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# Partial Fractions
### Partial Fractions
In solving algebra related problems and questions, we may sometimes deal with rational functions. A rational function is basically an algebraic polynomial fraction, in which we have polynomials on both the numerator and denominator.
Partial fractions is one of the simplest and most effective method in solving algebra related problems regarding rational functions.
In partial fractions, we separate the polynomials in our rational function into simpler form of polynomials.
Some of the applications of partial fractions include the solving of integration problems with rational functions, the binomial expansion and also the arithmetic series and sequences.
There are a few basic forms we need to memorize in partial fractions:
1.$$\enspace$$ The linear form:
$$\hspace{6em} {\small (ax + b)}$$
Example:
$${\large\frac{3x \ + \ 5}{(x \ + \ 1)(2x \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + \frac{B}{(2x \ + \ 7)} }$$
2.$$\enspace$$ The quadratic form of a linear factor:
$$\hspace{6em} {\small (cx \ + \ d)^{2}}$$
Example:
$$\frac{3x + 5}{(x + 1){(2x + 7)}^{2}} \equiv \frac{A}{(x + 1)} + {\small\boxed{\frac{B}{(2x + 7)} + \frac{C}{{(2x + 7)}^{2}}}}$$
3.$$\enspace$$ The quadratic form that cannot be factorized:
$$\hspace{6em} {\small (c{x}^{2} \ + \ d) }$$
Example:
$${\large\frac{3x \ + \ 5}{(x \ + \ 1)(2{x}^{2} \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + {\small\boxed{\frac{Bx \ + \ C}{(2{x}^{2} \ + \ 7)}}}}$$
After the polynomials in the denominator of the rational function is separated, make the denominators of the simpler terms to be the same. This is typically done by multiplying the denominators together .
To find each of the coefficients in the numerator (A, B or C), we can use substitution method or equating the coefficient method.
In the substitution method, we substitute a value of x that we freely choose in the left hand side numerator and the right hand side numerator and then find the coefficients one by one.
While in the equating the coefficient method, we expand the right hand side numerator and then compare each of the coefficients in the right hand side numerator with the left hand side numerator.
Both methods will be shown in the solution of the examples below. Give it a try and if you need any help, just look at the solution I have written. Cheers ! =) .
$$\\[1pt]$$
EXAMPLE:
$${\small 1.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{7{x}^{2} \ – \ 15x \ + \ 8}{(1 \ – \ 2x){(2 \ – \ x)}^{2} }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 2.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ x \ – \ 4{x}^{2} }{(3 \ – \ x)(2 \ + \ {x}^{2}) }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 3.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 27 }{(2x \ + \ 1)( {x}^{2} \ + \ 9 ) }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 4.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 10 x \ + \ 9 }{(2x \ + \ 1){( 2x \ + \ 3 )}^{2} }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 5.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 2x(5 \ – \ x) }{(3 \ + \ x){( 1 \ – \ x )}^{2} }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 6.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 7(a) and (b)
$$\\[1pt]$$
Let $${\small f(x) \ = \ {\large \frac{ 2 }{(2x \ – \ 1)( 2x \ + \ 1 ) }} }$$
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Using your answer to part (a), show that
$$\\[1pt]$$
$${\scriptsize {\Big( f(x) \Big)}^{2} \ = \ {\large \frac{ 1 }{ {(2x \ – \ 1)}^{2} }} \ – \ {\large \frac{ 1 }{ (2x \ – \ 1) }} }$$
$$\\[1pt]$$
$${\hspace{3em} \scriptsize \ + \ {\large \frac{ 1 }{ (2x \ + \ 1)}} \ + \ {\large \frac{ 1 }{ {(2x \ + \ 1)}^{2} }} . }$$
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 7.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 6(a)
$$\\[1pt]$$
Let $${\small f(x) \ = \ {\large \frac{ 5a }{(2x \ – \ a)( 3a \ – \ x ) }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 8.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 4(a)
$$\\[1pt]$$
Let $${\small f(x) \ = \ {\large \frac{ 15 \ – \ 6x }{(1 \ + \ 2x)( 4 \ – \ x ) }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 9.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 9(a)
$$\\[1pt]$$
Let $${\small f(x) \ = \ {\large \frac{ 14 \ – \ 3x \ + \ 2x^{2} }{(2 \ + \ x)( 3 \ + \ x^{2} ) }} }$$.
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 10.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 10
$$\\[1pt]$$
The variables $$x$$ and $$t$$ satisfy the differential equation:
$$\\[1pt]$$
$$\hspace{1.2em}{\large \frac{\mathrm{d}x}{\mathrm{d}t} } = x^{2} (1 + 2x)$$,
$$\\[1pt]$$
and $$x = 1$$ when $$t = 0$$.
$$\\[1pt]$$
Using partial fractions, solve the differential equation, obtaining an expression for $$t$$ in terms of $$x$$.
$$\\[1pt]$$
$$\\[1pt]$$
PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1.\enspace}$$ Express $${\small {\large \frac{7{x}^{2} \ – \ 3x \ + \ 2}{x({x}^{2} \ + \ 1) }} }$$ in partial fractions.
$${\small 2. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 6 }{(3 \ – \ 2x)({x}^{2} \ + \ 4 )}} }$$
Express $${\small f(x) }$$ in partial fractions.
$${\small 3. \enspace}$$ Express $${\small {\large \frac{2 \ – \ x \ + \ 8{x}^{2}}{(1 \ – \ x)(1 \ + \ 2x)(2 \ + \ x) }} }$$ in partial fractions.
$${\small 4. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ {x}^{2} \ + \ 3x \ + \ 3 }{(x \ + \ 1)(x \ + \ 3 )}} }$$
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Express f(x) in partial fractions.
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Hence show that,
$$\hspace{3em} {\small \displaystyle \int_{0}^{3} f(x) \ \mathrm{d}x = 3 \ – \ \frac{1}{2} \ln 2.}$$
$${\small 5. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ {x}^{2} \ – \ 8x \ + \ 9 }{(1 \ – \ x){(2 \ – \ x )}^{2}}} }$$
Express $${\small f(x) }$$ in partial fractions.
$${\small 6. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 2{x}^{2} \ – \ 7x \ – \ 1 }{(x \ – \ 2)({x}^{2} \ + \ 3 )}} }$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[12pt]{\small 7. \enspace}$$ Express $${\small {\large \frac{ x \ + \ 5 }{(x \ + \ 1)({x}^{2} \ + \ 3 )}} }$$ in the form:
$$\hspace{2em} {\large \frac{A}{( x \ + \ 1)} + \frac{Bx \ + \ C}{ ( {x}^{2} \ + \ 3) } }$$
$${\small 8. \enspace}$$ Express in partial fractions $${\small {\large \frac{{x}^{4}}{ {x}^{4} \ – \ 1 }} }$$
$${\small 9. \enspace}$$ Express in partial fractions $${\small {\large \frac{{x}^{3} \ + \ x \ – \ 1}{ {x}^{2} \ + \ {x}^{4} }} }$$
$$\\[10pt]{\small 10.\enspace}$$ Express in partial fractions
$$\hspace{2em}{\small {\large \frac{x \ + \ 5}{ {x}^{3} \ + \ 5{x}^{2} \ + \ 7x \ + \ 3 }} }$$
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
1. Mr will
X/(X^2+4^2)(X^2+b^2)
I was given to solve but I didn’t get it at all I love the way you solve just wish u got it more
2. Express (x+2)/x^2 (x-1) in partial fraction hence express (^3-x^2+x+2 )/x^2 (x-1)
3. Hi Mr Will
Solution for question no. 2
5x^2 + x + 6 / (3 – 2x) (x^2 + 4)
I got
A = 3
B = – 1
C = – 2
1. Yes, it’s correct!
Great job, Anonymous =)
Cheers,
Mr Will
2. 1/x^2(x+1)
4. Number 2
5. Integrate
No.1 1/x(x^n+1)dx
No.2(x^2+1)*(x^2+2)/(x^2+3)*(x^2+4)
6. May you please guide me more on the tips used to substitute for the value of x because it’s the most challenging part inorder to eliminate other coefficients . Thankyou.
7. Solutions for the practice questions
No.1
7x^2-3x +2 /x(x^+1)
I failed to get the value to substitute for x so as to eliminate A.
No.2
5x^2+6
(3-2x) (x^2+4)
Ans
Solving values for B and C was a problem
Mr Will am not at using computer ,therefore please some advise more especially when it comes at Mathematica.
From Judith, Have a Good Night
1. Hi Judith,
For no 1, there are two factors in the denominator: the linear factor, x and the quadratic factor that can’t be factorized, $${\small {x}^{2} \ + \ 1}$$.
$$\\[1pt]$$
Then we can use the appropriate forms,
$$\\[1pt]$$
$${\small {\large \frac{7{x}^{2} \ – \ 3x \ + \ 2}{x({x}^{2} \ + \ 1) } \equiv \frac{A}{x}+\frac{Bx \ + \ C}{ ({x}^{2} \ + \ 1) } } }$$
$$\\[1pt]$$
Next, multiply the denominators at the right hand side together.
$$\\[1pt]$$
$${\small {\large \frac{7{x}^{2} \ – \ 3x \ + \ 2}{x({x}^{2} \ + \ 1) } \equiv \frac{A({x}^{2} \ + \ 1) + (Bx \ + \ C)(x)}{x({x}^{2} \ + \ 1)} } }$$
$$\\[1pt]$$
Since the left hand side and the right hand side of the identity now have the same denominator, we can simply work on just the numerator.
$$\\[1pt]$$
$${\scriptsize 7{x}^{2} \ – \ 3x \ + \ 2 } \equiv {\small A({x}^{2} \ + \ 1) + (Bx \ + \ C)(x)}$$
$$\\[1pt]$$
We can use the substitution method,
$$\\[10pt]$$Let $${\small x = 0}$$,
$$\\[5pt] {\small 7(0)^2 – 3(0) + 2 = A(0^2 + 1) }$$
$$\\[5pt]\hspace{4.8 em} {\small 2 \ = \ A }$$
$$\\[12pt]\hspace{4.9 em} {\small A \ = \ 2 }$$
And we continue to substitute the x values to get B and C. Note that, you can pick any values of x arbitrarily. The idea is to get two simultaneous linear equations and solve for B and C.
$$\\[1pt]$$
$$\\[10pt]$$Let $${\small x = 1}$$,
$$\\[5pt] {\small 7(1)^2 – 3(1) + 2 = 2(1^2 + 1) \! + \! (B(1) + C)( 1 ) }$$
$$\\[5pt]\hspace{4.8 em} {\small 6 \ = \ 4 \ + B + C }$$
$$\\[12pt]\hspace{4.9 em} {\small 2 \ = \ B + C }$$
$$\\[10pt]$$Let $${\small x = -1}$$,
$$\\[5pt] {\small 7(-1)^2 – 3(-1) + 2 = 2({-1}^{2} + 1) \! + \! (B(-1) + C)( -1 ) }$$
$$\\[5pt]\hspace{4.8 em} {\small 12 \ = \ 4 \ + B \ – \ C }$$
$$\\[12pt]\hspace{4.9 em} {\small 8 \ = \ B \ – \ C }$$
By using substitution or elimination method, we can then find B and C,
$$\\[10pt] \hspace{2 em} {\small B \ = \ 5}$$
$$\\[10pt] \hspace{2 em} {\small C \ = \ -3}$$
$$\\[10pt]$$ Hence, f(x) in partial fractions is:
$$f(x) \ = \ \frac{2}{x}+\frac{5x \ – \ 3}{ ({x}^{2} \ + \ 1) }$$
$$\\[1pt]$$
As for no 2, you can try it again Judith, just remember you can use the x value that make the factor (3 – 2x) to be zero to find the coefficient A and choose any values of x to find the coefficient B and C.
Cheers,
Will
8. Hi Mr Will I am referring to the practice questions you gave after the presentation , so that I get to know what am practicing is write or not thankyou.
1. Hi Judith,
You can try the questions and maybe post your solutions here so I can take a look and see if there are any mistakes.
Cheers
9. please Mr Will solutions for the above question
1. Hi Judith,
which question are you referring about?
10. 5x/(x+3)(x-2)
1. $${\small {\large \frac{ 5x}{(x \ + \ 3)( x \ – \ 2 ) } \equiv \frac{A}{(x+3)}+\frac{B}{ (x \ – \ 2) } } }$$
$$\\[1pt]$$
$${\small {\large \frac{ 5x}{(x \ + \ 3)( x \ – \ 2 ) } \equiv \frac{A(x-2) \ + \ B(x+3)}{(x+3)(x \ – \ 2)} } }$$
$$\\[1pt]$$
$${\small 5x \equiv A(x-2) \ + \ B(x+3) }$$
$$\\[1pt]$$
$$\\[10pt]$$Let $${\small x = -3}$$,
$$\\[10pt]\hspace{1.2 em} {\small 5(-3) \ = \ A(-3 \ – \ 2) }$$
$$\\[10pt]\hspace{2.2 em} {\small -15 \ = \ -5A }$$
$$\hspace{3 em} {\small A \ = \ 3 }$$
$$\\[1pt]$$
$$\\[10pt]$$Let $${\small x = 2}$$,
$$\\[10pt]\hspace{1.2 em} {\small 5(2) \ = \ B(2 \ + \ 3) }$$
$$\\[10pt]\hspace{2.4 em} {\small 10 \ = \ 5B }$$
$$\hspace{3 em} {\small B \ = \ 2 }$$
$$\\[1pt]$$
$$\\[10pt]$$ Hence,
$$f(x) \ = \ \frac{3}{(x \ + \ 3)}+\frac{2}{ (x \ – \ 2) }$$
11. If the roots of an equation x^2 -5x- 7=0 are (alpha symbol) and (beta symbol) find the equation whose roots are:
(a).(alpha symbol)^2,(beta symbol)^2
(b).(alpha symbol +1),(beta symbol +1)
(c).(alpha symbol^2* beta symbol),(alpha symbol*beta symbol^2)
1. The quadratic equation in its general form:
$$\hspace{3em} {\small a{x}^{2} + bx + c \ = \ 0}$$
with its roots: $${\small \alpha \ \textrm{and} \ \beta}$$,
has the formula for:
$$\hspace{1.5em}$$ the sum of roots, $$\enspace {\small \alpha \ + \ \beta \ = \ -\frac{b}{a}}$$
$$\hspace{1.5em}$$ the product of roots, $$\enspace {\small \alpha \ \times \ \beta \ = \ \frac{c}{a}}$$
$$\\[1pt]$$
And also, the new quadratic equation can be found if you know the sum of roots (s.r) and the product of roots (p.r),
$$\enspace {\small {x}^{2} \ – \ \textrm{(s.r)}x \ + \ \textrm{(p.r)} \ = \ 0}$$
$$\\[1pt]$$
Now, back to the question,
$$\hspace{3em} {\small {x}^{2} – 5x – 7 \ = \ 0}$$
The sum of roots, $${\small \alpha \ + \ \beta \ = \ 5}$$
The product of roots, $${\small \alpha \ \times \ \beta = -7}$$
$$\\[1pt]$$
$$\enspace$$(a). $${\small {\alpha}^{2} \ + \ {\beta}^{2} \ = \ {(\alpha \ + \ \beta)}^2 \ – \ 2 \ \alpha \ \times \ \beta }$$
$$\hspace{3em} \ {\small = \ 5^2 \ – \ 2 (-7)}$$
$$\hspace{3em} \ {\small = \ 39 }$$
$$\\[1pt]$$
$$\hspace{1em} \ {\small {\alpha}^{2} \times \ {\beta}^{2} = {(\alpha \ \times \ \beta)}^{2} }$$
$$\hspace{3em} \ {\small = \ (-7)^2 }$$
$$\hspace{3em} \ {\small = \ 49 }$$
$$\\[1pt]$$
The new equation is $${\small x^2 \ – \ 39x \ + \ 49 = 0 }$$
$$\\[1pt]$$
$$\enspace$$(b). $${\small (\alpha + 1) \ + \ (\beta + 1) \ = \ \alpha \ + \ \beta \ + \ 2 }$$
$$\hspace{3em} \ {\small = \ 5 \ + \ 2 }$$
$$\hspace{3em} \ {\small = \ 7 }$$
$$\\[1pt]$$
$$\ {\small (\alpha + 1) \ \times \ (\beta + 1) = \alpha \times \beta + \alpha + \beta + 1 }$$
$$\hspace{3em} \ {\small = \ -7 \ + \ 5 \ + \ 1 }$$
$$\hspace{3em} \ {\small = \ -1 }$$
$$\\[1pt]$$
The new equation is $${\small x^2 \ – \ 7x \ – \ 1 = 0 }$$
$$\\[1pt]$$
$$\enspace$$(c). $${\small {\alpha}^{2}\beta \ + \ \alpha{\beta}^{2} \ = \ \alpha \beta (\alpha \ + \ \beta) }$$
$$\hspace{3em} \ {\small = \ -7 (5)}$$
$$\hspace{3em} \ {\small = \ -35 }$$
$$\\[1pt]$$
$$\hspace{1em} \ {\small {\alpha}^{2}\beta \ \times \ \alpha{\beta}^{2} = {(\alpha \ \times \ \beta)}^{3} }$$
$$\hspace{3em} \ {\small = \ (-7)^3 }$$
$$\hspace{3em} \ {\small = \ -243 }$$
$$\\[1pt]$$
The new equation is $${\small x^2 \ + \ 35x \ – \ 243 = 0 }$$
Cheers
|
Algebra and Trigonometry
Key Concepts
Algebra and TrigonometryKey Concepts
3.1Functions and Function Notation
• A relation is a set of ordered pairs. A function is a specific type of relation in which each domain value, or input, leads to exactly one range value, or output. See Example 1 and Example 2.
• Function notation is a shorthand method for relating the input to the output in the form $y=f( x ).y=f( x ).$ See Example 3 and Example 4.
• In tabular form, a function can be represented by rows or columns that relate to input and output values. See Example 5.
• To evaluate a function, we determine an output value for a corresponding input value. Algebraic forms of a function can be evaluated by replacing the input variable with a given value. See Example 6 and Example 7.
• To solve for a specific function value, we determine the input values that yield the specific output value. See Example 8.
• An algebraic form of a function can be written from an equation. See Example 9 and Example 10.
• Input and output values of a function can be identified from a table. See Example 11.
• Relating input values to output values on a graph is another way to evaluate a function. See Example 12.
• A function is one-to-one if each output value corresponds to only one input value. See Example 13.
• A graph represents a function if any vertical line drawn on the graph intersects the graph at no more than one point. See Example 14.
• The graph of a one-to-one function passes the horizontal line test. See Example 15.
3.2Domain and Range
• The domain of a function includes all real input values that would not cause us to attempt an undefined mathematical operation, such as dividing by zero or taking the square root of a negative number.
• The domain of a function can be determined by listing the input values of a set of ordered pairs. See Example 1.
• The domain of a function can also be determined by identifying the input values of a function written as an equation. See Example 2, Example 3, and Example 4.
• Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation. See Example 5.
• For many functions, the domain and range can be determined from a graph. See Example 6 and Example 7.
• An understanding of toolkit functions can be used to find the domain and range of related functions. See Example 8, Example 9, and Example 10.
• A piecewise function is described by more than one formula. See Example 11 and Example 12.
• A piecewise function can be graphed using each algebraic formula on its assigned subdomain. See Example 13.
3.3Rates of Change and Behavior of Graphs
• A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change is determined using only the beginning and ending data. See Example 1.
• Identifying points that mark the interval on a graph can be used to find the average rate of change. See Example 2.
• Comparing pairs of input and output values in a table can also be used to find the average rate of change. See Example 3.
• An average rate of change can also be computed by determining the function values at the endpoints of an interval described by a formula. See Example 4 and Example 5.
• The average rate of change can sometimes be determined as an expression. See Example 6.
• A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See Example 7.
• A local maximum is where a function changes from increasing to decreasing and has an output value larger (more positive or less negative) than output values at neighboring input values.
• A local minimum is where the function changes from decreasing to increasing (as the input increases) and has an output value smaller (more negative or less positive) than output values at neighboring input values.
• Minima and maxima are also called extrema.
• We can find local extrema from a graph. See Example 8 and Example 9.
• The highest and lowest points on a graph indicate the maxima and minima. See Example 10.
3.4Composition of Functions
• We can perform algebraic operations on functions. See Example 1.
• When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.
• The function produced by combining two functions is a composite function. See Example 2 and Example 3.
• The order of function composition must be considered when interpreting the meaning of composite functions. See Example 4.
• A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
• A composite function can be evaluated from a table. See Example 5.
• A composite function can be evaluated from a graph. See Example 6.
• A composite function can be evaluated from a formula. See Example 7.
• The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See Example 8 and Example 9.
• Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
• Functions can often be decomposed in more than one way. See Example 10.
3.5Transformation of Functions
• A function can be shifted vertically by adding a constant to the output. See Example 1 and Example 2.
• A function can be shifted horizontally by adding a constant to the input. See Example 3, Example 4, and Example 5.
• Relating the shift to the context of a problem makes it possible to compare and interpret vertical and horizontal shifts. See Example 6.
• Vertical and horizontal shifts are often combined. See Example 7 and Example 8.
• A vertical reflection reflects a graph about the $x- x-$ axis. A graph can be reflected vertically by multiplying the output by –1.
• A horizontal reflection reflects a graph about the $y- y-$ axis. A graph can be reflected horizontally by multiplying the input by –1.
• A graph can be reflected both vertically and horizontally. The order in which the reflections are applied does not affect the final graph. See Example 9.
• A function presented in tabular form can also be reflected by multiplying the values in the input and output rows or columns accordingly. See Example 10.
• A function presented as an equation can be reflected by applying transformations one at a time. See Example 11.
• Even functions are symmetric about the $y- y-$ axis, whereas odd functions are symmetric about the origin.
• Even functions satisfy the condition $f(x)=f(−x). f(x)=f(−x).$
• Odd functions satisfy the condition $f(x)=−f(−x). f(x)=−f(−x).$
• A function can be odd, even, or neither. See Example 12.
• A function can be compressed or stretched vertically by multiplying the output by a constant. See Example 13, Example 14, and Example 15.
• A function can be compressed or stretched horizontally by multiplying the input by a constant. See Example 16, Example 17, and Example 18.
• The order in which different transformations are applied does affect the final function. Both vertical and horizontal transformations must be applied in the order given. However, a vertical transformation may be combined with a horizontal transformation in any order. See Example 19 and Example 20.
3.6Absolute Value Functions
• Applied problems, such as ranges of possible values, can also be solved using the absolute value function. See Example 1.
• The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction. See Example 2.
• In an absolute value equation, an unknown variable is the input of an absolute value function.
• If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable. See Example 3.
3.7Inverse Functions
• If $g(x) g(x)$ is the inverse of $f(x), f(x),$ then $g(f(x))=f(g(x))=x. g(f(x))=f(g(x))=x.$ See Example 1, Example 2, and Example 3.
• Only some of the toolkit functions have an inverse. See Example 4.
• For a function to have an inverse, it must be one-to-one (pass the horizontal line test).
• A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.
• For a tabular function, exchange the input and output rows to obtain the inverse. See Example 5.
• The inverse of a function can be determined at specific points on its graph. See Example 6.
• To find the inverse of a formula, solve the equation $y=f(x) y=f(x)$ for $x x$ as a function of $y. y.$ Then exchange the labels $x x$ and $y. y.$ See Example 7, Example 8, and Example 9.
• The graph of an inverse function is the reflection of the graph of the original function across the line $y=x. y=x.$ See Example 10.
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What side do you read first on a graph?
What side do you read first on a graph?
To locate (2,5), move 2 units to the right on the x-axis and 5 units up on the y-axis. The order in which you write x- and y-coordinates in an ordered pair is very important. The x-coordinate always comes first, followed by the y-coordinate.
Do you read a line graph left to right?
A line chart (aka line plot, line graph) uses points connected by line segments from left to right to demonstrate changes in value. The horizontal axis depicts a continuous progression, often that of time, while the vertical axis reports values for a metric of interest across that progression.
How do you read a line graph example?
A data point on a line graph represents the quantity or a number that matches a particular time in the x-axis. In the example shown, the number of bicycles sold in the month of January is 50. Similarly, in the month of February 30 bicycles were sold. We can interpret this data for each month using the data point.
How do you describe a line on a graph?
The formal term to describe a straight line graph is linear, whether or not it goes through the origin, and the relationship between the two variables is called a linear relationship. Similarly, the relationship shown by a curved graph is called non-linear.
How do you properly graph?
1. Step 1: Identify the variables.
2. Step 2: Determine the variable range.
3. Step 3: Determine the scale of the graph.
4. Step 4: Number and label each axis and title the graph.
5. Step 5: Determine the data points and plot on the graph.
6. Step 6: Draw the graph.
How do you show greater than on a graph?
When graphing a linear inequality on a number line, use an open circle for “less than” or “greater than”, and a closed circle for “less than or equal to” or “greater than or equal to”.
What is the meaning of ≤?
≤ ≥ These symbols mean ‘less than or equal to’ and ‘greater than or equal to’ and are commonly used in algebra.
How do you make a line graph look better?
7 steps to make a professional looking line graph in Excel or…
1. Replace the legend with direct labels.
2. Remove gridlines or make them lighter.
3. Clean up the axes.
4. Consider selective data labels.
5. Add text that explains the message.
6. Increase font sizes so they are easy to read.
7. Use color to focus attention.
What are the 7 parts of a line graph?
What are the 7 parts of a graph?
• The Title. The title offers a short explanation of what is in your graph.
• The Source. The source explains where you found the information that is in your graph.
• X-Axis. Bar graphs have an x-axis and a y-axis.
• Y-Axis.
• The Data.
• The Legend.
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## NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations Ex 7.1
#### Question 1:
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
(i)
There will be as many ways as there are ways of filling 3 vacant places
in succession by the given five digits. In this case, repetition of digits is allowed. Therefore, the units place can be filled in by any of the given five digits. Similarly, tens and hundreds digits can be filled in by any of the given five digits.
Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125
(ii)
In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits. Therefore, the number of ways of filling the units place of the three-digit number is 5.
Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits.
Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60
#### Question 2:
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
There will be as many ways as there are ways of filling 3 vacant places
in succession by the given six digits. In this case, the units place can be filled by 2 or 4 or 6 only i.e., the units place can be filled in 3 ways. The tens place can be filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled by any of the 6 digits in 6 different ways, as the digits can be repeated.
Therefore, by multiplication principle, the required number of three digit even numbers is 3 × 6 × 6 = 108
#### Question 3:
How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
There are as many codes as there are ways of filling 4 vacant places in succession by the first 10 letters of the English alphabet, keeping in mind that the repetition of letters is not allowed.
The first place can be filled in 10 different ways by any of the first 10 letters of the English alphabet following which, the second place can be filled in by any of the remaining letters in 9 different ways. The third place can be filled in by any of the remaining 8 letters in 8 different ways and the fourth place can be filled in by any of the remaining 7 letters in 7 different ways.
Therefore, by multiplication principle, the required numbers of ways in which 4 vacant places can be filled is 10 × 9 × 8 × 7 = 5040
Hence, 5040 four-letter codes can be formed using the first 10 letters of the English alphabet, if no letter is repeated.
#### Question 4:
How many 5–digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
It is given that the 5-digit telephone numbers always start with 67.
Therefore, there will be as many phone numbers as there are ways of filling 3 vacant places by the digits 0 – 9, keeping in mind that the digits cannot be repeated.
The units place can be filled by any of the digits from 0 – 9, except digits 6 and 7. Therefore, the units place can be filled in 8 different ways following which, the tens place can be filled in by any of the remaining 7 digits in 7 different ways, and the hundreds place can be filled in by any of the remaining 6 digits in 6 different ways.
Therefore, by multiplication principle, the required number of ways in which 5-digit telephone numbers can be constructed is 8 × 7 × 6 = 336
#### Question 5:
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
When a coin is tossed once, the number of outcomes is 2 (Head and tail) i.e., in each throw, the number of ways of showing a different face is 2.
Thus, by multiplication principle, the required number of possible outcomes is 2 × 2 × 2 = 8
#### Question 6:
Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
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# Even and Odd Numbers
Sep 23, 2022
## Key Concepts
• Identify a group of objects that are even or odd
### Introduction
Count the number of ice creams and tell if the number is even or odd.
There are seven ice creams. Divide 7 ice creams into pairs to identify whether the number 7 is even or odd.
We cannot divide 7 into 2s because one ice cream is left in the above image. So, seven is an odd number.
Can you count the number of apples present in the image?
Yes, there are 8 apples, and we can divide 8 into 2s. So, 8 is an even number.
### Even Numbers
The objects which we can count by 2s are known as even numbers.
Count the squares by 2s.
Here we can count the squares by 2s like 2, 4, 6, and 8. So, 2, 4, 6 and 8 are even numbers.
Example:
Write whether the number of objects present in the image is even or odd
There are 8 objects present in the image. We can count the objects by 2s. So, 8 is even.
Example:
Write whether the number of objects present in the image is even or odd.
Solution:
There are 10 objects present in the image. We can count the objects by 2s. So, 10 is even.
Example:
Write whether the number of objects present in the image is even or odd.
Solution:
There are 16 objects present in the image. We can count the objects by 2s and here 8 pairs of shoes are there. So, 16 is an even number.
Example:
Write whether the number of objects present in the image is even or odd
Solution:
There are 8 objects present in the image. We can count the objects by 2s. So, 8 is even.
### Odd Numbers
The object which we cannot count by 2s is known as odd numbers.
Count the squares by 2s.
Here we cannot count by 2s. So, 3 and 5 are odd numbers.
Example:
Write whether the number of objects present in the image is even or odd.
Solution:
There are 7 objects present in the image. We cannot count the objects by 2s. So, 7 is odd.
Example:
Write whether the number of objects present in the image is even or odd.
Solution:
There are 5 objects present in the image. We cannot count the objects by 2s. So, 5 is odd.
Example:
Write whether the number of objects present in the image is even or odd.
Solution:
There are 7 objects present in the image. We cannot count the objects by 2s. So, 7 is odd.
Example:
Write whether the number of objects present in the image is even or odd.
Solution:
There are 9 objects present in the image. We cannot count the objects by 2s. So, 9 is odd.
Example:
Write whether the number of objects present in the image is even or odd.
Solution:
There are 11 objects present in the image. We cannot count the objects by 2s. So, 11 is odd.
## Exercise
• Write whether the number is even or odd.
• Write the equation for this model
• Count the number of squares present in the image and write if it is odd or even.
• Write whether the number of objects present in the image is even or odd.
• Count the number of squares present in the image and write if it is odd or even?
• Write the equation for this model.
• Write whether the number of objects present in the image is even or odd.
• Count the number of squares present in the image and write if it is odd or even?
• Write the equation for this model.
• Count the number of squares present in the image and write if it is odd or even?
### What we have learned
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
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# North Dakota - Grade 1 - Math - Numbers and Operations in Base Ten - Comparing Numbers - 1.NBT.3
### Description
Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.
• State - North Dakota
• Standard ID - 1.NBT.3
• Subjects - Math Common Core
### Keywords
• Math
• Numbers and Operations in Base Ten
## More North Dakota Topics
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
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# Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.14
## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.14
10th Maths Exercise 3.14 Solutions Question 1.
Write each of the following expression in terms of α + β and αβ.
Solution:
Ex 3.14 Class 10 Samacheer Question 2.
The roots of the equation 2x2 – 7x + 5 = 0 are α and β. Without solving the root find
Solution:
2x2 – 7x + 5 = x2 – $$\frac{7}{2} x+\frac{5}{2}$$ = 0
α + β = $$\frac{7}{2}$$
αβ = $$\frac{5}{2}$$
10th Maths Exercise 3.14 Question 3.
The roots of the equation x2 + 6x – 4 = 0 are α, β. Find the quadratic equation whose roots are
(i) α2 and β2
(ii) $$\frac{2}{\alpha} \text { and } \frac{2}{\beta}$$
(iii) α2β and β2α
Solution:
If the roots are given, the quadratic equation is x2 – (sum of the roots) x + product the roots = 0.
For the given equation.
x2 + 6x – 4 = 0
α + β = -6
αβ = -4
(i) α2 + β2 = (α + β)2 – 2αβ
= (-6)2 – 2(-4) = 36 + 8 = 44
α2β2 = (αβ)2 = (-4)2 = 16
∴ The required equation is x2 – 44x – 16 = 0.
(iii) α2β + β2α = αβ(α + β)
= -4(-6) = 24
α2β × β2α = α3β3 = (αβ)3 = (-4)3 = -64
∴ The required equation = x2 – 24x – 64 – 0.
10th Maths Exercise 3.14 Samacheer Kalvi Question 4.
If α, β are the roots of 7x2 + ax + 2 = 0 and if β – α = $$\frac{-13}{7}$$ Find the values of a.
Solution:
10th Maths Exercise 3.14 Solution Question 5.
If one root of the equation 2y2 – ay + 64 = 0 is twice the other then find the values of a.
Solution:
Let one of the root α = 2β
α + β = 2β + β = 3β
Given
a2 = 576
a = 24, -24
Exercise 3.14 Class 10 Samacheer Question 6.
If one root of the equation 3x2 + kx + 81 = 0 (having real roots) is the square of the other then find k.
Solution:
3x2 + kx + 81 = 0
Let the roots be α and α2
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# 8x 8 3ax 5ax 2a
This is a mathematical equation that can be used to solve for unknown variables. It is a quadratic equation, which means that it has two solutions. It is often used to solve problems in algebra and calculus. In this equation, 8x is the coefficient of the squared variable, 3ax is the coefficient of the linear variable, and 5ax is the constant term. The 2a is the coefficient of the squared variable.
To solve this equation, you will need to first identify the two solutions. To do this, you will need to use the quadratic formula. This formula will give you the two solutions of the equation. Using the quadratic formula, you can find the values of x that make the equation equal to zero. The quadratic formula is as follows:
x = -b +-√b^2 – 4ac/2a
## How To Use The Quadratic Formula
To use the quadratic formula, you will need to plug in the appropriate values for a, b, and c. In this case, a is 2, b is 3, and c is 5. Plugging these numbers into the equation, you will get the following: x= -3 ± √9 – 20/4. This equation simplifies to x = -3 ± √-11/4. Simplifying further, you will get the two solutions of x = 5 and x = -2.
## Interpreting The Solutions
The two solutions of x = 5 and x = -2 tell you that when you plug in either of these values into the equation, the equation will equal zero. This means that if you plug in either of these values into the equation, the equation will be true. The equation 8x 8 3ax 5ax 2a is equal to 0 when x is equal to 5 or -2. This means that the equation has been solved.
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# Preview Warm Up California Standards Lesson Presentation.
## Presentation on theme: "Preview Warm Up California Standards Lesson Presentation."— Presentation transcript:
Preview Warm Up California Standards Lesson Presentation
Warm Up Solve each proportion. = = 1. b = 10 2. y = 8 = = 3. 4. m = 52
30 3 9 = 56 35 y 5 = 1. b = 10 2. y = 8 4 12 p 9 = 56 m 28 26 = 3. 4. m = 52 p = 3
Preparation for MG1.2 Construct and read drawings and models made to scale.
California Standards
Vocabulary similar corresponding sides corresponding angles
Similar figures have the same shape, but not necessarily the same size.
Corresponding sides of two figures are in the same relative position, and corresponding angles are in the same relative position. Two figures are similar if the lengths of corresponding sides are proportional and the corresponding angles have equal measures.
43°
A is read as “angle A. ” ∆ABC is read as “triangle ABC
A is read as “angle A.” ∆ABC is read as “triangle ABC.” “∆ABC ~∆EFG” is read as “triangle ABC is similar to triangle EFG.” Reading Math
Additional Example 1: Identifying Similar Figures
Which rectangles are similar? Since the three figures are all rectangles, all the angles are right angles. So the corresponding angles are congruent.
Compare the ratios of corresponding sides to see if they are equal. length of rectangle J length of rectangle K width of rectangle J width of rectangle K 10 5 4 2 ? = 20 = 20 The ratios are equal. Rectangle J is similar to rectangle K. The notation J ~ K shows similarity. length of rectangle J length of rectangle L width of rectangle J width of rectangle L 10 12 4 5 ? = 50 48 The ratios are not equal. Rectangle J is not similar to rectangle L. Therefore, rectangle K is not similar to rectangle L.
Which rectangles are similar?
Check It Out! Example 1 Which rectangles are similar? 8 ft A 6 ft B C 5 ft 4 ft 3 ft 2 ft Since the three figures are all rectangles, all the angles are right angles. So the corresponding angles are congruent.
Check It Out! Example 1 Continued
Compare the ratios of corresponding sides to see if they are equal. length of rectangle A length of rectangle B width of rectangle A width of rectangle B 8 6 4 3 ? = 24 = 24 The ratios are equal. Rectangle A is similar to rectangle B. The notation A ~ B shows similarity. length of rectangle A length of rectangle C width of rectangle A width of rectangle C 8 5 4 2 ? = 16 20 The ratios are not equal. Rectangle A is not similar to rectangle C. Therefore, rectangle B is not similar to rectangle C.
Additional Example 2: Finding Missing Measures in Similar Figures
A picture 10 in. tall and 14 in. wide is to be scaled to 1.5 in. tall to be displayed on a Web page. How wide should the picture be on the Web page for the two pictures to be similar? Set up a proportion. Let w be the width of the picture on the Web page. width of a picture width on Web page height of picture height on Web page 14 w = 10 1.5 14 ∙ 1.5 = w ∙ 10 Find the cross products. 21 = 10w 10w 10 21 = Divide both sides by 10. 2.1 = w The picture on the Web page should be 2.1 in. wide.
Check It Out! Example 2 A painting 40 in. long and 56 in. wide is to be scaled to 10 in. long to be displayed on a poster. How wide should the painting be on the poster for the two pictures to be similar? Set up a proportion. Let w be the width of the painting on the Poster. width of a painting width of poster length of painting length of poster 56 w = 40 10 56 ∙ 10 = w ∙ 40 Find the cross products. 560 = 40w 40w 40 560 = Divide both sides by 40. 14 = w The painting displayed on the poster should be 14 in. long.
A T-shirt design includes an isosceles triangle with side lengths 4.5 in, 4.5 in., and 6 in. An advertisement shows an enlarged version of the triangle with two sides that are each 3 ft. long. What is the length of the third side of the triangle in the advertisement? side of small triangle base of small triangle side of large triangle base of large triangle = 3 ft x ft 4.5 in. 6 in. = Set up a proportion. 4.5 • x = 3 • 6 Find the cross products. 4.5x = 18 Multiply.
A T-shirt design includes an isosceles triangle with side lengths 4.5 in, 4.5 in., and 6 in. An advertisement shows an enlarged version of the triangle with two sides that are each 3 ft. long. What is the length of the third side of the triangle in the advertisement? x = = 4 18 4.5 Solve for x. The third side of the triangle is 4 ft long.
Check It Out! Example 3 A flag in the shape of an isosceles triangle with side lengths 18 ft, 18 ft, and 24 ft is hanging on a pole outside a campground. A camp t-shirt shows a smaller version of the triangle with two sides that are each 4 in. long. What is the length of the third side of the triangle on the t-shirt? side of large triangle side of small triangle base of large triangle base of small triangle = 24 ft x in. 18 ft 4 in. = Set up a proportion. 18 ft • x in. = 24 ft • 4 in. Find the cross products. 18x = 96 Multiply.
Check It Out! Example 3 Continued
A flag in the shape of an isosceles triangle with side lengths 18 ft, 18 ft, and 24 ft is hanging on a pole outside a campground. A camp t-shirt shows a smaller version of the triangle with two sides that are each 4 in. long. What is the length of the third side of the triangle on the t-shirt? x = 5.3 96 18 Solve for x. The third side of the triangle is about 5.3 in. long.
Lesson Quiz Use the properties of similar figures to answer each question. 1. Which rectangles are similar? A and B are similar. 2. Karen enlarged a 3 in. wide by 5 in. tall photo into a poster. If the poster is 2.25 ft wide, how tall is it? 3.75 ft 3. A rectangular house is 32 ft wide and 68 ft long. On a blueprint, the width is 8 in. Find the length on the blueprint. 17 in.
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# The Calendar Trick
This trick makes you look like you've got the most awesome mental power. All you need is a calendar which has the dates lined up under the days of the week, so the numbers are arranged something like this:
. . 1 2 3 4 5
6 7 8 9 10 11 12
13 14 15 16 17 18 19
20 21 22 23 24 25 26
27 28 29 30 31
### The "9 number" trick
• Ask a friend to draw a 3x3 box around ANY nine of the numbers on the calendar.
• Almost immediately you can say what the nine numbers all add up to!
• See how long it takes your friend to check the answer on a calculator. (Yawn...snore...zzzz...)
For instance, let's say the red numbers are chosen.
. . 1 2 3 4 5
6 7 8 9 10 11 12
13 14 15 16 17 18 19
20 21 22 23 24 25 26
27 28 29 30 31
THE SECRET: All you do is look at the number in the middle and multiply it by 9.
In this case the middle number is 22 and 22 x 9 = 198.
HANDY TIP: To multiply by 9 quickly, just multiply by 10 then subtract your number.
So to get 22 x 9 you multiply 22 x 10 =220 (easy!) and then subtract 22. With a bit of practise you can do this quickly in your head.
### The "20 number" trick
• Ask a friend to draw a 5x4 box around ANY twenty of the numbers on the calendar.
• Almost immediately you can say what all twenty numbers add up to!
For instance, let's say the green numbers are chosen.
###### Thanks to Louise Lennartsson who spotted a mistake in the calendar grid on the left (which we've corrected). We estimate that over 50,000 people had already viewed this page before she saw it!
. 1 2 3 4 5 6
7 8 9 10 11 12 13
14 15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30 31
THE SECRET: Add together the smallest and the largest numbers in the group. Multiply the answer by 10.
In this case the smallest number is 2 and the largest is 27, so 2 + 27 = 29. Then 29 x 10 = 290.
With the "20 number trick" if the calendar month is February occasionally it will not be possible to put a box round 20 numbers.
### And if you don't want to use a calendar....
These tricks work with ANY grid of numbered boxes, just as long as the numbers are continuous. Try them both on this grid:
13 14 15 16 17 18 19 20 21 22 23
24 25 26 27 28 29 30 31 32 33 34
35 36 37 38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64 65 66 67
68 69 70 71 72 73 74 75 76 77 78
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# Lesson 10
Interpretemos relaciones
## Warm-up: Verdadero o falso: Multiplicar y dividir (10 minutes)
### Narrative
The purpose of this True or False is for students to demonstrate understandings they have for the relationship between multiplication and division. Students will use this understanding in the lesson when they recognize multiplicative relationships between patterns.
### Launch
• Display one statement.
• “Hagan una señal cuando sepan si la afirmación es verdadera o no, y puedan explicar cómo lo saben” // “Give me a signal when you know whether the statement is true and can explain how you know.”
• 1 minute: quiet think time
### Activity
• Share and record answers and strategy.
• Repeat with each problem.
### Student Facing
En cada caso, decide si la afirmación es verdadera o falsa. Prepárate para explicar cómo razonaste.
• $$276 \div 3 = \frac{1}{3} \times 276$$
• $$276 \div 3 = \frac{276}{6}$$
• $$(276 \div 3)\times2=\frac{2}{3}\times 276$$
### Activity Synthesis
• “Para justificar su razonamiento, ¿cómo les puede ayudar la relación que hay entre la multiplicación y la división?” // “How can the relationship between multiplication and division help you justify your reasoning?” (I used the fact that dividing by 3 is the same as multiplying by $$\frac{1}{3}$$.)
## Activity 1: Completemos y relacionemos: 3 patrones (20 minutes)
### Narrative
The purpose of this activity is for students to practice interpreting relationships between patterns generated from two different rules. Students may need to generate patterns beyond the boxes provided. Encourage them to continue the pattern as needed. Students may describe the patterns and relationships in different, but accurate ways. Encourage them to notice as many relationships as they can and describe the relationships in whatever way makes sense to them. This is the first time students see two patterns where the numbers in one pattern are not multiples of the numbers in the other pattern.
This activity uses MLR7 Compare and Connect. Advances: Representing, Conversing.
• Groups of 2
### Activity
• 5 minutes: independent work time
• 2 minutes: partner discussion
MLR7 Compare and Connect
• “Creen una presentación visual que muestre lo que pensaron sobre las relaciones que hay entre cada grupo de patrones. Incluyan detalles, como notas, diagramas y dibujos, para ayudar a los demás a entender sus ideas” // “Create a visual display that shows your thinking about the relationships between each set of patterns. Include details such as notes, diagrams, and drawings to help others understand your thinking.”
• 2–5 minutes: independent or group work
• 3–5 minutes: gallery walk
### Student Facing
1. Completa los patrones para cada grupo de reglas.
2. ¿Qué relaciones hay entre los patrones de cada grupo de reglas? Prepárate para explicar cómo pensaste.
Grupo A
Regla 1: empezar en 0 y siempre sumar 3.
Regla 2: empezar en 0 y siempre sumar 6.
Grupo B
Regla 1: empezar en 4 y siempre sumar 3.
Regla 2: empezar en 9 y siempre sumar 6.
Grupo C
Regla 1: empezar en 0 y siempre sumar 5.
Regla 2: empezar en 0 y siempre sumar 3.
### Activity Synthesis
• “¿En qué se parecen y en qué son diferentes las formas en las que representamos las relaciones que hay entre los patrones?” // “What is the same and what is different in the way we represented the relationships between the patterns?” (Some of us used numbers and symbols, some of us wrote sentences.)
• 30 seconds: quiet think time
• 1 minute: partner discussion
• “¿En qué se parecen y en qué son diferentes el grupo A y el grupo B?” // “What is the same and what is different about Set A and Set B?” (They both add the same amount, but one set starts at 0. Set B has a lot more odd numbers in it. In set A, each number in rule 2 is double the number in rule 1, but in set B, each number in rule 2 is 1 more than double the number in rule 1.)
• “¿Para cuál pareja de reglas fue más retador observar y describir las relaciones?” // “For which pair of rules was it most challenging to notice and describe relationships?” (Set B because there was no multiple that worked to get from one pattern to the other. Set C because there weren’t doubles or halves.)
## Activity 2: Generemos patrones (15 minutes)
### Narrative
The purpose of this activity is for students to interpret more complex relationships in corresponding terms from patterns generated from two different rules. Both sets of rules generate patterns that have the same relationships between corresponding terms. Each of the terms in rule 2 is $$\frac{3}{2}$$ times greater than each of the corresponding terms in rule 1 and each of the terms in rule 1 is $$\frac{2}{3}$$ the corresponding term in rule 2. Some students may state these relationships in different ways. The relationships between these patterns build directly on the third pair of rules from the previous activity where the terms for rule 1 were $$\frac{5}{3}$$ the corresponding terms in rule 2.
Representation: Internalize Comprehension. Synthesis: Use multiple examples and non-examples to emphasize the relationship between corresponding terms in patterns.
Supports accessibility for: Attention, Memory, Conceptual Processing
### Launch
• Groups of 2
• “Ustedes y su compañero van a trabajar individualmente en algunos problemas sobre patrones. Cuando acaben, discutan su trabajo con su compañero” // “You and your partner will each complete some problems about patterns independently. After you’re done, discuss your work with your partner.”
### Activity
• 5–7 minutes: independent work time
• 3–5 minutes: partner discussion
• “Revisen su trabajo y hagan los ajustes necesarios basándose en lo que aprendieron durante la discusión con su compañero” // “Look back at your work and make any revisions based on what you learned from your discussion.”
• 1–2 minutes: independent work time
• Monitor for students who:
• recognize that each term in pattern 2 is $$1\frac{1}{2}$$ times the corresponding term in pattern 1
• recognize that each term in pattern 1 is $$\frac{2}{3}$$ the corresponding term in pattern 2
### Student Facing
Compañero A
1. Genera patrones para las dos reglas.
Regla 1: empezar en 0 y siempre sumar 4.
Regla 2: empezar en 0 y siempre sumar 6.
2. Compara tus patrones. ¿Qué relaciones observas?
3. ¿Qué número saldrá en el patrón 2 cuando el número 40 salga en los cuadros del patrón 1?
4. ¿Qué número saldrá en el patrón 1 cuando el número 120 salga en los cuadros del patrón 2?
Compañero B
1. Genera patrones para las dos reglas.
Regla 1: empezar en 0 y siempre sumar 2.
Regla 2: empezar en 0 y siempre sumar 3.
2. Compara tus patrones. ¿Qué relaciones observas?
3. ¿Qué número saldrá en el patrón 2 cuando el número 30 salga en los cuadros del patrón 1?
4. ¿Qué número saldrá en el patrón 1 cuando el número 60 salga en los cuadros del patrón 2?
### Activity Synthesis
• Ask previously selected students to share their thinking.
• Display or write the numbers in the patterns for partner A.
• “¿Cómo podemos representar la relación que hay entre los números de los patrones usando ecuaciones de multiplicación?” // “How can we represent the relationship between the numbers in the patterns with multiplication equations?” (Each number in rule 2 is $$\frac{3}{2}$$ the corresponding number in rule 1. Each number in rule 1 is $$\frac{2}{3}$$ the corresponding number in rule 2: $$6=\frac{3}{2}\times 4$$, $$4= \frac{2}{3}\times 6$$.)
• Display or write the numbers in the patterns for partner B.
• “¿Cómo podemos representar las relaciones que hay entre los números de los patrones usando ecuaciones de multiplicación?” // “How can we represent the relationships between the numbers in the patterns with multiplication equations?” ($$3=1\frac{1}{2}\times 2$$ , $$2=\frac{2}{3}\times 3$$.)
## Lesson Synthesis
### Lesson Synthesis
“Hoy observamos y explicamos relaciones entre patrones. Algunas de las relaciones involucraban fracciones” // “Today we noticed and explained relationships between patterns. Some of the relationships involved fractions.”
“¿Qué relaciones encontraron entre los patrones que estudiamos hoy?” // “What relationships did you find between the patterns we studied today?” (Sometimes I could multiply each term in one pattern by the same number to get the corresponding number in the other pattern. Sometimes that number was a fraction.)
Consider asking students to record their response in a math journal and then share their response with a partner.
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GreeneMath.com - Equations Lesson
# In this Section:
In this section, we start digging into the basics of algebra and introduce the equation. We learn how to tell the difference between an equation and an algebraic expression. When we have an equation, we have two algebraic expressions that are separated by an equals sign. This equation can be solved and the answer can be checked. When we have an algebraic expression, there is no equality symbol. We only gain a value for the algebraic expression if a value for the variable is given. In this case only, we can substitute in for the variable and obtain a value for the algebraic expression. We then move on and learn how to determine if a given value is a solution to an equation. We do this by replacing the variable in the equation with our proposed solution. If after simplifying, the left and the right side are equal, meaning the same value, then our proposed solution is an actual solution.
Sections:
# In this Section:
In this section, we start digging into the basics of algebra and introduce the equation. We learn how to tell the difference between an equation and an algebraic expression. When we have an equation, we have two algebraic expressions that are separated by an equals sign. This equation can be solved and the answer can be checked. When we have an algebraic expression, there is no equality symbol. We only gain a value for the algebraic expression if a value for the variable is given. In this case only, we can substitute in for the variable and obtain a value for the algebraic expression. We then move on and learn how to determine if a given value is a solution to an equation. We do this by replacing the variable in the equation with our proposed solution. If after simplifying, the left and the right side are equal, meaning the same value, then our proposed solution is an actual solution.
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# Integral Definition
## Introduction
The integral is a fundamental concept in mathematics that may be used in a wide variety of contexts. It is absolutely necessary for anybody who wants to go further into mathematics or subjects linked to mathematics, such as physics and engineering. In this article, we will talk about integral, how it works, the many types of integrals, and some popular uses of this powerful tool.
## Definition of Integral
The values of a function that are discovered during the integration process are called integrals. "Integration" refers to the procedure through which f(x) is derived from f'(x). Problems of displacement, velocity, area, volume, etc., that result from the combination of many, discrete data points may be described by the numbers assigned by an integral. We are able to identify the function "f" if we are given its derivative, which is denoted by the letter f'. In this situation, the function denoted by f is referred to as the antiderivative or integral of f'.
Example: Given: f(x) = x2.
Derivative of f(x) = f'(x) = 2x = g(x)
if g(x) = 2x, then anti-derivative of g(x) = ∫ g(x) = x2
## Types of Integrals
The following categories of challenges may be solved with the assistance of integral calculus:
1. The challenge of determining a function when only its derivative is known.
2. The challenge in determining the region enclosed by the graph of a function for a given set of constraints. Because of this, the integral calculus may be broken down into two distinct subfields.
Definite Integrals (the value of the integrals are definite)
Indefinite Integrals (the value of the integral is indefinite with an arbitrary constant, C)
### Definite Integrals
The region under a curve that is exactly between two given points defines a definite integral. The definite integral is denoted by the notation ∫baf(x)dx, where "a" represents the lower limit and "b" represents the upper limit, for a function f(x) that is defined with reference to the x-axis. In order to calculate the area under a curve that is between two limits, we first need to split the area into rectangles and then add up each of those rectangles. The area may be estimated with a higher level of accuracy if there is a greater number of rectangles used. To do this, we first divide the area into an unlimited number of rectangles, each of which has the same very tiny size, and then we sum up all of the areas. This is the foundational principle that underpins the concept of definite integrals.
When it comes to determining the value of a definite integral, we have two different formulae. The first formula is known as the "definite integral as a limit sum," while the second formula is known as the "fundamental theorem of calculus."
• ∫baf (x) d x = lim n → ∞ ∑ n r = 1 h f (a + r h) , where h = b - a/ n
• ∫ b a f ( x ) d x = F ( b ) − F ( a ) , where F'(x) = f(x)
### Indefinite Integrals
These are integrals for which there is no known limit value, hence the final integral value is indefinite. For integrating algebraic equations, trigonometric functions, logarithmic functions, and exponential functions, indefinite integrals are what you'll need to employ. Thus, g'(x) is the derivative solution, which when integrated gives the original function of g(x). Since the integration does not return the constant value of the initial expression, a constant letter 'c' is added to the response produced by the integral.
## Implementations and Applications of Integrals
Calculation may be accomplished in a wide variety of ways, some of the more common ones being functions, differentiation, and integration. Integrals have several uses in other areas of study, including mathematics, science, and engineering. In addition, we make heavy use of integral formulae whenever we need to compute the areas of irregular shapes or areas that are contained inside a two-dimensional space.
The following is a brief overview to the many uses of integrals:
Integrals and Their Applications in Mathematics
• Determine the area under simple curves.
• The space between the lines and arcs of a circle, an ellipse, or a parabola.
• Find the centroid of a region, such as the centroid of a triangle with rounded sides.
• Determine the region between two curves.
Integrals and Their Applications in Physical Theory
• Centre of gravity
• Center of mass
• The mass and momentum of satellites
• thrust
Integrals and Their Applications in Engineering Mathematics
• Assessment of double and triple integrals.
• The tracing of curves and the determination of the areas of cartesian and polar curves.
• Estimate of the length of the curve and the determination of the volume of revolution around the axis.
• Estimating the area of curve.
Integrals' Everyday Applications in a Variety of Situations
• Integrals are used in the field of medical research to find the development of germs in the laboratory by assigning factors such as a change in temperature and diet.
• In the subject of epidemiology, medical research is used to find out how quickly a disease is spreading, where it originated, and how to treat it in the most effective way possible. Integrals are used in the field of medicine to analyze the pace at which an infectious illness is spreading.
• Neurology is the study of the human nervous system, which is a highly complex network of neurons that transmit messages from the brain to the rest of the body and regulate all bodily functions. The voltage that is present in a neuron at a certain location may be calculated using integral calculus.
• As a survey requires a wide variety of questions with a wide range of potential answers, statisticians employ integrals to analyze survey data in order to enhance marketing tactics for various businesses.
• Calculus, in addition to having applications in engineering, medical science, and other fields of study of a similar kind, also has applications in music, particularly in the areas of harmonic and acoustic.
## Conclusion
Students who master integral calculus may expect to enjoy a number of benefits in their academic and professional lives. It is not only important for early academic success, but it also has the potential to open up a number of other career opportunities. The examples that have been shown show how adaptable and useful this topic may be in many contexts outside of the classroom. Students may utilize these abilities to their advantage while seeking for new jobs or expanding their existing profession if they have a basic understanding of the topic at hand. In conclusion, calculus is a powerful tool that every student need to consider mastering.
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## Elementary Linear Algebra 7th Edition
The solution of this system is unique and it is: $$x=\cos\theta-\sin\theta,\quad y= \cos\theta + \sin\theta.$$
Here we can treat sines and cosines as ordinary coefficients and we will solve this system as we would solve any system of two equations with two unknowns. Step 1: Multiply the first equation by $\sin\theta$ and the second equation by $\cos\theta$: \begin{align*} (\sin\theta\cos\theta)x+(\sin^2\theta)y=&\sin\theta\\ (-\sin\theta\cos\theta)x+(\cos^2\theta)y =& \cos\theta. \end{align*} Step 2: Add these two equations together to eliminate $x$ and find $y$: $$(\sin\theta\cos\theta)x +(-\sin\theta\cos\theta)x + (\sin^2\theta)y +(\cos^2\theta)y = \sin\theta+\cos\theta$$ which becomes $$y(\sin^2\theta+\cos^2\theta) = \sin\theta+\cos\theta.$$ Remember that for any $\theta$ we have that $\sin^2\theta+\cos^2\theta = 1$ so this simplifies to $$y=\sin\theta + \cos\theta$$ Step 3: Now let us do the similar thing as in step 1. Multiply the first equation by $\cos\theta$ and the second one by $\sin\theta$: \begin{align*} (\cos^2\theta)x+ (\sin\theta\cos\theta) y = &\cos\theta\\ (-\sin^2\theta)x+(\sin\theta\cos\theta)y = &\sin\theta \end{align*} Step 4: Now subtract these equations to eliminate $y$ and find $x$ : $$(\cos^2\theta)x-(-\sin^2\theta)x + (\sin\theta\cos\theta) y -(\sin\theta\cos\theta)y =\cos\theta- \sin\theta$$ This then gives $$x(\sin^2\theta+\cos^2\theta) = \cos\theta-\sin\theta.$$ Using the fact that $\sin^2\theta +\cos^2\theta = 1$ we again get simpler expression: $$x=\cos\theta-\sin\theta.$$
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10
Q:
# 14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.
A) 11/379 B) 21/628 C) 24/625 D) 26/247
Explanation:
Total no of ways = (14 – 1)! = 13!
Number of favorable ways = (12 – 1)! = 11!
So, required probability = $\left(\frac{\left(\mathbf{11}\mathbf{!}\mathbf{×}\mathbf{3}\mathbf{!}\right)}{\mathbf{13}\mathbf{!}}\right)$ = $\frac{39916800×6}{6227020800}$ = $\frac{\mathbf{24}}{\mathbf{625}}$
Q:
In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color?
A) 2/9 B) 5/9 C) 4/9 D) 0
Explanation:
Number of white marbles = 4
Number of Black marbles = 5
Total number of marbles = 9
Number of ways, two marbles picked randomly = 9C2
Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2
= 1/6 + 5/18
= 4/9.
0 175
Q:
A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?
A) 2/3 B) 1/8 C) 3/8 D) 3/4
Explanation:
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
5 210
Q:
Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?
A) 1/4 B) 1/6 C) 1/8 D) 4
Explanation:
Required probability is given by P(E) =
9 411
Q:
Two dice are rolled simultaneously. Find the probability of getting the sum of numbers on the on the two faces divisible by 3 or 4?
A) 3/7 B) 7/11 C) 5/9 D) 6/13
Explanation:
Here n(S) = 6 x 6 = 36
E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}
=> n(E)=20
Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.
23 768
Q:
A person starting with 64 rupees and making 6 bets, wins three times and loses three times, the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is:
A) A gain of Rs. 27 B) A loss of Rs. 37 C) A loss of Rs. 27 D) A gain of Rs. 37
Explanation:
As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).
The overall resultant will remain same.
So final amount with the person will be (in all cases):
64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27
Hence the final result is:
64 − 27 37
A loss of Rs.37
12 855
Q:
A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of the heart is?
A) 1/26 B) 1/13 C) 2/13 D) 1/52
Explanation:
Here in this pack of cards, n(S) = 52
Let E = event of getting a queen of the club or a king of the heart
Then, n(E) = 2
P(E) = n(E)/n(S) = 2/52 = 1/26
8 694
Q:
A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ?
A) 14/33 B) 14/55 C) 12/55 D) 13/33
Explanation:
Total number of chairs = (3 + 5 + 4) = 12.
Let S be the sample space.
Then, n(s)= Number of ways of picking 2 chairs out of 12
12×11/2×66
Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.
=> 8C8×7/2×28
Therefore required probability = 28/66 = 14/33.
13 1154
Q:
The probability that a bowler bowled a ball from a point will hit by the batsman is ¼. Three such balls are bowled simultaneously towards the batsman from that very point. What is the probability that the batsman will hit the ball ?
A) 37/64 B) 27/56 C) 11/13 D) 9/8
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Other Number Properties: Identities,
Inverses, Symmetry, etc.
(page 2 of 2)
If your textbook gets really ornate, you may have to delve into some of the more esoteric properties of numbers. For this, you need to know that "the identity" is whatever doesn't change your number at all, and "the inverse" is whatever turns your number into the identity.
For addition, "the identity" is zero, because adding zero to anything doesn't change anything. The "inverse" is the additive inverse: it's the same number, but with the opposite sign. For instance, suppose your number is –6, and you're adding. The identity is zero, and the inverse is 6, because –6 + 6 = 0.
For multiplication, "the identity" is one, because multiplying by one doesn't change anything. The "inverse" is the multiplicative inverse: the same number, but on the opposite side of the fraction line. For instance, suppose your number is –6, and you're multiplying. The identity is one, and the inverse is
–1/6, because (–6)( –1/6 ) = 1.
You also know (if you've done any equation solving) that you can do anything you want to an equation, as long as you do the same thing to both sides. This is the "property of equality".
The basic fact that you need for solving many equations, especially quadratics, is that, if p×q = 0, then must have either p = 0 or else q = 0. The only way you can multiply two things and end up with zero is if one (or both) of those two things was zero to start with. This is the "zero-product property".
And there are some properties that you use to solve word problems, especially where substitution is required. Anything equals itself: this is the "reflexive" (reflecting onto itself) property. Also, it doesn't matter which order the equality is in; if x = y, then y = x: this is the "symmetric" (they match) property. You can "cut out the middleman", so to speak; if x = y and y = z, then you can say that x = z: this is the "transitive" (moving across) property. Two numbers are either equal to each other or unequal; this is the "trichotomy" law (so called because there are three cases for two given numbers, a < b, a = b,
or
a > b). And you can plug in for variables, so if x = 3, then 4x = 12, because 4x = 4(3): this is the "substitution" property.
Here are some examples. Note: textbooks vary somewhat in the names they give these properties; you'll need to refer to the examples in your book to know the exact format you should use.
Determine which property was used.
• 1×7 = 7
They multiplied, and they didn't change anything: the multiplicative identity.
• –7y = –7y
This is obvious: anything equals itself. They used the reflexive property.
• If 10 = y, then y = 10.
When solving an equation, I might rearrange things so I end up with the variable on the left. But I only switched sides; I didn't actually change anything: the symmetric property.
• x + 0 = x
• If 2(a + b) = 3c, and a + b = 9, then 2(9) = 3c.
You might be torn here between the transitive property and the substitution property. If you look closely, what they did was substitute "9" for "a + b", so they used the substitution property.
• 2 = x, so 2 + 5 = x + 5
They did the backwards of solving an equation, but the point is that they were working with an equation. They changed the equation by adding equal things to both sides: the additive property of equality. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
• If x + 2 = 10, then x + 2 + (–2) equals what, and why?
They solved the equation by getting rid of the 2 from both sides. Since they added the same thing to both sides, they got x = 8 by the additive property of equality.
• (x – 3)(x + 4) = 0, so x = 3 or x = –4.
They set the quadratic equal to zero, factored, and then solved each factor: the zero-product property.
• 4x = 8, so x = 2
They solved the equation by dividing both sides by 4, or, which is the same thing, multiplying both sides by ( 1/4 ). In other words, they changed the equation by doing the same multiplying to both sides: the multiplicative property of equality.
• If x is not equal to y and not less than y, what must be true of x, and why?
• By the trichotomy law, there are only three possible relationships between x and y, and they've eliminated two of them. Then x > y, by the trichotomy law.
• x + (–x) = 0
• ( 3/3 )( 2/5 ) + ( 5/5 )( 4/3 ) = 6/1520/15
They converted to a common denominator by multiplying both fractions by a useful form of 1; remember that 3/3 and 5/5 are just 1! So they used the multiplicative identity.
• If 5x = 0, what is x, and why?
You can do this in either of two ways: multiply both sides by 1/5 (the multiplicative property of equality) and then get x = 0, or you could say that, since 5 doesn't equal zero, then x must equal zero (by the zero-product property).
• ( 2/3 )( 3/2 ) = 1
They multiplied, and they ended up with one: the multiplicative inverse.
• If 3x + 2 = y and y = 8, then 3x + 2 = 8.
You might be torn here between the transitive property and the substitution property. What they did here was "cut out the middleman" by removing the "y" in the middle, so they used the transitive property.
• If x = 14, what does x equal, and why?
To solve this, you would multiply both sides by a negative one, to cancel off the minus sign. So:
x = –14, by the multiplicative property of equality.
• If x = 3 and y = –4, then what does xy equal, and why?
By substitution (plugging in for the variables), you get (3)(–4). In other words:
xy = –12, by the substitution property.
• Can x < x? Why or why not?
• By the reflexive property, x = x. By the trichotomy law, if a = b then a cannot be less than b. So the answer is "no, by the reflexive property and the trichotomy law".
Don't let the seeming pointlessness of these questions bother you. Instead, view this stuff as "gimme" questions for the next test.
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# How do you simplify Ln(-e) - ln(-1/e)?
Mar 23, 2016
This expression is not valid because $\ln$ function is only defined for positive numbers.
#### Explanation:
This expression is not valid because $\ln$ is not defined for negative numbers but if it was:
$\ln \left(e\right) - \ln \left(\frac{1}{e}\right)$
then you could use a formula: $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$ which would lead to:
$\ln \left(e\right) - \ln \left(\frac{1}{e}\right) = \ln \left(\frac{e}{\frac{1}{e}}\right) = \ln \left(e \cdot e\right) = \ln \left({e}^{2}\right) = 2 \cdot \ln e = 2$
In the last transformation I used an identity: $\ln \left({a}^{b}\right) = b \cdot \ln a$
Aug 10, 2017
$\ln \left(- e\right) - \ln \left(- \frac{1}{e}\right) = 2$
#### Explanation:
In order to simplify this, we need to use the difference law:
$\log a - \log b \equiv \log \left(\frac{a}{b}\right)$
$\ln \left(- e\right) - \ln \left(- \frac{1}{e}\right) = \ln \left(\frac{- e}{- \frac{1}{e}}\right) = \ln \left({e}^{2}\right) = 2 \ln e = 2$
We could also use the exponent and addition laws:
$b \log a \equiv \log \left({a}^{b}\right)$ and $\log a + \log b \equiv \log \left(a b\right)$
$\ln \left(- e\right) - \ln \left(- \frac{1}{e}\right) = \ln \left(- e\right) + \ln \left({\left(- \frac{1}{e}\right)}^{-} 1\right)$
$= \ln \left(- e\right) + \ln \left(- e\right) = \ln \left({e}^{2}\right) = 2$
Note that taking the $\log$ of negative numbers is only valid over the complex field.
Aug 16, 2017
If we are working in $\mathbb{C}$ and using principal values of $\ln \left(z\right)$, then it simplifies to $2$
#### Explanation:
$\ln \left(- e\right) = 1 + \pi i$
$\ln \left(- \frac{1}{e}\right) = - 1 + \pi i$
$\ln \left(- e\right) - \ln \left(- \frac{1}{e}\right) = \left(1 + \pi i\right) - \left(- 1 + \pi i\right) = 2 + 0 i = 2$
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## Worksheet on complex roots
In this worksheet on complex roots we are going to see problems in complex roots.
Complex roots: While solving quadratic equations, we get sometimes unreal roots(imaginary roots).
While solving quadratic equations, we will find the value of the discriminant to find the nature of the roots.
Discriminant ∆= b² - 4 ac (1) ∆ > 0 but not a perfect square (2) ∆ > 0 but a perfect square (3) ∆ = 0 (4) ∆ < 0 but a perfect square Nature of roots Real,unequal and irrational Real,unequal and rational Real, equal and rational Complex roots
Examples:
1. Solve the equation :
x²/2 = 3x - 5
Solution:
x² = 2(3x-5)
x² = 6x -10
x² - 6x +10 = 0.
a = 1, b = -6, c = 10
The value of x is
x = -(-6) ± √[(-6)² - 4(1)(10)]
2(1)
x = [6 ± √(36-40)]/2
= [6 ± √(-4)]/2
= (6 ± 2i)/2
= 3 ± i
So x = 3 + i, or 3-i
2. Find the roots of the equation:
x + 5/x = 3
Solution:
Rewriting the equation,
x(x + 5/x ) = 3x
x² + 5 = 3x
x² - 3x + 5 = 0
Here a = 1, b = -3, c = 5.
So substituting in the formula
x = -(-3) ± √[(-3)² - 4 (1)(5)]/ 2(1)
= ( 23 ± √ (9-20)) / 2
= ( 3 ± √(-11)) / 2
= (3 ± √11 i) / 2
So x = ( 3 + √11 i)/2 or ( 3 - √11 i)/2
Worksheet on complex roots
Find the roots of the given quadratic equations:
1. x² - x + 1 = 0
(A) x = (1 ± i√3)/2
(B) x = 1, 2
(C) x = 1 ± i
2. x² + 3x + 5 = 0
(A) x+ √2, x-√2
(B) (-3±i√11)/2
(C) -x±i√7
3. -5/x = x-2
(A) x+ √2, x-√2
(B) (-3±i√11)/2
(C) x±4i
4. 3x²+ 10x + 9 =0
(A) (-5 ± i√2)/3
(B) (-3±i√11)/2
(C) x±4i
5. (-x+3) = 2/(x-2)
(A) (-5 ± i√2)/3
(B) (-3±i√11)/2
(C) (5±i√7)/2
Students can try to solve the problems given in 'Worksheet on complex roots' them selves. They can verify the answers and solutions with the solutions provided. If you are having any doubt you can contact us through mail, we will help you to clear your doubts.
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# Area of Polygon in Coordinate Geometry
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Sub Topics Polygon can be defined as packed or closed shape made by joining more than two line segments end to end. The sides of a Polygon never bisect each other. The Point where two lines meet is known as vertex of polygon. Now we will see how to find the Area of Polygon in coordinate Geometry? For finding the area of polygon we have to follow some steps which are given below: Step 1: First we take two axis in a plane which is named as ‘x’ and y- axis. Step 2: Then we draw a polygon on the plane. Step 3: Then we find the coordinates of given polygon. Step 4: If we have all the coordinates of a polygon then we can easily find the area of polygon by putting all the coordinates in formula shown below. The formula for finding the area of polygon by using the coordinates: Area of ploygon = $\frac{1}{2}$ $\begin{vmatrix} x_{1} & y_{1} \\ x_{2} & y_{2}\\ x_{3} & y_{3}\\ .& .\\ . & .\\ x_{n} & y_{n}\\ x_{1}& y_{1} \end{vmatrix}$ = $\frac{1}{2}$ | (x1 y2 - y1 x2) + (x2 y3 - y2 x3) + ....... + (xn y1 - yn x1) | Now we see how to find the area of polygon by using the above steps: A polygon is given then we can find the area of polygon by putting the values in the formula: So we can write it as: X Y P 5 9 Q 8 7 R 10 3 S 3 3 P 5 9 Multiply 5 by 9, we get 45 and multiply 8 by 9 we get 72; 5 * 9 = 45; 8 * 9 = 72; So we can write it as: 45 – 72 = -27; When we multiply 8 by 3 we get 24 and multiply 10 by 7 we get 70; 8 * 3 = 24; 10 * 7 = 70; So we can write it as: 24 – 70 = -46; When we multiply 10 by 3 we get 30 and multiply 3 by 3 we get 9; 10 * 3 = 30; 3 * 3 = 9; So we can write it as: 30 – 9 = 21; When we multiply 3 by 9 we get 27 and multiply 5 by 3 we get 15; 3 * 9 = 27; 5 * 3 = 15; So we can write it as: 27 – 15 = 12; Now add all the value and then divide the number by 2 we get the area of polygon. So the sum of all the values is: => -37 – 46 + 21 + 12; => -83 + 33; => -50;Area = $\frac{1}{2}$ |-50| = $\frac{50}{2}$ = 25Area of polygon is 25 square units. Then the area of polygon is: A = | -40/ 2| A = 20 inch2
## Area of a Regular Polygon
A Polygon is a closed figure with all sides connected end to end. A polygon is a plane figure which is bounded by an enclosed path composite of a finite sequence of a straight line segments. These segments are the sides and the Point where they meet is polygon vertex or the corners.
There are many types of polygon and one of them is Regular Polygon. A regular polygon is a polygon whose all angles and all sides are equal. A convex and a star can be a regular polygon. Like all other polygons, the lines of a regular polygon do not bisect each other. The point where the lines of the polygon meet is known as vertex of polygon and area is the space occupied by the polygon.
Some of its properties are as follows:
A regular n - sided polygon has the property of Rotational Symmetry (those symmetry which seems to be same after its rotation) of the ‘n’ order.
All vertices of the Circle lie on the common co cyclic points. We can say that regular polygon follows the cyclic property.
All sides of regular polygons are equal.
Its examples are Equilateral Triangle, square, pentagon, hexagon, octagon, and decagon. These all are the simple regular polygon and are convex. Those regular polygons with the same number of sides are called similar polygon.
Formula for Area of a regular polygon is:
Area = (1/2)(apothem)(perimeter):
Where area stands for the area of regular polygon, apothem stands for the length of radius of the inscribed circle, and perimeter is the sum of all the sides of the regular polygon.
Let’s understand it in a simpler way with the help of an illustration. We will consider a Square to understand it in a simpler form. Consider a side of the square as 10 yards and its apothem as 5 yards, place the described data in the above mentioned formula the area of regular polygon (square) will be 1/2*5*40 = 100 yards square.
## Area Of Polygon Given the length of a side
A Polygon is a geometrical figure on a plane formed with straight line segments. Let us see how we can use the length of side to find other sides and all other perimeters of the polygons.
Let us see some of these geometrical figure and shapes.
1. Area of polygon
To find the area of polygon having Coordinate (x1 , y1), (x2 , y2) , (x3 , y3) .... (xn , yn) Plane we use the following formula:
Area of Polygon = $\frac{1}{2}$ | (x1 y2 - y1 x2) + (x2 y3 - y2 x3) + ....... + (xn y1 - yn x1) |
When sides of Regular Polygons are given we can find the area of the polygons by using particular formula:
1. Square
If we have given the length of a side of a Square, we can find:
1. Area of Square
As all sides of a square are equal we can find area of the square by squaring one of the side i.e. (side)2
2. Perimeter of Square = 4(a)
The perimeter of a square is the total distance covered by sides of the square. It is the path that surrounds area of the square. The “perimeter” word is a Greek word in which "Peri" means "around" and the "meter" means “to measure". To find the perimeter of the square we simply add the length of each side together
As the sides of the square are equal in length therefore; Perimeter = 4* (side).
3. Diagonal of Square
To find the diagonal of square we simply multiply its side by the square root of 2.
Diagonal of Square = (side) * (sqrt (2))
1. Right angled triangle
Using Pythagorean Theorem, we can find the length of sides of the triangle using the following formula:
(Hypotenuse) 2 = (Base) 2 + (Perpendicular) 2
1. Area of a triangle
When we have given the lengths of the triangle we can find area of the triangle using Heron’s Formula:
According to Heron’s Formula:
Area of the triangle = √s (s-a) (s-b) (s-c)
Where a, b, and c are the sides of the triangle
And ‘s’ is the perimeter of the triangle.
s = $\frac{a+b+c}{2}$
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CCSS.Math.Content.1.OA.A.2 Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Counting Coins
CCSS.Math.Content.1.OA.C.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Read The Clock and Write the Time
CCSS.Math.Content.1.MD.B.3 Tell and write time in hours and half-hours using analog and digital clocks.
Which Sign Makes the Sentence True
CCSS.Math.Content.1.OA.B.3 Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
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# NCERT Solutions for Class 6 Maths Chapter 8 Decimals
## NCERT Solutions for Class 6 Maths Chapter 8 Decimals
Today in our article you will be told that when your ncert solutions for class 6 maths Chapter 8 Decimals will be released and you will be provided complete information about it. In our article, you will also be given complete information about the model paper. Also, you will be provided complete information about the model paper download process in our article. So please read it carefully till the end and stay connected with our website till the end.
ncert solutions for class 6 maths Chapter 8 Decimals
## Exercise 8.1
1. Write the following numbers in the given table.
Hundreds Tens Ones Tenths (100) (10) (1) (1 / 10)
Solutions:
Rows Hundreds Tens Ones Tenths a 0 3 1 2 b 1 1 0 4
2. Write the following decimals in the place value table.
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205.9
Solutions:
Hundreds Tens Ones Tenths 19.4 0 1 9 4 0.3 0 0 0 3 10.6 0 1 0 6 205.9 2 0 5 9
3. Write each of the following as decimals:
(a) Seven-tenths
(b) Two tens and nine-tenths
(c) Fourteen point six
(d) One hundred and two ones
(e) Six hundred point eight
Solutions:
(a) The decimal form of Seven-tenths is 7 / 10 = 0.7
(b) The decimal form of two tens and nine tenths is 20 + 9 / 10 = 20.9
(c) The decimal form of fourteen point six is 14.6
(d) The decimal form of one hundred and two ones is 100 + 2 = 102.0
(e) The decimal form of six hundred points eight is 600.8
4. Write each of the following as decimals:
(a) 5 / 10
(b) 3 + 7 / 10
(c) 200 + 60 + 5 + 1 / 10
(d) 70 + 8 / 10
(e) 88 / 10
(f)
(g) 3 / 2
(h) 2 / 5
(i) 12 / 5
(j)
(k)
Solutions:
(a) 5 / 10 = 0.5
(b) 3 + 7 / 10 = 3 + 0.7
= 3.7
(c) 200 + 60 + 5 + 1 / 10 = 265 + 0.1
= 265.1
(d) 70 + 8 / 10 = 70 + 0.8
= 70.8
(e) 88 / 10 = 80 / 10 + 8 / 10
= 8 + 0.8
= 8.8
(f)
(g) 3 / 2 = (2 + 1) / 2
= 2 / 2 + 1 / 2
= 1 + 0.5
= 1.5
(h) 2 / 5 = 0.4
(i) 12 / 5 = (10 + 2) / 5
= 10 / 5 + 2 / 5
= 2 + 0.4
= 2.4
(j)
(k)
5. Write the following decimals as fractions. Reduce the fraction to the lowest form.
(a) 0.6
(b) 2.5
(c) 1.0
(d) 3.8
(e) 13.7
(f) 21.2
(g) 6.4
Solutions:
(a) 0.6 = 6 / 10
= 3 / 5
(b) 2.5 = 25 / 10
= 5 / 2
(c) 1.0 = 1
= 1
(d) 3.8 = 38 / 10
= 19 / 5
(e) 13. 7 = 137 / 10
(f) 21.2 = 212 / 10
= 106 / 5
(g) 6.4 = 64 / 10
= 32 / 5
6. Express the following as cm using decimals.
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4 cm 2 mm
(e) 162 mm
(f) 83 mm
Solutions:
We know that
1 cm = 10 mm
1 mm = 1 / 10 cm
(a) 2 mm = 2 / 10 cm
= 0.2 cm
(b) 30 mm = 30 / 10 cm
= 3.0 cm
(c) 116 mm = 116 / 10 cm
= 11.6 cm
(d) 4 cm 2 mm = [(4 + 2 / 10)] cm
= 4.2 cm
(e) 162 mm = 162 / 10 cm
= 16.2 cm
(f) 83 mm = 83 / 10 cm
= 8.3 cm
7. Between which two whole numbers on the number line are the given numbers lie?
Which of these whole numbers is nearer the number?
(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.1
(f) 4.9
Solutions:
(a) 0.8 lies between 0 and 1
0.8 is nearer to 1
(b) 5.1 lies between 5 and 6
5.1 is nearer to 5
(c) 2.6 lies between 2 and 3
2.6 is nearer to 3
(d) 6.4 lies between 6 and 7
6.4 is nearer to 6
(e) 9.1 lies between 9 and 10
9.1 is nearer to 9
(f) 4.9 lies between 4 and 5
4.9 is nearer to 5
8. Show the following numbers on the number line.
(a) 0.2
(b) 1.9
(c) 1.1
(d) 2.5
Solutions:
(a) 0.2 lies between the points 0 and 1 on the number line. The space between 0 and 1 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 0.2 is the second point between 0 and 1
(b) 1.9 lies between points 1 and 2 on the number line. The space between 1 and 2 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 1.9 is the ninth point between 1 and 2
(c) 1.1 lies between points 1 and 2 on the number line such that the space between 1 and 2 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 1.1 is the first point between 1 and 2
(d) 2.5 lies between points 2 and 3 on the number line such that the space between 2 and 3 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 2.5 is the fifth point between 2 and 3
9. Write the decimal number represented by the points A, B, C, and D on the given number line.
Solutions:
(a) Point A represents 0.8 cm on the given number line.
(b) Point B represents 1.3 cm on the given number line
(c) Point C represents 2.2 cm on the given number line
(d) Point D represents 2.9 cm on the given number line
10. (a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
Solutions:
(a) The length of Ramesh notebook is 9 cm 5 mm
The length in cm is [(9 + 5 / 10)] cm
= 9.5 cm
(b) The length of a gram plant is 65 mm
Hence, the length in cm is 65 / 10
= 6.5 cm
## Exercise 8.2
1. Complete the table with the help of these boxes and use decimals to write the number.
Solutions:
Rows Ones Tenths Hundreds Number (a) 0 2 6 0.26 (b) 1 3 8 1.38 (c) 1 2 8 1.28
2. Write the numbers given in the following place value table in decimal form.
Rows Hundreds Tens Ones Tenths Hundredths Thousandths 100 10 1 1 / 10 1/ 100 1 / 1000 (a) 0 0 3 2 5 0 (b) 1 0 2 6 3 0 (c) 0 3 0 0 2 5 (d) 2 1 1 9 0 2 (e) 0 1 2 2 4 1
Solutions:
(a) 3 + 2 / 10 + 5 / 100
= 3 + 0.2 + 0.05
= 3.25
(b) 100 + 2 + 6 / 10 + 3 / 100
= 102 + 0.6 + 0.03
= 102.63
(c) 30 + 2 / 100 + 5 / 1000
= 30 + 0.02 + 0.005
= 30.025
(d) 200 + 10 + 1 + 9 / 10 + 2 / 1000
= 211 + 0.9 + 0.002
= 211.902
(e) 10 + 2 + 2 / 10 + 4 / 100 + 1 / 1000
= 12 + 0.2 + 0.04 + 0.001
= 12.241
3. Write the following decimals in the place value table.
(a) 0.29
(b) 2.08
(c) 19.60
(d) 148.32
(e) 200.812
Solutions:
(a) 0.29
= 0.2 + 0.09
= 2 / 10 + 9 / 100
(b) 2.08
= 2 + 0.08
= 2 + 8 / 100
(c) 19.60
= 19 + 0.60
= 10 + 9 + 6 / 10
(d) 148.32
= 148 + 0.3 + 0.02
= 100 + 40 + 8 + 3 / 10 + 2 / 100
(e) 200.812
= 200 + 0.8 + 0.01 + 0.002
=200 + 8 / 10 + 1 / 100 + 2 / 1000
Hundreds Tens Ones Tenths Hundredths Thousandths 0 0 0 2 9 0 0 0 2 0 8 0 0 1 9 6 0 0 1 4 8 3 2 0 2 0 0 8 1 2
4. Write each of the following as decimals.
(a) 20 + 9 + 4 / 10 + 1 / 100
(b) 137 + 5 / 100
(c) 7 / 10 + 6 / 100 + 4 / 1000
(d) 23 + 2 / 10 + 6 / 1000
(e) 700 + 20 + 5 + 9 / 100
Solutions:
(a) 20 + 9 + 4 / 10 + 1 / 100
= 29 + 0.4 + 0.01
= 29.41
(b) 137 + 5 / 100
= 137 + 0.05
= 137.05
(c) 7 / 10 + 6 / 100 + 4 / 1000
= 0.7 + 0.06 + 0.004
=0.764
(d) 23 + 2 / 10 + 6 / 1000
= 23 + 0.2 + 0.006
= 23.206
(e) 700 + 20 + 5 + 9 / 100
= 725 + 0.09
= 725.09
5. Write each of the following decimals in words.
(a) 0.03
(b) 1.20
(c) 108.56
(d) 10. 07
(e) 0.032
(f) 5.008
Solutions:
The following are the decimals in words
(a) 0.03 = zero point zero three
(b) 1.20 = one point two zero
(c) 108.56 = one hundred eight point five six
(d) 10.07 = ten point zero seven
(e) 0.032 = zero point zero three two
(f) 5.008 = five point zero zero eight
6. Between which two numbers in tenths place on the number line does each of the given number lie?
(a) 0.60
(b) 0.45
(c) 0.19
(d) 0.66
(e) 0.92
(f) 0.57
Solutions:
(a) 0.60 lies between 0 and 0.1 in tenths place
(b) 0.45 lies between 0.4 and 0.5 in tenths place
(c) 0.19 lies between 0.1 and 0.2 in tenths place
(d) 0.66 lies between 0.6 and 0.7 in tenths place
(e) 0.92 lies between 0.9 and 1.0 in tenths place
(f) 0.57 lies between 0.5 and 0.6 in tenths place
7. Write as fractions in the lowest terms.
(a) 0.60
(b) 0.05
(c) 0.75
(d) 0.18
(e) 0.25
(f) 0.125
(g) 0.066
Solutions:
(a) 0.60 = 60 / 100
= 6 / 10
= 3 / 5
(b) 0.05 = 5 / 100
= 1 / 20
(c) 0.75 = 75 / 100
= 3 / 4
(d) 0.18 = 18 / 100
= 9 / 50
(e) 0.25 = 25 / 100
= 1 / 4
(f) 0.125 = 125 / 1000
= 1 / 8
(g) 0.066 = 66 / 1000
= 33 / 500
## Exercise 8.3
1. Which is greater?
(a) 0.3 or 0.4
(b) 0.07 or 0.02
(c) 3 or 0.8
(d) 0.5 or 0.05
(e) 1.23 or 1.2
(f) 0.099 or 0.19
(g) 1.5 or 1.50
(h) 1.431 or 1.490
(i) 3.3 or 3.300
(j) 5.64 or 5.603
Solutions:
(a) 0.3 or 0.4
The whole parts for both the numbers are the same. We know that the tenth part of 0.4 is greater than that of 0.3
∴ 0.4 > 0.3
(b) 0.07 or 0.02
Both the numbers have the same parts up to the tenth place but the hundredth part of 0.07 is greater than that of 0.02
∴ 0.07 > 0.02
(c) 3 or 0.8
The whole part of 3 is greater than that of 0.8
∴ 3 > 0.8
(d) 0.5 or 0.05
The whole parts for both the numbers are the same. Here the tenth part of 0.5 is greater than that of 0.05
∴ 0.5 > 0.05
(e) 1.23 or 1.20
Here both the numbers have the same parts up to the tenth place. The hundredth part of 1.23 is greater than that of 1.20
∴ 1.23 > 1.20
(f) 0.099 or 0.19
The whole parts for both the numbers are the same. Here the tenth part of 0.19 is greater than that of 0.099
∴ 0.099 < 0.19
(g) 1.5 or 1.50
We may find that both numbers have the same parts up to the tenth place. Here 1.5 have no digit at a hundredth place. It represents that this digit is 0, which is equal to the digit at a hundredth place of 1.50.
∴ Both these numbers are equal
(h) 1.431 or 1.490
Here, both the numbers have the same parts up to the tenth place but the hundredth part of 1.490 is greater than that of 1.431
∴ 1.431 < 1.490
(i) 3.3 or 3.300
Here, both numbers have the same parts up to the tenth place. There are no digits at a hundredth and thousandth place of 3.3. It represents that these numbers are 0, which is equal to the digits at the hundredth and thousandth place of 3.300.
∴ Both these numbers are equal
(j) 5.64 or 5.603
Here both numbers have the same parts up to the tenth place but the hundredth part of 5.64 is greater than that of 5.603
∴ 5.64 > 5.603
2. Make five more examples and find the greater number from them.
Solutions:
Five more examples are
(a) 32.55 or 32.5
The whole parts for both the numbers are the same. The tenth part is also equal, but the hundredth part of 32.55 is greater than that of 32.5
Hence, 32.55 > 32.5
(b) 1 or 0.99
The whole part of 1 is greater than that of 0.99
∴ 1 > 0.99
(c) 1.09 or 1.093
Here both the numbers have the same parts up to the hundredth. But the thousandth part of 1.093 is greater than that of 1.09
∴ 1.093 > 1.09
(d) 2 or 1.99
The whole part of 2 is greater than that of 1.99
∴ 2 > 1.99
(e) 2.08 or 2.085
Here both the numbers have the same parts up to the hundredth. But the thousandth part of 2.085 is greater than that of 2.08
∴ 2.085 > 2.08
## Exercise 8.4
1. Express as rupees using decimals.
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90 paise
(e) 725 paise
Solutions:
We know that there are 100 paise in 1 rupee
(a) 5 paise = 5 / 100 rupees
= Rupess 0.05
(b) 75 paise = 75 / 100 rupees
= Rupees 0.75
(c) 20 paise = 20 / 100 rupees
= Rupees 0.20
(d) 50 rupees 90 paise = [(50 + 90 / 100)] rupees
= Rupees 50.90
(e) 725 paise = 725 / 100 rupees
= Rupees 7.25
2. Express as meters using decimals.
(a) 15 cm
(b) 6 cm
(c) 2 m 45 cm
(d) 9 m 7 cm
(e) 419 cm
Solutions:
We know that there are 100 cm in 1 meter
(a) 15 cm = 15 / 100 m
= 0.15 m
(b) 6 cm = 6 / 100 m
= 0.06 m
(c) 2 m 45 cm = [(2 + 45 / 100)] m
= 2.45 m
(d) 9 m 7 cm = [(9 + 7 / 100)] m
= 9.07 m
(e) 419 cm = 419 / 100 m
= 4.19 m
3. Express as cm using decimals
(a) 5 mm
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93 mm
Solutions:
We know that there are 10 mm in 1 cm
(a) 5 mm = 5 / 10 cm
= 0.5 cm
(b) 60 mm = 60 / 10 cm
= 6.0 cm
(c) 164 mm = 164 / 10 cm
= 16.4 cm
(d) 9 cm 8 mm = [(9 + 8 / 10)] cm
= 9.8 cm
(e) 93 mm = 93 / 10 cm
= 9.3 cm
4. Express as km using decimals.
(a) 8 m
(b) 88 m
(c) 8888 m
(d) 70 km 5 m
Solutions:
We know that there are 1000 metres in 1 km
(a) 8 m = 8 / 1000 km
= 0.008 km
(b) 88 m = 88 / 1000 km
= 0.088 km
(c) 8888 m = 8888 / 1000 km
= 8.888 km
(d) 70 km 5 m = [(70 + 5 / 1000)] km
= 70.005 km
5. Express as kg using decimals.
(a) 2 g
(b) 100 g
(c) 3750 g
(d) 5 kg 8 g
(e) 26 kg 50 g
Solutions:
We know that there are 1000 grams in 1 kg
(a) 2 g = 2 / 1000 kg
= 0.002 kg
(b) 100 g = 100 / 1000 kg
= 0.1 kg
(c) 3750 g = 3750 / 1000 kg
= 3.750 kg
(d) 5 kg 8 g = [(5 + 8 / 1000)] kg
= 5.008 kg
(e) 26 kg 50 g = [(26 + 50 / 1000)] kg
= 26.050 kg
## Exercise 8.5
1. Find the sum in each of the following:
(a) 0.007 + 8.5 + 30.08
(b) 15 + 0.632 + 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38
Solutions:
(a) Sum of 0.007 + 8.5 + 30.08
0.007
8.500
+ 30.080
__________
38.587
__________
(b) Sum of 15 + 0.632 + 13.8
15.000
0.632
+ 13.800
_________
29.432
__________
(c) Sum of 27.076 + 0.55 + 0.004
27.076
0.550
+ 0.004
_____________
27.630
______________
(d) Sum of 25.65 + 9.005 + 3.7
25.650
9.005
+ 3.700
__________
38.355
___________
(e) Sum of 0.75 + 10.425 + 2
0.750
10.425
+ 2.000
_________
13.175
__________
(f) Sum of 280.69 + 25.2 + 38
280.69
25.20
+ 38.00
__________
343.89
___________
2. Rashid spent ₹ 35.75 on Maths book and ₹ 32.60 on Science book. Find the total amount spent by Rashid.
Solutions:
Cost of Maths book = ₹ 35.75
Cost of Science book = ₹ 32.60
The total amount spent by Rashid is
35.75
+ 32.60
__________
68.35
___________
∴ The total amount of money spent by Rashid is ₹ 68.35
3. Radhika’s mother gave her ₹ 10.50 and her father gave her ₹ 15.80, find the total amount given to Radhika by the parents.
Solutions:
The amount is given by Radhika’s mother = ₹ 10.50
The amount is given by Radhika’s father = ₹ 15.80
The total amount given by her parents
10.50
+ 15.80
__________
26.30
___________
∴ The total amount of money given by Radhika’s parents is ₹ 26.30
4. Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.
Solutions:
Cloth of shirt = 3 m 20 cm
Cloth of trouser = 2 m 5 cm
The total length of cloth is
3.20
+ 2.05
________
5.25
_________
∴ The total length of cloth bought by Nasreen is 5.25 m
5. Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Solutions:
Distance walked by Naresh in the morning = 2 km 35 m
= [(2 + 35 /1000)] km
= 2.035 km
Distance walked by him in the evening = 1 km 7 m
= [(1 + 7 / 1000)] km
= 1.007 km
Total distance walked by Naresh is
2.035
+ 1.007
_______
3.042
________
∴ The total distance walked by Naresh is 3.042 km
6. Sunita traveled 15 km 268 m by bus, 7 km 7 m by car, and 500 m on foot in order to reach her school. How far is her school from her residence?
Solutions:
Distance travelled by bus = 15 km 268 m
= [(15 + 268 / 1000)] km
= 15.268 km
Distance travelled by car = 7 km 7 m
= [(7 + 7 / 1000)] km
= 7.007 km
Distance walked by Sunita = 500 m
= 500 / 1000
= 0.500 km
Total distance of school from her residence is
15.268
7.007
+ 0.500
________
22.775
________
∴ The total distance of the school from her residence is 22.775 km
7. Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar, and 10 kg 850 g flour. Find the total weight of his purchases.
Solutions:
Weight of rice = 5 kg 400 g
= [(5 + 400 / 1000)] kg
= 5.400 kg
Weight of sugar = 2 kg 20 g
= [(2 + 20 / 1000)] kg
= 2.020 kg
Weight of flour = 10 kg 850 g
= [(10 + 850 / 1000)] kg
= 10.850 kg
Total weight of his purchases is
5.400
2.020
+ 10.850
___________
18.270
____________
∴ The total weight of his purchases is 18.270 kg
## Exercise 8.6
1. Subtract:
(a) ₹ 18.25 from ₹ 20.75
(b) 202.54 m from 250 m
(c) ₹ 5.36 from ₹ 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg
Solutions:
(a) ₹ 20.75 – ₹ 18.75
20.75
– 18.25
__________
2.50
___________
₹ 2.50
(b) 250 m – 202.54 m
250.00
– 202.54
___________
47.46
____________
47.46 m
(c) ₹ 8.40 – ₹ 5.36
8.40
– 5.36
_________
3.04
_________
₹ 3.04
(d) 5.206 km – 2.051 km
5.206
– 2.051
__________
3.155
__________
3.155 km
(e) 2.107 kg – 0.314 kg
2.107
– 0.314
_________
1.793
__________
1.793 kg
2. Find the value of:
(a) 9.756 – 6.28
(b) 21.05 – 15.27
(c) 18.5 – 6.79
(d) 11.6 – 9.847
Solutions:
(a) 9.756
– 6.280
_________
3.476
_________
(b) 21.05
– 15.27
___________
5.78
____________
(c) 18.50
– 6.79
___________
11.71
___________
(d) 11.600
– 9.847
____________
1.753
____________
3. Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Solutions:
Money is given to shopkeeper = ₹ 50.00
Price of the book = ₹ 35.65
Money that Raju will get back from the shopkeeper will be the difference of these two
∴ Money left with Raju is
50.00
– 35.65
___________
14.35
___________
Hence, money left with Raju is ₹ 14.35
4. Rani had ₹ 18.50. She bought one ice cream for ₹ 11.75. How much money does she have now?
Solutions:
Money with Rani = ₹ 18.50
Price of an ice cream = ₹ 11.75
Now money left with Rani will be the difference of these two
Hence, money left with her is
18.50
– 11.75
__________
6.75
___________
∴ Money left with Rani is ₹ 6.75
5. Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?
Solutions:
Length of cloth = 20 m 5 cm
= 20.05 m
Length of cloth to make a curtain = 4 m 50 cm
= 4.50 m
The length of cloth left with Tina will be the difference between these two
Thus the length of cloth left with her is
20.05
– 4.50
________
15.55
________
∴ The length of the remaining cloth left with Tina is 15.55 m
6. Namita travels 20 km 50 m every day. Out of this, she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?
Solutions:
Total distance travelled by Namita = 20 km 50 m
= 20.050 km
Distance travelled by bus = 10 km 200 m
= 10.200 km
Distance traveled by auto = Total distance traveled – Distance traveled by bus
∴ Distance to be traveled by auto is
20.050
– 10.200
________
9.850
________
∴ Namita traveled 9.850 km by auto
7. Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Solutions:
Total weight of vegetables Aakash bought = 10.000 kg
Weight of onions = 3 kg 500 g
= 3.500 kg
Weight of tomatoes = 2 kg 75 g
= 2.075 kg
Weight of potatoes = Total weight of vegetables bought – (weight of onions + weight of tomatoes)
= 10.000 – (3.500 + 2.075)
3.500
+ 2.075
________
5.575
________
10.000
– 5.575
_________
4.425
_________
∴ 4.425 kg is the weight of the potatoes
I hope you like the information and notes given by us. This will help you a lot in your upcoming exams. Here we are giving ncert solutions to all of you students, through which all of you can score very well in your exams. In this post, we are giving you ncert solutions for class 6 maths. hope you benefit from it. You can download the pdf file of this ncert solutions.
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Precalculus Examples
Split Using Partial Fraction Decomposition
Step 1
Decompose the fraction and multiply through by the common denominator.
For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place .
For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place .
Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is .
Cancel the common factor of .
Cancel the common factor.
Rewrite the expression.
Cancel the common factor of .
Cancel the common factor.
Divide by .
Reorder and .
Simplify each term.
Cancel the common factor of .
Cancel the common factor.
Divide by .
Apply the distributive property.
Move to the left of .
Cancel the common factor of .
Cancel the common factor.
Divide by .
Apply the distributive property.
Move to the left of .
Move .
Step 2
Create equations for the partial fraction variables and use them to set up a system of equations.
Create an equation for the partial fraction variables by equating the coefficients of from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.
Create an equation for the partial fraction variables by equating the coefficients of the terms not containing . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.
Set up the system of equations to find the coefficients of the partial fractions.
Step 3
Solve the system of equations.
Solve for in .
Rewrite the equation as .
Subtract from both sides of the equation.
Replace all occurrences of with in each equation.
Replace all occurrences of in with .
Simplify the right side.
Simplify .
Simplify each term.
Apply the distributive property.
Multiply by .
Multiply by .
Solve for in .
Rewrite the equation as .
Move all terms not containing to the right side of the equation.
Add to both sides of the equation.
Divide each term in by and simplify.
Divide each term in by .
Simplify the left side.
Cancel the common factor of .
Cancel the common factor.
Divide by .
Simplify the right side.
Divide by .
Replace all occurrences of with in each equation.
Replace all occurrences of in with .
Simplify the right side.
Simplify .
Multiply by .
Subtract from .
List all of the solutions.
Step 4
Replace each of the partial fraction coefficients in with the values found for and .
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# How do you find the first and second derivative of y=1/(1+e^-x)?
Jan 9, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{- x}}{1 + {e}^{- x}} ^ 2 , \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{{e}^{- 2 x} - {e}^{- x}}{1 + {e}^{- x}} ^ 3$
#### Explanation:
$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left({e}^{- x}\right) = - {e}^{- x}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
There are 2 approaches to differentiating this function.
$\left(1\right) \text{ Using the quotient rule}$
$\left(2\right) \text{ expressing " y=(1+e^(-x))^-1" and use chain rule}$
I'll use approach (1) you could perhaps try approach (2). The result will be the same.
differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$
$\text{ Given " y=(g(x))/(h(x))" then}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$g \left(x\right) = 1 \Rightarrow g ' \left(x\right) = 0$
$h \left(x\right) = 1 + {e}^{- x} \Rightarrow h ' \left(x\right) = - {e}^{- x}$
$\textcolor{b l u e}{\text{------------------------------------------------------------}}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 + {e}^{-} x\right) .0 - 1. \left(- {e}^{-} x\right)}{1 + {e}^{-} x} ^ 2 = \frac{{e}^{-} x}{1 + {e}^{-} x} ^ 2$
$\text{ To find " (d^2y)/(dx^2)" differentiate } \frac{\mathrm{dy}}{\mathrm{dx}}$
differentiate using the $\textcolor{b l u e}{\text{quotient rule/chain rule}}$
$\text{here } g \left(x\right) = {e}^{-} x \Rightarrow g ' \left(x\right) = - {e}^{-} x$
$h \left(x\right) = {\left(1 + {e}^{-} x\right)}^{2} \Rightarrow h ' \left(x\right) = 2 \left(1 + {e}^{-} x\right) . \left(- {e}^{-} x\right) \rightarrow$
$\Rightarrow h ' \left(x\right) = - 2 {e}^{-} x \left(1 + {e}^{-} x\right)$
$\textcolor{b l u e}{\text{---------------------------------------------------------------}}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{{\left(1 + {e}^{-} x\right)}^{2} \left(- {e}^{-} x\right) - \left({e}^{-} x\right) . \left(- 2 {e}^{-} x \left(1 + {e}^{-} x\right)\right)}{1 + {e}^{-} x} ^ 4$
$= \frac{- {e}^{-} x {\left(1 + {e}^{-} x\right)}^{2} + 2 {e}^{- 2 x} \left(1 + {e}^{-} x\right)}{1 + {e}^{-} x} ^ 4$
$= \frac{{e}^{-} x \left(1 + {e}^{-} x\right) \left(2 {e}^{-} x - 1 - {e}^{-} x\right)}{1 + {e}^{-} x} ^ 4 \leftarrow \text{ factoring}$
=(e^-xcancel((1+e^-x))(e^-x-1))/(cancel((1+e^-x)^3
$= \frac{{e}^{- 2 x} - {e}^{-} x}{1 + {e}^{-} x} ^ 3$
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# How to Find Perimeter
## How to Find Perimeter of Triangles, Perimeter of Rectangles, and Other Polygons With Examples And Interactive Exercises
The perimeter of a polygon is the distance around the outside of the polygon. A polygon is 2-dimensional; however, perimeter is 1-dimensional and is measured in linear units. To help us make this distinction, look at our picture of a rectangular backyard. The yard is 2-dimensional: it has a length and a width. The amount of fence needed to enclose the backyard (perimeter) is 1-dimensional. The perimeter of this yard is the distance around the outside of the yard, indicated by the red arrow; It is measured in linear units such as feet or meters.
To find the perimeter of a polygon, take the sum of the length of each side. The polygons below are much smaller than a fenced-in yard. Thus, we use smaller units in our examples, such as centimeters and inches.
Example 1: Find the perimeter of a triangle with sides measuring 5 centimeters, 9 centimeters and 11 centimeters.
Solution: P = 5 cm + 9 cm + 11 cm = 25 cm
### Example 2: A rectangle has a length of 8 centimeters and a width of 3 centimeters. Find the perimeter.
Solution 1: P = 8 cm + 8cm + 3 cm + 3 cm = 22 cm
Solution 2: P = 2(8 cm) + 2(3 cm) = 16 cm + 6 cm = 22 cm
### In Example 2, the second solution is more commonly used. In fact, in mathematics, we commonly use the following formula for perimeter of a rectangle:
where is the perimeter, is the length and is the width.
In the next few examples, we will find the perimeter of other polygons.
### Example 3: Find the perimeter of a square with each side measuring 2 inches.
Solution: = 2 in + 2 in + 2 in + 2 in = 8 in
### Example 4: Find the perimeter of an equilateral triangle with each side measuring 4 centimeters.
Solution: = 4 cm + 4 cm + 4 cm = 12 cm
### Example 3: Find the perimeter of a square with each side measuring 2 inches.
Solution: This regular polygon has 4 sides, each with a length of 2 inches. Thus we get:
= 4(2 in) = 8 in
### Example 4: Find the perimeter of an equilateral triangle with each side measuring 4 centimeters.
Solution: This regular polygon has 3 sides, each with a length of 4 centimeters. Thus we get:
= 3(4 cm) = 12 cm
### Example 5: Find the perimeter of a regular pentagon with each side measuring 3 inches.
Solution: = 5(3 in) = 15 in
### Example 6: The perimeter of a regular hexagon is 18 centimeters. How long is one side?
Solution: = 18 cm
Let S represent the length of one side. A regular hexagon has 6 sides, so we can divide the perimeter by 6 to get the length of one side (S).
S = 18 cm ÷ 6
S = 3 cm
### Directions: Read each question below. Click once in an ANSWER BOX and type in your answer; then click ENTER. Your answers should be given as whole numbers greater than zero. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR.
1 Find the perimeter of a triangle with sides measuring 10 inches, 14 inches and 15 inches. ANSWER BOX: = in RESULTS BOX:
2 A rectangle has a length of 12 centimeters and a width of 4 centimeters. Find the perimeter. ANSWER BOX: = cm RESULTS BOX:
3 Find the perimeter of a regular hexagon with each side measuring 8 meters. ANSWER BOX: = m RESULTS BOX:
4 The perimeter of a square is 20 feet. How long is each side? ANSWER BOX: = ft RESULTS BOX:
5 The perimeter of a regular pentagon is 100 centimeters. How long is each side? ANSWER BOX: = cm RESULTS BOX:
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# stat341 / CM 361
Computational Statistics and Data Analysis is a course offered at the University of Waterloo
Spring 2009
Instructor: Ali Ghodsi
## Sampling (Generating Random numbers)
### Lecture of May 12th, 2009
In order to study statistics computationally, we need a good way to generate random numbers from various distributions using computational methods, or at least numbers whose distribution appears to be random (pseudo-random). Outside a computational setting, this is fairly easy (at least for the uniform distribution). Rolling a die, for example, produces numbers with a uniform distribution very well.
We begin by considering the simplest case: the uniform distribution.
#### Multiplicative Congruential Method
One way to generate pseudorandom numbers from the uniform distribution is using the Multiplicative Congruential Method. This involves three integer parameters a, b, and n, and a seed variable x0. This method deterministically (based on the seed) generates a sequence of numbers with a seemingly random distribution (with some caveats). It proceeds as follows:
$x_{i+1} = ax_{i} + b \mod{n}$
For example, with a = 13, b = 0, m = 31, x0 = 1, we have:
$x_{i+1} = 13x_{i} \mod{31}$
So,
\begin{align} x_{0} &{}= 1 \end{align}
\begin{align} x_{1} &{}= 13 \times 1 + 0 \mod{31} \\ &{}= 13 \end{align}
\begin{align} x_{2} &{}= 13 \times 13 + 0 \mod{31} \\ &{}= 14 \end{align}
etc.
This method generates numbers between 0 and m - 1, and by scaling this output by dividing the terms of the resulting sequence by m - 1, we create a sequence of numbers between 0 and 1. For correctly chosen values of a, b, and m, this method will generate a sequence of integers including all integers between 0 and m - 1.
Of course, not all values of a, b, and m will behave in this way, and will not be suitable for use in generating pseudorandom numbers. In practice, it has been found by a paper published in 1988 by Park and Miller, that a = 75, b = 0, and m = 231 - 1 = 2147483647 (the maximum size of a signed integer in a 32-bit system) are good values for the Multiplicative Congruential Method.
There are however some drawbacks to this method:
• No integer will ever be generated twice in a row (otherwise the method would generate that integer forever from that point onwards)
• Can only be used to generate pseudorandom numbers from the uniform distribution
#### General Methods
Since the Multiplicative Congruential Method can only be used for the uniform distribution, other methods must be developed in order to generate pseudorandom numbers from other distributions.
##### Inverse Transform Method
This method uses the fact that when the inverse of a distribution cumulative density function for a given distribution is applied to the uniform distribution, the resulting distribution is the distribution of the cdf. This is shown by this theorem:
Theorem:
If $U \sim~ Unif[0, 1]$ is a random variable and $X = F^{-1}(U)$, where F is continuous, and is the cumulative density function for some distribution, then the distribution of the random variable X is given by F(X).
Proof:
Recall that, if f is the pdf corresponding to F, then,
$F(x) = P(X \leq x) = \int_{-\infty}^x f(x)$
So F is monotonically increasing, since the probability that X is less than a greater number must be greater than the probability that X is less than a lesser number.
Note also that in the uniform distribution on [0, 1], we have for all a within [0, 1], $P(U \leq a) = a$.
So,
\begin{align} P(F^{-1}(U) \leq x) &{}= P(F(F^{-1}(U)) \leq F(x)) \\ &{}= P(U \leq F(x)) \\ &{}= F(x) \end{align}
Completing the proof.
Procedure (Continuous Case)
This method then gives us the following procedure for finding pseudorandom numbers from a continuous distribution:
• Step 1: Draw $U~ \sim~ Unif [0,1]$.
• Step 2: Compute $X = F^{-1}(U)$.
Example:
Suppose we want to draw a sample from $f(x) = \lambda e^{-\lambda x}$ where $x \gt 0$ (the exponential distribution).
We need to first find $F(x)$ and then its inverse, $F^{-1}$.
$F(x) = \int^x_0 \theta e^{-\theta u} du = 1 - e^{-\theta x}$
$F^{-1}(x) = \frac{-log(1-y)}{\theta}$
Now we can generate our random sample $i=1\dots n$ from $f(x)$ by:
$1)\ u_i \sim UNIF(0,1)$
$2)\ x_i = \frac{-log(1-u_i)}{\theta}$
The $x_i$ are now a random sample from $f(x)$.
The major problem with this approach is that we have to find $F^{-1}$ and for many distributions it is too difficult (or impossible) to find the inverse of $F(x)$. Further, for some distributions it is not even possible to find $F(x)$
Procedure (Discrete Case)
The above method can be easily adapted to work on discrete distributions as well.
In general in the discrete case, we have $x_0, \dots , x_n$ where:
\begin{align}P(X = x_i) &{}= p_i \end{align}
$x_0 \leq x_1 \leq x_2 \dots \leq x_n$
$\sum p_i = 1$
Thus we can define the following method to find pseudorandom numbers in the discrete case (note that the less-than signs from class have been changed to less-than-or-equal-to signs by me, since otherwise the case of $U = 1$ is missed):
• Step 1: Draw $U~ \sim~ Unif [0,1]$.
• Step 2:
• If $U \leq p_0$, return $X = x_0$
• If $U \leq p_0 + p_1$, return $X = x_1$
• ...
• In general, if $U \leq p_0 + \dots + p_k$, return $X = x_k$
Example (from class):
Suppose we have the following discrete distribution:
\begin{align} P(X = 0) &{}= 0.3 \\ P(X = 1) &{}= 0.2 \\ P(X = 2) &{}= 0.5 \end{align}
The cumulative density function for this distribution is then:
$F(x) = \begin{cases} 0, & \text{if } x \lt 0 \\ 0.3, & \text{if } 0 \leq x \lt 1 \\ 0.5, & \text{if } 1 \leq x \lt 2 \\ 1, & \text{if } 2 \leq x \end{cases}$
Then we can generate numbers from this distribution like this, given $u_0, \dots, u_n$ from $U \sim~ Unif[0, 1]$:
$x_i = \begin{cases} 0, & \text{if } u_i \leq 0.3 \\ 1, & \text{if } 0.3 \lt u_i \leq 0.5 \\ 2, & \text{if } 0.5 \lt u_i \leq 1 \end{cases}$
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# Difference between revisions of "1983 IMO Problems/Problem 6"
## Problem 6
Let $a$, $b$ and $c$ be the lengths of the sides of a triangle. Prove that
$a^2 b(a-b) + b^2 c(b-c) + c^2 a(c-a) \geq 0$.
Determine when equality occurs.
## Solution 1
By Ravi substitution, let $a = y+z$, $b = z+x$, $c = x+y$. Then, the triangle condition becomes $x, y, z > 0$. After some manipulation, the inequality becomes:
$xy^3 + yz^3 + zx^3 \geq xyz(x+y+z)$.
By Cauchy, we have:
$(xy^3 + yz^3 + zx^3)(z+x+y) \geq xyz(y+z+x)^2$ with equality if and only if $\frac{xy^3}{z} = \frac{yz^3}{x} =\frac{zx^3}{y}$. So the inequality holds with equality if and only if x = y = z. Thus the original inequality has equality if and only if the triangle is equilateral.
## Solution 2
Without loss of generality, let $a \geq b \geq c > 0$. By Muirhead or by AM-GM, we see that $a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$.
If we can show that $a^3 b + b^3 c+ c^3 a \geq a^3 c + b^3 a + c^3 b$, we are done, since then $2(a^3 b + b^3 c+ c^3 a ) \geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$, and we can divide by $2$.
We first see that, $(a^2 + ac + c^2) \geq (b^2 + bc + c^2)$, so $(a-c)(b-c)(a^2 + ac + c^2) \geq (a-c)(b-c)(b^2 + bc + c^2)$.
Factoring, this becomes $(a^3 - c^3)(b-c) \geq (a-c)(b^3 - c^3)$. This is the same as:
$(a^3 - c^3)(b-c) + (b^3 - c^3)(c-a) \geq 0$.
Expanding and refactoring, this is equal to $a^3 (b-c) + b^3(c-a) + c^3 (a-b) \geq 0$. (This step makes more sense going backwards.)
Expanding this out, we have
$a^3b + b^3 c + c^3 a \geq a^3 c + b^3 a + c^3 b$,
which is the desired result.
## Solution 3
Let $s$ be the semiperimeter, $\frac{a+b+c}{2}$, of the triangle. Then, $a=s-\frac{-a+b+c}{2}$, $b=s-\frac{a-b+c}{2}$, and $c=s-\frac{a+b-c}{2}$. We let $x=\frac{-a+b+c}{2},$ $y=\frac{a-b+c}{2}$, and $z=\frac{a+b-c}{2}.$ (Note that $x,y,z$ are all positive, since all sides must be shorter than the semiperimeter.) Then, we have $a=s-x$, $b=s-y$, and $c=s-z$. Note that $x+y+z=s$, so $$a=y+z,b=x+z,c=x+y.$$ Plugging this into $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq0$$ and doing some expanding and cancellation, we get $$2x^3z+2xy^3+2yz^3-2x^2yz-2xy^2z-2xyz^2\geq0.$$ The fact that each term on the left hand side has at least two variables multiplied motivates us to divide the inequality by $2xyz$, which we know is positive from earlier so we can maintain the sign of the inequality. This gives $$\frac{x^2}{y}-x+\frac{y^2}{z}-y-z+\frac{z^2}{x}\geq0.$$ We move the negative terms to the right, giving $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z.$$ We rewrite this as $$\sum_{cyc}\frac{x^2}{y}\geq\sum_{cyc}rx+(1-r)y.$$ where $r$ is any real number. (This works because if we evaulate the cyclic sum, then as long as the coefficients of $x$ and $y$ on the right sum to 1 the right side will be $x+y+z$.
Thus, we need to show that there exists a real number $r$ such that $\frac{x^2}{y}\geq rx+(1-r)y$ for all positive $x,y$. We claim that $r=2$ works. This becomes $\frac{x^2}{y}\geq2x-y$, and since $y$ is positive we can multiply by $y$ to yield $x^2\geq2xy-y^2$, or $(x-y)^2\geq0$, which is obviously true by the Trivial Inequality. Thus, we are done with part (a).
(To clarify how this works, we have $\frac{x^2}{y}\geq 2x-y$, $\frac{y^2}{z}\geq 2y-z$, and $\frac{z^2}{x}\geq 2z-x$, we add these inequalities to get $\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z.$)
Equality occurs if and only if $\frac{x^2}{y}=2x-y$, $\frac{y^2}{z}=2y-z$, and $\frac{z^2}{x}=2z-x$ at the same time, since if they are not all equal then the left side will be greater than the right side. From there, it is easy to see that the equality case is $x=y=z$, which is $a=b=c$.
~john0512
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Find Solution of the Differential Equation dy/dx=x+y
The exact solution of dy/dx=x+y is equal to |x+y+1| = Cex where C is a constant. In this post, we will learn how to solve the differential equation dy/dx =x+y.
Solve dy/dx=x+y
Question: Find the general solution of $\dfrac{dy}{dx}=x+y$.
Solution:
To solve the given differential equation, we will use the substitute method. Let us put
z=x+y.
Differentiating with respect to x,
$\dfrac{dz}{dx}=1+\dfrac{dy}{dx}$.
⇒ $\dfrac{dy}{dx}=\dfrac{dz}{dx}-1$.
So from the given equation, we obtain that
$\dfrac{dz}{dx}-1=z$
⇒ $\dfrac{dz}{dx}=1+z$
⇒ $\dfrac{dz}{1+z}=dx$
Integrating, $\int \dfrac{dz}{1+z}=\int dx$ +k
⇒ $\ln |1+z|=x+k$
⇒ $\ln |1+x+y|=x+k$ as z=x+y.
⇒ $|1+x+y|=Ce^x$ where C=ek.
So the general solution of dy/dx=x+y equals to |x+y+1| = Cex where C denotes an integral constant.
More Differential Equations:
Solve dy/dx = sin(x+y)
Solve dy/dx = cos(x+y)
FAQs
Q1: What is the general solution of dy/dx=x+y?
Answer: The general solution of the differential equation dy/dx=x+y is given by |x+y+1| = Cex where C is a constant.
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# 2019 AIME II Problems/Problem 6
## Problem
In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$, for some fixed $b\ge2$. A Martian student writes down $$3\log(\sqrt{x}\log x)=56$$ $$\log_{\log x}(x)=54$$ and finds that this system of equations has a single real number solution $x>1$. Find $b$.
## Solution 1
Using change of base on the second equation to base b, $$\frac{\log x}{\log \log_{b}{x}}=54$$ $$\log x = 54 \cdot \log\log_{b}{x}$$ $$x = (\log_{b}{x})^{54}$$ Note by dolphin7 - you could also just rewrite the second equation in exponent form. Substituting this into the $\sqrt x$ of the first equation, $$3\log_{b}{((\log_{b}{x})^{27}\log_{b}{x})} = 56$$ $$3\log_{b}{(\log_{b}{x})^{28}} = 56$$ $$\log_{b}{(\log_{b}{x})^{84}} = 56$$
We can manipulate this equation to be able to substitute $x = (\log_{b}{x})^{54}$ a couple more times: $$\log_{b}{(\log_{b}{x})^{54}} = 56 \cdot \frac{54}{84}$$ $$\log_{b}{x} = 36$$ $$(\log_{b}{x})^{54} = 36^{54}$$ $$x = 6^{108}$$
However, since we found that $\log_{b}{x} = 36$, $x$ is also equal to $b^{36}$. Equating these, $$b^{36} = 6^{108}$$ $$b = 6^3 = \boxed{216}$$
## Solution 2
We start by simplifying the first equation to $$3\log_{b}{(\sqrt{x}\log x)}=\log_{b}{(x^{\frac{3}{2}}\log^3x)}=56$$ $$x^\frac{3}{2}\cdot \log_b^3x=b^{56}$$ Next, we simplify the second equation to $$\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54$$ $$\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))$$ $$x=\log_b^{54}x$$ Substituting this into the first equation gives $$\log_b^{54\cdot \frac{3}{2}}(x)\cdot \log_b^3x=\log_b^{84}x=b^{56}$$ $$x=b^{b^{\frac{56}{84}}}=b^{b^{\frac{2}{3}}}$$ Plugging this into $x=\log_b^{54}x$ gives $$b^{b^{\frac{2}{3}}}=\log_b^{54}(b^{b^\frac{2}{3}})=b^{\frac{2}{3}\cdot 54}=b^{36}$$ $$b^{\frac{2}{3}}=36$$ $$b=36^{\frac{3}{2}}=6^3=\boxed{216}$$ -ktong
## Solution 3
Apply change of base to $$\log_{\log x}(x)=54$$ to yield: $$\frac{\log_b(x)}{\log_b(\log_b(x))}=54$$ which can be rearranged as: $$\frac{\log_b(x)}{54}=\log_b(\log_b(x))$$ Apply log properties to $$3\log(\sqrt{x}\log x)=56$$ to yield: $$3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}$$ Substituting $$\frac{\log_b(x)}{54}=\log_b(\log_b(x))$$ into the equation $\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}$ yields: $$\frac{1}{2}\log_b(x)+\frac{\log_b(x)}{54}=\frac{28\log_b(x)}{54}=\frac{56}{3}$$ So $$\log_b(x)=36.$$ Substituting this back in to $$\frac{\log_b(x)}{54}=\log_b(\log_b(x))$$ yields $$\frac{36}{54}=\log_b(36).$$ So, $$b^{\frac{2}{3}}=36\Rightarrow \boxed{b=216}$$
-Ghazt2002
## Solution 4
1st equation: $$\log (\sqrt{x}\log x)=\frac{56}{3}$$ $$\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}$$ 2nd equation: $$x=(\log x)^{54}$$ So now substitute $\log x=a$ and $x=b^a$: $$b^a=a^{54}$$ $$b=a^{\frac{54}{a}}$$ We also have that $$\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}$$ $$\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}$$ This means that $\frac{14}{27}a=\frac{56}{3}$, so $$a=36$$ $$b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}$$.
-Stormersyle
## Solution 5 (Substitution)
Let $y = \log _{b} x$ Then we have $$3\log _{b} (y\sqrt{x}) = 56$$ $$\log _{y} x = 54$$ which gives $$y^{54} = x$$ Plugging this in gives $$3\log _{b} (y \cdot y^{27}) = 3\log _{b} y^{28} = 56$$ which gives $$\log _{b} y = \dfrac{2}{3}$$ so $$b^{2/3} = y$$ By substitution we have $$b^{36} = x$$ which gives $$y = \log _{b} x = 36$$ Plugging in again we get $$b = 36^{3/2} = \fbox{216}$$
--Hi3142
## Solution 6 (Also Substitution)
This system of equations looks complicated to work with, so we let $a=\log_bx$ to make it easier for us to read.
Now, the first equation becomes $3\log(\sqrt x \cdot a) = 56 \implies \log(\sqrt{x}\cdot a)=\frac{56}3$.
The second equation, $\log_{\log(x)}(x)=54$ gives us $\underline{a^{54} = x}$.
Let's plug this back into the first equation to see what we get: $\log_b(\sqrt{a^{54}}\cdot a)=\frac{56}3$, and simplifying, $\log_b(a^{27}\cdot a^1)=\log_b(a^{28})=\frac{56}{3}$, so $b^{\frac{56}3}=a^{28}\implies \underline{b^{\frac 23}=a}$.
Combining this new finding with what we had above $a^{54} = (b^{\frac 23})^{54} = x\implies \mathbf{b^{36} =x}$.
Now that we've expressed one variable in terms of the other, we can plug this into either equation, say equation 1. Then we get $\log_b(\sqrt{b^{36}}\cdot\log_b(b^{36})=\frac{56}3\implies$$\log_b(b^{18}\cdot 36)=\frac{56}3\implies b^{\frac{56}3}=b^{18}\cdot 36$.
Finally, that gives us that $\frac{b^{\frac{56}3}}{b^{18}}=36\implies b^{\frac{56}{3}-18}=b^{\frac{56}{3}-\frac{54}{3}}=b^{\frac 23}=36\implies b=36^{\frac 32}=6^3$. Thus, $b=\boxed{216}$.
~BakedPotato66
## Solution 7 (Easy System of Equations)
Using change of base on the second equation, we have
$$\frac{\log_{b} x}{\log_{b} \log_{b} x} = 54$$
Using log rules on the first equation, we have
$$\frac{3}{2} \log_{b} x + 3 \log_{b} \log_{b} x = 56$$
We notice that $\log_{b} x$ and $\log_{b} \log_{b} x$ are in both equations. Thus, we set $m = \log_{b} x$ and $n = \log_{b} \log_{b} x$ and we have
$$\frac{3}{2} m + 3n = 56$$ $$\frac{m}{n} = 54$$
Solving this yields $m = 36$, $n = \frac{2}{3}$.
Now, $n = \log_{b} \log_{b} x = \log_{b} m = \log_{b} 36$, so we have $\log_{b} 36 = \frac{2}{3}$. Solving this yields $b = \boxed{216}$.
The second equation implies that $$\log_{\log_b x} x=54\implies (\log_b x)^{54}=x \implies \log_b x=x^{\frac{1}{54}}$$ The first equation implies that $$3\log_b(\sqrt{x} \log_b x)=56 \implies b^{\frac{56}{3}}=\sqrt{x} \log_b x$$ Substituting the first result into the second gives us $$b^{\frac{56}{3}}=x^{\frac{1}{2}}\cdot x^{\frac{1}{54}} \implies b=x^{\frac{1}{36}}.$$ Because $b^{36}=x,$ $\log_b x=36$ by the definition of a logarithm. Substituting this into the second equation, $$\log_{36} x=54 \implies x=36^{54}.$$ Finally, $$b=(36^{54})^{\frac{1}{36}}=36^{54\cdot\frac{1}{36}}=6^{2*\frac{3}{2}}=6^3=\boxed{216}.$$
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# Planar Graphs & Euler’s Theorem
A planar graph is a graph which can be drawn in the plane without any edges crossing. Some pictures of a planar graph might have crossing edges, but it’s possible to redraw the picture to eliminate the crossings.
There are also many practical applications with a graph structure in which crossing edges are a nuisance, including design problems for circuits, subways, utility lines. Two crossing connections normally means that the edges must be run at different heights.
When a planar graph is drawn with no crossing edges, it divides the plane into a set of regions, called faces. By convention, we also count the unbounded area outside the whole graph as one face.
Suppose that we have a graph with e edges, v nodes, and f faces. We know that the Handshaking theorem holds, i.e. the sum of the degrees of the vertices is 2e. For planar graphs, we also have a Handshaking theorem for faces: the sum of the face degrees is 2e.
For connected planar graphs, we have Euler’s formula: v − e + f = 2. We’ll prove that this formula works.
Before we try to prove Euler’s formula, let’s look at one special type of planar graph: trees.
A tree doesn’t divide the plane into multiple regions, because it doesn’t contain any cycles. In graph theory, a tree has only one face: the entire plane surrounding it. So Euler’s theorem reduces to v − e = 1,
i.e. e = v − 1. Let’s prove that this is true, by induction.
Proof by induction on the number of edges in the graph.
Base Case: If the graph contains no edges and only a single vertex, the formula is clearly true.
IHOP: Suppose the formula works for all trees with up to k vertices. Let T be a tree with k + 1 vertices.
We need to show that T has k edges.
Now, we find a vertex with degree 1 (only one edge going into it). To do this start at any vertex r and follow a path in any direction, without repeating edges. Because T has no cycles, this path can’t return to any vertex it has already visited. So it must eventually hit a dead end: the vertex at the end must have degree 1. Call it p.
Remove p and the edge coming into it, making a new tree T′ with k vertices. By the IHOP, T′ has k − 1 edges. So T has k edges. Therefore the formula holds for T.
We can now prove Euler’s Formula (v − e + f = 2) works in general, for any connected planar graph.
Proof by induction on the number of edges in the graph.
Base Case: If e = 0, the graph consists of a single node with a single face surrounding it. So we have 1 − 0 + 1 = 2 which is clearly right.
IHOP: Suppose the formula works for all graphs with no more than n edges. Let G be a graph with n + 1 edges.
Case 1: G doesn’t contain a cycle. So G is a tree and we already know the formula works for trees since we proved it! Check!
Case 2: G contains at least one cycle. Pick an edge p that’s on a cycle. Remove p to create a new graph G′.
Since the cycle separates the plane into two faces, the faces to either side of p must be distinct. When we remove the edge p, we merge these two faces. So G′ has one fewer faces than G.
Since G′ has n edges, the formula works for G′ by the IHOP.
That is, v′ − e′ + f′ = 2.
But v′ = v, e′ = e − 1, and f′ = f − 1.
Substituting, we find that
v − (e − 1) + (f − 1) = 2
So, v − e + f = 2 as desired.
## Exam Time!
Use Euler’s Planar Graph Theorem to prove whether the graph is planar or nonplanar.
http://www.science4all.org/le-nguyen-hoang/eulers-formula-and-the-utilities-problem/
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# What is a chi square test and why is it used?
Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis.
Also, why chi square test is used?
Chi-Square Test for Independence. This lesson explains how to conduct a chi-square test for independence. The test is applied when you have two categorical variables from a single population. It is used to determine whether there is a significant association between the two variables.
What is the purpose of using the chi square test?
Tests for Different Purposes. Chi square test for testing goodness of fit is used to decide whether there is any difference between the observed (experimental) value and the expected (theoretical) value. For example given a sample, we may like to test if it has been drawn from a normal population.
What is the chi square test used for and what does it tell you?
The Chi-square test is intended to test how likely it is that an observed distribution is due to chance. It is also called a “goodness of fit” statistic, because it measures how well the observed distribution of data fits with the distribution that is expected if the variables are independent.
## What is the P value in a chi square test?
The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 19.58. We use the Chi-Square Distribution Calculator to find P(Χ2 > 19.58) = 0.0001. Interpret results. Since the P-value (0.0001) is less than the significance level (0.05), we cannot accept the null hypothesis.
## How do you do a chi square test?
Calculate the chi square statistic x2 by completing the following steps:
• For each observed number in the table subtract the corresponding expected number (O — E).
• Square the difference [ (O —E)2 ].
• Divide the squares obtained for each cell in the table by the expected number for that cell [ (O – E)2 / E ].
• ## What does it mean when you have a small chi square value?
A very small chi square test statistic means that your observed data fits your expected data extremely well. A very large chi square test statistic means that the data does not fit very well.
## Why T test is used?
A t-test is an analysis of two populations means through the use of statistical examination; a t-test with two samples is commonly used with small sample sizes, testing the difference between the samples when the variances of two normal distributions are not known.
## What is a chi squared test?
A chi-squared test, also written as χ2 test, is any statistical hypothesis test where the sampling distribution of the test statistic is a chi-squared distribution when the null hypothesis is true. Chi-squared tests are often constructed from a sum of squared errors, or through the sample variance.
## What is an Anova test used for?
Analysis of Variance (ANOVA) is a statistical method used to test differences between two or more means. It may seem odd that the technique is called “Analysis of Variance” rather than “Analysis of Means.” As you will see, the name is appropriate because inferences about means are made by analyzing variance.
## What kind of test to use?
Types of Statistical TestsType of TestUsePaired T-testTests for the difference between two related variablesIndependent T-testTests for the difference between two independent variablesANOVATests the difference between group means after any other variance in the outcome variable is accounted for
## How do you know when to reject the null hypothesis?
Set the significance level, α, the probability of making a Type I error to be small — 0.01, 0.05, or 0.10. Compare the P-value to α. If the P-value is less than (or equal to) α, reject the null hypothesis in favor of the alternative hypothesis. If the P-value is greater than α, do not reject the null hypothesis.
## What is the chi square test of independence?
Home | Chi-Square Test of Independence. The Chi-Square test of independence is used to determine if there is a significant relationship between two nominal (categorical) variables. The frequency of each category for one nominal variable is compared across the categories of the second nominal variable.
## Can the result of a chi square test be negative?
An intuitive idea of the general shape of the distribution can also be obtained by considering this sum of squares. Since χ2 is the sum of a set of squared values, it can never be negative. The minimum chi squared value would be obtained if each Z = 0 so that χ2 would also be 0. There is no upper limit to the χ2 value.
## What is the degree of freedom in chi square test?
The Chi-Square Test. A statistical test that can test out ratios is the Chi-Square or Goodness of Fit test. Chi-Square Formula. Degrees of freedom (df) = n-1 where n is the number of classes. Let’s test the following data to determine if it fits a 9:3:3:1 ratio.
## What is the definition of the null hypothesis?
A null hypothesis is a type of hypothesis used in statistics that proposes that no statistical significance exists in a set of given observations. The null hypothesis attempts to show that no variation exists between variables or that a single variable is no different than its mean.
## What is the F test?
An F-test is any statistical test in which the test statistic has an F-distribution under the null hypothesis. It is most often used when comparing statistical models that have been fitted to a data set, in order to identify the model that best fits the population from which the data were sampled.
## How can the chi square test be used in genetics?
Genetic analysis often requires the interpretation of numbers in various phenotypic classes. In such cases, a statistical procedure called the χ2 (chi-square) test is used to help in making the decision to hold onto or reject the hypothesis.
## How do you calculate Chi Square in Excel?
Calculate the chi square p value Excel: Steps
• Step 1: Calculate your expected value.
• Step 2: Type your data into columns in Excel.
• Step 3: Click a blank cell anywhere on the worksheet and then click the “Insert Function” button on the toolbar.
• Step 4: Type “Chi” in the Search for a Function box and then click “Go.”
• ## What is the mean of the chi square distribution?
The Chi Square distribution is the distribution of the sum of squared standard normal deviates. The degrees of freedom of the distribution is equal to the number of standard normal deviates being summed. The mean of a Chi Square distribution is its degrees of freedom.
## What is the Z test?
A Z-test is any statistical test for which the distribution of the test statistic under the null hypothesis can be approximated by a normal distribution. Because of the central limit theorem, many test statistics are approximately normally distributed for large samples.
## What is the Fisher’s exact test?
Fisher’s exact test is a statistical significance test used in the analysis of contingency tables. Although in practice it is employed when sample sizes are small, it is valid for all sample sizes.
## What is chi square test of homogeneity?
Chi-Square Test of Homogeneity. This lesson explains how to conduct a chi-square test of homogeneity. The test is applied to a single categorical variable from two or more different populations. It is used to determine whether frequency counts are distributed identically across different populations.
## How do you determine the degrees of freedom?
just create an account. For instance, if a sample size were ‘n’ on a chi-square test, then the number of degrees of freedom to be used in calculations would be n – 1. To calculate the degrees of freedom for a sample size of N=9. subtract 1 from 9 (df=9-1=8).
Categories FAQ
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A line segment is bisected by a line with the equation 4 y + 9 x = 8 . If one end of the line segment is at (5 ,2 ), where is the other end?
Oct 19, 2016
The other end is at $\left(\frac{1295}{97} , \frac{554}{97}\right)$
Explanation:
Write the given line in slope-intercept form:
$y = - \frac{9}{4} x + 2$
Because bisector is perpendicular, the slope of the line segment will be the negative reciprocal of its bisector, $\frac{4}{9}$.
This reference gives us an equation for the distance from the point to the line. The distance from the point to the line is:
$d = | \frac{4 \left(2\right) + 9 \left(5\right) - 8}{\sqrt{{4}^{2} + {9}^{2}}} | = 45 \frac{\sqrt{97}}{97}$
The length of the line segment is twice this distance, $90 \frac{\sqrt{97}}{97}$.
From point $\left(5 , 2\right)$, we move to the right a distance, (x), and up a distance y , we know that y is $\frac{4}{9} x$, and we know that length of the hypotenuse formed by this right triangle is $90 \frac{\sqrt{97}}{97}$
${\left(90 \frac{\sqrt{97}}{97}\right)}^{2} = {x}^{2} + {\left(\frac{4}{9} x\right)}^{2}$
${90}^{2} / 97 = \frac{81}{81} {x}^{+} \frac{16}{81} {x}^{2}$
${x}^{2} = 81 \frac{{90}^{2}}{97} ^ 2$
$x = \frac{810}{97}$
$y = \frac{360}{97}$
The other end of the line is at point:
$\left(5 + \frac{810}{97} , 2 + \frac{360}{97}\right) = \left(\frac{1295}{97} , \frac{554}{97}\right)$
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# Negative Binomial Distribution Python Examples
In this post, you will learn about the concepts of negative binomial distribution explained using real-world examples and Python code. We will go over some of the following topics to understand negative binomial distribution:
• What is negative binomial distribution?
• What is difference between binomial and negative binomial distribution?
• Negative binomial distribution real-world examples
• Negative binomial distribution Python example
## What is Negative Binomial Distribution?
Negative binomial distribution is a discrete probability distribution representing the probability of random variable, X, which is number of Bernoulli trials required to have r number of successes. This random variable is called as negative binomial random variable. And, the experiment representing X number of Bernoulli trials required to product r successes is called as negative binomial experiment. Let’s understand the concept with the example of tossing a coin. Let’s say we want to continue flipping the coin until 3 heads come. This experiment of flipping a coin until 3 heads (r=3) come can be called as negative binomial experiment. And, the number of times the coin need to be flipped in each experiment represent the value of negative binomial random variable, X.
The negative binomial experiment would have the following properties:
• The experiment consists of X repeated trials until r successes occur.
• Each trial could result in only two outcomes – success or failure. In other words, the trial must be Bernoulli trial
• The probability of success on each trial is same as P
• The trials are independent of each other. Thus, in tossing coin experiment, head in one trial will not impact what appears in the next trial. It could be either head or tail.
The negative binomial distribution can be represented as the following when X represents the number of trials needed to get r successes where the probability of success in each trial is P.
$\Large B^{*}(X; r, P)$
.
The expected value / mean of the negative binomial distribution is defined as the expected / mean number of trials required to achieve r successes where the probability of success is P.
$\Large \mu_{X} = \frac{r}{P}$
.
### Alternative Definitions of Negative Binomial Distribution
Negative binomial distribution definitions vary with the definition of negative binomial random variable. Here are different definitions of the negative binomial random variable:
• Negative binomial random variable X represents number of successes before the binomial experiment results in k failures. The probability of success is P and failure is 1 – P.
$\Large \mu_{X} = k*\frac{P}{(1-P)}$
.
• Negative binomial random variable X represents number of failures before the binomial experiment results in r successes. The probability of success is P and failure is 1 – P.
$\Large \mu_{X} = r*\frac{1-P}{P}$
.
## Difference between Binomial & Negative Binomial Distribution
To understand the difference between binomial and negative binomial distribution, lets understand the difference between binomial and negative binomial experiment.
Binomial experiment is getting number of successes in N number of Bernoulli trials. The binomial random variable is number of successes. In binomial distribution, the number of trials are fixed.
Negative binomial experiment is about performing Bernoulli trials until r successes is achieved. The negative binomial random variable, X, is number of trials which are required to achieve r successes. In negative binomial distribution, the number of trials are not fixed.
In both the above cases, the following properties holds good:
• The trials have only two outcomes – success and failures. In other words, the trials are Bernoulli trials.
• The trials are independent of each other. The outcome from one trial does not impact the following trials.
• The probability of getting success in each trial is one and the same.
## Negative Binomial Distribution Real-world Examples
Here are some real-world examples of negative binomial distribution:
• Let’s say there is 10% chance of a sales person getting to schedule a follow-up meeting with the prospect in the phone call. The number of calls that the sales person would need to get 3 follow-up meetings would follow the negative binomial distribution. Thus, one will be able to calculate what is the probability that the sales person get 3 follow-up calls set in the 10th call. As per the alternative definitions, the number of rejections before which the sales person achieve 3 follow-up calls will follow the negative binomial distribution.
• Let’s say there is 30% chance that a basketball player will be able to get a perfect score in the free throw. The number of free throws that may be required to get the first score will follow the negative binomial distribution. Recall that this will also follow the geometric distribution. Geometric distribution, that way, is considered as the special case of negative binomial distribution. This example can also be read as the following – Number of free throw failures which will required to get the first perfect score will follow negative binomial distribution.
## Negative Binomial Distribution Python Example
Here is the Python code representing negative binomial distribution. Pay attention that Scipy.stats nbinom can be used to calculate probability distribution.
import numpy as np
from scipy.stats import nbinom
import matplotlib.pyplot as plt
#
# X = Discrete negative binomial random variable representing number of sales call required to get r=3 leads
# P = Probability of successful sales call
#
X = np.arange(3, 30)
r = 3
P = 0.1
#
# Calculate geometric probability distribution
#
nbinom_pd = nbinom.pmf(X, r, P)
#
# Plot the probability distribution
#
fig, ax = plt.subplots(1, 1, figsize=(8, 6))
ax.plot(X, nbinom_pd, 'bo', ms=8, label='nbinom pmf')
ax.plot(X, nbinom_pd, 'bo', ms=8, label='nbinom pmf')
plt.ylabel("Probability", fontsize="18")
plt.xlabel("X - No. of Sales Call", fontsize="18")
plt.title("Negative Binomial Distribution - No. of Sales Call Vs Probability", fontsize="18")
ax.vlines(X, 0, nbinom_pd, colors='b', lw=5, alpha=0.5)
Here is how the negative binomial distribution plot would look like:
## Conclusions
Here is the summary of what you learned in this post regarding negative binomial distribution:
• Negative binomial distribution is a discrete probability distribution which models the number of trials it will take to achieve r successes. Alternative definition is number of failures it would take to r successes.
• Negative binomial experiment is number of Bernoulli trials it will take to achieve r successes.
• Negative binomial random variable is number of Bernoulli trials to get r successes.
## Ajitesh Kumar
I have been recently working in the area of Data analytics including Data Science and Machine Learning / Deep Learning. I am also passionate about different technologies including programming languages such as Java/JEE, Javascript, Python, R, Julia, etc, and technologies such as Blockchain, mobile computing, cloud-native technologies, application security, cloud computing platforms, big data, etc. I would love to connect with you on Linkedin. Check out my latest book titled as First Principles Thinking: Building winning products using first principles thinking.
Posted in statistics. Tagged with .
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# How do you determine whether the function f(x)= 2x^3-3x^2-36x-7 is concave up or concave down and its intervals?
Aug 3, 2015
You can use the second derivative test.
#### Explanation:
The second derivative test allows you to determine the intervals on which a function is concave up or concave down by examining the sign of the second derivative around the inflexion point(s).
Inflexion point(s) are determined by making the second derivative equal to zero.
If the second derivative is positive on a given interval, then the function will be concave up on the same interval. Likewise, if the second derivative is negative on a given interval, the function will be concave down on said interval.
So, calculate the first derivative first - use the power rule
$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(2 {x}^{3} - 3 {x}^{2} - 36 x - 7\right)$
${f}^{'} = 6 {x}^{2} - 6 x - 36$
Next, calculate the second derivative. Once again, use the power rule
$\frac{d}{\mathrm{dx}} \left({f}^{'} \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(6 {x}^{2} - 6 x - 36\right)$
${f}^{' '} = 12 x - 6$
Find the inflexion point by solving ${f}^{' '} = 0$
$12 x - 6 = 0 \implies x = \textcolor{g r e e n}{\frac{1}{2}}$
Now look on how the second derivative behaves for values of $x$ smaller than $\frac{1}{2}$ and larger than $\frac{1}{2}$.
The two intervals you're going to use are
• $\left(- \infty , \frac{1}{2}\right)$
For this interval, ${f}^{' '}$ will be negative, which means that $f \left(x\right)$ is concave down.
• $\left(\frac{1}{2} , + \infty\right)$
For this interval, ${f}^{' '}$ will be positive, which implies that $f \left(x\right)$ is concave up.
So, $f \left(x\right)$ is concave down on $\left(- \infty , \frac{1}{2}\right)$ and concave up on $\left(\frac{1}{2} , + \infty\right)$. The point $\left(\frac{1}{2} , f \left(\frac{1}{2}\right)\right)$ is the only inflexion point on the graph of $f \left(x\right)$.
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# McGraw Hill My Math Grade 2 Chapter 3 Lesson 7 Answer Key Problem-Solving Strategy: Make a Model
All the solutions provided in McGraw Hill Math Grade 2 Answer Key PDF Chapter 3 Lesson 7 Problem-Solving Strategy: Make a Model will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 2 Answer Key Chapter 3 Lesson 7 Problem-Solving Strategy: Make a Model
Mark and Dan were in a walk-a-thon to raise money for their basketball team. Mark walked 35 blocks.
Dan walked 52 blocks. How many blocks did they walk in all?
1. Understand Underline what you know. Circle what you need to find.
2. Plan How will I solve the problem?
3. Solve make a model.
87 blocks
4. Check Is my answer reasonable? Explain.
Explanation:
Mark walked 35 blocks.
Dan walked 52 blocks
35 + 52 = 87
so, they walked 87 blocks.
Practice the Strategy
Sara, Mary, and Jonny are collecting shells at the beach. Sara finds 45 shells. Jonny finds 23 shells. Mary finds 15 shells. How many shells do they find?
1. Understand Underline what you know. Circle what you need to find.
2. Plan How will I solve the problem?
3. Solve I will ….
____ shells
4. Check Is my answer reasonable? Explain.
Explanation:
Sara finds 45 shells. Jonny finds 23 shells.
45 + 23 = 68
Mary finds 15 shells.
68 + 15 = 83
83 shells that they find
Apply the Strategy
Question 1.
Karenna ran 15 blocks Tuesday. She ran 20 blocks on Wednesday. How many blocks did she run in all?
____ blocks
35 blocks
Explanation:
Karenna ran 15 blocks Tuesday. She ran 20 blocks on Wednesday.
15 + 20 = 35
35 blocks that she run in all.
Question 2.
____ action figures
60 action figures
Explanation:
33 + 12 = 45
45 + 15 = 60
He will be having 60
Question 3.
Nora made 16 points in her basketball game. Jenna made 24 points and Alicia made 22. No one else scored on their team. What was their final score?
_____ points
62 points
Explanation:
16 + 24 = 40
40 + 22 = 62
62 was their final score.
Review the Strategies
Choose a strategy
• Write a number sentence.
• Find a pattern.
• Make a model.
Question 4.
Melissa has 23 pairs of earrings. She buys 2 more pairs. Then she is given 6 more pairs for her birthday. How many pairs of earrings does she have in all?
___ pairs
31 pairs.
Explanation:
Melissa has 23 pairs of earrings. She buys 2 more pairs.
23 + 2 = 25
Then she is given 6 more pairs for her birthday.
25+ 6 = 31
31 pairs of earrings that she have in all
Question 5.
Jessica made muffins for the bake sale. She made 24 blueberry muffins and 10 strawberry muffins. She also made 12 banana muffins. How many muffins did she make in all?
46 muffins.
Explanation:
She made 24 blueberry muffins and 10 strawberry muffins.
24 + 10 = 34
She also made 12 banana muffins.
34 + 12 = 46
46 muffins that she make in all.
Question 6.
Jacob and Luis collected glass bottles to recycle. The first day they collected 12 bottles. The second day they collected 15. The third day they collected 18. If this pattern continues, how many will they collect on the fourth day?
____ bottles
21 bottles
Explanation:
The first day they collected 12 bottles. The second day they collected 15.
3 x 4 = 12
3 x 5 = 15
3 x 6 = 18
3 x 7 = 21
so, he might be collected 21 bottles
### McGraw Hill My Math Grade 2 Chapter 3 Lesson 7 My Homework Answer Key
Underline what you know. Circle what you need to find. Make a model to solve.
Question 1.
Angela and Brittany count 8 windows in the first building, 5 in the next building, and 13 in the last building. How many windows did they count in all?
____ windows
26 windows
Explanation:
8 windows in the first building,
5 in the next building,
8 + 5 = 13
and 13 in the last building.
13 + 13 = 26
26 windows that they count in all.
Question 2.
Eric found 14 pennies in the couch, 6 pennies in his mom’s car, and 2 pennies under his bed. How many pennies did he find in all?
______ pennies
22 pennies
Explanation:
Eric found 14 pennies in the couch,
6 pennies in his mom’s car,
14 + 6 = 20
and 2 pennies under his bed.
20 + 2 = 22
22 pennies that he find in all.
Question 3.
The girls collected 57 bottle caps. The boys collected 42 caps. How many bottle caps did they collect in all?
_____ cups
99 bottle caps
Explanation:
The girls collected 57 bottle caps. The boys collected 42 caps.
57 + 42 = 99
99 bottle caps that they collect in all.
Explanation:
13 + 21 = 34
34 + 37 = 40
Fluency Practice
Question 1.
Explanation:
The sum of 14 and 3 is 17
14 + 3 = 17
Question 2.
Explanation:
The sum of 16 and 6 is 22
16 + 6 = 22
Question 3.
Explanation:
The sum of 13 and 8 is 21
13 + 8 = 21
Question 4.
Explanation:
The sum of 14 and 9 is 23
14 + 9 = 23
Question 5.
Explanation:
The sum of 15 and 6 is 21
15 + 6 = 21
Question 6.
Explanation:
The sum of 16 and 3 is 19
16 + 3 = 19
Question 7.
Explanation:
The sum of 11 and 9 is 20
11 + 9 = 20
Question 8.
Explanation:
The sum of 12 and 2 is 14
12 + 2 = 14
Question 9.
Explanation:
The sum of 14 and 0 is 14
14 + 0 = 14
Question 10.
Explanation:
The sum of 18 and 5 is 23
18 + 5 = 23
Question 11.
Explanation:
The sum of 19 and 1 is 20
19 + 1 = 20
Question 12.
Explanation:
The sum of 17 and 2 is 19
17 + 2 = 19
Fluency Practice
Question 1.
Explanation:
The sum of 13 and 6 is 19
13 + 6 = 19
Question 2.
Explanation:
The sum of 16 and 2 is 18
16 + 2 = 18
Question 3.
Explanation:
The sum of 13 and 8 is 21
13 + 8 = 21
Question 4.
Explanation:
The sum of 26 and 5 is 31
26 + 5 = 31
Question 5.
Explanation:
The sum of 13 and 1 is 14
13 + 1 = 14
Question 6.
Explanation:
The sum of 13 and 3 is 16
13 + 3 = 16
Question 7.
Explanation:
The sum of 21 and 9 is 30
21 + 9 = 30
Question 8.
Explanation:
The sum of 16 and 2 is 18
16 + 2 = 18
Question 9.
Explanation:
The sum of 11 and 0 is 11
11 + 0 = 11
Question 10.
Explanation:
The sum of 13 and 2 is 15
13 + 2 = 15
Question 11.
Explanation:
The sum of 13 and 4 is 17
13 + 4 = 17
Question 12.
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# Multivariate Function, Chain Rule / Multivariable Calculus
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## 1. Multivariate Function Definition
A multivariate function has several different independent variables. While “classroom” calculus usually deals with one variable, you’ll deal with their multivariate counterparts in applied sciences. While calculus commonly deals with functions with just one independent and dependent variable, multivariate functions are far more useful in everyday life. This is because most real-world processes depend on more than one parameter.
The functions we typically deal with have a single input and single output. “Input” and “output” refers to independent and dependent variables, typically denoted with x and y
## Notation
In calculus we write multivariate functions as having a dependent variable z and independent variables x and y:
z = f(x, y)
The inputs of a multivariate function the domain. The set of all possible outputs is the range
## 2. What is Multivariable Calculus?
Multivariable calculus (also called multivariate calculus) deals with functions of several variables. Many techniques can be transferred from single variable calculus, including finding derivatives and integrals.
Multivariable calculus is a huge field that usually covers an entire semester, usually after at least one full year of single variable calculus.
## Comparison of Single and Multivariable Calculus
Functions of one variable (left) are graphed on an x-y axis; The graph on the right is multivariate and is graphed on a x, y, z axis.
In single variable (univariate) calculus, the focus is on a function with a single input and a single output, like y = f(x). With multivariable calculus, more than one input is involved, such as z = x2 + y2.
You can also have multiple outputs, like:
• x = f(t),
• y = g(t),
• z = h(t).
While derivatives in single variable calculus give you a tangent line, in multivariate calculus the result is a tangent vector. The tangent vector is just your usual vector, extended to a 3-dimensional plane.
Another interesting difference is limits. In single variable calculus, you have a limit from the left and a limit from the right. When both of those limits exist and agree with each other, then we say there’s a limit. With multivariable calculus, this is a lot more challenging, because discontinuities don’t happen on a single line graph: they happen to 3D objects, which you can approach from multiple sides. Discontinuities are no longer simple broken lines, they often behave more like black holes in space, with the function sucked into (or blown out of) the point of discontinuity.
Hue-luminance plot of exp(1/z), centered on the essential singularity at zero. The function behaves differently depending on which direction you approach the function from. Credit: Functor Salad | Wikimedia Commons.
## 4. 3. The Multivariate Chain Rule
In the multivariate chain rule (or multivariable chain rule) one variable is dependent on two or more variables. The chain rule consists of partial derivatives
For the function f(x,y) where x and y are functions of variable t, we first differentiate the function partially with respect to one variable and then that variable is differentiated with respect to t. The chain rule is written as:
## Example
Let’s take a look at an example that shows how the chain rule works.
Example: The number of footballs a sporting goods store sells over time is x(t) t4 + 1; y(t) = 3t2 + 6 describes the price of footballs over time. Let z(t) = xy represent the revenue the store earns from the sale of footballs. How does revenue change with respect to time?
While there are several ways you could approach the solution, the easiest way to solve is to use the multivariate chain rule.
First, note that as a function of t, z is a function of a single variable. In this example, we have:
• Δ z / Δ x = y
• Δ z / Δ y = x
• dx/dt = 4t3
• dy/dt = 6t
Apply the chain rule formula from above to get the change in revenue over time:
dz/dt = (y)(4t3) + (x)(6t)
## References
Evans, M.; Hastings, N.; and Peacock, B. Statistical Distributions, 3rd ed. New York: Wiley, p. 5, 2000.
CITE THIS AS:
Stephanie Glen. "Multivariate Function, Chain Rule / Multivariable Calculus" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/multivariate-function/
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To solve this problem we need to start by turning the available information into equations. Let the number of animals be c, p, h, and r for cows, pigs, chickens, and rabbits respectively. Then we have:
c + p + h + r = 100
10c + 3p + h + 0.5r = 100
c > p
h > r
Let's also assume that there is at least 1 of each animal, since otherwise it would not be on the list. From the second equation we can see that 1 <= c <= 9 (i. e., the number of cows must be between 1 and 9). In fact, since c > p and p > 0, there must be at least 2 cows. This leads to 8 cases as follows:
Case 1: c = 2
2 + p + h + r = 100
20 + 3p + h + 0.5r = 100
Therefore, –2p + 0.5r = 18
Since the number of pigs is less than the number of cows, p = 1. So r = 40, and h = 57. This solution satisfies all the conditions of the problem. But is it unique?
Case 2: c = 3
3 + p + h + r = 100
30 + 3p + h + 0.5r = 100
Therefore, –2p + 0.5r = 27
Solving for r gives r = 54 + 4p. Since p > 0, the number of rabbits must be greater than 54. But if r > 54, then there cannot be more chickens than rabbits because that would put the total number of animals over 100. Therefore, Case 2 fails to give a solution.
Cases 3–8: c > 3
Combining the first two equations where c is known gives
–2p + 0.5r = (100 – c) – (100 – 10c) = 9c
Solving for r gives r = 18c + 4p. As c increases, r increases too, so we can apply the reasoning from Case 2 to show that r is never less than 50, so h cannot be greater than r. Therefore, none of the remaining cases lead to a solution.
The solution from Case 1 is unique. The list must have called for:
2 Cows @
\$10 ea. = \$20
1 Pig @
\$3 ea. = \$3
57 Chickens @
\$1 ea. = \$57
40 Rabbits @
50¢ ea. = \$20
100 animals for \$100
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# What is 6 of 15? Learn How to Master Fractions!
When it comes to studying math, fractions can be one of the most challenging topics for learners. Especially when dealing with complex fractions, like “What is 6 of 15?” In this article, we will guide you on how to master fractions.
## Why learn fractions?
Understanding fractions is essential in day-to-day life. Whether you’re dividing a pizza among friends, calculating the recipe portion for a dish, or measuring your carpentry materials, fractions play a crucial role in making these tasks easy and accurate.
Fractions can seem confusing at first, but once you know how to work with them, they can become your best friend. Here’s how to get started.
## What are fractions?
A fraction is a part of a whole value that is typically represented in mathematical terms by dividing the numerator by the denominator separated by a slash. The numerator represents the number of parts you have, while the denominator represents the total number of parts that make up a whole.
### Types of Fractions
Fractions can be classified into different types based on their values and how they are written. These include:
• Proper Fractions
• Improper Fractions
• Mixed Fractions
• Equivalent Fractions
• Unit Fractions
• Decimal Fractions
## What is meant by 6 of 15?
6 of 15 is a fraction that represents six parts of a whole value that is divided into fifteen equal parts.
When calculating fractions, it is essential to remember their representation, which parts are being represented, and the total number of parts they represent.
### Converting 6 of 15 to a Decimal
To convert 6 of 15 to a decimal, you can use the following formula:
Decimal = numerator ÷ denominator
In this case, the numerator is 6, and the denominator is 15. So, 6 ÷ 15 = 0.40 (rounded to two decimal places).
## How to Simplify Fractions
Mathematicians can simplify fractions to make them easier to work with by reducing all numbers belonging to the fraction to their lowest possible values.
### Finding the Greatest Common Factor (GCF)
The greatest common factor is the largest possible number that can divide two or more values without leaving any remainder that suits them.
To find the GCF of 6 and 15:
1. List down all the factors of the two numbers, 6 and 15:
• For 6: 1, 2, 3, 6
• For 15: 1, 3, 5, 15
2. Identify the common factors in both lists; in this case, the number 3.
3. The GCF of 6 and 15 is 3.
To simplify 6 of 15, divide both the numerator and denominator by their GCF, which is 3. The result is 2/5.
To add and subtract fractions, the denominators must be equal. If they are not, you will have to convert them to equivalent fractions with a common denominator.
### Converting Fractions to Equivalent Fractions
To convert fractions to equivalent fractions with a common denominator, follow these steps:
1. Determine the least common multiple (LCM) of the denominators. The LCM of 3 and 5 is 15.
2. Multiply both the numerator and denominator of each fraction with its respective missing factor to obtain fractions with the LCM denominator.
3. Add or subtract the fractions accordingly.
4. Simplify the fraction if necessary.
## Multiplying fractions
To multiply fractions, multiply the two numerators separately and then multiply the two denominators. The result is your new fraction.
### Multiplying 6 of 15 with 2 of 7
To multiply 6 of 15 with 2 of 7, follow these steps:
1. Multiply the two numerators: 6 x 2 = 12.
2. Multiply the two denominators: 15 x 7 = 105.
3. The answer is 12 of 105.
## Dividing Fractions
Dividing fractions is the same as multiplying them, but you must first flip the second fraction before multiplying them. This means that you switch the numerator and denominator of the second fraction.
### Dividing 6 of 15 with 2 of 7
To divide 6 of 15 with 2 of 7, follow these steps:
1. Flip the second fraction: 2 of 7 becomes 7 of 2
2. Multiply the first and second fractions: 6 of 15 x 7 of 2 = 42 of 30.
3. Simplify the fraction: 42 of 30 = 7 of 5.
## Summary
Fractions are an essential part of math that is commonly used in everyday life. To master fractions, you must first understand their representation, types, and how to work with them. You can simplify, add, subtract, multiply and divide fractions once you have mastered these basics.
### Q. What is a fraction?
A fraction is a part of a whole value that is typically represented in mathematical terms by dividing the numerator by the denominator separated by a slash.
### Q. How do you convert fractions to decimals?
To convert fractions to decimals, divide the numerator by the denominator.
### Q. What is the greatest common factor?
The greatest common factor is the largest possible number that can divide two or more values without leaving any remainder that suits them.
### Q. How do you simplify fractions?
To simplify fractions, find the greatest common factor of the numerator and denominator and divide both of the by that number.
### Q. Can you add and subtract fractions?
Yes, you can add and subtract fractions, but the denominators must be equal.
### Q. How do you multiply fractions?
To multiply fractions, multiply the two numerators together and the two denominators together.
### Q. How do you divide fractions?
To divide fractions, flip the second fraction’s numerator and denominator and then multiply the first fraction with the newly flipped second fraction.
## References
• Math is Fun. (2021). Fractions. Math is Fun. https://www.mathsisfun.com/fractions-menu.html
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NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3
# NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3
## NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3
NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3.
### Ex 15.3 Class 8 Maths Question 1.
Draw the graphs for the following tables of values, with suitable scales on the axes.
(a) Cost of apples
Number of apples 1 2 3 4 5 Cost (in ₹) 5 10 15 20 25
(b) Distance travelled by a car
Time (in hours) 6 a.m. 7 a.m. 8 a.m. 9 a.m. Distances (in km) 40 80 120 160
(i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m.?
(ii) What was the time when the car had covered a distance of 100 km since it’s start?
(c) Interest on deposits for a year.
Deposit (in ₹) 1000 2000 3000 4000 5000 Simple Interest (in ₹) 80 160 240 320 400
(i) Does the graph pass through the origin?
(ii) Use the graph to find the interest on ₹ 2500 for a year.
(iii) To get an interest of ₹ 280 per year, how much money should be deposited?
Solution:
(a) To draw a graph of given data, we represent ‘Number of apples’ on the x-axis and ‘Cost (in ₹)’ on the y-axis. The ordered pairs (1, 5), (2, 10), (3, 15), (4, 20) and (5, 25) are plotted as points and are joined by the line segments as shown in the figure. This gives the required graph.
(b) To draw time-distance graph, we represent the ‘Time (in hours)’ on the x-axis and the ‘Distances (in km)’ on the y-axis.
Scale: 2 units = 1 hour on x-axis and 1 unit = 10 km on y-axis.
Plot the points (6, 40), (7, 80), (8, 120) and (9, 160).
Join them by the line segments to get the required graph as shown below:
(i) We observe from the graph that the distance covered by the car during 7.30 a.m. to 8.00 a.m. is 20 km. ( 120 – 100 = 20 km)
(ii) The time when the car had covered 100 km since it’s start is 7.30 a.m.
(c) To draw principal-simple interest graph, we represent ‘deposits (in ₹)’ on the x-axis and the ‘simple interest (in ₹)’ on the y-axis.
Scale: 1 unit = ₹ 1000 on x-axis.
1 unit = ₹ 80 on y-axis.
Plot the points (1000, 80), (2000, 160), (3000, 240), (4000, 320) and (5000, 400).
Join these points to get the required graph.
(i) Yes, the graph passes through the origin.
(ii) Corresponding to the principal ₹ 2500 on the x-axis, we get the interest to be ₹ 200 on the y-axis.
(iii) Corresponding to the interest ₹ 280 per year on the y-axis, we get the principal to be ₹ 3500 on the x-axis.
### Ex 15.3 Class 8 Maths Question 2.
Draw a graph for the following.
(i)
Side of square (in cm) 2 3 3.5 5 6 Perimeter (in cm) 8 12 14 20 24
Is it a linear graph?
(ii)
Side of square (in cm) 2 3 4 5 6 Area (in cm2) 4 9 16 25 36
Is it a linear graph?
Solution:
(i) To draw the graph of the given data, we represent the ‘side of square (in cm)’ on the x-axis and the ‘perimeter (in cm)’ on the y-axis. Scale: 1 unit = 1 cm on x-axis and y-axis.
Plot the points (2, 8), (3, 12), (3.5, 14), (5, 20) and (6, 24). Join these points to get the required graph.
Yes, it is a linear graph.
(ii)
To draw the graph of the given data, we represent the ‘side of square (in cm)’ on the x-axis and the ‘area (in cm2)’ on the y-axis. Scale: 1 unit = 1 cm on x-axis and 1 unit = 1 cm2 on y-axis.
Plot the points (2, 4), (3, 9), (4, 16), (5, 25) and (6, 36). Join these points to get the required graph.
No, it is not a linear.
You can also like these:
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
NCERT Solutions for Maths Class 12
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# Solving 35x2+-83x+60 using the Quadratic Formula
For your equation of the form "ax2 + bx + c = 0," enter the values for a, b, and c:
a x2 + b x + c = 0
Reset
You entered:
35x2+-83x+60=0.
There are no solutions in the real number domain.
There are two complex solutions: x = 1.1857142857143 + 0.55530833501648i, and x = 1.1857142857143 - 0.55530833501648i,
where i is the imaginary unit.
## Here's how we found that solution:
You entered the following equation:
(1) 35x2+-83x+60=0.
For any quadratic equation ax2 + bx + c = 0, one can solve for x using the following equation, which is known as the quadratic formula:
(2)
In the form above, you specified values for the variables a, b, and c. Plugging those values into Eqn. 1, we get:
(3) $$x=--83\pm\frac{\sqrt{-83^2-4*35*60}}{2*35}$$
which simplifies to:
(4) $$x=--83\pm\frac{\sqrt{6889-8400}}{70}$$
Now, note that b2-4ac is a negative number. Specifically in our case, 6889 - 8400 = -1511.
(5) $$x=--83\pm\frac{\sqrt{-1511}}{70}$$
This means that our solution will require finding the square root of a negative number. There is no real number solution for this, so our solution will be a complex number (that is, it will involve the imaginary number i, defined as the square root of -1.).
Let's calculate the square root:
(6) $$x=--83\pm\frac{38.871583451154i}{70}$$
This equation further simplifies to:
(7) $$x=-\frac{--83}{70}\pm0.55530833501648i$$
Solving for x, we find two solutions which are both complex numbers:
x = 1.1857142857143 + 0.55530833501648i
and
x = 1.1857142857143 - 0.55530833501648i
Both of these solutions are complex numbers.
These are the two solutions that will satisfy the equation 35x2+-83x+60=0.
### Notes
What is a quadratic equation? Any equation that can be written in the form:
ax2 + bx + c = 0.
\ In this equation, a, b, and c are constants. X is unknown. The constants a and b, are referred to as coefficients. Furthermore, it is worth pointing out that a cannot be equal to 0. Otherwise, the equation ceases to be a quadratic equation, and becomes a linear equation.
Compared to solving a linear equation, solving a quadratic equation requires a few more steps. However, you have this handy-dandy quadratic equation calculator. All kidding aside, quadratic equations can be always solved using the quadratic formula, which is the same technique used by this quadratic equation calculator. Try it, and it will explain each of the steps to you. The quadratic formula is written:
When you compute a solution to a quadratic equation, you will always find 2 values for x, called "roots". These roots may both be real numbers or, they may both be complex numbers. Depending on the values of a, b, and c, both roots may be equal, meaning there will only be one solution for x.
Quadratic equations are an important part of mathematics. Quadratic equations are needed to compute answers in many real-world fields, including physics, biology and business.
In our equation, a cannot be zero. However, b can be zero, and so can c.
We this quadratic equation calculator is useful to you. We encourage you to try it with different values, and to read the explanation for how to reach your answer. But we totally understand if you just want to use it to find the answers you're looking for. Thank you for your interest in Quadratic-Equation-Calculator.com.
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> > > > Problem 5BSC
# use the Kruskal - Wallis test. Triathlon Times Jeff Parent
## Problem 5BSC Chapter 13.5
Elementary Statistics | 12th Edition
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Problem 5BSC
use the Kruskal - Wallis test?. Triathlon Times? Jeff Parent is a statistics instructor who participates in triathlons. Listed below are times (in minutes and seconds) he recorded while riding a bicycle for five laps through each mile of a 3-mile loop. Use a 0.05 significance level to test the claim that the samples are from populations with the same median. What do the data suggest? Mile 1 3:15 3:24 3:23 3:22 3:21 Mile 2 3:19 3:22 3:21 3:17 3:19 Mile 3 3:34 3:31 3:29 3:31 3:29
Step-by-Step Solution:
Solution 5BSC Step 1 Kruskal-Wallis test satisfies the following two requirements. 1. Sample data must be independent and drawn according to simple random sampling. 2. Each sample must contain minimum of five entries. By using the data from exercise, we see that the samples are drawn at random, where the three Mile loop are independent to each other. Hence the first requirement is satisfied. Now, we see that the sample size of Mile 1 is 5, sample size of Mile 2 is 5 and sample size of Mile 3 is 5. Hence the second requirement is satisfied. Therefore, the two requirements of Kruskal-Wallis test are satisfied. Step 2 By using = 0.05 level of significance we need t o test the claim that the samples are from populations with the same median. The Hypotheses can be expressed as H0 The samples are from population with the same median. H1 The samples are from population with the different median. All these three mile with Mile 1, Mile 2, Mile 3 samples are combined together to get a big sample. We first order the absolute values of the difference scores and assign rank from 1 through ‘n’ to the smallest through largest absolute values of the difference scores, and assign the mean rank when there are ties in the absolute values of the difference scores. Mile 1 Mile 2 Mile 3 3:15(1) 3:19(3.5) 3:34(15) 3:24(10) 3:22(7.5) 3:31(13.5) 3:23(7.5) 3:21(5.5) 3:29(11.5) 3:22(7.5) 3:17(2) 3:31(13.5) 3:21(5.5) 3:19(3.5) 3:29(11.5) The smallest value is assigned as the rank 1, here 3:15 is the smallest value and so on up to the largest value 3:34 is assigned as rank 15.
Step 3 of 3
##### ISBN: 9780321836960
This full solution covers the following key subjects: mile, times, use, test, participates. This expansive textbook survival guide covers 121 chapters, and 3629 solutions. The full step-by-step solution to problem: 5BSC from chapter: 13.5 was answered by , our top Statistics solution expert on 03/15/17, 10:30PM. This textbook survival guide was created for the textbook: Elementary Statistics, edition: 12. Since the solution to 5BSC from 13.5 chapter was answered, more than 581 students have viewed the full step-by-step answer. Elementary Statistics was written by and is associated to the ISBN: 9780321836960. The answer to “use the Kruskal - Wallis test?. Triathlon Times? Jeff Parent is a statistics instructor who participates in triathlons. Listed below are times (in minutes and seconds) he recorded while riding a bicycle for five laps through each mile of a 3-mile loop. Use a 0.05 significance level to test the claim that the samples are from populations with the same median. What do the data suggest? Mile 1 3:15 3:24 3:23 3:22 3:21 Mile 2 3:19 3:22 3:21 3:17 3:19 Mile 3 3:34 3:31 3:29 3:31 3:29” is broken down into a number of easy to follow steps, and 87 words.
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proving-square-root-of-3-is-irrational-number-sqrt-3-is-irrational-number-proof
# Interactive video lesson plan for: Proving Square Root of 3 is Irrational number | Sqrt (3) is Irrational number Proof
#### Activity overview:
Proving Square Root of 3 is Irrational number | Sqrt (3) is Irrational number Proof
http://www.learncbse.in/ncert-class-10-math-solutions/
http://www.learncbse.in/ncert-solutions-class-10th-maths-chapter-1-real-numbers/
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In this video, Irrationality theorem is explained and proof of sqrt(3) is irrational number is illustrated in detail.More emphasis is laid on rational numbers, co-primes, assumption and contradiction of assume statement.
Irrational Numbers Defination:
A number ‘s ’ is called irrational if it cannot be written in the form,p/q where p and q are integers and q is not 0.
Fundamental Theorem of Arithmetic.
Let p be a prime number. If p divides a^2, then p divides a, where a is a positive integer.
00:02 Prove that sqrt(3) is irrational.
00:16 Assume contrary of given statement.
That is sqrt(3) is rational number which can be expressed in the for of p/q.Where p and q are
02:11 We get p and q are not co-prime.This bring contradiction to our statement.
02:30 Contardiction to assumption
02:35 sqrt(3) is irrational
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CBSE solutions for class 10 maths Chapter 1 Real Numbers Exercise 1.4
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CBSE class 10 maths NCERT Solutions Chapter 1 Real Numbers
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NCERT solutions for CBSE class 10 maths Real Numbers
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proving Irrational Numbers
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2.5 Applying the Power Rule
3 min readjanuary 29, 2023
Welcome back to AP Calculus with Fiveable! We are now diving into one of the most valuable fundamental concepts in calculus: the Power Rule. This is the first of many derivative rules that you’re going to learn about!
⚡ The Power Rule
The Power Rule states that if $f(x) = x^n$, where $n$ is constant, then the derivative $f'(x)$ is given by:
$f'(x) = n \cdot x^{(n-1)}$
So, the Power Rule provides a shortcut for finding the derivative without using the limit definition of a derivative. Wasn’t doing that annoying?!
🏋️♂️ Practice Problems
Let’s work on a few questions to get the Power Rule down!
1. Given $f(x) = x^4$, find $f'(x)$.
2. Given $f(x) = \frac{1}{x^5}$, find $f'(x)$.
3. Given $f(x) = \sqrt{x}$, find $f'(x)$.
4. Given $f(x) = x^6 +2x^4-10$, find $f'(x)$.
💡 Before we reveal the answers, remember:
1. The Power Rule with fractions can be tricky! Sometimes rewriting the equation can help.
2. The derivative of any constant is zero.
👀 Answers to Practice Problems
Note how for many of these problems, the equations were rewritten before solving for the derivative using the power rule.
1. $f'(x) = 4 \cdot x^{(4-1)}$ $= 4x^3$
2. $f(x) = x^{-5}$
$f'(x) = -5 \cdot x^{(-5-1)}$ $= -5x^{-6}$ $= \frac{-5}{x^{6}}$
3. $f(x) = x^{1/2}$
$f'(x) = \frac{1}{2} \cdot x^\frac{-1}{2}$ $= \frac {1}{2\sqrt{x}}$
4. $f'(x)=6x^5 + 8x^3$
Yep! That’s it. This lesson was super short. Want to jump into the rest of the derivative rules you have to know? ⏭️
2.5 Applying the Power Rule
3 min readjanuary 29, 2023
Welcome back to AP Calculus with Fiveable! We are now diving into one of the most valuable fundamental concepts in calculus: the Power Rule. This is the first of many derivative rules that you’re going to learn about!
⚡ The Power Rule
The Power Rule states that if $f(x) = x^n$, where $n$ is constant, then the derivative $f'(x)$ is given by:
$f'(x) = n \cdot x^{(n-1)}$
So, the Power Rule provides a shortcut for finding the derivative without using the limit definition of a derivative. Wasn’t doing that annoying?!
🏋️♂️ Practice Problems
Let’s work on a few questions to get the Power Rule down!
1. Given $f(x) = x^4$, find $f'(x)$.
2. Given $f(x) = \frac{1}{x^5}$, find $f'(x)$.
3. Given $f(x) = \sqrt{x}$, find $f'(x)$.
4. Given $f(x) = x^6 +2x^4-10$, find $f'(x)$.
💡 Before we reveal the answers, remember:
1. The Power Rule with fractions can be tricky! Sometimes rewriting the equation can help.
2. The derivative of any constant is zero.
👀 Answers to Practice Problems
Note how for many of these problems, the equations were rewritten before solving for the derivative using the power rule.
1. $f'(x) = 4 \cdot x^{(4-1)}$ $= 4x^3$
2. $f(x) = x^{-5}$
$f'(x) = -5 \cdot x^{(-5-1)}$ $= -5x^{-6}$ $= \frac{-5}{x^{6}}$
3. $f(x) = x^{1/2}$
$f'(x) = \frac{1}{2} \cdot x^\frac{-1}{2}$ $= \frac {1}{2\sqrt{x}}$
4. $f'(x)=6x^5 + 8x^3$
Yep! That’s it. This lesson was super short. Want to jump into the rest of the derivative rules you have to know? ⏭️
|
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TEXT
# Reading: Solving Systems with Inverses
Nancy plans to invest \$10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Nancy invest in each bond? What is the best method to solve this problem? There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix.
## Finding the Inverse of a Matrix
We know that the multiplicative inverse of a real number is a–1, and $\displaystyle{a}{a}^{{-{1}}}={a}^{{-{{1}}}}{a}={(\frac{{1}}{{a}})}{a}={1}$. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A–1 equals the identity matrix. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by In where n represents the dimension of the matrix. The next two equations are the identity matrices for a 2 × 2 matrix and a 3 ×3 matrix, respectively. [latex-display]\displaystyle{I}_{{3}}=\left[\matrix{{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}}\right][/latex-display] The identity matrix acts as a 1 in matrix algebra. For example, AI = IA =A A matrix that has a multiplicative inverse has the properties [latex-display]\displaystyle\left[\matrix{{A}{A}^{{-{{1}}}}={I}\\{A}^{{-{{1}}}}{A}={I}}\right][/latex-display] A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, $\displaystyle{A}{A}^{{-{{1}}}}={A}^{{-{{1}}}}{A}={I}$, is a requirement. Not all square matrices have an inverse, but if Ais invertible, then A–1 is unique. We will look at two methods for finding the inverse of a 2 × 2matrix and a third method that can be used on both 2 × 2and 3 × 3matrices.
## The Identity Matrix and Multiplicative Inverse
The identity matrixIn,is a square matrix containing ones down the main diagonal and zeros everywhere else.
#### 2×2
[latex-display]\displaystyle{I}_{{2}}=\left[\matrix{{1}&{0}\\{0}&{1}}\right][/latex-display]
#### 3×3
[latex-display]\displaystyle{I}_{{3}}=\left[\matrix{{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}}\right][/latex-display] If is an n × n matrix and Bis an n × n matrix such that AB = BA = In, then B = A–1, the multiplicative inverse of a matrix A.
### Example 1
Given matrix A, show that AI = IA =A $\displaystyle{A}=\left[\matrix{{3}&{4}\\-{2}&{5}}\right]$
#### Solution
Use matrix multiplication to show that the product of A and the identity is equal to the product of the identity and A [latex-display]\displaystyle{A}{I}=\left[\matrix{{3}&{4}\\-{2}&{5}}\right]\left[\matrix{{1}&{0}\\{0}&{1}}\right]=\left[\matrix{{3}\cdot{1}+{4}\cdot{0}&{3}\cdot{0}+{4}\cdot{1}\\-{2}\cdot{1}+{5}\cdot{0}&-{2}\cdot{0}+{5}\cdot{1}}\right]=\left[\matrix{{3}&{4}\\-{2}&{5}}\right][/latex-display] [latex-display]\displaystyle{A}{I}={\left[\matrix{{1}&{0}\\{0}&{1}}\right]{\left[\matrix{{3}&{4}\\-{2}&{5}}\right]={\left[\matrix{{1}\cdot{3}+{0}\cdot-{2}&{1}\cdot{4}+{0}\cdot{5}\\{0}\cdot{3}+{1}\cdot-{2}&{0}\cdot{4}+{1}\cdot{5}}\right]={\left[\matrix{{3}&{4}\\-{2}&{5}}\right][/latex-display]
### How To
Given two matrices, show that one is the multiplicative inverse of the other.
1. Given matrix A of order n × n and matrix B of order n × n multiply AB
2. If AB = I, then find the product BA. If BA = I, then B = A–1 and A = B–1
### Example 2
1. Show that the given matrices are multiplicative inverses of each other. $\displaystyle{A}={\left[\matrix{{1}&{5}\\-{2}&-{9}}\right],{B}={\left[\matrix{-{9}&-{5}\\{2}&{1}}\right]$
2. Show that the following two matrices are inverses of each other $\displaystyle{A}={\left[\matrix{{1}&{4}\\-{1}&-{3}}\right],{B}={\left[\matrix{-{3}&-{4}\\{1}&{1}}\right]$
#### Solutions
1. Multiply A and B. If both products equal the identity, then the two matrices are inverses of each other. [latex-display]\displaystyle{A}{B}={\left[\matrix{{1}&{5}\\-{2}&{9}}\right]\cdot{\left[\matrix{-{9}&-{5}\\{2}&{1}}\right]={\left[\matrix{{1}{(-{9})}+{5}{({2})}&{1}{(-{5})}+{5}{({1})}\\-{2}{(-{9})}-{9}{({2})}&-{2}{(-{5})}-{9}{({1})}}\right]={\left[\matrix{{1}&{0}\\{0}&{1}}\right][/latex-display] A and B are inverses of each other.
2. $\displaystyle{A}{B}={\left[\matrix{{1}&{4}\\-{1}&-{3}}\right]\cdot{\left[\matrix{-{3}&-{4}\\{1}&{1}}\right]={\left[\matrix{{1}{(-{3})}+{4}{({1})}&{1}{(-{4})}+{4}{({1})}\\-{1}{(-{3})}+-{3}{({1})}&-{1}{(-{4})}+-{3}{({1})}}\right]={\left[\matrix{{1}&{0}\\{0}&{1}}\right]$
## Finding the Multiplicative Inverse Using Matrix Multiplication
We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication.
### Example 3
Use matrix multiplication to find the inverse of the given matrix. [latex-display]\displaystyle{A}={\left[\matrix{{1}&-{2}\\{2}&-{3}}\right][/latex-display]
#### Solution
For this method, we multiply by a matrix containing unknown constants and set it equal to the identity. [latex-display]\displaystyle{\left[\matrix{{1}&-{2}\\{2}&-{3}}\right]{\left[\matrix{{a}&{b}\\{c}&{d}}\right]={\left[\matrix{{1}&{0}\\{0}&{1}}\right][/latex-display] Find the product of the two matrices on the left side of the equal sign. [latex-display]\displaystyle{\left[\matrix{{1}&-{2}\\{2}&-{3}}\right]{\left[\matrix{{a}&{b}\\{c}&{d}}\right]={\left[\matrix{{1}{a}-{2}{c}&{1}{b}-{2}{d}\\{2}{a}-{3}{c}&{2}{b}-{3}{d}}\right][/latex-display] Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0. [latex-display]\displaystyle{1}{a}-{2}{c}={1} {R}_{{1}}[/latex-display] [latex-display]\displaystyle{2}{a}-{3}{c}={0} {R}_{{2}}[/latex-display] Using row operations, multiply and add as follows: $\displaystyle{(-{2})}{R}_{{1}}+{R}_{{2}}\rightarrow{R}_{{2}}$. Add the equations, and solve for c. [latex-display]\displaystyle{1}{a}-{2}{c}={1}[/latex-display] [latex-display]\displaystyle{0}+{1}{c}=-{2}[/latex-display] [latex-display]\displaystyle{c}=-{2}[/latex-display] Back-substitute to solve for a. [latex-display]\displaystyle{\left[\matrix{{a}-{2}{(-{2})}={1}\\{a}+{4}={1}\\{a}=-{3}}\right][/latex-display] Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity. [latex-display]\displaystyle{\left[\matrix{{1}{b}-{2}{d}={0}&{R}_{{1}}\\{2}{b}-{3}{d}={1}&{R}_{{2}}}\right][/latex-display] Using row operations, multiply and add as follows: $\displaystyle{(-{2})}{R}_{{1}}+{R}_{{2}}={R}_{{2}}$. Add the two equations and solve for [latex-display]\displaystyle{\left[\matrix{{1}{b}-{2}{d}={0}\\{0}+{1}{d}={1}\\{d}={1}}\right][/latex-display] Once more, back-substitute and solve for [latex-display]\displaystyle{\left[\matrix{{b}-{2}{({1})}={0}\\{b}-{2}={0}\\{b}={2}}\right][/latex-display] [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{-{3}&{2}\\-{2}&{1}}\right][/latex-display]
## Finding the Multiplicative Inverse by Augmenting with the Identity
Another way to find the multiplicative inverse is by augmenting with the identity. When matrix A is transformed into I, the augmented matrix I transforms into A–1 For example, given [latex-display]\displaystyle{A}={\left[\matrix{{2}&{1}\\{5}&{3}}\right][/latex-display] augment with the identity [latex-display]\displaystyle{\left[\matrix{{2}&{1}&{\mid}&{1}&{0}\\{5}&{3}&{\mid}&{0}&{1}}\right][/latex-display] Perform row operations with the goal of turning into the identity.
1. Switch row 1 and row 2. $\displaystyle{\left[\matrix{{5}&{3}&{\mid}&{0}&{1}\\{2}&{1}&{\mid}&{1}&{0}}\right]$
2. Multiply row 2 by –2 and add to row 1. $\displaystyle{\left[\matrix{{1}&{1}&{\mid}&-{2}&{1}\\{2}&{1}&{\mid}&{1}&{0}}\right]$
3. Multiply row 1 by –2 and add to row 2. $\displaystyle{\left[\matrix{{1}&{1}&{\mid}&-{2}&{1}\\{0}&-{1}&{\mid}&{5}&-{2}}\right]$
4. Add row 2 to row 1. $\displaystyle{\left[\matrix{{1}&{0}&{\mid}&{3}&-{1}\\{0}&-{1}&{\mid}&{5}&-{2}}\right]$
5. Multiply row 2 by –1 $\displaystyle{\left[\matrix{{1}&{0}&{\mid}&{3}&-{1}\\{0}&{1}&{\mid}&-{5}&{2}}\right]$
The matrix we have found is A–1 [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{{3}&-{1}\\-{5}&{2}}]\right[/latex-display]
## Finding the Multiplicative Inverse of 2 × 2 Matrices Using a Formula
When we need to find the multiplicative inverse of a 2 × 2 matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity. If A is a 2 × 2 matrix, such as [latex-display]\displaystyle{A}={\left[\matrix{{a}&{b}\\{c}&{d}}\right][/latex-display] the multiplicative inverse of is given by the formula [latex-display]\displaystyle{A}^{{-{{1}}}}=\frac{{1}}{{{a}{d}-{b}{c}}}{\left[\matrix{{d}&-{b}\\-{c}&{a}}\right][/latex-display] where ad – bc ≠ 0. If ad – bc = 0, then A has no inverse.
### Example 4
1. Use the formula to find the multiplicative inverse of $\displaystyle{A}={\left[\matrix{{1}&-{2}\\{2}&-{3}}\right]$
2. Use the formula to find the inverse of matrix A. Verify your answer by augmenting with the identity matrix. $\displaystyle{A}={\left[\matrix{{1}&-{1}\\{2}&{3}}\right]$
3. Find the inverse, if it exists, of the given matrix. $\displaystyle{A}={\left[\matrix{{3}&{6}\\{1}&{2}}\right]$
#### Solutions
1. Using the formula, we have
[latex-display]\displaystyle{{A}^{{-{{1}}}}=\frac{{1}}{{{({1})}{(-{3})}-{(-{2})}{({2})}}}{\left[\matrix{-{3}&{2}\\-{2}&{1}}\right][/latex-display] [latex-display]\displaystyle{A}^{{-{{1}}}}=\frac{{1}}{{-{3}+{4}}}{\left[\matrix{-{3}&{2}\\-{2}&{1}}\right][/latex-display] [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{-{3}&{2}\\-{2}&{1}}\right][/latex-display] Perform row operations with the goal of turning A into the identity. a) Multiply row 1 by –2 and add to row 2. [latex-display]\displaystyle{\left[\matrix{{1}&-{2}&{\mid}&{1}&{0}\\{0}&{1}&{\mid}&-{2}&{1}}\right][/latex-display] b) Multiply row 1 by 2 and add to row 1. [latex-display]\displaystyle{\left[\matrix{{1}&{0}&{\mid}&-{3}&{2}\\{0}&{1}&{\mid}&-{2}&{1}}\right][/latex-display] c) So, we have verified our original solution. [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{-{3}&{2}\\-{2}&{1}}]}[/latex-display] 2. $\displaystyle{A}^{{-{{1}}}}={\left[\matrix{\frac{{3}}{{5}}&\frac{{1}}{{5}}\-\frac{{2}}{{5}}&\frac{{1}}{{5}}}\right]$ We will use the method of augmenting with the identity. [latex-display]\displaystyle{\left[\matrix{{3}&{6}&{\mid}&{1}&{0}\\{1}&{3}&{\mid}&{0}&{1}}\right][/latex-display] a) Switch row 1 and row 2. [latex-display]\displaystyle{\left[\matrix{{1}&{3}&{\mid}&{0}&{1}\\{3}&{6}&{\mid}&{1}&{0}}\right][/latex-display] b) Multiply row 1 by −3 and add it to row 2. [latex-display]\displaystyle{\left[\matrix{{1}&{3}&{\mid}&{0}&{1}\\{0}&{0}&{\mid}&-{3}&{1}}\right][/latex-display] c) There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.
## Finding the Multiplicative Inverse of 3 × 3 Matrices
Unfortunately, we do not have a formula similar to the one for a 2 × 2 matrix to find the inverse of a 3 × 3 matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse. Given a 3 × 3 matrix [latex-display]\displaystyle{A}={\left[\matrix{{2}&{3}&{1}\\{3}&{3}&{1}\\{2}&{4}&{1}}\right][/latex-display] augment A with the identity matrix [latex-display]\displaystyle{A}{\mid}{I}={\left[\matrix{{2}&{3}&{1}&{\mid}&{1}&{0}&{0}\\{3}&{3}&{1}&{\mid}&{0}&{1}&{0}\\{2}&{4}&{1}&{\mid}&{0}&{0}&{1}}\right][/latex-display] To begin, we write the augmented matrix with the identity on the right and A on the left. Performing elementary row operations so that the identity matrix appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example.
### How To
Given a 3 × 3 matrix, find the inverse
1. Write the original matrix augmented with the identity matrix on the right.
2. Use elementary row operations so that
3. the identity appears on the left.
4. What is obtained on the right is the inverse of the original matrix.
5. Use matrix multiplication to show that $\displaystyle{A}{A}^{{-{{1}}}}={1}{\text{and}}{A}^{{-{{1}}}}{A}={I}$.
### Example 5
1. Given the matrix A, find the inverse. $\displaystyle{A}={\left[\matrix{{2}&{3}&{1}\\{3}&{3}&{1}\\{2}&{4}&{1}}\right]$
2. Find the inverse of the matrix. $\displaystyle{A}={\left[\matrix{{2}&-{17}&{11}\\-{1}&{11}&-{7}\\{0}&{3}&-{2}}\right]$
#### Solutions
1. Augment A with the identity matrix, and then begin row operations until the identity matrix replaces A. The matrix on the right will be the inverse of A. $\displaystyle{\left[\matrix{{2}&{3}&{1}&{\mid}&{1}&{0}&{0}\\{3}&{3}&{1}&{\mid}&{0}&{1}&{0}\\{2}&{4}&{1}&{\mid}&{0}&{0}&{1}}\right]{\stackrel{{\text{Interchange } {R}_{{2}}\text{ and } {R}_{{1}}\text{ } }}{\rightarrow}}{\left[\matrix{{3}&{3}&{1}&{\mid}&{0}&{1}&{0}\\{2}&{3}&{1}&{\mid}&{1}&{0}&{0}\\{2}&{4}&{1}&{\mid}&{0}&{0}&{1}}\right]$ Thus,$\displaystyle{A}^{{-{{1}}}}={B}={\left[\matrix{-{1}&{1}&{0}\\-{1}&{0}&{1}\\{6}&-{2}&-{3}}\right]$
2. $\displaystyle{A}^{{-{{1}}}}={\left[\matrix{{1}&{1}&{2}\\{2}&{4}&-{3}\\{3}&{6}&-{5}}\right]$
OpenStax, Precalculus, "Solving Systems with Inverses," licensed under a CC BY 3.0 license.
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# How do you solve -52+m=84?
Jul 15, 2016
$m = 136$
#### Explanation:
$\textcolor{b l u e}{\text{Shortcut method}}$
Take the 52 to the other side of the = and change its sign
$m = 84 + 52 = 136$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Using first principles from which the shot cut is derived}}$
Given:$\text{ } \textcolor{g r e e n}{- 52 + m = 84}$
Add $\textcolor{red}{52}$ to both sides
$\textcolor{g r e e n}{- 52 \textcolor{red}{+ 52} + m = 84 \textcolor{red}{+ 52}}$
But $- 52 + 52 = 0$
$0 + m = 84 + 52 = 136$
$m = 136$
|
## Bayes Theorem
In this section we concentrate on the more complex conditional probability problems we began looking at in the last section.
### Example 19
Suppose a certain disease has an incidence rate of 0.1% (that is, it afflicts 0.1% of the population). A test has been devised to detect this disease. The test does not produce false negatives (that is, anyone who has the disease will test positive for it), but the false positive rate is 5% (that is, about 5% of people who take the test will test positive, even though they do not have the disease). Suppose a randomly selected person takes the test and tests positive. What is the probability that this person actually has the disease?
There are two ways to approach the solution to this problem. One involves an important result in probability theory called Bayes’ theorem. We will discuss this theorem a bit later, but for now we will use an alternative and, we hope, much more intuitive approach.
Let’s break down the information in the problem piece by piece.
Suppose a certain disease has an incidence rate of 0.1% (that is, it afflicts 0.1% of the population). The percentage 0.1% can be converted to a decimal number by moving the decimal place two places to the left, to get 0.001. In turn, 0.001 can be rewritten as a fraction: 1/1000. This tells us that about 1 in every 1000 people has the disease. (If we wanted we could write P(disease)=0.001.)
A test has been devised to detect this disease. The test does not produce false negatives (that is, anyone who has the disease will test positive for it). This part is fairly straightforward: everyone who has the disease will test positive, or alternatively everyone who tests negative does not have the disease. (We could also say P(positive | disease)=1.)
The false positive rate is 5% (that is, about 5% of people who take the test will test positive, even though they do not have the disease). This is even more straightforward. Another way of looking at it is that of every 100 people who are tested and do not have the disease, 5 will test positive even though they do not have the disease. (We could also say that P(positive | no disease)=0.05.)
Suppose a randomly selected person takes the test and tests positive. What is the probability that this person actually has the disease? Here we want to compute P(disease|positive). We already know that P(positive|disease)=1, but remember that conditional probabilities are not equal if the conditions are switched.
Rather than thinking in terms of all these probabilities we have developed, let’s create a hypothetical situation and apply the facts as set out above. First, suppose we randomly select 1000 people and administer the test. How many do we expect to have the disease? Since about 1/1000 of all people are afflicted with the disease, 1/1000 of 1000 people is 1. (Now you know why we chose 1000.) Only 1 of 1000 test subjects actually has the disease; the other 999 do not.
We also know that 5% of all people who do not have the disease will test positive. There are 999 disease-free people, so we would expect (0.05)(999)=49.95 (so, about 50) people to test positive who do not have the disease.
Now back to the original question, computing P(disease|positive). There are 51 people who test positive in our example (the one unfortunate person who actually has the disease, plus the 50 people who tested positive but don’t). Only one of these people has the disease, so
P(disease | positive) $\approx\frac{1}{51}\approx0.0196$
or less than 2%. Does this surprise you? This means that of all people who test positive, over 98% do not have the disease.
The answer we got was slightly approximate, since we rounded 49.95 to 50. We could redo the problem with 100,000 test subjects, 100 of whom would have the disease and (0.05)(99,900)=4995 test positive but do not have the disease, so the exact probability of having the disease if you test positive is
P(disease | positive) $\approx\frac{100}{5095}\approx0.0196$
which is pretty much the same answer.
But back to the surprising result. Of all people who test positive, over 98% do not have the disease. If your guess for the probability a person who tests positive has the disease was wildly different from the right answer (2%), don’t feel bad. The exact same problem was posed to doctors and medical students at the Harvard Medical School 25 years ago and the results revealed in a 1978 New England Journal of Medicine article. Only about 18% of the participants got the right answer. Most of the rest thought the answer was closer to 95% (perhaps they were misled by the false positive rate of 5%).
So at least you should feel a little better that a bunch of doctors didn’t get the right answer either (assuming you thought the answer was much higher). But the significance of this finding and similar results from other studies in the intervening years lies not in making math students feel better but in the possibly catastrophic consequences it might have for patient care. If a doctor thinks the chances that a positive test result nearly guarantees that a patient has a disease, they might begin an unnecessary and possibly harmful treatment regimen on a healthy patient. Or worse, as in the early days of the AIDS crisis when being HIV-positive was often equated with a death sentence, the patient might take a drastic action and commit suicide.
As we have seen in this hypothetical example, the most responsible course of action for treating a patient who tests positive would be to counsel the patient that they most likely do not have the disease and to order further, more reliable, tests to verify the diagnosis.
One of the reasons that the doctors and medical students in the study did so poorly is that such problems, when presented in the types of statistics courses that medical students often take, are solved by use of Bayes’ theorem, which is stated as follows:
Bayes’ Theorem
$P(A|B)=\frac{P(A)P(B|A)}{P(A)P(B|A)+P(\bar{A})P(B|\bar{A})}$
In our earlier example, this translates to
$P(\text{disease}|\text{positive})=\frac{P(\text{disease})P(\text{positive}|\text{disease})}{P(\text{disease})P(\text{positive}|\text{disease})+P(\text{nodisease})P(\text{positive}|\text{nodisease})}$
Plugging in the numbers gives
$P(\text{disease}|\text{positive})=\frac{(0.001)(1)}{(0.001)(1)+(0.999)(0.05)}\approx0.0196$
which is exactly the same answer as our original solution.
The problem is that you (or the typical medical student, or even the typical math professor) are much more likely to be able to remember the original solution than to remember Bayes’ theorem. Psychologists, such as Gerd Gigerenzer, author of Calculated Risks: How to Know When Numbers Deceive You, have advocated that the method involved in the original solution (which Gigerenzer calls the method of “natural frequencies”) be employed in place of Bayes’ Theorem. Gigerenzer performed a study and found that those educated in the natural frequency method were able to recall it far longer than those who were taught Bayes’ theorem. When one considers the possible life-and-death consequences associated with such calculations it seems wise to heed his advice.
### Example 20
A certain disease has an incidence rate of 2%. If the false negative rate is 10% and the false positive rate is 1%, compute the probability that a person who tests positive actually has the disease.
Imagine 10,000 people who are tested. Of these 10,000, 200 will have the disease; 10% of them, or 20, will test negative and the remaining 180 will test positive. Of the 9800 who do not have the disease, 98 will test positive. So of the 278 total people who test positive, 180 will have the disease. Thus
$P(\text{disease}|\text{positive})=\frac{180}{278}\approx0.647$
so about 65% of the people who test positive will have the disease.
Using Bayes theorem directly would give the same result:
$P(\text{disease}|\text{positive})=\frac{(0.02)(0.90)}{(0.02)(0.90)+(0.98)(0.01)}=\frac{0.018}{0.0278}\approx0.647$
### Try it Now 5
A certain disease has an incidence rate of 0.5%. If there are no false negatives and if the false positive rate is 3%, compute the probability that a person who tests positive actually has the disease.
## Counting
Counting? You already know how to count or you wouldn’t be taking a college-level math class, right? Well yes, but what we’ll really be investigating here are ways of counting efficiently. When we get to the probability situations a bit later in this chapter we will need to count some very large numbers, like the number of possible winning lottery tickets. One way to do this would be to write down every possible set of numbers that might show up on a lottery ticket, but believe me: you don’t want to do this.
### Basic Counting
We will start, however, with some more reasonable sorts of counting problems in order to develop the ideas that we will soon need.
### Example 21
Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks) and five choices for a main course (hamburger, sandwich, quiche, fajita or pizza). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?
Solution 1: One way to solve this problem would be to systematically list each possible meal:
soup + hamburger soup + sandwich soup + quiche
soup + fajita soup + pizza salad + hamburger
Assuming that we did this systematically and that we neither missed any possibilities nor listed any possibility more than once, the answer would be 15. Thus you could go to the restaurant 15 nights in a row and have a different meal each night.
Solution 2: Another way to solve this problem would be to list all the possibilities in a table:
In each of the cells in the table we could list the corresponding meal: soup + hamburger in the upper left corner, salad + hamburger below it, etc. But if we didn’t really care what the possible meals are, only how many possible meals there are, we could just count the number of cells and arrive at an answer of 15, which matches our answer from the first solution. (It’s always good when you solve a problem two different ways and get the same answer!)
Solution 3: We already have two perfectly good solutions. Why do we need a third? The first method was not very systematic, and we might easily have made an omission. The second method was better, but suppose that in addition to the appetizer and the main course we further complicated the problem by adding desserts to the menu: we’ve used the rows of the table for the appetizers and the columns for the main courses—where will the desserts go? We would need a third dimension, and since drawing 3-D tables on a 2-D page or computer screen isn’t terribly easy, we need a better way in case we have three categories to choose form instead of just two.
So, back to the problem in the example. What else can we do? Let’s draw a tree diagram:
This is called a “tree” diagram because at each stage we branch out, like the branches on a tree. In this case, we first drew five branches (one for each main course) and then for each of those branches we drew three more branches (one for each appetizer). We count the number of branches at the final level and get (surprise, surprise!) 15.
If we wanted, we could instead draw three branches at the first stage for the three appetizers and then five branches (one for each main course) branching out of each of those three branches.
OK, so now we know how to count possibilities using tables and tree diagrams. These methods will continue to be useful in certain cases, but imagine a game where you have two decks of cards (with 52 cards in each deck) and you select one card from each deck. Would you really want to draw a table or tree diagram to determine the number of outcomes of this game?
Let’s go back to the previous example that involved selecting a meal from three appetizers and five main courses, and look at the second solution that used a table. Notice that one way to count the number of possible meals is simply to number each of the appropriate cells in the table, as we have done above. But another way to count the number of cells in the table would be multiply the number of rows (3) by the number of columns (5) to get 15. Notice that we could have arrived at the same result without making a table at all by simply multiplying the number of choices for the appetizer (3) by the number of choices for the main course (5). We generalize this technique as the basic counting rule:
Basic Counting Rule
If we are asked to choose one item from each of two separate categories where there are m items in the first category and n items in the second category, then the total number of available choices is m · n.
This is sometimes called the multiplication rule for probabilities.
### Example 22
There are 21 novels and 18 volumes of poetry on a reading list for a college English course. How many different ways can a student select one novel and one volume of poetry to read during the quarter?
There are 21 choices from the first category and 18 for the second, so there are 21 · 18 = 378 possibilities.
The Basic Counting Rule can be extended when there are more than two categories by applying it repeatedly, as we see in the next example.
### Example 23
Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks), five choices for a main course (hamburger, sandwich, quiche, fajita or pasta) and two choices for dessert (pie or ice cream). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?
There are 3 choices for an appetizer, 5 for the main course and 2 for dessert, so there are
3 · 5 · 2 = 30 possibilities.
### Example 24
A quiz consists of 3 true-or-false questions. In how many ways can a student answer the quiz?
There are 3 questions. Each question has 2 possible answers (true or false), so the quiz may be answered in 2 · 2 · 2 = 8 different ways. Recall that another way to write 2 · 2 · 2 is 23, which is much more compact.
### Try it Now 6
Suppose at a particular restaurant you have eight choices for an appetizer, eleven choices for a main course and five choices for dessert. If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?
## Permutations
In this section we will develop an even faster way to solve some of the problems we have already learned to solve by other means. Let’s start with a couple examples.
### Example 25
How many different ways can the letters of the word MATH be rearranged to form a four-letter code word?
This problem is a bit different. Instead of choosing one item from each of several different categories, we are repeatedly choosing items from the same category (the category is: the letters of the word MATH) and each time we choose an item we do not replace it, so there is one fewer choice at the next stage: we have 4 choices for the first letter (say we choose A), then 3 choices for the second (M, T and H; say we choose H), then 2 choices for the next letter (M and T; say we choose M) and only one choice at the last stage (T). Thus there are 4 · 3 · 2 · 1 = 24 ways to spell a code worth with the letters MATH.
In this example, we needed to calculate n · (n – 1) · (n – 2) ··· 3 · 2 · 1. This calculation shows up often in mathematics, and is called the factorial, and is notated n!
Factorial
n! = n · (n – 1) · (n – 2) ··· 3 · 2 · 1
### Example 26
How many ways can five different door prizes be distributed among five people?
There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be 5! = 5 · 4 · 3 · 2 · 1 = 120 ways.
Now we will consider some slightly different examples.
### Example 27
A charity benefit is attended by 25 people and three gift certificates are given away as door prizes: one gift certificate is in the amount of $100, the second is worth$25 and the third is worth $10. Assuming that no person receives more than one prize, how many different ways can the three gift certificates be awarded? Using the Basic Counting Rule, there are 25 choices for the person who receives the$100 certificate, 24 remaining choices for the $25 certificate and 23 choices for the$10 certificate, so there are 25 · 24 · 23 = 13,800 ways in which the prizes can be awarded.
### Example 28
Eight sprinters have made it to the Olympic finals in the 100-meter race. In how many different ways can the gold, silver and bronze medals be awarded?
Using the Basic Counting Rule, there are 8 choices for the gold medal winner, 7 remaining choices for the silver, and 6 for the bronze, so there are 8 · 7 · 6 = 336 ways the three medals can be awarded to the 8 runners.
Note that in these preceding examples, the gift certificates and the Olympic medals were awarded without replacement; that is, once we have chosen a winner of the first door prize or the gold medal, they are not eligible for the other prizes. Thus, at each succeeding stage of the solution there is one fewer choice (25, then 24, then 23 in the first example; 8, then 7, then 6 in the second). Contrast this with the situation of a multiple choice test, where there might be five possible answers — A, B, C, D or E — for each question on the test.
Note also that the order of selection was important in each example: for the three door prizes, being chosen first means that you receive substantially more money; in the Olympics example, coming in first means that you get the gold medal instead of the silver or bronze. In each case, if we had chosen the same three people in a different order there might have been a different person who received the $100 prize, or a different goldmedalist. (Contrast this with the situation where we might draw three names out of a hat to each receive a$10 gift certificate; in this case the order of selection is not important since each of the three people receive the same prize. Situations where the order is not important will be discussed in the next section.)
We can generalize the situation in the two examples above to any problem without replacement where the order of selection is important. If we are arranging in order r items out of n possibilities (instead of 3 out of 25 or 3 out of 8 as in the previous examples), the number of possible arrangements will be given by
n · (n – 1) · (n – 2) ··· (nr + 1)
If you don’t see why (n r + 1) is the right number to use for the last factor, just think back to the first example in this section, where we calculated 25 · 24 · 23 to get 13,800. In this case n = 25 and r = 3, so n r + 1 = 25 — 3 + 1 = 23, which is exactly the right number for the final factor.
Now, why would we want to use this complicated formula when it’s actually easier to use the Basic Counting Rule, as we did in the first two examples? Well, we won’t actually use this formula all that often, we only developed it so that we could attach a special notation and a special definition to this situation where we are choosing r items out of n possibilities without replacement and where the order of selection is important. In this situation we write:
Permutations
nPr = n · (n – 1) · (n – 2) ··· (nr + 1)
We say that there are nPr permutations of size r that may be selected from among n choices without replacement when order matters.
It turns out that we can express this result more simply using factorials.
${}_{n}{{P}_{r}}=\frac{n!}{(n-r)!}$
In practicality, we usually use technology rather than factorials or repeated multiplication to compute permutations.
### Example 29
I have nine paintings and have room to display only four of them at a time on my wall. How many different ways could I do this?
Since we are choosing 4 paintings out of 9 without replacement where the order of selection is important there are 9P4 = 9 · 8 · 7 · 6 = 3,024 permutations.
### Example 30
How many ways can a four-person executive committee (president, vice-president, secretary, treasurer) be selected from a 16-member board of directors of a non-profit organization?
We want to choose 4 people out of 16 without replacement and where the order of selection is important. So the answer is 16P4 = 16 · 15 · 14 · 13 = 43,680.
### Try it Now 7
How many 5 character passwords can be made using the letters A through Z
a. if repeats are allowed
b. if no repeats are allowed
## Combinations
In the previous section we considered the situation where we chose r items out of n possibilities without replacement and where the order of selection was important. We now consider a similar situation in which the order of selection is not important.
### Example 31
A charity benefit is attended by 25 people at which three $50 gift certificates are given away as door prizes. Assuming no person receives more than one prize, how many different ways can the gift certificates be awarded? Using the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are 25 · 24 · 23 = 13,800 ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get$50, so it doesn’t really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6:
ABC ACB BAC BCA CAB CBA
How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are 3 · 2 · 1 = 6 ways to do this; we didn’t really need to list them all, we can just use permutations!
So, out of the 13,800 ways to select 3 people out of 25, six of them involve Abe, Bea and Cindy. The same argument works for any other group of three people (say Abe, Bea and David or Frank, Gloria and Hildy) so each three-person group is counted six times. Thus the 13,800 figure is six times too big. The number of distinct three-person groups will be 13,800/6 = 2300.
We can generalize the situation in this example above to any problem of choosing a collection of items without replacement where the order of selection is not important. If we are choosing r items out of n possibilities (instead of 3 out of 25 as in the previous examples), the number of possible choices will be given by $\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}$, and we could use this formula for computation. However this situation arises so frequently that we attach a special notation and a special definition to this situation where we are choosing r items out of n possibilities without replacement where the order of selection is not important.
Combinations
${}_{n}{{C}_{r}}=\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}$
We say that there are nCr combinations of size r that may be selected from among n choices without replacement where order doesn’t matter.
We can also write the combinations formula in terms of factorials:
${}_{n}{{C}_{r}}=\frac{n!}{(n-r)!r!}$
### Example 32
A group of four students is to be chosen from a 35-member class to represent the class on the student council. How many ways can this be done?
Since we are choosing 4 people out of 35 without replacement where the order of selection is not important there are ${}_{35}{{C}_{4}}=\frac{35\cdot34\cdot33\cdot32}{4\cdot3\cdot2\cdot1}$ = 52,360 combinations.
### Try it Now 8
The United States Senate Appropriations Committee consists of 29 members; the Defense Subcommittee of the Appropriations Committee consists of 19 members. Disregarding party affiliation or any special seats on the Subcommittee, how many different 19-member subcommittees may be chosen from among the 29 Senators on the Appropriations Committee?
In the preceding Try it Now problem we assumed that the 19 members of the Defense Subcommittee were chosen without regard to party affiliation. In reality this would never happen: if Republicans are in the majority they would never let a majority of Democrats sit on (and thus control) any subcommittee. (The same of course would be true if the Democrats were in control.) So let’s consider the problem again, in a slightly more complicated form:
### Example 33
The United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee?
In this case we need to choose 10 of the 15 Republicans and 9 of the 14 Democrats. There are 15C10 = 3003 ways to choose the 10 Republicans and 14C9 = 2002 ways to choose the 9 Democrats. But now what? How do we finish the problem?
Suppose we listed all of the possible 10-member Republican groups on 3003 slips of red paper and all of the possible 9-member Democratic groups on 2002 slips of blue paper. How many ways can we choose one red slip and one blue slip? This is a job for the Basic Counting Rule! We are simply making one choice from the first category and one choice from the second category, just like in the restaurant menu problems from earlier.
There must be 3003 · 2002 = 6,012,006 possible ways of selecting the members of the Defense Subcommittee.
## Probability using Permutations and Combinations
We can use permutations and combinations to help us answer more complex probability questions
### Example 34
A 4 digit PIN number is selected. What is the probability that there are no repeated digits?
There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 · 10 · 10 · 10 = 104 = 10000 total possible PIN numbers.
To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 · 9 · 8 · 7, or notice that this is the same as the permutation 10P4 = 5040.
The probability of no repeated digits is the number of 4 digit PIN numbers with no repeated digits divided by the total number of 4 digit PIN numbers. This probability is $\frac{{}_{10}{{P}_{4}}}{{{10}^{4}}}=\frac{5040}{10000}=0.504$
In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket. In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player’s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is 48C6 = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player’s ticket, so the probability of winning the grand prize is: $\frac{{}_{6}{{C}_{6}}}{{}_{48}{{C}_{6}}}=\frac{1}{12271512}\approx0.0000000815$ ### Example 36 In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of$1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.
As above, the number of possible outcomes of the lottery drawing is 48C6 = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by 6C5 = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by 42C1 = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: 6C5 · 42C1 = 6 · 42 = 252. So the probability of winning the second prize is.
$\frac{\left({}_{6}{{C}_{5}}\right)\left({}_{42}{{C}_{1}}\right)}{{}_{48}{{C}_{6}}}=\frac{252}{12271512}\approx0.0000205$
### Try it Now 9
A multiple-choice question on an economics quiz contains 10 questions with five possible answers each. Compute the probability of randomly guessing the answers and getting 9 questions correct.
### Example 37
Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.
In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands, 52C5. This number will go in the denominator of our probability formula, since it is the number of possible outcomes.
For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Since there are four Aces and we want exactly one of them, there will be 4C1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48C4 ways to select the four non-Aces. Now we use the Basic Counting Rule to calculate that there will be 4C1 · 48C4 ways to choose one ace and four non-Aces.
Putting this all together, we have
$P(\text{oneAce})=\frac{\left({}_{4}{{C}_{1}}\right)\left({}_{48}{{C}_{4}}\right)}{{}_{52}{{C}_{5}}}=\frac{778320}{2598960}\approx0.299$
### Example 38
Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.
The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:
$P(\text{twoAces})=\frac{\left({}_{4}{{C}_{2}}\right)\left({}_{48}{{C}_{3}}\right)}{{}_{52}{{C}_{5}}}=\frac{103776}{2598960}\approx0.0399$
It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.
### Try it Now 10
Compute the probability of randomly drawing five cards from a deck of cards and getting three Aces and two Kings.
## Birthday Problem
Let’s take a pause to consider a famous problem in probability theory:
Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?
Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let’s see if we should listen to our intuition. Let’s start with a simpler problem, however.
### Example 39
Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people?
There are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves “What is the alternative to having at least one shared birthday?” In this case, the alternative is that there are no shared birthdays. In other words, the alternative to “at least one” is having none. In other words, since this is a complementary event,
P(at least one) = 1 – P(none)
We will start, then, by computing the probability that there is no shared birthday. Let’s imagine that you are one of these three people. Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday? There are 365 days in the year (let’s ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364/365. Now we move to the third person. What is the probability that this third person does not have the same birthday as either you or the second person? There are 363 days that will not duplicate your birthday or the second person’s, so the probability that the third person does not share a birthday with the first two is 363/365.
We want the second person not to share a birthday with you and the third person not to share a birthday with the first two people, so we use the multiplication rule:
$P(\text{nosharedbirthday})=\frac{365}{365}\cdot\frac{364}{365}\cdot\frac{363}{365}\approx0.9918$
and then subtract from 1 to get
P(shared birthday) = 1 – P(no shared birthday) = 1 – 0.9918 = 0.0082.
This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small. Let’s make our group a bit bigger.
### Example 40
Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people?
Continuing the pattern of the previous example, the answer should be
$P(\text{sharedbirthday})=1-\frac{365}{365}\cdot\frac{364}{365}\cdot\frac{363}{365}\cdot\frac{362}{365}\cdot\frac{361}{365}\approx0.0271$
Note that we could rewrite this more compactly as
$P(\text{sharedbirthday})=1-\frac{{}_{365}{{P}_{5}}}{{{365}^{5}}}\approx0.0271$
which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.
### Example 41
Suppose 30 people are in a room. What is the probability that there is at least one shared birthday among these 30 people?
Here we can calculate
$P(\text{sharedbirthday})=1-\frac{{}_{365}{{P}_{30}}}{{{365}^{30}}}\approx0.706$
which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday!
If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn’t studied probability!) You wouldn’t be guaranteed to win, but you should win more than half the time.
This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts. If you still don’t believe the math, you can carry out a simulation. Just so you won’t have to go around rounding up groups of 30 people, someone has kindly developed a Java applet so that you can conduct a computer simulation. Go to this web page: http://www-stat.stanford.edu/~susan/surprise/Birthday.html, and once the applet has loaded, select 30 birthdays and then keep clicking Start and Reset. If you keep track of the number of times that there is a repeated birthday, you should get a repeated birthday about 7 out of every 10 times you run the simulation.
### Try it Now 11
Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?
|
# 6.07 Simple interest
Lesson
It costs money to borrow money. The extra money that banks and other lenders charge us to borrow money is called interest. Interest may also refer to the additional money that is earned from investing money, such as into a savings accounts. There are different types of interest and in this lesson we are going to talk about simple interest.
### What is simple interest?
Simple interest, or flat rate interest, describes a method of calculating interest where the interest amount is fixed (i.e. it doesn't change). The interest charge is always based on the original amount borrowed (or invested), and does not take into account any interest earned along the way (that is, interest on interest is not included).
Many financial institutions express their interest rates per year. For example, an interest rate might be given as $3%$3% per year.
To calculate simple interest, three quantities are involved: the principal amount $P$P that is invested (or borrowed), the number of time periods $t$t, and the interest rate of $r$r per time period. Simple interest is then calculated using the formula
$I=Prt$I=Prt
Note that sometimes $n$n is used instead of $t$t to represent the number of time periods.
To find the total value of the investment or loan after a given time period, we add the interest to the principal amount $P$P.
Summary
Simple interest is calculated as
$I=Prt$I=Prt
where $P$P is the principal amount invested (or borrowed), $r$r is the interest rate per time period, and $t$t is the number of time periods.
The total amount or value $A$A, earned after $t$t interest periods, is then calculated as
$A=P+I$A=P+I
#### Worked examples
##### question 1
a) Calculate the simple interest on a loan of $\$8580$$8580 at 2%2% per year for 1010 years. Round your answer to the nearest cent. Think: We can substitute the values for the principal, interest rate and time periods. Do: II == PrtPrt == 8580\times2%\times108580×2%×10 == 8580\times0.02\times108580×0.02×10 == \1716$$1716
b) What is the total value of the loan after $10$10 years?
Think: The total value is the original loan plus the interest that has accrued over $10$10 years.
Do: $1716+8580=\$10296$1716+8580=$10296
##### question 2
The interest on an investment of $\$3600$$3600 over 1010 years is \2520.00$$2520.00. If the annual interest rate is $r$r, find $r$r as a percentage.
Think: What values do we know that we can substitute into the formula?
Do:
$I$I $=$= $Prt$Prt $2520$2520 $=$= $3600\times r\times10$3600×r×10 $2520$2520 $=$= $36000r$36000r $r$r $=$= $\frac{2520}{36000}$252036000 $r$r $=$= $0.07$0.07 $r$r $=$= $7%$7%
#### Practice question
##### Question 3
For a simple interest rate of $6%$6% per year , calculate the number of years $T$T needed for an interest of $\$1174.20$$1174.20 to be earned on the investment \1957$$1957.
Give your answer as a whole number of years.
Enter each line of working as an equation.
### Simple interest calculations for time periods other than years
When calculating simple interest for time periods that are not years, such as months, weeks or days, we need to make sure the interest rate and the time periods are expressed using the same period. For example, if the rate $r$r is given per year then the value for $t$t needs to be expressed in years too.
Remember!
$1$1 year $=$= $12$12 months $=$= $52$52 weeks $=$= $4$4 quarters (of $3$3 months each) $=$= $365$365 days
#### Practice questions
Calculate the simple interest generated on a loan of $\$3860$$3860 at a rate of 9%9% per year for 1313 months. 1. Round your answer to the nearest cent. ##### Question 5 Calculate the simple interest earned on an investment of \7000$$7000 at $1.8%$1.8% per quarter for $9$9 years.
Give your answer to the nearest cent.
##### Question 6
Calculate the simple interest earned on an investment of $\$53205320 at $6%$6% per year for $95$95 weeks.
Assume that a year has $52$52 weeks.
Give your answer to the nearest cent.
### Outcomes
#### 7.RP.3
Use proportional relationships to solve multistep ratio and percent problems.
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Solving Levers with the Teeter-Totter Principle.
9sep10
\begin{document}
\textbf{Question:} Where should the fulcrum be if a weight of $+2$ is
placed at $A$ and a weight of $-3$, a balloon for example, is at $B$?
What if it's $+3, -2$ instead?
\textbf{Solution:}
\subsection{Analyse the known version of this problem}
In the lesson on barycentric coordinates,
the weight $a$ and $A$ and $b$ at $B$ become the coefficients in
$C = \frac{2A+1B}{2+1} = \frac{2}{3}A + \frac{1}{3} B$.
The placement of the fulcrum $C$ is obvious. Substitute a variable for a
constant to conjecture the general rule. Let's use a
$\frac{1}{3} = t$ and write
$C =_1 A + t(B-A) =_2 A(1-t) + tB .$
Equation 1 places the fulcrum a third of the way from $A$ to $B$.
Equation 2 identifies this point's barycentric coordinates.
\subsection{Generalize and check the validity}
Writing $X = \frac{2A + (-3)B}{2-3}$ leads to rewriting
rewriting $X = A + 2(B-A)$. This places the fulcrum twice
as far from $B$ making $A$ three times as far from the fulcrum.
The lever equation you learned in high school physics becomes
$3 \times 2 = 2 \times 3$.
\subsection{Do it again to make sure.}
Exchanging the weights leads to a surprising result. Following the same
recipe leads to $Y = 3A + (-2)B = A + 2(A-B)$, which places the fulcrum
on the other side. The lever equation remains the same.
\end{document}
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# The population of a herd of sheep declines exponentially. If it takes 3 years for 15% to decline then how long will it take for half the population to decrease ?
Mar 14, 2017
It would take 12.8 years.
#### Explanation:
Let initial number of sheep $\textsf{= {S}_{0}}$
Let the number of sheep remaining after time t $\textsf{= {S}_{t}}$
The equation for exponential decay gives us:
$\textsf{{S}_{t} = {S}_{0} {e}^{- k t}}$
If the number of sheep have declined by 15% then the number remaining must be 85% of the original total.
So $\textsf{{S}_{t} = 085 {S}_{0}}$
Putting in the numbers:
$\textsf{0.85 \cancel{{S}_{0}} = \cancel{{S}_{0}} {e}^{- k 3}}$
Taking natural logs of both sides gives:
$\textsf{\ln \left(0.85\right) = - k 3}$
$\therefore$$\textsf{- 0.1625 = - k 3}$
$\therefore$$\textsf{k = \frac{0.1625}{3} = 0.05416 \textcolor{w h i t e}{x} {\text{yr}}^{-} 1}$
I won't go into the derivation here but it can be shown that the expression for 1/2 life in terms of the decay constant k is given by:
$\textsf{{t}_{\frac{1}{2}} = \frac{0.693}{k}}$
$\therefore$$\textsf{{t}_{\frac{1}{2}} = \frac{0.693}{0.05416} = 12.8 \textcolor{w h i t e}{x} \text{yr}}$
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Question
A boat covers 32 km upstream and 36 km downstream in 7 hours. Also, it covers 40 km upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and that of the stream.
Hint: Let the speed of stream and boat in still water to be ‘u’ and ‘v’ respectively. Get the downstream speed $=\left( u+v \right)km/hr$ and upstream$=\left( v-u \right)km/hr$. Then use (time taken by boat to 32 km upstream) + (time taken by boat to go 36 km downstream) = 7
Likewise for the next case, put$\dfrac{1}{\left( v+u \right)}=A\ and\ \dfrac{1}{\left( v-u \right)}=B$ in further steps.
We are given that a boat covers 32 km upstream and 36 km downstream in 7 hours. It also covers 40 km upstream and 48 km downstream in 9 hours. We have to find the speed of the boat in still water and that of the stream.
Let us consider the speed of a boat in still water to be $'v'km/hr$. Also, let us consider the speed of the stream to be $'u'km/hr$.
We know that whenever a boat goes downstream, then the direction of stream and direction of boat is in the same direction. Hence the stream supports the boat while going downstream. So, we get,
Downstream speed of boat = speed of boat in still water + speed of stream…………………(i)
We also know that whenever a boat goes upstream then the direction of stream and direction of boat is in the opposite direction.
Hence, the stream opposes the boat while going upstream. So, we get,
Upstream speed of boat = speed of boat in still water – speed of stream …………………..(ii)
As we have assumed the speed of boat in still water as $'v'km/hr$and speed of stream$'u'km/hr$, by putting these in equation (i) and (ii), we get,
Downstream speed of boat $=\left( v+u \right)km/hr$
and upstream speed of boat$=\left( v-u \right)km/hr$
Now let us consider the time taken by boat to go ${{D}_{1}}$ km downstream as td.
Since we know that $time=\dfrac{\text{distance}}{\text{speed}}$, therefore by putting the values of time = td, distance = ${{D}_{1}}$ km and downstream speed $=\left( v+u \right)km/hr$, we get,
$td=\dfrac{{{D}_{1}}km}{\left( v+u \right)km/hr}$
In case 1, as we are given that boat goes 36 km downstream we get,
$t{{d}_{1}}=\dfrac{36km}{\left( v+u \right)km/hr}...............(iii)$
In case 2, as we given that boat goes 48 km downstream,
$t{{d}_{2}}=\dfrac{48km}{\left( v+u \right)km/hr}...............(iv)$
Now, let us consider time taken by boat to go ${{D}_{2}}$km upstream as tu.
Since we know that time = $\dfrac{\text{Distance}}{\text{Speed}}$ , therefore by putting the value of time = tu, Distance = ${{D}_{2}}$km and upstream speed = (v – u)km/hr, we get
$tu=\dfrac{{{D}_{2}}km}{\left( v-u \right)km/hr}$
In case 1, as we are given that boat goes 32 km upstream we get,
$t{{u}_{1}}=\dfrac{32km}{\left( v-u \right)km/hr}.............\left( v \right)$
In case 2, as we are given that boat goes 40 km upstream, we get,
$t{{u}_{2}}=\dfrac{40km}{\left( v-u \right)km/hr}.............\left( vi \right)$
Now, in case 1, we are given that boat takes total 7 hours to go 32 km upstream and 36 downstream, therefore we get,
$t{{d}_{1}}+t{{u}_{1}}=7\ hours$
By putting the values of $t{{d}_{1}}\And t{{u}_{1}}$ from equation (iii) and equation (v) respectively, we get,
$\Rightarrow \dfrac{36}{\left( v+u \right)}+\dfrac{32}{\left( v-u \right)}=7.........\left( vii \right)$
Now in case 2, we are given that boat takes total 9 hours to go 40 km upstream and 40 km downstream therefore, we get,
$t{{d}_{2}}+t{{u}_{2}}=9\ hours$
By putting the value of $t{{d}_{2}}\And t{{u}_{2}}$from equation (iv) and equation (vi) respectively, we get,
$\dfrac{48}{\left( v+u \right)}+\dfrac{40}{\left( v-u \right)}=9........\left( viii \right)$
Let us take $\dfrac{1}{\left( v+u \right)}=A\ and\ \dfrac{1}{\left( v-u \right)}=B$. By applying this we get equation (vii) & (viii) as,
36A + 32B = 7…………….(a)
48A + 40B = 9…………….(b)
By dividing by 4 in equation (a), we get,
\begin{align} & \dfrac{1}{4}\left( 36A+32B \right)=\dfrac{7}{4} \\ & =9A+8B=\dfrac{7}{4}.............\left( {{a}_{1}} \right) \\ \end{align}
By dividing by 5 in equation (b) we get,
\begin{align} & \dfrac{1}{5}\left( 48A+40B \right)=\dfrac{9}{5} \\ & =\dfrac{48A}{5}+8B=\dfrac{9}{5}.............\left( {{b}_{1}} \right) \\ \end{align}
By subtracting equation $\left( {{a}_{1}} \right)$ from equation $\left( {{b}_{1}} \right)$, we get,
\begin{align} & \left( \dfrac{48A}{5}+8B \right)-\left( 9A+8B \right)=\dfrac{9}{5}-\dfrac{7}{4} \\ & =\dfrac{48A}{5}-9A=\dfrac{\left( 4\times 9 \right)-\left( 5\times 7 \right)}{20} \\ & =\dfrac{48A-45A}{5}=\dfrac{36-35}{20} \\ & =\dfrac{3A}{5}=\dfrac{1}{20} \\ \end{align}
By cross multiplying above equation we get,
\begin{align} & A=\dfrac{5}{20\times 3} \\ & \Rightarrow A=\dfrac{1}{4\times 3}=\dfrac{1}{12} \\ \end{align}
As we know that $A=\dfrac{1}{\left( v+u \right)}$ therefore we get,
\begin{align} & \dfrac{1}{\left( v+u \right)}=\dfrac{1}{12} \\ & or\ v+u=12.........\left( {{a}_{2}} \right) \\ \end{align}
By putting the value of A in equation $\left( {{a}_{1}} \right)$, we get,
\begin{align} & 9\left( \dfrac{1}{12} \right)+8B=\dfrac{7}{4} \\ & =\dfrac{3}{4}+8B=\dfrac{7}{4} \\ & or\ 8B=\dfrac{7}{4}-\dfrac{3}{4} \\ & 8B=\dfrac{4}{4}=1 \\ & B=\dfrac{1}{8} \\ \end{align}
As we know that $B=\dfrac{1}{\left( v-u \right)}$, therefore we get,
\begin{align} & \dfrac{1}{\left( v-u \right)}=\dfrac{1}{8} \\ & or\ \left( v-u \right)=8...........\left( {{b}_{2}} \right) \\ \end{align}
By adding equation $\left( {{a}_{2}} \right)$ & $\left( {{b}_{2}} \right)$, we get,
\begin{align} & \left( v+u \right)+\left( v-u \right)=12+8 \\ & =2v=20 \\ & v=\dfrac{20}{2}=10km/hr \\ \end{align}
By putting the value of v in equation $\left( {{a}_{2}} \right)$, we get,
\begin{align} & 10+u=12 \\ & \Rightarrow u=12-10 \\ & \Rightarrow u=2km/hr \\ \end{align}
Therefore we get speed of boat in still water (v) = 10km/hr and speed of stream (u) = 2km/hr.
Note: Students must remember this approach of taking $\dfrac{1}{v+4}=A\And \dfrac{1}{v-4}=B$ because if they won’t consider this, the question becomes very confusing and even unsolvable due to term $\left( {{v}^{2}}-{{u}^{2}} \right)$ . Also students should take care that upstream speed is (v – u)km/hr and not (u – v)km/hr where v is speed of boat in still water and u is speed of stream. Students can cross check their answer by putting the values of v and u in various equations and see if it is satisfying them or not.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Distributive Property for Multi-Step Equations
a(x + b) = c
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Distributive Property for Multi-Step Equations
Suppose you were a contestant on a game show where they gave you the answer and you had to think of the corresponding question. If the answer were, "M(N+K)=MN+MK\begin{align*}M(N+K)= MN+MK\end{align*} or M(NK)=MNMK\begin{align*}M(N-K)= MN-MK\end{align*}," what would you give for the question? How about, "What is the Distributive Property?" In this Concept, you will learn all about the Distributive Property and how to use it to solve equations with multiple steps.
Guidance
Solving Multi-Step Equations by Using the Distributive Property
When faced with an equation such as 2(5x+9)=78\begin{align*}2(5x+9)=78\end{align*}, the first step is to remove the parentheses. There are two options to remove the parentheses. You can apply the Distributive Property or you can apply the Multiplication Property of Equality. This Concept will show you how to use the Distributive Property to solve multi-step equations.
Example A
Solve for x\begin{align*}x\end{align*}: 2(5x+9)=78.\begin{align*}2(5x+9)=78.\end{align*}
Solution: Apply the Distributive Property: 10x+18=78.\begin{align*}10x+18=78.\end{align*}
Apply the Addition Property of Equality: 10x+1818=7818.\begin{align*}10x+18-18=78-18.\end{align*}
Simplify: 10x=60.\begin{align*}10x=60.\end{align*}
Apply the Multiplication Property of Equality: 10x÷10=60÷10.\begin{align*}10x \div 10 = 60 \div 10.\end{align*}
The solution is x=6\begin{align*}x=6\end{align*}.
Check: Does 10(6)+18=78?\begin{align*}10(6) + 18 = 78?\end{align*} Yes, so the answer is correct.
Example B
Solve for n\begin{align*}n\end{align*} when 2(n+9)=6n.\begin{align*}2(n+9)=6n.\end{align*}
Solution:
2(n+9)=5n2n+29=5n2n+18=5n2n+2n+18=2n+5n18=3n1318=133n6=n\begin{align*}&2(n+9)=5n\\ &2\cdot n+2\cdot 9=5n\\ &2n+18=5n\\ &-2n+2n+18=-2n + 5n\\ &18=3n\\ &\frac{1}{3}\cdot18=\frac{1}{3}\cdot 3n\\ &6=n \end{align*}
2(6+9)=5(6)2(15)=3030=30\begin{align*} &2(6+9)=5(6)\\ &2(15)=30\\ &30=30\\ \end{align*}
Example C
Solve for d\begin{align*}d\end{align*} when 3(d+15)18d=0.\begin{align*}3(d+15)-18d=0.\end{align*}
Solution:
3(d+15)18d=03d+31518d=03d+4518d=015d+45=015d+4545=04515d=4511515d=11545d=3\begin{align*}&3(d+15)-18d=0\\ &3\cdot d+3\cdot 15-18d=0\\ &3d+45-18d=0\\ &-15d+45=0\\ &-15d+45-45=0-45\\ &-15d=-45\\ &-\frac{1}{15}\cdot -15d=-\frac{1}{15}\cdot-45\\ &d=3\\ \end{align*}
3(3+15)18(3)=03(3+15)18(3)=03(18)18(3)=05454=00=0\begin{align*}&3(3+15)-18(3)=0\\ &3(3+15)-18(3)=0\\ &3(18)-18(3)=0\\ &54-54=0\\ &0=0\\ \end{align*}
Vocabulary
Distributive Property: For any real numbers M, N,\begin{align*}M, \ N,\end{align*} and K\begin{align*}K\end{align*}:
M(N+K)=MN+MKM(NK)=MNMK\begin{align*}&M(N+K)= MN+MK\\ &M(N-K)= MN-MK\end{align*}
The Addition Property of Equality: For all real numbers a,b,\begin{align*}a, b,\end{align*} and c\begin{align*}c\end{align*}:
If a=b\begin{align*}a = b\end{align*}, then a+c=b+c\begin{align*}a + c = b + c\end{align*}.
The Multiplication Property of Equality: For all real numbers a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*}:
If a=b\begin{align*}a = b\end{align*}, then a(c)=b(c).\begin{align*}a(c)= b(c).\end{align*}
Guided Practice
Solve for x\begin{align*}x\end{align*} when 3(2x+5)+2x=7.\begin{align*}3(2x+5)+2x=7.\end{align*}
Solution:
Step 1: Apply the Distributive Property.
3(2x+5)+2x=732x+35+2x=76x+15+2x=7\begin{align*}&3(2x+5)+2x=7\\ &3\cdot 2x+3\cdot 5+2x=7\\ &6x+15+2x=7\\ \end{align*}
Step 2: Combine like terms.
6x+15+2x=78x+15=7\begin{align*}&6x+15+2x=7\\ &8x+15=7\\ \end{align*}
Step 3: Isolate the variable and its coefficient by using the Addition Property.
8x+15=78x+1515=7158x=8\begin{align*} &8x+15=7\\ &8x+15-15=7-15\\ &8x=-8\\ \end{align*}
Step 4: Isolate the variable by applying the Multiplication Property.
8x=8188x=818188x=818x=1\begin{align*} &8x=-8\\ &\frac{1}{8}\cdot 8x=-8\cdot \frac{1}{8}\\ &\frac{1}{8}\cdot 8x=-8\cdot \frac{1}{8}\\ & x=-1 \end{align*}
Substitute x=1\begin{align*}x=-1\end{align*} into 3(2x+5)=7.\begin{align*}3(2x+5)=7.\end{align*}
3(2(1)x+5)+2(1)=3(2+5)2=3(3)2=92=7.\begin{align*}3(2(-1)x+5)+2(-1)=3(-2+5)-2=3(3)-2=9-2=7.\end{align*}
Therefore, x=1.\begin{align*}x=-1.\end{align*}
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Multi-Step Equations (15:01)
Note that in the next video there IS an error. At one point the teacher says that -135/6 is in simplest form. There is a common factor of 3 in both numerator and denominator. -135/6 can be simplified by that factor of 3 to 45/2, or 22.5.
In 1 – 22, solve the equation.
1. 3(x1)2(x+3)=0\begin{align*}3(x - 1) - 2(x + 3) = 0\end{align*}
2. 7(w+20)w=5\begin{align*}7(w + 20) - w = 5\end{align*}
3. 9(x2)=3x+3\begin{align*}9(x - 2) = 3x + 3\end{align*}
4. 2(5a13)=27\begin{align*}2 \left (5a - \frac{1}{3} \right ) = \frac{2}{7}\end{align*}
5. 29(i+23)=25\begin{align*}\frac{2}{9} \left (i + \frac{2}{3} \right ) = \frac{2}{5}\end{align*}
6. 4(v+14)=352\begin{align*}4 \left (v + \frac{1}{4} \right ) = \frac{35}{2}\end{align*}
7. 22=2(p+2)\begin{align*}22=2(p+2)\end{align*}
8. (m+4)=5\begin{align*}-(m+4)=-5\end{align*}
9. 48=4(n+4)\begin{align*}48=4(n+4)\end{align*}
10. 65(v35)=625\begin{align*}\frac{6}{5} \left (v- \frac{3}{5} \right ) = \frac{6}{25}\end{align*}
11. \begin{align*}-10(b-3)=-100\end{align*}
12. \begin{align*}6v + 6(4v+1)=-6\end{align*}
13. \begin{align*}-46=-4(3s+4)-6\end{align*}
14. \begin{align*}8(1+7m)+6=14\end{align*}
15. \begin{align*}0=-7(6+3k)\end{align*}
16. \begin{align*}35=-7(2-x)\end{align*}
17. \begin{align*}-3(3a+1)-7a=-35\end{align*}
18. \begin{align*}-2 \left (n+ \frac{7}{3} \right )=- \frac{14}{3}\end{align*}
19. \begin{align*}- \frac{59}{60} = \frac{1}{6} \left (- \frac{4}{3} r-5 \right )\end{align*}
20. \begin{align*}\frac{4y+3}{7} = 9\end{align*}
21. \begin{align*}(c+3)-2c-(1-3c)=2\end{align*}
22. \begin{align*}5m-3[7-(1-2m)]=0\end{align*}
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# Division of Integers
Related Topics:
Lesson Plans and Worksheets for Grade 7
Lesson Plans and Worksheets for all Grades
Examples, videos, and solutions to help Grade 7 students recognize that division is the reverse process of multiplication, and that integers can be divided provided the divisor is not zero.
### Lesson 12 Student Outcomes
• Students recognize that division is the reverse process of multiplication, and that integers can be divided provided the divisor is not zero.
• Students understand that every quotient of integers (with a non-zero divisor) is a rational number and divide signed numbers by dividing their absolute values to get the absolute value of the quotient. The quotient is positive if the divisor and dividend have the same signs and negative if they have opposite signs.
### Lesson 12 Summary
• The rules for dividing integers are similar to the rules for multiplying integers (when the divisor is not zero). The quotient is positive if the divisor and dividend have the same signs, and negative if they have opposite signs.
The quotient of any 2 integers (with a non-zero divisor) will be a rational number.
### NYS Math Module 2 Grade 7 Lesson 12 Classwork
Exercise 1: Recalling the Relationship Between Multiplication and Division
a. List examples of division problems that produced a quotient that is a negative number.
b. If the quotient is a negative number, what must be true about the signs of the dividend and divisor?
c. List your examples of division problems that produced a quotient that is a positive number.
d. If the quotient is a positive number, what must be true about the signs of the dividend and divisor?
Rules for Dividing Two Integers:
• A quotient is negative if the divisor and the dividend have _____ signs.
• A quotient is positive if the divisor and the dividend have ____ signs.
Exercise 2: Is the Quotient of Two Integers Always an Integer
Is the quotient of two integers always an integer? Use the work space below to create quotients of integers. Answer the question and use examples or a counterexample to support your claim.
Conclusion: Every quotient of two integers is always a rational number, but not always an integer.
Exercise 3: Different Representation of the Same Quotient
Are the answers to the three quotients below the same or different? Why or why not?
a) -14 ÷ 7
b) 14 ÷ -7
c) -(14 ÷ 7)
Exercise 4: Fact Fluency—Integer Division
Lesson 12 Problem Set
1. Find the missing values in each column:
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# 2007 AIME II Problems/Problem 9
## Problem
Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$
## Solution
### Solution 1
Several Pythagorean triples exist amongst the numbers given. $BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105$. Also, the length of $EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287$.
Use the Two Tangent Theorem on $\triangle BEF$. Since both circles are inscribed in congruent triangles, they are congruent; therefore, $EP = FQ = \frac{287 - PQ}{2}$. By the Two Tangent theorem, note that $EP = EX = \frac{287 - PQ}{2}$, making $BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]$. Also, $BX = BY$. $FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]$.
Finally, $FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}$. Also, $FP = FQ + PQ = \frac{287 - PQ}{2} + PQ$. Equating, we see that $\frac{805 - PQ}{2} = \frac{287 + PQ}{2}$, so $PQ = \boxed{259}$.
### Solution 2
By the Two Tangent Theorem, we have that $FY = PQ + QF$. Solve for $PQ = FY - QF$. Also, $QF = EP = EX$, so $PQ = FY - EX$. Since $BX = BY$, this can become $PQ = FY - EX + (BY - BX)$$= \left(FY + BY\right) - \left(EX + BX\right) = FB - EB$. Substituting in their values, the answer is $364 - 105 = 259$.
### Solution 3
Call the incenter of $\triangle BEF$ $O_1$ and the incenter of $\triangle DFE$ $O_2$. Draw triangles $\triangle O_1PQ,\triangle PQO_2$.
Drawing $BE$, We find that $BE = \sqrt {63^2 + 84^2} = 105$. Applying the same thing for $F$, we find that $FD = 105$ as well. Draw a line through $E,F$ parallel to the sides of the rectangle, to intersect the opposite side at $E_1,F_1$ respectively. Drawing $\triangle EE_1F$ and $FF_1E$, we can find that $EF = \sqrt {63^2 + 280^2} = 287$. We then use Heron's formula to get:
$$[BEF] = [DEF] = 11 466$$.
So the inradius of the triangle-type things is $\frac {637}{21}$.
Now, we just have to find $O_1Q = O_2P$, which can be done with simple subtraction, and then we can use the Pythagorean Theorem to find $PQ$.
## Solution 4
Why not first divide everything by its greatest common factor, $7$? Then we're left with much simpler numbers which saves a lot of time. In the end, we will multiply by $7$.
From there, we draw the same diagram as above (with smaller numbers). We soon find that the longest side of both triangles is 52 (64 - 12). That means:
$A = rs$ indicating $26(9)=r(54)$ so $r = 13/3$.
Now, we can start applying the equivalent tangents. Calling them $a$, $b$, and $c$ (with $c$ being the longest and a being the shortest),
$a+b+c$ is the semi perimeter or $54$. And since the longest side (which has $b+c$) is $52$, $a=2$.
Note that the distance $PQ$ we desired to find is just $c - a$. What is $b$ then? $b = 13$. And $c$ is $39$. Therefore the answer is $37$.
|
# Identify lowest common denominators
## Interactive practice questions
Find the lowest common denominator of $\frac{1}{2}$12 and $\frac{1}{7}$17.
To do this, we will look at the multiples of $2$2 and $7$7.
a
Fill in the table with the first ten multiples of $2$2.
$\times$×
$1$1 $2$2 $3$3 $4$4 $5$5 $6$6 $7$7 $8$8 $9$9 $10$10
$2$2 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
b
Fill in the table with the first ten multiples of $7$7.
$\times$×
$1$1 $2$2 $3$3 $4$4 $5$5 $6$6 $7$7 $8$8 $9$9 $10$10
$7$7 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
c
Hence, what is the lowest common denominator of $\frac{1}{2}$12 and $\frac{1}{7}$17?
Easy
Approx 2 minutes
Find the lowest common denominator of $\frac{1}{2}$12 and $\frac{6}{7}$67.
To do this, we will look at the multiples of $2$2 and $7$7.
Find the lowest common denominator of $\frac{4}{5}$45 and $\frac{11}{12}$1112.
Find the lowest common denominator of $\frac{1}{12}$112 and $\frac{1}{7}$17.
### Outcomes
#### NA4-4
Apply simple linear proportions, including ordering fractions
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# 5.5: Substitution
The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.
At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to see? We are looking for an integrand of the form $$f[g(x)]g′(x)dx$$. For example, in the integral
$∫(x^2−3)^3 2x \, dx. \label{eq1}$
we have
$f(x)=x^3$
and
$g(x)=x^2−3.$
Then
$g'(x)=2x.$
and
$f[g(x)]g′(x)=(x^2−3)^3(2x),$
and we see that our integrand is in the correct form. The method is called substitution because we substitute part of the integrand with the variable u and part of the integrand with du. It is also referred to as change of variables because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.
Substitution with Indefinite Integrals
Let $$u=g(x)$$,, where $$g′(x)$$ is continuous over an interval, let $$f(x)$$ be continuous over the corresponding range of g, and let $$F(x)$$ be an antiderivative of $$f(x).$$ Then,
\begin{align} ∫f[g(x)]g′(x)\,dx &=∫f(u)\,du \nonumber \\ &=F(u)+C \nonumber\\ &= F(g(x))+C \end{align} \nonumber
Proof
Let $$f$$, $$g$$, $$u$$, and $$F$$ be as specified in the theorem. Then
$\dfrac{d}{dx}F(g(x))=F′(g(x))g′(x)=f[g(x)]g′(x).$
Integrating both sides with respect to x, we see that
$∫f[g(x)]g′(x)\,dx=F(g(x))+C.$
If we now substitute $$u=g(x)$$, and $$du=g'(x)dx$$, we get
$∫f[g(x)]g′(x)\,dx=∫f(u)\,du=F(u)+C=F(g(x))+C.$
Returning to the problem we looked at originally, we let $$u=x^2−3$$ and then $$du=2x\,dx$$.
Rewrite the integral (Equation \ref{eq1}) in terms of $$u$$:
$∫(x^2−3)^3(2x\,dx)=∫u^3\,du.$
Using the power rule for integrals, we have
$∫u^3\,du=\dfrac{u^4}{4}+C.$
Substitute the original expression for $$x$$ back into the solution:
$\dfrac{u^4}{4}+C=\dfrac{(x^2−3)^4}{4}+C.$
We can generalize the procedure in the following Problem-Solving Strategy.
Problem-Solving Strategy: Integration by Substitution
1. Look carefully at the integrand and select an expression $$g(x)$$ within the integrand to set equal to u. Let’s select $$g(x)$$. such that $$g′(x)$$ is also part of the integrand.
2. Substitute $$u=g(x)$$ and $$du=g′(x)dx.$$ into the integral.
3. We should now be able to evaluate the integral with respect to u. If the integral can’t be evaluated we need to go back and select a different expression to use as u.
4. Evaluate the integral in terms of u.
5. Write the result in terms of x and the expression $$g(x).$$
Example $$\PageIndex{1}$$: Using Substitution to Find an Antiderivative
Use substitution to find the antiderivative of $$∫6x(3x^2+4)^4\,dx.$$
Solution
The first step is to choose an expression for u. We choose $$u=3x^2+4$$ because then $$du=6x\,dx$$ and we already have $$du$$ in the integrand. Write the integral in terms of $$u$$:
$∫6x(3x^2+4)^4dx=∫u^4\,du.$
Remember that du is the derivative of the expression chosen for u, regardless of what is inside the integrand. Now we can evaluate the integral with respect to u:
$∫u^4\,du=\dfrac{u^5}{5}+C=\dfrac{(3x^2+4)^5}{5}+C.$
Analysis
We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for C of 1, we let $$y=\dfrac{1}{5}(3x^2+4)^5+1.$$ We have
$y=\dfrac{1}{5}(3x^2+4)^5+1,$
so
$y′=(\dfrac{1}{5})5(3x^2+4)^46x=6x(3x^2+4)^4.$
This is exactly the expression we started with inside the integrand.
Exercise $$\PageIndex{1}$$
Use substitution to find the antiderivative of $$∫3x^2(x^3−3)^2dx.$$
Hint: Let $$u=x^3−3.$$
Solution
$∫3x^2(x^3−3)^2dx=\dfrac{1}{3}(x^3−3)^3+C$
Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.
Example $$\PageIndex{2}$$: Using Substitution with Alteration
Use substitution to find the antiderivative of $∫z\sqrt{z^2−5}\,dz.$
Solution
Rewrite the integral as $$∫z(z^2−5)^{1/2}\,dz.$$ Let $$u=z^2−5$$ and $$du=2z\,dz$$. Now we have a problem because $$du=2z\,dz$$ and the original expression has only $$z\,dz$$. We have to alter our expression for du or the integral in u will be twice as large as it should be. If we multiply both sides of the du equation by $$\dfrac{1}{2}$$. we can solve this problem. Thus,
$u=z^2−5$
$du=2z\,dz$
$\dfrac{1}{2}du=\dfrac{1}{2}(2z)\,dz=z\,dz.$
Write the integral in terms of u, but pull the $$\dfrac{1}{2}$$ outside the integration symbol:
$∫z(z^2−5)^{1/2}\,dz=\dfrac{1}{2}∫u^{1/2}\,du.$
Integrate the expression in $$u$$:
$$\dfrac{1}{2}∫u^{1/2}\,du=(\dfrac{1}{2})\dfrac{u^{3/2}}{\dfrac{3}{2}}+C$$
$$=(\dfrac{1}{2})(\dfrac{2}{3})u^{3/2}+C$$
$$=\dfrac{1}{3}u^{3/2}+C$$
$$=\dfrac{1}{3}(z^2−5)^{3/2}+C$$
Exercise $$\PageIndex{2}$$
Use substitution to find the antiderivative of $$∫x^2(x^3+5)^9\,dx.$$
Hint: Multiply the du equation by $$\dfrac{1}{3}$$.
Solution
$\dfrac{(x^3+5)^{10}}{30}+C$
Example $$\PageIndex{3}$$: Using Substitution with Integrals of Trigonometric Functions
Use substitution to evaluate the integral $$∫\dfrac{\sin t}{\cos^3t}\,dt.$$
Solution
We know the derivative of $$\cos t$$ is $$−\sin t$$, so we set $$u=\cos t$$. Then $$du=−\sin t\,dt.$$
Substituting into the integral, we have
$∫\dfrac{\sin t}{\cos^3t}\,dt=−∫\dfrac{du}{u^3}.$
Evaluating the integral, we get
$−∫\dfrac{du}{u^3}=−∫u^{−3}\,du=−(−\dfrac{1}{2})u^{−2}+C.$
Putting the answer back in terms of t, we get
$∫\dfrac{\sin t}{\cos^3t}\,dt=\dfrac{1}{2u^2}+C=\dfrac{1}{2\cos^2t}+C.$
Exercise $$\PageIndex{3}$$
Use substitution to evaluate the integral $∫\dfrac{\cos t}{\sin^2t}\,dt.$
Hint: Use the process from Example to solve the problem.
Solution
$$−\dfrac{1}{\sin t}+C$$
Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, $$u$$ should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of $$u$$. This technique should become clear in the next example.
Example $$\PageIndex{4}$$: Finding an Antiderivative Using u-Substitution
Use substitution to find the antiderivative of $∫\dfrac{x}{\sqrt{x−1}}\,dx.$
Solution
If we let $$u=x−1,$$ then $$du=dx$$. But this does not account for the x in the numerator of the integrand. We need to express x in terms of u. If $$u=x−1$$, then $$x=u+1.$$ Now we can rewrite the integral in terms of u:
$∫\dfrac{x}{\sqrt{x−1}}\,dx=∫\dfrac{u+1}{\sqrt{u}}\,du=∫\sqrt{u}+\dfrac{1}{\sqrt{u}}\,du=∫(u^{1/2}+u^{−1/2})\,du.$
Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result. Thus,
$$∫(u^{1/2}+u^{−1/2})\,du=\dfrac{2}{3}u^{3/2}+2u^{1/2}+C$$
$$=\dfrac{2}{3}(x−1)^{3/2}+2(x−1)^{1/2}+C$$
$$=(x−1)^{1/2}[\dfrac{2}{3}(x−1)+2]+C$$
$$=(x−1)^{1/2}(\dfrac{2}{3}x−\dfrac{2}{3}+\dfrac{6}{3})$$
$$=(x−1)^{1/2}(\dfrac{2}{3}x+\dfrac{4}{3})$$
$$=\dfrac{2}{3}(x−1)^{1/2}(x+2)+C.$$
Exercise $$\PageIndex{4}$$:
Use substitution to evaluate the indefinite integral $∫\cos^3t\sin t\,dt.$
Hint: Use the process from Example to solve the problem.
Solution
$−\dfrac{cos^4t}{4}+C$
## Substitution for Definite Integrals
Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.
Substitution with Definite Integrals
Let $$u=g(x)$$ and let $$g'$$ be continuous over an interval $$[a,b]$$, and let $$f$$ be continuous over the range of $$u=g(x).$$ Then,
$∫^b_af(g(x))g′(x)dx=∫^{g(b)}_{g(a)}f(u)\,du.$
Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if $$F(x)$$ is an antiderivative of $$f(x),$$ we have
$∫f(g(x))g′(x)\,dx=F(g(x))+C.$
Then
\begin{align} ∫^b_af[g(x)]g′(x)\,dx &= F(g(x))\bigg|^{x=b}_{x=a} \nonumber \\ &=F(g(b))−F(g(a)) \nonumber\\ &= F(u) \bigg|^{u=g(b)}_{u=g(a)} \nonumber\\ &=∫^{g(b)}_{g(a)}f(u)\,du \nonumber\end{align} \nonumber
and we have the desired result.
Example $$\PageIndex{5}$$: Using Substitution to Evaluate a Definite Integral
Use substitution to evaluate $∫^1_0x^2(1+2x^3)^5\,dx.$
Solution
Let $$u=1+2x^3$$, so $$du=6x^2dx$$. Since the original function includes one factor of $$x^2$$ and $$du=6x^2dx$$, multiply both sides of the du equation by $$1/6.$$ Then,
$du=6x^2\,dx$
$\dfrac{1}{6}du=x^2\,dx.$
To adjust the limits of integration, note that when $$x=0,u=1+2(0)=1,$$ and when $$x=1,u=1+2(1)=3.$$ Then
$∫^1_0x^2(1+2x^3)^5dx=\dfrac{1}{6}∫^3_1u^5\,du.$
Evaluating this expression, we get
$\dfrac{1}{6}∫^3_1u^5\,du=(\dfrac{1}{6})(\dfrac{u^6}{6})|^3_1=\dfrac{1}{36}[(3)^6−(1)^6]=\dfrac{182}{9}.$
Exercise $$\PageIndex{5}$$
Use substitution to evaluate the definite integral $∫^0_{−1}y(2y^2−3)^5\,dy.$
Hint: Use the steps from Example to solve the problem.
Solution
$$\dfrac{91}{3}$$
Example $$\PageIndex{6}$$: Using Substitution with an Exponential Function
Use substitution to evaluate $∫^1_0xe^{4x^2+3}\,dx.$
Solution
Let $$u=4x^3+3.$$ Then, $$du=8x\,dx.$$ To adjust the limits of integration, we note that when $$x=0,u=3$$, and when $$x=1,u=7$$. So our substitution gives
$∫^1_0xe^{4x^2+3}\,dx=\dfrac{1}{8}∫^7_3e^udu=\dfrac{1}{8}e^u|^7_3=\dfrac{e^7−e^3}{8}≈134.568$
Exercise $$\PageIndex{6}$$:
Use substitution to evaluate $∫^1_0x^2cos(\dfrac{π}{2}x^3)\,dx.$
Solution
$$\dfrac{2}{3π}≈0.2122$$
Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in Example.
Example $$\PageIndex{7}$$: Using Substitution to Evaluate a Trigonometric Integral
Use substitution to evaluate $∫^{π/2}_0\cos^2θ\,dθ.$
Solution
Let us first use a trigonometric identity to rewrite the integral. The trig identity $$\cos^2θ=\dfrac{1+\cos 2θ}{2}$$ allows us to rewrite the integral as
$∫^{π/2}_0\cos^2θdθ=∫^{π/2}_0\dfrac{1+\cos2θ}{2}\,dθ.$
Then,
$∫^{π/2}_0(\dfrac{1+\cos2θ}{2})dθ=∫^{π/2}_0(\dfrac{1}{2}+\dfrac{1}{2}\cos 2θ)\,dθ$
$=\dfrac{1}{2}∫^{π/2}_0\,dθ+∫^{π/2}_0\cos2θ\,dθ.$
We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let $$u=2θ.$$ Then, $$du=2\,dθ,$$ or $$\dfrac{1}{2}\,du=dθ$$. Also, when $$θ=0,u=0,$$ and when $$θ=π/2,u=π.$$ Expressing the second integral in terms of $$u$$, we have
$$\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}∫^{π/2}_0cos^2θ\,dθ=\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}(\dfrac{1}{2})∫^π_0\cos u \,du$$
$$=\dfrac{θ}{2}|^{θ=π/2}_{θ=0}+\dfrac{1}{4}sinu|^{u=θ}_{u=0}$$
$$=(\dfrac{π}{4}−0)+(0−0)=\dfrac{π}{4}$$
## Key Concepts
• Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable u and du for appropriate expressions in the integrand.
• When using substitution for a definite integral, we also have to change the limits of integration.
## Key Equations
• Substitution with Indefinite Integrals
$$∫f[g(x)]g′(x)dx=∫f(u)du=F(u)+C=F(g(x))+C$$
• Substitution with Definite Integrals
$$∫^b_af(g(x))g'(x)dx=∫^{g(b)}_{g(a)}f(u)du$$
## Glossary
change of variables
the substitution of a variable, such as u, for an expression in the integrand
integration by substitution
a technique for integration that allows integration of functions that are the result of a chain-rule derivative
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Resource Lesson Basic Trigonometry
Given below are three classic triangles with which you should become familiar.
The six basic right-triangle ratios that you need to learn are:
Using the triangles and ratios given above, you should be able to state the EXACT value for each of the trig functions requested. All answers should be expressed as common fractions NOT decimals.
sine cosine tangent cotangent secant cosecant
30º
37º
45º
53º
60º
Notice that three of these relationships are reciprocal functions:
Another group of functions are called co-functions. These functions quantify relationships between complementary angles.
Prior to the advent of scientific and graphing calculators, early trigonometry tables were usually set up based on these co-function relationships. Notice in the linked table how they display the fact that the co-functions for the complementary angles 37º and 53º are equal. The functions for the angles listed "down" the left side are across the top of the page while the functions for the angles listed "up" the right side are across the bottom of the page. For example, take a moment to see that the first column, "Sin," across the top is aligned with the column labeled "Cos" across the bottom.
On a unit circle, that is, a circle with radius equal to 1, the x-value of each co-ordinate represents the cosine of the central angle, the y-value of each co-ordinate represents the sine of the central angle.
Use the unit circle given above to determine the values for the sine and cosine of each quadrant angle given below. The remaining values for tangent, cotangent, secant, and cosecant can be calculated by using the functional relationships stated above. Once again, you should be able to state the EXACT value for each of the trig functions requested. All answers should be expressed as common fractions NOT decimals.
sine cosine tangent cotangent secant cosecant
0º
90º
180º
270º
360º
A radian is a numerical ratio for any central angle that compares the magnitude of the intercepted arc length to the length of the circle's radius. This tells us that when the central angle θ equals 1 radian, the arc length s equals the radius. The expression s = rθ represents this relationship. Note that when the arc length equals an entire circumference, s = rθ 2πr = rθ θ = 2π radians = 360º When measured in radians, as θ → 0, tan θ → sin θ which in turn approaches θ. This is called the small angle approximation and incurs an error of no greater than 0.1% for angles less than 5º. You can verify these relationships by examining the values for θ, sin θ, tan θ in Table 2. All of these angle values are often represented graphically when we speak of circular functions. Trigonometrically, we generally use the variable x when expressing angles in terms of radians and θ when expressing them in terms of degrees. Another useful set of trig identities are the called the Pythagorean Identities. These identities are based on line values drawn to a unit circle. The double-angle formulas for sine and cosine. sin 2θ = 2 sin θ cos θ cos 2θ = cos 2θ - sin 2θ When solving for a missing side or angle is a triangle, there are two important relationships that apply to any triangle that can make your job easier: the Law of Sines and the Law of Cosines. The Law of Sines, , can be used to solve for a missing side or angle in a general triangle when you know either two sides and an angle opposite one side, oronly one side and all of the angles The Law of Cosines can be used to solve for a missing side in a general triangle when you know the other two sides and their included angle.
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### Geometry of Determinant
##### 2. Determinant of 3 by 3 matrix
The determinant of a 3 by 3 matrix is
det[ a11 a12 a13 ] = a11a22a33 + a12a23a31 + a13a21a32 - a13a22a31 - a11a23a32 - a12a21a33. a21 a22 a23 a31 a32 a33
A simple way of memorizing the formula is to repeat the first and the second columns to the right of the matrix and then multiply as follows to get the six terms.
Similar to the 2 by 2 case, the absolute value of the determinant of a 3 by 3 matrix is the volume of a parallelogram spanned by the three column vectors. The claim may be verified in a way similar to the 2 by 2 case. However, this involves complicated 3 dimensional picture and we will not carry this out here. See this exercise for a very special case. We will provide a more general explaination later on. For the moment, we will simply accept the interpretation.
A consequence of the interpretation is that
det[u v w] = 0
u v w lie on a plane (because the volume is zero)
u v w are linearly dependent (see this discussion)
⇔ [u v w] is not invertible. (by this result)
This clearly generalizes the criterion for the invertibility of a 2 by 2 matrix. The criterion will still be true for general n by n matrices, as long as the absolute value of the general determinant still measures the volume of the parallelogram.
Example The volume of the tetrahedron with the vertices (1, -2, 1), (2, 1, 2), (0, -1, 4), (2, -3, 1) is one sixth of the volume of the parallelogram spanned by u = (2, 1, 2) - (1, -2, 1) = (1, 3, 1), v = (0, -1, 4) - (1, -2, 1) = (-1, 1, 3), w = (2, -3, 1) - (1, -2, 1) = (1, -1, 0), which is
(1/6)| det[ 1 -1 1 ] | = (1/6)|1×1×0 + (-1)×(-1)×1 + 1×3×3 - 1×1×1 - 1×(-1)×3 - (-1)×3×0| = 2. 3 1 -1 1 3 0
Similar to the 2 by 2 case, the sign of the determinant of a 3 by 3 matrix is also determined by the orientation of the three column vectors. Specifically,
u, v, w follows the right hand rule ⇒ det[u v w] ≥ 0
u, v, w follows the left hand rule ⇒ det[u v w] ≤ 0
The reason for the orientation to determines the sign is similar to the 2 by 2 case. We may construct a (somewhat more complicated) movie to deform the given arrangement into one of the two standard arrangements.
The extension of the determinant to matrices of bigger size should reflect the requirements that the absolute value is the volume of a parallelogram and the sign should be the orientation of the arrangement of the column vectors. We emphasis that the following naive extension to 4 by 4 matrices does not satisfy the requirements.
In fact, the determinant of a 4 by 4 matrix contains 24 terms. In general, the determinant of an n by n matrix contains n! = 1×2×3×...×n terms.
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# T/F every function is a relation
Every function is a relation, but not every relation is a function.
Hope this helps :)
Answer: Every function is a relation. However, not every relation is a function. In a function, there cannot be two lists that disagree on only the last element.
Hope this helps :)
## Related Questions
Functions of angles, I need help
The Third Choice
Step-by-step explanation:
In this question, we have to compare the sine of angle A and the cosine of angle B. As a review, the sine function is
and the cosine function is
.
For angle A, side BC is the opposite side, while side AB is the hypotenuse. [The hypotenuse of a right triangle is the side that is opposite to the 90 degree angle. In this case, AB is the hypotenuse because it is opposite of angle C, which measures 90 degrees.]
Now, we can determine the sine of angle A.
.
For angle B, side BC is the adjacent side, while side AB is the hypotenuse.
Using this information, we can determine the cosine of angle B.
As we can see, both the sine of angle A and the cosine of angle B have the same value. In other words, they are equal to each other. Therefore, the correct answer is the third choice.
The answer is the third one
an 8 by 18 ft wall has three 4 by 5 ft windows.the area of the three windows is what percent of the area of the entire wall (including the windows)?
Percentage of area of the three windows of the area of entire wall = 41.67%
Step-by-step explanation:
To find the percentage of area of the three window of the area of entire wall, we will need to find the area of the window (Length*breath) as well as the area of the entire wall (length*breath).
Area of the window = 4ft * 5ft = 20ft²
Area of three windows = Area of the window * 3
Area of the three windows = 20ft² * 3
Area of the three windows = 60ft²
Area of the entire wall = 8ft * 18ft
Area of the entire wall = 144ft²
Percentage of area of the three windows of the area of entire wall = area of the three windows / area of the entire wall * 100
= 60ft²/144ft² * 100
= 41.67%
area of windows = (3×4×5)
= 60
area of wall = 8×18
= 144
=(60÷144)×100
=0.4167 × 100
=41. 67% or
=42 %
Four out of every five visitors at an amusement park buy day passes what percent of the visitors buy passes?
The percentage of visitors who buy passes is 80%.
### What is the percentage?
A percentage is a value per hundredth. Percentages can be converted into decimals and fractions by dividing the percentage value by a hundred.
We know fractions can be converted into percentages by multiplying them by 100.
Given, Four out of every five visitors at an amusement park buy day passes.
∴ The percentage of people who buy passes is (4/5)×100 = 80%.
brainly.com/question/24159063
#SPJ5
80%
THis is because you are supposed to multiply 20 to the denomintaor and numerator.
4 times 20=80
~JZ
Hope it helps
Which functions are increasing?
Choose all answers that are correct.
the 2,3 they are the best answers .
The increasing functions will be going upward.
- The second graph has an increasing function.
- The third one is also increasing from 0 to the other ordered pair.
Thirteen out of twenty-one people surveyed prefer fresh fruit over canned fruit. What percent of people prefer fresh fruit?
The percent of the people who prefer fresh fruit is 61.9%
Step-by-step explanation:
Assume x% people prefer fresh fruit over canned fruit. Then x can be calculated as:
x= No of people who prefer fresh fruit * 100
Total no of surveyed people
x = 13 * 100
21
x = 61.9%
What is 86% written as a decimal
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# ACT Math : How to find a solution set
## Example Questions
2 Next →
### Example Question #11 : How To Find A Solution Set
Given the follow inequality, which of the following presents a range of possible answers for the inequality: –3 < 3x + 2 ≤ 3.5
(–3, 1/2)
( –2, 2)
(–1,1)
(–1, ½)
(½, 1)
(–1, ½)
Explanation:
If you plug in the outer limits of the given ranges, (–1, ½) is the only combination that fits within the given equation. It is important to remember that "<" means “less than,” and "≤" means “less than or equal.” For example, if you answered (–2,2), plugging in 2 would make the the expression equal 8, which is greater than 3.5. And plugging in –2 for x would make the expression equal –4, which is less than –3, not greater. However, plugging in the correct answer (–1, ½) gives you –1 as your lower limit and 3.5 as your upper limit, which satisfies the equation. Both limits of the data set must satisfy the equation.
### Example Question #12 : How To Find A Solution Set
If , what is the product of the largest and smallest integers that satisfy the inequality?
7
0
–10
–5
5
0
Explanation:
The inequality in the question possesses an absolute value; therefore, we most solve for the variable being less than positive 6 and greater than negative 6. Let's start with the positive solution.
Add 4 to both sides of the inequality.
Divide both sides of the inequality by 2.
Now, let's solve for the negative solution
Add 4 to both sides of the inequality.
Divide both sides of the inequality by 2.
Using these solutions we can write the following statement:
The smallest integer that satisfies this equation is 0, and the largest is 4. Their product is 0.
2 Next →
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Differences of Integers with the Same Sign
## Subtract positive numbers from positive numbers and negatives from negatives.
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Practice Differences of Integers with the Same Sign
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Differences of Integers with the Same Sign (D-4)
Remember Jamie and the money? Take a look once again.
Jamie earned ten dollars cutting grass. He owes his brother twelve dollars from a movie that they went to. Does Jamie still owe his brother money if he gives him the whole ten dollars he made? How much does he still owe?
In the last Concept, you solved this problem by using a number line. We could also solve it by subtracting integers with the same sign.
In this Concept, you will learn how to solve this problem in a different way.
### Guidance
In the last Concept, you learn how to subtract integers using a number line. You won’t always have time to draw a number line though, so this Concept will teach you how to subtract integers that have the same sign without the visual aid. Let’s begin.
First, let’s look at subtracting two positive numbers.
9 – 4 = ____
In this problem, if we use the language of losses and gains, we could say that we have a gain of nine and a loss of four. Because our loss is not greater than the gain, our answer is positive.
This is a key point. If the loss is greater than the gain, then our answer would be negative. In this example, the loss of four is not greater than the gain of nine, so our answer remains positive.
9 – 4 = 5
The answer is positive 5.
Here is a problem where we are still finding the difference between two positive numbers, but the loss is greater than the gain.
3 – 8 = _____
In this problem we start with a positive three or a gain of three. Then we have a loss of eight. The loss is greater than the gain that we started with.
3 – 8 = -5
Our answer is negative. It is a negative five.
Yes. Actually there is an easier way to think about subtracting any two integers. You can always think in terms of losses and gains, but if that is difficult, we can think of subtraction as being the opposite of addition-that is the key to making things simpler. Here is the hint.
What does this look like? How can we rewrite a subtraction problem as an addition problem?
3 – 8 = 3 + -8 = ____
The subtraction became addition.
Positive three plus a negative 8 is still a negative 5.
Our answer did not change even though our method of solving it did. The answer is still -5.
How can we find the difference of two negative numbers?
-6 – -3 = ____
We can find this difference in two ways. The first way is to think in terms of losses and gains. The second is to change subtraction to addition by adding the opposite. Let’s start with losses and gains.
If we think of this problem in terms of losses and gains, we start with a loss of 6.
-6
Next, we don’t add another loss, but we take away a loss. If you take away a loss, that is the same thing as a gain. So we have a gain of 3.
-6 combined with a gain of 3 = -3
Our answer is -3.
Now, let’s solve the problem by changing subtraction to addition by adding the opposite.
-6 – -3 = -6 + 3
We changed the subtraction to addition and added the opposite. The opposite of the given value of negative three is positive three. Now we can solve the addition problem.
-6 + 3 = -3
Notice that the answer is the same no matter which way you approach it. The answer is still -3.
Practice what you have learned by finding the differences of the following integer pairs.
5 – 10 = ____
Solution: - 5
14 – 7 = ____
Solution: 7
#### Example C
-4 – -8 = ____
Solution: 4
Here is the original problem once again.
Jamie earned ten dollars cutting grass. He owes his brother twelve dollars from a movie that they went to. Does Jamie still owe his brother money if he gives him the whole ten dollars he made? How much does he still owe?
First, let's work with the information that we have been given.
Jamie earned ten dollars.
\begin{align*}10\end{align*}
He owes 12 dollars.
\begin{align*}-12\end{align*}
We can put it together for an expression.
\begin{align*}10 - 12 = 2\end{align*}
Jamie still owes his brother two dollars.
### Vocabulary
Here are the vocabulary words in this Concept.
Sum
the result of an addition problem.
Difference
the result of a subtraction problem.
### Guided Practice
Here is one for you to try on your own.
The bank started out the day with a loss of 25 points. Over the course of the day, a twelve point loss was taken away. What was the final statistic at the end of the day?
To work on this problem, we can write each loss as a negative number.
The bank started with a loss of 25 points.
\begin{align*}-25\end{align*}
Over the course of the day, a 12 point loss was taken away.
\begin{align*}- (-12)\end{align*}
We can put it together for a complete expression.
\begin{align*}-25 - (-12)\end{align*}
The bank's ending statistic was still a loss of 13 points.
### Video Review
Here are videos for review.
### Practice
Directions: Subtract each pair of integers.
1. \begin{align*}14 - 19\end{align*}
2. \begin{align*}24 - 19\end{align*}
3. \begin{align*}-1 - 7\end{align*}
4. \begin{align*}-4 - 12\end{align*}
5. \begin{align*}-14 - 29\end{align*}
6. \begin{align*}-24 - (-19)\end{align*}
7. \begin{align*}9 - 11\end{align*}
8. \begin{align*}13 - (-1)\end{align*}
9. \begin{align*}23 - 19\end{align*}
10. \begin{align*}-31 - 15\end{align*}
11. \begin{align*}-18 - (-19)\end{align*}
12. \begin{align*}-14 - 6\end{align*}
13. \begin{align*}-74 - 39\end{align*}
14. \begin{align*}54 - (-29)\end{align*}
15. \begin{align*}64 - 99\end{align*}
### My Notes/Highlights Having trouble? Report an issue.
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Binary number facts for kids
Kids Encyclopedia Facts
Numbers 0 to 7 shown in binary form.
The binary numeral system is a way to write numbers using only two digits: 0 and 1. These digits are used in computers as a series of "off" and "on" switches.
Most people use ten different digits — 0 to 9 — to write any number. For example:
• 6
• 52
• 137
• 9,826
• 54,936
This is called the decimal number system or base ten system, which means that this number system has ten different digits to construct a number. It is a good system because people have 10 fingers to help them.
Computers do not use the decimal number system. This is because computers are built with electronic circuits, each part of which can be either on or off. Since there are only two options, they can only represent two different digits: 0 and 1. This is called the binary number system, or base two. ("Bi" means two.) All the numbers are constructed from the two digits 0 and 1. A digit in binary (that's a 0 or a 1) is also called a bit – short for binary digit.
Computers use this number system to add, subtract, multiply, divide, and do all their other math and data. They even save data in the form of bits.
A bit by itself can only mean zero or one, so to represent bigger numbers (and even represent letters), bits are grouped together into chunks. Eight bits make a byte, and computers use as many bytes as they need to store what we need them to. Modern computers have many billions of bytes of storage.
Why Do We Use Binary?
Woman shows her height in binary numbers at the Copernicus Science Centre in Warsaw. (Credit: Alexander Baxevanis)
In normal math, we don't use binary. We were taught to use our normal number system. Binary is much easier to do math in than normal numbers because you only are using two number-symbols — 1 and 0 — instead of ten number-symbols — 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
With just two number-symbols, you can count quite high using things that just go "on" or "off," "yes" or "no". For example: How high can you count with your fingers? Most people would say that they could count to 10. If you count on your fingers with binary, you can count to 31 with one hand! With two hands, using binary, you can count up to 1023!
Computers use binary because they can only read and store an "on" or "off" charge. So, using 0 as "off" and 1 as "on," we can use numbers in electrical wiring. Think of it like this — if you had one color for every math symbol (0 to 9), you'd have ten colors. That's a lot of colors to memorize, but you have done it anyway. If you were limited to only black and white, you'd only have two colors. It would be so much easier to memorize, but you would need to make a new way of writing down numbers. Binary is just that — a new way to record and use numbers.
Binary Notation
In school, you were taught place value. Numbers have a ones place, a tens place, a hundreds place, a thousands place, a ten-thousands place, and so on. Each column, or place, is worth ten times more than the place to its right. If you haven't seen it before, place value looks like this:
Decimal column 10,000 1000 100 10 1
Digit 5 4 9 3 6
Value 5 × 10,000 4 × 1000 9 × 100 3 × 10 6 × 1
So the decimal number 54,936 is equal to 5×10000 + 4×1000 + 9×100 + 3×10 + 6×1.
Binary also has columns, but instead of multiplying by ten, they multiply by two each time:
Binary column 128 64 32 16 8 4 2 1
Bit 1 0 1 1 0 1 0 1
Value 1 × 128 0 × 128 1 × 32 1 × 16 0 × 8 1 × 4 0 × 2 1 × 1
So the binary number 10110101 = 1×128 + 1×32 + 1×16 + 1×4 + 1×1 = 181 in decimal.
This method lets us read binary numbers, but how do we write them? One way is to write a list of all the numbers starting from one and work upward. Just as adding 1 to 9 in decimal carries over to make 10 and 1 + 99 makes 100, in binary when you add one to one, carry a one over to the next place on the left. Follow along with this table to see how that works.
Base-10 Binary
1 1
2 10
3 11
4 100
5 101
6 110
7 111
8 1000
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111
16 10000
You'll notice that the values 1, 2, 4, 8, and 16 only need one 1 bit and some 0 bits, because there's a column with that value, and we just have to set the bit in that column to 1.
Base-10 version Binary
1 1
2 10
4 100
8 1000
16 10000
Have you noticed a pattern in writing binary numbers? Study the table for 1 to 16 again until you understand why in binary,
"1 + 1 = 10" and "1 + 100 = 101"
Since we use the decimal system, you may have lots of practice reading decimal but none yet reading binary. Reading binary may seem difficult and time-consuming at first, but with practice, it will become easier.
Translating to Base-10
The binary number for 52 is 110100. How do you read a binary number?
1. First, we at the ones column. Since it has a 0 in it, we don't add anything to the total.
2. Next, we look at the twos column. Nothing, so we move on to the next column.
3. We have a 1 in the fours column, so we add 4 to the total (total is 4).
4. Skipping the eights column since it has a 0, we have come to a 1 in the sixteens column. We add 16 to the total (total is 20).
5. Last, we have a 1 in the thirty-twos column. We add this to our total (total is 52).
We're done! We now have the number 52 as our total. The basics of reading a base-2 number is to add each column's value to the total if there is a 1 in it. You don't have to multiply like you do in base-10 to get the total (like the 5 in the tens column from the above base-10 example). This can speed up your reading of base-2 numbers. Let's look at that in a table.
Binary digit Column Binary digit's value
0 1 0
0 2 0
1 4 4
0 8 0
1 16 16
1 32 32
Total 52
Finding a Mystery Number
Now let's look at another number. The binary number is 1011, but we don't know what it is. Let's go through the column-reading process to find out what the number is.
1. The ones column has a 1 in it, so we add 1 x 1 to the total (total is 1).
2. The twos column has a 1 in it, so we add 1 x 2 to the total (total is 3).
3. The fours column has a 0 in it, so we add 0 x 4 to the total (total is still 3).
4. The eights column has a 1 in it, so we add 1 x 8 to the total (total is 11).
There are no more columns, so the total is the answer. The answer is 11!
Storing Text
Computers store everything in binary, including text. To do this, every letter, every punctuation character, and, in fact, a very large number of the symbols people have ever used, has been given its own number in a system called Unicode.
For example, if your name is "George," then the computer can store that in binary just by storing the number for "G", then for "e", and so on. The most common symbols in American English, like letters without accents, can be stored with just one byte. Other symbols, like "£" and "¿", need more than one byte because they have been given a bigger number. A few examples:
• G is stored as 71, which is "0100 0111" in binary.
• e is stored as one hundred one, which is "0110 0101" in binary.
The whole word "George" looks like:
```0100 0111 0110 0101 0110 1111 0111 0010 0110 0111 0110 0101
```
While this might look strange, see if you can find the rest of the letters in the word and what their decimal representation is!
History
The binary numeral system, in a manuscript by Gottfried Wilhelm Leibniz, 1697
The modern binary number system is credited to Gottfried Leibniz in 1679.
Binary was invented by many people, but the modern binary number system is credited to Gottfried Leibniz, a German mathematician, in 1679.
In 1817, John Leslie (a Scottish mathematician) suggested that primitive societies may have counted with objects (like pebbles) before they even had words to describe the total number of objects involved. Next, they would have discovered that this pile of objects could be reduced into two piles of equal measures (leaving either 0 objects or 1 object leftover). This remainder (odd = 1 or even = 0) would then be recorded and one of the piles removed, while the second pile was then further divided into two sub piles. If you record the remainder left over after the original pile has been divided in two and continue repeating this process, you will ultimately be left with just either 2 or 3 objects. If you record the remainder leftover (odd = 1 or even = 0) at the end of each reduction you will eventually be left with a tally record of 1's and 0's which will be the binary representation of your original pile of objects. So instead of representing your original pile of objects with a repeating number, or marks, or tokens (which for large numbers could be quite long), you have reduced your pile of objects into a more compact binary number. If you need to recover the original number of objects from this summarized binary number, it is easy enough to do: start with the first tally mark and then double it and add one if the next binary number contains a 1. Continue the process until the end of the binary number is reached. So, binary counting may be both the oldest and the most modern way of counting.
Binary has been used in nearly everything electronic; from calculators to supercomputers. Machine code is binary digits.
Sources
Wikijunior
Binary number Facts for Kids. Kiddle Encyclopedia.
|
# Introduction
## Congruence
Congruence modulo n is a congruence relation, meaning that it is an equivalence relation that is compatible with the operations of addition, subtraction, and multiplication. Congruence modulo n is denoted :
$$a \equiv b [n]$$
The congruence relation may be rewritten as (k ∈ ℤ):
$$a = b + k \times n$$
### Practical example
$$12 \equiv 3 [9]$$
$$21 \equiv 3 [9]$$
$$-6 \equiv 3 [9]$$
Because :
$$12 = 3 + 1 \times 9$$
$$21 = 3 + 2 \times 9$$
$$-6 = 3 - 1 \times 9$$
In python, you can use the % symbol.
>>> 12 % 9 == 21 % 9 == -6 % 9 == 3
True
$$12 \equiv 3 [9]$$
You can add or substract any number. Example with +3 and -5 :
$$15 \equiv 6 [9]$$
$$7 \equiv -2 [9]$$
The last equivalence is equals too (sometimes it's easier to deal with positive numbers):
$$7 \equiv 7 [9]$$
## Multiplication
$$12 \equiv 3 [9]$$
You can multiply with any number € Z (you cannot make a division, ex: multiply by 1/2).
1. Example with * 3 :
$$36 \equiv 9 [9]$$
Which is equals to :
$$0 \equiv 0 [9]$$
$$36 - 9 * 4 = 9 - 9 * 1$$
2. Example with * -5 :
$$-60 \equiv -15 [9]$$
Which is equals to :
$$3 \equiv 3 [9]$$
$$-60 + 9 * 7 = -15 + 2 * 9$$
## Divsion (warning)
You can't divise a equivalence relation.
$$12 \equiv 2 [10]$$
Divide by 2 :
$$6 \not\equiv 1 [10]$$
## Modular inverse
The multiplicative inverse :
$$x * a \equiv 1 [n]$$
$$x * a + n * k = 1$$
$$x \equiv a^{-1} [n]$$
It may be efficiently computed by solving Bézout's equation a * x + n * k = 1 using the Extended Euclidean algorithm (used to compute the GCD - Greatest common divisor).
### Practical example
$$5 \times x \equiv 1 [34]$$
$$5 \times x = 1 + 34 \times k$$
$$1 = 5 \times 7 - 34$$
You can use the modular inverse in Python :
$$5 \equiv x^{-1} [34]$$
>>> pow(5, -1, 34)
7
>>> from Crypto.Util.number import inverse
>>> inverse(5, 34)
7
>>> from gmpy2 import invert
>>> invert(5, 34)
mpz(7)
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# NCERT Solutions for Maths Class 5 Chapter 11 - Areas And Its Boundary
The chapter Areas and Its Boundaries focuses on dealing with the areas of given shapes and their respective boundaries or perimeters. It helps the students to learn how to calculate the areas and perimeters of various shapes through formulae. It covers the topics: philosophy Areas of Shapes philosophy Perimeters of Shapes philosophy Formulas for Areas and Perimeters NCERT Math-Magic questions are answered in a simple and engaging manner. We have also related 'Learning Concepts and interactive worksheets with the solutions. Our 'Learning Beyond' segment caters to all the probable questions that a child might think out of curiosity. Download Chapter 11 Areas and Its Boundaries in PDF format for free here.
## Access Answers to NCERT Solutions for Maths Class 5 Chapter 11 - Areas And Its Boundary
### Area and its Boundary
Question 1 :
Draw a square of 9 square cm. Write A on it. Draw another square with double the side. Write B on it. Answer these:
1) The perimeter of square A is __________ cm.
2) The side of square B is __________ cm.
3) The area of square B is __________ square cm.
4) The area of square B is __________ times the area of square A.
5) The perimeter of square B is __________ cm.
6) The perimeter of square B is __________ times the perimeter of square A.
Draw two squares, one of side 3 cm and another of side 6 cm.
1) Perimeter of square A is:
Perimeter = 4 × side
= 4 × 3 = 12 cm
2) The side of square B is 6 cm.
3) The area of the square B is:
Area = side × side
= 6 × 6 = 36 square cm.
4) The area of the square A is:
Area = side × side
= 3 × 3 = 9 square cm.
Since 36 = 4 × 9, the area of square B is 4 times the area of square A.
5) Perimeter of square B is:
Perimeter = 4 × side
= 4 × 6 = 24 cm
6) Since 24 = 2 × 12, the perimeter of square B is two times the perimeter of square A.
Question 2 :
Parth and Gini bought aam paapad (dried mango slice) from a shop. Their pieces looked like these.
Both could not make out whose piece was bigger. Suggest some ways to find out whose piece is bigger. Discuss.
Do it by yourself. Draw squares of side 1 cm on both the pieces, and count the number of squares to find the bigger piece.
Question 3 :
A friend of Parth and Gini showed one way, using small squares.
The length of piece A is 6 cm. So, 6 squares of side 1 cm can be arranged along its length. The width of piece A is 5 cm. So, 5 squares can be arranged along its width.
a) Altogether how many squares can be arranged on it? _______
b) So, the area of piece A = _______ square cm
c) In the same way find the area of piece B.
d) Who had the bigger piece? How much bigger?
a) Six squares of side 1 cm can be placed along the length of A, and five squares can be placed along its width. To find the total number of squares, multiply 6 by 5.
6 × 5 = 30
Therefore, 30 squares can be placed on the piece A, as shown below.
b) Since 30 squares of side 1 cm can be placed on piece A, its area is 30 square cm.
c) The length of piece B is 11 cm. Therefore, 11 squares can be placed along its length.
The width of piece B is 3 cm. Therefore, 3 squares can be placed along its width.
The total number of squares that can be placed on piece B is:
11 × 3 = 33
Therefore, its area is 33 square cm.
d) The area of piece A is 30 square cm, and the area of piece B is 33 square cm. Therefore, piece B is 3 square cm bigger than piece A.
Hence, Gini had 3 cm bigger piece.
Question 4 :
This stamp has an area of 4 square cm. Guess how many such stamps will cover this big rectangle.
Question 5 :
Take a 15 cm long thread. Make different shapes by joining its ends on this sheet.
A) Which shape has the biggest area? How much? _________ What is the perimeter of this shape? _________
B) Which shape has the smallest area? How much? ________ What is the perimeter of this shape? _________ Also make a triangle, a square, a rectangle and a circle. Find which shape has biggest area and which has the smallest.
Do it by yourself. Answers may vary.
Question 6 :
a) Measure the yellow rectangle. It is ________ cm long.
b) How many stamps can be placed along its length? ________
c) How wide is the rectangle? ________ cm.
d) How many stamps can be placed along its width? ________
e) How many stamps are needed to cover the rectangle? ________
f) How close was your earlier guess? Discuss.
g) What is the area of the rectangle? ________ square cm.
h) What is the perimeter of the rectangle? ________ cm.
a) Use a ruler to measure the length of the rectangle. It is 14 cm long.
b) The rectangle is 14 cm long, and each side of the stamp is 2 cm. Divide 14 by 2 to find the number of stamps that can be placed along the length of the rectangle.
14 2 = 7
Therefore, 7 stamps can be placed along its length.
c) Use a ruler to measure the width of the rectangle. It is 8 cm wide.
d) The rectangle is 8 cm wide, and each side of the stamp is 2 cm. Divide 8 by 2 to find the number of stamps that can be placed along the width of the rectangle.
8 2 = 4
Therefore, 4 stamps can be placed along its width.
e) 7 stamps can be placed along the length of the rectangle, and 4 stamps can be placed along its width. To find the total number of stamps that can be placed on the rectangle, multiply 7 by 4.
7 × 4 = 28
Therefore, 28 stamps are needed to cover the rectangle.
f) Do it by yourself.
g) The length of the rectangle is 7 cm, and its width is 4 cm.
Area of a rectangle can be found by:
Area = Length × Width
Therefore, area of the rectangle is:
7 × 4 = 28 square cm.
h) Perimeter of a rectangle can be found by:
Perimeter = 2 (Length + Width)
Therefore, the perimeter of the rectangle is:
2(7 + 4) = 2 × 11 = 22 cm.
Question 7 :
a) Arbaz plans to tile his kitchen floor with green square tiles. Each side of the tile is 10 cm. His kitchen is 220 cm in length and 180 cm wide. How many tiles will he need?
220 cm is the length of the kitchen, and each side of the tile is 10 cm
Number of tiles that can be placed along its length
= length of kitchen length of each tile.
= 220 10 = 22
Width of the kitchen is 180 cm.
Number of tiles that can be placed along its breadth
= 180 10 = 18
Since 22 tiles can be placed along the length of the kitchen, and 18 tiles can be placed along its width, multiply 22 by 16 to get the total number of tiles required.
22 × 18 = 396
Therefore, 396 tiles are required to tile the kitchen floor.
Question 8 :
b) The fencing of a square garden is 20 m in length. How long is one side of the garden?
Length of the fencing is 20 m. Length of the boundary of a square is given by:
Perimeter = 4 × Side
⇒ Side = Perimeter ÷ 4
To find the length of the side, divide 20 by 4.
20 4 = 5
Therefore, one side of the garden is 5 m long.
Question 9 :
c) A thin wire 20 centimetres long is formed into a rectangle. If the width of this rectangle is 4 centimetres, what is its length?
Perimeter of a rectangle is found by:
Perimeter = 2(Length + Width)
⇒Length + Width = Perimeter ÷ 2
Perimeter of the rectangle is the length of the wire that is 20 cm, and its width is 4 cm. Therefore, we have
Length + 4 = 20 ÷ 2
⇒ Length + 4 = 10
⇒ Length = 10 – 4 = 6 cm
Therefore, the length of the rectangle is 6 cm.
Question 10 :
d) A square carrom board has a perimeter of 320 cm. How much is its area?
All sides of a square are of the same length.
AB = BC = CD = AD = Side
Perimeter of a square is given by:
Perimeter = 4 × Side
⇒ Side = Perimeter ÷ 4
Since the perimeter of the carrom board is 320 cm, we have
Side = 320 4 = 80 cm.
Area of a square is given by:
Area = Side × Side
Therefore, we have
Area of the carrom board = 80 × 80 = 6400 square cm.
Question 11 :
e) How many tiles like the triangle given here will fit in the white design? Area of design = ___________ square cm.
Observe the given design. There is 1 square and 4 triangles. Since the triangle is half of a cm square, 6 triangular tiles can fit in the design. Area of design = 3 square cm.
Question 12 :
f) Sanya, Aarushi, Manav, and Kabir made greeting cards. Complete the table for their cards:
Step 1: Consider Sanya's card:
Length = 10 cm, width = 8 cm. It is a rectangle.
Perimeter = 2(length + width) = 2(10 + 8) = 2 × 18 = 36 cm
Area = length × width = 10 × 8 = 80 square cm
Step 2: Consider Manav’s card:
Length = 11 cm, perimeter = 44 cm
Perimeter = 44 = 4 × 11 = 4 × length.
So, Manav’s card is a square.
Therefore,
Width = Length = 11 cm = Side of the square
Area = Side × Side = 11 × 11 = 121 square cm.
Step 3: Consider Arushi’s card:
Width = 8 cm, area = 80 square cm.
Area = Length × Width
⇒ Length = Area ÷ Width = 80 ÷ 8 = 10 cm. So, it is a rectangle.
Perimeter = 2 (length + width) = 2(10 + 8) = 2 × 18 = 36 cm
Step 4: Consider Kabir’s card:
Perimeter = 40 cm, area = 100 square cm.
Perimeter = 40 = 4 × 10
Area = 10 × 10 = 100 square cm.
So, Kabir’s card is a square with side 10 cm. Therefore,
Length = 10 cm, width = 10 cm.
Question 13 :
Take a thick paper sheet of length 14 cm and width 9 cm. You can also use an old postcard.
a) What is its area?
b) What is its perimeter?
c) Now cut strips of equal sizes out of it. Using tape join the strips, end to end, to make a belt. How long is your belt? _________
d) What is its perimeter _________
e) Whose belt is the longest in the class? ________
Do it by yourself.
Area = 14 × 9 = 126 square cm.
Perimeter = 14 + 9 + 14 + 9 = 46 cm.
Question 14 :
A) Make two squares of one square metre each. Divide your class in two teams. Ready to play! Try these in your teams:
a) How many of you can sit in one square metre? ________
b) How many of you can stand in it? ________
c) Which team could make more children stand in their square? How many? ________
d) Which team could make more children sit in their square? How many?
Do it by yourself with the help of your friends. Answers may vary.
Question 15 :
B) Measure the length of the floor of your classroom in metres. Also measure the width.
a) What is the area of the floor of your classroom in square metres? __________
b) How many children are there in your class? _________
c) So how many children can sit in one square metre? __________
d) If you want to move around easily then how many children do you think should be there in one square metre? _________
Do it by yourself as directed. Use a measuring tape to measure the length and width of the floor of your classroom. Answers may vary.
Question 16 :
Nasreena is a farmer who wants to divide her land equally among her three children — Chumki, Jhumri and Imran. She wants to divide the land so that each piece of land has one tree. Her land looks like this.
a) Can you divide the land equally? Show how you will divide it. Remember each person has to get a tree. Colour each person’s piece of land differently.
b) If each square on this page is equal to 1 square metre of land, how much land will each of her children get? ___________ square m Chumki, Jhumri and Imran need wire to make a fence.
c) Who will need the longest wire for fencing? __________
d) How much wire in all will the three need? ___________
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Bimedians and Diagonals in a Quadrilateral
Denote the midpoints of the sides $AB$, $BC$, $CD$, and $DA$ of quadrilateral $ABCD$ as $X$, $Y$, $Z$, $W$, respectively. Segments $XZ$ and $YW$ are known as the bimedians of the quadrilateral. We have an engaging statement
1. The bimedians of a quadrilateral are equal if and only if its diagonals are perpendicular.
2. The bimedians of a quadrilateral are perpendicular if and only if its diagonals are equal.
Proof
1. The bimedians of a quadrilateral are equal if and only if its diagonals are perpendicular.
2. The bimedians of a quadrilateral are perpendicular if and only if its diagonals are equal.
Proof
The easiest way to establish the statement is by employing vector techniques. Choose an arbitrary origin $O$ and associate with every point $P$ the vector $p=\overline{OP}$, denoted by the corresponding low case.
Thus we have, $x=\frac{a+b}{2}$, $y=\frac{b+c}{2}$, $z=\frac{c+d}{2}$, $w=\frac{a+d}{2}$. The statement at hand can be reformulated as
1. $|x - z|=|y - w|$ iff $(a - c)\perp (b-d)$.
2. $(x - z)\perp (y - w)$ iff $|a-c|=|b-d|$.
The absolute value of vector $m$ is defined as the square root of the scalar product: $|m|=\sqrt{m\cdot m}$; while the condition of orthogonality of two vectors $m$ and $n$ is also expressible in terms of the scalar product: $m\perp n$ iff $m\cdot n=0$. Using the scalar product we are able to once more reformulate our statement
1. $(x - z)(x - z)=(y - w)(y - w)$ iff $(a - c)(b-d)=0$.
2. $(x - z)(y - w)=0$ iff $(a-c)(a-c)=(b-d)(b-d)$.
Substituting $x$, $y$, $z$, $w$ from the definition reduces the above to
1. $(a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)$ iff $(a - c)(b-d)=0$.
2. $(a+b - c-d)(b+c-d-a)=0$ iff $(a-c)(a-c)=(b-d)(b-d)$.
All that remains is pure algebra. For example,
$(a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)$
without the squares that are certainly the same on both sides, appears as
$2(ab-ac-ad-bc-bd+cd)=2(bc-bd-ab-cd-ac+ad)$,
which is equivalent to $(a - c)(b-d)=0$.
Similarly, written explicitly, $(a+b - c-d)(b+c-d-a)=0$ gives
\begin{align}ab+ac & -ad-aa +bb+bc-bd-ab \\& -bc-cc+cd+ac -bd-cd+dd+ad \\& =2ac-aa+bb-2bd-cc+dd=0.\end{align}
which is equivalent to $(a-c)(a-c)=(b-d)(b-d)$.
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How do you solve x + 4y = 12 and 2x - 3y = 6 using substitution?
Mar 18, 2018
$x = \frac{60}{11} \mathmr{and} y = \frac{18}{11}$
Explanation:
Here,$x + 4 y = 12. . . \to \left(1\right) \mathmr{and} 2 x - 3 y = 6. . . \to \left(2\right)$
From $\left(1\right) \textcolor{red}{x = 12 - 4 y} \ldots \to \left(3\right) .$
substituting value of x in $e q {u}^{n} \left(2\right)$,we get
$2 \textcolor{red}{\left(12 - 4 y\right)} - 3 y = 6 \implies 24 - 8 y - 3 y = 6$
$\implies 24 - 6 = 8 y + 3 y \implies 18 = 11 y \implies y = \frac{18}{11}$
From(3),we have
$x = 12 - 4 \left(\frac{18}{11}\right) = \frac{132 - 72}{11} = \frac{60}{11}$
from (1) $x + 4 y = 12$
$\implies L H S = \frac{60}{11} + 4 \left(\frac{18}{11}\right) = \frac{60}{11} + \frac{72}{11} = \frac{132}{11} = 12 = R H S$
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# How do you simplify (5 + 2sqrt2)(8 - 3sqrt2)?
Jul 19, 2018
See a solution process below:
#### Explanation:
To multiply and simplify these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.
$\left(\textcolor{red}{5} + \textcolor{red}{2 \sqrt{2}}\right) \left(\textcolor{b l u e}{8} - \textcolor{b l u e}{3 \sqrt{2}}\right)$ becomes:
$\left(\textcolor{red}{5} \times \textcolor{b l u e}{8}\right) - \left(\textcolor{red}{5} \times \textcolor{b l u e}{3 \sqrt{2}}\right) + \left(\textcolor{red}{2 \sqrt{2}} \times \textcolor{b l u e}{8}\right) - \left(\textcolor{red}{2 \sqrt{2}} \times \textcolor{b l u e}{3 \sqrt{2}}\right)$
$40 - 15 \sqrt{2} + 16 \sqrt{2} - 6 {\left(\sqrt{2}\right)}^{2}$
$40 + \left(- 15 + 16\right) \sqrt{2} - \left(6 \times 2\right)$
$40 + 1 \sqrt{2} - 12$
$40 + \sqrt{2} - 12$
$40 - 12 + \sqrt{2}$
$28 + \sqrt{2}$
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## How do you do the row reduction method?
Row Reduction Method
1. Multiply a row by a non-zero constant.
2. Add one row to another.
3. Interchange between rows.
4. Add a multiple of one row to another.
5. Write the augmented matrix of the system.
6. Row reduce the augmented matrix.
7. Write the new, equivalent, system that is defined by the new, row reduced, matrix.
## What is Gauss Jordan method to find inverse?
Gauss Jordan’s Matrix Inversion method. In this method we shall find the inverse of a matrix without calculating the determinant. In this method we shall write the augmented matrix of a quare matrix by writing a unit matrix of same order as that of side by side.
How do you solve a row matrix?
There are three kinds of elementary matrix operations.
1. Interchange two rows (or columns).
2. Multiply each element in a row (or column) by a non-zero number.
3. Multiply a row (or column) by a non-zero number and add the result to another row (or column).
What is row reduction method in matrix?
Row reduction (or Gaussian elimination) is the process of using row operations to reduce a matrix to row reduced echelon form. This procedure is used to solve systems of linear equations, invert matrices, compute determinants, and do many other things.
### What is matrix reduction method?
More generally, a matrix is said to be in reduced form if. (i) The first nonzero entry in a row (if any) is 1, while all other entries of the column containing. that 1 are 0; (ii) The first nonzero entry in a row is to the right of the first nonzero entry in each row above; and.
### Which method is used to find inverse of a matrix?
Another method of finding the inverse is by augmenting with the identity. We can augment a 3 × 3 matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. Write the system of equations as AX=B A X = B .
What is Gauss-Jordan Row reduction?
Gauss-Jordan reduction is an extension of the Gaussian elimination algorithm. It produces a matrix, called the reduced row echelon form in the following way: after carrying out Gaussian elimination, continue by changing all nonzero entries above the leading ones to a zero.
How to find the inverse of a square matrix using row reduction?
An online calculator that calculates the inverse of a square matrix using row reduction is presented. To find the inverse A − 1 , we start with the augmented matrix [ A | I n] and then row reduce it. If matrix A is invertible, the row reduction will end with an augmented matrix in the form
#### How to find the inverse of a matrix in MATLAB?
In order to find the inverse of the matrix following steps need to be followed: Form the augmented matrix by the identity matrix. Perform the row reduction operation on this augmented matrix to generate a row reduced echelon form of the matrix.
#### How do you find the inverse of a – 1?
To find the inverse A − 1, we start with the augmented matrix [ A | I n] and then row reduce it. If matrix A is invertible, the row reduction will end with an augmented matrix in the form [ I n | A − 1] where the inverse A − 1 is the n × n on the right side of [ I n | A − 1]
How to find the inverse of a matrix using Gauss-Jordan method?
Steps to find the inverse of a matrix using Gauss-Jordan method: In order to find the inverse of the matrix following steps need to be followed: Form the augmented matrix by the identity matrix. Perform the row reduction operation on this augmented matrix to generate a row reduced echelon form of the matrix.
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| Home | | Advanced Mathematics |
## Chapter: Biostatistics for the Health Sciences: Defining Populations and Selecting Samples
Biostatistics for the Health Sciences: Defining Populations and Selecting Samples - Exercises questions answers
EXERCISES
2.1 Why does the field of inferential statistics need to be concerned about sam-ples? Give in your own words the definitions of the following terms that per-tain to sample selection:
a. Sample
b. Census
c. Parameter
d. Statistic
e. Representativeness
f. Sampling frame
g. Periodic effect
2.2 Describe the following types of sample designs, noting their similarities and differences. State also when it is appropriate to use each type of sample design.
a. Random sample
b. Simple random samples
c. Convenience/grab bag samples
d. Systematic samples
e. Stratified
f. Cluster
g. Bootstrap
2.3 Explain what is meant by the term parameter estimation.
2.4 How can bias affect a sample design? Explain by using the terms selection bias, response bias, and periodic effects.
2.5 How is sampling with replacement different from sampling without replacement?
2.6 Under what circumstances is it appropriate to use rejection sampling methods?
2.7 Why would a convenience sample of college students on vacation in Fort Lauderdale, Florida, not be representative of the students at a particular col-lege or university?
2.8 What role does sample size play in the accuracy of statistical inference? Why is the method of selecting the sample even more important than the size of the sample?
Exercises 2.9 to 2.13 will help you acquire familiarity with sample selection. These exercises use data from Table 2.2.
2.9 By using the random number table (Table 2.1), draw a sample of 10 height measurements from Table 2.2. This sample is said to have size 10, or n = 10. The rows and columns in Table 2.2 have numbers, which in combination are the “addresses” of specific height measurements. For example, the number defined by row 15, column 4 denotes the 154th height measurement, or 61.
TABLE 2.2. Heights in Inches of 400 Female Clinic Patients
Use two indices based on numbers from Table 2.1. Draw one random number to select the row between 1 and 50 and another to choose the column between 1 and 8. Use the rejection method. List the ten values you have selected by this process. What name is given to the kind of sample you have selected?
2.10 Again use Table 2.2 to select a sample, but this time select only one random number from Table 2.1. Start in the row determined by the index for that ran-dom number. Choose the first value from the first column in that row; then skip the next seven columns and select the second value from column 8. Con-tinue skipping seven consecutive values before selecting the next value. When you come to the end of the row, continue the procedure on the next row. What kind of sampling procedure is this? Can bias be introduced when you sample in this way?
2.11 From the 400 height measurements in Table 2.2, we will take a sample of ten distinct values by taking the first six values in row 1 and the two values in the last two columns in row 2 and the last two columns in row 3. Let these ten values comprise the sample. Draw a sample of size 10 by sampling with re-placement from these 10 measurements.
a. List the original sample and the sample generated by sampling with re-placement from it.
b. What do we call the sample generated by sampling with replacement?
2.12 Repeat the procedure of Exercise 2.11 five times. List all five samples. How do they differ from the original sample?
2.13 Describe the population and the sample for:
a. Exercise 2.9
b. The bootstrap sampling plan in Exercise 2.11
2.14 Suppose you selected a sample from Table 2.2 by starting with the number in row 1, column 2. You then proceed across the row, skipping the next five numbers and take the sixth number. You continue in this way, skipping five numbers and taking the sixth, going to the leftmost element in the next row when all the elements in a row are exhausted, until you have exhausted the table.
a. What is such a sample selection scheme called?
b. Could any possible sources of bias arise from using this scheme?
2.9 From Table 2.1, start in the first column and the third row and proceed across the row to generate the random numbers, going back to the first column on the next row when a row is completed. Placing a zero and a decimal point in front of the first digit of the number (we will do this throughout), we get for the first random number 0.69386. This random number picks the row. We multiply 0.69386 by 50, getting 34.693. We will always round up. This will give us integers between 1 and 50. So we take row 35. Now the next number in the table is used for the column. It is 0.71708. Since there are 8 columns, we multiply 0.71708 by 8 to get 5.7366 and round up to get 6. Now, the first sample from the table is (35, 6), the value in row 35 column 6. We look this up in Table 2.2 and find the height to be 61 inches.
For the second measurement, we take the next pair of numbers, 0.88608 and 0.67251. After the respective multiplications we have row 45 and column 6. We compare (45, 6) to our list, which consists only of (35, 6). Since this pair does not repeat a pair on the list, we accept it. The list is now (35, 6) and (45, 6) and the sam-ples are, respectively, 61 and 65.
For the third measurement the next pair of random numbers is 0.22512 and 0.00169, giving the pair (12, 1). Since this pair is not on the list, we accept and the list becomes (35, 6), (45, 6), and (12, 1), with corresponding measurements 61, 65, and 59.
The next pair is 0.02887 and 0.84072, giving the pair (2, 7). This is accepted since it does not appear on the list. The resulting measurement is 63.
The next pair is 0.91832 and 0.97489 giving the pair (46, 8). Again, we accept. The corresponding measurement is 59.
We are half way to the result. The list of pairs is (35, 6), (45, 6), (12, 1), (2, 7), and (46, 8), corresponding to the sample measurements 61, 65, 59, 63, and 59.
The next pair of random numbers is 0.68381 and 0.61725 (note at this point we had to move to row 4 column 1). The pair is (35, 5). This again is not on the list and the corresponding measurement is 66.
The next pair of random numbers is 0.49122 and 0.75836 corresponding to the pair (25, 7). This is not on the list so we accept it and the corresponding sample measurement is 55.
The next pair of random numbers is 0.58711 and 0.52551 corresponding to the pair (8, 5). This pair is again not on our list so we accept it. The sample measure-ment is 65.
The next pair of random numbers is 0.58711 and 0.43014, corresponding to the pair (30, 4). This pair is again not on our list so we accept it. The sample measure-ment is 64.
The next pair of random numbers is 0.95376 and 0.57402, corresponding to the pair (48, 5). This pair is again not on our list so we accept it. The sample measurement is 57.
We now have 10 samples. Since we only took 10 out of 400 numbers (50 rows by 8 columns), our chances of a rejection on any sample was small and we did not get one.
The resulting 10 pairs are (35, 6), (45, 6), (12, 1), (2,7), (46,8), (35, 5), (25,7), (8, 5), (30, 4) and (48, 5) and the corresponding sample of ten measurements is 61, 65, 59, 63, 59, 66, 55, 65, 64, and 57.
Despite the complicated mechanism we used to generate the sample, this consti-tutes what we call a simple random sample since each of the 400 samples has prob-ability 1/400 of being selected first and each of the remaining 399 has probability 1/399 of being selected second, given they weren’t chosen first, etc.
2.11 a. The original sample is 61, 55, 52, 59, 62, 66, 63, 60, 67, and 64. We then index these samples 1–10. Index 1 corresponds to 61, 2 to 55, 3 to 52, 4 to 59, 5 to 62, 6 to 66, 7 to 63, 8 to 60, 9 to 67, and 10 to 64. We use a table of random num-bers to pick the index. We will do this by running across row 21 of Table 2.1 to gen-erate the 10 indices. The random numbers on row 21 are:
22011 71396 95174 43043 68304 36773 83931 43631 50995 68130
This we interpret as 0.22011, 0.71396, 0.95174, 0.43043, 0.68304, 0.36773, 0.83931, 0.43631, 0.50995, and 0.68130. To get the index, we multiply these num-bers by 10 and round up to the next integer. The resulting indices are, respectively, 3, 8, 10, 5, 7, 4, 9, 5, 6, and 7. We see that indices 5 and 7 each repeated once and indices 1 and 2 did not occur. The corresponding sample is 52, 60, 64, 62, 63, 59, 67, 62, 66, and 63.
b. The name we give to sampling with replacement n times from a sample of size n is bootstrap sampling. The sample we obtained we call a bootstrap sample.
2.13 a. A population is a complete list of all the subjects you are interested in. For Exercise 2.9, it consisted of the 400 height measurements for the female clinic pa-tients. The sample is the chosen subset of the population, often selected at random. In this case it consisted of a random sample of 10 measurements corresponding to the female patients in specific rows and columns of the table. The resulting 10 pairs were (35, 6), (45, 6), (12, 1), (2,7), (46,8), (35, 5), (25,7), (8, 5), (30, 4), and (48, 5) and the corresponding sample of 10 measurements were 61, 65, 59, 63, 59, 66, 55, 65, 64, and 57.
b. For the bootstrap sampling plan in Exercise 2.11, the population is the same set of 400 height measurements in Table 2.2. The original sample is a subset of size 10 taken from this population in a systematic fashion, as described in Exercise 2.11. The bootstrap sample is then obtained by sampling with replacement from this original sample of size 10. The resulting bootstrap sample is a sample of size 10 that may have some of the original sample values repeated one or more times depending on the result of the random drawing. As shown in our solution, the indices 5 and 7 repeated once each.
2.14 a. This method of sampling is systematic sampling. It specifically is a periodic method.
b. Because of the cyclic nature of the sampling scheme, there is a danger of bias. If the data is also cyclic with the same period we could be sampling only the peak values (or only the trough values). In that case, the sample estimate of the mean would be biased on the high side if we sampled the peaks and on the low side if we sampled the troughs.
Related Topics
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# Properties of Whole Number
To understand the concept of Whole Numbers first we need to know what are Natural Numbers and then we will be able to understand about Properties of Whole Number.
Natural Numbers come naturally when we start counting like numbers 1,2,3,4 and hence they are called Natural Numbers.
## Introduction
Whole Numbers are the natural numbers along with ‘ 0’. Therefore the numbers 0, 1,2,3,4,5 are called Whole Numbers.
Before knowing the Properties of Whole Numbers we should understand the following:-
• The successor of a Whole Number
• The predecessor of a Whole Number
• Representation of Whole Numbers on a Number Line
The properties of Whole Number are better understood with respect to the different operations in maths for addition, subtraction, multiplication, and division. It helps in many problems in an easier manner with the use of these properties.
At Winaum Learning, we believe that maths is a science that deals with logic, shapes, numbers, patterns, and arrangement. It is a life skill and requires an in-depth understanding. Students at Winaum Learning Maths Program do maths with worksheets, practice tests, continuous revisions of the concepts, activities, and games.
## Closure Property
### 1. Closure Property in Addition :
If a and b are two whole numbers then a+b is also a whole number.
Example 1 : a= 4, b=7
then 4+7 = 11 , implying 4, 7, and 11 all are whole numbers.
Example 2 . a = 4, b = 6, a + b = 10
Now a and b are whole numbers, hence the result of addtition of a+b is also a whole number.
Therefore 4 + 6 = 10 , 10 is also a whole number.
Example 3. Suppose a = 5, b = 9,
Then a + b = 14
5 + 9 = 14
Implying a+b which equals 14 is also a whole number.
Example 4. If a = 8, b = 3, and a &b are whole numbers, then
a + b = 8 + 3 = 11
11 is also a whole number.
1. a = 6, b = 10, a + b = 16
6 + 10 = 16
To summarise: On the Addition of any two whole numbers, you will see that the result of addition is again a whole number.
### 2. Closure Property Multiplication:
If a and b are two whole numbers then a x b is also a whole number.
Example 1. If 4 and 7 are whole numbers then,
4 x 7 =28.
Implying 4, 7, and 28 all are whole numbers.
Example 2. Asssume a and b are whole numbers and if a = 3, b = 6,
=> a x b
=> 3 x 6 = 18
Implying 3, 6, and 18 all are whole numbers.
Example 3 . If a = 5, b = 11, and a &b are whole numbers, then
=> a x b
=> 5 x 11
=> 55
Implying that 5, 11, and the multiplication result 55 all are whole numbers.
Example 4. If a = 9, b = 2, and a & b are whole numbers, then
=> a x b
=> 9 x 2
=> 18
Implying that 9, 2, and the multiplication result 18 all are whole numbers.
Example 5. If a = 1, b = 8, and a &b are whole numbers, then
=>a x b
=>1 x 8 = 8
Implying that 1, 8, and the multiplication result 8 all are whole numbers.
To Summarise: As we look at the examples above when we multiply any two whole numbers, then you will see that the multiplication result is again a whole number.
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## Commutative Property
• ### Commutative Property in Addition
If a and b are two whole numbers, then
a + b = b + a
If we change the order, the sum of whole numbers remains unchanged.
The sum of two whole numbers remains the same even after interchanging the order of the numbers.
• Example 1
If a and b are two whole numbers and a = 5, b = 5,
Then as per this property a + b = b + a
5 + 5 = 5 + 5
10 = 10
So if we see above the answer on the left-hand side and the answer on the right-hand side, both answers are 10.
• Example 2: If a and b are two whole numbers and a = 7, b = 8.
Then as per this property a + b = b + a
7 + 8 = 8 + 7
15 = 15
So if we see above the answer on the left-hand side and the answer on the right-hand side, both answers are 15.
• Example 3 :
If a and b are two whole numbers and a = 3, b = 9.
Then as per this property a + b = b + a
a = 3, b = 9,
a + b = b + a
3 + 9 = 9 + 3
12 = 12
So if we see above the answer on the left-hand side and the answer on the right-hand side, both answers are 12.
• Example 4. If a and b are two whole numbers and a = 6, b = 8.
Then as per this property a + b = b + a
a = 6, b = 8
a + b = b + a
6 + 8 = 8 + 6
14 = 14
So if we see above the answer on the left-hand side and the answer on the right-hand side, both answers are 14.
• Example 5. If a and b are two whole numbers and a = 7, b = 5,
a + b = b + a =12
7 + 5 = 7 + 5
12 = 12
So if we see above the answer on the left-hand side and the answer on the right-hand side, both answers are 12.
• ### Commutative Property in Multiplication
For any two whole numbers a and b
a x b =b x a
This implies that the product of whole numbers remain unchanged even if change the order of numbers while multiplication.
6 x 12= 12x 6
72= 72
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## Associative Property
• ### Associative Property of Addition
If a, b, and c are three whole numbers then ( a+b ) + c = a+ ( b+c ). This implies even if we change the association of numbers, the result does not change.
For Example: a= 3, b=7, and c= 5, then
3+7+5= 3+(7+5)
10+5= 3+12
15=15
LHS= RHS
Even if the grouping of the numbers has changed both on the left-hand side and the right-hand side, the result does not change, it’s 15 on both sides.
• ### Associative Property of Multiplication
If a, b, and c are three whole numbers then
( ax b ) x c = ax ( b x c ). This implies even if we change the association of numbers , the result does not changes.
## Identity Element
The identity element of addition for whole numbers is 0 i.e on adding 0, the number shall remain unchanged. It implies that the sum of any whole number and zero ( 0 ) is the number itself.
General Property: a+0 = a
• Example 1 : Assume a= 17, then 17 + 0 = 7 ( i.e. number itself )
• Example 2 : Assume a = 125, then 125 + 0 = 125 ( i.e. number itself )
## Distributive Property of Multiplication
1. ### Distributive Property in Addition
The Distributive Property says that when we multiply the sum of two or more addends by a number, then it will give same answer as multiplying each addend individually by the number and then adding each of the multiplication results.
General Property: a × ( b + c ) = ( a × b ) + ( b × c )
Example 1: Let a = 3 , b= 5 and c = 8
3×( 5+8 ) = (3× 5)+(3×8)
3 × 13 = 15 + 24
39 = 39
LHS = RHS
Example 2 : Let a = 10, b = 7 and c = 12
10×(7+12)=(10×7)+(10×12)
10 × 19 = 70 + 120
190 = 190
LHS = RHS
Example 3 : Let a = 25, b = 11 and c = 40
25×(11+40) =(25×11)+(25×40)
25 × 51 = 275 + 1000
1275 = 1275
LHS = RHS
Example 4 : Let a = 100, b = 70 and c = 30
100×(70+30)=(100×70)+(100×30)
100 × 100 = 7000 + 3000
10000 = 10000
LHS = RHS
### 2. Distributive Property in Subtraction
According to distributive property multiplying the difference of two numbers by a number will give the same result as multiplying each number individually and then finding the difference of the products.
General Property: a × (b – c) = ( a × b )– ( a × c )
Example 1: Let a = 9 , b = 5 and c = 2
9×(5–2) = (9×5)– (9×2)
9 × 3 = 45 – 18
27 = 27
LHS = RHS
• Example 2. Let a = 11, b = 10 and c = 7
11 ×(10–7) = (11×10)–(11×7)
11 × 3 = 110 –77
33 = 33
LHS = RHS
• Example 3 : Let a = 50, b = 25 and c = 12
50×(25–12)=(50×25)–(50×12)
50 × 13 = 1250 – 600
650 = 650
LHS = RHS
• Example 4 : Let a = 100, b= 75 and c = 20
100×(75–20)=(100×75)–(100×20)
100 × 55 = 7500 – 2000
5500 = 5500
LHS = RHS
Here at Winaum Learning, the Distributive Property of Multiplication with addition and subtraction and other Properties of Whole Number are well understood by the students in an active learning environment with maths orals, written work, exercises, assessments, puzzles, and other math activities at Winaum Learning online maths classes. Book a Free Trial.
## Difference Between Counting Numbers and Whole Numbers
S.No
Counting Numbers
Whole Numbers
1
Natural numbers are defined as the basic counting numbers. The counting number or natural number are {1, 2, 3, 4, 5, …}.
Whole numbers are defined as the set of natural numbers, and it started with zero. The whole numbers are{0,1, 2, 3, 4, 5, …}.
2
Counting or Natural Numbers are represented using the letter “N”.
To represent Whole Numbers we use the letter “W”.
3
Counting number starts from 1.
Whole number starts from 0.
4
We can say that all the counting numbers are considered as whole numbers.
All the whole numbers are not considered as counting numbers as 0 is not a counting number.
## Conclusion
To summarise there are 5 Properties of Whole Number that are Closure Property, Commutative Property, Associative Property, Identity Element and Distributive Property. Above we have gone into details understanding of the concept by giving several examples for each property.
## Frequently Asked Questions on Whole Numbers
Whole numbers are the positive numbers starting from the number 0 and go up to infinity.
For children at an early age, it is easiest for them to understand whole numbers as they can do counting comfortably with it. The four main pillars of Mathematics -addition, subtraction, multiplication, and division can be understood well when the foundation of numbers and counting is clear to the kids.
Zero – “0” is the smallest number whole number.
All the numbers starting from 0 like 0,1,2,3 4,5… are whole numbers as they are not decimals or fractions.
• Whole numbers is a set of positive integers.
• 0 is also a whole number.
• All the whole numbers except 0 are natural numbers too.
Numbers that are negative, fractional numbers having denominators other than 1, and irrational numbers are the numbers that are not whole numbers.
No, whole numbers can’t be negative. In fact, it is a set of positive integers.
Yes, all the natural numbers are whole numbers.
Yes, all the whole numbers are real numbers.
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1. ## probability problem
Suppose that there are $\displaystyle 2$ teams: $\displaystyle A$ and $\displaystyle B$. Team $\displaystyle A$'s pitcher throws a strike $\displaystyle 50 \%$ of the time and a ball $\displaystyle 50 \%$ of the time (successive pitches are independent from one another), and the pitcher never hits the batter. Knowing this, Team $\displaystyle B$'s manager instructs the first batter not to swing at anything. Calculate the following probabilities:
(a) The batter walks on the fourth pitch.
So $\displaystyle P \{\text{batter walks on fourth pitch} \} = \left(\frac{1}{2} \right)^4 = \frac{1}{16}$
(b) The batter walks on the sixth pitch (so two of the first five must be strikes).
So $\displaystyle P \{\text{batter walks on sixth pitch} \} = \frac{5 \cdot 4}{2^{5}} = \frac{20}{32} = \frac{5}{8}$
(c) The batter walks
So this is the same as $\displaystyle P \{\text{pitcher throws four balls in a row} \} = \frac{1}{16}$
(d) The first batter scores while no one is out. Not sure how to do this one.
2. In order for the batter to be walked on the sixth pitch it must be true that among the first five pitches there must be two strikes and three balls while the sixth pitch must be a ball. Here is an example $\displaystyle \underbrace {BSBSB}_5\boxed{B}$.
But the first five places can be rearranged in $\displaystyle {5 \choose 3}$ ways.
Thus the probability is $\displaystyle {5 \choose 3} \left( {\frac{1}{2}} \right)^6$.
3. why is it not $\displaystyle \frac{5!}{3!}$? Doesnt order matter?
4. also for (c), isnt the probability 1/2 from the law of large numbers?
5. Originally Posted by shilz222
why is it not $\displaystyle \frac{5!}{3!}$? Doesnt order matter?
How many ways are there to arrrange the string "SSBBB"?
It is not $\displaystyle \frac {5!} {3!}$.
For (c): How many ways are there to arrrange the string "SSBBBB"?
Each of those strings has a probiblity $\displaystyle \left( {\frac{1}{2}} \right)^6$
6. $\displaystyle \binom{5}{3}$ and $\displaystyle \binom{6}{4}$ respectively.
7. and for part (d) the batter scores implies that he walks (assuming he does not hit the ball) right? so how would we do this?
8. Originally Posted by shilz222
and for part (d) the batter scores implies that he walks (assuming he does not hit the ball) right? so how would we do this?
Sorry, but I know so little about sports is general and baseball in particular, that I cannot answer your question.
9. yeah what that does mean exactly?
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# Probabilities of Compound Events Probability of Two Independent Events If two events, A and B are independent, then the probability of both events.
## Presentation on theme: "Probabilities of Compound Events Probability of Two Independent Events If two events, A and B are independent, then the probability of both events."— Presentation transcript:
Probabilities of Compound Events Probability of Two Independent Events If two events, A and B are independent, then the probability of both events occurring is the product of each individual probability. P(A and B) = P(A) * P(B)
Probabilities of Compound Events Find the probability of drawing a face card, replacing the card and then drawing an ace using a standard deck of playing cards P(A)= 12/52 P(B)= 4/52 P(A and B) = 12/52 * 4/52 = 3/169
Probabilities of Compound Events In a survey it was determined that 7 out of 10 shoppers do not use coupons and 3 out of 8 shoppers only buy items on sale. What is the probability that a random shopper will use a coupon and buy a item on sale. ( It has been determined that these two situations are independent.) P(A) = 1 - 7/10 = 3/10 P(B) = 3/8 P(A and B) = 3/10 * 3/8 = 9/80
Probabilities of Compound Events Probability of Two Dependent Events If two events, A and B, are dependent, then the probability of both events occurring is the product of each individual probability. P(A and B) = P(A) * P(B following A)
Probabilities of Compound Events In a bag there are 4 red, 6 green and 3 blue candies. Bob will pick 3 candies randomly from the bag (no replacement). Independent or dependent? What is the probability that Bob drew all blue candies? P(A, then B, then C) 3/13 * 2/12 * 1/11 = 1/286
Probabilities of Compound Events Probability of Mutually Exclusive Events Mutually exclusive: When two events cannot happen at the same time If two events, A and B are mutually exclusive, then the probability that either A or B occurs is the sum of their probabilities P(A or B) = P(A) + P(B)
Probabilities of Compound Events Betty has 6 pennies, 4 nickels and 5 dimes in her pocket. If Betty takes one coin out of her pocket, what is the probability that it is a nickel or a dime? P(A) = 4/15 P(B) = 5/15 P(A or B) = 4/15 + 5/15 = 9/15 = 3/5
Probabilities of Compound Events Probability of Inclusive Events Mutually Inclusive: When two events can happen at the same time If Two events, A and B are inclusive, then the probability that either A or B occurs is the sum of their probabilities decreased by the probability of both events occurring. P(A or B) = P(A) + P(B) – P(A and B)
Probabilities of Compound Events In a particular group of hospital patients, the probability of having high blood pressure is 3/8, the probability of having arteriosclerosis is 5/12, and the probability of having both is ¼. Mutually exclusive or inclusive? What is the probability that a patient has either HBP or arteriosclerosis?
Probabilities of Compound Events There are 6 children in an art class, 4 girls and 2 boys. Four children will be chosen at random to act as greeters for an art exhibit. What is the probability that at least 3 girls will be selected?
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# Chapter 14: Lines and Angles Exercise – 14.2
### Question: 1
In Figure, line n is a transversal to line l and m. Identify the following:
(i) Alternate and corresponding angles in Figure.(i)
(ii) Angles alternate to ∠d and ∠g and angles corresponding to ∠RQF and angle alternate to ∠PQE in Figure. (ii)
(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Figure. (iii)
(iv) Pairs of interior and exterior angles on the same side of the transversal in Figure. (iii)
### Solution:
(i) Figure (i)
Corresponding angles:
∠EGB and ∠GHD
∠HGB and ∠FHD
∠EGA and ∠GHC
∠AGH and ∠CHF
The alternate angles are:
∠EGB and ∠CHF
∠HGB and ∠CHG
∠EGA and ∠FHD
∠AGH and ∠GHD
(ii) Figure (ii)
The alternate angle to ∠d is ∠e.
The alternate angle to ∠g is ∠b.
The corresponding angle to ∠f is ∠c.
The corresponding angle to ∠h is ∠a.
(iii) Figure (iii)
Angle alternate to ∠PQR is ∠QRA.
Angle corresponding to ∠RQF is ∠ARB.
Angle alternate to ∠POE is ∠ARB.
(iv) Figure (ii)
Pair of interior angles are
∠a is ∠e.
∠d is ∠f.
Pair of exterior angles are
∠b is ∠h.
∠c is ∠g.
### Question: 2
In Figure, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively, If ∠CMQ = 60°, find all other angles in the figure.
### Solution:
Corresponding angles:
∠ALM = ∠CMQ = 60°
Vertically opposite angles:
∠LMD = ∠CMQ = 60°
Vertically opposite angles:
∠ALM = ∠PLB = 60°
Here,
∠CMQ + ∠QMD = 180° are the linear pair
= ∠QMD = 180° – 60°
= 120°
Corresponding angles:
∠QMD = ∠MLB = 120°
Vertically opposite angles
∠QMD = ∠CML = 120°
Vertically opposite angles
∠MLB = ∠ALP = 120°
### Question: 3
In Figure, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.
### Solution:
Given that,
∠LMD = 35°
∠LMD and ∠LMC is a linear pair
∠LMD + ∠LMC = 180°
= ∠LMC = 180° – 35°
= 145°
So, ∠LMC = ∠PLA = 145°
And, ∠LMC = ∠MLB = 145°
∠MLB and ∠ALM is a linear pair
∠MLB + ∠ALM = 180°
= ∠ALM = 180° – 145°
= ∠ALM = 35°
Therefore, ∠ALM = 35°, ∠PLA = 145°.
### Question: 4
The line n is transversal to line l and m in Figure. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.
### Solution:
Given that, l ∥ m
So,
The angle alternate to ∠13 is ∠7
The angle corresponding to ∠15 is ∠7
The angle alternate to ∠15 is ∠5
### Question: 5
In Figure, line l ∥ m and n is transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.
### Solution:
Given that,
∠1 = 40°
∠1 and ∠2 is a linear pair
= ∠1 + ∠2 = 180°
= ∠2 = 180° – 40°
= ∠2 = 140°
∠2 and ∠6 is a corresponding angle pair
So, ∠6 = 140°
∠6 and ∠5 is a linear pair
= ∠6 + ∠5 = 180°
= ∠5 = 180° – 140°
= ∠5 = 40°
∠3 and ∠5 are alternative interior angles
So, ∠5 = ∠3 = 40°
∠3 and ∠4 is a linear pair
= ∠3 + ∠4 = 180°
= ∠4 = 180° – 40°
= ∠4 = 140°
∠4 and ∠6 are a pair interior angles
So, ∠4 = ∠6 = 140°
∠3 and ∠7 are pair of corresponding angles
So, ∠3 = ∠7 = 40°
Therefore, ∠7 = 40°
∠4 and ∠8 are a pair corresponding angles
So, ∠4 = ∠8 = 140°
Therefore, ∠8 = 140°
So, ∠1 = 40°, ∠2 = 140°, ∠3 = 40°, ∠4 = 140°, ∠5 = 40°, ∠6 = 140°, ∠7 = 40°, ∠8 = 140°
### Question: 6
In Figure, line l ∥ m and a transversal n cuts them P and Q respectively. If ∠1 = 75°, find all other angles.
### Solution:
Given that, l ∥ m and ∠1 = 75∘
We know that,
∠1 + ∠2 = 180° → (linear pair)
= ∠2 = 180° – 75°
= ∠2 = 105°
here, ∠1 = ∠5 = 75° are corresponding angles
∠5 = ∠7 = 75° are vertically opposite angles.
∠2 = ∠6 = 105° are corresponding angles
∠6 = ∠8 = 105° are vertically opposite angles
∠2 = ∠4 = 105° are vertically opposite angles
So, ∠1 = 75°, ∠2 = 105°, ∠3 = 75°, ∠4 = 105°, ∠5 = 75°, ∠6 = 105°, ∠7 = 75°, ∠8 = 105°
### Question: 7
In Figure, AB ∥ CD and a transversal PQ cuts at L and M respectively. If ∠QMD = 100°, find all the other angles.
### Solution:
Given that, AB ∥ CD and ∠QMD = 100°
We know that,
Linear pair,
∠QMD + ∠QMC = 180°
= ∠QMC = 180° – ∠QMD
= ∠QMC = 180° – 100°
= ∠QMC = 80°
Corresponding angles,
∠DMQ = ∠BLM = 100°
∠CMQ = ∠ALM = 80°
Vertically Opposite angles,
∠DMQ = ∠CML = 100°
∠BLM = ∠PLA = 100°
∠CMQ = ∠DML = 80°
∠ALM = ∠PLB = 80°
### Question: 8
In Figure, l ∥ m and p || q. Find the values of x, y, z, t.
### Solution:
Give that, angle is 80°
∠z and 80° are vertically opposite angles
= ∠z = 80°
∠z and ∠t are corresponding angles
= ∠z = ∠t
Therefore, ∠t = 80°
∠z and ∠y are corresponding angles
= ∠z = ∠y
Therefore, ∠y = 80°
∠x and ∠y are corresponding angles
= ∠y = ∠x
Therefore, ∠x = 80°
### Question: 9
In Figure, line l ∥ m, ∠1 = 120° and ∠2 = 100°, find out ∠3 and ∠4.
### Solution:
Given that, ∠1 = 120° and ∠2 = 100°
∠1 and ∠5 a linear pair
= ∠1 + ∠5 = 180°
= ∠5 = 180° – 120°
= ∠5 = 60°
Therefore, ∠5 = 60°
∠2 and ∠6 are corresponding angles
= ∠2 = ∠6 = 100°
Therefore, ∠6 = 100°
∠6 and ∠3 a linear pair
= ∠6 + ∠3 = 180°
= ∠3 = 180° – 100°
= ∠3 = 80°
Therefore, ∠3 = 80°
By, angles of sum property
= ∠3 + ∠5 + ∠4 = 180°
= ∠4 = 180° – 80° – 60°
= ∠4 = 40°
Therefore, ∠4 = 40°
### Question: 10
In Figure, l ∥ m. Find the values of a, b, c, d. Give reasons.
### Solution:
Given that, l ∥ m
Vertically opposite angles,
∠a = 110°
Corresponding angles,
∠a = ∠b
Therefore, ∠b = 110°
Vertically opposite angle,
∠d = 85°
Corresponding angles,
∠d = ∠c
Therefore, ∠c = 85°
Hence, ∠a = 110°, ∠b = 110°, ∠c = 85°, ∠d = 85°
### Question: 11
In Figure, AB ∥ CD and ∠1 and ∠2 are in the ratio of 3: 2. Determine all angles from 1 to 8.
### Solution:
Given that,
∠1 and ∠2 are 3: 2
Let us take the angles as 3x, 2x
∠1 and ∠2 are linear pair
= 3x + 2x = 180°
= 5x = 180°
= x = 180°/5
= x = 36°
Therefore, ∠1 = 3x = 3(36) = 108°
∠2 = 2x = 2(36) = 72°
∠1 and ∠5 are corresponding angles
= ∠1 = ∠5
Therefore, ∠5 = 108°
∠2 and ∠6 are corresponding angles
= ∠2 = ∠6
Therefore, ∠6 = 72°
∠4 and ∠6 are alternate pair of angles
= ∠4 = ∠6 = 72°
Therefore, ∠4 = 72°
∠3 and ∠5 are alternate pair of angles
= ∠3 = ∠5 = 108°
Therefore, ∠5 = 108°
∠2 and ∠8 are alternate exterior of angles
= ∠2 = ∠8 = 72°
Therefore, ∠8 = 72°
∠1 and ∠7 are alternate exterior of angles
= ∠1 = ∠7 = 108°
Therefore, ∠7 = 108°
Hence, ∠1 = 108°, ∠2 = 72°, ∠3 = 108°, ∠4 = 72°, ∠5 = 108°, ∠6 = 72°, ∠7 = 108°, ∠8 = 72°
### Question: 12
In Figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.
### Solution:
Linear pair,
= ∠4 + 60° = 180°
= ∠4 = 180° – 60∘
= ∠4 = 120°
∠4 and ∠1 are corresponding angles
= ∠4 = ∠1
Therefore, ∠1 = 120°
∠1 and ∠2 are corresponding angles
= ∠2 = ∠1
Therefore, ∠2 = 120°
∠2 and ∠3 are vertically opposite angles
= ∠2 = ∠3
Therefore, ∠3 = 120°
### Question: 13
In Figure, if l ∥ m ∥ n and ∠1 = 60°, find ∠2
### Solution:
Given that,
Corresponding angles:
∠1 = ∠3
= ∠1 = 60°
Therefore, ∠3 = 60°
∠3 and ∠4 are linear pair
= ∠3 + ∠4 = 180°
= ∠4 = 180° – 60°
= ∠4 = 120°
∠3 and ∠4 are alternative interior angles
= ∠4 = ∠2
Therefore, ∠2 = 120°
### Question: 14
In Figure, if AB ∥ CD and CD ∥ EF, find ∠ACE
### Solution:
Given that,
Sum of the interior angles,
= ∠CEF + ∠ECD = 180°
= 130° + ∠ECD = 180°
= ∠ECD = 180° – 130°
= ∠ECD = 50°
We know that alternate angles are equal
= ∠BAC = ∠ACD
= ∠BAC = ∠ECD + ∠ACE
= ∠ACE = 70° – 50°
= ∠ACE = 20°
Therefore, ∠ACE = 20°
### Question: 15
In Figure, if l ∥ m, n ∥ p and ∠1 = 85°, find ∠2.
### Solution:
Given that, ∠1 = 85°
∠1 and ∠3 are corresponding angles
So, ∠1 = ∠3
= ∠3 = 85°
Sum of the interior angles
= ∠3 + ∠2 = 180°
= ∠2 = 180° – 85°
= ∠2 = 95°
### Question: 16
In Figure, a transversal n cuts two lines l and m. If ∠1 = 70° and ∠7 = 80°, is l ∥ m?
### Solution:
We know that if the alternate exterior angles of the two lines are equal, then the lines are parallel.
Here, ∠1 and ∠7 are alternate exterior angles, but they are not equal
= ∠1 ≠ ∠7 ≠ 80°
### Question: 17
In Figure, a transversal n cuts two lines l and m such that ∠2 = 65° and ∠8 = 65°. Are the lines parallel?
### Solution:
vertically opposite angels,
∠2 = ∠3 = 65°
∠8 = ∠6 = 65°
Therefore, ∠3 = ∠6
Hence, l ∥ m
### Question: 18
In Figure, Show that AB ∥ EF.
### Solution:
We know that,
∠ACD = ∠ACE + ∠ECD
= ∠ACD = 35° + 22°
= ∠ACD = 57° = ∠BAC
Thus, lines BA and CD are intersected by the line AC such that, ∠ACD = ∠BAC
So, the alternate angles are equal
Therefore, AB ∥ CD — 1
Now,
∠ECD + ∠CEF = 35° + 45° = 180°
This, shows that sum of the angles of the interior angles on the same side of the transversal CE is 180 degrees
So, they are supplementary angles
Therefore, EF ∥ CD ——- 2
From eq 1 and 2
We can say that, AB ∥ EF
### Question: 19
In Figure, AB ∥ CD. Find the values of x, y, z.
### Solution:
Linear pair,
= ∠x + 125° = 180°
= ∠x = 180° – 125°
= ∠x = 55°
Corresponding angles
= ∠z = 125°
= ∠x + ∠z = 180°
= ∠x + 125° = 180°
= ∠x = 180° – 125°
= ∠x = 55°
= ∠x + ∠y = 180°
= ∠y + 55° = 180°
= ∠y = 180° – 55°
= ∠y = 125°
### Question: 20
In Figure, find out ∠PXR, if PQ ∥ RS.
### Solution:
We need to find ∠PXR
∠XRS = 50°
∠XPR = 70°
Given, that PQ ∥ RS
∠PXR = ∠XRS + ∠XPR
∠PXR = 50° + 70°
∠PXR = 120°
Therefore, ∠PXR = 120°
### Question: 21
In Figure, we have
(i) ∠MLY = 2∠LMQ
(ii) ∠XLM = (2x - 10)° and ∠LMQ = (x + 30) °, find x.
(iii) ∠XLM = ∠PML, find ∠ALY
(iv) ∠ALY = (2x - 15) °, ∠LMQ = (x + 40) °, find x.
### Solution:
(i) ∠MLY and ∠LMQ are interior angles
= ∠MLY + ∠LMQ = 180°
= 2∠LMQ + ∠LMQ = 180°
= 3∠LMQ = 180°
= ∠LMQ = 180°/3
= ∠LMQ = 60°
(ii) ∠XLM = (2x - 10)° and ∠LMQ = (x + 30) °, find x.
∠XLM = (2x - 10) ° and ∠LMQ = (x + 30) °
∠XLM and ∠LMQ are alternate interior angles
= ∠XLM = ∠LMQ
= (2x - 10) ° = (x + 30) °
= 2x – x = 30° + 10°
= x = 40°
Therefore, x = 40°
(iii) ∠XLM = ∠PML, find ∠ALY
∠XLM = ∠PML
Sum of interior angles is 180 degrees
= ∠XLM + ∠PML = 180°
= ∠XLM + ∠XLM = 180°
= 2∠XLM = 180°
= ∠XLM = 180°/2
= ∠XLM = 90°
∠XLM and ∠ALY are vertically opposite angles
Therefore, ∠ALY = 90°
(iv) ∠ALY = (2x - 15) °, ∠LMQ = (x + 40) °, find x.
∠ALY and ∠LMQ are corresponding angles
= ∠ALY = ∠LMQ
= (2x - 15) °= (x + 40) °
= 2x – x = 40° + 15°
=x = 55°
Therefore, x = 55°
### Question: 22
In Figure, DE ∥ BC. Find the values of x and y.
### Solution:
We know that,
ABC, DAB are alternate interior angles
∠ABC = ∠DAB
So, x = 40°
And ACB, EAC are alternate interior angles
∠ACB = ∠EAC
So, y = 40°
### Question: 23
In Figure, line AC ∥ line DE and ∠ABD = 32°, Find out the angles x and y if ∠E = 122°.
### Solution:
∠BDE = ∠ABD = 32° – Alternate interior angles
= ∠BDE + y = 180°– linear pair
= 32° + y = 180°
= y = 180° – 32°
= y = 148°
∠ABE = ∠E = 32° – Alternate interior angles
= ∠ABD + ∠DBE = 122°
= 32° + x = 122°
= x = 122° – 32°
= x = 90°
### Question: 24
In Figure, side BC of ΔABC has been produced to D and CE ∥ BA. If ∠ABC = 65°, ∠BAC = 55°, find ∠ACE, ∠ECD, ∠ACD.
### Solution:
Corresponding angles,
∠ABC = ∠ECD = 55°
Alternate interior angles,
∠BAC = ∠ACE = 65°
Now, ∠ACD = ∠ACE + ∠ECD
= ∠ACD = 55° + 65°
= 120°
### Question: 25
In Figure, line CA ⊥ AB ∥ line CR and line PR ∥ line BD. Find ∠x, ∠y, ∠z.
### Solution:
Given that, CA ⊥ AB
= ∠CAB = 90°
= ∠AQP = 20°
By, angle of sum property
In ΔAPD
= ∠CAB + ∠AQP + ∠APQ = 180∘
= ∠APQ = 180° – 90° – 20°
= ∠APQ = 70°
y and ∠APQ are corresponding angles
= y = ∠APQ = 70°
∠APQ and ∠z are interior angles
= ∠APQ + ∠z = 180°
= ∠z = 180° – 70°
= ∠z = 110°
### Question: 26
In Figure, PQ ∥ RS. Find the value of x.
### Solution:
Given,
Linear pair,
∠RCD + ∠RCB = 180°
= ∠RCB = 180° – 130°
= 50°
In ΔABC,
∠BAC + ∠ABC + ∠BCA = 180°
By, angle sum property
= ∠BAC = 180° – 55° – 50°
= ∠BAC = 75°
### Question: 27
In Figure, AB ∥ CD and AE ∥ CF, ∠FCG = 90° and ∠BAC = 120°. Find the value of x, y and z.
### Solution:
Alternate interior angle
∠BAC = ∠ACG = 120°
= ∠ACF + ∠FCG = 120°
So, ∠ACF = 120° – 90°
= 30°
Linear pair,
∠DCA + ∠ACG = 180°
= ∠x = 180° – 120°
= 60°
∠BAC + ∠BAE + ∠EAC = 360°
∠CAE = 360° – 120° – (60° + 30°)
= 150°
### Question: 28
In Figure, AB ∥ CD and AC ∥ BD. Find the values of x, y, z.
### Solution:
(i) Since, AC ∥ BD and CD ∥ AB, ABCD is a parallelogram
= ∠ACD = 180° – 65°
= 115°
Opposite angles of parallelogram,
= ∠CAD = ∠CDB = 65°
= ∠ACD = ∠DBA = 115°
(ii) Here,
AC ∥ BD and CD ∥ AB
Alternate interior angles,
∠DCA = x = 40°
∠DAB = y = 35°
### Question: 29
In Figure, state which lines are parallel and why?
### Solution:
Let, F be the point of intersection of the line CD and the line passing through point E.
Here, ∠ACD and ∠CDE are alternate and equal angles.
So, ∠ACD = ∠CDE = 100°
Therefore, AC ∥ EF
### Question: 30
In Figure, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 75°, find ∠DEF.
### Solution:
Let, G be the point of intersection of the lines BC and DE
Since, AB ∥ DE and BC ∥ EF
The corresponding angles,
= ∠ABC = ∠DGC = ∠DEF = 100°
### Course Features
• Video Lectures
• Revision Notes
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• Mind Map
• Study Planner
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• Discussion Forum
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r
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Predictive Hacks
The Expected Payoff of a Dice Game
In hiring process of Data Scientists positions, it is quite common to include some interview questions about probabilities. This post is about a probability quiz. Let’s consider the following question:
Question
You have the option to throw a die up to three times. You will earn the face value of the die. You have the option to stop after each throw and walk away with the money earned. The earnings are not additive. What is the expected payoff of this game?
This is a tricky questions, since the expected payoff depends on player’s strategy. We will provide a solution assuming that we are dealing with a “rational” player who tries to maximize their expected payoff.
One Roll
Let’s assume that the player has only one roll. Then, the expected return would be:
$$Roll_1= (1+2+3+4+5+6)/6=3.5$$
Hence, the expected payoff of one roll is 3.5
Two Roll
Now, let’s assume that the player has the option of up to 2 rolls. The best strategy for them would be to roll again if their result is less than the expected value of the “One Roll” which is 3.5. In other words, if they got 4, 5, or 6 should keep it and if they got 1, 2 or 3 should try again. Thus, the expected payoff in this case is:
$$Roll_2= ( Roll_1 + Roll_1 + Roll_1 +4+5+6)/6= ( 3.5+ 3.5 + 3.5 +4+5+6)/6 = 4.25$$
Hence, the expected payoff of two roll is 4.25
Three Roll
Now, let’s assume that the player has the option of up to 3 rolls. The best strategy for them would be to roll again if their result is less than the expected value of the “Two Roll” which is 4.25. In other words, if they got 5 or 6 should keep it and if they got 1, 2, 3 or 4 should try again. Thus, the expected payoff in this case is:
$$Roll_3= ( Roll_2 + Roll_2 + Roll_2 +Roll_2+5+6)/6= ( 4.25+ 4.25 + 4.25 +4.25+5+6)/6 \cong 4.67$$
Hence, the expected payoff of three roll is 4.67, which is the answer to our problem! Recursively, we can answer this question for n>3. Clearly, as n is getting larger, the expected return will converge to the maximum value which is 6.
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# Nilpotent Matrix and Eigenvalues of the Matrix
## Problem 11
An $n\times n$ matrix $A$ is called nilpotent if $A^k=O$, where $O$ is the $n\times n$ zero matrix.
Prove the followings.
(a) The matrix $A$ is nilpotent if and only if all the eigenvalues of $A$ is zero.
(b) The matrix $A$ is nilpotent if and only if $A^n=O$.
## Hint.
Hint for (a)
1. $(\Rightarrow)$ Consider $A \mathbf{x}=\lambda \mathbf{x}$, where $\lambda$ is an eigenvalue of $A$ and $\mathbf{x}$ is an eigenvector corresponding to $\lambda$.
2. $(\Leftarrow)$ Consider triangulation or Jordan normal/canonical form of $A$. Or use Cayley-Hamilton theorem.
## Proof of (a).
$(\Rightarrow)$
Suppose the matrix $A$ is nilpotent. Namely there exists $k \in \N$ such that $A^k=O$. Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be the eigenvector corresponding to the eigenvalue $\lambda$.
Then they satisfy the equality $A\mathbf{x}=\lambda \mathbf{x}$. Multiplying this equality by $A$ on the left, we have
$A^2\mathbf{x}=\lambda A\mathbf{x}=\lambda^2 \mathbf{x} .$ Repeatedly multiplying by $A$, we obtain that $A^k \mathbf{x}=\lambda^k \mathbf{x}$. (To prove this statement, use mathematical induction.)
Now since $A^k=O$, we get $\lambda^k \mathbf{x}=0_n$, $n$-dimensional zero vector.
Since $\mathbf{x}$ is an eigenvector and hence nonzero by definition, we obtain that $\lambda^k=0$, and hence $\lambda=0$.
$(\Leftarrow)$
Now we assume that all the eigenvalues of the matrix $A$ are zero.
We prove that $A$ is nilpotent.
There exists an invertible $n\times n$ matrix $P$ such that $P^{-1} A P$ is an upper triangular matrix whose diagonal entries are eigenvalues of $A$.
(This is always possible. Study a triangularizable matrix or Jordan normal/canonical form.)
Hence we have
$P^{-1} A P= \begin{bmatrix} 0 & * & \cdots & * \\ 0 & 0 & \cdots & * \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix}.$
Then we have $(P^{-1}AP)^n=O$. This implies that $P^{-1} A^n P=O$ and thus $A^n=POP^{-1}=O$.
Therefore the matrix $A$ is nilpotent.
#### Another proof of $(\Leftarrow)$ using Cayley-Hamilton theorem
Suppose that all the eigenvalues of the matrix $A$ are zero.
Then the characteristic polynomial of the matrix $A$ is
$p(t)=\det(A-tI)=\pm t^n.$
Hence by the Cayley-Hamilton theorem says that
$p(A)=\pm A^n=O,$ the zero matrix.
Thus, $A$ is nilpotent.
Note also that this method also proves the part (b).
## Proof of (b).
If $A^n=O$, then by definition the matrix $A$ is nilpotent.
On the other hand, suppose $A$ is nilpotent. Then by Part (a), the eigenvalues of $A$ are all zero. Then by the same argument of the proof of part (a) $(\Leftarrow)$, we have $A^n=O$.
## Comment.
Part (b) implies the following.
Suppose that you are given $n \times n$ matrix $B$.
You calculate the power $B^n$, and if it is not zero, then the power $B^k$ is never going to be the zero matrix $O$ no matter how large the number $k$ is.
## Related Question.
Problem. Prove that every diagonalizable nilpotent matrix is the zero matrix.
See the post ↴
Every Diagonalizable Nilpotent Matrix is the Zero Matrix
for a proof of this problem.
### More from my site
• Determinant/Trace and Eigenvalues of a Matrix Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues. Show that (1) $$\det(A)=\prod_{i=1}^n \lambda_i$$ (2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$ Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix […]
• If Every Trace of a Power of a Matrix is Zero, then the Matrix is Nilpotent Let $A$ be an $n \times n$ matrix such that $\tr(A^n)=0$ for all $n \in \N$. Then prove that $A$ is a nilpotent matrix. Namely there exist a positive integer $m$ such that $A^m$ is the zero matrix. Steps. Use the Jordan canonical form of the matrix $A$. We want […]
• Find All the Eigenvalues of $A^k$ from Eigenvalues of $A$ Let $A$ be $n\times n$ matrix and let $\lambda_1, \lambda_2, \dots, \lambda_n$ be all the eigenvalues of $A$. (Some of them may be the same.) For each positive integer $k$, prove that $\lambda_1^k, \lambda_2^k, \dots, \lambda_n^k$ are all the eigenvalues of […]
• If Eigenvalues of a Matrix $A$ are Less than $1$, then Determinant of $I-A$ is Positive Let $A$ be an $n \times n$ matrix. Suppose that all the eigenvalues $\lambda$ of $A$ are real and satisfy $\lambda <1$. Then show that the determinant $\det(I-A) >0,$ where $I$ is the $n \times n$ identity matrix. We give two solutions. Solution 1. Let […]
• Diagonalize the Upper Triangular Matrix and Find the Power of the Matrix Consider the $2\times 2$ complex matrix $A=\begin{bmatrix} a & b-a\\ 0& b \end{bmatrix}.$ (a) Find the eigenvalues of $A$. (b) For each eigenvalue of $A$, determine the eigenvectors. (c) Diagonalize the matrix $A$. (d) Using the result of the […]
• Nilpotent Matrices and Non-Singularity of Such Matrices Let $A$ be an $n \times n$ nilpotent matrix, that is, $A^m=O$ for some positive integer $m$, where $O$ is the $n \times n$ zero matrix. Prove that $A$ is a singular matrix and also prove that $I-A, I+A$ are both nonsingular matrices, where $I$ is the $n\times n$ identity […]
• Is the Product of a Nilpotent Matrix and an Invertible Matrix Nilpotent? A square matrix $A$ is called nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix. (a) If $A$ is a nilpotent $n \times n$ matrix and $B$ is an $n\times n$ matrix such that $AB=BA$. Show that the product $AB$ is nilpotent. (b) Let $P$ […]
• Finite Order Matrix and its Trace Let $A$ be an $n\times n$ matrix and suppose that $A^r=I_n$ for some positive integer $r$. Then show that (a) $|\tr(A)|\leq n$. (b) If $|\tr(A)|=n$, then $A=\zeta I_n$ for an $r$-th root of unity $\zeta$. (c) $\tr(A)=n$ if and only if $A=I_n$. Proof. (a) […]
### 3 Responses
1. 07/26/2016
[…] We want to show that all eigenvalues are zero. (Review Nilpotent matrix and eigenvalues of the matrix) […]
2. 03/15/2017
[…] $A$ is nilpotent, all the eigenvalues of $A$ are $0$. (See the post “Nilpotent matrix and eigenvalues of the matrix” for the proof.) Hence the diagonal entries of $D$ are zero, and we have $D=O$, the zero […]
3. 07/10/2017
[…] the post ↴ Nilpotent Matrix and Eigenvalues of the Matrix for a proof of this […]
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##### Determinant/Trace and Eigenvalues of a Matrix
Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues. Show that (1) $$\det(A)=\prod_{i=1}^n \lambda_i$$...
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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2014 AMC 10B Problems/Problem 8"
## Problem
A truck travels $\dfrac{b}{6}$ feet every $t$ seconds. There are $3$ feet in a yard. How many yards does the truck travel in $3$ minutes?
$\textbf {(A) } \frac{b}{1080t} \qquad \textbf {(B) } \frac{30t}{b} \qquad \textbf {(C) } \frac{30b}{t}\qquad \textbf {(D) } \frac{10t}{b} \qquad \textbf {(E) } \frac{10b}{t}$
## Solution
Converting feet to yards and minutes to second, we see that the truck travels $\dfrac{b}{18}$ yards every $t$ seconds for $180$ seconds. We see that he does $\dfrac{180}{t}$ cycles of $\dfrac{b}{18}$ yards. Multiplying, we get $\dfrac{180b}{18t}$, or $\dfrac{10b}{t}$, or $\boxed{\textbf{(E)}}$.
## Solution 2
We set a proportion by letting the $x$ being the number of feet the truck travels in $3$ minutes.
$\frac{\frac{b}{6}}{t}=\frac{x}{180}$
$\frac{b}{6t}=\frac{x}{180}$
$\frac{180b}{6t}=x$
$x=\frac{30b}{t}$
Remember $x$ is the number of feet the truck travels, so we divide by $3$ to convert to yards.
$\frac{x}{3}=\frac{10b}{t}$, which corresponds to $\boxed{\text{(E)}}$
~savannahsolver
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# Coordinated Calculus
## Section2.5The Chain Rule
### Supplemental Videos.
The main topics of this section are also presented in the following videos:
• The Chain Rule
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unl.yuja.com/V/Video?v=7114272&node=34303241&a=51254766&autoplay=1
• The Chain Rule Examples
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unl.yuja.com/V/Video?v=7114274&node=34303311&a=122067046&autoplay=1
• The Chain Rule More Examples
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unl.yuja.com/V/Video?v=7114273&node=34303268&a=187839820&autoplay=1
In addition to learning how to differentiate a variety of basic functions, we have also been developing our ability to use rules to differentiate certain algebraic combinations of them.
### Example2.56.
State the rule(s) used to find the derivative of each of the following combinations of $$f(x) = \sin(x)$$ and $$g(x) = x^2\text{:}$$
\begin{equation*} s(x) = 3x^2 - 5\sin(x)\text{,} \end{equation*}
\begin{equation*} p(x) = x^2 \sin(x), \text{and} \end{equation*}
\begin{equation*} q(x) = \frac{\sin(x)}{x^2}\text{.} \end{equation*}
Hint.
Start by rewriting each function as a specific combination of $$f$$ and $$g\text{,}$$ and then ascertain what rules are necessary to find derivatives.
Finding $$s'$$ uses the sum and constant multiple rules, because $$s(x) = 3g(x) - 5f(x)\text{.}$$ Determining $$p'$$ requires the product rule, because $$p(x) = g(x) \cdot f(x)\text{.}$$ To calculate $$q'$$ we use the quotient rule, because $$q(x) =\frac{f(x)}{g(x)}\text{.}$$
Solution.
We first observe that $$f'(x)=\cos(x)$$ and $$g'(x)=2x\text{.}$$
Since $$s(x)=3g(x)-5f(x)\text{,}$$ we will use the sum and constant multiple rules to find $$s'(x)\text{.}$$ Doing so, we find that
\begin{align*} s'(x)=\mathstrut \amp 3g'(x)-5f'(x)\\ =\mathstrut \amp 3(2x)-5(\cos(x))\\ =\mathstrut \amp 6x-5\cos(x)\text{.} \end{align*}
Since $$p(x)=g(x)\cdot f(x)\text{,}$$ we will use the product rule to determine $$p'(x)\text{.}$$ Proceeding thus, we find
\begin{align*} p'(x)=\mathstrut \amp g'(x)f(x)+g(x)f'(x)\\ =\mathstrut \amp (2x)(\sin(x))+(x^2)(\cos(x))\\ =\mathstrut \amp 2x\sin(x)+x^2\cos(x)\text{.} \end{align*}
Since $$q(x)=\frac{f(x)}{g(x)}\text{,}$$ we will use the quotient rule to calculate $$q'(x)\text{.}$$ In doing so, we see that
\begin{align*} q'(x)=\mathstrut \amp \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\\ =\mathstrut \amp \frac{(\cos(x))(x^2)-(\sin(x))(2x)}{(x^2)^2}\\ =\mathstrut \amp \frac{x^2\cos(x)-2x\sin(x)}{x^4}\\ =\mathstrut \amp \frac{x\cos(x)-2\sin(x)}{x^3}\text{.} \end{align*}
There is one more natural way to combine basic functions algebraically, and that is by composing them. For instance, let’s consider the function
\begin{equation*} C(x) = \sin(x^2)\text{,} \end{equation*}
and observe that any input $$x$$ passes through a chain of functions. In the process that defines the function $$C(x)\text{,}$$ $$x$$ is first squared, and then the sine of the result is taken. We can represent this using an arrow diagram as follows:
\begin{equation*} x \longrightarrow x^2 \longrightarrow \sin(x^2)\text{.} \end{equation*}
It turns out we can express $$C$$ in terms of the elementary functions $$f$$ and $$g$$ that were used above in Example 2.56. Observe that $$x$$ is the input for the function $$g\text{,}$$ and the result is then used as the input for $$f\text{.}$$ We write
\begin{equation*} C(x) = f(g(x)) = \sin(x^2) \end{equation*}
and say that $$C$$ is the composition of $$f$$ and $$g\text{.}$$ We will refer to $$g\text{,}$$ the function that is first applied to $$x\text{,}$$ as the inner function, while $$f\text{,}$$ the function that is applied to the result, as the outer function.
Given a composite function $$C(x) = f(g(x))$$ that is built from differentiable functions $$f$$ and $$g\text{,}$$ how do we compute $$C'(x)$$ in terms of $$f\text{,}$$ $$g\text{,}$$ $$f'\text{,}$$ and $$g'\text{?}$$ In the same way that the rate of change of a product of two functions, $$p(x) = f(x) \cdot g(x)\text{,}$$ depends on the behavior of both $$f$$ and $$g\text{,}$$ it makes sense intuitively that the rate of change of a composite function $$C(x) = f(g(x))$$ will also depend on some combination of $$f$$ and $$g$$ and their derivatives. The rule that describes how to compute $$C'$$ in terms of $$f$$ and $$g$$ and their derivatives is called the chain rule.
But before we can learn what the chain rule says and why it works, we first need to be comfortable decomposing composite functions so that we can correctly identify the inner and outer functions, as we did in the example above with $$C(x) = \sin(x^2)\text{.}$$
### Example2.57.
For each function given below, identify its fundamental algebraic structure. In particular, is the given function a sum, product, quotient, or composition of basic functions? If the function is a composition of basic functions, state a formula for the inner function $$g$$ and the outer function $$f$$ so that the overall composite function can be written in the form $$f(g(x))\text{.}$$ If the function is a sum, product, or quotient of basic functions, use the appropriate rule to determine its derivative.
1. $$\displaystyle h(x) = \tan(2^x)$$
2. $$\displaystyle p(x) = 2^x \tan(x)$$
3. $$\displaystyle r(x) = (\tan(x))^2$$
4. $$\displaystyle m(x) = e^{\tan(x)}$$
5. $$\displaystyle w(x) = \sqrt{x} + \tan(x)$$
6. $$\displaystyle z(x) = \sqrt{\tan(x)}$$
Hint.
1. What is the input of tangent here?
2. Try using the product rule.
3. $$(\tan(x))^2=\tan(x)\cdot\tan(x)\text{,}$$ but can also be written as a composition. How?
4. Should $$e^x$$ be the inner function or the outer function?
5. The “$$+$$” indicates this is fundamentally a sum.
6. What is the input of the square root function here?
1. With $$g(x)=2^x$$ and $$f(x)=\tan(x)$$ we have $$h(x)=f(g(x))\text{.}$$
2. Use the product rule; $$p'(x)=2^x\ln(2)\tan(x)+2^x\sec^2(x)\text{.}$$
3. Use the product rule; $$r(x)=2\tan(x)\sec^2(x)\text{.}$$ Or, $$r(x)=f(g(x))$$ when $$g(x)=\tan(x)$$ and $$f(x)=x^2\text{.}$$
4. $$m(x)=f(g(x))$$ when $$g(x)=\tan(x)$$ and $$f(x)=e^x\text{.}$$
5. Use the sum rule; $$w'(x)=\frac{1}{2\sqrt{x}}+\sec^2(x)\text{.}$$
6. With $$g(x)=\tan(x)$$ and $$f(x)=\sqrt{x}\text{,}$$ we have $$z(x)=f(g(x))\text{.}$$
Solution.
1. $$\tan(2^x)$$ is the composition of $$\tan(x)$$ and $$2^x\text{.}$$ Specifically, with $$f(x)=\tan(x)\text{,}$$ $$g(x)=2^x\text{,}$$ and $$h(x)=\tan(2^x)\text{,}$$ we can write $$h(x)=f(g(x))\text{.}$$
2. $$2^x\tan(x)$$ is the product of $$2^x$$ and $$\tan(x)\text{.}$$ Using the product rule to differentiate $$p(x)=2^x\tan(x)\text{,}$$ we end up with
\begin{align*} p'(x)=\mathstrut \amp \frac{d}{dx}\left[2^x\tan(x)\right]\\ =\mathstrut \amp \frac{d}{dx}\left[2^x\right]\tan(x)+2^x\frac{d}{dx}\left[\tan(x)\right]\\ =\mathstrut \amp 2^x\ln(2)\tan(x)+2^x\sec^2(x)\text{.} \end{align*}
3. $$(\tan(x))^2$$ is the composition of $$x^2$$ and $$\tan(x)\text{.}$$ In particular, with $$f(x)=x^2\text{,}$$ $$g(x)=\tan(x)\text{,}$$ and $$r(x)=(\tan(x))^2\text{,}$$ we can write $$r(x)=f(g(x))\text{.}$$
Alternatively, we can recognize $$(\tan(x))^2$$ as the product of $$\tan(x)$$ with itself. Using the product rule to differentiate $$r(x)=(\tan(x))^2\text{,}$$ we find
\begin{align*} r'(x)=\mathstrut \amp \frac{d}{dx}\left[\tan(x)\tan(x)\right]\\ =\mathstrut \amp \frac{d}{dx}\left[\tan(x)\right]\tan(x)+\tan(x)\frac{d}{dx}\left[\tan(x)\right]\\ =\mathstrut \amp (\sec^2(x))\tan(x)+\tan(x)(\sec^2(x))\\ =\mathstrut \amp 2\tan(x)\sec^2(x)\text{.} \end{align*}
4. $$e^{\tan(x)}$$ is the composition of $$e^x$$ and $$\tan(x)\text{.}$$ Specifically, with $$f(x)=e^x\text{,}$$ $$g(x)=\tan(x)\text{,}$$ and $$m(x)=e^{\tan(x)}\text{,}$$ we can write $$m(x)=f(g(x))\text{.}$$
5. $$\sqrt{x}+\tan(x)$$ is the sum of $$\sqrt{x}=x^{\frac{1}{2}}$$ and $$\tan(x)\text{.}$$ Using the sum rule to find the derivative of $$w(x)=\sqrt{x}+\tan(x)\text{,}$$ we find
\begin{align*} w'(x)=\mathstrut \amp \frac{d}{dx}\left[\sqrt{x}+\tan(x)\right]\\ =\mathstrut \amp \frac{d}{dx}\left[x^{\frac{1}{2}}\right]+\frac{d}{dx}\left[\tan(x)\right]\\ =\mathstrut \amp \frac12x^{-\frac{1}{2}}+\sec^2(x)\\ =\mathstrut \amp \frac{1}{2\sqrt{x}}+\sec^2(x)\text{.} \end{align*}
6. $$\sqrt{\tan(x)}$$ is the composition of $$\sqrt{x}$$ and $$\tan(x)\text{.}$$ In particular, with $$f(x)=\sqrt{x}\text{,}$$ $$g(x)=\tan(x)\text{,}$$ and $$z(x)=\sqrt{\tan(x)}\text{,}$$ we can write $$z(x)=f(g(x))\text{.}$$
### Subsection2.5.1The Chain Rule
Often a composite function cannot be written in an alternate algebraic form. For instance, the function $$C(x) = \sin(x^2)$$ cannot be expanded or otherwise rewritten, so it presents no alternate approaches to taking the derivative. But some composite functions can be expanded or simplified, and these provide a way to explore how the chain rule works. One example of this was the function $$r(x)=(\tan(x))^2$$ in Example 2.57; another example is investigated below in Example 2.58.
#### Example2.58.
Let $$f(x) = -4x + 7$$ and $$g(x) = 3x - 5\text{.}$$ Determine a formula for $$C(x) = f(g(x))$$ and compute $$C'(x)\text{.}$$ How is $$C'$$ related to $$f$$ and $$g$$ and their derivatives?
Hint.
$$C$$ is also linear.
$$C(x)=-12x+27$$ and $$C'(x)=-12\text{.}$$
Solution.
By the rules given for $$f$$ and $$g\text{,}$$
\begin{align*} C(x) =\mathstrut \amp f(g(x))\\ =\mathstrut \amp f(3x-5)\\ =\mathstrut \amp -4(3x-5) + 7\\ =\mathstrut \amp -12x + 20 + 7\\ =\mathstrut \amp -12x + 27\text{.} \end{align*}
Thus, $$C'(x) = -12\text{.}$$ Noting that $$f'(x) = -4$$ and $$g'(x) = 3\text{,}$$ we observe that $$C'$$ appears to be the product of $$f'$$ and $$g'\text{.}$$
It may seem that Example 2.58 is too elementary to illustrate how to differentiate a composite function. Linear functions are the simplest of all functions, and composing linear functions yields another linear function. While this example does not illustrate the full complexity of a composition of nonlinear functions, at the same time we remember that any differentiable function is locally linear, and thus any function with a derivative behaves like a line when viewed up close. The fact that the derivatives of the linear functions $$f$$ and $$g$$ are multiplied to find the derivative of their composition turns out to be a key insight.
We now consider a composition involving a nonlinear function.
#### Example2.59.
Let $$C(x) = \sin(2x)\text{.}$$ Use the double angle identity to rewrite $$C$$ as a product of basic functions, and use the product rule to find $$C'\text{.}$$ Rewrite $$C'$$ in the simplest form possible.
Hint.
The double angle identity says $$\sin(2\theta)=2\sin(\theta)\cos(\theta)\text{.}$$
$$C'(x)=2\cos(2x)\text{.}$$
Solution.
Using the double angle identity for the sine function, we write
\begin{equation*} C(x) = \sin(2x) = 2\sin(x)\cos(x)\text{.} \end{equation*}
Applying the product rule and simplifying, we find
\begin{equation*} C'(x) = 2\left((\cos(x))\cos(x) + \sin(x)(-\sin(x))\right) = 2(\cos^2(x) - \sin^2(x))\text{.} \end{equation*}
Next, we recall that the double angle identity for the cosine function states
\begin{equation*} \cos(2x) = \cos^2(x) - \sin^2(x)\text{.} \end{equation*}
Substituting this result into our expression for $$C'(x)\text{,}$$ we now have that
\begin{equation*} C'(x) = 2 \cos(2x)\text{.} \end{equation*}
In Example 2.59, if we let $$g(x) = 2x$$ and $$f(x) = \sin(x)\text{,}$$ we observe that $$C(x) = f(g(x))\text{.}$$ Note that $$g'(x) = 2$$ and $$f'(x) = \cos(x)\text{,}$$ so we can view the structure of $$C'(x)$$ as
\begin{equation*} C'(x) = 2\cos(2x) = g'(x) f'(g(x))\text{.} \end{equation*}
In this example, as in the example involving linear functions, we see that the derivative of the composite function $$C(x) = f(g(x))$$ is found by multiplying the derivatives of $$f$$ and $$g\text{,}$$ but with $$f'$$ evaluated at $$g(x)\text{.}$$
Intuitively, it makes sense that these two quantities are involved in the rate of change of a composite function: if we ask how fast $$C$$ is changing at a given $$x$$ value, it clearly matters how fast $$g$$ is changing at $$x\text{,}$$ as well as how fast $$f$$ is changing at the value of $$g(x)\text{.}$$ It turns out that this structure holds for all differentiable functions
21
It is important to recognize that we have not proved the chain rule, instead we have given a reason you might believe the chain rule to be true. A key component of mathematics is verifying one’s intuition through formal proof. We will omit the proof of the chain rule, but – just like other differentiation rules – the chain rule can be proved formally using the limit definition of the derivative.
as is stated in the chain rule.
#### Chain Rule.
If $$g$$ is differentiable at $$x$$ and $$f$$ is differentiable at $$g(x)\text{,}$$ then the composite function $$C$$ defined by $$C(x) = f(g(x))$$ is differentiable at $$x$$ and
\begin{equation*} C'(x) = f'(g(x)) g'(x)\text{.} \end{equation*}
As with the product and quotient rules, it is often helpful to think verbally about what the chain rule says: “If $$C$$ is a composite function defined by an outer function $$f$$ and an inner function $$g\text{,}$$ then $$C'$$ is given by the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.”
It is helpful to clearly identify the inner function $$g$$ and outer function $$f\text{,}$$ compute their derivatives individually, and then put all of the pieces together by the chain rule.
#### Example2.60.
Use the chain rule to determine the derivative of the function
\begin{equation*} r(x) = (\tan(x))^2\text{.} \end{equation*}
Hint.
We saw this function earlier in Example 2.57.
As we saw in Example 2.57, $$r'(x)=2\tan(x)\sec^2(x)\text{.}$$
Solution.
The function $$r$$ is composite, with inner function $$g(x) = \tan(x)$$ and outer function $$f(x) = x^2\text{.}$$ Organizing the key information involving $$f\text{,}$$ $$g\text{,}$$ and their derivatives, we have
$$f(x) = x^2\text{,}$$ $$g(x) = \tan(x)\text{,}$$ $$f'(x) = 2x\text{,}$$ $$g'(x) = \sec^2(x)\text{,}$$ $$f'(g(x)) = 2\tan(x)\text{.}$$
Applying the chain rule, we find that
\begin{equation*} r'(x) = f'(g(x))g'(x) = 2\tan(x) \sec^2(x)\text{.} \end{equation*}
As a side note, we remark that $$r(x)$$ is usually written as $$\tan^2(x)\text{.}$$ This is common notation for powers of trigonometric functions: e.g. $$\cos^4(x)\text{,}$$ $$\sin^5(x)\text{,}$$ and $$\sec^2(x)$$ are all composite functions, with the outer function a power function and the inner function a trigonometric one.
#### Example2.61.
For each function given below, identify an inner function $$g$$ and outer function $$f$$ to write the function in the form $$f(g(x))\text{.}$$ Determine $$f'(x)\text{,}$$ $$g'(x)\text{,}$$ and $$f'(g(x))\text{,}$$ and then apply the chain rule to determine the derivative of the given function.
1. $$\displaystyle h(x) = \cos(x^4)$$
2. $$\displaystyle p(x) = \sqrt{ \tan(x) }$$
3. $$\displaystyle s(x) = 2^{\sin(x)}$$
4. $$\displaystyle z(x) = \cot^5(x)$$
5. $$\displaystyle m(x) = (\sec(x) + e^x)^9$$
Hint.
1. The outer function is $$f(x) = \cos(x)\text{.}$$
2. The outer function is $$f(x) = \sqrt{x}\text{.}$$
3. The outer function is $$f(x) = 2^x\text{.}$$
4. The outer function is $$f(x) = x^5\text{.}$$
5. The outer function is $$f(x) = x^9\text{.}$$
1. $$h'(x) = -4x^3\sin(x^4)\text{.}$$
2. $$h'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}}\text{.}$$
3. $$h'(x) = 2^{\sin(x)}\ln(2)\cos(x)\text{.}$$
4. $$h'(x) = -5\cot^4(x) \csc^2(x)\text{.}$$
5. $$h'(x) = 9(\sec(x)+e^x)^8 (\sec(x)\tan(x) + e^x)\text{.}$$
Solution.
1. The outer function is $$f(x) = \cos(x)$$ while the inner function is $$g(x) = x^4\text{.}$$ We know that
\begin{equation*} f'(x) = -\sin(x), g'(x) = 4x^3, \ \text{and} \ f'(g(x)) = -\sin(x^4)\text{.} \end{equation*}
Hence by the chain rule,
\begin{equation*} h'(x) = f'(g(x))g'(x) = -4x^3\sin(x^4)\text{.} \end{equation*}
2. The outer function is $$f(x) = \sqrt{x}$$ and the inner function is $$g(x) = \tan(x)\text{.}$$ We know that
\begin{equation*} f'(x) = \frac{1}{2\sqrt{x}}, g'(x) = \sec^2(x), \ \text{and} \ f'(g(x)) = \frac{1}{2\sqrt{\tan(x)}}\text{.} \end{equation*}
Thus by the chain rule,
\begin{equation*} h'(x) = f'(g(x))g'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}}\text{.} \end{equation*}
3. The outer function is $$f(x) = 2^x$$ while the inner function is $$g(x) = \sin(x)\text{.}$$ We know that
\begin{equation*} f'(x) = 2^x \ln(2), g'(x) = \cos(x), \ \text{and} \ f'(g(x)) = 2^{\sin(x)}\ln(2)\text{.} \end{equation*}
Thus by the chain rule,
\begin{equation*} h'(x) = f'(g(x))g'(x) = 2^{\sin(x)}\ln(2)\cos(x)\text{.} \end{equation*}
4. The outer function is $$f(x) = x^5$$ and the inner function is $$g(x) = \cot(x)\text{.}$$ We know that
\begin{equation*} f'(x) = 5x^4, g'(x) = -\csc^2(x), \ \text{and} \ f'(g(x)) = 5\cot^4(x)\text{.} \end{equation*}
Hence by the chain rule,
\begin{equation*} h'(x) = f'(g(x))g'(x) = -5\cot^4(x) \csc^2(x)\text{.} \end{equation*}
5. The outer function is $$f(x) = x^9$$ and the inner function is $$g(x) = \sec(x) + e^x\text{.}$$ We know that
\begin{equation*} f'(x) = 9x^8, g'(x) = \sec(x)\tan(x) + e^x, \ \text{and} \ f'(g(x)) = 9(\sec(x)+e^x)^8\text{.} \end{equation*}
Hence by the chain rule,
\begin{equation*} h'(x) = f'(g(x))g'(x) = 9(\sec(x)+e^x)^8 (\sec(x)\tan(x) + e^x)\text{.} \end{equation*}
### Subsection2.5.2Using Multiple Rules Simultaneously
The chain rule now joins the sum, constant multiple, product, and quotient rules in our collection of techniques for finding the derivative of a function through understanding its algebraic structure and the basic functions that constitute it. It takes practice to get comfortable applying multiple rules to differentiate a single function, but using proper notation and taking a few extra steps will help.
#### Example2.62.
Find a formula for the derivative of $$h(t) = 3^{t^2 + 2t}\sec^4(t)\text{.}$$
We first observe that $$h$$ is the product of two functions: $$h(t) = a(t) \cdot b(t)\text{,}$$ where $$a(t) = 3^{t^2 + 2t}$$ and $$b(t) = \sec^4(t)\text{.}$$ We will need to use the product rule to differentiate $$h\text{.}$$ And because $$a$$ and $$b$$ are composite functions, we will also need the chain rule. We therefore begin by computing $$a'(t)$$ and $$b'(t)\text{.}$$
Writing $$a(t) = f(g(t)) = 3^{t^2 + 2t}$$ and finding the derivatives of $$f$$ and $$g$$ with respect to $$t\text{,}$$ we have
$$f(t) = 3^t\text{,}$$ $$g(t) = t^2 + 2t\text{,}$$ $$f'(t) = 3^t \ln(3)\text{,}$$ $$g'(t) = 2t+2\text{,}$$ $$f'(g(t)) = 3^{t^2 + 2t}\ln(3)\text{.}$$
Thus by the chain rule, it follows that
\begin{equation*} a'(t) = f'(g(t))g'(t) = 3^{t^2 + 2t}\ln(3) (2t+2)\text{.} \end{equation*}
Turning next to the function $$b\text{,}$$ we write $$b(t) = r(s(t)) = \sec^4(t)$$ and find the derivatives of $$r$$ and $$s$$ with respect to $$t\text{.}$$
$$r(t) = t^4\text{,}$$ $$s(t) = \sec(t)\text{,}$$ $$r'(t) = 4t^3\text{,}$$ $$s'(t) = \sec(t)\tan(t)\text{,}$$ $$r'(s(t)) = 4\sec^3(t)\text{.}$$
By the chain rule,
\begin{gather*} b'(t) = r'(s(t))s'(t) \\ = 4\sec^3(t)\sec(t)\tan(t) \\ = 4 \sec^4(t) \tan(t). \end{gather*}
Now we are finally ready to compute the derivative of the function $$h\text{.}$$ Recalling that $$h(t) = 3^{t^2 + 2t}\sec^4(t)\text{,}$$ by the product rule we have
\begin{equation*} h'(t) = \frac{d}{dt}\left[3^{t^2 + 2t}\right]\sec^4(t)+3^{t^2 + 2t} \frac{d}{dt}\left[\sec^4(t)\right]\text{.} \end{equation*}
From our work above with $$a$$ and $$b\text{,}$$ we know the derivatives of $$3^{t^2 + 2t}$$ and $$\sec^4(t)\text{.}$$ Therefore
\begin{equation*} h'(t) = 3^{t^2 + 2t}\ln(3) (2t+2)\sec^4(t) + 3^{t^2 + 2t} 4\sec^4(t) \tan(t)\text{.} \end{equation*}
The above calculation may seem tedious. However, by breaking the function down into small parts and calculating derivatives of those parts separately, we are able to accurately calculate the derivative of the entire function.
#### Example2.63.
Differentiate each of the following functions. State the rule(s) you use, label relevant derivatives appropriately, and be sure to clearly identify your overall answer.
1. $$\displaystyle p(r) = 4\sqrt{r^6 + 2e^r}$$
2. $$\displaystyle m(v) = \sin(v^2) \cos(v^3)$$
3. $$\displaystyle \displaystyle h(y) = \frac{\cos(10y)}{1+e^{4y}}$$
4. $$\displaystyle s(z) = 2^{z^2 \sec (z)}$$
5. $$\displaystyle c(x) = \sin\left(e^{x^2}\right)$$
Hint.
1. Use the constant multiple rule first, followed by the chain rule.
2. Observe that $$m$$ is fundamentally a product of composite functions.
3. Note that $$h$$ is a quotient of composite functions.
4. The function $$s$$ is a composite function with outer function $$2^z\text{.}$$
5. It is possible for a function to be a composite function with more than two functions in the chain.
1. $$p'(r) = \frac{4(6r^5 + 2e^r)}{2\sqrt{r^6 + 2e^r}}\text{.}$$
2. $$m'(v) = 2v \cos(v^2)\cos(v^3)-3v^2 \sin(v^2)\sin(v^3)\text{.}$$
3. $$h'(y) = \frac{ [-10\sin(10y)](1+e^{4y}) - \cos(10y) [4e^{4y}]}{(1+e^{4y})^2}\text{.}$$
4. $$s'(z) = 2^{z^2\sec(z)} \ln(2) [2z\sec(z)+z^2 \sec(z)\tan(z)]\text{.}$$
5. $$c'(x) = \cos\left(e^{x^2}\right) \left[e^{x^2}\cdot 2x\right]\text{.}$$
Solution.
1. By the constant multiple rule, $$p'(r) = 4\frac{d}{dr}\left[\sqrt{r^6 + 2e^r}\right]\text{.}$$ Using the chain rule to complete the remaining derivative, we see that
\begin{equation*} p'(r) = 4 \frac{1}{2\sqrt{r^6 + 2e^r}} \frac{d}{dr}[r^6 + 2e^r] = \frac{4(6r^5 + 2e^r)}{2\sqrt{r^6 + 2e^r}}\text{.} \end{equation*}
2. Observe that by the product rule,
\begin{equation*} m'(v) = \frac{d}{dv}[\sin(v^2)]\cos(v^3) +\sin(v^2) \frac{d}{dv}[\cos(v^3)]\text{.} \end{equation*}
Applying the chain rule to differentiate $$\cos(v^3)$$ and $$\sin(v^2)\text{,}$$ we see that
\begin{align*} m'(v) =\mathstrut \amp [\cos(v^2) \cdot 2v]\cos(v^3) + \sin(v^2) [-\sin(v^3) \cdot 3v^2]\\ =\mathstrut \amp 2v \cos(v^2)\cos(v^3)-3v^2 \sin(v^2)\sin(v^3)\text{.} \end{align*}
3. By the quotient rule,
\begin{equation*} h'(y) = \frac{\frac{d}{dy}[\cos(10y)](1+e^{4y}) - \cos(10y) \frac{d}{dy}[1+e^{4y}]}{(1+e^{4y})^2}\text{.} \end{equation*}
Applying the chain rule to differentiate $$\cos(10y)$$ and $$e^{4y}\text{,}$$ it follows that
\begin{equation*} h'(y) = \frac{ [-10\sin(10y)](1+e^{4y}) - \cos(10y) [4e^{4y}]}{(1+e^{4y})^2}\text{.} \end{equation*}
4. By the chain rule, we have $$s'(z) = 2^{z^2\sec(z)} \ln(2) \frac{d}{dz}[z^2 \sec(z)]\text{.}$$ Then with the product rule, we find that
\begin{equation*} s'(z) = 2^{z^2\sec(z)} \ln(2) [2z\sec(z)+z^2 \sec(z)\tan(z)]\text{.} \end{equation*}
5. Here we have the composition of three functions, rather than just two. If we first apply the chain rule to the outermost function (the sine function), we find that
\begin{equation*} c'(x) = \cos\left(e^{x^2}\right) \frac{d}{dx}\left[e^{x^2}\right]\text{.} \end{equation*}
Next we again apply the chain rule to find $$e^{x^2}\text{,}$$ using $$e^x$$ as the outer function and $$x^2$$ as the inner function. We end up with
\begin{equation*} c'(x) = \cos\left(e^{x^2}\right) \left[e^{x^2}\cdot 2x\right]\text{.} \end{equation*}
The chain rule now adds substantially to our ability to compute derivatives. Whether we are finding the equation of the tangent line to a curve, the instantaneous velocity of a moving particle, or the instantaneous rate of change of a certain quantity, the chain rule is indispensable if the function under consideration is a composition.
#### Example2.64.
Use known derivative rules (including the chain rule) as needed to answer each of the following questions.
1. Find an equation for the tangent line to the curve $$y= \sqrt{e^x + 3}$$ at the point where $$x=0\text{.}$$
2. If $$\displaystyle s(t) = \frac{1}{(t^2+1)^3}$$ represents the position function of a particle moving horizontally along an axis at time $$t$$ (where $$s$$ is measured in inches and $$t$$ in seconds), find the particle’s instantaneous velocity at $$t=1\text{.}$$ Is the particle moving to the left or right at that instant?
22
You may assume that this axis is like a number line, with “left” being the negative direction, and “right” being the positive direction.
3. Suppose that $$f(x)$$ and $$g(x)$$ are differentiable functions and that the following information about them is known:
If $$C(x)$$ is a function given by the formula $$f(g(x))\text{,}$$ determine $$C'(2)\text{.}$$ In addition, if $$D(x)$$ is the function $$f(f(x))\text{,}$$ find $$D'(-1)\text{.}$$
Hint.
1. Let $$f(x) = \sqrt{e^x + 3}\text{.}$$ Find $$f'(x)$$ and $$f'(0)\text{.}$$
2. Recall that $$s'(t)$$ tells us the instantaneous velocity at time $$t\text{.}$$
3. Since $$C(x) = f(g(x))\text{,}$$ it follows $$C'(x) = f'(g(x))g'(x)\text{.}$$ What is $$C'(2)\text{?}$$ What is a formula for $$D'(x)\text{?}$$
1. $$y - 2 = \frac{1}{4}(x-0)\text{.}$$
2. $$s'(1) = -\frac{3}{8}$$ inches per second, so the particle is moving left at the instant $$t = 1\text{.}$$
3. $$C'(2) = -10 \text{;}$$ $$D'(-1) = -20\text{.}$$
Solution.
1. Let $$f(x) = \sqrt{e^x + 3}\text{.}$$ By the chain rule, $$f'(x) = \frac{e^x}{2\sqrt{e^x + 3}}\text{,}$$ and thus $$f'(0) = \frac{1}{4}\text{.}$$ Note further that $$f(0) = \sqrt{1 + 3} = 2\text{.}$$ The tangent line is therefore the line through $$(0,2)$$ with slope $$\frac{1}{4}\text{,}$$ which is
\begin{equation*} y - 2 = \frac{1}{4}(x-0)\text{.} \end{equation*}
2. Observe that $$s(t) = (t^2 + 1)^{-3}\text{,}$$ and thus by the chain rule, $$s'(t) = -3(t^2 + 1)^{-4}(2t)\text{.}$$ We therefore see that $$s'(1) = -\frac{6}{16} = -\frac{3}{8}$$ inches per second, so the particle is moving left at the instant $$t = 1\text{.}$$
3. Since $$C(x) = f(g(x))\text{,}$$ it follows $$C'(x) = f'(g(x))g'(x)\text{.}$$ Therefore, $$C'(2) = f'(g(2))g'(2)\text{.}$$ From the given table, $$g(2) = -1\text{,}$$ so applying this result and using the additional given information,
\begin{equation*} C'(2) = f'(-1) g'(2) = (-5)(2) = -10\text{.} \end{equation*}
For $$D(x) = f(f(x))\text{,}$$ the chain rule tells us that $$D'(x) = f'(f(x))f'(x)\text{,}$$ so $$D'(-1) = f'(f(-1))f'(-1)\text{.}$$ Using the given table, it follows that
\begin{equation*} D'(-1) = f'(2)f'(-1) = (4)(-5) = -20\text{.} \end{equation*}
### Subsection2.5.3The Composite Version of Basic Function Rules
As we gain more experience with differentiation, we will become more comfortable in simply writing down the derivative without taking multiple steps. This is particularly simple when the inner function is linear, since the derivative of a linear function is a constant.
#### Example2.66.
Use the chain rule to differentiate each of the following composite functions whose inside function is linear:
\begin{equation*} \frac{d}{dx} \left[ (5x+7)^{10} \right] = 10(5x+7)^9 \cdot 5\text{,} \end{equation*}
\begin{equation*} \frac{d}{dx} \left[ \tan(17x) \right] = 17\sec^2(17x), \ \text{and} \end{equation*}
\begin{equation*} \frac{d}{dx} \left[ e^{-3x} \right] = -3e^{-3x}\text{.} \end{equation*}
More generally, an excellent exercise for getting comfortable with the derivative rules is as follows. First write down a list of all the basic functions whose derivatives we know, and list the derivatives. Then write a composite function with the inner function being an unknown function $$u(x)$$ and the outer function being a basic function. Finally, write the chain rule for the composite function. The following example illustrates this for two different functions.
#### Example2.67.
To determine
\begin{equation*} \frac{d}{dx}[\sin(u(x))]\text{,} \end{equation*}
where $$u$$ is a differentiable function of $$x\text{,}$$ we use the chain rule with the sine function as the outer function. Applying the chain rule, we find that
\begin{equation*} \frac{d}{dx}[\sin(u(x))] = \cos(u(x)) \cdot u'(x)\text{.} \end{equation*}
This rule is analogous to the basic derivative rule that $$\frac{d}{dx}[\sin(x)] = \cos(x)\text{.}$$
Similarly, since $$\frac{d}{dx}[a^x] = a^x \ln(a)$$ whenever $$a \gt 0\text{,}$$ it follows by the chain rule that
\begin{equation*} \frac{d}{dx}[a^{u(x)}] = a^{u(x)} \ln(a) \cdot u'(x)\text{.} \end{equation*}
This rule is analogous to the basic derivative rule that $$\frac{d}{dx}[a^{x}] = a^{x} \ln(a)\text{.}$$
### Subsection2.5.4Summary
• A composite function is one where the input variable $$x$$ first passes through one function, and then the resulting output passes through another. For example, the function $$h(x) = 2^{\sin(x)}$$ is composite since $$x \longrightarrow \sin(x) \longrightarrow 2^{\sin(x)}\text{.}$$
• Given a composite function $$C(x) = f(g(x))$$ where $$f$$ and $$g$$ are differentiable functions, the chain rule tells us that
\begin{equation*} C'(x) = f'(g(x)) g'(x)\text{.} \end{equation*}
### Exercises2.5.5Exercises
#### 1.Mixing rules: chain, product, sum.
Find the derivative of
$$f(x) = e^{5 x} (x^2 + 5^{x})$$
$$f'(x) =$$
#### 2.Mixing rules: chain and product.
Find the derivative of
$$v(t) = t^6 e^{-ct}$$
Assume that $$c$$ is a constant.
$$v'(t) =$$
#### 3.Using the chain rule repeatedly.
Find the derivative of
$$y = \sqrt{e^{-5 t^2}+7}$$
$${dy\over dt} =$$
#### 4.Derivative involving arbitrary constants $$a$$ and $$b$$.
Find the derivative of
$$f(x) = axe^{-bx + 12}$$
Assume that $$a$$ and $$b$$ are constants.
$$f'(x) =$$
#### 5.Chain rule with graphs.
Use the graph below to find exact values of the indicated derivatives, or state that they do not exist. If a derivative does not exist, enter dne in the answer blank. The graph of $$f(x)$$ is black and has a sharp corner at $$x = 2\text{.}$$ The graph of $$g(x)$$ is blue.
Let $$h(x) = f(g(x))\text{.}$$ Find
1. $$h'(1) =$$
2. $$h'(2) =$$
3. $$h'(3) =$$
(Enter dne for any derivative that does not exist.)
#### 6.Chain rule with function values.
Given $$F(3)=1, F'(3)=6, F(4)=1, F'(4)=2$$ and $$G(1)=5, G'(1)=7, G(4)=3, G'(4)=5\text{,}$$ find each of the following. (Enter dne for any derivative that cannot be computed from this information alone.)
1. $$H(4)$$ if $$H(x)=F(G(x))$$
2. $$H'(4)$$ if $$H(x)=F(G(x))$$
3. $$H(4)$$ if $$H(x)=G(F(x))$$
4. $$H'(4)$$ if $$H(x)=G(F(x))$$
5. $$H'(4)$$ if $$H(x)=F(x)/G(x)$$
#### 7.A product involving a composite function.
Find the derivative of $$f(x)=2 x\sin(4 x)$$
$$f'(x) =$$
#### 8.Using the chain rule to compare composite functions.
Consider the basic functions $$f(x) = x^3$$ and $$g(x) = \sin(x)\text{.}$$
1. Let $$h(x) = f(g(x))\text{.}$$ Find the exact instantaneous rate of change of $$h$$ at the point where $$x = \frac{\pi}{4}\text{.}$$
2. Which function is changing most rapidly at $$x = 0.25\text{:}$$ $$h(x) = f(g(x))$$ or $$r(x) = g(f(x))\text{?}$$ Why?
3. Let $$h(x) = f(g(x))$$ and $$r(x) = g(f(x))\text{.}$$ Which of these functions has a derivative that is periodic? Why?
#### 9.Chain rule with an arbitrary function $$u$$.
Let $$u(x)$$ be a differentiable function. For each of the following functions, determine the derivative. Each response will involve $$u$$ and/or $$u'\text{.}$$
1. $$\displaystyle p(x) = e^{u(x)}$$
2. $$\displaystyle q(x) = u(e^x)$$
3. $$\displaystyle r(x) = \cot(u(x))$$
4. $$\displaystyle s(x) = u(\cot(x))$$
5. $$\displaystyle a(x) = u(x^4)$$
6. $$\displaystyle b(x) = u^4(x)$$
#### 10.More on using the chain rule with graphs.
Let functions $$p$$ and $$q$$ be the piecewise linear functions given by their respective graphs in Figure 2.68. Use the graphs to answer the following questions.
1. Let $$C(x) = p(q(x))\text{.}$$ Determine $$C'(0)$$ and $$C'(3)\text{.}$$
2. Find a value of $$x$$ for which $$C'(x)$$ does not exist. Explain your thinking.
3. Let $$Y(x) = q(q(x))$$ and $$Z(x) = q(p(x))\text{.}$$ Determine $$Y'(-2)$$ and $$Z'(0)\text{.}$$
#### 11.Applying the chain rule in a physical context.
If a spherical tank of radius 4 feet has $$h$$ feet of water present in the tank, then the volume of water in the tank is given by the formula
\begin{equation*} V = \frac{\pi}{3} h^2(12-h)\text{.} \end{equation*}
1. At what instantaneous rate is the volume of water in the tank changing with respect to the height of the water at the instant $$h = 1\text{?}$$ What are the units on this quantity?
2. Now suppose that the height of water in the tank is being regulated by an inflow and outflow (e.g., a faucet and a drain) so that the height of the water at time $$t$$ is given by the rule $$h(t) = \sin(\pi t) + 1\text{,}$$ where $$t$$ is measured in hours (and $$h$$ is still measured in feet). At what rate is the height of the water changing with respect to time at the instant $$t = 2\text{?}$$
3. Continuing under the assumptions in (b), at what instantaneous rate is the volume of water in the tank changing with respect to time at the instant $$t = 2\text{?}$$
4. What are the main differences between the rates found in (a) and (c)? Include a discussion of the relevant units.
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# Trigonometric Equations
Trigonometric Equations are those equations that involve the trigonometric functions of a variable. In this article, you will learn how to find solutions for the given equations. These equations consist of one or more unknown angles. Let us consider an example, cos m – sin2 m = 0, which is a trigonometric equation that does not satisfy all the values of m. Hence for such equations, you will have to either find the value of m or you will have to find the solution.
Previously, you have learned that the values sin x and cos x are repeated after an interval of 2π and tan x and these values repeat itself after the interval of π. Principal Solutions are those solutions that lie in the interval of [ 0, 2π ] of such given trigonometry equations. A trigonometric equation will also have a general solution expressing all the values which would satisfy the given equation and it is expressed in a generalized form in terms of ‘n’. The general representation of these equations comprises the formula of the trigonometric equation;
E1 ( sin m, cos m, tan m ) = E2 ( sin m, cos m, tan m )
Here,
E1 – is a rational function.
E2 – is a rational function.
Since sine, cosine, and tangent are the three major trigonometric functions, the solutions will be derived for these equations will comprise of only these three ratios. Although the solutions for the other three ratios such as secant, cosec, and cotangent can be obtained with the help of those solutions.
Let us consider a basic equation to understand this concept. Equation: sin m = 0 and 0, π, and 2π will be the principal solutions for this case since these values satisfy any given equation which lies in between the [ 0, 2π ]. If the values of the sin m = 0, then the value of m = 0, π, 2π, – π, -2π, -6π, etc of the given equation. Therefore, the general solution for sin m = 0 will be m = nπ, where n belongs to Integers.
## Solutions for Trigonometric Equations
Equations Solutions sin m = 0 m = nπ cos m = 0 m = (nπ + π/2) tan m = 0 m = nπ sin m = 1 m = (2nπ + π/2) = (4n+1)π/2 cos m = 1 m = 2nπ sin m = sin θ m = nπ + (-1)nθ, where θ ∈ [-π/2, π/2] cos m = cos θ m = 2nπ ± θ, where θ ∈ (0, π] tan m = tan θ m = nπ + θ, where θ ∈ (-π/2 , π/2] sin2 m = sin2 θ m = nπ ± θ cos2 m = cos2 θ m = nπ ± θ tan2 m = tan2 θ m = nπ ± θ
### Proofs for Solutions of Trigonometric Equations
Theorem 1: For any real number j and k, sin j = sin k implies that j = nπ + ( – 1 ) . n . k, where n Є Z
Proof: consider the equation, sin j = sin k. Now, let us try and find the general solution of this equation.
sin j = sin k
⇒ sin j – sin k = 0
⇒ 2 cos ( j + k ) / 2 sin ( j – k ) / 2 = 0
⇒ cos ( j + k ) / 2 = 0 or sin ( j – k ) / 2 = 0
Upon taking the common solution from both the conditions, we get:
j = n π + (-1)n k, where n ∈ Z
Theorem 2: For any real numbers j and k, cos j = cos k, implies j = 2nπ ± k, where n Є Z.
Proof: Similarly, the general solution of cos j = cos k will be:
cos j – cos k = 0
2 sin ( j + k ) / 2 sin ( k – j ) / 2 = 0
sin ( j + k ) / 2 = 0 or sin ( j – k ) / 2 = 0
( j + k ) / 2 = ( n * π ) or ( j – k ) / 2 = ( n * π )
On taking the common solution from both the conditions, we get:
j = 2 * n * π ± k, where n ∈ Z
Theorem 3: Prove that if and k are not odd multiple of π / 2, then tan j = tan y implies that j = nπ + k, wheren Є Z.
Solution:
Similarly to find the solution of equations involving tan x or other functions, we can use the conversion of trigonometric equations.
In other words, if tan x = tan y then;
$\frac{{sin j }}{cos k}$ = $\frac{{cos j }}{cos k}$
sin j * cos k = sin k * cos j
sin j cos k – sin k cos j = 0
sin ( j – k ) = 0 [By trigonometric identity]
Hence, j – k = ( n * π ) or j = ( n * π + k ), where n ∈ Z.
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## 1. Introduction
The Marching Squares algorithm is a computer graphics algorithm introduced in the 1980s that can be used for contouring. We can use the Marching Squares algorithm to draw the lines of constant:
• altitude on a topographical map
• temperature on a temperature heat map
• pressure on a contour map for a pressure field
In this tutorial, we’ll learn how the Marching Squares algorithm determines the contour from a grid of sample points of the image.
## 2. Contour Curves and Level Curves
Let’s spend a brief moment understanding the mathematically precise definition of contour lines.
### 2.1. Mathematical Definition
We have been drawing graphs of functions of one-variable such as or implicit functions like for so long, that we give the matter little thought. Plotting functions of two variables is a much more difficult task.
The trick to putting together a reasonable graph of a function of two or more variables is to find a way to cut down on the dimensions involved. One way to achieve this is to draw certain special curves that lie on the surface . These special curves, called contour curves, are obtained by intersecting the surface with horizontal planes for various values of the constant .
These curves in the -plane are called the level curves of the original function .
The level curve at height of any function is the curve in defined by the equation , where is a constant. In mathematical notation,
### 2.2. Examples
Consider the function defined as:
By definition, the level curve at height is:
Thus, we see that the level curves for are circles centered at the origin of radius . They are summarized in the table below:
The family of level curves, the “topographic map” of the surface is shown in the figure below:
If we stack all level curves together, we can get a feeling for the complete graph of . It’s a surface that looks like an inverted dish and is called a paraboloid:
For our discussion, we require the output contour to possess the following properties:
• It must separate the interior points from the exterior points.
• Thus, it must be an orientable and closed curve.
## 3. Essentials of the Algorithm
Consider the simplistic image of a cloud given below. We are interested to trace the contours of the cloud:
For simplicity, let’s assume that the image of a cloud is represented by a square matrix of pixels. Each pixel value represents a single sample of light, carrying brightness information. It ranges between and . Black is represented by and white by .
### 3.1. Sampling the Image
We divide the entire image domain into squares. Each square has dimensions by . We sample the image at each of the grid points:
Next, we iterate through the grid to determine the vertex state in comparison to an iso-value . Each grid vertex is now assigned a binary state ( or ). An iso-value serves as a threshold for determining the vertex state. Suppose we choose as the threshold. The vertex state is determined as follows:
• – Pixel value less than the iso-value
• – Pixel value greater than the iso-value
The resulting binary grid , thus obtained is:
### 3.2. Configuration of Each Sub-Grid
Next, a subgrid can be associated with a configuration. We iterate through . At each sub-grid, we form a binary code based on the vertex value by iterating clockwise from the top-left corner to the bottom-left corner.
For example, if a given subgrid has four corners:
its binary code is said to be . The equivalent of the binary code in the decimal number system is called the configuration of the sub-grid. In the above example, the sub-grid has the configuration Thus, we have a configuration associated with each sub-grid as follows:
### 3.3. Lookup Table for the Contours
Imagine a single square with the contours of an object cutting through this square. That is, the square does not lie wholly in the interior or exterior of the object. A square has edges and for each edge, we have possible choices – either the contour(boundary) intersects this edge, or it does not. This results in possible configurations.
The configuration of each sub-grid is matched with one entry in the contours lookup table below:
The contours lookup table is intuitive. As mentioned before, there are possible configurations.
Let’s consider the configuration :
We think of the purple region as the interior of an object, and the blue region as the exterior of the object. We denote the state by a black vertex and the state by a white vertex. We define a bipolar pair to be two vertices in opposite states.
Intuitively, the contour of an object must intersect the edge of a bipolar pair. Thus, the configuration must correspond to an object whose contour intersects the top and left edges of the square. In this fashion, the contours in each sub-grid are approximated, as shown in the figure below:
## 4. Algorithm
The Marching Squares algorithm can be divided into two steps:
1. Sampling the grid points and converting them to a binary matrix .
2. March the squares in and build a set of edges .
Let’s see the pseudocode:
### 4.1. Sampling the Grid
The input to the Marching Squares algorithm is some sample of the original matrix . The parameters and control the resolution of the sample. For example, if and , we select the sequence of pixel-values:
The result of this operation should yield a matrix of size rows and columns.
Each selected pixel value is compared with the iso-value and converted to a bit-value or . is used to hold the bit matrix. The actual -coordinates of each pixel in the sample , concerning the original matrix is stored as a named tuple in a matrix .
For example, if , , we would find , .
Let’s see the pseudocode now:
### 4.2. Marching the Squares In
Next, we march the squares in the binary grid . Each square in , beginning with its top-left corner, has vertices, in clockwise order : , , and . We compute the reference value of each square. We also compute the mid-points of each square edge: north, east, south, and west. Finally, we determine the contour by searching the reference value in the contours lookup table:
Here’s a topographical map of the Mont Blanc region in the Swiss Alps. The Marching Squares algorithm when executed on this image produces nice contours around the mountain peaks:
## 5. Conclusion
In this article, we learned about an interesting Computer Graphics algorithm called the Marching Squares. To summarize, it is based on two core ideas. First, the contours can be constructed piecewise within each grid square, without reference to other squares. Second, each isocontour segment in a grid square can be retrieved from a lookup table.
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Now we have to find out the values of algebraic expressions using the given variables.
Steps to find the value of the algebraic expression:
Step $$1$$: Understand the problem first and then fix the variable and write the algebraic expression.
Step $$2$$: Substitute each variable by the given numerical value to get an arithmetical expression.
Step $$3$$: Try to solve or simplify the arithmetical expression by the BIDMAS method.
Step $$4$$: Now, the final value you obtained is the required value of the expression.
Example:
Question I)
Consider the algebraic expression $13m-9n-8$ and the value of $$m=$$ 12 and $$n=$$ 4.
Now substitute the values of $$m$$ and $$n$$, we get
$\begin{array}{l}=13\left(12\right)-9\left(4\right)-8\\ \\ =156-36-8\\ \\ =156-44\\ \\ =112\end{array}$
Therefore, the value of the algebraic expression $13m-9n-8$ $$=$$ 112.
Question II)
Consider the algebraic expression ${m}^{2}-4{n}^{2}+4$ with the value of $$m=$$ 12 and $$n=$$ 4.
Now substitute the values of $$m$$ and $$n$$, we get
$\begin{array}{l}={12}^{2}-4\phantom{\rule{0.147em}{0ex}}\left({4}^{2}\right)+4\\ \\ =144-4\phantom{\rule{0.147em}{0ex}}\left(16\right)+4\\ \\ =144-64+4\\ \\ =148-64\\ \\ =84\end{array}$
Therefore, the value of the algebraic expression ${m}^{2}-4{n}^{2}+4$ $$=$$ 84.
Question III)
Consider the algebraic expression is $2{m}^{2}-{8n}^{2}+13$ with the value of $$m=$$ 12 and $$n=$$ 4.
Now substitute the values of $$m$$ and $$n$$, we get
$\begin{array}{l}=\phantom{\rule{0.147em}{0ex}}2\left({12}^{2}\right)\phantom{\rule{0.147em}{0ex}}-8\left({4}^{2}\right)+13\\ \\ =\phantom{\rule{0.147em}{0ex}}2\left(144\right)\phantom{\rule{0.147em}{0ex}}-8\phantom{\rule{0.147em}{0ex}}\left(16\right)\phantom{\rule{0.147em}{0ex}}+13\\ \\ =\phantom{\rule{0.147em}{0ex}}288\phantom{\rule{0.147em}{0ex}}-128+13\\ \\ =301-128\\ \\ =147\end{array}$
Therefore, the value of the algebraic expression $2{m}^{2}-{8n}^{2}+13$ $$=$$ 147.
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#
Describing Distributions
Overview
When describing or comparing distributions, it is important to discuss the following characteristics: shape, center, spread, and outliers.
1. Shape
There are five main possible shapes for the distribution, as shown below. If the distribution is not symmetric, decide whether it is skewed right or left using the following rule:
The data is skewed in the same direction as the tail.
For example, if the data were concentrated on the left side, leaving a tail to the right, then one would conclude that the distribution is skewed right.
SYMMETRIC (Bell Shaped) SYMMETRIC (Uniform) SKEWED LEFT SKEWED RIGHT SYMMETRIC, BIMODAL
2. Center
There are three measurements of center that can be used for this purpose, listed in order from most common to least common.
A) Median: The median is the value that splits the data in half so that 50% of the data lies to the left and 50% lies to the right.
• Notation: μ
• If the sample size is odd, take the middle number. If the sample size is even, take the average of the two middle numbers.
• The median is the most frequently used measure of center for describing distributions.
B) Mean: The mean is the average value.
• Notation:
• The mean is calculated by summing all the values together and then dividing this by the total number of data points.
C) Mode: The mode is the most frequently observed value in the data set.
• The mode is rarely used when describing distributions.
*For more information on measures of center, see the 'Measures of Center' help page.
There are three measurements of spread that can be used for this purpose.
A) Range: The range is the difference between the minimum and maxiumum values.
• Stating the minimum and maximum values is the most common way to describe the spread of distributions, especially when you are not given the dataset itself and must estimate through visual inspection of the graph.
B) Standard Deviation: Standard deviation measures the amount of dispersion from the mean.
• Formula: Sx$$\sqrt{Σ( x_i - \bar{x} )^2 \over n-1 }$$
• Nonresistant (best used for symmetric data)
C) Interquartile Range (IQR): The IQR measures the difference between the first and third quartiles.
• Formula: IQR = Q3 - Q1
• Resistant (works with data that is skewed and/or has outliers)
*For more information on IQR & quartiles, see the 'Five Number Summary' help page.
4. Outliers
Outliers are the extreme values in the data set. They can be estimated through visual inspection or calculated using the following formulas:
lower threshold = Q1 - (1.5) (IQR)
upper threshold = Q3 + (1.5) (IQR)
Any values below the lower threshold or above the upper threshold can be considered outliers.
*For more information on IQR & quartiles, see the 'Five Number Summary' help page.
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# Methods of computing square roots
Methods of computing square roots
There are several methods for calculating the principal square root of a nonnegative real number. For the square roots of a negative or complex number, see below.
## Rough estimation
Many of the methods for calculating square roots of a positive real number S require an initial seed value. If the initial value is too far from the actual square root, the calculation will be slowed down. It is therefore useful to have a rough estimate, which may be very inaccurate but easy to calculate. If S ≥ 1, let D be the number of digits to the left of the decimal point. If S < 1, let D be the negative of the number of zeros to the immediate right of the decimal point. Then the rough estimation is this:
If D is odd, D = 2n + 1, then use $\sqrt{S} \approx 2 \cdot 10^n.$
If D is even, D = 2n + 2, then use $\sqrt{S} \approx 6 \cdot 10^n.$
Two and six are used because they approximate the geometric means of the lowest and highest possible values with the given number of digits: $\sqrt{\sqrt{1 \cdot 10}} = \sqrt[4]{10} \approx 2 \,$ and $\sqrt{\sqrt{10 \cdot 100}} = \sqrt[4]{1000} \approx 6 \,.$
When working in the binary numeral system (as computers do internally), an alternative method is to use $2^{\left\lfloor D/2\right\rfloor}$ (here D is the number of binary digits).
## Babylonian method
Graph charting the use of the Babylonian method for approximating the square root of 100 (10) using starting values x0 = 50, x0 = 1, and x0 = −5. Note that using a negative starting value yields the negative root.
Perhaps the first algorithm used for approximating $\sqrt S$ is known as the "Babylonian method", named after the Babylonians,[1] or "Heron's method", named after the first-century Greek mathematician Hero of Alexandria who gave the first explicit description of the method.[2] It can be derived from (but predates by many centuries) Newton's method. The basic idea is that if x is an overestimate to the square root of a non-negative real number S then $\scriptstyle S/x\,$ will be an underestimate and so the average of these two numbers may reasonably be expected to provide a better approximation (though the formal proof of that assertion depends on the inequality of arithmetic and geometric means that shows this average is always an overestimate of the square root, as noted in the article on square roots, thus assuring convergence). This is a quadratically convergent algorithm, which means that the number of correct digits of the approximation roughly doubles with each iteration. It proceeds as follows:
1. Begin with an arbitrary positive starting value x0 (the closer to the actual square root of S, the better).
2. Let xn+1 be the average of xn and S / xn (using the arithmetic mean to approximate the geometric mean).
3. Repeat step 2 until the desired accuracy is achieved.
It can also be represented as:
$x_0 \approx \sqrt{S}.$
$x_{n+1} = \frac{1}{2} \left(x_n + \frac{S}{x_n}\right),$
$\sqrt S = \lim_{n \to \infty} x_n.$
This algorithm works equally well in the p-adic numbers, but cannot be used to identify real square roots with p-adic square roots; it is easy, for example, to construct a sequence of rational numbers by this method that converges to +3 in the reals, but to −3 in the 2-adics.
### Example
To calculate $\sqrt{S}$, where S = 125348, to 6 significant figures, use the rough estimation method above to get x0. The number of digits in S is D = 6 = 2·2 + 2. So, n = 2 and the rough estimate is
$x_0 = 6 \cdot 10^2 = 600.000. \,$
$x_1 = \frac{1}{2} \left(x_0 + \frac{S}{x_0}\right) = \frac{1}{2} \left(600.000 + \frac{125348}{600.000}\right) = 404.457.$
$x_2 = \frac{1}{2} \left(x_1 + \frac{S}{x_1}\right) = \frac{1}{2} \left(404.457 + \frac{125348}{404.457}\right) = 357.187.$
$x_3 = \frac{1}{2} \left(x_2 + \frac{S}{x_2}\right) = \frac{1}{2} \left(357.187 + \frac{125348}{357.187}\right) = 354.059.$
$x_4 = \frac{1}{2} \left(x_3 + \frac{S}{x_3}\right) = \frac{1}{2} \left(354.059 + \frac{125348}{354.059}\right) = 354.045.$
$x_5 = \frac{1}{2} \left(x_4 + \frac{S}{x_4}\right) = \frac{1}{2} \left(354.045 + \frac{125348}{354.045}\right) = 354.045.$
Therefore, $\sqrt{125348} \approx 354.045 \,.$
### Convergence
Let the relative error in xn be defined by
$\varepsilon_n = \frac {x_n}{\sqrt{S}} - 1$
and thus
$x_n = \sqrt {S} \cdot (1 + \varepsilon_n).$
Then it can be shown that
$\varepsilon_{n+1} = \frac {\varepsilon_n^2}{2 (1 + \varepsilon_n)}$
and thus that
$0 \leq \varepsilon_{n+2} \leq \min \left\{\frac {\varepsilon_{n+1}^2}{2}, \frac {\varepsilon_{n+1}}{2} \right\}$
and consequently that convergence is assured provided that x0 and S are both positive.
#### Worst case for convergence
If using the rough estimate above with the Babylonian method, then the worst cases are:
\begin{align} S & = 1; & x_0 & = 2; & x_1 & = 1.250; & \varepsilon_1 & = 0.250. \\ S & = 10; & x_0 & = 2; & x_1 & = 3.500; & \varepsilon_1 & < 0.107. \\ S & = 10; & x_0 & = 6; & x_1 & = 3.833; & \varepsilon_1 & < 0.213. \\ S & = 100; & x_0 & = 6; & x_1 & = 11.333; & \varepsilon_1 & < 0.134. \end{align}
Thus in any case,
$\varepsilon_1 \leq 2^{-2}. \,$
$\varepsilon_2 < 2^{-5} < 10^{-1}. \,$
$\varepsilon_3 < 2^{-11} < 10^{-3}. \,$
$\varepsilon_4 < 2^{-23} < 10^{-6}. \,$
$\varepsilon_5 < 2^{-47} < 10^{-14}. \,$
$\varepsilon_6 < 2^{-95} < 10^{-28}. \,$
$\varepsilon_7 < 2^{-191} < 10^{-57}. \,$
$\varepsilon_8 < 2^{-383} < 10^{-115}. \,$
Remember that rounding errors will slow the convergence. It is recommended to keep at least one extra digit beyond the desired accuracy of the xn being calculated to minimize round off error.
## Exponential identity
Pocket calculators typically implement good routines to compute the exponential function and the natural logarithm, and then compute the square root of S using the identity[citation needed]
$\sqrt{S} = e^{\frac{1}{2}\ln S}.$
The same identity is used when computing square roots with logarithm tables or slide rules.
## Method of bisecting intervals
A simple way to compute a square root is the high/low method, similar to the bisection method. This method involves guessing a number based on known squares, then checking if its square is too high or too low and adjusting accordingly.
To find the square root of 20, first note that the square of 5 is 25, and that the square of 4 is 16. As 20 is greater than 16 and less than 25, the square root of 20 must be in between 4 and 5. Guessing 4.5, as the average of 4 and 5, yields 20.25 and is too high. The next step is to guess 4.4, yielding 19.36 and is too low. Therefore, as before, the square root of 20 must be in between 4.4 and 4.5. Continue this pattern until the desired number of decimal places is achieved. For example:
4.452 = 19.8025 (too low)
4.472 = 19.9809 (too low, but close)
4.482 = 20.0704 (too high)
4.4752 = 20.025625 (too high)
4.4732 = 20.007729 (too high, but close)
4.4722 = 19.998784 (too low)
Now it is known that the square root of 20 is between 4.472 and 4.473, so the square root of 20 to the first three decimal places is 4.472.
## Bakhshali approximation
This method for finding an approximation to a square root was described in an ancient Indian mathematical manuscript called the Bakhshali manuscript. It is equivalent to two iterations of the Babylonian method beginning with N. The original presentation goes as follows: To calculate $\sqrt{S}$, let N2 be the nearest perfect square to S. Then, calculate:
$d = S - N^2 \,\!$
$P = \frac{d}{2N}$
$A = N + P\,\!$
$\sqrt{S} \approx A - \frac{P^2}{2A}$
This can be also written as:
$\sqrt{S} \approx N + \frac{d}{2N} - \frac{d^2}{8N^3 + 4Nd} = \frac{8N^4 + 8N^2 d + d^2}{8N^3 + 4Nd} = \frac{N^4 + 6N^2S + S^2}{4N^3 + 4NS}$
### Example
Find $\sqrt{9.2345}.$
$N=3\,\!$
$d = 9.2345 - 3^2 = 0.2345\,\!$
$P = \frac{0.2345}{2 \times 3} = 0.0391$
$A = 3 + 0.0391 = 3.0391\,\!$
$\sqrt{9.2345} \approx 3.0391 - \frac{0.0391^2}{2 \times 3.0391} \approx 3.0388$
## Digit-by-digit calculation
This is a method to find each digit of the square root in a sequence. It is slower than the Babylonian method (if you have a calculator that can divide in one operation), but it has several advantages:
• It can be easier for manual calculations.
• Every digit of the root found is known to be correct, i.e., it does not have to be changed later.
• If the square root has an expansion that terminates, the algorithm terminates after the last digit is found. Thus, it can be used to check whether a given integer is a square number.
Napier's bones include an aid for the execution of this algorithm. The shifting nth-root algorithm is a generalization of this method.
The algorithm works for any base, and naturally, the way it proceeds depends on the base chosen.
### Decimal (base 10)
Write the original number in decimal form. The numbers are written similar to the long division algorithm, and, as in long division, the root will be written on the line above. Now separate the digits into pairs, starting from the decimal point and going both left and right. The decimal point of the root will be above the decimal point of the square. One digit of the root will appear above each pair of digits of the square.
Beginning with the left-most pair of digits, do the following procedure for each pair:
1. Starting on the left, bring down the most significant (leftmost) pair of digits not yet used (if all the digits have been used, write "00") and write them to the right of the remainder from the previous step (on the first step, there will be no remainder). In other words, multiply the remainder by 100 and add the two digits. This will be the current value c.
2. Find p, y and x, as follows:
• Let p be the part of the root found so far, ignoring any decimal point. (For the first step, p = 0).
• Determine the greatest digit x such that $y=(20 \cdot p + x) \cdot x$ does not exceed c.
• Note: 20p + x is simply twice p, with the digit x appended to the right).
• Note: You can find x by guessing what c/(20·p) is and doing a trial calculation of y, then adjusting x upward or downward as necessary.
• Place the digit x as the next digit of the root, i.e., above the two digits of the square you just brought down. Thus the next p will be the old p times 10 plus x.
3. Subtract y from c to form a new remainder.
4. If the remainder is zero and there are no more digits to bring down, then the algorithm has terminated. Otherwise go back to step 1 for another iteration.
#### Examples
Find the square root of 152.2756.
1 2. 3 4
/
\/ 01 52.27 56
01 1*1 <= 1 < 2*2 x = 1
01 y = x*x = 1*1 = 1
00 52 22*2 <= 52 < 23*3 x = 2
00 44 y = (20+x)*x = 22*2 = 44
08 27 243*3 <= 827 < 244*4 x = 3
07 29 y = (240+x)*x = 243*3 = 729
98 56 2464*4 <= 9856 < 2465*5 x = 4
98 56 y = (2460+x)*x = 2464*4 = 9856
00 00 Algorithm terminates: Answer is 12.34
Find the square root of 2.
1. 4 1 4 2
/
\/ 02.00 00 00 00
02 1*1 <= 2 < 2*2 x = 1
01 y = x*x = 1*1 = 1
01 00 24*4 <= 100 < 25*5 x = 4
00 96 y = (20+x)*x = 24*4 = 96
04 00 281*1 <= 400 < 282*2 x = 1
02 81 y = (280+x)*x = 281*1 = 281
01 19 00 2824*4 <= 11900 < 2825*5 x = 4
01 12 96 y = (2820+x)*x = 2824*4 = 11296
06 04 00 28282*2 <= 60400 < 28283*3 x = 2
The desired precision is achieved:
The square root of 2 is about 1.4142
### Binary numeral system (base 2)
Inherent to digit-by-digit algorithms is a search and test step: find a digit, $\, e$, when added to the right of a current solution $\, r$, such that $\,(r+e)\cdot(r+e) \le x$, where $\, x$ is the value for which a root is desired. Expanding: $\,r\cdot r + 2re + e\cdot e \le x$. The current value of $\,r\cdot r$—or, usually, the remainder—can be incrementally updated efficiently when working in binary, as the value of $\, e$ will be a single bit, and the operations needed to compute $\,2\cdot r\cdot e$ and $\,e\cdot e$ can be replaced with faster bit shift operations. This gives rise to simple computer implementations:[3]
short isqrt(short num) {
short res = 0;
short bit = 1 << 14; // The second-to-top bit is set: 1L<<30 for long
// "bit" starts at the highest power of four <= the argument.
while (bit > num)
bit >>= 2;
while (bit != 0) {
if (num >= res + bit) {
num -= res + bit;
res = (res >> 1) + bit;
}
else
res >>= 1;
bit >>= 2;
}
return res;
}
Faster algorithms, in binary and decimal or any other base, can be realized by using lookup tables—in effect trading more storage space for reduced run time.[4]
## Vedic duplex method for extracting a square root
The Vedic duplex method is an ancient Indian method[citation needed] of extracting the square root. It is a variant of the digit by digit method for calculating the square root of a whole or decimal number one digit at a time.[5] The duplex is the square of the central digit plus double the cross-product of digits equidistant from the center. The duplex is computed from the quotient digits (square root digits) computed thus far, but after the initial digits. The duplex is subtracted from the dividend digit prior to the second subtraction for the product of the quotient digit times the divisor digit. For perfect squares the duplex and the dividend will get smaller and reach zero after a few steps. For non-perfect squares the decimal value of the square root can be calculated to any precision desired. However, as the decimal places proliferate, the duplex adjustment gets larger and longer to calculate. The duplex method follows the Vedic ideal for an algorithm, one-line, mental calculation. It is flexible in choosing the first digit group and the divisor. Small divisors are to be avoided by starting with a larger initial group.
In short, to calculate the duplex of a number, double the product of each pair of equidistant digits plus the square of the center digit (of the digits to the right of the colon).
Number => Calculation = Duplex
574 ==> 2(5·4) + 72 = 89
406,739 ==> 2(4·9)+ 2(0·3)+ 2(6·7) = 72+0+84 = 156
123,456 ==> 2(1·6)+ 2(2·5)+ 2(3·4) = 12 +20 +24 = 56
88,900,777 ==> 2(8·7)+2(8·7)+2(9·7)+2(0·0)+2(0·0) = 112+112+126+0+0 = 350
48329,03711 ==> 2(4·1)+2(8·1)+2(3·7)+2(2·3)+2(9·0)= 8+16+42+12+0 = 78
In a square root calculation the quotient digit set increases incrementally for each step.
Number => Calculation = Duplex:
1 ==> 12 = 1
14 ==>2(1·4) = 8
142 ==> 2(1·2) + 42 = 4 + 16 = 20
14,21 ==> 2(1·1) + 2(4·2) = 2 + 16 = 18
14213 ==> 6+8+4 = 18
142,135 ==> 10+24+4 = 38
1421356 ==> 12+40+12+1 = 65
1421,3562 ==> 4+48+20+6 = 78
142,135,623 ==> 6+16+24+10+9 = 65
142,1356,237 ==> 14+24+8+12+30 = 88
142,13562,373 ==> 6+56+12+4+36+25 = 139
### Example 1, by discussion
Consider the perfect square 2809 = 532. Use the duplex method to find the square root of 2,809.
• Set down the number in groups of two digits.
• Define a divisor, a dividend and a quotient to find the root.
• Given 2809. Consider the first group, 28.
• Find the nearest perfect square below that group.
• The root of that perfect square is the first digit of our root.
• Since 28 > 25 and 25 = 52, take 5 as the first digit in the square root.
• For the divisor take double this first digit (2 · 5), which is 10.
• Next, set up a division framework with a colon.
• 28: 0 9 is the dividend and 5: is the quotient.
• Put a colon to the right of 28 and 5 and keep the colons lined up vertically. The duplex is calculated only on quotient digits to the right of the colon.
• Calculate the remainder. 28: minus 25: is 3:.
• Append the remainder on the left of the next digit to get the new dividend.
• Here, append 3 to the next dividend digit 0, which makes the new dividend 30. The divisor 10 goes into 30 just 3 times. (No reserve needed here for subsequent deductions.)
• Repeat the operation.
• The zero remainder appended to 9. Nine is the next dividend.
• This provides a digit to the right of the colon so deduct the duplex, 32 = 9.
• Subtracting this duplex from the dividend 9, a zero remainder results.
• Ten into zero is zero. The next root digit is zero. The next duplex is 2(3·0) = 0.
• The dividend is zero. This is an exact square root, 53.
### Example 1, analysis and square root framework
Find the square root of 2809.
Set down the number in groups of two digits.
The number of groups gives the number of whole digits in the root.
Put a colon after the first group, 28, to separate it.
From the first group, 28, obtain the divisor, 10, since
28>25=52 and by doubling this first root, 2x5=10.
Gross dividend: 28: 0 9. Using mental math:
Divisor: 10) 3 0 Square: 10) 28: 30 9
Duplex, Deduction: 25: xx 09 Square root: 5: 3. 0
Dividend: 30 00
Remainder: 3: 00 00
Square Root, Quotient: 5: 3. 0
### Example 2
Find the square root of 2,080,180,881. Solution by the duplex method: this ten-digit square has five digit-pairs, so it will have a five-digit square root. The first digit-pair is 20. Put the colon to the right. The nearest square below 20 is 16, whose root is 4, the first root digit. So, use 2·4=8 for the divisor. Now proceed with the duplex division, one digit column at a time. Prefix the remainder to the next dividend digit.
divisor; gross dividend: 8) 20: 8 0 1 8 0 8 8 1
read the dividend diagonally up: 4 8 7 11 10 10 0 8
minus the duplex: 16: xx 25 60 36 90 108 00 81
actual dividend: : 48 55 11 82 10 00 08 00
minus the product: : 40 48 00 72 00 00 0 00
remainder: 4: 8 7 11 10 10 0 8 00
quotient: 4: 5, 6 0 9. 0 0 0 0
Duplex calculations:
Quotient-digits ==> Duplex deduction.
5 ==> 52= 25
5 and 6 ==> 2(5·6) = 60
5,6,0 ==> 2(5·0)+62 = 36
5,6,0,9 ==> 2(5·9)+2(6·0) = 90
5,6,0,9,0 ==> 2(5·0)+2(6·9)+ 0 = 108
5,6,0,9,0,0 ==> 2(5·0)+2(6·0)+2(0·9) = 0
5,6,0,9,0,0,0 ==> 2(5·0)+2(6·0)+2(0·0)+92 = 81
Hence the square root of 2,080,180,881 is exactly 45,609.
### Example 3
Find the square root of two to ten places. Take 20,000 as the beginning group, using three digit-pairs at the start. The perfect square just below 20,000 is 141, since 1412 = 19881 < 20,000. So, the first root digits are 141 and the divisor doubled, 2 x 141 = 282. With a larger divisor the duplex will be relatively small. Hence, the multiple of the divisor can be picked without confusion.
Dividend: 2.0000 : 0 0 0 0 0 0 0 0
Diagonal;Divisor: (282) : 1190 620 400 1020 1620 1820 750 1120
Minus duplex: : xxxx 16 16 12 28 53 74 59
Actual dividend: 20000 : 1190 604 384 1008 1592 1767 676 1061
Minus product: 19881 : 1128 564 282 846 1410 1692 564 846
Remainder: 119 : 62 40 102 162 182 75 112 215
Root quotient: 1.41 : 4 2 1 3 5 6 2 3
Ten multiples of 282: 282; 564; 846; 1128; 1410; 1692; 1974; 2256; 2538; 2820.
## A two-variable iterative method
This method is applicable for finding the square root of $0 < S < 3 \,\!$ and converges best for $S \approx 1$. This, however, is no real limitation for a computer based calculation, as in base 2 floating point and fixed point representations, it is trivial to multiply $S \,\!$ by an integer power of 4, and therefore $\sqrt{S}$ by the corresponding power of 2, by changing the exponent or by shifting, respectively. Therefore, $S \,\!$ can be moved to the range $\frac{1}{2} \le S <2$. Moreover, the following method does not employ general divisions, but only additions, subtractions, multiplications, and divisions by powers of two, which are again trivial to implement. A disadvantage of the method is that numerical errors accumulate, in contrast to single variable iterative methods such as the Babylonian one.
The initialization step of this method is
$a_0 = S \,\!$
$c_0 = S-1 \,\!$
$a_{n+1} = a_n - a_n c_n / 2 \,\!$
$c_{n+1} = c_n^2 (c_n - 3) / 4 \,\!$
Then, $a_n \rightarrow \sqrt{S}$ (while $c_n \rightarrow 0$).
Note that the convergence of $c_n \,\!$, and therefore also of $a_n \,\!$, is quadratic.
The proof of the method is rather easy. First, rewrite the iterative definition of $c_n \,\!$ as
$1 + c_{n+1} = (1 + c_n) (1 - c_n/2)^2 \,\!$.
Then it is straightforward to prove by induction that
$S (1 + c_n) = a_n^2$
and therefore the convergence of $a_n \,\!$ to the desired result $\sqrt{S}$ is ensured by the convergence of $c_n \,\!$ to 0, which in turn follows from $-1 < c_0 < 2 \,\!$.
This method was developed around 1950 by M. V. Wilkes, D. J. Wheeler and S. Gill[6] for use on EDSAC, one of the first electronic computers.[7] The method was later generalized, allowing the computation of non-square roots.[8]
## Iterative methods for reciprocal square roots
The following are iterative methods for finding the reciprocal square root of S which is $1/\sqrt{S}$. Once it has been found, find $\sqrt{S}$ by simple multiplication: $\sqrt{S} = S \cdot (1/\sqrt{S})$. These iterations involve only multiplication, and not division. They are therefore faster than the Babylonian method. However, they are not stable. If the initial value is not close to the reciprocal square root, the iterations will diverge away from it rather than converge to it. It can therefore be advantageous to perform an iteration of the Babylonian method on a rough estimate before starting to apply these methods.
• One method is found by applying Newton's method to the equation (1 / x2) − S = 0. It converges quadratically:
$x_{n+1} = \frac{x_n}{2} \cdot (3 - S \cdot x_n^2).$
$y_n = S \cdot x_n^2\, , \!$
$x_{n+1} = \frac{x_n}{8} \cdot (15 - y_n \cdot (10 - 3 \cdot y_n)).$
## Taylor series
If N is an approximation to $\sqrt{S}$, a better approximation can be found by using the Taylor series of the square root function:
$\sqrt{N^2+d} = \sum_{n=0}^\infty \frac{(-1)^{n}(2n)!d^n}{(1-2n)n!^2 4^nN^{2n-1}} = N + \frac{d}{2N} - \frac{d^2}{8N^3} + \frac{d^3}{16N^5} - \frac{5d^4}{128N^7} + \cdots$
As an iterative method, the order of convergence is equal to the number of terms used. With 2 terms, it is identical to the Babylonian method; With 3 terms, each iteration takes almost as many operations as the Bakhshali approximation, but converges more slowly. Therefore, this is not a particularly efficient way of calculation.
## Other methods
Finding $\sqrt{S}$ is the same as solving the equation $x^2 - S = 0\,\!$. Therefore, any general numerical root-finding algorithm can be used. Newton's method, for example, reduces in this case to the Babylonian method. Other methods are less efficient than the ones presented above.
A completely different method for computing the square root is based on the CORDIC algorithm, which uses only very simple operations (addition, subtraction, bitshift and table lookup, but no multiplication).
## Continued fraction expansion
Quadratic irrationals (numbers of the form $\frac{a+\sqrt{b}}{c}$, where a, b and c are integers), and in particular, square roots of integers, have periodic continued fractions. Sometimes what is desired is finding not the numerical value of a square root, but rather its continued fraction expansion. The following iterative algorithm can be used for this purpose (S is any natural number that is not a perfect square):
$m_0=0\,\!$
$d_0=1\,\!$
$a_0=\left\lfloor\sqrt{S}\right\rfloor\,\!$
$m_{n+1}=d_na_n-m_n\,\!$
$d_{n+1}=\frac{S-m_{n+1}^2}{d_n}\,\!$
$a_{n+1} =\left\lfloor\frac{\sqrt{S}+m_{n+1}}{d_{n+1}}\right\rfloor =\left\lfloor\frac{a_0+m_{n+1}}{d_{n+1}}\right\rfloor\!.$
Notice that mn, dn, and an are always integers. The algorithm terminates when this triplet is the same as one encountered before. The expansion will repeat from then on. The sequence [a0; a1, a2, a3, …] is the continued fraction expansion:
$\sqrt{S} = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3+\,\ddots}}}$
### Example, square root of 114 as a continued fraction
Begin with m0 = 0; d0 = 1; and a0 = 10 (102 = 100 and 112 = 121 > 114 so 10 chosen).
\begin{align} \sqrt{114} & = \frac{\sqrt{114}+0}{1} = 10+\frac{\sqrt{114}-10}{1} = 10+\frac{(\sqrt{114}-10)(\sqrt{114}+10)}{\sqrt{114}+10} \\ & = 10+\frac{114-100}{\sqrt{114}+10} = 10+\frac{1}{\frac{\sqrt{114}+10}{14}}. \end{align}
$m_{1} = d_{0} \cdot a_{0} - m_{0} = 1 \cdot 10 - 0 = 10 \,.$
$d_{1} = \frac{S-m_{1}^2}{d_0} = \frac{114-10^2}{1} = 14 \,.$
$a_{1} = \left\lfloor \frac{a_0+m_{1}}{d_{1}} \right\rfloor = \left\lfloor \frac{10+10}{14} \right\rfloor = \left\lfloor \frac{20}{14} \right\rfloor = 1 \,.$
So, m1 = 10; d1 = 14; and a1 = 1.
$\frac{\sqrt{114}+10}{14} = 1+\frac{\sqrt{114}-4}{14} = 1+\frac{114-16}{14(\sqrt{114}+4)} = 1+\frac{1}{\frac{\sqrt{114}+4}{7}}.$
Next, m2 = 4; d2 = 7; and a2 = 2.
$\frac{\sqrt{114}+4}{7} = 2+\frac{\sqrt{114}-10}{7} = 2+\frac{14}{7(\sqrt{114}+10)} = 2+\frac{1}{\frac{\sqrt{114}+10}{2}}.$
$\frac{\sqrt{114}+10}{2}=10+\frac{\sqrt{114}-10}{2}=10+\frac{14}{2(\sqrt{114}+10)} = 10+\frac{1}{\frac{\sqrt{114}+10}{7}}.$
$\frac{\sqrt{114}+10}{7}=2+\frac{\sqrt{114}-4}{7}=2+\frac{98}{7(\sqrt{114}+4)} = 2+\frac{1}{\frac{\sqrt{114}+4}{14}}.$
$\frac{\sqrt{114}+4}{14}=1+\frac{\sqrt{114}-10}{14}=1+\frac{14}{14(\sqrt{114}+10)} = 1+\frac{1}{\frac{\sqrt{114}+10}{1}}.$
$\frac{\sqrt{114}+10}{1}=20+\frac{\sqrt{114}-10}{1}=20+\frac{14}{\sqrt{114}+10} = 20+\frac{1}{\frac{\sqrt{114}+10}{14}}.$
Now, loop back to the second equation above.
Consequently, the simple continued fraction for the square root of 114 is
$\sqrt{114} = [10;1,2,10,2,1,20,1,2,10,2,1,20,1,2,10,2,1,20,\dots].\,$
Its actual value is approximately 10.67707 82520 31311 21....
### Generalized continued fraction
A more rapid method is to evaluate its generalized continued fraction. From the formula derived there:
$\sqrt{z} = \sqrt{x^2+y} = x+\cfrac{y} {2x+\cfrac{y} {2x+\cfrac{y} {2x+\ddots}}} = x+\cfrac{2x \cdot y} {2(2z-y)-y-\cfrac{y^2} {2(2z-y)-\cfrac{y^2} {2(2z-y)-\ddots}}}$
the square root of 114 is quickly found:
$\sqrt{114} = \sqrt{11^2-7} = 11-\cfrac{7} {22-\cfrac{7} {22-\cfrac{7} {22-\ddots}}} = 11-\cfrac{22 \cdot 7} {470+7-\cfrac{49} {470-\cfrac{49} {470-\ddots}}}.$
Furthermore, the fact that 114 is 2/3 of the way between 102=100 and 112=121 results in
$\sqrt{114} = \cfrac{\sqrt{1026}}{3} = \cfrac{\sqrt{32^2+2}}{3} = \cfrac{32}{3}+\cfrac{2/3} {64+\cfrac{2} {64+\cfrac{2} {64+\cfrac{2} {64+\ddots}}}},$
which is simply the aforementioned [10;1,2, 10,2,1, 20,1,2, 10,2,1, 20,1,2, ...] evaluated at every third term. Combining pairs of fractions produces
$\sqrt{114} = \cfrac{\sqrt{32^2+2}}{3} = \cfrac{32}{3}+\cfrac{64/3} {2050-1-\cfrac{1} {2050-\cfrac{1} {2050-\cfrac{1} {2050-\ddots}}}},$
which is now [10;1,2, 10,2,1,20,1,2, 10,2,1,20,1,2, ...] evaluated at the third term and every six terms thereafter.
## Pell's equation
Pell's equation and its variants yield a method for efficiently finding continued fraction convergents of square roots of integers. However, it can be complicated to execute, and usually not every convergent is generated. The ideas behind the method are as follows:
• If (p, q) is a solution (where p and q are integers) to the equation $p^2 = S \cdot q^2 \pm 1\!$, then $\frac{p}{q}$ is a continued fraction convergent of $\sqrt{S}$, and as such, is an excellent rational approximation to it.
• If (pa, qa) and (pb, qb) are solutions, then so is:
$p = p_a p_b + S \cdot q_a q_b\,\!$
$q = p_a q_b + p_b q_a\,\!$
(compare to the multiplication of quadratic integers)
• More generally, if (p1, q1) is a solution, then it is possible to generate a sequence of solutions (pn, qn) satisfying:
$p_{m+n} = p_m p_n + S \cdot q_m q_n\,\!$
$q_{m+n} = p_m q_n + p_n q_m\,\!$
The method is as follows:
• Find positive integers p1 and q1 such that $p_1^2 = S \cdot q_1^2 \pm 1$. This is the hard part; It can be done either by guessing, or by using fairly sophisticated techniques.
• To generate a long list of convergents, iterate:
$p_{n+1} = p_1 p_n + S \cdot q_1 q_n\,\!$
$q_{n+1} = p_1 q_n + p_n q_1\,\!$
• To find the larger convergents quickly, iterate:
$p_{2n} = p_n^2 + S \cdot q_n^2\,\!$
$q_{2n} = 2 p_n q_n\,\!$
Notice that the corresponding sequence of fractions coincides with the one given by the Hero's method starting with $\textstyle\frac{p_1}{q_1}$.
• In either case, $\frac{p_n}{q_n}$ is a rational approximation satisfying
$\left|\frac{p_n}{q_n} - \sqrt{S}\right| < \frac{1}{q_n^2 \cdot \sqrt{S}}.$
## Approximations that depend on IEEE representation
On computers, a very rapid Newton's-method-based approximation to the square root can be obtained for floating point numbers when computers use an IEEE (or sufficiently similar) representation.
This technique is based on the fact that the IEEE floating point format approximates base-2 logarithm. For example, you can get the approximate logarithm of 32-bit single precision floating point number by translating its binary representation as an integer, scaling it by 2 − 23, and removing a bias of 127, i.e.
$x_\text{int} \cdot 2^{-23} - 127 \approx \log_2(x).$
For example, 1.0 is represented by a hexadecimal number 0x3F800000, which would represent $1065353216 = 127 \cdot 2^{23}$ if taken as an integer. Using the formula above you get $1065353216 \cdot 2^{-23} - 127 = 0$, as expected from log 2(1.0). In a similar fashion you get 0.5 from 1.5 (0x3FC00000).
To get the square root, divide the logarithm by 2 and convert the value back. The following program demonstrates the idea. Note that the exponent's lowest bit is intentionally allowed to propagate into the mantissa. One way to justify the steps in this program is to assume b is the exponent bias and n is the number of explicitly stored bits in the mantissa and then show that
$(((x_\text{int} / 2^n - b) / 2) + b) \cdot 2^n = (x_\text{int} - 2^n) / 2 + ((b + 1) / 2) \cdot 2^n.$
float sqrt_approx(float z)
{
union
{
int tmp;
float f;
} u;
u.f = z;
/*
* To justify the following code, prove that
*
* ((((val_int / 2^m) - b) / 2) + b) * 2^m = ((val_int - 2^m) / 2) + ((b + 1) / 2) * 2^m)
*
* where
*
* val_int = u.tmp
* b = exponent bias
* m = number of mantissa bits
*
* .
*/
u.tmp -= 1 << 23; /* Subtract 2^m. */
u.tmp >>= 1; /* Divide by 2. */
u.tmp += 1 << 29; /* Add ((b + 1) / 2) * 2^m. */
return u.f;
}
The three mathematical operations forming the core of the above function can be expressed in a single line. An additional adjustment can be added to reduce the maximum relative error. So, the three operations, not including the cast, can be rewritten as
u.tmp = (1 << 29) + (u.tmp >> 1) - (1 << 22) + a;
where a is a bias for adjusting the approximation errors. For example, with a = 0 the results are accurate for even powers of 2 (e.g., 1.0), but for other numbers the results will be slightly too big (e.g.,1.5 for 2.0 instead of 1.414... with 6% error). With m = -0x4C000, the errors are between about -3.5% and 3.5%.
If the approximation is to be used for an initial guess for Newton's method to the equation (1 / x2) − S = 0, then the reciprocal form shown in the following section is preferred.
### Reciprocal of the square root
A variant of the above routine is included below, which can be used to compute the reciprocal of the square root, i.e., $x^{-{1\over2}}$ instead, was written by Greg Walsh, and implemented into SGI Indigo by Gary Tarolli.[9][10] The integer-shift approximation produced a relative error of less than 4%, and the error dropped further to 0.15% with one iteration of Newton's method on the following line.[11] In computer graphics it is a very efficient way to normalize a vector.
float invSqrt(float x)
{
float xhalf = 0.5f*x;
union
{
float x;
int i;
} u;
u.x = x;
u.i = 0x5f3759df - (u.i >> 1);
x = u.x * (1.5f - xhalf * u.x * u.x);
return x;
}
Some VLSI hardware implements inverse square root using a second degree polynomial estimation followed by a Goldschmidt iteration (see division (digital)).[12]
## Negative or complex square
If S < 0, then its principal square root is
$\sqrt {S} = \sqrt {\vert S \vert} \, \, i \,.$
If S = a+bi where a and b are real and b ≠ 0, then its principal square root is
$\sqrt {S} = \sqrt{\frac{\vert S \vert + a}{2}} \, + \, \sgn (b) \sqrt{\frac{\vert S \vert - a}{2}} \, \, i \,.$
This can be verified by squaring the root.[13][14] Here
$\vert S \vert = \sqrt{a^2 + b^2}$
is the modulus of S. The principal square root of a complex number is defined to be the root with the non-negative real part.
## Notes
1. ^ There is no direct evidence showing how the Babylonians computed square roots, although there are informed conjectures. (Square root of 2#Notes gives a summary and references.)
2. ^ Heath, Thomas (1921). A History of Greek Mathematics, Vol. 2. Oxford: Clarendon Press. pp. 323–324.
3. ^ Fast integer square root by Mr. Woo's abacus algorithm
4. ^ Integer Square Root function
5. ^ Vedic Mathematics: Sixteen Simple Mathematical Formulae from the Vedas, by Swami Sankaracarya (1880-1960), Motilal Banarsidass Indological Publishers and Booksellers, Varanasi, India, 1965; reprinted in Delhi, India, 1975, 1978. 367 pages.
6. ^ M. V. Wilkes, D. J. Wheeler and S. Gill, "The Preparation of Programs for an Electronic Digital Computer", Addison-Wesley, 1951.
7. ^ M. Campbell-Kelly, "Origin of Computing", Scientific American, September 2009.
8. ^ J. C. Gower, "A Note on an Iterative Method for Root Extraction", The Computer Journal 1(3):142–143, 1958.
9. ^ Rys (2006-11-29). "Origin of Quake3's Fast InvSqrt()". Beyond3D. Retrieved 2008-04-19.
10. ^ Rys (2006-12-19). "Origin of Quake3's Fast InvSqrt() - Part Two". Beyond3D. Retrieved 2008-04-19.
11. ^ Fast Inverse Square Root by Chris Lomont
12. ^ "High-Speed Double-Precision Computation of Reciprocal, Division, Square Root and Inverse Square Root" by José-Alejandro Piñeiro and Javier Díaz Bruguera 2002 (abstract)
13. ^ Abramowitz, Miltonn; Stegun, Irene A. (1964). Handbook of mathematical functions with formulas, graphs, and mathematical tables. Courier Dover Publications. p. 17. ISBN 0-486-61272-4. , Section 3.7.26, p. 17
14. ^ Cooke, Roger (2008). Classical algebra: its nature, origins, and uses. John Wiley and Sons. p. 59. ISBN 0-470-25952-3. , Extract: page 59
Wikimedia Foundation. 2010.
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# +26 Lesson 4 1 Segments And Midpoints
## Lesson 4.1: Segments and Midpoints
### Introduction
In this lesson, we will explore the concept of segments and midpoints in geometry. Understanding segments and midpoints is crucial for solving geometric problems and constructing various shapes. By the end of this lesson, you will have a solid grasp of these concepts and be able to apply them to real-world situations.
### What is a Segment?
A segment is a part of a line that consists of two endpoints and all the points between them. It can be thought of as a line segment with a specific length. The length of a segment is the distance between its endpoints, which can be calculated using the distance formula.
### How to Identify Segments
To identify a segment, you need to locate its two endpoints. Endpoints are specific points on a line or curve that mark the beginning and end of a segment. Once you have identified the endpoints, you can draw a line between them to represent the segment.
### Types of Segments
There are several types of segments that you may encounter in geometry:
• Line segment: a straight path between two points.
• Ray: a part of a line that has one endpoint and extends infinitely in one direction.
• Open segment: a segment that does not include its endpoints.
• Closed segment: a segment that includes its endpoints.
### What is a Midpoint?
A midpoint is the point that divides a segment into two equal parts. It is located exactly halfway between the two endpoints of the segment. The midpoint divides the segment into two congruent segments, meaning they have the same length.
### How to Find the Midpoint
To find the midpoint of a segment, you can use the midpoint formula. The formula states that the x-coordinate of the midpoint is the average of the x-coordinates of the endpoints, and the y-coordinate of the midpoint is the average of the y-coordinates of the endpoints.
### Properties of Segments and Midpoints
Segments and midpoints have several interesting properties:
• Segment Addition Postulate: If three points A, B, and C are collinear, then AB + BC = AC.
• Midpoint Theorem: If M is the midpoint of segment AB, then AM = MB.
• Segment Bisector: A line, ray, or segment that divides a segment into two congruent parts.
• Perpendicular Bisector: A line, ray, or segment that intersects a segment at its midpoint and forms right angles with it.
### Applications of Segments and Midpoints
The concepts of segments and midpoints have various real-world applications:
• Construction: Architects and engineers use segments and midpoints to accurately measure and construct buildings and structures.
• Navigation: GPS systems and map applications use segments and midpoints to calculate distances and provide directions.
• Surveying: Land surveyors use segments and midpoints to divide land into plots and determine property boundaries.
• Art and Design: Artists and designers use segments and midpoints to create balanced and visually appealing compositions.
### Examples and Problem Solving
Let's work through a few examples to solidify our understanding of segments and midpoints:
1. Find the length of the line segment AB with endpoints A(3, 4) and B(8, 9).
2. Given that M is the midpoint of segment AB with endpoints A(2, 5) and B(6, 1), find the coordinates of M.
3. If N is the midpoint of segment PQ and PN = 8, find the length of segment PQ.
### Conclusion
Segments and midpoints are fundamental concepts in geometry that allow us to understand and analyze shapes and structures. By mastering these concepts, you will be equipped with the necessary tools to solve geometric problems and apply your knowledge to various real-world scenarios. Practice identifying segments, finding midpoints, and applying the properties of segments and midpoints to enhance your geometric skills.
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In this lesson, some interesting properties of quadrilaterals, trapezoids, and angle bisectors of triangles will be explored. Each of these will be proven using congruence and similarity.
### Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
## Investigating Properties of a Quadrilateral
In the net, a quadrilateral, the segments divide the sides into eight congruent segments.
• Use the measuring tool to investigate how the segments divide each other inside the quadrilateral.
• Explore what happens when the vertices are moved!
Example
## Investigating the Midpoints of a Quadrilateral
The Triangle Midsegment Theorem gives a relationship between a midsegment and a side of a triangle. There too, is an exciting result for quadrilaterals, formed by the midpoints of the sides of a quadrilateral. Illustrated in the diagram are and which are midpoints of the sides of the quadrilateral
Show that is a parallelogram, and that and bisect each other.
### Hint
Draw a diagonal in quadrilateral and focus on the two triangles.
### Parallelogram
Draw diagonal of quadrilateral and focus on the two triangles and
According to the Triangle Midsegment Theorem, both and are parallel to the diagonal and they are half the length of That means these midsegments are parallel to each other, and they have the same length.
These relationships can be plotted on the diagram.
Similarly, and are also parallel and have the same length.
By definition, when the opposite sides of a quadrilateral are parallel, then it is a parallelogram. Therefore, the quadrilateral is a parallelogram.
### Bisecting Diagonals
To show that the diagonals and bisect each other, focus on two of the triangles formed by these diagonals.
These triangles contain the following properties.
Claim Justification
Proved previously
Alternate Interior Angles Theorem
Alternate Interior Angles Theorem
These claims can be shown in the diagram.
It can be seen that triangles and have two pairs of congruent angles, and the included sides are also congruent. According to the Angle-Side-Angle (ASA) Congruence Theorem, the triangles are congruent.
Corresponding parts of congruent triangles are congruent.
This completes the proof that and bisect each other.
Example
## Investigating Properties of a General Trapezoid
The following example discusses a property of a general trapezoid. On the diagram is a trapezoid and is parallel to the bases through the intersection of the diagonals.
Show that is the midpoint of
### Hint
Look for similar triangles.
### Solution
There are several pairs of similar triangles on the diagram. Using the scale factors of the similarity transformations between these triangles, the length of and can be expressed in terms of the length of the bases and Here is the outline of a possible approach.
• Step 1: Investigate triangles and
• Step 2: Investigate triangles and to express the length of in terms of the length of the bases and
• Step 3: Investigate triangles and to express the length of in terms of the length of the bases and
• Step 4: Compare the expressions for the length of and the length of
Here are the details.
### Step 1
Focus on the triangles formed by the bases and the diagonals of the trapezoid.
The following table contains some information about these triangles.
Claim Justification
Alternate Interior Angles Theorem
Alternate Interior Angles Theorem
This can be indicated on the diagram.
According to the Angle-Angle (AA) Similarity Theorem, this means that the two triangles are similar, so the corresponding sides are proportional.
### Step 2
Focus now on the left side of the trapezoid.
Since is parallel to a dilated image of is The scale factor can be written in two different ways.
In this equality can be replaced by
According to Step 1, the ratio is the same as the ratio Substituting this in the equation above gives an equation that can be solved for
Solve for
This calculation gave an expression for the length of in terms of the lengths of the bases of the trapezoid.
### Step 3
On the right of the trapezoid there are two more similar triangles, and
The scale factor of the dilation between these two triangles can be written two different ways.
As in Step 2, this equation can also be solved for In this case the second equation of the result of Step 1 can be used.
Simplify right-hand side
Solve for
### Step 4
The results of Step 2 and Step 3 show that the expressions for and are identical, so these two segments are congruent.
This completes the proof that is the midpoint of Move the point on the base of the trapezoid to see an illustration of the claim.
Explore
## Investigating Angle Bisectors of Triangles
The next part of this lesson focuses on triangles. The diagram shows a triangle with one of its angle bisectors drawn. Move the vertices of the triangle and find a relationship between the displayed segment measures.
Discussion
## Triangle Angle Bisector Theorem
The relationship stated in the following theorem can be checked on the previous applet for different triangles.
The angle bisector of an interior angle of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
In the figure, if is an angle bisector, then the following equation holds true.
### Proof
In consider the angle bisector that divides into two congruent angles. Let and be these congruent angles.
By the Parallel Postulate, a parallel line to can be drawn through Additionally, if is extended, it will intersect this line. Let be their point of intersection.
Let be the alternate interior angle to formed at Also, let be the corresponding angle to formed at
By the Corresponding Angles Theorem, is congruent to Remember that it is also known that is congruent to By the Transitive Property of Congruence, and are congruent angles.
Additionally, by the Alternate Interior Angles Theorem, is congruent to Using the Transitive Property of Congruence one more time, it can be said that and are also congruent angles.
This can be shown in the diagram.
Note that is divided by which is parallel to Therefore, by the Triangle Proportionality Theorem, divides the other two sides of this triangle proportionally.
The Converse Isosceles Triangle Theorem states that if two angles in a triangle are congruent, the sides opposite them are congruent. This means that is congruent to Therefore, by the definition of congruent segments, they have the same length. can be substituted for in the above proportion.
Pop Quiz
## Practice the Triangle Angle Bisector Theorem
Find the measurement of the segment as indicated in the applet.
Example
## Solving Problems With the Triangle Angle Bisector Theorem
In segment is the angle bisector of the right angle at and is perpendicular to The length of the legs and are 5 and 12, respectively.
Find the length of Write the answer in exact form as a fraction.
### Hint
Start with finding the length of the hypotenuse and the length of
### Solution
Mark the lengths which were given in the prompt onto the diagram.
The length of the hypotenuse of the triangle can be found using the Pythagorean Theorem.
As can be seen on the diagram, the length of is the sum of and When rearranged, this can be written as Furthermore, let represent as it is unknown. Then, since was just found, it can be said that
According to the Angle Bisector Theorem, an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle. In this case, that would indicate proportionality between the ratio of the two segments of the hypotenuse and the ratio of the altitude and base.
Substituting the expressions from the diagram gives an equation that can be solved for which represents the length of
Solve for
This gives the lengths of the segments on the hypotenuse.
Recall that the task is to find the length of Using similar logic as before, if is used to represent the length of then
Since both and are perpendicular to these segments are parallel. According to the Triangle Proportionality Theorem, this means that divides sides and proportionally.
Substituting the expressions from the diagram gives an equation that can be solved for which represents the length of
Solve for
The length of is
Discussion
## Converse Triangle Angle Bisector Theorem
According to the Angle Bisector Theorem, an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle. The converse of this statement is also true.
If a segment from a vertex of a triangle divides the opposite side in proportion to the sides meeting at then the segment is an angle bisector of the triangle.
Based on the figure, the following conditional statement holds true.
This theorem is the converse of the Triangle Angle Bisector Theorem.
### Proof
Consider and the segment that connects vertex with its opposite side. Let be the point of intersection of the segment from and Now, will be extended to a point such that equals Additionally, a segment from to will be constructed.
It is given that divides the opposite side in proportion to the sides meeting at
Because is equal to by the Substitution Property of Equality can be substituted for in the proportion.
Therefore, is a segment between two sides of that divides and proportionally. Then, by the Converse Triangle Proportionality Theorem it can be stated that is parallel to
It is seen that and are corresponding angles. By the Corresponding Angles Theorem, is congruent to Furthermore, and are alternate interior angles, and by the Alternate Interior Angles Theorem these two angles are also congruent.
Because by the Isosceles Triangle Theorem is congruent to
Since and are both congruent to by the Transitive Property of Congruence, it follows that and are congruent angles.
By the same property, since and are both congruent to they are congruent angles.
Therefore, by the definition of an angle bisector is an angle bisector of the triangle.
Example
## Solving Problems With the Converse Triangle Angle Bisector Theorem
On the diagram, the markers on line are equidistant, the circles are centered at and at and is the point of intersection of the circles.
Show that bisects
### Hint
Express the lengths of the line segments in terms of the distance between consecutive markers.
### Solution
The lengths of some line segments can be expressed in terms of the distance between consecutive markers.
Claim Justification
By counting the markers
By counting the markers
Segment is a radius of the circle centered at Counting markers shows that the radius of this circle is units long.
Segment is a radius of the circle centered at Counting markers shows that the radius of this circle is units long.
These measurements can be indicated on the diagram.
The ratio of two sides of the triangle can be simplified.
That equals the ratio of the two segments on the third side of the triangle.
According to the converse of the Angle Bisector Theorem, this relationship between the proportions means that bisects
Closure
## Analyzing Properties of a Quadrilateral
At the beginning of this lesson, the following net was investigated. Recall that the segments drawn inside the net, a quadrilateral, cut the sides into congruent parts.
Show that all segments are cut by the others into congruent parts.
### Hint
Use the knowledge that the segments connecting the midpoints of opposite sides of any quadrilateral bisect each other.
### Solution
In the first exercise of this lesson, it was proved that the segments connecting the midpoints of opposite sides of any quadrilateral bisect each other.
This statement can be used several times considering different quadrilaterals to prove the claim that all segments are cut by the others into congruent parts.
### Step 1
First, consider only the midpoints of the original quadrilateral and the segments connecting these midpoints, They intersect at the mark.
As it can be seen, these segments bisect each other.
### Step 2
The segments connecting the midpoints of opposite sides cut the original quadrilateral into two smaller quadrilaterals. Focus on the segments connecting the midpoints of opposite sides of this smaller quadrilateral.
As shown, these segments also bisect each other.
### Step 3
Next, focus on another quadrilateral that differs from the previous two smaller ones. Again, take note of the segments connecting the midpoints of opposite sides.
These segments also bisect each other.
### Step 4
Now, consider a quarter of the original quadrilateral. Mark the segments that connect the midpoints of its opposite sides.
Again, these bisect each other.
### Step 5
Using similar logic as in the previous steps, intersection points in similar positions can also be marked.
This shows that when considering every second segment in both directions, the segments cut each other into congruent pieces.
### Step 6
The other intersection points can also be found as midpoints of certain segments. Therefore, continuing this process shows that all segments are cut by the others into congruent pieces.
### Extra
The solution above was based on finding midpoints again and again. Similar arguments can be used to prove the claim for , segments as well. The following applet can be used to check that a similar statement is also true when the number of segments on the sides is not a power of
While this claim will not be proved here, it is a worthwhile concept to consider.
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# How to Round Numbers to the Nearest: Methods and Examples
Learn how to round numbers (decimals, fractions, integers) to the nearest number with methods and examples.
By Brian Dean- publish on march 08, 2023
When dealing with numbers in everyday life, it is often necessary to round them to the nearest whole number, tenth, hundredth, or even thousandth. Rounding numbers can make calculations easier and provide more manageable values. However, rounding can also introduce errors if not done correctly. In this article, we will explore various methods of rounding numbers to the nearest value and provide examples to illustrate their use.
## Understanding Rounding
Before diving into the methods of rounding numbers, it is essential to understand the concept of rounding. Rounding involves approximating a number to the nearest value based on a specific criterion, such as the nearest whole number or nearest tenth. When rounding, we consider the digit in the place we are rounding to, as well as the digit immediately to the right of it. If the digit to the right is 5 or greater, we round up; otherwise, we round down.
## Rounding to the Nearest Whole Number
The simplest method of rounding is rounding to the nearest whole number. To round a number to the nearest whole number, we look at the digit in the ones place. If it is 5 or greater, we round up to the next whole number. If it is less than 5, we round down to the previous whole number. For example:
4.6 rounds to 5
2.2 rounds to 2
9.9 rounds to 10
### Rounding to the Nearest Tenth
Rounding to the nearest tenth involves rounding a number to the nearest tenth decimal place. To round to the nearest tenth, we look at the digit in the tenths place. If it is 5 or greater, we round up to the next tenth. If it is less than 5, we round down to the previous tenth. For example:
4.63 rounds to 4.6
2.29 rounds to 2.3
9.95 rounds to 10.0
### Rounding to the Nearest Hundredth
Rounding to the nearest hundredth involves rounding a number to the nearest hundredth decimal place. To round to the nearest hundredth, we look at the digit in the hundredths place. If it is 5 or greater, we round up to the next hundredth. If it is less than 5, we round down to the previous hundredth. For example:
4.627 rounds to 4.63
2.294 rounds to 2.29
9.954 rounds to 9.95
### Rounding to the Nearest Thousandth
Rounding to the nearest thousandth involves rounding a number to the nearest thousandth decimal place. To round to the nearest thousandth, we look at the digit in the thousandths place. If it is 5 or greater, we round up to the next thousandth. If it is less than 5, we round down to the previous thousandth. For example:
4.6274 rounds to 4.627
2.2945 rounds to 2.295
9.9543 rounds to 9.954
### Rounding to Significant Figures
Rounding to significant figures involves rounding a number to a specified number of significant figures. Significant figures are the digits in a number that carry meaning, including all non-zero digits and any zeros between them. To round to a specific number of significant figures, we look at the digit in the place corresponding to the desired significant figure. If the digit to the right is 5 or greater, we round up; otherwise, we round down. For example:
To round 123.456 to two significant figures, we look at the digit in the tens place, which is 5. Since 5 is greater than or equal to 5, we round up to 120. If we were rounding to three significant figures, we would look at the digit in the ones place, which is 6. Since 6 is greater than or equal to 5, we round up to 123.
## Using Different Rounding Methods
Different rounding methods can be used depending on the situation and the desired level of precision. For example, when dealing with money, it may be necessary to round to the nearest cent or even the nearest hundredth of a cent. In contrast, when dealing with measurements, it may be sufficient to round to the nearest inch or millimeter.
It is essential to understand which rounding method to use in a particular situation to avoid introducing errors in calculations or measurements. For example, rounding a measurement to the nearest millimeter when a higher level of precision is required can result in incorrect values.
## Conclusion
Rounding numbers is a fundamental mathematical concept used in various applications. Understanding how to round numbers to the nearest whole number, tenth, hundredth, or thousandth can help make calculations and measurements more manageable. It is crucial to choose the appropriate rounding method based on the situation and the level of precision required to avoid errors.
## FAQs
### Is rounding numbers always necessary?
Rounding numbers is not always necessary, but it can make calculations and measurements more manageable.
### What is the purpose of rounding numbers?
The purpose of rounding numbers is to approximate a number to the nearest value based on a specific criterion.
### Can rounding numbers introduce errors?
Rounding numbers can introduce errors if not done correctly or if the wrong rounding method is used.
### What is the simplest method of rounding?
The simplest method of rounding is rounding to the nearest whole number.
### What is the difference between rounding to the nearest tenth and rounding to the nearest hundredth?
Rounding to the nearest tenth involves rounding a number to the nearest tenth decimal place, while rounding to the nearest hundredth involves rounding a number to the nearest hundredth decimal place.
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# Guess The Weight | Cat Rat Dog Puzzle
Guess The Weight Cat, Rat, Dog Puzzle is one of the most viral puzzles on Instagram, Facebook, Pinterest, and Twitter. In this puzzle, there is a picture with cartoons of Cat, Rat, and Dog. We have to guess the weight. We can also say that we have to find the weight of Cat Rat and Dog. So many people think. It is too difficult to solve this puzzle.
But this is a very easy puzzle. We can solve this puzzle in a few minutes. Let us reveal the answer with a proper explanation.
## Cat Rat Dog Puzzle Answer
The answer to Cat Rat Dog Puzzle is 27 KG
First, let us see what we have —
In the first case, We have Rat + Cat = 10
In the second case, We have Rat + Dog = 20
In the third case, We have Cat + Dog = 24
In the last case, We have Cat + Rat + Dog = ?
Now, We have to find the last case value.
Therefore, We will add First Case, Second Case, and Third Case
(Rat + Cat) + (Rat + Dog) + (Cat + Dog) = 10 + 20 + 24
Then, Rat + Rat + Dog + Dog + Cat + Cat = 54
So, 2Rat + 2Dog + 2Cat = 54
Then, 2(Rat + Dog + Cat) = 54
Now, 2 is multiplying on Left Hand Side. We will send 2 on Right Hand Side and divide 54 from 2
Finally, Rat + Dog + Cat = 54/2
Then, Rat + Dog + Cat = 27
We can also write this
Cat + Rat + Dog = 27 KG
So, The answer to Guess the weight puzzle is 27 KG.
## Related Video
This is a related video of the Cat Rat Dog Puzzle.
## FAQ
Ans : 27 Kg
Ans : 27 Kg
### What is the meaning of “ Guess The Weight ” in Hindi?
Ans : वजन का अनुमान लगाएं ( Hindi translation of Guess the weight )
Share on:
## 7 thoughts on “Guess The Weight | Cat Rat Dog Puzzle”
1. 27 KG
2. Let R, C and D be Rat, Cat and Dog respectively.
R+C=10KG — (1)
R+D=20KG — (2)
C+D=24KG — (3)
Solving simultaneously, Subtracting eqn2 from eqn3
C-R=4
C=R+4 — (4)
Substituting eqn4 in eqn1
R+R+4=10
2R=6
R=3
Substituting R=3 in eqn2
3+D=20
D=17
Substituting D=17 in eqn3
C+17=24
C=7
Hence,
R+D+C = 3+17+7 = 27KG
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33 5.3 Angular Acceleration
Angular acceleration is denoted by the Greek letter alpha (α). Angular acceleration represents the time rate of change in angular velocity. Another way to think about this is how quickly something is speeding up or slowing down.
acceleration (α) = Δω/Δt
The units are rads/s2 or degrees/s2. When velocity is increasing, the acceleration is in the same direction of rotation to increase the velocity. When the velocity is decreasing, there has to be acceleration in the opposite direction of travel acting as a break to decrease the velocity.
Acceleration has a direction. If the object is moving in the counterclockwise direction (+) and gaining velocity, acceleration is positive. If velocity is decreased, acceleration is negative. If the object is moving in the clockwise direction (-) and gaining velocity, acceleration is negative. If velocity is decreased, acceleration is positive.
Example 1: Calculating the Angular Acceleration and Deceleration of a Bike Wheel
Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in rad/s2. (b) If she now slams on the brakes, causing an angular acceleration of -87.3 rad/s2, how long does it take the wheel to stop?
Strategy for (a)
The angular acceleration can be found directly from its definition in $\boldsymbol{\alpha=\frac{\Delta\omega}{\Delta{t}}}$ because the final angular velocity and time are given. We see that Δω is 250 rpm and Δt is 5.00 s.
Solution for (a)
Entering known information into the definition of angular acceleration, we get
$\begin{array}{lcl} \boldsymbol{\alpha} & \boldsymbol{=} & \boldsymbol{\frac{\Delta\omega}{\Delta{t}}} \\ {} & \boldsymbol{=} & \boldsymbol{\frac{250\textbf{ rpm}}{5.00\textbf{ s.}}} \end{array}$
Because Δω is in revolutions per minute (rpm) and we want the standard units of rad/s2 for angular acceleration, we need to convert Δω from rpm to rad/s:
$\begin{array}{lcl} \boldsymbol{\Delta\omega} & \boldsymbol{=} & \boldsymbol{250\frac{\textbf{rev}}{\textbf{min}}\cdotp\frac{2\pi\textbf{ rad}}{\textbf{rev}}\cdotp\frac{1\textbf{ min}}{60\textbf{ sec}}} \\ {} & \boldsymbol{=} & \boldsymbol{26.2\textbf{ rads.}} \end{array}$
Entering this quantity into the expression for $\boldsymbol{\alpha}$, we get
$\begin{array}{lcl} \boldsymbol{\alpha} & \boldsymbol{=} & \boldsymbol{\frac{\Delta\omega}{\Delta{t}}} \\ {} & \boldsymbol{=} & \boldsymbol{\frac{26.2\textbf{ rad/s}}{5.00\textbf{ s}}} \\ {} & \boldsymbol{=} & \boldsymbol{5.24\textbf{ rad/s.}^2} \end{array}$
Strategy for (b)
In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular acceleration and solving for Δt, yielding
$\boldsymbol{\Delta{t}\:=}\boldsymbol{\frac{\Delta\omega}{\alpha}}.$
Solution for (b)
Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that Δω is -26.2 rad/s, and $\boldsymbol{\alpha}$ is given to be -87.3 rad/s2. Thus,
$\begin{array}{lcl} \boldsymbol{\Delta{t}} & \boldsymbol{=} & \boldsymbol{\frac{-26.2\textbf{ rad/s}}{-87.3\textbf{ rad/s}^2}} \\ {} & \boldsymbol{=} & \boldsymbol{0.300\textbf{ s.}} \end{array}$
Discussion
Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall—the velocity change is large in a short time interval.
If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in Figure 2. Thus, linear acceleration is called tangential acceleration at.
Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know that in circular motion centripetal acceleration, ac, refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular motion experiences centripetal acceleration, as seen in Figure 3. Thus, at and ac are perpendicular and independent of one another. Tangential acceleration at is directly related to the angular acceleration $\boldsymbol{\alpha}$ and is linked to an increase or decrease in the velocity, but not its direction.
Now we can find the exact relationship between linear acceleration at and angular acceleration $\boldsymbol{\alpha}$. Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined to be
$\boldsymbol{a_{\textbf{t}}\:=}\boldsymbol{\frac{\Delta{v}}{\Delta{t}}.}$
For circular motion, note that v=rω, so that
$\boldsymbol{a_{\textbf{t}}\:=}\boldsymbol{\frac{\Delta(r\omega)}{\Delta{t}}.}$
The radius r is constant for circular motion, and so Δ(rω)=r(Δω). Thus,
$\boldsymbol{a_{\textbf{t}}=r}\boldsymbol{\frac{\Delta\omega}{\Delta{t}}.}$
By definition, $\boldsymbol{\alpha=\frac{\Delta\omega}{\Delta{t}}}.$ Thus,
$\boldsymbol{a_{\textbf{t}}=r\alpha},$
or
$\boldsymbol{\alpha\:=}\boldsymbol{\frac{a_{\textbf{t}}}{r}.}$
These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a bike’s drive wheels, the greater the acceleration of the bike. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration $\boldsymbol{\alpha}$.
So far, we have defined three rotational quantities— θ, ω, and $\boldsymbol{\alpha}$. These quantities are analogous to the translational quantities x, v, and a. Table 1 displays rotational quantities, the analogous translational quantities, and the relationships between them.
Rotational Translational Relationship
$\boldsymbol{\theta}$ $\boldsymbol{x}$ $\boldsymbol{\theta=\frac{x}{r}}$
$\boldsymbol{\omega}$ $\boldsymbol{v}$ $\boldsymbol{\omega=\frac{v}{r}}$
$\boldsymbol{\alpha}$ $\boldsymbol{a}$ $\boldsymbol{\alpha=\frac{a_{\textbf{t}}}{r}}$
Table 1. Rotational and Translational Quantities.
Section Summary
• Uniform circular motion is the motion with a constant angular velocity $\boldsymbol{\omega=\frac{\Delta\theta}{\Delta{t}}}.$
• In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is $\boldsymbol{\alpha=\frac{\Delta\omega}{\Delta{t}}}.$
• Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as $\boldsymbol{a_{\textbf{t}}=\frac{\Delta{v}}{\Delta{t}}}.$
• For circular motion, note that v=rω, so that
$\boldsymbol{a_{\textbf{t}}\:=}\boldsymbol{\frac{\Delta(r\omega)}{\Delta{t}}}.$
• The radius r is constant for circular motion, and so Δ(rω)=rΔω. Thus,
$\boldsymbol{a_{\textbf{t}}=r}\boldsymbol{\frac{\Delta\omega}{\Delta{t}}}.$
• By definition, Δω/Δt=$\boldsymbol{\alpha}$. Thus,
$\boldsymbol{a_{\textbf{t}}=r\alpha}$
or
$\boldsymbol{\alpha=}\boldsymbol{\frac{a_{\textbf{t}}}{r}}.$
Problems & Exercises
4: Unreasonable Results
You are told that a basketball player spins the ball with an angular acceleration of 100 rad/s2. (a) What is the ball’s final angular velocity if the ball starts from rest and the acceleration lasts 2.00 s? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent?
Glossary
angular acceleration
the rate of change of angular velocity with time
change in angular velocity
the difference between final and initial values of angular velocity
tangential acceleration
the acceleration in a direction tangent to the circle at the point of interest in circular motion
Solutions
Problems & Exercises
4:
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# Chapter 3: Functions – Exercise 3.2
## Functions– Exercise 3.2 – Q.1
We have,
f(x) = x2 - 3x + 4
Now,
f(2x + 1) = (2x + 1)2 - 3(2x + 1) + 4
= 4x2 + 1 + 4x - 6x - 3 + 4
= 4x2 - 2x + 2
It is given that
f(x) = f(2x + 1)
⟹ x2 — 3x + 4 = 4x2 - 2x + 2
⟹ 0 = 4x2 - x2 - 2x + 3x + 2 - 4
⟹ 3x2 + x - 2 = 0
⟹ 3x2 + 3x - 2x - 2 = 0
⟹ 3x (x + 1) - 2 (x + 1) = 0
⟹ (x +1) (3x -2) = 0
⟹ x + 1 = 0 or 3x - 2 = 0
⟹ x = -1 or x = 2/3
## Functions– Exercise 3.2 – Q.2
We have,
f(x) = (x – a)2 (x – b)2
Now,
f(a + b) = (a + b – a)2 (a + b – b)2
= b2a2
⟹ f(a + 6) = a2b2
We have,
Hence, proved
## Functions– Exercise 3.2 – Q.4
We have,
∴ f[f(x)] = x Hence, proved
## Functions– Exercise 3.2 – Q.5
We have,
∴ f[f(x)] = x Hence, proved
We have,
## Functions– Exercise 3.2 – Q.7
We have,
Adding equation (i) and equation (ii), we get
## Functions– Exercise 3.2 – Q.8
We have,
∴ f(tan θ) = sin 2θ Hence, proved.
We have,
## Functions– Exercise 3.2 – Q.10
We have,
f(x) = (a - xn)1/n, a > 0
Now,
f{f(x)} = f(a - xn)1/n
= [a-{(a - xn)1/n}n]1/n
= [a - (a - xn)]1/n
= [a - a + xn]1/n
= (xn)1/n
= (x)n×1/n
= x
∴ f{f(x)} = x Hence, proved.
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# 2.1 Linear functions (Page 8/17)
Page 8 / 17
## Using tabular form to write an equation for a linear function
[link] relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.
w , number of weeks 0 2 4 6 P(w) , number of rats 1000 1080 1160 1240
We can see from the table that the initial value for the number of rats is 1000, so $b=1000.$
Rather than solving for $m,$ we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.
$P\left(w\right)=40w+1000$
If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using $\left(2,1080\right)$ and $\left(6,1240\right)$
Is the initial value always provided in a table of values like [link] ?
No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into $f\left(x\right)=mx+b,$ and solve for $b.$
A new plant food was introduced to a young tree to test its effect on the height of the tree. [link] shows the height of the tree, in feet, $x$ months since the measurements began. Write a linear function, $H\left(x\right),$ where $x$ is the number of months since the start of the experiment.
$x$ 0 2 4 8 12 $H\left(x\right)$ 12.5 13.5 14.5 16.5 18.5
$H\left(x\right)=0.5x+12.5$
Access this online resource for additional instruction and practice with linear functions.
## Key equations
slope-intercept form of a line $f\left(x\right)=mx+b$ slope point-slope form of a line $y-{y}_{1}=m\left(x-{x}_{1}\right)$
## Key concepts
• The ordered pairs given by a linear function represent points on a line.
• Linear functions can be represented in words, function notation, tabular form, and graphical form. See [link] .
• The rate of change of a linear function is also known as the slope.
• An equation in the slope-intercept form of a line includes the slope and the initial value of the function.
• The initial value, or y -intercept, is the output value when the input of a linear function is zero. It is the y -value of the point at which the line crosses the y -axis.
• An increasing linear function results in a graph that slants upward from left to right and has a positive slope.
• A decreasing linear function results in a graph that slants downward from left to right and has a negative slope.
• A constant linear function results in a graph that is a horizontal line.
• Analyzing the slope within the context of a problem indicates whether a linear function is increasing, decreasing, or constant. See [link] .
• The slope of a linear function can be calculated by dividing the difference between y -values by the difference in corresponding x -values of any two points on the line. See [link] and [link] .
• The slope and initial value can be determined given a graph or any two points on the line.
• One type of function notation is the slope-intercept form of an equation.
• The point-slope form is useful for finding a linear equation when given the slope of a line and one point. See [link] .
• The point-slope form is also convenient for finding a linear equation when given two points through which a line passes. See [link] .
• The equation for a linear function can be written if the slope $m$ and initial value $b$ are known. See [link] , [link] , and [link] .
• A linear function can be used to solve real-world problems. See [link] and [link] .
• A linear function can be written from tabular form. See [link] .
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
|
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2019 | May-Jun | (P2-9709/21) | Q#3
Question
Find the equation of the normal to the curve
at the point (3, 1).
Solution
We are given equation of the curve as;
We are required to find the equation of normal to the curve at the point (3, 1).
To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).
We need coordinates of the point on the curve (and normal).
We are already given coordinates of a point on the curve (and normal) as (3, 1).
Next we need to find slope of normal at (3, 1) in order to write its equation.
If a line is normal to the curve , then product of their slopes and at that point (where line is normal to the curve) is;
Therefore, if we can find slope of the curve C at point (3, 1) then we can find slope of the normal to the curve at this point.
Therefore, we first find expression for gradient of the curve.
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Therefore;
Rule for differentiation of is:
First we differentiate .
Rule for differentiation of is:
Next we differentiate .
To find from an implicit equation, differentiate each term with respect to , using the chain rule to differentiate any function as .
We utilize Product Rule to differentiate .
If and are functions of , and if , then;
If , then;
Let and , then;
Rule for differentiation of is:
For ;
Therefore;
Next we differentiate .
Rule for differentiation of is:
Lastly,we differentiate .
Rule for differentiation of is:
Hence;
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by
substituting x-coordinates of that point in the expression for gradient of the curve;
Therefore;
Hence;
With coordinates of a point on the tangent and its slope in hand, we can write equation of the tangent.
Point-Slope form of the equation of the line is;
|
# Examples of Recursion
### Towers of Hanoi
#### What are the Towers of Hanoi?
Is the end of the world near? At a monastery in the city of Hanoi, Vietnam, a group of monks has made it their life's work to solve the Towers problem, known due to its location as the Towers of Hanoi. Legend has it that the world will end when the monks finally solve the puzzle.
The puzzle is this. In the monastery are 3 pegs made of diamond. Resting on these pegs are 64 discs made of solid gold. None of the 64 discs are the same size; in fact, disc 1 is slightly larger in diameter than disc 2, which is slightly larger in diameter than disc 3, which is slightly larger in diameter than disc 4, etc. The initial configuration of the puzzle has all 64 discs piled in order of size on the first peg with the largest disc on the bottom.
Figure %: Example Towers Configuration
To solve the puzzle, all 64 discs must be moved to the third peg. Easy you say? The problem is that due to the fragility of the gold, you are not allowed to rest a larger disc on top of a smaller one, and only one disc may be removed from the pegs at any one time.
Figure %: Moving a single disc
There are many ways to solve this problem. The first is purely guess and check. Those who fear the end of the world may be hoping that the monks are using this approach. However, they're not; in fact, the monks all know the exact sequence of moves to solve the problem.
#### Solving Towers of Hanoi
##### One Disc
Let's simplify the problem to clarify our thinking. Let's imagine the Towers of Hanoi problem with only one disc.
Figure %: Towers Problem with 1 Disc
How do we solve this problem? Simple. We just move the disc on the first pole to the third pole.
Figure %: Towers Solution for 1 disc
##### Two Discs
Let's make the problem slightly bigger. Imagine two discs.
Figure %: Towers Problem with 2 Discs
How do we solve this problem? Simple, again.
1. Use the one disc solution to move the top disc to the intermediate pole.
2. Use the one disc solution to move the bottom disc to the final pole.
3. Use the one disc solution to move the top disc to the final pole.
Figure %: Towers Solution for 2 Discs
##### Three Discs
How about with three discs?
1. Use the two disc solution to move the top discs to the intermediate pole.
2. Use the one disc solution to move the bottom disc to the final pole.
3. Use the two disc solution to move the top discs to the final pole.
##### N Discs
So how about with N Discs?
1. Use the N - 1 disc solution to move the top discs to the intermediate pole.
2. Use the one disc solution to move the bottom disc to the final pole.
3. Use the N - 1 disc solution to move the top discs to the final pole.
Figure %: Towers Solution for N discs
And, voila! A recursive solution to solving the Towers of Hanoi! Note that the problem can be solved iteratively as well; however it makes much more intuitive sense recursively.
#### Coding Towers of Hanoi
Now that we know how to solve an n -disc problem, let's turn this into an algorithm we can use.
If there is one disc, then we move 1 disc from the source pole to the destination pole. Otherwise, we move n - 1 discs from the source pole to the temporary pole, we move 1 disc from the source pole to the destination pole, and we finish by moving the n - 1 discs from the temporary pole to the destination pole.
```
void TOH(int n, int p1, int p2, int p3)
{
if (n == 1) printf("Move top disc from %d to %d.\n", p1, p2);
else {
TOH(n-1, p1, p3, p2);
printf("Move top disc from %d to %d.\n", p1, p2);
TOH(n-1, p3, p2, p1);
}
}
```
Of course, we can simplify this to the following:
```
void TOH(int n, int p1, int p2, int p3)
{
if (n>1) TOH(n-1, p1, p3, p2);
printf("Move top disc from %d to %d.\n", p1, p2);
if (n>1) TOH(n-1, p3, p2, p1);
}
```
Pretty cool, huh? This example shows the power of recursion to turn what seems like a hard and intricate problem into something much more simple that can be solved in three lines of code.
#### Will the World End Tomorrow?
Actually, the whole story of the monks is just a legend. In fact, it isn't even an old legend. The story was created in 1883 by a mathematician named Edouard Lucas. He had invented an eight disc, three tower puzzle, and created the legend in order to sell his product.
That being said, what if the story were true? Should we be worried about the world ending when the monks solve the puzzle? After all, they live in the 21st century, too, and have access to the same information about recursion that we have.
Luckily, just as mathematics helps us solve the puzzle, it also helps prove that our grandchildren will still have a world to live in. In order to figure out how long it will take the monks to solve the puzzle, we need to write a recurrence relation, a recursive formula for describing the size of a recursive problem. Let's call our recursive formula T(n), where n is the number of discs.
As seen above, the base case for the Towers of Hanoi problem is when n is 1. This is when the monks just have to move one disc from one pole to another. So T(1) = 1 . For the recursive case where n! = 1 , we need a more complicated formula. The monks, in the recursive case, follow a three step procedure. They move n - 1 discs, then they move 1 disc, and then they move n - 1 discs. So T(n) = T(n - 1) + T(1) + T(n - 1) = 2T(n - 1) + 1 .
Now we know that to solve a Towers problem with n discs takes T(n) = 2T(n - 1) + 1 steps. It would be nice if we had a closed-form solution to this recurrence so that we could figure out exactly how long it will take. A closed form solution is a formula without recursion, meaning we can simply plug in numbers and get our answer.
Let's plug in some sizes and solve the Towers problem with those sizes to see if we can find a closed-form solution.
• T(1) = 1
• T(2) = 3
• T(3) = 7
• T(4) = 15
• T(5) = 31
• T(6) = 63
• T(7) = 127
• T(8) = 255
• T(9) = 511
• T(10) = 1023
Notice a pattern here? T(n) = 2n - 1 . To prove to yourself that this is true, try modifying your TOH code to count the number of disc moves. Create a global variable count, run your modified TOH code, and then print out count.
Now we can easily compute how long it would take the monks to solve their 64-disc Towers problem. 264 - 1 is approximately 18.45x1018 (note that if you actually tried to run the TOH code on your computer it would most likely take a very, very long time). If the monks could move a disc in a millisecond (an incredibly rate considering the size and weight of each disc), it would take them approximately 584,600,000 years to solve the puzzle. It appears the world is safe for now.
## Take a Study Break
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# Domain and Range Relations
“Domain” are numbers that you give to the function. “Range” means the numbers that the function gives back to you. The first thing one should know about domain and range is that domain is plotted on X-coordinates while Range is plotted on Y-coordinates. Even in cases where you have to find the domain and range from the existing graphs, the rule is the same. You can always think of functions as an existing machine where you put your number and you get a different number out. Some machines can take the numbers that you are giving them and some machines don’t. Some machines will just take out any number from the lot beyond your imagination while some are only known to produce specific numbers.
Domain and Range Mapping Diagrams
If there are two existing non-empty sets X and Y, and we have a relation R defined between them as a subset of all the cumulative elements X x Y, the subset is then called as the result of the ‘relation’ existing between the elements of the first set and the elements of the second set.
Find the Domain and Range of the Relation
In the figure given above, there is a relation from set X to Y. All the rectangular blocks are “related” to the triangular blocks with R A relation may have finite or infinite ordered pairs. If we take a relation from set X to Y, it is commonly referred to as ‘relation on X.’ The maximum number of relations that can be defined from set X (having m elements) to Y (having n elements) is equal to 2mn.
Domains are generally easy to find. Finding ranges sometimes can be complicated. In many textbooks, the word “image” is used rather than “range” and “pre-image” for the domain. The reason for that being “range” is used in two different ways in mathematics. It usually means image, the set of values that the function takes on. It is also used for “co-domain.”
Domain and Range Relations Examples
For example, think of graph y=3x+5. You can take any number that you want for x, so the domain for this number will consist of all the real numbers including negative infinity to positive infinity. But, if your function is y=x^2, your domain is still the set of real numbers. But for any real number, whose square results in a positive number. So, the range of the function would be non negative real numbers including zero.
The range is simply the set of all second components of the ordered pairs, with duplicates ignored, so {2; 1; 5; 10}. That eliminates A, B, and D. The domain is the set of all sets that you are allowed to choose from for the first component of the ordered pairs in an itemized list of the relational pairs. In short, if you think of a domain as “all possible inputs” and range as “all possible outputs,” you’ll have the right idea.
Domain and Range of each Relation
In all the branches of mathematics, an element in the domain is usually associated with another element of a co-domain. A co-domain, here is a set of all the allowed values while the range is the set of use values for the second component in the ordered pairs. So, here the range turns out to be a subset of the co-domain.
But for any relation, there are no absolute restrictions on how many elements can exist in a co-domain and one element in the domain can be related to – sometimes they can be zero. Therefore, any element of the domain may or may not be related to any element in the co-domain and similarly, there might not be a relational pair with the value of the first component. In such cases, you wouldn’t be knowing the element is in the domain just with a glance at the relational pairs.
How to Denote Domain Range Relation
Domain and range are usually denoted using interval notation, which could look like any of these for both the domain and range, but for this example, let’s just show domain: (smallest value in domain, the largest value in the domain) OR (smallest value in domain, the largest value in the domain) OR (smallest value in domain, the largest value in the domain) OR (smallest number in the domain, largest number in the domain).
|
#### Need Help?
Get in touch with us
# Numerical Patterns
Sep 19, 2022
## Key Concepts
• Numerical sequences
• Multiplication sequences
• Numerical patterns from the word problems
### Introduction
In this chapter, we will learn about numerical sequences, addition sequences, multiplication sequences, and word problems.
## Numerical sequences
Numerical sequence is a list of numbers arranged by a pattern or rule.
The pattern or rule of a sequence can be adding a number or subtracting a number.
Numerical sequence is one of the best concepts for the students to sharpen their numerical reasoning capability.
Example: Find the missing number
4, 11, 18, 25, _____.
Solution: In this example, the difference between each number is 7.
The rule for this sequence is to add 7 each time.
The missing number in the sequence: 25 + 7 = 32
Addition sequence is a sequence in which a particular number is added each time.
Example 1:
A shopkeeper sells balls. His sales keep increasing by 2 balls per day.
Using the numbers, the sequence is:
The rule to go from one number to the next number is ‘add 2’.
Example 2:
Maria has \$5. She will save \$10 each week. Steepen has \$8 and will also save \$10 each week.
Maria uses the rule “add 10” to create tables to see how much each will have saved after each week.
What relationship do you notice between the corresponding terms?
At the end of each week, Steepen has saved \$3 more than Maria.
Example 3:
Patterns on the Calendar: Possible increasing patterns include
• In each row numbers increase by 1.
• In each column numbers increase by 7.
• On the diagonal from left to right, numbers increase by 8.
• On the diagonal from right to left, numbers increase by 6.
### Multiplication sequences
A sequence made by multiplying by the same value each time.
Example 1: 2, 4, 8, 16, 32, 64, 128, 256, …
(Each number is 2 times the number before it)
Example 2: 3, 6, 12, 24, 48, …
(Each number is 2 times the number before it)
This below image indicates multiplication sequence
Example 3:
Jack is preparing for an exam. Each week he reads 20 pages in text book, 40 pages in work book.
He created a table to record his reading. Find the total number of pages that he will read from the text book and the work book after 5 weeks? Can you identify any relationship between reading the text book and the work book?
Solution:
Since jack reads 20 pages each week. Add 20 to find the next term for the total pages he reads.
And add 40 to find each term in the pattern for the total number of pages he reads.
Compare the corresponding terms in the patterns:
20×2 = 40
40×2 = 80
60×2 = 120
80×2= 160
100×2 = 200
So, the total number of work book pages is always 2 times the total number of text book pages.
Example 4:
Sheila and Patrick are making a table to compare gallons, quarts, and pints. Use the rule “add 4” to complete the column for the number of quarts. Then, use the rule “add 8” to complete the column for the number of pints.
Solution:
Compare the corresponding terms in the patterns:
4×2 = 8
8×2 = 16
12×2 = 24
16×2= 32
20×2 = 40
24×2= 48
So, the total number of pints is always 2 times the total number of quarts.
## Exercise
1. Consider the following sequence. What number should come next?
2. 1, 3, 5, 7…. b) 3, 6, 11, 18……
3. Consider the following sequence. What is the missing number?
4. 26, 21, …, 11, 6. b) 19, …, 31, 37
5. Find the missing terms in the following sequence:
8, _, 16, _, 24, 28, 32
• If hickory tree continues to grow 1 feet each year, how tall will the tree be after 10 years?
• Riya’s weight increases by 5 kgs every year. What will be Riya’s weight after 5 years?
• Find the next number in the sequence:
• 4, 8, 16, 32, ___ b) 5, 10, 20, 40, __
• What should come next:
• 2, 4, 9, 11, 16, … b) 30, 28, 25, 21, 16, …
• Each week Stem lifts weights twice and runs 4 times. Stem uses the rules “add 2” and “add 4” to complete the table. What relationship do you notice between the corresponding terms?
• Bridget is organizing her books and putting them on shelves. She put 72 books on the first shelf, 81 books on the second shelf, 90 books on the third shelf, and 99 books on the fourth shelf. If this pattern continues, how many books will Bridget put on the fifth shelf?
• Marcy is sorting pencils into boxes. She put 2 pencils in the first box, 4 pencils in the second box, 8 pencils in the third box, and 16 pencils in the fourth box. If this pattern continues, how many pencils will Marcy put in the fifth box?
### What have we learned
• Numerical sequences
• Multiplication sequences
• Numerical patterns from the word problems
### Concept Map
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
|
# A-level Mathematics/Edexcel/Decision 2/Assignment Problems
Assignment Problems (AP) (also known as Allocation Problems, although they will be known as the former throughout this article) are a common form of problem. A distinguishing feature of AP is that one agent is assigned to one and only one task. Examples in real-life situations could be:
• Assigning contracts
• Assigning Sales Personnel to Territories
An example of such a problem would be three lorries, A, B and C, and three loads that need to be transported, 1, 2 and 3. An example arrangement would be A → 1, B → 2 and C → 3. This means that Lorry A transports Load 1, Lorry B transports Load 2 and Lorry C transports Load 3.
## Balanced Problems
A balanced AP is one in which:
(number of agents) = (number of tasks)
Table 1
1 2 3
A 15 10 12
B 10 7 11
C 15 6 9
Consider again our lorry and load example. In this example, the number of agents (lorries) was 3 and the number of tasks (loads to be transported) is also equal to 3. Thus, it is a balanced problem. Initially, one will consider only balanced problems and how they are to be solved.
Upon expansion of the earlier problem, it was revealed that the cost for each lorry to transport each load is different. These are represented in Table 1.
The manager of the fleet of lorries wishes to minimise to total cost to complete all three jobs. How would one allocate the lorries to the loads? The table of numbers given is called the cost matrix of the problem.
### Formulation as a Linear Programming Problem
A Linear Programming Problem can easily be developed for AP. One would define $x$ij (where agent i is assigned task j) where:
$x$ij=1, when agent i is completing task j
$x$ij=0, when agent i does not complete task j
Thus, one can obtain the contraints for the Linear Programming Problem by considering each agent/task in turn:
As each agent is assigned to exactly one job, we can deduce:
for agent A: $x$A1 + $x$A2 + $x$A3 = 1
for agent B: $x$B1 + $x$B2 + $x$B3 = 1
for agent C: $x$C1 + $x$C2 + $x$C3 = 1
Similarly, since each job is assigned to exactly one agent, one can deduce:
for job 1: $x$A1 + $x$B1 + $x$C1 = 1
for job 2: $x$A2 + $x$B2 + $x$C2 = 1
for job 3: $x$A3 + $x$B3 + $x$C3 = 1
If we apply this to the aforementioned example, one can see that for lorry A, it can only carry one of loads 1, 2 or 3. So, $x$A1 could = 1, with $x$A2 and $x$A3 being zero, or $x$A2 = 1, with $x$A1 and $x$A3 being zero, etc.
The obtain the objective function for this example, we must consider the total costs.
For lorry A, costs will be: 15$x$A1 $+$ 10$x$A2 $+$ 12$x$A3
For lorry B, costs will be: 10$x$B1 $+$ 7$x$B2 $+$ 11$x$B3
For lorry C, costs will be: 15$x$C1 $+$ 6$x$C2 $+$ 9$x$C3
The means, therefore, that the objection function, Z, for this example is to minimise Z where:
Z = 15$x$A1 $+$ 10$x$A2 $+$ 12$x$A3 $+$ 10$x$B1 $+$ 7$x$B2 $+$ 11$x$B3 $+$ 15$x$C1 $+$ 6$x$C2 $+$ 9$x$C3
Finally, this means that one can conclude the LPP to be:
Minimise Z, where Z = 15$x$A1 $+$ 10$x$A2 $+$ 12$x$A3 $+$ 10$x$B1 $+$ 7$x$B2 $+$ 11$x$B3 $+$ 15$x$C1 $+$ 6$x$C2 $+$ 9$x$C3, subject to:
$x$A1 + $x$A2 + $x$A3 = 1
$x$B1 + $x$B2 + $x$B3 = 1
$x$C1 + $x$C2 + $x$C3 = 1
$x$A1 + $x$B1 + $x$C1 = 1
$x$A2 + $x$B2 + $x$C2 = 1
$x$A3 + $x$B3 + $x$C3 = 1
## Solution of the Balanced Assignment Problem
An algorithm was developed by Harold Kuhn in the 1950s, and was named the Hungarian Algorithm. Its name is a reference to Dénes Kőnig and Jenő Egerváry, a pair of prominent Hungarian Mathematicians whose techniques are applied in this algorithm.
### The Hungarian Algorithm
For the Hungarian Algorithm to work, the assignment problem must meet the following criteria:
• Each worker must be assigned, as we previously saw, to exactly one job.
• Each job must be assigned to exactly one worker.
• In the cost matrix, a constant can be added or subtracted from all cost values in any row or column without having any effect on the optimal assignments.
There are three stages in the implementation of the Hungarian Algorithm:
• Find the opportunity cost matrix
• Test for an optimal assignment. If one can be made, use it and stop.
• If an optimal assignment cannot be made, revise the opportunity cost matrix and return to the previous step.
To demonstrate these three steps, let us consider again the lorry example:
#### Finding the Opportunity Cost Matrix
The Opportunity Cost Matrix (OCM) is essentially a modified version of our initial Cost Matrix. In order to change the Cost Matrix into the OCM, one must carry out two stages.
1. Carry out row reduction
2. Carry out column reduction
Row reduction is the operation in which the smallest number in each row is subtracted from every number in that row. With our earlier example:
Cost Matrix
1 2 3
A 15 10 12
B 10 7 11
C 15 6 9
Cost Matrix with Row Minima
1 2 3 Row Minimum
A 15 10 12 10
B 10 7 11 7
C 15 6 9 6
Cost Matrix after Row Reduction
1 2 3
A 5 0 2
B 3 0 4
C 9 0 3
Next, one completes the column reduction. This is the same as row reduction, only with the smallest number in each column being subtracted from each value within that column. Thus:
Cost Matrix after Row Reduction
1 2 3
A 5 0 2
B 3 0 4
C 9 0 3
Cost Matrix with Column Minima
1 2 3
A 5 0 2
B 3 0 4
C 9 0 3
Column Minimum 3 0 2
Cost Matrix after Column Reduction
1 2 3
A 2 0 0
B 0 0 2
C 6 0 1
Thus, the completed Opportunity Cost Matrix is:
Opportunity Cost Matrix
1 2 3
A 2 0 0
B 0 0 2
C 6 0 1
#### Interpreting The Opportunity Cost Matrix
Since the aim of an assignment problem is to assign exactly one agent to exactly one task, one must match each agent up to the task which best minimises overall cost. After completing the row reduction and column reduction, it is logical to assume that the values that remain in the table are the minimum possible. Similarly, since the values in the table are the cost-values, the lowest values in the table are, surely, those of lowest cost. In order to find a solution, one must select one zero from each row and one zero from each column. However, one cannot select more than one value from any row or column, as it will result in one agent/task being assigned twice, which cannot occur in an Assignment Problem.
Again, let's consider the earlier example. If we were to select one zero from each column/row without any doubles occurring, the only possible solution is (with selected costs marked by asterices):
Opportunity Cost Matrix
1 2 3
A 2 0 0*
B 0* 0 2
C 6 0* 1
If we then assign this solution to our original cost matrix, one can see the following:
Cost Matrix
1 2 3
A 15 10 12*
B 10* 7 11
C 15 6* 9
Thus, the final assignment is:
A → 3 (cost 12)
B → 1 (cost 10)
C → 2 (cost 6)
Z = 12 + 10 + 6 = 28
It is interesting to see that, despite this being the cheapest solution, it doesn't actually utilise all of the lowest values within the table.
### Optimality of a Solution
An important issue with AP is the optimality of solution - when one obtains a solution, how can one be sure it truly is the best solution available? With the Hungarian algorithm, there is a fairly straightforward test for optimality that concerns drawing lines through the zeros in the Opportunity Cost Matrix. If the number of lines (horizontal and vertical only) is equal to the number of rows/columns, the solution obtained is optimal. If, however, the number of lines needed does not equal the number of rows/columns, the solution is not optimal, and can be improved. This is a somewhat perplexing technique, but one that is straightforward to implement and clear to interpret.
Consider the following OCM:
Opportunity Cost Matrix
1 2 3
A 1 0 2
B 5 0 0
C 6 3 1
It is fairly easy to see that it would require only two lines to draw through all of the zeros - one could use a horizontal line through row B, and a vertical line through column 2, for instance. 2≠3 (the number of rows/columns), thus an optimal solution cannot be made.
Consider, on the other hand, our lorry example. Is the solution that we obtained optimal?
Opportunity Cost Matrix
1 2 3
A 2 0 0
B 0 0 2
C 6 0 1
One can see that, in order to strike through each zero, one would need three lines (for instance, one through column 2, one through row B and one through row A). 3 = 3, thus we have an optimal solution that cannot be improved upon.
#### Revision of the Opportunity Cost Matrix
In the event of achieving a non-optimal assignment, one must revise the OCM. There are several steps in this process:
1. Check the costs uncovered by the drawn lines
2. Identify the smallest uncovered value
3. Subtract this value from all of the uncovered costs
4. Add this value to any costs that are at the intersection of any two drawn lines
Take our non-optimal solution from earlier. The smallest uncovered value was one. Subtracting one from each uncovered value gives:
Opportunity Cost Matrix
1 2 3
A 0 0 1
B 5 0 0
C 5 3 0
Adding one to the point of intersection of the two lines (in this example, one could have drawn them through row B and column 2, therefore the point of intersection is cell B2) gives:
Opportunity Cost Matrix
1 2 3
A 0 0 1
B 5 1 0
C 5 3 0
Again, only two lines (this time, they must be through row A and column 3) are needed. They intersect at cell A3. The smallest uncovered value is, again, 1. So, subtracting one from each uncovered cell and adding one to cell A3 gives:
Opportunity Cost Matrix
1 2 3
A 0 0 2
B 4 0 0
C 4 2 0
Finally, the minimum number of lines needed to draw through all of the zeros is 3, which is the number of rows/columns. This means any assignment made will be optimal.
## Solutions of Unbalanced Assignment Problems
Consider the following example:
A taxi company has four drivers (A, B, C, D), each currently at different locations. Three customers (1, 2, 3) have called the taxi company and need to be driven to their destinations. The total distance that would be travelled, by the driver, is represented in the cost matrix below:
Cost Matrix
1 2 3
A 11 19 17
B 21 15 13
C 15 18 21
D 18 15 17
The Taxi Company wishes to minimise the total distance travelled by its drivers. Find an optimal solution to this Assignment Problem.
Solution:
Initially, one must add a dummy column, in order to make this a 'balanced' problem. Let this column be called "Row 4". Since there isn't actually a fourth customer, all of the costs in this matrix will be zero. This will mean that the otherwise most expensive taxi in the matrix will result in being assigned the dummy - in other words, it won't be used in the problem.
First, let's carry out row and column reduction:
Cost Matrix with Row Minima
1 2 3 4 Row Minimum
A 11 19 17 0 0
C 15 18 21 0 0
D 18 15 17 0 0
... with Column Minima
1 2 3 4
A 11 19 17 0
B 21 15 13 0
C 15 18 21 0
D 18 15 17 0
Column Minimum 11 15 13 0
Opportunity Cost Matrix
1 2 3 4
A 0 4 4 0
B 10 0 0 0
C 4 3 8 0
D 7 4 0 0
As one can see, all of the zeros in the Opportunity Cost Matrix can only be struck through using 4 lines (for instance, 1 line vertically through Column 4, 3 lines horizontally through Rows A, B & D). Thus, we've already arrived at the optimal solution because 4 lines can be used to mark through the entire matrix. The only possible solution is:
A → 1
B → 3
C → 2
D → 4*
Z = 11 + 13 + 18 = 42
• Of course, customer 4 is in fact a 'dummy' - i.e., D doesn't actually transport any customer. Thus, the only three drivers needed are drivers A, B and C. Driver A will transport Passenger 1, Driver B will transport Passenger 3 and Driver C will transport Passenger 2.
## Maximisation Problems
All of the examples that have been considered so far have been Minimisation problems - in other words, they've been ones in which the aim of the problem was to minimise a cost. Assignment problems can, however, be used for the opposite reason - to maximise a gain or a profit. Consider the example of ice-cream seller. They know that, in certain towns, they will sell more icecream than others. The day, too, changes how much ice-cream s/he sells - they only drive the van on Monday, Wednesday and Saturday (M, W and S respectively). Taking into account that the ice-cream van can only visit one town at a time, which town should the ice-cream van visit on which day, in order to maximise his/her profit.
The following cost-matrix represents the earnings (in £100s) for each town (A, B, C) on each day (M, W, S):
Cost Matrix
M W S
A 10 7 11
B 10 12 12
C 9 13 11
If one were to carry out row and column reduction, as per usual, and then to come to a solution, it would be a solution that gave the minimum profit that the ice-cream van could make. In order to find the maximum profit, we first identify the largest value in the matrix. In this case, it is 13. We then subtract each number from 13 and assign the new number to the cell the original number was in. In other words, cell A1 (of current value 10) becomes $13 - 10 = 3$.
After completing this process for the whole matrix, we achieve the following:
Cost Matrix
M W S
A 3 6 2
B 3 1 1
C 4 0 2
We can now carry out row and column reduction, as we usually would. This, eventually, results in the following Opportunity Cost Matrix:
Opportunity Cost Matrix
M W S
A 0 6 1
B 0 1 0
C 1 0 1
Since the smallest number of lines that can be used to strike-through each of the zeros in this matrix is 3 (which is equal to the number of rows/columns), an optimal assignment can be made. The only possible assignment is:
A → M
B → S
C → W
On the cost matrix, this gives us:
Cost Matrix
M W S
A 10* 7 11
B 10 12 12*
C 9 13* 11
This means that:
On Monday, the van will visit town A.
On Wednesday, the van will visit town B.
On Saturday, the van will visit town C.
The total profit gives:
$Z = 10 + 12 + 13 = 35$
Thus the maximum total profit of the ice-cream van is £3500.
|
RS Aggarwal Solutions: Trigonometric Identities (Exercise 8B)
# RS Aggarwal Solutions: Trigonometric Identities (Exercise 8B) | Mathematics (Maths) Class 10 PDF Download
``` Page 1
1. co n a b m s si si s n, a b n co
2 2 2 2
m n a b If and prove that,
Sol:
We have
=
+
=
=
=
=
Hence,
2. n x a s e c bta n , y a b s e c ta If and prove that
2 2 2 2
x y a b .
Sol:
We have
=
Page 2
1. co n a b m s si si s n, a b n co
2 2 2 2
m n a b If and prove that,
Sol:
We have
=
+
=
=
=
=
Hence,
2. n x a s e c bta n , y a b s e c ta If and prove that
2 2 2 2
x y a b .
Sol:
We have
=
-
=
=
=
=
Hence,
3. If sin cos 1
x y
a b
and cos sin 1,
x y
a b
prove that
2 2
2 2
2.
x y
a b
Sol:
We have
Squaring both side, we have:
Again,
Now, adding (i) and (ii), we get:
4. If sec tan mand sec tan , n show that 1. m n
Sol:
We have
Again,
Now, multiplying (i) and (ii), we get:
Page 3
1. co n a b m s si si s n, a b n co
2 2 2 2
m n a b If and prove that,
Sol:
We have
=
+
=
=
=
=
Hence,
2. n x a s e c bta n , y a b s e c ta If and prove that
2 2 2 2
x y a b .
Sol:
We have
=
-
=
=
=
=
Hence,
3. If sin cos 1
x y
a b
and cos sin 1,
x y
a b
prove that
2 2
2 2
2.
x y
a b
Sol:
We have
Squaring both side, we have:
Again,
Now, adding (i) and (ii), we get:
4. If sec tan mand sec tan , n show that 1. m n
Sol:
We have
Again,
Now, multiplying (i) and (ii), we get:
=
=
=1
5. If cos cot e c mand sec cot , c o n show that 1. m n
Sol:
We have
=
=
6. If
3
cos x a and
3
sin , y b prove that
2 2
3 3
1.
x y
a b
Sol:
=
=
=
=
= 1
7. If tan sin mand tan sin , n prove that
2
2 2
16 . m n m n
Sol:
=
=
=
=
=
Page 4
1. co n a b m s si si s n, a b n co
2 2 2 2
m n a b If and prove that,
Sol:
We have
=
+
=
=
=
=
Hence,
2. n x a s e c bta n , y a b s e c ta If and prove that
2 2 2 2
x y a b .
Sol:
We have
=
-
=
=
=
=
Hence,
3. If sin cos 1
x y
a b
and cos sin 1,
x y
a b
prove that
2 2
2 2
2.
x y
a b
Sol:
We have
Squaring both side, we have:
Again,
Now, adding (i) and (ii), we get:
4. If sec tan mand sec tan , n show that 1. m n
Sol:
We have
Again,
Now, multiplying (i) and (ii), we get:
=
=
=1
5. If cos cot e c mand sec cot , c o n show that 1. m n
Sol:
We have
=
=
6. If
3
cos x a and
3
sin , y b prove that
2 2
3 3
1.
x y
a b
Sol:
=
=
=
=
= 1
7. If tan sin mand tan sin , n prove that
2
2 2
16 . m n m n
Sol:
=
=
=
=
=
=
=
=
=
= s
=
=
=
8. If cot tan mand sec cos n prove that
2 2
2 2
3 3
( ) ( ) 1 m n m n
Sol:
Now,
=
=
=
=
=
=
=
=
=
=
=
=
=
Page 5
1. co n a b m s si si s n, a b n co
2 2 2 2
m n a b If and prove that,
Sol:
We have
=
+
=
=
=
=
Hence,
2. n x a s e c bta n , y a b s e c ta If and prove that
2 2 2 2
x y a b .
Sol:
We have
=
-
=
=
=
=
Hence,
3. If sin cos 1
x y
a b
and cos sin 1,
x y
a b
prove that
2 2
2 2
2.
x y
a b
Sol:
We have
Squaring both side, we have:
Again,
Now, adding (i) and (ii), we get:
4. If sec tan mand sec tan , n show that 1. m n
Sol:
We have
Again,
Now, multiplying (i) and (ii), we get:
=
=
=1
5. If cos cot e c mand sec cot , c o n show that 1. m n
Sol:
We have
=
=
6. If
3
cos x a and
3
sin , y b prove that
2 2
3 3
1.
x y
a b
Sol:
=
=
=
=
= 1
7. If tan sin mand tan sin , n prove that
2
2 2
16 . m n m n
Sol:
=
=
=
=
=
=
=
=
=
= s
=
=
=
8. If cot tan mand sec cos n prove that
2 2
2 2
3 3
( ) ( ) 1 m n m n
Sol:
Now,
=
=
=
=
=
=
=
=
=
=
=
=
=
=
= RHS
Hence proved.
9. If
3 3
(cos sin ) (sec cos ) , e c a a n d b prove that
2 2 2 2
( ) 1 a b a b
Sol:
=
=
2
3
1
3
cos
a=
sin
=
=
=
=
=
=
=
=
= RHS
Hence, proved.
10. If 2sin 3cos 2, prove that 3sin 2cos 3.
Sol:
=
=
= 4+9
= 13
```
## Mathematics (Maths) Class 10
120 videos|463 docs|105 tests
## Mathematics (Maths) Class 10
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Question Video: Using Right Triangle Trigonometry to Solve Word Problems | Nagwa Question Video: Using Right Triangle Trigonometry to Solve Word Problems | Nagwa
# Question Video: Using Right Triangle Trigonometry to Solve Word Problems Mathematics • Third Year of Preparatory School
## Join Nagwa Classes
A kite, which is at a perpendicular height of 44 m, is attached to a string inclined at 60° to the horizontal. Find the length of the string accurate to one decimal place.
02:23
### Video Transcript
A kite, which is at a perpendicular height of 44 meters, is attached to a string inclined at 60 degrees to the horizontal. Find the length of the string accurate to one decimal place.
Let’s begin by drawing a sketch of this problem. We have a kite which is attached to a string. This string is inclined at an angle of 60 degrees to the horizontal and the perpendicular height of the kite. So that means the height of the kite that makes a right angle with the horizontal is 44 meters. We can now see that we have a right triangle formed by the horizontal, the vertical, and the string of the kite. We want to calculate the length of the string, so let’s label that as 𝑦 meters. We’re working with a right triangle, so we can approach this problem using trigonometry.
We’ll begin by labeling the three sides of the triangle in relation to the angle of 60 degrees. Next, we’ll recall the acronym SOHCAHTOA to help us decide which trigonometric ratio we need here. The side whose length we know is the opposite, and the side we want to calculate is the hypotenuse. So we’re going to be using the sine ratio. For an angle 𝜃 in a right triangle, this is defined as the length of the opposite divided by the length of the hypotenuse. We can then substitute the values for 𝜃, the opposite and the hypotenuse, into this equation giving sin of 60 degrees equals 44 over 𝑦.
We need to be careful because the unknown appears in the denominator of this fraction. Next, we solve this equation. As 𝑦 appears in the denominator, the first step is to multiply both sides of the equation by 𝑦, which gives 𝑦 sin 60 degrees is equal to 44. Next, we divide both sides of the equation by sin of 60 degrees, giving 𝑦 equals 44 over sin of 60 degrees. And then we evaluate on our calculators, which must be in degree mode, giving 50.806. The question asks for our answer accurate to one decimal place. So we round this value and include the units which are meters. The length of the string to one decimal place is 50.8 meters.
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# How To Calculate Displacement
Long summer days are perfect to go hiking. Let’s say you decide to pay a visit to a friend who lives 3 km away. You both meet and decide to have an ice cream in a store nearby, just 500 m away. After having a good time, you walk back to your friend’s house, and then go back home.
You sure did some exercise that day! Nevertheless, according to physics, your total displacement was… well, zero. This is because the magnitude of your displacement is defined as the difference between your starting and ending points, which in this case were the same: your house. On the other hand, the distance you travelled was indeed 7 km.
Don’t worry. It might sound absurd, but this definition of displacement is quite useful when considering physical phenomena such as body motion, potential energies, etc. Let’s discover why!
## How to calculate displacement
1. Strategically establish a coordinate system, for example, with its origin on the starting point of the trajectory.
2. Determine the coordinates of the starting and ending points of the trajectory: (xi, yi) and (xf, yf), respectively.
3. Calculate the displacement along the x- and y-axes by subtracting each corresponding coordinate pair: dx=xf-xi and dy=yf-yi, respectively.
4. Calculate the magnitude of the displacement using the Pythagorean theorem:
1. Express the direction of the displacement vector through the angle it forms with the x-axis using this equation:
## What is displacement
If you could attach a color smoke bomb to a bird’s tail and let it fly free, you would be able to observe the path it draws in the sky, like the following image shows. Physicists call the pathway followed by moving bodies their trajectory
Trajectories can have infinite shapes, and can be 1-, 2- or 3-dimensional. An example of a 1-dimensional trajectory is that of an object in free fall. All objects on the Earth tend to fall to its center due to the force of gravity. This force acts in a straight line connecting the object’s center of mass and the center of gravity of the Earth. Since the movement occurs along a straight line, we can say it is 1-dimensional.
An example of a 2D trajectory is that of an ant walking around to find food on your kitchen floor. The ant’s position changes in two directions parallel to the floor’s surface. If we use a coordinate system, we can call these directions the x- and the y-axes. Every direction the ant moves can be described with an x- and a y-component. Since our kitchen ant can’t fly, its trajectory is bound to the ground, so it only changes in two possible directions and is therefore 2-dimensional.
Lastly, the bird from our previous example draws a 3-dimensional trajectory in the sky. In this case its position changes in all possible directions. If you think about it in terms of its location on the map, the bird could fly to the north, east, south or west, but also change its altitude. If we use a coordinate system, the x- and y-axes could be used to describe motion along the four cardinal directions. Nevertheless, we would need a third axis, namely the z-axis, to describe the bird’s flying altitude.
In physics, when an object moves in space, we say it covers a certain distance. This is the length of the trajectory followed by the object. To measure the distance our kitchen ant travels, we would have to divide its entire trajectory in very small sections, measure the length of each section, as the following image shows, and then add all of the results together. The more sections we divide the pathway in, the more accurate the calculated distance travelled by the ant will be.
Knowing the distance travelled by a body in motion is very useful to calculate, for example, the instantaneous velocity at any point of the trajectory. Nevertheless, sometimes we only want to know what is the last position of the body after following a certain trajectory. In these cases we consider its displacement, rather than the distance it travelled.
Displacement is defined as a vector, whose magnitude is the distance between the final and initial positions of the moving body along a straight line connecting both points. The resulting vector, usually denoted as s, simply points from the start to the end of the trajectory. In the case of our kitchen ant, the displacement vector would look like this:
## How to calculate displacement
As mentioned in the previous section, the magnitude of the displacement vector is the distance between the starting and ending points in a straight line. To measure it, in our kitchen ant example, we can locate a coordinate system with its origin on the starting point, as the following image shows. This way, that point will have coordinates (0, 0). We can now measure the x- and y-components of the ant’s final position, namely xf and yf.
We now have the coordinates of the ant’s starting and ending points: (0, 0) and (xf, yf), respectively. If you examine the previous picture, and draw a straight line connecting the start and the end of the ant’s trajectory, you will notice two triangles are formed with the red lines. These triangles have a horizontal length of dx=xf-0, and a vertical height of dy=yf-0. They are also right triangles, which means one of its angles is 90°, for example, the one formed between the vertical red line and the x-axis. Finally, their hypotenuse is the magnitude of our displacement vector.
To calculate it, we may use equation 1, which is simply the definition of a right triangle’s hypotenuse or the Pythagorean theorem. This yields distance D=dx2+dy2. Now, remember that the difference between D and s is that the former is the magnitude of the displacement vector, and is therefore a scalar (it does not have a particular direction). To express the displacement vector’s direction, we can define angle in our coordinate system as the one formed between the displacement and the x-axis.
Again, considering the triangle formed by the vector and the coordinate system’s axes, we can calculate its tangent as: tan =dy/dx. Look at the next image for help. In order to solve for the angle , we need to take its inverse tangent or arctan, using equation 2. This way, we have calculated the displacement vector’s magnitude, D, and its direction, using angle .
Exercise 1: Your friend Lucy decides to go and have a cup of coffee in a store near her house. She goes out and walks around the park, as the following image shows.
Calculate the displacement vector (magnitude and direction) of Lucy’s motion around the town. Use the provided grid as help to calculate positions. All numbers are in meters. Keep in mind, Lucy’s trajectory does not start at the origin of the selected coordinate system, so you will have to calculate the horizontal and vertical distances before using equations 1 and 2.
1. What is the magnitude of Lucy’s displacement?
2. What is the angle the displacement vector forms with respect to a horizontal line?
3. What distance did Lucy’s travel?
1. 15,23 m
2. 23,2°
3. 20 m
## What is displacement useful for
In linear motion, the displacement vector is the root that allows us to formally calculate velocity as a vector. In this case, velocity is defined as the first time derivative of the displacement:
Objects certainly do not move along straight lines, but in free trajectories. So, in order to apply this definition of velocity, the actual trajectory is divided into very small straight lines, called differentials. That is why we write ds in equation 3 instead of just s. Consequently, since the distance travelled by the body is divided into differentials and we want to calculate its velocity, we have to divide the time needed to cover said distances into very small periods of time, called time differentials. These are denoted as dt in the previous equation.
In the examination of more advanced physical phenomena, the total change in an object’s position, which is measured through its displacement, is more important than the actual trajectory it follows. This happens, for example, when measuring the magnetic potential energy of a magnet inside a magnetic field.
A magnet produces its own magnetic field around it. Therefore, when it is introduced in an external magnetic field, both interact and forces are generated. This is similar to you putting two magnets close to each other: each magnet’s field interacts with the other’s, and you will probably feel a very strong force acting on both of them. In very general terms, our magnet will tend to move inside an external magnetic field due to said forces.
This is somewhat similar to you skydiving from an airplane. Before you jump, your body will tend to fall to the ground, but it does do so thanks to the aircraft carrying you. Since you have the potential to fall to the ground, which you will eventually use once you jump, physicists describe this as your gravitational potential energy.
In the case of our magnet, we can also associate the potential of moving inside an external magnetic field to a magnetic potential energy. Now, if you want to calculate the potential energy difference from one point to another and see if it increases or decreases as you move the magnet around, you will need to do so considering its displacement. This is because the magnetic potential depends solely on the magnet’s current position —as your gravitational potential depends only on your current altitude—, and not the trajectory it followed to get there in the first place.
A similar situation arises when calculating a charged particle’s electric potential difference when moved inside an electric field. The particle’s potential difference is a function of its displacement, and not of the distance it travels.
If you want to learn more about the difference between distance and displacement, go ahead and check out this cK-12 simulator, where you can explore these concepts further.
### Different Uses and Applications of organic chemistry
The uses and applications of organic chemistry range from life-saving pharmaceutical discoveries to the vibrant colors of fruits and vegetables we see daily. In this
### Ace organic chemistry- tips and tricks, cheat sheets, summary
Mastering organic chemistry takes time and practice. It might seem a daunting task at once if the right approach is not adopted. But in our
### Named reactions of organic chemistry- an overview
This article on organic reactions is a special one in our organic chemistry series. It will guide you through how chemical transformations unlock diversity at
### Organic Spectroscopy
Organic spectroscopy can be used to identify and investigate organic molecules. It deals with the interaction between electromagnetic radiation (EMR) and matter. These waves travel
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### LOCATING THE DECIMAL: MULTIPLICATION & DIVISION
There are plenty of instruments out there that don't have unit rods marked on the beam. The following techniques are one solution to solving the question, "Where should the decimal point go?"
##### MULTIPLICATION:
Here's a method to consider for locating the decimal point in problems of multiplication.
Rule: Simplify both numbers to read x.xxx... Take the net of the decimal point movement in both the multiplicand and in the multiplier. (This yields a number.) In the final analysis, undo by that number by moving in the opposite direction. This is probably best explained in the following examples:
Example 1: 31.68 x 0.00082
Put both numbers in the form x.xxx .....
Move the decimal in the multiplicand one place to the left (call it L1).
Move the decimal in the multiplier four places to the right (call it R4).
This gives : 3.168 x 8.2
Take the net of these : L1 plus R4 = R3.
In the final analysis remember to *undo* R3 by moving the decimal point L3.
Now, look again at the equation : The approximation is 3 x 8 = 24
Undo R3 by moving the decimal point L3. This leaves 0.024
The actual answer to the problem is : 31.68 x 0.00082 = 0.0259776
Example 2: 5326.879 x 0.00000079
Put both numbers in the form x.xxx .....
Think : L3 for the multiplicand and R7 for the multiplier
5.326879 x 7.9
Take the net of these : L3 plus R7 = R4
Remember to "undo" R4 at the end by moving the decimal point L4.
The approx answer is 5 x 8 = 40
Undo R4 by moving the decimal point L4. This leaves 0.0040
The actual answer to to the problem is : 5326.879 x 0.00000079 = 0.004208234
##### DIVISION:
Here's a method to consider for locating the decimal point in problems of division. The method is similar to that done for problems of multiplication in that we simplify both numbers to read x.xxx.... However, it's a little more complicated but once gotten used to it's very easy.
Rule: Simplify both numbers to read x.xxx... Take the net of the decimal point movement in the dividend and the *opposite* of the decimal point movement in the divisor. (This yields a number.) In the final analysis, undo by that number by moving in the opposite direction. Probably best explained by example. :)
Example 1: 0.68 ÷ 390
Put both numbers in the form x.xxx .....
Move the decimal in the dividend one place to the right (call it R1)
Move the decimal in the divisor two places to the left (call it L2)
The result is : 6.8 ÷ 3.9
Take the net of the decimal movement in the dividend and the *opposite* of the movement in the divisor. That is;
R1 plus R2 = R3
6.8 ÷ 3.9 is approximately 7 ÷ 4 = 1.75
Undo R3 by moving the decimal L3 : this equals 0.00175
The actual solution : 0.68 ÷ 390, is 0.00174358
Example 2: 0.073 ÷ 0.0054
Put both numbers in the form x.xxx .....
Think : R2 for the dividend and R3 for the divisor
The result is : 7.3 ÷ 5.4.
Take the net of the decimal movement in the dividend and the *opposite* of the movement in the divisor. That is;
R2 plus L3 = L1
7.3 ÷ 5.4 is approximately 7 ÷ 5 = 1.4
Undo L1 by moving the decimal R1 : this equals 14.00
Actual solution : 0.073 ÷ .0054 = 13.518...
Example 3: 897 ÷ 0.00061
Put both numbers in the form x.xxx .....
Think : L2 for the dividend and R4 for the divisor
8.97 ÷ 6.1
Take the net of the decimal movement in the dividend and the *opposite* of the movement in divisor. That is;
L2 plus L4 = L6
8.97 ÷ 6.1 is approximately 9 ÷ 6 = 1.5
Undo L6 by moving the decimal R6 : this equals 1500000.00
Actual solution : 897 ÷ 0.00061 = 1470491.803
Back to Multiplication
Back to Division
Back to Finding the Decimal
Gary Flom Atlanta Georgia USA
Email
gsflom[at]bellsouth[dot]net
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Readers of this blog should have noticed a trend by now: many of the GRE math concepts we’ve reviewed are likely straight out of your 9th grade algebra textbook. The straightforward nature of these concepts actually make them even more important to review, as relying on existing knowledge can lull you to sleep on test day (not to mention the points it will cost you on the GRE).
Up for today is the least common multiple, more affectionately known by many as simply the LCM.
## The basics of finding the least common multiple
Questions on the GRE won’t be this simple, but to get us started, let’s find the least common multiple of 12, 15 and 21 by rewriting each number into its prime factors:
12: 2 x 2 x 3
15: 3 x 5
21: 3 x 7
After writing out each number’s prime factors, we take the least number of prime numbers required to ‘build’ each number, and then multiply them together. We do not count the overlaps.
As such, we see that the LCM in this case is 2 x 2 x 3 x 5 x 7 = 420 .
Questions in which you’re asked to find the least common multiple are indicative of the task at hand. In the simplest terms, you’re simply trying to identify the smallest group of prime factors that are required to create the numbers in a given set.
Now that we’ve tackled a simple example, let’s review what a LCM question on the GRE might look like.
## Least common multiples on the GRE
Consider the example below:
If the least common multiple of two numbers is 45, and one of the numbers is 9, what is the other number?
A. 3
B. 5
C. 7
D. 9
E. 12
The wording of this question might seem to be a bit confusing, but let’s break this down in a similar fashion to the example we discussed earlier by simply listing out all the prime factors of 45.
45: 3 x 3 x 5
Take a look at the prime factorisation of 45. The question states that the other number is 9, which makes sense now that we see that two of the first three prime factors are 3. That leaves us with 5 as our only remaining factor, and also the answer to our question.
Questions involving least common multiples can seem tricky at times, especially since the people at ETS responsible for writing the GRE do everything they can to create complicated questions from simple concepts. However, applying the basics of how least common multiples work will give you the tools necessary to crack even the toughest questions on the GRE.
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1. Chapter 10 Class 11 Straight Lines
2. Serial order wise
3. Miscellaneous
Transcript
Misc 21 Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0. Let the given parallel lines be Line AB : 9x + 6y − 7 = 0 Line CD : 3x + 2y + 6 = 0 Multiplying equation of line CD by 3 3(3x + 2y + 6) = 3(0) 9x + 6y + 18 = 0 So equation of line CD is 9x + 6y + 18 = 0 Let the third line be PQ As PQ is equidistant from parallel lines AB & CD So, PQ parallel to the line AB & CD We need to find equation of line PQ Since Line PQ is parallel to the line AB & CD ∴ Slope of PQ = Slope of AB Finding slope of line AB 9x + 6y − 7 = 0 6y = 7 − 9x 6y = − 9x + 7 y = ( − 9𝑥 + 7)/6 y = (( − 9)/6)x + 7/9 y = (( − 3)/2)x + 7/9 The above equation of the form y = mx + c where m = slope of line ∴ Slope of AB = m = ( − 3)/2 Now, Slope of PQ = Slope of AB ∴ Slope of PQ = ( − 3)/2 Let equation of line PQ be y = mx + c where m is the slope of PQ So, m = ( − 3)/2 Putting values y = ( − 3)/2x + c 2y = -3x + c 2y + 3x − c = 0 3x + 2y − c = 0 Multiplying by 3 3(3x + 2y − c) = 3(0) 9x + 6y − 3c = 0 Also, given that Line PQ is equidistant from the line AB and line CD Distance between line AB & PQ = Distance between line CD & PQ Distance between line AB & PQ Distance between two parallel lines Ax + By + c1 = 0 & Ax + By + c2 = 0 is d = |𝐶_1 − 𝐶_2 |/√(𝐴^2 + 𝐵^2 ) Line AB: 9x + 6y − 7 = 0 The above equation is of the form Ax + By + c1 = 0 Here A = 9 , B = 6 & c1 = − 7 Distance between parallel line AB & PQ is d = | −3𝑐 − (−7)|/√(〖(9)〗^2 + 〖(6)〗^2 ) d = |−3𝑐 + 7|/√117 Distance between line CD & PQ Line CD : 9x + 6y + 18 = 0 The above equation is of the form Ax + By + c1 = 0 Here A = 9 , B = 6 & c1 = 18 d = |−3𝑐 − 18|/√117 Since, Distance between line AB & PQ = Distance between line CD & P |−3𝑐 + 7|/√117 = |−3𝑐 − 18|/√117 |−3𝑐+7| = |−3𝑐 −18| –3c + 7 = –3c – 18 or –3c + 7 = −(–3c − 18) Taking –3c + 7 = –3c − 18 –3c + 3c + 7 = −18 7 = − 18 not possible Hence , Equation of line PQ is 9x + 6y – 3c = 0 9x + 6y – 3((−11)/6) = 0 9x + 6y + 11/2 = 0 18x + 12y + 11 = 0 Which is required equation
Miscellaneous
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Rate Problems - SPECIAL MATH PROBLEMS - SAT Test Prep
## CHAPTER 9SPECIAL MATH PROBLEMS
### Lesson 4: Rate Problems
What Are Rates?
The word rate comes from the same Latin root as the word ratio. All rates are ratios. The most common type of rate is speed, which is a ratio with respect to time, as in miles per hour or words per minute, but some rates don”t involve time at all, as in miles per gallon. Rate units always haveper in their names: miles per gallon, meters per second, etc. Per, remember, means divided by, and is like the colon (:) or fraction bar in a ratio.
The Rate Pyramid
The name of any rate is equivalent to its formula. For instance, speed is miles per hour
or
Since this formula is similar to the “average” formula, you can make a rate pyramid.
This can be a great tool for solving rate problems. If a problem gives you two of the quantities, just put them in their places in the pyramid, and do the operation between them to find the missing quantity.
Example:
How long will it take a car to travel 20 miles at 60 miles per hour?
Simply fill the quantities into the pyramid: 20 miles goes in the distance spot, and 60 miles an hour goes in the speed spot. Now what? Just do the division the way the diagram says: 20 miles ÷ 60 miles per hour = 1/3 hour.
Whenever you work with formulas, you can check your work by paying attention to units. For instance, the problem above asks how long, so the calculation has to produce a time unit. Check the units in the calculation:
Two-Part Rate Problems
Rate problems are tougher when they involve two parts. When a problem involves, say, two people working together at different rates and times, or a two-part trip, you have to analyze the problem more carefully.
Example:
Toni bicycles to work at a rate of 20 miles per hour, then takes the bus home along the same route at a rate of 40 miles per hour. What is her average speed for the entire trip?
At first glance, it might seem that you can just average the two rates: miles per hour, since she is traveling the same distance at each of the two speeds. But this won”t work, because she isn”t spending the same time at each speed, and that is what”s important. But if that”s true, you might notice that she spends twice as much time going 20 miles per hour as 40 miles per hour (since it”s half as fast), so instead of taking the average of 20 and 40, you can take the average of two 20s and a 40:
miles per hour. Simple! But if that doesn”t make sense to you, think of it this way: Imagine, for simplicity”s sake, that her trip to work is 40 miles. (It doesn”t matter what number you pick, and 40 is an easy number to work with here.) Now the average speed is simply the total distance divided by the total time (as the pyramid says). The total distance, there and back, is 80 miles. The total time is in two parts. Getting to work takes her 40 miles ÷ 20 miles per hour = 2 hours. Getting home takes her 40 miles 40 miles ÷ per hour = 1 hour. So the totaltime of the trip is 3 hours. The average speed, then, must be 80 miles ÷ 3 hours = 26.67 miles per hour!
Concept Review 4: Rate Problems
For each of the following rates, write the formula of the rate and the corresponding “rate pyramid.”
1. Speed is miles per hour.
2. Efficiency is miles per gallon of fuel.
3. Typing speed is pages per minute.
Find the missing quantity, including the units, in each of these rate situations.
4. A train travels for 375 miles at 75 mph.
5. A car that gets 28 miles per gallon uses 4.5 gallons of fuel.
6. Harold can type 600 words in 5 minutes.
7. A landscaper who cuts 1.6 acres of grass per hour cuts an 8-acre lot.
8. A train leaves New York at 1:00 pm, going 50 mph, bound for Philadelphia, which is 90 miles away. If it makes no stops, at what time should it be expected to arrive?
9. Anne can paint a room in 2 hours, and Barbara can paint a room in 3 hours. When they work together, their work rate is the sum of their rates working separately. How long should it take them to paint a room if they work together?
SAT Practice 4: Rate Problems
1. Janice and Edward are editors at a newspaper. Janice can edit 700 words per minute and Edward can edit 500 words per minute. If each page of text contains 800 words, how many pages can they edit, working together, in 20 minutes?
2. Two cars leave the same point simultaneously, going in the same direction along a straight, flat road, one at 35 mph and one at 50 mph. After how many minutes will the cars be 5 miles apart?
3. What is the average speed, in miles per hour, of a sprinter who runs ¼ mile in 45 seconds?
(A) 11.25 mph
(B) 13.5 mph
(C) 20 mph
(D) 22 mph
(E) 25 mph
4. A car travels d miles in t hours and arrives at its destination 3 hours late. At what average speed, in miles per hour, should the car have gone in order to have arrived on time?
5. If , how many hours does it take a train traveling at miles per hour to travel ?
(C) x
6. In three separate 1-mile races, Ellen finishes with times of x minutes, y minutes, and z minutes. What was her average speed, in miles per hour, for all three races?
7. A hare runs at a constant rate of a mph, a tortoise runs at a constant rate of b mph, and . If they race each other for d miles, how many more hours, in terms of a, b, and d, will it take the tortoise to finish than the hare?
8. Sylvia drives 315 miles and arrives at her destination in 9 hours. If she had driven at an average rate that was 10 mph faster than her actual rate, how many hours sooner would she have arrived?
(A) 1.75
(B) 2.00
(C) 2.25
(D) 2.50
(E) 2.75
Concept Review 4
1. Speed = #miles ÷ hours
2. Efficiency =#miles ÷#gallons
3. Typing speed =#pages ÷ #minutes
4. 375 miles ÷ 75 mph = 5 hours for the trip.
5. 28 miles per gallon × 4.5 gallons = 126 miles the car can go before it runs out of fuel.
6. 600 words ÷ 5 minutes = 120 words per minute is Harold”s typing speed.
7. 8 acres ÷ 1.6 acres per hour = 5 hours for the job.
8. 90 miles ÷ 50 mph = 1.8 hours, or 1 hour 48 minutes for the entire trip. At 1 hour and 48 minutes after 1:00 pm, it is 2:48 pm.
9. Anne can paint one room in 2 hours, so her rate is ½ room per hour. Barbara can paint one room in 3 hours, so her rate is room per hour. When they work together, their rate is room per hour. So to paint one room would take one room ÷ room per hour = hours, or 1.2 hours, or 1 hour 12 minutes.
SAT Practice 4
1. 30 Working together, they edit words per minute. Since each page is 800 words, that”s 1,200 words per minute ÷ 800 words per page = 1.5 pages per minute. In 20 minutes, then, they can edit pages.
2. 20 Since the two cars are traveling in the same direction, their relative speed (that is, the speed at which they are moving away from each other) is mph. In other words, they will be 15 miles farther apart each hour. Therefore, the time it takes them to get 5 miles apart is 5 miles ÷ 15 miles per hour = 1/3 hour, which is equivalent to 20 minutes.
3. C Since there are seconds in an hour, hour. Speed = distance ÷ time = ¼ mile ÷ 45/3,600 hour = 3,600/180 = 20 miles per hour.
4. C To arrive on time, the car must take hours for the whole trip. To travel d miles in hours, the car must go miles per hour.
5. E According to the rate pyramid, time = distance ÷ speed = miles ÷ miles per hour = hours. Or you can pick a simple value for x, like 2, and solve numerically.
6. D Speed = miles ÷ hours. Her total time for the three races is minutes, which we must convert to hours by multiplying by the conversion factor (1 hour/60 minutes), which gives us hours. Since her total distance is 3 miles, her overall speed is 3 miles ÷ hours = 180/ miles per hour.
7. B If the hare”s rate is a mph, then he covers d miles in d/a hours. Similarly, the tortoise covers d miles in d/b hours. The difference in their finishing times, then, is .
8. B Sylvia”s speed is 315 miles ÷ 9 hours = 35 mph. If she were to go 10 mph faster, then her speed would be 45 mph, so her time would be 315 miles ÷ 45 mph = 7 hours, which is 2 hours sooner.
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# Math - 5th Grade. two digit by one digit multiplication fact families subtraction with regrouping
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## Transcription
1 Number and Operations Understand division of whole numbers N.MR N.MR N.MR Understand the meaning of division of whole numbers with and without remainders; relate division to and to repeated subtraction. understand relate-application Relate division of whole numbers with remainders relate - application to the form a = bq + r, e.g., 34 5 = 6 r 4, so 5 * = 34; note remainder (4) is less than divisor (5). Write mathematical statements involving division for given situations. Multiply and divide whole numbers write - application ontent division, whole numbers, remainders,, subtraction division, whole numbers, remainders, divisor, quotient, dividend factor, product, divisor, dividend, quotient, equation two digit by one digit multiplication fact families subtraction with regrouping two digit by one digit multiplication fact families addition with regrouping repeated subtraction variables identify the situation as multiplication or division factor x factor = product dividend / divisor = quotient I can recall basic multiplication and division facts. (4 x 3) I can solve basic division problems with or without remainders. (4 / 3 or 12 / 4) I can convert a fraction to a division problem. I understand the relationship between repeated subtraction and division. I can divide a four digit number by a two digit number with or without a remainder. I can recall basic multiplication and division facts. (4 x 3) I can solve basic division problems with or without remainders. (4 / 3 or 12 / 4) I understand the relationship between repeated subtraction and division. I can divide a four digit number by a two digit number with or without a remainder. I can identify if I have to multiply or divide. I can identify the missing part of the equation. I can correctly write a multiplication or division problem. N.FL Multiply a multi-digit number by a two-digit number; recognize and be able to explain common computational errors such as not accounting for place value. multiply - application recognize - knowledge explain - factor, place value, error two digit by one digit multiplication fact families addition and subtraction with regrouping I can recall basic multiplication and division facts. (4 x 3) I can multiply a two digit number by a one digit number. I can explain how to multiply numbers using place value. N.FL Solve applied problems involving multiplication and division of whole numbers. division, whole numbers Skills needed & Sequencing of Skills Apply knowledge of math facts to find product or quotient. xplain thought process used to solve problems I can choose the correct operation to solve the math problem. I can apply knowledge of math facts to solve the problem. I can explain the process I used to solve the problem involving whole numbers. N.FL Divide fluently up to a four-digit number by a twodigit number. divide - application division, estimation, remainders dividend divisor quotient knowledge of basic multiplication and division facts understand relationship between multiplication and division solve basic division problems with or without remainders divide two digit by one digit numbers solve division problems involving decimals to the hundredths place I can fluently solve basic multiplication and division facts. I can solve basic division problems with or without remainders. I can divide a two digit by one digit number. I can solve division problems involving decimals to the hundredths place. I can find quotients of whole numbers with up to four-digit dividends and two-digit divisors. = xtended = ore O:\lementary urriculum\working\lem Math\5th Grade\5th Grade Math Learning Targets.FINAL xlsx 1
2 ontent Find prime factorizations of whole numbers N.MR Find the prime factorization of numbers from 2 through 50, express in exponential notation, e.g., 24 = 2³ x 3', and understand that every whole number greater than 1 is either prime or can be expressed as a product of primes. factors, prime numbers, exponential notation, whole number, product, exponents understand the difference of prime numbers versus composite numbers break down number to its prime factors I can find prime number to 50. I can comprehend what a prime number vs. a composite number is. I can break down a number to prime factors. I can express the whole number in an equation as a product of prime numbers with exponents. Understand meaning of decimal and percentages N.M Understand the relative magnitude of ones, tenths, and hundredths and the relationship of each place value to the place to its right, e.g., one is 10 tenths, one tenth is 10 hundredths. place value, shift patterns recognize place value of digits in ones, tenths, hundredths place understand base ten values (.G. 1 whole = 10 tenths) understand that digits shifting to the right represent 1/10 the value of that digit and digits shifting to the left represent 10 x's as much as that digit explain patterns in the placement of the decimal point and the number of 0's when multiplying by powers of 10 I can identify the place value of numbers to the hundredths place. I understand the value of whole numbers, tenths, and hundredth in relation to each other. I understand and can explain that the placement of the decimal point effects the value of a number. N.M Understand percentages as parts out of 100, use % notation, and express a part of a whole as a percentage. percents, identify place value of digits to hundredths understand that a percentage is out of 100 or 1 whole be able to express a percentage as a fraction out of 100 or right as a decimal out of 1 whole I can identify place value of digits to hundredths. I can understand that a percentage is out of 100 or 1 whole. I can express a percentage as a fraction out of 100 or as a decimal out of one whole. I can relate to decimals to percents. Understand as division statements; find equivalent N.M Understand a fraction as a statement of division, e.g., 2 3 = 2/3, using simple and pictures to represent., parts, wholes, division Understand that are parts of a whole. Understand that a fraction is broken into parts and a total. Know when the divisor is bigger than the dividend, the divisor becomes your denominator. N.M Given two, e.g., 1/2 and 1/4, express them as with a common denominator, but not necessarily a least common denominator, e.g., 1/2 = 4/8 and 3/4 = 6/8; use denominators less than 12 or factors of 100., numerators, equivalent Understand that a fraction represents a part of a whole. Understand value of numerator vs. denominator. Find common multiples of two given numbers to 100. Use process of multiplication to convert denominators into common denominators. I know that a fraction is part of a whole number or equal to one. I understand the value of the numerator is part of the denominators. I can find common multiples of numbers to 100. I can multiply to convert denominators to common denominators. I know that the numerator must be multiplied by the same number the denominator was when making the denominators common. = xtended = ore O:\lementary urriculum\working\lem Math\5th Grade\5th Grade Math Learning Targets.FINAL xlsx 2
5 ontent xpress, interpret, and use ratios; find equivalences N.MR xpress and decimals as percentages and vice versa., decimals, percents, converting between all three xplain how and decimals represents parts of wholes. xplain the relationship between decimals and and how the can represent the same amount. Recognize how percentages relate to and decimals as parts of a whole. I can write a decimal as a fraction. I can explain how, decimals, and percents all represent parts in a whole. I can convert a decimal to a fraction to a percent. N.M xpress ratios in several ways given applied situations, e.g., 3 cups to 5 people, 3 : 5, 3/5; recognize and find equivalent ratios. recognize - knowledge ratios, equivalent, Measurement Know, and convert among, measurement units within a given system M.UN Recognize the equivalence of 1 liter, 1,000 ml and 1,000 cm³ and include conversions among liters, milliliters, and cubic centimeters. recognize - knowledge include - equivalence, volume, capacity, metric system, place value base ten place value to the thousandth and thousands place understanding of metric prefixes (centi, kilo, milli, etc ) know that equivalence means "equal to" I can recognize place value to the thousandths and thousands place. I know the meaning and value of the metric prefixes. I know that equivalence means equal to. M.UN Know the units of measure of volume: cubic centimeter, cubic meter, cubic inches, cubic feet, cubic yards, and use their abbreviations (cm³, m³, in³, ft³, yd³). volume, exponents Understand the difference between area and volume onvert between inches, feet, and yards onvert between centimeter and meter Understand what exponents are, how they are used, and what they mean I can state how many inches are in a foot, and how many feet in a yard. I can state how many centimeters are in a meter. I know what an exponent is. I know the value of an exponent. M.UN ompare the relative sizes of one cubic inch to one cubic foot, and one cubic centimeter to one cubic meter. compare - analysis volume, feet, meters, inches centimeter onvert between inches and feet onvert between centimeters and meters Formula for volume is length times width times height or the area of the base times the height I can convert between inches and feet, centimeters and meters. I can state the formula for area. I can state the formula for volume. M.UN onvert measurements of length, weight, area, volume, and time within a given system using easily manipulated numbers. = xtended = ore convert - area, volume, measurement, Using the standard and metric system, students can convert within the given unit of measurement 16 ounces in a pound 4 quarts in a gallon 12 inches in a foot 1000 grams in a kilogram Understand that as the unit of measurement gets smaller, the number of units gets larger I know how many ounces are in pound. I know how many quarts are in gallon, pints in a quart, cups in pints, etc. I understand that when I am measuring with smaller units, I will have more of them. O:\lementary urriculum\working\lem Math\5th Grade\5th Grade Math Learning Targets.FINAL xlsx 5
6 ontent Find areas of geometric shapes using formulas M.PS Represent relationships between areas of rectangles, triangles, and parallelograms using models. represent - area, rectangles, parallelograms, triangles, models two triangles make a rectangle or parallelogram demonstrate knowledge using a picture I can demonstrate how to divide a rectangle into two triangles. I can demonstrate how to divide a parallelogram into two triangles. Find areas of geometric shapes using formulas M.T Understand and know how to use the area formula of a triangle: A = ½bh (where b is length of the base and h is the height), and represent using models and manipulatives. represent - area, triangles, fraction 1/2 = divide by two, formula area of triangle = base times height divided by two create proof picture showing why the area of a triangle is divided by two and an rectangle/parallelogram is not I can state the formula for the area of a triangle. I can apply the formula for area of a triangle to a given situation. I can demonstrate my knowledge using a proof picture. M.T Understand and know how to use the area formula for a parallelogram: A = bh, and represent using models and manipulatives. represent - formula, area, parallelograms, length, width, multiplication area of parallelogram= base times height create proof picture showing why the area is found using the formula I can state the formula for the area of a parallelogram. I can apply the formula for area of parallelogram to a given situation. I can demonstrate my knowledge using a proof picture. Understand the concept of volume M.T Build solids with unit cubes and state their volumes. build - synthesis volume, 3-D shapes N build three-dimensional solids using cubes calculate volume I can build a model using models (prisms). I can calculate the volume of the model that I built. M.T Use filling (unit cubes or liquid), and counting or measuring to find the volume of a cube and rectangular prism. use - application volume, 3-D shapes, cube, prism N M.PS Solve applied problems about the volumes of rectangular prisms using multiplication and division and using the appropriate units. volume, prisms, division know the various volume formulas solve for missing variable identify correct unit used can label answer using appropriate measurement unit I can state the formula for volume of rectangular prism. I can identify what variable is missing in a situation. I can solve for the missing variable using multiplication or division. I can label my answer with the appropriate unit of measurement. = xtended = ore O:\lementary urriculum\working\lem Math\5th Grade\5th Grade Math Learning Targets.FINAL xlsx 6
7 ontent Geometry Know the meaning of angles, and solve problems G.TR G.GS Associate an angle with a certain amount of turning; know that angles are measured in degrees; understand that 90,180, 270, and 360 are associated respectively, with ¼, ½, and, ¾ and full turns. Measure angles with a protractor and classify them as acute, right, obtuse, or straight. Know the meaning of angles, and solve problems associate - measure - classify - application rotation, degrees, angles, protractors, fractional turns protractor, types of angles, degrees, Understand that from a point of origin, as you rotate around 360, each place you stop creates an angle. Know that 1/4 turn equals 90 (right angle), 1/2 turn equals 180 (straight angle), 3/4 turn equals 270 (reflex angle), and a full turn is 360. Understand that an angle is made up of two rays that meet at their beginning points. Know that an acute angle is less than 90 and an obtuse angle is 91 to 179, and a strait angle is 180 I know that an angle is made with two line segments or rays. I know where I begin is the point of origin or 0. I can rotate/turn clockwise or counterclockwise to create angles. I understand that 1/4 turn equals 90 (right angle). I understand that 1/2 turn equals 180 (straight angle). I understand that 3/4 turn equals 270 (reflex angle). I understand that rotation from any point back that same I know point what (point degrees of origin) make equals an obtuse, 360 /full acute, turn. right, and straight angle. I can use a protractor to measure the size of an angle. G.GS Identify and name angles on a straight line and vertical angles. identify - knowledge name - knowledge types of angles, straight line=180, vertical angles Know that two crossed lines are intersecting lines. Know that vertical angles are opposite angles created by two crossed lines. Be able to name the three given points on an angle created on a straight line. Identify the difference between an angle on a straight line and vertical angles. I can name an angle on a straight line using three given points. I can name the two vertical angles on intersecting lines. I can identify the difference between a n angle on a straight line and vertical angles. G.GS Find unknown angles in problems involving angles on a straight line, angles surrounding a point, and vertical angles. types of angles, straight line=180, vertical angles, angles in a circle (360) Know that a straight lines equals 180. Know that a turn around a circle equals 360. Know that vertical angles are opposite angles created by two crossed lines. Given a problem involving a straight line, vertical angles, or angles surrounding a point, know that the sum of the given angles must be equal to 180 or 360. I know that a straight line equals 180. I know that a full turn around a circle equals 360. I know that vertical angles are opposite angles created by two crossed lines. I can find the missing angle size in an equation involving straight lines, angles surrounding a point, or vertical angles. G.GS Know that angles on a straight line add up to 180 and angles surrounding a point add up to 360 ; justify informally by "surrounding" a point with angles. justify - evaluation types of angles, straight line=180, vertical angles, angles in a circle (360) A straight line equals a half circle. Angles found on a straight line must be equal or less than 180. A full turn around a circle equals 360. Any angles surrounding a point, know that the sum of the given angles must be equal to 180 or 360. I know that a straight lines makes half a circle. I know that an angle found on a straight line can't equal more than 180. I know that a full turn around a circle equals 360. I know that any angles surrounding a point must equal 360 or less. I can add or subtract 3-digit numbers. = xtended = ore O:\lementary urriculum\working\lem Math\5th Grade\5th Grade Math Learning Targets.FINAL xlsx 7
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