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Mister Exam
# Differential equation ylnydx+xdy=0
y() =
y'() =
y''() =
y'''() =
y''''() =
from to
### The solution
You have entered [src]
d
x*--(y(x)) + log(y(x))*y(x) = 0
dx
$$x \frac{d}{d x} y{\left(x \right)} + y{\left(x \right)} \log{\left(y{\left(x \right)} \right)} = 0$$
x*y' + y*log(y) = 0
Detail solution
Given the equation:
$$x \frac{d}{d x} y{\left(x \right)} + y{\left(x \right)} \log{\left(y{\left(x \right)} \right)} = 0$$
This differential equation has the form:
f1(x)*g1(y)*y' = f2(x)*g2(y),
where
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(y \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = - \frac{1}{x}$$
$$\operatorname{g_{2}}{\left(y \right)} = y{\left(x \right)} \log{\left(y{\left(x \right)} \right)}$$
We give the equation to the form:
g1(y)/g2(y)*y'= f2(x)/f1(x).
Divide both parts of the equation by g2(y)
$$y{\left(x \right)} \log{\left(y{\left(x \right)} \right)}$$
we get
$$\frac{\frac{d}{d x} y{\left(x \right)}}{y{\left(x \right)} \log{\left(y{\left(x \right)} \right)}} = - \frac{1}{x}$$
We separated the variables x and y.
Now, multiply the both equation sides by dx,
then the equation will be the
$$\frac{dx \frac{d}{d x} y{\left(x \right)}}{y{\left(x \right)} \log{\left(y{\left(x \right)} \right)}} = - \frac{dx}{x}$$
or
$$\frac{dy}{y{\left(x \right)} \log{\left(y{\left(x \right)} \right)}} = - \frac{dx}{x}$$
Take the integrals from the both equation sides:
- the integral of the left side by y,
- the integral of the right side by x.
$$\int \frac{1}{y \log{\left(y \right)}}\, dy = \int \left(- \frac{1}{x}\right)\, dx$$
Detailed solution of the integral with y
Detailed solution of the integral with x
Take this integrals
$$\log{\left(\log{\left(y \right)} \right)} = Const - \log{\left(x \right)}$$
Detailed solution of the equation
We get the simple equation with the unknown variable y.
(Const - it is a constant)
Solution is:
$$\operatorname{y_{1}} = y{\left(x \right)} = e^{\frac{C_{1}}{x}}$$
C1
--
x
y(x) = e
$$y{\left(x \right)} = e^{\frac{C_{1}}{x}}$$
Graph of the Cauchy problem
The classification
separable
1st exact
separable reduced
lie group
separable Integral
1st exact Integral
separable reduced Integral
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 0.6908192813300311)
(-5.555555555555555, 0.5958133031715364)
(-3.333333333333333, 0.4218748991197924)
(-1.1111111111111107, 0.07508464091707924)
(1.1111111111111107, nan)
(3.333333333333334, nan)
(5.555555555555557, nan)
(7.777777777777779, nan)
(10.0, nan)
(10.0, nan)
To see a detailed solution - share to all your student friends
To see a detailed solution,
share to all your student friends: |
# Exercise 1.1
Question 1
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
Sol :
(i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220
Sol :
(ii)196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii) 867 and 255
Sol :
(iii)867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51,
Therefore, HCF of 867 and 255 is 51.
Question 2
Show that any positive odd integer is of the form , or or where q is some integer.
Sol :
Let a be any positive integer and . Then, by Euclid’s algorithm,
for some integer , and r = 0, 1, 2, 3, 4, 5 because
Therefore, or or or or or
Also, , where k1 is a positive integer
, where k2 is an integer
, where k3 is an integer
Clearly, are of the form , where k is an integer.
Therefore, are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form , or ,
or
Question 3
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Sol :
HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid’s algorithm to find the HCF.
The HCF is 8.
Therefore, they can march in 8 columns each.
Question 4
Use Euclid’s division lemma to show that the square of any positive integer is either of form for some integer m.
[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Sol :
Let a be any positive integer and b = 3.
Then for some integer
And r = 0, 1, 2 because
Therefore, or or
Or,
Where are some positive integers
Hence, it can be said that the square of any positive integer is either of the form
Question 5
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m , 9m+1 or 9m+8
Sol :
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
a=3q or 3q+1 or 3q+2
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
a3=(3q)3=27q3=9(3q3)
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1,
a= (3q +1)3
a3 = 27q27q9q + 1
a3 = 9(3q3q+ q) + 1
a9m + 1
Where m is an integer such that m = (3q3q+ q)
Case 3: When a = 3q + 2,
a= (3q +2)3
a3 = 27q54q36q + 8
a3 = 9(3q6q4q) + 8
a9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1 ,
or 9m + 8.
Insert math as
$${}$$ |
Here we are going to see how to factor quadratic equation.A equation which is in the form of ax² + bx + c is known as quadratic equation. Here a,b and c are just numbers.
To factor a quadratic equation, we have to follow the steps given below.
Step 1 :
Multiply the coefficient of x2 by the constant term
Step 2 :
Split this number into two parts and the product of those parts must be equal to this number and simplified value must be equal to the middle term.
Step 3 :
Group them into linear factor
Let us look into some example problems to understand the above concept.
Example 1 :
Factor x² + 7 x + 12
Solution :
Step 1 :
By multiplying the coefficient of x2 by the constant term 12, we get 12.
Step 2 :
Now we need to spit this 12 as two parts, and the product of those parts must be equal to 12 and simplified value must be equal to the middle term (7).
= x2 + 3x + 4x + 12
Step 3 :
Grouping into linear factors
= x (x + 3) + 4 (x + 3)
= (x + 4) (x + 3)
Example 2 :
Factor y² - 16 y + 60
Solution :
Step 1 :
By multiplying the coefficient of y2 by the constant term 60, we get 60.
Step 2 :
Now we need to spit this 60 as two parts, and the product of those parts must be equal to 60 and simplified value must be equal to the middle term (-16).
Since the middle term is negative, both factors will be negative.
= y2 - 10y - 6y + 60
Step 3 :
Grouping into linear factors
= y (y - 10) - 6 (y - 10)
= (y - 10) (y - 6)
Example 3 :
Factor x² + 9x - 22
Solution :
Step 1 :
By multiplying the coefficient of x2 by the constant term -22, we get -22.
Step 2 :
Now we need to spit this -22 as two parts, and the product of those parts must be equal to -22 and simplified value must be equal to the middle term (9).
Since the last term is negative, the factors will be in the combination of positive and negative.
= x2 + 11x - 2x - 22
Step 3 :
Grouping into linear factors
= x (x + 11) - 2 (x + 11)
= (x + 11) (x - 2)
Example 4 :
Factor x² - 2x - 99
Solution :
Step 1 :
By multiplying the coefficient of x2 by the constant term -99, we get -99.
Step 2 :
Now we need to spit this -99 as two parts, and the product of those parts must be equal to -99 and simplified value must be equal to the middle term (-2).
Since the middle and last term are negative, the factors will be in the combination of positive and negative.
= x2 - 11x + 9x - 99
Step 3 :
Grouping into linear factors
= x (x - 11) + 9 (x - 11)
= (x - 11) (x + 9)
Example 5 :
Factor 3x² + 19x + 6
Solution :
Step 1 :
By multiplying the coefficient of x2 by the constant term 18, we get 18.
Step 2 :
Now we need to spit this 18 as two parts, and the product of those parts must be equal to 18 and simplified value must be equal to the middle term (19).
= 3x2 + 1x + 18x + 6
Step 3 :
Grouping into linear factors
= x (3x + 1) + 6 (3x + 1)
= (x + 6) (3x + 1)
Example 6 :
Factor 9x² - 16x + 7
Solution :
Step 1 :
By multiplying the coefficient of x2 by the constant term 63, we get 63.
Step 2 :
Now we need to spit this 63 as two parts, and the product of those parts must be equal to 63 and simplified value must be equal to the middle term (-16).
Since the middle term is negative, both factors will be negative.
= 9x2 - 9x - 7x + 7
Step 3 :
Grouping into linear factors
= 9x (x - 1) - 7 (x - 1)
= (x - 1) (9x - 7)
Example 7 :
Factor 2x² + 17x - 30
Solution :
Step 1 :
By multiplying the coefficient of x2 by the constant term -30, we get -30.
Step 2 :
Now we need to spit this -30 as two parts, and the product of those parts must be equal to -30 and simplified value must be equal to the middle term (17).
Since the last term is negative, the factors will be in the combination of positive and negative.
= 2x2 + 20x - 3x - 30
Step 3 :
Grouping into linear factors
= 2x (x + 10) - 3 (x + 10)
= (x + 10) (2x - 3)
Let us see the next example on "Factoring quadratics"
Example 8 :
Factor 18x² - x - 4
Solution :
Step 1 :
By multiplying the coefficient of x2 by the constant term -72, we get -72.
Step 2 :
Now we need to spit this -72 as two parts, and the product of those parts must be equal to -72 and simplified value must be equal to the middle term (-1).
Since the middle and last term are negative, the factors will be in the combination of positive and negative.
= 18x2 - 9x + 8x - 4
Step 3 :
Grouping into linear factors
= 9x (2x - 1) + 4 (2x - 1)
= (2x - 1) (9x + 4)
## Related topics
After having gone through the stuff given above, we hope that the students would have understood "Factoring quadratics"
If you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
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Algebra word problems
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Word problems on comparing rates
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Word problems on fractions
Word problems on mixed fractrions
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Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
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OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
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Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
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 |
General Solution to Linear Difference Equation: Is this correct
Notation:
Let $$Q_1[f(x)] = \lim _{h \rightarrow 1} \frac{f(x + h) - f(x) }{h}$$
And let
$$Q_1^{-1}\left[f(x)\right] = G(x)|Q_1\left[g(x)\right] = f(x)$$
Consider the equation
$$a_0(x) + a_1(x)f(x) + a_2(x)Q_1\left[f(x)\right] = 0$$
How do we express the general solution to this equation?
A natural first step is to divide the equation by $a_2(x)$ to find:
$$\frac{a_0(x)}{a_2(x)} + \frac{a_1(x)}{a_2(x)}f(x) + Q_1\left[f(x)\right] = 0$$
We can rename the coefficients (being that they are general) to:
$$b_0(x) + b_1(x) f(x) + Q_1\left[f(x)\right] = 0$$ Now subtract $b_0(x)$ from both sides: (I am attempting a Duhamel's formula style approach)
$$b_1(x)f(x) + Q_1\left[f(x)\right] = -b_0(x)$$
Now note that for a pair of functions there exists a product rule. Given $a(x)$ and $b(x)$
$$Q_1\left[a(x)b(x)\right] = Q_1\left[a(x)\right]Q_1\left[b(x)\right] + Q_1\left[a(x)\right]b(x) + a(x)Q_1\left[b(x)\right]$$
If we let $b(x) = f(x)$ then we note that we obtain a form of:
$$\left(Q_1\left[a(x)\right] + a(x)\right)Q_1\left[f(x)\right] + Q_1\left[a(x)\right]f(x)$$
At this point it is evident that to solve:
$$b_1(x)f(x) + Q_1\left[f(x)\right] = -b_0(x)$$
We need a function $\lambda(x)$ such that
$$\lambda(x)b_1(x)f(x) + \lambda(x)Q_1\left[f(x)\right] = Q_1\left[f(x)a(x)\right]$$
Therefore:
$$\lambda(x)b_1(x) = Q_1\left[a(x)\right]$$
$$\lambda(x) = Q_1\left[a(x)\right] + a(x)$$
Thus:
$$\lambda(x)b_1(x) + a(x) = \lambda(x)$$
Therefore:
$$\lambda(x) = \frac{a(x)}{1 - b_1(x)}$$
And thus both equations breakdown to:
$$\frac{b_1(x)}{1 - b_1(x)} a(x) = Q_1\left[a(x)\right]$$
Let $a(x) = 2^{g(x)}$ then it follows
$$Q_1\left[2^{g(x)}\right] = 2^{g(x)}\left(2^{Q_1\left[g(x)\right]} - 1\right)$$
Thus we are merely tasked with solving
$$\left(2^{Q_1\left[g(x)\right]} - 1\right) = \frac{b_1(x)}{1 - b_1(x)}$$
Which has solution
$$g(x) = Q_1^{-1}\left[\log_2\left\lbrace\frac{1}{1 - b_1(x)}\right\rbrace\right]$$
Thus the general solution has the form
$$2^\left({Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}\right)f(x) = Q_1^{-1}\left[-b_0(x)*\frac{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}}{1 - b_1(x)}\right]$$
Therefore
$$f = \frac{Q_1^{-1}\left[-b_0(x)*\frac{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}}{1 - b_1(x)}\right]}{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}}$$
Simplifying and substituting $\frac{a_0(x)}{a_2(x)}$ for $b_0(x)$ and $\frac{a_1(x)}{a_2(x)}$ for $b_1(x)$
We conclude that the general solution to:
$$a_0(x) + a_1(x)f(x) + a_2(x)Q_1\left[f(x)\right] = 0$$
is:
$$f = \frac{Q_1^{-1}\left[\frac{a_0(x)}{a_2(x)}*\frac{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - \frac{a_1(x)}{a_2(x)}}\right)\right]}}{\frac{a_1(x)}{a_2(x)} - 1}\right]}{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - \frac{a_1(x)}{a_2(x)}}\right)\right]}}$$
I think this is correct. But a second eye is always helpful
• *on a side note how do you make equations bigger as well as create left and right parenthesis that engulf the ENTIRE expression between them, resizing if need be? – frogeyedpeas Jul 8 '14 at 0:15
• For parentheses, try \left( and \right) (or \left[ and \right] or \left\{ and \right\}. – Mark Fischler Jul 8 '14 at 0:28
• I added the \left and \right for you but where you have powers of 2, it looks just as confusing (and you can't add any more brackets due to how many are there) – Jam Jul 8 '14 at 0:52
• thanks man! I think its an improvement from before – frogeyedpeas Jul 8 '14 at 0:56 |
Year 8 Interactive Maths - Second Edition
## Geometry
Geometry is used in astronomy, navigation, surveying and many practical occupations. Surveyors, architects and engineers often use geometrical drawings. Navigation by air and sea depends on accurate geometry.
In Year 7, some geometrical facts were considered. In this section, these basic facts will be revised as they will be used to learn further geometry and develop logical reasoning.
###### Recall that:
• The sum of adjacent angles forming a straight line is 180º.
• If the sum of two angles adds up to 180º, then the angles are said to be supplementary.
The supplement of a is b because a + b = 180º.
The supplement of b is a because a + b = 180º.
• If the sum of two angles adds up to 90º, then the angles are said to be complementary.
The complement of x is y because x + y = 90º.
The complement of y is x because x + y = 90º.
### Method for Finding the Value of a Pronumeral
To find the value of a pronumeral from given information, write an equation. Then solve the equation and write the answer in words.
#### Example 1
Use the information given in the diagram to find x. Give reasons for your answer.
##### Solution:
So, the value of x is 131º.
#### Example 2
Use the information given in the diagram to find x. Give reasons for your answer.
##### Solution:
So, the value of x is 36º.
#### Example 3
What is the complement of 42º?
##### Solution:
Let xº be the complement of 42º.
So, the complement of 42º is 48º.
#### Example 4
What is the supplement of 48º?
##### Solution:
Let xº be the supplement of 48º.
So, the supplement of 48º is 132º. |
# Fractions - Converting Mixed Number To Improper Fraction Calculator
## Converting Mixed Number To Improper Fraction Calculator
Let's investigate how to convert a Mixed Number To Improper Fraction. Mixed Numbers have two parts: an Integer number and a Fraction of a number. For example: 31/4. Below are the rules that define a correct mixed number.
• As stated, a mixed number must have an integer portion and a fraction portion.
• The integer portion of a mixed number can not be zero.
• The numerator of the fraction portion of the mixed number can not be zero.
• The denominator of the fraction portion of the mixed number can not be zero. Division by zero is undefined!
• The denominator of a fraction must be greater than numerator. In other word, it must be a Proper Fraction.
So how do we convert a Mixed Number to an Improper Fraction? Follow these simple steps:
• ## Step 1:
Multiply the integer number by the fraction's denominator.
• ## Step 2:
Take the product of the multiplication in Step 1, and add the fraction's numerator to it.
• ## Step 3:
The result of Step 2 is the numerator of the improper fraction and the fraction denominator is used for the improper fraction's denominator.
$\text{Integer} + \frac{\mathrm{Numerator}}{\mathrm{Denominator}} =$
$\frac{\mathrm{\left(Integer\right) x \left(Denominator\right) + Numerator}}{\mathrm{Denominator}}$
$\text{Example: }2\frac{3}{7} = \frac{\mathrm{2 x 7 + 3}}{7} = \frac{\mathrm{17}}{7}$
Please Read! These are instructions to input mixed number into the calculator: if you want to enter a whole number and a fraction (a mixed number) there must be a space between the integer and the fraction - for example type "13 1/2".
Mixed Number To Improper Fraction Calculator
Enter Mixed Number Here
Then Press The SUBMIT Button |
## Saturday, August 12, 2017
### Geometry: The Intersection Point of a Quadrilateral
Geometry: The Intersection Point of a Quadrilateral
The Setup
We are given four points, A, B, C, and D, as four Cartesian coordinates. We connect the four points, starting with A, in a clockwise form to form a quadrilateral. We will designate each points as the coordinates:
A: (ax, ay)
B: (bx, by)
C: (cx, cy)
D: (dx, dy)
Draw a line from one corner to the opposite corner. This results in two lines: AC and DB. The two lines (show above in lime green) meet at point P. The goal is determine the coordinates of P.
Slope
We know the equation of the line is y = m*x + b, where m is the slope and b is the y-intercept. Note that:
y = m*x + b
y – m*x = b
In geometry, the slope of a line containing two points is generally defined as:
m = (change in y coordinates)/(change in x coordinates) = Δy/Δx = (y2 – y1)/(x2 – x1)
The slope of AC: SAC = (cy – ay)/(cx – ax)
The slope of BD: SBD = (dy – by)/(dx –bx)
The Intercept
We can easily deduce that solving the general equation of the line y = m*x + b for the intercept yields b = y – m*x. If follows that:
The intercept of AC: IAC = cy – SAC * cx = ay – SAC *ax
The intercept of BD: IBD = by – SBD * bx = dy – IBD *dx
Finding the Intersection Point
Now that the slopes and intercepts are determined, we can form the following system of equations:
(I) y = SAC * x + IAC
(II) y = SBD * x + IBD
Solving for x and y will find our intersection point P (px, py). Subtracting (II) from (I) (see above):
0 = (SAC – SBD) * x + (IAC – IBD)
We can solve for x.
-(IAC – IBD) = (SAC – SBD) * x
(-1*IAC - -1*IBD) = (SAC – SBD) * x
(-IAC + IBD) = (SAC – SBD) * x
(IBD – IAC) = (SAC – SBD) * x
Hence:
x = px = (IBD – IAC)/(SAC – SBD)
It follows that we determine y by either equation (I) or (II):
y = py = SAC * px + IAC = SBD * px + IBD
Our desired coordinates of point P are found.
Eddie
This blog is property of Edward Shore, 2017.
x
### Numworks: Allowing Repeated Calculations in Python
Numworks: Allowing Repeated Calculations in Python Introduction Say we want the user to repeat a calculation or a routine for as lo... |
# Progression through the teaching of addition and subtraction
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## Transcription
1 Progression through the teaching of addition and subtraction
2 Maths has changed! The maths work your child is doing at school may look very different to the kind of sums you remember. This is because children are encouraged to work mentally, where possible, using personal jottings to help support their thinking. Formal calculations are introduced from Year 3 onwards. Children are then encouraged to use these methods for calculations they cannot solve in their heads.
3 Parents Meeting on: Progression through Calculations When faced with a problem, we want children to ask themselves. Do I need jottings? Shall I use a pencil and paper method? Can I do it in my head? Do I need to use a calculator? Lancashire Mathematics Team
4 How would you solve these calculations? = = = = 27 5 = = =
5 Laying the foundations for addition o Partitioning and subtraction o Rounding o Compensating o Counting on and back o Bridging through 10s, 100s, 1000s boundaries o Addition and subtraction facts Lancashire Mathematics Team
6 Addition + THE FOLLOWING ARE STANDARDS THAT WE EXPECT THE MAJORITY OF CHILDREN TO ACHIEVE. Reception and Year 1 Children are encouraged to develop a mental picture of the number system in their heads to use for calculation. They develop ways of recording calculations using pictures.
7 A number line is just a picture of how we work out some calculations in our heads!
8 They use number lines and practical resources to support calculation and teachers demonstrate the use of the number line = Year 1 Children then begin to use number lines to support their own calculations using a numbered line to count on in ones = Bead strings or bead bars can be used to illustrate addition including bridging through ten by counting on 2 then counting on 3.
9 Year 2 Children will begin to use empty number lines themselves starting with the larger number and counting on. First counting on in tens and ones = Then helping children to become more efficient by adding the units in one jump (by using the known fact = 7) = Followed by adding the tens in one jump and the units in one jump.
11 Year 3 Children will continue to use empty number lines with increasingly large numbers, including compensation where appropriate. Count on from the largest number irrespective of the order of the calculation =
12 Compensation = Year Children will begin to use informal pencil and paper methods (jottings) to support, record and explain mental methods building on existing mental strategies. Stage 1: Adding the most significant digits first, then moving to adding least significant digits ( ) 200 ( ) 11 (7 + 4) 140 (60 +80) (7 + 5) 352
13 Year 3 Moving to adding the least significant digits first in preparation for carrying (7 + 4) 12 (7+5) 80 ( ) 140 (60+80) (200+0) 352
14 Your turn! = (2 + 6) ( ) 118
15
16 Year 4 Children will in Year 4 be introduced to carrying above the line Using similar methods, children will: Add several numbers with different numbers of digits; Begin to add 2 or more 3-digit sums of money, with or without adjustment from pence to pounds; Know that the decimal points should line up under each other, particularly when adding or subtracting mixed amounts, e.g p
17 Year 5 Following formal addition methods with carrying above the line being introduced at Year 4, children should at Year 5, extend the carrying method to numbers with at least four digits Using similar methods, children will: Children would use rounding to estimate the answer to the calculation. So is about , which is approximately Add several numbers with different numbers of digits; Begin to add two or more decimal fractions with up to three digits and the same number of decimal points Know that decimal points should line up under each other, particularly when adding or subtracting mixed amounts, e.g 3.2m 280cm
18 Year 6 Children should extend the carrying method to numbers with any number of digits Using similar methods, children will: Add several numbers with different numbers of digits; Begin to add two or more decimal fractions with up to four digits and either one or two decimal places; Know that decimal points should line up under each other, particularly when adding or subtracting mixed amounts, e.g
19 Add Plus Altogether Addition Total Count on Increase Sum Make Vocabulary.
20 Subtraction - THE FOLLOWING ARE STANDARDS THAT WE EXPECT THE MAJORITY OF CHILDREN TO ACHIEVE. Reception and Year 1 Children are encouraged to develop a mental picture of the number system in their heads to use for calculation. They develop ways of recording calculations using pictures etc.
21 There are five frogs. If 2 frogs jumped into the lake how many would be left?
22 Year 1 They use number lines and practical resources to support calculation. Teachers demonstrate the use of the number line. 6 3 = The number line should also be used to show that 6-3 means the difference between 6 and 3 or the difference between 3 and 6 and how many jumps they are apart
23 Year 1 Children then begin to use numbered lines to support their own calculations - using a numbered line to count back in ones = Bead strings or bead bars can be used to illustrate subtraction including bridging through ten by counting back 3 then counting back = 8
24 Year 2 Children will begin to use empty number lines to support calculations. Counting back First counting back in tens and ones =
25 Then helping children to become more efficient by subtracting the units in one jump (by using the known fact 7 3 = 4) = Year Subtracting the tens in one jump and the units in one jump
27 Counting on If the numbers involved in the calculation are close together or near to multiples of 10, 100 etc, it can be more efficient to count on = Year =
28 Year 3 Children continue to use empty number lines with increasingly large numbers. Children are then taught this expanded method using partitioning. 89 = = 32 Initially, the children are taught using examples that do not need the children to exchange.
29 Year 3 From this the children move to exchanging; 71 = - 46 Step 1: Step 2: = 25 This would be recorded by the children as = 25
31 Year 4 Partitioning and decomposition 754 = 86 Step Step (adjust from T to U) Step (adjust from H to T) = 668
32 This would be recorded by the children as: = 668 Year 4 As decomposition this would look like this:
33 Common calculation errors!
34 Year 4 Children should: Be able to subtract numbers with different numbers of digits; Using this method, children should also begin to find the difference between 2 3-digit sums of money, with or without adjustment from the pence to pounds; Know that decimal points should line up under each other. For example: 8.95 = leading to = Alternatively, children can set the amounts to whole numbers, i.e and convert to pounds after the calculation.
35 Year 4 Where the numbers are involved in the calculation are close together or near to multiples of 10, 100 etc. counting on using a number line should be used =
36 Year 5 The teaching of subtraction continues from Year 4 where an expanded method will have been introduced. This expanded method uses partitioning Step 1: 754 = Step 2: (adjusting from T to U) Step 3: (adjusting from H to T) = This would be recorded by the children as = 468
37 Year 5 Decomposition Children would use rounding to estimate the answer to the calculation. So is about , which is approximately 500. Children should: Be able to subtract numbers with different numbers of digits Begin to find the number between two decimal fractions with up to three digits and the same number of decimal places this could be in the context of money or measures Know that decimal points should line up under each other.
38 Year 5 Where the numbers are involved in the calculation are close together or near to multiples of 10, 100 etc. counting on using a number line should be used =
39 Year 6 Decomposition Now using 4 digit numbers and beyond. Children should: be able to subtract numbers with different numbers of digits; Be able to subtract two or more decimal fractions with up to three digits and either one or two decimal places; this could be in the context of money or measures know that decimal points should line up under each other.
40 Year 6 Where the numbers are involved in the calculation are close together or near to multiples of 10, 100 etc. counting on using a number line should be used =
41
42 Subtraction Vocabulary How many are left? Take (away) Difference between How many have gone? Decrease 1 less How many fewer is... than less How many are left over? Count back
43 Key messages Children need to develop skills such as counting, partitioning and recombining numbers They need to build an awareness of the number system, value of numbers and number relationships They need to recall facts such as halving and doubling, number bonds and multiplication facts From all of these they learn to construct strategies that they can apply in many different areas. The questions at the forefront of their minds: Can I do it in my head? If not which method will help me?
44 Thank you for attending our workshop on the progression through addition and subtraction.
### Calculations at Holy Cross. Addition and Subtraction
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### Addition Subtraction Multiplication Division. size. The children develop ways of 5s and 10s. recording calculations using pictures etc.
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### PROGRESSION THROUGH CALCULATIONS FOR SUBTRACTION
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### Product of two natural numbers
YAP Von Bing
March 9, 2022
Let $$a$$ and $$b$$ be natural numbers, i.e., 1, 2, 3, etc. Let their product be $$n = ab$$ . You might agree that $$n$$ cannot be smaller than $$a$$ or $$b$$ , i.e., $$a \le n$$ and $$b \le n$$ . This is easy enough to see in numerous examples. $$2 \times 3 = 6$$ , and 6 is greater than both 2 and 3. $$1 \times 5 = 5$$ , and both 1 and 5 are less than or equal to the product 5. The second example shows that generally the inequality is not strict, i.e., we do not always have $$a < n$$ . How can we demonstrate the general fact, or $${\bf theorem}$$ ? How can we prove that for any natural numbers $$a$$ and $$b$$ with product $$n = ab$$ , both $$a \le n$$ and $$b \le n$$ hold? We cannot check every case, because there are infinitely many natural numbers.
Here is a visual approach. $$n$$ identical squares can be arranged in a rectangle with $$a$$ rows, each having $$b$$ squares. Clearly, the total $$n$$ cannot be less than the number of squares in a column, $$a$$ , or the number of squares in a row, $$b$$ . We may translate the pictorial argument into algebra.
$$n = ab = a + a(b-1) \ge a$$
The second equality is key, which reflects the separation of a column from the others. Since $$b-1 \ge 0$$ , $$a(b-1) \ge 0$$ , hence we are certain that $$n \ge a$$ . Both the picture and the algebra also show that in the special case $$b > 1$$ , indeed $$n > a$$ . You might want to make a similar argument for $$n \ge b$$ .
Let us suppose that $$a \le b$$ . Can you say something stronger than $$a \le n$$ ? For instance, can you find an expression $f(n)$ in terms of $$n$$ , such that $$a \le f(n) \le n$$ ? You might get a clue from concrete examples:
$$16 = 1 \times 16 = 2 \times 8 = 4 \times 4,$$
$$24 = 1 \times 24 = 2 \times 12 = 3 \times 8 = 4 \times 6$$
or the special case $$a=b$$ , where $$a = \sqrt{n}$$ .
It makes sense to try $$f(n) = \sqrt{n}$$ . Our examples work:
$$1, 2, 4 \le 4 = \sqrt{16}$$
$$1, 2, 3, 4 \le 4.9 \approx \sqrt{24}$$
Hence we are more confident of the $${\bf conjecture}: \$$
Let $$a \le b$$ be any natural numbers. Let $$n = ab$$ . Then $$a \le \sqrt{n}$$ .
We now give a proof of the conjecture. Suppose the first two sentences hold but the conclusion is false, i.e., $$a > \sqrt{n}$$ . Combining it with the first sentence gives $$b \ge a > \sqrt{n}$$ . Multiplying the two inequalities, we get $$ab > n$$ , which violates the second sentence. Assuming the conclusion is false has led to an absurdity. Therefore, the assumption is wrong, and the conclusion indeed follows. This concludes the proof. Now that the conjecture has been proved, it becomes a theorem.
The above is an example of $${\bf proof by contradiction}$$ , which was used by the ancient Greeks to prove a variety of facts about numbers, such as there are infinitely many prime numbers, and the more surprising fact that $$\sqrt{2}$$ is irrational, i.e., it cannot be expressed as the ratio of two natural numbers. The proofs of these facts are more intricate, but the underlying sturcture is exactly the same as here. In order to show that a conclusion follows from a premise, we assume that the conclusion is false. Then we combine the premise with the assumption logically to derive a false statement. Something has gone wrong. Since the derivation is logically correct, we must give the assumption up. Hence the conclusion must follow from the premise.
As practice, you may want to write down a proof by contradiction to show that with the same premise as above, $$b \ge \sqrt{n}$$ . This fact also follows directly from the previous fact. From $$a \le \sqrt{n}$$ , we know $$1/a \ge 1/\sqrt{n}$$ . Hence
$$b = n \times \frac{1}{a} \ge n \times \frac{1}{\sqrt{n}} = \sqrt{n}$$
We saw earlier that if $$a=b$$ , then both inequalities become equalities. Furthermore, if $$a<b$$ , then both inequalities are strict. In conclusion, we may make a more complete statement as follows:
$${\bf Theorem.}$$ Let $$a \le b$$ be any natural numbers, and let $$n = ab$$ . Then
$$a \le \sqrt{n} \le b$$
Moreover, if $$a<b$$ , then $$a < \sqrt{n} < b$$ . Otherwise, $$a=b = \sqrt{n}$$ .
Notice that the conjecture and theorem are general statements, i.e., they are about all possible values of $$a$$ and $$b$$ , as long as the values are natural numbers. The phrase “any natural numbers” should serve as a reminder of the generality. You may want to formulate analogous conjectures about three natural numbers $$a, b, c$$ and their product $$abc$$ , given $$a \le b \le c$$ , and then try to prove or disprove them. |
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# Solving Two Equations in Two Unknowns
Alignments to Content Standards: A-REI.C.5
Lisa is working with the system of equations $x + 2y = 7$ and $2x - 5y = 5$. She multiplies the first equation by 2 and then subtracts the second equation to find $9y = 9$, telling her that $y = 1$. Lisa then finds that $x = 5$. Thinking about this procedure, Lisa wonders
There are lots of ways I could go about solving this problem. I could add 5 times the first equation and twice the second or I could multiply the first equation by -2 and add the second. I seem to find that there is only one solution to the two equations but I wonder if I will get the same solution if I use a different method?
1. What is the answer to Lisa's question? Explain.
2. Does the answer to (a) change if we have a system of two equations in two unknowns with no solutions? What if there are infinitely many solutions?
## IM Commentary
The goal of this task is to help students see the validity of the elimination method for solving systems of two equations in two unknowns. That is, the new system of equations produced by the method has the same solution(s) as the initial system. This is a subtle and vital point, though students should already be familiar with implementing this procedure before working on this task.
It is not difficult to verify that a solution to the initial system of equations is also a solution to the new system. The key to the success of the elimination method, however, is that all steps in the algorithm are reversible. This is why a solution to the simpler system of equations is also a solution to the original system. This can be seen geometrically when taking a multiple of an equation since, for example, $x + 2y = 7$ and $2x + 4y = 14$ define the same line in the plane. The geometric intuition is lost, however, when equations are added or subtracted as this creates a new line, having the same point of intersection with the line defined by $2x - 5y = 5$.
## Solution
1. Fortunately the answer to Lisa's question is yes, she will get the same answer regardless of which (non-zero) multiples of the equations she takes in order to eliminate a variable. The key fact that makes this work is that the process is reversible so she can also manipulate the new, simpler system ($2x-5y=5$ and $9y=9$) to get back the original system. Concretely, suppose $x=a, y=b$ is a solution to $x+2y=7$ and $2x-5y=5$: for these equations, this means that $a = 5$ and $b = 1$ but this is not essential for the reasoning. So $2a-5b=5$ and $a+2b=7$. Lisa's first step is to multiply $x+2y=7$ by 2 to get $2x+4y=14$. Since $a+2b=7$ we also have $2a+4b=14$. The second step is to subtract $2x-5y=5$ from $2x+4y=14$ giving the equation $9y=9$. Since $2a-b=5$ and $2a+b=14$ the same reasoning applies to $a$ and $b$. We have just shown that if $x=a$ and $y=b$ is a solution to $2x-5y=5$ and $x+2y=7$ then it is also a solution to $9y=9$ and $x+2y=7$.
The same reasoning can be applied going back in the other direction. Briefly, if $x=a$ and $y=b$ is a solution to the pair of equations $9y=9$ and $x+2y=7$, then we can multiply the second equation by 2 and subtract the first equation to give $2x-5y=5$ and $x=a$ and $y=b$ will be a solution to this as well as to $x+2y=7$. This means that the original equations have the same solutions as the simpler, modified version. This argument will apply no matter how Lisa manipulates the equations, as long as the steps are reversible, that is as long as she does not multiply by 0 (undoing this would mean dividing by 0 which is not defined).
2. The method of elimination of variables works for any system of two equations in two unknowns. The reasoning in part (a) applies to any solution $x=a$ and $y=b$ to a system of two linear equations in two unknowns. If there are infinitely many solutions to one set of equations there will be infinitely many solutions to a new set obtained by adding and multiplying the original two equations (as long as the multiple is not 0). Similarly, if there are no solutions to a set of two linear equations in two unknowns there will also be no solutions to a new set obtained by adding and multiplying these (if there were a solution, it would solve the original two equations when the process is reversed as seen in the solution to part (a)). |
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## Equations and inequalities: solving quadratic equations
A quadratic equation is an equation where the power of the variable is at most 2. The following are examples of quadratic equations.
$\begin{array}{ccc}\hfill 2{x}^{2}+2x& =& 1\hfill \\ \hfill \frac{2-x}{3x+1}& =& 2x\hfill \\ \hfill \frac{4}{3}x-6& =& 7{x}^{2}+2\hfill \end{array}$
Quadratic equations differ from linear equations by the fact that a linear equation only has one solution, while a quadratic equation has at most two solutions. However, there are some special situations when a quadratic equation only has one solution.
We solve quadratic equations by factorisation, that is writing the quadratic as a product of two expressions in brackets. For example, we know that:
$\left(x+1\right)\left(2x-3\right)=2{x}^{2}-x-3.$
In order to solve:
$2{x}^{2}-x-3=0$
we need to be able to write $2{x}^{2}-x-3$ as $\left(x+1\right)\left(2x-3\right)$ , which we already know how to do. The reason for equating to zero and factoring is that if we attempt to solve it in a 'normal' way, we may miss one of the solutions. On the other hand, if we have the (non-linear) equation $f\left(x\right)g\left(x\right)=0$ , for some functions $f$ and $g$ , we know that the solution is $f\left(x\right)=0$ OR $g\left(x\right)=0$ , which allows us to find BOTH solutions (or know that there is only one solution if it turns out that $f=g$ ).
## Investigation : factorising a quadratic
1. $x+{x}^{2}$
2. ${x}^{2}+1+2x$
3. ${x}^{2}-4x+5$
4. $16{x}^{2}-9$
5. $4{x}^{2}+4x+1$
Being able to factorise a quadratic means that you are one step away from solving a quadratic equation. For example, ${x}^{2}-3x-2=0$ can be written as $\left(x-1\right)\left(x-2\right)=0$ . This means that both $x-1=0$ and $x-2=0$ , which gives $x=1$ and $x=2$ as the two solutions to the quadratic equation ${x}^{2}-3x-2=0$ .
1. First divide the entire equation by any common factor of the coefficients, so as to obtain an equation of the form $a{x}^{2}+bx+c=0$ where $a$ , $b$ and $c$ have no common factors. For example, $2{x}^{2}+4x+2=0$ can be written as ${x}^{2}+2x+1=0$ by dividing by 2.
2. Write $a{x}^{2}+bx+c$ in terms of its factors $\left(rx+s\right)\left(ux+v\right)$ . This means $\left(rx+s\right)\left(ux+v\right)=0$ .
3. Once writing the equation in the form $\left(rx+s\right)\left(ux+v\right)=0$ , it then follows that the two solutions are $x=-\frac{s}{r}$ or $x=-\frac{u}{v}$ .
4. For each solution substitute the value into the original equation to check whether it is valid
There are two solutions to a quadratic equation, because any one of the values can solve the equation.
Solve for $x$ : $3{x}^{2}+2x-1=0$
1. As we have seen the factors of $3{x}^{2}+2x-1$ are $\left(x+1\right)$ and $\left(3x-1\right)$ .
2. $\left(x+1\right)\left(3x-1\right)=0$
3. We have
$x+1=0$
or
$3x-1=0$
Therefore, $x=-1$ or $x=\frac{1}{3}$ .
4. We substitute the answers back into the original equation and for both answers we find that the equation is true.
5. $3{x}^{2}+2x-1=0$ for $x=-1$ or $x=\frac{1}{3}$ .
Sometimes an equation might not look like a quadratic at first glance but turns into one with a simple operation or two. Remember that you have to do the same operation on both sides of the equation for it to remain true.
You might need to do one (or a combination) of:
• For example,
$\begin{array}{ccc}\hfill ax+b& =& \frac{c}{x}\hfill \\ \hfill x\left(ax+b\right)& =& x\left(\frac{c}{x}\right)\hfill \\ \hfill a{x}^{2}+bx& =& c\hfill \end{array}$
• This is raising both sides to the power of $-1$ . For example,
$\begin{array}{ccc}\hfill \frac{1}{a{x}^{2}+bx}& =& c\hfill \\ \hfill {\left(\frac{1}{a{x}^{2}+bx}\right)}^{-1}& =& {\left(c\right)}^{-1}\hfill \\ \hfill \frac{a{x}^{2}+bx}{1}& =& \frac{1}{c}\hfill \\ \hfill a{x}^{2}+bx& =& \frac{1}{c}\hfill \end{array}$
• This is raising both sides to the power of 2. For example,
$\begin{array}{ccc}\hfill \sqrt{a{x}^{2}+bx}& =& c\hfill \\ \hfill {\left(\sqrt{a{x}^{2}+bx}\right)}^{2}& =& {c}^{2}\hfill \\ \hfill a{x}^{2}+bx& =& {c}^{2}\hfill \end{array}$
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
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# The distance covered for a race is 1km. When A and B run the race, A wins by 40 meters when A and C run, then A wins by 70 meters. When B and C run(a) B wins by 110m(b) C wins by 100m(c) B wins by 314m(d) C wins by 50m
Last updated date: 19th Jul 2024
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Hint: For solving this question you should know about the calculating distance by comparisons and ratios. In this problem we will find the comparison ratio between A and B and then we find the same distance ratio for A and C. And then with the relationship ratio of these we find the distance for which B and C run for winning.
According to our question it is given that the distance covered for a race is 1km. When A and B run the race, A wins by 40 meters, when A and C run, then A wins by 70 meters. Then find the right option for B and C run.
As we know that if two or more than two parameters are given to calculate any third parameter on the same scale then we find the third one by comparing and by taking ratios of those with one another.
So, if we see our question, it is given that if A and B run then A wins by 40 meters.
So, we can write it as,
$A:B=1000:960$
And when A and C run then A wins by 70 meters.
So, we can write $A:C=1000:930$
So, when B and C run,
B win by C $=\dfrac{960-930}{960}\times 1000$
B win by C $=\dfrac{30}{960}\times 1000$
B win by C $=31\dfrac{1}{4}$m
So, the correct answer is “Option c”.
Note: While solving these types of questions you should know about the taking right component with right in the ratios. Because we can only find the right answer if the ratio of two is right. And the comparison of that will be given a final right answer. |
Horizontal And Vertical Tangents For Parametric Curves
Recall that if we have a parametric curve in the form $x = f(t)$ and $y = g(t)$ and we want find a derivative for this curve, dy/dx, then it follows that:
(1)
\begin{align} \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \end{align}
# Horizontal Tangents
Recall that with a function f(x), horizontal tangents arise when f'(x) = 0, that is, the numerator of f'(x) is 0. This is similar for parametric curves.
For a parametric curve with a derivative defined by:
(2)
\begin{align} \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \end{align}
If $\frac{dy}{dt} = 0 \quad \mathbf{and} \quad \frac{dx}{dt} ≠ 0$, then a horizontal tangent exists for the specific value of t that satisfies this condition.
## Example 1
Find any horizontal tangents for the parametric curve defined by $x = 2t$ and $y = t^2$.
This is a rather simple example, but let's first differentiate to obtain dx/dt, and dy/dt:
(3)
Now let's set dy/dt = 0. We thus get $2t = 0$, so t = 0. When t = 0, dx/dt ≠ 0 since dx/dt is just a constant. Hence there exists a horizontal tangent at t = 0. We can plug this into our parametric equations to get the coordinates (0, 0) where this occurs.
# Vertical Tangents
When we have a function f(x), there were no such things as vertical tangents (although there was vertical asymptotes). Nevertheless, parametric curves are not necessarily functions, and vertical tangents can exist. If $\frac{dy}{dx} ≠ 0 \quad \mathbf{and} \quad \frac{dx}{dt} = 0$, then a vertical tangent exists for the specific value of t that satisfies this condition.
## Example 2
Find any vertical tangents for the parametric curve defined by $x = \cos \theta$ and $y = \sin \theta$.
Finding dx/dθ and dy/dθ, we obtain:
(4)
Hence dx/dθ = 0 when θ = 0 and π. Also note that dy/dθ ≠ 0 when θ = 0 and when θ = π, hence there exists two vertical tangents when θ = 0 and when θ = π. We can plug these into our parametric equations to get the xy-coordinates (1, 0) and (-1, 0)
# Exceptions
So what are we supposed to do if when dy/dt = 0, dx/dt = 0, or if when dx/dt = 0, dy/dt = 0? Well we have to use limits to determine what happens at these points.
## Example 3
Find any vertical tangents for the parametric curve defined by $x = r(\theta - \sin \theta)$ and $y = r(1 - \cos \theta)$.
First let's differentiate dx/dθ and dy/dθ to obtain dy/dx.
(5)
Therefore:
(6)
\begin{align} \frac{dy}{dx} = \frac{r \sin \theta}{r - r \cos \theta} \\ = \frac{r \sin \theta}{r(1 - \cos \theta)} \\ = \frac{\sin \theta}{1 - \cos \theta} \end{align}
So dx/dθ = 0 when θ = 0, π, 2π, … But notice that when θ = 0, π, 2π, …, then dy/dθ = 0, since sin(0) = 0, sin(π) = 0, …. So we need to use our knowledge of limits in order to figure out what is happening here. Let's just use 2π as an example,
(7)
\begin{align} \lim_{\theta \to 2\pi} \frac{dy}{dx} = \lim_{\theta \to 2\pi} \frac{\sin x}{1 - \cos x} \end{align}
We can use L'Hospital's Rule to evaluate this limit further, first from the right side:
(8)
\begin{align} \lim_{\theta \to 2\pi^+} \frac{\sin x}{1 - \cos x} \\ = \lim_{\theta \to 2\pi^+} \frac{\cos x}{\sin x} \\ = \infty \end{align}
Now let's look at the limit from the left:
(9)
\begin{align} \lim_{\theta \to 2\pi^-} \frac{\sin x}{1 - \cos x} \\ = \lim_{\theta \to 2\pi^+} \frac{\cos x}{\sin x} \\ = -\infty \end{align}
Since 2π exists as a point on our curve, it thus follows that when theta is 2π, a vertical tangent exists. |
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# Solving Systems of Linear Inequalities
Example 1
Graph the system of inequalities.
2x + y -4
6x + 3y > 0
Solution
Step 1 Solve the first inequality for y. Then graph the inequality.
To solve 2x + y -4 for y, subtract 2x from both sides. The result is y - 2x - 4.
To graph y - 2x - 4, first graph the equation y = - 2x - 4.
• Plot the y-intercept (0, -4). Then use the slope, to plot a second point at (1, -6).
For the inequality y - 2x - 4, the inequality symbol is “â€. This stands for “is less than or equal to.â€
• To represent “equal to,†draw a solid line through (0, - 4) and (1, - 6).
• To represent “less than,†shade the region below the line.
Step 2 Solve the second inequality for y. Then graph the inequality.
To solve 6x + 3y > 0 for y, do the following:
Subtract 6x from both sides. 3y > - 6x + 0
Divide both sides by 3. y > - 2x + 0
To graph y > - 2x + 0, first graph the equation y = 2x + 0.
• Plot the y-intercept (0, 0). Then use the slope, , to plot a second point at (1, -2).
For the inequality y > - 2x, the inequality symbol is “>â€. This stands for “is greater than.â€
• Since the inequality symbol “>†does not contain “equal to,†draw a dotted line through (0, 0) and (1, -2).
• To represent “greater than,†shade the region above the line.
Step 3 Shade the region where the two graphs overlap.
The shaded regions do not overlap.
So the system of inequalities has no solution.
Example 2
Graph the system of inequalities.
y x + 5
y < x - 3
Solution
Step 1 Solve the first inequality for y. Then graph.
The graph of y x + 5 is shown.
Step 2 Solve the second inequality for y. Then graph.
The graph of y < x - 3 is shown.
Step 3 Shade the region where the two graphs overlap. The solution is the region below the line y = x - 3 since it is where the graphs overlap.
The solution of the systems is the dark region where the graph overlaps.
Example 3
Given the following system of linear inequalities:
x > - 5
x 3
y < 4
y 2
a. Graph the system.
b. Find the coordinates of three points that satisfy all four inequalities.
c. Find the points of intersection of each pair of corresponding lines.
d. Find the area of the region whose points are the solution of the system.
Solution
a. Graph each inequality and shade the region where the graphs overlap.
b. Each point in the shaded region is a solution of the system. For example, (0, 0), ( -2, 1), and ( -4, 1) each satisfy the system.
c. The points of intersection are the corners of a rectangle.
The points are (3, 4), ( -5, 4), ( -5, -2), and (3, -2).
(Note: Only one corner (3, 2), is also a solution of the system.)
d. To find the area of a rectangle, multiply its length by its width.
The length of the rectangle is 8 units; the width is 6 units.
The area is length width · 8 units · 6 units = 48 square units. |
## Taking Down A Basic Proof
The first time I heard the idea that .999 = 1, it was explained to me like this:
1/3 = .333…
(1/3)3 = (.333…)3
1 = .999…
One divided by 3 equals point 3 repeating. If you multiply both sides by 3, you get 1 equals point 9 repeating.
When I first saw this proof, my mind was blown. I had heard that .999 repeating was exactly equal to 1 rather than almost, but I had never seen it proven. This was proof. I shared this with several people, and I was so proud to be “in the know”. But upon further examination, I see that there are a couple of problems with this seemingly easy math problem as I will demonstrate in this post.
The first is that 1 divided by 3 does not equal point 3 repeating. It’s almost equal. The explanation is long but educational. Bear with me on this one.
When we were kids first learning about division, we were taught a mathematical model that I’ll refer to as the Remainder Model. When we divided 10 by 3, we got “3 with a remainder of 1” (written as 3R1). If you wanted to check the math, you’d do it in reverse by multiplying by the number you divided by and then adding the remainder, so 3(3) + 1 gives you 10. This allowed us to divide uneven numbers exactly.
When we got older we replaced the Remainder Model with the Decimal Model. Instead of stopping at 3R1, we’d drop a zero from behind our original 3 (because it’s actually 3.000…) and then divide our remainder of 1 with a 0 (10) by 3 again, giving us 3.3. Now while we didn’t do this in school (at least I know *I* didn’t), we could have used both models to get an exact answer… all you’d have to do is carry the remainder as many times as you drop zeroes from behind the 3. Dividing 10 by 3 just a couple of times gave us 3.3R.1 (checked by multiplying 3(3.3) to get 9.9 and then adding the remainder of .1 to get 10), and doing it three times gave us 3.33R.01 and doing it ten times gave us 3.333333333R.000000001. Through inductive reasoning, we can see that if you divided 10 by 3 an infinite number of times you’d get 3.333…R.000…01. That remainder may get infinitely small, but it never disappears — no matter how many times you do the division, it will always carry a remainder.
Thus, 10 divided by 3 does not equal 3.333… but rather almost equals 3.333… because there’s a remainder. It’s easy to see why this is also true of dividing 1 by 3. In our proof above, we don’t actually get 1 = .999… but rather 1 = .999… with a remainder of .000…01.
I said above that there were “a couple of problems”, and this rounding error is just the first. The second one is with multiplying .333… by 3. “Why would that be a problem?” you may ask. Well, I don’t see it as a personal problem, but rather one for my opponents. There are those who would “reify” .333… (see my last post) and suggest that this problem is unsolvable because multiplication requires that both numbers have a finite end. After all, how else would you even start such a math problem? The answer is easy — through inductive reasoning. We know that any finite string of 3s multiplied by 3 will give us an equally long string of 9s, and there’s no reason to think this will change just because the string of 3s is infinitely long. But this is a problem for some people, and I’ve seen this proof less and less over time because mathematicians have realized that allowing this multiplication creates an obvious double standard. While this may seem a trivial problem compared to the first one, it is major enough to prevent many mathematicians from even offering this proof any more.
One final note: While I welcome criticism, please don’t tell me that my math is wrong unless you can suggest how it could have been done right. Don’t tell me that there is no remainder when you divide 1/3 unless you can demonstrate how it is done evenly. If you can’t offer an alternative explanation, there’s a good chance that there isn’t one.
Posted in Uncategorized | 5 Comments
## Reification
I’d like to answer an objection here, one that I hear commonly and that I’m sure I’ll hear again (even despite this post): people say that a number like 1.999…98 is impossible. If you aren’t familiar with that number, it is 1 point 9 repeating followed by a single 8, found by multiplying point 9 repeating by 2. I’ve explained how I came to this conclusion (inductive reasoning) and why it is consistent with what we know about math, but people like to think I’m being inconsistent with the definition of infinity.
You may have noticed the title of this post and thought, “What is reification?” Reification is a logical fallacy in which an abstract idea is treated like a physical object. For example, many people talk about feelings being “bottled up”, and suggest that letting out anger through venting will reduce the anger that you have. Studies have proven this wrong, because anger is not a physical object that can be reduced by “getting rid of it” through expressing it.
The problem with this objection to my method is that mathematicians who think that it’s impossible to “bookend” an infinite series of 9’s with a decimal point and an 8 will bookend other infinities without batting an eye. Consider the number of real numbers between 0 and 1. The obvious answer is “they’re infinite“, despite the existence of a floor and a ceiling to this series of numbers. One might argue that I’m arguing apples and oranges — the series between 0 and 1 is made up of numbers, not digits. So let’s convert that to digits. If you took all the real numbers between 0 and 1 and listed them from least to greatest, then drew a line crossing the first digit of each number, your result would be 01234567891, where every number except that final 1 would be repeated infinitely. Would this new number be consider a “real number”? “Sure!”, say proponents of Cantor’s diagonal argument, which creates a real number through almost identical means.
Do I agree that the number created is a real number? No, but that’s not the point. It’s not important to agree that a real number is made here, but instead to focus on the fact that the digits in this number are indisputably infinite, even though they end abruptly with other infinite series of digits and finally with one finite digit. This is because we’re not talking about a physical reality but a thought experiment. Numbers are just abstract ideas that we use to describe the relationships between mathematical models such as graphed lines, sets of objects, and sometimes just other numbers. While it may offend your sensibilities to try to grasp a concept like .000….01 (point zero repeating ending with a single one), there is no logical reason that such a number can’t exist.
One final note: While I welcome comments, please don’t tell me I’m wrong unless you can explain how I could have been right. If you can’t come up with an alternative explanation, there’s a good chance that there isn’t one.
Posted in Uncategorized | 8 Comments
## A Simple Proof
It’s almost too easy to explain why .999… does not equal 1. Even fools can see the reason right away — it appears to take a college education to get it wrong, as counter-intuitive as that may be.
If .999… = 1, and X – X = 0, then 1 – .999… should equal 0. It doesn’t. You get .000…01. How do I know this? Inductive reasoning.
1.00 – .99 = .01
1.00000 – .99999 = .00001
1.0000000000 – .9999999999 = .0000000001
1 minus point 9 repeating for any length equals point 0 repeating for one digit shorter than the string of 9s followed by a single 1.
There’s no reason to suppose that this pattern suddenly changes just because the string of 9s is infinitely long. For any finite length this is true, so for an infinite length it should still be true.
But let’s not stop there. If .999… = 1 then X squared = X squared. Does it? No.
.999 squared = .998001
.99999 squared = .9999800001
.999… squared = .999…980….0001
None of these equals 1 squared (which is 1). Again, I’ve demonstrated that any finite length of 9’s squared does not even come very close to being 1. It’s true that it gets closer as the string gets longer, but it will never change to become 1.000… No matter how long the number is, you’ll still be multiplying a 9 by a 9, and that will give you a 1, not a zero.
It’s even more revealing when you multiply both sides of the equation by 2.
.999…(2) = 1.999…98
1(2) = 2
Here you see that the distance between both sides doubles, which is consistent with any unbalanced equation. Take any two numbers, multiply both of them by 2, and you’ll always find that the difference between them also doubles (because if X-Y=Z, then 2X-2Y=2Z, where X and Y are any two numbers and Z is their difference).
There is literally nothing you can do to both sides of the equation .999… = 1 and have them still coming out equal. Many people assume that 9.999… = 10 (by multiplying both sides by 10) or that 3.999… = 4 (by adding 3 to both sides), but this is a logical fallacy known as “begging the question“… the conclusion is derived from prior belief that .999… = 1, not because it’s otherwise provable that these equations are true.
One final note: While I welcome criticism, please don’t tell me that I’m doing these problems wrong unless you can explain how they are done right. Don’t tell me that .999… multiplied by 2 does not equal 1.999…98 unless you can explain how those numbers are multiplied correctly and come out to a figure that supports your conclusion. If you can’t offer an alternative explanation, there’s a good chance that there isn’t one.
Posted in Uncategorized | 13 Comments
## Inductive Reasoning
Throughout this blog, I’ll refer many times to getting my answers through inductive reasoning, and it will probably benefit you to know what that is. Inductive Reasoning is, along with deductive reasoning, one of the two methods of figuring things out with logic. While deduction makes a more airtight case than induction, it is still an extremely useful tool that would be foolish to ignore. It is frequently used in mathematics, and may perhaps be necessary for any problem involving infinite numbers.
A common example of inductive reasoning is used in figuring out whether the sun will rise tomorrow. It rose this morning, and yesterday morning, and every morning of my life, and every recorded instance of mornings in history. Because the pattern fits so many intervals uninterrupted and without a single pattern break, it is said to be a strong inductive argument.
Mathematicians also use inductive reasoning to figure out, for example, what you get when you divide 1 by 3. We all know it’s (roughly) .333… , which is read as “point 3 repeating”. How does one reach this conclusion? We can’t keep dividing 10 by 3, dropping the remainder of 1, adding a 0, and dividing again by 3 over and over again. There isn’t enough time to do the whole problem, and it would be pointless to do it infinitely anyway because we all spot the pattern rather quickly. Because we know that this pattern will repeat every time without fail, we inductively reason that, no matter how many digits we add to the problem, we’ll get nothing but threes.
It’s important to note that this pattern doesn’t have a reason to break or interrupt — what is true about it at the beginning is true at the end, even if we keep it up into infinity. Strangely, some mathematicians would prefer to believe that the pattern does suddenly change when a number becomes infinitely long. Why? We’ll explore that in further blog posts, but the important thing to note here is that inductive reasoning will tell us what a pattern will do for any length of math problem, even an infinitely long one, and they are all strong arguments because there are an infinite number of examples in which each one works.
One final note: While I welcome criticism, please do not claim that I am wrong unless you can explain how I could have been right. If you don’t believe that inductive reasoning is useful for figuring out a problem like 1/3, please tell me how one does figure out such a problem. If you don’t have an alternative, there’s a good chance that there isn’t one.
Posted in Uncategorized | 8 Comments |
# Cube Root of Numbers
Last Updated: April 27, 2024
## Cube Root of Numbers
The cube root of a number is a value that, when multiplied by itself two more times, yields the original number. For instance, the cube root of 27, expressed as 3√27, is 3 because multiplying 3 by itself twice more (3 x 3 x 3) equals 27, or 3³. This process reveals the original value that was cubed, indicating that 27 is a perfect cube. The term “cube root” itself suggests the root cause of the cube. To determine the cube root of perfect cubes, the prime factorization method is typically employed, illustrating the concept’s significance akin to that of the square root.
## What is Cube Root?
The cube root of a number is the value that, when multiplied by itself three times (or cubed), results in the original number. It answers the question, “Which number, when cubed, gives the current value?” For example, the cube root of 8, denoted as , is 2, because 2×2×2=8. This concept is crucial in mathematics for finding the original base value of a cube number. Identifying the cube root is especially relevant for working with volumes in geometry, where understanding the dimensions of cubic shapes is necessary. The cube root symbol is represented as , with the number whose cube root is being sought placed inside the bracket.
## Cube Root Symbol
The cube root symbol is denoted as ³√. This notation is used to represent the cube root of a number, which is placed inside the bracket. For example, ³√27 signifies the cube root of 27.
## Cube Root Formula
The cube root formula is a way to find out which number was multiplied by itself three times to get a certain number, and it’s written with a special symbol, ³√. To find the cube root, you can start by breaking down the number into its prime factors. Then, use the cube root formula. If we have a number x, and x=y×y×y, then the cube root of x is simply ³√x=y. This means y is the cube root of x. If y turns out to be a whole number, then x is called a perfect cube.
Cuberoot of x= ³√x= ³√(y×y×y) = y
## How to Find Cube Root of a Number?
### Method 1: Prime Factorization
1. Break Down into Prime Factors: Start by breaking the number down into its prime factors. Prime factorization involves dividing the number into its basic building blocks—prime numbers that, when multiplied together, give the original number.
2. Group the Factors: Group these prime factors into sets of three identical numbers. Each set represents the cubed factor.
3. Multiply One from Each Group: Take one number from each group and multiply them together. This product is the cube root of the original number.
### Method 2: Using the Cube Root Symbol
• Apply the Cube Root Symbol: For simpler calculations or with the help of a calculator, you can directly apply the cube root symbol (³√x) to the number. This directly gives you the cube root without the need for factorization.
### Example
To find the cube root of 27:
1. Prime Factorize 27: 27=3×3×3
2. Group the Factors: There is one group of three 3s.
3. Take the Cube Root: The cube root of 27 is 3, because 3×3×3=27.
### Method 3: Estimation
• Estimate: For numbers that are not perfect cubes or when you’re without a calculator, estimate by finding the two closest perfect cube numbers it falls between. The cube root of your number will be between the cube roots of these two perfect cubes.
### Tips
• Perfect cubes are numbers like 8, 27, 64, etc., where the cube roots are whole numbers (2, 3, 4, respectively).
• For non-perfect cubes, the result may be a decimal or irrational number, which often requires a calculator for precise calculation.
## What is the cube root of 64?
The cube root of 64 is 4. This is because multiplying 4 by itself three times (4x4x4) equals 64.
## What is a cube root of 27?
The cube root of 27 is 3. This is determined by the fact that 3 multiplied by itself three times (3x3x3) equals 27.
Text prompt |
# Converting Standard Form To Slope Intercept
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Converting Standard Form To Slope Intercept – There are many forms employed to represent a linear equation, the one most frequently encountered is the slope intercept form. You may use the formula of the slope-intercept to identify a line equation when that you have the straight line’s slope and the y-intercept. This is the point’s y-coordinate at which the y-axis intersects the line. Read more about this particular line equation form below.
## What Is The Slope Intercept Form?
There are three basic forms of linear equations, namely the standard slope, slope-intercept and point-slope. Although they may not yield the same results when utilized in conjunction, you can obtain the information line that is produced faster using an equation that uses the slope-intercept form. Like the name implies, this form utilizes the sloped line and it is the “steepness” of the line is a reflection of its worth.
This formula can be used to find a straight line’s slope, the y-intercept, also known as x-intercept where you can apply different formulas available. The equation for a line using this specific formula is y = mx + b. The slope of the straight line is symbolized through “m”, while its y-intercept is indicated via “b”. Each point of the straight line is represented with an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” need to remain variables.
## An Example of Applied Slope Intercept Form in Problems
When it comes to the actual world in the real world, the slope intercept form is used frequently to represent how an item or issue changes over an elapsed time. The value provided by the vertical axis is a representation of how the equation deals with the intensity of changes over what is represented by the horizontal axis (typically the time).
A basic example of the use of this formula is to discover how many people live in a certain area as the years pass by. In the event that the population in the area grows each year by a predetermined amount, the point value of the horizontal axis increases one point at a time with each passing year and the point amount of vertically oriented axis will increase in proportion to the population growth by the set amount.
You may also notice the starting point of a question. The beginning value is located at the y-value of the y-intercept. The Y-intercept is the point where x is zero. By using the example of the above problem the beginning point could be the time when the reading of population begins or when time tracking begins , along with the changes that follow.
This is the location that the population begins to be monitored to the researchers. Let’s suppose that the researcher is beginning to perform the calculation or take measurements in 1995. Then the year 1995 will represent”the “base” year, and the x = 0 points would occur in the year 1995. So, it is possible to say that the 1995 population is the y-intercept.
Linear equations that use straight-line formulas can be solved in this manner. The starting point is expressed by the y-intercept and the rate of change is expressed through the slope. The main issue with this form generally lies in the interpretation of horizontal variables particularly when the variable is attributed to one particular year (or any other type of unit). The trick to overcoming them is to ensure that you comprehend the definitions of variables clearly. |
Decopose in simply fractions. 2/2x(6x+4)
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
We have to simplify the fraction: 2/(2x*(6x+4)) and write it as a sum of simplified fractions.
2/(2x*(6x+4))
=> 1/x*(6x + 4)
This can be written as (A / x) + (B / (6x + 4))
=> [A(6x + 4) + Bx]/ x*(6x + 4) = 1/x*(6x + 4)
=> 6Ax + 4A + Bx = 1
=> 4A = 1 and 6A + B = 0
=> A = 1/4 and B = -6*(1/4) = -3/2
The fraction 2/2x(6x+4) can be written as (1/4x) - 3/[2*(6x + 4)]
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
We'll have to get 2 irreducible fractions because of the 2 factors from denominator.
2/2x(6x+4) = 1/x(6x+4)
The final ratio 1/x(6x+4) is the result of addition or subtraction of 2 elementary fractions, as it follows:
1/x(6x+4) = A/x + B/(6x+4) (1)
We'll multiply by x(6x+4) both sides:
1 = A(6x+4) + Bx
We'll remove the brackets:
1 = 6Ax + 4A + Bx
We'll factorize by x to the right side:
1 = x(6A+B) + 4A
Comparing both sides, we'll get:
6A+B = 0
4A = 1 => A = 1/4
6/4 + B = 0
B = -3/2
We'll substitute A and B into the expression (1) and we'll get the algebraic sum of 2 elementary fractions:
2/2x(6x+4) = 1/x(6x+4) = 1/4x - 3/2(6x + 4)
neela | High School Teacher | (Level 3) Valedictorian
Posted on
2/2x(6x+4) = 1/2x(3x+2)
Let 1/2x((3x+2) = A/2x+B/(3x+2)
=> 1 = A(3x+2)+B(2x) (1)
Put = x= 0 in (1)
=> 1 = A*2. So A = 1/2.
Put x=-2/3 in (1)
=> 1 = 0+B*2* (-2/3) . S B = -3/4.
=> = 2/2x(6x+4) = 1/2x(3x+2) = 1/4x - 3/4(3x+2). |
2. ## can u see it this time?
If u can't read it, the question says "Supply the coordinates for each figure without introducing any new coordinates.
question (a) says: Triangle ABC has a right angle at A and angle B=Pie over 3.
question (b) says: BCi s a quarter circle with centre at (0,0). Line AD has equation y=x.
3. In part a, the slope of line BC is $\displaystyle \tan \left( {\frac{{2\pi }}{3}} \right)$.
4. Originally Posted by Raiden_11
If u can't read it, the question says "Supply the coordinates for each figure without introducing any new coordinates.
question (a) says: Triangle ABC has a right angle at A and angle B=Pie over 3.
The length of the base of the triangle is $\displaystyle a$.
Now, using the trig ratio for tangent, we see that
$\displaystyle \tan B = \frac {|AC|}{a}$ ........$\displaystyle |AC|$ means the length of the line connecting A and C
$\displaystyle \Rightarrow \tan \left( \frac { \pi}{3} \right) = \frac {|AC|}{a}$
$\displaystyle \Rightarrow \sqrt {3} = \frac {|AC|}{a}$
$\displaystyle \Rightarrow a = \frac {|AC|}{ \sqrt {3}} = \frac { \sqrt {3} |AC|}{3}$
So we have $\displaystyle B \left( \frac { \sqrt {3} |AC|}{3}, 0 \right)$
question (b) says: BCi s a quarter circle with centre at (0,0). Line AD has equation y=x.
Note that $\displaystyle r = |AD| = |AB|$
Also note that the angle the line y = x makes with the x-axis is $\displaystyle 45^{ \circ}$, or $\displaystyle \frac { \pi}{4}$
Using the trig ratio for sine we see that
$\displaystyle \sin \left( \frac { \pi}{4} \right) = \frac {|DE|}{|AD|} = \frac {|DE|}{r}$ .........E is the point where a vertical line from D cuts the x-axis, |DE| represents the length of that line
$\displaystyle \Rightarrow r = \frac {|DE|}{ \frac { \sqrt {2}}{2}} = \frac {2 |DE|}{ \sqrt {2}}$
So $\displaystyle B \left( 0, \frac {2 |DE|}{ \sqrt {2}} \right)$
These questions are a bit vague to me. I guess we could do the last one without introducing a new point, if we used the cosine rule or something like that
5. It appears to me as if the question (a) is asking for the coordinates of C.
Because line BC has slope $\displaystyle \tan \left( {\frac{{2\pi }}{3}} \right) = - \sqrt 3$ then say C(0,c) so $\displaystyle \frac{{c - 0}}{{0 - a}} = - \sqrt 3 \quad \Rightarrow \quad c = a\sqrt 3 \quad or\quad C:\left( {0,a\sqrt 3 } \right).$
Thus we have written the coordinates of C in terms of the given.
6. Originally Posted by Plato
It appears to me as if the question (a) is asking for the coordinates of C.
Because line BC has slope $\displaystyle \tan \left( {\frac{{2\pi }}{3}} \right) = - \sqrt 3$ then say C(0,c) so $\displaystyle \frac{{c - 0}}{{0 - a}} = - \sqrt 3 \quad \Rightarrow \quad c = a\sqrt 3 \quad or\quad C:\left( {0,a\sqrt 3 } \right).$
Thus we have written the coordinates of C in terms of the given.
oh yeah, silly me. and i suppose question (b) is asking for coordinate D right...or C? or both, yeah, both
well, i'll leave my post up there, just in case anybody wonders, "well, how does a and r relate to all of this?"
7. ## Here's another question guys....it seems simple but i think theres more to it.
Thanks for the help earlier can u help me answer this?
8. ## I really need help!
These are questions from the grade 12 geometry course i am taking.
1. Triangle ABC has medians CE and BD as shown in the diagram. Use a coordinate proof and a vector proof to prove BF : FD = 2 : 1
9. Coordinates prove:
We can assume that $\displaystyle A\in Oy,B,C\in Ox$ and $\displaystyle A(0,2a),B(2b,0),C(2c,0)$, for the simplicity of calculus.
Now $\displaystyle D(c,a),E(b,a)$. Then you can write the equations of $\displaystyle BD$ and $\displaystyle CE$, solve the system and you will find the coordinates of the point $\displaystyle F$. Then you can calculate the length of the segments $\displaystyle BF,FD$.
10. ## thanks....but
Thanks but can you show me how to solve it just so i can have a clear idea.
11. ## Also
Also,how did you get those points in the first place...can u show me in detail please. Much appreciated.
12. Given a triangle $\displaystyle ABC$ in plane we can get the $\displaystyle Ox$ and $\displaystyle Oy$ axis anywhere in that plane. So we can get $\displaystyle Ox$ axis passing through $\displaystyle B$ and $\displaystyle C$ and $\displaystyle Oy$ axis passing through $\displaystyle A$.
Now, if there are two points $\displaystyle M_1(x_1,y_1),M_2(x_2,y_2)$, then the midpoint $\displaystyle M$ of the segment $\displaystyle M_1M_2$ has the coordinates $\displaystyle \displaystyle \frac{x_1+x_2}{2}$ and $\displaystyle \displaystyle \frac{y_1+y_2}{2}$. Then $\displaystyle D(c,a), \ E(b,a)$.
The equation of the line passing through $\displaystyle M_1(x_1,y_1),M_2(x_2,y_2)$ is $\displaystyle \displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$ or
$\displaystyle \left|\begin{array}{lll} x & y & 1\\ x_1 & y_1 & 1\\ x_2 & y_2 & 1 \end{array}\right|=0$
Using one of these equations you can now write the equations of $\displaystyle BD$ and $\displaystyle CE$, then solve the system formed by these equations to find the coordinates of $\displaystyle F$.
Then you can use the distance formula between two points in plane:
$\displaystyle M_1M_2=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
to calculate $\displaystyle BF$ and $\displaystyle FD$
It is more clear now?
13. Vector prove:
Let $\displaystyle \displaystyle \frac{BF}{FD}=k$
Then $\displaystyle \displaystyle \overrightarrow{CF}=\frac{1}{1+k}\overrightarrow{C B}+\frac{k}{1+k}\overrightarrow{CD}$ and
$\displaystyle \displaystyle \overrightarrow{CE}=\frac{1}{2}(\overrightarrow{CB }+\overrightarrow{CA})=\displaystyle \frac{1}{2}\left(\overrightarrow{CB}+2\overrightar row{CD}\right)=\frac{1}{2}\overrightarrow{CB}+\ove rrightarrow{CD}$.
The vectors $\displaystyle \overrightarrow{CF}$ and $\displaystyle \overrightarrow{CE}$ are colinear, so
$\displaystyle \displaystyle \frac{\frac{1}{1+k}}{\frac{1}{2}}=\frac{\frac{k}{1 +k}}{1}\Rightarrow k=2$, so $\displaystyle \displaystyle \frac{BF}{FD}=2$
14. ## I have another problem
Thanks man, i really appreciate the help but i have two more problems. |
# Two charges of -5 C and -1 C are positioned on a line at points 3 and 5 , respectively. What is the net force on a charge of 2 C at -2 ?
Mar 25, 2017
$\text{Net Force is } - 3.97 \cdot {10}^{9} u n i t s$
$\text{The negative sign shows that charges at the points A and B }$$\text{attract the charge at the point of C.}$
#### Explanation:
$\text{Both of charge at the points A and B apply a force on}$
$\text{the charge at the point of C.}$
$\text{let "F_A" be a force that A applies C.}$
$\text{The charge at the point of A attracts the charge at C}$
$\textcolor{g r e e n}{{F}_{A}} = \frac{k \cdot {q}_{A} \cdot {q}_{C}}{{d}_{\text{AC}}^{2}} = - \frac{K \cdot 5 \cdot 2}{3 + 2} ^ 2 = - \frac{K \cdot 2.5}{25} = - \frac{2 K}{5}$
$\text{let "F_B" be a force that B applies C.}$
$\text{The charge at the point of B attracts the charge at C}$
$\textcolor{b l u e}{{F}_{B}} = \frac{k \cdot {q}_{B} \cdot {q}_{C}}{{d}_{\text{BC}}^{2}} = - \frac{K \cdot 1 \cdot 2}{5 + 2} ^ 2 = - \frac{K \cdot 2.1}{49} = - \frac{2 K}{49}$
$\textcolor{red}{{F}_{C}} = \textcolor{g r e e n}{{F}_{A}} + \textcolor{b l u e}{{F}_{B}}$
$\textcolor{red}{{F}_{C}} = - \frac{2 K}{5} - \frac{2 K}{49}$
$\textcolor{red}{{F}_{C}} = - 2 K \left(\frac{1}{5} + \frac{1}{49}\right)$
$\textcolor{red}{{F}_{C}} = - 2 K \left(\frac{49 + 5}{49 \cdot 5}\right)$
$\textcolor{red}{{F}_{C}} = - 2 K \frac{54}{245}$
$K = 9 \cdot {10}^{9}$
$\textcolor{red}{{F}_{C}} = - 2 \cdot 9 \cdot {10}^{9} \left(\frac{54}{245}\right)$
color(red)(F_C)=-3.97.10^9 " |
Goseeko blog
# What is integration?
IntroductionIntegration is the reverse process of differentiation. in other words It is also called anti-differentiation.
Integration calculus has its own application in economics, Engineering, Physics, Chemistry, business, commerce, etc.
The integral of a function is denoted by the sign
Let the function is y = f(x), So that its derivative is-
Then
Where c is the arbitrary constant.
For example,
A function,
Then, its derivative-
Or
Then
Here c is an arbitrary constant.
Some fundamental integrals-
## Methods of integration
Simple integration-
1. When the function is an algebraic function-
Some standard form are-
The integration of x^n will be as follows-
Example: Find the integral of-
Sol.
We know that-
Then
Example: Find the integral
Sol.
We know that-
Then
Example: Evaluate-
Sol.
## By substitution
Example: Evaluate the following integral-
Sol.
Let us suppose,
Then
Or
Substituting –
Logarithmic function-
Example: Evaluate the following integral-
Sol.
Let us suppose-
Now
## Integral of exponential function
Example: Evaluate-
Sol.
Let,
Now substituting-
Integration of product of two functions-
Suppose we have two function say- f(x) and g(x), then
The integral of product of these two functions is-
Note-
We chose the first function as method of ILATE-
Which is-
I – Inverse trigonometric function
L – Log function
A – Algebraic function
T- Trigonometric function
E- Exponential function
Example: Evaluate-
Sol.
Here according to ILATE,
First function = log x
Second function = x^n
We know that-
Then
On solving, we get- |
Square Roots - Examples, Exercises and Solutions
What are those mysterious square roots that often confuse students and complicate their lives? The truth is that to understand them, we need to grasp the concept of the inverse operation.
What is a square root?
When we solve an exercise like $5=25^2$, it's clear that $5$ times $5$ (that is, multiplying the number by itself) results in $25$. This is the concept of a power, or to be more precise, a square power, which to apply, we must multiply the figure or the number by itself.
The concept of "square root" refers to the inverse operation of squaring numbers.
That is, if we have $X^2=25$ and we want to find the value of $X$, what we need to do is perform an identical operation on both sides of the equation.
This operation is the square root.
So, we have: $\sqrt{X^2} = \sqrt{25}$ and the result is $X=5$.
Examples with solutions for Square Roots
Exercise #1
Choose the largest value
Step-by-Step Solution
Let's calculate the numerical value of each of the roots in the given options:
$\sqrt{25}=5\\ \sqrt{16}=4\\ \sqrt{9}=3\\$and it's clear that:
5>4>3>1 Therefore, the correct answer is option A
$\sqrt{25}$
Exercise #2
$\sqrt{441}=$
Step-by-Step Solution
The root of 441 is 21.
$21\times21=$
$21\times20+21=$
$420+21=441$
$21$
Exercise #3
$(\sqrt{380.25}-\frac{1}{2})^2-11=$
Step-by-Step Solution
According to the order of operations, we'll first solve the expression in parentheses:
$(\sqrt{380.25}-\frac{1}{2})=(19.5-\frac{1}{2})=(19)$
In the next step, we'll solve the exponentiation, and finally subtract:
$(19)^2-11=(19\times19)-11=361-11=350$
350
Exercise #4
$\sqrt{64}=$
8
Exercise #5
$\sqrt{36}=$
6
Exercise #6
$\sqrt{49}=$
7
Exercise #7
$\sqrt{121}=$
11
Exercise #8
$\sqrt{100}=$
10
Exercise #9
$\sqrt{144}=$
12
Exercise #10
$\sqrt{x}=1$
1
Exercise #11
$\sqrt{36}=$
6
Exercise #12
$\sqrt{16}=$
4
Exercise #13
$\sqrt{x}=2$
4
Exercise #14
$\sqrt{9}=$
3
Exercise #15
$\sqrt{4}=$ |
# AP Statistics : How to find confidence intervals for a mean
## Example Questions
### Example Question #1 : Confidence Intervals
Suppose you have a normally distributed variable with known variance. How many standard errors do you need to add and subtract from the sample mean so that you obtain 95% confidence intervals?
Explanation:
To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract . This is because 95% of the values drawn from a normally distributed sampling distribution lie within 1.96 standard errors from the sample mean.
### Example Question #1 : Confidence Intervals
An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $11,000. The standard deviation of the 24-car sample is$2,500.
Provide a 98% confidence interval for the true mean cost of repair.
Explanation:
Standard deviation for the samle mean:
Since n < 30, we must use the t-table (not the z-table).
The 98% t-value for n=24 is 2.5.
### Example Question #1 : Confidence Intervals
300 hundred eggs were randomly chosen from a gravid female salmon and individually weighed. The mean weight was 0.978 g with a standard deviation of 0.042. Find the 95% confidence interval for the mean weight of the salmon eggs (because it is a large n, use the standard normal distribution).
Explanation:
Because we have such a large sample size, we are using the standard normal or z-distribution to calculate the confidence interval.
Formula:
We must find the appropriate z-value based on the given for 95% confidence:
Then, find the associated z-score using the z-table for
Now we fill in the formula with our values from the problem to find the 95% CI.
### Example Question #1 : Confidence Intervals
A sample of observations of 02 consumption by adult western fence lizards gave the following statistics:
Find the confidence limit for the mean 02 consumption by adult western fence lizards.
Explanation:
Because we are only given the sample standard deviation we will use the t-distribution to calculate the confidence interval.
Appropriate Formula:
Now we must identify our variables:
We must find the appropriate t-value based on the given
t-value at 90% confidence:
Look up t-value for 0.05, 55 , so t-value= ~ 1.6735
90% CI becomes:
### Example Question #1 : Confidence Intervals
Subject Horn Length (in) Subject Horn Length (in) 1 19.1 11 11.6 2 14.7 12 18.5 3 10.2 13 28.7 4 16.1 14 15.3 5 13.9 15 13.5 6 12.0 16 7.7 7 20.7 17 17.2 8 8.6 18 19.0 9 24.2 19 20.9 10 17.3 20 21.3
The data above represents measurements of the horn lengths of African water buffalo that were raised on calcium supplements. Construct a 95% confidence interval for the population mean for horn length after supplments.
Explanation:
First you must calculate the sample mean and sample standard deviation of the sample.
Because we do not know the population standard deviation we will use the t-distribution to calculate the confidence intervals. We must use standard error in this formula because we are working with the standard deviation of the sampling distribution.
Formula:
To find the appropriate t-value for 95% confidence interval:
Look up in t-table and the corresponding t-value = 2.093.
Thus the 95% confidence interval is:
### Example Question #1 : Confidence Intervals And Mean
The population standard deviation is 7. Our sample size is 36.
What is the 95% margin of error for:
1) the population mean
2) the sample mean
1) 12.266
2) 3.711
1) 15.554
2) 3.656
1) 11
2) 3
1) 14.567
2) 4.445
1) 13.720
2) 2.287
1) 13.720
2) 2.287
Explanation:
For 95% confidence, Z = 1.96.
1) The population M.O.E. =
2) The sample standard deviation =
The sample M.O.E. = |
# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 3 - 5/6 = 13/6 = 2 1/6 ≅ 2.1666667
Spelled result in words is thirteen sixths (or two and one sixth).
### How do you solve fractions step by step?
1. Subtract: 3 - 5/6 = 3/1 - 5/6 = 3 · 6/1 · 6 - 5/6 = 18/6 - 5/6 = 18 - 5/6 = 13/6
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(1, 6) = 6. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 1 × 6 = 6. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - three minus five sixths = thirteen sixths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Fraction and a decimal
Write as a fraction and a decimal. One and two plus three and five hundredths
• Evaluate expression
Calculate the value of the expression z/3 - 2 z/9 + 1/6, for z = 2
Add this two mixed numbers: 1 5/6 + 2 2/11=
• Samuel
Samuel has 1/3 of a bag of rice and Isabella has a 1/2 bag of rice. What fraction of are bag of rice do they have altogether?
• Kelly
Kelly and Dan are collecting clothes for a clothing drive. Dan collected 1/2 as many clothes as Kelly did. If Kelly collected 3 bags of clothes, how many bags of clothes did Dan collect?
• Paper clips
Mrs. Bright is organizing her office supplies. There are 5 open boxes of paper clips in her desk drawer. Each box has 1/2 of the paper clips remaining. How many boxes of paper clips are left?
• Classroom
One-eighth of 9th class was interested in studying at a grammar school, at a business academy one sixth, at secondary vocational schools quarter, to SOU one third and the remaining three students were interested in the school of art direction. How many st
• Third of an hour
How many minutes is a third of an hour? Do you know to determine a third of the lesson hour (45min)?
• Juan is
Juan is making cookies. He makes 2 batches on Monday and 4 batches on Tuesday. He uses 3/4 cup of flour in each batch. How much flour does juan use?
• Boys and glasses
2/3 of the students in Sarah’s class are boys. Of the boys, 1/3 of them wear glasses. What fraction of the students wear glasses?
• Grandmother and grandfather
Grandmother baked cakes. Grandfather ate half, then quarter of the rest ate Peter and Paul ate half of rest. For parents left 6 cakes. How many cakes maked the grandmother?
• Expressions with variable
This is algebra. Let n represent an unknown number and write the following expressions: 1. 4 times the sum of 7 and the number x 2. 4 times 7 plus the number x 3. 7 less than the product of 4 and the number x 4. 7 times the quantity 4 more than the number |
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RD Chapter 15- Linear Inequations Ex-15.4 Interview Questions Answers
Related Subjects
Question 1 : Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.
Let ‘x’ be the smaller of the two consecutive odd positive integers. Then the other odd integer is x + 2.
Given:
Both the integers are smaller than 10 and their sum is more than 11.
So,
x + 2 < 10 and x + (x + 2) > 11
x < 10 – 2 and 2x + 2 > 11
x < 8 and 2x > 11 – 2
x < 8 and 2x > 9
x < 8 and x > 9/2
9/2 < x < 8
Note the odd positive integers lying between 4.5 and 8.
x = 5, 7 [Since, x is an odd integer]
x < 10 [From the given statement]
9/2 < x < 10
Note that, the upper limit here has shifted from 8 to 10. Now, x is odd integer from 4.5 to 10.
So, the odd integers from 4.5 to 10 are 5, 7 and 9.
Now, let us find pairs of consecutive odd integers.
Let x = 5, then (x + 2) = (5 + 2) = 7.
Let x = 7, then (x + 2) = (7 + 2) = 9.
Let x = 9, then (x + 2) = (9 + 2) = 11. But, 11 is greater than 10.
∴ The required pairs of odd integers are (5, 7) and (7, 9)
Question 2 : Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40.
2.
Given:
Both the natural numbers are greater than 10 and their sum is less than 40.
So,
x > 10 and x + x + 2 <40
x > 10 and 2x < 38
x > 10 and x < 38/2
x > 10 and x < 19
10 < x < 19
From this inequality, we can say that x lies between 10 and 19.
So, the odd natural numbers lying between 10 and 19 are 11, 13, 15 and 17. (Excluding 19 as x < 19)
Now, let us find pairs of consecutive odd natural numbers.
Let x = 11, then (x + 2) = (11 + 2) = 13
Let x = 13, then (x + 2) = (13 + 2) = 15
Let x = 15, then (x + 2) = (15 + 2) = 17
Let x = 17, then (x + 2) = (17 + 2) = 19.
x = 11, 13, 15, 17 [Since, x is an odd number]
∴ The required pairs of odd natural numbers are (11, 13), (13, 15), (15, 17) and (17, 19)
Question 3 : Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.
Let ‘x’ be the smaller of the two consecutive even positive integers. Then the other even integer is x + 2.
Given:
Both the even integers are greater than 5 and their sum is less than 23.
So,
x > 5 and x + x + 2 < 23
x > 5 and 2x < 21
x > 5 and x < 21/2
5 < x < 21/2
5 < x < 10.5
From this inequality, we can say that x lies between 5 and 10.5.
So, the even positive integers lying between 5 and 10.5 are 6, 8 and 10.
Now, let us find pairs of consecutive even positive integers.
Let x = 6, then (x + 2) = (6 + 2) = 8
Let x = 8, then (x + 2) = (8 + 2) = 10
Let x = 10, then (x + 2) = (10 + 2) = 12.
x = 6, 8, 10 [Since, x is even integer]
∴ The required pairs of even positive integer are (6, 8), (8, 10) and (10, 12)
Question 4 : The marks scored by Rohit in two tests were 65 and 70. Find the minimum marks he should score in the third test to have an average of at least 65 marks.
Given:
Marks scored by Rohit in two tests are 65 and 70.
Let marks in the third test be x.
So let us find minimum x for which the average of all three papers would be at least 65 marks.
That is,
Average marks in three papers ≥ 65 …(i)
Average is given by:
Average = (sum of all numbers)/(Total number of items)
= (marks in 1st two papers + marks in third test)/3
= (65 + 70 + x)/3
= (135 + x)/3
Substituting this value of average in the inequality (i), we get
(135 + x)/3 ≥ 65
(135 + x) ≥ 65 × 3
(135 + x) ≥ 195
x ≥ 195 – 135
x ≥ 60
This inequality means that Rohit should score at least 60 marks in his third test to have an average of at least 65 marks.
So, the minimum marks to get an average of 65 marks is 60.
∴ The minimum marks required in the third test is 60.
Question 5 : A solution is to be kept between 86o and 95oF. What is the range of temperature in degree Celsius, if the Celsius (C)/Fahrenheit (F) conversion formula is given by F = 9/5C + 32.
Let us consider F1 =86o F
And F2 =95o
We know, F = 9/5C + 32
F1 =9/5 C1 + 32
F1 –32 = 9/5 C1
C1 =5/9 (F1 – 32)
= 5/9 (86 – 32)
= 5/9 (54)
= 5 × 6
= 30o C
Now,
F2 =9/5 C2 + 32
F2 –32 = 9/5 C2
C2 =5/9 (F2 – 32)
= 5/9 (95 – 32)
= 5/9 (63)
= 5 × 7
= 35o C
The range oftemperature of the solution in degree Celsius is 30° C and 35° C.
Question 6 : A solution is to be kept between 30oC and 35oC. What is the range of temperature in degree Fahrenheit?
Let us consider C1 =30o C
And C2 =35o
We know, F = 9/5C + 32
F1 =9/5 C1 + 32
= 9/5 × 30 + 32
= 9 × 6 + 32
= 54 + 32
= 86o F
Now,
F2 =9/5 C2 + 32
= 9/5 × 35 + 32
= 9 × 7 + 32
= 63 + 32
= 95o F
The range oftemperature of the solution in degree Fahrenheit is 86° F and 95° F.
krishan |
# A triangle has sides with lengths: 16, 8, and 19. How do you find the area of the triangle using Heron's formula?
Dec 26, 2015
(see below for method)
Approximate area $63.2$
#### Explanation:
Heron's Formula tells us that if we are given a triangle with sides $a , b , c$ and a semi-perimeter $s = \frac{a + b + c}{2}$, the area of the triangle is:
color(white)("XXX")"Area"_triangle = sqrt(s(s-a)(s-b)(s-c))
For the given values $\left(a , b , c\right) = \left(16 , 8 , 19\right)$
$\textcolor{w h i t e}{\text{XXX}} s = \frac{43}{2}$
$\textcolor{w h i t e}{\text{XXX}} \left(s - a\right) = \frac{11}{2}$
$\textcolor{w h i t e}{\text{XXX}} \left(s - b\right) = \frac{27}{2}$
$\textcolor{w h i t e}{\text{XXX}} \left(s - c\right) = \frac{5}{2}$
Applying the formula (and using a calculator)
color(white)("XXX")"Area"_triangle ~~63.2 |
# AZ Merit Boot Camp 6th Grade Math - Higley Unified School ...
AZ Merit Boot Camp 6th Grade Math Ms. McClure San Tan Elementary Day 1 I can use my knowledge of the order of operations to create equivalent expressions.
Day 1 Expressions and Equations 6.EE.A.3 Equivalent expressions Must have the same value Contain variables
Multiplication is repeat addition Division is repeat subtraction Exponents are repeat multiplication Inverse Operations Multiplication & Division Addition & Subtraction Order of Operations: GEMS!
Grouping symbols Exponents Multiplication/division Subtraction/addition Day 1 Expressions and Equations 6.EE.A.3 Day 2
I can write and understand numerical expressions involving whole number exponents. I can determine the answer to expressions when given the specific value of a variable. I can identify when two expressions are equivalent. Day 2 Expressions and Equations
6.EE.A.1, 6.EE.A.2c, 6.EE.A.4 Use substitution to evaluate expressions when given the value of a variable. Parts of an expression Coefficient: The number before the variable Variable: Letter that stands for a number Constant: Number that stays the same
Terms: Separated by plus or minus signs Day 2 Expressions and Equations 6.EE.A.1, 6.EE.A.2c, 6.EE.A.4 Day 3 I can write expressions using numbers and letters (with the letters standing for numbers).
I can identify the parts of an expressions using mathematical words. I can understand that in 2(8+7), (8+7) can be thought of as two separate numbers or as 15. Day 3 Expressions and Equations 6.EE.A.2a and 6.EE.A.2b When
defining variables, we must be specific with the units We can either use GEMS or the distributive property to solve 3(5-2) Distribute algebraic expressions just as we distribute numerical expressions Use the distributive property to write an equivalent expression for 8n+16 as 8(n+2) because 8 is a common factor
Day 3 Expressions and Equations 6.EE.A.2a and 6.EE.A.2b Day 4 I can understand that solving an equation or inequality is like answering a question. I can use variables to represent numbers and write expressions when solving realworld problems. I can solve real-world problems and mathematical problems by writing and solving equations. I can write an inequality which has many solutions and represent these solutions on
a number line. I can use variables to represent two quantities in a real world problem and write an equation to express the quantities. I can use graphs and tables to show the relationship between dependent and independent variables. Day 4 Expressions and Equations 6.EE.B.5, 6.EE.B.6, 6.EE.B.7, 6.EE.B.8, and 6.EE.C.9
To solve an equation or inequality, determine the number(s) that, when substituted for the variable, makes the number sentence true To solve equations, we must isolate the variable, using inverse operations
When representing solutions to inequalities on number lines Closed circle includes the number (greater than or equal to/less than or equal to) Open circle does not include the number (greater than/less than) When your variable is to the left of the inequality symbol, the inequality symbol is pointing in the direction you shade on the number line The dependent variable depends on the independent variable In a table, the independent variable goes on the left On a graph, the independent variable goes along the bottom
In an equation, the dependent variable stands alone Day 4 Expressions and Equations 6.EE.B.5, 6.EE.B.6, 6.EE.B.7, 6.EE.B.8, and 6.EE.C.9 Day 5 I can divide two fractions. I can solve word problems involving the
division of fractions by fractions. Day 5 Number Sense 6.NS.A.1 When dividing a fraction by a fraction: KEEP, CHANGE, FLIP!
If a number is a whole number, change it to a fraction by giving it a denominator of 1 Day 5 Number Sense 6.NS.A.1 Day 6 I can divide multi-digit numbers. I can add, subtract, multiply, and divide
multi-digit numbers involving decimals. Day 6 Number Sense 6.NS.B.2 and 6.NS.B.3 When dividing decimals, we can multiply our dividend and our divisor by a power of 10 to get rid of the decimal in our divisor and divide
normally (or think of it as moving the decimal point over the same amount of times) When adding or subtracting decimals, be sure to line up the decimal point When multiplying decimals, move your decimal to the left in your product as many numbers as there are to the right of the decimal in the problem Day 6 Number Sense 6.NS.B.2 and 6.NS.B.3
Day 7 I can find the greatest common factor of two whole numbers less than or equal to 12. I can find the least common multiple of two whole numbers less than or equal to 12. I can use the distributive property to show the sum of two whole numbers 1-100 in different ways.
Day 7 Number Sense 6.NS.B.4 To find GCF and LCM, we can use the upside-down birthday cake method For GCF, we multiply the numbers weve taken out to the left For LCM, we multiply the L, or all numbers weve taken out
to the left as well as the numbers left underneath The greatest common factor will not be greater than both of the numbers The least common multiple will be at least as big as both of the numbers To use the distributive property, we find a common factor of our addends
For example, 5+15 = 5(1+3) Day 7 Number Sense 6.NS.B.4 Day 8 I can understand that positive and negative numbers are used to describe amounts of having opposite values. I can use positive and negative numbers to show amounts
in real-world situations and explain what the number 0 means in those situations. I can recognize opposite signs of a numbers as indicating places on opposite sides of 0 on a number line. I can understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane. Day 8 Number Sense 6.NS.C.5, 6.NS.C.6a, and 6.NS.C.6b
62 and 62 are opposites because they are on opposite sides of 0 and the same distance away on a number line We must be familiar with mathematical vocabulary in real-world situations such as deposit, withdraw, etc. to determine whether an amount is positive or negative We must be able to define 0 in these situations, for example 0 could represent no change in a bank account with a \$60 balance, or it could represent \$0
On the coordinate plane, (x, y) is located in Quadrant I (-x, y) is located in Quadrant II (-x, -y) is located in Quadrant III (x, -y) is located in Quadrant IV Day 8 Number Sense 6.NS.C.5, 6.NS.C.6a, and 6.NS.C.6b Day 8
Day 9 I can place integers and other numbers on a number line diagram. I can place ordered pairs on a coordinate plane. I can understand the distance between two numbers including positives and negatives on a number line. I can understand and explain what rational numbers mean in real-world situations. I can understand absolute value as they apply to real-world situations.
I can tell the difference between comparing absolute values and ordering positive and negative numbers. Day 9 Number Sense 6.NS.6c, 6.NS.C.7a, 6.NS.C.7b, 6.NS.C.7c, and 6.NS.C.7d We must be consistent with our scale on a number line
and on the coordinate plane The x-coordinate of an ordered pair takes you right (positive) or left (negative), the y-coordinate takes you up (positive), or down (negative) on the coordinate plane Rational numbers include integers, decimals, and fractions To determine the distance on a number line between numbers with opposite signs, add their absolute values Absolute value is a numbers distance away from 0, it
tells us the numbers magnitude (or size) Day 9 Number Sense 6.NS.6c, 6.NS.C.7a, 6.NS.C.7b, 6.NS.C.7c, and 6.NS.C.7d Day 9 Day 10
I can understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane. I can graph in all four quadrants of the coordinate plane to help me solve real-world and mathematical problems. I can determine the distance between points in the same first coordinate or the same second coordinate.
Day 10 Number Sense 6.NS.C.6b, 6.NS.C.8, and 6.NS.C.9 To determine the distance between points with the same first coordinate or the same second coordinate, look
at the coordinate thats different, find the absolute values of each, If they have different signs, add the absolute values If they have the same signs, Subtract the absolute values Day 10 Number Sense 6.NS.C.6b, 6.NS.C.8, and 6.NS.C.9
Day 10 Day 11 I can understand ratios and the language used to describe two amounts. I can make tables of equivalent ratios, find the missing values in the tables, plot those values on a coordinate plane, and use the
tables to compare ratios. Day 11 Ratios and Proportions 6.RP.A.1 and 6.RP.A.3a Ratios are comparisons between 2 amounts Remember: ORDER MATTERS! Common ratio language: For each, For every, etc.
Ratios can be written as fractions, with a colon (:) or with to between the two amounts Equivalent ratio tables can be thought of as repeat addition or multiplication by a common factor to help us find the missing values in the tables When making ratio tables, be sure to label what each column or row represents (ex. Miles and Hours) In order to compare ratios, they must have something in common such as the numerator or the denominator Another way we can compare is by dividing
When plotting values of ratio tables on the coordinate plane, think of each ratio as an ordered pair Day 11 Ratios and Proportions 6.RP.A.1 and 6.RP.A.3a Day 11
Day 12 I can understand how to find a rate when given a specific ratio. I can solve unit rate problems. I can find a percent of a quantity as rate per 100. I can solve problems involving finding the whole if I am given a part and the percent. I can use what I know about ratios to convert units of measurement.
Day 12 Ratios and Proportions 6.RP.A.2, 6.RP.A.3b, 6.RP.A.3d, and 6.RP.A.3c Be sure to label each part of your ratio or rate To find a unit rate, you must determine How many for 1/for each/for every/per, For example if Carlos drives 30 miles in
2 hours, his unit rate is 15 miles/hour Remember, is/of = %/100 = part/whole We can also change the percent to a decimal and multiply by the number to find a percent of a number If we know it takes 5 feet per string, we know it takes 60 inches per string because we can convert our measurement unit by multiplying or dividing by the appropriate amount. In this instance, we knew more inches would fit than feet because they are smaller, so we multiplied.
Day 12 Ratios and Proportions 6.RP.A.2, 6.RP.A.3b, 6.RP.A.3d, and 6.RP.A.3c Day 12
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# Derivative Of Sec 3 X
Derivative Of Sec 3 X. For example, we can use the derivative of sec x to differentiate h ( x) = sec ( 3 x + 4). Hence, we have ∫sec 3x tan 3x dx = (1/3) sec 3x. (u+v)' = u'+v' formula for calculating the derivative of a function product : We know that the derivative of sec a is sec a tan a.
The derivative rule for sec (x) is given as: We can find this derivative using the quotient rule: Date exercisable and expiration date (month/day/year) 7.
## Hence, we have ∫sec 3x tan 3x dx = (1/3) sec 3x.
Let f (x) = (sec x) 2. We know that the derivative of sec a is sec a tan a. To find the derivative of sec square x square with respect to x square, that is, d (sec 2 (x 2 ))/d (x 2 ), we will use the chain rule method of derivatives.
## What Is The Derivative Of (Sec X) 2?
On both the sides by using chain rule as follows d dx (x) = d dx (sec3y) 1 =. (u+v)' = u'+v' formula for calculating the derivative of a function product : Find dy/dx y=sec (3x) y = sec(3x) y = sec ( 3 x) differentiate both sides of the equation. We know that the derivative of sec a is sec a tan a.
### The Differentiation Or Derivative Of Secant Function With Respect To A Variable Is Equal To The Product Of Secant And Tangent Functions.
To differentiate, use the chain rule:
### Kesimpulan dari Derivative Of Sec 3 X.
First, a parser analyzes the mathematical function. Formula for calculating the derivative of a function sum : 3 d dx [sec(x)] 3 d d x [ sec ( x)] the derivative of sec(x) sec ( x). D⁄dxsec (x) = tan (x)sec (x) this derivative rule gives us the ability to quickly and directly differentiate sec (x). |
AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.4
AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.4 Textbook Questions and Answers.
AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.4
10th Class Maths 8th Lesson Similar Triangles Ex 8.4 Textbook Questions and Answers
Question 1.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Given : □ ABCD is a rhombus.
Let its diagonals AC and BD bisect each other at ‘O’.
We know that “the diagonals in a rhombus are perpendicular to each other”.
In △AOD; AD2 = OA2 + OD2 ………. (1)
[Pythagoras theorem]
In △COD; CD2 = OC2 + OD2 ………. (2)
[Pythagoras theorem]
In △AOB; AB2 = OA2 + OB2 ………. (3)
[Pythagoras theorem]
In △BOC; BC2 = OB2 + OC2 ………. (4)
[Pythagoras theorem]
Adding the above equations we get AD2 + CD2 + AB2 + BC2 = 2 (OA2 + OB2 + OC2 + OD2)
Question 2.
ABC is a right triangle right angled at B. Let D and E be any points on AB and BC respectively. Prove that AE2 + CD2 = AC2 + DE2.
Given: In △ABC; ∠B = 90°
D and E are points on AB and BC.
R.T.P.: AE2 + CD2 = AC2 + DE2
Proof: In △BCD, △BCD is a right triangle right angled at B.
∴ BD2 + BC2 = CD2 ……… (1)
[∵ Pythagoras theorem states that hypotenuse2 = side2 + side2]
In △ABE; ∠B = 90°
Adding (1) and (2), we get
BD2 + BC2 + AB2 + BE2 – CD2 + AE2
(BD2 + BE2) + (AB2 + BC2) = CD2 + AE2
DE2 + AC2 – CD2 + AE2 [Q.E.D.]
[∵ (i) In △DBE, ∠B = 90° and DE2 = BD2 + BE2
(ii) In △ABC, ∠B = 90° and AB2 + BC2]
Question 3.
Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
Given: △ABC, an equilateral triangle;
AD – altitude and the side is a units, altitude h units.
R.T.P: 3a2 = 4h2
Proof: In △ABD, △ACD
∠B = ∠C [∵ 60°]
∴ ∠BAD = ∠DAC [∵ Angle sum property]
Also, BA = CA
∴ △ABD s △ACD (by SAS congruence condition)
Hence, BD = CD = $$\frac{1}{2}$$BC = $$\frac{a}{2}$$ [∵ c.p.c.t]
Now in △ABD, AB2 = AD2 + BD2
[∵ Pythagoras theorem]
a2 = h2 + $$\left(\frac{a}{2}\right)^{2}$$
a2 = h2 + $$\frac{a^{2}}{4}$$
h2 = $$\frac{4 a^{2}-a^{2}}{4}$$
∴ h2 = $$\frac{3 a^{2}}{4}$$
⇒ 4h2 = 3a2 (Q.E.D)
Question 4.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR.
Given: In △PQR, ∠P = 90° and PM ⊥ QR.
R.T.P : PM2 = QM . MR
Proof: In △PQR; △MPR
∠P = ∠M [each 90°]
∠R = ∠R [common]
∴ △PQR ~ △MPR ……… (1)
[A.A. similarity]
In △PQR and △MQP,
∠P = ∠M (each 90°)
∠Q = ∠Q (common)
∴ △PQR ~ △MQP ……… (2)
[A.A. similarity]
From (1) and (2),
△PQR ~ △MPR ~ △MQP [transitive property]
∴ △MPR ~ △MQP
$$\frac{MP}{MQ}$$ = $$\frac{PR}{QP}$$ = $$\frac{MR}{MP}$$
[Ratio of corresponding sides of similar triangles are equal]
$$\frac{PM}{QM}$$ = $$\frac{MR}{PM}$$
PM . PM = MR . QM
PM2 = QM . MR [Q.E.D]
Question 5.
ABD is a triangle right angled at A and AC ⊥ BD.
Show that (i) AB2 = BC BD
(iii) AC2 = BC DC.
Given: In △ABD; ∠A = 90° AC ⊥ BD
R.T.P.:
i) AB2 = BC . BD
Proof: In △ABD and △CAB,
∠B = ∠B [common]
∴ △ABD ~ △CBA
[by A.A. similarity condition]
Hence, $$\frac{AB}{BC}$$ = $$\frac{BD}{AB}$$ = $$\frac{AD}{AC}$$
[∵ Ratios of corresponding sides of similar triangles are equal]
$$\frac{AB}{BD}$$ = $$\frac{BC}{AB}$$
⇒ AB . AB = BC . BD
∴ AB2 = BC . BD
ii) AD2 = BD . CD
∠D = ∠D (common)
∴ △ABD ~ △CAD [A.A similarity]
Hence, $$\frac{AB}{AC}$$ = $$\frac{BD}{AD}$$ = $$\frac{AD}{CD}$$
⇒ $$\frac{BD}{AD}$$ = $$\frac{AD}{CD}$$
AD2 = BD . CD [Q.E.D]
iii) AC2 = BC . DC
Proof: From (i) and (ii)
△ACB ~ △DCA
[∵ △BAD ~ △BCA ~ △ACD
Hence, $$\frac{AC}{DC}$$ = $$\frac{BC}{AC}$$ = $$\frac{AB}{AD}$$
$$\frac{AC}{DC}$$ = $$\frac{BC}{AC}$$
AC . AC = BC . DC
AC2 = BC . DC [Q.E.D]
Question 6.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Given: In △ABC; ∠C = 90°; AC = BC.
R.T.P.: AB2 = 2AC2
Proof: In △ACB; ∠C = 90°
Hence, AC2 + BC2 = AB2
[Square of the hypotenuse is equal to sum of the squares of the other two sides – Pythagoras theorem]
⇒ AC2 + AC2 = AB2 [∵ AC = BC given]
⇒ AB2 = 2AC2 [Q.E.D.]
Question 7.
‘O’ is any point in the interior of a triangle ABC.
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB, show that
i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Given: △ABC; O’ is an interior point of △ABC.
OD ⊥ BC, OE ⊥ AC, OF ⊥ AB.
R.T.P.:
i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
Proof: In OAF, OA2 = AF2 + OF2 [Pythagoras theorem]
⇒ OA2 – OF2 = AF2 …….. (1)
In △OBD,
OB2 = BD2 + OD2
⇒ OB2 – OD2 = BD2 …….. (2)
In △OCE, OC2 = CE2 + OE2
OC2 – OE2 = CE2 ……… (3)
Adding (1), (2) and (3) we get,
OA2 – OF2 + OB2 – OD2 + OC2 – OE2 = AF2 + BD2 + CE2
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ……… (4)
ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
In △OAE,
OA2 = AE2 + OF2 ……… (1)
⇒ OA2 – OE2 = AE2
In △OBF, OB2 = BF2 + OF2
OB2 – OF2 = BF2 ……… (2)
In △OCD, OC2 = OD2 + CD2
OC2 – OD2 = CD2 ……… (3)
Adding (1), (2) and (3) we get
OA2 – OE2 + OB2 – OF2 + OC2 – OD2 = AE2 + BF2 + CD2
⇒ OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AE2 + CD2 + BF2
⇒ AF2 + BD2 + CE2 = AE2 + CD2 + BF2 [From problem (i)]
Question 8.
A wire attached to vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Height of the pole AB = 18 m.
Length of the wire AC = 24 m.
Distance beween the pole and the stake be ‘d’ meters.
By Pythagoras theorem,
Hypotenuse2 = side2 + side2
242 = 182 + d2
d2 = 242 – 182 = 576 – 324 = 252
= $$\sqrt{36 \times 7}$$
∴ d = 6√7 m.
Question 9.
Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Let the height of the first pole AB = 6 m.
Let the height of the second pole CD = 11 m.
Distance between the poles AC = 12 m.
From the figure □ ACEB is a rectangle.
∴ AB = CE = 6 m
ED = CD – CE = 11 – 6 = 5 m
Now in △BED; ∠E = 90°; DE = 5 m; BE = 12 m
BD2 = BE2 + DE2
[hypotenuse2 = side2 + side2 – Pythagoras theorem]
= 122 + 52
= 144 + 25
BD2 = 169
BD = √l69 = 13m
∴ Distance between the tops of the poles = 13 m.
Question 10.
In an equilateral triangle ABC, D is a point on side BC such that BD = $$\frac{1}{3}$$ BC. Prove that 9AD2 = 7AB2.
In △ABE, ∠E = 90°
⇒ $$\overline{\mathrm{AB}}$$ is hypotenuse.
∴ AB2 = AE2 + BE2
AE2 = AB2 – BE2
⇒ AE2 = AB2 – $$\left(\frac{BC}{2}\right)^{2}$$
= AE2 = AB2 – $$\left(\frac{AB}{2}\right)^{2}$$ (∵ AB = BC)
⇒ AE2 = $$\frac{3}{4}$$AB2 ……… (1)
⇒ $$\overline{\mathrm{AD}}$$ is hypotenuse.
⇒ AD2 = AE2 + DE2
⇒ AE2 = AD2 + DE2
⇒ 28 AB2 = 36 AD2
⇒ 7 AB2 = 9 AD2
Hence proved.
Question 11.
In the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it. Prove that 8 AE2 = 3 AC2 + 5 AD2.
In △ABC, ∠B=90°
⇒ $$\overline{\mathrm{AD}}$$ is hypotenuse.
AC2 = AB2 + BC2
3AC2 = 3AB2 + 3BC2 …….. (1)
In △ABD, ∠B = 90°
∴ AD2 = AB2 + BD2 = AB2 + $$\left(\frac{BC}{3}\right)^{2}$$
⇒ AD2 = AB2 + $$\frac{\mathrm{BC}^{2}}{9}$$
⇒ 5 AD2 = 5 AB2 + $$\frac{5 \mathrm{BC}^{2}}{9}$$ …….. (2)
(1) + (2)
3 AC2 + 5 AD2 = 3 AB2 + 3 BC2 + 5 AB2 + $$\frac{5}{9} \mathrm{BC}^{2}$$
= 8AB2 + $$\frac{32}{9} \mathrm{BC}^{2}$$ ……… (3)
Now in △ABE, ∠B = 90°
⇒ $$\overline{\mathrm{AE}}$$ is hypotenuse.
⇒ AE2 = AB2 + BE2 = AB2 + $$\left(\frac{2}{3} BC\right)^{2}$$
= AB2 + $$\frac{4}{9} \mathrm{BC}^{2}$$
⇒ AE2 = 8AB2 + $$\frac{32}{9} \mathrm{BC}^{2}$$ ……… (4)
∴ RHS of (3) and (4) are equal.
∴ LHS of (3) and (4) are equal.
∴ 8 AE2 = 3 AC2 + 5 AD2.
Hence proved.
Question 12.
ABC is an isosceles triangle right angled at B. Equilateral triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of △ABE and △ACD.
$$\frac{\Delta \mathrm{ABE}}{\Delta \mathrm{ACD}}$$ = $$\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}$$
= $$\frac{a^{2}}{2 a^{2}}$$ = $$\frac{1}{2}$$ |
# Order of Operations
This page contains links to free math worksheets for Order of Operations problems. Click one of the buttons below to view a worksheet and its answer key. You can also use the 'Worksheets' menu on the side of this page to find worksheets on other math topics.
## Order of Operations
These order of operations worksheets mix basic arithmetic, including parentheses and exponents. If you are looking for order of operations worksheets that test your PEMDAS acumen, these math worksheets are a good start. You can also find order of operations worksheets with negative numbers and order of operations worksheets with comparisons on pages.
## Order of Operations (PEMDAS)
The order of operations are set of conventions used in math to decide what order operations need to be evaluated in to consistently get to the answer to a problem. These are also called precedence rules, and the occur in math problems as well as computer programming languages. Students make errors related to order of operations because we train them to read left-to-right, and a natural tendency is simply to evaluate a math program the same way. Left-to-right processing of a mathematical expression is the short road to whatever you get when you divide by zero. Just call it bad.
Instead, check with dear Aunt Sally. PEMDAS is a mnemonic tool used to help remember what operations to perform in what order. PEMDAS stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. We can remember this ordering with the phrase, 'Please excuse dear Aunt Salley.' By remembering this phrase, we know the order to evaluate terms in an expression. Anything inside parentheses is always evaluated first, even if it contains operations that are of lower precendence. Always work from the 'inside out' when dealing with expressions that have parentheses. Within a set of parathenses, the same rules for order of operations apply, so look for other parenthees and similarlly follow all the other rules below.Next, consider any terms that have exponents. The exponent is something that you might consider strongly attached to a term in an expression, much like a sign on a number. Following this, consider any multiplication or division operations. These operations are of equal precedence, so they can be evaluated in any order themselves. The same is true of the next set of operations, addition and subtraction. They can be evaluated in any order as long as you've done all of the preceding operations completely.
The order of operations worksheets in this section provide plenty of practice, and they gradually introduce each step in the PEMDAS mnemonic. If you work your way through all of them, you'll be an order of operations expert in no time. Aunt Sally would be proud. |
# Probability Of Multiple Events – Conditions, Formulas, and Examples
The probability of multiple events is an interesting topic discussed in mathematics and statistics. There are instances where we’re observing multiple events and want particular results – when this happens, knowing how to calculate the probability of multiple events comes in handy.
The probability of multiple events helps us measure our chances of getting the desired outcomes when two or more vents are occurring. The measured probability will heavily depend on whether the given events are independent or dependent.
Seeing that this is a more complex topic than the earlier topics of probability, make sure to refresher your knowledge on the following:
• Understand how we calculate probabilities of a single event.
• Review what complementary probabilities are.
Let’s begin by understanding when we apply the particular probability we’re discussing – and we can do so by studying the spinner shown in the next section.
What are multiple events in probability?
The probability of multiple events occurs when we’re trying to calculate the probability of observing two or more events. These include experiments where we’re observing different behaviors simultaneously, drawing cards with multiple conditions, or predicting the outcome of a multi-colored spinner.
Speaking of spinners, why don’t we observe the image shown above? From this, we can see that the spinner is divided into seven regions and distinguished by either the region’s colors or labels.
Here are examples of multiple events we can check from the spinners:
• Finding the probability of spinning a violet or an $a$.
• Finding the probability of spinning a blue or a $b$.
These two conditions will require us to calculate the probability of two events occurring at the same time.
Multiple events probability definition
Let’s dive right into the definition of multiple event probabilities and when they occur. The probability of multiple events measures the likelihood that two or more events occur at the same time. We sometimes lookout for the probability of when one or two outcomes happen and whether these outcomes overlap each other.
The probability will depend on an important factor: whether the multiple events are independent or not and whether they are mutually exclusive.
• Dependent events (also known as conditional events) are events where a given event’s outcomes are affected by the remaining events’ outcomes.
• Independent events are events where one event’s outcomes are not affected by the rest of the events’ outcomes.
Here are some examples of events that are dependent and independent of each other.
Dependent Events Independent Events Drawing two balls consecutively from the same bag. Finding one ball each from two bags. Picking two cards without replacement. Picking a card and rolling a die. Buying more lottery tickets to win the lottery. Winning the lottery and seeing your favorite show on a streaming platform.
Events can also be mutually exclusive– these are events where they can never happen simultaneously. Some examples of mutually exclusive are the chances of turning left or right at the same time. Ace and king cards from a deck are also mutually exclusive.
Knowing how to distinguish these two events will be extremely helpful when we learn how to evaluate the probabilities of two or more events that occur together.
How to find the probability of multiple events?
We’ll be using different approaches when finding the probability of multiple events occurring together depending on whether these events are dependent, independent, or mutually exclusive.
Finding the Probability of Independent Events
\begin{aligned}P(A \text{ and } B) &=P(A) \times P(B)\\P(A \text{ and } B \text{ and } C\text{ and }…) &=P(A) \times P(B) \times P(C) \times … \end{aligned}
When we’re working with independent events, we can calculate the probability occurring together by multiplying the respective probabilities of the events occurring individually.
Let’s say we have the following objects handy:
• A bag that contains $6$ red and $8$ blue chips.
• A coin is in your purse.
• A deck of cards is on your office table.
How do we find the probability that we get a red chip and toss the coin and get tails, and draw a card with a heart suit?
These three events are independent of each other, and we can find the probability of these events occurring together by first finding the probability that they occur independently.
As a refresher, we can find their independent probabilities by dividing the number of outcomes by the total number of possible outcomes.
Event Symbol Probability Getting a red chip $P(r)$ $P(r) = \dfrac{6}{14} = \dfrac{5}{7}$ Tossing the coin and get a tails $P(t)$ $P(t) = \dfrac{1}{2}$ Drawing a hearts $P(h)$ $P(h) = \dfrac{13}{52} = \dfrac{1}{4}$ \begin{aligned}P(r \text{ and }t \text{ and }h)&= P(r) \cdot P(t)\cdot P(h)\\&= \dfrac{5}{7}\cdot \dfrac{1}{2} \cdot \dfrac{1}{4}\\&= \dfrac{5}{56} \end{aligned}
Finding the Probability of Dependent Events
\begin{aligned}P(A \text{ and } B) &=P(A) \times P(B \text{ given } A)\\&= P(A)\times P(B|A)\\P(A \text{ and } B \text{ and } C) &=P(A) \times P(B \text{ given } A)\times P(C \text{ given } A\text{ and }B)\\&=P(A) \times P(B|A)\times P(C|A \text{ and } B) \end{aligned}
We can calculate for the probability of dependent events occurring together as shown above. Need a refresher on what $P(A|B)$ represents? It simply means the probability of $A$, once $B$ has happened. You’ll know more about conditional probability and be able to try out more complex examples here.
Let’s say we want to find out the probability of getting three jacks consecutively if we don’t return the drawn card each draw. We can keep in mind that three events are occurring in this situation:
• The probability of getting a jack on the first draw – we still have $52$ cards here.
• The probability of getting a second jack on the second draw (we now have $3$ jacks and $51$ cards).
• The third event is getting a third jack for the third row – $2$ jacks left and $50$ cards on the deck.
We can label these three events as $P(J_1)$, $P(J_2)$, and $P(J_3)$. Let’s work on the important components to calculate the probability of these three dependent events happening together.
Event Symbol Probability Drawing a jack the first time $P(J_1)$ $\dfrac{4}{52}= \dfrac{1}{13}$ Drawing a jack the second time $P(J_2|J_1)$ $\dfrac{4 -1}{52 -1} = \dfrac{1}{17}$ Drawing a jack the third time $P(J_3|J_1 \text{ and } J_2)$ $\dfrac{3-1}{51 -1} = \dfrac{1}{25}$ \begin{aligned}P(J_1) \times P(J_2 \text{ given } J_1)\times P(J_3 \text{ given } J_2\text{ and }J_1)&=P(J_1) \times P(J_2|J_1)\times P(J_3|J_1 \text{ and } J_2)\\&=\dfrac{4}{52}\cdot\dfrac{3}{51}\cdot\dfrac{2}{50}\\&= \dfrac{1}{13}\cdot \dfrac{1}{17}\cdot \dfrac{1}{25}\\&= \dfrac{1}{5525} \end{aligned}
Finding the Probability of Mutually Exclusive or Inclusive Events
We may also need to explore if the given events are mutually inclusive or exclusive to help us calculate the probability of multiple events where the outcome we’re looking for do not require all outcomes to occur altogether.
Here’s a table that summarizes the formula for mutually exclusive or inclusive events:
Type of Event Formula for the Probability Mutually Inclusive $P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)$ Mutually Exclusive $P(A \text{ or } B) = P(A) + P(B)$
Keep in mind that we’re now using “or” because we’re looking for the probabilities of events that occur individually or occur together.
These are all the concepts and formulas you’ll need to understand and solve problems that involve multiple events’ probability. We can go ahead and try out these examples shown below!
Example 1
A canvas bag contains $6$ pink cubes, $8$ green cubes, and $10$ purple cubes. One cube is removed from the bag and then replaced. Another cube is drawn from the bag, and repeat this one more time. What is the probability that the first cube is pink, the second cube is purple, and the third is another pink cube?
Solution
Keep in mind that the cubes are returned each time we draw another. Since the next draw’s probability is not affected by the first draw results, the three events are independent of each other.
When this happens, we multiply the individual probabilities to find the probability of having the outcome that we want.
Event Symbol Probability Drawing a pink cube in the first draw $P(C)$ $P(C_1) = \dfrac{6}{24}= \dfrac{1}{4}$ Drawing a purple cube in the second draw $P(C_2)$ $P(C_2) = \dfrac{10}{24}= \dfrac{5}{12}$ Drawing another pink cube in the third draw $P(C_3)$ $P(C_3) = \dfrac{6}{24}= \dfrac{1}{4}$ \begin{aligned}P(C_1 \text{ and }C_2\text{ and }C_3)&= P(C_1) \cdot P(C_2)\cdot P(C_3)\\&= \dfrac{1}{4}\cdot \dfrac{5}{12} \cdot \dfrac{1}{4}\\&= \dfrac{5}{192} \end{aligned}
This means that the probability of drawing a pink cube then a purple cube then another pink cube is equal to $\dfrac{5}{192}$.
Example 2
A book club of $40$ enthusiastic readers, $10$ prefers nonfiction books, and $30$ prefers fiction. Three book club members will be randomly selected to serve as the next book club meeting’s three hosts. What is the probability that all three members will prefer nonfiction?
Solution
When the first member is selected as the first host, we can no longer include them in the next random selection. This shows that the three outcomes are dependent on each other.
• For the first selection, we have $40$ members and $30$ nonfiction readers.
• For the second selection, we now have $40 -1 = 39$ members and $30- 1= 29$ nonfiction readers.
• Hence, for the third, we have $38$ members and $28$ nonfiction readers.
Event Symbol Probability Randomly selecting a nonfiction reader $P(N_1)$ $\dfrac{30}{40}= \dfrac{3}{4}$ Selecting another nonfiction reader $P(N_2|N_1)$ $\dfrac{29}{39}$ Selecting a nonfiction reader the third time $P(N_3|N_1 \text{ and } N_2)$ $\dfrac{28}{38} = \dfrac{14}{19}$ \begin{aligned}P(N_1) \times P(N_2 \text{ given } N_1)\times P(N_3 \text{ given }N_2\text{ and }N_1)&=P(N_1) \times P(N_2|N_1)\times P(N_3|N_1 \text{ and } N_2)\\&=\dfrac{30}{40}\cdot\dfrac{29}{39}\cdot\dfrac{28}{38}\\&= \dfrac{3}{4}\cdot \dfrac{29}{39}\cdot \dfrac{14}{19}\\&= \dfrac{203}{494} \end{aligned}
Hence, the probability of selecting three nonfiction readers is equal to $\dfrac{203}{494}\approx 0.411$.
Example 3
Let’s go back to the spinner that was introduced to us in the first section, and we can actually determine the probabilities of the following:
a. Spinning a violet or an $a$.
b. Spinning a blue or a red.
Solution
Let’s take note of the colors and labels found in each spinner.
Color $\rightarrow$Label $\downarrow$ Violet Green Red Blue Total $a$ $1$ $1$ $0$ $1$ $3$ $b$ $2$ $0$ $0$ $0$ $2$ $c$ $0$ $0$ $1$ $1$ $2$ Total $3$ $1$ $1$ $2$ $7$
Take note of the keyword “or” – this means that we account for the probability that either outcome occurs. For problems like this, it’s important to note whether the conditions are mutually exclusive or inclusive.
For the first condition, we want the spinner to land on either a violet region or a region labeled $a$, or both.
• There are $3$ violet regions and $3$ regions labelled $a$.
• There is a $1$ region where it’s both violet and labeled $a$.
This shows that the incident is mutually inclusive. Hence, we use $P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)$
\begin{aligned}P(V \text{ or } a) &= P(V) + P(a) – P(V \text{ and } a)\\&=\dfrac{3}{7} + \dfrac{3}{7} – \dfrac{1}{7}\\&= \dfrac{5}{7}\end{aligned}
a. This means that the probability is equal to $\dfrac{5}{7}$.
It’s impossible to land on a red region and a blue one all at the same time. This means that these two events are mutually exclusive. For these types of events, we add their individual probabilities.
b. This means that the probability is equal to $\dfrac{1}{7} + \dfrac{2}{7} = \dfrac{3}{7}$.
Practice Questions
1. A canvas bag contains $12$ pink cubes, $20$ green cubes, and $22$ purple cubes. One cube is removed from the bag and then replaced. Another cube is drawn from the bag, and repeat this one more time. What is the probability that the first cube is green, the second cube is purple, and the third is another green cube?
2. In a book club of $50$ enthusiastic readers, $26$ prefer nonfiction books, and $24$ prefer fiction. Three book club members will be randomly selected to serve as the three hosts of the next book club meeting
a. What is the probability that all three members will prefer fiction?
b. What is the probability that all three members will prefer nonfiction?
3. Using the same spinner from the first section, determine the probabilities of the following:
a. Spinning a green or an $a$.
b. Spinning a $b$ or a $c$.
1. $\dfrac{1100}{19683} \approx 0.056$
2.
a. $\dfrac{253}{2450} \approx 0.103$
b. $\dfrac{13}{98} \approx 0.133$
3.
a. $\dfrac{3}{7}$
b. $\dfrac{4}{7}$ |
# How do you solve the inequality abs(2-3x)>=2/3?
Jul 26, 2017
See a solution process below: $\left(- \infty , \frac{4}{9}\right]$; $\left[\frac{8}{9} , + \infty\right)$
#### Explanation:
The absolute value function takes any negative or positive term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
$- \frac{2}{3} \ge 2 - 3 x \ge \frac{2}{3}$
First, subtract $\textcolor{red}{2}$ from each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:
$- \frac{2}{3} - \textcolor{red}{2} \ge 2 - \textcolor{red}{2} - 3 x \ge \frac{2}{3} - \textcolor{red}{2}$
$- \frac{2}{3} - \left(\frac{3}{3} \times \textcolor{red}{2}\right) \ge 0 - 3 x \ge \frac{2}{3} - \left(\frac{3}{3} \times \textcolor{red}{2}\right)$
$- \frac{2}{3} - \frac{6}{3} \ge - 3 x \ge \frac{2}{3} - \frac{6}{3}$
$- \frac{8}{3} \ge - 3 x \ge - \frac{4}{3}$
Now, divide each segment by $\textcolor{b l u e}{- 3}$ to solve for $x$ while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:
$\frac{- \frac{8}{3}}{\textcolor{b l u e}{- 3}} \textcolor{red}{\le} \frac{- 3 x}{\textcolor{b l u e}{- 3}} \textcolor{red}{\le} \frac{- \frac{4}{3}}{\textcolor{b l u e}{- 3}}$
$\frac{\frac{- 8}{3}}{\frac{\textcolor{b l u e}{- 3}}{1}} \textcolor{red}{\le} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}} x}{\cancel{\textcolor{b l u e}{- 3}}} \textcolor{red}{\le} \frac{\frac{- 4}{3}}{\frac{\textcolor{b l u e}{- 3}}{1}}$
$\frac{- 8 \times 1}{3 \times \textcolor{b l u e}{- 3}} \textcolor{red}{\le} x \textcolor{red}{\le} \frac{- 4 \times 1}{3 \times \textcolor{b l u e}{- 3}}$
$\frac{- 8}{- 9} \textcolor{red}{\le} x \textcolor{red}{\le} \frac{- 4}{- 9}$
$\frac{8}{9} \textcolor{red}{\le} x \textcolor{red}{\le} \frac{4}{9}$
Or
$x \le \frac{4}{9}$; $x \ge \frac{8}{9}$
Or, in interval notation:
$\left(- \infty , \frac{4}{9}\right]$; $\left[\frac{8}{9} , + \infty\right)$ |
# Inequality – Prove the Bound for Square Roots of Consecutive Numbers
inequalitynumber-comparison
Question a) Prove that $$\frac{1}{90} < \sqrt{2024} – \sqrt{2023} <\frac{1}{88}$$
Question b) Is $$\sqrt{2024} – \sqrt{2023}$$ smaller or larger than $$\frac{1}{89}$$
No calculator is allowed, obviously and neither are methods of guessing the roots, we must move from the true inequality $$0<1$$ to that which we must prove.
Guessing that we must start by squaring both sides which is how i started.
$$\frac{1}{90^2} < 4047-2*\sqrt{2024}*\sqrt{2023} <\frac{1}{88^2}$$
i focus on the left side of the equation to first succesfuly prove that and get one side to one
$$1 < 90^2 (4047-2*\sqrt{2024}*\sqrt{2023})$$
This was were i got stuck maybe this is the intended transformation?
$$1 < 90^2 (2*45^2 -3 -2*\sqrt{45^2-1)(45^2-2)})$$
Now thanks to help i have a basic proof. We start by multiplying by the conjugate
$$\frac{\sqrt{2024} + \sqrt{2023}}{90} < (\sqrt{2024} – \sqrt{2023})(\sqrt{2024} + \sqrt{2023})<\frac{\sqrt{2024} + \sqrt{2023}}{88}$$
Simplify
$$\frac{\sqrt{45^2 -1} + \sqrt{45^2 -2}}{2*45} < 1 <\frac{\sqrt{45^2 -1} + \sqrt{45^2 -2}}{2*(45-1)}$$
Now simple proof says that in LL we have two numbers smaller than 45 added togheter in the numerator and larger than that in the denominator. While in RL we have that these must be larger than 44 and thus a denominator that is smaller than the numerator. QED LL is smaller than one and RL is larger and thus the original statement is true.
Any help is appreciated!
This was a question on a test on the chapter inequalites
My teacher should provide a solution in about one week.
Let $$x = \frac{1}{\sqrt{2024} - \sqrt{2023}}$$. Rationalizing the denominator gives:
$$x = \frac{\sqrt{2024} + \sqrt{2023}}{(\sqrt{2024} - \sqrt{2023})(\sqrt{2024} + \sqrt{2023})}$$ $$= \frac{\sqrt{2024} + \sqrt{2023}}{2024 - 2023}$$ $$= \sqrt{2024} + \sqrt{2023}$$
Note that $$44^2 = 1936$$ and $$45^2 = 2025$$. So each square root must be in this interval.
$$44 < \sqrt{2023} < 45$$ $$44 < \sqrt{2024} < 45$$
$$88 < \sqrt{2024} + \sqrt{2023} < 90$$ $$88 < x < 90$$
Since all values have the same sign, taking the reciprocals inverts the order:
$$\frac{1}{88} > \frac{1}{x} > \frac{1}{90}$$ $$\boxed{\frac{1}{90} < \sqrt{2024} - \sqrt{2023} < \frac{1}{88}}$$
If we bisect the interval $$(44, 45)$$, we get:
$$44.5^2 = 1980.25$$ $$44.5 < \sqrt{2023} < 45$$ $$44.5 < \sqrt{2024} < 45$$ $$89 < x < 90$$
$$\boxed{\frac{1}{90} < \sqrt{2024} - \sqrt{2023} < \frac{1}{89}}$$ |
graemethin
bannerdive
graemethin2
EASIER THAN YOU THINK...
# Linear Equations
The simplest graphs to master are straight line graphs.
Every straight line can be drawn using just two points, or a point and a gradient. These two bits of information can be found in a variety of ways and your first task is to learn how. You will learn to recognise the equations that produce straight lines. They contain an x and/or a y term and some numbers, but nothing else, nothing fancy … no xy terms or powers or radicals or other functions … just a simple x and/or y term. Two examples are on the graph at left. Such equations are called linear equations.
You will learn that linear equations come in a variety of forms, depending on the positions of the terms relative to the = sign. 2x – 3y + 6 = 0 is in the General Form, 2x – 3y = -6 is in the Standard Form, y = 2x + 7 is in the Gradient-Intercept Form, and x/4 – y/3 = 1 is in the Intercept Form. These are the main forms for linear equations. You will likely concentrate on the General Form and Gradient-Intercept Form during your schooling. Once you have learned to graph lines using each of these forms, you will then learn to calculate the equations for parallel and perpendicular lines, calculate where lines will intersect (see image at left), and calculate the distance from a point to a line.
This is quite a body of skills to master, but I encourage you to practise until you find this work almost intuitive. All your ‘straight line’ skills will form a foundation for other curve sketching skills, for the fundamentals of calculus, and for learning about sequences and series and other mathematical matters.
How to Find the Equation of a Parallel Line in 4-5 Lines of Work
So often, we make calculations more difficult than they need be. This is very true of the way in which we find the equations of lines parallel to a given line through some given point.
So that students understand that the two parallel lines will have the same gradient, we teach them to rearrange a linear equation into the gradient-intercept form first ... and then (often) convert it back to some other form at the end of the process. This 'double conversion' is useful in driving home the principle that parallel lines have identical gradients, but it is terribly wasteful and inefficient if you simply want to find the final equation!
In this video I show you how to identify where the gradient's information is 'stored' in a linear equation. I then show you how to use that knowledge to simplify the entire calculation to just 4-5 lines of (simple) work!
Elementary-Intermediate: If you are new to calculating the equations for parallel lines, please watch this longer video (26:44):
Advanced: If you already know how to perform this calculation the conventional way, and would like to see the simpler format, then you might like to watch this shorter video (12:24):
How to Find the Equation of a Perpendicular Line in 5-6 Lines of Work
As I shared in my videos concerning parallel lines, we often make calculations more difficult than they need be. This is also true of the way in which we find the equations of lines perpendicular to a given line through some given point.
So that students understand that two perpendicular lines will have gradients that are negative reciprocals of each other, we teach them to rearrange a linear equation into the gradient-intercept form first ... and then (often) convert it back to some other form at the end of the process. This 'double conversion' is useful in driving home the principle that perpendicular lines have gradients that are related in that way, but it is terribly wasteful and inefficient if you simply want to find the final equation!
In this video I show you how to identify where the gradient's information is 'stored' in a linear equation. I then show you how to use that information to simplify the entire calculation to just 5-6 lines of (simple) work!
Brilliant!!! This helps so much!!! So much easier than anything else I have found. This is now simple.
Thanks again for showing the simple way of doing this. I look forward to the next video!! Absolutely brilliant.
Once again, excellent! I’ve watched many others, that have not helped nearly as much as these. Thanks for making it so easy to understand. When things get complicated, it is easy to make a mistake, but your method of writing down the structure helps to prevent mistakes.
Thanks a million!!! I’ve watched many videos and read tutorials, but still could not get two in a row correct on Khan Academy. Your explanation was so clear and simple that it finally made sense to me and now I can get all of the problems correct!!! Thank you very much.
It all makes total sense now! Thanks very much for all the derivative videos!!! Very helpful. |
# Relative speed problem
A highway patrol plane is flying $$1$$ mile above a long, straight road, with constant ground speed of $$120$$ m.p.h. Using radar, the pilot detects a car whose distance from the plane is $$1.5$$ miles and decreasing at a rate of $$136$$ m.p.h.
### How fast is the car traveling along the highway? (Hint: You may give an exact answer, or use the fact that $$\sqrt5 \approx 2.2$$.)
I drew a triangle diagram depicting the position of the plane and car relative to the ground, where the height of the triangle is 1, the horizontal distance from the plane to the car is given by $$x$$, and the hypotenuse of the triangle $$y$$ is given by the Pythagorean theorem: $$y = \sqrt{x^2+1}$$.
At time $$t = 0$$, we know that $$y = 1.5$$ and $$x= \sqrt{y^2-1} = \frac{\sqrt{5}}{2}$$
By differentiating y, we get: $$\frac{dy}{dt} = \frac{2x}{2\sqrt{x^2+1}} \frac{dx}{dt} = \frac{x}{\sqrt{x^2+1}} \frac{dx}{dt} = -136 mph$$
Note that $$\frac{dy}{dt}$$ is negative since the distance between the car and the plane is decreasing at a rate of 136 mph.
When we subsitute $$x= \frac{\sqrt{5}}{2}$$, we get: $$\frac{dx}{dt} = \frac{\sqrt{x^2+1}}{x} *\left(-136 mph \right) = \frac{3}{\sqrt5} *\left(-136 mph \right) \approx \frac{-408}{2.2} \approx -185.45 mph$$
Given that we now know that $$\frac{dx}{dt}$$ is approximately equal to -185.45 mph and the velocity of the plane $$\left( \frac{dp}{dt}\right)$$ is given by 120 mph, what is the formula for the velocity of the car $$\left( \frac{dc}{dt}\right)$$?
What confuses me is the negative and positive signs of the velocities. For example, I got the following formula: $$\frac{dc}{dt} = \frac{dp}{dt} -\frac{dx}{dt}$$
However, this formula only produces the correct answer if $$\frac{dp}{dt} = -120 mph$$, making $$\frac{dc}{dt} = 65.45 mph$$
How do we know that the velocity of the plane is negative?
• Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site. Commented Nov 22, 2019 at 16:04
• logic and computations both seem correct, don't think of the velocity as something positive or negative but as same or opposite. If we assume the car is traveling along the x-axis, the plane will be traveling opposite to this direction as their relative distances (both x and y) are shrinking. Commented Nov 23, 2019 at 6:21
You defined $$x$$ as the horizontal distance between the airplane and car. As the airplane approaches the car, the distance shrinks, which makes $$\frac{dx}{dt}$$ negative.
The rate at which the horizontal distance shrinks depends on whether the airplane and car are heading in the same direction or opposite directions. If the airplane and car were heading in the same direction, the rate at which the horizontal distance shrinks would be the difference of the speeds of the airplane and car, which would be smaller than the speed of the airplane. Since $$\left|\frac{dx}{dt}\right| = 185.45~\text{mph}$$ exceeds the speed of the airplane, the airplane and car must be headed in opposite directions, so the rate at which the horizontal distance shrinks must equal the sum of the speeds of the airplane and car. Since $$\frac{dx}{dt} = -185.45~\text{mph}$$ the relative horizontal speed at which the airplane and car approach each other is $$\left|\frac{dx}{dt}\right| = 185.45~\text{mph}$$ Hence, $$185.45~\text{mph} = \left|\frac{dp}{dt}\right| + \left|\frac{dc}{dt}\right| = 120~\text{mph} + \left|\frac{dc}{dt}\right| \implies \left|\frac{dc}{dt}\right| = 65.45~\text{mph}$$ The reason the velocities of both the airplane and car are negative is that as the airplane and car approach each other, the distance between them shrinks.
$$\frac{dp}{dt} = -120~\text{mph}$$ since the speed of the plane causes the distance $$x$$ between the airplane and the car to decrease at a rate of $$120~\text{mph}$$. Similarly, $$\frac{dc}{dt} = -65.45~\text{mph}$$ since the speed of the car causes the distance $$x$$ between the airplane and the car to decrease at a rate of $$65.45~\text{mph}$$.
The problem statement is ambiguous.
Assuming both the airplane and the car are moving along parallel, perfectly horizontal straight lines (one along the road, which is assumed to be level, the other parallel to and above the road) and that those parallel lines are exactly $$1$$ mile apart, the horizontal distance $$x$$ between the car and the airplane must be decreasing at the rate of about $$182.46$$ miles per hour. (Note: if you're going to give an answer to five digits, don't make it depend on a factor that is accurate to only two digits.)
And that is all we know from the problem statement about the relative motion of the car and the aircraft. We don't know whether the car is approaching the aircraft from far ahead or catching up to it from behind.
If it's approaching from far ahead, the car's speed is approximately $$182.46 - 120 = 62.46 \mathrm{mph}.$$
If it's catching up from behind, the car's speed is approximately $$182.46 + 120 = 302.46 \mathrm{mph}.$$
We can reasonably rule out the second possibility if the car is a stock Ford Escort, but not if it is a specially-modified Bugatti Chiron.
The best assumption to make when a problem like this is set, if you must give the answer as one number with no further explanation, is that whoever set the problem did not even consider the second possibility. |
# Variation of Parameters
Author: John J Weber III, PhD Corresponding Textbook Sections:
• Section 4.6 -- Variation of Parameters
## Expected Educational Results
• Objective 13–1: I understand when variation of parameters is required to find particular solutions to nonhomogeneous equations.
• Objective 13–1: I can use variation of parameters to find particular solutions to nonhomogeneous equations.
## Variation of Parameters
### Definition: Particular Solution
A solution to a nonhomogeneous DE is called the particular solution, $\displaystyle y_p(t)$.
### Definition: Variation of Parameters
Variation of parameters is a more general method to find $y_p(t)$.
Variation of parameters is used for nonhomogeneous solutions when $f(t)$ contains factors or terms other than $P_m(t)$, $e^{rt}$, $\cos{(\beta t)}$, and $\sin{(\beta t)}$.
### Definition: General Solution
$\displaystyle y_g(t)=y_h(t)+y_p(t)$.
### Method of Variation of Parameters
NOTE: The terms determined by the method of variation of parameters must be linearly independent of the terms in $y_h(t)$.
1. Find two linearly independent solutions $\left\{y_1(t),y_2(t)\right\}$ to the corresponding homogeneous equation.
2. Let $y_p(t)=v_1(t)y_1(t)+v_2(t)y_2(t)$, for some functions $v_1(t)$ and $v_2(t)$.
3. Solve for $v_1(t)$:
• $v_1(t)=\int{\dfrac{-f(t)y_2(t)}{a W[y_1,y_2](t)}\,dt}$
4. Solve for $v_2(t)$:
• $v_2(t)=\int{\dfrac{f(t)y_1(t)}{a W[y_1,y_2](t)}\,dt}$
5. The particular solution is $y_p(t)=v_1(t)y_1(t)+v_2(t)y_2(t)$.
6. Write the general solution, $y(t)=y_h(t)+y_p(t)$.
#### Investigation 02
Solve the following DEs:
1. $\displaystyle y^{\,\prime\prime}+4y^{\prime}+4y=\sin{(x)}$
2. $\displaystyle y^{\,\prime\prime}+y^{\prime}=\sec{(x)}$
3. $\displaystyle y^{\,\prime\prime}+4y^{\prime}+4y=x^{-2}e^{-2x}$, $x>0$
4. $\displaystyle y^{\,\prime\prime}+4y=\tan{(2x)}$
5. $\displaystyle y^{\,\prime\prime}-2y^{\prime}+2y=e^x\sec{(x)}$
6. $\displaystyle y^{\,\prime\prime}+2y^{\prime}+y=e^{-x}\ln{(x)}$
7. $\displaystyle y^{\,\prime\prime}(t)-2y^{\,\prime}(t)+y(t)=\frac{e^t}{1+t^2}$ |
# Illustrative Mathematics Grade 8, Unit 1, Lesson 11: What Is the Same?
Learning Targets:
• I can decide visually whether or not two figures are congruent.
Related Pages
Illustrative Math
#### Lesson 11: What Is the Same?
Let’s decide whether shapes are the same.
Illustrative Math Unit 8.1, Lesson 11 (printable worksheets)
#### Lesson 11 Summary
The following diagrams describe how two figures are congruent if one can be lined up exactly with the other by a sequence of rigid transformations.
#### Lesson 11.1 Find the Right Hands
A person’s hands are mirror images of each other. In the diagram, a left hand is labeled. Shade all of the right hands.
#### Lesson 11.2 Are They the Same?
For each pair of shapes, decide whether or not they are the same.
#### Lesson 11.3 Area, Perimeter, and Congruence
1. Which of these rectangles have the same area as Rectangle R but different perimeter?
2. Which rectangles have the same perimeter but different area?
3. Which have the same area and the same perimeter?
4. Use materials from the geometry tool kit to decide which rectangles are congruent. Shade congruent rectangles with the same color.
#### Are you ready for more?
In square ABCD, points E, F, G, and H are midpoints of their respective sides. What fraction of square ABCD is shaded? Explain your reasoning.
#### Lesson 11 Practice Problems
1. If two rectangles have the same perimeter, do they have to be congruent? Explain how you know.
2. Draw two rectangles that have the same area, but are not congruent.
3. For each pair of shapes, decide whether or not it appears that the two shapes are congruent. Explain your reasoning.
4. a. Reflect Quadrilateral A over the x-axis. Label the image quadrilateral B. Reflect Quadrilateral B over the y-axis. Label the image C.
b. Are Quadrilaterals A and C congruent? Explain how you know.
5. The point (-2,-3) is rotated 90 degrees counterclockwise using center (0,0). What are the coordinates of the image?
6. Describe a rigid transformation that takes Polygon A to Polygon B.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Solved: 4. A newspaper finds that the mean number of typographical errors per page is five. Find the probabi
#### ByDr. Raju Chaudhari
Oct 15, 2020
1. A newspaper finds that the mean number of typographical errors per page is five. Find the probability that
a. exactly five typographical errors will be found on a page,
b. fewer than five typographical errors will be found on a page, and
c. no typographical errors will be found on a page.
#### Solution
Let $X$ denote the number of typographical errors per page. The mean number of typographical errors per page is five, i.e., $E(X)=\lambda = 5$.
The random variable $X$ follows Poisson distribution. That is, $X\sim P(5)$.
The probability mass function of Poisson distribution with $\lambda =5$ is
\begin{aligned} P(X=x) &= \frac{e^{-5}(5)^x}{x!},\\ &\quad \; x=0,1,2,\cdots \end{aligned}
a. The probability that exactly five typographical errors will be found on a page is
\begin{aligned} P(X=5) &= \frac{e^{-5}5^{5}}{5!}\\ &= 0.1755 \end{aligned}
b. The probability that fewer than five typographical errors will be found on a page is
\begin{aligned} P(X < 5) &=\sum_{x=0}^{4}P(X=x)\\ &=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)\\ &= \frac{e^{-5}5^{0}}{0!}+\frac{e^{-5}5^{1}}{1!}+\frac{e^{-5}5^{2}}{2!}+\frac{e^{-5}5^{3}}{3!}+\frac{e^{-5}5^{4}}{4!}\\ &= 0.0067+0.0337+0.0842+0.1404+0.1755\\ &= 0.4405 \end{aligned}
c. The probability that no typographical errors will be found on a page is
\begin{aligned} P(X=0) &= \frac{e^{-5}5^{0}}{0!}\\ &= 0.0067 \end{aligned} |
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# In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State?A) 7600B) 8000C) 8400D) Data inadequate.
Last updated date: 20th Jun 2024
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Hint:
First assume the number of candidates appeared in the examination as ‘x’. Then compute 6% of it as selected from state A, and also compute 7% of it as selected from state B. Then make a simple linear equation based on these two facts and one more fact given in the question. After solving the equation in a single variable, we can get the answer easily.
Complete step by step solution:
First, let us assume that the total number of candidates appeared in the examination as ‘x’ from each state A and B.
Now, 6% candidates got selected from the total appeared candidates, so selected candidates from state A, will be,
=$x \times \dfrac{6}{{100}} \\ \Rightarrow \dfrac{{6x}}{{100}} \\$ … (1)
Further, 7% candidates got selected from the total appeared candidates, so selected candidates from state B, will be,
=$x \times \dfrac{7}{{100}} \\ \Rightarrow \dfrac{{7x}}{{100}} \\$ …(2)
It is given in the question that from state B , 80 more candidates got selected than the state A.
Therefore from equation (1) and (2), we get
$\dfrac{{7x}}{{100}} = \dfrac{{6x}}{{100}} + 80$
Then we do transformation of variable term towards LHS, then
$\dfrac{{7x}}{{100}} - \dfrac{{6x}}{{100}} = 80$
Now we do some more simplification as follows to get the value of x,
$\Rightarrow \dfrac{x}{{100}} = 80 \\ \Rightarrow x = 8000 \\$
$\therefore$ The total number of candidates appearing in the examination from each state is 8000.
Thus the correct option is B.
Note:
Simple arithmetic based questions using the percentage value of some term, is very popular for many competitive examinations as well. Also, to enhance the mathematical logical and computational skill practice of such questions is very important. Also, here solving linear equations in a single variable is the case for solution. |
# Factors of 5
Last Updated: August 16, 2024
## Factors of 5
Factors of 5 are the numbers that can be multiplied together to produce the original number, 5. These factors are 1 and 5. When you divide 5 by any of these numbers, the result is a whole number with no remainder. Understanding the factors of a number is essential in various mathematical calculations, including simplifying fractions, finding common denominators, and solving algebraic equations. The simplicity of 5, being a prime number, means its factors are limited, making it a straightforward example for learning about the concept of factors in mathematics.
## What are the Factors of 5?
The factors of 5 are the numbers that can be multiplied together to result in 5. These numbers are 1 and 5. When 5 is divided by either of these factors, the result is a whole number with no remainder. This makes 5 a prime number, as it has exactly two distinct factors: 1 and itself. Understanding factors is crucial for various mathematical operations, including simplifying fractions and solving equations. The factors of 5 are straightforward due to its prime nature, illustrating the concept of factors clearly and simply.
## Factors Pairs of 5
Factor pairs of 5 are the sets of two numbers that, when multiplied together, equal 5. Since 5 is a prime number, it has only one factor pair:
• (1, 5)
This means 1 multiplied by 5 equals 5. Factor pairs are useful in various mathematical contexts, such as solving equations and understanding number properties. For the number 5, its single factor pair (1, 5) reflects its simplicity as a prime number.
## How to Calculate Prime Factors of 5?
Calculating the prime factors of a number involves breaking it down into the prime numbers that multiply together to give the original number. For the number 5, here’s a step-by-step guide:
### Step 1: Understand Prime Numbers
Prime numbers are natural numbers greater than 1 that have no divisors other than 1 and themselves. Examples include 2, 3, 5, 7, and so on.
### Step 2: Identify the Number
Recognize the number you want to factorize. In this case, it is 5.
### Step 3: Check for Smallest Prime Number
Start with the smallest prime number, which is 2. Determine if 2 can divide 5 without leaving a remainder. Since 5 is an odd number, 2 cannot divide it.
### Step 4: Move to the Next Prime Number
Proceed to the next prime number, which is 3. Check if 3 can divide 5. Since 5 divided by 3 leaves a remainder, 3 is not a factor of 5.
### Step 5: Check if the Number Itself is Prime
Since no smaller prime numbers can divide 5, check if 5 is a prime number. A prime number is only divisible by 1 and itself. Since 5 meets this criterion, it is a prime number.
### Step 6: Conclusion
Conclude that the prime factors of 5 are simply the number 5 itself.
## Factors of 5 : Examples
### Example 1: Basic Arithmetic
You need to divide a set of 5 pencils equally among 5 students.
### Example 2: Doubling a Recipe
A smoothie recipe calls for 5 bananas to serve 5 people. You need to adjust the recipe to serve 10 people.
Double the number of bananas using the factor 5. So, 5×2=10 bananas are needed for 10 people.
### Example 3: Distributing Stickers
You have 25 stickers and want to distribute them equally in 5 gift bags.
Divide the stickers by the number 5. Each bag gets 25÷5=5 stickers.
### Example 4: Organizing Teams
25 students need to be divided into teams, each containing 5 students.
Form teams by dividing the total number of students by 5. There are 25÷5=5 teams.
### Example 5: Calculating Multiples
You want to calculate the first five multiples of 5 to help students learn multiplication.
• 1st multiple: 5×1=5
• 2nd multiple: 5×2=10
• 3rd multiple: 5×3=15
• 4th multiple: 5×4=20
• 5th multiple: 5×5=25
## Factors of 5 : Tips
Understanding the factors of a number can simplify many mathematical tasks, from simplifying fractions to solving complex equations. The number 5, being a prime number, has unique characteristics that make it straightforward yet essential in various calculations. Here are some tips and tricks to help you master the factors of 5.
1. Recognize that 5 is a prime number, meaning it has no divisors other than 1 and itself.
2. To quickly check if 5 is a factor of a number, see if the number ends in 0 or 5. If it does, 5 is a factor.
3. Remembering the multiples of 5 (5, 10, 15, 20, etc.) can help quickly identify if 5 is a factor of a larger number.
4. Use the factor pair method by pairing 1 with 5. Since 5 is prime, these are the only pairs for 5.
5. When comparing two numbers, check if both are divisible by 5 to identify 5 as a common factor. For instance, both 20 and 25 have 5 as a common factor.
6. Use 5 to simplify fractions. For example, 15/20 can be simplified by dividing both the numerator and denominator by 5, resulting in 3/4.
7. To find the greatest common divisor (GCD) involving 5, list all factors and identify the largest common one. If 5 is a factor, it may be the GCD.
8. Use factors to determine the least common multiple (LCM). For 5, the LCM with another number is often a multiple of 5.
9. Remember that factors can also be negative. For 5, the negative factors are -1 and -5.
10. Create a factor tree for visualization, though simple for 5. Start with 5 and show its factors branching into 1 and 5.
## Are the factors of 5 always whole numbers?
Yes, when discussing factors in the context of integers, the factors of 5 are always whole numbers. They include both positive (1, 5) and negative (-1, -5) whole numbers.
## Can 5 be a factor of composite numbers?
Yes, 5 can be a factor of composite numbers. A composite number is a number that has more than two factors. For example, 10 and 15 are composite numbers that have 5 as one of their factors.
## Can the number 5 be a factor of zero?
Yes, 5 can be considered a factor of zero because any non-zero number multiplied by zero equals zero. Hence, 5 multiplied by zero equals zero.
## Can prime factorization be applied to 5?
Yes, prime factorization can be applied to 5. Since 5 is already a prime number, its prime factorization is simply 5 itself.
## Is 5 a common factor of 20 and 25?
Yes, 5 is a common factor of both 20 and 25. This is because 5 divides both numbers exactly: 20÷5=420÷5=4 and 25÷5=525÷5=5.
## Is 5 divisible by any other number?
No, 5 is not divisible by any number other than 1 and itself. As a prime number, its only divisors are 1 and 5.
## What is the factor tree of 5?
A factor tree of 5 is simple since 5 is a prime number. The factor tree consists of 5 itself, with no further branches as it cannot be factored further.
## What is the factor tree of 5?
A factor tree of 5 is simple since 5 is a prime number. The factor tree consists of 5 itself, with no further branches as it cannot be factored further.
Text prompt |
## What is the equation of the vertical asymptote for the graph below?
Vertical asymptotes can be found by solving the equation n(x) = 0 where n(x) is the denominator of the function ( note: this only applies if the numerator t(x) is not zero for the same x value). Find the asymptotes for the function . The graph has a vertical asymptote with the equation x = 1.
## What is the vertical asymptote in an equation?
A vertical asymptote is a vertical line that guides the graph of the function but is not part of it. It can never be crossed by the graph because it occurs at the x-value that is not in the domain of the function. … denominator then x = c is an equation of a vertical asymptote.
## How do you find the vertical asymptote of a graph?
Vertical asymptotes can be found by solving the equation n(x) = 0 where n(x) is the denominator of the function ( note: this only applies if the numerator t(x) is not zero for the same x value). Find the asymptotes for the function . The graph has a vertical asymptote with the equation x = 1.
## How do you write an equation for an asymptote?
Asymptote Equation
For Oblique asymptote of the graph function y=f(x) for the straight-line equation is y=kx+b for the limit x→+∞ x → + ∞ if and only if the following two limits are finite.
## What is the vertical asymptote of this function?
A vertical asymptote (or VA for short) for a function is a vertical line x = k showing where a function f(x) becomes unbounded. In other words, the y values of the function get arbitrarily large in the positive sense (y→ ∞) or negative sense (y→ -∞) as x approaches k, either from the left or from the right.
## How do you know if there are no vertical asymptotes?
Since the denominator has no zeroes, then there are no vertical asymptotes and the domain is “all x”. Since the degree is greater in the denominator than in the numerator, the y-values will be dragged down to the x-axis and the horizontal asymptote is therefore “y = 0”.
## What is a vertical asymptote on a graph?
Vertical asymptotes occur where the denominator becomes zero as long as there are no common factors. … If there are no vertical asymptotes, then just pick 2 positive, 2 negative, and zero. Put these values into the function f(x) and plot the points. This will give you an idea of the shape of the curve.
## How can you identify a function from a graph?
You can use the vertical line test on a graph to determine whether a relation is a function. If it is impossible to draw a vertical line that intersects the graph more than once, then each x-value is paired with exactly one y-value. So, the relation is a function.
## What is vertical and horizontal asymptote?
There are three kinds of asymptotes: horizontal, vertical and oblique. For curves given by the graph of a function y = ƒ(x), horizontal asymptotes are horizontal lines that the graph of the function approaches as x tends to +∞ or −∞. Vertical asymptotes are vertical lines near which the function grows without bound.
## How many vertical asymptotes can a function have?
You may know the answer for vertical asymptotes; a function may have any number of vertical asymptotes: none, one, two, three, 42, 6 billion, or even an infinite number of them! However the situation is much different when talking about horizontal asymptotes.
## How do you find the asymptotes of a function?
1. Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0.
2. Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote.
## What is an asymptote example?
An asymptote is a line that the graph of a function approaches but never touches. … In this example, there is a vertical asymptote at x = 3 and a horizontal asymptote at y = 1. The curves approach these asymptotes but never cross them.
## What is an asymptote in math?
Asymptote, In mathematics, a line or curve that acts as the limit of another line or curve. For example, a descending curve that approaches but does not reach the horizontal axis is said to be asymptotic to that axis, which is the asymptote of the curve.
## Can a function be defined at a vertical asymptote?
Regarding other aspects of calculus, in general, one cannot differentiate a function at its vertical asymptote (even if the function may be differentiable over a smaller domain), nor can one integrate at this vertical asymptote, because the function is not continuous there.
## Which has a vertical asymptote exponential or logarithmic?
A logarithmic function will have the domain as (0, infinity). The range of a logarithmic function is (−infinity, infinity). The logarithmic function graph passes through the point (1, 0), which is the inverse of (0, 1) for an exponential function. The graph of a logarithmic function has a vertical asymptote at x = 0.
## How do you find the asymptotes of an exponential function?
Exponential Functions
A function of the form f(x) = a (bx) + c always has a horizontal asymptote at y = c. For example, the horizontal asymptote of y = 30e6x – 4 is: y = -4, and the horizontal asymptote of y = 5 (2x) is y = 0. |
# Monotone Convergence Theorem: Examples, Proof
Not all bounded sequences converge, but if a bounded a sequence is also monotone (i.e. if it is either increasing or decreasing), then it converges. This fact, that every bounded monotone sequence converges is called the monotone convergence theorem [1].
For example, the sequence an = 2 -(4/n) converges by the theorem as it is both bounded above and monotone:
We can prove that a sequence converges using the theorem.
Example question: Prove that the following sequence converges [2]:
Solution: In order to apply the monotone convergence theorem, we have to show that the sequence is both monotone and bounded:
• The sequence is monotone decreasing because an + 1 < an
• The sequence is bounded below by zero (you can deduce this because the numerator is always smaller than the denominator, so if you try to head towards zero, you’ll just get smaller and smaller fractions, but they will never go past zero).
Therefore, by the monotone convergence theorem, this sequence converges.
## Proof of the Monotone Convergence Theorem
We can use the monotone convergence theorem to prove that bounded monotone sequences converge. But we can also prove the theorem itself. Specifically, we want to prove that the limit of a monotone bounded sequence exists. Several proofs exist, but this version is probably the most concise [3, 4]:
Suppose that an is an increasing sequence.
• Define S as the set of terms in an (S = an: n ∈ ℕ).
• We know that the limit superior L = sup(S) exists via the Completeness Axiom because this is a bounded set.
• Let ε > 0.
• Then L – ε is not an upper bound for S; by the definition of supremum (i.e. L is the least upper bound). Therefore, there must be a number N such that aN L – ε.
• Since an is increasing for all n ≥ N, it follows that an > L – ε.
• since L is an upper bound, we know that anL < L + ε.
• From the above, we can conclude that for all n ≥ N: L – ε < an < L + ε and so |anL| < ε.
## References
[1] Bakker, L. Math 341 Lecture #8 §2.4: The Monotone Convergence Theorem and a First Look at Infinite Series. Retrieved May 6, 2021 from: https://math.byu.edu/~bakker/M341/Lectures/Lec09.pdf
[2] Romik, D. (2011). Math 25 — Solutions to Homework Assignment #7. UC Davis, Spring 2011. Retrieved May 6, 2021 from: https://www.math.ucdavis.edu/~romik/teaching-pages/mat25-hw7solns.pdf
[3] More concise proof of part (a) of the monotone convergence theorem. Math 0450
Honors intro to analysis Spring, 2009. Retrieved May 6, 2021 from: http://www.math.pitt.edu/~sph/0450/0450-notes12.pdf
[4] Math 410 Section 2.3: The Monotone Convergence Theorem. Retrieved May 6, 2021 from: http://www.math.umd.edu/~immortal/MATH410/lecturenotes/ch2-3.pdf |
# LADR Assorted Solutions - Math 441/541 Assignment 1...
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Math 441/541 – Assignment 1: Solutions Chapter 1 3. Here we use the associativity of scalar multiplication and the fact that - 1 · v = - v . - ( - v ) = - 1 · ( - v ) = - 1 · ( - 1 · v ) = ( - 1 · - 1) v = 1 · v = v. 4. Let a F and v V and suppose av = 0. If a = 0 then we’re done. So assume a 6 = 0. Then we need to show that v = 0. Since a 6 = 0, we know from the fact that F is a field that a has a multiplicative inverse 1 a F , such that a · 1 a = 1. Hence we have 0 = av = 1 a · 0 = 1 a ( av ) = 1 a · a v = 1 · v = v. Thus, v = 0 as required. 5. a) { ( x 1 , x 2 , x 3 ) F 3 | x 1 + 2 x 2 + 3 x 3 = 0 } . 0 + 2(0) + 3(0) = 0, hence the zero vector 0 is in the set. If x = ( x 1 , x 2 , x 3 ) and y = ( y 1 , y 2 , y 3 ) are in the set, then x + y = ( x 1 + y 1 , x 2 + y 2 , x 3 + y 3 ), so check ( x 1 + y 1 ) + 2( x 2 + y 2 ) + 3( x 3 + y 3 ) = [ x 1 + 2 x 2 + 3 x 3 ] + [ y 1 + 2 y 2 + 3 y 3 ] = 0 + 0 = 0. Finally, if a F , then ax = ( ax 1 , ax 2 , ax 3 ) so we check ( ax 1 ) + 2( ax 2 ) + 3( ax 3 ) = a [ x 1 + 2 x 2 + 3 x 3 ] = a · 0 = 0 . So this set is a subspace. b) { ( x 1 , x 2 , x 3 ) F 3 | x 1 + 2 x 2 + 3 x 3 = 4 } . This set is not a subspace because 0 + 2(0) + 3(0) = 0 6 = 4, hence the zero vector is not an element of it. (in fact, the other two conditions fail as well!) c) { ( x 1 , x 2 , x 3 ) F 3 | x 1 x 2 x 3 = 0 } . The zero vector is in this set, and actually it is closed under scalar multiplication as well, but it is not closed under addition. Counterexample: (1 , 0 , 0) and (0 , 1 , 1) are both in the set, but the sum (1 , 1 , 1) is not. d) { ( x 1 , x 2 , x 3 ) F 3 | x 1 = 5 x 3 } . 0 = 5(0), hence the zero vector is in this set. Given vectors x and y in the set, we know that x 1 = 5 x 3 and y 1 = 5 y 3 , hence x 1 + y 1 = 5 x 3 +5 y 3 = 5( x 3 + y 3 ), so it is closed under addition. Similarly, if x 1 = 5 x 3 then ax 1 = a ( x 1 ) = a (5 x 3 ) = 5( ax 3 ), so it is also closed under scalar multi- plication. Thus it is a subspace. 6. The lattice Z 2 = { ( x, y ) R 2 | x, y Z } inside of R 2 is closed under addition and inverses (because the sum of any two integers is another integer and the additive inverse of an integer is an integer), but it is not closed under scalar multiplication. As an example (1 , 0) Z , but 1 2 · (1 , 0) = ( 1 2 , 0 ) 6∈ Z 2 . 1
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8. Let U 1 , U 2 , . . . , U n V be subspaces. Then U 1 U 2 ∩ · · · ∩ U n is a subspace of V too. Proof: Since each U i is a subspace we know that 0 U i for each i . Hence 0 U 1 ∩ · · · ∩ U n too. If x, y U 1 ∩ · · · ∩ U n , then specifically x, y U i for each i . Hence x + y U i for each i , which implies x + y U 1 ∩ · · · ∩ U n . Finally, if a F and x U 1 ∩ · · · ∩ U n , then specifically x U i for each i . Thus, ax U i for each i , which implies that ax U 1 ∩ · · · ∩ U n . 12. Addition of subspaces does have an additive identity, namely the zero subspace { 0 } . Since U + { 0 } = { u + 0 | u U } = { u | u U } = U . Given any subspaces U and W we know that U + W contains both U and W as subspaces. Hence, U + W = { 0 } implies that U = { 0 } and W = { 0 } . Hence, the only subspace with an additive inverse is the zero subspace { 0 } and its inverse is itself.
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## Chapter 11: SERIES COMPLETION
### Introduction
This involves recognizing the next term in a series and choosing one from the alternatives given.
Q. 7 , 12 , 19 , X , 39
A. X = 28 as the series follows the pattern: 7, 7+5, 12+7, 19+9, 28+11 where the number is added by the next odd number.
Q. 0, 6, 24, 60, 120, 210, X
A. X = 336 as series follows pattern 13-1 , 23-2 , 33-3 , 43-4....
Q. 4 , 6 , 12 , 14 , 28 , 30 , X
A. There are 2 series in this series 4, 12, 28 ... and 6, 14, 30 and both follow pattern +8, +16, +32, So we get missing value as 28+32 = 60
Q. 1, 4, 9, 16, 25, X
A. Series follows pattern: +3, +5, +7, +9 ...
Q. 20, 19, 17, X, 10, 5
A. Series follows pattern: -1, -2, -3, -4, -5
Q. 6, 11, 21, 36, 56...
A. Pattern: +5, +10, +15, +20, +25...
Q. 1, 6, 13, 22, 33, ...
A. Pattern: +5, +7, +9, +11 ...
Q. 1, 9, 17, 33, 49, 73, X
A. Pattern: + 4 * 2, + 4 * 2, + 4 * 4, + 4 * 4, + 4 * 6, + 4 * 6
Q. 0.5, 1.5, 4.5, 13.5, X
A. Pattern: + 1 , + 1 * 3, + 3 * 3, + 9 * 3 ; Multiply the previous value obtained by 3 each time.
Q. 19, 2, 38, 3, 114, 4, X
A. There are two series: 2, 3, 4, ... and 19, 38, 114, X. The second series follows patterns: * 2, * 3 , * 4, ... So answer is 114*4 = 456
### Choose the wrong alternative
Q. 7, 28, 63, 124, 215, 342, 611 ..
A. Series follows pattern: 23 - 1, 33-1, 43-1, 53-1 ... Now we have to check which term doesn't fit this pattern.
Q. 3, 8, 15, 24, 34, 48, 63 ...
A. Series pattern: next term - previous term = 5, 7, 9 ...
Q. 196, 169, 144, 121, 80
A. Pattern: 142, 132, 122, 112 ...
Q. 121, 143, 165, 186, 209
A. Pattern: 112 -02, 122-12, 132-22, 142 - 32, 152 - 42 . Thus odd term is 186.
Q. 1, 2, 4, 8, 16, 32, 64, 96
A. 96 as it is not double of previous
Q. 1, 3, 7, 15, 27, 63, 127
A. Series pattern: +2, +4, +8, +16 .... 27 doesn't follow this pattern.
Q. 6, 14, 30, 64, 126 ...
A. Pattern: +8, +16, +32, +64 and 64 breaks the pattern.
Q. 1236, 2346, 3456, 4566, 5686
A. 5686 breaks the pattern; The first three digits form continuous series so last number should be 5676.
Q. 5, 10, 40, 80, 320, 550, 2560
A. Series follows the pattern: * 2, * 4, * 2, * 4 ....
Q. 11, 2, 21, 3, 32, 4, 41, 5, 51, 6
A. There are two series: 2, 3, 4, 5.. 11 , 21 , 31 , 41 , 51. 32 breaks the second series.
### Alphabet Series
Alphabet series sums can be solved using this diagram as reference.
Q. ' ' , ' ', R, T, V, X, Z
A. Series starts from N and skips one alphabet.
Q. P, K, G, D, B, A
A. Series has the pattern: - 5, -4, -3 , -2 ...
Q. BDF, CFI, DHL, ....
A. The first term is moved one place ahead, second two places and third three places.
Q. YEB, WFD, UHG, SKI .....
A. The first term is moved back by 1 then 2 then 3 steps. The second and third terms are moved forward by 1 then 2 then 3 steps. Missing term :QOL
Q.
AB, DG, HM, MT, SB, ...
A. The first term is moved ahead in pattern +2, +3, +4 ... The second term is moved ahead in pattern: +4, +5, +6...
Q. P3, R5, T8, V12, ...
A. The first term is moved ahead by +1, +2, +3 and second term by +2, +3, +4 ... |
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# Set Operations
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## What are Set Operations?
Set operations can be defined as the operations that are performed on two or more sets to obtain a single set containing a combination of elements from both all the sets being operated upon. There are basically three types of operation on sets in Mathematics; they are:
• The Union of Sets (∪)
• The Intersection of Sets (∩)
• The Difference between Sets (ー)
Let us discuss the operation of set in Math in details:
### The Union of Sets (∪)
In Operation set in math, the union of sets can be described as the set that contains all the elements of all the sets on which the union operation is applied. The union of sets can be denoted by the symbol ∪. The set formed by the union of P and Q will contain all the elements of set P and set Q together. The union of sets can be interpreted as:
P ∪ Q = n(P) + n(Q)
Where n(P) represents, the cardinal number of set P and
n(Q) represents the cardinal number of set Q
Let’s take an example:
Set P- {1, 2, 3, 4, 5} and Set Q- {7, 8, 9, 10}
Therefore, P ∪ Q = {1, 2, 3, 4, 5, 7, 8, 9, 10}
### The Intersection of Sets (∩)
The intersection of sets is referred to as a set containing the elements that are common to all the sets being operated upon. It is denoted by the symbol ∩. The set that is formed by the intersection of both the sets will contain all the elements that are common in set P as well as Set Q.
Let’s take an example:
Set P= {1, 2, 3, 4} and Set Q= {3, 4, 5, 6}
Then, P ∩ Q = {3, 4}
### The Difference Between Sets (-)
The difference of two sets refers to a set that contains the elements of one set that are not present in the other set. It is denoted by the symbol -. Let us say there are two sets P and Q, then the difference between P and Q can be represented as P - Q.
Let’s take an example:
Set P= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and Set Q= {8, 9, 10}
Therefore, P - Q= {1, 2, 3,4, 5, 6, 7}, here we can see that P - Q contains all those elements that are present in P but not in Q.
## Set Operations Venn Diagrams
Venn diagrams are the pictorial representation of sets and set operations. Given below are some set operations Venn diagrams with the explanation.
### Venn Diagram for Union of Sets
The above diagram represents the union of two sets A and B.
### Venn Diagram for Intersection of Sets
The above diagram represents the Intersection of two sets A and B.
### Venn Diagram for Difference Between Sets
The above diagram represents the difference between the set A and B or A - B.
### Properties of Set Operations
There are certain properties of set operations; these properties are used for set operations proofs. The properties are as follows:
• Distributive Property
• Commutative Property
• Distributive Property states that:
If there are three sets P, Q and R then,
P ∩ (Q ∪ R) = (P ⋂ Q) ∪ (P ∩ R)
P ∪ (Q ∩ R) = (P ∪ Q) ∩ (P ∪ R)
• The Commutative Property states that:
If there are two sets P and Q then,
P ∪ Q = Q ∪ P
P ∩ Q = Q ∩ P
## Subset and Powerset
### What is a Subset?
A subset can be defined as a set whose elements are the members of another set. The subset is represented by the symbol ⊆. Let us say C is a subset of D then is represented as C ⊆ D. Let us take an example where set D ={1, 2, 3, 4, 5, 6, 7, 8} and set C= { 3, 4, 5, 6, 7}, as we can see here all the elements of set D are present in set C. Therefore, Set C is a subset of Set D.
### What is Powerset?
A Powerset is a set of all the subsets. If a set A= {1, 2, 3}, so the subsets of A are
{}
{1}
{2}
{3}
{1, 2}
{2, 3}
{1, 3}
{1, 2, 3}
And powerset of A which is denoted by P(A) is { {}, {1}, {2}, {3}, {1, 2}, {2, 3}, {3, 4}, {1, 2, 3}}
The number of element in a power set can be calculated by the formula 2\[^{n}\].
## FAQs on Set Operations
1. What is the Difference Between Powerset and Subset?
Ans: The difference between subset and powerset is:
A Subset is a set that contains all the elements that are a member of another set. Whereas a powerset is set of all the subsets. A Subset is denoted by the symbol ⊆, and powerset is denoted by P(). Let us take an example where set A is {1, 2} and set D is {2}. Therefore it can be said that set D is a subset of set A. Now, suppose we have a set A ={1, 2} then subsets of A can be {}, {1}, {2}, {1, 2} and so the powerset of A, i.e. P(A) = {{}, {1}, {2}, {1, 2}}.
2. What is the Complement of a Set?
Ans: In set operations math, consider a universal set U, and A is a subset of U, then the complement of a set A will contain all the elements of set U leaving the elements of set A. The complement of set A is denoted as A’. Let us take an example if Set U, i.e. an universal set which has the following elements U= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A is a subset of U, which has the following elements A= {2, 3, 4, 5, 6}, then complement of set A, i.e. A’ ={1, 7, 8, 9, 10}. |
Fixed Points
# Fixed Points
So far we have looked at the Bisection Method and Newton's Method for approximating roots of functions. We are about to introduce another root finding method know as the Fixed Point Method, but before we do so, we will need to learn about special types of points on functions known as fixed points which we define below.
Definition: The value $\alpha$ is called a Fixed Point of the function $g$ if $\alpha$ if $\alpha = g(\alpha)$, that is $g$ maps $\alpha$ back to itself. The equation $x_{n+1} = g(x_n)$ is called the Fixed Point Iteration.
The following very simple lemma gives us a way to determine the fixed points of a function.
Lemma 1: A value $x$ is a fixed point of the function $g$ if and only if $y = g(x)$ intersects the diagonal line $y = x$.
• Proof: $\Rightarrow$ Suppose that $x$ is a fixed point of the function $g$. Then $x = g(x)$. But $g(x) = y$, so $x = y$, and so $(x, g(x))$ lies on the line $y = x$.
• $\Leftarrow$ Suppose that $y = g(x)$ intersects the diagonal line $y = x$. Then since $y = x$, we have that $x = g(x)$ so $x$ is a fixed point of $g$. $\blacksquare$
Thus far we have not even mentioned whether a fixed point to a function is guaranteed to exist. Theorem 1 below gives us a condition that guarantees the existence fixed points while Theorem 2 gives us a condition (though not necessary) that guarantees the uniqueness of a fixed point on an interval.
Theorem 1 (Condition for The Existence of Fixed Points): If $g$ is a continuous function on the interval $[a, b]$ and if for all $x \in [a, b]$ we have that $g(x) \in [a, b]$ then there exists at least one fixed point in $[a, b]$.
• Proof: Let $g$ be a continuous function on $[a, b]$ such that if $x \in [a, b]$ then $g(x) \in [a, b]$. Consider the following two cases.
• Case 1: If $a = g(a)$ and/or $b = g(b)$, the $g$ has a fixed point on at least one of endpoints of the interval $[a, b]$ and we are done.
• Case 2: Suppose that $a \neq g(a)$ and $b \neq g(b)$. Then $g$ does not have a fixed point on either of the endpoints of $[a, b]$. Therefore we have that $g(a) > a$ (since $a < g(x) < b$) and $g(b) < b$ (since $a < g(x) < b$. Define the function $h$ as:
(1)
$$h(x) = g(x) - x$$
• We note that $h$ is a continuous function on $[a, b]$ since $y = g(x)$ and $y = x$ are both continuous on $[a, b]$. Now $h(a) = g(a) - x > 0$ and $h(b) = g(b) - b < 0$. By the Intermediate Value Theorem, there exists a root $\alpha \in (a, b)$ such that $h(\alpha) = 0$, and so $h(\alpha) = 0$ implies that $g(\alpha) - \alpha = 0$, thus $\alpha = g(\alpha)$ and so $\alpha$ is a fixed point of $g$. $\blacksquare$
Theorem 2 (Condition for The Uniqueness of a Fixed Point): If $g$ is a continuous function on the interval $[a, b]$ and if for all $x \in [a, b]$ we have that $g(x) \in [a, b]$. Furthermore, if $g$ is differentiable of $(a, b)$ and there exists a number $k \in \mathbb{R}$ such that $0 < k < 1$ such that for all $x \in (a, b)$ we have that $\mid g'(x) \mid ≤ k$ , then the fixed point $\alpha \in [a, b]$ is unique.
• Proof: We are already guaranteed a fixed point $\alpha \in [a, b]$ from Theorem 1. Suppose that there exists a $k \in \mathbb{R}$ such that $0 < k < 1$ and such that $\mid g'(x) ≤ k < 1$. Suppose that $\alpha$ and $\beta$ are both fixed points in $[a,b]$ such that $\alpha \neq \beta$. By the Mean Value Theorem, there exists a real number $\xi$ between $\alpha$ and $\beta$ (and therefore $\xi \in (a, b)$) such that:
(2)
\begin{align} \quad \frac{g(\alpha) - g(\beta)}{\alpha - \beta} = g'(\xi) \\ \quad g(\alpha) - g(\beta) = g'(\xi)(\alpha - \beta) \end{align}
• Since $\xi \in (a, b)$ we have that $\mid g'(\xi) \mid ≤ 1$. Furthermore, since $\alpha$ and $\beta$ are fixed points, we have that $\mid \alpha - \beta \mid = \mid g(\alpha) - g(\beta)$ and so:
(3)
\begin{align} \quad \mid \alpha - \beta \mid = \mid g(\alpha) - g(\beta) \mid = \mid g'(\xi) \mid \mid \alpha - \beta \mid < \mid \alpha - \beta \mid \end{align}
• But this is a contradiction which comes from assuming that $\alpha \neq \beta$. Thus, the fixed point $\alpha$ is unique. $\blacksquare$
## Example 1
Show that the function $f(x) = x^2$ has two fixed points. Show that $\mid f'(x) \mid ≥ 1$ for some $x \in (-1, 2)$
The function $f(x) = x^2$ intersects the line $y = x$ when $x = 0$ and $x = 1$ since $f(0) = 0^2 = 0$ and $f(1) = 1^2 = 1$. Therefore $x = 0$ and $x = 1$ are the two fixed points of $g$.
Now $f'(x) = 2x$. This function is strictly increasing on $[-1, 2]$, and $f(-1) = -2$ and $f(2) = 4$ are the absolute maximum and absolute minimum (respectively) on $[-1, 2]$. Clearly there exists $x \in (-1, 2)$ such that $\mid f'(x) \mid ≥ 1$, take $x = -0.5$ for example. Then $\mid f'(-0.5) \mid = \mid 2(-0.5) \mid = 1 ≥ 1$. |
# How do you simplify (5^(1/3))^6?
Jul 22, 2016
${5}^{\frac{1}{3} \times \frac{6}{1}} = {5}^{2} = 25$
#### Explanation:
The power law of indices states:
${\left({x}^{m}\right)}^{n} = {x}^{m n}$
When raising a power to another power, multiply the indices.
This is exactly what we have here.
${5}^{\frac{1}{3} \times \frac{6}{1}} = {5}^{2} = 25$
Jul 22, 2016
$25$
#### Explanation:
${\left({a}^{b}\right)}^{c} = {a}^{b \cdot c}$
Use the concept above to simplify the value.
${\left({5}^{\frac{1}{3}}\right)}^{6}$
$= {5}^{\left(\frac{1}{3} \cdot 6\right)}$
$= {5}^{\frac{6}{3}}$
$= {5}^{2}$
Now square $5$.
$= 25$ |
# Addition and Subtraction of Dissimilar Fractions – Fractions with Different Denominators
Okay. Here’s the thing. The addition or subtraction of dissimilar fractions (unlike fractions) is slightly different compared to the addition or subtraction of similar fractions (like fractions).
For the addition or subtraction of similar fractions, we are dealing only with the numerators. If you missed my previous lesson about it, you can review my previous discussion about the addition and subtraction of similar fractions.
Meanwhile, for the addition or subtraction of dissimilar fractions, it means that we will be dealing both with numerators and denominators.
For you to proceed with this lesson easily, please make sure that you have the full understanding of the following prerequisite topics:
1. Addition and subtraction of similar fractions (or like fractions).
2. Finding equivalent fractions.
3. Finding the “Least Common Denominator” or LCD.
Let’s start with the basic steps.
### How to Add Unlike Fractions (dissimilar fractions) Step-by-step.
To make it easy, I divided the procedure on how to add or subtract unlike fractions into two (2) basic steps as follows:
Step 1. Change all unlike fractions (dissimilar fractions) into like fractions (similar fractions).
In order to this, we need to change the denominators of the fractions into a common denominator. Remember that similar fractions has the same or common or like denominators.
The common denominator must be a multiple of all different denominators. Meaning to say, the common denominator can be divided by all the denominators without any remainder.
There are infinite numbers of common denominator for any fractions, but we will consider only the smallest or the lowest one. This is what do we called the least common denominator or LCD.
Finally, obtain the equivalent fractions of those fractions using the common denominator obtained.
Step 2. Proceed with the same procedure as adding or subtracting similar fractions.
We will follow the same procedure of adding or subtracting similar fractions as follows:
• Add or subtract the numerators.
• Copy the denominators.
• Convert the final answer in the lowest term.
Let us have an example.
### Adding Unlike Fractions – Examples
Example No.1 Find the sum of 1/4 and 1/2.
First step is to find the least common denominator of 1/4 and 1/2. It is the same as finding the least common multiple of 4 and 2.
Our least common denominator is 4 as shown above. The next to do next is to put 4 as our new denominators.
Then, we will find the equivalent fractions of our original fractions with 4 as the denominator.
1/4 will not change. 4 is already the denominator.
The equivalent fraction of 1/2 with 4 as the denominator is 2/4.
Now, we have 1/4 plus 2/4. We can easily add this now because they are similar fractions.
1/4 plus 2/4 is equal to 3/4. We just simply add the numerators.
Since our sum which is equal to 3/4 is already in the lowest term, 3/4 is the final answer.
Example No.2 Find the sum of 2/3 and 1/4.
The first thing that we want to do is to change those fractions into similar fractions.
In order to do that, we must find the LCD. The LCD is the least common multiple of 3 and 4.
In our solving above, the least common multiple of 3 and 4 is 12. We will use 12 as our new denominator.
The next thing to do is to find the equivalent fraction of 2/3 and 1/4 with 12 as the denominator.
The equivalent fraction of 2/3 with 12 as the denominator is 8/12.
The equivalent fraction of 1/4 with 12 as the denominator is 3/12.
Since we have already changed them into similar fractions, we can proceed with adding their numerators.
8/12 plus 3/12 is equal to 11/12.
11/12 is our answer.
More example of adding unlike fractions.
Example No.3 Find the sum of 1/6 and 3/4.
Example No.4 Find the sum of 4/5 and 3/4.
Example No.5 Find the sum of 1/6, 2/9, and 5/18.
### How to Subtract Unlike Fractions (dissimilar fractions) – Steps
First, we will change dissimilar fractions into similar fractions by finding the least common denominator.
Second, add them like similar fractions.
And then finally, convert the answer in the lowest term.
### Subtracting Unlike Fractions – Examples
Example No.6 Subtract 3/4 from 4/5.
What will be our least common denominator? The least common multiple of 5 and 4 is 20. Thus, 20 is our least common denominator.
The equivalent fraction of 4/5 with 20 as the denominator is 16/20.
The equivalent fraction of 3/4 with 20 as the denominator is 15/20.
Since we already have similar fractions, perform the subtraction.
16/20 minus 15/20 is equal to 1/20.
1/20 is in the lowest term.
Thus, 1/20 is the final answer.
Example No.7 Subtract 1/4 from 2/3.
First, solve for the least common denominator. The least common denominator is 12 as shown in our calculation.
The equivalent fraction of 2/3 with 12 as the denominator is 8/12.
While the equivalent fraction of 1/4 with 12 as the denominator is 3/12.
Thus, 8/12 minus 3/12 is equal to 5/12.
5/12 is already in the lowest term.
Thus, 1/20 is the final answer.
### Summary – Adding and Subtracting Unlike Fractions
1. To add or subtract dissimilar fractions (or unlike fractions), change dissimilar fractions into similar fractions (or like fractions). Then, add or subtract the fractions as similar fractions. The sum or difference must in the lowest term.
2. To change dissimilar fractions (or unlike fractions) into similar fractions (or like fractions0, we need to find the Least Common Denominator or LCD.
### Related Topics
If you have some questions or problems related to the above topic, please feel free to drop it in the comments section below.
Please do not forget to share this post with your friends, classmates, brothers, sisters, or siblings.
Enjoy!
## 11 comments
1. Chyke says:
I like the way you patiently explained every step that it is easily understandable to a child. I think parents will love this site. I am just seeing it for the first time. As a child, I wasn’t too good at math but then we didn’t have access to materials like this and patient teachers such as you. Or maybe we did have good teachers but were too resigned to follow what they taught.
1. Thank you Chyke. I really appreciate it. I love teaching kids and I love sharing my math knowledge to them.
2. Your explanation was very organized and systematic. Honestly, I was never much of a math wiz, but your use of pictures was very effective. The steps were also extremely clear. Great job!
1. Thank you. I love teaching. Please come back for more my interesting posts.
3. A Phill says:
As a father , I am teaching my daughter simple and basic maths and seeing your website and how you have painstakingly analysed every step that it is easily understandable to a child.
I am bookmarking your blog because I can learn more and transfer the skills to my girls. I use Kumon to teach them at the moment but when it comes to the logic I need to delve deeper just like you have done.
Growing up I loved maths and further maths but now into construction but not lost my love for the subject.
Thanks
1. Exactly. This is actually one my purpose – to help parents or kids to understand basic math topic better.
Regards.
4. Jukka says:
This was actually a really well thought out lesson, thank you for that! I was always decent with math in school so this was nothing new really but I can’t actually remember that we would have learned to find the LCD the way you showed here. It’s really simple and logical and I actually learned something new :).
1. You’re welcome. I tried to present all the lessons in its most basic approach.
5. Anne Santiago says:
This post contains a very detailed explanation in adding and subtracting fractions. It present an effective method on finding the LCD. I could say this is very helpful to pupils who just started learning fractions. I also want to share my idea on dealing with fractions but I think this is suitable for higher grade levels.
To add and subtract fractions successfully is to make the rules stick to your memory.
Rules are:
Same denominator:
Add both numerators then reduce. The result would be the final answer.
Different denominator (4 steps):
1. Multiply the numerator of first fraction to the denominator of second fraction. The result is the new numerator of first fraction.
2. Multiply the numerator of the second fraction to the denominator of first fraction. The result is the new numerator of second fraction.
3. Multiply both denominators. The result is the common denominator for two fractions.
4. Add the two new numerators. The result is the answer.
To make it stick to your memory:
Rules for subtraction:
Same denominator:
Subtract second numerator from first then reduce. The result would be the final answer.
Different denominator (4 steps):
1. Multiply the numerator of first fraction to the denominator of second fraction. The result is the new numerator of first fraction.
2. Multiply the numerator of the second fraction to the denominator of first fraction. The result is the new numerator of second fraction.
3. Multiply both denominators. The result is the common denominator for two fractions.
4. Subtract new second numerator from first new numerator. The result is now the answer.
To make it stick to your memory:
Same numerator:
Add two fractions 50 times.
Subtract two fractions 50 times.
Different denominator:
Add two fractions 100 times.
Subtract two fractions 100 times.
To check if your answer is right and your step by step solution is correct:
Use fraction calculator with button from http://www.fractioncalc.com. It is important that you follow the correct steps in adding unlike fraction. Unlike fractions are those fractions with different denominator.
The key here is to make the rules implanted into the minds of the students so that they will never forget.
6. Yuzak maiku says:
I am very very happy with this website.
Tanks.
7. Pingback: bahis siteleri |
NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1.
Board CBSE Textbook NCERT Class Class 7 Subject Maths Chapter Chapter 8 Chapter Name Comparing Quantities Exercise Ex 8.1 Number of Questions Solved 3 Category NCERT Solutions
## NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1
Question 1.
Find the ratio of:
(a) ₹ 5 to 50 paise
(b) 15 kg to 210 g
(c) 9 m to 27 cm
(d) 30 days to 36 hour
Solution:
(a) ₹ 5 : 50 paise
= ₹ 5 : 50 paise
= 5 × 100 paise : 50 paise
= 500 paise : 50 paise
= 10 : 1.
(b) 15 kg to 210g
= 15 kg : 210g
= 15 × 1000 g : 210g
= 15000 g : 210g = 500:7.
(c) 9 m to 27 cm
= 9m : 27 cm = 9 × 100 cm : 27 cm
= 900 cm : 27 cm
900 : 27 = 100 : 3.
(d) 30 days to 36 hours
= 30 days : 36 hours
30 × 24 hours : 36 hours
= 720 hours : 36 hours
= 720 : 36 = 20 : 1.
Question 2.
In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Solution:
Let x computers be needed for 24 students.
Hence, 12 computers will be needed for
24 students.
Aliter:
In a computer lab,
Every 6 students need 3 computers.
∴ Every 1 student needs $$\frac { 3 }{ 6 }$$ computers.
∴ Every 24 students need $$\frac { 3 }{ 6 }$$ × 24 = 3 × 4 = 12 computers
Question 3.
Population of Rajasthan = 570 lakhs
and population of UP = 1660 lakhs.
Area of Rajasthan =3 lakh km2
and area of UP = 2 lakh km2.
(i) How many people are there per km2 in both these States?
(ii) Which State is less populated?
Solution:
(i) Number of people per sq. km. in Rajasthan state
(ii) The Rajasthan State is less populated.
We hope the NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1, drop a comment below and we will get back to you at the earliest. |
# Slope
Slope refers to the Steepness of a Line
Basics
Slope refers to the steepness of a line.
Line 1 is steeper than Line 2.
Line 1’s slope is greater than Line 2’s slope.
A line is a flat straight figure extending infinitely through two points. (See Line L above.)
x-axis = horizontal
y-axis = vertical
x = horizontal coordinate
y = vertical coordinate
(x, y) = point representing horizontal and vertical coordinates on Line L
(o, b) = y-intercept = point where Line L crosses the y-axis
m = slope
Formulas (Mathematics Formula Sheet)
slope of a line
m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Formulas (NOT on Mathematics Formula Sheet)
slope as change
m = $\bf\displaystyle\frac{change\hspace{0.1cm}in\hspace{0.1cm}y}{change\hspace{0.1cm}in\hspace{0.1cm}x}$
Question
What is the slope of Line 1?
Answer
m = 2
Answer Process
slope as change
m = $\bf\displaystyle\frac{change\hspace{0.1cm}in\hspace{0.1cm}y}{change\hspace{0.1cm}in\hspace{0.1cm}x}$
m = $\bf\displaystyle\frac{10\hspace{0.1cm}boxes\hspace{0.1cm}up\hspace{0.1cm}positive\hspace{0.1cm}y\hspace{0.1cm}axis}{5\hspace{0.1cm}boxes\hspace{0.1cm}along\hspace{0.1cm}positive\hspace{0.1cm}x\hspace{0.1cm}axis}$ = $\bf\displaystyle\frac{10}{5}$ = 2
Question
What is the slope of Line 2?
Answer
m = $\bf\displaystyle\frac{1}{2}$
Answer Process
slope as change
m = $\bf\displaystyle\frac{change\hspace{0.1cm}in\hspace{0.1cm}y}{change\hspace{0.1cm}in\hspace{0.1cm}x}$
m = $\bf\displaystyle\frac{5\hspace{0.1cm}boxes\hspace{0.1cm}up\hspace{0.1cm}positive\hspace{0.1cm}y\hspace{0.1cm}axis}{10\hspace{0.1cm}boxes\hspace{0.1cm}along\hspace{0.1cm}positive\hspace{0.1cm}x\hspace{0.1cm}axis}$ = $\bf\displaystyle\frac{5}{10}$ = $\bf\displaystyle\frac{1}{2}$
Question
Using the points (3, 1) and (8, 3) in the graph below, calculate the slope of line L.
Answer
m = $\bf\displaystyle\frac{2}{5}$
Answer Process
(x1, y1) = (3, 1)
(x2, y2) = (8, 3)
slope of a line
m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
m = $\bf\displaystyle\frac{3-1}{8-3}$ = $\bf\displaystyle\frac{2}{5}$
Input Display Comment
blinker clears screen
3 – 1 8 – 3 $\bf\displaystyle\frac{3-1}{8-3}$ m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
$\bf\displaystyle\frac{2}{5}$ Answer
###### Slope
Question
Express the previous answer in decimal form.
Answer
m = 0.4
Answer Process
See Toggle.
$\bf\displaystyle\frac{2}{5}$ ⇔ 0.4
Input Display Comment
blinker clears screen
3 – 1 8 – 3 $\bf\displaystyle\frac{3-1}{8-3}$ m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
$\bf\displaystyle\frac{2}{5}$
0.4 Answer
(Toggle)
###### Slope
Question
Using the points (1, 2) and (4, 9) in the graph below, calculate the slope of line L in decimal form.
Answer
m = 2.333333333
Answer Process
(x1, y1) = (1, 2)
(x2, y2) = (4, 9)
slope of a line
m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
m = $\bf\displaystyle\frac{9-2}{4-1}$ = $\bf\displaystyle\frac{7}{3}$ = 2.333333333 after toggle
Input Display Comment
blinker clears screen
9 – 2 4 – 1 $\bf\displaystyle\frac{9-2}{4-1}$ m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
$\bf\displaystyle\frac{7}{3}$
2.333333333 Answer
(Toggle)
###### Slope
Question
Round the previous answer to the nearest tenth.
Answer
m = 2.3
Answer Process
See Rounding.
2.333333333 2.3
Question
On the graph below, plot the points (2, 3) and (6, 9), draw a line through them, and calculate the slope of the line in decimal form.
Answer
m = 1.5
Answer Process
(x1, y1) = (2, 3)
(x2, y2) = (6, 9)
slope of a line
m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
m = $\bf\displaystyle\frac{9-3}{6-2}$ = $\bf\displaystyle\frac{3}{2}$ = 1.5 after toggle
Input Display Comment
blinker clears screen
9 – 3 6 – 2 $\bf\displaystyle\frac{9-3}{6-2}$ m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
$\bf\displaystyle\frac{3}{2}$
1.5 Answer
(Toggle)
###### Slope
Question
On the graph below, plot the points (4, 1) and (10, 2) and calculate their slope in decimal form rounded to the nearest hundredth (without drawing a line).
Answer
m = 0.17
Answer Process
(x1, y1) = (4, 1)
(x2, y2) = (10, 2)
slope of a line
m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
m = $\bf\displaystyle\frac{2-1}{10-4}$ = $\bf\displaystyle\frac{1}{6}$ = 0.166666667 after toggle = 0.17 after rounding
Input Display Comment
blinker clears screen
2 – 1 10 – 4 $\bf\displaystyle\frac{2-1}{10-4}$ m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
$\bf\displaystyle\frac{1}{6}$
0.166666667 Answer
(before rounding)
###### Slope
Question
On the graph below, plot the points (3, 2.5) and (7, 6.25) and calculate their slope in decimal form to the nearest hundredth (without drawing a line).
Answer
m = 0.94
Answer Process
(x1, y1) = (3, 2.5)
(x2, y2) = (7, 6.25)
slope of a line
m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
m = $\bf\displaystyle\frac{6.25-2.5}{7-3}$ = 0.9375 =0.94 after rounding
Input Display Comment
blinker clears screen
6.25 – 2.5 7 – 3 $\bf\displaystyle\frac{6.25-2.5}{7-3}$ m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
0.9375 Answer
(before rounding)
###### Slope
Question
The following table shows the age in years and height in inches of a boy and a girl.
On the grid below the table, plot the x-y coordinates for the boy at ages 2 and 8 and calculate their slope in decimal form to the nearest tenth (without drawing a line).
Boy Girl Age (yr) Height (in) Age (yr) Height (in) 2 34.00 2 33.50 3 37.50 3 37.00 4 40.00 4 39.50 5 43.00 5 42.25 6 45.25 6 45.00 7 48.00 7 48.00 8 50.25 8 50.00 9 52.50 9 52.50 10 54.50 10 54.25
Answer
m = 2.7
Answer Process
(x1, y1) = (2, 34.00)
(x2, y2) = (8, 50.25)
slope of a line
m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
m = $\bf\displaystyle\frac{50.25-34.00}{8-2}$ = 2.708333333= 2.7 after rounding
Input Display Comment
blinker clears screen
50.25 – 34.00 8 – 2 $\bf\displaystyle\frac{50.25-34.00}{8-2}$ m = $\bf\displaystyle\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
2.708333333 Answer
(before rounding)
###### Slope
Practice – Questions
1. What is the slope of Line 1?
2. What is the slope of Line 2?
3. Using the points (3.5, 1.5) and (7, 3.25) in the graph below, calculate the slope of line L.
4. On the graph below, plot the points (4, 1) and (8, 2), draw a line through them, and calculate the slope of the line in decimal form.
5. On the graph below, plot the points (1, 4) and (2, 8) and calculate their slope (without drawing a line).
6. The following table shows the age in years and height in inches of a boy and a girl.
On the grid below the table, plot the x-y coordinates for the girl at ages 3 and 9 and calculate their slope in decimal form (without drawing a line).
Boy Girl Age (yr) Height (in) Age (yr) Height (in) 2 34.00 2 33.50 3 37.50 3 37.00 4 40.00 4 39.50 5 43.00 5 42.25 6 45.25 6 45.00 7 48.00 7 48.00 8 50.25 8 50.00 9 52.50 9 52.50 10 54.50 10 54.25
7. With reference to Question 6, express your answer as a fraction.
8. The following table shows the age in years and mileage for Car A and Car B.
On the grid below the table, plot the x-y coordinates for Car A at years 3 and 6 and calculate their slope (without drawing a line).
Car A Car B Age Mileage Age Mileage 1 10000 1 11000 2 22000 2 20500 3 33500 3 29725 4 39000 4 41000 5 48000 5 50500 6 62000 6 58750 7 71000 7 69900 8 83250 8 81000 9 90500 9 89900 10 99500 10 99500
9. The following table shows the age in years and mileage for Car A and Car B.
On the grid below the table, plot the x-y coordinates for Car B at years 2 and 9 and calculate their slope (without drawing a line).
Car A Car B Age Mileage Age Mileage 1 10000 1 11000 2 22000 2 20500 3 33500 3 29725 4 39000 4 41000 5 48000 5 50500 6 62000 6 58750 7 71000 7 69900 8 83250 8 81000 9 90500 9 89900 10 99500 10 99500
10. With reference to Question 9, round your answer to the nearest tenth.
Practice – Answers
1. m = 1
2. m = 2
3. m = 0.5
4. m = 0.25
5. m = 4
6. m = 2.583333333
7. $\bf\displaystyle m =\frac{31}{12}$
See Toggle.
8. m = 9500
9. m = 9914.285714
10. m = 9914.3 |
# 1957 AHSME Problems/Problem 37
## Problem
In right triangle $ABC, BC = 5, AC = 12$, and $AM = x; \overline{MN} \perp \overline{AC}, \overline{NP} \perp \overline{BC}$; $N$ is on $AB$. If $y = MN + NP$, one-half the perimeter of rectangle $MCPN$, then:
$[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair A = origin; pair M = (1,0); pair C = (2,0); pair P = (2,0.5); pair B = (2,1); pair Q = (1,0.5); draw(A--B--C--cycle); draw(M--Q--P); label("A",A,SW); label("M",M,S); label("C",C,SE); label("P",P,E); label("B",B,NE); label("N",Q,NW);[/asy]$
$\textbf{(A)}\ y = \frac {1}{2}(5 + 12) \qquad \textbf{(B)}\ y = \frac {5x}{12} + \frac {12}{5}\qquad \textbf{(C)}\ y =\frac{144-7x}{12}\qquad \\ \textbf{(D)}\ y = 12\qquad \qquad\quad\,\, \textbf{(E)}\ y = \frac {5x}{12} + 6$
## Solution
Because $AC=12$ and $AM=x$, $MC=12-x$. Let $MN=z$. Then, because $MNPC$ is a rectangle, $NP=12-x$ and $PC=z$, and so $BP=5-z$. By AA similarity, $\triangle AMN \sim \triangle NPB$. From this similarity, we can solve the following proportion for $z$: \begin{align*} \frac{BP}{PN} &= \frac{NM}{AM} \\ \frac{5-z}{12-x} &= \frac z x \\ 5x-xz &= 12z-xz \\ 5x &= 12z \\ z &= \frac{5x}{12} \end{align*} Because $y=MN+NP=z+12-x$, we can now substitute for $z$ and find $y$ in terms of $x$: \begin{align*} y &= z+12-x \\ &= \frac{5x}{12}-x+12 \\ &= \frac{-7x}{12}+\frac{144}{12} \\ &= \frac{144-7x}{12} \end{align*} Thus, our answer is $\boxed{\textbf{(C) }\frac{144-7x}{12}}$. |
# Quantitative Methods for Business Probability Distribution Regression Questions
Measures of central tendency and SpreadMeasures of central tendency:
Mean: Average of all observations (sum of all observations divided by the number of observations)
Mode: the most frequent observation
Median: the “middle” observation (50% of all observations have a value below the median and 50% of all observations have a
value above the median, the average between the middle two observations if we have an even number of observations)
When the median differs from the mean: Skewness
Range: The difference between the largest and the smallest observation
Variance: typical distance (squared) from the mean:
Standard deviation: square root of variance:
The Interquartile range: the difference between the third quartile and the first quartile.
2
Probability theory
1. If all events are equally likely (like for a dice, each side is equally likely) we compute
probability by the formula:
Number of outcomes when the event occurs / Number of possible outcomes
Typical question: Suppose you toss two fair (i.e., unbiased) six-faced dice.
What is the probability of getting the number 6 for both dices?
Solution: There are 6*6 possible outcomes. Why: 6 for the first dice and 6 for the second dice. How many
outcomes are there when the event (getting 6 on both) occurs? Only one: the outcome in which we get 6 on the
first dice and 6 on the second dice. Now we use the formula: 1/36.
Typical question: A card is drawn randomly from a deck of 52 cards. A) What is the probability that it is red? B)
What is the probability that it has the value 10? C) What is the probability that it is a red card with the value 10?
Solution: A) There are 52 cards. Half of the cards are red: the probability that it is red is thus 26/52 = 1/2. B) There
are four cards with the value 10: the probability a card has the value 10 is thus 4/52 = 1/13. C)There are only two
red cards with a value 10, so probability is 2/52 = 1/26.
3
Probability theory
2. What is Probability of A and B?
For example, what is the probability of getting the number 4 on the first dice and the number 5 on the
second dice?
If events are independent (they don’t influence each other, for example the outcome of one dice is
not influenced by the outcome of the other dice) then
P(A and B) = P(A)*P(B)
Typical question: There are 5 green books and 8 red books. Eric picks one book at random and then replaces it.
Anna picks another book at random and then replaces it. A) What is the probability that they both pick green
books? B) What is the probability that Eric picks a green book and Anna picks a red book?
Solution: A) The probability of picking a green book is 5/(5+8) = 5/13. The probability that Eric and Anna picks a
green book is P(Green)*P(Green) = (5/13)*(5/13) = 0.1479. B) The probability of picking a red book is 8/(5+8) =
8/13. the probability that Eric picks a green book and Anna picks a red book is (5/13)*(8/13) = 0.2367.
4
Probability theory
3. What is Probability of A or B?
It is P(A or B) = P(A) + P(B) – P(A and B).
A
B
Example: what is the probability a of getting ‘King’ OR a ‘red card?
• It is Pr(red) + Pr(King) -Pr(red AND king)
A and B
• = (26/52)+(4/52)-(2/52) = 28/52
Typical question: 22% of all firms in an industry have bought insurance. 15% have bought insurance and have a
risk consultant. 30% have a risk consultant. What proportion has insurance or a risk consultant?
Solution: P(I or R) = P(I)+P(R)-P(I and R) = 0.22+0.30- 0.15 = 0.52 – 0.15 = 0.37
Typical question: Suppose we flipped a coin and rolled a six faced unbiased dice. What is the probability of getting
a head on the coin OR the number 6 on the dice?
Solution: Use the formula: P(head) = 0.5, P(6) = 1/6. P(head and 6) = 0.5*(1/6). So, we get P(head or 6) = 0.5 +
(1/6) – 0.5*(1/6) = (6/12)+(2/12)-(1/12) = 7/12 = 0.58333.
5
Probability theory
4. Conditional Probability
The probability of A given that we know that B happened, P(A | B)
Formula (Bayes Rule): P(A | B) = P(A and B) / P(B)
Can also be written as: P(A | B) = P(B | A)P(A) / P(B)
Typical question: At the Warwick Foods factory, cookies are tested for crispness. 80% of th crispy cookies pass the
test but 10% of the not crispy cookies also pass the test. 80% of all cookies are crispy. What is the probability that
a cookie is crispy given that it passed the test?
Solution: P(crispy | pass the test) = P(pass the test and crispy ) / P(pass the test).
How to compute P(pass the test and crispy): 80% of the cookies are crispy. Out of these, 80% pass the test. Thus,
0.8*0.8 = 0.64 (64%) cookies are crispy and pass the test.
How to compute P(pass the test): 64% of cookies are crispy and pass the test. Some not crispy all pass the test.
What proportion are not crispy and pass the test: 20% are not crispy and 10% of these pass the test = 0.2*0.1 =
0.02. Overall, 0.8*0.8+0.2*0.1 = 0.64+0.02 = 0.66 pass the test.
Overall: P(crispy | pass the test) = 0.64 / 0.66 = 0.969697
6
Probability theory
Typical question: It is determined that 25 of every 100 marketing students and 600 of every 1,000 finance
students wear glasses. In a room there are 400 finance students and 100 marketing students. What is the
conditional probability that a student in this room without glasses is a finance student?
i.e., calculate Prob(Finance student | No glasses).
Solution: P(Finance student | No glasses) = P(Finance student and No glasses ) / P(No glasses).
How to compute P(Finance student and No glasses ). 400/500 = 80% of the students in the room are finance
students. 60% of these have glasses and 40% do not have glasses. Thus, P(Finance student and No glasses ) =
0.8*0.4 = 0.32.
How to compute P(No glasses): 32% are finance students and don’t have glasses. 20% are marketing students in
the room and 75% of them do not have glasses, i.e., P(marketing student and No glasses ) = 0.2*0.75 = 0.15. Thus,
P(No glasses) = 0.8*0.4+0.2*0.75 = 0.47.
Overall: P(Finance student | No glasses) = 0.8*0.4 / (0.8*0.4+0.2*0.75) = 0.32 / 0.47 = 0.681
7
Binomial Distribution
Suppose that on each trial either a success or a failure occurs. If the
probability of a success is p, what is the probability that we get exactly k
successes in n trials?
We calculate this using the Binomial distribution.
We can use the formula or the table.
The table lists the probability that we get r or more successes in n trials,
when the probability is p.
For example, the probability that
we get 4 or more successes in 8 trials
when p = 0.35 is the number
= 0.2935
8
9
Binomial Distribution
Typical question: The probability of a success is 0.35. In 9 trials, what is the probability of getting between
(and including) 3 and to (and including) 6 successes?
Solution: We can calculate this as follows: we need the red part (3,4,5 or 6). We can calculate this by first
calculating: P(3 or more). Then we calculate: P(7 or more).
Prob(between 3 and 6) = P(3 or more) – P(7 or more).
Successes
Hits
0
1
2
3
4
5
6
7
8
9
Pr(7 or more successes)
= see the table on next side
= 0.0112
Pr = P(3 or more successes) =
See the table on next side
= 0.6627
Prob(between 3 and 6) =
P(3 or more) – P(7 or more) = 0.6627-0.0112 =
0.5512
10
Binomial Distribution
Binomial Distribution
The probability of getting exactly k successes in n trials, when the probability of a
success is p, is
𝑛 !
𝑝 (1 − 𝑝)”#!
𝑘
𝑛
!!
=
, where n! = n*(n-1)*(n-2)….2*1. For example, 5! = 5*4*3*3*1.
𝑘 #! !\$# !
Note that 0! is defined as 1.
Here
Typical question: The probability of a success is 0.35. In 9 trials, what is the probability of getting exactly 4
successes?
Solution: We use the formula
𝑛 !
“!
%!
𝑝 (1 − 𝑝)”#! =
𝑝 ! (1 − 𝑝)”#! =
0.35& (1 − 0.35)%#&
!! “#! !
&! %#& !
𝑘
%!
%∗)∗*∗+∗’∗&∗,∗-∗.
%∗)∗*∗+
,1-&
=
0.35& (0.65)’ =
0.35& (0.65)’ =
0.35& (0.65)’ =
0.35& (0.65)’ = 0.219386
&!’!
(&∗,∗-∗.)(‘∗&∗,∗-∗.)
(&∗,∗-∗.)
(-&)
12
The Normal Distribution
0.1
Probability
is very useful because averages tend to
0.15
0.05
If X is normally distributed, the the probability
that X is larger or smaller than some number c, Average Restaurant Review on Yelp
only depends on the mean and the standard
deviation
0
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Average
Typical question: X is a normally distributed variable with average = 50 and standard deviation = 10. What is the
probability that x is above 55?
Solution: We standardize: (55-mean)/stdev = (55-50)/10 = 5/10 = 0.5.
We look up P(z>0.5) in the table.
We get P(z>0.5)=0.3085
A figure helps a lot:
0.51
13
13
Normal Distribution
Typical question: X is a normally distributed variable with average = 50 and standard deviation = 10. What is the
probability that x is below 45?
Solution:
We standardize: (45-mean)/stdev = (45-50)/10 = -5/10 = -0.5.
A figure helps a lot:
Because of symmetry
P(z < -0.5) = P(z > 0.5)
= 0.3085
-1
-0.5
0.5
1
14
Normal Distribution
Table
15
Confidence interval
Idea behind: Suppose you take a sample of 50 restaurants and calculate the mean review score in the sample. The
sample mean is 3.67 say. You know that the population mean is not exactly 3.67. The population mean could differ
from 3.67. How much could it differ? A confidence interval is one way of quantifying how much it could vary. You
say: the population mean will lie, 95 % of the time, between the lower limit and the upper limit. This is a 95%
confidence interval.
How to calculate it for a Sample mean:
The upper limit = sample mean + z*STEM
The lower limit = sample mean – z*STEM
Here z is the critical value:
z = 2 for an approximate 95% confidence interval, z = 1.96 for an exact 95% confidence interval
z = 3 for an approximate 99% confidence interval, z = 2.58 for an exact 99% confidence interval
STEM = standard deviation / sqrt(n), where n is sample size.
How to calculate it for a Proportion:
q = observed proportion in the sample.
The upper limit = q+ z*STEP
The lower limit = q- z*STEP
Here z is the critical value
STEP =
!×(\$%!)
16
Confidence interval
Typical question: An auditor of a small business has sampled 100 accounts. The sample mean is £435, and the
sample standard deviation is £86. Find an approximate 95% confidence interval for the average amount of all
accounts.
Solution: Sample size n = 100, sample mean m = 435, sample standard deviation s = 86, the confidence level is
approximate 95%, and the critical value z = 2.
STEM = s/sqrt(n) = 86 / sqrt(100) = 8.6.
A rough 95% confidence interval for the average amount of all accounts is
[mean-z*STEM, mean+z*STEM] = [435 – 2*8.6, 435 + 2*8.6] = [417.8, 452.2].
Typical question: A random sample of 100 preschool children in Coventry revealed that only 80 had been
vaccinated. Provide an approximate 95% confidence interval for the proportion vaccinated in Coventry.
Solution: : Sample size n = 100, sample proportion 80/100 = 0.8, the confidence level is approximate 95%, and
the critical value z = 2. We calculate STEP:
STEP =
(.*×(.+
\$((
=
(.\$,
(.-
= \$( = 0.04. Upper limit = 0.8+2*0.04 = 0.88. Lower limit = 0.8-2*0.04 = 0.72.
\$((
17
Confidence interval
Typical question: A University administrator wants to survey students about their evaluation of a new program.
The administrator believes that evaluations have a standard deviation of 1.2. How many students do they need to
sample to ensure that the width of an exact 99% confidence interval (critical value 2.58) is at most 0.4? Your
answer should be an integer. Use at least 4 decimals in all your computations.
Solution: We use the formula: 𝑛 =
+∗σ∗/ +
. σ is the standard deviation = 1.2. z is critical value = 2.58. w = width
0
of the confidence interval = 0.4. We insert this into the formula
2 ∗ 1.2 ∗ 2.58 +
𝑛=
= 15.48+ = 239.63
0.4
The answer should be an integer. Is the answer 239 or 240? It is 240, because if the sample size is smaller than
239.63, the confidence interval would be wider than at 239.63 .
18
Hypothesis Testing
General idea: We asses whether a “null hypothesis” can explain the data by calculating:
What is the probability that we would observe the data we do observe if the null hypothesis was true?
Specifically, we compute: what is the probability that we would observe a result as different from the null
hypothesis as we observe or even more different.
For example: we observe an increase of 10 on average in sample. We ask: can such a change occur by chance even
if there is no change in the population? We calculate: what is the probability of observing a change of 10 or more
if there has not been a change in the population (i.e., the null hypothesis is true)?
How do we calculate these probabilities? We use the normal approximation: we rely on the fact that averages
(and proportions) tend to be normally distributed.
19
Hypothesis Testing
How to compute a hypothesis test for a mean:
We compute: t = (m – Null Hypothesis)/STEM, where m is the sample mean.
STEM = standard deviation / sqrt(n)
We compare t to a critical value, z. The critical value depends on the significance value.
5 % significance value = critical value 2 (or exact 1.96)
1 % significance value = critical value 3 (or exact 2.58)
Reject the null hypothesis if t < -z or if t > z.
How to compute a hypothesis test for a Proportion:
We compute: t = (q – Null Hypothesis proportion)/STEP, where q is the sample proportion.
STEP =
1×(\$%1)
. (observe: p = the null hypothesis proportion, we use this one when we calculate STEP)
We compare t to a critical value, z. The critical value depends on the significance value.
5 % significance value = critical value 2 (or exact 1.96)
1 % significance value = critical value 3 (or exact 2.58)
Reject the null hypothesis if t < -z or if t > z.
20
Hypothesis Testing
Typical question: Robin suspects that a coin is biased. She throws it 25 times and gets head 15 times. Using this
data, test the null hypothesis that the coin is unbiased, i.e., test the null hypothesis that the probability of getting
of head is 0.5, using a significance level of 5%. What is the value of the test statistic (the “t-value” or “t-statistic”)?
Solution: Here STEP =
1×(\$%1)
=
(.2×(\$%(.2)
+2
=
(.+2
+2
= 0.1.
t = (q – Null Hypothesis proportion)/STEP, where q is the sample proportion
= (0.6-0.5)/ 0.1 = 1.
Because t = 1 < 1.96, the null hypothesis is NOT rejected. 21 Hypothesis Testing Typical question: A study of a new blood pressure medicine, with 81 patients, found that the average reduction in blood pressure was 10 with a standard deviation of 30. Test the null hypothesis that the medicine did not change blood pressure using a significance level of 5%. What is the value of the test statistic (the”t-value")? Round your answer to one decimal place. Use at least 4 decimals in all your computations. Can you reject the null hypothesis at a 5% significance level? Solution: STEM = stdev/ 𝑛 = 30/ 81 = 30/9 = 3.333. Test statistic = (m - null hypothesis) / STEM = (10-0)/ STEM = 10 / 3.333 = 3. This test statistic should be compared to the critical value, which is 1.96 (exact) or 2 (approximate). Because 3 > 1.96, we can reject the Null Hypothesis at a 5% significance level.
22
Hypothesis Testing
How to compute a hypothesis test for a difference between two means, if the null hypothesis is zero:
3 !”
!
We compute: t = #\$%&’
where STEDM =
(!”
(“”
+
)”
)!
We compare t to a critical value, z. The critical value depends on the significance value.
5 % significance value = critical value 2 (or exact 1.96)
1 % significance value = critical value 3 (or exact 2.58)
Reject the null hypothesis if t < -z or if t > z.
Typical question: There are 64 Traders in firm A. On average their profit was equal to 5.6 with a standard deviation equal
to 11.4. There are 78 Traders in firm B. On average their profit was equal to 4.3 with a standard deviation equal to 8.9. Test
the hypothesis that the average profits of the two firms is equal, using a significance level of 5%. Can you reject the null
hypothesis of equal means at a 5% significance level?
Solution: STEDM =
(!”
(”
+ )” =
)
**., ”
../”
+ 0, = 1.704
-.
!
“! !””
1.0!,.2
Test statistic = #\$%&’ = *.-3, = 0.7629.
This test statistic should be compared to the critical value, which is 1.96 (exact) or 2 (approximate). Because 0.7629 < 1.96, we CANNOT reject the Null Hypothesis at a 5% significance level. 23 Anova: idea behind the test A. What we test • C. We Calculate Within, Between and Total Variability ANOVA tests the null hypothesis H 0: μ 1 = μ 2 = … = μ K That is, “the group means are all equal” – • 2 A B The alternative hypothesis is 2 or, “the group means are not all equal” B. Idea Behind Test SSTot= ∑ ( x − x ) 2 C 2 SSBet = nA ( x A − x ) + nB ( x B − x ) + nC ( xC − x ) H1: μi ≠ μj for some i, j – 2 SSWith = ∑ ( x − x A ) + ∑ ( x − xB ) + ∑ ( x − xC ) 2 2 SSTot = SSBet + SSWith 40" 35" 30" D. The test is based on the F ratio: 25" 20" 15" 10" F= 5" 0" 0" 5" Picture of10"three 15"groups20"of data25" 30" Unlikely to have same means if: Large Variation Between Means & Small Variation Within Groups MSBet = SSBet / (k −1) MSWith = SSWith / (n − k) If F > critical value, null hypothesis is rejected.
24
Anova: how to compute it
1. Calculate sums and sum of squares
2. Calculate:
Tj2 T 2
SSBet = ∑ −
n
j nj
2
T
SSTot = S 2 −
n
SSWith = SSTot − SSBet
Sum of all observations (Total) = T
Sum of all squared observations = S2
3. Calculate the F ratio:
F=
MSBet = SSBet / (k −1)
MSWith = SSWith / (n − k)
4. Test the null hypothesis of no differences
in means, at significance level a.
If F > critical value, hypothesis is rejected.
CRsignif (dfbet, dfwith ) = CRsignif (k −1, n − k)
25
Typical question: Given the data below, test the null hypothesis that the averages are equal for all
groups, using an Anova and a significance level of 0.05. What is the F-ratio? Keep at least 4 decimals in
all computations and round the final answer to 1 decimal place.
Group 1
5
8
Group 2
3
10
Group 1
Solution:
SUMS
Group 3
9
9
Group 2
Group 4
8
12
Group 3
Group 4
Obs
Obs^2
Obs
Obs^2
Obs
Obs^2
Obs
Obs^2
5
25
3
9
9
81
8
64
8
64
10
100
9
81
12
144
13
89
13
109
18
162
20
208
From Table above, we can calculate:
• Total = 13+13+18+20 = 64 ; Sum of Squares (SS) = 89 + 109 + 162 + 208 = 568
• SS(between) = (13*13/2) +(13*13/2)+ (18*18/2)+(20*20/2) – (64*64/8)= 19
• SS(Total) = 568- (64*64/8) = 56.
• SS(within) = SS(total) – SS(between) = 56-19= 37.
• MS(Between) = 19/(4-1) = 6.333
• MS (Within) = 37/(8-4) = 9.25
‘#
0.222
• 𝐹 = ‘##\$%&\$\$’ = /.41 = 0.6847
&(%)(‘
Typical question: Can the null hypothesis be rejected at 5% significance?
F distribution, 0.05 significance level
k-1 = 4-1 = 3
n-k = 8-4 = 4
Critical value is 6.59
Because F ratio (0.7) is below 6.59, the null
hypothesis cannot be rejected.
Two Factor Anova: idea behind the test
Now 2 factors are varied (previously we considered one factor).
For example, we may have data on performance for firms using
two different hiring approaches and two different quality
management approaches:
!
!
!
!
!
Intuitive!
Quality!
Management!
Statistical!
Hiring!Approach!
!
More!
Educated!
6!
5!
7!
11!
14!
12!
Less!!
Educated!
8!
7!
9!
1!
3!
2!
!
!
!
!
!
Intuitive!
Quality!
Management!
Hiring!Approach!
!
Less!!
More!
Educated!
Educated!
!
!
xI,LE = 8
Statistical!
!
Main Effect of
Hiring Approach
MSH = SSH / (kH −1)
F=
2
MSWith
Main Effect of
Quality Approach
!
xS,ME =! 12.3 xS = 7.17
xLE = 5
xME = 9.17
!
2
Interaction Effect
F = MSH*Q = SSH*Q / (kH −1)(kQ −1)
MSWith
We can also examine the interaction effect: Does the benefit of a
statistical approach to quality management vary with how
educated the workforce is?
14″
8″
7″
10″
7″
6″
8″
Intui.ve”
6″
Sta.s.cal”
4″
4″
Intui0ve”
3″
Sta0s0cal”
2″
2″
1″
0″
0″
1″
2″
3″
0″
0″
1″
2″
3″
4.5″
7″
9″
4″
6″
8″
3.5″
6″
5″
5″
Intui1ve”
4″
Sta1s1cal”
3″
2
2.5″
2″
1.5″
1″
1″
0″
1″
2″
3″
Intui.ve”
Sta.s.cal”
2
2
Intui/ve”
3″
Sta/s/cal”
2″
0″
1″
2″
3″
Sta1s1cal”
3″
1″
0″
0″
0″
Intui1ve”
4″
2″
1″
0.5″
5″
0″
1″
2″
3″
0″
1″
2″
2
+nS,LE ( xS,LE − x ) + nS,ME ( xS,ME − x ) − SSH − SSQ
6″
4″
2
SSH*Q = nI. LE ( x I,LE − x ) + nI,ME ( x I,ME − x )
7″
5″
3″
2″
0″
SSQ = nI ( x I − x ) + nS ( xS − x )
No Interaction Effect: Effect of Factor 1 independent of factor 2
Total effect = Factor 1 effect + Factor 2 effect
9″
8″
12″
2
MSQ = SSQ / (kQ −1)
F=
MSWith
Given this data, we can examine the effect of each factor (Is it
better to hire more educated workers? Does a statistical quality
management approach lead to better results?)
Interaction Effect: Effect of Factor 1
depends on Factor 2
SSH = nLE ( x LE − x ) + nME ( x ME − x )
xI = 7
xI,ME = 6
xS,LE = 2
!
!
Generally: test main effects and interaction effects by computing
between and within variability
3″
28
Two Factor Anova: how to compute it
1. Calculate sums and sum of squares
2. Calculate Sums
T2
SSTot = S −
n
THj *THj T 2
TQj *TQj T 2
SSH = ∑
− , SSQ = ∑
n
n
n
n
Hj
Qj
j∈H
j∈Q
2
THiQj *THiQj T 2
SSH*Q = ∑
− SSH − SSQ
nij
n
i, j
Deduct these!
Interaction = variability
not due to main effects
SSWith = SSTot − SSH − SSQ − SSH*Q
3. Calculate the F ratios:
MSH = SSH / (kH −1)
FH =
MSWith = SSWith / (n − kH kQ )
FQ =
MSQ = SSQ / (kQ −1)
MSWith = SSWith / (n − kH kQ )
MSH*Q = SSH*Q / (kH −1)(kQ −1)
FH*Q =
MSWith = SSWith / (n − kH kQ )
4. Test hypotheses
If Fi > critical value, null-hypothesis is rejected.
Critical values:
CRH ,signif (kH −1, n − kH kQ )
CRQ,signif (kQ −1, n − kH kQ )
CRHQ,signif ((kH −1)(kQ −1), n − kH kQ )
29
Two Factor Anova: typical question
Deduct Ss_a
and SS_b
Solution:
1 = ka-1 = 2-1, 12 = n-ka*kb = 16-2*2
1 = kb-1 = 2-1, 12 = n-ka*kb = 16-2*2
1 = (ka-1)(kb-1) = (2-1)(2-1),
12 = n-ka*kb = 16-2*2
30
Mann-whitney test
Idea behind the test: If the null hypothesis that the two samples
come from the same distribution is true, the rank scores (ranging
from 1 = the lowest to 20 = the highest) should be evenly
distributed among the two samples.
It would be unlikely that all the low values are in sample 1 if the
null hypothesis is true.
For example: suppose we have 2 modules with 30 students in
each (thus 60 students overall). Suppose the height of the
students are drawn from the same distribution. How likely is it
that all the shortest 30 students end up in module 1? Not very
likely.
Note that if all the shortest 30 students end up in module 1, we
know that all the tallest 30 students end up in module 2. Thus,
we only need to look at one of the two samples to tell if the
distribution of the rank scores is uneven.
How to compute it:
1.
2.
Rank all scores (in both lists) from the lowest (Rank = 1) to the
highest (score N = total number of observations)
Sum the Ranks for each list. The rank sums are:
n1
n1
∑ R and ∑ R
3.
i,2
i,1
i=1
i=1
Calculate an U-value for each sample using the following
formulae:
n1
n (n +1)
U1 = ∑ Ri,1 − 1 1
2
i=1
n2
n (n +1)
U2 = ∑ Ri,2 − 2 2
2
i=1
Here n1 is the number of observations in list 1 and n2 is the number
of observations in list 2.
4. Compare the smaller U-value with the critical value in a Table for the
Mann-Whitney test. The calculated value must be equal to or smaller
than the table value for significance.
31
Mann-whitney test
Mann-whitney test
Typical question: Using the data below, test the hypothesis that there is no difference
between the two samples using a Mann-Whitney test and a significance level of 0.05. What
is the value of the test statistic (the lowest u value)? Can you reject the null-hypothesis?
Sample 1: 44 87 82 9 86 25 72 81
Sample 2: 73 13 49 12 57 15
Solution: Total 8+6 = 14 observations
Rank all observations from 1 (lowest) to 14 (highest).
Sample 1
Ranks
Sample 2
Ranks
44
6
87
14
82
12
9
1
86
13
25
5
73
10
13
3
49
7
12
2
57
8
15
4
The sum of ranks for sample 1 is: 71.
The U value for sample 1 is: 71 – 8*9/2 = 71 – 36 = 35
The sum of ranks for sample 2 is: 34.
The U value for sample 2 is: 34 – 6*7/2 = 34 – 21 = 13.
The smallest U value is: 13
72
9
81
11
Sum is 71
Sum is 34
The critical value is: 8
The smallest U value (13)
Is not smaller or equal to 8. Thus, we do
not reject the null hypothesi
33
Chi-square test
Critical values: df = (number rows -1)*(number columns-1)
Idea behind test: we want to test if two dimensions are independent (not associated). Null
hypothesis: not associated.
For example, is the proportion of students who pass the test independent of whether the
students come from Sweden?
From Sweden
Not from Sweden
Pass exam
20
45
Do not pass exam
10
15
There are 20+10+45+15 = 90 students.
30/90 are from Sweden.
65/90 pass the exam
If passing the exam was independent of being from Sweden, the proportion of students who
pass the exam AND are from Sweden would be:
P(pass exam)*P(Sweden) = (65/90)*(30/90)= .
The number of students, among the 90, who passed the exam and were from Sweden would
thus be (60/90)*(30/90) *90 = (2/9)*90 = 21 2/3. This is different from what we observe
(which is 20).
In this test we compare the expected numbers, in each “box”’, with the observed.
Expected in the box ”Sweden and pass exam” = (60/90)*(30/90) *90
This can be written as = (60*30/90)
To see if the observed differ from the expected we compute:
(observed-expected)^2 and divide with expected.
We do this for every “box” and sum these numbers.
We reject if the sum is large, because in this case the observed differ a lot from the expected
(and expected = expected if the null hypothesis of no association is true).
34
Chi-square test: typical question and how to compute it
35
Covariance
Definition:
cov(x, y) =
1
∑( x − x ) ( y − y )
n
Y
Y
Y
Average x
Correlation
Definition: rxy =
Stdev of y
Stdev of x
Y
Y
The correlation coefficient
is between -1 and +1
cov(x, y)
sx sy
Y
Y
Y
Y
Y
Y
Y
X
X
X
X
X
X
∑ xy − x * y
cov(x, y) =
Positive association:
Cov > 0
Negative
association
No
association
Cov < 0 Cov = 0 Simpler Formula: cov(x, y) = n Simpler Formulas: ∑x − x * x 2 Sx = n ∑y − y * y S = 2 ∑ xy − x * y y n n Positive association: Negative association No association Correlation > 0
Correlation < 0 Correlation =0 Perfect linear association and positive slope implies correlation coefficient = +1 150 Typical question: The data shows the performance (y) and amount of practice (x) for 4 individuals. Based on this data, compute the Covariance. Typical question: The data shows the performance (y) and amount of practice (x) for 4 individuals. Based on this data, compute the correlation coefficient. Solution: Solution: 100 50 0 0 50 100 150 36 Regression Problem: finding the best fitting line Typical question: The following table shows profitability (Y) and investment in online advertising (X) for five firms. What is the value of the slope (b) of the regression line, Y = a+bX? What is the value of the intercept? Firm 1 Advertising 3.0 Profitability 7.0 2 11.0 9.0 3 8.0 9.0 4 4.0 3.0 5 6.0 8.0 Solution: Computations can be done as follows: Solution ŷ = â + b̂x b̂ = SSxy â = y − b̂x SSx Computations: N SSxy = ∑ (xi − x )(yi − y ) i=1 # N &# N & % ∑ xi (% ∑ yi ( N \$ '\$ i=1 ' = ∑ xi yi − i=1 N i=1 obs 1 2 3 4 5 Sum Mean X 3 11 8 4 6 32.00 6.40 X*X 9 121 64 16 36 246.00 Y 7 9 9 3 8 36.00 7.20 X*Y 21 99 72 12 48 252.00 Y*Y 49 81 81 9 64 284.00 N SSx = ∑ (xi − x )2 i=1 # N &# N & % ∑ xi ( % ∑ xi ( N '\$ i=1 ' 2 \$ i=1 = ∑ xi − N i=1 37 Typical question: Given the data below compute the R-square coefficient. R-square coefficient How well does the line fit the data? It fits well if the sum of the squared distances is small obs 1 2 3 4 obs 1 2 3 4 Sum Mean ei2 = (yi − ŷi )2 N N ∑ e = ∑(y − ŷ ) 2 i i=1 i 2 i y 3 6 2 8 Solution: Computations can be done as follows: Squared distance between observation i and line: The line fits well if the sum of the squared distances x 2 3 4 7 x 2 3 4 7 16.000 4.000 x*x 4 9 16 49 78.000 y 3 6 2 8 19.000 4.750 x*y 6 18 8 56 88.000 y*y 9 36 4 64 113.000 First, we need to compute the slope (b) and the intercept (a). The formula for the !! slope is: ! = !!!". To calculate b we compute: ! ∑#" !! = 1 − %\$\$! & we need to compute ∑(%! and **& . # ∑(%! = ('! + (!! + ((! + ()! = (∑&)(∑() *+∗*+ = 88 − * . + = 88 − 76 = 12, and %%" = ∑' / − ) (∑&)(∑&) *+∗*+ !! */ * ) + = 78 − * . + = 78 − 64 = 14. Thus, the slope is !!!" = *. ≈ 0.857. ! 0.0016 + 4.44 + 7.563 + 0.46 = 12.464. (∑+)(∑+) Also, **& = ∑4 ! − % - & = The intercept is: 6 = (7 − !'̅ = 4.75 − 0.857 ∗ 4 ≈ 1.321. 113 − % Second, we compute the predicted y values ((
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# How to solve algebraic equations by long division method?
To Do :
We have to explain , how to solve algebraic equations by long division method.
Solution :
For example,
1. Divide the polynomial $6-t+$$t^5$ by the polynomial $t^2$$+5$.
Write both the dividend and the divisor in descending order of the powers of the variables.
Therefore, the dividend is $t^5$$-t+6$ and the divisor is $t^2$$+5$.
To make the division process simpler, represent the dividend as $t^5+0\times t^4+0\times t^3+0\times t^2-t+6$ because zero multiplied by anything will result in zero, and thus will not change the value of the expression.
The first step is to divide the first term of the dividend by the first term of the divisor. So, divide $t^5$ by $t^2$ to get $t^3$
The second step is to multiply the divisor by this $t^3$, that is, $t^3$ forms the first term of the quotient. Subtract the result, $t^5+0\times t^4+5\times t^3$gives the remainder as $-5t^3+0t^2-t+6$
The degree of the remainder is still greater than the degree of the divisor. Perform the above process till the degree of the remainder is less than the degree of the divisor.
$t^2+0t+5$ ) $\ t^5+0\times t^4+0\times t^3+0\times t^2-t+6$ ($t^3$ $\frac{t^5}{t^2}=t^3$
$t^5+0\times t^4+5t^3$
------------------
$\ t^2+0t+5$$\times t^2-t+6$
$\frac{-5t^3}{t^2}=-5t$
$-5t^3+0\times t^2-25t$
--------------------
$24t+6$
The degree of the new remainder 24t+6 is less than the degree of the divisor. So, stop the process here.
When the polynomial $6-t+t^5$ is divided by the polynomial $t^2+5$ , the quotient is $t^3-5t$ and the remainder is 24t+6.
Tutorialspoint
Simply Easy Learning |
# How do you evaluate 2((tan^-1(1/3))-tan^-1(-1/7)?
##### 1 Answer
Jul 3, 2016
$\frac{\pi}{4}$.
#### Explanation:
We have to use these Rules : R(1) : tan^-1x+tan^-1y=tan^-1{(x+y)/(1-xy)}; x,y>0, xy<1.
$R \left(2\right) : {\tan}^{-} 1 \left(- x\right) = - \left({\tan}^{-} 1 x\right) , x \in \mathbb{R}$.
The Given Exp. $= 2 {\tan}^{-} 1 \left(\frac{1}{3}\right) - {\tan}^{-} 1 \left(- \frac{1}{7}\right)$
$= {\tan}^{-} 1 \left(\frac{1}{3}\right) + {\tan}^{-} 1 \left(\frac{1}{3}\right) + {\tan}^{-} 1 \left(\frac{1}{7}\right) \ldots \ldots \ldots \ldots \ldots . . \left[R \left(2\right)\right]$
$= {\tan}^{-} 1 \left\{\frac{\frac{1}{3} + \frac{1}{3}}{1 - \frac{1}{3} \cdot \frac{1}{3}}\right\} + {\tan}^{-} 1 \left(\frac{1}{7}\right) \ldots \ldots \ldots \ldots \ldots . \left[R \left(1\right)\right]$
$= {\tan}^{-} 1 \left(\frac{2}{3} \cdot \frac{9}{8}\right) + {\tan}^{-} 1 \left(\frac{1}{7}\right)$
${\tan}^{-} 1 \left(\frac{3}{4}\right) + {\tan}^{-} 1 \left(\frac{1}{7}\right)$
$= {\tan}^{-} 1 \left\{\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{28}}\right\} \ldots \ldots \ldots \ldots \ldots . \left\{R \left(1\right)\right]$
$= {\tan}^{-} 1 \left\{\frac{\frac{25}{28}}{\frac{25}{28}}\right\}$
$= {\tan}^{-} 1 \left(1\right)$
$= \frac{\pi}{4}$. |
# How do you find the domain of (x+1)/(x^2 - 1)?
Apr 29, 2017
The domain of $\mathbb{R} - \left\{- 1 , 1\right\}$
#### Explanation:
We need
${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
Let rewrite the function by factorising
$\frac{x + 2}{{x}^{2} - 1} = \frac{x + 2}{\left(x + 1\right) \left(x - 1\right)}$
Let $f \left(x\right) = \frac{x + 2}{\left(x + 1\right) \left(x - 1\right)}$
As we cannot divide by $0$, $x \ne 1$ and $x \ne - 1$
Therefore,
the domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 1 , 1\right\}$ |
# 5.4 The Fundamental Theorem. The Fundamental Theorem of Calculus, Part 1 If f is continuous on, then the function has a derivative at every point in,
## Presentation on theme: "5.4 The Fundamental Theorem. The Fundamental Theorem of Calculus, Part 1 If f is continuous on, then the function has a derivative at every point in,"— Presentation transcript:
5.4 The Fundamental Theorem
The Fundamental Theorem of Calculus, Part 1 If f is continuous on, then the function has a derivative at every point in, and
First Fundamental Theorem: 1. Derivative of an integral.
2. Derivative matches upper limit of integration. First Fundamental Theorem: 1. Derivative of an integral.
2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. First Fundamental Theorem:
1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. New variable. First Fundamental Theorem:
1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. The long way: First Fundamental Theorem:
1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant.
The upper limit of integration does not match the derivative, but we could use the chain rule.
The lower limit of integration is not a constant, but the upper limit is. We can change the sign of the integral and reverse the limits.
The Fundamental Theorem of Calculus, Part 2 If f is continuous at every point of, and if F is any antiderivative of f on, then (Also called the Integral Evaluation Theorem) We already know this! To evaluate an integral, take the anti-derivatives and subtract.
Example 10 : The technique is a little different for definite integrals. We can find new limits, and then we don’t have to substitute back. new limit We could have substituted back and used the original limits.
Example 8: Wrong! The limits don’t match! Using the original limits: Leave the limits out until you substitute back. This is usually more work than finding new limits
Example 9 as a definite integral : Rewrite in form of ∫ u n du No constants needed- just integrate using the power rule.
Substitution with definite integrals Using a change in limits
Download ppt "5.4 The Fundamental Theorem. The Fundamental Theorem of Calculus, Part 1 If f is continuous on, then the function has a derivative at every point in,"
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Linear Inequalities in One Variable
The mathematical statement which says that one quantity is not equal to another quantity is called an inequality. As the word “inequality” indicates something that is not equal, an algebraic inequality refers to an algebraic expression that is not equal to another. In an inequality, two expressions are either less than (or equal to) or greater than (or equal to) another expression.
A linear inequality is a mathematical statement that relates a linear expression as either less than or greater than another. The following are some examples of linear inequalities..
For example: If m and n are two quantities such that m ≠ n; then any one of the following relations (conditions) will be true:
i.e., either
(i) m > n
(ii) m ≥ n
(iii) m < n
Or, m ≤ n
Each of the four conditions, given above, is an inequality.
A solution to a linear inequality is a real number that will produce a true statement when substituted for the variable. Linear inequalities have either infinitely many solutions or no solution. If there are infinitely many solutions, graph the solution set on a number line and/or express the solution using interval notation.
Two permissible rules:
Addition — Subtraction Rule: If the same number or expression is added to or subtracted from both sides of an inequation, the resulting inequation has the same solution (or solutions) as the original.
Multiplication — Division Rule:
(i) If both sides of an inequation are multiplied or divided by the same positive number, the resulting inequation has the same solution (or solutions) as the original.
(ii) If both sides of an inequation are multiplied or divided by the same negative number, the resulting inequation has the same solution (or solutions) as the original if the symbol of the inequality is reversed. Thus, the only difference between solving a linear equation and solving an inequation concerns multiplying or dividing both sides by a negative number. Therefore, always reverse the symbol of an inequation when multiplying or dividing by a negative number.
Consider the following statement:
“x is a number which when added to 2 gives a sum less than 6.”
The above sentence can be expressed as x + 2 < 6, where ‘<’ stands for “is less than”.
x + 2 < 6 is a linear inequality in one variable, x.
Clearly, any number less than 4 when added to 2 has a sum less than 6.
So, x is less than 4.
We say that the solutions of the inequality x + 2 < 6 are x < 4.
The form of a linear inequality in one variable is ax + b < c, where a, b and c are fixed numbers belonging to the set R.
If a, b and c are real numbers, then each of the following is called a linear inequality in one variable:
Similarly, ax + b > c (‘>’ stands for “is greater than”)
ax + b ≥ c (‘≥’ stands for “is greater than or equal to”)
ax + b ≤ c (‘≤’ stands for “is less than or equal to”)
are linear inequation in one variable. In an inequation, the signs ‘>’, ‘<’, ‘≥’ and ‘≤’ are called signs of inequality.
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Definition:
A factor of a number is always an exact divisor of that number. For eg: 4 is a factor of 24. This means that the number 24 is exactly divisible by 4 (Remainder of this division is 0). Also note that a factor of a number is always less than or equal to that number. For example: The factors of 12 are 1, 2, 3, 4, 6, 12. Each of these factors is less than or equal to the number 12.
Factors exist in pairs. For example: the factors of 18 are 1, 2, 3, 6, 9, 18. These factors can be paired as follows: 1 & 18, 2 & 9, 3 & 6. The product of each pair is 18. These pairs of factors are referred to as factor pairs.
Practice questions:
Q1: List the factors of 60.
A1: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
• Every number is a factor of itself.
For example: 24 x 1 = 24, hence 24 is a factor of itself
• 1 is a factor of every number
• Prime numbers have only two factors - the number itself and the number 1.
For example: the factors of 17 are 17 and 1.
• The number of factors of a given number are always finite.
For example: the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Common Factors
Consider the factors of 24 and factors of 32
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Factors of 32: 1, 2, 4, 8, 16, 32
If you notice there are some numbers that are factors of both 24 and 32 namely 1, 2, 4, 8. Such numbers that are factors of two or more numbers are referred to as common factors of those numbers.
Common factors can also be represented by Venn diagrams.
Highest Common Factor (HCF)
In the above example, the number 8 is the highest common factor to appear in both lists and is called the Highest Common Factor of 24 and 32. Highest Common Factor or HCF is sometimes also referred to as the Greatest Common Divisor.
The Highest Common Factor of two or more numbers is defined as the greatest number that divides each of those numbers exactly.
Practice questions:
Q1: List the common factors of 20 and 18.
A1: Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 18: 1, 2, 3, 6, 9, 18
The common factors of 20 and 18 are 1 and 2.
Q2: Find the Highest Common Factor of 24 and 64.
A2: The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
Factors of 64: 1, 2, 4, 8, 16, 32, 64.
The common factors of 24 and 64 are 1, 2, 4, 8. The HCF of 24 and 64 is 8.
Q3: Find the Highest Common Factor of 13 and 21.
A3: The factors of 21 are 1, 3, 7, 21
Factors of 13: 1, 13 (Note that 13 is a prime number)
The Highest Common Factor of 13 and 21 is 1.
Important facts about Highest Common Factors:
• The HCF of a given set of numbers can never be greater than any of the numbers in the set.
• Co-prime numbers are numbers who do not have any common factor between them other than 1. In other words, the HCF of co-prime numbers is always 1.
• The product of HCF and LCM of two numbers is equal to the product of those numbers.
For example: Consider two number 12 and 9.
The LCM of 12 and 9 is 36.
The HCF of 12 and 9 is 3.
The Product of 12 and 9 is 108.
The product of LCM and HCF is 36 x 3 = 108.
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# Applications of the Sine Law
Lesson
#### Example
At a picnic by the river two children wondered how tall the cliff opposite the river was. By taking some measurements, like the angles and distances shown, can you help them find the height of the cliff?
Goal: Find length $AB$AB.
In triangle $ABC$ABC we only have angle measurements before we can use trigonometry to find length $AB$AB we will need a side length. Length $BC$BC can be found using triangle $BCD$BCD.
In triangle $BCD$BCD, angle B is $180-69-47=64$1806947=64° (angles in a triangle sum to $180$180)
Using the sine rule:
$\frac{b}{\sin B}$bsinB $=$= $\frac{d}{\sin D}$dsinD $d$d is length $BC$BC, named d because it is opposite angle $D$D $\frac{143}{\sin64^\circ}$143sin64° $=$= $\frac{d}{\sin47^\circ}$dsin47° then we can rearrange to find length $d$d $\frac{143\sin47^\circ}{\sin64^\circ}$143sin47°sin64° $=$= $d$d $d$d $=$= $116.36$116.36 m
Now we are only halfway through. We still need to find the height of the cliff.
Using tan we can complete the question
$\tan39^\circ$tan39° $=$= $\frac{height}{116.36}$height116.36 $height$height $=$= $116.36\times\tan39^\circ$116.36×tan39° $height$height $=$= $94.2265$94.2265 ...
So the cliff is approximately $94$94m tall.
## Using the sine rule
The sine rule is useful when you want to find:
• a side length if you know 2 angles and another side
• an angle size if you know 1 angle and 2 corresponding side lengths
The best way to start problems involving applications of the sine rule is to label the angles of the triangle, label the corresponding sides and then use the most appropriate version of the sine rule.
The sine rule
$\frac{a}{\sin A}=\frac{b}{\sin B}$asinA=bsinB =$\frac{c}{\sin C}$csinC
we can also use the alternative form
$\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa=sinBb = $\frac{\sin C}{c}$sinCc
depending on what we are trying to use the law for.
The sine rule shows: that the lengths of the sides in a triangle are proportional to the
sines of the measures of the angles opposite them
Here are some worked examples.
##### Question 2
Mae observes a tower at an angle of elevation of $12^\circ$12°. The tower is perpendicular to the ground.
Walking $67$67 m towards the tower, she finds that the angle of elevation increases to $35^\circ$35°.
1. Calculate the angle $\angle ADB$ADB.
2. Find the length of the side $a$a.
3. Using the rounded value of the previous part, evaluate the height $h$h, of the tower.
##### Question 3
Lucy travelled on a bearing of $39^\circ$39° from point A to point B. She then travelled on a bearing of $159^\circ$159° for $17$17 km to point C which is east of point A.
1. Find the size of the angle $\angle CAB$CAB.
2. Find the size of the angle $\angle ABC$ABC
3. Solve for $x$x, the distance Lucy would have to travel due east to return to point A.
### Outcomes
#### 10D.T3.04
Solve problems involving the measures of sides and angles in acute triangles |
# How to Divide Integers Part 2
This is the second video (see first video) in our series on division of integers. In this video, we will learn division by solving more examples. Watch the Taglish math tutorial video above and then answer the Practice Exercises below.
Practice Exercises
1.) – 105 ÷ (-5)
2.) 34 ÷ (-17)
3.) – 45 ÷ 5
4.)- 18 ÷ -3
5.) -72 ÷ 9
1.) 21
2.) -2
3.) -9
4.) 6
5.) -8
# How to Divide Integers Part 1
This is the first video in our series on Division of Integers. In this video, we discuss the rules on dividing integers as well as give some examples. Watch the Taglish math tutorial video above and then answer the Practice Exercises below.
Practice Exercises
1.) – 14 ÷ 2
2.) 16 ÷ (-8)
3.) – 42 ÷ (- 7)
4.) – 88 ÷ 8
5.) 72 ÷ 8
1.) -7
2.) -2
3.) 6
4.) -11
5.) 9
# How to Multiply Integers Part 3
This is the third and the last part of our series on Multiplication of Integers (see first and second parts). In this video, we discuss more examples on how to multiply integers.
Watch the Taglish math tutorial video above and then answer the Practice Exercises below.
Practice Exercises:
1.) – 4 × 3 × (- 2) × 1
2.) (- 4) (- 1)(-8)
3.) (-1)(9)(-3) (-1)
4.)(-2)(-3)(1) (-5)(-1)
5.)(-7)(2)(1) (-2)(6)
1.) 24
2.) -32
3.) -27
4.) 30
5.) 168
# How to Multiply Integers Part 2
After learning the basics of multiplication of integers, we continue learning by discussing more examples in this video. Here, we apply the rules we have learned in the previous tutorial in multiplying more than two integers.
Watch the Taglish math tutorial video above and then answer the Practice Exercises below.
Practice Exercises: Convert the following subtraction expressions to an addition expression and then find the answer.
1.) – 3 × 4
2.) (- 7) (- 2)(-3)
3.) (1)(12)(-1)
4.) (- 3) × 3 × (-3)
5.) (- 4) × 2 × (-6) × 2
1.) -12
2.) -42
3.) -12
4.) 27
5.) 96
# How to Multiply Integers Part 1
In this video, we learn how to multiply positive and negative integers. We will also discuss the rules in multiplication of integers. Note that aside from the × sign, parentheses are also used in multiplication. For example, – 6 × 3 can also be written as (-6)(3).
Watch the video above and then answer the Practice Exercises below.
Practice Exercises: Convert the following subtraction expressions to an addition expression and then find the answer.
1.) – 8 × 4
2.) – 5 × – 4
3.) (7)(-12)
4.) 9 × 10
5.) – 12 × 0
1.) -32
2.) 20
3.) -84
4.) 90
5.) 0
# How to Subtract Integers Part 3
In this video, we summarize what we have learned in the first part and second part of this tutorial series on subtraction of integers. The main point of the discussion in the video is that any subtraction expression can be converted into an addition expression.
Watch the video above and then answer the Practice Exercises below.
Practice Exercises: Convert the following subtraction expressions to an addition expression and then find the answer.
1.) – 12 – 11
2.) 8 – 12
3.) 5 – (- 7)
4.)- 13 – (- 4)
5.) 0 – 21
1.) – 12 – 11 = – 12 + (-11) = -23
2.) 8 – 12 = 8 + (-12) = -4
3.) 5 – (- 7) = 5 + 7 = 12
4.)- 13 – (- 4) = – 13 + 4 = -9
5.) 0 – 21 = 0 + (-21) = -21
# How to Subtract Integers Part 2
In this video, we continue learning about subtraction of integers. We are going to use the rules that we have learned in the previous video to answer more exercises.
Watch the video above and then answer the Practice Exercises below.
Practice Exercises
1.) 6 – 10
2.) – 17 – 7
3.) 0 – (- 12)
4.) – 4 – (- 15)
5.) 0 – 21 Continue reading
# How to Subtract Integers Part 1
In this video, we are going to learn how to subtract integers. In doing this, we are going to use what we have learned in adding integers and convert subtraction sentences to addition sentences.
Watch the video above and then answer the Practice Exercises below.
Practice Exercises
1.) 4 – 8
2.) – 12 – 6
3.) 7 – (-9)
4.) – 3 – (- 12)
5.) 0 – 17 Continue reading
# How to Add Integers Part 3
In Part 1 and Part 2 of this series, we learned how to use the number line and positive and negative chips to add integers. In this video, we will learn how to add integers where the given has more than two addends. Watch the video above and then answer the Practice Exercises.
Practice Exercises
1.) 8 + 2 + (-22)
2.) 15 + (-4) + (-3)
3.) – 14 + (- 5) + (-3)
4.) 12 + (-3) + (-9)
5.) -11 + 2 + (-6)
# How to Add Integers Part 2
In the previous video, we learned how to use the number line and positive and negative chips to add integers. In this video, we continue our discussion. We will have more worked examples on adding integers. Watch the video above and answer the practice exercises.
Practice Exercises
1.) 25 + (-10)
2.) – 60 + (-4)
3.) – 34 + 15
4.) 37 + 9
5.) – 45 + 0 Continue reading
# How to Add Integers Part 1
In this video, we are going to use the number line to illustrate addition of integers. We are also going to represent the integers are positive and negative chips to illustrate addition.
Watch the video above and answer the Practice Exercises.
Practice Exercises
1.) 12 + (-5)
2.) -9 + (-4)
3.) -15 + 8
4.) 11 + 8
5.) 0 + (- 3)
6.) 6 + (-6)
# How to Divide Numbers with Decimals
After learning addition, subtraction, and multiplication of numbers with decimals, let us now learn how to divide numbers with decimals. The easy method of doing this is by multiplying both the dividend and divisors of the decimals being divided to eliminate the decimal point. Watch the Tagalog video above to learn this method and then answer the exercises below.
Practice Exercises
1.) 2.8 ÷ 0.2
2.) 12 ÷ 0.5
3.) 0.065 ÷ 0.13
4.) 8.04 ÷ 3
5.) 1.26 ÷ 0.3
# How to Multiply Numbers with Decimals
After learning how to add and subtract numbers with decimals, let us now learn how to multiply numbers with decimals. To know how to multiply numbers with decimals, watch the video above and then answer the exercises below.
Practice Exercises
1.) 2 × 2.5
2.) 3.1 × 1.2
3.) 4.52 × 10
4.) 5.21 × 0.01
4.) 8.21 × 3.26 Continue reading
# How to Subtract Numbers with Decimals
After learning how to add numbers with decimals, let us now learn how to subtract numbers with decimals. In subtracting numbers with decimals, the method is just to align the decimal points of the two numbers and then subtract them. Watch the video above to learn more and then answer the exercises below.
Practice Exercise
1.) 2.3 – 1.8
2.) 3.05 – 2.12
3.) 4.062 – 3.02
4.) 12.01 – 6.025
5.) 6.513 – 2.61 Continue reading
# How to Add Numbers with Decimals
In this video, we are going to learn how to add decimals. To add decimals, arrange the numbers vertically and align their decimal points. Watch the video above and answer the exercises below.
Practice Exercises
1.) 1.5 + 0.01
2.) 3.05 + 2.6 + 4.008
3.) 4.58 + 1.15 + 2.008
4.) 12.006 + 3.5 + 18.21
5.) 6.5 + 2.86 + 2.1 + 5 Continue reading
# Division of Fractions
After learning multiplication of fractions, we now learn division of fractions. This series is composed only of two videos. The first video involves division of fractions improper fractions and the second video on mixed fractions. To divide fractions, you just have to get the reciprocal of the divisor and then multiply it by the dividend. The result is the answer of the division.
DF1 Division of Fractions (Proper and Improper)
This video is a tutorial on how to divide proper and improper fractions.
DF2 Division of Fractions (Mixed Fractions)
This video is a tutorial on how to divide mixed fractions by proper fractions, division of mixed fractions by mixed fractions, as well as division of mixed fractions by whole numbers.
In the next series, we will be learning about operations on decimals. |
# What is a Statistic
This tutorial provides definitions and examples of a statistic. Further, it explains the relationship between a statistic and sampling distribution.
## The Definition of a Statistic
A statistic is a function of observable random variables, T=t(X1, X2, …, Xn), which does not depend on any unknown parameters.
script t is the function that we apply to X1, X2, …, Xn to define the statistic, which is denoted by capital T.
The intent of the use of a statistic T is to make inferences about the distribution of the set of random variables. Thus, if the variables are not observable or if the function t(X1, X2, …, Xn) depends on unknown parameters, T would not be useful in making such inferences.
## Example 1 of a statistic
Note that, the set of observable random variables need not be a random sample. For instance, 13 planes were in service, and the first 10 air conditioner failer times were as follows.
23,50,50,55,74,90,97,102,130,194
For this case, $$T = \sum_{i=1}^{10} y_i+2y_{10} = 1447$$ is a statistic. For sure, X1, X2, …, and Xn can be random samples, but they need not be random samples.
## Example 2 of a statistic
Let X1, X2, …, and Xn represent a random sample from a population. The sample mean $$\bar{X}$$ is a statistic with the function of t(X1, X2, …, Xn) = (X1, X2, …, Xn)/n. Often, the statistic of the sample mean $$\bar{X}$$ is written as follows.
$$\bar{X} = \sum_{i=1}^{n} \frac{X_i}{n}$$
Note that, in the function above, $$\bar{X}$$ uses the capital case of $$X$$. In contrast, when a random sample is observed, the value of $$\bar{X}$$ computed from the data is denoted by lowercase $$\bar{x}$$.
The sample mean $$\bar{x}$$ is useful because it can estimate the population mean and population variance. In particular, if X1, X2, …, Xn represent a random sample from a population f(x) with $$E(X) =\mu$$ and $$Var(X) =\sigma^2$$. Then, we can get:
$$E(\bar{X}) = \mu$$
$$Var(\bar{X}) = \frac{\sigma^2}{n}$$
## A statistic and sampling distribution
A statistic is also a random variable. The distribution of a statistic is referred to as a derived distribution or sampling distribution, in contrast to the population distribution.
Many important statistics can be expressed as a linear combination of independent normal random variables. For instance, if X1, X2, …, Xn denotes a random sample from $$N(\mu, \sigma^2)$$, then, we can get the sampling distribution of $$\bar{X}$$ as follows.
$$\bar{X} \sim N(\mu, \sigma^2/n)$$ |
# Thread: another system of equations
1. ## another system of equations
solve for x,y
xy = e
x^(ln(y))=1/e^2
2. Hello, jmedsy!
A fascinating problem . . .
Solve for $x,y$
. . . $xy \:=\: e\;\;\;\;[1]$
. . $x^{\ln y} \:=\:\frac{1}{e^2}\;\;[2]$
Take logs of equation [1]: . $\log(xy) \:=\:\ln(e) \quad\Rightarrow\quad \ln(x) + \ln(y) \:=\:1]$
Take logs of equation [2]: . $\ln\left(x^{\ln y}\right) \:=\:\ln\left(\frac{1}{e^2}\right) \quad\Rightarrow\quad \ln(x)\cdot\ln(y) \:=\:-2$
We have a system of equations: . $\begin{array}{cccc} \ln(x) + \ln(y) &=& 1 \\ \ln(x)\cdot\ln(y) &=&-2 \end{array}$
Substitute: . $\begin{array}{ccc}u &=& \ln(x) \\ v &=& \ln(y) \end{array} \quad\Rightarrow\quad \begin{array}{cccc}u + v &=& 1 & [3] \\ u\cdot v &=& -2 & [4] \end{array}$
From [3], we have: . $v \:=\:1-u\;\;[5]$
Substitute into [4]: . $u(1-u) \:=\:-2 \quad\Rightarrow\quad u^2 - u - 2 \:=\:0$
Factor: . $(u+1)(u-2) \:=\:0 \quad\Rightarrow\quad u \:=\:\text{-}1,\:2$
. . $u = \text{-}1\!:\;\text{Substitute into [5]: }\:v \:=\:1-(\text{-}1) \quad\Rightarrow\quad v \,=\,2$
. . $u = 2\!:\;\text{Substitute into [5]: }\:v \:=\:1-2 \quad\Rightarrow\quad v \,=\,\text{-}1$
We have: . $(u,v) \:=\:(\text{-}1,2),\:(2,\text{-}1)$
Back-substitute:
. . $(u,v) \,=\,(\text{-}1,2) \quad\Rightarrow\quad \begin{array}{cccc}\ln(x) \;=\;\text{-}1 & \Rightarrow & x \;=\;\frac{1}{e} \\ \ln(y) \;=\; 2 & \Rightarrow & y \;=\; e^2 \end{array}$
. . $(u,v) \,=\,(2,\text{-}1) \quad\Rightarrow\quad \begin{array}{cccc}\ln(x) \;=\;2 & \Rightarrow & x \;=\;e^2 \\ \ln(y) \;=\;-1 & \Rightarrow & y \;=\;\frac{1}{e} \end{array}$
Two solutions: . $\left(\frac{1}{e^2},\;e^2\right),\;\left(e^2,\;\fr ac{1}{e}\right)$
3. finally came up with the same thing, thanks |
Linear Functions & Inequalities Linear Equation:
Notes # Alpha 3
A linear equation has the form +Ax + By = C, where A and B are both not zero. The graph of a linear equation is always a line.
Solution of a Linear Equation: An ordered pair that makes the equation true. Each ordered pair corresponds to a point in the coordinate plane. Linear Function: A linear function is defined by f(x) = mx + b, where m and b are real numbers. Constant Function: A function f is a constant function if f(x) = b. The graph is a horizontal line. A constant function either has no zeros (b ≠ 0) or every value of x is a zero (b = 0). Linear Inequality: A linear inequality has the form Ax + By < C, Ax + By > C, Ax + By > C, or Ax + By < C, where A and B are both not zero. The graph of linear inequality consists of a boundary and the shading of a region. Ex A: Write an inequality that describes each graph.
#1)
#2)
y = -.8x
#3)
x = -3
-3x + y = 10
-3x + y = 5
x=3
Inequality:
Inequality:
Inequality:
Ex B: Graph each inequality or equation.
#1)
y = 4x + 2
#2)
2y < x – 5
#3)
y = |3x – 2|
Note: When graphing, you must label all x- and y-int. You must also label vertices. You must also write the equation on the curve. Shade as needed.
Linear Relations & Functions Page 1 of 2
Linear Functions & Inequalities Zeros of the f: values of x for which f(x) = 0. (zeros of a function are the x-intercepts.) X-intercept: The point at which a graph crosses the x-axis. In a linear function, the x
intercept will have coordinates −
b b ,0 . Thus, − is the x-intercept. m m
Ex C: Find the zero of each function.
#1)
f(x) = 0.5x + 4
#2)
Zero =
f(x) = 10x
Zero =
#3)
f(x) = 7x + 3
Note: You can either sub 0 for f(x) and solve for x. Or you can use the formula
Zero =
Ex D: Complete the following word problem.
Mike can spend up to \$40 per day plus \$0.35 per mile when renting a car to use on company business. The total cost of the daily rental (C) is a function of the total miles driven. a. Write a linear inequality that expresses the acceptable daily rental car cost.
b. Graph the inequality.
Note: When graphing make sure x is the independent variable and y is the dependant variable.
Linear Relations & Functions Page 2 of 2
Alpha 3 Notes Linear Functions & Inequalities
Alpha 3 Notes Linear Functions & Inequalities |
Which number has more digits and by how many?
Paula writes down all 2015 digit positive integers into one row, forming a long number. Peter writes down all 2016 digit numbers into a long row and deleted all 0s. Whose number has more digits and by how many?
I tried to count how many 0s there are, but there are too many and it takes too long. What is the fast way to approach this?
There are $9\cdot 10^{2014}$ different $2015$ digit positive integers. By concatenating these together in some fashion, we get a total of $2015\cdot (9\cdot 10^{2014})$ digits in sequence.
There are $9\cdot 10^{2015}$ different $2016$ digit positive integers. A tenth of those will have a zero in the second digit. Similarly a tenth of them will have a zero in the third digit, fourth digit, etc...
Concatenating the integers together, we have $2016\cdot 9\cdot 10^{2015}$ however by deleting a zero from the second digit of each where it occurs will remove $\frac{1}{10}\cdot 9\cdot 10^{2015}$ digits. Similarly for each other digit position (except for the first of course since a $k$-digit number cannot start with zero). This brings our new count to $2016\cdot 9\cdot 10^{2015}-2015\cdot 9\cdot 10^{2015}\cdot \frac{1}{10} = (20160-2015)\cdot 9\cdot 10^{2014}$
• @GerardL. Do you first understand how many different $2015$ digit numbers there are? There are $9\cdot 10^{2014}$ different $2015$ digit numbers. If you take two $2015$ digit numbers and concatenate them, how many digits long will the result be? There will be $2\cdot 2015$ digits. If you take three $2015$ digit numbers and concatenate them, how many digits long will the result be? There will be $3\cdot 2015$ digits. If you take as many $2015$ digit numbers as exist and concatenate them together, how many digits long will it be? $(9\cdot 10^{2014})\cdot 2015$ digits long. Mar 11, 2017 at 2:13
• Similarly in the concatenation of all possible $2016$ digit long numbers there will be $9\cdot 10^{2015}\cdot 2016$ digits total. After that, we figure out how many of the digits must be removed because they were zeroes and we recognize that a tenth of the digits "in the second position" need to be removed and a tenth of the digits "in the third position" need to be removed, etc... giving us a total of $2015\cdot 9\cdot 10^{2015}\cdot \frac{1}{10}$ digits needing to be removed, giving us the final result. Mar 11, 2017 at 2:15 |
# Math calculator solve for x
Math calculator solve for x is a software program that helps students solve math problems. Our website can solving math problem.
## The Best Math calculator solve for x
In this blog post, we discuss how Math calculator solve for x can help students learn Algebra. Solving the square is a mathematical technique used to find the value of a variable in a quadratic equation. The name comes from the fact that the technique can be used to draw a square on a graph, which can then be used to solve for the value of the variable. The most common way to solve the square is by using the Quadratic Formula, which states that the value of the variable is equal to the negative of the coefficient of the squared term, divided by twice the coefficient of the linear term. Solving the square can be a difficult process, but with practice it can become easier. In addition, there are many software programs and online calculators that can help to solve the square. With some patience and effort, anyone can learn how to solve the square.
Once this has been accomplished, the resulting equation can be solved for the remaining variable. In some cases, it may not be possible to use elimination to solve a system of linear equations. However, by understanding how to use this method, it is usually possible to simplify a system of equations so that it can be solved using other methods.
Let's say you're a cashier and need to figure out how much change to give someone from a \$20 bill. You would take the bill and subtract it from 20, which would give you the amount of change owed. So, if someone gave you a \$20 bill, you would give them back \$16 in change since 20-4 equals 16. You can use this same method to solve problems with larger numbers as well. For example, if someone gave you a \$50 bill, you would take the bill and subtract it from 50, which would give you the amount of change owed. So, if someone gave you a \$50 bill, you would give them back \$40 in change since 50-10 equals 40. As you can see, this method is simple yet effective when trying to figure out how much change to give someone. Give it a try next time you're stuck on a math problem!
This may seem like a lot of work, but the FOIL method can be a very helpful tool for solving trinomials. In fact, many algebra textbooks recommend using the FOIL method when solving trinomials. So next time you're stuck on a trinomial, give the FOIL method a try. You might be surprised at how helpful it can be.
No one likes doing math, but it's a necessary evil that we all have to deal with at some point in our lives. Fortunately, there's now an app that can take care of those pesky math word problems for us. All you have to do is take a picture of the problem and the app will provide the solution. The app uses Optical Character Recognition (OCR) to read the text from the image and then solves the problem using artificial intelligence. So far, it's been pretty accurate and has even managed to stump a few math experts. So if you're looking for a way to avoid doing math, this app is definitely worth checking out.
## Instant support with all types of math
It does all kinds of problems! And shows you the steps to solving them too. I used to think of it as a cheat sheet, now I think of it more as a tool. Also, a lot of other reviews said they had to pay for things. I have the full app for free and I didn't spend a penny on anything! It would be awesome if they made a "photo ELA" or other subjects too. It is 100% ad free. Not even little side ads at the top or bottom. It is great!!
Maya Ross
The best calculator ever. This is an absolute gem, the calculator can easily recognize many handwritings, even hurried scrawls. You can also input data manually with an awesome interface which is leagues ahead of other calculator apps allowing complex problems to be entered. But the best feature is its ability to show the steps of the calculation with great detail, most useful app for any student who wants to strengthen his/her understanding of both foundation level and complex math.
Xyla Gonzales
College algebra word problems with solutions Math homework answers app App that does your math Problem solvers phone number Online photo math |
##### Common Core Math For Parents For Dummies with Videos Online
In Common Core math, sixth graders apply their knowledge of factors and multiples to look for the greatest common factor (or GCF) of two numbers and the least common multiple (LCM) of two numbers.
The greatest common factor of two numbers is the largest number that is a factor of both numbers. The first step in thinking about GCF is to think about common factors. For example, think about the numbers 4 and 10. The factors of 4 are 1, 2, and 4. The factors of 10 are 1, 2, 5, and 10. The common factors of these numbers are 1 and 2. The greatest common factor is 2 because it's the biggest number on the list of common factors. So you can say that the GCF of 4 and 10 is 2.
The least common multiple is the smallest (or least) number that is a multiple of both numbers. The first step in thinking about LCM is to think about common multiples. For example, think again about the numbers 4 and 10. The multiples of 4 are 4, 8, 12, 16, 20, 24, and so on (notice that these are the numbers you say when you skip count by 4, starting at 4 — skip counting is another way to think about the multiples of a number). The multiples of 10 are 10, 20, 30, 40, and so on. If you continue both of these lists far enough, there turn out to be many numbers on both lists: 20, 40, 60, and 80 are all examples of common multiples of 4 and 10. The least common multiple is the smallest number on that list: 20. So you can say that the LCM of 4 and 10 is 20. |
Glossary
Numbers or algebraic expressions are factors (or divisors) of another number if they multiply to give that number. For example, 3 and 4 are factors of 12 as 3×4=12. This can be written algebraically as
$$x$$ and $$y$$ are factors of $$m$$, if $$m=xy$$
For polynomial expressions the same rule applies. For example, $$x-4$$ and $$x-2$$ are factors of the quadratic expression $$x^2-6x+8$$ because $$(x-4)(x-2)=x^2-6x+8$$.
According to the factor theorem, if $$p(x)$$ is a polynomial and $$p(a)=0$$ for some number $$a$$, then $$x-a$$ is a factor of $$p(x)$$. Conversely, if $$p(x)$$ is divisible by $$x-a$$ then $$p(a)=0$$.
This follows from the more general remainder theorem, which states that the remainder of the division of a polynomial $$p(x)$$ by a linear polynomial $$x-a$$ is equal to $$p(a)$$. This relationship is often stated in the form $$px=q(x)(x-a)+p(a)$$, where $$q(x)$$ is another polynomial, usually referred to as the quotient. It follows that, if $$p(a)=0$$, the remainder is $$0$$ and $$p(x)$$ is divisible by $$x-a$$.
The factor theorem can be used to obtain factors of a polynomial; for example, if $$p(x)=x^3-3x^2+5x-6$$, then it is easy to check that $$p(2)=2^3-3×2^2+5×2-6=0$$. So by the factor theorem $$x-2$$ is a factor of $$x^3-3x^2+5x-6$$.
To factorise a number or algebraic expression is to express it as a product; for example,
15 is factorised when expressed as a product: 15=3×5.
$$x^2-3x+2$$ is factorised when written as a product: $$x^2-3x+2=(x-1)(x-2)$$
A five-number summary is a method of summarising a data set using five statistics: the minimum value, the lower quartile, the median, the upper quartile and the maximum value. Box plots are a useful method of graphically depicting five-number summaries.
The fraction $$\frac ab$$ (written alternatively as a/b), where a and b are integers unequal to zero. For example, $$\frac35$$ refers to 3 of 5 equal parts of the whole.
In the fraction $$\frac ab$$ the number a is the numerator and the number b is the denominator.
Frequency, or observed frequency, is the number of times that a particular value occurs in a data set. For grouped data, it is the number of observations that lie in that group or class interval.
An expected frequency is the number of times that a particular event is expected to occur when a chance experiment is repeated a number of times. If the experiment is repeated n times, and on each of those times the probability that the event occurs is p, then the expected frequency of the event is np.
For example, suppose that a fair coin is tossed 5 times and the number of heads showing recorded. Then the expected frequency of ‘heads’ is 5/2. This example shows that the expected frequency is not necessarily an observed frequency, which in this case is any one of the numbers 0, 1, 2, 3, 4 or 5.
The relative frequency is given by the ratio $$\frac fn$$, where f is the frequency of occurrence of a particular data value or group of data values in a data set and n is the number of data values in the data set.
A frequency distribution is the division of a set of observations into a number of classes, together with a listing of the number of observations (the frequency) in that class. Frequency distributions can be displayed in the form of a frequency table, a two-way-table or in graphical form.
A frequency table lists the frequency (number of occurrences) of observations in different ranges, called class intervals.
The frequency distribution of the heights (in cm) of a sample of 46 people is displayed in the form of a frequency table below.
The information in a frequency table can also be displayed graphically in the form of a histogram or using a column graph.
A function $$f$$ assigns to each element of a set of input values (called the domain) precisely one element of a set of output values (called the range). In mathematical modelling, the independent variable is usually chosen as the input values for the function. The output values then represent the dependent variable.
Functions are usually defined by a formula for $$f(x)$$ in terms of $$x$$; for example, the formula $$f(x)=x^2$$ defines the ‘squaring function’ that maps each real number x to its square $$x^2$$. The graph of this function is shown below. |
We've updated our
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# Solving Linear Equations
### Learning Objectives
• Use properties of equality to isolate variables and solve algebraic equations
• Use the properties of equality and the distributive property to solve equations containing parentheses
• Clear fractions and decimals from equations to make them easier to solve
• Solve equations that have one solution, no solution, or an infinite number of solutions
Steps With an End In Sight
## Use properties of equality to isolate variables and solve algebraic equations
There are some equations that you can solve in your head quickly, but other equations are more complicated. Multi-step equations, ones that takes several steps to solve, can still be simplified and solved by applying basic algebraic rules such as the multiplication and addition properties of equality. In this section we will explore methods for solving multi-step equations that contain grouping symbols and several mathematical operations. We will also learn techniques for solving multi-step equations that contain absolute values. Finally, we will learn that some equations have no solutions, while others have an infinite number of solutions. First, let's define some important terminology:
• variables: variables are symbols that stand for an unknown quantity, they are often represented with letters, like x, y, or z.
• coefficient: Sometimes a variable is multiplied by a number. This number is called the coefficient of the variable. For example, the coefficient of 3x is 3.
• term: a single number, or variables and numbers connected by multiplication. -4, 6x and $x^2$ are all terms
• expression: groups of terms connected by addition and subtraction. $2x^2-5$ is an expression
• equation: an equation is a mathematical statement that two expressions are equal. An equation will always contain an equal sign with an expression on each side. Think of an equal sign as meaning "the same as." Some examples of equations are $y = mx +b$, $\frac{3}{4}r = v^{3} - r$, and $2(6-d) + f(3 +k) = \frac{1}{4}d$
The following figure shows how coefficients, variables, terms, and expressions all come together to make equations. In the equation $2x-3^2=10x$, the variable is $x$, a coefficient is $10$, a term is $10x$, an expression is $2x-3^2$.
There are some equations that you can solve in your head quickly. For example—what is the value of y in the equation $2y=6$? Chances are you didn’t need to get out a pencil and paper to calculate that $y=3$. You only needed to do one thing to get the answer: divide 6 by 2. Other equations are more complicated. Solving $\displaystyle 4\left( \frac{1}{3}t+\frac{1}{2}\right)=6$ without writing anything down is difficult! That’s because this equation contains not just a variable but also fractions and terms inside parentheses. This is a multi-step equation, one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules. Remember that you can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The addition property of equality and the multiplication property of equality explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you’ll keep both sides of the equation equal. If the equation is in the form $ax+b=c$, where x is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division.
### Example
Solve $3y+2=11$.
Answer: Subtract 2 from both sides of the equation to get the term with the variable by itself.
$\displaystyle \begin{array}{r}3y+2\,\,\,=\,\,11\\\underline{\,\,\,\,\,\,\,-2\,\,\,\,\,\,\,\,-2}\\3y\,\,\,\,=\,\,\,\,\,9\end{array}$
Divide both sides of the equation by 3 to get a coefficient of 1 for the variable.
$\begin{array}{r}\,\,\,\,\,\,\underline{3y}\,\,\,\,=\,\,\,\,\,\underline{9}\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9\\\,\,\,\,\,\,\,\,\,\,y\,\,\,\,=\,\,\,\,3\end{array}$
$y=3$
### Example
Solve $3x+5x+4-x+7=88$.
Answer: There are three like terms $3x$, $5x$, and $–x$ involving a variable. Combine these like terms. 4 and 7 are also like terms and can be added.
$\begin{array}{r}\,\,3x+5x+4-x+7=\,\,\,88\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x+4+7=\,\,\,88\end{array}$
The equation is now in the form $ax+b=c$, so we can solve as before.
$7x+11\,\,\,=\,\,\,88$
Subtract 11 from both sides.
$\begin{array}{r}7x+11\,\,\,=\,\,\,88\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-11\,\,\,\,\,\,\,-11}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x\,\,\,=\,\,\,77\end{array}$
Divide both sides by 7.
$\begin{array}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{7x}\,\,\,=\,\,\,\underline{77}\\7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,=\,\,\,11\end{array}$
$x=11$
Some equations may have the variable on both sides of the equal sign, as in this equation: $4x-6=2x+10$. To solve this equation, we need to “move” one of the variable terms. This can make it difficult to decide which side to work with. It doesn’t matter which term gets moved, $4x$ or $2x$, however, to avoid negative coefficients, you can move the smaller term.
### Examples
Solve: $4x-6=2x+10$
Answer: Choose the variable term to move—to avoid negative terms choose $2x$
$\,\,\,4x-6=2x+10\\\underline{-2x\,\,\,\,\,\,\,\,\,\,-2x}\\\,\,\,2x-6=10$
Now add 6 to both sides to isolate the term with the variable.
$\begin{array}{r}2x-6=10\\\underline{\,\,\,\,+6\,\,\,+6}\\2x=16\end{array}$
Now divide each side by 2 to isolate the variable x.
$\begin{array}{c}\frac{2x}{2}=\frac{16}{2}\\\\x=8\end{array}$
In this video, we show an example of solving equations that have variables on both sides of hte equal sign. https://youtu.be/f3ujWNPL0Bw
## The Distributive Property
As we solve linear equations, we often need to do some work to write the linear equations in a form we are familiar with solving. This section will focus on manipulating an equation we are asked to solve in such a way that we can use the skills we learned for solving multi-step equations to ultimately arrive at the solution. Parentheses can make solving a problem difficult, if not impossible. To get rid of these unwanted parentheses we have the distributive property. Using this property we multiply the number in front of the parentheses by each term inside of the parentheses.
### The Distributive Property of Multiplication
For all real numbers a, b, and c, $a(b+c)=ab+ac$. What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the steps we have already practiced to isolate the variable and solve the equation.
### Example
Solve for $a$. $4\left(2a+3\right)=28$
Answer: Apply the distributive property to expand $4\left(2a+3\right)$ to $8a+12$
$\begin{array}{r}4\left(2a+3\right)=28\\ 8a+12=28\end{array}$
Subtract 12 from both sides to isolate the variable term.
$\begin{array}{r}8a+12\,\,\,=\,\,\,28\\ \underline{-12\,\,\,\,\,\,-12}\\ 8a\,\,\,=\,\,\,16\end{array}$
Divide both terms by 8 to get a coefficient of 1.
$\begin{array}{r}\underline{8a}=\underline{16}\\8\,\,\,\,\,\,\,\,\,\,\,\,8\\a\,=\,\,2\end{array}$
$a=2$
In the video that follows, we show another example of how to use the distributive property to solve a multi-step linear equation. https://youtu.be/aQOkD8L57V0 In the next example, you will see that there are parentheses on both sides of the equal sign, so you will need to use the distributive property twice. Notice that you are going to need to distribute a negative number, so be careful with negative signs!
### Example
Solve for $t$. $2\left(4t-5\right)=-3\left(2t+1\right)$
Answer: Apply the distributive property to expand $2\left(4t-5\right)$ to $8t-10$ and $-3\left(2t+1\right)$ to$-6t-3$. Be careful in this step—you are distributing a negative number, so keep track of the sign of each number after you multiply.
$\begin{array}{r}2\left(4t-5\right)=-3\left(2t+1\right)\,\,\,\,\,\, \\ 8t-10=-6t-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$
Add $-6t$ to both sides to begin combining like terms.
$\begin{array}{r}8t-10=-6t-3\\ \underline{+6t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6t}\,\,\,\,\,\,\,\\ 14t-10=\,\,\,\,-3\,\,\,\,\,\,\,\end{array}$
Add 10 to both sides of the equation to isolate t.
$\begin{array}{r}14t-10=-3\\ \underline{+10\,\,\,+10}\\ 14t=\,\,\,7\,\end{array}$
The last step is to divide both sides by 14 to completely isolate t.
$\begin{array}{r}14t=7\,\,\,\,\\\frac{14t}{14}=\frac{7}{14}\end{array}\\$
[latex-display]t=\frac{1}{2}\\[/latex-display] We simplified the fraction $\frac{7}{14}\\$ into $\frac{1}{2}\\$
In the following video, we solve another multi-step equation with two sets of parentheses. https://youtu.be/StomYTb7Xb8
### Try It
[ohm_question]72650[/ohm_question]
Sometimes, you will encounter a multi-step equation with fractions. If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. This will clear all the fractions out of the equation. See the example below.
### Example
Solve $\frac{1}{2}x-3=2-\frac{3}{4}x$ by clearing the fractions in the equation first.
Answer: Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.
$\begin{array}{r}\frac{1}{2}x-3=2-\frac{3}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\end{array}$
Use the distributive property to expand the expressions on both sides. Multiply.
$\begin{array}{r}4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)-4\left(-\frac{3}{4}x\right)\\\\ \frac{4}{2}x-12=8-\frac{12}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\\ 2x-12=8-3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{array}$
Add 3x to both sides to move the variable terms to only one side. Add 12 to both sides to move the variable terms to only one side.
$\begin{array}{r}2x-12=8-3x\, \\\underline{+3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3x}\\ 5x-12=8\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$
Add 12 to both sides to move the constant terms to the other side.
$\begin{array}{r}5x-12=8\,\,\\ \underline{\,\,\,\,\,\,+12\,+12} \\5x=20\end{array}$
Divide to isolate the variable.
$\begin{array}{r}\underline{5x}=\underline{5}\\ 5\,\,\,\,\,\,\,\,\,5\\ x=4\end{array}$
$x=4$
Of course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve. In the following video, we show how to solve a multi-step equation with fractions. https://youtu.be/AvJTPeACTY0 Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find the easiest! Remember to check your answer by substituting your solution into the original equation.
### Try It
[ohm_question]52524[/ohm_question]
Sometimes, you will encounter a multi-step equation with decimals. If you prefer not working with decimals, you can use the multiplication property of equality to multiply both sides of the equation by a a factor of 10 that will help clear the decimals. See the example below.
### Example
Solve $3y+10.5=6.5+2.5y$ by clearing the decimals in the equation first.
Answer: Since the smallest decimal place represented in the equation is 0.10, we want to multiply by 10 to make 1.0 and clear the decimals from the equation.
$\begin{array}{r}3y+10.5=6.5+2.5y\,\,\,\,\,\,\,\,\,\,\,\,\\\\ 10\left(3y+10.5\right)=10\left(6.5+2.5y\right)\end{array}$
Use the distributive property to expand the expressions on both sides.
$\begin{array}{r}10\left(3y\right)+10\left(10.5\right)=10\left(6.5\right)+10\left(2.5y\right)\end{array}\\$
Multiply.
$30y+105=65+25y$
Move the smaller variable term, $25y$, by subtracting it from both sides.
$\begin{array}{r}30y+105=65+25y\,\,\\ \underline{-25y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-25y} \\5y+105=65\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$
Subtract 105 from both sides to isolate the term with the variable.
$\begin{array}{r}5y+105=65\,\,\,\\ \underline{\,\,\,\,\,\,-105\,-105} \\5y=-40\end{array}$
Divide both sides by 5 to isolate the y.
$\begin{array}{l}\underline{5y}=\underline{-40}\\ 5\,\,\,\,\,\,\,\,\,\,\,\,\,5\\ \,\,\,x=-8\end{array}$
$x=-8$
In the following video, we show another example of clearing decimals first to solve a multi-step linear equation. https://youtu.be/wtwepTZZnlY Here are some steps to follow when you solve multi-step equations.
### Solving Multi-Step Equations
1. (Optional) Multiply to clear any fractions or decimals. 2. Simplify each side by clearing parentheses and combining like terms. 3. Add or subtract to isolate the variable term—you may have to move a term with the variable. 4. Multiply or divide to isolate the variable. 5. Check the solution.
## Classify Solutions to Linear Equations
There are three cases that can come up as we are solving linear equations. We have already seen one, where an equation has one solution. Sometimes we come across equations that don't have any solutions, and even some that have an infinite number of solutions. The case where an equation has no solution is illustrated in the next examples.
## Equations with no solutions
### Example
Solve for x. $12+2x–8=7x+5–5x$
Answer: Combine like terms on both sides of the equation.
$\displaystyle \begin{array}{l}12+2x-8=7x+5-5x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\end{array}$
Isolate the x term by subtracting 2x from both sides.
$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\\\,\,\,\,\,\,\,\,\underline{-2x\,\,\,\,\,\,\,\,\,\,-2x\,\,\,\,\,\,\,\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4= \,5\end{array}$
This false statement implies there are no solutions to this equation. Sometimes, we say the solution does not exist, or DNE for short.
This is not a solution! You did not find a value for x. Solving for x the way you know how, you arrive at the false statement $4=5$. Surely 4 cannot be equal to 5! This may make sense when you consider the second line in the solution where like terms were combined. If you multiply a number by 2 and add 4 you would never get the same answer as when you multiply that same number by 2 and add 5. Since there is no value of x that will ever make this a true statement, the solution to the equation above is “no solution.” Be careful that you do not confuse the solution $x=0$ with “no solution.” The solution $x=0$ means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation. Also, be careful not to make the mistake of thinking that the equation $4=5$ means that 4 and 5 are values for x that are solutions. If you substitute these values into the original equation, you’ll see that they do not satisfy the equation. This is because there is truly no solution—there are no values for x that will make the equation $12+2x–8=7x+5–5x$ true.
Try solving these equations. How many steps do you need to take before you can tell whether the equation has no solution or one solution? a) Solve $8y=3(y+4)+y$ Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution. [practice-area rows="1"][/practice-area]
Solve $8y=3(y+4)+y$
First, distribute the 3 into the parentheses on the right-hand side.
$8y=3(y+4)+y=8y=3y+12+y$
Next, begin combining like terms.
$8y=3y+12+y = 8y=4y+12$
Now move the variable terms to one side. Moving the $4y$ will help avoid a negative sign.
$\begin{array}{l}\,\,\,\,8y=4y+12\\\underline{-4y\,\,-4y}\\\,\,\,\,4y=12\end{array}$
Now, divide each side by $4y$.
$\begin{array}{c}\frac{4y}{4}=\frac{12}{4}\\y=3\end{array}$
Because we were able to isolate y on one side and a number on the other side, we have one solution to this equation.
b) Solve $2\left(3x-5\right)-4x=2x+7$ Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution. [practice-area rows="1"][/practice-area]
Answer: Solve $2\left(3x-5\right)-4x=2x+7$. First, distribute the 2 into the parentheses on the left-hand side.
$\begin{array}{r}2\left(3x-5\right)-4x=2x+7\\6x-10-4x=2x+7\end{array}$
Now begin simplifying. You can combine the x terms on the left-hand side.
$\begin{array}{r}6x-10-4x=2x+7\\2x-10=2x+7\end{array}$
Now, take a moment to ponder this equation. It says that $2x-10$ is equal to $2x+7$. Can some number times two minus 10 be equal to that same number times two plus seven? Let's pretend $x=3$. Is it true that $2\left(3\right)-10=-4$ is equal to $2\left(3\right)+7=13$. NO! We don't even really need to continue solving the equation, but we can just to be thorough. Add $10$ to both sides.
$\begin{array}{r}2x-10=2x+7\,\,\\\,\,\underline{+10\,\,\,\,\,\,\,\,\,\,\,+10}\\2x=2x+17\end{array}$
Now move $2x$ from the right hand side to combine like terms.
$\begin{array}{l}\,\,\,\,\,2x=2x+17\\\,\,\underline{-2x\,\,-2x}\\\,\,\,\,\,\,\,0=17\end{array}$
We know that $0\text{ and }17$ are not equal, so there is no number that x could be to make this equation true. This false statement implies there are no solutions to this equation, or DNE (does not exist) for short.
### Algebraic Equations with an Infinite Number of Solutions
You have seen that if an equation has no solution, you end up with a false statement instead of a value for x. It is possible to have an equation where any value for x will provide a solution to the equation. In the example below, notice how combining the terms $5x$ and $-4x$ on the left leaves us with an equation with exactly the same terms on both sides of the equal sign.
### Example
Solve for x. $5x+3–4x=3+x$
Answer: Combine like terms on both sides of the equation.
$\displaystyle \begin{array}{r}5x+3-4x=3+x\\x+3=3+x\end{array}$
Isolate the x term by subtracting x from both sides.
$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,x+3=3+x\\\,\,\,\,\,\,\,\,\underline{\,-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-x\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,=\,\,3\end{array}$
This true statement implies there are an infinite number of solutions to this equation, or we can also write the solution as "All Real Numbers"
You arrive at the true statement “$3=3$.” When you end up with a true statement like this, it means that the solution to the equation is “all real numbers.” Try substituting $x=0$ into the original equation—you will get a true statement! Try $x=-\frac{3}{4}$, and it also will check! This equation happens to have an infinite number of solutions. Any value for x that you can think of will make this equation true. When you think about the context of the problem, this makes sense—the equation $x+3=3+x$ means “some number plus 3 is equal to 3 plus that same number.” We know that this is always true—it’s the commutative property of addition!
### Example
Solve for x. $3\left(2x-5\right)=6x-15$
Answer: Distribute the 3 through the parentheses on the left-hand side.
$\begin{array}{r}3\left(2x-5\right)=6x-15\\6x-15=6x-15\end{array}$
Wait! This looks just like the previous example. You have the same expression on both sides of an equal sign. No matter what number you choose for x, you will have a true statement. We can finish the algebra:
$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,6x-15=6x-15\\\,\,\,\,\,\,\,\,\underline{\,-6x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-6x\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-15\,\,=\,\,-15\end{array}$
This true statement implies there are an infinite number of solutions to this equation.
In the following video, we show more examples of attempting to solve a linear equation with either no solution or many solutions. https://youtu.be/iLkZ3o4wVxU In the following video, we show more examples of solving linear equations with parentheses that have either no solution or many solutions. https://youtu.be/EU_NEo1QBJ0
## Summary
Complex, multi-step equations often require multi-step solutions. Before you can begin to isolate a variable, you may need to simplify the equation first. This may mean using the distributive property to remove parentheses or multiplying both sides of an equation by a common denominator to get rid of fractions. Sometimes it requires both techniques. If your multi-step equation has an absolute value, you will need to solve two equations, sometimes isolating the absolute value expression first. We have seen that solutions to equations can fall into three categories:
• One solution
• No solution, DNE (does not exist)
• Many solutions, also called infinitely many solutions or All Real Numbers
And sometimes, we don't need to do much algebra to see what the outcome will be.
• Screenshot: Steps With an End In Sight. Provided by: Lumen Learning License: CC BY: Attribution.
• Solving Two Step Equations (Basic). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
• Solving an Equation that Requires Combining Like Terms. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
• Solve an Equation with Variable on Both Sides. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
• Solving an Equation with One Set of Parentheses. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
• Solving an Equation with Parentheses on Both Sides. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
• Solving an Equation with Fractions (Clear Fractions). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
• Solving an Equation with Decimals (Clear Decimals). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
### CC licensed content, Shared previously
• Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
• Ex 4: Solving Absolute Value Equations (Requires Isolating Abs. Value). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution. |
Home | | Maths 4th Std | Patterns in numbers: check if it is a multiple of nine
# Patterns in numbers: check if it is a multiple of nine
Cast out nines from a given number to check if it is a multiple of nine.
Patterns in numbers
Cast out nines from a given number to check if it is a multiple of nine.
EXAMPLE
Is 46908 multiple of 9?
46908 = 4 + 6 + 0 + 8
= 18
= 1 + 8
= 9
Let us Know
Any number or combination of digits in that number which add to 9 can be cast out from the given number. Then the sum of remaining digits of the number is divisible by 9 or multiple of 9.
In addition problem, we can check the sum by casting out nines.
EXAMPLE 1
Check the following numbers whether it is a multiple of 9 or not
24689 = 2 + 4 + 6 + 8 = 20 (It's not a multiple of 9.)
9108 = 0 (It's a multiple of 9.)
3165 = 1 + 5 = 6 (It's not a multiple of 9.)
EXAMPLE 2
3356 + 4729 = 8085
8 + 4 = 21
12 = 21
1 + 2 = 2 + 1
3 = 3
In subtraction problem, we can check the difference by casting out nines method.
(Remember that subtraction is nothing more than addition in reverse).
EXAMPLE
4897 − 2186 = 2711
4897 − 2186 = 2711
19 − 8 = 2
10 − 8 = 2
2 = 2
Let us Know
Think of a two digit number say 52, then subtract the reverse of its digits, 25 from 52.
Difference = 52 − 25 = 27
27 is a multiple of 9.
Activity
Tags : Patterns | Term 1 Chapter 3 | 4th Maths , 4th Maths : Term 1 Unit 3 : Patterns
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
4th Maths : Term 1 Unit 3 : Patterns : Patterns in numbers: check if it is a multiple of nine | Patterns | Term 1 Chapter 3 | 4th Maths |
# How Many Reflectional Symmetry Does The Regular Hexagon Have
Reflectional symmetry is a term used in geometry to describe the property of a shape that can be divided into two equal halves by a line, where one half is the mirror image of the other. In this article, we will explore the concept of reflectional symmetry in the context of the regular hexagon.
## What is a Regular Hexagon?
A regular hexagon is a six-sided polygon where all six sides are of equal length and all internal angles are 120 degrees. It is a type of equilateral polygon and is a two-dimensional shape with straight sides. The regular hexagon is a common shape in nature and can be found in honeycombs, snowflakes, and various architectural designs.
## Reflectional Symmetry in Regular Hexagons
When we talk about the reflectional symmetry of a regular hexagon, we are referring to the number of lines of symmetry that can be drawn to divide the hexagon into two equal halves, each being the mirror image of the other. To determine the number of reflectional symmetries in a regular hexagon, we can use a simple formula:
Number of reflectional symmetries = Number of sides ÷ 2
For a regular hexagon, the formula would be:
Number of reflectional symmetries = 6 ÷ 2 = 3
Therefore, a regular hexagon has three reflectional symmetries. This means that there are three different lines of symmetry that can be drawn to divide the hexagon into two equal halves.
## Understanding Reflectional Symmetry
To better understand the concept of reflectional symmetry in a regular hexagon, let’s consider the visual representation of the lines of symmetry. In the case of a regular hexagon, the lines of symmetry would pass through the following points:
• Two opposite vertices of the hexagon
• The midpoints of two opposite sides of the hexagon
• The midpoints of two non-adjacent sides of the hexagon
By visualizing these lines of symmetry, we can see how each one divides the regular hexagon into two mirror-image halves. This property is what defines the reflectional symmetry of the shape.
## Table of Reflectional Symmetry
For a more comprehensive understanding, let’s create a table to summarize the reflectional symmetry of the regular hexagon:
Line of SymmetryDescription
1Passes through two opposite vertices of the hexagon
2Passes through the midpoints of two opposite sides of the hexagon
3Passes through the midpoints of two non-adjacent sides of the hexagon
## Importance of Reflectional Symmetry
The concept of reflectional symmetry is not only important in geometry, but it also has practical applications in various fields. In art and design, understanding symmetry is essential for creating visually pleasing compositions. In architecture, symmetry plays a crucial role in the aesthetics of buildings and structures. Even in biology, symmetry is a defining characteristic of many organisms.
For the regular hexagon, knowing that it has three reflectional symmetries can inform the design process when using this shape in various contexts. Whether it’s in the creation of patterns, tessellations, or architectural elements, the knowledge of reflectional symmetry adds a layer of understanding and intentionality to the design work.
## Conclusion
In conclusion, the regular hexagon possesses three reflectional symmetries, meaning there are three different lines of symmetry that can be drawn to divide the hexagon into two equal, mirror-image halves. Understanding reflectional symmetry is not only fundamental to geometry but also has practical applications in art, design, and architecture.
By recognizing and utilizing the reflectional symmetry of the regular hexagon, we can gain a deeper appreciation for its geometric properties and leverage its aesthetic and structural potential in various creative and practical endeavors.
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Applications Using Radicals
Word problems involving radicals
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Applications Using Radicals
Applications Using Radicals
Mathematicians and physicists have studied the motion of pendulums in great detail because this motion explains many other behaviors that occur in nature. This type of motion is called simple harmonic motion and it is important because it describes anything that repeats periodically. Galileo was the first person to study the motion of a pendulum, around the year 1600. He found that the time it takes a pendulum to complete a swing doesn’t depend on its mass or on its angle of swing (as long as the angle of the swing is small). Rather, it depends only on the length of the pendulum.
License: CC BY-NC 3.0
The time it takes a pendulum to complete one whole back-and-forth swing is called the period of the pendulum. Galileo found that the period of a pendulum is proportional to the square root of its length: \begin{align*}T = a \sqrt{L}\end{align*}. The proportionality constant, \begin{align*}a\end{align*}, depends on the acceleration of gravity: \begin{align*}a = \frac{2 \pi}{\sqrt{g}}\end{align*}. At sea level on Earth, acceleration of gravity is \begin{align*}g = 9.81 \ m/s^2\end{align*} (meters per second squared). Using this value of gravity, we find \begin{align*}a = 2.0\end{align*} with units of \begin{align*}\frac{s}{\sqrt{m}}\end{align*} (seconds divided by the square root of meters).
License: CC BY-NC 3.0
Up until the mid \begin{align*}20^{th}\end{align*} century, all clocks used pendulums as their central time keeping component.
Real-World Application: Pendulums
Graph the period of a pendulum of a clock swinging in a house on Earth at sea level as we change the length of the pendulum. What does the length of the pendulum need to be for its period to be one second?
The function for the period of a pendulum at sea level is \begin{align*}T = 2 \sqrt{L}\end{align*}.
We start by making a table of values for this function:
\begin{align*}L\end{align*} \begin{align*}T = 2 \sqrt{L}\end{align*}
0 \begin{align*}T = 2 \sqrt{0} = 0\end{align*}
1 \begin{align*}T = 2 \sqrt{1} = 2\end{align*}
2 \begin{align*}y = 2 \sqrt{2} = 2.8\end{align*}
3 \begin{align*}y = 2 \sqrt{3} = 3.5\end{align*}
4 \begin{align*}y = 2 \sqrt{4} = 4\end{align*}
5 \begin{align*}y = 2 \sqrt{5} = 4.5\end{align*}
Now let's graph the function. It makes sense to let the horizontal axis represent the length of the pendulum and the vertical axis represent the period of the pendulum.
License: CC BY-NC 3.0
We can see from the graph that a length of approximately \begin{align*}\frac{1}{4}\end{align*} meters gives a period of 1 second. We can confirm this answer by using our function for the period and plugging in \begin{align*}T = 1 \ second\end{align*}:
\begin{align*}T & = 2 \sqrt{L} \Rightarrow 1 = 2 \sqrt{L}\end{align*}
\begin{align*}&\text{Square both sides of the equation:} && 1 = 4L\\ &\text{Solve for} \ L: && L = \frac{1}{4} \ meters\end{align*}
Real-World Application: TV Screens
“Square” TV screens have an aspect ratio of 4:3; in other words, the width of the screen is \begin{align*}\frac{4}{3}\end{align*} the height. TV “sizes” are traditionally represented as the length of the diagonal of the television screen. Graph the length of the diagonal of a screen as a function of the area of the screen. What is the diagonal of a screen with an area of \begin{align*}180 \ in^2\end{align*}?
Let \begin{align*}d =\end{align*} length of the diagonal, \begin{align*}x =\end{align*} width
Then 4 \begin{align*}\times\end{align*} height = 3 \begin{align*}\times\end{align*} width
Or, height = \begin{align*}\frac{3}{4}x\end{align*}.
License: CC BY-NC 3.0
The area of the screen is: \begin{align*}A =\end{align*} length \begin{align*}\times\end{align*} width or \begin{align*} A = \frac{3}{4} x^2\end{align*}
Find how the diagonal length relates to the width by using the Pythagorean theorem:
\begin{align*}x^2 + \left(\frac{3}{4} x \right)^2 & = d^2\\ x^2 + \frac{9}{16}x^2 & = d^2\\ \frac{25}{16}x^2 & = d^2 \Rightarrow x^2 = \frac{16}{25}d^2 \Rightarrow x = \frac{4}{5}d\end{align*}
Therefore, the diagonal length relates to the area as follows: \begin{align*} A = \frac{3}{4} \left( \frac{4}{5}d \right)^2 = \frac{3}{4} \cdot \frac{16}{25}d^2 = \frac{12}{25}d^2\end{align*}.
We can also flip that around to find the diagonal length as a function of the area: \begin{align*}d^2 = \frac{25}{12} A\end{align*} or \begin{align*}d = \frac{5}{2 \sqrt{3}} \sqrt{A}\end{align*}.
Now we can make a graph where the horizontal axis represents the area of the television screen and the vertical axis is the length of the diagonal. First let’s make a table of values:
\begin{align*}A\end{align*} \begin{align*}d = \frac{5}{2 \sqrt{3}} \sqrt{A}\end{align*}
0 0
25 7.2
50 10.2
75 12.5
100 14.4
125 16.1
150 17.6
175 19
200 20.4
License: CC BY-NC 3.0
From the graph we can estimate that when the area of a TV screen is 180 \begin{align*}in^2\end{align*} the length of the diagonal is approximately 19.5 inches. We can confirm this by plugging in \begin{align*}A = 180\end{align*} into the formula that relates the diagonal to the area: \begin{align*}d = \frac{5}{2\sqrt{3}} \sqrt{A} = \frac{5}{2 \sqrt{3}} \sqrt{180} = 19.4 \ inches\end{align*}.
Radicals often arise in problems involving areas and volumes of geometrical figures.
Real-World Application: Pool Dimensions
A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square feet. Find the dimensions of the pool and the area of the pool.
Make a sketch:
License: CC BY-NC 3.0
Let \begin{align*}x =\end{align*} the width of the pool. Then:
Area \begin{align*}=\end{align*} length \begin{align*}\times\end{align*} width
Combined length of pool and walkway \begin{align*}= 2x + 2\end{align*}
Combined width of pool and walkway \begin{align*}= x + 2\end{align*}
\begin{align*}\text{Area} = (2x + 2)(x + 2)\end{align*}
Since the combined area of pool and walkway is \begin{align*}400 \ ft^2\end{align*} we can write the equation
\begin{align*}(2x + 2)(x + 2) = 400\end{align*}
\begin{align*}\text{Multiply in order to eliminate the parentheses:} && 2x^2 + 4x + 2x + 4 & = 400\\ \text{Collect like terms:} && 2x^2 + 6x + 4 &= 400\\ \text{Move all terms to one side of the equation:} && 2x^2 + 6x - 396 & = 0\\ \text{Divide all terms by 2:} && x^2 + 3x - 198 & = 0\\ \text{Use the quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ && x & = \frac{-3 \pm \sqrt{3^2 - 4(1)(-198)}}{2(1)}\\ && x & = \frac{-3 \pm \sqrt{801}}{2} = \frac{-3 \pm 28.3}{2}\\ && x & = 12.65 \ feet\end{align*}
(The other answer is negative, so we can throw it out because only a positive number makes sense for the width of a swimming pool.)
So the dimensions of the pool are: \begin{align*}length = 12.65\end{align*} and \begin{align*}width = 25.3\end{align*} (since the width is 2 times the length)
That means that the area of just the pool is \begin{align*}A = 12.65 \cdot 25.3 \to 320 ft^2\end{align*}
Check by plugging the result in the area formula:
Area \begin{align*}= (2(12.65) + 2)(12.65 + 2) = 27.3 \cdot 14.65 = 400 \ ft^2.\end{align*}
The answer checks out.
Example
Example 1
The volume of a soda can is \begin{align*}355 \ cm^3\end{align*}. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.
Make a sketch:
License: CC BY-NC 3.0
Let \begin{align*}x =\end{align*} the radius of the cylinder base. Then the height of the cylinder is \begin{align*}4x\end{align*}.
The volume of a cylinder is given by \begin{align*}V = \pi R^2 \cdot h\end{align*}; in this case, \begin{align*}R\end{align*} is \begin{align*}x\end{align*} and \begin{align*}h\end{align*} is \begin{align*}4x\end{align*}, and we know the volume is 355.
Solve the equation:
\begin{align*}355 & = \pi x^2 \cdot(4x)\\ 355 & = 4 \pi x^3\\ x^3 & = \frac{355}{4 \pi}\\ x & = \sqrt[3]{\frac{355}{4 \pi}} = 3.046 \ cm\end{align*}
Check by substituting the result back into the formula:
\begin{align*}V = \pi R^2 \cdot h = \pi (3.046)^2 \cdot (4 \cdot 3.046) = 355 \ cm^3\end{align*}
So the volume is \begin{align*}355 \ cm^3\end{align*}. The answer checks out.
Review
1. If a certain model of a laptop has a diagonal of 15.4 inches and a length of 14.35 inches, find the width.
2. If a certain model of a laptop has a width of 12.78 inches and an area of 114.25 inches squared, find the diagonal.
3. The acceleration of gravity can also given in feet per second squared. It is \begin{align*}g = 32 \ ft/s^2\end{align*}at sea level.
1. Graph the period of a pendulum with respect to its length in feet.
2. For what length in feet will the period of a pendulum be 2 seconds?
4. The acceleration of gravity on the Moon is \begin{align*}1.6 \ m/s^2\end{align*}.
1. Graph the period of a pendulum on the Moon with respect to its length in meters.
2. For what length, in meters, will the period of a pendulum be 10 seconds?
5. The acceleration of gravity on Mars is \begin{align*}3.69 \ m/s^2\end{align*}.
1. Graph the period of a pendulum on the Mars with respect to its length in meters.
2. For what length, in meters, will the period of a pendulum be 3 seconds?
6. The acceleration of gravity on the Earth depends on the latitude and altitude of a place. The value of \begin{align*}g\end{align*} is slightly smaller for places closer to the Equator than places closer to the poles and the value of \begin{align*}g\end{align*} is slightly smaller for places at higher altitudes that it is for places at lower altitudes. In Helsinki the value of \begin{align*}g = 9.819 \ m/s^2\end{align*}, in Los Angeles the value of \begin{align*}g = 9.796 \ m/s^2\end{align*} and in Mexico City the value of \begin{align*}g = 9.779 \ m/s^2\end{align*}.
1. Graph the period of a pendulum with respect to its length for all three cities on the same graph.
2. Use the formula to find for what length, in meters, will the period of a pendulum be 8 seconds in each of these cities?
7. The aspect ratio of a wide-screen TV is 2.39:1.
1. Graph the length of the diagonal of a screen as a function of the area of the screen.
2. What is the diagonal of a screen with area \begin{align*}150 \ in^2\end{align*}?
For 8-10, rationalize the denominator.
1. The volume of a soup can is \begin{align*}452 \ cm^3\end{align*}. The height of the can is three times the radius of the base. Find the radius of the base of the cylinder.
2. The volume of a spherical balloon is \begin{align*}950 \ cm^3\end{align*}. Find the radius of the balloon. (Volume of a sphere \begin{align*}= \frac{4}{3} \pi R^3\end{align*}).
3. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is \begin{align*}180 \ in^2\end{align*}, what is the width of the frame?
Review (Answers)
To view the Review answers, open this PDF file and look for section 11.5.
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Vocabulary Language: English
TermDefinition
Period The period of a wave is the horizontal distance traveled before the $y$ values begin to repeat.
Radical Expression A radical expression is an expression with numbers, operations and radicals in it.
Simple Harmonic Motion Simple Harmonic Motion is periodically recurring motion, commonly modeled using a sin or a cosine function.
Volume Volume is the amount of space inside the bounds of a three-dimensional object.
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So I was trying to derive a way to approximate $\pi$ using my compass and straightedge, and then use algebra. I began by creating a $30$ degree angle:
So $CAD$ is $30$ degrees. Then, I imagined bisecting this angle into infinity. Notice if we draw a segment from $C$ to $D$, we get an isosceles. We can calculate for this length using the law of cosines: $x = \sqrt{2-2\cos(\frac{30}{2r}})$, where $r$ is a reiteration (another bisection). So, if I am correct in my assumption, $\pi$ should be about $x \cdot n$, if $n$ are the number of divisions in the circle. We can find $n$ easily enough: $n = 24 \cdot 2^{r-2}$. we know this becuase, as we bisect, we get a table of values: $\left \{ 24, 48, 96, \ldots \right \}$. Likewise, we find $\theta$ thusly: $\theta = \frac{30}{2^{r}}$. This table of values is found by dividing 360 by $n$: $\left \{ 30, 15, 7.5, ... \right \}$. So to find $x$, we use the law of cosine, take the square, and multiply by n. This gives me the equation: $\pi = \sqrt{2- 2 \cdot \cos \dfrac{30}{2^{r}}} \cdot 24 \cdot 2^{r-2}$ This makes sense from the way I constructed it, but not here: $\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos \frac{30}{2^{r}}} \cdot 24 \cdot 2^{r-2}$
This becomes: $\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos(30 \cdot 0)} \cdot 24 \cdot 2^{r-2} = 0$ Obviously, $\pi \neq 0$. So, I tested with two values of $r$ that my calculator could handle. For $r =17$, $\pi = 3.14159265358$ (correct to 11 decimal places.) However, at $r = 18$, $\pi = 3.1415$ (correct to only 4 decimal places.) So why does this equation get close to $\pi$, as it is supposed to, and then stop, and then appear to approach zero? Thanks for the help, I am not all that great at math, and would really appreciate it!
Note by Drex Beckman
5 years, 1 month ago
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Extremely interesting! How are you saying that limit is 0? (Hint: It's an indeterminate form, ($0 \cdot \infty$))
The correct limit is $\pi$, just like you wanted.
- 5 years, 1 month ago
Well, it seemed like as r approached infinity, the cosine would approach 0. of course, cos(0) = 1 and so we would get $0 * 24 * 2^{r-2}$. Since the limit is $\pi$, is there something wrong with my calculations? Since the precision seemed to degrade for higher r's. Thanks, I don't have the experience of ever learning limits in school, so I do not realize there is such a thing as indeterminate forms, but I was unsure what to do with the $0 * \infty$ case. I just assumed for any number, you would get zero. Thanks for the help! :)
- 5 years, 1 month ago
Oh okay, I'll try to explain then. Suppose you had two functions, $f(x)$ and $g(x)$,
$\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to \infty} g(x) = 0$. Now what is $\lim_{x \to \infty} f(x)\cdot g(x)$?
If you think about it, you really can't say, because it could be $0$ or $\infty$ or some value in between.
Why? It depends on the functions $f(x)$ and $g(x)$ themselves.
Let me give you some examples. Let $L = \lim_{x \to \infty} f(x)\cdot g(x)$.
1. $f(x) = x$ and $g(x) = \frac{1}{x} \Rightarrow L = 1$ as $f(x)g(x) = 1$ always.
2. $f(x) = x^2$ and $g(x) = \frac{1}{x} \Rightarrow L = \infty$ as $f(x)g(x) = x$ always.
3. $f(x) = x$ and $g(x) = \frac{1}{x^2} \Rightarrow L = 0$ as $f(x)g(x) = \frac{1}{x}$ always.
4. $f(x) = 5x$ and $g(x) = \frac{1}{x} \Rightarrow L = 5$ as $f(x)g(x) = 5$ always.
So, we've seen that the limit can be anything really. This is why we call $0 \cdot \infty$ an indeterminate form, it can 'evaluate' to anything.
Now, to your question, how do we evaluate $\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos \frac{30}{2^{r}}} \cdot 24 \cdot 2^{r-2}$?
Here, $f(x) = 24 \cdot 2^{r - 2}$ and $g(x) = \sqrt{2 - 2 \cdot \cos \frac{30}{2^r}}$ (Understand why.)
So what is L, here?
(Convert $30$ to $\frac{\pi}{6}$ radians)
L = $\lim_{r \rightarrow \infty} \sqrt{2- 2 \cdot \cos \frac{\pi}{6 \cdot 2^{r}}} \cdot 6 \cdot 2^{r}$
Let $t = \dfrac{1}{6\cdot 2^r}$.
L = $\lim_{t \rightarrow 0} \dfrac{\sqrt{2- 2 \cdot \cos \pi t }}{t} = \lim_{t \rightarrow 0} \dfrac{\sqrt{2}\sqrt{1 - \cos \pi t }}{t} = \lim_{t \rightarrow 0} \dfrac{\sqrt{2}\sqrt{2 \sin^2 \frac{\pi t}{2}}}{t} = \lim_{t \rightarrow 0} \dfrac{2 sin \frac{\pi t}{2}}{t} = \pi$
I used the well-known facts that $1 - \cos 2x = 2 \sin^2 x$ and $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
- 5 years, 1 month ago
Well explained.
It's a very common misconception to "Let $n \rightarrow \infty$ in a specific portion of the expression, while ignoring the rest of it".
Staff - 5 years ago
Thanks a lot, I think I understand it now! Very clear explanation, also! +1
- 5 years, 1 month ago
Thanks! Also, the reason your calculator drops off in precision is because of rounding off errors and approximate values for the cosines and sines.
- 5 years, 1 month ago
Right, that was what I suspected, but I put a bit too much faith in the arbitrary precision calculations, I guess. xD
- 5 years, 1 month ago |
# Question Video: Expressing a Definite Integral as a Limit of Riemann Sum Mathematics • Higher Education
Express β«_(0)^(2π) 3 sin 5π₯ dπ₯ as the limit of Riemann sums.
04:01
### Video Transcript
Express the definite integral from zero to two π of three times the sin of five π₯ with respect to π₯ as the limit of Riemann sums.
Weβre given the definite integral of a trigonometric function. And instead of evaluating this definite integral, we instead need to express this as the limit of a Riemann sum. So to answer this question, we need to recall how we would represent a definite integral as the limit of Riemann sums. We recall if π is an integrable function on a closed interval from π to π, then we know the definite integral from π to π of π of π₯ with respect to π₯ will be equal to the limit as π approaches infinity of its Riemann sum.
For the right Riemann sum, this will be the limit as π approaches infinity of the sum from π equals one to π of π of π₯ π times β³π₯, where β³π₯ is our interval width, π minus π all divided by π, and π₯ π will be our sample points, π plus π times β³π₯. We want to use this on the definite integral given to us in the question. First, our function π of π₯ will be our integrand three times the sin of five π₯. Next, our values of π and π will be the upper and lower limits of integration, in this case zero and two π.
So to use this, we first need to check that our function π is integrable on our closed interval of integration. In our case, we need to check that our function three sin of five π₯ is integrable on the closed interval from zero to two π. And in this case, the easiest way to do this is to notice that three sin of five π₯ is a trigonometric function. And trigonometric functions are continuous across their entire domain. In fact, the domain of this function is all real values of π₯. And a function being continuous on an interval means that it must be integrable on this interval. So in this case, our function π is integrable on any interval. So in particular, it will be integrable on the closed interval from zero to two π.
This justifies our use of this limit result. But before we can do that, we need to find expressions for β³π₯ and π₯ π. Letβs start with β³π₯. We know β³π₯ is our interval width π minus π all divided by π. We know π is two π and π is zero. So we get β³π₯ is two π minus zero all divided by π. And of course, we can simplify this to give us two π over π. Now that we found β³π₯, weβre ready to find our sample points π₯ π. We know that our sample points π₯ π will be equal to π plus π times β³π₯. We know π is zero and β³π₯ is two π over π. So we get π₯ π is equal to zero plus π times two π over π. And weβll write this as two π π divided by π.
The last thing we need to do to express this limit is find an expression for π evaluated at π₯ π. And to find this, we just need to substitute our expression for π₯ π into our expression for π of π₯. So we substitute two π π over π into three sin of five π₯. This gives us three sin of five times two π π over π. And we can simplify this since five multiplied by two π will give us 10π. So weβve shown that π evaluated at π₯ π is three sin of 10π times π divided by π.
Weβre now ready to evaluate our definite integral as the limit of a right Riemann sum. We can just write in our expression for π evaluated at each of our sample points π₯ π and our expression for β³π₯. This gives us the limit as π approaches infinity of the sum from π equals one to π of three sin of 10π times π divided by π multiplied by two π over π. And weβll just simplify this expression slightly. Instead of multiplying by two π by π at the end of our expression, weβll do this at the start of our expression. And we can also notice three multiplied by two can simplify to give us six.
And doing this gives us our final answer, the limit as π approaches infinity of the sum from π equals one to π of six π by π multiplied by the sin of 10π times π divided by π. And itβs also worth pointing out since we know our integrand is integrable, if we evaluated our definite integral by using techniques such as what we know about integrating trigonometric functions and if we were to evaluate this limit by using what we know about limit results, we would get exactly the same answer. |
# Introduction to Number Bases
In Clock Arithmetic and Modular Systems, we have learned about a different number system, a number system whose largest digit is 12. We observed that in that system, we can only use the numbers 1 through 12. We also noticed that 12 acts as 0 since 12 added to any number is equal to that number. If we change 12 to 0, we can only use 0 through 11 as digits.
The number system that we use everyday, the decimal number system, uses 0 through 9 as digits. In the decimal number system, if we add 1 to the largest digit which is 9, we add 1 to the number on next place value and write 0. For example, 9 + 1 = 10 and 10 means that 1 tens and 0 ones. In the decimal number system, 325 means 3 tens squared (or hundreds), 2 tens and 5 ones. Using the expansion notation, we have
$325 = 3(10^2) + 2(10^1) + 5(10^0)$.
From the pattern, we can see that every digit of a number in the decimal system is multiplied by a power of 10. That is why we call it the decimal system; the root word deci means 10. We are probably using base 10 because we have 10 fingers. If human’s fingers were 8, we would have probably chosen a number system with base 8.
If we apply the principle above to the clock number system, its base will be 12. Again, we use the digits 0 through 11 in this system. To distinguish this number system from base 10, we denote it by placing the base as a subscript. For instance, 693 to base 12 is written as $695_{12}$. In expanded form,
$6(12^2) + 9(12^1) + 3(12^0)$
The numbers 10 and 11 are considered as a single digit in base 12 but has 2 digits. This can be confusing. For example, we may interpret $101$ as
$10(12^1) + 1(12^0)$
and
$1(12^2) + 0(12^1) + 1(12^0)$.
In order to avoid confusion, we can use different symbols for both digits. We can use A for 10, and B for 11 (or any symbols we want). Using the letters, the first case can be written as $A1$ and the second as $101$.
The most familiar number system that uses this strategy is the hexadecimal system (base 16). The hexadecimal system uses 0-9 as digits, and letters A to E to 10-15. That means that in the hexadecimal system, $F14A$ means
$15(16^3) + 1(16^2) + 4(16^1) + 10(16^0)$
since F = 15 and A = 10.
Different number bases or number systems are used in many branches of mathematics. The binary system, for instance, is used by computers to transfer data. The hexadecimal system is used to denote colors. Some number systems, particularly those which are related to prime numbers, are used in cryptography.
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NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5
# NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5
## NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5 are the part of NCERT Solutions for Class 6 Maths. Here you can find the NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5.
### Ex 5.5 Class 6 Maths Question 1.
Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming a letter ‘L’.
(d) The letter V.
Solution:
(a) Yes, the adjacent edges of a table top are the models of perpendicular lines.
(b) No, the lines of a railway tracks are parallel to each other. So, they are not a model for perpendicular lines.
(c) Yes, the line segments forming the letter ‘L’ are the model for perpendicular lines.
(d) No, the line segments forming the letter ‘V’ are not a model for perpendicular lines.
### Ex 5.5 Class 6 Maths Question 2.
Let PQ be the perpendicular to the line segment XY. Let PQ and XY intersect in the point A. What is the measure of PAY?
Solution:
Since PQ XY.
PAY = 90°
### Ex 5.5 Class 6 Maths Question 3.
There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Solution:
The figures of the two set-squares in all geometry boxes are given below:
The measure of angles of set-square (a) are: 30°, 60° and 90°.
The measure of angles of set-square (b) are: 45°, 45° and 90°.
Yes, they have a common angle of measure 90°.
### Ex 5.5 Class 6 Maths Question 4.
Study the diagram. The line l is perpendicular to line m.
(a) Is CE = EG?
(b) Does PE bisects CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH
Solution:
(a) Yes.
Since, CE = 2 units and EG = 2 units
Hence, CE = EG.
(b) Yes, PE bisects CG because E is the mid-point of CG.
(c) The line segments for which PE is the perpendicular bisector are: DF and BH.
(d) (i) True (ii) True (iii) True
You can also like these:
NCERT Solutions for Maths Class 7
NCERT Solutions for Maths Class 8
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
NCERT Solutions for Maths Class 12 |
# Solve this
Question:
If $f(x)=\left\{\begin{array}{cl}\frac{x-4}{|x-4|}+a, & x<4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|}+b, & x>4\end{array}\right.$. Then $f(x)$ is continuous at $x=4$, then $a+b=$
Solution:
The function $f(x)=\left\{\begin{array}{ll}\frac{x-4}{|x-4|}+a, & x<4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|}+b, & x>4\end{array}\right.$ is continuous at $x=4$.
$\therefore f(4)=\lim _{x \rightarrow 4} f(x)$
$\Rightarrow f(4)=\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)$
$|x-4|= \begin{cases}-(x-4), & x<4 \\ x-4, & x \geq 4\end{cases}$
Now,
$f(4)=a+b$ ...(2)
$\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4}\left[\frac{x-4}{-(x-4)}+a\right]=\lim _{x \rightarrow 4} \frac{x-4}{-(x-4)}+\lim _{x \rightarrow 4} a=-1+a$ ......(3)
$\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4}\left[\frac{x-4}{(x-4)}+b\right]=\lim _{x \rightarrow 4} \frac{x-4}{(x-4)}+\lim _{x \rightarrow 4} b=1+b$ ....(4)
From (1), (2), (3) and (4), we get
$a+b=-1+a=1+b$
So,
$a+b=-1+a$
$\Rightarrow b=-1$
Also,
$a+b=1+b$
$\Rightarrow a=1$
$\therefore a+b=1+(-1)=0$
Thus, the value of a + b is 0.
If $f(x)=\left\{\begin{array}{cl}\frac{x-4}{|x-4|}+a, & x<4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|}+b, & x>4\end{array}\right.$. Then $f(x)$ is continuous at $x=4$, then $a+b=$ ___0___. |
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# National Curriculum: Ratio and Proportion Year 6 - Making Connections
Created on 15 October 2013 by ncetm_administrator
Updated on 05 January 2015 by ncetm_administrator
# Making Connections
Pupils should make rich connections across mathematical ideas to develop fluency, mathematical reasoning and competence in solving increasingly sophisticated problems. They should also apply their mathematical knowledge to science and other subjects.
(National Curriculum, September 2013, page 10)
## Connections within Mathematics
### Making connections to other topics within this year group
#### Number – Fractions
• use common factors to simplify fractions; use common multiples to express fractions in the same denomination
• associate a fraction with division and calculate decimal fraction equivalents (e.g. 0.375) for a simple fraction (e.g. ⅜)
• recall and use equivalences between simple fractions, decimals and percentages, including in different contexts
When working on ratio and proportion and/or fractions, including decimals and percentages, there are opportunities to make connections between them, for example:
In the same way that fractions describe the relationship between parts of a whole and the whole of which they are a part, proportion also expresses the relationship between parts and wholes. Ratio, on the other hand, describes the relationships between parts. When working with ratio, the whole can be inferred by understanding the total number of parts. In turn, once the total number of parts is known, any number of those parts can be expressed as a fraction or proportion of the whole. It is helpful to work through an example: imagine 10 counters of which 4 are red and 6 are blue. When comparing the red with the blue, we see that there are 4 red for every 6 blue; the ratio is 4:6. This can be simplified to 2:3. Proportion is usually represented as a fraction. Four out of ten counters are red, so, 4/10 are red. Six out of ten counters are blue, so, 6/10 are blue. These fractions can be simplified to ⅖ and ⅗. You can practise this with the children using counters, interlocking links or cubes or a collection of classroom items, for example, 8 pencils and 4 rubbers. 812 of the collection are pencils, this fraction can be simplified to ⅔. 412 are rubbers, this can be simplified to ⅓. Simplifying fractions covers the first objective in the above list.
The fractions made can also be converted to decimals and percentages, for example 0.4 or 40% of the counters are red and 0.6 or 60% of the counters are blue. 0.3 of 30% of the collection are pencils and 0.7 or 70% are rubbers.
Using the ITP Fractions gives opportunities to visually make comparisons between fractions, decimals, ratio and proportion.
Ratio and proportion is another ITP worth exploring with the children. In this ITP you can demonstrate comparisons between different ratios and proportions with volumes of liquids.
#### Number-Multiplication and division
When working on ratio and proportion and/or multiplication and division there are opportunities to make connections between them, for example:
Scaling an object or an amount down by multiplying by a fraction, for example, if sketching a tree or building in the school’s grounds, the height would need to be scaled down. You could ask the children to make an estimate of the height of the object by walking away from it until, when they bend down they can see its top from between their legs. They put a marker at this point and then measure from the marker to the base of the object. This will give a reasonable estimate of its height because the child will be looking at the top of the tree at an angle of approximately 45 degrees if viewed in this way. Therefore, the height of the tree will be the same as the distance from it. To make a sketch, this measurement needs to be scaled down. For example if it was 10m tall and they wanted to make a scale drawing showing the tree as 1m high, they would need to draw the tree at a scale of 1:10 (or multiply the height of the tree by a scale factor of 1/10).
Children could draw classroom objects to scale. For example, the height of a table measuring 50cm could be scaled down to (multiplied by scale factor) 1/5 to make the table10cm in the drawing.
#### Statistics
• interpret and construct pie charts and line graphs and use these to solve problems
When working on ratio and proportion and/or statistics there are opportunities to make connections between them, for example:
Using the ITP Data Handling is an effective way to explore comparisons using a pie chart. In the example below, the children could estimate the percentages of people of different ages and compare them. They could present the information as proportions using fraction and/or decimals. They could compare data as ratios, for example the ratio of those under 20 to those between 20 and 50 is 1:2.
### Year 5
#### Number-Fractions (including decimals and percentages)
• read and write decimal numbers as fractions [for example, 0.71 = 71100 ]
#### Non statutory guidance
• Pupils should be taught throughout that percentages, decimals and fractions are different ways of expressing proportions.
### Key Stage 3
#### Number
• interpret fractions and percentages as operators
#### Ratio, proportion and rates of change
• use scale factors, scale diagrams and maps
• express one quantity as a fraction of another, where the fraction is less than 1 and greater than 1
• use ratio notation, including reduction to simplest form
• divide a given quantity into two parts in a given part:part or part:whole ratio; express the division of a quantity into two parts as a ratio
• understand that a multiplicative relationship between two quantities can be expressed as a ratio or a fraction
## Cross-curricular and real life connections
Learners will encounter ratio and proportion:
Within the science curriculum there are opportunities to work with ratio and proportion. For example pupils should select the most appropriate ways to answer science questions using different types of scientific enquiry, including observing changes over different periods of time, noticing patterns, grouping and classifying things, carrying out comparative and fair tests and finding things out using a wide range of secondary sources of information. The children could, for example, construct pie charts or use ratio and proportion to compare groupings and classifications or the results of tests that they carry out.
Within the geography curriculum there are opportunities to connect with ratio and proportion, for example in the introduction of the Key Stage 2 Programme of Study it states that pupils should extend their knowledge and understanding beyond the local area to include the United Kingdom and Europe, North and South America. This will include the location and characteristics of a range of the world’s most significant human and physical features. Children could, for example, find and compare distances between countries or cities, compare population statistics, temperatures, lengths of rivers, heights of mountains. The results of any comparisons could be displayed in a pie chart.
See, for example:
There are also opportunities to work with ratio and proportion, linked to history, for example, ‘pupils should continue to develop a chronologically secure knowledge and understanding of British, local and world history, establishing clear narratives within and across the periods they study. The children could, for example, represent the lengths of the different periods in history and the rules of different British monarchs using pie charts’ (History Programme of Study). This would enable them to make comparisons using proportion as fractions or percentages.
See, for example:
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Worksheet Solution: Data Handling
# Worksheet Solution: Data Handling - Mathematics (Maths) Class 7
Q.1. Find the mode of the following data:
24, 26, 23, 26, 22, 25, 26, 28
Ans: Arranging the given data with the same value together, we get
22, 23, 24, 25, 26, 26, 26, 28
Here, 26 occurs the greatest number of times i.e. 3 times
Thus, the required mode = 26.
Q.2. Find the median of the following data:
8, 6, 10, 12, 14
Ans: Let us arrange the given data in increasing order,
6, 8, 10, 12, 14
n = 5 (odd)
Median = (n + 1/2)th term = 3rd term = 10
Thus, the required median = 10.
Q.3. A fair die is rolled, find the probability of getting a prime number.
Ans: Number on a die = 1, 2, 3, 4, 5, 6
n(S) = 6
Prime numbers = 2, 3, 5
n(E) = 3
Probability = n(E) / n(S) = 3 / 6 = 1 / 2
Thus the required probability = 1 / 2.
Q.4. Find the mean of the first 5 multiples of 3.
Ans: Five multiples of 3 are 3, 6, 9, 12 and 15
Mean =
= 45 / 5
= 9
Q.5. Find the mean and median of first five prime numbers.
Ans: First five prime numbers are: 2, 3, 5, 7 and 11
Mean =
= 28 / 5
= 5.6
Q.6. A bag contains 5 white and 9 red balls. One ball is drawn at random from the bag. Find the probability of getting
(a) a white ball
(b) a red ball
Ans: Total number of balls = 5 + 9 = 14 balls
(a) Number of white ball = 5
n(E) = 5
Probability of getting white ball = n(E)/n(S) = 5 / 14
(b) Number of red balls = 9
n(E) = 9
Probability of getting white ball = n(E)/n(S) = 9 / 14
Q.7. Find the range of the following data:
21, 16, 30, 15, 16, 18, 10, 24, 26, 20
Ans: Greatest number 30
Smallest number = 10
Range = 30 – 10 = 20
Q.8. Find the average of the numbers 8, 13, 15.
Ans: Average = Sum of the numbers / Total number of terms
= 8 + 13 + 15 / 3
= 36/3
= 12
Q.9. Find the median of the following data:
20, 14, 6, 25, 18, 13, 19, 10, 9, 12
Ans: Arranging the given data in increasing order, we get
6, 9, 10, 12, 13, 14, 18, 19, 20, 25
n = 10 (even)
Thus, the required median = 13.5
Q.10. The following bar graph shows the number of books sold by a publisher during the five consecutive years. Read the bar graph and answer the following questions:
(i) About how many books were sold in 2008, 2009 and 2012 years?
(ii) In which years were 575 books were sold?
(iii) In which years were the minimum number of books sold?
Ans:
(i)
(ii) In the year of 2012, maximum number of books i.e. 575 were sold.
(iii) Minimum number of books i.e. 150 were sold in the year 2008.
Q.11. The marks obtained (out of 10) by 80 students in a class test are given below:
Find the mode of the above data.
Ans: In the given frequency distribution table, we find that the observation 7 has maximum frequency, i.e., 20
Hence, the required mode = 7.
Q.12. State true of false.
(i) A data set may have more than one mode.
(ii) The data set 4, 8, 11, 7, 5, 0 has the mode 0.
(iii) The median can be value which is not equal to any of the observation.
(iv) It is impossible to get a sum of 14 of the numbers on both dice when a pair of dice is thrown together
(v) Mean of the observations can be lesser than each of the observations.
(vi) Mean of a data set is the value which divides the value into two equal parts.
(vii) The value of median changes if the value in a data set are arranged in ascending order instead of descending order.
(viii) Rolling the number 7 on a standard die is an impossible event.
(ix) Mean of the data is always from the given data.
(x) Mode of the data is always from the given data.
Ans: (i) - True
(ii) - False
(iii) - False
(iv) - True
(v) - False
(vi) - False
(vii) - False
(viii) - True
(ix) - False
(x) - True
Q.13. A batsman scored the following number of runs in six innings:
36, 35, 50, 46, 60, 55
Calculate the mean runs scored by him in an inning
Ans: 47
Q.14. Fill in the blanks:
(i) The difference between the highest and lowest values of observations in a set of data is called the_________
(ii) The representation of numbers using bars of uniform width, the length of the bars depicting the frequency is called the_________
(iii) When the raw data is arranged in ascending or descending order, it forms an_________
(vi) The probability of getting 1 in tossing of coin is_________
(v) The outcomes which have equal chances of occurrence are called_________.
(vi) Different forms of central tendency are _________and__________
(vii) On dividing the sum of observation by total number of observations, we obtain the_________
(viii) The possible results of an experiment are called_________
(ix) The measure of chance of happening something is called_________
(x) The probability of getting a tail in a throw of a dice is_________.
Ans: (i) - Range
(ii) - Bar graph
(iii) - array
(iv) - 0
(v) - equally likely
(vi) - Mean, median and Mode
(vii) - mean
(viii) - outcomes
(ix) - probability
(x) - 0
Q.15. The mean of 5 numbers is 25. If the four numbers 16, 26, 31, 32. Find the fifth numbers.
Ans: 20
The document Worksheet Solution: Data Handling | Mathematics (Maths) Class 7 is a part of the Class 7 Course Mathematics (Maths) Class 7.
All you need of Class 7 at this link: Class 7
## Mathematics (Maths) Class 7
121 videos|351 docs|45 tests
## FAQs on Worksheet Solution: Data Handling - Mathematics (Maths) Class 7
1. What is data handling in class 7?
Ans. Data handling in class 7 refers to the process of collecting, organizing, analyzing, and interpreting data. It involves various techniques such as creating tables, graphs, and charts to represent data in a meaningful way.
2. How can I collect data for a data handling project in class 7?
Ans. There are several ways to collect data for a data handling project in class 7. You can conduct surveys, interviews, or observations to gather information. You can also use existing data sources such as books, websites, or databases to collect relevant data.
3. What are the different methods of organizing data in class 7?
Ans. In class 7, data can be organized using various methods such as creating frequency tables, tally marks, or bar graphs. You can also use pictographs or line graphs to represent data. The choice of method depends on the type of data and the purpose of the analysis.
4. How can I analyze data in class 7?
Ans. Data analysis in class 7 involves examining the collected data to identify patterns, trends, or relationships. You can calculate measures of central tendency such as mean, median, or mode to summarize the data. You can also use measures of dispersion such as range or standard deviation to understand the spread of the data.
5. Why is data handling important for class 7 students?
Ans. Data handling is important for class 7 students as it helps develop critical thinking and analytical skills. It enables students to make sense of information, draw conclusions, and make informed decisions. Data handling also enhances their mathematical and statistical abilities, which are essential in various real-life situations.
## Mathematics (Maths) Class 7
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# Tag Archives: inequality sign change
## Why do we reverse/flip the inequality sign?
You have probably remembered in your high school or college days the following rule: If we multiply the numbers on both sides of the inequality by a negative number, then the inequality sign should be reversed.
For example, if we want to find the solution of the inequality $-\frac{1}{2}x > 8$, we multiply both sides by $-2$ and reverse the greater than sign giving us $x < -16$. Now, why did the $>$ sign became $<$?
Before we proceed with our discussion, let us first remember 2 important things we have learned in elementary mathematics:
1. The number line is arranged in such a way that the negative numbers are at the left hand side of $0$ and the positive numbers are at its right hand side such as shown in Figure 1.
2. If we have $2$ numbers $a$ and $b$, then $a > b$ if $a$ is at the right of $b$ on the number line. For example, in Figure 1, $2 > -1$ since $2$ is at the right of $-1$.
Figure 1 – The number line
If we generalize the statements above, suppose we have two numbers, say, $a$ and $b$ such that $a > b$, if we multiply them to a negative number $c$, instead of having $ac > bc$, the answer should be $ac < bc$.
For specific values, let’s choose $a = 2$ and $b = -1$ as shown in the diagram above and choose $c = -1$. Note that we will just use these values for discussion purposes, but we may take any value, as long as the conditions above hold. It would help, if we think of $a$ and $b$ as two points on the number line with $a$ as a blue point on the right $b$, a red point.
And note that before multiplying with a negative number, VALUE OF BLUE POINT > VALUE OF RED POINT.
Since $a$ and $b$ are variables, we need to multiply all the numbers on the number line by $-1$. This is to ensure that whatever values we choose for $a$ and $b$, we multiplied it by $-1$. If we multiply every number on the number line by $-1$, the geometric consequence would be a number line with negative numbers on the right hand side of $0$, and positive numbers at the left hand side of $0$ as shown in Figure 2.
Figure 2 – Afer multiplying all numbers on the number line by -1
But negative numbers should be at the left hand side of $0$ so we reverse the position of the number line, or rotate it 180 degrees with zero as the point of rotation. The resulting figure is shown in Figure 3.
Notice that the blue and red points changed order and that the blue point is now at the left of the red point. Therefore, VALUE OF BLUE POINTVALUE OF RED POINT. That is, why the inequality sign was reversed.
Summarizing, multiplying an inequality by a negative number is the same as reversing their order on the number line. That is, if $a, b$ and $c$ are real numbers, $a > b$ and $c<0$, then $ac < bc$.
Our summary above is actually a mathematical theorem. The proof of this theorem is shown below. It is a very easy proof, so, I suppose, that you would be able to understand it.
Theorem: If $a, b$ and $c$ are real numbers, with $a > b$ and $c<0$, then $ac < bc$.
Proof:
Subtracting $b$ from both sides, we have $a - b>0$.
Now, $a - b>0$ means $a - b$ is positive.
Since $c$ is negative, therefore, $c(a - b)$ is negative (negative multiplied by positive is negative)
Since $c(a - b)$ is negative, therefore, $c(a - b) < 0$.
Distributing $c$, we have $ac - bc < 0$.
Adding $bc$ to both sides, we have $ac < bc$ which is what we want to show .$\blacksquare$ |
# What is the formula of vertex of parabola?
## What is the formula of vertex of parabola?
The vertex formula helps to find the vertex coordinates of a parabola. The standard form of a parabola is y = ax2 + bx + c. The vertex form of the parabola y = a(x – h)2 + k.
How do you find the vertex when graphing?
Parabolas always have a lowest point (or a highest point, if the parabola is upside-down). This point, where the parabola changes direction, is called the “vertex”. If the quadratic is written in the form y = a(x – h)2 + k, then the vertex is the point (h, k).
What is a vertex in math on a graph?
A vertex (or node) of a graph is one of the objects that are connected together. The connections between the vertices are called edges or links. A graph with 10 vertices (or nodes) and 11 edges (links). For more information about graph vertices, see the network introduction.
### How do you solve a parabola equation?
The solution for any quadratic equation or parabola can be found by using a little algebra and the general formula for the quadratic equation, which is : x = -b ± sqrt(b^2 – 4ac) / 2a. Figure out the coefficients a, b, and c by looking at the given formula.
How do you calculate the vertex of a parabola?
In parabola, vertex is the measurement of the maximum or minimum point in a curve. Vertex of the parabola is defined by using the quadratic equation. Graphically, vertex of the parabola can be found by using the formula y = a(x – h) 2 + k Here, h,k are the vertex.
What is the equation for the vertex of a parabola?
But the equation for a parabola can also be written in “vertex form”: y = a ( x − h ) 2 + k. In this equation, the vertex of the parabola is the point ( h , k ) .
## What is the equation for finding vertex?
The “vertex” form of an equation is written as y = a(x – h)^2 + k, and the vertex point will be (h, k). Your current quadratic equation will need to be rewritten into this form, and in order to do that, you’ll need to complete the square.
How do you calculate vertex in Algebra?
Use the vertex formula for finding the x-value of the vertex. The vertex is also the equation’s axis of symmetry. The formula for finding the x-value of the vertex of a quadratic equation is x = -b/2a. Plug in the relevant values to find x. |
New Zealand
Level 6 - NCEA Level 1
# Using Equations - Blood Alcohol Content
Lesson
A formula is a convenient way of writing down the method for doing a calculation. It shows how some relevant quantities are related to one another without specifying the actual quantities.
If all but one of the quantities mentioned in a formula are known, then the remaining quantity can be calculated by the rules contained in the formula.
For example, we might have some quantities represented by the letters $A,m,x$A,m,x and $p$p, and a formula to show how they are related. The formula could be expressed in words: To obtain $A$A, multiply $x$x by $p$p and add the result to half of $m$m. This could be written as
$A=\frac{m}{2}+px$A=m2+px.
Note that the letters $p$p and $x$x are written next to each other. The convention is that letters next to each other are to be multiplied.
Now, suppose we are given the information that in a particular situation the quantities are $m=8$m=8, $p=1.5$p=1.5 and $x=3$x=3. The formula tells us how to calculate the value of $A$A. We substitute the given numbers for the letters in the formula. So, we have
$A=\frac{8}{2}+1.5\times3$A=82+1.5×3
We multiply $1.5$1.5 by $3$3 and add the result to half of $8$8 and obtain the answer: $A=8.5$A=8.5.
There are formulas (or formulae) for many different purposes. We consider two of these that are to do with alcohol consumption.
## Equivalent Standard Drinks
By describing an alcoholic drink in terms of the number of standard drinks it contains, we are able to compare its alcoholic content with that of other drinks. We can also use the number of standard drinks in further calculations to do with blood alcohol content.
The number of Equivalent Standard Drinks in a given drink depends on the volume of the drink and on the concentration of alcohol in the drink.
We use the letter $N$N for the number of Equivalent Standard Drinks; we use $V$V for the volume of the drink measured in litres, and the letter $A$A represents the concentration given as the percentage of alcohol in a litre of the drink. These are combined by the formula
$N=0.789\times V\times A$N=0.789×V×A
##### Example 1
What is the equivalent in standard drinks of a $375$375 mL bottle of beer with an alcohol concentration of $4.5%$4.5% ?
Before substituting the numbers into the formula, we have to express the volume in litres, not millilitres. So, we put $V=0.375$V=0.375 because there are $1000$1000 mL in $1$1 L. Now, the calculation is
$N=0.789\times0.375\times4.5$N=0.789×0.375×4.5
This works out to be $N=1.33$N=1.33 standard drinks, to two decimal places.
## Blood Alcohol Content
The Blood Alcohol Content (BAC) formula predicts what a person's blood alcohol content will be if they consume a given number of standard drinks over a certain number of hours. The BAC also depends on the person's weight and on whether the person is male or female.
There could be other factors that influence a particular person's blood alcohol content. The BAC formula gives an estimate based on the main factors. We give two versions of the formula, one for males and one for females.
$BAC_{\text{male}}=\frac{10N-7.5H}{6.8M}$BACmale=10N7.5H6.8M
$BAC_{\text{female}}=\frac{10N-7.5H}{5.5M}$BACfemale=10N7.5H5.5M
In both formulae, the letter $N$N stands for the number of standard drinks consumed, the letter $H$H stands for the number of hours over which the drinks were consumed, and $M$M stands for the person's mass in kilograms.
Note that if the person's weight is given in pounds, this number must be converted to kilograms before it can be used in the formula. The mass in pounds should be multiplied by $0.45$0.45 to convert it to kilograms, since $1$1lb $=$= $0.45$0.45kg.
##### Example 2
A female weighing $158$158 pounds consumes $3$3 standard drinks in $2.5$2.5 hours. What is her predicted blood alcohol content?
Her weight in kilograms is $158\times0.45=71.1$158×0.45=71.1. So, according to the formula,
$BAC_{\text{female}}$BACfemale $=$= $\frac{10\times3-7.5\times2.5}{5.5\times71.1}$10×3−7.5×2.55.5×71.1 $=$= $\frac{30-18.75}{391.05}$30−18.75391.05 $=$= $0.03$0.03
##### Example 3
To calculate the number of standard drinks we use the formula $N=0.789\times V\times A$N=0.789×V×A; where
$V$V is the volume of the drink in litres (L); and
$A$A is the percentage of alcohol (% alc/vol) in the drink.
1. Calculate the number of standard drinks in a $375$375 mL can of pre-mix drink that has an alcohol content of $7%$7% alc/vol to the nearest decimal place.
##### Example 4
To calculate the blood alcohol content (BAC) of a person we use the formula $BAC_{male}=\frac{10N-7.5H}{6.8M}$BACmale=10N7.5H6.8M for males and $BAC_{female}=\frac{10N-7.5H}{5.5M}$BACfemale=10N7.5H5.5M for females; where
$N$N is the number of standard drinks consumed;
$H$H is the number of hours of drinking; and
$M$M is the person's mass in kilograms.
1. Bill is an adult male that weighs $87$87kg and has consumed $4$4 standard drinks in $4$4 hours. Calculate his blood alcohol content correct to three decimal places.
##### Example 5
We want to calculate the BAC of a female with a mass of $53$53kg, who drinks three $425$425ml bottles of full strength beer with an alcohol content of $4.8%$4.8% alc/vol, over $3$3 hours.
1. First calculate the number of standard drinks correct to one decimal place. Remember we use the formula $N=0.789\times V\times A$N=0.789×V×A; where
$V$V is the volume of the drink(s) in litres (L); and
$A$A is the percentage of alcohol (% alc/vol) in the drink(s).
2. Using your answer to part (a), calculate her blood alcohol content (BAC) correct to three decimal places. Remember, we use the formula $BAC_{female}=\frac{10N-7.5H}{5.5M}$BACfemale=10N7.5H5.5M where;
$N$N is the number of standard drinks consumed;
$H$H is the number of hours of drinking; and
$M$M is the person's mass in kilograms.
### Outcomes
#### U5223
Use formulae and equations to solve problems in the workplace |
Study Guide
# Area, Volume, and Arc Length - Integrating with Polar Coordinates
## Integrating with Polar Coordinates
### Slicing Cold Pizza
First, let's forget about calculus and use our knowledge of fractions to answer the following question.
### Sample Problem
What is the area of the shaded region?
Think of the shaded region as a piece of cold pizza. The central angle of the slice is
The central angle of the entire pizza is 2π.
This means the slice is this fraction of the whole pizza:
Since the area of the whole pizza is
πr2 = π(1)2 = π,
the area of the piece is
.
We can use this idea to find the area of a piece of cold anchovy pizza for any pizza described by a polar function
r = f(θ).
First we slice the piece into tiny slices. The central angle of a tiny slice is Δθ.
The angle θ measures the position of a slice by its angle from the positive x-axis, going counterclockwise. If we look just at the slice at position θ, we can pretend this slice is part of a perfectly round pizza with radius f(θ).
The tiny slice would be the fraction
of the whole, perfectly round pizza. So the area of the tiny slice is approximately
Writing this a little more neatly, the π terms cancel out and we're left with
as the area of the tiny slice.
If we want to add up the areas of all the tiny slices and take the limit as the number of slices approaches ∞ to get an integral, we need to know the reasonable values of θ.
The original piece of cold pizza must go from some angle α to some angle β.
Using the values α and β as the limits of integration, the area of the original piece of pizza is
This is usually written as
Our cold, leftover slice of pizza came in handy for more than just satisfying our hunger pains at 4 AM. We used the cold slice to derive the area for a slice-like portion of a circle in polar coordinates.
When doing problems with polar coordinates, we can use the formula
All we have to do is put in the correct values for α and β and the correct function for f(θ).
We promised before, when talking about triangles, that we'd have the power to derive any area equation we could possibly need. If you forget this formula, don't panic. Just recreate it. Remember that each tiny slice is like a fraction of circle, so the area of the region is
Cancel the π's and move the fraction in front of the integral, and there's the formula:
Be Careful:
To find the area of a region in between two polar graphs, we can do it the hard way by adding areas directly. We like to work smarter instead of harder, though, by subtracting the smaller area from the larger area. |
Pizza Portions
My friends and I love pizza. Can you help us share these pizzas equally?
Doughnut
How can you cut a doughnut into 8 equal pieces with only three cuts of a knife?
Pies
Grandma found her pie balanced on the scale with two weights and a quarter of a pie. So how heavy was each pie?
Fraction Fascination
Age 7 to 11Challenge Level
We had some good solutions come in for this challenge.
Sophie and Emma, Keenan and Dylan from Crossacres Primary School wrote:
The smaller square: We traced the shape on tracing paper. Then we cut the triangle shapes out. We found out that the two big triangles fitted to make half (we had to flip one over). Because there were two of them, we halved a half which makes $\frac{1}{4}$.
Then we saw that the small triangle was half of a quarter which is $\frac{1}{8}$.
Then we turned all of the fractions into eighths and added them together and got $\frac{5}{8}$.
This means that the last triangle (the big one) is equal to $\frac{3}{8}$ because $\frac{5}{8}$ + $\frac{3}{8}$ = 1 whole.
and
You will need scissors and more than one copy of the shapes. First we cut out the four triangles that make up the small square. Then we took the two right-angled triangles so that they made an oblong. We put the oblong on top of another copy of the square.The two oblongs covered a half of the square, that makes them a quarter each. This means that the large triangle and small triangle together make up a half too. We then made a square using two of the small triangles. When we put this square on the original square we could see that it equals a quarter. That means that the small triangle is equal to one eighth because that is half of a quarter. The large triangle is equal to three eighths. Work this out by adding one quarter + one quarter + one eighth = five eighths and then to make 1 whole it is three eighths (the largest triangle).
When we looked at the second big square straight away we could see that it was made up of four of the smaller squares. That means that each of the fractions we first worked out would be four times smaller. We separated the large square into the four smaller squares and wrote down what each of the triangles was now worth. To work out the size of the rhombus we worked out
$\frac{1}{16}$ + $\frac{1}{16}$ + $\frac{1}{16}$ + $\frac{1}{16}$ = $\frac{4}{16}$.
Then we simplified the fraction to $\frac{1}{4}$.
Then we checked our answer by using four cut out triangles we placed them on the square and they covered one quarter of the shape.
Esme from St John's CEP School sent in the following:
For the first picture I drew two lines to make quarters on the picture. In the top half one triangle is half of the half (to make the fractions easy to understand I changed them into eighths) which is two eighths and as the top triangle is the same as the left hand side triangle you know that the left hand side triangle is also two eighths.
In the bottom right hand corner the triangle is half the quarter so the triangle is one eighth.
Then if you add up all the eighths you get five eighths that means that the remaining centre triangle is three eighths.
For the second picture I split it into sixteenths but I calculated it in to thirty-twoths so I could work out the answer more easily.
In the top right hand corner triangle it was half of two sixteenths which (in thirty-twoths) is two thirty-twoths. That is the same as each corner and the quarters of the big middle shape which means overall the middle shape is eight thirty-twoths.
In the centre of the right and left hand side each side's two sixteenths is one sixteenth of the two sixteenths. That means that each side triangle is two thirty-twoths.
Then if you add up the triangles you have calculated so far you get twenty thirty-twoths so the remaining shapes have twelve thirty-twoths all together so each of the remaining triangles is three thirty-twoths.
Lydia also from St. John's wrote:
I did this by looking at the bottom line and realising that the central triangle touched halfway, so I drew a vertical line from this point upwards. This then shows that the triangle on the right is half of the rectangle you have just created, therefore the triangle is a quarter of the entire square.
On the left hand side the central triangle's point touches halfway too and I drew the line horizontally across; now the square is in quarters. The top two quarters are split in half by a second triangle which makes that triangle a quarter and together with the triangle on the right a half. Half of one of the smaller squares (bottom left) is filled with a tiny right-angled triangle, and half a quarter is an eighth so the small triangle is one-eighth of the entire square. So now you have one-eighth and a half which makes five eighths. 8 minus 5 equals 3 so the central triangle is three as that is the remaining value!
Bottom left triangle = $\frac{1}{8}$; Top triangle = $\frac{2}{8}$; Triangle to the right = $\frac{2}{8}$; Central triangle = $\frac{3}{8}$
Here is a response from Esther and Tracy from Withernsea Primary School:
First we drew a vertical and horizontal line in order to split the square into quarters. We then looked at the triangle on the left of the square and decided that it was half of the left half of the square. This meant that it was a quarter of the whole square.
The triangle at the top of the square was also a quarter.
The smallest triangle was half the size of these two. Half of a quarter is an eighth.
So, the three triangles were a quarter, a quarter and an eighth.
To calculate the last triangle we turned the denominators into eighths. So, they became $\frac{2}{8}$, $\frac{2}{8}$ and $\frac{1}{8}$. This meant that the central triangle must have been $\frac{3}{8}$ as: $\frac{2}{8}$ + $\frac{2}{8}$ + $\frac{1}{8}$ + $\frac{3}{8}$ = $\frac{8}{8}$.
For the next problem we again divided the square into quarters. Looking at the top left hand square, we decided that the part of the central shape was a quarter of this square. This means that this must be one sixteenth of the whole shape. As there is are four of these triangles, they must take up four sixteenths of the whole square. So the central shape is four sixteenths or one quarter of the whole square.
Sakura, Ema and Wictoria at the VIenna International School in Austria we had sent in the following;
We first labelled the four triangles in size order A, B, C, D. Then we cut the triangles up and noticed that triangles B + C were equal sizes and together measured half the square. Therefore B was $\frac{1}{4}$ and C was $\frac{1}{4}$ of the total area.
Next we split the original diagram into four quadrants. We identified that triangle D was half of a quadrant and therefore equated to $\frac{1}{8}$ of the total area.
We then converted all of our fractions to eighths. We had $\frac{2}{8}$, $\frac{2}{8}$ + $\frac{1}{8}$ = $\frac{5}{8}$. Therefore we knew triangle A would be $\frac{3}{8}$ as the whole would add up to $\frac{8}{8}$.
We checked our answer by cutting the triangles further.
Zach sent in an excellent account of his method of finding solutions. He began by saying:
Start by labelling the triangles. Note that Triangles A, C & D are right angled, whilst Triangle B is isosceles. Triangle A = Triangle C
To read Zach's full solution, see one of the following:Fraction Fascination - Zach.docx or PDF Fraction Fascination - Zach .pdf
Thank you for these interesting solutions that were sent in. Please keep sending in reports on how you set about doing these challenges. |
# Is f(x)=(x^2-3x-2)/(x+1) increasing or decreasing at x=1?
Oct 8, 2016
Increasing
#### Explanation:
To determine if the graph is increasing or decreasing at a certain point, we can use the first derivative.
• For values in which $f ' \left(x\right) > 0$, $f \left(x\right)$ is increasing as the gradient is positive.
• For values in which $f ' \left(x\right) < 0$, $f \left(x\right)$ is decreasing as the gradient is negative.
Differentiating $f \left(x\right)$,
We have to use quotient rule.
$f ' \left(x\right) = \frac{u ' v - v ' u}{v} ^ 2$
Let $u = {x}^{2} - 3 x - 2$ and $v = x + 1$
then $u ' = 2 x - 3$ and $v ' = 1$
So $f ' \left(x\right) = \frac{\left(2 x - 3\right) \left(x + 1\right) - \left({x}^{2} - 3 x - 2\right)}{x + 1} ^ 2 = \frac{{x}^{2} + 2 x - 1}{x + 1} ^ 2$
Subbing in $x = 1$,
$f ' \left(x\right) = \frac{{1}^{2} + 2 \left(1\right) - 1}{1 + 1} ^ 2 = \frac{1}{2} , \therefore f ' \left(x\right) > 0$
Since the $f ' \left(x\right) > 0$ for $x = 1$, $f \left(x\right)$ is increasing at $x = 1$ |
High School Algebra: Aussie Fir Tree
The Aussie Fir Tree task is a culminating task for a 2-3 week unit on algebra that uses the investigation of growing patterns as a vehicle to teach students to visualize, identify and describe real world mathematical relationships. Students who demonstrate mastery of the unit are able to solve the Aussie Fir Tree task in one class period.
Suggested Use: Review the task and rubric before looking at the student work. Then, look at the student work and click on the red arrows to see an explanation of the student's performance on the task. Scroll down to the bottom of the student work to see suggested instructional next steps.
Student C (Performance below Standard)
Many students at this level are using recursive rules to extend the pattern. While this is helpful and easy to use for a small number of cases, it is cumbersome and prone to errors when trying to extend the pattern for larger numbers. Students need to see that it is more helpful to find more generalizable rules to solve problems for all cases.
One helpful way to do this is to ask students break down the pattern into simpler parts. "What do you see when you look at the pattern? How can you decompose the shape into parts? How does the tree trunk grow? How could we use algebra to describe how to find the tree trunk for any pattern number?" Breaking down the problem into simpler steps helps students manage the thinking in smaller chunks. But students in this stage need to learn questions to help them progress in their thinking, to develop strategies for finding a generalizable rule.
Students need to be encouraged to give more detail about what they see. So in class the teacher might ask, "That's interesting, can you tell me a bit more? Or where do you see the n in the diagram or the (n+1)?" The more detailed their descriptions usually the easier it is to quantify the ideas symbolically. Students should also be more descriptive in thinking about classes of numbers. In elementary school it is good to notice that numbers are odd or even, but by this grade level students should start to classify numbers as consecutive or consecutive odd numbers, multiples of . . ., powers of . . . , triangular numbers, etc. The types of patterns that students think about should be expanded.
Students need to have experiences thinking about types of linear patterns; those that are proportional and those that are not proportional. Take the work of student 692. The strategy of doubling from case 5 to case 10 works for proportional patterns, but not for patterns with a constant. Students can benefit by looking at two cases at the same time and comparing which one will work by doubling and which one won't. Drawing graphs of the situations can help to clarify this idea. |
# How do you increase 40m by 30%?
Dec 12, 2016
40m increased by 30% is 52m
#### Explanation:
To increase 40 by 30% you need to add "30% of 40m" to the original 40m.
To calculate 30% 0f 40m we know "Percent" or "%" means "out of 100" or "per 100", Therefore 30% can be written as $\frac{30}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Putting this together we can write:
$\frac{30}{100} \cdot 40 m = \frac{1200 m}{100} = 12 m$
Adding the 12m to the original 40m gives:
$40 m + 12 m = 52 m$ |
# How do you graph 3x+y= -2?
Jun 15, 2018
See a solution process below:
#### Explanation:
First, solve for two points which solve the equation and plot these points:
First Point: For $x = 0$
$\left(3 \cdot 0\right) + y = - 2$
$0 + y = - 2$
$y = - 2$ or $\left(0 , - 2\right)$
Second Point: For $y = 1$
$3 x + 1 = - 2$
$3 x + 1 - \textcolor{red}{1} = - 2 - \textcolor{red}{1}$
$3 x + 0 = - 3$
$3 x = - 3$
$\frac{3 x}{\textcolor{red}{3}} = - \frac{3}{\textcolor{red}{3}}$
$x = - 1$ or $\left(- 1 , 1\right)$
We can next plot the two points on the coordinate plane:
graph{(x^2+(y+2)^2-0.035)((x+1)^2+(y-1)^2-0.035)=0 [-10, 10, -5, 5]}
Now, we can draw a straight line through the two points to graph the line:
graph{(3x + y + 2)(x^2+(y+2)^2-0.035)((x+1)^2+(y-1)^2-0.035)=0 [-10, 10, -5, 5]} |
Select Board & Class
Basic Mathematics(Pre-requisite)
Differential calculus
Do you know what a physical quantity is? A physical quantity is any physical property that can be expressed in numbers. For example, time is a physical quantity as it can be expressed in numbers, but beauty is not as it cannot be expressed in numbers.
Scalar Quantities
• If a physical quantity can be completely described only by its magnitude, then it is a scalar quantity. To measure the mass of an object, we only have to know how much matter is present in the object. Therefore, mass of an object is a physical quantity that only requires magnitude to be expressed. Therefore, we say that mass is a scalar quantity.
• Some more examples of scalar quantities are time, area, volume, and energy.
• We can add scalar quantities by simple arithmetic means.
• It is difficult to plot scalar quantities on a graph.
Vector Quantities
• There are some physical quantities that cannot be completely described only by their magnitudes. These physical quantities require direction along with magnitude. For example, if we consider force, then along with the magnitude of the force, we also have to know the direction along which the force is applied. Therefore, to describe a force, we require both its magnitude and direction. This type of physical quantity is called a vector quantity.
Therefore, we can define vector quantity as the physical quantity that requires both magnitude and direction to be described.
• Some examples of vector quantities are velocity, force, weight, and displacement.
• Vector quantities cannot be added or subtracted by simple arithmetic means.
• Vector quantities can easily be plotted on a graph.
Scalars v/s Vectors
Scalars Vectors A scalar quantity has only magnitude. A vector quantity has both magnitude and direction. Scalars can be added, subtracted, multiplied, and divided just as ordinary numbers i.e., scalars are subjected to simple arithmetic operations. Vectors cannot be added, subtracted, and multiplied following simple arithmetic laws. Arithmetic division of vectors is not possible at all. Example: mass, volume, time, distance, speed, work, temperature Example: displacement, velocity, acceleration, force
Position Vector
Position vector of a point in a coordinate system is the straight line that joins the origin and the point.
Magnitude of the vector is the length of the straight line and its direction is along the angle θ from the positive x-axis.
Displacement Vector
Displacement vector is the straight line joining the initial and final positions.
Equality of Vectors
Two vectors and are said to be equal, if and only if they have the same magnitude and the same direction.
Scalars vs. Vectors
Scalars Vectors A scalar quantity has magnitude only. A vector quantity has both magnitude and direction. Scalar quantities can be added, subtracted, multiplied and divided just like ordinary numbers, i.e., scalars are subjected to simple arithmetic operations. Vectors cannot be added, subtracted or multiplied following simple arithmetic rules. Arithmetic division of vectors is not possible at all. Example: Mass, volume, time, distance, speed, work, temperature, etc. Example: Displacement, velocity, acceleration, force, etc.
Distance & Displacement:
Position Vector
The position vector of a point in a coordinate system is the straight line that joins the origin and the point.
The magnitude of a vector is the length of the straight line. Its direction is along the angle θ from the positive x-axis.
Displacement Vector
Displacement vector is the straight line joining the initial and the final position.
Equality of Vectors
Two vectors $\stackrel{\to }{A}$ and $\stackrel{\to }{B}$ are said to be equal only if they have the same magnitude and the same direction.
Negative vector
Negative vector is a vector whose magnitude is equal to that of a given vector, but whose direction is opposite to that of the given vector.
Zero vector
Zero vector is a vector whose magnitude is zero and have an arbitrary direction.
Resultant vector
The resultant vector of two or more vectors is a vector which produces the same effect as produced by the individual vectors together.
Multiplication of Vectors by Real Numbers
• Multiplication of a vector $\stackrel{\to }{A}$ with a positive number k only changes the magnitude of…
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# Stuck on this - In ABC right triangle AC= $2+\sqrt{3}$ and BC = $3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$
In ABC right triangle $$AC= 2+\sqrt{3}$$ and $$BC = 3+2\sqrt{3}$$. circle touches point C and D, Find the Area of $$AMD$$
Here's my strategy of solving this, I'm not sure if it's correct, if you find my explanation hard to understand you can just ignore and write the solution in your own way, thanks.
1. first area of main triangle, we know AC and CB so it'll be easy to calculate that,
2. to find the radius, we'll reflect triangle ABC on the left side of the circle, turning it into circle inscribed in isosceles triangle, and find it with the formula
3. to find the area of $$AMD$$ I'll subtract the area of sector $$OMD$$, triangle $$OAD$$ and triangle $$CDB$$ from triangle $$AMD$$,
4. $$DBC$$ is an isosceles triangle, so $$CB=DB$$, then to find the area, I split it into 2 right triangles(it becomes 90 30 60 triangle) and find its height. So we got the Area of $$DBC$$
5. Now similarly $$OAD$$ is isosceles, $$OD=OC=radius$$ of the circle which we "found" also, split this in 2 to get right triangles and then calculate with Pythagorean theorem to find the height so we get Area of $$OMD$$ too, maybe we could find angles with trigonometry? I don't know that, and if we get the angle of $$DOA$$ we could find the sector $$OMD$$ as well and subtract it to the main triangle so we get the area of $$AMD$$.
The area $$T$$ in question can be found as the difference between the area of $$\triangle AID$$ and a circular segment $$IDM$$: \begin{align} T&= \tfrac12\,|AI|\cdot|DI|\sin\angle DIA - \frac12\angle DIA\cdot|DI|^2 , \end{align}
where $$\angle DIA$$ is measured in radians.
To use the standard route to find the radius, consider this circle as inscribed in isosceles $$\triangle AFB$$, for which
\begin{align} |AC|& = 2+\sqrt3 ,\\ |BC|& = 3+2\sqrt3 ,\\ |AB| &=\sqrt{|AC|^2+|BC|^2} \\ &=\sqrt{28+16\sqrt3}=4+2\sqrt3 ,\\ |FB|&=2|BC|=6+4\sqrt3 ,\\ S_{\triangle AFB}&=|AC|\cdot|BC| =12+7\,\sqrt3 ,\\ \rho&=\tfrac12(2|AB|+|FB|) =|AB|+|BC|=7+4\sqrt3 , \end{align}
\begin{align} r&=|DI|=|MI|=|EI| \\ r &=\frac{S_{\triangle AFB}}{\rho} =\frac{12+7\sqrt3}{7+4\sqrt3} = \frac{(12+7\sqrt3)(7-4\sqrt3)}{(7+4\sqrt3)(7-4\sqrt3)} \\ &=\sqrt3 ,\\ \end{align}
hence, \begin{align} |AI|&=|AC|-|CI|=2 ,\\ \cos\angle DIA&=\frac{|DI|}{|AI|}=\frac{\sqrt3}2 ,\\ \angle DIA&=\frac \pi 6 , \end{align}
and the answer is
\begin{align} T&=\tfrac12\cdot2\cdot\sqrt3\cdot\tfrac12 -\tfrac12\cdot\tfrac{\pi}{6}\cdot(\sqrt3)^2 \\ &=\tfrac14(2\sqrt3-\pi) \approx 0.08062724 . \end{align}
• nice, but second way of solution gives different answer than first one I think? What I don't understand here is the area of $AID$ triangle, why is the height of the triangle being multiplied by hypotenuse? in your first solution Commented Jun 24, 2019 at 20:16
• @bartly ajames: $|AD|=|AI|\cdot\sin\angle DIA$.. Commented Jun 24, 2019 at 20:25
Clearly, $$\angle CAB = 60$$ because $$BC = 3+2\sqrt{3} = \sqrt{3} \cdot (2 + \sqrt{3}) = \sqrt{3} \cdot AC$$. Now, we need to find the lengths $$AM$$ and $$AD$$. Let $$r$$ be the radius of the circle. It is clear that $$OC = r$$, and $$AO = \sqrt{3}r$$ because of the 30-60-90 triangle $$AOD$$, as $$OD$$ is a tangent to $$AB$$. Therefore, $$r (1 + \sqrt{3}) = AC = 2 + \sqrt{3} \implies r = \frac{\sqrt{3}+1}{2}$$.
Now, we compute the area. Note that $$AM = AO - MO = (\sqrt{3}-1)r$$, and $$AD = \frac{\sqrt{3}}{3}r$$, so we have that $$[AMD] = AM \cdot AD \cdot \frac{\sqrt{3}}{4} = (\sqrt{3}-1)(\frac{\sqrt{3}}{3})(\frac{\sqrt{3}}{4})r^2 = \frac{1+\sqrt{3}}{8}$$.
• Area of $AMD$ is $\frac{\sqrt3}{2}-\frac{\pi}{4}$ according to test answers, is that correct? Commented Jun 24, 2019 at 17:12
$$\measuredangle B=\arctan\frac{2+\sqrt3}{3+2\sqrt3}=\arctan\frac{2+\sqrt3}{\sqrt3(2+\sqrt3)}=\arctan\frac{1}{\sqrt3}=30^{\circ}.$$
Thus, $$OC=(3+2\sqrt3)\tan15^{\circ}=(3+2\sqrt3)(2-\sqrt3)=\sqrt3(2+\sqrt3)(2-\sqrt3)=\sqrt3$$ and $$\measuredangle AOD=30^{\circ}.$$ Can you end it now?
• did you calculate that angle for my strategy? so does that mean i could get the answer my way? i didn't really do calculations Commented Jun 24, 2019 at 17:13
• @bartly ajames You don't need to get $S_{\Delta BCD}$ because to get $S_{ODBC}$ is much more easier. Commented Jun 24, 2019 at 17:37
Found this solution which looks the simplest to me:
• $$BC = 3 + 2\sqrt{3} = \sqrt{3}AC$$;
• $$BD = BC$$;
• $$AB = 2AC$$;
• $$AD = 1$$;
• $$OD = \sqrt{3}$$; (tg(<ABC) = AC/BC = AD/OD)
• $$S_{ADO} = \sqrt{3}/2$$; (by AD and OD)
• $$S_{MOD} = pi/4;$$
• $$S_{AMD} = \sqrt{3}/2 - pi/4;$$
p.s. pretty sure there might be better solution by using integrals and (or) function derivatives but that's not mine
In a video by Mathematics professor Michael Penn, this is solved first by finding out the hypotenuse of the triangle and writing it in the form $$a+b\sqrt{3}$$, where $$a, b \in \mathbb{R}$$. Then, he finds the area of the triangle ($$AOD$$) and the sector ($$EOD$$) as shown in Figure 1 and subtracts the area of the sector from the triangle to find the shaded area.
However, the shaded area can also be found out using calculus.
We need to find the equation of the line $$AB$$ and the circle. Before that, we need to find the radius of the circle.
In $$\Delta ABC$$, let $$\angle ABC = \theta$$. Then, \begin{align*} \tan \theta &= \frac{AC}{CB}=\frac{1}{\sqrt{3}} \\ \implies \theta &= \frac{\pi}{6} \end{align*} Let us consider $$\Delta COB$$ and $$\Delta DOB$$. It can be easily shown that $$\Delta COB \cong \Delta DOB$$. Hence, $$OB$$ bisects $$\angle ABC$$. $$\therefore \angle CBO = \frac{\pi}{12}$$ In $$\Delta CBO$$, \begin{align*} \tan{\frac{\pi}{12}} &= \frac{r}{3+2\sqrt{3}} \\ \therefore r &= \sqrt{3} \end{align*} Now, consider a cartesian coordinate system with point $$C$$ as the origin, as shown in Figure 2. Equation of line $$AB$$ is given by, $$$$y=f(x)=\frac{(12+7\sqrt{3})-x(2+\sqrt{3})}{(3+2\sqrt{3})} \tag{1}\label{eq:1}$$$$
The circle is centred at $$(0, \sqrt{3})$$. The equation of the circle is, $$$$y=g(x)=\sqrt{3} \pm \sqrt{3-x^2} \tag{2}\label{eq:2}$$$$ Here, the $$\pm$$ sign joins the upper half of the circle with the lower half to create the desired circle. But our purpose can be served with the upper half of the circle itself, i.e., the semicircle (The blue semicircular curve as shown in Figure 3. So, we can replace the $$\pm$$ with $$+$$ to reduce calculational labour.
If we solve for $$x$$ and $$y$$ in equations \eqref{eq:1} and \eqref{eq:2}, we will get the point $$D(x, y)$$ at which line $$AB$$ touches the circle (or semicircle).
We get $$x=\frac{\sqrt{3}}{2}$$ and $$y=\frac{3}{2}+\sqrt{3}$$. To compute the shaded area, we only need $$x$$. We will integrate $$f(x)$$ and $$g(x)$$ from $$x=0$$ to $$x=\sqrt{3}/2$$
Area under the line AB from $$x=0$$ to $$x=\sqrt{3}/2$$ is, \begin{align} I_{1} &= \int_{0}^{\sqrt{3}/2} f(x)dx \\ &= \frac{1}{3+2\sqrt{3}}\left(\frac{45\sqrt{3}+78}{8}\right) \\ &= \frac{7\sqrt{3}+12}{8} ~~\text{ sq. unit} \tag{3}\label{eq:3} \end{align}
Also, area under the upper semicircular curve from $$x=0$$ to $$x=\sqrt{3}/2$$ is, \begin{align} I_{2} &= \int_{0}^{\sqrt{3}/2} g(x)dx \\ &= \frac{3\sqrt{3}+12}{8} + \frac{\pi}{4} ~~\text{ sq. unit} \tag{4}\label{eq:4} \end{align}
Finally, the required shaded area is given by, $$$$I_{1} - I_{2} = \frac{7\sqrt{3}+12}{8} - \frac{3\sqrt{3}+12}{8} + \frac{\pi}{4}$$$$ $$$$\boxed{\therefore ~~\text{Reqd. Area} = \left( \frac{\sqrt{3}}{2} - \frac{\pi}{4} \right) ~~\text{ sq. unit}} \tag{5}\label{eq:5}$$$$ |
Albert Teen
YOU ARE LEARNING:
In some quadratic expressions, the coefficient of the squared term is more than one.
Have a look at this example
1
What is the coefficient for $x^2$ in this quadratic expression?
2
When $x^2$ has a coefficient, you factorise a quadratic expression into the form $(ax+b)(cx+d)$
First you want to find out what $ax$ and $cx$ might be.
3
$ax$ multiplied by $cx$ have to give $4x^2$. Which 2 of these options equal $4x^2$?
4
So there are two options for what $ax$ and $cx$ are in the factorised expression
You will come back to finding out which option is the correct one in a minute.
5
Now you look at $b$ and $d$. When they are multiplied, they have to equal $9$. Which 2 of these options fulfil that?
6
So there are also two options for what $b$ and $d$ are in the factorised expression
That means there are 4 options for what the factorised expression might be!
7
The correct factorised expression results in $4x^2 + 12x +9$ when you expand the brackets. Which one of the options is that?
8
So option C is the correct factorisation of $4x^2+12x+9$
The other options don't give you the correct expression when you expand the brackets.
Option A gives you $4x^2+15x+9$
Option B gives you $4x^2+13x+9$
Option D gives you $4x^2+20x+9$
Recap! Sometimes $x^2$ in a quadratic expression will have a coefficient in front of it. There are 3 steps to factorising the expression then.
1
First you find options for $ax$ and $cx$
If you multiply $ax$ and $cx$ is should equal the first term in the original expression.
Here the first term is $6x^2$, so the options are $2x \times 3x$ and $6x \times x$
2
Then you look at options for $b$ and $d$
If you multiply $b$ and $d$ it should equal the last term in the original expression.
Here the last term is $8$, so the options are $4 \times 2$ and $8 \times 1$
3
Finally, you can now expand all the options to find out which option is correct
In this example, that is the option where the middle term is $16x$ once you have expanded the brackets and simplified the expression.
What is the correct factorisation of $9x^2+11x+2$?
Factorise $2x^2+10x+8$ into the form $(ax + b)(cx+d)$
Now take a look at this example
1
Here one of the terms is negative
You need to factorise so that you will end up with $-12x$ as the middle term.
2
True or false? If $ax$, $cx$, $b$ and $d$ are all positive, all terms will also be positive once you have expanded the brackets.
3
So to get to a negative term when you expand the brackets, at least one of the terms in the factorised expression has to be _____________.
4
First you find the options for $ax$ and $cx$
They are either $4x \times x$ or $2x \times 2x$
5
Now, the options for $b$ and $d$ have to make $+9$ when multiplied. Pick all the options below that equal $+9$
6
So there are actually 4 options for $b$ and $d$
Two pairs of positives and two pairs of negatives.
You could also write the options as $\pm 3 \times \pm3$ and $\pm9 \times \pm1$ to save space.
7
You can exclude two of these pairs as correct options for $b$ and $d$. Which ones?
8
So now you are down to two options for $b$ and $d$. That gives you 4 options for factorised expressions. Which one is the correct one?
9
So when one of the terms in the original expression is negative, at least one of the terms in the factorised expression has to also be negative
Here both $b$ and $d$ are negative, because when they are multiplied they have to give the positive number $+9$
What is the correct factorisation of $6x^2+19x-7$?
Factorise $5x^2-22x+8$ into the form $(ax + b)(cx+d)$
Summary! There are 3 steps to factorising these harder quadratic expressions where there is a coefficient in front of $x^2$
1
First you find options for $ax$ and $cx$
If you multiply $ax$ and $cx$ is should equal the first term in the original expression.
2
Then you look at options for $b$ and $d$
If you multiply $b$ and $d$ it should equal the last term in the original expression.
3
Finally, you can now expand all the options to find out which option is correct
In this example, that is the option where the middle term is $-22x$ once you have expanded the brackets and simplified the expression.
4
When one of the terms in the original expression is negative, at least one of the terms in the factorised expression has to also be negative
If you only have positive terms, all terms will be positive when you expand the brackets. |
# Prominent Steps of How to Solve Ratios With Useful Examples
A ratio is one of the parts of a mathematical word practiced to match the number of amounts to the amount of other numbers. This is usually practiced in both math and expert conditions. This post on how to solve ratios has illustrated a few of the daily instances when one can practice ratios:
• If one measures the winnings on a play
• While one distributes a pack of sweets honestly among their friends
• When a person changes its Dollars to Pounds or vice-versa while going on vacation
• If one works how many glasses of beer they require for a party
• During the calculation of how much cost they need to must pay on their income
Ratios can be normally utilized to connect two quantities, though individuals might also be utilized to analyze multiple measures. Besides this, ratios are often involved in numeric values’ reasoning tests, where people can be performed in several methods. That is why one is capable of recognizing and planning the ratios; however, individuals are manifested.
## Numerous methods to understand how to solve ratios
Ratios can normally be given as two or more numeric terms classified with a colon, for instance, 9:2 or 1:5 or 5:3:1. Though these might also be presented in many other methods, three examples are expressed differently.
### Scaling a ratio
Ratios are very useful in a number of ways, and the basic reason for this is that it allows us to range the quantity. It indicates rising or reducing the quantity of anything. This is unusually beneficial for something such as scale maps or models, where very high amounts can be changed to enough fewer illustrations, which are yet perfect.
Scaling is additionally essential for raising or reducing the number of components into a chemical reaction or recipe. Ratios might be estimated higher or lower by multiplying each section of the ratio with the equivalent product. This is the most useful point for how to solve ratios. Let’s take an instance:
George requires to cook pancakes for nine mates, but his recipe only produces sufficient pancakes for three mates. What amount of the elements will he require to utilize?
Pancake Ingredients (works 3)
• 300ml milk
• 100g flour
• 2 large eggs
To check how to solve ratios, one needs first to recognize the ratio. It has a three-part degree, whereby:
• 300ml milk
• 100g flour
• 2 large eggs
= 300:100:2
Besides this, one requires to work how much they require to balance the ingredients with.
As the recipe is for three individuals, but George wants a recipe for severe nine persons.
Because 9/3 = 3, the required ratio must be estimated by three (it is seldom represented with 3). Now, he requires to multiply every ingredient of the ratio with 3:
• 300 x 3 = 900
• 100 x 3 = 300
• 2 x 3 = 6
Hence, to get adequate pancakes for nine persons, George will require 900ml milk, 300g flour, and 6 eggs. This is one method for how to solve ratios; now, let’s move to the other two methods that are listed below.
### Reducing the ratios
Seldom ratio can not be shown in its most manageable structure that addresses it more difficult to manage. For instance, if a person has 6 hens, and all together lay 42 eggs each day. It can be interpreted as the ratio 6:42 (or given as a portion that will show: 6/42).
Decreasing a ratio implies changing the ratio into a standard form, making it more accessible to practice. This is executed by dividing each quantity of numbers into a ratio with the highest number that it can divide by. Let’s take an example of it:
Stella has 17 birds, and all eat 68kg of seed per week. Sam has 11 birds, and all eat 55kg seed per week. Find out who has the greediest birds?
To answer how to solve ratios, one should first recognize and analyze these two ratios:
• Stella’s ratio = 17:68, explain it by dividing each number with 17, which provides a ratio as 1:4
• Sam’s ratio = 11:55, analyze it by dividing each number with 11, which provides a ratio as 1:5
It implies that Stella’s birds eat 4kg seed per week, while Sam’s birds eat 5kg seed per week. Hence, Sam’s birds are greedier.
### Analyzing unknown values from existing ratios
This is another method that ratios are individually beneficial because this allows the learners to work for unknown and new measures depending on a known (existing) ratio. There are several methods for determining these kinds of problems. Initiate with using the cross-multiplication.
Mandeep and Gabriel are going to get married. Both have estimated that all require 40 glasses of wine for the 80 guests. At the moment, both get to know that another 10 guests are coming to attend their marriage. Find out how much wine do both require in total?
Initially, one requires to work on the ratio of the glass of wine with guests. They practiced = 40 wine:80 guests.
Then analyze it as 1 wine:2 guest (also we can say that 0.5 glass wine/guest).
Both have 90 guests who are going to attend (80 + the extra 10 = 90). Therefore, one requires multiplying 90 with 0.5 = 45 glasses of wine. See for the contents in the sort of problem that can seldom demand the total order and the extra ordered. This is how to solve ratios effectively.
See also Math Vs Statistics: Top 9 Important Points One Should Know
### Things that you should remember while solving ratios
• Remember, one is studying the ratio of the best method. For instance, the ratio of colors can be represented as 3 reds to 9 blue can be represented as 3:9, not 9:3. The initial article in the statement arrives initially.
• Be accurate with understanding the contents. For instance, individuals often make errors with topics like “Sam has 10 animals and 5 birds. Determine the ratio of animals to birds.” It is fascinating to tell the ratio is 10:5, but it will be wrong as the problem demands the ratio of animals to birds. One requires determining the total number of pets (10 + 5 = 15). Therefore the right ratio will be 10:15 (or 2:3).
• Avoid putting off with decimals or units. The sources will be the equivalent, whether all connect to complete fractions, numbers, m2, or £. Assure one should take note of the units in the views and change them to similar units. For instance, if one requires a ratio of 100g to 0.50kg, then change each unit to either kilos or grams.
## Conclusion
To sum up the post on how to solve ratios, we can say that three different methods can be used to solve them. Besides these methods, some common mistakes can be done by learners. Therefore, try to remember these and avoid them while solving ratios. Ratios have significant uses in day-to-day lives that are beneficial to solve various daily problems. So, learn the methods to solve ratio problems and get the benefits of these to overcome daily numeric problems. If you think that you need help then you can tell us that I need help with my math homework. And, Get the best and affordable help with my homework math with the help of us.
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Use Long Division: Four Digits by Two Digits
Long division is a written method for solving division problems where the divisor is 2 digits. We can also use it to find remainders in division problems. Remainders are the amount left over after a division calculation.
Example 1 => 12 goes into 4 zero times, so we look at 49. How many times does 12 go into 49? 12 goes into 48 four times. 49 subtract 48 leaves a remainder of 1. Write the remaining 1 underneath the 9 and 8 with the 2 from above next to it to make 12. How many times does 12 go into 12? 12 goes into 12 once, with no remainder.
Example 2 => 14 goes into 2 zero times, so we look at 29. How many times does 14 go into 29? 14 goes into 28 twice. 29 subtract 28 leaves a remainder of 1. Write the remaining 1 underneath the 9 and 8 with the 7 from above next to it to make 17. How many times does 14 go into 17? 14 goes into 14 once. 17 subtract 14 leaves a remainder of 3.
Example 3 => 13 goes into 2 zero times, so we look at 28. How many times does 13 go into 28? 13 goes into 26 twice. 28 subtract 26 leaves a remainder of 2. Write the remaining 2 underneath the 8 and 6 with the 9 from above next to it to make 29. How many times does 13 go into 29? 13 goes into 26 twice. 29 subtract 26 leaves a remainder of 3. Write the remaining 3 underneath the 9 and 6 with the 9 from above next to it to make 39. How many times does 13 go into 39? 13 goes into 39 three times, with no remainder.
Example 4 => 12 goes into 1 zero times, so we look at 15. How many times does 12 go into 15? 12 goes into 12 once. 15 subtract 12 leaves a remainder of 3. Write the remaining 3 underneath the 5 and 2 with the 9 from above next to it to make 39. How many times does 12 go into 39? 12 goes into 36 three times. 39 subtract 36 leaves a remainder of 3. Write the remaining 3 underneath the 9 and 6 with the 8 from above next to it to make 38. How many times does 12 go into 38? 12 goes into 36 three times. 38 subtract 36 leaves a remainder of 2.
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# IT Project: Lesson Plan Template: Poll Everywhere
Endorsement Subject Class Unit Lesson Resources Standard Elementary Education Math 3rd Grade Math Basic Multiplication and Division Learning How to Multiply and Divide Textbook, Computer and Smartboard
MA 3.1.2 Operations: Students demonstrate the meaning of multiplication with whole numbers.
MA 3.1.2.a Represent multiplication as repeated addition using objects, drawings, words, and symbols (e.g., 3 x 4 = 4 + 4 + 4) MA 3.1.2.b Use objects, drawings, words and symbols to explain the relationship between multiplication and division (e.g., if 3 x 4 = 12 then 12 3 = 4.) MA 3.1.2.c Use drawings, words, and symbols to explain the meaning of the factors and product in a multiplication sentence (e.g., in 3 x 4 = 12, 3 and 4 are factors and 12 is the total or product. The first factor (3) tells how many sets while the second factor tells how many are in each set. Another way to say this is that 3 groups of 4 equals 12 total.) MA 3.1.2.d Use drawings, words, and symbols to explain the meaning of multiplication using an array (e.g., an array with 3 rows and 4 columns represents the multiplication sentence 3 x 4 = 12
## Preparing the Student Assessment of Student Learning
Lesson Sequence
Students should know how to multiply and divide whole numbers using numbers, symbols, words and drawings to show relationships and knowledge. A SmartBoard will be used to explain and show examples of the lesson. Students can practice examples on the SmartBoard. At the end of the lesson, a PollEverywhere quiz will be completed by the students via computer. Objective 1-Students will understand how to multiply whole numbers. Objective 2-Students will understand how to divide whole numbers. Objective 3-Students will understand the relationship between multiplication and division. Objective 4-Students will have a deeper understanding of the meaning and reasoning of multiplication and division. Objective 5-Students will be able to use symbols, words and objects as well as numbers to complete multiplication and division problems. The introductory lesson of multiplication and division is necessary. Students will need an access link to the PollEverywhere quiz for completion of the quiz. Students will use Knowledge to observe and recognize the differences between multiplication and division. Students will also be able to Understand by identifying sets within the problems. Students will Apply by solving the multiplication and division problems. Students will Analyze by calculating, explaining, comparing and differentiating the multiplication and division problems. Students will Evaluate their own and a peers work to ensure they have the right answer. Students will Create by designing their own problems using pictures and objects. The teacher will explain that they are going to be learning multiplication and division. This will begin with examples of numbers only problems. After explaining, allow the students to do practice problems. After teacher feels the students are sufficient in numbers only problems, move on to story problems and problems using pictures and objects. Once taught, have the students do more example problems until teacher again feels students are masters. At the completion of the lesson, have the students log on and take the quiz on multiplication and division problems using PollEverywhere. Teacher will then evaluate
the quizzes to ensure students have properly learned the objectives listed above. POLL QUIZ QUESTIONS 1. 2x2=? a. 4 b.7 c.2 d.1 2. 7x0=? a.7 b.1 c.0 d.10 3. 10/5=? a.4 b.50 c.2 d.5 4. 6/3=? a.9 b.3 c.2 d.18 5. 4x(?)=4 a.1 b.4 c.0 d.8 6. 50/(?)=5 a.20 b.10 c.0 d.1 7. Sally has 2 sets of apples. All together she has 6 apples. How many apples are in each set? a.2 b.1 c.4 d.3 8. Johnny has completed 3 good deeds. He has received 6 gold stars. How many gold stars does Johnny receive for 1 good deed? a.3 b.1 c.2 d.4 The teacher will grade the quiz on PollEverywhere involving multiplication and division
Measurement
of Success |
# How Do You Calculate Imperial Units of Measurement?
This entry uses the imperial system of measurement. For the metric system, click here.
Here, you are given some examples on what happens when you multiply numbers by different units of measurements.
Example 1
A room is $\text{}3\text{}\phantom{\rule{0.17em}{0ex}}\text{ft}$ wide and $\text{}4\text{}\phantom{\rule{0.17em}{0ex}}\text{ft}$ long. How many square feet is the room?
$3\phantom{\rule{0.17em}{0ex}}\text{ft}\cdot 4\phantom{\rule{0.17em}{0ex}}\text{ft}=3\cdot 4\cdot \text{ft}\cdot \text{ft}=12\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{2}$
Example 2
You are celebrating your birthday and you want to serve Kool-Aid from a small plastic pool in your garden. You fill the pool with $\text{}80.5\text{}\phantom{\rule{0.17em}{0ex}}\text{pt}$ Kool-Aid. How many cups of Kool-Aid is that?
$\begin{array}{llll}\hfill 80.5\phantom{\rule{0.17em}{0ex}}\text{pt}& =\left(80.5\cdot 2\right)\phantom{\rule{0.17em}{0ex}}\text{cups}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =161\phantom{\rule{0.17em}{0ex}}\text{cups}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Thus, you have 161 cups of Kool-Aid for your guests.
Example 3
What is the volume of a cube-shaped container with sides of $\text{}6\text{}\phantom{\rule{0.17em}{0ex}}\text{ft}$? Give your answer in gallons (gal).
First, you have to find the volume of the container. That is, $V=l\cdot b\cdot h$. Hence, the calculation is like this: $\begin{array}{llll}\hfill V& =6\phantom{\rule{0.17em}{0ex}}\text{ft}\cdot 6\phantom{\rule{0.17em}{0ex}}\text{ft}\cdot 6\phantom{\rule{0.17em}{0ex}}\text{ft}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =6\cdot 6\cdot 6\cdot \text{ft}\cdot \text{ft}\cdot \text{ft}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =216\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Now, you have to calculate into gallon. Note that $1$ ft3 approximately equals $7.48$ gal. $\begin{array}{llll}\hfill 216\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}& =125\cdot 7.48\phantom{\rule{0.17em}{0ex}}\text{gal}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1615.68\phantom{\rule{0.17em}{0ex}}\text{gal}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
The volume of the container is thus $1615.68$ gallons.
You are also going to experience situations with compound units which consists of several simple units. For example, the unit for speed is a compound unit, consisting of miles and hours (mi/h or mph).
Example 4
A car driver is driving 70 miles in $\text{}1.5\text{}$ hours. How fast has he driven on average per hour?
In this exercise, you need to know the formula for distance, velocity and time:
$s=v\cdot t.$
Since you are going to find the speed, you must first solve the formula for $v$: $\begin{array}{llll}\hfill s& =v\cdot t\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{s}{t}& =\frac{v\cdot t}{t}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill v& =\frac{s}{t}=\frac{70\phantom{\rule{0.17em}{0ex}}\text{mi}}{1.5\phantom{\rule{0.17em}{0ex}}\text{h}}\approx 46.7\phantom{\rule{0.17em}{0ex}}\text{mi/h}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Thus, the car driver drove $46.7$ mi/h on average. |
# Question Video: Using Laws of Logarithms to Verify Equality Statements Mathematics • 10th Grade
Is it true that log_(π) (π₯ + π¦) = log_(π) π₯ + log_(π) π¦?
02:52
### Video Transcript
Is it true that the logarithm base π of π₯ plus π¦ is equal to the logarithm base π of π₯ plus the logarithm base π of π¦?
In this question, weβre given a logarithmic equation involving values of π, π₯, and π¦, and we need to determine if this equation is true. We might be tempted to try and prove this is true by using the definition of a logarithm. However, since this isnβt one of our given rules of logarithms, itβs usually easier to check whether the statement is true for a few values first. So we want to choose some values of π, π₯, and π¦ to check if both sides of the equations are equal.
To do this, letβs recall some laws of logarithms to make both sides of the equation easy to evaluate. First, we recall for any base π thatβs a positive real number not equal to one, the log base π of π is equal to one. Therefore, if we chose π is equal to three, π₯ is equal to three, and π¦ is equal to three, both terms on the right-hand side of this equation would be of the form log base π of π. So we could evaluate both of these terms.
For the statement to be true, it would need to hold for all values of π, π₯, and π¦ which leave the equation well defined. So letβs check if this is true for π is three, π₯ is three, and π¦ is three. The right-hand side of our equation becomes log base three of three plus log base three of three, which is equal to one plus one, which of course is just equal to two. Letβs now check if the same is true on the left-hand side of the equation. When π is three, π₯ is three, and π¦ is three, the left-hand side of the equation becomes the log base three of three plus three, which we can simplify to give us the log base three of six. For the statement to be true, this value needs to be equal to two.
However, we can show that this is not true. For example, for any positive real number π, we know the log base π of π to the πth power is equal to π, since the logarithmic function of base π is the inverse function of π to the πth power. An application of this result tells us that the log base three of three squared is equal to two. And of course three squared is equal to nine. Itβs not equal to six.
Therefore, both sides of the equation are not equal for these values of π, π₯, and π¦, so the statement is false. However, for extra clarity, letβs also use our calculators to evaluate the log base three of six. To two decimal places itβs 1.63. Once again, this is not equal to the right-hand side of the equation. So we can say that no, it is not true that the log base π of π₯ plus π¦ is equal to the log base π of π₯ plus the log base π of π¦.
However, there is a statement thatβs very close to this which is true. Itβs called the product rule for logarithms. It states for any positive real numbers π, π₯, and π¦, where π is not equal to one, the log base π of π₯ times π¦ is equal to the log base π of π₯ plus the log base π of π¦, with the big difference being weβre taking the product of π₯ and π¦ inside of our logarithm. Weβre not taking the sum. In either case, we can show that no, it is not true that the log base π of π₯ plus π¦ is equal to the log base π of π₯ plus the log base π of π¦. |
# Solving Induction $\prod\limits_{i=1}^{n-1}\left(1+\frac{1}{i}\right)^{i} = \frac{n^{n}}{n!}$
I try to solve this by induction:
$$\prod_{i=1}^{n-1}\left(1+\frac{1}{i} \right)^{i} = \frac{n^{n}}{n!}$$
This leads me to: $$\prod_{i=1}^{n+1-1}\left(1+\frac{1}{i}\right)^{i} = \frac{(n+1)^{n+1}}{(n+1)!} = \frac{(n+1)^{n}(n+1)}{n!(n+1)}\\ \prod_{i=1}^{n+1-1}\left(1+\frac{1}{i}\right)^{i} = \frac{(n+1)^{n+1}}{(n+1)!}$$
I tried to solve this that way: $$\prod_{i=1}^{n-1}\left(1+\frac{1}{i}\right)^{i}\left(1+\frac{1}{n}\right)^{n} = \frac{(n+1)^{n+1}}{(n+1)!}$$
Which is equivalent to: $$\frac{n^{n}}{n!}\left(1+\frac{1}{n}\right)^{n} = \frac{(n+1)^{n}(n+1)}{n!(n+1)}$$
I'm not sure if every step is right, but now i can't solve this further. Please help :)
You are almost finished. Note that since $1+\frac{1}{n}=\frac{n+1}{n}$, we have $$\left(1+\frac{1}{n}\right)^n=\frac{(n+1)^n}{n^n}=\frac{(n+1)^{n+1}}{(n+1)n^n}.$$
Remark: The induction argument should be written up in a more formal style. Deal with the base case explicitly. Then do the induction step, showing that if the assertion holds for $n=k$, then it holds for $n=k+1$. Even though it is quite all right to do your "scratch" computations backwards, the writeup should be more direct.
So for the induction step, we assume that for a given $k$, we have $$\prod_{i=1}^{k-1}\left(1+\frac{1}{i} \right)^{i} = \frac{k^{k}}{k!}\tag{\ast}.$$ We want to show that $$\prod_{i=1}^{k}\left(1+\frac{1}{i} \right)^{i} = \frac{(k+1)^{k+1}}{(k+1)!}.$$
Note that $$\prod_{i=1}^{k}\left(1+\frac{1}{i} \right)^{i} =\left( \prod_{i=1}^{k-1}\left(1+\frac{1}{i} \right)^{i} \right) \left(1+\frac{1}{k}\right)^k .$$ By the induction assumption $(\ast)$, the right-hand side is equal to $$\frac{k^{k}}{k!}\left(1+\frac{1}{k}\right)^k.$$ Continue.
• Thank you very much, nice trick ;) You helped a beginner very much :) May 10 '12 at 15:41
• Thank you for the structure tip too. Of course i use the induction step and wrote it down better than i did it here :) The main problem is i don't know the english words for induction beginning, step and presumption May 10 '12 at 15:49
Hint $\$ The LHS and RHS both satisfy $\rm\:f(n+1)/f(n) = (1+1/n)^n,\,\ f(1) = 1.\:$ But it is trivial to prove by induction the uniqueness of solutions of such first-order difference equations, which yields the sought equality: LHS = RHS. As I often emphasize, uniqueness theorems provide powerful tools for proving equalities.
Note that the solution of such recurrences may be represented by (indefinite) products
$$\rm \left[\begin{eqnarray}\rm f(n\!+\!1) \,&=&\rm\, a_n\: f(n)\\ \rm f(1) \,&=&\,1 \end{eqnarray} \right]\,\iff\, f(n)\, =\, \prod_{k=1}^{n-1}\:\! a_k$$
Thus the uniqueness theorem yields that such products are well-defined. It is a gap in most courses that this fact is not proved (making many such inductive proofs circular). For more on "definitions by induction" see the award-winning Monthly exposition of Leon Henkin referenced here and here.
This is a special case of telescopy. For as below we can write the RHS as a product of its term ratios
$$\rm\ g(n)\ =\ \frac{g(n)}{\color{#c00}{g(n-1)}}\ \frac{\color{#c00}{g(n-1)}}{\color{#0a0}{g(n-2)}}\ \frac{\color{#0a0}{g(n-2)}}{\cdots }\ \cdots\ \frac{\cdots}{\color{brown}{g(3)}}\ \frac{\color{brown}{g(3)}}{\color{blue}{g(2)}}\ \frac{\color{blue}{g(2)}}{1}$$
Then the proof amounts to saying that both expressions are equal because they are both products of the same expression (here $\rm\:(1+1/i)^i\:).\:$ The usual ad-hoc induction proof of your problem amounts to essentially proving this uniqueness theorem in this specific case. But here, in fact, proving the general case is simpler than proving the special case, because it is much easier to see the telescopic cancellation without the obfuscating details of the special case. Further, one obtains a general proof that can be reused for all problems of this type. Who could ask for more?
• Sounds good :) Im a total beginner, i have no idea what you mean ;) But thank you May 10 '12 at 15:42
• Wow, nice :) Sry i accepted the answer above because it was the direct answer of my question, but this is amazing. I think i will use this stack more often May 10 '12 at 16:00
• @blang After you do a few more of these you might find it helpful to revisit this answer. It yields an algorithm for handling problems of this type - removing the need for any guesswork or intuitive leaps. It can be understood at high-school level if explained appropriately. You can find many more examples in my linked posts, both multiplicative and additive. May 10 '12 at 16:02 |
# Metric System Presentation
```Announcements
Two Youtube videos are posted and a homework
Office Hours
Competency:
II -- Apply quantitative skills to represent and
communicate scientific concepts.
Skills:
Convert between units algebraically
Understand how to use the metric prefixes
Questions to think about during this lesson:
What is measurement?
Why do we use standard units of measurement?
What is the International System of Units?
What are the advantages of using the SI?
What are the SI units?
This topic is about measurement which can be an observation or
a description that includes a number and a unit.
Long time ago, people measured quantities in many different
ways which included using their bodies such as arms, feet and
steps. These systems had many problems such as that some
people do not have the same length of arm and feet.
Making standard measurements proves to work well because
someone at Abaarso can measure a quantity and another person
in Hargeisa can also measure the same quantity in the same
amount of Kilograms. Experiments must be repeatable to make
accurate results and standard units allow us to repeat other
scientists' experiments. The system of measurement that is best
suited to scientific purpose is the metric system.
The metric system was developed in France near the end of the
eighteenth century and it uses the decimal basis to convert one
metric to another. Because the metric system is multiples of ten,
calculation can greatly be simplified by moving the decimal
place.
The metric system that officially scientists use is called the
International System of Units which is SI Units in short. This
lesson is going to distinguish between the physical quantities
and units of measurement. All the things or the quantities that
are measurable in the universe such as length are known as
physical quantities. The units used to measure length, for
instance, is meter which is the metric unit of measurement. In
the table below, the seven fundamental units of measurement
are shown with the seven physical quantities they measure and
also their symbols for each physical quantity and SI Unit.
Physical Quantity
Physical Quantity
SI Unit
SI Unit
Name
Symbol
Name
Abbreviation
Length
l
metre
m
Mass
m
kilogram
kg
Time
t
second
s
Electric Current
I
ampere
A
Temperature
T
Kelvin
K
Amount of Substance
n
mole
mol
Luminous Intensity
l
candela
cd
Physical quantity
Physical quantity
SI Unit
SI unit
Name
Symbol
Abbreviation
Name
m2
area
A
volume
V
m3
cubic meter
frequency
f
Hz
Hertz
velocity
v
acceleration
a
m s –1
m s –2
square meter
meters per second
meters per second squared
force
F
N
Newton
pressure
P or p
Pa
Pascal
power
P
W
Watt
energy
E
J
Joule
voltage
V
V
Volt
resistance
R
Ω
Ohm
conductance
G
S
Siemens
charge
Q
C
Coulomb
``` |
NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers
# NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers
Subject specialists have designed NCERT solutions for Maths Class 8 Chapter 1 which includes thorough solutions for reference. These solutions are updated according to the latest CBSE syllabus for class 8 2024-25 and are provided in easy language for understanding. Tips and tricks are also provided.
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These solutions are provided so a student can clear his doubts and get help with deep understanding of the concept. Also you can refer them to make the chapter notes and revisions notes. PDF of this can also be downloaded from website.
## Chapter 1 Rational Numbers Class 8 PDF
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1. Rational Numbers
## NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Summary
This chapter includes 2 exercises and the NCERT Solutions for Class 8 Maths provided here contains to the point answers for all the questions present in these exercises. The concepts present in the chapter are given below.
• Rational numbers are closed under the operations of addition, subtraction and multiplication.
• The operations addition and multiplication are
• Commutative for rational numbers.
• Associative for rational numbers.
• Rational number 0 is the additive identity for rational numbers.
• Rational number 1 is the multiplicative identity for rational numbers.
• Distributivity of rational numbers: For all rational numbers a, b and c, a(b + c) = ab + ac and a(b – c) = ab – ac
• Rational numbers can be represented on a number line.
• Between any two given rational numbers, there are countless rational numbers. The idea of mean helps us to find rational numbers between two rational numbers.
### Access Question and Answers to NCERT Class 8 Maths Chapter 1 Rational Numbers
1. Using appropriate properties, find: (i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6
Solution:
• -2/3 × 3/5 + 5/2 – 3/5 × 1/6
• = -2/3 × 3/5 – 3/5 × 1/6 + 5/2 (by commutativity)
• = 3/5 (-2/3 – 1/6) + 5/2
• = 3/5 ((-4 – 1)/6) + 5/2 (by distributivity)
• = 3/5 ((-5)/6) + 5/2
• = -15/30 + 5/2
• = -1/2 + 5/2
• = 4/2
• = 2
(ii) 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
Solution:
• 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
• = 2/5 × (- 3/7) + 1/14 × 2/5 – (1/6 × 3/2) (by commutativity)
• = 2/5 × (- 3/7 + 1/14) – 3/12
• = 2/5 × ((-6 + 1)/14) – 1/4
• = 2/5 × ((-5)/14) – 1/4
• = -10/70 – 1/4
• = -1/7 – 1/4
• = (-4 – 7)/28
• = -11/28
2. Write the additive inverse of each of the following: (i) 2/8
The additive inverse of 2/8 is -2/8.
(ii) -5/9
The additive inverse of -5/9 is 5/9.
(iii) -6/-5 = 6/5
The additive inverse of 6/5 is -6/5.
(iv) 2/-9 = -2/9
The additive inverse of -2/9 is 2/9.
(v) 19/-16 = -19/16
The additive inverse of -19/16 is 19/16.
3. Verify that: -(-x) = x for:
(i) x = 11/15
We have, x = 11/15.
The additive inverse of x is -x (as x + (-x) = 0).
Then, the additive inverse of 11/15 is -11/15 (as 11/15 + (-11/15) = 0).
The same equality, 11/15 + (-11/15) = 0, shows that the additive inverse of -11/15 is 11/15.
Or, -(-11/15) = 11/15, i.e., -(-x) = x.
(ii) x = -13/17
We have, x = -13/17.
The additive inverse of x is -x (as x + (-x) = 0).
Then, the additive inverse of -13/17 is 13/17 (as 13/17 + (-13/17) = 0).
The same equality (-13/17 + 13/17) = 0 shows that the additive inverse of 13/17 is -13/17.
Or, -(-13/17) = -13/17, i.e., -(-x) = x.
4. Find the multiplicative inverse of the following: (i) -13
The multiplicative inverse of -13 is -1/13.
(ii) -13/19
The multiplicative inverse of -13/19 is -19/13.
(iii) 1/5
The multiplicative inverse of 1/5 is 5.
(iv) -5/8 × (-3/7) = 15/56
The multiplicative inverse of 15/56 is 56/15.
(v) -1 × (-2/5) = 2/5
The multiplicative inverse of 2/5 is 5/2.
(vi) -1
The multiplicative inverse of -1 is -1.
5. Name the property under multiplication used in each of the following: (i) -4/5 × 1 = 1 × (-4/5) = -4/5
The property of multiplicative identity is used here.
(ii) -13/17 × (-2/7) = -2/7 × (-13/17)
The property of commutativity is used here.
(iii) -19/29 × 29/-19 = 1
The property of multiplicative inverse is used here.
6. Multiply 6/13 by the reciprocal of -7/16.
Solution:
Reciprocal of -7/16 = 16/-7 = -16/7
6/13 × (Reciprocal of -7/16) = 6/13 × (-16/7) = -96/91
7. Tell what property allows you to compute 1/3 × (6 × 4/3) as (1/3 × 6) × 4/3.
Solution: The Associativity Property allows you to compute 1/3 × (6 × 4/3) as (1/3 × 6) × 4/3. This property states that the way in which factors are grouped in a multiplication problem does not change the product.
8. Is 8/9 the multiplication inverse of -9/8? Why or why not?
Solution: No, 8/9 is not the multiplicative inverse of -9/8.The multiplicative inverse of a number x is a number y such that x × y = 1.In this case, if 8/9 were the multiplicative inverse of -9/8, then we should have:
8/9 × (-9/8) = 1However, the actual calculation shows:
8/9 × (-9/8) = -1 ≠ 1Therefore, 8/9 is not the multiplicative inverse of -9/8.
1. If 0.3 is the multiplicative inverse of x, why or why not?
Solution
• 0.3 = 3/10
• Multiplicative inverse means the product of the number and its inverse should be 1.
• 3/10 × 10/3 = 1
• Therefore, 0.3 is the multiplicative inverse of 10/3.
1. Write:
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
Solution
(i) The rational number that does not have a reciprocal is 0.
Reason: The reciprocal of 0 is 1/0, which is undefined.(ii) The rational numbers that are equal to their reciprocals are 1 and -1.
Reason:
• Reciprocal of 1 is 1/1 = 1
• Reciprocal of -1 is -1/(-1) = 1
(iii) The rational number that is equal to its negative is 0.
Reason: The negative of 0 is -0, which is still 0.
1. Fill in the blanks.
(i) Zero has _______reciprocal.
(ii) The numbers ______and _______are their own reciprocals
(iii) The reciprocal of – 5 is ________.
(iv) Reciprocal of 1/x, where x ≠ 0 is _________.
(v) The product of two rational numbers is always a ________.
(vi) The reciprocal of a positive rational number is _________.
Solution:
(i) Zero has no reciprocal.
(ii) The numbers -1 and 1 are their own reciprocals.
(iii) The reciprocal of -5 is -1/5.
(iv) Reciprocal of 1/x, where x ≠ 0, is x.
(v) The product of two rational numbers is always a rational number.
(vi) The reciprocal of a positive rational number is positive.
### Exercise 1.2 Page: 20
1. Represent these numbers on the number line:
(i) 7/4
(ii) -5/6
Solution
(i) 7/4
• Divide the line between the whole numbers into 4 parts.
• The rational number 7/4 lies 7 points away from 0 towards the positive number line.
(ii) -5/6
• Divide the line between the integers into 6 parts.
• The rational number -5/6 lies 5 points away from 0 towards the negative number line.
1. Represent -2/11, -5/11, -9/11 on a number line.
Solution
• Divide the line between the integers into 11 parts.
• The rational numbers -2/11, -5/11, and -9/11 lie at a distance of 2, 5, and 9 points away from 0 towards the negative number line, respectively.
1. Write five rational numbers which are smaller than 2.
Solution
• 2 can be written as 20/10
• The five rational numbers smaller than 2 are: 2/10, 5/10, 10/10, 15/10, 19/10
1. Find the rational numbers between -2/5 and 1/2.
Solution
• -2/5 = -20/50
• 1/2 = 25/50
• The rational numbers between -20/50 and 25/50 are: -18/50, -15/50, -5/50, -2/50, 4/50, 5/50, 8/50, 12/50, 15/50, 20/50
1. Find five rational numbers between:
(i) 2/3 and 4/5
(ii) -3/2 and 5/3
(iii) 1/4 and 1/2
Solution
(i) 2/3 and 4/5
• 2/3 = 40/60, 4/5 = 48/60
• The five rational numbers between 40/60 and 48/60 are: 41/60, 42/60, 43/60, 44/60, 45/60
(ii) -3/2 and 5/3
• -3/2 = -9/6, 5/3 = 10/6
• The five rational numbers between -9/6 and 10/6 are: -1/6, 2/6, 3/6, 4/6, 5/6
(iii) 1/4 and 1/2
• 1/4 = 6/24, 1/2 = 12/24
• The five rational numbers between 6/24 and 12/24 are: 7/24, 8/24, 9/24, 10/24, 11/24
1. Write five rational numbers greater than -2.
Solution
• -2 can be written as -20/10
• The five rational numbers greater than -2 are: -10/10, -5/10, -1/10, 5/10, 7/10
1. Find ten rational numbers between 3/5 and 3/4.
Solution
• 3/5 = 48/80, 3/4 = 60/80
• The ten rational numbers between 48/80 and 60/80 are: 49/80, 50/80, 51/80, 52/80, 54/80, 55/80, 56/80, 57/80, 58/80, 59/80
The main topics covered in this chapter include:
• 1.1 Introduction
• 1.2 Properties of Rational Numbers
• 1.2.1 Closure
• 1.2.2 Commutativity
• 1.2.3 Associativity
• 1.2.4 The role of zero
• 1.2.5 The role of 1
• 1.2.6 Negative of a number
• 1.2.7 Reciprocal
• 1.2.8 Distributivity of multiplication over addition for rational numbers.
• 1.3 Representation of Rational Numbers on the Number Line
• 1.4 Rational Numbers between Two Rational Numbers
### Access exercise-wise NCERT Solutions Class 8 Maths of this chapter here:
• 11 Questions (11 Short Answer Questions)
• 7 Questions (7 Short Answer Questions)
### NCERT Rational Numbers Class 8 Solutions
Numbers are the most basic block of Mathematics. In middle school, the students may have learned the different types of numbers including natural numbers, whole numbers, integers etc. this chapter deals with another set of numbers, the rational numbers. This chapter includes almost all the concepts that a student of Class 8 has to learn about rational numbers.
This chapter also provide the method of representing a rational number on a number line as well as the method of finding rational numbers between 2 rational numbers. Students must study this chapter to learn more about Rational Numbers and the concepts coming under them. By learning these solution students can easily score good marks in this chapter.
## NCERT Solutions for Class 8 Maths Chapter 1 FAQs
### What is the meaning of rational numbers according to these NCERT Solutions?
According to these NCERT Solutions, rational numbers can be represented in p/q form where q is not equal to zero. It comes under the type of real number. Any fraction with non-zero denominators is a rational number. Hence, we should say that ‘0’ is also a rational number, as we can represent it in many forms such as 0/1, 0/2, 0/3, etc. But, 1/0, 2/0, 3/0, etc. are not rational, since they give us infinite values.
### List out the important concepts discussed in these NCERT Solutions.
Main concepts including in this chapter are listed below: 1.1 Introduction 1.2 Properties of Rational Numbers 1.2.1 Closure 1.2.2 Commutativity 1.2.3 Associativity 1.2.4 The role of zero 1.2.5 The role of 1 1.2.6 Negative of a number 1.2.7 Reciprocal 1.2.8 Distributivity of multiplication over addition for rational numbers.
### Is INFINITE LEARN providing Solutions for Class 8 Maths Chapter 1?
Yes, INFINITE LEARN website comes with accurate and detailed solutions for all questions provided in the NCERT Textbook. INFINITE LEARN brings you NCERT Solutions for Class 8 Maths, made by our subject matter experts for smooth and easy understanding of concepts. These solutions comes with detailed step-by-step explanations of problems given in the NCERT Textbook. The NCERT Solutions of this chapter can be downloaded in the form of a PDF and it can be used as a quick revision tool.
### What is a rational number in Class 8?
In Class 8, a rational number is a number that can be expressed as a fraction where the numerator and denominator are integers, and the denominator is not zero. Rational numbers include integers, fractions, and terminating or repeating decimals.
### What is a short answer for a rational number?
A rational number is a number that can be expressed as a fraction where the numerator and denominator are integers, and the denominator is not zero. It can be written in the form a/b, where a and b are integers and b is not equal to zero.
### What are 8 rational numbers?
Eight rational numbers can be any numbers that can be expressed as fractions. For example: 1/2 -3/4 5 -7/3 2.5 (which is 5/2) -1 0 3/7
### What is an irrational number in Class 8?
In Class 8, an irrational number is a number that cannot be expressed as a fraction of two integers. These numbers have non-repeating and non-terminating decimal expansions. Examples of irrational numbers include √2, π (pi), and e (Euler's number).
### How to solve rational numbers in Class 8?
To solve rational numbers in Class 8, you can perform operations like addition, subtraction, multiplication, and division on fractions. Simplify fractions by finding the common factors between the numerator and denominator. Convert mixed numbers to improper fractions for easier calculations. Practice converting between fractions and decimals to enhance your understanding of rational numbers.
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# Jane Street Test
- - posted in
The questions below appeared on an online test administered by Jane Street Capital Management to assess whether a person is worthy of a phone interview.
### Question 1
Suppose we choose four numbers at random without replacement from the first 20 prime numbers. What is the probability that the sum of these four numbers is odd?
Solution By definition, an even number is an integer multiple of 2; that is, even numbers have the form $2n$, for some integer $n$. An odd number is one more than an even number; that is, odd numbers have the form $2n+1$ for some integer $n$.
If all four numbers are odd, then the sum is even. This follows from the fact that the sum of two odd numbers is even: $(2n+1) + (2k+1) = 2(n+k) + 2 = 2(n+k+1)$. And the sum of two even numbers is again even: $2n+2k = 2(n+k)$. Therefore, in order for the sum of the four primes chosen to be odd, the only even prime number, 2, must be among the four numbers we selected.
Here are two ways to compute the probability that 2 is one of the four numbers selected: First, consider the probability of not selecting 2: $$P(\text{2 is not among the chosen primes}) = \frac{19}{20} \frac{18}{19} \frac{17}{18} \frac{16}{17} = \frac{16}{20} = \frac{4}{5}.$$
Then the probability that 2 is among the four chosen prime numbers is $1 - \frac{4}{5}$, or $\frac{1}{5}$.
Alternatively, we could count the number of four-tuples that have a 2. This is the same as taking 4 times the number of ways to choose the other three numbers: $19\cdot 18\cdot 17$. (The factor of 4 appears since we could put the 2 in any of four positions). Now divide this by the total number of four-tuples, $20 \cdot 19\cdot 18\cdot 17$, to arrive at the same answer as above:
### Question 2
Suppose you have an 8 sided die with sides labeled 1 to 8. You roll the die once and observe the number $f$ showing on the first roll. Then you have the option of rolling a second time. If you decline the option to roll again, you win $f$ dollars. If instead you exercise the option to roll a second time, and if $s$ shows on the second roll, then you win $f + s$ dollars if $f+s < 10$ and 0 dollars if $f+s \geq 10$. What is the best strategy?
Solution
A strategy is given by specifying what to do in response to each possible first roll, $f$. The best strategy will maximize the expected winnings.
The expected winnings if we roll just once is simply the number that appears, $f$.
Let $S$ be the random variable representing the value showing on the second roll. This is uniformly distributed with $P(S=s) = 1/8$. Therefore, if we observe $f$ and then decide to roll twice, the expected winnings are
where $\chi$ denotes the characteristic function, that is, $\chi_{\mathrm{True}}=1$ and $\chi_{\mathrm{False}}=0$.
If we let $N= \min{8, 10 - f -1}$, then the expected value of rolling twice is
If we observe $f=1$ on the first roll, then
Since this is greater than 1, we should certainly exercise the option to roll twice if 1 shows up on the first roll.
Continuing this way, we have
Since this is larger than 4, we should roll twice when observing 4 on the first roll.
However, if $f = 5$, then
Since this is less than 5, we should not roll a second time when a 5 is observed on the first roll.
General strategy: Exercise the option to roll a second time if the first results in 1, 2, 3, or 4. Otherwise, decline the second roll and take the value shown on the first roll.
### Question 3
(The original question was for 3 coins, with respective probabilities 0.5, 0.3, and 0.2 of coming up heads. This version is a generalization to $n$ coins with arbitrary probabilities that sum to 1.)
Suppose you have a bag with $n$ coins, $C_1, C_2, \dots, C_n$, and coin $C_i$ has probability $p_i$ of coming up heads when flipped. Assume $p_1 + p_2 + \cdots p_n = 1$. Suppose you draw a coin from the bag at random and flip it and it comes up heads. If you flip the same coin it again, what is the probability it comes up heads?
Denote by $H_i$ the event that heads turns up on the $i$-th flip, and by a slight abuse of notation, let $C_i$ denote the event that we are flipping coin $C_i$. Then
Applying Bayes’ Theorem,
Assuming all coins are equally likely to have been drawn from the bag, $P(C_i)=\frac{1}{n}$. Therefore,
whence,
Therefore,
For the special case given in the original problem, we have
### Question 4
Suppose you have two urns that are indistinguishable from the outside. One of the urns contains 3 one-dollar coins and 7 ten-dollar coins. The other urn contains 5 one-dollar coins and 5 ten-dollar coins. Suppose you choose an urn at random and draw a coin from it at random.
You find that it is a $10 coin. Now you are given the option to draw again (without replacing the first coin) from either the same urn or the other urn. Should you draw from the same urn or switch? Solution The problem is uninteresting if we assume the draws are made with replacement. Clearly we should not switch urns in this case. Assume the second draw is made without replacing the first coin. Let$X$be a random variable denoting the value of the second draw. We compute the expected values$E(X | \text{ stay})$and$E(X | \text{ switch})$and whichever is higher determines our strategy. Let us call the urn that starts with 7 ten-dollar coins urn$A$, and the urn that starts with 5 ten-dollar coins urn$B$. Let$A_1$denote the event that urn$A$is chosen for the first draw and let$B_1$denote the event that urn$B$is chosen for the first draw. Let$T$denote the event that the first draw produces a ten-dollar coin. The probability that a ten-dollar coin appears on the first draw is Given that$T$occurred, the probability we were drawing from$A$is, by Bayes’ Theorem, Similarly the probability we were drawing from$B$, given$T$, is Now, if we decide to draw again from the same urn and that urn happens to be$A$, there will be 6 ten-dollar coins remaining, so our expected value of drawing again from$A$is If we decide to draw from the same urn and it happens to be$B$, there will be 4 ten-dollar coins remaining, so So, if we don’t switch urns, the expected value of the second draw is Suppose instead we decided to switch urns. In the event$A_1$that the first draw was from urn$A$, then urn$B$will contain its original contents: 5 ten-dollar coins and 5 one-dollar coins. Therefore, If we switch in the event$B_1$, then we choose from a pristine urn$A\$ having 7 ten-dollar coins and 3 one-dollar coins. Thus,
Therefore,
So, the expected value of the second draw if we switch urns is a little more than 6.83.
Conclusion: The best strategy is to switch urns before drawing the second coin. |
finding the slope and y intercept of a line pdf
# finding the slope and y intercept of a line pdf
Find the x-intercept and y-intercept of each line. Use the intercepts to graph the equation.Write an equation of a line in slope-intercept form to express the cost, y, of a large cheese pizza with x toppings. Multiple Choice Identify the choice that best completes the statement or answers the question. Find the slope of the line. 31 Find the x- and y-intercepts of the graph of the equation. Slope and Y intercept (math.com) Every straight line can be represented by an equation: y mx b. This is called the slope-intercept form.1.4 Equations of Lines and Modeling Find the slope and the y intercept of a line given the equation y mx b, or f(x) mx b. Graph a linear equation Using Slopes and Intercepts. x-intercept: the x-coordinate of the point at which a line crosses the x-axis.Write each equation in slope-intercept form and then find the slope and y -intercept. Goal 1 writing linear equations. In Lesson 2.3 you learned to find the slope and y-intercept of a line whose equation is given. In this lesson you will study the reverse process. When graphing a line we found one method we could use is to make a table of values. However, if we can identify some properties of the line, we may be able to make a graph much quicker and easier. One such method is nding the slope and the y-intercept of the equation.
c. Use the slope and y-intercept to find the equation of the line in slope-intercept form.41. OPEN ENDED Draw the graph of a line that has a y-intercept but no x- intercept. What is the slope of the line? 498 Chapter 9 Algebra: Linear Functions. Slope Intercept Form www.mathwarehouse.com/algebra/linearequation/slope- intercept-form.php.Practice. Find the equation of the line that has given slope and y- intercept.
The slope is defined as m (y2 y1)/(x2 x1) for any two points on the line. We call y mx b the Slope-Intercept Form of the linear equation.Finding The Equation of a Line: The Point-Slope Form The equation of a line that passes through (x1, y1) and has slope m is given by. 1 Finding Slope and y-intercept Find the slope and y-intercept of the graph of each function.Move 7 units up from (0, 0) since the slope is positive and 1 unit right . Repeat to find more points on the line. Write the equation of a line using slope and y-intercept.Graph a line using slope and y-intercept. When graphing a line we found one method we could use is to make a table of values. CHAPTER 6 Analyse Linear Relations 6.1 The Equation of a Line in Slope y- Intercept Form: y mx b The Equation of a Line in Slope y-Intercept Form: y mx b Vertical and Horizontal LinesThen, write the equation of each line. Solution: To find the slope, use two points on the graph. You can find the slope of a line by comparing any two of its points.The line crosses the y-axis at (0, 4). So, the y-intercept is 4. Find the slope and y-intercept of the line in each graph. Want some practice finding the y-intercept of a line? In this tutorial, youre given the slope of a line and a point on that lin.22 Day Vegan Diet PDF. Constitution Party Issues. Finding the Slope and y-Intercept of an Equation: Words. Symbols.y mx b. slope y-intercept. y 3x 1. Graph. EXAMPLE: Identify the slope and y-intercept of the line with the given equation. The slope of a line through the points (x1, y1) and (x2, y2) is as followsAdditional Example 2A: Using Slope-Intercept Form to Find Slopes and y- intercepts. Write each equation in slope-intercept form, and then find the slope and y-intercept. Warm Up 3. Find the slope and the y-intercept of the line represented by the table.Determine the initial value of a linear relationship. Use the slope- intercept form of an equation of a line. Download and Read Finding The Slope And Y Intercept Of A Line.Now, we will show you a new book enPDFd finding the slope and y intercept of a line that can be a new way to explore the knowledge. Finding the y-intercept of a line The Y-Intercept: the point where the graph of a line crosses the . theExample 3: Slope-Intercept Form of Linear Equations: y mx b (m stands for slope and b stands for y-intercept). Find the b (y-intercept) and write the equation of the line in slope- intercept form. m. b.Slope-Intercept Form: ymx b. We will also talk about x-intercept and y-intercept, parallel and perpendicular lines.Definition: The slope of a line measures the steepness of a line or the rate of change of the line. To find the slope of a line you need two points. As what we refer, find slope and y intercept of a line has several motives for you to pick as one of the sources. First, this is very connected to your problem now. This book also offers simple words to utter that you can digest the information easily from that book. 3. Graph a line whose equation is in the form Ax By C. 4. Find the equation of a line given its slope and yintercept. 5. Work with linear models in slopeintercept form. This is correct for a positive slope. Example 2. Find the slope of the line that passes through C(1, 4) and D(3, -2).This can be represented by the function 60x 15y 4,740. Find the x- and y-intercepts. Submit. More "finding slope and y intercept" pdf. Advertisement.7.1 The Slope-Intercept Form 7.1 OBJECTIVES 1. Find the slope and y intercept from the equation of a line 2. Finding the Slope and y Intercept. Increments Slope of a Line Parallel and Perpendicular Lines Equations of Lines Applications and why . . . Linear equations are used extensively in business and economic applications.In Exercises 2730, find the (a) slope and (b) y-intercept, and (c) graph the line. |
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