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# How much trigonometry is there in Calcuel 2
## Trigonometric quantities in right triangles
Cosine and sine play a key role in studying right triangles. Given a right triangle ABC with side lengths a, b, c and angles α, β, γ with γ = π / 2, we can assume by shifting, rotating and mirroring that A is the zero point, C on the positive x-axis and B is in the first quadrant. Then B = (b, a) is a point on the circle K.cwhose radius corresponds to the hypotenuse c of the triangle ABC. Hence applies
(b, a) = B = c (cos α, sin α), b = c cos α, a = c sin α.
The other trigonometric functions tangent, cotangent, secant and Kosenkan can also be illustrated with the help of right triangles. But first we want the addition theorems
cos (α + β) = cos α cos β - sin α sin β,
sin (α + β) = cos α sin β + sin α cos β for all α, β ∈ ℝ
prove with the help of right triangles.
###### Second proof of addition theorems
Let α, β ∈ ℝ. We first assume that α, β> 0 and α + β <π 2.="" everything="" else="" will="" follow="" from="" this.="" under="" these="" conditions,="" α="" +="" β="" is="" the="" angle="" of="" a="" right="" triangle.="" we="" construct="" such="" a="" triangle="" as="" follows,="" where="" with="" ∠="" abc="" we="" denote="" the="" angle="" of="" a="" triangle="" with="" corners="" a,="" b,="" c="" belonging="" to="" corner="">
Let P = (cos α, sin α). Let Q = (x0, 0) the point on the x-axis with ∠0QP = β. Then ∠Q0P = α. Finally, let A be the point on the straight line QP with ∠0AP = π / 2. Then ∠0PA = α + β. So the triangle 0PA is as desired.
We can now count on:
sin (α + β) = 0A = x0 sin β = (cos α + cot β sin β) sin β = sin α cos β + cos α sin β.
The cosine theorem is proven analogously (or deduced from the sine theorem with the help of Pythagoras' theorem).
We now consider more general angles. The addition theorems are clear because of sin 0 = cos (π / 2) = 0 and sin (π / 2) = cos 0 = 1, if one of the two angles is 0 or π / 2. If α, β ∈ [ 0, π / 2] with α + β ≥ π / 2, then let α * = π / 2 - α, β * = π / 2 - β. Then according to what has already been proven that
sin (α + β) = sin (π - (α + β)) = sin (α * + β *) = sin α * cos β * + cos α * sin β * = cos α sin β + sin α cos β = sin α cos β + cos α sin β.
We argue analogously for the cosine. We have thus shown the addition theorems for all α, β ∈ [0, π / 2] (this corresponds to points P and Q on the unit circle in the first quadrant). From this we can deduce the theorems for any angles α, β ∈ ℝ by adding
α = α1 + k1 π / 2, β = β1 + k2 π / 2 with k1, k2 ∈ ℤ and α1, β1 ∈ [0, π / 2]
write and apply the shift formulas.
### Identification of the trigonometric quantities
The six trigonometric quantities
cos α, sin α, tan α, cot α, sec α, csc α
appear in numerous geometric figures. We consider two typical examples. The identification of the quantities results from their definition by applying the ray theorem.
In the following we denote the length of the line between two points A and B of the plane with AB. If A = (x1, y1) and B = (x2, y2), so it is true
AB =.
In our first figure we consider a point P on the unit circle with the angle α ∈] 0, π / 2 [and the half-ray of the plane defined by the zero point 0 and P. This half-ray intersects the Cartesian coordinate grid at two points R and S. All six quantities are included in the resulting figure.
With the help of the figure we can illustrate many properties of the functions. For example, tan α = TR tends towards infinity if the angle α tends towards π / 2. Furthermore, R = S = (1, 1) and thus tan α = cot α, if α = π / 4.
In the second figure we consider the tangent of the unit circle K at P. It defines two points on the axes:
Combined, we get the following figure, which focuses on the four perhaps less obvious quantities tan α, cot α, sec α and csc α:
The figure can be analyzed further. The intersection of the two tangent lines drawn in green defines, for example, the bisector of α.
Our considerations also motivate the naming of the trigonometric functions tan, cot, sec and csc. The tangent and the cotangent measure the lengths of certain sections of tangents of the unit circle. Likewise, the secant and the cosecant measure the lengths of certain sections of secants of the unit circle. A “co” always refers to the complementary angle π / 2 - α. This also explains why the function defined by 1 / cos α is not referred to as “co” even though the cosine is used. In our triangles, the quantity 1 / cos α is assigned to the angle α, while 1 / sin α belongs to the complementary angle π / 2 - α. |
Study Materials for 2023 CFA®, FRM®, Actuarial, GMAT® and EA® Exams
# GMAT® Data Sufficiency Questions
### What is the Structure of GMAT Data Sufficiency Questions?
Data Sufficiency questions is one of the sections in Quantitative reasoning questions (the other one is Problem-solving questions). Data Sufficiency questions are designed to measure your ability to analyze a quantitative problem, ascertain which data is relevant, and decide the extent to which there is enough data to solve the problem.
A data sufficiency question consists of a question and two statements. To answer the question, one should first identify the statement that provides information relevant to the question and then eliminate all the other possible answers by using math knowledge and other everyday facts. There will be 14 to 15 questions on data sufficiency in each quantitative section.
## Tips for Data Sufficiency Questions?
• The quantitative takes 62 minutes to complete and includes 31 questions. Therefore, you do have on average 2 minutes to answer each question.
• There are 14 to 15 data sufficiency questions in every quantitative section.
• Determine whether the problem allows only a single value or a range of values. Note that the primary objective is to determine whether there is enough data to solve the problem.
• Avoid unnecessary assumptions about geometrical figures, as they are not necessarily drawn to scale.
4.5 Million
50 Thousand
# 1 Rated
## Division & Factoring
When integer y is divided by 2, is the remainder 1?
(1) $$(-1)^{(y + 2)} = -1$$
(2) $$y$$ is prime.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
(1) That $$(–1)^{(y + 2)} = -1$$, dictates that y + 2 must be odd and therefore y is odd since only an odd exponent can produce a value < 0. Furthermore, if an odd number is divided by 2, the remainder must be 1; SUFFICIENT.
(2) That y is prime does not definitively identify y as definitively even or odd, so the remainder could be either 0 or 1; NOT SUFFICIENT.
The correct answer is A; statement 1 alone is sufficient.
## Number Properties
Chiku received a $3.50 per hour raise this week. If last week she worked 40 hours per week at her old pay rate, how many fewer hours can she work this week and still guarantee that she makes more this week than she did last week? (1) She made$620 last week.
(2) Her raise was 20 percent greater than that of any of her coworkers.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
(1) If Chiku made 620 last week, it is possible to solve for her old pay rate, then her new pay rate, and then the number of hours she’d need to work to guarantee she makes more this week than she did last week; SUFFICIENT. (2) That her raise was greater than that of any of her coworkers introduces a new unknown variable; NOT sufficient. The correct answer is A; statement 1 alone is sufficient. ## Question 3 ## Percentages (Data Sufficiency) If 60 percent of the freshman class at Westown High is male, does the freshman class have more than 100 male students? (1) The freshman class has more than 150 students. (2) The freshman class has 63 more male students than female students. A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient. B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient. C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D) EACH statement ALONE is sufficient. E) Statements (1) and (2) TOGETHER are not sufficient. (1) Defining the total number of students in the freshman class as greater than 150 does not determine the number of male nor female students definitively, and so the question cannot be answered; NOT sufficient. (2) Given: • 60 percent of the freshman class is male. • The freshman class has 63 more male students than female students. Let’s use s to denote the total number of students in the freshman class, m to represent the number of male students, and f to represent the number of female students. From the information given in statement (2): \begin{align*}m&=f+63\\m&=0.60s\\f&=0.40s\end{align*} We can set up the following equation using m and f: $$0.60s=0.40s+63$$ Solving for s: \begin{align*}0.20s&=63\\s&=\frac{63}{0.20}\\s&=315\end{align*} Now, calculating the number of male students (m): \begin{align*}m&=0.60s\\m&=0.60\times315\\m&=189\end{align*} Therefore, the number of male students is 189, which is indeed more than 100. Thus, statement (2) alone is sufficient to determine that there are more than 100 male students. The correct answer is B: Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient. ## Question 4 ## Exponents & Radicals (Data Sufficiency) The amount of bacteria in a culture after some $$t$$ hours is given by the function $$f(t)=pe^{kt}$$ where $$p$$ is a constant. What is the number of bacteria after 8 hours? (1) There were approximately 275 bacteria after 2 hours (2) $$k=0.16$$ A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient. B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient. C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D) EACH statement ALONE is sufficient. E) Statements (1) and (2) TOGETHER are not sufficient. The correct answer is: C) (1) From the function, $$f(t)=pe^{kt}$$, when $$t=2, f(t)=275$$. The equation becomes $$pe^{2k}=275$$. This is one equation in two unknowns, hence, we cannot solve it. More information is required; NOT SUFFICIENT. (2) When $$k=0.16$$, we have $$f(t)=pe^{0.16t}$$. In this case, $$p$$ is unknown, hence more information is required; NOT SUFFICIENT. Considering the two cases, we have $$pe^{2k}=275$$ and $$f(t)=pe^{0.16t}$$ which reduced to $$pe^{2(0.16)}=275$$. We solve the equation \begin{align*} pe^{2(0.16)} &=275\\ pe^{3.2} &=275\\ \ln \left(pe^{3.2}\right) &=\ln 275\\ \ln p + 3.2 &=\ln 275\\ \ln p &=\ln 275 – 3.2 = 2.41677\\ p& =e^{2.41677}=11.21\end{align*}. We now solve the equation When $$t=8$$, we have $$f(8)=11.21e^{0.16(8)} =11.21e^{1.28}=40.32$$ The correct answer is C; BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. ## Question 5 ## 3D Geometry (Data Sufficiency) A cylindrical layer cake has two vanilla layers surrounding a marzipan layer, represented by the shaded region, as shown below. Is the cake split into three layers of equal volume? (1) The height of marzipan layer is equal to the radius of the whole cake. (2) The volume of the entire cake is 81π cubic inches. A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient. B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient. C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D) EACH statement ALONE is sufficient. E) Statements (1) and (2) TOGETHER are not sufficient. The correct answer is: E) Note that the volume for a cylinder can be found from the formula $$πr^2 × height$$ where r = radius and the height is perpendicular to the radius. (1) That the height for the marzipan layer is equal to the radius of the cake means that the volume of the marzipan layer can be found from the simplified formula $$πr^3$$, but the total cake volume remains dependent on the actual height of the cake; NOT sufficient. (2) Represent the information algebraically as $$πr^2 × height = 81$$, but no information is provided about the height in relation to the radius; NOT sufficient. (Together) The information together still does not relate the height to the radius. For instance the radius could be 3 and the ratio of the marzipan layer to the others could be equal as each would have a volume of 27π. Alternatively, the radius could = 1 and the marzipan layer would be significantly smaller than the other two layers. The correct answer is E; both statements together are still not sufficient. ## Question 6 ## Inequalities At Rounders Grocer, orange slices cost2 a pound, pineapple chunks cost $3 a pound, and cut watermelon cost$5 a pound. If Sally buys enough of these three fruits from Rounders to make five pounds of fruit salad, and at least one pound of each, which fruit did she buy the most of by weight?
(1) Sally spent less than $5 on orange slices. (2) Sally spent more than$18 on her fruit salad.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
First note that Sally must buy at least one pound of each of the fruits, which would result in a sum fixed cost of $10. (1) If Sally spent less than$5 on orange slices, then she spent must have spent more on one of the other two fruits, but it is unclear which; NOT sufficient
(2) If Sally spent more than $18 then she must spend more than$8 on the two-variable pounds of fruit, which could only be accomplished by buying two more pounds of cut watermelon; Sufficient
The correct answer is B; statement 2 alone is sufficient.
## Plane Geometry
The Embeyay Expressway and Emefay Expressway intersect at a perpendicular junction coming from Ulster City and Finster Town, respectively, while a direct road connecting the two municipalities is entirely straight. How much further would a motorist traveling the expressways between Ulster City to Finster Town have to drive in comparison to another motorist who used the direct road?
(1) The distance from Ulster City to the Embeyay Expressway and Emefay Expressway junction is 12 kilometers.
(2) The distance from Ulster City to Finster Town on the direct road is 15 kilometers.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
Recognize that a right triangle can be formed with the junction creating the 90° angle and the two municipalities at the other two vertices.
Therefore, only two of the distances between the locations will be necessary to find the third by way of the Pythagorean Theorem.
(1) This provides the distance for one of the two legs of the right triangle only; NOT sufficient.
(2) This provides the distance for the hypotenuse of the right triangle only; NOT sufficient.
(Together) Complete the theorem as $$12^2+b^2 =15^2$$ to find that the shorter leg distance is 9. From there, 12 + 9 = 21 is 6 kilometers further than 15 kilometers; SUFFICIENT.
## Descriptive Statistics
A supermarket display of canned corn is shaped like a pyramid with one can on top and two more cans in each row below. If the display is only one can deep for the entire pyramid, what is the median number of cans in a row in the pyramid?
(1) There are 100 cans in the pyramid.
(2) The range of cans per row is 18.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
Remember that if there is a constant difference in terms that the average for the sequence = median of the sequence.
The formula for average is sum divided by the number of terms, so in this case the total number of cans = median × number of rows.
And the total number of cans = x + (x + 2) + (x + 4) … up to the total number of rows.
Therefore, two equations are already theoretically known, and only one additional equation is needed to solve for all of the values involved.
(1) If 100 is the total number of cans, then it would be possible to count from 1 + 3 + 5 + 7… to determine the total number of rows, and thereby the median, without completing the process; SUFFICIENT.
(1)If the range in cans per row is 18, then it would be possible to determine that the number of cans in the last row is 19 and from there the median, without completing the process; SUFFICIENT
The correct answer is D; each statement alone is sufficient.
## Coordinate Geometry
If line $$d$$ in the coordinate plane has the equation $$y = mx + b$$, where $$m$$ and $$b$$ are constants, what is the slope of line $$d$$?
(1) Line $$d$$ intersects the line with equation $$y = 6x + 2$$ at the point (1, 8).
(2) Line $$d$$ is parallel to the line with equation $$y = (2 – m)x + b – 4$$.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
Note that the slope of a line with equation $$y = mx + b$$ is$$m$$.
(1) A line passing through the point (1, 8) can have any value for its slope, so it is impossible to determine the slope of line $$d$$.
For example, the line $$y = x + 7$$ intersects $$y = 6x + 2$$ at (1, 8) with a slope of 1, while the line $$y = 2x + 6$$ intersects $$y = 6x + 2$$ at (1, 8) with a slope of 2; NOT sufficient.
(2) Parallel lines have the same slope, and it is possible to solve for the slope as $$m = 2 – m$$, $$2m = 2$$, and $$m = 1$$; SUFFICIENT.
## Rates, Work, and Combined Time
Cars enter a parking garage at a certain rate. At the same time, cars leave the parking garage at a different rate. At what rate, in cars per minute, is the number of cars in the garage changing?
(1) Cars enter the garage at a rate of 13.5 cars per 4.5 minutes.
(2) Cars leave the garage at a rate of 5 cars per minute.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
(1) While this rate can be converted to $$\frac {13.5}{4.5} = 3$$ cars/minute entering the garage, it says nothing about the rate at which cars leave; NOT sufficient.
(2) This statement says nothing about the rate at which cars enter the garage; NOT sufficient.
(Together) With the rates at which cars both enter and leave the garage, the target question can be answered; SUFFICIENT.
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How do you solve 16-x^2>0 using a sign chart?
May 11, 2017
The solution is $x \in \left(- 4 , + 4\right)$
Explanation:
We need
${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
We factorise the inequality
$16 - {x}^{2} > 0$
$\left(4 + x\right) \left(4 - x\right) > 0$
Let
$f \left(x\right) = \left(4 + x\right) \left(4 - x\right)$
We can build the sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$+ 4$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$4 + x$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$4 - x$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$
Therefore,
$f \left(x\right) > 0$ when $x \in \left(- 4 , + 4\right)$ |
1. Class 12
2. Important Question for exams Class 12
Transcript
Ex 8.1, 10 Find the area bounded by the curve 𝑥2=4𝑦 and the line 𝑥=4𝑦 – 2 Let AB represent the line 𝑥=4𝑦−2 AOB represent the curve 𝑥2=4𝑦 First we find Points A and B Points A & B are the intersection of curve and line We know that, 𝑥=4𝑦−2 Putting in equation of curve , we get 𝑥2=4𝑦 4𝑦−22=4𝑦 16 𝑦2+4−16𝑦=4𝑦 16 𝑦2−16𝑦−4𝑦+4=0 16 𝑦2−20𝑦+4=0 4 4 𝑦2−5𝑦+1=0 4 𝑦2−5𝑦+1=0 4 𝑦2−4𝑦−𝑦+1=0 4𝑦 𝑦−1−1 𝑦−1=0 4𝑦−1 𝑦−1=0 So, y = 14 , y = 1 As Point A is in 2nd Quadrant ∴ A = −1 , 14 & Point B is in 1st Quadrant ∴ B = 2 , 1 Finding required area Required Area = Area APBQ – Area APOQBA = −12𝑦1𝑑𝑥− −12𝑦2𝑑𝑥 Required Area = −12 𝑥 + 24𝑑𝑥− −12 𝑥24𝑑𝑥 = 14 −12 𝑥+2𝑑𝑥 − 14 −12 𝑥2 𝑑𝑥 = 14 𝑥22+2𝑥−12− 14 𝑥33−12 = 14 22 − −122+2 2− −1− 14 23 − −133 = 14 4 − 12+2 3− 14 8 + 13 = 14 32+6− 14 93 = 14 32+6−3 = 14 32+3 = 14 92 = 98 ∴ Required Area = 𝟗𝟖 Square units
Class 12
Important Question for exams Class 12 |
# How do you solve \sqrt { x + 6} = 6- 2\sqrt { 5- x }?
Aug 20, 2017
$x \approx 2.66$
#### Explanation:
For starters, you know that you must have
$x + 6 \ge 0 \implies x \ge - 6$
and
$5 - x \ge 0 \implies x \le 5$
That is the case because the expressions under the two square roots must be positive when working with real numbers.
So, right from start, you know that the solution interval cannot include any value of $x$ that is not part of
$x \in \left[- 6 , 5\right] \text{ } \textcolor{b l u e}{\left(1\right)}$
Moreover, you need to have
$6 - 2 \sqrt{5 - x} \ge 0$
$6 \ge 3 \sqrt{5 - x}$
This simplifies to
$2 \ge \sqrt{5 - x} \text{ } \textcolor{b l u e}{\left(2\right)}$
Now, rearrange the equation as
$\sqrt{x + 6} + 2 \sqrt{5 - x} = 6$
Square both sides of the equation to get
${\left(\sqrt{x + 6} + 2 \sqrt{5 - x}\right)}^{2} = {6}^{2}$
${\left(\sqrt{x + 6}\right)}^{2} + 2 \cdot \sqrt{x + 6} \cdot 2 \cdot \sqrt{5 - x} + {\left(2 \sqrt{5 - x}\right)}^{2} = 36$
This is equivalent to
$x + 6 + 4 \sqrt{\left(x + 5\right) \left(5 - x\right)} + 4 \left(5 - x\right) = 36$
Rearrange the equation to isolate the new square root term on one side of the equation
$4 \sqrt{\left(x + 6\right) \left(5 - x\right)} = 36 - x - 6 - 20 + 4 x$
$\sqrt{\left(x + 6\right) \left(x - 5\right)} = \frac{10 + 3 x}{4}$
Next, square both sides of the equation again to get
${\left(\sqrt{\left(x + 6\right) \left(x - 5\right)}\right)}^{2} = {\left(\frac{10 + 3 x}{4}\right)}^{2}$
$\left(x + 6\right) \left(5 - x\right) = \frac{100 + 60 x + 9 {x}^{2}}{16}$
This will be equivalent to
$- {x}^{2} - x + 30 = \frac{100 + 60 x + 9 {x}^{2}}{16}$
$- 16 {x}^{2} - 16 x + 480 = 9 {x}^{2} + 60 x + 100$
$25 {x}^{2} + 76 x - 380 = 0$
Use the quadratic formula to find the two roots of this quadratic equation
${x}_{1 , 2} = \frac{- 76 \pm \sqrt{{76}^{2} - 4 \cdot 25 \cdot \left(- 380\right)}}{2 \cdot 25}$
${x}_{1 , 2} \approx \frac{- 76 \pm 209}{50} \implies \left\{\begin{matrix}{x}_{1} \approx \frac{- 76 - 209}{50} \approx - 5.7 \\ {x}_{2} \approx \frac{- 76 + 209}{50} \approx 2.66\end{matrix}\right.$
Now, notice that both solutions satisfy $\textcolor{b l u e}{\left(1\right)}$, but that only one satisfies $\textcolor{b l u e}{\left(2\right)}$, since
$2 \textcolor{red}{\cancel{\textcolor{b l a c k}{\ge}}} \sqrt{5 - \left(- 5.7\right)}$
$2 \textcolor{red}{\cancel{\textcolor{b l a c k}{\ge}}} \sqrt{10.7}$
This means that $x \approx - 5.7$ is an extraneous solution, i.e. it does not satisfy the original equation.
Therefore, you can say that $x \approx 2.66$ is the only solution to the original equation. |
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Sigma summation
We could easily have many more forces to add. While towing your boat, the stream might get stronger and a stormy wind may appear, so you’ll need help from nearby strangers:
$$\vec F_\text{result}= \vec F_\text{you}+ \vec F_\text{friend}+ \vec F_\text{stream}+ \vec F_\text{stranger1}+ \vec F_\text{stranger2}+ \vec F_\text{stranger3}+ \vec F_\text{stranger4}+ \vec F_\text{wind}$$
We start having a lot of terms to add up here. Let’s invent a simple symbol to mean ‘add them all, no-matter how many there are’. Let’s pick the Greek uppercase letter Sigma $\sum$ and write:
$$\vec F_\text{result}=\sum\vec F$$
Some would call this: sigma notation.[1,2]
If we in the future want to know how much force only the four strangers aided with, we can quickly invent a way of writing that we only want those four added up. We could for example quickly number all the terms: $\underbrace{\vec F_\text{you}}_1+ \underbrace{ \vec F_\text{friend}}_2+ \underbrace{ \vec F_\text{stream}}_3+ \underbrace{ \vec F_\text{stranger1}}_4+ \underbrace{ \vec F_\text{stranger2}}_5+ \underbrace{ \vec F_\text{stranger3}}_6+ \underbrace{ \vec F_\text{stranger4}}_7+ \underbrace{ \vec F_\text{wind}}_8$ – let’s call this numbering indexes1 – and then indicate that we only want to sum up those between number 4 to 7 (from index 4 to 7) in for example this way:
$$\vec F_\text{strangers}=\sum_4^7\vec F$$
There is of course no need to write these numbers when we want to sum up all terms.
References:
1. Summation Notation’ (web page), Columbia University, www.columbia.edu/itc/sipa/math/summation.html (accessed Aug. 28th, 2019)
2. The Summation Symbol’ (article), HEC Montreal, www.hec.ca/en/cams/help/topics/The_summation_symbol.pdf (accessed Aug. 28th, 2019)
3. Online Etymology Dictionary’ (dictionary), Douglas Harper, www.etymonline.com |
# Question: How many bracelet with no luck can be formed from 7 different colored beads?
Contents
It would be 7! = 5040 diffrent necklaces.
## How many bracelets with no lock can be formed from 6 different colored beads?
= 720 distinct ways (permutations) to arrange the 6 different beads.
2520. 5040.
## How many necklaces can be formed with 8 different Colour beads?
The number of ways in which 8 different beads be strung on a necklace is. 2500. 2520.
## How many arrangements of beads are possible in a bracelet if there are 6 different designs of beads?
Since there are 6! linear arrangements of six distinct beads, the number of distinguishable circular arrangements is 6! 6=5!
THIS IS EXCITING: What were beads used for in Africa?
## How many necklaces can you make with 6 beads of 3 colors?
The first step is easy: the number of ways to colour 6 beads, where each bead can be red, green or blue, is 36 = 729.
## How many ways can 5 beads be arranged in a circular bracelet?
Therefore the total number of different ways of arranging 5 beads is 242=12 . Now, we can select 5 beads from 8 different beads in 56 ways. In each way we can arrange them in 12 ways. Therefore the total number of ways in which 5 beads, chosen from 8 different beads be threaded on a circular ring is 672.
## What is circular permutation?
Circular permutation is the total number of ways in which n distinct objects can be arranged around a fix circle.
## What are the number of ways in which 10 beads can be arranged to form a necklace 9 !/ 2 9 10 10?
Answer: This is called a cyclic permutation. The formula for this is simply (n-1)!/2, since all the beads are identical. Hence, the answer is 9!/2 = 362880/2 = 181440.
## How many necklaces can be formed with 6 white and 5 red beads if each necklace is unique how many can be formed?
5! but correct answer is 21.
## How many ways can 7 boys seat around circle?
Since in this question we have to arrange persons in a circle and 7 persons have to be arranged in a circle so that every person shall not have the same neighbor. Hence there are 360 ways to do the above arrangement and therefore the correct option is A.
## How many different bangles can be formed from 8 different colored heads?
How many different bangles can be formed from 8 different colored beads? Answer: 5,040 bangles .
## How many ways can 6 differently Coloured beads be threaded on a string?
Assuming that the beads are different, the first bead can be picked in 6 ways. Then the second bead can be picked in 5 ways. And the third bead can be picked in 4 ways, etc. Multiplying these together, we get 6*5*4*3*2*1 = 720 ways.
## How many necklaces can be made by using 10 round beads all of a different colors?
There are 10 beads of distinct colours; say, A, B, C, D, E, F, G, H, I and J. If no restriction is imposed then, there are (10!) = 3628800 ways to put these ten distinctly coloured beads into a necklace. |
Estimate the radius, diameter, and area of the circle. answer choices . For purposes of hand-calculation, pi is assumed to be 3.14 when solving these problems. Play. The exclusive pages contain a lot of pdf worksheets in finding area, circumference, arc length, and area of sector. Circumference = Π×diameter = 10 Π. radius = diameter ÷ 2 = 10 ÷ 2 = 5. answer choices . Therefore, Diameter $$= 2 \times$$ Radius These worksheets are a great supplement to any unit on circles and or any sub lesson plan! Students will be given a measure to represent the diameter, radius or circumference of a circle. Diameter is a straight line segment passing through the center of circle, the endpoints of which touches the circumference of circle. You can see here that this is one part of a circle. Use = 3.14 to calculate your answers. If my radius is 10ft, what is my diameter? Radius = D/2. Click hereto get an answer to your question ️ If the diameter of circle is 10 cm, then find the radius of circle. Representing inequalities: Fill in the gaps. 12m 32 km 2. Example 1: Radius = 6 cm. The radius of a circle is 5.8 m. Calculate its area. Tags: Question 6 . Which circle has a radius of 16 km? Find the area of a circle given its radius, diameter, or circumference. 7.1 Estimating Areas. What is the diameter of Sue's wheel? Practice worksheet on the radius, diameter, circumference, and area of a circle. Show Answer. Formula for Area with Diameter. Find lines of symmetry, identify symmetrical shapes, and sketch symmetrical figures. 0 times. Next Perimeter of a Semi-Circle Practice Questions. Estimate the area of circle B. Problem 2. Values of π. π is equal to approximately 3.14. So, circumference equals (I’m just gonna rewrite the formula to help us follow our work), $$C=\pi \times d$$, equals pi times diameter. View solution. 3. Solved Examples. Let's consider an example to look at the use of these relations. If the diameter of a circle is 142.8 millimeters, then what is the radius? Let's find the circumference of the circle. Circumference = 2 × π × radius Practice finding the radius and diameter of a circle using both vocabulary and visuals. A circle’s circumference is approximately 76 cm. The students should use a ruler to measure the diameter of each circle to the nearest inch. ... Share practice link. Author: Jenny Beech. Either the diameter or the radius is shown on each shape. The formula for the circumference of a circle is $$C=\pi \times d$$, or it can be written as $$C=2\times \pi \times r$$. Area = Π (radius)² = Π (5)² = 25Π. To play this quiz, ... Is the red line a radius or diameter of the circle? Practice that skill now. Find the area of the circle. Quiz: Want To Find Out What Type Of Guy Is Right For You? To play this quiz, please finish editing it. Formulas: 1) Circumference C = or C= 2) Diameter, d = 2r and radius , 3) Area of a circle , A = π r2 -----1) . Find the perimeter of the table View solution Find the area of the semicircle whose radius is 4 cm. Radius = 18/2 = 9 cm . Area of . 6 inches. Assume that the radius of a circle is 21 cm. 1. Question 34: In the adjoining figure, and is mid-point of . The Diameter of a Circle. Finish Editing. Which circle has a radius of 9 cm? Each circle picture shows a radius or diameter. Calculate its radius. Logged in members can use the Super Teacher Worksheets filing cabinet to save their favorite worksheets. Enter the radius, diameter, circumference or area of a Circle to find the other three. Students will be given a measure to represent the diameter, radius or circumference of a circle. ... AB is a diameter of a circle and P is a variable point on the circumference of the circle. This measurement is called the radius. I’m going to use the formula with the diameter for this one. 6 a. Radius and Diameter of a Circle Radius and Diameter 1. These exercises are curated for students of grade 4 through high school. Diameter. Share practice link. The diameter is always twice the radius, so either form of the equation works. Great vocabulary and definitions. For simplicity, we use = 3.14. 5. Students answer the short answer questions about circles, circumference, and pi. The formula for the radius can be written as $$r=\frac{d}{2}$$, and the formula for diameter can be written as $$d=2r$$. The shape of the top of a table in a restaurant is that of a segment of a circle with the centre O and ∠ B O D = 9 0 ∘, B O = O D = 6 0 cm. Answers should be given in one of the following 3 formats to be considered correct: separated by "and" (A and B), separated by 1 space (A B), or separated by 2 spaces (A B). The radius of a circle is . For complete access to thousands of printable lessons click the button or the link below. Search for: Author: Jenny Beech. Can You Pass This Basic World History Quiz? Can your students solve these brain bafflers? Divide 9 inches by 2. 7th - 8th grade. 5th - 7th grade. Circumference = 2Π × radius = 2Π × 4 = 8Π. Previous Arc Length Practice Questions. Area of a Circle Calculator. In this lesson you will explore the properties of circles. Share practice link. 1) . How are radius and diameter related? Answers should be given in one of the following 3 formats to be considered correct: separated by "and" (A and B), separated by 1 space (A B), or separated by 2 spaces (A B). Use the formula pi-r-squared to calculate the area of each circle. Diameter – Diameter is a line segment that starts at one side of circle, passes through the center of the circle and touches the other end of the circle. 'AB' is the line segment which connects the point 'A' and point 'B' passing through the center 'O'. Students will:•Identify and describe the diameter, radius, chord, and circumference of a circle.•Investigate and describe the relationship between: diameter and radiusdiameter and chordradius and circumference Foundation Maths - Circles 3. Fun maths practice! Payton has a yo-yo with a diameter of 4 cm, what is the radius of the yo-yo? Practice. $$\pi$$ (Pi) is a mathematical constant which is the ratio of the circumference of a circle to its diameter. Circumference And Area Of Circle Quiz! Similarly, the formula for the area of a circle is tied to π and the radius: Area of a circle: A = πr 2 Q. What Do You Know About Gluconeogenesis? Also, figure out which circles are the largest, which are the smallest, and which are the same size. Pi. Sue has a bicycle wheel with a radius of 6.5 in. The diameter of a circle is 3.8 ft. answer choices . To change a diameter to a radius you DIVIDE by 2. radius, sector, arc, tangent, chord, radii, diameter, circumference. Now to find the circumference of a circle, we will need to use a formula. The … Find the diameter and radius of each circle. The size of the circle changes when the length of the radius varies. If you're seeing this message, it means we're having trouble loading external resources on our website. Diameter = 2 × radius. SURVEY . Our calculation is correct! The radius of the circle is. This quiz is incomplete! However, you may see other more precise values for π, such as 3.142 or 3.14159. In the last lesson, we learned that the formula for circumference of a circle is: . Method 1 144π – 100π = 44π. Which Harry Potter Hogwarts House Do You Belong To Quiz! The d represents the measure of the diameter, and r represents the measure of the radius. Circumference of a circle: C = πd = 2πr. ... Finding the radius and diameter from the area of a circle. Radius. This type of activity is known as Practice. The formula is πr 2. This quiz is incomplete! Answer 3: Practice that skill now. 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Question 34: in the RESULTS BOX to indicate whether your answer is correct or.., perimeter, area, accurate to 1 decimal place or html formats purposes of hand-calculation, is! = 144π segment that joins the center of the surface of the circle use these for peer sessions. Also, figure out the circumference a diameter of 37 inches students should use ruler... The endpoints of which touches the circumference, accurate to 1 decimal place ) area radius and diameter of a circle practice a.. The diameter of a circle is the radius of 15 inches three times the diameter for this one questions. Centre point '\ ( O\ ) ' order radius and diameter of a circle practice find the area of a is. For students to solve the problems, such as 3.142 or 3.14159 area and circumference of a circle questions... For peer practice sessions, small group instruction, or circumference of circle type Guy! Circumference = 2Π × radius = 2Π × 4 radius and diameter of a circle practice 8Π precise values for,... 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In this page 10th grade geometry solution2 we are going to see solutions of some practice questions.
(2) In the figure AP = 3 cm, AR = 4.5 cm, AQ = 6 cm, AB = 5 cm and AC = 10 cm. Find the length of AD.
Solution:
From the given information we get, (AB/AP) = (AC/AQ)
In triangle ABC, (AB/AP) = (AC/AQ)
By using converse of “Thales theorem” PQ is parallel to BC.
RD = x
In triangle ABD, PR is parallel to BD
(5/3) = (4.5 + x)/4.5
(5 x 4.5)/3 = 4.5 + x
7.5 = 4.5 + x
x = 7.5 – 4.5
x = 3
Here we need to find the length of AD = 4.5 + x
= 4.5 + 3
= 7.5 cm
(3) E and F are points on the sides PQ and PR respectively, of a triangle PQR. For each of the following cases. Verify EF is parallel to QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Solution:
First let us draw the picture for the above details
To verify whether EF is parallel to QR we have to check the condition (PE/EQ) = (PF/FR) (3.9/3) = (3.6/2.4) 1.3 ≠ 1.5 So the sides EF and QR are not parallel.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Solution:
First let us draw the picture for the above details
To verify whether EF is parallel to QR we have to check the condition (PE/EQ) = (PF/FR) (4/4.5) = (8/9) 0.88 = 0.88 So the sides EF and QR are parallel.
(4) In the figure,AC is parallel to BD and CF is parallel to DF, if OA = 12 cm, AB = 9 cm, OC = 8 cm and EF= 4.5 cm, then find FQ.
Solution:
In triangle OBD, AC is parallel to BD
By using “Thales theorem” we get,
(OA/AB) = (OC/CD)
(12/9)= (8/CD)
CD = (8 x 9)/12
= 72/12
= 6 cm
In triangle ODF, the sides CE and DF are parallel, by using “Thales theorem” we get,
(OC/CD)= (OE/EF)
(8/6) = (OE/4.5)
OE = (8 x 4.5)/6
= 36/6
= 6 cm
So, OF = OE + EF
= 6 + 4.5
= 10.5 cm |
How do you simplify 9.594 div 0.06?
Apr 9, 2017
A way to get round the decimal place in the initial division expression.
$159.9$
Explanation:
Note that $9.594$ is the same as $9594 \times \frac{1}{1000}$
Note that $0.06$ is the same as $6 \times \frac{1}{100}$
So $9.594 \div 0.006$ is the same as $\left[9594 \div 6\right] \times \left[\frac{1}{1000} \div \frac{1}{100}\right]$
$\left(9594 \div 6\right) \times \frac{1}{10}$
I will do the division first then multiply by the $\frac{1}{10}$ at the very end
$\text{ } 9594$
$\textcolor{m a \ge n t a}{1000} \times 6 \to \underline{6000} \leftarrow \text{ subtract}$
$\text{ } 3594$
$\textcolor{m a \ge n t a}{\textcolor{w h i t e}{1} 500} \times 6 \to \underline{3000} \leftarrow \text{ subtract}$
$\text{ } 594$
$\textcolor{m a \ge n t a}{\textcolor{w h i t e}{10} 90} \times 6 \to \underline{\textcolor{w h i t e}{0} 540} \leftarrow \text{ subtract}$
$\text{ } \textcolor{w h i t e}{0} 54$
$\textcolor{m a \ge n t a}{\textcolor{w h i t e}{100} 9} \times 6 \to \text{ "ul(54) larr" subtract}$
$\text{ } 0$
As we have 0 we may now stop the division
$9594 \div 6 = \textcolor{m a \ge n t a}{1599}$
Now we multiply by the $\frac{1}{10}$ giving:
$159.9$
~~~~~~~~~~~~~~~~~~~~~~~~~~~
If the division had not finished with a 0 at that point then we would have gone into decimal values.
Aug 11, 2017
$= 159.6$
Explanation:
The rule with dividing BY a decimal is simple ... DON'T!!
You can change any decimal into a whole number by multiplying by a power of $10$
Write the division as a fraction:
$\frac{9.594}{0.06}$
Multiply the numerator and the denominator by the same number
$\frac{9.594}{\textcolor{b l u e}{0.06}} \textcolor{red}{\times \frac{100}{100}} \text{ } \leftarrow \left[\textcolor{red}{\frac{100}{100} = 1}\right]$
This changes the denominator into a whole number:
$= \frac{959.4}{\textcolor{b l u e}{6}} \text{ } \leftarrow$ divide as usual
$6 | \underline{9 {\text{ "^3 5" "^5 9." }}^{5} 4}$
$\textcolor{w h i t e}{\times} 1 \textcolor{w h i t e}{x . x} 5 \textcolor{w h i t e}{\times x} 9. \textcolor{w h i t e}{x .} 6 \text{ } \leftarrow$ details below
$= 159.6$
$9 \div 6 = 1 \text{ carry } 3$
$35 \div 6 = 5 \text{ carry } 5$
$59 \div 6 = 9 \text{ carry } 5$
$54 \div 6 = 9$ , no remainder |
# ANALYZING STUDENT WORK
STRATEGY
Although analyzing student work isn’t always easy, the benefits are invaluable. By studying our students’ approach to solving problems, we can discover new ways to teach a familiar topic, learn more about a particular child’s strengths and weaknesses, and plan modifications or adjustments in our current instruction.
Let’s walk through a student’s solution to Problem 4 of the Problem Set of Grade 5, Module 3, Lesson 15:
Step 1: The student drew a tape diagram. She knew to label the whole as 20 minutes, parts 1 and 2 with the number of minutes Gavin spent on those questions, and part 3 with a “?” to represent the unknown.
Step 2: The tape diagram helped her decide to subtract the amount of time Gavin spent on questions 1 and 2 from the total time spent on the quiz.
Step 3: Common units were formed and 20 was renamed. The student recognized the need for common units in this subtraction problem and was able to correctly determine the equivalent fractions. Why was the value of 20 renamed? She recognized that it would be too challenging to subtract the mixed fractions from 20, so 20 was decomposed into 19 and 1. The number 1 was renamed 20/20. Now, she is able to subtract.
Step 4: The student subtracted the first two mixed numbers in step 3. To do so, she first subtracted whole numbers, 19 and 9, to get a value of 10 and then subtracted the fractions to find a difference. Finally, she added 10 and 5/20.
Step 5: Because there weren’t enough twentieths in 5/20 to subtract 16/20, she unbundled a 1 to make 9 and 20/20, which became 9 25/20 when she added 5/20.
Step 6: After subtracting the two mixed numbers in step 5 the student remembered the answer needed to be written in minutes and seconds. She knew there are 60 seconds in a minute. Renaming the fraction as sixtieths made it easier to determine the final answer, which she wrote as a statement: Gavin had 6 minutes 27 seconds for question 3.
In analyzing this work and breaking down what this student was thinking in each part of the solution, we learned more about the knowledge and skills needed to successfully answer this problem. This student needed to know how to apply the “Read, Draw, Write” process, rename units, subtract mixed numbers, rename a fraction greater than one, and decompose or unbundle units in a way that simplifies problem-solving. The analysis here helps us anticipate where a student might get held up and allows us to customize our lessons to fit the needs of our students.
Looking at these steps also reveals alternative ways to solve the same problem, such as the following:
Although many of the steps in this alternate approach are the same as the steps detailed above, the strategy is different. Encouraging and accepting multiple strategies is a wonderful practice to incorporate into your math classroom because it addresses different learning styles. It also allows students to create their own strategy, which increases motivation to solve problems and strengthens their mathematical thinking when asked to explain or justify their work. Celebrating multiple solutions in our classrooms also creates a sense of excitement and fosters positive feelings toward math.
Planning the time to analyze student work has major impacts on our instruction and the success of our students. This can also be a great exercise to do in collaboration with colleagues. You may gain a different perspective, discover new ideas, and even increase your own understanding of mathematics.
This post is by Dawn Pensack who is a Grades 5–6 Eureka Math writer. She taught math for 10 years (mostly in Grade 6).
By: Pia Mohsen
### PROBLEM SOLVING THE RDW WAY
By: Lisa Watts-Lawton |
By third grade we hope that most of our third graders know most of their addition and subtraction facts from memory. But usually they don’t. So do we spend precious math time drilling and killing them on those facts? NO, there is a better way. Teach them the Properties of Addition! Yes, these properties are actually smart strategies to add more efficiently especially with multi digit numbers.
Click READ MORE below to see the entire post.
The Zero or Identity Property of Addition (ZPA) is a quick and fairly easy property to teach: any number added to zero is that number. Or stated another way: any addend added to zero is that addend. I have my students practice this property and 2 others on these mats shown in the picture. Though they are part of a paid resource, a white board with headings works just as well.
Practicing with the Properties of Addition
The Commutative Property of Addition (CPA) is a MUST. If a student is stuck with 7 + 5, then maybe the student knows 5 + 7! The Commutative Property states that addends can be added in any order. Again I have my students practice this property on the mats and using unifix cubes. Watch the video below to see a demonstration.
Watch as my student works with the Commutative Property of Addition
Understanding the Commutative Property also helps with mental math or counting up. If you are presented with 3 + 8, you can start with the larger number and count up 3 to get to 11. I can do this because the CPA allows me to start with any addend. Sometimes younger students are stuck and are daunted by having to count up from a smaller number. Start with the larger number! Working closely with the CPA, is the Associative Property of Addition.
The Associative Property of Addition (APA) allows me to group any addends and add them. So why not group digits that add up to 10! Why 10? Because it has a zero! (see Zero Property of Addition). See the example below to see how the (APA) really helps with adding multi digit numbers or with column addition. I still remember my math teacher from high school giving us that tip! Look for combinations of 10s.
Associative Property of Addition is great for finding pairs that make 10.
So, it may be a while before third graders or even fourth graders are automatic with the addition and subtraction facts. But knowing the Properties of Addition students have a strategy for adding when their memory fails.
You can see all my products for the Properties of Addition HERE.
The websites were accessed using Chrome and Safari. mathplayground.com does support mobile devices such as tablets and phones. fun4thebrain.com requires flash to work with iOS, so it does not work properly with an iPad or iPhone. However, paid iOS and Android apps are available for those games as well. So, if your students have access to Chrome Books or other laptops, these sites are easily accessible and full of learning!
fun4thebrain.com
Nicely designed website with many games for students to chose from that will appeal to both boys and girls. The site also has the link to iOS apps for the iPhone and Android apps. However, those are paid apps.
www.fun4thebrain.com
www.mathplayground.com
Another great website with many games for students to chose from. The best part of this website is that at the bottom of each game screen is either the Common Core State Math Standard or the Mathematical Practice it supports. So if parents or administrators are in doubt that these games will support the CCSS, you have some evidence.
www.mathplayground.com
Click the links below to get more ideas from Math Tip Monday bloggers!
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Calculate the value
Simplify the expression
$29 ^{ 2 } +2 \times 29+1$
$900$
Calculate the value
$\color{#FF6800}{ 29 } ^ { \color{#FF6800}{ 2 } } + 2 \times 29 + 1$
Calculate power
$\color{#FF6800}{ 841 } + 2 \times 29 + 1$
$841 + \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 29 } + 1$
Multiply $2$ and $29$
$841 + \color{#FF6800}{ 58 } + 1$
$\color{#FF6800}{ 841 } \color{#FF6800}{ + } \color{#FF6800}{ 58 } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
Find the sum
$\color{#FF6800}{ 900 }$
$900$
Calculation using the perfect square formula
$29 ^ { 2 } + 2 \times 29 + \color{#FF6800}{ 1 }$
Present as the shape of the power
$29 ^ { 2 } + 2 \times 29 + \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$
$29 ^ { 2 } + \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 29 } + 1 ^ { 2 }$
Change the term in the middle
$29 ^ { 2 } + \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 29 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } + 1 ^ { 2 }$
$\color{#FF6800}{ 29 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 29 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } ^ { \color{#FF6800}{ 2 } }$
Organize the expression using $A^{2} ± 2AB + B^2 = (A ± B)^{2}$
$\left ( \color{#FF6800}{ 29 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( \color{#FF6800}{ 29 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { 2 }$
Add $29$ and $1$
$\color{#FF6800}{ 30 } ^ { 2 }$
$\color{#FF6800}{ 30 } ^ { \color{#FF6800}{ 2 } }$
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$\color{#FF6800}{ 900 }$
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New Rochelle High School Geometry Summer Assignment
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2 Evaluating Algebraic Expressions Evaluate Numerical Expressions - Numerical expressions often contain more than one operation. To evaluate them, use the rules for order of operations shown below. Order of Operations Step 1 Evaluate expressions inside grouping symbols. Step Evaluate all powers. Step Do all multiplication and/or division from left to right. Step 4 Do all addition and/or subtraction from left to right. Evaluate Algebraic Expressions - Algebraic expressions may contain more than one operation. Algebraic expressions can be evaluated if the values of the variables are known. First, replace the variables with their values. Then use the order of operations to calculate the value of the resulting numerical expression. Evaluate Numerical Expressions Evaluate Numerical Expressions a. [ + (1 ) ] Evaluate x + 5(y ) if x = and y = 1. [ + (1 ) ] = ( + 4 ) Divide 1 by. x + 5(y ) = ( + 16) Find 4 squared. + 5(1 ) Replace x with and y with 1. = (18) Add and (1 ) Evaluate. = 54 Multiply and (9) Subtract from Multiply 5 and 9. 5 Add 8 and 45. The solution is What is the value of 5xy if x and y? (1) 90 () 90 () 180 (4) 180. If y x 5x 6, what is the value of y when x? (1) () 4 () 4 (4) 4. Evaluate the expression [1 + (5 ) ]. 4. Evaluate the expression 5x (y + z) if x 6, y 8, and z. 5. Evaluate the expression (10x) + 100a if x and 4 a. 5
3 Solving Equations Multi-Step Equations - To solve equations with more than one operation, often called multi-step equations, undo operations by working backward. Reverse the usual order of operations as you work. Equations With Variables on Each Side - To solve an equation with the same variable on each side, first use the Addition or the Subtraction Property of Equality to write an equivalent equation that has the variable on just one side of the equation. Then solve the equation. Equations With Grouping Symbols - When solving equations that contain grouping symbols, first use the Distributive Property to eliminate grouping symbols. Then solve. Multi-Step Equations Equations With Variables on Each Side Solve 5x + =. Solve 5y 8 = y x + = Subtract from each side. 5y 8 y = y + 1 y Subtract y from each side. 5x = 0 Simplify. y 8 = 1 Simplify. 5x = Divide each side by 5. y = Add 8 from each side. x = 4 Simplify. y = 0 Simplify. Equations With Grouping Symbols y = 0 y = 10 Divide each side by. Simplify. Solve 4(a 1) = 10(a 5). 8a 4 = 10a + 50 Distributive Property 18a = 54 Simplify. 18a 8a a = 10a a Add 10a to each side. Divide each side by a 4 = 50 Simplify. a = Simplify. 18a = Add 4 to each side. 6. If 1x 8x 1 10, then x (1) () () 1 (4) 1 7. If x x 10, then x (1) 4 7 () 7 1 () 5 (4) 9 8. Solve x 1 9x Solve x 5 8x Solve 5 6 x x.
4 Factoring Factor x + bx + c - To factor a trinomial of the form x + bx + c, find two integers, m and p, whose sum is equal to b and whose product is equal to c. Factoring x + bx + c x + bx + c = (x + m)(x + p), where m + p = b and mp = c Factor x + 7x + 10 Factor x + 6x 16 In this trinomial, b = 7 and c = 10. Factors of 10, 5 7 Since + 5 = 7 and 5 = 10, let m = and p = 5. x + 7x + 10 = (x + 5)(x + ) Factor x 8x + 7 In this trinomial, b = 6 and c = 16. This means m + p is positive and mp is negative. Make a list of the factors of 16, where one factor of each pair is positive. In this trinomial, b = 8 and c = 7. Notice that m + p is negative and mp is positive, so m and p are both negative. Since 7 + ( 1) = 8 and ( 7)( 1) = 7, m = 7 and p = 1. Therefore, m = and p = 8. x 8x + 7 = (x 7)(x 1) x + 6x 16 = (x )(x + 8) 11. Which is a factor of Sum of Factors 1, y y 0? (1) y 6 () y 6 () y (4) y 1. Which product is a factored form of x x 1? x x x x Factors of 16 Sum of Factors 1, , 16 15, 8 6, 8 6 (1) () () x 6 x 1 (4) x 1 x 6 1. Factor x 8x Factor x 1x x 17x 60 4
5 Solving Quadratic Equations Solve Equations by Factoring - Factoring and the Zero Product Property can be used to solve many equations of the form x + bx + c = 0. Solve x + 6x = 7 x + 6x = 7 Original equation x + 6x 7 = 0 Rewrite equation so that one side equals 0. (x 1)(x + 7) = 0 Factor. x 1 = 0 or x + 7 = 0 Zero Product Property x = 1 x = 7 Solve each equation. The solution set is { 7, 1}, the roots are 7 and If x x 0, which is the greater of the two roots? (1) () () (4) 17. If x 11x 0 0, which is the smaller of the two roots? (1) 6 () 5 () 5 (4) Solve y y Solve x x Solve x x 10. 5
6 Slope Find Slope - The slope of a line is the ratio of change in the y-coordinates (rise) to the change in the x- coordinates (run) as you move in the positive direction. Find the slope of the line that passes through Find the value of r so that the line through (10, r) and (, 5) and (4, ). (, 4) has a slope of. 7 Let (, 5) = (x 1, y 1 ) and (4, ) = (x, y ). m = y y 1 x x 1 m = y y 1 x x 1 Slope formula 7 = 4 r 10 = 5 4 ( ) = 7 7 y =, y 1 = 5, x = 4, x 1 = 7 = 4 r 7 Slope formula m = 7, y = 4, y 1 = r, x =, x 1 = 10 Simplify. Simplify. ( 7) = 7(4 r) Cross multiply. = 1 14 = 8 7r Distributive Property 14 = 7r Subtract 8 from each side. = r Divide each side by What is the slope of line l shown in the. What is the slope of a line through points 4, 6,8? accompanying diagram? and (1) () 5 5 () (4) 5 (1) 5 () 5 () 5 5 (4) 5. Determine the slope of the line through the points A, and 1,7 B? 4. Determine the slope of the line through the points A 4, and 0,6 B? 5. If points, and, 5 x are on a line whose slope is 7, what is the value of x? 6
7 Equations of Lines Slope-Intercept Form y = mx + b, where m is the slope and b is the y-intercept Write an Equation Given the Slope and a Point Write an equation of the line that passes through (, 1) Write an Equation Given Two Points Write an equation of the line that passes through with a slope of 1. (1, ) and (, ). 4 The line has slope 1. Replace m with 1 and (x, y) with (, 1) 4 4 Find the slope m. To find the y-intercept, replace m with in the slope-intercept form. its computed value and (x, y) with (1, ) in the slope- y = mx + b Slope-intercept form intercept form. Then solve for b. 1 = 1 4 ( ) + b m = 1 4, y = 1, and x = m = y y 1 x x 1 1 = 1 + b Multiply. m = 1 Slope formula y =, y 1 =, x =, x 1 = 1 1 = b Add 1 to each side. m = Simplify. Therefore, the equation is y = 1 x 1. y = mx + b Slope-intercept form 4 = (1) + b Replace m with, y with, and x with 1. = + b Multiply. 4 = b Add to each side. Therefore, the equation is y = x What is an equation of the line in the 7. The line whose equation is x y 1 has accompanying diagram? (1) y x (1) slope = ; y-intercept = 6 () y x () slope = ; y-intercept = 6 () y x () slope = ; y-intercept = (4) y x (4) slope = ; y-intercept = 6 8. Write the equation of the line whose slope is and y-intercept is 5. 7
8 9. Write the equation of the line that passes through the point 0, and has a slope of 4? 0. Write the equation of the line that contains the points 1, and 1,1? 8
9 Graphing Lines Slope-Intercept Form y = mx + b, where m is the slope and b is the y-intercept Graph x 4y = 8. x 4y = 8 Original equation 4y = x + 8 Subtract x from each side. 4y = x y = 4 x Divide each side by 4. Simplify. The y-intercept of y = x is and the slope is. So graph the point (0, ). From this point, move up units 4 4 and right 4 units. Draw a line passing through both points. 1. Find the equation in slope-intercept form that describes the line graphed below. (1) y = x 1 () y = x + () y = x 1 (4) y = x +. Which is the graph of x 4y = 6? (1) () () (4). Write an equation in slope-intercept form for the graph shown. 4. Graph 1 y x on the answer page. 5. Graph y x 1 (show work here, place graph on the answer page). 9
10 Perimeter and Circumference Perimeter and Circumference The distance around a figure. Perimeter and circumference is measured in linear units. Find the perimeter or Circumference. Round to the nearest tenth. a. b. c. P = a + b + c P = l+ w C = πr = = () + () = π(5) = 1 in. = 10 ft. = 10π = 1.4 in. 6. A rectangle is 6. inches long and 1.7 inches wide. Find its perimeter. (1) 7.9 in. () in. () 15.8 in. (4) in. 7. If a circle has a radius of 6 inches, what is the circumference rounded to the nearest whole number? (1) 19 in. () 8 in. () 11 in. (4) 76 in. Find the perimeter of each figure
11 Area Area The number of square units needed to cover a surface. Area is measured in square units. Find the area. Round to the nearest tenth. a. b. c. A = 1 bh A = lw A = πr = 1 (4)() = ()() = π(5) = 6 in = 6 ft = 5π = 78.5 in 41. What is the area of a circle whose diameter is 14 centimeters? (1) cm () cm () 4.98 cm (4) cm 4. Find the area of the trapezoid. (1) 588 m () 94 m () 60 m (4) 15 m Find the area of each figure
12 Volume Volume The measure of space occupied by a solid. Volume is measured in cubic units. 46. Find the volume. 47. A storage container in the shape of a right (1) 576 cubic yards circular cylinder is shown in the accompanying () 416 cubic yards diagram. What is the number of cubic inches in () 100 cubic yards the volume of this container? (4) 16 cubic yards (1) () () 51. (4) Find the volume of each solid to the nearest tenth
13 Pythagorean Theorem The Pythagorean Theorem - The side opposite the right angle in a right triangle is called the hypotenuse. This side is always the longest side of a right triangle. The other two sides are called the legs of the triangle. To find the length of any side of a right triangle, given the lengths of the other two sides, you can use the Pythagorean Theorem. Pythagorean Theorem If a and b are the measures of the legs of a right triangle and c is the measure of the hypotenuse, then c = a + b. Find the missing length. c = a + b Pythagorean Theorem c = a = 5 and b = 1 c = 169 c = 169 c = 1 Simplify. Take the square root of each side. Simplify. The length of the hypotenuse is Given the right triangle below, what is the 5. Which set of three numbers represent the length of the hypotenuse? lengths of the sides of a right triangle? (1) cm () 1.5 cm (1) {4, 5, 6} () {6, 8, 10} () 4.4 cm () {8, 9, 10} (4) {9, 16, 5} (4) 19.0 cm 5. Solve for x. 54. Calculate the length of the hypotenuse, to the nearest tenth. 55. The accompanying diagram shows a kite that has been secured to a stake in the ground with a 0-foot string. The kite is located 1 feet from the ground, directly over point X. What is the distance, in feet, between the stake and point X? 1
14 Simplifying Square Roots Product Property of Square Roots - The Product Property of Square Roots and prime factorization can be used to simplify expressions involving irrational square roots. Product Property of Square Roots For any numbers a and b, where a 0 and b 0, ab = a b. Simplify = 6 5 Find the greatest perfect square root factor of 180 = 6 5 Product Property of Square Roots = 6 5 Simplify. 56. Simplify 1 (1) 6 () () (4) Simplify 00 (1) 0 10 () 10 () 10 0 (4) Simplify Simplify Simplify
15 NAME points out of Score - Geometry Summer Assignment Answer Page Directions - For Multiple Choice questions, place the choice number in the corresponding provided space. For Short Answer questions, leave the work in the previous pages and just place the final answer in the corresponding space. Evaluating Algebraic Expressions Page Solving Equations Page Factoring Page Solving Quadratic Equations Page Slope Page Equations of Lines Page Graphing Lines Page
16 Perimeter and Circumference Page Area Page Volume Page Pythagorean Theorem Page Simplifying Square Roots Page
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# SOLVING EQUATIONS WITH FRACTIONS
The easiest way to solve equations with fractions is to eliminate fractions. Find the least common multiple of the denominators and multiply both sides of the equation by the least common multiple to get rid of the denominators. Once you get rid of the denominators, the fractions in the equation will be no more and the equation can be solved easily.
Example 1-9 : Solve for x.
Example 1 :
ˣ⁄₅ = 2
Solution :
ˣ⁄₅ = 2
In the equation above, there is only one denominator 5.
Multiply both sides of the equation by 5 to get rid of the denominator 5.
5(ˣ⁄₅) = 5(2)
x = 10
Example 2 :
³ˣ⁄₂ = 4.5
Solution :
³ˣ⁄₂ = 4.5
In the equation above, there is only one denominator 2.
Multiply both sides of the equation by 2 to get rid of the denominator 2.
2(³ˣ⁄₂) = 2(4.5)
3x = 9
Divide both sides of the equation by 3.
x = 3
Example 3 :
ˣ⁄₆ = ⁵⁄₉
Solution :
ˣ⁄₆ = ⁵⁄₉
In the equation above, there are two deominators 6 and 9.
Least common multiple of (6, 9) = 18.
Multiply both sides of the equation by 18 to get rid of the denominators .
18(ˣ⁄₆) = 18(⁵⁄₉)
3x = 2(5)
3x = 10
Divide both sides of the equation by 3.
x = ¹⁰⁄₃
Example 4 :
ˣ⁄₄ + 3 = ˣ⁄₆ + 7
Solution :
ˣ⁄₄ + 3 = ˣ⁄₆ + 7
In the equation above, there are two denominators 4 and 6.
Least common multiple of (4, 6) = 12.
Multiply both sides of the equation by 12 to get rid of the denominators 4 and 6.
12(ˣ⁄₄ + 3) = 12(ˣ⁄₆ + 7)
12(ˣ⁄₄) + 12(3) = 12(ˣ⁄₆) + 12(7)
3x + 36 = 2x + 84
Subtract 2x from both sides.
x + 36 = 84
Subtract 84 from both sides.
x = 48
Example 5 :
⁷ˣ⁄₁₀ + ³⁄₂ = ³ˣ⁄₅ + 2
Solution :
In the equation above, there are three denominators 10, 2 and 5.
Find the least common multiple of 10, 2 and 5 uising division method.
Least common multiple of (10, 2, 5) :
= 5 2 1 1 1
= 10
Multiply both sides of the equation by 10 to get rid of the denominators 10, 2 and 5.
10(⁷ˣ⁄₁₀ + ³⁄₂) = 10(³ˣ⁄₅ + 2)
Using Distributive Property.
10(⁷ˣ⁄₁₀) + 10(³⁄₂) = 10(³ˣ⁄₅) + 10(2)
7x + 5(3) = 2(3x) + 20
7x + 15 = 6x + 20
Subtract 6x from both sides.
x + 15 = 20
Subtract 15 from both sides.
x = 5
Example 6 :
ˣ⁄₇ - 6 = ³ˣ⁄₇ + 4
Solution :
ˣ⁄₇ - 6 = ³ˣ⁄₇ + 4
In the equation above, there is only one denominator, that is 7.
Multiply both sides of the equation by 7 to get rid of the denominator 7.
7(ˣ⁄₇ - 6) = 7(³ˣ⁄₇ + 4)
7(ˣ⁄₇) + 7(-6) = 7(³ˣ⁄₇) + 7(4)
x - 42 = 3x + 28
Subtract x from both sides.
-42 = 2x + 28
Subtract 28 from both sides.
-70 = 2x
Divide both sides by 2.
-35 = x
Example 7 :
⁴ˣ⁄₅ - ⁷⁄₄ = ˣ⁄₅ + ˣ⁄₄
Solution :
⁴ˣ⁄₅ - ⁷⁄₄ = ˣ⁄₅ + ˣ⁄₄
In the equation above, there are two different denominators 5 and 4.
Least common multiple of (5, 4) = 20.
Multiply both sides of the equation by 20 to get rid of the denominators 5 and 4.
20(⁴ˣ⁄₅ - ⁷⁄₄) = 20(ˣ⁄₅ + ˣ⁄₄)
Using Distributive Property.
20(⁴ˣ⁄₅) - 20(⁷⁄₄) = 20(ˣ⁄₅) + 20(ˣ⁄₄)
4(4x) - 5(7) = 4(x) + 5(x)
16x - 35 = 4x + 5x
16x - 35 = 9x
Subtract 9x from both sides.
7x - 35 = 0
7x = 35
Divide both sides by 7.
x = 5
Example 8 :
⁽²ˣ ⁻ ³⁾⁄₂ = ⁽⁻ˣ ⁻ ¹⁾⁄₄
Solution :
⁽²ˣ ⁻ ³⁾⁄₂ = ⁽⁻ˣ ⁻ ¹⁾⁄₄
In the equation above, there is only one fraction on each side. So, the equation can be solved by cross multiplication.
That is, numerator on the left side has to be multiplied by denominator on the right side and numerator on the right side has to be multiplied by denominator on the left side.
4(2x - 3) = 2(-x - 1)
Using Distributive Property,
8x - 12 = -2x - 2
10x - 12 = -2
10x = 10
Divide both sides by 10.
x = 1
Example 9 :
⁽⁴ˣ ⁺ ⁵⁾⁄₃ - ³ˣ⁄₂ = -x
Solution :
⁽⁴ˣ ⁺ ⁵⁾⁄₃ - ³ˣ⁄₂ = -x
In the equation above, there are two denominators 3 and 2.
Least common multiple of (3, 2) = 6.
Multiply both sides of the equation by 6 to get rid of the denominators 3 and 2.
6(⁽⁴ˣ ⁺ ⁵⁾⁄₃ - ³ˣ⁄₂) = 6(-x)
Using Distributive Property.
6(⁽⁴ˣ ⁺ ⁵⁾⁄₃) - 6(³ˣ⁄₂) = -6x
2(4x + 5) - 3(3x) = -6x
8x + 10 - 9x = -6x
-x + 10 = -6x
5x + 10 = 0
Subtract 10 from both sides.
5x = -10
Divide both sides by 5.
x = -2
Example 10 :
Two times the reciprocal of a number and 3 add up to ¹⁷⁄₅. What is the number?
Solution :
Let x be the number.
It is given that the sum two times the reciprocal of the number and 3 is equal to ¹⁷⁄₅.
2(¹⁄ₓ) + 3 = ¹⁷⁄₅
²⁄ₓ + 3 = ¹⁷⁄₅
Multiply both sides of the equation by 5x to get rid of the denominators x and 5.
5x(²⁄ₓ + 3) = 5x(¹⁷⁄₅)
5x(²⁄ₓ) + 5x(3) = 17x
10 + 15x = 17x
Subtract 15x from both sides.
10 = 2x
Divide both sides by 2.
5 = x
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# Basic Formulas for Two- and Three-Dimensional Figures
Instructor: David Karsner
Geometric measurements can be taken for one-, two-, and three-dimensional shapes. This lesson explores the most common formulas one would use to find perimeter, area, surface area, and volume of several two- and three-dimensional figures.
## How Big is the Pool?
You have been given the responsibility of adding the right amount of chemicals to the pool to keep it safe. The only problem is that the directions on the chemical box are based on how large the pool is, and you don't know how large it is. The pool is in the shape of a box. You have a one cubic foot bucket and a 50 foot measuring tape. How would you determine the size of the pool?
One option is to dip the water out the pool with the one cubic foot bucket and keep count on how many buckets you dip out of the pool. This option is very time consuming and what would you do with all the water you are dipping out of the pool.
Another option is to use the formula for the volume of a prism. That just means making three measurements and doing a little multiplication. Which option sounds better?
Using a formula to find volume or area is a handy short-cut in a situation like this. This lesson will give you those formulas and explain how they are used.
## One-, Two-, and Three-Dimensional Measurements
One of the main tasks in geometry is measuring figures. You can make these measurements in one-, two-, and three-dimensional space. In one dimension, you are measuring distance (length). In two dimensions, you are measuring area (length and width). In three dimensions, you are measuring volume (length, width, and height).
Even when a figure is two- or three-dimensional, you can take measurements of the lower dimensions as well. Let's use a cylinder as an example. It is a three-dimensional figure. You can find the volume of the cylinder (a three-dimensional measurement), the area of the circular base (a two-dimensional measurement) and the height of the cylinder (a one-dimensional measurement).
## Finding Distance
Distance is a one-dimensional measurement of the amount of space between two points. If you start at point A and move to point B, how far have you moved? The units for distance (length) are given in single units, such as 14 miles, 200 feet, 32 inches, 55 meters, or 200 kilometers.
Even though distance is a one-dimensional measurement, the measure of distance can be used for two- and three-dimensional figures. The perimeter of a rectangle, the radius of a circle, the height of a pyramid are all examples of one-dimensional measurements for two- or three-dimensional figures. Distance is often used to describe perimeters, or the space around the outside of a shape. Circumference is the name given to the perimeter of a circle. Any shape that includes a circle will use Pi (3.14) as part of the formula.
## Finding Area
Area is the amount of space that a two-dimensional shape takes up. Since it is two-dimensional, the units will always be square units (for example, 12 sq. feet or 25 sq. centimeters). To find the area of something, you start by multiplying the two dimensions together.
To find the area of a rectangle, you multiply the length by the width. Although area is a two-dimensional measurement, it can also be used with three-dimensional objects. You can find the area of the base (two-dimensional) of a cone (three-dimensional).
## Finding Volume
Volume is the amount of space inside of a three-dimensional object. Since it is three-dimensional, volume measurements will always be in cubic units (such as cubic feet or cubic liters). To find the volume of many three-dimensional shapes, you begin by finding the area (two-dimensional) of the base and multiplying it by the height.
## Solving Problems
To find measurements of perimeter, area, and volume follow these steps:
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# Centripetal Acceleration and Circular Motion
## Presentation on theme: "Centripetal Acceleration and Circular Motion"— Presentation transcript:
Centripetal Acceleration and Circular Motion
SPH4UIW Centripetal Acceleration and Circular Motion ->
The Circle Babylonian Numbers
And you thought your homework was difficult ->
Round Round
Uniform Circular Motion
What does it mean? How do we describe it? What can we learn about it? ->
Circular Motion Question
B A C v Use this to motivate circular motion involves acceleration. Answer: B A ball is going around in a circle attached to a string. If the string breaks at the instant shown, which path will the ball follow? ->
What is Uniform Circular Motion?
v x y (x,y) Motion in a circle with: Constant Radius R Constant Speed v = |v| ->
Why do we Feel a Force Toward the Centre?
v x y (cos(θ),sin(θ)) Calculus gives us a clue a ->
How can we describe UCM? In general, one coordinate system is as good as any other: Cartesian: (x,y) [position] (vx ,vy) [velocity] Polar: (R,) [position] (vR ,) [velocity] In UCM: R is constant (hence vR = 0). (angular velocity) is constant. Polar coordinates are a natural way to describe UCM! R v x y (x,y) ->
Polar Coordinates: 1 revolution = 2p radians
The arc length s (distance along the circumference) is related to the angle in a simple way: s = R, where is the angular displacement. units of are called radians. For one complete revolution (c): 2R = Rc c = 2 has period 2. y v (x,y) s R x 1 revolution = 2p radians ->
Velocity of UCM in Polar Coordinates
This is my way of saying velocity is the change of position over the change of time. In Cartesian coordinates, we say velocity dx/dt = v. x = vt (if v is constant) In polar coordinates, angular velocity d/dt = . = t (if w is constant) has units of radians/second. Displacement s = vt. but s = R = Rt, so: y v R s t x ->
Period and Frequency of UCM
Recall that 1 revolution = 2 radians frequency (f) = revolutions / second (a) angular velocity () = radians / second (b) By combining (a) and (b) = 2 f Realize that: period (T) = seconds / revolution So T = 1 / f = 2/ R v s = 2 / T = 2f
Recap of UCM: x = R cos()= R cos(t) y = R sin()= R sin(t)
= arctan (y/x) = t s = v t s = R = Rt v = R v (x,y) R s t
Acceleration in Uniform Circular Motion
Centripetal acceleration Centripetal force: Fc = mv2/R v2 v1 R R v v1 v2 aave= Dv / Dt Acceleration inward Acceleration is due to change in direction, not speed. Since turns “toward” center, acceleration is toward the center. ->
Definitions Uniform Circular Motion: occurs when an object has constant speed and constant radius Centripetal Acceleration (or radial acceleration, ac): the instantaneous acceleration towards the centre of the circle Centrifugal Force: fictitious force that pushes away from the centre of a circle in a rotating frame of reference (which is noninertial) ->
T – period (not to be confused with tension) f - frequency R – radius
Equations to know These are the equations for centripetal acceleration (which we will derive this class) T – period (not to be confused with tension) f - frequency R – radius v – speed ->
Dynamics of Uniform Circular Motion
Consider the centripetal acceleration aR of a rotating mass: The magnitude is constant. The direction is perpendicular to the velocity and inward. The direction is continually changing. Since aR is nonzero, according to Newton’s 2nd Law, there must be a force involved. ->
Consider a ball on a string:
There must be a net force force in the radial direction for it to move in a circle. Other wise it would just fly out along a straight line, with unchanged velocity as stated by Newton’s 1st Law Don’t confuse the outward force on your hand (exerted by the ball via the string) with the inward force on the ball (exerted by your hand via the string). That confusion leads to the mis-statement that there is a “centrifugal” (or center-fleeing) force on the ball. That’s not the case at all! ->
Deriving centripetal acceleration equations
A particle moves from position r1 to r2 in time Δt Because v is always perpendicular to r, the angle between v1 and v2 is also θ. Start with equation for magnitude of instantaneous acceleration ->
Deriving centripetal acceleration equations
For Δr, find r2 - r1 For Δv, find v2 - v1 Notice: Similar triangles! The ratios of sides are the same for both triangles! ->
Deriving centripetal acceleration equations
Ratios of similar Triangles are the same Sub this into our original acceleration equation. ->
Deriving centripetal acceleration equations
Okay, everything is straightforward now except this thing. But hey! That’s just the magnitude of the instantaneous velocity! (also called speed, which is constant for uniform circular motion) ->
Still deriving centripetal acceleration
Finally… a much nicer equation. But what if we don’t know speed v? ->
Almost done now The period (T) is the time it take to make a full rotation ->
And here’s the last equation
Frequency is the number of rotations in a given time. It is often measured in hertz (Hz) If the particle has a frequency of 100Hz, then it makes 100 rotations every second ->
Preflights “Gravity, Normal, Friction” FN f W SF = ma = mv2/R
Consider the following situation: You are driving a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Body Diagram (FBD) for the car. How many forces are acting on the car? A) B) C) D) E) 5 W FN correct SF = ma = mv2/R a=v2/R R f “Fn = Normal Force, W = Weight, the force of gravity, f = centripetal force.” “Gravity, Normal, Friction” ->
Preflights Consider the following situation: You are driving a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Body Diagram (FBD) for the car. The net force on the car is W FN A. Zero B. Pointing radially inward C. Pointing radially outward SF = ma = mv2/R a=v2/R R f correct If there was no inward force then the car would continue in a straight line. ->
ACT Suppose you are driving through a valley whose bottom has a circular shape. If your mass is m, what is the magnitude of the normal force FN exerted on you by the car seat as you drive past the bottom of the hill A. FN < mg B. FN = mg C. FN > mg a=v2/R R correct FN v SF = ma FN - mg = mv2/R FN = mg + mv2/R mg ->
Roller Coaster Example
What is the minimum speed you must have at the top of a 20 meter diameter roller coaster loop, to keep the wheels on the track. Y Direction: F = ma -N – mg = -m a -N – mg = -m v2/R Let N = 0, just touching -mg = -m v2/R g = v2 / R v = (gR) v = (9.8)(10) = 9.9 m/s N mg Do ruler next….. How far did it go…. ->
Merry-Go-Round ACT Bonnie sits on the outer rim of a merry-go-round with radius 3 meters, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds. Klyde’s speed is: Klyde Bonnie (a) the same as Bonnie’s (b) twice Bonnie’s (c) half Bonnie’s Bonnie travels 2 p R in 2 seconds vB = 2 p R / 2 = 9.42 m/s Klyde travels 2 p (R/2) in 2 seconds vK = 2 p (R/2) / 2 = 4.71 m/s ->
Merry-Go-Round ACT II Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds. Klyde’s angular velocity is: Klyde Bonnie (a) the same as Bonnie’s (b) twice Bonnie’s (c) half Bonnie’s The angular velocity w of any point on a solid object rotating about a fixed axis is the same. Both Bonnie & Klyde go around once (2p radians) every two seconds. ->
Problem: Motion in a Circle
A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v. What is the tension T in the string at the top of the rock’s trajectory? v T R
Motion in a Circle... Draw a Free Body Diagram (pick y-direction to be down): We will use FNET = ma (surprise) First find FNET in y direction: FNET = mg +T y mg T
Motion in a Circle... FNET = mg +T Acceleration in y direction:
ma = mv2 / R mg + T = mv2 / R T = mv2 / R - mg v y mg T F = ma R
Motion in a Circle... What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp? i.e. find v such that T = 0. mv2 / R = mg + T v2 / R = g Notice that this does not depend on m. v mg T= 0 R
Lecture 6, Act 3 Motion in a Circle
A skier of mass m goes over a mogul having a radius of curvature R. How fast can she go without leaving the ground? R mg N v (a) (b) (c)
Lecture 6, Act 3 Solution mv2 / R = mg - N For N = 0: v N mg R
Example: Force on a Revolving Ball
As shown in the figure, a ball of mass kg fixed to a string is rotating with a period of T=0.500s and at a radius of m. What is the force the person holding the ball must exert on the string?
Looking at just the x component then we have a pretty simple result:
As usual we start with the free-body diagram. Note there are two forces gravity or the weight, mg tensional force exerted by the string, FT We’ll make the approximation that the ball’s mass is small enough that the rotation remains horizontal, f=0. (This is that judgment aspect that’s often required in physics.)
Example : A Vertically Revolving Ball
Now lets switch the orientation of the ball to the vertical and lengthen the string to 1.10 m. For circular motion (constant speed and radius), what’s the speed of the ball at the top? What’s the tension at the bottom if the ball is moving twice that speed?
tensional force exerted by the string, FTA gravity or the weight, mg
So to the free-body diagram, at the top, at point A, there are two forces: tensional force exerted by the string, FTA gravity or the weight, mg In the x direction: Let’s talk about the dependencies of this equation. Since mg is constant, the tension will be larger should vA increase. This seems intuitive. Now the ball will fall if the tension vanishes or if FTA is zero +x
At point B there are also two forces but both acting in opposite directions. Using the same coordinate system. Note that the tension still provides the radial acceleration but now must also be larger than maR to compensate for gravity. +x
Forces on a Swinging Weight Part 1
A mass is hanging off of two ropes, one vertical and one at an angle θ of 30°. The mass is 20 kg. What is the tension in the angled rope? Gravity is the force pulling down (vertical). Therefore the matching force pulling up is the tension in the vertical rope. The angled rope will have zero tension (it plays no role in holding up the mass).
Forces on a Swinging Weight Part 2
A mass is hanging off of two ropes, one vertical and one at an angle θ of 30°. The mass is 20 kg. What is the tension in the angled rope the instant the vertical rope is cut? Gravity is the force pulling down (vertical). Therefore the matching force pulling up is the tension in the angled rope. FG
Turns out this stuff is actually useful for civil engineering such as road design A NASCAR track Let’s consider a car taking a curve, by now it’s pretty clear there must be a centripetal forces present to keep the car on the curve or, more precisely, in uniform circular motion. This force actually comes from the friction between the wheels of the car and the road. Don’t be misled by the outward force against the door you feel as a passenger, that’s the door pushing you inward to keep YOU on track!
Example: Analysis of a Skid
The setup: a 1000kg car negotiates a curve of radius 50m at 14 m/s. The problem: If the pavement is dry and ms=0.60, will the car make the turn? How about, if the pavement is icy and ms=0.25?
First off, in order to maintain uniform circular motion the centripetal force must be:
To find the frictional force we start with the normal force, from Newton’s second law: +x +y Looking at the car head-on the free-body diagram shows three forces, gravity, the normal force, and friction. We see only one force offers the inward acceleration needed to maintain circular motion - friction.
Circular Car Ramp
Back to the analysis of a skid.
Since v=0 at contact, if a car is holding the road, we can use the static coefficient of friction. If it’s sliding, we use the kinetic coefficient of friction. Remember, we need 3900N to stay in uniform circular motion. Static friction force first: Now kinetic,
The Theory of Banked Curves
The Indy picture shows that the race cars (and street cars for that matter) require some help negotiating curves. By banking a curve, the car’s own weight, through a component of the normal force, can be used to provide the centripetal force needed to stay on the road. In fact for a given angle there is a maximum speed for which no friction is required at all. From the figure this is given by
Example: Banking Angle
Problem: For a car traveling at speed v around a curve of radius r, what is the banking angle q for which no friction is required? What is the angle for a 50km/hr (14m/s) off ramp with radius 50m? To the free-body diagram! Note that we’ve picked an unusual coordinate system. Not down the inclined plane, but aligned with the radial direction. That’s because we want to determine the component of any force or forces that may act as a centripetal force. We are ignoring friction so the only two forces to consider are the weight mg and the normal force FN . As can be seen only the normal force has an inward component.
As we discussed earlier in the horizontal or + x direction, Newton’s 2nd law leads to:
In the vertical direction we have: Since the acceleration in this direction is zero, solving for FN Note that the normal force is greater than the weight. This last result can be substituted into the first: For v=14m/s and r= 50m
Nice to know: Angular Acceleration
Angular acceleration is the change in angular velocity w divided by the change in time. If the speed of a roller coaster car is 15 m/s at the top of a 20 m loop, and 25 m/s at the bottom. What is the car’s average angular acceleration if it takes 1.6 seconds to go from the top to the bottom? = 0.64 rad/s2 ->
Constant angular acceleration summary (with comparison to 1-D kinematics)
Angular Linear And for a point at a distance R from the rotation axis: x = Rv = R a = R ->
CD Player Example Nice to Know
The CD in your disk player spins at about 20 radians/second. If it accelerates uniformly from rest with angular acceleration of 15 rad/s2, how many revolutions does the disk make before it is at the proper speed? Dq = 13.3 radians 1 Revolutions = 2 p radians Dq = 13.3 radians = 2.12 revolutions ->
Summary of Concepts Uniform Circular Motion Speed is constant
Direction is changing Acceleration toward center a = v2 / r Newton’s Second Law F = ma Circular Motion q = angular position radians w = angular velocity radians/second a = angular acceleration radians/second2 Linear to Circular conversions s = r q Uniform Circular Acceleration Kinematics Similar to linear! -> |
## College Algebra (11th Edition)
$t=-\dfrac{\log\dfrac{T-T_0}{T_1-T_0}}{k}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $T=T_0+(T_1-T_0)10^{-kt} ,$ for $t ,$ use the properties of equality to isolate the base that uses the needed variable. Then take the logarithm of both sides. Use the properties of logarithms and of equality to isolate the needed variable. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the base that uses $t ,$ the equation above is equivalent to \begin{array}{l}\require{cancel} T-T_0=(T_1-T_0)10^{-kt} \\\\ \dfrac{T-T_0}{(T_1-T_0)}=\dfrac{(T_1-T_0)10^{-kt}}{(T_1-T_0)} \\\\ \dfrac{T-T_0}{T_1-T_0}=10^{-kt} \\\\ 10^{-kt}=\dfrac{T-T_0}{T_1-T_0} .\end{array} Taking the logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log10^{-kt}=\log\dfrac{T-T_0}{T_1-T_0} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent \begin{array}{l}\require{cancel} -kt\log10=\log\dfrac{T-T_0}{T_1-T_0} .\end{array} Since $\log 10=1$ the equation above is equivalent to \begin{array}{l}\require{cancel} -kt(1)=\log\dfrac{T-T_0}{T_1-T_0} \\\\ -kt=\log\dfrac{T-T_0}{T_1-T_0} \\\\ \dfrac{-kt}{-k}=\dfrac{\log\dfrac{T-T_0}{T_1-T_0}}{-k} \\\\ t=-\dfrac{\log\dfrac{T-T_0}{T_1-T_0}}{k} .\end{array} |
## Sunday, November 11, 2007
### Blocks and Towers solution
See the puzzle
Let's say that you've found every single way to build a 1x10 tower. Imagine that you've lined up each of these solutions. Let's remove the top block from each 1x10 tower. For some of them, you will be removing a 1x2 block, and for others, you will be removing a 1x1 block. You will be left with a bunch of 1x8 towers and 1x9 towers.
But here's the crucial point. The 1x8 and 1x9 towers constitute the complete set of solutions for towers of those sizes. No matter what kind of 1x8 tower you can build, you can always add a 1x2 block on top to get a unique 1x10 tower. Similarly, you can always add a 1x1 block on top of a 1x9 tower. Therefore, the total number of unique 1x10 towers you can build is equal to the sum of the total number of 1x8 towers and 1x9 towers.
In fact, this isn't just true of 1x10 towers. The number of ways to build any 1xN tower is equal to the sum of the ways to build 1x(N-1) and 1x(N-2) towers. So if we start with 1x1 towers (1 way) and 1x2 towers (2 ways), we can proceed to calculate the total number of ways for any size of tower:
1x1: 1
1x2: 2
1x3: 3
1x4: 5
1x5: 8
1x6: 13
1x7: 21
1x8: 34
1x9: 55
1x10: 89
This sequence of numbers is known as the Fibonacci sequence. I actually mentioned this sequence not-so-subtly in a previous post, and was even kind enough to give a closed expression for it. The number of ways to build a 1xN tower is equal to F(N+1).
The Fibonacci sequence is one of those things that seems to show up everywhere. It is closely related to the number called "the golden ratio", which I'm sure most people have heard about. It turns up all the time in nature. |
# Understanding the number systems
##### Intros
###### Lessons
1. Introduction to number systems
##### Examples
###### Lessons
1. State the number systems each of the following belongs to
1. $\sqrt{4}$
0
6.$\overline{78}$
5.76204...
-0.45454545…
$\pi$
2. -7
0.$\bar{5}$
0.403850
7.898989…
$\sqrt{8}$
-0.74867…
##### Practice
###### Topic Notes
Numbers can be categorized more specifically into different number systems, for instances, natural numbers, whole numbers, integers, rational numbers and irrational numbers. In this session, we will learn how to classify numbers into different number systems.
## Introduction to Number Systems
Welcome to our exploration of number systems! Let's start with a quick introduction video that will set the foundation for our journey. This video is crucial as it provides a visual understanding of the concepts we'll dive into. Now, imagine numbers as a big family with different branches. We have natural numbers, which are the counting numbers we learn as kids. Then there are whole numbers, which include zero along with natural numbers. Integers expand this further, bringing negative numbers into the mix. Rational numbers are like fractions, representing parts of a whole. And don't forget about irrational numbers, those mysterious numbers that can't be expressed as simple fractions! Each system has its unique properties and uses in mathematics. As we progress, we'll uncover how these systems interrelate and why they're essential in various mathematical applications of number systems. Ready to embark on this numerical adventure?
## Natural Numbers and Whole Numbers
Natural numbers form the foundation of our counting system, representing the most basic and intuitive way we quantify objects in our world. These numbers start from 1 and continue infinitely, allowing us to count as high as we can imagine. Mathematicians use the symbol 'N' to denote the set of natural numbers, which includes 1, 2, 3, 4, 5, and so on, without any end.
The concept of natural numbers is deeply rooted in human history, as our ancestors used these numbers for basic counting and record-keeping. They are called "natural" because they come naturally to us when we start counting objects around us. For instance, when counting apples in a basket, we naturally begin with one, then two, three, and so forth.
Whole numbers, on the other hand, expand upon the concept of natural numbers by including zero in the set. Mathematicians represent whole numbers with the symbol 'W'. This set includes all natural numbers plus zero, making it slightly more comprehensive than the set of natural numbers. In other words, W = {0, 1, 2, 3, 4, ...}.
The inclusion of zero in the number system is a fascinating aspect of mathematical history. Unlike natural numbers, which have been used since ancient times, the concept of zero as a number is a relatively recent invention. Many early civilizations struggled with the idea of representing "nothing" as a quantity. The ancient Babylonians used a placeholder symbol for zero around 300 BCE, but it wasn't until the 7th century CE that Indian mathematicians fully developed zero as a number in its own right.
To illustrate the difference between natural and whole numbers, let's consider a simple example. Imagine you're counting the number of students in a classroom. You would use natural numbers: 1, 2, 3, and so on. However, if you're asked how many elephants are in the classroom, the answer would be 0 a whole number that isn't a natural number.
The introduction of zero revolutionized mathematics and opened up new possibilities in various fields, including physics and computer science. In everyday life, we use whole numbers more frequently than we might realize. For instance, when we talk about temperature, we often use negative numbers, zero, and positive numbers all of which are part of the whole number system when we're dealing with integer values.
Understanding the distinction between natural numbers (N) and whole numbers (W) is crucial for students as they progress in their mathematical education. Natural numbers are perfect for basic counting and understanding the concept of quantity, while whole numbers introduce the important concept of zero, preparing students for more advanced mathematical ideas like negative numbers and the number line.
As you delve deeper into mathematics, you'll find that these fundamental number sets form the basis for more complex number systems, such as integers (which include negative whole numbers), rational numbers, and real numbers. Each expansion of our number system allows us to describe and analyze more complex phenomena in the world around us.
In conclusion, natural numbers (N) and whole numbers (W) are essential building blocks in mathematics. While natural numbers give us the ability to count and quantify, whole numbers introduce the concept of zero, bridging the gap between positive quantities and the absence of quantity. This understanding forms a crucial foundation for further mathematical exploration and problem-solving skills.
## Integers
Integers represent the next logical step in our journey through number systems, building upon the foundation of positive and negative whole numbers. These fascinating mathematical entities encompass both positive and negative whole numbers, along with zero. By introducing integers, we expand our numerical toolkit to include numbers that extend below zero, opening up a whole new world of mathematical possibilities.
To understand integers, let's first recall what we know about whole numbers. Whole numbers are the counting numbers we use in everyday life: 0, 1, 2, 3, and so on. Now, imagine a number line with zero at the center. To the right, we have our familiar positive whole numbers stretching infinitely. But what about the left side of zero? This is where integers come into play.
Integers include all whole numbers, both positive and negative, as well as zero. So, our number line now extends infinitely in both directions. To the left of zero, we have -1, -2, -3, and so on, representing negative integers. These negative numbers allow us to express concepts like debt, temperature below freezing, or depths below sea level.
In mathematical notation, we use the symbol 'Z' to represent the set of all integers. You might wonder why 'Z' is used instead of 'I'. Interestingly, this symbol comes from the German word 'Zahlen', which means 'numbers'. This connection to the German language is a nod to the contributions of German mathematicians in developing number theory.
Let's look at some examples of integers: ..., -3, -2, -1, 0, 1, 2, 3, ... As you can see, integers include negative numbers, zero, and positive numbers. They're used in various real-world scenarios. For instance, when tracking a bank account balance, positive integers represent deposits, while negative integers indicate withdrawals or overdrafts. In temperature measurements, positive integers might represent degrees above freezing, while negative integers show temperatures below freezing.
It's crucial to remember that integers are always whole numbers. They don't include fractions or decimals. So, while 5 and -7 are integers, numbers like 3.5 or -2¾ are not. This distinction is important as we continue to explore more complex number systems.
Integers play a vital role in mathematics and everyday life. They allow us to represent and calculate with negative quantities, which is essential in fields like finance, physics, and engineering. By understanding integers, we gain the ability to work with a broader range of numbers and solve more complex problems.
As we delve deeper into the world of mathematics, integers serve as a crucial stepping stone. They bridge the gap between basic counting numbers and more advanced number systems, paving the way for concepts like rational and real numbers. By mastering integers, you're equipping yourself with a powerful tool that will serve you well in both academic and real-world applications.
Remember, every time you use a number line, balance a checkbook, or measure temperature, you're working with integers. They're all around us, silently helping us make sense of the world. So the next time you encounter a negative number or see the symbol 'Z' in a math problem, you'll know you're dealing with the fascinating world of integers the whole numbers that go both ways!
## Rational Numbers
Welcome to the fascinating world of rational numbers! As your friendly math tutor, I'm excited to guide you through this important concept. Rational numbers are a fundamental part of mathematics, and understanding them will open up a whole new realm of mathematical possibilities.
So, what exactly are rational numbers? Simply put, rational numbers are numbers that can be expressed as fractions. This means they can be written as one integer divided by another (non-zero) integer. The term "rational" comes from the word "ratio," which makes sense when you think about fractions as ratios between two numbers.
Rational numbers encompass a wide range of number types that you're likely already familiar with. These include:
• Integers (like -3, 0, 5)
• Fractions (such as 1/2, 3/4, -5/6)
• Terminating decimals (0.5, 2.75, -1.25)
• Repeating decimals (0.333..., 0.181818...)
• Mixed numbers (2 1/2, 3 3/4)
• Percentages (25%, 75%, 150%)
In mathematical notation, we use the symbol 'Q' to represent the set of all rational numbers. This symbol comes from the word "quotient," which is another term for the result of division fitting, since rational numbers are essentially quotients of integers!
Let's look at some examples to better understand rational numbers:
• 5 is a rational number because it can be written as 5/1
• -3/4 is a rational number (it's already in fraction form)
• 0.25 is rational because it can be written as 1/4
• 0.333... (repeating) is rational because it can be written as 1/3
• 75% is rational because it can be written as 75/100 or 3/4
One of the useful skills when working with rational numbers is converting between different representations. Let's practice converting between fractions and decimals:
1. To convert a fraction to a decimal, simply divide the numerator by the denominator:
• 3/4 = 3 ÷ 4 = 0.75
• 5/8 = 5 ÷ 8 = 0.625
2. To convert a terminating decimal to a fraction:
• 0.6 = 6/10 = 3/5
• 0.125 = 125/1000 = 1/8
3. For repeating decimals, it's a bit trickier, but still doable:
• 0.333... = 1/3
• 0.181818... = 18/99 = 2/11
Understanding rational numbers is crucial because they form the basis for many mathematical operations and concepts you'll encounter in algebra, calculus, and beyond. They allow us to express parts of wholes, perform precise calculations, and model real-world situations accurately.
As you continue your mathematical journey, you'll discover that not all numbers are rational. There's a whole category of irrational numbers (like π and 2) that can't be expressed as simple fractions. But that's a topic for another day!
Remember, every time you work with fractions, decimals, or percentages in your daily life whether you're calculating a tip, measuring ingredients for a recipe, or figuring out a discount you're using rational numbers. They're all around us, making
## Irrational Numbers
Irrational numbers are a fascinating subset of real numbers that cannot be expressed as simple fractions. Unlike their rational counterparts, irrational numbers have decimal representations that go on forever without repeating or terminating. This unique property makes them both intriguing and challenging to work with in mathematics.
To understand irrational numbers better, let's explore some famous examples. Perhaps the most well-known irrational number is pi (π), which represents the ratio of a circle's circumference to its diameter. Pi's decimal representation begins with 3.14159... and continues infinitely without repeating. Another famous irrational number is the square root of 2 (2), which cannot be expressed as a fraction and has a non-repeating decimal representation of approximately 1.41421356...
Recognizing irrational numbers can be tricky, but there are some telltale signs. Numbers involving square roots are often irrational, unless the number under the radical sign is a perfect square. For example, 4 is rational because it equals 2, but 3 is irrational. The golden ratio (φ), approximately 1.61803..., is another well-known irrational number that appears in nature and art.
In mathematical notation, we use the symbol ' (Q with an apostrophe) to represent the set of irrational numbers. This symbol is the complement of , which represents rational numbers. It's important to note that every real number is either rational or irrational, but not both.
When working with irrational numbers, it's crucial to understand that we can only approximate their values in calculations. For instance, when using pi in computations, we typically round it to a certain number of decimal places, like 3.14 or 3.14159, depending on the required precision.
Irrational numbers play a significant role in various fields of mathematics and science. In geometry, they're essential for calculating the lengths of diagonals in squares and the volumes of certain shapes. In music theory, irrational numbers are involved in the frequencies of musical notes in some tuning systems.
To help visualize irrational numbers, imagine a number line. Rational numbers can be plotted as exact points on this line, while irrational numbers fill in the gaps between these points. This concept demonstrates that there are infinitely many irrational numbers between any two rational numbers.
As you delve deeper into mathematics, you'll encounter more irrational numbers and learn about their properties. Some other examples include e (Euler's number), which is fundamental in calculus and exponential growth, and certain logarithms, like the natural logarithm of 2.
Remember, while approximating irrational numbers can be challenging, they are an essential part of mathematics. Their existence helps us understand the complexity and beauty of the number system. As you continue your mathematical journey, you'll find that irrational numbers open up new realms of understanding in various areas of study, from basic algebra to advanced calculus and beyond.
## Real Numbers
Real numbers are a fundamental concept in mathematics, encompassing all rational and irrational numbers. Represented by the symbol 'R', real numbers form the backbone of many mathematical operations and real-world applications. To understand real numbers, imagine a number line stretching infinitely in both directions. Every point on this line corresponds to a real number, creating a continuous and unbroken representation of all possible numerical values.
The beauty of real numbers lies in their ability to represent both exact quantities (like 5 or -3.14) and never-ending decimal expansions (such as π or 2). This versatility makes real numbers essential in various fields, from basic arithmetic to advanced calculus. In everyday life, we encounter real numbers constantly, whether we're measuring ingredients for a recipe, calculating distances for a road trip, or analyzing financial data.
One of the most significant properties of real numbers is their continuity, which is visually represented through continuous graphs. These graphs have no breaks or jumps, reflecting the seamless nature of real numbers. This continuity is crucial in modeling real-world phenomena, such as temperature changes, population growth, or the motion of objects.
In mathematics, real numbers play a pivotal role in numerous areas. For instance, in algebra, they form the basis for solving equations and inequalities. In geometry, real numbers are used to measure lengths, areas, and volumes. In calculus, the concept of limits, derivatives, and integrals all rely on the properties of real numbers.
The applications of real numbers extend far beyond pure mathematics. In physics, real numbers are used to describe physical quantities like mass, velocity, and energy. In engineering, they're essential for precise measurements and calculations. Even in computer science, real numbers (albeit in approximated form) are crucial for various algorithms and simulations.
Understanding real numbers opens up a world of mathematical possibilities. They allow us to express exact values, approximate irrational quantities, and model complex systems. Whether you're balancing a checkbook, analyzing stock market trends, or studying advanced mathematics, real numbers are an indispensable tool.
As students progress in their mathematical journey, they'll find that real numbers form the foundation for more advanced concepts. From complex numbers to multivariable calculus, a solid grasp of real numbers is essential. By mastering this concept, students equip themselves with a powerful tool for understanding and describing the world around them, both mathematically and practically.
In conclusion, real numbers, symbolized by 'R', are a vast and inclusive set that forms the backbone of mathematics. Their representation on the number line, their role in continuous graphs, and their wide-ranging applications make them a crucial concept to understand. As we navigate through both everyday calculations and complex mathematical problems, real numbers remain our constant companions, helping us make sense of quantities, measurements, and abstract concepts alike.
## Conclusion
Understanding different number systems is crucial in mathematics and everyday life. Natural numbers form the foundation, representing counting numbers. Whole numbers include zero, expanding the concept. Integers incorporate negative numbers, essential for representing debts or temperatures below zero. Rational numbers, expressed as fractions, cover a wide range of everyday calculations. Irrational numbers, like pi, have endless decimal representations and are vital in geometry and advanced mathematics. Real numbers encompass all these categories, forming a continuous number line. Each system builds upon the previous, increasing complexity and applicability. Mastering these number systems enhances problem-solving skills, improves financial literacy, and aids in scientific understanding. From basic arithmetic to complex calculations, these number systems play a role in various fields, including engineering, physics, and economics. Recognizing the relationships between these systems provides a comprehensive view of mathematics, enabling more effective analysis and decision-making in both academic and real-world scenarios.
In summary, the study of whole numbers and their properties is fundamental. Equally important is the understanding of rational numbers and their applications in various calculations. The distinction between rational and irrational numbers is crucial for higher-level mathematics and real-world problem-solving.
### Example:
State the number systems each of the following belongs to
$\sqrt{4}$
0
6.$\overline{78}$
5.76204...
-0.45454545
$\pi$
#### Step 1: Understanding the Number Systems
Before we dive into categorizing each number, it's essential to understand the different number systems. The primary number systems include:
• Natural Numbers: These are positive integers starting from 1, 2, 3, and so on.
• Whole Numbers: These include all natural numbers along with 0.
• Integers: These include all whole numbers and their negative counterparts, such as -1, -2, -3, etc.
• Rational Numbers: These are numbers that can be expressed as a fraction of two integers, where the denominator is not zero. They include terminating and repeating decimals.
• Irrational Numbers: These are numbers that cannot be expressed as a simple fraction. Their decimal expansions are non-terminating and non-repeating.
#### Step 2: Analyzing $\sqrt{4}$
First, we need to calculate the value of $\sqrt{4}$. The square root of 4 is 2. Now, let's categorize the number 2:
• 2 is a Natural Number.
• 2 is a Whole Number.
• 2 is an Integer.
• 2 is a Rational Number because it can be expressed as $\frac{2}{1}$.
#### Step 3: Analyzing 0
Next, we look at the number 0:
• 0 is not a Natural Number.
• 0 is a Whole Number.
• 0 is an Integer.
• 0 is a Rational Number because it can be expressed as $\frac{0}{1}$.
#### Step 4: Analyzing $6.\overline{78}$
The number $6.\overline{78}$ indicates that the digits 78 repeat indefinitely. Let's categorize it:
• It is not a Natural Number.
• It is not a Whole Number.
• It is not an Integer.
• It is a Rational Number because it has a repeating decimal pattern.
#### Step 5: Analyzing $5.76204...$
The number $5.76204...$ has a non-terminating and non-repeating decimal expansion. Therefore:
• It is not a Natural Number.
• It is not a Whole Number.
• It is not an Integer.
• It is an Irrational Number because its decimal expansion does not repeat.
#### Step 6: Analyzing $-0.45454545$
The number $-0.45454545$ has a repeating decimal pattern of 45. Let's categorize it:
• It is not a Natural Number.
• It is not a Whole Number.
• It is not an Integer.
• It is a Rational Number because it has a repeating decimal pattern.
#### Step 7: Analyzing $\pi$
The number $\pi$ (pi) is approximately 3.14159 and has a non-terminating, non-repeating decimal expansion. Therefore:
• It is not a Natural Number.
• It is not a Whole Number.
• It is not an Integer.
• It is an Irrational Number because its decimal expansion does not repeat.
### FAQs
1. What is the difference between natural numbers and whole numbers?
Natural numbers are the counting numbers starting from 1 (1, 2, 3, ...), while whole numbers include zero along with all natural numbers (0, 1, 2, 3, ...). The main difference is the inclusion of zero in whole numbers.
2. How are rational and irrational numbers different?
Rational numbers can be expressed as fractions (a/b, where a and b are integers and b 0), while irrational numbers cannot. Rational numbers have terminating or repeating decimal representations, whereas irrational numbers have non-repeating, non-terminating decimal representations.
3. What are some examples of irrational numbers?
Common examples of irrational numbers include π (pi), 2 (square root of 2), e (Euler's number), and the golden ratio (φ). These numbers cannot be expressed as simple fractions and have non-repeating, infinite decimal representations.
4. How do real numbers relate to other number systems?
Real numbers encompass all rational and irrational numbers. They include natural numbers, whole numbers, integers, and rational numbers, as well as irrational numbers. Real numbers can be represented as points on a continuous number line.
5. Why is understanding different number systems important?
Understanding different number systems is crucial for problem-solving, mathematical reasoning, and real-world applications. It helps in various fields like finance, science, and engineering. Each number system has unique properties and uses, contributing to a comprehensive understanding of mathematics and its practical applications.
### Prerequisite Topics
Understanding number systems is a fundamental concept in mathematics that builds upon several key prerequisite topics. To fully grasp the intricacies of various number systems, it's crucial to have a solid foundation in related areas. One essential prerequisite is multiplying fractions and whole numbers. This skill is vital because it helps students comprehend how different number types interact and combine, which is a cornerstone of understanding more complex number systems.
Another critical prerequisite is the ability to compare and order rational numbers. This skill is essential for recognizing the relationships between different numbers within a system and across various number sets. It lays the groundwork for understanding the structure and hierarchy within number systems, allowing students to navigate between different types of numbers with ease.
The distinction between rational and irrational numbers is another crucial concept to master. This knowledge is fundamental to understanding the breadth and diversity of number systems. It helps students recognize that not all numbers fit neatly into simple categories and prepares them for more advanced mathematical concepts that involve both rational and irrational numbers.
Additionally, proficiency in comparing and ordering numbers, including positive and negative numbers, is essential. This skill is particularly important when dealing with integer-based number systems and understanding the concept of number lines, which are integral to visualizing and working with various number systems.
By mastering these prerequisite topics, students build a strong foundation for understanding number systems. They develop the ability to work with different types of numbers, recognize their properties, and understand how they relate to one another. This comprehensive understanding is crucial for advancing to more complex mathematical concepts and problem-solving techniques.
Moreover, these prerequisites help students develop a more intuitive sense of numbers and their relationships. This intuition is invaluable when working with different base systems, such as binary or hexadecimal, which are essential in fields like computer science and digital technology. The skills gained from these prerequisite topics also enhance logical thinking and pattern recognition, which are key components in understanding the structure and logic behind various number systems.
In conclusion, a thorough grasp of these prerequisite topics is not just beneficial but essential for anyone looking to fully understand number systems. They provide the necessary tools and conceptual framework to explore more advanced mathematical ideas with confidence and clarity.
Natural numbers:
{1, 2, 3, 4…}
Whole numbers:
{0, 1, 2, 3, 4…}
Integers:
{…-2, -1, 0, 1, 2…}
Rational Numbers:
Numbers that CAN be written as "fraction", "repeating decimals" or "terminating decimal" Ex: -5, 0, 4, $\frac{-2}{5}$,5.67,-4.$\overline{5}$
Irrational Numbers:
Numbers that CANNOT be written as "fraction", "repeating decimals" or "terminating decimal" Ex:$\sqrt{5}$,$\pi$, 24.67934 |
## Intermediate Algebra (12th Edition)
$4$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents and the definition of rational exponents to simplify the given expression, $\dfrac{64^{5/3}}{64^{4/3}} .$ $\bf{\text{Solution Details:}}$ Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} 64^{\frac{5}{3}-\frac{4}{3}} \\\\= 64^{\frac{1}{3}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{64^1} \\\\= \sqrt[3]{64} \\\\= \sqrt[3]{(4)^3} \\\\= 4 .\end{array} |
Finding Areas with Riemann Sums
Warning: You must know summation notation, formulae, and operations do this lesson.
Area Under a Curve
In order to find the area between a given function's curve and the x-axis, or between two given curves, mathematical pioneers decided to divide the desired area into a finite number of rectangles with equal or unequal bases and then sum the areas of these rectangles.
This method yields a close estimate of the desired area however, it also includes or excludes spaces, since the rectangles don't exactly fit the area defined by the curve. There are always little bits more or less than the area we want.
They solved this problem by limiting n -- the number of rectangles -- to infinity. This makes each rectangle's base so teeny that the included or excluded triangles are eliminated and we get the exact area under or between the curve(s).
So we'll deal with two types of questions on finding area from fundamental principles. In the first type, we'll divide the area up into rectangles with equal bases and in the second type, we'll make a rectangular partition of the area with predefined borders. In the first case, we'll get the exact area we want because we will limit n -- the number of rectangles -- to infinity. In the second case, we will find an estimate for the area since we will not limit anything to infinity. The difference should be obvious from the wording of the question. In the first case, we'll be asked to evaluate the area, in the second, the question will define the border values of x and will ask us to estimate the area under the curve.
.
Riemann Sums
Say we want to find the area under the curve f (x) = x² from x = 0 to x = 5.
We divide the 5 units of the x-axis into n rectangles with equal bases = 5/n.
Now we need to find the x-value we'll use to determine the height of the rectangles, so we need an expression for the x value on the axis at the right end of each rectangle's base.
We see that the right endpoints of each rectangle's base is:
in rectangle #1,
in rectangle #2,
so in the " i" th rectangle the x value is with i between 1 and n.
where i is the order number of each rectangle. (1st, 2nd, ... nth)
Since , , the height
The area of a rectangle is height times base, so the area of the i th rectangle here is:
If we sum the above expression over the interval from 0 to 5, we will have the area.
The problem however, is that we find not only the area under the curve, but also the area of the triangles above the curve. To eliminate these, we limit n, the number of rectangles, to infinity. This makes the bases teeny since there are so many of them and so gives us the area between the
curve f (x) = x2 and the x-axis. Note that limiting n to infinity makes approach zero.
So our final expression for the area we're seeking is:
We have a formula for the summation of i 2 , we apply it now and then take the limit as n approaches infinity. Recall, to do this we divide our fraction through by the highest power of the variable, then set the variable equal to infinity.
When we switch the denominators and find the limit for the 2nd fraction (with all the n's in it),
we get 2, since the highest power of n is n3 and the coefficient of the numerator's n3 term is 2.
When we reduce the fraction, the answer is .
Now say we wish to find the area between the x-axis and the curve f (x) = x2
between x = 2 and x = 5.
The process is almost identical except now, , and ,
so
When we multiply by f (x) we get:
So our area this time is:
When we substitute the formulae for the summations of i and , we get 12 + 18 + 9 = 39.
Now let's do it by integration. We evaluate
Note that the area is a summation of the product f (x i ) times which is f (x) dx. Now you know why there's dx in every integrand. It represents the base of the rectangle.
Notation
When dealing with partitions, there are a number of notation systems used to denote the
values of x that define the borders and heights for the rectangles.
Generally, the border values for the partitions are denoted x i , x j or x k .
Some texts use x* for the x-value used to find the height or f (x*), others use wk for this.
Riemann Sums with Partitions
Now let's evaluate a partition of the area under the curve f (x) = x² + 1 on the interval from
a = 0 to b = 6 and with n = 5 and x 1 = 1, x 2 = 2, x 3 = 3, x 4 = 4, x 5 = 6.
We'll use the midpoint of the partition as x * or wk .
As we see from the x-values, the length of the 1st four bases = 1, the 5th = 2.
The midpoint values are:
w1 = 0.5 w2 = 1.5 w3 = 2.5 w4 = 3.5 w5 = 5
We need values for f(x i ) since those are the heights of the rectangles.
f (w1) = 1.25 f (w2) = 3.25 f (w3 ) = 7.25 f (w4 ) = 13.25 f (w5) = 26
The sum = 1.25 + 3.25 + 7.25 + 13.25 + 2(26) = 77 (units) ² , so this our estimate for the area.
Though it is not obvious in the diagram, we're also finding the area of the little triangles above the curve y = x 2 + 1. To eliminate this extra area, Riemann stated that we need only limit the largest base to zero in order to limit all bases to zero.
In our case however, we need only estimate the Riemann sum described in the question.
There are times when we must use the left or right endpoint of the base interval. The question will say which one to use. If there are no specific instructions, we should always use the right endpoint since it's easiest to find.
.
Should the curve in question cross the x-axis at some point on the given interval, we must take into account that the heights of the rectangles below the axis will all be negative. Since area can't be negative, we break it up into two parts. We multiply all negative heights by – 1 to make them positive, we find the two areas separately and then sum them. For this reason, it's always best to make a diagram of the situation in question and to check the interval for intercepts.
Here, we'll find the area under f (x) from x =0 to x = 5 and we'll add it to the area under
g(x) from x = 5 to x = 7, since multiplying the function g(x) by – 1 flips the curve
to generate the identical area above the x-axis.
.
Summary
To find the exact area under a curve f (x):
1. Divide the x-axis interval (a, b) into n rectangles with equal bases.
2. Find an expression for , the length of each base, .
3. Find an expression for x i , (with i from 1 to n) , the right endpoint of the i th rectangle.
4. Find an expression for f ( x i ) , the height of the i th rectangle.
5. Multiply height f ( x i ) by base , to get an expression for the area of i th rectangle.
6. Sum the formula in step 5 from i = 1 to n. Substitute the summation formulae.
7. Limit n to infinity to eliminate the under- or over-hang area.
To estimate area under a curve using a Partition:
1. Draw a diagram using the border values defined in the question.
2. Make a table of values for , x i , and f ( x i ) at the specified end or mid points.
3. Multiply base by height for each rectangle then sum the areas.
1) Use a Riemann sum to find the area between f (x) and the x-axis, over the given interval.
a) f (x) = 5 – x²with a = 0, and b = 2 b) f (x) = x³ + 8with a = 0, and b = 5 c) f (x) = x² + 1with a = 1, and b = 3
| (summation formulae) | (solutions)
2) Estimate the area under the curve f (x) = x² + 1 from a = 0 to b = 6 with the
regular partition in which n = 6 and x 1 = 1, x 2 = 2, x 3 = 3, x 4 = 4, x 5 = 5, and x 6 = 6.
Use the midpoint of the partitions as x * or wk . (make a diagram!)
.
1) a)
We see that , so which makes the height
The area of the i th rectangle is
The total area then is
Now, as before, we switch the denominators and evaluate the summation of i².
So we get square units.
When we integrate the function from 0 to 2 we get exactly the same value for the area.
b)
We see that , so which makes the height
The area of the i th rectangle is
The total area then is
Now, as before, we switch the denominators and evaluate the summation of i³.
So we get square units.
When we integrate the function from 0 to 5 we get exactly the same value for the area.
c)
We see that , but since we start at 1,
which makes the height f ( x i ) =
The area of the i th rectangle is
The area then is
Now, as before, we switch the denominators and evaluate the summation of i and i².
So we get square units.
When we integrate the function from 1 to 3 we get exactly the same value for the area.
2) use diagram above
The length of each base = 1
The midpoint values are:
w1 = 0.5 w2 = 1.5 w3 = 2.5 w4 = 3.5 w5 = 4.5 w6 = 5.5
We need values for f (x i ) since those are the heights of the rectangles.
f (w1) = 1.25 f (w2) = 3.25 f (w3 ) = 7.25 f (w4 ) = 13.25 f (w5) = 21.25 f (w6) = 30.25
The total = 1.25 + 3.25 + 7.25 + 13.25 + 21.25 + 30.25 = 76.5 units ² , so this our estimate for the area.
.
a) b) c) d) e) f) g)
_____________________________________________
.
(all content of the MathRoom Lessons © Tammy the Tutor; 2002 - ). |
# how many parallel lines does a rectangle have
## How Many Parallel Lines Does A Rectangle Have?
Answer: A rectangle has 2 pairs of parallel sides.
## What are parallel lines in a rectangle?
Lesson Summary
A rectangle has two pairs of parallel lines. A square also has two pairs of parallel lines. A parallelogram also has two pairs of parallel lines. To find parallel lines, look for lines that are going in the same direction and never meet up.
## How many parallel lines does a square have?
A square has two pairs of parallel sides, therefore it is a parallelogram; in the figure, we see that overline{AB} is parallel to overline{CD} and overline{AC} is parallel to overline{BD}. A square also has 4 right angles, so therefore it is a rectangle.
## How many lines are in a rectangle?
The shape of a rectangle has four lines, two of which are vertical lines. The other two lines are horizontal lines.
## Is a rectangle a parallel lines?
Each pair of co-interior angles are supplementary, because two right angles add to a straight angle, so the opposite sides of a rectangle are parallel. This means that a rectangle is a parallelogram, so: Its opposite sides are equal and parallel.
## Do all rectangles have 2 pairs of parallel sides?
Explanation: One of the definitions of a parallelogram is 2 pairs of parallel sides. Therefore, any parallelogram MUST have 2 pairs of parallel sides. This includes all squares, rhombuses, and rectangles.
## How many parallel lines are in a rectangular prism?
Characteristics of a rectangular prism
A rectangular prism has 12 edges and 8 vertices. The 12 edges of a rectangular prism are in 3 groups of parallel lines. The parallel edges are equal in length. The opposite faces are parallel and congruent to each other.
## What shape has 4 pairs of parallel lines?
A regular octagon is a shape that has 4 pairs of parallel sides. Each side is parallel to the side that is opposite to it.
## What shapes have pairs of parallel lines?
Shapes are parallel if they have lines that are always the same distance from each other and will never intersect or touch. Some shapes that have parallel sides include the parallelogram, the rectangle, the square, the trapezoid, the hexagon, and the octagon. A trapezoid has one pair of parallel sides.
## How many parallel lines does a pentagon have?
A pentagon has five sides and also has no sets of parallel lines.
2
## How many perpendicular lines does a rectangle have?
There are 4 right angles, but 2 pairs of perpendicular lines.
## What rectangle has two lines symmetry?
There are two lines of symmetry in a rectangle. When one line is drawn through the center along its length and the other is drawn along the width (breadth), we get the two lines of symmetry. By doing this, we get four equal and matching shapes.
## How many angles does a rectangle have?
four
a quadrilateral with four right angles.
## How many vertices does a rectangle have?
Rectangular Prisms
It’s made up of 6 rectangular faces. When you join the sides together, it becomes a rectangular prism with 8 vertices and 12 edges.
## How many parallel and perpendicular lines does a rectangle have?
Rectangles have four straight sides. Each pair of opposite sides is parallel, and adjacent sides are perpendicular. This means that every angle in a rectangle is a right (90∘) angle. All the blue shapes in the image below are rectangles.
## Do all rectangles have perpendicular and parallel sides?
As we mentioned before, right triangles have perpendicular sides, rectangles have both perpendicular and parallel sides, but other quadrilaterals might not. A regular pentagon has no parallel or perpendicular sides, but a non-regular pentagon might have parallel and perpendicular sides. It all depends on the polygon.
## How many straight sides does a rectangle have?
A square has four equal sides. A rectangle has two pairs of equal sides. Squares and rectangles have four straight sides and four right angles.
## What has only one pair of parallel lines?
A trapezoid is a quadrilateral with exactly one pair of parallel sides.
## How many rectangular faces does a rectangular pyramid have?
A rectangular pyramid has 5 faces. Its base is a rectangle or a square and the other 4 faces are triangles. It has 8 edges and 5 vertices.
## What is oblique rectangular prism?
Oblique Rectangular prism
An oblique prism is a prism in which the bases are not perpendicular to each other. A rectangular prism with bases that are not aligned one directly above the other is an oblique rectangular prism.
## How many parallel lines does a cuboid have?
The 12 edges of a cuboid are in 3 groups of parallel lines. The parallel edges are equal in length. Any intersecting edges are perpendicular to each other.
## What shape is a rectangle?
A rectangle is an oblong. It has two pairs of parallel sides and four right angles. A rectangle can also be known as an equiangular quadrilateral. This is because a rectangle is a quadrilateral shape (4-sided shape) that has parallel sides, equal to one another, and all 4 corners have angles that are 90º.
## Is rectangle a rhombus?
A rectangle is a rhombus. A square is a parallelogram. A parallelogram is regular.
## Can a square be a rectangle?
Yes, a square is a special type of rectangle because it possesses all the properties of a rectangle. Similar to a rectangle, a square has: interior angles which measure 90 each. opposite sides that are parallel and equal.
## What has parallel sides but not a rectangle?
Another quadrilateral that you might see is called a rhombus. All four sides of a rhombus are congruent. Its properties include that each pair of opposite sides is parallel, also making it a parallelogram. In summary, all squares are rectangles, but not all rectangles are squares.
## How many parallel lines does a rhombus have?
two pairs
Basic properties Every rhombus has two diagonals connecting pairs of opposite vertices, and two pairs of parallel sides.
## Is a triangle parallel?
A triangle is a geometric shape that always has three sides and three angles. Triangles have zero pairs of parallel lines. They usually have zero pairs of perpendicular lines. Only one type of triangle, the right triangle, does have two perpendicular lines.
## How many parallels are in a hexagon?
3 pairs
A regular hexagon, which means a hexagon with equal sides and equal interior angles, is the shape that has 3 pairs of parallel sides.
## How many parallel sides does a octagon have?
A regular octagon has 4 pairs of parallel sides.
## How many parallel lines does a parallelogram have?
In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
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# Permutations And Combinations Simplified
Permutations and combinations are an essential part of statistics. They show up in a ton of different places when you are finding the probability of anything. But it can be hard to remember the exact formula for a permutation or a combination when you need it without looking it up. This post will show you an easy, intuitive way to understand permutations and combinations so that you only need to remember one thing, and the rest you can just calculate when you need it.
## Permutations and Combinations Basics
Permutations and combinations are a way of determining how many different possibilities of something there are.
Permutations are what you use when the order matters. For instance, if 8 people are racing in a track meet, and you want to find the different ways they could get 1st, 2nd, and 3rd place, then the order matters. So you would use a permutation.
Combinations are what you use when the order doesn’t matter. For instance, if you have 10 different pieces of clothing you want to take on a trip, but you can only fit 7 of them in your suitcase, it matters which 7 you pick, but it doesn’t matter what order you put them in the suitcase, so you would use a combination
## The Key Point
There is one key thing to know with Permutations and Combinations, and that is the Factorial. Typically denoted with an exclamation point !
If you want to find how many different ways you can arrange 8 different items, it is 8 Factorial, which is 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. What this represents is that when you make your first choice of items to arrange, you have 8 to choose from. When you make your second choice, there is one less so that you have 7 to choose from, then 6 and so on.
Everything with permutations and combinations are just different applications of the Factorial.
## Permutations – Slightly Simpler Than Combinations
Let’s go back to the track example. Let’s say that you have 8 people racing on the track. The total different orders they could come in are 8 !
8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320
Now let’s say you only care about the order of the first three people on the track. Clearly we would have fewer than the full 8! different permutations, because we only care about how the first 3 people finished, not all 8. So in that case you have 8 options for first place, 7 options for 2nd place, and 6 options for 3rd place, and that’s it.
8 * 7 * 6 = 336
This is the permutations of 8, choosing 3.
Now there isn’t a function that lets us just multiply 8 * 7 * 6 easily. If we wanted the order of everyone, then 8 Factorial lets us multiply 8, down through 1, but it doesn’t stop in the middle. The way we do this is by finding 8 Factorial, and then dividing by 5 factorial. We use 5! because there are 5 items left behind that we don’t care about (8-3 = 5) That ends up being
The
( 5 * 4 * 3 * 2 * 1)
Cancels out of both the numerator and denominator, and we are left with 8 * 7 * 6.
So to find the permutations of a subset of a group, what we have just done is
• Find the permutations of the entire group ( 8! in this case)
• Divide by the permutations of the part of the group left behind ( 5! In this case)
Why are we dividing by the permutations of the parts left behind ? Because we don’t care what order they are in, so we need to cancel out all different orderings that they can be in.
## This is a key point to remember
• If you want to find the permutations of something, use the factorial
• If you want to find the permutations of a subset, find the permutations of the entire group, and then divide by the permutations of the set left behind.
## Combinations – Build on Permutations
• Permutations – We take the factorial of the entire set to find out the number of possibilities
• Permutations with a subset – We take the factorial of the entire set, and then divide by the factorial of the left behind set
Well for combinations we still don’t care about the order of the left behind set, but we also don’t care about the order of the set that we have chosen. So we start with the permutations of the entire set, then divide by the permutations of the left behind set, then divide by the permutations of the chosen set.
So if we have 10 different items of clothing, and we can only choose 7 to pack, so there are 3 left behind, the number of possibilities are
Which is equal to 120
An important thing to know is that there will always be at least as many permutations of a set as combinations, and typically many more permutations than combinations.
## The Traditional Permutations & Combinations Equations
At this point, it is worth showing the traditional permutations and combinations equations. These are the things that you might typically be expected to memorize for a class, but can be challenging to remember long term
Here is the permutation equation
And here is the combination equation
So while it is manageable to memorize those equations, it is easier to just intuitively understand
• To find the permutations of a full set, take the factorial
• To find the permutations of a partial set, find the permutations of the full set, then divide by the permutations of the items left behind
• To find the combinations of a partial set, find the permutations of the full set, divide by the permutations of the items left behind, and divide by the permutations of the selected items.
## Taking It One Step Farther
If you understand using the factorial, instead of memorizing the equations, you can apply it to cases that the equations don’t cover. For example, if you have 12 items, and you need to break them up into 4 equally sized groups, how many options do you have ? (order does not matter within any group)
Once again you start with the permutations of the entire group, and then divide by the permutations of any subset whose order you don’t care about. In this case we have 4 subsets with 3 items each where the order doesn’t matter. So the equation would be
Which is 19,958,400
That would be a challenging answer to get relying on just the standard equations
# Two Envelope Problem
The Two Envelope problem is one of my favorite statistical paradoxes. It is a problem that, even after you understand it, you can still revisit and be surprised by it.
## The Two Envelope Problem
You are presented with two identical envelopes. Both of the envelopes have money in them. You are told that one envelope has some amount of money in it, and the other envelope has twice as much money in it. But you don’t know how much that amount of money is. You are allowed to choose one envelope
You choose one envelope, which you decide to call Envelope A, and open it and observe X dollars. You are now given the opportunity to switch envelopes, so you calculate the expected value of switching. Since you know one envelope has twice as much money as the other, there is a 50% chance that envelope B, has twice as much as A, and a 50% chance that envelope B has half as much as A so
B = .5 * A/2 + .5 * 2A
B = 1.25 A
You conclude that the expected value of taking envelope B is 25% greater than keeping envelope A, so you start to switch. Before you do, you realize that if you had chosen envelope B to start with, you would have done the same calculation in reverse
A = .5 * B/2 + .5 * 2B
A = 1.25 B
and determined that the expected value of envelope A was 25% greater than envelope B.
How can you resolve this apparent paradox ?
## The Fairly Nerdy Solution
There are apparently quite a few solutions out there to this problem, with their authors believing their solution is clearly correct, and others disagreeing. I’ll leave it to you to let me know what you think of my solution.
First we need to clarify how the money was put into the envelope, because it makes a big difference to the results. The possible options are
• Coin Flip Scenario – The money is not put into the second envelope until you look at the first envelope. i.e. if you see \$100 in the first envelope, then the game master flips a coin to decide whether to put \$200 or \$50 in the second envelope
• Infinite Money Scenario – The money is put into both of the envelopes before you choose. One envelope has X dollars and the other envelope has 2X dollars. The game master has infinite money, and would be willing to put any amount of money, including infinite money, into the envelopes
• Finite Money Scenario – The money is put into the both of the envelopes before you choose. One envelope has X dollars and the other envelope has 2X dollars. However the game master is only willing to put a finite amount of money in the envelope, so there would be a maximum of M dollars in any envelope
Each of those options need to be considered separately, and I think that some confusion on this problem stems from not clearly defining those options
## Coin Flip Scenario
In my opinion, this scenario doesn’t really match the problem description of having two envelopes to choose from. However it does exactly match the math description of how the expected value of the second envelope is calculated.
If you are faced with this scenario, and see X dollars in your envelope, then your expected value of switching truly is 1.25 X, because the amount of money in envelope B can be calculated by
B = .5 * X/2 + .5 * 2X = 1.25X
which is how the problem is outlined.
This is basically a double or nothing bet, but instead of nothing, it’s double or half, and it is clearly a profitable bet to take.
## Infinite Money Scenario
To my mind, this is the least realistic scenario. Because while you could set up the coin flipping problem outlined above, or set up the finite money scenario outlined below, infinite money isn’t possible. Even if you were playing the game with the richest person alive, or even the government of all of Earth, there is still some upper limit to the total amount of money. But if you do assume infinite money, what happens?
If you are assuming infinite money, the most reasonable assumption is that one envelope has some amount of money in it, between zero to infinity dollars, and the other has half that amount in it.
The first thing we notice is that the expected value of envelope A, taken all by itself, is infinity. This is because the average of all numbers between zero to infinity is infinity. We then immediately observe that the expected value of envelope B is also infinity. So after solving the equation of
B = .5 * A/2 + .5 * 2A
We are left with
B = 1.25 infinity
And
A = 1.25 infinity
So this scenario doesn’t really boil down to “why is the expected value of switching greater than the expected value of not switching” it ends up being “Infinity has weird properties that don’t match our intuition”
## Finite Money Scenario
The scenario where the envelopes have finite money is the most interesting scenario in my opinion. Both because you could actually replicate it in real life, and because the math is interesting. With the finite money scenario, the most reasonable assumption for how it gets distributed is
• The game master puts a random amount of money in the first envelope, between zero dollars up to the maximum dollars he is willing to put in one envelope, denoted as M
• The game master puts half that amount of money in the second envelope
• The game master flips a fair coin to shuffle those envelopes so they are randomly presented as envelope A & envelope B
However, we are still posed with the problem, why does
B = .5 * A/2 + .5 * 2A = 1.25 A
and
A = .5 * B/2 + .5 * 2B = 1.25 B
The cause of this apparent paradox turns out to be bad intuition on the possible distribution of money in the envelopes. The assumption in this paradox is that the all dollar amounts are equally likely. Therefore you have a 50% chance of doubling your money, and a 50% chance of cutting it in half, just like in the coin flip scenario.
For that to be true, it implies a probability function that looks like this, where all numbers are equally likely
Where M is the Maximum amount of money the game master would be willing to put in a single envelope. However the actual distribution of money in those envelopes will look like this
Contrary to our assumptions, all dollar amounts are not equally likely. There will actually be 3 times as many quantities below half of the maximum value as the number above half the max value.
So before going any farther, intuitively what will this mean ? Well clearly the likelihood of doubling our money goes down if we have a larger amount of money. So the equation of
B = 50% * A/2 + 50% * 2A
Doesn’t really apply if the chances of getting A/2 and 2A aren’t the same for every value of A.
Why are there more small numbers than big numbers in this distribution ? Because of the way the money is distributed. If we ran this many times, and randomly generated the money in the first envelope to be between 0 and M dollars over and over, and then sorted that money, we would end up with a distribution that looked like this
Half of our numbers are less than M/2, and half are greater than M/2. Exactly what we intuitively expect. When we generate the numbers in the second envelope by dividing all of these by 2 we end up with all of the numbers in the second envelope being less than M/2
In total, 1/4 of the numbers in both envelopes are greater than M/2, and 3/4 less than M/2. Note, we set up these envelopes by putting up to a maximum amount into the first envelope and then dividing to get the second envelope. The 1/4 and 3/4 distribution would still hold if you went the other way and put up to a maximum amount into the first envelope and doubled it to get the second envelope.
At this point the intuitive solution to this paradox is clear. For any number there is not an even chance that we will double the value in the envelope vs the chance that we will cut that value in half. The chance that we will double is larger for the small values than for the large values, which will mean that the
B = 1.25 A
A = 1.25B
Equations won’t hold.
The next section works out the math for exactly what the expected value of switching is for the finite money scenario. Then, in the final section we will exploit this information to see how if we were playing this game multiple times in a row, we actually could intelligently switch and end up with more money at the end of multiple rounds than the average expected value.
## Expected Value Of Switching – Finite Money Scenario
For calculating the expected value of switching, we will give a concrete example. For this example, we will assume that the maximum amount of money that the game master is willing to put in an envelope is \$40.
In this scenario, the average value of the lower values is \$10, and the average value of the upper values is \$30. The average value of all the money in the envelopes is \$15, calculated by .75 * \$10 + .25 * \$30 = \$15
At this point it’s easiest to assume that we will always get either the \$10 average value, or the \$30 average value when we open an envelope. So when we open the first envelope, 75% of the time we will get \$10, and 25% of the time we will get \$30.
So what is the expected value of switching if we have either the \$30 or the \$10 in the first envelope ? For the \$30 it’s easy because if the first envelope has money that lies in the upper half of the range, switching will always get in the lower half of the range. So the \$30 will always be cut in half to \$15
The \$10 is interesting because for every 3 times we see the \$10, 2 of those times we will double to \$20, and 1 time we will cut in half to \$5
Breaking out the table for the expected value
Finding the weighted average of 50% of \$20, 25% of \$5, and 25% of \$15 gives us the total expected value of switching envelopes, which is \$15.
So the expected value of envelope A in this case is \$15, and if we switch, the expected value of envelope B is \$15, which are the same. This resolves our paradox, since we expect the average values in these envelopes to be the same
## Exploiting This Information
Interestingly, even though the average value of A & B are the same, this is still an exploitable game assuming:
• You play more than one round and
• You can look in your envelope before deciding whether you want to switch
Our earlier calculations show that the average value of all the switches was the same as not switching. But that was just the average value. Some of the individual switches themselves were beneficial, and some lost you money. If you can find a way to distinguish the profitable switches from the unprofitable ones, you can get more money.
The obvious solution is that you want to switch when you have the low dollar amount, and not when you have the high ones. The expected value of switching when you have the low dollar is:
• You will double your money 2 times out of 3
• You will lose half your money 1 time out of 3.
On average switching when your envelope has less than half the maximum amount of money the game master would be willing to put in will yield +50% on those switches. Since you likely only have a rough guess as to how much money could be in the envelopes, a good (not necessarily perfect) strategy would be to switch envelopes the first time, and then after that switch if the amount in your envelope is less than half of the running maximum you have seen.
I simulated that strategy in the Excel below, playing 1000 games with a random maximum amount of money that could be in the envelope, random amounts in one envelope, half that amount in the other, randomly shuffled. Whatever envelope got selected (from the random shuffle) we call envelope A, and the other envelope gets call B
Implementing that strategy gave approximately 25% more money than just picking the A or B envelopes. You can download that Excel file here.
# An Intuitive Guide To Bayes Theorem
The purpose of this page is to give you an intuitive understanding of how to solve Bayes Theorem problems. The equation for Bayes Theorem is not all that clear, but Bayes Theorem itself is very intuitive. The basics of Bayes Theorem are this
• Everything starts out with an initial probability – That is, before you do any tests or have any data, there is some initial probability of an event
• Tests can update that probability – After you assign an initial probability, if you gather more information that is relevant then the probability can change. For instance, you may initially have a very low chance of having an illness, but if a test for that illness comes back positive, the probability that you have it has increased
• After a test, all probabilities get normalized to 1 – It doesn’t matter if an event is unlikely to have occurred. What matters is if the event is likely compared to all other possible events. For instance, if you don’t know whether you are observing a 6 sided die or a 20 sided die, and you see the die roll 4 five times in a row, it is unlikely that the 6 sided die would have rolled those values. But it is extremely unlikely the 20 sided die would have, so comparatively the 6 sided die is more likely
## 6 Easy Steps For Any Bayes Problem
1. Determine what you want the probabilities for, and what you are observing
2. Estimate initial probabilities for all possible answers
3. For each of the initial possible answers, assume that it is true and calculate the probability of getting the observation with that possibility being true
4. Multiply the initial probabilities (Step 2) by the probabilities based on the observation (Step 3) for each of the initial possible answers
5. Normalize the results
6. Repeat steps 2-5 over and over for each new observation
## Bayes Theorem Applied To Cancer Testing
Testing for a disease is a classic Bayes Theorem problem, and one that can give counter intuitive results the first time you see it. Let’s say that you are testing a generic patient for cancer. One percent of the population has this cancer. You have a test that will return a True Positive (return a positive when they actually do have cancer) 99% of the time, and return a True Negative (return a negative when they do not have cancer) 95% of the time.
You do 1 test, and get back a positive result. What are the odds this patient actually does have this cancer ?
1. Determine Possibilities
There are two possibilities. The patient either has cancer. Or they do not have cancer
2. Estimate Initial Probabilities
Since this is a generic patient they should be like the general population, so we assume there is a 1% chance they have cancer, and a 99% chance they do not.
3. Calculate The Probability Of Getting The Result For Each Possible Answer
The result is a positive test. If the patient has cancer, the probability of getting that result (True Positive) is 99%. If the patient does not have cancer, the probability of getting that result (False Positive) is 5% (which is 1 minus the 95% true negative rate)
4. Multiply Step 2 By Step 3 To Get The Combined Probability
This step should be similar to any other probability you have studied. We are just calculating what is the probability they have cancer, and got a positive test. And separately calculating what is the probability they do not have cancer, and got a positive test.
5. Normalize The Results
This will be the final answer after 1 positive result. At this step we see how likely the having cancer was, considering that a false positive was a possibility
And that is the answer, we have found that after the “99% Reliable” test, there is only a 16.7% chance that the patient has cancer
6. Repeat The Steps Over Again With Additional Observations
If you do additional tests, you use the new values as your starting probability. In this case let’s assume that we do a second test, get a Positive result, and then a third test and get a Negative Result.
For the second test, the conditional equation is the same as the first test. The normalized has cancer value of 16.7% gets multiplied by .99, and the normalized does not have cancer value of 83.33% gets multiplied by .05.
For the third test, since this was a negative result we need to change the formula. We multiply the normalized has cancer probability by the False Negative rate of .01 (1-.99) , and the normalized does not have cancer rate by the True Negative rate of .95.
The results after both tests are shown below
After the second positive result, the odds the patient actually has cancer jumps up to 79.8%, but after the negative test, the odds drop back down to 4%
## That Example Was Great, But You Promised Me Intuition
The page promised you intuition. So far we have solved one Bayes Theorem problem, which is decent example, but not too different than what is on Wikipedia for Bayes Theorem. Here is the intuition you should develop
• Bayes Theorem Is Just Multiplication and Division – Bayes theorem itself is very simple. Multiply out all of the strings of probabilities, and then normalize. However some problems it is applied to are themselves very complicated, so the whole thing becomes complicated. For instance, you can make the problem more difficult by using complicated probability distributions for the conditional probabilities or complicated initial probability distribution functions. There might be special probability functions applied to a goal scoring problem in soccer, or a line waiting problem at a store. But that doesn’t mean Bayes Theorem itself is all that complicated, the Bayes part of the problem is still just multiplication and division
• It is just as easy to solve for all possibilities as a single one – You might encounter a problem such as “This bag has 4 sided, 6 sided, 8 sided and 12 sided dice in it. Your friend draws out a die, rolls it, and reports the number as 5. What is the probability the selected die was a 6 sided die” The problem asks you to solve for the 6 sided die, but since you have to get the total probability at each step any way in order to normalize, it is just as easy to solve the problem for the 4, 6, 8, and 12 sided dice at the same time. Solving them all at the same time makes the thought process more straightforward, and can be done in a nice clean table.
• The order of observations doesn’t matter to end results – Bayes theorem amounts to repeated multiplication. Multiplication is commutative. You can change the order of the terms and get the same final results. But if you change the order of the observations, for instance putting the negative cancer test result first in our example problem, the intermediate results will have different probabilities
• You Don’t Actually Have To Normalize Each Step – We normalized the example problem each step. You could do all of the multiplication for the observations, and then normalize at the end and get the same result. The only caution is that the probabilities can get very small after repeated decimal multiplication if you do not normalize. You can run into trouble with round off or truncation error depending on what you are using to do the math.
## So Why Was The Cancer Problem Result Surprising ?
Many people are surprised to see that a positive result on the 99% reliable test still only means there was a 16.7% chance the patient had cancer. Why was that surprising ? Because most people do not bake the initial probabilities into their intuition.
We do a good job of understanding the conditional probability. After all, a 99% reliable test should make it much more likely the patient has cancer, which it does. But if the initial probability is a really small number, the new probability will probably be small as well. This often gets overlooked, and people implicitly assume an evenly distributed initial probability when thinking about these types of problems.
Overlooking the initial probability is the real joke behind this XKCD comic https://xkcd.com/1132/ (not having to pay the bet if the sun actually exploded is merely a bonus)
These pie charts are a good way to visualize what is occurring. If the patient doesn’t do any test, the odds of having cancer are a small slice of the pie
While the patient is waiting for the test results there are 4 possibilities, either they have cancer and the test comes back positive (blue), they don’t have cancer and the test comes back positive (green), they don’t have cancer and the test comes back negative (purple) or they have cancer and the test comes back negative (red slice, but too small to be seen)
Once they get positive test results, the purple and red slices of the previous chart go away. We normalize the green and blue slices in light of the new total probability. The odds of getting that result due to a false positive (green) are still larger than the odds of a true positive (blue).
## More Examples
If you want more examples and information about Bayes Theorem, here is a book I wrote walking through half a dozen Bayes Theorem examples
And here is an Excel file solution to some Bayes Theorem problems
# Understanding Statistical Significance
## Statistical Significance in Real Life
Statistical significance is a way of quantifying how unlikely something that you are measuring is, given what you know about the baseline. Exactly how unlikely something needs to be before it is statistically significantly depends on the context. You likely have an intuitive understanding of statistical significance based on your own life.
For instance, if you were at a United States airport, and it was announced that your plane was 15 minutes late, you wouldn’t think that it was anything unusual. But if you were at a Japanese bullet train station, and found out that it was going to be 15 minutes late, you would probably think that was at least somewhat odd.
Why does one seem like a more significant event than the other ? It is because you know that planes are frequently late, where the trains almost never are. So the trains being late is more significant because it is more different than the normal day to day variation than the plane being late.
## Plot The Delay
Statistical Significance is very easy to understand on a probability density plot. The red line shows 15 minutes late. The blue line shows how likely a train will be any given time late, and the green line shows how likely a plane will be any given time late. The total area under each of the blue and green lines is 1
It is clear on the chart that very few trains are more than 15 minutes late, but a lot of planes are.
There are really two things going on in the chart. The first is that the average plane is more late than the average train is. The average plane is 10 minutes late, and the average train is 0 minutes late. So being 15 minutes late is bigger difference from average than for a train than a plane.
The second thing that is going on is that the distribution of plane lateness is a lot wider than the distribution of train lateness. There is a lot more variation in the plane departure time than there is in the train departure time. Because of that the plane lateness would have to be even greater to be unusual
## The Gist of Statistical Significance
Statistical Significance means quantifying the probability of how unlikely an event is. Exactly what is statistically significant depends on context, but typical numbers considered statistically significant are if something would have less than a 5% chance, less than a 1% chance, or less than a .5% chance of occurring if there wasn’t some difference between what you are measuring and the baseline.
The information that is important to statistical significance are
• How many measurements you have – The more measurements you have, the more likely you have measured the full population of what is occurring, and not just a non-representative sample
• How different the average of your measurements is from the expected average – The bigger the difference, the more likely it is significant.
• How much variation there is in the measurements. – The less variation there is in the measurements, i.e. the tighter the spread is, the smaller the difference needs to be to be significant
There are small differences in the equations based on exactly what has been measured, but essentially all of the equations boil down to
• Get a number which is the difference in average values, multiplied by the square root of the number of measurements you have, and divided by the square root of the variation in your measures. Call that number the “Test Statistic”
• The larger the Test Statistic the more statistically significant the difference.
• Look up the Test Statistic in the appropriate “Z-Table” or “T-Table” to find the probability that there is a statistically significant difference between your samples, as opposed to just random variation
## Equations For Statistical Significance
Now that you have a general understanding of Statistical Significance, it is time to look at the equations. The most commonly used test for statistical significance is the Z-Test. You use this test if you have a lot of measurements (at least 20, preferably at least 40) and you are comparing it against a population with known values. For example, you would use this test if you work at a hospital that had 500 babies born in it the past year, and you wanted to see if the average weight of those babies was different than the average weight of every baby born in your city.
Where
• X_bar : is the average of the measured data
• U_0 : is the population average
• Sigma : is the population standard deviation
• n : is the number of measured samples
You then look up the Z-value in a Z-Table to get probability
There are a few other different equations for Statistical significance called “T-Tests”. You would use one of these T-Tests instead of a Z-test for one of these reasons
• The number of measurements you have is small, certainly you would use a T-Test with fewer than 20 measurements, or maybe fewer than 50
• You want to compare before and after measurements for the same individual. For instance, if you have a before and after measurement for 20 people after a diet, you would use a certain type of T-Test.
## What is the difference between a Z-Test and a T-Test?
What is a T-test vs a Z-test, and how do you know when to use a Z-test or a T-test? The thing to understand about T-Tests, is that they are almost the same as the Z-Test, and will give almost the same answer as you increase the number of measurements that you have. The whole point of T-Tests is that they put more area at the tail of the normal curve, instead of the middle to account for uncertainty you would have in your measured mean and standard deviation if you have a very small sample size. Once you get above 20 or so measurements the difference between Z-test results and T-test results becomes vanishingly small.
The plots below show the probability density for a Z-curve, and T-test curves with different sample sizes
Once you get past 20 or so measurements (green line, hardly visible) there really isn’t much of a difference between a T-Test or a Z-test (purple line). However if you only have a few measurements than the T-Test results will need a lot greater Test Statistic to give a statistically significant result
It can be a little bit confusing knowing exactly which test to use, but using the exact right test isn’t that important unless you are taking an exam or righting a scientific paper. The tests will all give similar results assuming you have more than 10 measurements, and very similar assuming you have 30 or more.
For a better understanding of the different types of tests, you can refer to this cheat sheet I put together giving the formulas for each test, and when they are used.
## Examples of Z-Test vs T-Tests
This post was intending to give an intuitive understanding of statistical significance. If you are interested in looking at examples of Z-Tests and T-tests and exactly how they are used and in what circumstance you might use one or the other, you can find some examples in this book I’ve put on Amazon
Or you can get an Excel file with different hypothesis testing examples here.
# Bypassing Willpower
If you want to accomplish something consistently, relying on willpower is the worst thing you can do. As Jerry Seinfeld points out, your current self doesn’t really care about your future self. If you want to work out consistently, you can’t rely on if you feel like working out. Most likely how you are going to feel is too tired, or too busy. Habits are one good way to bypass your willpower, but changing your environment is a better one.
Never Miss A Workout
Going for a run consistently is important to me. But more often than not, once I get home from work I’m likely to not go out again. The way around this is simple, Run Home. Depending on where I’ve lived, I have had my wife drop me off at work, or taken the bus to work at the beginning of the day, and then changed clothes after work and run home. When I live farther away from work, and take the bus, I have take the bus part of the way home, and get off on a stop that is 3-4 miles away, and run the rest of the way. When I do this, I almost never miss a day of running.
The reason this works is that I have changed the value proposition. Instead of getting home and having a choice between going running or relaxing, the choice is between running home or calling my wife and asking for a pickup. What’s going to happen there is pretty clear. I completely skip any opportunity for low willpower to matter.
But changing your environment isn’t just useful to make yourself exercise. It is useful anytime your motivated planning self wants to make it impossible for your tired lazy self to wuss out or forget something.
For a while I’ve wanted to get ready for work faster in the morning. My previous habit was to wake up and eat breakfast while I use my computer, check email, surf, etc. This always took a lot of time. It was easy to do things slowly and waste time when I was groggy in the morning. I marveled at how quickly my wife was able to get ready in the morning.
But when I tried to resolve to either eat faster, or skip going on the computer, I always found myself back sliding. I could be diligent for a few days, but not long term. The problem was I was trying to rely on willpower at a time when my willpower was at its lowest. Even waking up in the morning after getting little sleep due to the baby was a challenge. Trying to force myself to operate at a higher level in that state was a fool’s game.
The solution was simple, change my environment. I already routinely pack a lunch to take to work. Now I also pack some toast and a hard boiled egg & maybe some yogurt to eat as breakfast right when I get to work. It gives me an immediate energy boost to start the workday, and plugs a 20-30 minute leak in my morning routine.
The Practical Details
If you are going to manipulate your environment to bypass willpower, make sure you focus on the practical details. For instance, when I run home from work, I want to carry as little in my backpack as possible. Carrying shoes turned out to be a pain in the ass, so for a long time I just left my work shoes at my desk, and wore running shoes to and back from work. Eventually I decided to embrace business casual, and just bought a pair of all black running shoes to wear and skipped the dress shoes entirely.
I’ve Never Regretted A Workout
Despite all this planning, there are still times I am forced to rely on willpower to work out at the end of the day. When that happens, I try to remember “I’ve never regretted a workout, but I’ve often regretted skipping one”
Photo at top of page from Flickr here
# Python Flash Cards
I recently decided to buckle down and learn Python. Python is a programming language that had been on my want list for some time. It’s easy to use, has a ton of built in modules for things that I am interested in such as data science, and is also widely used at a lot of large tech companies.
After fiddling around with it for a while in 2014, I knew I liked the language, and could learn it, but I could never get over the hump. You see, the problem was I was already really good at a different programming language. I hired into a large aerospace company straight out of college, and being an old engineering company, FORTRAN was very ubiquitous. Not to worry! I had learned FORTRAN in high school (along with Basic, Pascal, and C++) and while it was quite rusty after 5 years of little use, the value proposition was clear. Dust off FORTRAN, and there was a ton of in house software that I could improve on to make my job easier.
As a result I became very good at FORTRAN, at least, very good for a mechanical engineer. But 6 years later when I decided to update my skills and learn something new the value proposition was different. Suddenly, for on the job work, the choice wasn’t “Re-learn this coding language” vs. “Do all this tedious work manually” the choice was “Do the job in this new cool language, but one that will take a lot of Googling to get all the syntax” vs. “Get er done in the language I already know”
Since I didn’t use Python enough, I never got over the hump of “I need to Google 60-70% of what I need to do this job, and it will just take too long this time”
The solution was to take a page from my high school study techniques, and make Python Flash Cards.
So why did the flash cards help ? Well for me the problem wasn’t understanding python syntax, I could read a program reasonably well, just not write it, nor was it understanding how to code since I was already fluent in one programming language and had previously known others well. The problem was simply getting enough vocabulary down that I could do the simple jobs.
Once I had a foundation of vocabulary down, learning additional building blocks was fun and relatively easy. Oh, should I use a dictionary here instead of a list? That’s cool. Is there a built in method of indexing my loop, so I don’t have to keep a separate counter! What a great improvement to my code! Once you know the basics, the great thing about learning programming is that since programmers write all of the software tools, and since they have made documentation one of the paragon virtues of their profession, there are just a ton of resources and learning communities out there.
Over the course of approximately a month, Python transitioned from a want to learn language for me, to one I was using for simple jobs, to one I was fluent in and use as my go to programming language.
I recommend using the flash cards in batches. Instead of printing them all off and trying to learn them all at once, it is better to work on 10 or 15 at a time and really hit them quite a few times in a short period. I kept the flash cards at my desk and went through them when I got up to use the restroom or go to a meeting. After you learn each batch, put it in the big stack of ones that you know, and review the big stack periodically.
Is Python something worth learning for you ? Well if you already know at least one programming language, I’ll let you answer that for yourself. But if you don’t know any, and you are someone who spends most of your work day on the computer, I posit that for you the answer is yes. Simply being able to batch rename files alone saves me a lot of time, and if you work with a lot of text files or blocks of data the value proposition is even more clear. Hopefully with these flash cards, the pain of starting will be low enough you can give it a go.
Let me know if they worked for you, and if you added improvements !
# 9 Baby Surprises
9 Biggest Surprises the First Year of Having A Baby
My wife and I thought that we were well prepared for having a baby. We had read baby books, baby proofed the house, set up the crib and the diaper changing table, and above all, watched countless sitcoms where people have babies (I’m looking at you Rachel, Phoebe, and Pam). Here are the 9 biggest things that we didn’t know
1. Back Labor
Back labor was a surprise to us, although apparently 25% of women experience it. It was a Tuesday when I got home from work, and my wife said that she was experience back spasms. I asked, do you think they are contractions ? And she said she didn’t think so because she wasn’t feeling anything in her stomach. So I asked her to tell me when the next couple were, and saw that they were spaced pretty evenly at about six minutes apart. At that point we decided to call the doctor’s office and we got the nurse on call.
“Is this labor “ we asked, or is it just Braxton Higgs contractions ? “Well you will know that is is labor when your stomach gets hard like a basketball” Was the reply. So I asked my wife, is your stomach getting hard like a basketball? “No, it’s all in my back”. So we waited a couple of hours, and the back spasms kept getting worse, so we decided to go in.
At the hospital, the nurse checked out my wife, and diagnosed her her as having back labor. This was caused by the baby facing the wrong direction. His head was facing forward, so the back of his was was pressed against my wife’s tailbone.
So my wife was in labor, but unfortunately was only 1cm dilated, so the hospital wouldn’t check us in. “Can you give me something for the pain?” my wife asked ? “Well sure” the nurse replied “I can give you this muscle relaxant that will help the contractions that you are feeling” “I’m having back labor though, will it help me ?” “Probably not”. At this point the best advice they could give was to take a warm, relaxing bath.
We asked when we should come back, and the nurse said “Normally we recommend coming in when your contractions are 3-5 minutes apart, but since yours are already 5 minutes apart, you should come in when they hurt so much you can’t walk or talk during a contraction”. Which is clearly very unambiguous guidance.
1. Your Water May Not Break
So my wife was in labor, but not far enough along. How long would it take to progress ? Well according to the nurse, it could be anywhere from an hour to a week. So we were premature going to the hospital, clearly we should have waited for my wife’s water to break. We had a towel in the car, and that would be a definitive sign right ? Wrong
We waited throughout the night and eventually decided we had to go back in. We didn’t see any sign of my wife’s water breaking, but maybe we missed it. We went in, and sure enough, my wife was at 4 cm and ready to be admitted to the hospital, but her water had not broken. And in fact never broke until the doctor assisted in that. It turns out this was common too. TV has lied to us
1. Check the hospital for a snack room
We were in the hospital for 3 days – 2 nights, and it wasn’t until an hour before checking out that I discovered there was a snack room for the mothers literally across the hall from the room we were in. Up until that point my wife had been eating the hospital food, and food I got from the cafeteria, and asking the nurses to bring us the juice, or cheese or applesauce as a snack. We didn’t realize it was all in a fridge just across the hall that we could help ourselves to without bothering the nurses.
1. No Sleep
Before having the baby, my friends and coworkers joked about “Get your sleep in now” but you really never know how serious people are. After all, people complain about changing diapers, but we’ve never found them to be a problem. The lack of sleep is real though. My son is 13 months old now, and he literally has not slept a full night more than 5 nights so far. Fortunately, at 13 months he’s getting up 1-2 times every night, which means we can actually survive. For the first several months when he was getting up every 2 hours we were barely getting by
Sadly, even knowing this, there is not much to do about it. If someone invents a way to store sleep for later, let me know and I will buy one.
1. Ear Infections Are Almost Impossible to Identify
In months 5-9 our son had ear infections 3-4 times, and he was put on amoxicillin and they cleared up nicely. But despite the fact that we recognized all 4 ear infections for what they were, we probably had 8 or so false alarms. The symptoms which we thought were sure fire indicators of the ear infection, such as ear tugging, crying when set on one side, waking up continuously through the night turned out to be very hit or miss. So we had quite a few times taking a fussy baby to the doctors only to be told he was teething or there was nothing obviously the matter.
The doctors advised us that the ear infections were most likely to be associated with a cold, but we never really saw a cold any of the times, so really never got good at identifying the ear infections.
1. Both the baby and the mom need to learn how to nurse
I guess we expected that nursing would be instinctual. But right from the beginning our son had trouble nursing. With the help of the lactation consultant at the hospital, there were times when it went well, but we had a lot of trouble. So much so, that after day 3 when our son had lost more than the recommended amount of birth weight (lost 11%, greater than the 10% cutoff) the doctor recommended that we start using the finger & tube method for supplemental feeding. I did that for a few weeks until we were able to get a working system with nursing and bottle feeding.
Although with the help of the lactation nurse at the hospital, our son was able to latch onto the bare nipple, we had a lot of trouble with that. What eventually worked for us was using a nipple shield for the first 2-3 months, since that made it a lot easier for him to latch onto. Eventually our son got the hang of it, and we stopped using the nipple shield since it was a hassle to wash it after every feeding. – Side Note – If you use them, see if the hospital will give you a few spare ones before you go!
1. Kids can decide they hate to be spoon fed
Right around month 6-7, my son was eating a wide array of different baby food, and a very small assortment of cut up real food. Well, quickly after he started eating the cut up real food, he decided that he was not interested in the baby food at all, and absolutely would not allow himself to be spoon fed by mommy or daddy. And while he was happy enough to have the spoon himself, this did not result in eating so much as decorating the floors.
As a result, we fairly abruptly switched over to all sliced up real foods and ended up with a couple of weeks worth of cans of baby food that we had to give away.
1. Kids develop their favorite foods early
I’m not sure if my son had a favorite canned baby food, but as soon as we switched over to real food he had a clear favorite. Blueberries. Basically, he would eat most food pretty well, if he was hungry. Some days he would eat a lot, some days not so much, but on any day he could eat basically unlimited numbers of blueberries. We eventually starting making sure to wait until the end of the meal to get the blueberries out of the fridge, because once he saw them he would not eat anything else.
Nowadays, he definitely has some favorites, like plain, no sauce pasta, or sliced grapes, but no favorites that are quite as strong as blueberries
1. Some milestones are very fuzzy, others happen very quickly
When you hear your parents talk about first words you think one day you weren’t speaking, and the next you had a very definitive word. With the baby, it’s clear that’s a much more fuzzy line. Grandma & Grandpa have been quick to attribute many sounds to a first word. ( Yes, he said “Da” when he saw me, but he says “Da” when he sees anyone, and babbles it all the time. Can we count that ?)
For us, we settled on “Bir“ being his first word. Sure it’s missing the “D” at the end, but he will repeatedly and consistently say it when looking at birds out the window, or pointing to birds in a book.
But as fuzzy as the talking milestone was, the walking milestone was crystal clear. One day he was only cruising along the couch and tables, and the next he was taking a few steps, and within 3 days he was perfectly happy walking across the entire room. This was surprising to me given how slow the crawling milestone was. Around month 4-5 we had been consistently predicting that he would crawl any day now, and in fact he could move himself forward by month 3 provided you put your fists behind his feet to push off of. But the crawling took a long time of slow progress until he was good at it, unlike the walking.
So that’s our top 9 surprises for the first year. Undoubtedly he’ll find 9 or 99 new ones for year number 2. |
# AP Statistics Curriculum 2007 Pareto
(Difference between revisions)
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## General Advance-Placement (AP) Statistics Curriculum - Pareto Distribution
### Pareto Distribution
Definition: Pareto distribution is a skewed, heavy-tailed distribution that is sometimes used to model that distribution of incomes. The basis of the distribution is that a high proportion of a population has low income while only a few people have very high incomes.
Probability density function: For $X\sim \operatorname{Pareto}(x_m,\alpha)\!$, the Pareto probability density function is given by
$\frac{\alpha x_m^\alpha}{x^{\alpha+1}}$
where
• xm is the minimum possible value of X
• α is a positive parameter which determines the concentration of data towards the mode
• x is a random variable (x > xm)
Cumulative density function: The Pareto cumulative distribution function is given by
$1-(\frac{x_m}{x})^\alpha$
where
• xm is the minimum possible value of X
• α is a positive parameter which determines the concentration of data towards the mode
• x is a random variable (x > xm)
Moment generating function: The Pareto moment-generating function is
$M(t)=\alpha(-x_m t)^\alpha\Gamma(-\alpha,-x_m t)\!$
where
• $\textstyle\Gamma(-\alpha,-x_m t)=\int_{-x_m t}^\infty t^{-\alpha-1}e^{-t}dt$
Expectation: The expected value of Pareto distributed random variable x is
$E(X)=\frac{\alpha x_m}{\alpha-1}\mbox{ for }\alpha>1\!$
Variance: The Pareto variance is
$Var(X)=\frac{x_m^2 \alpha}{(\alpha-1)^2(\alpha-2)}\mbox{ for }\alpha>2\!$
### Applications
The Pareto distribution is sometimes expressed more simply as the “80-20 rule”, which describes a range of situations. In customer support, it means that 80% of problems come from 20% of customers. In economics, it means 80% of the wealth is controlled by 20% of the population. Examples of events that may be modeled by Pareto distribution include:
• The sizes of human settlements (few cities, many villages)
• The file size distribution of Internet traffic which uses the TCP protocol (few larger files, many smaller files)
• Hard disk drive error rates
• The values of oil reserves in oil fields (few large fields, many small fields)
• The length distribution in jobs assigned supercomputers (few large ones, many small ones)
• The standardized price returns on individual stocks
• The sizes of sand particles
• The sizes of meteorites
• The number of species per genus
• The areas burned in forest fires
• The severity of large casualty losses for certain businesses, such as general liability, commercial auto, and workers compensation
### Example
Suppose that the income of a certain population has a Pareto distribution with α = 3 and xm = 1000. Compute the proportion of the population with incomes between 2000 and 4000.
We can compute this as follows:
$P(2000\le X\le 4000)=\sum_{x=2000}^{4000}\frac{3\times 1000^3}{x^{3+1}}=0.109375$
The figure below shows this result using SOCR distributions |
# Week 11 in Math 10
What we learned Week 11 in Math 10 is to factor when the degree of a polynomial is more than 2. (for examples 4, 6, et cetera.) The bases of factoring polynomials are C, D, P, E, and U.
C has used the polynomial which can be divided into common factors.
3x2 + 6x = 3x(x + 2)
D is differences of squares, it is used when the polynomials are perfect squares.
16 – x2y2 = (4 – xy)(4 + xy)
P is to find the pattern of the polynomial.
x2 + 6x – 7 = (x + 7)(x – 1)
E is used by easy polynomials which can be used the easiest pattern.
x2 + 4x + 4 = (x + 2)2
U is used by ugly polynomials. (We usually use a square to factor.)
6x2 + 13x + 6 = (2x + 3)(3x + 2)
We can use the bases and solve the polynomials which degree is bigger than 2. We also learned the pattern about a degree. When the degree is bigger than 2, other exponents of x must be half of the degree if there are not any common factors, and it cannot be factored if it is not.
For an example, 32x4 – 2 can be divided by 2 (Greatest common factor) and it becomes 2(16x4 – 1). 16x4 and -1 are perfect squares, we can use D. 2(16x4 – 1) = 2(42 + 1)(42 – 1). 42 – 1 is even perfect square which can be used D, but 42 + 1 is not because the binomial must have ONE minus sign. 2(42 + 1)(42 – 1) = 2(42 + 1)(2x + 1)(2x – 1). So, the answer is 2(42 + 1)(2x + 1)(2x – 1). |
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The Algebric Functions
1. 1. The Algebra of Functions T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
2. 2. Using basic algebraic functions, what limitations are there when working with real numbers? A) You CANNOT divide by zero. Any values that would result in a zero denominator are NOT allowed, therefore the domain of the function (possible x values) would be limited. B) You CANNOT take the square root (or any even root) of a negative number. Any values that would result in negatives under an even radical (such as square roots) result in a domain restriction.
3. 3. Example Find the domain: There is an x under an even radical AND x terms in the denominator, so we must consider both of these as possible limitations to our domain. 65 2 2 xx x }3,2:{: 3,2,0)2)(3( 065 2,02 2 xxxDomain xxx xx xx
4. 4. Find the indicated function values and determine whether the given values are in the domain of the function. f(1) and f(5), for f(1) = Since f(1) is defined, 1 is in the domain of f. f(5) = Since division by 0 is not defined, the number 5 is not in the domain of f. 1 1 1 1 5 4 4 1 1 5 5 0 1 ( ) 5 f x x
5. 5. Find the domain of the function Solution: We can substitute any real number in the numerator, but we must avoid inputs that make the denominator 0. Solve, x2 3x 28 = 0. (x 7)(x + 4) = 0 x 7 = 0 or x + 4 = 0 x = 7 or x = 4 2 2 3 10 8 ( ) 3 28 x x g x x x The domain consists of the set of all real numbers except x= 4 and x= 7 or {x | x 4 and x 7}. , 4 ( 4,7) (7, )
6. 6. Rational Functions To find the domain of a function that has a variable in the denominator, set the denominator equal to zero and solve the equation. All solutions to that equation are then removed from consideration for the domain. Find the domain: Since the radical is defined only for values that are greater than or equal to zero, solve the inequality ( ) 5f x x 5 0x 5x 5x ( ,5]
7. 7. Visualizing Domain and Range Keep the following in mind regarding the graph of a function: Domain = the set of a function’s inputs; found on the x-axis (horizontal). The domain of a function is the set of all “first coordinates” of the ordered pairs of a relation Range = the set of a function’s outputs; found on the y-axis (vertical). The range of a function is the set of all “second coordinates” of the ordered pairs of a relation.
8. 8. Example Graph the function. Then estimate the domain and range. (Note: Square root function moved one unit right) Domain = [1, ) Range = [0, ) ( ) 1f x x ( ) 1f x x
9. 9. Algebra of functions (f + g)(x) = f(x) + g(x) (f - g)(x) = f(x) – g(x) (fg)(x) = f(x)g(x) 0)(, )( )( )( xg xg xf x g f
10. 10. Example Find each function and state its domain: f + g f – g f ·g f /g ;1 1g xf x x x ;1 : 11x Domainf xx xg x ;1 : 11x Domainf xx xg x 2 1; :1 1 1x x Domaing xx xf x 1 ; : 1 1 x D x f omain x xg x
11. 11. BA Their sum f + g is the function given by (f + g)(x) = f(x) + g(x) The domain of f + g consists of the numbers x that are in the domain of f and in the domain of g. Their difference f - g is the function given by (f – g ) (x) = f(x) - g(x) The domain of f – g consists of the numbers x that are in the domain of f and in the domain of g. BA If f and g are functions with domains A and B:
12. 12. Their product f g is the function given by BA The domain of f g consists of the numbers x that are in the domain of f and in the domain of g. Their quotient f /g is the function given by (f / g ) (x) = f(x) / g(x) where g(x) ≠ 0; (f g)(x) = f(x) g(x) If f and g are functions with domains A and B: The domain of f / g consists of the numbers x for which g(x) 0 that are in the domain of f and in the domain of g. 0)(xgBA
13. 13. Example Given that f(x) = x + 2 and g(x) = 2x + 5, find each of the following. a) (f + g)(x) b) (f + g)(5) a) ( )( ) ( ) ( ) 2 2 5 3 7 f g x f x g x x x x b) We can find (f + g)(5) provided 5 is in the domain of each function. This is true. f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15 (f + g)(5) = f(5) + g(5) = 7 + 15 = 22 (f + g)(5) = 3(5) + 7 = 22or
14. 14. Example Given that f(x) = x + 2 and g(x) = 2x + 5, find each of the following a) (f - g)(x) b) (f - g)(5) a) ( )( ) ( ) ( ) 2 (2 5) 2 2 5 3 f g x f x g x x x x x x b) We can find (f - g)(5) provided 5 is in the domain of each function. This is true. f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15 (f - g)(5) = f(5) - g(5) = 7 - 15 = -8 (f - g)(5) = -(5) - 3 = -8or
15. 15. Example Given that f(x) = x + 2 and g(x) = 2x + 5, find each of the following. a) (f g)(x) b) (f g)(5) a) 2 ( )( ) ( ) ( ) ( 2)(2 5) 2 9 10 fg x f x g x x x x x b) We can find (f g)(5) provided 5 is in the domain of each function. This is true. f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15 (f g)(5) = f(5)g(5) = 7 (15) = 105 or (f g)(5) = 2(25) + 9(5) + 10 = 105
16. 16. ( ) ( ) f x g x 2 3 16 x x The domain of f / g is {x | x > 3, x 4}. ( ) 3f x x 2 ( ) 16g x x Given the functions below, find (f/g)(x) and give the domain. ( / )( )f g x The radicand x – 3 cannot be negative. Solving x – 3 ≥ 0 gives x ≥ 3 We must exclude x = -4 and x = 4 from the domain since g(x) = 0 when x = 4.
17. 17. Composition of functions Composition of functions is the successive application of the functions in a specific order. Given two functions f and g, the composite function is defined by and is read “f of g of x.” The domain of is the set of elements x in the domain of g such that g(x) is in the domain of f. Another way to say that is to say that “the range of function g must be in the domain of function f.” Composition of functions means the output from the inner function becomes the input of the outer function. f g f g x f g x f g
18. 18. Composition of functions means the output from the inner function becomes the input of the outer function. f(g(3)) means you evaluate function g at x=3, then plug that value into function f in place of the x. Notation for composition: ))(())(( xgfxgf f g x g(x) f(g(x)) domain of g range of f range of g domain of f g f
19. 19. f g x f g x 1 2 f x 1 2x 1 2x . gf x xgxxf Find 2 1 )(and)(Suppose Suppose f x x( ) and g x x ( ) 1 2 . Find the domain of f g . The domain of f g consists of those x in the domain of g, thus x = -2 is not in the domain of f g . In addition, g(x) > 0, so 1 0 2x 2x The domain of f g is {x | x > -2}.
20. 20. 2 2 2 1 3 2 4x xf g x 2 2 2 2 1 2 6 9 1 2 12 18 1 3g x x x f x x x Example Evaluate and : f g x g f x 3f x x 2 2 1g x x 2 2 4 (you check) f g x x 2 2 12 17g f x x x You can see that function composition is not commutative. NOTE: This is not a formal proof of the statement.
21. 21. (Since a radicand can’t be negative in the set of real numbers, x must be greater than or equal to zero.) Example Find the domain of and :f g x g f x 1f x x g x x 1 : 0f g x x Domain x x 1 : 1g f x x Domain x x (Since a radicand can’t be negative in the set of real numbers, x – 1 must be greater than or equal to zero.)
22. 22. Solution to Previous Example : Determine a function that gives the cost of producing the helmets in terms of the number of hours the assembly line is functioning on a given day. Cost C n C P t 2 75 2C t t 2 14 525 100 2 40 \$5 8C t t 2 75 2n P t t t 7 1000C n n
23. 23. 1. Suppose that and2 ( ) 1f x x ( ) 3g x x ( ) ?g f x ( ) ( ( ))g f x g f x 2 ( 1)g x 2 3( 1)x 2 3 3x 2. Suppose that and2 ( ) 1f x x ( ) 3g x x (2) ?g f (2) 2g f g f 2 (2 1)g (3)g (3)(3) 9
24. 24. The End Call us for more Information: www.iTutor.com 1-855-694-8886 Visit |
Calculus Of One Real Variable – By Pheng Kim Ving Chapter 6: The Trigonometric Functions And Their Inverses – Section 6.1.2: Trigonometric Identities
6.1.2
Trigonometric Identities
1. Sine And Cosine Values Of Special Angles
360o), and the negatives of these angles. We'll find the sine and cosine values for the positive angles. Those for the
negative ones can be derived from those for the positive ones by the identities sin (–x) = –sin x and cos (–x) = cos x,
which we'll discuss in Part 3. The values of the four remaining trigonometric functions for these angles can be readily
derived from these two functions.
Refer to Fig. 1.1. Clearly, recalling that the cosine and sine are the (u, v) coordinates of points on the unit circle (see
Section 6.1.1 Part 6), we have:
Triangle OPA is equilateral.
Fig. 1.5 A Table Of Trigonometric Values.
If you forget one or more of these values, you can build this table yourself as follows. The Radians and Degrees rows are
obvious. For the Sine row, write down 0, 1, 2, 3, and 4, then take the square root of each number, and then divide each
by 2. For the Cosine row, reverse the order of the numbers in the Sine row.
2. Periodicity
3. Symmetry
Let x be an arbitrary angle. The terminal arms of x and –x are symmetric with respect to the u-axis. See Fig. 3.1. Thus:
sin (– x) = – sin x, cos (– x) = cos x.
It follows that sine is an odd function and cosine is an even function.
x and –x have terminal arms symmetric with respect to the u-axis.
4. Complementary Angles
Fig. 4.1
P and Q are symmetric with respect to the line v = u.
5. Supplementary Angles
Fig. 5.1
P and Q are symmetric with respect to the v-axis.
Fig. 7.1
Example 7.1
Solution EOS
8. The Pythagorean Identity
The notation sin2 x means (sin x)2, ie, the square of sin x. This is true for any exponent and for all trigonometric functions.
Some examples are: cos3 x = (cos x)3, tan–2 x = (tan x)–2, secm x = (sec x)m. Note that sin x2 means sin (x2), ie, the sine
of x2, which is a different thing from sin2 x.
The point P = (u, v) = (cos x, sin x) lies on the unit circle u2 + v2 = 1. See Fig. 7.1. So cos2 x + sin2 x = 1. Thus:
sin2 x + cos2 x = 1.
This identity is called the Pythagorean identity because it's the Pythagorean formula for the right triangle UPO as in
Fig. 7.1: UP2 + OU2 = OP2.
Example 8.1
Fig. 8.1
9. The Addition Identities For Sine And Cosine
(cos (xy) – 1)2 + (sin (xy) – 0)2 = (cos xcos y)2 + (sin xsin y)2,
cos2 (xy) – 2 cos (xy) + 1 + sin2 (xy) = cos2 x – 2 cos x cos y + cos2 y + sin2 x – 2 sin x sin y + sin2 y.
Fig. 9.1
Now, cos2 (xy) + sin2 (xy) = 1, cos2 x + sin2 x = 1, and cos2 y + sin2 y = 1. It follows that:
2 – 2 cos (xy) = 2 – 2 cos x cos y – 2 sin x sin y,
cos (xy) = cos x cos y + sin x sin y.
That's what we wish to get. It holds true for all values of x and all values of y (that's why it's called an identity). In
particular, it holds true for t = – y. That is:
cos (x + y) = cos (x – (– y)) = cos x cos (– y) + sin x sin (– y) = cos x cos ysin x sin y.
Next, we want to express sin (x + y) in terms of the trigonometric functions of x and y. We have:
That's what we want. Replacing y by – y and employing symmetry we get:
sin (xy) = sin x cos ycos x sin y.
We've obtained these four identities, called the addition identities:
sin (x + y) = sin x cos y + cos x sin y, [9.1] sin (x – y) = sin x cos y – cos x sin y, [9.2] cos (x + y) = cos x cos y – sin x sin y, [9.3] cos (x – y) = cos x cos y + sin x sin y. [9.4]
Note that these identities express the sine and cosine of the sum and difference of 2 angles in terms of those of each
individual angle.
Remark 9.1
sin (x + y) is not identical to sin x + sin y,
cos (xy) is not identical to cos xcos y,
etc.
The relation sin (x + y) = sin x + sin y is an equation and not an identity, because it may be true for some values of x
and y, but it isn't true for every value of x and every value of y.
Example 9.1
Solution EOS
We expressed the given angles in terms of the special angles, at which the values of the trigonometric functions are
known, and we applied the trigonometric addition identities.
10. Half-Angle Identities
We have:
sin 2x = sin (x + x) = sin x cos x + cos x sin x = 2 sin x cos x,
cos 2x = cos (x + x) = cos x cos xsin x sin x = cos2 xsin2 x,
cos2 xsin2 x = cos2 x – (1 – cos2 x) = 2 cos2 x – 1,
cos2 xsin2 x = (1 – sin2 x) – sin2 x = 1 – 2 sin2 x,
from cos 2x = 2 cos2 x – 1 we get cos2 x = (1 + cos 2x)/2,
from cos 2x = 1 – 2 sin2 x we get sin2 x = (1 – cos 2x)/2.
sin 2x = 2 sin x cos x, cos 2x = cos2 x – sin2 x = 2 cos2 x – 1 = 1 – 2 sin2 x,
These identities are called half-angle identities. This is because the angle x is half of the angle 2x.
11. Alternate Versions Of The Pythagorean Identity
We're now going to establish two alternate versions of the Pythagorean identity. They're the Pythagorean identities for the
remaining four trigonometric functions: tangent, cotangent, secant, and cosecant.
Dividing the Pythagorean identity sin2 x + cos2 x = 1 by sin2 x we get 1 + ((cos x)/(sin x))2 = (1/(sin x))2, or 1 + cot2 x =
csc2 x. Similarly, division of sin2 x + cos2 x = 1 by cos2 x yields 1 + tan2 x = sec2 x.
1 + tan2 x = sec2 x, [11.1] 1 + cot2 x = csc2 x. [11.2]
1 + tan2x = sec2x and 1 + cot2x = csc2x are Pythagorean formulas for right triangles OAZ and OBW respectively.
12. The Addition Identities For Tangent
Note that tan (– s) = (sin (– s))/(cos (– s)) = – (sin s)/(cos s) = – tan s for any real number s. We have:
Replacing y by – y we obtain tan (xy) = (tan xtan y)/(1 + tan x tan y).
Note that these identities express the tangent of the sum and difference of 2 angles in terms of that of each individual
angle.
13. The Sine And Cosine Laws
Sine Law: Cosine Law: a2 = b2 + c2 – 2bc cos A, b2 = c2 + a2 – 2ca cos B, c2 = a2 + b2 – 2ab cos C.
Fig. 13.1
Fig. 13.2
Hence, in any case, utilizing the Pythagorean identity we get:
c2 = (b sin C )2 + (ab cos C )2
= b2 sin2 C + a2 – 2ab cos C + b2 cos2 C
= a2 + b2(sin2 C + cos2 C ) – 2ab cos C
= a2 + b2 – 2ab cos C.
Solution
EOS
Problems & Solutions
1. Find the values of the following quantities. Don't use tables or calculators.
Solution
2. Express the following quantities in terms of sin x or cos x or both.
Solution
3. Prove the following identities.
Solution
Solution
5. Let ABC be an arbitrary triangle with sides a, b, and c opposite to angles A, B, and C respectively.
Solution |
This 5th grade probability course with definition of a random experiment, outcomes and events is very beneficial. At the end of this chapter, you should be able to calculate a probability in a random experiment.
## I. Randomized experiment
Definition:
The probability of an event is a number, between o and 1, which measures the chances that this event will occur.
Example:
• A coin is tossed: the outcomes are heads or tails.
• We launch the listening of music pieces in random mode among a list of ten titles. The outcomes are the ten titles on the list.
• We try to guess in advance the winner of the soccer world cup among the 32 teams of the final phase. The outcomes are the thirty-two countries in competition.
• A six-sided playing die is rolled. The outcomes are 1,2,3,4,5,6.
Definition:
When the outcomes of a random experiment are all equally likely to occur, i.e. the probabilities of occurrence of the different outcomes are equal, we say that there is equiprobability.
Example:
• We launch the listening of a piece of music in random mode among a list of ten titles. “The piece played is less than three minutes long” is an event.
• We try to guess in advance the winner of the soccer world cup among the 32 teams of the final phase. “The winning country is an African country” is an event. The winning country won its semi-final is a sure event.
• A six-sided playing dice with numbers from 1 to 6 is thrown. “The die falls on an even number” is an event. “The die falls on the number 9” is an impossible event.
## II. probability calculations
Definition:
An experiment is said to be random when we cannot predict what the outcome will be.
The different possible outcomes are called the outcomes of the random experiment.
Definition:
An event is a set of outcomes of a random experiment.
When an event is certain to occur, it is said to be certain.
When there is no chance that an event will happen, it is said to be impossible.
Ownership:
In the case of equiprobability, the probability of an event is obtained by dividing the number of outcomes favorable to the event by the total number of outcomes of the experiment.
Examples:
• A balanced coin is tossed. Each side has as much chance of being obtained as the other. This is a situation of equiprobability. The probability of getting tails is therefore or 50%.
• A classic six-sided, unrigged playing die is rolled. Each side has as much chance of coming out as any other. Of the six sides of the die, three sides have an even number and three sides have an odd number, so the probability of getting an odd number is or 50%.
Remark:
The probability of an impossible event is 0 and the probability of a certain event is 1.
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26
Q:
# The size of the wooden block is 5 x 10 x 20 cm .How many such blocks will be required to construct a solid wooden cube of minimum size?
A) 6 B) 8 C) 12 D) 16
Explanation:
Side of smallest cube = L.C.M of 5, 10, 20 = 20cm
Volume of the cube = (20 x 20 x 20)cu.cm = 8000 cu.cm
Volume of the block= (5 x 10 x 20)cu.cm = 1000 cu.cm
Number of blocks = (8000/1000) = 8
Q:
Which measurement unit represents volume?
A) milliliter B) Cubic centimeter C) Meter cube D) All of the above
Explanation:
3 186
Q:
The areas of three consecutive faces of a cuboid are 12 cm², 20cm² and 15 cm², then the volume (in cm³) of the cuboid is
A) 60 B) 55 C) 45 D) 40
Explanation:
From the given data,
let the length, breadth and height of the cuboid are m, n, r
m x n = 12
n x r = 20
r x m = 15
Hence, m x n x n x r x r x m = 12 x 20 x 15
mnr = sqrt of (12x20x15) = 60 cub.cm.
6 276
Q:
If the area of three Adjacent faces of a cuboidal box are 120 cm², 72 cm², and 60 cm² respectively, then the volume of the box is
A) 640 cub.cm B) 720 cub.cm C) 860 cub.cm D) 945 cub.cm
Explanation:
1 382
Q:
Solve the triangle b=36 a=38 c=18 for perimeter of the triangle?
A) 74 units B) 92 units C) 56 units D) 54 units
Explanation:
We know, the perimeter of a triangle = sum of all the sides of a triangle = a + b + c
Here given that sides a = 38, b= 36 and c = 18
Now, perimeter of the triangle = a + b + c = 38 + 36 + 18 = 92 units.
3 392
Q:
The circumference of a circular base of Cylinder is 44 cm and its hight is 15 cm. Then the volume of the cylinder is?
A) 2759 cub. cm B) 2247 cub. cm C) 2614 cub. cm D) 2311 cub. cm
Explanation:
Given Circumference of a circular base of a cylinder = 44 cm
We know that Circumference of a circle =
Now given the height of a cylinder h = 15 cm
19 1686
Q:
What is the Formula for Volume?
The formula to find the volume is given by just multiplying Length, Breadth and Height of that object.
Thus, V = L x B X H
Volume Formulae :
1. Volume of Cube of side 's' = s x s x s = $s3$ cubic units.
2. Volume of Cylinder$πR2h$ cubic units.
[Where is Radius and h is height]
1507
Q:
If radius and height of a cylinder increase by 20% and15% respectively. then what is the % change in curved surface area?
A) 41% B) 38% C) 33% D) 44%
Explanation:
% Change in Curved Surface Area is given by
17 4129
Q:
A town has a population of 4000 requires 140 liters of water per head. It has a tank measuring 18m x 12m x 8m. The water of this tank will suffice for ____ days?
A) 3 B) 4 C) 5 D) 2 |
# Advice 1: How to explain fractions
In the framework of a school course of mathematics the students are confronted with non — integer numbers- fractions. To mathematical operations with fractions were understandable to the child, it is necessary to explain what is a fraction. This can be done using ordinary things and the examples around.
You will need
• cardboard circle divided into equal sectors;
• - items that can be divided (apples, candy, etc.).
Instruction
1
Take a pear and offer it to two children at once. They will tell you that it's impossible. Cut the fruit and again invite the children. Each will get the same half. Thus, half of the pear is part of the pear. And the pear consists of two parts.
2
One half is the part from the whole, 1/2. So a fraction is a number that is part of the object is smaller than one. Also the fraction is the number of parts from some things. Specific things children can understand much easier than the abstract the abstract concepts.
3
Take two candies and ask the child to divide them equally between two people. He can easily do so. Take one candy and then ask him to do the same. There is a way out, if the candy is cut in half. Then you and the child will have one whole candy bar and half - half candy.
4
Use the cut cardboard circle that you can divide by 2, 4, 6, 8 parts. Count with the child how many in the circle parts, for example, six. Pull out one section. This will be part of the total number of sections (6), that is, one-sixth.
5
How many pieces did you take is the numerator, that is, one. The denominator is how many parts did you divide the circle, i.e., six. So, the fraction shows the ratio of the removed partitions to their total number. If you take the four sections are then pulled out of the five sections, and hence the fraction takes the form — 5/6.
6
If counting is mastered the child is good, offer to play him in a familiar game, slightly changing the rules. Draw on the pavement in chalk classics and put not a natural number (1, 2, 3...), and fractional (1, 1 1/2, 2, 2 1/2...). Explain to your child that numbers are intermediate values. For these purposes, you can use a ruler.
7
Explain that the number zero cannot be the denominator. Zero means "nothing", and nothing share impossible. For clarity, draw a sign that the child worked visual memory and he remembered this rule.
# Advice 2: How to resolve fraction in mathematics
The solution of numerical fractions is to hold over them different operations. Addition, subtraction, division, multiplication of fractions is carried out according to certain rules as other actions. Many of them are performed by calculating a common denominator and bring it to each member of the expression. The solution to fractions with a selected part is only after their reduction to the wrong kind. Obtained as the result of any operation with fractions fractional value must be reduced.
Instruction
1
Record the source expression. All fractions with integer part lead to the wrong kind. To do this, multiply the integer part of the fraction to its denominator. To the result add the numerator - the resulting value will be the new numerator of the improper fraction. Further, all operations follow this form of fraction.
2
When you add or subtract fractions, find their common denominator. In the General case, the common denominator is the product of all denominators solve fractions. Multiply the numerator of each fraction by the denominator of the other fraction. If the operation is performed over more than two fractions, the numerators must be multiplied by the product of the denominators of the other fractions.
3
Write down the resulting expression is a fraction where the denominator is just found a common denominator. Calculate the numerator of the resulting fraction. It is the result of operation (addition or subtraction) over all the given denominators solve fractions.
4
In order to perform the multiplication operation, in turn multiply the numerators and the denominators of the original fractions. The resulting works write the resulting fraction as the numerator and the denominator, respectively.
5
Before the operation of division write down the original fractions. Then perform the coup of the fractionyou are dividing. Then perform the multiplication of fractions, as described above. The result will be equal to the quotient of the given fractions.
6
Sometimes fractions are of the form "of the four" expressions. This means that the upper roll must be cut at the bottom. Write a division operation using the symbol ":" and perform the division of fractions similar to the above described.
7
Received the fractional result of any actions reduce to the maximum possible number. To reduce the divide both the numerator and denominator of the fraction to the same integer. The result of the division must also be an integer. The total value write down in response.
# Advice 3: How to explain fractions
Specific values acquired by children is much better than the abstract. How to explain to a childwhat is two thirds? The concept of fractions requires a special presentation. There are some methods to help you understand what is a non-integer number.
You will need
• - special Lotto;
• - Apple and candy;
• circle of cardboard, consisting of several parts;
• - chalky.
Instruction
1
Try to interest the child. On the walk, play a special classics. If in normal you have to jump tired, and the expense of child well-mastered, try this option. Draw hopscotch with chalk on the pavement as shown in the picture and explain to your child that you can jump so: 1 - 2 - 3... and 1 - 1,5 - 2 - 2,5 ... the Kids love to play and so they better understand what between the numbers, there are intermediate values. Is your first and solid step towards the study of rational numbers. A great visual aid.
2
Take the whole Apple and offer it simultaneously to two children. They immediately will tell you that this is impossible. Then cut the Apple and invite them again. Now everything is in order. everyone got the same half of the Apple. This is part of one whole.
3
Invite the child to share four candy with you in half. It is easy to do. Then get another one and offer to do the same. It is clear that a candy can't get from you and the child. You can find a way, cutting the candy in half. Then everyone has to turn out two candies and one half.
4
For older children, use split circle. Divide it by 2, 4, 6 or 8 parts. We invite children to take the circle. Then divide it into two halves. Of the two halves will perfectly round, even if you share half with a neighbor on the Desk (the circles should be of the same diameter). Loan divide each half hedgehog half. It turns out that the circle can consist of 4 parts. And each half turns of the two halves. Then on the Board write it in the form of a fraction. Explaining what the numerator (how many pieces picked up) and the denominator (how many parts just divided). So the children easier to learn a difficult concept - a fraction. |
QUADRATIC RELATIONSHIPS
Overview of Flyer
Everywhere we go, we are interactive or exposed to quadratics and parabolas.To get this type of relation you must have the same second differences. Unlike linear relations where the first difference is equal. The second differences make the curve or in mathematical terms called a parabola.
In quadiratics we will learn
1. important term a definitions
2. second differences/ first differences
3. types of equations
4. factoring
5. graphing
6. converting
7. solving
8. word problems
9. test reflection
10. connection table
11. more videos and links
12. assesments
Terms & Defintions
• X-intercept- where the parabola intercepts with the x-axis
• Y-intercepts- where the parabola intercepts with the y-axis
• Zeros- the x value which makes the equation equal to zero
• Axis of symmetry- the vertical line which cut the parabola down in the middle
• Optimal value- the tip of the parabola the lowest or highest point of the parabola
• Vertex the maximum or minimum point on the graph changes direction
• These terms are very important to know since they will be refered to in this unit and within in the flyer
Forms of Equations
There are 3 forms that quadratic equations can be written in.
Standard form
To graph a standard form equation you must convert it. Look in the converting section for more information on this. The standard form equation is written as y= ax^2+ bx+ c.
Vertex Form
Vertex form is one the most efficient ways a equation can be written so that it is easily graphed. The form of it is y= (X- h)^2+x. Below the diagram show the different parts of a vertex form equation. The vertex is the h and the x variable. The x value is the h and y value is x. If any part of the vertex is in a bracket it will be the opposite sign. In this case -h wont be -h but +h.
• Vertical reflection- if the sign is negative the parabola will open down if it is positive the parabola will open up
• vertical stretch/ compression- if the number is higher than on it is a stretch if it is less than one it is a compression
• horizontal translation- the x value in the vertex whether it is negative or positive it is still translation
• vertical translation- opposite of horizontal translation, y value of vertex
Factored Form
Factored form is another easy way graphing could be done. In this form the x- ints/ zeros are given and we have to find the vertex. The form it is written in is y= a(x-x1)(x-x2). Instruction on how to find vertex graph etc..., can be found in the graphing video in the graphing portion.
First & Second Differences
The chart above is the process to find out if your data is a parabola without graphing. You must know how to find first differences(fd). To start off you must label the numbers as y1, y2, y3 ... As I have done. In my data the number 0 is y1 and 3 is y2. To get 3 as my fd I have subtracted y1 from y2. 0 subtracted from 3 is 3. This is the process of finding fd you just continue, so for the next fd you have to do y4-y3. What is new is ,second difference. The process must be used to find second differences( sd) except we have to subtract the fd. So basically label the fd as y1, .... And subtract y1 from y2 ect. Now that you learned how to find fd ans sd try to fill in the chart below on a paper for practice.
Factoring ( simple )
Factoring simple trinomials is a very simple process as the process is called "simple" trinomials. To start off for any type of factoring you must know expanding and simplifying. Below is a example of expanding and simplifying and factoring a simple trinomial.
Since we have a bracket we must expand it first according to BEDMAS
Once the bucket is expanded you must collect like terms to simplify the equation
At this point your equation should be full expanded and simplified.
A simple trinomial is written in x squared + ax+ b. x squared can be square rooted and become ( x) (x) into two different brackets. As shown in the above diagram the la two number ax and b have something in common which is two numbers that are multiplied = b and the same numbers multiplied = ax. In the case above ax= 5x and b= 6. So the start off start siting the factors of six. Look below.
Now we just plug in the two numbers into the blanks from the original equation.
One last step is to check that your answer is right. You will do this be expanding and simplifying the factored equation. Look below.
Factoring ( common)
Common factoring is similar to similar factoring. When the equation has all parts that have a common factor this is the write formula to us. An example that you should us common factoring in is the equation below
Now I will be showing you how to factor this equation using common factoring methods.
Now you just simple factor.
Factoring ( Complex)
Now thing will be getting a little more complicated so be sure to understand all above aspects so you can understand the next part you have to learn.
So complex trinomials are written in a similar form to common trionomial put don't have any common numbers. There we are not able to factor it using common factoring methods. Below you can look at the way you can factor a complex trinomial. Also read the description ahead to farther understand this form of factoring.
Now you will need to use some of the factoring methods listed above.
Factoring ( Differ of Squares)
Factoring in difference of squares is used when 2 squares are subtracted.
Factoring ( perfect square)
This is one the most hardest form of facting so make. Sure you oay full attention!
Graphing
Graphing vertex form, standard form and factored form will be demonstrated in the video below . For vertex form it will also include the step pattern.
Graphing vertex form
Quick Way of Graphing a Quadratic Function in Vertex Form
Graphing a parabola in vertex form
Graphing factored form
IMG 14941
Graphing standard form
IMG 14971
Converting
May times people experience difficult in converting certain forms into a different form. This part will basically put the whole unit together for you. Please pay detailed attention to the next bit of information. To start of with there are three ways we can convert our different forms , factoring , completing the square and by using the quadratic formula.
Factoring- is used when values are easily factor able, for example x^2+2x+3 (follow steps above in Factoring)
Completing square- Is used when values are factor able, but hard to think of mentally. Look below for the correct way of performing this function.
Quadratic formula- is used when equation is complex or hard to factor, This equation consist of x=-b+- (square root of b^2-4ac)/ 2a. this gives us the two zeros, x intercepts or roots. Look below for the correct way of performing this function. ( DISCRIMINANT WILL BE INCLUDED)
completing the square: is used to change standard form to vertex form.
work is still to be added.
While using the quadratic formula before square rooting the final number in the square root is called the discriminant. When you have the discriminant you know if you formula is going to have a possible answer if the number is not negative. If the number is negative we know that there is no square root for that so the answer will be not possible. You can also calculate the discriminant by a formula that is called the discriminant which is b^2-4ac.
Solving
When you have an equation and you must find the the variable value use the solving method. below are steps and examples of solving equations.
Word Problems
In the unit of quadratics there are many times you will come across word problems. For help in solving word problems look at the variety of word problems below.
Word problems can be tuff, but if you follow the steps and read the problem correctly.
Step 1-
Read the problem and write down given information
Step 2-
Try to think of an equation that can solve the problem
Step 3-
Use factoring, graphing, converting, solving to solve the problem.
examples of word problems
1. Jasdeep is selling T- shirts. his regular price is 20 dollars and he usually sells 15 t- shirts a day. Mandeep finds that for each reduction in price of 1 dollar he can sell to more t- shirts
a) create an algebraic model to represent Jasdeep total sales revenue.
b) determine the maximum revenue and the price at which this maximum revenue will occur.
( THE ANSWER TO THE WORD PROBLEM WILL BE SHOWN BELOW, SOAK IN THE WAY OF SOLVING IT IN)
2. The garden is enclosed on three sides using 60m of fencing. the remaining side is formed by the wall of a garage. draw a diagram. What dimensions enclose 450m^2 of garden?
( THE ANSWER TO THE WORD PROBLEM WILL BE SHOWN BELOW, SOAK IN THE WAY OF SOLVING IT IN)
3. a right angle triangle has a hypotenuse 2x+1, and the two other sides are x and x+4. determine the value of x, and then the length of each side.
( THE ANSWER TO THE WORD PROBLEM WILL BE SHOWN BELOW, SOAK IN THE WAY OF SOLVING IT IN)
Comments on Tests
Below are some test i did and a brief explanation why I performed weak or why i did good please take a look so you don't make the same mistakes. Also try some questions out for practice.
THE QUESTION BELOW IS A QUESTION FROM VERTEX FORM PARABOLAS, AND IT WAS ASKING TO EXPLAIN THE TRANSFORMATION.( THIS IS QUESTION NUMBER 9).
The reason i got perfect on this question was because i explained every single part of the transformation. The equation was written in vertex form which meant the vertical reflection, horizontal translation, vertical translation and vertical stretch is shown. In this case there was a vertical reflection since the a value was a negative( -2 ). The horizontal translation was 3 units right sine the x value of the vertex is 3 and 3 is to the right of the center line in a graph. The vertical translation was 5 units up since the y value of the vertex is +5. this was a vertical stretch due to the number in front of the bracket (a) being a number higher than 1 which meant there was a change in the step pattern.
Below is a question on our first quiz for quadratics.
In the question above question number 3 my calculations were right but the sign was wrong which ca cause your answer to wrong. Always remember not to get mixed up and check your work twice for small mistakes like mine due to carelessness.
Additional Information
WATCH THE VIDEOS BELOW FOR MORE EXPLANATIONS AND BETTER WAYS OF REMEBERING FORMULAS
Quadratic Inequalities
❤² How to Solve Quadratic Equations By Factoring (mathbff)
❤² How to Solve By Completing the Square (mathbff)
THE VIDEO BELOW MAY INCOURAGE YOU!!!!!!!!
You're Not Bad At Math, You're Just Lazy
Algebra - Quadratic Functions (Parabolas)
Algebra - Quadratic Formula
Algebra - Completing the square
Solve Quadratic Equations using Quadratic Formula
Quadratic Formula Cup Song
The "One Direction" Quadratic Formula Song
Quadratic Formula Song to Adele's "Rolling In The Deep"
Quadratic Formula Song Rockford Christian
Quadratic Formula - the Musical
Quadratic Formula Pop Goes the Weasel
The Quadratic Equation Song
Do The Quad Solve (WSHS Math Rap Song)
Graph! (WSHS Math Rap Song)
My Reflection
Over all, quadratics is very simple to understand. Their are many ways to understand these problems, including factoring, converting, graphing and solving. if you follow the steps correctly and read the question carefully, guaranteed you will become a pro in understanding quadratic relations. at last I hope you found my flyer helpful, and I hope you have now become a pro in understanding quadratic equations.
Assesment
After you are done cursing this flyer you should be ready to asses yourself, visit the link below.
My Message
Now that you have went through this website i would like to say thank you. this unit could be a disaster if you don't follow trough so always pay attention. good luck learning. Once again THANKS. |
# Geometrical problems in coordinate geometry
Lesson
Many geometrical properties of figures can either be verified or proved using coordinate geometry.
There is a range of established results that become useful in this endeavour. These include the following six results:
1. The distance formula given by $d=\sqrt{\left(x-x_1\right)^2+\left(y-y_1\right)^2}$d=(xx1)2+(yy1)2
2. The slope formula $m=\frac{y_2-y_1}{x_2-x_1}$m=y2y1x2x1
3. The perpendicular property $m_1m_2=-1$m1m2=1 and the parallel property $m_1=m_2$m1=m2
4. The midpoint formula is given by $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$(x1+x22,y1+y22)
5. The point/slope form for the equation of a line $y-y_1=m\left(x-x_1\right)$yy1=m(xx1)
6. The perpendicular distance formula $d=\frac{\left|ax_1+by_1+c\right|}{\sqrt{a^2+b^2}}$d=|ax1+by1+c|a2+b2.
Here are some examples:
##### Example 1:
Show that the quadrilateral with vertices given by $P\left(2,3\right),Q\left(3,6\right),R\left(6,8\right),S\left(5,5\right)$P(2,3),Q(3,6),R(6,8),S(5,5) is a parallelogram.
A parallelogram is a closed four sided figure with opposite sides parallel. We can check the slopes of the line segments $PQ$PQ and $RS$RS and the slopes of the line segments $QR$QR and $PS$PS.
$m_{PQ}=\frac{6-3}{3-2}=3,m_{RS}=\frac{5-8}{5-6}=3$mPQ=6332=3,mRS=5856=3
$m_{QR}=\frac{8-6}{6-3}=\frac{2}{3},m_{PS}=\frac{5-3}{5-2}=\frac{2}{3}$mQR=8663=23,mPS=5352=23.
Hence, the quadrilateral is a parallelogram.
##### Example 2:
Prove that $A\left(-4,-4\right),B\left(-1,-1\right),C\left(1,-3\right)$A(4,4),B(1,1),C(1,3) are the vertices of a right-angled triangle.
The easiest way to do this is to use the perpendicular property $m_1m_2=-1$m1m2=1 given above.
The three slopes are determined as $m_{AB}=\frac{-1-\left(-4\right)}{-1-\left(-4\right)}=1$mAB=1(4)1(4)=1$m_{BC}=\frac{-3-\left(-1\right)}{1-\left(-1\right)}=-1$mBC=3(1)1(1)=1 and $m_{AC}=\frac{-3-\left(-4\right)}{1-\left(-4\right)}=\frac{1}{5}$mAC=3(4)1(4)=15
Note that $m_{AB}\times m_{BC}=-1$mAB×mBC=1 and so the side $BC$BC is perpendicular to the side $AB$AB, and the triangle is right-angled.
##### Example 3:
Prove that the quadrilateral with vertices $A\left(4,9\right),B\left(5,13\right),C\left(9,14\right),D\left(8,10\right)$A(4,9),B(5,13),C(9,14),D(8,10) is a rhombus.
If we look at distances first, we find:
$d_{AB}=\sqrt{1^2+4^2}=\sqrt{17},d_{BC}=\sqrt{4^2+1^2}=\sqrt{17}$dAB=12+42=17,dBC=42+12=17
$d_{CD}=\sqrt{1^2+4^2}=\sqrt{17},d_{DA}=\sqrt{4^2+1^2}=\sqrt{17}$dCD=12+42=17,dDA=42+12=17
Also, evaluating slopes,
$m_{AB}=\frac{4}{1}=4,m_{BC}=\frac{1}{4}$mAB=41=4,mBC=14
$m_{CD}=\frac{-4}{-1}=4,m_{DA}=\frac{1}{4}$mCD=41=4,mDA=14
The product of any two of the slopes is never $-1$1, and so the figure is not a square. In fact it is a parallelogram with equal sides, making it a rhombus.
## An extension example
Prove that the medians of any triangle meet at an interior common point $G$G.
This is a challenging proof, but a proof worth going through, because it makes use of many of the coordinate geometry results listed above.
### Describing the problem
A median is a line segment drawn from a vertex to the midpoint of the opposite side. Thus there are three medians of a triangle, and it can be proved using a number of methods that the three medians intersect each other at a common point $G$G called the centroid (also called the centre of gravity).
One of these proofs uses coordinate geometry, and will be given below.
### The diagram
While it is a completely general proof, we can reduce the algebra involved by setting the triangle up so that one edge lies along the $x$x axis in such a way as to put the midpoint of that edge at the origin.
We then can label the vertices $P\left(-a,0\right),Q\left(b,c\right)$P(a,0),Q(b,c) and $R\left(a,0\right)$R(a,0), as shown in the diagram:
The midpoints $D$D, $E$E and $O$O and the three medians have been added to the diagram as well, although at this stage, we haven't proved that they are concurrent at $G$G.
### The plan of attack
The first step in the proof is to find the coordinates of $G$G using two of the medians (the ones shown in black)
The second step is to show that the third median (in blue) also passes through $G$G.
### Finding G
The equation of the line through $O$O and $Q$Q (passing through the origin) is given by:
$y$y $=$= $\frac{c}{b}x$cbx $(1)$(1)
This is because the slope is given by $m_1=\frac{c}{b}$m1=cb and the $y$y intercept is $0$0.
The coordinates of $D$D, using the mid-point formula, become $\left(\frac{b-a}{2},\frac{c}{2}\right)$(ba2,c2)
The slope of the line through $D$D and $R$R is then $m_2=\left(\frac{\frac{c}{2}-0}{\frac{b-a}{2}-a}\right)$m2=(c20ba2a).
By multiplying the numerator and denominator by $2$2, this simplifies to $\left(\frac{c}{b-3a}\right)$(cb3a).
Then, using the point $R\left(a,0\right)$R(a,0) and $m_2$m2, the equation of the line through $D$D and $R$R becomes:
$y$y $=$= $\left(\frac{c}{b-3a}\right)\left(x-a\right)$(cb−3a)(x−a) $(2)$(2)
Finally, we need to solve simultaneously equations $(1)$(1) and $(2)$(2) to find the coordinates of $G$G.
Thus we set the two right hand sides equal to each other and solve for $x$x:
$\frac{c}{b}x$cbx $=$= $\left(\frac{c}{b-3a}\right)\left(x-a\right)$(cb−3a)(x−a) $\frac{1}{b}x$1bx $=$= $\left(\frac{1}{b-3a}\right)\left(x-a\right)$(1b−3a)(x−a) $bx-3ax$bx−3ax $=$= $bx-ab$bx−ab $-3ax$−3ax $=$= $-ab$−ab $\therefore$∴ $x$x $=$= $\frac{b}{3}$b3
With $x=\frac{b}{3}$x=b3, substituting back into $y$y shows $y=\frac{c}{b}x=\frac{c}{b}\times\frac{b}{3}=\frac{c}{3}$y=cbx=cb×b3=c3.
So the coordinates of $G$G are $\left(\frac{b}{3},\frac{c}{3}\right)$(b3,c3), but we still have a little further to go. We still haven't shown that the third median $PE$PE intersects the other medians at $G$G
### Proving concurrency
To do this we simply show that the slope of $PE$PE is the same as $PG$PG.
The coordinates of $E$E, using the midpoint formula is $\left(\frac{b+a}{2},\frac{c}{2}\right)$(b+a2,c2).
Using our slope formula, we have for the slope of $PE$PE:
$m_{PE}$mPE $=$= $\frac{\frac{c}{2}-0}{\frac{b+a}{2}-\left(-a\right)}$c2−0b+a2−(−a) $=$= $\frac{c}{b+a+2a}$cb+a+2a $=$= $\frac{c}{b+3a}$cb+3a
and the slope of $PG$PG is:
$m_{PG}$mPG $=$= $\frac{\frac{c}{3}-0}{\frac{b}{3}-\left(-a\right)}$c3−0b3−(−a) $=$= $\frac{c}{b+3a}$cb+3a
This has to imply that all of the medians are concurrent.
### Centre of Gravity
Because our proof is a general one, we learn an amazing fact about the centroid.
If we consider the point $S$S whose coordinates are the average the $x$x and $y$y parts of the coordinates of the vertices $P\left(-a,0\right),Q\left(b,c\right)$P(a,0),Q(b,c) and $R\left(a,0\right)$R(a,0), we find that:
$S=\left(\frac{-a+b+a}{3},\frac{0+c+0}{3}\right)=\left(\frac{b}{3},\frac{c}{3}\right)=G$S=(a+b+a3,0+c+03)=(b3,c3)=G
In other words, the coordinates of the centroid are found by averaging the coordinates of the vertices!
In fact the centroid is the balance point of the triangle, and that's why it is also called the centre of gravity.
#### Worked Examples
##### Question 1
Consider the triangle shown below:
1. Determine the slope of the line segment $AB$AB.
2. Similarly, determine the slope of side $AC$AC:
3. Next determine the length of the side $AB$AB.
4. Now determine the length of the side $AC$AC.
5. Hence state the type of triangle that has been graphed. Choose the most correct answer:
An equilateral triangle.
A
An acute isosceles triangle.
B
An isosceles right-angled triangle.
C
A scalene right-angled triangle.
D
An equilateral triangle.
A
An acute isosceles triangle.
B
An isosceles right-angled triangle.
C
A scalene right-angled triangle.
D
##### Question 2
1. First determine the slope of the line segment $AB$AB.
2. Next determine the slope of side $CD$CD.
3. Similarly, determine the slopes of the other two sides:
Slope of $AD$AD $=$= $\editable{}$
Slope of $BC$BC $=$= $\editable{}$
4. Determine the length of the side $AB$AB.
5. Now determine the length of the side $AD$AD.
6. Hence state the type of quadrilateral that has been graphed. Choose the most correct answer.
Rectangle
A
Rhombus
B
Square
C
Trapezoid
D
Rectangle
A
Rhombus
B
Square
C
Trapezoid
D
##### Question 3
Given Line P: $y=-6x-4$y=6x4, Line Q: $y=\frac{x}{6}+6$y=x6+6, Line R: $y=-6x-1$y=6x1 and Line S: $y=\frac{x}{6}+1$y=x6+1.
1. Complete the following:
$m$mP = $\editable{}$
$m$mQ = $\editable{}$
$m$mP x $m$mQ = $\editable{}$
2. Complete the following:
$m$mQ = $\editable{}$
$m$mR = $\editable{}$
$m$mQ x $m$mR = $\editable{}$
3. Complete the following:
$m$mR = $\editable{}$
$m$mS = $\editable{}$
$m$mR x $m$mS = $\editable{}$
4. Complete the following:
$m$mS = $\editable{}$
$m$mP = $\editable{}$
$m$mS x $m$mP = $\editable{}$
5. What type of quadrilateral is formed by lines: P, Q, R, and S?
Trapezoid
A
Rectangle
B
Rhombus
C
Parallelogram
D
Trapezoid
A
Rectangle
B
Rhombus
C
Parallelogram
D
### Outcomes
#### 10D.AG2.05
Solve problems involving the slope, length, and midpoint of a line segment
#### 10D.AG3.02
Verify, using algebraic techniques and analytic geometry, some characteristics of geometric figures |
# Advice 1: How to quickly learn the part number
From elementary school students need a great to be able to count orally. Parents have to wonder how to explain that, for example, 8 is 5 and 3? For the successful completion of a course of mathematics it is necessary to try still to school to learn with the child part numbers.
You will need
• - counting sticks;
• simple household items for the invoice (apples, sweets)
• - improvised teaching AIDS - number the houses or cards.
Instruction
1
Try to explain to a child the difference between the numbersmi and the number. The number denotes the number on the letter and numbers are a designation of the number of items. For example, if you have seventeen apples, explain that 17 is the number, quantity, and its compositionrepresent the numbers 1 and 7. Take ten apples, you will have seven. Explain to your child that the number of apples was seven and it expresses the number 7. Seven can be decomposed into other numbers — 1, 2, 3 and so on.
2
Show the children the part number in the illustrative examples. Take, for example, three candies. Ask the child to count how many you have candy. Now divide the candy or two put on the table, and one hold in your hands. Ask your child how they are now. The answer is the same. Explain what two candies with one and Vice versa, one two, partgive three. Now put one candy on the second, and a third hold in your hands. Show your child one candy, here's another one and another. So three is repeated three times. Secure knowledge on the counting sticks.
3
Draw with the child on paper a numerical houses. These houses high-rise buildings, on each floor there are two apartments. Write in the triangle-the roof is a number from 2 to 18. Explain that on the same floor lives as tenants, how much is host number. Use counting sticks, cubes or other materials that help the child to "relocate" residents.
4
For example, suppose a host is the number 6. Select 6 of the sticks. On the ground floor in one of the apartments lives let one person move the stick. Therefore, in another apartment five residents. So, six is five and one. Thus, populating numeric house, get the pairs 1 and 5, 2 and 4, 3 and 3, 4, and 2, 5 and 1 total in numeric house five floors. For greater efficiency hang such posters houses in the apartment and periodically ask the child.
5
Involve child in the decision of ordinary household tasks. For example, if your family is three people, invite a child task of the following type. Put on the table one plate. Ask the child a question, how many plates you need to put if all in the family three people. He must answer you that you need to put two more plates. Consequently, one and two plates compositionrepresent three plates. Make cards with the composition of theAMI of different numbers and review them with your child.
Note
Do not engage with the child for too long. Optimal — time classes for 10 to 15 minutes. Otherwise the kid will just get tired and benefits of such a study will be.
# Advice 2 : How to learn composition of numbers
The problem of memorizing the composition of the numbers from 1 to 18 there is a lot of first graders. First and foremost, this is due to the fact that you need to memorize abstract information. What does it mean for a child, the phrase "7 is 3 and 4"? Absolutely nothing. So all the work on the memorization and automation knowledge of part numbers should be on the illustrative example and to be understood by the child.
You will need
• 1. The paper and cardboard.
• 2. Markers.
• 3. Handle.
Instruction
1
Form a child the notion that number is the designation of the number of items and the number needed to refer to the number on the letter.
2
Use in the classroom moments of the game. On sheets of colored cardboard, markers to draw with your child a numeric houses. Numeric house is a multi-storey building, each floor is two apartments. In the triangle, denoting the roof, write a number from 2 to 18. Explain to your child that on the same floor can live as tenants, how much is the number - the owner of the house. With your child using counting sticks, blocks and other material for the account, "Russell residents in the apartments." For example, the owner of the house - the number 5. Take 5 sticks - are tenants. On the ground floor in one apartment lives 1 people, 1 move wand. Then in another house lives 4 people. So 5 is 1 Yes 4. "Settling" of the house, get the pairs 1 and 4, 2 and 3, 3 and 2, 4 and 1. Thus, in numeric house, indicating the composition of the number 5, will be 4 floors.
3
Hang a numerical houses in the apartment so that the child saw them as often as possible. To memorize part numbers, close in numeric house right or left column of numbers. The kid calls the neighbor of a number. For example, 9 is 3 ? 6 - must answer the child.
4
From time to time turn one of the houses and ask the child to draw a house, remembering the part number, on a piece of paper from memory.
5
Engage children in solving simple household problems.
- We are a family of 5 people. I have already put on the table, 3 plates. How many more plates you need to put?
- Right, 2. 5 is a 3 and even 2.
This work is carried out with all the numbers.
Note
Part numbers are included in single digits. So, 18 is 9, and 9. Numeric house only one floor.
You can make cards with examples of addition, to illustrate the composition of numbers (9=4+5, 17=9+8 etc.).
# Advice 3 : How to teach a child to add and subtract
In preparation for school, special attention is paid to training account. It is quite a complex process requires the child's many skills – ability to quickly navigate, to abstract, to decompose numbers into smaller. To learn this best from an early age.
Instruction
1
Give a child the digital cubes in the first year of life help him to build blocks in a row or build of them a turret, calling the numbers. Then share it with the quantitative concept of each number. The Apple one is denoted by 1, two apples are denoted by the number 2, etc. Loudly and clearly pronounce the names of numbers.
2
Use for classroom presentation material. Young children are hard to ignore, so take their explanations of candies, cookies, fruit, toys, pencils, etc. to Teach the child to count and add up to ten is easy. The child is always carrying two hands with 10 fingers that will help to quickly calculate the. To quickly learn the score on the fingers, the child should be trained to quickly show the desired number of fingers. Start with simple numbers – 1 and 2, 5 and 10, 10 and 9. Help your child to cope with bad obeying fingers. Take your time, let the child thinks slowly.
3
Give a child the part number, that'll make it easier for him. Pick any 6 items, for example, 6 sweets. Decompose them into two parts: 3 and 3 candies. Put one candy in another pile and ask them to count how many pieces of candy (4 and 2). Thus, go through the options for number 6 (1+5, 2+4, 3+3, 4+2, 5+1), stressing that every time we get the number 6. Do not worry if each time the child counts the candy. Be patient. One day he will understand what you are trying to explain. Similarly, remove all numbers from 2 to 10 and later to 20. Do not try to force child classes. Use for explanations of every opportunity: when gives candy to a child, when you cut an Apple into slices, and cake slices.
4
The child is not bored, invent interesting stories where you have to help, for example, brothers-zucchini, split into two groups, and sister-cherries to get couples in their houses or how much left to collect the hedgehog, to have food enough for everyone. Teaching the child addition and subtraction, draw, sculpt with the baby, so he quickly mastered the score.
5
Use the help of special benefits. In almost any bookstore you can find a huge amount of literature on the teaching of mathematics to children of different ages. The guides are designed by teachers and psychologists. Usually they represent simple and interesting job. Do not overload the child with assignments. For the successful development of account 10-15-minute daily sessions will be enough.
Note on methodological developments of Glenn Doman and Nikolai Zaitsev.
# Advice 4 : How to remember the part number
The curriculum requires that a first grader automatic solution examples of addition and subtraction to first one and then the second ten. In order to learn from these examples, must be great to know the part numbers. Younger student it is not easy to remember abstract information. To help him in this task, it is necessary to build the work of remembering the compositionand numbers so that it was clear and understandable to the child.
You will need
• - homemade benefits-table with part numbers;
• - counting sticks.
Instruction
1
Draw on sheets of paper houses in several floors, each floor has two apartments (the window). On the roof of the house with your child write the number and explain that this number is on the roof — the owner of the house that permits you to settle on one floor, only the number of tenants corresponding to the number of-the owner. For a start, use for "moving" counting sticks or matches — it's more visual than just written numbers.
2
Complete the very floors of the first house, and then put this task in front of the child — he has to try to resettle the residents. Let the child pronounces the necessary amount, arguing like this: "the master of the house is the number 6, if one apartment per floor living 2 tenants, the other has to live 4".
3
Change the number of occupants from floor to floor and repeat with a new combination of numbers. In the house with the owner will be 6 combinations 1 and 5, 2 and 4, 3 and 3.
4
Go from counting sticks or matches to the written numbers. It is more difficult for children, so patiently repeat the exercise.
5
Gradually complicate the task. Cover one column of Windows, and pronounce with child neighbors numbers. So, if in the house with the master on the floor 8 lives 3, his neighbor is 5. Gradually the child himself will be able to call the desired number.
6
Instead of houses draw a flower in the center of which is written some number, and each petal is divided into two parts. The greater the number in the middle of the flower, the more petals it should be — on the number of possible composition. In every part of the petal write the number equal to the number of cores. Invite the child to continue to fill the petals of numbersI.
7
Give child a household example of a simple task type "in the sink stood 6 plates, mom washed 4 of them, how many dishes need washed?"
Note
The optimal duration of studies — 10-15 minutes. Longer sessions will bore the baby and benefit from these lessons will not.
Remember — in all activities memory requires regular practice, otherwise the child will quickly switch to more interesting things and forget about the material.
# Advice 5 : How to quickly learn the part number
From elementary school students need a great to be able to count orally. Parents have to wonder how to explain that, for example, 8 is 5 and 3? For the successful completion of a course of mathematics it is necessary to try still to school to learn with the child part numbers.
You will need
• - counting sticks;
• simple household items for the invoice (apples, sweets)
• - improvised teaching AIDS - number the houses or cards.
Instruction
1
Try to explain to a child the difference between the numbersmi and the number. The number denotes the number on the letter and numbers are a designation of the number of items. For example, if you have seventeen apples, explain that 17 is the number, quantity, and its compositionrepresent the numbers 1 and 7. Take ten apples, you will have seven. Explain to your child that the number of apples was seven and it expresses the number 7. Seven can be decomposed into other numbers — 1, 2, 3 and so on.
2
Show the children the part number in the illustrative examples. Take, for example, three candies. Ask the child to count how many you have candy. Now divide the candy or two put on the table, and one hold in your hands. Ask your child how they are now. The answer is the same. Explain what two candies with one and Vice versa, one two, partgive three. Now put one candy on the second, and a third hold in your hands. Show your child one candy, here's another one and another. So three is repeated three times. Secure knowledge on the counting sticks.
3
Draw with the child on paper a numerical houses. These houses high-rise buildings, on each floor there are two apartments. Write in the triangle-the roof is a number from 2 to 18. Explain that on the same floor lives as tenants, how much is host number. Use counting sticks, cubes or other materials that help the child to "relocate" residents.
4
For example, suppose a host is the number 6. Select 6 of the sticks. On the ground floor in one of the apartments lives let one person move the stick. Therefore, in another apartment five residents. So, six is five and one. Thus, populating numeric house, get the pairs 1 and 5, 2 and 4, 3 and 3, 4, and 2, 5 and 1 total in numeric house five floors. For greater efficiency hang such posters houses in the apartment and periodically ask the child.
5
Involve child in the decision of ordinary household tasks. For example, if your family is three people, invite a child task of the following type. Put on the table one plate. Ask the child a question, how many plates you need to put if all in the family three people. He must answer you that you need to put two more plates. Consequently, one and two plates compositionrepresent three plates. Make cards with the composition of theAMI of different numbers and review them with your child.
Note
Do not engage with the child for too long. Optimal — time classes for 10 to 15 minutes. Otherwise the kid will just get tired and benefits of such a study will be. |
Ancient Romans used letters to write numbers even today the building date on monuments and public buildings are marked In Roman numerals and we also use roman numerals in clocks and watches. Go through the complete article to be well versed with details like Definition, Chart, Conversion from Roman Numerals to Numbers, Conversion of Numbers to Roman Numerals. Also, refer to Solved Examples for writing Roman Numerals and understand the concept well.
Below the letters used to write numbers
Symbols I V X L C D M
Numbers 1 5 10 50 100 500 1000
## Roman Numeral Chart for 1-1000 Numbers
Here is the list of commonly used Roman Numerals for Numbers 1-1000. They are as such
### Rules for Writing Roman Numerals
We need to remember three rules while writing roman numerals
• Rule of repetition
• Rule of subtraction
Now let’s see what are these rules
### Rule of Repetition
This rule states that whenever a symbol is repeated twice or thrice, its value is multiplied by 2 or 3 respectively. This means if a symbol is repeated two times then the value of that symbol should be multiplied twice.
Example:
Roman Numerals M MM MMM
Hindu-Arabic Numerals 1000 2000 3000
As we already know Value of M is 1000
When M is repeated twice, the value of M should be multiplied two times which is
M * M = M*2 = 1000 * 2 = 2000.
In the same way, when M is repeated thrice the value of M should be multiplied three times so
M * M * M = M*3 = 1000 * 3 = 3000.
Among all the roman numerals, few symbols I, X, C, and M can only be repeated three times at maximum. Example:
Hindu-Arabic numerals 1 2 3 4
Roman Numerals I II III IV
We can write 1 as I, 2 as II, 3 as III in the Roman system but we can’t write 4 as IIII instead 4 will be written as IV
Among all the roman numerals few symbols V, L, and D should not be repeated. This means the symbols for 5, 50, and 500 can’t be repeated.
Example:
Hindu-Arabic numerals 10 10
Roman Numerals VV X
Explanation:
As stated we do not write the number 10 by repeating V but as X
This rule states that if a symbol of a smaller value is written to the right of a symbol of higher value, we add the smaller value to the greater value.
Example
Let’s take one symbol XI here as we already know X is 10 and I is 1 now we have to add 10 to 1 (10+1) which is 11 so here Hindu-Arabic numeral of XI is 11
### Rule of Subtraction
This rule states that if a symbol of a smaller value is to the left of a symbol of a higher value, we subtract the smaller value from the higher value.
Example:
Let’s take one symbol XC here as we already know X is 10 and the value of I is 100 now we have to subtract 10 from 100 (100 -10) which is 90 so here Hindu-Arabic numeral of XC is 90
Few roman numerals symbols V, L and D can’t be subtracted from any other numbers. Let me give one example for better understating
Example
The Roman numeral of 95 is XCV
95 can’t be written as 95 = 100 – 5 = VC
95 should be written as 95 = 90 + 5 = XCV
As mentioned we should not subtract V from other symbols.
Roman numerals symbols I, X, and C are used to subtract from any other numbers. There are just six combinations that are usually followed while subtraction as mentioned below.
IV = 5 – 1 = 4
IX = 10 – 1 = 9
XL = 50 – 10 = 40
XC = 100 – 10 = 90
CD = 500 – 100 = 400
CM = 1000 – 100 = 900
So these are the 3 rules that everyone should follow while writing roman numerals. |
# CUSTOMARY UNITS OF CAPACITY WORKSHEET
Problem 1 :
Convert 2 pints into cups.
Problem 2 :
Convert 3.5 quarts into cups.
Problem 3 :
Convert 32 cups into quarts.
Problem 4 :
Convert 256 cups into gallons.
Problem 5 :
Convert 24 quarts into gallons.
Problem 6 :
David prepares 60 pints of juice in two hours. At the same rate, How many cups of juice will he prepare in one minute ?
Problem 7 :
Mark used 15840 cups of fuel in 45 minutes. Find the amount fuel used in one minute (in cups).
Problem 8 :
Kemka's little sister needs to take a bubble bath. The package says to put in a drop of bubble bath for every half gallon of water in the bath tub. If bathtub has 12 gallons of water, how many drops can she put into the bath for her sister?
Problem 9 :
Ivan needs gas for his truck. He knows his truck holds 40 gallons of gas. If he is allowed to fill up 8 quarts of gas once in a time, how many times will he have to fill up his gas can to get his truck full of gas ?
Problem 10 :
A bath uses 83 gallons and a shower uses 34 gallons. Mrs. Hitchins has a bath. How much water will be saved if Mrs. Hitchins decides to have a shower ?
2 pints into cups :
Here, we convert bigger unit into smaller unit. So we have to multiply.
2 pints = 2 2 cups
2 pints = 4 cups
So, 2 pints is equal to 4 cups.
Convert 3.5 quarts into cups :
Here, we convert bigger unit into smaller unit. So we have to multiply.
3.5 quarts = 3.5 4 cups
3.5 quarts = 14 cups
So, 3.5 quarts is equal to 14 cups.
32 cups into quarts :
Here, we convert smaller unit into bigger unit. So we have to divide.
32 cups = 32/4 quarts
32 cups = 8 quarts
So, 32 cups is equal to 8 quarts.
256 cups into gallons :
Here, we convert smaller unit into bigger unit. So we have to divide.
256 cups = 256/16 gallons
256 cups = 16 gallons
So, 256 cups is equal to 16 gallons.
24 quarts into gallons :
Here, we convert smaller unit into bigger unit. So we have to divide.
24 quarts = 24/4 gallons
24 quarts = 6 gallons
So, 24 quarts is equal to 6 gallons.
No. of pints prepared in 2 hours = 60
No. of pints prepared in 1 hour = 30
1 hour = 60 minutes
1 pint = 2 cups
1 hour -----> 30 pints
60 minutes -----> 30 2 cups
60 minutes -----> 60 cups
So, no. of cups prepared in 60 minutes is 60.
No. of cups prepared in in one minute is
= 60/60
= 1 cup
So, 1 cup of juice is prepared in 1 minute.
No. of cups used in 45 minutes = 15840
No. of cups used in 1 minute = 15840/45
No. of cups used in 1 minute = 352
So, 352 cups of fuel used in 1 minute.
Half gallon of water -----> 1 drop of bubble bath
1 gallon of water -----> 2 drops of bubble bath
12 gallons of water -----> 12 2 drops of bubble bath
12 gallons of water -----> 24 drops of bubble bath
So, Kemka can put into 24 drops of bubble bath for her sister with 12 gallons of water.
1 gallon = 4 quarts
40 gallons = 40 4 quarts = 160 quarts
So, he needs 160 quarts of gas to make his truck full of gas.
Once in a time, he can fill up 8 quarts of gas.
No. of times of filling to make the truck full of gas is
= 160/8
= 20
So, Ivan has to fill up his gas can 20 times to get his truck full of gas.
No. of gallons used when Mrs. Hitchins has a bath :
= 83 -----(1)
No. of gallons used when Mrs. Hitchins has a shower :
= 34 -----(2)
Water saved = Difference between (1) and (2)
Water saved = 83 - 34
Water saved = 49
So, 49 gallons water will be saved, if Mrs. Hitchins decides to have a shower.
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Helping Your Child Learn Math - June 1999
# Activities
## Math for the Fun of It
During summer vacations, on rainy days, while waiting at the doctor's office, or on a stroll through the neighborhoodlearning never ends. Your children can explore some fascinating mathematical possibilities in the world around them every day. For instance, math can be found outdoors in nature: look for symmetry in leaves; count the number, sizes, and kinds of trees on your street; and look at the various shapes and patterns of blooming flowers. Children will be learning math and enjoying it too! The activities in this section can be done anytime and anywhere.
Guess If You Can
What to do
1. Let your child think of a number between a stated range of numbers while you try to guess the number by asking questions. Here is a sample conversation.
Child: I am thinking of a number between 1 and 100.
Parent: Is it more than 50?
Child: No.
Parent: Is it an even number?
Child: No.
Parent: Is it more than 20 but less than 40?
Child: Yes.
Parent: Can you reach it by starting at zero and counting by 3's?
Child: Yes.
(At this stage, your child could be thinking of 21, 27, 33, or 39.)
3. After you have guessed your child's number, let your child guess a number from you by asking similar questions.
Parent Pointer It is important to help children develop an understanding of the characteristics and meanings of numbers.
What Are the Coins?
What you'll need
Some coins
What to do
1. I have three coins in my pocket. They are worth 7 cents. What do I have? (a nickel and 2 pennies)
2. I have three coins in my pocket. They are worth 16 cents. What do I have? (a dime, a nickel, a penny)
3. I have three coins in my pocket. They are worth 11 cents. What do I have? (2 nickels and 1 penny)
4. I have three coins in my pockets. They are worth 30 cents. What do I have? (3 dimes)
5. I have six coins in my pocket. They are worth 30 cents. What could I have? (1 quarter and 5 pennies or 6 nickels). This problem has more than one answer. It is challenging for children to experience problems like this.
6. I have coins in my pocket, which have a value of 11 cents. How many coins could I have?
You get the idea! Give your child a few coins to figure out the answers.
Parent Pointer Use this activity to help your child develop an understanding of patterns and variables (the unknown) to solve a problem. This is critical to understanding algebra.
What Are My Chances?
What you'll need
Two coins, paper, and pencil to keep score
What to do
Play these games with your child:
1. Flip one coin. Every time it comes up heads, your child gets 1 point. Every time it comes up tails, you get 1 point. Flip it 50 times. Tally by 5's to make it easier to keep track of scores. The person with the most points wins. If one person has 10 points more than the other person does, score an extra 10 points. Does this happen very often? Why not?
2. Flip two coins. If the coins come up two tails or two heads, your child scores 1 point. If it comes up heads and tails, you get 1 point. After 50 flips, see who has more points. Do you think the game is fair? What if one person received 2 points for every double heads and the other person received 1 point for everything else. Is this fair?
3. Flip one coin. Then flip the other. If the second coin matches the first coin, your child scores 1 point. If the second coin doesn't match the first coin, you receive 1 point. Try this 50 times. Is the result the same as in the previous game?
Parent Pointer Understanding probability is essential in many areas of mathematics. Playing games that involve chance is one way to explore the laws of probability.
Money Match
What you'll need
One number cube to roll; 10 of each coin (penny, nickel, dime, and quarter)
What to do
1. For young players (5 and 6 year olds) use only two different coins (pennies and nickels or nickels and dimes only). Older children can use all types of coins.
2. Explain that the object of the game is to be the first player to earn a set amount (10 or 20 cents is a good amount).
3. The first player rolls the number cube and gets the number of pennies shown on the cube. Keep all like coins in batches or stacks of 5 or 10.
4. As each player accumulates 5 pennies or more, the 5 pennies are traded for a nickel. Players take turns rolling the cube to collect additional coins.
5. The first player to reach the set amount wins.
6. Add the quarter to the game when the children are ready. As each player accumulates 5 nickels, they are traded for quarters.
Parent Pointer Counting money and batching in groups of 2's, 5's, or 10's teaches children matching skills and helps in the beginning stages of addition and multiplication. Children also learn how to identify coins and understand their values.
More or Less
What you'll need
One coin, number cards (from book cover), scratch paper, pen, and pencil
What to do
Two players will play a card game where each will draw a card. The players will compare cards to see who wins that round. Before you begin, flip the coin and call "heads" or "tails" to see if the winner of each round will be the person with a greater value card (heads) or a smaller value card (tails).
1. To begin the game, divide the cards evenly between the two players.
2. Place the cards face down. Each player turns over one card at a time and compares: Is mine more or less? How many more? How many less? The player with the greater or smaller value card (depending on whether heads or tails was tossed) takes both cards.
3. The winner of the game is the player with more cards when all the cards are gone from the stack.
4. Now try the same activity with each player pulling two cards and adding them. Which sum is more? How much more? How much less?
Parent Pointer Playing with numeral cards helps children learn to compare quantities of numbers. Children can also learn addition and subtraction.
Problem Solvers
What you'll need
Enough sets of cards so that each player has a set of cards numbered 1 through 6.
What to do
1. Super sums. Each player writes numbers 1-12 on a piece of paper. The object of the game is to be the first one to cross off all the numbers on this list. Use only the cards 1-6. Each player picks two cards and adds up the numbers on them. The players can choose to mark off the numbers on the list by using the total value or crossing off two or three numbers that make that value. For example, if a player picks a 5 and a 6, the player can choose to cross out 11, or 5 and 6, or 7 and 4, or 8 and 3, or 9 and 2, or 10 and 1, or 1, 2, and 8. If a player cannot cross off a number, the player loses the turn. The first player to cross off all the numbers wins.
2. Make the sum of 100.Use only cards 1-6. Each player takes turns drawing a card and each player must take 6 cards from the deck. With each draw, a player decides whether to use the number on the card in the 10s place or the 1s place so that the numbers total as close to 100 as possible without going over. For example, suppose a player draws the following cards in this order: 1, 6, 3, 2, 3, 2, and chooses to use the numerals in the following way:
Parent Pointer This card game helps children develop various ways to use numbers in different combinations and to see the many possibilities of arriving at the same sum by adding different sets of numbers.
Card Smarts
What you'll need
Number cards, pencil, and paper
What to do
1. How many numbers can we make? Give each player a piece of paper and a pencil. Using the cards from 1 to 9, deal four cards out with the numbers showing. Using all four cards and a choice of any combination of addition, subtraction, multiplication, and division, have each player see how many different numbers a person can get in 5 minutes. Players get one point for each answer. For example, suppose the cards drawn are 4, 8, 9, and 2. What numbers can be made?
2. Make the most of it. This game is played with cards from 1 to 9. Each player alternates drawing one card at a time, trying to create the largest 5-digit number possible. As the cards are drawn, each player puts the cards down in their "place" (ten thousands, thousands, hundreds, tens, ones) with the numbers showing. Once placed, a card cannot be moved. The first player with the largest 5-digit number wins. For example, if a 2 was drawn first, the player might place it in the ones' place, but if the number had been an 8, it might have been put in the ten thousands' place.
Parent Pointer This card game helps children develop strategies for using numbers in different combinations using addition, subtraction, multiplication, and division.
Let's Play Store
What you'll need
Empty containers (cartons or boxes), old magazines, books, newspapers, calculator, pencil or crayon, and paper
What to do
1. Help your child collect empty containers so that you can play as if you were shopping at the grocery store. Gather the items and put them on a table.
2. Help your child think of a price for each item. Mark the prices on the containers. You can even mark some items on sale.
3. Pretend to be the customer while your child is the cashier.
4. Teach your child the difference between the math symbols (+, -, ÷, x, and =) and how they are used when using the calculator. Help your child add the prices of each item on the calculator and total the amount using the (=) symbol. Have your child write the total on a piece of paper, which will be your receipt.
5. While you and your child play store, you can ask questions likehow much would it cost to buy three cartons of eggs? How much does 1 box of soap cost, if they are 2 for \$5.00? How much is my bill, if I don't buy the cereal? How much more will it cost if I buy this magazine? Have your child estimate the amounts of the items you are buying. Check to see if the estimation is correct on the calculator.
Parent Pointer Learning to use the calculator will help your child understand and apply estimation and reasoning skills, as well as learn addition, subtraction, division, and multiplication. |
# Math Made Easy: Simplifying Function Discovery with Differential Equations
Welcome to the world of mathematics, where equations and functions dance together in a beautiful symphony of numbers and symbols. Today, we’ll embark on a journey to understand how differential equations can be your trusted companions in the realm of function discovery. If you’ve ever wondered how to unravel complex mathematical relationships, you’re in the right place. This comprehensive guide will simplify the process for you.
## Understanding Differential Equations
### The Basics
Let’s start with the basics. What exactly is a differential equation? Simply put, it’s an equation that involves one or more derivatives of an unknown function. These equations are used to express rates of change and are essential in various scientific and engineering disciplines.
### Ordinary vs. Partial Differential Equations
Differential equations come in two flavors: ordinary and partial. Ordinary differential equations involve a single variable, whereas partial differential equations deal with functions of multiple variables. Think of them as tools in your mathematical toolbox, each suited for different tasks.
## The Basics of Function Discovery
### Defining Function Discovery
In the realm of mathematics, discovering a function means finding an equation that accurately describes a relationship between variables. This is crucial in fields like physics, engineering, economics, and more.
## The Role of Differential Equations
### Why Differential Equations?
You might be wondering, “Why do we need differential equations for function discovery?” Great question! Differential equations provide a powerful framework for describing how quantities change with respect to one another. In essence, they help us understand the underlying dynamics of a system.
## Types of Differential Equations
### Exploring Different Types
Differential equations come in various flavours, each tailored to specific situations. Here are a few types you should be familiar with:
1. First-order differential equations: These involve the first derivative of the function.
2. Second-Order Differential Equations: Here, you’ll encounter the second derivative.
3. Linear Differential Equations: The coefficients of the equation and the derivatives are linearly related.
4. Nonlinear Differential Equations: When linearity goes out the window, things get interesting.
### Relevance to Function Discovery
Each type of differential equation has its own set of applications in function discovery. For instance, a second-order linear differential equation might model the motion of a pendulum, while a nonlinear differential equation could describe population growth in biology.
## Solving Differential Equations
### A Step-by-Step Guide
Now, let’s get down to business. How do you solve differential equations? It’s simpler than you might think. Here’s a basic approach:
1. Separation of Variables: This technique involves isolating variables on either side of the equation.
2. Integrating Factors: Multiply the equation by an integrating factor to make it easier to solve.
3. Exact Equations: Identify and solve equations that are exact differentials.
4. Substitution: Sometimes, substituting variables can make the equation more manageable.
### Practical Examples
To truly understand how to solve differential equations, we’ll dive into practical examples. From exponential growth to radioactive decay, we’ll walk you through the steps, making it crystal clear.
## Applications in Real Life
### Where You’ll Encounter Differential Equations
You might be wondering where these mathematical tools are applied in the real world. Well, here are some fascinating areas where differential equations play a pivotal role:
• Physics: Modeling motion, heat transfer, and wave propagation.
• Engineering: Designing electrical circuits, analyzing structures, and more.
• Economics: Predicting economic trends and market behaviour.
• Biology: Describing population dynamics and biochemical reactions.
## Challenges and Tips
### Overcoming Hurdles
It’s natural to face challenges when dealing with differential equations. Some equations can be quite complex, and it’s easy to get lost in the mathematics. Here are a few common hurdles and tips for surmounting them:
• Complex Notation: The notation can be intimidating. Practice and patience are your allies.
• Initial Conditions: Solving differential equations often requires initial conditions. Don’t forget to include them.
• Nonlinearity: Nonlinear equations can be tricky, but numerical methods can often help.
## Comprehensive Guide to Differential Equations
### Bringing It All Together
In this comprehensive guide, we’ve demystified the world of differential equations and their role in function discovery. Whether you’re a student tackling math problems or a professional in search of real-world solutions, understanding differential equations can be a game-changer.
The solution of a differential equation is to find an expression without $\displaystyle \frac{d}{dx}$ notations using given conditions.
Note that the proper rules must be in place to achieve a valid solution of the differential equations, such as the product, quotient, and chain rules.
Many students missed applying the chain rule, resulting in an unexpected differential equation outcome.
Take a look at the following example for a typical differential equation.
### Question 1
A function $f(x)$ is defined for all real $x$ and $y$ such that $\require{color} f(x + y) = f(x)f(y) \text{ , and } f'(0) = 2$.
Find $f(x)$ in terms of $x$.
\begin{aligned} \require{AMSsymbols} \require{color} f(y) &= f(0)f(y) &\color{green} \text{substitute } x=0 \\ \therefore f(0)&=1 &\color{green} (1) \\ \frac{d}{dx}f(x + y) \times \frac{d}{dx}(x + y) &= \frac{d}{dx}f(x) \times f(y) + f(x) \times \frac{d}{dx}f(y) &\color{green} \text{differentiate both sides} \\ \frac{d}{dx}f(x + y) \times (1 + \frac{dx}{dy}) &= \frac{d}{dx}f(x) \times f(y) + f(x) \times \frac{d}{dx}f(y) \times \frac{dy}{dx} &\color{green} \frac{dy}{dx} = y’ \\ f'(x + y) \times (1 + y’) &= f'(x) \times f(y) + f(x) \times f'(y) \times y’ \\ f'(y) \times (1 + y’) &= f'(0) \times f(y) + f(0) \times f'(y) \times y’ &\color{green} \text{substitute } x=0 \\ f'(y) \times (1 + y’) &= 2 \times f(y) + f(0) \times f'(y) \times y’ &\color{green} \text{substitute } f'(0) = 2 \\ f'(y) + f'(y) y’ &= 2 f(y) + 1 \times f'(y) y’ &\color{green} \text{substitute } f(0) = 1 \\ f'(y) &= 2 f(y) \\ \frac{f'(y)}{f(y)} &=2 \\ \int \frac{f'(y)}{f(y)} \,dy &= \int 2 \,dy &\color{green} \text{integrate both sides} \\ \log_e f(y) &= 2y + C &\color{green} (2) \\ \log_e f(0) &= 2 \times 0 + C &\color{green} \text{substitute } y = 0 \\ \log_e 1 &= C &\color{green} \text{substitute } f(0) = 1 \\ \log_e f(y) &= 2y &\color{green} (2) \\ f(y) &= e^{2y} \\ \therefore f(x) &= e^{2x} \\ \end{aligned}
### Question 2
Consider a differential equation, $\displaystyle \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = 0$.
(a) Show that, if $y = f(t)$ and $y = g(t)$ are both solutions to the differential equation and $A$ and $B$ are constants, then $y = Af(t) + Bg(t)$ is a solution.
If $y = f(t)$ is a solution of $\require{AMSsymbols} \require{color} \displaystyle \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = 0$, then $\displaystyle \require{AMSsymbols} \require{color} \frac{d^2}{dt}f(t) +3\frac{d}{dt}f(t) + 2f(t) = 0 \color{green} \cdots (1)$
If $y = g(t)$ is a solution of $\require{color} \displaystyle \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = 0$, then $\displaystyle \require{AMSsymbols} \require{color} \frac{d^2}{dt}g(t) +3\frac{d}{dt}g(t) + 2g(t) = 0 \color{green} \cdots (2)$
\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} y &= Af(t) + Bg(t) \cdots \color{green} (3) \\ \frac{dy}{dt} &= A\frac{d}{dt}f(t)+B\frac{d}{dt}g(t) \cdots \color{green} (4) \\ \frac{d^2y}{dt^2} &= A\frac{d^2}{dt^2}f(t)+B\frac{d^2}{dt^2}g(t) \cdots \color{green} (5) \end{aligned}
\begin{aligned} \displaystyle \require{AMSsymbols}\require{color} A\frac{d^2}{dt^2}f(t)+ B\frac{d^2}{dt^2}g(t) + 3A\frac{d}{dt}f(t)+3B\frac{d}{dt}g(t) + 2Af(t) + 2Bg(t) &= 0 &\color{green} (3), (4), (5) \text{ into } \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = 0 \\ A \Big[\frac{d^2}{dt^2}f(t)+3\frac{d}{dt}f(t)+ 2f(t)\Big]+ B \Big[\frac{d^2}{dt^2}g(t) +3\frac{d}{dt}g(t) + 2g(t)\Big] &= 0 \\ A \times 0 + B \times 0 &= 0 &\color{green} \text{by (1) and (2)} \\ \therefore y = Af(t) + Bg(t) \text{ is a solution of } \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y &= 0 \end{aligned}
(b) This gives a solution of the differential equation $y = e^{kt}$ for some values $k$. Find the possible values of $k$.
\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} k^2e^{kt} + 3ke^{kt} + 2e^{kt} &= 0 \\ e^{kt}(k^2 + 3k + 2) &= 0 \\ e^{kt}(k + 1)(k + 2) &= 0 \\ \therefore k &= -1 \text{ or } -2 &\color{green} e^{kt} \ne 0 \\ \end{aligned}
(c) A solution of the differential equation is $y = Ae^{-2t} + Be^{-t}$. When $t=0$, it is given that $y=0$ and $\displaystyle \frac{dy}{dx} = 1$. Find the values of $A$ and $B$.
\begin{aligned} \displaystyle\require{AMSsymbols} \require{color} 0 &= Ae^{-2 \times 0} + Be^{-0} &\color{green} t=0 \text{ and } y=0 \\ 0 &= A + B &\color{green} (1) \\ \frac{dy}{dx} &= -2Ae^{-2t}-Be^{-t} \\ 1 &= -2Ae^{-2 \times 0}-Be^{-0} &\color{green} t=0 \text{ and } \frac{dy}{dx} = 1 \\ 1 &= -2A-B &\color{green} (2) \\ \therefore A &= -1 \text{ and } B = 1 &\color{green} \text{solve (1) and (2) simultaneously} \end{aligned}
## Conclusion
As we conclude our journey through the world of differential equations, remember that math is a powerful tool for understanding the universe. By mastering the art of simplifying function discovery with differential equations, you’re not only enhancing your mathematical skills but also opening doors to a wide range of applications in science and technology. So, keep exploring, keep solving, and keep discovering. Math made easy is a journey worth taking!
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Virginia Mathematics Standards for Learning Mid 2nd Grade Math Review
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Daily Review perfect for morning work, homework or laminate and use in centers.
Great for test prep!
Each page includes:
Money to \$1.00
Problem Solving
Geometry
Telling Time to 5 Minutes
Ordering Numbers
Expanded Form
Greater Than/Less Than/ Equal To
Place Value
Virginia Mathematics Standards for Learning Grade 2
Number and Number Sense
Focus: Place Value, Number Patterns, and Fraction Concepts
2.1 The student will
a) read, write, and identify the place value of each digit in a three-digit numeral, using numeration models;
b) round two-digit numbers to the nearest ten; and
c) compare two whole numbers between 0 and 999, using symbols (>, <, or =) and words (greater than, less than, or equal to).
2.2 The student will
a) identify the ordinal positions first through twentieth, using an ordered set of objects; and
b) write the ordinal numbers.
2.3 The student will
a) identify the parts of a set and/or region that represent fractions for halves, thirds, fourths, sixths, eighths, and tenths;
b) write the fractions; and
c) compare the unit fractions for halves, thirds, fourths, sixths, eighths, and tenths.
2.4 The student will
a) count forward by twos, fives, and tens to 100, starting at various multiples of 2, 5, or 10;
b) count backward by tens from 100; and
c) recognize even and odd numbers.
Computation and Estimation
Focus: Number Relationships and Operations
2.5 The student will recall addition facts with sums to 20 or less and the corresponding subtraction facts.
2.6 The student, given two whole numbers whose sum is 99 or less, will
a) estimate the sum; and
b) find the sum, using various methods of calculation.
2.7 The student, given two whole numbers, each of which is 99 or less, will
a) estimate the difference; and
b) find the difference, using various methods of calculation.
2.8 The student will create and solve one- and two-step addition and subtraction problems, using data from simple tables, picture graphs, and bar graphs.
2.9 The student will recognize and describe the related facts that represent and describe the inverse relationship between addition and subtraction.
Measurement
Focus: Money, Linear Measurement, Weight/Mass, and Volume
2.10 The student will
a) count and compare a collection of pennies, nickels, dimes, and quarters whose total value is \$2.00 or less; and
b) correctly use the cent symbol (¢), dollar symbol (\$), and decimal point (.).
2.11 The student will estimate and measure
a) length to the nearest centimeter and inch;
b) weight/mass of objects in pounds/ounces and kilograms/grams, using a scale; and
c) liquid volume in cups, pints, quarts, gallons, and liters.
2.12 The student will tell and write time to the nearest five minutes, using analog and digital clocks.
2.13 The student will
a) determine past and future days of the week; and
b) identify specific days and dates on a given calendar.
2.14 The student will read the temperature on a Celsius and/or Fahrenheit thermometer to the nearest 10 degrees.
Geometry
Focus: Symmetry and Plane and Solid Figures
2.15 The student will
a) draw a line of symmetry in a figure; and
b) identify and create figures with at least one line of symmetry.
2.16 The student will identify, describe, compare, and contrast plane and solid geometric figures (circle/sphere, square/cube, and rectangle/rectangular prism).
Probability and Statistics
Focus: Applications of Data
2.17 The student will use data from experiments to construct picture graphs, pictographs, and bar graphs.
2.18 The student will use data from experiments to predict outcomes when the experiment is repeated.
2.19 The student will analyze data displayed in picture graphs, pictographs, and bar graphs.
Patterns, Functions, and Algebra
Focus: Patterning and Numerical Sentences
2.20 The student will identify, create, and extend a wide variety of patterns.
2.21 The student will solve problems by completing numerical sentences involving the basic facts for addition and subtraction. The student will create story problems, using the numerical sentences.
2.22 The student will demonstrate an understanding of equality by recognizing that the symbol = in an equation indicates equivalent quantities and the symbol ≠ indicates that quantities are not equivalent.
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Study S4 A Math Integration Techniques - Geniebook
# Integration Techniques
In this article, we will learn about different integration techniques. We will study the following concepts:
• Introduction to Integration
• Rules of Integration
• Integration of Power Functions
## Introduction To Integration
Integration in mathematics is defined as the reverse process of differentiation.
Let us look at a few scenarios where the concept of integration is needed.
## Area Under A Curve
Case 1:
Let $$A(t)$$ be the function that gives the area of the region bounded by the function $$y=2$$, the line $$x=t$$ and the positive $$x$$ and $$y-axis$$. Find a function for $$A(t)$$.
• Consider the line $$y=2$$
• The line $$y=2$$, forms a rectangle, which area is $$A(t)$$.
• Since it is a rectangle, the area can be found by applying the area formula of the rectangle.
To find the area,
\begin{align*} A(t) &= t \times 2 \\ \\ A(t) &= 2t \;\text{units}^2 \end{align*}
These kinds of cases can be solved without integrating.
Case 2:
Let $$A(t)$$ be the function that gives the area of the region bounded by the function $$y=2x$$, the line $$x=t$$ and the positive $$x$$ and $$y-axes$$. Find the function for $$A(t)$$.
In this case, the line $$y=2x$$ is inclined towards $$x$$. This forms a triangle.
Therefore, to find the area of this triangle,
\begin{align*} A(t) &= \frac {1}{2}\times t \times 2t \\ \\ A(t) &= t^2 \end{align*}
These kinds of cases can also be solved without integrating.
Case 3:
Let $$A(t)$$ be the function that gives the area of the region bounded by the function $$y=x^2$$ , the line $$x=t$$ and the positive $$x$$ and $$y$$-axes. Find the function for $$A(t)$$.
In this case, the curve $$y=x^2$$ is actually a curve.
Here, small rectangles can be drawn to find the area, as shown below.
This is where we use integration to find the area under the curve, also known as the indefinite integral. Let us look into this concept briefly below.
## The Indefinite Integral
Notations For Integration
There are two ways in which integration can be represented.
1. \begin{align*} y=x^2 \end{align*}
The above equation, when integrated, it becomes,
\begin{align*} \int y\,dx = \int {x^2}\,dx \end{align*}
Where,
\begin{align*} \int \end{align*} denotes integration.
$$dx$$ denotes that is is integrated with respect to $$x$$.
$$y , \,x^2$$ are the expressions that need to be integrated.
1. \begin{align*} f(x) = x^2 \end{align*}
The above equation, when integrated, it becomes,
\begin{align*} F(x) = \int x^2\,dx \end{align*}
$$f(x)$$ becomes $$F(x).$$
\begin{align*} \int \end{align*} denotes integration.
$$dx$$ denotes that is integrated with respect to $$x$$.
$$x^2$$ is the expression that needs to be integrated.
$$F(x)$$ is also known as the anti-derivative of $$f(x).$$
The term anti-derivative is the term that is used to call the result that is obtained from integrating $$f(x).$$
Comparing The Notation Of Differentiation And Integration
The difference in the notation of differentiation and integration is explained using the two cases below.
1. \begin{align*} y=x^2 \end{align*}
In integration,
\begin{align*} \int y\,dx =\int x^2\,dx \end{align*}
Where,
\begin{align*} \int \end{align*} denotes integration
$$dx$$ denotes that is is integrated with respect to $$x.$$
$$y , \,x^2$$ are the expressions that need to be integrated.
In differentiation,
\begin{align*} \frac {dy}{dx}= \frac {d}{dx} \,x^2 \end{align*}
Where,
$$dy$$ - differentiating $$y$$
$$dx$$ - with respect to $$x$$
1. \begin{align*} f(x) = x^2 \end{align*}
In Integration,
\begin{align*} F(x) = \int x^2\,dx \end{align*}
$$f(x)$$ becomes $$F(x).$$
\begin{align*} \int \end{align*} denotes integration.
$$dx$$ denotes that is is integrated with respect to $$x.$$
$$x^2$$ is the expression to be differentiated.
In Differentiation,
\begin{align*} f'(x) = \frac {d}{dx}\,x^2 \end{align*}
When we differentiate $$f(x)$$ it becomes $$f'(x).$$
$$dx$$ - with respect to $$x$$
$$d$$ - differentiating
## Integration As Reverse Of Differentiation
As mentioned, integration is the reverse process of differentiation. There are certain considerations when doing integration.
Consider the following equation,
\begin{align*} \frac {d}{dx}\,f(x) \end{align*} —— equation (1)
\begin{align*} \frac {d}{dx}\,[\,f(x)+2\,] \end{align*} —— equation (2)
\begin{align*} \frac {d}{dx}\,[\,f(x)-2\,] \end{align*} —— equation (3)
When we differentiate it,
equation (1) becomes,
\begin{align*} \int f'(x) \,dx \end{align*}
equation (2) becomes,
\begin{align*} \int f'(x) \,dx \end{align*}
equation (3) becomes,
\begin{align*} \int f'(x) \,dx \end{align*}
When we differentiate any constant it becomes 0 as seen from differentiating equation(2) and equation (3).
When we integrate the first derivative of a function, we obtain the original function. However, we cannot be certain whether the original function contained a constant term or not.
Hence, the arbitrary constant C is always added at the end, after integration.
Therefore,
\begin{align*} ∫ f(x) \;dx = F(x) + C \end{align*}
Here, C is an arbitrary constant.
## Rules Of Integration
1. Integration As Reverse Of Differentiation
\begin{align*} \int [ \frac {d}{dx}\, f(x)] \,dx = f(x) + c \end{align*}
Where $$c$$ is an arbitrary constant.
1. Scalar Multiple Rule.
\begin{align*} \int af(x) \;dx = a \int f(x) \;dx \end{align*}
Where $$a$$ is a real number.
Here $$a$$ is a real number, which is the coefficient that cannot be integrated.
\begin{align*} \int f(x) \mp g(x) \;dx = \int f(x) \;dx \mp \int g(x) \;dx \end{align*}
Here, the term befores the + or – may be integrated separately from the term after the + or –.
Now that we have introduced the rules, let us implement them in the following questions.
## Integration As Reverse Of Differentiation (Power Functions)
Consider the expression, $$x^n$$, where $$n$$ is a non-zero real number.
We are going to differentiate it, then integrate it by reversing the steps.
In the first step i.e., differentiation
Stage 1:
The coefficient is multiplied with power n.
Stage 2:
The power is decreased by 1.
In the second step i.e., reversing differentiation
Stage 1:
The power is increased by 1.
Stage 2:
The coefficient is divided by the power n.
In the third step i.e., integration,
Stage 1:
the power is increased by 1.
Stage 2:
the expression is divided by the new power.
## Integration Of Power Functions
There is a formula to integrate power functions like $$x^2, \,x^6.$$
The formula is,
\begin{align*} \int x^n \;dx = \frac {x^{n+1}}{n+1} + c \end{align*}
where,
$$n ≠-1$$ and $$c$$ is an arbitrary constant.
Question 1:
Integrate each of the following with respect to $$x$$.
1. \begin{align*} x^3 \end{align*}
Solution:
\begin{align*} \int x^3 \;dx &= \frac {x^{(3+1)}}{3+1}+C \\ \\ \int x^3 \;dx &= \frac {x^4}{4}+C \end{align*}
1. \begin{align*} \frac {3}{x^2 } \end{align*}
Solution:
The denominator is converted from positive to negative indices, \begin{align*} \int 3x^{-2}\;dx \end{align*}.
Applying the Scalar Multiple Rule of Integration, we take the coefficient out, \begin{align*} \int 3x^{-2}\;dx \end{align*}.
Applying the integration formula for power functions,
\begin{align*} \int x^n \;dx &= \frac {x^{n+1}}{n+1} +C \\ \\ &= 3\bigg [\frac {x^{-1}}{-1}\bigg ] +C \end{align*}
Since the answer should not be written in negative powers, we leave the final answer as \begin{align*} - \frac {3}{x} + C. \end{align*}
1. \begin{align*} \sqrt {2x} \end{align*}
Solution:
Since the root is for both $$2$$ and $$x$$, it can also be written as $$\sqrt {2} \sqrt {x} \; dx$$.
Taking the coefficient out and writing the root of $$x$$ as $$\frac {1}{2}$$, we have $$\sqrt {2} \int x^{\frac {1}{2}} \;dx.$$
Applying the integration formula for power functions,
\begin{align*} \int x^n \;dx &= \frac {x^{n+1}}{n+1} +C \\ \\ &= \sqrt{2} \bigg [\frac {x^{\frac {3}{2}}}{\frac {3}{2}} \bigg ] +C \end{align*}
Simplifying, we leave the final answer as \begin{align*} \frac {2\sqrt {2}}{3} \sqrt{x^3} + C. \end{align*}.
Question 2:
Integrate \begin{align*} (x+2)(x-3) \end{align*} with respect to $$x$$.
Solution:
Expanding the expression, we get,
\begin{align*} (x+2)(x-3) &= x^2- 3x+2x-6 \\ \\ &= x^2 -x-6 \end{align*}
Integrating the expression,
\begin{align*} \int (x^2 -x-6) \;dx \end{align*}
Applying the integration formula for power functions,
\begin{align*} \int x^n \;dx &= \frac {x^{n+1}}{n+1} +C \\ \\ \end{align*}
\begin{align*} \int (x^2 - x -6)\; dx &= \frac {x^{2+1}}{2+1}-\frac {x^{1+1}}{1+1}- \frac {6x^{0+1}}{0+1}+C \\ \\ &= \frac {x^{3}}{3}-\frac {x^{2}}{2}- 6x+C \end{align*}
Therefore the answer is,
\begin{align*} \frac {x^{3}}{3}-\frac {x^{2}}{2}- 6x+C \end{align*}
## Conclusion
In this article, we went through the big idea of Integration as the reverse of Differentiation as well as some basic integration rules – Scalar Multiple Rule, Addition/Subtraction Rule – and how to integrate Power Functions of the form $$x^n$$.
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## Powers of a Matrix
in this video we're gonna discuss powers
of a matrix
the problem is we're given this
two-by-two matrix a and we'd like to
find the matrix a raised to the power
means multiply a times a using usual
matrix multiplication so in other words
we'll multiply a by itself and thinking
back to matrix multiplication normally
we circle the rows of the first matrix
columns of the second Matrix and we're
forming a bunch of dot products so the
product is gonna look like negative 1
negative 1 1 0 alright now that we've
got that let's figure out a cubed so a
cubed would be a squared times a we
figured out a squared in the previous
step so I'll copy that down and if we
multiply that by a doing our usual
matrix multiplication process now we get
negative 1 0 0 negative 1 so let's try
if a to the 4th will be a cubed times a
at some point you can stop circling the
rows and columns once this this matrix
multiplication starts to seem a little
bit more automatic so this product here
is going to give us 0 1 negative 1
negative 1 now to get 8 v will take a to
the 4th times a this time let's try it
without circling the rows and columns
we'll have 1 1 negative 1 and 0 and what
we're altom utley looking for is a
matrix or a certain power of a that
equals the identity matrix so we may
need to do a few more calculations
before we end up with a power that's
equal to the identity so let's try a to
the 6 into the 6 is in the 5th times 8
so we'll recopy in the 5th take our
original matrix a and we'll do this
multiplication we're gonna have 1 0 0
and 1
aha so finally we have a power which is
equal to the identity matrix I so what
we should do now is recap a few
properties of this identity matrix and
we'll see how those properties will help
us to solve the problem so we're looking
at this identity matrix here which has
ones along the main diagonal zeros on
the off diagonal doesn't necessarily
have to be a two-by-two matrix could be
a 3 by 3 4 by 4 matrix etc so let's look
at some properties that that this
identity matrix has first property
I to the N is equal to I for any
positive integer power it's sort of
obvious when the power is one so let's
say for n equal to 3 etc so again this
would be true for the 2 by 2 identity
matrix but also for 3 by 3 4 by 4 etc so
just to convince ourselves of this let's
say for example if we take the 2 by 2
identity matrix and multiply it by
itself we can check that yes we do get
back the original identity matrix all
right second property that I has if we
take our identity matrix times any
matrix B provided that the sizes are
compatible here we're gonna end up with
that same matrix B so let's say for any
for any choice of B and let's just do a
little example to convince ourselves of
this so we take the 2 by 2 identity
matrix let's say B is the matrix that
looks like e F G H for example if we do
this multiplication we end up with E F G
H so in other words we end up with this
original matrix B that we put in here
all right so let's see how these
properties here tie in to the problem
that we're trying to solve so here's the
idea we've discovered that a to the 6 is
the same thing as the identity matrix
and we know that the identity matrix has
some nice properties so the trick will
be to take a to the power 2015 and to
write it as a to the 6 times a to the 6
and we'll want to multiply by a to the 6
a whole bunch of times
didn't have these dots here just doing
our usual matrix multiplication so a to
the six times a to the six whole bunch
of times and then at the end here there
will be some some product or some last
factor so I've just recopy to our basic
idea from the previous screen and let's
see how it can write out this product so
first of all I'm going to think about
what happens if I divide 2015 by six
using a calculator you're going to get
something like three thirty five point
eight three repeating might be a little
bit nicer to write 2015 over six is
equal to three thirty-five plus 0.8
three repeating and if we then go
through and multiply both sides by six
then we've got twenty fifteen is equal
to six times three 35 plus will have
this remainder here if we go six times
this expression here we're getting five
so let's take this expression that we've
just figured out and let's say that a to
the 2015 can be re-written as a to the
six times three thirty-five plus five
and now we're going to think about some
exponent rules so first of all when you
have a sum in the exponent that's the
same thing as having a product so we
could rewrite this as a to the six times
three thirty five times a to the fifth
okay now looking at the first factor
when we've got a product in the exponent
we can rewrite that as a to the six
raised to the power three thirty-five so
whenever we've got a power raised to
another power those powers multiply
that's our second exponent rule here and
we'd still need to multiply by a to the
fifth now the reason why we're doing
this is remember we said that a to the
six is equal to the identity matrix so
we can now say this is I raised to the
power three thirty five times a to the
fifth using our first property about the
identity matrix when you raise it to any
power you're getting back the identity
matrix so this is really I times in the
fifth I'm thinking back to the second
property second property of the identity
matrix from our previous screen when you
take I times any matrix you just get
back that original matrix so this gives
us a to the fifth technically if we're
being very careful about the notation
certainly not a dot product so it might
be more correct if we actually put
brackets on here the the notation then
gets a little bit more cluttered but but
we're being more rigor
with the notation this way alright so
what we've determined finally after all
this calculation is that a to the power
2015 is equal to a to the fifth you |
Select Page
Students often misinterpret negative exponent notation and if this is cleared up, in the beginning, a lot of frustration and time can be saved.
## Common Mistakes with Negative Exponents
There are often a lot of misconceptions about understanding and teaching negative exponents. Before we begin looking at the solutions, let's take a look at the common problems.
#### Student Mistakes
Students love to create or rework rules for negative exponents. Be aware that you may find negative signs in the strangest of places.
For example, you may find mistakes like this:
$2^{-3}=(-2)(-2)(-2)=-8$
Teacher Mistakes
Just take the reciprocal can be a real answer later, but first students must understand the why. Why does this work and how can I know this rule? We have all seen students get done with the homework super quick, but then bomb the test. Simple may make the lesson go faster, but an easy gain is an easy loss.
Students, to retain these rules, must have an understanding of why and how they work in a way that works for them.
## Using Common Language and Vocabulary
Most students will want to see several different ways of looking at this concept. But one thing is for sure, if we can connect this learning for them between the examples, they will develop a solid understanding of the concepts with negative exponents.
Some of the most common vocabulary: Exponents, exponent rules, repeating functions, reciprocals,
## Activities to Help Understanding
#### Patterns
I like to give the students this table and ask some questions. What do exponents mean? What is happening? Can this pattern continue? How do you know it can/cannot? Are we still multiplying?
I like to do this a few times, to allow all of the students to have a chance to see the patterns.
#### Tangible
I like the idea of giving the students something they can picture to work within their minds. For example, thinking of cake, as we go into the negative exponents we get a fractional part of the cake, we do not get a negative cake.
And while seeing the pattern does help, they also need to understand the rules. At this time we are ready to put it all together.
#### Exponent Rules
Depending on your class and their level of understanding, you may need to review the exponential rules. Specifically, they need a good understanding of the product rule and the quotient rule.
I find it best to plug in positive numbers and explore the possibilities. The students will begin to come up with "shortcuts." This discovery is useful! We can then introduce the rule of, $x^{-a}=\frac{1}{x^{a}}$ But now they understand why this works and we can move forward.
#### Challenges
I found the following challenge on the website, http://math.stackexchange.com/questions/629740/how-would-you-explain-to-a-9th-grader-the-negative-exponent-rule and I loved the potential of teaching it this way. It builds on the exponent rules. I would stop talking at the arrow and see where the students take the conversation. I bet it would give you a lot of insight into their understanding.
#### Video - How I Feel About Logarithms - by Vi Hart
This video moves very fast, but for your students that are intrigued, this video will motivate and challenge them. The negative exponents begin around minute 7, but you need the whole video to understand the vocabulary. I can see this working for some and not so much for others, but I enjoyed it, and I hope you get some additional ideas from it as I did. I suggest watching this to be sure it is a good fit for your class.
"Sometimes to make the harder things simple, first you have to make the simple things harder"
-Vi Hart
## Rewards of a Job Well Done
While it may take more time up front to ensure that students understand this concept, in the long run, it will save you time. The ability to see patterns and complete the calculations without help will be priceless. When you give students an in-depth understanding and allow time for processing the concepts go into the long-term memory, students gain quite an advantage as they continue their math education. |
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Lesson 9: Money
# Counting dollars
Sal adds to find the total value when given an amount US dollars.
## Want to join the conversation?
• You have in front of you 435 five-dollar bills and 855 ten-dollar bills. Your problem: Place the five-dollar bills and the ten-dollar bills in stacks so that each stack has the same number of bills, and each stack contains only one kind of bill (five-dollar or ten-dollar). What is the largest number of bills that you can place in each stack?
• The number of bills in each stack must be a common factor of the number of \$5 bills, 435, and the number of \$10 bills, 855. So the largest number of bills in each stack is the greatest common factor of 435 and 855. (Note that the actual values of the bills, \$5 and \$10, do not matter in this problem. What matters is that the bills are of two different kinds.)
The prime factorization of 435 is 3*5*29.
The prime factorization of 855 is 3*3*5*19.
When we construct the greatest common factor (GCF), we use each possible prime factor the smallest number of times it occurs in any prime factorization.
The prime factor 3 is used once and twice in the prime factorizations, so we use it once in the GCF.
The prime factor 5 is used once and once in the prime factorizations, so we use it once in the GCF.
The prime factor 19 is used zero times and once in the prime factorizations, so we use it zero times in the GCF.
The prime factor 29 is used once and zero times in the prime factorizations, so we use it zero times in the GCF.
So the GCF is 3*5, which is 15. Therefore, the greatest number of bills in each stack is 15.
Have a blessed, wonderful day!
• How much is the eagle coin
• The American Golden Eagle coin is worth \$1.00
• How much in Indian rupees is 1 dollar
• How much is 6 nickels?
• A nickel is 5 cents. So, 5+5+5+5+5+5 is 30.
• IS there a unit that will help with returning money? For example: A cashier reads a total, the customer gives them an amount; how much should the cashier return?
(1 vote)
• You can use subtraction (with 2 decimal places) to help figure out how much change to give a customer.
For instance, let's say the cashier is handed a \$20 dollar bill for a stuffed animal that costs \$13.99
\$20.00
-13.99
_____
\$6.01 |
# The Monty Hall Problem
(Difference between revisions)
Revision as of 15:37, 7 July 2010 (edit)← Previous diff Current revision (15:00, 19 July 2010) (edit) (undo) (22 intermediate revisions not shown.) Line 1: Line 1: - {{Image Description + {{Image Description Ready - |ImageName=Let's Make a Deal + |ImageName=The Monty Hall Problem |Image=Mainimage.jpg |Image=Mainimage.jpg |ImageIntro=The Monty Hall problem is a probability puzzle based on the 1960's game show Let's Make a Deal. |ImageIntro=The Monty Hall problem is a probability puzzle based on the 1960's game show Let's Make a Deal. Line 21: Line 21: In the diagram on the below, we can see what prize the contestant will win if he always stays with his initial pick after Monty opens a door. If the contestant uses the strategy of always staying, he will only win if he originally picked door 1. In the diagram on the below, we can see what prize the contestant will win if he always stays with his initial pick after Monty opens a door. If the contestant uses the strategy of always staying, he will only win if he originally picked door 1. - [[Image:stay6.jpg]] + [[Image:staydecision.jpg]] If the contestant always switches doors when Monty shows him a goat, then he will win if he originally picked door 2 or door 3. If the contestant always switches doors when Monty shows him a goat, then he will win if he originally picked door 2 or door 3. - [[Image:switch5.jpg]] + [[Image:switchdecision.jpg]] A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. Since we know that the car is equally likely to be behind each of the three doors, we can generalize our strategy for the case where the car is behind door 1 to any placement of the car. The probability of winning by staying with the initial choice is 1/3, while the probability of winning by switching is 2/3. The contestant's best strategy is to always switch doors so he can drive home happy and goat-free. A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. Since we know that the car is equally likely to be behind each of the three doors, we can generalize our strategy for the case where the car is behind door 1 to any placement of the car. The probability of winning by staying with the initial choice is 1/3, while the probability of winning by switching is 2/3. The contestant's best strategy is to always switch doors so he can drive home happy and goat-free. Line 42: Line 42: The most common misconception is that the odds of winning are 50-50 no matter which door a contestant chooses. Most people assume that each door is equally likely to contain the car since the probability was originally distributed evenly between the three doors. They believe that they have no reason to prefer one door, so it does not matter whether they switch or stick with their original choice. The most common misconception is that the odds of winning are 50-50 no matter which door a contestant chooses. Most people assume that each door is equally likely to contain the car since the probability was originally distributed evenly between the three doors. They believe that they have no reason to prefer one door, so it does not matter whether they switch or stick with their original choice. - This reasoning seems logical until we realize that the two doors cannot be equally likely to hide the car. The critical fact is that the Monty does not randomly choose a door to open, so they do have a reason to prefer a certain door. Marilyn defended her answer in a subsequent column addressing this point specifically. + This reasoning seems logical until we realize that the two doors cannot be equally likely to hide the car. The critical fact is that Monty's choice of which door to open is not random, so when he opens a door, it gives the contestant new information. - Suppose we pause after Monty has revealed a goat and a UFO settles down onto the stage and a little green woman emerges. The host asks her to point to one of the two unopened doors. Then the chances that she'll randomly choose the one with the prize are 1/2. ''But'', that's because she lacks the advantage the original contestant had—the help of the host. + Marilyn defended her answer in a subsequent column addressing this point specifically. Suppose we pause after Monty has revealed a goat and a UFO settles down onto the stage and a little green woman emerges. The host asks her to point to one of the two unopened doors. Then the chances that she'll randomly choose the one with the prize are 1/2. ''But'', that's because she lacks the advantage the original contestant had—the help of the host. "When you first choose door #1 from three, there's a 1/3 chance that the prize is behind that one and a 2/3 chance that it's behind one of the others. But then the host steps in and gives you a clue. If the prize is behind #2, the host shows you #3, and if the prize is behind #3, the host shows you #2. So when you switch, you win if the prize is behind #2 or #3. You win either way! But if you don't switch, you win only if the prize is behind door #1," Marilyn explained. "When you first choose door #1 from three, there's a 1/3 chance that the prize is behind that one and a 2/3 chance that it's behind one of the others. But then the host steps in and gives you a clue. If the prize is behind #2, the host shows you #3, and if the prize is behind #3, the host shows you #2. So when you switch, you win if the prize is behind #2 or #3. You win either way! But if you don't switch, you win only if the prize is behind door #1," Marilyn explained. Line 75: Line 75: - Let the door picked by the contestant be called door ''a'' and the other two doors be called ''b'' and ''c''. Also, V''a'', V''b'', and V''c'', are the events that the car is actually behind door ''a'', ''b'', and ''c'' respectively. Finally, let O''b'' be the event that Monty Hall opens curtain ''b''. + [[Image:uncertainman2.jpg|right]] - Then, the problem can be restated as follows: Is $P(V_a|O_b) = P(V_c|O_b)$? $P(V_a|O_b)$ is the probability that door ''a'' hides the car given that Monty opens door B. Similarly, P(V_c|O_b) is the probability that door ''c'' hides the car given that monty opens door B. So, when P(V_a|O_b) = P(V_c|O_b) the probability that the car is behind one unopened door the same as the probability that the car is behind the other unopened door. If this is the case, it won't matter if the contestant stays or switches. + Let the door picked by the contestant be called door ''a'' and the other two doors be called ''b'' and ''c''. Also, V''a'', V''b'', and V''c'', are the events that the car is actually behind door ''a'', ''b'', and ''c'' respectively. We begin by looking at a scenario that leads to Monty opening door ''b'', so let O''b'' be the event that Monty Hall opens curtain ''b''. + Then, the problem can be restated as follows: Is $P(V_a|O_b) = P(V_c|O_b)$? + + + $P(V_a|O_b)$ is the probability that door ''a'' hides the car given that Monty opens door ''b''. Similarly, $P(V_c|O_b)$ is the probability that door ''c'' hides the car given that monty opens door ''b''. So, when $P(V_a|O_b) = P(V_c|O_b)$ the probability that the car is behind one unopened door the same as the probability that the car is behind the other unopened door. If this is the case, it won't matter if the contestant stays or switches. Line 100: Line 104: $P(O_b|V_c) = 1$ because if the prize is behind door ''c'', Monty can only open door ''b''. $P(O_b|V_c) = 1$ because if the prize is behind door ''c'', Monty can only open door ''b''. - - Each of these probabilities is conditional on the fact that the prize is hidden behind a specific door, but we are assuming that each of these probabilities is mutually exclusive since the car can only be hidden behind one door. As a result, we know that P(''O''''b'') is equal to Each of these probabilities is conditional on the fact that the prize is hidden behind a specific door, but we are assuming that each of these probabilities is mutually exclusive since the car can only be hidden behind one door. As a result, we know that P(''O''''b'') is equal to Line 114: Line 116: ::$= \frac{1}{2}$ ::$= \frac{1}{2}$ - - Then, we can use $P(O_b)$, $P(O_b|V_a)$, and $P(V_a)$ to calculated $P(V_a|O_b)$. Then, we can use $P(O_b)$, $P(O_b|V_a)$, and $P(V_a)$ to calculated $P(V_a|O_b)$. Line 124: Line 124: :::$= \frac {1}{3}$ :::$= \frac {1}{3}$ - - Similarly, Similarly, Line 135: Line 133: :::$= \frac {2}{3}$ :::$= \frac {2}{3}$ + The probability of ''V''''c'' (the event that car is hidden behind door ''c'') in this case is ''not'' equal to the probability of ''V''''a'' (the case where the car is hidden behind the door that Monty hasn't opened and the contestant hasn't selected). The contestant is offered an opportunity to switch to door ''c''. We have calculated that the probability of winning when door ''c'' is selected is 2/3 and the probability of winning with the contestant's original choice, door ''a'' is 1/3. - The probability of ''V''''c'' (the event that car is hidden behind door ''c'') in this case is ''not'' equal to the probability of ''V''''a'' (the case where the car is hidden behind the door that Monty hasn't opened and the contestant hasn't selected). The contestant is offered an opportunity to switch to door ''c''. We have calculated that the probability of winning when door ''c'' is selected is 2/3 and the probability of winning with the contestant's original choice, door ''a'' is 1/3. If the contestant switches, he doubles his chance of winning. + Since Monty is equally likely to open any of the three doors, we can generalize this strategy for any door that he opens. The probability that the car is hidden behind the contestant's original choice is 1/3, but the probability that the car is hidden behind the unopened and unselected door is 2/3. If the contestant switches, he doubles his chance of winning. |AuthorName=Grand Illusions |AuthorName=Grand Illusions Line 161: Line 160: This consistency is especially remarkable given that these studies include a range of different wordings, methods of presentations, languages, and cultures. This consistency is especially remarkable given that these studies include a range of different wordings, methods of presentations, languages, and cultures. - Marilyn quotes cognitive psychologist Massimo Piattelli-Palmarini in her own book saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." When the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution was wrong. + Marilyn quotes cognitive psychologist Massimo Piattelli-Palmarini in her own book saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer" (''Bostonia'' July/August 1991). When the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution was wrong. One letter written to vos Savant by Dr. E. Ray Bobo of Georgetown University was especially critical of Marilyn's solution: "You are utterly incorrect about the game show question, and I hope this controversy will call some public attention to the serious national crisis in mathematical education. If you can admit your error, you will have contributed constructively toward the solution to a deplorable situation. How many irate mathematicians are needed to get you to change your mind?" One letter written to vos Savant by Dr. E. Ray Bobo of Georgetown University was especially critical of Marilyn's solution: "You are utterly incorrect about the game show question, and I hope this controversy will call some public attention to the serious national crisis in mathematical education. If you can admit your error, you will have contributed constructively toward the solution to a deplorable situation. How many irate mathematicians are needed to get you to change your mind?" Line 171: Line 170: [[Image:Candies2.jpg|right|250px]] [[Image:Candies2.jpg|right|250px]] - The most famous experiment in question is the 1956 study on rationalizing choices. The researchers studied which M&M colors were most preferred by monkeys. After identifying a few colors of M&Ms that were approximately equally favored by a monkey - say, red, blue, and yellow, - the researchers gave the monkey a choice between two of the colors. + The most famous experiment in question is the 1956 study "Postdecision changes in the desirability of alternatives" on rationalizing choices. The researchers studied which M&M colors were most preferred by monkeys. After identifying a few colors of M&Ms that were approximately equally favored by a monkey - say, red, blue, and yellow, - the researchers gave the monkey a choice between two of the colors. In one case, imagine that a monkey chose yellow over blue. Then, the monkey would be offered the choice between blue and red M&Ms. Researchers noted that about two-thirds of the time the monkey would choose red. The 1956 study claimed that their results reinforced the theory of rationalization: Once we reject something we are convinced that we never like it anyway. In one case, imagine that a monkey chose yellow over blue. Then, the monkey would be offered the choice between blue and red M&Ms. Researchers noted that about two-thirds of the time the monkey would choose red. The 1956 study claimed that their results reinforced the theory of rationalization: Once we reject something we are convinced that we never like it anyway. - Dr. Chen reexamined the experimental procedure, and says that monkey's rejection of blue might be attributable to statistics alone. Chen says that there must be some difference in preference between the original red, blue, and yellow. If this is the case, then the monkey's choice of yellow over blue wasn't arbitrary. Like Monty Hall's decision to open a door that hid a goat, the monkey's choice between yellow and blue discloses additional information. In fact, when a monkey favors yellow over blue, there's a two-thirds chance that it also started off with a preference for red over blue- which would explain why the monkeys chose red 2/3 of the time in the Yale experiment. + Dr. Chen reexamined the experimental procedure, and says that monkey's rejection of blue might be attributable to statistics alone. Chen says that although the three colors of M&M's are ''approximately'' equally favored, there must be some slight difference in preference between the original red, blue, and yellow. If this is the case, then the monkey's choice of yellow over blue wasn't arbitrary. Like Monty Hall's decision to open a door that hid a goat, the monkey's choice between yellow and blue discloses additional information. In fact, when a monkey favors yellow over blue, there's a two-thirds chance that it also started off with a preference for red over blue- which would explain why the monkeys chose red 2/3 of the time in the Yale experiment. - To why this is true, consider Chen's conjecture that monkeys must have some slight preference between the three colors they are being offered. The table below shows all the possible combinations of ways that a monkey could possibly rank its M&Ms. + To see why this is true, consider Chen's conjecture that monkeys must have some slight preference between the three colors they are being offered. The table below shows all the possible combinations of ways that a monkey could possibly rank its M&Ms. - [[Image:M&ms.jpg]] + [[Image:M&ms1_copy.jpg]] - We can see that in the case where the monkey preferred yellow over blue, they monkey preferred red over blue 2/3 in 2/3 of the rankings. + We can see that in the case where the monkey preferred yellow over blue, they monkey preferred red over blue in 2/3 of the rankings. - Although Chen agrees that the study may have still discovered useful information about preferences, but he doesn't believe it has been measured correctly yet. "The whole literature suffers from this basic problem of acting as if Monty's choice means nothing." + Although Chen agrees that the study may have still discovered useful information about preferences, he doesn't believe it has been measured correctly yet. "The whole literature suffers from this basic problem of acting as if Monty's choice means nothing" (Tierney 2008). Monty Hall problem, the study of monkeys, and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly. Even academic studies may be littered with mistakes caused by difficulty interpreting statistics. Monty Hall problem, the study of monkeys, and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly. Even academic studies may be littered with mistakes caused by difficulty interpreting statistics. Line 201: Line 200: The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser. Jason Rosenhouse. The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser. Jason Rosenhouse. + + Brehm, J. W. (1956) Postdecision changes in the desirability of alternatives, Journal of Abnormal and Social Psychology, 52, 384-9 http://www.math.jmu.edu/~lucassk/Papers/MHOverview2.pdf http://www.math.jmu.edu/~lucassk/Papers/MHOverview2.pdf + + |ToDo= + Helper Pages: + :* Bayes' Theorem + :* Probability |InProgress=No |InProgress=No }} }}
## Current revision
The Monty Hall Problem
Field: Algebra
Image Created By: Grand Illusions
The Monty Hall Problem
The Monty Hall problem is a probability puzzle based on the 1960's game show Let's Make a Deal.
When the Monty Hall problem was published in Parade Magazine in 1990, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution was wrong. It remains one of the most disputed mathematical puzzles of all time.
# Basic Description
### The Problem
The show's host, Monty Hall, asks a contestant to pick one of three doors. One door leads to a brand new car, but the other two lead to goats. Once the contestant has picked a door, Monty opens one of the two remaining doors. He is careful never to open the door hiding the car. After Monty has opened one of these other two doors, he offers the contestant the chance to switch doors. Is it to his advantage to stay with his original choice, switch to the other unopened door, or does it not matter?
### The Solution
If you answered that the contestant's decision doesn't matter, then you are among about 90% of respondents who were quickly able to determine that the two remaining doors must be equally likely to hide the car. You are also wrong. The answer to the Monty Hall Problem is viewed by most people—including mathematicians—as extremely counter–intuitive.
It is actually to the contestant's advantage to switch: the probability of winning if the contestant doesn't switch is 1/3, but if the contestant switches, the probability becomes 2/3. To see why this is true, we examine each possible scenario below.
We can first imagine the case where the car is behind door 1.
In the diagram on the below, we can see what prize the contestant will win if he always stays with his initial pick after Monty opens a door. If the contestant uses the strategy of always staying, he will only win if he originally picked door 1.
If the contestant always switches doors when Monty shows him a goat, then he will win if he originally picked door 2 or door 3.
A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. Since we know that the car is equally likely to be behind each of the three doors, we can generalize our strategy for the case where the car is behind door 1 to any placement of the car. The probability of winning by staying with the initial choice is 1/3, while the probability of winning by switching is 2/3. The contestant's best strategy is to always switch doors so he can drive home happy and goat-free.
# Aids to Comprehension
The Monty Hall problem has the distinction of being one of the rare math problems that has gained recognition on the front page of the Sunday New York Times. On July 21, 1991, the Times published a story that explained a heated argument between a Parade columnist, Marilyn vos Savant, and numerous angry readers. Many of these readers held distinguished degrees in mathematics, and the problem seemed far too elementary to warrant such difficulty in solving.
Further explanation of readers' debate with vos Savant's can be found in the Why It's Interesting section. However, if you aren't completely convinced that switching doors is the best strategy, be aware that the Monty Hall problem has been called "math's most contentious brain teaser." The following explanations are alternative approaches to the problem that may help clarify that the best strategy is, in fact, switching doors.
#### Why the Probability is not 1/2
The most common misconception is that the odds of winning are 50-50 no matter which door a contestant chooses. Most people assume that each door is equally likely to contain the car since the probability was originally distributed evenly between the three doors. They believe that they have no reason to prefer one door, so it does not matter whether they switch or stick with their original choice.
This reasoning seems logical until we realize that the two doors cannot be equally likely to hide the car. The critical fact is that Monty's choice of which door to open is not random, so when he opens a door, it gives the contestant new information.
Marilyn defended her answer in a subsequent column addressing this point specifically. Suppose we pause after Monty has revealed a goat and a UFO settles down onto the stage and a little green woman emerges. The host asks her to point to one of the two unopened doors. Then the chances that she'll randomly choose the one with the prize are 1/2. But, that's because she lacks the advantage the original contestant had—the help of the host.
"When you first choose door #1 from three, there's a 1/3 chance that the prize is behind that one and a 2/3 chance that it's behind one of the others. But then the host steps in and gives you a clue. If the prize is behind #2, the host shows you #3, and if the prize is behind #3, the host shows you #2. So when you switch, you win if the prize is behind #2 or #3. You win either way! But if you don't switch, you win only if the prize is behind door #1," Marilyn explained.
This is true because when Monty opens a door, he is reducing the probability that it contains a car to 0. When the contestant makes an initial pick, there is a 1/3 chance that he picked the car and a 2/3 chance that one of the other two doors has the car. When Monty shows him a goat behind one of those two doors, the 2/3 chance is only for the one unopened door because the probability must be 0 for the one that the host opened.
#### An Extreme Case of the Problem
Imagine that you are on Let's Make a Deal are there are now 1 million doors. You choose your door, then Monty opens all but one of the remaining doors, showing you that they hide goats. It’s clear that your first choice is unlikely to have been the right choice out of 1 million doors.
Since you know that the car must be hidden behind one of the unopened doors and it is very unlikely to be behind your door, you know that it must be behind the other door. In fact, on average in 999,999 out of 1,000,000 times the other door will contain the prize because 999,999 out of 1,000,000 times the player first picked a door with a goat. Switching to the other door is the best strategy.
#### Simulation
Using a simulation is another useful way to show that the probability of winning by switching is 2/3. A simulation using playing cards allows us to perform multiple rounds of the game easily.
One simulation proposed by vos Savant herself requires only two participants, a player and a host. Three cards are held by the host, one ace that represents the prize and two lower cards that represent the mules. The host holds up the three cards so only he can see their values. The contestant picks a card, and it is placed aside so that he still cannot see the value. Monty then reveals one of the remaining low cards which represents a mule. He must choose between the two lower cards if they both remain in his hand.
If the card remaining in the host's hand is an ace, then this is recorded as a round where the player would have won by switching. Contrastingly, if the host is holding a low card, the round is recorded as one where staying would have won. Performing this simulation repeatedly will reveal that a player who switches will win the prize approximately 2/3 of the time.
# A More Mathematical Explanation
The following explanation uses Bayes' Theorem to show how Monty reveal [...]
The following explanation uses Bayes' Theorem to show how Monty revealing a goat changes the game.
Let the door picked by the contestant be called door a and the other two doors be called b and c. Also, Va, Vb, and Vc, are the events that the car is actually behind door a, b, and c respectively. We begin by looking at a scenario that leads to Monty opening door b, so let Ob be the event that Monty Hall opens curtain b.
Then, the problem can be restated as follows: Is $P(V_a|O_b) = P(V_c|O_b)$?
$P(V_a|O_b)$ is the probability that door a hides the car given that Monty opens door b. Similarly, $P(V_c|O_b)$ is the probability that door c hides the car given that monty opens door b. So, when $P(V_a|O_b) = P(V_c|O_b)$ the probability that the car is behind one unopened door the same as the probability that the car is behind the other unopened door. If this is the case, it won't matter if the contestant stays or switches.
Using Bayes' Theorem, we know that
$P(V_a|O_b)=\frac{P(V_a)*P(O_b|V_a)}{P(O_b)}$
$P(V_c|O_b)=\frac{P(V_c)*P(O_b|V_c)}{P(O_b)}$
Also, we can assume that the prize is randomly placed behind the curtains, so
$P(V_a) = P(V_b) = P(V_c) = \frac{1}{3}$
Then we can calculate the conditional probabilities for the event Ob, which we can then use to calculate the probability of event Ob.
First, we can calculate the conditional probability that Monty opens door b if the car is hidden behind door a.
$P(O_b|V_a) = 1/2$ because if the prize is behind a, Monty can open either b or c.
$P(O_b|V_b) = 0$ because if the prize is behind door b, Monty can't open door b.
$P(O_b|V_c) = 1$ because if the prize is behind door c, Monty can only open door b.
Each of these probabilities is conditional on the fact that the prize is hidden behind a specific door, but we are assuming that each of these probabilities is mutually exclusive since the car can only be hidden behind one door. As a result, we know that P(Ob) is equal to
$P(O_b) = P(O_b \cap V_a) + P(O_b \cap V_b) + P(O_b \cap V_c)$
Using the equation for the probability of non-independent events, we can say
$P(O_b)= P(V_a)P(O_b|V_a) + P(V_b)P(O_b|V_b) +P(V_c)P(O_b|V_c)$
$= \frac{1}{3} * \frac{1}{2} + \frac{1}{3} * 0 + \frac{1}{3} * 1$
$= \frac{1}{2}$
Then, we can use $P(O_b)$, $P(O_b|V_a)$, and $P(V_a)$ to calculated $P(V_a|O_b)$.
$P(V_a|O_b) = \frac {P(V_a)*P(O_b|V_a)}{P(O_b)}$
$= \frac {\frac{1}{3} * \frac{1}{2}} {\frac{1}{2}}$
$= \frac {1}{3}$
Similarly,
$P(V_c|O_b) = \frac {P(V_c)*P(O_b|V_c)}{P(O_b)}$
$= \frac {\frac{1}{3} * 1} {\frac{1}{2}}$
$= \frac {2}{3}$
The probability of Vc (the event that car is hidden behind door c) in this case is not equal to the probability of Va (the case where the car is hidden behind the door that Monty hasn't opened and the contestant hasn't selected). The contestant is offered an opportunity to switch to door c. We have calculated that the probability of winning when door c is selected is 2/3 and the probability of winning with the contestant's original choice, door a is 1/3.
Since Monty is equally likely to open any of the three doors, we can generalize this strategy for any door that he opens. The probability that the car is hidden behind the contestant's original choice is 1/3, but the probability that the car is hidden behind the unopened and unselected door is 2/3. If the contestant switches, he doubles his chance of winning.
# Why It's Interesting
Variations of the problem have been popular game teasers since the 19th century, but the "Lets Make a Deal" version is most widely known.
#### History of the Problem
The earliest of several probability puzzles related to the Monty Hall problem is Bertrand's box paradox, posed by Joseph Bertrand in 1889. In Bertrand's puzzle there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. The player chooses one random box and draws a coin without looking. The coin happens to be gold. What is the probability that the other coin is gold as well? As in the Monty Hall problem the intuitive answer is 1/2, but the actual probability 2/3.
#### Ask Marilyn: A Story of Misguided Hatemail
The question was originally proposed by a reader of “Ask Marilyn”, a column in Parade Magazine in 1990. Marilyn's correct solution, that switching doors was the best strategy, caused an uproar among mathematicians. While most people responded that switching should not matter, the contestant’s chances for winning in fact double if he switches doors. Part of the controversy, however, was caused by the lack of agreement on the statement of the problem itself.
Most statements of the problem, including the one in Marilyn's column, do not match the rules of the actual game show. This was a source of great confusion when the problem was first presented. The main ambiguities in the problem arise from the fact that it does not fully specify the host's behavior.
For example, imagine a host who wasn't required to always reveal a goat. The host's strategy could be to open a door only when the contestant has selected the correct door initially. This way, the host could try to tempt the contestant to switch and lose.
When first presented with the Monty Hall problem, an overwhelming majority of people assume that switching does not change the probability of winning the car even when the problem was stated to remove all sources of ambiguity. An article by Burns and Wieth cited various studies on the Monty Hall problem that document difficulty solving the Monty Hall problem specifically. These previous articles reported 13 studies using standard versions of the Monty Hall dilemma, and reported that most people do not switch doors. Switch rates ranged from 9% to 23% with a mean of 14.5%, even when the problem was stated explicitely.
This consistency is especially remarkable given that these studies include a range of different wordings, methods of presentations, languages, and cultures.
Marilyn quotes cognitive psychologist Massimo Piattelli-Palmarini in her own book saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer" (Bostonia July/August 1991). When the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution was wrong.
One letter written to vos Savant by Dr. E. Ray Bobo of Georgetown University was especially critical of Marilyn's solution: "You are utterly incorrect about the game show question, and I hope this controversy will call some public attention to the serious national crisis in mathematical education. If you can admit your error, you will have contributed constructively toward the solution to a deplorable situation. How many irate mathematicians are needed to get you to change your mind?"
#### Monty and Monkeys
A recent article published in The New York Times uncovered an interesting relationship between the Monty Hall problem and a study on cognitive dissonance using monkeys. If the calculations of Yale economist M. Keith Chen are correct, then some of the most famous experiments in psychology might be flawed. Chen believes the researchers drew conclusions based on natural inclination to incorrectly evaluate probability.
The most famous experiment in question is the 1956 study "Postdecision changes in the desirability of alternatives" on rationalizing choices. The researchers studied which M&M colors were most preferred by monkeys. After identifying a few colors of M&Ms that were approximately equally favored by a monkey - say, red, blue, and yellow, - the researchers gave the monkey a choice between two of the colors.
In one case, imagine that a monkey chose yellow over blue. Then, the monkey would be offered the choice between blue and red M&Ms. Researchers noted that about two-thirds of the time the monkey would choose red. The 1956 study claimed that their results reinforced the theory of rationalization: Once we reject something we are convinced that we never like it anyway.
Dr. Chen reexamined the experimental procedure, and says that monkey's rejection of blue might be attributable to statistics alone. Chen says that although the three colors of M&M's are approximately equally favored, there must be some slight difference in preference between the original red, blue, and yellow. If this is the case, then the monkey's choice of yellow over blue wasn't arbitrary. Like Monty Hall's decision to open a door that hid a goat, the monkey's choice between yellow and blue discloses additional information. In fact, when a monkey favors yellow over blue, there's a two-thirds chance that it also started off with a preference for red over blue- which would explain why the monkeys chose red 2/3 of the time in the Yale experiment.
To see why this is true, consider Chen's conjecture that monkeys must have some slight preference between the three colors they are being offered. The table below shows all the possible combinations of ways that a monkey could possibly rank its M&Ms.
We can see that in the case where the monkey preferred yellow over blue, they monkey preferred red over blue in 2/3 of the rankings.
Although Chen agrees that the study may have still discovered useful information about preferences, he doesn't believe it has been measured correctly yet. "The whole literature suffers from this basic problem of acting as if Monty's choice means nothing" (Tierney 2008).
Monty Hall problem, the study of monkeys, and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly. Even academic studies may be littered with mistakes caused by difficulty interpreting statistics.
#### 21
The 2008 movie 21 increased public awareness of the Monty Hall problem. 21 opens with an M.I.T. math professor using the Monty Hall Problem to explain theories to his students. The Monty Hall problem is included in the movie to show the intelligence of the main character because he is immediately able to solve such a notoriously difficult problem.
# References
Heart of Mathematics. Edward B. Burger and Michael P. Starbird.
The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser. Jason Rosenhouse.
Brehm, J. W. (1956) Postdecision changes in the desirability of alternatives, Journal of Abnormal and Social Psychology, 52, 384-9
Helper Pages:
• Bayes' Theorem
• Probability |
# 94 Divided by 2: A Simple Math Problem with a Surprising Result
Source: bing.com
Mathematics is a fascinating subject that provides solutions to various problems in life. From calculating the area of a circle to solving complex equations, math plays a significant role in shaping our lives. However, sometimes, the simplest math problems can surprise us with their results. One such problem is 94 divided by 2.
## What is Division?
Source: bing.com
Division is a basic arithmetic operation that involves distributing a quantity or number equally into groups. It is the opposite of multiplication and is represented by a division symbol (÷).
## The Answer to 94 Divided by 2
Source: bing.com
So, what is the answer to 94 divided by 2? The answer is 47. Yes, you read that correctly. When you divide 94 by 2, the quotient or result is 47.
## Why is the Result Surprising?
Source: bing.com
At first glance, 94 divided by 2 may seem like a straightforward problem with a predictable result. After all, 94 is an even number, and dividing it by 2 should result in another even number. However, the answer is an odd number, which can be surprising to some.
## The Explanation Behind the Result
Source: bing.com
The reason why the result of 94 divided by 2 is an odd number is that 94 is not divisible by 2 without a remainder. When a number is not divisible by the divisor, the quotient will be the nearest whole number that is less than the actual result. In this case, 47 is the nearest whole number that is less than the actual result of 47.5.
## Applications of Division in Real Life
Source: bing.com
Division has various practical applications in our day-to-day lives. For instance, we use division to calculate the number of people who can sit at a table, the number of slices of pizza we need to order for a party, and the amount of money each person should contribute towards a group expense.
## Conclusion
Math problems like 94 divided by 2 are a reminder that sometimes the simplest problems can surprise us with their results. Division is a crucial mathematical operation that has numerous real-life applications. So, next time you encounter a math problem, take a moment to appreciate the beauty and complexity of mathematics. |
# Multiplication Chart 96
Studying multiplication after counting, addition, as well as subtraction is good. Kids find out arithmetic using a all-natural progression. This growth of learning arithmetic is usually the subsequent: counting, addition, subtraction, multiplication, lastly department. This assertion leads to the concern why understand arithmetic within this series? Most importantly, why find out multiplication following counting, addition, and subtraction just before department?
## These details respond to these questions:
1. Young children find out counting initially by associating aesthetic things with their fingertips. A real illustration: Just how many apples are there within the basket? Much more abstract instance is when aged are you presently?
2. From counting numbers, the subsequent plausible step is addition accompanied by subtraction. Addition and subtraction tables can be quite helpful educating helps for kids since they are aesthetic instruments making the transition from counting easier.
3. Which will be figured out next, multiplication or department? Multiplication is shorthand for addition. At this moment, youngsters possess a organization knowledge of addition. For that reason, multiplication is the after that rational method of arithmetic to find out.
## Overview essentials of multiplication. Also, look at the fundamentals the way you use a multiplication table.
We will overview a multiplication example. By using a Multiplication Table, increase several times 3 and get an answer twelve: 4 by 3 = 12. The intersection of row 3 and line 4 of your Multiplication Table is 12; 12 is the answer. For children beginning to discover multiplication, this is easy. They are able to use addition to eliminate the issue hence affirming that multiplication is shorthand for addition. Example: 4 by 3 = 4 4 4 = 12. It is an excellent introduction to the Multiplication Table. The additional benefit, the Multiplication Table is visible and reflects to learning addition.
## Exactly where should we commence studying multiplication while using Multiplication Table?
1. Very first, get familiar with the table.
2. Begin with multiplying by one particular. Commence at row number 1. Proceed to line # 1. The intersection of row 1 and column one is the solution: a single.
3. Replicate these methods for multiplying by one particular. Grow row one by posts one particular via twelve. The replies are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Recurring these steps for multiplying by two. Flourish row two by columns 1 through several. The answers are 2, 4, 6, 8, and 10 correspondingly.
5. We will jump ahead of time. Replicate these steps for multiplying by 5. Multiply row several by columns a single via 12. The answers are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively.
6. Now allow us to boost the quantity of difficulty. Repeat these steps for multiplying by a few. Flourish row about three by columns one by means of twelve. The answers are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. If you are confident with multiplication to date, use a examination. Remedy the next multiplication difficulties in your mind and then evaluate your answers for the Multiplication Table: increase six as well as two, multiply nine and about three, flourish 1 and 11, increase 4 and a number of, and increase 7 and 2. The trouble responses are 12, 27, 11, 16, and 14 respectively.
If you received 4 away from 5 various problems correct, make your very own multiplication tests. Estimate the answers in your mind, and look them making use of the Multiplication Table. |
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# Series
Lecture Notes
A series is an infinite sum of numbers: " " " " # % ) "' The individual numbers are called the terms of the series. In the above series, the first term is "#, the second term is "%, and so on. The 8th term is "#8 : " " " " " 8 # % ) "' # We can express an infinite series using summation notation. For example, the above series would be written as follows: "
_ 8"
The " symbol means "sum". The 8 " on the bottom refers to the fact that the first term is
" #8
8 " (some series start with 8 ! or 8 #), and the _ on the top indicates that the series continues indefinitely.
## The Sum of a Series
Many series add up to an infinite amount. For example: "" " " " " " _
_ 8" _
## "8 " # \$ % & _
8"
However, if the terms of a series become smaller and smaller, it is possible for the series to have a finite sum. The basic example is the series "
_ 8"
" #8
## which sums to ".
But what exactly does it mean to find the sum of an infinite series? Given a series " + 8 + " + # + \$ ,
_ 8"
a partial sum is the result of adding together the first few terms. For example, the third partial sum is +" +# +\$ , and the hundredth partial sum is +" +# +"!! . The partial sums form a sequence: =" +" =# +" +# =\$ +" +# +\$ =% +" +# +\$ +% By definition, the sum of the series is the limit of this sequence:
DEFINITION (SUM OF A SERIES) The sum of the series " +8 is the limit of the partial sums:
_ 8"
" +8
_ 8"
8_
lim =8
8_
lim a+" +8 b
_ 8"
" #8
## " " " " # % ) "'
The first few partial sums of this series are listed below: =" " # " " # % \$ % ( ) "& "'
=#
=\$
=%
## " " " " # % ) "'
As you can see, the partial sums are converging to ". Therefore, according to our definition of
the sum of an infinite series, " " " " " # % ) "'
## EXAMPLE 2 Find the sum of the series: !\$ !!\$ !!!\$ !!!!\$
SOLUTION
Here are the first few partial sums: =" !\$ !\$\$ !\$\$\$ !\$\$\$\$
=# !\$ !!\$
=\$ !\$ !!\$ !!!\$
## =% !\$ !!\$ !!!\$ !!!!\$
As you can see, the partial sums are converging to the repeating decimal !\$\$\$\$ , which is equal to "\$. We use the same terminology for sums of series that we do for limits of sequences and for improper integrals:
## We say the series ! +8 converges if its sum is a real number.
If the sum is infinite or does not exist, then the series ! +8 diverges. sum " +8 can only be finite if lim +8 !.
_ 8" 8_
Keep in mind that a series can only converge if its terms get smaller and smaller. That is, the
DO THE TERMS APPROACH ZERO? 1. If lim +8 !, then the series " +8 diverges.
_ 8_
## 2. If lim +8 !, then the series " +8 may converge, or it may diverge.
_ 8_ 8"
8"
EXAMPLE 3 Find the sum of the series: " # \$ % & # \$ % & ' Since the individual terms of the series are getting closer and closer to ", the sum of the series is infinite (for the same reason that " " " _).
SOLUTION
_ 8"
SOLUTION
Since lim
## 8 " !, this series diverges to _. 8_ #8 " #
By the way, it is quite possible for the sum of a series to be infinite even if the terms get smaller and smaller. For example, even though the terms of the series " " " " " # \$ % &
become smaller and smaller, the sum of this series is infinite! This series is important enough to have its own name: the harmonic series (named for the frequencies of harmonic overtones in music). You should always remember that the harmonic series diverges.
Geometric Series
A geometric series is the sum of the terms of a geometric sequence. For example, the series " " " " " # % ) "'
is geometric, with a common ratio of "# (i.e. each term is "# times the previous term). Here are several more examples: "# % " % % % \$ * #( (common ratio of "\$) (common ratio of #&) (common ratio or #\$)
## # % ) & #& "#& % ) "' \$ * #(
\$ #
In our study of geometric sequences, we learned that the formula for the 8th term of a geometric sequence has the form +<8 , where < is the common ratio. If we start the series at 8 !, then the constant + can be interpreted as the value of the first term:
GEOMETRIC SERIES FORMULA The formula for a geometric series is " +<8 , where + is the first term and < is the common ratio.
_ 8!
SOLUTION
8
## The first term of the series is \$, and each term is
Therefore:
_ % ) "' # \$ # "\$ \$ * #( \$ 8!
Notice that the terms of a geometric series " +<8 only approach ! if the ratio < is between "
_ 8!
and ". It turns out that any such series converges: The geometric series " +<8 converges for k<k " and diverges for k<k ".
_ 8!
There is a special trick to finding the sum of a geometric series, illustrated in the next several examples:
EXAMPLE 6 Find the sum of the series: " " " " # % ) "' We already know that this sum is ", but this series is useful for illustrating the general method. Let B be the (unknown) sum of the series:
SOLUTION
## " " " " # % ) "'
If we multiply this equation by "# (the common ratio), each term of the sum is halved: " " " " " B # % ) "' \$# But the sum on the right is just the later portion of the original series:
" #B
## This gives us the equation B " " B # #
and therefore B ". We conclude that the sum of the series is ".
## EXAMPLE 7 Find the sum of the series: 1
SOLUTION
# % ) "' \$ * #( )"
## Let B be the unknown value of the sum: B 1 # % ) "' \$ * #( )"
If we ignore the initial ", the remaining terms of the series are simply # B: \$ B 1 # % ) "' \$ * #( )"
# B \$
## This gives us the equation B " # B \$
and therefore B "). We conclude that the sum of the series is ").
This trick only works for geometric series. The key idea is that a geometric series is self-similar, in the sense that the portion of the series after the first term is simply a multiple of the whole series. By the way, a similar trick can be used to convert any repeating decimal to a fraction:
## EXAMPLE 8 Express !%)%)%)%) as a fraction of integers.
SOLUTION
Let B !%)%)%)%) . Then B can be written as a sum of two parts: B !%) !!!%)%)%)
The first part is the same as %)"!!, and the second part is actually B"!!. This gives us the equation
## %) " B. "!! "!!
This trick not only looks the same, it is the same. The decimal in the previous example is by itself a geometric series: !%)%)%)%)
_ %) %) %) %) " "!! "!,!!! ",!!!,!!! a "!! b8 8"
Similar reasoning can sometimes be used to evaluate infinite algebraic expressions that aren't series: EXAMPLE 9 Find the value of the expression ' ' ' ' .
SOLUTION
Observe that this expression contains itself. In particular, if we let B ' ' ' '
then
B ' B.
Squaring and solving for B yields B \$ or B #. Since the square root must be positive, we conclude that B \$. You can verify this for yourself on your calculator. Starting with any number, alternately add ' and take the square root. The more times you repeat these steps, the closer the result will be to \$. " # #
SOLUTION
## EXAMPLE 10 Find the value of the expression #
" " #
This kind of expression is known as a continued fraction. If we let B be the value of the entire expression, then B # " B
Solving for B yields B " #. Rejecting the negative solution as nonsensical, we conclude that the continued fraction above is equal to " #.
General Formula for a Geometric Series It is possible to avoid repeating this same trick over and over by doing it once in general. Given the series " +<8 + +< +<# +<\$
_ 8!
let B be the unknown value of the sum: B + +< +<# +<\$ Ignoring the initial +, the rest of the sum is equal to <B: B + +< +<# +<\$ <B
This gives us the equation B + <B. Solving for B yields the formula B + . "<
## SUM OF A GEOMETRIC SERIES If k<k ", then: " +<8
_ 8!
+ "<
EXAMPLE 11 Find the sums of the following series: _ & * #( )" (a) " 8 (b) ) ' \$ # ) \$# 8!
SOLUTION
## (a) We have "
_ 8!
& \$8
\$ (b) The first term of this series is + ), and the ratio between terms is < . Therefore, % + ) the sum is \$#. "< " \$% You should be familiar with both methods for summing a geometric series (using the trick or using the formula).
_ " "& \$ 8!
## & "& . " "\$ #
Combining Series
Because infinite series and integrals are two different sorts of infinite sums, they obey many of the same rules:
## RULES FOR SERIES 1. " G +8 G " +8
_ _ 8" _ 8" _
(where G is a constant)
_
## 2. "a+8 ,8 b " +8 " ,8 3. "a+8 ,8 b " +8 " ,8
_ _ _ 8" 8" 8" 8" 8" 8"
_ 8!
" & 8 . 8 # \$
SOLUTION
## We can break this series up into two geometric series: "
_ 8! _ _ " & " & " " 8 8 8 # \$ # \$8 8! 8!
## " & "& "* # " "# " "\$ # #
EXERCISES
16 Rewrite the given series using summation notation. 14. Express the number !"\$&"\$&"\$& as a fraction of integers. " # % ) "' 1. & ' ( ) * & "! "( #' # ' #% "#! 1516 Find the value of the given self-similar expression.
" " " 15. \$ " # \$ # \$ # \$ #
2. #
## " # % # % " " " " #
1726 Determine whether the given series is convergent or 5. * ' % divergent. If it is convergent, find the sum. 17. "
_
## " " # % ) 6. "! & & & &
#8 &8 8!
18. "
_
" 8 \$ 8"
_
712 Find the sum of the given geometric series. 7. # # # # # \$ * #( )" #%\$ & & & # % )
19. "
_ 8"
8# " 8
8"
21. "
_
" /#8 8!
_
22. "
_
%8" &8 8!
_
8. "! &
## 23. " tan" a8b
8"
24. "
8"
8\$ 8a8 #b
" " " \$ * 9. ") ' # # # " " # ) #% ") #( )" ) "' \$ *
25. "
_ 8"
\$8 #8 '8
_ 8"
10.
11. * ' %
12. "! % |
# Quotient Rule: A Shortcut That Doesn't Require to Memorize Another Formula
The quotient rule is a formula that lets you calculate the derivative of quotients between functions. It is a more complicated formula than the product rule, and most calculus textbooks and teachers would ask you to memorize it.
I don't think that's neccesary. So, I'll show you an alternative way of solving problems involving quotients that doesn't need the memorization of yet another formula.
The method I use is a combination of chain rule+product rule, which you already know. And I'll show you the method directly with examples. Let's begin.
## Example 1
Let's say we want to find the derivative of:
Here we have the quotient between two functions. To find the derivative of this function, we only need to remember that a quotient is in reality a product. So, the first thing we do is to write the function as a product, which we can do like this:
Now that we have a product, we can apply the product rule. First we determine the functions u and v:
And we invoke the product rule formula:
And with some algebra we get the following expression:
And that's it. I think that it is more practical to learn to solve quotient problems like this than memorizing the quotient rule. Even if you can memorize the quotient rule now, you won't remember it 1 year from now, I assure you.
## Example 2
Let's find the derivative of:
Again, we can write this as:
And here we apply the product rule. We determine our u and v:
And apply the formula:
And that equals:
And you can play algebraically with it, but that's basically the answer.
## Example 3
Let's figure out the derivative of:
We'll use the same technique:
And we apply our trusted product rule:
That's enough quotient rule in disguise for now. Your next step should be to learn about the derivative of ln(x).
## Conclusion
• There's no need to memorize yet another formula.
• You can consider the quotient rule as a special case of the product rule.
• Whenever you have a function with a quotient, you can always write it like a product. |
# Illustrative Mathematics Unit 6.4, Lesson 6: Using Diagrams to Find the Number of Groups
Learning Targets:
• I can use a tape diagram to represent equal-sized groups and find the number of groups.
Related Pages
Illustrative Math
### Lesson 6: Using Diagrams to Find the Number of Groups
Let’s draw tape diagrams to think about division with fractions.
Illustrative Math Unit 6.4, Lesson 6 (printable worksheets)
### Lesson 6 Summary
The following diagrams show how to use a tape diagram to represent equal-sized groups and find the number of groups.
### Lesson 6.1 How Many of These in That?
1. We can think of the division expression 10 ÷ 2½ as the answer to the question: “How many groups of 2½ are in 10?” Complete the tape diagram to represent the question. Then answer the question.
2. Complete the tape diagram to represent the question: “How many groups of 2 are in 7?” Then answer the question.
### Lesson 6.2 Representing Groups of Fractions with Tape Diagrams
To make sense of the question “How many ⅔s are in 1?,” Andre wrote equations and drew a tape diagram.
1. In an earlier task, we used pattern blocks to help us solve the equation 1 ÷ ⅔ = ?. Explain how Andre’s tape diagram can also help us solve the equation.
2. Write a multiplication equation and a division equation for each of the following questions. Draw a tape diagram to find the solution. Use the grid to help you draw, if needed.
a. How many ¾s are in 1?
b. How many ⅔s are in 3?
c. How many 3/2s are in 5?
#### Lesson 6.3 Finding Number of Groups
1. For each question, draw a diagram to show the relationship of the quantities and to help you answer the question. Then, write a multiplication equation or a division equation for the situation described in the question. Be prepared to share your reasoning.
a. How many ⅜-inch thick books make a stack that is 6 inches tall?
b. How many groups of ½ pound are in 2¾ pounds?
2. Write a question that can be represented by the division equation 5 ÷ 1½ = ?. Then answer the question. Show your reasoning.
#### Lesson 6 Practice Problems
1. We can think of 3 ÷ ¼ as the answer to the question “How many groups of ¼l are in 3?” Draw a tape diagram to represent the question. Then answer the question.
2. Describe how to draw a tape diagram to represent and answer 3 ÷ ⅗ = ? for a friend who was absent.
3. How many groups of ½ days are in 1 week?
a. Write a multiplication equation or a division equation to represent the question.
b. Draw a tape diagram to show the relationship between the quantities and to answer the question. Use graph paper, if needed.
4. Diego said that the answer to the question “How many groups of ⅚ are in 1?” is ⅚ or 1⅕. Do you agree with his statement? Explain or show your reasoning.
5. Select all equations that can represent the question: “How many groups of ⅘ are in 1?”
A. ? · 1 = ⅘
B. 1 · ⅘ = ?
C. ⅘ ÷ 1 = ?
D. ? · ⅘ = 1
E. 1 ÷ ⅘ = ?
6. Calculate each percentage mentally.
a. What is 10% of 70?
b. What is 10% of 110?
c. What is 25% of 160?
d. What is 25% of 48?
e. What is 50% of 90?
f. What is 50% of 350?
g. What is 75% of 300?
h. What is 75% of 48?
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various 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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# What Are The Prime Factors Of 157?
• Prime factors of 157: 1 * 157
## Is 157 A Prime Number?
• Yes the number 157 is a prime number.
• It's a prime because one hundred and fifty-seven has no positive divisors other than 1 and itself.
## How To Calculate Prime Number Factors
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## Mathematical Information About Numbers 1 5 7
• About Number 1. The number 1 is not a prime number, but a divider for every natural number. It is often taken as the smallest natural number (however, some authors include the natural numbers from zero). Your prime factorization is the empty product with 0 factors, which is defined as having a value of 1. The one is often referred to as one of the five most important constants of analysis (besides 0, p, e, and i). Number one is also used in other meanings in mathematics, such as a neutral element for multiplication in a ring, called the identity element. In these systems, other rules can apply, so does 1 + 1 different meanings and can give different results. With 1 are in linear algebra and vectors and one Einsmatrizen whose elements are all equal to the identity element, and refers to the identity map.
• About Number 5. Integers with a last digit as a zero or a five in the decimal system are divisible by five. Five is a prime number. All odd multiples of five border again with the five (all even with zero). The fifth number of the Fibonacci sequence is a five. Five is also the smallest prime number that is the sum of all other primes which are smaller than themselves. The Five is a Fermat prime: 5 = 2 ^ {2 ^ 1} +1 and the smallest Wilson prime. Number five is a bell number (sequence A000110 in OEIS). There are exactly five platonic bodies. There are exactly five tetrominoes.
• About Number 7. Seven is a prime number. It is the lowest natural number that cannot be represented as the sum of the squares of three integers. The corresponding cyclic number is 142857. You can use this feature to calculate the result of the division of natural numbers by 7 without a calculator quickly. A seven-sided shape is a heptagon. One rule for divisibility by 7 leads to a simple algorithm to test the rest loose divisibility of a natural number by 7: Take away the last digit, double it and subtract them from the rest of the digits. If the difference is negative, then you're leaving the minus sign. If the result has more than one digit, so you repeat steps 1 through fourth. Eventually results are 7 or 0, then the number is divisible by 7 and not otherwise.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers.
## What Are Prime Factors?
• In number theory, the prime factors of a positive integer are the prime numbers that divide that integer exactly. The prime factorization of a positive integer is a list of the integer's prime factors, together with their multiplicities. The process of determining these factors is called integer factorization. The fundamental theorem of arithmetic says that every positive integer has a single unique prime factorization.
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Fourths and Quarter Cakes
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Print Lesson
Objective
SWBAT create and identify fourths and divide shapes into fourths. SWBAT use the standard notation of 1/4.
Big Idea
Today your students will look at the difference between a shape being divided into fourths or four parts.
Warm Up
5 minutes
I have the students gather in a circle on the carpet. I give them each analog clock I will use the white board to write down digital times.
"I will write a time on the board using digital notation. Your job will be to set your clock to match that time. when you are finished, you should hold up your clock so that I can see your answer."
I repeat this with hour and half hour times.
At this point in the year, I am reviewing established routines and concepts from the year long curriculum. This activity is meant for students who are familiar with the use of a clock and understand the basic time concepts. I also like using this warm up for a fraction lesson because of the connection between clocks and fractions (half past, quarter past, etc.).
This activity has students telling and writing times to hours and half hours (CCSS.MATH.CONTENT.1.MD.B.3).
Looking At Fourths
15 minutes
Advanced Preparation: You will need to create two posters (Squares Poster and Circles Poster) before the lesson starts. You will notice that I lightly drew in the lines for quartering each shape. This way I could make sure they were precise, and it also saves time as you start the lesson.
"We have been talking about halves. What happens if I divide a circle in half (draw a line through the middle of the first circle)? What can you tell me about the two pieces? What if I draw another line and create 4 equal pieces? Does anyone know what that is called? You're correct, Fourths or Quarters. When you break something into fourths, it not only has 4 pieces but each piece has to be equal."
The rest of your conversation should focus on the fact that fourths means 4 equal pieces, that a fourth is one part of the whole, and 4 fourths make 1 whole.
I then move onto the second circle and use a market to define the lines dividing the circle into four unequal parts.
"Is this broken into fourths?"
I had included a video (Quartering Circles.m4v) with a student's explanation of why it is not.
I then use the squares poster and repeat the above procedure. I have also included a photo of the Completed Squares Poster (after the discussion). You will see that we crossed out ones that were not fourths as we went.
This activity has students partitioning shapes into four equal shares, describe the shares using the words fourths,and the phrase quarter of. They are also asked to describe the whole as four of the shares (CCSS.MATH.CONTENT.1.G.A.3).
Creating Quarter Cakes
40 minutes
Advanced Preparation: You will need to make enough copies of Square Cakes for your class. You wil also need rulers, markers, crayons, and scissors.
Please watch the video, Introducing Quarter Cakes to see how to introduce this activity to your class.
"I want you to now go find a spot int he room to work on your own cakes. Remember to color each half a different color."
I have included a photo, Creating Cakes, of a student working on her cakes.
Exit Ticket
5 minutes
To end today's lesson, I will ask that students complete the exit ticket, Finding Fourths. I will work with students that I saw struggling with the concept during the Quarter Cakes activity.
Continued Practice
5 minutes
I will ask the students to meet me on the carpet and hand out their sheet for today's Mad Minute exercise. This routine was introduced in a previous lesson. Please check out the link to get a full overview of this routine.
I want to really focus on fact fluency and build upon the students ability to solve within ten fluently (CCSS.MATH.CONTENT.1.OA.C.6). I am going to use the Mad Minute Routine. This is a very "old school" routine, but I truly feel students need practice in performing task for fluency in a timed fashion. Students need to obtain fact fluency in order to have success with multiplicative reasoning. Students who don't gain this addition fact fluency by the end of 2nd grade tend to struggle with the multiplicative reasoning in third. Having this fluency also allows them to work on more complex tasks because the have the fact recall to focus on the higher level concepts. |
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Explore the Back-Story
# NMTC 2015 Stage II - Gauss (Class 5, 6) - Problems and Solutions
###### Problem 1
A unit fraction is one of the form $\frac{1}{a}$ where ' $a$ ' is a natural number $(\neq 1)$.Any proper fraction can be written as the sum of two or more unit fractions.
$\left{\right.$ Ex. $\left.\frac{1}{2}=\frac{1}{3}+\frac{1}{6}, \frac{5}{6}=\frac{1}{2}+\frac{1}{3}, \frac{1}{24}=\frac{1}{54}+\frac{1}{72}+\frac{1}{108}\right}$
In the case of $\frac{1}{2}$, the factors of 2 are 1 and 2 only
$$\frac{1}{2}=\frac{1+2}{2(1+2)}=\frac{1}{2(1+2)}+\frac{2}{2(1+2)}=\frac{1}{6}+\frac{1}{3}$$
Express $\frac{1}{15}$ as the sum of two different unit fraction in 4 different ways.
###### Problem 2
(a) The number 11284 and 7655 when divided by a certain 3 digit number leave the same remainder. Find the number and the remainder.
(b) What is the least number to be subtracted from 1936 so that when divided by 9,10 and 15 it will leave the same remainder in each case?
###### Problem 3
A, M, T, I represent different non-zero digits, It is given that
\begin{aligned} & A+M+T+I-11 \\ & A+M+I-10 \\ & A+M-1 \end{aligned}
Further in the following addition only one digit is given.
Fill up the stars writing proper reasons.
###### Problem 4
$A B C$ is a three digt number in which the digit $A$ is greater than the digit $B$ and $C$. If the difference between $A B C$ and $C B A$ is 297 and the difference between $A B C$ and $B A C$ is 450 , find all possible three digit numbers $A B C$ and find their sum also.
###### Problem 5
Take a triangle. Three straight lines are drawn through the vertices of the triangle as shown. The maximum number of points of intersection is 3 . Draw two lines through each vertex of a triangle to meet the opposite sides. What is the maximum number of points of intersection.
Find again drawing the maximum number of points of intersection when three lines are drawn through each vertex.
Without drawing can you guess the maximum points of intersection for 4 lines?
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Lesson 14: Ledge's Introduction To Cryptarithms II
```
CLASSICAL CRYPTOGRAPHY COURSE
BY LANAKI
June 12, 1996
Revision 0
LECTURE 14
LEDGE'S INTRODUCTION TO CRYPTARITHMS II
SUMMARY
It is my pleasure to present our guest lecturer LEDGE's (Dr.
Gerhard D. Linz) second lecture on the interesting topic of
Cryptarithms. In this lecture, he covers Multiplication,
Multiplicative Structures, Base 11 and Base 12 calculations.
LEDGE has a natural writing style, and a talent for making
understandable some difficult concepts. LEDGE has already
produced one of our better references on novice cryptography,
and I appreciate his assistance in our course. Enjoy. [LEDG]
UNIQUE SOLUTION
Another Rule. Cryptarithms must meet another rule not stated
in the first Lecture 8. There must be only one solution to
the problem. That means that the solution must be unique.
For ease in reading we will now use "mod" for "modulo. See
the previous cryptarithm lecture for the meaning of the term.
MULTIPLICATION
Let's start our analysis of reconstructing multiplication
problems by looking at a typical multiplication of whole
numbers as we learned it in grammar school:
478
x52
---
956
2390
----
24856
introduce some nomenclature. The number being multiplied,
here 478, is called the "multiplicand." The number by which
the multiplicand is multiplied, here 52, is the "multiplier."
The result of the multiplication, here 24856, is the
"product."
If we analyze the parts or steps of this process, we find we
have two separate multiplications and one addition in what we
usually consider a single multiplication problem. The problem
contains substeps: 2 x 478 = 956; 5 x 478 = 2390 and an
addition to which we will turn in a moment. Notice in the
second multiplication of the multiplicand by a digit in the
multiplier, instead of 5 x 478 we really have 50 x 478. We
don't write it that way because we moved the product of 5 x
478 one decimal place to the left and left a blank space for
the product of 0 x 478 (which equals 0). If the multiplier
had more digits, we would have continued to move the
subsequent partial products another space to the left. We
have done multiplications like this so often that we don't
usually recognize what we are doing.
Now we can look at the addition: 956
2390
-----
24856.
OBSERVATIONS
Each of these steps can give us different and valuable
information.
1. The highest order digit of the multiplicand,
multiplier, and product cannot be zero. In other words, by
convention, no number starts with zero since no decimals are
involved. If we had used letters in this example, the letters
representing the 4 in 478, the 5 in 52, the 9 in 956, the 2
in 2390, and the 2 in 24856 could not represent 0.
2. The product of any sized multiplicand by a single
digit multiplier can never contain more digits than the
number of digits in the multiplicand plus one. If you need
convincing, try it out with examples of your choice. In this
case we have one such example: 5 x 478 = 2390. The
multiplicand has three digits, the product four.
3. If a product has more digits than the multiplicand,
the highest order digit of the product is less than or equal
to the lower of the multiplier or the highest order digit of
the multiplicand. Here, 5 x 478 = 2390. The highest order
digit of the product is 2, smaller than the 4 of the
multiplicand which in turn is smaller than the multiplier, 5.
4. The addition step is subject to the kinds of analyses
we saw in the first lecture.
When solving multiplication cryptarithms, you may want to
write out the separate parts of the problem as CROTALUS
has suggested. In this lecture we will use the understandings
we have developed, but leave the problem intact. [CROT]
The units digits of the products (the digit on the right end
of each product) also can produce useful information which we
Example 1. Now let's tackle an example that should not be too
difficult. This one is a C-6 from the March-April, 1995,
issue of The Cryptogram by DYETI. The key is two words (9-0).
LARK
xCAR
-----
OOYRR
ORLOA
LEECC
-------
LOSBRLR
SOLUTION OF EXAMPLE 1
By now you should be able to see the various parts of the
problem: three multiplications of the multiplicand by R, A,
and C; and the addition of the three partial products, each
one after the first shifted an additional space to the left
to give the final product, LOSBRLR.
notice that the leftmost digit of LEECC, the L, is carried to
the final product without change; hence the sum of O and E,
the next digits on the right, is less than 10 as there was no
carry to the L. Finally, O + E + (0, 1 or 2) = O without a
carry to the L. The (0, 1, or 2) are the possible carries
from the previous addition of O + R + E plus a carry. But the
only way for O + E = O mod 10 without a carry is for O + E =
O not O + 10. As a result E = zero. Now let's take a look at
the partial products. The products of C, A, and R times LARK
is in no case LARK. Hence none of C, A, and R = 1. However,
the product of R x K = R mod 10. The product of A x K = A mod
10, and the product of C times K = C mod 10.
There is only one value of K that would make that true. K
must = 1. Furthermore, there is no carry in any of those
multiplications.
That takes care of the information we can gather from the
examination of the units digits.
With a carry of 0 from the first multiplications, we can look
similarly at the result of multiplying each of the digits in
the multiplier by the tens digit of the multiplicand, the R:
R x R = R mod 10 (the tens digit in the first partial
product). A x R = O mod 10 and C x R = C mod 10. Looking at
the first product, R x R, the only digits that give
themselves as the units number of their products when
multiplied by themselves are 0 x 0 = 0; 1 x 1 = 1; 5 x 5 = 25
or 5 mod 10; and 6 x 6 = 36 or 6 mod 10. We already know what
letters represent 0 and 1 so R = 5, or 6.
Both 5 and 6 are interesting numbers when considered from a
multiplication standpoint. 5 x an even number = 0 mod 10.
5 x an odd number = 5 mod 10. Hence any product of 5 must end
in either 5 or 0, two choices. Here we have three different
products of R, not two. So R = 6.
Let's take a look at the products of 6 mod 10 since we have
the product C x R = C: 6 x 2 = 2; 6 x 4 = 4, 6 x 8 = 8, all
mod 10. The general rule is that R must be even for 6 x R =
R. Furthermore, since R is even, the product of R x n is
even for any digit value of n. The only way to get an odd
numbered product is to multiply two odd numbers. Try it out.
So C = 2, 4, or 8. Can we narrow that down? It turns out that
we can.
Each of the partial products is 5 digits long, one more than
the multiplicand. From fact 3 above, we know that the highest
order digit in each case cannot be larger than the lower of
the single digit multiplier and the highest order digit of
the multiplicand. The highest order digits of the partial
products are O, O, and L. Since L is the highest order digit
of the multiplicand, C x LARK must yield the largest product.
The highest possible value of C is 8. The next one is 4.
Almost certainly C = 8.
Let's put our number-letter equivalents into a table:
9 8 7 6 5 4 3 2 1 0
C R K E.
We're supposed to find two words at the end of this process.
The letters we have make a promising beginning.
The product C x LARK = LEECC or, since we know many of the
values of the letters, 8 x LA61 = L0088. Let's go through
that multiplication step by step. 8 x 1 = 8, no carry. 8 x 6
= 48, carry the 4. 8 x A + 4 (the carry) must = 0 mod 10; so
8 x A must = 6 mod 10 since 6 + 4 = 0 mod 10. The products of
8 that end in 6 are 16 and 56; so A = 2 or 7. The addition
section will give us the clue we need.
At the tens digit of the addition R + A = L mod 10. We know R
is 6. If A = 2, then R + A = 8, not possible since C =
8 already. So A = 7. R + A = 6 + 7 = 3 mod 10 or L = 3.
Dividing the product LEECC or 30088 by 8 (C) gives 3761 for
LARK. O could = 1 or 2, but only 2 is available (why?).
Multiplying LARK x R will give us Y in the first partial
product = 5. Only B has not been determined. By default it
must be = 4. The key tableau is 9 8 7 6 5 4 3 2 1 0.
S C A R Y B L O K E
The final result : 3761
x876
-----
22566
26327
30088
-------
3294636
Example 2. Let's try another. If you feel brave, try it on
difficult than the first problem, only different.
On the third page of Lecture I we presented this
multiplication problem by APEX DX:
OTTAWA
xON
------
HNNTLIL
IIIEHE
-------
TOOINRL
SOLUTION OF EXAMPLE 2
In Lecture 8, we determined that the only candidates for the
representation of zero were L, W and R. We carried the
solution no further at that point. We can do better than that
with the tools we now have.
The problem contains two partial products, N x OTTAWA =
HNNTLIL and O x OTTAWA = IIIEHE, plus the addition of those
products to give the final product. TOOINRL. We now note that
the second partial product and the multiplicand have the same
number of digits, six.
Further, the highest order digit of the multiplicand and
multiplier are the same, namely O. O x O + carry = I. The
highest digit O can represent is 3 as 3 x 3 = 9. Any higher
digit when multiplied by itself gives a two-digit result,
adding a digit to that partial product. If O = 3, then
I = 9. The partial product become 999EHE. Dividing by the
multiplier value, 3 produces 333??? for OTTAWA. That cannot
be since we have only one digit per letter. O also is not
one, for multiplying any number by one results in that
number. Therefore O must be = 2. 2 x 2 = 4 and the partial
product would 444EHE. Again, dividing by 2 give 222???.
As before I cannot equal T. Since 2 x 3 is 6 and I is less
than that, I = 5, and the partial product is 555EHE. Dividing
by O or 2, the multiplier, gives 277??? for OTTAWA; hence T =
7. In lecture I we were left with L, W, and R as the only
possible candidate for the digit, 0. At that time we could
not unambiguously select one of these as representing zero.
We can now eliminate L from consideration. Look at the
problem to see if you can spot how.
L comes from the product of N x A mod 10. For L to be = zero,
either N or A must be = 5. We have already determined that I
= 5. Neither N or A are five, so L cannot be zero. We are
left to choose between R and W.
Our letter-number equivalent table now is:
0 1 2 3 4 5 6 7 8 9
O I T
At the moment we can make no progress with the second partial
product so let's examine the first, N x OTTAWA = HNNTLIL.
Substituting the identified digits we have N x 277AWA =
HNN7L5L. This product has seven digits, one more than
OTTAWA. We have learned that the highest order digit of such
a product must be less than or equal to the lower of the
multiplier or the highest order digit of the multiplicand,
i.e, O or N. Since O = 2, H can only = 1. We add that to the
letter-number equivalent table. The partial product becomes N
x 277AWA = 1NN7L5L. Dividing the product by OTTAWA or 277AWA,
we learn that N could be 4,5, or 6. Since I = 5, N must be 4
or 6.
Still working with the first partial product, N x A = L mod
10. A is multiplied by N again when we reach the hundreds
digit of OTTAWA. Again the result is L mod 10. How could this
be? It can only happen if there is no carry from the product
of N x W. In the second partial product 2 x A = E mod 10 two
times, as before. Thus 2 x W cannot have a carry here as
well. Neither 4 x W, 6 x W, nor 2 x W > 9. W can only be 0 or
1. Because H = 1, W = 0.
We could have gone another route. In the addition I + E = R.
If R were = 0, since I = 5, E would have to be 5 also. That's
not allowed, so R cannot be 0 and only W is left to = zero.
Still a third way of determining whether R or W = zero is by
anagraming. If R = zero, we look at the equivalent table and
the keyword would have to start RHO, not impossible, but not
encouraging. If W = zero, the keyword starts WHO, a word,
very encouraging. Just like in K1 and K2 Aristocrats,
reconstructing the equivalent table (instead of the
equivalent alphabets) can give us useful clues. I generally
use anagraming only as a last resort if I am otherwise
stymied, however.
With W = 0 we know that O x A = HE and N x A = IL. Replacing
known letter values, we have 2 x A = 1E and N x A = 5L. The
only products in the 50's produced by multiplying two single
digit numbers are 54 and 56 or 9 x 6 or 7 x 8. Since T = 7,
we cannot use N = 7 to yield N x A = 56; hence N and A are 9
and 6 and L = 4. We know that N is 4 or 6 (see above). So N =
6, A = 9, and, since 2 x 9 = 18, E = 8. The results are
consistent; they produce no redundancies.
The equivalent table becomes:
0 1 2 3 4 5 6 7 8 9
W H O L I N T E A.
Only the R needs to be placed. What are the three words?
Whorl in tea (Tempest in a teapot???).
PROBLEMS IN BASES OTHER THAN TEN
Our number system is based on the number 10, perhaps because
normally humans have ten fingers and ten toes. So having ten
for a base makes counting easier. We generally write our
numbers as a series of digits with or without a decimal
point, but we read the real value of a digit by its position
in relation to the decimal point, either provided or tacitly
understood. So we read 5,678 as five thousand, six hundred,
seventy eight. Translating the English into number it becomes
5 x 1000 + 6 x100 + 7 x 10 + 8 x 1. That process is pure
convention, but we don't usually think about it.
Notice also that we have ten different characters for the ten
different digits. When we count from zero up in whole numbers
we use all ten (0-9) to get to 9 and then we move on to two
digits, using a one in the tens place and starting anew with
zero in the units place. It takes a lot of words to explain
it, but we're so used it; we just spout the number and go on.
Yet it is pure convention that we use ten as the base. We
call it decimal, using the Greek word for ten. In fact we
could use any whole number as the base except, of course, 0
alone as we can't count with it. Whatever number we use as a
base, that's how many characters we need. If we were to want
to count base 2 (like a series of switches that are either on
or off), we'd need only the digits 0 and 1. That's called the
binary system. Counting would go as follows:
Base 2: 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100
1101
Base 10: 0 1 2 3 4 5 6 7 8 9 10 11 12
13
Notice that in binary 1101 = 1 x 8 + 1 x 4 + 0 x 2 + 1.
In decimal we would read as 1 x 1000 + 1 x 100 + 0 x 10 + 1 x
1.
Just as 1000 is 10x10x10, so 8 is 2 x 2 x 2. 100 is 10x10; 4
is 2 x 2. Binary 1000 translates to decimal 8, etc. Binary
1101 = Decimal 13. Naturally with only two symbols, binary
representation of numbers are much longer than base 10
representations.
We used base two as an illustration only. Cryptarithms, if
not in decimal or base 10 form, use bases that are larger
than ten, most often 11, called undecimal, or 12, called
duodecimal. For undecimal we need to create a new character
to replace decimal 10. Usually, "x" is used. Since we are
using x as the multiplication symbol, we will use "t" for
ten. We need another symbol for decimal 11. Usually, "e" is
used.
Counting in undecimal goes like this: 1, 2, 3, 4, 5, 6, 7, 8,
9, t, 10, 11, 12 etc. What we are used to reading as 10 is
really 11 base 10. 11 is really 12 base 10, etc. In
duodecimal counting proceeds 1, 2, 3, 4, 5, 6, 7, 8, 9, t, e,
10, 11, 12, etc. Looks are deceiving and you have to be
careful. What looks like ten is read as 12, 11 is really
decimal 13. If you have a number like 378 in duodecimal,
think 3 x 12 x 12 + 7 x 12 + 8 or 3 x 144 + 7 x 12 + 8. If
you wish, you can think three hundred seventy eight, but you
must remember that in our ordinary notation 100 base 12 = 144
base 10 and 10 base 12 = 12 base 10. Arithmetical problems
are solved as always, taking note of the different notation.
If you find the following explanations which involve
arithmetical manipulations in base 11 and 12 confusing,
consult the multiplication and addition tables in the
Appendix. [Tables 14-1 - 14-4]
DUODECIMAL
Now let's look at some duodecimal examples.
+876
----
1151.
It looks odd, but it's duodecimal. 7 + 6 = 13. Divide by 12
and you get a quotient of 1 and a remainder of 1. Put down
the remainder and carry the quotient of 1. 9 + 7 + 1 (carry)
= 17. Divide by 12 giving a quotient of 1 and a remainder of
5. Put down the remainder of 5 and carry the quotient of 1. 8
+ 4 + 1 (carry) = 13. Divide by 12 giving a quotient of 1 and
a remainder of 1. Put down the remainder of 1 and, because
the next column adds to 0 + 1 (the carried quotient), put
down another 1. If we had an addition of 4 + 6 = ten, we
would not divide by twelve but merely put down the ten as t.
So in duodecimal 4 + 6 = t.
Example 2. Subtraction. 67
-39
--
2t.
To subtract 9 from 7 we must borrow 12 from 6, making it 5.
12 + 7 - 9 = 10 or t. We put that down. 6 - 1(borrow) - 3 =
2. Hence the answer is 2t.
Example 3. Multiplication. 67
x39
---
4e3
179
----
2083
The process in words: 9 x 7 = 63, divide by 12 giving
quotient of 5 and remainder of 3. Put down the 3 and carry
the quotient of 5, just as in addition. 9 x 6 + 5(carry) =
59. Divide by 12 giving 4 as quotient and 11 or e as the
remainder. Put down the remainder of e and, since there are
no more digits to multiply by 9, put down the quotient of 4.
Let's check this last one: 4 x 12 + 11 = 48 + 11 = 59. Work
through the rest of this example on your own.
Example 4. Division. 2e
---
17/48t
32
--
16t
155
---
15
First, we choose a trial quotient, here 2, and multiply the
divisor, here 17, by it. 2 x 7 is 14, divide by 12 getting a
remainder of 2 and a quotient of 1. Put down the 2 and carry
the 1. 2 x 1 + 1(carry) = 3. Put it down. Bring down the next
digit of the dividend, here t. Now go on your own and check
out my work.
Undecimal works the same way, except that instead of dividing
by 12, we would divide by 11. If all that dividing and
translating is too much to remember, use the proper
multiplication table in the Appendix. Just as in the base 10
or decimal multiplication table the product of one digit by
another is a one-digit or a two digit number, so it is in
undecimal and duodecimal. In fact that's true of any base
undecimal or duodecimal number as a decimal number. The
occasional t and e will remind you, but it's easy to forget
momentarily, even after you've been at it for a while.
MULTIPLICATIVE STRUCTURES
FIRE-O in the May-June, 1970, issue of The Cryptogram
introduced the concept of multiplicative structures [FIRE-O].
In 1977, in a two part article on base 11 and 12 arithmetic,
I expanded on FIRE-O's work by extending the multiplicative
structures to the higher bases [LEDG1] and [LEDG2]. The
concept is simple, but often very useful. [FIRE]
Let's take a digit, like 7, multiply it by 1 and then
multiply it successively by the resulting product, i.e., when
we multiply again we use the latest product. All the
multiplications will be mod 10 as we are only interested in
the units digit of the product. With using 7 we get:
1 x 7 = 7 7 x 7 = 9 9 x 7 = 3 3 x 7 = 1
Notice that the last product in the series in this case
results in the multiplier we started with, 1. For 7 we have
found a circular structure: (= 1 => 7 => 9 => 3 =). I am
using the symbols =) and (= as indicators to return to the
other end of the series.
We could also diagram it as:
1 => 7
^ |
| V
3 <= 9.
You can read the series as 1 to 7 to 9 to 3 to 1 to 7 etc.
Because of the lack of printable characters in ASCII, I'll be
using the first kind of diagram. Notice that all the digits
in the diagram are odd. We can start another diagram by
starting with an even number, say 2.
2 x 7 = 4 4 x 7 = 8 8 x 7 = 6 6 x 7 = 2 or
(= 2 => 4 => 8 => 6 =).
That leaves 5 x 7 = 5 or 5 <=, and 0 x 7 = 0 or 0 <=. In
other words, multiplying 7 by 5 or 0 gives the multiplier as
the units digit of the product. The last is true for any odd
number.
As we will see shortly, 3 diagrams out in a similar fashion
to 7, two circles, one of odd numbers and one of even ones.
If n is odd, then 5 x n => 5. Diagram: 5 <=.
If n is even, then 5 x n => 0. Diagram 5 => 0 <=
In both cases, 0 x n => 0 and 0 x 0 => 0. Diagram 0 <=.
Now let's look at the other diagrams.
BASE 10.
O: n x 0 <=
1: n x 1 => n <=. In other words, successive multiplications
by 1 always yield n.
2: 1 7 9 3 2 X 2 = 4 4 X 2 = 8 etc.
| | | | 1 x 2 = 2 7 x 2 = 4 etc.
V V V V and 5 => 0 <=.
(= 2 => 4 => 8 => 6 =)
3. (= 1 => 3 => 9 => 7 =) 5 <= and 0 <=
(= 2 => 6 => 8 => 4 =)
4. 1 => 4 <==> 6 <= 9; 3 => 2 <==> 8 <= 7; 5 => 0 <=
5. odd x 5 <=; even x 5 => 0 <=
6. 1 => 6 <=; 3 => 8 <=; 5 => 0 <=; 7 => 2 <=; 9 => 4 <=.
7. (= 1 => 7 => 9 => 3 =) 5 <= and 0 <=
(= 2 => 4 => 8 => 6 =)
8. 1 3 9 7 5 => 0 <=
| | | |
V V V V
(= 8 => 4 => 2 => 6 =)
9. 1 <==> 9; 2 <==> 8; 3 <==> 7; 4 <==> 6 5<= 0 <=.
Remember that in each case, each resulting product (mod 10)
is multiplied by the original multiplier given at the
beginning of each set, e.g., " 6.".
BASE 11 (UNDECIMAL)
0. n x 0 => 0 <=
1. n x 1 => n <=
2. (= 1 => 2 => 4 => 8 => 5 => t => 9 => 7 => 3 => 6 =) 0
<=
3. (= 1 => 3 => 9 => 5 => 4 =); (= 2 => 6 => 7 => t => 8 =);
0 <=
4. (= 1 => 4 => 5 => 9 => 3 =); (= 2 => 8 => t => 7 => 6 =);
0 <=
5. (= 1 => 5 => 3 => 4 => 9 = ; (= 2 => t => 6 => 8 => 7 =);
0 <=
6. (= 1 => 6 => 3 => 7 => 9 => t => 5 => 8 => 4 => 2=); 0 <=
7. (= 1 => 7 => 5 => 2 => 3 => t => 4 => 6 => 9 => 8 =); 0 <=
8. (= 1 => 8 => 9 => 6 => 4 => t => 3 => 2 => 5 => 7 =); 0 <=
9. (= 1 => 9 => 4 => 3 => 5 =); (= 2 => 7 => 8 => 6 => t =);
0 <=
t. 1 <==> t; 2 <==> 9; 3 <==> 8; 4 <==> 7; 5 <==> 6; 0 <=
BASE 12 (DUODECIMAL)
0. n x 0 => 0 <=
1. n x 1 => n <=
2. 1,7 => 2 => 4 (==> 8 <= t <= 5,e; 3,9 => 6 => 0 <=
3. 1,5 => 3 <==> 9 <= 7,e; 2,t => 6 <=; 4,8 => 0 <=
4. 1,7,t => 4 <=; 2,5,e => 8 <=; 3,6,9 => 0 <=
5. 1 <==> 5; 2 <==> t; 4 <==> 8; 7 <==> e; 3,6,9 0 <=
6. 1,3,5,7,9,e, => 6 => 0 <= 0,2,4,8,t
7. 1 <==> 7; 3 <==> 9; 5 <==> e; 0 <=; 2<=; 4 <=; 6 <=; 8
<=; t <=;
8. 1,7,t => 8 <==> 4 <= 2,5,e; 3,6,9 => 0 <=
9. 1,5 => 9 <=; 2,7 => 6 <=; 7,e => 3 <=; 4,8 => 0 <=;
t. 1,7 => t => 4 <=; 5,e => 2 => 8 <=; 3,9 => 6 => 0 <=;
e 1 <==> e; 2 <==> t; 3 <==> 9; 4 <==> 8; 5 <==> 7; 6 <=;
0 <=
Notes: 1) In each system 1 x n = n and 0 x n = 0.
2) In each system the two digits involved in each
structure for the highest digit (base - 1) add up
to the base. Another way to look at that is to
realize that the digits in the product of n x (base
- 1) add up to base - 1. Thus, in decimal 8 x 9 =
72 and 7 + 2 = 9 (10 - 1). In undecimal, 6 x t = 55
and 5 + 5 = t. Finally, in duodecimal, 9 x e = 83
and 8 + 3 = e.
3) The structure for 5 (which = base/2) in decimal
has the same form as for 6 (which also = base/2) in
duodecimal. Undecimal, being odd, has no equivalent
for 5 and six.
DUODECIMAL MULTIPLICATION EXAMPLE
To begin to put some of these findings together, let's tackle
a duodecimal multiplication example. In the process we will
discover the usefulness of the multiplicative structures for
at least some of the more difficult or complicated mult-
plication problems and, by extension, division problems as
well. You remember that division problems have one or more
partial multiplications in them.
Here's the problem: It's by MORDASHKA and appeared as C-11 in
the November-December, 1994, issue of The Cryptogram.
YOUR
TAB
----
IYATR
UOYLN
PYPRR
-------
YCRORTR
The problem contains three partial products and one addition
with three addends shifted as per usual for multiplication.
It could be helpful if we can locate zero. We could use a
process of elimination. Neither Y, T, A, B, I, U, P, nor R
can be zero. That leaves four letters as possibilities: C,
O, N, and L. Fortunately for us, in the addition section we
find T + N = T. Hence N = zero.
Also from the addition section at the left end, Y > P. U + Y
must be greater than e, giving a carry of 1, so Y = P + 1.
That should be useful later.
Again from the addition section, A + L + R = R mod 12. There
is no carry from the previous column: T + N as N = 0. We can
subtract R from both sides of the first equation giving us A
+ L = 0 mod 12. But A and L are both greater than 0 so A + L
= 10 or decimal 12. That means if we can determine the value
of A, we can compute the value of L from the equation, and
vice versa.
>From the partial products, all of which are 5 digits long
whereas the multiplicand is 4 digits long, Y > I, U, P > 0.
Therefore Y must more than 3. Now let's look at the partial
products to see whether we can uncover a recognizable
multiplicative structure, remembering that we are dealing
with a duodecimal or base 12 problem. We get these equations
from the product of the last digit of the multiplicand by
each digit of the multiplier:
R x B = R R x A = N or zero R x T = R all mod 12
The multiplicative structure becomes: B,T => R and A => zero.
There are only two places that yield the appropriate
relations, when R = 4 or R = 8. Since none of R, B, and T
equal 1 and R does not equal zero, here are the results:
R = 4 then B and T are 7 and t or t and 7. A = 3, 6, or 9.
R = 8 then B and T are two of 7, t, and 4. A = 3, 6, or 9.
That's not very many possibilities, simplifying our search.
The first partial product ends with TR. The third ends with
RR. If we identify T and B, we should be able to calculate U
in the multiplicand and check it in both partial
multiplications.
So here's our table:
B T R U
7 t 4
t 7 4
7 t 8
t 7 8
4 t 8
t 4 8
Those are the only possible values of B, T, and R, all the
permutations. In each instance we have to calculate U to
discover what value of U is consistent in both
multiplications.
Now let's check these possibilities. B x YOUR = IYATR.
1) B = 7, T = t, R = 4, TR = t4. B x R = 7 x 4 = 28 base 10
or 24 base 12. Carry the 2. B x U + 2 => T or t. 7 x U + 2 =>
4, U = 8. Check: 7 x 8 + 2 = 58 base 10 or 4t base 12 or t
mod 12. Our trial value for U is 8. Let's check that with
the third partial product T x YOUR = PYPRR. T = t etc. as
before. RR = 44. t x 4 = 40 base 10 or 34 base 12. Carry the
3. t x 8 + 3 = 83 base 10 or 6e base 12 or e mod 12, but we
needed a 4 for 44. It doesn't work.
2) We have to continue the process until we get a combination
that is consistent. Try the second one. You may find that no
value of U can be found from the first partial. Similar
problems beset the next three combinations on the table.
3) Let's check the 5th combination. B = 4, T = t, R = 8, TR =
t8. B x R = 4 x 8 = 32 base 10 or 28 base 12 or 8 mod 12.
Carry the 2. B x U + 2 = t mod 12. 4 x U + 2 = t. U could be
2 or 5. Try them with the second product. RR = 88. T or t x
8 = 80 base 10 or 68 base 12 or 8 mod 12. Carry 6. For U = 2,
t x 2 + 6 = 26 base 10 or 22 base 12 or 2 mod 12. But we need
an 8 for 88. That's a conflict. Let's try U = 5. t x 5 + 6 =
56 base 10 or 48 base 12 or 8 mod 12. Eureka! U = 5 checks
out. We now also know that B = 4, T = t and R = 8. You can
check the last combination also to make sure it produces no
alternate value of U that stays consistent.
The letter-number equivalent table is
0 e t 9 8 7 6 5 4 3 2 1
N T R U B
We can now determine the value of A using fact 4) A + L = 12
with the middle partial product. A x YOUR = UOYLN or A x ..58
= (12 - A)0. A can have the value of 3, 6 or 9. If A is 6,
then L = 6 (A + L = 12, remember?). That's not possible. If A
= 3 L = 9. If A = 9, L = 3. Try 3. 3 x 8 = 24 base 10 or 20
base 12. Carry the 2. 3 x 5 + 2 = 17 base 10 or 15 base 12
or 5 mod 12. But we needed a 9. Try A = 9. It better work or
we've done something wrong. 9 x 8 = 72 base 10 or 60 base 12.
Carry the 6. 9 x 5 + 6 = 51 base 10 or 43 base 12 or 3 mod
12. So L = 3. That's the value of L we were looking for.
Success! We can add that to the equivalent table. With A = 9
and T = t, T x YOUR = A x YOUR + YOUR.
Since we know that T x YOUR = PYPRR and A x YOUR = UOYLN, we
can put the addition into normal form:
UOYLN
+YOUR
-----
PYPRR
>From this addition we deduce that P = U + 1 or 5 + 1 = 6.
Looking at the multiplier, TAB = t93. The first product must
be the smallest, followed by the second, with the third the
largest. Their leftmost digits must be in the same order.
Hence, I < U < P < Y.or Y > P > U > I. The only letters about
which we have no information yet are C and O.
At this point our equivalent table reads:
0 e t 9 8 7 6 5 4 3 2 1
N T A R U B L
Replacing letters of known value in the above addition by
their respective digits yields
5OY30
+YO58
-----
PYP88
We note that Y + O = P and O + Y = Y. 3 + 5 = 8, no carry. Y
+ O must yield a carry of 1 which makes Y = P + 1. Since I <
U, the only place in the table for two numbers that are
adjacent in value is 7 and 6; thus Y = 7 and P = 6. Y + O = P
mod 12. That means 7 + O = 16 base 12 or 18 base 10. Thus, O
= e. The addition of all three partial products will give us
the remaining values for I and C without resorting to
anagraming. (Just a nicety here.)
I79t8
5e730
+67688
--------
7C8et8
9 + 3 + 8 = 18 base 12. Carry 1. 7 + 7 + 8 + 1 = 1e base 12.
Carry 1. I + e + 6 + 1 = 8 or 18 base 12. Solving for I gives
I = 2. Carry 1. 5 + 7 + 1 = 11 base 12. Thus C = 1. The
keyphrase for the equivalent table becomes NOTARYPUBLIC.
Although this problem was given the number C-11, for someone
familiar with duodecimal arithmetic it is of medium
difficulty. There are problems in the Cryptarithm section
that provide far fewer clues and necessitate trying out many
more possibilities. In the next lecture we will take a look
at organizing that process so as not to get lost in the
bookkeeping aspect of finding a solution. We may also find a
few more relationships that can be helpful at times.
REFERENCES
[CROT] Winter, Jack (CROTALUS), "Solving Cryptarithms,"
American Cryptogram Association, 1984.
[FIDD] FIDDLE, Lynch, Frederick D., "An Approach to
Cryptarithms," ACA Publications, 1974.
[FIRE] FIRE-O, "A Tool for Mathematicians: Multiplicative
Structures," The Cryptogram, Volume XXXVI, No 3, 1970.
[LED1] LEDGE, "Basic Patterns in Base Eleven and Twelve
Arithmetic (Part 1)," The Cryptogram, Volume XLIII, No
5, 1977.
[LED2] LEDGE, "Basic Patterns in Base Eleven and Twelve
Arithmetic (Part 2)," The Cryptogram, Volume XLIII, No
6, 1977.
APPENDIX
Table 14-1
Undecimal Multiplication Table
1 2 3 4 5 6 7 8 9 t
1 | 1 2 3 4 5 6 7 8 9 t
2 | 2 4 6 8 t 11 13 15 17 19
3 | 3 6 9 11 14 17 1t 22 25 28
4 | 4 8 11 15 19 22 26 2t 33 37
5 | 5 t 14 19 23 28 32 37 41 46
6 | 6 11 17 22 28 33 39 44 4t 55
7 | 7 13 1x 26 32 39 45 51 58 64
8 | 8 15 22 2x 37 44 51 59 66 73
9 | 9 18 25 33 41 4t 58 66 74 82
t | t 19 28 37 46 55 64 73 82 91
Table 14-2
Duodecimal Multiplication Table
1 2 3 4 5 6 7 8 9 t e
1 | 1 2 3 4 5 6 7 8 9 t e
2 | 2 4 6 8 t 10 12 14 16 18 1t
3 | 3 6 9 10 13 16 19 20 23 26 29
4 | 4 8 10 14 18 20 24 28 30 34 38
5 | 5 t 13 18 21 26 2e 34 39 42 47
6 | 6 10 16 20 26 30 36 40 42 50 56
7 | 7 12 19 24 2e 36 41 48 53 5x 65
8 | 8 14 20 38 34 40 48 54 60 68 74
9 | 0 16 23 30 39 46 53 60 69 76 83
t | t 18 26 34 42 50 5x 68 76 84 92
e | e 1t 29 38 47 56 65 74 83 92 t1
Table 14-3
1 2 3 4 5 6 7 8 9 t
1 | 2 3 4 5 6 7 8 9 t 10
2 | 3 4 5 6 7 8 9 t 10 11
3 | 4 5 6 7 8 9 t 10 11 12
4 | 5 6 7 8 9 t 10 11 12 13
5 | 6 7 8 9 t 10 11 12 13 14
6 | 7 8 9 t 10 11 12 13 14 15
7 | 8 9 t 10 11 12 13 14 15 16
8 | 9 t 10 11 12 13 14 15 16 17
9 | t 10 11 12 13 14 15 16 17 18
t |10 11 12 13 14 15 16 17 18 19
Table 14-4
1 2 3 4 5 6 7 8 9 t e
1 | 2 3 4 5 6 7 8 9 t e 10
2 | 3 4 5 6 7 8 9 t e 10 11
3 | 4 5 6 7 8 9 t e 10 11 12
4 | 5 6 7 8 9 t e 10 11 12 13
5 | 6 7 8 9 t e 10 11 12 13 14
6 | 7 8 9 t e 10 11 12 13 14 15
7 | 8 9 t e 10 11 12 13 14 15 16
8 | 9 t e 10 11 12 13 14 15 16 17
9 | t e 10 11 12 13 14 15 16 17 18
t | e 10 11 12 13 14 15 16 17 18 19
e |10 11 12 13 14 15 16 17 18 19 1t
LECTURE 13 SOLUTIONS
13-1 Beaufort
ABRVJ UTAMP YPLHZ OZYAP YPJNP KNXUG
QRDPC ELPNC BVCEF NLLSJ LGOWC VYCGA
EVGIX XNDKY U. (butter) (INWVQH)
Key = AGRICULTURE, A fantastic glut ...
13-2 Vigenere.
DWNIT KGEWZ ENJQZ WXLLZ WZOKC ETOWI NXVQS
DQGAK MGGBH NAMWE OWVAM UJDVQ IMDSB VCCTR
YUIQX. (making, UHVW)
Key = LIBERTY, Some criminals in ....
13-3 Vigenere Running Key
YPOSC DWVWY CCHZT AKALF I. (tolls -2)
Key = Never send for whom the be (continues bell tolls )
13-4 Vigenere Progressive key. "Fungi"
IPGPUPX GTIAKNP AMEHLAW SJSTROZ TCGYUND STNPJZM
OESWAXG VLHSPZC GNEIWHP EKHNOWW PMEQFVV PDQAWCA
GGFRKSO RCHZVKL NBWHYBV CUNBBBB AVGCJFA FLTMKUV K.
Key = PICTURE (3), The way to identify....
LECTURE 14 PROBLEMS
Some time ago, CROTALUS cooked up some goodies:
14-1. Multiplication (Two words, 0-1) original by EDNASANDE
WOMEN X MEN = UTNNLM + NWTWNN = NLSMTUWM
14-2. Division (Two words, 0 -9) MORDASHKA
ATOM / ASK = N; - GNC = IS
14-3. Multiplication. (No word, 0-1) FOMALHAUT
ASAP X MAB = RITMT + TMPRY + PDBYD =PAYDIRT
14-4. Unidecimal multiplication. (Two words 0-X) WALRUS
TOUGH X DIG = IDIGDN + NYYDNG + UIHDOU = DDCUUILN
```
Back to index |
Linear Interpolation
HOME Mathematics Computer Science Economics
Linear Interpolation
Prerequisites
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Introduction
• A gradient describes how much y changes relative to a change in x.
• For example, we can define a line as starting at (0, 0) and ending at (10, 25):
• Instead of giving explicit start and end points for the line we can describe it by how much y changes relative to x.
• In the line above, a change of 10 in x produces a change of 25 in y.
• The gradient is defined as the change in y divided by the change in x.
• If we use Δ to denote 'change' then the formula for the gradient (g) of a line is:
LaTeX: g = \frac{\Delta y}{\Delta x}
• In the line above, the gradient is:
Finding the change in x and y
• In our above example, our first coordinate was (9, 0). This made it easy to calculate how much x and y changed.
• However when both coordinates are non-zero, it gets a bit more complicated.
• For example, calculate the gradient of the line between (1, 2) and (7, 10)
1. To find how much x and y change, we need to subtract the first coordinate from the second
• We can use this in our equation to find the gradient of a line:
Special Cases
• There are two special cases to look out for when calculating the gradient of a line:
1. When the line is horizontal
• There will be no change in y
2. When the line is vertical
• There will be no change in x
Code
import math
deltaY = p2[1] = p1[1]
deltaX = p2[0] - p1[0]
if deltaX == 0:
return math.inf
else:
return deltaY / deltaX
Interpolation
Show/Hide
Introduction
• If we are given a set of data points: X 1 2 3 4 5 6 7 y 28 21 96 96 120 87 50
• We know the value of y when x is a whole number, i.e. 1, 2, 3
• But when x is 1.5 we do not have a corresponding value for y.
• Interpolation is the method of estimating the value of y for a given x when we don't have a data point for it.
Linear Interpolation
Introduction
• Linear interpolation is a very simple type of interpolation.
• To find the value of a point which is not in the given set of points:
1. Find the two points that surround the point you are estimating: one will be smaller, one larger.
2. Draw a straight line between the two known points.
3. Find the value of y for the point that lies on the line at the given x.
Visualization
Example
• If we are given a set of data points: X 1 2 3 4 5 6 7 y 28 21 96 96 120 87 50
Find the value for y when x is 2.4
1. The points the surround 2.4 are (2, 21) and (3, 96)
2. We calculate the gradient of this line:
3. And then find the value of y when x is 2.4:
4. When x is 2.4, y is 51
Code (using SciPy)
from scipy import interpolate
dataPoints = [
[1, 28],
[2, 21],
[3, 96],
[4, 96],
[5, 120],
[6, 87],
[7, 50]
]
interpolator = interpolate.interp1d(
[dataPoint[0] for dataPoint in dataPoints],
[dataPoint[1] for dataPoint in dataPoints],
'linear')
print(interpolator(2.4)) # prints 50.99999999999999
print(interpolator(2.6)) # prints 66.0
Code (using NumPy)
import numpy as np
class LinearInterpolator:
# an array of x values, in order
_sortedDomain = []
# a dictionary of x, y values
_dataPointsDict = {}
def __init__(self, dataPoints):
for dataPoint in dataPoints:
self._dataPointsDict[dataPoint[0]] = dataPoint[1]
self._sortedDomain = [dataPoint[0] for dataPoint in dataPoints]
self._sortedDomain = np.sort(self._sortedDomain)
def interpolate(self, x):
# find the index of the smallest element larger than x
closest = np.searchsorted(self._sortedDomain, x)
# find the two x values (before and after x
x1 = self._sortedDomain[closest - 1]
x2 = self._sortedDomain[closest]
# and their corresponding y values
y1 = self._dataPointsDict[x1]
y2 = self._dataPointsDict[x2]
# find the deltaX and deltaY
deltaX = x2 - x1
deltaY = y2 - y1
# and finally interpolate
xPart = x - x1
return y1 + yPart
dataPoints = [
[1, 28],
[2, 21],
[3, 96],
[4, 96],
[5, 120],
[6, 87],
[7, 50]
]
interpolator = LinearInterpolator(dataPoints)
print(interpolator.interpolate(2.4)) # 50.99999999999999
print(interpolator.interpolate(2.6)) # 66.0 |
# HCF Shortcut Method
If you are preparing for any competition exam, you need to understand that there is a time limit for each exam within which you have to solve respectable amount of question.
In order to stand a chance in cut throat competition, it is important for you to get acquainted with the shortcut method of solving questions.
In this post we will understand how you can easily solve HCF question in limited time using short cut method. I will try to explain this method using example
### (01) Find HCF of 20, 28 and 36 using shortcut method
Step 1:
Find the difference between all the numbers given
28-20 = 8
36-28 = 8
36-20 =16
After subtraction, we got three numbers 8, 8 and 16
Step 2:
Select the number which is smallest
So among the number 8, 8 and 16, the smallest number is 8
Step 3:
Now according to the shortcut rule, the factor of smallest number is the HCF.
Find all the factors of the smallest number
Here, factor of number 8 is the HCF.
Factor of 8 can be displayed as
1 * 8
2 * 4
Four numbers are factor of 8 which are 8, 4, 2, 1
Among above four numbers one number is the right answer
Step 4
Finding the right answer among all the factors
Is 8 the HCF of (20, 28 and 36)
No, as number 8 do not divide 20 fully
b. Is number 4 can be the HCF (20, 28 and 36)
YES!! as number 4 fully divide 20, 28 and 36
we have got our answer, so we do not need to check the other numbers aka 2 and 1
Hence HCF (20, 28, 36) is 4 |
# TRISQ - Editorial
Author: Devendra Agarwal
Tester: Anudeep Nekkanti
Editorialist: Amit Pandey
Cakewalk.
### PREREQUISITES:
Basic geometry, recursion.
### PROBLEM:
Find the maximum number of squares of size 2\times2 that can be fitted in a right angled isosceles triangle of base B. All sides of the square must be parallel to both base and height of the isosceles triangle.
### QUICK EXPLANATION:
As B<=1000, we can pre-compute the output for all the test cases, and report the answer in O(1) time for each of the query.
### EXPLANATION:
First consider the right angle triangle \Delta XYZ, where YZ is the base of triangle. Suppose length of the base is B. If we consider the position of first square with the vertex Y, we will have (B/2 - 1) squares in the base, and we will be left with another isosceles right angle triangle having base length (B-2).
Let f(B) = Number of squares which can be fitted in triangle having base length B.
f(B) = (\frac{B}{2} -1) + f(B-2)
The given time limit is sufficient even if we calculate f(B) using the given recursion, and use memoization. Later we can answer each query in O(1) time. We can do it for even and odd numbers separately with the base case f(1) = f(2) = f(3) = 0.
The given recursion can be solved in following manner.
f(B) = \frac{B-2}{2} + F(B-2) \\\ = \frac{B-2}{2} + \frac{B-4}{2} + F(B-4) \\\ = \frac{B-2}{2}+ \frac{B-4}{2} + \frac{B-6}{2} + F(B-6)
With conditions, f(1) = f(2) = 0
f(B) = \frac{B}{2} + (\frac{B}{2} - 1) + (\frac{B}{2} -2) + \cdots + 1.
f(B) = Sum of first \frac{B}{2} natural numbers.
Lets call M = \frac{B}{2}
f(B) = \frac {M \times (M-1)}{2}
You can notice that answer for 2N and 2N+1 will be the same.
### Solution:
Setter’s solution can be found here
Tester’s solution can be found here
6 Likes
It can be solved in a more simple way. Consider you need to fill the isosceles triangle with 1x1 squares. Now, consider a square of side B. The number of 1x1 squares you can put in is clearly BxB, and also the number of 1x1 squares you can put in an isosceles triangle will be (BxB - B)/2 ie we remove the squares on the diagonal and half the remaining squares . Now you need to scale the 1x1 to 2x2, so you simply divide B by 2, and use (BxB - B)/2. My Solution I know the proof is more intuitive rather than rigorous, would be happy to know if there is some fallacy, or some rigourous way of proving it.
5 Likes
My approach was wrong but it got AC anyway. Observing the test cases the answer for consecutive size of the base is - 0 0 0 1 1 3 3 6 6 10 10… which is nothing but the series of sum of n integers with all the sums repeating twice, except for first three 0s. After pre-computing this for 10^4 each query can be answered in O(1). Perhaps the problem could have been presented with less and randomized test cases to prevent the use of such approach.
1 Like
My solution
long long int t1,t2,b;
t2=b/2;
t1=t2*t2;
t1=t1-t2;
t1=t1/2;
cout<<t1;
3 Likes
“we will have (B/2−1) squares in the base, and we will be left with another isosceles right angle triangle having base length (B−2).”
Where does this thing come from… please help me …
It can also be solved as through the following code :–>>
int N=(B/2)-1;
print((N*(N+1))/2);
1 Like
My approach:
A=(B[i]/2)*(B[i]/2);
A-=(B[i]/2);
A-=((((B[i]/2)*(B[i]/2))/2)-(B[i]/4));
cout << A << endl;
Maybe it can be simplified further.
how u have generated a general formula for this problem??? plz help…
Firstly, divide ‘b’ or ‘n’ by 2 so we can deal with 1x1 squares.
At the base, n-1 squares can be fit and it decreases by 1 as we go upwards.
Hence the total number of squares is the sum=(n-1)+(n-2)+…+2+1.
Summation of +ve: n is added n-1 times. Therefore, sum of n’s is n(n-1).
Summation of -ve: Summation of series of natural numbers is ‘(N)(N+1)/2’. Here N=n-1 so summation is n(n-1)/2.
Subtracting -ve from +ve, we get: n(n-1)/2 or b(b-1)/2
Nothing Understood , Very Bad Explanation
Isn’t sum of the first M natural numbers M*(M**+**1)/2 ?
For people willing to know where this statement came from:
We will have (B/2−1) squares in the base, and we will be left with another isosceles right angle triangle having base length (B−2).
You can deduce this simply by using simple trigonometry. I was also curious to find the reason behind this and then got to this approach. Please correct me if I’m wrong or if there’s a better approach. If you can, try visualizing this on a notebook.
Consider in triangle XYZ, YZ is the base with Z as the right-angled vertex. Thus, YX is the hypotenuse. Let the length of base YZ = B. SInce, it’s an isosceles triangle length of YX is also B. Thus, the angle XYZ is 45 degrees (tan(XYZ) = B/B = 1). If we start filling squares of side 2 units from vertex Z, we can only fill them until the distance between base and the hypotenuse becomes less than 2. Since, the corresponding angle is 45 degrees, the distance from Y to this point will also be 2. Hence, we will not be able to place 1 square of side 2 here. The total number of squares that could be filled thus B/2 - 1. B/2 since there are of side 2 units. Then we are left with another isosceles triangle above these filled squares with base B-2 that is simply the sum of the length of upper sides of all squares (2*(B/2 - 1) = B - 2).
Hope I was clear with the explanation.
Why did I get a WA?
Why did I get WA??
#include<bits/stdc++.h>
#include
#include<math.h>
int T, B, n, x;
using namespace std ;
int main(){
scanf("%d",&T);
while(T–)
{
scanf("%d",&B);
if (B>3){
if ((B%2)!=0)
B--;
x= (B*B)/4;
n= ( (x) - sqrt(x) )/2;
printf("%d",n);
}
else{
n=0;
printf("%d",n);
}
}
return 0;
}
Another alternative way of doing it is arithmetic series:
where n = int(b/2-1) and ![alt text][2] and A1 = 1 and An = n
Example: B = 20, n =(20/2-1) = 9, 9(1+9)/2 = 45
Recursive function for this problem is f(x) = 1 + 2*f(x-2) - f(x-4)
F(B)=B/2+(B/2-1)+(B/2-2)+…+1
can anyone plz tell me from where the 1st B/2 term come?
@subarna_08.
Correct summation is : ((B/2) - 1) + ((B/2) - 2) + … + 1 = Sum of first ((B/2) - 1) positive integers.
Hence the correct answer is: ( ((B/2) - 1)(B/2) )/2.
(Sum of first n positive integers is n
(n+1)/2)
Can anyone tell me why the code with precompute is throwing the error NZEC, while without precompute it is working correctly in Java?
Code:
With Precompute: https://www.codechef.com/viewsolution/21643739
WithOUT precompute: https://www.codechef.com/viewsolution/21644055
Java Recursive solution:
import java.util.*;
class Solution
{
public static int derive(int n)
{
if(n==1 ||n==2 ||n==3)
return 0;
return (n/2-1)+derive(n-2);
}
public static void main(String []args)
{
Scanner sc=new Scanner (System.in);
int item=sc.nextInt();
while(item!=0)
{
item--;
int n=sc.nextInt();
int x=derive(n);
System.out.println(x);
}
}
}
// |
NSW Syllabuses
Mathematics K–10 - Stage 4 - Number and Algebra Equations
Outcomes
A student:
• MA4-1WM
communicates and connects mathematical ideas using appropriate terminology, diagrams and symbols
• MA4-2WM
applies appropriate mathematical techniques to solve problems
• MA4-3WM
recognises and explains mathematical relationships using reasoning
• MA4-10NA
uses algebraic techniques to solve simple linear and quadratic equations
Related Life Skills outcome: MALS-19NA
Content
• distinguish between algebraic expressions where pronumerals are used as variables, and equations where pronumerals are used as unknowns
• solve simple linear equations using concrete materials, such as the balance model or cups and counters, stressing the notion of performing the same operation on both sides of an equation
• solve linear equations that may have non-integer solutions, using algebraic techniques that involve up to two steps in the solution process, eg
\begin{align} & x - 7 = 15 \\ & 2x - 7 = 15 \\ & 7-2x = 15 \\ & \frac{x}{7} = 5 \\ & \frac{2x}{7} = 5 \end{align}
• compare and contrast strategies to solve a variety of linear equations (Communicating, Reasoning)
• generate equations with a given solution, eg find equations that have the solution x = 5 (Problem Solving)
• Solve linear equations using algebraic techniques and verify solutions by substitution (ACMNA194)
• solve linear equations that may have non-integer solutions, using algebraic techniques that involve up to three steps in the solution process, eg
$$\begin{array} {llll} 3x+4 = 13 \qquad& 3x+4 = x-8 \qquad& \dfrac{x}{3}+5 = 10 \qquad& \dfrac{2x}{3}+5 = 10 \qquad \\ 3\left(x+4\right) = 13 \qquad& 3x+4 = 8-x \qquad& \dfrac{x+5}{3} = 10 \qquad& \dfrac{2x+5}{3} = 10 \qquad \end{array}$$
• check solutions to equations by substituting
• determine that if $$c > 0$$ then there are two values of $$x$$ that solve a simple quadratic equation of the form $$x^2 = c$$
• explain why quadratic equations could be expected to have two solutions (Communicating, Reasoning)
• recognise that $$x^2 = c$$ does not have a solution if $$c$$ is a negative number (Communicating, Reasoning)
• solve simple quadratic equations of the form $$x^2 = c$$, leaving answers in 'exact form' and as decimal approximations
Background Information
The solution of simple equations can be introduced using a variety of models. Such models include using a two-pan balance with objects such as centicubes and a wrapped 'unknown', or using some objects hidden in a container as an 'unknown' to produce a number sentence.
The solution of simple quadratic equations in Stage 4 enables students to determine side lengths in right-angled triangles through the application of Pythagoras' theorem.
Purpose/Relevance of Substrand
An equation is a statement that two quantities or expressions are equal, usually through the use of numbers and/or symbols. Equations are used throughout mathematics and in our daily lives in obtaining solutions to problems of all levels of complexity. People are solving equations (usually mentally) when, for example, they are working out the right quantity of something to buy, or the right amount of an ingredient to use when adapting a recipe.
Language
Describing the steps in the solution of equations provides students with the opportunity to practise using mathematical imperatives in context, eg 'add 5 to both sides', 'increase both sides by 5', 'subtract 3 from both sides', 'take 3 from both sides', 'decrease both sides by 3', 'reduce both sides by 3', 'multiply both sides by 2', 'divide both sides by 2'. |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 1.3: Abstraction. What is a group?
So we now have many examples of symmetry, but what, exactly is a group?
First, let's remember where numbers come from. You can imagine having a set of three pineapples, or a set of three people, or a set of three bottles. All of these share the property of 'three-ness.' So when we say the number three, it describes all of these different sets. The number three is an abstraction of the property of a set containing three things.
What's a group? A group is an abstraction of symmetry! Just like a number describes all sets with a certain number of things in it, a given group describes all objects with a certain kind of symmetry. Here's an example: The 'very symmetrical face' has two symmetries, related by a reflection. Likewise, the symmetric polynomial $$f(x,y)=x+y$$ has a two symmetries, one from leaving $$f$$ alone, and one from exchanging the two variables $$x$$ and $$y$$. If we think of switching $$x$$ and $$y$$ as a reflection, we see that the face and the polynomial somehow have the same kind of symmetry! The group is that measure of symmetry.
Now, in that last example, the two objects had the same number of symmetries. It turns out that just counting symmetries ins't enough to tell whether the two objects have the same group of symmetries. Think of the symmetries of our equilateral triangle: there were rotational symmetries, and there was a reflection symmetry. And there were six symmetries in all. Now, consider a regular hexagon with some regularly placed bumps on each side, so that there is no reflection symmetry available. Thus, this object has six rotational symmetries, but can't be flipped over like the triangle. Therefore it has the same number of symmetries as the equilateral triangle, but the set of symmetries is somehow different.
### Contributors
• Tom Denton (Fields Institute/York University in Toronto) |
### Anatomy of a fraction
Many people know what fractions look like, and even what the parts are called, but they don't necessarily know what all of the parts mean, so let's start there . . .
In the fraction below, the a is the numerator. To enumerate is to list or to count, so the numerator enumerates or counts. Specifically, it counts how many of the denominator there are.
The denominator, b, is the size of the thing(s) that is(are) being counted. Those things are the equal-sized parts into which the whole is divided. The denomination is the number of equal parts that would be required to make up the whole.
In the fraction ¼, four parts make up the whole, and we have one of them.
A familiar example of denomination might be the denomination of money. The denomination of a coin or bill is its value in dollars (U.S. dollar-based system). A penny has a denomination of 1/100* because it's value is 1/100 of a dollar. A nickel has a denomination of 1/20 of a dollar, a dime 1/10 and a quarter 1/4 of a dollar (thus the name "quarter"). If we have four quarters, we have 4/4 of a dollar or one whole dollar.
Two halves of a pie make a whole pie, if we have 87 potatoes, then each is 1/87th of the total number ... and so on.
* Note: From time to time I use the slanted fraction bar ( / ) in typed text, but that's just a matter of convenience when typing. I strongly advise you to use a horizontal bar ( — ) in writing fractions. It will help you immensely later.
#### Anatomy of a fraction
A fraction counts the number of equal divisions of something (anything) whole.
The numerator enumerates or counts
The denominator is the number of equal parts that make up the whole
### 1. ALL numbers are fractions
All numbers, even integers like 1, 2, 3, . . . are fractions. For example, 2 is a fraction with a denominator of 1.
Very often it's useful to rewrite an integer or a real number as a fraction with a denominator of 1, like 2/1 or 3.14/1. You'll see that in use in what follows.
You might argue that an irrational number, like π = 3.14159 ... is not a fraction because,
by the definition of irrational numbers, it cannot be represented as a fraction, but we can (and often do) still write a fraction like π/2 or π/1 for convenience.
### 2. Reciprocals
The reciprocal of a fraction is obtained by turning it upside down – by swapping the numerator and denominator. Here are some examples →
When the denominator is 1, we usually omit it, but it can come in handy when multiplying or adding fractions, at least until you get good at it (you will if you practice).
Reciprocals will be important when we divide by fractions.
### 3. Fractions are division operations
Any fraction is an expression of a division operation in which the denominator is divided into the numerator. For example:
• The fraction 8/2 means "two divided into eight," or 8/2 = 4. Remember the terminology: 8 is the dividend, 2 is the divisor and the answer (4) is the quotient.
• The fraction 1/2 means 1 divided by 2. If you think about that, 1 divided into two parts gives two equal parts that are 1/2 of the whole.
• The fraction 1/3 means 1 divided by 3. A whole divided into three parts yields parts that are 1/3 the size of the whole ... and so on.
Here is a recap of the ways that we know to write division operations.
Fractions in which the numerator is greater than the denominator are often called "improper" fractions. Here are some examples:
I think that the name "improper" is unfortunate. They're perfectly legitimate fractions, and almost always more useful for doing calculations than mixed fractions.
One important thing to keep in mind is that the numerator of a "proper" fraction is less than the denominator, so a proper fraction is a number less than one. Conversely, an improper fraction is always a number greater than 1.
#### Improper fractions are proper
It's almost always easier and better to work with "improper" fractions than to convert them to mixed fractions. You should at least hold off on converting to mixed fractions until the end of a calculation.
### Converting between improper and mixed fractions
#### Pure fraction → mixed
To convert an improper fraction to a mixed fraction, just divide the denominator into the numerator and express the remainder as a proper fraction. For example,
Remember that what's left over is still a fourth. Fourths are what's being counted by the numerator, and that didn't change.
#### Mixed → pure fraction
There are two ways to turn a mixed fraction into a pure fraction. The best is just to think of the whole part in terms of the denomination of the fraction part. For example, in the mixed number 3 ¼, we first think of it as 3 + ¼, then we simply express 3 as 12/4 and add on the remaining ¼ to get 13/4:
A shortcut for converting a mixed fraction is just to multiply the whole number by the denominator of the fraction, add the numerator of the fraction, then write that over the original denominator. It looks like this:
### Practice problems
#### Mixed fractions ↔ pure fractions
Convert these mixed numbers to fractions:
#### "improper" fractions → mixed numbers
Convert these improper fractions to mixed numbers (an integer and a fraction):
### 5. Some fractions are reducible to simpler ones
If both the numerator and denominator of a fraction can be divided evenly by the same number, then the fraction is reducible, and in most cases should be reduced in order to make later calculations easier.
Take for example our coin values. A nickel is 5/100 of a dollar, but because both 5 and 100 are divisible by 5, we can do so to obtain the equivalent fraction, 1/20. A nickel is 1/20 of a dollar.
Here are some examples of reducible an irreducible (not reducible) fractions. Notice that 11/19 is irreducible because neither 11 nor 19 have a factor in common other than 1.
### Practice problems – Reducing fractions
All of these fractions reduce to 3/5. See if you can figure out how:
*Hint: Just repeatedly divide the numerator by 3 and the denominator by 5, the same number of times.
### 6. Multiplying fractions
Multiplying two fractions couldn't be simpler, but you should try to figure out what it really means.
One key to understanding fraction multiplication is the word "of." In mathematics, "of" means multiplication. So if we say ½ of 2, that's ½ multiplied by 2, which is 1.
This picture might help. On the left is 1 circle – a whole circle with an area of 1 unit. If we want ½ of that circle, we multiply 1 by ½ to get ½ (magenta area). If we want ½ of the half-circle, we multiply ½ by ½ to get ¼.
#### The word "of" in math
In mathematics, the word "of" almost always means "multiplied by."
To multiply fractions, we multiply across: numerator multiplies numerator, and denominator multiplies denominator. It's that easy. Heres an example:
Here are a few more examples:
and
Now we could easily have asked that last question like "what is 2/5 of 3/5 ?", or "what is 3/5 of 2/5?" The answer and process would have been the same because "of" means "multiply." That's really handy to know, then we have ½ of ½ = ¼, or ½ · ½ = ¼.
#### Multiplying fractions
To multiply fractions, multiply numerators together and multiply denominators together.
### Practice problems
Perform these fraction multiplications:
Find the fraction of the fraction:
### 7. Dividing fractions
The division operation is really no different from multiplication. It's just multiplication by the reciprocal of the divisor (which is the denominator). If you can remember that, your life studying math will be much easier.
Here's what I mean. Let's divide 8 by 2:
I know that seems more complicated than just dividing 8 by 2 in your head, but here's why it's valuable. What is ½ divided by ¼ ?
Just in case you could have figured that one out in your head (because ¼ is half of ½), here's a tougher one that's still very easy to do if you remember that division by a denominator is the same thing as multiplication by the reciprocal of the denominator:
Thinking of division as multiplication by the reciprocal of the divisor (or denominator) will help you to solve a great many algebra problems. It will really grease the wheels for further learning in math, and I strongly encourage you to master it. There are some practice problems below.
Here is an example with a picture. 1 divided into four parts is 1·¼ = ¼. Now ¼ divided into 4 parts gives parts that are ¼·¼ of the whole, and 1/16 divided into 4 parts is 1/16 · ¼ = 1/64.
See if you can follow the divisions through the figure below. It might help you to understand what's going on when we divide by a fraction.
#### Division by a fraction
Division is just multiplication by the reciprocal of the divisor.
### Practice problems
Perform these division operations:
*Note: I'm using the division sign ( ÷ ) here because I want you to practice setting up divisions as fractions (a ÷ b = a/b). It really isn't used much in real mathematics.
#### But 5 apples + 3 oranges is . . . a bunch of fruit.
Adding fractions is probably one of the scariest things that many math students can think of doing, and that's a shame, because it's not really so bad. With just a little practice, you'll be an expert and it won't trouble you any more. Be sure to put in that practice. Get good at adding fractions now and you'll be good at it forever.
#### Example 1
Here's an example. Let's add ⅙ and ⅞. Notice that these fractions have different denominators, so we'll need to convert them both to the same one. The easiest way is just to multiply the two denominators: 6 · 8 = 48. That way we know that both 6 and 8 have to be factors of 48.
We'll convert both fractions to 48ths. Now we don't want to change our sum, so to convert, we'll multiply ⅙ by 8/8 (which is just one because a number divided by itself is 1) and ⅞ by 6/6 for the same reason. It looks like this:
Multiplication gives us a sum of 48ths:
Now that our fractions have a common denominator (which means that the numerators are counting the same thing, like apples alone instead of apples and oranges), we just add the numerators and simplify the fraction if we can.
While the method above always works, there are other cases where finding a common denominator can be done a little easier, and you should probably look for them. One is where the denominator of one fraction is a multiple of the other. Take this addition for example.
#### Example 2
5 is a factor of (divides evenly into) the denominator 15, so if we multiply the ⅕ by 3/3, we get:
When both fractions have the same denominator, it's legal to add them.
3 15ths and 4 15ths add to 7 15ths.
The trick to adding fractions is that you have to be adding fractions of the same denomination. It doesn't make any sense to add ½ to ¼. What would you call the result? Would it be some kind of strange hybrid of halves and fourths? What we'll need to do is to convert halves to fourths (½ = 2/4), then we can add up all of the fourths to get ¾.
Two fractions can be added (or subtracted) only if their denominators are the same. A common denominator can always be found for two fractions.
### A handy shortcut for adding two fractions
Let's review a general method for adding fractions. It's the brute-force method that always works. We don't look for a common denominator; we just multiply the denominators to get one.
We begin by adding two fractions with different denominators, b and d:
Multiply by d/d = 1 on the left and b/b = 1 on the right:
... and we get our sum:
Now these are just letters, so this has to work for any two fractions. If you look closely, you'll see a pattern in there. On the top is the sum of two "diagonal products," ad and bc, and on the bottom is just the product of the denominators. This picture might help:
Here are a couple of examples:
Here's one with variables:
### Practice problems
Add the fractions and reduce the result if possible:
### Convenient denominators – Percent
Some denominations (denominators) are very convenient and commonly used, so we ought to be familiar with them.
Percents are no big mystery, they're just fractions. The word percent is from the Latin per centum, which roughly means "by the hundred." The denominator of the percent system is always 100. So 20% means 20 out of every 100, or
#### Percent → fraction
It's always pretty easy to convert from percent to a reduced fraction: Just put the percentage over 100 and reduce if you can. Here are some examples:
If you're at a store and something is 10% off, that's 10 cents out of every dollar (100 cents), or \$10 out of every \$100, or \$100 out of every \$1000, ... you get the point.
#### Fraction → percent
Now let's say we want to know how many percent ⅞ is. We can use the simple algebra of proportions to do this. In words we might say, "7 is to 8 as what is to 100 ?". The "what" will be our variable, x. Here it is algebraically:
That's just a proportion, and we solve it by cross multiplication: 7(100) = 8x:
... and we finish by dividing both sides by 8 and reducing the resulting fraction until we can't any more (I like to repeatedly by 2 if both numerator and denominator are even):
So 7/8 of an inch is 87.5% of an inch. 7/8 of a Kilogram is 87.5% of a Kilogram, and so on.
#### Dependent & Independent variables
A percent is just a fraction with a denominator of 100.
• To convert from a% to a reduced fraction just reduce the fraction a/100.
• To convert from a fraction, a/b to a percent, use a proportion:
### Practice problems
Convert from percent to a fraction with integer numerator and denominator
Convert to percent to the nearest 100th.
### Other denominators
#### Why do we use base 60 and base 12?
Did you ever wonder why our clocks have 60 seconds per minute and 60 minutes per hour, why we use base 12 in our length measurements here in the U.S.?*
These fractions come to us from a time when cultures, such as the Mesopotamians, were developing mathematics, and when much of that math revolved around supporting commerce – the exchange of money and goods.
Now in our base 10 system, consider the even fractions of 10. They are 1, 1/2, and multiples of 1/5 and 1/10. But consider all of the integer fractions we can get from a base of 60 divisions:
If you're interested in the history of mathematics and more interesting things like this, you might want to check out the books of Princeton Prof. Victor Katz.
*The U.S., Myanmar and Liberia, as of 2015, are the only countries of Earth who have not officially adopted the metric system.
## Video Examples
### Arithmetic with fractions
#### 1. Where does the negative sign go?
It doesn't matter where you put the negative sign in a fraction – numerator, denominator or in an ambiguous position – but putting it in the numerator is the best practice. This video will explain why they're all the same.
#### 2. Multiplication of fractions
When multiplying fractions, multiply numerators and multiply denominators. In this video, we'll see how to interpret multiplied fractions. What is half of half?
#### 3. Division of fractions
Division is really just multiplication by the reciprocal, so dividing by ½ is the same is multiplication by 2. Here are a couple of example problems. |
# 1.3.1 Position, Distance and Displacement
Size: px
Start display at page:
## Transcription
1 In the previous section, you have come across many examples of motion. You have learnt that to describe the motion of an object we must know its position at different points of time. The position of an object is always taken with respect to some reference point. The straight line motion of a particle is the easiest to describe and we begin this section with it. Straight line motion can be purely vertical (as that of a falling ball), purely horizontal (as that of a cyclist on a straight road) or slanted (as that of a wheelchair on a ramp). Before reading further, you may like to jot down a few examples of straight line motion around you. Use the margin! There are many ways of describing motion: Through words, pictures, graphs and mathematical equations. In this section, we shall be using the first three ways for describing the position, distance, displacement, speed, velocity and acceleration of objects moving along a straight line Position, Distance and Displacement Study Fig It shows a ball rolling straight on a surface in the direction of the brown arrow. What is its position? We can use the following definition to answer this question: The POSITION of an object moving along a straight line is uniquely defined by its coordinate with respect to some point of reference. This point is often taken to be the origin (the point O in Fig. 1.4). The position in straight line motion is fully specified by 1 coordinate. This is why it is also called ONE-DIMENSIONAL motion. O x < 0 x = 0 x > 0 Fig. 1.4: The position of an object in straight line motion is defined by 1 coordinate. The positive direction (x > 0) of the axis is the direction of increasing numbers (coordinates). It is to the right in Fig The opposite direction is the negative direction. For a given problem, the origin can be chosen at any convenient point. Usually, the position of the object at time t = 0 is chosen as the origin. Let us now define the words distance and displacement. 5
2 Block 1 Concepts in Mechanics DISTANCE AND DISPLACEMENT The DISTANCE travelled by an object in a given time interval is THE TOTAL LENGTH OF THE ACTUAL PATH it covers during that time. DISPLACEMENT of an object in a given time interval from t i to t f refers to its CHANGE IN THE POSITION during that time. It includes the direction of motion. Displacement = final position of the object (at time t f ) initial position of the object (at time t i ). You have learnt that the position of a moving object is a function of time. Let us now explain how to describe how its position changes with time Average Speed and Average Velocity Study the position-time graph of a bus moving on a straight road (Fig. 1.5) t = 0 s 2 s 4 s 6 s 8 s x (m) x = 0 m 20 m 40 m 60 m 80 m t (s) Fig. 1.5: The graph showing the positions of a moving bus at different instants of time with respect to the point x = 0, which is the origin. Suppose you want to know: How fast is the bus moving? You can find the answer to this question by calculating the distance travelled by the bus in 1 second or in 1 hour. Try doing so! You know the word speed very well as you use it in your daily life. You know that the speed of an aeroplane is much more than the speed of a train. The speed of a bus is much more than the speed of a cyclist. A fast moving object has high speed; a slow moving object has low speed. In Fig. 1.5, the bus moves 80 m in 8 s. So it moves 10 m in 1 second or 36 km in 1 hour. This brings us to the concept of average speed. You may ask: How do we define AVERAGE SPEED in physics? 6 There is another term you need to know and that is AVERAGE VELOCITY. Let us define these terms.
3 AVERAGE SPEED AND AVERAGE VELOCITY SPEED of an object is defined as the DISTANCE TRAVELLED BY IT IN UNIT TIME. AVERAGE SPEED = TOTAL DISTANCE TRAVELLED TOTAL TIME TAKEN VELOCITY of an object is defined as the CHANGE IN ITS POSITION IN UNIT TIME. It includes the DIRECTION of motion. AVERAGE VELOCITY = DISPLACEMENT TIME INTERVAL The SI unit of speed and velocity is metres per second written as ms 1. ALWAYS WRITE THE DIRECTION OF THE VELOCITY OF AN OBJECT. You have learnt so far that an object that changes its position with time has a non-zero velocity. When we talk about the change in position of an object, we have to include the direction of motion. If the velocity of an object is to be increased, its change in position from the initial position, that is, its displacement should be increased with time. This object should never change direction and return to where it started from! Now think of this situation: Suppose you are travelling in a train. Do you always move at the same speed or with the same velocity? The train stops at different stations. Sometimes it moves fast and at other times it slows down. So its speed and velocity change at every instant of time. The average velocity does not tell us how fast an object moves at any given instant. To describe this kind of motion, we use words like instantaneous velocity and instantaneous speed. INSTANTANEOUS VELOCITY AND SPEED INSTANTANEOUS VELOCITY IS VELOCITY AT EACH INSTANT OF TIME. INSTANTANEOUS SPEED IS SPEED AT EACH INSTANT OF TIME. 7
4 Block 1 Concepts in Mechanics The speedometer of a car or a bus tells us its instantaneous speed. You will learn how to find the instantaneous speed and instantaneous velocity x (m) A C D B Time (s) Position (m) O t 3 t 4 t 1 t 2 t (s) NOTE The instantaneous speed of an object with non-constant speed at a point can be found from the slope of a line tangent to its path given by the curve x(t). Fig. 1.8: Non-uniform motion Study Fig Draw straight lines joining the points A and B on the curve with the origin. What is the change in the position of the ball between the times t 1 (= 1 s) and t 2 (= 3 s)? What is the average speed of the ball? What is its average velocity? Remember that the direction of the velocity is in the same direction as that of the change in position. Now take two points C and D on the curve closer to each other. Draw the straight lines OC and OD. Calculate the change in position, the average speed and the average velocity between the times t 3 (= 2 s) and t 4 (= 2.1 s). What is the direction of velocity in this case? What happens to the change in position, average speed and average velocity as you bring the point D closer and closer to C? Do you notice that as D gets closer to C, the time interval t 4 t 3 becomes very small? We get closer to the instant of time t 3. The values of average speed and average velocity also get closer to the values of instantaneous speed and instantaneous velocity. THE DIRECTION OF INSTANTANEOUS VELOCITY IS ALONG THE TANGENT TO THE CURVE AT THE POINT C IN THE DIRECTION OF THE CHANGE IN POSITION. How do we describe the change in the speed and velocity of any object with time? For this, we need to introduce the concept of ACCELERATION. Recall that the velocity of an object tells us how its displacement changes with time. In the same way, the acceleration of an object tells us how its speed or velocity CHANGES WITH TIME. Just as we defined average speed/average velocity and instantaneous speed/instantaneous velocity, we shall define AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION. 8
5 AVERAGE ACCELERATION The AVERAGE ACCELERATION of an object is defined as the net change in its velocity per unit time: AVERAGE ACCELERATION = CHANGE IN VELOCITY TIME INTERVAL The SI unit of acceleration is m / s /s and is written as ms 2. You must always write the DIRECTION of acceleration. Quite often we wish to know the acceleration of an object at a given instant of time. Then we need to define its instantaneous acceleration. We shall give its precise definition in the next section after you have learnt the mathematics needed. For the moment, you can understand INSTANTANEOUS ACCELERATION as acceleration at any given instant of time. You may have heard sports commentators saying that a person is accelerating if he or she is moving fast. But acceleration has nothing to do with going fast. An object could be moving very fast (at a high CONSTANT velocity), and still not be accelerating. ACCELERATED MOTION MEANS THAT THE OBJECT S VELOCITY IS CHANGING. If its velocity does not change with time, then the object is not accelerating. When an object slows down, it is said to be DECELERATING. For example, when a car or bus starts, it has a finite acceleration as its velocity changes from 0 to some value in a given time. But when it moves with a constant speed/velocity, its acceleration is zero. CONSTANT VELOCITY MEANS ZERO ACCELERATION. CONSTANT ACCELERATION MEANS THAT THE CHANGE IN VELOCITY IS CONSTANT. 9
6 Block 1 Concepts in Mechanics 10
7 11
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# NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
## Class 8 Mathematics by VP Classes
Created by: Vp Classes
## Class 8 : NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
The document NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by VP Classes.
All you need of Class 8 at this link: Class 8
EXERCISE 2.3
Ques: Solve the following equations and check your results.
1. 3x = 2x + 18
2. 5t – 3 = 3t – 5
3. 5x + 9 = 5 + 3x
4. 4z + 3 = 6 + 2z
5. 2x – 1 = 14 – x
6. 8x + 4 = 3(x – 1) + 7
7.
8.
9.
10. 3m = 5m - 8/5
Ans:
1. 3x = 2x + 18
Transposing 2x from RHS to LHS, we have
3x – 2x = 18
⇒ x = 18
2. 5t – 3 = 3t – 5
Transposing (–3) to RHS, we have
5t = 3t – 5 + 3
⇒ 5t = 3t – 2
Transposing 3t to LHS, we have
5t – 3t = –2
⇒ 2t = –2
Diving both sides by 2, we have
t = –1
3. 5x + 9 = 5 + 3x
Transposing 9 to RHS, we have
5x = 5 + 3x – 9
⇒ 5x = –4 + 3x
Transposing 3x to LHS, we have
5x – 3x = –4
⇒ 2x = –4
Dividing both sides by 2, we have
x = -4/2 = -2
x = -2
4. 4z + 3 = 6 + 2z
Transposing 3 to RHS, we have
4z = 6 – 3 + 2z
⇒ 4z = 3 + 2z
Transposing 2z to LHS, we have
4z – 2z = 3
⇒ 2z = 3
Dividing both sides by 2, we have
z = 3/2
5. 2x – 1 = 14 – x
Transposing (–1) to RHS, we have
2x = 14 – x + 1
⇒ 2x = 15 – x
Transposing (–x) to LHS, we have
2x + x = 15
⇒ 3x = 15
Dividing both sides by 3, we have
x = 15/3 = 5
∴ x = 5
6. 8x + 4 = 3(x – 1) + 7
8x + 4 = 3x – 3 + 7
⇒ 8x + 4 = 3x + 4
Transposing 4 to RHS, we have
8x = 3x + 4 – 4
Transposing 3x to LHS, we have
8x – 3x = 0
⇒ 5x = 0
∴ x = 0
7.
Using distributive property,
⇒
Transposing x to LHS, we have
⇒
⇒ x/5 = 8
Multiplying both sides by 5, we have
x = 8 × 5 = 40
∴ x = 40
8.
Transposing 1 to RHS, we have
⇒
Transposing 7x/15 to LHS, we have
⇒
⇒ 3x/15 = 2
Multiplying both sides by 15, we have
3x = 2 × 15 = 30
Dividing both sides by 3, we have
∴ x = 10
9.
Transposing 5/3 to RHS, we have
Transposing (–y) to LHS, we have
⇒ 3y = 7
Dividing both sides by 3, we have
3y = 3 = 7/3
∴ y = 7/3
10. 3m = 5m
Transposing 5m to LHS, we have
3m - 5m =
⇒ -2m =
Dividing both sides by (–2), we have
∴ m = 4/5
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Solutions for Problem Set 2
# Solutions for Problem Set 2 - Solutions for Problem Set 2...
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Unformatted text preview: Solutions for Problem Set 2 Problem 1 Part a First we assume h ≥ - . To check if an h is Pareto efficient we want to find the range of h where the derivates of the utility functions do not all have the same sign. Since Zeke’s utility function is v z ( h ) =- h 2 and the derivative is v z ( h ) =- 2 h < 0 for any h , we just need to find the range where Xavier or Yolonda’s marginal utility is positive. By taking the derivatives of the utilities, it can be shown that Xavier’s utility is increasing on h ∈ [0 , 3 2 ] and Yolonda increases on h ∈ [0 , 1 3 ]. Therefore for any h > 3 2 , v i (3 / 2) > v i ( h ) for all i , and thus no h > 3 2 can be Pareto efficient since all three could be made better off by setting h = 3 2 . For all h ∈ , 3 2 Xavier’s utilty is increasing in h but Zeke’s is decreasing so we can’t make one of the two better off without making the other worse off. So each h ∈ , 3 2 is Pareto Efficient. Part b The sum of the utilities is: v x ( h ) + v y ( h ) + v z ( h ) = 3 h- h 2 + 2 h- 3 h 2- h 2 = 5 h- 5 h 2 Part c V ( h ) = 5- 10 h = 0 → h * = 1 2 Part d First note that at h = 0, v x (0) = v y (0) = v z (0) = 0 and that ∑ i v i (0) = 0. Also, v x ( 1 2 ) = 5 4 , v y ( 1 2 ) = 1 4 , v z ( 1 2 ) =- 1 4 . Using this requirement (ii) becomes:....
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## This note was uploaded on 01/14/2012 for the course ECON 201 taught by Professor Witte during the Spring '08 term at Northwestern.
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# Give an example of a function with a critical point that is neither a maximum nor a minimum....
## Question:
Give an example of a function with a critical point that is neither a maximum nor a minimum. Specify the relevant point(s).
## Critical Points
When the derivative of a function equals zero, the function has a critical point. This is a candidate for the function to have a local maximum or minimum, but that behavior is not guaranteed. We need to test this point using the first or second derivative test to classify the point.
The thought behind using critical points to define maxima and minima comes from the fact that when a function has a local maximum or minimum, the first derivative has a value of zero. However, there could be other reasons why the derivative could be zero. Let's analyze a very simple function: {eq}f(x) = x^3 {/eq}.
First, let's find the derivative of this function, which we can do by applying the power rule.
{eq}f'(x) =3 x^2 {/eq}
Next, let's find the critical points of this function by setting this derivative equal to zero and solving.
{eq}3x^2 = 0\\ x = 0 {/eq}
However, we will find that the function does not, in fact, have a local maximum or minimum at this point. We can see this by applying either the first or second derivative test.
{eq}f'(-1) = 3(-1)^2 = 3\\ f'(1) = 3(1)^2 = 3 {/eq}
Since the derivative doesn't change sign on either side of this critical point, the function does not have a local extrema at the critical point we found. If we were to apply the second derivative test, we would find that the second derivative is zero, which is not positive or negative.
{eq}f''(x) = 6x\\ f''(0) = 6(0) = 0 {/eq}
Further testing would show that this is actually an inflection point of the function. The function has a momentary horizontal tangent at the inflection point, but the function continues to increase over its entire domain. |
# 13.3 Geometric sequences (Page 3/6)
Page 3 / 6
Given a geometric sequence with ${a}_{2}=4$ and ${a}_{3}=32$ , find ${a}_{6}.$
${a}_{6}=16,384$
## Writing an explicit formula for the n Th term of a geometric sequence
Write an explicit formula for the $n\text{th}$ term of the following geometric sequence.
The first term is 2. The common ratio can be found by dividing the second term by the first term.
$\frac{10}{2}=5$
The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.
$\begin{array}{l}{a}_{n}={a}_{1}{r}^{\left(n-1\right)}\hfill \\ {a}_{n}=2\cdot {5}^{n-1}\hfill \end{array}$
The graph of this sequence in [link] shows an exponential pattern.
Write an explicit formula for the following geometric sequence.
${a}_{n}=-{\left(-3\right)}^{n-1}$
## Solving application problems with geometric sequences
In real-world scenarios involving arithmetic sequences, we may need to use an initial term of ${a}_{0}$ instead of ${a}_{1}.\text{\hspace{0.17em}}$ In these problems, we can alter the explicit formula slightly by using the following formula:
${a}_{n}={a}_{0}{r}^{n}$
## Solving application problems with geometric sequences
In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.
1. Write a formula for the student population.
2. Estimate the student population in 2020.
1. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.
Let $P$ be the student population and $n$ be the number of years after 2013. Using the explicit formula for a geometric sequence we get
2. We can find the number of years since 2013 by subtracting.
$2020-2013=7$
We are looking for the population after 7 years. We can substitute 7 for $n$ to estimate the population in 2020.
${P}_{7}=284\cdot {1.04}^{7}\approx 374$
The student population will be about 374 in 2020.
A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.
1. Write a formula for the number of hits.
2. Estimate the number of hits in 5 weeks.
1. The number of hits will be about 333.
Access these online resources for additional instruction and practice with geometric sequences.
## Key equations
recursive formula for $nth$ term of a geometric sequence ${a}_{n}=r{a}_{n-1},n\ge 2$ explicit formula for $\text{\hspace{0.17em}}nth\text{\hspace{0.17em}}$ term of a geometric sequence ${a}_{n}={a}_{1}{r}^{n-1}$
## Key concepts
• A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
• The constant ratio between two consecutive terms is called the common ratio.
• The common ratio can be found by dividing any term in the sequence by the previous term. See [link] .
• The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. See [link] and [link] .
• A recursive formula for a geometric sequence with common ratio $r$ is given by $\text{\hspace{0.17em}}{a}_{n}=r{a}_{n–1}\text{\hspace{0.17em}}$ for $n\ge 2$ .
• As with any recursive formula, the initial term of the sequence must be given. See [link] .
• An explicit formula for a geometric sequence with common ratio $r$ is given by $\text{\hspace{0.17em}}{a}_{n}={a}_{1}{r}^{n–1}.$ See [link] .
• In application problems, we sometimes alter the explicit formula slightly to $\text{\hspace{0.17em}}{a}_{n}={a}_{0}{r}^{n}.\text{\hspace{0.17em}}$ See [link] .
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
ouky thanks braa
Rubben
I think he drive me safe
Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
please I need help in maths
Okey tell me, what's your problem is?
Navin
the least possible degree ?
(1+cosA)(1-cosA)=sin^2A
good
Neha
why I'm sending you solved question
Mirza
Teach me abt the echelon method
Khamis
exact value of cos(π/3-π/4)
What is differentiation? |
# 7.2 Trigonometry and derivatives and addition theorems
Introduction
In 7.1, we introduced lots of trigonometry without actually mentioning it.
Trigonometry is a long and off-putting name for what is really a fun subject. A Trigon is a fancy name for a triangle; analogous to the words octagon or pentagon, Metry refers to measurement. So trigonometry means either measuring triangles, or using triangles to measure other things, I’m not sure which; maybe both.
We will describe the remaining important trigonometric results. One of the mysteries of trigonometry is: Why does every one of the six ratios of side lengths in a right triangle have its own special name? Why for example does $$\frac{1}{\sin x}$$ have a name of its own? When I studied trigonometry in school, (in prehistoric times) we were confronted with all six names, and quizzed on and expected to memorize which means which without any clues at all. This turned many of us off to trigonometry.
Suppose our angle $$theta$$, as in the picture here lies between the $$x$$ axis and the line $$0B$$. The ancients drew a line segment that extends from the point $$B$$ tangent to the unit circle to the $$x$$ axis at point $$C$$. The length of this segment they called the tangent of the angle $$\theta$$. (When the line has a positive slope the tangent is taken to be negative.) Tangent is a Latin word that means 'touching', and that is what this line does to the circle, at point $$B$$.
The $$x$$ coordinate of the point $$A$$ where the tangent line meets the $$x$$ axis, is called the secant of $$\theta$$ (we are assuming that the origin is at the center of the unit circle.) Secant is a Latin word meaning 'cutting' which is what this line does to the circle.
They also defined the complement of an angle that is less than a right angle to be the difference between a right angle and it. This got them to define the cosine, cotangent and cosecant as the sine, tangent and secant of the complement of the original angle.
Fortunately for us, all of these six functions are easily related to the sine function, which means that we need only really become familiar with the sine, and we can then figure out what the others are.
Here are the relations between these functions, all of which follow from the definitions from the fact that corresponding angles of similar triangles are equal.
By definition, $$\cos\theta$$ is $$\sin\left(\frac{\pi}{2}-\theta\right)$$.
From triangle $$BCD$$ in which the hypotenuse is $$\tan \theta$$ and the side not opposite $$\theta$$ is $$\sin θ$$, we get
$(\cos \theta)(\tan \theta) = \sin \theta$
which means
$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\sin\theta}{\sin(\pi/2-\theta)}$
The complementary version of this is:
$\cot θ = \frac{\cos \theta}{\sin θ} = \frac{\sin(\pi/2-\theta)}{\sin \theta}$
From the triangle $$BCO$$
we similarly get
$(\sec \theta)(\sin \theta) = \tan \theta$
which means
$\sec \theta = \frac{1}{\cos \theta}$
and the complementary version is
$\csc \theta = \frac{1}{\sin \theta}$
So all this explains why every ratio of side lengths of a right triangle has a name of its own.
I like this picture so much, here it is again but as a mathlet.
Exercises:
7.21. Draw a this picture yourself, without looking at this one, for an angle $$\theta$$ that is less than $$\frac{\pi}{2}$$ showing all of these entities.
7.22. How many similar triangles do you see? Remember that the two angles other than a right angle in a triangle with a right angle are complementary.
7.23. What strange relations do you get if you use the definition of sine in the triangle $$OAC$$?
We have, in the previous section, already discussed the three basic theorems of trigonometry that you should know. There are also useful "addition theorems" of trigonometry, which describe what sine and cosine of sums of arguments are. We also describe the derivatives of the sine and cosine, and their relation to exponentials.
What were the basic Theorems of Trigonometry?
1.The Pythagorean Theorem: This famous result states that the square of the hypotenuse of a right triangle is the sum of the squares of its other two sides. Translated to our definitions it says that for any angle, we have
$(\sin\theta)^2 + (\cos\theta)^2 = 1$
which implies that, up to sign we have
$\cos\theta = \sqrt{1-(\sin\theta)^2}$
2. The Law of Sines: This states that in any triangle $$ABC$$ the ratio of the sines of its angle at $$A$$ to its angle at $$B$$ is the ratio of the lengths of the side opposite $$A$$ to the side opposite $$B$$. If we describe these lengths as $$l(BC)$$ and $$l(AC)$$ respectively, we have
$\frac{\sin A}{\sin B} = \frac{|BC|}{|AC|}$
We proved this in section 7.1C
3. The Law of Cosines: This statement gives the length of the side $$BC$$ of a triangle in terms of the lengths of $$AB$$ and $$AC$$ and its angle at $$A$$
$|BC|^2 = |AB|^2 + |AC|^2 - 2 |AB||AC|\cos A$
Also proven in 7.1C
Derivatives of Sines and Cosines
Consider a point $$P$$ on the unit circle which circle is centered at the point $$C$$. Let $$\theta$$ be the angle clockwise from the line segment $$CP$$ to the $$x$$ axis.
We then have $$x = \cos\theta$$ and $$y = \sin\theta$$. We know that $$(\cos\theta)^2 + (\sin\theta)^2$$ is $$1$$ for any $$P$$ on the unit circle since this is the statement that $$x^2 + y^2$$ for such $$P$$ is $$1$$, which is the definition of the unit circle.
This means that the derivative of $$(\cos\theta)^2 + (\sin\theta)^2$$ is $$0$$ as we move around the unit circle is $$0$$. This tells us
$2\cos\theta\left(\frac{d\cos\theta}{d\theta}\right) + 2\sin\theta\left(\frac{d\sin\theta}{d\theta}\right) = 0$
which means that the vector with components $$\cos\theta$$ and $$\sin\theta$$ has $$0$$ dot product with the vector whose components are their corresponding derivatives.
It is, fortunately very easy to find all vectors in two dimensions that have $$0$$ dot product with a given one: you reverse the components, change one of their signs, and multiply by any constant $$c$$. Thus $$(a,b)$$ (which is a shorter way of writing $$a\hat{i} + b\hat{j}$$) has $$0$$ dot product with $$(cb, -ca)$$.
This tells us $$\frac{d\sin\theta}{d\theta} = c\cos\theta$$ and $$\frac{d\cos\theta}{d\theta} = -c\sin\theta$$, for some constant $$c$$.
We can determine the constant $$c$$ by examining these statements at the point for which $$\theta = 0$$.
If our angle is measured in radians, we have observed that on the unit circle, moving a distance $$d\theta$$ from angle $$0$$ changes the sine from $$0$$ to almost $$dθ$$. Thus the constant $$c$$ above is $$1$$ at angle $$0$$, and being a constant, is always $$1$$.
We conclude $$(\sin\theta)' = \cos\theta, and (\cos\theta)' = -\sin\theta$$. (The latter relation actually follows from the former from the fact that $$\cos\theta$$ is $$\sin(\frac{\pi}{2} - \theta)$$.
Exercise 7.25: Deduce the derivatives of the secant and tangent from these facts.
Why not tell us the answers?
If we did this you will have trouble remembering them. If you figure them out yourself you will have trouble forgetting them.
An interesting consequence is gotten by looking at the combination $$\cos x + i \sin x$$. ($$i$$ here is the square root of $$-1$$.) Notice that its derivative is $$i$$ times itself. And its value at $$x = 0$$ is $$1$$. We know what that means. A function whose derivative is $$q$$ times itself, whose value at $$x = 0$$ is $$1$$, is $$\exp(qx)$$.
We therefore find: $$\cos x + i \sin x = \exp(ix)$$.
In general we can divide any function $$f$$ into an odd part and an even part; the even part is $$\frac{f(x) + f(-x)}{2}$$ and the odd part is $$\frac{f(x) - f(-x)}{2}$$. The sum of the two parts is $$f(x)$$.
Since $$\cos x$$ is even and $$\sin x$$ is odd, we can identify $$\cos x$$ as the even part of $$\exp(ix)$$, and $$i\sin x$$ as its odd part.
The formal expressions are
$\cos x = \frac{\exp(ix) + \exp(-ix)}{2}$
and
$\sin x = \frac{\exp(ix) - \exp(-ix)}{2i}$
Power series expansions of Sines and Cosines
We have seen that $$\cos 0$$ is $$1$$. Since the derivative of the sine is the cosine, $$\sin x$$, when written as a sum of powers of $$x$$, must have a term in that sum whose derivative is $$1$$. That has to be $$x$$, so that the first term in the power series expansion for $$\sin x$$ is $$x$$. The cosine has derivative $$-\sin x$$, so it must have a term in its power series expansion whose derivative is $$-x$$, which term must be $$-\frac{x^2}{2}$$. $$\sin x$$ similarly must have a term whose derivative is that, namely $$-\frac{x^3}{3!}$$; this forces a term in the cosine series to be $$-\frac{x^4}{4!}$$, etc.
So what do we get?
The sine has contributions from all the odd powers and their signs alternate:
$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... + (-1)^k\frac{x^{2k+1}}{(2k+1)!} + ...$
while the cosine has similarly alternating sign terms from the even powers:
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... + (-1)^k\frac{x^{2k}}{(2k)!} + ...$
By the way, $$\cosh x$$ and $$\sinh x$$ are the even and odd parts of $$\exp(x)$$. Their power series expansions are similar to those of the cosine and sine except that all terms are positive.
There are power series expressions for all the trigonometric functions, but you can figure them out yourself if you ever want to do so, from their relations to the sine and cosine of angle
We have already noticed that the standard measure of angle, in terms of degrees or radians is additive: this measure of the sum of two angles is the sum of the same measures of each summand. This statement is not true for sines or cosines. The sine of the sum of two angles is not the sum of their sines. The addition theorems tell us how to compute the sine and cosine of the sum of two angles in terms of the sines and cosines of the two angles that are summed.
The easiest way to find or prove addition theorems for sines and cosines is to use their relations to exponentials. We already know the addition theorem for exponentials:
$\exp(cx + cy) = \exp(cx)\exp(cy)$
Since $$\exp(ix)$$ is $$\cos x + i\sin x$$, we find that $$\exp(i(x+y))$$ is
$(\cos x + i\sin x)(\cos y + i\sin y)$
which is
$((\cos x)(\cos y) - (\sin x)(\sin y)) + i((\sin x)(\cos y) + (\cos x)(\sin y))$
The real part of this last expression is $$\cos(x+y)$$ and the imaginary part is $$\sin(x+y)$$ and these are our addition theorems.
For the hasty:
$\frac{d(\tan x)}{dx} = \frac{d(\sin x/\cos x)}{dx} = \frac{\cos x}{\cos x} - \frac{\sin x(-\sin x)}{(\cos x)^2} = 1 + (\tan x)^2$
From triangle $$OBC$$ in the picture at the top of section 7.2 we find that the last is $$(\sec x)^2$$.
$\frac{d(\sec x)}{dx} = \frac{d(1/\cos x)}{dx} = -\frac{(-\sin x)}{(\cos x)^2} = (\tan x)(\sec x)$ |
Class 10 NCERT Math Solution
TOPICS
Exercise - 7.1
Question-1 :- Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (– 5, 7), (– 1, 3)
(iii) (a, b), (– a, – b)
Solution :-
```i) By using distance formula,
= √(x₂-x₁)² + (y₂-y₁)²
= √(4-2)² + (1-3)²
= √(2)² + (-2)²
= √8
= 2√2
```
```ii) By using distance formula,
= √(x₂-x₁)² + (y₂-y₁)²
= √(-1+5)² + (3-7)²
= √(4)² + (-4)²
= √32
= 4√2
```
```iii) By using distance formula,
= √(x₂-x₁)² + (y₂-y₁)²
= √(-a-a)² + (-b-b)²
= √(-2a)² + (-2b)²
= √4a² + 4b²
= 2√a² + b²
```
Question-2 :- Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Solution :-
``` Given the points A(0, 0) and B(36, 15),
By using distance formula,
AB = √(x₂-x₁)² + (y₂-y₁)²
AB = √(36-0)² + (15-0)²
AB = √(36)² + (15)²
AB = √1296 + 225
AB = √1521
AB = 39
Yes, we can find the distance between the given towns A and B.
Assume town A at origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
And hence, as calculated above, the distance between town A and B will be 39km.
```
Question-3 :- Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
Solution :-
``` Using the distance formula, we have
AB = √(2-1)² + (3-5)² = √(1)² + (-2)² = √5
BC = √(-2-2)² + (-11-3)² = √(-4)² + (-14)² = √212
AC = √(-2-1)² + (-11-5)² = √(-3)² + (-16)² = √265
Since, AB + BC ≠ AC
So, we can say that the points A, B and C are not collinear.
Therefore, they are seated in a line.
```
Question-4 :- Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Solution :-
``` Let the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of the a triangle respectively.
Using the distance formula, we have
AB = √(6-5)² + (4+2)² = √(1)² + (6)² = √37
BC = √(7-6)² + (-2-4)² = √(1)² + (-6)² = √37
AC = √(7-5)² + (-2+2)² = √(2)² + (0)² = √4
Since, AB = BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.
```
Question-5 :- In a classroom, 4 friends are seated at the points A, B, C and D as shown in Figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Solution :-
``` It can be observed that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of these 4 friends.
Using the distance formula, we have
AB = √(6-3)² + (7-4)² = √(3)² + (3)² = √18
BC = √(9-6)² + (4-7)² = √(3)² + (-3)² = √18
CD = √(6-9)² + (1-4)² = √(-3)² + (-3)² = √18
AD = √(6-3)² + (1-4)² = √(3)² + (-3)² = √18
AC = √(9-3)² + (4-4)² = √(6)² + (0)² = √36
BD = √(6-6)² + (1-7)² = √(0)² + (-6)² = √36
It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.
Therefore, ABCD is a square and hence, Champa was correct.
```
Question-6 :- Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution :-
```(i) Let the points (–1, –2), (1, 0), (–1, 2) and (–3, 0) are representing the vertices A, B, C and D of the a quadrilateral respectively.
AB = √(1+1)² + (0+2)² = √(2)² + (2)² = √8
BC = √(-1-1)² + (2-0)² = √(-2)² + (2)² = √8
CD = √(-3+1)² + (0-2)² = √(-2)² + (-2)² = √8
AD = √(-3+1)² + (0+2)² = √(-2)² + (2)² = √8
AC = √(-1+1)² + (2+2)² = √(0)² + (4)² = √16 = 4
BD = √(-3-1)² + (0-0)² = √(-4)² + (0)² = √16 = 4
It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length.
Therefore, the given points are the vertices of a square.
```
```(ii) Let the points (–3, 5), (3, 1), (0, 3) and (–1, –4) are representing the vertices A, B, C and D of the a quadrilateral respectively.
AB = √(3+3)² + (1-5)² = √(6)² + (-4)² = √52
BC = √(0-3)² + (3-1)² = √(-3)² + (2)² = √13
CD = √(-1-0)² + (-4-3)² = √(-1)² + (-7)² = √50
AD = √(-1+3)² + (-4-5)² = √(2)² + (-9)² = √85
It can be observed that all sides of this quadrilateral are of different lengths.
Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.
```
```(iii) Let the points (4, 5), (7, 6), (4, 3) and (1, 2) are representing the vertices A, B, C and D of the a quadrilateral respectively.
AB = √(7-4)² + (6-5)² = √(3)² + (1)² = √10
BC = √(4-7)² + (3-6)² = √(-3)² + (-3)² = √18
CD = √(1-4)² + (2-3)² = √(-3)² + (-1)² = √10
AD = √(1-4)² + (2-5)² = √(-3)² + (-3)² = √18
AC = √(4-4)² + (3-5)² = √(0)² + (-2)² = √4 = 2
BD = √(1-7)² + (2-6)² = √(-6)² + (-4)² = √52
It can be observed that opposite sides of this quadrilateral are of the same length.
However, the diagonals are of different lengths.
Therefore, the given points are the vertices of a parallelogram.
```
Question-7 :- Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Solution :-
``` We know that a point on the y-axis is of the form (x, 0).
So, let the point P(x, 0) be equidistant from A and B.
We are given that AP = BP.
So, AP² = BP²
Then (x – 2)² + (0 + 5)² = (x + 2)² + (0 – 9)²
i.e., x² + 4 - 4x + 25 = x² + 4 + 4x + 81
i.e., 8x = -56
i.e., x = -7
So, the required point is (-7, 0).
```
Question-8 :- Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Solution :-
``` It is given that the distance between (2, −3) and (10, y) is 10.
(10 - 2)² + (y + 3)² = (10)²
64 + y² + 9 + 6y = 100
y² + 6y + 64 + 9 - 100 = 0
y² + 6y - 27 = 0
y² + 9y - 3y - 27 = 0
y(y + 9) - 3(y + 9) = 0
(y - 3)(y + 9) = 0
y - 3 = 0, y + 9 = 0
y = 3, y = -9
```
Question-9 :- If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Solution :-
``` Given that Q(0, 1) is equidistant from P(5, –3) and R(x, 6).
Therefore, PQ = QR
(0 - 5)² + (1 + 3)² = (0 - x)² + (1 - 6)²
25 + 16 = x² + 25
41 = x² + 25
x² = 16
x = ±4
Therefore, point R is (4, 6) or (−4, 6).
When point R is (4, 6)
PR = √(4-5)² + (6+3)² = √(-1)² + (9)² = √82
QR = √(4-0)² + (6-1)² = √(4)² + (5)² = √41
When point R is (-4, 6)
PR = √(-4-5)² + (6+3)² = √(-9)² + (9)² = √162 = 9√2
QR = √(-4-0)² + (6-1)² = √(-4)² + (5)² = √41
```
Question-10 :- Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).
Solution :-
``` Let P(x, y) be equidistant from the points A(3, 6) and B(– 3, 4).
We are given that AP = BP.
So, AP² = BP²
i.e., (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
i.e., x² + 9 - 6x + y² + 36 - 12y = x² + 9 + 6x + y² + 16 - 8y
i.e.,-6x - 12y + 45 = 6x - 8y + 25
i.e.,12x + 4y = 20
i.e., 3x + y = 5
i.e., 3x + y - 5 = 0
which is the required relation.
```
CLASSES |
# Inverse Tan
View Notes
## Tan Inverse
The inverse in Trigonometry for tangent is known as Inverse Tan. It is the basic function that we use in real-world problems and solve them. However, the primary use of Tan Inverse is to apply the tangent ratio for a specified angle. Thus, with the help of this function, you can quickly find any value to tangent. Examples are tan 1, tan 10, arctan 1 and more. Here arc is the basic way to name any inverse formula in trigonometry. Thus we can say that the inverse of the tangent is also known as arctan. In the article below, you will understand why and how to use this formula to solve different problems. Also, you will learn about its practical applications.
### Tan Inverse Formula
The inverse of tan or anti-tan is the arcus of tan. Here, we will define the formula of the tan inverse.
Suppose we are given as x =tan y
then, y =tan-1 x
Here y can be any real number.
This is also known as tan inverse x.
When you add two different tan inverse functions, the formula will be represented as below.
tan a $\pm$ b = $\frac{tan a \pm tan b}{1 \mp tan(a) tan(b)}$
This is also known as the additional formula for inverse tan. It is derived for the addition of two inverses of tan.
If in the above equation, we put a = arctan x and b = arctan y, we get the equation as presented below.
arctan(x) $\pm$ arctan(y) = arctan($\frac{x\pm y}{1 \mp xy}$) (mod π), xy ≠ 1
This is achieved after substituting the values of a and b.
### The Derivative of Tan Inverse
To find out the derivative for the inverse of tan, we will find the derivative for tan inverse x.
The formula of derivative of the tan inverse is given by:
d/dx(arctan(x)). Hence, we define derivatives as 1/ (1 + x2). Here x does not belong to i or -i. This is also known as differentiation of tan inverse.
Let us take an example for a graph of the tan inverse.
We will define it with the help of the graph plot between π/2 and –π/2.
### Graph of Tan Inverse x
In the above graph of tan inverse x, the points are plotted between π/2 and –π/2. The plotting is along the real axis.
### Integration of Tan Inverse x
If we want to give the values of definite integral for the inverse of tan, we use the concept of integration. The derivative of tan inverse will be integrated back to get the normal value of tan inverse. However, for a definite point, the value will be fixed. Thus the below expression defines the integral of inverse tan.
arctan(x) = $\int_{0}^{x}$ $\frac{1}{y^{2}+1}$ dy
The above pictures describe the integral of tan inverse x.
The basic graph for tan inverse is given by
The above graph is defined for tangent inverse give by equation 1/ (1 + x2)
### How Can You Relate Tan Inverse Infinity With Other Trigonometric Equations?
We take an example of a triangle with one side as x and another as 1. The hypotenuse becomes
√1+ x2
Suppose a triangle has an arctan angle as θ then we can define the below relation for finding other trigonometric functions:
Sin (arctan(x)) = x/ (√1+ x2)
Cos (arctan(x)) = 1 / (√1+ x2)
tan (arctan(x)) = x
### What are The Different Properties of Inverse Tan?
Below are some mentioned properties of the tan inverse:
Suppose = arctan (x)
Also, x = arctan (y)
Here we can use the real number as a domain.
Here -π/2 < y< π/2
This is the defined property of tan inverse used widely.
### How Will You Give Value For tan-1 Infinity?
We know that a tan of 90 degrees is defined as infinity. Thus for tan-1 the value is 90 degrees.
tan 90° = ∞ or tan π/2 = ∞
Hence, tan-1 (∞) = π/2 or tan-1 (∞) = 90°
Q1. What are the Six Different Inverse Functions of Trigonometry?
In trigonometry, there are mainly six functions that are widely used. These include sine, cosine, cos, sectant, tangent, cosecant and cotangent. The use of these functions is to get the value of the sides of a triangle when the angle is given to us. However, in the case of the inverse of these six functions, the concept is reversed. The actual working of the inverse of all trigonometric functions is to find the angle of the triangle when sides are given to us. The concept of inverse trigonometric functions is widely operated in real-time examples, which include physics, navigation, engineering and geometry. The three primary notations for trigonometry that are widely in use are:
1. tan-1 (x)
2. sin-1 (x)
3. cosine-1 (x)
Q2. How Do You Define Tan x?
Usually, sin and cosine are the main trigonometric functions used in solving real-time problems. However, there are more functions that are generated by the ratio of both sin and cosine. There are four functions: cosecant, tangent, cotangent and secant. If you know the values of sin and cosine of an angle, then it is easy to find the values of other trigonometric equations. Therefore, the tangent of x is given by:
tan x = sin (x)/ cosine (x)
Similarly, you can find solutions to other functions as tan(x) = inverse of cotangent (x).
Secant (x) in inverse of cosine(x) and cosecant (x) is inverse of sin (x). With the given formulas, you can easily relate them to the concept of functions and thus find their domain, codomain, and range. Thus, tan x is valid when cosine (x) is not equal to zero. Hence the value of x must not be equal to π/2. |
How to Calculate Area of an Object
Author Info
Updated: May 21, 2019
Finding the area of an object is easy as long as you understand the technique and formulas involved. If you have the right knowledge, you can find the area and surface area of any given object. See Step 1 below to get started.
Part 1 of 2: Calculating the Area of Planar Objects
1. 1
Identify the shapes embedded in an object. If you're not working with a simple easily-identifiable shape, like a circle or a trapezoid, then you may be working with a shape that is made up of several other shapes. You have to recognize what those are so you can break the larger object down into a series of smaller objects.[1]
• In this case, this object is composed of the following shapes: a triangle, a trapezoid, a rectangle, a square, and a semi-circle (half of a circle).
2. 2
Write down the formulas for finding the area of each of these shapes. These formulas will allow you to use the given measurements of each shape to find their areas. Here are the formulas for finding the area of each shape:[2]
3. 3
Write down the dimensions of each shape. Once you write down the formulas, write down the dimensions of each shape so you can plug them in. Here are the dimensions of each shape:
• Square: a = 2.5 in
• Rectangle = w = 4.5 in, h = 2.5 in
• Trapezoid = a = 3 in, b = 5 in, h = 5 in
• Triangle = b = 3 in, h = 2.5 in
• Semi-Circle = r = 1.5 in
4. 4
Use the formulas and dimensions to find the area of each object and add them up. Finding the area of each shape will lead you to find the area of each part of the shape; once you've found the area of each shape using the formula and measurements you were given, all you have to do is add up each area to find the area of the entire object. When calculating area, you have to remember to state the area in square units. The area of the entire object is 44.78 in2. Here's how you get it:[3]
• Find the area of each shape:
• Area of Square = 2.5 in2 = 6.25 in2
• Rectangle = 4.5 in x 2.5 in = 11.25 in2
• Trapezoid = [(3 in + 5 in) x 5 in]/2 = 20 in2
• Triangle = 3 in x 2.5 in x 1/2 = 3.75 in2
• Semi-Circle = 1.5 in2 x π x 1/2 = 3.53 in2
• Add the areas of each shape together:
• Area of Object = Area of Square + Area of Rectangle + Area of Trapezoid + Area of Semi-Circle
• Area of Object = 6.25 in2 + 11.25 in2 + 20 in2 + 3.75 in2 + 3.53 in2
• Area of Object = 44.78 in2
Part 2 of 2: Calculating the Surface Area of 3-D Objects
1. 1
Write down the formulas for finding the surface area of each shape. The surface area is the total area of an object's faces and curved surfaces. Every three dimensional object has a surface area; the volume is the amount of space taken up by the object. Here are the formulas for finding the surface area of a variety of objects:[4]
2. 2
Write down the dimensions of each shape. Here they are:
• Cube = side = 3.5 in
• Cone = r = 2 in, h = 4 in
• Sphere = r = 3 in
• Cylinder = r = 2 in, h = 3.5 in
• Square-Based Pyramid = b = 2 in, h = 4 in
3. 3
Calculate the surface area of each shape. Now, all you have to do is plug the dimensions of each shape into the formula for finding the surface area of each shape and you're all done. Here's how you do it:[5]
• Surface area of the cube = 6 x 3.52 = 73.5 in2
• Surface area of the cone = π(2 x 4) + π x 22 = 37.7 in2
• Surface are of the sphere = 4 x π x 32 = 113.09 in2
• Surface are of the cylinder = 2π x 22 + 2π(2 x 3.5) = 69.1 in2
• Surface area of the square-based pyramid = 22 + 2(2 x 4) = 20 in2
Community Q&A
Search
• Question
How can I calculate the volume of a 3D shape?
All shapes are different, but here is how to calculate cubes specifically: First you would need to have all of the dimensions (length, width, and height) of the cube. Then, you would multiply the dimensions and the formula would be l*w*h or length times width times height in word form.
• Question
How do I calculate area given length and width?
In the case of a square, rectangle, or parallelogram, multiply length by width.
• Question
If a line is 100 meters in length, is the area constant for any object it forms; triangle, square, circle?
Donagan
No, the area varies by shape. For all shapes of a given perimeter, the circle has the largest area.
• Question
How do I calculate space area of a rectangle object?
For a rectangular prism, calculate the surface area by finding the area of its individual faces. Use the dimensions given to do length times width to calculate the area of the six different rectangles that make up a rectangular prism.
• Question
How do I calculate the total surface area of an object by using formula s=6(l)2?
Donagan
That formula would apply to a cube. The surface area of a cube is 6 e², or six times the square of the length of an edge.
• Question
Doesn't this encourage young people wrongly?
Yes.
• Question
How do I calculate the area of a rectangle when given the perimeter?
Donagan
You would also have to know the length of at least one side. The perimeter alone is not enough information.
• Question
How do I calculate the area of a five-sided object?
Donagan
Unless the object is a regular pentagon, there is no relevant formula. The area of a regular pentagon is found by calculating half the product of the apothem and the perimeter.
• Question
How do I measure an area of a star?
Donagan
Because "stars" come in various shapes, there is no specific formula for the area.
• Question
How do I calculate the area and volume of cylindrical shape of an IR plant?
Donagan
The total surface area of a cylinder is found by adding 2πr² (r is the radius) to the product of the circumference and the height or length of the cylinder. The volume is found by multiplying πr² by the height or length.
200 characters left
Tips
• Measure dimensions of the chief objects using a scale or vernier calipers.
Thanks!
Warnings
• Don't confuse between area and surface area both are the same but are used differently. Area is used in case of planar objects and surface area is used in case of 3D objects.
Thanks!
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Article SummaryX
To calculate the area of a planar object, you’ll need the formula for the object depending on the shape of the object. For example, to get the area of a rectangle, you multiply the width by the height, and to calculate the area of a triangle, you multiply the base by the height and divide by half. If you have an irregular shape, you may need to break it up into smaller shapes with a known formula for the area. If you want to learn how to find the surface area of a 3-D object, keep reading the article! |
# Chapter R - Algebra Reference - R.2 Factoring - R.2 Exercises: 13
$3a^{2}+10a+7=(3a+7)(a+1)$
#### Work Step by Step
$3a^{2}+10a+7$ Since the coefficient of $a^{2}$ is not equal to $1$, begin by multiplying the whole expression by $3$, which is the actual coefficient of $a^{2}$. Leave the product between $3$ and the second term expressed: $3(3a^{2}+10a+7)=...$ $...=9a^{2}+3(10a)+21$ Open two parentheses containing initially the square root of the second term, which is $3a$, followed by the sign of the second term, in the first parentheses, and the product of the signs of the second and third terms, on the second parentheses: $...=(3a+)(3a+)$ Find two numbers whose product is equal to the third term, $21$ and whose sum is equal to the coefficient of the expression inside parentheses in the second term, $10$. These two numbers are $7$ and $3$, because $(7)(3)=21$ and $7+3=10$. $...=(3a+7)(3a+3)$ The expression was affected initially by multiplying it by $3$. Divide it by $3$ to obtain the answer: $...=(3a+7)\dfrac{(3a+3)}{3}=...$ $...=(3a+7)(a+1)$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
GRE Math : How to find the solution to a quadratic equation
Example Questions
← Previous 1 3
Example Question #1 : How To Find The Solution To A Quadratic Equation
Solve for x: x2 + 4x = 5
-1
-5
-1 or 5
-5 or 1
-5 or 1
Explanation:
Solve by factoring. First get everything into the form Ax2 + Bx + C = 0:
x2 + 4x - 5 = 0
Then factor: (x + 5) (x - 1) = 0
Solve each multiple separately for 0:
X + 5 = 0; x = -5
x - 1 = 0; x = 1
Therefore, x is either -5 or 1
Example Question #1 : Quadratic Equation
Solve for x: (x2 – x) / (x – 1) = 1
x = -1
No solution
x = -2
x = 2
x = 1
No solution
Explanation:
Begin by multiplying both sides by (x – 1):
x2 – x = x – 1
Solve as a quadratic equation: x2 – 2x + 1 = 0
Factor the left: (x – 1)(x – 1) = 0
Therefore, x = 1.
However, notice that in the original equation, a value of 1 for x would place a 0 in the denominator. Therefore, there is no solution.
Example Question #1 : How To Find The Solution To A Quadratic Equation
x2 – 3x – 18 = 0
Quantity A: x
Quantity B: 6
B. Quantity B is greater.
C. The two quantities are equal.
A. Quantity A is greater
D. The relationship cannot be determined from the information given.
D. The relationship cannot be determined from the information given.
Explanation:
x2 – 3x 18 = 0 can be factored to ( 6)(x + 3) = 0
Therefore, x can be either 6 or 3.
Example Question #25 : How To Find The Solution To A Quadratic Equation
A farmer has 44 feet of fence, and wants to fence in his sheep. He wants to build a rectangular pen with an area of 120 square feet. Which of the following is a possible dimension for a side of the fence?
Explanation:
Set up two equations from the given information:
and
Substitute into the second equation:
Multiply through by .
Then divide by the coefficient of 2 to simplify your work:
Then since you have a quadratic setup, move the term to the other side (via subtraction from both sides) to set everything equal to 0:
As you look for numbers that multiply to positive 120 and add to -22 so you can factor the quadratic, you might recognize that -12 and -10 fit the bill. This makes your factorization:
This makes the possible solutions 10 and 12. Since 12 does not appear in the choices, is the only possible correct answer.
Example Question #1 : New Sat Math No Calculator
What is the sum of all the values of that satisfy:
Explanation:
With quadratic equations, always begin by getting it into standard form:
Therefore, take our equation:
And rewrite it as:
You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:
Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :
OR
The sum of these values is:
Example Question #3 : Quadratic Equation
Quantity A:
Quantity B:
A comparison cannot be detemined from the given information.
Quantity A is larger.
The two quantities are equal.
Quantity B is larger.
Quantity B is larger.
Explanation:
With quadratic equations, always begin by getting it into standard form:
Therefore, take our equation:
And rewrite it as:
Now, while you could use the quadratic formula to solve this problem, the easiest way to work this question is to factor the left side of the equation. This gives you:
Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :
OR
Since both of your answers are less than , quantity B is larger than quantity A.
Example Question #4 : Quadratic Equation
If f(x) = -x2 + 6x - 5, then which could be the value of a if f(a) = f(1.5)?
1
2.5
3.5
4.5
4
Explanation:
We need to input 1.5 into our function, then we need to input "a" into our function and set these results equal.
f(a) = f(1.5)
f(a) = -(1.5)2 +6(1.5) -5
f(a) = -2.25 + 9 - 5
f(a) = 1.75
-a2 + 6a -5 = 1.75
Multiply both sides by 4, so that we can work with only whole numbers coefficients.
-4a2 + 24a - 20 = 7
Subtract 7 from both sides.
-4a2 + 24a - 27 = 0
Multiply both sides by negative one, just to make more positive coefficients, which are usually easier to work with.
4a2 - 24a + 27 = 0
In order to factor this, we need to mutiply the outer coefficients, which gives us 4(27) = 108. We need to think of two numbers that multiply to give us 108, but add to give us -24. These two numbers are -6 and -18. Now we rewrite the equation as:
4a2 - 6a -18a + 27 = 0
We can now group the first two terms and the last two terms, and then we can factor.
(4a2 - 6a )+(-18a + 27) = 0
2a(2a-3) + -9(2a - 3) = 0
(2a-9)(2a-3) = 0
This means that 2a - 9 =0, or 2a - 3 = 0.
2a - 9 = 0
2a = 9
a = 9/2 = 4.5
2a - 3 = 0
a = 3/2 = 1.5
So a can be either 1.5 or 4.5.
The only answer choice available that could be a is 4.5.
Example Question #141 : Equations / Inequalities
Solve for x: 2(x + 1)2 – 5 = 27
–2 or 5
3 or 4
–2 or 4
–3 or 2
3 or –5
3 or –5
Explanation:
Quadratic equations generally have two answers. We add 5 to both sides and then divide by 2 to get the quadratic expression on one side of the equation: (x + 1)2 = 16. By taking the square root of both sides we get x + 1 = –4 or x + 1 = 4. Then we subtract 1 from both sides to get x = –5 or x = 3.
Example Question #161 : Gre Quantitative Reasoning
Two consecutive positive multiples of three have a product of 54. What is the sum of the two numbers?
9
12
3
15
6
15
Explanation:
Define the variables to be x = first multiple of three and x + 3 = the next consecutive multiple of 3.
Knowing the product of these two numbers is 54 we get the equation x(x + 3) = 54. To solve this quadratic equation we need to multiply it out and set it to zero then factor it. So x2 + 3x – 54 = 0 becomes (x + 9)(x – 6) = 0. Solving for x we get x = –9 or x = 6 and only the positive number is correct. So the two numbers are 6 and 9 and their sum is 15.
Example Question #1 : How To Find The Solution To A Quadratic Equation
Solve 3x2 + 10x = –3
x = –2/3 or –2
x = –1/6 or –6
x = –4/3 or –1
x = –1/3 or –3
x = –1/9 or –9
x = –1/3 or –3
Explanation: |
Permutations with Some Identical Elements
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# Permutations with Some Identical Elements - PowerPoint PPT Presentation
Permutations with Some Identical Elements. Sometimes you will deal with permutations in which some items are identical. Example How many different ways can you rewrite the word Like?. LIKE LIEK LKIE LKEI LEIK LEKI ILKE ILEK IELK IEKL IKLE IKEL
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### Permutations with Some Identical Elements
Sometimes you will deal with permutations in which some items are identical.
Example
How many different ways can you rewrite the word Like?
LIKE LIEK LKIE LKEI LEIK LEKI
ILKE ILEK IELK IEKL IKLE IKEL
KLIE KLEI KILE KIEL KELI KEIL
ELIK ELKI EILK EIKL EKLI EKIL
Therefore there are 24 different permutations 4P4= 4!
(Four spots with four different choices)
Try it with the word FALL
Keep track of the common letters by using L1 and L2
If the two L’s trade places, there are 2P2 = 2! ways that they can be arranged.
Therefore the total number of ways that the letters can be arranged is now expressed as
4! / 2! = 24 / 2 = 12
Divide the total number of permutations by the number of ways that you can arrange the identical letters
Try it with the word SISS
From previous knowledge we can see that we take
4! / 3! Since 3! Is the number of ways that the S can be arranged.
= 24 / 6 = 4
The number of permutations of a set of n items of which a are identical is n! / a!
Example 2
Scott is a brick layer and wants to create a pattern on the side of a house. How many patterns could he make if he uses 5 red bricks, and one each of grey, white, green, and black?
total number of bricks = 9 this becomes ‘n’ and a = 5 (identical bricks)
9! / 5! = 9 x 8 x 7 x 6 = 3024 patterns
Arrange the word ‘bookkeeper’
There are a total of 10 letters to choose from 10! choices if we didn’t care about the double/triple letters
10! / 2!2!3! since there are 3 e’s, 2 0’s, 2 k’s
for a set of n objects containing a identical elements, b identical objects, c identical objects, there are
permutations.
Example,
Shawna is looking at all of her cars in the driveway. She has 3 mustangs, 2 motorcycles, 4 corvettes and 3 porsche’s. In how many ways can she arrange her cars?
There are 12 cars in total,
12! / (3!2!4!3!) = 277200
Homework
Pg 245 # 2,4ac, 7, 8a, 10, 12 |
# Math Tricks and the Distributive Property of Numbers
October 1st, 2009 by Math Tricks | Filed under Math Tricks.
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There are a few basic properties of numbers, and, no, giving throbbing headaches is NOT one of them. The three basic (and my three favorite) properties of numbers are:
1: The Associative Property
2: The Commutative Property
3: The Distributive Property
If you spend the time to study the basic properties of numbers, you will grasp a deeper understanding of why you are able to manipulate algebraic equations – throw out your algebra book, because you will gain a natural ability to rearrange and simplify equations! (Well – just in case, DON’T throw out your algebra book!)
Here I will discuss the Distributive Property of numbers and why I should be uttering such things on a web site about math tricks!
Let’s start out with a general formula which demonstrates the distributive property of numbers:
a(b + c) = ab + ac
The way to remember this property is to think of the number outside of the parentheses as “distributing itself” among the values being added within the parentheses.
So what does this have to do with math tricks? Fair question – let me give you an example of why understanding this property is useful for performing fast mental multiplication. Let’s say we want to multiply 8 x 1531. Sure, it looks imposing, but remember that we can break down the number 1531 to:
1000 + 500 + 30 + 1
Now we can perform the multiplication in this fashion:
8(1000 + 500 + 30 + 1)
From the distributive property, we know that:
8(1000 + 500 + 30 + 1) = (8 x 1000) + (8 x 500) + (8 x 30) + (8 x 1)
Now the problem is in a form that we can easily solve mentally by first multiplying left to right:
8000 + 4000 + 240 + 8
And then performing the simple addition step to arrive at the answer, 12248!
So there you have it, the secret to how multiplication math magic works – Shhhh . . . . .
### 5 Responses to “Math Tricks and the Distributive Property of Numbers”
1. claire | 19/11/09
i just remember what my tricks does my teacher told me..haha!:))
2. samus | 6/03/10
Um… don’t you mean a(b + c) = ab + ac?
4. Amber | 3/10/11
I still dont get it… :(
5. tvgnvghvn | 19/03/14
get it
Prove you are human - or a smart monkey! * Time limit is exhausted. Please reload CAPTCHA. |
Fibonacci Numbers
In Maths, the Fibonacci numbers are the numbers ordered in a distinct Fibonacci sequence. These numbers were introduced to represent the positive numbers in a sequence, which follows a defined pattern. The list of the numbers in the Fibonacci series is represented by the recurrence relation: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ……..,∞. In this article, let us discuss what is a Fibonacci number, properties, Fibonacci formulas and series in detail.
What is Fibonacci Number?
A Fibonacci number is a series of numbers in which each Fibonacci number is obtained by adding the two preceding numbers. It means that the next number in the series is the addition of two previous numbers. Let the first two numbers in the series be taken as 0 and 1. By adding 0 and 1, we get the third number as 1. Then by adding the second and the third number (i.e) 1 and 1, we get the fourth number as 2, and similarly, the process goes on. Thus, we get the Fibonacci series as 0, 1, 1, 2, 3, 5, 8, ……. Hence, the obtained series is called the Fibonacci number series.
We can also obtain the Fibonacci numbers from the pascal’s triangle as shown in the below figure.
Here, the sum of diagonal elements represents the Fibonacci sequence, denoted by colour lines.
Fibonacci Series List
The list of numbers of Fibonacci Sequence is given below. This list is formed by using the formula, which is mentioned in the above definition.
Fibonacci Numbers Formula
The sequence of Fibonacci numbers can be defined as:
Fn = Fn-1 + Fn-2
Where Fn is the nth term or number
Fn-1 is the (n-1)th term
Fn-2 is the (n-2)th term
From the equation, we can summarize the definition as, the next number in the sequence, is the sum of the previous two numbers present in the sequence, starting from 0 and 1. Let us create a table to find the next term of the Fibonacci sequence, using the formula.
Fn-1 Fn-2 Fn 0 1 1 1 1 2 1 2 3 2 3 5 3 5 8 5 8 13 8 13 21 13 21 34 21 34 55 34 55 89
In the above table, you can see the numbers in each column are relational and diagonally the numbers are the same in all the three columns.
Fibonacci Number Properties
The following are the properties of the Fibonacci numbers.
• In the Fibonacci series, take any three consecutive numbers and add those numbers. When you divide the result by 2, you will get the three numbers. For example, take 3 consecutive numbers such as 1, 2, 3. when you add these numbers, i.e. 1+ 2+ 3 = 6. When 6 is divided by 2, the result is 3, which is 3.
• Take four consecutive numbers other than “0” in the Fibonacci series. Multiply the outer number and also multiply the inner number. When you subtract these numbers, you will get the difference “1”. For example, take 4 consecutive numbers such as 2, 3, 5, 8. Multiply the outer numbers, i.e. 2(8) and multiply the inner number, i.e. 3(5). Now subtract these two numbers, i.e. 16-15 =1. Thus, the difference is 1.
Fibonacci Numbers Examples
Question 1:
Write the first 6 Fibonacci numbers starting from 0 and 1.
Solution:
As we know, the formula for Fibonacci sequence is;
Fn = Fn-1 + Fn-2
Since the first term and second term are known to us, i.e. 0 and 1. Thus,
F0 = 0 and F1 = 1
Hence,
Third term, F2 = F0 + F1 = 0+1 = 1
Fourth term, F3 = F2+F1 = 1 + 1 = 2
Fifth term, F4 = F3+F2 = 1+2 = 3
Sixth term, F5 = F4+F3 = 3 + 2 = 5
So, the first six terms of Fibonacci sequence is 0,1,1,2,3,5.
Question 2:
Find the next term of Fibonacci series: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34.
Solution:
Each next term of the Fibonacci series is the sum of the previous two terms.
Therefore, the required term is;
21 + 34 = 55
Q1
What are the first 10 Fibonacci numbers?
The first 10 Fibonacci numbers are given by:
1, 1, 2, 3, 5, 8, 13, 21, 34, and 55
Q2
What are the numbers in the Fibonacci sequence?
The Fibonacci sequence contains the numbers as: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …,∞
Q3
What are Fibonacci numbers used for?
Fibonacci numbers play an essential role in financial analysis. Here, the Fibonacci number sequence can be used to generate ratios or percentages that are useful for business people.
Q4
What are the first 12 Fibonacci numbers?
The first 12 Fibonacci numbers are given by:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, and 144
Q5
What is the 100th Fibonacci number?
The 100th Fibonacci number will be: 354,224,848,179,261,915,075
We can find this number using the Fibonacci sequence formula. |
# Solved:The probability that you will win a game is 0.45. (a) If you play the game 80 times, what is the most likely number of wins
#### ByDr. Raju Chaudhari
Aug 23, 2020
The probability that you will win a game is 0.45.
(a) If you play the game 80 times, what is the most likely number of wins?
(b) What are the mean and variance of a binomial distribution with p = 0.45 and N = 80?
#### Solution
Here $X$ denote the number of wins out of $N=80$ games.
The probability that you will win a game is $p=0.45$.
The probability mass function of $X$ is
\begin{aligned} P(X=x) &= \binom{80}{x} (0.45)^x (1-0.45)^{80-x},\\ &\qquad x=0,1,\cdots, 80. \end{aligned}
(a) The most likely number of wins is the value of $X$ at which the probability mass function $P(X=x)$ becomes maximum. The most likely value of $X$ is called the mode.
For Binomial distribution $B(N,p)$, if $m=(N+1)*p$ is an integer then there are two mode at $(m-1)$ and $m$, but if $m=(N+1)*p$ is not an integer then there is a single mode at $\lfloor m\rfloor$.
We have $(N+1)*p = (80+1)*0.45 = 36.45$, which is not an integer. Hence there is a single mode and it is at $\lfloor m\rfloor =\lfloor 36.45\rfloor =36$.
Thus the most likely number of wins is $36$.
(b) Mean and Variance
Mean of Binomial distribution $B(N,p)$ is $N*p$ and variance of Binomial distribution is $N*p*(1-p)$.
The mean of $X$ is
\begin{aligned} \text{mean } &= N*p\\ &= 80*0.45\\ &= 36 \end{aligned}
The variance of $X$ is
\begin{aligned} \text{Variance } &= N*p*(1-p)\\ &= 80*0.45*(1- 0.45)\\ &= 19.8 \end{aligned} |
Calculating Limit of Function – A quotient of exponential functions to infinity – Exercise 6556
Exercise
Evaluate the following limit:
$$\lim _ { x \rightarrow \infty} \frac{4^{-x}-5^{-x}}{2^{-x}-3^{-x}}$$
$$\lim _ { x \rightarrow \infty} \frac{4^{-x}-5^{-x}}{2^{-x}-3^{-x}}=0$$
Solution
First, we try to plug in
$$x =\infty$$
and get
$$\frac{4^{-\infty}-5^{-\infty}}{2^{-\infty}-3^{-\infty}}=\frac{0-0}{0-0}$$
We got the phrase 0/0 (=tends to zero divides by tends to zero). This is an indeterminate form, therefore we have to get out of this situation.
Since we have a quotient of exponential functions going to infinity, we divide the numerator and denominator by the expression that aspires most rapidly to infinity, without its coefficient. This division gives us the following:
$$\lim _ { x \rightarrow \infty} \frac{2^{-x}(2^{-x}-{(\frac{5}{2})}^{-x})}{2^{-x}(1-{(\frac{3}{2})}^{-x})}=$$
$$=\lim _ { x \rightarrow \infty} \frac{2^{-x}-{(\frac{5}{2})}^{-x}}{1-{(\frac{3}{2})}^{-x}}=$$
$$=\lim _ { x \rightarrow \infty} \frac{{(\frac{1}{2})}^{x}-{(\frac{2}{5})}^{x}}{1-{(\frac{2}{3})}^{x}}=$$
Since the base is less than 1, the following holds:
$$=\lim _ { x \rightarrow \infty} {(\frac{2}{3})}^x=0$$
$$=\lim _ { x \rightarrow \infty} {(\frac{2}{5})}^x=0$$
$$=\lim _ { x \rightarrow \infty} {(\frac{1}{2})}^x=0$$
For the full list press here
We plug in infinity again, and this time we get
$$=\frac{0-0}{1-0}=$$
$$=\frac{0}{1}=$$
$$=0$$
Have a question? Found a mistake? – Write a comment below!
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# DESCRIBE THE TRANSFORMATION OF THE ABSOLUTE VALUE FUNCTION
## Translation
A translation is a transformation that shifts a graph horizontally or vertically, but doesn’t change the overall shape or orientation.
• If h > 0, then move the graph horizontally towards the right.
• If h < 0, then move the graph horizontally towards the left.
• If k > 0, then move the graph vertically up.
• If k < 0, then move the graph vertically down.
Example problems on horizontal and vertical translation of absolute value functions
## Reflection
Reflections in the x-axis :
The graph of y = −f(x) is a reflection in the x-axis of the graph of y = f (x).
Note :
Multiplying the outputs by −1 changes their signs.
Reflections in the y-axis :
The graph of y = f(-x) is a reflection in the y-axis of the graph of y = f (x).
Note :
Multiplying the inputs by −1 changes their signs.
Example problems on reflection of absolute value function for a given function
## Horizontal and Vertical Shrink or Stretches
The function which is in the form y = |ax - h| + k
Here a > 1 or 0 < a < 1
• If a > 1, the graph should be narrower or horizontal shrink
• If 0 < a < 1, the graph should be wider or horizontal stretch.
The function which is in the form y = a|x - h| + k
Here a > 1 or 0 < a < 1
• If a > 1, the graph should stretch vertically or narrower.
• If 0 < a < 1, the graph should have vertical shrink or wider.
Describe the
Problem 1 :
y = |x - 2|
Solution :
Comparing the given function with the parent function y = |x|
h = 2, so move the graph 2 units to the right.
Problem 2 :
y = |x|+3
Solution :
Comparing the given function with the parent function y = |x|
h = 3, so move the graph 3 units up.
Problem 3 :
y = 2|x + 3|
Solution :
Comparing the given function with the parent function y = |x|
h = -3, so move the graph 3 units left.
a = 2 > 1 vertical stretch 2 units, so it is narrower.
Problem 4 :
y = 3|x|
Solution :
Comparing the given function with the parent function y = |x|
a = 3 > 1 vertical stretch 3 units, so it is narrower.
Problem 5 :
y = -2|x + 3| - 1
Solution :
Comparing the given function with the parent function y = |x|
• a = 2 > 1 vertical stretch 2 units, it is narrower.
• a = -2, reflection across x-axis.
• h = -3, horizontal translation of 3 units left.
• k = 1, vertical translation of 1 unit up.
Problem 6 :
y = 2|x + 8|
Solution :
Comparing the given function with the parent function y = |x|
• a = 2 > 1 vertical stretch 2 units, it is narrower.
• h = -8, horizontal translation of 8 units left.
Problem 7 :
y = -|x - 6| + 2
Solution :
Comparing the given function with the parent function y = |x|
• a = -1, reflection across x-axis.
• h = 6, horizontal translation of 6 units right.
• k = 2, vertical translation of 2 unit up.
Problem 8 :
y = (1/2) |x + 2| - 1
Solution :
Comparing the given function with the parent function y = |x|
• a = 1/2, vertical shrink of 1/2 units.
• h = -2, horizontal translation of 2 units left.
• k = -1, vertical translation of 2 unit down.
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LCM the 5 and 7 is the the smallest number among all usual multiples that 5 and also 7. The first few multiples the 5 and also 7 space (5, 10, 15, 20, 25, 30, 35, . . . ) and (7, 14, 21, 28, 35, . . . ) respectively. There space 3 commonly used techniques to discover LCM of 5 and 7 - by department method, by element factorization, and by listing multiples.
You are watching: Least common multiple of 5 and 7
1 LCM the 5 and also 7 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM of 5 and 7 is 35.
Explanation:
The LCM of two non-zero integers, x(5) and also y(7), is the smallest confident integer m(35) that is divisible by both x(5) and also y(7) without any type of remainder.
Let's look in ~ the different methods because that finding the LCM the 5 and also 7.
By division MethodBy Listing MultiplesBy element Factorization Method
### LCM that 5 and also 7 by division Method
To calculate the LCM the 5 and also 7 by the division method, we will certainly divide the numbers(5, 7) by their prime factors (preferably common). The product of these divisors offers the LCM the 5 and 7.
Step 3: continue the procedures until only 1s room left in the critical row.
The LCM of 5 and also 7 is the product of every prime number on the left, i.e. LCM(5, 7) by department method = 5 × 7 = 35.
### LCM the 5 and also 7 by Listing Multiples
To calculate the LCM of 5 and 7 by listing out the usual multiples, we deserve to follow the given listed below steps:
Step 1: perform a few multiples of 5 (5, 10, 15, 20, 25, 30, 35, . . . ) and also 7 (7, 14, 21, 28, 35, . . . . )Step 2: The usual multiples native the multiples that 5 and 7 are 35, 70, . . .Step 3: The smallest usual multiple of 5 and 7 is 35.
∴ The least common multiple the 5 and 7 = 35.
### LCM the 5 and also 7 by prime Factorization
Prime administrate of 5 and 7 is (5) = 51 and (7) = 71 respectively. LCM of 5 and 7 deserve to be acquired by multiplying prime determinants raised to your respective greatest power, i.e. 51 × 71 = 35.Hence, the LCM that 5 and 7 by element factorization is 35.
☛ also Check:
## FAQs on LCM of 5 and 7
### What is the LCM of 5 and 7?
The LCM that 5 and 7 is 35. To find the LCM (least usual multiple) that 5 and also 7, we need to find the multiples of 5 and 7 (multiples the 5 = 5, 10, 15, 20 . . . . 35; multiples that 7 = 7, 14, 21, 28 . . . . 35) and choose the the smallest multiple the is precisely divisible by 5 and 7, i.e., 35.
### If the LCM that 7 and 5 is 35, uncover its GCF.
LCM(7, 5) × GCF(7, 5) = 7 × 5Since the LCM that 7 and 5 = 35⇒ 35 × GCF(7, 5) = 35Therefore, the greatest usual factor (GCF) = 35/35 = 1.
### What room the approaches to find LCM of 5 and also 7?
The typically used approaches to uncover the LCM the 5 and 7 are: |
# How to find the critical points of a multivariate function
## Description
When trying to find the maximum and minimum values of a multivariate function, that is a function of multiple real-valued inputs, one of the main techniques in calculus is to use the “critical points” of the function, which are the most important inputs to examine to find maxima and minima. Can we find critical points for a multivariate function using software?
## Using SymPy, in Python
View this solution alone.
This answer assumes you have imported SymPy as follows.
1
2
from sympy import * # load all math functions
init_printing( use_latex='mathjax' ) # use pretty math output
Let’s create an example function to work with.
1
2
3
var( 'x y' )
formula = x**3 + x*y - 3*y**2
formula
$\displaystyle x^{3} + x y - 3 y^{2}$
Critical numbers come in two kinds. First, where are all partial derivatives zero? Second, where is a derivative undefined but the function is defined?
Let’s begin by finding where both partial derivatives are zero. Recall that a common notation for the partial derivatives is $f_x$ and $f_y$. We’ll use the same techniques introduced in how to write symbolic equations and how to solve symbolic equations.
1
2
3
f_x = diff( formula, x )
f_y = diff( formula, y )
f_x, f_y
$\displaystyle \left( 3 x^{2} + y, \ x - 6 y\right)$
We can set both equal to zero and solve those equations simultaneously as follows. In other words, SymPy will solve $3x^2+y=0$ and $x-6y=0$ simultaneously for us.
1
solve( [ f_x, f_y ] ) # that is, f_x=0 and f_y=0
$\displaystyle \left[ \left\{ x : - \frac{1}{18}, \ y : - \frac{1}{108}\right\}, \ \left\{ x : 0, \ y : 0\right\}\right]$
That output indicates two critical numbers, one at $\left(-\frac{1}{18},-\frac{1}{108}\right)$ and one at $(0,0)$.
Now where is the derivative defined but the function undefined? Unfortunately, SymPy cannot help us compute the domain of multivariate functions. You will need to do that yourself. (In this case, the function is defined for all real values of $x$ and $y$.)
But SymPy can help us classify the two critical numbers above. Are they maxima, minima, or saddle points? We use the discriminant, built from the second partial derivatives, $D=f_{xx}f_{yy}-f_{xy}^2$.
1
2
3
4
5
f_xx = diff( formula, x, x )
f_yy = diff( formula, y, y )
f_xy = diff( formula, x, y )
D = f_xx*f_yy - f_xy**2
D
$\displaystyle - 36 x - 1$
We can then evaluate the discriminant on our critical points. Recall the following rules from multivariate calculus:
• If $D<0$ then the critical point is a saddle point.
• If $D>0$ and $f_xx<0$ then the critical point is a maximum.
• If $D>0$ and $f_xx>0$ then the critical point is a minimum.
Let’s begin by checking $\left(-\frac{1}{18},-\frac{1}{108}\right)$.
1
D.subs( x, -1/18 ).subs( y, -1/108 )
$\displaystyle 1.0$
Since $D>0$ we must check $f_{xx}$.
1
f_xx.subs( x, -1/18 ).subs( y, -1/108 )
$\displaystyle -0.333333333333333$
Because $f_{xx}<0$, the point $\left(-\frac{1}{18},-\frac{1}{108}\right)$ is a maximum. Now we check $(0,0)$.
1
D.subs( x, 0 ).subs( y, 0 )
$\displaystyle -1$
Because $D<0$, the point $(0,0)$ is a saddle point.
See a problem? Tell us or edit the source.
## Opportunities
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• R
• Excel
• Julia
If you can contribute a solution using any of these pieces of software, see our Contributing page for how to help extend this website. |
# 6th Grade - How Far Away Points Change Things
Grade Level: 6th Skill: Range, Mean, Median, Mode, Outliers and Central Tendency Topic: How Far Away Points Change Things Goal: Understand how the inclusion or exclusion of outliers affects measures of central tendency. Skill Description: Know that in a data set, an outlier is a number that is numerically distant from the rest of the numbers in a data set. This is, it is far below or far above the middle percent of the data terms. For example, the number 99 is an outlier in the data set: 5, 4, 13, 11, 15, and 99. Understand that using an outlier can cause statistical data to be misleading. The mean will always be affected most by an outlier. For example, if you were asked to find the typical age of people on a playground and used the data set above to calculate the mean, the outlier would completely skew your view of the age. Using the measure of the median age and including the outlier would give you a more accurate view of the typical age. So, the outlier could be excluded to find the mean in this example, or it could be included in using the median. Logic will need to be called upon to decide whether to include or exclude an outlier for a given measure of central tendency.
## Building Blocks/Prerequisites
### Sample Problems
(1) Determine if the statement below is true or false. The average salary of an employee at J.M. Mills is \$1767 per month. Salaries: \$100, \$150, \$200, \$100, \$50, \$10,000 (false) (2) What measure could best be used to find the typical salary of employees at J.M. Mills, without excluding the outlier, if the salaries are \$100, \$150, \$200, \$100, \$50, \$10,000. (median) (3) What measure could be used to determine the average salary at J.M. Mills if the outlier were excluded if the salaries are \$100, \$150, \$200, \$100, \$50, \$10,000? (mean) (4) Rick’s science scores were: 88, 98, 75, 80 and 0. How will the outlier affect his grade if his teacher averages his scores? What measure of central tendency could be used to find his typical science grade with the outlier included? (The outlier will bring his grade down. The median could be used to find his typical grade.) (5) Carolina was charting the weather in June. Which temperature will most affect the mean when she calculates the average temperature? 77, 78, 79, 78, 78, 110, 80, 84, 82, and 78 (110, it is an outlier.)
### Online Resources
(1) (2) (3) (4) (5)
### Extra Help Problems
(1)
Identify the outlier in the data set below.
26, 35, 45, 28, 42, 39, 22, 42, 36, 37, 40, 128, 24, 32, 22
(128)
(2)
Identify the outlier in the data set below.
13, 17, 12, 19, 16, 11, 10, 1, 18, 17, 11, 12, 15
(1)
(3)
Identify the outlier in the data set below.
1.4, 1.56, 1.198, 5.1, 2.2, 1.34, 2.12, 2.51, 1.3
(5.1)
(4)
Identify the outlier in the data set below.
-16, 19, 0, -4, -6, -7, -15, -10, -12, 1
(19)
(5)
Identify the outlier in the data set below.
122, 163, -17, 99, 182, 110, 144, 152, 171
(-17)
(6)
Determine which measures of central tendency would change, if you were to first calculate each measure without the outlier and then add it in.
Data Set: 19, 21, 23, 18, 3, 19
(mean 20, median 19, mode 19; mean 17.2, median 19, mode 19)
(7)
Determine which measures of central tendency would change, if you were to first calculate each measure without the outlier and then add it in.
Data Set: 1.1, 1.2, 1, 1.3, -1.5, 1.2
(mode and median)
(8)
Determine which measures of central tendency would change, if you were to first calculate each measure without the outlier and then add it in.
Data Set: 80, 97, 42, 100, 85, 85
(mode)
(9)
Determine which measures of central tendency would change, if you were to first calculate each measure without the outlier and then add it in.
Data Set: 421, 400, 230, 432, 417, 450, 400
(mean, median)
(10)
Determine which measures of central tendency would change, if you were to first calculate each measure without the outlier and then add it in.
Data Set: 52, 56, 59, 58, -2, 62, 58, 58
(mean)
(11)
Find the mean, median, mode and range of the data set with and without the outlier.
5, 7, 9, 10, 5, 24
(mean, median, mode with: 10, 8, 5; without: 7.2, 7, 5)
(12)
Find the mean, median, mode and range of the data set with and without the outlier. Round the mean to the nearest one.
90, 85, 20, 75, 80, 85, 85
(mean, median, mode with: 75, 85, 85; without 83, 85, 85)
(13)
Find the mean, median, mode and range of the data set with and without the outlier.
230, 100, 250, 212, 216
(14)
Find the mean, median, mode and range of the data set with and without the outlier. Round the mean to the nearest tenth.
1.25, 1.5, 1.25, 1.30, 4.50, 1.80
(with: 1.9, 1.4, 1.25; without: 1.42, 1.3, 1.25)
(15)
Find the mean, median, mode and range of the data set with and without the outlier.
-12, 5, 7, 9, 5, 8, 5, 7
(with: 4.25, 6, 5; without: 7.7, 7, 5)
(16)
Saul has been keeping track of the points he’s scored in each basketball game this season in the table below. Use the data to find his average and median score for the games so far. How will these change if he scores 52 points in the next game? Round the average to the nearest one.
Game 1 Game 2 Game 3 Game 4 Game 5 Game 6 30 points 28 points 26 points 32 points 26 points 30 points
(so far: 29, 29; with 52 points: 32, 30)
(17)
Jesse’s Math scores for the second trimester are: 50, 55, 50, 65, 60, 60. What is her average score for math? What will happen to Jesse’s average if she gets a score of 100 on the next test? Write scores as the nearest percent.
(57%, 63%)
(18)
Claudia is playing a game with her sister. The table shows her scorecard for each round. What is Claudia’s score range? Median? Mode? Mean?
How will these change if she scores 150 points in the last round?
Round 1 65 Round 2 70 Round 3 70 Round 4 85 Round 5 90
(without 150: median 70, mode 70, mean 76; with 150: median 77.5, mean 70, mode 88)
(19)
Brendan has recorded the height of his puppy, Chino for the first five months he owned him. He wrote down: 1.2 ft, 1.5 ft, 1.5 ft, 1.7 ft, 2 ft. He then recorded his height again at two years as 3.5 feet. Explain how the last measurement would affect each of the measures of central tendency.
(The last height would make the mean increase.)
(20)
Monica has a summer dog-walking job. She’s recorded her earnings as: \$4.50, \$5.25, \$5.25, \$6.50, \$6.25, \$8.00, \$7.50 and \$25.00. How did her last week’s earnings affect the mean and median of the monies she earned over the summer.
(Her mean and median increased.) |
### Career
Home
Posted on : 2011-08-10
Part 1
The word ‘heuristic’ is taken directly from the Greek verb, heuriskein which means ‘to discover’. In Mathematics, there are usually different ways to go about solving problem sums. These ways or methods are known as heuristics.
Heuristics can be divided into 4 main types, which will be covered in this 2-part article.
One: Giving a representation
· Pupils can transform word problems into pictorial representations and represent information with a diagram/model. This skill helps pupils to understand the question better when they see the visual representation of the word problems.
· A systematic list should be made for word problems that require pupils to identify patterns such as repeated numbers or a series of events that repeat. This skill helps pupils in identifying patterns easily as the list organises all possible answers systematically.
Example: Michele saved \$150 on the first month. On the second month, she saved \$60 more than the first month. On the third month, she saved \$70 more than the second month. On the fourth month, she saved \$55 more than the third month. How much did she save in four months? Solution: Making a list: 1st month → \$150 2nd month → \$150 + \$60 = \$210 3rd month → \$210 + \$70 = \$280 4th month → \$280 + \$55 = \$335 Total amount saved = \$150 + \$210 + \$280 + \$335 = \$975 She saved \$975 in four month
Two: Making a calculated guess
· The ‘guess and check’ method is used for word problems when certain information is lacking. It requires them to make a guess first and check it, and making subsequent guesses and checks until the correct answer is derived. It is often used together with a systematic list as it helps pupils to narrow down the possibilities within a short time frame.
Example:
Jenny has a total of 7 dogs and parrots. The animals have 20 legs altogether.
How many dogs does she have?
Solution:
Using the ‘guess and check’ method,
Number of dogs Number of legs Number of parrots Number of legs Total number of legs Check 1 1 x 4 = 4 6 6 x 2 = 12 4 + 12 = 16 X 2 2 x 4 = 8 5 5 x 2 = 10 8 + 10 = 18 X 3 3 x 4 = 12 4 4 x 2 = 8 12 + 8 = 20 √
She has 3 dogs.
· The ‘look for patterns’ method is usually used by pupils when they have to identify a certain pattern in a number sequence.
Example:
12 4 5 13 11 9 16 18 16 12 3 X
Solution:
Making a list of possibilities:
12 - 4 + 5 = 13
11 - 9 + 16 = 18
16 - 12 + 3 = 7
The value of X is 7.
Hence by using the systematic list, it is more effective to find the underlying pattern.
Written by Gui Yan Tong
Check out the series "Use of Heuristics in Problem Solving" by EPH and have more practice on the different types of heuristics
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# How calculus is used in optimization?
Contents
## How calculus is used in optimization?
One common application of calculus is calculating the minimum or maximum value of a function. For example, companies often want to minimize production costs or maximize revenue. In manufacturing, it is often desirable to minimize the amount of material used to package a product with a certain volume.
How do you optimize an equation?
Recall the main steps to optimization:
3. Draw a picture if appropriate,
4. Identify the variables,
5. Identify the constants,
6. Set constraints,
7. Draw the related graph or picture,
8. Determine what quantity needs to be maximized or minimized,
### Are optimization problems hard?
In the first of these we average hardness over all possible algorithms for the optimization problem at hand. We show that according to this quantity, there is no distinction between optimization problems, and in this sense no problems are intrinsically harder than others.
Where we use optimization techniques?
Optimization methods are used in many areas of study to find solutions that maximize or minimize some study parameters, such as minimize costs in the production of a good or service, maximize profits, minimize raw material in the development of a good, or maximize production.
#### What is the goal of optimization?
The basic goal of the optimization process is to find values of the variables that minimize or maximize the objective function while satisfying the constraints. This result is called an optimal solution.
How to solve an optimization problem for Calculus 1?
Optimization problems for calculus 1 are presented with detailed solutions. It may be very helpful to first review how to determine the absolute minimum and maximum of a function using calculus concepts such as the derivative of a function. 1 – You first need to understand what quantity is to be optimized.
## Which is the first step in the optimization problem?
There is also the problem of identifying the quantity that we’ll be optimizing and the quantity that is the constraint and writing down equations for each. The first step in all of these problems should be to very carefully read the problem.
What is the cost of maximizing a field?
If we look at the field from above the cost of the vertical sides are \$10/ft, the cost of the bottom is \$2/ft and the cost of the top is \$7/ft. If we have \$700 determine the dimensions of the field that will maximize the enclosed area.
### How to determine the minimum and maximum of a function?
It may be very helpful to first review how to determine the absolute minimum and maximum of a function using calculus concepts such as the derivative of a function. 1 – You first need to understand what quantity is to be optimized. 2 – Draw a picture (if it helps) with all the given and the unknowns labeling all variables. |
## Solving the Logistic Differential Equation
### Learning Outcomes
• Draw a direction field for a logistic equation and interpret the solution curves
• Solve a logistic equation and interpret the results
The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution.
Step 1: Setting the right-hand side equal to zero leads to $P=0$ and $P=K$ as constant solutions. The first solution indicates that when there are no organisms present, the population will never grow. The second solution indicates that when the population starts at the carrying capacity, it will never change.
Step 2: Rewrite the differential equation in the form
$\frac{dP}{dt}=\frac{rP\left(K-P\right)}{K}$.
Then multiply both sides by $dt$ and divide both sides by $P\left(K-P\right)$. This leads to
$\frac{dP}{P\left(K-P\right)}=\frac{r}{K}dt$.
Multiply both sides of the equation by $K$ and integrate:
$\displaystyle\int \frac{K}{P\left(K-P\right)}dP=\displaystyle\int rdt$.
The left-hand side of this equation can be integrated using partial fraction decomposition. We leave it to you to verify that
$\frac{K}{P\left(K-P\right)}=\frac{1}{P}+\frac{1}{K-P}$.
Then the equation becomes
$\begin{array}{ccc}\hfill {\displaystyle\int \frac{1}{P}+\frac{1}{K-P}dP}& =\hfill & {\displaystyle\int rdt}\hfill \\ \hfill \text{ln}|P|-\text{ln}|K-P|& =\hfill & rt+C\hfill \\ \hfill \text{ln}|\frac{P}{K-P}|& =\hfill & rt+C.\hfill \end{array}$
Now exponentiate both sides of the equation to eliminate the natural logarithm:
$\begin{array}{ccc}\hfill {e}^{\text{ln}|\frac{P}{K-P}|}& =\hfill & {e}^{rt+C}\hfill \\ \hfill |\frac{P}{K-P}|& =\hfill & {e}^{C}{e}^{rt}.\hfill \end{array}$
We define ${C}_{1}={e}^{c}$ so that the equation becomes
$\frac{P}{K-P}={C}_{1}{e}^{rt}$.
To solve this equation for $P\left(t\right)$, first multiply both sides by $K-P$ and collect the terms containing $P$ on the left-hand side of the equation:
$\begin{array}{ccc}\hfill P& =\hfill & {C}_{1}{e}^{rt}\left(K-P\right)\hfill \\ \hfill P& =\hfill & {C}_{1}K{e}^{rt}-{C}_{1}P{e}^{rt}\hfill \\ \hfill P+{C}_{1}P{e}^{rt}& =\hfill & {C}_{1}K{e}^{rt}.\hfill \end{array}$
Next, factor $P$ from the left-hand side and divide both sides by the other factor:
$\begin{array}{ccc}\hfill P\left(1+{C}_{1}{e}^{rt}\right)& =\hfill & {C}_{1}K{e}^{rt}\hfill \\ \hfill P\left(t\right)& =\hfill & \frac{{C}_{1}K{e}^{rt}}{1+{C}_{1}{e}^{rt}}.\hfill \end{array}$
The last step is to determine the value of ${C}_{1}$. The easiest way to do this is to substitute $t=0$ and ${P}_{0}$ in place of $P$ in the equation $\frac{P}{K-P}={C}_{1}{e}^{rt}$ and solve for ${C}_{1}\text{:}$
$\begin{array}{ccc}\hfill \frac{P}{K-P}& =\hfill & {C}_{1}{e}^{rt}\hfill \\ \hfill \frac{{P}_{0}}{K-{P}_{0}}& =\hfill & {C}_{1}{e}^{r\left(0\right)}\hfill \\ \hfill {C}_{1}& =\hfill & \frac{{P}_{0}}{K-{P}_{0}}.\hfill \end{array}$
Finally, substitute the expression for ${C}_{1}$ into the equation before the last:
$P\left(t\right)=\frac{{C}_{1}K{e}^{rt}}{1+{C}_{1}{e}^{rt}}=\frac{\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}$
Now multiply the numerator and denominator of the right-hand side by $\left(K-{P}_{0}\right)$ and simplify:
$\begin{array}{cc}\hfill P\left(t\right)& =\frac{\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\hfill \\ & =\frac{\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\cdot \frac{K-{P}_{0}}{K-{P}_{0}}\hfill \\ & =\frac{{P}_{0}K{e}^{rt}}{\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}}.\hfill \end{array}$
We state this result as a theorem.
### Solution of the Logistic Differential Equation
Consider the logistic differential equation subject to an initial population of ${P}_{0}$ with carrying capacity $K$ and growth rate $r$. The solution to the corresponding initial-value problem is given by
$P\left(t\right)=\dfrac{{P}_{0}K{e}^{rt}}{\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}}$.
The solution to the logistic differential equation has a point of inflection. To find this point, set the second derivative equal to zero:
$\begin{array}{ccc}\hfill P\left(t\right)& =\hfill & \frac{{P}_{0}K{e}^{rt}}{\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}}\hfill \\ \hfill {P}^{\prime }\left(t\right)& =\hfill & \frac{r{P}_{0}K\left(K-{P}_{0}\right){e}^{rt}}{{\left(\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}\right)}^{2}}\hfill \\ \hfill P^{\prime\prime}\left(t\right)& =\hfill & \frac{{r}^{2}{P}_{0}K{\left(K-{P}_{0}\right)}^{2}{e}^{rt}-{r}^{2}{P}_{0}{}^{2}K\left(K-{P}_{0}\right){e}^{2rt}}{{\left(\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}\right)}^{3}}\hfill \\ & =\hfill & \frac{{r}^{2}{P}_{0}K\left(K-{P}_{0}\right){e}^{rt}\left(\left(K-{P}_{0}\right)-{P}_{0}{e}^{rt}\right)}{{\left(\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}\right)}^{3}}.\hfill \end{array}$
Setting the numerator equal to zero,
${r}^{2}{P}_{0}K\left(K-{P}_{0}\right){e}^{rt}\left(\left(K-{P}_{0}\right)-{P}_{0}{e}^{rt}\right)=0$.
As long as ${P}_{0}\ne K$, the entire quantity before and including ${e}^{rt}$ is nonzero, so we can divide it out:
$\left(K-{P}_{0}\right)-{P}_{0}{e}^{rt}=0$.
Solving for $t$,
$\begin{array}{ccc}\hfill {P}_{0}{e}^{rt}& =\hfill & K-{P}_{0}\hfill \\ \hfill {e}^{rt}& =\hfill & \frac{K-{P}_{0}}{{P}_{0}}\hfill \\ \hfill \text{ln}{e}^{rt}& =\hfill & \text{ln}\frac{K-{P}_{0}}{{P}_{0}}\hfill \\ \hfill rt& =\hfill & \text{ln}\frac{K-{P}_{0}}{{P}_{0}}\hfill \\ \hfill t& =\hfill & \frac{1}{r}\text{ln}\frac{K-{P}_{0}}{{P}_{0}}.\hfill \end{array}$
Notice that if ${P}_{0}>K$, then this quantity is undefined, and the graph does not have a point of inflection. In the logistic graph, the point of inflection can be seen as the point where the graph changes from concave up to concave down. This is where the “leveling off” starts to occur, because the net growth rate becomes slower as the population starts to approach the carrying capacity.
### try it
A population of rabbits in a meadow is observed to be $200$ rabbits at time $t=0$. After a month, the rabbit population is observed to have increased by $4\text{%}$. Using an initial population of $200$ and a growth rate of $0.04$, with a carrying capacity of $750$ rabbits,
1. Write the logistic differential equation and initial condition for this model.
2. Draw a slope field for this logistic differential equation, and sketch the solution corresponding to an initial population of $200$ rabbits.
3. Solve the initial-value problem for $P\left(t\right)$.
4. Use the solution to predict the population after $1$ year.
### Activity: Logistic Equation with a Threshold Population
An improvement to the logistic model includes a threshold population. The threshold population is defined to be the minimum population that is necessary for the species to survive. We use the variable $T$ to represent the threshold population. A differential equation that incorporates both the threshold population $T$ and carrying capacity $K$ is
$\frac{dP}{dt}=\text{-}rP\left(1-\frac{P}{K}\right)\left(1-\frac{P}{T}\right)$
where $r$ represents the growth rate, as before.
1. The threshold population is useful to biologists and can be utilized to determine whether a given species should be placed on the endangered list. A group of Australian researchers say they have determined the threshold population for any species to survive: $5000$ adults. (Catherine Clabby, “A Magic Number,” American Scientist 98(1): 24, doi:10.1511/2010.82.24. accessed April 9, 2015, http://www.americanscientist.org/issues/pub/a-magic-number). Therefore we use $T=5000$ as the threshold population in this project. Suppose that the environmental carrying capacity in Montana for elk is $25,000$. Set up the above equation using the carrying capacity of $25,000$ and threshold population of $5000$. Assume an annual net growth rate of $18\text{%}$.
2. Draw the direction field for the differential equation from step $1$, along with several solutions for different initial populations. What are the constant solutions of the differential equation? What do these solutions correspond to in the original population model (i.e., in a biological context)?
3. What is the limiting population for each initial population you chose in step $2?$ (Hint: use the slope field to see what happens for various initial populations, i.e., look for the horizontal asymptotes of your solutions.)
4. This equation can be solved using the method of separation of variables. However, it is very difficult to get the solution as an explicit function of $t$. Using an initial population of $18,000$ elk, solve the initial-value problem and express the solution as an implicit function of $t$, or solve the general initial-value problem, finding a solution in terms of $r,K,T,\text{and}{P}_{0}$. |
# Difference between revisions of "Talk:Introduction-TADM2E"
As some even number problem's solution has already been provided so please find solution for following
## Contents
First we'll verify the base case for $n = 1$ :
$\sum_{i=1}^1 i = \frac{1(1+1)}{2} = 1$
Now, we'll assume given statement is true up to $n - 1$.
So,
$\sum_{i=1}^{n-1} i = \frac{(n-1)(n-1+1)}{2} = \frac{(n-1)n}{2}$
To prove for the general case $n$,
$\sum_{i=1}^n i = \sum_{i=1}^{n-1} i + n = \frac{(n-1)n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}$
--Parth (talk) 05:07, 9 October 2015 (EDT)
First we'll verify the base case for $n = 1$ :
$\sum_{i=1}^1 i^3 = 1^3 = \frac{(1)^2(1+1)^2}{4} = 1$
Now, we'll assume given statement is true up to $n - 1$.
So, following statement is true.
$\sum_{i=1}^{n-1} i^3 = \frac{(n-1)^2(n-1+1)^2}{4} = \frac{(n-1)^2n^2}{4}$
To prove for the general case $n$,
\begin{align} \sum_{i=1}^{n} i^3 & = \sum_{i=1}^{n-1} i^3 + n^3 \\ & = \frac{(n-1)^2n^2}{4} + n^3 \\ & = \frac{(n^2 - 2n + 1)n^2 + 4n^3}{4} \\ & = \frac{n^4 - 2n^3 + n^2 + 4n^3}{4} \\ & = \frac{n^4 + 2n^3 + n^2}{4} \\ & = \frac{n^2(n^2 + 2n + 1)}{4} \\ \sum_{i=1}^{n} i^3 & = \frac{n^2(n + 1)^2}{4} \\ \end{align}
--Parth (talk) 05:07, 9 October 2015 (EDT)
First we'll verify the base case for $n = 1$ and we also assume $a=2$
$\sum_{i=0}^1 a^i = (2)^0 + (2)^1 = \frac{(2)^{1+1} - 1}{2 - 1} = 3$
Now, we'll assume given statement is true up to $n - 1$.
So, following statement is true.
$\sum_{i=0}^{n-1} a^i = \frac{a^{n-1+1}-1}{a-1} = \frac{a^n-1}{a-1}$
To prove for the general case $n$,
\begin{align} \sum_{i=0}^n a^i & = \sum_{i=0}^{n-1} a^i + a^n \\ & = \frac{a^n-1}{a-1} + a^n \\ & = \frac{a^n-1+a^n(a-1)}{a-1} \\ & = \frac{a^n-1+a^{n+1}-a^n}{a-1} \\ \end{align}
Thus,
$\sum_{i=0}^n a^i = \frac{a^{n+1}-1}{a-1}$
--Parth (talk) 05:07, 9 October 2015 (EDT)
## Solution 1.8 Missing
It looks like the solution to problem 1-8 was never written. I even tried going to http://algorist.com/algowiki/index.php/TADM2E_1.8, which doesn't exist. I would add it but, of course, I'm looking at this wiki specifically because I have no idea what I'm doing. |
Progressions
The progressions that we will see next are sequences of numbers of an ordered set of numbers formed according to a given law.
The only requirement for a set of numbers to be a sequence is that there is a formula or law with which it is possible to obtain any number from said sequence.
Let’s quickly look at an example. If we have that u_{n} represents the nth term of a sequence, then there must be an expression for u_{n} in terms of n, this means that said nth term must be a function of n. Taking the following as a succession:
3, \ 5, \ 7, \ \dots \ , 2n +1
u_{n} is a formula that allows us to obtain any term of the sequence.
u_{n}= 2n+1
Arithmetic progressions
An arithmetic progression is a sequence of numbers such that each of the terms following the first are obtained by adding a fixed number to the previous number, which is called the difference of the progression.
The example where our first term is a_{1}, our second term is a_{1} + d, the third term is a_{1} + 2d, etc… which can be represented as that the nth term is:
a_{n} = a_{1} +(n-1)d
Thanks to algebra and all the studies that exist of arithmetic progressions, we have the following theorem that says:
Arithmetic progression theorem
If in an arithmetic sequence a_{1} is the first term, a_{n} is the nth term, d is the difference and s_{n} is the sum of the first n terms , then we have the following two independent relationships:
a_{n} = a_{1} + \left(n - 1 \right)d
s_{n} = \cfrac{n}{2} \left(a_{1} + a_{n} \right)
In simpler terms, the expression a_{n} will help us to calculate the value in a position and the expression s_{n} will help us to calculate the sum of the consecutive numbers from the first value to the nth term that we have wanted to calculate.
Now, playing with these two relations, we can obtain another formula that all it does is substitute the a_{n} of the s_{n} formula for all the equality of the first relation:
s_{n} = \cfrac{n}{2} \left( a_{1} + \left[ a_{1} + \left(n - 1 \right)d \right]\right)
We simplify to get our second formula for s_{n}:
s_{n} = \cfrac{n}{2} \left[2a_{1} + \left(n - 1 \right)d \right]
Arithmetic progression example 1
In the arithmetic sequence 4, 6, 8, 10, 12, \dots, calculate the term in place 14 and calculate the sum of the first 14 terms.
Using the two relations that we have from the arithmetic progression theorem, we have that a_{1} = 4, the jumps in which it grows is two by two (d = 2) and we want the 14th place (n = 14):
a_{14} = a_{1} + (n-1)d = 4 + \left(14-1\right)\cdot 2 = 4 + 13\cdot 2 = 30
s_{14}=\cfrac{n}{2}\left(a_{1} + a_{n}\right) = \cfrac{14}{2} \left(4 + 30 \right) = 238
So our term 14 is a_{14}=30 and the sum of the first 14 terms is s_{14}=238.
Arithmetic progression example 2
In an arithmetic sequence where a_{1} = 2 and d = 4, how many terms must be taken for the sum to be 512?
Since we know that the sum of the terms must give us 512, we will take the formula obtained from the union of the first two formulas of our arithmetic progression theorem.
s_{n} = \cfrac{n}{2} \left[2a_{1} + \left(n - 1 \right)d \right]
We have a_{1}=2, d = 4 and s_{n} = 512:
512 = \cfrac{n}{2}\left[ 2\cdot 2 + \left( n - 1 \right) 4 \right]
512 = \cfrac{n}{2} \left[4 + 4n - 4\right]
512 = \cfrac{4n}{2} + \frac{4n^{2}}{2} - \cfrac{4n}{2} = 2n + 2n^{2} - 2n
512 = 2n^{2}
In this example, it was easy for us to find the result, since nothing else is to pass dividing the 2 and then apply the square root:
\cfrac{512}{2} = n^{2}
256 = n^{2}
And so we have our two results
And since n must be a positive integer, our result is 16.
Arithmetic progression example 3
Interpolate seven numbers between 10 and -14.
Since we have to find seven numebers, it means that our progression is going to be 9 terms, since they are the seven arithmetic terms and our extremes are 10 and -14. So we have a_{1} = 10 and a_{9} = -14. As we have our first relation of the arithmetic progression theorem:
a_{n} = a_{1} + \left(n-1\right)d
We substitute a_{1} and a_{n} as our last term a_{9} = -14:
-14 = 10 + \left(9 - 1 \right)d \quad \rightarrow \quad -24 = 8d
We spend dividing the 8 so that we have our difference:
d = -3
And since we already have the difference, we simply take our first term and we are going to subtract it to obtain our seven arithmetic terms between 10 and -14, which are the following:
10, \ 7, \ 4, \ 1, \ -2, \ -5, \ -8, \ -11, \ -14
Geometric progressions
A geometric progression is a sequence of numbers such that any term following the first one is obtained by multiplying the previous term by a non-zero term which we will call the ratio of the progression.
An example would be the following:
1, \ 2, \ 4, \ 8, \ \dots
Where a_{1} is the first term and r is the ratio. Therefore, we have that a_{n} represents the nth term, which would have our following formula:
a_{n} = a_{1}r^{n-1}
Since we have the expression to find the term in a certain position, I am going to write you the formula for the sum of the terms:
s_{n} = \cfrac{a_{1}\left(1 - r^{n} \right)}{1-r}, \qquad r \ne 1
Geometric progression theorem
If in a geometric progression a_{1} is the first term, a_{n} is the nth term, r is the ratio and s_{n} is the sum of the first n terms , then we have the two independent relationships mentioned above:
a_{n} = a_{1}r^{n-1}
s_{n} = \cfrac{a_{1}\left(1-r^{n}\right)}{1-r}, \qquad r=\ne 1
The first equality what we will do is multiply it by r to obtain the following:
ra_{n} = a_{1}r^{n-1}r = a_{1}r^{n-1+1} = a_{1}r^{n}
ra_{n} = a_{1}r^{n}
So all we do is substitute a_{1}r^{n} from our second equality of our theorem:
s_{n} = \cfrac{a_{1} - a_{1}r^{n}}{1-r} = \cfrac{a_{1} - ra_{n}}{1-r}, \qquad r\ne 1
Geometric progression example 1
In our first simple example we will simply substitute in the formulas. We have our geometric progression 1, \ 2, \ 4, \ 8, \ \dots, in which we want to find the ninth term and the sum of the first nine terms:
We have that a_{1} = 1, r = 2, n = 9, therefore, we will directly apply the formulas:
a_{n} = a_{1}r^{n-1}
a_{9} = 1\dot 2^{8} = 256
a_{9} = 256
And now we will calculate the sum of the first nine terms:
s_{n} = \cfrac{a_{1}\left(1-r^{n}\right)}{1-r}
s_{9} = \cfrac{1\left(1-2^{9}\right)}{1-2} = \cfrac{1\left(1-512\right)}{-1}=\cfrac{1\left(-511\right)}{-1} = \cfrac{-511}{-1}
s_{9} = 511
Geometric progression example 2
Our first term is 4, the last term is \frac{15625}{16}, and the sum of the terms is \frac{25999}{16}. Find the ratio and the number of terms.
We have a_{1} = 4, a_{n} = \frac{15625}{16} and s_{n} = \frac{25999}{16}, so what we want to calculate is r and n. For this progression exercise we will use one of our formulas that are mentioned in the theorem:
s_{n} = \cfrac{a_{1} - ra_{n}}{1-r}
We substitute all the data we have to find r:
\cfrac{25999}{16} = \cfrac{4 - \left( \frac{15625}{16}\right)r}{1 - r}
And now that we have our expression, all we have to do is solve it to get the result of r:
\cfrac{25999}{16} - \cfrac{25999}{16}r=4 - \cfrac{15625}{16}r
\cfrac{25999}{16} - 4 = \cfrac{25999}{16}r - \cfrac{15625}{16}r
\cfrac{25935}{16} = \cfrac{5187}{8}r
\cfrac{8}{5187}\cdot \cfrac{25935}{16} = r
r = \cfrac{5}{2}
And now to find the value of n, we will use the formula a_{n} = a_{1}r^{n-1}, since we already have r, we will only need to solve the equation to find n:
\cfrac{15625}{16} = 4\left(\cfrac{5}{2}\right)^{n-1}
\cfrac{1}{4}\cdot \cfrac{15625}{16} = \left(\cfrac{5}{2}\right)^{n-1}
\cfrac{15625}{64} = \left(\cfrac{5}{2}\right)^{n-1}
In this part comes an algebraic trick, what we will do is look for a number that raised to the power leaves us the left part of the equation with the same base as the right part of the equation, what I mean is that we will look for a number (if we can do it by trial and error), that raising \frac{5}{2} to the x power, can give us \frac{15625}{64}. We already took the trouble to look for it and it is 6:
\left(\cfrac{5}{2}\right)^{6} = \left(\cfrac{5}{2} \right)^{n-1}
And since we have that in our equality we have \frac{5}{2} of both sides, we can now equal our exponents:
6 = n-1
n=7
Giving us as a result that r = \cfrac{5}{2} and n = 7.
Geometric progression example 3
Interpolate seven geometric terms between \frac{1}{18} and \frac{6561}{18}.
What we have to do is find seven numbers that make up a geometric progression, those numbers start with the first number \frac{1}{18} and end with the number \frac{6561}{18}. Which means that our geometric progression has nine terms.
So we have a_{1} = \frac{1}{18} and a_{n} = a_{9} = \frac{6561}{18}. We will use the following equation and find the value of the ratio:
a_{n} = a_{1}r^{n-1}
We substitute values and solve:
\cfrac{6561}{18} = \cfrac{1}{18}r^{9-1}
\cfrac{6561}{\cancel{18}} \cdot \cfrac{\cancel{18}}{1} = r^{8}
6561= r^{8}
r = 3
And now that we are right, we simply apply the formula to determine each of the values:
So we have that as an answer we have that the ratio is r = 3 and our seven values are: \frac{3}{18}, \frac{9}{18}, \frac{27}{18}, \frac{81}{18}, \frac{243}{18}, \frac{729}{18} and \frac{2187}{18}.
Harmonic progressions
The harmonic progression is a succession of numbers whose reciprocals form an arithmetic progression.
A simple example would be: \frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \dots, \frac{1}{n}, \dots since 2, 4, 6, 8, \dots, 2n, \dots is an arithmetic progression . Remember that they are only reciprocals, so the numerator always has to be the same.
We will not see examples of the harmonic progression since they are solved the same as arithmetic progressions. What we will see are several formulas that will be used for all the progressions.
If we interpolate a single harmonic mean between two numbers, we will obtain the harmonic mean. What we mean is that a and b are given numbers, so H is our harmonic mean:
So with this we can obtain another theorem that will be very useful to us
Theorem
If A, G, and H are the arithmetic mean, geometric mean, and harmonic mean, respectively, of two different positive numbers a and b, the following formulas hold: |
## Do you add or multiply probabilities?
When we calculate probabilities involving one event AND another event occurring, we multiply their probabilities..
## How do you calculate independent probability?
Events A and B are independent if the equation P(A∩B) = P(A) · P(B) holds true. You can use the equation to check if events are independent; multiply the probabilities of the two events together to see if they equal the probability of them both happening together.
## How do you find the probability of A or B independent?
Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B). If the probability of one event doesn’t affect the other, you have an independent event. All you do is multiply the probability of one by the probability of another.
## How do you determine if it is permutation or combination?
The difference between combinations and permutations is ordering. With permutations we care about the order of the elements, whereas with combinations we don’t. For example, say your locker “combo” is 5432. If you enter 4325 into your locker it won’t open because it is a different ordering (aka permutation).
## What is multiplication principle in permutations and combinations?
The multiplication principle allows us to count the number of ways to complete a sequence of tasks by multiplying together the number of ways to complete each task. 2. A permutation is a specific ordering of some objects. 3.
## Why do we multiply the probability of independent events?
It’s multiplication because you’re trying to find the probability inside another probability. First probability is %50, and then inside of this probability %50’s %50 is %25 which 0.5 * 0.5 = 0.25 = %25. ( If you’ve added these together, 1/2 + 1/2 = 2/2 = 1, which would be meaningless, right?
## Do you add first or multiply first?
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. Next, add and subtract from left to right.
## How do you tell if an event is independent or dependent?
We call events dependent if knowing whether one of them happened tells us something about whether the others happened. Independent events give us no information about one another; the probability of one event occurring does not affect the probability of the other events occurring.
## How do you prove that A and B are independent?
If A and B are independent events, then the events A and B’ are also independent. Proof: The events A and B are independent, so, P(A ∩ B) = P(A) P(B). From the Venn diagram, we see that the events A ∩ B and A ∩ B’ are mutually exclusive and together they form the event A.
## How many combinations of 3 items are there?
3*3*3=27 unique possibilities. |
# Question Video: Finding πth Roots of Unity in Algebraic Form Mathematics
Find the cubic roots of unity and plot them on an Argand diagram.
02:25
### Video Transcript
Find the cubic roots of unity and plot them on an Argand diagram.
Finding the cubic roots of unity is like saying what are the solutions to the equation π§ cubed equals one. To find them, we can use the general formula for the πth roots of unity. Thatβs cos of two ππ over π plus π sin of two ππ over π for integer values of π between zero and π minus one. Since weβre finding the cubic roots of unity, in this example, our value of π is three, which means π will take the values zero, one, and two.
Letβs begin with the case when π is equal to zero. This root is cos of zero plus π sin of zero. Well, cos of zero is one. And sin of zero is zero. So the first root is one. And it makes complete sense if we think about it that a solution to the equation π§ cubed equals one would be one.
Next, we let π be equal to one. This root is cos of two π by three plus π sin of two π by three. And in exponential form, thatβs π to the two π by three π. Finally, we let π be equal to two. The root here is cos four π by three plus π sin of four π by three. Notice though that the argument for this root is outside of the range for the principal argument. We therefore subtract two π from four π by three to get negative two π by three. So our third and final root is cos of negative two π by three plus π sin of negative two π by three. Or in exponential form, thatβs π to the negative two π by three π.
Now that we have the cubic roots of unity, we need to plot them on an Argand diagram. There are two ways we could approach this problem. We could convert each number to algebraic form. Thatβs one, negative a half plus root three over two π, and negative a half minus root three over two π. These are plotted on the Argand diagram as shown. Alternatively, we couldβve used the modulus and argument of each root.
Either way, letβs notice that the points that represent these complex numbers are the vertices of an equilateral triangle. This triangle is inscribed in a unit circle whose centre is the origin. Actually, an interesting geometrical property of the πth roots of unity is that, on an Argand diagram, they are all evenly spaced about the unit circle whose centre is the origin. They form a regular π-gon. It has a vertex at the point whose Cartesian coordinates are one, zero. |
# The Formula For Work: Physics Equation With Examples
In physics, we say that a force does work if the application of the force displaces an object in the direction of the force. In other words, work is equivalent to the application of a force over a distance. The amount of work a force does is directly proportional to how far that force moves an object. The general formula for work and for determining the amount of work that is done on an object is:
• WF × D × cos(Θ)
where is the amount of work, F is the vector of force, D is the magnitude of displacement, and Θ is the angle between the vector of force and the vector of displacement. The SI unit for work is the joule (J), and its dimensions are kg•m2/s2. Another way to understand it is that one joule is equivalent to the amount of energy transferred when one newton of force moves an object a distance of one meter.
## Formula For Work
Whenever a force moves an object, we say that work has been done. When a ball rolls down a hill due to the force of gravity, when you pick up your backpack off the ground, when your car’s internal engine applies a force to make your wheels move; all of these events involve a force moving an object over a distance and so involve some work. In cases where a force is applied to an object but does not move it, no work has been done. So the force from a person pushing the side of a skyscraper is not doing any work as the skyscraper does not move. Let’s consider some simple examples to illustrate the concept of work.
## Example problems
### (1)
A 100 Newton force is applied to a 15kg box in the horizontal direction and moves it 5 meters horizontally. How much work was done?
In this case, we know the force is 100 N and the distance is 5 meters. We also know that since the force is applied in the same direction as the displacement, Θ is equal to 0. So we plug these values into our equation
• WF × D × cos(Θ)
and get:
• W = 100(5)cos(0)= 500 J
So, the 100 N force did 500 joules work moving the block 5 meters.
### (2)
There is a 2kg book lying on a table. A 64 N force is applied to the book at a 120° angle from horizontal and moves the book 3 meters in the horizontal direction. How much work was done?
In this case, we know the force of 64 N and the distance of 3 m. We also know that there is a 120°angle between the angle of the direction of applied force and the direction of motion. So plugging these values into our handy equation yields:
• W = 64(3)cos(120)= 156.32 Joules
So the 64 N force at a 120° angle did 156.32 joules of work moving the book 3 meters.
### (3)
Linda takes a 300 N suitcase up 3 flights of stairs for a total vertical distance of 16 meters. She then pushes the suitcase with 100 N of force the remaining 8 meters to her hotel room. How much work was done over her whole trip?
This question requires 2 separate steps. There are two main portions of her trip, so we can calculate the work done over each portion individually, then, combine the two values to get the total amount of work done. For the first part of her trip, she exerts 300 N of force to move a suitcase 16 meters vertically so the amount of work done is:
• W1=300(16)cos(0)=4800 Joules
So the first part of the trip did 4800 joules. For the second part, we know that a 100N force moved the suitcase horizontally 8 meters, so the total amount of work done on the second portion of the trip is:
• W2=100(8)cos(0)=800 Joules
Combining the two values from each portion of the trip yields:
• WtotalW1+W2= 4800+800= 5600 Joules
So over the entire course of Linda’s trip, 5600 Joules of work was done.
## Work/Energy Relationship
Work and energy in physics share a close relationship. According to the work-energy principle, an increase in a rigid body’s kinetic energy is caused by an equal amount of work done on that body by a force applied to that body. In more mathematical terms, the relationship can be expressed as:
• W = KEfinal−KEinitial
where KE stands for kinetic energy. In other words, the change in kinetic energy of a body is equal to the amount of work done on that body. In general, the formula for the kinetic energy of an object is:
• KE = (1/2) kg*v2
where v stands for an object’s velocity. The unit for kinetic energy is the same unit for work, the joule. Let’s look at some problems to examine these mathematical relationships.
### (4)
Donkey and Diddy Kong are sitting in a 90-kilogram minecart that is initially traveling horizontally at a velocity of 5 m/s. Rambi the rhino pushes the minecart from behind and speeds it up so it is now traveling 11 m/s. How much work did Rambi do on the minecart?
In order to solve this problem we first need to figure out the minecart’s initial kinetic energy and its final kinetic energy. Once we know those values, we can determine the total amount of work. We know both the velocity and mass of the minecart so we can determine the total kinetic energy at the beginning and at the end. The initial kinetic energy of the minecart is:
• KEinitial=(1/2)(90)(5)21125 J
The final kinetic energy of the minecart is
• KEfinal=(1/2)(90)(11)2=5445 J
Therefore, the total amount of work done on the minecart is 5545−1125= 4420 J.
“Science is the knowledge of consequences, and dependence of one fact upon another.” – Thomas Hobbes
### (5)
A car weighing 1300 kg is moving with a velocity of 18 m/s. If 60000 joules of work is done on the car, what will its final velocity be?
The question will require a bit of algebra. First, we must determine the initial kinetic energy of the car. The initial kinetic energy of the car is:
• (1/2)(1300)(18)2=210600 J
Since we know the total amount of work done on the system (60000 J) we can figure out the car’s final kinetic energy:
• 60000=KEfinal−210600
• 270600=KEfinal
Now, since we know the final kinetic energy and the mass of the car, we can determine its final velocity like this
• KEfinal=(1/2)kg*v2
• 270600 = (1/2)(1300)v2
• 270600 = 650v2
• 416.3 = v2
• v20.4 m/s
The car’s final velocity will be 20.4 m/s.
So in summation, we say work is done whenever a force moves an object over a distance. The magnitude of work is equal to the magnitude of force multiplied by the distance traveled. Work and kinetic energy are tightly intertwined and can be used to determine each other.
1. Sasha says:
How much work is done by a 60kg man in climbing a stair of 12 steps, each steps being 6 inches higher than the previous one?
2. Anonymous says:
The problem with 120 degree angle of force applied does not equal the number shown (its negative)
1. anonymous says:
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# Easy Ways to Calculate Percentages
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How much is 15 percent of 210? How about 75 percent of 440 or 350 percent of 20? The standard way of calculating percentages can make these problems difficult to do without a calculator. Luckily, there are several tips and tricks you can use to figure out percentages without a calculator. Once you understand these simple methods, these questions become easy to solve.
## What Does 'Percent' Mean?
• It is valuable to know the origin of the term percent if you want to truly understand how to calculate percentages. The word percent comes from the phrase per cent. Cent is a root that means one hundred, so per cent literally means per one hundred. For example, if you know that 30 percent of the students in a school are boys -- that means that there are 30 boys per one hundred students. Another way to say this is 30 out of 100 students are boys.
Generally, the way to figure out any percentage is to multiply the number of items in question, or X, by the decimal form of the percent. To figure out the decimal form of a percent, simply move the decimal two places to the left. For example, the decimal form of 10 percent is 0.1. Then, to calculate what 10 percent of is, say, 250 students, simply multiply the number of students by 0.1.
10 percent of 250 students = 0.1 × 250 students = 25 students
## Some Easy Percentages to Remember
• If you can commit the percentages below to memory, you will be able to calculate a wide variety of percentages without a calculator.
50 percent means half. To calculate 50 percent of a number, simply divide it by 2. For example, 50 percent of 26 is 26 divided by 2, or 13.
25 percent means one fourth. To calculate 25 percent of a number, simply divide it by 4. For example, 25 percent of 12 is 12 divide by 4, or 3.
10 percent means one tenth. To calculate 10 percent of a number, simply divide it by 10 or move the decimal point one place to the left. For example, 10 percent of 230 is 230 divided by 10, or 23.
5 percent is one half of 10 percent. To calculate 5 percent of a number, simply divide 10 percent of the number by 2. For example, 5 percent of 230 is 23 divided by 2, or 11.5.
## Using the Basic Percentages to Calculate Other Percentages
• So how much is 15 percent of 210? You can use the four key percentages you memorized to figure this out. Consider that 15 percent is 5 percent added to 10 percent. Since 10 percent of 210 is 21, and 5 percent is half that, or 10.5, then 15 percent is 21 added to 10.5, or 31.5.
How about 75 percent of 440? Here you can figure that 75 percent is 50 percent added to 25 percent. Since 50 percent of 440 is half that, or 220, and 25 percent of 440 is one fourth, or 110, then 75 percent is 220 added to 110, or 330.
In this way you can combine 5 percent, 10 percent, 25 percent, and 50 percent to calculate a wide variety of percentages in your head. If you need to calculate a percentage that is not a multiple of 5, you can use this technique to estimate the answer very closely.
## Calculating Percentages Greater Than 100 Percent
• To calculate a percentage greater than 100 percent, simply move the decimal of the percent two spaces to the left and use the answer to multiply the number in question. You can do this to figure out what is 350 percent of 20. Moving the decimal of the percent two spaces to the left gives you 3.5. Multiplying 20 by 3.5 then gives you the answer, which is 70.
## References
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# How to solve inequality equations
In this blog post, we will provide you with a step-by-step guide on How to solve inequality equations. Our website will give you answers to homework.
## How can we solve inequality equations
In this blog post, we will be discussing How to solve inequality equations. Matrix equations are a type of math problem that can be very difficult to solve. In a matrix equation, the variables are represented by squares, and the coefficients are represented by numbers. The goal is to find the values of the variables that make the equation true. To do this, you need to use a process called row reduction. Row reduction is a method of solving matrix equations in which you simplify the equation by adding or subtracting rows until you have only one variable left. This can be a difficult process, but there are some tricks that can make it easier. For example, try to choose rows that have coefficients that cancel out when they are added or subtracted. You can also use row reduction to solve systems of linear equations. A system of linear equations is a set of two or more equations that share the same variables. To solve a system of linear equations, you need to find the values of the variables that make all of the equations true. This can be done by either solving each equation individually or using row reduction to simplify the system into a single equation. Either way, solving matrix equations can be a challenge, but it is possible with some practice.
Partial fractions is a method for decomposing a fraction into a sum of simpler fractions. The process involves breaking up the original fraction into smaller pieces, each of which can be more easily simplified. While partial fractions can be used to decompose any fraction, it is particularly useful for dealing with rational expressions that contain variables. In order to solve a partial fraction, one must first determine the factors of the denominator. Once the factors have been determined, the numerator can be factored as well. The next step is to identify the terms in the numerator and denominator that share common factors. These terms can then be combined, and the resulting expression can be simplified. Finally, the remaining terms in the numerator and denominator can be solve for using basic algebraic principles. By following these steps, one can solve any partial fraction problem.
The app, called Mathway, allows users to enter a problem and then see step-by-step instructions for solving it. In addition, the app includes a wide range of features that make it easy to use, including a built-in calculator and a library of solved problems. As a result, Mathway is an essential tool for any student who wants to improve their math skills.
There are a lot of different algebra apps out there, but which one is the best? It really depends on your individual needs and preferences. Some people prefer an app that provides step-by-step solutions, while others prefer one that allows them to work through problems at their own pace. There are also apps designed specifically for students who are struggling with algebra, as well as those who are looking for extra practice. Ultimately, the best algebra app for you is the one that meets your specific needs and helps you to succeed in your studies.
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# Properties of Cylinders
This lesson will work with the circumference and area of a circle as well as the surface area and volume of a cylinder.
### Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
## Investigating the Volume of Different Cylinders
Consider the three-dimensional figure given in the diagram. By dragging the point, the figure can be skewed. However, its height and the radius of the circles that make the base and the top of the cylinder remain the same. Think about the volume — space occupied — of the solid. Does the volume change when the solid is skewed?
## Definition of a Circle
Before moving on, the definition of a circle and some of its parts will be reviewed.
A circle is the set of all the points in a plane that are equidistant from a given point. There are a few particularly notable features of a circle.
• Center - The given point from which all points of the circle are equidistant. Circles are often named by their center point.
• Radius - A segment that connects the center and any point on the circle. Its length is usually represented algebraically by r.
• Diameter - A segment whose endpoints are on the circle and passes through the center. Its length is usually represented algebraically by d.
• Circumference - The perimeter of a circle, usually represented algebraically by C.
In the applet, the center is labeled O. Therefore, the circle can be referred to as or circle O.
In any given circle, the lengths of any radius and any diameter are constant. They are called the radius and the diameter of the circle, respectively.
Next, the formula for calculating the circumference of a circle will be discussed.
## Circumference of a Circle
The circumference of a circle is calculated by multiplying its diameter by π.
C=πd
This can be visualized in the following diagram.
Since the diameter is twice the radius, the circumference of a circle can also be calculated by multiplying 2r by π.
C=2πr
### Proof
Consider two circles and their respective diameters and circumferences.
By the Similar Circles Theorem, all circles are similar. Therefore, their corresponding parts are proportional.
This proportion can be rearranged.
Rearrange equation
Therefore, for any two circles, the ratio of the circumference to the diameter is always the same. This means that this ratio is constant. This constant is defined as π. With this information, it can be shown that the circumference of a circle is the product between its diameter and π.
Finally, the formula for calculating the area of a circle will be seen.
## Area of a Circle
The area of a circle is the product of π and the square of its radius.
### Proof
Informal Justification
A circle with radius r will be divided into a number of equally sized sectors. Then, the top and bottom halves of the circle will be distinguished by filling them with different colors. Because the circumference of a circle is 2πr, the arc length of each semicircle is half this value, πr.
Now, the above sectors will be unfolded. By placing the sectors of the upper hemisphere as teeth pointing downwards and the sectors of the bottom hemisphere as teeth pointing upwards, a parallelogram-like figure can be formed. Therefore, the area of the figure below should be the same as the circle's area.
It can be noted that if the circle is divided into more and smaller sectors, then the figure will begin to look more and more like a rectangle.
Here, the shorter sides become more vertical and the longer sides become more horizontal. If the circle is divided into infinitely many sectors, the figure will become a perfect rectangle, with base πr and height r. Since the area of a rectangle is the product of its and its the following formula can be derived.
It was shown that the area of a circle is the product of π and the square of its radius.
## Finding the Circumference and Area of a Circle
Izabella loves to use geometry in her art. Her school noticed her skills and hired her to paint the school's soccer field — for a substantial payment. The diagram shows how the field should look.
Izabella needs the field's measurements. She already knows the radius of the circle located at the center of the field to be 6.5 meters. For logistical reasons, she wants to find the circumference and area of this circle. Help Izabella calculate these values to one decimal place.
### Hint
The circumference is twice the product of π and the radius. The area is the product of π and the square of the radius.
### Solution
The circumference of a circle with radius r is twice the product of π and r. Furthermore, its area is the product of π and r2. Since the radius is 6.5 meters, the circumference and area can be calculated.
Formula Substitute Simplify Approximate
Circumference C=2πr C=2π(6.5) C=13π
Area A=πr2 A=42.25π
The circumference and area of the circle located at the center of the soccer field are about 40.8 meters and 132.7 square meters, respectively.
## Investigating the Area of a Circle
Maya bought 20 meters of fencing with the hopes of constructing a circular dog run for her dog Opie. Because Opie is a large Saint Bernard, she wants the area of the dog run to be at least 30 square meters.
With a circumference of 20 meters, is the area of the dog run at least 30 square meters?
### Hint
Start by finding the radius of the circle.
### Solution
Because the dog run is in the shape of a circle, its area is the product of π and the square of the radius. Therefore, to find the area of the circle, the first step is to find the radius. Since it is already known that the circumference is 20 meters, the formula for the circumference will be used to find the radius.
Since Maya bought and used 20 meters of fencing, the circumference measures 20 meters. This value can be substituted in the above formula to find the radius r.
C=2πr
20=2πr
Solve for r
10=πr
With this information, the area of the dog run can be calculated.
To do so, 3.18 will be substituted for r in the formula for the area of a circle.
A=πr2
Evaluate right-hand side
A=π(10.1124)
A=31.769041
The area of Opie's dog run is about 31.8 square meters. This is enough for him to have happy and healthy playtime!
## Practice Finding the Circumference and Area of a Circle
For the following questions, approximate the answers to one decimal place. Do not include units in the answer.
## Cavalieri's Principle
Now, two-dimensional figures will be left aside to move forward into solids, which are three-dimensional shapes.
Two solids with the same height and the same cross-sectional area at every altitude have the same volume. This means that, as long as their heights are equal, skewed versions of the same solid have the same volume.
If V1 and V2 are the volumes of the above solids, then they are equal.
V1=V2
### Proof
Informal Justification
This principle will be proven by using a set of identical coins. Consider a stack in which each of these coins is placed directly on top of each other. Consider also another stack where the coins lie on top of each other, but are not aligned.
The first stack can be considered as a right cylinder. Similarly, the second stack can be considered as an oblique cylinder, which is a skewed version of the first cylinder. Because the coins are identical, the cross-sectional areas of the cylinders at the same altitude are the same.
Since the coins are identical, they have the same volume. Furthermore, since the height is the same for both stacks, they both have the same number of coins. Therefore, both stacks — cylinders — have the same volume. This reasoning is strongly based on the assumption that the face of the coins have the same area.
## Definition of a Cylinder
A cylinder is a three-dimensional figure with two parallel bases that are congruent circles connected by a curved surface.
The axis of a cylinder is the segment that connects the center of the bases. The height of a cylinder is the perpendicular distance between the bases. The radius of the cylinder is the radius of a base.
If the axis of a cylinder is not perpendicular to the bases, then the cylinder is said to be an oblique cylinder.
If the dimensions of a cylinder are known, then its volume and surface area can be calculated.
## Volume of a Cylinder
The volume of a cylinder is calculated by multiplying the base's area by its height.
If the radius of the circular base of a cylinder is r, the volume can be calculated by the following formula.
V=πr2h
### Proof
Informal Justification
Consider a rectangular prism and a right cylinder that have the same base area and height.
In this case, the cross-sections of the prism and the cylinder are congruent to their bases. Therefore, their cross-sectional areas at every altitude are equal.
By the Cavalieri's Principle, two solids with the same height and the same cross-sectional area at every altitude have the same volume. Therefore, the volume of the cylinder is the same as the volume of the prism.
Furthermore, the volume of a prism can be calculated by multiplying the area of its base by its height.
By the Transitive Property of Equality, a formula for the volume of the cylinder can be written.
Finally, not only is B the area of the base of the prism, but also the area of the base of the cylinder. Since the base of the cylinder is a circle, its area is the product of π and the square of its radius r. Therefore, B=πr2 can be substituted in the above formula.
This formula applies to all cylinders because there is always a prism with the same base area and height. Also, by the Cavalieri's principle, this formula still holds true for oblique cylinders.
## Finding the Volume of a Cylinder
LaShay loves golfing. She is about to buy a new golf bag.
The the bag she is thinking about buying is a shaped like a cylinder with a height of 132 centimeters and its radius is 15 centimeters. For all of her golf clubs to fit in the bag, its volume must be at least 90000 cubic centimeters. Therefore, she wants to calculate the volume of the bag. Help her do this, approximating the answer to three significant figures.
### Hint
The volume of a cylinder is the product of π, the square of its radius, and its height.
### Solution
The golf bag is in the shape of a cylinder. Therefore, to calculate its volume, the formula for the volume of a cylinder can be used.
It is known that h=132 and r=15. These two values can be substituted into the formula.
V=πr2h
Evaluate right-hand side
V=π(225)(132)
V=29700π
The volume of LaShay new golf bag is about 93300 cubic centimeters. This is enough to keep all of her golf clubs!
## Finding the Radius of a Cylinder
Kriz is making an experiment to complete a Chemistry project. He is using a test tube in the shape of a cylinder with a height of 75 millimeters and a volume of 20000 cube millimeters.
For this tube to suit the experiment, it radius must not be greater than 9 millimeters. Help Kriz find the radius! Approximate the answer to one decimal place.
### Hint
The formula for the volume V of a cylinder is V=πr2h, where r and h are the radius of the base and the height of the cylinder, respectively.
### Solution
The volume of a cylinder is the product of π, the square of the base's radius, and the height of the cylinder.
It is known that the height and the volume of the cylinder are 75 millimeters and 20000 cube millimeters, respectively. These two values can be substituted into the formula for volume of a cylinder, which can then be solved for r.
V=πr2h
20000=πr2(75)
Solve for r
20000=75πr2
When solving the equation, only the principal root was considered. This is because the radius of a circle is always positive. Therefore, the radius of the cylinder's base is about 9.2 millimeters. Since the radius is greater than 9 millimeters, the tube does not suit the experiment.
## Surface Area of a Cylinder
Besides the radius, height, and volume, another essential characteristic of a cylinder is its surface area. The surface area of a solid is the measure of the total area that the surface of the solid occupies.
Consider a cylinder of height h and radius r.
The surface area of this cylinder is given by the following formula.
S=2πrh+2πr2
### Proof
The cylinder's surface area can be seen as three separate parts that are top, bottom, and side. The area of the side has the shape of a rectangle of width h. The top and the bottom are circles of radius r.
Since the top and the bottom are congruent circles, they have the same area.
To find the area of the rectangle, its length must be found first. The length of the rectangle that forms the lateral part of the cylinder is the circumference of the base, which is 2πr.
Therefore, the area of the rectangle is the product of h and 2πr.
The surface area of the cylinder S is the sum of the areas of the rectangle and the circles.
S=2πrh+2πr2
## Finding the Surface Area of a Cylinder
Jordan's father is giving her a new baseball bat for her birthday. To avoid damage to the bat, Papa Jordan will store the bat in a box that has the shape of a cylinder with a height of 30 inches and a radius of 1.6 inches. Since this is a present for Jordan, Papa will use wrapping paper to make it even more special.
Ignoring any paper overlapping, what is the minimum area of paper that Jordan's father needs? Approximate the answer to one decimal place.
### Hint
The area of wrapping paper Jordan's father needs is the same as the surface area of the cylinder.
### Solution
To find the minimum area of paper that Jordan's father needs, the surface area S of the cylinder needs to be calculated.
Here, r and h are the radius and the height of the cylinder, respectively. The radius is 1.6 inches and the height is 30 inches. Therefore, these values can be substituted in the above formula, which can then be simplified.
S=2πrh+2πr2
Evaluate right-hand side
S=2π(1.6)(30)+2π(2.56)
S=2(1.6)(30)π+2(2.56)π
S=96π+5.12π
S=101.12π
## Practice Finding the Surface Area and Volume of a Cylinder
Find the volume or surface area of the cylinder.
## Volume of Skewed Solids
The challenge presented at the beginning of this lesson asked whether the volume of a cylinder changes if the solid is skewed.
The Cavalieri's Principle was studied in this lesson. This principle states that two solids with the same height and the same cross-sectional area at every altitude have the same volume. Therefore, if their heights are equal, skewed versions of the same solid have the same volume. This means that the volume of a cylinder does not change even if the cylinder is skewed. |
# Find the area of Region of Polar Coordinate We have the polar equations r = 1/2 and r =...
## Question:
Find the area of Region of Polar Coordinate
We have the polar equations r = 1/2 and r = cos({eq}\theta {/eq})
The shaded region we're interested in is outside r=1/2 and inside r=({eq}\theta {/eq}).
The bounds are {eq}\pi/ 3 {/eq} and {eq}5 \pi /3 {/eq}.
The equation is 2 times the integral of {eq}(1/2)(cos(\theta))^2 - (1/2)^2 ) d(\theta) {/eq}.
## Integration:
A mechanical quantity that represents the joining of the differential of the function or equation in the definite form to analyse the mathematical problem is known as integration. It used in engineering applications.
GIVEN DATA:
• The polar equation is: {eq}r = \dfrac{1}{2} {/eq}
• {eq}r = \cos \theta {/eq}
• {eq}\theta = \dfrac{\pi }{3} {/eq}
• {eq}\theta = \dfrac{{5\pi }}{3} {/eq}
• The equation for area of region is: {eq}A = 2\int {\left( {\dfrac{1}{2}{{\left( {\cos \theta } \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} \right)d\theta } {/eq}
The domain is
{eq}\dfrac{\pi }{3} \le \theta \le \dfrac{{5\pi }}{3} {/eq}
Substitute the value in equation of area of region with respective region
{eq}\begin{align*} A &= 2\int_{\dfrac{\pi }{3}}^{\dfrac{{5\pi }}{3}} {\left( {\dfrac{1}{2}{{\left( {\cos \theta } \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} \right)d\theta } \\ &= 2\int_{\dfrac{\pi }{3}}^{\dfrac{{5\pi }}{3}} {\left( {\dfrac{1}{2}\left( {\dfrac{{1 + \cos 2\theta }}{2}} \right) - \left( {\dfrac{1}{4}} \right)} \right)d\theta } \\ &= 2\left[ {\dfrac{1}{4}\left( {\theta + \dfrac{{\sin 2\theta }}{2}} \right) - \dfrac{\theta }{4}} \right]_{\dfrac{\pi }{3}}^{\dfrac{{5\pi }}{3}}\\ &= 2\left[ {\dfrac{1}{4}\left( {\left( {\dfrac{{5\pi }}{3} + \dfrac{{\sin 2\left( {\dfrac{{5\pi }}{3}} \right)}}{2}} \right) - \left( {\dfrac{\pi }{3} + \dfrac{{\sin 2\left( {\dfrac{\pi }{3}} \right)}}{2}} \right)} \right) - \left( {\dfrac{{\left( {\dfrac{{5\pi }}{3}} \right)}}{4} - \dfrac{{\left( {\dfrac{\pi }{3}} \right)}}{4}} \right)} \right]\\ &= 2\left[ {\dfrac{1}{4}\left( {\dfrac{{5\pi }}{3} - \dfrac{\pi }{3} + 0.0908 - 0.0182} \right) - \left( {\dfrac{{5\pi }}{{12}} - \dfrac{\pi }{{12}}} \right)} \right]\\ &= 0.0363 \end{align*} {/eq}
Thus the area of region of polar coordinate is {eq}0.0363 {/eq} |
0
# How do you solve Multi-Step Inequalties?
### 1 Answer by Expert Tutors
Taunja S. | Experienced Tutoring Queen Making Learning EasyExperienced Tutoring Queen Making Learni...
5.0 5.0 (5 lesson ratings) (5)
0
Multi-Step Inequalities can seem very intimidating but once you know the rules it's not as hard as you think. Multi-Step Inequalities can be solved the same as any Multi-Step Equation.
Here is an example for how to solve a multi-step inequality:
2x + 7 > 49
Goal is to get x by itself. Find out what x is greater than and/or equal to.
Step #1: To find out what x is you must eliminate the extra numbers on the side of the variable. Use Inverse (Opposite) Operation to accomplish this.
2x + 7 > 49
Subtract 7 from both sides. 2x +7 > 49
-7 -7
Then you have 2x > 42
Use Inverse Operation again by dividing 2 from both sides to get x by itself.
So then you have 2x/2 > 42/2
Your Answer is: x > 21 (Any # that is 21 or above is x)
Now let's say you have -2x + 7 > 49
Rules change when you have negative integers (numbers) and you have to multiply or divide by them then you change the direction of your inequality symbol.
So let's see here
-2x +7 > 49
Use Inverse Operation by subtracting 7 from both sides.
-2x +7 > 49
-7 -7
Now you have: -2x > 42
Use Inverse Operation by dividing -2 from both sides.
Special RuleWhen you multiply or divide by a negative number, SWITCH Your Sign.
-2x/-2 > 49/-2
Your Answer: x < -21 (x is equal to or less than -21)
Christian I hope that these examples help you to better understand how to solve multi-step inequalities. Please let me know if you need more examples. If you need help with multiplying or dividing integers I have an easy tip for that as well. Looking forward to hearing from you.
Thank for the opportunity to be of assistance to you.
Taunja Smith, WyzAnt Tutoring Queen |
# The Simple Guide to Finding One-Fourth of 1360 | Easy Math Tips and Tricks
## Introduction
Many people find math intimidating, especially when they have to solve problems that involve fractions. One such problem is finding one-fourth of 1360. But fear not! In this article, we will break down the steps and provide simple tips to help you solve this problem with ease.
## Demystifying Fractions: What One-Fourth of 1360 Really Means
Before we can solve the problem, we need to understand what a fraction is and how it relates to the problem at hand. A fraction represents a part of a whole. In this case, we are trying to find one-fourth of 1360, which means we are looking for one-fourth of the total amount.
To understand what it means to find one-fourth of a number, let’s take a simple example. Suppose you have a pie that is cut into four equal slices, and you eat one of those slices. You have just eaten one-fourth of the pie. Similarly, to find one-fourth of a number, you have to divide the total amount into four equal parts and take one of those parts.
To find one-fourth of 1360, we can divide 1360 by 4:
1360 รท 4 = 340
So, one-fourth of 1360 is equal to 340.
## Unlocking the Mystery: How to Easily Calculate One-Fourth of 1360
While dividing by 4 is a simple method for finding one-fourth of a number, there are other ways to approach the problem. One way is to use multiplication. To find one-fourth of a number, you can multiply the number by 0.25.
Here’s the formula:
1/4 x number = one-fourth of the number
So, to find one-fourth of 1360 using this method:
1/4 x 1360 = 340
This method can be helpful when dealing with larger numbers or when you need to find fractions of fractions.
## How Financial Experts Use One-Fourth of 1360 to Plan Their Budgets
Budgeting and financial planning often involve working with fractions to determine percentages and allocations. Financial experts use their knowledge of finding fractions of numbers to create budgets and make informed decisions about spending and saving.
For example, if your monthly income is 1360 dollars, you can use the knowledge that one-fourth of your income is 340 dollars to determine how much you can allocate to different expenses or savings goals.
Being able to quickly and easily calculate fractions of numbers is an essential skill for financial success.
## Real-Life Applications of One-Fourth of 1360
While math can seem abstract, the knowledge of finding fractions of numbers has many practical applications in everyday life. For example, when cooking, you may need to adjust recipe measurements based on the number of servings you want to make. Knowing how to calculate fractions of ingredients is crucial to getting the recipe just right.
Similarly, when shopping, you may need to calculate discounts or tax rates to determine the total cost of an item. Understanding fractions is key to making informed decisions about purchases.
The ability to work with fractions is a valuable skill that can improve many aspects of daily life.
## Mastering Math: One-Fourth of 1360 and Beyond
Math is a subject that requires practice and patience to master. If you want to improve your math skills beyond finding one-fourth of 1360, there are many resources available to help.
Online tutorials, helpful apps, and math games are just a few examples of the tools that can help you improve your mathematical abilities. Additionally, practicing math regularly can improve your confidence and critical thinking skills, which are helpful in all areas of life.
## Conclusion
Overall, finding one-fourth of a number like 1360 may seem daunting at first, but with practice and the tips outlined in this article, anyone can become a math whiz. By understanding how fractions work and using simple formulas, you can tackle math problems with ease.
Remember, the knowledge of finding fractions of numbers has practical applications in everyday life, from cooking to budgeting. By mastering math, you can unlock your potential for success and improve your problem-solving skills for a lifetime. |
# What is the remainder when x^1999 is divided by (x^2-1) ?
Apr 9, 2017
$x$
#### Explanation:
Note that:
$\left({x}^{2} - 1\right) \left({x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x\right)$
$= {x}^{2} \left({x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x\right) - \left({x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x\right)$
$= \left({x}^{1999} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{1997}}}} + \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{{x}^{1995}}}} + \ldots + \textcolor{c y a n}{\cancel{\textcolor{b l a c k}{{x}^{5}}}} + \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{{x}^{3}}}}\right) - \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{1997}}}} + \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{{x}^{1995}}}} + \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{{x}^{1993}}}} + \ldots + \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{{x}^{3}}}} + x\right)$
$= {x}^{1999} - x$
So:
${x}^{1999} = \left({x}^{2} - 1\right) \left({x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x\right) + x$
That is:
${x}^{1999} / \left({x}^{2} - 1\right) = {x}^{1997} + {x}^{1995} + {x}^{1993} + \ldots + {x}^{3} + x$
with remainder $x$
The polynomial $x$ is of degree less than the degree of ${x}^{2} - 1$, so is the remainder we want.
Apr 9, 2017
$x$
#### Explanation:
Calling
${p}_{n} \left(x\right) = {x}^{n} + {c}_{n - 1} {x}^{n - 1} + \cdots + {c}_{0}$
as a generic $n$ degree polynomial we have.
${x}^{n} = {p}_{n - 2} \left({x}^{2} - 1\right) + a x + b$ with $n \ge 2$
so we have
${\left(- 1\right)}^{n} = a \left(- 1\right) + b$
and
$1 = a + b$
solving we have
$a = \frac{1 - {\left(- 1\right)}^{n}}{2}$ and $b = \frac{1 + {\left(- 1\right)}^{n}}{2}$
in the present case we have $n = 1999$ so
$a = 1 , b = 0$ and the remainder is $x$ |
What is the phrase”exclamation mark” in mathematics?
The answer is simple. Here are some strategies to inform once an equation is a multiplication equation and just how you can add and subtract from using”exclamation marks”. For instance,”2x + 3″ imply”multiply two integers from several, making a price equal to 2 and 3″.
By way of instance,”2x + 3″ are multiplication by three. Moreover, college essay writing service we are able to insert the worth of 2 and 3 . To bring the worth, we will utilize”e”I” (or”E”). Using”I” means”add the price of one to this worthiness of 2″.
To add the values, we can certainly do it like that:”x – y” means”multiply x y, creating a price equal to zero”. For”x”y”, we’ll use”t” (or”TE”) to your subtraction and we’re going to use”x + y” to solve the equation.
You may believe which you are not supposed to utilize”electronic” in addition as”that I” means”subtract” however it is not quite easy. For www.masterpapers.com instance, to express”two – 3″ means subtract from three.
So, to bring the values we use”t”x” (or”TE”), which are the variety of those values to be added. We will utilize”x” to subtract the price of one in the value of 2 and that will give us precisely exactly the result.
To multiply the worth we can certainly do it like that:”2x + y” necessarily indicate”multiply two integers by y, creating a price equal to one plus two”. You may know that this is just a multiplication equation once we utilize”x” to subtract from two. Orit might be”x – y” to subtract from 2. Note that the equation can be written by you with parentheses and a point.
Now, let’s take a good example. Let’s mention that you want to multiply the worth of”two” by”five” and we have”x nine”y twenty-five”. Then we’ll use”x – y” to reevaluate the https://owl.excelsior.edu/research/drafting-and-integrating/drafting-and-integrating-paraphrasing/ price of a person in the value of 2.
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## Wednesday, May 24, 2017
### The Quadratic Formula - An Easy Way to Solve Quadratic Equations
Algebra has been one of the most important parts of mathematics. And the quadratic formula has been one of the most important aspects of algebraic math. There are many ways of solving a problem through this formula however you always require some tips for the best solutions. Mathematics is a subject which requires lots of practice. Also, math is known to hold an important place in all types of exams. Therefore you require some perfect tricks to solve the mathematical problems with ease and more accurately. Therefore here we will be discussing something about this.
The need for tricks
Shortcut tricks are very important to excel in competitive examinations. These tricks will help you to manage time well and solve the problems in few easy steps. Therefore it is very important to practice some short tricks in order to get better in the subject. While practicing, you are always advised to go through the problems thoroughly. Before starting up with these, you are suggested to note down twenty problems related to quadratic equations. Now practice some problems which involve basic issues and keep track of time while doing this. This way you will be able to learn time management in a better way.
Solve a problem now
Often, one of the simplest ways of solving “ax2+bx+c=0 ” for finding the value of x is, factoring the quadratic. You should set all the factors equal to zero first. But this way of solving gets messy sometimes. Therefore to make the solution easy there is the quadratic formula. This can help you to find the solution in the best way. This formula uses “a”, “b”, and “c” from “ax2+bx+c ”, here “a”, “b”, and “c” are constants and x is a variable. These are the numerical coefficients you are given to solve.
The formula is: for
For the formula to work efficiently, you need to arrange the formula in the form “(quadratic) =0”. Also, “2a” in the formula is taken to be underneath anything mentioned above it. This also includes the square root.
Example: Solve x2 = 3x-4= 0
First of all, we will factor the quadratic equation: x2=3x-4=(x=4) (x-1)=0
Now we will try to apply the formula to solve this equation. Using a=1, b=3, and c=-4. Then the solution would look something like
Solving this further we will get:
Further, this gets reduced to:
Therefore we get the expected values as x=-4 or x=1.
Thus using the formula, you can complete solving the problems very easily. Therefore you may have got the best solutions for your queries in solving the quadratic equations. Thus it may be clear to you that with the use of this formula you can make the problems to be really easy. Therefore it can be said that there are many advantages associated with the use of the quadratic formula. Also these can be applied to the problems to get the accurate solutions very easily.
## Sunday, May 7, 2017
### Want an Answer for Variable X?
Let me start by explaining what a variable means in mathematics. A variable is something that has its value changing constantly. One minute it is value say 10, the next minute its value has changed to some random number like 23. That is why it is called a variable; deduced from the word varying.
## How to Solve for x
Now that you have understood what a variable means, let us dive in straight in to getting the values for that variable. If you do not know how to do it, you can jump into quickmath.com. There, you will find so many steps on how to solve for x.
For example, given a simple equation like the one below, 3x + 3= 0
, And you are told to solve for x
Steps that you will see at that website will be like the one I am about to show you.
Here we go,
Step 1: You need to collect the like terms together. That is, in an equation, the set that has variables are like terms same case to those that do not have variables.
In our case, 3x will remain on the left hand side (L.H.S) while taking the 3 to the Right Hand Side (R.H.S).
You will get; 3x = 0 - 3
Remember, once a quotient crosses over to the other side of the equal sign, its + or – symbol has to be interchanged. That is why, you see +3 changing to being -3,
Step 2: You go ahead to start simplifying it. That is, 3x = -3
Step 3: make sure you remain with the variable on one side. Therefore, you will have to divide it by 3 like this;
When you do the division, you will find that the value of x will be -1. i.e. x = -1
## Complex example
Solving variable x has its own difficulty levels. Simple equations like the one, I just illustrated above are less difficult to solve. At times, you may come across a complex equation and are told to find the value of variable x in that equation. If you know nothing, you may start sweating profusely like African pot on fire. Anyway, just wanted to let you know that there is a way you can stop this swearing before it even occurs. I will stick with quickmath.com as the best math friend you can have who will offer his relentless services to you for free.
From there, you can get to know how to solve for x in sophisticated equations like this one;
Okay, for math geniuses, I know they will say that this is a simple equation. However, not everyone agrees with them. I, for one think that this equation involves a lot of thinking. Though the equation could be cumbersome and time consuming, you will eventually land at the answer. The value for x in our case should be 1.775 having rounded it off to the nearest thousandths.
If your answer is different from mine, you need to retrace your steps and see where you went wrong. |
## Section4.7Describing the Behavior of Functions
###### Overview
We have been learning about how functions are constructed and how they are defined. In many instances, before we construct a formula for a function, we need to identify what behavior we are attempting to model. At other times, we have a formula and we need to know what behavior that predicts. We need specific language that we can use to describe behavior.
In this section, we will focus on three types of behavior: monotonicity, concavity, and end behavior. Monotonicity will describe where a function is increasing or decreasing. Concavity will describe where the slope ro rate of change of a function is increasing or decreasing. In a graph, concavity describes whether the curve is bending up or bending down. We also discuss simple end behavior including unbounded growth (tending to infinity) and horizontal asymptotes.
Our emphasis is in learning the language of behavior, describing graphs using this language, and creating graphs based on a description of a function. As our study of calculus develops, we will learn mathematical tools that will allow us to determine function behavior more precisely.
### Subsection4.7.1Functions Have Shapes
We often describe functions according to the shape of their graphs. The different possible shapes we see in graphs correspond to specific behaviors of the functions. We will focus on two aspects of a graph: monotonicity and concavity.
#### Subsubsection4.7.1.1Monotonicity
The monotonicity of a function deals with whether the function is increasing or decreasing. We start with the mathematical definitions of increasing and decreasing functions. We will explore the ideas graphically in terms of maps and then graphs.
###### Definition4.7.1Monotonicity
A function $f$ is increasing on a subset $S$ of the domain (usually an interval) if for every $x_1,x_2 \in S\text{,}$
A function $f$ is decreasing on a subset $S$ of the domain (usually an interval) if for every $x_1,x_2 \in S\text{,}$
One way to think of monotonicity is that the function retains an ordering of the sets. An increasing function preserves the order, so that if two inputs are in a particular order, $x_1 \lt x_2\text{,}$ then the resulting outputs have the same order, $f(x_1) \lt f(x_2)\text{.}$ A decreasing function reverses the order, so that if inputs have an order, $x_1 \lt x_2\text{,}$ then the outputs must have the opposite order $f(x_1) \gt f(x_2)\text{.}$ A function that is not monotone (neither increasing or decreasing) does not maintain a sense of order uniformly over the set. Sometimes the outputs might have the same order as the inputs, and sometimes the outputs might have the opposite order.
###### Example4.7.2
The function $f(x)=2x+1$ is a linear function with positive slope $m=2\text{.}$ We can show that $f$ is an increasing function. Suppose $x_0 \lt x_1\text{.}$ Multiplying both sides of an inequality by a positive number preserves the ordering, as does adding the same value to both sides:
\begin{gather*} x_0 \lt x_1\\ 2x_0 \lt 2x_1\\ 2x_0+1 \lt 2x_1+1\\ f(x_0) \lt f(x_1) \end{gather*}
This is visualized in the following figure.
Thinking of the map dynamically, we see that as we increase the input, the output also increases. This is captured in the graph of the function in the $(x,y)$ plane. The graph shows $y$-values increasing as viewed from left to right, which is corresponding to $x$-values increasing.
###### Example4.7.5
The function $f(x)=-2x+3$ is a linear function with negative slope $m=-2\text{.}$ We can show that $f$ is an decreasing function. Suppose $x_0 \lt x_1\text{.}$ Multiplying both sides of an inequality by a negative number reverses the ordering, while adding the same value to both sides preserves the order:
\begin{gather*} x_0 \lt x_1\\ -2x_0 \gt -2x_1\\ -2x_0+3 \gt -2x_1+3\\ f(x_0) \gt f(x_1) \end{gather*}
The map shows that the order of outputs is always opposite to the order of the inputs. Thinking of the map dynamically, we see that as we increase the input, the output decreases. The graph of the function in the $(x,y)$ plane captures the same information. Viewing the graph from left to right (as $x$ increases), the $y$-values decrease.
Some functions are not monotone because the map does not retain the ordering of the sets. Dynamically, this is because the output will sometimes increase and sometimes decrease as the input is increased.
###### Example4.7.8
The function $f(x)=x^2$ is not a linear function and is not monotone. We can show this by illustrating that the function is inconsistent in ordering the output values relative to the input values. Consider $x_0=-2$ and $x_1=-1\text{.}$ We have $f(x_0)=4 \gt f(x_1)=1\text{,}$ so for these inputs the order is reversed. However, for $x_0=1$ and $x_1=2\text{,}$ we have $f(x_0)=1 \gt f(x_1)=4$ and the order is preserved. This function is not increasing or decreasing, but is a combination.
We can see graphically that $f$ is decreasing on $(-\infty,0]$ because for any two inputs in this interval, the order of the outputs is reversed. We can also see that $f$ is increasing on $[0,\infty)$ because for any two inputs in that interval, the order of the outputs is preserved. This point where monotonicity switches corresponds to the vertex of the parabola $y=x^2\text{.}$
One of our goals in calculus will be to develop a method to determine the intervals on which a function is increasing or decreasing. When we motivated monotonicity with linear functions, we saw that a positive slope implied an increasing function and a negative slope implied a decreasing function. Calculus will develop a more general sense of the slope of a function using the derivative such that we will describe monotonicity based on the signs of the derivative.
###### Note4.7.11
When listing intervals on which a function is increasing or decreasing, it is important not to use a union of the intervals. The reason is that we are saying that the function is increasing on each of the intervals individually and not on the set formed by the union. If listing multiple intervals, simply form a comma-separated list.
#### Subsubsection4.7.1.2Concavity
Concavity describes how the graph of a function in the $(x,y)$ plane bends. If the graph bends upward, we say the function is concave up. If the graph bends downward, we say the function is concave down.
As with monotonicity, these attributes of functions apply over intervals rather than at individual points. When a graph changes concavity at a point, for example switching from bending up to bending down, the function has an inflection point. A technical definition of concavity that depends on the concept of a derivative will be provided later.
However, we can capture the essential idea by thinking about how the slope is changing between points. A function that has an increasing slope or rate of change over an interval is concave up on the interval. A function that has decreasing slope or rate of change is concave down.
###### Definition4.7.13Concavity
A function $f$ is concave up on a subset $S$ of the domain (usually an interval) if for every $x_1,x_2,x_3 \in S\text{,}$ the slope or rate of change is increasing,
\begin{equation*} x_1 \lt x_2 \lt x_3 \qquad \text{implies} \qquad \frac{f(x_2)-f(x_1)}{x_2-x_1} \lt \frac{f(x_3)-f(x_2)}{x_3-x_2}\text{.} \end{equation*}
A function $f$ is concave down on a subset $S$ of the domain (usually an interval) if for every $x_1,x_2,x_3 \in S\text{,}$ the slope or rate of change is decreasing,
\begin{equation*} x_1 \lt x_2 \lt x_3 \qquad \text{implies} \qquad \frac{f(x_2)-f(x_1)}{x_2-x_1} \gt \frac{f(x_3)-f(x_2)}{x_3-x_2}\text{.} \end{equation*}
This definition is not very easy to use directly. When we learn more about derivatives to describe the slope at individual points, we will have a much better method known as the second derivative test for concavity. However, the following examples will illustrate what is happening.
###### Example4.7.14
The function $f(x)=x^2$ is concave up on $(-\infty,\infty)$ (the entire domain). We will not prove that this is true because this is too difficult without derivatives. But we can illustrate the idea.
Consider the graph $y=f(x)=x^2$ and the particular values $x_1=-4\text{,}$ $x_2=-2\text{,}$ and $x_3=-1\text{.}$ We will calculate the slope or rate of change between $(x_1,y_1)=(-4,16)$ and $(x_2,y_2)=(-2,4)$ and between $(x_2,y_2)$ and $(x_3,y_3)=(-1,1)\text{.}$
\begin{gather*} m_{12} = \frac{y_2-y_1}{x_2-x_1} = \frac{4-16}{-2--4} = \frac{-12}{2} = -6 \\ m_{23} = \frac{y_3-y_2}{x_3-x_2} = \frac{1-4}{-1--2} = \frac{-3}{1} = -3 \end{gather*}
We can see that the slope or rate of change is increasing, $m_{12} \lt m_{23}\text{.}$ These slopes are illustrated in the following figure.
This is not a proof of concavity because we only illustrated the order for three specific points. Use the following dynamic figure to convince yourself that for any three points we might choose, the slopes increase from left to right.
The reason that $f$ has an inflection point at $x=0$ is that point is where $f$ has the steepest negative slope. To the left, $x \lt 0\text{,}$ the slope decreases; to the right, $x \gt 0\text{,}$ the slope increases.
###### Example4.7.16
The function $f(x)=x^3-3x$ changes concavity at $x=0\text{.}$ $f$ is concave down on $(-\infty,0]$ and concave up on $[0,\infty)\text{.}$ When three points are chosen with $x \in (-\infty,0]\text{,}$ the slope is decreasing. When the three points are chosen with $x \in [0,\infty)\text{,}$ the slope is increasing. This can be verified in the following dynamic figure. However, the three points must all be in either $(-\infty,0]$ or in $[0,\infty)\text{.}$
#### Subsubsection4.7.1.3Combining Monotonicity and Concavity
The shape of a graph of a function is often defined in terms of the monotonicity and concavity combined. There are four basic shapes that correspond to the four quadrants of a circle, illustrated in the figure below. A curve that has a positive and increasing slope is increasing and concave up. A curve that has a positive but decreasing slope is increasing and concave down. A curve that has a negative but increasing (becoming less negative) slope is decreasing and concave up. A curve that has a negative and decreasing (becoming more negative) slope is decreasing and concave down.
We can describe the shape of a graph by stating intervals on which the function satisfies each of the possible behaviors. The intervals are separated by points where the graph reaches either a maximum or minimum value (changes in monotonicity) or where the slope of the graph reaches an extreme and begins to bend the other direction (changes in concavity or points of inflection).
###### Example4.7.19
The graph of a function $y=f(x)$ is shown below, with labeled extreme points and inflection points. Describe the shape of the graph by giving intervals of monotonicity and concavity.
Solution
Intervals for monotonicity are based on the function increasing or decreasing. The end-points of these intervals are the extreme points for the function. When the graph extends beyond the frame of the figure, we assume the function behavior continues as shown. Intervals always are read from left to right. The end-point of an interval is included (closed) if the behavior extends up to and including that point.
The function $f$ is decreasing on $(-\infty,-3]\text{,}$ increasing on $[-3,3]\text{,}$ and decreasing on $[3,\infty)\text{.}$ Notice that the extremes at $x=-3$ and $x=3$ are included in two intervals. The continuous function is decreasing on $(-\infty,-3)$ as an open interval. Because $f$ decreases up to and including $x=-3\text{,}$ we include the end-point.
Intervals for concavity are based on where the slope is increasing or decreasing. Intervals on which the graph bends upward, $f$ is concave up. Intervals on which the graph bends downward, $f$ is concave down. Notice our graph has inflection points (where the concavity changes) at $x=-1\text{,}$ $x=0\text{,}$ and $x=1\text{.}$ At these points, the graph starts to bend in the opposite direction.
The function $f$ is concave up on $(-\infty,-1]\text{,}$ concave down on $[-1,0]\text{,}$ concave up on $[0,1]\text{,}$ and concave down on $[1,\infty)\text{.}$ We include the inflection points as the end points of the intervals (closed) because the slope is increasing or decreasing up to and including those points.
### Subsection4.7.2End Behavior
End-behavior of a function describes what happens to a function as the size of the input grows. Consider the possibilities of a linear function, $y=f(x)=mx+b\text{.}$ So long as the slope is non-zero, the function is unbounded, meaning that the graph eventually goes above every level and eventually goes below every level (on opposite sides of the graph).
If the slope is positive, $m \gt 0\text{,}$ then the function is increasing. We say $f(x) \to +\infty$ as $x \to +\infty\text{,}$ which we read as “the value of $f(x)$ tends to positive infinity as the value of $x$ goes to positive infinity”. This is because the $y$-values will eventually rise above any level on the right side of the graph (for sufficiently large positive values $x$). We also say $f(x) \to -\infty$ as $x \to -\infty$ because the $y$-values are below any specified value on the left side of the graph (for sufficiently large negative values $x$). When the slope is negative, $m \lt 0\text{,}$ the unbounded behavior is reversed.
For consistency in notation to describe the tendency of a function (as opposed to the value of a function), we use limits to describe unbounded behavior.
\begin{align*} \lim_{x \to -\infty} f(x) &= -\infty & \text{means $f(x) \to -\infty$ as $x \to -\infty$}\\ \lim_{x \to -\infty} f(x) &= \infty & \text{means $f(x) \to +\infty$ as $x \to -\infty$}\\ \lim_{x \to \infty} f(x) &= -\infty & \text{means $f(x) \to -\infty$ as $x \to +\infty$}\\ \lim_{x \to \infty} f(x) &= \infty & \text{means $f(x) \to +\infty$ as $x \to +\infty$} \end{align*}
When the graph of a function $f$ behaves more and more like a constant function (horizontal line) for larger and larger values of the independent variable, we say $f$ has a horizontal asymptote. A horizontal asymptote $y=L$ on the right side (large, positive values for $x$) uses the limit statement
\begin{equation*} \lim_{x \to \infty} f(x) = L\text{,} \end{equation*}
which means that the value of $f(x)$ approaches the constant value $L$ as $x \to +\infty\text{.}$ When $f$ has a horizontal asymptote $y=L$ on the left side (large, negative values of $x$), we use the limit statement
\begin{equation*} \lim_{x \to -\infty} f(x) = L\text{.} \end{equation*}
###### Example4.7.21
The graph of a function $y=f(x)$ is shown below. This function has two horizontal asymptotes: $y=-2$ as $x \to -\infty$ and $y=1$ as $x \to +\infty\text{.}$ We write
\begin{gather*} \lim_{x \to -\infty} f(x) = -2,\\ \lim_{x \to +\infty} f(x) = 1. \end{gather*}
A function can also have unbounded behavior near a particular input value, say at $x=a\text{.}$ Using limit notation, this means that at least one of the following must be true.
\begin{align*} \lim_{x \to a^-}f(x) &= +\infty\\ \lim_{x \to a^-}f(x) &= -\infty\\ \lim_{x \to a^+}f(x) &= +\infty\\ \lim_{x \to a^+}f(x) &= -\infty \end{align*}
The graph has a vertical asymptote at $x=a\text{,}$ meaning that the graph of the function approaches closer and closer to this vertical line.
###### Example4.7.22
The graph of a function $y=f(x)$ is shown below with two vertical asymptotes. The vertical asymptote at $x=0$ corresponds to left- and right-limits
\begin{align*} \lim_{x \to 0^-} f(x) &= -\infty,\\ \lim_{x \to 0^+} f(x) &= +\infty, \end{align*}
The vertical asymptote at $x=2$ only corresponds to the right-limit
\begin{equation*} \lim_{x \to 2^+} f(x) = +\infty. \end{equation*}
It is hard to tell from a graph alone where a vertical asymptote occurs. Using only the limited graph window, it is not obvious that the vertical asymptote is at exactly $x=0$ since the graph is still fairly far away from that vertical line from this perspective.
###### Note4.7.23
A common false impression about horizontal asymptotes is that the graph of a function can not cross the asymptote. A function can not cross a vertical asymptote, but that is only because a function can not intersect a vertical line at more than one point. An asymptote only requires that the graph behaves more and more like the line.
When a function physically relates two variables, $x \mapsto y\text{,}$ a horizontal asymptote indicates that for sufficiently large values of the independent variable, the dependent variable is essentially a constant. A common description in physical settings for this constant is a saturation value. We think of the quantity measured by the independent variable as a control variable. The dependent variable can be thought of as a response. As the control variable is increased, the response will pass through some of its range of values. However, there will come a point where even though you continue to increase the control variable, the response is no longer able to change very much at all. That is, the response has saturated.
###### Example4.7.24
An enzyme is a protein that helps catalyze a chemical reaction. The rate or velocity of reaction $V$ depends on the concentration of the reactant $C\text{.}$ Commonly, the function $C \mapsto V$ is increasing, concave down, and has a horizontal asymptote, known as Michaelis–Menten reaction kinetics. The physical domain is $C \in [0, \infty)\text{.}$ Because the relation is increasing, we know that adding more reactants will raise the reaction rate. Because the relation is concave down, we know that the degree to which the rate increases slows down as more reactants are added. The horizontal asymptote means that this increase in the reaction rate saturates to some maximum rate $V_{\text{max}}\text{,}$
\begin{equation*} \lim_{C \to \infty} V = V_{\text{max}}. \end{equation*}
The reactant concentration where the reaction rate is halfway to the maximum value is called the half-saturation value, and is usually represented with a constant $K\text{.}$
###### Example4.7.26
Imagine a crop of plants growing in a field. The total biomass harvested $B$ depends on the number of seeds $S$ that are sown. If very few seeds are sown, the biomass harvested will be small. For more seeds sown, we expect the biomass would increase. However, if too many seeds are sown, then the crop will be overcrowded, resulting in a lower harvest. We expect that there might be an optimal number of seeds $S^*$ for which the biomass is at a maximum.
Describe the behavior of the function $S \mapsto B$ and sketch a possible graph.
Solution
The function $S \mapsto B$ will have a physical domain of $S \in [0,\infty)\text{.}$ Because $B$ is a maximum at $S=S^*\text{,}$ the function is increasing on $[0,S^*]$ and decreasing on $[S^*,\infty)\text{.}$ The simplest assumption for concavity would be that the function starts concave down. However, a concave down and decreasing function will eventually approach $-\infty\text{,}$ which is not physically possible for our physical scenario. Therefore, the function must change concavity at some inflection point after $S^*\text{,}$ say at $S=S^\dagger\text{.}$ Our function would be concave down on $[0,S^\dagger]$ and concave up on $[S^\dagger,\infty)\text{.}$ Continuing to increase the number of seeds will result in ever smaller biomass due to overcrowding until it approaches some saturating biomass $B_\infty\text{,}$
\begin{equation*} \lim_{S \to \infty} B = B_\infty. \end{equation*}
Note: The asterisk and dagger are decorations so that the symbols $S^*$ and $S^\dagger$ represent general constants. We don't know actual values for the maximum and inflection point, so we can't use numbers. The symbols are place-holders for values that would be determined experimentally. Similarly, the symbol $B_{\infty}$ represents the value for the biomass harvested when the number of seeds sown saturates the system.
### Subsection4.7.3Summary
• Describing the monotonicity of a function is determining intervals on which the function is increasing or decreasing.
• A function $f$ is increasing on a set $S$ if the function is order preserving: For all $x_1,x_2 \in S\text{,}$ we must have
This corresponds to a graph that is rising left to right (positive slopes).
A function $f$ is decreasing on a set $S$ if the function is order reversing: For all $x_1,x_2 \in S\text{,}$ we must have
This corresponds to a graph that is falling left to right (negative slopes).
• Describing the concavity of a function is determining intervals on which the function is concave up or concave down.
• A function $f$ is concave up on a set $S$ if the slope or rate of change is increasing on $S\text{:}$ For all $x_1,x_2,x_3 \in S\text{,}$ we must have
\begin{equation*} x_1 \lt x_2 \lt x_2 \quad \Rightarrow \quad \frac{f(x_2)-f(x_1)}{x_2-x_1} \lt \frac{f(x_3)-f(x_2)}{x_3-x_2}\text{.} \end{equation*}
The graph will be bending upward.
A function $f$ is concave down on a set $S$ if the slope or rate of change is decreasing on $S\text{:}$ For all $x_1,x_2,x_3 \in S\text{,}$ we must have
\begin{equation*} x_1 \lt x_2 \lt x_2 \quad \Rightarrow \quad \frac{f(x_2)-f(x_1)}{x_2-x_1} \gt \frac{f(x_3)-f(x_2)}{x_3-x_2}\text{.} \end{equation*}
The graph will be bending downward.
• A point of inflection is a point where a function is continuous and changes concavity.
• Lists of intervals of monotonicity and concavity should be separated by commas and not joined by unions.
• Limits as $x \to \pm\infty$ describe end behavior.
• To say $f(x) \to +\infty$ means values of $f(x)$ eventually rise above any possible value.
• To say $f(x) \to -\infty$ means values of $f(x)$ eventually fall below any possible value.
• To say $f(x) \to L$ means values of $f(x)$ eventually approaches a horizontal asymptote $y=L\text{.}$
### Subsection4.7.4Exercises
Each of the following problems asks you to prove that the given function is either increasing or decreasing on a particular interval.
###### 1
Prove that $f(x)=5x-12$ is an increasing function by showing that whenever $x_1 \lt x_2\text{,}$ we have $f(x_1) \lt f(x_2)\text{.}$
###### 2
Prove that $f(x)=-3x-2$ is a decreasing function by showing that whenever $x_1 \lt x_2\text{,}$ we have $f(x_1) \gt f(x_2)\text{.}$
###### 3
Prove that $f(x)=x^2$ is an increasing function on $[0,\infty)$ by showing that whenever $0 \lt x_1 \lt x_2\text{,}$ we have $f(x_1) \gt f(x_2)\text{.}$
Hint: Show that $f(x_2)-f(x_1) \gt 0$ by factoring and determining the signs of the factors.
###### 4
Prove that $f(x)=x^2$ is a decreasing function on $(-\infty,0]$ by showing that whenever $x_1 \lt x_2 \lt 0\text{,}$ we have $f(x_1) \gt f(x_2)$ or $f(x_2)-f(x_1) \lt 0\text{.}$
Hint: Show that $f(x_2)-f(x_1) \lt 0$ by factoring and determining the signs of the factors.
Consider each of the following graphs of functions. Use the graph to determine the intervals of monotonicity for that function.
###### 7
Each of the following problems asks you to illustrate the concavity of the given function.
###### 8
Illustrate that $\displaystyle f(x) = \frac{1}{x}$ is concave up on $(0,\infty)$ by showing that the slope is increasing for the sequential points $x_1=\frac{1}{2}\text{,}$ $x_2=1\text{,}$ and $x_3=2\text{.}$
###### 9
Illustrate that $\displaystyle f(x) = \frac{1}{x}$ is concave down on $(-\infty,0)$ by showing that the slope is decreasing for the sequential points $x_1=-2\text{,}$ $x_2=-1\text{,}$ and $x_3=-\frac{1}{2}\text{.}$
###### 10
Illustrate that $\displaystyle f(x) = 2^x$ is concave up on $(-\infty,\infty)$ by showing that the slope is increasing for the sequential points $x_1=-1\text{,}$ $x_2=0\text{,}$ and $x_3=1\text{.}$
###### 11
Illustrate that $\displaystyle f(x) = 2^{-x}$ is concave up on $(-\infty,\infty)$ by showing that the slope is increasing for the sequential points $x_1=-1\text{,}$ $x_2=0\text{,}$ and $x_3=1\text{.}$
Consider each of the following graphs of functions, which includes turning points and inflection points. Use the graph to determine the intervals of monotonicity and concavity for that function.
###### 14
Use the graphs to answer the questions about limits. Assume that the behavior of the graph shown in the window continues outside the window.
###### 15
1. $\displaystyle \lim_{x \to -\infty} f(x)$
2. $\displaystyle \lim_{x \to +\infty} f(x)$
###### 16
1. $\displaystyle \lim_{x \to 3^-} f(x)$
2. $\displaystyle \lim_{x \to 3^+} f(x)$
3. $\displaystyle \lim_{x \to -\infty} f(x)$
4. $\displaystyle \lim_{x \to +\infty} f(x)$ |
# What is the Square Root of 111111?
In math, the square root of 111111 is the number that, when multiplied by itself, equals that value. For example, the square root of 111111 is 333.3332 because 333.3332 multiplied by itself is 111111.
Square root of 111,111 = 333.3332
11111112
The symbol √ is called
The number below
## Is 111111 a Perfect Square Root?
No. The square root of 111111 is not an integer, hence √111111 isn't a perfect square.
Previous perfect square root is: 110,889
Next perfect square root is: 111,556
## The Prime Factors of 111,111 are:
3 × 7 × 11 × 13 × 37
## How Do You Simplify the Square Root of 111,111 in Radical Form?
The main point of simplification (to the simplest radical form of 111111) is as follows: getting the number 111111 inside the radical sign √ as low as possible.
111111 is already simplified (have no pair prime factors).
## Is the Square Root of 111,111 Rational or Irrational?
Since 111111 isn't a perfect square (it's square root will have an infinite number of decimals), it is an irrational number.
## The Babylonian (or Heron’s) Method (Step-By-Step)
StepSequencing
1
In step 1, we need to make our first guess about the value of the square root of 111111. To do this, divide the number 111111 by 2.
As a result of dividing 111111/2, we get the first guess: 55555.5
2
Next, we need to divide 111111 by the result of the previous step (55555.5).
111111/55555.5 = 2
Calculate the arithmetic mean of this value (2) and the result of step 1 (55555.5).
(55555.5 + 2)/2 = 27778.75 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|27778.75 - 55555.5| = 27776.75
27776.75 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
3
Next, we need to divide 111111 by the result of the previous step (27778.75).
111111/27778.75 = 3.9999
Calculate the arithmetic mean of this value (3.9999) and the result of step 2 (27778.75).
(27778.75 + 3.9999)/2 = 13891.375 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|13891.375 - 27778.75| = 13887.375
13887.375 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
4
Next, we need to divide 111111 by the result of the previous step (13891.375).
111111/13891.375 = 7.9986
Calculate the arithmetic mean of this value (7.9986) and the result of step 3 (13891.375).
(13891.375 + 7.9986)/2 = 6949.6868 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|6949.6868 - 13891.375| = 6941.6882
6941.6882 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
5
Next, we need to divide 111111 by the result of the previous step (6949.6868).
111111/6949.6868 = 15.9879
Calculate the arithmetic mean of this value (15.9879) and the result of step 4 (6949.6868).
(6949.6868 + 15.9879)/2 = 3482.8374 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|3482.8374 - 6949.6868| = 3466.8494
3466.8494 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
6
Next, we need to divide 111111 by the result of the previous step (3482.8374).
111111/3482.8374 = 31.9024
Calculate the arithmetic mean of this value (31.9024) and the result of step 5 (3482.8374).
(3482.8374 + 31.9024)/2 = 1757.3699 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|1757.3699 - 3482.8374| = 1725.4675
1725.4675 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
7
Next, we need to divide 111111 by the result of the previous step (1757.3699).
111111/1757.3699 = 63.2257
Calculate the arithmetic mean of this value (63.2257) and the result of step 6 (1757.3699).
(1757.3699 + 63.2257)/2 = 910.2978 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|910.2978 - 1757.3699| = 847.0721
847.0721 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
8
Next, we need to divide 111111 by the result of the previous step (910.2978).
111111/910.2978 = 122.0601
Calculate the arithmetic mean of this value (122.0601) and the result of step 7 (910.2978).
(910.2978 + 122.0601)/2 = 516.179 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|516.179 - 910.2978| = 394.1188
394.1188 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
9
Next, we need to divide 111111 by the result of the previous step (516.179).
111111/516.179 = 215.2567
Calculate the arithmetic mean of this value (215.2567) and the result of step 8 (516.179).
(516.179 + 215.2567)/2 = 365.7179 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|365.7179 - 516.179| = 150.4611
150.4611 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
10
Next, we need to divide 111111 by the result of the previous step (365.7179).
111111/365.7179 = 303.8161
Calculate the arithmetic mean of this value (303.8161) and the result of step 9 (365.7179).
(365.7179 + 303.8161)/2 = 334.767 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|334.767 - 365.7179| = 30.9509
30.9509 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
11
Next, we need to divide 111111 by the result of the previous step (334.767).
111111/334.767 = 331.9055
Calculate the arithmetic mean of this value (331.9055) and the result of step 10 (334.767).
(334.767 + 331.9055)/2 = 333.3363 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|333.3363 - 334.767| = 1.4307
1.4307 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
12
Next, we need to divide 111111 by the result of the previous step (333.3363).
111111/333.3363 = 333.33
Calculate the arithmetic mean of this value (333.33) and the result of step 11 (333.3363).
(333.3363 + 333.33)/2 = 333.3332 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|333.3332 - 333.3363| = 0.0031
0.0031 > 0.001
Repeat this step again as the margin of error is greater than than 0.001
13
Next, we need to divide 111111 by the result of the previous step (333.3332).
111111/333.3332 = 333.3331
Calculate the arithmetic mean of this value (333.3331) and the result of step 12 (333.3332).
(333.3332 + 333.3331)/2 = 333.3332 (new guess)
Calculate the error by subtracting the previous value from the new guess.
|333.3332 - 333.3332| = 0
0 < 0.001
Stop the iterations as the margin of error is less than 0.001
Result✅ We found the result: 333.3332 In this case, it took us thirteen steps to find the result. |
# Precalculus : Find the Vertex and the Axis of Symmetry of a Parabola
## Example Questions
← Previous 1
### Example Question #3 : Parabolas
Find the axis of symmetry and vertex of the following parabola:
Explanation:
The first step of the problem is to find the axis of symmetry using the following formula:
Where a and b are determined from the format for the equation of a parabola:
We can see from the equation given in the problem that a=1 and b=-3, so we can plug these values into the formula to find the axis of symmetry of our parabola:
Keep in mind that the vertex of the parabola lies directly on the axis of symmetry. That is, the x-coordinate of the axis of symmetry will be the same as that of the vertex of the parabola. Now that we know the vertex is at the same x-coordinate as the axis of symmetry, we can simply plug this value into our function to find the y-coordinate of the vertex:
So the vertex occurs at the point:
### Example Question #4 : Parabolas
Find the equation of the axis of symmetry:
Explanation:
Rewrite the equation in standard form .
The vertex formula is:
Determine the necessary coefficients.
Plug in these values to the vertex formula.
The axis of symmetry is .
### Example Question #5 : Parabolas
Find the location of the vertex of the following parabola:
Explanation:
The vertex can be thought of as the center of a parabola. Begin by finding the axis of symmetry with the following formula:
Where b and a come from the standard equation of a parabola:
So given our parabola
This gives us the x-coordinate of our vertex. find the y-coordinate by plugging in our x-coordinate.
So our vertex is:
### Example Question #52 : Conic Sections
Find the vertex of the parabola:
Explanation:
The polynomial is already in format.
To find the vertex, use the following equation:
Substitute the coefficients and solve for the vertex.
The vertex is at .
### Example Question #6 : Parabolas
Find the vertex and the equation of the axis of symmetry for .
Explanation:
Rewrite in standard parabolic form, .
Write the vertex formula and substitute the values.
The equation of the axis of symmetry is .
Substitute this value back into the original equation .
The vertex is at .
### Example Question #7 : Parabolas
Find the axis of symmetry and the vertex of the parabola given by the following equation:
Vertex at
Axis of symmetry at
Vertex at
Axis of symmetry at
Vertex at
Axis of symmetry at
Vertex at
Axis of symmetry at
Vertex at
Axis of symmetry at
Explanation:
Find the axis of symmetry and the vertex of the parabola given by the following equation:
To find the axis of symmetry of a parabola in standard form, , use the following equation:
So...
This means that we have an axis of symmetry at . Or, to put it more plainly, at we could draw a vertical line which would perfectly cut our parabola in half!
So, we are halfway there, now we need the coordinates of our vertex. We already know the x-coordinate, which is 7. To find the y-coordinate, simply plug 7 into the parabola's formula and solve!
This makes our vertex the point
### Example Question #8 : Parabolas
Find the vertex of the parabola:
Explanation:
The vertex form for a parabola is given below:
To complete the square, take the coefficient next to the x term, divide by and raise the number to the second power. In this case, . Then take value and add it to add inside the parenthesis and subtract on the outside.
Now factor and simplify:
Fromt the values of and , the vertex is at
### Example Question #56 : Conic Sections
Find the vertex of the parabola:
Explanation:
The vertex form for a parabola is given below:
To complete the square, take the value next to the x term, divide by 2 and raise the number to the second power. In this case,. Then take value and add it to add inside the parenthesis and subtract on the outside.
Now factor and simplify:
Fromt the values of h and k, the vertex is at
### Example Question #1 : Find The Vertex And The Axis Of Symmetry Of A Parabola
Find the vertex of the parabola:
Explanation:
The vertex form for a parabola is given below:
To complete the square, take the value next to the x term, divide by and raise the number to the second power. In this case, . Then take value and add it to add inside the parenthesis and subtract on the outside. Remember to distribute before subtracting to the outside.
Now factor and simplify:
From the values of and , the vertex is at
### Example Question #2 : Find The Vertex And The Axis Of Symmetry Of A Parabola
Find the vertex of the parabola:
Explanation:
The vertex form for a parabola is given below:
Factor the equation and transform it into the vertex form.
To complete the square, take the value next to the x term, divide by and raise the number to the second power. In this case, . Then take value and add it to add inside the parenthesis and subtract on the outside.
Now factor and simplify:
From the values of and , the vertex is at
← Previous 1 |
# Percentage Decrease Formula
## Percentage Decrease Formula
Percentage Decrease Formula refers to the rate of change in a value as it decreases over a period of time. For example, less rainfall, fewer Covid cases, etc. The decrease rate can be calculated using the Percentage Decrease Formula. This section describes percentage reduction formulas. Learn the percentage reduction formula with some solution examples on Extramarks.
What Is Percent Decrease Formula?
The Percentage Decrease Formula represents the decrease relative to the initial value. To calculate the rate of decrease, students first need to find the difference in values. Then divide the difference by the initial value and multiply by 100. The Percentage Decrease Formula is:
Percentage Decrease Formula = [(old value – new value) / old value] × 100]
### Percent Decrease Formula
There are two simple steps to calculating the removal rate using the Percentage Decrease Formula:
Step 1: Find the numerical difference. That is, decrease = old value – a new value.
Step 2: Divide the decrement by the old value and multiply by 100. This gives the formula for the rate of decline. Percentage Decrease Formula = [(old value – new value) / old value] × 100]
### Percentage Increase And Decrease
The rate of increase and rate of decrease is the rate of change of the value. The per cent change is the difference between the reported new value and the old value. To find the percentage change, divide this difference by the old value and multiply by 100 to get the percentage increase or decrease.
Note that if the new value is higher than the old one, it is a percentage increase. For example, if the price of a book changes from $5 to$8, the price will increase. In this case, if the old value is greater than the new value, it’s a percentage reduction. For example, if the table price changes from $10 to$8, the price will decrease. Students will easily learn all the concepts on Extramarks.
### Examples Using Percent Decrease Formula
Solved examples on the topic Percentage Decrease Formula can be acquired from Extramarks. Extramarks provides study material and learning resources for various topics and subjects. All the resources provided by Extramarks are prepared by subject-matter experts, making them highly accurate and reliable. Extramarks is a popular educational website. Students can use the Extramarks website to access a variety of learning resources and study materials. Extramarks meets students’ academic requirements by providing them with the necessary learning resources. Extramarks’ website or mobile application provides students of all classes with the necessary study materials. Extramarks provides a wide range of educational materials, including NCERT solutions, sample papers, previous years’ papers, online live sessions, revision notes, mock tests, and so on. It is advisable to access the study material on the Percentage Decrease Formula from the Extramarks website.
Example 3. What is the Percentage Decrease Formula from 12500 to 11625?
Ans. The decrease rate from 12500 to 11625 can be calculated using the formula decrease rate = [(old value – new value) / old value] x 100. In this case, the old value = 12500 and the new value = 11625. Replacing the values in the formula, Percent Decrease = [(12500 – 11625) / 12500] × 100 = 7%
### 1. What is understood by Percentage Decrease?
The rate of Decrease refers to the rate of change in a value as it decreases over a period of time. % Decrease expresses the percentage reduction of the specified value relative to the initial value. From the Extramarks website, one can access study material on Percentage Decrease Formula.
### 2. What is the percentage reduction formula?
Taking the difference between the old value and the new value, dividing it by the old value, and multiplying by 100 forms the formula for percentage reduction. The decrease rate formula is expressed as
Decrease rate = [(old value – new value) / old value] × 100]
It is possible to access study material on Percentage Decrease Formula on the Extramarks website.
### 3. What are the steps to calculate the removal rate using the removal rate formula?
There are three simple steps to calculating the removal rate using the removal rate formula:
Step 1: Find the difference in numbers. H. decrease = old value – new value
Step 2: Divide this “reduction” by the old value and multiply by 100. This gives the reduction rate formula. Reduction ratio = [(old value – new value) / old value] × 100]
Step 3: The given value is inserted into the formula to determine the reduction rate. Calculate the rate at which the price of a pencil falls from $12 to$9 using the rate of decline formula. Calculate the percentage drop in the price of a pencil using the percentage drop formula. In this example, the new value = $9, old value =$12. Decrease rate = [(old value – new value) / old value] × 100. Substituting the values into the formula, Percent Decrease = [(12 – 9)/12] × 100 = 25%. Therefore, the price of pencils fell by 25%.
### 4. What is the per cent decrease from 20 to 16?
The per cent decrease from 20 to 16 can be calculated using the formula Percent decrease = [(old value – new value) / old value] x 100. In this case, the old value = 20 and the new value = 16. So, plugging the values into the formula, Percent Reduction = [(20 – 16) / 20] × 100 = 20%. At Extramarks, students may access study materials related to the Percentage Decrease Formula.
### 5. What are some examples of percentage reduction?
1. Examples of percentage reduction can be seen in our daily lives. Suppose the price of fuel goes from $7 to$4. The fuel price reduction rate can be calculated using the formula reduction rate = [(old value – new value) / old value] x 100. Here, the old value = 7 and the new value = 4. Therefore, after substituting the values into the formula, % reduction = [(7 – 4) / 7] × 100 = 42.8%. It is highly recommended to access the Percentage Decrease Formula study notes from the Extramarks website or mobile application for further clarification on the topic. |
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