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# 7.4: Partial Fractions
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##### Learning Objectives
• Integrate a rational function using the method of partial fractions.
• Recognize simple linear factors in a rational function.
• Recognize repeated linear factors in a rational function.
• Recognize quadratic factors in a rational function.
We have seen some techniques that allow us to integrate specific rational functions. For example, we know that
$\int \dfrac{du}{u}=\ln |u|+C \nonumber$
and
$\int \dfrac{du}{u^2+a^2}=\dfrac{1}{a}\tan^{−1} \left(\dfrac{u}{a}\right)+C.\nonumber$
However, we do not yet have a technique that allows us to tackle arbitrary quotients of this type. Thus, it is not immediately obvious how to go about evaluating
$\int \dfrac{3x}{x^2−x−2}\,dx.\nonumber$
However, we know from material previously developed that
$\int \left(\dfrac{1}{x+1}+\dfrac{2}{x−2}\right)\,dx=\ln |x+1|+2\ln |x−2|+C.\nonumber$
In fact, by getting a common denominator, we see that
$\dfrac{1}{x+1}+\dfrac{2}{x−2}=\dfrac{3x}{x^2−x−2}.\nonumber$
Consequently,
$\int \dfrac{3x}{x^2−x−2}\,dx=\int \left(\dfrac{1}{x+1}+\dfrac{2}{x−2}\right)\,dx.\nonumber$
In this section, we examine the method of partial fraction decomposition, which allows us to decompose rational functions into sums of simpler, more easily integrated rational functions. Using this method, we can rewrite an expression such as:
$\dfrac{3x}{x^2−x−2}\nonumber$
as an expression such as
$\dfrac{1}{x+1}+\dfrac{2}{x−2}.\nonumber$
The key to the method of partial fraction decomposition is being able to anticipate the form that the decomposition of a rational function will take. As we shall see, this form is both predictable and highly dependent on the factorization of the denominator of the rational function. It is also extremely important to keep in mind that partial fraction decomposition can be applied to a rational function $$\dfrac{P(x)}{Q(x)}$$ only if $$deg(P(x))<deg(Q(x))$$. In the case when $$deg(P(x))≥deg(Q(x))$$, we must first perform long division to rewrite the quotient $$\dfrac{P(x)}{Q(x)}$$ in the form $$A(x)+\dfrac{R(x)}{Q(x)}$$, where $$deg(R(x))<deg(Q(x))$$. We then do a partial fraction decomposition on $$\dfrac{R(x)}{Q(x)}$$. The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form $$\int \dfrac{P(x)}{Q(x)}\,dx$$, where $$deg(P(x))≥deg(Q(x)).$$
##### Example $$\PageIndex{1}$$: Integrating $$\displaystyle \int \frac{P(x)}{Q(x)}\,dx$$, where $$deg(P(x))≥deg(Q(x))$$
Evaluate
$\int \dfrac{x^2+3x+5}{x+1}\,dx. \nonumber$
Solution
Since $$deg(x^2+3x+5)≥deg(x+1),$$ we perform long division to obtain
$\dfrac{x^2+3x+5}{x+1}=x+2+\dfrac{3}{x+1}. \nonumber$
Thus,
$\int \dfrac{x^2+3x+5}{x+1}\,dx=\int \left(x+2+\dfrac{3}{x+1}\right)\,dx=\dfrac{1}{2}x^2+2x+3\ln |x+1|+C. \nonumber$
Visit this website for a review of long division of polynomials.
##### Exercise $$\PageIndex{1}$$
Evaluate
$\int \dfrac{x−3}{x+2}\,dx. \nonumber$
Hint
Use long division to obtain $$\dfrac{x−3}{x+2}=1−\dfrac{5}{x+2}. \nonumber$$
$x−5\ln |x+2|+C \nonumber$
To integrate $$\displaystyle \int \dfrac{P(x)}{Q(x)}\,dx$$, where $$deg(P(x))<deg(Q(x))$$, we must begin by factoring $$Q(x)$$.
## Nonrepeated Linear Factors
If $$Q(x)$$ can be factored as $$(a_1x+b_1)(a_2x+b_2)…(a_nx+b_n)$$, where each linear factor is distinct, then it is possible to find constants $$A_1,A_2,…A_n$$ satisfying
$\dfrac{P(x)}{Q(x)}=\dfrac{A_1}{a_1x+b_1}+\dfrac{A_2}{a_2x+b_2}+⋯+\dfrac{A_n}{a_nx+b_n}. \label{eq:7.4.1}$
The proof that such constants exist is beyond the scope of this course.
In this next example, we see how to use partial fractions to integrate a rational function of this type.
##### Example $$\PageIndex{2}$$: Partial Fractions with Nonrepeated Linear Factors
Evaluate $$\displaystyle \int \dfrac{3x+2}{x^3−x^2−2x}\,dx.$$
Solution
Since $$deg(3x+2)<deg(x^3−x^2−2x)$$, we begin by factoring the denominator of $$\dfrac{3x+2}{x^3−x^2−2x}$$. We can see that $$x^3−x^2−2x=x(x−2)(x+1)$$. Thus, there are constants $$A$$, $$B$$, and $$C$$ satisfying Equation \ref{eq:7.4.1} such that
$\dfrac{3x+2}{x(x−2)(x+1)}=\dfrac{A}{x}+\dfrac{B}{x−2}+\dfrac{C}{x+1}. \nonumber$
We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,
$\dfrac{3x+2}{x(x−2)(x+1)}=\dfrac{A(x−2)(x+1)+Bx(x+1)+Cx(x−2)}{x(x−2)(x+1)}. \nonumber$
Now, we set the numerators equal to each other, obtaining
$3x+2=A(x−2)(x+1)+Bx(x+1)+Cx(x−2).\label{Ex2Numerator}$
There are two different strategies for finding the coefficients $$A$$, $$B$$, and $$C$$. We refer to these as the method of equating coefficients and the method of strategic substitution.
Strategy one: Method of Equating Coefficients
Rewrite Equation $$\ref{Ex2Numerator}$$ in the form
$3x+2=(A+B+C)x^2+(−A+B−2C)x+(−2A). \nonumber$
Equating coefficients produces the system of equations
\begin{align*} A+B+C &=0 \\[4pt] −A+B−2C &= 3 \\[4pt] −2A &=2. \end{align*}
To solve this system, we first observe that $$−2A=2⇒A=−1.$$ Substituting this value into the first two equations gives us the system
$$B+C=1$$
$$B−2C=2$$.
Multiplying the second equation by $$−1$$ and adding the resulting equation to the first produces
$$−3C=1,$$
which in turn implies that $$C=−\dfrac{1}{3}$$. Substituting this value into the equation $$B+C=1$$ yields $$B=\dfrac{4}{3}$$. Thus, solving these equations yields $$A=−1, B=\dfrac{4}{3}$$, and $$C=−\dfrac{1}{3}$$.
It is important to note that the system produced by this method is consistent if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.
Strategy two: Method of Strategic Substitution
The method of strategic substitution is based on the assumption that we have set up the decomposition correctly. If the decomposition is set up correctly, then there must be values of $$A, B,$$ and $$C$$ that satisfy Equation $$\ref{Ex2Numerator}$$ for all values of $$x$$. That is, this equation must be true for any value of $$x$$ we care to substitute into it. Therefore, by choosing values of $$x$$ carefully and substituting them into the equation, we may find $$A, B$$, and $$C$$ easily. For example, if we substitute $$x=0$$, the equation reduces to $$2=A(−2)(1)$$. Solving for $$A$$ yields $$A=−1$$. Next, by substituting $$x=2$$, the equation reduces to $$8=B(2)(3)$$, or equivalently $$B=4/3$$. Last, we substitute $$x=−1$$ into the equation and obtain $$−1=C(−1)(−3).$$ Solving, we have $$C=−\dfrac{1}{3}$$.
It is important to keep in mind that if we attempt to use this method with a decomposition that has not been set up correctly, we are still able to find values for the constants, but these constants are meaningless. If we do opt to use the method of strategic substitution, then it is a good idea to check the result by recombining the terms algebraically.
Now that we have the values of $$A, B,$$ and $$C,$$ we rewrite the original integral:
$\int \dfrac{3x+2}{x^3−x^2−2x}\,dx=\int \left(−\dfrac{1}{x}+\dfrac{4}{3}⋅\dfrac{1}{x−2}−\dfrac{1}{3}⋅\dfrac{1}{x+1}\right)\,dx. \nonumber$
Evaluating the integral gives us
$\int \dfrac{3x+2}{x^3−x^2−2x}\,dx=−\ln |x|+\dfrac{4}{3}\ln |x−2|−\dfrac{1}{3}\ln |x+1|+C. \nonumber$
In the next example, we integrate a rational function in which the degree of the numerator is not less than the degree of the denominator.
##### Example $$\PageIndex{3}$$: Dividing before Applying Partial Fractions
Evaluate $$\displaystyle \int \dfrac{x^2+3x+1}{x^2−4}\,dx.$$
Solution
Since $$deg(x^2+3x+1)≥deg(x^2−4),$$ we must perform long division of polynomials. This results in
$\dfrac{x^2+3x+1}{x^2−4}=1+\dfrac{3x+5}{x^2−4} \nonumber$
Next, we perform partial fraction decomposition on $$\dfrac{3x+5}{x^2−4}=\dfrac{3x+5}{(x+2)(x−2)}$$. We have
$\dfrac{3x+5}{(x−2)(x+2)}=\dfrac{A}{x−2}+\dfrac{B}{x+2}. \nonumber$
Thus,
$3x+5=A(x+2)+B(x−2). \nonumber$
Solving for $$A$$ and $$B$$ using either method, we obtain $$A=11/4$$ and $$B=1/4.$$
Rewriting the original integral, we have
$\int \dfrac{x^2+3x+1}{x^2−4}\,dx=\int \left(1+\dfrac{11}{4}⋅\dfrac{1}{x−2}+\dfrac{1}{4}⋅\dfrac{1}{x+2}\right)\,dx. \nonumber$
Evaluating the integral produces
$\int \dfrac{x^2+3x+1}{x^2−4}\,dx=x+\dfrac{11}{4}\ln |x−2|+\dfrac{1}{4}\ln |x+2|+C. \nonumber$
As we see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrational function. The trick is to convert the nonrational function to a rational function through a substitution.
##### Example $$\PageIndex{4}$$: Applying Partial Fractions after a Substitution
Evaluate $$\displaystyle \int \dfrac{\cos x}{\sin^2x−\sin x}\,dx.$$
Solution
Let’s begin by letting $$u=\sin x.$$ Consequently, $$du=\cos x\,dx.$$ After making these substitutions, we have
$\int \dfrac{\cos x}{\sin^2x−\sin x}\,dx=\int \dfrac{du}{u^2−u}=\int \dfrac{du}{u(u−1)}. \nonumber$
Applying partial fraction decomposition to $$\dfrac{1}{u(u−1)}$$ gives $$\dfrac{1}{u(u−1)}=−\dfrac{1}{u}+\dfrac{1}{u−1}.$$
Thus,
$\int \dfrac{\cos x}{\sin^2x−\sin x}\,dx=−\ln |u|+\ln |u−1|+C=−\ln |\sin x|+\ln |\sin x−1|+C. \nonumber$
##### Exercise $$\PageIndex{2}$$
Evaluate $$\displaystyle \int \dfrac{x+1}{(x+3)(x−2)}\,dx.$$
Hint
$\dfrac{x+1}{(x+3)(x−2)}=\dfrac{A}{x+3}+\dfrac{B}{x−2} \nonumber$
$\dfrac{2}{5}\ln |x+3|+\dfrac{3}{5}\ln |x−2|+C \nonumber$
## Repeated Linear Factors
For some applications, we need to integrate rational expressions that have denominators with repeated linear factors—that is, rational functions with at least one factor of the form $$(ax+b)^n,$$ where $$n$$ is a positive integer greater than or equal to $$2$$. If the denominator contains the repeated linear factor $$(ax+b)^n$$, then the decomposition must contain
$\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+⋯+\dfrac{A_n}{(ax+b)^n}. \label{eq:7.4.2}$
As we see in our next example, the basic technique used for solving for the coefficients is the same, but it requires more algebra to determine the numerators of the partial fractions.
##### Example $$\PageIndex{5}$$: Partial Fractions with Repeated Linear Factors
Evaluate $$\displaystyle \int \dfrac{x−2}{(2x−1)^2(x−1)}\,dx.$$
Solution
We have $$deg(x−2)<deg((2x−1)^2(x−1)),$$ so we can proceed with the decomposition. Since $$(2x−1)^2$$ is a repeated linear factor, include
$\dfrac{A}{2x−1}+\dfrac{B}{(2x−1)^2} \nonumber$
in the decomposition in Equation \ref{eq:7.4.2}. Thus,
$\dfrac{x−2}{(2x−1)^2(x−1)}=\dfrac{A}{2x−1}+\dfrac{B}{(2x−1)^2}+\dfrac{C}{x−1}. \nonumber$
After getting a common denominator and equating the numerators, we have
$x−2=A(2x−1)(x−1)+B(x−1)+C(2x−1)^2. \label{Ex5Numerator}$
We then use the method of equating coefficients to find the values of $$A, B,$$ and $$C$$.
$x−2=(2A+4C)x^2+(−3A+B−4C)x+(A−B+C). \nonumber$
Equating coefficients yields $$2A+4C=0$$, $$−3A+B−4C=1$$, and $$A−B+C=−2$$. Solving this system yields $$A=2, B=3,$$ and $$C=−1.$$
Alternatively, we can use the method of strategic substitution. In this case, substituting $$x=1$$ and $$x=1/2$$ into Equation $$\ref{Ex5Numerator}$$ easily produces the values $$B=3$$ and $$C=−1$$. At this point, it may seem that we have run out of good choices for $$x$$, however, since we already have values for $$B$$ and $$C$$, we can substitute in these values and choose any value for $$x$$ not previously used. The value $$x=0$$ is a good option. In this case, we obtain the equation $$−2=A(−1)(−1)+3(−1)+(−1)(−1)^2$$ or, equivalently, $$A=2.$$
Now that we have the values for $$A, B,$$ and $$C$$, we rewrite the original integral and evaluate it:
\begin{align*} \int \dfrac{x−2}{(2x−1)^2(x−1)}\,dx &=\int \left(\dfrac{2}{2x−1}+\dfrac{3}{(2x−1)^2}−\dfrac{1}{x−1}\right)\,dx \\[4pt] &=\ln |2x−1|−\dfrac{3}{2(2x−1)}−\ln |x−1|+C. \end{align*}
##### Exercise $$\PageIndex{3}$$
Set up the partial fraction decomposition for
$\int \dfrac{x+2}{(x+3)^3(x−4)^2}\,dx. \nonumber$
(Do not solve for the coefficients or complete the integration.)
Hint
Use the problem-solving method of Example $$\PageIndex{5}$$ for guidance.
$\dfrac{x+2}{(x+3)^3(x−4)^2}=\dfrac{A}{x+3}+\dfrac{B}{(x+3)^2}+\dfrac{C}{(x+3)^3}+\dfrac{D}{(x−4)}+\dfrac{E}{(x−4)^2} \nonumber$
## The General Method
Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let’s outline the basic method in the following problem-solving strategy.
##### Problem-Solving Strategy: Partial Fraction Decomposition
To decompose the rational function $$P(x)/Q(x)$$, use the following steps:
1. Make sure that $$deg(P(x))<deg(Q(x)).$$ If not, perform long division of polynomials.
2. Factor $$Q(x)$$ into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic that has no real zeros.
3. Assuming that $$deg(P(x))<deg(Q(x)$$, the factors of $$Q(x)$$ determine the form of the decomposition of $$P(x)/Q(x).$$
1. If $$Q(x)$$ can be factored as $$(a_1x+b_1)(a_2x+b_2)…(a_nx+b_n)$$, where each linear factor is distinct, then it is possible to find constants $$A_1,A_2,...A_n$$ satisfying $\dfrac{P(x)}{Q(x)}=\dfrac{A_1}{a_1x+b_1}+\dfrac{A_2}{a_2x+b_2}+⋯+\dfrac{A_n}{a_nx+b_n}. \nonumber$
2. If $$Q(x)$$ contains the repeated linear factor $$(ax+b)^n$$, then the decomposition must contain $\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+⋯+\dfrac{A_n}{(ax+b)^n}. \nonumber$
3. For each irreducible quadratic factor $$ax^2+bx+c$$ that $$Q(x)$$ contains, the decomposition must include $\dfrac{Ax+B}{ax^2+bx+c}. \nonumber$
4. For each repeated irreducible quadratic factor $$(ax^2+bx+c)^n,$$ the decomposition must include $\dfrac{A_1x+B_1}{ax^2+bx+c}+\dfrac{A_2x+B_2}{(ax^2+bx+c)^2}+⋯+\dfrac{A_nx+B_n}{(ax^2+bx+c)^n}. \nonumber$
5. After the appropriate decomposition is determined, solve for the constants.
6. Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.
Now let’s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic $$ax^2+bx+c$$ is irreducible if $$ax^2+bx+c=0$$ has no real zeros—that is, if $$b^2−4ac<0.$$
##### Example $$\PageIndex{6}$$: Rational Expressions with an Irreducible Quadratic Factor
Evaluate
$\int \dfrac{2x−3}{x^3+x}\,dx.\nonumber$
Solution
Since $$deg(2x−3)<deg(x^3+x),$$ factor the denominator and proceed with partial fraction decomposition. Since $$x^3+x=x(x^2+1)$$ contains the irreducible quadratic factor $$x^2+1$$, include $$\dfrac{Ax+B}{x^2+1}$$ as part of the decomposition, along with $$\dfrac{C}{x}$$ for the linear term $$x$$. Thus, the decomposition has the form
$\dfrac{2x−3}{x(x^2+1)}=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x}.\nonumber$
After getting a common denominator and equating the numerators, we obtain the equation
$2x−3=(Ax+B)x+C(x^2+1).\nonumber$
Solving for $$A,B,$$ and $$C,$$ we get $$A=3, B=2,$$ and $$C=−3.$$
Thus,
$\dfrac{2x−3}{x^3+x}=\dfrac{3x+2}{x^2+1}−\dfrac{3}{x}.\nonumber$
Substituting back into the integral, we obtain
\begin{align*} \int \dfrac{2x−3}{x^3+x}\,dx &=\int \left(\dfrac{3x+2}{x^2+1}−\dfrac{3}{x}\right)\,dx \nonumber \\[4pt] &=3\int \dfrac{x}{x^2+1}\,dx+2\int \dfrac{1}{x^2+1}\,dx−3\int \dfrac{1}{x}\,dx & & \text{Split up the integral} \\[4pt] &=\dfrac{3}{2}\ln ∣x^2+1∣+2 \tan^{−1}x−3\ln |x|+C. & & \text{Evaluate each integral} \end{align*}
Note: We may rewrite $$\ln ∣x^2+1∣=\ln (x^2+1)$$, if we wish to do so, since $$x^2+1>0.$$
##### Example $$\PageIndex{7}$$: Partial Fractions with an Irreducible Quadratic Factor
Evaluate $$\displaystyle \int \dfrac{\,dx}{x^3−8}.$$
Solution: We can start by factoring $$x^3−8=(x−2)(x^2+2x+4).$$ We see that the quadratic factor $$x^2+2x+4$$ is irreducible since $$2^2−4(1)(4)=−12<0.$$ Using the decomposition described in the problem-solving strategy, we get
$\dfrac{1}{(x−2)(x^2+2x+4)}=\dfrac{A}{x−2}+\dfrac{Bx+C}{x^2+2x+4}. \nonumber$
After obtaining a common denominator and equating the numerators, this becomes
$1=A(x^2+2x+4)+(Bx+C)(x−2). \nonumber$
Applying either method, we get $$A=\dfrac{1}{12},B=−\dfrac{1}{12},$$ and $$C=−\dfrac{1}{3}.$$
Rewriting $$\int \dfrac{\,dx}{x^3−8},$$ we have
$\int \dfrac{\,dx}{x^3−8}=\dfrac{1}{12}\int \dfrac{1}{x−2}\,dx−\dfrac{1}{12}\int \dfrac{x+4}{x^2+2x+4}\,dx. \nonumber$
We can see that
$\int \dfrac{1}{x−2}\,dx=\ln |x−2|+C,\nonumber$
but
$\int \dfrac{x+4}{x^2+2x+4}\,dx \nonumber$
requires a bit more effort. Let’s begin by completing the square on $$x^2+2x+4$$ to obtain
$x^2+2x+4=(x+1)^2+3. \nonumber$
By letting $$u=x+1$$ and consequently $$du=\,dx,$$ we see that
\begin{align*} \int \dfrac{x+4}{x^2+2x+4}\,dx &=\int \dfrac{x+4}{(x+1)^2+3}\,dx & & \text{Complete the square on the denominator} \\[4pt] &=\int \dfrac{u+3}{u^2+3}\,du & & \text{Substitute }u=x+1,\,x=u−1,\text{ and } du=dx \\[4pt] &=\int \dfrac{u}{u^2+3}du+\int \dfrac{3}{u^2+3}du & & \text{Split the numerator apart} \\[4pt] &=\dfrac{1}{2}\ln ∣u^2+3∣+\dfrac{3}{\sqrt{3}}\tan^{−1}\dfrac{u}{\sqrt{3}}+C & & \text{Evaluate each integral} \\[4pt] &=\dfrac{1}{2}\ln ∣x^2+2x+4∣+\sqrt{3}\tan^{−1}\left(\dfrac{x+1}{\sqrt{3}}\right)+C & & \text{Rewrite in terms of }x\text{ and simplify} \end{align*}
Substituting back into the original integral and simplifying gives
$\int \dfrac{ \,dx}{x^3−8}=\dfrac{1}{12}\ln |x−2|−\dfrac{1}{24}\ln |x^2+2x+4|−\dfrac{\sqrt{3}}{12}\tan^{−1}\left(\dfrac{x+1}{\sqrt{3}}\right)+C. \nonumber$
Here again, we can drop the absolute value if we wish to do so, since $$x^2+2x+4>0$$ for all $$x$$.
##### Example $$\PageIndex{8}$$: Finding a Volume
Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of $$f(x)=\dfrac{x^2}{(x^2+1)^2}$$ and the x-axis over the interval $$[0,1]$$ about the y-axis.
Solution
Let’s begin by sketching the region to be revolved (see Figure $$\PageIndex{1}$$). From the sketch, we see that the shell method is a good choice for solving this problem.
The volume is given by
$V=2π\int ^1_0x⋅\dfrac{x^2}{(x^2+1)^2}\,dx=2π\int ^1_0\dfrac{x^3}{(x^2+1)^2}\,dx. \nonumber$
Since $$deg((x^2+1)^2)=4>3=deg(x^3),$$ we can proceed with partial fraction decomposition. Note that $$(x^2+1)^2$$ is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get
$\dfrac{x^3}{(x^2+1)^2}=\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{(x^2+1)^2}. \nonumber$
Finding a common denominator and equating the numerators gives
$x^3=(Ax+B)(x^2+1)+Cx+D. \nonumber$
Solving, we obtain $$A=1, B=0, C=−1,$$ and $$D=0.$$ Substituting back into the integral, we have
$V=2π\int _0^1\dfrac{x^3}{(x^2+1)^2}\,dx=2π\int _0^1\left(\dfrac{x}{x^2+1}−\dfrac{x}{(x^2+1)^2}\right)\,dx=2π\left(\dfrac{1}{2}\ln (x^2+1)+\dfrac{1}{2}⋅\dfrac{1}{x^2+1}\right)\Big|^1_0=π\left(\ln 2−\tfrac{1}{2}\right). \nonumber$
##### Exercise $$\PageIndex{4}$$
Set up the partial fraction decomposition for $\int \dfrac{x^2+3x+1}{(x+2)(x−3)^2(x^2+4)^2}\,dx. \nonumber$
Hint
Use the problem-solving strategy.
$\dfrac{x^2+3x+1}{(x+2)(x−3)^2(x^2+4)^2}=\dfrac{A}{x+2}+\dfrac{B}{x−3}+\dfrac{C}{(x−3)^2}+\dfrac{Dx+E}{x^2+4}+\dfrac{Fx+G}{(x^2+4)^2} \nonumber$
## Key Concepts
• Partial fraction decomposition is a technique used to break down a rational function into a sum of simple rational functions that can be integrated using previously learned techniques.
• When applying partial fraction decomposition, we must make sure that the degree of the numerator is less than the degree of the denominator. If not, we need to perform long division before attempting partial fraction decomposition.
• The form the decomposition takes depends on the type of factors in the denominator. The types of factors include nonrepeated linear factors, repeated linear factors, nonrepeated irreducible quadratic factors, and repeated irreducible quadratic factors.
## Glossary
partial fraction decomposition
a technique used to break down a rational function into the sum of simple rational functions
This page titled 7.4: Partial Fractions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax. |
# How do you solve x^2-x=12?
Feb 3, 2015
We can use the Sum-Product method.
Bring everything to one side:
${x}^{2} - x = 12 \to {x}^{2} - x - 12 = 0$
We now have a quadratic equation of the form
$a {x}^{2} + b x + c = 0$ where $a = 1 , b = - 1 \mathmr{and} c = - 12$
We find two numbers that will give $c = - 12$ as a product and $b = - 1$ as a sum (or difference).
We can try $1 \cdot 12 , 2 \cdot 6 , 3 \cdot 4$
$3 \mathmr{and} 4$ will fit, with a $-$sign to the $4$, as that would make the sum $3 - 4 = - 1$ and the product $3 \cdot - 4 = - 12$
Now we can rewrite the equation:
$\left(x + 3\right) \left(x - 4\right) = 0$
Which leaves us with two possibilities:
$\left(x + 3\right) = 0 \to x = - 3$ OR $\left(x - 4\right) = 0 \to x = 4$
$x = - 3 \mathmr{and} x = 4$ |
Only one out of fifty people can figure out these top three mistakes in mathematics. Can you figure them out?
###### Mistake #1: Order of Operations
What is the answer to 8 + 8 ÷ 8 + 8 x 8 – 8 = ?
Step 1. 8 + 8 ÷ 8 + 8 x 8 – 8 =
Step 2. 16 ÷ 8 + 8 x 8 – 8 =
Step 3. 2 + 8 x 8 – 8 =
Step 4. 10 x 8 – 8 =
Step 5. 80 – 8 = 72
If you said 72, you are wrong! There is a mistake in step 2 and in step 3.
The answer is actually 65 because the multiplication and division should be done before the addition and subtraction.
Step 1. 8 + 8 ÷ 8 + 8 x 8 – 8
Step 2. 8 + (8 ÷ 8) + (8 x 8) – 8
Step 3. 8 + 1 + 64 – 8 = 65 (Correct)
PEMDAS is Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction. The order of the letters in PEMDAS tells you which part to calculate first. This is also known as the Order of Operations.
###### Mistake #2: Division by Zero
0 ÷ 2 = 0 so what is the answer to 2 ÷ 0 = ?
If you said 0, you are wrong!
The answer is actually undefined because you cannot divide by zero.
Step 2. Multiply both sides by a, so ab = a²
Step 3. Subtract b2 from both sides, so ab – b² = a² – b²
Step 4. Factor both sides, so b(a-b) = (a+b)(a-b)
Step 5. Divide both sides by a – b, so b = a + b
Step 6. Remember that a = b, so b = 2b
Step 7. Divide both sides by b, so 1 = 2
Obviously, one does not equal two! This shows the problems when you try to divide by zero. In step 5, a – b is zero because of a = b. Therefore, there was a division by zero in this step.
###### Mistake #3: Forgotten Parentheses
What is the answer to (-4)² =?
If you said negative 16 was the square of negative 4, you are wrong!
The answer is actually 16 not -16.
Here is an incorrect way to solve this problem: -4² = -(4)(4) = -16
And here is the correct way: (-4)² = (-4)(-4) = 16
The mistake is forgetting about the parentheses. Most people forget about the parentheses and just answer -16. |
Assignment 1
The purpose of this activity is to explore binary operations of functions and the composition of functions.
Let
f(x) = x/5 - 4
g(x) = -3x + 2
f(x) is the function shown in purple and g(x) is the function shown in red. The slope of f(x) is 1/5 and the intercept is -4. The slope of g(x) is -3 and the intercept is 2. We can verify this information by looking at the graphs of these equations show below.
Now explore the graph of h(x), where h(x) = f(x) + g(x). Notice how this graph relates to the original functions.
Notice the slope of this new function appears to be very similar to that of g(x). Algebraically, this function can be shown as h(x) = -(14/5)x -2. The slope is -14/5 and the y - intercept is -2.
Let's now explore the graph of h'(x), where
h'(x) = f(x).g(x).
Here we can see that the graph of two linear functions multiplied together is in this case a parabola.
Let's now explore the graph of h''(x) where h''(x)
= f(x)/g(x).
Here we notice that the graph is a hyperbola.
Lastly, look at the graph of h'''(x), where h'''(x) = f(g(x)).
The compositions of these two functions, f of g of x, is a linear function.
Now, let f(x) and g(x) be two different functions and explore these graphs again.
Let f(x) = 5x + 2, and let g(x) = x/2 + 1.
Below are the graphs of these two equations. f(x) is the purple graph and g(x) is the red graph.
h(x) = f(x) + g(x) is shown by the blue function. Again, notice how the process of addition causes the original equation with the largest absolute value to dominate.
h'(x) = f(x).g(x) Again, when we multiply two linear equations together, we get a parabola.
h''(x) = f(x)/g(x) Dividing one linear equation by another yields a hyperbola.
h'''(x) = f(g(x)) The composition of two linear functions, f of g of x, creates a new function, also linear.
Summary of Findings
When two linear functions are added together, the result is a third linear function. The slopes and y-intercepts of the functions follow the basic properties of addition when added together, respectively.
When two linear functions are multiplied together, the resulting function will be a parabola, provided that the slope of neither function is 0. In that case, the resulting function will either be a linear function (if one function has zero slope) or a constant function (if both functions have zero slopes.) Likewise, the resulting function when one linear function is divided by another will be a hyperbola, provided neither function has a zero slope. If a linear function with a non zero slope is divided by a function with a zero slope, the resulting function will also be linear.
The composition of two linear functions always yields another linear function. |
# Difference of Two Sets
How to find the difference of two sets?
If A and B are two sets, then their difference is given by A - B or B - A.
If A = {2, 3, 4} and B = {4, 5, 6}
A - B means elements of A which are not the elements of B.
i.e., in the above example A - B = {2, 3}
In general, B - A = {x : x
B, and x A}
If A and B are disjoint sets, then A – B = A and B – A = B
Solved examples to find the difference of two sets:
1. A = {1, 2, 3} and B = {4, 5, 6}.
Find the difference between the two sets:
(i) A and B
(ii) B and A
Solution:
The two sets are disjoint as they do not have any elements in common.
(i) A - B = {1, 2, 3} = A
(ii) B - A = {4, 5, 6} = B
2. Let A = {a, b, c, d, e, f} and B = {b, d, f, g}.
Find the difference between the two sets:
(i) A and B
(ii) B and A
Solution:
(i) A - B = {a, c, e}
Therefore, the elements a, c, e belong to A but not to B
(ii) B - A = {g)
Therefore, the element g belongs to B but not A.
3. Given three sets P, Q and R such that:
P = {x : x is a natural number between 10 and 16},
Q = {y : y is a even number between 8 and 20} and
R = {7, 9, 11, 14, 18, 20}
(i) Find the difference of two sets P and Q
(ii) Find Q - R
(iii) Find R - P
(iv) Find Q – P
Solution:
According to the given statements:
P = {11, 12, 13, 14, 15}
Q = {10, 12, 14, 16, 18}
R = {7, 9, 11, 14, 18, 20}
(i) P – Q = {Those elements of set P which are not in set Q}
= {11, 13, 15}
(ii) Q – R = {Those elements of set Q not belonging to set R}
= {10, 12, 16}
(iii) R – P = {Those elements of set R which are not in set P}
= {7, 9, 18, 20}
(iv) Q – P = {Those elements of set Q not belonging to set P}
= {10, 16, 18}
Set Theory
Sets
Objects Form a Set
Elements of a Set
Properties of Sets
Representation of a Set
Different Notations in Sets
Standard Sets of Numbers
Types of Sets
Pairs of Sets
Subset
Subsets of a Given Set
Operations on Sets
Union of Sets
Intersection of Sets
Difference of two Sets
Complement of a Set
Cardinal number of a set
Cardinal Properties of Sets
Venn Diagrams
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## Recent Articles
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Guided Lessons
# Arrays for Fraction Products
Illustrating products for fractions with arrays is fun! Use this lesson plan to teach your students to visualize products as arrays when faced with multiplying fraction factors.
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Students will be able to illustrate fraction factor products with arrays.
(5 minutes)
• Show your students the following equation: 2/3 x ?/? = 6/12 and have them think, pair, and share their thoughts on the missing factor fraction.
• Call on students to share out their ideas. Note relevant academic terms for future reference and confirm the missing factor to be 3/4.
• Draw a rectangle where:
• The vertical side is labeled 3/4
• The horizontal side is labeled 2/3
• The center of the rectangle is labeled 6/12
• Tell your class that today’s lesson will be on how to draft an array for fraction products when both factors are also fractions. An array is an illustration strategy, great for displaying products and sums of repeat addition.
(10 minutes)
• On grid paper show your class a rectangle where:
• The vertical side is labeled 1/2
• The horizontal side is labeled 1/4
• Below the rectangle is the equation: 1/2 x 1/4 = 1/8
• Show your students how to divide the rectangle into two horizontal halves and shade one (1/2 total).
• Next, divide the rectangle vertically into four sections and shade in one part (1/4, you will notice some overlap).
• Explain that the overlap, 1/8, is an illustration of the product of 1/2 x 1/4 in the form of an array.
(10 minutes)
• Guide your class through the same procedure for the following expressions:
• 3/4 x 1/5
• 1/6 x 2/7
(10 minutes)
• Answer any clarifying questions and hand out grid paper to your students.
• Have your class solve the following exercises with arrays:
• 3/4 x 2/9
• 5/6 x 4/5
• 6/7 x 2/5
Support
• Using unit fractions (i.e. 1/3, 1/4, 1/2, 1/5,) for factors makes easier arrays.
Enrichment
• Challenge students to draft arrays with mixed number factors by adding whole numbers to their fraction factors.
• It’s convenient to snap pictures of student work at different stages, upload photos and import them into word processing documents like word or Google Sheets for online reference.
• Drafting student responses during reflection discussions via computer and projector for your class to see makes for an interactive lesson closure setup.
(5 minutes)
• Show your students an array (i.e. for 2/3 x 3/5) and have them explain the equation it represents.
(15 minutes)
• Have your students share out exercise answers with the whole class. If they need help, they can "phone a friend" for assistance or "hand off" to a peer.
• Discuss: What can we learn from arrays for fraction products? Note student responses for future reference.
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# Factor Patterns at Your Fingertips
Take a look at the interactive model below (and here). Most of the numbers in the array are shaded orange, but several are blue. What is special about these blue values? They are the factors of 32, the largest number in the array.
Try dragging the red point to change the dimensions of the array. You’ll see that the pattern of blue and orange changes, with the blue-shaded numbers indicating the factors of the largest number in the array.
If the goal of studying factors is to factor actual numbers, then this model is horrible—it does the work for you! But mastering factoring, while important, is not nearly as interesting as exploring the mathematical relationships between numbers and their factors. And that’s the benefit of this Sketchpad model: It makes the factoring itself easy and allows students to focus on the numerical and visual patterns of the blue and orange-shaded numbers.
Here is a partial list of questions and observations that students might make while exploring the interactive model above:
• The number 1 is always shaded blue.
• The number in the bottom-right corner of the array is always shaded blue.
• Our array has 11 rows, and I see that the number 11 is shaded blue. In general, if we have n rows in our grid, then the number n will be shaded blue.
• What arrays have just two of their numbers shaded blue?
• There are only two ways to display the factors of prime numbers in the array—either as a single row of circles or a single column of circles.
• If the array has at least 2 columns and 2 rows, then the number it represents isn’t prime.
• If the number of rows and columns in the array are equal and prime, the array will contain exactly three numbers shaded blue.
• We found a way to create arrays with exactly four numbers shaded blue. Drag the red point to form a single row of numbers. Make sure the largest number in the row is prime. Then drag the red point straight up to create a prime number of columns. That does the trick.
• When our array contains an even number of rows, the rows in the upper half of the array are filled entirely with orange circles. Only the number we’re factoring is shaded blue. Why is that?
• We can pair every number that is shaded blue with a partner. For example, the factors of 18 are 1, 2, 3, 6, 9, and 18. Let’s pair 1 and 18, 2 and 9, and 3 and 6 together. In each pair, the product of the numbers is 18.
• What happens when we pair the factors of 25? Its factors are 1, 5, and 25. We can pair 1 and 25 together, but can we pair 5 with itself?
• In most of our arrays, there are an even number of circles shaded blue. But in some cases, the number of blue circles is odd. Is there a way to predict whether there will be an even or odd number of blue circles?
• We dragged the red point so that the numbers from 1-20 all appeared in a single row. We wanted to find other ways to display those 20 circles in the array. The numbers in blue—1, 2, 4, 5, 10, and 20—gave us a big hint. Since 2 is a factor of 20, we can make a 2 x 10 array. Similarly, we can make a 4 x 5, a 5 x 4, and a 10 x 2 array.
• If the number of columns is even, the number 2 is always shaded blue. If the number of columns is odd, the number 2 alternates between orange and blue as I drag the red point up to add more rows.
• We created a game. We scrolled our sketch window so that you can only see the bottom row of circles. Your challenge is to make an educated guess about the total number of circles in the array. |
# In a triangle ABC right angled at B, find the value of tan A
Last Updated : 12 Oct, 2021
Trigonometry is the branch of mathematics that deals with the relationship of sides with angles in a triangle. Trigonometry is based on the principle that “If two triangles have the same set of angles then their sides are in the same ratio”. Side length can be different but side ratios are the same. Trigonometry is very important in physics and mathematics, it is used to find, the height of towers, or distance between stars, and in GPS Navigation System.
### Trigonometric Functions
Trigonometric functions are also called circular functions or trigonometric ratios. The relationship of angles and sides is represented by these trigonometric functions. There are six trigonometric functions Sine, Cosine, Tangent, Cosecant, Secant, Cotangent. The sides representations for the six ratios are,
• sin A = Perpendicular / Hypotenuse
• cos A = Base / Hypotenuse
• tan A = Perpendicular / Base
• cot A = Base / Perpendicular
• sec A = Hypotenuse / Base
• cosec A = Hypotenuse / Perpendicular.
Here, A is the angle opposite to the perpendicular side. Let’s learn what the Perpendicular, base, and hypotenuse of a right-angled triangle are,
1. Perpendicular: The side in front of the angle is perpendicular. In this case, the side in front of 30° is called it’s perpendicular.
2. Base: A base is one of the sides which touches angle, but hypotenuse can never be considered as base.
3. Hypotenuse: It is a side opposite to 90°. it is the largest side.
Note: Perpendicular and base changes as angle changes. In a triangle, a side is perpendicular for an angle, but the same side is a base for another angle, but the hypotenuse remains the same because it is a side opposite to angle 90°.
As shown in the above diagram for the same triangle if considered angle 30° the perpendicular is the side PQ, but if considered angle 60° the perpendicular is side QR.
### In a triangle ABC right angled at B, find the value of tan A
Tan – The tan of an angle A is the ratio of lengths of perpendicular to the base.
Tan A = Perpendicular / Base
Let us see how a triangle ABC right angled at B looks like.
Since Tan (A) = perpendicular / Base.
In a triangle right angled at B, Tan (A) = BC / AB.
### Sample Problems
Question 1: In a right angle triangle, angle A is 60°, and the base is 3m. Find the length of perpendicular.
Solution:
Given: Base = 3m
Tan 60 = √3
P/B = √3
P/3 = √3/1
p = 3√3
Question 2: In a right angle triangle, angle A is 30°, and the Perpendicular is 3m. Find the length of the Base.
Solution:
Given: perpendicular = 3m
Tan 30° = 1/√3
P/B = 1/√3
3/B = 1/√3
B = 3√3
Question 3: In a right angle triangle, for an angle A, the Perpendicular is 9√3m and the Base is 9m, find the angle A.
Solution:
Given: perpendicular = 9√3m, Base = 9m.
Tan A = 9√3 / 9
Tan A = √3
Tan (60°) = √3
Angle A = 60°
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GCSE Maths Geometry and Measure Area
Area Of A Trapezium
# Area Of A Trapezium
Here we will learn about finding the area of a trapezium, including compound area questions, questions with missing side lengths and questions involving unit conversion.
There are also area of a trapeziumworksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What is the area of a trapezium?
The area of a trapezium is the amount of space inside the trapezium. It is measured in units squared ( cm^{2}, m^{2}, mm^{2} etc.)
A trapezium (also known as a trapezoid) is a quadrilateral with exactly one pair of parallel sides. The parallel sides are indicated using small arrows.
The plural of trapezium is trapezia or trapeziums.
An isosceles trapezium is a trapezium where the base angles are equal and therefore the left and right side lengths are also equal.
The area of a trapezium is calculated by:
2. Dividing by 2
3. Multiplying by the height.
This can also be summarised with the following formula:
\text { Area }_{\text {Trapezium }}=\frac{a+b}{2} \times h
Where a and b are the parallel sides
Where h is the perpendicular height of the trapezium
## How to find the area of a trapezium
In order to find the area of a trapezium
1. Find the sum of the parallel sides.
2. Divide by 2 .
3. Multiply by the perpendicular height of the trapezium.
Another way of calculating the area is by using the formula below:
\text { Area }_{\text {Trapezium }}=\frac{a+b}{2} \times h
Where a and b are the parallel sides
Where h is equal to the perpendicular height of the trapezium
## Related lessons on area
Area of a trapezium is part of our series of lessons to support revision on area. You may find it helpful to start with the main area lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
## Area of a trapezium examples
### Example 1: finding area given the parallel side lengths and height
Find the area of the trapezium below:
1. Find the sum of the parallel sides.
4+6=10
2Divide by 2.
10\div2=5
3Multiply by the perpendicular height of the trapezium.
In this case the perpendicular height is 3 .
5\times3=15
\text { Area }=15 \mathrm{~m}^{2}
### Example 2: finding the area of a trapezium requiring converting units
Find the area of the trapezium below:
Find the sum of the parallel sides.
Divide by 2.
Multiply by the perpendicular height of the trapezium.
### Example 3: worded question
Sarah is looking to design the flower bed below at the front of her house. She would like to cover the entire flower bed with rose bushes. Each bush needs 1.5m^{2} to allow the roots to spread out evenly. What is the maximum number of rose bushes she can plant on her flower bed?
Find the sum of the parallel sides.
Divide by 2.
Multiply by the perpendicular height of the trapezium.
### Example 4: area of compound shapes
Find the area of the shaded region below:
Find the sum of the parallel sides.
Divide by 2.
Multiply by the perpendicular height of the trapezium.
### Example 5: calculating height given the area
The area of the trapezium below is 128m^{2} . Calculate the value of h in the trapezium below:
Rearrange the area of a trapezium formula
Substitute in the known values
### Example 6: calculating base length given the area
The area of a trapezium is 322 square millimetres and its height is 14mm . If the length of one of the parallel sides is 27mm , what is the length of the other parallel side?
Let’s start out by sketching out a trapezium using what is given in the question:
Rearrange the area of a trapezium formula
Substitute in the known values
### Common misconceptions
• Using incorrect units for the answer
A common error is to forget to include squared units when asked to calculate area.
• Forgetting to convert measures to a common unit
Before calculating the area of a trapezium, pupils must look at the units given in the question. If different units are given e.g. length = 4m and width = 3cm pupils must convert them either both to cm or both to m .
• Using length of the non-parallel sides when calculating area and not the height
Sometimes in a question we are given additional lengths which are not needed in our calculations. Sometimes we are given the length of the non-parallel sides of the trapezium. We must be careful to not mistake these measurements for height.
### Practice area of a trapezium questions
1. Find the area of the trapezium below:
20.5m^{2}
36m^{2}
67.5m^{2}
47.5m^{2}
Add the parallel sides 9 and 7 together. Divide the sum by 2 . Times by the perpendicular height of 4.5 .
2. Find the area of the trapezium below:
5085cm^{2}
5085m^{2}
2055.6cm^{2}
2055.6m^{2}
Convert 0.68m into cm . Add the 45cm and 68cm together. Divide by 2 . Multiply by 90 .
3. Sally is looking to design the garden below at the back of her house. She would like to cover the entire garden with grass. An individual carpet of grass covers an area of 4 square metres. How many rolls of grass will be needed to cover the entire garden?
42.5
10
11
165
Add the parallel sides 11 and 6 together. Divide the sum by 2 . Times by the perpendicular height of 5 . Divide by 4 .
4. Find the area of the shaded region below:
224m^{2}
84.5m^{2}
44m^{2}
55m^{2}
Find the area of the rectangle by multiplying its length times width. (5 x 11) . Split the hexagon in half and find the area of each trapezium. To do this add the parallel sides ( 11 and 15 ), divide by 2 , and multiply by the perpendicular height which in this case is 6.5 . As there are two congruent trapeziums, multiply the area of the trapezium by 2 and add on the area of the rectangle to get the compound area.
5. The area of the trapezium below is 68m^{2} . Calculate the value of h in the trapezium below:
8.5m^{2}
60m^{2}
16m
8.5m
Start with the area of a trapezium formula \text { Area }_{\text {Trapezium }}=\frac{a+b}{2} \times h and rearrange to get the formula h=\frac{2 A}{(a+b)} . Substitute the values of the area and the parallel sides and solve for the height.
6. Calculate the value of a in the trapezium below:
55mm
13mm
45mm
14mm
Start with the area of a trapezium formula \text { Area }_{\text {Trapezium }}=\frac{a+b}{2} \times h and rearrange to get the formula a=\frac{2 A}{h}-b . Make sure to isolate a by moving the other variables to the other side of the equal sign and changing to the opposite sign. Substitute the known values and solve for the length of the parallel side.
### Area of a trapezium GCSE questions
1. Find the area of the trapezium below.
(3 marks)
Finding sum of parallel sides (12+18=30)
(1)
Dividing by 2 or multiplying by height
(1)
75m^{2} (cao with units)
(1)
2. (a) Calculate the value of x in the trapezium below.
(b) The above trapezium is enlarged by a scale factor of 2 . What is the area of the new trapezium?
(7 marks)
(a)
Evidence of adding the parallel sides 11+17 = 28
(1)
Dividing by 2
(1)
70\div14 seen
(1)
5m seen (cao including units)
(1)
(b)
2^{2} oe seen
(1)
70\times4 seen
(1)
280m^{2}
(1)
3. A painter is looking to paint his deck. Below is a blueprint of the deck. A can of paint costs £5.50 and covers an area of 1.5cm^{2} . How much would it cost to paint the entire deck?
(5 marks)
Adding parallel sides of trapezium and dividing by 2 or multiplying by height
(1)
50cm^{2} seen as area of trapezium
(1)
\frac{1}{2}bh seen or correct values substituted or 24cm^{2} seen with some calculations
(1)
Evidence of adding areas or dividing total are by 1.5 or multiplying by £5.50
(1)
£271.33
(1)
## Learning checklist
You have now learned how to:
• Calculate and compare the area of trapeziums including using standard units
• Calculate the area of trapeziums and related composite shapes
## Still stuck?
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# 4 Quarts of Water
Question: If you had an infinite supply of water and a 5 quart and 3 quart pails, how would you measure exactly 4 quarts? and What is the least number of steps you need?
Update: There is a better solution in the comments section which requires only 6 steps.
Answer: This question is very simple actually. Since we can’t hold 4 quarts in the 3 quart pail, we have to look to filling up the 5 quart pail with exactly 4 quarts. Lets count the steps as we move along
1. Fill 3 quart pail ( 5p – 0, 3p – 3)
2. Transfer to 5 quart pail (5p – 3, 3p – 0)
3. Fill 3 quart pail ( 5p – 3, 3p – 3)
4. Transfer to 5 quart pail (5p – 5, 3p – 1)
5. Empty 5 quart pail (5p – 0, 3p – 1)
6. Transfer to 5 quart pail (5p – 1, 3p – 0)
7. Fill 3 quart pail ( 5p – 1, 3p – 3)
8. Transfer to 5 quart pail (5p – 4, 3p – 0) We are done!!!
That was easy right. Now for those who are mathematical and need everything solved in terms of a formula, here comes a little more mathematical solution.
Now the general steps are to fill up the 3 quart pail and keep transferring to the 5 quart pail (empty if full) until we hit 4 quarts. Therefore, the total amount of water we filled in the 3 quart pail must be equal to 4 more than a multiple of 5 (since we discard 5 quarts of water at a time). From this, we can derive this formula.
5n + 4 = 3m, where n and m are arbitrary positive integers
n represents the number of times we had to empty the 5 quart pail and m represents the number of times we had to fill up the 3 quart pail.
Now all we have to do is solve for the lowest set of positive integer solutions for {n,m}. {1, 3} is the lowest set. Some of the other solutions are {4, 8}, {7, 13}, {10, 18} and so on.
Hope you enjoyed the nerdy mathematical solution.
If you have any questions, please feel free to send me an email atย support@mytechinterviews.com. If you have any interview questions which you feel would benefit others, I would love to hear about it.
If you're looking for some serious preparation for your interviews, I'd recommend this book written by a lead Google interviewer. It has 189 programming questions and solutions:
## 12 Responses
1. Pawan says:
6 steps
1. Fill 5p
2. Transfer 3quart of 5 to 3p
3. Make empty 3p
4. Transfer remaining 2quart of 5p to 3p
5. Fill 5p
6. Transfer 1quart of 5p to 3p =>>Ans remaining 4quart in 5p
2. Anonymous says:
1. Fill 5 and 3 quart pail simultaneously
2. Empty 3 quart pail once its filled, and now let it fill for remaining 2 quart (i.e. when 3 quart pail if full, we have 2 quartz in 5 quart pail)
3. Now, by the time we fill 3 quart pail with 2 quartz of water, we have 5 quartz pail full. So now empty the 5 quarts pail.
4. Now, 3 quart pail is getting filled with 1 more quartz of water and simultaneously 5 quart pail is also getting filled. When 3 quart pail is full, shift the entire water in 5 quart pail (Since, right now 5 quart pail has 1 quart of water and 3 quartz pail has 3 quartz of water)
3. theAmitom says:
***** Best Solution ***** ๐
1. Fill 5 quart pail & Transfer to 3 quart pail (5p = 2, 3p = 3)
2. Throw water of 3 quart pail & Transfer 5 quart pail water to 3 quart pail (5p = 0, 3p = 2)
3. Fill 5 quart pail & Transfer to 3 quart pail (5p = 4, 3p = 3)
You will be left with 4 quarts in 5 quart pail ๐
4. RJ says:
Shrenik’s answer assumes that the pail is a perfect cylinder or prism, whose sides make a 90 degree angle with the base. If that is the case, then that answer could possibly work (if you are completely accurate when you’re tilting). But most pails are tapered, having a wider top and a narrower bottom. And in those cases, that method would not work.
5. Avin says:
Containers need not to be symmetrical. So minimum #steps will be 6 by the solution provided by Michael.
6. Gaurav says:
I am impressed with Shrenik’s solution…but I have my own alternative:
1. Fill 3 quart pail and transfer to 5 quart pail
2. Fill 3 quart pail again and while transferring, now, when 5 quart pail is full, retain the water in 3 quart pail (1 quart), whilst emptying the 5 quart pail.
3. Now transfer 1 quart from 3 quart pail to 5 quart pail.
4. Fill 3 quart pail and transfer to 5 quart pail.
Maybe transferring the contents is also considered one step!
7. db says:
I’m with Shrenik. I was going to leave that answer but he/she already did. Its the fastest and its the simplest. Such logic hasn’t landed me a job anywhere near google or microsoft though…
9. Rog says:
Thinking a little outside the box, or pail as it were…
1. Place the 3 quart pail in 5 quart pail.
2. Fill the 5 quart pail, whilst holding the 3 quart pail such that both rims are coplanar.
3. Remove the 3 quart pail and transfer the contents (2 quarts) to it.
4. Fill the 5 quart pail.
5. Transfer the contents to the 3 quart pail until full (1 quart).
You are now left with 4 quarts in the 5 quart pail.
10. Doggie says:
I guess that could be a solution provided you talk in through with the interviewers and they don’t mind that answer. My guess is that they will likely make an excuse at that point to make sure you can’t use that logic. ๐
11. How about using the diagonal-divides-in-half property of a symmetrical container? For example if you fill a pail to the brim and then tilt it to the point where the surface of the water overlaps the diagonal plane of the pail from the bottom to the top, the pail will be exactly 1/2 full.
1. Fill 5 quart pail, tilt
2. Fill 3 quart pail, tilt
3. Transfer contents of 3 quart to 5 quart pail
12. I’m pretty certain there is a faster solution to this, but I’m not sure that I can mathematically prove that it always works.
1. Fill the 5 quart jug (5p = 5, 3p = 0)
2. Transfer to 3 quart jug (5p = 2, 3p = 3)
3. Throw away 3 quart jug (5p = 2, 3p = 0)
4. Transfer to 3 quart jug (5p = 0, 3p = 2)
5. Fill 5 quart jug (5p = 5, 3p = 2)
6. Transfer to 3 quart jug (5p = 4, 3p = 3)
Hope that proves interesting.
XHTML: These are some of the tags you can use: `<a href=""> <b> <blockquote> <code> <em> <i> <strike> <strong>` |
• # Revision:Further Vectors
## Vector equations of a line
Lines can be written in several different ways:
There’s the c4 way:
The other c4 way:
Also the Cartesian form:
## Vector equations of a plane
Similar to the c4 line equation:
Similar again: I used numbers for the sake of notation. 2 i’s won’t look pretty!
Here’s a new one:
Or:
Or:
## Vector product
• The definition of the vector product is where is the unit vector perpendicular to both a and b in the direction given by the right hand screw rule.
• For vectors in component form, the vector product may be calculated using either of the following determinants, the second of which is given in the formula booklet:
or
• The vector product of any 2 parallel vectors is zero
• The vector product is anti-commutative as
• Area of a triangle is , of a parallelogram is , and representing adjacent sides.
## Intersections
#### Point of intersection of a line and a plane
• Express the position vector of a general point on the line as a single vector, e.g.
• substitute this vector into the normal form of the equation of a plane and find , e.g.
etc.
• Substitute your value of back into the equation of the line.
#### Line of intersection of 2 intersecting planes
• Find the direction vector of the line by using the fact that the line is perpendicular to both normal vectors, i.e.
• Find any point , common to both planes by using any value (usually 0) for x, y or z
• Equation of line is
## Angles
• The direction vector of a line is given by d in the equation r= d
• The direction of the normal to a plane is given by n in the equation r.n = a.n
• The angle between two planes is the angle between their 2 normals, n_1 and n_2. This is usually found by using the scalar product, a.b =
• The angle between two lines is the angle between their 2 direction vectors, d_1 and d_2.
• The angle between a line and a plane is found by finding the angle between d and n then subtracting from 90 degrees. Alternatively, if is the angle between d and n then as you may use
## Distances
• The distance of a plane r.n = p from the origin is or where the sign is unimportant.
• To find the distance between two planes, first find their distances from the origin. If the two p-values are the same sign, the planes are on the same side of the origin so subtract the 2 distances. If the two p-values are different signs, the origin lies between them so add the positive distances.
#### Distance between a plane and a point, A
1)*Express the equation in the form
• If the point is , quote the result from the formula booklet giving:
Distance =
2)*Find the equation of the plane through the given point A, parallel to the original plane, using r.n = a.n
• Find the distance between these 2 planes.
3)*Locate any point, P on the plane and find the vector
• If is the angle between and n, then the required distance h is so h =
4)*Find the co-ordinates of F, the foot of the perpendicular from A to the plane and find AF. This method should only really be used if F is specifically asked for because it’s a bit long winded. Anyway, F can be found as follows:
• As AF is perpendicular to the plane, then n, so = a + n
• Proceed as for and intersection of a line and a plane
#### Distance between a point A, from a line L
1)*Locate any point, P on L and find the vector
• If is the angle between and d, then the required distance h is so h =
2)* Find the co-ordinates of F, the foot of the perpendicular from A to L and find AF. F can be found as follows:
• As F lies on L, express the position vector of F as a single vector involving
• Use the fact that to find
• As is perpendicular to L, then so use this to find and hence F
#### Distance between 2 skew lines
• Locate any point, A on and any point B on and find the vector = ba
• Find the common normal, i.e. the vector which is perpendicular to both and . This is given by n = and may be simplified if appropriate
• If is the angle between and n, then the required distance h is so h =
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# Unit 1 Overview: Exploring One-Variable Data
Josh Argo
Jed Quiaoit
Josh Argo
Jed Quiaoit
## What is Statistics?
is all about data. We collect sets of data, analyze our data and ultimately, use our data sets to make inferences about larger sets of individuals in our population.
We are going to be focusing on univariate data or one-variable data in this unit. This is data that only has one aspect of it that is being measured. Among our sets of , we will divide our data sets into two different types: quantitative and categorical. ๐
### Quantitative Data
Have you ever wondered what the average AP score was? Or perhaps the average number of bananas purchased at the grocery store per bunch? Both of these are examples of because each individual is assigned a quantity. Whether it is assigning each test taker an AP score, or each banana bunch purchased, each individual being measures is assigned a number. One of the big giveaways for is that we can take the , or the average, of the data set. In other words, is average-able. ๐ฒ
EXAMPLE: You have taken 5 exams in your math class and you want to know your average score. The scores on the exams are as follows:
Exam 1: 80
Exam 2: 90
Exam 3: 70
Exam 4: 85
Exam 5: 75
To find the average, you need to add up all of the exam scores and then divide by the total number of exams. In this case, the total score is 80 + 90 + 70 + 85 + 75 = 400, and the total number of exams is 5.
Therefore, the average exam score is 400 / 5 = 80; in this example, your average exam score is 80.
This is a very simple example and in practice, you may encounter more complex problems that involve larger datasets and more variables. However, the basic principle of finding the average by summing the values and dividing by the count remains the same.
๐ก uses means, or averages, to make inference!
### Categorical Data
On the flip side, we have . Have you ever asked a group of people whether they liked coffee? What about what their favorite vegetable is? How about if they prefer ๐ฉ or ๐ช for dessert? Each of these types of surveys would be examples of . The reason why is because each individual chooses a category: do you fall into the ๐ฉ or ๐ช category? Because of this separation of data, it is impossible to calculate the average dessert preference. After all, it would not make sense to make a statement like "the average dessert preference is a cookie." Instead, we typically measure categorical datasets using measures like proportions. It makes a lot more sense to make a statement like, "the of people who prefer cookies is 0.65."
Here are some examples of statements outlining using proportions:
1. "In a survey of 100 people, 50% identified as male and 50% identified as female."
2. "In a sample of 300 customers, 20% reported having a positive experience with the company's customer service, while 80% reported a negative experience."
3. "Of the 1000 people surveyed, 30% reported having a bachelor's degree, while 70% reported having a high school diploma or lower level of education."
4. "Of the 200 products reviewed, 40% received a rating of 4 or 5 stars, while 60% received a rating of 3 stars or lower."
5. "In a study of 500 students, 25% reported experiencing bullying at school, while 75% reported not experiencing bullying."
In each of these examples, the of individuals or items in each category is described using percentages. This allows us to see the relative frequency or prevalence of each category within the data. โ๏ธ
๐ก uses percentages, or proportions, to make inference.
## What's the Point of Statistics?
In practice, is used in a wide range of fields, including business, economics, biology, psychology, social sciences, and many others. It is a powerful tool for understanding and interpreting real-world phenomena, and is used to inform decision-making, policy-making, and research in a variety of contexts. ๐
• Collecting data through surveys, experiments, or other methods
• Describing and summarizing data using measures such as , median, mode, and standard deviation
• Visualizing data using graphs and plots
• Testing hypotheses and making inferences about population parameters based on sample data
• Building statistical models to predict outcomes or understand relationships between variables
Even beyond this course, there are many different branches of , including , , , and more. Each of these areas has its own set of techniques and approaches for analyzing data! โ๏ธ
## Context of Data
One of the major things that is going to feel very different for this course as opposed to other mathematics courses you have taken in the past is the way in which you record your answers. In an Algebra or Calculus course, it is sufficient to say "x = 5" when that is your answer. In AP , it is a good idea to go ahead and get in a habit of tying your answer to whatever the specific context of the problem you are working on. Instead of simply saying, "x = 5" make your answer more specific by saying things like "the average number of bananas per bunch is 5." ๐งฉ
๐ก Our goal in is not just to find the correct answer, but to communicate our findings to our audience so that the answer is useful in making further predictions.
## Describing Data
Perhaps the biggest concept and skill of this first unit is being able to describe data. In , this consists of four main parts: , , , and . It is also important to include context in your answer. ๐
For example, if we had a set of data regarding the amount of bananas per bunch purchased, a model response may look like the following: "The number of bananas purchased was 5 bananas, There was one outlier when a customer purchased a bunch of 12 bananas. The of our data distribution was fairly symmetric. The range of bananas per bunch was 10, with the largest bunch being 12 and the smallest bunch being 2."
In , this process may look different. It is usually more valuable with context data to discuss which category was most likely to happen and which was least likely to happen. For example, a description could look like this: "Our most likely outcome was people who prefer donuts with a of 0.45 and our least likely outcome was people who prefer cookies with a of 0.15." ๐จโ๐ณ
Sometimes it is also beneficial with to discuss raw counts rather than proportions. However, it is more likely that the AP exam will ask you to describe a distribution of a set rather than a set. For more information on content from Unit 1, check the link below! ๐โโ๏ธ
๐ฅ Watch: AP Stats - Unit 1 Streams
# Key Terms to Review (13)
Categorical Data
: Categorical data refers to data that can be divided into categories or groups based on qualitative characteristics.
Center
: The center refers to the middle or average value of a data set. It represents the typical or central value around which the data tends to cluster.
Descriptive Statistics
: Descriptive statistics involves organizing, summarizing, and presenting data in a meaningful way to describe its main features.
Inferential Statistics
: Inferential statistics involves using sample data to make inferences or draw conclusions about a population.
Mean
: The mean is the average of a set of numbers. It is found by adding up all the values and dividing by the total number of values.
Outliers
: Outliers are extreme values that significantly differ from other values in a dataset. They can greatly affect statistical analyses and should be carefully examined.
Predictive Modeling
: Predictive modeling involves using historical data and statistical algorithms to make predictions about future outcomes.
Proportion
: A proportion is a fraction or percentage that represents the relationship between a part and a whole in a population or sample.
Quantitative Data
: Quantitative data refers to numerical information that can be measured or counted. It involves quantities and can be analyzed using mathematical methods.
Shape
: In statistics, shape refers to the overall appearance or form of a distribution. It describes how the data is distributed and can be characterized by its symmetry, skewness, or modality.
: Spread refers to how much variability or dispersion exists within a data set. It measures how far apart the values are from each other.
Statistics
: Statistics is the study of collecting, analyzing, interpreting, presenting, and organizing data. It involves using mathematical techniques to make sense of information and draw conclusions.
Univariate Data
: Univariate data refers to a type of data that consists of only one variable. It focuses on analyzing and describing patterns within a single set of observations.
# Unit 1 Overview: Exploring One-Variable Data
Josh Argo
Jed Quiaoit
Josh Argo
Jed Quiaoit
## What is Statistics?
is all about data. We collect sets of data, analyze our data and ultimately, use our data sets to make inferences about larger sets of individuals in our population.
We are going to be focusing on univariate data or one-variable data in this unit. This is data that only has one aspect of it that is being measured. Among our sets of , we will divide our data sets into two different types: quantitative and categorical. ๐
### Quantitative Data
Have you ever wondered what the average AP score was? Or perhaps the average number of bananas purchased at the grocery store per bunch? Both of these are examples of because each individual is assigned a quantity. Whether it is assigning each test taker an AP score, or each banana bunch purchased, each individual being measures is assigned a number. One of the big giveaways for is that we can take the , or the average, of the data set. In other words, is average-able. ๐ฒ
EXAMPLE: You have taken 5 exams in your math class and you want to know your average score. The scores on the exams are as follows:
Exam 1: 80
Exam 2: 90
Exam 3: 70
Exam 4: 85
Exam 5: 75
To find the average, you need to add up all of the exam scores and then divide by the total number of exams. In this case, the total score is 80 + 90 + 70 + 85 + 75 = 400, and the total number of exams is 5.
Therefore, the average exam score is 400 / 5 = 80; in this example, your average exam score is 80.
This is a very simple example and in practice, you may encounter more complex problems that involve larger datasets and more variables. However, the basic principle of finding the average by summing the values and dividing by the count remains the same.
๐ก uses means, or averages, to make inference!
### Categorical Data
On the flip side, we have . Have you ever asked a group of people whether they liked coffee? What about what their favorite vegetable is? How about if they prefer ๐ฉ or ๐ช for dessert? Each of these types of surveys would be examples of . The reason why is because each individual chooses a category: do you fall into the ๐ฉ or ๐ช category? Because of this separation of data, it is impossible to calculate the average dessert preference. After all, it would not make sense to make a statement like "the average dessert preference is a cookie." Instead, we typically measure categorical datasets using measures like proportions. It makes a lot more sense to make a statement like, "the of people who prefer cookies is 0.65."
Here are some examples of statements outlining using proportions:
1. "In a survey of 100 people, 50% identified as male and 50% identified as female."
2. "In a sample of 300 customers, 20% reported having a positive experience with the company's customer service, while 80% reported a negative experience."
3. "Of the 1000 people surveyed, 30% reported having a bachelor's degree, while 70% reported having a high school diploma or lower level of education."
4. "Of the 200 products reviewed, 40% received a rating of 4 or 5 stars, while 60% received a rating of 3 stars or lower."
5. "In a study of 500 students, 25% reported experiencing bullying at school, while 75% reported not experiencing bullying."
In each of these examples, the of individuals or items in each category is described using percentages. This allows us to see the relative frequency or prevalence of each category within the data. โ๏ธ
๐ก uses percentages, or proportions, to make inference.
## What's the Point of Statistics?
In practice, is used in a wide range of fields, including business, economics, biology, psychology, social sciences, and many others. It is a powerful tool for understanding and interpreting real-world phenomena, and is used to inform decision-making, policy-making, and research in a variety of contexts. ๐
• Collecting data through surveys, experiments, or other methods
• Describing and summarizing data using measures such as , median, mode, and standard deviation
• Visualizing data using graphs and plots
• Testing hypotheses and making inferences about population parameters based on sample data
• Building statistical models to predict outcomes or understand relationships between variables
Even beyond this course, there are many different branches of , including , , , and more. Each of these areas has its own set of techniques and approaches for analyzing data! โ๏ธ
## Context of Data
One of the major things that is going to feel very different for this course as opposed to other mathematics courses you have taken in the past is the way in which you record your answers. In an Algebra or Calculus course, it is sufficient to say "x = 5" when that is your answer. In AP , it is a good idea to go ahead and get in a habit of tying your answer to whatever the specific context of the problem you are working on. Instead of simply saying, "x = 5" make your answer more specific by saying things like "the average number of bananas per bunch is 5." ๐งฉ
๐ก Our goal in is not just to find the correct answer, but to communicate our findings to our audience so that the answer is useful in making further predictions.
## Describing Data
Perhaps the biggest concept and skill of this first unit is being able to describe data. In , this consists of four main parts: , , , and . It is also important to include context in your answer. ๐
For example, if we had a set of data regarding the amount of bananas per bunch purchased, a model response may look like the following: "The number of bananas purchased was 5 bananas, There was one outlier when a customer purchased a bunch of 12 bananas. The of our data distribution was fairly symmetric. The range of bananas per bunch was 10, with the largest bunch being 12 and the smallest bunch being 2."
In , this process may look different. It is usually more valuable with context data to discuss which category was most likely to happen and which was least likely to happen. For example, a description could look like this: "Our most likely outcome was people who prefer donuts with a of 0.45 and our least likely outcome was people who prefer cookies with a of 0.15." ๐จโ๐ณ
Sometimes it is also beneficial with to discuss raw counts rather than proportions. However, it is more likely that the AP exam will ask you to describe a distribution of a set rather than a set. For more information on content from Unit 1, check the link below! ๐โโ๏ธ
๐ฅ Watch: AP Stats - Unit 1 Streams
# Key Terms to Review (13)
Categorical Data
: Categorical data refers to data that can be divided into categories or groups based on qualitative characteristics.
Center
: The center refers to the middle or average value of a data set. It represents the typical or central value around which the data tends to cluster.
Descriptive Statistics
: Descriptive statistics involves organizing, summarizing, and presenting data in a meaningful way to describe its main features.
Inferential Statistics
: Inferential statistics involves using sample data to make inferences or draw conclusions about a population.
Mean
: The mean is the average of a set of numbers. It is found by adding up all the values and dividing by the total number of values.
Outliers
: Outliers are extreme values that significantly differ from other values in a dataset. They can greatly affect statistical analyses and should be carefully examined.
Predictive Modeling
: Predictive modeling involves using historical data and statistical algorithms to make predictions about future outcomes.
Proportion
: A proportion is a fraction or percentage that represents the relationship between a part and a whole in a population or sample.
Quantitative Data
: Quantitative data refers to numerical information that can be measured or counted. It involves quantities and can be analyzed using mathematical methods.
Shape
: In statistics, shape refers to the overall appearance or form of a distribution. It describes how the data is distributed and can be characterized by its symmetry, skewness, or modality.
: Spread refers to how much variability or dispersion exists within a data set. It measures how far apart the values are from each other.
Statistics
: Statistics is the study of collecting, analyzing, interpreting, presenting, and organizing data. It involves using mathematical techniques to make sense of information and draw conclusions.
Univariate Data
: Univariate data refers to a type of data that consists of only one variable. It focuses on analyzing and describing patterns within a single set of observations. |
Video: Simplifying Quotients of Fractions Containing Monomials
Simplify the expression (4𝑥²𝑦³/10𝑥²𝑦)/(2𝑦⁴/3𝑥⁴).
02:26
Video Transcript
Simplify the expression four 𝑥 squared 𝑦 cubed divided by 10𝑥 squared 𝑦 all divided by two 𝑦 to the fourth power divided by three 𝑥 to the fourth power.
To begin, let’s go ahead and rewrite this. This way, it’s a little bit easier to see that we’re dividing two fractions. And when dividing two fractions, we actually multiply by the second fraction’s reciprocal. So we change the division to multiplication. And then, we’re multiplying by the reciprocal. So we flip the second fraction. And when multiplying fractions, we multiply the numerators together and the denominators together. So we can cancel anything on the numerators with anything on the denominators. Two can go into itself once. And it can also go into four twice. These 𝑥 squares can cancel.
Before cancelling anymore, let’s see what we have left. On the numerator, we have two 𝑦 cubed three and 𝑥 to the fourth. So two times three gives us six. And we have 𝑦 cubed. And we also have 𝑥 to the fourth. On the denominator, we have 10. And we have 𝑦. And we have 𝑦 to the fourth. So 𝑦 times 𝑦 to the fourth is where we add their exponents. So 𝑦 would be 𝑦 to the first power. And one plus four would give us five. So the six tenths can reduce. So how come we didn’t catch that before? Well, two could’ve went into 10. And that’s where it would’ve simplified.
So back to where we were, six tenths, that does reduce. We can simplify them both by two, making three-fifths. And then we have 𝑦 to the third on the numerator and 𝑦 to fifth on the denominator. We can think of this two ways. When dividing with like bases, we subtract their exponents. And three minus five would be negative two. But when we have a negative exponent, if it’s on the numerator, we can move it to the denominator and make it positive, and vice versa. If it would’ve been negative on the denominator, we could’ve moved it to the numerator to make it positive.
Another way to think about this is if 𝑦 cubed is on the top, that means there are three 𝑦s on the top. And 𝑦 to the fifth is on the bottom. So there will be five 𝑦s on the bottom. Three of them would cancel. And there would be two left on the bottom. And then lastly, we have an 𝑥 to the fourth power. And there are no 𝑥s in the denominator. So we can’t simplify anymore. Therefore, three 𝑥 to the fourth power divided by five 𝑦 squared will be our final answer. |
# Ohm s Law and Simple DC Circuits
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1 Ohm s Law and Simple DC Circuits 2EM Object: Apparatus: To confirm Ohm s Law, to determine the resistance of a resistor, and to study currents, potential differences, and resistances in simple direct current (DC) parallel and series circuits. Two multimeters, various resistors, connecting wires, solderless breadboarding system, and potentiometer. FOREWORD Ohm s Law states a simple relation among the potential difference V across a conductor, the current I through it, and the resistance R the conductor presents to the current. A mathematical statement of Ohm s Law is: (1) R = V/I If V and I are measured in volts and amperes, respectively, the resistance R will be given in Ohms. This relation is so well established that it hardly needs verification by students in beginning Physics courses. However, the experience gained in setting up the necessary circuitry to verify Ohm s Law and its application to simple DC circuits justifies an experiment in which laboratory skills are stressed. Particularly in this experiment, the student will learn how to determine the resistance of a resistor, the proper use of a voltmeter and ammeter, how a potentiometer works, and will develop some basic wiring skills. The validity of Ohm s law can be easily demonstrated if a circuit is set up whereby V and I can be varied for a conductor of resistance R. Figure 1 shows the schematic of such a circuit. A + + B R S + V Figure 1. Schematic of a circuit that can be used to verify Ohm s Law. In this schematic, B represents the battery or source of electrical potential difference, R represents the resistor, V represents the voltmeter used to measure the potential difference across the resistor, A represents the ammeter used to measure the current through the resistor, and S represents a SPST switch. Note the polarity of the battery and the meters. The potential difference V can be varied by changing the size of the battery and/or by using a potential divider, and the resulting current I can be measured. 2EM 1
2 Measurements will be made on individual resistors and also on resistors connected in series and in parallel. Figure 2 shows two resistors R 1 and R 2 with series and parallel connections. R 1 R 2 (a) (b) R 1 R 2 Resistors in series Resistors in parallel Figure 2. Two resistors connected in (a) series and (b) parallel. The equivalent resistance R T of two resistors in series is given by: (2) R T = R 1 + R 2 When two resistors are connected in parallel the equivalent resistance is given by: (3) 1 = R t R 1 R 2 PROCEDURE Figure 3 shows the solderless breadboarding system we will be using in this experiment and Figure 4 shows just the breadboard and connections beneath the cover plate. Figure 3. Solderless breadboarding system used in this experiment. 2EM 2 Figure 4. Breadboard and internal connections.
3 Figure 5 shows the resistance color code. 1st Band 2nd Band COLOR BANDS 1st 2nd 3rd 4th Black Brown Red Orange Yellow Green ,000 Blue 6 6 1,000,000 Purple 7 7 Grey 8 8 White 9 9 Gold 10 5% Silver % 20% rd Band 4th Band Figure 5. Resistance color code. Part I. Determining the Resistance of a Resistor In your component box, you will find some resistors painted blue and some resistors painted orange. In this part of the experiment, we will determine the resistance of the blue resistors, the orange resistors, the blue and orange resistors in series, and the blue and orange resistors in parallel. Using a blue resistor, wire the circuit shown in Figure 6. The small device with the knob is the potential divider. Note the polarity of the meters. For this part of the experiment, you will need to measure electrical potential differences in the range of volts and electric currents in the range of ma. Be sure to have your lab instructor check your wiring before you connect the final wire (see. Figure 6.) I final wire + V R slider potential divider Figure 6. Circuit used to determine the resistance of a resistor. The potential divider is placed across a 5 volt DC source and then used to tap off 0 to 5 volts potential difference. A schematic representation of a potential divider is shown in Figure 7. 2EM 3
4 b slider + V a Figure 7. Schematic representation of a potential divider. As you move the slider from point a to point b, you tap off 0 to 5 volts. The voltmeter will read 0 volts at a, vary from 0 to 5 volts as you move the slider from a to b, and read 5 volts at point b. The potential divider you are using today varies slightly from the above and is shown in Figure 8. b slider + V a Figure 8. Schematic representation of the potential divider used in this experiment. An assembly diagram of the circuit to be wired (schematic shown in Figure 6) is shown in Figure 9. + V R I b b Potential Divider a Slider + 5 volts a 0 volts Figure 9. Assembly diagram for the schematic shown in Figure 6. 2EM 4
5 Vary the potential difference applied to the circuit by turning the knob of the potential divider. Record the current I for at least eight different values of the potential difference V between 0 and 4 volts. Repeat this process for the orange resistor, the blue and orange resistors in series, and finally, the blue and orange resistors in parallel. On one sheet of graph paper, plot V vs. I for each of these four cases. Write the equation of the plot for each of these four cases. From the equation of these plots, determine the resistance of the blue resistor, orange resistor, blue and orange resistors in series, and the blue and orange resistors in parallel. Using the determined value of your resistors, verify that for resistors in series equivalent resistance is given by: (2) R T = R 1 + R 2 Using the determined value of your resistors, verify that for resistors in parallel, the equivalent resistance is given by: (3) 1 = R t R 1 R 2 Part II. Investigating Resistors in Series Wire the circuit shown in Figure 10 on your solderless breadboard. R 1 =100Ω R 2 =200Ω Figure 10. Two resistors in series. We are interested in the electrical potential difference (V) across each element of the circuit and the electric current (I) through each element of the circuit. Since you have only one voltmeter and ammeter, you will need to move them around. Figure 11 shows where we would like to take measurements and what we will call them. Note the polarity of the meters. V 1 V 2 I 1 I 2 I T R 1 =100Ω R 2 =200Ω V Total V Power Supply Figure 11. Measurements to be made with two resistors in series. To keep this simple, wire the circuit as shown in Figure 10, then use the voltmeter to determine the potential difference at all the sites indicated in Figure 11. Next, move the ammeter around to the sites shown and measure the current. Record your measurements on the Data and Calculation Sheet and then perform the calculations indicated. 2EM 5
6 Part III. Investigating resistors in parallel. Wire the circuit shown in Figure 12 on your solderless breadboard. R 1 =100Ω R 2 =200Ω Figure 12. Two resistors in parallel. We are interested in the electrical potential difference (V) across each element of the circuit and the electric current (I) through each element of the circuit. Since you have only one voltmeter and ammeter, you will need to move them around. Figure 13 shows where we would like to take measurements and what we will call them. Note the polarity of the meters. V 1 R 1 =100Ω R 2 =200Ω V 2 V Total I 1 I 2 I Total VPower Supply Figure 13. Measurements to be made with two resistors in parallel. To keep this simple, wire the circuit as shown in Figure 12, then use the voltmeter to determine the potential difference at all the sites indicated in Figure 13. Next, move the ammeter around to the sites shown and measure the current. Record your measurements on the Data and Calculation Sheet and then perform the calculations indicated. 2EM 6
7 NAME SECTION DATE DATA AND CALCULATION SUMMARY R blue R orange R blue and R orange R blue and R orange in series in parallel V (Volts) I (ma) I (ma) I (ma) I (ma) Equation for the plot of V vs. I for R blue : Equation for the plot of V vs. I for R orange : Equation for the plot of V vs. I for R blue and R orange in series : Equation for the plot of V vs. I for R blue and R orange in parallel : Experimental values for the resistors R blue = R blue and R orange in series = R orange = R blue and R orange in parallel = Verify that resistors in series add as R 1 + R 2 = R T Verify that resistors in series add as 1/R 1 + 1/R 2 = 1/R T 2EM 7
8 Part II. Data Calculations V 1 = I 1 = R 1 = 100Ω R 1 = V 1 /I 1 = P = I 1 V 1 = V 2 = I 2 = R 2 = 200Ω R 2 = V 2 /I 2 = P 2 = I 2 V 2 = V T = I T = R T = V T /I T = P T = I T V T = V PS = Summary Statements: Describe how potential differences behave in a series circuit: Describe how the current behaves in a series circuit: Describe how the resistance behaves in a series circuit: Describe how the power behaves in a series circuit: Describe how Ohm s Law applies to each resistive element: Describe how Ohm s Law applies to the entire circuit: 2EM 8
9 Part III. Data Calculations V 1 = I 1 = R 1 = 100Ω R 1 = V 1 /I 1 = P = I 1 V 1 = V 2 = I 2 = R 2 = 200Ω R 2 = V 2 /I 2 = P 2 = I 2 V 2 = V T = I T = R T = V T /I T = P T = I T V T = V PS = Summary Statements: Describe how the potential differences behave in a parallel circuit: Describe how the current behaves in a parallel circuit: Describe how the resistance behaves in a parallel circuit: Describe how the power behaves in a parallel circuit: Describe how Ohm s Law applies to each resistive element: Describe how Ohm s Law applies to the entire circuit: 2EM 9
10 2EM 10
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# Bayes's theorem
Let's start considering the following example:
A screw factory has two machines, the $$M1$$, which is old, and does $$75\%$$ of all the screws, and the $$M2$$, newer but small, that does $$25\%$$ of the screws. The $$M1$$ does $$4\%$$ of defective screws, while the $$M2$$ just does $$2\%$$ of defective screws. If we choose a screw at random: what is the probability that ir turns out to be defectively?
As we have seen, we could solve the problem using the theorem of the total probability. Remember that it was the same as to use a tree to solve it.
Let's look now at the problem from another point of view. If we know that a screw is defective: what is the probability that it has been made by machine $$M1$$?
In other words, we are wondering for the conditional probability $$P(M1/D)$$. On the one hand, for the definition of conditional probability, we have that:
$$P(M1/D)=\dfrac{P(M1\cap D)}{P(D)}$$$On the other hand, if we represent our problem in a tree, we see that we can compute $$P(M1\cap D)$$, since it is the probability of the colored branch: to be made by $$M1$$ and to be defective. Remember that the theorem of total probability is: $$P(D) = P(M1)\cdot P(D/M1) + P(M2)\cdot P(D/M2)$$$ So we obtain: $$P(M1/D)=\dfrac{P(M1)\cdot P(D/M1)}{P(M1)\cdot P(D/M1) + P(M2)\cdot P(D/M2)}$$$In our example, $$P(M1/D)=\dfrac{0,75\cdot 0,04}{0,75\cdot 0,04 + 0,25\cdot 0,02}=0,857$$$
What we have done to solve this problem can be generalized through the Bayes' theorem.
• Bayes' Theorem:
Let $$A_1,A_2,\ldots, A_n$$ be a partition of the sample space and $$B$$ uan event associated with the same experiment. Then, we have that: $$P(A_i/B)=\dfrac{P(A_i)\cdot P(B/A_i)}{P(A_1)\cdot P(B/A_1)+\ldots+P(A_n)\cdot P(B/A_n)}$$$Let's see how to apply it. We have three boxes with light bulbs. The first one contains $$10$$ bulbs, with $$4$$ of them broken; in the second one there are $$6$$ light bulbs, and only one broken, and in the third one there are three broken light bulbs out of eight. Now we want to ask ours: If we take a broken bulb what is the probability that comes from box $$1$$? Let's remember that $$C1$$, $$C2$$, $$C3$$ represent boxes $$1$$, $$2$$ and $$3$$. Also $$F =$$ "broken bulb", for what $$\overline{F}=$$ "not broken bulb". Now, we are only interested in the top branch of our tree. We are interested in $$P(C1/F)$$. From Bayes' theorem, $$P(C1/F)=\dfrac{P(C1)\cdot P(F/C1)}{P(C1)\cdot P(F/C1)+P(C2)\cdot P(F/C2)+P(C3)\cdot P(F/C3)}$$$
In our case, $$P(C1/F)=\dfrac{\dfrac{1}{3}\cdot \dfrac{4}{10}}{\dfrac{1}{3}\cdot \dfrac{4}{10} + \dfrac{1}{3}\cdot \dfrac{1}{6}+\dfrac{1}{3}\cdot \dfrac{3}{8}}=\dfrac{\dfrac{4}{30}}{\dfrac{113}{360}}=\dfrac{48}{113}=0,425$$$that is to say, $$42,5\%$$. Suppose we have $$250$$ doctors from Europe meeting in a conference. Among these $$115$$ are Germans; $$65$$, French, and $$70$$ Englishmen. We also know that, $$75\%$$ of the Germans, $$60\%$$ of the French and $$65\%$$ of the Englishmen are in favour of using a new vaccine for the flu. In order to decide whether the vaccine is finally used they agree on the following: among all the doctors they select at random three doctors, who answer if they are in favour or not (with replacement). Remember that with replacement means that a doctor can be selected all three times (or two times). The vaccine is approved out of these three picks, at least two agree on using the vaccine. We are now interested in: If we pick a doctor randomly and he is in favor of using the vaccine: what is the probability that he is French? Let's consider the following events: $$A =$$ "German doctor", $$F =$$ "French doctor", $$I =$$ "English doctor", and $$V =$$ "to be in favor of the vaccine" (and therefore, $$\overline{V}=$$ "o be against the vaccine"). We represent our problem in a tree. We are looking for $$P(F/V)$$. From Bayes' theorem, $$P(F/V)=\dfrac{P(F)\cdot P(V/F)}{P(A)\cdot P(V/A)+P(F)\cdot P(V/F)+P(I)\cdot P(V/I)}$$$
In our case, $$P(F/V)=\dfrac{\dfrac{65}{250}\cdot 0,6}{\dfrac{115}{250}\cdot 0,75 + \dfrac{65}{250}\cdot 0,6+\dfrac{70}{250}\cdot 0,65} =0,228$$\$ |
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The Trig solver is a very basic tool for solving differential equations. It takes a pair of input values and the equation to be solved, and outputs the solution. The input values can be any kind of number - real numbers, complex numbers, or even other trigonometric functions. The most important part of a trigonometric solver is the input function - it takes in two values and produces one output value. A simple function would look like this: f(x,y) = x² + y² The output value will be whatever value that f(x,y) equals when the input values x and y are both equal to 0. If x = 0 and y = 0, then both the input values are equal to zero. Therefore, f(0,0) = 1. That's why this function outputs 1 as its solution when x = y = 0. An example of an input function might look like this: f(x,y) = sin(x)/cos(y) * cos(2*pi*x/3) + sin(2*pi*y/3) * sin(2*pi*x/3) In this example, we have three pieces of information: x , y , and pi . When we solve for f(x,y), we get three different solutions depending on
If someone can find a solution to homelessness, for example, it could drastically improve the quality of life for many people. So if you have the skills and knowledge required to solve such problems, you should pursue them! One of the best ways to get involved in solving hard problems is by joining an organization like Engineers Without Borders (EWB). EWB works on difficult and important projects around the world, including building schools in remote areas and helping farmers improve their crops. You can work with other engineers and solve real-world challenges that matter to people everywhere.
Solving quadratics by factoring is a method that uses the quadratic formula to solve for a root of a given quadratic expression. Factoring is often used to simplify a polynomial in terms of factoring one of the factors out. This can be done by dividing both sides by the factor and simplifying as much as possible. The result is then multiplied by the original expression, which can be reduced using the quadratic formula. To simplify a quadratic expression, one must first factor out the highest power term, then factor out all lower powers of x, and finally divide both sides by x-a. By doing this step-by-step, it will become fairly simple to solve for any given value of x that makes sense. And since the values are in standard form, it is easy to see what part of equation needs to be factored out.
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# 8.4: Special Right Triangles
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Identify and use the ratios involved with isosceles right triangles.
• Identify and use the ratios involved with 30-60-90 triangles.
## Review Queue
Find the value of the missing variable(s). Simplify all radicals.
1. Do the lengths 6, 6, and \begin{align*}6 \sqrt{2}\end{align*} make a right triangle?
2. Do the lengths \begin{align*}3,3 \sqrt{3},\end{align*} and 6 make a right triangle?
Know What? The Great Giza Pyramid is a pyramid with a square base and four isosceles triangles that meet at a point. It is thought that the original height was 146.5 meters and the base edges were 230 meters.
First, find the length of the edge of the isosceles triangles. Then, determine if the isosceles triangles are also equilateral triangles. Round your answers to the nearest tenth.
You can assume that the height of the pyramid is from the center of the square base and is a vertical line.
## Isosceles Right Triangles
There are two types of special right triangles, based on their angle measures. The first is an isosceles right triangle. Here, the legs are congruent and, by the Base Angles Theorem, the base angles will also be congruent. Therefore, the angle measures will be \begin{align*}90^\circ, 45^\circ,\end{align*} and \begin{align*}45^\circ\end{align*}. You will also hear an isosceles right triangle called a 45-45-90 triangle. Because the three angles are always the same, all isosceles right triangles are similar.
Investigation 8-2: Properties of an Isosceles Right Triangle
Tools Needed: Pencil, paper, compass, ruler, protractor
1. Construct an isosceles right triangle with 2 in legs. Use the SAS construction that you learned in Chapter 4.
2. Find the measure of the hypotenuse. What is it? Simplify the radical.
3. Now, let’s say the legs are of length \begin{align*}x\end{align*} and the hypotenuse is \begin{align*}h\end{align*}. Use the Pythagorean Theorem to find the hypotenuse. What is it? How is this similar to your answer in #2?
\begin{align*}x^2 + x^2 & = h^2\\ 2x^2 & = h^2\\ x \sqrt{2} & = h\end{align*}
45-45-90 Corollary: If a triangle is an isosceles right triangle, then its sides are in the extended ratio \begin{align*}x : x : x \sqrt{2}\end{align*}.
Step 3 in the above investigation proves the 45-45-90 Triangle Theorem. So, anytime you have a right triangle with congruent legs or congruent angles, then the sides will always be in the ratio \begin{align*}x : x : x \sqrt{2}\end{align*}. The hypotenuse is always \begin{align*}x \sqrt{2}\end{align*} because that is the longest length. This is a specific case of the Pythagorean Theorem, so it will still work, if for some reason you forget this corollary.
Example 1: Find the length of the missing sides.
a)
b)
Solution: Use the \begin{align*}x : x : x \sqrt{2}\end{align*} ratio.
a) \begin{align*}TV = 6\end{align*} because it is equal to \begin{align*}ST\end{align*}. So, \begin{align*}SV = 6 \sqrt{2}\end{align*} .
b) \begin{align*}AB = 9 \sqrt{2}\end{align*} because it is equal to \begin{align*}AC\end{align*}. So, \begin{align*}BC = 9 \sqrt{2} \cdot \sqrt{2} = 9 \cdot 2 = 18\end{align*}.
Example 2: Find the length of \begin{align*}x\end{align*}.
a)
b)
Solution: Again, use the \begin{align*}x : x : x \sqrt{2}\end{align*} ratio, but in these two we are given the hypotenuse. We need to solve for \begin{align*}x\end{align*} in the ratio.
a) \begin{align*}12 \sqrt{2} = x \sqrt{2}\!\\ 12 = x\end{align*}
b) \begin{align*}x \sqrt{2} = 16\!\\ x = \frac{16}{ \sqrt{2}} \cdot \frac{ \sqrt{2}}{\sqrt{2}} = \frac{16 \sqrt{2}}{2} = 8 \sqrt{2}\end{align*}
In part b, we rationalized the denominator. Whenever there is a radical in the denominator of a fraction, multiply the top and bottom by that radical. This will cancel out the radical from the denominator and reduce the fraction.
## 30-60-90 Triangles
The second special right triangle is called a 30-60-90 triangle, after the three angles. To construct a 30-60-90 triangle, start with an equilateral triangle.
Investigation 8-3: Properties of a 30-60-90 Triangle
Tools Needed: Pencil, paper, ruler, compass
1. Construct an equilateral triangle with 2 in sides.
2. Draw or construct the altitude from the top vertex to the base for two congruent triangles.
3. Find the measure of the two angles at the top vertex and the length of the shorter leg.
The top angles are each \begin{align*}30^\circ\end{align*} and the shorter leg is 1 in because the altitude of an equilateral triangle is also the angle and perpendicular bisector.
4. Find the length of the longer leg, using the Pythagorean Theorem. Simplify the radical.
5. Now, let’s say the shorter leg is length \begin{align*}x\end{align*} and the hypotenuse is \begin{align*}2x\end{align*}. Use the Pythagorean Theorem to find the longer leg. What is it? How is this similar to your answer in #4?
\begin{align*}x^2 + b^2 & = (2x)^2\\ x^2 + b^2 & = 4x^2\\ b^2 & = 3x^2\\ b & = x \sqrt{3}\end{align*}
30-60-90 Corollary: If a triangle is a 30-60-90 triangle, then its sides are in the extended ratio \begin{align*}x : x \sqrt{3} : 2x\end{align*}.
Step 5 in the above investigation proves the 30-60-90 Corollary. The shortest leg is always \begin{align*}x\end{align*}, the longest leg is always \begin{align*}x \sqrt{3}\end{align*}, and the hypotenuse is always \begin{align*}2x\end{align*}. If you ever forget this corollary, then you can still use the Pythagorean Theorem.
Example 3: Find the length of the missing sides.
a)
b)
Solution: In part a, we are given the shortest leg and in part b, we are given the hypotenuse.
a) If \begin{align*}x = 5\end{align*}, then the longer leg, \begin{align*}b = 5 \sqrt{3}\end{align*}, and the hypotenuse, \begin{align*}c = 2(5) = 10\end{align*}.
b) Now, \begin{align*}2x = 20\end{align*}, so the shorter leg, \begin{align*}f = 10\end{align*}, and the longer leg, \begin{align*}g = 10 \sqrt{3}\end{align*}.
Example 4: Find the value of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
a)
b)
Solution: In part a, we are given the longer leg and in part b, we are given the hypotenuse.
a) \begin{align*}x \sqrt{3} = 12\!\\ x = \frac{12}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{12 \sqrt{3}}{3} = 4 \sqrt{3}\!\\ \text{Then, the hypotenuse would be}\!\\ y = 2 \left ( 4 \sqrt{3} \right ) = 8 \sqrt{3}\end{align*}
b) \begin{align*}2x = 15 \sqrt{6}\!\\ x = \frac{15 \sqrt{6}}{2}\!\\ \text{The, the longer leg would be}\!\\ y = \left ( \frac{15 \sqrt{6}}{2} \right ) \cdot \sqrt{3} = \frac{15 \sqrt{18}}{2} = \frac{45 \sqrt{2}}{2}\end{align*}
Example 5: Find the measure of \begin{align*}x\end{align*}.
Solution: Think of this trapezoid as a rectangle, between a 45-45-90 triangle and a 30-60-90 triangle.
From this picture, \begin{align*}x = a + b + c\end{align*}. First, find \begin{align*}a\end{align*}, which is a leg of an isosceles right triangle.
\begin{align*}a = \frac{24}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{24 \sqrt{2}}{2} = 12 \sqrt{2}\end{align*}
\begin{align*}a = d\end{align*}, so we can use this to find \begin{align*}c\end{align*}, which is the shorter leg of a 30-60-90 triangle.
\begin{align*}c = \frac{12 \sqrt{2}}{ \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{12 \sqrt{6}}{3} = 4 \sqrt{6}\end{align*}
\begin{align*}b = 20\end{align*}, so \begin{align*}x = 12 \sqrt{2} + 20 + 4 \sqrt{6}\end{align*}. Nothing simplifies, so this is how we leave our answer.
Know What? Revisited The line that the vertical height is perpendicular to is the diagonal of the square base. This length (blue) is the same as the hypotenuse of an isosceles right triangle because half of a square is an isosceles right triangle. So, the diagonal is \begin{align*}230 \sqrt{2}\end{align*}. Therefore, the base of the right triangle with 146.5 as the leg is half of \begin{align*}230 \sqrt{2}\end{align*} or \begin{align*}115 \sqrt{2}\end{align*}. Do the Pythagorean Theorem to find the edge.
\begin{align*}edge = \sqrt{\left (115 \sqrt{2} \right )^2 + 146.5^2} \approx 218.9 \ m\end{align*}
In order for the sides to be equilateral triangles, this length should be 230 meters. It is not, so the triangles are isosceles.
## Review Questions
1. In an isosceles right triangle, if a leg is \begin{align*}x\end{align*}, then the hypotenuse is __________.
2. In a 30-60-90 triangle, if the shorter leg is \begin{align*}x\end{align*}, then the longer leg is __________ and the hypotenuse is ___________.
3. A square has sides of length 15. What is the length of the diagonal?
4. A square’s diagonal is 22. What is the length of each side?
5. A rectangle has sides of length 4 and \begin{align*}4 \sqrt{3}\end{align*}. What is the length of the diagonal?
6. A baseball diamond is a square with 90 foot sides. What is the distance from home base to second base? (HINT: It’s the length of the diagonal).
For questions 7-18, find the lengths of the missing sides.
1. Do the lengths \begin{align*}8 \sqrt{2}, 8 \sqrt{6}\end{align*}, and \begin{align*}16 \sqrt{2}\end{align*} make a special right triangle? If so, which one?
2. Do the lengths \begin{align*}4 \sqrt{3}, 4 \sqrt{6}\end{align*} and \begin{align*}8 \sqrt{3}\end{align*} make a special right triangle? If so, which one?
3. Find the measure of \begin{align*}x\end{align*}.
4. Find the measure of \begin{align*}y\end{align*}.
5. What is the ratio of the sides of a rectangle if the diagonal divides the rectangle into two 30-60-90 triangles?
6. What is the length of the sides of a square with diagonal 8 in?
For questions 25-28, it might be helpful to recall #25 from section 8.1.
1. What is the height of an equilateral triangle with sides of length 3 in?
2. What is the area of an equilateral triangle with sides of length 5 ft?
3. A regular hexagon has sides of length 3 in. What is the area of the hexagon? (Hint: the hexagon is made up a 6 equilateral triangles.)
4. The area of an equilateral triangle is \begin{align*}36 \sqrt{3}\end{align*}. What is the length of a side?
5. If a road has a grade of \begin{align*}30^\circ\end{align*}, this means that its angle of elevation is \begin{align*}30^\circ\end{align*}. If you travel 1.5 miles on this road, how much elevation have you gained in feet (5280 ft = 1 mile)?
6. Four isosceles triangles are formed when both diagonals are drawn in a square. If the length of each side in the square is \begin{align*}s\end{align*}, what are the lengths of the legs of the isosceles triangles?
1. \begin{align*}4^2 + 4^2 = x^2\!\\ {\;} \quad \ \ 32 = x^2\!\\ {\;} \qquad x = 4 \sqrt{2}\end{align*}
2. \begin{align*}3^2 + z^2 = 6^2 \qquad \left( 3 \sqrt{3} \right)^2 + 9^2 = y^2\!\\ {\;} \qquad z^2 = 27 \qquad \qquad \quad \ \ 108=y^2\!\\ {\;} \qquad \ z = 3 \sqrt{3} \qquad \qquad \quad \ \ y = 6 \sqrt{3}\end{align*}
3. \begin{align*}x^2 + x^2 = 10^2\!\\ {\;} \ \ \quad 2x^2 = 100\!\\ {\;} \qquad x^2 = 50\!\\ {\;} \qquad \ x = 5 \sqrt{2}\end{align*}
4. Yes, \begin{align*}6^2 + 6^2 = \left(6 \sqrt{2} \right)^2 \rightarrow 36 + 36 = 72\end{align*}
5. Yes, \begin{align*}3^2 + \left(3 \sqrt{3} \right)^2 = 6^2 \rightarrow 9 + 27 = 36\end{align*}
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## Tuesday, July 24, 2012
### Subtracting Integers
In one of the middle school sessions at TMC12 we were talking about integers and how students struggle especially with subtraction. We were brainstorming different strategies (chips, number lines, rules, etc.) This is the distance strategy I shared with the group.
When working with students it is important for them to understand that subtraction is more than just "take away". Yes, we can figure 25 - 3 as "You have 25 candies in your hand and you eat 3. What do you have left?" but that limits us when we expand to include integers.
Using a number line students can also recognize that the distance between 25 and 3 is also 22. Continue building on what students know.
13 - 4 = 13 take away 4 but also the distance between 13 and 4
So what happens when you switch it?
4 - 13
Many students will intuitively notice the result would have to be negative. What is the distance between 4 and 13? It is the same as 13 and 4 only it is negative because the smaller number is first.
*Students have to understand the above to be able to move forward.
9 - (-2) = the distance between 9 and -2, notice 9 is larger than -2 = 11
-2 - 9 = the distance between -2 and 9, notice the -2 is smaller than 9 = -11
-3 - (-8) = the distance between -3 and -8, notice the -3 is larger than -8 = 5
If students have trouble determining the distance, provide a number line. Counting on to determine how far apart numbers are is another strategy that some students might need practice with.
-135 - 124 =
(the distance from -135 to 0 is 135, the distance from 0 to 124 is 124 so the distance from -135 to 124 is 135 + 124)
1. Hmmm... I never thought of it this way! Thanks for the share!
2. Cool, thanks for sharing! If you use "distance" to describe subtraction, how do you distinguish this from the definition of absolute value? So, for example, when you say 4 - 13 means finding the distance between 4 and 13, how would students see the difference between that question and abs(4 - 13)?
1. When we were first talking about it at TMC12, I was saying distance and direction but that seemed to get confusing. From my experience, absolute value is something that students continue to struggle with and the more opportunities to discuss the better. I can absolutely see this conversation of absolute value naturally following the subtraction. Any other suggestions are welcome and appreciated. Thanks for commenting.
2. Another analogy for subtracting integers that I've found helpful is the "Hot and Cold Cubes" story from IMP, Year 1. http://bit.ly/M821Xq has a copy of the story, but the basic idea is that you can use a hot cube to represent a positive number and a cold cube to represent a negative number. In this analogy, one hot cube raises the temperature 1 degree and one cold cube decreases it by 1 degree. So both adding hot cubes and removing cold cubes raises the temperature while removing hot cubes or adding cold cubes decreases the temperature.
3. Well. It does. Since absolute value is just the pure distance, no sign, it shows why abs(4-13) is the same as abs(13-4), because they are the same distance, just different directions. I don't think the "directions" are confusing. In fact, I think they're fairly intuitive, especially when you need to reverse directions.
4. This video - https://www.teachingchannel.org/videos/teaching-subtracting-integers - addresses the hot cube / cold cube thing that Anna mentioned. I think the teacher in the video does a great job with the lesson - but I find it sort of amusing that The Teaching Channel refers to this chef's endeavors as a "Real-World Scenario." Right... because we all use hot cubes in our kitchens all the time.
3. I use the distance in my classroom too and the students do pick up on it! Whenever I teach adding and subtracting integers I have them all sing row your boat too |
## 6.3 Mole-Mass Conversions
### Learning Objective
1. Convert quantities between mass units and mole units.
Example 3 in Section 6.2 "Atomic and Molar Masses" stated that the mass of 2 mol of U is twice the molar mass of uranium. Such a straightforward exercise does not require any formal mathematical treatment. Many questions concerning mass are not so straightforward, however, and require some mathematical manipulations.
The simplest type of manipulation using molar mass as a conversion factor is a mole-mass conversionThe conversion from moles of material to the mass of that same material. (or its reverse, a mass-mole conversion). In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).
We established that 1 mol of Al has a mass of 26.98 g (Example 3 in Section 6.2 "Atomic and Molar Masses"). Stated mathematically,
1 mol Al = 26.98 g Al
We can divide both sides of this expression by either side to get one of two possible conversion factors:
The first conversion factor can be used to convert from mass to moles, and the second converts from moles to mass. Both can be used to solve problems that would be hard to do “by eye.”
### Note
The algebra skills we are using here are the same skills that we used in Chapter 1 "Chemistry, Matter, and Measurement" to perform unit conversions.
### Example 5
What is the mass of 3.987 mol of Al?
Solution
The first step in a conversion problem is to decide what conversion factor to use. Because we are starting with mole units, we want a conversion factor that will cancel the mole unit and introduce the unit for mass in the numerator. Therefore, we should use the conversion factor. We start with the given quantity and multiply by the conversion factor:
Note that the mol units cancel algebraically. (The quantity 3.987 mol is understood to be in the numerator of a fraction that has 1 in the unwritten denominator.) Canceling and solving gives
Our final answer is expressed to four significant figures.
### Skill-Building Exercise
1. How many moles are present in 100.0 g of Al? (Hint: you will have to use the other conversion factor we obtained for aluminum.)
Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure 6.2 "A Simple Flowchart for Converting between Mass and Moles of a Substance" is a chart for determining what conversion factor is needed, and Figure 6.3 "A Flowchart Illustrating the Steps in Performing a Unit Conversion" is a flow diagram for the steps needed to perform a conversion.
Figure 6.2 A Simple Flowchart for Converting between Mass and Moles of a Substance
It takes one mathematical step to convert from moles to mass or from mass to moles.
Figure 6.3 A Flowchart Illustrating the Steps in Performing a Unit Conversion
When performing many unit conversions, the same logical steps can be taken.
### Example 6
A biochemist needs 0.00655 mol of bilirubin (C33H36N4O6) for an experiment. How many grams of bilirubin will that be?
Solution
To convert from moles to mass, we need the molar mass of bilirubin, which we can determine from its chemical formula:
33 C molar mass: 33 × 12.01 g = 396.33 g 36 H molar mass: 36 × 1.01 g = 36.36 g 4 N molar mass: 4 × 14.00 g = 56.00 g 6 O molar mass: 6 × 16.00 g = 96.00 g Total: 584.69 g
The molar mass of bilirubin is 584.69 g. (We did this calculation in Example 4 in Section 6.2 "Atomic and Molar Masses".) Using the relationship
1 mol bilirubin = 584.69 g bilirubin
we can construct the appropriate conversion factor for determining how many grams there are in 0.00655 mol. Following the steps from Figure 6.3 "A Flowchart Illustrating the Steps in Performing a Unit Conversion":
The mol bilirubin unit cancels. The biochemist needs 3.83 g of bilirubin.
### Skill-Building Exercise
1. A chemist needs 457.8 g of KMnO4 to make a solution. How many moles of KMnO4 is that?
### To Your Health: Minerals
For our bodies to function properly, we need to ingest certain substances from our diets. Among our dietary needs are minerals, the noncarbon elements our body uses for a variety of functions, such developing bone or ensuring proper nerve transmission.
The US Department of Agriculture has established some recommendations for the RDIs of various minerals. The accompanying table lists the RDIs for minerals, both in mass and moles, assuming a 2,000-calorie daily diet.
Mineral Male (age 19–30 y) Female (age 19–30 y)
Ca 1,000 mg 0.025 mol 1,000 mg 0.025 mol
Cr 35 µg 6.7 × 10−7 mol 25 µg 4.8 × 10−7 mol
Cu 900 µg 1.4 × 10−5 mol 900 µg 1.4 × 10−5 mol
F 4 mg 2.1 × 10−4 mol 3 mg 1.5 × 10−4 mol
I 150 µg 1.2 × 10−6 mol 150 µg 1.2 × 10−6 mol
Fe 8 mg 1.4 × 10−4 mol 18 mg 3.2 × 10−4 mol
K 3,500 mg 9.0 × 10−2 mol 3,500 mg 9.0 × 10−2 mol
Mg 400 mg 1.6 × 10−2 mol 310 mg 1.3 × 10−2 mol
Mn 2.3 mg 4.2 × 10−5 mol 1.8 mg 3.3 × 10−5 mol
Mo 45 mg 4.7 × 10−7 mol 45 mg 4.7 × 10−7 mol
Na 2,400 mg 1.0 × 10−1 mol 2,400 mg 1.0 × 10−1 mol
P 700 mg 2.3 × 10−2 mol 700 mg 2.3 × 10−2 mol
Se 55 µg 7.0 × 10−7 mol 55 µg 7.0 × 10−7 mol
Zn 11 mg 1.7 × 10−4 mol 8 mg 1.2 × 10−4 mol
This table illustrates several things. First, the needs of men and women for some minerals are different. The extreme case is for iron; women need over twice as much as men do. In all other cases where there is a different RDI, men need more than women.
Second, the amounts of the various minerals needed on a daily basis vary widely—both on a mass scale and a molar scale. The average person needs 0.1 mol of Na a day, which is about 2.5 g. On the other hand, a person needs only about 25–35 µg of Cr per day, which is under one millionth of a mole. As small as this amount is, a deficiency of chromium in the diet can lead to diabetes-like symptoms or neurological problems, especially in the extremities (hands and feet). For some minerals, the body does not require much to keep itself operating properly.
Although a properly balanced diet will provide all the necessary minerals, some people take dietary supplements. However, too much of a good thing, even minerals, is not good. Exposure to too much chromium, for example, causes a skin irritation, and certain forms of chromium are known to cause cancer (as presented in the movie Erin Brockovich).
### Concept Review Exercises
1. What relationship is needed to perform mole-mass conversions?
2. What information determines which conversion factor is used in a mole-mass conversion?
1. The atomic or molar mass is needed for a mole-mass conversion.
2. The unit of the initial quantity determines which conversion factor is used.
### Key Takeaway
• It is possible to convert between moles of material and mass of material.
### Exercises
1. What is the mass of 8.603 mol of Fe metal?
2. What is the mass of 0.552 mol of Ag metal?
3. What is the mass of 6.24 × 104 mol of Cl2 gas?
4. What is the mass of 0.661 mol of O2 gas?
5. What is the mass of 20.77 mol of CaCO3?
6. What is the mass of 9.02 × 10−3 mol of the hormone epinephrine (C9H13NO3)?
7. How many moles are present in 977.4 g of NaHCO3?
8. How many moles of erythromycin (C37H67NO13), a widely used antibiotic, are in 1.00 × 103 g of the substance?
9. Cortisone (C21H28O5) is a synthetic steroid that is used as an anti-inflammatory drug. How many moles of cortisone are present in one 10.0 mg tablet?
10. Recent research suggests that the daily ingestion of 85 mg of aspirin (also known as acetylsalicylic acid, C9H8O4) will reduce a person’s risk of heart disease. How many moles of aspirin is that?
1. 480.5 g
2. 4.42 × 106 g
3. 2,079 g
4. 11.63 mol
5. 2.77 × 10−5 mol |
# How do you solve x^3+x^2+4x+4>0 using a sign chart?
##### 1 Answer
Feb 4, 2017
The answer is x in ]-1, +oo[
#### Explanation:
Let $f \left(x\right) = {x}^{3} + {x}^{2} + 4 x + 4$
Before, we need to find the factors of $f \left(x\right)$
$f \left(- 1\right) = - 1 + 1 - 4 + 4 = 0$
Therefore, $\left(x + 1\right)$ is a factor
To find the other factors, we perform a long division
$\textcolor{w h i t e}{a a a a}$${x}^{3} + {x}^{2} + 4 x + 4$$\textcolor{w h i t e}{a a a a}$$|$$x + 1$
$\textcolor{w h i t e}{a a a a}$${x}^{3} + {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a}$$|$${x}^{2} + 4$
$\textcolor{w h i t e}{a a a a a}$$0 + 0 + 4 x + 4$
$\textcolor{w h i t e}{a a a a a a a a a a a a}$$4 x + 4$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$0 + 0$
Therefore,
$f \left(x\right) = \left(x + 1\right) \left({x}^{2} + 4\right)$
$\forall x \in \mathbb{R} , \left({x}^{2} + 4\right) > 0$
So, we can build the sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
Therefore,
$f \left(x\right) > 0$ when x in ]-1, +oo[ |
# Math Expressions Grade 4 Student Activity Book Unit 7 Lesson 11 Answer Key Decimals Greater Than 1
This handy Math Expressions Student Activity Book Grade 4 Answer Key Unit 7 Lesson 11 Decimals Greater Than 1 provides detailed solutions for the textbook questions.
## Math Expressions Grade 4 Student Activity Book Unit 7 Lesson 11 Decimals Greater Than 1 Answer Key
Discuss Symmetry Around the Ones
Question 1.
Discuss symmetries and relationships you see in the place value chart.
Explanation:
Decimal numbers have 2 parts. A Whole number and a fractional part (or) decimal part. The two parts (Whole number and Decimal parts) are separated by a point (.) called the decimal point. so, their relationship is that before the decimal point whole numbers are placed as ones, tens, or hundreds, and after the decimal point decimal parts are placed as tenths, or hundredths.
Question 2.
Is it easier to see place value patterns in a or b? Discuss why.
a. 500 50 5 .5 .05
Explanation:
Here both represent the same, the place value of fraction parts consists of a sequence of decimal (base ten)digits.
b. 500 50 5 0.5 0.05
Explanation:
In the above given 0.5 and 0.05 we see that the whole number part is zero .this is because the value of a proper fraction is always less than one whole so, 500 50 5 0.5 0.05 is easier even though 500 50 5 .5 .05 are equivalent.
Use your Decimal Secret Code Cards to make numbers on the frame.
t
Write Numbers in Decimal Form
Read and write each mixed number as a decimal.
Question 3.
3$$\frac{1}{10}$$ ________________
3.1
Explanation:
Given, mixed fraction 3 (1/10), so, by separating the parts we get,3+(1/10), we know that 1/10 is 0.1 that is 3+0.1=3.1.
Question 4.
5$$\frac{7}{100}$$ ________________
5.07,
Explanation:
Given, mixed fraction 5 (7/100), so, by separating the parts we get,5+(7/100), we know that 7/100=0.07, that is 5+0.07=5.07.
Question 5.
2$$\frac{46}{100}$$ ________________
2.46,
Explanation:
Given, mixed fraction 2 (46/100), so, by separating the parts we get,2+(46/100), we know that 46/100 =0.46, that is 2+0.46=2.46.
Question 6.
28$$\frac{9}{10}$$ _________________
28.9,
Explanation:
Given, the mixed fraction 28 (9/10), so, by separating the parts we get, 28+(9/10), we know that 9/10=0.9 is 28+0.9=28.9
Read and write each decimal as a mixed number.
Question 7.
12.8 ________________
12$$\frac{8}{10}$$,
Explanation:
Given,12.8 that is, 1 decimal place so the denominator is 10. so, 8/10, so the whole number remains the same that is 12 (8/10).
Question 8.
3.05 ________________
3$$\frac{5}{100}$$,
Explanation:
Given,3.05 that is,2 decimal places so the denominator is 100. so,5/100, so the whole number remains the same is 3 (5/100).
Question 9.
4.85 _________________
4$$\frac{85}{00]$$,
Explanation:
Given, 4.85 that is,2 decimal places so the denominator is 100. so, 85/100, So, the whole number remains the same that is,4 (85/100).
Question 10.
49.7 _________________
49$$\frac{7}{10}$$,
Explanation:
Given,49.7, that is,1 decimal place so the denominator is 10, so, 7/10.so, the whole number remains the same that is,49 (7/10).
Read each word name. Then write a decimal for each word name.
Question 11.
sixty-one hundredths
0.61,
Explanation:
Given, sixty-one hundredths are written as,61/100, can be written in decimal form as, 0.61.
Question 12.
six and fourteen hundredths
6.14,
Explanation:
Given, six and fourteen-hundredths are written as 614/100, can be written in decimal form as,6.14.
Question 13.
seventy and eight tenths
70.8.
Explanation:
Given, seventy and eight-tenths wrote as,708/100, can be written in decimal form as 70.8.
Question 14.
fifty-five and six hundredths
55.06.
Explanation:
Given, fifty-five and six-hundredths wrote as, 5506/100, can be written in decimal form as 55.06.
Expanded Form
Write each decimal in expanded form.
Question 15.
8.2 _________________
8+(2/10).
Explanation:
Given, 8.2 in decimal form, 8 is the whole number and 2 is the decimal part which has 1 decimal place so, the denominator is 10 which is 2/10 which means the expanded form is 8+(2/10).
Question 16.
17.45 _________________
10+7+(4/10)+(5/100).
Explanation:
Given,17.45 in decimal form, 17 is a whole number that is 10+7 and 45 is the decimal part that is, (4/10)+(5/100) which means, the expanded form is 10+7+(4/10)+(5/100).
Question 17.
106.24 _________________
100+00+6+(2/10)+(4/100).
Explanation:
Given, 106.24 in decimal form, 106 is a whole number that is 100+00+6 and 24 is the decimal part that is, (2/10)+(4/100) which means, the expanded form is 100+00+6+(2/10)+(4/100).
Question 18.
50.77 _________________
50+0+(7/10)+(7/100).
Explanation:
Given,50.77 in decimal form,50 is the whole number that is 50+0 and 77 is the decimal part that is (7/10)+(7/100) which means, the expanded form is 50+0+(7/10)+(7/100).
Question 19.
312.09 _________________
300+10+2+(0/10)+(9/100).
Explanation:
Given, 312.09 in decimal form, 312 is a whole number that is 300+10+2 and 09 is the decimal part is (0/10)+(9/100), which means, the expanded form is 300+10+20(0/10)+(9/100).
Question 20.
693.24 _________________
600+90+3+(2/10)+(4/100).
Explanation:
Given, 693.24 in decimal form,693 is a whole number that is 600+90+3 and 24 is a decimal part is (2/10)+(4/100), which means, the expanded form is 600+90+3+(2/10)+(4/100).
Solve.
Question 21.
There are 100 centimeters in 1 meter. A snake crawls 3 meters and 12 more centimeters. What decimal represents the number of meters the snake crawls? |
# 7th Grade Math Geometry Project Choice Board Worksheets Boards Create Free Kindergarten Algebra 5th
Comfort Melina October 23, 2020 Math Worksheet
First, the Basics! The x axis of a graph refers to the horizontal line while the y axis refers to the vertical line. Together these lines form a cross and the point where they both meet is called the origin. The value of the origin is always 0. So if you move your pencil from the origin to the right, you are drawing a line across the positive values of the x axis, i.e., 1, 2, 3 and so on. From the origin to the left, you’re moving across the negative values of the x axis, i.e., -1, -2, -3 and so on. If you go up from the origin, you are covering the positive values of the y axis. Going down from the origin, will take you to the negative values of the y axis.
Today we all know that benefits of math are considerable. Math is not a subject one learns by reading the problems and solutions. American children have very little practice with multi-step problems, and very few opportunities to think their way in to and through problems that don’t look like ’all the others’. With a packed curriculum and the increased emphasis on testing, our children are taught tons of procedures – but procedures disconnected from when to use them, and why. Sustained thinking – the key ingredient to math success – is painfully absent in too many math classes.
When a child learns to relate math to everyday questions, he will be great at it from the simplest addition all the way to trigonometry. To convert percentages, decimals and fractions is thus one essential skill. How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom’s delicious apple pie is gone. How many kids in school have done their homework? Again this can be answered in several ways: in percentages, 70%; or in ratio, 7:10; Both of these mean out of ten kids in class there are seven good ones who did and three not-so-good ones who didn’t. The bottom line is that kids learn math much better when it makes sense.
Track Record, Another advantage of these math worksheets is that kids and parents will be able to keep them to serve as their references for review. Since worksheets are easy to correct, students will be able to identify the items and areas that they had mistakes so that they will be able to correct those deficiencies. Keeping record is really a good thing; As a parent, you will be able to go back through them and assess their strong and weak areas. Keeping track you will be able to track your child’s progress as empirical evidence.
### Year 5 Maths Revision Worksheets Pdf
Nov 06, 2020
#### Did U Hear About Math Worksheet Answers
Nov 06, 2020
##### Unit Circle Worksheet Math 36 Answers
Nov 07, 2020
###### Ks2 Maths Fractions Worksheets
Nov 06, 2020
Rather than using worksheets, a better method is to use individual size white boards and have the child writing entire facts many times. Having a child writing 9 x 7 = 7 x 9 = 63 while saying ”nine times seven is the same as seven times nine and is equal to sixty-three” is many times more successful than a worksheet with 9 x 7 = ___ and the student just thinks the answer once and then writes that answer on the duplicate problems. I will admit that there is one type of worksheet that I used in the past and found relatively beneficial, although it had a different kind of flaw. For my Basic Math, Pre-Algebra, and Algebra classes, I had several books of ”self-checking” worksheets. These worksheets had puns or puzzle questions at the top, and as the students worked the problems they were given some kind of code for choosing a letter to match that answer. If they worked the problems correctly, the letters eventually answered the pun or riddle. Students enjoyed these worksheets, but there are a couple problem areas even with these worksheets. Some students would get the answer to the riddle early and then work backward from letter to problem answer, so they weren’t learning or practicing anything.
I remember that with my Mom everything was somehow connected to math. She made me count the buttons in my shirt as she dressed me up, asked questions that demanded answers that are related to sums, like how many pair of shoes do you have? How many buttons are there on your Daddy’s shirt? Count all the furniture in the living room and several math games. All my toys were one way or the other math related. I had puzzles, and tons of things Mom had me do as games on daily basis at home to get me ready for kindergarten! In fact, she continued guiding me towards being math friendly throughout kindergarten and first grade during which time 1st grade math worksheets was my constant companion.
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# Coin Toss Probability Formula
Reviewed by:
Last updated date: 19th Sep 2024
Total views: 359.7k
Views today: 5.59k
## What is the Coin Toss Probability Formula?
Let's look at a few things about flipping a coin before studying the coin toss probability formula. There are two potential consequences when flipping a coin: heads or tails. We don't know which way the coin will land on a given toss, but we do know it will either be Head or Tail. Tossing a coin, on the other hand, is a random experiment since you know the set of outcomes but not the exact outcome for each random experiment execution.
The probability formula for a coin flip can be used to calculate the probability of some experiment. This page discusses the concept of coin toss probability along with the solved examples.
### Results of Coin Toss Probability
Each result has a predetermined likelihood, which remains constant from trial to trial. Heads and tails share the same chance of 1/2 when it comes to coins. In addition, there are cases where the coin is skewed, resulting in varying odds for heads and tails. In this part, we'll look at probability distributions with just two potential outcomes and fixed probabilities that add up to one. Binomial distributions are the name for certain types of distributions.
Here are some examples of problems involving coin toss chances.
If a coin is flipped, there are two potential outcomes: a ‘head' (H) or a ‘tail' (T), and it is difficult to determine whether the toss will end in a ‘head' or a ‘tail.'
Assuming the coin is equal, then the coin probability is 50% or 1/2
This is because you know the result would be either head or tail, and both are equally probable.
The Probability for Equally Likely Outcomes is:
Total number of favorable outcomes and Total number of possible outcomes
Where,
Total number of possible outcomes = 2
1. Coin toss probability formula for heads
If the favorable outcome is head (H).
A number of favorable outcomes = 1.
Therefore, P (getting heads) = $\frac{number \, of \, favorable \, outcomes}{total \, number \, of \, possible \, outcomes} = \frac{1}{2}$
1. Coin toss probability formula for tails.
If the favorable outcome is tail (T).
A number of favorable outcomes = 1.
Therefore, P (getting heads) = $\frac{number \, of \, favorable \, outcomes}{total \, number \, of \, possible \, outcomes} = \frac{1}{2}$
### Coin Probability Problems
1. A Coin is Tossed Thrice at Random. What is the Probability of Getting
2. The Same Face?
Solution:
The possible outcomes of a given event are {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
So, the total number of outcomes = 8.
1. Number of favorable outcomes for event E
= Number of outcomes appears at least one head
= 4 (as HHH, HHT, HTH, HTT, THH, THT, TTH are having at least one head).
= 7/8
So, by definition, P(F) = 7/8
1. Number of favorable outcomes for event E
= Number of outcomes having the same face
= 2 (as HHH, TTT has the same face).
= 2/8
= 1/4
So, by definition, P(E) = ¼
1. What is the Probability of Getting a Head When Tossing a Coin?
Solution:
When a single coin is tossed, the possible outcomes can be {H, T}.
Thus, the total number of possible outcomes = 2
So the number of favorable outcomes = 1
Therefore,
the probability of getting head is,
P(H) = Number of Favorable Outcomes/Total Number of Possible Outcomes
= 1/2
So, by definition P(H) = ½
3. Two Coins are Tossed Randomly 150 Times and it is Found That Two Tails Appeared 60 Times, One Tail Appeared 74 Times and No Tail Appeared 16 Times.
If two coins are tossed at random, then what is the probability of,
1. Getting 2 Tails
2. Getting 1 Tail
3. Getting 0 Tail
Solution:
Total number of trials = 150
Number of times 2 tails appear = 60
Number of times 1 tail appears = 74
Number of times 0 tail appears = 16
Let T1, T2, and T3 be the events of getting 2 tails, 1 tail and 0 tail respectively.
1. P(getting 2 tails)
The number of times two tails appear is 60.
= P(T1) = Number of Times two Tails Appears/Total Number of Possible Outcomes
= 60/150
= 0.40
The probability of getting 2 tails is 0.40
1. P(getting 1 tail)
The number of times one tail appears is 74
= P(T2) = Number of Times one Tail AppearsTotal Number of Possible Outcomes
= 74/150
= 0.4933
The probability of getting one tail is 0.4933
1. P(getting 0 tail)
The number of times zero tail appear is 16
= P(T3) = Number of Times Zero Tail AppearsTotal Number of Possible Outcomes
= 16/150
= 0.1067
The probability of getting zero tail is 0.1067
Note:
Remember while tossing 2 coins simultaneously, the only possible outcomes are T1, T2, T3,
= P (T1) + P (T2) + P (T3)
= 0.40 + 0.4933 + 0.1067
= 1
## FAQs on Coin Toss Probability Formula
1. What is the Sample Space When Four Coins are Tossed?
Four coins are tossed simultaneously, then the sample space is,
Tossed coins = 4
Hence, The number of faces = 24= 16
{HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
2. Where Did the Coin Toss Originate From?
The view of a random result as an act of divine will is the historical basis of coin-flipping.
The Romans named coin flipping Navia Aut Caput ("ship or head") since certain coins had a ship on one hand and the emperor's head on the other. This was known as cross and pile in England.
3. Who Won the Super Bowl 2020 Coin Toss?
In the 2020 year, the 49ers won the Super Bowl LIV coin toss after picking tails.
4. What is the probability of getting heads if you flip a coin 3 times?
On flipping a coin 3 times the probability of getting 3 heads, we get total eight outcomes as {HHH, THH, HTH, HHT, TTH, THT, HTT, TTT}
Total outcomes are - 8 and among these three heads has one outcome only.
Now, use the coin toss probability formula and apply the values below:
P (getting three heads) =number of favorable outcomestotal number of possible outcomes = 18
Therefore, P (getting three heads) = 0.125. |
FutureStarr
What Is 25 Percent of 35
## What Is 25 Percent of 35
What are some of the best number tricks you've ever learned? As so many of my middle-school students can tell you, it's 2/5ths of 3/5ths of a whole, or 23/64. I used to share my strategies with them and they'd laugh, but one day I realized that I should probably share them with you. You'll even find a fun quiz to test your competence at number tricks below.
Sale Price = \$20 (answer). This means the cost of the item to you is \$20. You will pay \$20 for a item with original price of \$25 when discounted 20%. In this example, if you buy an item at \$25 with 20% discount, you will pay 25 – 5 = 20 dollars.
What is 35 percent of 80? Write 35% as 35100. 35100 of 80 = 35100 × 80. Therefore, the answer is 28. If you are using a calculator, simply enter 35÷100×80 which will give you 28 as the answer. (Source: www.seniorcare2share.com)
### Calculation
If something costs \$35 and is on sale for 25% off, then how much would it cost? Here we will show you how to calculate how much you save (discount) and how much you have to pay if something you want to buy is regularly \$35, but is currently on sale for 25 percent off.
Here is a Percentage Calculator to solve similar calculations such as 35 is 25 percent of what number. You can solve this type of calculation with your values by entering them into the calculator's fields, and click 'Calculate' to get the result and explanation. (Source: percentage-calculator.net)
### Value
via GIPHY
How to calculate percentage Determine the whole or total amount of what you want to find a percentage for. Divide the number that you wish to determine the percentage for. Multiply the value from step two by 100.
Similarly, you can multiply the 1% answer by any number to find any percentage value. For example, to find 80% of 4,500, multiply the 45 by 80 to get 3,600. (Source: www.seniorcare2share.com)
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# Multiplying 2 fractions: 5/6 x 2/3
CCSS.Math:
## Video transcript
We're asked to multiply 5/6 times 2/3 and then simplify our answer. So let's just multiply these two numbers. So we have 5/6 times 2/3. Now when you're multiplying fractions, it's actually a pretty straightforward process. The new numerator, or the numerator of the product, is just the product of the two numerators, or your new top number is a product of the other two top numbers. So the numerator in our product is just 5 times 2. So it's equal to 5 times 2 over 6 times 3, which is equal to-- 5 times 2 is 10 and 6 times 3 is 18, so it's equal to 10/18. And you could view this as either 2/3 of 5/6 or 5/6 of 2/3, depending on how you want to think about it. And this is the right answer. It is 10/18, but when you look at these two numbers, you immediately or you might immediately see that they share some common factors. They're both divisible by 2, so if we want it in lowest terms, we want to divide them both by 2. So divide 10 by 2, divide 18 by 2, and you get 10 divided by 2 is 5, 18 divided by 2 is 9. Now, you could have essentially done this step earlier on. You could've done it actually before we did the multiplication. You could've done it over here. You could've said, well, I have a 2 in the numerator and I have something divisible by 2 into the denominator, so let me divide the numerator by 2, and this becomes a 1. Let me divide the denominator by 2, and this becomes a 3. And then you have 5 times 1 is 5, and 3 times 3 is 9. So it's really the same thing we did right here. We just did it before we actually took the product. You could actually do it right here. So if you did it right over here, you'd say, well, look, 6 times 3 is eventually going to be the denominator. 5 times 2 is eventually going to be the numerator. So let's divide the numerator by 2, so this will become a 1. Let's divide the denominator by 2. This is divisible by 2, so that'll become a 3. And it'll become 5 times 1 is 5 and 3 times 3 is 9. So either way you do it, it'll work. If you do it this way, you get to see the things factored out a little bit more, so it's usually easier to recognize what's divisible by what, or you could do it at the end and put things in lowest terms. |
# Difference between revisions of "2016 AMC 10A Problems/Problem 18"
## Problem
Each vertex of a cube is to be labeled with an integer $1$ through $8$, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$
## Solution 1
First of all, the adjacent faces have the same sum $(18$, because $1+2+3+4+5+6+7+8=36$, $36/2=18)$, so now consider the $opposite \text{ } sides$ (the two sides which are parallel but not in same face of the cube); they must have the same sum value too. Now think about the extreme condition 1 and 8 , if they are not sharing the same side, which means they would become end points of $opposite \text{ } sides$, we should have $1+X=8+Y$, but no solution for $[2,7]$, contradiction.
Now we know $1$ and $8$ must share the same side, which sum is $9$, the $opposite \text{ } side$ also must have sum of $9$, same thing for the other two parallel sides.
Now we have $4$ parallel sides $1-8, 2-7, 3-6, 4-5$. thinking about $4$ end points number need to have sum of $18$. It is easy to notice only $1-7-6-4$ vs $8-2-3-5$ would work.
So if we fix one direction $1-8 ($or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$
Now, the problem is same as the problem to arrange $4$ points in a two-dimensional square. which is $4!/4$=$\boxed{\textbf{(C) }6.}$
## Solution 2
Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to one, then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x
$3,7,8:$ Does not work. $4,6,8:$ Works. $5,6,7:$ Does not work. $5,6,8:$ Works. $5,7,8:$ Does not work. $6,7,8:$ Works.
So our answer is $3\cdot 2=\boxed{\textbf{(C) }6.}$
## Solution 3
We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$, it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same face, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly, if $8$ and $7$ were on the same face, the only way to get the sum is with $1$ and $2$. This means that $6$ and $7$ are not on the same edge as $8$, or in other words they are diagonally across from it on the same face, or on the other end of the cube.
Now we look at three cases, each yielding two solutions which are reflections of each other:
1) $6$ and $7$ are diagonally opposite $8$ on the same face. 2) $6$ is diagonally across the cube from $8$, while $7$ is diagonally across from $8$ on the same face. 3) $7$ is diagonally across the cube from $8$, while $6$ is diagonally across from $8$ on the same face.
This means the answer is $3\cdot 2=\boxed{\textbf{(C) }6.}$ |
# How do you solve the following system?: 3x-y=4 , 3x - 5y = 2
May 18, 2018
Lettuce solve $3 x$
${1}^{s t} \text{ equation}$
$3 x - y = 4$
Transfer $y$
$3 x = 4 + y$
Just remember this value as in the next equation we have $3 x$ too
color(blue)(x=(4+y)/3
${2}^{n d} \text{ equation}$
$3 x - 5 y = 2$
Put value of $3 x$
$4 + y - 5 y = 2$
$4 - 4 y = 2$
Factorize $4$ outta it
$4 \left(1 - y\right) = 2$
Transfer $4$
$1 - y = \frac{2}{4}$
Transfer 1
$- y = \frac{1}{2} - 1$
Common Denominator
$- y = \frac{1}{2} - \frac{2}{2}$
$\cancel{-} y = \cancel{-} \frac{1}{2}$
color(Red)(y=1/2
$x = \frac{4 + y}{3}$
You get
$x = \frac{4 + \frac{1}{2}}{3}$
Common denominator
$x = \frac{\frac{8}{2} + \frac{1}{2}}{3}$
$x = \frac{9}{2} \div 3$
Invert
$x = {\cancel{9}}^{3} / 2 \times \frac{1}{\cancel{3}}$
You end up with
color(red)(x=3/2 |
Maths
Question 1
1. (i) Use the data in this file “Censusdata.xls” to develop quotas for this survey by age and gender for one of the regions**. Briefly explain how you calculated the quotas.
We choose Dublin city as our region for analysis, first we get the quotas for the region by age as follows:
First we add up the total for the regions, this is to say we add the males and females at their age groups as follows:
total male and female
percentage
0-9 years
51406
1/34
Maths
10%
10-19 years
54297
11%
20-29 years
119454
24%
30-39 years
84043
17%
2/34
Maths
40-49 years
62527
12%
50-59 years
49866
10%
60-69 years
38851
8%
70-79 years
30064
3/34
Maths
6%
80 years+
15703
3%
total
506211
After getting the total we divide each total of males and females with the sum total of the region and then get the percentage level of this ratio.
Quotas by gender:
For the selected sample of Dublin city it is easy to get the quota of males and females, this is done by simply adding up all the females and the males as follows then getting the percentage value of each as follows:
4/34
Maths
total
percentage
male
248087
49%
female
258124
51%
total
5/34
Maths
506211
1. (ii) What other variables do you think the quotas should be broken down by for this survey
In this region the survey can be broken down into regions, this can consider the quota by the regions that are given.
1. (iii) Explain briefly why you feel quota sampling may or may not be suitable for this survey
Quota sampling is one of the methods that can be used to obtain a sample from a population, the greatest weakness of this type of sampling is that it is not randomly selected and therefore the issue of probability in choosing an element for the sample is not used. For this survey therefore it would not be appropriate to use quota sampling as it does not use random selection of elements of the sample.
Question 2
(i)
The best technique to select a sample is through a random sample, random sampling involves the use of random numbers to select the sample. The first step is to assign a numbers to the sample, this involves for example assigning each company from number 1 to the population size number, in our case we assign the numbers 1 to 50.
6/34
Maths
The second step is to come up with random numbers, this random numbers can be generated using a calculator or computer, the random numbers produced will in our case be a two number random number and for this reason we will consider the first two numbers of every random number produced, for example when we get the random number 0.3899003 then we will consider 0.38 which will give us the random number 38, if we get a random that exceeds our sample size for example when we have 0.9877 then we reject this random number.
When 20 valid random samples are produced the next step is to match the random samples to the sample, this involves the recording of the sample by selecting those elements of the data that correspond to the random number generated. The results from these steps give us the random sample.
In our case we will have the following random numbers generated using a computer,
31, 6, 30, 27, 13, 4, 5, 38, 46, 47, 42, 26, 32, 37, 25, 36, 34, 17, 3 and 10
The above random numbers correspond to the following elements in the data as shown in the table below which shows the random sample collected:
sample
random numbers
7/34
Maths
value
1
31
company
31
36000
2
6
company
6
8/34
Maths
23000
3
30
company
30
29000
4
27
company
27
3000
9/34
Maths
5
13
company
13
49000
6
4
company
4
41000
7
10/34
Maths
5
company
5
46000
8
38
company
38
20000
9
46
11/34
Maths
company
46
40000
10
47
company
47
30000
11
42
company
12/34
Maths
42
13000
12
26
company
26
7000
13
32
company
32
13/34
Maths
1000
14
37
company
37
45000
15
25
company
25
22000
14/34
Maths
16
36
company
36
17000
17
34
company
34
6000
18
15/34
Maths
17
company
17
34000
19
3
company
3
38000
20
10
16/34
Maths
company
10
24000
(ii) Using the sample of 20 tax returns calculate
sales
1
36000
2
23000
17/34
Maths
3
29000
4
3000
5
49000
6
41000
7
46000
18/34
Maths
8
20000
9
40000
10
30000
11
13000
12
7000
19/34
Maths
13
1000
14
45000
15
22000
16
17000
17
6000
18
20/34
Maths
34000
19
38000
20
24000
total
524000
mean
26200
21/34
Maths
The total for the selected sample is 524000, to get the mean of this data we divide the total by the sample size which in this case will be 524,000 / 20 = 26200 this means that the mean or average is equal to 26200.
The standard deviation is derived from the following formula: the formula below gives the variance and to get the standard deviation we find the square root of the variance:
∑ (X –X’)2
S2 = ____________
N -1
Where X’ is the mean and x is the sample elements, N is the sample size which in this case is equal to 20, the table below shows the calculation of the standard deviation
sales
X -X’
(X-X’)2
22/34
Maths
1
36000
9800
96040000
2
23000
-3200
10240000
3
23/34
Maths
29000
2800
7840000
4
3000
-23200
538240000
5
49000
24/34
Maths
22800
519840000
6
41000
14800
219040000
7
46000
19800
25/34
Maths
392040000
8
20000
-6200
38440000
9
40000
13800
190440000
26/34
Maths
10
30000
3800
14440000
11
13000
-13200
174240000
27/34
Maths
12
7000
-19200
368640000
13
1000
-25200
635040000
14
28/34
Maths
45000
18800
353440000
15
22000
-4200
17640000
16
17000
29/34
Maths
-9200
84640000
17
6000
-20200
408040000
18
34000
7800
30/34
Maths
60840000
19
38000
11800
139240000
20
24000
-2200
4840000
31/34
Maths
total
524000
4.273E+09
mean
26200
2.249E+08
32/34
Maths
standard deviation
14996.84
14996.842
4.273 X 109
S2 = ____________
20 -1
S2 = 2.249 X 108
S = 14996.842
33/34
Maths
Quartile range:
The range of the data is given by the maximum amount minus the minimum amount, in our case the minimum amount is 1000 and the maximum amount is 49000, therefore our range is equal to 49,000 -1,000 = 48,000
From our sample that was generated randomly our mean sales value is 26200 with a standard deviation value equal to 14996.84 and our range of this data is equal to 48,000. this shows that we have a very large deviation of data from the mean as depicted by the standard deviation and the range.
34/34 |
# Question: What Is The GCF Of 36 And 24?
## What is the HCF of 24 36 and 48?
find gcf (24, 36 and 48) = .
step 3The common prime factor or the product of all common prime factors between all integers is the highest common factor of this group of integers (24, 36 & 48)..
## Whats the HCF of 24 and 60?
i) The HCF of 24 and 60 is found by comparing the prime factors and multiplying the numbers common to both lists. So the HCF of 24 and 60 is 2 x 2 x 3 = 12. ii) The LCM of 24 and 60 is found by multiplying together all the prime factors of both numbers.
## What is the HCF of 36?
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36. Therefore, common factor of 24 and 36 = 1, 2, 3, 4, 6, 8 and 12. Highest common factor (H.C.F) of 24 and 36 = 12.
## What is the GCF of 35 and 49?
Answer and Explanation: The greatest common factor of 35 and 49 is 7. To find the greatest common factor, start by listing the factors for each of the numbers.
## What is the GCF of 18 and 63?
The factors of 63 are 63, 21, 9, 7, 3, 1. The common factors of 18 and 63 are 9, 3, 1, intersecting the two sets above. In the intersection factors of 18 ∩ factors of 63 the greatest element is 9. Therefore, the greatest common factor of 18 and 63 is 9.
## What is the prime factorization of 24 and 36?
The prime factorization of 24 is: 2 x 2 x 2 x 3. The prime factorization of 36 is: 2 x 2 x 3 x 3. The prime factors and multiplicities 24 and 36 have in common are: 2 x 2 x 3.
## What is the HCF of 24 32?
8Thus, the highest common factor of 24 and 32 is 8. Hence, HCF of 24 and 32= 8.
## What is the GCF of 18 and 27?
The factors of 18 are 1, 2, 3, 6, 9, 18. The factors of 27 are 1, 3, 9, 27. The common factors of 18 and 27 are 1, 3 and 9. The greatest common factor of 18 and 27 is 9.
## What is the GCF and LCM of 36 and 24?
The GCF of 24 and 36 is 12. ‘GCF’ means ‘greatest common factor’.
## What is the GCF of 24 36 and 60?
The common factors for 24,36,60 24 , 36 , 60 are 1,2,3,4,6,12 1 , 2 , 3 , 4 , 6 , 12 . The GCF (HCF) of the numerical factors 1,2,3,4,6,12 1 , 2 , 3 , 4 , 6 , 12 is 12 .
## What is the LCM of 24 18 and 36?
The least common multiple 72 is a product of common & odd prime factors between the integers which is divisible by each one an integer of this same group. The step by step work for LCM of 18, 24 and 36 may useful to understand how to find LCM for two or three numbers.
## What is the LCM of 3 18 and 24?
Least common multiple (LCM) of 18 and 24 is 72.
## What is the GCF of 15 and 25?
5The Greatest Common Factor of 15and25 is 5. (the number itself and 1 do not count). The factor(s) of 25 is 5 .
## What is the GCF of 18 and 30?
We found the factors and prime factorization of 18 and 30. The biggest common factor number is the GCF number. So the greatest common factor 18 and 30 is 6.
## What is the GCF of 24 and 32?
The greatest common factor of 24 and 32 is 8.
## What is the GCF of 24 and 18?
Consider the numbers 18, 24, and 36. Their greatest common factor is 6. Another method of finding the greatest common factor is by using prime factorization.
## What’s the GCF of 24?
Example 4: Find the GCF of 24 and 36. The common factors of 24 and 36 are 1, 2, 3, 4, 6 and 12. The greatest common factor of 24 and 36 is 12. Example 5: Find the GCF of 56 and 63.
## What is the LCM of 24 36 and 72?
Answer: LCM of 24 and 36 is 72. 2. What are the Factors of 24?
## What is the LCM of 24 and 30?
Least common multiple (LCM) of 24 and 30 is 120.
## What is the HCF of 36 and 84?
23. Find the highest common factor of 36 and 84. H.C.F. = 22 x 3 = 12.
## What is the HCF of 20 28 and 36?
The common factor of 20, 28 and 36 is 2(occurring twice). Thus, HCF of 20, 28 and 36 is 2 × 2 = 4. |
North Carolina High School Mathematics
Math I Unpacking Document
The Real Number System N-RN
Common Core Cluster
Extend the properties of exponents to rational exponents.
Common Core Standard / Unpacking
What does this standard mean that a student will know and be able to do?
N-RN.1 Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define to be the cube root of 5 because we want to hold, so must equal 5.
N-RN.2 Rewrite expressions involving radicals and rational exponents using the properties of exponents. / N-RN.1 In order to understand the meaning of rational exponents, students can initially investigate them by considering a pattern such as:
What is the pattern for the exponents? They are reduced by a factor of each time. What is the pattern of the simplified values? Each successive value is the square root of the previous value. If we continue this pattern, then .
Once the meaning of a rational exponent (with a numerator of 1) is established, students can verify that the properties of integer exponents hold for rational exponents as well. For example,
since
since
Ex. Use an example to show why holds true for expressions involving rational exponents like or .
N-RN.2Students should be able to use the properties of exponents to rewrite expressions involving radicals as expressions using rational exponents. At this level, focus on fractional exponents with a numerator of 1.
Ex. Simplify the following.
N-RN.2 Students should be able to use the properties of exponentstorewrite expressions involving rational exponents as expressions using radicals. At this level, focus on fractional exponents with a numerator of 1.
Ex. Simplify the following.
Quantities* N-Q
Common Core Cluster
Reason quantitatively and use units to solve problems.
Common Core Standard / Unpacking
What does this standard mean that a student will know and be able to do?
N-Q.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.
N-Q.2 Define appropriate quantities for the purpose of descriptive modeling.
N-Q.3 Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. / N-Q.1Use units as a tool to help solve multi-step problems. For example, students should use the units assigned to quantities in a problem to help identify which variable they correspond to in a formula. Students should also analyze units to determine which operations to use when solving a problem. Given the speed in mph and time traveled in hours, what is the distance traveled? From looking at the units, we can determine that we must multiply mph times hours to get an answer expressed in miles: (Note that knowledge of the distance formula is not required to determine the need to multiply in this case.)
N-Q.1Based on the type of quantities represented by variables in a formula, choose the appropriate units to express the variables and interpret the meaning of the units in the context of the relationships that the formula describes.
Ex. When finding the area of a circle using the formula , which unit of measure would be appropriate for the radius?
1. square feet
2. inches
3. cubic yards
4. pounds
Ex. Based on your answer to the previous question, what units would the area be measured in?
N-Q.1When given a graph or data display, read and interpret the scale and origin. When creating a graph or data display, choose a scale that is appropriate for viewing the features of a graph or data display. Understand that using larger values for the tick marks on the scale effectively “zooms out” from the graph and choosing smaller values “zooms in.” Understand that the viewing window does not necessarily show the x- or y-axis, but the apparent axes are parallel to the x- and y-axes. Hence, the intersection of the apparent axes in the viewing window may not be the origin. Also be aware that apparent intercepts may not correspond to the actual x- or y-intercepts of the graph of a function.
Ex. What scale would be appropriate for making a histogram of the following data that describes the level of lead in the blood of children (in micrograms per deciliter) who were exposed to lead from their parents’ workplace?
10, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 23, 24, 25, 27, 31, 34, 34, 35, 35, 36, 37, 38, 39, 39, 41, 43, 44, 45, 48, 49, 62, 73
N-Q.2 Define the appropriate quantities to describe the characteristics of interest for a population. For example, if you want to describe how dangerous the roads are, you may choose to report the number of accidents per year on a particular stretch of interstate. Generally speaking, it would not be appropriate to report the number of exits on that stretch of interstate to describe the level of danger.
Ex. What quantities could you use to describe the best city in North Carolina?
Ex. What quantities could you use to describe how good a basketball player is?
N-Q.3Understand that the tool used determines the level of accuracy that can be reported for a measurement. For example, when using a ruler, you can only legitimately report accuracy to the nearest division. If I use a ruler that has centimeter divisions to measure the length of my pencil, I can only report its length to the nearest centimeter.
Ex. What is the accuracy of a ruler with 16 divisions per inch?
Seeing Structure in Expressions A-SSE
Common Core Cluster
Interpret the structure of expressions.
Common Core Standard / Unpacking
What does this standard mean that a student will know and be able to do?
A-SSE.1Interpret expressions that represent a quantity in terms of its context.
a.Interpret parts of an expression, such as terms, factors, and coefficients.
b.Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret as the product of P and a factor not depending on P.
A-SSE.2 Use the structure of an expression to identify ways to rewrite it. For example, see x4 – y4 as (x2)2 – (y2)2, thus recognizing it as a difference of squares that can be factored as (x2 – y2)(x2 + y2). / A-SSE.1a. Students manipulate the terms, factors, and coefficients in difficult expressions to explain the meaning of the individual parts of the expression. Use them to make sense of the multiple factors and terms of the expression. For example, consider the expression 10,000(1.055)5. This expression can be viewed as the product of 10,000 and 1.055 raised to the 5th power. 10,000 could represent the initial amount of money I have invested in an account. The exponent tells me that I have invested this amount of money for 5 years. The base of 1.055 can be rewritten as (1 + 0.055), revealing the growth rate of 5.5% per year. At this level, limit to linear expressions, exponential expressions with integer exponents, and quadratic expressions.
Ex. The expression 20(4x) + 500 represents the cost in dollars of the materials and labor needed to build a square fence with side length x feet around a playground. Interpret the constants and coefficients of the expression in context.
A-SSE.1b Students group together parts of an expression to reveal underlying structure.For example, consider the expression that represents income from a concert wherep is the price per ticket. The equivalent factored form,, shows that the income can be interpreted as the price times the number of people in attendancebased on the price charged. At this level, limit to linear expressions, exponential expressions with integer exponents, and quadratic expressions.
Ex. Without expanding, explain how the expression can be viewed as having the structure of a quadratic expression.
A.SSE.2 Students rewrite algebraic expressions by combining like terms or factoring to reveal equivalent forms of the same expression.
Ex. Expand the expression to show that it is a quadratic expression of the form .
Seeing Structure in Expressions A-SSE
Common Core Cluster
Write expressions in equivalent forms to solve problems.
Common Core Standard / Unpacking
What does this standard mean that a student will know and be able to do?
A-SSE.3Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.
1. Factor a quadratic expression to reveal the zeros of the function it defines.
/ A-SSE.3a Students factor quadratic expressions and find the zeros of thequadratic function they represent. Zeroes are the x-values that yield a y-value of 0. Students should also explain the meaning of the zeros as they relate to the problem. For example, if the expression x2– 4x +3 represents the path of a ball that is thrown from one person to another, then the expression (x – 1)(x – 3) represents its equivalent factored form. The zeros of the function, (x – 1)(x – 3) = y would be x= 1 and x= 3, because an x-value of 1 or 3 would cause the value of the function to equal 0. This also indicates the ball was thrown after 1 second of holding the ball, and caught by the other person 2 seconds later. At this level, limit to quadratic expressions of the form ax2 + bx + c.
Ex. The expression is the income gathered by promoters of a rock concert based on the ticket price, m. For what value(s) of m would the promoters break even?
Arithmetic with Polynomials and Rational Expressions A-APR
Common Core Cluster
Perform arithmetic operations on polynomials
Common Core Standard / Unpacking
What does this standard mean that a student will know and be able to do?
A-APR.1Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. / A-APR.1 The Closure Property means that when adding, subtracting or multiplying polynomials, the sum, difference, or product is also a polynomial. Polynomials are not closed under division because in some cases the result is a rational expression rather than a polynomial. At this level, limit to addition and subtraction of quadratics and multiplication of linear expressions.
A-APR.1 Add, subtract, and multiply polynomials. At this level, limit to addition and subtraction of quadratics and multiplication of linear expressions.
Ex. If the radius of a circle is kilometers, what would the area of the circle be?
Ex. Explain why does not equal.
Creating Equations* A-CED
Common Core Cluster
Create equations that describe numbers or relationships
Common Core Standard / Unpacking
What does this standard mean that a student will know and be able to do?
A-CED.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
A-CED.2 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
A-CED.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.
A-CED.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrangeOhm’s law V = IR to highlight resistance R. / A-CED.1From contextual situations, write equations and inequalities in one variable and use them to solve problems. Include linear and exponential functions. At this level, focus on linear and exponential functions.
Ex. The Tindell household contains three people of different generations. The total of the ages of the three family members is 85.
1. Find reasonable ages for the three Tindells.
2. Find another reasonable set of ages for them.
3. One student, in solving this problem, wrote C + (C+20)+ (C+56) = 85
1. What does C represent in this equation?
2. What do you think the student had in mind when using the numbers 20 and 56?
3. What set of ages do you think the student came up with?
Ex. A salesperson earns \$700 per month plus 20% of sales. Write an equation to find the minimum amount of sales needed to receive a salary of at least \$2500 per month.
Ex. A scientist has 100 grams of a radioactive substance. Half of it decays every hour. Write an equation to find how long it takes until 25 grams are left.
A-CED.2Given a contextual situation, write equations in two variables that represent the relationship that exists between the quantities. Also graph the equation with appropriate labels and scales. Make sure students are exposed to a variety of equations arising from the functions they have studied. At this level, focus on linear, exponential and quadratic equations.Limit to situations that involve evaluating exponential functions for integer inputs.
Ex. In a woman’s professional tennis tournament, the money a player wins depends on her finishing place in the standings. The first-place finisher wins half of \$1,500,000 in total prize money. The second-place finisher wins half of what is left; then the third-place finisher wins half of that, and so on.
1. Write a rule to calculate the actual prize money in dollars won by the player finishing in nth place, for any positive integer n.
2. Graph the relationship that exists between the first 10 finishers and the prize money in dollars.
3. What pattern do you notice in the graph? What type of relationship exists between the two variables?
A-CED.3 Use constraints which are situations that are restricted to develop equations and inequalities and systems of equations or inequalities. Describe the solutions in context and explain why any particular one would be the optimal solution. Limit to linear equations and inequalities.
Ex. The Elite Dance Studio budgets a maximum of \$100 per month for newspaper and yellow pages advertising. The news paper charges \$50 per ad and requires at least four ads per month. The phone company charges \$100 dollars for half a page and requires a minimum of two advertisements per month. It is estimated that each newspaper ad reaches 8000 people and that each half page of yellow page advertisement reaches 15,000 people. What combination of newspaper and yellow page advertising should the Elite Dance Studio use in order to reach the maximum number of people? What assumptions did you make in solving this problem? How realistic do you think they are?
A-CED.4 Solve multi-variable formulas or literal equations, for a specific variable. Explicitly connect this to the process of solving equations using inverse operations. Limit to formulas which are linear in the variable of interest or to formulas involving squared or cubed variables.
Ex. If , solve for T2
Reasoning with Equations and Inequalities A-REI
Common Core Cluster
Understanding solving equations as a process of reasoning and explain the reasoning
Common Core Standard / Unpacking
What does this standard mean that a student will know and be able to do?
A-REI.1Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. / A-REI.1 Relate the concept of equality to the concrete representation of the balance of two equal quantities. Properties of equality are ways of transforming equations while still maintaining equality/balance. Assuming an equation has a solution, construct a convincing argument that justifies each step in the solution process with mathematical properties.
Ex. Solve 5(x+3)-3x=55 for x. Use mathematical properties to justify each step in the process.
Common Core Cluster
Solve equations and equalities in one variable.
Common Core Standard / Unpacking
What does this standard mean that a student will know and be able to do?
A-REI.3Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. / A-REI.3 Solve linear equations in one variable, including coefficients represented by letters.
Ex. Solve, Ax +B =C for x. What are the specific restrictions on A?
Ex. What is the difference between solving an equation and simplifying an expression?
Ex. Grandma’s house is 20 miles away and Johnny wants to know how long it will take to get there using various modes of transportation.
1. Model this situation with an equation where time is a function of rate in miles per hour.
2. For each mode of transportation listed below, determine the time it would take to get to Grandma’s.
Mode of Transportation / Rate of Travel in mph / Time of Travel hrs.
bike / 12mph
car / 55mph
walking / 4mph
A-REI.3 Solve linear inequalities in one variable, including coefficients represented by letters.
Ex. A parking garage charges \$1 for the first half-hour and \$0.60 for each additional half-hour or portion thereof. If you have only \$6.00 in cash, write an inequality and solve it to find how long you can park.
Ex. Compare solving an inequality in one variable to solving an equation in one variable, also compare solving a linear inequality to solving a linear equation.
Common Core Cluster
Solve systems of equations.
Common Core Standard / Unpacking
What does this standard mean that a student will know and be able to do?
A-REI.6Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. / A.REI.6Solve systems of equations exactly by using the substitution method and solve systems of equations by using the elimination method (sometimes called linear combinations).
Ex. Solve the system by elimination, checking your solution by graphing using technology.
3x + 2y = 6
x - 4y = 2
Ex. Solve the system by substitution, checking your solution by graphing using technology.
-3x + 5y = 6
2x + y = 6
A.REI.6 Solve systems of equations approximately by using graphs. Graph the system of linear functions on the same coordinate plane and find the point of intersection. This point is the solution to the system because it is the one point that makes all equations in the system true. Equations may be in standard or slope-intercept form.
Ex. The equations y = 18 + .4m and y = 11.2 + .54m give the lengths of two different springs in centimeters, as mass is added in grams, m, to each separately.
1. Graph each equation on the same set of axes.
2. What mass makes the springs the same length?
3. What is the length at that mass?
4. Write a sentence comparing the two springs.
Common Core Cluster
Represent and solve equations and inequalities graphically.
Common Core Standard / Unpacking |
# How do you write an equation in point-slope form for the given (0, 6), (3,0)?
Mar 19, 2018
$y - 0 = - 2 \left(x - 3\right)$
#### Explanation:
$\text{the equation of a line in "color(blue)"point-slope form}$ is.
•color(white)(x)y-y_1=m(x-x_1)
$\text{where m is the slope and "(x_1,y_1)" a point on the line}$
$\text{to calculate m use the "color(blue)"gradient formula}$
•color(white)(x)m=(y_2-y_1)/(x_2-x_1)
$\text{let "(x_1,y_1)=(0,6)" and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 0\right)$
$\Rightarrow m = \frac{0 - 6}{3 - 0} = \frac{- 6}{3} = - 2$
$\text{use either of the given points as a point on the line}$
$\text{using "(3,0)" then}$
$y - 0 = - 2 \left(x - 3\right)$
$\Rightarrow y = - 2 \left(x - 3\right) \leftarrow \textcolor{red}{\text{in point-slope form}}$ |
# On Understanding Probability Puzzles
The theory of probability tries to assign a number to our belief that something is “true”. Having its roots in gambling, probability is an intricate part of everyone’s life in this age. The insurance industry, finance industry, casinos, health care industry, double blind experiments, election exit polling etc are just a few instances where probability plays a crucial role. Before going into the abstract theory, one of the best ways to understand probability is to see it from an experimental perspective. Let’s consider the following concrete example.
— 1. Girl born on Tuesday problem —
Problem 1 If a couple has their first girl child born on a Tuesday, what is the probability that their second child is also a girl?
You might wonder the presence of Tuesday in the question. Just like most other probability puzzles, this one is ill posed. And just like most other probability puzzles, there are numerous ways in which to make the problem well posed. Once this is done, will land up with a unique answer. And different people, depending upon how they make the problem well posed typically land up with different and correct answers. The funny part is I have seen people arguing very passionately about how their solution is the “correct” one. Let’s first simulate a thought experiment in our mind. We need to generate an experiment whose outcomes contain our events of interest. Let’s ignore the Tuesday information for a moment and assume for simplicity that everyone has just two children. Consider the following experiment.
There are four outcomes of the above experiments which are ${(B,B),(B,G),(G,B),(G,G)}$. Therefore, for this interpretation of Problem 1, the solution for the puzzle without the Tuesday information is ${\frac{1}{4}}$. Now, let’s put back in the Tuesday for our puzzle. The setup for our thought experiment is as follows.
Solution: For this new experiment, the outcomes ${(C[1],D[1],C[2],D[2])}$ corresponding to two pairs of gender and day of the week can take ${2\times 7 \times 2 \times 7 = 196}$ values. Now, we are already told that the first child is a girl born on Tuesday. Therefore, our sample space size is reduced to ${1\times 1\times 2\times 7}$. The event of interest to us, i.e., both girls, can take ${1\times 1\times 1 \times 7}$ values corresponding to ${(Girl,Tuesday,Girl, Anyday)}$. Therefore, under this interpretation, our answer is ${1/2}$. $\Box$
I am sure there are other ways to interpret the question. But once you have made the interpretation and set up your experiment, you should get a unique answer. Let’s change the problem a bit as follows.
Problem 2 If a couple has a girl child born on a Tuesday, what is the probability that their second child is also a girl?
Solution: In this case, we are only told that the couple has a girl child born on a Tuesday. It could be the first girl child born on a Tuesday or the second girl child born on a Tuesday. Let’s compute the sample space size first. Out of all the 196 outcomes, only ${2\times 7 + 2\times 7 -1 = 27}$ outcomes have a girl child born on a Tuesday. And out of all the 196 possible outcomes, only ${1\times 7 + 1\times 7 -1 = 13}$ outcomes have both girls, of which one is born on a Tuesday. Therefore, the required probability is ${13/27}$. $\Box$
— 2. The Month Hall problem —
One of my favorite probability puzzles is the Monty Hall problem. Here is the description.
Problem 3 Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
Solution: Let’s set up our experiment for this problem as follows.
The above experiment returns the probability of winning. If one is not switching doors after being shown a goat door, then one sees that you win if ${InitialPick = CarDoor}$ which happens with probability ${1/3}$. Therefore, with switching strategy, the probability of winning is the complementary event with value ${2/3}$. In other words, with the switching strategy, you win if you made the bad choice initially. And with probability ${2/3}$, you are bound to make a bad choice at the beginning. To appreciate this statement better, consider an extreme Monty Hall problem with 1000 doors and the host shows you 998 goats. You are asked if you want to change your initial pick. In this case, your chances of making a bad choice initially is ${999/1000}$, i.e., your chances of winning with the swap strategy is ${999/1000}$. The point being that the host is not opening a door for you at random. S/he knows what is behind each door and one is essentially exploiting this information with the swap strategy. $\Box$
— 3. The concentric circles and chord problem —
Let’s consider a more serious looking problem. I have posed this problem to a couple of my friends and have never got the same answer.
Problem 4 Consider two concentric circles of radii 1 and 2 respectively. If you pick a chord on the outer circle at random, what is the probability that the chord will touch or intersect the inner circle?
Solution: Here is one way to make the problem well posed.
For this interpretation of Problem 4, one gets a unique answer which is ${1/3}$. I am aware of at least two other ways to make the problem well posed and which will give you (correct) answers ${1/2}$ and ${1/4}$. The corresponding experiments are different ways to make the phrase “pick at random” precise. For example, one can pick a number ${L}$ uniformly from the interval ${\left[ 0,4 \right]}$, an angle ${\theta}$ uniformly from ${\left[ 0,\pi \right]}$ and consider a chord of length ${L}$ and orientation ${\theta}$. Or one can sample a point ${P}$ uniformly inside the disc of radius ${2}$ and consider the (almost) unique chord for which ${P}$ is the mid point. $\Box$
To summarize, do not panic if a probability puzzle seems out of reach for you. Stay calm and run a thought simulation of the problem in your mind and things will become clear. As an exercise, try to solve the following problems by making them well-posed.
— 4. Exercise —
Problem 5 Let ${A}$ be the ordered array of integers ${\{1,\dots,N\}}$ where ${N}$ is unknown but surely between 1 and 100. A person picks up an element of ${A}$ at random and it turns out to be 10. What is the expected value of ${N}$?
As slightly related problem is the following.
Problem 6 Let ${A}$ be the ordered array of integers ${\{1,\dots,N\}}$ where ${N\ge 1}$. A person picks up an element of ${A}$ at random and it turns out to be 10. What is the expected value of ${N}$?
Problem 7 If the probability of seeing a bus at a particular intersection in a 20 minute window is 0.9, what is the probability of seeing a bus at the same intersection in a 5 minute window?
Problem 7 seems to be very popular in the puzzle world. It is quite interesting and most people who ask this don’t realize the subtlety because they themselves heard it from someone else or saw it in a book and never thought deeply about it. Again, think of an experiment and simulate it in your mind. You can come with with at least three different interpretations for the underlying experiment and hence at least three correct answers.
The following two sections are a bit more technical and not really in the puzzle category. I will assume knowledge of some basic results in probability.
— 5. The Brownian motion —
Let’s try to understand the one-dimensional Brownian motion, denoted by ${X_t}$, from an experimental point of view. The formal construction for the Brownian motion can be found here for example. And you will get a good historical perspective of the Brownian motion here. One of the basic results of Paley, Wiener and Zygmund in Brownian motion is the following
Theorem 1 (Paley, Wiener, Zygmund) Almost surely, the Brownian motion is nowhere differentiable, i.e., the map ${t\rightarrow X_t}$ is almost surely nowhere differentiable.
We will take this fact for granted but lets see what it means experimentally. Lets have two indices, ${\omega}$ and ${t}$. The index ${\omega }$ represents the index of our experiment output and the index ${t}$ represents the time index for for an instance of the experiment. For example, say I have 100 petri dishes with water and I place a pollen grain in the middle of each one of them at ${t=0}$ and let it go. In this case, my ${\omega }$ ranges from 1 to 100 and for each ${\omega }$, the time ranges from 0 to ${\infty}$. There are various interesting questions which one could ask. For example,
• What is the probability that the pollen grain will hit the boundary of the petri dish?
• What is the expected time for the pollen grain to hit the boundary?
• What is the probability that the pollen grain, starting from the middle of the petri dish, will hit a particular point on the boundary?
• What is the mean and variance of the location of the pollen grain in time?
One could of course conduct millions of physical or numerical experiments and empirically compute these numbers. The theory of Brownian motion is an attempt to model this experimental setup into a rigorous mathematical framework so that these numbers can be computed without too much effort. The way this is achieved is by considering (uncountably large) number of petri dishes, placing a pollen grain in the middle of each one of them and letting it go. If ${\Sigma = \left[ 0,\infty \right)}$ and ${\omega \in \mathbb{R}}$ is the index for the ${\omega^{\rm th}}$ petri dish, then the underlying experimental space is of size ${\mathbb{R}^{\Sigma}}$. Now, it is a highly nontrivial task to show that one can indeed think of ${\mathbb{R}^{\Sigma}}$ as a probability space, i.e., a measure space with a finite measure called the Wiener measure. That is the content of Kolmogorov Extension theorem which can be found in any graduate level probability textbook. For the Brownian motion, one can consider a single experimental output, ${X_t}$, to be an element of ${\mathbb{R}^{\Sigma}}$ for some index ${\omega }$. In other words, ${B(\omega ,t)\in \mathbb{R}^{\Sigma}}$ denotes the position of the pollen grain in the ${\omega^{\rm th}}$ petri dish at time ${t}$.
So the statement ${t\rightarrow X_t}$ is almost surely nowhere differentiable has the following experimental setup. Let ${N}$ be a large integer and let ${\omega_1,\dots,\omega_N}$ be ${N}$ petri dishes with a pollen grain in the middle of each one of them at ${t=0}$. Let the experiment run for time ${t\in \Sigma}$. Let ${D_N}$ be the number of petri dishes such that the corresponding pollen grain trajectory ${X_t}$ is differentiable at time ${t}$. Then, ${D_N/N\rightarrow 0}$ as ${N\rightarrow \infty}$.
Construct an experiment to solve the following.
Problem 8 We saw above that for each ${t}$, the brownian motion ${X_t}$ is almost surely not differentiable. Does this imply that almost surely every ${t}$ is a point of nondifferentiability? In other words, does { for all t } and { almost surely } commute?
— 6. Compressive Sensing —
Let me conclude by discussing the experimental setup in the compressive sensing (CS) scenario. I will stick to the notation used in this Candes-Tao paper. In compressive sensing, the underlying setup is the following. One has a signal known to be sparse in some particular basis. The question is then to quantify the minimum number of “random” samples that need to be taken to guarantee that one can reconstruct the original signal with high “probability”. The problem reduces to “solving” an equation of the form ${Ax=B}$. See these slides of Terry Tao for a high level overview. The nontrivial thing about CS is in making all the above quotes mathematically precise. The two versions of results in the CS literature are the uniform results and the non-uniform results. I will now explain the difference when seen from an experimental point of view. Here is the statement taken from Tao’s blog. In the theorem below, the signal ${f:\mathbb{Z}/N\mathbb{Z}\rightarrow \mathbb{C}}$, thought of as a complex vector of size ${N}$. And ${\Lambda\subset \mathbb{Z}/N\mathbb{Z}}$ is the set of observed frequencies.
Theorem 2 Suppose ${\Lambda}$ is a random set with ${\Lambda\gg S\log N}$. Then any given ${S}$-sparse function ${f}$ will be recovered exactly by ${l^1}$ minimisation with overwhelming probability. If one makes the stronger hypothesis ${|{\Lambda}|\gg S\log^4 N}$ , then with overwhelming probability {\rm all ${S}$-sparse} functions will be recovered exactly by ${l^1}$ minimisation. (Again, ${N}$ is not required to be prime.)
The big difference between these two results from an experimental point of view is the following. In the stronger hypothesis case, one allows for both ${A}$ and ${x}$ to be random. In the weaker hypothesis case, ${x}$ is fixed and ${A}$ is chosen to be random. The weaker hypothesis experiment is
The stronger hypothesis experiment is
As one can see, the underlying sample space has dimensions
• ${\sum_{i=1}^k{N\choose i}\leq k{N\choose k}}$ in the weaker hypothesis case
• ${{N\choose |{\Lambda}|}\left(\sum_{i=1}^k{N\choose i}\right)\leq {N\choose |{\Lambda}|}k{N\choose k}}$ in the stronger hypothesis case
Therefore, the sample space is exponential in ${k}$ for the weak case and is exponential in ${|{\Lambda}|+k}$ for the strong case. The proof for the strong case for Gaussian ensembles is relatively easy because of exponential tail bounds for minimum and maximum singular values. A simple union bound argument suffices. See Section 4.1 of this paper. But because of this large dimensionality, the proof for the Fourier measurement case becomes quite nontrivial. See CT and RV for more technical details. |
# Chapter 4 - Vector Spaces - 4.7 Coordinates and Change of Basis - 4.7 Exercises - Page 210: 34
$$P^{-1}= \left[ \begin {array}{cccccccccc} {\frac {12}{157}}&{\frac {45}{157}}&-{\frac {17}{157}}&-{\frac {1}{157}}&-{\frac {4}{157}} \\ {\frac {32}{157}}&-{\frac {37}{157}}&{ \frac {7}{157}}&{\frac {47}{314}}&{\frac {31}{314}} \\ {\frac {5}{314}}&-{\frac {99}{314}}&{ \frac {3}{157}}&{\frac {287}{628}}&{\frac {49}{628}} \\ {\frac {10}{157}}&-{\frac {41}{157}}&{ \frac {12}{157}}&{\frac {103}{314}}&-{\frac {59}{314}} \\ -{\frac {7}{157}}&{\frac {13}{157}}&{ \frac {23}{157}}&-{\frac {25}{314}}&{\frac {57}{314}}\end {array}\right] .$$
#### Work Step by Step
Given $$B=\{(1,0,0,0,0),(0,1,0,0,0),(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1) \} , \\ B^{\prime}= \{(2,4,-2,1,0),(3,-1,0,1,2),(0,0,-2,4,5), (2,-1,2,1,1),(0,1,2,-3,1) \} .$$ To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix $$\left[B^{\prime} B\right]= \left[ \begin {array}{cccccccccc} 2&4&-2&1&0&1&0&0&0&0 \\ 3&-1&0&1&2&0&1&0&0&0\\ 0&0&-2&4 &5&0&0&1&0&0\\ 2&-1&2&1&1&0&0&0&1&0 \\ 0&1&2&-3&1&0&0&0&0&1\end {array} \right] .$$ Using Gauss-Jordan elimination to obtain the transition matrix $$\left[\begin{array}{ll}{I_{5}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccccccc} 1&0&0&0&0&{\frac {12}{157}}&{\frac {45}{157}}&-{\frac {17}{157}}&-{\frac {1}{157}}&-{\frac {4}{157}} \\ 0&1&0&0&0&{\frac {32}{157}}&-{\frac {37}{157}}&{ \frac {7}{157}}&{\frac {47}{314}}&{\frac {31}{314}} \\ 0&0&1&0&0&{\frac {5}{314}}&-{\frac {99}{314}}&{ \frac {3}{157}}&{\frac {287}{628}}&{\frac {49}{628}} \\ 0&0&0&1&0&{\frac {10}{157}}&-{\frac {41}{157}}&{ \frac {12}{157}}&{\frac {103}{314}}&-{\frac {59}{314}} \\ 0&0&0&0&1&-{\frac {7}{157}}&{\frac {13}{157}}&{ \frac {23}{157}}&-{\frac {25}{314}}&{\frac {57}{314}}\end {array} \right] .$$ So, we have $$P^{-1}= \left[ \begin {array}{cccccccccc} {\frac {12}{157}}&{\frac {45}{157}}&-{\frac {17}{157}}&-{\frac {1}{157}}&-{\frac {4}{157}} \\ {\frac {32}{157}}&-{\frac {37}{157}}&{ \frac {7}{157}}&{\frac {47}{314}}&{\frac {31}{314}} \\ {\frac {5}{314}}&-{\frac {99}{314}}&{ \frac {3}{157}}&{\frac {287}{628}}&{\frac {49}{628}} \\ {\frac {10}{157}}&-{\frac {41}{157}}&{ \frac {12}{157}}&{\frac {103}{314}}&-{\frac {59}{314}} \\ -{\frac {7}{157}}&{\frac {13}{157}}&{ \frac {23}{157}}&-{\frac {25}{314}}&{\frac {57}{314}}\end {array}\right] .$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
## CONTENTS
Grade 6 is a benchmark year for every child. Almost on the threshold of junior school, Grade 6 is the bridge between the foundations of elementary school and new concepts of junior school. This bridge makes it imperative for every Grade 6 students’ mathematics to be concrete and clear. Entering the world of ratios with the understanding of dimensions and basic fractions and decimals from Grade 5, Grade 6 elaborates on the applications of the basic concepts including the metric system, probability, fractions while introducing new facets of probability and graphs.
## Whole numbers
In mathematics whole numbers are positive integers or nonnegative integers if you include zero. There is no fraction or decimal parts in whole numbers they also cannot be negative. Whole numbers are important because they are basic counting numbers.
Place values in whole numbers
Word names for numbers
Roman numerals
## Decimals
A number that has a decimal point followed by digits that show a value smaller than one. Understanding decimal numbers is important to our daily lives as it relates to money, fractional amounts or metric measures.
What decimal number is illustrated?
Decimal place values
Word names for decimal numbers
Put decimal numbers in order
Inequalities with decimals
Round decimals
Round whole numbers and decimals: find the missing digit
Decimal number lines
## Integers
Integers are like whole numbers which means they do not have fraction or decimal part but unlike whole numbers they can be negative.
Understanding integers
Integers on number lines
Graph integers on horizontal and vertical number lines
Compare and order integers
In the primary grades, students develop an understanding of part-whole concepts – they learn that two or more parts can be combined to create a whole (addition) or they can be separated from a whole (Subtraction).
Add and subtract whole numbers up to millions
Add and subtract whole numbers: word problems
Estimate sums and differences of whole numbers
Estimate sums and differences: word problems
Add and subtract decimals: word problems
Estimate sums and differences of decimals
Maps with decimal distances
## Multiplication
The multiplication of numbers can be seen as repeated addition. For example, 3×4 = 4+4+4. It is important to learn the basic operations for future math works.
Multiply whole numbers
Multiply whole numbers: word problems
Multiply numbers ending in zeroes
Multiply numbers ending in zeroes: word problems
Multiply three or more numbers
Multiply three or more numbers: word problems
Estimate products
## Division
Separation of partitioning of objects from a set for an equal share without any disparity in the fixed number of groups. The resultant will be a whole number quotient and a remainder if the number is not divisible further.
Divisibility rules
Divide by two-digit numbers
Divide by two-digit numbers: word problems
Division patterns with zeroes
Divide numbers ending in zeroes: word problems
Estimate quotients
## Multiply and divide decimals
Multiplying and dividing decimals is like the normal multiplication and division with one small difference aka decimals points. Just need to do the multiplication without the decimal then insert the decimal in the right spot.
Multiply a decimal by a one-digit whole number
Multiply a decimal by a multi-digit whole number
Multiply decimals and whole numbers: word problems
Multiply two decimals using grids
Multiply decimals
Estimate products of decimal numbers
Inequalities with decimal multiplication
Divide decimals by whole numbers
Divide decimals by whole numbers: word problems
Multiply and divide decimals by powers of ten
Division with decimal quotients
Inequalities with decimal division
## Exponents
The exponent of a number tells us how many times to multiply that number with itself. It is written as a small number to the right and above the base number.
Write multiplication expressions using exponents
Evaluate exponents
Find the missing exponent or base
Exponents with decimal bases
## Number theory
The study of whole numbers and their properties is called Number theory. Number theory is a large and interesting area in mathematics, including studying Prime numbers, rational numbers and so on.
Prime or composite
Identify factors
Prime factorization
Prime factorization with exponents
Greatest common factor
Least common multiple
GCF and LCM: word problems
## Fractions and mixed numbers
Fractions help us to understand how many equal parts of a whole we have.
Fractions and mixed numbers review
Understanding fractions: word problems
Equivalent fractions review
Write fractions in lowest terms
Fractions of a group: word problems
Least common denominator
Compare fractions using models
Compare fractions with like and unlike denominators
Compare fractions: word problems
Convert between improper fractions and mixed numbers
Convert between decimals and fractions or mixed numbers
Graph and order fractions on number lines
Order fractions
Put a mix of decimals, fractions and mixed numbers in order
We can add or subtract fractions like the normal numbers if their denominators are equal. If the denominators are not equal, we have to change it by multiplying or dividing but we have to apply the same to the top (numerators).
1. K.1
Add and subtract fractions with like denominators using number lines
2. K.2
Add and subtract fractions with like denominators
Add and subtract fractions with like denominators: word problems
Inequalities with addition and subtraction of like fractions
Add and subtract mixed numbers with like denominators
Add and subtract mixed numbers with like denominators: word problems
Estimate sums and differences of mixed numbers
## Multiply fractions
Multiplying the fractions is quite easy, you have to just multiply the numerators and denominators to each other and at the end just simplify the fraction if needed.
Fractions of a number: word problems
Multiply unit fractions by whole numbers using number lines
Multiply unit fractions by whole numbers using models
Multiples of fractions
Multiply fractions by whole numbers using number lines
Multiply fractions by whole numbers using models
Multiply fractions by whole numbers I
Multiply fractions by whole numbers II
Estimate products of fractions and whole numbers
## Mixed operations
Mixing different basic operations to solve questions and teaching the order of operations which tells us the order to solve steps in expressions with more than one operation.
Add, subtract, multiply or divide two whole numbers
Add, subtract, multiply or divide two whole numbers: word problems
Evaluate numerical expressions involving whole numbers
Add, subtract, multiply or divide two decimals
Add, subtract, multiply or divide two decimals: word problems
Evaluate numerical expressions involving decimals
Add, subtract or multiply two fractions
Add, subtract or multiply two fractions: word problems
## Rational numbers
A Rational number is made by dividing two integers (integer is a number with no fraction or decimal part). Most of the numbers we use in everyday life are rational numbers. If you can write a number as a simple fraction, then it is a rational number.
Compare rational numbers
Put rational numbers in order
## Problem solving and estimation
A question that needs a solution. In mathematics some problems use words, so you need to learn how to interpret them into mathematical expressions and find the appropriate answer to the question. Problem solving is important skill in life which helps you to tackles the problem in life and find the best solution for them. Sometimes when we try to find an answer to our question, we cannot find the exact answer, so we estimate it. Estimation is the process of finding a value that is close enough to the right answer.
Estimate to solve word problems
Multi-step word problems
Word problems with extra or missing information
Guess-and-check word problems
Distance/direction to starting point
Use logical reasoning to find the order
## Ratios and rates
A ratio shows the relative sizes of two or more values, in other words a Ratio compares values. A ratio says how much of one thing there is compared to another thing. For example, if there is 1 boy and 3 girls, we can write the ratio like 1:3 (for every 1 boy there are 3 girls.)
Write a ratio
Write a ratio: word problems
Identify equivalent ratios
Write an equivalent ratio
Ratio tables
Unit rates and equivalent rates
Compare ratios: word problems
Do the ratios form a proportion?
Solve the proportion
Scale drawings: word problems
## Percents
“Percent” comes from the Latin Per Centum. The Latin word Centum means 100. When we say Percent, we mean per 100, so 1 percent means 1 per 100. We use the symbol % to show the percent. For example, 50% means 50 per 100. Understanding percentages is important for math skills and real life, for example stores advertise discounts on their products by using percents, like 30% off on math books.
What percentage is illustrated?
Convert between percents, fractions and decimals
Compare percents to each other and to fractions
Compare percents and fractions: word problems
Percents of numbers and money amounts
Percents of numbers: word problems
Find what percent one number is of another
Find what percent one number is of another: word problems
## Units of measurement
A quantity used as a standard of measurement; it is how much makes up “1” of the measurement. So, 1 second is a unit of time or the basic unit of length in metric is meter so 1 meter is a unit of length.
Estimate metric measurements
Convert and compare metric units
Metric mixed units
Convert square and cubic units of length
Convert between cubic metres and litres
Compare temperatures above and below zero
## Money
Teaching students about how to add and subtract money values and understanding the prices and rounding them up. It is an important practice for real life problems like understanding the prices and how money relates to real world.
Find the number of each type of coin
Add and subtract money amounts: word problems
Multiply money by whole numbers
Multiply money: word problems
Divide money amounts
Divide money amounts: word problems
## Consumer math
Consumer math is about learning spending money skills by using basic math skills such as basic operations, percent and other skills. Consumer math is an important skill for everyday life, you will learn how to calculate sale prices, tax and interests.
Sale prices
Which is the better coupon?
Unit prices
Unit prices with fractions and decimals
Percents – calculate tax, tip, mark-up and more
Simple interest
## Time
Time is the ongoing sequence of events taking place. The past, present and future. We can measure time using clocks. Time has different units like second, minutes, hours, days and so on.
Elapsed time
Time units
Find start and end times
Convert between 12-hour and 24-hour time
## Coordinate plane
The plane containing X axis and Y axis is called coordinate plane. Cartesian coordinated can be used to pinpoint where we are on a map or graph. We can mark a point on a graph by how far along and how far up it is, the point (10,6) is 10 units along and 6 units up.
Objects on a coordinate plane
Graph points on a coordinate plane
Coordinate planes as maps
Follow directions on a coordinate plane
## Expressions and properties
Numbers, symbols and operators grouped together that show the value of something is called an expression. A variable is a symbol for a number we do not know yet.
Write variable expressions
Write variable expressions: word problems
Evaluate variable expressions with whole numbers
Evaluate multi-variable expressions
Evaluate variable expressions with decimals
Identify terms and coefficients
Sort factors of expressions
Properties of multiplication
Solve for a variable using properties of multiplication
Identify equivalent expressions
## One-variable equations
An equation says that two things are equal, it will have an equal sign “=”. A variable is a symbol for a number we do not know yet. A single variable equation (one-variable equation) is an equation in which there is only one variable used. Note that the variable can be used multiple times or used on either side of the equation; all that matters is that the variable remains the same.
Does x satisfy an equation?
Which x satisfies an equation?
Write an equation from words
Model and solve equations using algebra tiles
Write and solve equations that represent diagrams
Solve one-step equations with whole numbers
Solve one-step equations with decimals
Solve one-step equations: word problems
## Two-variable equations
An equation says that two things are equal, it will have an equal sign “=”. A variable is a symbol for a number we do not know yet. Two-variable equation is like single variable equation but there are two variables, to solve these types of equations you have to rewrite in such a way to eliminate one of the variables and then solve for the remaining variable.
Does (x, y) satisfy an equation?
Identify independent and dependent variables
Solve word problems involving two-variable equations
Complete a table for a two-variable relationship
Write a two-variable equation
Identify the graph of an equation
## Two-dimensional figures
Two-dimensional geometry or plane geometry is about flat shapes like triangles and circles. Two-dimensional figures have only two dimensions such as width and height but no thickness. It also known as “2D”.
Identify and classify polygons
Measure angles
Types of angles
Estimate angle measurements
Classify triangles
Identify trapezoids
Find missing angles in triangles and quadrilaterals
Sums of angles in polygons
Name angles
Parts of a circle
## Symmetry and transformations
When two or more parts are identical after a flip, slide or turn we say it has symmetry. The simplest symmetry is reflection symmetry, sometimes called line symmetry or mirror symmetry. Other types of symmetries are Rotational symmetry and point symmetry. When we change a shape by using Turn, flip, slide or resize it is called transformation.
Symmetry
Reflection, rotation and translation
Translations: graph the image
Reflections: graph the image
Rotations: graph the image
Similar and congruent figures
Find side lengths of similar figures
## Three-dimensional figures
Having three dimensions such as Height, Width and Depth, like any real-world object is a three-dimensional figure. Three-dimensional geometry is about solid shapes like spheres or cubes. It is also known as “3D”.
Identify polyhedra
Which figure is being described?
Nets of three-dimensional figures
Front, side and top view
## Geometric measurement
Geometric measurement is studying the properties of shapes by measuring them, like finding the Area or Perimeter of a shape.
Perimeter
Area of squares and rectangles
Area of triangles
Area of parallelograms and trapezoids
Area of compound figures
Area between two rectangles
Area and perimeter of figures on grids
Area and perimeter: word problems
Rectangles: relationship between perimeter and area
Compare area and perimeter of two figures
Volume of cubes and rectangular prisms
Surface area of cubes and rectangular prisms
Volume of triangular prisms
Surface area of triangular prisms
Relate volume and surface area
## Data and graphs
A collection of facts, such as numbers, measurements or observations is called data. We can create a table with the data. A diagram of values, usually shown as lines is called graph.
Interpret pictographs
Create pictographs
Interpret stem-and-leaf plots
Create stem-and-leaf plots
Interpret line plots
Create line plots
Create frequency tables
Interpret bar graphs
Create bar graphs
Interpret double bar graphs
Create double bar graphs
Interpret histograms
Create histograms
Circle graphs with fractions
Interpret line graphs
Create line graphs
Interpret double line graphs
Create double line graphs
Choose the best type of graph
## Statistics
Statistics is a branch of mathematics dealing with the collection, analysis, interpretation, and presentation of data. Statistics is a strong tool in everyday life to get answers about data and make concrete decisions.
Calculate mean, median, mode and range
Interpret charts to find mean, median, mode and range
Mean, median, mode and range: find the missing number
Identify representative, random and biased samples
## Probability
Probability is the chance of something happening or how likely it is that some event will happen. Probability is a number between 0 (not happening) to 1 (certainly happening).
Combinations
Probability of one event
Make predictions
Probability of opposite, mutually exclusive and overlapping events
Compound events – find the number of outcomes by counting |
# Are Quadrilaterals always 360 degrees?
## Are Quadrilaterals always 360 degrees?
You would find that for every quadrilateral, the sum of the interior angles will always be 360°. Since the sum of the interior angles of any triangle is 180° and there are two triangles in a quadrilateral, the sum of the angles for each quadrilateral is 360°.
Is the sum of the interior angles of a quadrilateral is 360?
A quadrilateral has 4 angles. The sum of its interior angles is 360 degrees.
What is the interior angles of a quadrilateral?
90° (for square and rectangle)
### How many degrees are in the interior of a quadrilateral?
360°
Why do the interior angles add up to 360?
A special rule exists for regular polygons: because they are equiangular, the exterior angles are also congruent, so the measure of any given exterior angle is 360/n degrees. As a result, the interior angles of a regular polygon are all equal to 180 degrees minus the measure of the exterior angle(s).
What shapes interior angles add up to 360?
The General Rule
Shape Sides Sum of Interior Angles
Triangle 3 180°
Pentagon 5 540°
Hexagon 6 720°
## Is an oval 360 degrees?
The maximum length of an arc is 23,040, or 360 degrees, which would make an entire oval or circle. The minimum length is 2,304, or 36 degrees, one-tenth the circumference of a circle.
Quadrilaterals are composed of two triangles. Seeing as we know the sum of the interior angles of a triangle is 180°, it follows that the sum of the interior angles of a quadrilateral is 360°.
How do you find the angle of a quadrilateral?
Subtract the sum of the three angles from 360, to get the final angle. For example, a quadrilateral with the angles 40, 45, and 115 degrees, you would get a fourth angle of 160 degrees (360 – 200 = 160).
### Do all shapes have 360 degree rotational symmetry?
Regular polygons have a degree of rotational symmetry equal to 360 divided by the number of sides. An object has rotational symmetry if that figure is itself after you rotate it less than 180 degrees. If it is itself after exactly 180 degrees no more no less then that figure has point symmetry.
What type of angle is 360?
Angles that are 180 degrees (θ = 180°) are known as straight angles. Angles between 180 and 360 degrees (180°< θ < 360°) are called reflex angles. Angles that are 360 degrees (θ = 360°) are full turn.
What are the measures of the interior angles of a quadrilateral?
The sum of the measures of the interior angles of a quadrilateral is 360°. In the diagram shown below, find m ∠A. Because the figure ABCD is a closed figure and it is covered by four segments, it is quadrilateral. Substitute m∠B = 105°, m∠C = 113°, m∠D = 75°. Simplify.
What shapes interior angles add up to 360? Shape Sides Sum of Interior Angles Triangle 3 180° Quadrilateral 4 360 ° Pentagon 5 540° Hexagon 6 720°
## Do all polygons have 360 degrees?
Thereof, do all shapes have 360 degrees? Polygons – Quadrilaterals. So, the sum of the interior angles of a quadrilateral is 360 degrees. All sides are the same length (congruent) and all interior angles are the same size (congruent). To find the measure of the interior angles, we know that the sum of all the angles is 360 degrees (from above)
What is the sum of the angles in a convex quadrilateral?
The Quadrilateral Sum Conjecture tells us the sum of the angles in any convex quadrilateral is 360 degrees. Remember that a polygon is convex if each of its interior angles is less that 180 degree. In other words, the polygon is convex if it does not bend “inwards”. |
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# Chapter One: Vectors in R3
## 1 The Cartesian coordinate system
A point P in the space (resp. in the plane) can be associated to an ordered 3-tuple of
real numbers (x1 , x2 , x3 ) (resp. an ordered pair (x1 , x2 )) via the Cartesian coordinate
system. The Cartesian coordinate system consists of three mutually perpendicular
number lines (resp. two mutually perpendicular number lines, if it is in the plane),
called the x-axis, y-axis and the z-axis, whose origins coincides at a single point,
called the origin of the coordinate system. Click here to see the picture and the
explanation of the Cartesian coordinate system. Hence, we write the space as R3 =
{(x1 , x2 , x3 ); x1 , x2 , x3 ∈ R} (resp. the plane as R2 = {(x1 , x2 ); x1 , x2 ∈ R}), where
R is the set of real numbers. In this course, we usually do not distinguish the point
P and the ordered 3-tuple of real numbers (x1 , x2 , x3 ).
2 Vectors in R3
A vector is a line segment with a direction. A point P = (x1 , x2 , x3 ) uniquely cor-
responds to v = OP ~ , which visualized geometrically as the arrow pointing from the
origin O to the point P . In this way, there is one-to-one correspondence between
the vectors v in R3 and the points P ∈ R3 (as well as an ordered 3-tuple of real
numbers (x1 , x2 , x3 )). In this course, we usually do not distinguish the point P ,
an ordered 3-tuple of real numbers (x1 , x2 , x3 ), as well as the vector v. We also
use R3 to denote the set of vectors v in the space. On R3 , we can do addition
v + w = (x1 + y1 , x2 + y2 , x3 + y3 ), if v = (x1 , x2 , x3 ), w = (y1 , y2 , y3 ), and scalar
multiplication λv = (λx1 , λx2 , λx3 ) if v = (x1 , x2 , x3 ). The set R3 forms a vector
space with the addition and scalar multiplication defined above.
## Definition 1.1 We define the length of the vector v = (x1 , x2 , x3 ) to be
q
kvk = x21 + x22 + x23 .
## We say that v is a unit vector if it has length 1, i.e. kvk = 1.
1
3 The dot product in R3
Definition 1.2 The dot product of two vectors v = (x1 , x2 , x3 ), w = (y1 , y2 , y3 ) is
defined as
v · w = x 1 y1 + x2 y2 + x3 y3 .
## Some properties of dot product:
(a) kvk2 = v · v;
(b) v · w = kvkkwk cos θ, where θ is angle between v and w;
(c) v · w = 0 if and only if v and w are orthogonal;
(d) |v · w| ≤ kvkkwk (Schwarz inequality);
(e) v · w = w · v;
(f) u · (v + w) = u · v + u · w;
(g) kv + wk ≤ kvk + kwk (the triangle inequality).
## 4 The cross product in R3
Definition 1.3 The cross product of two vectors v = (x1 , x2 , x3 ), w = (y1 , y2 , y3 )
is the vector given by
v × w = (x2 y3 − x3 y2 , x3 y1 − x1 y3 , x1 y2 − x2 y1 ).
## (a) v × w is orthogonal to v and w if the vectors are linearly independent, and
satisfies the right-hand rule(Click here to see the right-hand rule);
(b) v × w = −w × v, in particular v × v = 0;
(c) If the vectors are both nonzero, then
kv × wk = kvkkwk sin θ,
## where θ is angle between v and w;
(d) v × w = 0 if and only if v and w are linearly dependent;
(e) |u · (v × w)| is the volume of a parallelepiped spanned by the vectors u, v
and w.
2
Let i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1). Then we have (called the right-hand
rule)(Click here to see the right-hand rule)
(1.1) i × j = k, k × i = j, j × k = i.
We can use (1.1) to calculate the cross product. For example, to calculate v × w
for v = (1, 2, 3), w = (2, 5, 1), we write v = i + 2j + 3k, w = 2i + 5j + k, then, using
(1.1) and the fact that i × i = j × j = k × k = 0, we have
## v×w = (i + 2j + 3k) × (2i + 5j + k)
= 5i × bf j + i × k + 4j × i + 2j × k + 6k × i + 15k × j
= 5k − j − 4k + 2i + 6j − 15i
= −13i + 5j + k.
## 5 Differentiation and integration for vector valued
functions
A vector valued function u(t) in R3 assignes, for each t ∈ [a, b] a unique vector value
u(t) ∈ R3 . Hence
u(t) = (x1 (t), x2 (t), x3 (t)), t ∈ [a, b],
where x1 , x2 , x3 are functions of variable t. We define the diifferentiation of u(t) as
du(t)
u0 (t) = = (x01 (t), x02 (t), x03 (t)).
dt
We recall the following differentiation rules for vector valued functions:
## (cu(t))0 = cu0 (t)
(f (t)u(t))0 = f (t)u0 (t) + f 0 (t)u(t)
(u(t) · v(t))0 = u(t) · v0 (t) + u0 (t) · v(t)
(u(t) × v(t))0 = u(t) × v0 (t) + u0 (t) × v(t)
[u(f (t))]0 = f 0 (t)u0 (f (t)).
3
Similarily, we define the integration of u(t) over [a, b] as
!
Z b Z b Z b Z b
u(t)dt = x1 (t)dt, x2 (t)dt, x3 (t)dt .
a a a a |
# Warm Up #1. We have 22 students in this class, how many ways could Ms. Fraser group everyone so that all the groups are equal? What if we had one more.
## Presentation on theme: "Warm Up #1. We have 22 students in this class, how many ways could Ms. Fraser group everyone so that all the groups are equal? What if we had one more."— Presentation transcript:
Warm Up #1. We have 22 students in this class, how many ways could Ms. Fraser group everyone so that all the groups are equal? What if we had one more student? What if we had one less? #2. A large restaurant can seat 164 people, how could they be seated so that each table has the same number of people?
Warm Up September 8 Take out a sheet of looseleaf Number warm up is your title Number from 1-8 Mil.Hun. ThouTen ThouThousandsHundredsTensOnes
PRIME AND COMPOSITE NUMBERS LESSON 2
PRIME NUMBERS A Prime Number can be divided evenly only by 1, or itself. EXAMPLES: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
COMPOSITE NUMBERS A Composite Number can be divided evenly by numbers other than 1 or itself. EXAMPLES: 4,6,8,9,10,12,14,15,16,18,20,21,22,24,25, 26,27,28,30,32,33,34,35,36,38,39,40
FACTORS A number may be made by multiplying two or more other numbers together. The numbers that are multiplied together are called factors. EXAMPLE:
FACTORS A number may be made by multiplying two or more other numbers together. The numbers that are multiplied together are called factors. EXAMPLE: 3 X 4 = 12 3 X 4 = 12 FACTOR PRODUCT
FACTORS Therefore PRIME numbers have 2 factors: one and itself. COMPOSITE numbers have more than two factors. The number ONE only has one factor (1), therefore it is not prime nor composite but SPECIAL.
Try these Is the number prime or composite? a) 9b) 7c) 23d) 24 e) 57f) 144 List all of the factors of the composite numbers
GCF GCF – GREATEST COMMON FACTOR The GCF is the highest factor that two or more numbers have in common. EXAMPLE: The factors of 18 are {1,2,3,6,9,18} The factors of 24 are {1,2,3,4,6,8,12,24}
GCF GCF – GREATEST COMMON FACTOR The GCF is the highest factor that two or more numbers have in common. EXAMPLE: The factors of 18 are {1,2,3,6,9,18} The factors of 24 are {1,2,3,4,6,8,12,24} The GCF is 6.
Try these Find the Greatest Common Factor. ( GCF) a) 28, 49b) 32, 48c) 24, 36
PRIME FACTORS Factors that are prime numbers. When a composite number is written as a product of all of its prime factors, we have the prime factorization of the number. EXAMPLE: the number 72 can be written as a product of primes as: 72 =2x2x2x3x3. The expression "2x2x2x3x3" is said to be the prime factorization of 72.
TRY THIS ONE Find the PRIME factors of 32 SOLUTION:
TRY THIS ONE Find the PRIME factors of 32 SOLUTION: 32 4 x 8 2 x 2 x 2 x 2 x 2 = 2 5
Try these Express the number as a product of its prime factors. a) 42
LCM LCM – LOWEST COMMON MULTIPLE A common multiple is a number that is a multiple of two or more numbers. common multiplecommon multiple The common multiples of 3 and 4 are: 0, 12, 24,….
LCM The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both. least common multipleleast common multiple EXAMPLE: Multiples of 3: 6,9,12,15,18,21,24,… Multiples of 4: 8,12,16,20,24,…. LCM of 3 and 4 is:
The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both. least common multipleleast common multiple EXAMPLE: Multiples of 3: 6,9,12,15,18,21,24,… Multiples of 4: 8,12,16,20,24,…. LCM of 3 and 4 is: 12
Try these Find the Lowest Common Multiple. ( LCM ) a) 18, 27b) 10, 25
1. Two companies are using the telephone book to choose winners for some promotional prizes. One company phones every 36th name and the second company phones every 24th name. What is the position of the first name that wins both prizes?
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Place Value
View Notes
Number System:
A number system is a method of representing numbers. It is also known as the numeration scheme, which determines a set of values to describe a quantity. The position of a digit in a number determines the value of the digit in it. For instance, 5 in 350 represents 5 tens, or 50; but 5 in 5,006 represents 5,000. Kids need to know that while the same digit can be present in many numbers, its value depends on where it is in the number.
Place value is most often taught to children through a place value chart as given below.Â
Ten Millions Millions Hundred Thousands Ten Thousands Thousands Hundreds Tens Ones Decimal Point Tenths Hundredths Thousandths 1 2 4 5 6 7 8 9 . 1 2 3
This will also only appear with letters on them to denote Millions, Hundreds of Thousands, Ten Thousand, Thousands, Hundreds, Tens, Ones, Tenths, Hundredths, and so on, per place.
Example
In the number 1329:
• 1 holds the thousands position.
• 3 holds the hundreds position.
• 2 holds the tens position.
• 9 holds the ones position.
More on Place Value
Whole numbers are arranged in groups of three, called periods. Each period features a hundred(s), ten(s) and one(s) position.
Number 123 456 789 below as an example. Here, 8 holds the tens position, 5 holds the ten-thousands position, and so on.
Decimal numbers are also organized in groups of threes called periods. Each period features a ten(th), hundred(th), and thousand(th) position as explained in the below example.
Place Value and the Face Value
The digit's place value is the product of the digit's face value and its place value, while a digit's face value is the digit itself.
In this example, we will calculate the face value and place a value of 6 in 6, 45,100.
The face value of 6 in 6, 45,100 is ‘6’
The place value of the digit is obtained by multiplying the face value of the digit and the value of its place so, the place value of 6 in 6,45,100 is 6 x 1,00,000 = 6,00,000 (6 Lakh).
Â
Expanded Notation:
We express each digit of a number to its position value in an extended form. Let's see the extended 29,1233 number notation.
In three different ways, this can be expanded:
    1.  2 ten thousands + 9 thousands + 1 hundred + 2 tens + 3 ones
      2.  (2 x 10,000) + (9 x 1,000) + (1 x 100) + (2 x 10) + (3 x 1)
      3.  20000 + 9000 + 100 + 20 + 3
Standard form of 60000+4000+40+6 is 64,046
Â
The Indian System and the International System
Indian System:Â
The first period, consisting of three place values (ones, tens, and hundreds), is one in the Indian system of numeration, starting from the right. There are thousands in the next period, consisting of two-position values (thousands and ten thousand). Lakhs, consisting of two-position values (lakhs and ten lakhs), and then crores and so on is the third cycle from the right. This numeration scheme is often referred to as the Hindu-Arabic system of numeration. To divide the cycles, we use commas, which help us read and write large numbers. The first comma comes from the right after three digits (i.e., after one period) in the Indian scheme, and the next comma comes after the next two digits (i.e., after the thousand periods) and then after every two digits, and so on.
Indian Place Value Chart
Crores (C) Crores (C) Crores (C) Lakhs (L) Thousands(Th) Thousands(Th) Ones Ones Ones Ten crores(TC) Crores (C) Ten Lakhs (TL) Lakhs(L) Ten Thousands (TTh) Thousands (Th) Hundreds(H) Tens(T) Ones(O) 10,00,00,000 1,00,00,000 10,00,000 1,00,000 10,000 1,000 100 10 1
Consider an example of the International Numeration System:
In the Indian numeration scheme,
92357385 = 9,23,57,3855
Likewise, 2930625 will be translated as 29,30,625 in the Indian method of numeration.
The first cycle, consisting of three place values (ones, tens, and hundreds), is one in the International Numeration Scheme, beginning from the right. Thousands belong to the next period, consisting of three place values (one thousand, ten thousand, and one hundred thousand) and then millions and then billions.
International Place Value Chart
Billions (B) Billions (B) Billions (B) Millions (M) Millions (M) Millions (M) Thousands(Th) Thousands(Th) Thousands(Th) Ones Ones Ones Hundred Billions(HB) Ten Billions (TB) One Billion (B) Hundred Millions(HM) Ten Millions (TM) One Million (M) Hundred Thousands(HTH) Ten Thousand (TTh) Thousands(Th) Hundreds (H) Tens (T) Ones(0) 100,000,000,000 10,000,000,000 1,000,000,000 100,000,000 10,000,000 1,000,000 100,000 10,000 1,000 100 10 1
Comparison of Two Number Systems:
Crores (C) Crores (C) Crores (C) Lakhs (L) Thousands(Th) Thousands(Th) Ones Ones Ones Ten crores(TC) Crores (C) Ten Lakhs (TL) Lakhs(L) Ten Thousands (TTh) Thousands(Th) Hundreds(H) Tens(T) Ones(O) 10,00,00,000 1,00,00,000 10,00,000 1,00,000 10,000 1,000 100 10 1
Billions (B) Billions (B) Billions (B) Millions (M) Millions (M) Millions (M) Thousands(Th) Thousands(Th) Thousands(Th) Ones Ones Ones Hundred Billions(HB) Ten Billions (TB) One Billion (B) Hundred Millions(HM) Ten Millions (TM) One Million (M) Hundred Thousands(HTH) Ten Thousands(TTh) Thousands(Th) Hundreds (H) Tens (T) Ones(0) 100,000,000,000 10,000,000,000 1,000,000,000 100,000,000 10,000,000 1,000,000 100,000 10,000 1,000 100 10 1
From above we can conclude that:
100 thousands = 1 lakh
1 million = 10 lakhs
10 millions = 1 crore
100 millions= 10 crores
Place Value for Decimals
Place value of chart looks like a figure below:
Below are some of the mentioned points which need to be captured while reading the writing of any number system.
• Separate the period by using commas when writing a number in the Indian numeral system or the international numeral system.
• In each period, read the number and say the name of the period after each number in the period.
• Skip the time containing zeros at all the locations when reading a number.
1.How to Make a Place Value Chart?
Answer: To form an Indian place value chart, draw 4 columns showing the 4 periods: ones, thousands, lakhs, and crores. Draw sub-columns for each time to display various position values: ones, tens, hundreds, thousands, ten thousand, lakhs, ten lakhs, crores, and ten crores.Â
Â
Similarly, draw 3 columns representing the 3 periods: ones, thousands, and millions, to create an International Place Value Map. Draw sub-columns for each time to display the different values of the place: ones, tens, hundreds, thousands, ten thousand, hundreds of thousands, millions, ten million, hundreds of millions.
Â
For example, the number 26984 is regarded in the Indian place value chart as follows.
 |
# The net monthly profit, in dollars, from the sale of a certain item, is given by the formula P(x)...
## Question:
The net monthly profit, in dollars, from the sale of a certain item, is given by the formula {eq}P(x) = 10^6\left [ 1 + (x-1)e^{0.001x} \right ], {/eq} where x is the number of items sold. Find the number of items that yield the maximum profit. At full capacity, the factory can produce 2000 items per month.
## Finding Maximum Profit from Sales of Item
The net monthly profit in dollars from the sale of a certain item is given as an explicit function of the number of items sold. Using a first derivative and second derivative test from Calculus, if applicable, we find the number of items sold that lead to a maximum value for the profit. The profit function contains the product of a linear and exponential term. The first derivative of such a function requires the use of the product rule for derivatives. Using the first and second derivative is an excellent tool used to find points of local maximum and minimum of a function.
To find the point of maximum of the profit P(x) we need to find its critical points. To that end we calculate its first derivative using the product rule for derivatives from Calculus as follows:
{eq}P'(x) = 10^6 \left[ 0 + (x-1)(0.001)e^{0.001x} + e^{0.001x}(1) \right] \qquad (1) {/eq}
Since P'(x) exists for all values of x, we find critical points of P(x) by solving P'(x) = 0 from (1) to get
{eq}10^6 \left[ 0 + (x-1)(0.001)e^{0.001x} + e^{0.001x}(1) \right] = 0 \implies e^{0.001x} \left[ 0.001(x-1)+1 \right] = 0 \qquad (2) {/eq}
From (2) we get the equation
{eq}0.001(x-1)+1 = 0 \implies 1 = (1-x)(0.001) \implies 1000 = 1-x \implies x = -999. {/eq}
Since a negative number of items is not possible, therefore no meaningful critical points exist. Hence a local maximum of the profit function P(x) does not exist. |
# 2.2 Histograms, frequency polygons, and time series graphs (Page 6/15)
Page 6 / 15
Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table.
Data Value (# cars) Frequency Relative Frequency Cumulative Relative Frequency
What does the frequency column in [link] sum to? Why?
65
What does the relative frequency column in [link] sum to? Why?
What is the difference between relative frequency and frequency for each data value in [link] ?
The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value.
What is the difference between cumulative relative frequency and relative frequency for each data value?
To construct the histogram for the data in [link] , determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling.
Answers will vary. One possible histogram is shown:
Construct a frequency polygon for the following:
1. Pulse Rates for Women Frequency
60–69 12
70–79 14
80–89 11
90–99 1
100–109 1
110–119 0
120–129 1
2. Actual Speed in a 30 MPH Zone Frequency
42–45 25
46–49 14
50–53 7
54–57 3
58–61 1
3. Tar (mg) in Nonfiltered Cigarettes Frequency
10–13 1
14–17 0
18–21 15
22–25 7
26–29 2
Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger.
Depth of Hunger Frequency
230–259 21
260–289 13
290–319 5
320–349 7
350–379 1
380–409 1
410–439 1
Find the midpoint for each class. These will be graphed on the x -axis. The frequency values will be graphed on the y -axis values.
Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men?
Life Expectancy at Birth – Women Frequency
49–55 3
56–62 3
63–69 1
70–76 3
77–83 8
84–90 2
Life Expectancy at Birth – Men Frequency
49–55 3
56–62 3
63–69 1
70–76 1
77–83 7
84–90 5
Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births.
Sex/Year 1855 1856 1857 1858 1859 1860 1861 Female 45,545 49,582 50,257 50,324 51,915 51,220 52,403 Male 47,804 52,239 53,158 53,694 54,628 54,409 54,606 Total 93,349 101,821 103,415 104,018 106,543 105,629 107,009
Sex/Year 1862 1863 1864 1865 1866 1867 1868 1869 Female 51,812 53,115 54,959 54,850 55,307 55,527 56,292 55,033 Male 55,257 56,226 57,374 58,220 58,360 58,517 59,222 58,321 Total 107,069 109,341 112,333 113,070 113,667 114,044 115,514 113,354
Sex/Year 1871 1870 1872 1871 1872 1827 1874 1875 Female 56,099 56,431 57,472 56,099 57,472 58,233 60,109 60,146 Male 60,029 58,959 61,293 60,029 61,293 61,467 63,602 63,432 Total 116,128 115,390 118,765 116,128 118,765 119,700 123,711 123,578
The following data sets list full time police per 100,000 citizens along with homicides per 100,000 citizens for the city of Detroit, Michigan during the period from 1961 to 1973.
Year 1961 1962 1963 1964 1965 1966 1967 Police 260.35 269.8 272.04 272.96 272.51 261.34 268.89 Homicides 8.6 8.9 8.52 8.89 13.07 14.57 21.36
Year 1968 1969 1970 1971 1972 1973 Police 295.99 319.87 341.43 356.59 376.69 390.19 Homicides 28.03 31.49 37.39 46.26 47.24 52.33
1. Construct a double time series graph using a common x -axis for both sets of data.
2. Which variable increased the fastest? Explain.
3. Did Detroit’s increase in police officers have an impact on the murder rate? Explain.
#### Questions & Answers
Uttam
How can I calculate the Class Mark, Relative frequency and the cumulative frequency on a frequency table?
what is the important in business planning and economics
explain the limitation and scope of statistics
mahelt
statistics is limited to use where data can be measured quantitatively. statistics scope is wider such as in economic planning, medical science etc.
Gurpreet
can you send me mcq type questions
Yas
Umar
which books are best to learn applied statistics for data science/ML
Gurpreet
A population consists of five numbers 2,3,6,8,11.consists all possible samples of size two which can be drawn with replacement from this population. calculate the S.E of sample means
A particular train reaches the destination in time in 75 per cent of the times.A person travels 5 times in that train.Find probability that he will reach the destination in time, for all the 5 times.
0.237
Amresh
explain that this answer-0.237
umesh
p(x=5)= 5C0 p^5 q^0 solve this
Amresh
umesh
ok
umesh
5C0=1 p^5= (3/4)^5 q^0=(1/4)^0
Amresh
Hint(0.75 in time and 0.25 not in time)
kamugi
what is standard deviation?
It is the measure of the variation of certain values from the Mean (Center) of a frequency distribution of sample values for a particular Variable.
Dominic
what is the number of x
10
Elicia
Javed Arif
Jawed
how will you know if a group of data set is a sample or population
population is the whole set and the sample is the subset of population.
umair
if the data set is drawn out of a larger set it is a sample and if it is itself the whole complete set it can be treated as population.
Bhavika
hello everyone if I have the data set which contains measurements of each part during 10 years, may I say that it's the population or it's still a sample because it doesn't contain my measurements in the future? thanks
Alexander
Pls I hv a problem on t test is there anyone who can help?
Peggy
What's your problem Peggy Abang
Dominic
Bhavika is right
Dominic
what is the problem peggy?
Bhavika
hi
Sandeep
Hello
hi
Bhavika
hii Bhavika
Dar
Hi eny population has a special definition. if that data set had all of characteristics of definition, that is population. otherwise that is a sample
Hoshyar
three coins are tossed. find the probability of no head
three coins are tossed consecutively or what ?
umair
umair
or .125 is the probability of getting no head when 3 coins are tossed
umair
🤣🤣🤣
Simone
what is two tailed test
if the diameter will be greater than 3 cm then the bullet will not fit in the barrel of the gun so you are bothered for both the sides.
umair
in this test you are worried on both the ends
umair
lets say you are designing a bullet for thw gun od diameter equals 3cm.if the diameter of the bullet is less than 3 cm then you wont be able to shoot it
umair
In order to apply weddles rule for numerical integration what is minimum number of ordinates
excuse me?
Gabriel
why?
didn't understand the question though.
Gabriel
which question? ?
We have rules of numerical integration like Trapezoidal rule, Simpson's 1/3 and 3/8 rules, Boole's rule and Weddle rule for n =1,2,3,4 and 6 but for n=5?
John
Someone should help me please, how can I calculate the Class Mark, Relative frequency and the cumulative frequency on a frequency table?
IJOGI
geometric mean of two numbers 4 and 16 is:
10
umair
really
iphone
quartile deviation of 8 8 8 is:
iphone
sorry 8 is the geometric mean of 4,16
umair
quartile deviation of 8 8 8 is
iphone
can you please expalin the whole question ?
umair
mcq
iphone
h
iphone
can you please post the picture of that ?
umair
how
iphone
hello
John
10 now
John
how to find out the value
can you be more specific ?
umair
yes
KrishnaReddy
what is the difference between inferential and descriptive statistics
descriptive statistics gives you the result on the the data like you can calculate various things like variance,mean,median etc. however, inferential stats is involved in prediction of future trends using the previous stored data.
umair
if you need more help i am up for the help.
umair
Thanks a lot
Anjali
Inferential Statistics involves drawing conclusions on a population based on analysis of a sample. Descriptive statistics summarises or describes your current data as numerical calculations or graphs.
fred
my pleasure😊. Helping others offers me satisfaction 😊
umair
inferential statistics the results of the statistical analysis of the sample data of the population are used for generalization or decision making about the population why descriptive statistics, the analyzed data are presented without generalization or decision making about the population.
IJOGI |
# Math Tips and Tricks
Nowadays calculators are so prevalent that most students wouldn’t know what to do if they didn’t have one to help them work out their calculations. However, they might find themselves in a situation where they don’t have access to a calculator or aren’t allowed to use one, and in these cases, knowing some math “tricks” can come in very handy.
TIP #1
When multiplying or dividing with numbers ending in zeros, drop the zeros and do the multiplication/division problem, then add the zeros back.
Examples:
400 x 200 = 4 x 2 = 8 (and add back the 4 zeros) = 8000
400/20 = 4/2 = 2 (and add back one zero) = 20
TIP #2
Any number whose digits add up to a number that is divisible by three is also divisible by three.
Example:
123 – the sum of 1 + 2 + 3 is 6, which is divisible by 3, so 123 is divisible by 3 (it is equivalent to 41).
TIP #3
When you multiply decimals, initially ignore the decimals and then add them back to your solution.
Example
1.2 x 3
First drop the decimal and think “12 x 3”, which is 36.
Reinsert the decimal place one spot from the right to obtain 3.6
TIP #4
When multiplying numbers whose tens digits are the same and the ones digits add up to ten, multiply the tens digit by the next whole number, multiply the ones digits together and combine the two products.
Example:
61 x 69
First, multiply the tens digit (6) by the next whole number (7). You get 42.
Second, multiply the ones digits together (1 x 9). You get 09 (a leading zero is added for single-digit products).
Third, combine these products together to get the answer of 4209.
TIP #5
When multiplying a 2-digit number by 11, add the digits of the number together and then insert the sum between the two digits to obtain the answer.
Example:
12 x 11
1 + 2 = 3
Insert the 3 between the 1 and the 2 to get the answer of 132.
TIP #6
When finding the fourth power of a number, think of it as being the square of a square.
Example:
3 raised to the fourth power
This is the same as the square of three squared. Three squared is nine and nine squared is 81.
TIP #7
When squaring any number ending in five, multiply the tens digit by the next whole number and then add on 25.
Example:
35 squared
First, multiply the tens digit (3) by the next whole number (4). You get 12. |
# 1st Class Mathematics Comparison of Numbers Introduction
## Introduction
Category : 1st Class
### Introduction
Comparison enables us to identity what is greater or smaller between two objects and what is greatest and smallest among more than two objects.
For example, if we compare&, it is clear thatis bigger number but if we compare
& then
is the smallest andis the biggest number.
Whilecomes between&therefore, it is greater or bigger than but smaller or less than.
Comparison of Numbers
Let us see the given below examples:
It means thatis greater thanand '>'is the sign of 'greater than' or bigger than 'or more than'
(a) is greater than
(b) is bigger than
(c) is more than
is also written asand read asis less than
Therefore, it is clear that the mouth of the sign '>' is open to the side of the greater or bigger number.
Howis a greater number betweenand 3?
For that we apply vertical line trick:
When countedis more than
Hence Take another example:
Which one is bigger betweenand?
Clearlyis more than
or
and or
It means the numberhas two digits and the numberhasdigit. Therefore,is greater than Take another example:
If we compareand Then:
or
has three digits andhas two digits. Therefore,is greater than.
If the two numbers have equal number of digits then the digits at extreme left are compared and the number with greater digit is greater.
If we compareandthen:
or
To find this result we compare the two numbers.
While comparing two digit numbers, if extreme left digits are equal or same then we compare the next digits to the right and so on. In this case also the number with greater digit is greater.
When we compare and or
Therefore,
1. Which one of the given below options is false?
(a)1 > 9
(b) 4 > 3
(c) 5 > 3
(d) 9 > 8
Explanation
Option (a) is correct because:
Hence,
Option (b) is incorrect as it is true. Let's check:
Hence,
Option (c) is incorrect as it is true. Let's check:
Hence,
Option (d) is incorrect as it is true. let us check:
Hence,
14 > 10 will be read as____.
(a) 10 is greater than 14
(b) 14 is less than 10
(c) 10 is bigger than 14
(d) 14 is greater than 10
Expiation
is read as "14 is greater than 10"
Smaller Numbers
See the following examples:
It meansis less than 3 and '<' is the sign of 'less than' or 'smaller than'.
(a) is smaller than
(b) is less than
is also written as. So it is clear that the mouth of the sign '<' is closed to the side of the smaller number. Howis smaller in?
For this, Let's see the vertical line trick:
When countedis less than
Hence,
Take another example:
Which one is smaller when andare compared?
Here we have
Clearlyis less than
Hence, or
Other Rules
Rule 1: If a number has less digits than another, it is smaller of the two. If we haveand Then or
Let us take another example:
If we compareand
Then or
Rule2: If the two numbers have equal number of digits then the digits on extreme left are compared and the number with smaller digit is smaller. Whenandare compared thenis true.
To know how is it possible just see the trick given below:
Thus And hence,
Rule3: While comparing two digit numbers, if extreme left digits are same then we compare the second digit from left. In this case also, the number with smaller digit is smaller.
When we compareand
Then
or
Trick:
Here
Hence,
Just see the numbers inside the egg given below and find out which one is the smallest?
(a) 15
(b) 16
(c) 10
(d) 9
Explanation
The only single digit number given. Rest of the options is incorrect because they are double digit numbers and therefore, bigger than
Which one is wrong?
(a)
(b)
(c)
(d)
Explanation
Option (d) is correct because it is wrong.
Let's check
Hence,or
Option (a) is incorrect because it is true.
Let's check:
Hence, Option (b) is incorrect because it is true. In fact 2 digits number 10 is greater than 5.
Option (c) is incorrect because it is true Let's check:
Hence, or
Equal Numbers
Equal means neither greater nor less than.
Let's consider
and
These numbers are equal because if one or more zeroes is put before any number, the number remains unchanged. Thusandcan be written as:
or
Where '=' is the sign of equal
is equal to
Similarly take another example:
andis equal and they can be written as
or
Which one of the following pair of numbers is not equal?
(a) 2 and 02
(b) 02 and 2
(c) 002 and 2
(d) 20 and 2
Explanation
is a double digit number andis a single digit number. So.
Choose the incorrect statement.
(a) 05 and 5 are equal
(b) 60 and 06 are equal
(c) 07 and 007 are equal
(d) 08 and 8 are equal |
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# 2017 AMC 8 Problems/Problem 19
## Problem
For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
## Solution 1
Factoring out $98!+99!+100!$, we have $98!(1+99+99*100)$ which is $98!(10000)$ Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$. The $19$ is because of all the multiples of $5$. Now $10,000$ has $4$ factors of $5$, so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$.
## Solution 2
The number of $5$'s in the factorization of $98! + 99! + 100!$ is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by $5$, until you can't divide by $5$ anymore. Factorizing $98! + 99! + 100!$, you get $98!(1+99+9900)=98!(10000)$. To find the number of trailing zeroes in 98!, we do $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22$. Now since $10000$ has 4 zeroes, we add $22 + 4$ to get $\boxed{\textbf{(D)}\ 26}$ factors of $5$.
~ pi_is_3.14 |
# 9.0: Prelude to Hypothesis Testing
CHAPTER OBJECTIVES
By the end of this chapter, the student should be able to:
• Differentiate between Type I and Type II Errors
• Describe hypothesis testing in general and in practice
• Conduct and interpret hypothesis tests for a single population mean, population standard deviation known.
• Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown.
• Conduct and interpret hypothesis tests for a single population proportion
One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of \$60,000 per year.
Figure $$\PageIndex{1}$$: You can use a hypothesis test to decide if a dog breeder’s claim that every Dalmatian has 35 spots is statistically sound. (Credit: Robert Neff)
A statistician will make a decision about these claims. This process is called "hypothesis testing." A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analysis of the data, to reject the null hypothesis. In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests.
Hypothesis testing consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion. To perform a hypothesis test, a statistician will:
• Set up two contradictory hypotheses.
• Collect sample data (in homework problems, the data or summary statistics will be given to you).
• Determine the correct distribution to perform the hypothesis test.
• Analyze sample data by performing the calculations that ultimately will allow you to reject or decline to reject the null hypothesis.
• Make a decision and write a meaningful conclusion.
To do the hypothesis test homework problems for this chapter and later chapters, make copies of the appropriate special solution sheets. See Appendix E.
### Glossary
Confidence Interval (CI)
an interval estimate for an unknown population parameter. This depends on:
• The desired confidence level.
• Information that is known about the distribution (for example, known standard deviation).
• The sample and its size.
Hypothesis Testing
Based on sample evidence, a procedure for determining whether the hypothesis stated is a reasonable statement and should not be rejected, or is unreasonable and should be rejected. |
# How many beats are half notes?
## How many beats are half notes?
two beats
The half note gets two beats; so you will play the note and count to two. The dotted half note gets three beats and the duration with four beats is the whole note.
## How many beats were in each measure?
Common Time Signatures In this time signature there are 4 beats possible in each measure, and the quarter note represents one beat. The top number of the time signature tells how many beats are in each measure, and the bottom number tells which note will represent one beat.
What is half note beat?
A whole note gets 4 beats, a half note gets 2 beats, and a quarter note gets 1 beat.
How do you count half beats?
The top number always denotes the number of beats in a measure, and the bottom always signifies what note gets the beat. If the bottom number is an 8, then you should count eighth notes. If the bottom number is a 2, then you should count half notes.
### How many beats does a whole note tied to a half note receive?
A WHOLE NOTE = 4 Beats A HALF NOTE = 2 Beats A QUARTER NOTE = 1 Beat In the following example, you can see that in each measure, there are 4 beats . 1 Whole Note = 4 Beats . 2 Half Notes = 4 Beats . 4 Quarter Notes = 4 Beats .
### How many counts is a half note equal to?
A B sixteenth rest one quarter of a beat of silence How many quarter notes equal one whole n four How many half notes equal one whole note two How many eighth notes equal one whole no eight
How many sixteen notes equal a half note?
By adding the flag it becomes a note of half that value – an eighth note. By adding another flag, it becomes half as long as an eighth note – a sixteenth note. It takes two sixteenth notes to equal one eighth note. It takes four sixteenth notes to equal one quarter note.
How do many half notes equal to quarter notes?
A hollow half note equals two filled in quarter notes. If you play a note that lasts for all four beats of the measure, you’re playing a whole note. Like the half note, the whole note’s notehead is hollow, but its shape is slightly different – more oval than round. Whole notes hold out for all four counts. |
Lesson Explainer: Resolution of Forces | Nagwa Lesson Explainer: Resolution of Forces | Nagwa
# Lesson Explainer: Resolution of Forces Mathematics
In this explainer, we will learn how to solve problems about the resolution of a force into two directions.
Force is a vector quantity and so a force can be represented by an arrow in the direction of the force with a length proportional to the magnitude of the force.
The direction of a force can be expressed in terms of a coordinate system. Probably the most intuitive example of such a system is a two-dimensional system with perpendicular axes. It is a common convention to label these axes and , as shown in the following figure.
The direction of a force may be parallel to one of the axes of the coordinate system. This is shown in the following figure for a force for which the point of action is the origin of the system.
When the line of action of a force is parallel to one axis of such a coordinate system, it is necessarily perpendicular to the other axis of the system.
The direction of force can however make an arbitrary angle with lines parallel to either of the axes of the system, as shown in the following figure.
The line of action of the force shown makes an angle with the -axis of the system and makes an angle with the -axis of the system.
A force acting in an arbitrary direction can be expressed in terms of two components. Each component is parallel to one of the axes of the system and perpendicular to the other axis. The directions of these components are, therefore, perpendicular to each other. The perpendicular components of a force and the force itself are shown in the following figure.
The magnitudes of the perpendicular components of a force can be determined from trigonometric rules for right triangles. Consider the following figure of a right triangle which has an angle .
The ratios of the lengths of the opposite and adjacent sides to the length of the hypotenuse are determined by the following equations: and
A force, , and its perpendicular components form a right triangle, as shown in the following figure.
The arrows representing the force and its components are assumed to be in a coordinate system where angle is the angle from the -axis of the system.
The ratios of the magnitudes of the components of the force to the force are, therefore, given by hence, and hence,
The perpendicular components of a force are often represented acting at the point at which the force acts. The following figure shows that representing the components of the force in this way is equivalent to representing the force as the sum of its perpendicular components.
Let us look at an example of resolving a force into perpendicular components.
### Example 1: Resolving a Force into Two Perpendicular Components
Resolve a force of 81 N into two perpendicular components and as shown in the figure. Give your answer correct to two decimal places.
The component is given by
To two decimal places, .
The component is given by
To two decimal places, .
It is worth noting that the magnitudes of the components of a force are dependent on the coordinate system chosen to represent force vectors.
For example, a force that acts along the -axis of a coordinate system has two nonzero components in another coordinate system, as shown in the figure for a coordinate system that is rotated by an angle to the coordinate system in which is parallel to the -axis.
The lines of action of the components and are along the - and -axes of this rotated coordinate system.
Let us look at an example of resolving a force into perpendicular components relative to an arbitrary direction.
### Example 2: Resolving a Particle’s Weight into the Directions Parallel and Perpendicular to the Plane
A body weighing 72 N is placed on a plane that is inclined at to the horizontal. Resolve its weight into two components and , where is the component in the direction of the plane and is the component normal to the plane.
The following figure shows the forces acting on the body, where is the 72 N weight of the body.
The components of the weight parallel to and perpendicular to the plane are perpendicular to each other; hence, and
Let us now look at another such example.
### Example 3: Resolving the Weight of a Body on an Inclined Plane
A particle weighing 69 N is placed on a plane inclined at an angle to the horizontal, where . Resolve the weight of the particle into two components, and , where is parallel to a line of greatest slope and is perpendicular to .
The following figure shows the force of the weight of the particle and its components parallel to and perpendicular to the plane. A section of the plane that corresponds to a right triangle is shown.
The question states that
From this, we can see that the sides of the triangle opposite to and adjacent to the angle have lengths in the ratio . In a right triangle with the sides having this ratio of lengths, the ratio of the side adjacent to to the length of the hypotenuse is , and the ratio of the side opposite to the length of the hypotenuse is .
From this we see that
The component of the weight of the particle perpendicular to the plane, , is rotated through an angle from the direction of the weight of the particle, so has a magnitude given by
Conversely, the component of the weight of the particle parallel to the plane, , has a magnitude given by
Thus far we have considered components of a force where the components are perpendicular to each other. We can consider a force as consisting of two nonperpendicular components. The following figure shows a force and two nonperpendicular components and , where
For a force , there are infinite pairs of components that sum to . and are only one example of such a pair of components.
For nonperpendicular components of a force, it is necessary to use different trigonometric rules to those of right triangles.
The components and can be represented as acting from the same point as , as shown in the following figure.
Lines from the head of to the head of and from the head of to the head of can be drawn. These lines complete a parallelogram, as shown in the following figure.
The line from the head of to the head of is parallel to . The line from the head of to the head of is parallel to . The line of action of is a line from which coangles can be defined for the parallelogram, as shown in the following figure.
The parallelogram consists of two similar triangles, for which the unknown angle, , is given by as shown in the following figure.
Let us consider one of the triangles of the parallelogram, as shown in the following figure.
The lengths of the sides of a triangle are related to the angles of the triangle by the sine rule where , , and are the lengths of the sides opposite to the angles , , and .
For , , and ,
Let us look at an example of resolving a force that has nonperpendicular components.
### Example 4: Resolving a Force into Two Components Shown in a Diagram
A force of magnitude 41 N acts due south. It is resolved into two components as shown on the diagram. Find the magnitudes of and . Give your answer to two decimal places.
A parallelogram can be drawn with side lengths proportional to and . A straight line can be drawn from the vertex at which these sides meet to the opposite vertex of the parallelogram. This line has a length of 41 length units, as shown in the following figure.
If we consider one of the triangles of this parallelogram, it contains an angle , as shown in the following figure.
The angle is given by
The magnitudes of and can be determined by applying the sine rule in the triangle:
We have, therefore, that
To two decimal places, .
We also have that
To two decimal places, .
Let us look at another such example.
### Example 5: Resolving Forces in a Real-Life Context
The diagram shows a body of weight 69 N suspended by 2 light, inextensible strings, and . Both strings make an angle of with the horizontal. Resolve the weight of the body into two components in the direction of and in the direction of . Give your answers to the nearest newton.
The strings are at the same angle from the horizontal and so the components and have equal magnitude. The line of action of the weight of the body is perpendicular to .
The following figure shows two right triangles. For each triangle, the non-right angles are and .
The angle between each component and the weight is also and so is given by
A parallelogram can be defined that has a vertex at and another vertex vertically below proportional to the weight, where the lengths of each side of the parallelogram are proportional to the magnitude of either component. A triangle of this parallelogram is shown in the following figure.
From this, we can see that
To the nearest newton, .
Each component has a magnitude of 57 N.
Now let us summarize what we have learned in these examples.
### Key Points
• A two-dimensional force, , can be resolved into components, and . If these components are perpendicular to each other in a coordinate system used to represent the force, then these components have magnitudes given by and where is the angle between and the -axis of the coordinate.
• A force can be resolved into components that are not perpendicular to each other. A parallelogram can be formed that has a diagonal of length proportional to the magnitude of the force and sides that have lengths proportional to the magnitudes of the components of the force that act at angles to the force proportional to the internal angles of the parallelogram. The lengths of the sides of the parallelogram can be determined using the sine rule where is a triangle of the parallelogram with side lengths , , and , where , , and are lengths of the sides of the triangle opposite the angles , , and . |
# Interpolation Formula
The method of finding new values for any function using the set of values is done by interpolation. The unknown value on a point is found out using this formula. If linear interpolation formula is concerned then it should be used to find the new value from the two given points. If compared to Lagrange’s interpolation formula, the “n” set of numbers should be available and Lagrange’s method is to be used to find the new value.
The following is Linear Interpolation Formula
$\large y=y_{1}+\frac{(x-x_{1})}{(x_{2}-x_{1})} \times (y_{2}-y_{1})$
The La-grange’s Interpolation Formula is given as,
$y=\frac{(x-x_{1})(x-x_{2})……(x-x_{n})}{(x_{o}-x_{1})(x_{0}-x_{2})….(x_{0-}-x_{n})}y+\;\frac{(x-x_{0})(x-x_{n})….(x-x_{n})}{(x_{1}-x_0)(x_{1}-x_{2}….((x_{1}-x_{n})}y_{1}\;+\frac{(x-x_{1})(x-x_{2})….(x-x_{n})}{(x_{o}-x_{1})(x_{0}-x_{2})…..(x_{o}-x_{n})}y_{n}$
Some solved problems on interpolation are given below:
### Solved Examples
Question 1: Using the interpolation formula, find the value of y at x = 8 given some set of values (2, 6), (5, 9) ?
Solution:
The known values are,$x_{0}=8, x_{1}=2, x_{2}=5, y_{1}=6, y_{2}=9$
$y=y_{1}+\frac{(x-x_{1})}{(x_{2}-x_{1})} \times (y_{2}-y_{1})$
$y=6+(\frac{(8-2)}{(5-2)} \times (9-6)$
y = 6 + 6
y = 12
More topics in Interpolation Formula Quadratic Interpolation Formula Linear Interpolation Formula Lagrange Interpolation Formula
Related Formulas LCM Formula N Choose K Formula Logarithm Formula Margin of Error Formula Prism Formula Percentile Formula Perimeter of Rhombus Formula Relative Standard Deviation Formula |
# Factors of 45
Understanding Prime and Composite Numbers When studying mathematics, one of the fundamental concepts that students learn is that of prime and composite numbers. In this article, we will delve into the factors of 45, a composite number, and explore the various ways in which it can be broken down into its component parts.
## Factors Calculator
Enter Number
Factors of 45: 1, 3, 5, 9, 15 and 45
Negative Factors of 45: -1, -3, -5, -9, -15 and -45
Prime Factors of 45: 3, 5
Prime Factorization of 45: 3 × 3 × 5
Factors of 45 in Pairs: (1, 45), (3, 15) and (5, 9)
Negative Pairs Factors of 45: (-1, -45), (-3, -15) and (-5, -9)
Prime Numbers
A prime number is a whole number greater than 1 that can only be divided evenly by 1 and itself. Some examples of prime numbers include 2, 3, 5, 7, and 11. Prime numbers are important in mathematics as they are used in many mathematical formulas and algorithms, and are also used in the study of number theory.
Composite Numbers
A composite number is a whole number greater than 1 that can be divided evenly by a number other than 1 or itself. In other words, a composite number is not a prime number. Some examples of composite numbers include 4, 6, 8, 9, and 10.
## What are the factors of 45?
The method of calculating the factors of 45 is as follows. First, each number can be divided by one and by itself.
Consequently, 1 and 45 are the factors of 45.
By dividing a number by 1, 2, 3, 4… we can discover all its factors.
(i) 45 ÷ 1 = 45
This division gives the remainder 0 and so is divisible by 45. So please put them 1 and 45 in your factor list.
1, …. 45
(ii) 45 ÷ 2 = 22.5
This division gives the remainder 22.5, not being thoroughly divided. So we will not write 2 and 22.5 on the list.
(iii) 45 ÷ 3 = 15
This division gives the remainder 0 and so is divisible by 15. So please put them 3 and 15 in your factor list.
1, 3 …. 15, 45
(iv) 45 ÷ 4 = 11.25
This division gives the remainder 11.25, not being thoroughly divided. So we will not write 4 and 11.25 on the list.
(iv) 45 ÷ 5 = 9
This division gives the remainder 0 and so is divisible by 15. So please put them 3 and 15 in your factor list.
1, 3, 5 …. 9, 15, 45
(v) Since we don’t have any more numbers to calculate, we are putting the numbers so far.
So 1, 3, 5, 9, 15 and 45 are factors of 45.
## Factor pairs of 45
1 x 45 = 45
3 x 15 = 45
5 x 9 = 45
So, (1, 45), (3, 15) and (5, 9) are factor pairs of 45
## Factor pairs of -45
-1 x -45 = 45
-3 x -15 = 45
-5 x -9 = 45
So, (-1, -45), (-3, -15) and (-5, -9) are negative pair factors of 45
## Prime Factorization of 45
45 ÷ 3 = 15
15 ÷ 3 = 5
5 ÷ 5 = 1
Therefore, 3 × 3 × 5 are Prime factorization of 45.
## Factor tree of 45
`````` 45
/ \
3 15
/ \
3 5
`````` |
# Adding by a Form of 0 (Part 4)
In my previous post, I wrote out a proof (that an even number is an odd number plus 1) that included the following counterintuitive steps:
$2k = (2k - 1) + 1 = ([2k - 1 - 1] + 1) + 1$
A common reaction that I get from students, who are taking their first steps in learning how to write mathematical proofs, is that they don’t think they could produce steps like these on their own without a lot of coaching and prompting. They understand that the steps are correct, and they eventually understand why the steps were necessary for this particular proof (for example, the conversion from $2k-1$ to $[2k - 1 -1]+1$ was necessary to show that $2k-1$ is odd).
Not all students initially struggle with this concept, but some do. I’ve found that the following illustration is psychologically reassuring to students struggling with this concept. I tell them that while they may not be comfortable with adding and subtracting the same number (net effect of adding by 0), they should be comfortable with multiplying and dividing by the same number because they do this every time that they add or subtract fractions with different denominators. For example:
$\displaystyle \frac{2}{3} + \frac{4}{5} = \displaystyle \frac{2}{3} \times 1 + \frac{4}{5} \times 1$
$= \displaystyle \frac{2}{3} \frac{5}{5} + \frac{4}{5} \times \frac{3}{3}$
$= \displaystyle \frac{10}{15} + \frac{12}{15}$
$= \displaystyle \frac{22}{15}$
In the same way, we’re permitted to change $2k-1$ to $2k-1 + 0$ to $2k -1 - 1 + 1$.
Hopefully, connecting this proof technique to this familiar operation from 5th or 6th grade mathematics — here in Texas, it appears in the 5th grade Texas Essential Knowledge and Skills under (3)(H) and (3)(K) — makes adding by a form of 0 in a proof somewhat less foreign to my students.
## One thought on “Adding by a Form of 0 (Part 4)”
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# Orthocentre
The 3 altitudes of a triangle intersect at a common point called the orthocentre. Its location depends on the type of triangle; unlike the median, it can be located either inside, outside, or on the vertex of a triangle.
### Acute Triangle
The orthocentre is located on the inside of the triangle.
### Right Triangle
The orthocentre is located on the vertex of the right angle.
### Obtuse Triangle
The orthocentre is located on the outside of the triangle.
## Steps to Calculate Orthocentre
1. Calculate the slope of at least 2 side lengths of the triangle using the formula $$m = \cfrac{y2 - y1}{x2 - x1}$$
2. Using the slopes of the side lengths, find the slopes of the altitudes by calculating the perpendicular slopes using the formula $$m = \cfrac{-1}{m}$$
3. Use the slope-point formula $$y - y₁ = m(x - x₁)$$ to determine the altitudes. $$m$$ represents the slope of the altitude and $$(x₁, y₁)$$ represents the point it connects to
4. Set the altitudes equal to each other to solve for the x-coordinate of the orthocentre
5. Plug the x-value into one of the altitude formulas to solve for the y-coordinate
Example
Find the orthocentre of the following triangle:
We can find the slopes of 2 side lengths of the triangle. In this instance, we will find the slopes of sides $$\text{AB}$$ and $$\text{BC}$$ with $$\text{A}$$ acting as Point 1, $$\text{B}$$ acting as Point 2 and $$\text{C}$$ acting as Point 3:
m(ᴀʙ) $$= \cfrac{y₂ - y₁}{x₂ - x₁}$$
m(ᴀʙ) $$= \cfrac{4 - 1}{3 - 1}$$
m(ᴀʙ) $$= \cfrac{3}{2}$$
m(ʙᴄ) $$= \cfrac{y₃ - y₂}{x₃ - x₂}$$
m(ʙᴄ) $$= \cfrac{1 - 4}{5 - 3}$$
m(ʙᴄ) $$= \cfrac{-3}{2}$$
Now that we have the slopes, we can determine the slopes of the altitudes by calculating the inverse slopes:
m(altitude)AB $$= \cfrac{-1}{3/2}$$
m(altitude)AB $$= \cfrac{-2}{3}$$
m(altitude)BC $$= \cfrac{-1}{-3/2}$$
m(altitude)BC $$= \cfrac{2}{3}$$
Since we have the slopes of the altitudes, we can plug these values into the slope-point formula to determine the altitudes For side length $$\text{AB}$$, we will be using Point $$\text{C}$$. For side length $$\text{BC}$$, we will be using Point $$\text{A}$$:
altitudeAB: $$y - 1 = \cfrac{-2}{3}(x - 5)$$
altitudeAB: $$y - 1 = \cfrac{-2}{3}x + \cfrac{10}{3}$$
altitudeAB: $$y = \cfrac{-2}{3}x - \cfrac{10}{3} + \cfrac{3}{3}$$
altitudeAB: $$y = \cfrac{-2}{3}x + \cfrac{13}{3}$$
altitudeBC: $$y - 1 = \cfrac{2}{3}(x - 1)$$
altitudeBC: $$y - 1 = \cfrac{2}{3}x - \cfrac{2}{3}$$
altitudeBC: $$y = \cfrac{2}{3}x - \cfrac{2}{3} + \cfrac{3}{3}$$
altitudeBC: $$y = \cfrac{2}{3}x + \cfrac{1}{3}$$
Now that we have both altitudes, we can determine the location of the orthocentre. We can set the altitudes equal to each other to solve for $$x$$.
$$\cfrac{-2}{3}x + \cfrac{13}{3} = \cfrac{2}{3}x + \cfrac{1}{3}$$
$$\cfrac{-2}{3}x - \cfrac{2}{3}x = \cfrac{1}{3} - \cfrac{13}{3}$$
$$(\cfrac{-3}{4})(\cfrac{-4}{3}x) = (\cfrac{-3}{4})(\cfrac{-12}{3})$$
$$x = \cfrac{36}{12} = 3$$
Now we can plug the x-coordinate into one of the altitude formulas to solve for $$y$$:
$$y = (\cfrac{-2}{3})(3) + \cfrac{13}{3}$$
$$y = \cfrac{-6}{3} + \cfrac{13}{3}$$
$$y = \cfrac{7}{3}$$
Therefore, we can determine that the location of the orthocentre is $$(3, \(\cfrac{7}{3}$$)\).
Determine the orthocentre of △$$\text{CRT}$$ with coordinates $$\text{C}(1,8)$$, $$\text{R}(3, 12)$$ and $$\text{T}(6,2)$$. |
Chapter 4 Matrices , Determinants , Linear Equations
This chapter includes two parts: matrices , determinants, and systems of linear algebraic equations .
In the previous part, the basic concepts of matrices and determinants were described, with emphasis on the properties and basic operations of various types of matrices. In addition, the methods for finding eigenvalues and eigenvectors of matrices were also introduced, as well as related content, such as similarity transformation, etc. ;In the part of linear equations, it focuses on the solution of n equations with n unknowns , and also briefly discusses the structure of the solution . Finally, it also briefly explains the integer coefficient linear equations and linear inequalities .
§1 Matrix and determinant
1. Matrix and its rank
[ Matrix and Square Matrix ] Number field (Chapter 3, § 1 ) The m × n numbers a ij ( i =1,2, , m ; j =1,2, , n ) on F are determined by A rectangular array of positions, called an m × n matrix . Denoted as
A =
The horizontal row is called the row, the vertical row is called the column, a ij is called the element of the i -th row and the j -th column of the matrix, and the matrix A is abbreviated as ( a ij ) or ( a ij ) m n .
An n - by - n matrix is also calledsquare matrix of order n , and a 11 , a 12 , ..., a nn are called the elements ofthe main diagonal ofmatrix A.
A matrix whose number of rows m and number of columns are both finite is called a finite matrix . Otherwise, it is called an infinite matrix .
[ Linear correlation of vectors is linearly independent ] For a set of vectors x 1 , x 2 ,..., x m in n -dimensional space , if there is a set of numbers k i ( i = 1 , 2 ,... , m ) , so that
k 1 x 1 + k 2 x 2 + + k m x m =0
If established, the set of vectors is said to be linearly dependent on F , otherwise the set of vectors is said to be linearly independent on F.
Discussion of Linear Dependency of Vector Groups:
A sufficient and necessary condition for the linear correlation of 1 ° vector groups x 1 , x 2 , ..., x m is that at least one vector x i can be represented by a linear combination of other vectors, that is
2 ° The set of vectors containing zero vectors must be linearly dependent .
In the 3 ° vector group x 1 , x 2 ,..., x m , if two vectors are equal: x i = x j ( i j ), then the vector group is linearly related .
4 ° If the vector groups x 1 , x 2 ,..., x r are linearly correlated, the vector group formed by adding several vectors is still linearly correlated; if the vector groups x 1 , x 2 ,..., x m are linearly independent, Then the vector group composed of any part of the vector is also linearly independent .
5 ° If x 1 , x 2 , . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ a linear combination of .
[ Row Vector and Column Vector · Matrix Rank ] An n -dimensional vector composed of elements in any row of the matrix is called a row vector , and is written as
a i =( a i 1 , a i 2 ,..., a in ) ( i =1,2,..., m )
The m -dimensional vector formed by the elements of any column of the matrix is called a column vector, denoted as
( j =1 ,2 ,..., n )
In the formula, t represents transposition, that is, the row (column) is converted into a column (row) .
If r of the n column vectors of matrix A are linearly independent ( r n ) , and all column vector groups whose number is greater than r are linearly correlated, then the number r is called the column rank of matrix A. Similarly, matrix A can be defined row rank .
The column rank and row rank of a matrix A must be equal, it is also called the rank of the matrix, denoted as rank A = r .
The rank of a matrix is also equal to the maximum order of subforms (see this section, two) in that matrix that are not equal to zero .
Second, the determinant
1. The determinant and its Laplace expansion theorem
[ Nth order determinant ] Let
is a number determined by n 2 numbers a ij ( i , j =1,2,..., n ) arranged in the form of an n -order square matrix , and its value is n ! sum of terms
In the formula, k 1 , k 2 ,..., k n is a sequence obtained by exchanging the element order of the sequence 1, 2,..., n for k times, and the Σ number represents the pair of k 1 , k 2 ,... ., k n take all the permutations of 1, 2, ..., n and sum up, then the number D is called the corresponding determinant of the n -order square matrix . For example, the fourth-order determinant is 4 ! The sum of terms of the form , and where a 13 a 21 a 34 a 42 corresponds to k = 3, that is, the symbol of the front end of the term should be
( -1 ) 3 .
If the n -order square matrix A = ( a ij ) , then the corresponding determinant D of A is written as
D =| A |=det A = det( a ij )
If the corresponding determinant of matrix A is D = 0 , A is called a singular matrix, otherwise it is called a non-singular matrix .
[ label set ] any k elements i 1 , i 2 ,..., i k in the sequence 1,2 ,..., n satisfy
1 i 1 < i 2 <...< i kn (1)
i 1 , i 2 ,..., i k form a subcolumn of {1,2,..., n } with k elements and {1,2,..., n } with k elements The totality of the sub-columns satisfying (1) is denoted as C ( n , k ), obviously C ( n , k ) has a total of sub-columns . Therefore C ( n , k ) is a label set with elements (see twentieth Chapter 1, §1 , ii), C ( n , k )The elements of σ , τ ,..., σ∈ C ( n , k ) are denoted by
σ ={ i 1 , i 2 ,..., i k }
is a subsequence of {1,2,..., n } that satisfies (1) . If τ ={ j 1 , j 2 ,..., j k } C ( n , k ), then σ = τ means i 1 = j 1 , i 2 = j 2 ,..., i k = j k .
[ Subform , main subform , cosubform , algebraic cosubform ]
Take any k rows and k columns ( 1 kn - 1 ) from the n -order determinant D , and the k -order determinant formed by the elements at the intersection of the k-row and k-column is called the k - order child of the determinant D formula, write
, σ , τC ( n , k )
If the selected k rows and k columns are the i 1 , i 2 ,..., i k rows and i 1 , i 2 ,..., i k columns, respectively, then the resulting k -order subformula is called Main subform . That is, when σ = τC ( n , k ) , it is the main subform .
The n - k order determinant obtained by removing k rows (σ) and k columns ( τ ) from the determinant D is called the cofactor of the subformula, denoted as .
If σ ={ i 1 , i 2 ,..., i k } , τ ={ j 1 , j 2 ,..., j k } , then
is the algebraic cofactor of the subform .
In particular, when k = 1 , σ ={ i } , τ ={ j }, the subform is an element a ij , the cofactor of a ij is written as A ij , the algebraic cofactor of a ij is written as A ij , that is
and have ( 2 )
or ( 3 )
[ Laplace expansion theorem ] If any k rows ( 1 kn -1 ) are taken in the n -order determinant D , then all the k -order sub-formulas and their respective algebras contained in these selected rows The sum of the products of cofactors is equal to the determinant D , that is, for any σ∈ C ( n , k ) , 1 kn -1 ,
( 4 )
where ∑ represents the summation of all elements in the label set C ( n , k ) .
Laplace's theorem is done for rows, and similar results are obtained for columns
( 5 )
Furthermore
( 6 )
( 7 )
Obviously ( 2 ), ( 3 ) are special cases of ( 6 ), ( 7 ), respectively .
[ Laplace's identity ] Let A =( a ij ) m n , B =( b ij ) m n ( mn ), and let l = , all n -order subformulas of A are U 1 , U 2 , ... , U l , and the corresponding n -order subformulas of B are V 1 , V 2 , ..., V l , then
det( A τ B )=
2. The nature of the determinant
1 ° ï A 1 A 2 L A m ï = ï A 1 ï ï A 2 ï L ï A m ï
ï A m ï = ï A ï m , ï kA ï = k n ï A ï
In the formula, A 1 , A 2 , L , and A m are all square matrices of order n , and k is any complex number .
2 ° After the row and column are interchanged , the value of the determinant remains unchanged, that is,
| |=| A |
where represents the transposed matrix of A (see § 2 of this chapter ) .
3 ° Interchange any two rows (or columns) of the determinant, change the sign of the determinant . For example
=
4 ° Multiplying a row (or column) of the determinant by the number α is equivalent to multiplying the determinant by the number α . For example
= a
5 ° Multiply one row (or column) element of the determinant by the number α and add it to the corresponding element in another row (or column), the value of the determinant does not change . For example
=
6 ° If a row (or column) of the determinant is all zero, then the determinant is equal to zero .
A determinant is zero if two rows (or columns) of the determinant have identical or proportional elements .
A determinant is zero if a row (or column) of elements in the determinant is a linear combination of corresponding elements in some other row (or column) .
7 ° If all elements of a row (or column) in the determinant can be expressed as the sum of two terms, the determinant can be expressed by the sum of two determinants of the same order . For example
= +
3. Several special determinants
[ diagonal determinant ]
=
[ triangle determinant ]
=
[ Second-order determinant ]
[ third-order determinant ]
= + +
memory method
The value of the determinant is equal to the sum of the products of the elements on the solid lines minus the sum of the products of the elements on the dotted lines .
[ fourth-order determinant ]
= - + -
= - +
+ - +
Note that the fourth-order and higher-order determinants cannot use the memory method of the third-order determinant, and should be expanded according to the Laplace expansion theorem by the method of step-by-step reduction .
[ Vandermonde determinant ]
=
where Õ is the product of all pairs ( i , j ) ( 1 j < i n ) .
[ Reciprocal symmetric determinant ]
= |
## Precalculus (6th Edition) Blitzer
Two square roots of $9$ using DeMoivre’stheorem are $3$ and $-3$.
The roots of a complex number represented in polar form can be found by using DeMoivre’s theorem. DeMoivre’stheorem for finding complex roots states that every complex number has two distinct complex square roots, three distinct complex cube roots, four distinct complex fourth roots, and so on. Each root has the same modulus. \begin{align} & r=z_{k}^{n} \\ & {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \end{align} Here, $k$ is the number of distinct $n\text{th}$ roots and $\theta$ is in radians. There are exactly two square roots of $9$ as per DeMoivre’s theorem. First write the provided number in rectangular form as follows: $9=9+0i$ Convert the above expression into polar form as follows: \begin{align} & 9=r\left( \cos \theta +i\sin \theta \right) \\ & =9\left( \cos 0+i\sin 0 \right) \end{align} Apply DeMoivre’s theorem to find the two roots of $9$. ${{z}_{k}}=\sqrt[2]{1}\left[ \cos \left( \frac{0+2\pi k}{2} \right)+i\sin \left( \frac{0+2\pi k}{2} \right) \right]$ Here, $k=0,1$. Insert the above values of $k$ to find two distinct square roots of the provided number. For $k=0$, \begin{align} & {{z}_{0}}=\sqrt[2]{9}\left[ \cos \left( \frac{0+2\pi \left( 0 \right)}{2} \right)+i\sin \left( \frac{0+2\pi \left( 0 \right)}{2} \right) \right] \\ & =3\left[ \cos 0+i\sin 0 \right] \\ & =3 \end{align} For $k=1$, \begin{align} & {{z}_{1}}=\sqrt[2]{9}\left[ \cos \left( \frac{0+2\pi \left( 1 \right)}{2} \right)+i\sin \left( \frac{0+2\pi \left( 1 \right)}{2} \right) \right] \\ & =3\left[ \cos \left( \pi \right)+i\sin \left( \pi \right) \right] \end{align} Substitute the values of $\cos \left( \pi \right)$ and $\sin \left( \pi \right)$. Then, \begin{align} & {{z}_{1}}=3\left[ -1.0+i\left( 0 \right) \right] \\ & =-3 \end{align} The two square roots of the complex number $9$ are $3$ and $-3$. |
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Classical Geometry Puzzle: Finding the Radius
This is another look at a puzzle from Mind Your Decisions. The problem is to find the radius of the following circle:
We are told that line AD and BC are perpendicular and the lengths of some parts of chords, but not much more! First I’ll look at my attempt to solve this. It’s not quite as “nice” as the solution in the video as it requires the use of a calculator, but it still does the job.
Method 1, extra construction lines:
These are the extra construction lines required to solve this problem. Here is the step by step thought process:
1. Find the hypotenuse of triangle AGC.
2. Use the circle theorem angles in the same segment are equal to show that angle CBD = angle CAG.
3. Therefore triangle AGC and GBD are similar, so length BG = 4. We can now use Pythagoras to find length BD.
4. We can find length CD by Pythagoras.
5. Now we have 3 sides of a triangle, CDB. This allows use to find angle BDC using the cosine rule.
6. Now we the circle theorem angles in the same segment are equal to show that angle BDC = angle BEC.
7. Now we use the circle theorem angles in a semi circle are 90 degrees to show ECB = 90.
8. Now we have a right angled triangle BCE where we know both an angle and a side, so can use trigonometry to find the length of BE.
9. Therefore the radius is approximately 4.03.
Method 2, creating a coordinate system
This is a really beautiful solution – which does not require a calculator (and which is discussed in the video above). We start by creating a coordinate system based around point G at (0,0). Because we have perpendicular lines we can therefore create coordinates for A, B and C. We also mark the centre of the circle as (p,q).
Next we create 3 equations by substituting in our coordinates:
Next we can do equation (3) – equation (1) to give:
Next we can substitute this value for p into equations (1) and (3) and equate to get:
Lastly we can substitute both values for p and q into equation (1) to find r:
We get the same answer as before – though this definitely feels like a “cleaner” solution. There are other ways to solve this – but some of these require the use of equations you may not already know (such as the law of sines in a circumcircle, or the equation for perpendicular chords and radius). Perhaps explore any other methods for solving this – what are the relative merits of each approach?
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here. |
# Base Method of Squaring a number
### Base Method of Squaring a number:
Base method of squaring is very much easy and quick technique of Vedic mathematics. In this method we need to consider two types of base, which are
• Actual base: 10, 100, 1000, 10000 and so on
• Working base: 20, 30, 40, 200, 300, 400 etc. Working base taken by multiplying the actual base with integers. If we have to find square of 22 then the working base must be 20(10×2), but actual base is 10.
I know these things are little confusing, but don‟t worry after some example it will clear doubts about this method. Let us take some example:-
Example 1: Find the square of 24
Solution: Here actual base is 10, but we cannot take 10 as working base. We need to take 20 as working base since (10×2=20).
Since 24 is more than 20, so we need to add 4 to 24, and we get (24+4)=28, now since our working base is 20 and 20 is obtain by multiplying actual base which is 10 with 2, we need to multiply 28 with 2. So, (28×2) =56, and “56” will be the LHS of the final answer.
Now for RHS we need to find the square of 4 which is 16. Now here is the confusion is whether to put 16 or 6 in RHS. We need to put only “6” in RHS, because our actual base containing only one zero.
Remember, if actual base is 100 we need to put 2 digits in the RHS, if 1000 we put 3 digits in RHS. Now what happened to the “1” in 16. We simply add 1 to 56 in LHS. So, final answer is 576.
Example 2: Find square 296
Solution: Here actual base is 100 and working base is 300, since 296 is closer to 300 than 200. Step by step solution method is analyzed below,
• 1 st of all find out how much less is 296 from 300, so 296 is 4 less from 300 so we need to subtract 4 from 296 which is 292.
• Now multiply 292 with 3 since our actual base is multiply with 3 to get working base.
• (292×3)=876, which will be our LHS.
• Now for RHS we need two digit, since our actual base has two zero.
• Squaring 4 we get the RHS, i.e. 16 is in RHS.
• So our final answer is 87616.
Example 3: Find square of 104.
Solution: Here our actual base is also our working base, since 104 is closer to 100. Step by step detailed analysis is given below,
• 1 st of all add 4 to 104 to get LHS which is 108, since 104 is more than 100 by 4.
• Now for RHS square the number 4 and we get 16 as RHS.
• So final answer is 10816.
Checklist:
1. Check for actual and working base, if actual base itself close to the
number takes it as working base.
2. Select the base which is closer to the number.
3. Put the digit equal to the number of zeros in actual base.
4. If we get 2 digit number of actual base 10 for RHS then adds the unit
place digit to the LHS. Same rule is applied for 100, 1000 base and
so on.
40 |
```8-3 Factoring x2 + bx + c
Preview
Warm Up
California Standards
Lesson Presentation
8-3 Factoring x2 + bx + c
Warm Up
1. Which pair of factors of 8 has a sum of
9? 1 and 8
2. Which pair of factors of 30 has a sum of
–17? –2 and –15
Multiply.
3. (x + 2)(x + 3) x2 + 5x + 6
4. (r + 5)(r – 9)
r2 – 4r – 45
8-3 Factoring x2 + bx + c
California
Standards
11.0 Students apply basic factoring
techniques to second- and simple thirddegree polynomials. These techniques
include finding a common factor for all terms
in a polynomial, recognizing the difference of
two squares, and recognizing perfect squares
of binomials.
8-3 Factoring x2 + bx + c
In Chapter 7, you learned how to multiply two
binomials using the Distributive Property or the
FOIL method. In this lesson, you will learn how to
factor a trinomial into two binominals.
8-3 Factoring x2 + bx + c
Notice that when you multiply (x + 2)(x + 5), the
constant term in the trinomial is the product of
the constants in the binomials.
(x + 2)(x + 5) = x2 + 7x + 10
8-3 Factoring x2 + bx + c
Use this fact to factor some trinomials into binomial
factors. Look for two integers (positive or negative)
that are factors of the constant term in the trinomial.
Write two binomials with those integers, and then
multiply integers to check.
If no two factors of the constant term work, we say
the trinomial is not factorable.
8-3 Factoring x2 + bx + c
(
+
(x +
)(
+
)(x +
) Write two sets of parentheses.
) The first term is x2, so the variable
terms have a coefficient of 1.
The constant term in the trinomial is 36.
Try factors of 36 for the constant terms in the binomials.
(x + 1)(x + 36) = x2 + 37x + 36
(x + 2)(x + 18) = x2 + 20x + 36
(x + 3)(x + 12) = x2 + 15x + 36
8-3 Factoring x2 + bx + c
The factors of x2 + 15x + 36 are (x + 3)(x + 12).
x2 + 15x + 36 = (x + 3)(x + 12)
Check (x + 3)(x + 12)
= x2 + 12x + 3x + 36 Use the FOIL method.
= x2 + 15x + 36
The product is the
original trinomial.
8-3 Factoring x2 + bx + c
Remember!
When you multiply two binomials, multiply:
First terms
Outer terms
Inner terms
Last terms
8-3 Factoring x2 + bx + c
Check It Out! Example 1a
x2 + 10x + 24
(
+
(x +
)(
+
)
)(x +
)
Write two sets of parentheses.
The first term is x2, so the variable
terms have a coefficient of 1.
The constant term in the trinomial is 24.
Try factors of 24 for the constant terms in the binomials.
(x + 1)(x + 24) = x2 + 25x + 24
(x + 2)(x + 12) = x2 + 14x + 24
(x + 3)(x + 8) = x2 + 11x + 24
(x + 4)(x + 6) = x2 + 10x + 24
8-3 Factoring x2 + bx + c
Check It Out! Example 1a Continued
x2 + 10x + 24
The factors of x2 + 10x + 24 are (x + 4)(x + 6).
x2 + 10x + 24 = (x + 4)(x + 6)
Check (x + 4)(x + 6)
= x2 + 4x + 6x + 24 Use the FOIL method.
= x2 + 10x + 24
The product is the
original trinomial.
8-3 Factoring x2 + bx + c
Check It Out! Example 1b
x2 + 7x + 12
(
+
(x +
)(
+
)
)(x +
)
Write two sets of parentheses.
The first term is x2, so the variable
terms have a coefficient of 1.
The constant term in the trinomial is 12.
Try factors of 12 for the constant terms in the binomials.
(x + 1)(x + 12) = x2 + 13x + 12
(x + 2)(x + 6) = x2 + 8x + 12
(x + 3)(x + 4) = x2 + 7x + 12
8-3 Factoring x2 + bx + c
Check It Out! Example 1b Continued
Factor each trinomial and check.
x2 + 7x + 12
The factors of x2 + 7x + 12 are (x + 3)(x + 4).
x2 + 7x + 12 = (x + 3)(x + 4)
Check (x + 3)(x + 4)
= x2 + 4x + 3x + 12
= x2 + 7x + 12
Use the FOIL method.
The product is the
original trinomial.
8-3 Factoring x2 + bx + c
The method of factoring used in Example 1 can be
made more efficient. Look at the product of (x + a)
and (x + b).
x2
ab
(x + a)(x + b) = x2 + ax + bx + ab
ax
bx
= x2 + (a + b)x + ab
The coefficient of the middle term is the sum of a
and b. The third term is the product of a and b.
8-3 Factoring x2 + bx + c
8-3 Factoring x2 + bx + c
When c is positive, its factors have the same sign.
The sign of b tells you whether the factors are
positive or negative. When b is positive, the
factors are positive and when b is negative, the
factors are negative.
8-3 Factoring x2 + bx + c
Additional Example 2A: Factoring x2 + bx + c
When c is Positive
x2 + 6x + 5
(x +
)(x +
)
b = 6 and c = 5; look for factors of 5
whose sum is 6.
Factors of 5 Sum
1 and 5
6
The factors needed are 1 and 5.
(x + 1)(x + 5)
Check (x + 1)(x + 5) = x2 + 5x + x + 5
= x2 + 6x + 5
Use the FOIL method.
The product is the
original trinomial.
8-3 Factoring x2 + bx + c
Additional Example 2B: Factoring x2 + bx + c
When c is Positive
x2 + 6x + 9
b = 6 and c = 9; look for factors of 9
(x + )(x + )
whose sum is 6.
Factors of 9 Sum
1 and 9
10
3 and 3
6 The factors needed are 3 and 3.
(x + 3)(x + 3)
Check (x + 3)(x + 3 ) = x2 + 3x + 3x + 9 Use the FOIL method.
The product is the
2
= x + 6x + 9
original trinomial.
8-3 Factoring x2 + bx + c
Additional Example 2C: Factoring x2 + bx + c
When c is Positive
x2 – 8x + 15
(x +
)(x +
)
b = –8 and c = 15; look for
factors of 15 whose sum is –8.
Factors of –15 Sum
–1 and –15 –16
–3 and –5 –8 The factors needed are –3 and –5 .
(x – 3)(x – 5)
Check (x – 3)(x – 5 ) = x2 – 5x – 3x + 15 Use the FOIL method.
The product is the
2
= x – 8x + 15
original trinomial.
8-3 Factoring x2 + bx + c
Check It Out! Example 2a
x2 + 8x + 12
(x +
)(x +
Factors of 12
1 and 12
2 and 6
)
Sum
13
8
b = 8 and c = 12; look for factors of
12 whose sum is 8.
The factors needed are 2 and 6 .
(x + 2)(x + 6)
Check (x + 2)(x + 6 ) = x2 + 6x + 2x + 12 Use the FOIL method.
= x2 + 8x + 12
The product is the
original trinomial.
8-3 Factoring x2 + bx + c
Check It Out! Example 2b
x2 – 5x + 6
(x +
)(x+
)
b = –5 and c = 6; look for
factors of 6 whose sum is –5.
Factors of 6 Sum
–1 and –6 –7
–2 and –3 –5 The factors needed are –2 and –3.
(x – 2)(x – 3)
Check (x – 2)(x – 3) = x2 – 3x – 2x + 6 Use the FOIL method.
= x2 – 5x + 6
The product is the
original trinomial.
8-3 Factoring x2 + bx + c
Check It Out! Example 2c
x2 + 13x + 42
(x +
)(x +
)
b = 13 and c = 42; look for
factors of 42 whose sum is
13.
Factors of 42 Sum
1 and 42 43
2 and 21 23
6 and 7 13 The factors needed are 6 and 7.
(x + 6)(x + 7)
Check (x + 6)(x + 7) = x2 + 7x + 6x + 42 Use the FOIL method.
= x2 + 13x + 42 The product is the
original trinomial.
8-3 Factoring x2 + bx + c
Check It Out! Example 2d
x2 – 13x + 40
b = –13 and c = 40; look for factors
of 40 whose sum is –13.
(x + )(x+ )
Factors of 40
–2 and –20
–4 and –10
–5 and –8
Sum
–22 The factors needed are –5 and –8.
–14
–13
(x – 5)(x – 8)
Check (x – 5)(x – 8) = x2 – 8x – 5x + 40 Use the FOIL method.
= x2 – 13x + 40
The product is the
original polynomial.
8-3 Factoring x2 + bx + c
When c is negative, its factors have opposite
signs. The sign of b tells you which factor is
positive and which is negative. The factor
with the greater absolute value has the same
sign as b.
8-3 Factoring x2 + bx + c
Additional Example 3A: Factoring x2 + bx + c When
c is Negative
Factor each trinomial.
x2 + x – 20
(x +
)(x +
)
Factors of –20 Sum
–1 and 20
19
–2 and 10
8
–4 and 5
1
(x – 4)(x + 5)
b = 1 and c = –20; look for
factors of –20 whose sum is
1. The factor with the greater
absolute value is positive.
The factors needed are 5 and
–4.
8-3 Factoring x2 + bx + c
Additional Example 3B: Factoring x2 + bx + c When
c is Negative
Factor each trinomial.
x2 – 3x – 18
(x +
)(x +
Factors of –18
1 and –18
2 and – 9
3 and – 6
)
Sum
–17
– 7
– 3
(x – 6)(x + 3)
b = –3 and c = –18; look for
factors of –18 whose sum is
–3. The factor with the
greater absolute value is
negative.
The factors needed are 3 and
–6.
8-3 Factoring x2 + bx + c
If you have trouble remembering the rules for
which factor is positive and which is negative,
you can try all the factor pairs and check their
sums.
8-3 Factoring x2 + bx + c
Check It Out! Example 3a
x2 + 2x – 15
(x +
)(x +
)
b = 2 and c = –15; look for
factors of –15 whose sum is 2.
The factor with the greater
absolute value is positive.
Factors of –15 Sum
–1 and 15
14
–3 and 5
2 The factors needed are –3 and
5.
(x – 3)(x + 5)
Check (x – 3)(x + 5) = x2 + 5x – 3x – 15 Use the FOIL method.
= x2 + 2x – 15
The product is the
original polynomial.
8-3 Factoring x2 + bx + c
Check It Out! Example 3b
x2 – 6x + 8
(x +
)(x +
Factors of 8
–1 and –6
–2 and –4
)
Sum
–7
–6
(x – 2)(x – 4)
b = –6 and c = 8; look for
factors of 8 whose sum is –6.
The factors needed are –4
and –2.
Check (x – 2)(x – 4) = x2 – 4x – 2x + 8
= x2 – 6x + 8
Use the FOIL method.
The product is the
original polynomial.
8-3 Factoring x2 + bx + c
Check It Out! Example 3c
x2 – 8x – 20
(x +
)(x +
)
Factors of –20 Sum
1 and –20 –19
2 and –10 –8
(x – 10)(x + 2)
b = –8 and c = –20; look for
factors of –20 whose sum is
–8. The factor with the
greater absolute value is
negative.
The factors needed are –10
and 2.
Check (x – 10)(x + 2) = x2 + 2x – 10x – 20
= x2 – 8x – 20
Use the FOIL method.
The product is the
original polynomial.
8-3 Factoring x2 + bx + c
A polynomial and the factored form of the
polynomial are equivalent expressions. When
you evaluate these two expressions for the
same value of the variable, the results are the
same.
8-3 Factoring x2 + bx + c
Factor y2 + 10y + 21. Show that the original
polynomial and the factored form have the
same value for y = 0, 1, 2, 3, and 4.
y2 + 10y + 21
(y + )(y + )
Factors of 21 Sum
1 and 21
22
3 and 7
10
(y + 3)(y + 7)
b = 10 and c = 21; look for factors
of 21 whose sum is 10.
The factors needed are 3 and 7.
8-3 Factoring x2 + bx + c
Evaluate the original polynomial and the
factored form for y = 0, 1, 2, 3, and 4.
y
y2 + 10y + 21
y
(y + 7)(y + 3)
0
02 + 10(0) + 21 = 21
0
(0 + 7)(0 + 3) = 21
1
12 + 10(1) + 21 = 32
1
(1 + 7)(1 + 3) = 32
2
22 + 10(2) + 21 = 45
2
(2 + 7)(2 + 3) = 45
3
32 + 10(3) + 21 = 60
3
(3 + 7)(3 + 3) = 60
4
42 + 10(4) + 21 = 77
4
(4 + 7)(4 + 3) = 77
The original polynomial and the factored form
have the same value for the given values of n.
8-3 Factoring x2 + bx + c
Check It Out! Example 4
Factor n2 – 7n + 10. Show that the original
polynomial and the factored form have the
same value for n = 0, 1, 2, 3, and 4.
n2 – 7n + 10
b = –7 and c = 10; look for factors
(n + )(n + )
of 10 whose sum is –7.
Factors of 10 Sum
–1 and –10 –11 The factors needed are –2 and
–5.
–2 and –5 – 7
(n – 5)(n – 2)
8-3 Factoring x2 + bx + c
Check It Out! Example 4 Continued
Evaluate the original polynomial and the
factored form for n = 0, 1, 2, 3, and 4.
n
n2 –7n + 10
n
(n – 5)(n – 2 )
0
02 – 7(0) + 10 = 10
0
(0 – 5)(0 – 2) = 10
1
12 – 7(1) + 10 = 4
1
(1 – 5)(1 – 2) = 4
2
22 – 7(2) + 10 = 0
2
(2 – 5)(2 – 2) = 0
3
32 – 7(3) + 10 = –2
3
(3 – 5)(3 – 2) = –2
4
42 – 7(4) + 10 = –2
4
(4 – 5)(4 – 2) = –2
The original polynomial and the factored form
have the same value for the given values of n.
8-3 Factoring x2 + bx + c
Lesson Quiz: Part I
Factor each trinomial.
1. x2 – 11x + 30 (x – 5)(x – 6)
2. x2 + 10x + 9
(x + 1)(x + 9)
3. x2 – 6x – 27
(x – 9)(x + 3)
4. x2 + 14x – 32
(x + 16)(x – 2)
8-3 Factoring x2 + bx + c
Lesson Quiz: Part II
5. Factor n2 + n – 6. Show that the original
polynomial and the factored form have the
same value for n = 0, 1, 2, 3 ,and 4.
(n + 3)(n – 2)
``` |
# Know Your Division: Divide by 2 (1)
In this worksheet, students divide by two.
Key stage: KS 1
Curriculum topic: Number: Multiplication and Division
Curriculum subtopic: Use Multiplication/Division Facts (2, 5, 10)
Difficulty level:
### QUESTION 1 of 10
In this worksheet, you must divide by 2.
Example
Here are 6 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils?
6 ÷ 2 = 3
3 pencil boxes are needed.
Here are 4 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
Here are 2 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
Here are 8 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
Here are 4 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
Here are 16 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
Here are 8 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
Here are 10 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
Here are 12 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
Here are 14 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
Here are 18 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
• Question 1
Here are 4 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
2
EDDIE SAYS
4 ÷ 2 = 2
2 x 2 = 4
• Question 2
Here are 2 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
1
EDDIE SAYS
2 ÷ 2 = 1
2 x 1 = 2
• Question 3
Here are 8 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
4
EDDIE SAYS
8 ÷ 2 = 4
2 x 4 = 8
• Question 4
Here are 4 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
2
EDDIE SAYS
4 ÷ 2 = 2
2 x 2 = 4
• Question 5
Here are 16 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
8
EDDIE SAYS
16 ÷ 2 = 8
2 x 8 = 16
• Question 6
Here are 8 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
4
EDDIE SAYS
8 ÷ 2 = 4
2 x 4 = 8
• Question 7
Here are 10 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
5
EDDIE SAYS
10 ÷ 2 = 5
2 x 5 = 10
• Question 8
Here are 12 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
6
EDDIE SAYS
12 ÷ 2 = 6
2 x 6 = 12
• Question 9
Here are 14 pencils.
2 pencils fit into each pencil box.
How many pencil boxes can be filled with these pencils? (Just write the number)
7
EDDIE SAYS
14 ÷ 2 = 7
2 x 7 = 14
• Question 10
Here are 18 chicks.
2 chicks fit into one nest.
How many nests can be filled with these chicks? (Just write the number)
9
EDDIE SAYS
18 ÷ 2 = 9
2 x 9 = 18
---- OR ----
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# Difference between revisions of "2011 USAMO Problems/Problem 5"
## Problem
Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that $\angle Q_1 BC = \angle ABP$, $\angle Q_1 CB = \angle DCP$, $\angle Q_2 AD = \angle BAP$, $\angle Q_2 DA = \angle CDP$. Prove that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if $\overline{Q_1 Q_2} \parallel \overline{CD}$.
## Solution
First note that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if the altitudes from $Q_1$ and $Q_2$ to $\overline{AB}$ are the same, or $|Q_1B|\sin \angle ABQ_1 =|Q_2A|\sin \angle BAQ_2$. Similarly $\overline{Q_1 Q_2} \parallel \overline{CD}$ iff $|Q_1C|\sin \angle DCQ_1 =|Q_2D|\sin \angle CDQ_2$.
If we define $S =\frac{|Q_1B|\sin \angle ABQ_1}{|Q_2A|\sin \angle BAQ_2}\cdot\frac{|Q_2D|\sin \angle CDQ_2}{|Q_1C|\sin \angle DCQ_1}$, then we are done if we can show that S=1.
By the law of sines, $\frac{|Q_1B|}{|Q_1C|}=\frac{\sin\angle Q_1CB}{\sin\angle Q_1BC}$ and $\frac{|Q_2D|}{|Q_2A|}=\frac{\sin\angle Q_2AD}{\sin\angle Q_2DA}$.
So, $S=\frac{\sin \angle ABQ_1}{\sin \angle BAQ_2}\cdot\frac{\sin \angle CDQ_2}{\sin \angle DCQ_1}\cdot\frac{\sin \angle BCQ_1}{\sin \angle CBQ_1}\cdot\frac{\sin \angle DAQ_2}{\sin \angle ADQ_2}$
By the terms of the problem, $S=\frac{\sin \angle PBC}{\sin \angle PAD}\cdot\frac{\sin \angle PDA}{\sin \angle PCB}\cdot\frac{\sin \angle PCD}{\sin \angle PBA}\cdot\frac{\sin \angle PAB}{\sin \angle PDC}$. (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.)
Rearranging yields $S= \frac{\sin \angle PBC}{\sin \angle PCB}\cdot\frac{\sin \angle PDA}{\sin \angle PAD}\cdot\frac{\sin \angle PCD}{\sin \angle PDC}\cdot\frac{\sin \angle PAB}{\sin \angle PBA}$.
Applying the law of sines to the triangles with vertices at P yields $S=\frac{|PC|}{|PB|}\frac{|PA|}{|PD|}\frac{|PD|}{|PC|}\frac{|PB|}{|PA|}=1$. |
# Increasing and bounded sequence proof
Prove that the sequence $a_n= 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln(n)$ is increasing and bounded above. Conclude that it’s convergent.
This what I got so far
Proof:
Part 1: Proving $a_n$ is increasing by induction.
Base: $a_1=1$
$a_2=1+\frac 12= \frac 32$
$a_1≤a_2$
So the base case is established.
Induction step: We assume that $a_{n-1}≤a_n$. We will show that $a_n≤a_{n+1}$. Since
$a_{n-1}≤a_n$
$$1+ \frac 12+ \frac 13+\cdots+ \frac{1}{(n-1)}-\ln(n-1) \leq 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln n$$
How should I continue?
• Hint: $\ln(n)-\ln(n-1)=\int_{n-1}^n \frac{1}{x}\,dx$. Commented Sep 20, 2013 at 16:48
• In reference to Pocho, make a drawing of the function y=1/x and draw rectangles with base 1 and the width as y-values on top of it. Look at the difference of the area of each rectangle and the area under the curve passing through that rectangle. Commented Sep 20, 2013 at 16:52
• I've just extended my answer by adding a proof of boundedness. Commented Sep 21, 2013 at 18:06
This sequence is NOT increasing and is in fact DECREASING.
Proof:
$$a_n = 1+\frac{1}{2} + \cdots + \frac{1}{n} - \ln(n)$$
$$a_{n+1}=1+\frac{1}{2} + \cdots + \frac{1}{n}+\frac{1}{n+1} - \ln(n+1)$$
so
$$a_{n+1}=a_n + (\ln(n) +\frac{1}{n+1} - \ln(n+1)) = a_n + \frac{1}{n+1} - (\ln(n+1) - \ln(n))$$
$$\ln(n+1)-\ln(n) = \int_n^{n+1}\frac{1}{t}dt > \frac{1}{n+1}$$
To see the last line just draw a graph of $\frac{1}{t}$ between $n$ and $n+1$ and draw a box of width $1$ and height $\frac{1}{n+1}$ (i.e. the box touches $\frac{1}{t}$ at $t=\frac{1}{n+1}$).
NOTE: Your base case was not done properly since you forgot to subtract $\ln(1)$ and $\ln(2)$ (granted $\ln(1) = 0$) if you subtracted $\ln(2) \approx 0.693147 > 0.5$ you would have seen the base case not to hold.
I think you should have been asked to show that $$1+\frac12+\frac13+\cdots+\frac1n-\ln(n+1) \tag 1$$ increases with $n$. The previous term is $$1+\frac12+\frac13+\cdots+\frac{1}{n-1} - \ln n. \tag 2$$ Then the problem is to show that when you subtract $(2)$ from $(1)$, the difference is nonnegative. You get $$\frac1n - \ln(n+1) + \ln n = \int_n^{n+1} \frac1n - \frac1x \, dx. \tag 3$$ It is easy to show that that is nonnegative.
To show that $(1)$ is bounded above, first notice that $(1)$ is equal to the sum of expressions like the one in $(3)$: $$\sum_{k=1}^n \int_k^{k+1} \frac1k - \frac1x\, dx. \tag 4$$ Then $$[\text{expression in (4)}] \ge \sum_{k=1}^n \int_k^{k+1} \frac1k - \frac{1}{k+1}\, dx = \sum_{k=1}^n \frac1k - \frac{1}{k+1},$$ and this is a telescoping sum, and all its terms are non-negative, and it adds up to $1$.
• I don't understand step (1) and (2), is (1) is my $a_n$? it doesn't look exactly the same Commented Sep 21, 2013 at 13:35
• It's not the same, since it has $n+1$ as the argument to the logarithm where you had $n$. As I said, if you were assigned an exercise asking you to prove that your $a_n$ is increasing, then whoever wrote the exercise probably ought instead to have asked you to show that the sequence in $(1)$ is bounded and increasing. Commented Sep 21, 2013 at 17:54
• I just fixed an error: it should be $\displaystyle\int_n^{n+1}$, not $\displaystyle\int_{n-1}^n$. ${}\qquad{}$ Commented Sep 21, 2013 at 17:58
@Patrick's answer shows that the sequence is decreasing. However, we can show boundedness and covergence in a single step since the terms of the sequence are non-negative. I only provide a skeleton solution, which will need fleshing out with a limit argument.
We can begin by rewriting the sequence as: \begin{equation*} \sum_{k=1}^n \frac{1}{k} -\log{n} = \sum_{k=1}^n \frac{1}{k} -\int_{1}^n \frac{1}{x} d{x} = \overbrace{\sum_{k=1}^{n-1} \int_{k}^{k+1} \Bigl( \frac{1}{k} -\frac{1}{x} \Bigr) d{x} +\frac{1}{n}}^\circledast. \end{equation*} The integral's limits provide the inequalities $k\le x \le {k+1}$, with which we obtain the following bounds on the integrand: $$0 \le \frac{1}{k} - \frac{1}{x} \le \frac{1}{k} -\frac{1}{k+1} = \frac{1}{k(k+1)} < \frac{1}{k^2}.$$ Then, after substituting our new integrand $\tfrac{1}{k^2}$, we obtain: \begin{equation*} 0 \le \sum_{k=1}^n \frac{1}{k} -\log{n} =\overbrace{\sum_{k=1}^{n-1} \int_{k}^{k+1} \Bigl( \frac{1}{k} -\frac{1}{x} \Bigr) d{x} +\frac{1}{n}}^\circledast \le \overbrace{\sum_{k=1}^{n-1} \frac{1}{k^2}}^{\spadesuit} +\frac{1}{n}. \end{equation*}
By comparison with the convergent series $\spadesuit$, we find that the original sequence converges. |
What is a solution to the differential equation dy/dt=e^t(y-1)^2?
Apr 4, 2018
The General Solution is:
$y = 1 - \frac{1}{{e}^{t} + C}$
Explanation:
We have:
$\frac{\mathrm{dy}}{\mathrm{dt}} = {e}^{t} {\left(y - 1\right)}^{2}$
We can collect terms for similar variables:
$\frac{1}{y - 1} ^ 2 \setminus \frac{\mathrm{dy}}{\mathrm{dt}} = {e}^{t}$
Which is a separable First Order Ordinary non-linear Differential Equation, so we can "separate the variables" to get:
$\int \setminus \frac{1}{y - 1} ^ 2 \setminus \mathrm{dy} = \int {e}^{t} \setminus \mathrm{dt}$
Both integrals are those of standard functions, so we can use that knowledge to directly integrate:
$- \frac{1}{y - 1} = {e}^{t} + C$
And we can readily rearrange for $y$:
$- \left(y - 1\right) = \frac{1}{{e}^{t} + C}$
$\therefore 1 - y = \frac{1}{{e}^{t} + C}$
$y = 1 - \frac{1}{{e}^{t} + C}$
Apr 4, 2018
$y = - \frac{1}{{e}^{t} + C} + 1$
Explanation:
This is a separable differential equation, which means it can be written in the form:
$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot f \left(y\right) = g \left(x\right)$
It can be solved by integrating both sides:
$\int \setminus f \left(y\right) \setminus \mathrm{dy} = \int \setminus g \left(x\right) \setminus \mathrm{dx}$
In our case, we first need to separate the integral into the right form. We can do this by dividing both sides by ${\left(y - 1\right)}^{2}$:
$\frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{1}{y - 1} ^ 2 = {e}^{t} \cancel{{\left(y - 1\right)}^{2} / {\left(y - 1\right)}^{2}}$
$\frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{1}{y - 1} ^ 2 = {e}^{t}$
Now we can integrate both sides:
$\int \setminus \frac{1}{y - 1} ^ 2 \setminus \mathrm{dy} = \int \setminus {e}^{t} \setminus \mathrm{dt}$
$\int \setminus \frac{1}{y - 1} ^ 2 \setminus \mathrm{dy} = {e}^{t} + {C}_{1}$
We can solve the left hand integral with a substitution of $u = y - 1$:
$\int \setminus \frac{1}{u} ^ 2 \setminus \mathrm{du} = {e}^{t} + {C}_{1}$
$\int \setminus {u}^{-} 2 \setminus \mathrm{du} = {e}^{t} + {C}_{1}$
${u}^{-} \frac{1}{- 1} + {C}_{2} = {e}^{t} + {C}_{1}$
Resubstituting (and combining constants) gives:
$- \frac{1}{y - 1} = {e}^{t} + {C}_{3}$
Multiply both sides by $y - 1$:
$- 1 = \left({e}^{t} + {C}_{3}\right) \left(y - 1\right)$
Divide both sides by ${e}^{t} + {C}_{3}$:
$- \frac{1}{{e}^{t} + {C}_{3}} = y - 1$
$y = - \frac{1}{{e}^{t} + C} + 1$ |
# Symmetric group of an equilateral triangle.
In Symmetry of regular polygons, we found all the symmetries of regular polygons. Today, we will look at what happens when we define the composition of such symmetries and the properties this set will have.
To begin with, recall that for the equilateral triangle we had the symmetries R0, R120, R240, F1, F2 and F3, where Ra is a rotation about the center of the triangle by an angle of a degrees, and Fa is a flip over the vertex labeled a. As a reminder of how we labeled the vertices, we have a picture of our triangle below. Note that the labels are number counterclockwise.
Our goal now is to figure out what happens if we perform two of these actions in a row, so let’s look at an example. Suppose that we rotate the triangle by 120 degrees, we then have
We then rotate the triangle again another 120 degrees and the triangle now ends up like
Note that if we had started out rotating 240 degrees, the resulting picture would be the same. In this way, we say that R120oR120=R240, that is the composition of the two rotations of 120 degrees is a rotation of 240 degrees.
As another example, if we flip over the vertex labeled 1, we end up with
If we then flip over the vertex 1 again, we end up with
Which is the same as doing nothing, so we would say F1oF1=R0. In order to find the composition of different symmetries, we note that while would say the left of the right these means we first put the triangle in the the right one, then put it into the second one. To see this, we want to find R120oF1, so, we first flip the triangle over the first vertex and get
If we now take the resulting picture and rotate it 120 degrees, we will end up with the picture
This is the same end result as if we had flipped the original triangle over the 2 vertex, so we say that R120oF1=F2. If we now find F1oR120, we then rotate 120 degrees and get the picture
After this we will flip the triangle over the 1 vertex. Note that by the 1 vertex, however, we are referring to the numbering in the original picture. That is, the vertex 2 is now where the vertex 1 was, so this is the vertex we flip over. Doing so, we end up with
We no see that this is the same as flipping over the vertex 3, so we get that F1oR120=F3. I want to point out that the reason I took the time to discuss the order of symmetries is because if we change the order in which we perform the symmetries, we end up with different answers in this case. In the classroom with students I would do these examples, then I would leave it up to the students to find the rest of the compositions. In order to keep track of the results, I would have the students fill out a Cayley (multiplication) table. When using the table, the top row is the right entry and the left column is the left entry when determining the order to multiply. Therefore, we would do the top symmetry first followed by the left symmetry. I give the completed table below.
Now that we have the multiplication table, we can look at some properties of this set up composition. There a few things to note;
1. The composition of two symmetries is again a symmetry.
2. The composition of symmetries (AoB)oC=Ao(BoC).
3. There is a symmetry, R0, such that R0oA=AoR0=A.
4. For every symmetry, A, there is a symmetry, B, such that AoB=BoA=R0.
5. There exists symmetries A and B such that AoB is not equal to BoA.
The property 1 states that the set of symmetries is closed under the binary operation of composition, 2 states that associativity holds, 3 states that an identity exists, 4 states that every symmetry has an inverse and 5 states that the composition is not commutative in general. The first four properties tell us that these symmetries form what is called a group. I would note that if we look at the integers under addition, then these four properties also hold where the identity is 0 and the inverse of any number k is -k since k+(-k)=0. With the integers, however, we do have that a+b=b+a, so the fact that we don’t have commutativity in group of symmetries is something distinct that I like to point out for students. In fact, this is the smallest example of a group that does not have the commutative property.
Together with the fact that this process allows students to actively participate in the process and play with objects, I feel this is a great way to introduce people to sets of number and operations other than addition and multiplication. I have success with this project in general education math classes as well as upper level courses. There is about it that provides a fun insight and different view point to students that don’t spend a lot of time with math, whereas it provides ample opportunity for math majors to pursue deeper questions. I hope you had fun with this, and that, if you try this in class, your students enjoy it as well.
Next time in the symmetry series we will look at the symmetry groups of all polygons. While we took the time to look at compositions of symmetries individually, but in the next post we will look at things in a more general way. This will allow us to find what the symmetric group will look like for any polygon. I hope you’ll join me for that journey as well in a couple days. Make sure to follow the blog to be updated when posts are made.
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# Find the area of the quadrilateral whose vertices taken in order, are $\left( { - 4, - 2} \right),\left( { - 3, - 5} \right),\left( { - 3, - 2} \right){\text{ and }}\left( {2,3} \right)$.
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Hint- To solve this question we will use the formula given by $\Delta = \dfrac{1}{2}\left| {{y_1}({x_2} - {x_3}) + {y_2}({x_3} - {x_1}) + {y_3}({x_1} - {x_2})} \right|$ , we will divide the quadrilateral into two triangles and then apply the formula.
Complete step-by-step solution -
Given that vertices of quadrilateral are $\left( { - 4, - 2} \right),\left( { - 3, - 5} \right),\left( { - 3, - 2} \right){\text{ and }}\left( {2,3} \right)$
Let $A\left( { - 4, - 2} \right),B\left( { - 3, - 5} \right),C\left( { - 3, - 2} \right){\text{ and D}}\left( {2,3} \right)$ be the vertices of the quadrilateral.
As we know that the area of the quadrilateral ABCD= area of $\Delta ABC$ + area of $\Delta ACD$
Area of $\Delta ABC$ is given by
$= \dfrac{1}{2}\left| {{y_1}({x_2} - {x_3}) + {y_2}({x_3} - {x_1}) + {y_3}({x_1} - {x_2})} \right|$
In $\Delta ABC$
${x_1} = - 4,{x_2} = - 3,{x_3} = 3,{y_1} = - 2,{y_2} = - 5,{y_3} = - 2$
Substituting above values in the formula, we get
$= \dfrac{1}{2}\left| { - 2( - 3 - 3) - 5(3 + 4) - 2( - 4 + 3)} \right| \\ = \dfrac{1}{2}\left| { - 2( - 6) - 5(7) - 2( - 1)} \right| \\ = \dfrac{1}{2}\left| {12 - 35 + 2} \right| \\ = \dfrac{1}{2}\left| { - 21} \right| \\ = 10.5{\text{ sq}}{\text{.units}} \\$
Area of $\Delta ACD$ is given by
$= \dfrac{1}{2}\left| {{y_1}({x_2} - {x_3}) + {y_2}({x_3} - {x_1}) + {y_3}({x_1} - {x_2})} \right|$
In $\Delta ACD$
${x_1} = - 4,{x_2} = 3,{x_3} = 2,{y_1} = - 2,{y_2} = - 2,{y_3} = 3$
Substituting above values in the formula, we get
$= \dfrac{1}{2}\left| { - 2( - 3 - 2) - 2(2 + 4) - 3( - 4 - 3)} \right| \\ = \dfrac{1}{2}\left| { - 2 - 12 - 21} \right| \\ = \dfrac{1}{2}\left| { - 35} \right| \\ = 17.5{\text{ sq}}{\text{.units}} \\$
Therefore, the area of the quadrilateral is = $10.5 + 17.5 = 28{\text{ sq}}{\text{. units}}$
Hence, the area of the quadrilateral is $28.0{\text{ sq}}{\text{. units}}$
Note- The given problem is the coordinate geometry problem and the vertices of the quadrilateral are given. In order to solve such questions, try to break the quadrilateral in two triangles and apply the formula to calculate the area of the triangle. |
# What is cos (2 arcsin (3/5))?
Jul 21, 2015
$\frac{7}{25}$
#### Explanation:
First consider that : $\epsilon = \arcsin \left(\frac{3}{5}\right)$
$\epsilon$ simply represents an angle.
This means that we are looking for color(red)cos(2epsilon)!
If $\epsilon = \arcsin \left(\frac{3}{5}\right)$ then,
$\implies \sin \left(\epsilon\right) = \frac{3}{5}$
To find $\cos \left(2 \epsilon\right)$ We use the identity : $\cos \left(2 \epsilon\right) = 1 - 2 {\sin}^{2} \left(\epsilon\right)$
$\implies \cos \left(2 \epsilon\right) = 1 - 2 \cdot {\left(\frac{3}{5}\right)}^{2} = \frac{25 - 18}{25} = \textcolor{b l u e}{\frac{7}{25}}$
Jul 22, 2015
We have:
$y = \cos \left(2 \arcsin \left(\frac{3}{5}\right)\right)$
I will do something similar to Antoine's method, but expand on it.
Let $\arcsin \left(\frac{3}{5}\right) = \theta$
$y = \cos \left(2 \theta\right)$
$\theta = \arcsin \left(\frac{3}{5}\right)$
$\sin \theta = \frac{3}{5}$
Using the identity $\cos \left(\theta + \theta\right) = {\cos}^{2} \theta - {\sin}^{2} \theta$, we then have:
$\cos \left(2 \theta\right) = \left(1 - {\sin}^{2} \theta\right) - {\sin}^{2} \theta = 1 - 2 {\sin}^{2} \theta$
(I didn't remember the result, so I just derived it)
$= 1 - 2 {\left\{\sin \left[\arcsin \left(\frac{3}{5}\right)\right]\right\}}^{2}$
$= 1 - 2 {\left(\frac{3}{5}\right)}^{2}$
$= \frac{25}{25} - 2 \left(\frac{9}{25}\right)$
$= \frac{25}{25} - \frac{18}{25} = \textcolor{b l u e}{\frac{7}{25}}$ |
In mathematics, a synthetic division is a well-known technique in algebraic equations to do polynomial division in a simplified way. It is a technique to divide the polynomials with the help of the coefficients of the polynomials.
In algebra, this process is used to simplify the process of polynomial division with the system of coefficients. Polynomials are algebraic equations in which the variables are raised to the power and take its product by coefficients.
In this article, we’ll learn and explain the term synthetic division with the help of examples.
In algebra, the synthetic division is a method for dividing polynomials with the help of its factors. This algebraic equation technique is also known as the short division method. The main purpose of synthetic division is to simplify the process of polynomial long division.
It is the shortened method to find the quotient and remainder of the division of polynomials as the polynomial division is used frequently in algebra for various purposes. When the linear factors are available then this technique is used to get the results in a few calculations by dividing the coefficients of a polynomial by the linear factor.
The formula of synthetic division is:
In mathematics, the synthetic division is a faster way to divide the polynomial as compared to the polynomial long division method. Here are a few steps to perform synthetic division.
### First step
First of all, arrange the terms of the given polynomials from least to greatest order. Keep one thing in mind, if there is any missing term while writing a term from larger to smaller write the missing power of x in the polynomial filled with zeros. Such as
A polynomial expression 3x4 – 3x2 + 2x3 + 1 divided by x – 2 is given then write it in descending order and write the missing term.
3x4 – 3x2 + 2x3 + 1 will be written as 3x4 + 2x3 – 3x2 + 0x + 1
Dividend = 3x4 + 2x3 – 3x2 + 0x + 1
Divisor = x – 2
Write the constant coefficient of the polynomial expression inside the division symbol.
### Second step
Now write the divisor outside the symbol of division. Such as the given divisor is x – 2 then you have to take this divisor equal to zero and find the value of x that will be written outside the notation of division. Such as x – 2 = 0 → x = 2
### Third step
Multiply the divisor “2” by the leading coefficient of the given polynomial “3”, and write the result under the next coefficient in the polynomial.
### Fourth step
Add the multiplied result to the 2nd coefficient of the expression and write the sum below it while writing the leading coefficient as it is.
### Fifth step
Continue to multiply the term and add it to the next term until you have covered all the coefficients. In other words, repeat the step third and fourth until the last coefficient.
### Sixth step
Write the final result, with the remainder on top and the coefficients of the quotient below.
Q(x) = 3x3 + 8x2 + 13x + 26
R = 53
Hence
## Synthetic DivisionBenefits
Synthetic division is a well-known method in algebraic expressions that has many benefits over other methods of dividing polynomials.
## Examples of synthetic division
Here are a few examples of synthetic division.
Example 1
Divide the polynomial P(x) = 5x3 + 3x4 + 4x5 – 2x + 6x2 + 12 by (x – 3).
Solution
Step 1: Arrange the polynomial in descending order and write their coefficients.
P(x) = 4x5 + 3x4 + 5x3 + 6x2– 2x + 12
Coefficients of polynomial = 4, 3, 5, 6, -2, 12
Step 2:Now take the linear factor and find the value of the unknown.
x – 3 = 0
x = 3
Step 3:Multiply the divisor “3” by the leading coefficient of the given polynomial “4”, and write the result under the next coefficient in the polynomial.
Step 4:Add the multiplied result to the 2nd coefficient of the expression and write the sum below it while writing the leading coefficient as it is.
Step 5:Now repeat steps 3 & 4 to complete the division.
Step 6: Write the final result, with the remainder on top and the coefficients of the quotient below.
Q(x) = 4x4 + 15x3 + 41x2 + 129x + 385
R = 1167
Hence,
P(x)/ (x – a) = Q(x) + R/ (x – a)
[5x3 + 3x4 + 4x5 – 2x + 6x2 + 12] / (x – 3) = [4x4 + 15x3 + 41x2 + 129x + 385] + 1167/(x – 3)
A synthetic division calculator is an alternative way to solve the problems of polynomial division without involving into lengthy calculations.
Example 2
Divide the polynomial P(y) = 2y2 – 4y3 – y – 11 by (y – 1).
Solution
Step 1: Arrange the polynomial in descending order and write their coefficients.
P(y) = – 4y3 + 2y2– y – 11
Coefficients of polynomial = -4, 2, -1, -11
Step 2:Now take the linear factor and find the value of the unknown.
y – 1 = 0
y = 1
Step 3:Multiply the divisor “1” by the leading coefficient of the given polynomial “-4”, and write the result under the next coefficient in the polynomial.
Step 4: Add the multiplied result to the 2nd coefficient of the expression and write the sum below it while writing the leading coefficient as it is.
Step 5: Now repeat steps 3 & 4 to complete the division.
Step 6: Write the final result, with the remainder on top and the coefficients of the quotient below.
Q(x) = -4y2 – 2y – 3
R = -14
Hence,
P(x)/ (x – a) = Q(x) + R/ (x – a)
[-4y3 + 2y2 – y – 11] / (y – 1) = [-4y2 – 2y – 3] – 14/(y – 1)
## Conclusion
Now you can take assistance from this post to learn the basics and calculations of the synthetic division to calculate the quotient and remainder of polynomial division. There is also some solved examples available above to understand the topic briefly.
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Partial fraction decomposition (PFD) is a method for solving simultaneous equations. It gives the solution of A * B = C in terms of A and B, and C = A * B. If we have two equations, A * B = C and A + B = C, then PFD gives us an equation of the form (A * B) - (A + B) = 0. The PFD algorithm solves the system by finding a solution to the following equation: A(B - C) = 0 This can be expressed as a simpler equation in terms of partial fractions as: B - C / A(B - C) = 0 This solution is called a "mixed" or "mixed-order" solution. Mixed-order solutions typically have less accuracy than higher-order solutions, but are much faster to compute. The PFD solver computes mixed-order solutions based on an interpolation scheme that interpolates between values of a function at points where it crosses zero. This scheme makes the second derivative zero on these points, and therefore the interpolant will be quadratic on these points. These points are computed iteratively so that they become increasingly accurate while computing time is reduced. Typically, linear systems like this are solved by double-differencing or Taylor's series expansion to approximate the second derivative term at
Elimination equations are a type of math problem in which you have to find the solution that leaves the least number of equations. They are often used when you have to find the minimum or maximum value for one variable after another variable has been changed. There are four types of elimination equations: Linear: One variable is raised to a power, and the other variables are multiplied by it. For example, if one variable is raised to the power 3 and another to the power 8, then the resulting equation would be (3x8) = 64. The solution would be 32. Square: Two variables are multiplied. For example, if one variable is squared (or raised to 4) and another is squared (or raised to 4), then their resulting product is 16. The solution would be 8. Cubed: Three variables are multiplied. For example, if one variable is cubed (i.e., raised to 8) and another is cubed (i.e., raised to 8), then their resulting product is 56. The solution would be 40. To solve an elimination equation, you first need to identify which equation needs solving. Then you need to identify all of the variables involved in that equation and their values at each step in your problem, such as x1 = 1, x2 = 2, x3 = 4, … . This will allow you to
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# Marbles And Ratio Math
### The angles of a triangle are in the ratio 1 3 8.
Marbles and ratio math.
The required ratio is given by. If the ratio of the red marbles to the blue marbles is the same for both john and jane then john has how many more blue marbles than jane. Kevin has a jar where he keeps 15 marbles out of which 5 are red 3 are green 4 are blue and the rest are pink. In other words 12 cups of flour and 8 cups of milk.
Find the ratio of pink marbles to all the marbles. Write the ratio of the number of blue marbles to the total number of marbles in terms of r b and w. 2 4 12. So 11 x 3 33 and 4 7 x 3 is 12 21 4 3 12 7 3 21.
If there are 18 boys in the class how many girls are in that class. Jane has 20 marbles all of them either red or blue. The ratio is still the same so the pancakes should be just as yummy. The perimeter of a rectangle is equal to 280 meters.
Solution the total number of marbles is r b w the total ratio of blue marbles to the total number of marbles is r r b w. To make pancakes for a lot of people we might need 4 times the quantity so we multiply the numbers by 4. We get the ratio from john. There are r red marbles b blue marbles and w white marbles in a bag.
Find the area of the rectangle. If there are 42 marbles in the bag how many of the marbles are red. So the ratio of flour to milk is 3. The ratio of its length to its width is 5 2.
Write the ratio of the number of blue marbles to the total number of marbles in terms of r b and w. The ratio s total starts at 11 4 7 the total of the ratio 4 7 and in order for the grand total to be 33 you need to multiply 11 by 3 and therefor the ratio that was added to make it. There are a total of 15 marbles in the jar out of which 3 are pink marbles. In level 1 the problems ask for a specific ratio such as noah drew 9 hearts 6 stars and 12 circles.
There are r red marbles b blue marbles and w white marbles in a bag. Let x blue marbles for jane 20 x red marbles for jane.
### Problem Sums Part V Maths Sums Math Primary
Source : pinterest.com |
# Problem of the Week Problem E and Solution Medians
## Problem
In $$\triangle ABC$$, $$\angle ABC = 90^{\circ}$$. A median is drawn from $$A$$ to side $$BC$$, meeting $$BC$$ at $$M$$ such that $$AM=5$$. A second median is drawn from $$C$$ to side $$AB$$, meeting $$AB$$ at $$N$$ such that $$CN=2\sqrt{10}$$.
Determine the length of the longest side of $$\triangle ABC$$.
Note: In a triangle, a median is a line segment drawn from a vertex of the triangle to the midpoint of the opposite side.
## Solution
Since $$AM$$ is a median, $$M$$ is the midpoint of $$BC$$. Then $$BM=MC$$. Let $$BM=MC=y$$.
Since $$CN$$ is a median, $$N$$ is the midpoint of $$AB$$. Then $$AN=NB$$. Let $$AN=NB=x$$.
Since $$\angle B=90^{\circ}$$, $$\triangle NBC$$ is a right-angled triangle. Using the Pythagorean Theorem, \begin{align*} NB^2+BC^2&=CN^2\\ x^2+(2y)^2&=(2\sqrt{10})^2\\ x^2+4y^2&=40\tag{1}\end{align*}
Since $$\angle B=90^{\circ}$$, $$\triangle ABM$$ is a right-angled triangle. Using the Pythagorean Theorem, \begin{align*} AB^2+BM^2&=AM^2\\ (2x)^2+y^2&=5^2\\ 4x^2+y^2&=25 \tag{2}\end{align*}
Adding equations $$(1)$$ and $$(2)$$, we get $$5x^2+5y^2=65$$ or $$x^2+y^2=13$$.
The longest side of $$\triangle ABC$$ is the hypotenuse $$AC$$. Using the Pythagorean Theorem, \begin{aligned} AC^2&=AB^2+BC^2\\ &=(2x)^2+(2y)^2\\ &=4x^2+4y^2\\ &=4(x^2+y^2)\end{aligned}
Since $$x^2 + y^2 =13$$, we have $$AC^2 = 4(13)$$. And since $$AC>0$$, $$AC=2\sqrt{13}$$ follows.
Therefore, the length of the longest side of $$\triangle ABC$$ is $$2\sqrt{13}$$.
Note: The solver could have instead solved a system of equations to find $$x=2$$ and $$y=3$$, and then proceed to solve for the longest side. The above approach was provided to expose the solver to alternate way to think about the solution of this problem. |
# Periodic continued fraction
In mathematics, an infinite periodic continued fraction is a continued fraction that can be placed in the form
${\displaystyle x=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{\quad \ddots \quad a_{k}+{\cfrac {1}{a_{k+1}+{\cfrac {\ddots }{\quad \ddots \quad a_{k+m-1}+{\cfrac {1}{a_{k+m}+{\cfrac {1}{a_{k+1}+{\cfrac {1}{a_{k+2}+{\ddots }}}}}}}}}}}}}}}}}}$
where the initial block of k + 1 partial denominators is followed by a block [ak+1ak+2,...ak+m] of partial denominators that repeats ad infinitum. For example, ${\displaystyle {\sqrt {2}}}$ can be expanded to a periodic continued fraction, namely as [1,2,2,2,...].
The partial denominators {ai} can in general be any real or complex numbers. That general case is treated in the article convergence problem. The remainder of this article is devoted to the subject of simple continued fractions that are also periodic. In other words, the remainder of this article assumes that all the partial denominators ai (i ≥ 1) are positive integers.
## Purely periodic and periodic fractions
Since all the partial numerators in a regular continued fraction are equal to unity we can adopt a shorthand notation in which the continued fraction shown above is written as
{\displaystyle {\begin{aligned}x&=[a_{0};a_{1},a_{2},\dots ,a_{k},a_{k+1},a_{k+2},\dots ,a_{k+m},a_{k+1},a_{k+2},\dots ,a_{k+m},\dots ]\\&=[a_{0};a_{1},a_{2},\dots ,a_{k},{\overline {a_{k+1},a_{k+2},\dots ,a_{k+m}}}]\end{aligned}}}
where, in the second line, a vinculum marks the repeating block.[1] Some textbooks use the notation
{\displaystyle {\begin{aligned}x&=[a_{0};a_{1},a_{2},\dots ,a_{k},{\dot {a}}_{k+1},a_{k+2},\dots ,{\dot {a}}_{k+m}]\end{aligned}}}
where the repeating block is indicated by dots over its first and last terms.[2]
If the initial non-repeating block is not present – that is, if k = -1, a0 = am and
${\displaystyle x=[{\overline {a_{0};a_{1},a_{2},\dots ,a_{m-1}}}],}$
the regular continued fraction x is said to be purely periodic. For example, the regular continued fraction for the golden ratio φ – given by [1; 1, 1, 1, ...] – is purely periodic, while the regular continued fraction for the square root of two – [1; 2, 2, 2, ...] – is periodic, but not purely periodic.
## As unimodular matrices
Such periodic fractions are in one-to-one correspondence with the real quadratic irrationals. The correspondence is explicitly provided by Minkowski's question-mark function. That article also reviews tools that make it easy to work with such continued fractions. Consider first the purely periodic part
${\displaystyle x=[0;{\overline {a_{1},a_{2},\dots ,a_{m}}}],}$
This can, in fact, be written as
${\displaystyle x={\frac {\alpha x+\beta }{\gamma x+\delta }}}$
with the ${\displaystyle \alpha ,\beta ,\gamma ,\delta }$ being integers, and satisfying ${\displaystyle \alpha \delta -\beta \gamma =1.}$ Explicit values can be obtained by writing
${\displaystyle S={\begin{pmatrix}1&0\\1&1\end{pmatrix}}}$
which is termed a "shift", so that
${\displaystyle S^{n}={\begin{pmatrix}1&0\\n&1\end{pmatrix}}}$
and similarly a reflection, given by
${\displaystyle T\mapsto {\begin{pmatrix}-1&1\\0&1\end{pmatrix}}}$
so that ${\displaystyle T^{2}=I}$. Both of these matrices are unimodular, arbitrary products remain unimodular. Then, given ${\displaystyle x}$ as above, the corresponding matrix is of the form [3]
${\displaystyle S^{a_{1}}TS^{a_{2}}T\cdots TS^{a_{m}}={\begin{pmatrix}\alpha &\beta \\\gamma &\delta \end{pmatrix}}}$
and one has
${\displaystyle x=[0;{\overline {a_{1},a_{2},\dots ,a_{m}}}]={\frac {\alpha x+\beta }{\gamma x+\delta }}}$
as the explicit form. As all of the matrix entries are integers, this matrix belongs to the modular group ${\displaystyle SL(2,\mathbb {Z} ).}$
A quadratic irrational number is an irrational real root of the quadratic equation
${\displaystyle ax^{2}+bx+c=0}$
where the coefficients a, b, and c are integers, and the discriminant, b2 − 4ac, is greater than zero. By the quadratic formula every quadratic irrational can be written in the form
${\displaystyle \zeta ={\frac {P+{\sqrt {D}}}{Q}}}$
where P, D, and Q are integers, D > 0 is not a perfect square (but not necessarily square-free), and Q divides the quantity P2 − D (for example (6+8)/4). Such a quadratic irrational may also be written in another form with a square-root of a square-free number (for example (3+2)/2) as explained for quadratic irrationals.
By considering the complete quotients of periodic continued fractions, Euler was able to prove that if x is a regular periodic continued fraction, then x is a quadratic irrational number. The proof is straightforward. From the fraction itself, one can construct the quadratic equation with integral coefficients that x must satisfy.
Lagrange proved the converse of Euler's theorem: if x is a quadratic irrational, then the regular continued fraction expansion of x is periodic.[4] Given a quadratic irrational x one can construct m different quadratic equations, each with the same discriminant, that relate the successive complete quotients of the regular continued fraction expansion of x to one another. Since there are only finitely many of these equations (the coefficients are bounded), the complete quotients (and also the partial denominators) in the regular continued fraction that represents x must eventually repeat.
## Reduced surds
The quadratic surd ${\displaystyle \zeta ={\frac {P+{\sqrt {D}}}{Q}}}$ is said to be reduced if ${\displaystyle \zeta >1}$ and its conjugate ${\displaystyle \eta ={\frac {P-{\sqrt {D}}}{Q}}}$ satisfies the inequalities ${\displaystyle -1<\eta <0}$. For instance, the golden ratio ${\displaystyle \phi =(1+{\sqrt {5}})/2=1.618033...}$ is a reduced surd because it is greater than one and its conjugate ${\displaystyle (1-{\sqrt {5}})/2=-0.618033...}$ is greater than −1 and less than zero. On the other hand, the square root of two ${\displaystyle {\sqrt {2}}=(0+{\sqrt {8}})/2}$ is greater than one but is not a reduced surd because its conjugate ${\displaystyle -{\sqrt {2}}=(0-{\sqrt {8}})/2}$ is less than −1.
Galois proved that the regular continued fraction which represents a quadratic surd ζ is purely periodic if and only if ζ is a reduced surd. In fact, Galois showed more than this. He also proved that if ζ is a reduced quadratic surd and η is its conjugate, then the continued fractions for ζ and for (−1/η) are both purely periodic, and the repeating block in one of those continued fractions is the mirror image of the repeating block in the other. In symbols we have
{\displaystyle {\begin{aligned}\zeta &=[{\overline {a_{0};a_{1},a_{2},\dots ,a_{m-1}}}]\\[3pt]{\frac {-1}{\eta }}&=[{\overline {a_{m-1};a_{m-2},a_{m-3},\dots ,a_{0}}}]\,\end{aligned}}}
where ζ is any reduced quadratic surd, and η is its conjugate.
From these two theorems of Galois a result already known to Lagrange can be deduced. If r > 1 is a rational number that is not a perfect square, then
${\displaystyle {\sqrt {r}}=[a_{0};{\overline {a_{1},a_{2},\dots ,a_{2},a_{1},2a_{0}}}].}$
In particular, if n is any non-square positive integer, the regular continued fraction expansion of n contains a repeating block of length m, in which the first m − 1 partial denominators form a palindromic string.
## Length of the repeating block
By analyzing the sequence of combinations
${\displaystyle {\frac {P_{n}+{\sqrt {D}}}{Q_{n}}}}$
that can possibly arise when ζ = (P + D)/Q is expanded as a regular continued fraction, Lagrange showed that the largest partial denominator ai in the expansion is less than 2D, and that the length of the repeating block is less than 2D.
More recently, sharper arguments [5][6][7] based on the divisor function have shown that L(D), the length of the repeating block for a quadratic surd of discriminant D, is given by
${\displaystyle L(D)={\mathcal {O}}({\sqrt {D}}\ln {D})}$
where the big O means "on the order of", or "asymptotically proportional to" (see big O notation).
### Canonical form and repetend
The following iterative algorithm [8] can be used to obtain the continued fraction expansion in canonical form (S is any natural number that is not a perfect square):
${\displaystyle m_{0}=0\,\!}$
${\displaystyle d_{0}=1\,\!}$
${\displaystyle a_{0}=\left\lfloor {\sqrt {S}}\right\rfloor \,\!}$
${\displaystyle m_{n+1}=d_{n}a_{n}-m_{n}\,\!}$
${\displaystyle d_{n+1}={\frac {S-m_{n+1}^{2}}{d_{n}}}\,\!}$
${\displaystyle a_{n+1}=\left\lfloor {\frac {{\sqrt {S}}+m_{n+1}}{d_{n+1}}}\right\rfloor =\left\lfloor {\frac {a_{0}+m_{n+1}}{d_{n+1}}}\right\rfloor \!.}$
Notice that mn, dn, and an are always integers. The algorithm terminates when this triplet is the same as one encountered before. The algorithm can also terminate on ai when ai = 2 a0,[9] which is easier to implement.
The expansion will repeat from then on. The sequence [a0; a1, a2, a3, ...] is the continued fraction expansion:
${\displaystyle {\sqrt {S}}=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}+\,\ddots }}}}}}}$
#### Example
To obtain 114 as a continued fraction, begin with m0 = 0; d0 = 1; and a0 = 10 (102 = 100 and 112 = 121 > 114 so 10 chosen).
{\displaystyle {\begin{aligned}{\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{aligned}}}
${\displaystyle m_{1}=d_{0}\cdot a_{0}-m_{0}=1\cdot 10-0=10\,.}$
${\displaystyle d_{1}={\frac {S-m_{1}^{2}}{d_{0}}}={\frac {114-10^{2}}{1}}=14\,.}$
${\displaystyle a_{1}=\left\lfloor {\frac {a_{0}+m_{1}}{d_{1}}}\right\rfloor =\left\lfloor {\frac {10+10}{14}}\right\rfloor =\left\lfloor {\frac {20}{14}}\right\rfloor =1\,.}$
So, m1 = 10; d1 = 14; and a1 = 1.
${\displaystyle {\frac {{\sqrt {114}}+10}{14}}=1+{\frac {{\sqrt {114}}-4}{14}}=1+{\frac {114-16}{14({\sqrt {114}}+4)}}=1+{\frac {1}{\frac {{\sqrt {114}}+4}{7}}}.}$
Next, m2 = 4; d2 = 7; and a2 = 2.
${\displaystyle {\frac {{\sqrt {114}}+4}{7}}=2+{\frac {{\sqrt {114}}-10}{7}}=2+{\frac {14}{7({\sqrt {114}}+10)}}=2+{\frac {1}{\frac {{\sqrt {114}}+10}{2}}}.}$
${\displaystyle {\frac {{\sqrt {114}}+10}{2}}=10+{\frac {{\sqrt {114}}-10}{2}}=10+{\frac {14}{2({\sqrt {114}}+10)}}=10+{\frac {1}{\frac {{\sqrt {114}}+10}{7}}}.}$
${\displaystyle {\frac {{\sqrt {114}}+10}{7}}=2+{\frac {{\sqrt {114}}-4}{7}}=2+{\frac {98}{7({\sqrt {114}}+4)}}=2+{\frac {1}{\frac {{\sqrt {114}}+4}{14}}}.}$
${\displaystyle {\frac {{\sqrt {114}}+4}{14}}=1+{\frac {{\sqrt {114}}-10}{14}}=1+{\frac {14}{14({\sqrt {114}}+10)}}=1+{\frac {1}{\frac {{\sqrt {114}}+10}{1}}}.}$
${\displaystyle {\frac {{\sqrt {114}}+10}{1}}=20+{\frac {{\sqrt {114}}-10}{1}}=20+{\frac {14}{{\sqrt {114}}+10}}=20+{\frac {1}{\frac {{\sqrt {114}}+10}{14}}}.}$
Now, loop back to the second equation above.
Consequently, the simple continued fraction for the square root of 114 is
${\displaystyle {\sqrt {114}}=[10;{\overline {1,2,10,2,1,20}}].\,}$ (sequence A010179 in the OEIS)
114 is approximately 10.67707 82520. After one expansion of the repetend, the continued fraction yields the rational fraction ${\displaystyle {\frac {21194}{1985}}}$ whose decimal value is approx. 10.67707 80856, a relative error of 0.0000016% or 1.6 parts in 100,000,000.
#### Generalized continued fraction
A more rapid method is to evaluate its generalized continued fraction. From the formula derived there:
${\displaystyle {\sqrt {z}}={\sqrt {x^{2}+y}}=x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {y}{2x+\ddots }}}}}}=x+{\cfrac {2x\cdot y}{2(2z-y)-y-{\cfrac {y^{2}}{2(2z-y)-{\cfrac {y^{2}}{2(2z-y)-\ddots }}}}}}}$
and the fact that 114 is 2/3 of the way between 102=100 and 112=121 results in
${\displaystyle {\sqrt {114}}={\cfrac {\sqrt {1026}}{3}}={\cfrac {\sqrt {32^{2}+2}}{3}}={\cfrac {32}{3}}+{\cfrac {2/3}{64+{\cfrac {2}{64+{\cfrac {2}{64+{\cfrac {2}{64+\ddots }}}}}}}}={\cfrac {32}{3}}+{\cfrac {2}{192+{\cfrac {18}{192+{\cfrac {18}{192+\ddots }}}}}},}$
which is simply the aforementioned [10;1,2, 10,2,1, 20,1,2] evaluated at every third term. Combining pairs of fractions produces
${\displaystyle {\sqrt {114}}={\cfrac {\sqrt {32^{2}+2}}{3}}={\cfrac {32}{3}}+{\cfrac {64/3}{2050-1-{\cfrac {1}{2050-{\cfrac {1}{2050-\ddots }}}}}}={\cfrac {32}{3}}+{\cfrac {64}{6150-3-{\cfrac {9}{6150-{\cfrac {9}{6150-\ddots }}}}}},}$
which is now ${\displaystyle [10;1,2,{\overline {10,2,1,20,1,2}}]}$ evaluated at the third term and every six terms thereafter.
## Notes
1. ^ Pettofrezzo & Byrkit 1970, p. 158.
2. ^ Long 1972, p. 187.
3. ^
4. ^ Davenport 1982, p. 104.
5. ^
6. ^
7. ^
8. ^
9. ^
## References
• Beceanu, Marius (5 February 2003). "Period of the Continued Fraction of sqrt(n)" (PDF). Theorem 2.3. Archived from the original (PDF) on 21 December 2015. Retrieved 3 May 2022.
• Long, Calvin T. (1972). Elementary Introduction to Number Theory (3 Sub ed.). Waveland Pr Inc. LCCN 77-171950. |
# Monthly Archives: January 2018
## Mobile Algebra
Following the introductory use of structure and emoji math to introduce systems, my teaching partner and I continued with mobiles as suggested by the authors of “An Emoji is Worth a Thousand Variables.” EDC has this great website, SolveMe Mobiles, that has 200 mobile puzzles like this:
Each shape in this mobile has a value (or weight) and the total value (or weight) in this mobile is 60 (units). Go ahead and solve the mobile.
This mobile represents a system of four unknowns. Using traditional algebra symbols it might look like this:
A couple of those equations have just one variable, so it may not be quite as intimidating to look at the traditional symbols. On the other hand, the mobile shapes are just so accessible to everyone!
We needed to move our students away from systems that had one variable defined for them, though, and the SolveMe site, as great as it is, always includes some kind of hint. So we started to make up our own mobiles.
As first, students used a lot of educated guessing to solve the mobiles. Then there was a breakthrough.
Take a closer look at the left-hand mobile.
Students realized that they could “cross off” the same shapes on equal branches and the mobile would stay balanced. In the example above, you can “cross off” two triangles and one square. Whatever remains is equivalent, though it no longer totals 36. Therefore, two triangles equals one square. Using that relationship, some students then substituted two triangles for the one square in the left branch. Then they had a branch of 6 triangles with a total of 18. So, each triangle is worth 3. Other students used the same relationship to substitute one square for the two triangles in the right branch, resulting in a branch of 3 squares with a total of 18. So, each square is worth 3.
We were floored. We had never discussed the idea of substitution, but here it was, naturally arising from students reasoning about the structure in the mobile.
Looking closer at the center mobile, students used the same “cross out” method to find the relationship that 2 triangles equals 3 squares. If we’d been teaching the substitution method in a more traditional way, kids would have been pushed to figure out how much 1 triangle (or 1 square) was worth before making the substitution step. We knew substitution was happening here, but we didn’t invent this approach so we just followed closely to see where our students took us. Since 2 triangles equals 3 squares, some kids substituted 3 squares for the two triangles on the right branch of the mobile. Others made two substitutions of 6 squares for the 4 triangles on the left branch. Either way the result was a branch of 7 squares that totaled 14. It seemed quite natural to them.
What would you do with this one?
Next up: Moving to traditional symbols. The final (?) post of this saga.
Filed under #CCSS, BMTN, problem solving, teaching
## Emoji math
The October 2017 issue of Mathematics Teacher included the article “An Emoji is Worth a Thousand Variables,” by Tony McCaffrey and Percival G. Matthews. (Note that you need to be an NCTM member to access the article without purchasing it.) The authors introduced their students to systems by using sets of equations comprised of emojis, similar to those puzzles that are found on Facebook. Lots of people, including those who say they hate math or they aren’t good at math or, “I’m not a math person” will do puzzles like these. They get lots of likes, answers posted in the comments, and shares – probably because this doesn’t look much like math. Take a moment and solve the puzzle.
What if, instead of the emoji puzzle, I had posted this puzzle:
The two are actually equivalent, but the abstract nature of the second representation is enough to make our students who see themselves as not math people shut down.
My fall term teaching partner and I used the emoji approach with our students. We hadn’t discovered the website yet, so we gave our students this short worksheet. Many of our students had struggled with math prior to coming to Baxter Academy. They were in self-contained or pull-out special education settings or in pre-algebra classes in middle school. A few probably had something closer to algebra 1, but they certainly hadn’t solved systems of three or more equations. They had no problems understanding what the emoji puzzles were asking of them. They weren’t put off by the number of equations or the number of icons. They were able to explain their solution process clearly and with great detail.
In a follow-up exercise, asking students to make connections between the emoji representation and the more traditional representation, nearly 80% of my students saw and could articulate the direct connection between the icons and the variables.
If the emoji system is more engaging and more accessible, then why don’t we use more of them to introduce systems of equations? Is it because emoji systems seem to “dumb down” the mathematics? The authors of the article make the case against that view:
“The algebra represented in [the emoji system] is not dumbed down at all. Notice that the puzzle presents a linear system in three variables … First-year algebra students are generally not exposed to three-variable systems; indeed, when McCaffrey checked all the first-year algebra texts in his school’s faculty library, none included systems of three variables. Although McCaffrey’s students had never seen three-variable systems before this class, most found the puzzle intuitive enough to solve.”
It is important to transition from emojis to more formal algebra, but it’s not important to start with the most abstract representation – the one that leaves too many of our students behind.
Filed under #CCSS, BMTN, teaching
## Using Structure to Solve Equations
Last November at the ATMNE 2017 fall conference, I attended a session where Gail Burrill highlighted some TI-Nspire documents that helped kids build concepts about expressions and equations. Sure, they’re targeted at middle level, but you use the tools that your students need, right, not where you wish they were?
Anyway, the one that really caught my eye was called “Using Structure to Solve Equations.” It was the perfect activity at the perfect time because my students were kind of struggling with solving (what I thought) were some fairly simple equations. Their struggle was about trying to remember the steps they knew someone had showed them rather than trying to reason their way through the equation. Just before learning about this activity, I had found myself using these strategies. For example, to help a student solve an equation like 3x + 8 = 44, I found myself covering up the 3x and asking the student, “What plus 8 equals 44?” When they came up with 36, I would follow up with, “So 3x must equal 36. What times 3 equals 36?” This approach helped several students – they stopped trying to remember and began to reason.
There were too many students who were still trying to remember, though. Their solutions to equations involving parentheses like 6(x – 4) – 7 = 5 or 8(x+ 9) + 2(x + 9) = 150 included classic distributive property mistakes. The “Using Structure …” activity was just like my covering up part of the equation and asking the question. But each student could work on their own equation and at their own pace. The equations were randomly generated so each kid had something different to work with. Working through the “Using Structure … ” activity helped kids to stop and think for a moment rather than employing a rote procedure.
In a follow-up to the activity, I asked kids to explain why 8(x+ 9) + 2(x + 9) = 150 could be thought of as 10(x + 9) = 150. My favorite response was because “8 bunches of (x + 9) and 2 bunches of (x + 9) makes 10 bunches of (x + 9).” From there, they saw it as trivial to say that x = 6. These were students who had previously struggled with solving equations like 3x + 8 = 44 because they couldn’t remember the procedure. After “Using Structure …” they didn’t have to remember, they just had to think!
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# 6.2: Exterior Angles in Convex Polygons
Difficulty Level: At Grade Created by: CK-12
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What if you were given a twelve-sided regular polygon? How could you determine the measure of each of its exterior angles
### Exterior Angles in Convex Polygons
Recall that an exterior angle is an angle on the outside of a polygon and is formed by extending a side of the polygon.
As you can see, there are two sets of exterior angles for any vertex on a polygon. It does not matter which set you use because one set is just the vertical angles of the other, making the measurement equal. In the picture above, the color-matched angles are vertical angles and congruent. The Exterior Angle Sum Theorem stated that the exterior angles of a triangle add up to \begin{align*}360^\circ\end{align*}. Let’s extend this theorem to all polygons.
Watch the second half of this video.
#### Investigation: Exterior Angle Tear-Up
Tools Needed: pencil, paper, colored pencils, scissors
1. Draw a hexagon like the hexagons above. Color in the exterior angles as well.
2. Cut out each exterior angle and label them 1-6.
3. Fit the six angles together by putting their vertices together. What happens?
The angles all fit around a point, meaning that the exterior angles of a hexagon add up to \begin{align*}360^\circ\end{align*}, just like a triangle. We can say this is true for all polygons.
Exterior Angle Sum Theorem: The sum of the exterior angles of any polygon is \begin{align*}360^\circ\end{align*}.
Proof of the Exterior Angle Sum Theorem:
Given: Any \begin{align*}n-\end{align*}gon with \begin{align*}n\end{align*} sides, \begin{align*}n\end{align*} interior angles and \begin{align*}n\end{align*} exterior angles.
Prove: \begin{align*}n\end{align*} exterior angles add up to \begin{align*}360^\circ\end{align*}
NOTE: The interior angles are \begin{align*}x_1, x_2, \ldots x_n\end{align*}.
The exterior angles are \begin{align*}y_1, y_2, \ldots y_n\end{align*}.
Statement Reason
1. Any \begin{align*}n-\end{align*}gon with \begin{align*}n\end{align*} sides, \begin{align*}n\end{align*} interior angles and \begin{align*}n\end{align*} exterior angles. Given
2. \begin{align*}x_n^\circ\end{align*} and \begin{align*}y_n^\circ\end{align*} are a linear pair Definition of a linear pair
3. \begin{align*}x_n^\circ\end{align*} and \begin{align*}y_n^\circ\end{align*} are supplementary Linear Pair Postulate
4. \begin{align*}x_n^\circ+ y_n^\circ=180^\circ\end{align*} Definition of supplementary angles
5. \begin{align*}(x_1^\circ+x_2^\circ+\ldots+x_n^\circ)+(y_1^\circ+ y_2^\circ+\ldots+ y_n^\circ)=180^\circ n\end{align*} Sum of all interior and exterior angles in an \begin{align*}n-\end{align*}gon
6. \begin{align*}(n-2)180^\circ=(x_1^\circ+ x_2^\circ+\ldots+x_n^\circ)\end{align*} Polygon Sum Formula
7. \begin{align*}180^\circ n=(n-2)180^\circ+(y_1^\circ+ y_2^\circ+\ldots+ y_n^\circ)\end{align*} Substitution PoE
8. \begin{align*}180^\circ n=180^\circ n-360^\circ+(y_1^\circ+ y_2^\circ+\ldots+ y_n^\circ)\end{align*} Distributive PoE
9. \begin{align*}360^\circ=(y_1^\circ+ y_2^\circ+\ldots+ y_n^\circ)\end{align*} Subtraction PoE
#### Solving for Unknown Angle Measurments
What is \begin{align*}y\end{align*}?
\begin{align*}y\end{align*} is an exterior angle, as well as all the other given angle measures. Exterior angles add up to \begin{align*}360^\circ\end{align*}, so set up an equation.
\begin{align*}70^\circ + 60^\circ + 65^\circ + 40^\circ + y & = 360^\circ\\ y & = 125^\circ\end{align*}
#### Measuring Exterior Angles
What is the measure of each exterior angle of a regular heptagon?
Because the polygon is regular, each interior angle is equal. This also means that all the exterior angles are equal. The exterior angles add up to \begin{align*}360^\circ\end{align*}, so each angle is \begin{align*}\frac{360^\circ}{7} \approx 51.43^\circ\end{align*}.
#### Calculating the Sum of Exterior Angles
What is the sum of the exterior angles in a regular 15-gon?
The sum of the exterior angles in any convex polygon, including a regular 15-gon, is \begin{align*}360^\circ\end{align*}.
#### Earlier Problem Revisited
The exterior angles of a regular polygon sum to \begin{align*}360^\circ\end{align*}. The measure of each exterior angle in a dodecagon (twelve-sided regular polygon) is \begin{align*}\frac{360^\circ}{12} = 30^\circ\end{align*}.
### Examples
Find the measure of each exterior angle for each regular polygon below:
#### Example 1
12-gon
For each, divide by \begin{align*}360^\circ\end{align*} and by the given number of sides.
\begin{align*}30^\circ\end{align*}
#### Example 2
100-gon
\begin{align*}3.6^\circ\end{align*}
#### Example 3
36-gon
\begin{align*}10^\circ\end{align*}
### Review
1. What is the measure of each exterior angle of a regular decagon?
2. What is the measure of each exterior angle of a regular 30-gon?
3. What is the sum of the exterior angles of a regular 27-gon?
Find the measure of the missing variables:
1. The exterior angles of a quadrilateral are \begin{align*}x^\circ, 2x^\circ, 3x^\circ,\end{align*} and \begin{align*}4x^\circ.\end{align*} What is \begin{align*}x\end{align*}?
Find the measure of each exterior angle for each regular polygon below:
1. octagon
2. nonagon
3. triangle
4. pentagon
5. 50-gon
6. heptagon
7. 34-gon
8. Challenge Each interior angle forms a linear pair with an exterior angle. In a regular polygon you can use two different formulas to find the measure of each exterior angle. One way is \begin{align*}\frac{360^\circ}{n}\end{align*} and the other is \begin{align*}180^\circ - \frac{(n-2)180^\circ}{n}\end{align*} (\begin{align*}180^\circ\end{align*} minus Equiangular Polygon Formula). Use algebra to show these two expressions are equivalent.
9. Angle Puzzle Find the measures of the lettered angles below given that \begin{align*}m \ || \ n\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
Exterior Angle Sum Theorem Exterior Angle Sum Theorem states that the exterior angles of any polygon will always add up to 360 degrees.
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# How do you solve 5x + 3y = -8 and 4x + 6y = 8 using matrices?
Jun 27, 2018
The solution is $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 4 \\ 4\end{matrix}\right)$
#### Explanation:
The equation in matrix form is
$A X = B$
where
The matrix $A = \left(\begin{matrix}5 & 3 \\ 4 & 6\end{matrix}\right)$
$X = \left(\begin{matrix}x \\ y\end{matrix}\right)$
and
$B = \left(\begin{matrix}- 8 \\ 8\end{matrix}\right)$
The solution is
$X = {A}^{-} 1 B$
The inverse of $A$ is
${A}^{-} 1 = \frac{1}{|} \left(5 , 3\right) , \left(4 , 6\right) | \left(\begin{matrix}6 & - 3 \\ - 4 & 5\end{matrix}\right)$
$= \frac{1}{18} \left(\begin{matrix}6 & - 3 \\ - 4 & 5\end{matrix}\right)$
Therefore,
$X = \frac{1}{18} \left(\begin{matrix}6 & - 3 \\ - 4 & 5\end{matrix}\right) \left(\begin{matrix}- 8 \\ 8\end{matrix}\right)$
$= \frac{1}{18} \left(\begin{matrix}- 72 \\ 72\end{matrix}\right)$
$= \left(\begin{matrix}- 4 \\ 4\end{matrix}\right)$
The solution is
$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 4 \\ 4\end{matrix}\right)$ |
# ARITHMETIC MEAN OF UNGROUPED DATA
Arithmetic mean (AM) is one of the measures of central tendency which can be defined as the sum of all observations divided by the number of observations.
Formula to find arithmetic mean :
= Sum of all observations / Number of observations
Example 1 :
The marks obtained by 10 students in a test are 15, 75, 33, 67, 76, 54, 39, 12, 78, 11. Find the arithmetic mean.
Solution :
Mean = Total marks of 10 students / 10
= (15 + 75 + 33 + 67 + 76 + 54 + 39 + 12 + 78 + 11) / 10
= 460 / 10
= 46
Example 2 :
Find the mean of 2, 4, 6, 8, 10 , 12, 14, 16.
Solution :
Mean = Sum of given numbers / 8
= (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16) / 8
= 72 / 8
= 9
Example 3 :
John studies for 4 hours, 5 hours and 3 hours respectively on three consecutive days. How many hours does he study daily on an average?
Solution :
The average study time of John
= Total number of study hours / Number of days for which he studied
= (4 + 5 + 3) / 3
= 12 / 3
= 4 hours
Thus, we can say that John studies for 4 hours daily on an average.
Example 4 :
A batsman scored the following number of runs in six innings:
36, 35, 50, 46, 60, 55
Calculate the mean runs scored by him in an inning.
Solution :
To find the mean, we find the sum of all the observations and divide it by the number of observations.
Mean = Total runs / Number of innings
= (36 + 35 + 50 + 46 + 60 + 55) / 6
= 47
Thus, the mean runs scored in an inning are 47.
Example 5 :
The ages in years of 10 teachers of a school are:
32, 41, 28, 54, 35, 26, 23, 33, 38, 40
What is the mean age of these teachers?
Solution :
Mean age of the teachers
= Sum of age of teachers / Number of teachers
= (23 + 26 + 28 + 32 + 33 + 35 + 38 + 40 + 41 + 54) /10
= 350 / 10
= 35 years
Example 6 :
Following table shows the points of each player scored in four games:
Player Game 1 Game 2 Game 3 Game 4 ABC 1408 16811 106Did not play 10413
(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
(iii) B played in all the four games. How would you find the mean?
(iv) Who is the best performer?
Solution :
(i) Mean score of A = (14 + 16 + 10 + 10) / 4
= 12.5
Mean score of A per game is 12.5
(ii) To find the mean number of points per game for C, we have to divide the total points by 3.
Because he didn't participate in game 3. Total number of games he played is 3.
(iii) Mean score of B = (0 + 8 + 6 + 4) / 4
= 18/4
= 4.5
Mean score of B per game is 4.5
(iv) To choose the best performer, we have to find the mean score of each player.
Mean score of C = (8 + 11 + 13) / 3
= 32/3
= 10.6
Mean score of A per game is 12.5
Mean score of B per game is 4.5
Mean score of C per game is 10.6
Hence C is the best performer.
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# Three easy steps to find all global extrema of a function
In this post we discuss global extrema of a real-valued 1-variable function over all reals.
### Step 1: find all stationary points of a given function
By definition a stationary point of a function $$f(x)$$ is a solution of $$f'(x)=0$$. In this step many students waste their time by evaluating the 2nd derivative $$f”(x)$$ at each stationary point to determine if it is a local extremum, maximum or minimum. We shall detect global extrema simply by comparing $$f(x)$$ at all critical points, which is usually easier than $$f”(x)$$.
### Step 2: find all points where the 1st derivative is undefined
This important step is often missed. For example, the derivative of $$f(x)=|x|$$ is never zero.
However, we should certainly consider the point $$x=0$$, where the derivative of $$f(x)=|x|$$ is undefined. Otherwise we miss the global minimum of $$f(x)=|x|$$ at $$x=0$$ over all reals.
### We should consider all critical points and possibly more
All points where $$f'(x)=0$$ or $$f'(x)$$ is undefined are usually called critical. Sometimes we might not be sure if the derivative $$f'(x)$$ is well-defined or not.
A common mistake in sketching the graph $$y=|\cos x|$$ is to draw neighbourhoods of the points $$x=\frac{\pi}{2}+\pi n$$ for any integer $$n$$, where $$\cos x=0$$, either as smooth arcs in $$y=\cos^2 x$$ above or as vertical cusps in $$y=\sqrt{|\cos x|}$$ below.
In fact, the neighbourhoods of these points locally look like $$y=|x|$$, see the picture at the beginning of the post. Indeed, $$y=cos x$$ looks like $$y=x$$ around the points $$x=\frac{\pi}{2}+\pi n$$. So the derivative of $$f(x)=|\cos x|$$ is actually undefined at $$x=\frac{\pi}{2}+\pi n$$ for any integer $$n$$.
However we don’t need to justify this conclusion if we are interested only in extreme values of $$f(x)=|\cos x|$$. We may simply consider all “potentially critical” points, namely $$x=\frac{\pi}{2}n$$ for any integer $$n$$, where $$f'(x)=0$$ (at local extrema $$x=\pi n$$ of $$y=\cos x$$) or where $$f'(x)$$ might be undefined for $$f(x)=|\cos x|$$.
### Step 3: compare the function values at all critical points
Finally we should simply compare the values of $$f(x)$$ at all (potentially) critical points and choose the points that have the largest and smallest values.
For instance, $$f(x)=|\cos x|$$ over all real x has the largest value 1at the infinitely many global maxima $$x=\pi n$$ and the smallest value 0 at the infinitely many global minima $$x=\frac{\pi}{2}+\pi n$$ for any integer $$n$$. Here is the full justification of the extreme values: $$0\leq|\cos x|\leq 1$$ for all $$x$$.
If we are interested in extreme values of a function over a closed interval, not over the whole real line, then there is an extra step that will be discussed in one of our future posts.
The traditional riddle below is Problem 3.5 in our current MAT 2014 course, which was modified from original Question 1B in past exam MAT 2007.
• Riddle 18: Find the greatest value of $$(4\cos^4(5x-6)-3)^2$$ over all real $$x$$.
• How to submit: to write your full answer, submit a comment.
• Hint: you can write a full solution to this riddle in 1-2 lines.
• Warning: using derivatives is possible, but is hard to justify.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: Carlo has solved the problem, see this attempt and the final solution.
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# Finding local extrema of functions: 3 mistakes and 3 tips
This post discusses common mistakes and gives tips how to easily find local extrema of real-valued 1-variable functions.
### 1st mistake: any local extremum = a stationary point?
This myth has been dethroned in riddle 14. Actually, the function $$f(x)=|x|$$ has a local (even global) minimum at $$x=0$$, because $$|x|\geq 0$$ for all $$x$$. However, the derivative $$f'(x)=\left\{ \begin{array}{l} 1 \mbox{ for } x>0,\\ -1 \mbox{ for } x<0 \end{array} \right.$$ is undefined at $$x=0$$.
On the other hand, the function $$f(x)=x^3$$ from the previous post has a stationary point at $$x=0$$, which is not a local extremum. So the concepts of a stationary point and a local extremum are independent in the sense that one doesn’t follow from another.
### 1st tip: find all points where the derivative is undefined
The examples above imply that we should study all point where $$f'(x)$$ is undefined to find local extrema of $$f(x)$$. If the derivative $$f'(x)$$ is a fraction, we may start from
finding all points where the denominator of $$f'(x)$$ vanishes.
For example, the function $$f(x)=\sqrt{x}$$ is well-defined for all $$x\geq 0$$, but the derivative $$f'(x)=\frac{1}{2\sqrt{x}}$$ is not defined at $$x=0$$, where the tangent line to $$y=\sqrt{x}$$ is vertical.
### 2nd mistake: sufficient or necessary conditions of extrema
Recall that a stationary point of a function $$f(x)$$ is by definition a solution of the equation $$f'(x)=0$$. Here is a theorem stating when a stationary point $$x=a$$ is a local extremum.
Sufficient conditions of a local extremum. For a stationary point $$x=a$$ of a function $$f(x)$$ with a well-defined derivative $$f'(x)$$ for all $$x$$, if $$f”(a)<0$$ then $$x=a$$ is a local maximum, if $$f”(a)>0$$ then $$x=a$$ is a local minimum.
The conditions above are sufficient, but are not necessary as the example of $$f(x)=|x|$$ at $$x=0$$ shows. If $$f”(a)=0$$, then more analysis is needed. This is a typical question of singularity theory, which is richer for more than 1 variable.
### 2nd tip: check your rule for basic shapes of $$\pm x^2$$.
Many students often forget which inequality $$f”(a)<0$$ or $$f”(a)>0$$ corresponds to a local maximum or a local minimum. The simple trick is to remember the basic shapes of $$x^2$$ and $$-x^2$$. Namely, the positive parabola $$f(x)=x^2$$ has $$f”(0)=2>0$$ and a local minimum at $$x=0$$. Similarly, the negative parabola $$f(x)=-x^2$$ has $$f”(0)=-2<0$$ and a local maximum at $$x=0$$.
### 3rd mistake: does $$f”(a)=0$$ mean that $$x=a$$ is a point of inflection?
We have seen dozens of student scripts wrongly claiming that “$$f”(0)=0 \Rightarrow x=0$$ is a point of inflection”. Here is a proper geometric definition of a point of inflection of $$f(x)$$: if the graph $$y=f(x)$$ goes from one side of its the tangent line at $$x=a$$ to another side in a small neighbourhood of $$x=a$$, then $$x=a$$ is a point of inflection.
Analytically, if $$L(x)=f(a)+f'(a)(x-a)$$ is the tangent line to $$y=f(x)$$, then the difference $$f(x)-L(x)$$ changes its sign at a point of inflection $$x=a$$. For instance, at a stationary point $$x=a$$ the tangent line $$L(x)=f(a)$$ is horizontal and the same difference $$f(x)-f(a)$$ keeps its sign around $$x=a$$. Hence a point of inflection can not be a local extremum of $$f(x)$$.
For example, $$x=0$$ is a point of inflection of $$f(x)=x^3$$, because $$y=x^3$$ intersects the tangent line $$y=0$$ and goes from the lower half-plane to the upper half-plane.
### 3rd tip: use well-known inequalities for justifying extrema
The function $$f(x)=x^4$$ has a local minimum, but not a point of inflection at $$x=0$$. The sufficient conditions from the theorem above are not satisfied as $$f'(0)=0=f”(0)$$. However, the justification is even easier: $$x=0$$ is a global minimum of $$f(x)=x^4$$ over all real $$x$$ as $$x^4\geq 0$$.
• Riddle 15: state conditions when $$x=a$$ is a point of inflection of $$f(x)$$.
• How to submit: to write your full answer, simply submit a comment.
• Hint: give sufficient conditions in derivatives of $$f(x)$$ without proof.
• Warning: not all conditions are equalities, $$f'(a)=0$$ isn’t enough.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
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# Traffic jams and 3 common mistakes on turning points
This post discusses the commonly confused concept of a turning point.
### 1st mistake: stationary = turning?
Let $$f(x)$$ be the position of a moving car in the $$x$$-axis (on a highway if you wish). Then the derivative $$f'(x)$$ is the speed (or the length of the velocity vector) of the car.
The solutions of $$f'(x)=0$$ are the points where the speed is 0, so the car is stationary. Hence the solutions of $$f'(x)=0$$ are called stationary points of the function $$f(x)$$.
If a car comes to a stationary point (a stop), it doesn’t mean that the car will make a U-turn. You have certainly been in a traffic jam, where a car stops for a while and then starts moving again in the same direction. Despite this overwhelming real-life practice, stationary points are often (and wrongly) called turning points even at university.
### 2nd mistake: stationary points = extrema?
Another common mistake is to confuse stationary points with local extrema. We have seen hundreds of scripts claiming something like “$$f'(a)=0$$, hence $$x=a$$ is a minimum (or a maximum)”. The simple counter-example is $$f(x)=x^3$$. Indeed, $$f'(x)=3x^2$$, so $$x=0$$ is a stationary point. However, $$x=0$$ is neither a local minimum nor a local maximum.
A stationary point and a local extremum are different concepts. A stationary point is analytically defined as a solution of $$f'(x)=0$$. A local extremum is geometrically defined below.
### A proper definition of a local maximum
A point $$x=a$$ is called a (strict) local maximum of $$f(x)$$ if $$f(x)<f(a)$$ for all points $$x\neq a$$ sufficiently close to the point $$a$$. Notice that the inequality $$f(x)<f(a)$$ can’t hold at $$x=a$$ and is not required over the whole domain of $$f(x)$$.
For example, $$f(x)=\cos x$$ has a local maximum at $$x=0$$, because $$\cos x<1$$ for all $$x\neq 0$$ over $$-2\pi<x<2\pi$$.
### A proper definition of a local minimum
Similarly, a point $$x=a$$ is called a (strict) local minimum of $$f(x)$$ if $$f(x)>f(a)$$ for all points $$x\neq a$$ sufficiently close to the point $$a$$.
The word strict refers to the strict inequality $$f(x)>f(a)$$. Non-strict local extrema allow the condition $$f(x)\geq f(a)$$ for all $$x$$ sufficiently close to the point $$a$$. For example, $$x=0$$ can be considered as a non-strict local minimum of $$f(x)=\left\{ \begin{array}{l} x \mbox{ for } x\geq 0,\\ 0 \mbox{ for } x\leq 0. \end{array}\right.$$
Local minima and maxima can be called local extrema. The word extremum means either a minimum or a maximum. A turning point of a function $$f(x)$$ is the same concept as a strict local extremum. Indeed, the graph $$y=f(x)$$ “turns” (or makes a U-turn) at any strict local extremum of $$f(x)$$. However, experts usually say a local extremum, not a “turning point”.
### 3rd mistake: extreme point = extreme value?
The final common mistake is to confuse extrema with extreme values by writing, for instance, “$$f(x)=x^2+1$$ has the minimum $$f(0)=1$$“. The point $$x=0$$ is indeed a local minimum of $$f(x)=x^2+1$$. However, the value $$f(0)=1$$ is called a local minimum value, not a local minimum point. So a point and the value of a function at this point are very different.
• Riddle 14: for a function $$f(x)$$, should any local extremum be a stationary point?
• How to submit: to write your full answer, simply submit a comment to this post.
• Hint: the derivative $$f'(a)$$ should be well-defined at any stationary point $$x=a$$.
• Warning: justify that any extremum has $$f'(a)=0$$ or give a counter-example.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: IS has solved the riddle giving the counter-example f(x)=|x|.
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# Mistakes with square roots: easy, common, potential
We congratulate all readers of our blog on Chinese New Year!
In this post we discuss common mistakes with square roots of real numbers. As usual in mathematics, we start from the definition: the square root $$\sqrt{x}$$ is well-defined only for a non-negative real number $$x\geq 0$$ and equals a unique number $$y$$ such that $$x=y^2$$.
The graph of the square root function $$y=\sqrt{x}$$ is shown in the picture above. The function is always increasing and has the vertical tangent at the origin. If you understand derivatives, you may check that $$y'(x)\to+\infty$$ as $$0<x\to 0$$. We discuss only real numbers here. The square root of a complex number is another (interesting and more advanced) story.
### Easy mistake: simplifying $$\sqrt{x^2}$$
Rather surprisingly, it is a very popular myth that $$\sqrt{x^2}=x$$. At least, this is the first answer we got at many summer schools and also from 2nd year maths undergraduates at a top UK university. Our next question was to check if the identity holds for $$x=-1$$. Then most students realised that $$\sqrt{x^2}=x$$ fails for $$x=-1$$.
Some students were still in doubt and claimed that $$\sqrt{1}$$ is $$1$$ or $$-1$$. We knew that all students have calculators at hand and asked them to compute $$\sqrt{1}$$ with their calculators. The expected correct answer $$\sqrt{1}=1$$ left no chance for the myth that $$\sqrt{x^2}=x$$ can hold for all real $$x$$.
Indeed, $$\sqrt{x^2}=|x|=\left\{\begin{array}{c} x \mbox{ for } x\geq 0,\\ -x \mbox{ for } x<0. \end{array} \right.$$ At one of the summer schools we were told that the absolute value $$|x|$$ is learned in the UK only at A-levels (at the age of 16-18). So there is little chance to get the correct identity $$\sqrt{x^2}=|x|$$. Later is better than never, so just in case we have sketched the graph $$y=|x|$$ on the left.
### Common mistake: squaring both sides
In the riddle from the post about success at Oxbridge interviews we asked to find all real $$x$$ such that $$x\geq\sqrt{3x-2}$$. Unfortunately, public attempts 1 and 2 (and more private attempts by e-mail) started from squaring both sides: $$x^2\geq 3x-2$$.
The resulting inequality is not equivalent to the original one, because $$x=-1$$ satisfies $$x^2\geq 3x-2$$, but not $$x\geq\sqrt{3x-2}$$. So the popular myth that $$\sqrt{x^2}=x$$ has led to serious consequences that squaring both sides is a safe operation, but it is not!
When a given equation or inequality contains functions that are not well-defined for all real numbers, experts start from writing the domain when the problem makes sense. Best students always remember that a good style to finish a solution is to substitute answers back into an original equation.
However, this final check should not be considered as a part of a solution, but only as an opportunity to develop self-criticism, a key mathematical skill. Indeed, if we have infinitely many answers (as in the case of an inequality), we can not substitute back all our answers or such a substitution could be too hard. So if a final check fails, our solution fails, but the final check is ok.
For the inequality $$x\geq\sqrt{3x-2}$$, we first write that square root makes sense only when $$3x-2\geq 0,\; x\geq\frac{2}{3}$$. In this domain squaring both sides produces the equivalent inequality $$x^2\geq 3x-2,\; x^2-3x+2\geq 0,\; (x-1)(x-2)\geq 0$$.
The last inequality has the solutions $$x\leq 1$$ and $$x\geq 2$$. If we remember the domain $$x\geq\frac{2}{3}$$, the final answer is $$\frac{2}{3}\leq x\leq 1$$ and $$x\geq 2$$. Just in case, we may check that the boundary values $$\frac{2}{3},1,2$$ satisfy the original inequality $$x\geq\sqrt{3x-2}$$.
### Potential confusion: Q6(ii) in STEP I exam 2007
Here is the exact quote: “Given that $$x^3-y^3=(x-y)^4$$ and that $$x-y=d\neq 0$$, show that $$3xy=d^3-d^2$$. Hence show that $$2x=d\pm d\sqrt{\frac{4d-1}{3}}$$.”
Even without trying to solve the problem, any professional mathematician could spot within a few seconds that the expression $$\frac{4d-1}{3}$$ under the square root can be negative. Indeed, the only given restriction $$d\neq 0$$ allows $$d=-2$$ when we get the complex roots $$x=-1\pm i\sqrt{3}$$.
It is a good exercise in complex numbers to check that the pairs $$(x,y)=(-1+i\sqrt{3},1+i\sqrt{3})$$ and $$(x,y)=(-1-i\sqrt{3},1-i\sqrt{3})$$ actually satisfy the given equation $$x^3-y^3=(x-y)^4$$, so there is no mistake in the problem. However, complex numbers are not in the STEP I syllabus and we suspect that the case of complex roots was probably missed in the original problem.
The STEP examiners’ solution contains the identity $$\pm\sqrt{9d^2+12(d^3-d^2)}=\pm d\sqrt{12d-3}$$, which is luckily correct for all real d, but only if we use complex numbers with the sign $$\pm$$ in both sides. For real numbers, as was probably expected, we should write $$\sqrt{|9d^2+12(d^3-d^2)|}=|d|\sqrt{|12d-3|}$$, simply because the left hand side is not negative and so should the expressions under the square roots.
Unfortunately, there are other STEP problems when the sign $$\pm$$ doesn’t help. So one of the homework problem from our STEP I course has a parametric equation that can’t be solved by careless squaring both sides. The average mark for this homework is less than 16/20.
Just in case the picture above shows the curve $$x^3-y^3=(x-y)^4$$ excluding the line $$y=x$$ of easy solutions. The straight line $$x-y=d=\frac{1}{4}$$ meets the curve at the red point where $$x=\frac{1}{8}$$, hence $$y=-\frac{1}{8}$$. If $$d<\frac{1}{4}$$, the straight line $$x-y=d$$ doesn’t intersect the curve (in the real domain) and has only complex intersection points $$(x,y)$$, which are invisible on the usual plane.
• Riddle 12: find all real $$x,y$$ when the inequality $$\frac{x+y}{2}\geq\sqrt{xy}$$ holds.
• How to submit: to write your full answer, submit a comment.
• Warning: justify that you found all (not only some) real solutions.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
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# Mistakes in trigonometry: easy, common, unexpected
We congratulate all Oxbridge candidates who have received their offers in January 2014!
In this post we discuss common mistakes in solving trigonometric equations.
### Easy mistake: basic values often forgotten
If you remember basic values of trigonometric functions such as $$\sin\frac{\pi}{6}$$, $$\cos\frac{\pi}{4}$$, $$\tan\frac{\pi}{3}$$ without using a calculator, then at a real exam you will save time for more interesting problems.
A good exercise is to check the following basic values (for any integer n): $$\sin(n\pi)=0$$, $$\sin(\frac{\pi}{2}+n\pi)=(-1)^n=\cos(n\pi)$$, $$\tan(\frac{\pi}{4}+n\frac{\pi}{2})=(-1)^n=\cot(\frac{\pi}{4}+n\frac{\pi}{2})$$. You may check these identities for n=0, n=1 (possibly for n=2, n=3) and then use periodicity.
A good style to finish your solution to a trigonometric equation is to substitute your answers back into the original equation. If something is wrong, please don’t write “something is wrong”, but try to correct your computations. In our distance courses we give more specific tips to locate a computational error in a long chain of equations.
### Common mistake: not all solutions are found
We have seen hundreds of scripts that contain “$$\sin x=0\Rightarrow x=0$$“. This logical implication fails, because x=π also satisfies the equation $$\sin x=0$$. Here is the full correct solution “$$\sin x=0, x=n\pi$$ for any integer n”.
We wouldn’t write the arrow $$\Rightarrow$$, because this logical implication doesn’t say that all found values are correct. For instance, the implication “$$\sin x=0 \Rightarrow x=n\frac{\pi}{2}$$ for any integer n” is also logically correct. So this confusing notation will be a topic for one of our future posts.
Since trigonometric functions are periodic, a trigonometric equation usually has infinitely many solutions. To make problems simpler, the authors of MAT (entrance exams to Oxford and Imperial College London) often specify a short range of expected solutions, usually from 0 to 2π.
### Unexpected mistake: Q1 in STEP III exam 2007
Here is the exact quote: “The four real numbers θ1, θ2, θ3 and θ4 lie in the range $$0\leq \theta_i< 2\pi$$ and satisfy the equation $$p\cos(2\theta)+\cos(\theta-\alpha)+p=0$$, where p and $$\alpha$$ are independent of θ. Show that θ1234=nπ for some integer n.”
Let us consider the simple value of the parameter p=0. Then the given equation becomes $$\cos(\theta-\alpha)=0$$ and has only two real solutions in the range $$0\leq\theta<2\pi$$. For instance, if $$\alpha=\frac{\pi}{6}$$, then the two solutions of $$\cos(\theta-\frac{\pi}{6})=0$$ in the range $$0\leq\theta<2\pi$$ are
$$\theta=\frac{2\pi}{3}$$ and $$\theta=\frac{5\pi}{3}$$.
There is no way to assign these two values $$\frac{2\pi}{3}$$ and $$\frac{5\pi}{3}$$ to 4 variables whose sum θ1234 is a multiple of π. Indeed, if the value $$\frac{2\pi}{3}$$ is taken k (of 4) times, then the value $$\frac{5\pi}{3}$$ is taken 4-k times. Then $$k\frac{2\pi}{3}+(4-k)\frac{5\pi}{3}=n\pi$$, hence $$2k+5(4-k)=3n$$ and $$20=3(k+n)$$, which is impossible for integer k,n.
A brute-force alternative is to consider all 5 cases for possible values of the 4 angles $$0\leq\theta_1\leq\theta_2\leq\theta_3\leq\theta_4<2\pi$$ as follows.
(1) If $$\theta_1=\theta_2=\theta_3=\theta_4=\frac{2\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{8\pi}{3}$$.
(2) If $$\theta_1=\theta_2=\theta_3=\frac{2\pi}{3}$$ and $$\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{11\pi}{3}$$.
(3) If $$\theta_1=\theta_2=\frac{2\pi}{3}$$ and $$\theta_3=\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{14\pi}{3}$$.
(4) If $$\theta_1=\frac{2\pi}{3}$$ and $$\theta_2=\theta_3=\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{17\pi}{3}$$.
(5) If $$\theta_1=\theta_2=\theta_3=\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{20\pi}{3}$$.
In all 5 cases (1)-(5) above the sum $$\theta_1+\theta_2+\theta_3+\theta_4$$ is not a multiple of π, which disproves the statement from Q1 STEP III exam 2007. The STEP examiners’ solution to this problem is unfortunately too short and doesn’t consider exceptional cases when a denominator can be zero.
Rather surprisingly, neither the problem nor the solution were corrected in the official STEP resources for more than 6.5 years (checked on 24th January 2014). Fortunately, we have modified this problem and carefully considered all exceptional cases in our STEP III course.
• Riddle 11: solve the equation $$\sin x=a$$ for any real parameter a.
• How to submit: to write your full answer, submit a comment.
• Hint: there is a short formula for x depending on an integer n.
• Warning: a full answer should cover all possible values of a.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: K Wright has solved the problem, see attempts 1 and 2.
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# Success in Oxbridge interviews: 3 mistakes and 3 tips
We wish good luck to all Oxbridge candidates at their interviews in December!
### Mistake 1 : are numerical answers important?
Numerical answers have very little value in mathematics and are even less important in real life. Indeed, computing the page rank of a single web page is probably worth about 1 pence. However, the page rank algorithm (with many others) can quickly process the whole World Wide Web, which turned Google into a multi-billion company.
At a proper mathematics interview you are expected to demonstrate your thinking process. If interviewers are interested only in your numerical answers, they can get them in a much quicker and cheaper way via an on-line test.
### Tip 1 : never believe and learn how to prove rigorously
A few years ago one student argued that he should have got a full mark for his solution that gave a correct numerical answer justified only by the following phrase: “I believe that my answer is correct”. Students who are interested in beliefs will probably be welcomed in theology. Mathematicians are always expected to prove new results, for instance by using previously proved theorems.
### Mistake 2 : the easy ways to make a bad impression are
• being nervous, e.g. if you didn’t sleep well before
• talking too quiet or looking physically weak or tired
• reflecting on your performance during the interview
• asking inappropriate questions, e.g. how have I done?
Oxbridge interviews are usually in the morning, so you need to have a regular sleeping pattern at least a few weeks before your interview, not only at the very last night. Read our advice on physical exercise. There is no time and no point to think or to ask if you have done well before all interviews are finished.
### Tip 2 : read interview stories from past students
We have collected a few useful stories from our past successful students about their Oxbridge interviews and share this first hand experience with all our current students. For instance, all Cambridge candidates and some Oxford candidates sit a written test before oral interviews.
Some of our students happily announce shortly after their interviews that they have done well, e-mail us their questions and later receive rejections. In most these cases we clearly see that the questions were quite easy. If you get easy questions, you will not be told that you are considered as a weak candidate, because the job of interviewers is to keep all candidates happy.
One of our best students last year e-mailed that he could hardly complete any question and only with a lot of hints from his interviewers. However his questions were much harder than from other students. Actually, he received a standard Cambridge offer with a grade 1 in both STEP II and III, then gained two grades S after completing our STEP courses.
### Mistake 3 : avoiding proper feedback on your progress
We regularly encounter over-optimistic candidates who are self-studying without any expert advice on their progress. Then the first (and often last) feedback will be a “yes/no” from university admissions. The MAT examiners and Oxbridge admissions tutors guard hard all exam scores and STEP candidates can hope to receive only their numerical mark. So the current entrance exams at top UK universities provide little feedback to students after months of self-studies.
The education systems outside Europe and North America are quite opposite. Almost all exams are oral and students naturally get a lot of feedback. Oral exams are often harder and harsher, so it is a proper training for real life. That is why the key value of our distance courses is the detailed feedback on regular homework, where questions are usually harder than in past exams.
### Tip 3 : don’t train for a test, but aim higher
A pupil told a kung-fu master: “I have been training really hard for many months, but I still can not break through that board”. The master watched his attempt and then said: “If you hit the board, you will never break through it. You should hit beyond the board.”
Similarly in any learning, if students are trained only for a specific test, they are likely to fail, read
BBC Education: most A level grade predictions wrong. The winners always aim higher (much higher than their target) and that is why they often win even if something goes wrong.
The usual feedback from our students on the distance course for Oxbridge interviews: “your Oxbridge questions are much harder than anything I have done before at school”. Here is our powerful learning principle: the harder the training, the easier the exam!
• Riddle 9: Find all real solutions of the inequality $$x\geq\sqrt{3x-2}$$.
• How to submit: to write your full answer, submit a comment.
• Hint: remember that $$\sqrt{x^2}\neq x$$, read answers to this riddle.
• Warning: the very first step in most common attempts is wrong.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: Paul solved the riddle and won a prize, read our comment.
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# Mistakes in sketching graphs: easy, common, unexpected
Easy mistake: an axis can’t have 2 directions.
Several our students persistently started to sketch graphs from the diagram above. By definition, an axis is an infinite straight line with one direction, so we need to know a direction of increase of the variable along the axis. In computer programming the axes have different directions, because we usually type on a screen from top to bottom starting from the top left corner (the origin).
So the vertical $$y$$-axis is oriented downwards and the parabola $$y=(x-8)^2$$ has the reversed shape on such a diagram. However, if two directions are specified on the line of the $$y$$-coordinate, then we can not know how to draw even the simple parabola $$y=(x-8)^2$$. By the standard convention in mathematics the horizontal $$x$$-axis is oriented to the right hand side and the vertical $$y$$-axis is oriented upwards.
Common mistake: corners/cusps vs smooth points.
Many students often draw the graph $$y=2\sin^2 x$$ with corners and cusps at the points
$$x=\frac{\pi}{2}+\pi n$$ for any integer n, where the graph meets the $$x$$-axis. Starting from the familiar graph $$y=\sin x$$ and before scaling it by factor 2, the students probably reflect the negative parts to the upper half-plane and get the following picture with cusps (“acute corners”). This reflection should give right-angled (not acute) corners, because $$y=\sin x$$ meets the $$x$$-axis at angles $$\pm\frac{\pi}{4}$$ since the gradient is $$y'(x)=\cos x=\pm 1$$ at $$x=\frac{\pi}{2}+\pi n$$.
However, $$y=2\sin^2 x$$ has neither corners nor cusps, because the derivative $$y'(x)=2\sin x \cos x =\sin 2x$$ is well-defined everywhere. Indeed, $$y'(x)=sin 2x=0$$ at $$x=\frac{\pi}{2}+\pi n$$, so the graph the $$x$$-axis smoothly touches $$y=2\sin^2 x$$ at all the points $$x=\frac{\pi}{2}+\pi n$$ as in the correct picture below.
Actually, best students know the formula $$2\sin^2 x=1-\cos 2x$$ and sketch the simpler graphs $$y=\cos x, y=\cos 2x, y=1-\cos 2x$$.
Unexpected mistake: Q3(v) in the MAT paper 2011.
Q3 in the MAT paper 2011 is about the cubic parabola $$y(x)=x^3-x$$ and its tangent line $$y=m(x-a)$$ having a slope $$m>0$$ and meeting the $$x$$-axis at a point $$x=a\leq 1$$.
Through any point (a,0) for $$a<-1$$ we can draw 3 tangent lines to $$y=x^3-x$$. One the them touches the cubic parabola at a point $$0<x<1$$ and has a negative slope. However, there are two different tangent lines with a positive slope. Only one of them was shown in the original problem as in the picture above. No restrictions apart from $$m>0, a\leq -1$$ were given in the problem. Hence we may have the second tangent line for $$b<a<-1$$ in our picture below.
Part (ii) asks to prove that $$a=\frac{2b^3}{3b^2-1}$$. This formula also works for both tangent lines: if $$b=-2$$, then $$a=-\frac{16}{11}>b=-2$$.
Part (iii) asks to find an approximate value of b when $$a=-10^6$$. The examiners’ solution of part (iii) considers only one possibility when $$a=\frac{2b^3}{3b^2-1}$$ has a large (absolute) value because of a small denominator, which implies that $$b\approx -\frac{1}{\sqrt{3}}$$ (if negative). However, the second possibility is that b also has a large (absolute) value. For instance, if we set $$b=\frac{3}{2}a$$, then $$\frac{2b^3}{3b^2-1}=\frac{\frac{27}{4}a^3}{\frac{27}{4}a^2-1}=a+\frac{a}{\frac{27}{4}a^2-1}$$ is approximately equal to $$a$$ when $$a$$ has a large absolute value. So another tangent line at $$b\approx -\frac{3}{2}10^6$$ meets the x-axis at the same point $$a=-10^6$$.
Part (iv) asks to show that the tangent line meets the cubic parabola at the second point $$c=-2b$$. This formula also works for both tangent lines: if $$b=-2$$, then the tangent line $$y=11\Big(x+\frac{16}{11}\Big)=11x+16$$ meets $$y(x)=x^3-x$$ at $$x=c=4$$ where $$y(4)=60=11\cdot 4+16$$.
Part (v) asks to find the largest possible area of the region R bounded above by the tangent line and bounded below by $$y=x^3-x$$. For both tangent lines, the region R is finite because both
tangent lines eventually intersect the cubic parabola at $$c=-2b$$. The examiners’ solution to part (v) says: “We can see that as $$a$$ increases then the tangent line rises and so the area of R increases. So the area is greatest when $$a=-1$$“.
This claim holds only for the first tangent line when $$-1\leq b<0$$. However, our last picture with the second tangent line above shows a much larger area when $$b<-1$$. Actually both tangent lines coincide when $$a=-1$$, so one family of tangent lines for $$a<-1<b<0$$ is joining another family of tangent lines for $$b<a<-1$$. The picture in the problem with only one tangent line for $$a<b$$ degenerates in the case $$a=b=-1$$ when two points A,B merge and can hardly be used as a reference for part (v).
It is rather surprising that Q3(v) in the MAT paper 2011 and its solution have not been amended in Oxford MAT 2011 solutions and in Imperial MAT 2011 for almost 2 years after the actual exam: checked on 20th September 2013. You can find more details on this problem in our web tutorial Tangent lines and areas bounded by cubic parabolas.
Mathematics is a wonderful subject, because students can check all arguments and find more efficient solutions themselves. That is why we keep learning from our smart students.
• Riddle 6: does $$x^2+y^2=1$$ define a function $$y(x)$$ for $$-1\leq x\leq 1$$?
• How to submit: to write your full answer, simply submit a comment.
• Hint: sketch the curve $$x^2+y^2=1$$ on the plane, you may try to express $$y(x)$$.
• Warning: a function $$y(x)$$ should have a single value of $$y$$ over $$-1\leq x\leq 1$$.
• Restriction: only the first fully correct public answer will be rewarded.
• Prize: free 1-year access to one of our interactive web tutorials.
• Update: Carlo has solved the problem, see the comment.
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Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.6 Questions and Answers.
## Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6
Question 1.
Evaluate:
i. $$\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right]\left[\begin{array}{lll} {[2} & -4 & 3 \end{array}\right]$$
ii. $$\left[\begin{array}{lll} 2 & -1 & 3 \end{array}\right]\left[\begin{array}{l} 4 \\ 3 \\ 1 \end{array}\right]$$
Solution:
i. \begin{aligned} \left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right]\left[\begin{array}{lll} 2 & -4 & 3 \end{array}\right] &=\left[\begin{array}{lll} 3(2) & 3(-4) & 3(3) \\ 2(2) & 2(-4) & 2(3) \\ 1(2) & 1(-4) & 1(3) \end{array}\right] \\ &=\left[\begin{array}{ccc} 6 & -12 & 9 \\ 4 & -8 & 6 \\ 2 & -4 & 3 \end{array}\right] \end{aligned}
ii. $$\left[\begin{array}{lll} 2 & -1 & 3 \end{array}\right]\left[\begin{array}{l} 4 \\ 3 \\ 1 \end{array}\right]$$
= [2(4)-1(3)+ 3(1)]
= [8 – 3 + 3] = [8]
Question 2.
If A = $$\left[\begin{array}{cc} 1 & -3 \\ 4 & 2 \end{array}\right]$$ B = $$\left[\begin{array}{cc} 4 & 1 \\ 3 & -2 \end{array}\right]$$, = show that AB ≠ BA.
Solution:
From (i) and (ii), we get
AB ≠ BA
Question 3.
If A = $$\left[\begin{array}{ccc} -1 & 1 & 1 \\ 2 & 3 & 0 \\ 1 & -3 & 1 \end{array}\right]$$ ,B = $$\left[\begin{array}{lll} 2 & 1 & 4 \\ 3 & 0 & 2 \\ 1 & 2 & 1 \end{array}\right]$$ state whether AB = BA? Justify your answer.
Solution:
From (i) and (ii), we get
AB ≠ BA
Question 4.
Show that AB = BA, where
i. A = $$\left[\begin{array}{rrr} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{array}\right]$$ , B = $$\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{array}\right]$$
ii. A = $$\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$$, B = $$\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$$
Solution:
From (i) and (ii), we get
AB = BA
From (i) and (ii), we get
AB = BA
[Note: The question has been modified.]
Question 5.
If A = $$\left[\begin{array}{cc} 4 & 8 \\ -2 & -4 \end{array}\right]$$, prove that A2 = 0.
Solution:
A2 = A.A
= $$\left[\begin{array}{cc} 4 & 8 \\ -2 & -4 \end{array}\right]\left[\begin{array}{cc} 4 & 8 \\ -2 & -4 \end{array}\right]$$
= $$\left[\begin{array}{cc} 16-16 & 32-32 \\ -8+8 & -16+16 \end{array}\right]$$
= $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$ = 0
Question 6.
Verify A(BC) = (AB)C in each of the following cases:
i. A = $$=\left[\begin{array}{cc} 4 & -2 \\ 2 & 3 \end{array}\right]$$, B = $$\left[\begin{array}{cc} -1 & 1 \\ 3 & -2 \end{array}\right]$$ and C = $$\left[\begin{array}{cc} 4 & 1 \\ 2 & -1 \end{array}\right]$$
ii. A = $$\left[\begin{array}{ccc} 1 & -1 & 3 \\ 2 & 3 & 2 \end{array}\right]$$, B = $$\left[\begin{array}{cc} 1 & 0 \\ -2 & 3 \\ 4 & 3 \end{array}\right]$$ and C = $$\left[\begin{array}{cc} 1 & 2 \\ -2 & 0 \\ 4 & -3 \end{array}\right]$$
Solution:
From (i) and (ii), we get
A(BC) = (AB)C.
Question 7.
Verify that A(B + C) = AB + AC in each of the following matrices:
i. A = $$\left[\begin{array}{cc} 4 & -2 \\ 2 & 3 \end{array}\right]$$, B = $$\left[\begin{array}{cc} -1 & 1 \\ 3 & -2 \end{array}\right]$$ and C = $$=\left[\begin{array}{cc} 4 & 1 \\ 2 & -1 \end{array}\right]$$
ii. A = $$\left[\begin{array}{ccc} 1 & -1 & 3 \\ 2 & 3 & 2 \end{array}\right]$$, B = $$\left[\begin{array}{cc} 1 & 0 \\ -2 & 3 \\ 4 & 3 \end{array}\right]$$ and C = $$\left[\begin{array}{cc} 1 & 2 \\ -2 & 0 \\ 4 & -3 \end{array}\right]$$
Solution:
From (i) and (ii), we get
A(B + C) = AB + AC.
[Note: The question has been modified.]
Question 8.
If A = $$\left[\begin{array}{cc} 1 & -2 \\ 5 & 6 \end{array}\right]$$, B = $$\left[\begin{array}{cc} 3 & -1 \\ 3 & 7 \end{array}\right]$$, find AB – 2I, where I is unit matrix of order 2.
Solution:
Question 9.
If A = $$\left[\begin{array}{ccc} 4 & 3 & 2 \\ -1 & 2 & 0 \end{array}\right]$$, B = $$\left[\begin{array}{cc} 1 & 2 \\ -1 & 0 \\ 1 & -2 \end{array}\right]$$, show that matrix AB is non singular.
Solution:
im
∴ AB is non-singular matrix.
Question 10.
If A = , find the product (A + I)(A – I).
Solution:
[Note : Answer given in the textbook is $$\left[\begin{array}{ccc} 9 & 6 & 4 \\ 15 & 32 & -2 \\ 35 & -7 & 29 \end{array}\right]$$
However, as per our calculation it is $$\left[\begin{array}{ccc} 10 & 10 & 4 \\ 25 & 39 & 2 \\ 35 & 7 & 22 \end{array}\right]$$. ]
Question 11.
If A = $$\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right]$$, B = $$\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]$$, find α, if A2 = B.
Solution:
A2 = B
∴ By equality of matrices, we get
α2 = 1 and α + 1 = 2
∴ α = ± 1 and α = 1
∴ α = 1
Question 12.
If A = $$\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]$$, show that A2 – 4A is scalar matrix.
Solution:
A2 – 4A = A.A – 4A
Question 13.
If A = $$\left[\begin{array}{cc} 1 & 0 \\ -1 & 7 \end{array}\right]$$, find k so that A2 – 8A – kI = O, where I is a unit matrix and O is a null matrix of order 2.
Solution:
A2 – 8A – kI = O
∴ A.A – 8A – kI = O
∴ by equality of matrices, we get
1 – 8 – k = 0
∴ k = -7
Question 14.
If A = $$\left[\begin{array}{cc} 8 & 4 \\ 10 & 5 \end{array}\right]$$, B = $$\left[\begin{array}{cc} 5 & -4 \\ 10 & -8 \end{array}\right]$$, show that (A+B)2 = A2 + AB + B2.
Solution:
We have to prove that (A + B)2 = A2 + AB + B2,
i.e., to prove A2 + AB + BA + B2 = A2 + AB + B2,
i.e., to prove BA = 0.
BA = $$\left[\begin{array}{cc} 5 & -4 \\ 10 & -8 \end{array}\right]\left[\begin{array}{cc} 8 & 4 \\ 10 & 5 \end{array}\right]$$
$$\left[\begin{array}{cc} 40-40 & 20-20 \\ 80-80 & 40-40 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$
Question 15.
If A = $$\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]$$, prove that A2 – 5A + 7I = 0, where I is unit matrix of order 2.
Solution:
A2 – 5A + 7I = 0 = A.A – 5A + 7I = 0
Question 16.
If A = $$\left[\begin{array}{cc} 3 & 4 \\ -4 & 3 \end{array}\right]$$ and B = $$\left[\begin{array}{cc} 2 & 1 \\ -1 & 2 \end{array}\right]$$, show that (A + B)(A – B) = A2 – B2.
Solution:
We have to prove that (A + B)(A – B) = A2 – B2,
i.e., to prove A2 – AB + BA – B2 = A2 – B2,
i.e., to prove – AB + BA = 0,
i.e., to prove AB – BA.
From (i) and (ii), we get AB = BA
Question 17.
If A = $$\left[\begin{array}{cc} 1 & 2 \\ -1 & -2 \end{array}\right]$$, B = $$\left[\begin{array}{cc} 2 & a \\ -1 & b \end{array}\right]$$ and (A + B)2 = A2 + B2, find the values of a and b.
Solution:
Given, (A + B)2 = A2 + B2
∴ A2 + AB + BA + B2 = A2 + B2
∴ AB + BA = 0
∴ AB = -BA
∴ by equality of matrices, we get
– 2 + a = 0 and 1 + b = 0
a = 2 and b = -1
[Note: The question has been modified.]
Question 18.
Find matrix X such that AX = B,
where A = $$\left[\begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array}\right]$$ and B = $$\left[\begin{array}{c} -3 \\ -1 \end{array}\right]$$
Solution:
Let X = $$\left[\begin{array}{c} a \\ b \end{array}\right]$$
But AX = B
∴ $$\left[\begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{a} \\ \mathrm{b} \end{array}\right]=\left[\begin{array}{r} -3 \\ -1 \end{array}\right]$$
∴ $$\left[\begin{array}{c} a-2 b \\ -2 a+b \end{array}\right]=\left[\begin{array}{l} -3 \\ -1 \end{array}\right]$$
By equality of matrices, we get
a – 2b = -3 …(i)
-2a + b = -l …(ii)
By (i) x 2 + (ii), we get
-3b =-7
∴ b = $$\frac{7}{3}$$
Substituting b = $$\frac{7}{3}$$ in (i), we get
a – 2 ($$\frac{7}{3}$$) = -3
∴ a = -3 + $$\frac{14}{3}=\frac{5}{3}$$
∴ X = $$\left[\begin{array}{l} \frac{5}{3} \\ \frac{7}{3} \end{array}\right]$$
Question 19.
Find k, if A = $$\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]$$ and A2 = KA – 2I
Solution:
A2 = kA – 2I
∴ AA + 2I = kA
∴ $$\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]=\left[\begin{array}{ll} 3 k & -2 k \\ 4 k & -2 k \end{array}\right]$$
∴ By equality of matrices, we get
3k = 3
∴ k = 1
Question 20.
Find x, if $$\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 2 & 5 \end{array}\right]\left[\begin{array}{c} 1 \\ -2 \\ 3 \end{array}\right]$$ = 0
Solution:
∴ [6 + 12x + 14] =[0]
∴ By equality of matrices, we get
∴ 12x + 20 = 0
∴ 12x =-20
∴ x = $$\frac{-5}{3}$$
Question 21.
Find x and y, if $$\left\{4\left[\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 0 & 2 \end{array}\right]-\left[\begin{array}{ccc} 3 & -3 & 4 \\ 2 & 1 & 1 \end{array}\right]\right\}\left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right]=\left[\begin{array}{l} x \\ y \end{array}\right]$$
Solution:
∴ By equality of matrices, we get
x = 19 andy = 12
Question 22.
Find x, y, z if
$$\left\{3\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \\ 2 & 2 \end{array}\right]-4\left[\begin{array}{cc} 1 & 1 \\ -1 & 2 \\ 3 & 1 \end{array}\right]\right\}\left[\begin{array}{l} 1 \\ 2 \end{array}\right]=\left[\begin{array}{c} x-3 \\ y-1 \\ 2 z \end{array}\right]$$
Solution:
∴ By equality of matrices, we get
x – 3 = -6,y – 1 = 0, 2z = -2
∴ x = – 3, y = 1, z = – 1
Question 23.
If A = $$\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$$ show that A2 = $$=\left[\begin{array}{cc} \cos 2 \alpha & \sin 2 \alpha \\ -\sin 2 \alpha & \cos 2 \alpha \end{array}\right]$$
Solution:
Question 24.
If A = $$\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right]$$, B = $$\left[\begin{array}{cc} 0 & 4 \\ 2 & -1 \end{array}\right]$$
show that AB ≠ BA, but |AB| = |A| . |B|.
Solution:
AB = $$\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ 2 & -1 \end{array}\right]$$
Now, |AB| = $$\left|\begin{array}{cc} 4 & 2 \\ 10 & 7 \end{array}\right|$$ = 28 – 20 = 8
|A| = $$\left|\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right|$$ = 5 – 6 = -1
|B| = $$\left|\begin{array}{cc} 0 & 4 \\ 2 & -1 \end{array}\right|$$ = 0 – 8 = -8
∴ |A| . |B| = (-1).(-8) = 8 = |AB|
∴ AB ≠ BA, but |AB| = |A|.|B|
Question 25.
Jay and Ram are two friends in a class. Jay wanted to buy 4 pens and 8 notebooks, Ram wanted to buy 5 pens and 12 notebooks. Both of them went to a shop. The price of a pen and a notebook which they have selected was 6 and ₹ 10. Using matrix multiplication, find the amount required from each one of them.
Solution:
Let A be the matrix of pens and notebooks and B be the matrix of prices of one pen and one notebook.
The total amount required for each one of them is obtained by matrix AB.
∴ Jay needs ₹ 104 and Ram needs ₹ 150. |
Question
# Given the matrices A and B shown below , solve for X in the equation -frac{1}{3}X+frac{1}{2}A=B A=begin{bmatrix}-10 & 4 8 & 8 end{bmatrix}, B=begin{bmatrix}9 & -1 4& 2 end{bmatrix}
Matrices
Given the matrices A and B shown below , solve for X in the equation $$-\frac{1}{3}X+\frac{1}{2}A=B$$
$$A=\begin{bmatrix}-10 & 4 \\8 & 8 \end{bmatrix}, B=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}$$
2020-12-10
Step 1
the given matrices are:
$$A=\begin{bmatrix}-10 & 4 \\8 & 8 \end{bmatrix}, B=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}$$
we have to solve for X in the equation $$-\frac{1}{3}X+\frac{1}{2}A=B$$
Step 2
the given matrices are: $$A=\begin{bmatrix}-10 & 4 \\8 & 8 \end{bmatrix} \text{ and }B=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}$$
$$-\frac{1}{3}X+\frac{1}{2}A=B$$
$$\frac{-1}{3}X+\frac{1}{2}\begin{bmatrix}-10 & 4 \\8 & 8 \end{bmatrix}=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}$$
$$\frac{-1}{3}X+\begin{bmatrix}-\frac{10}{2} & \frac{4}{2} \\\frac{8}{2} & \frac{8}{2} \end{bmatrix}=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}$$
$$\frac{-1}{3}X+\begin{bmatrix}-5 & 2 \\4 & 4 \end{bmatrix}=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}$$
$$\frac{-1}{3}X=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}-\begin{bmatrix}-5 & 2 \\4 & 4 \end{bmatrix}$$
$$\frac{-1}{3}X=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}+\begin{bmatrix}5 & -2 \\-4 & -4 \end{bmatrix}$$
$$\frac{-1}{3}X=\begin{bmatrix}9+5 & -1-2 \\4-4& 2-4 \end{bmatrix}$$
$$\frac{-1}{3}X=\begin{bmatrix}14 & -3 \\0& -2 \end{bmatrix}$$
$$X=-3\begin{bmatrix}14 & -3 \\0& -2 \end{bmatrix}$$
$$X=\begin{bmatrix}14\times(-3) & -3\times(-3) \\0\times(-3)& -2\times(-3) \end{bmatrix}$$
$$X=\begin{bmatrix}-42 & 9 \\0 & 6 \end{bmatrix}$$
Step 3
therefore the matrix X is $$\begin{bmatrix}-42 & 9 \\0 & 6 \end{bmatrix}$$
therefore the solution of X for the equation $$-\frac{1}{3}X+\frac{1}{2}A=B \text{ is } X=\begin{bmatrix}-42 & 9 \\0 & 6 \end{bmatrix}$$ |
# Video: AQA GCSE Mathematics Higher Tier Pack 5 β’ Paper 1 β’ Question 16
π΄π΅, πΆπ·, and πΈπΉ are straight lines. π΄π΅ is parallel to πΆπ·. Find the value of π¦.
05:00
### Video Transcript
π΄π΅, πΆπ·, and πΈπΉ are straight lines. π΄π΅ is parallel to πΆπ·. Find the value of π¦.
We can see that π¦ has been used in an expression for the size of one of the angles in this diagram. We were told in the question that the line π΄π΅ is parallel to the line πΆπ·. So in order to answer this question, weβre going to need to use some facts about angles in parallel lines. Notice that the expressions weβve been given for two other angles in this diagram are in terms of a second variable π₯. So weβre probably going to need to calculate the value of π₯ before we can calculate the value of π¦. Iβm going to introduce the letters πΊ and π» onto the diagram to represent the points where the line πΈπΉ crosses π΄π΅ and πΆπ·.
Letβs look first of all at the angles π΄πΊπ» and πΆπ»πΈ. And we notice that theyβre enclosed within an πΉ shape. Itβs backwards. But itβs still an πΉ shape. The proper term for angles which are enclosed within an πΉ shape is corresponding angles. So we can conclude that angle π΄πΊπ» and πΆπ»πΈ are corresponding angles. Corresponding angles are equal to each other. So angle π΄πΊπ» is equal to angle πΆπ»πΈ. This means that we can add the expression of two π₯ plus 27 degrees to the top part of our diagram.
Now letβs look at angles π΄πΊπΉ and π΄πΊπ». Thatβs the angle I have marked in pink and the angle I have marked in orange. These two angles lie on a straight line. And we know that the sum of angles on a straight line is 180 degrees. So angle π΄πΊπΉ plus angle π΄πΊπ» equals 180 degrees. We can form an equation by adding the expressions for these two angles and setting it equal to 180. We have four π₯ plus three plus two π₯ plus 27 equals 180.
We can simplify this equation by grouping like terms. Firstly, four π₯ plus two π₯ is six π₯ and three plus 27 is 30. We are now in a position to be able to solve this equation. The first step is to subtract 30 from each side of the equation. On the left-hand side, weβre left with six π₯ and on the right-hand side, 180 minus 30 is 150.
Next, we need to divide both sides of this equation by six. On the left-hand side six π₯ divided by six is just π₯ and on the right-hand side, 150 divided by six is 25. You can work this out with a bit of logic if you remember that four lots of 25 are 100. So six lots of 25 are 150 or you can use a short division method. There are no sixes is in one. So we carry the one into the next column. There are two sixes in 15 as two sixes are 12 with a remainder of three. And there are five sixes in 30 with no remainder, giving our answer of 25.
So weβve found the value of π₯. But how would this help us with the question which was to find the value of π¦? There are two ways that we could now work out the value of π¦. Firstly, we could note that angle π΄πΊπ» and π΅πΊπ» are on a straight line, which means the sum of these two angles must be 180 degrees. As we know the value of π₯, we can work out the size of angle π΄πΊπ» and then form an equation to find the value of π¦. Or we can use the fact that angles π΄πΊπΉ and π΅πΊπ» are vertically opposite angles. Theyβre formed by the intersection of two straight lines. And we know that vertically opposite angles are equal. Letβs use this method.
Angle π΄πΊπΉ is four π₯ plus three degrees. And we know the value of π₯ is 25. So we can work out the size of this angle by substituting 25 for π₯. Four multiplied by 25 is 100 and adding three gives 103. So angle π΄πΊπΉ is 103 degrees. But as vertically opposite angles are equal, angle π΅πΊπ» is also equal to 103 degrees. So we can form an equation: five π¦ minus two equals 103.
To solve for π¦, we must first add two to each side, giving five π¦ equals 105, and then divide both sides of the equation by five to give π¦ equals 21. You can see that 105 divided by five is 21, either using a short division method or remembering that five times 20 is 100. So five times 21 will be 105.
We found the value of π¦. π¦ is equal to 21. |
How do you solve 2/7x-3=-8?
Jan 19, 2017
$x = - \frac{35}{2}$
Explanation:
First note that $\frac{2}{7} x = \frac{2 x}{7}$
To eliminate the fraction, multiply ALL terms on both sides of the equation by 7, the denominator of the fraction.
$\left({\cancel{7}}^{1} \times \frac{2 x}{\cancel{7}} ^ 1\right) - \left(7 \times 3\right) = \left(7 \times - 8\right)$
$\Rightarrow 2 x - 21 = - 56$
$2 x \cancel{- 21} \cancel{+ 21} = - 56 + 21$
$\Rightarrow 2 x = - 35$
To solve for x, divide both sides by 2
$\frac{\cancel{2} x}{\cancel{2}} = \frac{- 35}{2}$
$\Rightarrow x = - \frac{35}{2}$
$\textcolor{b l u e}{\text{As a check}}$
Substitute this value of x into the left side of the equation and if it equals the right side then it is the solution.
$x = - \frac{35}{2} \to \left({\cancel{2}}^{1} / {\cancel{7}}^{1} \times - {\cancel{35}}^{5} / {\cancel{2}}^{1}\right) - 3$
$= - 5 - 3 = - 8 = \text{ right side}$
$\Rightarrow x = - \frac{35}{2} \text{ is the solution}$ |
## Arithmetic Reasoning Flash Card Set 114175
Cards 10 Topics Adding & Subtracting Radicals, Averages, Defining Exponents, Exponent to a Power, Greatest Common Factor, Prime Number, Proportions, Simplifying Radicals
#### Study Guide
To add or subtract radicals, the degree and radicand must be the same. For example, $$2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$$ but $$2\sqrt{2} + 2\sqrt{3}$$ cannot be added because they have different radicands.
###### Averages
The average (or mean) of a group of terms is the sum of the terms divided by the number of terms. Average = $${a_1 + a_2 + ... + a_n \over n}$$
###### Defining Exponents
An exponent (cbe) consists of coefficient (c) and a base (b) raised to a power (e). The exponent indicates the number of times that the base is multiplied by itself. A base with an exponent of 1 equals the base (b1 = b) and a base with an exponent of 0 equals 1 ( (b0 = 1).
###### Exponent to a Power
To raise a term with an exponent to another exponent, retain the base and multiply the exponents: (x2)3 = x(2x3) = x6
###### Greatest Common Factor
The greatest common factor (GCF) is the greatest factor that divides two integers.
###### Prime Number
A prime number is an integer greater than 1 that has no factors other than 1 and itself. Examples of prime numbers include 2, 3, 5, 7, and 11.
###### Proportions
A proportion is a statement that two ratios are equal: a:b = c:d, $${a \over b} = {c \over d}$$. To solve proportions with a variable term, cross-multiply: $${a \over 8} = {3 \over 6}$$, 6a = 24, a = 4.
The radicand of a simplified radical has no perfect square factors. A perfect square is the product of a number multiplied by itself (squared). To simplify a radical, factor out the perfect squares by recognizing that $$\sqrt{a^2} = a$$. For example, $$\sqrt{64} = \sqrt{16 \times 4} = \sqrt{4^2 \times 2^2} = 4 \times 2 = 8$$. |
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# Write an Equation of a Circle
May 22, 2023
## Distance Between Ant Two Points in the X-y Plane
Let P( x1, y1 ) and Q( x2, y2) be any two points in a plane, as shown in the figure.
Hence, the distance ‘d’ between the points P and Q is
d =√(x2−x1)2+(y2−y1)2
This is called the distance formula
1. Find the distance between two points A(4, 3) and B(8, 6)
Solution: Compare these points with ( x1 , y1 ), ( x2 , y2)
x1=4, y1=3, x2=8, y2 =6
d =√(x2−x1)2+(y2−y1)2
=√(8−4)2+(6−3)2
=√42+32
=√16+9
=√25
d = 5 Units
2. Find the distance between two points A(3, 1) and B(6, 4)
Solution: Compare these points with ( x1 , y1 ), ( x2 , y2)
x1=3, y1 =1, x2=6, y2 =4
d =√(x2−x1)2+(y2−y1)2
=√(6−3)2+(4−1)2
=√32+32
=√9+9
=√18
=√9×2
d =3
2–√2
Units ≈4.24 Units
Center: The point in the plane that all points of the circle are equidistant to.
Radius: The line that represents the distance from any given point on the circle to the center.
Since the radius can be expressed as the distance between the center and all points around it, we can use the distance formula to make an equation for the circle.
Write an equation of a circle with a radius r and a center at the origin.
Let (x, y) represent any point on a circle with the center at the origin and radius r.
By the Pythagorean Theorem,
x2+y2= r
This is the equation of a circle with a radius r and a center at the origin.
Example 1
a. Write the equation of the circle shown.
Solution:
The radius is 3, and the center is at the origin.
x2 + y2= r2 Equation of circle.
x2 + y2= 3232 Substitute.
x2 + y2=9 Simplify.
The equation of the circle is x2+y2=92
b. Write the standard equation of the circle shown on the graph.
You can see that the center of the circle is at the origin (0, 0), and the radius is 2 units. Thus,
x2+y2=r2x2+y2=r2
Standard equation of a circle going through the origin
x2+y2=22×2+y2=22
Substitute.
x2+y2=4×2+y2=4
Simplify.
c. Write the standard equation of the circle shown on the graph.
You can see that the center of the circle is at the origin (0,0), and the radius is 20 units. Thus,
x2+y2=r2x2+y2=r2
Standard equation of a circle going through the origin
x2+y2=202 x2+y2=202
Substitute.
x2+y2=400×2+y2=400
Simplify.
Standard Equation of a Circle
CIRCLES CENTERED AT (h, k)
You can write the equation of any circle if you know its radius and the coordinates of its center.
Suppose a circle has a radius r and center (h, k).
Let (x, y) be a point on the circle.
The distance between (x, y) and (h, k) is r, so by the Distance Formula,
(x−h)2+(y−k)2 −−−−−−−−−−−−−−−−√(x−h)2+(y−k)2
=r
Square both sides to find the standard equation of a circle.
KEY CONCEPT
The standard equation of a circle with center (h, k) and radius r is:
(x−h)2+(y−h)2=r2(x−h)2+(y−h)2=r2
Example 2:
Write the standard equation of a circle with center and radius.
1. Center (0, -9) and radius 4.2
1. Center (2, 0), radius 2.5.
1. Center (-2, 5), radius 7
1. Write the standard equation of the circle shown on the graph.
1. Center (0, -9) and radius 4.2.
Solution:
(x−h)2+(y−h)2=r2(x−h)2+(y−h)2=r2
Standard equation of a circle
(x−0)2+(y−(−9))2=4.22(x−0)2+(y−(−9))2=4.22
Substitute.
x2+(y+9)2=x2+(y+9)2=17.64
Simplify.
2. Write the standard equation of a circle with a center (2, 0) and radius of 2.5.
Solution:
h=2, k=0 and r=2.5
(x−h)2+(y−h)2=r2(x−h)2+(y−h)2=r2
Standard equation of a circle.
(x−2)2+(y−0)2=2.52(x−2)2+(y−0)2=2.52
Substitute.
(x−2)2+y2=6.25(x−2)2+y2=6.25
Simplify.
3. Write the standard equation of a circle with center (-2, 5) and radius 7.
Solution:
h=-2, k=5 and r=7
(x−h)2+(y−h)2=r2(x−h)2+(y−h)2=r2
Standard equation of a circle.
(x−(−2))2+(y−5)2=72(x−(−2))2+(y−5)2=72
Substitute.
(x+2)2+(y−5)2=49(x+2)2+(y−5)2=49
Simplify.
4. You can see that the center of the circle is at the point (2, 3), and the radius is 2 units. Thus,
(x−a)2+(y−b)2=r2(x−a)2+(y−b)2=r2
Standard equation of a circle.
(x−2)2+(y−3)2=22(x−2)2+(y−3)2=22
Substitute.
(x−2)2+(y−3)2=4(x−2)2+(y−3)2=4
Simplify.
Example 3:
1. The point (-5, 6) is on a circle with center (-1, 3). Next, write the standard equation of the circle.
1. The point (3, 4) is on a circle whose center is (1, 4).
Write the standard equation of the circle.
1. The point (-1, 2) is on a circle whose center is (2, 6). Write the standard equation of the circle
Solutions:
a. Step 1:
To write the standard equation, you need to know the values of h, k, and r.
To find r, find the distance between the center and the point (-5, 6) on the circle.
r =
(x−h)2+(y−k)2 −−−−−−−−−−−−−−−−√ (x−h)2+(y−k)2
Distance formula
r =
[−5−(−1)]2+(6−3)2−−−−−−−−−−−−−−−−−−√ [−5−−1]2+(6−3)2
=
(−5+1)2+(3)2−−−−−−−−−−−−−√(−5+1)2+(3)2
Simplify
=
(−4)2+32−−−−−−−−−√(−4)2+32
=
16+9−−−−−√16+9
=
25−−√25
∴r = 5
1. The point (3, 4) is on a circle whose center is (1, 4). Write the standard equation of the circle.
Solution:
To write the standard equation, you need to know the values of h, k, and r.
To find r, find the distance between the center and the point (3, 4) on the circle.
h = 1, k =4, x=3 and y= 4
r =
(x−h)2+(y−k)2 −−−−−−−−−−−−−−−−√ (x−h)2+(y−k)2
Distance formula
r =
(3−1)2+(4−4)2−−−−−−−−−−−−−−−√ (3−1)2+(4−4)2
=
22+02−−−−−−√22+02
Simplify
=
4–√4
r =2
Substitute (h, k) = (1, 4) and r = 2 into the standard equation of a circle.
(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
Standard equation of a circle.
(x−1)2+(y−4)2=22(x−1)2+(y−4)2=22
Substitute.
(x−1)2+(y−4)2=4(x−1)2+(y−4)2=4
Simplify.
The standard equation of the circle is
(x−1)2+(y−4)2=4(x−1)2+(y−4)2=4
3. The point (-1, 2) is on a circle whose center is (2, 6). Write the standard equation of the circle.
Solution:
To write the standard equation, you need to know the values of h, k, and r.
To find r, find the distance between the center and the point (-1, 2) on the circle.
h =1, k =4, x=3 and y= 4
r =
(−1−2)2+(2−6)2 −−−−−−−−−−−−−−−−−√ (−1−2)2+(2−6)2
Distance formula
r =
(−3)2+(−4)2−−−−−−−−−−−−√ (−3)2+(−4)2
=
9+16−−−−−√9+16
Simplify
=
25−−√25
r = 5 Units
Substitute (h, k) = (2, 6) and r = 5 into the standard equation of a circle.
(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
Standard equation of a circle.
(x−2)2+(y−6)2=52(x−2)2+(y−6)2=52
Substitute.
(x−2)2+(y−6)2=25(x−2)2+(y−6)2=25
Simplify.
The standard equation of the circle is
(x−2)2+(y−6)2=25(x−2)2+(y−6)2=25
Example 4:
1. The equation of a circle is (x−4)2+(y+2)2=36(x−4)2+(y+2)2=36. Graph the circle.
1. The standard equation of a circle is (x+2)2+(y−2)2=4(x+2)2+(y−2)2=4. Graph the circle.
1. The standard equation of a circle is x2+(y−1)2=16×2+(y−1)2=16. Graph the circle.
1. The standard equation of a circle is (x+2)2+(y−3)2=25(x+2)2+(y−3)2=25. Graph the circle.
1. The equation of a circle is (x−4)2+(y+2)2=36(x−4)2+(y+2)2=36. Graph the circle.
Solution:
Rewrite the equation to find the center and radius.
(x−4)2+(y+2)2=36(x−4)2+(y+2)2=36
(x−4)2+[(y−(−2)]2=62(x−4)2+[(y−−2]2=62
The center is (4, -2), and the radius is 6.
Use a compass to graph the circle.
2. The equation of a circle is (x+2)2+(y−2)2=4(x+2)2+(y−2)2=4. Graph the circle.
Solution:
Rewrite the equation to find the center and radius.
(x+2)2+(y−2)2=4(x+2)2+(y−2)2=4
(x−(−2)2+(y−2)2=22(x−(−2)2+(y−2)2=22
The center is (-2, 2), and the radius is 2.
Use a compass to graph the circle.
3. The standard equation of a circle is
x2+(y−1)2=16×2+(y−1)2=16
Graph the circle.
Solution:
Rewrite the equation to find the center and radius.
x2+(y−1)2=16×2+(y−1)2=16
(x−0)2+(y−1)2=42(x−0)2+(y−1)2=42
The center is (0, 1), and the radius is 4.
Use a compass to graph the circle.
4. The standard equation of a circle is
(x+2)2+(y−3)2=25(x+2)2+(y−3)2=25
Graph the circle.
Solution:
Rewrite the equation to find the center and radius.
[x−(−2)]2+(y−3)2=25[x−−2]2+(y−3)2=25
[x−(−2)]2+(y−3)2=52[x−−2]2+(y−3)2=52
The center is (-2, 3), and the radius is 5.
Use a compass to graph the circle.
Use graphs of circles
Example 5:
Earthquakes
The epicenter of an earthquake is the point on Earth’s surface directly above the earthquake’s origin. A seismograph can be used to determine the distance to the epicenter of an earthquake. Seismographs are needed in three different places to locate an earthquake’s epicenter. Use the seismograph readings from locations A, B, and C to find the epicenter of an earthquake.
• The epicenter is 7 miles away from A(-2, 2.5).
• The epicenter is 4 miles away from B(4, 6).
• The epicenter is 5 miles away from C(3, -2.5).
Solution:
The set of all points equidistant from a given point is a circle, so the epicenter is located on each of the following circles.
• A with center (-2, 2.5) and radius 7
• B with center (4, 6) and radius 4
• C with center (3, -2.5) and radius 5
To find the epicenter, graph the circles on a graph where units are measured in miles. Find the point of intersection of all three circles.
• The epicenter is at about (5, 2).
1. Write the equation of a circle shown below.
1. Write the standard equation of a circle with center (2, -4) and radius 5.
1. The point (2, 3) is on a circle with center (4, 1). Write the standard equation of the circle.
1. The standard equation of a circle is (x−4)2+(y−1)2=1(x−4)2+(y−1)2=1. Graph the circle.
1. Determine whether the given equation defines a circle. If the equation defines a circle, rewrite the equation in standard form.
x2−8x+16+y2+2y+4=25×2−8x+16+y2+2y+4=25
Solutions:
Solution 1:
The radius is 4, and the center is at the origin.
x2+y2=r2x2+y2=r2
x2+y2=42×2+y2=42
x2+y2=16×2+y2=16
The equation of the circle is
x2+y2=16×2+y2=16
Solution 2:
h=2, k=-4 and r=5
Standard equation of a circle
(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
(x−2)2+[y−(−4)]2=52(x−2)2+[y−−4]2=52
(x−2)2+(y+4)2=25(x−2)2+(y+4)2=25
Solution 3:
Step 1: Point (2, 3) center (4, 1).
Point (x, y) center (h, k).
r =
(x−h)2+(y−k)2−−−−−−−−−−−−−−−√(x−h)2+(y−k)2
=
(2−4)2+(3−1)2−−−−−−−−−−−−−−−√(2−4)2+(3−1)2
=
(−2)2+(2)2−−−−−−−−−−√(−2)2+(2)2
=
4+4−−−−√4+4
=
8–√8
= 2
2–√2
r≈2.8
Step 2: Substitute (h, k) = (4, 1) and r =2.8 into the standard equation of a circle.
is
(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
(x−4)2+(y−1)2=2.82(x−4)2+(y−1)2=2.82
(x−4)2+(y−1)2=7.84(x−4)2+(y−1)2=7.84
Solution:
Rewrite the equation to find the center and radius.
(x−4)2+(y−1)2=1(x−4)2+(y−1)2=1
(x−4)2+(y−1)2=12(x−4)2+(y−1)2=12
(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
The center is (4, 1), and the radius is 1. Use a compass to graph the circle.
5. Determine whether the given equation defines a circle. If the equation defines a circle, rewrite the equation in standard form.
x2−8x+16+y2+2y+4=25×2−8x+16+y2+2y+4=25
Solution:
x2−8x+16+y2+2y+4=25×2−8x+16+y2+2y+4=25
x2−8x+42+y2+2y+22=52×2−8x+42+y2+2y+22=52
x2−2.x.4+42+y2+2.y.1+22=52×2−2.x.4+42+y2+2.y.1+22=52
(x−4)2+(y−2)2=52(x−4)2+(y−2)2=52
By using algebraic identity formulas
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 […] |
# 2.2 Graphs of linear functions (Page 9/15)
Page 9 / 15
Access these online resources for additional instruction and practice with graphs of linear functions.
## Key concepts
• Linear functions may be graphed by plotting points or by using the y -intercept and slope. See [link] and [link] .
• Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See [link] .
• The y -intercept and slope of a line may be used to write the equation of a line.
• The x -intercept is the point at which the graph of a linear function crosses the x -axis. See [link] and [link] .
• Horizontal lines are written in the form, $f\left(x\right)=b.$ See [link] .
• Vertical lines are written in the form, $x=b.$ See [link] .
• Parallel lines have the same slope.
• Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See [link] .
• A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x - and y -values of the given point into the equation, $f\left(x\right)=mx+b,$ and using the $b$ that results. Similarly, the point-slope form of an equation can also be used. See [link] .
• A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope. See [link] and [link] .
• A system of linear equations may be solved setting the two equations equal to one another and solving for $x.$ The y -value may be found by evaluating either one of the original equations using this x -value.
• A system of linear equations may also be solved by finding the point of intersection on a graph. See [link] and [link] .
## Verbal
If the graphs of two linear functions are parallel, describe the relationship between the slopes and the y -intercepts.
The slopes are equal; y -intercepts are not equal.
If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y -intercepts.
If a horizontal line has the equation $f\left(x\right)=a$ and a vertical line has the equation $x=a,$ what is the point of intersection? Explain why what you found is the point of intersection.
The point of intersection is $\left(a,a\right).$ This is because for the horizontal line, all of the $y$ coordinates are $a$ and for the vertical line, all of the $x$ coordinates are $a.$ The point of intersection will have these two characteristics.
Explain how to find a line parallel to a linear function that passes through a given point.
Explain how to find a line perpendicular to a linear function that passes through a given point.
First, find the slope of the linear function. Then take the negative reciprocal of the slope; this is the slope of the perpendicular line. Substitute the slope of the perpendicular line and the coordinate of the given point into the equation $y=mx+b$ and solve for $b.$ Then write the equation of the line in the form $y=mx+b$ by substituting in $m$ and $b.$
## Algebraic
For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular:
$\begin{array}{l}4x-7y=10\hfill \\ 7x+4y=1\hfill \end{array}$
a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size
what is the importance knowing the graph of circular functions?
can get some help basic precalculus
What do you need help with?
Andrew
how to convert general to standard form with not perfect trinomial
can get some help inverse function
ismail
Rectangle coordinate
how to find for x
it depends on the equation
Robert
whats a domain
The domain of a function is the set of all input on which the function is defined. For example all real numbers are the Domain of any Polynomial function.
Spiro
foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24.
difference between calculus and pre calculus?
give me an example of a problem so that I can practice answering
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
I want to learn about the law of exponent
explain this
what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created? |
# Subtracting from 5
11 teachers like this lesson
Print Lesson
## Objective
Students will be to able solve subtraction equations.
#### Big Idea
We have 10 fingers, so when learning to subtract starting with a number less than 10 or equal to 10, students can use their fingers to help them. Manipulatives and drawings are helpful too!
## Problem of the Day
5 minutes
I start each math lesson with a Problem of the Day. I use the procedures outlined here on Problem of the Day Procedures.
Today's Problem of the Day:
There were 8 bats in the tree. When some bugs came by, 4 flew away. How many bats are left in the tree?
I set this problem up with some structure to help the students organize their thinking. I give a blank number sentence to encourage students to write an equation. I also include a bat set to infinite cloner. If you do not have a SMART Board, you can use the PDF and manipulatives, pictures or students drawings.
Since we do this whole group, I have one student come up and do this problem. I remind the student to check their work when they are finished and have the class tell if they agree or disagree by showing a thumbs up or thumbs down.
## Presentation of Lesson
25 minutes
I start this by singing Five Little Speckled Frog. See it with a cute video here. I then write the number 5 on the board.
What number is this? Today we are going to look at what happens when we take away from 5. Let's start by using what we already have on our bodies that can help us count. Put up 5 fingers. If I have 5 and I take away 0, how many fingers do I still have up?
I show it on my fingers and write the equation on the board. 5 - 0 = 5 We continue with all of the other possibilities 5-1, 5-2, 5-3, 5-4, and 5-5.
When we are finished, I tell students that they are going to be practicing taking away from 5 on a Take Away From 5 Worksheet.
We are going to work on this paper together. When you get to your seat, do not touch your cup of counters. You need to get out a pencil and put your name on your paper. When your name is on your paper hold your pencil in the air, that will let me know that you are ready to start.
I use the procedures outlined here on the Paper Procedures. Prior to this lesson, I placed a plastic cup at each students' place containing four color tiles.
Use your color tiles to model each equation. Write the numbers to complete the equations.
The first thing the directions tell you to do is use the color tiles to model each equation. You will notice that the last question does not have an equation to model. For this question, you will need to use the color tiles to come up with your own way to subtract from 5. Check it out here.
We work through the first 5 questions together. For the last question, I show one way to do it on the SMARTBoard. I then allow the students to come up with their own way. I walk around and make sure that students are correctly counting writing their equations. When students are finished with their paper, they can put it in the basket and get their center.
## Practice
20 minutes
The centers for this week are:
I quickly circulate to make sure students are engaged and do not have any questions about how to complete the centers. Because of the trouble during our addition unit, I decided not to introduce any subtraction centers this early in the subtraction unit.
I pull 3 groups (5-10 minutes depending on need). We are working on writing equations in all groups. I verbally give the group an equation. I have them write it and then solve it using manipulatives. With students who are able to do this easily, I also have them try with drawing pictures instead of manipulatives.
Prior to clean up, I check in with the other tables to see how the centers are going. My students have been struggling with getting cleaned up quickly and quietly after centers. Lately I have been using counting down from 20 slowly instead of a clean up song. Counting backwards is as critical as counting up. Students need to be able to know the number that comes before, as well as after, any given number (w/i 10, w/i 20, etc.). Counting back is a critical strategy for subtraction.
The students like to count backwards with me as they clean up and I can lengthen or reduce the clean up time based on how students are doing and how much time we have.
## Closing
5 minutes
To close, I put a student's paper on the document camera a project it on the SMART Board and have that student explain their work. I have the class read the equation using the words minus and equals. I mention positive things noticed during centers as well as something that needs to be better next time.
I review what we did during our whole group lesson. "Today we learned about taking away from 5. Tomorrow we are going to continue this series of lessons where where we practice subtracting from a given number!" |
# Square Root and Cube Root
Square Root and Cube Root Concepts is used in almost every topic of mathematics. So, candidates can learn Square Root and Cube Root Elementary Algebra solving techniques, symbol, formula, tricks etc from this page. Also, Square Root and Cube Root Table PDF is also shared here that will be useful for the candidates during preparation of competitive examination and can help in solving the question in time.
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## Square Root and Cube Root
Square Root and Cube Root Symbol:
The square root is denoted by the symbol ‘√’, whereas the cube root is denoted by ‘∛’.
Examples:
√4 = √(2 × 2) = 2
∛27 = ∛(3 × 3 × 3) = 3
Square Root and Cube Root Chart:
Number Square root (√) Cube root (∛) 1 1.000 1.000 2 1.414 1.260 3 1.732 1.442 4 2.000 1.587 5 2.236 1.710 6 2.449 1.817 7 2.646 1.913 8 2.828 2.000 9 3.000 2.080 10 3.162 2.154 11 3.317 2.224 12 3.464 2.289 13 3.606 2.351 14 3.742 2.410 15 3.873 2.466
Square Root Formula:
When a value is multiplied by itself to give the original number then that number is a square root. Square Root is represented by the symbol √.
For example, 4 and -4 are the square roots of 16.
The formula to represent the square root is given below:
Solved Example: What is the square root of 144?
Solution:
The factors of 144 are given as,
144 = 12×12
√144 =√12×12
√144= 12
Solved Example: What is the square root of 80?
Solution:
The factors of 80 are given as,
80= 4×4×5
√80 =√4×4×5
√80=4√5
Cube Root Definition:
The cube root of any number say ‘a’, is the number ‘b’, which satisfy the equation given below:
b= a
This can be represented as:
a = ∛b
Cube Root Formula:
The cube root formula is used to give the cube root value of any number.
For Example:
5 Cube = 53 = 125
Cube Root of 125 = 3√125 = 5
Thus, the cube root of 125 is 5.
Solved Example: What is the cube root of 1728?
Solution:
The factors of 1728 are given as,
1728 = 12 × 12 × 12
3√1728 = 3√(12 × 12 × 12)
3√1728 = 12
Cube Root by Prime Factorisation Method:
We can find the cube-root of a number by the method of prime factorisation.
For Example:
2744= 2 × 2× 2 × 7 ×7 × 7= (2 × 7 )3
Therefore, the cube root of 2744 = ∛2744 = 2 × 7 = 14
Solve Here: Chain Rule Aptitude Questions and Answers
Easy Tricks to Find Square Roots and Cube Roots:
Finding Square Root:
Above 100:
1032 = 10609
Step 1. Add the number to the ones digit:
103 + 3 = 106
Step 2. Square the ones digit number (if the result is a single digit put a 0 in front of it):
32 = 09
Step 3. Place the result from Step 2 next to the result from Step 1: 10609
Below 100:
972 = 9409
Step 1. Subtract the number from 100: 100- 97 = 3
Step 2. Subtract the number (from Step 1) from original number : 97-3 =94
Step 3. Square the result from Step 1 (if the result is a single digit put a 0 in front of it): 3= 09
Step 4. Place the result from Step 3 next to the result from Step 2: 9409
Below 50:
482 = 2304
Step 1. Subtract the number from 50: 50-48=2
Step 2. Subtract the result (from Step 1) from 25: 25-2 =23
Step 3. Square the result from Step 1 if the result is a single digit put a 0 in front of it ) : 22 = 04
Step 4. Place the result from Step 3 next to the result from Step 2: 2304
Above 50:
532 = 2809
Step 1. Add 25 to the ones digit: 25 + 3 = 28
Step 2. Square the ones digit number ( if the result is a single digit put a 0 in front of it ) : 32 = 09
Step 3. Place the result from Step 2 next to the result from Step 1 : 2809
Finding Cube Root:
Remembering Units Digits:
Remember cubes of 1 to 10 and unit digits of these cubes.
1 = 1
2 = 8
3 = 7
4 = 4
5 = 5
6 = 6
7 = 3
8 = 2
9 = 9
10 = 0
Whenever unit digit of a number is 9, the unit digit of the cube of that number will also be 9. Similarly, if the unit digit of a number is 9, the unit digit of the cube root of that number will also be 9.
Deriving Cube Root From Remaining Digits
Find the cube root of 474552.
Unit digit of 474552 is 2. So we can say that unit digit of its cube root will be 8.
Now we find cube root of 447552 by deriving from remaining digits.
Let us consider the remaining digits leaving the last 3 digits. i.e. 474.
Since 474 comes in between cubes of 7 and 8.
So the ten’s digit of the cube root will definitely be 7
i.e. cube root of 474552 will be 78.
Solve Here: Number Series Question
Properties of Square Root:
• If the unit digit of a number is 2, 3, 7, and 8 then its square root is not a natural number.
• If a number ends in an odd number of zeros, then its square root is not a natural number.
• The square root of an even number is even and that of an odd number is odd.
• Negative numbers have no squares root in a set of real numbers.
Properties of Cube Root:
• Cube root of all the odd numbers is an odd number.
• Cube root of all the even natural numbers is even.
• The cube root of a negative integer always results in negative.
The details mentioned above about Square Root and Cube Root Topic is well written by the team members of reruitmentresult.com Individuals must go through it and prepare well for their examination.
Something That You Should Put an Eye On
Problem on HCF and LCM Profit And Loss Problems with Solutions Percentage Aptitude Questions Simple Interest Aptitude Questions Time And Distance Questions Compound Interest Aptitude Questions Decimal Fraction Important Formulas Logarithms Formulas |
The Paradox of Translating a Parabola
by
Sarah Major
The following is based on problem #3 given in Exploration 2, which states:
Examine graphs for the parabola
for different values of a, b, and c. (a, b, c can be any rational numbers).
Fix the values for a and b, vary c. Make at least 5 graphs on the same axes as you vary c.
Try an animation for the same range.
What is happening mathematicatically?
Can you prove this is a translation and that the shape of the parabola does not change?
For the sake of analyzing the mathematics behind these parabolas, I will use the simplest form possible by assigning a=1 and b=1. c will range from -2 to 2. The following is the graphs of those equations using Graphing Calculator 4.0:
From these graphs, we can see that the value for c determines where the y-intercept for the graph is. These locations make it easy to see what is visually happening with the graph. Each time c increases by 1, the y-intercept increases by the same amount. This, in turn, means that all of the points are "shifted up" by the same amount. That is, all of the points on the graphs are translated from (x,y) to (x,y+k). But why does adding or subtracting a certain number of units from the equation of a parabola cause the graph to vertically move that same number of units, and why that translation in particular?
Let's break this problem down mathematically. To make the notation easy, let's consider an equation, f(x). Now, say it has an initial point of (x1,y1), where: y1=f(x1).
Now say we are adding a constant, k (which we will say is positive), to our function so that, once added, we have y=f(x)+k. If we plug in our initial point into this new function, we get y1=f(x1)+k. If we plug in our initial point, we get a different y-value. We get the equation y2=f(x1)+k, or y2=y1+k if we substitute.
After this occurs our initial point, (x1,y1), is translated to (x1,y2), and in this case, y2=y1+k. As we can see, the x-value for the second function has stayed the same, so there is no change in the horizontal value, or x-value, at all. However, the y-value has changed by k units, so it has been moved up k units (since k is positive). If k was negative, then we would have a value subtracted from the original function, so the graph would move downwards. See below for a diagram of the process we just went through:
Therefore, since we have successfully generalized this concept and showed how it mathematically works, we can safely say that adding a value to a function translates the function to points that are that many units above the original points while subtracting from a function translates all of the points down that many units.
From the above, we can see that in the context of our original equations, that each graph is 1 unit higher than the one before it because we are adding 1 unit to the whole function. However, how do we know that the shape of the graph is not changing since we are not able to see the same amount of each graph? This answer comes from finding the area contained within the parabola.
The formula for finding the area contained by a parabola (commonly used to find the area under a parabolic arch) is:
where b is the base and h is the height. The height goes between the vertex of the parabola and the base while the base is made from the intersection of a horizontal line segment intersecting both sides of the parabola. See the following diagram made with Geometer's Sketchpad 4.0:
Why does this formula work? It comes from the construction of a rectangle in this area of a parabola. Let's change the diagram to that of a rectangle:
In this diagram, b times h would be the area of the rectangle. In essence, the formula says that the area in this part of the parabola is 2/3 of the area of the rectangle.
How do we know that this formula works? What other mathematical concept do we use to find the area under a graph? The first thing that should come to mind is the use of integration. However, when we integrate, we are finding the area UNDER the curve in reference to the x-axis. In the case of our graph above, the x-axis constitutes the TOP of our area. This means that if we integrate this particular parabola, we will get a negative area. The absolute value of this value will be the actual area that we are trying to calculate. Let's see what we get when we integrate this particular parabola:
For:
```
```
And what would the area be using our formula?
So the formula works. Will we have the same area if we add 4? Be mindful that we must choose one of our parameters, either base or height, to be the same as our first parabola to make the formula useful.
For:
So area is preserved when the parabola is shifted vertically.
Is there another way to see if the shape of a parabola is preserved if we move it vertically? This answer can come from looking at the definition of a parabola. We define a parabola as the set of points that are equidistant from the directrix and the focus. If we know that these distances are preserved when the parabola is shifted vertically, we'll know that the shape is preserved. But how do we find the directrix and focus of our parabola?
Most students are used to working with the vertex form of a parabola equation, which is:
where a is the stretch or shrink of the graph and (h,k) is the vertex. a is also the same a that appears in the general form of the parabola.
A form that students are usually not as familiar with is the conic form:
where, like the other form, (h,k) is the vertex, but p is the vertical distance from the vertex to the focus, making it also the vertical distance from the vertex to the directrix.
Is there a way to find p without having to convert the equation into the conic form? There is a way to find the equivalent of it in either the vertex form or the general form for the parabola:
```
```
So, essentially, 1/a = 4p. Therefore, if we want to find p using either the general form or the vertex form, we simply find the reciprocal of the value corresponding to a and divide it by 4.
In the context of the parabolas that we have been using, a = 1, so p = 1/4. We can then find the focus, which will be 1/4 units above the vertex, and the directrix, which will be 1/4 below the vertex. First, we must put the equation into vertex form to find the vertex:
```
```
So the vertex is (-0.5,-2.25). From this equation, a=1, so p=0.25. Since p is the distance from the vertex to the focus and also from the vertex to the directrix, we can use this to find the point that is our focus and the line that is our directrix. The focus would be (-0.5,-2) and our directrix would be the line y=-2.50. Let's graph these along with our original parabola:
One can visually see that the vertex of this parabola is equidistant from the vertex and the directrix. We can also pick some random points on the parabola to find the distance from them to our directrix and focus so that when we translate our parabola, we can see if ths distance is preserved and thus confirming that the shape and size is also preserved. Let's pick three points, two on the left side of the parabola and one on the right side. These points will be (-2,0), (-3,4), and (0.5,-1.25):
We first need to calculate the distance from each of the points to both the focus and the directrix. The distance from each point to the focus is easy because we can just use the distance formula:
The distance from the point to the directrix, however, is a bit more complicated because we are not given a certain point on the directrix in which to calculate the distance. We must find the path of least distance from the points to the line that is the directrix. We can do this by using linear algebra.
To do this, we first pick two points on the line, (x1,y1) and (x2,y2). Then we must find the vector perpindicular to the line, which will be given by:
Next we find the vector from the point in question, which we will call for the moment (x0,y0), to the first point on the line. This is given by:
Then, to find the distance from our initial point to the line, we simply project r onto v, which we will call d, by doing the following:
The two points on the line we choose do not matter. They are just specifically so that a vector perpindicular to it can be found. For my calculations, I will call the distance from a point to the focus df and the distance from a point to the directrix dd. Therefore, we get the following distances:
For (-2,0): df = 2.5, dd = 2.5
For (-3,4): df = 6.5, dd = 6.5
For (0.5,-1.25): df = 1.25, dd = 1.25
Just like the definition of a parabola states, the two distances for each point are the same. Therefore, it is not necessary to calculate them both, but for the sake of our investigation, we will to make sure they are the same.
Now, we need to shift the parabola vertically, say four units, and do the same calculations from the corresponding points on the new parabola to its new focus and directrix. Let's first shift our parabola. If we do a vertical translation of four units up, our new equation will be:
And the graph will be as follows:
Let's put our equation in vertex form so that we can find the vertex:
So the vertex is (-0.5,1.75). Since the a value is still the same, p is still 0.25, which is a good sign for what we are trying to show. We can now graph our focus and directrix. The focus will be the point (-0.5,2) and the directrix will be the line y = 1.5:
Now we need to find our corresponding points and make sure they lie on the graph of the parabola. Since the graph was shifted vertically up 4 units, our points should be (-2,4), (-3,8), and (0.5,2.75):
They all do appear on the parabola. Now just to find the distances from each point to the focus and from each point to the directrix:
For (-2,4): df = 2.5, dd = 2.5
For (-3,8): df = 6.5, dd = 6.5
For (0.5,2.75): df = 1.25, dd = 1.25
So, the distances for a translated parabola that has been shifted vertically are the same. Thus, the size and shape of the parabola has been preserved. |
Table of Contents (Long Version)
## Chapter 1. From Formula to Program
1.1 A Small Example
1.2 Sines and Cosines
1.3 Max's and Min's
1.4 Formulas that Involve Integers
We grow up in mathematics thinking that the formula'' is a ticket to solution space. Want the surface area of a sphere? Use A = 4pi r2. Have the cosine of some angle $\theta \in [0,\pi/2]$ and want $\cos(\theta/2)$? Use $\cos(\theta/2) = \sqrt{\frac{1+\cos(\theta)}{2}}.$ Want the minimum value $\mu$ of the quadratic function $q(x) = x^{2} + bx +c$ on the interval $[L,R]$? Use $\mu = \left\{ \begin{array}{ll} q(-b/2) & \mbox{if } L \leq -b/2 \leq R \\ & \\ \mbox{min}\{q(L),q(R)\} & \mbox{otherwise} \end{array}. \right.$ Want to know if year $y$ is a leap year? Use the rule that $y$ is a leap year if it is a century year divisible by 400 or a non-century year divisible by 4.
Sometimes the application of a formula involves a simple substitution. Thus, the surface area of a one-inch ball bearing is 4\pi(1/2)^{2} = pi square inches. Sometimes we have to check things before choosing the correct option'' within a formula. Since 0 <= 3 <= 10, the minimum value of q(x) = x^{2} -6x + 5 on the interval [0,10] is q(3) = -4. Sometimes we must clarify the assumptions upon which the formula rests. Thus, if a century year means something like 1400, 1500, and 2000, then 1900 was not a leap year.
Writing programs that use simple formulas like the above are a good way to begin our introduction to computational science. We'll see that the mere possession of a formula is just the start of the problem-solving process. The real ticket to solution space, if you travel by computer, is the program. And that's what we have to learn to write.
## Chapter 2. Numerical Exploration
2.1 Discovering limits
2.2 Confirming Conjectures
2.3 The Floating Point Terrain
2.4 Interactive Frameworks
All work and no play does not a computational scientist make. It is essential to be able to {\em play} with computational idea before moving on to its formal codification and development. This is very much a comment about the role of intuition. A computational experiment can get our mind moving in a creative direction. In that sense, merely watching what a program does is no different then watching a chemistry experiment unfold: it gets us to think about concepts and relationships. It builds intuition.
The chapter begins with a small example to illustrate this point. The area of a circle is computed as a limit of regular polygon areas. We discover'' pi by writing and running a sequence of programs.
Sometimes our understanding of an established result is solidified by experiments that confirm its correctness. In Section 2.2 we check out a theorem from number theory that says 3^{2k+1} + 2^{k} is divisible by 7 for all positive integers k.
To set the stage for more involved computational trips'' into mathematics and science, we explore the landscape of floating point numbers. The terrain is finite and dangerous. Our aim is simply to build a working intuition for the limits of floating point arithmetic. Formal models are not developed. We're quite happy just to run a few well chosen computational experiments that show the lay of the land and build an appreciation for the inexactitude of real arithmetic.
The design of effective problem-solving {\em environments} for the computational scientist is a research area of immense importance. The goal is to shorten the path from concept to computer program. We have much to say about this throughout the text, In Section 2.4 we develop the notion of an interactive framework that fosters the exploration of elementary computational ideas.
## Chapter 3. Elementary Graphics
3.1 Grids
3.2 Rectangles and Ovals
3.3 Granularity
• It is hard to overstate the importance of graphics to computational science. Three reasons immediately come to mind:
• In most large applications, the amount of numerical data that makes up the answer'' is just too much for the human mind to assimilate in tabular form.
• The visual display of data often permits the computational scientist to spot patterns that would otherwise be hidden.
• Many computations have geometric answers and it is more effective to show the answer than to describe it in numerical terms.
To build an appreciation for computer graphics we need to do computer graphics. In this chapter we get started with a handful of ThinkPascal graphics procedures that are utilized throughout the text. Elementary graphical problems are solved that involve grids, rectangles, and ovals. A fringe benefit of the chosen applications is that they give us an opportunity to build up our iteration expertise. Visual patterns involve repetition and repetition requires the writing of loops. Screen granularity provides another setting for exploring the interplay between continuous and the discrete mathematics.
## Chapter 4. Sequences
4.1 Summation
4.2 Recursions
In Chapter 2 we played with the sequence of regular n-gon areas $\{a_{n}\}$ where $A_{n} = \frac{n}{2} \sin\left( \frac{2\pi}{n} \right).$ We numerically discovered'' that $\lim_{n\rightarrow \infty} A_{n} = \pi \:,$ a fact that is consistent with our geometric intuition.
In this chapter we build our n-th term expertise'' by exploring sequences that are specified in various ways. At the top of our agenda are sequences of sums like $S_{n} = 1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^{2}}.$ Many important functions can be approximated by very simple summations, e.g., $\exp(x) \approx 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}.$ The quality of the approximation depends upon the value of $x$ and the integer $n$.
Sometimes a sequence is defined {\em recursively}. The $n$-term may be specified as a function of previous terms, e.g., $f_{n} = \left\{ \begin{array}{ll} 1 & \mbox{ if n = 1 or 2 }\\ & \\ f_{n-1} + f_{n-2} & \mbox{ if n \geq 3} \end{array} \right.$ Sequence problems of this variety give us the opportunity to practice some difficult formula-to-program transitions.
## Chapter 5. Random Simulations
5.1 Generating Random Reals and Integers
5.2 Estimating Probabilities and Averages
5.3 Monte Carlo
Many phenomena have a random, probabilistic aspect: the role of the dice, the diffusion of a gas, the number of customers the enter a bank between noon and 12:05. Some special tools are needed to simulate events like these with the computer. This section is about random number generation and how to write programs that answer questions about random phenomena.
## Chapter 6. Fast, Faster, Fastest
6.1 Benchmarking
6.2 Efficiency
How fast a program runs is usually of interest and so the intelligent acquisition of timing data, called {\em benchmarking}, is important. Benchmarking serves many purposes:
• It can be used to identify program bottlenecks. \item It can be used to the quantify how hard it is to solve a problem.
• It can be used to determine whether one solution process is more favorable than another.
• It can be used to calibrate the performance of a particular machine architecture.
However, in this chapter we merely illustrate the mechanics of benchmarking and show how it can be used to assess efficiency improvements as a program undergoes development.
Two examples are used to illustrate the design of efficient code. The plotting of an ellipse is used to show how to remove redundant arithmetic and function evaluation in a loop context. The computation of a rational approximation to $pi$ is used to show how the reduction of a doubly-nested fragment to a single loop can result in an order-of-magnitude speed-up.
Behind all the discussion is a quiet, but very important ambition: to build an aesthetic appreciation for the fast program. Programs that run fast are creations of beauty. This is widely accepted in practical settings where time is money. But in addition, program efficiency is something to revel in for its own sake. It should be among the aspirations for every computational scientist who writes programs.
## Chapter 7. Exponential Growth
7.1 Powers
7.2 Binomial Coefficients
There are a number of reasons why the built-in sin function is so handy. To begin with, it enables us to compute sines without having a clue about the method used. It so happens that the design of an accurate and efficient sine function is somewhat involved. But by taking the black box'' approach, we are able to be effective sin-users while being blissfully unaware of how the built-in function works. All we need to know is that sin-expects a real input value and that it returns the sine of that value interpreted in radians.
Another advantage of sin can be measured in keystrokes and program readability. Instead of disrupting the real business'' of a program with lengthy compute-the-sine fragments, we merely invoke sin as required. The resulting program is shorter and reads more like traditional mathematics.
A programming language like ThinkPascal always comes equipped with a library of built-in functions. The designers of the language determine the library's content by anticipating who will be using the language. If that group includes scientists and engineers, then invariably there will be built-in functions for the sine, cosine, log, and exponential functions because they are of central importance to work in these areas.
It turns out that if you need a function that is not part of the built-in function library, then you can write your own. The art of being able to write efficient, carefully organized functions is an absolutely essential skill for the computational scientist because it suppresses detail and permits a higher level of algorithmic thought.
To illustrate the mechanics of function writing we have chosen a set of examples that highlight a number of important issues. On the continuous side we look at powers, exponentials, and logs. These functions are monotone increasing and can be used to capture different rates of growth. Factorials and binomial coefficients are important for counting combinations. We bridge the continuous/discrete dichotomy through a selection of problems that involve approximation.
## Chapter 8.Patterns
8.1 Encapsulation
8.2 Hierarchy
Procedures hide computational detail and in that regard they are similar to functions. The procedures discussed in this chapter draw objects. Once such a procedure is written, it can be used as a black box.''
Writing and using procedures that draw geometric patterns is symbolic of what engineers and scientists do. Geometric patterns are defined by parameters and deciding what the "right'' parameters are requires a geometric intuition. Similar is the design of an alloy that requires a metallurgist's intuition or the building of a model to predict crop yield that requires a biologist's intuition. What are to be the constituent metals? What are the factors effecting the growth? Once the parameters are identified, construction is possible by setting their value. A pattern is drawn. An alloy is mixed. A model is formulated. Optimality can then pursued: What choice of parameter values renders the most pleasing pattern, the strongest alloy, the most accurate model of crop yield?
Our use of graphics procedures to shed light on the processes of engineering design and scientific discovery begins in this chapter. We start by showing how to "package'' the computations that produce the pattern. It's an occasion to practice the writing of clear specifications that define what a piece of software can do. Patterns can be built upon other, more elemental patterns, a fact that we use to motivate the design of procedure hierarchies. Optimization issues are discussed further in Chapters 13, 23, and 24.
## Chapter 9. Proximity
9.1 Distance
9.2 Inclusion
9.3 Collinearity
Questions of proximity are of central importance in computational science. How {\em near} is a given mechanical system to wild oscillation? How {\em near} is a given fluid flow to turbulence? How {\em near} is a given configuration of molecules to a state of minimal energy? How near is one digitized picture to another? The key word is near'' and the recognition that a distance function'' is required to measure nearness.'' The notion of distance is familiar to us in geometric settings:
• What is the distance between two points in the xy plane?
• What is the distance from a point to a line segment?
• What is the distance from a point to a polygon?
Our plan is to cut our nearness'' teeth on planar distance problems of this variety, illustrating the distinction between constrained and unconstrained optimization and the complicated boundary between exact mathematics and practical computation.
In the geometric setting, extreme nearness "turns into'' inclusion.'' Instead of asking how near one rectangle is to another, we may ask whether one rectangle is inside another. Questions like this have yes/no answers. Distance questions, on the other hand, have a continuity about them and culminate in the production of a single, nonnegative real number.
The problem of when three points are collinear gives us a snapshot of just how tricky it can be to handle a yes/no geometric question. In theory, three points either line up or they do not. In practice, fuzzy data and inexact arithmetic muddy the waters. For example, we may be using a telescope and a computer to determine the precise moment when both members of a binary star system line up. But both tools have limited precision. Stars and numbers that are too close together are impossible to resolve, and so the computational scientist formulates a distance function that can be used to investigate how near the astronomical situation is to exact collinearity.
## Chapter 10. Roots
10.1 Quadratic Equations
10.2 The Method of Bisection
10.3 The Method of Newton
Our first experiences with root-finding are typically with easy'' functions that permit exact, closed-form solutions like the quadratic equation: $ax^{2} +bx+c=0 \quad \Rightarrow \quad x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}, \:a\neq 0$ However, even the implementation of such a math book formula involves interesting computational issues.
In practical problems we are rarely able to express roots in closed form and this pushes us once again into the realm of the approximate. Just as we had to develop the notion of approximate collinearity to make computational progress'' in Section 9.3, so must we develop the notion of an approximate root. Two definitions are presented and exploited in the methods of bisection and Newton that we develop. The discussion of these implementations lead to some larger software issues. For example, these root finders expect the underlying function to be specified in a certain way. This may require the repackaging'' of an existing implementation that does not meet the required specification. The exercise of modifying "your'' software so that it can interact with "someone else's'' software is typical in computational science, where so many techniques are embodied in existing software libraries.
We use the development of a modest Newton method root-finder to dramatize the difference between a math book implementation of a formula and a finished, usable piece of software. It is absolutely essential for the computational scientist to appreciate the difficulties associated with software development.
## Chapter 11. Area
11.1 Triangulation
11.2 Tiling
11.3 Integration
The breaking down of large complex problems into smaller, solvable subproblems is at the heart of computational science. The calculation of area is symbolic of this enterprise and makes an interesting case study. When the region is simple, there may be a formula, e.g., $A = base\cdot height$. Otherwise, we may have to partition the region into simpler regions for which there are area formulas. For example, we can cover a polygon with triangles and then sum their areas. Other times we may have to resort to approximation. If we can pack (without overlap) N, h-by-h tiles inside a shape that is bounded by curves, then Nh^{2} approximates its area. A variation of this idea, with limits, leads to the concept of integration in the calculus. By exploring these limits we obtain yet another glimpse of the boundary between exact mathematics and approximate calculation.
## Chapter 12. Encoding Information
12.1 Notation and Representation
12.2 Place Value
This chapter is about the representation of information, a term already familiar to us. For example, a real number has a floating point representation when stored in a real variable. Numbers stored in integer variables have a different kind of representation. In this chapter we advance our understanding of the representation "idea'' by looking at several "conversion problems'' that involve numbers as strings of characters. A greater appreciation for the place-value notation is obtained.
## Chapter 13. Visualization
13.1 Exploratory Environments
13.2 Coordinate Systems
In Chapter 2 we developed the notion of numerical exploration, the main idea being that we could get a handle on difficult mathematical questions through computer experimentation. This is one of the most important aspects of computational science. To dramatize further this point, we enlist the services of computer graphics. Our geometric intuition and our ability to visualize go hand-in-hand. Both are essential in many problem-solving domains and graphics can lend a real helping hand.
We start by developing a handful of graphical tools that permit the construction of simple exploratory environments. Sometimes the computer visualization of a problem or a task is an end in itself. On other occasions, it merely sets the stage for analytical work. Regardless of how it is used, a "visualization system'' is driven by many behind-the-scenes computations that permit the suppression of mundane detail. Typical among these are the coordinate transformations that take us from "problem space'' to "screen space.'' The intelligent handling of these transformations leads to some important software issues.
## Chapter 14. Points in the Plane
14.1 Centroids
14.2 Max's and Min's
All of the programs that we have considered so far involve relatively few variables. We have seen problems that involve a lot of data, but there was never any need to store it "all at once.'' This will now change. Tools will be developed that enable us to store a large amount of data that can be accessed during program execution. We introduce this new framework by considering various problems that involve sets of points in the plane. If these points are given by $(x_{1},y_{1}),\ldots,(x_{n},y_{n})$, then we may ask:
• What is their centroid?
• What two points are furthest apart?
• What point is closest to the origin (0,0)?
• What is the smallest rectangle that contains all the points?
The usual {\tt readln}/{\tt writeln} methods for input and output are not convenient for problems like this. The amount of data is too large and too geometric.\footnote{In this chapter we'll be making extensive use of {\tt cGetPosition}, {\tt cDrawDot}, {\tt cDrawBigDot}, {\tt DrawAxes}, {\tt cMoveTo}, and {\tt cLinetO}. These procedures are declared in {\tt DDCodes} and are developed in Section 13.2. Throughout this chapter the underlying coordinate transformation is a non-issue and will be fixed with the $xy$ origin at the screen center and 10 pixels per unit $xy$ distance}
## Chapter 15. Tables
15.1 Set-Up
15.2 Plotting
15.3 Efficiency Issues
15.4 Look-Up
Suppose it costs one dollar to evaluate a function f(x) and that a given fragment calls f 1000 times. If each function call involves a different value of x, then $1000 must be spent on f evaluations. However, if only 10 different x values are involved, then there is a$10 solution to the f-evaluation problem:
• precompute' the 10 necessary f- evaluations and store them in an array. (This costs $10.) • extract the necessary f-values from the array during the execution of the fragment. Storing x-values and f-values in a pair of arrays is just a method for representing a table in the computer. A plotting environment is developed that allows us to display in a window the values in a table. Although the plotting tools that we offer are crude, they are good enough to build an appreciation for plotting as a vehicle that builds intuition about a function's behavior. The setting up of a table is an occasion to discuss several efficiency issues that have to do with function evaluation. A sine/cosine example is used to show how to exploit recursive relations that may exist between table entries. The "parallel '' construction of the entries in a table using array-level operations is also discussed Once a table is set up, there is the issue of looking up values that it contains. The methods of linear search and binary search are discussed. The "missing'' data problem is handled by linear interpolation. ## Chapter 16. Divisors 16.1 The Prime Numbers 16.2 The Apportionment Problem A division problem need not have a "happy ending.'' Quotients like$1 \div 0$are not defined. Ratios like 1/3 have no finite base-10 expansion. Real numbers like$\sqrt{2}$cannot be obtained by dividing one integer into another. Integers like 7 have no proper divisors. Etc, Etc. Division is hard. That's why we learn it last in grade school. That's why the IRS permits rounding to the nearest dollar. That's why base-60 systems were favored by the Maya and the Babylonians. (More numbers divide 60 than 10, and this permits a simpler arithmetic life.) Yes, division is by far the most interesting of the four arithmetic operations. But the idea of division transcends the purely numerical. Geometry and combinatorics are filled with partitioning problems. How can a polygon be divided into a set of triangles? How many ways can a set of m objects be divided into n non-empty subsets? In this chapter we consider a pair of representative division problems. One is purely arithmetic and involves the prime numbers. A prime number is an integer that has no divisors except itself and one. They figure in many important applications. Our treatment of the prime numbers in Section 16.1 is designed to build intuition about integer divisibility. The second division problem we consider also involves the integers, but it is essentially a partitioning problem with social constraints. This is the problem of apportionment, which in its most familiar form is this: how can 435 Congressional districts be divided among 50 states? Few division problems have such far-reaching ramifications and that alone is reason enough to study the computation. the algorithms that solve the apportionment But the apportionment problem is a good place to show how reasonable methods may differ in the results that they produce, a fact of life in computational science. ## Chapter 17. The Second Dimension 17.1 ij Thinking 17.2 Operations 17.3 Tables In Two Dimensions 17.4 Bit Maps As we have said before, the ability to think at the array level is very important in computational science. This is challenging enough when the arrays involved are linear, i.e., one-dimensional. Now we consider the two-dimensional array using this chapter to set the stage for more involved applications that involve this structure. Two-dimensional array thinking is essential in application areas that involve image processing. (A digitized picture is a 2-dimensional array.) Moreover, many 3-dimensional problems are solved by solving a sequence of 2-dimensional, "cross-section'' problems. We start by considering some array set-up computations in Section 17.1. The idea is to develop an intuition about the parts of a 2-dimensional array: its rows, its columns, and its subarrays. Once an array is set up, it can be searched and its entries manipulated. Things are not too different from the 1-dimensional array setting, but we get additional row/column practice in Section 17.2 by considering a look-for-the-max problem and also a mean/standard deviation calculation typical in data analysis. Computations that involve both 1- and 2-dimensional arrays at the same time are explored through a cost/purchase order/inventory application. Using a 2-dimensional array to store a finite snapshot of a 2-dimensional continuous function f(x,y) is examined in Section 17.3. In the last section we present the 2-dimensional boolean array as a vehicle for representing some familiar patterns of "yes-no'' data. ## Chapter 18. Polygons 18.1 Points 18.2 Line Segments 18.3 Triangles and Rectangles 18.4 N-gons Plane geometry is filled with hierarchies. For example, each side of a polygon is a line segment. In turn, each line segment is defined by two points, and each point is defined by two real numbers. Problem solving in this domain is made easier by using records. With records, the data that defines a problem can be "packaged'' in a way that facilitates our geometric thinking. ## Chapter 19. Special Arithmetics 19.1 The Extra Long Integer 19.2 The Rational 19.3 The Complex In this chapter we push out from the constraints of computer arithmetic. We have no way to represent exactly$100!$or even 1/3 or$\sqrt{-1}$. To address these constraints we develop three {\em environments}. The first is for very long integer arithmetic and will permit us to compute very large integers. The idea is to use an array to represent an integer. Functions are developed that permit the manipulation of integers that are stored in this fashion and enable us to compute exactly things like $100! = \begin{array}{l} \rule{0pt}{10pt}\\ 33262154439441526816992388562667004907159682643812146859296389521759999322991\\ 56089414639761565182825369792082722375825118521091686400000000000000000000000 \end{array}$ Quotients of integers give us the rational numbers. Unfortunately, even simple rational numbers like 1/3 have no exact (base-2) floating point representation. Clearly, the thing to do is to represent a rational number as a pair of integers in the computer. By doing this and developing arithmetic functions that can operate on rational numbers, we can compute exactly rational numbers like $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{10} = \frac{7381}{2580}$ Our third extended arithmetic system deals more with a shortcoming of the real numbers than with real floating point arithmetic. The problem is that the square root of a negative real number is not real, but complex. But if we let$i$stand for$\sqrt{-1}$, then many interesting doors are opened. For starters, square roots like$\sqrt{-36}$have a complex representation, e.g.,$6i$. General complex numbers have the form$a+bi$where$a$and$b$are real. We develop an environment that supports their representation and manipulation. The display of complex numbers in the complex plane enables us to acquire a geometric intuition about their behavior in certain computational settings. ## Chapter 20. Polynomials 20.1 Representation and Operations 20.2 Evaluation 20.3 Quotients A polynomial is a function of the following form: $p(x) = a_{0} + a_{1}x + a_{2}x^{2} + \cdots + a_{n}x^{n}.$ If$a_{n}$is nonzero, then the {\em degree} of$p(x)$is$n$. An$n$-degree polynomial has exactly$n$roots. Some of these roots may be complex, but if the coefficients are real, then the complex roots come in conjugate pairs. For quadratics (degree = 2), cubics (degree = 3) and quartics (degree = 4), there are closed formulae for the roots. A famous theorem by Galois states that no such recipes exist for polynomials having degree$\geq 5$. Polynomials are widely used because \begin{itemize} \item They have a tractable algebra with many interesting and useful properties. \item They are easy to integrate and differentiate. \item They can be used to approximate more complicated functions. \item They are used to build rational functions. \end{itemize} In this chapter we build an appreciation for these things and generally develop an ability to work with this important family of functions. ## Chapter 21. Permutations 21.1 Shifts and Shuffles 21.2 Sorting 21.3 Representation The when data is re-ordered it undergoes a {\em permutation}. An ability to compute with permutations and to reason about them is very important in computational science. Unfortunately, it's an activity that is very prone to error because it often involves intricate subscripting. So we start gently by discussing two very straightforward but important permutations: the shift and the perfect shuffle. These operations play a key role in many signal processing applications. Sorting is by far the most important permutation that arises in applications and three elementary methods are discussed in \S21.2: bubble sort, insertion sort, and merge sort. These methods are developed and compared in the context of real arrays. When arrays or records are to be sorted, other issues come to the fore and these are discussed in \S21.3. An important undercurrent throughout the chapter is the concept of data motion. On most advanced machines, program efficiency is a function of how much data flows back and forth between the computer's memory and processing units. The volume of arithmetic is almost a side issue. Because programs that implement permutations deal almost exclusively with moving data, they are good for rounding out our intuition about efficiency. ## Chapter 22. Optimization 22.1 Shortest Path 22.2 Best Design 22.3 Smallest Ellipse Optimization problems involve finding the best'' of something. The search for the optimum ranges over a set called the {\em search space} and the notion of best is quantified through an {\em objective function}. One example encountered in \S8.3 is to find the closest point on a line$L$to a given point$P$. Thus,$L$is the search space and Euclidean distance is the objective function. Using the calculus, a formula can be given that explicitly specifies the optimal point. This, however, is not typical. In practice, explicit recipes give way to algorithms and exact solutions give way to approximations. Suboptimal solutions are happily accepted if they are cheap to compute and good enough.'' To clarify these points we describe three different applications in this chapter. Our goal is to show how one goes about solving complicated optimization problems and build an appreciation for their role in computational science. In \S22.1 we consider the traveling salesperson problem where the aim is to find the shortest roundtrip path that visits each of$n$given points exactly once. The search space is huge, consisting of$(n-1)!$possible itineraries. A brute force search plan that considers every possibility is out of the question except for very small values of$n\$. But with an appropriately chosen computational rule-of-thumb called a heuristic, we show that it is not necessary to scan the entire search space. A good, low-mileage itinerary can be produced relatively cheaply.
In \S22.2 we use a small engineering design problem to discuss the important role that constraints play in optimization and how there is often more than one natural choice for an objective function. The problem is to build a 10-sprocket bicycle with a desirable range of gear ratios. As in the traveling salesperson problem, the number of possibilities to consider is huge, although finite. Constraints reduce the size of the search space and but extra care must be exercised to stay within the set of allowable solutions. The application is small as engineering design problems go, but rich enough in complexity to illustrate once again the key role of heuristics.
The last problem we consider is that of enclosing a given set of points with the smallest possible ellipse. In contrast to the previous two problems, this is a continuous optimization problem with a genuinely infinite search space. We set up a graphical environment that facilitates the search for the optimum ellipse.
## Chapter 23. Divide and Conquer
23.1 Recursion vs. Iteration
23.2 Repeated Halving
23.2 Mesh Refinement
The family of divide and conquer algorithms have a very central role to play in computational science. These algorithms involve the repeated subdivision of the original problem into successively smaller parts. The solutions of the smaller parts are then glued together'' in hierarchical fashion to obtain the overall solution. There are many variations of this theme and we cover several major examples in this chapter.
We have already met the divide and conquer idea. The method of bisection discussed in \S10.x divides'' the current bracketing interval in half and conquers'' that half known to include a root. The method of merge sort that we discussed in \S21.x proceeds by dividing the given list in halve, sorting (conquering) the two halves, and merging the results.
A new technique is required to carry out the divide and conquer solution strategy in its most powerful form, and that is the recursive procedure. Simply put, a recursive procedure (or function) calls itself. This capability is supported in Pascal and is of fundamental importance.
In \S23.1 we introduce the mechanics of recursion and develop a recursive function for exponentiation. The example is not very convincing because the nonrecursive algorithm is such a simple alternative, but it does permit a comparison of recursion and iteration. Binary squaring, merge sort, and other repeated halving'' computations are discussed in \S23.2. Dynamic mesh generation is developed in the last section. Many important applications in computational science involve solving complicated equations over complicated regions. A general solution strategy is to partition the region into the union of smaller, simpler, subregions. The problem, or more likely, a simplified version of the problem is then solved on each subregion. The mesh is often generated recursively. To give a sense of this enterprise, we show how to approximate a curve in the plane with a polygonal line whose break points are recursively determined.
## Chapter 24. Models and Simulation
24.1 Prediction and Intuition
24.2 The Effect of Dimension
24.3 Building Models from Data
Scientists use models to express what they know. The level of precision and detail depends upon several factors including the mission of the model, the traditions of the parent science, and the mathematical expertise of the model-builder. When a model is implemented as a computer program and then run, a computer simulation results. This activity is at the heart of computational science and we have dealt with it many times before. In this closing chapter we focus more on the model/simulation interface'' shedding light on how simulations are used, what makes them computationally intensive, and how they are tied up with data acquisition.
Suppose a physicist builds a complicated model that explains what happens to a neutron stream when it bombards a lead shield. A simulation based upon this model could be used to answer a design question: How thick must the shield be in order to make it an effective barrier from the safety point of view? The simulation acts as a predictor. The computer makes it possible to see what the underlying mathematics says.'' Alternatively, the physicist may just be interested in exploring how certain model parameters effect the simulation outcome. In this setting the simulation has a more qualitative, intuition-building role to play. The precise value of the numerical output is less important than the relationships that produce it. In \S24.1 we examine these two roles that computer simulation can play using Monte Carlo, which we introduced in \S6.3.
The time required to carry out a simulation on a grid usually depends strongly upon the grid's dimension. In \S24.2 we build an appreciation for this by experimenting with a family of one, two, and three dimensional problems.
In the last section we discuss the role that data plays in model-building. The least squares fitting of a line to a set of points in the plane illustrates that a model's parameters can sometimes be specified as a solution to an optimization problem. A ray tracing application shows how a two-dimensional density model can be obtained by gathering lots of data from one-dimensional snapshots. |
Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.
# Trailing Zeroes
Let's say you have this question:
## How many trailing zeroes does 100! have?
### Count the tens
The first method that comes to mind is to count the tens or count the 10's and the 2*5's.
1 2 3 4 5 6 7 8 9 10, so in 10! there are 2 zeroes. The problem with that method is it takes a lot of time, and it's completely unnecessary, because we only did the first 10, and you have 90 more numbers to look at.
### Count the fives
An important idea to understand is that 5 is a limiting factor. In other words, there are fewer 5's than 2's. That's important because in order to multiply to 10 you need a 2 and a 5.
The reason there are more 2's is that every other number has 2 as a factor. Every fifth number has a 5 as a factor. So, we can say that for every 5 in the factorial, there's a 2. So, all we have to do is count the 5's, and we can count the 0's.
So, what you have to do is count the numbers with one 5 (e.g. 5,10,15,20,30,35,40, 45, 55, etc. ), then add 1 for each number with two 5s (e.g. 25, 50, 75, 100), then add 1 for each number with three 5s (e.g. 125, 250, 375, 500), etc. So, our answer is:
### 100/5 + 100/25 + 100/125 = 20 + 4 + 0 (there's no 125 in the first 100 positive integers) = 24
So, there are 24 trailing zeroes in 100!
## How many trailing zeroes does 133! have?
This is similar, but I just want to demonstrate that you only take the integer part of each quotient.
133/5 + 133/25 + 133/125 = 26.6 + 5.32 + 1.064 (ignore the decimals) = 26 + 5 + 1 = 32
David Witten |
# GMAT Math: Possibilities for Variables
How do we know what the possible values for a variable on the GMAT are?
First of all, consider this DS problem
1) Is P > Q?
Statement #1: P = 5*Q
Statement #2: P = Q + R
For folks who are not as comfortable with math, or are a bit rusty with math, may fall into any one of a number of traps here. Nothing is specified about any of these three variables, and on the GMAT, when nothing is specified, we have to assume that every category of number is possible. This means we have to consider
a) 1 (the great exception to many rules)
b) 0 (another great exception to many rules
c) positive integers (1, 2, 3, 4, etc.)
d) negative integers (-1, -2, -3, -4, etc.)
e) positive fractions (1/2, 1/10, etc.)
f) negative fractions (-1/2, -1/10, etc.)
For a mathematic statement in variables to be true, it must be true for every category of number. For example, the commutative law (a + b = b + a and c*d = d*c) works for every single number on the number line, all infinity of them. Statements like that are rare indeed.
In the problem above, if you fall into the trap of thinking only in terms of positive numbers, you may go astray. If we specified that P & Q & R must be positive integers, then each statements would be sufficient on its own, and the answer would be D. BUT, as it stands, those variables could be anything.
If P and Q are positive, then P = 5*Q implies P > Q, but if P and Q are negative, then P = 5*Q implies Q > P (for example, consider P = -10 and Q = -2). Statement #1 is not sufficient.
If R is positive, then adding R makes something bigger, but if R is negative, then adding R makes something smaller. Statement #2 is not sufficient.
Even if we combine both statements — if we pick three positive numbers (e.g. P = 5, Q = 1, and R = 4) we get a “yes” answer to the prompt; but, if we pick, say P = Q = R = 0, then the both math equations work, and the answer to the prompt would be “no.” Nothing is sufficient. Answer is E precisely because there are too many possibilities to determine anything.
Whenever you see variables, and no constraints are explicitly specified in the problem, you must consider all possible cases!
Here’s a challenging DS practice question:
## Author
• Mike served as a GMAT Expert at Magoosh, helping create hundreds of lesson videos and practice questions to help guide GMAT students to success. He was also featured as “member of the month” for over two years at GMAT Club. Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike’s Youtube video explanations and resources like What is a Good GMAT Score? and the GMAT Diagnostic Test. |
## Thursday, 21 January 2010
### Prime Factorisation Homework
Textbook p9 #2:
The Prime factorisation of a number is 2^4 X 3^5 X 7^2 X 11.
Write down 3 factors of the number that are greater than 100.
(Hint: 16 X 11 = 176, which is a factor.)
My Answer: 2673, 539 and 23814
2^4 X 3^5 X 7^2 X 11= 2X2X2X2X3X3X3X3X3X7X7X11
2X2X2X2=16
3X3X3X3X3=243
7X7=49
Since the hint given is taking 16 times 11 to get the factor, then I guessed other factors could be calculated using 243 and 49 times 11 respectively.
So we have, 243 X 11=2673 and 49 X 11=539
Then I used the calculator to calculate the answer for 2X2X2X2X3X3X3X3X3X7X7X11 which is 2095632 and then I used the calculator to double check that both 2673 and 539 are factors for 2095632, which means my guess was correct.
Now, I have to find one more factor for this question. I used 2X2X2 (Please note that I used 2X2X2X2 to times 11 the previous time but now I am using 2X2X2 to times 11)times 11 and the answer is 88. So from here I know 88 is a factor of 2095632, but since it is not more than 100, I used 2095632 to divide it and got the result 23814.
After doing the homework, I realise that using the prime factorization, we are able to get the original number given and to find the factors and the highest common factors. :)
Textbook p9 #26:
The Prime factorisation of two numbers are 2 X 3^2 X 7^3 X 13 and3 X 7^2 X 13^3 X 17 .
Write down 3 common factors of the numbers.
My Answer: 3, 7 and 13
Prime number 3, 7 and 13 are present in the two prime factorisations and so these three numbers are the 3 common factors. :)
Lai Ziying
## Wednesday, 20 January 2010
### Maths Homework 19.1.2010
```Textbook: Brainworks (p8)
Q24(b) A mathematician proposed that "Every even number greater than 2 can be expressed as a sum of two prime numbers." Do you agree? Why?```
Yes, I do agree because I was able to express the even numbers (which I anyhow pick for myself to try out) as a sum of two prime numbers. Then I went online and found the information below.
Why Does Every Even Number Greater Than 2 Can Be Expressed As A Sum Of 2 Prime Numbers?
Even numbers have several ways to be the sum of two prime numbers. For example, the number 36 can be shown as '5+31', '7+29', or '17+19'. By looking at these examples, it appears that the higher the even number, the more pair of prime numbers there are that add up to it.
There is an assumption that says that every even number can be expressed as the sum of two prime numbers at least in one way. This assumption is called as Goldbach's conjecture, after Christian Goldbach, a Prussian mathematician who propounded it. it remains to be one of the most ancient problems that is yet not completely solved in the field of mathematics. The conjecture basically states that every even integer that is greater than the number 2 can be expressed as the sum of two primes.
Cited from: http://www.blurtit.com/q710414.html
Lai Ziying
## Tuesday, 12 January 2010
### 12 January: Numbers as a Language
After doing research on the Internet, I have learnt the history of the development of numeration systems. I choose the Roman numeration system to present 2010 is because the Roman numeration system is more pragmatic with respect to mathematics than the other numeration systems. Being pragmatic is how I feel about my learning is going to be in SST in 2010.
Lai Ziying
## Monday, 11 January 2010
### Communication Homework
1 Why is communication important?
We need so socialise with the others through communication and it also enables us to get connected with one another. Also, through communication, we will be able to understand the people around us as well as the world. Without communication, we are able to achieve nothing but become a loner.
2 What is/are your favourite form/s of communication? Why?
My favourite way of communicating with others is talking. I feel that I am able to express and receive messages more freely and clearly through talking. When we communicate by talking over the phone, we are able to ensure that the person gets the message on time as sometimes people check their online social networks and handphone messages only at a certain time and this may delay the person from getting the message right away.
3 How do you decide which form of communication to use in a situation?
I will look at the form of the message and the audience that I want to pass the message to and choose the form of communication that will ensure that the people get the message in time.
4 What difficulties do you face in communicating with others?
I don’t think I face any difficulties in communicating with others as long as I know what I want to say and how to go about expressing it in a way that people understand.
Lai Ziying
## Saturday, 9 January 2010
CD里播着：“很爱很爱你，所以愿意让你飞到更幸福的地方去。”
（赖同学我本人不认同这句话的后两个部分，我觉得那是情绪低谷时人们才会说的话）
（的确，依照我自己的生活经历而言，我曾经无数次地在博客中写道自己成长了，但随着这样的字眼出现的次数增加，我发现，我们只是在某一瞬间，在一个阶段后发觉自己长大了，但其实，我们也会觉醒，自己还在继续成长。路还很长，成长的脚步不间断。）
（有时候摸索，有时候迷惘，有时候绝望中带着坚强。有句话说，最坚强的人不是没有眼泪的人，而是含着泪奔跑的人。曾经有那么一个时刻，不，是许多那样的时刻，我害怕自己落队，害怕自己跟不上生活的脚步，而对于未来的“隐隐约约”，我有着强烈的担忧。但那些都是过程罢了，有担忧才有思考，有思考才有实践。不要忧愁，因为生活本身就让我们对它有信心，对我们自己有信心。时间到了，我们会依照自己的需要活出最好的自己。失望，但不绝望！）
It's just the beginning, we still have a long way to go... Enjoy! It may be tough, but beautiful :)
*(Life is tough but beautiful is my primary school teacher's motto)
101
## Wednesday, 6 January 2010
### My Personal Reflection On the Amazing Race
I think our school value "Expanding Our Learning Networks" means that our learning should not be just limited from the books or the school compound but anywhere. Such as we went around the Clementi Neighbourhood today and through out the whole activity, we observed, led, thought and so on. Through the whole process, we built up our bond within the class and learnt new things outside the school compound.This school value teaches me that we are able to learn new things from anywhere.
This value is important because it will teach us to be more observant of the things around us and we will be more creative and innovative in thinking. This value alslo suggests that we should not limit our learning network.
Lai Ziying
## Tuesday, 5 January 2010
### My Reflection On the Team Building Challenge
I think "Forging Excellence" means as a school, we should play our indivisual part and when all our parts are being played well, we are forging excellence. So the excellence is not for each indivusual, but to the whole school as a team. Since this "excellence" needs to be "forged", each one of us matters and is important. We should aim for our goal and live up to excellence and then when our dreams come true, that's when we are an excellent and united team as a school.
I think my interest and enthusiasm in learning will help me to success. Plus, I know what I want and I am willing to work towards my goal! No matter how tough it is going to be. Because I know I will pull through it all!
I think I should try to understand my friends around me and find an effective way to communicate with them. I have to make them feel that I respect them so that they may respect me in the same way and pay attention to my opinions.
I can contribute to the success of our class by sharing my thoughts with my classmates and help them if I am able to. Besides sharing point of views and helping one another, I will do whatever I can to make our class an united team so that we will be stronger!
Lai Ziying |
## Muzzle Velocity
Two cardboard target discs are mounted in a straight line with a 3 m gap between the discs’ flat faces. The discs both spin clockwise about their center point at a speed of 1,500 rpm. If a rifle is mounted directly in line with the targets and fires a bullet towards the discs such that the resulting bullet hole in the second disc is displaced 40° with respect to the hole made in the first disc, what is the approximate muzzle velocity in m/s? Ignore friction and gravity.
Hint
Muzzle velocity is the projectile/bullet speed right when it leaves the gun barrel’s end.
Hint 2
First, convert the 1,500 rotations per minute to rotations per second.
$$1,500\:\frac{rotations}{min}\cdot \frac{min}{60sec}=25\:rotations\:per\:sec$$$Muzzle velocity is the projectile/bullet speed right when it leaves the gun barrel’s end. First, convert the 1,500 rotations per minute to rotations per second. $$1,500\:\frac{rotations}{min}\cdot \frac{min}{60sec}=25\:rotations\:per\:sec$$$
To complete a full rotation, the disc needs to rotate 360°. Since the second disc displaced 40°, then:
$$\frac{40^{\circ}}{360^{\circ}}=0.11\:of\:a\:complete\:rotation$$$Therefore, the time it takes the second disc to rotate 40° is: $$\frac{0.11rotation}{25\:rotations\:per\:sec}=0.00444444\:sec$$$
Since the bullet travels 3m in this time, the muzzle velocity is:
$$\frac{3m}{0.004444sec}= 675\:m/s$$\$
675 m/s |
## FACTORING POLYNOMIALS USING ALGEBRAIC IDENTITIES
On this page "factoring polynomials using algebraic identities",we are going see clear explanation of factoring trinomials with two different variables.
## Factoring polynomials using algebraic identities (a+b)²and (a-b)²
Example 1:
Factorize 9x² - 24xy + 16y²
Solution:
Step 1:
To factor these kind of expressions first we try to write the first and last term as squares.
Step 2:
Since it is possible,we have to split the middle term as shown the below picture.
We write the middle term 24xy as the multiple of 2 x first term x last term
Step 3:
Comparing these terms with the algebraic identity.
(3x - 4y) (3x -4y) are the factors
Example 2:
Factorize 4x² + 12xy + 9y²
Solution:
Step 1:
To factor these kind of expressions first we try to write the first and last term as squares.
= 4x² + 12xy + 9y²
= 2² x² + 12 xy + 3²y²
= (2x)² + 12 xy + (3y)²
Step 2:
Since it is possible,we can to split the middle term as the multiple of 2 times product of first and last term
= (2x)² + 2 (2x) (3y) + (3y)²
Step 3:
Comparing these terms with the algebraic identity.
a² + 2ab + b² = (a + b)²
= (2x + 3y)²
(2x + 3y)(2x + 3y) are the factors
Example 3:
Factorize 16a² - 8a + 1
Solution:
Step 1:
To factor these kind of expressions first we try to write the first and last term as squares.
= 16a² - 8a + 1
= 4² a² - 8a + 1²
= (4a)² - 8a + (1)²
Step 2:
Since it is possible,we can to split the middle term as the multiple of 2 times product of first and last term
= (4a)² - 2 (4a) (1) + (1)²
Step 3:
Comparing these terms with the algebraic identity.
a² - 2ab + b² = (a - b)²
= (4a - 1)²
## Factoring polynomials using algebraic identities a²- b²
Example 4:
Factorize 16a² - 9b²
Solution:
= 16a² - 9b²
= 4² a² - 3²b²
= (4a)² - (3b)²
The above algebraic expression exactly matches the identity a² - b²
Formula for a² - b² is (a + b) (a - b). In the above expression we have "4a" instead of "a" and "3b" instead of "b".
= (4a + 3b) (4a - 3b)
Example 5:
Factorize (a + b)² - (a - b)²
Solution:
Let x = a + b and y = a - b
= x² - y²
The above algebraic expression exactly matches the identity a² - b²
Formula for a² - b² is (a + b) (a - b). In the above expression we have "x" instead of "a" and "y" instead of "b".
= (x + y) (x - y)
= [(a + b) + (a - b)] [(a + b) - (a -b)]
= (a + b + a - b) (a + b - a + b)
= 2a (2b)
= 4 a b
## Factoring polynomials using algebraic identities a³+b³and a³-b³
Example 6:
Factorize 8x³ - 125y³
Solution:
Step 1:
Let us try to write the numbers 8 and 125 in term of cube.
8 = 2³ and 125 = 5³
= 2³ x³ - 5³y³
= (2x)³ - (5y)³
Step 2:
The above algebraic expression exactly matches the identity a³ - b³.
Formula for a³ - b³ is (a - b) (a² + ab + b²)
= (2x - 5y) [ (2x)² + (2x) (5y) + (5y)²]
= (2x - 5y) [ 4x² + 10xy + 25y²]
Example 7:
Factorize 27x³ + 64y³
Solution:
Step 1:
Let us try to write the numbers 27 and 64 in term of cube.
27 = 3³ and 64 = 4³
= 3³ x³ + 4³y³
= (3x)³ + (4y)³
Step 2:
The above algebraic expression exactly matches the identity a³ + b³.
Formula for a³ + b³ is (a + b) (a² - ab + b²)
= (3x + 4y) [ (3x)² - (3x) (4y) + (4y)²]
= (3x + 4y) [ 9x² - 12xy + 16y²]
## Related topics
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
0
## Leapfrog Division IV
Published on Thursday, October 29, 2015 in , , ,
While I normally do my Leapfrog Division posts about a year apart, I though I'd wrap up this mental division series just 1 week after the previous entry.
In this post, you'll learn to mentally divide by numbers ending in 2!
STARTING POINTS: This is the most advanced technique of all of the Leapfrog Division posts, so you should be familiar with and practice the the previous techniques. Not only does this employ the basic ideas taught in the original Leapfrog Division post, but also the subtraction from 9 idea used in Leapfrog Division II, AND both the doubling and halfway-comparison concepts from Leapfrog Division III. If you comfortable with all of these concepts, then you're ready to move on to this version.
This version also introduces a new idea to the Leapfrog Division series: The stopping rule. In the previous versions, you could stop either when you realized the numbers were going to repeat, or when you didn't need any further precision. While the same is true here, in this version, you'll also need to stop when you get a quotient of 5, and a remainder of 0. You'll understand this better, including the exceptions, as you work your way through the technique.
THE TECHNIQUE: For our first example, we'll use 1732. As in the technique for dividing by numbers ending in 1, we will always start by reducing the numerator by 1, giving us 1632. Similar to the method for dividing by numbers ending in 8, we're also going to compare numbers to the half of the denominator. Anytime the numerator is greater than or equal to half of the original denominator, we'll reduce it by 1.
In the case of 32, half of that is 16, so we ask if the current numerator is greater than or equal to 16. In this example, it's currently 16 (exactly equal to half!), so we reduce by 1 more, giving us a division problem of 1522. As with all the Leapfrog Division techniques, we're now going to round the denominator to the nearest multiple of 10, and then divide it by 10. So, the problem becomes 153.
You shouldn't be surprised that we're going to divide this out using quotients and remainders. Starting with 152, we get:
• 15 ÷ 3 = 5 (remainder 0)
Naturally, you write the quotient down right away, with a 0 and a decimal in front, as in 0.5. Ordinarily, the stopping rule might tell us to stop here. After all, we have a quotient of 5, and a remainder of 0. However, since 1732 (our original problem) isn't exactly 0.5, there's definitely more numbers to calculate, so we'll continue. We should still keep the stopping rule in mind for later, however.
From here, you're going to use the doubling idea as taught in Leapfrog Division III, in which you double numbers, but only keep the ones (units) digit. 5 doubles to 0, because 5 × 2 = 10, and we only keep the ones digit, which is 0. Next, as in Leapfrog Division II, you're going to subtract the quotient from 9. In this example, 9 - 0 = 9, so the new quotient is now 9. Leapfrogging the remainder, 0, to the front of the quotient, we know have 09, or simply 9.
Before dividing, we need to ask whether 9 is greater than or equal to 16. It isn't, so we don't decrease the number at this stage. After that question, only then do we do the division again:
• 9 ÷ 3 = 3 (remainder 0)
We write the 3 down, giving us 0.53 so far. 3 doubled becomes 6, and 9 - 6 = 3. Leapfrogging the remainder of 0 in front, we have 03, or just 3. Is 3 greater than or equal to 16? No, so there's no decrease this time, either. The next division problem yields:
• 3 ÷ 3 = 1 (remainder 0)
So far, our mental work has given us an answer of 0.531. Repeat once more. Double the 1 to get 2, subtract 9 - 2 to get 7, and put the remainder 0 in front of the 7, giving us a new divisor of 07. Is 7 greater than or equal to 16? No, so there's no decrease. Moving on to the next division:
• 7 ÷ 3 = 2 (remainder 1)
Now we have 0.5312. 2 doubled becomes 4, and 9 - 4 = 5. Leapfrogging the 1 in front gives us 15. Is 15 greater than or equal to 16? No, so there's no decrease. Obviously, we can move on to the next division:
• 15 ÷ 3 = 5 (remainder 0)
At this point, with 0.53125 as our current answer, you'll note we have a quotient of 5 and a remainder of 0. This means that the stopping rule kicks in. So, our final result is 0.53125, which is exactly what 1732 equals!
Now that you understand the steps, let's work out 1922 as a second example. We start by reducing 19 by 1, which is ALWAYS the first step, giving us 1822. Half of 22 is 11, and 18 is greater than 11, so we decrease it by 1 again, leaving us with 1722. Rounding the denominator down and dividing by 10, our starting problem should be 172. We start there, and work through the problem this way:
• 17 ÷ 2 = 8 (remainder 1)
(8 doubles to 6, 9-6=3, 1 makes it 13, which is MORE than 11, so 13 - 1 = 12.)
• 12 ÷ 2 = 6 (remainder 0)
(6 doubles to 2, 9-2=7, 0 makes it 07, which is less than 11.)
• 7 ÷ 2 = 3 (remainder 1)
(3 doubles to 6, 9-6=3, 1 makes it 13, which is MORE than 11, so 13 - 1 = 12.)
Since we're already back to dividing by 12, you can see that this is going to repeat. Writing down just the quotients, we get the correct answer of 1922 ≈ 0.863!
TIPS: Yes, this has more steps than any of the other approaches taught in the Leapfrog Division series, and it's not difficult to confuse the steps of the various versions. The solution, as always, is practice, practice, practice!
You may have noticed that I referred to this as the last post in the Leapfrog Division series. Why is that? Because using the 4 different techniques I've taught, you can actually handle dividing by most numbers with just a little adjustment. How do you handle numbers ending in...
So, every number except those ending in 5 or 0 are covered. That's good incentive to practice these techniques, as you can amaze many people with the ability to handle almost every division problem thrown at you. I hope you've found this series enjoyable and useful. |
# EXPRESSIONS AND EQUATIONS
Expressions and equations :
This is one of the important basic stuff of algebra in math.
Students who would like to learn solving word problems in math must study this basic stuff.
## Writing expressions for the given verbal phrases
Let us see, how to write verbal phrase into expressions.
If there is any unknown value, in the given verbal phrase, replace it by some English alphabet.
## Writing expressions - Practice questions
Question 1 :
Write the given verbal phrase into mathematical expression
"The sum of 5 times a number and 8"
5x + 8
Question 2 :
Write the given verbal phrase into mathematical expression
"2 times the sum of 7 times a number and 4"
2(7x + 4)
Question 3 :
Write the given verbal phrase into mathematical expression
"7 less than 4 times a number"
4x - 7
Question 4 :
Write the given verbal phrase into mathematical expression
"One fifth of sum of 3 times a number and 9"
( 3x + 9 ) / 5
Question 5 :
Write the given verbal phrase into mathematical expression
"7 less than 3 times the sum of a number and 6"
3(x + 6) - 7
## Evaluating expressions - Practice questions
Question 1 :
Evaluate the given expression for x = 3 and y = 5.
3x + 2y
Plug x = 3 and y = 5 in the given expression
3(3) + 2(5) = 9 + 10 = 19
Question 2 :
Evaluate the given expression for m = 5 and n = 2.
m² + 2mn²
Plug m = 5 and n = 2 in the given expression
5² + 2(5)(2)² = 25 + 40 = 65
Question 3 :
Evaluate the given expression for s = 5.
s² + 7s - 2
Plug s = 5 in the given expression
5² + 7(5) - 2 = 25 + 35 - 2 = 58
Question 4 :
Evaluate the given expression for m = 1/3.
18m² + 3m + 7
Plug m = 1/3 in the given expression
18(1/3)² + 3(1/3) + 7 = 2 + 1 + 7 = 10
Question 5 :
Evaluate the given expression for m = 13.
m² + m - 54
Plug m = 13 in the given expression
13² + 13 -54 = 169 + 13 - 54 = 128
## Solving word problems using equations - Examples
Example 1 :
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool ?
Solution :
Let the breadth be x m.
The length will be (2x + 2) m.
Perimeter of swimming pool = 2(l + b) = 154 m
2(2x + 2 + x) = 154
2(3x + 2) = 154
Dividing both sides by 2,
3x + 2 = 77
Subtracting 2 on both sides, we get
3x + 2 - 2 = 77 - 2
3x = 75
On dividing 3 on both sides
x = 25
2x + 2 = 2 × 25 + 2 = 52
Hence, the breadth and length of the pool are 25 m and 52 m respectively.
Let us look at the next word problem on"Expressions and equations".
Example 2 :
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution :
Let one number be x.
Therefore, the other number will be x+ 15.
According to the question,
x + x + 15 = 95
2x + 15 = 95
Subtract 15 on both sides
2x = 95 - 15
2x = 80
Divide by 2 on both sides
x = 40
x + 15 = 40 + 15 = 55
Hence, the numbers are 40 and 55.
Let us look at the next word problem on"Expressions and equations".
Example 3 :
Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Solution :
Let the common ratio between these numbers be x. Therefore, the numbers will be 5x and 3x respectively.
Difference between these numbers = 18
5x - 3x= 18
2x= 18
Divide by 2 on both sides
x = 9
First number = 5x= 5 × 9 = 45
Second number = 3x= 3 × 9 = 27
Hence the required numbers are 27, 45.
Let us look at the next word problem on"Expressions and equations".
Example 4 :
Three consecutive integers add up to 51. What are these integers?
Solution :
Let three consecutive integers be x, x + 1, and x+ 2.
Sum of these numbers = x + x + 1 + x + 2 = 51
3x+ 3 = 51
Subtract by 3 on both sides
3x = 51 - 3
3x = 48
Divide by 3 on both sides
x = 16
x + 1 = 17
x + 2 = 18
Hence, the consecutive integers are 16, 17, and 18.
Let us look at the next word problem on"Expressions and equations".
Example 5 :
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution :
Let the three consecutive multiples of 8 be 8x, 8(x + 1), 8(x + 2).
Sum of these numbers = 8x + 8(x+ 1) + 8(x+ 2) = 888
8(x + x + 1 +x + 2) = 888
8 (3x + 3) = 888
Dividing by 8 both sides
3x+ 3 = 111
Subtract 3 on both sides
3x + 3 - 3 = 111 - 3
3x= 108
Divide by 3 on both sides
x = 36
First multiple = 8x = 8 × 36 = 288
Second multiple = 8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296
Third multiple = 8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304
Hence, the required numbers are 288, 296, and 304.
We hope that the students would have understood the stuff given on "Expressions and equations"
Apart from "Expressions and equations", if you need any other stuff in math, please use our google custom search here.
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Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Circle Theorems
This article explains circle theorems, including tangents, sectors, angles and proofs (with thanks to Revision Maths).
## Isosceles Triangle
Two Radii and a chord make an isosceles triangle.
## Perpendicular Chord Bisection
The perpendicular from the centre of a circle to a chord will always bisect the chord (split it into two equal lengths).
## Angles Subtended on the Same Arc
Angles formed from two points on the circumference are equal to other angles, in the same arc, formed from those two points.
## Angle in a Semi-Circle
Angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle. So c is a right angle.
### Proof
We can split the triangle in two by drawing a line from the centre of the circle to the point on the circumference our triangle touches.
We know that each of the lines which is a radius of the circle (the green lines) are the same length. Therefore each of the two triangles is isosceles and has a pair of equal angles.
But all of these angles together must add up to 180°, since they are the angles of the original big triangle.
Therefore x + y + x + y = 180, in other words 2(x + y) = 180.
and so x + y = 90. But x + y is the size of the angle we wanted to find.
## Tangents
A tangent to a circle is a straight line which touches the circle at only one point (so it does not cross the circle- it just touches it).
A tangent to a circle forms a right angle with the circle’s radius, at the point of contact of the tangent.
Also, if two tangents are drawn on a circle and they cross, the lengths of the two tangents (from the point where they touch the circle to the point where they cross) will be the same.
## Angle at the Centre
The angle formed at the centre of the circle by lines originating from two points on the circle’s circumference is double the angle formed on the circumference of the circle by lines originating from the same points. i.e. a = 2b.
### Proof
You might have to be able to prove this fact:
OA = OX since both of these are equal to the radius of the circle. The triangle AOX is therefore isosceles and so ∠OXA = a
Similarly, ∠OXB = b
Since the angles in a triangle add up to 180, we know that ∠XOA = 180 – 2a
Similarly, ∠BOX = 180 – 2b
Since the angles around a point add up to 360, we have that ∠AOB = 360 – ∠XOA – ∠BOX
= 360 – (180 – 2a) – (180 – 2b)
= 2a + 2b = 2(a + b) = 2 ∠AXB
## Alternate Segment Theorem
This diagram shows the alternate segment theorem. In short, the red angles are equal to each other and the green angles are equal to each other.
### Proof
You may have to be able to prove the alternate segment theorem:
We use facts about related angles
A tangent makes an angle of 90 degrees with the radius of a circle, so we know that ∠OAC + x = 90.
The angle in a semi-circle is 90, so ∠BCA = 90.
The angles in a triangle add up to 180, so ∠BCA + ∠OAC + y = 180
Therefore 90 + ∠OAC + y = 180 and so ∠OAC + y = 90
But OAC + x = 90, so ∠OAC + x = ∠OAC + y
Hence x = y
## Cyclic Quadrilaterals
cyclic quadrilateral is a four-sided figure in a circle, with each vertex (corner) of the quadrilateral touching the circumference of the circle. The opposite angles of such a quadrilateral add up to 180 degrees.
Area of Sector and Arc Length
If the radius of the circle is r,
Area of sector = πr2 × A/360
Arc length = 2πr × A/360
In other words, area of sector = area of circle × A/360
arc length = circumference of circle × A/360
# Useful Formulas
What is a problem? A problem = a fact + a judgment. That is a simple formula that tells us something about the way the world works. Maths is full of formulas, and that can intimidate some people if they don’t understand them or can’t remember the right one to use.
However, formulas should be our friends, as they help us to do sometimes complex calculations accurately and repeatably in a consistent and straightforward way. The following is a list of the most useful ones I’ve come across while teaching Maths to a variety of students at a variety of ages and at a variety of stages in their education.
## Averages
• The mean is found by adding up all the values and dividing the total by how many there are, eg the mean of the numbers 1-10 is 5.5, as 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55, and 55 ÷ 10 = 5.5.
• The mode is the most common value (or values), eg the mode of 1, 2, 2, 3, 4, 5 is 2.
• The median of an odd number of values sorted by size is the one in the middle, eg the median of the numbers 1-5 is 3. The median of an even number of values is the mean of the two numbers in the middle, eg the median of the numbers 1-10 is 5.5, as 5 and 6 are the numbers in the middle, and 11 ÷ 2 = 5.5.
• The range is the highest value minus the lowest, eg the range of the numbers 1-10 = 10 – 1 = 9.
## Geometry
• Angles around a point add up to 360º
• Angles on a straight line add up to 180º
• Opposite angles are equal, ie the two pairs of angles opposite each other when two straight lines bisect (or cross) each other
• Alternate angles are equal, ie the angles under the arms of a ‘Z’ formed by a line (or ‘transversal’) bisecting two parallel lines
• Corresponding angles are equal, ie the angles under the arms of an ‘F’ formed by a line (or ‘transversal’) bisecting two parallel lines
• Complementary angles add up to 90º
• Any straight line can be drawn using y = mx + c, where m is the gradient and c is the point where the line crosses the y-axis (the ‘y-intercept’)
• The gradient of a straight line is shown by δy/δx (ie the difference in the y-values divided by the difference in the x-values of any two points on the line, usually found by drawing a triangle underneath it)
## Polygons
• Number of diagonals in a polygon = (n-3)(n÷2) where n is the number of sides
• The sum of the internal angles of a polygon = (n-2)180º, where n is the number of sides
• Any internal angle of a regular polygon = (n-2)180º ÷ n, where n is the number of sides
## Rectangles
• Perimeter of a rectangle = 2(l + w), where l = length and w = width
Note that this is the same formula for the perimeter of an L-shape, too.
• Area of a rectangle = lw, where l = length and w = width
## Trapeziums
• Area of a trapezium = lw, where l = average length and w = width (in other words, you have to add both lengths together and divide by two in order to find the average length)
## Triangles (Trigonometry)
• Area of a triangle = ½bh, where b = base and h = height
• Angles in a triangle add up to 180º
• Pythagoras’s Theorem: in a right-angled triangle, a² + b² = c², ie the area of a square on the hypotenuse (or longest side) is equal to the sum of the areas of squares on the other two sides
Circles
• Circumference of a circle = 2πr, where r = radius
• Area of a circle = πr², where r = radius
• π = 3.14 to two decimal places and is sometimes given as 22/7
## Spheres
• Volume of a sphere = 4/3πr³, where r = radius
• Surface area of a sphere = 4πr², where r = radius
## Cuboids
• Volume of a cuboid = lwh, where l is length, w is width and h is height
• Surface area of a cuboid = 2(lw + lh + wh), where l is length, w is width and h is height
## Number Sequences
• An arithmetic sequence (with regular intervals) = xn ± k, where x is the interval (or difference) between the values, n is the value’s place in the sequence and k is a constant that is added or subtracted to make sure the sequence starts at the right number, eg the formula for 5, 8, 11, 14…etc is 3n + 2
• The sum of n consecutive numbers is n(n + 1)/2, eg the sum of the numbers 1-10 is 10(10 + 1)/2 = 110/2 = 55
## Other
• Speed = distance ÷ time
• Profit = sales – cost of goods sold
• Profit margin = profit ÷ sales
• Mark-up = profit ÷ cost of goods sold |
# Show that I_n = -1/n \ cosx \ sin^(n-1)x+(n-1)/n \ I_(n-2) where I_n=int \ sin^nx \ dx , and n ge 2?
Sep 24, 2017
We want to show that:
${I}_{n} = - \frac{1}{n} \setminus \cos x \setminus {\sin}^{n - 1} x + \frac{n - 1}{n} \setminus {I}_{n - 2}$
Where ${I}_{n} = \int \setminus {\sin}^{n} x \setminus \mathrm{dx}$, and $n \ge 2$.
We can write the integral as:
${I}_{n} = \int \setminus \left(\sin x\right) \left({\sin}^{n - 1} x\right) \setminus \mathrm{dx}$ provided $n \ge 2$
We can use integration by parts:
Let $\left\{\begin{matrix}u & = {\sin}^{n - 1} x & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = \left(n - 1\right) {\sin}^{n - 2} x \cos x \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = \sin x & \implies v & = - \cos x\end{matrix}\right.$
Then plugging into the IBP formula:
$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$
gives us:
$\int \setminus \left({\sin}^{n - 1} x\right) \left(\sin x\right) \setminus \mathrm{dx} = \left({\sin}^{n - 1} x\right) \left(- \cos x\right) - \int \setminus \left(- \cos x\right) \left(\left(n - 1\right) {\sin}^{n - 2} x \cos x\right) \setminus \mathrm{dx}$
$\therefore {I}_{n} = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) \int \setminus {\cos}^{2} x \setminus {\sin}^{n - 2} x \setminus \mathrm{dx}$
Using the identity, ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ we have:
${I}_{n} = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) \int \setminus \left(1 - {\sin}^{2} x\right) \setminus {\sin}^{n - 2} x \setminus \mathrm{dx}$
$\setminus \setminus \setminus = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) \int \setminus {\sin}^{n - 2} x - {\sin}^{n} x \setminus \mathrm{dx}$
$\setminus \setminus \setminus = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) \left({I}_{n - 2} - {I}_{n}\right)$
$\setminus \setminus \setminus = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) {I}_{n - 2} - \left(n - 1\right) {I}_{n}$
${I}_{n} + \left(n - 1\right) {I}_{n} = - \cos s x \setminus {\sin}^{n - 1} x + \left(n - 1\right) {I}_{n - 2}$
$\therefore n {I}_{n} = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) {I}_{n - 2}$
$\therefore {I}_{n} = - \frac{1}{n} \cos s x \setminus {\sin}^{n - 1} x + \frac{n - 1}{n} {I}_{n - 2}$ QED |
# HSPT Math : How to find surface area
## Example Questions
← Previous 1 3
### Example Question #31 : Cylinders
A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
156.25 in2
78.125π in2
312.5 in2
39.0625π in2
78.125π in2
Explanation:
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
### Example Question #1 : How To Find Surface Area
The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If r and h represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to r in terms of h?
r = h2/(h + 2)
r = h2 + 2h
r = 2h/(h – 2)
r = 2h2 + 2
r = h/(2h – 2)
r = 2h/(h – 2)
Explanation:
We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.
surface area of cylinder = surface area of bases + lateral surface area
The bases of the cylinder will be two circles with radius r. Thus, the area of each will be πr2, and their combined surface area will be 2πr2.
The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2πr, and the height is h, so the lateral area is 2πrh.
surface area of cylinder = 2πr2 + 2πrh
Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is πr2, and the height is h, so the volume of the cylinder is πr2h.
volume = πr2h
Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.
2πr2 + 2πrhπr2h
First, let's factor out 2πr from the left side.
2πr(h) = πr2h
We can divide both sides by π.
2r(h) = r2h
We can also divide both sides by r, because the radius cannot equal zero.
2(h) = rh
Let's now distribute the 2 on the left side.
2r + 2h = rh
Subtract 2r from both sides to get all the r's on one side.
2h = rh – 2r
rh – 2r = 2h
Factor out an r from the left side.
r(h – 2) = 2h
Divide both sides by h – 2
r = 2h/(h – 2)
The answer is r = 2h/(h – 2).
### Example Question #2 : How To Find Surface Area
What is the surface area of a cylinder with a radius of 4 and a height of 7?
225π
96π
164π
108π
88π
88π
Explanation:
Surface area of a cylinder = 2πr2 + 2πrh = 2 * 4π + 2 * 4 * 7 * π = 32π + 56π = 88π
### Example Question #1 : How To Find The Surface Area Of A Cube
What is the surface area of a cube with a volume of 1728 in3?
1728 in2
144 in2
432 in2
72 in2
864 in2
864 in2
Explanation:
This problem is relatively simple. We know that the volume of a cube is equal to s3, where s is the length of a given side of the cube. Therefore, to find our dimensions, we merely have to solve s3 = 1728. Taking the cubed root, we get s = 12.
Since the sides of a cube are all the same, the surface area of the cube is equal to 6 times the area of one face. For our dimensions, one face has an area of 12 * 12 or 144 in2. Therefore, the total surface area is 6 * 144 = 864 in2.
### Example Question #1 : How To Find Surface Area
If the volume of a cube is 216 cubic units, then what is its surface area in square units?
36
216
108
64
54
216
Explanation:
The volume of a cube is given by the formula V = , where V is the volume, and s is the length of each side. We can set V to 216 and then solve for s.
In order to find s, we can find the cube root of both sides of the equaton. Finding the cube root of a number is the same as raising that number to the one-third power.
This means the length of the side of the cube is 6. We can use this information to find the surface area of the cube, which is equal to . The formula for surface area comes from the fact that each face of the cube has an area of , and there are 6 faces in a cube.
surface area =
The surface area of the square is 216 square units.
### Example Question #5 : How To Find Surface Area
You have a cube with sides of 4.5 inches. What is the surface area of the cube?
Explanation:
The area of one side of the cube is:
A cube has 6 sides, so the total surface area of the cube is
### Example Question #3 : How To Find Surface Area
A cube has a surface area of 24. If we double the height of the cube, what is the volume of the new rectangular box?
Explanation:
We have a cube with a surface area of 24, which means each side has an area of 4. Therefore, the length of each side is 2. If we double the height, the volume becomes .
### Example Question #1 : How To Find The Surface Area Of A Cube
A cube has a surface area of 10m2. If a cube's sides all double in length, what is the new surface area?
640m2
80m2
20m2
320m2
40m2
40m2
Explanation:
The equation for surface area of the original cube is 6s2. If the sides all double in length the new equation is 6(2s)2 or 6 * 4s2. This makes the new surface area 4x that of the old. 4x10 = 40m2
### Example Question #1 : How To Find The Surface Area Of A Prism
A rectangular prism has a volume of 70 m3. If the length, width, and height of the prism are integers measured in meters, which of the following is NOT a possible measure of the surface area of the prism measured in square meters?
178
214
118
280
174
280
Explanation:
Since the volume is the product of length, width, and height, and each of these three dimensions are integers, it is important to know the factors of the volume. 70 = (2)(5)(7). This implies that each of these factors (and only these factors with the exception of 1) will show up in the three dimensions exactly once. This creates precisely the following five possibilities:
2, 5, 7
SA = 2((2)(5)+(2)(7)+(5)(7)) = 118
1, 7, 10
SA = 2((1)(7)+(1)(10)+(7)(10)) = 174
1, 5, 14
SA = 2((1)(5)+(1)(14)+(5)(14)) = 178
1, 2, 35
SA = 2((1)(2)+(1)(35)+(2)(35)) = 214
1, 1, 70
SA = 2((1)(1)+(1)(70)+(1)(70)) = 282
### Example Question #1 : How To Find The Surface Area Of A Prism
The three sides of a rectangular box all have integer unit lengths. If each of the side lengths is greater than one unit, and if the volume of the box is 182 cubic units, what is the surface area of the box in square units?
262
236
181
264
182
262
Explanation:
Let's call the side lengths of the box l, w, and h. We are told that l, w, and h must all be integer lengths greater than one. We are also told that the volume of the box is 182 cubic units.
Since the volume of a rectangular box is the product of its side lengths, this means that lwh must equal 182.
(l)(w)(h) = 182.
In order to determine possible values of l, w, and h, it would help us to figure out the factors of 182. We want to express 182 as a product of three integers each greater than 1.
Let's factor 182. Because 182 is even, it is divisible by 2.
182 = 2(91).
91 is equal to the product of 7 and 13.
Thus, 182 = 2(7)(13).
This means that the lengths of the box must be 2, 7, and 13 units.
In order to find the surface area, we can use the following formula:
surface area = 2lw + 2lh + 2hw.
surface area = 2(2)(7) + 2(2)(13) + 2(7)(13)
= 28 + 52 + 182
= 262 square units. |
# Difference between revisions of "2020 AMC 10A Problems/Problem 16"
The following problem is from both the 2020 AMC 12A #16 and 2020 AMC 10A #16, so both problems redirect to this page.
## Problem
A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$. The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$. (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth$?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$
## Solution 1
### Diagram
$[asy] size(10cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); filldraw((arc((0,0), 0.4, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.4, 90, 180)); filldraw((arc((1,0), 0.4, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.4, 180, 270)); filldraw((arc((1,1), 0.4, 180, 270))--(1,1)--cycle, gray); draw(arc((0,1), 0.4, 270, 360)); filldraw(arc((0,1), 0.4, 270, 360)--(0,1)--cycle, gray); [/asy]$
Diagram by Shurong.ge Using Asymptote
Note: The diagram only represents a small portion of the given $2020 * 2020$ square.
### Solution
We consider an individual one-by-one block.
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$, the area covered by the circles should be $0.5$. Because of this, and the fact that there are four circles, we write
$$4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}$$
Solving for $d$, we obtain $d = \frac{1}{\sqrt{2\pi}}$, where with $\pi \approx 3$, we get $d = \frac{1}{\sqrt{6}}$, and from here, we simplify and see that $d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}$ ~Crypthes
$\textbf{Note:}$ To be more rigorous, note that $d<0.5$ since if $d\geq0.5$ then clearly the probability is greater than $\frac{1}{2}$. This would make sure the above solution works, as if $d\geq0.5$ there is overlap with the quartercircles. $\textbf{- Emathmaster}$
## Solution 2
As in the previous solution, we obtain the equation $4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}$, which simplifies to $\pi d^2 = \frac{1}{2} = 0.5$. Since $\pi$ is slightly more than $3$, $d^2$ is slightly less than $\frac{0.5}{3} = 0.1\bar{6}$. We notice that $0.1\bar{6}$ is slightly more than $0.4^2 = 0.16$, so $d$ is roughly $\boxed{\textbf{(B) } 0.4}.$ ~emerald_block
~IceMatrix |
Introduction
Perimeter is the total length or total distance covered along the boundary of a closed shape. It can also be defined as the length of the outline of a shape.
We know that perimeter of a closed figure is the distance around it and the area is the part of the plane or region occupied by the closed figure. Watch the video to explore more about perimeter and area.
If you live in an apartment, you will notice that the apartment has a main gate and outer walls which form the ‘fence’ of the building. Similarly, an independent house has fencing all around it to separate it from others. In both the cases, the length of the fence around the building is the perimeter of the outer area.
Concepts
The chapter ‘Perimeter and Area’ covers the following concepts:
Squares and Rectangles
TA square has all its sides equal in length and a rectangle has its opposite sides equal. A square is a special kind of a rectangle whose all sides are equal in length. Now, let us have a look at the two rectangles below: one is vertical, and the other is horizontal.
What can you see? We can see that in the first figure, the length is the longest side and in the second figure breadth is the longest side. Is there any special rule to decide as to which side will be the length and breadth? No! there is no such rule.
Also, instead of saying length and breadth, we can say side 1 and side 2 or adjacent sides.
So, the perimeter of a rectangle = 2 (Side 1 + Side 2)
Or in other words,
the perimeter of a rectangle = 2 × the sum of adjacent sides
Area of a rectangle = Side 1 x Side 2
Or in other words,
Area of a rectangle = the product of adjacent sides
The amount of surface enclosed by a closed figure is called its area. Now let us consider the two rectangles below:
What is the perimeter of the two rectangles? The orange rectangle has a perimeter of 24 mm and the purple rectangle also has a perimeter of 24 mm. What about their areas? Do they have the same area?
Area of the orange rectangle = 33 mm
Area of the purple rectangle = 20 mm2
They do not have the same area. So, rectangles with the same perimeter can have different areas. Rectangles having the same area can have different perimeters.
Observe the shapes 1, 2, 3 and 4 in the figure.
How do we find the perimeter of any figure? Perimeter is the length of the boundary of any shape. Here each shape is made of squares of 1 unit. Hence, if the sides of a shape are made of 12 sides of squares, then its perimeter will be 12 units. How about the area?
We know that area is the total space occupied inside a shape. Hence, we must count the number of squares inside the shape. So, if a shape has 12 squares in total, then its area is 12 sq. units.
Area of a square = side × side
Area of a rectangle = length × breadth
Perimeter of a rectangle = 2 × (length + breadth)
Perimeter of a square = 4 × length of one side
Remember, to find the area between two rectangles:
Area of the remaining part of the rectangle = Area of the big rectangle – Area of the small rectangle
When the area and a side of a rectangle are given, find the missing side by using the formula of the area of a rectangle and then find the perimeter.
When the perimeter and a side are given, find the missing side by using the formula of the perimeter of a rectangle and then find the area.
Consider this example: Manoj walks around a rectangular park with area 920 m2 and length 40 m. Manoj wants to know how long he walks once around the park. Let us help Manoj to find the distance he walks once around the park.
To find the distance around the park we must first find the breadth of the park.
We know that area of a rectangle = length × breadth
920 m2 = 40 m × breadth
= 23 m
We know that the perimeter of the rectangular park is the total distance around the park. To find the distance covered by Manoj in one round, we need to find the perimeter of the park.
Perimeter = 2 × (40 m + 23 m)
= 2 × 63 m
= 126 m
Hence, Manoj covers 126 m by walking once around the park.
Rectangles
A rectangle can be divided into congruent parts. Figures having the same shape and size are known as congruent figures.
To check whether the parts are congruent or not, superimpose one part onto the other, if they match, they are said to be congruent. Congruent parts have the same size and area.
If we superimpose one figure over the other and they fit onto each other, then they are said to be congruent shapes. They will have the same shape and size.
Consider the triangles ABC and PQR.
Both have the same size and shape and hence, they are congruent. So, this can be expressed as
∆ABC ≅ ∆PQR
Take a rectangle and cut the rectangle along its diagonal to get two triangles. Superimpose one triangle onto the other.
Note: To superimpose, we will have to rotate the other triangle.
Are they exactly the same in size?
We can see that after superimposing, both triangles fit perfectly which implies that they are congruent to each other. Therefore, these two triangles have the same shape and size.
A rectangle of length 5 cm and breadth 4 cm is divided into two parts as shown below. Trace the rectangle on a sheet of paper and cut the rectangle along the line shown below to divide it into two parts.
Superimpose one part on the other, see if they match. (You may have to rotate one of them)
Are they congruent? Yes! The two parts are congruent to each other.
Now, take a rectangle and cut the rectangle along its diagonal to get two triangles. Superimpose one triangle on the other. Are they exactly the same in size?
We can see that after superimposing both the triangles fit perfectly which implies that they are congruent to each other. Therefore, these two triangles have the same shape and size. Since two parts of the triangles are congruent, they have the same area.
Together these two triangles form a rectangle.
We can also write this as,
Area of triangle(A) + Area of triangle(B)
= Area of rectangle
The triangular parts have the same area so,
Area of rectangle
= 2 × area of triangle (A)
We can also write it as,
Area of each triangle
= 1 2 × area of triangle
We can also cut the same rectangle in some other shapes which are congruent to each other.
A rectangle of length 6 cm and breadth 4 cm is divided into two parts. Trace the rectangle on another paper and cut off the rectangle along EF to divide it into two parts. Superimpose one part onto the other, see if they match. (You may have to rotate them)
Are they congruent? Yes!
The two parts are congruent to each other. So, the area of one part is equal to the area of the other part.
Therefore, the area of each congruent part
= 1 2 × the area of the triangle
= 1 2 × (6 × 4) cm2
= 12 cm2
Parallelograms
To calculate the area of the parallelogram we use the formula:
Area of a parallelogram = base × height
The area of the given parallelogram
= DC × AE
The parallelogram shown below has an area of 80 units2.
Find the missing base. How do we find the missing base? We know that the perimeter of a parallelogram = base × height
80 = base × height
80 = b × height
Hence, base (b) = 80 5 = 16 units.
Base of a parallelogram = area of the parallelogram height of the parallelogram
Height of a parallelogram = area of the parallelogram height of the parallelogram
Finding the missing dimensions in parallelograms requires us to work backwards.
We can figure out a missing dimension if we have been given the area and another measurement, such as the area and the height or the area and the base.
The area of a parallelogram is determined by multiplying the base and the height.
If you know the area and the base or the area and height, you can use this formula to find the missing dimension.
Triangle
Observe the given triangles. What do you notice?
Any side of a triangle can be the base. The diagrams above show the length of the base (b) and the height (h) of the triangles.
We can see that both ∆ ABC and ∆ DEF are congruent to each other.
Now, place both the triangles in such a way that their corresponding sides AB and ED are joined as shown below.
The new figure thus formed is a parallelogram. Compare the area of each of the triangles to the area of the parallelogram. The base and the height of the triangles are the same as the base and the height of the parallelogram, respectively.
The parallelogram is made of two equal triangles.
We have,
Area of parallelogram = base x height
Therefore,
Area of each triangle
= 1 2 × (Area of the parallelogram)
= 1 2 × (base x height)
= 1 2 × (b x h)
Observe the figure given below:
The triangles are on the base AB = 6 cm. What can you say about the height of each of the triangles corresponding to the base AB?
From the given figure, we can see that all triangles have the same base and the same altitude. Hence, they are equal in area.
All congruent triangles are equal in area but the triangles equal in area need not be congruent.
A gardener wants to know the cost of covering the whole of a triangular garden with grass.
We know that area is the total surface enclosed by a closed figure. In this case, we need to know the area of the triangular region.
Find the area of right - angled triangle ABC right angled at B, with one leg of length 20 cm and hypotenuse as 29 cm. Let us find the area of right - angled triangle ABC.
We are given,
• Height/Perpendicular (AB) = 20 cm
• Hypotenuse (AC) = 29 cm
• Base (BC) = ?
Let us use Pythagoras theorem to find the base (BC),
According to Pythagoras theorem, hypotenuse2 = perpendicular2 + base2
Base2 = hypotenuse2 - perpendicular2
(BC)2 = (AC)2 – (AB)2
(BC)2 = 292 – 202
(BC)2 = 841 – 400
(BC)2 = 441
BC = √441 = 21
So, the base BC is 21 cm.
We know that,
Area of a triangle = 1 2 × (base x height)
Area of right - angled triangle ABC = 1 2 × (BC × AB)
= 1 2 × (20 × 21)
= 210 cm2
Hence, the area of the right - angled triangle ABC is 210 cm2.
A triangular base of a prism has height 10 cm and base 15 cm; what will be the area of the triangular base? We know that,
Area of a triangle = 1 2 × (base x height)
Area of a triangular base = 1 2 × (20 × 15)
= 1 2300
= 150 cm2
Hence, the area of a triangular base of a prism is 150 cm2.
Remember, altitude of a triangle is the perpendicular drawn from a vertex of the triangle to its opposite side (this side is the base of the triangle). Also, the altitude, which is known as the height of the triangle, makes a right angle with the base. The picture below shows an altitude of a triangle.
When a triangle is a right triangle, the altitude or the height is one of the legs if the other leg is the base. If the triangle is obtuse, then an altitude may be outside the triangle. If the triangle is acute, then every altitude will be inside the triangle. The red lines below are all altitudes.
There are 2 types of problems in finding the missing dimensions of a triangle.
Type 1: To find the altitude of a triangle when the area and base of the triangle are given, use the following formula:
Altitude of a triangle = Area × 2 Base
Type 2: To find the base of a triangle when the area and altitude of the triangle are given, use the following formula:
Base of a triangle = Area × 2 Base
Circles
The distance from one point on a circle through the centre to another point on the circle is called the diameter. It is also the longest distance across the circle and is twice the radius.
We know that a circle is a closed figure. The length of the boundary or the perimeter of a circle is called the circumference. To calculate the circumference of the circle we need to know the value of a special number π (pi). The ratio of the circumference of any circle to its diameter is always a constant value, 22 7, which is the value of pi.
For convenience in calculation, we round off the value of ‘pi’ to 3.14.
Circumference of the circle is the length of the boundary of the circle. If we cut open the circle and straighten it, the length of the boundary will be the measure of the circumference of the circle.
Consider the following table. We are given the radius, diameter, circumference and the ratio of circumference to diameter for 4 different circles.
We know that π = 22 7 = 3.14
So, we can say that c d = π, where ‘C’ represents the circumference of the circle and ‘d’
represents its diameter. Or we can write it as C = π × d.
We know that diameter (d) of a circle is twice the radius (r), i.e., d = 2 × r.
So, C = π × d
C = π × 2r or C = 2πr
We can conclude that:
The perimeter of any shape is the total length of all its sides. Since a circle has no sides, its perimeter (also known as its circumference) is the length of the boundary of the circle.
A semicircle is half a circle, so we'll need to divide the formula for the circumference of a circle by two.
The circumference of a circle is 2πr or πd. So, the circumference of a semicircle is 1 2 × πd or πr, where ‘r’ is the radius.
But if we divide the circumference of the circle by two, we get the measure of only the curved part of the boundary of the semicircle. The measure of the straight part of the boundary is not included in that.
Since a semicircle, unlike a circle, has a straight side, we need to consider that if we are going to find its total perimeter.
That side is its diameter. So, we just need to add the diameter to the formula we have already determined. So, perimeter of the semicircle is given by πr + 2r or πr + d.
We know how to calculate the circumference of a circle if radius or diameter is given. But how to calculate the radius/diameter when the circumference is given? We know that circumference of a circle is:
C = 2πr when the radius is given, and C = πd when the diameter is given.
Hence, we can write r = C or d = C π
Where C = circumference, r = radius, d = diameter
A farmer digs a flower bed of radius 2 m at the centre of a field. He needs to purchase fertiliser. If 1 kg of fertiliser is required for 1 square metre area, how much fertiliser should he purchase?
What will be the cost of polishing a circular table-top of radius 3 m at the rate of ₹10 per square metre?
In such cases, we need to find the area of the circular region.
Let us find the area of a circle, using a graph paper. Draw a circle of radius 4 cm on a graph paper. Find the area by counting the number of squares enclosed.
As the edges are not straight, we get a rough estimate of the area of the circle by this method. The amount of space inside the circle is called the area of the circle.
We calculate the area of the circle using the formula:
Area of circle = πr2
where ‘r’ is the radius of the circle and π = 22 7
This formula is useful for measuring the space occupied by a circular field or a plot. Suppose you have to buy a table cloth, then with the formula you can calculate how much cloth is required to be bought.
Let us consider a problem.
If the circumference of a circular sheet is 154 m, find its radius. Also, find the area of the sheet.
How to calculate the radius or diameter?
We use the following formula to calculate the circumference of a circle:
C = 2πr when the radius is given, and C = πd when the diameter is given.
Hence, we can write r = C or d = C π
Where C = circumference, r = radius, d = diameter
When we know the radius of the circle, we can use the formula A = πr2 to calculate the area of the circle.
What is a ring and how does it look?
A ring is a circular disk with a circular hole in it. The outer and inner circles that define the ring are concentric which means that they share a common centre.
The dimensions R1 and R2 are the radii of the outer circle and the inner circle respectively. The area of a circular ring can be found by subtracting the area of the smaller (inner) circle from that of the larger (outer) circle.
Area of a ring
= Area of the larger circle – Area of the smaller circle
= πR12 – πR22
= π(R12 – R22)
To solve problems on finding the area of the path between 2 circles, follow the strategy given below.
• Identify or find the radius of the outer and inner circles.
• Calculate the area of the inner circle and the outer circle.
• Subtract the area of the inner circle from the area of the outer circle to find the area between the 2 circles.
To compare the area of a circle and the area of a rectangle when the circumference of the circle is equal to the perimeter of the rectangle:
• To find the area of the circle: Using the circumference calculate the radius and then find the area
• To find the area of the rectangle: Using the perimeter and the given one side, calculate the other side and then find the area.
Perimeter and Area
Let us learn how to convert units to other units while measuring areas.
You can find that this square of side 1 cm can be divided into 100 squares, each side 1 mm.
Area of a square of side 1 cm = Area of 100 squares of each side 1mm.
Therefore, 1 cm2 = 100 × 1 mm2 or
1 cm2 = 100 mm2
Similarly, 1 m2 = 1 m × 1 m
= 100 cm × 100 cm (As 1 m = 100 cm)
= 1000 cm2
In the metric system, areas of land are also measured in hectares [written “ha” in short]. A square of side 100 m has an area of 1 hectare.
So, 1 hectare = 100 × 100 m2
= 10,000 m2
You must have observed that quite often, in gardens or parks, some space is left all around in the form of a path or in between as cross paths.
A framed picture also has some space left all around it.
We need to find the areas of such pathways or borders when we want to find the cost of making them.
To solve problems on finding the area between 2 rectangles follow the given strategy
• Find the area of the outer rectangle and the inner rectangle
• Subtract the area of the inner rectangle from the area of the outer rectangle to find the area between them.
Common Errors
The following are topics in which students make common mistakes when dealing with perimeter and area:
• 1. Area of a triangle
• 2. Area of a triangle
• 3. Use the value of pi as directed in the question
• 4. Calculate the area of the shaded portion
Area Of A Triangle
Consider the given triangle. What is its area?
Sometimes, while calculating the area of a triangle we forget to multiply by half.
Remember:
Area Of A Triangle
Identify the correct base and height pair in a triangle. Consider the given figure. Calculate the area of the triangle ABC.
Here for triangle ABC, the correct pair of height and base,
• Base = 12 m
• Height = 3 m
Use The Value Of Pi As Directed In The Question
Usually, we take 4 values of pi,
• 3.14
• In terms of π
• Calculator value
• 22 2
However, in some questions, the value of pi to be considered is given:
1 .The radius of a circular pipe is 10 cm. What length of tape is required to wrap once around the pipe (π = 3.14)?
2. Find the perimeter of the given shape. (Take π = 22 7)
In the first question, we will use the value of π as 3.14 and in the second question, we use pi as 22/7. Why so? Because it is given in the question for easier calculation. Hence, be very careful while solving such questions.
Remark: So, we must use the value of π as mentioned in the question.
Calculate The Area Of The Shaded Portion
Observe the given question:
Two crossroads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m, and parallel to its sides. Find the area of the roads.
While solving such questions make sure that we subtract the area which is common in both the shapes.
Conclusion
The following table gives the length, breadth, and perimeter of some rectangles.
Fill in the blanks to find A, B, C, D, and E. |
# Calculus Applets
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## L'Hopital's Rule
Suppose that we want to find the value of , when f (a) = g (a) = 0. One method is to use L'Hopital's Rule, which says: if f (x) and g (x) are differentiable functions and f (a) = g (a) = 0, then , if the limit on the right exists. In other words, we can find the original limit by finding the limit of the ratio of the derivatives of the numerator and denominator functions. The following applet helps to explain why this works.
Try the following:
1. In the graph on the left, the applet shows a graph of . This is not defined for x = 0, but it clearly seems to have a limit there. If we let f (x) = e2x - 1 and g (x) = x, then we can use L'Hopital's Rule to find the limit via finding the derivatives of the numerator and denominator functions: . Note that we do not take the derivative of the ratio using the quotient rule, but rather separately find the derivatives of the numerator and denominator functions, then find the limit of their ratio. Notice that, while the original ratio is undefined for x = 0, the ratio of the derivatives is defined, hence we can evaluate the limit just by substituting x = 0.
Why does this work? The graph on the right shows f (x) and g (x). If you click the zoom in button several times, local linearity makes the curves look like straight lines. Hence the functions can be approximated by their tangent lines at x = 0 (i.e., f (x) ≈ 2x and g(x) ≈ x) , and the ratio can be approximated by the ratio of these tangent lines. In the limit, this equals the ratio of the slopes of the tangent lines, which is just the ratio of the derivatives of the numerator and denominator functions.
2. Select the second example from the drop down menu. This shows another example, . Again, if you click on Zoom In a few times, the graphs look like straight lines and the ratio can be evaluated just by using the ratio of the values of the derivatives.
3. Select the third example, showing . . In this case, the limit on the right does not exist, so L'Hopital's Rule cannot be used in this case.
4. L'Hopital's Rule also works if f (a) = g (a) = ∞, and when . |
# Data interpretation line graph
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Line graph data interpretation questions are asked in various competitive exams like SSC, IBPS clerk mains, SBI PO prelims, RRB, etc., all the concepts of line graph in Data Interpretation are explained from basics with example questions and solution below. By learning the concepts and practicing different types of data interpretation line graph tricks, questions help in mastering the line graph topics.
## What is the data interpretation line graph?
An updated version of the bar chart is a line graph. Usually, by joining the uppermost points of the bars we can form a line. Similarly, with other bars repeating the same process we can form a line chart.
Line graphs are progressively self-evident, exact and precise than graphs. Line graphs are very useful for determining trends, to study slopes, rates of change and for illustrating comparisons with respect to some time series.
### Construction of a line graph
Line graphs are represented on a special kind of paper called Graph Paper. This graph paper consists of various horizontal and vertical lines that form a number of squares inside the graph paper consist of two straight lines drawn at right angles intersecting each other at a point ‘o’ as shown in the figure. This point is called the origin.
The horizontal line is called x-axis represented by xox’. The distance measured from the right side of origin is considered as positive and distance measured left side of origin is taken as negative.
The vertical line is called the y-axis represented by yoy’. Here distance measured from the top side of origin is considered from origin to bottom is considered as negative from the graph we can see from quadrants formed by intersecting x-axis and y-axis.
Generally, for line graph in Data interpretation, we consider the first quadrant which is always positive unless negative quantities are given.
The pair of variables represented on the graph are denoted as (x , y). Here x is the value of an independent variable (x-coordinate) and y is the value of the dependent variable (y-coordinate).
Generally, line graphs show how a quantity of dependent variable changes with the value of the independent variable.
As there are represented on x-y axis line graphs are also called x-y charts. Data represented on x-y co-ordinates may be in different forms based on time series like years, half-year, quarter, month, week, hour, day etc.
Generally, dates related to economics and business will be in the form of time series. Let’s see various types of data representations that are possible to determine on x-y coordinates in line graph data interpretation.
### Types of graphs by data representation in the line graph (time series):
1. Single dependent. Variable line graph
Over a certain period of time, these graphs show changes in a single variable.
1. More than one dependent variable line graph:
For representing two or more dependent variables these graphs show two or more lines.
1. Speed time graph:
This is a special case in line graph where the speed of a body against time is shown on respective axes.
### Data interpretation line graph example with solution
Given below is a line graph represents data related to imports and exports of a company in the months of Jan to June(rupees in lacks). Observe the data given in the graph carefully and answer the questions asked.
1. What was the average import from Jan to June?
Sol:
All the imports of the company month-wise are (in lacks)
Jan=5
Feb=7.5
March=12.5
April=7.5
May=12.5
Jun=15
Average import from Jan to June =sum of total imports (in lacks)/number of months import done
=5+7.5+12.5+7.5+12.5+15/6
=60/6
=10(lacks)
1. Percentage decrease in exports from April to May
Sol:
Exports of the company in the months of April are 7.5 lacks.
Exports of the company in the month of May are 12.5 lacks.
Percentage decrease in exports=(12.5-7.5/7.5)×100
=66.66%
1. What was the difference in export between February and March?
Sol:
The exports of the company in February are 7.5 lacks.
The exports of the company in March are 17.5 lacks.
The difference in exports between February and March is (12.5-7.5) lack= 5 lacks |
# If x=16 is one of the root of the quadratic equation x^{2}-2\left(7x+16\right)=0,find the other...
## Question:
If {eq}x = 16 {/eq} is one of the root of the quadratic equation {eq}x^{2} - 2\left(7x + 16\right) = 0, {/eq} find the other root.
## Quadratic Equation and its Roots:
{eq}\\ {/eq}
Any quadratic equation has two roots or solutions, they may be real or complex in nature depending on the value of coefficients. Let us assume that we have two roots {eq}(\alpha) {/eq} and {eq}(\beta) {/eq} of the standard quadratic equation. Then there are two properties related to the sum and product of roots which are quite useful when we know anyone of the root and we have to determine the other one directly without using the standard formula.
\begin{align} &ax^{2} + bx + c = 0 \\[0.2cm] &x = \alpha, \; \beta \\[0.2cm] &\alpha + \beta = - \biggr( \dfrac {b}{a} \biggr) \\[0.2cm] &\alpha \times \beta = \biggr( \dfrac {c}{a} \biggr) \end{align}
{eq}\\ {/eq}
The described quadratic equation is given below:
\begin{align} x^{2} - 2 (7x + 16) &= 0 \\[0.2cm] x^{2} - 14x - 32 &= 0 &\left(\text{ Equation 1 } \right) \end{align}
We all know the standard form of a quadratic equation and its sum of roots property that is already mentioned in the context section:
\begin{align} ax^{2} + bx + c &= 0 &\left(\text{ Equation 2 } \right) \end{align}
Now, compare the coefficients of equations {eq}1{/eq} and {eq}2{/eq}:
$$a = 1, \; b - 14, \; c = -32$$
Let us assume that we have two roots of the given equation {eq}(\alpha \; \text {and} \; \beta) {/eq}:
\begin{align} \alpha + \beta &= - \biggr( \dfrac {b}{a} \biggr) \\[0.2cm] \alpha + 16 &= - \biggr( \dfrac {-14}{1} \biggr) \\[0.2cm] \alpha + 16 &= 14 \\[0.2cm] \alpha &=\boxed {-2 } \end{align}
Hence, the other root of the equation is {eq}x = -2 {/eq}. |
# USING SIMILAR TRIANGLES TO FIND SLOPE WORKSHEET
## About "Using similar triangles to find slope worksheet"
Using similar triangles to find slope worksheet :
Worksheet given in this section is much useful to the students who would like to practice problems on similar triangles and slope.
## Using similar triangles to find slope worksheet - Questions
1. In the diagram given below, using similar triangles, prove that the slope between the points D and F is the same as the slope between the points A and C.
2. Suppose that we label two other points on line ℓ as P and Q. Would the slope between these two points be different than the slope we found in the above activity ? Explain.
3. Find the slope of the line AB using the similar triangles as a guide.
## Using similar triangles to find slope worksheet - Answers
Question 1 :
In the diagram given below, using similar triangles, prove that the slope between the points D and F is the same as the slope between the points A and C.
Step 1 :
Draw the rise and run for the slope between points D and F. Label the intersection as point E. Draw the rise and run for the slope between points A and C. Label the intersection as point B.
Step 2 :
Write expressions for the slope between D and F and between A and B.
Slope between D and F : FE / DE
Slope between A and B : CB / AB
Step 3 :
Extend DE and AB across our drawing. DE and AB are both horizontal lines, so they are parallel.
Line l is a transversal that intersects parallel lines.
Step 4 :
Because DE and AB are parallel lines and is a transversal that intersects DE and AB,
m∠FDE and m∠CAB are corresponding angles and they are congruent.
m∠FED and m∠CBA are right angles and they are congruent.
Step 5 :
By Angle–Angle Similarity, triangle ABE and triangle CDF are similar triangles.
Step 6 :
Because triangle ABE and CDF are similar, the lengths of corresponding sides of similar triangles are proportional.
FE / CB = DE / AB
Step 7 :
Recall that you can also write the proportion so that the ratios compare parts of the same triangle :
FE / DE = CB / AB
Step 8 :
The proportion we wrote in step 8 shows that the ratios we wrote in step 2 are equal. So, the slope of line ℓ is constant.
Hence, the slope between the points D and F is the same as the slope between the points A and C.
Question 2 :
Suppose that we label two other points on line ℓ as P and Q. Would the slope between these two points be different than the slope we found in the above activity ? Explain.
No
The slope of the line is constant, so the slope between the points P and Q would be the same. Moreover, not only the two points P and Q, between any two points on ℓ, the slope would be same.
Question 3 :
Find the slope of the line AB using the similar triangles as a guide.
Step 1 :
Slope is a ratio between the change in y and the change in x. That is y/x.
Step 2 :
Both triangles rise 2 places (y) and run 3 places (x). So the slope is 2/3.
After having gone through the stuff given above, we hope that the students would have understood "Using similar triangles to find slope worksheet".
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