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# MATH 110 Automotive Worksheet #4
## Full text
(1)
### Ratios
The math name for a fraction is ratio. It is just a comparison of one quantity with another quantity that is similar. As an automotive technician, you will use ratios a lot.
You can use ratios when you convert from one unit to another. Since 12 inches = 1 foot, the conversion ratio can be written as a fraction
### or
You can see that these ratios relate inches and feet.
Example 1: Using a ratio to convert from 2.5 feet into an equivalent number of inches, we use the first ratio.
## )
### ⎛
So 2.5 feet equals 30 inches. Notice the feet units cancel (because
feet feet
### =
; that is, anything divided by itself is equal to 1). Always arrange the number you want to convert and the unit conversion ratio so that the units you don’t want cancel out and you are left with the units you do want in the numerator.
Example 2: Using a ratio to convert from 42 inches into an equivalent number of feet, we use the second ratio.
## )
### ⎛
So 42 inches equals 3.5 feet.
Notice that in each case, you use the conversion ratio that forces the units to cancel (divide) out. The units tell you which form of the conversion ratio you must use.
### Ratios in the Automotive Industry
Ratios are often written with a colon “:”
For instance, to express the windings of a coil with 1600 turns and 400 turns as a ratio you would say 1600:400, which can then be reduced to 4:1 since you can divide 400 into both sides of the ratio.
A gear ratio compares the number of teeth on different gears. It is the comparison of the number of teeth on the input gear to the number of teeth on the output gear (the input gear is the one the power is applied to, also called the driver gear. The output gear is also called the driven gear). To find the ratio, you should divide the number of teeth of the input gear by the number of teeth of the output gear. This is usually
(2)
stated as (the answer for the division):1. If the gear ratio of A to B is 3:1, then gear A has three times as many teeth as gear B.
### =
The principle behind gears is pretty simple. Gears with unequal numbers of teeth alter the speed between the input and out put gears. Say the gear ratio of the input gear to the output gear is 1:3 (like 10 teeth for gear A and 30 teeth for gear B). Then for every complete revolution of the input gear the output gear turns 1/3 the way around. This means you are slowing down the action (this is referred to in engineering terms as “Stepping Down”.) If we reverse everything so that the A has 30 teeth and B has 10, then the opposite happens and we “Step Up”-- then for every 1 turn of the input gear the output gear would turn 3 times and the ratio is now 3:1. This means the formula for the gear ratio can also be written as
### =
Example 3: The input gear has 75 teeth and the output gear has 10 teeth. What is the gear ratio? In this figure the gear ratio A to B is 12:18
or, after you reduce it by dividing both sides by 12 (the smaller of the two
numbers), you get a ratio of 1:1.5
Input Output Here the input gear turns faster than the output gear. Stepping down has the advantage of
producing more power although at a slower rate.
Input
Output Here the output gear turns faster than the input gear. Stepping up produces a much faster output speed, but mechanically delivers less power.
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Solution: From the first gear ratio formula listed above, the ratio of the number of teeth in the input to the number of teeth in the output is
### 75 =
. So the gear ratio is 7.5:1.
Example 4: The gear ratio of the teeth on input gear A to output gear B is 6:1. How many times faster does gear B move than gear A?
Solution: From the second gear ratio formula listed above, the ratio of the number of revolutions of A to B is 1:6. That is, for every complete turn of A, B turns 6 times. Therefore, B turns 6 times faster than A.
The transmission gear ratio refers to the number of times the speed is reduced by the transmission. It compares the speed of the crankshaft to the speed of the driveshaft. A transmission ratio of 3:1 means the speed is reduced three times by the transmission (the RPMs of the crankshaft is three times the RPMs of the driveshaft).
To find the transmission gear ratio, first find the gear ratio of each set of gears that are meshed together. Then multiply the first numbers of the gear ratios together and compare to the product of the second numbers. Then reverse the numbers. In the figure below, the arrows show the direction the power is applied.
Example 5: In low speed, gears A and B, and gears D and F are meshed. A has 14 teeth, B has 28 teeth, D has 16 teeth and F has 26 teeth. What is the transmission gear ratio?
(4)
Solution: First find the gear ratio for each set of meshed gears. For A and B, the gear ratio is 14:28 or 1:2. For gears D and F, the gear ratio is 16:26 or after dividing both sides by 16 gives, 1:1.625. To get the transmission gear ratio multiply the first numbers in each ratio together and do the same for the second set of numbers then compare. That is
1:2 1:1.625
or 1x1:2x1.625 which gives us 1:3.25. You then reverse the numbers to get 3.25:1. This means the transmission reduces the speed by a factor of 3.25 (the driveshaft spins 3.25 times slower than the crankshaft).
Example 6: In second gear, gears A and B and gears C and E are meshed. A has 14 teeth, B has 28 teeth, C has 21 teeth and E has 21teeth. What is the transmission gear ratio?
Solution: The gear ratio of each set of gears is: For A and B: 1:2. For C and E: 1:1. Then we multiply the numbers from each ratio: 1x1:2x1 which gives us. 1:2. Then reverse the numbers to get the transmission gear ratio: 2:1. This means the crankshaft spins 2 times faster than the driveshaft.
The axle or differential ratio refers to the number of times the speed of the driveshaft is reduced by the ring gear and pinion. It compares the speed of the driveshaft to the speed of the rear axle shaft. An axle ratio of 4:1 means the RPM’s of the driveshaft is 4 times greater than the RPM’s of the rear axle shaft. To find the axle ratio, find the ratio of the number of teeth of the ring gear to that of the pinion gear. See the figure:
Example 7: The pinion gear has 12 teeth and the ring gear has 42 teeth. What is the axle ratio?
Solution: The axle ratio is the ratio of the number of teeth of the ring gear to that of the pinion gear. So we have 42:12. Divide both sides by 12 to reduce it. That gives us 42/12:1 or 3.5:1. This means the driveshaft turns 3.5 times faster than the rear axle shaft does.
The total gear reduction is the ratio of the RPM of the crankshaft to the RPM of the rear axle shaft. In finding the total gear reduction, both the transmission gear ratio and the axle (differential) ratios are used. Multiply the first numbers from both ratios and then multiply the second set of numbers. Then compare them.
Example 8: A transmission gear ratio is 2:1 and the axle ratio is 3:1. What is the total gear reduction ratio?
multiply
multiply
Ring gear — mounted to the rear axle. Pinion gear —
attached to the driveshaft.
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Solution: The total gear reduction ratio is 3x2:1x1, or 6:1.
Another type of ratio you will run across in your work is the compression ratio. The compression ratio is a comparison to the amount of space (the volume in cubic inches or cubic centimeters-- cc) in the cylinder when the piston is at the bottom of the stroke (indicated by A in the drawing), and the amount of space when the piston is at the top of the stroke (indicated by B in the drawing). For example, if there is 8 times as much space when the piston is at the bottom of the stroke as when the piston is at the top of the stroke, the compression ratio is 8:1.
Example 9: There are 33 cubic inches of space when the piston is at the bottom of the stroke and 4 cubic inches of space when the piston is at the top of the stroke. What is the compression ratio?
Solution: The compression ratio is 33:4 or after dividing both sides by 4, 8.25:1.
### Proportions
A proportion is two ratios set equal to one another. A proportion is a fraction equal to a fraction. For example, 3:1 = 12:4 is the same thing as
### 3 =
. The above problems can be done using proportions.
Example 10: An input gear turns three revolutions while the output gear only turns one revolution. The larger gear has 27 teeth. How many teeth does the smaller gear have?
Solution: The gear ratio is 1:3 since the input gear turns three times as fast as the output gear. The larger gear with 27 teeth has to be the output gear since it turns more slowly than the input gear. We can set the gear ratio equal to the ratio of the input gear’s teeth to that of the output gear’s teeth: 1:3 = X:27, where X represents the unknown number of teeth in the input gear. This proportion can be written as two equal fractions:
(6)
### 1 =
. You can solve this for X by cross multiplying. To do this, multiply the denominators of each fraction by the numerators of the other and set them equal to each other leaving: (1)(27) = 3(X). This can be solved for X by dividing both sides by 3:
### =
Here you could have saved a step by just multiplying the 1 by the 27. That would leave you
### =
which reduces to X = 9. So the smaller input gear has 9 teeth.
### Homework Problems
1. Convert 57 inches to feet. State the answer two ways: just feet (in decimal form, like 2.3 feet), and feet and inches (like 3 feet 4 inches).
2. 1 inch = 25.4 millimeters. How many millimeters are there in 2.7 inches?
3. An input gear has 55 teeth and the output gear has 25 teeth. What is the gear ratio of these gears? Which gear turns faster and how much faster is it compared to the other gear?
4. The gear ratio of two gears A and B is 6.2:1. How many times slower does gear A move compared to gear B?
Refer to the transmission diagram earlier in this document for the next two problems.
5. What is the transmission gear ratio when the car is in second gear (intermediate speed), assuming gear A has 12 teeth, gear B has 36 teeth, gear C has 30 teeth and gear E has 24 teeth?
6. In high speed gear there are internal splines that mesh with the teeth on the back of the gear A. This gives direct drive. What is the transmission gear ratio?
7. A ring gear has 42 teeth and the pinion gear has 16 teeth. What is the axle ratio, and how much slower does the rear axle shaft spin compared to the driveshaft?
8. The second gear transmission gear ratio of a truck is 1.9:1. The rear axle ratio is 4.6:1. Find the total gear reduction. Also, how much faster does the crankshaft turn than the rear axle shaft? 9. A sport truck with a 2.3 Liter turbocharged diesel engine has a cylinder volume at bottom dead
center of 605 cc. At top dead center, the volume is 30 cc. What is the compression ratio?
10. A gear with 15 teeth turns as 160 RPM. It is driving a gear of 25 teeth. Find the RPM of the driven gear.
Updating...
## References
Related subjects : |
9.1 Square Roots
# 9.1 Square Roots
## 9.1 Square Roots
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. 9.1 Square Roots Pre-Algebra
2. Vocabulary • Square root: If , then m is the square root of n. • Radical sign: represents the positive square root. represents the negative square root. • Perfect square: Any number that has an integer square roots. • Example: ,
3. Perfect Squares These are perfect squares. They have the same length and the same width. Area = length ∙ width
4. Example 1: Finding Square Roots • What are the square roots of 81? • 92=81 and (-9)2=81. • The square roots of 81 are 9 and -9. • What are the square roots of 49?
5. Example 2: Evaluating Square Roots Why? When computing the square root with the radical sign, you will have only one answer-the positive square root
6. Example 3: Solving a Square Root Equation • To find the minimum speed a pole vaulter must run, you must use the following equation s = 8 • S is the pole vaulter’s speed in feet per second,and h is the height vaulted in feet. • If the pole vaulter vaults over a height of 25 feet, find the pole vaulter’s minimum speed. Write equation for speed of apole vaulter. Substitute 25 for h. Evaluate square root. Multiply.
7. Example 5: Solving Equations Using Square Roots • Solve the equation. When solving for a variable, you will have two answers-the positive and negative square root. a.x2= 25 x2= 25 Write original equation. Use definition of square root. Evaluate square root.
8. Example 5: Solving Equations Using Square Roots • Solve the equation. b.h2+ 5 = 54 h2+ 5 = 54 Write original equation. Undo the addition of 5. Simplify. Use definition of square root. Evaluate square root.
9. You Try! • Solve the equation. 1. 2. x2= 81 x2= 1 x = ±9 x = ±1
10. 12x2= 108 x = ±3 x2– 5 = – 1 x = ±2 3. 4.
11. Think about it!! Can you find ? Why or Why not? |
# 1992 AHSME Problems/Problem 16
## Problem
If $$\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}$$ for three positive numbers $x,y$ and $z$, all different, then $\frac{x}{y}=$
$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$
## Solution 1
$\fbox{E}$ We have $\frac{x+y}{z} = \frac{x}{y} \implies xy+y^2=xz$ and $\frac{y}{x-z} = \frac{x}{y} \implies y^2=x^2-xz \implies x^2-y^2=xz$. Equating the two expressions for $xz$ gives $xy+y^2=x^2-y^2 \implies x^2-xy-2y^2=0 \implies (x+y)(x-2y)=0$, so as $x+y$ cannot be $0$ for positive $x$ and $y$, we must have $x-2y=0 \implies x=2y \implies \frac{x}{y}=2$.
## Solution 2
We cross multiply the first and third fractions and the second and third fractions, respectively, for $$x(x-z)=y^2$$ $$y(x+y)=xz$$ Notice how the first equation can be expanded and rearranged to contain an $(x+y)$ term. $$x^2-xz=y^2$$ $$x^2-y^2=xz$$ $$(x+y)(x-y)=xz$$ We can divide this by the second equation to get $$\frac{(x+y)(x-y)}{y(x+y)}=\frac{xz}{xz}$$ $$\frac{x-y}{y}=1$$ $$\frac{x}{y}-1=1$$ $$\frac{x}{y}=2 \rightarrow \boxed{E}$$
## Solution 3
Since $$\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = \frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n},$$ we can say that $$\frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} = \frac {y+(x+y)+x}{(x-z)+z+y} = \boxed{2}.$$ $\implies \boxed{(E)}$.
$\textbf{Proof of the used property:}$
Let $$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = ... = \frac{a_n}{b_n} = r.$$
Therefore, \begin{align*} a_1 &= b_1r,\\ a_2 &= b_2r,\\ ...,\\ a_n &= b_nr \end{align*} so $$\frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n} = \frac{b_1r + b_2r + ... + b_nr}{b_1 + b_2 + ... + b_n}$$ $$= \frac{(b_1 + b_2 + ... + b_n)(nr)}{(b_1 + b_2 + ... + b_n)(n)} = r$$ --- NamelyOrange |
# Inter 1st Year Maths 1A Matrices Solutions Ex 3(a)
Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(a) will help students to clear their doubts quickly.
## Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(a)
I.
Question 1.
Write the following as a single matrix.
(i) $$\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]+\left[\begin{array}{lll} 1 & 0 & 0 \end{array}\right]$$
(ii) $$\left[\begin{array}{ccc} 3 & 9 & 0 \\ 1 & 8 & -2 \end{array}\right]+\left[\begin{array}{ccc} 4 & 0 & 2 \\ 7 & 1 & 4 \end{array}\right]$$
(iii) $$\left[\begin{array}{c} 0 \\ 1 \\ -1 \end{array}\right]+\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$$
(iv) $$\left[\begin{array}{cc} -1 & 2 \\ 2 & -2 \\ 3 & 1 \end{array}\right]-\left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ -2 & 1 \end{array}\right]$$
Solution:
Question 2.
If A = $$\left[\begin{array}{cc} -1 & 3 \\ 4 & 2 \end{array}\right]$$, B = $$\left[\begin{array}{cc} 2 & 1 \\ 3 & -5 \end{array}\right]$$, X = $$\left[\begin{array}{ll} x_{1} & x_{2} \\ x_{3} & x_{4} \end{array}\right]$$ and A + B = X, then find the values of x1, x2, x3 and x4.
Solution:
A + B = X
∴ x1 = 1, x2 = 4, x3 = 7, x4 = -3
Question 3.
If A = $$\left[\begin{array}{ccc} -1 & -2 & 3 \\ 1 & 2 & 4 \\ 2 & -1 & 3 \end{array}\right]$$, B = $$\left[\begin{array}{ccc} 1 & -2 & 5 \\ 0 & -2 & 2 \\ 1 & 2 & -3 \end{array}\right]$$ and C = $$\left[\begin{array}{ccc} -2 & 1 & 2 \\ 1 & 1 & 2 \\ 2 & 0 & 1 \end{array}\right]$$ then find A + B + C.
Solution:
Question 4.
If A = $$\left[\begin{array}{ccc} 3 & 2 & -1 \\ 2 & -2 & 0 \\ 1 & 3 & 1 \end{array}\right]$$, B = $$\left[\begin{array}{ccc} -3 & -1 & 0 \\ 2 & 1 & 3 \\ 4 & -1 & 2 \end{array}\right]$$ and X = A + B then find X.
Solution:
Question 5.
If $$\left[\begin{array}{cc} x-3 & 2 y-8 \\ z+2 & 6 \end{array}\right]=\left[\begin{array}{cc} 5 & 2 \\ -2 & a-4 \end{array}\right]$$ then find the values of x, y, z and a.
Solution:
Given $$\left[\begin{array}{cc} x-3 & 2 y-8 \\ z+2 & 6 \end{array}\right]=\left[\begin{array}{cc} 5 & 2 \\ -2 & a-4 \end{array}\right]$$
∴ x – 3 = 5 ⇒ x = 3 + 5 = 8
2y – 8 = 2 ⇒ 2y = 8 + 2 = 10 ⇒ y = 5
z + 2 = -2 ⇒ z = -2 – 2 = -4
a – 4 = 6 ⇒ a = 4 + 6 = 10
II.
Question 1.
If $$\left[\begin{array}{ccc} x-1 & 2 & 5-y \\ 0 & z-1 & 7 \\ 1 & 0 & a-5 \end{array}\right]=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 7 \\ 1 & 0 & 0 \end{array}\right]$$ then find the values of x, y, z and a.
Solution:
Given $$\left[\begin{array}{ccc} x-1 & 2 & 5-y \\ 0 & z-1 & 7 \\ 1 & 0 & a-5 \end{array}\right]=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 7 \\ 1 & 0 & 0 \end{array}\right]$$
∴ x – 1 = 1 ⇒ x = 1 + 1 = 2
5 – y = 3 ⇒ y = 5 – 3 = 2
z – 1 = 4 ⇒ z = 4 + 1 = 5
a – 5 = 0 ⇒ a = 5
Question 2.
Find the trace of $$\left[\begin{array}{ccc} 1 & 3 & -5 \\ 2 & -1 & 5 \\ 1 & 0 & 1 \end{array}\right]$$
Solution:
Trace of $$\left[\begin{array}{ccc} 1 & 3 & -5 \\ 2 & -1 & 5 \\ 1 & 0 & 1 \end{array}\right]$$ = Sum of the diagonal elements
= 1 – 1 + 1
= 1
Question 3.
If A = $$\left[\begin{array}{rrr} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 4 & 5 & -6 \end{array}\right]$$ and B = $$\left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right]$$ find B – A and 4A – 5B.
Solution:
Question 4.
If A = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right]$$ and B = $$\left[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right]$$ find 3B – 2A.
Solution: |
APNotes-Chap07 – ap gov note
University: Temple University
Course: Government in American Society (POLS 8101)
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Students shared 11 documents in this course
AP Statistics – Chapter 7 Notes: Sampling Distributions
7.1 – What is a Sampling Distribution?
Parameter – A parameter is a number that describes some characteristic of the population
Statistic – A statistic is a number that describes some characteristic of a sample
Symbols
used
Sample
Statistic
Population
Parameter
Proportions
ˆ
p
p
Means
x
Sampling Distribution – the distribution of all values taken by a statistic in all possible samples of the same
size from the same population
A statistic is called an unbiased estimator of a parameter if the mean of its sampling distribution is equal to
the parameter being estimated
Important Concepts for unbiased estimators
The mean of a sampling distribution will always equal the mean of the population for any sample size
The spread of a sampling distribution is affected by the sample size, not the population size.
Specifically, larger sample sizes result in smaller spread or variability.
7.2 – Sample Proportions
Choose an SRS of size n from a large population with
population proportion p having some characteristic of
interest.
Let 𝑝 be the proportion of the sample having that
characteristic. Then the mean and standard deviation
of the sampling distribution of 𝑝 are
Mean: 𝜇𝑝
=𝑝 Std. Dev.: 𝜎𝑝
=√𝑝(1−𝑝)
𝑛
With the Z-Statistic: 𝑧 = 𝑝
−𝑝
√𝑝(1−𝑝)
𝑛
CONDITIONS FOR NORMALITY
The 10% Condition
Use the formula for the standard deviation of
ˆ
p
only
when the size of the sample is no more than 10% of
the population size (𝑛 ≤ 1
10𝑁).
The Large Counts Condition
We will use the normal approximation to the
sampling distribution of
ˆ
p
for values of n and p that
satisfy
10np
and
(1 ) 10np
.
7.3 – Sample Means
Suppose that
x
is the mean of a sample from a large
population with mean
and standard deviation
.
Then the mean and standard deviation of the
sampling distribution of
x
are
Mean: 𝜇𝑥=𝜇 Std. Dev.: 𝜎𝑥=𝜎
√𝑛
With the Z-Statistic: 𝑧 = 𝑥−𝜇
𝜎√𝑛
CONDITIONS FOR NORMALITY
If an SRS is drawn from a population that has the
normal distribution with mean
and standard
deviation
, then the sample mean
x
will have the
normal distribution
( , )Nn
for any sample size.
Central Limit Theorem
If an SRS is drawn from any population with mean
and standard deviation
, when n is large
( 30)n
, the sampling distribution of the sample
mean
x
will have the normal distribution
( , )Nn
.
You are watching: AP Statistics – Chapter 7 Notes: Sampling Distributions 7 – What is a. Info created by GBee English Center selection and synthesis along with other related topics. |
# How do you find the zeros, if any, of y=x^2-3x+1 using the quadratic formula?
May 6, 2018
The solutions are $\setminus \frac{3 \pm \sqrt{5}}{2}$
#### Explanation:
$a {x}^{2} + b x + c = 0$
the quadratic formula states that the solutions, if any, are
${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$
The quantity $\setminus \Delta = {b}^{2} - 4 a c$ is called Determinant, and you can tell the number of solutions by its sign:
1. If $\Delta > 0$, then the square root is well defined and not null, so you have two solutions.
2. If $\Delta = 0$, the square root is zero as well. So, adding or subtracting it makes no difference, and you have two coincident solutions.
3. If $\Delta < 0$, the square root is not defined, and you have no solutions (at least using real numbers).
In your case, $a = 1$,$b = - 3$ and $c = 1$. So, you have
$\Delta = {\left(- 3\right)}^{2} - 4 \cdot 1 \cdot 1 = 9 - 4 = 5$, so you have two solutions, namely
${x}_{1 , 2} = \setminus \frac{- \left(- 3\right) \setminus \pm \setminus \sqrt{5}}{2 \cdot 1} = \setminus \frac{3 \pm \sqrt{5}}{2}$ |
# ICSE Solutions for Selina Concise Chapter 2 Compound Interest (using formula) Class 9 Maths
### Exercise 3(A)
1. Find the amount and the compound interest on Rs. 12,000 in 3 years at 5% compounded annually.
Given : P= Rs. 12,000; n = 3 years and r = 5%
= Rs. 13,891.50
C.I. = Rs. 13,891.50 – Rs. 12,000 = Rs. 1,891.50.
2. Calculate the amount of Rs. 15,000 is lent at compound interest for 2 years and the rates for the successive years are 8% and 10% respectively.
Given : P = Rs. 15,000; n = 2 years; r1 = 8 % and r2 = 10%
= Rs. 17,820.
3. Calculate the compound interest accrued on Rs. 6,000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.
Given : P = Rs. 6,000; n = 3 years; r= 5%; r= 8% and r3 = 10%
= Rs. 7,484.40
∴ C.I. = Rs. 7,484.40 – Rs. 6,000 = Rs. 1,484.40.
4. What sum of money will amount to Rs. 5,445 in 2 years at 10% per annum compound interest ?
Given: P= Rs. 5,445 ; n = 2 years and r = 10%
= Rs. 4,500
5. On what sum of money will the compound interest for 2 years at 5% per annum amount to Rs. 768.75?
Given : C.I.= Rs. 768.75; n= 2 years and r = 5%
= Rs. 7,500
6. Find the sum on which the compound interest for 3 years at 10% per annum amounts to Rs. 1,655.
Given : C.I. = Rs. 1,655; n = 3 years and r = 10%
= Rs. 5,000
7. What principal will amount to Rs. 9,856 in two years, if the rates of interest for successive years are 10% and 12% respectively ?
Given : A = Rs. 9,856 ; n = 2 years ; r1 = 10 % and r2 = 12%
= Rs. 8,000
8. On a certain sum, the compound interest in 2 years amounts to Rs. 4,240. If the rate of interest for the successive years is 10% and 15% respectively, find the sum.
⇒ (P + 4240) = P(1.265)
⇒ P = Rs. 16000
The sum is Rs.16,000.
9. At what per cent per annum will Rs.6,000 amount to Rs.6,615 in 2 years when interest is compounded annually?
At 5% per annum the sum of Rs. 6,000 amounts to Rs. 6,615 in 2 years when the interest is compounded annually.
10. At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years ?
Let Principal = Rs. y
Then Amount= Rs 1.44y
n= 2 years
On solving, we get r = 20 %
11. At what rate per cent will a sum of Rs. 4,000 yield Rs.1,324 as compound interest in 3 years ?
Given : P = Rs. 4,000, C.I. = Rs. 1,324 and n = 3 years
Now, A = P + I
⇒ A = Rs. (4,000 + 1,324) = Rs. 5,324
Thus, the rate of interest is 10%.
12. A person invests Rs5,000 for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to Rs6,272. Calculate :
(i) the rate of interest per annum.
(ii) the amount at the end of the third year.
Given: P = Rs. 5,000; A = Rs. 6,272 and n = 2 years.
= Rs. 7,024.64
13. In how many years will Rs. 7,000 amount to Rs. 9,317 at 10% per annum compound interest ?
Given : P = Rs. 7,000; A = Rs. 9,317 and r = 10%.
On comparing, n = 3 years
14. Find the time, in years, in which Rs. 4,000 will produce Rs. 630.50 as compound interest at 5% compounded annually.
Given : P= Rs. 4,000; C.I.= Rs. 630.50 and r = 5%
On comparing, n = 3 years
15. Divide Rs. 28,730 between A and B so that when their shares are lent out at 10% compound interest compounded per year, the amount that A receives in 3 years is the same as what B receives in 5 years.
Let share of A = Rs. y
share of B = Rs (28,730 – y)
rate of interest= 10%
According to question,
Amount of A in 3 years= Amount of B in 5 years
Therefore, share of A = Rs. 15,730
Share of B = Rs. 28,730 – Rs.15,730 = Rs. 13,000
16. A sum of Rs 44,200 is divided between John and Smith, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10% per annum compound interest, they will receive equal amounts on reaching 16 years of age.
(i) What is the share of each out of Rs44,200 ?
(ii) What will each receive, when 16years old ?
(i) Let share of John = Rs y
share of Smith = Rs (44,200 – y)
rate of interest= 10%
According to question,
Amount of John in 4 years = Amount of Smith in 2 years
= Rs. 29,282
17. The simple interest on a certain sum of money and at 10% per annum is Rs. 6,000 in 2 years, Find:
1. the sum.
2. the amount due to the end of 3 years and at the same rate of interest compounded annually.
3. the compound interest earned in 3 years.
(i) I = Rs. 6,000, T = 2 years and R = 10%
= Rs. 39,930
(iii) C.I. earned in 3 years = A – P = Rs. (39,930 – 30,000) = Rs. 930.
18. Find the difference between compound interest and simple interest on Rs. 8,000 in 2 years and at 5% per annum.
Given: P = Rs. 8,000, R = 5%, T = 2 years
For simple interest,
C.I. = A - P
= Rs. (8,820 – 8,000)
= Rs. 820
Now, C.I. – S.I. = Rs. (820 – 800) = Rs. 20.
Thus, the difference between the compound interest and the simple interest is Rs. 20.
### Exercise 3(B)
1. The difference between simple interest and compound interest on a certain sum is Rs. 54.40 for 2 years at 8 per cent per annum. Find the sum.
Let principal (P) = x
R = 8%
T = 2 years
X = Rs. 8500
Thus, principal sum = Rs. 8500
2. A sum of money, invested at compound interest, amounts to Rs. 19,360 in 2 years and to Rs. 23,425.60 in 4 years. Find the rate per cent and the original sum of money.
(for 2 years) A = Rs. 19360
T = 2 years
Let P = X
X = Rs. 16,000
Thus, sum = Rs. 16000
3. A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money becomes twenty-seven times of itself at the same rate of interest p.a.
Let principal = X, A = 3X, T = 8 years, R = ?
Case I,
T = 24
Time = 24 years.
4. On what sum of money will compound interest (payable annually) for 2 years be the same as simple interest on Rs. 9,430 for 10 years, both at the rate of 5 per cent per annum ?
P = Rs. 9430
R = 5%
T = 10 years
Thus, principal from = Rs. 46,000
5. Kamal and Anand each lent the same sum of money for 2 years at 5% at simple interest and compound interst respectively. Anand recived Rs. 15 more than Kamal. Find the amount of money lent by each and the interest received.
Let principal = Rs. 100, R = 5% T = 2 years
= Rs. 615
6. Simple interest on a sum of money for 2 years at 4% is Rs. 450. Find compound interest of the same sum and at the same rate for 2 years.
SI = Rs. 450
R = 4%
T = 2 years
P = ?
CI = A - P = 6084 - 5625 = Rs. 459
7. Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 per cent per annum by Rs. 228. Find the sum.
Let principal (P), R = 4%, T = 4 years
P = Rs. 96000
Thus, Principal = Rs. 96000
8. Compound interest on a certain sum of money at 5% per annum for two years is Rs. 246. Calculate simple interest on the same sum for 3 years at 6% per annum.
CI = Rs. 246, R = 5%, T = 2 years
CI = A – P
9. A certain sum of money amounts to Rs. 23,400 in 3 years at 10% per annum simple interest. Find the amount of the same sum in 2 years and at 10% p.a. compound interest.
Let the sum (principle) = X
Given Amount = 23400, R = 10% and T = 3 years
A = 21780
The amount of the same sum in 2 years and at 10% p.a. compound interest is 21780.
10. Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying Rs. 12,600 at the end of the first year and Rs. 17,640 at the end of the second year. Find the sum borrowed.
For the payment of Rs. 12,600 at the end of first year :
A = Rs. 12,600; n = 1 year and r = 5%
∴ Sum borrowed = Rs. (12,000 + 16,000 ) = Rs. 28,000.
### Exercise 3(C)
1. If the interest is compounded half-yearly, calculate the amount when principal is Rs. 7,400; the rate of interest is 5% per annum and the duration is one year.
Given: P = Rs. 7,400; r = 5% p.a. and n = 1 year
Since the interest is compounded half-yearly,
= Rs. 7,774.63
2. Find the difference between the compound interest compounded yearly and half-yearly on Rs. 10,000 for 18 months at 10% per annum.
(i) When interest is compounded yearly :
Given : P = Rs. 10,000 ; n = 18 months = 1.5 year and r = 10% p.a.
For 1 year
= Rs. 11,576.25
C.I.= Rs.11,576.25 – Rs.10,000 = Rs. 1,576.25
Difference between both C.I. = Rs. 1,576.25 – Rs. 1,550 = Rs. 26.25
3. A man borrowed Rs.16,000 for 3 years under the following terms:
20% simple interest for the first 2 years.
20% C.I. for the remaining one year on the amount due after 2 years, the interest being compounded half-yearly.
Find the total amount to be paid at the end of the three years.
For the first 2 years
⇒ A = 27,104
The total amount to be paid at the end of the three years is Rs. 27,104.
4. What sum of money will amount to Rs. 27,783 in one and a half years at 10% per annum compounded half yearly ?
⇒ P = 24,000
The sum of Rs. 24,000 amount Rs. 27,783 in one and a half years at 10% per annum compounded half yearly.
5. Ashok invests a certain sum of money at 20% per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If Geeta gets Rs. 33 more than Ashok in 18 months, calculate the money invested.
(i) For Ashok (interest is compounded yearly) :
Let P = Rs. y; n = 18 months = 1.5 year and r = 20% p.a.
For 1 year
Money invested by each person=Rs. 3,000.
6. At what rate of interest per annum will a sum of Rs. 62,500 earn a compound interest of Rs. 5,100 in one year? The interest is to be compounded half yearly.
⇒ r = 8
The rate of interest is 8%.
7. In what time will Rs. 1,500 yield Rs. 496.50 as compound interest at 20% per year compounded half-yearly ?
Given: P=Rs. 1,500; C.I.= Rs. 496.50 and r = 20%
Since interest is compounded semi-annually
8. Calculate the C.I. on Rs. 3,500 at 6% per annum for 3 years, the interest being compounded half-yearly.
Do not use mathematical tables. Use the necessary information from the following:
(1.06)3 =1.191016; (1.03)3 = 1.092727
(1.06)6 =1.418519; (1.03)6 = 1.194052
Given : P = Rs. 3,500; r = 6% and n = 3 years
Since interest is being compounded half-yearly
= 3,500[(1.03)6 – 1]
= 3,500[1.194052 – 1]
= 3,500 × 0.194052
= Rs. 679.18
9. Find the difference between compound interest and simple interest on Rs.12,000 and in 1.5 years at 10% compounded yearly.
= Rs. 13,860
∴ C.I. = Rs. 13,860 – Rs. 12,000 = Rs. 1,860
∴Difference between C.I. and S.I.
= Rs. 1,860 – Rs. 1,800 = Rs. 60.
10. Find the difference between compound interest and simple interest on Rs. 12,000 and in 1.5 years at 10% compounded half-yearly.
A = Rs. 13,891.50
C.I. = Rs. 13,891.50 – Rs. 12,000 = Rs. 1,891.50
∴ Difference between C.I. and S.I = Rs. 1,891.50 – Rs. 1,800 = Rs. 91.50.
### Exercise 3(D)
1.The cost of a machine is supposed to depreciate each year at 12% of its value at the beginning of the year. If the machine is valued at Rs. 44,000 at the beginning of 2008, find its value :
(i) at the end of 2009.
(ii) at the beginning of 2007.
Cost of machine in 2008 = Rs. 44,000
Depreciation rate = 12%
(i) ∴ Cost of machine at the end of 2009
= Rs. 50,000
2. The value of an article decreases for two years at the rate of 10% per year and then in the third year it increases by 10%. Find the original value of the article, if its value at the end of 3 years is Rs. 40,095.
Let X be the value of the article.
The value of an article decreases for two years at the rate of 10% per year.
The value of the article at the end of the 1st year is
X – 10% of X = 0.90X
The value of the article at the end of the 2nd year is
0.90X – 10% of (0.90X) = 0.81X
The value of the article increases in the 3rd year by 10%.
The value of the article at the end of 3rd year is
0.81x + 10% of (0.81x) = 0.891x
The value of the article at the end of 3 years is Rs. 40,095.
0.891X = 40,095
⇒ X = 45,000
The original value of the article is Rs. 45,000.
3. According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64,000. The census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2009 did the population of this town reach 74,088 ?
Population in 2009 (P) = 64,000
Let after n years its population be 74,088(A)
Growth rate= 5% per annum
On comparing, we get,
n = 3 years
4. The population of a town decreased by 12% during 1998 and then increased by 8% during 1999. Find the population of the town, at the beginning of 1998, if at the end of 1999 its population was 2,85,120.
Let the population in the beginning of 1998 = P
The population at the end of 1999 = 2,85,120(A)
r1 = – 12% and r2 = + 8%
5. A sum of money, invested at compound interest, amounts to Rs. 16,500 in 1 year and to Rs. 19,965 in 3 years. Find the rate per cent and the original sum of money invested.
Let sum of money be Rs P and rate of interest = r %
Money after 1 year = Rs. 16,500
Money after 3 years = Rs. 19,965
For 1 year
6. The difference between C.I. and S.I. on Rs. 7,500 for two years is Rs. 12 at the same rate of interest per annum. Find the rate of interest.
Given : P = Rs. 7,500 and Time(n) = 2 years
Let rate of interest = y%
⇒ y2 = 16
⇒ y = 4 %
7. A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 10 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.
Let Principal be Rs y and rate= r%
According to 1st condition
Amount in 10 years = Rs 3
On comparing, we get
n = 10×3 = 30 years
8. Mr. Sharma borrowed a certain sum of money at 10% per annum compounded annually. If by paying Rs.19,360 at the end of the second year and Rs. 31,944 at the end of the third year he clears the debt; find the sum borrowed by him.
At the end of the two years the amount is
⇒ P = Rs. 40,000
Mr. Sharma borrowed Rs. 40,000.
9. The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at 10% for a year is Rs. 15. Find the sum of money lent out.
Let sum of money be Rs. y
To calculate S.I.
⇒ y/400 = 15 ⇒ y = Rs. 6,000.
10. The ages of Pramod and Rohit are 16 years and 18 years respectively. In what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years?
Let Rs.X and Rs.Y be the money invested by Pramod and Rohit respectively such that they will get the same sum on attaining the age of 25 years.
Pramod will attain the age of 25 years after 25 – 16 = 9 years
Rohit will attain the age of 25 years after 25 -18 = 7 years
Pramod and Rohit should invest in 400 : 441 ratio respectively such that they will get the same sum on attaining the age of 25 years.
### Exercise 3(E)
1. Simple interest on a sum of money for 2 years at 4% is Rs .450. Find compound interest on the same sum and at the same rate for 1 year, if the interest is reckoned half yearly.
1st case
Given: S.I. = Rs 450 ; Time = 2 years and Rate = 4%
= Rs. 5852.25
∴ C.I. = 5,852.25 – 5,625 = Rs. 227.25
2. Find the compound interest to the nearest rupee on Rs. 10,800 for 2.5 years at 10% per annum.
Given : P = Rs. 10,800 ; Time = 2.5 years and Rate = 10% p.a.
For 2 years
∴ Rs.13,721 – Rs.10,800 = Rs.2,921
3. The value of a machine, purchased two years ago, depreciates at the annual rate of 10%. If its present value is Rs.97,200, find:
1. Its value after 2 years.
2. Its value when it was purchased.
(i) Present value of machine(P) = Rs.97,200
Depreciation rate = 10%
4. Anuj and Rajesh each lent the same sum of money for 2 years at 8% simple interest and compound interest respectively. Rajesh received Rs. 64 more than Anuj. Find the money lent by each and interest received.
Let the sum of money lent by both Rs. y
For Anuj
P = Rs. y ; rate = 8% and time = 2 years
5.Calculate the sum of money on which the compound interest (payable annually) for 2 years be four times the simple interest on Rs. 4,715 for 5 years, both at the rate of 5% per annum.
Given : Principal = Rs.4,715; time = 5 years and rate= 5% p.a.
6. A sum of money was invested for 3 years, interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted to Rs. 4,950, find the sum invested.
Given : C.I. for the 2nd year = Rs. 4,950 and rate = 15%
The sum invested is Rs.30,000.
7 . A sum of money is invested at 10% per annum compounded half yearly. If the difference of amounts at the end of 6 months and 12 months is Rs.189, find the sum of money invested.
Let the sum of money be Rs. y
and rate = 10% p.a. compounded half yearly
y = 3600.
8. Rohit borrows Rs. 86,000 from Arun for two years at 5% per annum simple interest. He immediately lends out this money to Akshay at 5% compound interest compounded annually for the same period. Calculate Rohit’s profit in the transaction at the end of two years.
P = Rs. 86,000; time = 2 years and rate = 5% p.a.
To calculate S.I
Profit = C.I. - S.I. = Rs.8,815 - Rs.8,600 = Rs.215
9. The simple interest on a certain sum of money for 3 years at 5% per annum is Rs.1,200. Find the amount and the compound interest due on this sum of money at the same rate and after 2 years. Interest is reckoned annually.
Let Rs.X be the sum of money.
Rate = 5 % p.a. Simple interest = Rs.1,200, n = 3 years.
⇒ A = 8,000( 1.1025 )
⇒ A = 8,820
C.I. = A – P
⇒ C.I. = 8,820 – 8,000
⇒ C.I. = 820.
The amount due after 2 years is Rs. 8,820 and the compound interest is Rs. 820.
10. Nikita invests Rs.6,000 for two years at a certain rate of interest compounded annually. At the end of first year it amounts to Rs.6,720. Calculate:
(a) The rate of interest.
(b) The amount at the end of the second year. |
Find the distance of the point (-1,-5,-10) from the point of intersection of the line $$\overrightarrow r = (2\hat i - \hat j + 2\hat k)+\lambda (3\hat i + 4\hat j + 2\hat k)$$ and the plane $$\overrightarrow r = (\hat i - \hat j + \hat k)=5.$$
Toolbox:
• Distance between two points is $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 1:
The equation of the line is $\overrightarrow r=2\hat i-\hat j+2\hat k+\lambda(3\hat i+4\hat j+2\hat k)$-----(1)
The equation of the plane is $\overrightarrow r.(\hat i-\hat j+\hat k)=5$------(2)
On solving equ(1) and (2)
(i.e)$[2\hat i-\hat j+2\hat k+\lambda(3\hat i+4\hat j+2\hat k)].(\hat i-\hat j+\hat k)=5$
On simplifying we get
$(2\hat i-\hat j+2\hat k).(\hat i-\hat j+\hat k)+\lambda(3\hat i+4\hat j+2\hat k).(\hat i-\hat j+\hat k)=5$
Multiplying by applying dot product we get
$(2+1+2)+\lambda(3-4+2)=5$
Step 2:
On simplifying we get
$5+\lambda=5$
Therefore $\lambda=0$
The point of intersection of the line and the plane is $(2\hat i-\hat j+2\hat k)$
The other point is $(-1,-5,-10)=-\hat i-5\hat j-10\hat k$
Step 3:
Distance between two points is $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Substituting for $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ we get,
$\sqrt {(2-(-1))^2+(-1+5)^2+(2-(-10))^2}$
$\Rightarrow \sqrt{3^2+4^2+12^2}$
$\Rightarrow \sqrt{9+16+144}$
$\Rightarrow \sqrt{169}$
$\Rightarrow 13$
Hence the required distance is $13$ |
# Volume and Surface Area of a Pyramid
Formula of volume and surface area of a pyramid are used to solve the problems step-by-step with the detailed explanation.
Worked-out examples on volume and surface area of a pyramid:
1. A right pyramid on a square base has four equilateral triangles for its four other faces, each edge being 16 cm. Find the volume and the area of whole surface of the pyramid.
Solution:
Let the square WXYZ be the base of the right pyramid and its diagonal WY and XZ intersect at O. If OP be perpendicular to the plane of the square at O, then OP is the height of the right pyramid.
By question, lateral faces of the pyramid are equilateral triangles; hence,
PW = WX = XY = YZ = ZW = 16 cm.
Now, from the right-angled ∆ WXY we get,
WY² = WX² + XY²
or, WY² = 16² + 16²
or, WY² = 256 + 256
or, WY² = 512
or, WY = √512
Therefore, WY = 16√2
Therefore, WO = 1/2 ∙ WY = 8√2
Again OP is perpendicular to the plane of the square WXYZ at O; hence, OP ┴ OW.
Therefore, from the eight angled triangle POW we get,
OP² + OW² = PW²
or, OP² = PW² - OW²
or, OP² = 16² - (8√2)²
or, OP² = (8√2)²
Therefore, OP = 8√2
Now, draw OEWX; then, OE = 1/2 XY = 8 cm.
Join PE,
Clearly, PE is the slant height of the right pyramid.
Since OPPE,
Hence from the right angle triangle POE we get,
PE² = OP² + OE²
or, PE² = (8√2)² + 8²
or, PE² = 128 + 64
or, PE² = 192
Therefore, PE = 8√3
Therefore, the required volume of a right pyramid = 1/3 × (area of the square WXYZ) × OP
= 1/3 × 16² × 8√2 cu. cm. = 1/3 ∙ 2048√2 cu. cm.
And area of its whole surface
= 1/2 (perimeter of square WXYZ) × PE + area of square WXYZ.
= [1/2 ∙ 4 ∙ 16 ∙ 8√3 + 16²] sq. cm.
= 256(√3 + 1) sq. cm.
2. The base of a right pyramid is a regular hexagon each of whose sides is 8 cm. and the lateral faces are isosceles triangles whose two equal sides are 12 cm. each.
Find the volume of the pyramid and the area of all its faces.
Solution:
Let O be the centre of the regular hexagon ABCDEF, the base of the right pyramid and P, the vertex of the pyramid. Join PA, PB, OB and PM where M is the mid-point of AB.
Then, OP is the height and PM, the slant height of the pyramid.
According to the question, AB = 8 cm. and
PA = PB = 12 cm; hence, AM = 1/2 ∙ AB = 4 cm.
Clearly, PMAB, hence from the right angled ∆ PAM we get ,
AM² + PM² = PA²
or, PM² = PA² - AM²
or, PM² = 12² - 4²
or, PM² = 144 - 16
or, PM² = 128
Therefore, PM = 8√2
Again, OP is perpendicular to the plane of the hexagon ABCDEF at O; hence OPOB.
Therefore, from the right angled ∆ POB we get ,
OP² + OB² = PB²
OP² = PB² - OB²
or, OP² = 12² - 8² (Since OB = AB = 8 cm)
or, OP² = 144 - 64
or, OP² = 80
Therefore, OP = 4√5.
Now, the area of the base of the pyramid = area of the regular hexagon ABCDEF
= {(6 ∙ 8²)/4} cot (π/6) [Since, the area of regular polygon of n sides = {(na²)/4} cot (π/n), a being the length of a side].
= 96√3 sq. cm.
Therefore, the required volume of the pyramid
= 1/3 × ( area of the the hexagagon ABCDEF) × OP
= 1/3 × 96√3 × 4√5 cu. cm.
= 128 √15 cu.cm.
And the area of all its faces
= area of the slant surfaces + area of the base
= 1/2 × perimeter of the base × slant height + area of hexagon ABCDEF
= [1/2 × 6 × 8 × 8√2 + 96√3] sq. cm.
= 96 (2√2 + √3] sq. cm.
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Mathematics questions
This Mathematics questions helps to fast and easily solve any math problems. We will also look at some example problems and how to approach them.
The Best Mathematics questions
Mathematics questions can be found online or in math books. Simple solutions math is a method of teaching that focuses on breaking down complex problems into small, manageable steps. By breaking down problems into smaller pieces, students can better understand the concepts behind the problem and find more creative solutions. Simple solutions math also encourages students to think outside the box and approach problems from different angles. As a result, Simple solutions math can be an effective way to teach problem-solving skills. In addition, Simple solutions math can help to improve test scores and grades. Studies have shown that students who use Simple solutions math outperform those who do not use the method. As a result, Simple solutions math is an effective tool for helping students succeed in school.
How to solve using substitution is best explained with an example. Let's say you have the equation 4x + 2y = 12. To solve this equation using substitution, you would first need to isolate one of the variables. In this case, let's isolate y by subtracting 4x from both sides of the equation. This gives us: y = (1/2)(12 - 4x). Now that we have isolated y, we can substitute it back into the original equation in place of y. This gives us: 4x + 2((1/2)(12 - 4x)) = 12. We can now solve for x by multiplying both sides of the equation by 2 and then simplifying. This gives us: 8x + 12 - 8x = 24, which simplifies to: 12 = 24, and therefore x = 2. Finally, we can substitute x = 2 back into our original equation to solve for y. This gives us: 4(2) + 2y = 12, which simplifies to 8 + 2y = 12 and therefore y = 2. So the solution to the equation 4x + 2y = 12 is x = 2 and y = 2.
A radical is a square root or any other root. The number underneath the radical sign is called the radicand. In order to solve a radical, you must find the number that when multiplied by itself produces the radicand. This is called the principal square root and it is always positive. For example, the square root of 16 is 4 because 4 times 4 equals 16. The symbol for square root is . To find other roots, you use division. For example, the third root of 64 is 4 because 4 times 4 times 4 equals 64. The symbol for the third root is . Sometimes, you will see radicals that cannot be simplified further. These are called irrational numbers and they cannot be expressed as a whole number or a fraction. An example of an irrational number is . Although radicals can seem daunting at first, with a little practice, they can be easily solved!
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# What is the instantaneous velocity of an object moving in accordance to f(t)= (t^2sin(t-pi),tcost) at t=pi/3 ?
Jan 2, 2016
$0.172261$
#### Explanation:
The instantaneous velocity is equal to $f ' \left(\frac{\pi}{3}\right)$.
$x \left(t\right) = {t}^{2} \sin \left(t - \pi\right)$
To find $x ' \left(t\right)$, use the product rule.
$x ' \left(t\right) = 2 t \sin \left(t - \pi\right) + {t}^{2} \cos \left(t - \pi\right)$
We also know that
$y \left(t\right) = t \cos t$
Again, differentiate with the product rule.
$y ' \left(t\right) = \cos t - t \sin t$
The derivative of the entire parametric equation is found as follows:
$f ' \left(t\right) = \frac{y ' \left(t\right)}{x ' \left(t\right)} = \frac{\cos t - t \sin t}{2 t \sin \left(t - \pi\right) + {t}^{2} \cos \left(t - \pi\right)}$
Find $f ' \left(\frac{\pi}{3}\right)$.
$f ' \left(\frac{\pi}{3}\right) = \frac{\cos \left(\frac{\pi}{3}\right) - \frac{\pi}{3} \sin \left(\frac{\pi}{3}\right)}{2 \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{3} - \pi\right) + {\left(\frac{\pi}{3}\right)}^{2} \cos \left(\frac{\pi}{3} - \pi\right)}$
$\approx 0.172261$ |
#### Transcript Document
```COURSE: JUST 3900
TIPS FOR APLIA
Chapters 7-9:
Test Review
Developed By:
John Lohman
Michael Mattocks
Aubrey Urwick
Chapter 7: Key Concepts
This chapter is on the distribution of sample means.
Sampling error is the natural discrepancy, or difference,
between a sample statistic and its corresponding
population parameter.
The distribution of sample means, or sampling
distribution, is the collection of sample means for all of
the possible random samples of a particular size (n) that
can be obtained from a population.
Chapter 7: Key Concepts
The mean of a sampling distribution, or expected value
of M, should always be equal to the population mean (µ).
The standard error of M (σM) is the standard deviation of
the distribution of sample means. It provides a measure
of how much distance is expected on average between a
sample mean (M) and the population mean (µ).
Standard Error of M: 𝜎𝑀 =
𝜎
𝑛
Chapter 7: Key Concepts
The central limit theorem states that for any population
with mean µ and standard deviation σ, the distribution of
sample means for sample size n will have a mean of µ
𝜎
and a standard deviation of 𝜎𝑀 = and will approach a
𝑛
normal distribution as n approaches infinity.
In other words, a sampling distribution has the following
characteristics:
A mean of µ. (Expected value of M)
A standard deviation of 𝜎𝑀 =
Is normal in shape if n is at least 30, or if the population
distribution is normal in shape.
𝜎
.
𝑛
(Standard error of M)
Chapter 7: Practice
Question 1: A population has a mean of µ = 100 and a
standard deviation of σ = 15.
a)
b)
c)
d)
For samples of size n = 5, what is the expected value and the
average difference between M and µ for the distribution of
sample means?
If the population distribution is not normal, describe the shape
of the distribution of sample means based on n = 5.
For samples of size n = 45, what is the expected value and the
average difference between M and µ for the distribution of
sample means?
If the population distribution is not normal, describe the shape
of the distribution of sample means based on n = 45.
Chapter 7: Practice
a)
Expected Value of M: µ = 100
Standard Error of M: 𝜎𝑀 =
b)
𝜎
𝑛
=
15
5
= 6.71
The distribution of sample means does not satisfy either
criterion to be normal. It would not be a normal distribution.
If the population distribution was normal, or if the sample size was at least 30,
the sampling distribution would be normal.
Chapter 7: Practice
c)
Expected Value of M: µ = 100
Standard Error of M: 𝜎𝑀 =
𝜎
𝑛
=
15
45
= 2.24
Notice how the expected value stays the same regardless of sample size, while
the standard error decreases as sample size increases.
d)
Because the sample size is greater than 30, the distribution of
sample means is a normal distribution.
Chapter 7: Key Concepts
Notice the relationship between sample size n and standard error 𝜎𝑀 .
As n increases, 𝜎𝑀 decreases.
Chapter 7: Key Concepts
Again, as sample size increases, standard error decreases.
Chapter 7: Key Concepts
Standard Deviation (σ)
1
5
10
15
20
Standard Error
1
𝜎𝑀 =
100
5
𝜎𝑀 =
100
10
𝜎𝑀 =
100
15
𝜎𝑀 =
100
20
𝜎𝑀 =
100
0.10
0.50
1.00
1.50
2.00
As standard deviation increases, so does standard error of M.
Chapter 7: Key Concepts
The relationship between sample size and standard error is an important concept.
Understanding this relationship can save you valuable time on the test.
Chapter 7: Practice
Question 2: If we have a sampling distribution with a
standard error of 𝜎𝑀 = 5 and sample size of n = 15, what
would the standard error be if we increased our sample
size to n = 30?
a)
b)
c)
d)
𝜎𝑀
𝜎𝑀
𝜎𝑀
𝜎𝑀
= 7.12
= 6.57
= 3.54
= 5.64
Chapter 7: Practice
c) 𝜎𝑀 = 3.54
𝜎
𝜎𝑀 =
𝑛
𝜎
5=
15
𝜎 = 19.36
𝜎𝑀 =
𝜎
19.36
=
= 3.54
𝑛
30
There are two ways to arrive at this answer.
1) Find the standard deviation, then re-calculate standard error for n = 30, or
2) Eliminate all of the other answer choices based on our rule: as sample size
increases standard error decreases. In this case, answer choice c) is the only
answer that is less than 5.
Chapter 7: Sampling
Distributions and Probability
Very similar to what we did in chapter 6, we can
calculate the probability of selecting a specific sample of
size n by calculating the z-score and looking up the
probability in the unit normal table.
𝑧=
𝑀−𝜇
𝜎𝑀
Chapter 7: Practice
Question 3: What is the probability of obtaining a sample
mean greater than M = 60 for a random sample of n = 16
scores selected from a normal population with a mean of
µ = 65 and a standard deviation of σ = 20? What if we
changed our sample size to n = 5?
Chapter 7: Practice
Find 𝜎𝑀 .
𝜎
𝑛
=
20
16
=
20
4
= 5.00
Find the z-score.
𝜎𝑀 =
𝑧=
𝑀−𝜇
𝜎𝑀
=
60−65
5
=
−5
5
= −1.00
Look up z = -1.00 in the unit normal table. (Column B)
p(z > -1.00) = 0.8413 (or 84.13%)
Chapter 7: Practice
Find 𝜎𝑀 .
𝜎
𝑛
=
20
5
20
= 2.24 = 8.94
Find the z-score.
𝜎𝑀 =
𝑧=
𝑀−𝜇
𝜎𝑀
=
60−65
8.94
−5
= 8.94 = −0.56
Look up z = -0.56 in the unit normal table. (Column B)
p(z > -0.56) = 0.7123 (or 71.23%)
Chapter 8: Key Concepts
Chapter 8 covers hypothesis testing.
Hypothesis tests allow us to make generalizations from
samples to populations about whether our treatment has
an effect.
When there appears to be a treatment effect, one of two
things has happened.
The discrepancy between µ and M is the result of sampling error
or,
Chapter 8: Key Concepts
There are 4 steps to hypothesis testing:
1)
State the hypothesis.
1)
2)
We state both the null hypothesis H0, which states that there is no
change, and the alternative hypothesis H1, which states that there
is a change.
Set criteria for a decision.
1)
Next, we divide our distribution up into two sections:
1)
2)
2)
3)
Sample means close to the null hypothesis.
Sample means very different from the null hypothesis.
We use the alpha level, or level of significance, to mark the critical
region, which is composed of sample values that are very unlikely
to be obtained if the null is true.
These boundaries are generally set at α = 0.05, α = 0.01, or
α = 0.001
Chapter 8: Key Concepts
Critical region
for 2-tail test at
α = 0.05.
α Level
z-Score
.05
+/- 1.96
.01
+/- 2.58
.001
+/- 3.30
Chapter 8: Key Concepts
Critical
region
for 1-tail
test at
α = 0.05.
α Level
z-Score
.05
+/- 1.65
.01
+/- 2.33
.001
+/- 3.10
Chapter 8: Key Concepts
There are 4 steps to hypothesis testing:
3)
Compute the sample statistic.
1)
4)
𝑧=
𝑀−𝜇
𝜎𝑀
=
𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛 −ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑧𝑒𝑑 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑀 𝑎𝑛𝑑 𝜇
Make a decision.
1)
2)
3)
Does our sample statistic fall in the critical region?
If yes, reject the null. The treatment has an effect.
If no, fail to reject the null. The treatment does not have an effect.
Chapter 8: Practice
Question 1: What combination of factors will most likely
lead to rejecting the null hypothesis?
a)
b)
c)
d)
n = 30; α = 0.05
n = 5; α = 0.05
n = 30; α = 0.01
n = 5; α = 0.01
Chapter 8: Practice
a)
b)
c)
d)
n = 30; α = 0.05
n = 5; α = 0.05
n = 30; α = 0.01
n = 5; α = 0.01
A larger sample size leads to a smaller
Standard error, which in turn gives us
A larger z-score, increasing the likelihood
That it will fall in the critical region.
The larger alpha level increases the size of
the critical region, making it more likely that
our z-score will fall in this region.
Chapter 8: Practice
Question 2: On average, what would we expect our zscore to equal if the null hypothesis is true?
Chapter 8: Practice
z = 0, indicating that there was no change in µ and the treatment
Remember that our null hypothesis states that µ doesn’t change, and the µ for
the unit normal table (as with all z-distributions) is 0.
Chapter 8: Practice
Question 3: State the null and alternative hypotheses for
a one-tailed test with µ = 50. We expect our treatment to
have a positive effect.
a)
b)
c)
d)
H0: µ = 50
H1: µ ≠ 50
H0: µ = 0
H 1: µ ≠ 0
H0: µ > 50
H1: µ ≤ 50
H0: µ ≤ 50
H1: µ > 50
Chapter 8: Practice
a)
b)
c)
d)
H0: µ = 50
H1: µ ≠ 50
H 0: µ = 0
H 1: µ ≠ 0
H0: µ > 50
H1: µ ≤ 50
H0: µ ≤ 50
H1: µ > 50
I often find it less confusing to state the alternative
hypothesis first in a one-tailed test. We know that if
there’s an effect, µ will get larger, µ > 0. However, if µ
doesn’t get larger, it will either stay the same or get
smaller, µ ≤ 0.
Chapter 8: Key Concepts
Occasionally, hypothesis tests can lead to error.
There are two types of error in hypothesis testing: Type I
and Type II.
Type I error occurs when a researcher rejects a null
hypothesis that is actually true. In these cases, the
sample statistic falls in the critical region, not because of
a treatment effect, but as a result of sampling error.
Type I error is also known as a false positive.
The probability of a Type I error is equal to the alpha
level, or level of significance.
Chapter 8: Key Concepts
Type II error, on the other hand, occurs when a
researcher fails to reject a false null hypothesis.
This means that a treatment effect really exists, but the
hypothesis test fails to detect it. This is often the case
with very small treatment effects.
Chapter 8: Key Concepts
α
1-β
1-α
β
Chapter 8: Practice
Question 4: Which of the following will decrease the risk
of a type I error?
a)
b)
c)
d)
e)
f)
Increasing the sample size (n)
Decreasing the alpha from α = 0.05 to α = 0.01
Moving from a one- to a two-tailed test
All of the above
None of the above
Some of the above
Chapter 8: Practice
a)
b)
c)
d)
e)
f)
Increasing the sample size (n)
Decreasing the alpha from α = 0.05 to α = 0.01
Moving from a one- to a two-tailed test
All of the above
None of the above
Some of the above
Increasing n reduces 𝜎𝑀 , making it less likely that the effect is due to chance.
Decreasing alpha makes it more difficult to get into the critical region, reducing
the chance of getting a type I error. Here, we reduced the probability of a type I
error from 5% to 1%.
Just like when we decrease alpha, moving from a one- to a two-tailed test
reduces the probability of getting a type I error.
Chapter 8: Practice
Question 5: When are type II errors likely to occur?
Chapter 8: Practice
When the treatment effect is very small.
Remember: Type II errors occur when we fail to reject a false null hypothesis.
In other words, our treatment has an effect, but for some reason our z-score
did not fall in the critical region.
Chapter 8: Key Concepts
However, hypothesis testing doesn’t tell the whole story.
It tells us whether there is a significant treatment effect,
but not the size of the effect.
To find effect size, we calculate Cohen’s d.
𝑀𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝑀𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 −𝜇𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝜎
𝐶𝑜ℎ𝑒𝑛′ 𝑠 𝑑 =
Effect sizes are summarized in the following chart.
=
Cohen’s d is not influenced by sample size.
Chapter 8: Practice
Question 6: A researcher selects a sample from a
population with µ = 45 and σ = 8. A treatment is
administered to the sample and, after treatment, the
sample mean is found to be M = 47. What is the size of
the treatment effect?
Chapter 8: Practice
𝑀𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 −𝜇𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝜎
𝐶𝑜ℎ𝑒𝑛′ 𝑠 𝑑 =
This is a small effect.
=
47−45
8
2
8
= = 0.25
Chapter 8: Key Concepts
Power refers to the probability that a hypothesis test will
correctly reject a false null hypothesis. That is, power is
the likelihood that the test will identify a treatment effect
if one really exists.
Like hypothesis testing, power is a 4-step process:
1)
Calculate standard error.
2)
𝜎𝑀 =
𝜎
𝑛
Locate boundary of critical region.
M = µ + (Critical z-score * 𝜎𝑀 )
Chapter 8: Key Concepts
3)
Calculate z-score for the difference between the treated
sample mean for the critical region boundary and the
population mean with the treatment effect.
4.
𝑧=
𝑀−𝜇
𝜎𝑀
Interpret power of the hypothesis test.
Find probability associated with your z-score.
Chapter 8: Key Concepts
Chapter 8: Practice
Question 7: What is the power of a hypothesis test if the
probability of a type II error is β = 0.6046?
Chapter 8: Practice
Power: 1 − β = 1 − 0.6046 = 0.3954 or 39.54%
Chapter 8: Practice
Question 8: What 3 factors increase power?
Chapter 8: Practice
1)
2)
3)
Sample size is increased.
Alpha is increased (e.g., from .01 to .05).
You go from a 2- to a 1-tail test.
Chapter 9: Practice
Question 1: Which of the following is a fundamental
difference between the t statistic and a z-score?
a)
b)
c)
d)
The t statistic uses the sample mean in place of the population
mean.
The t statistic uses the sample variance in place of the
population variance.
The t statistic computes the standard error by dividing the
standard deviation by n-1 instead of dividing by n.
All of the above are differences between z and t.
Chapter 9: Practice
a)
b)
c)
d)
The t statistic uses the sample mean in place of the population
mean.
The t statistic uses the sample variance in place of the
population variance.
The t statistic computes the standard error by dividing the
standard deviation by n-1 instead of dividing by n.
All of the above are differences between z and t.
Without the population standard deviation or variance, we cannot calculate
a z-score.
Chapter 9: Practice
Question 2: A sample of n = 25 is selected from a
population with a mean of µ = 50. A treatment is
administered to the individuals in the sample and, after
treatment, the sample has a mean of M = 56 and a
variance of s = 5.
If all other factors are held constant and the sample size is
increased to n = 25, is the sample sufficient to conclude that the
treatment has a significant effect? (Use a two-tailed test with
α = 0.05)
Chapter 9: Practice
Step 1: State hypotheses
H0: Treatment has no effect. (µ = 50)
H1: Treatment has an effect. (µ ≠ 50)
Chapter 9: Practice
Step 2: Set Criteria for Decision (α = 0.05)
t Critical: ± 2.064
Chapter 9: Practice
df = 24
t Distribution with
α = 0.05
Critical region
t = - 2.064
Critical region
t = + 2.064
Chapter 9: Practice
a)
b)
Step 3: Compute sample statistic
𝑠𝑀 =
𝑡=
𝑠
𝑛
𝑀−𝜇
𝑠𝑀
=
5
25
5
5
=
56−50
1
= = 1.00
=
6
1
= 6.00
Chapter 9: Practice
df = 24
t Distribution with
α = 0.05
Critical region
t = - 2.064
Critical region
t = + 2.064
t = 6.00
Chapter 9: Practice
Step 4: Make a decision
For a Two-tailed Test:
If -2.064 < tsample < 2.064, fail to reject H0
If tsample ≤ -2.064 or tsample ≥ 2.064, reject H0
tsample (6.00) > tcritical (2.064)
Thus, we reject the null and conclude that the treatment has an
effect.
Chapter 9: Practice
Question 3: A sample of n = 16 is selected from a
population. A treatment is administered to the sample
and, after treatment, the sample is found to be M = 86
with a standard deviation of s = 8. A confidence interval
is constructed and the interval spans μlower = 81.738 and
μupper = 90.262. How confident are we that our µ falls
within this interval?
Chapter 9: Practice
1)
Find the corresponding t-statistics for μupper and μlower.
𝑠
𝑛
=
8
16
8
𝑠𝑀 =
= 4 = 2.00
𝑡=
𝑀−𝜇
𝑠𝑀
=
81.738−86
2
=
−4.262
2
𝑡=
𝑀−𝜇
𝑠𝑀
=
90.262−86
2
=
4.262
2
= −2.131
= 2.131
Chapter 9: Practice
df = 15
Middle ?%
of t distribution
?% in the lower tail
t = - 2.131
?% in the upper tail
t = + 2.131
Chapter 9: Practice
2)
Use the t-distribution to find the alpha level (percentage
between both tails)
100% - 5% = 95%
Chapter 9: Practice
df = 15
Middle 95%
of t distribution
2.5% in the lower tail
t = - 2.131
2.5% in the upper tail
t = + 2.131
Chapter 9: Practice
|
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# Algebra Expressions
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Algebra Expressions Year 9
Note 1: Expressions We often use x to represent some number in an equation. We refer to the letter x as a variable. e.g. x + 5 means ‘a number with 5 added on’ x – 7 means ‘a number with 7 subtracted from it’
Note 1: Expressions A few Rules: A number should be written before a letter. y x 2 = 2y e.g. Terms should be written in alphabetical order. xyz and 3b x 4a = 12ab e.g. We don’t use the x or ÷ signs in algebra instead we write it like this: 5 x y = 5y x ÷ 9 = x e.g. 9
Activity: Expressions Match each algebraic expression with the phrase Five times a number x – 8 3 + x Three plus a number A number multiplied by seven x x Half of a number 2 2 A number plus six 7x 3x + 1 x A number divided by nine 5x 9 A number minus eight x Three times a number plus one x + 6 9
Activity: Expressions Write an equivalent algebraic expression for each phrase Twelve times a number 12x One plus a number 1 + x A number multiplied by three 3x x A quarter of a number 4 A number plus one hundred x + 100 x A number divided by nineteen 19 A number minus four x – 4 IWB Ex 11.01 Eight times a number minus one 8x – 1 Pg 275-276
Note 2: Substitution • We replace the variable (letter) with a number and calculate the answer. • Algebra follows the same rules as BEDMAS! e.g. If a = 2, b = -3, c = 5 then calculate: a + 5 3b a + b + c = 2 + 5 = 3 x -3 = 2 + -3 + 5 = 7 = -9 = 2 – 3 + 5 = 4
Note 2: Substitution Remember: Do multiplication and division before addition and subtraction Anything in brackets is worked out first A number in front of the bracket means multiply The fraction line means divide e.g. If d = 3, e = 7, f = -2 then calculate: 14− f 4 14 − − 2 4(d + e) 5def – 2e = 4 = 4(3 + 7) = 5 x 3 x 7 x -2 – 2 x 7 2 = 14 + = 4 x 10 = -210 – 14 4 = 40 = -224 = 4
Note 2: Substitution Remember: Do multiplication and division before addition and subtraction Anything in brackets is worked out first A number in front of the bracket means multiply The fraction line means divide e.g. If d = 3, e = 7, f = -2 then calculate: 3 −e 7 7 2(d - e) 5f – 2d 3× 7 − 7 2 = 2(3 - 7) = 5 x -2 – 2 x 3 = 7 2 2 = 2 x (-4) = -10 – 6 = 21− 7 = 2 x 16 7 = 32 = -16 = 2
Note 2: Substitution e.g. Number of tyres = 5x the number of cars b Number of tyres = 5 x 60 c 5 x 40 = 200 = 300
Starter - = 5 x 3 = 5 x -4 = 3 x -4 = 15 = -20 = -12 = 2 x -4 = 12 x 3 = 2 x -4 = -8 = 36 = -8 = 6 x 2 = 3 x 2 = 3 x 2 x -4 = 12 = 6 = -24
Note 3: Formulas A formula is a mathematical rule that explains how to calculate some quantity. e.g. John baby sits for his neighbours. He charges a set fee of \$10 plus \$5 for every hour (h), that he baby sits. A formula to calculate this charge is given by: Charge = \$10 + \$5h Use the formula to calculate the amount John charges if he baby sits for: 5 hours 3 hours = 10 + 5 x 5 = 10 + 5 x 3 = \$ 35 = \$ 25
Note 3: Formulas e.g. If John receives \$65 how long did he baby sit for? Charge = \$10 + \$5h \$65 = \$10 + \$5 x h 65 = 10 + 5h 65-10 = 10-10 + 5h 55 = 5h 5 5 11 = h IWB John baby sat for 11 hours Ex 12.01 Pg 299-302 Ex 12.02 Pg 305-308
Starter Multiply the base length by the height and divide by 2 5× 8 = 20 cm 2 2
Note 4: Multiplying Algebraic Expressions Rules: Multiply the numbers in the expression (these are written first) Write letters in alphabetical order
Note 5: Adding & Subtracting Algebraic Expressions Write in simplest form: y + y + y + y = 4y a + a + a + a + a = 5a a + a + a + a + a + b = 5a + b a + a + a + a + a + b + b + b = 5a + 3b x + y + x + y + x + y + x = 4x + 3y x + y + y + y + y - x = 4y
Note 5: Adding & Subtracting Algebraic Expressions Write in simplest form: y + 3y = 4y 3x + 5x = 8x 5x – 2x + 3x = 6x 5x – 4x = x 10x + x + 19x = 30x ALPHA 12p + 3r – 2p + 3r = 10p + 6r Ex 11.06 Pg 161
Note 6: Like and Unlike Terms Like terms are terms that contain the exact same variables (letters) or combinations of letters. e.g. Like Terms 2x, 5x, 25x, -81x, 13x, 0.5x… Remember: If we had written these xy, 2xy, -4xy, ½xy, -100xy,… terms properly (in alphabetical order), it 2abc, 4bac, 6 cab, 9abc, ….. would be more obvious that they are like terms.
Note 6: Like and Unlike Terms Like terms are terms that contain the exact same variables (letters) or combinations of letters. e.g. Unlike Terms 2, 2x, 3y 3a, 7ab, 8b 2p, 4r, 10s
Note 6: Like and Unlike Terms Expressions can have a mixture of LIKE and UNLIKE terms. e.g. 3x + 5y 2a – 7b + 5a + 8b = 7a + b Like terms can be grouped together and simplified Unlike terms cannot be simplified e.g. 5a + 2b – 3a + 6c + 4b 2a + 6b + 6c
Note 6: Like and Unlike Terms 2x + 10y 8x + 8y 4x + 1 6p + 9q x + 1 2x + 2 alpha x Ex 11.07 Pg 162 2x – 7 Ex 11.08 Pg 164
Starter Find the perimeter of this shape in terms of x x The perimeter is the sum of all side lengths P = x + (2x + 1) + (2x + 3) + (4x + 1) + (5x – 2) = 14x + 3 If the perimeter is 31 cm. What is the value of x and which side is the longest? P = 14x + 3 28 = 14x 31 = 14x + 3 14 14 x = 2
Note 7: Powers (exponents) Recall: When a variable (letter) is multiplied by itself many times, we use powers e.g. Write the following in index form: 3 p p x p x p = ________________ q 5 q x q x q x q x q = __________ s t 3 4 s x s x s x t x t x t x t = _______ p q s p x p x q x q x s x s = ________ 2 2 2 s t 3 3 s x t x t x t x s x s = __________
Note 7: Powers (exponents) Substituting: Evaluate the following when a =2, b = 7 and c = -3 2 2 2 2 a = 2 5a = 5 x 2 = 4 = 5 x 4 = 20 (b + c) 2 = (7 - 3) 2 2 = (4) 2 2 2 2 = 16 2a c = 2 x 2 x (-3) = 2 x 4 x 9 (5a) 2 = (5 x 2) = 72 2 2 = 10 = 100
Note 7: Powers (exponents) Multiplying: Simplify: b x b 3 4 = (b x b x b) x (b x b x b x b) = b 7 When multiplying power expressions with the same base, we add the powers. e.g. e x e g x g 5m x 4m 3 3 6 2 8 = e 2+6 = g 8+1 = 20m 3+3 = e 8 = g 9 = 20m 6
Note 7: Powers (exponents) Dividing: IWB c 5 c× c× c× c× c Ex 13.08 Pg 346 Simplify: = = c Ex 13.09 Pg 349 3 c 2 c× c PUZZLE Pg 350 When dividing power expressions with the same base, we subtract the powers. f e.g. 6 g ÷ g 25q 7 6s 8 7 f 3 5q 3 4s 2 = g 7-1 = f 6-3 = 5q 7-3 = 3s 6 = g 6 = f 3 = 5q 4 2
Starter How would you calculate 7 x 83 in your head? 7 x 80 + 7 x 3 560 + 21 = 581 What we have done in our head can also look like this: x x 7 (80 + 3) = 7 x 80 + 7 x 3 = 581
Note 8: Expanding Brackets To expand (remove) brackets: – Multiply the outside term by everything inside the brackets – Simplify where possible The e.g. Expand: Distributive Law a.) 4(x + y) = 4x + 4y b.) 2(x – y) = -2x + 2y − c.) 5(x – y + 2z) = 5x - 5y + 10z
Try These! e.g. Expand: a.) 8(x + y) = 8x + 8y f.) -8(x + y) = -8x – 8y b.) 4(x – y) = 4x - 4y g.) 5(x – 3y) = 5x – 15y c.) 2(x – y) = 2x - 2y h.) -(x – 2y) = -x + 2y d.) 3(-x + y) = -3x + 3y i.) -7(-x + 7y) = 7x – 49y e.) 9(x + y + z) = 9x + 9y + 9z j.) -4(3x - y + 5z) = -12x + 4y – 20z
Lets do some more e.g. Expand: a.) 8(x + 4) = 8x + 32 f.) -8(x + 8) = -8x – 64 b.) 4(x – 2y) = 4x – 8y g.) 5(5x – 3) = 25x – 15 c.) 2(x – 10) = 2x - 20 h.) -(x – 11) = -x + 11 d.) 3(2x + 9) = 6x + 27 i.) -7(x + 12) = -7x – 84 e.) -(x + 2y - z) = -x – 2y + z j.) -2(x - y + 14) = -2x + 2y – 28 |
# Find the determinant by using elementary row operations
I'm having a problem finding the determinant of the following matrix using elementary row operations. I know the determinant is -15 but confused on how to do it using the elementary row operations. Here is the matrix
$$\begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix}$$
Thank you
Note that the determinant of a lower (or upper) triangular matrix is the product of its diagonal elements. Using this fact, we want to create a triangular matrix out of your matrix \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} So, I will start with the last row and subtract it from the second row to get \begin{bmatrix} 2 & 3 & 10 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix} Now, I want to get rid of the $2$ in the first row. I thus multiply the last row by $2$ and subtract it from the first row to obtain: \begin{bmatrix} 0 & 1 & 16 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix} Finally, I subtract the second row from the first one to obtain \begin{bmatrix} 0 & 0 & 15 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix} We now have $$\det \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} = \det \begin{bmatrix} 0 & 0 & 15 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}$$
Now, I will transform the RHS matrix to an upper diagonal matrix. I can exchange the first and the last rows. Exchanging any two rows changes the sign of the determinant, and therefore $$\det \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} = -\det \begin{bmatrix} 1 & 1 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 15 \end{bmatrix}$$
The matrix on the RHS is now an upper triangular matrix and its determinant is the product of its diagonal elements, which is $15$. With the minus sign, the $\det$ of our initial matrix is thus $-15$. |
# Cube of Difference
In this chapter you will learn formula for cube of difference and along with the solved problems at the end.
## Cube of Difference definition
The formula for cube of difference is given as;
\mathtt{( a-b)^{3} =a^{3} -\ b^{3} -3ab.( a-b)}
The formula can also be written as;
\mathtt{( a-b)^{3} =a^{3} -\ b^{3} -3a^{2} b\ +3ab^{2}}
The cube of difference of two terms is given by difference of cube of individual terms and addition of \mathtt{-3a^{2} b\ \&\ 3ab^{2}}
## Proof of cube of difference formula
The expression given is \mathtt{( a-b)^{3}}
Rewriting the expression;
\mathtt{\Longrightarrow \ ( a-b)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( a-b) \ ( a-b)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( a-b)\left( a^{2} -2ab+b^{2}\right)}
Multiplying the expression;
\mathtt{\Longrightarrow \ a^{3} -2a^{2} b+ab^{2} -a^{2} b+2ab^{2} -b^{3}}\\\ \\ \mathtt{\Longrightarrow \ a^{3} -\ b^{3} -3a^{2} b\ +3ab^{2}}
Hence, we get the formula;
\mathtt{( a-b)^{3} =a^{3} -\ b^{3} -3a^{2} b\ +3ab^{2}}
## Proof of cube of difference formula
Let the given expression is \mathtt{( 5-2)^{3}}
Finding value using simple calculation
\mathtt{\Longrightarrow \ ( 5-2)^{3}}\\\ \\ \mathtt{\Longrightarrow \ 3^{3}}\\\ \\ \mathtt{\Longrightarrow \ 27}
Hence, 27 is the value of given expression.
Now let’s find value using the formula;
\mathtt{( a-b)^{3} =a^{3} -\ b^{3} -3a^{2} b\ +3ab^{2}}
Putting the values;
\mathtt{\Longrightarrow \ ( 5-2)^{3}}\\\ \\ \mathtt{\Longrightarrow \ 5^{3} -\ 2^{3} -3( 5)^{2}( 2) \ +3( 5)( 2)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 125-8-150+60}\\\ \\ \mathtt{\Longrightarrow \ 27}
The value of given expression is 27.
In both the above methods we got the same value, hence the formula is valid.
## Cube of Difference – Solved Problems
Example 01
Expand \mathtt{( 5x-3)^{3}}
Solution
The expression is in form of \mathtt{( a-b)^{3}}
We will use the formula;
\mathtt{( a-b)^{3} =a^{3} -\ b^{3} -3a^{2} b\ +3ab^{2}}
Where;
a = 5x
b = 3
Putting the values we get;
\mathtt{\Longrightarrow \ ( 5x-3)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 5x)^{3} -\ 3^{3} -3( 5x)^{2}( 3) \ +3( 5x)( 3)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 125x^{3} -27-225x^{2} +\ 135x}\\\ \\ \mathtt{\Longrightarrow \ 125x^{3} -225x^{2} +135x-27}
Hence, the above expression is expanded form of given question.
Example 02
Expand \mathtt{\left( 2x-y^{2}\right)^{3}}
Solution
The expression is in form of \mathtt{( a-b)^{3}}
We will use the formula;
\mathtt{( a-b)^{3} =a^{3} -\ b^{3} -3a^{2} b\ +3ab^{2}}
Where;
a = 2x
b = \mathtt{y^{2}}
Putting the values;
\mathtt{\Longrightarrow \ \left( 2x-y^{2}\right)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 2x)^{3} - \left( y^{2}\right)^{3} -3( 2x)^{2}\left( y^{2}\right) \ +3( 2x)\left( y^{2}\right)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 8x^{3} -y^{6} -6x^{2} y^{2} +\ 6xy^{4}}
Hence, the above expression is expanded form of given question.
Example 03
Find the value of \mathtt{27^{3}} using cube of difference formula.
Solution
The number can be written as;
\mathtt{( 27)^{3} \Longrightarrow ( 30-3)^{3}}
\mathtt{( 30-3)^{3}} is in form of expression \mathtt{( a-b)^{3}}
Where;
a = 30
b = 3
We will use the formula;
\mathtt{( a-b)^{3} =a^{3} -\ b^{3} -3a^{2} b\ +3ab^{2}}
Putting the values;
\mathtt{\Longrightarrow ( 30-3)^{3} \ }\\\ \\ \mathtt{\Longrightarrow \ ( 30)^{3} -\ ( 3)^{3} -3( 30)^{2}( 3) \ +3( 30)( 3)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 27000-27-8100+\ 810}\\\ \\ \mathtt{\Longrightarrow \ 19683}
Hence, the value of given expression is 19683.
Example 04
Find the value of \mathtt{( 99)^{3}} using cube of difference formula.
Solution
The number can be written as;
\mathtt{( 99)^{3} \ \Longrightarrow ( 100-1)^{3} \ }
\mathtt{( 100-1)^{3}} is in form of expression \mathtt{( a-b)^{3}}
Where;
a = 100
b = 1
We will use the formula;
\mathtt{( a-b)^{3} =a^{3} -\ b^{3} -3a^{2} b\ +3ab^{2}}
Putting the values;
\mathtt{\Longrightarrow ( 100-1)^{3} \ }\\\ \\ \mathtt{\Longrightarrow \ ( 100)^{3} -\ ( 1)^{3} -3( 100)^{2}( 1) \ +3( 100)( 1)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 1000000-1-30000+300}\\\ \\ \mathtt{\Longrightarrow \ 970299}
Hence, 970299 is the value of given expression.
Example 05
Find the value of \mathtt{x^{3} -64y^{3}}
If x – 4y = 5 and x . y = 12
Solution
It’s given that, x – 4y = 5
Taking cube on both sides;
\mathtt{( x-4y)^{3} =5^{3}}
\mathtt{( x-4y)^{3}} is in form of \mathtt{( a-b)^{3}}
Where;
a = x
b = 4y
We will apply the below formula;
\mathtt{( a-b)^{3} =a^{3} -\ b^{3} -3ab( a-b)}
Using the formula, we get;
\mathtt{x^{3} -64y^{3} -3.x.4y( x-4y) \ =\ 5^{3}}\\\ \\ \mathtt{x^{3} -64y^{3} -12xy( x-4y) \ =\ 5^{3}}\\\ \\ \mathtt{x^{3} -64y^{3} -12.12( 5) \ =125}\\\ \\ \mathtt{x^{3} -64y^{3} -720=125}\\\ \\ \mathtt{x^{3} -64y^{3} \ =\ 845}
Hence, 845 is the value of required expression.
Next chapter : Sum of cube formula |
Precalculus
# 10.5Conic Sections in Polar Coordinates
Precalculus10.5 Conic Sections in Polar Coordinates
## Learning Objectives
In this section, you will:
• Identify a conic in polar form.
• Graph the polar equations of conics.
• Define conics in terms of a focus and a directrix.
Figure 1 Planets orbiting the sun follow elliptical paths. (credit: NASA Blueshift, Flickr)
Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets’ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits.
In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body’s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system.
## Identifying a Conic in Polar Form
Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola $x=2+ y 2 x=2+ y 2$ shown in Figure 2.
Figure 2
In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus $P(r,θ) P(r,θ)$ at the pole, and a line, the directrix, which is perpendicular to the polar axis.
If $F F$ is a fixed point, the focus, and $D D$ is a fixed line, the directrix, then we can let $e e$ be a fixed positive number, called the eccentricity, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points $P P$ such that $e= PF PD e= PF PD$ is a conic. In other words, we can define a conic as the set of all points $P P$ with the property that the ratio of the distance from $P P$ to $F F$ to the distance from $P P$ to $D D$ is equal to the constant $e. e.$
For a conic with eccentricity $e, e,$
• if $0≤e<1, 0≤e<1,$ the conic is an ellipse
• if $e=1, e=1,$ the conic is a parabola
• if $e>1, e>1,$ the conic is an hyperbola
With this definition, we may now define a conic in terms of the directrix, $x=±p, x=±p,$ the eccentricity $e, e,$ and the angle $θ. θ.$ Thus, each conic may be written as a polar equation, an equation written in terms of $r r$ and $θ. θ.$
## The Polar Equation for a Conic
For a conic with a focus at the origin, if the directrix is $x=±p, x=±p,$ where $p p$ is a positive real number, and the eccentricity is a positive real number $e, e,$ the conic has a polar equation
$r= ep 1±ecosθ r= ep 1±ecosθ$
For a conic with a focus at the origin, if the directrix is $y=±p, y=±p,$ where $p p$ is a positive real number, and the eccentricity is a positive real number $e, e,$ the conic has a polar equation
$r= ep 1±esinθ r= ep 1±esinθ$
## How To
Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity.
1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
2. Identify the eccentricity $e e$ as the coefficient of the trigonometric function in the denominator.
3. Compare $e e$ with 1 to determine the shape of the conic.
4. Determine the directrix as $x=p x=p$ if cosine is in the denominator and $y=p y=p$ if sine is in the denominator. Set $ep ep$ equal to the numerator in standard form to solve for $x x$ or $y. y.$
## Example 1
### Identifying a Conic Given the Polar Form
For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.
1. $r= 6 3+2sinθ r= 6 3+2sinθ$
2. $r= 12 4+5cosθ r= 12 4+5cosθ$
3. $r= 7 2−2sinθ r= 7 2−2sinθ$
## Try It #1
Identify the conic with focus at the origin, the directrix, and the eccentricity for $r= 2 3−cosθ . r= 2 3−cosθ .$
## Graphing the Polar Equations of Conics
When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine $e e$ and, therefore, the shape of the curve. The next step is to substitute values for $θ θ$ and solve for $r r$ to plot a few key points. Setting $θ θ$ equal to $0, π 2 ,π, 0, π 2 ,π,$ and $3π 2 3π 2$ provides the vertices so we can create a rough sketch of the graph.
## Example 2
### Graphing a Parabola in Polar Form
Graph $r= 5 3+3cosθ . r= 5 3+3cosθ .$
### Analysis
We can check our result with a graphing utility. See Figure 4.
Figure 4
## Example 3
### Graphing a Hyperbola in Polar Form
Graph $r= 8 2−3sinθ . r= 8 2−3sinθ .$
## Example 4
### Graphing an Ellipse in Polar Form
Graph $r= 10 5−4cosθ . r= 10 5−4cosθ .$
### Analysis
We can check our result using a graphing utility. See Figure 7.
Figure 7 $r= 10 5−4cosθ r= 10 5−4cosθ$ graphed on a viewing window of $[ –3,12,1 ] [ –3,12,1 ]$ by and
## Try It #2
Graph $r= 2 4−cosθ . r= 2 4−cosθ .$
## Defining Conics in Terms of a Focus and a Directrix
So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.
## How To
Given the focus, eccentricity, and directrix of a conic, determine the polar equation.
1. Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of $y, y,$ we use the general polar form in terms of sine. If the directrix is given in terms of $x, x,$ we use the general polar form in terms of cosine.
2. Determine the sign in the denominator. If $p<0, p<0,$ use subtraction. If $p>0, p>0,$ use addition.
3. Write the coefficient of the trigonometric function as the given eccentricity.
4. Write the absolute value of $p p$ in the numerator, and simplify the equation.
## Example 5
### Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix
Find the polar form of the conic given a focus at the origin, $e=3 e=3$ and directrix $y=−2. y=−2.$
## Example 6
### Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix
Find the polar form of a conic given a focus at the origin, $e= 3 5 , e= 3 5 ,$ and directrix $x=4. x=4.$
## Try It #3
Find the polar form of the conic given a focus at the origin, $e=1, e=1,$ and directrix $x=−1. x=−1.$
## Example 7
### Converting a Conic in Polar Form to Rectangular Form
Convert the conic $r= 1 5−5sinθ r= 1 5−5sinθ$ to rectangular form.
## Try It #4
Convert the conic $r= 2 1+2cosθ r= 2 1+2cosθ$ to rectangular form.
## Media
Access these online resources for additional instruction and practice with conics in polar coordinates.
## 10.5 Section Exercises
### Verbal
1.
Explain how eccentricity determines which conic section is given.
2.
If a conic section is written as a polar equation, what must be true of the denominator?
3.
If a conic section is written as a polar equation, and the denominator involves $sinθ, sinθ,$ what conclusion can be drawn about the directrix?
4.
If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph?
5.
What do we know about the focus/foci of a conic section if it is written as a polar equation?
### Algebraic
For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity.
6.
$r= 6 1−2cosθ r= 6 1−2cosθ$
7.
$r= 3 4−4sinθ r= 3 4−4sinθ$
8.
$r= 8 4−3cosθ r= 8 4−3cosθ$
9.
$r= 5 1+2sinθ r= 5 1+2sinθ$
10.
$r= 16 4+3cosθ r= 16 4+3cosθ$
11.
$r= 3 10+10cosθ r= 3 10+10cosθ$
12.
$r= 2 1−cosθ r= 2 1−cosθ$
13.
$r= 4 7+2cosθ r= 4 7+2cosθ$
14.
$r(1−cosθ)=3 r(1−cosθ)=3$
15.
$r(3+5sinθ)=11 r(3+5sinθ)=11$
16.
$r(4−5sinθ)=1 r(4−5sinθ)=1$
17.
$r(7+8cosθ)=7 r(7+8cosθ)=7$
For the following exercises, convert the polar equation of a conic section to a rectangular equation.
18.
$r= 4 1+3sinθ r= 4 1+3sinθ$
19.
$r= 2 5−3sinθ r= 2 5−3sinθ$
20.
$r= 8 3−2cosθ r= 8 3−2cosθ$
21.
$r= 3 2+5cosθ r= 3 2+5cosθ$
22.
$r= 4 2+2sinθ r= 4 2+2sinθ$
23.
$r= 3 8−8cosθ r= 3 8−8cosθ$
24.
$r= 2 6+7cosθ r= 2 6+7cosθ$
25.
$r= 5 5−11sinθ r= 5 5−11sinθ$
26.
$r(5+2cosθ)=6 r(5+2cosθ)=6$
27.
$r(2−cosθ)=1 r(2−cosθ)=1$
28.
$r(2.5−2.5sinθ)=5 r(2.5−2.5sinθ)=5$
29.
$r= 6secθ −2+3secθ r= 6secθ −2+3secθ$
30.
$r= 6cscθ 3+2cscθ r= 6cscθ 3+2cscθ$
For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci.
31.
$r= 5 2+cosθ r= 5 2+cosθ$
32.
$r= 2 3+3sinθ r= 2 3+3sinθ$
33.
$r= 10 5−4sinθ r= 10 5−4sinθ$
34.
$r= 3 1+2cosθ r= 3 1+2cosθ$
35.
$r= 8 4−5cosθ r= 8 4−5cosθ$
36.
$r= 3 4−4cosθ r= 3 4−4cosθ$
37.
$r= 2 1−sinθ r= 2 1−sinθ$
38.
$r= 6 3+2sinθ r= 6 3+2sinθ$
39.
$r(1+cosθ)=5 r(1+cosθ)=5$
40.
$r(3−4sinθ)=9 r(3−4sinθ)=9$
41.
$r(3−2sinθ)=6 r(3−2sinθ)=6$
42.
$r(6−4cosθ)=5 r(6−4cosθ)=5$
For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix.
43.
Directrix: $x=4;e= 1 5 x=4;e= 1 5$
44.
Directrix: $x=−4;e=5 x=−4;e=5$
45.
Directrix: $y=2;e=2 y=2;e=2$
46.
Directrix: $y=−2;e= 1 2 y=−2;e= 1 2$
47.
Directrix: $x=1;e=1 x=1;e=1$
48.
Directrix: $x=−1;e=1 x=−1;e=1$
49.
Directrix: $x=− 1 4 ;e= 7 2 x=− 1 4 ;e= 7 2$
50.
Directrix: $y= 2 5 ;e= 7 2 y= 2 5 ;e= 7 2$
51.
Directrix: $y=4;e= 3 2 y=4;e= 3 2$
52.
Directrix: $x=−2;e= 8 3 x=−2;e= 8 3$
53.
Directrix: $x=−5;e= 3 4 x=−5;e= 3 4$
54.
Directrix: $y=2;e=2.5 y=2;e=2.5$
55.
Directrix: $x=−3;e= 1 3 x=−3;e= 1 3$
### Extensions
Recall from Rotation of Axes that equations of conics with an $xy xy$ term have rotated graphs. For the following exercises, express each equation in polar form with $r r$ as a function of $θ. θ.$
56.
$xy=2 xy=2$
57.
$x 2 +xy+ y 2 =4 x 2 +xy+ y 2 =4$
58.
$2 x 2 +4xy+2 y 2 =9 2 x 2 +4xy+2 y 2 =9$
59.
$16 x 2 +24xy+9 y 2 =4 16 x 2 +24xy+9 y 2 =4$
60.
$2xy+y=1 2xy+y=1$
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Analytic Geometry: Cylindrical Coordinates
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Cylindrical Coordinates
Cylindrical coordinates are obtained by using polar coordinates in a plane, and then adding a z-axis perpendicular to the plane passing through the pole. This gives three coordinates (r,theta,z) for any point.
To transform from rectangular Cartesian coordinates (x,y,z) to cylindrical ones and back, use the following formulas:
x = r cos(theta),
y = r sin(theta),
z = z,
r = ±sqrt(x2+y2),
theta = arctan(y/x),
z = z.
The sign of r is determined by which of the values of the arctangent function is chosen:
Sign of x Sign of y Quadrant of theta Sign of r
++ I+
++ III
+ II+
+ IV
I
III+
+ II
+ IV+
The quadrant of theta can always be chosen to make r positive, if it is so desired.
### Common Uses
The most common use of cylindrical coordinates is to give the equation of a surface of revolution. If the z-axis is taken as the axis of revolution, then the equation will not involve theta at all.
Examples:
• A paraboloid of revolution might have equation
z = r2.
This is the surface you would get by rotating the parabola z = x2 in the xz-plane about the z-axis. The cartesian coordinate equation of the paraboloid of revolution would be z = x2 + y2.
• A right circular cylinder of radius a whose axis is the z-axis has equation
r = R.
• A a sphere with center at the origin and radius R will have equation
r2 + z2 = R2.
• A right circular cone with vertex at the origin and axis the z-axis has equation
z = m r.
As another kind of example, a helix has the following equations:
r = R,
z = a theta.
Sometimes a change from rectangular to cylindrical coordinates makes computing difficult multiple integrals simpler.
Compiled by Robert L. Ward. |
# What is a Quadratic Equation? (with pictures)
D.R. Satori
D.R. Satori
A quadratic equation consists of a single variable with three terms in the standard form: ax2 + bx + c = 0. The first quadratic equations were developed as a method used by Babylonian mathematicians around 2000 BC to solve simultaneous equations. Quadratic equations can be applied to problems in physics involving parabolic motion, path, shape, and stability. Several methods have evolved to simplify the solution of such equations for the variable x. Any number of quadratic equation solvers, in which the values of the quadratic equation coefficients can be entered and automatically calculated, can be found online.
The three methods most commonly used to solve quadratic equations are factoring, completing the square, and the quadratic formula. Factoring is the simplest form of solving a quadratic equation. When the quadratic equation is in its standard form, it is easy to visualize if the constants a, b, and c are such that the equation represents a perfect square. First, the standard form must be divided through by a. Then, half of, what is now, the b/a term must be equal to twice, what is now, the c/a term; if this is true, then the standard form can be factored into the perfect square of (x ± d)2.
If the solution of a quadratic equation is not a perfect square and the equation cannot be factored in its present form, then a second solution method — completing the square — can be used. After dividing through by the a term, the b/a term is divided by two, squared, and then added to both sides of the equation. The square root of the perfect square can be equated to the square root of all the remaining constants on the right hand side of the equation in order to find x.
The final method of solving the standard quadratic equation is by directly substituting the constant coefficients (a, b, and c) into the quadratic formula: x = (-b±sqrt(b2-4ac))/2a, which was derived by the method of completing the squares in the generalized equation. The discriminant of the quadratic formula (b2 - 4ac) appears under a square root sign and, even before the equation is solved for x, can indicate the type and number of solutions found. The type of solution depends on whether the discriminant is equal to the square root of a positive or negative number. When the discriminant is zero, there is only one positive root. When the discriminant is positive, there are two positive roots, and when the discriminant is negative, there are both positive and negative roots.
## You might also Like
anon253895
The Diagonal Sum Method for solving quadratic equations ax^2 + bx + c = 0.
Concept of the method: Directly find 2 real roots, in the form of 2 fractions, knowing their product (c/a) and sum (-b/a).
Rule of signs for real roots:
1. When a and c have different signs, roots have opposite signs.
2. When a and c have same sign, roots have same sign.
a. If a and b have different signs, both roots are positive (+)
b. If a and b have same sign, both roots are negative (-).
Rule of the diagonal sum of a root pair.
Given a pair of real roots: (c1/a1, c2/a2).
Their product is equal to (c/a), meaning: c1c2 = c and: a1a2 = a. The numerators of a root pair are a factor-pair of c. The denominators are a factor-pair of a.
Their sum: (c1/a1 + c2/a2) = (c1a2 + c2a1)/a1a2 = -b/a. The sum (c1a2 + c2a1) is called the diagonal sum. From there comes the rule:
"The diagonal sum of a true root-pair must equal to (-b). If it equals to (-b), the solution is the negative of this pair."
SOLVING PROCESS
1. When a = 1. Solving x^2 + bx + c = 0.
The diagonal sum reduces to the sum of the 2 real roots. Solving is very fast, and doesn't need factoring, due to the Rule of signs
Example 1. Solve x^2 + 75x + 216 = 0. Both roots are negative. Write factor pairs of c = 216. They are: (-1, -216),(-2, -108),(-3, -72)...This sum is -3 - 72 = -75 = -b. The 2 real roots are -3 and -72.
1. When a is not 1. The DS Method directly selects the probable root pairs from the quotient (c/a).
a. If a and c are both prime numbers: the number of probable root pair is limited to one, except when 1 (or -1) is a real root.
Example 2. Solve 11x^2 - 142x - 13 = 0. Roots have different signs. There is unique probable root pair: (-1/11, 13/1). Its DS is 143- 1 = 142 = -b. The 2 real roots are -1/11 and 13.
b. If a and c are small numbers and may have one (or 2) factors. Most of the factorable quadratic equations given in classes/books belong to this case. In this case, we can write down all of the probable root pairs, then find the one that has a DS equal to -b (or b). Number of trials is always fewer than 4.
Example 3. Solve 10x^2 + 31x + 13 = 0. Both roots are negative. The constant a = 10 has 2 factor pairs: (1, 10) and (2, 5). There are 3 probable root pairs:
(-1/10, -13/1),(-1/2, -13/5),(-1/5, -13/2). The DS of the second pair is: -5 - 26 = -31 = -b. The 2 real roots are -1/2 and -13/5.
anon251448
Beyond the four existing methods (graphing, completing the square, factoring, formula) there is a new method called Diagonal Sum Method.
Its concept is direct, finding the two real roots in the form of two fractions knowing their sum (-b/a) and product (c/a). It is very fast when a = 1 and when the constants a, c are prime/small numbers. It is a trial and error method, same as the factoring one, but it reduces in half the number of permutations.
It saves the time used to solve the 2 binomials for x. It is considered as a shortcut of the factoring method. |
Monkey Math
The Fun Way to Learn Simple Addition!
Age: 4 +
This learning maths game by ‘Learning Mates’ has a total of 16 pieces. There is a little monkey that is on a stand and his head rocks back and forth depending on what numbers are placed on each handle that he is holding. He is 7 .25 inches tall and 9.5 inches wide. Plus there are 15 “banana bunches” or number tokens. Of these 15, there are 2 of each of numbers: 1, 2, 3, 4, 5 and 1 of each of numbers: 6, 7, 8, 9, and 10. Also included is an instruction booklet showing how to make learning simple addition a whole banana bunch of fun!
When the monkey is not balanced with equal numbers on either hand, the side with the greater number hangs down low and the other hand will be up. While his hand hangs low, his eyes are wild looking (crossed) since the eye sockets are not lining up with his eyes. That means it’s time to try again. When he’s balanced, you can see his eyes are looking straight at you.
When a number is placed on the right hand, you have to balance the bananas by putting the equal “number” of bananas /tokens on the other hand. For example, if you put a single bunch of 10 bananas on the right hand, you will have to put 2 or 3 of the other numbers together to get the total on the left hand up to 10 bananas. E.g. to make 10 – you can put 8 and 2 or 7 and 3 or 5, 4 and 1 etc. When you do have equal numbers showing, the monkey’s arms are balanced and you can see his eyes looking straight at you.
This helps children start learning about numbers and how they add up to other numbers. It helps teach about balance on a scale and helps them begin to grasp beginning math concepts in a FUN way. This game also introduces number recognition, sorting and counting.
How to make maximum use of ‘Monkey Math’!!!
1. For younger children you can start with using the banana bunches that are in pairs. Numbers 1,2,3,4 and 5 can be used to pair with each other or match together for number recognition.
Get your child to find the same numbers and pair them.
2. The banana bunches can also be used to reinforce number recognition.
a. You can call out a number and your child can pick the number up for you.
b. You can show the number to your child and your child says the number name to you.
3. Your child can use the banana bunches to arrange the numbers in sequence from 1 -10.
a. Can be used to arrange numbers from the smallest to biggest
b. Or biggest to smallest.
This gives a better visual perception as the child is able to see the numbers of bananas increasing or decreasing in quantity as he lays the bananas.
4. To begin using the Monkey balance use the banana bunches that are in pairs first so your child just hangs the exact same pair to balance the monkey’s arms before moving on to other number combinations.
Review by : Nafisa Juzer, www.r2learntoys.com.
Sometimes, it might prove to be a challenging task to sustain a child’s interest in learning maths. Came across any interesting product to help children learn numbers better? Do write in to us! |
# What is Mathematics for essay?
## What is Mathematics for essay?
Math Essay: Mathematics is generally defined as the science that deals with numbers. It involves operations among numbers, and it also helps you to calculate the product price, how many discounted prizes here, and If you good in maths so you can calculate very fast.
## What is the importance of mathematics?
Mathematics provides an effective way of building mental discipline and encourages logical reasoning and mental rigor. In addition, mathematical knowledge plays a crucial role in understanding the contents of other school subjects such as science, social studies, and even music and art.
Why is maths a Favourite subject?
My favourite subject is Maths as I love to play with numbers and solve mathematical problems. Maths gives me a lot of satisfaction and boosts my energy and thinking capacity while studying. I love the number game and can solve problems for hours at a stretch without getting bored.
Beginning in the 6th century BC with the Pythagoreans, with Greek mathematics the Ancient Greeks began a systematic study of mathematics as a subject in its own right. Around 300 BC, Euclid introduced the axiomatic method still used in mathematics today, consisting of definition, axiom, theorem, and proof.
### Is 0 an even number?
So what is it – odd, even or neither? For mathematicians the answer is easy: zero is an even number. Because any number that can be divided by two to create another whole number is even.
Is the number 1 even?
The number five can be divided into two groups of two and one group of one. Even numbers always end with a digit of 0, 2, 4, 6 or 8. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30 are even numbers. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31 are odd numbers.
2
## Which is the greatest odd number?
Explanation: Here, factors of 8642 are {1, 8642, 2, 4321, 29, 298, 58, 149} in which odd factors are {1, 4321, 29, 149} and the greatest odd factor among all odd factors is 4321.
What is the largest odd number?
Approach: Largest N-digit even number will be (10n) – 2 because the series for different values of N will be 8, 98, 998, 9998, ….. Similarly, largest N-digit odd number will be (10n) – 1 for the series 9, 99, 999, 9999, …..
### What is an odd number?
Odd numbers are whole numbers that cannot be divided exactly into pairs. Odd numbers, when divided by 2, leave a remainder of 1. 1, 3, 5, 7, 9, 11, 13, 15 … are sequential odd numbers. Odd numbers have the digits 1, 3, 5, 7 or 9 in their ones place.
### What is the formula of odd number?
The total of any set of sequential odd numbers beginning with 1 is always equal to the square of the number of digits, added together. If 1,3,5,7,9,11,…, (2n-1) are the odd numbers, then; Sum of first odd number = 1. Sum of first two odd numbers = 1 + 3 = 4 (4 = 2 x 2).
How many odd numbers are there?
There are 25 odd numbers from 1 to 50 while there are 50 in between 1 and 100. In case of numbers from 1 to 1000, there are 500 odd numbers and 500 even numbers. A few odd numbers list include numbers like: -5, -3, -1, 1, 3, 5 , 7 , 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, etc.
## How do you represent an odd number?
If is an integer (a whole number), then the expression represents an even number, because even numbers are the multiples of 2. The expressions 2 n − 1 and 2 n + 1 can represent odd numbers, as an odd number is one less, or one more than an even number.
## Why is 2n even?
Multiplying any whole number by 2 gives an even number, so 2n is an even number.
Is 99 an odd number?
99 is an odd number. Related links: Is 99 a composite number?
### Is 11 an odd number?
All the numbers ending with 1,3,5,7 and 9 are odd numbers. For example, numbers such as 11, 23, 35, 47 etc. are odd numbers.
### Is 49 an odd number?
49 is an odd number.
Is 49 a even number?
49 is not an even number.
## Is 32 a odd number?
32 is not an odd number.
## Why 49 is an odd number?
How is 49 an odd number? An odd number is any integer (a whole number) that cannot be divided by 2 evenly. That means the number is an odd number because it cannot be divided by 2 without a remainder.
Is 51 an even number?
51 is not an even number.
### Is 50 an even or odd number?
50 is an even number.
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The Pythagorean Theorem
Learning Objective(s)
· Use the Pythagorean Theorem to find the unknown side of a right triangle.
· Solve application problems involving the Pythagorean Theorem.
Introduction
A long time ago, a Greek mathematician named Pythagoras discovered an interesting property about right triangles: the sum of the squares of the lengths of each of the triangle’s legs is the same as the square of the length of the triangle’s hypotenuse. This property—which has many applications in science, art, engineering, and architecture—is now called the Pythagorean Theorem.
Let’s take a look at how this theorem can help you learn more about the construction of triangles. And the best part—you don’t even have to speak Greek to apply Pythagoras’ discovery.
The Pythagorean Theorem
Pythagoras studied right triangles, and the relationships between the legs and the hypotenuse of a right triangle, before deriving his theory.
The Pythagorean Theorem If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. This relationship is represented by the formula:
In the box above, you may have noticed the word “square,” as well as the small 2s to the top right of the letters in . To square a number means to multiply it by itself. So, for example, to square the number 5 you multiply 5 • 5, and to square the number 12, you multiply 12 • 12. Some common squares are shown in the table below.
Number Number Times Itself Square 1 12 = 1 • 1 1 2 22 = 2 • 2 4 3 32 = 3 • 3 9 4 42 = 4 • 4 16 5 52 = 5 • 5 25 10 102 = 10 • 10 100
When you see the equation , you can think of this as “the length of side a times itself, plus the length of side b times itself is the same as the length of side c times itself.”
Let’s try out all of the Pythagorean Theorem with an actual right triangle.
This theorem holds true for this right triangle—the sum of the squares of the lengths of both legs is the same as the square of the length of the hypotenuse. And, in fact, it holds true for all right triangles.
The Pythagorean Theorem can also be represented in terms of area. In any right triangle, the area of the square drawn from the hypotenuse is equal to the sum of the areas of the squares that are drawn from the two legs. You can see this illustrated below in the same 3-4-5 right triangle.
Note that the Pythagorean Theorem only works with right triangles.
Finding the Length of the Hypotenuse
You can use the Pythagorean Theorem to find the length of the hypotenuse of a right triangle if you know the length of the triangle’s other two sides, called the legs. Put another way, if you know the lengths of a and b, you can find c.
In the triangle above, you are given measures for legs a and b: 5 and 12, respectively. You can use the Pythagorean Theorem to find a value for the length of c, the hypotenuse.
The Pythagorean Theorem. Substitute known values for a and b. Evaluate. Simplify. To find the value of c, think about a number that, when multiplied by itself, equals 169. Does 10 work? How about 11? 12? 13? (You can use a calculator to multiply if the numbers are unfamiliar.) 13 = c The square root of 169 is 13.
Using the formula, you find that the length of c, the hypotenuse, is 13.
In this case, you did not know the value of c—you were given the square of the length of the hypotenuse, and had to figure it out from there. When you are given an equation like and are asked to find the value of c, this is called finding the square root of a number. (Notice you found a number, c, whose square was 169.)
Finding a square root takes some practice, but it also takes knowledge of multiplication, division, and a little bit of trial and error. Look at the table below.
Number x Number y which, when multiplied by itself, equals number x Square root y 1 1 • 1 1 4 2 • 2 2 9 3 • 3 3 16 4 • 4 4 25 5 • 5 5 100 10 • 10 10
It is a good habit to become familiar with the squares of the numbers from 0‒10, as these arise frequently in mathematics. If you can remember those square numbers—or if you can use a calculator to find them—then finding many common square roots will be just a matter of recall.
For which of these triangles is ? A) B) C) D) Show/Hide Answer
Finding the Length of a Leg
You can use the same formula to find the length of a right triangle’s leg if you are given measurements for the lengths of the hypotenuse and the other leg. Consider the example below.
Example Problem Find the length of side a in the triangle below. Use a calculator to estimate the square root to one decimal place. a = ? b = 6 c = 7 In this right triangle, you are given the measurements for the hypotenuse, c, and one leg, b. The hypotenuse is always opposite the right angle and it is always the longest side of the triangle. To find the length of leg a, substitute the known values into the Pythagorean Theorem. Solve for a2. Think: what number, when added to 36, gives you 49? Use a calculator to find the square root of 13. The calculator gives an answer of 3.6055…, which you can round to 3.6. (Since you are approximating, you use the symbol .) Answer
Which of the following correctly uses the Pythagorean Theorem to find the missing side, x? A) B) x + 8 = 10 C) D) Show/Hide Answer
Using the Theorem to Solve Real World Problems
The Pythagorean Theorem is perhaps one of the most useful formulas you will learn in mathematics because there are so many applications of it in real world settings. Architects and engineers use this formula extensively when building ramps, bridges, and buildings. Look at the following examples.
Example Problem The owners of a house want to convert a stairway leading from the ground to their back porch into a ramp. The porch is 3 feet off the ground, and due to building regulations the ramp must start 12 feet away from the base of the porch. How long will the ramp be? Use a calculator to find the square root, and round the answer to the nearest tenth. To solve a problem like this one, it often makes sense to draw a simple diagram showing where the legs and hypotenuse of the triangle lie. a = 3 b = 12 c = ? Identify the legs and the hypotenuse of the triangle. You know that the triangle is a right triangle since the ground and the raised portion of the porch are perpendicular—this means you can use the Pythagorean Theorem to solve this problem. Identify a, b, and c. Use the Pythagorean Theorem to find the length of c. 12.4 = c Use a calculator to find c. The square root of 153 is 12.369…, so you can round that to 12.4. Answer The ramp will be 12.4 feet long.
Example Problem A sailboat has a large sail in the shape of a right triangle. The longest edge of the sail measures 17 yards, and the bottom edge of the sail is 8 yards. How tall is the sail? Draw an image to help you visualize the problem. In a right triangle, the hypotenuse will always be the longest side, so here it must be 17 yards. The problem also tells you that the bottom edge of the triangle is 8 yards. Setup the Pythagorean Theorem. a = 15 15 • 15 = 225, so a = 15. Answer The height of the sail is 15 yards.
Summary
The Pythagorean Theorem states that in any right triangle, the sum of the squares of the lengths of the triangle’s legs is the same as the square of the length of the triangle’s hypotenuse. This theorem is represented by the formula . Put simply, if you know the lengths of two sides of a right triangle, you can apply the Pythagorean Theorem to find the length of the third side. Remember, this theorem only works for right triangles. |
# Thread: Arithmetic mean and Geometric mean
1. ## Arithmetic mean and Geometric mean
Q) A and B are two numbers such that there G.M is 20% lower than their A.M .Find the ratio between the two numbers.
Thanks,
Ashish
2. Originally Posted by a69356
Q) A and B are two numbers such that there G.M is 20% lower than their A.M .Find the ratio between the two numbers.
Thanks,
Ashish
$\displaystyle \sqrt{a b} = \frac{4}{5} \left( \frac{a + b}{2} \right)$
$\displaystyle \Rightarrow ab = \frac{4}{25} (a + b)^2 \Rightarrow 25 ab = 4a^2 + 8 ab + 4 b^2 \Rightarrow 25 = 4 \left( \frac{a}{b}\right) + 8 + 4 \left(\frac{b}{a}\right)$.
Let the ratio be $\displaystyle x = \frac{a}{b}$. Then:
$\displaystyle 25 = 4x + 8 + \frac{4}{x}$.
Your job is to solve for x.
3. Hello, Ashish!
My solution is similar to Mr. F's . . .
$\displaystyle A$ and $\displaystyle B$ are two numbers such that there G.M is 20% lower than their A.M.
Find the ratio between the two numbers.
Let $\displaystyle a$ and $\displaystyle b$ be the two numbers.
We have: .$\displaystyle \sqrt{ab} \:=\:\frac{4}{5}\left(\frac{a+b}{2}\right) \quad\Rightarrow\quad 5\sqrt{ab} \:=\:2(a+b)$
Square both sides: .$\displaystyle 25ab \;=\;4a^2 + 8ab + 4b^2 \quad\Rightarrow\quad 4x^2 - 17ab + 4b^2 \:=\:0$
. . which factors: .$\displaystyle (a - 4b)(4a - b) \:=\:0$
Hence, we have: .$\displaystyle \begin{Bmatrix}a - 4b \:=\:0 & \Rightarrow & \dfrac{a}{b} \:=\: 4 \\ \\[-2mm] 4a - b \:=\:0 & \Rightarrow & \dfrac{a}{b} \:=\:\dfrac{1}{4} \end{Bmatrix}$
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# two numbers A and B their gm is 20% less than ap find the ratio
Click on a term to search for related topics. |
RELATIVE SPEEED AND TRAIN QUESTIONS
Speed has no sense of direction unlike the velocity. Relative speed is the speed of one object as observed from another moving object. Questions on train are the classic examples of relative speed and in all these questions it is assumed that trains move parallel to each other – whether in the same direction or the opposite direction. Thus, we shall see how the relative speed is calculated and using it we come to know the time taken by the trains to cross each other and some other like aspects.
Important Formulas – Problems on Trains
1. x km/hr = (x×5)/18 m/s
1. y m/s = (y×18)/5 km/hr
1. Speed = distance/time, that is, s = d/t
1. velocity = displacement/time, that is, v = d/t
1. Time taken by a train x meters long to pass a pole or standing man or a post
= Time taken by the train to travel x meters.
1. Time taken by a train x meters long to pass an object of length y meters
= Time taken by the train to travel (x + y) metres.
1. Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2,
then their relative speed = (v1 – v2) m/s
1. Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s ,
then their relative speed = (v1+ v2) m/s
1. Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then
The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds
1. Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then
The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds
1. Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,
A’s speed: B’s speed = √q: √p
Solved Examples
Level 1
1.A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train? A. 190 metres B. 160 metres C. 200 metresAnswer : Option C D. 120 metres
Explanation :
Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s
Time taken to cross, t = 18 s
Distance Covered, d = vt = (400/36)× 18 = 200 m
Distance covered is equal to the length of the train = 200 m
2.A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m? A. 120 sec B. 99 s C. 89 s D. 80 s
Explanation :
v = 240/24 (where v is the speed of the train) = 10 m/s
t = (240+650)/10 = 89 seconds
3.Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is A. 10.8 s B. 12 s C. 9.8 s D. 8 s
Explanation :
Distance = 140+160 = 300 m
Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s
Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s
4.A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train? A. 79.2 km/hr B. 69 km/hr C. 74 km/hr D. 61 km/hr
Explanation :
Let x is the length of the train and v is the speed
Time taken to move the post = 8 s
=> x/v = 8
=> x = 8v — (1)
Time taken to cross the platform 264 m long = 20 s
(x+264)/v = 20
=> x + 264 = 20v —(2)
Substituting equation 1 in equation 2, we get
8v +264 = 20v
=> v = 264/12 = 22 m/s
= 22×36/10 km/hr = 79.2 km/hr
5.Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is A. 2 : 3 B. 2 :1 C. 4 : 3 D. 3 : 2
Explanation :
Ratio of their speeds = Speed of first train : Speed of second train
= √16: √ 9
= 4:3
6.Train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? A. 320 m B. 190 m C. 210 m D. 230 m
Explanation :
Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s
time = 9s
Total distance covered = 270 + x where x is the length of other train
(270+x)/9 = 500/9
=> 270+x = 500
=> x = 500-270 = 230 meter
7.Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet? A. 10.30 a.m B. 10 a.m. C. 9.10 a.m. D. 11 a.m.
Explanation :
Assume both trains meet after x hours after 7 am
Distance covered by train starting from P in x hours = 20x km
Distance covered by train starting from Q in (x-1) hours = 25(x-1)
Total distance = 110
=> 20x + 25(x-1) = 110
=> 45x = 135
=> x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am
8.Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is A. 42 B. 36 C. 28 D. 20
Explanation :
Distance covered = 120+120 = 240 m
Time = 12 s
Let the speed of each train = v. Then relative speed = v+v = 2v
2v = distance/time = 240/12 = 20 m/s
Speed of each train = v = 20/2 = 10 m/s
= 10×36/10 km/hr = 36 km/hr
Level 2
1.A train, 130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is A. 270 m B. 245 m C. 235 m D. 220 m
Explanation :
Assume the length of the bridge = x meter
Total distance covered = 130+x meter
total time taken = 30s
speed = Total distance covered /total time taken = (130+x)/30 m/s
=> 45 × (10/36) = (130+x)/30
=> 45 × 10 × 30 /36 = 130+x
=> 45 × 10 × 10 / 12 = 130+x
=> 15 × 10 × 10 / 4 = 130+x
=> 15 × 25 = 130+x = 375
=> x = 375-130 =245
2.A train has a length of 150 meters. It is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train. A. 182 km/hr B. 180 km/hr C. 152 km/hr D. 169 km/hr
Explanation :
Length of the train, l = 150m
Speed of the man, Vm= 2 km/hr
Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr
Relative Speed = Speed of train, Vt – Speed of man (As both are moving in the same direction)
=> 180 = Vt – 2 => Vt = 180 + 2 = 182 km/hr
3.Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively. If they cross each other in 23 seconds, what is the ratio of their speeds? A. Insufficient data B. 3 : 1 C. 1 : 3 D. 3 : 2
Explanation :
Let the speed of the trains be x and y respectively
length of train1 = 27x
length of train2 = 17y
Relative speed= x+ y
Time taken to cross each other = 23 s
=> (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y)
=> 4x = 6y => x/y = 6/4 = 3/2
4.A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the jogger? A. 46 B. 36 C. 18 D. 22
Explanation :
Distance to be covered = 240+ 120 = 360 m
Relative speed = 36 km/hr = 36×10/36 = 10 m/s
Time = distance/speed = 360/10 = 36 seconds
5.A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is A. None of these B. 280 meter C. 240 meter D. 200 meter
Explanation :
Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s
Length of the train = speed × time taken to cross the man = 15×20 = 300 m
Let the length of the platform = L
Time taken to cross the platform = (300+L)/15
=> (300+L)/15 = 36
=> 300+L = 15×36 = 540 => L = 540-300 = 240 meter
6.A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train? A. 62 m B. 54 m C. 50 m D. 55 m
Explanation :
Let x is the length of the train in meter and v is its speed in kmph
x/9 = (v-2) (10/36) — (1)
x/10 = (v-4) (10/36) — (2)
Dividing equation 1 with equation 2
10/9 = (v-2)/(v-4) => 10v – 40 = 9v – 18 => v = 22
Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m
7.A train is traveling at 48 kmph. It crosses another train having half of its length, traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform? A. 500 m B. 360 m C. 480 m D. 400 m
Explanation :
Speed of train1 = 48 kmph
Let the length of train1 = 2x meter
Speed of train2 = 42 kmph
Length of train 2 = x meter (because it is half of train1’s length)
Distance = 2x + x = 3x
Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s
Time = 12 s
Distance/time = speed => 3x/12 = 25
=> x = 25×12/3 = 100 meter
Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s
Speed of train1 = 48 kmph = 480/36 = 40/3 m/s
Distance = 200 + y where y is the length of the platform
=> 200 + y = 45×40/3 = 600
=> y = 400 meter
8.A train, 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)? A. 440 m B. 500 m C. 260 m D. 430 m
Explanation :
Distance = 800+x meter where x is the length of the tunnel
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s
Distance/time = speed
(800+x)/60 = 65/3 => 800+x = 20×65 = 1300
=> x = 1300 – 800 = 500 meter
9.Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is : A. 19 m B. 2779 m C. 1329 m D. 33 m
Explanation :
Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s
Time = 5 s
Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 2779 m = length of the fast train
APPSC GROUP 1 Notes brings Prelims and Mains programs for APPSC GROUP 1 Prelims and APPSC GROUP 1 Mains Exam preparation. Various Programs initiated by APPSC GROUP 1 Notes are as follows:- For any doubt, Just leave us a Chat or Fill us a querry––
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## A Multiplication Based Logic Puzzle
### 972 Happy Birthday, Andy!
Today is my brother Andy’s birthday. I know Andy can solve these puzzles so I’ve made him a puzzle cake with factors from 1 to 16. Adding extra factor possibilities complicates the puzzle and makes it a little more difficult to read as a multiplication table, but it is still solvable. Since these puzzles have only one solution and are solved by logic and not by guessing and checking, I added a clue right in the center of the cake to ensure a unique solution. Happy birthday, Andy!
Print the puzzles or type the solution in this excel file: 10-factors-968-977
Now I’ll share a little about the number 972 which is the 13th Achilles number. All of the exponents in its prime factorization are greater than 1, yet the greatest common factor of those exponents is still 1. The previous Achilles number, 968, and 972 are the closest two Achilles numbers so far.
I think 972 has some interesting representations when written in some other bases:
It’s 33030 in BASE 4 because 3(4⁴) + 2(4³) + 0(4²) + 3(4) + 0(1) = 3(256 + 64 + 4) = 3(324) = 972
363 in BASE 17 because 3(17²) + 6(17) + 3(1) = 972
300 in BASE 18 because 3(18²) = 3(324) = 972
RR in BASE 35 (R is 27 base 10) because 27(35) + 27(1) = 27(36) = 972
R0 in BASE 36 because 27(36) + 0(1) = 27(36) = 972
• 972 is a composite number.
• Prime factorization: 972 = 2 × 2 × 3 × 3 × 3 × 3 × 3, which can be written 972 = 2²× 3⁵
• The exponents in the prime factorization are 2 and 5. Adding one to each and multiplying we get (2 + 1)(5 + 1) = 3 × 6 = 18. Therefore 972 has exactly 18 factors.
• Factors of 972: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, 243, 324, 486, 972
• Factor pairs: 972 = 1 × 972, 2 × 486, 3 × 324, 4 × 243, 6 × 162, 9 × 108, 12 × 81, 18 × 54 or 27 × 36
• Taking the factor pair with the largest square number factor, we get √972 = (√324)(√3) = 18√3 ≈ 31.1769
Here are a few of the MANY possible factor trees for 972: |
## A botanist wishes to estimate the mean number of seeds for a certain fruit. She samples 18 specimens and finds the average number of seeds i
Question
A botanist wishes to estimate the mean number of seeds for a certain fruit. She samples 18 specimens and finds the average number of seeds is 48 with a standard deviation of 14. Construct a one-sided upper-bound 80% confidence interval for the mean number of seeds for the species using the appropriate critical value to three decimal places.
in progress 0
5 months 2021-08-25T07:43:17+00:00 1 Answers 0 views 0
The 80% confidence interval for the mean number of seeds for the species is of 50.848 seeds.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
T interval
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 18 – 1 = 17
80% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level(as we want the one-sided upper-bound confidence interval) of . So we have T = 0.863
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The upper end of the interval is the sample mean added to M. So it is 48 + 2.848 = 50.848 seeds.
The 80% confidence interval for the mean number of seeds for the species is of 50.848 seeds. |
# Weitzenbock by Sanchez
### Solution 1
Let $A=(0,2),\,$ $B=(-2u,0),\,$ $C=(2v,0),\,$ where $u=\cot B,\,$ $v=\cot C.\,$ Obviously, $u+v\gt 0.\,$ We have $[ABC]=2(u+v),\,$ $[NAB]=u^2+1,\,$ $[PAC]=v^2+1,\,$ $[MBC]=(u+v)^2,\,$ implying
$[MBNAPC]=[2+u^2+v^2+(u=v)^2]+2(u+v).$
Hence, we need to prove that $2\sqrt{3}(u+v)\le 2+u^2+v^2+(u+v)^2.\,$ Denote $u+v=2s\,$ and $uv=p.\,$ Our inequality becomes $2\sqrt{3}s\le 4s^2-p+1,\,$ or, equivalently, $p\le 4s^2-2\sqrt{3}s+1.,$ But $p\le s^2,\,$ hence, suffice it to show that $s^2\le 4s^2-2\sqrt{3}s+1,\,$ i.e., $(\sqrt{3}s-1)^2\ge 0,\,$ which is true.
Equality iff $\displaystyle u=v=\frac{1}{\sqrt{3}},\,$ i.e., when $\Delta ABC\,$ is equilateral.
### Solution 2
Denote $BC=a,\,$ $AC=b,\,$ $AB=c,\,$ $\displaystyle p=\frac{a+b+c}{2},\,$ $S=[ABC].\,$ Using Heron's formula, $S=\sqrt{p(p-a)(p-b)(p-c)},\,$
(1)
### Acknowledgment
Leo Giugiuc has kind posted a problem from the Peru Geometrico facebook group (originally by Miguel Ochoa Sanchez) at the CutTheKnotMath facebook page, along with his solution (Solution 1). Solution 2 is by Marian Cucoanes; Solution 3 is by Rachid Moussaoui. |
Mister Exam
# Inequality with the module Step by Step
For example, you have entered (calculator here):
$$- \left|{x - 3}\right| + \left|{x^{2} - 1}\right| > 7$$
### Rapid solution
$$\left(-\infty < x \wedge x < - \frac{3 \sqrt{5}}{2} - \frac{1}{2}\right) \vee \left(x < \infty \wedge - \frac{1}{2} + \frac{3 \sqrt{5}}{2} < x\right)$$ $$x\ in\ \left(-\infty, - \frac{3 \sqrt{5}}{2} - \frac{1}{2}\right) \cup \left(- \frac{1}{2} + \frac{3 \sqrt{5}}{2}, \infty\right)$$
### Detail solution
Given the inequality:
$$- \left|{x - 3}\right| + \left|{x^{2} - 1}\right| > 7$$
To solve this inequality, we must first solve the corresponding equation:
$$- \left|{x - 3}\right| + \left|{x^{2} - 1}\right| = 7$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x - 3 \geq 0$$
$$x^{2} - 1 \geq 0$$
or
$$3 \leq x \wedge x < \infty$$
we get the equation
$$- (x - 3) + \left(x^{2} - 1\right) - 7 = 0$$
after simplifying we get
$$x^{2} - x - 5 = 0$$
the solution in this interval:
$$x_{1} = \frac{1}{2} - \frac{\sqrt{21}}{2}$$
but x1 not in the inequality interval
$$x_{2} = \frac{1}{2} + \frac{\sqrt{21}}{2}$$
but x2 not in the inequality interval
2.
$$x - 3 \geq 0$$
$$x^{2} - 1 < 0$$
The inequality system has no solutions, see the next condition
3.
$$x - 3 < 0$$
$$x^{2} - 1 \geq 0$$
or
$$\left(1 \leq x \wedge x < 3\right) \vee \left(x \leq -1 \wedge -\infty < x\right)$$
we get the equation
$$- (3 - x) + \left(x^{2} - 1\right) - 7 = 0$$
after simplifying we get
$$x^{2} + x - 11 = 0$$
the solution in this interval:
$$x_{3} = - \frac{1}{2} + \frac{3 \sqrt{5}}{2}$$
$$x_{4} = - \frac{3 \sqrt{5}}{2} - \frac{1}{2}$$
4.
$$x - 3 < 0$$
$$x^{2} - 1 < 0$$
or
$$-1 < x \wedge x < 1$$
we get the equation
$$\left(1 - x^{2}\right) - \left(3 - x\right) - 7 = 0$$
after simplifying we get
$$- x^{2} + x - 9 = 0$$
the solution in this interval:
$$x_{5} = \frac{1}{2} - \frac{\sqrt{35} i}{2}$$
but x5 not in the inequality interval
$$x_{6} = \frac{1}{2} + \frac{\sqrt{35} i}{2}$$
but x6 not in the inequality interval
$$x_{1} = - \frac{1}{2} + \frac{3 \sqrt{5}}{2}$$
$$x_{2} = - \frac{3 \sqrt{5}}{2} - \frac{1}{2}$$
$$x_{1} = - \frac{1}{2} + \frac{3 \sqrt{5}}{2}$$
$$x_{2} = - \frac{3 \sqrt{5}}{2} - \frac{1}{2}$$
This roots
$$x_{2} = - \frac{3 \sqrt{5}}{2} - \frac{1}{2}$$
$$x_{1} = - \frac{1}{2} + \frac{3 \sqrt{5}}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$\left(- \frac{3 \sqrt{5}}{2} - \frac{1}{2}\right) + - \frac{1}{10}$$
=
$$- \frac{3 \sqrt{5}}{2} - \frac{3}{5}$$
substitute to the expression
$$- \left|{x - 3}\right| + \left|{x^{2} - 1}\right| > 7$$
$$- \left|{\left(- \frac{3 \sqrt{5}}{2} - \frac{3}{5}\right) - 3}\right| + \left|{-1 + \left(- \frac{3 \sqrt{5}}{2} - \frac{3}{5}\right)^{2}}\right| > 7$$
2
/ ___\ ___
23 |3 3*\/ 5 | 3*\/ 5 > 7
- -- + |- + -------| - -------
5 \5 2 / 2
one of the solutions of our inequality is:
$$x < - \frac{3 \sqrt{5}}{2} - \frac{1}{2}$$
_____ _____
\ /
-------ο-------ο-------
x2 x1
Other solutions will get with the changeover to the next point
etc.
$$x < - \frac{3 \sqrt{5}}{2} - \frac{1}{2}$$
$$x > - \frac{1}{2} + \frac{3 \sqrt{5}}{2}$$ |
# Vectors
Page 1 / 10
## Introduction
This chapter focuses on vectors. We will learn what is a vector and how it differs from everyday numbers. We will also learn how to add, subtract and multiply them and where they appear in Physics.
Are vectors Physics? No, vectors themselves are not Physics. Physics is just a description of the world around us. To describe something we need to use a language. The most common language used to describe Physics is Mathematics. Vectors form a very important part of the mathematical description of Physics, so much so that it is absolutely essential to master the use of vectors.
## Scalars and vectors
In Mathematics, you learned that a number is something that represents a quantity. For example if you have 5 books, 6 apples and 1 bicycle, the 5, 6, and 1 represent how many of each item you have.
These kinds of numbers are known as scalars .
Scalar
A scalar is a quantity that has only magnitude (size).
An extension to a scalar is a vector, which is a scalar with a direction. For example, if you travel 1 km down Main Road to school, the quantity 1 km down Main Road is a vector. The “ 1 km ” is the quantity (or scalar) and the “ down Main Road ” gives a direction.
In Physics we use the word magnitude to refer to the scalar part of the vector.
Vectors
A vector is a quantity that has both magnitude and direction.
A vector should tell you how much and which way .
For example, a man is driving his car east along a freeway at $100\phantom{\rule{2pt}{0ex}}\mathrm{km}·\mathrm{h}{}^{-1}$ . What we have given here is a vector – the velocity. The car is moving at $100\phantom{\rule{2pt}{0ex}}\mathrm{km}·\mathrm{h}{}^{-1}$ (this is the magnitude) and we know where it is going – east (this is the direction). Thus, we know the speed and direction of the car. These two quantities, a magnitude and a direction, form a vector we call velocity.
## Notation
Vectors are different to scalars and therefore have their own notation.
## Mathematical representation
There are many ways of writing the symbol for a vector. Vectors are denoted by symbols with an arrow pointing to the right above it. For example, $\stackrel{\to }{a}$ , $\stackrel{\to }{v}$ and $\stackrel{\to }{F}$ represent the vectors acceleration, velocity and force, meaning they have both a magnitude and a direction.
Sometimes just the magnitude of a vector is needed. In this case, the arrow is omitted. In other words, $F$ denotes the magnitude of the vector $\stackrel{\to }{F}$ . $|\stackrel{\to }{F}|$ is another way of representing the magnitude of a vector.
## Graphical representation
Vectors are drawn as arrows. An arrow has both a magnitude (how long it is) and a direction (the direction in which it points). The starting point of a vector is known as the tail and the end point is known as the head .
## Directions
There are many acceptable methods of writing vectors. As long as the vector has a magnitude and a direction, it is most likely acceptable. These different methods come from the different methods of expressing a direction for a vector.
## Relative directions
The simplest method of expressing direction is with relative directions: to the left, to the right, forward, backward, up and down.
## Compass directions
Another common method of expressing directions is to use the points of a compass: North, South, East, and West. If a vector does not point exactly in one of the compass directions, then we use an angle. For example, we can have a vector pointing $40{}^{\circ }$ North of West. Start with the vector pointing along the West direction: Then rotate the vector towards the north until there is a $40{}^{\circ }$ angle between the vector and the West. The direction of this vector can also be described as: W $40{}^{\circ }$ N (West $40{}^{\circ }$ North); or N $50{}^{\circ }$ W (North $50{}^{\circ }$ West)
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Draw a frame of reference with house A as the origin and write down the positions of houses B, C, D and E. |
# Areas And Perimeters Of Shapes Pdf
• and pdf
• Tuesday, December 22, 2020 9:46:15 AM
• 1 comment
File Name: areas and perimeters of shapes .zip
Size: 1818Kb
Published: 22.12.2020
The area of the circle is 1. Lesson 11 Polygons; Surface Area. Jan 10, - One glance helps students remember the key components of calculating area and perimeter in grade 3.
Comparing Numbers. Daily Math Review.
## Perimeter, Area, and Volume
Discover Shopee marketplace. So, as an equation, this looks like: When solving problems involving circles, it is important to think about what you are trying to determine. Program to calculate area of an Circle inscribed in a Square. One area of knowledge in geometry is simple identification of shapes, and learning the names for shapes with a certain number of sides is a rote activity.
The lesson utilizes a short video to solidify the relationship while maintaining a hands-on experience. Area of a sector formula. They approximate the area using squares These activities can be used practice or review and will get your students engaged in practicing finding the area and circumference of a circle.
In this lesson we first look at how to read three circle diagrams. Anyone caught forms additional arches. In this example, we are dividing the code using the Object Oriented Programming. Purpose of use. AREA 5 11 m. Explain your reasoning. Reading Three Circle Diagrams.
Improve your math knowledge with free questions in "Area and circumference of circles" and thousands of other math skills. The length of this line. Circle time is an important daily activity in the preschool classroom. Thus, the diameter of a circle is twice as long as the radius.
Area and Perimeter Area is a measure of the amount of space a two dimensional shape takes up, that is the space that is enclosed by its boundary. The formula used to find the area of a circlular sector - a pie-shaped part of a circle. The circumradius is defined as the radius of a circle that passes through all the vertices of a polygon. Arcs of a Circle. Example 1: Find the area of a circle whose circumference is 22 cm.
Activity 2: Possible journal entry or small. Preschool circle time. Small circle circulations pulmonary. The area of a circle is the number of square units inside that circle. Correct Answer — d cm. Practice and Problem Solving: D 1. Plot an easy triangle. Tap on PRINT, PDF or IMAGE button to print or download this grade-6 geometry worksheet to practice how to find the area and perimeter of 2-dimensional shapes such as square, rectangle, right angle triangle, circle, parallelogram, trapezoid, rhombus and pentagon.
Introduction 2 2. Printable circles for if you wish to do a discovery activity with your students on deriving the formula. Below are six versions of our grade 6 math worksheet on finding the area of a circle when given the radius or diameter; standard units of measurement are used.
The distance around a circular region is also known as its circumference. One example of a light-related circadian rhythm is sleeping at night and being awake during the day. Find the area of each circle. The following Java program to print the area of a circle has been written in five simple different ways, static method, using constructor, Interface, inheritance with sample outputs for each program.
Cages are now required for hammer, discus and. Solution The following diagram shows the circle drawn on squared paper. This Circle Worksheet is great for practicing solving for the circumference, area, radius and diameter of a circle. Catering to the learning needs of students in grade 5 through grade 8, these. This area-of-a-circle worksheet has an explanation and example at the top of the page. Example: Find the area of a circle whose circumference is 88 cm. If the triangle was a right triangle, it would be pretty easy to compute the area of the triangle by finding one-half the product of the base and the height.
Area of Sector and Arc Length. Finding the Area of a Circular Region. Use this circle calculator to find the area, circumference, radius or diameter of a circle. IED Activity 5. Area of a Circle. Staying in your local area means stay in the village, town, or part of the city where you live. ProcedureTeacher's activity Good morning class! Checking of attendance Who are absent today? Very Good! Now bring out your assignment. There's currently no direct support for a circle bound selector but you can follow these steps to implement it yourself.
Connect-the-dots with letters. Finding the Area of Composite Shapes - Set 2. I use the written words of the song on a chart as a shared reading text. Calculate the area of the shape. A chord is a segment whose endpoints are on a circle. Mathematicslearning MathematicsTeaching RashmiKathuria onlineteaching.
The diagram is not to scale. Sign in and start exploring all the free, organizational tools for your email. The World of Dante is a multi-media research tool intended to facilitate the study of the Divine Comedy through a wide range of offerings.
Options are numerous: you can choose metric or customary units or both, you can include or not include. Formula for the Area of a Circle. The teacher will demonstrate the necessary steps in order to.
He can go to the book area to look at books. Part Two: Triangle Area. Read the instructions to answer each of the following questions. These math worksheets are randomly created by our math worksheet generators, so you have an endless supply of quality math worksheets at your disposal. The distance across a circle through its center is called its diameter. The circumference of a circle is To determine these values, let's first take a closer look at the area and circumference formulas.
Easily calculate the area with these step-by-step examples. Notice that this formula uses the radius, so we will have to convert when we are given the diameter instead. Instead, almost all statements about area in Euclid, for example, is to say that one.
Even if they aren't being released, allow warm-ups with implements only in the cage, the circle or on the runway. Area of a Circle Worksheet attached. With this pizza circle sheet, your student will practice finding the radius, diameter, circumference and area of a circle. Name: You are given the radius of each circle. Learning Outcome The students are able to understand the concept of this activity through paper cutting. The distance from the center of a circle to any point along the circumference of a circle is called the radius.
The distance around a circle on the other hand is called the circumference c. This is the first area I set up every year. Since these are all the same, you can multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared. A 5 r2 5 10 2 cm2 Now we need to find of cm2 to determine the area of the sector formed by this central angle.
Worksheet Area. Complete the chart below and answer the following. Example BE is a diameter of the circle. The area of a polygon is the number of square units inside the polygon. Therefore, the area is,. Example 1 Estimate the area of a circle of radius 3 cm. Teachers and homeschooling moms can now. How long will it take. So here are the definitions of the two regions The above figure […].
## Free Math 7 Worksheets
Discover Shopee marketplace. So, as an equation, this looks like: When solving problems involving circles, it is important to think about what you are trying to determine. Program to calculate area of an Circle inscribed in a Square. One area of knowledge in geometry is simple identification of shapes, and learning the names for shapes with a certain number of sides is a rote activity.
The perimeter of a polygon or any other closed curve, such as a circle is the distance around the outside. The area of a simple, closed, planar curve is the amount of space inside. The volume of a solid 3 D shape is the amount of space displaced by it. Some formulas for common 2 -dimensional plane figures and 3 -dimensional solids are given below. The answers have one, two, or three dimensions; perimeter is measured in linear units , area is measured in square units , and volume is measured in cubic units. Names of standardized tests are owned by the trademark holders and are not affiliated with Varsity Tutors LLC. Media outlet trademarks are owned by the respective media outlets and are not affiliated with Varsity Tutors.
## Perimeter Worksheets
Area and Perimeter Formula are used to calculate the Area and Perimeter for the plane figure i,. The formulas are given for the following. Area and Perimeter of Right angle triangle b. Area and Perimeter of triangle c. Area and Perimeter of Square d.
This Percent Worksheet is great for practicing converting between percents, decimals, and fractions. You may select six different types of percentage conversion problems with three different types of Early geometry perimeter Worksheet.
The worksheets are very varied, and include:. Each worksheet is randomly generated and thus unique. The answer key is automatically generated and is placed on the second page of the file. You can generate the worksheets either in html or PDF format — both are easy to print. To get the worksheet in html format, push the button " View in browser " or " Make html worksheet ".
- Мы на месте.
Мидж налила себе стакан воды, надеясь, что это поможет ей успокоиться. Делая маленькие глотки, она смотрела в окно. Лунный свет проникал в комнату сквозь приоткрытые жалюзи, отражаясь от столешницы с затейливой поверхностью. Мидж всегда думала, что директорский кабинет следовало оборудовать здесь, а не в передней части здания, где он находился.
По приезде группу сразу же разделили. Все они подверглись проверке на полиграф-машине, иными словами - на детекторе лжи: были тщательно проверены их родственники, изучены особенности почерка, и с каждым провели множество собеседований на всевозможные темы, включая сексуальную ориентацию и соответствующие предпочтения. Когда интервьюер спросил у Сьюзан, не занималась ли она сексом с животными, она с трудом удержалась, чтобы не выбежать из кабинета, но, так или иначе, верх взяли любопытство, перспектива работы на самом острие теории кодирования, возможность попасть во Дворец головоломок и стать членом наиболее секретного клуба в мире - Агентства национальной безопасности.
Он перезагрузил монитор, надеясь, что все дело в каком-то мелком сбое. Но, ожив, монитор вновь показал то же. Чатрукьяну вдруг стало холодно. У сотрудников лаборатории систем безопасности была единственная обязанность - поддерживать ТРАНСТЕКСТ в чистоте, следить, чтобы в него не проникли вирусы. Он знал, что пятнадцатичасовой прогон может означать только одно: зараженный файл попал в компьютер и выводит из строя программу.
Лиланд Фонтейн решил, что с него довольно этого зрелища. - Выключите, - приказал. - Выключите эту чертовщину.
Чатрукьян заколебался. - Коммандер, мне действительно кажется, что нужно проверить… - Фил, - сказал Стратмор чуть более строго, - ТРАНСТЕКСТ в полном порядке. Если твоя проверка выявила нечто необычное, то лишь потому, что это сделали мы. А теперь, если не возражаешь… - Стратмор не договорил, но Чатрукьян понял его без слов. Ему предложили исчезнуть.
Когда распался последний силиконовый чип, громадная раскаленная лава вырвалась наружу, пробив верхнюю крышку и выбросив на двадцать метров вверх тучу керамических осколков, и в то же мгновение насыщенный кислородом воздух шифровалки втянуло в образовавшийся вакуум. Сьюзан едва успела взбежать на верхнюю площадку лестницы и вцепиться в перила, когда ее ударил мощный порыв горячего ветра. Повернувшись, она увидела заместителя оперативного директора АНБ; он стоял возле ТРАНСТЕКСТА, не сводя с нее глаз. Вокруг него бушевала настоящая буря, но в его глазах она увидела смирение. Губы Стратмора приоткрылись, произнеся последнее в его жизни слово: Сьюзан.
Он ничего не мог с собой поделать. Она была блистательна и прекрасна, равной ей он не мог себе даже представить. Его жена долго терпела, но, увидев Сьюзан, потеряла последнюю надежду. Бев Стратмор никогда его ни в чем не обвиняла. Она превозмогала боль сколько могла, но ее силы иссякли.
Стратмор медленно повернулся. Он являл собой печальное зрелище. - Это сделаю я, - сказал он, встал и, спотыкаясь, начал выбираться из-за стола. Сьюзан, чуть подтолкнув, усадила его на место. - Нет! - рявкнула .
Так продолжалось несколько недель. За десертом в ночных ресторанах он задавал ей бесконечные вопросы. Где она изучала математику.
#### Meditation the first and last freedom osho pdf
11.04.2021 at 03:37
#### Nonlinear programming analysis and methods pdf
11.04.2021 at 14:08
#### Linear and nonlinear differential equations pdf
13.04.2021 at 23:01
1. Mohammad K. 31.12.2020 at 02:08 |
# What are are the tests of divisibility of various numbers?
Jun 12, 2018
See below
#### Explanation:
Well, I'll list all the criteria I know of:
• A number is divisible by $2$ if it's even, i.e. if its last digit is $0 , 2 , 4 , 6$ or $8$
• A number is divisible by $3$ if the sum of its digits is divisible by $3$. This operation can be recursive. For example, we want to check if $69873$ is divisible by $3$. The sum of the digits is $33$. Again, $33$ is divisible by $3$ if the sum of its digits is, which is $3 + 3 = 6$. Thus, $69873$ is divisible by $3$.
• A number is divisible by $4$ if its last two digits are (including numbers ending by $00$). For example, $345876345 \textcolor{g r e e n}{44}$ is divisible by $4$ because $44$ is, while $348576 \textcolor{red}{33}$ isn't because $33$ isn't
• A number is divisible by $5$ if its last digit is $0$ or $5$.
• $6$ doesn't really have its criterion. You can tell if a number is divisible by $6$ if it divisible by $2$ and $3$ at the same time. This technique works for every number without a proper criterion, for example you can tell that a number is divisible by $14$ if it's divisible by $2$ and $7$.
• $7$ is a bit complicated: assuming your number has more than two digits, you must consider the number without the last digit, and subtract twice the last digit from the truncated number. You proceed until you get a two digit number. For example, start with $\textcolor{red}{9567} \textcolor{g r e e n}{6}$. This number is divisible by $7$ if $\textcolor{red}{9567} - 2 \cdot \textcolor{g r e e n}{6}$ is, which is $\textcolor{red}{955} \textcolor{g r e e n}{5}$. We perform another step to get to $\textcolor{red}{955} - 2 \cdot \textcolor{g r e e n}{5} = \textcolor{red}{94} \textcolor{g r e e n}{5}$. Last step: we have $\left(94\right) \textcolor{g r e e n}{-} 2 \cdot \left(5\right) = \textcolor{red}{84}$, which has two digits and is indeed divisible by $7$. So, $95676$ is divisible by $7$ as well.
• A number is divisible by $8$ if its last three digits are (including numbers ending by $000$).
• A number is divisible by $9$ if the sum of its digits is divisible by $9$.
• A number is divisible by $10$ if it ends with $0$. Actually, a number is divisible by ${10}^{n}$ if it ends with $n$ zeroes.
• $11$ requires some calculation as well: you must sum all the even-positioned digits, and subtract (absolute values) all the odd-positioned digits. If it is divisible by $11$ (or it is $0$), then so is the starting number. For example, start with $\textcolor{red}{6} \textcolor{g r e e n}{2} \textcolor{red}{5} \textcolor{g r e e n}{8} \textcolor{red}{3} \textcolor{g r e e n}{4}$. This leads to $- 6 + 2 - 5 + 8 - 3 + 4 = 0$. So, $625834$ is divisible by $11$
• $13$ works like $7$, except how have to sum four times the last digit to the rest of the number and check if the number is divisible by $13$. For example, start with $\textcolor{red}{730} \textcolor{g r e e n}{6}$. You have $\textcolor{red}{730} + 4 \cdot \textcolor{g r e e n}{6} = \textcolor{red}{75} \textcolor{g r e e n}{4}$. Again, you have $\textcolor{red}{75} + 4 \cdot \textcolor{g r e e n}{4} = 91$, which is divisible by $13$.
• $17$ is similar again, but you need to subtract five times the last digit to the rest of the number and check if the number is divisible by $17$ (absolute values again). For example, start with $\textcolor{red}{258} \textcolor{g r e e n}{4} \setminus \to \textcolor{red}{258} - 2 \cdot \textcolor{g r e e n}{4} = \setminus \textcolor{g r e e n}{23} \textcolor{g r e e n}{8}$. Finally, you have $\setminus \textcolor{g r e e n}{23} - 5 \cdot \textcolor{g r e e n}{8} = - 17$. Thus, $2584$ is divisible by $17$
• A number is divisible by $25$ if it ends with $00 , 25 , 50$ or $75$. |
## Methods of Calculation of Areas in Surveying | Simpson’s Rule
#### Calculation of Areas in Surveying | Simpson’s Rule
In one of my previous articles, I discussed Midpoint Ordinate Rule and Average Ordinate Rule in detail with an example and listed out various important methods used for the calculation of areas in Surveying. In this article, we will deal with the next important method (rule) i.e. Simpson’s Rule along with a numerical example used for the calculation of areas in the field of Surveying.
#### Here are the five important rules (Methods) used for the calculation of areas in Surveying:
1. Midpoint ordinate rule
2. Average ordinate rule
3. Simpson’s rule
4. Trapezoidal rule
5. Graphical rule
#### Simpson’s Rule
Statement
It states that, sum of first and last ordinates has to be done. Add twice the sum of remaining odd ordinates and four times the sum of remaining even ordinates. Multiply to this total sum by 1/3rd of the common distance between the ordinates which gives the required area.
Where O1, O2, O3, …. On are the lengths of the ordinates
d = common distance
n = number of divisions
#### Note:
This rule is applicable only if ordinates are odd, i.e. even number of divisions.
If the number of ordinates are even, the area of last division maybe calculated separated and added to the result obtained by applying Simpson’s rule to two remaining ordinates.
## Methods for Calculation of Areas in Surveying | Average Ordinate Rule
#### Calculation of Areas in Surveying | Average Ordinate Rule
In one of my previous articles, I discussed Midpoint Ordinate Rule in detail with an example and listed out various important methods used for the calculation of areas in Surveying. In this article, we will deal with the next important method (rule) used for the calculation of areas in the field of Surveying.
#### Here are the five important rules (Methods) used for the calculation of areas in Surveying:
1. Midpoint ordinate rule
2. Average ordinate rule
3. Simpson’s rule
4. Trapezoidal rule
5. Graphical rule
#### Average Ordinate Rule
The rule states that (to the average of all the ordinates taken at each of the division of equal length multiplies by baseline length divided by number of ordinates).
O1, O2, O3, O4….On ordinate taken at each of division.
L = length of baseline
n = number of equal parts (the baseline divided)
d = common distance
#### Area = [(O1+ O2+ O3+ …. + On)*L]/(n+1)
Here is an example of a numerical problem regarding the calculation of areas using Average Ordinate Rule
## Different Methods for the Calculation of Areas in Surveying
#### Different methods for the calculation of Areas in the field of Surveying
In this article, we will list out different methods to calculate the areas in Surveying and also study each of the method in depth… We will also explain each method with a suitable example for your better understanding…
#### Here are the five important rules (Methods) used for the calculation of areas in Surveying:
1. Midpoint ordinate rule
2. Average ordinate rule
3. Simpson’s rule
4. Trapezoidal rule
5. Graphical rule
We will now move on with our discussion on the first rule “Midpoint ordinate rule” with a suitable example.
#### Midpoint-ordinate rule
The rule states that if the sum of all the ordinates taken at midpoints of each division multiplied by the length of the base line having the ordinates (9 divided by number of equal parts).
In this, base line AB is divided into equal parts and the ordinates are measured in the midpoints of each division.
Area = ([O1 +O2 + O3 + …..+ On]*L)/n
L = length of baseline
n = number of equal parts, the baseline is divided
d = common distance between the ordinates
#### Example of the area calculation by midpoint ordinate rule
The following perpendicular offsets were taken at 10m interval from a survey line to an irregular boundary line. The ordinates are measured at midpoint of the division are 10, 13, 17, 16, 19, 21, 20 and 18m. Calculate the are enclosed by the midpoint ordinate rule. |
How do you factor 5x^2- 35x?
Mar 30, 2018
$5 x \left(x - 7\right)$
Explanation:
take out common factors
$\left(1\right) \text{ take " hcf(5,35) =color(red)(5)" out}$
$5 {x}^{2} - 35 x = \textcolor{red}{5} \left({x}^{2} - 7 x\right)$
$\left(2\right) \text{ take "hcf(x^2,x)=color(blue)(x )" out}$
$5 {x}^{2} - 35 x = \textcolor{red}{5} \left({x}^{2} - 7 x\right) = \textcolor{red}{5} \textcolor{b l u e}{x} \left(x - 7\right)$
Mar 30, 2018
$5 x \left(x - 7\right)$
Explanation:
We got:
$5 {x}^{2} - 35 x$
Take the greatest common factor out, which is $5$.
$= 5 \left({x}^{2} - 7 x\right)$
Notice how $x$ still remains in both factors inside, and so we can take it out, and we get:
$= 5 x \left(x - 7\right)$
Mar 30, 2018
$5 x \left(x - 7\right)$
Explanation:
Both terms are divisible by $5$, and $x$, so you factor those out of the polynomial. As for how this affects any solutions of this, because both factors equal $0$ (or whatever is on the other side of the equation, but I'll use $0$), $5 x = 0$, so $x = 0$, and $x - 7 = 0$, so $x = 7$ |
# The Pythagorean Number Values
Part II : Translating Letters into Numbers: The Pythagorean Number Values
Pythagorean Numerology
The Pythagorean system of numerology is the most widely used form of numerology. This is most likely due to the fact that it is very straightforward and easy to learn.
Before we get started, be sure you have read Part I of this guide and have a full understanding of the mathematics behind numerology.
In the previous section, we discussed how numbers with two or more digitsĀ are reduced to a single digit in Pythagorean numerology. This holds true in every case except with the numbers 11 and 22. These two numbers are considered master vibrations are not reduced. If you come up with 11 or 22 for one of your answers in the follow exercise do not reduce it further.
The Pythagorean Number Values
The following chart shows the numbers assigned to each letter in Pythagorean numerology and is the basis for many numerology readings. To read the chart, find the letter you are looking for and look at the corresponding number at the top of the column containing the letter. For example, the letter A would be 1, the letter B would be 2, the letter C would be 3, etc.
1 2 3 4 5 6 7 8 9 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z X
The following exercise will teach you to translate letters into numbers. This is one of the most basic components of numerology, so be sure you have a full understanding of it before moving on.
To begin with, write down your full name (first, middle and last). In Pythagorean numerology, it is important that you use your name as it appears on your birth certificate. Then assign a number to each letter in your name. To do that, locate the letter in the above chart, and then look at the number at the top of the column containing that letter. That number is the number you would assign to that letter. Do that for each of the letters in your full name. For our example, we will use a fictional character named John Alan Smith.
J O H N A L A N S M I T H 1 6 8 5 1 3 1 5 1 4 9 2 8
The next step is to add up all these numbers and then reduce the result to a single digit. You will want to be sure to add each name separately, then add those sums together to get the total for the whole name, then if necessary reduce to a single digit. We add the parts of the name separately instead of adding the whole name straight across because not only does the total number for the name have meaning, but each part of the name also has some bearing on who we are.
John = 1+6+8+5 = 20 = 2+0 = 2
Alan = 1+3+1+5 = 10 = 1+0 = 1
Smith = 1+4+9+2+8 = 24 = 2+4 = 6
Adding those totals together we get:
2+1+6 = 9
This means that in our example, John’s birthname reduces to a 9.
Exercise
Translate the following names into a single digit number using the Pythagorean chart:
a. Lucy Anne Anderson
b. Robert Jeffery Lewis
c. Dana Lee Johnson
CLICK HERE FOR THE ANSWERS (opens in a new window)
Continue to the Part II – Translating Letters Into Numbers : The Chaldean Number Values |
# NYT Digits Game Answers Today: How to Improve Your Score 14th July 2024
NYT Digits Game Answers: The New York Times’ Digits game has taken the world by storm, challenging players to use their mathematical skills to reach a target number with a limited set of digits. This game is not only entertaining, but it also helps to sharpen our mental math skills. However, some players may find it challenging to reach the target number within the given time frame. In this article, we will provide you with some tips and tricks to improve your score and increase your chances of winning the game.
## Understanding the Rules of Digits Game
Before we dive into the tips and tricks, let’s have a quick look at the rules of the Digits game. The game gives you six random digits that you have to use to reach a target number. You can use any mathematical operation, such as addition, subtraction, multiplication, and division, to get to the target number. You have 60 seconds to solve each puzzle, and the game gets harder as you progress to higher levels.
## Tips and Tricks to Improve Your Score
Here are some tips and tricks that you can use to improve your score in the Digits game:
Before you start solving the puzzle, take a moment to plan your strategy. Try to break down the target number into smaller, more manageable numbers. For example, if the target number is 100, you can break it down into 50 + 50 or 25 x 4. This will help you to focus on smaller numbers, making it easier to find the solution within the given time.
### Use Parentheses
Using parentheses can help you to group numbers together and simplify the equation. For example, if the target number is 20, you can use (10 – 5) x 2 to get to the solution. This will simplify the equation and make it easier to solve.
### Utilize the Commutative and Associative Properties
The commutative and associative properties of addition and multiplication can be used to simplify equations and reach the target number. For example, if the target number is 45, you can use 25 + 10 + 10 to get to the solution, or you can use (10 x 5) – 5 to get the same result. This will help you to save time and reach the target number faster.
### Use Multiplication and Division
Multiplication and division are often faster than addition and subtraction when solving the Digits game. For example, if the target number is 36, you can use 6 x 6 to get to the solution, which is faster than adding or subtracting numbers.
### Practice, Practice, Practice
Like any other skill, practice makes perfect. The more you play the Digits game, the better you will become at solving the puzzles. You can also try to challenge yourself by setting a time limit or trying to solve the puzzle with fewer moves.
## Today’s 14th July 2024 Digits Answers
• 64
• 25 + 10 = 35
• 35 – 3 = 32
• 32 x 2 = 64
• 124
• 25 x 5 = 125
• 125 – 3 = 122
• 122 + 2 = 124
• 296
• 15 – 3 = 12
• 25 x 12 = 300
• 300 – 4 = 296
• 351
• 15 x 6 = 90
• 90 x 4 = 360
• 360 – 9 = 351
• 477
• 25 x 19 = 475
• 5 – 3 = 2
• 475 + 2 = 477
## NYT Digits, Puzzle 1 Answer for May 6, 2024
1. 25 × 4 = 100
2. 100 − 10 = 90
3. 90 − 5 = 85
4. 85 + 1 = 86
## NYT Digits, Puzzle 1 Answer for May 3, 2024
The target is 51. There are two operations to get to 3-stars for the first puzzle.
1. 10 × 5 = 50
2. 50 + 1 = 51
## NYT Digits, Puzzle 2 Answer for May 3, 2024
The target is 155. There are three operations to get to 3-stars for the second puzzle.
1. 10 × 9 = 90
2. 90 × 2 = 180
3. 180 – 25 = 155
## NYT Digits, Puzzle 3 Answer for May 3, 2024
The target is 252. There are two operations to get to 3-stars for the third puzzle.
1. 25 + 11 = 36
2. 36 × 7 = 252
## NYT Digits, Puzzle 4 Answer for May 3, 2024
The target is 347. There are four operations to get to 3-stars for the fourth puzzle.
1. 9 × 6 = 54
2. 54 – 4 = 50
3. 50 × 7 = 350
4. 350 – 3 = 347
## NYT Digits, Puzzle 5 Answer for May 3, 2024
The target is 412. There are three operations to get to 3-stars for the fifth puzzle.
1. 20 – 3 = 17
2. 25 × 17 = 425
3. 425 – 13 = 412
### Can I use decimals in the Digits game?
No, the Digits game only allows whole numbers.
### Can I use the same digit twice in one equation?
Yes, you can use the same digit twice in one equation.
### How many levels are there in the Digits game?
The Digits game has an infinite number of levels, and the difficulty increases as you progress.
## Final Words
The Digits game is not only a fun way to pass the time, but it also helps to improve your mental math skills. By following the tips and tricks mentioned in this article, you can improve your score and increase your chances of winning the game. Remember to practice regularly and challenge yourself to become a Digits game expert.
Hi, I'm Selva a full-time Blogger, YouTuber, Affiliate Marketer, & founder of Coding Deekshi. Here, I post about programming to help developers.
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# Radical problem solver
Radical problem solver can help students to understand the material and improve their grades. We can solving math problem.
## The Best Radical problem solver
We'll provide some tips to help you choose the best Radical problem solver for your needs. distance = sqrt((x2-x1)^2 + (y2-y1)^2) When using the distance formula, you are trying to find the length of a line segment between two points. The first step is to identify the coordinates of the two points. Next, plug those coordinates into the distance formula and simplify. The last step is to take the square root of the simplify equation to find the distance. Let's try an example. Find the distance between the points (3,4) and (-1,2). First, we identify the coordinates of our two points. They are (3,4) and (-1,2). Next, we plug those coordinates into our distance formula: distance = sqrt((x2-x1)^2 + (y2-y1)^2)= sqrt((-1-3)^2 + (2-4)^2)= sqrt(16+4)= sqrt(20)= 4.47 Therefore, the distance between the points (3,4) and (-1,2) is 4.47 units.
There are many ways to solve problems involving interval notation. One popular method is to use a graphing calculator. Many graphing calculators have a built-in function that allows you to input an equation and then see the solution in interval notation. Another method is to use a table of values. This involves solving the equation for a few different values and then graphing the results. If the graph is a straight line, then the solution is simple to find. However, if the graph is not a straight line, then the solution may be more complicated. In either case, it is always important to check your work to make sure that the answer is correct.
Once the equation has been factored, you can solve each factor by setting it equal to zero and using the quadratic formula. Another method for solving the square is to complete the square. This involves adding a constant to both sides of the equation so that one side is a perfect square. Once this is done, you can take the square root of both sides and solve for the variable. Finally, you can use graphing to solve the square. To do this, you will need to plot the points associated with the equation and then find the intersection of the two lines. Whichever method you choose, solving the square can be a simple process as long as you have a strong understanding of algebra.
The next step is to use matrix operations to simplify the matrix. Finally, the solution to the system of linear equations can be found by solving the simplified matrix. By using a matrix to represent a system of linear equations, it is possible to solve the equation quickly and efficiently.
## Math solver you can trust
My little brother first suggested this app to me. I tried it and it's amazing. He doesn't do algebra or anything like that yet as he's in the third grade but I do! And I love it! A good Recommendation to both students and teachers struggling or taking long with math
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Solving quadratic inequalities Arithmetic math problems Math equations algebra Solve for an angle in right triangles Maclaurin series solver |
# Paint Calculation: How Much Paint Do I Need
At the finishing stage of your residential building project, this question will come to your mind “How much paint do I need?”.
So, in this post, I’ll show you the process of paint calculation for walls and ceilings.
You’ll also learn to calculate the required putty quantity.
After learning this, you can calculate the required paint quantity for your whole project.
So, let’s learn.
## How Much Paint Is Needed For The Walls Of A Bedroom?
I’ll show you the paint calculation process of a bedroom’s wall here.
Once you’ve learned the process, you’ll be able to calculate the required paint quantity for all of the rooms of an apartment, and eventually for the whole project.
Let’s say, this is our bedroom to be painted:
To calculate paint quantity…
### First, Get The Gross Painting Area Of Walls.
The formula is,
= Length of walls × Height
Here,
Length of walls is,
= 2 × (10′ + 12′)
= 44 feet.
And, the height is = 9′-6″
So the gross painting area is,
= 44 × 9′-6″
= 418 square feet.
But, Here is a beam in the bedroom:
In some rooms, you’ll get beams like this.
And, you need to add the beam bottom portion to the gross painting area.
In our example, it is.
= 5″ × 12′
= 5 square feet.
So, the total gross painting area of the bedroom’s walls is,
= 418 + 5
= 423 square feet.
### Next, Calculate The Area Of Openings.
Here we have two openings — a window and a door.
The area of the window opening is,
= 5′-0″ × 4′-6″
= 22.50 square feet.
And, the area of the door opening is,
= 3′-4″ × 7′-0″
= 23.31 square feet.
So, the total opening area is,
= 22.50 + 23.31
= 45.81 square feet.
### Now, Calculate The Net Painting Area
The formula is,
= Gross painting area — Opening area
= 423 — 45.81
= 377.19 square feet.
Say, 378 square feet.
This is the area you need to calculate paint for.
### Finally, Calculate How Much Paint You Need For The Area.
The formula is,
= Net painting area ÷ Coverage area of paint per liter.
The coverage of paint varies depending on the surface condition and manufacturer.
Generally, one-liter paint covers somewhere 40 to 50 square feet. However, you should check the manufacturer’s information.
For the sake of our calculation, Let’s take 45 square feet per liter.
So, the required paint quantity is,
= 378 ÷ 45
= 8.4 liters.
This is for one coat paint only.
To get the smooth paint finish, you need at least two coats of paint.
So, for two coats, the required paint quantity is,
= 2 × 8.4
= 16.80 liters.
This is the required paint quantity we need for our example bedroom.
But,
## How Much Paint Is Needed For The Ceiling?
Obviously, you aren’t going to use the same paint on walls and ceiling.
The paint on the ceiling is often different than the walls.
So, let’s learn the process of paint calculation for the ceiling.
Actually, the process is the same as the wall paint.
You first need to…
### Calculate The Gross Painting Area.
The formula is,
= Length × Width
= 10′-0″ × 12′-0″
= 120 square feet.
This is actually the net painting area for our ceiling as we don’t have any openings in the ceiling.
[NOTE: But sometimes, you’ll have some columns’ portion and beam portions to be deducted from the gross painting area.
For sake of the simplicity of this post I just didn’t put them in our example.
I think you got this.]
So, you just need to…
### Calculate the required paint quantity…
…For 120 square feet of painting area.
And, the formula is,
= Net painting area ÷ Coverage area of paint per liter.
Let’s assume the coverage area of paint per liter is 45 square feet.
So, the required paint quantity for one coat is,
= 120 ÷ 45
= 2.66 liters.
And for two coats,
= 2 × 2.66
= 5.33 liters.
As you now know the process of paint calculation, let’s see…
## How To Calculate The Required Quantity Of Putty
Before painting, you need to apply putty on a new surface.
So, it’s time to learn the process of putty calculation.
The process is the same as the paint calculation.
Let’s say, you need to calculate the required putty quantity for the walls of our example bedroom.
For that:
### First, Calculate The Net Painting Area
We calculated this before, at the time of paint calculation.
And that is 378 square feet.
### Now, Calculating The Required Putty Quantity
And, the formula is,
= Net painting area ÷ Coverage area of putty per kg.
The coverage area of putty is 12 — 15 square feet/kg for two coats and 8 — 12 square feet/kg for three coats.
Let’s say, you’re going to apply three coats of putty on the surface and the coverage area of putty is 10 square feet per kg.
So, the required putty quantity is,
= 378 ÷ 10
= 37.80 kg.
That’s it!
Conclusion
Calculating paint quantity is easy.
Right?
You just need to get the net painting area and divide the area with the coverage area of paint per liter.
Most of the manufacturers show the coverage area of the paint on their product’s label. You can get the information from there. |
# 2.4: 14. Universal derivation
## 14.1 An example: the Meno
In one of Plato’s dialogues, the Meno, Socrates uses questions and prompts to direct a young slave boy in the process of making a square that has twice the area of a given square, by using the diagonal of the given square as a side in the new square. Socrates draws a square 1 foot on a side in the dirt. The young boy at first just suggests that to double its area, the two sides of the square should be doubled, but Socrates shows him that this would result in a square that is four times the area of the given square; that is, a square of the size four square feet. Next, Socrates takes this 2×2 square, which has four square feet, and shows the boy how to make a square double its size.
Socrates: Tell me, boy, is not this a square of four feet that I have drawn?
Boy: Yes.
Socrates: And now I add another square equal to the former one?
Boy: Yes.
Socrates: And a third, which is equal to either of them?
Boy: Yes.
Socrates: Suppose that we fill up the vacant corner?
Boy: Very good.
Socrates: Here, then, there are four equal spaces?
Boy: Yes.[12]
So what Socrates has drawn at this point looks like:
Suppose each square is a foot on a side. Socrates will now ask the boy how to make a square that is of eight square feet, or twice the size of their initial 2×2 square. Socrates has a goal and method in drawing the square four times the size of the original.
Socrates: And how many times larger is this space than the other?
Boy: Four times.
Socrates: But it ought to have been twice only, as you will remember.
Boy: True.
Socrates: And does not this line, reaching from corner to corner, bisect each of these spaces?
By “spaces”, Socrates means each of the 2×2 squares. Socrates has now drawn the following:
Boy: Yes.
Socrates: And are there not here four equal lines that contain this new square?
Boy: There are.
Socrates: Look and see how much this new square is.
Boy: I do not understand.
After some discussion, Socrates gets the boy to see that where the new line cuts a small square, it cuts it in half. So, adding the whole small squares inside this new square, and adding the half small squares inside this new square, the boy is able to answer.
Socrates: The new square is of how many feet?
Boy: Of eight feet.
Socrates: And from what line do you get this new square?
Boy: From this. [The boy presumably points at the dark line in our diagram.]
Socrates: That is, from the line which extends from corner to corner of the each of the spaces of four feet?
Boy: Yes.
Socrates: And that is the line that the educated call the “diagonal”. And if this is the proper name, then you, Meno’s slave, are prepared to affirm that the double space is the square of the diagonal?
Boy: Certainly, Socrates.
For the original square that was 2×2 feet, by drawing a diagonal of the square we were able to draw one side of a square that is twice the area. Socrates has demonstrated how to make a square twice the area of any given square: make the new square’s sides each as large as the diagonal of the given square.
It is curious that merely by questioning the slave (who would have been a child of a Greek family defeated in battle, and would have been deprived of any education), Socrates is able to get him to complete a proof. Plato takes this as a demonstration of a strange metaphysical doctrine that each of us once knew everything and have forgotten it, and now we just need to be helped to remember the truth. But we should note a different and interesting fact. Neither Socrates nor the slave boy ever doubts that Socrates’s demonstration is true of all squares. That is, while Socrates draws squares in the dirt, the slave boy never says, “Well, Socrates, you’ve proved that to make a square twice as big as this square that you have drawn, I need to take the diagonal of this square as a side of my new square. But what about a square that’s much smaller or larger than the one you drew here?”
That is in fact a very perplexing question. Why is Socrates’s demonstration good for all, for any, squares?
## 14.2 A familiar strangeness
We have saved for last the most subtle issue about reasoning with quantifiers: how shall we prove something is universally true?
Consider the following argument. We will assume a first order logical language that talks about numbers, since it is sometimes easier to imagine something true of everything in our domain of discourse if we are talking about numbers.
All numbers evenly divisible by eight are evenly divisible by four.
All numbers evenly divisible by four are evenly divisible by two.
_____
All numbers evenly divisible by eight are evenly divisible by two.
Let us assume an implicit translation key, and then we can say that the following is a translation of this argument.
x (F x →G x )
x (G x →H x )
_____
x (F x →H x )
This looks like a valid argument. Indeed, it may seem obvious that it is valid. But to prove it, we need some way to be able to prove a universal statement.
But how could we do such a thing? There are infinitely many numbers, so surely we cannot check them all. How do we prove that something is true of all numbers, without taking an infinite amount of time and creating an infinitely long proof?
The odds are that you already know how to do this, although you have never reflected on your ability. You most likely saw a proof of a universal claim far back in grade school, and without reflection concluded it was good and proper. For example, when you were first taught that the sum of the interior angles of a triangle is equivalent to two right angles, you might have seen a proof that used a single triangle as an illustration. It might have gone something like this: assume lines AB and CD are parallel, and that two other line segments EF and EG cross those parallel lines, and meet on AB at E. Assume also that the alternate angles for any line crossing parallel lines are equal. Assume that a line is equivalent to two right angles, or 180 degrees. Then, in the following picture, b’=b, c’=c, and b’+c’+a=180 degrees. Thus, a+b+c=180 degrees.
Most of us think about such a proof, see the reasoning, and agree with it. But if we reflect for a moment, we should see that it is quite mysterious why such a proof works. That’s because, it aims to show us that the sum of the interior angles of any triangle is the same as two right angles. But there are infinitely many triangles (in fact, logicians have proved that there are more triangles than there are natural numbers!). So how can it be that this argument proves something about all of the triangles? Furthermore, in the diagram above, there are infinitely many different sets of two parallel lines we could have used. And so on.
This also touches on the case that we saw in the Meno. Socrates proves that the area of a square A twice as big as square B does not simply have sides twice as long as the sides of B; rather, each side of A must be the length of the diagonal of B. But he and the boy drew just one square in the dirt. And it won’t even be properly square. How can they conclude something about every square based on their reasoning and a crude drawing?
In all such cases, there is an important feature of the relevant proof. Squares come in many sizes, triangles come in many sizes and shapes. But what interests us in such proofs is all and only the properties that all triangles have, or all and only properties that all squares have. We refer to a triangle, or a square, that is abstract in a strange way: we draw inferences about, and only refer to, its properties that are shared with all the things of its kind. We are really considering a special, generalized instance.
We can call this special instance the “arbitrary instance”. If we prove something is true of the arbitrary triangle, then we conclude it is true of all triangles. If we prove something is true of the arbitrary square, then we conclude it is true of all squares. If we prove something is true of an arbitrary natural number, then we conclude it is true of all natural numbers. And so on.
## 14.3 Universal derivation
To use this insight, we will introduce not an inference rule, but rather a new proof method. We will call this proof method “universal derivation” or, synonymously, “universal proof”. We need something to stand for the arbitrary instance. For a number of reasons, it is traditional to use unbound variables for this. However, to make it clear that the variable is being used in this special way, and that the well-formed formula so formed is a sentence, we will use a prime—that is, the small mark “”—to mark the variable. Let α be any variable. Our proof method thus looks like this.
Where α does not appear in any open proof above the beginning of the universal derivation.
Remember that an open proof is a subproof that is not completed.
We will call any symbolic term of this form (x′, y′, z…) an “arbitrary term”, and it is often convenient to describe it as referring to the arbitrary object or arbitrary instance. But there is not any one object in our domain of discourse that such a term refers to. Rather, it stands in for an abstraction: what all the things in the domain of discourse have in common.
The semantics of an arbitrary instance is perhaps less mysterious when we consider the actual syntactic constraints on a universal derivation. One should not be able to say anything about an arbitrary instance α′ unless one has done universal instantiation of a universal claim. No other sentence should allow claims about α′. For example, you cannot perform existential instantiation to an arbitrary instance, since we required that existential instantiation be done to special indefinite names that have not appeared yet in the proof. But if we can only makes claims about α′ using universal instantiation, then we will be asserting something about α′ that we could have asserted about anything in our domain of discourse. Seen in this way, from the perspective of the syntax of our proof, the universal derivation hopefully seems very intuitive.
This schematic proof has a line where we indicate that we are going to use α′ as the arbitrary object, by putting α′ in a box. This is not necessary, and is not part of our proof. Rather, like the explanations we write on the side, it is there to help someone understand our proof. It says, this is the beginning of a universal derivation, and α′ stands for the arbitrary object. Since this is not actually a line in the proof, we need not number it.
We can now prove our example above is valid.
Remember that our specification of the proof method has a special condition, that α′ must not appear earlier in an open proof (a proof that is still being completed). This helps us avoid confusing two or more arbitrary instances. Here, there is no x appearing above our universal derivation in an open proof (in fact, there is no other arbitrary instance appearing in the proof above x), so we have followed the rule.
## 14.4 Two useful theorems: quantifier equivalence
Our definition of “theorem” remains the same for the first order logic and for the propositional logic: a sentence that can be proved without premises. However, we now have a distinction when it comes to the semantics of sentences that must be true. Generally, we think of a tautology as a sentence that must be true as a function of the truth-functional connectives that constitute that sentence. That is, we identified that a tautology must be true by making a truth table for the tautology. There are, however, sentences of the first order logic that must be true, but we cannot demonstrate this with a truth table. Here is an example:
x (F x v ¬F x )
This sentence must be true. But we cannot show this with a truth table. Instead, we need the concept of a model (introduced briefly in section 17.6) to describe this property precisely. But even with our intuitive semantics, we can see that this sentence must be true. For, we require (in our restriction on predicates) that everything in our domain of discourse either is, or is not, an F.
We call a sentence of the first order logic that must be true, “logically true”. Just as it was a virtue of the propositional logic that all the theorems are tautologies, and all the tautologies are theorems; it is a virtue of our first order logic that all the theorems are logically true, and all the logically true sentences are theorems. Proving this is beyond the scope of this book, but is something done in most advanced logic courses and texts.
Here is a proof that x(Fx v ¬Fx).
Let us consider another example of a logically true sentence that we can prove, and thus, practice universal derivation. The following sentence is logically true.
(( x (F x G x ) ^ x (F x H x )) x (F x (Gx ^H x ))
Here is a proof. The formula is a conditional, so we will use conditional derivation. However, the consequence is a universal sentence, so we will need a universal derivation as a subproof.
Just as there were useful theorems of the propositional logic, there are many useful theorems of the first order logic. Two very useful theorems concern the relation between existential and universal claims.
( x F x ↔ ¬ x ¬F x )
( x F x ↔ ¬∃ x ¬F x )
Something is F just in case not everything is not F. And, everything is F if and only if not even one thing is not F.
We can prove the second of these, and leave the first as an exercise.
## 14.5 Illustrating invalidity
Consider the following argument:
x (H x →G x )
¬H d
_____
¬G d
This is an invalid argument. It is possible that the conclusion is false but the premises are true.
Because we cannot use truth tables to describe the semantics of quantifiers, we have kept the semantics of the quantifiers intuitive. A complete semantics for first order logic is called a “model”, and requires some set theory. This presents a difficulty: we cannot demonstrate that an argument using quantifiers is invalid without a semantics.
Fortunately, there is a heuristic method that we can use that does not require developing a full model. We will develop an intuitive and partial model. The idea is that we will come up with an interpretation of the argument, where we ascribe a meaning to each predicate, and a referent for each term, and where this interpretation makes the premises obviously true and the conclusion obviously false. This is not a perfect method, since it will depend upon our understanding of our interpretation, and because it requires us to demonstrate some creativity. But this method does illustrate important features of the semantics of the first order logic, and used carefully it can help us see why a particular argument is invalid.
It is often best to create an interpretation using numbers, since there is less vagueness of the meaning of the predicates. So suppose our domain of discourse is the natural numbers. Then, we need to find an interpretation of the predicates that makes the first two lines true and the conclusion false. Here is one:
Hx: x is evenly divisible by 2
Gx: x is evenly divisible by 1
d: 3
The argument would then have as premises: All numbers evenly divisible by 2 are evenly divisible by 1; and, 3 is not evenly divisible by 2. These are both true. But the conclusion would be: 3 is not evenly divisible by 1. This is false. This illustrates that the argument form is invalid.
Let us consider another example. Here is an invalid argument:
x (Fx→G x )
F a
_____
G b
We can illustrate that it is invalid by finding an interpretation that shows the premises true and the conclusion false. Our domain of discourse will be the natural numbers. We interpret the predicates and names in the following way:
Fx: x is greater than 10
Gx: x is greater than 5
a: 15
b: 2
Given this interpretation, the argument translates to: Any number greater than 10 is greater than 5; 15 is greater than 10; therefore, 2 is greater than 5. The conclusion is obviously false, whereas the premises are obviously true.
In this exercise, it may seem strange that we would just make up meanings for our predicates and names. However, as long as our interpretations of the predicates and names follow our rules, our interpretation will be acceptable. Recall the rules for predicates are that they have an arity, and that each predicate of arity n is true or false (never both, never neither) of each n things in the domain of discourse. The rule for names is that they refer to only one object.
This illustrates an important point. Consider a valid argument, and try to come up with some interpretation that makes it invalid. You will find that you cannot do it, if you respect the constraints on predicates and names. Make sure that you understand this. It will clarify much about the generality of the first order logic. Take a valid argument like:
x (Fx→G x )
F a
_____
G a
Come up with various interpretations for a and for F and G. You will find that you cannot make an invalid argument.
In summary, an informal model used to illustrate invalidity must have three things:
1. a domain of discourse;
2. an interpretation of the predicates; and
3. an interpretation of the names.
If you can find such an informal model that makes the premises obviously true and the conclusion obviously false, you have illustrated that the argument is invalid. This may take several tries: you can also sometimes come up with interpretations for invalid arguments that make all the premises and the conclusion true; this is not surprising, when you remember the definition of valid (that necessarily, if the premises are true then the conclusion is true—in other words, it is not enough that the conclusion just happens to be true).
## 14.6 Problems
1. Prove the following. These will require universal derivation. (For the third, remember that the variables used in quantifiers are merely used to indicate the place in the following expression that is being bound. So, if we change the variable nothing else changes in our proof or use of inference rules.) The last three are challenging. For these last three problems, do not use the quantifier negation rules.
1. Premises: xFx, x (Fx ↔ Gx). Conclusion: xGx.
2. Premises: x(Fx → Gx). Conclusion: x(¬Gx → ¬Fx).
3. Premises: x(Fx ↔ Hx), y(Hy ↔ Gy). Conclusion: z(Fz ↔ Gz).
4. Conclusion: (x(¬Fx v Gx) → x(Fx → Gx)).
5. Conclusion: (x(Fx Gx) → (xFx ↔ xGx)).
6. Conclusion: (¬∃xFx x¬Fx).
7. Conclusion: (¬xFx x¬Fx).
8. Conclusion: (xFx ¬x¬Fx).
2. Create a different informal model for each of the following arguments to illustrate that it is invalid.
1. Premises: x(Fx → Gx), ¬Ga. Conclusion: ¬Fb.
2. Premises: x(Fx v Gx), ¬Fa. Conclusion: Gb.
3. Premises: x(Fx → Gx), xFx. Conclusion: Gc.
3. In normal colloquial English, write your own valid argument with at least two premises and with a conclusion that is a universal statement. Your argument should just be a paragraph (not an ordered list of sentences or anything else that looks like formal logic). Translate it into first order logic and prove it is valid.
4. Do we have free will? Much of the work that philosophers have done to answer this question focuses on trying to define or understand what free will would be, and understand the consequences if we do not have free will. Doubts about free will have often been raised by those who believe that physics will ultimately explain all events using deterministic laws, so that everything had to happen one way. Here is a simplified version of such an argument.
Every event is caused by prior events by way of natural physical laws. Any event caused by prior events by way of natural physical laws could not have happened otherwise. But, if all events could not have happened otherwise, then there is no freely willed event. We conclude, therefore, that there are no freely willed events.
Symbolize this argument and prove it is valid. You might consider using the following predicates:
Fx: x is an event.
Gx: x is caused by prior events by way of natural physical laws.
Hx: x could have happened otherwise.
Ix: x is a freely willed event.
(Hint: this argument will require universal derivation. The conclusion can be had using modus ponens, if you can prove: all events could not have happened otherwise.) Do you believe that this argument is sound?
[12] These passages are adapted from the Benjamin Jowett translation of the Meno. Versions of this translation are available for free on the web. Students hoping to read other works by Plato should consider Cooper and Hutchinson (1997). |
## What is the Circumference of a Circle and How to Measure it?
Circumference Calculator
#### What is the circumference of a circle?
The perimeter of a circle is known as its circumference.
Trace the outline of a few circular objects of different sizes like a CD, a circular bottle lid, a frisbee, a round bowl and a key ring on paper and cut them along their circumference. Then, do the following:
1. Measure the diameter of each circle by folding in half.
2. Measure the circumference of each circle using a string.
3. Record both readings in a table as below.
As you will notice, the circumference of each object is about 3 times its diameter.
More precisely, the circumference of a circle is
22 7
times its diameter.
The value
22 7
is represented by π and read as pi.
The approximate value of π is 3.14.
#### What is the formula to calculate circumference of a circle?
Circumference = π × Diameter
Or,
Circumference = π × 2 × Radius = 2 × π × Radius
Practice Unlimited Circumference Questions
#### 1. Find the circumference of the circles below.
a)
Diameter = 8 cm
Circumference = π × 8 cm = 3.14 × 8 cm = 25.12 cm
b)
Radius = 21 cm
Circumference
= 2 × π × 21 cm
= 2 ×
22 17
× 213 cm
132 cm
When calculation is easy, we use
22 7
instead of 3.14 as the value of π
#### 2. Find the perimeters of the following figures. Round off your answers to 2 decimal places.
a) A semicircle of diameter 17.6 cm.
Perimeter of a semicircle
= Half the circumference of the circle + diameter
= (
1 2
× π × 17.6) + 17.6
= (
1 2
× 3.14 × 17.6) + 17.6
= 27.632 + 17.6
= 45.232
≈ 45.23 cm
b) A quadrant of radius 5 cm.
Perimeter of a quadrant
= A quarter of the circumference of circle + (2 × radius)
= (
1 24
× 21 × 3.14 × 5) + (2 × 5)
= 7.85 + 10
= 17.85 cm
c) A three-quarter circle of radius 84 cm.
Perimeter of
3 4
circle
3 4
of the circumference of the circle + (2 × radius)
= (
3 24
× 21 ×
22 17
× 8412) + (2 × 84)
= (
3 12
× 1 ×
2211 1
× 12) + 168
= 396 + 168
= 564 cm
#### 3. Can a pizza of radius 12 cm fit into a plate that has a circumference of 100 cm?
Circumference of pizza = 2 × 3.14 × 12 = 75.36 cm
75.36 cm < 100 cm
Yes, the pizza can fit into the plate.
#### 4. The spoke of a wheel is 35 cm long. How far does the wheel travel in 1 rotation? Give your answer in metres.
Spoke of the wheel = Radius of the wheel
1 rotation of the wheel
= Circumference of the wheel
= 2 ×
22 17
× 355 cm
= 220 cm
= 2.2 m
The wheel travels 2.2 m in 1 rotation. |
# Prove? cotx/(cscx-1)=(cscx+1)/cotx
Remember that $1 + {\cot}^{2} x = {\csc}^{2} x$.
#### Explanation:
$\cot \frac{x}{\csc x - 1} = \frac{\csc x + 1}{\cot} x$
Remember that $1 + {\cot}^{2} x = {\csc}^{2} x$. This becomes useful if we multiply the terms with $\csc x$ to get them squared:
$\cot \frac{x}{\csc x - 1} \left(\frac{\csc x + 1}{\csc x + 1}\right) = \frac{\csc x + 1}{\cot} x$
$\frac{\cot x \csc x + \cot x}{{\csc}^{2} x - 1} = \frac{\csc x + 1}{\cot} x$
We can now use ${\csc}^{2} x - 1 = {\cot}^{2} x$
$\frac{\cot x \csc x + \cot x}{{\cot}^{2} x} = \frac{\csc x + 1}{\cot} x$
$\frac{\csc x + 1}{\cot x} = \frac{\csc x + 1}{\cot} x$ |
Title
College Algebra
Answer/Discussion to Practice Problems
Tutorial 23B: Rational Inequalities
Answer/Discussion to 1a I used the test-point method, but you could also use the sign graph of factors method.
Step 1: Write the rational inequality in standard form.
This quadratic inequality is already in standard form.
Step 2: Factor the numerator and denominator and find the values of x that make these factors equal to 0 to find the boundary points.
Numerator:
*Set numerator = 0 and solve
Denominator:
*Set denominator = 0 and solve
-7 and 2 are boundary points.
Below is a graph that marks off the boundary points -7 and 2 and shows the three sections that those points have created on the graph. Note that open holes were used on those two points since our original inequality did not include where it is equal to 0 and 2 makes the denominator 0. Note that the two boundary points create three sections on the graph: , , and .
Step 4: Test a point in each test interval found in step 3 to see which interval(s) is part of the solution set.
You can choose ANY point in an interval to represent that interval. Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get. Keep in mind that our original problem is . Since we are looking for the quadratic expression to be LESS THAN 0, that means we need our sign to be NEGATIVE. From the interval , I choose to use -8 to test this interval: (I could have used -10, -25, or -10000 as long as it is in the interval)
*Chose -8 from 1st interval to plug in for x
Since 1/10 is positive and we are looking for values that cause our quadratic expression to be less than 0 (negative), would not be part of the solution.
From the interval , I choose to use 0 to test this interval. (I could have used -6, -5, or -4 as long as it is in the interval)
*Chose 0 from 2nd interval to plug in for x
Since -7/2 is negative and we are looking for values that cause our expression to be less than 0 (negative), would be part of the solution.
From the interval , I choose to use 3 to test this interval. (I could have used 10, 25, or 10000 as long as it is in the interval)
*Chose 3 from 3rd interval to plug in for x
Since 10 is positive and we are looking for values that cause our quadratic expression to be less than 0 (negative), would not be part of the solution.
Interval notation: Graph: *Open interval indicating all values between -7 and 2 *Visual showing all numbers between -7 and 2 on the number line
Answer/Discussion to 1b I used the test-point method, but you could also use the sign graph of factors method.
Step 1: Write the rational inequality in standard form.
*Inv. of add. 2 is sub. 2 *Rewrite 2nd fraction with LCD of x - 2
Step 2: Factor the numerator and denominator and find the values of x that make these factors equal to 0 to find the boundary points.
Numerator:
*Set numerator = 0 and solve
Denominator:
*Set denominator = 0 and solve
-5 and 2 are boundary points.
Below is a graph that marks off the boundary points -5 and 2 and shows the three sections that those points have created on the graph. Note that there is an open hole at -2. Since that is the value that cause the denominator to be 0, we cannot include where x = 2. Since our inequality includes where it is equal to 0, and -5 causes only the numerator to be 0 there is a closed hole at -5. Note that the two boundary points create three sections on the graph: , , and .
Step 4: Test a point in each test interval found in step 3 to see which interval(s) is part of the solution set.
You can choose ANY point in an interval to represent that interval. Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get. Keep in mind that our inequality is > 0. Since we are looking for the quadratic expression to be GREATER THAN OR EQUAL TO 0, that means we need our sign to be POSITIVE (OR O). From the interval , I choose to use -6 to test this interval: (I could have used -10, -25, or -10000 as long as it is in the interval)
*Chose -6 from 1st interval to plug in for x
Since 1/8 is positive and we are looking for values that cause our quadratic expression to be greater than or equal to 0 (positive or 0), would be part of the solution.
From the interval , I choose to use 0 to test this interval. (I could have used -4, -3, or -2 as long as it is in the interval)
*Chose 0 from 2nd interval to plug in for x
Since -5/2 is negative and we are looking for values that cause our expression to be greater than or equal to 0 (positive or 0), would not be part of the solution.
From the interval , I choose to use 3 to test this interval. (I could have used 10, 25, or 10000 as long as it is in the interval)
*Chose 3 from 3rd interval to plug in for x
Since 8 is positive and we are looking for values that cause our quadratic expression to be greater than or equal to 0 (positive or 0), would be part of the solution.
Interval notation: Graph: *An closed interval indicating all values less than or equal to -5 and an open interval indicating all values greater then 2 *Visual showing all numbers less than or equal to -5 and indicating all values greater then 2
Last revised on Jan. 2, 2010 by Kim Seward.
All contents copyright (C) 2002 - 2010, WTAMU and Kim Seward. All rights reserved. |
# How do you find vertical, horizontal and oblique asymptotes for F(x) = (x - 1)/(x - x^3)?
Dec 20, 2016
Vertical asymptotes at $x = 0 , x = - 1 \mathmr{and} x = 1$
Horizontal asymptote at $y = 0$
#### Explanation:
Vertical asymptotes will occur when the denominator equals $0$.
$x - {x}^{3} = 0$
$x \left(1 - {x}^{2}\right) = 0$
$x \left(1 + x\right) \left(1 - x\right) = 0$
$x = 0 , - 1 \mathmr{and} 1$
Since the degree of the numerator is lesser than that of the denominator, there will be a horizontal asymptote at $y = 0$ and no slant asymptote.
Hopefully this helps!
Dec 21, 2016
vertical asymptotes at x =0 and x = - 1
horizontal asymptote at y = 0
#### Explanation:
Factorising f(x)
$f \left(x\right) = \frac{x - 1}{x \left(1 - {x}^{2}\right)} = \frac{x - 1}{x \left(1 - x\right) \left(1 + x\right)}$
$= \frac{- \left(1 - x\right)}{x \left(1 - x\right) \left(1 + x\right)} = \frac{- {\left(\cancel{1 - x}\right)}^{1}}{x \left(\cancel{1 - x}\right) \left(1 + x\right)} = \frac{- 1}{x \left(1 + x\right)}$
Hence there is an excluded value at x = 1
Indicating a hole at x = 1 in f(x). This hole is not in the simplified f(x).
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
solve: $x \left(1 + x\right) = 0 \Rightarrow x = 0 \text{ or } x = - 1$
$\Rightarrow x = 0 \text{ and " x=-1" are the asymptotes}$
Horizontal asymptotes occur as
${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$
divide terms on numerator/denominator by x
$f \left(x\right) = \frac{- \frac{1}{x}}{\frac{x}{x} + {x}^{2} / x} = \frac{- \frac{1}{x}}{1 + \frac{1}{x}}$
as $x \to \pm \infty , f \left(x\right) \to \frac{0}{1 + 0}$
$\Rightarrow y = 0 \text{ is the asymptote}$
Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0, denominator-degree 2) Hence there are no oblique asymptotes.
graph{(-1)/(x(1+x) [-20, 20, -10, 10]} |
iGCSE (2021 Edition)
# 7.10 Solving equations using radians
Lesson
Solving trigonometric equations in radians uses the same steps as when we solved them using degrees:
1. Find the related positive acute angle (by using exact values or the calculator).
2. Draw a unit circle and use the ASTC rule to determine the relevant quadrants.
3. Determine the value of the angles in the relevant quadrants that satisfy the equation in the given domain.
For harder equations, remember to simplify first. This may involve using trigonometric identities or algebraic techniques like factorising.
Remember!
Always pay careful attention to the domain in which the angle can lie.
Remember to modify the domain for equations with compound angles.
#### Practice questions
##### Question 1
If $\cos\theta=0.9063$cosθ=0.9063, find the value of $\theta$θ correct to $2$2 decimal places, where $0$0$\theta$θ < $2\pi$2π.
1. Find the acute angle $\theta$θ that solves the equation.
2. Now find ALL solutions to $\theta$θ in the range $0$0 to $2\pi$2π (separate alternative solutions by a comma). Round your solutions to $2$2 decimal places.
##### Question 4
Solve $\tan^2\left(x\right)+2\tan x+1=0$tan2(x)+2tanx+1=0 over the interval $[$[$0$0, $2\pi$2π$)$).
##### Question 5
Solve $\sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}$sin(x+π3)=12 where $0\le x\le2\pi$0x2π.
##### Question 6
Solve $\sin x=0$sinx=0 over the interval $[$[$-4\pi$4π, $4\pi$4π$]$]. |
69 minus 1 percent
This is where you will learn how to calculate sixty-nine minus one percent (69 minus 1 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 69 minus 1 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 69 of something.
69
(100%)
1 percent means 1 per hundred, so for each hundred in 69, you want to subtract 1. Thus, you divide 69 by 100 and then multiply the quotient by 1 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(69 ÷ 100) × 1
= 0.69
We made a pink square that we put on top of the image shown above to illustrate how much 1 percent is of the total 69:
The dark blue not covered up by the pink is 69 minus 1 percent. Thus, we simply subtract the 0.69 from 69 to get the answer:
69 - 0.69
= 68.31
The explanation and illustrations above are the educational way of calculating 69 minus 1 percent. You can also, of course, use formulas to calculate 69 minus 1%.
Below we show you two formulas that you can use to calculate 69 minus 1 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
69 - ((69 × 1/100))
69 - 0.69
= 68.31
Formula 2
Number × (1 - (Percent/100))
69 × (1 - (1/100))
69 × 0.99
= 68.31
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
70 minus 1 percent
Here is the next percent tutorial on our list that may be of interest. |
# ORDERING RATIONAL AND IRRATIONAL NUMBERS
Example 1 :
Order √22, (π + 1) and 4½ from least to greatest. Then graph them on the number line.
Solution :
Step 1 :
First approximate √22.
√22 is between 4 and 5. Since we don’t know where it falls between 4 and 5, we need to find a better estimate for √22.
So that we can compare it to 4½.
Since 22 is closer to 25 than 16, use squares of numbers between 4.5 and 5 to find a better estimate of √22.
4.52 = 20.25
4.62 = 21.16
4.72 = 22.09
4.82 = 23.04
Since 4.72 = 22.09, an approximate value for √22 is 4.7.
That is,
√22 ≈ 4.7 ----(1)
Step 2 :
An approximate value of π is 3.14. So an approximate value of π+1 is 4.14.
That is,
π + 1 ≈ 4.14 ----(2)
Step 3 :
The value of 4½ is 4.5.
That is,
4½ = 4.5 ----(3)
Step 4 :
Comparing (1), (2) and (3), we can order the numbers from least to greatest as given below.
π + 1, 4½ and √22
Step 5 :
Read the numbers from left to right to place them on a number line in order from least to greatest.
Example 2 :
Order 4√2, 2√3, 3√2, √17, 3√3 and 5 from least to greatest.
Solution :
Key Concept :
Most of the given real numbers are irrational numbers.
So, square the given real numbers and order them from least to greatest.
Step 1 :
Take square to the number 4√2.
(4√2)2 = (4)2(√2)2
(4√2)2 = (16)(2)
(4√2)2 = 32 ----(1)
Step 2 :
Take square to the number 2√3.
(2√3)2 = (2)2(√3)2
(2√3)2 = (4)(3)
(2√3)2 = 12 ----(2)
Step 3 :
Take square to the number 3√2.
(3√2)2 = (3)2(√2)2
(3√2)2 = (9)(2)
(3√2)2 = 18 ----(3)
Step 4 :
Take square to the number √17.
(√17)2 = 17 ----(4)
Step 5 :
Take square to the number 3√3.
(3√3)2 = (3)2(√3)2
(3√3)2 = (9)(3)
(3√3)2 = 27 -----(5)
Step 6 :
Take square to the number 5.
(5)2 = 25 ----(6)
Step 7 :
Comparing (1), (2), (3), (4), and (6), we can write the squares of the given irrational numbers from least to greatest as given below.
12, 17, 18, 25, 27, 32
In the above order, write the corresponding real number to its square to write the given real numbers in the order from least to greatest.
2√3, √17, 3√2, 5, 3√3, 4√2
Example 3 :
Order 2, √5, 61/3, 42/3 from least to greatest
Solution :
Step 1 :
First convert the radical in √5 as exponent.
Then, we have
√5 = 51/2
Step 2 :
Now, the given numbers are
2, 51/2, 61/3, 42/3
In the numbers listed above, we find the exponents 1/2, 1/3 and 2/3.
All the exponents are fractions and the denominators of the fractions are 2 and 3.
The least common multiple of the denominators 2 and 3 is 6
Step 3 :
Now take exponent 6 to all of the numbers in step 2.
26, (51/2)6, (61/3)6, (42/3)6
Simplify.
26, 53, 62, 44
64, 125, 36, 256
Step 4 :
Order the numbers from least to greatest.
36, 64, 125, 256
Step 5 :
Using the order of numbers in step 4, write the given rational and irrational numbers in the order from least to greatest.
61/3, 2, √5, 42/3
Kindly mail your feedback to v4formath@gmail.com
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# Domain, range, piecewise-defined functions
## • Use the notions of domain and range. • Interpret a graph of a piecewise-defined function. • Graph step functions.
1 The restaurant
WHAT: A function that represents a context is described, and students are asked to state a reasonable domain and range for the context F-IF.B.5.
WHY: The purpose of this task is to get students thinking about the domain and range of a function representing a particular context. Often when a function is being used to model a context, the expression for the function has a larger domain and range than is reasonable for the context. Asking students to focus on a function for which there is no formula focuses attention on the context itself. In order to complete this task, students will need to reason abstractly about the quantities in the context MP.2.
2 Finding the domain
WHAT: The function $f(x) = 2/(x – 3)$ is defined. Students evaluate $f(11)$ and attempt to evaluate $f(3)$. Then students are asked to list the algebraic operations needed to evaluate an input and describe restrictions that each operation places on the domain of the function F-IF.A.1.
WHY: The purpose of this task is to introduce the idea of the domain of a function by linking it to the evaluation of an expression defining the function. By thinking through the evaluation step by step, students isolate the exact point where a given input results in an undefined output. In part (d), any domain that excludes $x=3$ is possible. It is conventional when given a function defined by an expression to take the domain to be the largest possible, but it is worth pointing that this is a convention, not a mathematical fact. As students gain a mature understanding of functions they learn that the domain is something that is specified when you define the function, it does not come already attached. The function in this task can be broken down into two simple operations. More complicated variations are possible. Students must make use of structure to evaluate operations in the correct order. Also, stepping through the operations like this lays the groundwork for students to be able to make use of structure in future work MP.7.
3 The Parking Lot
WHAT: Students are asked to graph the cost of parking in a garage where the cost increases each half-hour or fraction thereof F-IF.C.7b.
WHY: This task provides an example of a step function arising from a real-world context F-IF.C.7b. The idea that the price “jumps” \\$.50 each half-hour or fraction thereof gives rise to why a single linear function is not an appropriate model for this problem MP.4. This task lends itself to a discussion about domain and range F-IF.A.1 and why the graph requires “hollow dots” and “solid dots” to indicate jumps.
4 Bank Account Balance
WHAT: Students are presented with a piecewise-defined graph that shows the balance in a bank account over a seven-day period. The graph is shown as a continuous function. Because the bank account balance changes discontinuously, the graph of a step function is a more accurate representation.
WHY: This task provides students with an opportunity to explain why the function shown in the graph is not the most appropriate model for the given context MP.2. The second part of this task is open-ended and allows for various models MP.4 (a step function F-IF.C.7b or bar graph are two possibilities). This could lend itself to a class discussion where students explain why their model best represents the given situation MP.3. |
# Bayes' theorem facts for kids
Kids Encyclopedia Facts
In probability theory and applications, Bayes' theorem shows the relation between a conditional probability and its reverse form. For example, the probability of a hypothesis given some observed pieces of evidence, and the probability of that evidence given the hypothesis. This theorem is named after Thomas Bayes ( or "bays") and is often called Bayes' law or Bayes' rule.
## Formula
The equation used is:
$P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}.$
Where:
• P(A) is the prior probability or marginal probability of A. It is "prior" in the sense that it does not take into account any information about B.
• P(A|B) is the conditional probability of A, given B. It is also called the posterior probability because it is derived from (or depends upon) the specified value of B.
• P(B|A) is the conditional probability of B given A. It is also called the likelihood.
• P(B) is the prior or marginal probability of B, and acts as a normalizing constant.
In many scenarios, P(B) is calculated indirectly using the formula $P(B) = P(B|A)P(A) + P(B|A^c)P(A^c)$, which simply states that the probability of B is the sum of the conditional probabilities based on whether A has occurred or not.
## Example
A simple example is as follows: There is a 40% chance of rain on Sunday. If it rains on Sunday, there is a 10% chance of rain on Monday. If it did not rain on Sunday, then there is an 80% chance that it will rain on Monday.
"Raining on Sunday" is event A, and "Raining on Monday" is event B.
• P( A ) = 0.40 = Probability of rain on Sunday.
• P( A ) = 0.60 = Probability of no rain on Sunday.
• P( B | A ) = 0.10 = Probability of rain on Monday, if it rained on Sunday.
• P( B | A ) = 0.90 = Probability of no rain on Monday, if it rained on Sunday.
• P( B | A ) = 0.80 = Probability of rain on Monday, if it did not rain on Sunday.
• P( B |A` ) = 0.20 = Probability of no rain on Monday, if it did not rain on Sunday.
The first thing we would normally calculate is the probability of it raining on Monday: This would be the sum of the probability of "Rain on Sunday and rain on Monday" and "No rain on Sunday and rain on Monday":
$0.40\times0.10 + 0.60\times0.80 = 0.52 = 52%$ chance
However, we were asked to calculate the probability that it rained on Sunday, given that it rained on Monday, then this is where Bayes' theorem comes in. It allows us to calculate the probability of an earlier event, given the result of a later event.
The equation used is:
$P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}.$
In our case, "Rain on Sunday" is event A, and "Rain on Monday" is event B.
• P(B|A) = 0.10 = Probability of rain on Monday, if it rained on Sunday.
• P(A) = 0.40 = Probability of rain on Sunday.
• P(B) = 0.52 = Probability of rain on Monday.
So, to calculate the probability it rained on Sunday, given that it rained on Monday, we use the formula:
$P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}.$
or:
$P(A|B) = \frac{0.10*0.40}{0.52} = .0769$
In other words, if it rained on Monday, then there is a 7.69% chance it rained on Sunday.
## Intuitive explanation
To calculate the probability of it having rained on Sunday, given that it rained on Monday, we can take the following steps:
• We know that it rained on Monday. Therefore, the total probability is P(B).
• The probability it rained on Sunday is P(A).
• The probability it rained on Monday, given that it rained on Sunday is P(B|A).
• The probability of raining on Sunday and raining Monday is P(A)*P(B|A).
• Therefore, the total probability of it having rained on Sunday, given that it rained on Monday, is the chance of it raining on Sunday and Monday divided by the total chance of it having rained on Monday.
Therefore,
$P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}.$
Another way to see this, which shows where Bayes' theorem comes from, is to consider the probability P(AB) that it rains on both Sunday and Monday. This can be calculated in two different ways, which give the same answer for P(AB):
$P(A)\, P(B|A) = P(B)\, P(A|B)$
In this regard, Bayes' theorem is just another way to write that equation. |
How Notation Obscures Patterns
This is another stab for me at what continues to prove to be a complicated topic: How our mess of mathematical notation obfuscates key patterns. This is also a rough draft of thoughts, not meant to be a polished product.
In Algebra II recently, the unit is radical notation and rational exponents. That means that logarithms are around the corner. This is the kernel of where our traditional notation goes from somewhat annoying to utterly slapdash. The underlying relationships are admittedly difficult, but the notation makes them so much worse.
But instead, I’ll keep it simpler. Today in class I reviewed how to add, multiply, and divide fractions, so I’ll reflect on my thoughts there.
First, let’s multiply fractions. Groovy. $$\frac{4}{5} \cdot \frac{3}{7}=\frac{12}{35}$$. Multiply across the top, multiply across the bottom. Simple enough rule, but… why? Why does it work?
Set that aside. Let’s divide fractions. $$\frac{4}{5} \div \frac{3}{7}=\frac{4}{5} \cdot \frac{7}{3}=\frac{28}{15}$$. In high school, we say, “Dividing by a fraction is equivalent to multiplying by its reciprocal.” In middle school, that’s given as “Keep-Change-Flip” (okay, stop screaming, Nix The Tricksers).
It is less obvious why this works.
Next, let’s add some fractions. $$\frac{4}{5} + \frac{3}{7}=\frac{28}{35}+\frac{15}{35}=\frac{43}{35}$$. It’s not at all clear why this works. It’s a confusing bunch of steps that can be condensed to the dreaded “Butterfly Method” (or “Bowtie Method”).
Next up: Converting from mixed numbers to improper fractions. $$6\frac{3}{7}=\frac{6}{1}+\frac{3}{7}=\frac{42+3}{7}=\frac{45}{7}$$. This has a trick, too: Multiply the integer by the denominator, add the product to the numerator, then put it over the denominator. But if we see a mixed number as a sum of an integer and a fraction, this is just another example of adding fractions (with unlike denominators).
Now… imagine our notation were different. A small tweak for now: Allow / to be used like -, that is, as a unary operator. /4 is 1/4, just as -4 is 0-4.
No more denominators. I’ll do the multiplication we just did, but no fractions and no binary division.
$$(4 \times /5) \times (3 \times /7) = 4 \times /5 \times 3 \times /7$$. Compare this to $$(4 + -5) + (3 + -7) = 4 + -5 + 3 + -7$$. Here’s a key feature: When dealing with real numbers, order doesn’t matter within addition, and it doesn’t matter within multiplication. We only need parentheses when subtraction or division are involved.
So $$4 + -5 + 3 + -7 = 4 + 3 + -5 + -7 = (4 + 3) + -(5 + 7)$$ and $$4 \times /5 \times 3 \times /7 = 4 \times 3 \times /5 \times /7 = (4 \times 3) \times /(5 \times 7)$$. Entirely parallel. No need for a special rule on how to multiply fractions: If you know how to deal with $$(4 – 5) + (3 – 7)$$, you know how to deal with $$(4 \div 5) \times (3 \div 7)$$.
Because I do think it’s difficult to read this new notation (I have suggestions for that, but that’s a separate item), I’ll tweak it to make it closer to what we’re used to: I’ll use $$\div$$ instead of $$\times /$$ for division (as I did at the end of the last paragraph). The parallel pattern persists: $$4 – 5 + 3 – 7 = 4 + 3 – 5 – 7 = (4 + 3) – (5 + 7)$$ and $$4 \div 5 \times 3 \div 7 = 4 \times 3 \div 5 \div 7 = (4 \times 3) \div (5 \times 7)$$.
Consider dividing fractions: $$(4 \div 5) \div (3 \div 7)$$. Its parallel within addition is: $$(4 – 5) – (3 – 7) = 4 – 5 – 3 + 7$$. When we get rid of the parentheses, we “distribute” the negative sign. What we’re really doing is distributing a coefficient of -1, but … notation!
So what happens when we distribute the division sign? If negating a negative results in a positive, what does dividing by a division do? In standard mathematics terms, that doesn’t even make any sense, but let’s just pretend. If $$-(-x) = x$$, then perhaps $$/(/x) = x$$ as well?
Indeed, the multiplicative inverse of the multiplicative inverse of a number is the original number, just as the additive inverse of the additive inverse of a number is the original number.
So… hold on, this is where things got weird for me when I noticed them. But it works! $$(4 \div 5) \div (3 \div 7) = 4 \div 5 \div 3 \times 7$$. And since each division sign only applies to its neighbor to the right (that is, because this is shorthand for $$4 \times /5 \times /3 \times 7$$), we can rearrange these items as we please. This is why $$(4 \div 5) \div (3 \div 7) = (4 \times 7) \div (5 \times 3)$$.
No need for “multiply by the reciprocal”: If someone can understand why $$(4 – 5) – (3 – 7) = (4 + 7) – (5 + 3)$$, they can understand why $$(4 \div 5) \div (3 \div 7) = (4 \times 7) \div (5 \times 3)$$. Completely parallel structures.
Next up is the monster: Adding fractions with unlike denominators. Note that I don’t mention subtraction. Subtraction is fairly trivial. We could get rid of subtraction today (from high school mathematics) without otherwise changing our notation just by always writing negatives instead.
Rewrite $$\frac{4}{5} + \frac{3}{7}$$ to $$(4 \div 5) + (3 \div 7)$$ and compare it to $$(4 \times 5) + (3 \times 7)$$. What are we doing when we “add” fractions? We’re creating a single expression involving division and two numbers.
Step back one more pace to $$(4 \times 5) + (3 \times 5) = (4 + 3) \times 5$$. This works because $$4 \times 5$$ and $$3 \times 5$$ have a common factor. We can’t do anything similar with $$(4 \times 5) + (3 \times 7)$$ because there’s no common factor. If we wanted to write something like $$(a + b) \times c$$ for this case, we would need to introduce a common factor.
I don’t think I’ve ever seen anyone deliberately introduce a common factor for that purpose, but maybe I’ve just missed it. Regardless, though, that’s what we do when adding fractions: We effectively introduce a common factor (of sorts). Here’s the whole path: $$(4 \div 5) + (3 \div 7) =$$ $$(4 \div 5 \div 7 \times 7) + (3 \div 7 \div 5 \times 5) =$$ $$(4 \times 7 \div (5 \times 7)) + (3 \times 5 \div (7 \times 5)) =$$ $$(28 \div 35) + (15 \div 35) =$$ $$(28 + 15) \div 35 = 43 \div 35$$.
At first blush, this might look like more work than the traditional method, but if so, that might be more a factor of its alienness. I feel like the underlying mathematics is more consistent and logical and, more to the point, that this allows us to see ways in which division works like multiplication, as well as how the multiplication/division layer work like the addition/subtraction layer.
A more radical approach would be to get rid of division entirely. Just as we can do all of our subtraction using negation (e.g., $$5 – 3 = 5 + -3)$$, we can do all of our division using reciprocals (e.g., $$5/3 = 5 \times /3$$. Just as $$-(-4) = 4$$, $$/(/4) = 4$$. Just as we can distribute negation across a polynomial (e.g. $$-(4 + 3) = -4 + -3$$), we can distribute reciprocals: $$/(4 \times 3) = /4 \times /3$$. The common symbols even lend themselves to this: Just as a + sign consists of two overlapping – signs, a $$\times$$ sign consists of two overlapping / signs.
I’m not sure where I sit on such a radical approach. I personally would be eager for it, but there’s a serious argument to be made about not just throwing current notation out entirely. Plus, as a teacher, I’m fully aware of how entrenched many of my colleagues are, and how my eagerness would be generally met by their lack of it.
My bigger point here, though, is that until I started digging into alternative notation, I didn’t see these parallels myself. I knew that multiplication generally acted like addition, but I didn’t notice how completely parallel the processes were.
And throwing in the exponential layer? I’ve started to dig here, but I’m still a long way off from something I find satisfying. I do know, though, that our current notation is a total mess. And messy notation isn’t conducive for clear understanding.
Clio Corvid
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# Scale Drawings
Scale drawings allow us to accurately recreate real-life objects with fixed scale factors. But how can we accomplish this? Let''s find out:
## An example of a scale drawing
Suppose our teacher asked us to draw a "model" of our school with all of its landmarks proportionally shown.
First, we need a scale factor. We also need to find the proportional distances between these landmarks.
Let''s say our scale factor is
1 cm:100 m
.
If we know that the playground is 500 meters away from the music room, we know that in our scale model, we can draw this distance as 5 cm. We would continue this process until the entire scale drawing was complete.
## More examples of scale drawings
If we wanted to create a scale drawing of a zoo with a scale factor of
1 cm: 400 m
, then how far apart are the lion''s cage and the snake area if they are 6cm apart in our drawing?
All we need to do is multiply
6*400
to get a total distance of 2,400 meters.
If an enlarged scale drawing of a microchip has a length of 6 cm but the actual length is 2mm, what is the scale factor of our drawing?
First, convert the lengths to the same unit: The scale drawing has a length of 60 mm.
60/2 = 30
. The scale factor is therefore 30:1.
## Topics related to the Scale Drawings
Graphing Scale and Origin
Scale Factor
## Flashcards covering the Scale Drawings
Basic Geometry Flashcards
Common Core: High School - Geometry Flashcards
## Practice tests covering the Scale Drawings
Common Core: High School - Geometry Diagnostic Tests
Intermediate Geometry Diagnostic Tests
## Pair your student with a stellar math tutor today
If you''ve been searching for an appropriate math tutor for your student, look no further than Varsity Tutors. We carefully interview tutors before they work with students, ensuring your student is always working with a tutor who understands the subjects they need help with. Reach out today and get your student started with private math tutoring.
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# Solving Rational Inequalities
View Notes
## Introduction to Solving Rational Inequalities
A key approach towards solving rational inequalities depends upon determining the critical values of the rational expression that divides the number line into distinctive open intervals. The critical values are the zeros (0s) of both the numerator as well as the denominator. Note that the zeros of the denominator make the rational expression undefined, so they should be immediately excluded or dismissed as a probable solution. However, zeros of the numerators also require to be verified for its possible inclusion to the entire solution.
## Different Ways of How to Solve Rational Inequalities
Struggling with the ways of how to find rational inequalities? Well! There are different ways of solving rational inequalities which are as below:
• Solving rational inequalities algebraically.
• Solving inequalities with rational expressions.
• Solving rational inequalities by graphing.
### How to Solve Rational Inequalities
Below are the summarized steps in order to find rational inequalities and solve them.
Step 1: Write the expression of inequality as one quotient on the left and zero (0) on the right.
Step 2: identify the critical points–the points where the rational expression will either be undefined or zero.
Step 3: Use the critical points for dividing the number line into intervals.
Step 4: Test the value of each interval. The number line displays the sign of each factor of the numerator as well as the denominator in each interval. The number line also shows the sign of the quotient.
Step 5: identify the intervals where inequality is appropriate.
Step 6: Write the solution in the form of interval notation.
### Rational
This is what a rational expression seems like having a ratio of two polynomials.
Sometimes we would require solving rational inequalities like these:
Symbol Words Example > greater than (x + 1)/(3 − x) > 2 < less than x/(x + 7) < −3 ≥ greater than or equal to (x−1)/(5−x) ≥ 0 ≤ less than or equal to (3 − 2x)/(x − 1) ≤ 2
### Solved Example
Example: Solve and simplify the given rational inequality:
(x2 - 3x - 4 )/(x2 - 8x + 16) < 0
Solution:
Step 1: Factor out both the numerator and denominator in order to find their zeros. In factored form, we will obtain:
(x+1) (x - 4) / (x - 4) (x - 4) < 0
Step 2: identify the zeros of the rational inequality by establishing each factor equal to zero
Step 3: Solve for x. We get:
Zeros of the numerators: –1 and 4
Zeros of the denominators: 4
Step 4: Consider taking the zeros as critical numbers in order to divide the number line into distinct intervals.
Step 5: Test the validity of each interval by choosing a test value and assessing them into the original rational inequality. Note that the ones in yellow are the numbers we selected.
If you observe, the only interval providing a true statement is (- 1, 4).
Additionally, the zeros of the numerator don’t check with the original rational inequality so we should disregard them.
Therefore, the final answer is just (−1,4).
### Did You Know?
• When solving a rational inequality, we must first write the inequality with 0 on the right and only one quotient on the left.
• Critical points are identified for using them to divide the number line into intervals.
• Once identifying the critical points, factors of the numerator and denominator, and quotient in each interval are found. This will identify the interval, or intervals, which consists of all the solutions of rational inequality.
Conclusion: Solution of rational inequality is written in interval notation. We should be very careful writing the notation as it helps to identify if or not the endpoints are included.
Q1. What is a Rational Inequality?
Answer: A rational inequality is an inequality which consists of a rational expression. While solving a rational inequality, we can use distinct methods as we use when solving linear inequalities. We particularly must remember that when we multiply or divide by a negative number, the inequality sign should reverse.
Moreover, when solving a rational inequality, we must carefully consider what value might make rational expression undefined and thus should be excluded.
Q2. What is the Difference Between a Linear Equation and a Rational Inequality?
Answer: When we solve an equation and the outcome is x=7, we are known that there is only one solution, which is 7.
On the other hand, when we solve inequality and the outcome is x > 7, we know that there are many solutions. We graph the outcome solving rational inequalities by graphing in order to better help show all the solutions, and we begin with 7. Seven becomes a critical point and then we plan if to shade to the left or right of it. The numbers to the right of 7 are greater than 7, so we shade to the right.
Q3. What is a Critical Point?
Answer: A critical point is a number that makes the rational expression zero (0) or undefined. |
Posted In: KS3 , Year 11 maths
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In maths, an equation is a statement that two things are equal. In an equation, there are two expressions, one on each side of the equal to sign.
The followings are examples of equations:
8 + 2 = 10
7 * 5 = 35
12 - 5 = 7
15 / 3 = 5
Variables in an equation
Variables are the unknown in an equation. An equation could contain one or more unknowns. In the equation 8 + x = 10, x is the unknown variable. We can now solve this equation to determine the value of x as shown below:
8 + x = 10
x = 10 - 8
Therefore, x = 2
This process of finding the value of the unknown is called solving the equation.
Practice questions
Question: Find the value of x in the following equations.
1. 12 - x = 7
2. 13 * x = 26
3. 5 + x = 8
( Answer: 5, 2, 3 )
Simultaneous equations
Simultaneous equations are two equations with two unknowns. They are called simultaneous equations because they must be solved together in order to determine the unknowns.
For example:
Given two simultaneous equations x + y = 10 and x - y = 6, find the values of x and y.
Equation 1: x + y = 10
Equation 2: x - y = 6
Add the two equations to eliminate the unknown y
2x = 16
x= 16/2 = 8
Now replace the value of x in equation 1
8 + y = 10
y= 10 - 8 = 2
Solving simultaneous equations using substitution
Sometimes you can use substitution to solve two simultaneous equations to determine the value of unknowns.
For example:
Given two simultaneous equations x + y = 12 and x = 2y, find the values of x and y.
Equation 1: x + y = 12
Equation 2: x = 2y
Replace the value of x in equation 1
2y + y = 12
3y = 12
y = 4
Now replace the value of y in equation x
x + 4 = 12
x = 8
Practice questions
Question: Find the value of x and y in the following equations: 3x + y = 7 and 3x - y = 5
( Answer: Equation 1: 3x + y = 7
Equation 2: 3x - y = 5
6x = 12
x = 2
Replace the value of x in equation 1
3 * 2 + y = 7
6 + y = 7
y= 1
)
Question: Find the value of x and y in the following equations: y - 2x = 1 and 2y - 3x = 5
( Answer: Equation 1: y - 2x = 1
Equation 2: 2y - 3x = 5
Rearrange equation 1: y = 1 + 2x
Substitute in equation 2: 2 (1 + 2x) - 3x = 5
2 + 4x - 3x = 5
2 + x = 5
x = 3
Use value of x in equation 1
y= 2 * 3 + 1 = 7
)
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# 3.2 Absolute value
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples.Objectives of this module: understand the geometric and algebraic definitions of absolute value.
## Overview
• Geometric Definition of Absolute Value
• Algebraic Definition of Absolute Value
## Absolute value—geometric approach
The absolute value of a number $a$ , denoted $|a|$ , is the distance from $a$ to 0 on the number line.
Absolute value speaks to the question of "how far," and not "which way." The phrase how far implies length, and length is always a nonnegative (zero or positive) quantity. Thus, the absolute value of a number is a nonnegative number. This is shown in the following examples:
$|4|=4$
$|-4|=4$
$|0|=0$
$-|5|=-5$ .
The quantity on the left side of the equal sign is read as "negative the absolute value of 5." The absolute value of 5 is 5. Hence, negative the absolute value of 5 is $-5$ .
$-|-3|=-3$ .
The quantity on the left side of the equal sign is read as "negative the absolute value of $-3$ ." The absolute value of $-3$ is 3. Hence, negative the absolute value of $-3$ is $-\left(3\right)=-3$ .
## Algebraic definition of absolute value
The problems in the first example may help to suggest the following algebraic definition of absolute value. The definition is interpreted below. Examples follow the interpretation.
## Absolute value—algebraic approach
The absolute value of a number $a$ is
$|a|=\left\{\begin{array}{rrr}\hfill a& \hfill \text{if}& \hfill a\ge 0\\ \hfill -a& \hfill \text{if}& \hfill a<0\end{array}$
The algebraic definition takes into account the fact that the number $a$ could be either positive or zero $\left(\ge 0\right)$ or negative $\left(<0\right)$ .
1. If the number $a$ is positive or zero $\left(\ge 0\right)$ , the first part of the definition applies. The first part of the definition tells us that if the number enclosed in the absolute bars is a nonnegative number, the absolute value of the number is the number itself.
2. If the number $a$ is negative $\left(<0\right)$ , the second part of the definition applies. The second part of the definition tells us that if the number enclosed within the absolute value bars is a negative number, the absolute value of the number is the opposite of the number. The opposite of a negative number is a positive number.
## Sample set a
Use the algebraic definition of absolute value to find the following values.
$|8|$ .
The number enclosed within the absolute value bars is a nonnegative number so the first part of the definition applies. This part says that the absolute value of 8 is 8 itself.
$|8|=8$
$|-3|$ .
The number enclosed within absolute value bars is a negative number so the second part of the definition applies. This part says that the absolute value of $-3$ is the opposite of $-3$ , which is $-\left(-3\right)$ . By the double-negative property, $-\left(-3\right)=3$ .
$|-3|=3$
## Practice set a
Use the algebraic definition of absolute value to find the following values.
$|7|$
7
$|9|$
9
$|-12|$
12
$|-5|$
5
$-|8|$
$-8$
$-|1|$
$-1$
$-|-52|$
$-52$
$-|-31|$
$-31$
## Exercises
For the following problems, determine each of the values.
$|5|$
5
$|3|$
$|6|$
6
$|14|$
$|-8|$
8
$|-10|$
$|-16|$
16
$-|8|$
$-|12|$
$-12$
$-|47|$
$-|9|$
$-9$
$|-9|$
$|-1|$
1
$|-4|$
$-|3|$
$-3$
$-|7|$
$-|-14|$
$-14$
$-|-19|$
$-|-28|$
$-28$
$-|-31|$
$-|-68|$
$-68$
$|0|$
$|-26|$
26
$-|-26|$
$-|-\left(-8\right)|$
$-8$
$-|-\left(-4\right)|$
$-|-\left(-1\right)|$
$-1$
$-|-\left(-7\right)|$
$-\left(-|4|\right)$
4
$-\left(-|2|\right)$
$-\left(-|-6|\right)$
6
$-\left(-|-42|\right)$
$|-|-3||$
3
$|-|-15||$
$|-|-12||$
12
$|-|-29||$
$|6-|-2||$
4
$|18-|-11||$
$|5-|-1||$
4
$|10-|-3||$
$|-\left(17-|-12|\right)|$
5
$|-\left(46-|-24|\right)|$
$|5|-|-2|$
3
${|-2|}^{3}$
$|-\left(2\cdot 3\right)|$
6
$|-2|+|-9|$
${\left(|-6|+|4|\right)}^{2}$
100
${\left(|-1|-|1|\right)}^{3}$
${\left(|4|+|-6|\right)}^{2}-{\left(|-2|\right)}^{3}$
92
$-{\left[|-10|-6\right]}^{2}$
$-{\left[-{\left(-|-4|+|-3|\right)}^{3}\right]}^{2}$
$-1$
A Mission Control Officer at Cape Canaveral makes the statement "lift-off, $T$ minus 50 seconds." How long before lift-off?
Due to a slowdown in the industry, a Silicon Valley computer company finds itself in debt $2,400,000$ . Use absolute value notation to describe this company's debt.
$|-2,400,000|$
A particular machine is set correctly if upon action its meter reads 0 units. One particular machine has a meter reading of $-1.6$ upon action. How far is this machine off its correct setting?
## Exercises for review
( [link] ) Write the following phrase using algebraic notation: "four times $\left(a+b\right)$ ."
$4\left(a+b\right)$
( [link] ) Is there a smallest natural number? If so, what is it?
( [link] ) Name the property of real numbers that makes $5+a=a+5$ a true statement.
( [link] ) Find the quotient of $\frac{{x}^{6}{y}^{8}}{{x}^{4}{y}^{3}}$ .
( [link] ) Simplify $-\left(-4\right)$ .
4
how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Eleven fifteenths of two more than a number is eight.
Please keep in mind that it's not allowed to promote any social groups (whatsapp, facebook, etc...), exchange phone numbers, email addresses or ask for personal information on QuizOver's platform. |
There are four natural subsets associated with a matrix. We have met three already: the null space, the column space and the row space. In this section we will introduce a fourth, the left null space. The objective of this section is to describe one procedure that will allow us to find linearly independent sets that span each of these four sets of column vectors. Along the way, we will make a connection with the inverse of a matrix, so Theorem FS will tie together most all of this chapter (and the entire course so far).
## Left Null Space
Definition LNS (Left Null Space) Suppose $A$ is an $m\times n$ matrix. Then the left null space is defined as $\lns{A}=\nsp{\transpose{A}}\subseteq\complex{m}$.
The left null space will not feature prominently in the sequel, but we can explain its name and connect it to row operations. Suppose $\vect{y}\in\lns{A}$. Then by Definition LNS, $\transpose{A}\vect{y}=\zerovector$. We can then write
\begin{align*} \transpose{\zerovector} &=\transpose{\left(\transpose{A}\vect{y}\right)} &&\text{ Definition LNS}\\ &=\transpose{\vect{y}}\transpose{\left(\transpose{A}\right)} &&\text{ Theorem MMT}\\ &=\transpose{\vect{y}}A &&\text{ Theorem TT} \end{align*}
The product $\transpose{\vect{y}}A$ can be viewed as the components of $\vect{y}$ acting as the scalars in a linear combination of the rows of $A$. And the result is a "row vector", $\transpose{\zerovector}$ that is totally zeros. When we apply a sequence of row operations to a matrix, each row of the resulting matrix is some linear combination of the rows. These observations tell us that the vectors in the left null space are scalars that record a sequence of row operations that result in a row of zeros in the row-reduced version of the matrix. We will see this idea more explicitly in the course of proving Theorem FS.
Example LNS: Left null space.
## Computing Column Spaces
We have three ways to build the column space of a matrix. First, we can use just the definition, Definition CSM, and express the column space as a span of the columns of the matrix. A second approach gives us the column space as the span of some of the columns of the matrix, but this set is linearly independent (Theorem BCS). Finally, we can transpose the matrix, row-reduce the transpose, kick out zero rows, and transpose the remaining rows back into column vectors. Theorem CSRST and Theorem BRS tell us that the resulting vectors are linearly independent and their span is the column space of the original matrix.
We will now demonstrate a fourth method by way of a rather complicated example. Study this example carefully, but realize that its main purpose is to motivate a theorem that simplifies much of the apparent complexity. So other than an instructive exercise or two, the procedure we are about to describe will not be a usual approach to computing a column space.
Example CSANS: Column space as null space.
This example motivates the remainder of this section, so it is worth careful study. You might attempt to mimic the second approach with the coefficient matrices of Archetype I and Archetype J. We will see shortly that the matrix $L$ contains more information about $A$ than just the column space.
## Extended echelon form
The final matrix that we row-reduced in Example CSANS should look familiar in most respects to the procedure we used to compute the inverse of a nonsingular matrix, Theorem CINM. We will now generalize that procedure to matrices that are not necessarily nonsingular, or even square. First a definition.
Definition EEF (Extended Echelon Form) Suppose $A$ is an $m\times n$ matrix. Add $m$ new columns to $A$ that together equal an $m\times m$ identity matrix to form an $m\times(n+m)$ matrix $M$. Use row operations to bring $M$ to reduced row-echelon form and call the result $N$. $N$ is the extended reduced row-echelon form of $A$, and we will standardize on names for five submatrices ($B$, $C$, $J$, $K$, $L$) of $N$.
Let $B$ denote the $m\times n$ matrix formed from the first $n$ columns of $N$ and let $J$ denote the $m\times m$ matrix formed from the last $m$ columns of $N$. Suppose that $B$ has $r$ nonzero rows. Further partition $N$ by letting $C$ denote the $r\times n$ matrix formed from all of the non-zero rows of $B$. Let $K$ be the $r\times m$ matrix formed from the first $r$ rows of $J$, while $L$ will be the $(m-r)\times m$ matrix formed from the bottom $m-r$ rows of $J$. Pictorially, \begin{equation*} M=[A\vert I_m] \rref N=[B\vert J] = \left[\begin{array}{c|c}C&K\\\hline 0&L\end{array}\right] \end{equation*}
Example SEEF: Submatrices of extended echelon form.
Theorem PEEF (Properties of Extended Echelon Form) Suppose that $A$ is an $m\times n$ matrix and that $N$ is its extended echelon form. Then
1. $J$ is nonsingular.
2. $B=JA$.
3. If $\vect{x}\in\complex{n}$ and $\vect{y}\in\complex{m}$, then $A\vect{x}=\vect{y}$ if and only if $B\vect{x}=J\vect{y}$.
4. $C$ is in reduced row-echelon form, has no zero rows and has $r$ pivot columns.
5. $L$ is in reduced row-echelon form, has no zero rows and has $m-r$ pivot columns.
Notice that in the case where $A$ is a nonsingular matrix we know that the reduced row-echelon form of $A$ is the identity matrix (Theorem NMRRI), so $B=I_n$. Then the second conclusion above says $JA=B=I_n$, so $J$ is the inverse of $A$. Thus this theorem generalizes Theorem CINM, though the result is a "left-inverse" of $A$ rather than a "right-inverse."
The third conclusion of Theorem PEEF is the most telling. It says that $\vect{x}$ is a solution to the linear system $\linearsystem{A}{\vect{y}}$ if and only if $\vect{x}$ is a solution to the linear system $\linearsystem{B}{J\vect{y}}$. Or said differently, if we row-reduce the augmented matrix $\augmented{A}{\vect{y}}$ we will get the augmented matrix $\augmented{B}{J\vect{y}}$. The matrix $J$ tracks the cumulative effect of the row operations that converts $A$ to reduced row-echelon form, here effectively applying them to the vector of constants in a system of equations having $A$ as a coefficient matrix. When $A$ row-reduces to a matrix with zero rows, then $J\vect{y}$ should also have zero entries in the same rows if the system is to be consistent.
## Four Subsets
With all the preliminaries in place we can state our main result for this section. In essence this result will allow us to say that we can find linearly independent sets to use in span constructions for all four subsets (null space, column space, row space, left null space) by analyzing only the extended echelon form of the matrix, and specifically, just the two submatrices $C$ and $L$, which will be ripe for analysis since they are already in reduced row-echelon form (Theorem PEEF).
Theorem FS (Four Subsets) Suppose $A$ is an $m\times n$ matrix with extended echelon form $N$. Suppose the reduced row-echelon form of $A$ has $r$ nonzero rows. Then $C$ is the submatrix of $N$ formed from the first $r$ rows and the first $n$ columns and $L$ is the submatrix of $N$ formed from the last $m$ columns and the last $m-r$ rows. Then
1. The null space of $A$ is the null space of $C$, $\nsp{A}=\nsp{C}$.
2. The row space of $A$ is the row space of $C$, $\rsp{A}=\rsp{C}$.
3. The column space of $A$ is the null space of $L$, $\csp{A}=\nsp{L}$.
4. The left null space of $A$ is the row space of $L$, $\lns{A}=\rsp{L}$.
The first two conclusions of this theorem are nearly trivial. But they set up a pattern of results for $C$ that is reflected in the latter two conclusions about $L$. In total, they tell us that we can compute all four subsets just by finding null spaces and row spaces. This theorem does not tell us exactly how to compute these subsets, but instead simply expresses them as null spaces and row spaces of matrices in reduced row-echelon form without any zero rows ($C$ and $L$). A linearly independent set that spans the null space of a matrix in reduced row-echelon form can be found easily with Theorem BNS. It is an even easier matter to find a linearly independent set that spans the row space of a matrix in reduced row-echelon form with Theorem BRS, especially when there are no zero rows present. So an application of Theorem FS is typically followed by two applications each of Theorem BNS and Theorem BRS.
The situation when $r=m$ deserves comment, since now the matrix $L$ has no rows. What is $\csp{A}$ when we try to apply Theorem FS and encounter $\nsp{L}$? One interpretation of this situation is that $L$ is the coefficient matrix of a homogeneous system that has no equations. How hard is it to find a solution vector to this system? Some thought will convince you that any proposed vector will qualify as a solution, since it makes all of the equations true. So every possible vector is in the null space of $L$ and therefore $\csp{A}=\nsp{L}=\complex{m}$. OK, perhaps this sounds like some twisted argument from Alice in Wonderland. Let us try another argument that might solidly convince you of this logic.
If $r=m$, when we row-reduce the augmented matrix of $\linearsystem{A}{\vect{b}}$ the result will have no zero rows, and all the leading 1's will occur in first $n$ columns, so by Theorem RCLS the system will be consistent. By Theorem CSCS, $\vect{b}\in\csp{A}$. Since $\vect{b}$ was arbitrary, every possible vector is in the column space of $A$, so we again have $\csp{A}=\complex{m}$. The situation when a matrix has $r=m$ is known by the term full rank, and in the case of a square matrix coincides with nonsingularity (see exercise FS.M50).
The properties of the matrix $L$ described by this theorem can be explained informally as follows. A column vector $\vect{y}\in\complex{m}$ is in the column space of $A$ if the linear system $\linearsystem{A}{\vect{y}}$ is consistent (Theorem CSCS). By Theorem RCLS, the reduced row-echelon form of the augmented matrix $\augmented{A}{\vect{y}}$ of a consistent system will have zeros in the bottom $m-r$ locations of the last column. By Theorem PEEF this final column is the vector $J\vect{y}$ and so should then have zeros in the final $m-r$ locations. But since $L$ comprises the final $m-r$ rows of $J$, this condition is expressed by saying $\vect{y}\in\nsp{L}$.
Additionally, the rows of $J$ are the scalars in linear combinations of the rows of $A$ that create the rows of $B$. That is, the rows of $J$ record the net effect of the sequence of row operations that takes $A$ to its reduced row-echelon form, $B$. This can be seen in the equation $JA=B$ (Theorem PEEF). As such, the rows of $L$ are scalars for linear combinations of the rows of $A$ that yield zero rows. But such linear combinations are precisely the elements of the left null space. So any element of the row space of $L$ is also an element of the left null space of $A$. We will now illustrate Theorem FS with a few examples.\\
Example FS1: Four subsets, \#1.
Example FS2: Four subsets, \#2.
The next example is just a bit different since the matrix has more rows than columns, and a trivial null space.
Example FSAG: Four subsets, Archetype G.
Example COV and Example CSROI each describes the column space of the coefficient matrix from Archetype I as the span of a set of $r=3$ linearly independent vectors. It is no accident that these two different sets both have the same size. If we (you?) were to calculate the column space of this matrix using the null space of the matrix $L$ from Theorem FS then we would again find a set of 3 linearly independent vectors that span the range. More on this later.
So we have three different methods to obtain a description of the column space of a matrix as the span of a linearly independent set. Theorem BCS is sometimes useful since the vectors it specifies are equal to actual columns of the matrix. Theorem BRS and Theorem CSRST combine to create vectors with lots of zeros, and strategically placed 1's near the top of the vector. Theorem FS and the matrix $L$ from the extended echelon form gives us a third method, which tends to create vectors with lots of zeros, and strategically placed 1's near the bottom of the vector. If we don't care about linear independence we can also appeal to Definition CSM and simply express the column space as the span of all the columns of the matrix, giving us a fourth description.
With Theorem CSRST and Definition RSM, we can compute column spaces with theorems about row spaces, and we can compute row spaces with theorems about row spaces, but in each case we must transpose the matrix first. At this point you may be overwhelmed by all the possibilities for computing column and row spaces. diagram CSRST is meant to help. For both the column space and row space, it suggests four techniques. One is to appeal to the definition, another yields a span of a linearly independent set, and a third uses Theorem FS. A fourth suggests transposing the matrix and the dashed line implies that then the companion set of techniques can be applied. This can lead to a bit of silliness, since if you were to follow the dashed lines twice you would transpose the matrix twice, and by Theorem TT would accomplish nothing productive.
Column Space and Row Space Techniques
Although we have many ways to describe a column space, notice that one tempting strategy will usually fail. It is not possible to simply row-reduce a matrix directly and then use the columns of the row-reduced matrix as a set whose span equals the column space. In other words, row operations do not preserve column spaces (however row operations do preserve row spaces, Theorem REMRS). See exercise CRS.M21. |
Subtraction of 3-digit numbers | Grade 3 | Learning Concepts
# Operations
## 3-Digit Subtraction for Class 3 Maths
Here the student will recall the properties of subtraction and learn about 3-digit subtraction. The subtraction with borrowing and the subtraction without borrowing are explained with examples.
The students will learn to
• Deducting the result of subtraction using addition
• Calculate subtraction of 3- digit subtraction with borrowing or 3- digit subtraction without borrowing.
• Identifying minuend and subtrahend.
• Checking the result of subtraction using addition
The learning concept has explained subtraction to class 3 students with examples, illustrations, and a concept map. At the end of the page, two printable 3- digit subtraction worksheets for class 3 with solutions are attached for the students.
Download the worksheets and solutions to assess our knowledge of the concept.
• Subtraction is an operation used to calculate the difference between two numbers.
• The symbol used to denote subtraction is ‘−‘.
• Subtraction is the process of taking away one number or amount from another.
• The number from which another number is subtracted is called the minuend and the number which is subtracted is called the subtrahend.
• The result is called the difference.
• The subtrahend is always smaller than the minuend.
### Examples:
1. 352 - 153
2. 756 - 545
1. We can subtract 153 from 352 as:
So, 100 + 100 - 1 = 200 - 1 = 199
Hence, 352 - 153 = 199
2. We can subtract 545 from 756 as:
So, 100 + 100 + 10 + 1 = 211
Hence, 756 - 545 = 211
## 3-Digit Subtraction Without Borrowing
• Start subtracting digits from the ones place.
• Subtract the digits at the tens place
• Subtract the digits at the hundreds place
Let us discuss the subtraction of the three-digit number without borrowing by examples.
### Examples:
1. 353 - 151
2. 756 - 545
1. 353-151 =
Hence, 353 - 151 = 202
2. 756 - 545 =
Hence, 756 - 545 = 211
## 3-Digit Subtraction With Borrowing
• Start subtracting digits from the ones place. If the minuend is lesser than the subtrahend then subtraction is not possible. We need to borrow ten from the minuend at tens place and add with the minuend at ones’ place and then do the subtraction.
• Subtract digits at tens place. If the minuend is lesser than the subtrahend then subtraction is not possible. need to borrow ten from the minuend at hundreds place and add with the minuend at ones’ place and then do the subtraction.
• Repeat the process
Let us discuss the subtraction of the three-digit number with borrowing by example.
### Examples:
1. 353 - 154
2. 756 - 548
1. 353 - 154
Hence, 353 - 154 = 199
2. 756 - 548
Hence, 756 - 548 = 208
The result of any subtraction can easily be checked using addition. To check whether the result of the subtraction (difference) is correct or not, one should add the subtrahend with the difference. If the sum results in the minuend, then the subtraction is correct.
Here, some examples are given below.
### Examples:
Find the difference and check by addition:
1. 605 - 263
2. 689 - 354
1. 605 - 263
So, the difference is 342.
Now, to check the result is correct, add the difference with subtrahend and check if the sum is the minuend.
Hence, it is clear that the answer of the subtraction is correct.
2. 689 - 354
So, the difference is 335.
Now, to check the result is correct, add the difference with subtrahend and check if the sum is the minuend.
Hence, it is clear that the answer of the subtraction is correct.
• - |
# Decimals on a Number Line Worksheet 4th Grade (with answer key + PDF)
Another type of fraction is a decimal, in which a dot separates the whole number from the fractional portion. With the exception of how they are represented, fractions and decimals on a number line look very similar. We can perform precise calculations with the aid of decimals where the differences in minor values are crucial.
## What are the “Decimals on a Number Line Worksheet 4th Grade ”?
This worksheet will explore some of the decimals on a number line.
## What are Decimals on a Number Line?
A decimal number denotes a portion of a whole. It is, in other words, the interval between two integers. The distance between them can be divided into tenths by using a number line from 0 to 1. When asked to represent 0.3 on a number line, we would place a dot on the third tenth.
## Instructions on how to use the “Decimals on a Number Line Worksheet 4th Grade”
Study the concept and examples given and try to solve the exercises below.
## Conclusion
On a number line, decimal numbers are represented using the same method as whole numbers. For decimal numbers, tenths are represented by equally dividing the adjacent whole numbers to represent them. The hundredths are represented by further dividing each of the tenths into equal intervals.
If you have any inquiries or feedback, please let us know.
## Decimals on a Number Line Worksheet 4th Grade (with answer key + PDF)
A dot is used to demarcate the whole number from the fractional portion in decimals, another type of fraction. Except for the representation method, the representation of decimals on a number line is very similar to that of fractions.
Representing Decimals on Number Line
Plotting decimal numbers on a number line is referred to as representing decimals on a number line. A dot is used to denote the separation of the integral and fractional parts of decimal numbers. It is possible for the integral part to be less than 0 or more than 0. The decimals are positioned between the integers on a number line.
How to Represent Decimals on Number Line?
On a number line, fractional representation and decimal representation are very similar. On a number line, the left side of 0 represents the negative region, and the right side represents the positive region. Let’s use the methods listed below to represent 0.3 on a number line.
Step 1: Mark 0 as the reference point on a number line.
Step II: The second step is to identify and mark the integers that the decimal lies between. Between 0 and 1 is 0.3.
Step III: We will now divide the integers 0 and 1 by 10 to complete this step.
Step IV: From this point on, we’ll take as many steps to the right of 0 as there are digits to the right of the decimal point. We will move three spaces to the right for 0.3.
Another example: We will represent 2.83 on the number line. We know that 2.83 lies between 2.8 and 2.9.
## Worksheet
Decimal Number Lines
1. Write the missing decimal values on each of the arrows.
1. Write the missing decimal values on each of the arrows.
1. Draw arrows to correctly place each number on the number line below.
1. Draw arrows to correctly place each number on the number line below. |
# Factoring the Difference of Two Squares
Hello, and welcome to this video on factoring the difference of two squares. When factoring polynomials, there are a few special patterns you’ll want to be on the lookout for. The one we’ll be talking about in this video is the difference of two squares.
When you see a binomial in the form $$a^{2}-b^{2}$$, then you know you are looking at a difference of two squares, and it will factor to $$(a+b)(a-b)$$. The reason there’s no middle term is because FOILing these binomials will result in the outer and inner terms canceling each other out.
$$a^{2}-ab+ab-b^{2}$$
$$a^{2}-b^{2}$$
Let’s look at a few examples.
$$x^{2}-4$$
We know this is an example of a difference of two squares because it is a binomial with a subtraction sign between the two terms, and both terms are perfect squares. To factor this binomial, take the positive square root of both terms (ignoring the minus sign).
$$\sqrt{x^{2}}=x$$
$$\sqrt{4}=2$$
Remember, a difference of squares binomial factors to $$(a+b)(a-b)$$. In this case, $$a=x$$ and $$b=2$$, so the factored form is:
$$(x+2)(x-2)$$
If you wanted to check your answer, you can always FOIL it and see if you get the original expression.
Let’s try another one.
$$x^{2}-36$$
First, check that this matches the pattern $$a^{2}-b^{2}$$. It does, so take the square root of both terms.
$$a=\sqrt{x^{2}}=x$$
$$b=\sqrt{36}=6$$
Now, write out the factored form of the binomial.
$$(x+6)(x-6)$$
Let’s try one that’s slightly harder.
$$4y^{2}-16$$
First, check that this matches the pattern $$a^{2}-b^{2}$$. It looks different from our other examples, but this does follow this pattern because $$4y^{2}$$ is a perfect square.
Take the positive square root of both terms.
$$a=\sqrt{4y^{2}}=2y$$
Now we know that $$2y$$ is the positive square root of $$4y^{2}$$ because we’re going to take the square root of our coefficient 4, $$\sqrt{4}=2$$, and then multiply it by $$y^{2}$$, which is $$y$$. So that’s how we simplify this.
$$b=\sqrt{16}=4$$
Then, create the factored form of the binomial.
$$(2y+4)(2y-4)$$
Let’s work through one more example before we go.
$$16a^{4}-49$$
First, make sure it follows the pattern $$a^{2}-b^{2}$$. Since it does, take the positive square root of both terms. So remember, to take the square root of an expression like this $$(16a^{4})$$, take the square root of the coefficient first. $$\sqrt{16}=4$$, and then $$\sqrt{a^{4}}=a^{2}$$, so we’ll have $$4a^{2}$$.
$$a=\sqrt{16a^{4}}=4a^{2}$$
$$b=\sqrt{49}=7$$
Then, write the factored form of the binomial.
$$(4a^{2}+7)(4a^{2}-7)$$
And there you have it! Now you should be more comfortable recognizing difference of squares problems and how to factor them. I hope this video was helpful. Thanks for watching, and happy studying!
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by Mometrix Test Preparation | This Page Last Updated: October 13, 2023 |
# Q: Why does the leading digit 1 appear more often than other digits in all sorts of numbers? What’s the deal with Benford’s Law?
Mathematician: Benford’s Law (sometimes called the Significant-Digit Law) states that when we gather numbers from many different types of random sources (e.g. the front pages of newspapers, tables of physical constants at the back of science textbooks, the heights of randomly selected animals picked from many different species, etc.), the probability that the leading digits (i.e. the left most non-zero digits) of one of these numbers will be $d$ is approximately equal to
$log_{10}(1 + \frac{1}{d})$.
That means the probability that a randomly selected number will have a leading digit of 1 is
$log_{10}(1 + \frac{1}{1}) = 0.301$
which means it will happen about 30.1% of the time, whereas the probability that the first two leading digits will be 21 is given by
$log_{10}(1 + \frac{1}{21}) = 0.020$
which means it will occur about 2.0% of the time. Note that if we were writing our numbers in a base $b$ other than 10 (i.e. decimal) we would simply replace the $log_{10}$ in the formulas above with $log_{b}$. Benford’s Law indicates that in base 10, the most likely leading digit for us to see is 1, the second most likely 2, the third most 3, the fourth most likely 4, and so on. But why should this be true, and to what sorts of sources of random numbers will it apply?
Some insight into Benford’s Law can be gleaned from the following mathematical fact: If there exists some universal distribution that describes the probability that numbers sampled from any source will have various leading digits, it must be the formula given above. The reason for this is because if such a formula works for all sources of data, then when we multiply all numbers produced by our source by any constant, the distribution of the likelihood of leading digits must not change. This is the property of “scale invariance”. Now notice that if we have a number whose leading digit is 5, 6, 7, 8, or 9, and we multiply that number by 2, the new leading digit will always be 1. But since this operation is not allowed to change the probability of leading digits, that means that the probability of having a leading digit of 1 must be the same as the probability of having a leading digit of any of 5, 6, 7, 8 or 9. This property is satisfied by the formula given above, since
$log_{10}(1 + \frac{1}{1}) =$ $log_{10}(1 + \frac{1}{5}) + log_{10}(1 +\frac{1}{6}) + log_{10}(1 + \frac{1}{7})$ $+ log_{10}(1 + \frac{1}{8}) + log_{10}(1 + \frac{1}{9})$
Of course, there is nothing special about multiplying the numbers from our random source by 2, so a similar property must hold regardless of what we multiply our numbers by. As it turns out, the formula for Benford’s Law is the only formula such that the distribution does not change no matter what positive number we multiply the output of our random source by.
There is a problem with the preceding argument, however, since it has been empirically verified that not all data sources satisfy Benford’s Law in practice, so the existence of a universal law for leading digits seems to contradict the available evidence. On the other hand though, a great deal of data has been collected (e.g. by Benford himself) indicating that the law holds to pretty good accuracy for many different types of data sources. In fact, it seems that the law was first discovered due to a realization (by astronomer Simon Newcomb in 1881) that the pages of logarithm tables at the back of textbooks are not equally well worn. What was noticed was that the earlier tables (with numbers starting with the digit 1) tended to look rather dirtier than the later ones. So the question remains, how can we justify all of these empirical observations of the law in a more rigorous mathematical way?
First of all, an important point to note is that when we sample values from some common probability distributions (like the exponential distribution and the log normal distribution) the leading digits that you get already come close to satisfying Benford’s Law (see the graphs at the bottom of the article). Hence, we should already expect the law to approximately hold in some real world scenarios. More importantly though, as was demonstrated by the probabilist Theodore Hill, if our process for sampling points actually involves sampling from multiple sources (which cover a variety of different scales, without favoring any scale in particular), and then group together all the points that we get from all of the sources, the distribution of leading digits will tend towards satisfying Benford’s Law. For the technical details and restrictions, check out Hill’s original 1996 paper.
Perhaps the best way to quickly convince yourself that Hill’s result is true is to look at the graphs found below. Various probability distributions are shown (the first five that I happened to think of) together with the frequency of leading digits (from 1 to 9) that I got when sampling 100,000 points from that distribution (where the frequencies are depicted by the blue bars). For each, the pink line shows what we would expect to get if Benson’s Law held perfectly. As you can see, for some distributions we get a good fit (e.g. the exponential and log normal distributions) whereas for others the fit is poorer (e.g. the uniform, normal and laplace distributions). What the third graph in each table shows is the distribution of leading digits that we get when, instead of sampling just from one copy of each distribution, we sample from 9 different copies (with equal probability), each of which has a different variance (in most cases chosen to be proportional to the values 1 up to 9). Hence, what we are doing is sampling from multiple distributions each of which is the same except for a scaling factor, and then pooling those samples together, at which point we calculate the probability of the various leading digits (how often is 1 the first non-zero digit, how often is two the first non-zero digit, etc). The result in every case shown is that this leads to a distribution of leading digits that fits Benford’s Law quite well.
To conclude, when we are dealing with data that is combined from multiple sources such that those sources have no systematic bias in how they are scaled, we can expect that the distribution of leading digits will approximate Benford’s Law. This will tend to apply to sources like numbers pulled each day from the front page of a newspaper, because the values found in this way will come from all different distributions (some will represent oil prices, others real estate prices, others populations, and so on).
Besides just being generally bizarre and interesting, Benford’s Law has lately found some real world applications. For certain types of financial data where Benford’s Law applies, fraud has actually been detected by noting that results made up out of thin air will generally be non-random and will not satisfy the proper distribution of leading digits.
Distribution Uniform Probability Density Function Distribution of Leading Digit Of Samples Distribution of Leading Digit Taking Samples From 9 Such Distributions With Different Variances
Distribution Normal Probability Density Function Distribution of Leading Digit Of Samples Distribution of Leading Digit Taking Samples From 9 Such Distributions With Different Variances
Distribution Laplace Probability Density Function Distribution of Leading Digit Of Samples Distribution of Leading Digit Taking Samples From 9 Such Distributions With Different Variances
Distribution Log Normal Probability Density Function Distribution of Leading Digit Of Samples Distribution of Leading Digit Taking Samples From 9 Such Distributions With Different Variances
Distribution Exponential Probability Density Function Distribution of Leading Digit Of Samples Distribution of Leading Digit Taking Samples From 9 Such Distributions With Different Variances
This entry was posted in -- By the Mathematician, Math. Bookmark the permalink. |
Thursday, January 12, 2017
Number of 2 by 2 matrices with all elements in {0,1,..,n} and its permanent equal to the sum of all elements
OEIS sequence A280934 is defined as: $a(n) =$ the number of 2 by 2 matrices with all elements integers in the set $\{0,1,..,n\}$ and its permanent is equal to the sum of all elements.
Consider a 2 by 2 matrix $\left(\begin{array}{cc}a & c \\ d & b\end{array}\right)$. The sum of all elements equal to the permanent is written as $a+b+c+d = ab+cd$. This can be rewritten as $(a-1)(b-1) + (c-1)(d-1) = 2$. In other words, $a(n) =$ the number of 2 by 2 matrices with all elements integers in the set $\{-1,..,n-1\}$ and its permanent is equal to 2. We will consider this alternative formulation in the rest of this note. Suppose $n > 3$ and at least one of members of the matrix has value $n-1$. Let us count how many of such matrices has permanent 2, i.e. $ab+cd = 2$.
First consider the case $a = n-1$. If $b = -1$, then $cd = n+1$. Note that $c \neq 1$ and $c \neq n+1$ since $-1\leq c,d\leq n-1$. Since $n> 3$, any nontrivial factor of $n+1$ is less than $n-1$. Thus setting $c>1$ equal to any nontrivial factor of $n+1$ and $d = (n+1)/c$ will result in a matrix with permanent 2. Thus there are $\tau(n+1)-2$ such matrices. Here $\tau(n)$ denotes the number of divisors of $n$. If $b = 0$, then $cd = 2$ and there are 2 cases: $(c,d) = (1,2)$ and $(c,d) = (2,1)$. If $b=1$, then $cd = 3-n$ and there are 2 cases as well: $(c,d) = (-1,n-3)$ and $(c,d) = (n-3,-1)$. Note that $cd \geq 1-n$ since $-1\leq c,d\leq n-1$. This means that if $b > 1$ then $ab+cd \geq n-1 > 2$. Thus we have a total of $\tau(n+1)+2$ cases. Note that in all cases, exactly one element of the matrix is of value $n-1$.
By symmetry, the same analysis applies to $b=n-1$, $c=n-1$ and $d=n-1$ as well. Since in all the cases only one element of the matrix is of value $n-1$, there is no double counting. Thus the number of matrices with at least one member equal to $n-1$ and permanent 2 is $4\tau(n+1)+8$. This implies the following recurrence relation for $a(n)$:
$$a(n) = a(n-1) + 4\tau(n+1)+8 \quad\mbox{for}\quad n> 3$$. |
# Find the second order derivatives of the functions given in $$\sin (\log x)$$
$\begin{array}{1 1} \large\frac{\sin(\log x)}{x} \\ \large\frac{\cos(\log x)}{x} \\ -\Large\frac{[\cos(\log x)+\cos(\log x)]}{x^2} \\-\Large\frac{[\sin(\log x)+\cos(\log x)]}{x^2} \end{array}$
Toolbox:
• $y=f(x)$
• $\large\frac{dy}{dx}$$=f'(x) • \large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big) • \large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$
Step 1:
$y=\sin (\log x)$
Differentiating with respect to $x$
$\large\frac{dy}{dx}=$$\cos (\log x).\large\frac{d}{dx}$$(\log x)$
$\quad\;=\cos(\log x).\large\frac{1}{x}$
$\quad\;=\large\frac{\cos(\log x)}{x}$
Step 2:
$\Large\frac{d^2y}{dx^2}=\frac{x.\Large\frac{d}{dx}[\cos(\log x)]-\cos(\log x).1}{x^2}$
$\quad\;=\Large\frac{[-x\sin (\log x).\Large\frac{1}{x}]-\cos (\log x).1]}{x^2}$
$\quad\;=-\Large\frac{[\sin(\log x)+\cos(\log x)]}{x^2}$ |
## List of Mathematics Grade 8 Whole Numbers Questions and Answers for Revision
Find a list of Mathematics Grade 8 Whole Numbers Questions and Answers for Revision in pdf format:
Here are some questions and answers for “Whole Numbers” mathematics for grade 8 in the South African CAPS curriculum:
1. What are whole numbers? A: Whole numbers are numbers that are positive integers, meaning they are greater than zero and do not include any fractional or decimal parts. Whole numbers range from 1, 2, 3, and so on.
2. How are whole numbers used in real-life situations? A: Whole numbers are used in a variety of real-life situations, such as counting objects, measuring distances, calculating the cost of items, determining the quantity of items in stock, and more.
3. What are some properties of whole numbers? A: Some properties of whole numbers include being commutative under addition and multiplication, being associative under addition and multiplication, having an identity element under addition and multiplication, and having inverse elements under addition.
4. How do you add whole numbers? A: To add whole numbers, you simply line up the numbers and add the digits in each place value column, starting from the rightmost column. If the sum of the digits in a column is greater than or equal to 10, you carry over the excess to the next place value column.
5. How do you subtract whole numbers? A: To subtract whole numbers, you line up the numbers and subtract the subtrahend from the minuend, starting from the rightmost place value column. If the result in a column is negative, you borrow from the next place value column.
6. How do you multiply whole numbers? A: To multiply whole numbers, you can use the standard algorithm, which involves multiplying each digit in one number by each digit in the other number and then adding the products. You can also use long multiplication, which involves breaking one of the numbers into parts and multiplying each part by the other number.
7. How do you divide whole numbers? A: To divide whole numbers, you can use the standard algorithm, which involves dividing the dividend by the divisor to find the quotient. You can also use long division, which involves dividing the dividend by the divisor and finding the quotient and remainder.
### Application Questions:
Copy and complete the following statements by giving answers to these questions, without doing any calculations with the given numbers.
(a) Is 8 Å~ 117 more than 2 000 or less than 2 000? than 2 000
(b) Is 27 Å~ 88 more than 3 000 or less than 3 000? than 3 000
(c) Is 18 Å~ 117 more than 3 000 or less than 3 000? than 3 000
(d) Is 47 Å~ 79 more than 3 000 or less than 3 000? than 3 000
What you have done when you tried to give answers to questions (a) to (d), is called estimation. To estimate is to try to get close to an answer without actually doing the calculations with the given numbers.
(a) The numbers 1 000, 2 000, 3 000, 4 000, 5 000, 6 000, 7 000, 8 000, 9 000 and 10 000 are all multiples of a thousand. In each case, write down the multiple of 1 000 that you think is closest to the answer. The numbers you write down are called estimates.
(b) In some cases you may achieve a better estimate by adding 500 to your estimate, or subtracting 500 from it. If so, you may add or subtract 500.
(c) If you wish, you may write what you believe is an even better estimate by adding or subtracting some hundreds. |
# Unit symbols & use
## Using units in your math
It is highly important that you attach symbols to numerical answers in Physics problems because the units change the meaning of your value significantly.
For example, let's say you get in trouble with your parents and they say you are restricted form using your phone. You ask "How long?" and they respond "5." This answer does not satisfy you, because there is a big difference between 5 minutes, 5 hours, 5 days and 5 weeks, all of which are units of time.
You should also understand that these units get altered when we perform functions like multiplication, division, and power functions (like squaring and finding the square root).
For example, we all know that when finding the area of a room 10 feet wide and 14 feet long we and up with an area of 140 square feet. The reason is because when we multiplied the number values we also multiplied the units and the result of ft x ft = ft2.
When units do not match we end up with a derived unit, which is simply a unit expressed by combining other units. That might sound complicated, but you are used to using some of these already.
For example, we all know that when finding the speed of a car that travels 300 miles in 5 hours we divide 300 miles by 5 hours and we get the answer of 60 miles per hour. When we write this algebraically it will look like this:
Miles per hour is a derived unit because it is a combination of 2 other units.
Derived units will be easier to process when performing math functions if you write them in fractional form as the example answer above has the mi on top and the h of the bottom of a fraction instead of writing an abbreviation like mph or using a slash notation like mi/h.
Identical units can also cancel out or be reduced through division or multiplication of fractions.
For example, when finding the amount of distance a car will travel in 5 hours if moving at an average speed of 60 mi/h we perform the following operations:
You can see that the hours in the bottom of the derived unit fraction cancel the hours in the time being multiplied and leave an approriate unit of miles for the distance.
In complex processes we will perform several operations at once and may end up with some very long winded derived units.
Insert example and practice here.
## Unit Symbols
The following is a list of symbols for common units we may use throughout the year. There are many more units possible, but these should cover pretty much everything we need. I do not bother with common prefixes or variations (ex. I will not list feet, yards, inches or centimeters).
The following are additional units just for AP Physics 2 |
# Multiplying Polynomials
### Multiplication of a Number by a Polynomial
It is another polynomial that has the same degree. The coefficients are the product of the coefficients of the polynomial and the number.
3 · (2x3 − 3x2 + 4x − 2) =
= 6x3 − 9x2 + 12x − 6
### Multiplication of a Monomial by a Polynomial
The monomial is multiplied by each and every one of the monomials that form the polynomial.
3x2 · (2x3 − 3x2 + 4x − 2) =
= 6x5 − 9x4 + 12x3 − 6x2
### Multiplication of Polynomials
P(x) = 2x2 − 3 Q(x) = 2x3 − 3x2 + 4x
Multiply each monomial from the first polynomial by each of the monomials in the second polynomial.
P(x) · Q(x) = (2x2 − 3) · (2x3 − 3x2 + 4x) =
= 4x5 − 6x4 + 8x3 − 6x3 + 9x2 − 12x =
Add the monomials of the same degree:
= 4x5 − 6x4 + 2x3 + 9x2 − 12x
The multiplication of polynomials is another polynomial whose degree is the sum of the degrees of the polynomials that are to be multiplied.
The polynomials can also be multiplied as follows:
#### Example
Multiply the polynomials using two different methods:
P(x) = 3x4 + 5x3 − 2x + 3 and Q(x) = 2x2 − x + 3
P(x) · Q(x) = (3x4 + 5x3 − 2x + 3) · (2x2 − x + 3) =
= 6x6 − 3x5 + 9x4 + 10x5 − 5x4 + 15x3
− 4x3 + 2x2 − 6x + 6x2 − 3x + 9 =
= 6x6 + 7x5 + 4x4 + 11x3 + 8x2 − 9x + 9 |
(7.2) Number, operation, and quantitative reasoning. The student adds, subtracts, multiplies, or divides to solve problems and justify solutions. The student.
Presentation on theme: "(7.2) Number, operation, and quantitative reasoning. The student adds, subtracts, multiplies, or divides to solve problems and justify solutions. The student."— Presentation transcript:
(7.2) Number, operation, and quantitative reasoning. The student adds, subtracts, multiplies, or divides to solve problems and justify solutions. The student is expected to: (C) use models to add, subtract, multiply, and divide integers and connect the actions to algorithms. 7.2C Addition and Subtraction of Integers
7.2C INSTRUCTIONAL ACTIVITY #1 Addition and subtraction of integers can be modeled using different colors of tiles to represent integers. For example, use blue tiles to represent positive integers and red tiles to represent negative integers. EXAMPLE 1: Use color tiles to model the addition problem ˉ7 + 3. Place 7 red tiles to represent ˉ7, then place 3 blue color tiles below them to represent 3. A tile representing a positive 1 and a negative 1 can be combined to represent a zero pair. Remove the tiles representing the zero pairs. The remaining tiles represent ˉ7 + 3, therefore ˉ7 + 3 = -4.
7.2C INSTRUCTIONAL ACTIVITY #1 Addition and subtraction of integers can be modeled using different colors of tiles to represent integers. For example, use blue tiles to represent positive integers and red tiles to represent negative integers. EXAMPLE 2: Use color tiles to model the subtraction problem ˉ4 - ˉ3. Place 4 red tiles to represent ˉ4. Remove 3 red tiles representing the subtraction of ˉ3. The remaining tile represents ˉ4 - ˉ3, therefore ˉ4 - ˉ3 = ˉ1.
7.2C INSTRUCTIONAL ACTIVITY #1 Addition and subtraction of integers can be modeled using a number line. Always begin the model at zero. Draw an arrow to the right to represent a positive integer or draw an arrow to the left to represent a negative integer as the first number in the problem. Next Draw an arrow to the right to represent adding a positive integer or draw an arrow to the left to represent adding a negative integer. or Draw an arrow to the right to represent subtracting a negative integer because you are adding the opposite of the negative integer. 0-10-9-8-7-6-5-4-3-210987654321 Example: ˉ2 - ˉ2 can be changed to addition and becomes ˉ2 + 2.
7.2C INSTRUCTIONAL ACTIVITY #1 Use the number line below to model the addition problem ˉ7 + 3. 0-10-9-8-7-6-5-4-3-210987654321 ˉ7 +3 Begin to model the addition problem at 0. Since ˉ7 is a negative integer, it can be modeled with an arrow ˉ7 units to the left of 0. Draw an arrow from 0 to ˉ7, which is 7 units to the left of 0. Since 3 is a positive integer, it can be modeled with an arrow 3 units to the right of ˉ7. Draw an arrow from ˉ7 to ˉ4, which is 3 units to the right of ˉ7. The model shows that the sum of ˉ7 and 3 is ˉ4.
7.2C INSTRUCTIONAL ACTIVITY #1 Use the number line below to model the addition problem ˉ6 + ˉ2. Start to model the addition problem at 0. 0-10-9-8-7-6-5-4-3-210987654321 ˉ6 ˉ2 Since ˉ6 is a negative integer, it can be modeled with an arrow 6 units to the left of 0. Draw an arrow from 0 to ˉ6, which is 6 units to the left of 0. Since ˉ2 is a negative integer, and is being added to ˉ6, it can be modeled with an arrow 2 units to the left of ˉ6. Draw an arrow from ˉ6 to ˉ8, which is 2 units to the left of ˉ6. The model shows that the sum of ˉ6 and ˉ2 is ˉ8.
7.2C INSTRUCTIONAL ACTIVITY #1 Use the number line below to model the addition problem ˉ5 - 2. Start to model the addition problem at 0. 0-10-9-8-7-6-5-4-3-210987654321 ˉ5 2 Since ˉ5 is a negative integer, it can be modeled with an arrow 5 units to the left of 0. Draw an arrow from 0 to ˉ5, which is 5 units to the left of 0. Since 2 is a positive integer, and is being subtracted from ˉ5, it can be modeled with an arrow 2 units to the left of ˉ5. Draw an arrow from ˉ5 to ˉ7, which is 2 units to the left of ˉ5. The model shows the difference between ˉ5 and 2 is ˉ7.
7.2C INSTRUCTIONAL ACTIVITY #1 Use the number line below to model the addition problem ˉ4 - ˉ3. Start to model the addition problem at 0. 0-10-9-8-7-6-5-4-3-210987654321 ˉ4 ˉ3ˉ3 Since ˉ4 is a negative integer, it can be modeled with an arrow 4 units to the left of 0. Draw an arrow from 0 to ˉ4, which is 4 units to the left of 0. Since ˉ3 is a negative integer that is being subtracted from ˉ4, it can be modeled with an arrow 3 units to the right of ˉ4 to represent subtracting a negative integer because you are adding the opposite of the negative integer or finding the difference between ˉ4 and ˉ3. Draw an arrow from ˉ4 to ˉ1, which is 3 units to the right of ˉ4. The model shows the difference between ˉ4 and ˉ 3 is ˉ1.
7.2C INSTRUCTIONAL ACTIVITY #1 Addition and subtraction of integers can be modeled using sketches to represent the integers. Use to represent +1 and to represent ˉ1 to model the addition problem ˉ6 + 2. Start to model the addition problem by representing ˉ6 and adding 2. Form zero pairs by grouping a representation of a ˉ 1 and +1. Mark off and remove the zero pairs that are equal to zero. Four tiles representing ˉ1s remain. The model shows that the sum of ˉ6 and 2 is ˉ4.
7.2C STUDENT ACTIVITY #1 Problem #1: Lyle scored 15 points in a board game. On his next turn he lost 5 points, and then he lost 6 more points. On his final turn Lyle gained 9 points. Determine Lyles final score by modeling the problem on a number line. Be sure to mark moves above the number line by using arrows and identifying the direction and the amount of the move. Start at 0 on the number line. Lyles beginning score is 15. Represent this score with the integer _______. To add 15, count 15 places to the right from 0. On his next turn Lyle lost 5 points. Represent this with the integer _______. To add -5, count 5 places to the left from 15 to _______. On his next turn Lyle lost 6 points. Represent this with the integer _______. To add -6, count 6 places to the left from 10 to _______. On his final turn Lyle gained 9 points. Represent this with the integer _______. To add 9, count 9 places to the right from 4 to _______. Represent Lyles final score with the expression _______ +_______ +_______ +_______. Lyles final score was _______. 0246810121416 15 +15 -5 10 -6 4 +9 13 15 -5 -6 9 13
7.2C STUDENT ACTIVITY #1 Problem #2: Leslie made \$12 babysitting on Friday night. On Saturday afternoon she spent \$4 for a movie ticket and \$3 for popcorn and a cold drink. On Saturday night she made another \$5 babysitting. Determine the amount of money Leslie had after babysitting on Saturday night drawing a sketch to model the problem using to represent +1 and to represent -1. Be sure to identify and remove zero pairs in your model. Write an explanation for each step in your model below each sketch. Step 1: Identify my positive and negative integers and use colored dots as symbols to identify them. Step 2: Combine my zero pairs. Step 3: Remove tiles representing zero pairs. Step 4: Remaining tiles represent the amount of money Leslie had after babysitting on Saturday night. \$10
7.2C STUDENT ACTIVITY #1 Problem #2: Leslie made \$12 babysitting on Friday night. On Saturday afternoon she spent \$4 for a movie ticket and \$3 for popcorn and a cold drink. On Saturday night she made another \$5 babysitting. Draw a number line to represent the amount of money Leslie had after babysitting on Saturday night. Be sure to mark moves above the number line by using arrows and identifying the direction and the amount of the move. 0246810121416 The following expression represents how much money Leslie had after babysitting on Saturday night. _______ +_______ +_______ +_______. Leslie had \$_____ after babysitting on Saturday night. +12 -4 -3 +5 12 -4 -3 9 10
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# Two marbles are drawn, without replacement, from an urn containing 4 red marbles, 5 white marbles, and 2 blue marbles. Determine the probability that:
a. Both are red.
b. The first marble drawn is red, and the second is blue.
c. The first marble drawn is blue, and the second is red.
d. One of the marbles is red and the other is blue.
Probability is the # of possible outcomes desired / # of possible total outcomes. In problems like this when you have consecutive draws, you find the probability of each individual draw and then multiply your results.
a. there are 4 reds to possible draw from the urn, there are 11 total marbles. The probability of drawing 1 red is 4/11.
After the 4/11 chance that you did get a red, now what is the probability of getting a second red?
*Note, the was no replacing so you have 1 red in your hand and 3 in the urn.
Now there is a 3/10 probability of getting the second red marble. 10 because you have one of the 11 in your hand so there are only 10 left in the urn.
so we have 4/11 and 3/10, multiply and we get 12/110 which reduces to 6 /55
b. same idea just keep your colors and number of marbles straight.
P(red) = 4/11
P(b after a red has been drawn) = 2/10, 2 blue still in the urn but only 10 marbles since you have 1 red in your hand.
Multiply 4/11 and 2/10 which equals 8/110 = 4/55
c. P(b) = 2/11, P(red after a blue has been drawn) = 4/10, multiply 8/110 = 4/55
d. this one is different, you have to account for either one being red or blue. so a red then blur or a blue then a red. We have already found these probabilities in b. and c. Since both situations need to be counted then we add them together. 4/55 +4/55 = 8/55
Why don't we multiply? Because that would give us a smaller probability and if you think about the problem, out chances increase if the order doesn't matter. (yes there is a mathematical explanation but I like to use common sense when possible)
a) (4/11)(3/10) = 6/55
b) (4/11)(2/10) = 4/55
c) (2/11)(4/10) = 4/55
d) 2(4/55) = 8/55
Remember: P(A and B)=P(A)P(B) for independent events (which marble drawings are)
P(A or B)=P(A)+P(B) for disjoint events
1. P(red and red) =(4/11)(3/10)=12/110
2. P(red and blue)=(4/11)(2/10)=8/110
3. P(blue and red)=(2/11)(4/10)=8/110
4. P(red or blue)=P(red and blue)+P(blue and red)=16/110 |
# The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3,
Question:
The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.
Solution:
Let the numerator and denominator of the fraction be $x$ and $y$ respectively. Then the fraction is $\frac{x}{y}$
The sum of the numerator and denominator of the fraction is 12. Thus, we have
$x+y=12$
$\Rightarrow x+y-12=0$
If the denominator is increased by 3, the fraction becomes $\frac{1}{2}$. Thus, we have
$\frac{x}{y+3}=\frac{1}{2}$
$\Rightarrow 2 x=y+3$
$\Rightarrow 2 x-y-3=0$
So, we have two equations
$x+y-12=0$
$2 x-y-3=0$
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
$\frac{x}{\mid \times(-3)-(-1) \times(-12)}=\frac{-y}{\mid \times(-3)-2 \times(-12)}=\frac{1}{\mid \times(-1)-2 \times 1}$
$\Rightarrow \frac{x}{-3-12}=\frac{-y}{-3+24}=\frac{1}{-1-2}$
$\Rightarrow \frac{x}{-15}=\frac{-y}{21}=\frac{1}{-3}$
$\Rightarrow \frac{x}{15}=\frac{y}{21}=\frac{1}{3}$
$\Rightarrow x=\frac{15}{3}, y=\frac{21}{3}$
$\Rightarrow x=5, y=7$
Hence, the fraction is $\frac{5}{7}$. |
# FAQ: What Is Algebraic Expression In Math?
## What is an algebraic expression in math terms?
An expression containing variables, numbers, and operation symbols is called an algebraic expression. is an example of an algebraic expression. Each expression is made up of terms. A term can be a signed number, a variable, or a constant multiplied by a variable or variables.
## What is algebraic expression and equation?
An expression is a number, a variable, or a combination of numbers and variables and operation symbols. An equation is made up of two expressions connected by an equal sign.
## What are the types of algebraic expression?
There are 3 main types of algebraic expressions which include:
• Monomial Expression.
• Binomial Expression.
• Polynomial Expression.
## How do you identify an algebraic expression?
Algebraic expressions are combinations of variables, numbers, and at least one arithmetic operation. For example, 2x+4y−9 is an algebraic expression. Term: Each expression is made up of terms. A term can be a signed number, a variable, or a constant multiplied by a variable or variables.
## Is 5 an algebraic expression?
The expressions in which the numbers, or variables, or both, are connected by operational signs (+, – etc.) are called algebraic expressions. For example 5, 4x, a+b, x−y etc.
You might be interested: Question: What Is Math Bingo?
## What is algebra formula?
Here is a list of Algebraic formulas – a2 – b2 = (a – b)(a + b) (a + b)2 = a2 + 2ab + b2. a2 + b2 = (a + b)2 – 2ab. (a – b)2 = a2 – 2ab + b2.
## How do you write an algebraic expression?
If x is 2, then the expression 9 + x has a value of 11. If x is 6, then the expression has a value of 15. So 9 + x is an algebraic expression. In the next few examples, we will be working solely with algebraic expressions.
## What is the name of a term without variable in an algebraic expression?
Constants. Constants are the terms in the algebraic expression that contain only numbers. That is, they’re the terms without variables. We call them constants because their value never changes, since there are no variables in the term that can change its value.
## What is a basic expression?
Expressions are basically the building blocks of Statements, in that every BASIC statement is made up of keywords (like GOTO, TO, STEP) and expressions. So expressions include not just the standard arithmetic and boolean expressions (like 1 + 2), but also lvalues (scalar variables or arrays), functions, and constants.
## How many terms are there in expression?
A Term is either a single number or a variable, or numbers and variables multiplied together. So, now we can say things like “that expression has only two terms “, or “the second term is a constant”, or even “are you sure the coefficient is really 4?”
## What is like terms in algebraic expression?
In algebra, like terms are terms that have the same variables and powers. The coefficients do not need to match. Unlike terms are two or more terms that are not like terms, i.e. they do not have the same variables or powers. |
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Math Help > Basic Algebra > Polynomial > Polynomial Remainder Theorem
The polynomial remainder theorem states, If f(x) is a polynomial, then the remainder obtained by dividing f(x) by x-r equals f(r) In other words, x-r is a factor of f(x)-f(r) To show this is true, we will first show that x-r is a factor of axn-arn, where a is any real number, and n is a nonnegative integer. Then, since the polynomial f(x)-f(r) is composed of a sum of numbers of the form axn-arn, and x-r divides every term of that sum, then it follows that x-r divides f(x)-f(r). Before we begin, perhaps a definition of "remainder" is needed. As you recall, in division of integers, if r is the remainder of the division p/d, then there is an integer, q, such that qd+r=p, and 0 <= r < d. In division of polynomials the same idea holds, except the quotient and remainders are both polynomials. In division of polynomials, if r is the remainder of the division p/d (both polynomials), then there is a polynomial, q, such that qd+r=p, and 0 <= O(r) < d. Here O(r) means the "order" or "degree" of polynomial r. Proof that (x-r) | (axn-arn) for all real a and nonnegative integer n. If (x-r) | (xn-rn) then (x-r) | (axn-arn), so I'll just show the former. If n=0, then x-r is a factor of xn-rn, because x0-r0=0, and x-r is a factor of 0. Now let's suppose x-r is a factor of xn-rn. We will show x-r is a factor of xn+1-rn+1. First, it is clear that (x-r) | (xn-rn)(x-r). I will write a series of expressions that are all equivalent to (xn-rn)(x-r). (x-r) | (xn-rn)(x-r) (x-r) | xn+1-xnr-rnx+rn+1 by multiplying the terms together (x-r) | xn+1-xnr+2rn+1-rnx-rn+1 by adding and subtracting 2arn+1. (x-r) | xn+1-xnr+rnr+rnr-rnx-rn+1 by expressing 2arn+1 as arnr+arnr. (x-r) | xn+1-r(xn-rn)-(x-r)(rn)-rn+1 by gathering terms, and factoring r and (x-r) (x-r) | (xn+1-rn+1) - r(xn-rn) - (x-r)(rn) by rearranging terms (x-r) | r(xn-rn) (x-r) | (x-r)(rn) (x-r) | (xn+1-rn+1) (x-r)|a-b-c, (x-r)|b, (x-r)|c, so (x-r)|a. This proves that (x-r) | (xn-rn) for all nonnegative integer n. Therefore (x-r) | (axn-arn) for all real a and nonnegative integer n. Since the polynomial f(x)-f(r) is composed of a sum of numbers of the form axn-arn, and x-r divides every term of that sum, then it follows that x-r divides f(x)-f(r). In other words, If f(x) is a polynomial, then the remainder obtained by dividing f(x) by x-r equals f(r)
### Related pages in this website
Number Theory
Synthetic Division
Descartes' Rule Of Signs
Polynomial
The webmaster and author of this Math Help site is Graeme McRae. |
# Formulas for the Arithmetic Questions on the GRE Quantitative Reasoning Test
If you’re preparing for the GRE Test, and you want to get the best score possible in the Quantitative Reasoning section, you have come to the right place! At Union Test Prep, we believe in you and want you to succeed. Below, we have the essential formulas you’ll need to use for the arithmetic questions found on the GRE Quantitative Reasoning Test. Check out our GRE blogs for formula charts that relate to the other typesof Quantitative Reasoning problems on the GRE.
And if you are a bit rusty with math, it doesn’t matter—you can also check our other resources:
Study guide for the GRE Quantitative Reasoning Test
Practice questions for the GRE Quantitative Reasoning Test
Flashcards for the GRE Quantitative Reasoning Test
Remember that this chart only shows formulas for one part of the test: Arithmetic. Access our other three formula charts for the GRE Quantitative Reasoning section, here:
Algebra
Geometry
Data Analysis
## Arithmetic Formulas for the GRE Quantitative Reasoning Section
Formula Symbols Comment
$$\dfrac{a}{b} + \dfrac{c}{d}= \dfrac{(a\cdot d) + (c \cdot b)}{b\cdot d}$$ a, b, c, d = any real number simplify (if possible)
$$\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{a\cdot c}{b\cdot d}$$ a, b, c, d = any real number simplify (if possible)
$$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a \cdot d}{b \cdot c}$$ a, b, c, d = any real number simplify (if possible)
$$a \frac{b}{c} = \dfrac{(a\cdot c) + b}{c}$$ a, b, c = any real number simplify (if possible)
$$a \cdot b \% = a \cdot \frac{b}{100}$$ a = any real number
b% = any percent
simplify (if possible)
$$\% = \dfrac{\vert{b-a}\vert}{b} \cdot 100 = \dfrac{c}{b} \cdot 100$$ % = % increase or decrease
a = new value
b = original value
c = amount of change
$$x^a \cdot x^b = x^{a+b}$$ a, b, x = any real number
$$\dfrac{x^a}{x^b} = x^{a-b}$$ a, b, x = any real number $$x \ne 0$$
$$(x^a)^b = x^{a\cdot b}$$ a, b, x = any real number
$$(x \cdot y )^a = x^a \cdot y^a$$ a, x, y = any real number
$$x^1 = x$$ x = any real number
$$x^0= 1$$ x = any real number $$x \ne 0$$
$$x^{-a} = \dfrac{1}{x^a}$$ a, x = any real number $$x \ne 0$$ |
# Thread: optimization for a box
1. ## optimization for a box
Find the dimensions of a rectangular box of maximum volume if the total surface area is given as 64cm^2.
Find the dimensions of a rectangular box of maximum volume if the total surface area is given as 64cm^2.
Let the base be m and height be n. I am assuming the problem indicates square base is used.
$V = mmn = m^2n$
Assuming the box has an open top, the surface area is
$m^2 + 4mn = 64$
Solve for n from the surface area, then plug the expressions into the volume formula.
$n = \frac{64-m^2}{4m}$
$V = m^2 \frac{64-m^2}{4m} = \frac{64m-m^3}{4}$
Find the first derivative of volume:
$V' = 8 - 0.75 m^2 = 0
$
Then solve for m.
The story changes if this is a closed top box.
Find the dimensions of a rectangular box of maximum volume if the total surface area is given as 64cm^2.
So the volume of a box is
$V=lwh$
Subject to the surface area
$S=2lw+2lh +2wh=64$
$F(l,w,h,\lambda) =lhw-\lambda(2lw+2lh+2wh-64)$
$\nabla F=0$
$hw-\lambda(2w+2h)=0$
$lh-\lambda(2l+2h)=0$
$lw-\lambda(2l+2w)=0$
$0=(2lw+2lh+2wh-64)$
Now multiply the first equation by l and the 2nd by -w and add them together
$lwh-\lambda(2lw+2lw)=0$
$-lwh+\lambda(2lw+2wh)=0$
$-2lw+2wh=0 \iff 2w(-l+h)=0 \iff l=h$
Doing something similar you can show that $w=l=h$
Plugging into the last equation we get.
$2l^2+2l^2+2l^2-64=0 \iff 6l^2=64 \iff l^2 =\frac{32}{3} \iff l=4 \sqrt{\frac{2}{3}}$
So the maximal volume will be a cube.
I hope this helps
4. Nice going, TheEmptySet... I was way off here. A calculus subforum for single-variable people like me? Just kidding. |
1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Examples
Transcript
Example 47 Find intervals in which the function given by f(𝑥) =310𝑥4 – 45𝑥3– 3𝑥2 + 365𝑥 + 11 is (a) strictly increasing (b) strictly decreasing f𝑥 = 310𝑥4 – 45𝑥3– 3𝑥2 + 365𝑥 + 11 Step 1: Finding f’𝑥 f’𝑥 = 310 × 4𝑥3 – 45 × 3𝑥2 – 3 × 2x + 365 + 0 f’𝑥 = 1210𝑥3– 125𝑥2– 6x + 365 f’𝑥 = 65𝑥3− 125𝑥2– 6x + 365 f’𝑥 = 6𝑥35−2𝑥25−𝑥+65 f’𝑥 = 6𝑥3 − 2𝑥2− 5𝑥 + 65 = 65 𝑥3−2𝑥2−5𝑥+6 = 65𝑥−1𝑥2−𝑥−6 = 65 𝑥−1𝑥2−3𝑥+2𝑥−6 = 65 𝑥−1𝑥𝑥−3+2𝑥−3 = 65 𝑥−1𝑥+2𝑥−3 Hence f’𝑥 = 65𝑥−1𝑥+2𝑥−3 Step 2: Putting f’𝑥=0 65𝑥−1𝑥+2𝑥−3 = 0 𝑥−1𝑥+2𝑥−3 = 0 Hence x = –2 , 1 & 3 Step 3: Plotting point on real line Thus, we get four disjoint intervals i.e. −−2 ,−2, 1 ,1 , 3, 3 , uc1 ⇒ f𝑥 is strictly decreasing on the interval 𝑥 ∈−−𝟐& 𝟏 , 𝟑 f𝑥 is strictly increasing on the interval 𝑥 ∈−𝟐,𝟏 & 𝟑 , uc1
Examples |
# How to Use Trigo R-Formula
Trigonometry (Trigo) R-formula is a must-know formula in O-levels A-Math. In this revision note, you will learn the Trigo R-formula and how to apply it to solve common O-levels A-Math questions.
Before you read on, you might want to download this entire revision notes in PDF format to print it out, or to read it later.
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## R−Formula
The R−Formula allows an expression involving the sum (or difference) of sine and cosine functions to be expressed as a single trigonometric function in sine or cosine. In general,
a sin θ ± b cos θ = R sin(θ ± a)
a cos θ ± b sin θ = R sin(θ ∓ a)
where a > 0, b > 0, α is acute, R = √(a2 + b2) and α= tan−1(b/a).
## R-Formula Trigo Example
The diagram below shows an L-shaped rod, XYZ, placed against an upright wall. Given that XY = 2 m, YZ = 8 m and angle OZY = θ°.
(i) Show that OZ = (2 sin θ + 8 cos θ) m.
(ii) Express OZ in the form R sin (θ – α), where α is acute.
(iii) Hence state the maximum value of OZ and the value of θ for which OZ is maximum.
(iv) Find the value of θ for which OZ = 6 m.
(i) Show that OZ = (2 sin θ + 8 cos θ) m.
Let W be on YZ such that XW be parallel to OZ. Let A be on XW and B be on OZ such that YB is perpendicular to OB. Hence angle XWY = angle OZY = θ (corresponding angles, XW // OZ)
In the right−angled triangle XYW,
Angle YXW = 180° − angle XWY − 90° = 90° − θ (sum of angles in triangle = 180°)
In right−angled triangle XYA,
Angle XYA = 180° − angle YXW − 90° = θ (sum of angles in triangle = 180°)
sin∠XYA = sin θ =AX/2
AX = 2 sin θ = OB
In right−angled triangle YBZ,
cos∠YZB = cos θ = BZ/8
BZ = 8 cos θ
OZ = OB + BZ = 2 sin θ + 8 cos θ (shown)
(ii) Express OZ in the form R sin (θ – α), where α is acute.
Using the Trigo R-Formula,
α = tan-1(8/2) = 76.0° (1 dp)
2 sin θ + 8 cos θ = √(22 + 82) sin(θ + 76.0°) = √68 sin (θ + 76.0°) (ans)
(iii) Hence state the maximum value of OZ and the value of θ for which OZ is maximum.
−1 ≤ sin(θ + 76.0°) ≤ 1
−√68 ≤ √68 sin(θ + 76.0°) ≤ √68
The maximum value of OZ is √68 = 8.25 m (3 sf ) (ans)
OZ is maximum when sin(θ + 76.0°) = 1.
θ + 76.0° = 90°
θ = 14.0° (ans)
(iv) Find the value of θ for which OZ = 6 m.
√68 sin(θ + 76.0°) = 6
sin(θ + 76.0°) = 6/√68
θ + 76.0° = 46.6861°
θ = 29.3° (1 dp) (ans)
Note: While it can be inferred that θ may lie in the 1st or 2nd quadrant, there is no need to find the basic angle and obtain the angle in the 2nd quadrant since θ is an acute angle based on the context and diagram
## Check out our exam guide on other topics here!
Secondary Math Revision Notes
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# Calculus 3 : Vectors and Vector Operations
## Example Questions
### Example Question #51 : Vectors And Vector Operations
Find the angle between the two vectors. Round to the nearest degree.
Explanation:
In order to find the angle between the two vectors, we follow the formula
and solve for
Using the vectors in the problem, we get
Simplifying we get
To solve for
we find the
of both sides and get
and find that
### Example Question #51 : Vectors And Vector Operations
Find the angle between the two vectors. Round to the nearest degree.
Explanation:
In order to find the angle between the two vectors, we follow the formula
and solve for
Using the vectors in the problem, we get
Simplifying we get
To solve for
we find the
of both sides and get
and find that
### Example Question #52 : Vectors And Vector Operations
Find the angle between the two vectors. Round to the nearest degree.
Explanation:
In order to find the angle between the two vectors, we follow the formula
and solve for
Using the vectors in the problem, we get
Simplifying we get
To solve for
we find the
of both sides and get
and find that
### Example Question #53 : Vectors And Vector Operations
Calculate the angle between the vectors and , and express the measurement of the angle in degrees.
Explanation:
The angle between the vectors and is given by the following equation:
where represents the cross product of the vectors and , and and represent the respective magnitudes of the vectors and .
We are given the vectors and . Calculate , and , and then substitute these results into the formula for the angle between these vectors, as shown:
,
,
and
.
Hence,
The principal angle for which is . Hence, the angle between the vectors and measures .
### Example Question #55 : Vectors And Vector Operations
Find the angle between the gradient vector and the vector where is defined as:
Explanation:
Find the angle between the gradient vector and the vector where is defined as:
_____________________________________________________________
Compute the gradient by taking the partial derivative for each direction:
At we have:
____________________________________________________________
The angle between two vectors and can be found using the dot product:
We wish to find the angle between the two vectors:
Compute the dot product between and
Therefore the dot product is:
Compute the magnitude of
Compute the magnitude of
Now put it all together:
### Example Question #1 : Distance Between Vectors
Given the vectors
find the distance between them, .
Explanation:
To find the distance between the vectors
we do the following calculation:
### Example Question #1 : Distance Between Vectors
Find the distance between the two vectors
Explanation:
To find the distance between the two vectors
we make the following calculation
### Example Question #1 : Distance Between Vectors
Find the Euclidian distance between the two vectors:
Explanation:
The Euclidian distance between two vectors is:
Plugging in the numbers given, we have:
### Example Question #4 : Distance Between Vectors
Find the distance between if and .
Explanation:
Write the formula to find the magnitude of the vector .
Substitute the points into the equation assuming and .
### Example Question #5 : Distance Between Vectors
Calculate the length of line segment AB given A(5, 2, 0) and B(6, 0, 3): |
# Question Video: Simplifying Algebraic Expressions Mathematics • 7th Grade
Simplify (2πΒ³ β 4πΒ² β 9) + (5πΒ³ β 3π + 1).
02:11
### Video Transcript
Simplify two π cubed minus four π squared minus nine plus five π cubed minus three π plus one.
Before we start this question, it is important to remember two things. Firstly, we can only group or collect terms with the same exponent. In this question, we can group the two π cubed and the five π cubed and also the negative nine and positive one.
Thereβs only one π squared term and, likewise, only one π term. Therefore, these two terms cannot be simplified.
It is also important to remember our rules when two signs are touching. If we have two positive signs, our resultant sign is an addition. A positive and a negative results in a subtraction or negative sign. And two negative signs result in a positive or addition sign.
Grouping the π cubed terms gives us two π cubed plus five π cubed. As two plus five is equal to seven, two π cubed plus five π cubed equals seven π cubed.
As previously mentioned, there is only one π squared and one π term. Therefore, negative four π squared and negative three π need to be in the answer.
Finally, we need to collect the numbers negative nine plus positive one. As we have two positive signs touching each other, the resultant sign is positive or addition, leaving us with negative nine plus one. This is equal to negative eight.
The simplified version of two π cubed minus four π squared minus nine plus five π cubed minus three π plus one is seven π cubed minus four π squared minus three π minus eight. |
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## Lesson Tutor : Lesson Plans : Algebra Grade 9: translating math sentences or expressions into equations.
/ Lesson Tutor : Lesson Plans : Algebra Grade 9: translating math sentences or expressions into equations.
A Beginning Look at Basic Algebra – Lesson 1
by Elaine Ernst Schneider
Objective(s): By the end of this lesson the student will be able to:
Define these terms: variable; algebraic expression; signs of operation; order of operations.
Pre-Class Assignment:
Resources/Equipment/Time Required:
Outline:
Algebra provides the basics for all higher math. You will work with numbers and letters (variables) to form sentences (expressions) that you can solve. The best way to learn math is by practicing it, so each lesson will include exercises using the skills learned.
A place to begin:
Letters in math are called variables. They can stand for different numbers at different times.
A mathematical sentence is called an expression. It can include numbers, variables, signs of operation, and symbols of inclusion.
Signs of operation tell you what to do to the sentence. The four operations are addition, subtraction, multiplication, and division.
Symbols of inclusion are parentheses ( ) and brackets [ ].
An important caution:
Be very neat in your calculations. Many an algebra problem is missed because the student misread what he or she had written or did not “line up” the column correctly for subtraction or division. Always double check operations. You don’t want to miss a problem because you added incorrectly.
Let’s Get Started:
To “evaluate” an expression means to find its value, or to solve it. The first rule to learn about algebra is “what to do when.” The order in which an expression’s operations are done can completely change the answer.
When evaluating an algebraic expression, first look for the symbols which show the innermost work. That can be expressed by use of parentheses or brackets. If BOTH parentheses and brackets are present, the parentheses are usually the innermost and should be worked first.
Here is an example:
24 + [46 – (2 X 11)]
24 + [46 – 22]
24 + 24
48
Now it’s time for you to try a few.
EXERCISE:
9 – (4 X 2)
(9 – 4) X 2
(9 – 4) X (2 X 1)
48 – [42 – (3 X 9)]
63 – [8/2 + (14 – 10)]
(Note: 8/2 is the same as 8 divided by 2, just like in fractions.)
[800/ (200 X 4)]
28 + [ 10 – (4 + 2) ]
(11-5) X (10 + 14)
125 / ( 5 X 5) (Remember from number 5? / = divided by.)
[28 – (4 X 5)] – 4 |
# 2019 AMC 8 Problems/Problem 1
## Problem 1
Ike and Mike go into a sandwich shop with a total of $30.00$ to spend. Sandwiches cost $4.50$ each and soft drinks cost $1.00$ each. Ike and Mike plan to buy as many sandwiches as they can, and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how many items will they buy?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$
## Solution 1
We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$. Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they bought a total of $9$ items. Therefore, the answer is $\boxed{D = 9 }$
- SBose
## Solution 2 (Using Algebra)
Let $s$ be the number of sandwiches and $d$ be the number of sodas. We have to satisfy the equation of $$4.50s+d=30$$ In the question, it states that Ike and Mike buys as many sandwiches as possible. So, we drop the number of sodas for a while. We have: $$4.50s=30$$ $$s=\frac{30}{4.5}$$ $$s=6R30$$ We don't want a remainder so the maximum number of sandwiches is $6$. The total money spent is $6\cdot 4.50=27$. The number of dollar left to spent on sodas is $30-27=3$ dollars. $3$ dollars can buy $3$ sodas leading us to a total of $6+3=9$ items. Hence, the answer is $\boxed{(D) = 9}$
- by interactivemath |
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If you took algebra in high school, you probably still have nightmares about using the quadratic formula. It's one of the most complex formulas in high school math. Understanding when and how to use it can seem impossible.
The truth is, the quadratic equation is simpler to solve than it looks. In fact, after using our quadratic formula calculator, you'll wonder why you were ever afraid of it in the first place.
Once you have the equation in standard form, enter the values of a, b, and c into the calculator to generate an instant result. In the guide below, go over some of the most important questions to help simplify one of the most commonly heard of formulas today.
## What is a quadratic equation?
Second-order equations (also called quadratic equations) are equations with one variable in which the highest order term is of the second order.
The quadratic formula is used to solve any second-order algebraic equation.
### Examples
3x2 = 3 is second-order.
0 = 5x2 + 3x + 2 is also second-order.
But 4x3 + 2x2 + x+1 = 0 is not and neither is 2x+1 = 0.
When solving a second-order equation, it's hard to know where to begin. Until you learn about the quadratic formula, that is.
With this formula in your arsenal, you can solve any second-order equation in a matter of seconds.
The formula looks like this:
#### x = (-b ± √(b2 - 4ac)) / 2a
If this is the first time you've seen it since high school, it probably looks like gibberish. But don't worry, we're going to walk you through it step by step.
By the time you finish this guide, you'll be an expert.
## How do you use the quadratic formula?
As mentioned above, you can use the quadratic formula any time you have a second-order algebraic equation in one variable.
### Step One
First, manipulate your equation until you have a zero on one side. You want to put in standard form, which looks like this:
#### ax2 + bx + c = 0
So for example, if your equation looks like this:
2x2 + 3x - 2 = x+3
Rearrange it so it looks like:
2x2 + 2x - 5 = 0
So, in our case, a = 2, b = 2, and c = -5.
### Step Two
Finally, plug these values into the equation like this:
### Solve
x = (-2 ± √(4 - 4 * 2 * (-5))
4
x = -2 ± √4 - -40
4
x = -2 ± √44
4
x = -2 ± 2√11
4
x = -2 ± 2√11
44
x = -1 ± √11
22
This gives you x = -2.158 and x = 1.158. Luckily, you can solve the equation even quicker with a quadratic formula calculator.
## How many solutions should I get?
### Two solutions
Quadratic equations (usually) have two correct answers. This occurs when b2-4ac, known as the discriminant, is greater than zero.
Since you are taking the square root of the discriminant, any number greater than zero can have two outcomes. The ± sign in the formula gives you the two answers, one for + and one for -.
### One solution
Some quadratic equations only have one answer. This occurs when b2-4ac, known as the discriminant, is equal to zero. And therefore,
### = -b2a
Don't worry if this happens. The square root of zero is zero, which means x has only one possible answer.
### No Solutions
It's possible for your equation to have no solutions. This occurs when the discriminant (b2-4ac) is less than zero.
Because you can't take the square root of a negative number, your equation will have no real solutions. It will, however, have "imaginary" solutions". Unless you're in an academic setting, you don't need to worry about those.
## When can you skip the quadratic formula?
The quadratic equation is quick and easy to solve, especially if you use our quadratic formula calculator. But sometimes, you can solve your equation even faster using other methods.
### C Term = 0
For example, if you have the equation:
x2 + 4x = 0
In other words, your c term is equal to zero and you can see right away that one solution is x = 0. You then divide both sides by x, leaving you with:
x + 4 = 0
So, x = -4 is your other solution.
You could have arrived at the same answer using the quadratic formula, but it may have taken you longer.
### Factoring
Furthermore, if your second-order equation is easy to factor or
a perfect square,
that's often a better way to solve it.
For example, if your equation were:
x2 - 3x + 2 = 0
You could factor it to:
(x-1)(x-2) = 0
And your solutions would be x = 1 and x = 2.
But of course, you need to have experience factoring polynomials. If that's not something you know how to do, you can always use the quadratic equation.
## How does that look in a graph?
Consider a graph of f(x). The zeros will be the values of x at which the graph crosses the x-axis.
### One Solution
If the quadratic equation only yields one answer, then the function only has one zero. That means the graph touches the x-axis and "bounces" right back.
### No Solutions
If the quadratic equation doesn't yield any real answers, then f(x) has no real zeros. That means that the graph never touches the x-axis. It's either entirely above or entirely below it.
### Practice, Practice, Practice
Now that you understand the quadratic equation, it's time to use it. After all, the best way to understand a new concept is to apply it over and over again.
The quadratic equation may look intimidating, but once you get the hang of it, you'll realize how simple it is.
So get out there and start using it. You have the power to solve any quadratic equation in the world.
If you found this guide helpful, please share it with your friends. For other helpful math tools and solvers, head over to our math section and try an exponent calculator. |
# 5.7 Proving That Figures Are Special Quadrilaterals
## Presentation on theme: "5.7 Proving That Figures Are Special Quadrilaterals"— Presentation transcript:
5.7 Proving That Figures Are Special Quadrilaterals
Objective: After studying this section, you will be able to prove that a quadrilateral is: A rectangle A kite A rhombus A square An isosceles triangle
Proving that a quadrilateral is a rectangle
F G Show that the quadrilateral is a parallelogram first then use one of the methods to complete the proof. 1. If a parallelogram contains at least one right angle, then it is a rectangle (reverse of the definition). 2. If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. You can prove that a quadrilateral is a rectangle without first showing that it is a parallelogram 3. If all four angles of a quadrilateral are right angles, then it is a rectangle.
Proving that a quadrilateral is a kite
To prove that a quadrilateral is a kite, either of the following methods can be used. 1. If two disjoint pairs of consecutive sides of a quadrilateral are congruent, then it is a kite (reverse of the definition). 2. If one of the diagonals of a quadrilateral is the perpendicular bisector of the other diagonal, then the quadrilateral is a kite.
Proving that a quadrilateral is a rhombus
J O K M Show that the quadrilateral is a parallelogram first then apply either of the following methods. 1. If a parallelogram contains a pair of consecutive sides that are congruent, then it is a rhombus (reverse of the definition). 2. If either diagonals of a parallelogram bisects two angles of the parallelogram, then it is a rhombus. You can prove that a quadrilateral is a rhombus without first showing that it is a parallelogram 3. If the diagonals of a quadrilateral are perpendicular bisectors of each other, then the quadrilateral is a rhombus.
Proving that a quadrilateral is a square
The following method can be used to prove the NPRS is a square. 1. If a quadrilateral is both a rectangle and a rhombus, then it is a square (reverse of the definition).
Proving that a trapezoid is an isosceles
B C 1. If the nonparallel sides of a trapezoid are congruent, then it is isosceles (reverse of the definition). 2. If the lower or the upper base angles of a trapezoid are congruent, then it is isosceles. 3. If the diagonals of a trapezoid are congruent, then it is isosceles.
What is the most descriptive name for quadrilateral ABCD with vertices
A = (-3, -7), B = (-9, 1), C = (3, 9), and D = (9, 1)?
Given: ABCD is a parallelogram
Prove: ABCD is a rhombus C
Given: GJMO is a parallelogram
Prove: OHKM is a rectangle G H J K
Summary Homework: worksheet
Write a description of each of three special quadrilaterals without using the names of the quadrilaterals. Each description should include sufficient properties to establish the quadrilateral’s identity. Homework: worksheet |
FutureStarr
A Increase by 30 Percent
# Increase by 30 Percent
via GIPHY
This is an easy enough calculation. The number 60, multiplied by 1. 3, which is a 30% increase.
### Increase
The first step in increasing a number by a percentage is to convert the percent to a decimal. The easiest way to do this is move the decimal point two points to the left. For example, 30 percent as a decimal is 0.3, and 50 percent as a decimal is 0.5. If you have a calculator with a percent key (%), enter your number and press % to convert the percent to a decimal (you may have to press the = key on some models). Another way to work out the decimal is to remember that 100 percent is 1, because it is the whole of something. This means 50 percent is one-half (0.5), 25 percent is one-quarter (0.25), 75 percent is three-quarters (0.75) and so on. The concept of percent increase is basically the amount of increase from the original number to the final number in terms of 100 parts of the original. An increase of 5 percent would indicate that, if you split the original value into 100 parts, that value has increased by an additional 5 parts. So if the original value increased by 14 percent, the value would increase by 14 for every 100 units, 28 by every 200 units and so on. To make this even more clear, we will get into an example using the percent increase formula in the next section.
In some math problems, you might know the percent increase or decrease and the new amount, and need to work out the original amount. For example, you know a bed with a \$280 sale price has been reduced by 30 percent. To work out the original price of the bed, you have to establish what percentage of the original price the sale price is. The original price is 100 percent and 30 percent has been taken off, so the sale price is 70 percent of the original price. Divide the sale price (280) by the numerical value of 70 percent, or 0.7, to work out the original price. The answer is 400, so you know the original price of the bed was \$400. (Source: sciencing.com)
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# Consider the following statementsStatement-1: Differentiation and integration are processes involving limits.Statement-2: The process of differentiation and integration are inverse of each other.Choose the correct option.(a) Statement-1 is true(b) Statement-2 is true(c) Both statements are true(d) Both statements are false
Last updated date: 13th Jul 2024
Total views: 60.9k
Views today: 1.60k
Answer
Verified
60.9k+ views
Hint: To find the correct option, write the basic formulas involved with the integration and differentiation of any function and then check the validity of the options.
To solve this question, we will check the validity of both the statements.
We will begin by writing the formulas involved by integration and differentiation of any function to check the first statement.
Consider any general function $f\left( x \right)$.
To find the differentiation the function $f\left( x \right)$at$x=a$, we will use the formula $f'\left( a \right)=\underset{h\to \infty 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$.
Thus, we observe that differentiation of any function involves the limit.
To find the integral of the function $f\left( x \right)$ over an interval $[a,b]$, we will subdivide the interval $[a,b]$into$n$ subintervals of equal width, $\vartriangle x$, and from each interval choose a point ${{x}_{i}}^{*}$. Then the integral of the function is$\int\limits_{a}^{b}{f\left( x \right)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{f\left( {{x}_{i}}^{*} \right)}\vartriangle x$.
Thus, we observe that the integral of any function involves limit.
Hence, we observe that Statement-1 is correct.
Now, we will check the second statement which says that the process of integration and differentiation are inverse of each other.
One must know that differentiation forms an algebraic expression that helps in the calculation of the gradient of a curve at any point. While, integration is used to calculate the area under any curve.
Hence, we observe that differentiation tries to lower down the function into simpler parts from which it has originated. While, integration tries to combine the smaller fragments from which the function is obtained.
Thus, integration and differentiation are opposite processes.
Hence, statement-2 is correct.
So, both statements are correct.
Option (c) is the correct answer.
Note: One must clearly know the basic formulas involved with the differentiation and integration formulas. |
Question
## Introduction
I’m going to let you in on a secret: math is not that difficult. In fact, it’s pretty straightforward if you just follow the rules and use your brain. That’s why today, I’m going to teach you about the difference between any two consecutive natural numbers using only one example and no more than three steps of logic.
## The difference between any two consecutive natural numbers is the number that comes immediately after them.
The difference between any two consecutive natural numbers is the number that comes immediately after them. For example, consider the following:
• The difference between 2 and 3 is 1.
• The difference between 3 and 4 is 1.
• The difference between 4 and 5 is 2 (or -1).
## You can find the difference between two natural numbers by counting how many numbers come before them.
Let’s say you want to find the difference between any two consecutive natural numbers. You can do this by counting how many numbers come before them, then subtracting that number from both sides of your equation. For example, suppose we want to know what one plus three is (4). The first thing we do is count how many numbers come before 1 and 3: there are 2 integers between them (1 and 2). So our equation looks like this:
1 + 3 – 2 = 4
## If you divide the smaller number by 2, you get the other one.
If you divide the smaller number by 2, you get the other one.
The smaller number is the first half of the larger number.
The larger number is the second half of the larger number.
This means that any two consecutive natural numbers will always have a difference of 1 between them!
## It’s simple math!
The difference between any two consecutive natural numbers is simply the difference in their absolute values. For example, if you think about what the difference between 1 and 2 is, it’s easy to see that this would be -1. If we take another example of 10 and 11, then the difference between them would be -1 as well.
You can easily calculate these differences yourself with a calculator or in your head by taking advantage of simple subtraction rules:
• If there are no zeroes involved in either number (e.g., 1), then just subtract one from another (10 – 9 = 1).
• If only one number contains zeroes (e.g., 10), then simply subtract 0 from any other number(s) before doing so (11 – 00 = 11).
There you have it. The difference between any two consecutive natural numbers is the number that comes immediately after them. It’s simple math!
1. # What Is The Difference Between Any Two Consecutive Natural Numbers
Numbers are a fundamental part of our lives. We use them every day to manage our finances, plan our days, and more. But what do the different types of numbers mean? In this blog post, we will explore the difference between any two consecutive natural numbers. From basic math to more complicated concepts, this article will have something for everyone. So don’t miss it!
## What is a number?
Numbers are symbols that denote amounts, sizes, or quantities. There are many different numbers, and each has a specific meaning. For example, the number “two” can mean two people, two rooms in a house, or two cigarettes. The number “ten” can mean ten cookies, ten books, or ten minutes.
There are many different types of numbers. Some numbers are cardinal (meaning they can be used to count), while others are ordinal (meaning they can be used to rank things). Cardinal numbers include one, two, three, and so on. Ordinal numbers include first, second, third, and so on.
There is also the zero symbol. Zero is not a number in and of itself- it’s just a placeholder for when no number exists yet! Numbers can also have negative values (like -5), which means five less than what was originally written.
Numbers come in different sizes and shapes- some are square like 4 or 5 inches on each side, while others are round like a coin or an apple. Numbers can also be written in various fonts to make them look nicer or more official- like bold text for 10 or italics for 15.
## What are the different types of numbers?
There are a lot of different types of numbers, and it can be tough to tell the difference between them. In this article, we will discuss the different types of numbers and how to identify them.
Consecutive Numbers:
When two natural numbers are consecutive, they are said to be consecutive. For example, 2, 3, 4 are consecutive. This means that the next number in the sequence is immediately followed by the number before it. For example, if you were to ask someone what the next number in the sequence is, they would say 5.
Disjunctive Numbers:
A disjunctive number is a number that consists of two parts. The first part lists all of the possible values for the second part and it doesn’t have to be a whole number. For example, -5, -3, 2 can all be considered disjunctive numbers because they list three possible values for x: 0, 1, 2. The second part is always a whole number and it doesn’t have to follow any specific pattern.
Polarity Numbers:
Polarity numbers are similar to disjunctive numbers but there is one important difference: one side of the polarity number always has to be a whole number while the other side can be any value. For example, -4 can also be considered a polarity number because it has an even side (4) and a odd side (-3). Again, x doesn’t have to be a whole number and it can be either positive or negative.
Real Numbers:
Real numbers are the simplest type of number and they are the numbers that we experience every day. Everything that we see in life, from the size of a mountain to the temperature outside, is a real number. There are a lot of different real numbers, but the most common ones are integers (like 2, 3, 5, 10).
Integers:
An integer is a real number that is written using digits. For example, 2 is an integer because it is written as 2 with two digits after the decimal point. There are a lot of different integers, but the most common ones are whole numbers (like 2, 3, 5, 10).
Decimals:
A decimal is a number that is written using fractional digits. For example, 4.5 is a decimal because it is written as 4.5 with three decimal points after the 1. In other words, 4.5 represents 0.45 and not 4500 like a whole number would represent. There are a lot of different decimals, but the most common ones are fractions (like 0.33, .79).
## The difference between two consecutive natural numbers
There is a big difference between two consecutive natural numbers. For example, the difference between 2 and 3 is 1, while the difference between 3 and 4 is 2. There are many other examples like this where one number is larger than the other.
## The sum of two consecutive natural numbers
The sum of two consecutive natural numbers is always greater than the sum of the two original natural numbers. This is because the sequence of addition continues indefinitely and eventually produces a larger number than any individual number in the sequence. For example, if you add 2 and 3, you get 5, which is greater than both 2 and 3. If you add 4 and 5, you get 10, which is greater than both 4 and 5.
## The product of two consecutive natural numbers
The product of two consecutive natural numbers is the sum of their digits. For example, the product of 3 and 4 is 5 because 3 + 4 = 5.
## The difference between three consecutive natural numbers
There is a big difference between any two consecutive natural numbers. For example, the difference between 3 and 4 is 2, while the difference between 5 and 6 is 3. The reason for this is that when we add 1 to either of these numbers, we get the number that follows it (in this case, 2). However, when we add 2 to either of these numbers, we instead get the number that preceded it (in this case, 1). In other words, the difference between any two consecutive natural numbers is always greater than 1.
## The sum of three consecutive natural numbers
The sum of three consecutive natural numbers is 12. The difference between any two consecutive natural numbers is 4.
## The product of three consecutive natural numbers
The difference between any two consecutive natural numbers is always a number. The product of three consecutive natural numbers is the sum of their differences, or 3 + 2 + 1.
## The difference between four consecutive natural numbers
There is a big difference between four consecutive natural numbers. The first number is 1, the second number is 2, the third number is 3, and the fourth number is 4. Here’s how those numbers compare:
1 is different from 2 because 1 is larger than 2.
2 is different from 3 because 2 is smaller than 3.
3 is different from 4 because 3 is larger than 4.
## The sum of four consecutive natural numbers
The sum of four consecutive natural numbers is 16. The difference between any two consecutive natural numbers is 4.
2. 🤔 Have you ever wondered what the difference is between any two consecutive natural numbers? Well, you’re in luck! We’re here to explain the difference between any two consecutive natural numbers and why it’s important.
Natural numbers are numbers that start from 1 and go up indefinitely. They are also known as counting numbers or positive integers. So, the difference between any two consecutive natural numbers is always 1.
Let’s say we have two consecutive natural numbers, 3 and 4. The difference between them is the number that lies in between them, which is 1. If we have two consecutive natural numbers, 5 and 6, the difference between them is also 1.
It doesn’t matter what two consecutive natural numbers you choose, the difference between them will always be 1! This is an important concept in mathematics, especially in sequences and series.
As you can see, understanding the difference between any two consecutive natural numbers is pretty simple. Knowing this piece of information can help you solve a variety of math problems, including problems related to sequences and series. 🤓
So, next time you come across any two consecutive natural numbers, remember that the difference between them is always 1. 🤓 |
# Coordinate Plane Distances Worksheet (With answer key +PDF)
The distance formula, which is derived from the Pythagorean Theorem, is employed to determine the separation between two points in the plane. The distance between two points in an XY plane is calculated using the distance formula, which is used in coordinate geometry or Euclidean geometry. The x-coordinate, also known as the abscissa, is a point’s separation from the y-axis. The y-coordinate, also known as the ordinate, is the distance between a point and the x-axis.
## What is the “Coordinate Plane Distances Worksheet (With answer key +PDF)”?
This worksheet will explore some of the Coordinate Plane Distances.
What is Distance on Coordinate Plane?
When two points on a coordinate plane share the same x- or y-coordinate, we can calculate their distance by adding up their respective units. To assist in calculating the distance, we can use the values on the x- and y-axes.
This worksheet will help you to understand the distance on the coordinate plane. It provides clear information about the particular topic.
Instructions on how to use the “Coordinate Plane Distances Worksheet (With answer key +PDF).”
Study the concept and examples given and try to solve the given exercises below.
## Conclusion
Plotting points and graphing lines can both be done on the coordinate plane. With the aid of this system, we can create and understand algebraic concepts as well as visually describe algebraic relationships.
If you have any inquiries or feedback, please let us know.
## Coordinate Plane Distances Worksheet (With answer key +PDF)
### A straight line connects any two points with the shortest distance. The distance formula can be used to determine this distance.
EXAMPLE:
FINDING THE DISTANCE BETWEEN TWO POINTS
## Worksheet
1. Calculate the distance on the coordinate plane using the formula.
1. Reflection
What do you think about solving the Coordinate Plane Distances? Did you find it difficult to follow the pattern in each step? |
Sales Toll Free No: 1-800-481-2338
# How to Find the Reciprocal of a Negative Number?
TopThe term fraction number means that number is written in form of $\frac{a}{b}$, where 'a' and 'b' are whole numbers and b $\neq$ 0. Now, if we simply look at any whole number, say 5. We observe that number 5 can also be written in the form of $\frac{5}{1}$. Now, we will compare the above number with the format of the fraction and find that a = 5 and b = 1. So. we come to a conclusion that every whole number can also be written as fraction number.
In order to find the reciprocal of any fraction number, we will simply write the numerator in the place of denominator and the denominator is written in the place of numerator. Now, what if given number is in the form of negative number. Let us see how to find the reciprocal of a negative number. Let us consider a negative number say -9. We say that -9 can also be read in the form of $\frac{-9}{1}$. Now, if we try to write the number in its reciprocal form, we come to the conclusion that above given number can be written in form of its reciprocal as follows:
Reciprocal of $\frac{-9}{1}$ = $\frac{1}{-9}$
We observe that this form of writing any number is not the standard form. So, we say that in order to convert it to standard form, we will multiply the numerator and denominator by (-1). Thus, the resultant number is $\frac{-1}{9}$.
The number so formed is not the fraction, as numerator is not the whole number but an integer. Thus, we say that it is in the form of $\frac{p}{q}$, where 'p' and 'q' are integers. So, it is a rational number. |
SAT II Math II : Other 2-Dimensional Geometry
Example Questions
Example Question #1 : Other 2 Dimensional Geometry
Which of the following describes a triangle with sides of length 9 feet, 4 yards, and 180 inches?
The triangle is acute and scalene.
The triangle is right and scalene.
The triangle is acute and isosceles, but not equilateral.
The triangle is right and isosceles, but not equilateral.
The triangle is acute and equilateral.
The triangle is right and scalene.
Explanation:
3 feet make a yard, so 9 feet is equal to 3 yards. 36 inches make a yard, so 180 inches is equal to yards. That makes this a 3-4-5 triangle. 3-4-5 is a well-known Pythagorean triple; that is, they have the relationship
and any triangle with these three sidelengths is a right triangle. Also, since the three sides are of different lengths, the triangle is scalene.
The correct response is that the triangle is right and scalene.
Example Question #2 : Other 2 Dimensional Geometry
Which of the following describes a triangle with sides of length two yards, eight feet, and ten feet?
The triangle is right and scalene.
The triangle is right and isosceles.
The triangle is acute and scalene.
The triangle is acute and isosceles.
The triangle cannot exist.
The triangle is right and scalene.
Explanation:
Two yards is equal to six feet. The sidelengths are 6, 8, and 10, which form a well-known Pythagorean triple with the relationship
The triangle is therefore right. Since no two sides have the same length, it is also scalene.
Example Question #3 : Other 2 Dimensional Geometry
The above figure shows a square garden (in green) surrounded by a dirt path feet wide throughout. Which of the following expressions gives the distance, in feet, from one corner of the garden to the opposite corner?
Explanation:
The sidelength of the garden is feet less than that of the entire lot - that is, . Since the garden is square, the path from one corner to the other is a diagonal of a square, which has length times the sidelength. This is
feet.
Example Question #4 : Other 2 Dimensional Geometry
The above figure shows a square garden (in green) surrounded by a dirt path six feet wide throughout. Which of the following expressions gives the distance, in feet, from one corner of the garden to the opposite corner?
Explanation:
The sidelength of the garden is less than that of the entire lot - that is, . Since the garden is square, the path from one corner to the other is a diagonal of a square, which has length times the sidelength. This is
feet.
Example Question #5 : Other 2 Dimensional Geometry
A circle is inscribed inside a square that touches all edges of the square. The square has a length of 3. What is the area of the region inside the square and outside the edge of the circle?
Explanation:
Solve for the area of the square.
Solve for the area of the circle. Given the information that the circle touches all sides of the square, the diameter is equal to the side length of the square.
This means that the radius is half the length of the square:
Subtract the area of the square and the circle to determine the area desired.
Example Question #1 : Other 2 Dimensional Geometry
Figure is not drawn to scale
is a diameter of the circle; its length is ten; furthermore we know the following:
Give the length of (nearest tenth) |
# 5th Grade Division Problems – An Easier Way to Divide Large Numbers (Video Tutorials included)!
### Long DivisionThe Bigger the Numbers – The More Room for Errors!
Long division is an awkward system that confuses many students. Even when children understand the concept behind long division, they still struggle if they can not write neatly and in straight columns. Beyond that – children struggle further when they estimate incorrectly, because they have to erase and start again. The whole system of long division is outdated and confusing.
### Area Division
###### If you are following these 8-video posts in order, then you have already read my explanation about Area Division. Please skip directly to “On to the Videos” below.
Area Division is an alternative to long division. I have been teaching elementary and middle school children for over a quarter of a century. Long division confuses more children than almost any other educational concept. The entire process is confusing and quite frustrating for young minds. As I will demonstrate, children do not have to estimate correctly every time when they do area division. And, if they make a mistake, they simply create another area box and continue to solve the problem.
### Break it Down
Break it Down is a strategy that can be used with Area Division. However, this is only used when a child does not know their multiplication fact, or if they estimate wrong. If a child knows their multiplication facts and can estimate correctly there is no need for the Break it Down strategy.
Break it Down works particularly well for students that do not know all their multiplication facts. The video below shows how children can use the “Break it Down” strategy with any division problem.
### This is my Fourth of Eight Video-Posts
1. An Introduction to Area Division
2. Dividing a 4-digit number by a 1-digit number (ie: 4,224 ÷ 8)
3. Dividing a 3-digit number by a 2-digit number (ie: 738 ÷ 82)
4. Dividing a 4-digit number by a 2-digit number (ie: 2,952 ÷ 82)
5. Dividing a 3-digit number by a 2-digit number with decimal representation of the remainder (ie: 533 ÷ 82 = 6.5) – Coming Soon
6. Dividing a 2-digit number by a tenth-place decimal (ie: 35 ÷ 0.8 = 43.75) – Coming Soon
7. Dividing a decimal by another decimal (ie: 28.8 ÷ 0.8 = 36) – Coming Soon
In the summer of 2017, I decided to develop an alternative to long division. That alternative is called “Area Division”. I will teach you Area Division in a series of video-posts.
The video below is the second progression for learning my latest strategy, Area Division. In order to fully understand this new strategy, it is important to start with my first video-post (An Introduction to Area Division) and work your way up. If your child is already in 5th or 6th grade, this may seem too easy. However, it is important that they understand the basic concept of how this new strategy works before they jump to the problems they’re doing in class. If they take their time and watch each consecutive video, you will understand this new method of division. Then you will reap the rewards of success!
### How to use these Video-Posts
First – Children are introduced to a concept with a “WATCH ME” demonstration video.
Then – Children continue developing their skills by working side-by-side with me. They are given a problem and work through that problem, while they watch the “WORK WITH ME” video.
Finally – Children perfect their skills by completing problems by themselves. Then they will watch the “ON YOUR OWN” videos. These last videos encourage the child’s independent work, and then review the problem making sure that the child gains understanding.
### On to the Video Lessons
These Video-Posts are designed in a similar manner to how I teach a concept in my class. Therefore, I will be addressing the children directly for the rest of this post.
### Baby Llamas For Sale
1 – Watch Me
You are raising baby llamas and sell them to llama farms around the world. Your llamas are the cutest. You made \$3,431 by selling 73 llamas. How much do each of your baby llamas cost?
This one is easy. Click on the link below to watch how this problem is solved. Pay close attention. The next challenge will be very similar to this one.
2 – Work with Me
Your parents just buy a travel-home. Now it’s time to hit the road. You’re heading across the states. It’s 2,924 miles from California to New York. If you got from California to New York in 43-hours, how fast did you drive?
This one is a little harder.
You will need the following materials:
• A blank piece of paper
• A pencil
Watch and complete this challenge with me. I am taking you through a step-by-step process. Together, we will solve this problem.
Pay close attention. The next challenge will be very similar to this one.
### Grand Prix Motorcycle Race
You and your family go to the Grand Prix Motorcycle Race every year. Each competitor must ride his or her bike 9,732 mile in just 12-days. How many miles must the racers ride their motorcycles each day?
You will need the following materials:
• A blank piece of paper
• A pencil
This problem is solved just like the earlier one. If your estimate is incorrect, you must make another area box and continue to solve the problem.
Don’t worry if you make a mistake – Some of our biggest leaps in learning come when we make a mistake then see what we did wrong and fix it.
• Once you have completed this challenge, click the video below:
• Keep your paper with you while you watch the video.
• If you make a mistake, pause the video and fix your mistake.
• That’s the fastest way to learn!
### Whale eating Krill
This whale is huge, but it eats tiny krill. In order to survive a whale like this must eat 40-million krill per day. That’s around 8,016 pounds of krill a day. About how many pounds of krill do whales eat per hour?
Gather the following materials:
• A blank piece of paper
• A pencil
Solve this problem just as you did in the earlier one. If you estimate incorrectly, make another area box and continue to solve the problem.
Don’t worry if you make a mistake – Some of our biggest leaps in learning come when we make a mistake then see what we did wrong and fix it.
• Once you have completed this challenge, click the video below:
• Keep your paper with you while you watch the video.
• If you have made a mistake, pause the video and fix your mistake.
• That’s the fastest way to learn!
You are a scientist studying the deadly frogs of South America. These Frogs have beautiful colors, but they are deadly. Scientists believe that the are deadly because of the insects that they eat, which include tiny beetles, termites, and fruit flies.
There are 59 frog eating 8,024 insects. If they all eat the same amount of insects, how many insects does each frog eat?
Gather the following materials:
• A blank piece of paper
• A pencil
Solve this problem just as you did in the earlier one. If you estimate incorrectly, make another area box and continue to solve the problem.
Don’t worry if you make a mistake – Some of our biggest leaps in learning come when we make a mistake then see what we did wrong and fix it.
• Once you have completed this challenge, click the video below:
• Keep your paper with you while you watch the video.
• If you make a mistake, pause the video and fix your mistake.
• That’s the fastest way to learn!
Your next challenge will be in the Video-Blog-Post, Dividing a 6-digit number by a 3-digit number with decimal representation of the remainder (ie: 750,864 ÷ 824). The concept builds from this one, so it is important that you understand all the videos that you have just worked through.
Have a great day – Brian McCoy
##### On another note –
Read my post entitled: Free Learning Games Kids – Do they really motivate children to learn?
#### Novels by Brian McCoy –
– A Children’s Fantasy Book intended for readers in 2nd through 4th grades.
– an Action & Adventure Novel intended for readers in Middle School
https://www.amazon.com/RED-Brian-D-McCoy/dp/1520676468/ref=sr_1_1?ie=UTF8&qid=1503101676&sr=8-1&keywords=REd+by+brian+mcCoy
##### Want more Tutorials?
I’ve created an Educational Fantasy Game called, TeachersDungeon. It is set to the Common Core Educational Standards, and is web-based, so it can be played on any device. Many of the questions are accompanied by tutorials like the ones you saw here. |
# MAXIMUM AND MINIMUM VALUES OF QUADRATIC FUNCTIONS WORKSHEETS
Maximum and minimum values of quadratic functions worksheets :
Here we are going to see some practice questions on finding maximum and minimum values of quadratic functions.
## Maximum and minimum values of quadratic functions worksheets - Practice questions
(1) Find the minimum or maximum value of the quadratic equation given below.
f(x) = 2x² + 7x + 5
(2) Find the minimum or maximum value of the quadratic equation given below.
f(x) = -2x² + 6x + 12
(3) Find the minimum or maximum value of the quadratic equation given below.
f(x) = -5x² + 30x + 200
(4) Find the minimum or maximum value of the quadratic equation given below.
f(x) = 3x² + 4x + 3
## Find the maximum and minimum value of quadratic function - Solution
Question 1 :
Find the minimum or maximum value of the quadratic equation given below.
f(x) = 2x² + 7x + 5
Solution :
Since the coefficient of x2 is positive, the parabola is open upward.
So, the function will have only the minimum value.
x-coordinate of minimum value = -b/2a
y-coordinate of minimum value = f(-b/2a)
a = 2 b = 7 and c = 5
x-coordinate = -7/2(2) ==> -7/4
y-coordinate = f(-7/4)
f(x) = 2x² + 7x + 5
f(-7/4) = 2(-7/4)² + 7(-7/4) + 5
= 2(49/16) - (49/4) + 5
= (49/8) - (49/4) + 5
= (49 - 98 + 40) / 8
= -9/8
Hence the minimum value is -9/8.
Question 2 :
Find the minimum or maximum value of the quadratic equation given below.
f(x) = -2x² + 6x + 12
Solution :
Since the coefficient of xis negative, the parabola is open downward.
So, the function will have only the maximum value.
x-coordinate of maximum value = -b/2a
y-coordinate of maximum value = f(-b/2a)
a = -2 b = 6 and c = 12
x-coordinate = -6/2(-2) ==> 6/4 ==> 3/2
y-coordinate = f(3/2)
f(x) = -2x² + 6x + 12
f(3/2) = -2(3/2)² + 6(3/2) + 12
= -2(9/4) + 3(3) + 12
= -9/2 + 9 + 12
= -9/2 + 21
= (-9 + 42)/2
= 33/2
Hence the maximum value is 33/2.
Question 3 :
Find the minimum or maximum value of the quadratic equation given below.
f(x) = -5x² + 30x + 200
Solution :
Since the coefficient of xis negative, the parabola is open downward.
So, the function will have only the maximum value.
x-coordinate of maximum value = -b/2a
y-coordinate of maximum value = f(-b/2a)
a = -5, b = 30 and c = 200
x-coordinate = -30/2(-5) ==> 30/10 ==> 3
y-coordinate = f(3)
f(x) = -5x² + 30x + 200
f(3) = -5(3)² + 30(3) + 200
= -5(9) + 90 + 200
= -45 + 290
= 245
Hence the maximum value is 245.
Question 4 :
Find the minimum or maximum value of the quadratic equation given below.
f(x) = 3x² + 4x + 3
Solution :
Since the coefficient of xis positive, the parabola is open upward.
So, the function will have only the minimum value.
x-coordinate of minimum value = -b/2a
y-coordinate of minimum value = f(-b/2a)
a = 3, b = 4 and c = 3
x-coordinate = -4/2(3) ==> -4/6 ==> -2/3
y-coordinate = f(-2/3)
f(x) = -3x² + 4x + 3
f(-2/3) = 3(-2/3)² + 4(-2/3) + 3
= 3(4/9) - (8/3) + 3
= 12/9 - 8/3 + 3
= 4/3 - 8/3 + 3
= (4 - 8 + 9)/3
= (13 - 8)/3
= 5/3
Hence the minimum value is 245.
After having gone through the stuff given above, we hope that the students would have understood "Maximum and minimum values of quadratic functions worksheets".
Apart from the stuff given on this web page, if you need any other stuff in math, please use our google custom search here.
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Profit and loss shortcuts
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Remainder when 2 power 256 is divided by 17
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Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Video: AQA GCSE Mathematics Foundation Tier Pack 4 • Paper 3 • Question 14
Calculate the size of angle 𝑥.
02:28
### Video Transcript
Calculate the size of angle 𝑥.
There are few things to be aware of before we start this question. Firstly, it says that the diagram is not drawn accurately. This means that we can’t measure the angle with a protractor. We have to use our angle properties. It is also important to note that the line marked in orange appears to be a straight line. However, we cannot assume this, unless we’ve been told it is definitely a straight line. In this case, it is not. This means that we can’t use the angle property that angles on a straight line add up to 180 degrees.
The angle marked with a square or box is a right angle and is therefore equal to 90 degrees. The angle property that we will use to solve this question is that angles in a circle or at a point sum or add up to 360 degrees. The five angles inside the circle must add up to 360. We can write this as an equation: 𝑥 plus 102 plus 50 plus 90 plus 42 is equal to 360. Our first step is to simplify the equation by adding 102, 50, 90, and 42. These four numbers sum to 284. Therefore, 𝑥 plus 284 is equal to 360.
Our final step to calculate the value of 𝑥 is to subtract 284 from both sides of the equation. 284 minus 284 is equal to zero. Therefore, we have 𝑥 on the left-hand side. 360 minus 284 is equal to 76. This means that the size of angle 𝑥 is 76 degrees. We can substitute this value back into our initial equation to check that our answer is correct. 76 plus 102 plus 50 plus 90 plus 42 is indeed equal to 360. |
# Basic Mathematics & Statistics Concepts Quiz
CharismaticRuby
·
Start Quiz
Study Flashcards
## 10 Questions
### How are variations expressed?
As absolute differences or relative differences
Probability
### What is differentiation?
The process of finding the rate of change of a function at a given point
### What is a matrix?
A rectangular array of numbers, symbols, or expressions, organized in rows and columns
Square matrices
### What are the four basic operations in arithmetic?
Addition, subtraction, multiplication, and division
### What is the formula for calculating the mean of a set of numbers?
Sum of all the numbers divided by the number of items
### Name two common types of graphs and charts used to represent data.
Line graphs and bar charts
### How is the median of a set of numbers determined?
It is the middle value when the numbers are arranged in ascending or descending order
Measurement
## Basic Mathematics & Statistics
Mathematics and statistics are essential tools in various fields, including science, engineering, and business. In this article, we will explore the basic concepts of arithmetic, graphs and charts, mean and median, measurement, probability, variants, set theory, matrix and determinants, and differentiation.
### Arithmetic
Arithmetic is the branch of mathematics that deals with the properties and relationships of numbers, including addition, subtraction, multiplication, and division. It is the foundation for more advanced mathematical concepts, such as algebra and geometry.
### Graphs and Charts
Graphs and charts are visual representations of data that help to identify, analyze, and interpret patterns, trends, and relationships. Common types of graphs and charts include line graphs, bar charts, pie charts, and scatter plots.
### Mean and Median
Mean and median are measures of central tendency, used to describe the typical value of a set of numbers. The mean is the sum of all the numbers divided by the number of items, while the median is the middle value when the numbers are arranged in ascending or descending order.
### Measurement
Measurement is the process of assigning numbers to objects or quantities to describe their size, shape, or amount. It involves determining the magnitude of a quantity by comparing it with a known standard or unit of measurement.
### Probability
Probability is the branch of mathematics that deals with the likelihood of events occurring. It involves calculating the chances of specific outcomes, such as rolling a dice or drawing cards from a deck.
### Variations
Variations are changes or differences in a set of numbers or values. They can be expressed as absolute differences or relative differences, and understanding variations helps in analyzing and comparing different sets of data.
### Set Theory
Set theory is the branch of mathematics that deals with the properties and relationships of collections of objects, called sets. It forms the foundation of modern mathematics and has applications in various fields, including computer science and data analysis.
### Matrix and Determinants
A matrix is a rectangular array of numbers, symbols, or expressions, organized in rows and columns. Determinants are scalar values associated with square matrices, used to calculate various properties and solve systems of linear equations.
### Differentiation
Differentiation is the process of finding the rate of change of a function at a given point, often used to study the behavior of functions and solve optimization problems. It involves calculating the derivative of a function, which represents the slope of the tangent line at a specific point.
In conclusion, understanding the basic concepts of mathematics and statistics is essential for various applications in science, engineering, and business. Mastering these topics can help you make informed decisions and solve problems more effectively.
Test your knowledge of fundamental concepts in mathematics and statistics, including arithmetic, graphs and charts, mean and median, measurement, probability, variations, set theory, matrix and determinants, and differentiation.
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# Trigonometry help
Trigonometry help can be found online or in mathematical textbooks. We will give you answers to homework.
## The Best Trigonometry help
Math can be a challenging subject for many learners. But there is support available in the form of Trigonometry help. Differential equations are equations that describe the relationship between a quantity and a change in that quantity. There are many types of differential equations, which can be classified into two main categories: linear and nonlinear. One example of a differential equation is the equation y = x2, which describes the relationship between the height and the width of a rectangle. In this case, x represents height and y represents width. If we want to find out how high or how wide a rectangle will be, we can find the height or width by solving this equation. For example, if we want to know what the height of a rectangle will be, we simply plug in an x value and solve for y. This process is called “back substitution” because it makes use of back-substitution. For example, if we want to know what the width of a rectangle will be, we plug in an x value and solve for y. Because differential equations describe how one quantity changes when another quantity changes, solving them can often be used to predict what will happen to one variable if another variable changes or is kept constant. In addition to predicting what will happen in the future, differential equations can also be used to simulate how systems behave in the past or present. Because these simulations involve using estimates of past values as inputs into models instead of actual values from the past, they are often referred as
The rule of equilateral triangles is that the three sides must be equal. Two-to-one correspondence: When you take two shapes like a circle and a square and connect them with lines to form a rectangle. The rule of rectangle is that the four sides must be equal. Three-to-one correspondence: When you take three shapes like an ellipse and a triangle and connect them with lines to form a parallelogram. The rule of parallelogram is that the five sides must be equal. Two-way correspondence: When you have two shapes like a circle and a square or two rectangles or two triangles and then connect them with lines to form another shape such as an equilateral triangle or rectangle or ellipse or parallelogram. Addition math problems: Adding numbers from one set to another set is the most common arithmetic math problem you will encounter in school. You can do it by adding sets
It is important that you use the same units for both sides of the equation (e.g., cm or inches). Next, we need to identify one side as the hypotenuse, which is the longest side of the triangle. In this case, it is going to be a long side that measures 5 cm (or 5 inches). Finally, we need to multiply all three sides by their corresponding integers, so that they become equal lengths: 5 + 3 = 8 cm (or 8 inches). The right triangle has been solved.
A step by step calculator is a calculator that guides you through every step of a calculation, making it easy to follow along. You can enter numbers one by one, or enter an equation and the calculator will automatically calculate everything else. One example of a step by step calculator is an online mortgage calculator. This tool will guide you through each step of your mortgage application, showing you how much you could potentially save with different loan terms. There are many different types of calculators out there, but they all work the same way: They help you solve problems by giving you step-by-step instructions for how to solve them. These calculators can be very helpful if you aren't sure how to solve a certain problem in math or if you need to calculate something like an interest rate or when your taxes are due.
As a college student who struggles with algebra like. horrendously. this has been a very helpful tool. Although you have to pay to get the step by step solving instructions, I usually don't need them and subsequently this has aided me in trial and error. It is very useful!!! It's kind of like having Sheldon Cooper breathing down your neck to ensure you get things correct, which is a massive compliment, dudes.
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### Nathalie Bryant
Table of values solver Problem solving helper Factorial of zero How to solve radical equations Input output rule solver Radical calculator solver |
# Get Answers to all your Questions
#### Provide solution RD Sharma maths class 12 chapter 10 differentiation exercise 10.6 question 5 maths textbook solution
Answer: $\inline \frac{d y}{d x}=\frac{y^{2} \cot x}{1-y \log \sin x}$
Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.
Given: $\inline y=\left ( \sin x \right )^{\left ( \sin \right )^{..... \infty }}$
Solution:
Here it is given that,
$\inline y=\left ( \sin x \right )^{\left ( \sin \right )^{..... \infty }}$
This can be written as:
$\inline \log y=\log \left ( \sin x \right )^{y}$
Taking log on both sides, we get:
$\inline \log y=\log \left ( \sin x \right )^{y}$
$\inline \therefore \log y=y\log \left ( \sin x \right )$ …(1)
Differentiating (1) w.r.t x,
\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=y \frac{d}{d x}(\log \sin x)+\log \sin x \cdot \frac{d}{d x}(y) \\ &\frac{1}{y} \frac{d y}{d x}=y \frac{1}{\sin x}(\cos x)+\log \sin x \frac{d y}{d x} \\ &\frac{d y}{d x}\left(\frac{1}{y}-\log \sin x\right)=y \cot x \\ &\frac{d y}{d x}\left(\frac{1-y \log \sin x}{y}\right)=y \cot x \\ &\therefore \frac{d y}{d x}=\frac{y^{2} \cot x}{1-y \log \sin x} \end{aligned}
Hence, it is proved. |
# How do you find the polar coordinates of the point?
Mar 14, 2018
I am assuming that you want to convert a point in Cartesian Coordinate form to Polar Coordinate Form.
I will use an example in the explanation below.
#### Explanation:
Let us say that we want to convert a point $\left(0 , 2\right)$ in Cartesian Form to Polar Form.
Let us analyze the image below:
We have a right-triangle and hence we can use Pythagoras Theorem to define a relationship:
${x}^{2} + {y}^{2} = {r}^{2}$
$\tan \theta = \frac{y}{x}$
We can see that
$\cos \theta = \frac{x}{r}$
Hence,
color(blue)(x=r cos theta color(red)([ 1 ]
$\sin \theta = \frac{y}{r}$
Hence,
color(blue)(y = r sin theta color(red)([ 2 ]
Observe that, if we are given the point $\left(x , y\right)$, then we can define $\left(r , \theta\right)$ and express $\left(x , y\right)$ in terms of $\left(r \cos \theta , r \sin \theta\right)$.
Now, let us work on the point in Cartesian coordinate form, in our example:
$\left(0 , 2\right)$
Let us plot this point and look at the graph:
Now, we are in a position to convert the $\left(0 , 2\right)$ to equivalent Polar form using color(red)([ 1 ] and color(red)([ 2 ]
Polar form is color(green)((r, theta)
rArr color(blue)((2, pi/2)
Hope it helps. |
## Reverse Percentages in Numeracy Aptitude Tests
Many who are asked to complete numerical aptitude test are frequently caught out on reverse percentage questions. In fact, only few follow right steps to work out problems in such questions correctly. In simple terms, reverse percentages are mathematical functions used to solve problems related to finding out unknown quantities after they have increased or decreased. For example, you know that the price of house was £320,000 in 2013 which represented 10% increase from 2012. You would use reverse percentage to find the value of house in 2012. If you are not acquainted or used to these concepts it may be hard to know how to multiply, divide or what operation to take to get to the right answer. In the below tutorial you will find practice examples that will help you to understand how to work out these kinds of problems correctly in your numerical reasoning tests.
How to work out reverse percentage increase
Reverse percentage increase is one of the most important topics to master for your numeracy aptitude tests. It referrers to calculating algebraic functions related to finding value X after it had increased by certain percentage amount. For example, assume that amount X raised by 30% to new figure 70. To find for X you need to apply following formula 70 / (1 + 0.30). To demonstrate this on real test example consider the below question and inspect the figure below.
If the iPhones sales in September rose by 15% from previous month then what was the number of iPhones sold during the August?
As indicated above, to solve this you need to make use of reverse percentage increase that is 200 / (1 + 0.15). However, some test takers that are not acquainted with this method may calculate 200 x (1 – 0.15) which is the incorrect way. Note that this approach rather refers to calculating percentage decrease which is to be applied to situations where you are asked to reduce given amount by certain proportion. Hence, pay attention and read carefully what the question is asking you to do so that you use the correct calculation.
How to work out reverse percentage decrease
When you are required to work out reverse percentage decrease for example to solve for value M that has decreased say by 25% over some period of time to value N which equals to 90 then the formula to use to solve for N would be formulated in following way 90 / (1 – 0.25) = 120. To illustrate this on real example consider the question below.
If the BlackBerry trades went down by 20% from August to September then what was the number of BlackBerry sales in August?
If you inspect the above graph you can see that in September trading company sold 100 BlackBerry iPhones. Hence, you may set up your equation as follows 100 / (1 – 0.20) to calculate sales for August. Fairly simple isn’t it.
In your numerical reasoning test the faster and quicker way to calculate the above is to mentally subtract given percentage, in the above case 0.20 from 1, and only perform calculation of 100 / 0.80 on your calculator. Similarly, for reverse percentage increase add together in your head 1 and 0.20 and then calculate on your calculator only 100 / 1.20. Remember, time is precious in numerical reasoning tests and hence attempt to make as many calculations in your head as possible. |
# Math in Focus Grade 6 Chapter 7 Lesson 7.5 Answer Key Real-World Problems: Algebraic Expressions
Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 7 Lesson 7.5 Real-World Problems: Algebraic Expressions to score better marks in the exam.
## Math in Focus Grade 6 Course 1 A Chapter 7 Lesson 7.5 Answer Key Real-World Problems: Algebraic Expressions
### Math in Focus Grade 6 Chapter 7 Lesson 7.5 Guided Practice Answer Key
Complete.
Question 1.
Raoul is y years old. Kayla is 6 years older than Raoul and Isaac is 4 years younger than Raoul.
a) Find Kayla’s age.
Kayla is years old.
y + 6 years
Explanation:
Raoul is y years old.
Kayla is 6 years older than Raoul and Isaac is 4 years younger than Raoul.
So, Kayla is y + 6 years old.
b) Find Isaac’s age.
Isaac is years old.
y – 4 years
Explanation:
Raoul is y years old.
Kayla is 6 years older than Raoul and Isaac is 4 years younger than Raoul.
So, Isaac is y – 4 years old.
c) If y = 12, find the sum of Raoul’s age and Isaac’s age.
When y =12,
Isaac’s age:
=
= years old
Sum of Raoul’s age and Isaac’s age:
+ =
The sum of Roul’s age and Isaac’s age is years.
20 years.
Explanation:
When y =12,
Isaac’s age:
12 – 4 = 8 years old
Sum of Raoul’s age and Isaac’s age:
12 + 8 = 20
The sum of Roul’s age and Isaac’s age is 20 years.
Question 2.
A pickup truck uses 1 gallon of gas for every 14 miles traveled,
a) How far can it travel on 3p gallons of gas?
1 gallon → miles
3p gallons → = miles
It can travel miles on 3p gallons of gas.
42 miles
Explanation:
1 gallon → 14 miles
3p gallons → 3p • 14 = 42 p miles
It can travel 42 p miles on 3p gallons of gas.
b) How many gallons of gas have been used after the pickup truck has traveled v miles? Evaluate this expression when v = 56.
14 miles → gallon
v miles → ÷ = gallons
gallons have been used.
When v = 56,
=
=
4 groups
Explanation:
14 miles → 4 gallon
v miles → 56 ÷ 14 = 4 gallons
4 gallons have been used.
When v = 56,
56 = 14 x 4
= 4 groups
Question 3.
There were three questions in a mathematics test. Salma earned m points for the first question and twice the number of points for the second question,
a) How many points did she earn for the first two questions?
+ =
She earned points for the first two questions.
3m points
Explanation:
m + 2m = 3m
She earned 3m points for the first two questions.
b) If she received a total of 25 points on the test, how many points did she earn for the third question?
She earned points for the third question.
10 points
Explanation:
Salma received a total of 25 points on the test,
Third question: 25 – 3m = 25 – (3 •5)
= 25 – 15
= 10
c) If m = 5, find the points she earned for each question.
First question: m = 5
Second question: 2m = 2 •
=
Third question: 25 – 3m = 25 – (3 • )
= 25 –
=
She earned points for the first question, points for the second question and points for the third question.
10 points
Explanation:
First question: m = 5
Second question: 2m = 2 • 5 = 10
Third question: 25 – 3m = 25 – (3 •5)
= 25 – 15
= 10
She earned 5 points for the first question, 10 points for the second question and 10 points for the third question.
### Math in Focus Course 1A Practice 7.5 Answer Key
Question 1.
Jenny is x years old. Thomas is 3 times as old as she is. Jenny is 5 years older than Alexis.
a) Find Alexis’s age in terms of x.
x+5
Explanation:
Jenny age x years
Thomas age 3x
Alexis are x + 5
b) Find Thomas’s age in terms of x.
Explanation:
Jenny is x years old.
Thomas is 3 times as old as she is.
Thomas age 3x.
c) If x = 12, how much older is Thomas than Jenny?
36 years
Explanation:
Thomas is 3x = 3(12) = 36 years
Question 2.
A van travels from Town A to Town B. It uses 1 gallon of gas for every 24 miles traveled.
a) How many gallons of gas does the van use if it travels 3x miles?
$$\frac{3x}{24}$$ gallons of gas
Explanation:
1 gallon = 24 miles
3x miles = $$\frac{3x}{24}$$ gallons of gas.
b) The van uses 2y gallons of gas for its journey from Town A to Town B. Find the distance between Town A and Town B.
48y miles
Explanation:
1 gallon = 24 miles
2y gallons = 2y x 24 miles
= 48y miles
Question 3.
Brian bought x apples and some oranges. Brian bought 3 more oranges than apples.
a) Find the total number of fruit Brian bought in terms of x.
2x + 3
Explanation:
x apples
x + 3 oranges
the total number of fruit Brian bought in terms of x is
= x + x +3 = 2x + 3
b) Find the total amount of money, in cents, that Brian spent on the fruit. Give your answer in terms of x.
80x + 150 cents
Explanation:
2x ( 40 ) + 3 ( 50)
= 80x + 150 cents
c) If Brian could have bought exactly 12 pears with the amount of money that was spent on the apples and oranges, find the cost of each pear, in cents, in terms of x.
$$\frac{80x+150}{12}$$
Explanation:
12p = 80x + 150 cents
p = $$\frac{80x+150}{12}$$
Question 4.
A rectangle has a width of x centimeters and a perimeter of 8x centimeters. A square has sides of length $$\frac{1}{4}$$ that of the length of the rectangle.
a) Find the length of the rectangle.
Explanation:
8x = 2(x + l)
=> 8x/2= x + l
=> 4x = x + l
=> 4x – x = l
length = 3x
b) Find the perimeter of the square.
Explanation:
A square has sides of length $$\frac{1}{4}$$
the perimeter of the square = 4 x $$\frac{3x}{4}$$ = 3x
c) Find how many centimetres greater the rectangle’s perimeter is than the square’s perimeter if x = 4.
20 centimetres greater the rectangle’s perimeter is than the square’s perimeter.
Explanation:
rectangle’s perimeter = 8x
square’s perimeter = 3x
if x = 4.
rectangle’s perimeter = 8x = 8 x 4 = 32 cm
square’s perimeter = 3x = 3 x = 12 cm
32 – 12 = 20 cm
20 centimetres greater the rectangle’s perimeter is than the square’s perimeter
d) Find how many square centimetres greater the rectangle’s area is than the square’s area if x = 4.
39 sq cm
Explanation:
Square area = side x side
= $$\frac{3x}{4}$$ x $$\frac{3x}{4}$$
= $$\frac{3x}{4}$$ x $$\frac{3x}{4}$$ substitute x = 4
= $$\frac{3 x 4}{4}$$ x $$\frac{3 x 4}{4}$$ = 9 sq cm
Rectangle area = length x width
= 3x . x = 3 x 4 . 4 = 48 sq cm
48 – 9 = 39 sq cm many square centimeters greater the rectangle’s area is than the square’s area.
Question 5.
Jose bought 4 comic books and 2 nonfiction books. The 4 comic books cost him 8y dollars. If the cost of one nonfiction book is (3 + 7y) dollars more expensive than the cost of one comic book, find
a) the cost of the 2 nonfiction books in terms of y.
(15y + 6)/2
Explanation:
4 comic books = 8y
1 comic book = 8y/4 = y/2
one notification book = y/2 + (3 + 7y) = (15y + 6)/2
b) the total amount that Jose spent on the books if y = 4.
$98 Explanation: 8y + 15y + 6 if y = 4 = 8.4 + 15.4 + 6 = 32 + 60 + 6 = 98$
Question 6.
Wyatt has (2x – 1) one-dollar bills and (4x + 2) five-dollar bills. Susan has 3x dollars more than Wyatt.
a) Find the total amount of money that Wyatt has in terms of x.
22x – 9
Explanation:
Wyatt has (2x – 1) one-dollar bills and (4x + 2) five-dollar bills
2x-1 + 5(4x + 2)
= 2x – 1 + 20x + 10
= 22x – 9
b) Find the number of pens that Wyatt can buy if each pen costs 50 cents.
44x -1 8 pens
Explanation:
Wyatt has (2x – 1) one-dollar bills and (4x + 2) five-dollar bills
2x-1 + 5(4x + 2)
= 2x – 1 + 20x + 10
= 22x – 9/(1/2)
= (22x – 9 ) x 2
= 44x -1 8 pens can buy
c) If x = 21, find how much money Susan will have now if Wyatt gives her half the number of five-dollar bills that he has.
516$Explanation: Susan has 3x dollars more than Wyatt Wyatt has 22x – 9 2x – 1 + (5(4x + 2))/2 = (4x – 2 + 20x + 10)/2 = (24x – 8)/2 now Susan has x = 21 = 2x – 1+(4x + 2)/2 + 3x = 2x-1 + (4 x 21 +2 )/2 + 3 x 21 = 2×21 -1 + 86/2 + 63 = 41 + 43 + 63 = 147$ (to be re calculated)
Brain @ Work
Question 1.
Find the perimeter of the figure in terms of x, given that all the angles in the figure are right angles. If x = 5.5, evaluate this expression. |
# Math Circle Session 5: Powers, Limits, and Fractals
Yesterday saw the next installment of our math circle with primary school children. Again, following but extending Rozhkovskaya’s book, we used fractals as an interesting way in to consolidate (and in the case of the younger pupils, introduce) powers of numbers, to explore the children’s intuition about infinite series, and to make some pretty pictures!
We started by producing a fractal tree. The rule is simple: after Year 1, the trunk has grown. At the end of Year 2, two branches have grown from that, after Year 3, two further branches from each existing branch, and so on. Children were quickly able to tell us that the number of branches doubled each year, some took great pleasure in calculating or reciting powers of two. The template for this exercise was taken from Rozhkovskaya’s book, and is quite clever: lines are drawn across the page heights of 8 squares, 12 squares, 14 squares, 15 squares, and 15.5 squares, and each successive year’s branches should be drawn to reach the corresponding height. A bird is drawn flying around 18 squares up. The question is posed: will the tree ever reach the bird?
The children had mixed views on whether the tree will reach the bird. The most common view was that it must. One of the older children was able to provide a line of reasoning: “Each time, the height of the tree increases. Since it is always increasing, even by small amounts, it must eventually reach any height.” We then looked at how much gap there is between the height of the tree and a line of height 16 as the years progress: 16 at the beginning, 8 at the end of Year 1, 4 at the end of Year 2, and so on. Children could see that there was always a positive gap between the height of the tree and the height 16, for any finite number of years. However, this clashed with several children’s intuition, and it took quite a lot of discussion before it was generally accepted that an infinite series can sum to a finite limit. One of the youngest children in the group asked some very probing questions, such as “is infinity a number”, which led to some useful side discussions. Random banter between children later in the session about “hacking” their siblings’ passwords led onto a consolidation of this discussion by asking the question “If you had to press an infinite number of keys on a keyboard, could you do it in a finite amount of time? No? What if the first key took 8 seconds to press, the next 4, the next 2, and so on…”
We then moved onto a practical activity of constructing a sequence of approximations to the Sierpinski triangle by sticking little white triangles on coloured paper (pictured). One of the children noticed quickly, and was able to clearly explain, that the number of white triangles was increasing in powers of three. At this point we ran out of time, and our session ended. I intend to pick up and use the Sierpinski triangle again soon in math circle, through Pascal’s triangle modulo 2. |
## ParthKohli 3 years ago How do I solve $$\text{SYSTEM OF EQUATIONS?}$$ Look below to see the tutorial.
1. ParthKohli
All of us probably know one-variable equations. But, the two-variable linear equations may confuse people. In this tutorial, solving two-variable systems would be explained by substitution and elimination. $$\LARGE \color{MidnightBlue}{\text {SUBSTITUTION}}$$ $$\Large \color{purple}{\rightarrow 0.45x +0.65y = 18.55 }$$ $$\Large \color{purple}{\rightarrow x + y = 35 }$$ Okay, now let's take any of the variables. Let's say x for the sake of convenience. Now, we just have to solve for x by subracting y from both sides, and we get: $$\Large \color{purple}{\rightarrow x = 35 - y }$$ Now, we'll replace x with 35 - y, and we get: $$\Large \color{purple}{\rightarrow 0.45(35 - y) + 0.65y = 18.55 }$$ Now, we can solve the equation easily. $$\Large \color{MidnightBlue}{\text{ELIMINATION} }$$ This one is again easy. You have to seek the possible ways. Let's solve the same system again! So, let's multiply the equation 'x + y = 35' by 0.45(both sides). We get: $$\Large \color{purple}{\rightarrow 0.45x + 0.45y = 15.75 }$$ Subtract both equations: $$\Large \color{purple}{\rightarrow (0.45x - 0.45x) + (0.65y - 0.45y) = 19.25 }$$ $$\Large \color{purple}{\rightarrow 0.20y = 19.25 }$$ $$\Large \color{purple}{\rightarrow y = 96.25 }$$ Now, we can just solve for x because we know the value of y.
2. kevinkeegan
ok...
3. Kreshnik
@ParthKohli Is this "Tutorial" called "Collecting medals" ? lol
4. ParthKohli
Eh?
5. satellite73
i think we need a separate section for these
6. ParthKohli
@satellite73 Exactly what I think.
7. Romero
So we can copy and paste a link and tell the asker to come back if he/she still had more questions after reading the tutorials @satellite73
8. amistre64
your tutorial is fine as is, but its missing a few important pointers. The "why" it works would be helpful. And also if you could explain the 3 possible results that can occur. The notion that we can find infinite or no consistent solution is something that tends to confuse people. And the knowledge of why these outcomes are possible explains those mysteries.
9. 2bornot2b
@ParthKohli excellent work! I appreciate it.
10. ParthKohli
@amistre64 I'd certainly take care of that in the future. @2bornot2b Thank you :)
11. 2bornot2b
This tutorial might turn out to be quite helpful for someone searching the internet for "how to solve simultaneous equations". @amistre64 I believe parthkohli prepared it for high school level, therefore he decided to exclude that part.
12. ParthKohli
@2bornot2b I don't know if it's high school or not, but considering that the person knows one-variable linear equations, I posted it.
13. inkyvoyd
You forgot one important thing. First of all, you said system of equations. You need to limit that down to first degree 2 varibles determined. Second of all, you forgot cramers rule. Other than that, good enough.
14. inkyvoyd
*2 imporant things
15. inkyvoyd
Did you know that you can reduce AES encryption down to a "system of equations"? You claimed to solve that, xD
16. ParthKohli
@inkyvoyd Yes, about the Cramer's rule, I have written on the top line that I will explain it using only substitution and elimination. You see, these are two simple methods. It'd have taken hours if I'd used the cramer's rule and graphs.
17. inkyvoyd
Actually, cramer's rule was the fastest way to program the system of equation's solution.
18. FoolForMath
^^ No.
19. lgbasallote
isnt's cramer's rule linear algebra o.O
20. lgbasallote
i like gauss-jordan elimination for the record :p
21. inkyvoyd
Well, my graphing calculator says so.
22. FoolForMath
Computing determinant is itself tedious.
23. inkyvoyd
Well, programming a computer algebra system is more tedious -.-
24. FoolForMath
When you will study numerical analysis you will understand what I meant :)
25. inkyvoyd
I will never study numerical analysis, and I will never understand zzz
26. inkyvoyd
(Neither will my Ti-84Plus SE)
27. lgbasallote
since they said the good comments already i'll be the "simon cowell" from a tutor to another one. 1) You should solve the SUBSTITUTION and ELIMINATION methods thoroughly not just end in the middle 2) You did not state WHY you multiplied 0.45 in the ELIMINATION method. 3) Inconsistency. In the ELIMINATION method you solved for y. You did not solve for y in the SUBSTITUTION method though. and again, you cut in the middle 4) You did not prove that the results of the SUBSTITUTION method and the ELIMINATION method were the same. that are all my bad comments...otherwise it was good :) deserved the medals ^_~
28. inkyvoyd
Want to see my tutorial? The answer is always no. I just bother people until they see it, then they just give medals and leave. :/ @lgbasallote
29. inkyvoyd
They never provide feedback!
30. lgbasallote
one more thing... "Okay, now let's take any of the variables. Let's say x for the sake of convenience. Now, we just have to solve for x by subracting y from both sides" be more specific...you did not state which you used, if first equation or second equation..although that is pretty obvious when looked at the next step, these things should still not be omitted. @inkyvoyd honestly...a lot of us do not know what you're saying :p haha no offense
31. inkyvoyd
What I'm saying? look at my link! LOOOOL
32. inkyvoyd |
# Multiplying and dividing with integers
You also have to pay attention to the signs when you multiply and divide. There are two simple rules to remember:
When you multiply a negative number by a positive number then the product is always negative.
When you multiply two negative numbers or two positive numbers then the product is always positive.
This is similar to the rule for adding and subtracting: two minus signs become a plus, while a plus and a minus become a minus. In multiplication and division, however, you calculate the result as if there were no minus signs and then look at the signs to determine whether your result is positive or negative. Two quick multiplication examples:
$3\cdot (-4)=-12$
3 times 4 equals 12. Since there is one positive and one negative number, the product is negative 12.
$(-3)\cdot (-4)=12$
Now we have two negative numbers, so the result is positive.
Turning to division, you may recall that you can confirm the answer you get by multiplying the quotient by the denominator. If you answer is correct then the product of these two numbers should be the same as the numerator. For example,
$\frac{12}{3}=4$
In order to check whether 4 is the correct answer, we multiply 3 (the denominator) by 4 (the quotient):
$3\cdot 4=12$
What happens when you divide two negative numbers? For example,
$\frac{(-12)}{(-3)}=\: ?$
For the denominator (-3) to become the numerator (-12), you would have to multiply it by 4, therefore the quotient is 4.
So, the quotient of a negative and a positive number is negative and, correspondingly, the quotient of a positive and a negative number is also negative. We can conclude that:
When you divide a negative number by a positive number then the quotient is negative.
When you divide a positive number by a negative number then the quotient is also negative.
When you divide two negative numbers then the quotient is positive.
The same rules hold true for multiplication.
## Video lesson
Calculate the following expressions
$(-4)\cdot (-12),\: \: \: \: \frac{-12}{3}$ |
# MAP 8th Grade Math : Statistics and Probability
## Example Questions
### Example Question #1 : Map 8th Grade Math
The scatter plot provided displays a group of students' test scores versus the number of missing assignments the students have. Based on plot, select the best answer that describes the direction of the points.
A negative, linear association
A positive, nonlinear association
A positive, linear association
A negative, non linear association
A negative, linear association
Explanation:
The data points in the scatter plot move up the y-axis as the x-axis decreases; thus the data points show a negative association. Also, the data points do not curve, or go up and down, but gradually decreased; thus the scatter plot shows a linear association. We could even draw a "best fit" line:
### Example Question #2 : Statistics And Probability
A middle school teacher conducted a survey of the grade class and found that students were athletes and of those students drink soda. There were students that were not athletes, but drank soda. Last, they found that students did not have a curfew nor were on honor roll. Given this information, how many students don't drink soda?
Explanation:
To help answer this question, we can construct a two-way table and fill in our known quantities from the question.
The columns of the table will represent the students who are athletes or are not athletes and the rows will contain the students who drink soda or do not drink soda. The first bit of information that we were given from the question was that students were athletes; therefore, needs to go in the "athlete" column as the row total. Next, we were told that of those students, drinks soda; therefore, we need to put in the "athlete" column and in the "drinks soda" row. Then, we were told that students were not athletes, but drink soda, so we need to put in the "not an athlete" column and the "drinks soda" row. Finally, we were told that students are not athletes or soda drinkers, so needs to go in the "not an athlete" column and "doesn't drink soda" row. If done correctly, you should create a table similar to the following:
Our question asked how many students don't drink soda. We add up the numbers in the "doesn't drink soda" row to get the total, but first we need to fill in a gap in our table, students who were athletes, but don't drink soda. We can take the total number of students who are athletes, , and subtract the number of students who drink soda,
This means that students who are athletes, don't drink soda.
Now, we add up the numbers in the "doesn't drink soda" row to get the total:
This means that students don't drink soda.
### Example Question #1 : Statistics And Probability
Select the answer choice with a data set that includes an outlier. |
# Counting Problems
Rating
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The authors
Matt W.
## Basics on the topicCounting Problems
After this lesson, you will be able to compute percentages by using the formula, (Part/Whole) x 100 = Percent.
The lesson begins by teaching you how to identify the numerical values for the terms part and whole from a given context. It leads you to learn the formula, (Part/Whole) x 100 = Percent. It concludes with solving counting problems.
Learn how to compute percentages by helping Terence with his counting problems.
This video includes key concepts, notation, and vocabulary such as the terms whole and part that are used to find the percent in the formula, (Part/Whole) x 100 = Percent. Whole represents the total number of outcomes and part represents the number of outcomes that fit the criteria you’re looking for.
Before watching this video, you should already be familiar with solving the equation Quantity = Percent x Whole for one of the missing values, converting percents into fractions and decimals (and the other way around), and computations with fractions.
After watching this video, you will be prepared to learn more about probability and combinatorics.
Common Core Standard(s) in focus: 7.RP.A.1, 7.RP.A.2c, 7.RP.A.3 A video intended for math students in the 7th grade Recommended for students who are 12 - 13 years old
### TranscriptCounting Problems
Terence has a habit of embarassing himself in front of his classmate Magnolia. This time, not only has he just shut a tentacle in his locker door, but in the shock of the moment, he forgot his locker combination! Let's see if we can save Terence some embarrassment by helping him out of this Counting Problem. Terence's locker combination consists of three different numbers: 12, 49, and 37. And Terence knows that the combination ends with the number 49. If each number is used only once, what percent of all possible locker combinations involving 12, 49, and 37, end with the number 49? To answer that question, we need to find out two things: Our whole, our part, then, we'll be able to calculate a percent. Let's start by determining the whole. What do you think represents our whole, or total, in this problem? Since we know the numbers 12, 49, and 37 are each used once, our whole will be all the possible locker combinations containing only those numbers. Given this list of all possible outcomes, we can see that there are 1, 2, 3, 4, 5, 6 possible combinations which use our 3 numbers. These 6 combinations represent the whole. If we are looking for the percent of combinations that end with the number 49, what do you think is the "part" in this problem? Take a look at the table and count how many combinations end in the number 49. Looks like these two fit the bill, so 2 will be our part. Two combinations out of 6 equals two-sixths, or one-third. That means only about 33.3 percent of all possible combinations end with the number 49. Magnolia doesn't seem fazed by the whole tentacle incident, so she and Terence head off to their next class. But it looks like there are just three open seats left, which means there are only a few possible seating arrangements for them to choose from. In what percent of these do Terence and Magnolia end up sitting next to each other? Just like before, we need to identify the part and the whole in order to calculate a percent. So, what do you think represents our "whole" in this scenario? Looking at these different seating arangements, we can see there are 6 possible outcomes total, so that will represent our whole. Now, what should the "part" be in this problem? We want to know the percent of seating arrangements that have Terence and Magnolia next to each other. So the part is the number of outcomes that match this description. Looking in our chart, we can see that Terence is next to Magnolia in 1, 2, 3, 4 out of 6 possible options. Now that we know the part and the whole, we can calculate the percent. 4 out of 6 possible outcomes is four-sixths, or two-thirds. Rounding off our decimal, that means in about 66.7 percent of all seating arrangements, Terence and Magnolia end up next to each other. So, while Terence snags a seat, let's review counting problems. When dealing with counting problems, remember that you're really just working with parts, wholes, and percents. Given a list of all possible outcomes, count the total number of outcomes to find the whole. That number goes in your denominator. Then, figure out how many outcomes fit the criteria you're looking for. This number is the part, so it goes in the numerator. Finally, you can calculate the percent. Back in class with Terence and Magnolia, I don't think that's the outcome that Terence was hoping for. But maybe that new kid is nice? |
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# Find the distance between the following pairs of points:$\left( i \right){\text{ }}\left( {2,3} \right),\left( {4,1} \right) \\ \left( {ii} \right){\text{ }}\left( { - 5,7} \right),\left( { - 1,3} \right) \\ \left( {iii} \right){\text{ }}\left( {a,b} \right),\left( { - a, - b} \right) \\$
Last updated date: 24th Jul 2024
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Hint – Whenever we are given two points, we can easily find the distance between them using the concept of distance formula. Use this same concept to find distance for all three options.
Whenever we are given two pair of point such that the points are $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ then the distance between these points can be given as $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$……………. (1)
Now the first pair of point is $\left( i \right){\text{ }}\left( {2,3} \right),\left( {4,1} \right)$
Using equation (1) we get
Distance = $\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {1 - 3} \right)}^2}}$
Distance= $\sqrt {{2^2} + {{\left( { - 2} \right)}^2}}$
$\Rightarrow \sqrt {4 + 4} = 2\sqrt 2$ m
Now the second pair of point is $\left( {ii} \right){\text{ }}\left( { - 5,7} \right),\left( { - 1,3} \right)$
Using equation (1) we get
Distance = $\sqrt {{{\left( { - 1 - \left( { - 5} \right)} \right)}^2} + {{\left( {3 - 7} \right)}^2}}$
Distance= $\sqrt {{4^2} + {{\left( { - 4} \right)}^2}}$
$\Rightarrow \sqrt {16 + 16} = 4\sqrt 2$ m
Now the second pair of point is $\left( {iii} \right){\text{ }}\left( {a,b} \right),\left( { - a, - b} \right)$
Using equation (1) we get
Distance = $\sqrt {{{\left( { - a - \left( a \right)} \right)}^2} + {{\left( { - b - b} \right)}^2}}$
Distance= $\sqrt {{{\left( { - 2a} \right)}^2} + {{\left( { - 2b} \right)}^2}}$
$\Rightarrow \sqrt {4{a^2} + 4{b^2}} = 2\sqrt {{a^2} + {b^2}}$ m
Note – Whenever we face such type of problems, the key concept is compare the given points whose distance is to be calculated with any general point $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ and then applying the distance formula to obtain the right distance between them. |
## Presentation on theme: "Unit 2 – Quadratic, Polynomial, and Radical Equations and Inequalities"— Presentation transcript:
Chapter 5 – Quadratic Functions and Inequalities 5.6 – The Quadratic Formula and the Discriminant
5.6 – The Quadratic Formula and the Discriminant
In this section we will learn how to: Solve quadratic equations by using the Quadratic Formula Use the discriminant to determine the number and type of roots of a quadratic equation.
5.6 – The Quadratic Formula and the Discriminant
Exact solutions to some quadratic equations can be found by graphing, factoring or by using the Square Root Property. Completing the square can be used to solve any quadratic equation, but it can be tedious if the equations have fractions/decimals Fortunately, a formula exists that can be used to solve any quadratic equation of the form ax2 + bx + c = 0
5.6 – The Quadratic Formula and the Discriminant
ax2 + bx + c = 0
5.6 – The Quadratic Formula and the Discriminant
The solutions of a quadratic equation of the form ax2 + bx + c = 0, where a ≠ 0, are given by the following formula: x= -b ± √b2-4ac 2a
5.6 – The Quadratic Formula and the Discriminant
Example 1 Solve x2 – 8x = 33 by using the Quadratic Formula
5.6 – The Quadratic Formula and the Discriminant
Example 2 Solve x2 – 34x +289 = 0 by using the Quadratic Formula When the value of the radicand is 0, there is only one rational root
5.6 – The Quadratic Formula and the Discriminant
Example 3 Solve x2 – 6x + 2 = 0 by using the Quadratic Formula.
5.6 – The Quadratic Formula and the Discriminant
Example 4 Solve x = 6x by using the Quadratic Formula
5.6 – The Quadratic Formula and the Discriminant
The expression b2 – 4ac is called the discriminant The value of the discriminant can be used to determine the number and type of roots of a quadratic equation.
5.6 – The Quadratic Formula and the Discriminant
Consider ax2 + bx + c = 0, where a, b, and c are rational numbers Value of Discriminant Type and number of roots Example of Graph of Related Function b2 – 4ac > 0; b2 – 4ac is a perfect square 2 real, rational roots b2 – 4ac is not a perfect square 2 real, irrational roots b2 – 4ac = 0 1 real, rational root b2 – 4ac < 0 2 complex roots
5.6 – The Quadratic Formula and the Discriminant
The discriminant can help you check the solutions of a quadratic equation. Your solutions must match in number and in type to those determined by the discriminant
5.6 – The Quadratic Formula and the Discriminant
Example 5 Find the value of the discriminant for each quadratic equation. Describe the number and type of roots for the equation. x2 + 3x + 5 = 0 x2 – 11x + 10 = 0
5.6 – The Quadratic Formula and the Discriminant
Method Can be Used When to Use Example Graphing sometimes Use only if an exact answer is not required Factoring Use if the constant term is 0 or if the factors are easily determined x2 – 3x = 0 Square Root Property Use for equations in which a perfect square is equal to a constant (x + 13)2 = 9 Completing the Square always Useful for equations of the form x2 + bx + c = 0 when b is even x2 + 14x – 9 = 0 Quadratic Formula Useful when other methods fail or are too tedious 3.4x2 – 2.5x = 0
5.6 – The Quadratic Formula and the Discriminant
HOMEWORK Page 281 #15 – 45 odd |
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## Algebra 2
### Course: Algebra 2>Unit 3
Lesson 2: Greatest common factor
# Greatest common factor of monomials
Learn how to find the GCF (greatest common factor) of two monomials or more.
### What you should be familiar with before this lesson
A monomial is an expression that is the product of constants and nonnegative integer powers of $x$, like $3{x}^{2}$. A polynomial is a sum of monomials.
You can write the complete factorization of a monomial by writing the prime factorization of the coefficient and expanding the variable part. Check out our Factoring monomials article if this is new to you.
### What you will learn in this lesson
In this lesson, you will learn about the greatest common factor (GCF) and how to find this for monomials.
## Review: Greatest common factors in integers
The greatest common factor of two numbers is the greatest integer that is a factor of both numbers. For example, the GCF of $12$ and $18$ is $6$.
We can find the GCF for any two numbers by examining their prime factorizations:
• $12=2\cdot 2\cdot 3$
• $18=2\cdot 3\cdot 3$
Notice that $12$ and $18$ have a factor of $2$ and a factor of $3$ in common, and so the greatest common factor of $12$ and $18$ is $2\cdot 3=6$.
## Greatest common factors in monomials
The process is similar when you are asked to find the greatest common factor of two or more monomials.
Simply write the complete factorization of each monomial and find the common factors. The product of all the common factors will be the GCF.
For example, let's find the greatest common factor of $10{x}^{3}$ and $4x$:
• $10{x}^{3}=2\cdot 5\cdot x\cdot x\cdot x$
• $4x=2\cdot 2\cdot x$
Notice that $10{x}^{3}$ and $4x$ have one factor of $2$ and one factor of $x$ in common. Therefore, their greatest common factor is $2\cdot x$ or $2x$.
1) What is the greatest common factor of $9{x}^{2}$ and $6x$?
2) What is the greatest common factor of $12{x}^{5}$ and $8{x}^{3}$?
3) What is the greatest common factor of $5{x}^{7}$, $30{x}^{4}$, and $10{x}^{3}$?
## A note on the variable part of the GCF
In general, the variable part of the GCF for any two or more monomials will be equal to the variable part of the monomial with the lowest power of $x$.
For example, consider the monomials $6{x}^{5}$ and $4{x}^{2}$:
• Since the lowest power of $x$ is ${x}^{2}$, that will be the variable part of the GCF.
• You could then find the GCF of $6$ and $4$, which is $2$, and multiply this by ${x}^{2}$ to obtain $2{x}^{2}$, the GCF of the monomials!
This is especially helpful to understand when finding the GCF of monomials with very large powers of $x$. For example, it would be very tedious to completely factor monomials like $32{x}^{100}$ and $16{x}^{88}$!
## Challenge Problems
4*)What is the greatest common factor of $20{x}^{76}$ and $8{x}^{92}$?
5*) What is the greatest common factor of $40{x}^{5}{y}^{2}$ and $32{x}^{2}{y}^{3}$?
## What's next?
To see how we can use these skills to factor polynomials, check out our next article on factoring out the greatest common factor!
## Want to join the conversation?
• I keep mixing up GCF and LCM, anyone has a good way to memorize the difference?
• What is the difference between binomials and monomials.
• A binomial has 2 terms (2 items being added or subtracted).
Examples: 3x^3y + 6xy; 7 - 5y
A monomial has 1 term.
Examples of monomials: 4; 5ab^2; 7x/8
Hope this helps.
• I still don't understand the difference between monomials and polynomials. When you add 3x+3x it's 6x. That's not a polynomial. That's a monomial!
• I'm not exactly sure what your asking but I know this much:
A monomial is a one term polynomial.
A bionomial is a two term polynomial.
A trinomial is a three term polynomial.
• Why is it so hard to understand GCF when its used in a lot of ways
• IIt is hard to learn at first, because you are not used to looking at two different numbers and instead of t usual math operations, such as adding subtracting, your breaking them down. into smaller numbers and finding out what they have in common.(AKA greatest common factor)
• Why do we have to show 3 factors multiplied together, for example, one of the example questions I got wrong was because I showed the factors of 12 by 6 by 2, but the correct way of factoring the 12 is by 2 by 2 by 3. Why?
• You need to keep factoring until you get the prime factors of the numbers, so you can easily find the gcf of various numbers.
• what is the gcf 4(12)+4(8)
• Multiply: 48 + 32
Now find the GCF.
Or, factors the numbers down to prime factors and find all the common factors.
The GCF = 16
• What is the difference between binomials and monomials
• A monomial is a polynomial with one term (such as x or 3 or y^2). A binomial is a polynomial with two terms (such as 3x + 2 or x^2 + 3x). A trinomial is next with 3 terms (x^2+4x+5).
• When factoring, are you able to have more than two terms? for example 24x3, you could write that as 12x x 2x2 but you could also write it as 3x x 4x x 2x. Does it still work in that case?
• Notice that in your examples you can still go further:
12 x 2 becomes 6x2 x 2, which in turn becomes:
3x2 x 2 x 2
Which is the stopping point, because every number is prime. Prime numbers can't be factored out further since they aren't divisible by anything other than 1 and themselves.
3 x 4 x 2 => 4 is not a prime, we can factor it out further:
3 x 2x2 x 2
As you can see, we arrived at the same expression. Hope that helps!
• When factoring the coefficient, do you use the prime factorization or the GCF of those two numbers? |
# 2.2 Factorisation of Quadratic Expression
(A) Factorisation quadratic expressions of the form
ax2 + bx + c, b = 0 or c = 0
1. Factorisation of quadratic expressions is a process of finding two linear expressions whose product is the same as the quadratic expression.
2. Quadratic expressions ax2 + c and ax2 + bx that consist of two terms can be factorised by finding the common factors for both terms.
Example 1:
Factorise each of the following:
(a) 2x2+ 6
(b) 7p2– 3p
(c) 6x2– 9x
Solution:
(a) 2x2+ 6 = 2 (x2 + 3) ← (2 is common factor)
(b) 7p2– 3p = p (7p – 3) ← (p is common factor)
(c) 6x2– 9x = 3x (2x – 3) ← (3x is common factor)
(B) Factorisation of quadratic expressions in the form ax2c , where a and c are perfect squares
Example 2:
(a) 9p2– 16
(b) 25x2– 1
(c) $\frac{1}{4}-\frac{1}{25}{x}^{2}$
Solution:
(a) 9p2– 16 = (3p)2 – 42= (3p – 4) (3p + 4)
(b) 25x2– 1 = (5x)2 – 12= (5x – 1) (5x + 1)
(c)
$\begin{array}{l}\frac{1}{4}-\frac{1}{25}{x}^{2}={\left(\frac{1}{2}\right)}^{2}-{\left(\frac{1}{5}x\right)}^{2}\\ \text{}=\left(\frac{1}{2}-\frac{1}{5}x\right)\left(\frac{1}{2}+\frac{1}{5}x\right)\end{array}$
(C) Factorisation quadratic expressions in the form ax2 + bx + c, where a ≠ 0, b ≠ 0 and c ≠ 0
Example 3:
Factorise each of the following
(a) 3y2+ 2y – 8
(b) 4x2– 12x + 9
Solution:
(a)
Factorise using the Cross Method
3y2+ 2y – 8 = (3y – 4) (y + 2)
(b)
4x2– 12x + 9 = (2x – 3) (2x – 3) |
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