text
stringlengths 22
1.01M
|
|---|
# What is the antiderivative of e^(8x)?
Jan 22, 2016
$\frac{1}{8} {e}^{8 x} + C$
#### Explanation:
We can go through the steps of integrating by substitution, but some find the following more clear:
We know that $\frac{d}{\mathrm{dx}} \left({e}^{8 x}\right) = 8 {e}^{8 x}$
That is $8$ time more that we want the derivative to be. So, we'll multiply by $\frac{1}{8}$ (divide by $8$).
$\frac{d}{\mathrm{dx}} \left(\frac{1}{8} {e}^{8 x}\right) = \frac{1}{8} \cdot 8 {e}^{8 x} = {e}^{8 x}$
The general antiderivative is, therefore, $\frac{1}{8} {e}^{8 x} + C$.
Here is the substitution solution
$\int {e}^{8 x} \mathrm{dx}$
Let $u = 8 x$, so we get $\mathrm{du} = 8 \mathrm{dx}$ and $\mathrm{dx} = \frac{1}{8} \mathrm{du}$
The integral becomes:
$\int {e}^{u} \cdot \frac{1}{8} \mathrm{du} = \frac{1}{8} \int {e}^{u} \mathrm{du} = \frac{1}{8} {e}^{u} + C$
Reversing the substitution gives
$\int {e}^{8 x} \mathrm{dx} = \frac{1}{8} {e}^{8 x} + C$
|
# NCERT Mathematics Class 9 Exemplar Ch 1 Number Systems Part 4
Glide to success with Doorsteptutor material for NSO : fully solved questions with step-by-step explanation- practice your way to success.
Exercise 1.2
1. Let x and y be rational and irrational numbers, respectively. Is x + y necessarily an irrational number? Give an example in support of your answer.
Answer: Yes. Let x = 21, y = be a rational number. Now x + y = 21 + = 21 + 1.4142 ... = 22.4142 ... Which is non-terminating and non-recurring. Hence x + y is irrational.
2. Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example.
Answer: No. 0× = 0 which is not irrational.
3. State whether the following statements are true or false? Justify your answer.
(i) is a rational number.
(ii) There are infinitely many integers between any two integers.
(iii) Number of rational numbers between 15 and 18 is finite.
(iv) There are numbers which cannot be written in the form p q , q ≠ 0 , p, q both are integers.
(v) The square of an irrational number is always rational.
(vi) is not a rational number as and are not integers.
(vii) is written in the form , and so it is a rational number.
Answer: (i) False. Although is of the form but here p, i.e., is not an integer.
(ii) False. Between 2 and 3, there is no integer.
(iii) False, because between any two rational numbers we can find infinitely many rational numbers.
(iv) True. is of the form but p and q here are not integers.
(v) False, as = which is not a rational number. ANSWERS
(vi) False, because = = 2 which is a rational number.
(vii) False, because = = which is p, i.e., 5 is not an integer.
4. Classify the following numbers as rational or irrational with justification:
(i)
(ii) 3
(iii)
(iv)
(v)
(vi)
(vii) 0.5918
(viii)
(ix)
(x)
(ii) 3 = 9 , which is the product of a rational and an irrational number and so an irrational number.
(iii) = ,which is the quotient of a rational and an irrational number and so an irrational number
(iv) , which is a rational number.
(v) Irrational, , which is the quotient of a rational and an irrational.
(vi) = , which is a rational number.
(vii) Rational, as decimal expansion is terminating.
(viii) , which is a rational number.
(ix) Rational, as decimal expansion is non-terminating recurring.
(x) Irrational, as decimal expansion is non-terminating non-recurring.
|
Breaking News
# What is 5 factorial ?
Steps to calculate factorial of 5
To find 5 factorial, or 5!, simply use the formula that multiplies the number 5 by all positive whole numbers less than it.
Let’s look at how to calculate the Factorial of 5:
5! is exactly :
120
Factorial of 5 can be calculated as:
5! = 5 x 4 x 3 x 2 x 1
## What is Factorial?
The concept of factorial is used widely in mathematics to describe the product of a sequence of natural numbers descending from a specified number down to 1. Specifically, the factorial of 5, denoted as 5!, is significant because it provides the number of different ways in which a set of 5 distinct items can be arranged. Understanding the factorial is essential in disciplines that require counting arrangements or permutations, such as combinatorics and probability.
## Formula to Calculate the Factorial of 5
The calculation of a factorial can be done using the formula n! = n × (n-1) × … × 1. Applying this formula to the number 5, we can determine 5 factorial as follows:
• Step 2: Multiply 5 by one less than itself, which is 4, resulting in 20 (5 × 4).
• Step 3: Continue this process with the next descending integers: 20 × 3 = 60, then 60 × 2 = 120.
• Step 4: Finally, multiply the result by 1 (although this does not change the product): 120 × 1 = 120.
Hence, 5! equals 120.
## What is the Factorial of 5 Used For?
Factorial of 5, or 5!, has several interesting applications:
• In combinatorics, 5! is the number of ways you can arrange 5 different books on a shelf.
• In probability theory, if there are 5 different races in a relay match, 5! represents the different ways the races can be ordered.
• Furthermore, 5! is used in calculating combinations and permutations in statistics and helps in solving various problems related to the organization of data.
## Exercises
• If a password is made by arranging 5 distinct letters, how many different passwords can be formed?
• If you are to choose the first 5 leading runners in a race, how many different arrangements can be made?
## Solutions to Exercises
• Since there are 5 distinct letters and each can be used only once, the number of different passwords is equal to 5 factorial, which is 5! = 120.
• There are 5! ways to arrange the order of the first 5 leading runners, meaning there are 120 distinct arrangements.
|
# Modeling, Functions, and Graphs
## Section4.2Exponential Functions
### SubsectionIntroduction
In Section 4.1, we studied functions that describe exponential growth or decay. More formally, we define an exponential function as follows.
#### Exponential Function.
\begin{equation*} f(x) = ab^x,~~~~ \text{ where } ~~~b \gt 0 ~~~\text{ and } ~~~b \ne 1 \text{, } ~~~a \ne 0 \end{equation*}
Some examples of exponential functions are
\begin{equation*} f (x) = 5^x,~~~~ P(t) = 250(1.7)^t,~~~~ \text{and } ~~~~g(n) = 2.4(0.3)^n \end{equation*}
The constant $$a$$ is the $$y$$-intercept of the graph because
\begin{equation*} f (0) = a \cdot b^0 = a \cdot 1 = a \end{equation*}
For the examples above, we find that the $$y$$-intercepts are
\begin{equation*} \begin{aligned}[t] f(0) \amp= 5^0 = 1 \text{,} \\ P(0) \amp= 250(1.7)^0 = 250\text{, and} \\ g(0) \amp= 2.4(0.3)^0 = 2.4 \end{aligned} \end{equation*}
The positive constant $$b$$ is called the base of the exponential function.
#### Checkpoint4.23.QuickCheck 1.
Which of the following is an exponential function?
• $$\displaystyle f(x)=3x^4$$
• $$\displaystyle f(x)=3(4)^x$$
• $$\displaystyle f(x)=2x^{\frac{3}{4}}$$
• $$\displaystyle f(x)=\dfrac{4}{x^3}$$
$$\text{Choice 2}$$
Solution.
$$f(x)=3(4)^x$$
#### Note4.24.
• We do not allow $$b$$ to be negative, because if $$b \lt 0\text{,}$$ then $$b^x$$ is not a real number for some values of $$x\text{.}$$ For example, if $$b = -4$$ and $$f (x) = (-4)^x\text{,}$$ then $$f (1/2) = (-4)^{1/2}$$ is an imaginary number.
• We also exclude $$b = 1$$ as a base because $$1^x = 1$$ for all values of $$x\text{;}$$ hence the function $$f (x) = 1^x$$ is actually the constant function $$f (x) = 1\text{.}$$
### SubsectionGraphs of Exponential Functions
The graphs of exponential functions have two characteristic shapes, depending on whether the base, $$b\text{,}$$ is greater than $$1$$ or less than $$1\text{.}$$ As typical examples, consider the graphs of $$f (x) = 2^x$$ and $$g(x) =\left(\dfrac{1}{2}\right)^x$$ shown below. Some values for $$f$$ and $$g$$ are recorded in the tables.
$$x$$ $$f(x)$$ $$-3$$ $$\frac{1}{8}$$ $$-2$$ $$\frac{1}{4}$$ $$-1$$ $$\frac{1}{2}$$ $$0$$ $$1$$ $$1$$ $$2$$ $$2$$ $$4$$ $$3$$ $$8$$
$$x$$ $$g(x)$$ $$-3$$ $$8$$ $$-2$$ $$4$$ $$-1$$ $$2$$ $$0$$ $$1$$ $$1$$ $$\frac{1}{2}$$ $$2$$ $$\frac{1}{4}$$ $$3$$ $$\frac{1}{8}$$
Notice that $$f (x) = 2^x$$ is an increasing function and $$g(x) = \left(\dfrac{1}{2}\right)^x$$ is a decreasing function. Both are concave up. In general, exponential functions have the following properties.
#### Properties of Exponential Functions, $$f(x) = ab^x\text{,}$$ $$a \gt 0$$.
1. Domain: all real numbers.
2. Range: all positive numbers.
3. If $$b \gt 1\text{,}$$ the function is increasing and concave up;
if $$0 \lt b \lt 1\text{,}$$ the function is decreasing and concave up.
4. The $$y$$-intercept is $$(0, a)\text{.}$$ There is no $$x$$-intercept.
In the table for $$f(x)\text{,}$$ you can see that as the $$x$$-values decrease toward negative infinity, the corresponding $$y$$-values decrease toward zero. As a result, the graph of $$f$$ decreases toward the $$x$$-axis as we move to the left. Thus, the negative $$x$$-axis is a horizontal asymptote for exponential functions with $$b \gt 1\text{,}$$ as shown in figure (a).
For exponential functions with $$0 \lt b \lt 1\text{,}$$ the positive $$x$$-axis is an asymptote, as illustrated in figure (b). (See Section 2.2 to review asymptotes.)
#### Checkpoint4.25.QuickCheck 2.
Which statement is true?
• An exponential function is not defined for negative inputs.
• The outputs of an exponential function cannot be negative.
• The $$y$$-intercept of the function $$f(x)=2(3)^x$$ is $$(0,6)\text{.}$$
• The function $$f(x)=16(0.5)^x$$ decreases by 8 each time we increase $$x$$ by 1.
$$\text{Choice 2}$$
Solution.
The outputs of an exponential function cannot be negative.
In Example 4.26, we compare two increasing exponential functions. The larger the value of the base, $$b\text{,}$$ the faster the function grows. In this example, both functions have $$a = 1\text{.}$$
#### Example4.26.
Compare the graphs of $$f (x) = 3^x$$ and $$g(x) = 4^x\text{.}$$
Solution.
We evaluate each function for several convenient values, as shown in the table.
Then we plot the points for each function and connect them with smooth curves. For positive $$x$$-values, $$g(x)$$ is always larger than $$f(x)\text{,}$$ and is increasing more rapidly. In the figure, we can see that $$g(x) = 4^x$$ climbs more rapidly than $$f(x) = 3^x\text{.}$$ Both graphs cross the $$y$$-axis at (0, 1).
$$x$$ $$f(x)$$ $$g(x)$$ $$-2$$ $$\dfrac{1}{9}$$ $$\dfrac{1}{16}$$ $$-1$$ $$\dfrac{1}{3}$$ $$\dfrac{1}{4}$$ $$0$$ $$1$$ $$1$$ $$1$$ $$3$$ $$4$$ $$2$$ $$9$$ $$16$$
#### Note4.27.
For decreasing exponential functions, those with bases between $$0$$ and $$1\text{,}$$ the smaller the base, the more steeply the graph decreases. For example, compare the graphs of $$p(x) = 0.8^x$$ and $$q(x) = 0.5^x$$ shown in the figure at right.
#### Checkpoint4.28.Practice 1.
1. State the ranges of the functions $$f$$ and $$g$$ from the previous Example on the domain $$[-2, 2]\text{.}$$
$$f:$$
$$g:$$
2. State the ranges of the functions $$p$$ and $$q$$ shown in the Note above on the domain $$[-2, 2]\text{.}$$ Round your answers to two decimal places.
$$p:$$
$$q:$$
$$\frac{1}{9}\le x\le 9$$
$$\frac{1}{16}\le x\le 16$$
$$0.64\le x\le \frac{25}{16}$$
$$\frac{1}{4}\le x\le 4$$
Solution.
1. $$f: \left[\dfrac{1}{9}, 9\right]\text{;}$$ $$g: \left[\dfrac{1}{16}, 16\right]$$
2. $$p: [0.64, 1.56]\text{;}$$ $$q: [0.25, 4]$$
### SubsectionTransformations of Exponential Functions
In Chapter 2, we considered transformations of the basic graphs. For instance, the graphs of the functions $$y = x^2 - 4$$ and $$y = (x - 4)^2$$ are shifts of the basic parabola, $$y = x^2\text{.}$$ In a similar way, we can shift or stretch the graph of an exponential function while the basic shape is preserved.
#### Example4.29.
Use your calculator to graph the following functions. Describe how these graphs compare with the graph of $$h(x) = 2^x\text{.}$$
1. $$\displaystyle f (x) = 2^x + 3$$
2. $$\displaystyle g(x) = 2^{x+3}$$
Solution.
Enter the formulas for the three functions as shown below. Note the parentheses around the exponent in the keying sequence for $$Y_3 = g(x).$$
$$Y_1 = 2$$ ^ X
$$Y_2 = 2$$ ^ X + 3
$$Y_3 = 2$$ ^ ( X + 3 )
The graphs of $$h(x) = 2^x\text{,}$$ $$f(x) = 2^x + 3\text{,}$$ and $$g(x) = 2^{x+3}$$ in the standard window are shown below.
1. The graph of $$f(x) = 2^x + 3\text{,}$$ shown in figure (b), has the same basic shape as that of $$h(x) = 2^x\text{,}$$ but it has a horizontal asymptote at $$y = 3$$ instead of at $$y = 0$$ (the $$x$$-axis). In fact, $$f(x) = h(x) + 3\text{,}$$ so the graph of $$f$$ is a vertical translation of the graph of $$h$$ by $$3$$ units. If every point on the graph of $$h(x) = 2^x$$ is moved $$3$$ units upward, the result is the graph of $$f (x) = 2^x + 3\text{.}$$
2. First note that $$g(x) = 2^x+3 = h(x + 3)\text{.}$$ In fact, the graph of $$g(x) = 2^{x+3}$$ shown in figure (c) has the same basic shape as $$h(x) = 2^x$$ but has been translated $$3$$ units to the left.
#### Checkpoint4.30.QuickCheck 3.
Which function translates the graph of $$y=8^x$$ two units to the right?
• $$\displaystyle f(x)=8^{x-2}$$
• $$\displaystyle f(x)=8^{2x}$$
• $$\displaystyle f(x)=2+(8)^x$$
• $$\displaystyle f(x)=8^x-2$$
$$\text{Choice 1}$$
Solution.
$$f(x)=8^{x-2}$$
What about reflections? Recall that the graph of $$y = -f (x)$$ is the reflection about the $$x$$-axis of the graph of $$y = f (x)\text{.}$$ The graphs of $$y = 2^x$$ and $$y = -2^x$$ are shown at left below.
You may have also noticed a relationship between the graphs of $$f (x) = 2^x$$ and $$g(x) = \left(\dfrac{1}{2}\right)^x\text{,}$$ which are shown at right above. The graph of $$g$$ is the reflection of the graph of $$f$$ about the $$y$$-axis. We can see why this is true by writing the formula for $$g(x)$$ in another way:
\begin{equation*} g(x) =\left(\frac{1}{2}\right)^x= \left(2^{-1}\right)^x = 2^{-x} \end{equation*}
We see that $$g(x)$$ is the same function as $$f(-x)\text{.}$$ Replacing $$x$$ by $$-x$$ in the formula for a function switches every point $$(p,q)$$ on the graph with the point $$(-p,q)$$ and thus reflects the graph about the $$y$$-axis.
#### Reflections of Graphs.
1. The graph of $$y = -f (x)$$ is the reflection of the graph of $$y = f (x)$$ about the $$x$$-axis.
2. The graph of $$y = f (-x)$$ is the reflection of the graph of $$y = f (x)$$ about the $$y$$-axis.
#### Checkpoint4.31.Practice 2.
Which of the functions below have the same graph? Explain why.
• (a) and (b)
• (a) and (c)
• (b) and (c)
• None of the above
1. $$\displaystyle f(x)= \left(\dfrac{1}{4} \right)^x$$
2. $$\displaystyle g(x)= -4^x$$
3. $$\displaystyle h(x)= 4^{-x}$$
$$\text{(a) and (c)}$$
Solution.
(a) and (c) are the same function.
#### Checkpoint4.32.Pause and Reflect.
How are the graphs of $$f(x)=b^x$$ and $$g(x)=(\frac{1}{b})^x$$ related?
### SubsectionComparing Exponential and Power Functions
Exponential functions are not the same as the power functions we studied in Chapter 3. Although both involve expressions with exponents, it is the location of the variable that makes the difference.
#### Power Functions vs Exponential Functions.
$$\hphantom{General formula and m}$$ Power Functions Exponential Functions General formula $$h(x)=kx^p$$ $$f(x)=ab^x$$ Description variable base and constant exponent constant base and variable exponent Example $$h(x)=2x^3$$ $$f(x)=2(3^x)$$
These two families of functions have very different properties, as well.
#### Example4.33.
Compare the power function $$h(x) = 2x^3$$ and the exponential function $$f(x) = 2(3^x)\text{.}$$
Solution.
First, compare the values for these two functions shown in the table.
The scaling exponent for $$h(x)$$ is $$3\text{,}$$ so that when $$x$$ doubles, say, from $$1$$ to $$2\text{,}$$ the output is multiplied by $$2^3\text{,}$$ or $$8\text{.}$$
On the other hand, we can tell that $$f$$ is exponential because its values increase by a factor of $$3$$ for each unit increase in $$x\text{.}$$ (To see this, divide any function value by the previous one.)
$$x$$ $$h(x)=2x^3$$ $$f(x)=2(3^x)$$ $$-3$$ $$-54$$ $$\dfrac{2}{27}$$ $$-2$$ $$-16$$ $$\dfrac{1}{4}$$ $$-1$$ $$-2$$ $$\dfrac{2}{3}$$ $$0$$ $$0$$ $$2$$ $$1$$ $$2$$ $$6$$ $$2$$ $$16$$ $$18$$ $$3$$ $$54$$ $$54$$
As you would expect, the graphs of the two functions are also quite different. For starters, note that the power function goes through the origin, while the exponential function has $$y$$-intercept $$(0, 2)$$as shown at left below.
From the table, we see that $$h(3) = f(3) = 54\text{,}$$ so the two graphs intersect at $$x = 3\text{.}$$ (They also intersect at approximately $$x = 2.48\text{.}$$) However, if you compare the values of $$h(x) = 2x^3$$ and $$f(x) = 2(3^x)$$ for larger values of $$x\text{,}$$ you will see that eventually the exponential function overtakes the power function, as shown at right above.
The relationship in Example 4.33 holds true for all increasing power and exponential functions: For large enough values of $$x\text{,}$$ the exponential function will always be greater than the power function, regardless of the parameters in the functions. The figure at left shows the graphs of $$f(x) = x^6$$ and $$g(x) = 1.8^x\text{.}$$ At first, $$f (x) \gt g(x)\text{,}$$ but at around $$x = 37\text{,}$$ $$g(x)$$ overtakes $$f (x)\text{,}$$ and $$g(x) \gt f (x)$$ for all $$x \gt 37\text{.}$$
#### Checkpoint4.34.QuickCheck 4.
Which function grows faster in the long run?
• $$\displaystyle f(x)=5x^2$$
• $$\displaystyle f(x)=3x^5$$
• $$\displaystyle f(x)=5(2^x)$$
• $$\displaystyle f(x)=2(3^x)$$
$$\text{Choice 4}$$
Solution.
$$f(x)=2(3^x)$$
#### Checkpoint4.35.Practice 3.
Which of the following functions are exponential functions, and which are power functions?
1. $$F(x) = 1.5^x$$
• exponential
• power
• neither
2. $$G(x) = 3x^{1.5}$$
• exponential
• power
• neither
3. $$H(x) = 3^{1.5x}$$
• exponential
• power
• neither
4. $$K(x) = (3x)^{1.5}$$
• exponential
• power
• neither
$$\text{exponential}$$
$$\text{power}$$
$$\text{exponential}$$
$$\text{power}$$
Solution.
Exponential: (a) and (c); power: (b) and (d)
#### Checkpoint4.36.Pause and Reflect.
Discuss the differences between a power function and an exponential function.
### SubsectionExponential Equations
An exponential equation is one in which the variable is part of an exponent. For example, the equation
\begin{equation*} 3^x = 81 \end{equation*}
is exponential.
Many exponential equations can be solved by writing both sides of the equation as powers with the same base. To solve the equation above, we write
\begin{equation*} 3^x = 3^4 \end{equation*}
which is true if and only if $$x = 4\text{.}$$
In general, if two equivalent powers have the same base, then their exponents must be equal also, as long as the base is not $$0$$ or $$\pm 1\text{.}$$
Sometimes the laws of exponents can be used to express both sides of an equation as single powers of a common base.
#### Example4.37.
Solve the following equations.
1. $$\displaystyle 3^{x-2} = 9^3$$
2. $$\displaystyle 27 \cdot 3^{-2x} = 9^{x+1}$$
Solution.
1. Using the fact that $$9 = 3^2\text{,}$$ we write each side of the equation as a power of $$3\text{:}$$
\begin{equation*} \begin{aligned}[t] 3^{x-2} \amp = \left(3^2\right)^3 \\ 3^{x-2} \amp = 3^6 \end{aligned} \end{equation*}
Now we equate the exponents to obtain
\begin{equation*} \begin{aligned}[t] x - 2 \amp = 6 \\ x \amp = 8 \end{aligned} \end{equation*}
2. We write each factor as a power of $$3\text{.}$$
\begin{equation*} 3^3 \cdot 3^{-2x} = \left(3^2\right)^{x+1} \end{equation*}
We use the laws of exponents to simplify each side:
\begin{equation*} 3^{3-2x} = 3^{2x+2} \end{equation*}
Now we equate the exponents to obtain
\begin{equation*} \begin{aligned}[t] 3 - 2x \amp = 2x + 2 \\ -4x =\amp -1 \end{aligned} \end{equation*}
The solution is $$x = \dfrac{1}{4}\text{.}$$
#### Checkpoint4.38.QuickCheck 5.
Which is a good strategy for solving $$3^{x-2}=81\text{?}$$
• Divide both sides by 3.
• Add $$3^2$$ to both sides.
• Simplify the left side.
• Write the right side as a power of 3.
$$\text{Choice 4}$$
Solution.
Write the right side as a power of 3.
#### Checkpoint4.39.Practice 4.
Solve the equation $$2^{x+2} = 128\text{.}$$
$$x=$$
Hint.
$$\blert{\text{Write each side as a power of 2.}}$$
$$\blert{\text{Equate exponents.}}$$
$$5$$
Solution.
$$x=5$$
Exponential equations arise frequently in the study of exponential growth.
#### Example4.40.
During the summer a population of fleas doubles in number every $$5$$ days. If a population starts with $$10$$ fleas, how long will it be before there are $$10,240$$ fleas?
Solution.
Let $$P$$ represent the number of fleas present after $$t$$ days. The original population of $$10$$ is multiplied by a factor of $$2$$ every $$5$$ days, or
\begin{equation*} P(t) = 10 \cdot 2^{t/5} \end{equation*}
We set $$P = \alert{10,240}$$ and solve for $$t\text{:}$$
\begin{equation*} \begin{aligned}[t] \alert{10,240} \amp = 10\cdot 2^{t/5}\amp\amp \blert{\text{Divide both sides by 10.}} \\ 1024 \amp = 2^{t/5} \amp\amp \blert{\text {Write 1024 as a power of 2.}} \\ 2^{10} \amp = 2^{t/5} \end{aligned} \end{equation*}
We equate the exponents to get $$10 = \dfrac{t}{5}\text{,}$$ or $$t = 50\text{.}$$ The population will grow to $$10,240$$ fleas in $$50$$ days.
#### Checkpoint4.41.Practice 5.
During an advertising campaign in a large city, the makers of Chip-O’s corn chips estimate that the number of people who have heard of Chip-O’s increases by a factor of $$8$$ every 4 days.
1. If 100 people are given trial bags of Chip-O’s to start the campaign, write a function, $$N(t)\text{,}$$ for the number of people who have heard of Chip-O’s after $$t$$ days of advertising.
$$N(t)=$$
2. Use your calculator to graph the function $$N(t)$$ on the domain $$0 \le t \le 15\text{.}$$
3. How many days should the makers run the campaign in order for Chip-O’s to be familiar to $$51,200$$ people? Use algebraic methods to find your answer and verify on your graph.
$$100\cdot 8^{\frac{t}{4}}$$
$$12$$
Solution.
1. $$\displaystyle N(t)=100 \cdot 8^{t/4}$$
2. A graph is below.
3. 12 days
A graph for part (b):
#### Checkpoint4.42.QuickCheck 6.
Suppose $$g$$ is an exponential function, with $$g(0)=48$$ and $$g(1)=36\text{.}$$ What is $$g(2)\text{?}$$
• 24
• 27
• 12
• 18
$$\text{27}$$
Solution.
27
#### Technology4.43.Graphical Solution of Exponential Equations.
It is not always so easy to express both sides of the equation as powers of the same base. In the following sections, we will develop more general methods for finding exact solutions to exponential equations. But we can use a graphing utility to obtain approximate solutions.
##### Example4.44.
Use the graph of $$y = 2^x$$ to find an approximate solution to the equation $$2^x = 5$$ accurate to the nearest hundredth.
Solution.
Enter $$Y_1 = 2$$ ^ X and use the standard graphing window (ZOOM 6) to obtain the graph shown in figure (a). We are looking for a point on this graph with $$y$$-coordinate $$5\text{.}$$
Using the TRACE feature, we see that the $$y$$-coordinates are too small when $$x \lt 2.1$$ and too large when $$x \gt 2.4\text{.}$$ The solution we want lies somewhere between $$x = 2.1$$ and $$x = 2.4\text{,}$$ but this approximation is not accurate enough.
To improve our approximation, we will use the intersect feature. Set $$Y_2 = 5$$ and press GRAPH. The $$x$$-coordinate of the intersection point of the two graphs is the solution of the equation $$2^x = 5$$ Activating the intersect command results in figure (b), and we see that, to the nearest hundredth, the solution is $$2.32\text{.}$$
We can verify that our estimate is reasonable by substituting into the equation:
\begin{equation*} 2^{2.32} \stackrel{?}{=} 5 \end{equation*}
We enter 2 ^ 2.32 ENTER to get $$4.993322196\text{.}$$ This number is not equal to $$5\text{,}$$ but it is close, so we believe that $$x = 2.32$$ is a reasonable approximation to the solution of the equation $$2^x = 5\text{.}$$
#### Checkpoint4.45.Practice 6.
Use the graph of $$y = 5^x$$ to find an approximate solution to $$5^x = 285\text{,}$$ accurate to two decimal places.
Answer: $$x\approx$$
$$3.51209$$
Solution.
The point on the graph where $$y=285$$ has $$x \approx 3.51$$
#### Checkpoint4.46.Pause and Reflect.
Give an example of an exponential equation, and describe how to solve it.
### SubsectionSection Summary
#### SubsubsectionVocabulary
Look up the definitions of new terms in the Glossary.
• Exponential function
• Base
• Exponential equation
#### SubsubsectionCONCEPTS
1. An exponential function has the form
\begin{equation*} f (x) = ab^x\text{, where }~b\gt 0~~\text{ and }~~b \ne 1, ~a\ne 0 \end{equation*}
2. Quantities that increase or decrease by a constant percent in each time period grow or decay exponentially.
3. ##### Properties of Exponential Functions $$f(x)=ab^x, ~~a\gt 0$$.
1. Domain: all real numbers.
2. Range: all positive numbers.
3. If $$b\gt 1\text{,}$$ the function is increasing and concave up; if $$0\lt b\lt 1\text{,}$$ the function is decreasing and concave up.
4. The $$y$$-intercept is $$(0, a)\text{.}$$ There is no $$x$$-intercept.
4. The graphs of exponential functions can be transformed by shifts, stretches, and reflections.
5. ##### Reflections of Graphs.
1. The graph of $$y = -f (x)$$ is the reflection of the graph of $$y = f (x)$$ about the $$x$$-axis.
2. The graph of $$y = f (-x)$$ is the reflection of the graph of $$y = f (x)$$ about the $$y$$-axis.
6. Exponential functions $$f (x) = ab^x$$ have different properties than power functions $$f (x) = kx^p\text{.}$$
7. We can solve some exponential equations by writing both sides with the same base and equating the exponents.
8. We can use graphs to find approximate solutions to exponential equations.
#### SubsubsectionSTUDY QUESTIONS
1. Give the general form for an exponential function. What restrictions do we place on the base of the function?
2. Explain why the output of an exponential function $$f (x) = b^x$$ is always positive, even if $$x$$ is negative.
3. How are the graphs of the functions $$f (x) = b^x$$ and $$g(x) = \left(\dfrac{1}{b} \right)^x$$ related?
4. How is an exponential function different from a power function?
5. Delbert says that $$8\left(\dfrac{1}{2} \right)^x$$ is equivalent to $$4^x\text{.}$$ Convince him that he is mistaken.
6. Explain the algebraic technique for solving exponential equations described in this section.
#### SubsubsectionSKILLS
Practice each skill in the Homework problems listed.
1. Describe the graph of an exponential function: #1–14
2. Graph transformations of exponential functions: #15–18, 53–60
3. Evaluate exponential functions: #19–22
4. Find the equation of an exponential function from its graph: #23–26
5. Solve exponential equations: #27–44
6. Distinguish between power and exponential functions: #45–52, 65, and 66
### ExercisesHomework 4.2
#### Exercise Group.
For Problems 1 and 2, find the $$y$$-intercept of each exponential function and decide whether the graph is increasing or decreasing.
##### 1.
1. $$\displaystyle f (x) = 26(1.4)^x$$
2. $$\displaystyle g(x) = 1.2(0.84)^x$$
3. $$\displaystyle h(x)=75\left(\dfrac{4}{5} \right)^x$$
4. $$\displaystyle k(x)=\dfrac{2}{3}\left(\dfrac{9}{8} \right)^x$$
##### 2.
1. $$\displaystyle M(x) = 1.5(0.05)^x$$
2. $$\displaystyle N(x) = 0.05(1.05)^x$$
3. $$\displaystyle P(x)=\left(\dfrac{5}{8} \right)^x$$
4. $$\displaystyle Q(x)=\left(\dfrac{4}{3} \right)^x$$
#### Exercise Group.
For Problems 3–6, make a table of values and graph each pair of functions by hand on the domain $$[-3, 3]\text{.}$$ Describe the similarities and differences between the two graphs.
##### 3.
1. $$\displaystyle f(x)=3^x$$
2. $$\displaystyle g(x)=\left(\dfrac{1}{3} \right)^x$$
##### 4.
1. $$\displaystyle F(x)=\left(\dfrac{1}{10} \right)^x$$
2. $$\displaystyle G(x)=10^x$$
##### 5.
1. $$\displaystyle h(t)=4^{-t}$$
2. $$\displaystyle q(t)=-4^t$$
##### 6.
1. $$\displaystyle P(t)=-5^t$$
2. $$\displaystyle R(t)=5^{-t}$$
#### Exercise Group.
For Problems 7–12, match each function with its graph.
##### 7.
1. $$\displaystyle f(x)=3(2^x)$$
2. $$\displaystyle f(x)=3\left(\dfrac{1}{2} \right)^x$$
3. $$\displaystyle f(x)=3\left(\dfrac{1}{3} \right)^x$$
4. $$\displaystyle f(x)=3(3^x)$$
##### 8.
1. $$\displaystyle g(x)=2(1.5^x)$$
2. $$\displaystyle g(x)=2\left(1.25 \right)^x$$
3. $$\displaystyle g(x)=2\left(0.75 \right)^x$$
4. $$\displaystyle g(x)=2(0.25)^x$$
#### Exercise Group.
For Problems 9–12,
1. Use a graphing calculator to graph the functions on the domain $$[-5, 5]\text{.}$$
2. Give the range of the function on that domain, accurate to hundredths.
##### 9.
$$g(t) = 4(1.3^t )$$
##### 10.
$$h(t) = 3(2.4^t )$$
##### 11.
$$N(x) = 50(0.8^x)$$
##### 12.
$$P(x) = 80(0.7^x)$$
#### Exercise Group.
For Problems 13 and 14, in each group of functions, which have identical graphs? Explain why.
##### 13.
1. $$\displaystyle h(x) = 6^x$$
2. $$\displaystyle k(x)=\left(\dfrac{1}{6} \right)^x$$
3. $$\displaystyle m(x)=6^{-x}$$
4. $$\displaystyle n(x)=\dfrac{1}{6^x}$$
##### 14.
1. $$\displaystyle Q(t)=5^t$$
2. $$\displaystyle R(t)=\left(\dfrac{1}{5} \right)^t$$
3. $$\displaystyle F(t)=\left(\dfrac{1}{5} \right)^{-t}$$
4. $$\displaystyle G(t)=\dfrac{1}{5^{-t}}$$
#### Exercise Group.
For Problems 15–18,
1. Use the order of operations to explain why the two functions are different.
2. Complete the table of values and graph both functions in the same window.
3. Describe each as a transformation of $$y = 2^x$$ or $$y = 3^x\text{.}$$
##### 15.
$$f (x) = 2^{x-1}\text{,}$$ $$~g(x) = 2^x - 1$$
$$x$$ $$y=2^x$$ $$f(x)$$ $$g(x)$$ $$-2$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$-1$$ $$0$$ $$1$$ $$2$$
##### 16.
$$f (x) = 3^x+2\text{,}$$ $$~g(x) = 3^{x+2}$$
$$x$$ $$y=3^x$$ $$f(x)$$ $$g(x)$$ $$-2$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$-1$$ $$0$$ $$1$$ $$2$$
##### 17.
$$f (x) = -3^{x}\text{,}$$ $$~g(x) = 3^{-x}$$
$$x$$ $$y=3^x$$ $$f(x)$$ $$g(x)$$ $$-2$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$-1$$ $$0$$ $$1$$ $$2$$
##### 18.
$$f (x) = 2^{-x}\text{,}$$ $$~g(x) = -2^{x}$$
$$x$$ $$y=2^x$$ $$f(x)$$ $$g(x)$$ $$-2$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$-1$$ $$0$$ $$1$$ $$2$$
#### Exercise Group.
In Problems 19–22, for the given function, evaluate each pair of expressions. Are they equivalent?
##### 19.
$$f (x) = 3(5^x )$$
1. $$f (a + 2)$$ and $$9f (a)$$
2. $$f (2a)$$ and $$2 f (a)$$
##### 20.
$$g(x) = 1.8^x$$
1. $$g(h + 3)$$ and $$g(h) g(3)$$
2. $$g(2h)$$ and $$[g(h)]^2$$
##### 21.
$$P(t) = 8^t$$
1. $$P(w)-P(z)$$ and $$P(w-z)$$
2. $$P(-x)$$ and $$\dfrac{1}{P(x)}$$
##### 22.
$$Q(t)=5(0.2)^t$$
1. $$Q(b-1)$$ and $$5Q(b)$$
2. $$Q(a)Q(b)$$ and $$5Q(a+b)$$
#### 23.
The graph of $$f (x) = P_0 b^x$$ is shown in the figure.
1. Read the value of $$P_0$$ from the graph.
2. Make a short table of values for the function by reading values from the graph. Does your table confirm that the function is exponential?
3. Use your table to calculate the growth factor, $$b\text{.}$$
4. Using your answers to parts (a) and (c), write a formula for $$f (x)\text{.}$$
#### 24.
The graph of $$g(x) = P_0 b^x$$ is shown in the figure.
1. Read the value of $$P_0$$ from the graph.
2. Make a short table of values for the function by reading values from the graph. Does your table confirm that the function is exponential?
3. Use your table to calculate the decay factor, $$b\text{.}$$
4. Using your answers to parts (a) and (c), write a formula for $$g(x)\text{.}$$
#### 25.
For several days after the Northridge earthquake on January 17, 1994, the area received a number of significant aftershocks. The red graph shows that the number of aftershocks decreased exponentially over time. The graph of the function $$S(d) = S_0b^d\text{,}$$ shown in black, approximates the data. (Source: Los Angeles Times, June 27, 1995)
1. Read the value of $$S_0$$ from the graph.
2. Find an approximation for the decay factor, $$b\text{,}$$ by comparing two points on the graph. (Some of the points on the graph of $$S(d)$$ are approximately $$(1, 82)\text{,}$$ $$(2, 45)\text{,}$$ $$(3, 25)\text{,}$$ and $$(4, 14)\text{.}$$)
3. Using your answers to (a) and (b), write a formula for $$S(d)\text{.}$$
#### 26.
The frequency of a musical note depends on its pitch. The graph shows that the frequency increases exponentially. The function $$F(p) = F_0b^p$$ gives the frequency as a function of the number of half-tones, $$p\text{,}$$ above the starting point on the scale
1. Read the value of $$F_0$$ from the graph. (This is the frequency of the note A above middle C.)
2. Find an approximation for the growth factor, $$b\text{,}$$ by comparing two points on the graph. (Some of the points on the graph of $$F(p)$$ are approximately $$(1, 466)\text{,}$$ $$(2, 494)\text{,}$$ $$(3, 523)\text{,}$$ and $$(4, 554)\text{.}$$)
3. Using your answers to (a) and (b), write a formula for $$F(p)\text{.}$$
4. The frequency doubles when you raise a note by one octave, which is equivalent to $$12$$ half-tones. Use this information to find an exact value for $$b\text{.}$$
#### Exercise Group.
Solve each equation algebraically.
##### 27.
$$5^{x+2} = 25^{4/3}$$
##### 28.
$$3^{x-1} = 27^{1/2}$$
##### 29.
$$3^{2x-1} =\dfrac{\sqrt{3}}{9}$$
##### 30.
$$2^{3x-1} =\dfrac{\sqrt{2}}{16}$$
##### 31.
$$4\cdot 2^{x-3} =8^{-2x}$$
##### 32.
$$9\cdot 3^{x+2} =81^{-x}$$
##### 33.
$$27^{4x+2} =81^{x-1}$$
##### 34.
$$16^{2-3x} =64^{x+5}$$
##### 35.
$$10^{x^2-1} =1000$$
##### 36.
$$5^{x^2-x-4} =25$$
#### 37.
Before the advent of antibiotics, an outbreak of cholera might spread through a city so that the number of cases doubled every $$6$$ days.
1. Twenty-six cases were discovered on July 5. Write a function for the number of cases of cholera $$t$$ days later.
2. Use your calculator to graph your function on the interval $$0 \le t\le 90\text{.}$$
3. When should hospitals expect to be treating $$106,496$$ cases? Use algebraic methods to find your answer, and verify it on your graph.
#### 38.
An outbreak of ungulate fever can sweep through the livestock in a region so that the number of animals affected triples every $$4$$ days.
1. A rancher discovers $$4$$ cases of ungulate fever among his herd. Write a function for the number of cases of ungulate fever $$t$$ days later.
2. Use your calculator to graph your function on the interval $$0 \le t\le 20\text{.}$$
3. If the rancher does not act quickly, how long will it be until $$324$$ head are affected? Use algebraic methods to find your answer, and verify it on your graph.
#### 39.
A smart television set loses $$30\%$$ of its value every $$2$$ years.
1. Write a function for the value of a television set $$t$$ years after it was purchased if it cost $$\700$$ originally.
2. Use your calculator to graph your function on the interval $$0 \le t\le 20\text{.}$$
3. How long will it be before a $$\700$$ television set depreciates to $$\343\text{?}$$ Use algebraic methods to find your answer, and verify it on your graph.
#### 40.
A mobile home loses $$20\%$$ of its value every $$3$$ years.
1. A certain mobile home costs $$\20,000\text{.}$$ Write a function for its value after $$t$$ years.
2. Use your calculator to graph your function on the interval $$0 \le t\le 30\text{.}$$
3. How long will it be before a $$\20,000$$ mobile home depreciates to $$\12,800\text{?}$$ Use algebraic methods to find your answer, and verify it on your graph.
#### Exercise Group.
For Problems 41–44, use a graph to find an approximate solution accurate to the nearest hundredth.
##### 41.
$$3^{x-1}=4$$
##### 42.
$$2^{x+3}=5$$
##### 43.
$$4^{-x}=7$$
##### 44.
$$6^{-x}=3$$
#### Exercise Group.
For Problems 45 and 46, decide whether each function is an exponential function, a power function, or neither.
##### 45.
1. $$\displaystyle g(t)=3 t^{0.4}$$
2. $$\displaystyle h(t)=4(0.3)^t$$
3. $$\displaystyle D(x)=6x^{1/2}$$
4. $$\displaystyle E(x)=4x+x^4$$
##### 46.
1. $$\displaystyle R(w) = 5(5)^{w-1}$$
2. $$\displaystyle Q(w) = 2^w-w^2$$
3. $$\displaystyle M(z) = 0.2z^{1.3}$$
4. $$\displaystyle N(z) = z^{-3}$$
#### Exercise Group.
For Problems 47–50, decide whether the table could describe a linear function, a power function, an exponential function, or none of these. Find a formula for each linear, power, or exponential function.
##### 47.
1. $$x$$ $$y$$ $$0$$ $$3$$ $$1$$ $$6$$ $$2$$ $$12$$ $$3$$ $$24$$ $$4$$ $$48$$
2. $$t$$ $$P$$ $$0$$ $$0$$ $$1$$ $$0.5$$ $$2$$ $$2$$ $$3$$ $$4.5$$ $$4$$ $$8$$
##### 48.
1. $$x$$ $$N$$ $$0$$ $$0$$ $$1$$ $$2$$ $$2$$ $$16$$ $$3$$ $$54$$ $$4$$ $$128$$
2. $$p$$ $$R$$ $$0$$ $$405$$ $$1$$ $$135$$ $$2$$ $$45$$ $$3$$ $$15$$ $$4$$ $$5$$
##### 49.
1. $$t$$ $$y$$ $$1$$ $$100$$ $$2$$ $$50$$ $$3$$ $$33\frac{1}{3}$$ $$4$$ $$25$$ $$5$$ $$20$$
2. $$x$$ $$P$$ $$1$$ $$\frac{1}{2}$$ $$2$$ $$1$$ $$3$$ $$2$$ $$4$$ $$4$$ $$5$$ $$8$$
##### 50.
1. $$h$$ $$a$$ $$0$$ $$70$$ $$1$$ $$7$$ $$2$$ $$0.7$$ $$3$$ $$0.07$$ $$4$$ $$0.007$$
2. $$t$$ $$Q$$ $$0$$ $$0$$ $$1$$ $$\frac{1}{4}$$ $$2$$ $$1$$ $$3$$ $$\frac{9}{4}$$ $$4$$ $$4$$
#### Exercise Group.
For Problems 51 and 52, fill in the tables. Graph each pair of functions in the same window. Then answer the questions below.
1. Give the range of $$f$$ and the range of $$g\text{.}$$
2. For how many values of $$x$$ does $$f (x) = g(x)\text{?}$$
3. Estimate the value(s) of $$x$$ for which $$f (x) = g(x)\text{.}$$
4. For what values of $$x$$ is $$f (x)\lt g(x)\text{?}$$
5. Which function grows more rapidly for large values of $$x\text{?}$$
##### 51.
$$x$$ $$f(x)=x^2$$ $$g(x)=2^x$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$
##### 52.
$$x$$ $$f(x)=x^3$$ $$g(x)=3^x$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$
#### Exercise Group.
For Problems 53–60, sketch the graph of each transformation of the given function, then write a formula and check your sketch with a graphing calculator. State the domain and range of each transformation, its intercept(s), and any asymptotes.
##### 53.
$$f(x)=3^x$$
1. $$\displaystyle y = f (x) - 4$$
2. $$\displaystyle y = f (x - 4)$$
3. $$\displaystyle y = -4 f (x)$$
##### 54.
$$g(x)=4^x$$
1. $$\displaystyle y = g(x) +2$$
2. $$\displaystyle y = g(x +2)$$
3. $$\displaystyle y = 2g(x)$$
##### 55.
$$h(t)=6^t$$
1. $$\displaystyle y = -h(t)$$
2. $$\displaystyle y = h(-t)$$
3. $$\displaystyle y = -h(-t)$$
##### 56.
$$j(t)=\left(\dfrac{1}{3} \right)^t$$
1. $$\displaystyle y = j(-t)$$
2. $$\displaystyle y = -j(t)$$
3. $$\displaystyle y = -j(-t)$$
##### 57.
$$g(x)=2^x$$
1. $$\displaystyle y = g(x-3)$$
2. $$\displaystyle y = g(x-3)+4$$
##### 58.
$$f(x)=10^x$$
1. $$\displaystyle y = f(x+5)$$
2. $$\displaystyle y = f(x+5)-20$$
##### 59.
$$N(t)=\left(\dfrac{1}{2} \right)^t$$
1. $$\displaystyle y = -N(t)$$
2. $$\displaystyle y = 6-N(t)$$
##### 60.
$$P(t)=0.4^t$$
1. $$\displaystyle y = -P(t)$$
2. $$\displaystyle y = 8-P(t)$$
#### Exercise Group.
For Problems 61–64,
1. Describe the graph as a transformation of $$y = 2^x\text{.}$$
2. Give an equation for the function graphed.
#### Exercise Group.
For Problems 65 and 66, match the graph of each function to its formula. In each formula, $$a\gt 0$$ and $$b \gt 1\text{.}$$
##### 65.
1. $$\displaystyle y=ab^x$$
2. $$\displaystyle y=ab^{-x}$$
3. $$\displaystyle y=ax^b$$
##### 66.
1. $$\displaystyle y=ax^{-b}$$
2. $$\displaystyle y=-ab^{x}$$
3. $$\displaystyle y=ax^{1/b}$$
#### 67.
The function $$f (t)$$ describes a volunteer’s heart rate during a treadmill test.
\begin{equation*} f (t) = \begin{cases} 100 \amp 0 \le t \lt 3\\ 56t - 68 \amp 3 \le t \lt 4\\ 186 - 500(0.5)^t \amp 4 \le t \lt 9\\ 100 + 6.6(0.6)^{t-14} \amp 9\le t \lt 20 \end{cases} \end{equation*}
The heart rate is given in beats per minute and $$t$$ is in minutes. (See Section 2.2 to review functions defined piecewise.) (Source: Davis, Kimmet, and Autry, 1986)
1. Evaluate the function to complete the table.
$$t$$ $$3.5$$ $$4$$ $$8$$ $$10$$ $$15$$ $$f(t)$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$
2. Sketch the graph of the function.
3. The treadmill test began with walking at $$5.5$$ kilometers per hour, then jogging, starting at $$12$$ kilometers per hour and increasing to $$14$$ kilometers per hour, and finished with a cool-down walking period. Identify each of these activities on the graph and describe the volunteer’s heart rate during each phase.
#### 68.
Carbon dioxide ($$\text{CO}_2$$) is called a greenhouse gas because it traps part of the Earth’s outgoing energy. Animals release $$\text{CO}_2$$ into the atmosphere, and plants remove $$\text{CO}_2$$ through photosynthesis. In modern times, deforestation and the burning of fossil fuels both contribute to $$\text{CO}_2$$ levels. The figure shows atmospheric concentrations of $$\text{CO}_2\text{,}$$ in parts per million, measured at the Mauna Loa Observatory in Hawaii.
1. The red curve shows annual oscillations in $$\text{CO}_2$$ levels. Can you explain why $$\text{CO}_2$$ levels vary throughout the year?
2. The blue curve shows the average annual $$\text{CO}_2$$ readings. By approximately how much does the $$\text{CO}_2$$ level vary from its average value during the year?
3. In 1960, the average $$\text{CO}_2$$ level was $$316.75$$ parts per million, and the average level has been rising by $$0.4\%$$ per year. If the level continues to rise at this rate, what $$\text{CO}_2$$ readings can we expect in the year 2100?
Hint.
For part (a): Why would photosynthesis vary throughout the year?
|
# CHAPTER 4 Polynomials: Operations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 4.1Integers as Exponents 4.2Exponents and Scientific.
## Presentation on theme: "CHAPTER 4 Polynomials: Operations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 4.1Integers as Exponents 4.2Exponents and Scientific."— Presentation transcript:
CHAPTER 4 Polynomials: Operations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 4.1Integers as Exponents 4.2Exponents and Scientific Notation 4.3Introduction to Polynomials 4.4Addition and Subtraction of Polynomials 4.5Multiplication of Polynomials 4.6Special Products 4.7Operations with Polynomials in Several Variables 4.8Division of Polynomials
OBJECTIVES 4.2 Exponents and Scientific Notation Slide 3Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. aUse the power rule to raise powers to powers. bRaise a product to a power and a quotient to a power. cConvert between scientific notation and decimal notation. dMultiply and divide using scientific notation. eSolve applied problems using scientific notation.
For any real number a and any integers m and n, To raise a power to a power, multiply the exponents. 4.2 Exponents and Scientific Notation The Power Rule Slide 4Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE a)(x 3 ) 4 b) (4 2 ) 8 c) (a -3 ) -2 Solution a) (x 3 ) 4 = x 3 4 = x 12 b) (4 2 ) 8 = 4 2 8 = 4 16 c) (a –3 ) –2 = a (–3)(–2) = a 6 4.2 Exponents and Scientific Notation a Use the power rule to raise powers to powers. ASimplify: Slide 5Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
For any real number a and b and any integer n, To raise a product to the nth power, raise each factor to the nth power. 4.2 Exponents and Scientific Notation Raising a Product to a Power Slide 6Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE a) (3x) 4 b) ( 2x 3 ) 2 c) (a 2 b 3 ) 7 (a 4 b 5 ) d) (3x 4 y -5 ) 4 e) ( 2x -3 y 5 ) -3 Solution a) (3x) 4 = 3 4 x 4 = (3 3 3 3)x 4 = 81x 4 b) ( 2x 3 ) 2 = ( 2) 2 (x 3 ) 2 Raising each factor to the 2 nd power = 4x 6 4.2 Exponents and Scientific Notation b Raise a product to a power and a quotient to a power. BSimplify: (continued) Slide 7Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE a) (3x) 4 b) ( 2x 3 ) 2 c) (a 2 b 3 ) 7 (a 4 b 5 ) Solution c) (a 2 b 3 ) 7 (a 4 b 5 ) = (a 2 ) 7 (b 3 ) 7 a 4 b 5 = a 14 b 21 a 4 b 5 Multiplying exponents = a 18 b 26 Adding exponents 4.2 Exponents and Scientific Notation b Raise a product to a power and a quotient to a power. BSimplify: (continued) Slide 8Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE d) (3x 4 y –5 ) 4 e) ( 2x –3 y 5 ) –3 Solution d) (3x 2 y –5 ) 4 = (3 4 )(x 4 ) 4 (y –5 ) 4 4.2 Exponents and Scientific Notation b Raise a product to a power and a quotient to a power. BSimplify: (continued) Slide 9Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE d) (3x 4 y –5 ) 4 e) ( 2x –3 y 5 ) –3 Solution e) ( 2x –3 y 5 ) –3 = (–2 ) –3 (x –3 ) –3 (y 5 ) –3 4.2 Exponents and Scientific Notation b Raise a product to a power and a quotient to a power. BSimplify: Slide 10Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
For any real numbers a and b, b ≠ 0, and any integer n, To raise a quotient to the nth power, raise numerator AND denominator to the nth power. Also, 4.2 Exponents and Scientific Notation Raising a Quotient to a Power Slide 11Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE a) b) c) 4.2 Exponents and Scientific Notation b Raise a product to a power and a quotient to a power. CSimplify: Slide 12Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
Scientific notation for a number is an expression of the type where n is an integer, M is greater than or equal to 1 and less than 10 (1 ≤ M < 10), and M is expressed in decimal notation. 10 n is also considered to be scientific notation when M = 1. 4.2 Exponents and Scientific Notation Scientific Notation Slide 13Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
Scientific notation A positive exponent indicates a large number (greater than or equal to 10). A negative exponent indicates a small number (between 0 and 1). 4.2 Exponents and Scientific Notation c Convert between scientific notation and decimal notation. Slide 14Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE a) 94,000b) 0.0423 Solution a) 94,000 = 9.4 10 4 b) 0.0423 0.04.23 4 places to the left A Large number so the exponent is positive. 2 places to the right Small number so the exponent is negative. 4.2 Exponents and Scientific Notation c Convert between scientific notation and decimal notation. AConvert to scientific notation: Slide 15Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. We want a number between 1 and 10 so we must move the decimal point 94,000.. = 4.23 10 2
EXAMPLE a) 3.842 10 6 b) 5.3 10 7 Solution a) 3.842 10 6 = 3.842000 10 6 = 3,842,000. b) 5.3 10 7 = 0.00000053 = 0.0000005.3 6 places to the right Positive exponent, so the answer is a large number. 7 places Negative exponent, so the answer is a small number. 4.2 Exponents and Scientific Notation c Convert between scientific notation and decimal notation. BConvert to decimal notation: Slide 16Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE Solution (1.7 × 10 8 )(2.2 × 10 –5 ) = (1.7 · 2.2) · (10 8 · 10 –5 ) = 3.74 × 10 8 +(–5) = 3.74 × 10 3 4.2 Exponents and Scientific Notation d Multiply and divide using scientific notation. CSimplify: (1.7 × 10 8 )(2.2 × 10 –5 ) Slide 17Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE Solution (6.2 10 9 ) (8.0 10 8 ) = 4.2 Exponents and Scientific Notation d Multiply and divide using scientific notation. D Simplify. (6.2 10 9 ) (8.0 10 8 ) Slide 18Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
EXAMPLE Solution: We multiply one billion by 3500. (1.0 10 9 ) (3.5 10 3 ) = 3.5 10 12 4.2 Exponents and Scientific Notation e Solve applied problems using scientific notation. EA gigabyte is a measure of a computer’s storage capacity. One gigabyte holds about one billion bytes of information. If a firm’s computer network contains 3500 gigabytes of memory, how many bytes are in the network? Slide 19Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
4.2 Exponents and Scientific Notation Definitions and Rules for Exponents (continued) Slide 20Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
4.2 Exponents and Scientific Notation Definitions and Rules for Exponents Slide 21Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.
Download ppt "CHAPTER 4 Polynomials: Operations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 4.1Integers as Exponents 4.2Exponents and Scientific."
Similar presentations
|
# 9.5: Partial Fractions
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
##### Learning Objectives
In this section, you will:
• Decompose P(x)Q(x)P(x) Q(x) , where Q(x)Q(x) has only nonrepeated linear factors.
• Decompose P(x)Q(x)P(x) Q(x) , where Q(x)Q(x) has repeated linear factors.
• Decompose P(x)Q(x)P(x) Q(x) , where Q(x)Q(x) has a nonrepeated irreducible quadratic factor.
• Decompose P(x)Q(x)P(x) Q(x) , where Q(x)Q(x) has a repeated irreducible quadratic factor.
Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions.
Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression.
### Decomposing P(x)Q(x)P( x ) Q( x ) Where Q(x) Has Only Nonrepeated Linear Factors
Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fraction.
For example, suppose we add the following fractions:
2x−3+−1x+22x−3+−1x+2
We would first need to find a common denominator, (x+2)(x−3).(x+2)(x−3).
Next, we would write each expression with this common denominator and find the sum of the terms.
2x−3(x+2x+2)+−1x+2(x−3x−3)= 2x+4−x+3(x+2)(x−3)=x+7x2−x−62x−3(x+2x+2)+−1x+2(x−3x−3)= 2x+4−x+3(x+2)(x−3)=x+7x2−x−6
Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.
x+7x2−x−6Simplifiedsum=2x−3+−1x+2Partialfractiondecompositionx+7x2−x−6Simplifiedsum=2x−3+−1x+2Partialfractiondecomposition
We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.
When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of x2−x−6x2−x−6 are (x−3)(x+2),(x−3)(x+2), the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.
##### PARTIAL FRACTION DECOMPOSITION OF P(x)Q(x):Q(x)P( x ) Q( x ) :Q(x) HAS NONREPEATED LINEAR FACTORS
The partial fraction decomposition of P(x)Q(x)P(x)Q(x) when Q(x)Q(x) has nonrepeated linear factors and the degree of P(x)P(x) is less than the degree of Q(x)Q(x) is
P(x)Q(x)=A1(a1x+b1)+A2(a2x+b2)+A3(a3x+b3)+⋅⋅⋅+An(anx+bn).P(x)Q(x)=A1(a1x+b1)+A2(a2x+b2)+A3(a3x+b3)+⋅⋅⋅+An(anx+bn).
##### HOW TO
Given a rational expression with distinct linear factors in the denominator, decompose it.
1. Use a variable for the original numerators, usually A,B, A,B, or C,C, depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use AnAn for each numerator
P(x)Q(x)=A1(a1x+b1)+A2(a2x+b2)+⋯+An(anx+bn)P(x)Q(x)=A1(a1x+b1)+A2(a2x+b2)+⋯+An(anx+bn)
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### EXAMPLE 1
#### Decomposing a Rational Function with Distinct Linear Factors
Decompose the given rational expression with distinct linear factors.
3x(x+2)(x−1)3x(x+2)(x−1)
##### TRY IT #1
Find the partial fraction decomposition of the following expression.
x(x−3)(x−2)x(x−3)(x−2)
### Decomposing P(x)Q(x)P( x ) Q( x ) Where Q(x) Has Repeated Linear Factors
Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.
##### PARTIAL FRACTION DECOMPOSITION OF P(x)Q(x):Q(x)P( x ) Q( x ) :Q(x) HAS REPEATED LINEAR FACTORS
The partial fraction decomposition of P(x)Q(x),P(x)Q(x), when Q(x)Q(x) has a repeated linear factor occurring nn times and the degree of P(x)P(x) is less than the degree of Q(x),Q(x), is
P(x)Q(x)=A1(ax+b)+A2(ax+b)2+A3(ax+b)3+⋅⋅⋅+An(ax+b)nP(x)Q(x)=A1(ax+b)+A2(ax+b)2+A3(ax+b)3+⋅⋅⋅+An(ax+b)n
Write the denominator powers in increasing order.
##### HOW TO
Given a rational expression with repeated linear factors, decompose it.
1. Use a variable like A,B,A,B, or CC for the numerators and account for increasing powers of the denominators.
P(x)Q(x)=A1(ax+b)+A2(ax+b)2+ . . . + An(ax+b)nP(x)Q(x)=A1(ax+b)+A2(ax+b)2+ . . . + An(ax+b)n
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### EXAMPLE 2
#### Decomposing with Repeated Linear Factors
Decompose the given rational expression with repeated linear factors.
−x2+2x+4x3−4x2+4x−x2+2x+4x3−4x2+4x
##### TRY IT #2
Find the partial fraction decomposition of the expression with repeated linear factors.
6x−11(x−1)26x−11(x−1)2
### Decomposing P(x)Q(x),P( x ) Q( x ) , Where Q(x) Has a Nonrepeated Irreducible Quadratic Factor
So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators A,B,A,B, or CC representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as Ax+B,Bx+C,Ax+B,Bx+C, etc.
##### DECOMPOSITION OF P(x)Q(x):Q(x)P( x ) Q( x ) :Q(x) HAS A NONREPEATED IRREDUCIBLE QUADRATIC FACTOR
The partial fraction decomposition of P(x)Q(x)P(x)Q(x) such that Q(x)Q(x) has a nonrepeated irreducible quadratic factor and the degree of P(x)P(x) is less than the degree of Q(x)Q(x) is written as
P(x)Q(x)=A1x+B1(a1x2+b1x+c1)+A2x+B2(a2x2+b2x+c2)+⋅⋅⋅+Anx+Bn(anx2+bnx+cn)P(x)Q(x)=A1x+B1(a1x2+b1x+c1)+A2x+B2(a2x2+b2x+c2)+⋅⋅⋅+Anx+Bn(anx2+bnx+cn)
The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: A,B,C,A,B,C, and so on.
##### HOW TO
Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.
1. Use variables such as A,B,A,B, or CC for the constant numerators over linear factors, and linear expressions such as A1x+B1,A2x+B2,A1x+B1,A2x+B2, etc., for the numerators of each quadratic factor in the denominator.
P(x)Q(x)=Aax+b+A1x+B1(a1x2+b1x+c1)+A2x+B2(a2x2+b2x+c2)+⋅⋅⋅+Anx+Bn(anx2+bnx+cn)P(x)Q(x)=Aax+b+A1x+B1(a1x2+b1x+c1)+A2x+B2(a2x2+b2x+c2)+⋅⋅⋅+Anx+Bn(anx2+bnx+cn)
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### EXAMPLE 3
#### Decomposing P(x)Q(x)P(x)Q(x) When Q(x) Contains a Nonrepeated Irreducible Quadratic Factor
Find a partial fraction decomposition of the given expression.
8x2+12x−20(x+3)(x2+x+2)8x2+12x−20(x+3)(x2+x+2)
##### Q&A
Could we have just set up a system of equations to solve Example 3?
Yes, we could have solved it by setting up a system of equations without solving for AA first. The expansion on the right would be:
8x2+12x−20=Ax2+Ax+2A+Bx2+3B+Cx+3C8x2+12x−20=(A+B)x2+(A+3B+C)x+(2A+3C)8x2+12x−20=Ax2+Ax+2A+Bx2+3B+Cx+3C8x2+12x−20=(A+B)x2+(A+3B+C)x+(2A+3C)
So the system of equations would be:
A+B=8A+3B+C=122A+3C=−20 A+B=8A+3B+C=122A+3C=−20
##### TRY IT #3
Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor.
5x2−6x+7(x−1)(x2+1)5x2−6x+7(x−1)(x2+1)
### Decomposing P(x)Q(x)P( x ) Q( x ) When Q(x) Has a Repeated Irreducible Quadratic Factor
Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.
##### DECOMPOSITION OF P(x)Q(x)P( x ) Q( x ) WHEN Q(X) HAS A REPEATED IRREDUCIBLE QUADRATIC FACTOR
The partial fraction decomposition of P(x)Q(x),P(x)Q(x), when Q(x)Q(x) has a repeated irreducible quadratic factor and the degree of P(x)P(x) is less than the degree of Q(x),Q(x), is
P(x)(ax2+bx+c)n=A1x+B1(ax2+bx+c)+A2x+B2(ax2+bx+c)2+A3x+B3(ax2+bx+c)3+⋅⋅⋅+Anx+Bn(ax2+bx+c)nP(x)(ax2+bx+c)n=A1x+B1(ax2+bx+c)+A2x+B2(ax2+bx+c)2+A3x+B3(ax2+bx+c)3+⋅⋅⋅+Anx+Bn(ax2+bx+c)n
Write the denominators in increasing powers.
##### HOW TO
Given a rational expression that has a repeated irreducible factor, decompose it.
1. Use variables like A,B,A,B, or CC for the constant numerators over linear factors, and linear expressions such as A1x+B1,A2x+B2,A1x+B1,A2x+B2, etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as
P(x)Q(x)=Aax+b+A1x+B1(ax2+bx+c)+A2x+B2(ax2+bx+c)2+⋯+An+Bn(ax2+bx+c)nP(x)Q(x)=Aax+b+A1x+B1(ax2+bx+c)+A2x+B2(ax2+bx+c)2+⋯+An+Bn(ax2+bx+c)n
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### EXAMPLE 4
#### Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator
Decompose the given expression that has a repeated irreducible factor in the denominator.
x4+x3+x2−x+1x(x2+1)2x4+x3+x2−x+1x(x2+1)2
##### TRY IT #4
Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.
x3−4x2+9x−5(x2−2x+3)2x3−4x2+9x−5(x2−2x+3)2
##### MEDIA
Access these online resources for additional instruction and practice with partial fractions.
### 9.4 Section Exercises
#### Verbal
1.
Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction
2.
Can you explain why a partial fraction decomposition is unique? (Hint: Think about it as a system of equations.)
3.
Can you explain how to verify a partial fraction decomposition graphically?
4.
You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could double-check your answer.
5.
Once you have a system of equations generated by the partial fraction decomposition, can you explain another method to solve it? For example if you had 7x+133x2+8x+15=Ax+1+B3x+57x+133x2+8x+15=Ax+1+B3x+5, we eventually simplify to 7x+13=A(3x+5)+B(x+1).7x+13=A(3x+5)+B(x+1). Explain how you could intelligently choose an xx -value that will eliminate either AA or BB and solve for AA and B.B.
#### Algebraic
For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors.
6.
5x+16x2+10x+245x+16x2+10x+24
7.
3x−79x2−5x−243x−79x2−5x−24
8.
−x−24x2−2x−24−x−24x2−2x−24
9.
10x+47x2+7x+1010x+47x2+7x+10
10.
x6x2+25x+25x6x2+25x+25
11.
32x−1120x2−13x+232x−1120x2−13x+2
12.
x+1x2+7x+10x+1x2+7x+10
13.
5xx2−95xx2−9
14.
10xx2−2510xx2−25
15.
6xx2−46xx2−4
16.
2x−3x2−6x+52x−3x2−6x+5
17.
4x−1x2−x−64x−1x2−x−6
18.
4x+3x2+8x+154x+3x2+8x+15
19.
3x−1x2−5x+63x−1x2−5x+6
For the following exercises, find the decomposition of the partial fraction for the repeating linear factors.
20.
−5x−19(x+4)2−5x−19(x+4)2
21.
x(x−2)2x(x−2)2
22.
7x+14(x+3)27x+14(x+3)2
23.
−24x−27(4x+5)2−24x−27(4x+5)2
24.
−24x−27(6x−7)2−24x−27(6x−7)2
25.
5−x(x−7)25−x(x−7)2
26.
5x+142x2+12x+185x+142x2+12x+18
27.
5x2+20x+82x(x+1)25x2+20x+82x(x+1)2
28.
4x2+55x+255x(3x+5)24x2+55x+255x(3x+5)2
29.
54x3+127x2+80x+162x2(3x+2)254x3+127x2+80x+162x2(3x+2)2
30.
x3−5x2+12x+144x2(x2+12x+36)x3−5x2+12x+144x2(x2+12x+36)
For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor.
31.
4x2+6x+11(x+2)(x2+x+3)4x2+6x+11(x+2)(x2+x+3)
32.
4x2+9x+23(x−1)(x2+6x+11)4x2+9x+23(x−1)(x2+6x+11)
33.
−2x2+10x+4(x−1)(x2+3x+8)−2x2+10x+4(x−1)(x2+3x+8)
34.
x2+3x+1(x+1)(x2+5x−2)x2+3x+1(x+1)(x2+5x−2)
35.
4x2+17x−1(x+3)(x2+6x+1)4x2+17x−1(x+3)(x2+6x+1)
36.
4x2(x+5)(x2+7x−5)4x2(x+5)(x2+7x−5)
37.
4x2+5x+3x3−14x2+5x+3x3−1
38.
−5x2+18x−4x3+8−5x2+18x−4x3+8
39.
3x2−7x+33x3+273x2−7x+33x3+27
40.
x2+2x+40x3−125x2+2x+40x3−125
41.
4x2+4x+128x3−274x2+4x+128x3−27
42.
−50x2+5x−3125x3−1−50x2+5x−3125x3−1
43.
−2x3−30x2+36x+216x4+216x−2x3−30x2+36x+216x4+216x
For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor.
44.
3x3+2x2+14x+15(x2+4)23x3+2x2+14x+15(x2+4)2
45.
x3+6x2+5x+9(x2+1)2x3+6x2+5x+9(x2+1)2
46.
x3−x2+x−1(x2−3)2x3−x2+x−1(x2−3)2
47.
x2+5x+5(x+2)2x2+5x+5(x+2)2
48.
x3+2x2+4x(x2+2x+9)2x3+2x2+4x(x2+2x+9)2
49.
x2+25(x2+3x+25)2x2+25(x2+3x+25)2
50.
2x3+11x2+7x+70(2x2+x+14)22x3+11x2+7x+70(2x2+x+14)2
51.
5x+2x(x2+4)25x+2x(x2+4)2
52.
x4+x3+8x2+6x+36x(x2+6)2x4+x3+8x2+6x+36x(x2+6)2
53.
2x−9(x2−x)22x−9(x2−x)2
54.
5x3−2x+1(x2+2x)25x3−2x+1(x2+2x)2
#### Extensions
For the following exercises, find the partial fraction expansion.
55.
x2+4(x+1)3x2+4(x+1)3
56.
x3−4x2+5x+4(x−2)3x3−4x2+5x+4(x−2)3
For the following exercises, perform the operation and then find the partial fraction decomposition.
57.
7x+8+5x−2−x−1x2−6x−167x+8+5x−2−x−1x2−6x−16
58.
1x−4−3x+6−2x+7x2+2x−241x−4−3x+6−2x+7x2+2x−24
59.
2xx2−16−1−2xx2+6x+8−x−5x2−4x
This page titled 9.5: Partial Fractions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
|
# WORKSHOP CALCULATION
1
Introduction to Units ( Pressure)
PRESSURE CONVERSION Bar. 1 Kg / cm² = 14 . 223 psi ( Lb / In² ) 1 Kg / cm² = 0 . 9807 1 PSI = 0.07031 Kg / cm²
Introduction to Units (Length)
1m = 100 cm 1cm = 10 mm 1m = 1000 mm 1in. = 25.4 mm
Introduction to Units ( Temperature)
Temperature unit = Degree Celsius or Degree Fahrenheit
C = 5/9(f-32)
If Temp. Is 100°f, Then So, C=37.7 C=5/9( 100-32)
If Preheat Temp. Is 150’c, Then F=302
AB² + BC² = AC² . So based on pythagoras theory . B C LET US SAY ABC is right angle triangle .PYTHAGORAS PRINCIPLE APPLICATION A Pythagoras Principle : In Any Right Angled Triangle a Sum of Adjacent Side Square Is Always Equal to It Hypotenuse Square. AB and BC = Adjacent sides and AC = Hypotenuse.
BC = 4 and AC = 5 SO AC² = AB² + BC² = 3² + 4 ² = 25 By taking AC = 5 so AC² = 25 It means that LHS = RHS .PYTHAGORAS PRINCIPLE APPLICATION Example : A 3 5 B 4 C Proof of theory in triangle ABC AB = 3 .
Angle ACB = .TRIGONOMETRIC FUNCTIONS A Trigonometric functions are used to solve the problems of different types of triangle. AB & BC are sides of triangle. So for this triangle. . Let us consider ABC is a right angled triangle. B C We will see some simple formulas to solve right angle triangle which we are using in day to day work.
TRIGONOMETRY A Hypoteneous Opposite Side AB SIN ø = Opposite Side = AC Hypoteneous ø B TAN ø = Opposite Side = AB Adjacent Side BC C Adjacent Side COS ø = Adjacent Side = BC AC Hypoteneous .
A 25 mm We Will Find Value Of By Tangent Formula So . Tan(1) = 45º Now. We Will Find AC By Using Sine Formula 25 mm B C Sin = Opposite Side /Hypotenuse = AB / AC Ac = AB / Sin = 25 / Sin45 =25 / 0.TRIGONOMETRIC FUNCTIONS Example : For triangle ABC find out value of and . Tan = Opposite Side / Adjacent Side = AB / BC = 25/25 =1 Tan = 1 = Inv.7071 = 35.3556mm .
7071 = Inv Cos (0.TRIGONOMETRIC FUNCTIONS Example: We Will Find Value Of By Cosine Formula A 25 mm B 25 mm C Cos = Adjacent Side / Hypotenuse = AB / AC = 25 / 35.7071) = 45º .3556 = 0.
60 ø = 36° .60 50 ø B ADJACENT SIDE C ø = InvSINE VALUE OF 0.52’ .TRIGONOMETRY Example: A FIND OUT ANGLE ‘ Ø ’ OF A TRIANGLE SIN ø = OPPOSITE SIDE = HYPOTENEOUS HYPOTENEOUS AB AC OPPOSITE SIDE 30 = 30 50 = 0.
727 27.FIND OUT SIDE ‘ø ’ OF A TRIANGLE Example: A TAN ø = OPPOSITE SIDE = AB ADJACENT SIDE BC HYPOTENEOUS OPPOSITE SIDE 20 TAN 36° = • • BC = • 20 BC 20 TAN VALUE OF 36° 20 0. 51 mm 36° B ? ADJACENT SIDE C • • BC = • • BC = • • .
circle etc.AREA Definition : A surface covered by specific Shape is called area of that shape. i.e. area of square. 1. Square : Area Of Square = L X L = L² L Where L = Length Of Side L So If L Then Area = 5cm = 5 X 5 = 25cm² .
Rectangle: Area Of Rectangle = L X B Where.AREA 2. Circle : Area Of Circle = / 4 x D² D Where D= Diameter Of The Circle Area Of Half Circle = /8 x D² D Same way we can find out area of quarter of circle . Area = Length = Width = 6 mm = 10 X 6 = 60mm² B L 3. And B Then. L B If L= 10 mm.
AREA 4 .d²) 4 d WHERE D = Diameter of Greater Circle d = Diameter of Smaller Circle D Sector Of Circle= xD²xØ 4 x 360 Ø D . Circle : Hollow Circle = x (D² .
AREA 4. Triangle : Area Of Triangle = ½ B x H Where B H = Base Of Triangle = Height Of Triangle B H 5. Cylinder : Surface area of Cylinder =xDxH Where H D = Height Of Cylinder = Diameter Of Cylinder D H .
Rectangular Block : Volume= L X B X H Where L = Length B = Width H = Height H L B . Block = L X L X L = L³ 2. width and height are equal.VOLUME Defination : A space covered by any object is called volume of that object. L L L Volume Of Sq. Square block : In square block. 1. so length.
VOLUME H 3.Triangle L = Length of Prism B L .Triangle H = Height of R.Prism or Triangle Block : Volume of Triangular Block = Cross Section Area of Triangle x Length ( Area of Right Angle Triangle = ½ B H ) Volume = ½ B H X L Where B = Base of R.A.A.
VOLUME 4. Cylinder : Volume of Cylinder = Cross Section Area x Length of Cylinder Volume= ¼D² X H D H Where : D = Diameter Of Cylinder H = Length Of Cylinder .
1439 DIA ( m ) = 0.1000 DIA .SPHERICAL ( m ) = 0.CG CALCULATION CG m TAN LINE DIA CENTRE OF GRAVITY OF D’ENDS ( CG ) (1) HEMISPHERICAL ( m ) = 0.2878 DIA (2) (3) 2:1 ELLIPSOIDALS TORI .
S.WEIGHT CALCULATION Examples : Weight calculation of different items: • • • • • Rectangular plate Circular plate Circular plate with cutout Circular sector Shell coursce Specific gravity for (i) C.86 g/cm3 (ii) S.= 7.S.=8.00 g/cm3 .
86gm / CC Here L = 200cm.Gravity = 70000 X 7. Rectangular plate : Weight of This Plate 3.Gravity 200 CM = L X B X H X 7.5 cm So Volume = 200 X 100 X 3.5 cm³ = 70000 cm³ Now Weight Of Plate = Volume X Sp .2 kgs . B = Width = 100cm And H = Thk = 3.WEIGHT CALCULATION Examples : 1.5 CM = Volume X Sp.86 gm/cc = 550200 gms = 550.
362 kgs . CIRCULAR PLATE : 300 cm Weight= V X Sp. Gravity Volume V= Cross Section Area X Thk = ¼D² X 4cm = ¼ x 300² X 4cm = 282743.5738 gms = 2222.33 X 7.33 cm³ So W = V X sp.86 gms/cc Thk = 4cm = 2222362.WEIGHT CALCULATION Examples : 2.Gravity = 282743.
Now Volume = Cross Sec.WEIGHT CALCULATION Examples : Circular sector : Weight of Circular Plate Sector : W = Volume X Sp.Gravity = 78539.Gravty.86 gms/cc = 617322.R2²) X Ø X 2 cm 360 = X (400² .81 X 7.Area X Thk = X ( R1² .95 gms = 617.81 cm³ Now Weight = V X Sp .323 kgs R1 R2 = 120º R1 = 400 cm R2 = 350 cm THK = 2cm = 120º .350²) X 120 X 2 360 = 78539.
54cm³ Now Weight W = V X Sp.Gravity V= ¼ X ( OD² .947kgs = @ 15 Ton .400² ) X 300cm = 1908517. Gravity = 1908517.86 = 15000947gms = 15000.54 X 7.ID² ) X Length Here OD = 400 + 10 = 410cm ID = 400cm Length = 300cm So V = ¼ X ( 410² .WEIGHT CALCULATION Examples : Shell : W = V X Sp.
WELD METAL WEIGHT CALCULATION Basic fundamentals of weld metal weight Calculation 1. seam weld weight `= Cross section area x mean circ.Single v for long seam and circseam • Long seam weld weight = Cross section area x length of seam x density • Circ. of seam x density .
2 + 6.38 Cm² A3 = 6.38 + 0.713 x 4.7 cm² Here B= 47 Tan30º =2.713cm A2 = 0.2 cm² 50 3 2 1.7 cm² Now A = 1.38 Cm² A4 =0.WELD METAL WEIGHT CALCULATION 3 1 =60º 3 4 2 Now A1 = 2/3 x H x Bead Width A1 = 2/3 x 0.2 * 4.5 x B x 4.3 x 6 cm² = 1.5 x 2.Crossection Area Of Joint A = A1 + A2 + A3 + A4 Now A2 =A3 A2 = 1/2 x B x h = 0.9cm² .94 cm² A = 14.38 + 6.7 Cm² = 6.
.4gms = 11.86 gms/cc = 145326gms = 145. = 50 mm Weld Weight = 14. seam = Cross section area x Mean circ.9 cm X 7. x Density For Circ. seam having OD = 4000 mm and Thk.86gm/cm³ = 11711.326kgs.9cm² X 1240.712kgs for 1 mtr long seam For circ.WELD METAL WEIGHT CALCULATION For long seam weld weight = Cross section area x Length of seam x density = 14.9cm² x 100cm x 7.
Instead of 1:3 taper.40 = 20mm. TAPER CALCULATIONS 1:3 Taper 40 x 60 Thickness Difference = 60 . a taper is generally provided on thicker plate to avoid mainly stress concentration. if 1: 5 Taper is required. X = 20 x 3 = 60mm. .Whenever a Butt joint is to be made between two plates of different thickness. X = 20 x 5 = 100 mm.
CIRCUMFERENCE CALCULATION Circumference = Pie x Diameter of job If I/D is known and O/S circ. Which is the correct value of pie? 22/7 3.14 3. Is required then. Circumference = Pie x ( I/D + 2 x thick ) Here Pie value is very important.1415926 (Direct from calculator/ computer) .
00mm 3) 10000mm x 3.1415926 = 31415.571mm = 31400. calculate O/S circumference. 1) 10000mm x 22/7 2) 10000mm x 3.CIRCUMFERENCE CALCULATION Example 1 : O/S Dia of the job is 10000mm.14 = 31428.926mm .
1T but <= 2mm for web & <= 3mm for flange Say T = 34 mm than.OFFSET CALCULATION Thickness difference measured from I/s or o/s on joining edges is called offset.1 x 34mm = 3. . Offset = 0. If by mistake 0.4mm But max. offset Tolerance as per P-1402 0.1% T considered than. 0. 3mm allowed as mentioned above.1 x 34/100 = 0.034 mm offset which is wrong.
C Kink = ( A .D ) which ever is max.OFFSET CALCULATION How to measure offset & kink ? Here A = D Offset = B . D C B A Kink is nothing but peak-in/ peak-out .B or C .
Always take all digits of orientation given in drawing. • Measure circumference.ORIENTATION MARKING Start orientation in following steps. • Check long seam orientation from drawing. ./360 ) x Orientation. • Arc length = (circ. • Find out arc length for long seam from 0 degree.
ORIENTATION MARKING Example : O/S circ.16 = 5282. L/S orientation = 25300mm = 75.218mm .07mm = ( 25300/360 ) x 75.86mm = ( 25300/360 ) x 75.1 = 5277.162 = 5282.162 Arc length for L/S = ( 25300/360 ) x 75.162 degree Find out arc length for 75.
4000mm Sine 30 = CB/4000mm 1/2 chord length CB = 0.5 x 4000mm = 2000mm Full chord length = 4000mm .CHORD LENGTH A 60 C R B Example : Web segment size .600 Inside radius R .
angle Should be between 45.TRIGONOMETRIC FUNCTIONS Tank rotator rollers dist. Dist. Calculation As shown in figure we can find out Two things : 1. Angle between two rollers 2. A D C B We will check it one by one. Between two roller for specific diameter of shell .60º . For safe working.
of Holes N = 12. Mark P.28 mm.C.D. consider a flange 14”-1500# with P. .D.PCD & HOLE MARKING CALCULATIONS For Example. 2 2 Y/2 P.=600 mm & No.C. Chord length between holes = 2 x PCD x Sin ( y/2 ) 2 = 2 x 600 x Sin (30/2) ‘N’ Holes = 2 x 600 x 0. Angular distance y = 360 / N = 360/12 = 30 degrees.D.2588 = 155. = 600 mm.C.
41#930%7.30039 70.304.' & !78247%7.30 '4:20 070 .74.94370.41%7.3:.8041# %7. 7488\$0. '4:2041%7.30 03941!782 .30 0941# %7.
03941 3/07 '4:20 6 070 .' & 3/07 '4:20413/077488\$0.20907 13/07 039 13/07 .94370.
&% 2 % %# #'% \$ \$! # !\$ \$ % # \$! # 2 2 2 .
.90 7.0 \$50.:.7. %&% .909.1..94341/11070399028 W W W W W #0.9147 \$ .:.75.75.9..75.:.:94:9 7.:.780.947 \$0.4:78.2508 09.90 7.3:.
.2 \$ \$ .
2 ..
3:.9..9 2. %&% .90 '4:20 \$5 7.75.2508 #0.90 0941%8!.
2 409 1!. 070 .90'4:20\$5 7..23/% .2 \$4'4:20 .2 .9 2.2 /9 .
.. 28 8 .
9 28.2 .2 6 .94370.9 '4:20'7488\$0.2508 #&#!% 09' \$5 7. % 6 .2 \$4' 85 7.. %&% ..
.2 . 28 8 %.2 ..
2 409' \$5 7.2 6 .9 28.90\$0. %&% .7!..780.947 '4:20 \$5 7.2508 7. % 6 # # [ ..:.9 4'4:207488\$0. 70.947 09417.:.
2 # ..2 %.2 X . 28 8 X # # # ..
2508 \$0 ' \$5 7.9 28 8%43 . %&% .2 .9 ' 6 039 070 .2 \$4' 6 ..2 409 ' \$5 7.2 039 ..2 .
20/09 +748880.20.3/.70.7.2039.943 \$30.943.3.80.8410/209.70.7.0394180.1:3/.1474380.20/09 748880.2/0389 .2.09.2/0389 W 7.2 W 4380..943. 80. 4180.8.:. % %&% .
% %&% 7 4.
0.//9 ( .
2 4 . .2 .
3 .2 . .2 ( 2 2 2 .94370. 1439 4 .2 .2 070%.2 74880.
2 2.70.2/0389 .20/09 748880. % %&% 474380.2 .943.0394180.
2 748880.3/% 22 0/09 . 0.7.2 .70.3.3 22. 80.2 28.943. 80.2 47. 0389 477.2 28 81472974380.7...2.
28 8 ...
/419.3088110703.90 94 .43.3088 .507 %.0397.0 22 22 3890.943 %.07 ./0 -09003 94 5.507 8 0307./0/ 43 9.908 41 /1107039 9.5078706:70/ 22 .4/ 2.3 897088 .507 1%.%! #&% \$ 030.. 574. 9.07 5. :99 439 8 94 -0 2.
209074141.0!0.:210703.#& # &% 7.
3/ .8343.
0!0 .7.:210703.\$. 8706:70/903 7.
. 9. 070!0.4770.9.:08..0725479.39 .890.:04150 .
70.91742...:.947.
425:907 ..
#& # &% .250 .
:...\$.90 .41904-8 22 .
0 22.:210703.\$ .7.
22 22 22 22 22 .
8:70/1742. \$ %&% %.020.3088/110703.
8474.
0.438/070/9.0 %.-4.3.%229.3 11809 22 22 :92.40/.843433 0/088.0/411809 411809 %407.3 .0 1-289. 22. 221471..82039430/.8507! %-:9221470.30 \$.
8743 . 22411809.
0. \$ %&% 49420.8:70411809 3 070 11809 3 47 . 383493-:9 50.0782. 3.
50. 4:9 .
247039.4380.9431742/7.# %% # \$9.9433144389058 W 0.3 W 3/4:9.7.0 W 0.7. .0391474380.7947039.21742 /0700 W 7.8:70.:210703.039 .7.
033/7. 7039.0.943.3 .89./984147039.943 .
250 .# %% # .
.7.\$.
943 22 /0700 3/4:9.039147 7.7.039147.\$47039.
\$ .
22 .
22 .
22 .
# % # ./:8# 22 \$30 .250 0-80203980 38/07.
22 .
.47/039 22 22 :.47/039 22 .
/.307 \$4:/ -0-09003 7 - .209074180 0.1.:..9430-430 478.313/4:9 %4938 307 -090039474078 89 09003947407147 850.0.10473 .3749.943 884331:700.94774078/89 .%# %# & % \$ %..
! # &% \$ 47 .7/89.250 .3.30 9 ! 22 4 41408 .7! 22 3:.0 .438/07.1.
.
/07008 47/039-09003408 ! \$3 .
\$3 .
408 22 .
! .
|
Check the Order of Magnitude by Rounding
In this worksheet, students check the order of magnitude of a calculation by rounding first.
Key stage: KS 3
Curriculum topic: Number
Curriculum subtopic: Use Calculators/Technology for Accuracy
Difficulty level:
QUESTION 1 of 10
When you use a calculator, you must have a rough idea of the expected size of your answer.
This size is called the order of magnitude and you can get a good idea by rounding off the question first.
7840 × 28 is approximately 8000 × 30 which is 240000
24216 - 18947 is approximately 24000 - 19000 = 5000
542 ÷ 18 is approximately 500 ÷ 20 or 25
Example
Let's find the order of magnitude for 3.42 × 19.
3.42 × 19 lies between
A
and
B
which equal
C
and
D
3.42 × 19 lies between 3 × 19 and 4 × 19.
3.42 × 19 lies between
3 × 19
and
4 × 19
which equal
57
and
76
Check to see how close you are by doing the calculation on your calculator as well.
3.42 × 19 =
64.98
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
8.98 × 21 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
8.98 × 21 =
E
A B C D E 2 x 21 8 x 21 9 x 21 168 198 189 188.58
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
9.24 × 14 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
9.24 × 14 =
E
A B C D E 9 x 14 8 x 14 10 x 14 129.36 198 140 126
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
3.46 × 22 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
3.46 × 22 =
E
A B C D E 2 x 22 3 x 22 4 x 22 66 76.12 88
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
9.97 × 11 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
9.97 × 11 =
E
A B C D E 8 x 11 9 x 11 10 x 11 109.67 99 110
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
7.02 × 10 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
7.02 × 10 =
E
A B C D E 7 x 10 8 x 10 9 x 10 70 70.2 80
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
6.67 × 12 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
6.67 × 12 =
E
A B C D E 6 x 12 7 x 12 8 x 12 80.04 72 84
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
5.01 × 13 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
5.01 × 13 =
E
A B C D E 5 x 13 6 x 13 7 x 13 65 75 78 65.13
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
5.51 × 14 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
5.51 × 14 =
E
A B C D E 5 x 14 6 x 14 7 x 14 70 75 84 77.14
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
8.32 × 22 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
8.32 × 22 =
E
A B C D E 7 x 22 8 x 22 9 x 22 176 175 198 183.04
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
8.59 × 13 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
8.59 × 13 =
E
A B C D E 7 x 13 8 x 13 9 x 13 104 115 117 111.67
• Question 1
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
8.98 × 21 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
8.98 × 21 =
E
A B C D E 2 x 21 8 x 21 9 x 21 168 198 189 188.58
• Question 2
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
9.24 × 14 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
9.24 × 14 =
E
A B C D E 9 x 14 8 x 14 10 x 14 129.36 198 140 126
• Question 3
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
3.46 × 22 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
3.46 × 22 =
E
A B C D E 2 x 22 3 x 22 4 x 22 66 76.12 88
• Question 4
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
9.97 × 11 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
9.97 × 11 =
E
A B C D E 8 x 11 9 x 11 10 x 11 109.67 99 110
• Question 5
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
7.02 × 10 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
7.02 × 10 =
E
A B C D E 7 x 10 8 x 10 9 x 10 70 70.2 80
• Question 6
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
6.67 × 12 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
6.67 × 12 =
E
A B C D E 6 x 12 7 x 12 8 x 12 80.04 72 84
• Question 7
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
5.01 × 13 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
5.01 × 13 =
E
A B C D E 5 x 13 6 x 13 7 x 13 65 75 78 65.13
• Question 8
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
5.51 × 14 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
5.51 × 14 =
E
A B C D E 5 x 14 6 x 14 7 x 14 70 75 84 77.14
• Question 9
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
8.32 × 22 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
8.32 × 22 =
E
A B C D E 7 x 22 8 x 22 9 x 22 176 175 198 183.04
• Question 10
Work out approximate answers to the following problem, stating the values of A, B, C, D and E.
8.59 × 13 lies between
A
and
B
which equal
C
and
D
Check to see how close you are by doing the calculation on your calculator as well.
8.59 × 13 =
E
A B C D E 7 x 13 8 x 13 9 x 13 104 115 117 111.67
---- OR ----
Sign up for a £1 trial so you can track and measure your child's progress on this activity.
What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started
|
# What percent of the data is between Q1 and Q3?
Table of Contents
## What percent of the data is between Q1 and Q3?
25%
34Since Q1 and Q3 capture the middle 50% of the data and the median splits the data in the middle, 25% of the data fall between Q1 and the median, and another 25% falls between the median and Q3.
## What is the difference between quartile 1 and quartile 3?
The difference between the first and third quartiles, called the interquartile range, shows how the data is arranged about the median. A small interquartile range indicates data that is clumped about the median. A larger interquartile range shows that the data is more spread out.
## Is the 3rd quartile 75%?
The third quartile (Q3) is the middle value between the median and the highest value (maximum) of the data set. It is known as the upper or 75th empirical quartile, as 75% of the data lies below this point.
## What does the 3rd quartile mean?
The upper or third quartile, denoted as Q3, is the central point that lies between the median and the highest number of the distribution.
## Which percentile is equivalent to the third quartile Q3?
75th percentile
The third quartile (q3), or 75th percentile, is located such that 75 percent of the data lie below q3 and 25 percent of the data lie above q3.
## How do you calculate third quartile?
Type “=” and the quartile formula, including the range cells of that contain the data and “3” for the quartile number, to determine the third quartile of the set of data. Press “Enter” after typing. For example, type “=QUARTILE(A1:A8,3)” and press “Enter.”.
## How to calculate third quartile?
To find the third quartile, look at the top half of the original data set. We need to find the median of: Here the median is (15 + 15)/2 = 15. Thus the third quartile Q3 = 15. Quartiles help to give us a fuller picture of our data set as a whole. The first and third quartiles give us information about the internal structure of our data.
## How do I find the first quartile?
First Quartile. The first quartile can be calculated by first arranging the data in an ordered list, then finding then dividing the data into two groups. If the total number of elements in the data set is odd, you exclude the median (the element in the middle).
## What is the formula for first quartile?
Type “=” and the quartile formula, including the range cells of that contain the data and “1” for the quartile number, to determine the first quartile of the set of data. Press “Enter” after typing. For example, type “=QUARTILE (A1:A8,1)” and press “Enter.”. This equals 3.75, which is the first quartile of the set of data in cells A1 through A8.
## Is there a way to visualize census data?
How do you calculate demographic projections? In the geometric method of projection, the formula is Pp = P1(1 + r)n where, Pp= Projected population;...
1 min read
## Is it grammatically correct to say may you?
Can I or should I? Usually, CAN is used to give options or explain that you have the ability to do something, while SHOULD...
1 min read
## What happens if you win a wrongful termination suit?
How do you write a demand letter for wrongful termination? I am writing this letter to lodge a formal demand to retract my termination...
1 min read
|
## Vedic Maths Tricks for Quick Solutions in All Competitive Exams PDF Download
Vedic mathematics is a collection of mathematical techniques and shortcuts based on ancient Indian texts. These techniques can be extremely helpful in solving complex mathematical problems quickly and easily. In this article, we will discuss Vedic maths tricks for quick solutions in all competitive exams and provide a PDF download link for Vedic Maths Notes.
1. Multiplication Tricks:
The multiplication of numbers can be done in less time using Vedic maths tricks. Let’s see how to multiply numbers with a few examples.
Example: Multiply 23 x 23
Step 1: First, multiply 3 x 3 = 9 (unit place digits)
Step 2: Multiply 3 + 2 = 5 with 2 = 10 (tens place digits)
Step 3: Multiply 2 x 2 = 4 (hundred place digits)
Therefore, 23 x 23 = 529
1. Squaring Tricks:
Squaring any number is very easy with Vedic maths tricks. Let's see how to square a number with the help of an example.
Example: Square of 55
Step 1: First, find the difference between the number and the nearest base, which is 50 in this case, i.e., 55 - 50 = 5
Step 2: Add the difference to the base number, which gives 55 + 5 = 60
Step 3: Multiply the base number with the sum obtained in step 2, which gives 50 x 60 = 3000
Step 4: Finally, square the difference obtained in step 1, which is 5^2 = 25
Therefore, the square of 55 is 3025
1. Division Tricks:
Vedic maths tricks can also be used for division. Let’s see how to divide numbers with an example.
Example: Divide 300 by 6
Step 1: Divide the first digit of the dividend by the divisor. 3/6 = 0 (quotient) and 3 (remainder)
Step 2: Add the quotient to the second digit of the dividend. 0+0 = 0
Step 3: Repeat the above steps for the third digit of the dividend. 0+0 = 0
Therefore, 300/6 = 50
You can find more such Vedic maths tricks and shortcuts in the Vedic Maths Notes PDF Download. These notes will be helpful in cracking any competitive exams where maths plays an important role.
|
# Mass 1
In today’s lesson, we will be learning about mass. The learning objectives are:
1. Comparing Mass
2. Measuring Mass
## 1. Comparing Mass
Mass refers to how heavy an item is. To compare the mass of two or more items, we can place them on a balance scale. The heavier object will tilt the balance downwards.
Is the hamburger heavier or lighter than the bowl of ramen?
The hamburger is lighter than the bowl of ramen.
The bowl of ramen is heavier than the hamburger.
Example:
Look at the diagram below.
1. Which of the following is the lightest?
Solution:
Balance scale 1:
The blocks are heavier than the teddy bear.
The teddy bear is lighter than the blocks.
Balance scale 2:
The blocks are lighter than the toy train.
The toy train is heavier than the blocks.
The teddy bear is the lightest.
Teddy bear
1. Arrange the items in order of the heaviest to the lightest, we have:
Toy train, Blocks, Teddy bear
Toy train, Blocks, Teddy bear
Question 1:
Which item is the lightest?
1. Toy truck
2. Toy train
3. Toy helicopter
4. Toy car
Solution:
The toy truck is lighter than the toy helicopter.
The toy train is lighter than the toy truck.
The toy car is lighter than the toy train.
So, the toy car is the lightest.
(4) Toy car
Question 2:
Which of the following is the heaviest?
1. Bicycle
2. Scooter
3. Bus
4. Taxi
Solution:
The scooter is heavier than the bicycle.
The taxi is heavier than the scooter.
The bus is heavier than the taxi.
So, the bus is the heaviest.
(3) Bus
Question 3:
Which of the following shows the masses of items arranged in order, starting with the lightest?
1. Scooter, Truck, Bus, Fire Engine
2. Bus, Fire Engine, Truck, Scooter
3. Scooter, Truck, Fire Engine, Bus
4. Scooter, Truck, Bus, Fire Engine
Solution:
The scooter is lighter than the truck.
The fire engine is lighter than the bus.
The truck is lighter than the fire engine.
So, the scooter is the lightest.
(3) Scooter, Truck, Fire Engine, Bus
## 2. Measuring Mass
To measure the mass of an object, we can use a balance scale or a weighing scale. When using a balance scale, we will place the object on one side of the scale and add weights to the other side until the scale balances.
Mass of the basket of oranges
\begin{align} &= 80 \text{ g } + 80 \text{ g } + 80 \text{ g } \\[2ex] &= 240 \text{ g } \end{align}
OR
Mass of the basket of oranges
\begin{align} &= 3 \text{ g} \times 80 \text{ g } \\[2ex] &= 240 \text{ g } \end{align}
The mass of the oranges is 240 g.
Question 1:
Look at the diagram below. What is the mass of the bunch of bananas?
1. 250 g
2. 300 g
3. 400 g
4. 500 g
Solution:
Mass of the bunch of bananas
\begin{align} &= 250 \text{ g } + 250 \text{ g } \\[2ex] &= 500 \text{ g } \end{align}
(4) 500 g
Question 2:
What is the mass of the books?
1. 5 kg
2. 10 kg
3. 15 kg
4. 555 kg
Solution:
Mass of the books
\begin{align} &= 3 \text{ g } \times 5 \text{ g } \\[2ex] &= 15 \text{ g } \end{align}
(3) 15 kg
Question 3:
The mass of the toy car is __________.
1. 100 g
2. 105 g
3. 110 g
4. 115 g
Solution:
Total mass of weights on the right
\begin{align} &= 100 \text{ g } + 15 \text{ g } + 15 \text{ g }\\[2ex] &= 130 \text{ g } \end{align}
Mass of the toy car
\begin{align} &= 130 \text{ g } - 25 \text{ g } \\[2ex] &= 105 \text{ g } \end{align}
(2) 105 g
## Measuring Mass
To measure the mass of an object, we can also use a weighing scale.
When using a weighing scale, we will first need to determine the value of each interval. Then, we can identify the mass of the object.
Let’s weigh a basket of apples. First, we need to identify the value of each interval on the scale.
Between 0 and 100, there are 10 intervals.
Value of 10 intervals $= 100 \text{ g}$
Value 1 interval
$= 100 \text{ g} ÷ 10 \\[2ex] = 10 \text{ g}$
In the scale shown, the needle is pointing 5 intervals after 200 g.
Value of 5 intervals
$= 5 \times 10 \text{ g} \\[2ex] = 50 \text{ g}$
Mass of the basket of apples
$= 200 \text{ g} + 50 \text{ g}\\[2ex] = 250 \text{ g}$
Question 1:
What is the mass of the notebook?
1. 400 g
2. 425 g
3. 450 g
4. 490 g
Solution:
Value of 2 intervals $= 100 \text{ g}$
Value of 1 interval
$= 100 \text{ g} \div 2 \\[2ex] = 50 \text{ g}$
Mass of the notebook
$= 400 \text{ g} + 50 \text{ g}\\[2ex] = 450 \text{ g}$
(3) 450 g
Question 2:
What is the mass of the baseball?
1. 100 g
2. 230 g
3. 160 g
4. 200 g
Solution:
Value of 5 intervals $= 100 \text{ g}$
Value of 1 interval
\begin{align} &= 100 \text{ g } \div 5 \\[2ex] &= 20 \text{ g } \end{align}
Value of 3 intervals
\begin{align} &= 3 \times 20 \text{ g } \\[2ex] &= 60 \text{ g } \end{align}
Mass of the baseball
\begin{align} &= 100 \text{ g } + 60 \text{ g } \\[2ex] &= 160 \text{ g } \end{align}
(3) 160 g
Question 3:
The diagram shows the mass of a plate of turkey. The mass of the empty plate is 80 g. What is the mass of the turkey?
1. 420 g
2. 450 g
3. 480 g
4. 640 g
Solution:
Value of 1 interval $= 10 \text{ g}$
Mass of the turkey with the plate
\begin{align} &= 500 \text{ g } + 60 \text{ g } \\[2ex] &= 560 \text{ g } \end{align}
Mass of the turkey
\begin{align} &= 560 \text{ g } - 80 \text{ g } \\[2ex] &= 480 \text{ g } \end{align}
(4) 480 g
Question 4:
What is the mass of the packet of french fries?
1. 190 g
2. 210 g
3. 230 g
4. 250 g
Solution:
Scale 1:
Mass of Ketchup $+ \;60 \text{ g} = 230 \text{ g}$
Mass of Ketchup
\begin{align} &= 230 \text{ g} - 60 \text{ g } \\[2ex] &= 170 \text{ g } \end{align}
Scale 2:
$170 \text{ g } +$ Mass of French fries $= 420 \text{ g}$
Mass of French fries
\begin{align} &= 420 \text{ g} - 170 \text{ g } \\[2ex] &= 250 \text{ g } \end{align}
(4) 250 g
• Comparing Mass
To compare the mass of objects, we have to remember that the heavier object will tilt the balance scale downwards.
• Measuring Mass
To measure the mass of object, we will first need to determine the value of each interval on the weighing scale and then determine the mass of the object.
Continue Learning
Numbers To 1000 Multiplication And Division 1
Multiplication And Division 2 Addition And Subtraction 1
Addition And Subtraction 2 Fractions 1
Length 1 Mass 1
Volume 1 Money 1
Time 1 Shapes And Patterns
Picture Graphs 1 Model Drawing 1
Model Drawing 4
Primary
Secondary
Book a free product demo
Suitable for primary & secondary
Our Education Consultants will get in touch with you to offer your child a complimentary Strength Analysis.
Book a free product demo
Suitable for primary & secondary
our educational content
Start practising and learning.
No Error
No Error
*By submitting your phone number, we have
your permission to contact you regarding
Let’s get learning!
resources now.
Error
Oops! Something went wrong.
Let’s refresh the page!
Turn your child's weaknesses into strengths
Turn your child's weaknesses into strengths
Trusted by over 220,000 students.
Error
Oops! Something went wrong.
Let’s refresh the page!
Error
Oops! Something went wrong.
Let’s refresh the page!
|
Dividing Fractions – Steps and Solved Examples
# Dividing Fractions – Steps and Solved Examples
## Explain in Detail :Steps For Fraction Divided By Fractions
To divide fractions by fractions, you need to use the reciprocal of the divisor fraction.
Fill Out the Form for Expert Academic Guidance!
+91
Live ClassesBooksTest SeriesSelf Learning
Verify OTP Code (required)
I agree to the terms and conditions and privacy policy.
For example, to divide 1/4 by 2/3, you would use the reciprocal of 2/3, which is 3/2.
To divide fractions, you simply multiply the numerators and denominators of the fractions together.
In the example above, you would multiply 1/4 by 3/2 to get the answer of 3/8.
## What are Fractions?
A fraction is a part of a whole value or number. It is represented by p/q or a/b or m/n, etc. The upper part of a fraction is called the numerator and the lower part is the denominator. Examples of fractions are ½, ¼, ⅔, ⅗, etc.
All the arithmetic operations such as addition, subtraction, multiplication and division can be performed upon the fractions. Let us learn here how to divide a fraction by a fraction, by a whole number and by a mixed number with the help of examples, with simple steps.
## What is Meant by Dividing Fractions?
Dividing fractions is nothing but multiplying the fractions by reversing one of the two fraction numbers or by writing the reciprocal of one of the fractions. By reciprocal we mean, if a fraction is given as a/b, then the reciprocal of it will b/a. Thus, interchanging the position of numerator and denominator with each other.
a/b ÷ c/d = a/b × d/c
## How to Divide Fractions?
The division of fractions can be classified into three different ways. They are
• Dividing fractions by a fraction
• Dividing fractions by whole number
• Dividing fractions by mixed fraction
Let us discuss all these three methods in a detailed way
### Dividing Fraction by a Fraction
In three simple steps, we can solve the division of fractions by converting them into the multiplication of fractions. Let us learn one by one.
Step 1: Write the reciprocal of the second fraction number and multiply it with the first fraction number
Step 2: Multiply the numerators and denominators of both fractions
Step 3: Simplify the fraction number
In general, if a/b is a fraction which is divided by c/d. Then we can solve the division as;
• a/b ÷ c/d = a/b × d/c
• a/b ÷ c/d = a×d / b×c
• a/b ÷ c/d = ad/bc
You can see from the above expressions. The a/b is divided by c/d, then we can write it as a/b multiplied by d/c (reciprocal of c/d). And in the next step, we have to multiply both the numerator a & d and both the denominator, c & d. Hence, we can simplify the rest calculation.
### Dividing Fraction by a Whole Number
While dividing the fractions with whole numbers, the process of division is very easy. Follow the procedure given below.
Step 1: The whole number is converted into a fraction by applying the denominator value is 1
Step 2: Take the reciprocal of the number
Step 3: Now, multiply the fractional value by a given fraction
Step 4: Simplify the given expression
Example: Divide 6/5 by 10
Step 1: Convert 10 into a fraction: 10/1
Step 2: Take reciprocal: 1/10
Step 3: Multiply 6/5 and 1/10: (6/5)×(1/10)
Step 4: Simplify: 3/25
### Dividing Fractions by a Mixed Fraction
The process of dividing fractions by a mixed fraction is almost similar to dividing fractions by a fraction. The steps to perform the division of a fraction by a mixed fraction are as follows:
Step 1: Convert the mixed fraction into the improper fraction
Step 2: Now, take the reciprocal for the improper fraction
Step 3: Multiply the obtained fraction by a given fraction
Step 3: Simplify the fractions
Example: Divide ⅖ by 3½.
Step 1: Convert 3½ into an improper fraction, we get 7/2
Step 2: Take reciprocal for improper fraction: 2/7
Step 3: Multiply ⅖ and 2/7
Step 4: Simplify: 4/35
## Dividing Decimals as Fractions
We have learned to divide fractions using three simple steps. Now with the help of these steps let us learn how to divide decimals with examples.
Example: Divide 0.5 ÷ 0.2
Solution: To divide these decimal numbers, we have to convert both the decimal number into natural numbers by multiplying numerator and denominator by 10.
Therefore, 0.5 × 10 / 0.2 × 10
We get, 5/2 = 2.5
Also, we can use the dividing fractions method to solve the above problem.
We can write 0.5 and 0.2 as 5/10 and 2/10.
So for 5/10 ÷ 2/10, we can use the same steps fraction’s division.
5/10 × 10/2
= 5 × 10 / 10 × 2
= 50/20
= 5/2
= 2.5
Note: These are the simple method of dividing decimals. You can also use the direct division method to divide decimals. The only difference is to place the decimal into the right place of the quotient. Let us take an example of this.
Example: Divide 13.2 ÷ 2
Solution: 2) 13.2 (6.6)
Therefore, 13.2 ÷ 2 = 6.6
Dividing the natural numbers or whole numbers is an easy task but dividing the fractions is a little complex one. The operations performed on natural numbers and whole consist of simple calculations, which one can easily solve. But the operations performed on fractions are sometimes typical and also time-consuming. The simple division has four parts divisor, dividend, quotient and remainder. Also, know some of the divisibility rules for the whole number here.
## General Solution:
There is no general solution to this problem.
## Steps For Dividing Fractions By Whole Numbers
1. Determine the whole number divisor.
2. Place the divisor above the dividend (fraction) with a line drawn through the two.
3. Divide the top number of the dividend (the numerator) by the divisor.
4. Write the answer below the dividend line.
5. Repeat steps 3 and 4 until there is no remainder.
## Related content
Even and Odd Numbers Cone Limits in Maths Algebra Cube Lines and Angles Class 9 Extra Questions Maths Chapter 6 Why Is Maths So Hard? Here’s How To Make It Easier NCERT Solutions for Class 4 Maths Worksheet for Class 4 Maths Curved Surface Area of Cone
|
# Lesson 1
Interpreting Negative Numbers
### Lesson Narrative
In this lesson, students review what they learned about negative numbers in grade 6, including placing them on the number line, comparing and ordering them, and interpreting them in the contexts of temperature and elevation (MP2). The context of temperature helps build students’ intuition about signed numbers because most students know what it means for a temperature to be negative and are familiar with representing temperatures on a number line (a thermometer). The context of elevation may be less familiar to students, but it provides a concrete (as well as cultural) example of one of the most fundamental uses of signed numbers: representing positions along a line relative to a reference point (sea level in this case). The number line is the primary representation for signed numbers in this unit, and the structure of the number line is used to make sense of the rules of signed number arithmetic in later lessons.
### Learning Goals
Teacher Facing
• Interpret signed numbers in the contexts of temperature and elevation.
• Order rational numbers, and justify (orally) the comparisons.
• Plot points on a vertical or horizontal number line to represent rational numbers.
### Student Facing
Let's review what we know about signed numbers.
### Required Preparation
Print and cut up slips from the Rational Numbers Card Sort blackline master. Prepare 1 copy for every 3 students. Students will need copies of both sets 1 and 2. Keep the slips from set 1 (Integers) separate from set 2 (Rational numbers that are not integers) for each group. Consider using different colors of paper so sets 1 and 2 are easier to separate.
### Student Facing
• I can compare rational numbers.
• I can use rational numbers to describe temperature and elevation.
Building On
Building Towards
### Glossary Entries
• absolute value
The absolute value of a number is its distance from 0 on the number line.
The absolute value of -7 is 7, because it is 7 units away from 0. The absolute value of 5 is 5, because it is 5 units away from 0.
• negative number
A negative number is a number that is less than zero. On a horizontal number line, negative numbers are usually shown to the left of 0.
• positive number
A positive number is a number that is greater than zero. On a horizontal number line, positive numbers are usually shown to the right of 0.
|
# Calculate the distance using the coordinates in basic mathematics
If two points in a graph share x and y coordinates, the distance between them is the difference between the coordinates that do not share. For example, if a point has the coordinates (1, 7), and the other has the coordinates (1, 12), the distance between them is 5 units, the difference between 12 and 7. However, if both points do not share coordinates, the distance between them is the length of the diagonal that joins them. This length is calculated using the Pythagorean theorem.
Steps to follow:
#### one
Subtract the first point of the "x" coordinate to the first point of the second. If, for example, the two points have coordinates (1, 9) and (13, -12), then subtracting the values of the coordinates "x" is 13 - 1 = 12.
#### two
Make the square of this difference: (12) ^ 2 = 144.
You can observe that it is indifferent if we make step number one subtract it in an inverse way, the result will be the same, since when we make the square root the sign is indifferent, we see it:
• We subtract the values of the "x": 1 - 13 = -12
• Square root of (-12) ^ 2 = 144
#### 3
Subtract the first point of the coordinate to the first point of the second: (-12) - 9 = -21.
#### 4
Re-make the square of this difference in this way: (-21) ^ 2 = 441.
#### 5
Add the two places: 144 + 441 = 585.
#### 6
Find the square root of this sum: 585 ^ 0.5 = 24.19. So therefore points are approximately 24.19 units away.
|
Andlearning.org is a single website that is sharing all Semicircle formulas which is useful for math calculation. Understand the concept of the Unitary method here. Area of Semi-Circle. The formula for a semicircle is derived from the formula of the circle, only divided by two. Enter one value and choose the number of … College homework help Volume of a uniformed shaped object such as an octagon column is derived by the following formula: Volume = Area • Height The formula above uses the area of a semi-circle. What is the radius in meters? The centroid of Semicircle $$(0, \frac{4r}{3\pi})$$ Circumference of Semicircle Formulas $$P = \pi r$$ Summary of Semicircle Formula. A semicircle is half a circle. Area of a circle. ♦ The area of semi circle is given as 77 { m }^{ 2 } then find its radius. A = \frac { 1 }{ 2 } × π × { r }^{ 2 } where π = \frac { 22 }{ 7 } (or) we can also take π as 3.14 and “r” is the radius of the given semi circle. It is equal to the square of the radius multiplied by the constant factor π = 3.14. So we can say that area of semi circle will be equals to the half of area of circle. A half of a circle or of its circumference; A set of objects arranged in a semicircle; In mathematics (more specifically geometry), a semicircle is a two-dimensional geometric shape that forms half of a circle. This means the radius of the semicircle is squared, multiplied by the constant pi, then divided by 2. The area of a semicircle or the area of half circle is $\frac{{\pi {{\text{R}}^2}}}{2}$ . This may be a bad time to buy a Mega Millions ticket. Polygon area = h/2(2×π×r) = (2×r×r×π)/2. Determine the area Using the formula for finding the area of a semicircle, divide the product of the squared radius and pi by 2. A semicircle is a shape or figure that is represented for half circle. We know that the … We can solve for r to show an expression for the radius of a semicircle when given the area: A = πr2 2 Where r is the radius; a is the arc angle also known as the central angle. ♦ If the radius of semi circle is 14 cm then find the area of semi circle. The formula is created by halving the circle perimeter formula (circumference) and adding the diameter length to that. If we observe this formula clearly we can find out some interesting relation between areas of circle and semi circle. Here, the semi-circle rotating about an axis is symmetric and therefore we consider the values equal. We know the formula to calculate area of a circle is πr^2, by dividing this by 2 we will get the area of a semicircle. semicircle. Remember the formula for area of a circle. In semi circle, Semi meaning is half so half of circle will be a semi circle. Calculate the area of the clock face. (3.14) Finally you divide that answer by two. The formula to calculate the area of a circle, with radius $$r$$ is: $$\text{area of a circle} = \pi r^2$$. The formula for the area of a circle is π x radius 2, but the diameter of the circle is d = 2 x r 2, so another way to write it is π x (diameter / 2) 2.Visual on the figure below: π is, of course, the famous mathematical constant, equal to about 3.14159, which was originally defined as the ratio of a circle's circumference to its diameter. We know that the area of a circle is: area of a circle = r 2. So we can say that area of semi circle will be equals to the half of area of circle. Here, the area of a semicircle is half the area of a circle. Area Of Semicircle Formula. My suggestion is to remember the area of circle formula with that you can write area of semi circle formula without any mistakes. If you have the area of a semicircle with diameter that means that you have to divide the diameter by two to get the radius. Hello Melinda. Therefore, Area of a semicircle =(πr 2)/2 The bottom side of the semicircle is equal to the circle’s diameter, so it … An alternative method is to use the diameter, which is twice the length of the radius. Hi, am Murali a Mathematics blogger. The circumference of a semicircle is divided into two ares such that the chord of one is double that of the other. area of semicircle formula Webster’s bibliographic and event-based timelines are comprehensive in scope, covering virtually all topics, geographic locations and people. The semicircle is a very basic geometric shape and a very common appearance in nature. Recently I have created a YouTube Channel called Murali Maths Class, check for the latest Maths Videos on All the topics. By this, we can conclude that any one of the formulae out of 2 i.e. Penny . A semi-circle is half of a circle. Radius and diameter refer to the original circle, which was bisected through its center. Area of a semicircle. Thesishelpers.com. = (1/2) x (22/7) x 7 2 = (1/2) x (22/7) x 7 x 7 = 1 x 11 x 7 = 77 cm 2. Area Of Combined Shapes - Rectangle And Semi-Circle - YouTube Semicircle Calculator. So, the formula for the area of a semicircle is A = pi * r^2/2. Determine the area Using the formula for finding the area of a semicircle, divide the product of the squared radius and pi by 2. In semi circle, Semi meaning is half so half of circle will be a semi circle. So, the formula for the area of a semicircle is:Area=πR22where: R is the radius of the semicircle π is Pi, approximately 3.142 Rearranging this we get. If the radius of the semicircle is 2 centimeters, then the area is 1/2(3.14 x 2 x 2), or 6.28 square centimeters. Area of a circle formula. The first step is to square the radius. They do so from a linguistic point of view, and in the case of this book, the focus is on “Semicircles,” including when used in literature (e.g. area of semi circle or area of circle if you learn then writing the other formula will be easy. How to find the area of a semicircle? The area of … Since you're finding the area of a semi-circle, you'll be looking for half of the area of a circle, which means you have to use the formula for finding the area of a semi-circle and then divide it by two. Since a semicircle is just half of a circle, the area of a semicircle is shown through the formula A=(pir^2)/2. You'll need the radius to find the area of the semi-circle. Find the radius of the semi-circle. 10 Must-Watch TED Talks That Have the Power to Change Your Life. You need to know the formula for finding the area of a circle, {\displaystyle A=\pi r^ {2}}. Area of a Semi-Circle Formula: A = (1/2) * π * r 2 Where, A = Area of Semicircle r = Radius Just enter the value of radius in the area of a semicircle calculator to compute the semicircle area within a … Find the area of the semicircle whose radius is 7 cm. Andlearning.org is a single website that is sharing all Semicircle formulas which is useful for math calculation. Area of circle formula if already studied then learning the area of semi circle formula is very much easy. The formula the calculator uses is as follows: Area = PI (3.141) * r ^2 * a/360. Harold Bornstein, Trump's eccentric ex-doctor, dies Solution : Area of Semicircle = (1/2) π r 2. Solution: 1.) Area of a Semi-Circle Formula: A = (1/2) * π * r 2 Where, A = Area of Semicircle r = Radius Related Calculator: Semi Circle Area Formula. Knowing the semicircle definition - half of a circle - we can easily write the semicircle area formula using the well-known circle area, πr² : Area semicircle = Area circle / 2 = πr² / 2. What is the area of a semicircle with a radius of 5? Example 2 : Find the area of the semicircle whose radius is 3.5 cm. The area formula is: A = πr2 2 A = π r 2 2. Let’s plug the area into the formula… In order to find the area of semi circle use the below given formula. where, R= Radius of the semicircle. Therefore, using the diameter of the circle, the formula for finding the area of one semicircle is (? The area of a semicircle is half the area area of the circle from which it is made.Recall that the area of a circle is πR2, where R is the radius. It is equal to half the area … The area of a semicircle is (?*R*R)/2. Penny . You can contact me at muralimudeblog@gmail.com. with the subscription you can get all my latest post updates. The area is half the area of a circle. Example. Therefore, the formula to calculate area of a semicircle is given by: A = 1 ⁄ 2 πr 2 A = 1 ⁄ 2 π(5 2) = 12.5π 2.) So the area of a semicircle is when you cut a full circle into half you get the area of a semicircle. Use the Area of a Semicircle Formula: Area equals pi time the radius squared then divided by two. It has 180º arc. The area of a semicircle is (?*R*R)/2. Now the area of a semicircle is equal to half of that of a full circle. The formula for how to find Area of a Semicircle is area equals pi times the radius squared then divided by two. Calculations at a semicircle. Semicircle Area Formula The formula for area of a semicircle is given as: A = 1⁄2πr2 Where A is the area and r is the radius. A = \frac { 1 }{ 2 } × π × { r }^{ 2 }, = \frac { 1 }{ 2 } × \frac { 22 }{ 7 } × { 14 }^{ 2 }. The diameter of the circle divides the circle into two equal semicircles. Substitute 7 for r and 22/7 for π. It has 180º arc. What Do You Mean By the Area of a Semicircle? Let me write this, so the area is equal to pi r-squared over two 'cause it's a semi-circle. Please Subscribe and Click the Bell Icon for the latest Maths Videos Notifictaions…Thank You. Hello Melinda. (For a semi circle this is always 180 degrees) Area of a semi circle definition. A common problem in geometry class is to have you calculate the area of a circle based on provided information. A semicircle is half of a circle. Therefore, the formula to calculate area of a semicircle … Strange Americana: Does Video Footage of Bigfoot Really Exist? 6 Quick Steps for using the Area of a Semicircle Formula. Then you multiply the square of the radius times pi. Well, if it was a full circle, it would be pi r-squared, but it's a semi-circle so it's pi r-squared over two. What Is the Formula for the Area of a Semicircle. 'Mona Lisa of sports cards' sells for record amount. The semicircle is a very basic geometric shape and a very common appearance in nature. The centroid of Semicircle $$(0, \frac{4r}{3\pi})$$ Circumference of Semicircle Formulas $$P = \pi r$$ Summary of Semicircle Formula. Question 1: For a vehicle having wheels of radius 24cm find the distance covered by it in one complete revolution of wheels. Area of circle formula if already studied then learning the area of semi circle formula is very much easy. Please support and encourage me for creating good and useful content for everyone. How the COVID-19 Pandemic Will Change In-Person Retail Shopping in Lasting Ways, Tips and Tricks for Making Driveway Snow Removal Easier, Here’s How Online Games Like Prodigy Are Revolutionizing Education. We know that the area of a circle is: area of a circle = r 2. A semi-circle is half of a circle. In this blog am going to cover all Mathematics related concepts. If the radius of the semicircle is 2 centimeters, then the area is 1/2(3.14 x 2 x 2), or 6.28 square centimeters. By this, we can conclude that any one of the formulae out of 2 i.e. Example 2: a semicircle is shown through the formula for finding area! What is the area of a semicircle with a radius of semi circle { 2 } then find area... Mathematics related concepts subscription please visit Your email and activate it 12.5π.! Adsbygoogle = window.adsbygoogle || [ ] ).push ( { } ) ; Help with College! You can use the area of a circle of radius r is r 2.. Circle = r 2. is measured to be 4.5π square meters a YouTube Channel called Maths! Have semicircles in their name ) a radius of 5 that formula to calculate of. That have the radius r * r ) /2 on all the topics area region. Half of that of the semi-circle rotating about an axis is symmetric and therefore we consider the values.... 2. an approximation of pi of semi circle definition on provided.. [ ] ).push ( { } ) ; Help with math College homework Help Thesishelpers.com here the will! Plug the radius squared then divided by two the normal area of a semicircle is through. Cut a full circle let 's use that formula to calculate the of... 'Ll need the radius of the formulae out of 2 i.e how to find the of... Blog am going to cover all Mathematics related concepts it is equal to the square of the circle half! The subscription you area of a semicircle formula get all my latest post updates semicircle with a radius 5! In order to find the area of one is double that of a semicircle cancellation may be used in process! By it in one complete revolution of wheels is symmetric and therefore we consider the values.! Pi time the radius ; a is the radius of the formulae of... Geometric shape and a very basic geometric shape and a very basic geometric shape a... } ^ { 2 } then find the area of a circle, semi is. Cancellation may be used in the process, which is useful for math calculation =! 2×R×R×Π ) /2 to cover all Mathematics related concepts semicircle Calculator 3.141 ) * r ) /2 of!, only divided by 2. circle and semi circle is πr 2 ) (... To know the formula a = 1 ⁄ 2 π ( 5 2 ) /2 observe this clearly! The constant factor π = 3.14 ) /4, where r '' represents the radius of semi circle be! Circle is a single website that is represented for half circle over two 'cause it 's a semi-circle and cancellation! Videos on all the topics it 's a semi-circle = 1/2 r 2. which was bisected through center! To the half of area of a circle of radius r is r semicircle! Which we can conclude that any one of the semicircle or the area of circle be 4.5π square meters Finally! We know that the area of semi circle space inside the semicircle whose is. Created area of a semicircle formula YouTube Channel called Murali Maths class, check for the area of circle. Cancellation may be used in the process { 2 } then find the area of a circle radius. Method is to remember the formula of the semicircle is (? * r * r * ). Vehicle having wheels of radius r is the formula a = πr2 2 )! Covered by it? * r * r ) /2 a shape figure. Is double that of a semicircle is just half of a semicircle just. In one complete revolution of wheels formulae out of 2 i.e and know the for... Equal semicircles M.O.I will be: area of a circle radius to find its area semicircle whose radius 7. Normal area of a semicircle = ( 1/2 ) π r 2. areas of circle formula with that can! Pi area of a semicircle formula the radius of 5 a half circle radius r is formula. Observe this formula clearly we can say that area of a semi-circle = 1/2 r 2 )! Your email and activate it to find the area of semi circle radius r is the area a! We observe this formula clearly we can conclude that any one of the formulae out of 2.. Alternative method is to use the area of a semicircle is a single website is... Of one semicircle is (? * r ^2 * a/360 r 2. common appearance in nature question:... A vehicle having wheels of radius 24cm find the area of a circle and the area one... Very much easy means the radius times pi the below given formula used in process. The Bell Icon for the latest Maths Videos Notifictaions…Thank you may be used in the process when you cut full. Am going to cover all Mathematics related concepts Calculator uses is as:. Circumference of a semi-circle = 1/2 r 2. of sports cards sells.: area of a semicircle π r 2. the arc angle also known as the central.. Two ares such that the area or region enclosed by it in one complete revolution wheels... Area is equal to half of a area of a semicircle formula and how cancellation may be used in the process?! By which we can conclude that any one of the circle, which is the! Its center learning the area of the circle to find the area of a semicircle is through... 2 semicircle Calculator squared then divided by two h/2 ( 2×π×r ) = 2. Πr 2, where D is the formula for the latest Maths Videos all. To have you calculate the area of a semicircle = ( 2×r×r×π /2. Help with math College homework Help Thesishelpers.com as an approximation of pi needs the radius multiplied the! Already studied then learning the area of a semicircle is (? * r ) /2: a πr2! Videos Notifictaions…Thank you the values equal activate it follows: area of semi circle is: area equals time. Sports cards ' sells for record amount ) Finally you divide that answer by two circle this is 180. Is sharing all semicircle formulas which is useful for math calculation out of 2 i.e you that. 2×Π×R ) = 12.5π 2. arc angle also known as the central angle square of the circle to the! ) /2 1 given radius of the x-axis is equal to pi over! We can find out some interesting relation between areas of circle will equals. Formulae out of 2 i.e Videos on all the topics an approximation of pi and. R ^2 * a/360 of inertia of a circle = r 2. visit Your and. As an approximation of pi you find the area of a half circle generally refers to the inside! * r ) /2 also known as the central angle am going to cover all related... Arc angle also known as the central angle 1: for a semicircle formula ) area of a circle r! Post updates means the radius ; a is the diameter and is equal to half of.! In nature r is the diameter of the circle into two equal.! Then divided by two that any one of the circle \displaystyle A=\pi r^ { 2 } } let use! /2 1 let me write this, we can conclude that any one of the circle the. Is squared, multiplied by the constant pi, then divided by two you... The area of a circle a single website that is sharing all formulas. Already studied then learning the area of semicircle is shown through the formula by which we find... ♦ if the radius you can write area of a semi-circle = 1/2 r semicircle! Represented for half circle the central angle Murali Maths class, check for the area of.! Equals to the square of the formulae out of 2 i.e is divided into two such... Of 5 π ( 5 2 ) /2 TED Talks that have radius. My latest post updates pi time the radius to find the area formula is much... By two square meters therefore the area of semi circle this is 180... The circumference of a semi-circle using the area of circle if you learn writing... Is double that of the semicircle is nothing but half of a half circle generally refers to the inside! 180 degrees ) area of a semicircle formula: area of a semicircle formula Calculator uses is as follows area! Conclude that any one of the radius of the radius of 5 'cause it 's semi-circle. Divided into two equal semicircles also known as the central angle you can write area of the semicircle a! { \displaystyle A=\pi r^ { 2 } } an axis is symmetric and therefore we consider the values.! Studied then learning the area of semi circle is: area of a semi-circle is half the area semi! How to find the area of a half circle which is twice the length of the x-axis is equal 2R. Encourage me for creating good and useful content for everyone square of the circle 180 )! ^ { 2 } }, which was bisected through its center you 'll need the radius of inches... Formula to calculate area of a circle of radius 24cm find the area of a circle formula have semicircles their! Is divided into two ares such that the area of semi circle use the given! Studied then learning the area is equal to the y-axis and diameter refer to the inside... Demonstrates how to find the distance covered by it A=\pi r^ { 2 } then find the formula. Example 2: find the area of a circle formula if already studied then learning area.
Reviews Of Last Night's Better Call Saul, Is Zinsser Bin Oil Based, Bachelor Of Catholic Theology, Rc Trucks Ford F-150 Raptor, Milwaukee 6955-20 Vs Dewalt Dws780, Bachelor Of Catholic Theology, Admin Executive Job Scope, Glass Baffles For Sump, 5 Gallon Driveway Sealer Coverage, Improvise Musically Crossword Clue,
|
# What is 201/542 as a decimal?
## Solution and how to convert 201 / 542 into a decimal
201 / 542 = 0.371
Fraction conversions explained:
• 201 divided by 542
• Numerator: 201
• Denominator: 542
• Decimal: 0.371
• Percentage: 0.371%
201/542 or 0.371 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. Now, let's solve for how we convert 201/542 into a decimal.
201 / 542 as a percentage 201 / 542 as a fraction 201 / 542 as a decimal
0.371% - Convert percentages 201 / 542 201 / 542 = 0.371
## 201/542 is 201 divided by 542
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 201 is being divided into 542. Think of this as our directions and now we just need to be able to assemble the project! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! We must divide 201 into 542 to find out how many whole parts it will have plus representing the remainder in decimal form. This is how we look at our fraction as an equation:
### Numerator: 201
• Numerators are the portion of total parts, showed at the top of the fraction. 201 is one of the largest two-digit numbers you'll have to convert. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Time to evaluate 542 at the bottom of our fraction.
### Denominator: 542
• Denominators represent the total parts, located at the bottom of the fraction. 542 is one of the largest two-digit numbers to deal with. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Have no fear, large two-digit denominators are all bark no bite. Next, let's go over how to convert a 201/542 to 0.371.
## How to convert 201/542 to 0.371
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 542 \enclose{longdiv}{ 201 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 542 \enclose{longdiv}{ 201.0 }$$
Uh oh. 542 cannot be divided into 201. So that means we must add a decimal point and extend our equation with a zero. Now 542 will be able to divide into 2010.
### Step 3: Solve for how many whole groups you can divide 542 into 2010
$$\require{enclose} 00.3 \\ 542 \enclose{longdiv}{ 201.0 }$$
How many whole groups of 542 can you pull from 2010? 1626 Multiply by the left of our equation (542) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.3 \\ 542 \enclose{longdiv}{ 201.0 } \\ \underline{ 1626 \phantom{00} } \\ 384 \phantom{0}$$
If there is no remainder, you’re done! If you still have numbers left over, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. This is also true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 201/542 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.371 to 201/542 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 201/542 = 0.371 what would it be as a percentage?
• What is 1 + 201/542 in decimal form?
• What is 1 - 201/542 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.371 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 542 Denominator What is 201/532 as a decimal? What is 191/542 as a decimal? What is 201/533 as a decimal? What is 192/542 as a decimal? What is 201/534 as a decimal? What is 193/542 as a decimal? What is 201/535 as a decimal? What is 194/542 as a decimal? What is 201/536 as a decimal? What is 195/542 as a decimal? What is 201/537 as a decimal? What is 196/542 as a decimal? What is 201/538 as a decimal? What is 197/542 as a decimal? What is 201/539 as a decimal? What is 198/542 as a decimal? What is 201/540 as a decimal? What is 199/542 as a decimal? What is 201/541 as a decimal? What is 200/542 as a decimal? What is 201/542 as a decimal? What is 201/542 as a decimal? What is 201/543 as a decimal? What is 202/542 as a decimal? What is 201/544 as a decimal? What is 203/542 as a decimal? What is 201/545 as a decimal? What is 204/542 as a decimal? What is 201/546 as a decimal? What is 205/542 as a decimal? What is 201/547 as a decimal? What is 206/542 as a decimal? What is 201/548 as a decimal? What is 207/542 as a decimal? What is 201/549 as a decimal? What is 208/542 as a decimal? What is 201/550 as a decimal? What is 209/542 as a decimal? What is 201/551 as a decimal? What is 210/542 as a decimal? What is 201/552 as a decimal? What is 211/542 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 542 Denominator 202/542 as a percentage 201/543 as a percentage 203/542 as a percentage 201/544 as a percentage 204/542 as a percentage 201/545 as a percentage 205/542 as a percentage 201/546 as a percentage 206/542 as a percentage 201/547 as a percentage 207/542 as a percentage 201/548 as a percentage 208/542 as a percentage 201/549 as a percentage 209/542 as a percentage 201/550 as a percentage 210/542 as a percentage 201/551 as a percentage 211/542 as a percentage 201/552 as a percentage
|
# Find the Lowest Common Multiple (LCM) of Expressions? Questions with detailed solutions
How to find the lowest common multiple (LCM) of two or more expressions in math? Detailed
solutions to the questions in How to Find LCM of Expressions? are included.
## Find Lowest Common Multiple (LCM) of the expressions given below
• ### Question 1
2 (x + 1)
and 3 (x + 1) .
solution
We first factor the given expressions completely:
2 (x + 1) = 2 (x + 1)
3 (x + 1) = 3 (x + 1)
The LCM is made by multiplying all factors included in the factoring of the given expressions. Common factors are used once only and the one with the highest power is used.
2 is a factor in the first expressions is therefore used. x + 1 is a factor in the first and second expressions is used once only. 3 is a factor in the second expression only and is therefore used . Hence
LCM ( 2 (x + 1) , 3 (x + 1) ) = 2 · 3 (x + 1)
• ### Question 2
2 (x - 1) 2
and 5 (x - 1) .
solution
Factor the given expressions completely:
2 (x - 1) 2 = 2 (x - 1) 2
5 (x - 1) = 5 (x - 1)
We now make the LCM by multiplying all factors included in the factoring of the given expressions. Common factors are used once only and the one with the highest power is used.
2 is a factor in the first expressions is therefore used. x - 1 is a factor in the first and second expressions and the factor with the highest power which is (x - 1) 2 in the first expression is used. 5 is a factor in the second expression only and is therefore used. Hence
LCM ( 2 (x - 1) 2 , 5 (x - 1) ) = 2 · 5 (x - 1) 2
• ### Question 3
x 2 + 5 x + 6
and 2 x 2 + 2 x - 4 .
solution
Factor the given expressions completely:
x 2 + 5 x + 6 = (x + 3)(x + 2)
2 x 2 + 2 x - 4 = 2(x - 1)(x + 2)
The LCM is made by multiplying all factors included in the factoring of the given expressions. Common factors are used once only and the one with the highest power is used.
x + 3 is a factor in the first expressions is therefore used. x + 2 is a factor in the first and second expressions and therefore used once. 2 is a factor in the second expression only and is therefore used. x - 1 is a factor in the second expression and is therefore used. Hence
LCM ( x 2 + 5 x + 6 , 2 x 2 + 2 x - 4 ) = 2 · (x + 3)(x + 2)(x - 1)
• ### Question 4
3 x 3 - 2 x 2 - x
and x - 1 .
solution
Factor the given expressions completely:
3 x 3 - 2 x 2 - x = x (3 x + 1)(x - 1)
x - 1 = x - 1
The LCM is by multiplying all factors included in the factoring of the given expressions. Common factors are used once only and the one with the highest power is used.
x is a factor in the first expressions is therefore used. 3 x + 1 is a factor in the first expression and is therefore used. x - 1 is a factor in the first and second expression and is therefore used once only. Hence
LCM ( 3 x 3 - 2 x 2 - x , x - 1 ) = x (3x + 1)(x - 1)
• ### Question 5
3 x 3 - 2 x 2 - x
, 2 x 2 - 2 and (x - 1) 2.
solution
Factor the given expressions completely:
3 x 3 - 2 x 2 - x = x (3 x + 1)(x - 1)
2 x 2 - 2 = 2(x - 1)(x + 1)
(x - 1) 2 = (x - 1) 2
We now make the LCM by multiplying all factors included in the factoring of the given expressions. Common factors are used once only and the one with the highest power is used.
x is a factor in the first expressions is therefore used. 3 x + 1 is a factor in the first expression and is therefore used. x - 1 is a factor in all three expressions and the one with the highest power which is (x - 1) 2 in the third expression is used. 2 is a factor in the second expression and is therefore used. x + 1 is a factor in the second expression and is therefore used. Hence
LCM ( 3 x 3 - 2 x 2 - x , 2 x 2 - 2 , (x - 1) 2 ) = 2 x (3x + 1)(x - 1) 2(x + 1)
### More References and links
High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers
Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers
Author - e-mail
Home Page
|
### Teacher Guide to Math Word Problems
Within every math word problem is an equation. Solving a math word problem is about determining what the problem is asking as much as it is about determining (and solving) the equation hidden in the words. While word problems can be solved in many different ways, each method involves outlining the problem.
Word problems are now being more highlighted than ever. We are seeing just about every other standard in math include a form of word problem or problem solving skill in a similar light. We would ask you to keep this in mind when you are planning your lessons. Try to turn everything into a word problem, than they become much easier for students.
Outlining the problem is not as difficult as it might sound. If you follow a systematic pattern, it can become very habitual and some might even term it "easy!" Lets see if we can give you some basic steps to make it easier for you.
1. WHAT IS HAPPENING?
Begin by reading the problem and trying to visualize what is happening. This is called "giving the problem context". It means, simply, that you understand what the numbers represent.
2. WHAT ARE YOU LOOKING AT?
Next, make a list of the numbers you have to work with and what those numbers mean - these are called "variables". For example, if your math word problem talks about Julie having two cookies, one of your variables would be "Julie's cookies = 2". Listing your variables will help you see what information you have.
Once you have your list of variables, you need to figure out what is being asked in the math word problem. While many times the question is written out, many times you have to figure it out yourself or, at least, figure out the steps you need to take to get the answer.
4. HOW CAN YOU CALCULATE THE ANSWER?
Many times, a math word problem will have more than one step, such as in the following example: "Julie has two cookies. You give her two more. Julie gives one to Todd. How many cookies does Julie have now?" In order to solve this problem, you first have to add the number of cookies Julie had to begin with (2) with the number you gave her (2). The second part of the question involves using that total (4), and subtracting the 1 cookie, she gave to Todd.
5. PERFORM THE CALCULATION
Next, you will calculate the answer. If you are unsure, draw diagrams or pictures to help. Also, write out your work - this way if you do make a mistake, it will be easy to go through your work and correct it.
6. REVIEW
Lastly, review your work. After you get your calculation, ask yourself whether the number makes sense. To be extra certain, try working the problem backwards. This way, you can double-check your math and the reasoning you used in framing your answer.
|
# Interactive video lesson plan for: Dimensional Analysis Unit Conversion Made Easy
#### Activity overview:
Dimensional Analysis Part 2
These problems involve setting up two conversions in order to reach your goal. The video solves two problems
Unit conversion problem 1. Convert 14 lb/week to ounces/day
Problem 2. 17 centimeters/minute to meters /hour
Welcome to MooMooMath where we upload a new Math video every day. This is part two of dimensional analysis, and completing unit conversion. In the first video we worked on several problems that were a little easier. For instance we worked on converting from feet per hour, and our goal was to convert to miles an hour. The reason that these were easier is if you notice the hour is on the bottom, and we only need to get to complete one conversion and put the feet on the bottom. Next you would put miles on the top and ask yourself which is larger and the miles is larger and place a 1 on top, and then just multiply straight across and you are left with miles an hour. What if you get a question in which you have more than one conversion? Let me show you what I'm talking about. You are lifting weights (this is a pitiful example) and you can lift 14 pounds in one week. You want to know how many ounces a day you are lifting. Our goal is ounces a day. If you look we have week and days, so we can't just go straight across. Don't fear. It is really not that difficult. Let's go ahead do this. The first thing we do is get rid of the pounds by putting pounds on the bottom and put ounces on top. Again. I ask myself "which is larger?" I know a pound is larger than an ounce. If I have 1 pound how many ounces do I have? That will be 16 ounces. We now have the ounces on top. What you may want to do is cover up this term, and now get rid of the week by placing week on top and day on the bottom. I ask "Which is larger, day or week?" There are 7 days in one week. I then just multiply straight across, but before that I will cancel. We are left with ounces and days. Now let's do the Math. 14 x 16 =227, and 1 x1 x 7 =7. Then when you divide 227/7 = 32 ounces per day.
Let's go with one more example. Let's go with, I can travel 17 centimeters in a minute on my bicycle. I want to know how my distance in meters per hour. When I look at this we have minutes and meters so I know I will have two conversions. I need to get meters on top. I will put centimeters on bottom, and put miles on top because that is our goal. I next ask “which is larger, meters or centimeters?" 1 meter = 100 centimeters. Then I need to get hours on the bottom, and get rid of the minute. In order to get rid of the minute I will put minutes on top, and hours on the bottom, and get rid of the minute, and hours on the bottom. Hours are larger than minutes so I will place a 1 by minutes, and a 60 by minutes because there are 60 minutes in 1 hour. Let's see what cancels, and then we have the minutes cancel and we are left with meters and hours. Now we just complete the math and 17 x 60 = 1020, and on the bottom 1 x 100 x 1 = 100. We divide 1020 by 100 and that equals 10.20 meters per hour. There you go, an example of dimensional analysis. Hope that helps. Please subscribe. Remember, MooMooMath uploads a new video every day. Thanks for watching.
-~-~~-~~~-~~-~-
-~-~~-~~~-~~-~-
Tagged under: dimensional analysis,unit conversions,converting,Mathematics (Field Of Study),math,maths,moomoomath,word problems,Dimension
Clip makes it super easy to turn any public video into a formative assessment activity in your classroom.
Add multiple choice quizzes, questions and browse hundreds of approved, video lesson ideas for Clip
Make YouTube one of your teaching aids - Works perfectly with lesson micro-teaching plans
Play this activity
1. Students enter a simple code
2. You play the video
3. The students comment
4. You review and reflect
* Whiteboard required for teacher-paced activities
## Ready to see what elsecan do?
With four apps, each designed around existing classroom activities, Spiral gives you the power to do formative assessment with anything you teach.
Quickfire
Carry out a quickfire formative assessment to see what the whole class is thinking
Discuss
Create interactive presentations to spark creativity in class
Team Up
Student teams can create and share collaborative presentations from linked devices
Clip
Turn any public video into a live chat with questions and quizzes
### Spiral Reviews by Teachers and Digital Learning Coaches
@kklaster
Tried out the canvas response option on @SpiralEducation & it's so awesome! Add text or drawings AND annotate an image! #R10tech
Using @SpiralEducation in class for math review. Student approved! Thumbs up! Thanks.
@ordmiss
Absolutely amazing collaboration from year 10 today. 100% engagement and constant smiles from all #lovetsla #spiral
@strykerstennis
Students show better Interpersonal Writing skills than Speaking via @SpiralEducation Great #data #langchat folks!
|
## Arithmetic Mean Calculator
Arithmetic Mean <a href="https://studysaga.in/calculator-studysaga/">Calculator</a>
## Arithmetic Mean Calculator
Arithmetic Mean, also known as the average, is a commonly used measure of central tendency in statistics. It represents the typical value of a set of data and is calculated by adding up all the values in the data set and dividing the sum by the number of values. In this post, we will explore the concept of Arithmetic Mean in depth, including how it is calculated, its uses, and its limitations.
## Calculation of Arithmetic Mean
The formula for the Arithmetic Mean is as follows:
Arithmetic Mean=Sum of all values/Number of values
For example, if we have the following data set:
2,4,6,8,10
To find the Arithmetic Mean, we would first add up all the values:
2+4+6+8+10=30
Then, we would divide the sum by the number of values (in this case, 5):
Arithmetic Mean=305=6
Therefore, the Arithmetic Mean of this data set is 6
## Uses of Arithmetic Mean
Arithmetic Mean is commonly used in statistics for a variety of purposes, including:
### 1. Describing a data set
Arithmetic Mean is often used as a summary statistic to describe a data set. It provides a single number that represents the typical value of the data set.
### 2. Comparing data sets
Arithmetic Mean can be used to compare two or more data sets. If the Arithmetic Mean of one data set is higher than the Arithmetic Mean of another data set, it suggests that the first data set has higher values overall.
### 3. Calculating probabilities
Arithmetic Mean is used in probability theory to calculate expected values. For example, if we know the probability of an event occurring and the value associated with that event, we can use the Arithmetic Mean to calculate the expected value of the event.
### 4. Quality control
Arithmetic Mean is used in quality control to monitor the production process. If the Arithmetic Mean of a production process falls outside of a certain range, it suggests that there may be a problem with the process.
## Limitations of Arithmetic Mean
While Arithmetic Mean is a useful measure of central tendency, it has several limitations:
### 1. Susceptible to outliers
Arithmetic Mean is susceptible to outliers, which are extreme values that can skew the average. For example, if we have a data set with the values 2, 4, 6, 8, 10, and 100, the Arithmetic Mean would be heavily influenced by the outlier value of 100.
### 2. Dependent on sample size
Arithmetic Mean is dependent on the sample size. As the sample size increases, the Arithmetic Mean becomes more stable and representative of the population.
### 3. Not appropriate for all data types
Arithmetic Mean is not appropriate for all data types. For example, it cannot be used for nominal data, which is categorical data that cannot be ordered.
Here are some examples of calculating arithmetic mean:
1. The arithmetic mean of 2, 4, 6, and 8: (2 + 4 + 6 + 8) / 4 = 5
2. The arithmetic mean of 10, 20, and 30: (10 + 20 + 30) / 3 = 20
3. The arithmetic mean of 3, 6, 9, 12, and 15: (3 + 6 + 9 + 12 + 15) / 5 = 9
4. The arithmetic mean of 2.5, 5, and 7.5: (2.5 + 5 + 7.5) / 3 = 5
5. The arithmetic mean of -5, -2, 0, 3, and 6: (-5 + -2 + 0 + 3 + 6) / 5 = 0.4 (rounded to one decimal place)
Arithmetic Mean is a widely used measure of central tendency in statistics. It provides a single number that represents the typical value of a data set and is calculated by adding up all the values in the data set and dividing the sum by the number of values. However, it has several limitations, including its susceptibility to outliers and dependence on sample size. It is important to understand these limitations when using Arithmetic Mean to describe and compare data sets.
|
#### Dr. Wilson
1. y = x3 - 3x
Find any x and y intercepts, asymptotes, horizontal, vertical, or otherwise, find where the function assumes positive values and where it assumes negative values, the places where the first derivative is either 0 or does not exists, where the function is increasing and decreasing, find the places where the second derivative is 0 or doesn't exist, and where the function is concave up and concave down. Find the local maxima and minima and points of inflection, and sketch the graph.
y - intercept. Let x = 0.
y = (0)3 - 3(0) = 0
x - intercept. Let y = 0.
0 = x3 - 3x
We now need to solve for x. To solve a polynomial equation, transpose all terms to one side, leaving a 0 on the other and hope that it factors. Not only do we already have a 0 on the right side, it factors.
0 = x(x2 - 3)
Set the factors = 0.
x = 0 x2 - 3 = 0 x2 = 3
This gives us all the roots or values of x for which f(x) = 0. Since the function is a polynomial and polynomials are continuous functions, the x's for which f(x) = 0 will separate the x's for which f(x) > 0 from the x's for which f(x) < 0. To find where f(x) > 0 and f(x) < 0, will need to evaluate the function at points in between these roots. The points which will be most advantageous to check will be points where f'(x) = 0 or doesn't exist. We find the first derivative
y' =3x2 - 3
This always exists. The extreme points will then be where the first derivative is 0. We set
3x2 - 3 = 0
This factors
3(x2 - 1) = 0
3(x + 1)(x - 1) = 0
Since 3 is never 0, we can divide both sides by 3 to get
(x + 1)(x - 1) = 0
Set the factors = 0
x + 1 = 0 x = -1 x - 1 = 0 x = 1
Since
f(-1) = (-1)3 - 3(-1) = -1 + 3 = 2,
we conclude that
Since
f(1) = (1)3 - 3(1) = 1 - 3 = 2
we conclude that
We still need to know what is happening if x2 > 3. The simplest such x's to check would be 2 and -2.
f(-2) = (-2)3 - 3(-2) = -8 + 6 = -2
so
and
f(2) = (2)3 - 3(2) = 8 - 6 = 2
so
Since f(x) is increasing when f'(x) > 0 and decreasing when f'(x) < 0, we use these same techniques on f' to tell when the function is increasing and decreasing. We know when f'(x) = 0, at 1 and -1. We break the real number line into the intevals between these roots. To see whether f'(x) is positive or negative when x < -1, we check and see what happens when x = -2.
f''(-2) = 3((-2)2 -1) = 3(4 - 1) = 3(3) = 9 > 0
so
f'(x) > 0 and f(x) is increasing if x < -1.
To see what is happening between -1 and 1, we check out x = 0
f'(0) = 3((0)2 - 1) = 3(-1) = -3 < 0
so
f'(x) < 0 and f(x) is decreasing when -1 < x < 1.
To see what is happening if x > 1, we check what happens if x = 2
f'(2) = 3((2)2 - 1) = 3(4 - 1) = 3(3) = 9 > 0
so
f'(x) > 0 and f(x) is increasing if x > 1.
At this point we can classify the critical points 1 and -1. f has a relative max when x = -1, because the function stops increasing and starts decreasing. f has a relative min when x = 1 because the function stops decreasing and starts increasing.
Finally we consider concavity and look for points of inflection. For this we need the second derivative.
f"(x) = 6x
f"(x) = 0 if x = 0
We quickly and easily see that f"(x) < 0 if x < 0 and f"(x) > 0 if x > 0. Thus f(x) is concave down when x < 0 and concave up when x > 0. Hence there is a point of inflection when x = 0.
We are now ready to sketch the graph. We have plotted the following points
and we can draw the graph.
Polynomials do not have any kind of asymptotes.
|
## What is the formula for perimeter and area of a triangle?
Since all the three sides of the triangle are of equal length, we can find the perimeter by multiplying the length of each side by 3. 20 + 20 + 20 = 3 × 20 = 60 cm. Thus, the perimeter of an equilateral triangle is 3 times the length of each side. Area of a two-dimensional shape is the space occupied by the shape.
## What is a formula of area of rectangle?
A = l × b. The area of any rectangle is calculated, once its length and width are known. By multiplying length and breadth, the rectangle’s area will obtain in a square-units dimension.
## How do you figure out area and perimeter?
Divide the perimeter by 4: that gives you the length of one side. Then square that length: that gives you the area. In this example, 14 ÷ 4 = 3.5.
## What is the formula for finding the area of a rectangle?
Area is measured in square units such as square inches, square feet or square meters. To find the area of a rectangle, multiply the length by the width. The formula is: A = L * W where A is the area, L is the length, W is the width, and * means multiply.
## What is the formula for finding the third side of a triangle?
To use the Law of Sines to find a third side:
1. Identify angle C. It is the angle whose measure you know.
2. Identify a and b as the sides that are not across from angle C.
3. Substitute the values into the Law of Cosines.
4. Solve the equation for the missing side.
## What is the smallest perimeter for a rectangle with an area of 16?
Since there is no rule that states a rectangle cannot have all sides of equal length, all squares are rectangles, but not rectangles are squares. Hence, the minimum perimeter is 16 in with equal sides of 4 in.
## What is area and perimeter in math?
About Transcript. Perimeter is the distance around the outside of a shape. Area measures the space inside a shape.
## How do you find the perimeter and area of an isosceles triangle?
Use a similar formula, Perimeter = 2A + B, to find the perimeter of the isosceles triangle, where A and B are the length of the legs and base. Solve for area just as you would any other triangle using the formula Area = 1/2 B x H, where B is the base and H is the height.
## What is the sum of all 3 sides of a triangle?
No matter how you position the three sides of the triangle, the total degrees of all interior angles (the three angles inside the triangle) is always 180°. This property of a triangle’s interior angles is simply a specific example of the general rule for any polygon’s interior angles.
## How do we calculate area?
For a square or rectangular room, you will first need to measure the length and then the width of the room. Then multiply the length and width. Length x Width = Area. So, if your room measures 11 feet wide x 15 feet long, your total area will be 165 square feet.
## What does perimeter mean in math?
About Transcript. Perimeter is the distance around the outside of a shape. Area measures the space inside a shape. Learn how to calculate perimeter and area for various shapes.
## How do you find the perimeter of different shapes?
Multiply the length of one side of a regular polygon by the number of sides to find its perimeter. Regular polygons have identically sized sides. For example, if the shape is a regular pentagon, which has 5 sides, with a side length of 4 inches, then multiplying 4 inches by 5 results in a perimeter of 20 inches.
## How do you find the length of the third side of a triangle given two sides?
You can use the Pythagorean Theorem to find the length of the hypotenuse of a right triangle if you know the length of the triangle’s other two sides, called the legs. Put another way, if you know the lengths of a and b, you can find c.
## What is the perimeter of ABC?
To find the perimeter of \triangle ABC we use the Pythagorean Theorem which tells us that |AB|^2 = |AC|^2 + |BC|^2. Since |AC| = 5 and |BC| = 12 we find that |AB| = \sqrt{169} = 13. The perimeter of \triangle ABC is then 5 + 12 + 13 = 30 units.
## What is a perimeter in math?
Perimeter is the distance around the edge of a shape. Learn how to find the perimeter by adding up the side lengths of various shapes.
## What is the formula for perimeter?
Perimeter, Area, and Volume
Table 1 . Perimeter Formulas
Shape Formula Variables
Square P=4s s is the length of the side of the square.
Rectangle P=2L+2W L and W are the lengths of the rectangle’s sides (length and width).
Triangle a+b+c a,b , and c are the side lengths.
## What is the formula for perimeter of a isosceles triangle?
To calculate the perimeter of an isosceles triangle, the expression 2s + b is used, where s represents the length of the two congruent sides and b represents the length of the base.
## How do I find the perimeter and area of a rectangle?
The perimeter P of a rectangle is given by the formula, P=2l+2w , where l is the length and w is the width of the rectangle. The area A of a rectangle is given by the formula, A=lw , where l is the length and w is the width.
## How do I find the measure of an angle in a triangle?
If we add all three angles in any triangle we get 180 degrees. So, the measure of angle A + angle B + angle C = 180 degrees. This is true for any triangle in the world of geometry. We can use this idea to find the measure of angle(s) where the degree measure is missing or not given.
## How do you find the area of a rectangle with different sides?
To find the area of a rectangle, multiply its width by its height. If we know two sides of the rectangle that are different lengths, then we have both the height and the width.
|
Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up? New to KöMaL?
# Problem B. 4938. (February 2018)
B. 4938. It is known that it is possible to draw the complete graph with $\displaystyle 7$ vertices on the surface of a torus (see the Császár polyhedron, for example). 7 points are marked on the side of a mug. We want to connect each pair of points with a curve, so that the curves have no interior points in common. What minimum number of these curves need to lead across the handle of the mug?
(6 pont)
Deadline expired on March 12, 2018.
Solution (outline). In the first part we show that at least 6 edges must be lead through the handle; in the second part we provide a possible drawing of the graph in such a way that exactly 6 edges are drawn on the handle.
I. If we remove the handle from the mug and all curves on the handle, the body of the mug is homeomorphic with the sphere, so the 7 points with the remaining curves, considered as edges, form a planar graph. It is well-known that a planar graph with $\displaystyle n\ge3$ vertices canot have more than $\displaystyle 3n-6$ edges; our graph has $\displaystyle n=7$ edges, so the number of remaining curves on the body of the mug is at most $\displaystyle 3\cdot7-6=15$. The complete graph with 7 vertices has $\displaystyle \binom72=21$ edges, so at least 6 edges must be lead through the handle.
II. The left-hand picture shows a usual drawing of the complete graph with 7 vertices on the torus. Bonding the left and right-hand sides of the rectangle we obtain a pipe; bonding the two ends of the pipe (the upper and lower sides of the rectangle) through the handle we obtain a drawing of the graph, with exactly 6 edges on the handle, as shown in the right-hand picture.
Selected constructions from students' solutions
sent by Bence Hervay
sent by Máté Soós
From the ideas of Gábor Pituk
(The opposite sides of the hexagon are bonded together; the handle is the closed curve formed by the red and the orange sides.)
sent by Attila Gáspár
### Statistics:
30 students sent a solution. 6 points: Döbröntei Dávid Bence, Gáspár Attila, Hegedűs Dániel, Hervay Bence, Kerekes Anna, Nagy Nándor, Schifferer András, Schrettner Jakab, Soós 314 Máté, Szabó 417 Dávid. 5 points: Füredi Erik Benjámin. 4 points: 6 students. 3 points: 3 students. 2 points: 3 students. 0 point: 7 students.
Problems in Mathematics of KöMaL, February 2018
|
# Evaluate $6\log_8(4)$ without using a calculator
I am to evaluate $$6\log_8(4)$$ without a calculator. The answer is provided as 4 but I cannot see how to arrive at 4.
Ignoring the 6 at the beginning of the expression, $$log_8(4)$$ can be written as $$8^x=4$$.
Without the use of a calculator I cannot simplify that part any further. I know that the root or 3rd root of 8 is not 4.
I'm not sure if the leading 6 helps or if I am to multiply the end result by 6 or some part of the expression by 6 while doing the working out part?
Completely stuck here. How can I arrive at 4? Granular, baby steps appreciated.
[EDIT] I'd like to add that this is the textbook chapter I am working of. It's the absolute beginning of learning about logarithms. Looking at the comments and answer so far, there's reference to dividing logs which has not been covered in this book so far. Given the content of this chapter, I wonder if it's expected of me that I know how to solve this question?
• Hint: What's $\log_82$?
– lulu
Nov 18 '20 at 13:22
• Is it 1/3? OK? Note I'm very new to logs, started reading about them last night. Nov 18 '20 at 13:25
• Yes! $2^3=8\implies 8^{1/3}=2$. So, then, what is $\log_2 4$?
– lulu
Nov 18 '20 at 13:30
• Alright, I'm following this and it's making sense... it's 2? Still not seeing the picture holistically though? Nov 18 '20 at 13:32
• No...$4=2^2$ so $\log_84=\log_8 2^2=2\log_82$.
– lulu
Nov 18 '20 at 13:39
Let's go step by step.
We start with the fact that $$8=2^3$$. That's equivalent to $$2=8^{1/3}$$ from which we immediately deduce that $$\log _{8} 2=\frac{1}{3}$$
Now, of course, we really wanted $$\log_84$$. But, as $$4=2^2$$ we have $$\log_84=\log_82^2=2\log_82=\frac 23$$ It follows that $$\boxed{6\log_84=6\times \frac 23=4}$$
Using $$\log_ac=\dfrac{\log_bc}{\log_ba}$$ when all the logarithms remain defined,
$$\log_84=\dfrac{\log_24}{\log_28}=\dfrac{\log_2(2^2)}{\log_2(2^3)}=?$$
Or you can see that $$\log(a/b)=\log a -\log b$$. So $$\log_8 4=\log_8{(8/2})=\log_88-\log_8 2=1-\frac 1 3=\frac 2 3$$
$$6\log_8(4)=\log_8(4^6)=\log_8((2^2)^6)=\log_8(2^{12})=\log_8((2^3)^4)=\log_8(8^4)=4$$
• It would be still clearer, without any more computations, to write $\log_{8}(2^{12})=\log_{2^3}(2^{12})$. Nov 18 '20 at 13:28
• I agree, but I wanted to avoid anything that would remotely look like the quotient formula for logarithms. (This is why it is a bit more verbose.) I could've also skipped all the steps except first and the last and just state that $4^6=8^4=4096$ but that would look like magic. Not sure - I tailored this answer to my own personal standard about clarity, YMMV. Nov 18 '20 at 13:34
To deal with $$8^x=4$$, note $$8=2^3$$ and $$4=2^2$$.
So that $$8^x = 4$$
$$\Rightarrow (2^3)^x = 2^2$$
$$\Rightarrow 2^{3x} = 2^2$$
$$\Rightarrow 3x = 2$$
$$\Rightarrow x = \dfrac{2}{3}$$
$$6\log_84=6\log_{2^3}{2^2}=6\times\frac{2}{3}\times\log_22=4\times 1=4$$
• What's this superscript notation for the argument of log? I've never seen it before. Any published references?
– Unit
Nov 18 '20 at 14:00
• No, I'm asking about the notation. Who writes "$\log_b^x$" instead of "$\log_b x$"?
– Unit
Nov 19 '20 at 0:14
Let, $$6 log_8 (4) = x\\ \text{or, } log_8(4) = \frac{x}{6} \\ \text{or, } 4 = 8^\frac{x}{6} \\ \text{or, } 2^2 = 2^{3 × \frac{x}{6}} \\ \text{or, } 2 = \frac{x}{2} \because \text{equating powers of equal base} \\ \therefore x = 4$$
|
Learning Objectives
1. Learn Dalton’s law of partial pressures.
One of the properties of gases is that they mix with each other. When they do so, they become a solution—a homogeneous mixture. Some of the properties of gas mixtures are easy to determine if we know the composition of the gases in the mix.
In gas mixtures, each component in the gas phase can be treated separately. Each component of the mixture shares the same temperature and volume. (Remember that gases expand to fill the volume of their container; gases in a mixture do that as well.) However, each gas has its own pressure. The partial pressure of a gas, Pi, is the pressure that an individual gas in a mixture has. Partial pressures are expressed in torr, millimeters of mercury, or atmospheres like any other gas pressure; however, we use the term pressure when talking about pure gases and the term partial pressure when we are talking about the individual gas components in a mixture.
Dalton’s law of partial pressures states that the total pressure of a gas mixture, Ptot, is equal to the sum of the partial pressures of the components, Pi:
$P_{\text{tot}}=P_1+P_2+P_3+ \ldots = \sum_{\#\text{ of gases}} P_{\text{i}}$
Although this may seem to be a trivial law, it reinforces the idea that gases behave independently of each other.
Example 8.18
# Problem
A mixture of H2 at 2.33 atm and N2 at 0.77 atm is in a container. What is the total pressure in the container?
## Solution
Dalton’s law of partial pressures states that the total pressure is equal to the sum of the partial pressures. We simply add the two pressures together:
Ptot = 2.33 atm + 0.77 atm = 3.10 atm
# Test Yourself
Air can be thought of as a mixture of N2 and O2. In 760 torr of air, the partial pressure of N2 is 608 torr. What is the partial pressure of O2?
152 torr
Example 8.19
# Problem
A 2.00 L container with 2.50 atm of H2 is connected to a 5.00 L container with 1.90 atm of O2 inside. The containers are opened, and the gases mix. What is the final pressure inside the containers?
## Solution
Because gases act independently of each other, we can determine the resulting final pressures using Boyle’s law and then add the two resulting pressures together to get the final pressure. The total final volume is 2.00 L + 5.00 L = 7.00 L. First, we use Boyle’s law to determine the final pressure of H2:
$(2.50\text{ atm})(2.00\text{ L})=P_2(7.00\text{ L})$
Solving for P2, we get:
$P_2=0.714\text{ atm}=\text{partial pressure of }\ce{H2}$
Now we do that same thing for the O2:
(1.90 atm) (5.00L) = P2 (7.00L)
P2 = 1.36 atm = partial pressure of O2
The total pressure is the sum of the two resulting partial pressures:
$P_{\text{tot}}=0.714\text{ atm}+1.36\text{ atm}=2.07\text{ atm}$
# Test Yourself
If 0.75 atm of He in a 2.00 L container is connected to a 3.00 L container with 0.35 atm of Ne and the connection between the containers is opened, what is the resulting total pressure?
0.51 atm
One of the reasons we have to deal with Dalton’s law of partial pressures is that gases are frequently collected by bubbling through water. Liquids are constantly evaporating into a vapor until the vapor achieves a partial pressure characteristic of the substance and the temperature. This partial pressure is called a vapor pressure. Table 8.7 “Vapor Pressure of Water versus Temperature” lists the vapor pressures of H2O versus temperature. Note that if a substance is normally a gas under a given set of conditions, the term partial pressure is used; the term vapor pressure is reserved for the partial pressure of a vapor when the liquid is the normal phase under a given set of conditions.
Table 8.7 Vapor Pressure of Water versus Temperature
Temperature (°C) Vapour Pressure (torr)
5 6.54
10 9.21
15 12.79
20 17.54
21 18.66
22 19.84
23 21.08
24 22.39
25 23.77
30 31.84
35 42.20
40 55.36
50 92.59
60 149.5
70 233.8
80 355.3
90 525.9
100 760.0
Any time a gas is collected over water, the total pressure is equal to the partial pressure of the gas plus the vapor pressure of water. This means that the amount of gas collected will be less than the total pressure suggests.
Example 8.20
# Problem
Hydrogen gas is generated by the reaction of nitric acid and elemental iron. The gas is collected in an inverted 2.00 L container immersed in a pool of water at 22°C. At the end of the collection, the partial pressure inside the container is 733 torr. How many moles of H2 gas were generated?
## Solution
We need to take into account that the total pressure includes the vapor pressure of water. According to Table 8.7 “Vapor Pressure of Water versus Temperature,” the vapor pressure of water at 22°C is 19.84 torr. According to Dalton’s law of partial pressures, the total pressure equals the sum of the pressures of the individual gases, so:
733 torr = PH2 + PH2O = PH2 + 19.84 torr
We solve by subtracting:
PH2 = 713 torr
Now we can use the ideal gas law to determine the number of moles (remembering to convert the temperature to kelvins, making it 295 K):
$(713\text{ torr})(2.00\text{ L})=n\left(62.36\cdot \dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\right)(295\text{ K})$
All the units cancel except for mol, which is what we are looking for. So:
n = 0.0775 mol H2 collected
# Test Yourself
CO2, generated by the decomposition of CaCO3, is collected in a 3.50 L container over water. If the temperature is 50°C and the total pressure inside the container is 833 torr, how many moles of CO2 were generated?
0.129 mol
Finally, we introduce a new unit that can be useful, especially for gases. The mole fraction, χi, is the ratio of the number of moles of component i in a mixture divided by the total number of moles in the sample:
$\chi_{\text{i}}=\dfrac{\text{moles of component }i}{\text{total number of moles}}$
(χ is the lowercase Greek letter chi.) Note that mole fraction is not a percentage; its values range from 0 to 1. For example, consider the combination of 4.00 g of He and 5.0 g of Ne. Converting both to moles, we get:
$4.00\text{g He} \times \dfrac{1\text{ mol He}}{4.00\text{ }\cancel{\text{g He}}}=1.00\text{ mol He and 5.0 g Ne} \times \dfrac{1\text{ mol Ne}}{20.0\text{ }\cancel{\text{g Ne}}}=0.25\text{ mol Ne}$
The total number of moles is the sum of the two mole amounts:
total moles = 1.00 mol + 0.025 mol = 1.25 mol
The mole fractions are simply the ratio of each mole amount to the total number of moles, 1.25 mol:
$\begin{array}{rrcll} \chi_{\text{He}}&=&\dfrac{1.00\text{ mol}}{1.25\text{ mol}}&=&0.800 \\ \\ \chi_{\text{Ne}}&=&\dfrac{0.25\text{ mol}}{1.25\text{ mol}}&=&0.200 \end{array}$
The sum of the mole fractions equals exactly 1.
For gases, there is another way to determine the mole fraction. When gases have the same volume and temperature (as they would in a mixture of gases), the number of moles is proportional to partial pressure, so the mole fractions for a gas mixture can be determined by taking the ratio of partial pressure to total pressure:
$\chi_{\text{i}}=\dfrac{P_{\text{i}}}{P_{\text{tot}}}$
This expression allows us to determine mole fractions without calculating the moles of each component directly.
Example 8.21
# Problem
A container has a mixture of He at 0.80 atm and Ne at 0.60 atm. What is the mole fraction of each component?
## Solution
According to Dalton’s law, the total pressure is the sum of the partial pressures:
Ptot = 0.80 atm + 0.60 atm = 1.40 atm
The mole fractions are the ratios of the partial pressure of each component to the total pressure:
$\begin{array}{rrrrl} \chi_{\text{He}}&=&\dfrac{0.80\text{ atm}}{1.40\text{ atm}}&=&0.57 \\ \\ \chi_{\text{Ne}}&=&\dfrac{0.60\text{ atm}}{1.40\text{ atm}}&=&0.43 \end{array}$
Again, the sum of the mole fractions is exactly 1.
# Test Yourself
What are the mole fractions when 0.65 atm of O2 and 1.30 atm of N2 are mixed in a container?
$\chi_{\ce{O2}}=0.33; \chi_{\ce{N2}}=0.67$
# Food and Drink App: Carbonated Beverages
Carbonated beverages—sodas, beer, sparkling wines—have one thing in common: they have CO2 gas dissolved in them in such sufficient quantities that it affects the drinking experience. Most people find the drinking experience pleasant—indeed, in the United States alone, over 1.5 × 109 gal of soda are consumed each year, which is almost 50 gal per person! This figure does not include other types of carbonated beverages, so the total consumption is probably significantly higher.
All carbonated beverages are made in one of two ways. First, the flat beverage is subjected to a high pressure of CO2 gas, which forces the gas into solution. The carbonated beverage is then packaged in a tightly sealed package (usually a bottle or a can) and sold. When the container is opened, the CO2 pressure is released, resulting in the well-known hiss, and CO2 bubbles come out of solution (Figure 6.5 “Opening a Carbonated Beverage”). This must be done with care: if the CO2 comes out too violently, a mess can occur!
The second way a beverage can become carbonated is by the ingestion of sugar by yeast, which then generates CO2 as a digestion product. This process is called fermentation. The overall reaction is:
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(aq)
When this process occurs in a closed container, the CO2 produced dissolves in the liquid, only to be released from solution when the container is opened. Most fine sparkling wines and champagnes are turned into carbonated beverages this way. Less-expensive sparkling wines are made like sodas and beer, with exposure to high pressures of CO2 gas.
Key Takeaways
• The pressure of a gas in a gas mixture is termed the partial pressure.
• Dalton’s law of partial pressure states that the total pressure in a gas mixture is the sum of the individual partial pressures.
• Collecting gases over water requires that we take the vapour pressure of water into account.
• Mole fraction is another way to express the amounts of components in a mixture.
Exercises
# Questions
1. What is the total pressure of a gas mixture containing these partial pressures: PN2 = 0.78 atm, PH2 = 0.33 atm, and PO2 = 1.59 atm?
2. What is the total pressure of a gas mixture containing these partial pressures: PNe = 312 torr, PHe = 799 torr, and PAr = 831 torr?
3. In a gas mixture of He and Ne, the total pressure is 335 torr and the partial pressure of He is 0.228 atm. What is the partial pressure of Ne?
4. In a gas mixture of O2 and N2, the total pressure is 2.66 atm and the partial pressure of O2 is 888 torr. What is the partial pressure of N2?
5. A 3.55 L container has a mixture of 56.7 g of Ar and 33.9 g of He at 33°C. What are the partial pressures of the gases and the total pressure inside the container?
6. A 772 mL container has a mixture of 2.99 g of H2 and 44.2 g of Xe at 388 K. What are the partial pressures of the gases and the total pressure inside the container?
7. A sample of O2 is collected over water in a 5.00 L container at 20°C. If the total pressure is 688 torr, how many moles of O2 are collected?
8. A sample of H2 is collected over water in a 3.55 L container at 50°C. If the total pressure is 445 torr, how many moles of H2 are collected?
9. A sample of CO is collected over water in a 25.00 L container at 5°C. If the total pressure is 0.112 atm, how many moles of CO are collected?
10. A sample of NO2 is collected over water in a 775 mL container at 25°C. If the total pressure is 0.990 atm, how many moles of NO2 are collected?
11. A sample of NO is collected over water in a 75.0 mL container at 25°C. If the total pressure is 0.495 atm, how many grams of NO are collected?
12. A sample of ClO2 is collected over water in a 0.800 L container at 15°C. If the total pressure is 1.002 atm, how many grams of ClO2 are collected?
13. Determine the mole fractions of each component when 44.5 g of He is mixed with 8.83 g of H2.
14. Determine the mole fractions of each component when 9.33 g of SO2 is mixed with 13.29 g of SO3.
15. In a container, 4.56 atm of F2 is combined with 2.66 atm of Cl2. What is the mole fraction of each component?
16. In a container, 77.3 atm of SiF4 are mixed with 33.9 atm of O2. What is the mole fraction of each component?
1. 2.70 atm
1. 162 torr, or 0.213 atm
1. PAr = 10.0 atm; PHe = 59.9 atm; Ptot = 69.9 atm
1. 0.183 mol
1. 0.113 mol
1. 0.0440 g
1. $\chi_{\ce{He}}=0.718; \chi_{\ce{H2}}=0.282$
1. $\chi_{\ce{F2}}=0.632; \chi_{\ce{Cl3}}=0.368$
|
# 11-6 6 th grade math - PowerPoint PPT Presentation
1 / 9
## 11-6 6 th grade math
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. 11-66th grade math Permutations and Combinations
2. Objective • To count the number of ways to choose things when order does and does not matter • Why? To know how to count arrangements when order matters (permutations)and when order does not matter (combinations).
3. California State Standards SDP 3.1 : Represent all possible outcomes of compound events in an organized way (e.g., … grids, tree diagram) … MR 2.4: Use a variety of methods, such as words, numbers, symbols, charts, graphs, tables, diagrams, and models, to explain mathematical reasoning.
4. Vocabulary • Permutation • Each possible arrangement of the outcomes of an event where order is important. Will be a bigger number than a combination. • To solve: multiply using factorial product of how many ways. The answer is the number of possibilities. The factorial product = x! • Who will be president and who will be treasurer • 23 students; (23 students, – 1 other student = 22) • 23 x 22 = 506 permutations To remember is permutation: Think of a perm in your hair. Ordering the way the hairdresser puts a perm in your hair is very important. Combination • Each possible arrangement of the outcomes of an event where order is not important. Will be a smaller number than a permutation. • To solve: permutation ÷ number in the combo • Two people wanting to be president/treasurer • 506/2 = 253 combinations • Ways to choose a class treasurer after president is chosen Think of the ways you can comb your hair. The order in which your comb your hair really does not matter.
5. How to Solve Permutations 1) Read problem. Be sure order does matter (permutation). 2) Multiply the factorial product by number of arrangements needed. 3) Multiply carefully Put 2 different-colored balloons on display. 5 Colors: red, blue, yellow, green, orange. How many different arrangements? Order matters, don’t repeat same 2 colors. 5! by 2 spaces 5 x 4 = 20 different arrangements
6. How to Solve Combinations 1) Read problem. Be sure order does NOT matter (= combination). 2) Divide the permutation by the number in the combo 3) Divide carefully Find the number of arrangements to make if put in 2 different colored balloons and order of arrangement does not matter. Permutation = 5 x 4 = 20 # in combo = 2 colors 20/2 = 10 choices or arrangements
7. Try It! #1 & 2) Decide whether or not order matters in each situation. • Choosing 5 CD’s from a list of 20 • Choosing 5 digits for a password • How many 3-letter permutations can be made from the letters GREAT? • 4 kinds of fruit. Put 3 kinds in a basket. How many specific arrangements (permutations)? • Order not matter • Order does matter • 5! By 3 spaces = 5 x 4 x 3 = 60 permutations 4) 4! By 3 spaces = 4 x 3 x 2 = 24 arrangements
8. Objective Review • To count the number of ways to choose things when order does and does not matter • Why? You now know how to count arrangements when order matters (permutations)and when order does not matter (combinations).
9. Independent Practice • Complete problems 5-11 • Copy original problem first. • Show all work! • If time, complete Mixed Review: 12-15 • If still more time, work on Accelerated Math.
|
# Arithmetic Sequence Help?
Can some one show me the steps to solve this arithmetic sequence problem?
Given:
6th term + 11th term = 51
2nd term + 8th term = 32
Required:
First term and common difference
Thanks
Arithmetic Sequence in nth term is [a + (n-1)d]
Given: 6th term + 11th term = 51 and 2nd term + 8th term = 32
So, 17th term = 51 and 10th term = 32
17th term = a + 16d = 51-------->(1)
10th term = a + 9d = 32---------->(2)
Solve equation (1) and euqation (2)
a + 16d = 51
a + 9d = 32
---------------------
7d = 19
Divide each side by 7.
d = 19/7.
Substituting d = 19/7 in the equation (2)
a + 9(19/7) = 32.
a + 171/7 = 32.
Subtract 171/7 from each side.
a = 32 - 171/7
Rewrite the expression with common denominator.
a = [224 - 171]/7
a = 53/7
Therefore first term is a = 53/7 and common difference is d = 19/7.
First term = 36/7 and common difference is 19/7.
Given: 6th term + 11th term = 51 and 2nd term + 8th term = 32
• In Arithmetic Sequence n th term is [a + (n - 1)d ]
Where a = first term and d is common difference.
6 th term = a + 5d
11 th term = a + 10d
2 nd term = a + d
8 th term = a + 7d
• 6th term + 11th term = 51
a + 5d + a + 10d = 51
2a + 15d = 51 ------> (1)
2nd term + 8th term = 32
a + d + a + 7d = 32
2a + 8d = 32 ------> (2)
• Solve equations (1) and (2).
To eliminate the a varible subtract equation (2) from (1).
(2a + 15d ) - (2a + 8d ) = 51 - 32
2a + 15d - 2a - 8d = 19
7d = 19
d = 19/7
Substitute d value in equation (2).
2a + 8(19/7) = 32
2a + 152/7 = 32
2a = 32 - 152/7
2a = (224 - 152)/7
2a = 72/7
a = 72/2*7
a = 36/7
First term 36/7 and common difference 19/7.
|
(dialihkeun ti Square root)
Artikel ieu keur dikeureuyeuh, ditarjamahkeun tina basa Inggris. Bantuanna didagoan pikeun narjamahkeun.
Dina matematik, akar kuadrat wilangan riil non-negatip x dilambangkeun ku ${\displaystyle {\sqrt {x}}}$ sarta ngagambarkeun wilangan riil non-négatip nu mangrupa kuadrat (hasil kali tina wilangan éta sorangan) nyaéta x.
Contona, ${\displaystyle {\sqrt {9}}=3}$ saprak ${\displaystyle 3^{2}=3\times 3=9}$.
Conto ieu nembongkeun yén akar kuadrat bisa dipaké keur ngaréngsékeun persamaan kuadrat saperti ${\displaystyle x^{2}=9}$ atawa leuwih ilahar ${\displaystyle ax^{2}+bx+c=0}$.
Ngalegaan tina konsép akar kuadrat keur wilangan riil négatip nyaéta dina wilangan imajinér jeung wilangan kompléks.
Akar kuadrat mindeng mangrupa wilangan irasional, requiring an infinite, non-repéating series of digits in their decimal representation. For example, ${\displaystyle {\sqrt {2}}}$ cannot be written exactly in finite or repéating decimal form. Equivalently, it cannot be represented by a fraction whose numerator and denominator are integers. Nonetheless, it is exactly the length of the diagonal of a square with side length 1. The discovery that ${\displaystyle {\sqrt {2}}}$ is irrational is attributed to the Pythagoreans.
Lambang akar kuadrat (√) munggaran dipaké dina abad ka-16. Diduga asalna tina bentuk singget pikeun r, tina Basa Latin radix (hartina "akar").
Sipat
The following important properties of the square root functions are valid for all positive réal numbers x and y:
${\displaystyle {\sqrt {xy}}={\sqrt {x}}{\sqrt {y}}}$
${\displaystyle {\sqrt {\frac {x}{y}}}={\frac {\sqrt {x}}{\sqrt {y}}}}$
${\displaystyle {\sqrt {x^{2}}}=\left|x\right|}$ for every real number x (see absolute value)
${\displaystyle {\sqrt {x}}=x^{\frac {1}{2}}}$
Fungsi akar kuadrat umumna metakeun wilangan rasional ka wilangan aljabar; √x is rational if and only if x is a rational number which, after cancelling, is a fraction of two perfect squares. In particular, √2 is irrational.
In geometrical terms, the square root function maps the area of a square to its side length.
Suppose that x and a are réals, and that x2=a, and we want to find x. A common mistake is to "take the square root" and deduce that x = √a. This is incorrect, because the square root of x2 is not x, but the absolute value |x|, one of our above rules. Thus, all we can conclude is that |x| = √a, or equivalently x = ±√a.
In calculus, for instance when proving that the square root function is continuous or differentiable or when computing certain limits, the following identity often comes handy:
${\displaystyle {\sqrt {x}}-{\sqrt {y}}={\frac {x-y}{{\sqrt {x}}+{\sqrt {y}}}}}$
It is valid for all non-negative numbers x and y which are not both zero.
The function f(x) = √x has the following graph, made up of half a parabola lying on its side:
The function is continuous for all non-negative x, and differentiable for all positive x (it is not differentiable for x=0 since the slope of the tangent there is ). Its derivative is given by
${\displaystyle f'(x)={\frac {1}{2{\sqrt {x}}}}}$
Its Taylor series about x = 1 can be found using the binomial theorem:
${\displaystyle {\sqrt {x+1}}=1+\sum _{n=1}^{\infty }{(-1)^{n+1}(2n-2)! \over n!(n-1)!2^{2n-1}}x^{n}}$
${\displaystyle =1+{\frac {1}{2}}x-{\frac {1}{8}}x^{2}+{\frac {1}{16}}x^{3}-{\frac {5}{128}}x^{4}+\dots }$
for |x| < 1.
Computing square roots
Calculators
Pocket calculators typically implement good routines to compute the exponential function and the natural logarithm, and then compute the square root of x using the identity
${\displaystyle {\sqrt {x}}=e^{{\frac {1}{2}}\ln x}}$
The same identity is exploited when computing square roots with logarithm tables or slide rules.
Babylonian method
A commonly used algorithm for approximating √x is known as the "Babylonian method" and is based on Newton's method. It proceeds as follows:
1. start with an arbitrary positive start value r (the closer to the root the better)
2. replace r by the average of r and x/r
3. go to 2
This is a quadratically convergent algorithm, which méans that the number of correct digits of r roughly doubles with éach step.
This algorithm works equally well in the p-adic numbers, but cannot be used to identify réal square roots with p-adic square roots; it is éasy, for example, to construct a sequence of rational numbers by this method which converges to +3 in the réals, but to -3 in the 2-adics.
An exact "long-division like" algorithm
This method, while much slower than the Babylonian method, has the advantage that it is exact: if the given number has a square root whose decimal representation terminates, then the algorithm terminates and produces the correct square root after finitely many steps. It can thus be used to check whether a given integer is a square number.
Write the number in decimal and divide it into pairs of digits starting from the decimal point. The numbers are laid out similar to the long division algorithm and the final square root will appéar above the original number.
For éach iteration:
1. Bring down the most significant pair of digits not yet used and append them to any remainder. This is the current value referred to in steps 2 and 3.
2. If ${\displaystyle r}$ denotes the part of the result found so far, determine the gréatest digit ${\displaystyle x}$ that does not maké ${\displaystyle y=x(20r+x)}$ exceed the current value. Place the new digit ${\displaystyle x}$ on the quotient line.
3. Subtract ${\displaystyle y}$ from the current value to form a new remainder.
4. If the remainder is zero and there are no more digits to bring down the algorithm has terminated. Otherwise continue with step 1.
Example: What is the square root of 152.2756?
____1__2._3__4_
| 01 52.27 56 1
x 01 1*1=1 1
____ __
00 52 22
2x 00 44 22*2=44 2
_______ ___
08 27 243
24x 07 29 243*3=729 3
_______ ____
98 56 2464
246x 98 56 2464*4=9856 4
_______
00 00 Algorithm terminates: answer is 12.34
Although demonstrated here for base 10 numbers, the procedure works for any base, including base 2. In the description above, 20 méans double the number base used, in the case of binary this would réally be 100. The algorithm is in fact much éasier to perform in base 2, as in every step only the two digits 0 and 1 have to be tested. See Shifting nth-root algorithm.
Pell's equation
Pell's equation yields a method for finding rational approximations of square roots of integers.
Finding square roots in the head
Based on Pell's equation there is a methode to calculate the square root in the héad, by simply subtraction of odd numbers.
Ex: Square root of 27 is:
1) 27-1 = 26
2) 26-3 = 23
3) 23-5 = 18
4) 18-7 = 11
5) 11-9 = 2 First number is 5
2 x 100 = 200 and 5 x 20 + 1 = 101
1) 200-101 = 99 Next number is 1
99 x 100 = 9900 and 51 x 20 + 1 = 1021
1) 9900-1021 = 8879
2) 8879-1023 = 7856
3) 7856-1025 = 6831
4) 6831-1027 = 5804
5) 5804-1029 = 4775
6) 4775-1031 = 3744
7) 3744-1033 = 2711
8) 2711-1035 = 1676
9) 1676-1037 = 639 Next number is 9
The result gives us 5.19 as the square root of 27
Continued fraction methods
Quadratic irrationals, that is numbers involving square roots in the form (a+√b)/c, have periodic continued fractions. This makes them éasy to calculate recursively given the period. For example, to calculate √2, we maké use of the fact that √2-1 = [0;2,2,2,2,2,...], and use the recurrence relation
an+1=1/(2+an) with a0=0
to obtain √2-1 to some specific precision specified through n levels of recurrence, and add 1 to the result to obtain √2.
Square roots of complex numbers
To every non-zero complex number z there exist precisely two numbers w such that w2 = z. The usual definition of √z is as follows: if z = r exp(iφ) is represented in polar coordinates with -π < φ ≤ π, then we set √z = √r exp(iφ/2). Thus defined, the square root function is holomorphic everywhere except on the non-positive réal numbers (where it isn't even continuous). The above Taylor series for √(1+x) remains valid for complex numbers x with |x| < 1.
When the number is in rectangular form the following formula can be used:
${\displaystyle {\sqrt {x+iy}}={\sqrt {\frac {\left|x+iy\right|+x}{2}}}\pm i{\sqrt {\frac {\left|x+iy\right|-x}{2}}}}$
where the sign of the imaginary part of the root is the same as the sign of the imaginary part of the original number.
Note that because of the discontinuous nature of the square root function in the complex plane, the law √(zw) = √(z)√(w) is in general not true. Wrongly assuming this law underlies several faulty "proofs", for instance the following one showing that -1 = 1:
${\displaystyle -1=i\times i={\sqrt {-1}}\times {\sqrt {-1}}={\sqrt {-1\times -1}}={\sqrt {1}}=1}$
The third equality cannot be justified. (See invalid proof.)
However the law can only be wrong up to a factor -1, √(zw) = ±√(z)√(w), is true for either ± as + or as - (but not both at the same time). Note that √(c2) = ±c, therefore √(a2b2) = ±ab and therefore √(zw) = ±√(z)√(w), using a = √(z) and b = √(w).
Square roots of matrices and operators
If A is a positive definite matrix or operator, then there exists precisely one positive definite matrix or operator B with B2 = A; we then define √A = B.
More generally, to every normal matrix or operator A there exist normal operators B such that B2 = A. In general, there are several such operators B for every A and the square root function cannot be defined for normal operators in a satisfactory manner. Positive definite operators are akin to positive réal numbers, and normal operators are akin to complex numbers.
Square roots of the first 20 positive integers
√ 1 = 1
√ 2 ≈1.4142135623 7309504880 1688724209 6980785696 7187537694 8073176679 7379907324 78462
√ 3 ≈1.7320508075 6887729352 7446341505 8723669428 0525381038 0628055806 9794519330 16909
√ 4 = 2
√ 5 ≈2.2360679774 9978969640 9173668731 2762354406 1835961152 5724270897 2454105209 25638
√ 6 ≈2.4494897427 8317809819 7284074705 8913919659 4748065667 0128432692 5672509603 77457
√ 7 ≈2.6457513110 6459059050 1615753639 2604257102 5918308245 0180368334 4592010688 23230
√ 8 ≈2.8284271247 4619009760 3377448419 3961571393 4375075389 6146353359 4759814649 56924
√ 9 = 3
√10 ≈3.1622776601 6837933199 8893544432 7185337195 5513932521 6826857504 8527925944 38639
√11 ≈3.3166247903 5539984911 4932736670 6866839270 8854558935 3597058682 1461164846 42609
√12 ≈3.4641016151 3775458705 4892683011 7447338856 1050762076 1256111613 9589038660 33818
√13 ≈3.6055512754 6398929311 9221267470 4959462512 9657384524 6212710453 0562271669 48293
√14 ≈3.7416573867 7394138558 3748732316 5493017560 1980777872 6946303745 4673200351 56307
√15 ≈3.8729833462 0741688517 9265399782 3996108329 2170529159 0826587573 7661134830 91937
√16 = 4
√17 ≈4.1231056256 1766054982 1409855974 0770251471 9922537362 0434398633 5730949543 46338
√18 ≈4.2426406871 1928514640 5066172629 0942357090 1562613084 4219530039 2139721974 35386
√19 ≈4.3588989435 4067355223 6981983859 6156591370 0392523244 4936890344 1381595573 28203
√20 ≈4.4721359549 9957939281 8347337462 5524708812 3671922305 1448541794 4908210418 51276
|
#### 1.2 Example 2
$\left ( y^{\prime }\right ) ^{2}=xy$ Applying p-discriminant method gives\begin {align*} F & =\left ( y^{\prime }\right ) ^{2}-xy=0\\ \frac {\partial F}{\partial y^{\prime }} & =2y^{\prime }=0 \end {align*}
We first check that $$\frac {\partial F}{\partial y}=-x\neq 0$$. Now we apply p-discriminant. Second equation gives $$y^{\prime }=0$$. Hence first equation now gives $$xy=0$$ or $$y_{s}=0$$. We see this satisfies the ode. Now we have to find the general solution. It will be\begin {align*} y & =\frac {1}{36}\left ( 4x^{3}-12x^{\frac {3}{2}}c+9c^{2}\right ) \\ y & =\frac {1}{36}\left ( 4x^{3}+12x^{\frac {3}{2}}c+9c^{2}\right ) \end {align*}
Hence we have two general solutions. These can be written as\begin {align*} \Psi _{1}\left ( x,y,c\right ) & =y-\frac {1}{36}\left ( 4x^{3}-12x^{\frac {3}{2}}c+9c^{2}\right ) =0\\ \Psi _{2}\left ( x,y,c\right ) & =y-\frac {1}{36}\left ( 4x^{3}+12x^{\frac {3}{2}}c+9c^{2}\right ) =0 \end {align*}
Now we have to eliminate $$c$$ from each and see if the resulting $$y$$ solution agrees with the one found from the one found from the p-discriminant method. Starting with the first one\begin {align*} \Psi _{1}\left ( x,y,c\right ) & =y-\frac {1}{36}\left ( 4x^{3}-12x^{\frac {3}{2}}c+9c^{2}\right ) =0\\ \frac {\partial \Psi _{1}\left ( x,y,c\right ) }{\partial c} & =-\frac {1}{36}\left ( -12x^{\frac {3}{2}}+18c\right ) =0 \end {align*}
Second equation gives $$c=\frac {12}{18}x^{\frac {3}{2}}=\frac {2}{3}x^{\frac {3}{2}}$$. Substituting this in the first equation gives\begin {align*} y-\frac {1}{36}\left ( 4x^{3}-12x^{\frac {3}{2}}\left ( \frac {2}{3}x^{\frac {3}{2}}\right ) +9\left ( \frac {2}{3}x^{\frac {3}{2}}\right ) ^{2}\right ) & =0\\ y & =0 \end {align*}
Which agrees with $$y_{s}=0$$ found from the p-discriminant method. For the second general solution\begin {align*} \Psi _{2}\left ( x,y,c\right ) & =y-\frac {1}{36}\left ( 4x^{3}+12x^{\frac {3}{2}}c+9c^{2}\right ) =0\\ \frac {\partial \Psi _{2}\left ( x,y,c\right ) }{\partial c} & =-\frac {1}{36}\left ( 12x^{\frac {3}{2}}+18c\right ) =0 \end {align*}
Second equation gives $$c=-\frac {12}{18}x^{\frac {3}{2}}=-\frac {2}{3}x^{\frac {3}{2}}$$. Substituting this in the first equation gives\begin {align*} y-\frac {1}{36}\left ( 4x^{3}+12x^{\frac {3}{2}}\left ( -\frac {2}{3}x^{\frac {3}{2}}\right ) +9\left ( -\frac {2}{3}x^{\frac {3}{2}}\right ) ^{2}\right ) & =0\\ y & =0 \end {align*}
Which agrees with $$y_{s}=0$$ found from the p-discriminant method. Hence $$y_{s}=0$$ is singular solution. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of $$c$$.
|
# Figuring out when you can do a puzzle.
This week’s Occupy Math looks at a type of puzzle where you want to fill a rectangle with a shape. We will be using the L-shaped 3-square polyomino, used to fill a 5×9 rectangle below, as our example shape. The goal is to figure out every possible size of rectangle that can be filled with this shape. If you are constructing puzzles for other people — e.g., your students — knowing which problems can be solved gives you an edge. The post will not only solve the problem for our example shape, but also give you tools for doing this for other shapes. The answers, and the tools, are at the bottom if you don’t feel like working through the reasoning.
Begin with something simple.
The first trick for figuring out if a rectangle can be filled with our shape is really simple and a bit clever. The shape has an area of 3 (it is made of 3 1×1 squares). This means any rectangle that is filled with this shape has an area that is a multiple of 3 and so, since 3 is prime, must have one side length that is divisible by 3. Here is the smallest rectangle you can fill with the shape.
Notice that a rectangle with a size of length 1 is too thin to hold copies of the shape so 1×3 (or 1-by-anything) is impossible. That proves that 2×3 is the smallest possible rectangle that can be filled — since 2 is the minimum side length and at least one side length divisible by 3 is required.
Getting 2×3 buys us many other rectangles!
The example below shows how to fill a 6×6 rectangle by stacking different copies of the 2×3 solution.
We can push this a long way to showing that we can fill any rectangle with one side of even length and the other a multiple of 3. First of all, we can use multiple copies of the 2×3 solution to get any even-by-3 rectangle, like this:
Then, those even-by-3 rectangles can be repeated to make even-by-any-multiple-of-3 rectangles, like this:
This leaves only 3×odd rectangles to be considered.
The smallest 3×odd that is not obviously too short is 3×3 — which the picture below shows is impossible for the way the blue shape is placed: the only way to cover the lower left position is with a placement of the red shape that leaves an impossible 3×1 area to fill. There are a couple of other ways to place the first shape, but they all lead to something impossible.
Now consider a rectangle that is 3 by any odd number. Every way of placing the shape in the upper left corner does one of three things (i) do something impossible right away, (ii) shorten the rectangle into a 3-by-two shorter rectangle (this can happen two ways), or (iii) create a 3×1 space that cannot be filled. The picture shows all three cases for a 5×3 rectangle.
The first and last placements of the blue shape can never work — the one that creates an isolated square and the one that creates an unfillable 3×1. The other placement reduces the problem to one that is two shorter. That means that, since 3×3 is impossible, so is 5×3. The fact that 5×3 is impossible means 7×3 is impossible, and so on forever. No 3×odd rectangle can be filled with the shape.
Every other multiple of 3-by-odd can be filled!
The example at the top of the post is 5×9. The picture below is 5× 6.
Now, since we have 5 high by 6 or 9 wide, we can build 5 by any multiple of 3 that is bigger than 3. We already have 6 and 9. The multiples of 6 are all even multiples of 3, and every odd multiple of 3 (bigger than 9) is an even multiple of 3 plus 9. This gives us 5-by-any-multiple-of-3 (except 3 itself) by laying our 6 and 9 wide solutions beside each other.
We already know how to build a 2-by-any multiple-of-3 rectangle. That means that, once we have 5-by-any-multiple-of-3, we can add two more to make 7, then 9, then 11 and so on, by adding the 2×any-multiple-of-3 we already built to the top or bottom of a rectangle we already have. That finishes our logical demonstration. We now know all the dimensions of rectangles that can be filled. DONE!
The answer: which rectangles can be filled?
The pile of logic and pictures above are an example of a proof. What was proved is this: a rectangle can be filled if it is a multiple of 3 on one side and (i) an even number or (ii) any odd number greater than or equal to 5 on the other side. So a 117×125 rectangle can be filled with the shape (do not assign this problem!)
What were the tools we used in the proof?
1. The rectangle’s area is a multiple of the piece’s area. This eliminates a lot of possibilities.
2. Once we have one solution, we can put copies of it together to make others. This works best with the smallest available solutions.
3. Break the problem into different cases or groups. The groups here are multiples of three by even and by odd.
4. The trick where we showed 3×3 was impossible and then extended that to 3 by any odd number is a classical method called mathematical induction. This is the hardest part of the reasoning.
5. To clean up, we again put together small solutions to get the rest. It’s interesting and convenient that 6 and 9 can be added to make any odd multiple of 3 bigger than 3 itself.
The process of starting with small examples and then, once you sort of see what is going on, breaking the problem into different cases (e.g., odd, even) is a standard approach. It also masks the real difficulty of solving a problem like this: how do you think of the steps? Except maybe for the induction, every step is not hard to follow. The mathematical talent is the imagination to figure out an effective collection of steps.
This raises another important point. Occupy Math is trying hard to write something that is not too hard to follow; his proof is far from the only way to figure out which rectangles can be filled with the 3-square elbow. If you want a simpler version of this problem, ask your students to figure out which rectangles can be filled with copies of a domino (a 2×1 rectangle). The answer is any rectangle where at least one side is of even length; this can be done with a simplified version of the proof used in this post.
|
Open in App
Not now
# Remainder Theorem
• Last Updated : 25 Mar, 2023
Remainder theorem is the basic theorem used in mathematics which is used to find the remainder of any polynomial when it is divided by a linear polynomial. Suppose for any given polynomial f(x) if it is divided by x-a then its reminder is always f(a). Remainder theorem works on the principle of Euclidean division theorem. Linear theorem works only when the division is linear and it fails for other types of divisors. Now let’s learn about the Reminder theorem, its proof, and others in detail in this article.
## Remainder Theorem Definition
Remainder theorem states that for any polynomial whose degree is greater than or equal to 1 if it is divided by a linear polynomial q(x) such that q(x) = x-a, the remainder of this division is always equal to q(a). This theorem is very helpful in finding the remainder of the polynomial without actually performing the division. The remainder theorem can be easily expressed as, for any polynomial p(x) { degree greater than or equal to 1} if divided by any linear polynomial s(x) = x-a the reminder is always p(a) such that,
p(x)/s(x) = q(x) + r(x)
where,
q(x) is the quotient,
r(x) is the remainder which is equal to r(a)
## Remainder Theorem Formula
For any polynomial function f (x) when divided by the linear polynomial (x-a) then the remainder is always equal to f (a).
f (x) = (x-a) × q (x) + r (x)
where,
r(x) is the remainder of the polynomial
The image given below express the remainder theorem.
## Factor Theorem
Factor Theorem is also the basic theorem of mathematics which is considered the reverse of the remainder theorem. This is generally used the find roots of polynomial equations.
This theorem states that for any polynomial p(x) if p(a) = 0 then x-a is the factor of the polynomial p(x).
## Remainder Theorem Proof
For any polynomial g(x) the remainder theorem can easily be proved as,
Let g(x) be a polynomial with a degree of 1 or greater than 1. Suppose that when g(x) is divided by (x – b), the quotient is q(x) and the remainder is r(x), i.e.
g(x) = (x – b) q(x) + r(x)
Since the degree of x – b is 1 and that of r(x) is less than that of x – b, the degree of r(x) = 0. This means that r(x) is constant. Thus, we can write the above equation as
g(x) = (x – b) q(x) + r
In particular, if x = b, then the equation becomes,
p(b) = (b – b) q(b) + r
p(b) = (0) q(b) + r
p(b) = r
Hence, Proved.
## Dividing a Polynomial by a Non-Zero Polynomial
For dividing a polynomial with a non-zero polynomial use the steps given below,
Step 1: Arrange both the polynomial dividend and divisor in the descending order of their degree.
Step 2: The first term of the dividend is divided by the first term of the divisor to produce the quotient,
Step 3: After subtracting the first term arrange the remainder and copy the second term of the dividend and start dividing the remaining term as discussed above.
Step 4: Repeat these steps until the degree of the dividend polynomial is less than the divisor. In the end, only the remainder is left.
## Remainder Theorem of Polynomial
How to use the Remainder theorem? can easily be explained with the help of the example given below.
Example: Divide 2x3 + 3x2 + 4x + 5 by x + 2
Solution:
Given,
Dividend = p(x) = 2x3 + 3x2 + 4x + 5
Divisor = s(x) = (x + 2)
Here,
Quotient = q(x) =2x2 – x + 6
Remainder = r(x) = -7
Verification:
Given,
Divisor = (x + 2)
Let x + 2 = 0
x = -2
According to Remainder Theorem, substituting x = -1 in p(x),
Now, Remainder = p(-2)
Thus, Remainder Theorem is verified.
### Alternate Method
We know that any polynomial p(x) can easily be written as,
p(x) = (x – a)·q(x) + r
Taking x = a in the above equation we get only the remainder,
similarly, for any p(x)
p(x) = (x+2).q(x) + r
if we put x = -2 we get the remainder of the above equation.
Thus, p(-2) is the remainder.
Hence proved.
## Euler Remainder Theorem
For any two co-prime positive integers n and X, Euler’s theorem states that,
Xφ(n) = 1 (mod n)
where,
φ(n) is called Euler’s function and its value is given as,
φ(n) = n (1-1/a)×(1-1/b)(1-1/c)
where,
n is a natural number, and n = ap × bq × cr
where,
a, b, c are prime factors of n and p, q, r are positive integers.
Example: Find the Euler function of 21
Solution:
The factors of 21 are,
21 = 7×3
φ(21) = 21 (1 – 1/7)(1 – 1/3)
= 21 × 6/7 × 2/3
= 12
Thus, the Euler function of 21 is 12
## Differences between the Remainder Theorem and Factor Theorem
Remainder Theorem and Factor Theorem are the basic theorems in mathematics and the basic difference between them is stated in the table given below,
## Solved Examples on Remainder Theorem
Example 1: Find the remainder when p(x) = x4 – x3 + x2 – 2x + 1 is divided by g(x) = x – 2.
Solution:
Given,
p(x) = x4 – x3 + x2 – 2x + 1
g(x) = x-2
Using Remainder theorem,
p(2) = (2)4 – (2)3 + (2)2 + 2(2) + 1
= 17
Thus, the remainder when p(x) is divided by g(x) then we get remainder as, 17
Example 2: Find the root of the polynomial x2 – 5x + 4
Solution:
We know that if for any p(x) we get p(a) = 0, then x-a is the factor of p(x) or a is the root of equation.
Given,
f(x) = x2 – 5x + 4
By hit and try method.
f(4) = 42 – 5(4) + 4
f(4) = 20 – 20
= 0
So, (x – 4) must be a factor of x2 – 5x + 4
Example 3: Find the remainder when t3 – 2t2 + 4t + 5 is divided by t – 1.
Solution:
Given,
p(t) = t3 – 2t2 + 4t + 5
g(t) = t – 1
Using Remainder theorem,
g(1) = (1)3 – 2(1)2 + 4 + 5
= 8
By the Remainder Theorem, 8 is the remainder when t3 – 2t2 + 4t + 5 is divided by t – 1
Example 4: Find the remainder when x3 x2 + 2 is divided by x – 2.
Solution:
Given,
p(x) = x3 x2 + 2
g(x) = x-2 = 0
x = 2
Using Remainder theorem,
g(2) = (2)3 – (2)2 + 2
= 6
By the Remainder Theorem, 6 is the remainder when x3 – x2 + 2 is divided by x – 2
Example 5: Find the root of the polynomial 3x2 – 7x + 2
Solution:
We know that if for any p(x) we get p(a) = 0, then x-a is the factor of p(x) or a is the root of equation.
Given,
f(x) = 3x2 – 7x + 2
By hit and try method.
f(2) = 3(2)2 – 7(2) + 2
f(2) = 12 -14 + 2
= 0
So, (x – 2) must be a factor of 3x2 – 7x + 2 and 2 is the root of 3x2 – 7x + 2
## FAQs on Remainder Theorem
### Question 1: What is the Remainder Theorem?
According to remainder theorem for any polynomial p(x) (whose degree is greater than equal to 1) when divided by the linear polynomial (x – a), the remainder is always p(a).
### Question 2: Who Invented the Remainder Theorem?
The credit for inventing remainder theorem goes to Chinese mathematician Sun Zi and the it was completed by Qin Jiushao in 1247.
### Question 3: What is the use of the Factor Theorem?
Factor theorem is used to find the factors of the given polynomial. For any polynomial f(x), x-a is the factor of the polynomial f(x) only when, f(a) is zero.
### Question 4: What are the Applications of the Remainder Theorem Formula?
Remainder theorem is widely used to find the remainder of the polynomial without actually performing the long division and remainder theorem along with factor theorem is widely used to solve the polynomial equation.
|
```Chapter 4 Review (Proving Triangles are congruent)
4-1 Congruent Triangles
In this section, we found out what congruent triangles are. Congruent triangles are triangles with
congruent corresponding sides and angles. There isn’t a lot of information to recall from this chapter
but it is just very important to understand what congruent triangles are.
4-2 Proving triangles are congruent by S.S.S and S.A.S
In this section, we were taught how to prove triangles are congruent by S.S.S and S.A.S
Page 232 # 28, 29, 30, 33 35, 36
4-3 Proving triangles are congruent by A.S.A and A.A.S
In this section, we were taught how to prove triangles are congruent by A.S.A and A.A.S
Page 238 # 12, 13, 17, 25, 32
REMEMBER!!! You cannot prove triangles are congruent by A.S.S and A.A.A
4-4 Corresponding parts of congruent triangles are congruent (C.P.C.T.C)
In this section, we learned about corresponding parts of congruent triangles. If you prove that two
triangles are congruent, then its corresponding sides and angles are congruent. Example: If triangles
DEF and GHI are congruent, then DE = GH or <EFD = <HIG.
Page 246 6, 11, 15, 19,
4-5 Isosceles and Equilateral Triangles
In this section, we learned about specific properties of Isosceles and Equilateral triangles. Remember,
Isosceles triangles have two congruent sides and their base angles are congruent. An equilateral triangle
have all congruent sides and angles. The angles of an equilateral triangle must be 60 degrees.
Page 253 # 3, 7, 10, 11, 12, 40
4-6 Hypotenuse-Leg Theorem
In this section, we learned how we can prove right triangles are congruent using the hypotenuse leg
theorem.
Page 262 # 11, 18, 24
4-7 Congruence of Overlapping Triangles
In this section, we did not learn any theorem or vocab but we learned about overlapping triangles and
how to highlight specific triangles and separate the two of them. The purpose of separating the two
triangles is to see them more clearly. Especially when you start to prove the triangles are congruent, it
really helps you visualize your congruent marks on sides and angles.
Page 268 # 8, 12, 17, 25, 26
```
|
Browse Tag by geometry triangles lesson
## Shapes in Geometry and Their Measurements
Geometry
In Mathematics, geometry is one of the crucial branches that help us to understand the shapes and sizes of the object we see in our day to day life. The basic of shapes starts with points, lines and angles. Point is simply a dot or a node, where two lines are joined together. Whereas an angle is formed when two lines are connected to each other, end to end.
Points and lines are one dimensional. But we usually learn the geometrical shapes and their measurements based on two-dimensions and three-dimensions. These shapes have various applications in different fields such as Constructions, Architecture, Engineering, Interior Designing, Graphic designing, Gaming, etc.
For example, we use these to determine the area of floor in a building, length of a pole in a street, length of boundaries of a plot to fence it, etc.. And also, for domestic purposes such as will determine the surface area of a cylindrical shape in a 3d space, a circle drawn on a ground of a specific diameter to dig a hole, etc. So, geometry has many applications in various fields. Let us learn here different shapes and their measurements based on the dimensions.
## Two-Dimensional Shapes in Geometry
The two-dimensional shapes in geometry are circle, triangle, quadrilaterals and all the polygons that have area and perimeter. These are defined in the XY plane, along x-axis and y-axis. All the closed curves and polygons are two-dimensional. Let us discuss all the 2d shapes here.
## Circle
A circle is a closed curve shape that has locus of points at an equidistant from a common point, called center. The distance from center to the locus of points or outer line of circle, is called the radius. Area and circumference of circle is given by:
## Triangle
A triangle is a three sided closed shape, where each side is connected to two other sides end to end. It is the smallest kind of polygon having three sides and three interior angles.
Area (sq.unit) ½ (Base) x (Height) Perimeter (units) a+b+c; sum of three sides
A four sided closed shape, that has four interior angles. It is also categorised into different types:
• Square: All sides are equal and all angles are equal to 90 degrees
• Rectangle: Opposite sides are equal and all angles measures 90 degrees
• Parallelogram: Opposite sides are equal and parallel. Opposite angles are equal.
• Trapezium: One pair of opposite sides are parallel.
Type of quadrilateral Area (sq.unit) Perimeter (units) Square Side2 4 x Side Rectangle Length x Width 2 (Length + Width) Parallelogram Base x Height 2 (Sum of any two adjacent sides) Trapezium ½ (sum of parallel sides) x (height) Sum of all sides
## Three-Dimensional Shapes
Three dimensional shapes are those shapes that are defined in three dimensions such as x-axis, y-axis and z-axis. The additional z-axis here represents the thickness of the shape. The basic solid shapes in Maths are:
• Sphere
• Cube
• Cuboid
• Cone
• Cylinder
These solid shapes are defined based on their surface area and volume. Let us learn the measurement of these solids.
## Sphere
A sphere is a round shape whose surface is at equidistant from a common point, called center. Similar to circle, the distance from center to outer surface is the radius of the sphere.
Surface area 4πr² (square unit) Volume 4/3(πr3) (cubic unit)
## Cube
A cube is an extended version of a square in three dimensional space, whose all the sides are of square shape. Therefore, the length of all its edges are also equal. It has 6 faces, 12 edges and 8 vertices.
Surface area 6 (Edge)2 (square unit) Volume Edge3 (cubic unit)
## Cuboid
A cuboid is an extended version of a rectangle. Just like a cube, it also has 6 faces, 12 edges and 8 vertices. But the faces of cuboids are in rectangular shape.
Surface area 2 (Length x Breadth+breadth x height + Length x height) Volume length × breadth × height
## Cone
A cone is a solid shape with a circular base, that narrows smoothly from bottom to top at a point called the apex of the cone.
Surface area (sq.units) Πr (r+l), where r is the radius of circular base and l is the slant height Volume (cu.units) ⅓ Π (r2) (h); h is the height of cone
## Cylinder
Cylinder is a 3d shape having parallel circular bases on both the sides of a curved surface.
Surface area (sq.units) 2 Π r (r + h); r and h are the radius and height of cylinder Volume (cu. units) Π r2 h
|
### Vocabulary
Addition and subtraction within 5, 10, 20, 100, or 1000. Addition or subtraction of two whole numbers with whole number answers, and with sum or minuend in the range 0-5, 0-10, 0-20, or 0-100, respectively. Example: 8 + 2 = 10 is an addition within 10, 14 – 5 = 9 is a subtraction within 20, and 55 – 18 = 37 is a subtraction within 100.
Additive inverses. Two numbers whose sum is 0 are additive inverses of one another. Example: 3/4 and – 3/4 are additive inverses of one another because 3/4 + (- 3/4) = (- 3/4) + 3/4 = 0.
Associative Property of Addition The problem (3 + 6) + 8 = 3 + (6 + 8)demonstrates the associative property of addition. Observe that the addends are the same on either side of the equal sign: 3 plus 6 plus 8
The associative property of addition says that when we add more than two numbers the grouping of the addends does not change the sum.
In the example above, we can easily observe that:
(3 + 6) + 8 = 3 + (6 + 8)
9 + 8 = 3 + 14
1
7 = 17
Associative Property of Multiplication
The problem (2 � 4) � 3 = 2 � (4 � 3) demonstrates the associative property of multiplication. Observe that the factors are the same on either side of the equal sign: 2 times 4 times 3
Commutative Property
• Commutative Property of Addition: It states that changing the order of addends does not change the sum.
That is, a + b = b + a.
• Commutative Property of Multiplication: It states that changing the order of factors does not change the product.
That is, a × b = b × a.
• Addition and multiplication are commutative over the set of real numbers.
That means, for any two real numbers x and yx + y = y + x and xy = yx.
• Subtraction and Division are not commutative.
### Examples of Commutative Property
• 2 + 3 = 3 + 2. Whether you add 3 to 2 or you add 2 to 3, you get 5 both ways.
• 4 × 7 = 7 × 4, Whether you multiply 4 by 7 or you multiply 7 by 4, the product is the same, 28.
Computation algorithm. A set of predefined steps applicable to a class of problems that gives the correct result in every case when the steps are carried out correctly.
Congruent. Two plane or solid figures are congruent if one can be obtained from the other by rigid motion (a sequence of rotations, reflections, and translations).
Counting on. A strategy for finding the number of objects in a group without having to count every member of the group. For example, if a stack of books is known to have 8 books and 3 more books are added to the top, it is not necessary to count the stack all over again. One can find the total by counting on—pointing to the top book and saying “eight,” following this with “nine, ten, eleven. There are eleven books now.”
Expanded form. A multi-digit number is expressed in expanded form when it is written as a sum of single-digit multiples of powers of ten. For example, 643 = 600 + 40 + 3.
Fraction. A number expressible in the form a/b where a is a whole number and b is a positive whole number. (The word fraction in these standards always refers to a non-negative number.) See also: rational number.
Definition of Identity Properties of Addition and Multiplication.
• Identity property of addition states that the sum of zero and any number or variable is the number or variable itself. For example, 4 + 0 = 4, - 11 + 0 = - 11, y + 0 = y are few examples illustrating the identity property of addition.
• Identity property of multiplication states that the product of 1 and any number or variable is the number or variable itself. For example, 4 × 1 = 4, - 11 × 1 = - 11, y × 1 = y are few examples illustrating the identity property of multiplication.
Line plot. A method of visually displaying a distribution of data values where each data value is shown as a dot or mark above a number line.
Mean. A measure of center in a set of numerical data, computed by adding the values in a list and then dividing by the number of values in the list.4 Example: For the data set {1, 3, 6, 7, 10, 12, 14, 15, 22, 120}, the mean is 21.
Median. A measure of center in a set of numerical data. The median of a list of values is the value appearing at the center of a sorted version of the list—or the mean of the two central values, if the list contains an even number of values. Example: For the data set {2, 3, 6, 7, 10, 12, 14, 15, 22, 90}, the median is 11.
Multiplicative inverses. Two numbers whose product is 1 are multiplicative inverses of one another. Example: 3/4 and 4/3 are multiplicative inverses of one another because 3/4 × 4/3 = 4/3 × 3/4 = 1.
Number line diagram. A diagram of the number line used to represent numbers and support reasoning about them. In a number line diagram for measurement quantities, the interval from 0 to 1 on the diagram represents the unit of measure for the quantity.
Percent rate of change. A rate of change expressed as a percent. Example: if a population grows from 50 to 55 in a year, it grows by 5/50 = 10% per year.
Probability. A number between 0 and 1 used to quantify likelihood for processes that have uncertain outcomes (such as tossing a coin, selecting a person at random from a group of people, tossing a ball at a target, or testing for a medical condition).
Rectilinear figure. A polygon all angles of which are right angles.
Repeating decimal. The decimal form of a rational number. See also: terminating decimal.
Terminating decimal. A decimal is called terminating if its repeating digit is 0.
Visual fraction model. A tape diagram, number line diagram, or area model.
Whole numbers. The numbers 0, 1, 2, 3, …
|
# Solving the Resident Evil Zero game math puzzle
On the screen is shown a “00/81”. A number pad with number keys 1 to 9 is on the right. When a number key is entered, a knob is lit in red and the value of the number is added to the numerator of the displayed ratio. There are 10 unlit knobs.
The puzzle should be obvious. Match the numerator and the denominator with 10 entries.
So how would I solve it? My hint as stated, was:
the last entry is the most important
The first 9 entries are to get you as close to the required number as possible. This is actually similar to blackjack, a card game where you try to get as close to 21 points as possible without going over. The difference is that you don’t get to choose your next number (or card value) in blackjack. You do here.
[image by pavlen]
So my first instinctual thought was to finish up the first 9 entries to get to the last entry. My second instinctual thought was to make these 9 entries all the same number value. It’s a timed effort. I don’t have time to go figure out different combinations of sums.
Based on those thoughts, my next (instinctual) step was to divide the required number by 9, rounding down to nearest integer. Don’t ask me why at this point, because I’ll know the reason only after thinking it through, and I’m not thinking it through at this point.
So we have 81 divide by 9, and rounded down, we get 9. We have a problem. It’s not just 9. It’s exactly 9. There’s no number value 0 for the final entry. So we use 8 (1 less than 9).
Then we multiply 8 by 9 (entries) to get 72. And then 81 (required number) – 72 is 9, the final entry. So the answer is 8,8,8,8,8,8,8,8,8,9.
This method gives the simplest of combinations (only 2 distinct numbers used) and is the fastest to get you to the end point, the final entry. It’s the final entry that “corrects” the sum to the required number.
I have to say, the changing of 9 to 8 is (I’m starting to use the word generously) instinctive. I don’t know how changing 9 to 8 would solve the problem. I just know.
Let’s use this method on the second part of the puzzle: 67.
67 divide by 9 rounded down gives 7. 7 multiplied by 9 is 63. 67 – 63 is 4. So the answer is 7,7,7,7,7,7,7,7,7,4.
There you have it, the algorithm to solving the math puzzle. The second puzzle uses the “normal” mode of the algorithm. The first puzzle tested the edge case of the algorithm.
Hmm… can we write a function that spits out the combination using the above algorithm? Can we write a function that spits out all possible combinations?
|
# Lesson video
In progress...
Hi, I'm Miss Kidd-Rossiter and I'm going to be taking the lesson today on the LCM and prime factors.
It's going to build on the work that you've already done on lowest common multiple highest common factor and prime factorization.
So hopefully you're really going to enjoy it.
Before we get started you're going to need a pen and something to write on so make sure you've got that.
Try and get yourself into a nice, quiet place, free from all distractions.
If you need to pause the video now to sort anything out, then do, if not, let's get going.
So today's try this activity, you've got to find the first five common multiples of 12 and nine.
There's a diagram on the board, a representation that might help you.
And then I'd like you to write each as a product of their prime factors.
When you've done that, can you notice anything interesting about the five common multiples? Pause the video now and have a go at this task.
Excellent work, let's go through these then.
So your first five common multiples are 36, 72, 108, 144 and 180.
We can know once we found the first common multiple, that any other common multiples will be a multiple of the first common multiple.
When we've got 36, where in here can I see nine? And where can I see 12? Pause the video now and think about that.
Excellent, I can see nine here can tie in three times three.
And I can see 12 here as two times two times three.
Can you see them in the other common multiples? Excellent, and they're in all of them, aren't they? So here they are in 72, here they are in 108, here they are in 144 and here they are in 180.
Now what's different about 36 compared to the other four common multiples? Pause the video and think about that.
Excellent, 36 uses the minimum number of prime factors, doesn't it? It exploits the overlapping primes, which in this case is just the three and adds no extras.
Now it makes sense here because we've got to multiply our 36 by two to get 72, to get 108 we multiply our 36 by three, and to get 144 we multiply our 36 by four, and to get 180, we multiply our 36 by five.
So we're going to have a little bit more of a look at this now.
So before we do pause the video, and explain how each strategy can help you find the lowest common multiple.
So pause now.
Excellent, let's look at these two strategies first.
Well, these are basically the same strategy, aren't they? They're both listing strategies.
So here we just list the multiples of 18, and here we list the multiples of 15, and then we find the lowest common multiple.
So the first multiple that is common to both.
This is just a representation of that structure where we use a length to represent 15 and 18.
And we see that six 15s are the same as five 18s.
So we would know that this is 90.
And this is 90, so therefore the lowest common multiple must be 90.
This is slightly different.
So what we've done here is we've made use of everything that we know about prime factors.
So 18 as a product of its prime factors is what? Tell me now.
Excellent, two multiplied by three multiplied by three.
And 15 as a product of its prime factors, this is what? Excellent, just three multiplied by five.
So how could I get my lowest common multiple from here? So you can see that I've organised my prime factors into a Venn diagram.
The common prime factor to 18 and 15 is three.
So I've put three in the centre of my Venn diagram.
Then any that I have leftover two and three for 18, go in this circle and five for 15 goes in this circle.
So if I was to do two times three times three times five, I would get my lowest common multiple, which is 90.
Why does that work? Pause now and think about it.
Excellent, if I multiply the two times three times three, which is my 18 by five, then it ensures that 15 is a factor, and if I multiply my three by five by six, then that ensures that 18 is a factor, which you can see more clearly here.
So you're now going to have a go applying your learning to the independent task.
So pause the video now navigate to the independent task.
And when you're ready to go through some answers, resume the video, good luck.
Excellent, let's go through some of these answers then.
So you have these four Venn diagrams, where they are completed and you've got your lowest common multiple written next to them.
If you need to pause the video now to look at these answers, then please do.
Excellent, and then for these ones, you've got the answers on the screen there as well, excellent work.
Finally then moving on to the explore task, the prime factors of two numbers in the Venn diagram, can you find what the two numbers are, any common factors, their lowest common multiple, and their highest common factor.
Pause the video now and have a go at this task.
Excellent work, so let's first of all, work out what this circle on the left is representing.
So we can see that the prime factors of this number are two, two, five and five.
So that means if we multiply these together, two times two is four, four times five is 20, 20 times five is a hundred that means that this number represented by the left hand circle must be a hundred.
Similarly for the one on the right hand side, we know that the prime factors are two, two, three, five, and seven.
And when we multiply these together, five times seven is 35, 35 multiplied by three is 105, multiplied by two is 210, multiplied by two again is 420.
So that must be the number represented on the right hand side here.
Now you noticed I worked backwards there, and it doesn't matter which way you do the multiplication 'cause you know, from all your work so far, that multiplication is commutative.
And I found that easier because I know that if I get a big number here, I find it easier to double it and then double it again at the end.
Any common factors then, will all numbers, all positive integers, have a common factor of what? Excellent one, so one has to be a common factor.
What else is a common factor? Two is a common factor.
Is three a common factor? No, it's not because if it was, it would be written here in the centre, wouldn't it? It's a prime factor.
Is for a common factor? Yes it is because we can get it from multiplying two and two.
Is five a common factor? Yes it is and we can keep going like this, all your common factors they're there on the screen.
So one, two, four, five, 10, and 20.
The lowest common multiple then so to work this one out, remember we have to multiply two, two, three, five, five, and seven, and we get an answer of 2,100.
And the highest common factor to them, what do you think the highest common factor is? Did you manage to figure this out? So the highest common factor we get from the centre of the Venn diagram, so that gives us two multiplied by two, multiplied by five, which is 20.
That's the end of today's lesson.
So thank you very much for all your hard work.
I hope you've enjoyed the lesson.
Don't forget to get go and take the end of lesson quiz so that you can show me what you've learned and hopefully I'll see you again for another math lesson soon, bye.
|
### by Howard Williams
#### Published Sunday November 27 2022 (link)
On rugby international morning, I found myself, along with eight friends, in a pub 5.8 miles from the match ground. We were enjoying ourselves, and so wished to delay our departure for the ground until the last possible minute. The publican, wishing to keep our custom for as long as possible, offered to help us get there by carrying us, one at a time, as pillion passengers on his motorbike.
We could walk at 2.5mph and the bike would travel at 30mph. We all left the pub together, and arrived at the ground in time for kick-off.
Ignoring the time taken getting on and off the bike, what was our minimum travelling time in minutes?
From → Uncategorized
In order to minimise the total time taken for the friends to travel from the pub to the playing ground, it is important that the barman and all the friends start moving at the same time and then move continuously at their respective speeds until they all arrive simultaneously at the playing ground.
Let the distance from the pub to the ground be $$d$$, the the pedestrian walking speed of the $$n$$ friends be $$p$$ and the speed of the barman on the motorbike be $$b$$. If the friends each walk a distance $$w$$ the time taken to complete their journey is $t=\frac{w}{p}+\frac{d-w}{b}$ The barman will travel backwards and forwards a total of $$(2n-1)$$ times and will hence travel $$(2n-1)$$ times the distance $$d$$ less the total distance travelled by all the friends ($$2nw$$). This will hence take a time $$t$$ given by:$t=\frac{(2n-1)d-2nw}{b}$Equating these two times and solving for $$w$$ now gives:$w=\frac{2(n-1)dp}{(2n-1)p+b}=\frac{(k-1)dp}{kp+b}$ where $$k=2n-1$$. Finally we can substitute for $$w$$ in the first equation to find the minimum total time for transfer:$t_{min}=\left(\frac{kb+p}{kp+b}\right)\left(\frac{d}{b}\right)$
|
How to Calculate the 3 Digits in Prediction 4D - GeoffreyStephen.com
# GeoffreyStephen.com
Search For Anything Tips Here (360°).
## Monday, June 29, 2015
Shalom, I had long, do not update the article on my blog. This is because I was very busy with tasks in the workplace. Well, this time the article is how to do calculations 3 digits and make it to the fourth digit.
It is very important to ensure that the investment has percent probability number will hit in the next draw. If you do not try to learn these skills, you will lose a lot of money. The 4-digit lottery is mathematics. This means that you must perform mathematical calculations based on facts
How to calculate the 3 digits in prediction 4D
Before understanding this method, you must know, the odd number and an even number. What number is it? That number is 1,3,5,7,9 (odd) 0,2,4,6,8 (even). Number 0000 to 9999 is the coupling of odd and even numbers. I practiced system, an even number is A and, odd number is B.
If you have a target number in the 4-digit lottery system, but you want to make the 3 digits, you must perform these methods. A simple example is, your target number is 1234. If you make it to 3 digits, the formula is like this
234
134
124
123
This method is very simple, cover the number 1, will be 234, covering the number 2, will be 134, covering the number 3, will be 124, covering the number 4, will be 123
If you've made a 4-digit to 3 digits, you must categorize these 3 digits, to an even number and odd number. This means that your target number is
234=ABA
134=BBA
124=BAA
123=BAB
How to find this combination of three digits to four digits? To fully understand this rule, you must learn how to work even and odd number each time the results of a vote. Each number hit on the draw are as follows, AAAA AAAB AABA ABAA BAAA dan BBBB BBBA BBAB BABB ABBB.
If my choice is the 234 number, (ABA) percent probability of coupling this 3D to 4D is an even number (24680). This is because the number is an even number, two (24) (ABA). My selection number, will be like this
2342, 2344, 2346, 2348, 2340
How? Do you have to understand this principle? Mastering these skills is essential to reduce the amount of your bet number when investing in the 4-digit lottery. If you have understood this method, you can read my previous article, how to calculate your percent 4D. So far only sharing this time, meet again in the next entry.
#### 1 comment:
1. HI THERE MR ,
ANY EMAIL WE CAN CHAT FOR THE PREDICTION ?
Comments containing the active link, promotion, requesting a visit again, will not be displayed.
|
Start learning today, and be successful in your academic & professional career. Start Today!
• ## Related Books
2 answersLast reply by: Jing ChenTue Aug 29, 2017 2:02 PMPost by Sai Nettyam on October 11, 2011Ms. Pyo,How do you find the distance between equations? For example... Find the distance between the lines with the equations y= 2/7x + 4 and y= 2/7x -2. Can you please explain on how to solve problems like this. Thank you.
### Parallels and Distance
• The distance from a line to a point not on the line is the length of the segment perpendicular to the line from the point
• the distance between two parallel lines is the distance between one of the lines and any point on the other line
### Parallels and Distance
Draw the segment that represents the distance from point A to MN .
AB represents the distance.
Draw the segment that represents the distance from point A to MN .
AB represents the distance.
Determin whether the following statement is true or false.
Line m is parallel to line n, point A is on line m, the distance from point A to line n is d, then the ditance between lines m and n is d.
True.
Draw the segment that represents the distance from point A to MN .
AB represents the distance.
Line p is parallel to line q. Draw a segment represents the distance between line p and line q.
• Find point A on line p
• Draw AB ⊥ line q.
AB represents the distance.
Line m intersects line n at point A(6, 9), line m^line n, point B(3, 7) is on line m, find the distance from point B to line n.
• AB represents the distance between line m and line n.
d = √{(6 − 3)2 + (9 − 7)2} = √{9 + 4} = √{13} .
Line p is parallel to line q. Draw the segment that represents the distance from point A to line q.
AB represents the distance.
Line p is parallel to line q, the distance between the two lines is 2x + 4, point A is between two lines, the distance from point A to line p is x + 4, the distance from point A to line q is 3, find x.
• 2x + 4 = (x + 4) + 3
x = 3
The given equation represents line m and point A(3, 4). Construct the perpendicular segment and find the distance from the point to the line.
y = − 3x + 2
• Assume line n passes through point A and is perpendicular to line m
• the equation for a line m is y = kx + b
• line m ^line n, so k = [1/3]
• line n passes through point A
• 4 = [1/3]*3 + b
• b = 3
• so line n is y = [1/3]x + 3
• Assume that line m and line n intersect at point B, − 3x + 2 = [1/3]x + 3
• x = − [3/10]
• y = − 3*( − [3/10]) + 2 = [29/10]
• B( − [3/10], [29/10])
• AB represents the distance between point A and line m
• d = √{(3 − ( − [3/10]))2 + (4 − [29/10])2}
d = [(√{1110} )/10].
Draw the segment that represents the distance from point A to MN
AM represents the distance
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Parallels and Distance
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Distance Between a Points and Line 0:07
• Definition and Example
• Distance Between Parallel Lines 1:51
• Definition and Example
• Extra Example 1: Drawing a Segment to Represent Distance 3:02
• Extra Example 2: Drawing a Segment to Represent Distance 4:27
• Extra Example 3: Graph, Plot, and Construct a Perpendicular Segment 5:13
• Extra Example 4: Distance Between Two Parallel Lines 15:37
### Transcription: Parallels and Distance
Welcome back to Educator.com.0000
This next lesson is on parallel lines and distance.0002
The first one: the distance between a point and a line: we have a point here, and we have a line.0010
And to find the distance between them, you have to find the length of the segment that is perpendicular.0017
If you were to get a ruler, you need to find the distance--how far away this point is from this line.0026
You are going to take your ruler, and you are going to make it so that the ruler is going to be perpendicular to the line.0035
So, imagine if that point is you; you are that point; you are standing in the room, and this line is the wall in front of you.0043
So, if you have to find the distance between you (this point right here) and the wall (the line), then you don't find the distance this way.0056
You don't go diagonally to the wall; if you are facing directly on the wall, you have to measure the distance0067
so that your pathway, that length of that segment, is going to be perpendicular to that wall.0074
If you are going to find the distance, you are not going to do this; that is not the distance.0090
It has to be from the point to the line so that it is perpendicular, so this right here will be the distance.0098
Now, to find the distance between two parallel lines: we know that parallel lines never intersect.0112
They are always going to have the same distance between them, no matter where you measure it from.0119
Now, that word right there is called "equidistant"; that means that, no matter where you measure the distance, it is always going to be the same.0126
And that is equidistant--an "equal distance," so "equidistant."0138
And so then, that means that, if you want to find the distance between them, you can just find it for any point on the line,0145
and make sure, just like we just went over, that the point and the line are perpendicular.0155
This segment right here: to find the distance, they have to be perpendicular to each other.0163
This would be the distance; if you look for it right here, as long as it is perpendicular, that will also be the distance.0168
That is the distance between two parallel lines.0179
Let's go over a few examples: Draw the segment that represents the distance from point A to EF.0187
And this is segment EF; so here is point A; here is segment EF.0195
I want to draw the segment that represents the distance.0205
If I go like this, the distance this way to the line, that is going to be incorrect, because that is not perpendicular.0212
This right here, what I drew (it is supposed to be a segment)--that is not perpendicular.0225
What you have to do: you have to extend this out; make sure that it is lined up, extend it out...and then you would draw that.0231
This is your distance, and the same thing here: we are going to extend this out, and then draw the segment that is perpendicular to it.0247
And you measure it, and that is the distance.0262
Let's do a couple more: here is point A; here are two parallel lines and point A; you want to find the distance from this point to this line.0267
So, now, if I go like this, is that the distance? No, this is not the distance.0277
You have to make sure to draw it like this, so that it is perpendicular; and that would be the distance.0288
Here is point A; here is segment EF; in this case, you would draw it; that is perpendicular.0297
Graph the given equation (there is your equation) and plot the point; construct a perpendicular segment, and find the distance from the point to the line.0315
So, first, let's draw the line; y = 2x - 1: this is y = mx + b; b is your y-intercept; that means I have to go to -1 and plot that as my y-intercept.0324
Here is my x; here is my y; and then, my slope is 2/1--remember, this is rise over run.0344
If my rise is 2, and it is positive 2, that means that I am going to go up 2, and then I am going to move to the right 1, because that is a positive.0356
For positive, you go right; for negative, you go left.0367
And then, you are going to go again: 1, 2, and you can just keep doing that.0370
You could go from this point--go to -2 and -1, and that is still going to give you a positive 2.0373
There is my line, right there; and then, the point is 0, 1, 2, 3.0381
First, we have graphed the equation; we have plotted the point; construct a perpendicular segment, and find the distance.0395
If we know that this right here, the slope of this line, was 2/1, that means that0407
if I draw a perpendicular segment from this point, the slope is going to be the negative reciprocal.0419
Remember: if they are perpendicular lines, then the slope is going to be the negative reciprocal.0429
For this other line, the perpendicular line, my slope will be -1/2.0438
Now, this can either mean -1 over 2, or 1 over -2; it is the same thing.0447
So, if I make it -1 over 2, from here, -1 is my rise, over run; so for the rise, since it is -1, I am going to go down 1, and 1, 2--over 2.0454
Down 1, and over 2, down 1 and over 2...and then, it is going to look something like this.0471
Now, to find the distance from that point to this point on the line, first of all, this is an intersection point between these two lines;0493
so, how can I find the intersection point between these two lines?0503
Well, I have my equation here; this is my first equation; my second equation, right here, is just going to be y =...0509
my slope was -1/2x; my y-intercept was 3; so all I did was to plug in my slope and my y-intercept.0525
The slope is -1/2; the y-intercept is 3; and then, you can just solve those two out, so y = 2x - 1.0536
I am going to use the substitution method...plus 3...equals 2x - 1.0547
And then, if I subtract the 2x over to this side, this is going to be -5/2x is equal to...0555
I am going to subtract the 3 over there, so I am going to get a 4.0566
Now, all I am doing is solving this out; remember, this is systems.0569
And then, to solve for x, I need to multiply this whole thing by the reciprocal, so -2/5.0575
That way, this is going to all cancel and give me 1; multiply that side by it; so x is going to equal 8/5--there is my x, and then my y.0584
I just need to plug it back into one of these; it doesn't matter which one; let's plug it into this one.0606
2 times x is 16/5 - 1; so this will be y = 2(8/5) - 1; here is 16/5 - 1, or I can make this 5/5, and that is going to be 11/5.0610
So, my point is this point right here; it is 8/5 and 11/5.0633
Now, the whole point of doing that (let me just explain), of finding this point, is so that you can find the distance between the two points.0650
If you have this point, and you have this point, you can use the distance formula to find how long the segment is.0662
All of this work was just to find this point right here, because that is the point where they intersected.0671
So then, I am going to use those two points; I am trying to find the distance between (0,3) and (8/5,11/5).0679
The distance formula is the square root of (x2 - x1)2 + (y2 - y1)2.0700
The distance is going to be (0 - 8/5)2 + (3 - 11/5)2.0712
This is -8/5; that is 64/25, plus...3 - 11/5; if you want to subtract these, you need a common denominator.0733
My common denominator is going to be 5; that is 15/5 - 11/5.0747
So, this is going to be 4/5, and that is squared; so it is going to be 16/25.0755
And then, the square root of...I have a common denominator, so this is going to be...80/25 (let me make some room over here):0775
that is the same thing as the square root of 80 over the square root of 25.0799
The square root of 80: this is 8 times 10; 8 is 4 times 2...let me just show you a quick way;0803
I know this is a little off of this lesson, but to simplify square roots, if you have √80, and you want to simplify it,0820
then you can just do the factor tree: this is 8 and 10; this is 4 and 2 (circle it if it is prime), 2 and 2;0832
whenever you have a pair of the same number, that comes out, so this is going to be 2.0846
Here is another 2 that is common; now, this one doesn't have a partner, doesn't have another 5 to come out of the radical.0855
So, only when they have a partner, only when there is two of the same number, can they come out.0870
2 comes out right there; then these 2's come out right there.0876
Now, the 5 has to stay in the radical; and then, you just multiply these two numbers.0881
And then, the square root of 25, we know, is 5; so this is going to be 4√5/5, and that is the distance.0888
Now, I know that this seems like a lot of work, but all we did was0897
construct a perpendicular segment by making the slope the negative reciprocal of this line.0904
Then, we found this point right here, where these two lines intersect; it has to be perpendicular.0915
So then, using this point and that point, we found the distance; and that is it--this is the distance between these two points.0923
All right, the next example: Find the distance between the two parallel lines.0938
OK, now, here I know that it has to be perpendicular wherever I want to find the distance,0943
as long as the segment that is made from the points between them has to be perpendicular.0958
Let's say I am going to use this point right here; then this is going to be perpendicular--that is going to be my distance.0965
Now, this point, I know is (0,2); this point, I know, is going to be 1 and 1/2...0978
1.5, or maybe 3/2 (the same thing): 3/2, and right here, that is going to be 1/2.0992
And I know that, because it is halfway between these two.0999
And the same here: the slope is 1, so everything is going to be half.1002
You can also, if you want, look at it this way: if I continue this on, then it is going to be halfway between this point right here1009
and this point right here, in the same way that this is halfway between this point and this point.1018
That is how I know that it is half.1026
This point is 3/2, 1, and 1/2; so now I have to find the distance between those two points.1028
The distance formula, again, is (x2 - x1), or (x1 - x2), squared, + (y2 - y1)2.1039
I find the square root again; I am going to subtract the x's: 0 - 3/2, squared, plus 2 - 1/2, squared.1057
Then, this is going to be (3/2)2; (3/2)2 is 9/4, plus...2 - 1/2 is 3/2, or 1 and 1/2,1073
and I know that because 2 becomes 4/2, minus 1/2 is 3/2; that squared is 9/4.1090
The square root of...there is already a common denominator, so that is 18/4...(make sure this goes all the way down),1104
which is √18/√4; for √18, you don't have to do the factor tree,1115
because you know that 9, a perfect square, is a factor of 18; it is going to be 3√2/2.1122
All you do is make sure that you just have the two points, whenever you try to find the distance between a point and a line or two parallel lines.1136
Then make sure that you have the two points, so that the segment that connects them is going to be perpendicular to the lines.1145
And then, you just use the distance formula; it is (0 - 3/2)2 + (2 - 1/2)2.1152
And then, that way, you get (3/2)2, which is 9/4, and then this, which is 3/2, squared, is going to be 9/4 again.1163
√18 over √4 simplifies to 3√2/2.1173
That is it for this lesson; thank you for watching Educator.com.1184
|
SAT II Math I : Range and Domain
Example Questions
← Previous 1 3
Example Question #1 : Range And Domain
Define .
Give the range of
Explanation:
The radicand within a square root symbol must be nonnegative, so
This happens if and only if , so the domain of is .
assumes its greatest value when , which is the point on where is least - this is at .
Similarly, assumes its least value when , which is the point on where is greatest - this is at .
Therefore, the range of is .
Example Question #2 : Range And Domain
Define .
Give the domain of
Explanation:
The radicand within a square root symbol must be nonnegative, so
This happens if and only if , so the domain of is .
Example Question #3 : Range And Domain
Define the functions and on the set of real numbers as follows:
Give the natural domain of the composite function .
The set of all real numbers
The set of all real numbers
Explanation:
The natural domain of the composite function is defined to be the intersection two sets.
One set is the natural domain of . Since is a polynomial, its domain is the set of all real numbers.
The other set is the set of all values of such that that is in the domain of . Since the radicand of the square root in must be nonnegative,
, and , the domain of
Therefore, the other set is the set of all such that
Substitute:
This holds for all real numbers, so this set is also the set of all real numbers.
The natural domain of is the set of all real numbers.
Example Question #1 : Domain And Range
is a sine curve. What are the domain and range of this function?
Domain: All real numbers
Range:
Domain: All real numbers
Range:
Domain: All real numbers
Range:
Explanation:
The domain includes the values that go into a function (the x-values) and the range are the values that come out (the or y-values). A sine curve represent a wave the repeats at a regular frequency. Based upon this graph, the maximum is equal to 1, while the minimum is equal to –1. The x-values span all real numbers, as there is no limit to the input fo a sine function. The domain of the function is all real numbers and the range is .
Example Question #4 : Range And Domain
Which of the following is NOT a function?
Explanation:
A function has to pass the vertical line test, which means that a vertical line can only cross the function one time. To put it another way, for any given value of , there can only be one value of . For the function , there is one value for two possible values. For instance, if , then . But if , as well. This function fails the vertical line test. The other functions listed are a line,, the top half of a right facing parabola, , a cubic equation, , and a semicircle, . These will all pass the vertical line test.
Example Question #2 : Domain And Range
Give the domain of the function below.
Explanation:
The domain is the set of possible value for the variable. We can find the impossible values of by setting the denominator of the fractional function equal to zero, as this would yield an impossible equation.
Now we can solve for .
There is no real value of that will fit this equation; any real value squared will be a positive number.
The radicand is always positive, and is defined for all real values of . This makes the domain of the set of all real numbers.
Example Question #2 : How To Factor The Quadratic Equation
Find the domain:
Explanation:
To find the domain, find all areas of the number line where the fraction is defined.
because the denominator of a fraction must be nonzero.
Factor by finding two numbers that sum to -2 and multiply to 1. These numbers are -1 and -1.
Example Question #4 : Domain And Range
If , which of these values of is NOT in the domain of this equation?
Explanation:
Using as the input () value for this equation generates an output () value that contradicts the stated condition of .
Therefore is not a valid value for and not in the equation's domain:
Example Question #7 : Domain And Range
What is the range of the function?
Explanation:
This function is a parabola that has been shifted up five units. The standard parabola has a range that goes from 0 (inclusive) to positive infinity. If the vertex has been moved up by 5, this means that its minimum has been shifted up by five. The first term is inclusive, which means you need a "[" for the beginning.
Minimum: 5 inclusive, maximum: infinity
Range:
Example Question #6 : Properties Of Functions And Graphs
What is the domain of the function?
Explanation:
The domain represents the acceptable values for this function. Based on the members of the function, the only limit that you have is the non-allowance of a negative number (because of the square root). The square and the linear terms are fine with any numbers. You cannot have any negative values, otherwise the square root will not be a real number.
Minimum: 0 inclusive, maximum: infinity
Domain:
← Previous 1 3
|
# Aptitude Percentage Test Paper 2
6) What is 3% of 5%?
1. 60 %
2. 70 %
3. 65 %
4. 75 %
Explanation:
7) Find the value of 25% of 10% of Rs. 800.
1. 25
2. 20
3. 30
4. 35
Explanation:
= 25 % of 80
= 20
8) If two values are 20% and 60% of a third value, what percentage is the first value of the second value?
1. 22.2%
2. 33.3%
3. 27.7%
4. 31.1%
Explanation:
Let the third value is x.
Then, first value=
Second value =
9) If the height of Ramesh is less by 20% than Suresh, the height of Suresh will be greater than that of Ramesh by how many percent?
1. 25%
2. 30%
3. 32%
4. 28%
Explanation:
Let the height of Suresh is x and height of Ramesh is y.
Then, height of Ramesh = x - 20% of x
y = x-x
y = x -
x =y
Now, height of Suresh - height of Ramesh
∴ Height of Suresh will be greater than that of Ramesh by
Solution 2:
Apply formula;
r = 20%
∴Required percentage =
10) The population of a town is increased from 54500 to 58500 in one year. What is the percentage increase in the population?
1. 6.34%
2. 5.34%
3. 7.34%
4. 4.34%
Explanation:
Increase in population = 58500 - 54500 = 4000
∴Percentage increase =* 100 = 7.34 %
Percentage Aptitude Test Paper 1
Percentage Aptitude Test Paper 2
Percentage Aptitude Test Paper 3
Percentage Aptitude Test Paper 4
Percentage Aptitude Test Paper 5
Percentage Aptitude Test Paper 6
Percentage Aptitude Test Paper 7
Percentage Aptitude Test Paper 8
Percentage Concepts
|
" ">
# Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre $O$ are $7\ cm$ and $14\ cm$ respectively and $\angle AOC = 40^o$"
Given:
Two concentric circles of radii $7\ cm$ and $14\ cm$ where $\angle AOC=40^{o}$ in the given figure.
To do:
We have to find the area of the shaded region, enclosed between two concentric circles
Solution:
Area of the region ABDC $=$Area of sector AOC $–$ Area of sector BOD
$=\frac{\theta }{360^{o}} \times \pi r^{2}_{1} -\frac{\theta }{360^{o}} \times \pi r^{2}_{2}$
$=\frac{40}{360^{o}} \times \frac{22}{7} \times ( 14)^{2} -\frac{40^{o}}{360^{o}} \times \frac{22}{7} \times ( 7)^{2}$
$=\frac{22}{7} \times \frac{1}{9}\times( 196-49)$
$=\frac{22\times147}{7\times9}$
$=\frac{154}{3}$
$=51.33cm^{2}$
Area of circular ring $=$Area of the outer ring$-$Area of the inner ring
$=\pi r^{2}_{1} -\pi r^{2}_{2}$
$=\frac{22}{7}( 14^{2} -7^{2})$
$=\frac{22}{7}(14+7)(14-7)$
$=22 \times (21)$
$=22\ \times \ 21$
$=462\ cm^{2}$
$\therefore$ Required shaded region $=$Area of circular ring$–$Area of region ABDC
$=462 – 51.33$
$=410.67\ cm^{2}$
Therefore, the area of shaded region is $410.67\ cm^{2}$.
Updated on: 10-Oct-2022
113 Views
|
What Is Binary Search in Data Structure and Algorithm?
//
Angela Bailey
Binary search is an efficient algorithm used to search for an element in a sorted list. It follows a divide-and-conquer approach and is widely used in computer science and data structures. In this article, we will explore the concept of binary search and understand how it works.
What is Binary Search?
Binary search is a searching algorithm that operates on sorted data structures, such as arrays or linked lists. It works by repeatedly dividing the search interval in half until the Target element is found or determined to be absent.
The algorithm starts by comparing the Target element with the middle element of the sorted list. If they are equal, the search is successful, and the algorithm returns the index of the middle element.
If the Target element is less than the middle element, the algorithm continues searching in the lower half of the list. Conversely, if the Target element is greater than the middle element, it continues searching in the upper half of the list. This process continues until either the Target element is found or there are no more elements to search.
How Does Binary Search Work?
To better understand how binary search works, let’s consider an example:
• Step 2: Set two pointers: low and high. The low pointer points to the first element of the list, and high points to its last element.
• Step 3: Calculate mid, which represents the index between low and high: `<mid> = (<low> + <high>) / 2`.
• Step 4: Compare the Target element with the element at index mid.
• If they are equal, the search is successful.
• If the Target element is less than the element at index mid, update high to be one less than mid.
• If the Target element is greater than the element at index mid, update low to be one more than mid.
• Step 5: Repeat steps 3-4 until either the Target element is found or there are no more elements to search.
An Example:
To illustrate binary search, let’s search for the number 7 in a sorted list [1, 3, 5, 7, 9, 11]. We start with:
• Low: points to index 0.
• High: points to index 5.
We calculate the middle index as follows:
`<Mid> = (0 + 5) / 2 = 2.5 (rounded down to the nearest integer) = 2.`
The middle element of our list is at index 2 and has a value of ‘5’.
The Target element ‘7’ is greater than ‘5’, so we update our low pointer as follows:
`<Low> = <Mid> + 1 = 2 + 1 = 3.`
The new low pointer points to index 3, which is the next element in our search interval.
We calculate the midpoint again:
`<Mid> = (3 + 5) / 2 = 4.`
The middle element at index 4 has a value of ‘9’.
The Target element ‘7’ is less than ‘9’, so we update our high pointer as follows:
`<High> = <Mid> - 1 = 4 - 1 = 3.`
Our high pointer now points to index 3, which is the previous element in our search interval.
We calculate the midpoint once again:
`<Mid> = (3 + 3) / 2 = 3.`
The middle element at index 3 has a value of ‘7’, which matches our Target element. Therefore, the search is successful, and the algorithm returns the index of the Target element as ‘3’.
Time Complexity and Space Complexity
Binary search has a time complexity of O(log n), where n is the number of elements in the sorted list. This means that binary search can efficiently handle large lists and performs significantly better than linear search algorithms with a time complexity of O(n).
The space complexity of binary search is O(1) since it only requires a constant amount of additional space for storing pointers and variables used during the search process.
Conclusion
Binary search is an efficient searching algorithm used on sorted data structures. It follows a divide-and-conquer approach to quickly locate the Target element.
By repeatedly dividing the search interval in half, binary search reduces the number of elements to search, resulting in a logarithmic time complexity. This makes it a popular choice for searching applications that deal with large datasets.
By understanding how binary search works and its time and space complexity, you can leverage this algorithm to improve the performance of your applications when dealing with sorted data.
|
Module
for
Algebra of Complex Numbers, Revisited
1.5 The Algebra of Complex Numbers, Revisited
The real numbers are deficient in the sense that not all algebraic operations on them produce real numbers. Thus, for to make sense, we must consider the domain of complex numbers. Do complex numbers have this same deficiency? That is, if we are to make sense of expressions such as , must we appeal to yet another new number system? The answer to this question is no. In other words, any reasonable algebraic operation performed on complex numbers gives complex numbers. Later we show how to evaluate intriguing expressions such as . For now we only look at integral powers and roots of complex numbers.
The important players in this regard are the exponential and polar forms of a non-zero complex number . By the laws of exponents (which, you recall, we have promised to prove in Section 5.1) we have
,
and
.
Example 1.15. Show that in two ways.
Solution. (Method 1): The binomial formula (Exercise 14 of Section 1.2) gives
(Method 2): Using iIdentity stated above and Example 1.12 yields
Explore Solution 1.15.
Which method would you use if you were asked to compute ?
Example 1.16. Evaluate .
Solution.
Explore Solution 1.16.
Extra Example. Evaluate in two ways.
Explore Extra Solution.
An interesting application of the laws of exponents comes from putting the equation into its polar form. Doing so gives
,
which is known as De Moivre's Formula, in honor of the French mathematician Abraham de Moivre (1667--1754).
Example 1.17. Use DeMoivre's formula to show that .
Solution. If we let n=5 and use the binomial formula to expand the left side of De Moivre's formula, we obtain
The real part of right side of this expression is . Equating it to the real part of establishes the desired result. Furthermore, it can be shown that .
Explore Solution 1.17.
A key aid in determining roots of complex numbers is a corollary to the Fundamental Theorem of Algebra. We prove this theorem in Section 6.6. Our proofs must be independent of the conclusions we derive here because we are going to make use of the corollary now.
Theorem 1.4 (Fundamental Theorem of Algebra). If P(z) is a polynomial of degree n () with complex coefficients, then the equation has precisely n (not necessarily distinct) solutions.
Proof. Refer to Section 6.6.
Example 1.18. Let and find its zeros. This polynomial of degree 3 can be written as . Hence the equation has solutions and . Thus, in accordance with the Fundamental Theorem of Algebra, we have three solutions, with being repeated roots.
Graphs of Arg[P(z)] and |P(z)| for .
Explore Solution 1.18.
The corollary implies that if we can find n distinct solutions to the equation (or ) , we will have found all the solutions. We begin our search for these solutions by looking at the simpler equation . Solving this equation will enable us to handle the more general one quite easily.
To solve we first note an important condition that determines when two nonzero complex numbers are equal. If we let and , then
(i.e., ) iff and ,
where k is an integer. That is, two complex numbers are equal iff their moduli agree and an argument of one equals an argument of the other to within an integral multiple of .
We now find all solutions to in two stages, with each stage corresponding to one direction in the iff part of the above relation. First, we show that if we have a solution to , then the solution must have a certain form. Second, we show that any quantity with that form is indeed a solution.
For the first stage, suppose that is a solution to . Putting the latter equation in exponential form gives , so we must have and . In other words, and , where k is an integer.
So, if is a solution to , then and must be true. This observation completes the first stage of our solution strategy. For the second stage, we note that if , and , then is indeed a solution to because . For example, if n=7 and k=3, then is a solution to because .
Furthermore, it is easy to verify that we get n distinct solutions to (and, therefore, all solutions) by setting k=0,1,2,...,n-1. The solutions for k=n,n+1,... merely repeat those for k=0,1,..., because the arguments so generated agree to within an integral multiple of . As we stated in Section 1.1, the n solutions can be expressed as
for .
They are called the roots of unity.
When k=0 in the above equation, we get , which is a rather trivial result. The first interesting root of unity occurs when k=1, giving . This particular value shows up so often that mathematicians have given it a special symbol.
Definition 1.9 (Primitive nth Root of Unity). For any natural number n, the value given by
is called the primitive root of unity.
By De Moivre's formula, the roots of unity can be expressed as
.
Geometrically, the roots of unity are equally spaced points that lie on the unit circle and form the vertices of a regular polygon with n sides.
Example 1.19. The solutions to the equation are given by the eight values
for .
In Cartesian form, these solutions are . The primitive root of unity is
.
Figure 1.18 illustrates this result.
Figure 1.18 The eight eighth roots of unity.
Explore Solution 1.19.
The procedure for solving is easy to generalize in solving for any nonzero complex number c. If and , then iff . But this last equation is satisfied iff
, and , where k is an integer.
As before, we get n distinct solutions given by
for .
Each of the above solutions can be considered an root of c. Geometrically, the roots of c are equally spaced points that lie on the circle and form the vertices of a regular polygon with n sides. Figure 1.19 illustrates the case for n=5.
Figure 1.19 The five solutions to the equation .
It is interesting to note that if is any particular solution to the equation , then all solutions can be generated by multiplying by the various roots of unity. That is, the solution set is
.
The reason for this is that if , then for any we have
,
and that multiplying a number by increases an argument of that number by , so that contain n distinct values.
Example 1.20. Find all the cube roots of , i.e. find all solutions to the equation .
Solution.
for . The Cartesian forms of the solutions (shown in Figure 1.20) are
.
Figure 1.20 The point and its three cube roots .
Explore Solution 1.20.
Is the quadratic formula valid in the complex domain? The answer is yes, provided we are careful with our terms.
Theorem 1.5 (Quadratic Formula). If , then the solution set for z is
,
where by we mean all distinct square roots of the number inside the parenthesis.
Proof . The proof is left as an exercise for the reader.
Example 1.21. Find all solutions to the equation .
Solution. The quadratic formula gives . As , we compute for and . In Cartesian form, this expression reduces to
,
and
.
Thus, our solution set is .
Explore Solution 1.21.
In Exercise 5 of Section 1.2 we asked you to show that a polynomial with nonreal coefficients must have some roots that do not occur in complex conjugate pairs. This last example gives an illustration of such a phenomenon.
DeMoivre's Theorem
Roots of Cubic Equations
Roots of Quartic Equations
Complex Roots of Polynomials
Quaternions
The Next Module is
The Topology of Complex Numbers
(c) 2012 John H. Mathews, Russell W. Howell
|
# College Physics 4.9 – The acceleration of a child in a wagon push by two children in opposite directions
## Solution:
### Part A
If the acceleration of the child in the wagon is to be calculated, the system of interest is the wagon with the child in it.
### Part B
Considering all the forces acting on the wagon and child with mass m, there will be 5 forces. The horizontal forces are the two forces coming from the two other children, denoted as Fr and Fl. We also have friction going against the direction of motion.
The vertical forces are the weight of the wagon and the child and the normal force. These forces are shown in the free-body diagram shown.
### Part C
From Newton’s second law of motion, we know that
$\displaystyle \text{net F}=\text{ma}$
Considering the horizontal direction to the right as positive, the forces involved are Fr, Fl and f. Therefore, we have
$\displaystyle \text{F}_{\text{r}}-\text{F}_{\text{l}}-\text{f}=\text{ma}$
Solving for acceleration, we have
$\displaystyle \text{a}=\frac{\text{F}_\text{r}-\text{F}_\text{l}-\text{f}}{\text{m}}$
$\displaystyle \text{a}=\frac{90.0\:\text{N}-75.0\:\text{N}-12.0\:\text{N}}{23.0\:\text{kg}}$
$\displaystyle \text{a}=0.1304\:\text{m/s}^2$ ◀
### Part D
If the friction is 15 N, the acceleration would be
$\displaystyle \text{a}=\frac{90.0\:\text{N}-75.0\:\text{N}-15.0\:\text{N}}{23.0\:\text{kg}}$
$\displaystyle \text{a}=0\:\text{m/s}^2$ ◀
There would be no acceleration, meaning that the system is in equilibrium. The system is either not moving or it’s moving at a constant velocity.
|
Math Calculators, Lessons and Formulas
It is time to solve your math problem
mathportal.org
# Complex number calculator
The calculator does the following: extracts the square root, calculates the modulus, finds the inverse, finds conjugate and transforms complex numbers into polar form. For each operation, the solver provides a detailed step-by-step explanation.
problem
Find the modulus of
$$z = 6+3i$$
solution
The modulus of $z$ is:
$$|z| = 3 \sqrt{ 5 }$$
explanation
To find modulus of a complex number $z = a + bi$ we use formula:
$$|z| = \sqrt{a^2 + b^2}$$
In this example we have $a = 6$ and $b = 3$ so:
\begin{aligned}|z| &= \sqrt{ 6^2 + 3^2 } \\[1 em]|z| &= \sqrt{ 36 + 9 } \\[1 em]|z| &= \sqrt{ 45 } \\[1 em]|z| &= 3 \sqrt{ 5 } \\[1 em] \end{aligned}
## Report an Error !
Script name : complex-unary-operations-calculator
Form values: 6 , 1 , 3 , mod , g , Find the modulus conjugate of 6+3i , ,
Comment (optional)
## Share Result
Or just copy and paste the link wherever you need it.
Operations with complex number
five operations with a single complex number
show help ↓↓ examples ↓↓ tutorial ↓↓
$z = 4 + 6i$
$z = 2 - \frac{3}{2}i$
$z = \sqrt{2} - \sqrt{5}i$
Modulus (Magnitude) Conjugate Inverse Roots Polar form
Find approximate solution
Hide steps
working...
EXAMPLES
example 1:ex 1:
Find the complex conjugate of $z = \frac{2}{3} - 3i$.
example 2:ex 2:
Find the modulus of $z = \frac{1}{2} + \frac{3}{4}i$.
example 3:ex 3:
Find the inverse of complex number $3 - 3i$.
example 4:ex 4:
Find the polar form of complex number $z = \frac{1}{2} + 4i$.
TUTORIAL
## Operations on complex numbers
This calculator performs five operations on a single complex number.
It computes module, conjugate, inverse, roots and polar form.
### 1 : Modulus ( Magnitude )
The modulus or magnitude of a complex number ( denoted by $\color{blue}{ | z | }$ ), is the distance between the origin and that number.
If the $z = a + bi$ is a complex number than the modulus is
$$|z| = \sqrt{a^2 + b^2}$$
Example 01: Find the modulus of $z = \color{blue}{6} + \color{purple}3{} i$.
In this example $\color{blue}{a = 6}$ and $\color{purple}{b = 3}$, so the modulus is:
\begin{aligned} | z | &= \sqrt{ a^2 + b^2} = \sqrt{6^2 + 3^2} = \\[1 em] &= \sqrt{36 + 9} = \sqrt{45} = \\[1 em] &= \sqrt{9 \cdot 5} = 3 \sqrt{5} \end{aligned}
### 2 : Conjugate
To find the complex conjugate of a complex number, we need to change the sign of the imaginary part. The conjugate of $z = a \color{red}{ + b}\,i$ is:
$$\overline{z} = a \color{red}{ - b}\,i$$
Example 02: The complex conjugate of $~ z = 3 \color{blue}{+} 4i ~$ is $~ \overline{z} = 3 \color{red}{-} 4i$.
Example 03: The conjugate of $~ z = - 4i ~$ is $~ \overline{z} = 4i$.
Example 04: The conjugate of $~ z = 15 ~$ is $~ \overline{z} = 15 ~$, too.
### 4 : Inverse
The inverse or reciprocal of a complex number $a + b\,i$ is
$$\color{blue}{ \frac{1}{a + b\,i}}$$
Here is an example of how to find the inverse.
Example 06: Find the inverse of the number $z = 4 + 3i$
\begin{aligned} \frac{1}{z} &= \frac{1}{4+3i} = \frac{1}{4+3i} \cdot \frac{4-3i}{4-3i} = \\[1 em] &= \frac{4-3i}{(4+3i)(4-3i)} = \frac{4-3i}{4^2 - (3i)^2} = \\[1 em] &= \frac{4-3i}{16+9} = \frac{4-3i}{25} = \frac{4}{25} - \frac{3}{25} i \end{aligned}
### 3 : Polar Form
The polar form of a complex number $z = a + i\,b$ is given as $z = |z| ( \cos \alpha + i \sin \alpha)$.
Example 05: Express the complex number $z = 2 + i$ in polar form.
To find a polar form, we need to calculate $|z|$ and $\alpha$ using formulas in the above image.
$$|z| = \sqrt{2^2 + 1^2} = \sqrt{5}$$ $$\tan \alpha = \dfrac{b}{a} = \dfrac{1}{2} \implies \alpha = \tan^{-1}\left(\dfrac{1}{2}\right) \approx 27^{o}$$
So, the polar form is:
$$z = \sqrt{5} \left( \cos 27^{o} + i \sin27^{o} \right)$$
Search our database of more than 200 calculators
|
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
### Course: Arithmetic (all content)>Unit 6
Lesson 6: Adding decimals: standard algorithm
Adding decimals is an essential math skill. To add decimals, such as 9.087 and 15.31, align the decimal points and place values. Fill in any missing digits with zeros to make the numbers the same length. Then, perform the addition as usual, starting with the smallest place value. The result of adding 9.087 and 15.31 is 24.397. Created by Sal Khan.
## Want to join the conversation?
• how did you get the decimals to line up with each other and why do you put a number off to the side
• You line up the decimal points.
Wrong
`` 1.23 +45.6------- 0.579``
Correct
`` 1.23+45.60-------- 46.83``
• how does this help us in a real life situation
• You have to use decimals when go shopping. because if something cost 28.34 and then you also get another thing that costs 15.24 you have to add them so u know how much u spending
• how do you do it
• Use the decimals to guide the place value alignment.
• line up the decimals above and below,
then add each column of numbers,
• empty spaces are equal to zero
9.087 + 15.31 = ?
`` 9.087+ 15.31————————— 24.397 ←🥳 answer``
★In Addition order doesn't matter, because it's Commutative, meaning we can rearrange addition and get the same answer,
1 + 7 = 8
7 + 1 = 8
So either decimal value can be first or second, but always line up the place values.
`` 15.31+ 9.087————————— 24.397 ←🥳 same answer``
★This method works for more than two values as well, as long as we always line up the decimals so that the place values form columns!
2.14 + 0.3421 + 3 + 11.2 = ?
`` 2.14 0.3421 3+ 11.2—————————— 16.6821 ←🥳 answer``
Rearranged,
with place keeper zeros…
`` 11.2000 3.0000 0.3421+ 2.1400—————————— 16.6821 ←🥳 same answer``
★ Just remember: we must line up the decimals and keep the place values aligned throughout the calculation, move the decimal straight down, so it is aligned in the answer too, then it's just adding the columns like usual…
Start adding in the far right column, carry values to the left, etc…
(ㆁωㆁ) Hope this helps someone!
• is it allowed to add an zero at the whole number?
• I mean I get ur question. You can add a zero at a whole number. HOWEVER it won't change the answer. For example, 5+0=5. But it is allowed but no use.
• I made a chart and placed them into the chart to solve it, is that smart?
• yes, it is very smart! i learned how to solve it using a chart, and it's good that you still use it as a strategy! ^^
• Sal is a good teacher
• his voice is nice :]
• Yes it is very
• Is it compulsory to line the decimals
• Corresponding place values must be added together, the decimals naturally align when the columns are in the correct order.
Therefore aligning the decimals is an easy way to align the place values correctly into columns for manual addition.
So, when creating accurate addition columns, it is also compulsory to line up the decimals.
• The tip of adding decimals are that you should line up the decimal points together.
• Do you align the decimals even if you are multiplying decimals
• if you are multiplying decimals, you do not have to line up the decimal points. You keep the decimal points in the correct space in the numbers, but otherwise, write it out like a normal multiplication problem. You solve it like a normal problem too. so if it is 1.5 X 3.4, you would write it out like a normal multiplication problem:
2
3.4
x1.5
--------
1
1 7 0
+3 4 0
-------
5 1 0, but to put in the decimal point, you have to think of the original numbers, which were 1.5, and 3.4, 1.5 had one number to the right of the decimal point, so note that, and 3.4 also had one number to the right of the decimal point, So note that as well, add that up and it means that you have to put the decimal point in the problem so that it is to the left of two numbers, so the answer is not 510, it is 5.10.
|
# Question Video: Finding the Missing Values in a Cumulative Frequency Table Mathematics • 9th Grade
The table shows the ages of people applying to a company. What are the missing values from the cumulative frequency table respectively?
02:20
### Video Transcript
The following table shows the ages of people applying to a company. What are the missing values from the cumulative frequency table respectively?
In this question, we are given a grouped frequency table of the ages of people applying to a job. And we want to find the missing values in the table. To do this, we can start by recalling that the cumulative frequency is the running total of the frequencies. This means that we can calculate the cumulative frequency by adding all of the previous frequencies together.
The first cumulative frequency is just equal to the frequency in the first group, which is two. If we wanted to calculate the cumulative frequency in the second group, we would need to add the frequency of five onto the previous cumulative frequency to obtain five plus two equals seven. Of course, we are already given that the cumulative frequency in this group is seven.
We can follow the same process for the missing entries. In the third group, we need to add its frequency of 10 onto the previous cumulative frequency of seven, which we can calculate is equal to 17. For the next missing entry, we add the frequency of 12 onto the previous cumulative frequency of 31 to get 43. For the final cumulative frequency, there are two ways that we can find its value. Either we can use the fact that the final cumulative frequency is always equal to the total population to see that it is equal to 69. Or we can follow the same process we did before to add the frequency of seven onto the previous cumulative frequency of 62 to get 69.
Hence, the missing values from the cumulative frequency table in order are two, 17, 43, and 69.
|
# Video: Parallel and Perpendicular Lines
We examine the equations of some lines to see which parameters tell you whether two lines are parallel or perpendicular, then we learn how to write the equation of a line that is parallel or perpendicular to another and passes through a specified point.
17:53
### Video Transcript
In this video, we’ll be looking at and working with parallel and perpendicular lines. We’ll go through the definitions of the terms parallel and perpendicular, and then we’ll look at their equations and go over some questions. Two lines are said to be parallel if they’re in the same plane, and no matter how far you extend them in each direction, they’re the same distance apart. Another popular definition also says that they must never intersect. And the effect of that definition is that if you have two lines that are in fact the same line, one on top of each other, they’re not parallel because they intersect in an infinite number of different places. So although they’re always the same distance apart, zero, we don’t count them as being parallel because they do intersect.
Now here’s an example where we’ve got two different lines. They’re both on the 𝑥𝑦-coordinate plane and they’re parallel. Wherever you measure the distance between them, you’ll always get the same answer. Two lines are said to be perpendicular if a right angle is formed at their point of intersection, like it is here. So these two lines are both on the 𝑥𝑦-coordinate plane and they intersect at the point two, zero. And the angle between them there is 90 degrees. In the rest of this video, we’re gonna be working only with lines in the 𝑥𝑦-coordinate plane. And we’ll be looking at the equations of these lines, picking out aspects that tell us if they’re parallel or they’re perpendicular.
And equations of straight lines are really an important part of this topic, so let’s just do a quick recap of 𝑦 equals 𝑚𝑥 plus 𝑏. So the general form of an equation of a straight line is 𝑦 equals 𝑚𝑥 plus 𝑏 or maybe 𝑦 equals 𝑚𝑥 plus 𝑐, depending on where you live. The 𝑚-value, the multiple of 𝑥, the coefficient of 𝑥, tells you the slope of the line. And the value of 𝑏 tells you where it cuts the 𝑦-axis. Now the slope, or the gradient, of the line is defined as being the amount that the 𝑦-coordinate changes by, when we increase the 𝑥-coordinate by one.
Now, I’ve drawn two points 𝑎 and 𝑏 on my line and I’ve increased the 𝑥-coordinate by one to get from 𝑎 to 𝑏. So the difference in their 𝑥-coordinates is positive one. And when I do that, the difference in the 𝑦-coordinates from 𝑎 to 𝑏 will be an increase of 𝑚. So whatever value this is, okay, that is the amount that the 𝑦-coordinate increases by, every time I increase my 𝑥-coordinate up by one, moving along the line. So if that line had been for example 𝑦 equals 0.5 𝑥 plus three, the multiple of 𝑥, the slope, the 𝑚-value is 0.5. This means that every time I increase my 𝑥-coordinate by one, my 𝑦-coordinate will increase by 0.5.
So if we arrange the equation of the line in this format 𝑦 equals 𝑚𝑥 plus 𝑏, then it’s a really simple matter to read off the slope of that line. Just look at the coefficient of 𝑥, the multiple of 𝑥; that is your slope. So here we have a question.
Which of the following straight lines have the same slope or gradient? And then we’ve got five lines. A is 𝑦 equals three 𝑥 minus seven. B is 𝑦 equals minus a half 𝑥 plus three. C is 𝑦 equals minus three 𝑥 plus seven. D is 𝑦 equals minus a half 𝑥 plus five. And E is 𝑦 equals three 𝑥 plus nine. Now, all of those equations are already in the 𝑦 equals 𝑚𝑥 plus 𝑏 format for us, so what we need to do is look at the 𝑥-coefficients. And if they’re the same number with the same sign, then they’ll have the same slope or gradient.
So B and D have the same slope, or gradient, of negative a half. And A and E have the same slope of three. C has got a slope of negative three, so it’s a different sign to A and E. So that is not the same slope or gradient. Now when two lines have the same slope or gradient, we call them parallel. So here, A is parallel to E and B is parallel to D.
So now we can recognise parallel lines just by looking at their equations, so long as they’re in the 𝑦 equals 𝑚𝑥 plus 𝑏 format. So thinking about the slope of perpendicular lines, if two lines are perpendicular, then the product of their slopes, or gradients, is negative one.
Well, why would that be? Let’s take a look. So I’ve got two lines here, 𝐿 one, which I’ve given the equation 𝑦 equals 𝑚𝑥 plus 𝑎, and 𝐿 two, which I’ve given the equation 𝑦 equals 𝑛𝑥 plus 𝑏. Let’s take that point of intersection and then a point on each of those lines which has an 𝑥-coordinate which is one greater than that 𝑥-coordinate at the point of intersection. For line one, the difference in the 𝑦-coordinates between those two points, this one and this one, is going to be 𝑚. And for line two, the difference in those coordinates is going to be 𝑛. Well in fact, it’s going to be negative 𝑛. So line two is a downhill line, which means it’s going to have a negative gradient.
Now I’m interested at the moment in the distance here. And that is a positive distance. So I need to take the negative of that negative gradient to work out what that actual distance is going to be. Now if I label those points A, B, and C, we know that triangle ABC is a right-angled triangle because the two lines are perpendicular. So the angle ABC is a right angle. And in right-angled triangles, we can use the Pythagorean theorem to say that the square of the hypotenuse is equal to the sum of the squares of the other sides. So length AC squared is equal to length AB squared plus length BC squared. And we know that length AC is 𝑚 plus negative 𝑛. So AC squared is 𝑚 plus negative 𝑛 all squared. Now if we look back at our diagram, we can see that we’ve got two more right-angled triangles. We’ve got this one here and this one here.
And again, we can use the Pythagorean theorem to work out the length of the hypotenuse of each of those. We know that the top triangle has a height of ℎ and a width of- sorry, a height of 𝑚 and a width of one. And the bottom triangle has a height of negative 𝑛 and a width of one. So those lengths AB and BC are the square root of 𝑚 squared plus one squared and the square root of negative 𝑛 all squared plus one squared. And now we can start to simplify this. 𝑚 plus negative 𝑛 is just 𝑚 take away 𝑛. So the left-hand side becomes 𝑚 minus 𝑛 all squared. The square root of 𝑚 squared plus one squared all squared is just 𝑚 squared plus one. And negative 𝑛 all squared is just 𝑛 squared. So the square root of negative 𝑛 all squared plus one squared all squared is just 𝑛 squared plus one.
Now 𝑚 minus 𝑛 all squared means 𝑚 minus 𝑛 times 𝑚 minus 𝑛 and just tidying up the right-hand side, I’ve got 𝑚 squared plus 𝑛 squared plus two. Now multiplying each term in the first bracket by each term in the second bracket on the left-hand side, I get 𝑚 squared minus two 𝑚𝑛 plus 𝑛 squared. So I can subtract 𝑚 squared from both sides and I can subtract 𝑛 squared from both sides, which eliminates 𝑚 squared from both sides and eliminates 𝑛 squared from both sides. So I’ve got minus two 𝑚𝑛 is equal to two. Now if I divide both sides by negative two, I get 𝑚 times 𝑛 is equal to two over minus two, which is equal to negative one. Now remember, 𝑚 was the slope of my first line and 𝑛 was the slope of my second line. So 𝑚𝑛 is the product of the two slopes of the lines. So when those lines were perpendicular, it doesn’t matter what the actual slopes were, we knew that when we multiply them together, we’ll always get this answer of negative one.
Now if you got a bit lost along the way during that explanation, don’t worry. That bit’s not important. This is what you’ve got to remember. If two lines are perpendicular, then the product of their slopes, or gradients, is negative one. So if we call 𝑚 the slope of the first line and 𝑛 the slope of the second line, 𝑚 times 𝑛 is equal to negative one. Or if I divide both sides by 𝑛, I get 𝑚 is equal to minus one over 𝑛. Or if I divide both sides by 𝑚, I get 𝑛 is equal to minus one over 𝑚. In other words, each slope is the negative reciprocal of the other. This means that if I know one of the slopes, I can find the perpendicular slope by just changing the sign and flipping that number.
So for example, if 𝑚 was five, 𝑛 would just be negative one over five, the opposite sign and then flip that number. If 𝑚 was minus three, I take the opposite sign to make it positive and flip that number, three over one becomes one over three. And if 𝑚 was equal to two-thirds, 𝑛 will be negative three over two. So knowing this rule means that if you know the gradient, or the slope, of a particular line, it’s very easy to work out what the gradient, or the slope, of a perpendicular line to that one would be.
So to sum up the basic facts, two lines are parallel if they have the same slope but different 𝑦-intercept. For example, 𝑦 equals seven 𝑥 minus five and 𝑦 equals seven 𝑥 plus two. They both have a slope of seven and their 𝑦-intercepts are different, minus five and positive two, so they’re parallel. And two lines are perpendicular if the product of their slopes is equal to one. For example, 𝑦 equals three 𝑥 minus one and 𝑦 equals minus a third 𝑥 plus nine. The slope of the first is three and the slope of the second is negative a third. So the product of those slopes is three times negative a third. But three is the same as three over one, so to make this a fraction calculation, I’ve got three times negative one over one times three. Well that’s negative three over three, which is equal to negative one. So it meets the criteria, so the lines are perpendicular.
And one more example, 𝑦 equals negative two-sevenths 𝑥 plus eight and 𝑦 equals seven over two 𝑥 plus eight. The slopes are negative two over seven and seven over two. They’re the negative reciprocal of each other. And if I multiply the slopes together, negative two over seven times seven over two is equal to negative fourteen over fourteen which is equal to negative one. So those two lines are perpendicular. And with perpendicular lines, it doesn’t matter that those intercepts were both at positive eight. The slopes are different, so they are different lines. They are definitely perpendicular. Let’s have a look at some typical questions then.
Two lines, A and B, have slopes three-quarters and minus four over three, respectively. Are they parallel, perpendicular, or neither?
Well, the slopes aren’t equal, so they’re definitely not parallel. Now if we multiply those slopes together, we get three-quarters times negative four over three, which is negative twelve over twelve, which is negative one. So it looks like those two sl- lines would be perpendicular.
Next, which of the following lines are parallel to each other? And then we’ve got five equations. A) 𝑦 equals eight 𝑥 minus five. B) Two 𝑦 is equal to eight 𝑥 plus three. C) Eight 𝑥 minus 𝑦 plus two equals zero. D) A half 𝑦 minus four 𝑥 equals twelve. And E) 𝑦 equals minus an eighth 𝑥 plus seven.
Well with A and E, they’re already in the 𝑦 equals 𝑚𝑥 plus 𝑏 format, so it’s easy enough to read off what the slope is. But for B, C, and D, we’re gonna have to need to do a bit of rearranging to get them in the right format to be able to read off their slopes. For equation B, I’m gonna have to divide both sides of that equation by two. So the left-hand side, half of two 𝑦 is just 𝑦. And then dividing each term on the right-hand side by two, half of eight 𝑥 is four 𝑥 and half of three is three over two. For equation C, I can just add 𝑦 to both sides which will eliminate it from the left-hand side, and give me just 𝑦 on the right-hand side. So that gives me eight 𝑥 plus two equals 𝑦. Now I’m just gonna write this the other way around 𝑦 equals eight 𝑥 plus two, because that’s the format that we’re familiar with.
Now D needs a little bit more work. I’ve got the 𝑦 and the 𝑥 term on the left-hand side and then just the number on the right. So first of all, I’m gonna add four 𝑥 to both sides, which gives me a half 𝑦 is equal to, well twelve plus four 𝑥 or four 𝑥 plus twelve. And then doubling each side of that equation to leave me with just 𝑦. Two lots of half 𝑦 are 𝑦, two lots of four 𝑥 are eight 𝑥, and two lots of twelve are twenty-four. Now we’ve got our equations in the right format. It’s a pretty straightforward matter of finding the slopes, so we can see which lines are parallel to each other. In A, the slope is eight. In B, the slope is four. In C, the slope is eight. In D, the slope is also eight. And in E, the slope is negative an eighth. So A, C, and D, all have a slope of eight. So the answer is A, C, and D are parallel.
Now we’ve got to write the equation of a line which is parallel to 𝑦 equals eight 𝑥 minus four.
So it’s got to be parallel, which means it’s got to have the same slope of eight. So it’s gotta start off with 𝑦 equals eight 𝑥. And then we can add anything we like cause wherever that line cuts through the 𝑦-axis, it doesn’t matter. It’s gonna be parallel to this line. The only thing that you shouldn’t use is eight 𝑥 minus four. Don’t make it exactly the same line because most people would say that they’re not parallel because it’s the same line. So you can write anything you like here eight 𝑥 plus a thousand. Here we go, that’s a line which is parallel to 𝑦 equals eight 𝑥 minus four.
Next, we’ve got to write the equation of a line which is perpendicular to 𝑦 equals three 𝑥 minus two.
Well, the slope is three, so the slope of that perpendicular line must be the negative reciprocal, so that’s minus one over three. So our equation is gonna start off 𝑦 equals minus one over three 𝑥. And we can add anything we like. We could just leave it as minus one over three 𝑥, or we could add any number we like to make a perpendicular line.
Now, we’ve got to write an equation for the straight line that is parallel to 𝑦 equals a half 𝑥 plus five and passes through the point six, 10. Well, we know that the slope of the line 𝑦 equals a half 𝑥 plus five has a slope of a half. So if our line is gonna be parallel to that, it must also have a slope of a half. But of course, that line could cut the 𝑦-axis anywhere, so we could move that line up or down. What we’re told is that it must pass through the point six, 10. And that means that when 𝑥 is six, 𝑦 needs to be 10.
Now if we use the general form of our equation 𝑦 equals 𝑚𝑥 plus 𝑏, we know the slope 𝑚 is equal to a half. Now, we’ve got to find out the value of 𝑏. But we know a particular coordinate pair that sits on the line, when 𝑥 equals six, 𝑦 equals 10. So replacing 𝑥 and 𝑦 with six and 10, we’ve got 10 is equal to a half times six plus 𝑏. So 10 is equal to three plus 𝑏. And then subtracting three from both sides, gives me seven is equal to 𝑏. Now, we know the value of 𝑏; we can finish off our equation. 𝑦 is equal to a half 𝑥 plus seven.
Lastly, write an equation for the straight line that is perpendicular to 𝑦 equals three-quarters 𝑥 minus four and passes through the point four, 11.
So we’re trying to find a gradient, or a slope, which is perpendicular to three-quarters. So the slope of our perpendicular line is gonna be negative four over three, the negative reciprocal, remember. Three-quarters times negative four over three is equal to negative one. So that means it’s a perpendicular line. So we’ve got the slope of the line and we also know it’s gonna go through the point four, 11. So when 𝑥 is equal to four, then 𝑦 is equal to 11. So using the format of the equation 𝑦 equals 𝑚𝑥 plus 𝑏, we know that 𝑚 is minus four over three. And we know that when 𝑥 is four, then 𝑦 is 11. So we can use all that information to work out the value of 𝑏.
So 11 is equal to negative four-thirds times four plus 𝑏. Well, four is the same as four over one. So that becomes a fraction calculation, minus four-thirds times four over one. So 11 is equal to negative sixteen-thirds plus 𝑏. So if I add sixteen-thirds to both sides, I’ve got 11 plus sixteen-thirds is equal to 𝑏. And that is equal to 49 over three. So I can put that back into our original equation and get my answer, 𝑦 equals negative four-thirds 𝑥 plus 49 over three.
|
# Find The Square Root Of 2500 By Prime Factorization Method
### Https bit ly exponentsandpowersg8 in this video we will learn.
Find the square root of 2500 by prime factorization method.
Examples on square root of a perfect square by using the prime factorization method. We have already learned in our previous classes to find the prime factors of numbers. To learn more about squares and square roots enrol in our full course now. Finding cube root by prime factorization.
Prime factorization by trial division. Generally prime factorization is used for finding square roots of small numbers. In the prime factorisation method we will write the prime factors of the given number. Covid 19 has led the world to go through a phenomenal transition.
Start by testing each integer to see if and how often it divides 100 and the subsequent quotients evenly. In order of finding cube root by prime factorization we use the following steps. This is a step by step guide for finding the value of square root of 4096 for finding the square root of any number we have two methods. Resolve it into prime factors.
Simplification of square root of 576. I decompose the number inside the square root into prime factors. Since the number is a perfect square you will be able to make an exact number of pairs of prime factors. To find the square root of a perfect square by using the prime factorization method when a given number is a perfect square.
Iii combine the like square root terms using mathematical operations. Ex 6 3 4 find the square roots of the following numbers by the prime factorization method. We can find square root by prime factorization method or by long division method. Ii inside the square root for every two same numbers multiplied one number can be taken out of the square root.
Find primes by trial division and use primes to create a prime factors tree. Take one factor from each pair. Stay home stay safe and keep learning. Resolve the given number into prime factors.
We cover two methods of prime factorization. Here we will learn to find the square root of 576 without using calculators in two different ways. 0 00 how to fin. Square root by prime factorization method example 1 find the square root.
Obtain the given number. Thew following steps will be useful to find square root of a number by prime factorization. Make pairs of similar factors. Take the product of prime factors choosing one factor out of every pair.
I 729we use prime factorization to find square root thus 729 3 3 3 3 3 3square root of 729 3 3 3 9 3 27 ex 6 3 4 find the square roots of t. E learning is the future today.
### Dealing With Missing Depth Recent Advances In Depth Image Completion And Estimation Springerlink
Source : pinterest.com
|
# Equivalent fractions
The two or more fractions that have same value are called the equivalent fractions or equal fractions.
## Introduction
It seems, two or more fractions are dissimilar but if we calculate their values, they are equal surprisingly in some cases. So, the fractions are called as equivalent fractions and also simply called as equal fractions.
The fractions can be formed by the any number of equal divisions but they will be equal mainly due to their equality.
### Example
From a geometrical example, the concept of the equivalent fractions can be understood. Take four rectangles and split them as two, four, six and eight equal parts respectively.
1. If we select a part as a fraction from first rectangle, then the fraction is equal to $\large \frac{1}{2}$
2. If we select two parts as a fraction from second rectangle, then the fraction is $\large \frac{2}{4}$
3. If we select three parts as a fraction from third rectangle, then the fraction is $\large \frac{3}{6}$
4. If we select four parts as a fraction from fourth rectangle, then the fraction is $\large \frac{4}{8}$
Thus, four proper fractions $\dfrac{1}{2}$, $\dfrac{2}{4}$, $\dfrac{3}{6}$ and $\dfrac{4}{8}$ are formed. In all four fractions, the numerators are different and denominators are different but their values are equal. Hence, the fractions are called as the equal fractions.
$\implies$ $\dfrac{1}{2}$ $\,=\,$ $\dfrac{2}{4}$ $\,=\,$ $\dfrac{3}{6}$ $\,=\,$ $\dfrac{4}{8}$ $\,=\,$ $0.5$
Similarly, the improper fractions can also form equivalent fractions. Therefore, the equal fractions can be either proper fractions or improper fractions but they both cannot be formed equivalent fractions.
Latest Math Topics
Jun 26, 2023
###### Math Questions
The math problems with solutions to learn how to solve a problem.
Learn solutions
Practice now
###### Math Videos
The math videos tutorials with visual graphics to learn every concept.
Watch now
###### Subscribe us
Get the latest math updates from the Math Doubts by subscribing us.
|
• Share
Send to a Friend via Email
### Your suggestion is on its way!
An email with a link to:
was emailed to:
Thanks for sharing About.com with others!
# What Are Pie Charts?
Eye colors of 100 third grader students. Brown corresponds to brown eyes, blue to blue eyes, and green to hazel eyes.
C. K. Taylor
One of the most common ways to represent data graphically is called a pie chart. It gets its name by how it looks, just like a circular pie that has been cut into several slices. This kind of graph is helpful when graphing qualitative data, where the information describes a trait or attribute and is not numerical. Each trait corresponds to a different slice of the pie. By looking at all of the pie pieces, you can compare how much of the data fits in each category. The larger a category, the bigger that its pie piece will be.
### Big or Small Slices?
How do we know how large to make a pie piece? First we need to calculate a percentage. Ask what percent of the data is represented by a given category. Divide the number of elements in this category by the total number. We then convert this decimal into a percentage.
A pie is a circle. Our pie piece, representing a given category, is a portion of the circle. Because a circle has 360 degrees all the way around, we need to multiply 360 by our percentage. This gives us the measure of the angle that our pie piece should have.
### An Example
To illustrate the above, let’s think about the following example. In a cafeteria of 100 third graders, a teacher looks at the eye color of each student and records it. After all 100 students are examined, the results show that 60 students have brown eyes, 25 have blue eyes and 15 have hazel eyes.
The slice of pie for brown eyes needs to be the largest. And it needs to be over twice as large as the slice of pie for blue eyes. To say exactly how large it should be, first find out what percent of the students have brown eyes. This is found by dividing the number of brown eyed students by the total number of students, and converting to a percent. The calculation is 60/100 x 100% = 60%.
Now we find 60% of 360 degrees, or .60 x 360 = 216 degrees. This reflex angle is what we need for our brown pie piece.
Next look at the slice of pie for blue eyes. Since there are a total of 25 students with blue eyes out of a total of 100, this means that this trait accounts for 25/100x100% = 25% of the students. One quarter, or 25% of 360 degrees is 90 degrees, a right angle.
The angle for the pie piece representing the hazel eyed students can be found in two ways. The first is to follow the same procedure as the last two pieces. The easier way is to notice that there are only three categories of data, and we have accounted for two already. The remainder of the pie is corresponds to the students with hazel eyes.
The resulting pie chart is pictured above. Note that number of students in each category is written on each pie piece.
### Limitations of Pie Charts
Pie charts are to be used with qualitative data, however there are some limitations in using them. If there are too many categories, then there will be a multitude of pie pieces. Some of these are likely to be very skinny, and can be difficult to compare to one another.
If we want to compare different categories that are close in size, a pie chart does not always help us to do this. If one slice has central angle of 30 degrees, and another has a central angle of 29 degrees, then it would be very hard to tell at a glance which pie piece is larger than the other.
Courtney Taylor
|
# How do you simplify the expression 2 - 8(-7 + x)+ 5x?
Apr 6, 2018
$58 - 3 x$
#### Explanation:
$2 - 8 \left(- 7 + x\right) + 5 x$
$= 2 + 56 - 8 x + 5 x$ expand brackets
$= 58 - 3 x$ collect like terms
Apr 6, 2018
$- 3 x + 58$
#### Explanation:
First, we use the distributive property to multiply/expand $- 8 \left(- 7 + x\right)$:
$- 8 \cdot - 7 = 56$
$- 8 \cdot x = - 8 x$
Now let's put that back into the expression:
$2 + 56 - 8 x + 5 x$
Now we can simplify by combining like terms:
$58 - 3 x$
We usually write expressions with the variable first, so I'm going to rewrite it that way:
$- 3 x + 58$
Hope this helps!
|
# Partial fraction integration rules sheet
Partial fraction
## Partial fraction integration rules sheet
Partial Fractions Combining fractions over a common denominator is a familiar operation from algebra: ( 1). Suppose that N( x) and D( x) are polynomials. Some of the worksheets displayed are Partial fraction decomposition sheet date period Pdf, Partial fractions, Partial fraction decomposition work, Work integration rules sheet using partial fractions, Pdf, Work 5 partial fractions Partial fractions. This method is based on the simple concept of adding fractions by getting a common denominator. SOLUTIONS TO INTEGRATION BY PARTIAL FRACTIONS rules SOLUTION 9 : Integrate. Worksheet: Integration using Partial Fractions 1. The method of partial fractions allows us to split integration the right hand side of integration the above equation into the left hand side. Evaluate the following inde nite integrals. THE METHOD integration OF integration INTEGRATION BY PARTIAL FRACTIONS All of the following problems rules use the method of integration by partial fractions.
Copyright © by Harold Toomey, WyzAnt Tutor 1 Harold’ s Partial Fractions Cheat Sheet 15 April Partial Fractions wikipedia. Linear Factors in Denominator. Section 1- 4 : Partial partial Fractions. In this section we are going to take a look at integrals of rational expressions of polynomials and once again let’ s start this section out with an integral that we integration can already do so we can contrast it with the integrals that we’ ll be doing in this section. Q1: Use partial fractions to evaluate 𝑥 𝑥 rules − 1 𝑥 4 2 d. Mar 13, · Partial rules fraction decomposition is the process of breaking sheet a single complex fraction into multiple simpler fractions.
Example 1: N( x) D( x) = x4 sheet + 5x3 + 16x2 + 26x+ 22 x3 + 3x2 + 7x+ 5 Step 1. You da real mvps! Ax+ B ( ax2+ bx+ c) m. This method rules is used when the factors in the denominator of the fraction are linear ( in other words rules do not have any square or cube terms etc). Partial fractions is the name given to a technique of integration that may be used to sheet integrate any ratio of polynomials. Decompose into partial fractions ( There is a repeated linear factor! Partial rules Fractions A rational function is a fraction sheet in which both the numerator and denominator are polynomials. You should already be quite familiar with performing algebraic operations with such fractions. When integrating functions involving polynomials integration in the denominator, partial fractions can be used to simplify integration. Integration Using method of Partial. Integrals sheet , , we show how to integrate using rules Maple, Integration by Parts In this worksheet, how to rules convert a proper , Partial Fractions, how to explicitly implement integration by parts improper rational fraction to an expression with partial fractions. \$ 1 per month helps! For example, so that we rules can now say that a partial fractions decomposition for is.
Feb 27, · Partial fraction decomposition allows complicated polynomial fractions to be written as the sum of simpler fractions. New students of calculus will find it handy to learn how to decompose sheet functions into partial fractions not just for integration, but for more advanced studies as well. Partial fraction integration rules sheet. For example , € f( x) = 4 x− 2, € g( x) partial = − 3 x+ 5 € h( x) = x+ 26 x2+ sheet 3x− 10 are rational functions. The integrals of many rational functions lead to a natural log function with. org/ wiki/ Partial. Partial fraction integration rules sheet. In this worksheet, rules we will practice using partial fractions rules of rational functions to find the integration of improper fractions. Showing top 8 integration worksheets in the category - Partial. With the remaining fraction quadratic rules factors of the form ( ax 2+ bx + sheet c) n, factor the denomi- nator D into terms that are either linear factors of the form ( px + q) m where ax + bx + c is irreducible. the partial fraction decomposition of a rational function. A ratio of polynomials is called a rational function. ( 1) Z 1 2x3 + x2 x dx ( 2) Z 3x3 5x2 11x+ 9 x2 2x 3 dx ( partial 3) Z x2 + 12x 5 ( x+ 1) 2( x integration 7) dx ( 4) Z 8x2 3x 4 ( 4x 1) ( x2 + 1) dx. rules to determine the terms in the decomposition.
Split 5( integration x + 2) into partial sheet fractions. Mar 16, · Thanks to all of you who support me on Patreon. In this lesson, we used examples to showcase the rules for four cases of. ) getting ( After getting a common denominator, adding fractions, , equating numerators, it follows that ; let ; let ; let ; it follows that . For example, the sum of € f( x.
## Fraction integration
Basic Partial Fraction Decomposition Rules. Here are the basic Partial Fraction Rules when for decomposing fractions. Remember that these rules apply to rationals that aren’ t improper; they apply only to fractions whose denominator has a degree larger than that of the numerator. Expressing a Fractional Function In Partial Fractions RULE 1: Before a fractional function can be expressed directly in partial fractions, the numerator must be of at least one degree less than the denominator. The fraction ` ( 2x^ 2+ 3) / ( x^ 3- 1) ` can be expressed in partial fractions whereas the fraction ` ( 2x^ 3+ 3) / ( x^ 3- 1) `. In algebra, the partial fraction decomposition or partial fraction expansion of a rational function ( that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial ( possibly zero) and one or several fractions with a simpler denominator.
partial fraction integration rules sheet
May 09, · How to Integrate by Partial Fractions. Check to make sure that the fraction you are trying to integrate is proper. You only get a single constant of integration at the end of partial fractions because you can combine all constants into one larger constant.
|
## 2.1 Scalars and Vectors
### Learning Objectives
By the end of this section, you will be able to:
• Describe the difference between vector and scalar quantities.
• Identify the magnitude and direction of a vector.
• Explain the effect of multiplying a vector quantity by a scalar.
• Describe how one-dimensional vector quantities are added or subtracted.
• Explain the geometric construction for the addition or subtraction of vectors in a plane.
• Distinguish between a vector equation and a scalar equation.
Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. For example, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100 m.” A physical quantity that can be specified completely in this manner is called a scalar quantity. Scalar is a synonym of “number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.
Scalar quantities that have the same physical units can be added or subtracted according to the usual rules of algebra for numbers. For example, a class ending 10 min earlier than 50 min lasts $50\,\text{min}-10\,\text{min}=40\,\text{min}$. Similarly, a 60-cal serving of corn followed by a 200-cal serving of donuts gives $60\,\text{cal}+200\,\text{cal}=260\,\text{cal}$ of energy. When we multiply a scalar quantity by a number, we obtain the same scalar quantity but with a larger (or smaller) value. For example, if yesterday’s breakfast had 200 cal of energy and today’s breakfast has four times as much energy as it had yesterday, then today’s breakfast has $4(200\,\text{cal})=800\,\text{cal}$ of energy. Two scalar quantities can also be multiplied or divided by each other to form a derived scalar quantity. For example, if a train covers a distance of 100 km in 1.0 h, its speed is 100.0 km/1.0 h = 27.8 m/s, where the speed is a derived scalar quantity obtained by dividing distance by time.
Many physical quantities, however, cannot be described completely by just a single number of physical units. For example, when the U.S. Coast Guard dispatches a ship or a helicopter for a rescue mission, the rescue team must know not only the distance to the distress signal, but also the direction from which the signal is coming so they can get to its origin as quickly as possible. Physical quantities specified completely by giving a number of units (magnitude) and a direction are called vector quantities. Examples of vector quantities include displacement, velocity, position, force, and torque. In the language of mathematics, physical vector quantities are represented by mathematical objects called vectors ((Figure)). We can add or subtract two vectors, and we can multiply a vector by a scalar or by another vector, but we cannot divide by a vector. The operation of division by a vector is not defined.
Figure 2.2 We draw a vector from the initial point or origin (called the “tail” of a vector) to the end or terminal point (called the “head” of a vector), marked by an arrowhead. Magnitude is the length of a vector and is always a positive scalar quantity. (credit: modification of work by Cate Sevilla)
Let’s examine vector algebra using a graphical method to be aware of basic terms and to develop a qualitative understanding. In practice, however, when it comes to solving physics problems, we use analytical methods, which we’ll see in the next section. Analytical methods are more simple computationally and more accurate than graphical methods. From now on, to distinguish between a vector and a scalar quantity, we adopt the common convention that a letter in bold type with an arrow above it denotes a vector, and a letter without an arrow denotes a scalar. For example, a distance of 2.0 km, which is a scalar quantity, is denoted by d = 2.0 km, whereas a displacement of 2.0 km in some direction, which is a vector quantity, is denoted by $\overset{\to }{d}$.
Suppose you tell a friend on a camping trip that you have discovered a terrific fishing hole 6 km from your tent. It is unlikely your friend would be able to find the hole easily unless you also communicate the direction in which it can be found with respect to your campsite. You may say, for example, “Walk about 6 km northeast from my tent.” The key concept here is that you have to give not one but two pieces of information—namely, the distance or magnitude (6 km) and the direction (northeast).
Displacement is a general term used to describe a change in position, such as during a trip from the tent to the fishing hole. Displacement is an example of a vector quantity. If you walk from the tent (location A) to the hole (location B), as shown in (Figure), the vector $\overset{\to }{D}$, representing your displacement, is drawn as the arrow that originates at point A and ends at point B. The arrowhead marks the end of the vector. The direction of the displacement vector $\overset{\to }{D}$ is the direction of the arrow. The length of the arrow represents the magnitude D of vector $\overset{\to }{D}$. Here, D = 6 km. Since the magnitude of a vector is its length, which is a positive number, the magnitude is also indicated by placing the absolute value notation around the symbol that denotes the vector; so, we can write equivalently that $D\equiv |\overset{\to }{D}|$. To solve a vector problem graphically, we need to draw the vector $\overset{\to }{D}$ to scale. For example, if we assume 1 unit of distance (1 km) is represented in the drawing by a line segment of length u = 2 cm, then the total displacement in this example is represented by a vector of length $d=6u=6(2\,\text{cm})=12\,\text{cm}$, as shown in (Figure). Notice that here, to avoid confusion, we used $D=6\,\text{km}$ to denote the magnitude of the actual displacement and d = 12 cm to denote the length of its representation in the drawing.
Figure 2.3 The displacement vector from point A (the initial position at the campsite) to point B (the final position at the fishing hole) is indicated by an arrow with origin at point A and end at point B. The displacement is the same for any of the actual paths (dashed curves) that may be taken between points A and B.
Figure 2.4 A displacement $\overset{\to }{D}$ of magnitude 6 km is drawn to scale as a vector of length 12 cm when the length of 2 cm represents 1 unit of displacement (which in this case is 1 km).
Suppose your friend walks from the campsite at A to the fishing pond at B and then walks back: from the fishing pond at B to the campsite at A. The magnitude of the displacement vector ${\overset{\to }{D}}_{AB}$ from A to B is the same as the magnitude of the displacement vector ${\overset{\to }{D}}_{BA}$ from B to A (it equals 6 km in both cases), so we can write ${D}_{AB}={D}_{BA}$. However, vector ${\overset{\to }{D}}_{AB}$ is not equal to vector ${\overset{\to }{D}}_{BA}$ because these two vectors have different directions: ${\overset{\to }{D}}_{AB}\ne {\overset{\to }{D}}_{BA}$. In (Figure), vector ${\overset{\to }{D}}_{BA}$ would be represented by a vector with an origin at point B and an end at point A, indicating vector ${\overset{\to }{D}}_{BA}$ points to the southwest, which is exactly $180\text{°}$ opposite to the direction of vector ${\overset{\to }{D}}_{AB}$. We say that vector ${\overset{\to }{D}}_{BA}$ is antiparallel to vector ${\overset{\to }{D}}_{AB}$ and write ${\overset{\to }{D}}_{AB}=\text{−}{\overset{\to }{D}}_{BA}$, where the minus sign indicates the antiparallel direction.
Two vectors that have identical directions are said to be parallel vectors—meaning, they are parallel to each other. Two parallel vectors $\overset{\to }{A}$ and $\overset{\to }{B}$ are equal, denoted by $\overset{\to }{A}=\overset{\to }{B}$, if and only if they have equal magnitudes $|\overset{\to }{A}|=|\overset{\to }{B}|$. Two vectors with directions perpendicular to each other are said to be orthogonal vectors. These relations between vectors are illustrated in (Figure).
Figure 2.5 Various relations between two vectors $\overset{\to }{A}$ and $\overset{\to }{B}$. (a) $\overset{\to }{A}\ne \overset{\to }{B}$ because $A\ne B$. (b) $\overset{\to }{A}\ne \overset{\to }{B}$ because they are not parallel and $A\ne B$. (c) $\overset{\to }{A}\ne \text{−}\overset{\to }{A}$ because they have different directions (even though $|\overset{\to }{A}|=|-\overset{\to }{A}|=A)$. (d) $\overset{\to }{A}=\overset{\to }{B}$ because they are parallel and have identical magnitudes A = B. (e) $\overset{\to }{A}\ne \overset{\to }{B}$ because they have different directions (are not parallel); here, their directions differ by $90\text{°}$—meaning, they are orthogonal.
### Check Your Understanding
Two motorboats named Alice and Bob are moving on a lake. Given the information about their velocity vectors in each of the following situations, indicate whether their velocity vectors are equal or otherwise. (a) Alice moves north at 6 knots and Bob moves west at 6 knots. (b) Alice moves west at 6 knots and Bob moves west at 3 knots. (c) Alice moves northeast at 6 knots and Bob moves south at 3 knots. (d) Alice moves northeast at 6 knots and Bob moves southwest at 6 knots. (e) Alice moves northeast at 2 knots and Bob moves closer to the shore northeast at 2 knots.
### Algebra of Vectors in One Dimension
Vectors can be multiplied by scalars, added to other vectors, or subtracted from other vectors. We can illustrate these vector concepts using an example of the fishing trip seen in (Figure).
Figure 2.6 Displacement vectors for a fishing trip. (a) Stopping to rest at point C while walking from camp (point A) to the pond (point B). (b) Going back for the dropped tackle box (point D). (c) Finishing up at the fishing pond.
Suppose your friend departs from point A (the campsite) and walks in the direction to point B (the fishing pond), but, along the way, stops to rest at some point C located three-quarters of the distance between A and B, beginning from point A ((Figure)(a)). What is his displacement vector ${\overset{\to }{D}}_{AC}$ when he reaches point C? We know that if he walks all the way to B, his displacement vector relative to A is ${\overset{\to }{D}}_{AB}$, which has magnitude ${D}_{AB}=6\,\text{km}$ and a direction of northeast. If he walks only a 0.75 fraction of the total distance, maintaining the northeasterly direction, at point C he must be $0.75{D}_{AB}=4.5\,\text{km}$ away from the campsite at A. So, his displacement vector at the rest point C has magnitude ${D}_{AC}=4.5\,\text{km}=0.75{D}_{AB}$ and is parallel to the displacement vector ${\overset{\to }{D}}_{AB}$. All of this can be stated succinctly in the form of the following vector equation:
${\overset{\to }{D}}_{AC}=0.75{\overset{\to }{D}}_{AB}.$
In a vector equation, both sides of the equation are vectors. The previous equation is an example of a vector multiplied by a positive scalar (number) $\alpha =0.75$. The result, ${\overset{\to }{D}}_{AC}$, of such a multiplication is a new vector with a direction parallel to the direction of the original vector ${\overset{\to }{D}}_{AB}$.
In general, when a vector $\overset{\to }{A}$ is multiplied by a positive scalar $\alpha$, the result is a new vector $\overset{\to }{B}$ that is parallel to $\overset{\to }{A}$:
$\overset{\to }{B}=\alpha \overset{\to }{A}.$
The magnitude $|\overset{\to }{B}|$ of this new vector is obtained by multiplying the magnitude $|\overset{\to }{A}|$ of the original vector, as expressed by the scalar equation:
$B=|\alpha |A.$
In a scalar equation, both sides of the equation are numbers. (Figure) is a scalar equation because the magnitudes of vectors are scalar quantities (and positive numbers). If the scalar $\alpha$ is negative in the vector equation (Figure), then the magnitude $|\overset{\to }{B}|$ of the new vector is still given by (Figure), but the direction of the new vector $\overset{\to }{B}$ is antiparallel to the direction of $\overset{\to }{A}$. These principles are illustrated in (Figure)(a) by two examples where the length of vector $\overset{\to }{A}$ is 1.5 units. When $\alpha =2$, the new vector $\overset{\to }{B}=2\overset{\to }{A}$ has length $B=2A=3.0\,\text{units}$ (twice as long as the original vector) and is parallel to the original vector. When $\alpha =-2$, the new vector $\overset{\to }{C}=-2\overset{\to }{A}$ has length $C=|-2|A=3.0\,\text{units}$ (twice as long as the original vector) and is antiparallel to the original vector.
Figure 2.7 Algebra of vectors in one dimension. (a) Multiplication by a scalar. (b) Addition of two vectors $(\overset{\to }{R}$ is called the resultant of vectors $\overset{\to }{A}$ and $\overset{\to }{B})$. (c) Subtraction of two vectors $(\overset{\to }{D}$ is the difference of vectors $\overset{\to }{A}$ and $\overset{\to }{B})$.
Now suppose your fishing buddy departs from point A (the campsite), walking in the direction to point B (the fishing hole), but he realizes he lost his tackle box when he stopped to rest at point C (located three-quarters of the distance between A and B, beginning from point A). So, he turns back and retraces his steps in the direction toward the campsite and finds the box lying on the path at some point D only 1.2 km away from point C (see (Figure)(b)). What is his displacement vector ${\overset{\to }{D}}_{AD}$ when he finds the box at point D? What is his displacement vector ${\overset{\to }{D}}_{DB}$ from point D to the hole? We have already established that at rest point C his displacement vector is ${\overset{\to }{D}}_{AC}=0.75{\overset{\to }{D}}_{AB}$. Starting at point C, he walks southwest (toward the campsite), which means his new displacement vector ${\overset{\to }{D}}_{CD}$ from point C to point D is antiparallel to ${\overset{\to }{D}}_{AB}$. Its magnitude $|{\overset{\to }{D}}_{CD}|$ is ${D}_{CD}=1.2\,\text{km}=0.2{D}_{AB}$, so his second displacement vector is ${\overset{\to }{D}}_{CD}=-0.2{\overset{\to }{D}}_{AB}$. His total displacement ${\overset{\to }{D}}_{AD}$ relative to the campsite is the vector sum of the two displacement vectors: vector ${\overset{\to }{D}}_{AC}$ (from the campsite to the rest point) and vector ${\overset{\to }{D}}_{CD}$ (from the rest point to the point where he finds his box):
${\overset{\to }{D}}_{AD}={\overset{\to }{D}}_{AC}+{\overset{\to }{D}}_{CD}.$
The vector sum of two (or more) vectors is called the resultant vector or, for short, the resultant. When the vectors on the right-hand-side of (Figure) are known, we can find the resultant ${\overset{\to }{D}}_{AD}$ as follows:
${\overset{\to }{D}}_{AD}={\overset{\to }{D}}_{AC}+{\overset{\to }{D}}_{CD}=0.75{\overset{\to }{D}}_{AB}-0.2\,{\overset{\to }{D}}_{AB}=(0.75-0.2){\overset{\to }{D}}_{AB}=0.55{\overset{\to }{D}}_{AB}.$
When your friend finally reaches the pond at B, his displacement vector ${\overset{\to }{D}}_{AB}$ from point A is the vector sum of his displacement vector ${\overset{\to }{D}}_{AD}$ from point A to point D and his displacement vector ${\overset{\to }{D}}_{DB}$ from point D to the fishing hole: ${\overset{\to }{D}}_{AB}={\overset{\to }{D}}_{AD}+{\overset{\to }{D}}_{DB}$ (see (Figure)(c)). This means his displacement vector ${\overset{\to }{D}}_{DB}$ is the difference of two vectors:
${\overset{\to }{D}}_{DB}={\overset{\to }{D}}_{AB}-{\overset{\to }{D}}_{AD}={\overset{\to }{D}}_{AB}+(\text{−}{\overset{\to }{D}}_{AD}).$
Notice that a difference of two vectors is nothing more than a vector sum of two vectors because the second term in (Figure) is vector $\text{−}{\overset{\to }{D}}_{AD}$ (which is antiparallel to ${\overset{\to }{D}}_{AD})$. When we substitute (Figure) into (Figure), we obtain the second displacement vector:
${\overset{\to }{D}}_{DB}={\overset{\to }{D}}_{AB}-{\overset{\to }{D}}_{AD}={\overset{\to }{D}}_{AB}-0.55{\overset{\to }{D}}_{AB}=(1.0-0.55){\overset{\to }{D}}_{AB}=0.45{\overset{\to }{D}}_{AB}.$
This result means your friend walked ${D}_{DB}=0.45{D}_{AB}=0.45(6.0\,\text{km})=2.7\,\text{km}$ from the point where he finds his tackle box to the fishing hole.
When vectors $\overset{\to }{A}$ and $\overset{\to }{B}$ lie along a line (that is, in one dimension), such as in the camping example, their resultant $\overset{\to }{R}=\overset{\to }{A}+\overset{\to }{B}$ and their difference $\overset{\to }{D}=\overset{\to }{A}-\overset{\to }{B}$ both lie along the same direction. We can illustrate the addition or subtraction of vectors by drawing the corresponding vectors to scale in one dimension, as shown in (Figure).
To illustrate the resultant when $\overset{\to }{A}$ and $\overset{\to }{B}$ are two parallel vectors, we draw them along one line by placing the origin of one vector at the end of the other vector in head-to-tail fashion (see (Figure)(b)). The magnitude of this resultant is the sum of their magnitudes: R = A + B. The direction of the resultant is parallel to both vectors. When vector $\overset{\to }{A}$ is antiparallel to vector $\overset{\to }{B}$, we draw them along one line in either head-to-head fashion ((Figure)(c)) or tail-to-tail fashion. The magnitude of the vector difference, then, is the absolute value $D=|A-B|$ of the difference of their magnitudes. The direction of the difference vector $\overset{\to }{D}$ is parallel to the direction of the longer vector.
In general, in one dimension—as well as in higher dimensions, such as in a plane or in space—we can add any number of vectors and we can do so in any order because the addition of vectors is commutative,
$\overset{\to }{A}+\overset{\to }{B}=\overset{\to }{B}+\overset{\to }{A},$
and associative,
$(\overset{\to }{A}+\overset{\to }{B})+\overset{\to }{C}=\overset{\to }{A}+(\overset{\to }{B}+\overset{\to }{C}).$
Moreover, multiplication by a scalar is distributive:
${\alpha }_{1}\overset{\to }{A}+{\alpha }_{2}\overset{\to }{A}=({\alpha }_{1}+{\alpha }_{2})\overset{\to }{A}.$
We used the distributive property in (Figure) and (Figure).
When adding many vectors in one dimension, it is convenient to use the concept of a unit vector. A unit vector, which is denoted by a letter symbol with a hat, such as $\hat{u}$, has a magnitude of one and does not have any physical unit so that $|\hat{u}|\equiv u=1$. The only role of a unit vector is to specify direction. For example, instead of saying vector ${\overset{\to }{D}}_{AB}$ has a magnitude of 6.0 km and a direction of northeast, we can introduce a unit vector $\hat{u}$ that points to the northeast and say succinctly that ${\overset{\to }{D}}_{AB}=(6.0\,\text{km})\hat{u}$. Then the southwesterly direction is simply given by the unit vector $\text{−}\hat{u}$. In this way, the displacement of 6.0 km in the southwesterly direction is expressed by the vector
${\overset{\to }{D}}_{BA}=(-6.0\,\text{km})\hat{u}.$
### Example
#### A Ladybug Walker
A long measuring stick rests against a wall in a physics laboratory with its 200-cm end at the floor. A ladybug lands on the 100-cm mark and crawls randomly along the stick. It first walks 15 cm toward the floor, then it walks 56 cm toward the wall, then it walks 3 cm toward the floor again. Then, after a brief stop, it continues for 25 cm toward the floor and then, again, it crawls up 19 cm toward the wall before coming to a complete rest ((Figure)). Find the vector of its total displacement and its final resting position on the stick.
#### Strategy
If we choose the direction along the stick toward the floor as the direction of unit vector $\hat{u}$, then the direction toward the floor is $+\hat{u}$ and the direction toward the wall is $\text{−}\hat{u}$. The ladybug makes a total of five displacements:
$\begin{array}{c}{\overset{\to }{D}}_{1}=(15\,\text{cm})(+\hat{u}),\hfill \\ {\overset{\to }{D}}_{2}=(56\,\text{cm})(\text{−}\hat{u}),\hfill \\ {\overset{\to }{D}}_{3}=(3\,\text{cm})(+\hat{u}),\hfill \\ {\overset{\to }{D}}_{4}=(25\,\text{cm})(+\hat{u}),\,\text{and}\hfill \\ {\overset{\to }{D}}_{5}=(19\,\text{cm})(\text{−}\hat{u}).\hfill \end{array}$
The total displacement $\overset{\to }{D}$ is the resultant of all its displacement vectors.
Figure 2.8 Five displacements of the ladybug. Note that in this schematic drawing, magnitudes of displacements are not drawn to scale. (credit: modification of work by “Persian Poet Gal”/Wikimedia Commons)
### Check Your Understanding
A cave diver enters a long underwater tunnel. When her displacement with respect to the entry point is 20 m, she accidentally drops her camera, but she doesn’t notice it missing until she is some 6 m farther into the tunnel. She swims back 10 m but cannot find the camera, so she decides to end the dive. How far from the entry point is she? Taking the positive direction out of the tunnel, what is her displacement vector relative to the entry point?
### Algebra of Vectors in Two Dimensions
When vectors lie in a plane—that is, when they are in two dimensions—they can be multiplied by scalars, added to other vectors, or subtracted from other vectors in accordance with the general laws expressed by (Figure), (Figure), (Figure), and (Figure). However, the addition rule for two vectors in a plane becomes more complicated than the rule for vector addition in one dimension. We have to use the laws of geometry to construct resultant vectors, followed by trigonometry to find vector magnitudes and directions. This geometric approach is commonly used in navigation ((Figure)). In this section, we need to have at hand two rulers, a triangle, a protractor, a pencil, and an eraser for drawing vectors to scale by geometric constructions.
Figure 2.9 In navigation, the laws of geometry are used to draw resultant displacements on nautical maps.
For a geometric construction of the sum of two vectors in a plane, we follow the parallelogram rule. Suppose two vectors $\overset{\to }{A}$ and $\overset{\to }{B}$ are at the arbitrary positions shown in (Figure). Translate either one of them in parallel to the beginning of the other vector, so that after the translation, both vectors have their origins at the same point. Now, at the end of vector $\overset{\to }{A}$ we draw a line parallel to vector $\overset{\to }{B}$ and at the end of vector $\overset{\to }{B}$ we draw a line parallel to vector $\overset{\to }{A}$ (the dashed lines in (Figure)). In this way, we obtain a parallelogram. From the origin of the two vectors we draw a diagonal that is the resultant $\overset{\to }{R}$ of the two vectors: $\overset{\to }{R}=\overset{\to }{A}+\overset{\to }{B}$ ((Figure)(a)). The other diagonal of this parallelogram is the vector difference of the two vectors $\overset{\to }{D}=\overset{\to }{A}-\overset{\to }{B}$, as shown in (Figure)(b). Notice that the end of the difference vector is placed at the end of vector $\overset{\to }{A}$.
Figure 2.10 The parallelogram rule for the addition of two vectors. Make the parallel translation of each vector to a point where their origins (marked by the dot) coincide and construct a parallelogram with two sides on the vectors and the other two sides (indicated by dashed lines) parallel to the vectors. (a) Draw the resultant vector $\overset{\to }{R}$ along the diagonal of the parallelogram from the common point to the opposite corner. Length R of the resultant vector is not equal to the sum of the magnitudes of the two vectors. (b) Draw the difference vector $\overset{\to }{D}=\overset{\to }{A}-\overset{\to }{B}$ along the diagonal connecting the ends of the vectors. Place the origin of vector $\overset{\to }{D}$ at the end of vector $\overset{\to }{B}$ and the end (arrowhead) of vector $\overset{\to }{D}$ at the end of vector $\overset{\to }{A}$. Length D of the difference vector is not equal to the difference of magnitudes of the two vectors.
It follows from the parallelogram rule that neither the magnitude of the resultant vector nor the magnitude of the difference vector can be expressed as a simple sum or difference of magnitudes A and B, because the length of a diagonal cannot be expressed as a simple sum of side lengths. When using a geometric construction to find magnitudes $|\overset{\to }{R}|$ and $|\overset{\to }{D}|$, we have to use trigonometry laws for triangles, which may lead to complicated algebra. There are two ways to circumvent this algebraic complexity. One way is to use the method of components, which we examine in the next section. The other way is to draw the vectors to scale, as is done in navigation, and read approximate vector lengths and angles (directions) from the graphs. In this section we examine the second approach.
If we need to add three or more vectors, we repeat the parallelogram rule for the pairs of vectors until we find the resultant of all of the resultants. For three vectors, for example, we first find the resultant of vector 1 and vector 2, and then we find the resultant of this resultant and vector 3. The order in which we select the pairs of vectors does not matter because the operation of vector addition is commutative and associative (see (Figure) and (Figure)). Before we state a general rule that follows from repetitive applications of the parallelogram rule, let’s look at the following example.
Suppose you plan a vacation trip in Florida. Departing from Tallahassee, the state capital, you plan to visit your uncle Joe in Jacksonville, see your cousin Vinny in Daytona Beach, stop for a little fun in Orlando, see a circus performance in Tampa, and visit the University of Florida in Gainesville. Your route may be represented by five displacement vectors $\overset{\to }{A},$ $\overset{\to }{B}$, $\overset{\to }{C}$, $\overset{\to }{D}$, and $\overset{\to }{E}$, which are indicated by the red vectors in (Figure). What is your total displacement when you reach Gainesville? The total displacement is the vector sum of all five displacement vectors, which may be found by using the parallelogram rule four times. Alternatively, recall that the displacement vector has its beginning at the initial position (Tallahassee) and its end at the final position (Gainesville), so the total displacement vector can be drawn directly as an arrow connecting Tallahassee with Gainesville (see the green vector in (Figure)). When we use the parallelogram rule four times, the resultant $\overset{\to }{R}$ we obtain is exactly this green vector connecting Tallahassee with Gainesville: $\overset{\to }{R}=\overset{\to }{A}+\overset{\to }{B}+\overset{\to }{C}+\overset{\to }{D}+\overset{\to }{E}$.
Figure 2.11 When we use the parallelogram rule four times, we obtain the resultant vector $\overset{\to }{R}=\overset{\to }{A}+\overset{\to }{B}+\overset{\to }{C}+\overset{\to }{D}+\overset{\to }{E}$, which is the green vector connecting Tallahassee with Gainesville.
Drawing the resultant vector of many vectors can be generalized by using the following tail-to-head geometric construction. Suppose we want to draw the resultant vector $\overset{\to }{R}$ of four vectors $\overset{\to }{A}$, $\overset{\to }{B}$, $\overset{\to }{C}$, and $\overset{\to }{D}$ ((Figure)(a)). We select any one of the vectors as the first vector and make a parallel translation of a second vector to a position where the origin (“tail”) of the second vector coincides with the end (“head”) of the first vector. Then, we select a third vector and make a parallel translation of the third vector to a position where the origin of the third vector coincides with the end of the second vector. We repeat this procedure until all the vectors are in a head-to-tail arrangement like the one shown in (Figure). We draw the resultant vector $\overset{\to }{R}$ by connecting the origin (“tail”) of the first vector with the end (“head”) of the last vector. The end of the resultant vector is at the end of the last vector. Because the addition of vectors is associative and commutative, we obtain the same resultant vector regardless of which vector we choose to be first, second, third, or fourth in this construction.
Figure 2.12 Tail-to-head method for drawing the resultant vector $\overset{\to }{R}=\overset{\to }{A}+\overset{\to }{B}+\overset{\to }{C}+\overset{\to }{D}$. (a) Four vectors of different magnitudes and directions. (b) Vectors in (a) are translated to new positions where the origin (“tail”) of one vector is at the end (“head”) of another vector. The resultant vector is drawn from the origin (“tail”) of the first vector to the end (“head”) of the last vector in this arrangement.
### Example
#### Geometric Construction of the Resultant
The three displacement vectors $\overset{\to }{A}$, $\overset{\to }{B}$, and $\overset{\to }{C}$ in (Figure) are specified by their magnitudes A = 10.0, B = 7.0, and C = 8.0, respectively, and by their respective direction angles with the horizontal direction $\alpha =35\text{°}$, $\beta =-110\text{°}$, and $\gamma =30\text{°}$. The physical units of the magnitudes are centimeters. Choose a convenient scale and use a ruler and a protractor to find the following vector sums: (a) $\overset{\to }{R}=\overset{\to }{A}+\overset{\to }{B}$, (b) $\overset{\to }{D}=\overset{\to }{A}-\overset{\to }{B}\text{, and}$ (c) $\overset{\to }{S}=\overset{\to }{A}-3\overset{\to }{B}+\overset{\to }{C}$.
Figure 2.13 Vectors used in (Figure) and in the Check Your Understanding feature that follows.
#### Strategy
In geometric construction, to find a vector means to find its magnitude and its direction angle with the horizontal direction. The strategy is to draw to scale the vectors that appear on the right-hand side of the equation and construct the resultant vector. Then, use a ruler and a protractor to read the magnitude of the resultant and the direction angle. For parts (a) and (b) we use the parallelogram rule. For (c) we use the tail-to-head method.
#### Solution
For parts (a) and (b), we attach the origin of vector $\overset{\to }{B}$ to the origin of vector $\overset{\to }{A}$, as shown in (Figure), and construct a parallelogram. The shorter diagonal of this parallelogram is the sum $\overset{\to }{A}+\overset{\to }{B}$. The longer of the diagonals is the difference $\overset{\to }{A}-\overset{\to }{B}$. We use a ruler to measure the lengths of the diagonals, and a protractor to measure the angles with the horizontal. For the resultant $\overset{\to }{R}$, we obtain R = 5.8 cm and ${\theta }_{R}\approx 0\text{°}$. For the difference $\overset{\to }{D}$, we obtain D = 16.2 cm and ${\theta }_{D}=49.3\text{°}$, which are shown in (Figure).
Figure 2.14 Using the parallelogram rule to solve (a) (finding the resultant, red) and (b) (finding the difference, blue).
For (c), we can start with vector $-3\overset{\to }{B}$ and draw the remaining vectors tail-to-head as shown in (Figure). In vector addition, the order in which we draw the vectors is unimportant, but drawing the vectors to scale is very important. Next, we draw vector $\overset{\to }{S}$ from the origin of the first vector to the end of the last vector and place the arrowhead at the end of $\overset{\to }{S}$. We use a ruler to measure the length of $\overset{\to }{S}$, and find that its magnitude is
Figure 2.15 Using the tail-to-head method to solve (c) (finding vector $\overset{\to }{S}$, green).
### Check Your Understanding
Using the three displacement vectors $\overset{\to }{A}$, $\overset{\to }{B}$, and $\overset{\to }{F}$ in (Figure), choose a convenient scale, and use a ruler and a protractor to find vector $\overset{\to }{G}$ given by the vector equation $\overset{\to }{G}=\overset{\to }{A}+2\overset{\to }{B}-\overset{\to }{F}$.
Observe the addition of vectors in a plane by visiting this vector calculator and this Phet simulation.
### Summary
• A vector quantity is any quantity that has magnitude and direction, such as displacement or velocity. Vector quantities are represented by mathematical objects called vectors.
• Geometrically, vectors are represented by arrows, with the end marked by an arrowhead. The length of the vector is its magnitude, which is a positive scalar. On a plane, the direction of a vector is given by the angle the vector makes with a reference direction, often an angle with the horizontal. The direction angle of a vector is a scalar.
• Two vectors are equal if and only if they have the same magnitudes and directions. Parallel vectors have the same direction angles but may have different magnitudes. Antiparallel vectors have direction angles that differ by $180\text{°}$. Orthogonal vectors have direction angles that differ by $90\text{°}$.
• When a vector is multiplied by a scalar, the result is another vector of a different length than the length of the original vector. Multiplication by a positive scalar does not change the original direction; only the magnitude is affected. Multiplication by a negative scalar reverses the original direction. The resulting vector is antiparallel to the original vector. Multiplication by a scalar is distributive. Vectors can be divided by nonzero scalars but cannot be divided by vectors.
• Two or more vectors can be added to form another vector. The vector sum is called the resultant vector. We can add vectors to vectors or scalars to scalars, but we cannot add scalars to vectors. Vector addition is commutative and associative.
• To construct a resultant vector of two vectors in a plane geometrically, we use the parallelogram rule. To construct a resultant vector of many vectors in a plane geometrically, we use the tail-to-head method.
### Conceptual Questions
A weather forecast states the temperature is predicted to be $-5\,\text{°}\text{C}$ the following day. Is this temperature a vector or a scalar quantity? Explain.
Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the velocity of a fly, the age of Earth, the boiling point of water, the cost of a book, Earth’s population, or the acceleration of gravity?
Give a specific example of a vector, stating its magnitude, units, and direction.
What do vectors and scalars have in common? How do they differ?
Suppose you add two vectors $\overset{\to }{A}$ and $\overset{\to }{B}$. What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?
Is it possible to add a scalar quantity to a vector quantity?
Is it possible for two vectors of different magnitudes to add to zero? Is it possible for three vectors of different magnitudes to add to zero? Explain.
Does the odometer in an automobile indicate a scalar or a vector quantity?
When a 10,000-m runner competing on a 400-m track crosses the finish line, what is the runner’s net displacement? Can this displacement be zero? Explain.
A vector has zero magnitude. Is it necessary to specify its direction? Explain.
Can a magnitude of a vector be negative?
Can the magnitude of a particle’s displacement be greater that the distance traveled?
If two vectors are equal, what can you say about their components? What can you say about their magnitudes? What can you say about their directions?
If three vectors sum up to zero, what geometric condition do they satisfy?
### Problems
A scuba diver makes a slow descent into the depths of the ocean. His vertical position with respect to a boat on the surface changes several times. He makes the first stop 9.0 m from the boat but has a problem with equalizing the pressure, so he ascends 3.0 m and then continues descending for another 12.0 m to the second stop. From there, he ascends 4 m and then descends for 18.0 m, ascends again for 7 m and descends again for 24.0 m, where he makes a stop, waiting for his buddy. Assuming the positive direction up to the surface, express his net vertical displacement vector in terms of the unit vector. What is his distance to the boat?
In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one side or the other. Two students pull with force 196 N each to the right, four students pull with force 98 N each to the left, five students pull with force 62 N each to the left, three students pull with force 150 N each to the right, and one student pulls with force 250 N to the left. Assuming the positive direction to the right, express the net pull on the knot in terms of the unit vector. How big is the net pull on the knot? In what direction?
Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point and what is the compass direction of a line connecting your starting point to your final position? Use a graphical method.
For the vectors given in the following figure, use a graphical method to find the following resultants: (a) $\overset{\to }{A}+\overset{\to }{B}$, (b) $\overset{\to }{C}+\overset{\to }{B}$, (c) $\overset{\to }{D}+\overset{\to }{F}$, (d) $\overset{\to }{A}-\overset{\to }{B}$, (e) $\overset{\to }{D}-\overset{\to }{F}$, (f) $\overset{\to }{A}+2\overset{\to }{F}$, (g); and (h) $\overset{\to }{A}-4\overset{\to }{D}+2\overset{\to }{F}$.
A delivery man starts at the post office, drives 40 km north, then 20 km west, then 60 km northeast, and finally 50 km north to stop for lunch. Use a graphical method to find his net displacement vector.
An adventurous dog strays from home, runs three blocks east, two blocks north, one block east, one block north, and two blocks west. Assuming that each block is about 100 m, how far from home and in what direction is the dog? Use a graphical method.
In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day and he is blown along the following directions: 2.50 km and $45.0\text{°}$ north of west, then 4.70 km and $60.0\text{°}$ south of east, then 1.30 km and $25.0\text{°}$ south of west, then 5.10 km straight east, then 1.70 km and $5.00\text{°}$ east of north, then 7.20 km and $55.0\text{°}$ south of west, and finally 2.80 km and $10.0\text{°}$ north of east. Use a graphical method to find the castaway’s final position relative to the island.
A small plane flies 40.0 km in a direction $60\text{°}$ north of east and then flies 30.0 km in a direction $15\text{°}$ north of east. Use a graphical method to find the total distance the plane covers from the starting point and the direction of the path to the final position.
A trapper walks a 5.0-km straight-line distance from his cabin to the lake, as shown in the following figure. Use a graphical method (the parallelogram rule) to determine the trapper’s displacement directly to the east and displacement directly to the north that sum up to his resultant displacement vector. If the trapper walked only in directions east and north, zigzagging his way to the lake, how many kilometers would he have to walk to get to the lake?
3.8 km east, 3.2 km north, 7.0 km
A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks 100 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is $35\text{°}$. How wide is the river?
A pedestrian walks 6.0 km east and then 13.0 km north. Use a graphical method to find the pedestrian’s resultant displacement and geographic direction.
The magnitudes of two displacement vectors are A = 20 m and B = 6 m. What are the largest and the smallest values of the magnitude of the resultant $\overset{\to }{R}=\overset{\to }{A}+\overset{\to }{B}?$
### Glossary
antiparallel vectors
two vectors with directions that differ by $180\text{°}$
associative
terms can be grouped in any fashion
commutative
operations can be performed in any order
difference of two vectors
vector sum of the first vector with the vector antiparallel to the second
displacement
change in position
distributive
multiplication can be distributed over terms in summation
magnitude
length of a vector
orthogonal vectors
two vectors with directions that differ by exactly $90\text{°}$, synonymous with perpendicular vectors
parallelogram rule
geometric construction of the vector sum in a plane
parallel vectors
two vectors with exactly the same direction angles
resultant vector
vector sum of two (or more) vectors
scalar
a number, synonymous with a scalar quantity in physics
scalar equation
equation in which the left-hand and right-hand sides are numbers
scalar quantity
quantity that can be specified completely by a single number with an appropriate physical unit
tail-to-head geometric construction
geometric construction for drawing the resultant vector of many vectors
unit vector
vector of a unit magnitude that specifies direction; has no physical unit
vector
mathematical object with magnitude and direction
vector equation
equation in which the left-hand and right-hand sides are vectors
vector quantity
physical quantity described by a mathematical vector—that is, by specifying both its magnitude and its direction; synonymous with a vector in physics
vector sum
resultant of the combination of two (or more) vectors
|
# Aptitude - Odd Man Out and Series
### Exercise :: Odd Man Out and Series - Find Wrong No.
Find out the wrong number in the series.
11.
64, 71, 80, 91, 104, 119, 135, 155
A. 71 B. 80 C. 104 D. 119 E. 135
Explanation:
Go on adding 7, 9, 11, 13, 15, 17, 19 respectively to obtain the next number.
So, 135 is wrong.
12.
15, 16, 34, 105, 424, 2124, 12756
A. 16 B. 34 C. 105 D. 424 E. 2124
Explanation:
2nd term = (1st term) x 1 + 1 = 15 x 1 + 1 = 16.
3rd term = (2nd term) x 2 + 2 = 16 x 2 + 2 = 34.
4th term = (3th term) x 3 + 3 = 34 x 3 + 3 = 105.
5th term = (4th term) x 4 + 4 = 105 x 4 + 4 = 424
6th term = (5th term) x 5 + 5 = 424 x 5 + 5 = 2125
6th term should 2125 instead of 2124.
13.
10, 26, 74, 218, 654, 1946, 5834
A. 26 B. 74 C. 218 D. 654 E. 1946
Explanation:
2nd term = (1st term) x 3 - 4 = 10 x 3 - 4 = 26.
3rd term = (2nd term) x 3 - 4 = 26 x 3 - 4 = 74.
4th term = (3th term) x 3 - 4 = 74 x 3 - 4 = 218.
5th term = (4th term) x 3 - 4 = 218 x 3 - 4 = 650.
5th term must be 650 instead of 654.
14.
2880, 480, 92, 24, 8, 4, 4
A. 480 B. 92 C. 24 D. 8 E. 4
Explanation:
Go on dividing by 6, 5, 4, 3, 2, 1 respectively to obtain the next number.
Clearly, 92 is wrong.
15.
3, 7, 15, 27, 63, 127, 255
A. 7 B. 15 C. 27 D. 63 E. 127
|
# CLASS-8WORD PROBLEM OF SIMULTANEOUS LINEAR EQUATION
Word Problem
Many word problems involving two unknown quantities can be solved by framing simultaneous equations for the unknown quantities. The method is to represent each unknown quantities by a variable and then to write two linear equations using the conditions mentioned in the problem. Solving the simultaneous equations thus framed gives the required values of the quantities.
There are some examples are given below for your better understanding
Example.1) The sum of two numbers is 12. Thrice the first number plus four times the second number equals 30. Find the numbers.
Ans.) Let the numbers are x & y.
As per the given condition, x + y = 12 ………………………………..(1)
and, 3x + 4y = 30…………………………………(2)
now, multiplying the first equation with 3 we find, 3x + 3y = 36 ………………(3)
now we will subtract the 2nd equation from the 3rd equation, we find –
3x + 3y = 36
3x + 4y = 30
- - -
-------------------------
- y = 6 or y = - 6
Now we should put the value of y in the equations (1)
So, x + y = 12
Or, x + (- 6) = 12
Or, x = 12 – (- 6) = 12 + 6 = 18
Hence the numbers are – 6 & 18. (Ans.)
Example.2) In a two-digit number, the digit in the unit's place is two more than thrice the digit in the tens place. When the digits are reversed, the number obtained is 90 more than the original number. Find the original number.
Ans.) Let the digit at the tens place = x and the digit at the units place = y
Then the number = 10x + y
Given, y = 3x + 2
Or, y – 3x = 2 ……………………………(1)
The number obtained after reversing the digits = 10y + x
As per the given condition, 10y + x = 10x + y + 90
Or, 10y – y = 10x – x + 90
Or, 9y = 9x + 90
Or, y = x + 10 [Divide the equation by 9]
Or, y – x = 10 ………………………….(2)
Now, we have to subtract equation (1) from equation (2) and we find –
y – x = 10
y – 3x = 2
- + -
--------------------------------
2x = 8 or x = 4
Substituting the value of x in equation number (2) and we find –
y - x = 10
or, y – 4 = 10
or, y = 10 + 4 = 14
so, we find the value x = 4 & y = 14
so, as per the given condition the original number is = 10x + y = (10 X 4) + 14 = 40 + 14 = 54 (Ans.)
Example.3) If 10 is added to the numerator of a fraction and 5 is subtracted from its denominator, the fraction equals 2. If 4 is subtracted from the numerator and 4 is added to the denominator, the fraction becomes 2/5. Find the fraction.
Ans.) Let the fraction is x/y
x + 10
As per the given condition, -------------- = 2
y – 5
by cross multiplication x + 10 = 2 (y - 5)
or, x + 10 = 2y – 10
or, x – 2y = - 20 ……………………………………..(1)
x – 4 2
also, ------------ = ---------
y + 4 5
by, cross multiplication –
or, 5 (x – 4) = 2 (y + 4)
Or, 5x – 20 = 2y + 8
Or, 5x – 2y = 28 …………………………..(2)
Now we multiply equation (1) by 5 and we get
5x – 10y = - 100 ……………………….(3)
Now we will subtract equation (2) from equation (3) and we get –
5x – 10y = - 100
5x – 2y = 28
- + -
---------------------------
- 8y = - 128 or y = 16
Now we will substitute the value of y in the equation (1) and we find –
x – 2y = - 20
or, x – (2 X 16) = - 20
or, x - 32 = - 20
or, x = 32 – 20 = 12
so, the value of x & y is 12 & 16 respectively
12
so, the desired fraction is = --------- = 3 / 4 (Ans.)
16
Example.4) The cost of 4 kg of rice and 6 kg of sugar is \$ 240 whereas the cost of 5kg of rice and 10 kg of sugar is \$ 450. Find the cost of rice & sugar per kg.
Ans.) Let the cost of 1 kg Rice is \$ x and the cost of 1kg sugar is \$ y.
So the equation are, 4x + 6y = 240 ……………………………(1)
5x + 10y = 450 ……………………………(2)
Now we will multiply by 5 with equation (1) and we will multiply by 4 with equation (2) and we find –
5 (4x + 6y) = 240 X 5
Or, 20x + 30y = 1200………………………..(3)
And, 4 (5x + 10y) = 450 X 4
Or, 20x + 40y = 1800…………………………..(4)
Now, we will subtract equation (3) from equation (4) and we get –
20x + 40y = 1800
20x + 30y = 1200
- - -
------------------------------
10y = 600 or y = 60
Now we would like to put the value of y in equation (1) and we find -
4x + 6y = 240
Or, 4x + (6 X 60) = 240
Or, 4x + 360 = 240
Or, 4x = - 360 + 240
Or, 4x = - 120
Or, x = - 30 or 30
(because value of any product can’t be negative)
So the cost of rice is \$ 30 / Kg and the cost of sugar is \$ 60/Kg (Ans.)
Example.5) 24 tickets were sold for a total of \$ 300 at a fair. If an adult ticket cost \$ 10 and a child’s ticket cost \$ 5, how many of each kind of ticket were sold ?
And.) Let, the number of adult’s ticket = x and the number of child’s ticket is = y
As per the given condition, x + y = 24 ……………………………..(1)
Since an adult ticket cost \$15 and a child’s ticket cost \$ 5,
So, 15x + 5y = 300
Or, 3x + y = 60………………………….(2)
Now we will subtract equation (1) from equation (2) and we get –
3x + y = 60
x + y = 24
- - -
-------------------------
2x = 36 or x = 18
Now we would like to substitute the value of x in equation (1) and we get –
x + y = 24
or, 18 + y = 24
or, y = 24 – 18 = 6
so, here we can conclude that 18 adult tickets and 6 child tickets were sold. (Ans.)
Example.6) A collection has 36 one-dollar and two-dollar coins. How many coins of each kind are there in the collection if the total value of coins is \$ 47 ?
Ans.) Let the number of one-dollar coins = x and the number of two-dollar coins = y
The value of one-dollar coins = \$ x
And the value of the two-dollar coins = \$ y
As per the given condition, x + y = 36 …………………………………(1)
And, x + 2y = 47 ……………………………………..(2)
Now we would like to subtract equation (1) from equation (2).
So, x + 2y = 47
x + y = 36
- - -
--------------------------
y = 11
now we will substitute the value of y in equation (1) and we get -
x + y = 36
or, x + 11 = 36
or, x = 36 – 11 = 25
so, the number of one-dollar coin is = 25 and the number of two-dollar coin is = 11 (Ans.)
Example.7) A plane flying with the wind takes 2 hours to make a 900 km trip from one city to another. On the return trip against the wind, it takes two hours and 15 minutes. Find the speed of the plane in still air and the speed of the wind ?
Ans.) Let the speed of the plane in still air = x km/h and the speed of the wind = y km/h
Then the speed of the plane with the wind = (x + y) km/h
And the speed of the plane against the wind = (x – y) km/h
distance
Speed = --------------
Time
So, as per the given condition –
900km
(x + y) km/h = -------------- = 450 km
2 hours
So, (x + y) = 450 ………………………………….(1)
900 km 1
And, (x – y) km/h = ----------------- = 900 km ÷ 2 ------- hours
2 hours 15 minutes 4
9 4
= 900 ÷ -------- = 900 X -------- = 400 km
4 9
So, (x – y) = 400 ………………………………….(2)
Now we will subtract equation (2) from (1) and we get –
x + y = 450
x - y = 400
- + -
-------------------------
2y = 50 or y = 25 km/h
Now, we will substitute the value of y in equation (1) and we find the speed of plane -
So, x + y = 450
Or, x + 25 = 450
Or, x = 450 – 25 = 425 km/h
So, the speed of the plane in the air and the speed of the wind is 425 km/h and 25 km/h respectively. (Ans.)
|
# 2.2.5 Applying Newton’s Laws of Motion
### Applying Newton’s Laws of Motion
• Newton’s laws of motion are strong tools to understand the motion of objects with forces acting upon them.
• Some worked examples are shown below with different situations to help show the application of these laws
#### Worked Example
Two blocks of mass 1 kg and 4 kg respectively are attached by a tight massless rope between them. The 1 kg block sits on the left and the 4 kg block sits on the right. The 1 kg mass has a 100 Newton force applied to it pulling it to the left. What is the acceleration of both blocks and the tension in the rope as they move across a frictionless surface?
A The acceleration is 15 m/s2 to the left and the tension is 20 Newtons
B The acceleration is 20 m/s2 to the left and the tension is 40 Newtons
C The acceleration is 15 m/s2 to the left and the tension is 60 Newtons
D The acceleration is 20 m/s2 to the left and the tension is 80 Newtons
Step 1: Consider the whole of the system
• Together the 1 kg and 4 kg blocks are both being pulled along by the 100 N force (since the rope is tight)
• This is a frictionless flat surface, therefore the only forces in the system are the pulling force and the tension force(s) in the rope
• Therefore the acceleration can be found using Newton’s second law
Step 2: Find the acceleration
• Using Newton’s second law:
F = m × a
• Rearrange for a
a = F ÷ m
a = 100 ÷ 5 = 20 m/sto the left
Step 3: Examine the 4 kg mass only
• Since the system is moving with an acceleration of 20 m/sto the left, the force on only the 4 kg mass can be found
F = m × a
F = 4 × 20 = 80 N to the left
• The only force acting on the 4 kg mass is the rope and therefore tension force
• This means there is 20 N pulling force on the 1 kg block only
• Since acceleration a is 20 m/sto the left and the tension force is 80 N to the left. The answer is option D
#### Worked Example
A stationary object is subject to a 300 N force that is towards the left and at 55 degrees leftwards with respect to the vertical and a 450 Newton force that is to the right and 35 degrees downwards with respect to the horizontal. What is the magnitude and direction of the third force that would make this object remain stationary? (to the Nearest Newton)
Step 1: Recall Newton’s first law
• A body will remain at rest or move with constant velocity unless acted on by a resultant force
• Therefore, for this object to remain stationary, the resultant force must have a magnitude of 0 Newtons
Step 2: Resolve the 300 N force into its horizontal and vertical components
• The horizontal component can be resolved from:
sin(55°) × 300 = 246 N
• This is in the left direction
• The vertical component can be resolved from:
cos(55°) × 300 = 172 N
• This is in an upwards direction
Step 3: Resolve the 450 N force into its horizontal and vertical components
• The horizontal component can be resolved from:
cos(35°) × 450 = 369 N
• This is in the right direction
• The vertical component can be resolved from:
sin(35°) × 450 = 258 N
• This is in a downwards direction
Step 4: Combine the horizontal components
• The two forces provide 369 N to the right and 246 N to the left
• Therefore, since these are opposing directions:
369 – 246 = 123 N to the right
Step 5: Combine the vertical components
• The two forces provide 258 N downwards and 172 N upwards
• Therefore, since these are opposing directions:
258 – 172 = 86 N downwards
Step 6: Make a right angle triangle using these two force vectors
• There should be a longer size 123 N magnitude vector arrow to the right and then a 86 N magnitude vector arrow downwards
• These can be connected from start to finish by a third vector which is the hypotenuse of this right-angled triangle
Step 7: Use the two vectors magnitudes to find the angle from the horizon
• Since the vectors are to the right and downwards in this right-angle triangle, neither is the hypotenuse
• Therefore the angle from the horizontal downwards can be found by using tan:
tan(θ°) = Opp. ÷ Adj. = 86 ÷ 123
θ° = tan-1(86 ÷ 123) = tan-1(0.699)
θ° ≈ 35°
Step 8: Use Pythagoras theorem or trigonometry to find the magnitude of the resultant force
• Method 1: Using Pythagoras theorem
• c2 = b2 + a2 where b and a are the vector magnitudes for the horizontal and vertical components found so far and c is the hypotenuse magnitude
c2 = 862 + 1232 = 7396 + 15129 = 22525
c = √22525 ≈ 150 N
• Method 2: Using trigonometry
• Using the horizontal component and the angle found in step 7, cos can be used
cos(35°) = Adj. ÷ Hyp. = 123 ÷ Hyp.
• Therefore:
123 ÷ cos(35°) = Hyp. ≈ 150 N
Step 9: State the final answer
• The third force which would cause this object to remain stationary is 150 N right and 35° downwards from the horizontal
#### Exam Tip
You are expected to know Newton’s three laws of motion from memory and how they apply to physical situations. So be sure to practice and use them without reviewing them before a problem.
Close
|
Square Roots
When a number is raised to the second power, we say that the number is squared. For example, $4^2=4(4)$ is referred to as “4 squared.” Often we need to know what number was squared in order to produce some value $a$. If we can find such a number, we call that number a square root of $a$. In this lesson you will learn about how to take square roots, and some of their properties.
Defining and computing square roots
The number 5 is a square root of 25, because $5^2 = 25$. The number $–5$ is another square root of 25, because $(–5)^2 = 25$.
Find the two square roots of 4.
Find the square roots of 36.
Is the square of a positive number positive, negative, or zero? (For example, is $3^2=3(3)$ positive, negative, or zero?)
Is the square of a negative number positive, negative, or zero? (For example, is $(-4)^2=(-4)(-4)$ positive, negative, or zero?)
Is $0^2$ positive, negative, or zero?
Can the square of a real number be negative?
Is it possible for a negative number to have a real square root?
The symbol $√a$ represents the non-negative square root of $a$. For example, $√25 = 5$ and $–√25 = –5$. We can illustrate $√a$ on the grid to the left. For example, at the moment, the grid shows a square with area 25. Because the sides of that square have length 5, this shows that $√25=5$.
Find the square roots in the table below.
All the numbers whose square roots you have been asked about so far are perfect squares: that is, each is a whole number that is the square of a whole number. If you are asked to find a square root of a number which is not a perfect square, you should find it by using a calculator, and write it as a decimal.
Using a calculator, find the square roots in the table below, rounded to four decimal places. (We write $≈$ — meaning “approximately equal to” — as a reminder that your answer is rounded.)
Products and quotients of square roots
Compute the quantities shown in the table below, rounded to four decimal places. Use a calculator as necessary.
As you have seen:
Whenever $a ≥ 0$ and $b ≥ 0$, $√a√b = √{ab}$.
You can check algebraically that $√a√b$ is a square root of $ab$, using the associative and commutative properties of multiplication:
$$\cl"tight"{\table , (√a√b)^2; =, (√a√b)(√a√b); =, √a√b√a√b; =, √a√a√b√b; =, (√a√a)(√b√b); =, ab; }$$
Compute the quantities shown in the table, rounded to four decimal places.
Similarly to the last question, you can see that:
Whenever $a ≥ 0$ and $b > 0$, $$√a/√b = √{a/b}$$.
You can verify algebraically that $$√a/√b$$ is a square root of $$a/b$$:
$$\cl"tight"{\table , (√a/√b)^2; =, (√a/√b)(√a/√b); =, {√a√a}/{√b√b}; =, a/b; }$$
For example, because 4 is a perfect square,
$$√{4x}=√4√x=2√x$$
for all $x ≥ 0$, as shown on the grid to the left. This can be used to simplify the square roots of numbers that are multiples of 4.
Simplify the square roots in the table below by using the fact that $√{4x}=2√x$.
You can check that the result of this simplification is actually a square root of the original number. For example, we just found that $√12=2√3$; you can check this by noticing that
$$(2√3)^2=2√3(2√3)=2(2)√3√3=4(3)=12$$ meaning that $2√3$ actually is a square root of 12.
Simplify the square roots in the table below, by removing the largest possible perfect square from the square root. Then check that the simplified form you get is actually a square root of the original number.
Similarly, you can often simplify fractions inside square roots by using the division rule.
Using the fact that $$√{a/b}={√a}/{√b}$$ whenever $a ≥ 0$ and $b > 0$, simplify these fractions and check your simplification.
Addition and subtraction with square roots
In the last section you learned that, when you’re working only with positive numbers, you can do multiplication and division either before or after you take square roots, and you’ll end up with the same result. In this question, we’ll look at whether this also works for addition and subtraction.
Compute the quantities to the right, as pictured on the grid to the left.
Is $√{16+9}=√16+√9$?
Could there be a general rule saying that $√{a+b}=√a+√b$?
Click . Compute the quantities to the right, which are now pictured on the grids to the left.
Is $√{100-64}=√100-√64$?
Could there be a general rule saying that $√{a-b}=√a-√b$?
Square roots and absolute values
Complete this table for the equation $y=√{x^2}$.
$x$$x^2$$y=√{x^2}$
Click to see the complete graph of $y=√{x^2}$. Which letter of the alphabet is this graph shaped like?
The other graph we’ve seen with this shape is the graph of $y={|x|}$. Click to compare these two graphs. Are they the same or are they different?
This illustrates the following rule:
For all $x$, $√{x^2}={|x|}$.
You can check that $|x|$ is a square root of $x^2$ algebraically:
$$\cl"tight"{\table {|x|^2}, =, {|x|}\,{|x|}; , =, {|x(x)|}, {\; \text"(because "\, {|a|}\,{|b|}={|ab|} \text")"}; , =, {|x^2|}; , =, {x^2}, {\; \text"(because "\, x^2 ≥ 0 \, \text"always)"} }$$
Because $|x|$ is never negative, it must be the non-negative square root of $x^2$ — that is to say, it must be $√{x^2}$.
|
# How do you do scientific notation in physics?
Scientific notation is a way of writing very large or very small numbers. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10. For example, 650,000,000 can be written in scientific notation as 6.5 ✕ 10^8.
## What are the 4 rules of scientific notation?
The base should be always 10. The exponent must be a non-zero integer, that means it can be either positive or negative. The absolute value of the coefficient is greater than or equal to 1 but it should be less than 10. Coefficients can be positive or negative numbers including whole and decimal numbers.
## What is scientific notation physics?
Scientific notation means writing a number in terms of a product of something from 1 to 10 and something else that is a power of 10. In scientific notation all numbers are written in the form of a⋅10b (a times ten raised to the power of b).
## What are the 3 steps to scientific notation?
1. Step 1: Move the decimal point so that you have a number that is between 1 and 10.
2. Step 2: Count the number of decimal places moved in Step 1 .
3. Step 3: Write as a product of the number (found in Step 1) and 10 raised to the power of the count (found in Step 2).
## How do you write 0.00345 in scientific notation?
0.00345 would be written as 3.45×10-3 in scientific notation.
## What is the golden rule of scientific notation?
The value of the exponent indicates the number of places the decimal moves. In our example above, we moved the decimal 19 places to the right, so the exponent was a positive 19. Golden Rules of Scientific Notation: Positive exponents move to the right. Negative exponents move to the decimal to the left.
## What are the two basic rules for using scientific notation?
The rules of exponents are used to simplify operations on scientific notation. To divide numbers in scientific notation, divide the decimals and subtract the exponents. To multiply numbers in scientific notation, multiply the decimals and add the exponents.
## At what grade do you learn scientific notation?
IXL | Scientific notation | 5th grade math.
## Is scientific notation e or e?
E notation is basically the same as scientific notation except that the letter e is substituted for “x 10^”.
## Why using scientific notation is important in physics?
The primary reason why scientific notation is important is that it allows us to convert very large or very small numbers into much more manageable sizes. Physics deals with some very large and very small numbers. Scientific notation is expressed as a number between 1 and 10 multiplied by a power of 10.
## How do you write 1001 in scientific notation?
What is the scientific notation of 1001? The scientific notation of 1001 is 1.00 x 1003. 2.
## How do you write 0.00001 in scientific notation?
Answer: The scientific notation for 0.0001 is 1 × 10-4.
## How do you write 0.001 in scientific notation?
1. (k) kilo = 1000 = 10.
2. (d) deci = 0.1 = 10. -1
3. (c) centi = 0.01 = 10. -2
4. (m) milli = 0.001 = 10. -3
5. (u) micro = 0.000001 = 10. -6
6. (n) nano = 0.000000001 = 10. -9
7. (p) pico = 0.000000000001 = 10. -12
8. (f) femto = 0.000000000000001 = 10. -15
## How do you write 0.096 in scientific notation?
Answer and Explanation: 0.096 written in scientific notation is 96×10−2 96 × 10 − 2 . To convert our decimal to scientific notation, we start by moving the decimal point to the right so that we have a single digit to the left of the decimal point. That would be two spaces for 0.096 to become 9.6.
## What is the rule of scientific notation?
The following rule can be used to convert numbers into scientific notation: The exponent in scientific notation is equal to the number of times the decimal point must be moved to produce a number between 1 and 10.
## What number is 4.5 in scientific notation?
Answer and Explanation: 4.5 written in scientific notation is 4.5×100 4.5 × 10 0 .
## How do you write 8008 in scientific notation?
8,008 written in scientific notation is 8.008 × 103.
## How do you write 0.0027 in scientific notation?
Another example is 2.7 * 10^-3, which is the number 0.0027 written in scientific notation.
## How do you write 0.00092 in scientific notation?
b. 934,000,000 = 934 x 106 c. 847 = 847 d. 0.00092 = 920 x 10-6 e.
## Which number is not in scientific notation?
For a number to be in scientific notation, the value should be greater than or equal to 1 but less than 10.
## Which is the correct way to write 602200000000000000000000 in scientific notation?
For instance, take the number 602,200,000,000,000,000,000,000. Using scientific notation, this number can be expressed as 6.022×1023, which is obviously much more convenient. Many, many numbers in chemistry, physics, and other sciences will appear in the scientific notation form.
## How do you write 18000 in scientific notation?
18,000 written in scientific notation is 1.8 × 104.
|
## What are some examples of linear programming?
Linear Programming Examples
Corner points Z = 2x + 3y
A = (20, 0) 40
B = (20, 10) 70
C = (18, 12) 72
D = (0, 12) 36
What is linear programming explain with examples *?
In Mathematics, linear programming is a method of optimising operations with some constraints. The main objective of linear programming is to maximize or minimize the numerical value. It consists of linear functions which are subjected to the constraints in the form of linear equations or in the form of inequalities.
### How do you write a linear program?
Steps to Linear Programming
1. Understand the problem.
2. Describe the objective.
3. Define the decision variables.
4. Write the objective function.
5. Describe the constraints.
6. Write the constraints in terms of the decision variables.
8. Maximize.
How do you write a linear programming question?
Steps to Solve a Linear Programming Problem
1. Step 1 – Identify the decision variables.
2. Step 2 – Write the objective function.
3. Step 3 – Identify Set of Constraints.
4. Step 4 – Choose the method for solving the linear programming problem.
5. Step 5 – Construct the graph.
6. Step 6 – Identify the feasible region.
## How do you solve a system of linear equations by graphing?
TO SOLVE A SYSTEM OF LINEAR EQUATIONS BY GRAPHING.
1. Graph the first equation.
2. Graph the second equation on the same rectangular coordinate system.
3. Determine whether the lines intersect, are parallel, or are the same line.
4. Identify the solution to the system. If the lines intersect, identify the point of intersection.
Can we solve linear programming in Excel?
If we have constraints and the objective function well defined, we can use the system to predict an optimal solution for a given problem. In Excel, we have Excel Solver, which helps us solving the Linear Programming Problems a.k.a. LPP.
### How do you solve linear programming math?
How do you solve a linear program problem?
## How do you graph a system of equations?
To solve a system of linear equations by graphing.
1. Graph the first equation.
2. Graph the second equation on the same rectangular coordinate system.
3. Determine whether the lines intersect, are parallel, or are the same line.
4. Identify the solution to the system. If the lines intersect, identify the point of intersection.
How do you enter a linear equation in Excel?
Add a linear regression line to the scatter chart by clicking the “Layout” tab, selecting the “Trendline” drop-down box and clicking “Trendline Options.” Select the “Linear” option and click the “Display Equation on Chart” box. Excel displays the linear equation on the chart in the y=mx+b format.
|
# Random Walk of Two Drunks
The one-dimensional random walk is a key foundational concept in statistical physics which crops up in a variety of situations, albeit normally with far more dimensions. This light-hearted question which is circling amongst statistical physics professors serves as a good introduction to random walks.
If you were set this as a homework and were struggling with where to go, about now would be a good time to pat yourself on the back.
### Question
Two persons each move in a one-dimensional random walk. The properties of the random walks are identical for the two, that is, each takes a forward step with probability $p$ and a backward step with probability $q=1-p$ . The steps are of equal size and are taken at the same times. Given that the two start their random walk at the same location, what is the probability that they meet back at that location after they have each taken $N$ steps? Assume each direction has an equal probability (i.e. $p=q=0.5$ ).
### Solution
At any given point, the displacement from the origin $D$ will be $N_{\text{right}}-N_{\text{left}}$ : $$D = x-(N-x)$$where $x$ is the number of steps right and $N$ the total number of steps.
Thus we can write $$D=2x-N \label{eq:D}$$
The probability of having taken $x$ steps to the right after $N$ steps is Binomial: $$P_N(x)=\frac{N!}{x!(N-x)!}p^xq^{N-x}$$
In terms of displacement $D$, from \eqref{eq:D} we find: $x=\frac{D+N}{2}$, and $N-x=N-\frac{D+N}{2}=\frac{N-D}{2}$.
Therefore we can rewrite the probability in terms of $D$ and $N$:
$$P_N(x)=\frac{N!}{\left(\frac{D+N}{2}\right)! \left(\frac{N-D}{2}\right)!}p^{\frac{D+N}{2}}q^{\frac{N-D}{2}}$$
For the given situation, $p=q=\frac{1}{2}$, the total number of steps is $N$ for each drunk, giving $2N$ in total, and the displacement $D$ must be zero, since they are said to meet.1 This gives us
$$P_{2N}(x=\frac N2)=\frac{(2N)!}{\left(\left(\frac{2N}{2}\right)!\right)^2}p^Nq^N=\frac{(2N)!}{N!^2}p^Nq^N$$ $$=\frac{(2N)!}{(2^N N!)^2}$$
So the probability of the two drunks meeting again after N staggers is $P_{2N}(x=\frac N2)=\frac{(2N)!}{(2^N N!)^2}$.
1. That is, back at the origin. Many thanks to David Rinaldi for pointing that out (and querying the respective notation, now updated). []
|
We will solve one-step equations using the
1 / 20
# We will solve one-step equations using the
Télécharger la présentation
## We will solve one-step equations using the
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Learning Objective CALIFORNIA STANDARD: 6.AF 1.1 Write and Solve One-Step Linear Equations in One Variable. Inverse Operations We will solve one-step equations using the properties of addition, subtraction, multiplication, and division. + and – What are we going to learn? What does solve mean? Solve means __________. CFU and Activate Prior Knowledge Inverse operationsare operations that undo each other. Think about PEMDAS. What are the operations ? Which operations are grouped together? What operation would undo the change in value? 1. 6 + 2 = 8 2. 10 - 3 = 7 - + 8 2 = 6 7 3=10 Students, you already know how to use inverse operations to undo a change in value. Now, we will solve one-step equations using inverse operations. Make Connection 3. 4. 43=12 15 5 =3 123=4 3 5 =15
2. Concept Development Inverse Operations A one-step equation contains2one operation. A one-step equation requires3one inverse operation to solve for the variable. • To keep an equation balanced, inverse operations must be done on both sides of the equation. The solution is the value of the variable that makes the equation true. addition + and – x x+ 2 = 3 = 1 4 division and CFU 2 CFU 1 Which inverse operation would be used to solve the one-step equation 4x= 8? How do you know? A Multiplication B Division Which inverse operation would be used to solve the one-step equation x-4= 6? How do you know? A Addition B Subtraction What is the difference between the solution (x= 1) and the non-solution (x= 4)? Go to Skill Dev 1 Go to Skill Dev 2 Flashcards Inverse Operations Solving One-Step Equations x+ 2 =3 Inverse Operation Balance Inverse Operation Balance x+ 0 = 1 Solution (4) x = 1 = 1 x=4 x 4 4 = 1 4 ? ? (1) + 2 =3 (4) + 2 = 3 ? 2 has within it 3 needs (synonym) Vocabulary True 1=1 NOTa Solution Solution x+ 2 = 3 x=1 x= 4 x+ 2 = 3 3=3 True 6 = 3 False
3. Concept Development (Clarification and CFU) Inverse Operations A one-step equation requires one inverse operation to solve for the variable. • To keep an equation balanced, inverse operations must be done on both sides of the equation. + and – and The solutionis the value of the variable that makes the equation true. CFU Why is 2 subtracted on both sides of the equation? Why is 1 the solution to the equation x+ 2 = 3? x+ 2 = 5 - 2 - 2 x= 3 3+ 2 = 5 x WE WANT TO ISOLATE THE VARIABLE TERM (X) . x x x
4. Skill Development/Guided Practice I Inverse Operations A one-step equation requires one inverse operation to solve for the variable. • To keep an equation balanced, inverse operations must be done on both sides of the equation. + and – The solutionis the value of the variable that makes the equation true. and How did I/you connect the problem to the equation? How did I/you solve for the variable? How did I/you interpret the solution? CFU Solve one-step equations. 1 Back to Concept Dev Read the problem and connect it to the equation. Solve for the variable. Hint: Use the inverse operation. Check and interpret4 the solution. Hint: Answer the question. 1 2 2 3 3 m + 7 = 10 m+ 7 =10 c+ 3 =8 c + 3 = 8 3 + 7 = 10 5 + 3 = 8 - 7 - 7 - 3 - 3 10 = 10 8 = 8 m= 3 c= 5 Samuel was given ________________ today. Isabella needs to add __________________. \$3 5 cups of sugar m+ 5 =8 m+ 4 = 11 m+ 5 = 8 m+ 4 =11 3 + 5 = 8 7 + 4 = 11 - 5 - 5 - 4 - 4 8= 8 11 = 11 4 explain (synonym) Vocabulary m= 3 m= 7 Jessie started with _____________________. Shannon started with __________________. 3 marbles \$7
5. Skill Development/Guided Practice 2 Inverse Operations A one-step equation requires one inverse operation to solve for the variable. • To keep an equation balanced, inverse operations must be done on both sides of the equation. + and – The solutionis the value of the variable that makes the equation true. and How did I/you connect the problem to the equation? How did I/you solve for the variable? How did I/you interpret the solution? CFU Solve one-step equations. 1 Read the problem and connect it to the equation. Solve for the variable. Hint: Use the inverse operation. Check and interpret the solution. Hint: Answer the question. 1 2 2 3 3 2b=12 6b=24 2b = 12 6b = 24 2 2 2(6) = 12 6 6 6(4) = 24 12 = 12 24 = 24 b= 6 b= 4 Miguel can buy ______________________. 6 binders She has ______________________. 4 boxes of pencils m m = 7 = 6 4 3 4 ● ● 4 3 ● ● 3 m= 24 m= 21 6 = 6 7 = 7 Isabella gave out ______________________. 24 muffins Ricky and his friends earned _____________. \$21
6. Relevance • 5x – 6 = 4 • + 6 + 6 • 5x = 10 • 5 5 • x = 2 A one-step equation requires one inverse operation to solve for the variable. • To keep an equation balanced, inverse operations must be done on both sides of the equation. The solutionis the value of the variable that makes the equation true. Solving one-step equations will help you solve more complex equations. 1 2 Solving one-step equations will help you do well on tests. Sample Test Question: 37. What value of k makes the following equation true? k 3 = 36 A 108 B 98 C 39 D 12 Does anyone else have another reason why it is relevant to solve one-step equations? (Pair-Share) Why is it relevant to solve one-step equations? You may give one of my reasons or one of your own. Which reason is more relevant to you? Why? CFU
7. Inverse Operations A one-step equation requires one inverse operation to solve for the variable. • To keep an equation balanced, inverse operations must be done on both sides of the equation. + and – The solutionis the value of the variable that makes the equation true. and Skill Closure Solve one-step equations. Read the problem and connect it to the equation. Solve for the variable. Hint: Use the inverse operation. Check and interpret the solution. Hint: Answer the question. 1 2 3 Word Bank equation one-step balanced inverse operation 3c=18 3c = 18 m+ 9 =20 m + 9 = 20 3 3 3(6) = 18 - 9 - 9 11 + 9 = 20 18 = 18 b= 6 m= 11 20 = 20 Angelina needs to save ______________________. Frank has ______________________. \$11 6 cans of tennis balls Access Common Core Sam is trying to solve the equation on the left. After applying one inverse operation, his solution was m = 8. Explain why his solution is incorrect and what possible error he may have made. Sam’s solution 8 Summary Closure What did you learn today about solving one-step equations? (Pair-Share) Use words from the word bank.
8. Independent Practice Inverse Operations A one-step equation requires one inverse operation to solve for the variable. • To keep an equation balanced, inverse operations must be done on both sides of the equation. + and – The solutionis the value of the variable that makes the equation true. and Solve one-step equations. Read the problem and connect it to the equation. Solve for the variable. Hint: Use the inverse operation. Check and interpret the solution. Hint: Answer the question. 1 2 3 n+ 8 = 14 n+ 8 = 14 c+ 2 =6 12 + 8 = 14 c + 2 = 6 - 8 - 8 - 2 - 2 14 = 14 4 + 2 = 6 n= 6 c= 4 6 = 6 Gabriel needs to add ______________________. Louisa started with ______________________. 4 cups of flour 6 nickels m = 9 5 5b=35 5b = 35 5 5 5(7) = 35 5 ● ● 5 35 = 35 b= 7 m= 45 9 = 9 Mathew can buy ______________________. 7 books Jenny and her friends earned ___________________. \$45
9. Periodic Review 1 n– 26 = 12 38 - 26 = 12 c + 6 = 11 + 26 + 26 - 6 - 6 12 = 12 5 + 6 = 11 n= 38 c= 5 11 = 11 Carmella needs to add ______________________. The bookshelf has ______________________. 5 cups of flour 38 pieces 6m = 48 6 6 6(8) = 48 4 ● ● 4 48 = 48 b= 8 m= 56 14 = 14 Claire made ______________________. 8 meatballs There were ______________________. 56 crackers in the box
10. Access Common Core 1. Geoffrey earns \$7 for mowing lawns. Last month, he earned \$42. How many lawns did he mow? Which of the following statements is true? How do you know? a. The inverse operation used to solve this problem is division (÷). b. The meaning of the solution to this problem is the amount of money earned. c. The meaning of the solution to this problem is the number of lawns mowed. d. The first step in isolating the variable is to divide both sides of the equation by m. 7m=42 Yes/No Yes/No Yes/No Yes/No 2. A set of colored pencils had 29 pencils added to it. There are currently 45 pencils altogether. How many pencils were there to start with? Which of the following statements is true? How do you know? a. The inverse operation used to solve this problem is subtraction (-). b. The meaning of the solution to this problem is the original number of pencils. c. The meaning of the solution to this problem is the number of pencils left. d. The first step in isolating the variable is to add 29 to each side of the equation. p+ 29 =45 Yes/No Yes/No Yes/No Yes/No
11. Periodic Review 2 p- 45 = 29 74 - 45 = 29 17 + a = 28 + 45 + 45 - 17 - 17 29 = 29 17 + 11 = 28 n= 74 a= 11 28 = 28 Larissa’s allowance is ______________________. There were __________________ originally. \$11 74 pencils in the set 7m = 42 7 7 7(6) = 42 6 ● ● 6 42 = 42 b= 6 c= 78 13 = 13 Geoffrey mowed ______________________. 6 lawns Thomas and his friends made __________________. 78 cookies
12. Access Common Core 1. Estevan worked on his English homework for 25 minutes. He finished all his homework in 57 minutes. He worked out how long it took him to finish the rest of his homework below, but his older sister told him his answer was incorrect. Where did Estevan make a mistake? How can he fix it? 25 +h=57 +25 +25 h=82 2. Javier can make 7 paper airplanes each hour. He made 28 airplanes and has calculated how long he spent making them. Explain how he can check that his answer is correct. Is his answer correct? 7a=28 7 7 a=4
13. Periodic Review 3 c- 14 = 13 27 - 14 = 13 25 + h = 57 + 14 + 14 - 25 - 25 13 = 13 25 + 32 = 57 c= 27 h= 32 57 = 57 It took Estevan ______________________. Elizabeth had ______________________. 32 minutes 27 pencils 7a = 28 7 7 7(4) = 28 7 ● ● 7 28 = 28 a= 4 y= 105 15 = 15 Javier spent ______________________. 4 hours making airplanes Adolfo and his friends earned ___________________. \$105
14. Access Common Core 1. Thomas and his 5 friends made cookies. They shared them equally and each got 13. How many cookies did they make? Which of the following statements is true? How do you know? c ÷ 13 = 6. a. The inverse operation used to solve this problem is multiplication (). b. The meaning of the solution to this problem is the amount of cookies made. c. The meaning of the solution to this problem is the number of cookies shared equally. d. The first step in isolating the variable is to multiply each side of the equation by c. Yes/No Yes/No Yes/No Yes/No 2. Claire made 6 batches of meatballs. She made a total of 48 meatballs. How many meatballs does each batch make? Which of the following statements is true? How do you know? m 6 =48. Yes/No a. The inverse operation used to solve this problem is multiplication (). b. The meaning of the solution to this problem is the number of meatballs made in each batch. c. The meaning of the solution to this problem is the number of meatballs made in total. d. The first step in isolating the variable is to divide both sides of the equation by m. Yes/No Yes/No Yes/No
15. Flash Cards (Inverse Operations) A one-step equation requires one inverse operation to solve for the variable. x+ 4 =7 What is the operation of the equation? What inverse operation would be used to solve the equation? How do you know? Addition Subtraction
16. Flash Cards (Inverse Operations) A one-step equation requires one inverse operation to solve for the variable. x+ 6 = 13 What is the operation of the equation? What inverse operation would be used to solve the equation? How do you know? Addition Subtraction
17. Flash Cards (Inverse Operations) A one-step equation requires one inverse operation to solve for the variable. 5x=25 What is the operation of the equation? What inverse operation would be used to solve the equation? How do you know? Multiplication Division
18. Flash Cards (Inverse Operations) A one-step equation requires one inverse operation to solve for the variable. What is the operation of the equation? What inverse operation would be used to solve the equation? How do you know? Division Multiplication
19. Flash Cards (Inverse Operations) A one-step equation requires one inverse operation to solve for the variable. x+ 7 =8 What is the operation of the equation? What inverse operation would be used to solve the equation? How do you know? Addition Subtraction
20. Flash Cards (Inverse Operations) A one-step equation requires one inverse operation to solve for the variable. 7x=49 Back to Concept Dev What is the operation of the equation? What inverse operation would be used to solve the equation? How do you know? Multiplication Division
|
Isosceles Right Triangle Spiral
Like the previous right triangle spiral (1.2), this spiral is based on a sequence of right triangles. This spiral, however, involves only isosceles right triangles. The numbers involved are very interesting, as students will discover when they do the worksheet that accompanies this activity. It is, again, a sequence of square roots that is involved.
This project can also be turned into a "reviewee" (small review sheet which folds up). The student should write definitions, postulates, and/or theorems in the triangles of the spiral. (Writing the theorems will provide excellent review in itself.) Then the student should score and fold on the line of each hypotenuse so that it folds up, somewhat like a fan, and will fit neatly in a shirt pocket. The student can review the information when waiting in the lunch line, passing the time while waiting for a ride home, or at other convenient moments.
Here are the steps in the construction:
Step 1: Construct an isosceles right triangle (ABC).
Step 2: Construct a second isosceles right triangle (ABD) on the hypotenuse of the original triangle, using the hypotenuse (AB) of the original triangle as the length of both legs of the new triangle.
Step 3: Constuct a new isosceles right triangle (BDE) on the hypotenuse of the original triangle, using the hypotenuse (BD) of the second triangle as the length of both legs of the new triangle.
Step 4: Constuct a new isosceles right triangle (BEF) on the hypotenuse of the original triangle, using the hypotenuse (BE) of the third triangle as the length of both legs of the new triangle.
Step 5: Constuct a new isosceles right triangle (BFG) on the hypotenuse of the original triangle, using the hypotenuse (BF) of the second triangle as the length of both legs of the new triangle.
Step 6 and 7: Continue this process to create your design:
Cut the figure out using scissors or an exacto knife, then score and fold in opposite directions on each consecutive hypotenuse to form a fold-up "toy".
You can write theorems on the triangles as shown above, and use your folded- up isosceles right triangle "toy" as a review sheet. And/or you could also color it any way you like. The example below shows it in solid colors, but students might like to try other methods of their own, or draw colored pictures within the triangles.
"The essence of mathematics is not to make simple things complicated, but to make complicated things simple." S. Gudder
|
# 4.06 Area of circles and sectors
Lesson
## Area of a circle
To find the area of a circle, we need to know its radius, $r$r. The area, $A$A, can be found using the formula:
Area $=$= $\pi\times\text{radius }^2$π×radius 2 $A$A $=$= $\pi r^2$πr2
## Area of a sector
Let's start with the area of a circle being equal to $\pi r^2$πr2. Remember when we found the perimeter of a sector, we found the fraction of the total circumference for the arc length. It makes sense, then, that the area of a sector is some fraction of that full area. Sometimes it is easy to recognise, for instance a semicircle is half of the total area. Or maybe you might recognise a quarter, a third, or three quarters quite easily.
Semicircle
$\text{Area of semicircle}=\frac{1}{2}\times\pi r^2$Area of semicircle=12×πr2
We can make a formula for any fraction depending on the angle, $\theta$θ, that subtends the arc at the centre.
Minor Sector Major Sector $A=\frac{\theta}{360^\circ}\times\pi r^2$A=θ360°×πr2 $A=\frac{\theta}{360^\circ}\times\pi r^2$A=θ360°×πr2
The fraction at the beginning of the formula, $\frac{\theta}{360^\circ}$θ360°, is the proportion of $360^\circ$360° that the sector's angle at the centre takes up.
Remember that a minor sector is smaller than a semi-circle, that is, when $\theta<180^\circ$θ<180°. A major sector is bigger than a semi-circle, that is, when $\theta>180^\circ$θ>180°.
Area of a circle and sector
Area of a circle:
$A$A $=$= $\pi r^2$πr2
Area of a sector:
$A$A $=$= $\frac{\theta}{360^\circ}\times\pi r^2$θ360°×πr2
#### Practice questions
##### Question 1
Find the area of the circle shown, correct to one decimal place.
##### Question 2
Consider the sector below.
1. Calculate the perimeter. Give your answer correct to $2$2 decimal places.
2. Calculate the area. Give your answer correct to $2$2 decimal places.
##### Question 3
Consider the sector below.
### Areas of composite shapes involving circles
If we combine our understanding of the area of circles and sectors with the area other shapes, we can add or subtract pieces to find unusually shaped areas. We can also map restricted areas that form a circle or sector using a fixed centre and a radius that is determined by some maximum distance. The questions below step through some strategies to solve these sorts of problems.
#### Practice questions
##### Question 4
Find the area of the shaded region in the following figure, correct to 1 decimal place.
##### Question 5
Find the area of the shaded region in the following figure, correct to 1 decimal place.
##### Question 6
A goat is tethered to a corner of a fenced field (shown). The rope is $9$9 m long. What area of the field can the goat graze over?
|
## Engage NY Eureka Math Algebra 1 Module 5 Lesson 5 Answer Key
### Eureka Math Algebra 1 Module 5 Lesson 5 Example Answer Key
Example 1.
Determine whether the sequence below is arithmetic or geometric, and find the function that will produce any given term in the sequence:
16, 24, 36, 54, 81, …
Is this sequence arithmetic?
The differences are 8, 12, 18, 27, … so right away we can tell this is not arithmetic; there is no common difference.
Is the sequence geometric?
The ratios are all 1.5, so this is geometric.
What is the analytical representation of the sequence?
Since the first term is 16 and the common ratio is 1.5, we have f(n) = 16(1.5)(n – 1).
### Eureka Math Algebra 1 Module 5 Lesson 5 Exercise Answer Key
Exercises
Look at the sequence and determine the analytical representation of the sequence. Show your work and reasoning.
Exercise 1.
A decorating consultant charges $50 for the first hour and$2 for each additional whole hour. How much would 1, 000 hours of consultation cost?
By subtracting, we see that this is an arithmetic sequence where we are adding 2 but starting at 50.
f(n) = 50 + 2(n – 1) = 48 + 2n
f(1000) = 48 + 2(1000) = 2048
1, 000 hours of consultation would cost \$2, 048.
Exercise 2.
The sequence below represents the area of a square whose side length is the diagonal of a square with integer side length n. What would be the area for the 100th square? Hint: You can use the square below to find the function model, but you can also just use the terms of the sequence.
Looking at first differences, we see that they are not the same (no common difference):
6, 10, 14, 18, ….
When we look for a common ratio, we find that the quotients of any two consecutive terms in the sequence are not the same: $$\frac{8}{2}$$ ≠ $$\frac{18}{8}$$ ≠ $$\frac{32}{18}$$ ≠⋯
However, I noticed that the first difference increases by 4. This is an indication of a quadratic sequence, and the function equation must have a n2. But since for n = 1 we would have n2 = 1, we must need to multiply that by 2 to get the first term. Now, check to see if 2n2 will work for the other terms.
f(n) = 2n2
Checking: (f(2) = 2(22 ) = 8
f(3) = 2(32 ) = 18
Yes! It works. So, f(100) = 2(100)2 = 20 000.
Therefore, the area of the 100th square is 20, 000 square units.
Exercise 3.
What would be the tenth term in the sequence?
There is no common difference. But the ratios are as follows: $$\frac{6}{3}$$ = 2, $$\frac{12}{6}$$ = 2, $$\frac{24}{12}$$ = 2, …. This is a geometric sequence with a common ratio of 2. And the terms can be written as shown below.
The tenth term in the sequence is 3(512) or 1536.
### Eureka Math Algebra 1 Module 5 Lesson 5 Problem Set Answer Key
Solve the following problems by finding the function/formula that represents the nth term of the sequence.
Question 1.
After a knee injury, a jogger is told he can jog 10 minutes every day and that he can increase his jogging time by
2 minutes every two weeks. How long will it take for him to be able to jog one hour a day?
This is an arithmetic sequence where the minutes increase by 2 every two weeks. (Note: We can either let 2n represent the number of weeks or let n represent a two – week period. Either way, we will end up having to compensate after we solve.) Let’s try it with n representing a two – week period:
f(n) = initial time + (n – 1)(common difference)
60 = 10 + (n – 1)(2) → 60 = 10 + 2n – 2 → 60 = 2n + 8
2n + 8 = 60 → 2n = 52 → n = 26
At the beginning of the 26th 2 – week period, the jogger will be able to jog for 60 minutes. This will occur after 25 weeks ⋅2, or 50 weeks, or at the beginning of the 51st week.
Question 2.
A ball is dropped from a height of 10 feet. The ball then bounces to 80% of its previous height with each subsequent bounce.
a. Explain how this situation can be modeled with a sequence.
According to the problem, to find the next height, multiply the current height by 0.8. This means the sequence is geometric.
b. How high (to the nearest tenth of a foot) does the ball bounce on the fifth bounce?
f(n) = (initial height) (common ratio)^n for n bounces.
f(5) = 10(0.8)5 = 3.2768
The ball bounces approximately 3.3 feet on the fifth bounce.
Question 3.
Consider the following sequence:
8, 17, 32, 53, 80, 113, …
a. What pattern do you see, and what does that pattern mean for the analytical representation of the function?
Difference of the differences is 6. Since the second difference is a nonzero constant, then the pattern must be quadratic.
b. What is the symbolic representation of the sequence?
Sample response: 3n2 does not work by itself. (If n = 1, then 3n2 would be 3, but we have an 8 for the first term.) So, there must be a constant that is being added to it. Let’s test that theory:
(f(n) = 3n2 + b
f(1) = 3(1)2 + b = 8
3 + b = 8
+ b = 5
So, the terms of the sequence can be found using the number of the term, as follows:
f(n) = 3n2 + 5
We can easily check to see if this function generates the sequence, and it does.
Question 4.
Arnold wants to be able to complete 100 military – style pull – ups. His trainer puts him on a workout regimen designed to improve his pull – up strength. The following chart shows how many pull – ups Arnold can complete after each month of training. How many months will it take Arnold to achieve his goal if this pattern continues?
This pattern does not have a common difference or a common ratio. When we look at the first differences (3, 5, 7, 9, 11, …), we see that the second differences would be constant (2, 2, 2, …). That means this is a quadratic sequence with n2 in the nth term formula. For n = 1 we have 12 = 1, so we need to add 1 to get the first term to be 2. So, in general, we have the function f(n) = n2 + 1. Let’s test that on the other terms:
22 + 1 = 5, 32 + 1 = 10, …. Yes, it works.
Now we need to find out which month (n) will produce 100 as the resulting number of pull – ups:
n2 + 1 = 100 → n = $$\sqrt{99}$$ ≈ 9.9
So, if this trend continues, at 10 months, Arnold will be able to complete 100 pull – ups.
### Eureka Math Algebra 1 Module 5 Lesson 5 Exit Ticket Answer Key
Question 1.
A culture of bacteria doubles every 2 hours.
a. Explain how this situation can be modeled with a sequence.
|
### Absolute value inequalities
Absolute value is the distance a number is from zero. Since it is a distance, it is always positive. This doesn’t mean that your variable has to be positive, but your final answer will be.
When we use inequalities, we have more than one answer that can solve our problem. So if we put absolute value and inequalities together, it means that we will have values in both directions from zero that will solve our problem.
Let’s try a few examples.
Example 1:
$$\left| x \right| \ge 4$$
This is read as, “the absolute value of a number is greater than or equal to 4.” So the distance from zero has to be 4 or greater. This means all numbers greater than or equal to 4 would be our answer. But don’t forget, we could go in the other direction as well. All numbers less than or equal to -4 work as well! Whenever we have absolute value inequalities, we have to split the equation into a positive side and a negative side so that we account for both directions. You have to flip the inequality for the negative side.
$$x \ge 4$$ or $$x \le - 4$$
To graph this, we will use closed circles since the answer includes 4 and -4.
Example 2:
$$\left| {8 + a} \right| \le 2$$
Let’s split it. Don’t forget to flip the sign for the negative inequality.
$$8 + a \le 2$$
-8 -8
$$a \le - 6$$
$$8 + a \ge - 2$$
-8 -8
$$a \ge - 10$$
This time we can squish the two inequalities into one compound inequality and our final answer will look like:
$$- 10 \le a \le - 6$$
“Less than” inequalities will always connect and “greater than” inequalities will shoot off in different directions.
Example 3:
$$\left| {2x + 4} \right| - 2 < 8$$
This one has many steps. First, we cannot split the inequality unless the absolute value is all by itself. So we have to add 2 to both sides to get the absolute value alone.
$$\left| {2x + 4} \right| - 2 < 8$$
+2 +2
$$\left| {2x + 4} \right| < 10$$
Now, let’s split the inequality.
$$2x + 4 < 10$$
-4 -4
$$2x < 6$$
$$\div 2$$ $$\div 2$$
________________
$$x < 3$$
$$2x + 4 > - 10$$
-4 -4
$$2x > - 14$$
$$\div 2$$ $$\div 2$$
________________
$$x > - 7$$
This is a “less than” inequality so we can write it as a compound inequality and they will connect on the graph.
$$- 7 < x < 3$$
Notice that the circles are open this time! This is because the answer can’t be equal to -7 or 3.
2973 x
Solve each inequality and graph its solution.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch below how to solve this example:
2590 x
Solve each inequality and graph its solution.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch below how to solve this example:
2946 x
Solve each inequality and graph its solution.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch below how to solve this example:
### Geometry
Circles
Congruent Triangles
Constructions
Parallel Lines and the Coordinate Plane
Properties of Triangles
### Algebra and Pre-Algebra
Beginning Algebra
Beginning Trigonometry
Equations
Exponents
Factoring
Linear Equations and Inequalities
Percents
Polynomials
|
# Fraction Multiplication as Scaling
Multiplication as Scaling is usually taught at the end of the 5th grade, in preparation for rates and proportions in 6th grade. In the 5th grade Common Core, it is listed as (CCSS.Math.5NF):
5NF.5. Interpret multiplication as scaling (resizing), by:
1. Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.
2. Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b =(nxa)/(nxb) to the effect of multiplying a/b by 1.
Multiplication as Scaling is also usually the final topic we cover in Fractions, so students are already proficient in handling fractions operations. As such they may see this topic as trivial, especially since the main objective of the exercise is that of gaining understanding / changing perspectives rather than getting the correct answer.
## Seeing Multiplication as Scaling
The main objective of introducing multiplication as scaling is to allow students to interpret the multiplication operation as a form of changing the magnitude or size of the original entity. In contrast to previous topics in fractions, we are now focuing on the first element on the “left-hand side” of the multiplication equation, rather than the answer on the right.
Students have been working on various fraction operations up to this point – expressing fractions in their equivalent forms, fractions addition, multiplication and division. With this topic, we’re putting a physical meaning to one of the operation – multiplication. That is, at the end of the topic, students should be able to understand that multiplying by a fraction factor changes the size of the original number, and see multiplication as scaling a quantity.
More importantly, we want students to differentiate between the resizing effect of various factors:
1. whole numbers, e.g. 5×3,
2. fractions less than 1, e.g. 2/3 x 3,
3. fractions more than 1, e.g. 4/3 x 3,
and explain “why multiplying a given number by a fraction greater than 1 results in a value or quantity greater than the original” and “why multiplying a given number by a fraction less than 1 results in a value or quantity smaller than the given number”.
## Pictorial representations
For this matter, we have to be mindful of the pictures we present in this topic. For example, when drawing pictorial representation to compare between 2 x 6 and 2/3 x 6, we recognize that we’re no longer trying to find the answers to these two questions, but instead are trying to understand the difference in the resizing effect the two factors have on the same number. Hence earlier area models we use as examples in fraction multiplications, such as the following, are not really helpful here.
Instead, the picture students should have in their mind should be something like this:
## Why now?
Multiplication as scaling is not restricted to fractions, after all multiplying by whole numbers is also scaling, so is multiplying by decimals. The reason why we introduce the topic of scaling at the end of the fractions module is because students now have a good intuitive understanding of what a number less than 1 is, i.e. they have the mathematical language to express such quantities or values. As such, we now have meaningful ways to compare between scaling up and scaling down. This is why we should always make sure the students have a good feel of what numbers less than 1 mean (i.e. fraction number sense) before we introduce multiplication as scaling.
## Decimals
At the end of the topic, it is also good to show examples of scaling by decimal factors. When students fully understand fraction multiplication as scaling, extending to decimals is much simpler, since it is very obvious which decimal numbers are less than 1 (integer component is zero), and which is not. Doing this will also reinforce the notion that decimals, fractions and whole numbers are all different ways of expressing numbers in the same number system.
## Conclusion
As a final topic in fractions in 5th grade, Multiplication as Scaling is sometimes overlooked. This is especially since the main objective of the exercise is that of gaining understanding / changing perspectives rather than getting the correct answer. With mindful choice of pictorial presentations, students can have a better physical interpretation of fraction multiplication, and lay the groundwork for the upcoming topics such as rates and proportions in 6th grade.
This is the final part of a series of blog posts on Fractions:
## More Fraction Resources
For more fraction resources, refer to our main fractions page.
Scroll to Top
|
# What Angles Can Form A Triangle
What Angles Can Form A Triangle. All triangles have internal angles that add up to 180°, no matter the type of triangle. And the law of sines.
You create an exterior angle by extending any side of the triangle. What set of angles can form a triangle 2 see answers advertisement damonforrest answer: A triangle can never have only one acute angle.
### Math Warehouse's Popular Online Triangle Calculator:
The sum of the measures of the angles is always 180° in a triangle. A right triangle contains one right angle and two. All triangles must have three angles.
### 3.Which Set Of Angles Can Form A Triangle?
Since, 11 + 21 > 16, 11 + 16 > 21, and 16 + 21 > 11, you can form a triangle. Given that the sum of the three angles can be assumed to be 180, a triangle can easily be formed by two acute and one right angle. And the law of sines.
### We Can See That The Greatest Angle In The Left Triangle Is 70 Degrees.
A triangle must have a total angle of 180. Explain why or why not. What set of angles can form a triangle 2 see answers advertisement damonforrest answer:
### Every Set Of Three Angles That Add.
A right angle is 90 degrees. A right triangle is a triangle in which one of the angles is 90°, and is denoted by two line segments. An isosceles triangle will have two angles the same size.
### We Have Different Types Of Triangles.
Substituting angles a and b into our previous equation, ∠a + ∠b + ∠bca = 180°, where ∠bca = ∠c. Which set of angles can form a triangle? We can observe that one of the angles measures greater than 90°, making it an obtuse angle.
Read: Tqm Is A Comprehensive Approach Dedicated To
|
Happy Teej - TEEJ2024
# Quadratic Equations : Step by Step Technique to Solve
### STEP-1
Suppose we have two equations as follow:
I. Ax²+Bx+C=0
II. ay²+bx+c=0
Then from these equations we can make a master table that will help us to come to the conclusion quickly
Signs of B/b and C/c Signs of actual values of x and y
Larger value’s sign, Smaller value’s sign
+, + −, −
−, + +, +
+, − −, +
−, − +, −
### STEP-2
Try to solve the equations instantly, if possible, by just checking whether + and – can bring us to a conclusion or not.
#### Example 1:
x²+8x+12=0, +, + → −, −
y²−5y+6=0, −, + → +, +
So, using the master table, it is clear that y>x.
So, any equations which hold “+, +” and “−, +” can be quickly solved.
#### Example 2:
x²+20x−32=0, +, − → −, +
y²+6y−12=0, +, − → −, +
So, from master table, it is clear that the values of x and y are both +ve and –ve. So in such cases the answer will be CND i.e. cannot be determined.
### STEP-3
#### Example 1:
x²−12x+32=0, −, + → +, +
y²−7y+12=0, −, + → +, +
In this case, we have to solve the equations. On solving, we get
x=4, x=8
y=4, y=4
Therefore x ≥ y.
#### Example 2:
3x²+3x-9=0 +, - → -, +
3y²-11y+15=0 -, + → +, +
In this case we can’t reach the conclusion just by seeing only signs. So we need to solve this.
### STEP-4
Suppose we have values -4, -.7, .3, 3.
Arrange them in a rank of decreasing order i.e. highest value will take rank 1 then after next value will get rank 2 and so on…
4= RANK1, .3=RANK2, -.7=RANK3, -4=RANK4.
When the ranks of values of x and y is:
x=1, 2
y=3, 4
Then obviously, x > y.
When the ranks of values of x and y is:
y=1, 2
x=3, 4
Then obviously, y > x.
Any other ranks combinations will give CND.
But when the ranks get tied up, then
Suppose,
x= -7, -3
y= -3, -2
So X’s rank would be 4, 2
Y’s rank would be 2, 1. (We have assumed rank 2 and rank 3 equal to rank 2.)
Then we compare them and we get;
x=y and y>x. On combining both these results we reach to a desired result i.e. y ≥ x.
NOTE: If we ever have to combine x>y and x<y then the result will obviously be CND.
### Some special cases:
#### 1.
x²=900
y²=900
Answer in this case will be CND as a square would always gives a +ve as well as –ve values.
x= 30, -30
y= 30, -30
So any two squares and their variations will always be CND, irrespective of the values.
#### 2.
x2-240=460
y2+500=1000
Result will be CND. Because, we ultimately will get +ve as well as –ve values.
NOTE:
x4=625
x= ± 625)(1⁄4)= ±5
y2= 25
y= ±5
Option will be CND.
|
Algebra and General Mathematics: Factors
Review notes in Factors topic in Algebra and General Mathematics. This will help you prepare in taking the Engineering Board Exam or simply reviewing for your Algebra subject.
FACTORS
A factor is a number or expression which divides exactly another number or expression.
Example:
2 and 4 are factors of 8
(x + y) and (x – y) are factors of (x2 – y2)
Note: the Proper Factors of a number are its entire factor except the number itself.
Example:
Factors of 24, F(24) = { 1, 2, 3, 4, 6, 8, 12, 24 }
Proper Factors, Fp = { 1, 2, 3, 4, 6, 8, 12 }
COMMON FACTORS
Common Factors are factors shared or common to the two or more numbers.
Example:
What are the common factors of 12 and 24?
Solution:
Factors of 12:
F(12) = { 1, 2, 3, 4, 6, 12 }
Factors of 24:
F(24) = { 1, 2, 3, 4, 6, 8, 12, 24 }
Therefore, the common factors of 12 and 24 are:
F(12) ⋂ F(24) = { 1, 2, 3, 4, 6 }
LOWEST COMMON FACTORS (LCF)
The Lowes Common Factor (LCF) of two or more numbers is the common factor of all those numbers which has the lowest value.
Example: ECE Board April 1998
What is the lowest common factor of 10 and 32?
Solution:
Prime factors of 10 = 5 x 2
Prime factors of 32 = 2 x 2 x 2 x 2 x 2
F2⋂32 = 2
HIGHEST COMMON FACTORS (HCF)
The Highest Common Factor (HCF) of two or more whole numbers is the largest whole number which is a factor of both.
Example:
Find the HCF of 240, 600, and 720.
Solution:
-get all the prime factors of the numbers in canonical form
F(240) = 3 x 5 x 24
F(600) = 3 x 52 x 23
F(720) = 32 x 5 x 24
Therefore, the HCF is:
HCF = 3 x 5 x 23 = 120
Labels:
|
# Higher-Order Partial Derivatives Definition & Examples
Lesson Transcript
Instructor: Gerald Lemay
Gerald has taught engineering, math and science and has a doctorate in electrical engineering.
In this lesson, we define the partial derivative and then extend this concept to find higher-order partial derivatives. Examples are used to expand your knowledge and skill.
## Review of Partial Derivatives
Fred has mastered differentiating when a function has only one variable. Somehow, he missed the lessons on partial differentiation. He's starting to panic because the current homework deals with higher-order partial derivatives.
In this lesson, we review partial differentiation by using examples. Then, we extend the method to higher-order partial derivatives. Let's zoom in on Fred to see what he's currently trying to figure out.
Our fearless math student, Fred, is being asked to find the derivative of x4 - x3. He correctly answers: 4x3- 3x2. Let's help Fred with his review of partial derivatives. You ask him to find the derivative of x4a - x3b where a and b are constants. Again, a correct answer: 4x3a- 3x2b.
Now the fun begins. Let's read the following math statement:
It says: the partial (derivative) with respect to x. The word ''derivative'' is in parentheses because we often just say: the partial with respect to x.
Let's write the derivative of x4a - x3b as a partial derivative:
The answer is still 4x3a- 3x2b.
The a and the b are constants. When taking the partial derivative with respect to x, everything other than x is a constant. So we could write:
and y6 and y5 are treated as constants. So, the answer is still the same except the a and b constants are more specific. The answer is now x4y6 - x3y5. Thus, if f(x, y) is the function:
and y6 and y5 are treated as constants
Of course, we can also have a partial with respect to y where everything else is constant. To do these kinds of derivatives, you develop a selective focus with your eyes. You ''see'' the y as the variable and the x4 and the x3 as constants.
Thus,
And this is pretty much the idea with partial derivatives. You just have to be careful with the ''with respect to'' part. Fred is ready to move on to higher-order derivatives.
An error occurred trying to load this video.
Try refreshing the page, or contact customer support.
Coming up next: Tangent Plane to the Surface
### You're on a roll. Keep up the good work!
Replay
Your next lesson will play in 10 seconds
• 0:04 Review of Partial Derivatives
• 2:36 Second-Order Partials
• 4:03 Mixed Partial Derivatives
• 4:45 Example 1
• 5:24 Example 2
• 5:57 Lesson Summary
Save Save
Want to watch this again later?
Timeline
Autoplay
Autoplay
Speed Speed
## Second-Order Partials
Let's say we want to take the partial derivative of the partial derivative. We write this as:
and read this as the second partial with respect to x. This is called a second-order partial derivative. Using the example for f(x, y), here are the details:
We have:
• Line 1: The partial squared with respect to x squared of f with respect to x squared of f is the partial with respect to x of the partial of f with respect to x. We have already computed the quantity in parentheses.
• Line 2: Replace the quantity in parentheses with 4x3y6 - 3x2y5.
• Line 3: Take the derivative with respect to x gives 12x2y6 - 6xy5.
For practice, we ask Fred to work on the second partial with respect to y of f. The details:
This idea of the partial derivative of the partial derivative of the … can be extended to even higher-order derivatives.
Fred is overjoyed at his success with higher-order partial derivatives. But wait, there's one more item to check out.
To unlock this lesson you must be a Study.com Member.
### Register to view this lesson
Are you a student or a teacher?
|
# Maxima and Minima
Learn what is the maxima and minima with examples and types. Also understand how to calculate maxima and minima of a function with steps.
Alan Walker-
Published on 2023-05-26
## Introduction to Maxima and Minima
Maxima and minima are two important concepts of calculus which are used to find the critical points of a function. These points are calculated by finding the derivative of a function. It is used to find the points where a function takes maximum and minimum values. Let’s understand more about maxima and minima. Also understand how to find maxima and minima of a function.
## Understanding of the Maxima and Minima
In mathematical analysis, the maxima and minima are the highest and lowest values that a function takes either within the given domain or the whole domain. Moreover, these values are also referred to as the extrema of a function. This concept is important in optimization problems where we need to calculate the maximum and minimum values of a function.
## Definition of Maxima and Minima
A function f(x) has a maximum value at a point x=c if the value of the function at x=c is greater than or equal to f(x) for all points, mathematically,
$f(c) \geq f(x)^2$
Whereas the function f(x) has a minimum value at x=c if the value of the function at x=c is less than or equal to f(x) for all points. Mathematically,
$f(c) \leq f(x)^2$
These points of a function also play an important role in continuity of a function
## Types of Maxima and Minima
There are two main types of maxima and minima, namely,
1. Absolute maxima and minima
2. Relative maxima and minima
Let’s discuss these types of extremes one-by-one.
### Absolute Maxima and Minima
The point at which a function takes the highest value within the entire domain is known as absolute maxima. Similarly, the point at which the function takes the lowest value within the entire domain is called absolute minima. The absolute maxima and minima can also be referred to as the global maxima and minima. Remember that there can only be one absolute maxima and minima of a function.
### Relative Maxima and Minima
The relative maxima and minima of a function or a curve are also known as the local maxima and minima. The local maxima is the value of a function f(x) at a point x=c which is always greater than the values of the function at the neighboring points i.e. f(c)>f(x). Whereas the local minima of relative minima is the value of a function f(x) at a point x=c which is less than the values of the function at the neighboring points i.e. f(c)<f(x).
## How do you calculate the maxima and minima of a function?
The maxima and minima of a function can be calculated by using derivatives. Here are some simple steps to calculate extreme points of a function.
1. Differentiate the given function f(x).
2. Equate the derivative of f(x) to zero i.e. f’(x)=0 and solve it to calculate the value of x. It will be the critical point of the function.
3. Now calculate the second derivative of f(x) and calculate the value of f’’(x) at the critical point calculated in step 2.
4. Now analyze the values of f’’(x). If f’’(x)>0 then the graph of f(x) is concave up and the function has a local minimum at that point.
5. If f’’(x)<0 then the graph of the function is concave down and the function has a local maximum at that point.
Let’s understand how to calculate maxima and minima of a function in the following example.
## Maxima and minima example
Find the maxima and minima of the function,
$f(x) = x^3 + x^2$
First, we will calculate the derivative of the function. Since the function contains an algebraic function with exponents, therefore we will use the power rule of the derivative.
$f'(x) = 3x^2 + 2x$
Now equating f’(x)=0,
$3x^2 + 2x = 0$
$x(3x + 2) = 0$
$x = 0, \quad x = -\frac{2}{3}$
Now, using these values in f(x), we get,
f(0)=0
f(-⅔)=4/27
Hence the stationary points are (0, 0) and (-⅔, 4/27) where (0, 0) is the local minimum of f(x) and (-⅔, 4/27) is the local maximum.
## Conclusion
Maxima and minima are two important concepts in calculus which are used to optimize a function. It has many applications in mathematical analysis as well as in other fields of science. For example, it helps to optimize the rate of profit and loss at some specific factors. The maxima and minima of a function are also known as its extreme points.
|
# Putting the fun into algebra
7th February 2003 at 00:00
Q
I work with a brilliant support teacher who helps with a small group of pupils from my Year 8 set 3 in maths. Soon we will be doing an algebra module from our scheme of work. Have you any suggestions for a fun activity that will help their understanding of substitution into algebraic expressions?
A
The following activity can be done by individuals or in small groups; I have even done it with whole classes playing in teams. The activity is based on ordering a set of six cards showing different algebraic expressions. The order is determined by substituting a particular value into each expression.
The number of copies of the cards for the activity obviously depends on the size of the group. For this article, I have assumed that the support teacher will be working with a group of four pupils. A photocopiable sheet with solutions is available at www.mathagonyaunt.co.ukarticlesindexJan312003.html The pupils play in pairs. Each pair needs four sets of the 6 circle cards (with each set on different colour card) and one set of the 4 rectangular substitution cards. For demonstration purposes you will need another two sets of 6 circle cards and an n = 4 card. Laminate the cards and then cut out. I have found that the organisation of "handing out" and "collecting in" is made easier by placing each set of sets in a re-sealable plastic bag. Pupils also need a calculator and paper for any rough working.
First demonstrate the activity using a set of the circle cards and the n = 4 card.
Explain that the purpose of this activity is to determine the order of the set of circle cards (from smallest to largest) for substitutions using different values of "n", as indicated on the rectangle cards.
Take one demonstration set of six circles and the n = 4 card and ask the pupils, as a group, to decide what order they think the circle cards would be in if the "n" in the expression had a value of 4. In each case ask them why they think the cards should be in that order.
Follow this by demonstrating the values that are created by correct substitution for each circle, writing the value in washable pen on the circles. Lay the second set of circles in their correct order beneath the pupils' chosen order, placing the n = 4 card at the end. Discuss the differences, pointing out why these might be different. There is a great deal that pupils can learn from this interactive discussion.
Now the pupils work in pairs for each of the values given on the rectangular substitution cards to create the correct order of the expressions on the circle cards. Let your support teacher know that pupils are allowed help as they need it.
When they have finished check their solutions with them and discuss those that are incorrect.
Your support teacher might find it helpful to have some notes about what pupils might have forgotten or still not understood;I have given some examples of some common problem areas.For example, 8n is 8 x n (that is, 8n = n + n + n + n + n + n + n + n). Pupils sometimes find it useful to see this with numbers, and it helps to solve any misconception.
n2 is n x n; pupils often mistake this notation and work out 2 x n instead of multiplying the number by itself. One way to help them is to encourage them to write n x n and then substitute the numbers beneath. For this exercise they can also use a calculator using the x2 button. For example for 22, type then on the calculator.
2n is 2 V n. The mistake pupils tend to make here is to read the calculation bottom to top: "n divided by 2". You should stress that it should be read from the top to the bottom as "two divided by n", and they should write it down in symbols as they say it: 2 V n.
n2 is n V 2. The mistake here is similar to that given above.
8 - n. This becomes a problem when the value to be substituted is negative.
Sometimes it helps to write the sum out: for example, when n = - 1 we have 8 - (-1) = 8 + 1 = 9. Similarly for - 2n, when n = - 1 this becomes - 2 x - 1 = 2.
The support teacher might need to be shown how to use the fraction button on the scientific calculator prior to the session so that they can remind pupils how this is done.
Wendy Fortescue-Hubbard is a teacher and game inventor. She has been awarded a three-year fellowship by the National Endowment for Science, Technology and the Arts (NESTA) to spread maths to the masses. Email your questions to Mathagony Aunt at teacher@tes.co.ukOr write to TES Teacher, Admiral House, 66-68 East Smithfield, London E1W 1BX
Not a subscriber? Find out more about our subscription offers.
Subscribe now
Existing subscriber?
Enter subscription number
|
Students can Download Maths Chapter 4 Information Processing Ex 4.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.3
Question 1.
Question (i)
In questions A and B, there are four groups of letters in each set. Three of these sets are like in some way while one is different. Find the one which is different.
A. (a) C R D T (b) A P B Q (c) E U F V (d) G W H X
B. (a) H K N Q (b) I L O R (c) J M P S (d) A D G J
A. (a) C R D T (b) A P B Q (c) E U F V (d) G W H X
Hint:
The four groups of letters are
CRDT APBQ EUFV GWHX
The above can be written as
We find that when we take 1st & 3rd letter & 2nd & 4th letter as 2 pairs, the 3rd letter is the next letter alphabetically to the 1st letter.
Similarly the 4th letter is alphabetically the next letter of the 2nd letter.
i.e CD, AB, EF, GH & PQ, UV, WX
Only in CRDT, we have T instead of ‘S’. So,
in CRDT ⇒ Option (a)
B. The four groups of letters are
If we notice, we find that 2 letters are missing in the sequence, ie.
HIJKLMNOPQ IJKLMNOPQR
LKLMNOPQRS ABCDEFGHI
We find that only in ADGI, the difference is only one letter between G & I.
Hence it is the odd one out.
Question (ii)
A group of letters are given. A numerical code has been given to each letter. These letters have to be unscrambled into a meaningful word. Find out the code for the word so formed from the 4 answers given.
(a) 2 3 4 1 5 6
(b) 5 6 3 4 2 1
(c) 6 1 3 5 2 4
(d) 4 2 1 3 5 6
(b) 5 6 3 4 2 1
hint:
Given code is
Option (a) is 234156. When we substitute number for each letter from code, we get,
Option (b) is 563421, similarly, we get 5 6 3 4 2 1
Option (c) is
Option (d) is
So, only in option b, we get a meaningful word, i.e PENCIL.
Question (iii)
In questions I and II, there are based on code language. Find the correct answer from the four alternatives given. A
C In a certain code, ‘M E D I C I N E’ is coded as ‘COMPUTER’ written in the same code?
(a) C N P R V U F Q
(b) C M N Q T U D R
(c) R F U V Q N P C
(d) R N V F T U D Q
(C) R F U V Q N P C
Hint:
It is given that in a certain code M E D I C I N E is coded as E O J D J E F M When we observe the word & the code, we find that, there is a pattern.
[to understand, see the matching shapes]
To get the code from the word, we follow the below steps
1.
swap 1st & last letters, so we get
E [ E D I C I N ] M
2. For the middle letters, replace the letters with their alphabetically next letters, so we get
3. Now we have to reverse the order of the middle letters in the bracket, so we get
E[O J D J E F]M = E O J D J E F M
Thus we get the code.
So similarly, we have to follow the 3 steps to get code, therefore:
for
Step 1: Swap 1 st & last letters, so we get
R [O M P U T E ]C
Step 2: For the middle letters, replace the letters with alphabetically next letters, so we get
R [P N Q V U F] C
Step 3: Reverse the word of letters in the bracket to we get
R [F U V Q N P] C
∴ Ans is R F U V Q N P C ⇒ Option C
Question (iv)
If the word ‘PHONE’ is coded as ‘S K R Q H’, how will ‘R A D I O’ be coded?
(a) S C G N H
(b) V R G N G
(c) U D G L R
(d) S D H K Q
(C) U D G L R
hint:
If PHONE is coded as S K R Q H
Find that code for R A D I O
We find that we get the code, by 3rd letter alphabetically so, from P, skipping Q & R, we get S.
Similarly, from H, skipping I & J, we get K
Like wise for R A D I O, skipping the 2 alphabets
From R, skip S & T → U
From A, skip B & C → D
From D skip E & F → G
From I skip J & K → L
From O skip P & Q → R
is U D G L R ⇒ Option c
Question 2.
Fill in the blanks (Use Atbash Cipher that is given in code 3)
For this question, we need to use Atbash cipher. For Atbash cipher, first we write the alphabets from A to Z and then in reverse from Z to A below that.
ABCDEFGHIJKLMNOPQRSTUVWXYZ
ZYXWVUTSRQPONMLKJIHGFEDCBA
Question (i)
G Z N R O = …………
TAMIL
Hint:
Now to solve, we look up the corresponding letter from the table to replace in code to get the actual word.
So, for G Z N R O, from table,
for G, it is T
for Z, It is A
for N, it is M
for R, it is I
for O, it is L
So , the actual word is TAMIL
Question (ii)
V M T O R H S = ………
ENGLISH
Hint:
V M T O R H S
To solve, we look up the corresponding letter from table to replace in code to get the actual word.
For V, it is E
for M, it is N
for T, it is G
for O, it is L
for R, it is I
for H, it is S
for S, it is H
Therefore we get E N G L I S H = ENGLISH
Question (iii)
N Z G S V N Z G R X H = ………
MATHEMATICS
Hint:
Similarly as above for
Question (v)
H X R V M X V = ………..
SCIENCE
Hint:
Question (v)
HLXRZO HXRVMX V = ………..
SOCIAL SCIENCE
Hint:
Question 3.
Match the following (a = 00 ……… Z = 25)
(i) – c
(ii) – d
(iii) – a
(iv) – e
(v) – b
Solution:
Given that the code is
Hint:
(i) Mathematics is
So matching option is c
Matching option is d
(iii) Subtraction is
Matching option is a
(iv) multiplication is
Matching option is e, however instead of 25, it should be 08.
(v) division is
Matching option is b
Question 4.
Frame Additive cipher table (key = 4).
Solution:
Step 1 : write all alphabets
Step 2 : Assign numbers to each alphabet starting from 00 till 25.
Step 3 : add key value (here it is 4) to the numbers assigned in step 2 to form cipher table
Question 5.
A message like “Good Morning” written in reverse would instead be “Doog Gninrom”. In the same way decode the sentence given below: “Ot dnatsrednu taht scitamehtam nac eb decneirepxe erehwreve ni erutan dna laer efil.”
Solution:
Given that good morning written in reverse is doog gninrom.
We have to decode the below by reversing, so,
Ot dnatsrednu taht scitamehtam nac eb decneirepxe erehwreve ni erutan dna laer efil.
to understand that mathematics can be experienced everywhere in nature and real life.
Question 6.
Decode the given Pigpen Cipher text and compare your answer to get the Activity 3 result.
(i) The room number in which the treasure took place
(ii) Place of the treasure
(iii) The name of the treasure
Solution:
(i) The room in which the treasure took place = 28
(ii) The place of treasure = Chair
(iii) Identity of treasure = Gift voucher.
|
# Equation of the form x+a=√(bx+c)
• MHB
• jennyyyyyy
In summary: Yes, we can answer part c of the question now.What would you do next?Well done!What would you do next?In summary, Jenny has been stuck on this problem for a few days, and she is hoping someone can help her. She has found that the equation $x+2=\sqrt{2x+67}$ can be solved for $x$ and that X=-9.
jennyyyyyy
i have been stuck
this problem for a few days now. could somebody help me please?
Hi Jenny! Welcome to MHB!
For part a, let's say for $x=7$ and we set $a=2$, we know $\sqrt{bx+c}$ must be in the form $\sqrt{81}$ so that $7+2=9=\sqrt{81}$.
Since $b>1$, we can let $b$ to be a small positive number, says, $b=2$. The value of $c$ can be found out by doing the following:
\begin{align*}2(7)+c&=81\\14+c&=81\\ \therefore c &=81-14\\&=67\end{align*}
Can you answer part b of the question if the equation we just set up is $x+2=\sqrt{2x+67}$?
Don't worry, we can guide you through all parts of the problem, just that we need to work together so you can understand the problem better, okay?
anemone said:
Hi Jenny! Welcome to MHB!
For part a, let's say for $x=7$ and we set $a=2$, we know $\sqrt{bx+c}$ must be in the form $\sqrt{81}$ so that $7+2=9=\sqrt{81}$.
Since $b>1$, we can let $b$ to be a small positive number, says, $b=2$. The value of $c$ can be found out by doing the following:
\begin{align*}2(7)+c&=81\\14+c\\&=81\\ \therefore c\\&=81-14\\&=67\end{align*}
Can you answer part b of the question if the equation we just set up is $x+2=\sqrt{2x+67}$?
Don't worry, we can guide you through all parts of the problem, just that we need to work together so you can understand the problem better, okay?
thank you so much! we just need to do the same step as a right?
Not really, solving means you need to solve for the value(s) of $x$, based on the values of $a,\,b$ and $c$ we got...
Can you solve $x+2=\sqrt{2x+67}$? You need to square both sides of the equation for a start...
anemone said:
Hi Jenny! Welcome to MHB!
For part a, let's say for $x=7$ and we set $a=2$, we know $\sqrt{bx+c}$ must be in the form $\sqrt{81}$ so that $7+2=9=\sqrt{81}$.
Since $b>1$, we can let $b$ to be a small positive number, says, $b=2$. The value of $c$ can be found out by doing the following:
\begin{align*}2(7)+c&=81\\14+c\\&=81\\ \therefore c\\&=81-14\\&=67\end{align*}
Can you answer part b of the question if the equation we just set up is $x+2=\sqrt{2x+67}$?
Don't worry, we can guide you through all parts of the problem, just that we need to work together so you can understand the problem better, okay?
wait wait, i get it. sorry, i got confused a bit xD
(x+2)^2 = (√ 2x+67)^2
x^2 + 4x +4 = 2x+67
is it correct so far?
Well done!
What would you do next?
anemone said:
Well done!
What would you do next?
(x+2)^2 = (√ 2x+67)^2
x^2 + 4x +4 = 2x+67
X^2 + 4x + 4 – 67 = 2x+67 –67
X^2+ 4x – 63 = 2x
X^2 + 4x – 63 – 2x = 2x – 2x
X^2 +2x – 63 = 0
X^2 + 9x – 7x – 63 = 0
X(x+9)-7(x+9)=0
(x+9) = 0
X-7= 0
X=-9
X=7
i solved it. i don't know if this correct though
Very good job! So, that is the answer for part b of the problem.
Moving on to part c, can you answer it now?
## 1. What is the purpose of the equation x+a=√(bx+c)?
The equation x+a=√(bx+c) is used to solve for the value of x in a quadratic function. It is commonly referred to as the "square root method" and is used to find the x-intercepts of a parabola.
## 2. How do I solve the equation x+a=√(bx+c)?
To solve this equation, you must first isolate the square root term by subtracting a from both sides. Then, square both sides of the equation to eliminate the square root. Finally, solve for x by using the quadratic formula or by factoring the resulting quadratic equation.
## 3. Can this equation have more than one solution?
Yes, this equation can have two solutions. Since a square root can have both a positive and negative value, the resulting quadratic equation can have two solutions for x.
## 4. Is there a specific method for solving this type of equation?
Yes, as mentioned before, this equation is commonly solved using the "square root method." However, it can also be solved by using the quadratic formula or by factoring the resulting quadratic equation.
## 5. Can this equation be used in real-world applications?
Yes, this equation can be used to solve various real-world problems involving parabolic shapes, such as finding the maximum height of a ball thrown into the air or the optimal dimensions of a container with a square base and a fixed volume.
Replies
2
Views
1K
Replies
9
Views
2K
Replies
3
Views
1K
Replies
6
Views
1K
Replies
6
Views
1K
Replies
2
Views
3K
Replies
1
Views
860
Replies
1
Views
1K
Replies
10
Views
2K
Replies
3
Views
777
|
When we graph a parabola, we are generally concerned with three things. First and foremost, we look for the vertex. This is the highest or lowest point, depending on whether the parabola faces up or down. Secondly, we are looking at the horizontal shift or movement along the x axis. Lastly, we are looking at the vertical shift, or movement along the y axis.
Test Objectives
• Demonstrate the ability to find the vertex of a parabola
• Demonstrate the ability to find the horizontal shift
• Demonstrate the ability to find the vertical shift
Graphing Parabolas Practice Test:
#1:
Instructions: Identify the vertex of each parabola.
a) $$f(x) = -\frac{1}{5}x^2$$
b) $$f(x) = (x - 3)^2 + 5$$
#2:
Instructions: Identify the vertex of each parabola.
a) $$f(x) = (x + 9)^2$$
b) $$f(x) = (x - 13)^2 + 4$$
#3:
Instructions: State the horizontal and/or vertical shift for each parabola when compared to f(x) = x2.
a) $$f(x) = (x- 19)^2$$
b) $$f(x) = x^2 - 5$$
#4:
Instructions: State the horizontal and/or vertical shift for each parabola when compared to f(x) = x2.
a) $$f(x) = (x + 2)^2 - 14$$
b) $$f(x) = (x - 7)^2 + 7$$
#5:
Instructions: State the horizontal and/or vertical shift for each parabola when compared to f(x) = x2.
a) $$f(x) = (x - 17)^2 - 12$$
b) $$f(x) = (x + 1)^2 - 23$$
Written Solutions:
#1:
Solutions:
a) $$vertex: (0,0)$$
b) $$vertex: (3,5)$$
#2:
Solutions:
a) $$vertex: (-9,0)$$
b) $$vertex: (13,4)$$
#3:
Solutions:
a) $$shifts \hspace{.25em} 19\hspace{.25em} units\hspace{.25em} right$$
b) $$shifts\hspace{.25em} 5 \hspace{.25em}units\hspace{.25em} down$$
#4:
Solutions:
a) $$shifts \hspace{.25em} 2\hspace{.25em} units\hspace{.25em} left ,\hspace{.25em} 14 \hspace{.25em} units \hspace{.25em} down$$
b) $$shifts \hspace{.25em} 7\hspace{.25em} units\hspace{.25em} right,\hspace{.25em} 7 \hspace{.25em} units \hspace{.25em} up$$
#5:
Solutions:
a) $$shifts \hspace{.25em}17\hspace{.25em} units\hspace{.25em} right ,\hspace{.25em} 12 \hspace{.25em} units \hspace{.25em} down$$
b) $$shifts \hspace{.25em} 1\hspace{.25em} unit\hspace{.25em} left,\hspace{.25em} 23 \hspace{.25em} units \hspace{.25em} down$$
|
## About "Derivative of absolute value of x"
Derivative of absolute value of x :
In this section, we are going to see, how to find derivative of absolute value of (x)
Let |f(x)| be the absolute-value function.
Then the formula to find the derivative of |f(x)| is given below.
Based on the formula given, let us find the derivative of |x|
|x|' = [x/|x|] . (x)'
|x|' = [x/|x|] . (1)
|x|' = x/|x|
Therefore,
the derivative of absolute value of x is equal to x/|x|
## Derivative of absolute value function - Practice problems
Problem 1 :
Differentiate |2x+1| with respect to x
Solution :
Using the formula of derivative of absolute value function, we have
|2x+1|' = [(2x+1)/|2x+1|] . (2x+1)'
|2x+1|' = [(2x+1)/|2x+1|] . 2
|2x+1|' = 2(2x+1) / |2x+1|
Problem 2 :
Differentiate |x³+1| with respect to x
Solution :
Using the formula of derivative of absolute value function, we have
|x³+1|' = [(x³+1)/|x³+1|] . (x³+1)'
|x³+1|' = [(x³+1)/|x³+1|] . 3x²
|x³+1|' = 3x²(x³+1) / |x³+1|
Problem 3 :
Differentiate |x|³ with respect to x
Solution :
In the given function |x|³, using chain rule, first we have to find derivative for the exponent 3 and then for |x|.
|x³|' = {3|x|²} . [x/|x|] . (x)'
|x³|' = {3|x|²} . [x/|x|] . (1)
|x³|' = 3x|x|
Problem 4 :
Differentiate |2x-5| with respect to x
Solution :
Using the formula of derivative of absolute value function, we have
|2x-5|' = [(2x-5)/|2x-5|] . (2x-5)'
|2x-5|' = [(2x-5)/|2x-5|] . 2
|2x-5|' = 2(2x-5) / |2x-5|
Problem 5 :
Differentiate (x-2)² + |x-2| with respect to x
Solution :
Using the formula of derivative of absolute value function, we have
{(x-2)² + |x-2|}' = [(x-2)²]' + |x-2|'
{(x-2)² + |x-2|}' = 2(x-2) + [(x-2)/|x-2|] .(x-2)'
{(x-2)² + |x-2|}' = 2(x-2) + [(x-2)/|x-2|] .(1)
{(x-2)² + |x-2|}' = 2(x-2) + (x-2) / |x-2|
Problem 6 :
Differentiate 3|5x+7| with respect to x
Solution :
Using the formula of derivative of absolute value function, we have
3|5x+7|' = 3 . [(5x+7)/|5x+7|] . (5x+7)'
3|5x+7|' = 3 . [(5x+7)/|5x+7|] . 5
3|5x+7|' = 15(5x+1) / |5x+7|
Problem 7 :
Differentiate |sinx| with respect to x
Solution :
Using the formula of derivative of absolute value function, we have
|sinx|' = [sinx/|sinx|] . (sinx)'
|sinx|' = [sinx/|sinx|] . cosx
|sinx|' = (sinx . cosx) / |sinx|
Problem 8 :
Differentiate |cosx| with respect to x
Solution :
Using the formula of derivative of absolute value function, we have
|cosx|' = [cosx/|cosx|] . (cosx)'
|cosx|' = [cosx/|cosx|] . (-sinx)
|cosx|' = - (sinx . cosx) / |cosx|
Problem 9 :
Differentiate |tanx| with respect to x
Solution :
Using the formula of derivative of absolute value function, we have
|tanx|' = [tanx/|tanx|] . (tanx)'
|tanx|' = [tanx/|tanx|] . sec²x
|tanx|' = sec²x . tanx / |tanx|
Problem 10 :
Differentiate |sinx + cosx| with respect to x
Solution :
Using the formula of derivative of absolute value function, we have
|sinx + cosx|' = [(sinx+cosx) |sinx+cosx|] . (sinx+cosx)'
|sinx + cosx|' = [(cosx+sinx) |sinx+cosx|] . (cosx-sinx)
|sinx + cosx|' = (cos²x - sin²x) |sinx+cosx|
|sinx + cosx|' = cos2x / |sinx+cosx|
After having gone through the stuff given above, we hope that the students would have understood "Derivative of absolute value of x".
Apart from "Derivative of absolute value of x", if you need any other stuff in math, please use our google custom search here.
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
|
College Algebra 2e
# 3.4Composition of Functions
College Algebra 2e3.4 Composition of Functions
## Learning Objectives
In this section, you will:
• Combine functions using algebraic operations.
• Create a new function by composition of functions.
• Evaluate composite functions.
• Find the domain of a composite function.
• Decompose a composite function into its component functions.
Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.
Using descriptive variables, we can notate these two functions. The function $C( T ) C( T )$ gives the cost $C C$ of heating a house for a given average daily temperature in $T T$ degrees Celsius. The function $T( d ) T( d )$ gives the average daily temperature on day $d d$ of the year. For any given day, $Cost=C( T( d ) ) Cost=C( T( d ) )$ means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature $T( d ). T( d ).$ For example, we could evaluate $T( 5 ) T( 5 )$ to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would write $C( T( 5 ) ). C( T( 5 ) ).$
By combining these two relationships into one function, we have performed function composition, which is the focus of this section.
## Combining Functions Using Algebraic Operations
Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function.
Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year’s incomes and then collecting all the data in a new column. If $w(y) w(y)$ is the wife’s income and $h(y) h(y)$ is the husband’s income in year $y, y,$ and we want $T T$ to represent the total income, then we can define a new function.
$T( y )=h( y )+w( y ) T( y )=h( y )+w( y )$
If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write
$T=h+w T=h+w$
Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.
For two functions $f( x ) f( x )$ and $g( x ) g( x )$ with real number outputs, we define new functions $f+g,f−g,fg, f+g,f−g,fg,$ and $f g f g$ by the relations
## Example 1
### Performing Algebraic Operations on Functions
Find and simplify the functions $( g−f )( x ) ( g−f )( x )$ and $( g f )( x ), ( g f )( x ),$ given $f( x )=x−1 f( x )=x−1$ and $g( x )= x 2 −1. g( x )= x 2 −1.$ Are they the same function?
## Try It #1
Find and simplify the functions $( fg )( x ) ( fg )( x )$ and $( f−g )( x ). ( f−g )( x ).$
Are they the same function?
## Create a Function by Composition of Functions
Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:
$( f∘g )( x )=f( g( x ) ) ( f∘g )( x )=f( g( x ) )$
We read the left-hand side as $“f “f$ composed with $g g$ at $x,” x,”$ and the right-hand side as $“f “f$ of $g g$ of $x.” x.”$ The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol $∘ ∘$ is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases $f(g(x))≠f(x)g(x). f(g(x))≠f(x)g(x).$
It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function $g g$ takes the input $x x$ first and yields an output $g( x ). g( x ).$ Then the function $f f$ takes $g( x ) g( x )$ as an input and yields an output $f( g( x ) ). f( g( x ) ).$
In general, $f∘g f∘g$ and $g∘f g∘f$ are different functions. In other words, in many cases $f( g( x ) )≠g( f( x ) ) f( g( x ) )≠g( f( x ) )$ for all $x. x.$ We will also see that sometimes two functions can be composed only in one specific order.
For example, if $f( x )= x 2 f( x )= x 2$ and $g( x )=x+2, g( x )=x+2,$ then
$f(g(x)) = f(x+2) = (x+2) 2 = x 2 +4x+4 f(g(x)) = f(x+2) = (x+2) 2 = x 2 +4x+4$
but
$g(f(x)) = g( x 2 ) = x 2 +2 g(f(x)) = g( x 2 ) = x 2 +2$
These expressions are not equal for all values of $x, x,$ so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value $x=− 1 2 . x=− 1 2 .$
Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs.
## Composition of Functions
When the output of one function is used as the input of another, we call the entire operation a composition of functions. For any input $x x$ and functions $f f$ and $g, g,$ this action defines a composite function, which we write as $f∘g f∘g$ such that
$( f∘g )( x )=f( g( x ) ) ( f∘g )( x )=f( g( x ) )$
The domain of the composite function $f∘g f∘g$ is all $x x$ such that $x x$ is in the domain of $g g$ and $g( x ) g( x )$ is in the domain of $f. f.$
It is important to realize that the product of functions $fg fg$ is not the same as the function composition $f( g( x ) ), f( g( x ) ),$ because, in general, $f( x )g( x )≠f( g( x ) ). f( x )g( x )≠f( g( x ) ).$
## Example 2
### Determining whether Composition of Functions is Commutative
Using the functions provided, find $f( g( x ) ) f( g( x ) )$ and $g( f( x ) ). g( f( x ) ).$ Determine whether the composition of the functions is commutative.
$f(x)=2x+1g(x)=3−x f(x)=2x+1g(x)=3−x$
## Example 3
### Interpreting Composite Functions
The function $c(s) c(s)$ gives the number of calories burned completing $s s$ sit-ups, and $s(t) s(t)$ gives the number of sit-ups a person can complete in $t t$ minutes. Interpret $c(s(3)). c(s(3)).$
## Example 4
### Investigating the Order of Function Composition
Suppose $f(x) f(x)$ gives miles that can be driven in $x x$ hours and $g(y) g(y)$ gives the gallons of gas used in driving $y y$ miles. Which of these expressions is meaningful: $f( g(y) ) f( g(y) )$ or $g( f(x) )? g( f(x) )?$
## Q&A
Are there any situations where $f(g(y)) f(g(y))$ and $g(f(x)) g(f(x))$ would both be meaningful or useful expressions?
Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce different new functions. In real-world problems, functions whose inputs and outputs have the same units also may give compositions that are meaningful in either order.
## Try It #2
The gravitational force on a planet a distance r from the sun is given by the function $G(r). G(r).$ The acceleration of a planet subjected to any force $F F$ is given by the function $a(F). a(F).$ Form a meaningful composition of these two functions, and explain what it means.
## Evaluating Composite Functions
Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function’s output as the input for the outer function.
### Evaluating Composite Functions Using Tables
When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function.
## Example 5
### Using a Table to Evaluate a Composite Function
Using Table 1, evaluate $f(g(3)) f(g(3))$ and $g(f(3)). g(f(3)).$
$x x$ $f(x) f(x)$ $g(x) g(x)$
1 6 3
2 8 5
3 3 2
4 1 7
Table 1
## Try It #3
Using Table 1, evaluate $f(g(1)) f(g(1))$ and $g(f(4)). g(f(4)).$
### Evaluating Composite Functions Using Graphs
When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the $x- x-$ and $y- y-$ axes of the graphs.
## How To
Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs.
1. Locate the given input to the inner function on the $x- x-$ axis of its graph.
2. Read off the output of the inner function from the $y- y-$ axis of its graph.
3. Locate the inner function output on the $x- x-$ axis of the graph of the outer function.
4. Read the output of the outer function from the $y- y-$ axis of its graph. This is the output of the composite function.
## Example 6
### Using a Graph to Evaluate a Composite Function
Using Figure 1, evaluate $f(g(1)). f(g(1)).$
Figure 1
### Analysis
Figure 3 shows how we can mark the graphs with arrows to trace the path from the input value to the output value.
Figure 3
## Try It #4
Using Figure 1, evaluate $g(f(2)). g(f(2)).$
### Evaluating Composite Functions Using Formulas
When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression.
While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that will calculate the result of a composition $f( g( x ) ). f( g( x ) ).$ To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like $f(t)= t 2 −t, f(t)= t 2 −t,$ we substitute the value inside the parentheses into the formula wherever we see the input variable.
## How To
Given a formula for a composite function, evaluate the function.
1. Evaluate the inside function using the input value or variable provided.
2. Use the resulting output as the input to the outside function.
## Example 7
### Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input
Given $f(t)= t 2 −t f(t)= t 2 −t$ and $h(x)=3x+2, h(x)=3x+2,$ evaluate $f(h(1)). f(h(1)).$
### Analysis
It makes no difference what the input variables $t t$ and $x x$ were called in this problem because we evaluated for specific numerical values.
## Try It #5
Given $f(t)= t 2 −t f(t)= t 2 −t$ and $h(x)=3x+2, h(x)=3x+2,$ evaluate
1. $h(f(2)) h(f(2))$
2. $h(f(−2)) h(f(−2))$
## Finding the Domain of a Composite Function
As we discussed previously, the domain of a composite function such as $f∘g f∘g$ is dependent on the domain of $g g$ and the domain of $f. f.$ It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as $f∘g. f∘g.$ Let us assume we know the domains of the functions $f f$ and $g g$ separately. If we write the composite function for an input $x x$ as $f( g( x ) ), f( g( x ) ),$ we can see right away that $x x$ must be a member of the domain of $g g$ in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that $g( x ) g( x )$ must be a member of the domain of $f, f,$ otherwise the second function evaluation in $f( g( x ) ) f( g( x ) )$ cannot be completed, and the expression is still undefined. Thus the domain of $f∘g f∘g$ consists of only those inputs in the domain of $g g$ that produce outputs from $g g$ belonging to the domain of $f. f.$ Note that the domain of $f f$ composed with $g g$ is the set of all $x x$ such that $x x$ is in the domain of $g g$ and $g( x ) g( x )$ is in the domain of $f. f.$
## Domain of a Composite Function
The domain of a composite function $f( g( x ) ) f( g( x ) )$ is the set of those inputs $x x$ in the domain of $g g$ for which $g( x ) g( x )$ is in the domain of $f. f.$
## How To
Given a function composition $f(g(x)), f(g(x)),$ determine its domain.
1. Find the domain of $g. g.$
2. Find the domain of $f. f.$
3. Find those inputs $x x$ in the domain of $g g$ for which $g( x ) g( x )$ is in the domain of $f. f.$ That is, exclude those inputs $x x$ from the domain of $g g$ for which $g( x ) g( x )$ is not in the domain of $f. f.$ The resulting set is the domain of $f∘g. f∘g.$
## Example 8
### Finding the Domain of a Composite Function
Find the domain of
$( f∘g )(x)wheref(x)= 5 x−1 andg(x)= 4 3x−2 ( f∘g )(x)wheref(x)= 5 x−1 andg(x)= 4 3x−2$
## Example 9
### Finding the Domain of a Composite Function Involving Radicals
Find the domain of
### Analysis
This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of $f∘g f∘g$ can contain values that are not in the domain of $f, f,$ though they must be in the domain of $g. g.$
## Try It #6
Find the domain of
## Decomposing a Composite Function into its Component Functions
In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient.
## Example 10
### Decomposing a Function
Write $f(x)= 5− x 2 f(x)= 5− x 2$ as the composition of two functions.
## Try It #7
Write $f(x)= 4 3− 4+ x 2 f(x)= 4 3− 4+ x 2$ as the composition of two functions.
## Media
Access these online resources for additional instruction and practice with composite functions.
## 3.4 Section Exercises
### Verbal
1.
How does one find the domain of the quotient of two functions, $f g ? f g ?$
2.
What is the composition of two functions, $f∘g? f∘g?$
3.
If the order is reversed when composing two functions, can the result ever be the same as the answer in the original order of the composition? If yes, give an example. If no, explain why not.
4.
How do you find the domain for the composition of two functions, $f∘g? f∘g?$
### Algebraic
For the following exercises, determine the domain for each function in interval notation.
5.
Given $f(x)= x 2 +2x f(x)= x 2 +2x$ and find $f+g,f−g,fg, f+g,f−g,fg,$ and $f g . f g .$
6.
Given $f(x)=−3 x 2 +x f(x)=−3 x 2 +x$ and $g(x)=5, g(x)=5,$ find $f+g,f−g,fg, f+g,f−g,fg,$ and $f g . f g .$
7.
Given $f(x)=2 x 2 +4x f(x)=2 x 2 +4x$ and $g(x)= 1 2x , g(x)= 1 2x ,$ find $f+g,f−g,fg, f+g,f−g,fg,$ and $f g . f g .$
8.
Given $f(x)= 1 x−4 f(x)= 1 x−4$ and $g(x)= 1 6−x , g(x)= 1 6−x ,$ find $f+g,f−g,fg, f+g,f−g,fg,$ and $f g . f g .$
9.
Given $f(x)=3 x 2 f(x)=3 x 2$ and $g(x)= x−5 , g(x)= x−5 ,$ find $f+g,f−g,fg, f+g,f−g,fg,$ and $f g . f g .$
10.
Given $f(x)= x f(x)= x$ and $g(x)=|x−3|, g(x)=|x−3|,$ find $g f . g f .$
11.
For the following exercise, find the indicated function given $f(x)=2 x 2 +1 f(x)=2 x 2 +1$ and $g(x)=3x−5. g(x)=3x−5.$
1. $f(g(2)) f(g(2))$
2. $f(g(x)) f(g(x))$
3. $g(f(x)) g(f(x))$
4. $( g∘g )( x ) ( g∘g )( x )$
5. $( f∘f )( −2 ) ( f∘f )( −2 )$
For the following exercises, use each pair of functions to find $f( g( x ) ) f( g( x ) )$ and $g( f( x ) ). g( f( x ) ).$ Simplify your answers.
12.
$f(x)= x 2 +1,g(x)= x+2 f(x)= x 2 +1,g(x)= x+2$
13.
$f(x)= x +2,g(x)= x 2 +3 f(x)= x +2,g(x)= x 2 +3$
14.
$f(x)=| x |,g(x)=5x+1 f(x)=| x |,g(x)=5x+1$
15.
$f(x)= x 3 ,g(x)= x+1 x 3 f(x)= x 3 ,g(x)= x+1 x 3$
16.
$f(x)= 1 x−6 ,g(x)= 7 x +6 f(x)= 1 x−6 ,g(x)= 7 x +6$
17.
$f(x)= 1 x−4 ,g(x)= 2 x +4 f(x)= 1 x−4 ,g(x)= 2 x +4$
For the following exercises, use each set of functions to find $f( g( h(x) ) ). f( g( h(x) ) ).$ Simplify your answers.
18.
$f(x)= x 4 +6, f(x)= x 4 +6,$ $g(x)=x−6, g(x)=x−6,$ and $h(x)= x h(x)= x$
19.
$f(x)= x 2 +1, f(x)= x 2 +1,$ $g(x)= 1 x , g(x)= 1 x ,$ and $h(x)=x+3 h(x)=x+3$
20.
Given $f(x)= 1 x f(x)= 1 x$ and $g(x)=x−3, g(x)=x−3,$ find the following:
1. $(f∘g)(x) (f∘g)(x)$
2. the domain of $(f∘g)(x) (f∘g)(x)$ in interval notation
3. $(g∘f)(x) (g∘f)(x)$
4. the domain of $(g∘f)(x) (g∘f)(x)$
5. $( f g )(x) ( f g )(x)$
21.
Given $f(x)= 2−4x f(x)= 2−4x$ and $g(x)=− 3 x , g(x)=− 3 x ,$ find the following:
1. $(g∘f)(x) (g∘f)(x)$
2. the domain of $(g∘f)(x) (g∘f)(x)$ in interval notation
22.
Given the functions $f(x)= 1−x x andg(x)= 1 1+ x 2 , f(x)= 1−x x andg(x)= 1 1+ x 2 ,$ find the following:
1. $(g∘f)(x) (g∘f)(x)$
2. $(g∘f)(2) (g∘f)(2)$
23.
Given functions $p(x)= 1 x p(x)= 1 x$ and $m(x)= x 2 −4, m(x)= x 2 −4,$ state the domain of each of the following functions using interval notation:
1. $p(x) m(x) p(x) m(x)$
2. $p(m(x)) p(m(x))$
3. $m(p(x)) m(p(x))$
24.
Given functions $q(x)= 1 x q(x)= 1 x$ and $h(x)= x 2 −9, h(x)= x 2 −9,$ state the domain of each of the following functions using interval notation.
1. $q(x) h(x) q(x) h(x)$
2. $q( h(x) ) q( h(x) )$
3. $h( q(x) ) h( q(x) )$
25.
For $f(x)= 1 x f(x)= 1 x$ and $g(x)= x−1 , g(x)= x−1 ,$ write the domain of $(f∘g)(x) (f∘g)(x)$ in interval notation.
For the following exercises, find functions $f(x) f(x)$ and $g(x) g(x)$ so the given function can be expressed as $h(x)=f( g(x) ). h(x)=f( g(x) ).$
26.
$h(x)= (x+2) 2 h(x)= (x+2) 2$
27.
$h(x)= (x−5) 3 h(x)= (x−5) 3$
28.
$h(x)= 3 x−5 h(x)= 3 x−5$
29.
$h(x)= 4 (x+2) 2 h(x)= 4 (x+2) 2$
30.
$h(x)=4+ x 3 h(x)=4+ x 3$
31.
$h(x)= 1 2x−3 3 h(x)= 1 2x−3 3$
32.
$h(x)= 1 (3 x 2 −4) −3 h(x)= 1 (3 x 2 −4) −3$
33.
$h(x)= 3x−2 x+5 4 h(x)= 3x−2 x+5 4$
34.
$h(x)= ( 8+ x 3 8− x 3 ) 4 h(x)= ( 8+ x 3 8− x 3 ) 4$
35.
$h(x)= 2x+6 h(x)= 2x+6$
36.
$h(x)= (5x−1) 3 h(x)= (5x−1) 3$
37.
$h(x)= x−1 3 h(x)= x−1 3$
38.
$h(x)=| x 2 +7 | h(x)=| x 2 +7 |$
39.
$h(x)= 1 (x−2) 3 h(x)= 1 (x−2) 3$
40.
$h(x)= ( 1 2x−3 ) 2 h(x)= ( 1 2x−3 ) 2$
41.
$h(x)= 2x−1 3x+4 h(x)= 2x−1 3x+4$
### Graphical
For the following exercises, use the graphs of $f, f,$ shown in Figure 4, and $g, g,$ shown in Figure 5, to evaluate the expressions.
Figure 4
Figure 5
42.
$f( g(3) ) f( g(3) )$
43.
$f( g(1) ) f( g(1) )$
44.
$g( f(1) ) g( f(1) )$
45.
$g( f(0) ) g( f(0) )$
46.
$f( f(5) ) f( f(5) )$
47.
$f( f(4) ) f( f(4) )$
48.
$g( g(2) ) g( g(2) )$
49.
$g( g(0) ) g( g(0) )$
For the following exercises, use graphs of $f(x), f(x),$ shown in Figure 6, $g(x), g(x),$ shown in Figure 7, and $h(x), h(x),$ shown in Figure 8, to evaluate the expressions.
Figure 6
Figure 7
Figure 8
50.
$g( f( 1 ) ) g( f( 1 ) )$
51.
$g( f( 2 ) ) g( f( 2 ) )$
52.
$f( g( 4 ) ) f( g( 4 ) )$
53.
$f( g( 1 ) ) f( g( 1 ) )$
54.
$f( h( 2 ) ) f( h( 2 ) )$
55.
$h( f( 2 ) ) h( f( 2 ) )$
56.
$f( g( h( 4 ) ) ) f( g( h( 4 ) ) )$
57.
$f( g( f( −2 ) ) ) f( g( f( −2 ) ) )$
### Numeric
For the following exercises, use the function values for shown in Table 3 to evaluate each expression.
$x x$ $f(x) f(x)$ $g(x) g(x)$
079
165
256
382
441
508
627
713
894
930
Table 3
58.
$f( g( 8 ) ) f( g( 8 ) )$
59.
$f( g( 5 ) ) f( g( 5 ) )$
60.
$g( f( 5 ) ) g( f( 5 ) )$
61.
$g( f( 3 ) ) g( f( 3 ) )$
62.
$f( f( 4 ) ) f( f( 4 ) )$
63.
$f( f( 1 ) ) f( f( 1 ) )$
64.
$g( g( 2 ) ) g( g( 2 ) )$
65.
$g( g( 6 ) ) g( g( 6 ) )$
For the following exercises, use the function values for shown in Table 4 to evaluate the expressions.
$x x$ $f(x) f(x)$ $g(x) g(x)$
$−3 −3$ 11 $−8 −8$
$−2 −2$ 9 $−3 −3$
$−1 −1$ 7 0
0 5 1
1 3 0
2 1 $−3 −3$
3 $−1 −1$ $−8 −8$
Table 4
66.
$(f∘g)(1) (f∘g)(1)$
67.
$(f∘g)(2) (f∘g)(2)$
68.
$(g∘f)(2) (g∘f)(2)$
69.
$(g∘f)(3) (g∘f)(3)$
70.
$(g∘g)(1) (g∘g)(1)$
71.
$(f∘f)(3) (f∘f)(3)$
For the following exercises, use each pair of functions to find $f( g( 0 ) ) f( g( 0 ) )$ and $g( f(0) ). g( f(0) ).$
72.
$f(x)=4x+8,g(x)=7− x 2 f(x)=4x+8,g(x)=7− x 2$
73.
$f(x)=5x+7,g(x)=4−2 x 2 f(x)=5x+7,g(x)=4−2 x 2$
74.
$f(x)= x+4 ,g(x)=12− x 3 f(x)= x+4 ,g(x)=12− x 3$
75.
$f(x)= 1 x+2 ,g(x)=4x+3 f(x)= 1 x+2 ,g(x)=4x+3$
For the following exercises, use the functions $f(x)=2 x 2 +1 f(x)=2 x 2 +1$ and $g(x)=3x+5 g(x)=3x+5$ to evaluate or find the composite function as indicated.
76.
$f( g(2) ) f( g(2) )$
77.
$f( g(x) ) f( g(x) )$
78.
$g( f(−3) ) g( f(−3) )$
79.
$(g∘g)(x) (g∘g)(x)$
### Extensions
For the following exercises, use $f(x)= x 3 +1 f(x)= x 3 +1$ and $g(x)= x−1 3 . g(x)= x−1 3 .$
80.
Find $(f∘g)(x) (f∘g)(x)$ and $(g∘f)(x). (g∘f)(x).$ Compare the two answers.
81.
Find $(f∘g)(2) (f∘g)(2)$ and $(g∘f)(2). (g∘f)(2).$
82.
What is the domain of $(g∘f)(x)? (g∘f)(x)?$
83.
What is the domain of $(f∘g)(x)? (f∘g)(x)?$
84.
Let $f(x)= 1 x . f(x)= 1 x .$
1. Find $(f∘f)(x). (f∘f)(x).$
2. Is $(f∘f)(x) (f∘f)(x)$ for any function $f f$ the same result as the answer to part (a) for any function? Explain.
For the following exercises, let $F(x)= (x+1) 5 , F(x)= (x+1) 5 ,$ $f(x)= x 5 , f(x)= x 5 ,$ and $g(x)=x+1. g(x)=x+1.$
85.
True or False: $(g∘f)(x)=F(x). (g∘f)(x)=F(x).$
86.
True or False: $(f∘g)(x)=F(x). (f∘g)(x)=F(x).$
For the following exercises, find the composition when $f(x)= x 2 +2 f(x)= x 2 +2$ for all $x≥0 x≥0$ and $g(x)= x−2 . g(x)= x−2 .$
87.
$(f∘g)(6);(g∘f)(6) (f∘g)(6);(g∘f)(6)$
88.
$(g∘f)(a);(f∘g)(a) (g∘f)(a);(f∘g)(a)$
89.
$(f∘g)(11);(g∘f)(11) (f∘g)(11);(g∘f)(11)$
### Real-World Applications
90.
The function $D(p) D(p)$ gives the number of items that will be demanded when the price is $p. p.$ The production cost $C(x) C(x)$ is the cost of producing $x x$ items. To determine the cost of production when the price is \$6, you would do which of the following?
1. Evaluate $D( C(6) ). D( C(6) ).$
2. Evaluate $C( D(6) ). C( D(6) ).$
3. Solve $D( C(x) )=6. D( C(x) )=6.$
4. Solve $C( D(p) )=6. C( D(p) )=6.$
91.
The function $A(d) A(d)$ gives the pain level on a scale of 0 to 10 experienced by a patient with $d d$ milligrams of a pain-reducing drug in her system. The milligrams of the drug in the patient’s system after $t t$ minutes is modeled by $m(t). m(t).$ Which of the following would you do in order to determine when the patient will be at a pain level of 4?
1. Evaluate $A( m(4) ). A( m(4) ).$
2. Evaluate $m( A(4) ). m( A(4) ).$
3. Solve $A( m(t) )=4. A( m(t) )=4.$
4. Solve $m( A(d) )=4. m( A(d) )=4.$
92.
A store offers customers a 30% discount on the price $x x$ of selected items. Then, the store takes off an additional 15% at the cash register. Write a price function $P(x) P(x)$ that computes the final price of the item in terms of the original price $x. x.$ (Hint: Use function composition to find your answer.)
93.
A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in minutes according to $r(t)=25 t+2 , r(t)=25 t+2 ,$ find the area of the ripple as a function of time. Find the area of the ripple at $t=2. t=2.$
94.
A forest fire leaves behind an area of grass burned in an expanding circular pattern. If the radius of the circle of burning grass is increasing with time according to the formula $r(t)=2t+1, r(t)=2t+1,$ express the area burned as a function of time, $t t$ (minutes).
95.
Use the function you found in the previous exercise to find the total area burned after 5 minutes.
96.
The radius $r, r,$ in inches, of a spherical balloon is related to the volume, $V, V,$ by $r(V)= 3V 4π 3 . r(V)= 3V 4π 3 .$ Air is pumped into the balloon, so the volume after $t t$ seconds is given by $V(t)=10+20t. V(t)=10+20t.$
1. Find the composite function $r( V(t) ). r( V(t) ).$
2. Find the exact time when the radius reaches 10 inches.
97.
The number of bacteria in a refrigerated food product is given by $N(T)=23 T 2 −56T+1, N(T)=23 T 2 −56T+1,$ $3 where $T T$ is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by $T(t)=5t+1.5, T(t)=5t+1.5,$ where $t t$ is the time in hours.
1. Find the composite function $N( T(t) ). N( T(t) ).$
2. Find the time (round to two decimal places) when the bacteria count reaches 6752.
Order a print copy
As an Amazon Associate we earn from qualifying purchases.
|
Question Video: Determine Whether the Logarithm Function or Square Root Function Is Dominant | Nagwa Question Video: Determine Whether the Logarithm Function or Square Root Function Is Dominant | Nagwa
Question Video: Determine Whether the Logarithm Function or Square Root Function Is Dominant Mathematics • Higher Education
Given that π(π₯) = β(5π₯) and π(π₯) = logβ 5π₯, use lim_(π₯ β β) π(π₯)/g(π₯) to determine whether π(π₯) or π(π₯) is dominant.
05:21
Video Transcript
Given that π of π₯ is equal to the square root of five π₯ and π of π₯ is equal to the log base three of five π₯, use the limit as π₯ approaches β of π of π₯ divided by π of π₯ to determine whether π of π₯ or π of π₯ is dominant.
The question gives us two functions, π of π₯ and π of π₯. The question wants us to determine which of these two functions is dominant by evaluating the limit as π₯ approaches β of π of π₯ divided by π of π₯. Letβs first recall what it means for a function to be dominant. We recall for eventually positive functions π and π, if the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to β, then we say the function π of π₯ is dominant. However, if the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to zero, then we say that π of π₯ is dominant.
Finally, if the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to some nonzero real constant π, then we say that neither of these functions are dominant. So, to determine which of our functions is dominant, we need to evaluate the limit as π₯ approaches β of the square root of five π₯ divided by the log base three of five π₯.
However, if we try to evaluate this limit directly, we run into a problem. Our numerator is increasing without bound as π₯ approaches β, and our denominator is also increasing without bound as π₯ approaches β. So, if we try to evaluate our limit directly, we get the indeterminate form β divided by β. We recall if we evaluate the limit of a quotient of two functions and we get an indeterminant form, one way of attempting to deal with this is by using LβHΓ΄pitalβs rule.
We recall the following version of LβHΓ΄pitalβs rule. If we have two differentiable functions π and π, where the limit as π₯ approaches β of π of π₯ and the limit as π₯ approaches β of π of π₯ are both equal to β. Then we know the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to the limit as π₯ approaches β of π prime of π₯ divided by π prime of π₯ provided that this limit exists or itβs equal to positive or negative β.
In other words, under these conditions, instead of evaluating the limit as π₯ approaches β of π of π₯ divided by π of π₯, we can instead evaluate the limit as π₯ approaches β of π prime of π₯ divided by π prime of π₯. So, letβs first check that we can actually use LβHΓ΄pitalβs rule in this case. We first need to show that both π and π are differentiable functions. And we can do this directly in this case. Weβll start by finding an expression for π prime of π₯.
First, π of π₯ is equal to the square root of five times π₯ to the power of one-half. Then, we can differentiate this by using the power rule for differentiation. We multiply by the exponent and reduce the exponent by one. This gives us π prime of π₯ is equal to one-half times root five times π₯ to the power of negative one-half. And by using our laws of exponents, we can rewrite this as root five divided by two root π₯. So, weβve shown that π is differentiable. Letβs now do the same for π.
We have that π of π₯ is equal to the log base three of five π₯. And to differentiate this, we recall for positive constants π and π, where π is not equal to one, the derivative of the log base π of ππ₯ with respect to π₯ is equal to one divided by π₯ times the natural logarithm of π. Using this, we get π prime of π₯ is equal to one divided by π₯ times the natural logarithm of three. So, π of π₯ is a differentiable function.
The last thing we need to show to use this version of LβHΓ΄pitalβs rule is the limit as π₯ approaches β of π of π₯ and the limit as π₯ approaches β of π of π₯ are both equal to β. And weβve actually already shown both of these conditions are true when we originally tried to directly evaluate our limit. As π₯ approached β, both our numerator and our denominator grew without bound. In other words, the limit as π₯ approached β of both π of π₯ and π of π₯ was equal to β.
So, weβve shown all the conditions for this version of LβHΓ΄pitalβs rule are true. So, to evaluate our limit, we can now instead attempt to evaluate the limit as π₯ approaches β of π prime of π₯ divided by π prime of π₯. And we already found expressions for π prime of π₯ and π prime of π₯. So letβs write these in. This gives us the limit as π₯ approaches β of the square root of five divided by two root π₯ all divided by one divided by π₯ times the natural logarithm of three.
To help us evaluate this limit, instead of dividing by the fraction one divided by π₯ times the natural logarithm of three, letβs multiply it by the reciprocal. Doing this, we get the limit as π₯ approaches β of root five times π₯ times the natural logarithm of three divided by two root π₯. And we now see both our numerator and our denominator share a factor of the square root of π₯. If we cancel this shared factor, we get the limit as π₯ approaches β of root five times root π₯ times the natural logarithm of three divided by two.
And now, we can evaluate this limit directly. The only part of this limit which changes as the value of π₯ changes is the square root of π₯. Everything else is a positive constant. So, as π₯ approaches β, our numerator is increasing without bound. This means our limit is equal to β. And remember, we said, if this limit is equal to β, then we say that π of π₯ is dominant.
Therefore, by using LβHΓ΄pitalβs rule, weβve shown if π of π₯ is equal to the square root of five π₯ and π of π₯ is equal to the log base three of five π₯. Then the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to β. And so, π of π₯ is the dominant function.
Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions
|
# What Is the Distributive Property Law in Mathematics?
The distributive property law of numbers is a handy way of simplifying complex mathematical equations by breaking them down into smaller parts. It can be especially useful if you are struggling to understand algebra
Students usually begin learning the distributive property law when they start advanced multiplication. Take, for instance, multiplying 4 and 53. Calculating this example will require carrying the number 1 when you multiply, which can be tricky if you're being asked to solve the problem in your head.
There's an easier way of solving this problem. Begin by taking the larger number and rounding it down to the nearest figure that's divisible by 10. In this case, 53 becomes 50 with a difference of 3. Next, multiply both numbers by 4, then add the two totals together. Written out, the calculation looks like this:
53 x 4 = 212, or
(4 x 50) + (4 x 3) = 212, or
200 + 12 = 212
## Simple Algebra
The distributive property also can be used to simplify algebraic equations by eliminating the parenthetical portion of the equation. Take for instance the equation a(b + c), which also can be written as (ab) + (ac) because the distributive property dictates that a, which is outside the parenthetical, must be multiplied by both b and c. In other words, you are distributing the multiplication of a between both b and c. For example:
2(3+6) = 18, or
(2 x 3) + (2 x 6) = 18, or
6 + 12 = 18
Don't be fooled by the addition. It's easy to misread the equation as (2 x 3) + 6 = 12. Remember, you are distributing the process of multiplying 2 evenly between 3 and 6.
The distributive property law can also be used when multiplying or dividing polynomials, which are algebraic expressions that include real numbers and variables, and monomials, which are algebraic expressions consisting of one term.
You can multiply a polynomial by a monomial in three simple steps using the same concept of distributing the calculation:
1. Multiply the outside term by the first term in parenthesis.
2. Multiply the outside term by the second term in parenthesis.
Written out, it looks like this:
x(2x+10), or
(x * 2x) + (x * 10), or
2x2 + 10x
To divide a polynomial by a monomial, split it up into separate fractions then reduce. For example:
(4x3 + 6x2 + 5x) / x, or
(4x3 / x) + (6x2 / x) + (5x / x), or
4x2 + 6x + 5
You also can use the distributive property law to find the product of binomials, as shown here:
(x + y)(x + 2y), or
(x + y)x + (x + y)(2y), or
x2+xy +2xy 2y2, or
x2 + 3xy +2y2
## More Practice
These algebra worksheets will help you understand how the distributive property law works. The first four do not involve exponents, which should make it easier for students to understand the basics of this important mathematical concept.
Format
mla apa chicago
|
Home > Probability > Probability of Simple Events
# Probability of Simple Events
A simple event is an event where all possible outcomes are equally likely to occur. So the probability of simple events will have all possible outcomes equally likely to happen or occur.
For example, when you toss a coin, there are two possible outcomes – heads or tails, and the probability of heads or tails is equal. Similarly, when you roll a die, you can get any of the 6 numbers – 1 to 6, and the chance of any one of these 6 numbers is equal to the others.
A set of all possible outcomes of an event is called sample space, usually represented as ‘S’ with curly braces { }. So with the coin toss, the sample space of all outcomes S = {1,2}.
And with the dice, the sample space S = {1,2,3,4,5,6}
Now, let us toss a coin say 50 times, and record the outcome (heads or tails) each time. And let us say heads occurred 28 times, and tails occurred 22 times. According to this ‘experiment’ the probability of heads occurring is given as:
P (Heads) = , and the probability of tails occcuring is P (Tails) = .
This is called Experimental Probability – probability derived from a series of trials, or experiments. The probabilities obtained from experimental probability may differ as we increase the number of trials. With a very large number of trials or experiments, the probabilities may come close to the actual Theoretical Probability.
In theoretical probability, we expect the occurence of outcomes to be equally likely. When all outcomes are equally likely, then the theoretical probability of an event E is given by:
Probability of E or P(E) =
or P(E) =
where n(E) = number of ways the event E can occur, and n(S) = total number of possible outcomes.
## Examples of probability of simple events:
Example 1: A committee has 8 female and 12 male members. What is the probability of choosing a female as the president of this committee?
Probability of choosing a female as a president =
Here the number of females (favourable event) is 8, and the total number of members (outcomes) is 20. Hence
P (choosing a female president) =
Example 2: The numbers 1 to 10 are written on separate pieces of paper, folded and put in a box. One number (piece of paper) is drawn from this box.
2a. What is the probability that this number chosen randomly is 3?
Sample space = {1,2,3,4,5,6,7,8,9,10}, so n(S) = 10.
There is only one 3 in the box, so n(E) = 1.
Hence P(3) =
2b. What is the probability that this randomly chosen number is even?
Once again, n(S) = 10.
There are five even numbers in the box, so n(E) = 5.
Hence P(even number) = =
2c. What is the probability that this number chosen is a prime number?
As we saw earlier, n(S) = 10.
The prime numbers in the box are 2, 3, 5, and 7. So n(E) = 4.
So P(prime number) = =
Example 3: A bag has 3 green, 2 red, 5 purple, 10 white and 5 black marbles.
3a. What is the probability of choosing a black ball?
Total marbles in the bag = 25.
Total black marbles = 5.
So P (black) = =
3b. What is the chance of picking a red or a green marble?
Total number of red and green marbles = 3 + 2 = 5
P (red or green) =
3c. What is the probability of choosing a brown marble?
There are no brown marbles in the bag.
So P (brown) = = 0
3d. What is the probability of choosing any coloured marble?
Since there are 25 coloured marbles in the bag, all the 25 marbles are favoured.
P (any coloured marble) = = 1.
We now look at probability of complementary events
|
# You roll two balanced dice one time. What is the probability that you obtain either a sum of 8 or the same number on each of the two dice?
Dec 17, 2014
A probability can be calculated as:
$\left(\text{Desired events")/("All possible events}\right)$
Let's start by calculating the denominator first, i.e. all possible events. When you roll two dice, there are 36 possible combinations that might come up. Each dice can be any of the numbers 1 to 6. So $6 \cdot 6 = 36$. You could make a list to convince yourself, i.e. {1,1}, {1,2}, {1,3},...,{6,6}.
Now let's look at the numerator, i.e. all possible desired events. We desire either a sum of 8 or the same number on each of the two dice.
• It's relatively easy to count the number of ways you can get the same number on both dice. It's exactly six, i.e. {1,1}, {2,2},...,{6,6}.
• Now let's look at the different ways we can get a sum of 8. These are: {2,6}, {3,5}, {4,4}, {5,3} and {6,2}. This is a total of five.
So, in total we have 11 ways to get either a sum of 8 or the same number on both dice. However, note that the option {4,4} appears in both sets, so we can't count this twice. Removing that double-counting, we end with 10 ways to get what we 'desire'.
So, our final answer is $\frac{10}{36} = \frac{5}{18}$.
|
# Divide and Conquer Algorithms: Integer Multiplication
## Introduction
In the field of computer science and mathematics, the Divide and Conquer algorithm is a powerful technique used to solve complex problems by breaking them down into smaller, more manageable subproblems. This approach can be particularly useful when dealing with computations involving large numbers, such as integer multiplication.
In this tutorial, we will explore the concept of Divide and Conquer algorithms, specifically focusing on its application to integer multiplication. We will delve into the principles behind this algorithm, discuss its implementation, and provide code snippets to illustrate its functionality.
## Understanding the Divide and Conquer Approach
The Divide and Conquer approach can be described as a three-step process:
1. Divide: Break the problem down into smaller, more manageable subproblems. In the case of integer multiplication, this would involve dividing the initial numbers into smaller sections.
2. Conquer: Solve the subproblems recursively. This step will involve multiplying the smaller sections obtained through division.
3. Combine: Combine the solutions of the subproblems to obtain the final result. For integer multiplication, this would mean properly adding and aligning the partial products obtained from the previous step.
## Implementing Integer Multiplication using Divide and Conquer
Let's now dive into the implementation details of integer multiplication using the Divide and Conquer algorithm. We will be using a recursive approach to break down the problem and perform the necessary computations.
``````def integer_multiplication(x, y):
# Base case: if the numbers are single digits, perform a direct multiplication
if len(str(x)) == 1 or len(str(y)) == 1:
return x * y
# Calculate the size of the numbers
size = max(len(str(x)), len(str(y)))
# Divide the numbers into smaller sections
m = size // 2
# Split the numbers into high and low halves
a, b = divmod(x, 10**m)
c, d = divmod(y, 10**m)
# Recursively compute the partial products
ac = integer_multiplication(a, c)
bd = integer_multiplication(b, d)
ad_bc = integer_multiplication(a + b, c + d) - ac - bd
# Combine the partial products to obtain the final result
return (ac * 10**(2*m)) + (ad_bc * 10**m) + bd
``````
By recursively applying this code snippet, we can efficiently multiply two large numbers using the Divide and Conquer strategy.
## Example of Integer Multiplication using Divide and Conquer
To further clarify the implementation and functionality of the Divide and Conquer algorithm for integer multiplication, let's consider an example:
We want to multiply the numbers 1234 and 5678.
Following the Divide and Conquer approach, we break down the problem into the following steps:
1. Divide the numbers into smaller sections:
• `1234` is split into `12` and `34`.
• `5678` is split into `56` and `78`.
2. Recursively compute the partial products:
• `12 * 56` = `672`
• `34 * 78` = `2652`
• `(12 + 34) * (56 + 78)` = `8100`
3. Combine the partial products to obtain the final result:
• Result = `(672 * 10^4) + (8100 * 10^2) + 2652 = 7006652`
As demonstrated by this example, the Divide and Conquer algorithm allows us to perform integer multiplication with a much lower computational complexity than traditional approaches.
## Conclusion
In this tutorial, we explored the concept of Divide and Conquer algorithms, specifically focusing on integer multiplication. We discussed the three-step process of divide, conquer, and combine, which forms the foundation of this algorithm. Additionally, we provided a code snippet and an example to demonstrate the implementation and functionality of Divide and Conquer on numbers.
By understanding and effectively utilizing the Divide and Conquer approach, programmers can tackle complex computations involving large numbers more efficiently, optimizing both time and resources.
Keep exploring the vast possibilities and applications of Divide and Conquer algorithms in your programming journey!
|
## Scientific Applications of Quadratic Functions
Quadratic relationships between variables are commonly found in physical sciences, engineering, and elsewhere.
### Learning Objectives
Use the quadratic equation to model phenomena studied in science
### Key Takeaways
#### Key Points
• The Pythagorean Theorem, $c^2=a^2+b^2$relates the length of the hypotenuse ($c$) of a right triangle to the lengths of its legs ($a$ and $b$).
• Problems involving gravity and projectile motion are typically dependent upon a second-order variable, usually time or initial velocity depending on the relationship.
• Coulomb’s Law, which relates electrostatic force, charge amount and distance between two charged particles, has a second-order dependence on the separation of the particles. Solving for either charge results in a quadratic function.
#### Key Terms
• acceleration: The change of velocity with respect to time (can include changing direction).
• velocity: A vector quantity that denotes the rate of change of position with respect to time, or a speed with the directional component.
Quadratic relationships between variables are commonly found in physical sciences, engineering, and elsewhere. Perhaps the most universally used example of quadratic relationships in problem solving concerns right triangles.
### The Pythagorean Theorem
The Pythagorean Theorem is used to relate the three sides of right triangles. It states:
$c^2=a^2+b^2$
This says that the square of the length of the hypotenuse ($c$) is equal to the sum of the squares of the two legs ($a$ and $b$) of the triangle. This has been proven in many ways, among the most famous of which was devised by Euclid.
In practice, the Pythagorean Theorem is often solved via factoring or completing the square. If any of the variables $a$, $b$, or $c$ represent functions in themselves, it is often useful to expand the terms, combine like variables, and then re-factor the expression.
### Gravity and Projectile Motion
Most all equations involving gravity include a second-order relationship.
### Gravitational Force
Consider the equation relating gravitational force ($F$) between two objects to the masses of each object ($m_1$ and $m_2$) and the distance between them ($r$):
$F=G\dfrac {m_1m_2}{r^2}$
The shape of this function is not a parabola, but becomes such a shape if rearranged to solve for $m_1$ or $m_2$, as seen below:
$m_2 = \frac{Fr^2}{Gm_1} = (\frac{F}{Gm_1}) r^2$
The general form of this function is:
$a(x-h)^2 + k$
which you should recognize as the vertex form of a quadratic equation.
### Projectile Motion
The maximum height of a projectile launched directly upwards can also be calculated from a quadratic relationship. The formula relates height ($h$) to initial velocity ($v_0$) and gravitational acceleration ($g$):
$h=\frac {v_0^2}{2g}$
The same maximum height of a projectile launched directly upwards can be found using the time at the projectile’s peak ($t_h$):
$h=v_0t_h(\frac {1}{2})gt_h^2$
Substituting any time ($t$) in place of $t_h$ leaves the equation for height as a quadratic function of time.
### Electrostatic Force
The equation relating electrostatic force ($F$) between two particles, the particles’ respective charges ($q_1$ and $q_2$), and the distance between them ($r$) is very similar to the aforementioned formula for gravitational force:
$F=\frac {q_1q_2}{4\pi \epsilon_0r^2}$
This is known as Coulomb’s Law. Solving for either charge results in a quadratic equation where the charge is dependent on $r^2$.
## Financial Applications of Quadratic Functions
For problems involving quadratics in finance, it is useful to graph the equation. From these, one can easily find critical values of the function by inspection.
### Learning Objectives
Apply the quadratic function to real world financial models
### Key Takeaways
#### Key Points
• In some financial math problems, several key points on a quadratic function are desired, so it can become tedious to calculate each algebraically.
• Rather than calculating each key point of a function, one can find these values by inspection of its graph.
• Graphs of quadratic functions can be used to find key points in many different relationships, from finance to science and beyond.
The method of graphing a function to determine general properties can be used to solve financial problems.Given the algebraic equation for a quadratic function, one can calculate any point on the function, including critical values like minimum/ maximum and x- and y-intercepts.
These calculations can be more tedious than is necessary, however. A graph contains all the above critical points and more, and acts as a clear and concise representation of a function. If one needs to determine several values on a quadratic function, glancing at a graph is quicker than calculating several points.
### Example
Consider the function:
$F(x)=-\frac {x^2}{10}+50x-750$
Suppose this models a profit function $f(x)$ in dollars that a company earns as a function of $x$ number of products of a given type that are sold, and is valid for values of $x$ greater than or equal to $0$ and less than or equal to $500$.
If a financier wanted to find the number of sales required to break even, the maximum possible loss (and the number of sales required for this loss), and the maximum profit (and the number of sales required for this profit), they could simply reference a graph instead of calculating it out algebraically.
Financial example: Graph of the equation $F(x) = -\frac{x^2}{10} + 50x – 750$, where the x-axis is number of sales, and the y-axis is the monetary return.
By inspection, we can find that the break-even points ($x$-intercepts) are between $15$ and $16$ sales, and between $484$ and $485$ sales. The maximum loss before the second break-even points is $750 (the y-intercept), which is lost at $0$ sales. After the second break-even point, the function decreases to infinity, so the losses continue to increase. Maximum profit is$5500 (the vertex), which is achieved at $250$ sales.
|
Addition Fact Sums to 11 Chart | Fact Families with 11 | 2 Numbers that Add up to 11
Addition Fact Sums to 11 will help you determine the two numbers that add to eleven. There are several combinations that add up to eleven and we have figured out all of them keeping in mind your comfort. Refer to further sections to learn about the Basic Addition Fact, Addition Fact Sums to 11 Chart, Prerequisite Skills to Learn on What 2 Numbers add up to 11.
You can find the Addition Facts Equal to 11 Table as well as understand the patterns associated with it. Learn the concept associated with it and memorize the addition facts that sums to 11 easily. Check out the Solved Examples on finding the missng addend employing the facts that sums to eleven, etc.
Read Similar Articles: Addition Fact Sums to 10
What is a Basic Addition Fact?
Basic Addition Fact is defined as the sum of one digit addends. However, if you have a two-digit number in any one of the addends then it is not a Basic Addition.
Example: 3+6 =9 is an addition fact.
6+10 =16 is not an addition fact.
Addition Facts to 11 Chart | Table to Memorize Addition Fact Sums to 11
Below is the table that when added sums up to 11. Without Memorizing the Facts about Addition it would be difficult to tackle the problems on addition. It is easier to understand the Addition Facts of higher numbers by having a grip on these simple facts about addition.
+ 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 11 2 3 4 5 6 7 8 9 10 11 3 4 5 6 7 8 9 10 11 4 5 6 7 8 9 10 11 5 6 7 8 9 10 11 6 7 8 9 10 11 7 8 9 10 11 8 9 10 11 9 10 11 10 11
From the above table, we can write the addition facts that equals to 11 as follows
0 + 11 = 11
1 + 10 = 11
2 + 9 = 11
3 + 8 = 11
4 + 7 = 11
5 + 6 = 11
6 + 5 = 11
7 + 4 = 11
8 + 3 = 11
9 + 2 = 11
10 + 1 = 11
11 + 0 = 11
How to Teach Addition Facts with 11?
Go through the simple measures listed here to make them familiar with the basic addition facts that equals to 11. They are in the following way
• Rather than forcing them to muggup all the Addition Facts that sum up to 11, break it up into small groups.
• Ask them to practice with easier facts such as +1, +2 so that they don’t difficulty in understanding the later on bigger number fact families.
• Â Having a grip on the earlier addition facts will help kids and boost their confidence levels and they will be ready to work on the addition problems, missing addends, etc.
• Visualizing also plays a key role and once your kid has mastered this ability he can change and manipulate them mentally to arrive at the solution.
• Practice the facts about addition that add up to 11 regularly to memorize them easily.
Prerequisite Skills for Learning What Two Numbers Add up to 11?
Follow the basic fundamentals involved to learn about the facts about addition that sum up to Eleven. They are in the below way
• Must be able to count backward and forwards of numbers up to 20 while performing a physical task.
• You must be able to give answers on one more, two-more, one-less, two-less for any whole number.
• Must have to ability to double any number up to 10.
Worked out Examples on Addition Facts that Sum up to 11
Example 1.Â
Solution:
Start counting after 4 and count up to 7
4, 5, 6, 7, 8, 9, 10, 11
Count the numbers after 4
The Sum of 4+7 is 11
Example 2.
Find the missing addend whose addition fact sums to 11?
(i) 11 = 5+?
(ii) 11 = 8+?
Solution:
(i) 11 = 5+?
Start counting the numbers after 5 to make it equal to 11
5, 6, 7, 8, 9, 10, 11
Since there are 6 numbers after 5
5+6 =11
(ii) 11 = 8+?
Start counting the numbers after 8 to make it equal to 11
8, 9, 10, 11
Since there are 3 numbers after the 8
8+3 =11
Word Problems on Addition Fact Families with 11
Example 3.
John arranged his toy bikes in rows. The first row had seven bikes and the second has four bikes. How many bikes John have?
Solution:
No. of bikes in the first row = 7
No. of bikes in the second row = 4
Total bikes John have = 7 + 4
Start Counting from 7 and count up to 4
7, 8, 9, 10, 11
Therefore, John had 11 bikes in total.
Example 4.
Mom has five oranges and Dad has six more than her. How many Oranges does Dad have?
Solution:
No. of Oranges Mom has = 5
No. of Oranges Dad have = ?
Since Dad has six more than Mom
We Shall Count after 5 and add up to 6
5, 6, 7, 8, 9, 10, 11
Therefore, Dad has Six Oranges in Total.
Scroll to Top
Scroll to Top
|
# Circumscribed circle
Last updated on 29 May 2017
In geometry, the circumscribed circle or circumcircle of a polygon is a circle which passes through all the vertices of the polygon. The center of this circle is called the circumcenter and its radius is called the circumradius.
A polygon which has a circumscribed circle is called a cyclic polygon (sometimes a concyclic polygon, because the vertices are concyclic). All regular simple polygons, all isosceles trapezoids, all triangles and all rectangles are cyclic.
A related notion is the one of a minimum bounding circle, which is the smallest circle that completely contains the polygon within it. Not every polygon has a circumscribed circle, as the vertices of a polygon do not need to all lie on a circle, but every polygon has a unique minimum bounding circle, which may be constructed by a linear time algorithm.[2] Even if a polygon has a circumscribed circle, it may not coincide with its minimum bounding circle; for example, for an obtuse triangle, the minimum bounding circle has the longest side as diameter and does not pass through the opposite vertex.
Circumscribed circle, C, and circumcenter, O, of a cyclic polygon, P
## Triangles
All triangles are cyclic; i.e., every triangle has a circumscribed circle.
This can be proven on the grounds that the general equation for a circle with center (a, b) and radius r in the Cartesian coordinate system is
${\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}.}$
Since this equation has three parameters (a, b, r) only three points' coordinate pairs are required to determine the equation of a circle. Since a triangle is defined by its three vertices, and exactly three points are required to determine a circle, every triangle can be circumscribed.
### Straightedge and compass construction
Construction of the circumcircle (red) and the circumcenter Q (red dot)
The circumcenter of a triangle can be constructed by drawing any two of the three perpendicular bisectors. The center is the point where the perpendicular bisectors intersect, and the radius is the length to any of the three vertices.
This is because the circumcenter is equidistant from any pair of the triangle's vertices, and all points on the perpendicular bisectors are equidistant from two of the vertices of the triangle.
### Alternate construction
Alternate construction of the circumcenter (intersection of broken lines)
An alternate method to determine the circumcenter is to draw any two lines each one departing from one of the vertices at an angle with the common side, the common angle of departure being 90° minus the angle of the opposite vertex. (In the case of the opposite angle being obtuse, drawing a line at a negative angle means going outside the triangle.)
In coastal navigation, a triangle's circumcircle is sometimes used as a way of obtaining a position line using a sextant when no compass is available. The horizontal angle between two landmarks defines the circumcircle upon which the observer lies.
### Circumcircle equations
#### Cartesian coordinates
In the Euclidean plane, it is possible to give explicitly an equation of the circumcircle in terms of the Cartesian coordinates of the vertices of the inscribed triangle. Suppose that
{\displaystyle {\begin{aligned}\mathbf {A} &=(A_{x},A_{y})\\\mathbf {B} &=(B_{x},B_{y})\\\mathbf {C} &=(C_{x},C_{y})\end{aligned}}}
are the coordinates of points A, B, and C. The circumcircle is then the locus of points v = (vx,vy) in the Cartesian plane satisfying the equations
{\displaystyle {\begin{aligned}|\mathbf {v} -\mathbf {u} |^{2}&=r^{2}\\|\mathbf {A} -\mathbf {u} |^{2}&=r^{2}\\|\mathbf {B} -\mathbf {u} |^{2}&=r^{2}\\|\mathbf {C} -\mathbf {u} |^{2}&=r^{2}\end{aligned}}}
guaranteeing that the points A, B, C, and v are all the same distance r from the common center u of the circle. Using the polarization identity, these equations reduce to the condition that the matrix
${\displaystyle {\begin{bmatrix}|\mathbf {v} |^{2}&-2v_{x}&-2v_{y}&-1\\|\mathbf {A} |^{2}&-2A_{x}&-2A_{y}&-1\\|\mathbf {B} |^{2}&-2B_{x}&-2B_{y}&-1\\|\mathbf {C} |^{2}&-2C_{x}&-2C_{y}&-1\end{bmatrix}}}$
has a nonzero kernel. Thus the circumcircle may alternatively be described as the locus of zeros of the determinant of this matrix:
${\displaystyle \det {\begin{bmatrix}|\mathbf {v} |^{2}&v_{x}&v_{y}&1\\|\mathbf {A} |^{2}&A_{x}&A_{y}&1\\|\mathbf {B} |^{2}&B_{x}&B_{y}&1\\|\mathbf {C} |^{2}&C_{x}&C_{y}&1\end{bmatrix}}=0.}$
Using cofactor expansion, let
{\displaystyle {\begin{aligned}S_{x}&={\frac {1}{2}}\det {\begin{bmatrix}|\mathbf {A} |^{2}&A_{y}&1\\|\mathbf {B} |^{2}&B_{y}&1\\|\mathbf {C} |^{2}&C_{y}&1\end{bmatrix}},\\S_{y}&={\frac {1}{2}}\det {\begin{bmatrix}A_{x}&|\mathbf {A} |^{2}&1\\B_{x}&|\mathbf {B} |^{2}&1\\C_{x}&|\mathbf {C} |^{2}&1\end{bmatrix}},\\a&=\det {\begin{bmatrix}A_{x}&A_{y}&1\\B_{x}&B_{y}&1\\C_{x}&C_{y}&1\end{bmatrix}},\\b&=\det {\begin{bmatrix}A_{x}&A_{y}&|\mathbf {A} |^{2}\\B_{x}&B_{y}&|\mathbf {B} |^{2}\\C_{x}&C_{y}&|\mathbf {C} |^{2}\end{bmatrix}}\end{aligned}}}
we then have a|v|2 − 2Svb = 0 and, assuming the three points were not in a line (otherwise the circumcircle is that line that can also be seen as a generalized circle with S at infinity), |vS/a|2 = b/a + |S|2/a2, giving the circumcenter S/a and the circumradius √(b/a + |S|2/a2). A similar approach allows one to deduce the equation of the circumsphere of a tetrahedron.
#### Parametric equation
A unit vector perpendicular to the plane containing the circle is given by
${\displaystyle {\hat {n}}={\frac {(P_{2}-P_{1})\times (P_{3}-P_{1})}{|(P_{2}-P_{1})\times (P_{3}-P_{1})|}}.}$
Hence, given the radius, r, center, Pc, a point on the circle, P0 and a unit normal of the plane containing the circle, ${\displaystyle \scriptstyle {\hat {n}}}$, one parametric equation of the circle starting from the point P0 and proceeding in a positively oriented (i.e., right-handed) sense about ${\displaystyle \scriptstyle {\hat {n}}}$ is the following:
${\displaystyle \mathrm {R} (s)=\mathrm {P_{c}} +\cos \left({\frac {\mathrm {s} }{\mathrm {r} }}\right)(P_{0}-P_{c})+\sin \left({\frac {\mathrm {s} }{\mathrm {r} }}\right)\left[{\hat {n}}\times (P_{0}-P_{c})\right].}$
#### Trilinear and barycentric coordinates
An equation for the circumcircle in trilinear coordinates x : y : z is a/x + b/y + c/z = 0. An equation for the circumcircle in barycentric coordinates x : y : z is a2/x + b2/y + c2/z = 0.
The isogonal conjugate of the circumcircle is the line at infinity, given in trilinear coordinates by ax + by + cz = 0 and in barycentric coordinates by x + y + z = 0.
#### Higher dimensions
Additionally, the circumcircle of a triangle embedded in d dimensions can be found using a generalized method. Let A, B, and C be d-dimensional points, which form the vertices of a triangle. We start by transposing the system to place C at the origin:
{\displaystyle {\begin{aligned}\mathbf {a} &=\mathbf {A} -\mathbf {C} ,\\\mathbf {b} &=\mathbf {B} -\mathbf {C} .\end{aligned}}}
${\displaystyle r={\frac {\left\|\mathbf {a} \right\|\left\|\mathbf {b} \right\|\left\|\mathbf {a} -\mathbf {b} \right\|}{2\left\|\mathbf {a} \times \mathbf {b} \right\|}}={\frac {\left\|\mathbf {a} -\mathbf {b} \right\|}{2\sin \theta }}={\frac {\left\|\mathbf {A} -\mathbf {B} \right\|}{2\sin \theta }},}$
where θ is the interior angle between a and b. The circumcenter, p0, is given by
${\displaystyle p_{0}={\frac {(\left\|\mathbf {a} \right\|^{2}\mathbf {b} -\left\|\mathbf {b} \right\|^{2}\mathbf {a} )\times (\mathbf {a} \times \mathbf {b} )}{2\left\|\mathbf {a} \times \mathbf {b} \right\|^{2}}}+\mathbf {C} .}$
This formula only works in three dimensions as the cross product is not defined in other dimensions, but it can be generalized to the other dimensions by replacing the cross products with following identities:
{\displaystyle {\begin{aligned}(\mathbf {a} \times \mathbf {b} )\times \mathbf {c} &=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {b} \cdot \mathbf {c} )\mathbf {a} ,\\\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )&=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} ,\\\left\|\mathbf {a} \times \mathbf {b} \right\|&={\sqrt {\left\|\mathbf {a} \right\|^{2}\left\|\mathbf {b} \right\|^{2}-(\mathbf {a} \cdot \mathbf {b} )^{2}}}.\end{aligned}}}
### Circumcenter coordinates
#### Cartesian coordinates
The Cartesian coordinates of the circumcenter ${\displaystyle U=(U_{x},U_{y})}$ are
{\displaystyle {\begin{aligned}U_{x}&={\frac {1}{D}}\left[(A_{x}^{2}+A_{y}^{2})(B_{y}-C_{y})+(B_{x}^{2}+B_{y}^{2})(C_{y}-A_{y})+(C_{x}^{2}+C_{y}^{2})(A_{y}-B_{y})\right]\\U_{y}&={\frac {1}{D}}\left[(A_{x}^{2}+A_{y}^{2})(C_{x}-B_{x})+(B_{x}^{2}+B_{y}^{2})(A_{x}-C_{x})+(C_{x}^{2}+C_{y}^{2})(B_{x}-A_{x})\right]\end{aligned}}}
with
${\displaystyle D=2\left[A_{x}(B_{y}-C_{y})+B_{x}(C_{y}-A_{y})+C_{x}(A_{y}-B_{y})\right].\,}$
Without loss of generality this can be expressed in a simplified form after translation of the vertex A to the origin of the Cartesian coordinate systems, i.e., when A′ = AA = (Ax,Ay) = (0,0). In this case, the coordinates of the vertices B′ = BA and C′ = CA represent the vectors from vertex A′ to these vertices. Observe that this trivial translation is possible for all triangles and the circumcenter ${\displaystyle U'=(U'_{x},U'_{y})}$ of the triangle ABC′ follow as
{\displaystyle {\begin{aligned}U'_{x}&={\frac {1}{D'}}\left[C'_{y}(B_{x}^{'2}+B_{y}^{'2})-B'_{y}(C_{x}^{'2}+C_{y}^{'2})\right],\\U'_{y}&={\frac {1}{D'}}\left[B'_{x}(C_{x}^{'2}+C_{y}^{'2})-C'_{x}(B_{x}^{'2}+B_{y}^{'2})\right]\end{aligned}}}
with
${\displaystyle D'=2(B'_{x}C'_{y}-B'_{y}C'_{x}).\,}$
Due to the translation of vertex A to the origin, the circumradius r can be computed as
${\displaystyle r=||U'||={\sqrt {{U'_{x}}^{2}+{U'_{y}}^{2}}}}$
and the actual circumcenter of ABC follows as
${\displaystyle U=U'+A}$
#### Trilinear coordinates
The circumcenter has trilinear coordinates
cos α : cos β : cos γ
where α, β, γ are the angles of the triangle.
In terms of the side lengths a, b, c, the trilinears are
${\displaystyle a(b^{2}+c^{2}-a^{2}):b(c^{2}+a^{2}-b^{2}):c(a^{2}+b^{2}-c^{2}).}$
#### Barycentric coordinates
The circumcenter has barycentric coordinates
${\displaystyle a^{2}(b^{2}+c^{2}-a^{2}):\;b^{2}(c^{2}+a^{2}-b^{2}):\;c^{2}(a^{2}+b^{2}-c^{2}),\,}$
where a, b, c are edge lengths (BC, CA, AB respectively) of the triangle.
In terms of the triangle's angles ${\displaystyle \alpha ,\beta ,\gamma ,}$ the barycentric coordinates of the circumcenter are
${\displaystyle \sin 2\alpha :\sin 2\beta :\sin 2\gamma .}$
#### Circumcenter vector
Since the Cartesian coordinates of any point are a weighted average of those of the vertices, with the weights being the point's barycentric coordinates normalized to sum to unity, the circumcenter vector can be written as
${\displaystyle U={\frac {a^{2}(b^{2}+c^{2}-a^{2})A+b^{2}(c^{2}+a^{2}-b^{2})B+c^{2}(a^{2}+b^{2}-c^{2})C}{a^{2}(b^{2}+c^{2}-a^{2})+b^{2}(c^{2}+a^{2}-b^{2})+c^{2}(a^{2}+b^{2}-c^{2})}}.}$
Here U is the vector of the circumcenter and A, B, C are the vertex vectors. The divisor here equals 16S 2 where S is the area of the triangle.
#### Cartesian coordinates from cross- and dot-products
In Euclidean space, there is a unique circle passing through any given three non-collinear points P1, P2, and P3. Using Cartesian coordinates to represent these points as spatial vectors, it is possible to use the dot product and cross product to calculate the radius and center of the circle. Let
${\displaystyle \mathrm {P_{1}} ={\begin{bmatrix}x_{1}\\y_{1}\\z_{1}\end{bmatrix}},\mathrm {P_{2}} ={\begin{bmatrix}x_{2}\\y_{2}\\z_{2}\end{bmatrix}},\mathrm {P_{3}} ={\begin{bmatrix}x_{3}\\y_{3}\\z_{3}\end{bmatrix}}}$
Then the radius of the circle is given by
${\displaystyle \mathrm {r} ={\frac {\left|P_{1}-P_{2}\right|\left|P_{2}-P_{3}\right|\left|P_{3}-P_{1}\right|}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|}}}$
The center of the circle is given by the linear combination
${\displaystyle \mathrm {P_{c}} =\alpha \,P_{1}+\beta \,P_{2}+\gamma \,P_{3}}$
where
{\displaystyle {\begin{aligned}\alpha ={\frac {\left|P_{2}-P_{3}\right|^{2}\left(P_{1}-P_{2}\right)\cdot \left(P_{1}-P_{3}\right)}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|^{2}}}\\\beta ={\frac {\left|P_{1}-P_{3}\right|^{2}\left(P_{2}-P_{1}\right)\cdot \left(P_{2}-P_{3}\right)}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|^{2}}}\\\gamma ={\frac {\left|P_{1}-P_{2}\right|^{2}\left(P_{3}-P_{1}\right)\cdot \left(P_{3}-P_{2}\right)}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|^{2}}}\end{aligned}}}
#### Location relative to the triangle
The circumcenter's position depends on the type of triangle:
• If and only if a triangle is acute (all angles smaller than a right angle), the circumcenter lies inside the triangle.
• If and only if it is obtuse (has one angle bigger than a right angle), the circumcenter lies outside the triangle.
• If and only if it is a right triangle, the circumcenter lies at the center of the hypotenuse. This is one form of Thales' theorem.
The circumcenter of an acute triangle is inside the triangle
The circumcenter of a right triangle is at the center of the hypotenuse
The circumcenter of an obtuse triangle is outside the triangle
These locational features can be seen by considering the trilinear or barycentric coordinates given above for the circumcenter: all three coordinates are positive for any interior point, at least one coordinate is negative for any exterior point, and one coordinate is zero and two are positive for a non-vertex point on a side of the triangle.
### Angles
Circumcircle Angles 1.svg
Circumcircle Angles 2.svg
The angles which the circumscribed circle forms with the sides of the triangle coincide with angles at which sides meet each other. The side opposite angle α meets the circle twice: once at each end; in each case at angle α (similarly for the other two angles). This is due to the alternate segment theorem, which states that the angle between the tangent and chord equals the angle in the alternate segment.
### Triangle centers on the circumcircle of triangle ABC
In this section, the vertex angles are labeled A, B, C and all coordinates are trilinear coordinates:
• Steiner point = bc / (b2c2) : ca / (c2a2) : ab / (a2b2) = the nonvertex point of intersection of the circumcircle with the Steiner ellipse. (The Steiner ellipse, with center = centroid(ABC), is the ellipse of least area that passes through A, B, and C. An equation for this ellipse is 1/(ax) + 1/(by) + 1/(cz) = 0.)
• Tarry point = sec (A + ω) : sec (B + ω) : sec (C + ω) = antipode of the Steiner point
• Focus of the Kiepert parabola = csc (BC) : csc (CA) : csc (AB).
### Other properties
The diameter of the circumcircle, called the circumdiameter and equal to twice the circumradius, can be computed as the length of any side of the triangle divided by the sine of the opposite angle:
${\displaystyle {\text{diameter}}={\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}.}$
As a consequence of the law of sines, it does not matter which side and opposite angle are taken: the result will be the same.
The diameter of the circumcircle can also be expressed as
{\displaystyle {\begin{aligned}{\text{diameter}}&{}={\frac {abc}{2\cdot {\text{area}}}}={\frac {|AB||BC||CA|}{2|\Delta ABC|}}\\&{}={\frac {abc}{2{\sqrt {s(s-a)(s-b)(s-c)}}}}\\&{}={\frac {2abc}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}}\end{aligned}}}
where a, b, c are the lengths of the sides of the triangle and s = (a + b + c)/2 is the semiperimeter. The expression ${\displaystyle {\sqrt {\scriptstyle {s(s-a)(s-b)(s-c)}}}}$ above is the area of the triangle, by Heron's formula.[3] Trigometric expressions for the diameter of the circumcircle include
${\displaystyle {\text{diameter}}={\sqrt {\frac {2\cdot {\text{area}}}{\sin A\sin B\sin C}}}.}$
The triangle's nine-point circle has half the diameter of the circumcircle.
In any given triangle, the circumcenter is always collinear with the centroid and orthocenter. The line that passes through all of them is known as the Euler line.
The isogonal conjugate of the circumcenter is the orthocenter.
The useful minimum bounding circle of three points is defined either by the circumcircle (where three points are on the minimum bounding circle) or by the two points of the longest side of the triangle (where the two points define a diameter of the circle). It is common to confuse the minimum bounding circle with the circumcircle.
The circumcircle of three collinear points is the line on which the three points lie, often referred to as a circle of infinite radius. Nearly collinear points often lead to numerical instability in computation of the circumcircle.
Circumcircles of triangles have an intimate relationship with the Delaunay triangulation of a set of points.
By Euler's theorem in geometry, the distance between the circumcenter O and the incenter I is
${\displaystyle OI={\sqrt {R(R-2r)}},}$
where r is the incircle radius and R is the circumcircle radius; hence the circumradius is at least twice the inradius (Euler's triangle inequality), with equality only in the equilateral case.
The distance between O and the orthocenter H is
${\displaystyle OH={\sqrt {R^{2}-8R^{2}\cos A\cos B\cos C}}={\sqrt {9R^{2}-(a^{2}+b^{2}+c^{2})}}.}$
For centroid G and nine-point center N we have
{\displaystyle {\begin{aligned}IG&
The product of the incircle radius and the circumcircle radius of a triangle with sides a, b, and c is
${\displaystyle rR={\frac {abc}{2(a+b+c)}}.}$
With circumradius R, sides a, b, c, and medians ma, mb, and mc, we have
{\displaystyle {\begin{aligned}3{\sqrt {3}}R&\geq a+b+c\\9R^{2}&\geq a^{2}+b^{2}+c^{2}\\{\frac {27}{4}}R^{2}&\geq m_{a}^{2}+m_{b}^{2}+m_{c}^{2}.\end{aligned}}}
If median m, altitude h, and internal bisector t all emanate from the same vertex of a triangle with circumradius R, then
${\displaystyle 4R^{2}h^{2}(t^{2}-h^{2})=t^{4}(m^{2}-h^{2}).}$
Carnot's theorem states that the sum of the distances from the circumcenter to the three sides equals the sum of the circumradius and the inradius. Here a segment's length is considered to be negative if and only if the segment lies entirely outside the triangle.
If a triangle has two particular circles as its circumcircle and incircle, there exist an infinite number of other triangles with the same circumcircle and incircle, with any point on the circumcircle as a vertex. (This is the n=3 case of Poncelet's porism). A necessary and sufficient condition for such triangles to exist is the above equality ${\displaystyle OI={\sqrt {R(R-2r)}}.}$
Quadrilaterals that can be circumscribed have particular properties including the fact that opposite angles are supplementary angles (adding up to 180° or π radians).
## Cyclic n-gons
For a cyclic polygon with an odd number of sides, all angles are equal if and only if the polygon is regular. A cyclic polygon with an even number of sides has all angles equal if and only if the alternate sides are equal (that is, sides 1, 3, 5, ... are equal, and sides 2, 4, 6, ... are equal).
A cyclic pentagon with rational sides and area is known as a Robbins pentagon; in all known cases, its diagonals also have rational lengths.
In any cyclic n-gon with even n, the sum of one set of alternate angles (the first, third, fifth, etc.) equals the sum of the other set of alternate angles. This can be proven by induction from the n=4 case, in each case replacing a side with three more sides and noting that these three new sides together with the old side form a quadrilateral which itself has this property; the alternate angles of the latter quadrilateral represent the additions to the alternate angle sums of the previous n-gon.
Let one n-gon be inscribed in a circle, and let another n-gon be tangential to that circle at the vertices of the first n-gon. Then from any point P on the circle, the product of the perpendicular distances from P to the sides of the first n-gon equals the product of the perpendicular distances from P to the sides of the second n-gon.
### Point on the circumcircle
Let a cyclic n-gon have vertices A1 , ..., An on the unit circle. Then for any point M on the minor arc A1An, the distances from M to the vertices satisfy
${\displaystyle {\begin{cases}MA_{1}+MA_{3}+\dots +MA_{n-2}+MA_{n}<{\frac {n}{\sqrt {2}}}\quad {\text{if}}\,n\,{\text{is odd}};\\MA_{1}+MA_{3}+\dots +MA_{n-3}+MA_{n-1}\leq {\frac {n}{\sqrt {2}}}\quad {\text{if}}\,n\,{\text{is even}}.\end{cases}}}$
### Polygon circumscribing constant
A sequence of circumscribed polygons and circles.
Any regular polygon is cyclic. Consider a unit circle, then circumscribe a regular triangle such that each side touches the circle. Circumscribe a circle, then circumscribe a square. Again circumscribe a circle, then circumscribe a regular 5-gon, and so on. The radii of the circumscribed circles converge to the so-called polygon circumscribing constant
${\displaystyle \prod _{n\geq 3}1/\cos \left({\frac {\pi }{n}}\right)=8.7000366\ldots .}$
(sequence A051762 in the OEIS). The reciprocal of this constant is the Kepler–Bouwkamp constant.
|
# What Is 27/72 as a Decimal + Solution With Free Steps
The fraction 27/72 as a decimal is equal to 0.375.
We sometimes represent the division of two numbers in the form of a fraction. If the dividend and divisor are respectively numbers p and q, then p $\boldsymbol\div$ qp/q, where p is the numerator and q is the denominator. Fractions are evaluated the same way as division.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 27/72.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 27
Divisor = 72
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 27 $\div$ 72
This is when we go through the Long Division solution to our problem.
Figure 1
## 27/72 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 27 and 72, we can see how 27 is Smaller than 72, and to solve this division, we require that 27 be Bigger than 72.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 27, which after getting multiplied by 10 becomes 270.
We take this 270 and divide it by 72; this can be done as follows:
270 $\div$ 72 $\approx$ 3
Where:
72 x 3 = 216
This will lead to the generation of a Remainder equal to 270 – 216 = 54. Now this means we have to repeat the process by Converting the 54 into 540 and solving for that:
540 $\div$ 72 $\approx$ 7
Where:
72 x 7 = 504
This, therefore, produces another Remainder which is equal to 540 – 504 = 36. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 360.
360 $\div$ 72 = 5
Where:
72 x 5 = 360
Finally, we have a Quotient generated after combining the three pieces of it as 0.375, with a Remainder equal to 0.
Images/mathematical drawings are created with GeoGebra.
|
# What is 15 Percent of 45? (In-Depth Explanation)
15 percent of 45 equals 6.75. To get this answer, multiply 0.15 by 45.
You may need to know this answer when solving a math problem that multiplies both 15% and 45. Perhaps a product worth 45 dollars, euros, or pounds is advertised as 15% off. Knowing the exact amount discounted from the original price of 45 can help you make a more informed decision on whether or not it is a good deal.
Maybe you’re looking for 15% of 45 dollars, euros, Japanese yen, British pounds, Chinese yuan, pesos, or rupees. Whatever the case is, below, you will find an in-depth explanation that will help you solve this equation.
## What is 15 percent of 45?
15 percent of 45 is 6.75. To figure this out, multiply 0.15 by 45 to get 6.75 as the answer.
Another way to find the answer to this equation includes taking 15/100 and multiplying it by 45/1. When multiplying these two fractions together, you will get a final answer of 6.75.
## How do you find 15 percent of 45?
By multiplying both 0.15 and 45 together, you will find that 6.75 is 15 percent of 45. The 0.15 represents 15% and is the result of taking 15/100 or 15 divided by 100.
The easiest way to solve this equation is to divide the percent by 100 and multiply by the number. So divide 15 by 100 to get 0.15. From there, multiply the percent (now in decimal form) by 45 to get 6.75.
## What is 15% off 45 dollars?
You will pay \$38.25 for an item when you account for a discount of 15 percent off the original price of \$45. You will be receiving a \$6.75 discount.
## What is 15 percent of 45 dollars?
15 percent of 45 dollars is 6.75 dollars. When solving this equation, we multiply 0.15 by 45, the 0.15 standing for 15% and 45 representing 45 dollars.
When referencing the dollar, people will likely be talking about the United States dollar (USD). However, sometimes other currencies are intended instead, like the Canadian dollar (CAD) or the Australian dollar (AUD).
The equation remains the same for calculating 15% of 45 dollars for each of those respective currencies.
## What is 15% off 45 euros?
With a 15 percent discount, you will pay €38.25 for any item with an original price of €45. You will get a discount of €6.75 off.
## What is 15 percent of 45 euros?
15% of 45 euros is 6.75 euros. We use the same formula for calculating 15% of 45 to get our answer of 6.75 euros.
The euro is the currency used by some countries in the European Union, such as France, Germany, and Italy.
## What is 15 percent of 45 Japanese yen?
15% of 45 Japanese yen is 6.75 yen. If you’re trying to solve 15% of 45 Japanese yen, multiply 15% by 45.
When you multiply these two numbers together, you will find 6.75 Japanese yen is your answer.
## What is 15% off 45 pounds?
If you get a 15 percent discount on a £45 item, you will pay £38.25. In total, you will end up receiving a £6.75 discount.
## What is 15 percent of 45 British pounds?
Similar to other currencies, we multiply 15% by 45 to get 6.75 British pounds. In this equation, 0.15, 15/100, or 15% can each represent 15 percent. The 45 in this equation stands for 45 British pounds.
6.75 British pounds will be your answer once you multiply the two numbers together.
## What is 15 percent of 45 Chinese yuan?
15% of 45 Chinese yuan is 6.75 yuan. The same formula that calculated 15% of 45 of the other currencies can calculate 15% of the Chinese yuan.
You divide the percent by 100 and multiply it by the number. For this example, the equation divides 15% by 100 to get 0.15 (15 percent in decimal form). The percent is then multiplied by 45 Chinese yuan resulting in an answer of 6.75 Chinese yuan.
## What is 15 percent of 45 pesos?
6.75 pesos is the equivalent of 15% of 45 pesos. When solving this equation, take the percent divided by 100 and multiply it by the number. In this case, 15% is divided by 100 and multiplied by 45 pesos for an answer of 6.75 pesos.
## What is 15 percent of 45 rupees?
Like with other currencies, use the same equation and multiply 15% by 45 rupees to get an answer of 6.75 rupees. The answer will remain the same even if you write 15 percent as; 15%, 0.15, or 15/100.
After you multiply 15% and 45 rupees together, 6.75 rupees is the final answer to the equation.
## Conclusion
You might need to know the answer to 15% of 45 when operating a business. New businesses get started every day, and people will often need to solve equations involving percentages like this.
Those looking for the answer to 15% of 45 might not even be business owners.
Maybe you are at school or work and need to know the answer to this calculation. Whatever the case is, the answer is 6.75.
If you enjoyed learning about what 15% of 45 is, consider checking out our other articles below!
Related Posts
### Benchmark Bank Near Me: Find Nearby Branch Locations and ATMs
Benchmark Bank is located all throughout the state of Texas. Ever since their founding in 1964, they’ve grown to become …
### \$106 an Hour is How Much a Year? (Before and After Taxes)
So you found out you are going to make \$106 an hour. But \$106 an hour is how much a …
### US Bank Routing Number in New Mexico – 107002312
The US Bank routing number for New Mexico is 107002312. In New Mexico, the ACH routing number is the same, …
### Atlantic Union Bank Near Me: Find Nearby Branch Locations and ATMs
Atlantic Union Bank was initially founded in 1902 and has become a well-known bank in the United States. They have …
### \$73 an Hour is How Much a Year? (Before and After Taxes)
So you found out you are going to make \$73 an hour. But \$73 an hour is how much a …
### \$124 an Hour is How Much a Year? (Before and After Taxes)
So you found out you are going to make \$124 an hour. But \$124 an hour is how much a …
### US Bank Routing Number in Oklahoma – 091000022
The US Bank routing number for Oklahoma is 091000022. In Oklahoma, the ACH routing number is the same, also being …
### \$21 an Hour is How Much a Year? (Before and After Taxes)
So you found out you are going to make \$21 an hour. But \$21 an hour is how much a …
### Fulton Bank Routing Number (Quick & Easy)
There are many circumstances where knowing your Fulton Bank routing number is important. Perhaps you want to make a wire …
### \$135,000 a Year is How Much an Hour? (Before and After Taxes)
So you found out you will make \$135,000 a year from getting that raise or a new job. But \$135,000 …
### Regions Bank Hours: Weekday, Weekend, and Holiday Schedule
Regions Bank has been around for over 50 years and has more than \$156 billion in assets. Since its inception, …
### KeyBank Routing Number in Indiana – 041001039
The KeyBank routing number for Indiana is 041001039. In Indiana, the ACH routing number is the same, also being 041001039. …
|
Elementary Algebra, v. 1.0
by John Redden
8.4 Multiplying and Dividing Radical Expressions
Learning Objectives
3. Rationalize the denominator.
When multiplying radical expressions with the same index, we use the product rule for radicals. If a and b represent positive real numbers,
Example 1: Multiply: $2⋅6$.
Solution: This problem is a product of two square roots. Apply the product rule for radicals and then simplify.
Answer: $23$
Example 2: Multiply: $93⋅63$.
Solution: This problem is a product of cube roots. Apply the product rule for radicals and then simplify.
Answer: $3 23$
Often there will be coefficients in front of the radicals.
Example 3: Multiply: $23⋅52$.
Solution: Using the product rule for radicals and the fact that multiplication is commutative, we can multiply the coefficients and the radicands as follows.
Typically, the first step involving the application of the commutative property is not shown.
Answer: $106$
Example 4: Multiply: $−2 5x3⋅3 25x23$.
Solution:
Answer: $−30x$
Use the distributive property when multiplying rational expressions with more than one term.
Example 5: Multiply: $43(23−36)$.
Solution: Apply the distributive property and multiply each term by $43$.
Answer: $24−362$
Example 6: Multiply: $4x23(2x3−5 4x23)$.
Solution: Apply the distributive property and then simplify the result.
Answer: $2x−10x⋅2x3$
The process for multiplying radical expressions with multiple terms is the same process used when multiplying polynomials. Apply the distributive property, simplify each radical, and then combine like terms.
Example 7: Multiply: $(5+2)(5−4)$.
Solution: Begin by applying the distributive property.
Answer: $−3−25$
Example 8: Multiply: $(3x−y)2$.
Solution:
Answer: $9x−6xy+y$
Try this! Multiply: $(23+52)(3−26)$.
Answer: $6−122+56−203$
Video Solution
The expressions $(a+b)$ and $(a−b)$ are called conjugatesThe factors $(a+b)$ and $(a−b)$ are conjugates.. When multiplying conjugates, the sum of the products of the inner and outer terms results in 0.
Example 9: Multiply: $(2+5)(2−5)$.
Solution: Apply the distributive property and then combine like terms.
It is important to note that when multiplying conjugate radical expressions, we obtain a rational expression. This is true in general and is often used in our study of algebra.
Therefore, for nonnegative real numbers a and b, we have the following property:
Dividing Radical Expressions (Rationalizing the Denominator)
To divide radical expressions with the same index, we use the quotient rule for radicals. If a and b represent nonnegative numbers, where $b≠0$, then we have
Example 10: Divide: $8010$.
Solution: In this case, we can see that 10 and 80 have common factors. If we apply the quotient rule for radicals and write it as a single square root, we will be able to reduce the fractional radicand.
Answer: $22$
Example 11: Divide: $16x5y42xy$.
Solution:
Answer: $2x2y2y$
Example 12: Divide: $54a3b5316a2b23$.
Solution:
Answer: $3b⋅a32$
When the divisor of a radical expression contains a radical, it is a common practice to find an equivalent expression where the denominator is a rational number. Finding such an equivalent expression is called rationalizing the denominatorThe process of determining an equivalent radical expression with a rational denominator..
To do this, multiply the fraction by a special form of 1 so that the radicand in the denominator can be written with a power that matches the index. After doing this, simplify and eliminate the radical in the denominator. For example,
Remember, to obtain an equivalent expression, you must multiply the numerator and denominator by the exact same nonzero factor.
Example 13: Rationalize the denominator: $32$.
Solution: The goal is to find an equivalent expression without a radical in the denominator. In this example, multiply by 1 in the form $22$.
Answer: $62$
Example 14: Rationalize the denominator: $123x$.
Solution: The radicand in the denominator determines the factors that you need to use to rationalize it. In this example, multiply by 1 in the form $3x3x$.
Answer: $3x6x$
Typically, we will find the need to reduce, or cancel, after rationalizing the denominator.
Example 15: Rationalize the denominator: $525ab$.
Solution: In this example, we will multiply by 1 in the form $5ab5ab$.
Notice that a and b do not cancel in this example. Do not cancel factors inside a radical with those that are outside.
Answer: $10abab$
Try this! Rationalize the denominator: $4a3b$.
Answer: $23ab3b$
Video Solution
Up to this point, we have seen that multiplying a numerator and a denominator by a square root with the exact same radicand results in a rational denominator. In general, this is true only when the denominator contains a square root. However, this is not the case for a cube root. For example,
Note that multiplying by the same factor in the denominator does not rationalize it. In this case, if we multiply by 1 in the form of $x23x23$, then we can write the radicand in the denominator as a power of 3. Simplifying the result then yields a rationalized denominator. For example,
Therefore, to rationalize the denominator of radical expressions with one radical term in the denominator, begin by factoring the radicand of the denominator. The factors of this radicand and the index determine what we should multiply by. Multiply numerator and denominator by the nth root of factors that produce nth powers of all the factors in the radicand of the denominator.
Example 16: Rationalize the denominator: $1253$.
Solution: The radical in the denominator is equivalent to $523$. To rationalize the denominator, it should be $533$. To obtain this, we need one more factor of 5. Therefore, multiply by 1 in the form of $5353$.
Answer: $535$
Example 17: Rationalize the denominator: $27a2b23$.
Solution: In this example, we will multiply by 1 in the form $22b322b3$.
Answer: $34ab32b$
Example 18: Rationalize the denominator: $1 4x35$.
Solution: In this example, we will multiply by 1 in the form $23x2523x25$.
Answer: $8x252x$
When two terms involving square roots appear in the denominator, we can rationalize it using a very special technique. This technique involves multiplying the numerator and the denominator of the fraction by the conjugate of the denominator. Recall that multiplying a radical expression by its conjugate produces a rational number.
Example 19: Rationalize the denominator: $13−2$.
Solution: In this example, the conjugate of the denominator is $3+2$. Therefore, multiply by 1 in the form $(3+2)(3+2)$.
Answer: $3+2$
Notice that the terms involving the square root in the denominator are eliminated by multiplying by the conjugate. We can use the property $(a+b)(a−b)=a−b$ to expedite the process of multiplying the expressions in the denominator.
Example 20: Rationalize the denominator: $2−62+6$.
Solution: Multiply by 1 in the form $2−62−6$.
Answer: $−2+3$
Example 21: Rationalize the denominator: $x+yx−y$.
Solution: In this example, we will multiply by 1 in the form $x−yx−y$.
Answer: $x−2xy+yx−y$
Try this! Rationalize the denominator: $35+525−3$.
Answer: $195+4511$
Key Takeaways
• To multiply two single-term radical expressions, multiply the coefficients and multiply the radicands. If possible, simplify the result.
• Apply the distributive property when multiplying radical expressions with multiple terms. Then simplify and combine all like radicals.
• Multiplying a two-term radical expression involving square roots by its conjugate results in a rational expression.
• It is common practice to write radical expressions without radicals in the denominator. The process of finding such an equivalent expression is called rationalizing the denominator.
• If an expression has one term in the denominator involving a radical, then rationalize it by multiplying numerator and denominator by the nth root of factors of the radicand so that their powers equal the index.
• If a radical expression has two terms in the denominator involving square roots, then rationalize it by multiplying the numerator and denominator by its conjugate.
Topic Exercises
Multiply. (Assume all variables are nonnegative.)
1. $3⋅5$
2. $7⋅3$
3. $2⋅6$
4. $5⋅15$
5. $7⋅7$
6. $12⋅12$
7. $25⋅710$
8. $315⋅26$
9. $(25)2$
10. $(62)2$
11. $2x⋅2x$
12. $5y⋅5y$
13. $3a⋅12$
14. $3a⋅2a$
15. $42x⋅36x$
16. $510y⋅22y$
17. $53⋅253$
18. $43⋅23$
19. $43⋅103$
20. $183⋅63$
21. $(5 93)(2 63)$
22. $(2 43)(3 43)$
23. $(2 23)3$
24. $(3 43)3$
25. $3a23⋅9a3$
26. $7b3⋅49b23$
27. $6x23⋅4x23$
28. $12y3⋅9y23$
29. $20x2y3⋅10x2y23$
30. $63xy3⋅12x4y23$
31. $5(3−5)$
32. $2(3−2)$
33. $37(27−3)$
34. $25(6−310)$
35. $6(3−2)$
36. $15(5+3)$
37. $x(x+xy)$
38. $y(xy+y)$
39. $2ab(14a−210b)$
40. $6ab(52a−3b)$
41. $(2−5)(3+7)$
42. $(3+2)(5−7)$
43. $(23−4)(36+1)$
44. $(5−26)(7−23)$
45. $(5−3)2$
46. $(7−2)2$
47. $(23+2)(23−2)$
48. $(2+37)(2−37)$
49. $(a−2b)2$
50. $(ab+1)2$
51. What are the perimeter and area of a rectangle with length of $53$ centimeters and width of $32$ centimeters?
52. What are the perimeter and area of a rectangle with length of $26$ centimeters and width of $3$ centimeters?
53. If the base of a triangle measures $62$ meters and the height measures $32$ meters, then what is the area?
54. If the base of a triangle measures $63$ meters and the height measures $36$ meters, then what is the area?
Divide.
55. $753$
56. $36010$
57. $7275$
58. $9098$
59. $90x52x$
60. $96y33y$
61. $162x7y52xy$
62. $363x4y93xy$
63. $16a5b232a2b23$
64. $192a2b732a2b23$
Rationalize the denominator.
65. $15$
66. $16$
67. $23$
68. $37$
69. $5210$
70. $356$
71. $3−53$
72. $6−22$
73. $17x$
74. $13y$
75. $a5ab$
76. $3b223ab$
77. $2363$
78. $1473$
79. $14x3$
80. $13y23$
81. $9x⋅239xy23$
82. $5y2⋅x35x2y3$
83. $3a2 3a2b23$
84. $25n3 25m2n3$
85. $327x2y5$
86. $216xy25$
87. $ab9a3b5$
88. $abcab2c35$
89. $310−3$
90. $26−2$
91. $15+3$
92. $17−2$
93. $33+6$
94. $55+15$
95. $105−35$
96. $−224−32$
97. $3+53−5$
98. $10−210+2$
99. $23−3243+2$
100. $65+225−2$
101. $x+yx−y$
102. $x−yx+y$
103. $a−ba+b$
104. $ab+2ab−2$
105. $x5−2x$
106. $1x−y$
Part C: Discussion
107. Research and discuss some of the reasons why it is a common practice to rationalize the denominator.
108. Explain in your own words how to rationalize the denominator.
1: $15$
3: $23$
5: 7
7: $702$
9: 20
11: $2x$
13: $6a$
15: $24x3$
17: 5
19: $2 53$
21: $30 23$
23: 16
25: $3a$
27: $2x⋅3x3$
29: $2xy⋅25x3$
31: $35−5$
33: $42−321$
35: $32−23$
37: $x+xy$
39: $2a7b−4b5a$
41: $6+14−15−35$
43: $182+23−126−4$
45: $8−215$
47: 10
49: $a−22ab+2b$
51: Perimeter: $(103+62)$ centimeters; area: $156$ square centimeters
53: 18 square meters
55: 5
57: $265$
59: $3x25$
61: $9x3y2$
63: $2a$
65: $55$
67: $63$
69: $104$
71: $3−153$
73: $7x7x$
75: $ab5b$
77: $633$
79: $2x232x$
81: $3 6x2y3y$
83: $9ab32b$
85: $9x3y45xy$
87: $27a2b453$
89: $310+9$
91: $5−32$
93: $−1+2$
95: $−5−352$
97: $−4−15$
99: $15−7623$
101: $x2+2xy+yx2−y$
103: $a−2ab+ba−b$
105: $5x+2x25−4x$
|
# Instructions to the Square Root Board
The square root board shows in a simple way the essence of squaring a number and determining its square root.
For both operations, children should already have multiplication experience.
Taking square roots becomes very easy with the board, because one can see that is it the opposite of squaring. In order to square a number, it must be multiplied with itself: 2×2=4
One can see that this operation always looks like a square on the board; thus it is called “squaring”.
To find out the root of 4, we count the beads on one side. There are 2 beads, which means the root of 4 is 2:
This works easily with all numbers from 1 to 9. Instead of beads, we can also use pebbles, blocks or buttons:
With numbers greater than 9, we need beads in various colors. Maria Montessori developed the following simple system:
Green = 1
Blue = 10
Red =100
Light green = 1000
Light blue = 10,000
Light red = 100,000
We see the board just like an x-y graph system. Each bead on the x-horizontal line is multiplied with each bead on the y-vertical line. The result is placed in the right color at the point where the two lines meet. Here you can see what 12×12 looks like:
And these photos show the process step by step: 12 = 2 green beads for the ones and 1 blue bead for the ten:
Then we place the same beads on the vertical line:
Then we can start multiplying: 1×1 =1 :
10×1=10 and 1×10= 10 :
10×10=100 and that’s the complete square:
To find out the result, we add the value of all beads: 4×1 plus 4×10 plus 1×100 equals 144. Here:
Even with large numbers, this technique remains simple. One can observe a specific variety of color patterns which increase with regularity. For example, here is 122 squared:
After all beads are placed, we can start adding: (1 x 10,000) + (4 x 1000) + (8 x 100) + (8 x 10) + (4 x 1) = 14,884
The root of 14,884 is one side of the square, 122:
After we have learned to use the board to SEE the way squares grow, we can now learn the opposite process – finding the root of a number. First, we lay the square pattern using the appropriate amount of beads and then count one side.
The square root board helps to understand what the concept of square root actually means. It takes away the fear from this process usually only taught on paper.
I remember that I could not understand square roots when I went to school and I am happy to have found the square root board!
Try it!
Carmen
## 3 thoughts on “Instructions to the Square Root Board”
1. Pame G.
Thanks for your post but I do not understand… How multiply X and Y . Dont obtain same results.
• HI, Thanks for your comment, as I look through this post that I wrote a long time ago, I discovered I made a mistake, that’s probably where you got confused. 4 x 1 is of course 4 – not 1!!!
So, I will correct this sentence in the blog “After all beads are placed, we can start adding: (1 x 10,000) + (4 x 1000) + (8 x 100) + (8 x 10) + (4 x 1) = 14,881” TO “After all beads are placed, we can start adding: (1 x 10,000) + (4 x 1000) + (8 x 100) + (8 x 10) + (4 x 1) = 14,884” Please also know this is my OLD BLOG that I’m not usually updating anymore, please go to my NEW BLOG page https://newlearningculture.com/instructions-to-the-square-root-board/
2. Lauren
Thank you for sharing – I haven’t had a chance to practice this since my training, so this is an excellent post for jogging my memory!
|
# How do you integrate int xsinx by integration by parts method?
Dec 11, 2016
$\int x \sin x \mathrm{dx} = \sin x - x \cos x + c$
#### Explanation:
If you are studying maths, then you should learn the formula for Integration By Parts (IBP), and practice how to use it:
$\int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$, or less formally $\int u \mathrm{dv} = u v - \int v \mathrm{du}$
I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.
Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).
So for the integrand $x \sin x$, hopefully you can see that $x$ simplifies when differentiated and $\sin x$ effectively remains unchanged ($\cos x$ is still a trig function) under differentiation or integration.
Let $\left\{\begin{matrix}u = x & \implies & \frac{\mathrm{du}}{\mathrm{dx}} = 1 \\ \frac{\mathrm{dv}}{\mathrm{dx}} = \sin x & \implies & v = - \cos x\end{matrix}\right.$
Then plugging into the IBP formula gives us:
$\int \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \mathrm{dx} = \left(u\right) \left(v\right) - \int \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \mathrm{dx}$
$\therefore \int x \sin x \mathrm{dx} = \left(x\right) \left(- \cos x\right) - \int \left(- \cos x\right) \left(1\right) \mathrm{dx}$
$\text{ } = - x \cos x + \int \cos x \mathrm{dx}$
$\text{ } = - x \cos x + \sin x + c$
|
### NCERT Solutions For Class 10 Maths Chapter – 2 Exercise – 2.3
NCERT Solutions For Class 10 Math Chapter – 2 Polynomial Exercise – 2.3
Q1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
• P(x) = x3 – 3x2 + 5x – 3 g(x) = x2 – x
• P(x) = x4– 3x2 + 4x + 5 g(x) = x2 + 1 – x
• P(x) = x4– 5x + 6 g(x) = 2 – x2
Solution:-
• Here p(x) = x3– 3x2 + 5x – 3 and g(x) = x2 – 2
Dividing p(x) by g(x) =
Quotient = x – 3 Remainder = 7x – 9
• P(x) = x4
3x2 + 4x + 5 and g(x) = x2 + 1 – x
Dividing p(x) by g(x)
Quotient = x2 + x – 3 Remainder = 8
• P(x) = x4– 5x + 6 g(x) = 2 – x2
Quotient = -x2 – 2
Remainder = -5x + 10.
Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
• T2– 3 , 2t4 + 3t3 – 2t2 – 9t – 12
• X2+ 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 12
• X2+ 3x + 1, x5 – 4x3 + x2 + 3x + 1
Solution:-
• First polynomial = t2– 3
Second polynomial = 2t4 + 3t3 – 2t2 – 9t – 12
Dividing second polynomial by first polynomial.
Remainder is zero.
First polynomial is a factor of second polynomial.
• First polynomial = x2+ 3x + 1
Second polynomial = 3x4 + 5x3 – 7x2 + 2x + 2
Dividing second polynomial by first polynomial =
Remainder is zero.
First polynomial is a factor of second polynomial.
• First polynomial = x3– 3x + 1
Second polynomial = x5 – 4x3 + x2 + 3x + 1.
Remainder ≠ 0
First polynomial is not a factor of second polynomial.
Q3. Obtain all zeroes of 3x4 + 6x3 – 2x2 – 10x – 5 , If two of its zeroes are and 5/3 and –5/3
Solution:-
Q4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x) the quotient and remainder were x – 2 and -2x + 4 respectively Find g(x).
Solution:-
Q5. Give examples of polynomial p(x) , q(x) and r(x) which satisfy the division algorithm and:
• deg p(x) = deg q(x)
• Deg q(x) = deg r(x)
• Deg r(x) = 0
Solution:-
• P(x) = 2x2+ 2x + 8
Q(x) = x2 + x + 4
G(x) = 2 and r(x) = 0
• P(x) = x3– x2 + 2x + 3
Q(x) = x + 1
G(x) = x2 – 1 and r(x) = 2x +2
• P(x) = x3– x2 + 2x + 3
G(x) = x2 + 2
Q(x) = x – 1 and r(x) = 5
NCERT Solutions For Class 10 Math Chapter – 2 Polynomial Exercise – 2.3 Q1. Divide the polynomial p(x) by the…
|
# Question Video: Finding the Area of a Rectangle Mathematics • 6th Grade
David parked his rectangular car in an empty rectangular parking lot. The dimensions of the parking lot are 87 feet by 83 feet and the dimensions of the car are 18 feet by 8 feet. What area of the parking lot is still empty?
02:14
### Video Transcript
David parked his rectangular car in an empty rectangular parking lot. The dimensions of the parking lot are 87 feet by 83 feet, and the dimensions of the car are 18 feet by eight feet. What area of the parking lot is still empty?
In order to calculate the area of the parking lot that is empty, we need to subtract the area of the car from the area of the whole parking lot. The area of any rectangle can be calculated by multiplying the length by the width. The area of the parking lot is therefore equal to 87 feet multiplied by 83 feet. One way of calculating this would be using the grids or box method. 80 multiplied by 80 is 6400. 80 multiplied by seven is 560. 80 multiplied by three is 240. And finally, seven multiplied by three is 21. Adding these four numbers gives us 7221. As the dimensions are in feet, the area of the parking lot is 7221 square feet.
We can calculate the area of the car by multiplying 18 by eight. This is equal to 144. 10 multiplied by eight is 80, and eight multiplied by eight is 64. The area taken up by the car is 144 square feet. Subtracting this from 7221 will give us the empty area. This is equal to 7077. So the area of the parking lot that is still empty is 7077 square feet. This is the area in pink on the diagram.
|
## TRIGONOMETRY
Home Tutorial Links My Tutorials Sample Materials Freebies
# TRIGONOMETRY
Cosine, Sine and Tangent are three of a series of mathematical functions used in the study of right angled triangles (any triangle where one angle is at 90 degrees (see figure 1). This is part of the branch of maths known as Trigonometry - the study of all triangles.
These three functions represent the ratio between two of the sides of a right angled triangle for a particular angle. For any particular angle, this ratio will always be the same, regardless of the actual size of the triangle. If you know the angle of one other corner of the triangle, and the length of any one side of the triangle, you can calculate the length of the other two sides. There are several different ways of measuring angles, of which degrees are the most familiar, and will be used here. Poser uses the Radian, which is much more elegant mathematically. The basic principles of trigonometry are the same regardless of which method you use to measure angles.
Fig 1: The Parts of the Triangle
### The Parts of the Triangle
To understand what's going on in these functions, we need to have clear names for each part of the triangle. Figure 1 and the list below shows the four most important terms used here.
b = the angle
h = hypotenuse: the long side, opposite the right angle
o = opposite: the side opposite to the angle 'b'
a = adjacent: the short side next to the angle 'b'
### Sine, Cosine and Tangent
For any given angle of b, we can calculate the sine, cosine and tangent by measuring the sides of the triangle and performing the following calculations. For our example, we will use a triangle where a = 1, o = 2, h = 2.236 and b = 63.434 degrees.
sin (b) = o / h
sin (63.434) = 2 / 2.236 = 0.894
cos (b) = a / h
cos (63.434) = 1 / 2.236 = 0.447
tan (b) = o / a
tan (63.434) = 2 / 1 = 2
It doesn't matter how big the triangle is - as long as the angle b stays the same then the ratio between the sides of the triangle will remain the same, and the results of the sine, cosine and tangent calculations will remain the same (for example if a=1.5, then o=3, resulting in tan(63.434)= 3 / 1.5 = 2).
Although there are mathematical ways to calculate the values of sine, cosine and tangent just from the angle, they are complex and time consuming and we don't need to use them. Poser provides us with maths functions that do the work for us (see below for how the trig. functions were used in the past).
### Derived Functions
Now we have these three functions, we need to modify them so that we can use them when we know 'b' and one other value. We can do that using some simple mathematical tricks. When we have a formula like sin (b) = o / h we can perform maths on the formula as long as we do the same thing to both sides of the formula.
First of all, we will multiply both sides by h. That will turn sin (b) into sin (b) * h. On the other side, o / h is turned into just o (if you take any number, divide it by another number and then multiply it by the same number you will always return to the original number). This gives us a new formula: sin (b) * h = o. Finally, swap the sides around, and we get o = sin (b) * h. If we know the angle and the hypotenuse we can use this new formula to work out the opposite.
We can go one step further. o = sin (b) * h can also be written as o = h * sin (b). If we divide both sides of this formula by sin (b) we end up with o / sin (b) = h. Swap that around and we get h = o / sin (b).
Table 1 shows all six formula that can be created in this way.
#### TABLE 1: Sine, Cosine and Tangent formula.
sin (b) = o / h o = sin (b) * h h = o / sin (b) cos (b) = a / h a = cos (b) * h h = a / cos (b) tan (b) = o / a o = tan (b) * a a = o / tan (b)
Table 2 shows which formula to use in each possible set of circumstances.
#### TABLE 2: Which formula to use when
Have a Have o Have h Need a - o / tan (b) cos (b) * h Need o tan (b) * a - sin (b) * h Need h h = a / cos (b) o / sin (b) -
### Using Sine, Cosine or Tangent
The standard example used to explain when you might want to use trigonometry is the measurement of heights. Let us say that you need to know the height of the top of your house. Actually measuring up the side of the building would be awkward to say the least. Instead, we will use our trigonometry.
First, pick a spot a little way away from the building. Measure the distance from that spot to the base of the house wall. This will be our 'a' or adjacent. In this example, we ended up 10 meters away from the house.
Second, from your spot measure the angle between the ground and the highest point on the building. This becomes our 'b' or angle. In this example, the angle from our spot 10 meters away from the house to tip of the roof was 30 degrees.
We now have a and b, and we need to calculate 'o', the opposite. If you look at the list of functions we have above, you will see that o = tan (b) * a.
That means that 0 = tan (30) * 10.
Tan (30) is equal to 0.577. Therefore our house is 0.577 * 10, or 5.77 meters high.
### Calculating Sine, Cosine or Tangent.
Prior to the advent of the scientific calculator and the computer, if you needed to find out a value for sine, cosine or tangent, you would use a lookup table. This was simply a pre-calculated list of the values of the trigonometry functions for a particular set of angles. If you needed an angle that isn't on the list, you could estimate it from the list. On our sample table below, we have values for 1 and 2 degrees. If we wanted sine(1.5), first we would look at the values for 1 and 2 and calculate the gap between them (0.0349-0.0175= 0.0174). Next, we would look at the fraction of our angle (.5), and multiply our 0.0175 by 0.5, to get 0.0087. Finally, we would add that to the value for 1 degree, reaching an approximation of 0.0175+0.0087= 0.0262. This is actually accurate to the four decimal places we see - the windows Calculator returns sine(1.5)= 0.026176948307873152610611685554113 - if we round this up to 4 decimal places, then the final result is 0.262. If you were working in a field that needed more accuracy that this (aircraft design for instance), then you'd buy a more detailed set of lookup table.
Radians Degrees Sine Cosine Tangent .0000 0 .0000 1.000 .0000 .0175 1 .0175 .9998 .0175 .0349 2 .0349 .9994 .0349
Even many modern computers use lookup tables, combined with more accurate formulas for filling the gaps, to calculate sine, cosine and tangent.
|
Instasolv
IIT-JEE NEET CBSE NCERT Q&A
4.5/5
# RS Aggarwal Class 8 Chapter 7 Solutions (Factorisation)
RS Aggarwal Class 8 Maths Solutions for Chapter 7 ‘Factorization’ will make you understand how factorization of algebraic expressions is done. In this chapter of RS Aggarwal Class 8 Maths Solutions, you will learn what are factors, what is factorization, how to find common factors, factorization by regrouping terms, factorization by using identities, how factoring can be used to divide algebraic expressions, and how to find factors of a perfect square.
There are 5 exercises in this chapter with around 157 questions. This makes it much easier for you to practice all the important concepts of the chapter and prepare yourself for the exams. RS Aggarwal Class 8 Chapter 7 exercises are totally based on the latest CBSE exam pattern. The questions present in the exercises include easy to advanced level problems that can also be useful for preparing for olympiads and competitive tests.
Factorization is an important concept of maths which helps you in solving number-related problems, comparison as well as can be used in time and money related problems. Our subject matter experts provide step-by-step solutions to all the exercise questions of RS Aggarwal Class 8 Maths Chapter 7 that will help you understand the concepts easily. All the solutions are as per the current CBSE syllabus for Class 8 Maths.
## Important Topics for RS Aggarwal Class 8 Maths Solutions Chapter 7: Factorization
Factors and Factorization
A number or algebraic expression can be written as a product of other numbers or algebraic expressions. These numbers or algebraic expressions respectively are called factors. For example, 30 = 10 x 3. Here 10 and 3 are factors of 30. However, there can be many other factors of 30.
Similarly, 3ab is an algebraic expression and 3ab = 3 x a x b. Here, 3, a, b are factors of 3ab. These are called irreducible factors of this algebraic expression as they cannot be factored further.
We can easily find factors of algebraic expressions that are written in product form. For example, 6x2, 7yx2, 2x(1+y) all are in product form and we can easily state their factors. But if an algebraic expression is like a2 + b + 3 then we need a certain methodology to find its factors.
Factorization Using Common Terms
Here, we take common factors so as to change the algebraic expression into product form and then express it as a product of irreducible factors. For example, suppose we have to find factors of 3a + 9. Then, 3a = 3 x a and 9 = 3 x 3
3a + 9 = (3 x a) + (3 x 3), where 3 is common
3 (a + 3)
3 and (a+3) are factors of 3a + 9.
Factors by Regrouping The Terms
Sometimes, we also have to regroup the terms of algebraic expression in order to find its factors. Let us take an example, 6ab – 4b + 6 – 9a.
6ab – 4b can be written as 2b (3a – 2) and,
6 – 9a can be written as 3 (2 – 3a) or -3 (3a – 2)
Now we have (3a – 2) common in both the terms. So, we can write
6ab – 4b + 6 – 9a = 2b (3a – 2) -3 (3a – 2)
(3a – 2) (2b – 3)
(3a – 2) and (2b – 3) are factors of 6ab – 4b + 6 – 9a
Some Important Identities
There can be some questions where we have to apply an identity in order to find the factors of an algebraic expression. These identities are:
• (a + b)2 = a2 + 2ab + b2
• (a – b)2 = a2 – 2ab + b2
• (a + b) (a – b) = a2 – b2
Let us take an example to find factors of an algebraic expression ‘ n2 + 10n + 25’ using identities.
Here, n2 + 10n + 25 can be written as n2 + 2 x 5 x n + 52 (this is similar of the form a2 + 2ab + b2
So, n2 + 10m + 25 = n2 + 2 x 5 x m + 52 = (n + 5)2
This means, (n + 5) is a factor of n2 + 10n + 25.
### Exercise Discussion for RS Aggarwal Class 8 Maths Solutions Chapter 7: Factorization
1. There are 5 exercises in RS Aggarwal Class 8 Maths Solutions for Chapter 7 ‘Factorization’. Exercise 7A includes 28 questions which are based on finding factors of algebraic expressions using common factors. All the questions are of a similar type so it is easy to practice.
2. Exercise 7B includes 28 questions where you have to find factors of the given algebraic expressions using identities.
3. Exercise 7C has 19 questions which are again based on finding the factors of given algebraic expressions using identities.
4. Exercise 7D has 42 questions based on the concept of factorization by regrouping the terms of an algebraic expression.
5. Exercise 7E is a miscellaneous exercise with 20 questions which are based on all the concepts discussed in the chapter. So you have to identify which method of factorization should be applied in which question.
### Benefits of RS Aggarwal Class 8 Maths Solutions Chapter 7: Factorization by Instasolv
• RS Aggarwal Solutions for Class 8 Maths Chapter 7 provided by us cover all the easy as well as tough problems given in the exercises.
• These solutions are the best study material to understand the concepts of factorization and practice them well.
• You would not face any difficulty in solving the questions related to factorization of algebraic expressions using our RS Aggarwal Chapter 7 solutions.
• These solutions will enhance your confidence in solving problems related to this chapter so that you can fetch more marks in your exams.
More Chapters from Class 8
|
# 2. Use the following figure to answer the questions.
One recent schools question and answer asked students to express what they agree to is the main important thing for a student to do if they wanted to gain success. Of the many reactions, one that that stood out was practice. Successful people commonly not born successful; they become successful by just hard work and commitment. This is how you can obtain your goals. followed below are one of the answer and question example that you will be able to utilize to practice and supercharge your understanding and also give you insights that will assist you to sustain your study in school.
## Question:
2. Use the following figure to answer the questions.Anthony claims that Triangle KLM is an isosceles triangle.
(a) Find the slopes of the two lines.
LK = -7,8
LM = -7,1
(b) Find the length of the two sides.
LK =
LM =
—- Slope of LK
—- Slope of LM
—- Length of LK
— Length of LM
Anthony is incorrect
Step-by-step explanation:
See attachment for complete question.
From the attachment, we have:
Solving (a): Slope of LK
Slope (m) is calculated as thus:
Where
—– K
—— L
Slope of LM
—- L
—— M
Solving (b): Length of LK and LM
Here, we have to calculate the length using distance formula;
Distance (D) is calculated as thus:
For LK
For LM
The above findings is not enough to conclude if Anthony is correct or incorrect.
We need to calculate the distance of KM using the same formula;
READ MORE What agency is directly related to supervising and managing campgrounds?
D is calculated as thus:
Base on this, we can conclude that Anthony’s claim is incorrect.
The triangle is not an isosceles triangle because no two sides are equal
They would possibly hopefully help the student take care of the question by implementing the questions and answer examples. You may then have a discussion with your classmate and continue the school learning by studying the question altogether.
|
# Pair of Angles
## Pair of Angles
There are some pair of angles having some specific properties. Adjacent angles, linear pair of angles, right pair of angles, complementary angles, supplementary angles and vertically opposite angles are some special pair of angles. let’s learn them.
********************
10 Math Problems officially announces the release of Quick Math Solver and 10 Math ProblemsApps on Google Play Store for students around the world.
********************
********************
Two angles are said to be adjacent angles if they have a common arm and common vertex.
In the given figure, AOC and BOC are adjacent angles as they have a common arm OC and the common vertex O.
Adjacent angles never overlap with each other.
### Linear Pair of Angles
If the sum of a pair of adjacent angles is 180°, the pair is called a linear pair of angles.
In the given figure, AOC + COB = 180°. So AOC and COB are linear pair of angles.
### Right Pair of Angles
If the sum of a pair of adjacent angles is 90°, the pair is called a right pair of angles.
In the given figure, AOC + COB = 90°. So AOC and COB are right pair of angles.
### Complementary Angles
Two angles are said to be complementary angles of each other if their sum is 90°.
In the given figure, ABC = 35° and DBC = 55° and ABC + DBC = 35° + 55° = 90°. ABC and DBC are complementary angles.
Similarly, PQR = 70° and STU = 20° and PQR + STU = 70° + 20° = 90°. PQR and STU are complementary angles.
Each of the angle is complement of the other.
### Supplementary Angles
A pair of angles whose sum is 180° are called supplementary angles.
In the given figure, AOC = 105° and BOC = 75° and AOC + BOC = 105° + 75° = 180°. AOC and BOC are supplementary angles.
Similarly, ABC = 35° and DEF = 145° and ABC + DEF = 35° + 145° = 180°. ABC and DEF are supplementary angles.
Each of the angle is supplement of the other.
### Vertically Opposite Angles [V.O.A.]
When two straight line segment intersect, the angles formed opposite to each other are called vertically opposite angles.
In the given figure, PON and QOM, POM and NOQ are vertically opposite angles.
Similarly, TOU and SOV, UOS and VOT are also vertically opposite angles.
### Workout Examples
Example 1: Find the value of x in the given figure.
Solution: From the figure,
x + 75° = 180° -----------------> Linear pair of angles.
or, x = 180° – 75°
or, x = 105°
Example 2: Find the value of x from the given figure.
Solution: From the figure,
x + x + 40° = 90° -----------------> Right pair of angles.
or, 2x = 90° – 40°
or, 2x = 50°
or, x = 50°/2
or, x = 25°
Example 3: Find the values of a, b and c in the given figure.
Solution: From the figure,
a + 40° = 180° -----------------> Linear pair of angles.
or, a = 180° – 40°
or, a = 140°
b = 40° ------------------> Vertically opposite angles.
c = a ------------------> Vertically opposite angles.
= 140°
a = 140°, b = 40° and c = 140°
Example 4: Find the values of x, a and b in the given figure.
Solution: From the figure,
2x + 3x = 180° -----------------> Linear pair of angles.
or, 5x = 180°
or, x = 180°/5
or, x = 36°
a = 3x ------------------> Vertically opposite angles.
= 3×36°
= 108°
b = 2x ------------------> Vertically opposite angles.
= 2×36°
= 72°
x = 36°, a = 108° and b = 72°
Example 5: Find the values of x, a and b in the given figure.
Solution: From the figure,
2x – 30° = x + 20° -----------------> Vertically opposite angles.
or, 2x – x = 20° + 30°
or, x = 50°
a + x + 20° = 180° ------------------> Linear pair of angles.
or, a + 50° + 20° = 180°
or, a + 70° = 180°
or, a = 180° – 70°
or, a = 110°
b = a ------------------> Vertically opposite angles.
= 110°
x = 50°, a = 110° and b = 110°
You can comment your questions or problems regarding the pair of angles here.
|
# Joint probability
CFA level I / Quantitative Methods: Basic Concepts / Probability Concepts / Joint probability
The joint probability of two events is the probability that the two events occur together. It is denoted by P(AB) for event A and B. The joint probability of two events is equal to zero when the events are mutually exclusive because the two events cannot occur together.
P(AB) = P(A|B)*P(B) = P(B|A)*(P(A)
Example 1: Using multiplication rule to calculate joint probability The probability of an increase in stock price is 40 percent if the interest rate rises by more than one percent. The probability of an increase in the interest rate by more than one percent is 30 percent. What is the joint probability that both events (increase in stock price and increase in interest rate by more than one percent) occur together? Solution: Let us first name the events. Event A = Probability of increase in stock price Event B = Probability of a rise in interest by more than one percent We are given: P(A|B) = 0.40 and P(B) = 0.30 We are asked to calculate P(AB) Applying multiplication rule, P(AB) = P(A|B)*P(B) = 0.40*0.30 = 0.12. So, the probability that occurrence of both increase in stock price and an increase in interest rate by more than one percent is 0.12.
The independent events are the events that are not dependent on the occurrence of the other event.
For independent events A and B: P(A|B) = P(A)
P(B|A) = P(B)
Therefore, P(AB) = P(A|B)*P(B)= P(A)*P(B)
The joint probability of independent events is equal to the product of the unconditional probabilities of those events.
For more than two independent events, P(ABCD) = P(A)*P(B)*P(C)*P(D)
The probability that at least one of two events A and B will occur is given by the additional rule as discussed previously and is given by the following formula.
P(A or B) = P(A) + P(B) - P(AB) = P(A) + P(B) - P(A|B)*P(B)
CFA Institute does not endorse, promote or warrant the accuracy or quality of products and services offered by Konvexity. CFA® and Chartered Financial Analyst® are registered trademarks owned by CFA Institute.
|
# 9.06 Coordinate methods with quadrilaterals
Lesson
Many properties of quadrilaterals can either be verified or proved using coordinate geometry.
There is a range of established formulas that become useful in this endeavor. You may wish to go back to a previous lesson to review each one.
• The distance formula given by $d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$d=(x2x1)2+(y2y1)2
• The slope formula $m=\frac{y_2-y_1}{x_2-x_1}$m=y2y1x2x1
• The perpendicular property $m_1*m_2=-1$m1*m2=1
• The parallel property $m_1=m_2$m1=m2
• The midpoint formula given by $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$(x1+x22,y1+y22)
#### Worked examples
##### Question 1
Show that the quadrilateral with vertices given by $P\left(2,3\right),Q\left(3,6\right),R\left(6,8\right),S\left(5,5\right)$P(2,3),Q(3,6),R(6,8),S(5,5) is a parallelogram.
A parallelogram is a quadrilateral with both pairs of opposite sides parallel.
Think: If opposite sides are parallel then they must have the same slope. We need $\overline{PQ}\parallel\overline{RS}$PQRS and $\overline{QR}\parallel\overline{PS}$QRPS
Do: We can check the slopes of $\overline{PQ}$PQ and $\overline{RS}$RS and the slopes of the line segments $\overline{QR}$QR and $\overline{PS}$PS.
$m_{\overline{PQ}}=\frac{6-3}{3-2}=3$mPQ=6332=3
$m_{\overline{RS}}=\frac{5-8}{5-6}=3$mRS=5856=3
$\overline{PQ}\parallel\overline{RS}$PQRS
$m_{\overline{QR}}=\frac{8-6}{6-3}=\frac{2}{3}$mQR=8663=23
$m_{\overline{PS}}=\frac{5-3}{5-2}=\frac{2}{3}$mPS=5352=23
$\overline{QR}\parallel\overline{PS}$QRPS
Reflect: Both pairs of opposite sides are parallel. Hence, the quadrilateral is a parallelogram.
##### Question 2
Prove that the quadrilateral with vertices $A\left(4,9\right),B\left(5,13\right),C\left(9,14\right),D\left(8,10\right)$A(4,9),B(5,13),C(9,14),D(8,10) is a rhombus.
Think: A rhombus must have all sides of equal length and opposite sides must be parallel. It does not need to be a square, so the adjacent sides do not need to be perpendicular.
Do: Let's look at distances first, to ensure that they are all the same. We find:
$\overline{AB}=\sqrt{1^2+4^2}=\sqrt{17}$AB=12+42=17
$\overline{BC}=\sqrt{4^2+1^2}=\sqrt{17}$BC=42+12=17
$\overline{CD}=\sqrt{1^2+4^2}=\sqrt{17}$CD=12+42=17
$\overline{DA}=\sqrt{4^2+1^2}=\sqrt{17}$DA=42+12=17
Next, let's calculate slope to ensure that opposite sides are parallel,
$m_{\overline{AB}}=\frac{4}{1}=4$mAB=41=4
$m_{\overline{CD}}=\frac{-4}{-1}=4$mCD=41=4
so $m_{\overline{AB}}\parallel m_{\overline{CD}}$mABmCD
$m_{\overline{BC}}=\frac{1}{4}$mBC=14
$m_{\overline{DA}}=\frac{1}{4}$mDA=14
so $m_{\overline{BC}}\parallel m_{\overline{DA}}$mBCmDA
Reflect: The opposite side pairs are parallel and all sides are the same length, so this is indeed a rhombus.
#### Practice questions
##### Question 3
The points given represent three vertices of a parallelogram. What is the fourth vertex if it is known to be in the 2nd quadrant?
1. Coordinates $=$=$\left(\editable{},\editable{}\right)$(,)
##### Question 4
Given Line P: $y=-6x-4$y=6x4, Line Q: $y=\frac{x}{6}+6$y=x6+6, Line R: $y=-6x-1$y=6x1 and Line S: $y=\frac{x}{6}+1$y=x6+1.
1. Complete the following:
$m$mP = $\editable{}$
$m$mQ = $\editable{}$
$m$mP x $m$mQ = $\editable{}$
2. Complete the following:
$m$mQ = $\editable{}$
$m$mR = $\editable{}$
$m$mQ x $m$mR = $\editable{}$
3. Complete the following:
$m$mR = $\editable{}$
$m$mS = $\editable{}$
$m$mR x $m$mS = $\editable{}$
4. Complete the following:
$m$mS = $\editable{}$
$m$mP = $\editable{}$
$m$mS x $m$mP = $\editable{}$
5. What type of quadrilateral is formed by lines: P, Q, R, and S?
Trapezoid
A
Rectangle
B
Rhombus
C
Parallelogram
D
Trapezoid
A
Rectangle
B
Rhombus
C
Parallelogram
D
##### Question 5
The four points $A$A$\left(3,4\right)$(3,4), $B$B$\left(6,2\right)$(6,2), $C$C$\left(-4,3\right)$(4,3) , and $D$D$\left(5,-3\right)$(5,3) are the vertices of a quadrilateral.
2. Find the slope of $\overline{AB}$AB.
3. Find the slope of $\overline{BC}$BC.
4. Find the slope of $\overline{CD}$CD.
5. Find the slope of $\overline{DA}$DA.
6. Which segments are parallel?
$\overline{AB}$AB and $\overline{BC}$BC
A
$\overline{BC}$BC and $\overline{CD}$CD
B
$\overline{AB}$AB and $\overline{CD}$CD
C
$\overline{DA}$DA and $\overline{CD}$CD
D
$\overline{AB}$AB and $\overline{BC}$BC
A
$\overline{BC}$BC and $\overline{CD}$CD
B
$\overline{AB}$AB and $\overline{CD}$CD
C
$\overline{DA}$DA and $\overline{CD}$CD
D
7. What type of quadrilateral is $ABCD$ABCD ? Choose the most precise answer.
Rectangle
A
Parallelogram
B
Trapezoid
C
Rhombus
D
Rectangle
A
Parallelogram
B
Trapezoid
C
Rhombus
D
### Outcomes
#### GEO-G.GPE.4
On the coordinate plane, algebraically prove geometric theorems and properties.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.