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Derivative of x^4 by First Principle, Power Rule
The derivative of x^4 is equal to 4x3 which can by proved by first principle and power rule. The formula of the derivative of x4 is given below.
$\dfrac{d}{dx}$(x4) = 4x3.
Let us now learn how to differentiate x to the power 4 by the power rule and the first principle of derivatives.
Derivative of x^4 by Power Rule
Recall the power rule of derivatives: the derivative of xn by the power rule is
$\dfrac{d}{dx}(x^n)=nx^{n-1}$.
Putting n=4, we will get the derivative of x4 which is equal to
$\dfrac{d}{dx}(x^4)=4x^{4-1}=4x^3$.
Derivative of x^4 by First Principle
The derivative of f(x) by the first principle is as follows: $\frac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$.
In this formula, we put f(x)=x4. So the differentiation of x4 using the first principle will be calculated as follows:
$\dfrac{d}{dx}(x^4)$ $=\lim\limits_{h \to 0} \dfrac{(x+h)^4-x^4}{h}$.
Expanding (x+h)4 using the formula (a+b)4=a4+b4+4a3b+6a2b2+4ab3, we get that
$\dfrac{d}{dx}(x^4)$ $=\lim\limits_{h \to 0}$ $\dfrac{x^4 + h^4 + 4x^3 h+6x^2h^2+4xh^3 -x^4}{h}$
= $\lim\limits_{h \to 0}$ $\dfrac{h^4 + 4x^3 h+6x^2h^2+4xh^3}{h}$
= $\lim\limits_{h \to 0}$ $\dfrac{h(h^3 + 4x^3 +6x^2h+4xh^2)}{h}$
= $\lim\limits_{h \to 0} h^3 + 4x^3 +6x^2h+4xh^2$
= $0^3 + 4x^3 +6x^2 \cdot 0+4x \cdot 0^2$
= 4x3.
So the derivative of x^4 by the first principle is 4x3.
Read Also: Derivative of 1/x by First Principle
Question: Find the derivative of sin4x.
Let z=sinx. Then by the chain rule of derivatives,
$\dfrac{d}{dx}(\sin^4 x)=\dfrac{d}{dz}(z^4) \cdot \dfrac{dz}{4x}$
= $4z^3 \cdot \cos x$
= 4sin3x cosx as z=sinx.
So the derivative of sin4x is equal to 4sin3x cosx.
Derivative of ax
Derivative of 1/(1+x)
Derivative of e3x
Derivative of $\sqrt{e^x}$
FAQs
Q1: What is the derivative of x4?
Answer: The derivative of x4 is equal to 4x3, that is, d/dx(x4) = 4x3.
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# What is the least common multiple of 9 and 15?
Feb 26, 2018
$45$
#### Explanation:
First we need to write out the prime factors of $9$ and $15$.
$9 : 3 \times 3$
$15 : 3 \times 5$
Now we group them together:
$9 : \left(3 \times 3\right)$
$15 : \left(3\right) \times \left(5\right)$
Next we take the largest groups of each number:
$9$ has two $3$s while $15$ has 1 $5$.
We multiply the largest groups together:
$L C M \left(9 , 15\right) = 3 \times 3 \times 5 = 45$
$\frac{45}{15} = 3$, $\frac{45}{9} = 5$
Feb 28, 2018
$\lcm \left(9 , 15\right) = 45$
#### Explanation:
listing the multiples
multiples of $9 : \text{ } \left\{9 , 18 , 27 , 36 , \textcolor{red}{45} , 54 , 63 , 72 , 81 , \textcolor{red}{90} , 99 , \ldots\right\}$
multiples of $15 : \text{ } \left\{15 , 30 , \textcolor{red}{45} , 60 , 75 , \textcolor{red}{90} , 105 , . .\right\}$
common multiples$\text{ } \left\{45 , 90 , \ldots\right\}$
$\lcm \left(9 , 15\right) = 45$
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How do you integrate int (lnx)^2 by integration by parts method?
2 Answers
Oct 10, 2016
$x {\left(\ln x\right)}^{2} - 2 x \ln x + 2 x + C$
Explanation:
$I = \int {\left(\ln x\right)}^{2} \mathrm{dx}$
Integration by parts takes the form: $\int u \mathrm{dv} = u v - \int v \mathrm{du}$
So, for the given integral, let:
$\left\{\begin{matrix}u = {\left(\ln x\right)}^{2} \text{ "=>" "du=(2lnx)/xdx \\ dv=dx" "=>" } v = x\end{matrix}\right.$
Plugging these into the integration by parts formula, this becomes:
$I = x {\left(\ln x\right)}^{2} - \int x \left(\frac{2 \ln x}{x}\right) \mathrm{dx}$
$I = x {\left(\ln x\right)}^{2} - 2 \int \ln x \mathrm{dx}$
To solve this integral, we will reapply integration by parts:
$\left\{\begin{matrix}u = \ln x \text{ "=>" "du=1/xdx \\ dv=dx" "=>" } v = x\end{matrix}\right.$
Thus, this becomes:
$I = x {\left(\ln x\right)}^{2} - 2 \left[x \ln x - \int x \left(\frac{1}{x}\right) \mathrm{dx}\right]$
$I = x {\left(\ln x\right)}^{2} - 2 x \ln x + 2 \int \mathrm{dx}$
$I = x {\left(\ln x\right)}^{2} - 2 x \ln x + 2 x + C$
Oct 10, 2016
$\therefore \int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} - 2 x \ln x + 2 x + c$
Explanation:
Remember the formula for IBP: $\int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$
Let $u = \ln x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$
Let $\frac{\mathrm{dv}}{\mathrm{dx}} = \ln x \implies v = \int \ln \mathrm{dx} = x \ln x - x$ (see additional notes)
Substitute into the IBP equation:
$\int \left(\ln x\right) \left(\ln x\right) \mathrm{dx} = \left(\ln x\right) \left(x \ln x - x\right) - \int \left(x \ln x - x\right) \frac{1}{x} \mathrm{dx}$
$\therefore \int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} - x \ln x - \int \left(\ln x - 1\right) \mathrm{dx}$
$\therefore \int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} - x \ln x - \left(x \ln x - x - x\right) + c$
$\therefore \int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} - x \ln x - x \ln x + 2 x + c$
$\therefore \int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} - 2 x \ln x + 2 x + c$
Additional Notes:
How do you find $\int \ln \mathrm{dx} = x \ln x - x$. Firstly its is extremely helpful to learn this result, but if you can't or you need to prove it then you need to use IBP again:
Let $u = \ln x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$
Let $\frac{\mathrm{dv}}{\mathrm{dx}} = 1 \implies v = x$
Substitute into the IBP equation:
$\int \left(\ln x\right) \left(1\right) \mathrm{dx} = \left(\ln x\right) \left(x\right) - \int \left(x \cdot \frac{1}{x}\right) \mathrm{dx}$
$\therefore \int \ln x \mathrm{dx} = x \ln x - \int \left(1\right) \mathrm{dx}$
$\therefore \int \ln x \mathrm{dx} = x \ln x - x$ $\left(+ c\right)$
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# How do you solve the inequality 2abs(7-x)+4>1 and write your answer in interval notation?
Jul 23, 2018
$\left[\frac{11}{2} , \frac{17}{2}\right]$
#### Explanation:
Isolate the absolute value (in red):
$2 \setminus \textcolor{red}{| 7 - x |} + 4 \setminus > 1$
$2 \setminus \textcolor{red}{| 7 - x |} \setminus > 1 - 4$
$\setminus \textcolor{red}{| 7 - x |} \setminus > \setminus \frac{- 3}{2}$
Expand absolute value:
$- \frac{3}{2} \setminus < 7 - x \setminus < - \left(- \frac{3}{2}\right)$
$- \frac{3}{2} \setminus < 7 - x \setminus < \frac{3}{2}$
$- \frac{3}{2} - 7 \setminus < - x \setminus < \frac{3}{2} - 7$
Note the sign change here. This happens when you divide by negatives.
$\frac{3}{2} + 7 \setminus \textcolor{\to m a \to}{\setminus >} x$ and $x \setminus \textcolor{\to m a \to}{\setminus >} - \frac{3}{2} + 7$
Effectively, this means $- \frac{3}{2} + 7 \setminus < x \setminus < \frac{3}{2} + 7$
Simplifying new inequality:
$- \frac{3}{2} + 7 \setminus < x \setminus < \frac{3}{2} + 7$
$- \frac{3}{2} + \frac{14}{2} \setminus < x \setminus < \frac{3}{2} + \frac{14}{2}$
$\frac{11}{2} \setminus < x \setminus < \frac{17}{2}$
In interval notation:
$\left[\frac{11}{2} , \frac{17}{2}\right]$ (brackets are used when you have $\setminus < , \setminus >$ signs).
Jul 23, 2018
The solution is $x \in \left(- \infty , + \infty\right)$
#### Explanation:
Let
$7 - x = 0$
$\implies$, $x = 7$
$\forall x \in \mathbb{R}$
$2 | 7 - x | > 0$
And
$2 | 7 - x | + 4 > 1$
The solution is
$x \in \left(- \infty , + \infty\right)$
graph{2|7-x|+3 [-21.99, 29.33, -11.31, 14.36]}
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Finding cost price, selling price, profit % , loss%
Chapter 8 Class 8 Comparing Quantities
Concept wise
Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month
### Transcript
Example 7 Meenu bought two fans for Rs 1200 each. She sold one at a loss of 5% and the other at a profit of 10%. Find the selling price of each. Also find out the total profit or loss. Given, Number of fans purchased = 2 Cost Price of 1 fan = Rs 1200 ∴ Total Cost Price (CP) = Number of bulbs purchased × Cost Price of 1 Bulb = 2 × 1200 = 2400 ∴ CP = Rs 2400 Total Selling Price (SP) = Selling Price of First fan + Selling Price of Second fan Selling Price of First Fan Cost Price = 1200 Loss Percentage = 5% Now, Loss Percentage = 𝐿𝑜𝑠𝑠/(𝐶𝑜𝑠𝑡 𝑃𝑟𝑖𝑐𝑒) × 100 5 = 𝐿𝑜𝑠𝑠/1200 × 100 (5 × 1200)/100 = loss 5 × 12 = Loss 60 = Loss ∴ Loss = 60 Selling Price of Second Fan Cost Price = 8000 Profit Percentage = 10% Now, Profit Percentage = 𝑃𝑟𝑜𝑓𝑖𝑡/(𝐶𝑜𝑠𝑡 𝑃𝑟𝑖𝑐𝑒) × 100 10 = 𝑃𝑟𝑜𝑓𝑖𝑡/1200 × 100 10 = 𝑃𝑟𝑜𝑓𝑖𝑡/12 10 × 12 = Profit 120 = Profit ∴ Profit = 120 We know that, Loss = Cost Price − Selling Price 60 = 1200 − Selling Price Selling Price = 1200 − 60 ∴ Selling Price = 1140 We know that, Profit = Selling Price − Cost Price 120 = Selling Price − 1200 120 + 1200 = Selling Price 1320 = Selling Price ∴ Selling Price = 1320 ∴ Selling Price of first fan = Rs 1140 & Selling Price of Second fan = Rs 1320 Putting the values in equation (1), Total Selling Price (SP) = Selling Price of first fan + Selling Price of Second fan = 1140 + 1320 = 2460 Now, CP = Rs 2400 & SP = Rs 2460 Since SP > CP ∴ Meenu made a Profit Profit = SP − CP = 2460 − 2400 = 60 Thus, Selling Price of first fan = Rs 1140 Selling Price of second fan = Rs 1320 & Profit = Rs 60
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Solve approximation x5
2
In this chapter, we will develop certain techniques that help solve problems stated in words. These techniques involve rewriting problems in the khung of symbols. For example, the stated problem
"Find a number which, when added to lớn 3, yields 7"
may be written as:
3 + ? = 7, 3 + n = 7, 3 + x = 1
and so on, where the symbols ?, n, & x represent the number we want khổng lồ find. We điện thoại tư vấn such shorthand versions of stated problems equations, or symbolic sentences. Equations such as x + 3 = 7 are first-degree equations, since the variable has an exponent of 1. The terms khổng lồ the left of an equals sign cosplay the left-hand thành viên of the equation; those to lớn the right trang điểm the right-hand member. Thus, in the equation x + 3 = 7, the left-hand thành viên is x + 3 & the right-hand thành viên is 7.
Bạn đang xem: Solve approximation x5
SOLVING EQUATIONS
Equations may be true or false, just as word sentences may be true or false. The equation:
3 + x = 7
will be false if any number except 4 is substituted for the variable. The value of the variable for which the equation is true (4 in this example) is called the solution of the equation. We can determine whether or not a given number is a solution of a given equation by substituting the number in place of the variable & determining the truth or falsity of the result.
Example 1 Determine if the value 3 is a solution of the equation
4x - 2 = 3x + 1
Solution We substitute the value 3 for x in the equation and see if the left-hand thành viên equals the right-hand member.
4(3) - 2 = 3(3) + 1
12 - 2 = 9 + 1
10 = 10
Ans. 3 is a solution.
The first-degree equations that we consider in this chapter have at most one solution. The solutions lớn many such equations can be determined by inspection.
Example 2 Find the solution of each equation by inspection.
a.x + 5 = 12b. 4 · x = -20
Solutions a. 7 is the solution since 7 + 5 = 12.b.-5 is the solution since 4(-5) = -20.
SOLVING EQUATIONS USING ADDITION và SUBTRACTION PROPERTIES
In Section 3.1 we solved some simple first-degree equations by inspection. However, the solutions of most equations are not immediately evident by inspection. Hence, we need some mathematical "tools" for solving equations.
EQUIVALENT EQUATIONS
Equivalent equations are equations that have identical solutions. Thus,
3x + 3 = x + 13, 3x = x + 10, 2x = 10, and x = 5
are equivalent equations, because 5 is the only solution of each of them. Notice in the equation 3x + 3 = x + 13, the solution 5 is not evident by inspection but in the equation x = 5, the solution 5 is evident by inspection. In solving any equation, we transform a given equation whose solution may not be obvious khổng lồ an equivalent equation whose solution is easily noted.
The following property, sometimes called the addition-subtraction property, is one way that we can generate equivalent equations.
If the same quantity is added lớn or subtracted from both membersof an equation, the resulting equation is equivalent lớn the originalequation.
In symbols,
a - b, a + c = b + c, and a - c = b - c
are equivalent equations.
Example 1 Write an equation equivalent to
x + 3 = 7
by subtracting 3 from each member.
Solution Subtracting 3 from each thành viên yields
x + 3 - 3 = 7 - 3
or
x = 4
Notice that x + 3 = 7 and x = 4 are equivalent equations since the solution is the same for both, namely 4. The next example shows how we can generate equivalent equations by first simplifying one or both members of an equation.
Example 2 Write an equation equivalent to
4x- 2-3x = 4 + 6
by combining lượt thích terms và then by adding 2 lớn each member.
Combining lượt thích terms yields
x - 2 = 10
Adding 2 lớn each member yields
x-2+2 =10+2
x = 12
To solve an equation, we use the addition-subtraction property khổng lồ transform a given equation khổng lồ an equivalent equation of the khung x = a, from which we can find the solution by inspection.
Example 3 Solve 2x + 1 = x - 2.
We want to lớn obtain an equivalent equation in which all terms containing x are in one member and all terms not containing x are in the other. If we first add -1 to (or subtract 1 from) each member, we get
2x + 1- 1 = x - 2- 1
2x = x - 3
If we now địa chỉ cửa hàng -x to (or subtract x from) each member, we get
2x-x = x - 3 - x
x = -3
where the solution -3 is obvious.
The solution of the original equation is the number -3; however, the answer is often displayed in the form of the equation x = -3.
Since each equation obtained in the process is equivalent to lớn the original equation, -3 is also a solution of 2x + 1 = x - 2. In the above example, we can check the solution by substituting - 3 for x in the original equation
2(-3) + 1 = (-3) - 2
-5 = -5
The symmetric property of equality is also helpful in the solution of equations. This property states
If a = b then b = a
This enables us to interchange the members of an equation whenever we please without having to be concerned with any changes of sign. Thus,
If 4 = x + 2thenx + 2 = 4
If x + 3 = 2x - 5then2x - 5 = x + 3
If d = rtthenrt = d
There may be several different ways khổng lồ apply the addition property above. Sometimes one method is better than another, và in some cases, the symmetric property of equality is also helpful.
Example 4 Solve 2x = 3x - 9.(1)
Solution If we first địa chỉ cửa hàng -3x khổng lồ each member, we get
2x - 3x = 3x - 9 - 3x
-x = -9
where the variable has a negative coefficient. Although we can see by inspection that the solution is 9, because -(9) = -9, we can avoid the negative coefficient by adding -2x and +9 to lớn each member of Equation (1). In this case, we get
2x-2x + 9 = 3x- 9-2x+ 9
9 = x
from which the solution 9 is obvious. If we wish, we can write the last equation as x = 9 by the symmetric property of equality.
SOLVING EQUATIONS USING THE DIVISION PROPERTY
Consider the equation
3x = 12
The solution khổng lồ this equation is 4. Also, lưu ý that if we divide each thành viên of the equation by 3, we obtain the equations
whose solution is also 4. In general, we have the following property, which is sometimes called the division property.
If both members of an equation are divided by the same (nonzero)quantity, the resulting equation is equivalent khổng lồ the original equation.
In symbols,
are equivalent equations.
Example 1 Write an equation equivalent to
-4x = 12
by dividing each thành viên by -4.
Solution Dividing both members by -4 yields
In solving equations, we use the above property to lớn produce equivalent equations in which the variable has a coefficient of 1.
Example 2 Solve 3y + 2y = 20.
We first combine like terms khổng lồ get
5y = 20
Then, dividing each member by 5, we obtain
In the next example, we use the addition-subtraction property và the division property khổng lồ solve an equation.
Example 3 Solve 4x + 7 = x - 2.
Solution First, we địa chỉ -x & -7 to lớn each member to get
4x + 7 - x - 7 = x - 2 - x - 1
Next, combining lượt thích terms yields
3x = -9
Last, we divide each thành viên by 3 lớn obtain
SOLVING EQUATIONS USING THE MULTIPLICATION PROPERTY
Consider the equation
The solution to lớn this equation is 12. Also, lưu ý that if we multiply each thành viên of the equation by 4, we obtain the equations
whose solution is also 12. In general, we have the following property, which is sometimes called the multiplication property.
If both members of an equation are multiplied by the same nonzero quantity, the resulting equation Is equivalent lớn the original equation.
In symbols,
a = b và a·c = b·c (c ≠ 0)
are equivalent equations.
Example 1 Write an equivalent equation to
by multiplying each member by 6.
Solution Multiplying each member by 6 yields
In solving equations, we use the above property to produce equivalent equations that are không tính phí of fractions.
Example 2 Solve
Solution First, multiply each thành viên by 5 to get
Now, divide each member by 3,
Example 3 Solve
.
Solution First, simplify above the fraction bar khổng lồ get
Next, multiply each member by 3 to obtain
Last, dividing each member by 5 yields
FURTHER SOLUTIONS OF EQUATIONS
Now we know all the techniques needed to lớn solve most first-degree equations. There is no specific order in which the properties should be applied. Any one or more of the following steps listed on page 102 may be appropriate.
Steps lớn solve first-degree equations:Combine lượt thích terms in each thành viên of an equation.Using the addition or subtraction property, write the equation with all terms containing the unknown in one member and all terms not containing the unknown in the other.Combine lượt thích terms in each member.Use the multiplication property to remove fractions.Use the division property to obtain a coefficient of 1 for the variable.
Example 1 Solve 5x - 7 = 2x - 4x + 14.
Solution First, we combine lượt thích terms, 2x - 4x, to yield
5x - 7 = -2x + 14
Next, we showroom +2x & +7 to lớn each thành viên and combine like terms to get
5x - 7 + 2x + 7 = -2x + 14 + 2x + 1
7x = 21
Finally, we divide each thành viên by 7 to obtain
In the next example, we simplify above the fraction bar before applying the properties that we have been studying.
Example 2 Solve
Solution First, we combine like terms, 4x - 2x, to lớn get
Then we add -3 to each thành viên and simplify
Next, we multiply each thành viên by 3 lớn obtain
Finally, we divide each member by 2 to lớn get
SOLVING FORMULAS
Equations that involve variables for the measures of two or more physical quantities are called formulas. We can solve for any one of the variables in a formula if the values of the other variables are known. We substitute the known values in the formula và solve for the unknown variable by the methods we used in the preceding sections.
Example 1 In the formula d = rt, find t if d = 24 & r = 3.
Solution We can solve for t by substituting 24 for d and 3 for r. That is,
d = rt
(24) = (3)t
8 = t
It is often necessary to solve formulas or equations in which there is more than one variable for one of the variables in terms of the others. We use the same methods demonstrated in the preceding sections.
Example 2 In the formula d = rt, solve for t in terms of r and d.
Solution We may solve for t in terms of r & d by dividing both members by r to yield
from which, by the symmetric law,
In the above example, we solved for t by applying the division property lớn generate an equivalent equation. Sometimes, it is necessary to lớn apply more than one such property.
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## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.1
NCERT Solutions of Chapter 11 Conic Sections Exercise 11.1 is provided here which will help you solving difficult questions easily and completing your homework in no time. NCERT Solutions for Class 11 Maths are prepared by Studyrankers experts that are detailed and correct so you can improve your score in the examinations.
1. Find the equation of the circle with centre (0, 2) and radius 2.
Here h = 0, k = 2 and r = 2. Therefore, the required equation of the circle is
(x – 0)2 + (y – 2)2 = (2)2
or x 2 + y 2 – 4y + 4 = 4
or x 2 + y 2 – 4y = 0
2. Find the equation of the circle with centre (–2, 3) and radius 4.
Here h = –2, k = 3 and r = 4. Therefore, the required equation of the circle is
[x – (–2)]2 + (y – 3)2 = (4)2
or (x + 2)2 + (y – 3)2 = 16
or x 2 + 4x + 4 + y 2 – 6y + 9 = 16.
3. Find the equation of the circle with centre (1/2 , 1/4) and radius 1/12.
Equation of the circle is
(x – 1/2)2 + (y – 1/4)2 = 1/122 = 1/144
=> x2 + y2 – x – y/2 + 1/4 + 1/16 = 1/144
=> 36x 2 + 36y 2 – 36x – 18y + 11 = 0.
4. Find the equation of the circle with centre (1, 1) and radius √2.
Centre of circle is (1, 1), radius = √2
Equation of circle is
(x – 1)2 + (y – 1)2 = (√2)2 =2
or x2 + y2 – 2x – 2y + 2 = 2
or x2 + y– 2x – 2y = 0.
5. Find the equation of the circle with centre (–a, –b) and radius √a2 + b2
Centre of circle is (–a, –b), radius =√a2 + b2
∴ Equation of the circle is
(x + a)2 + (y + b)2 = a2 – b2
Or x+ y2 + 2xa + 2yb + a2 + b2 = a2 – b2
Or x2 + y2 + 2ax + 2by + 2b2 = 0.
6. Find the centre and radius of the circle.
(x + 5)+ (y – 3)2 = 36
Comparing the equation of the circle
(x + 5)2 + (y – 3)2 = 36
with (x – h)2 + (y – k)2 = 2
∴ –h = 5 or h = –5, k = 3, r 2 = 36, r = 6
∴ Centre of the circle is (–5, 3) and radius = 6
7. Find the centre and radius of the circle.
x2 + y2 – 4x – 8y – 45 = 0
The given equation is
x2 + y2 – 4x – 8y – 45 = 0
or (x2 – 4x) + (y2 – 5y) = 45
Now completing the squares with in the parenthesis, we get
(x 2 – 4x + 4) + (y 2 – 8y + 16) = 4 + 16 + 45
or (x – 2)2 + (y – 4)2 = 65
Therefore, the given circle has centre at (2, 4) and radius √65.
8. Find the centre and radius of the circle.
x2 + y2 – 8x + 10y – 12 = 0.
The given equation is
x2 + y2 – 8x + 10y – 12 = 0
or (x 2 – 8x) + (y 2 + 10) = 12
or (x 2 – 8x + 16) + (y 2 + 10y + 25) = 12 + 16 + 25
or (x – 4)2 + (y + 5)2 = 53
Therefore, the given circle has centre at (4, –5) and radius √53.
9. Find the centre and radius of the given circle
2x 2 + 2y 2 – x = 0.
Equation of circle is 2x 2 + 2y 2 – x = 0
=> x2 + y2 – x/2 = 0 => (x2 – x/2) + y2 = 0
=> (x2 – x/2 + 1/16) + y2 = 1/16
=> (x – 1/4)2 + y2 = 1/16
Centre is (1/4 , 0)and radius is 1.
10. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Let the equation of the circle be
(x – h)2 + (y – k)2 = r2 … (i)
The points (4, 1) and (6, 5) lies on it
∴ (4 – h)2 + (1 – k)= r2
=> h2 + k– 8h – 2k + 17 = r2 …..(ii)
and (6 – h)2 + (5 – k)2 = r2 …..(iii)
The centre (h, k) lies on
4x + y = 16
4h + k = 16 … (iv)
Subtracting (iii) from (ii),
∴ 4h + 8k – 44 = 0 Þh + 2k = 11 … (v)
Multiplying (v) by 4, 4h + 8k = 44
Subtracting eqn (iv) from it
7k = 44 – 16 = 28 ∴ k = 4
From (v) h + 8 = 11 ∴ h = 3
Putting h = 3, k = 4 in (ii)
9 + 16 – 24 – 8 + 17 = r2
=> 42 – 32 = r 2
∴ r 2 = 10
∴ Equation of the circle is
(x – 3)2 + (y – 4)2 = 10
=> x 2 + y 2 – 6x – 8y + 15 = 0
11. Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.
Let the equation of the circle be
(x – h) 2 + (y – k) 2 = r 2 … (i)
Since, the points (2, 3) and (–1, 1) lies on it.
∴ (2 – h) 2 + (3 – k) 2 = r 2
h 2 + k 2 – 4h – 6k + 13 = r 2 … (ii)
Centre (h, k) lies on x – 3y – 11 = 0
h – 3k – 11 = 0 … (iii)
Subtracting (ii) from (i) 6h + 4 k – 11 = 0 … (iv)
Multiply eqn (iii) by 6 6h – 18 k – 66 = 0 … (v)
Subtracting (v) from (iv) 22k + 55 = 0
∴ k = -(55/22) = -(5/2)
from (iii) h = 3k + 11 = -(15/2) + 11 = 7/2
Put the value of h and k in (2 – h) 2 + (3 – k) 2 = r2
(2 – 7/2)2 + (3 + 5/2)2 = r2
=> r2 = 9/4 + 121/4 130/4 = 65/2
∴ Equation of the circle
(x – 7/2)+ (y + 5/2)2 = 65/2
=> x2 + y2 – 7x + 5y + 49/4 + 25/4 – 65/2 = 0
=> x2 + y2 – 7x + 5y – 14 = 0
12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Let the equation of the circle be
(x – h)2 + (y – k)2 = r2 … (i)
r = 5 ∴ r 2 = 25
Centre lies on x -axis is k = 0
Equation (i) becomes (x – h) 2 + y 2 = 25 (2, 3) lies on it
∴ (2 – h) 2 + 9 = 25 => (2 – h) 2 = 16,
∴ 2 – h = ±4 => h = –2, 6
When h = –2, equation of circle
(x + 2)2 + y 2 = 25
=> x2 + y2 + 4x – 21 = 0
When h = 6, (x – 6)2 + y 2 = 25,
2 + y 2 – 12x + 11 = 0
Thus, required circles are x 2 + y 2 + 4x – 21 = 0
and x2 + y 2 – 12x + 11 = 0
13. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
a, b are the intercepts made by the circle on the co-ordinate axes at A and B, C the mid point of AB is the centre of the circle
∴ centre (a/2 , b/2);
=
∴ Equation of the circle is
(x – a/2)2 + (y – b/2)2 =
x2 + y2 – ax – by + a2/4 + b2/4 = (a2 + b2)4
=> x2 + y2 – ax – by = 0
14. Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
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# Tutor profile: Catherine L.
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Catherine L.
Tutor for 5 Years
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## Questions
### Subject:Geometry
TutorMe
Question:
The measure of an angle is 14 degrees. What is the complementary angle? What is the supplementary angle?
Inactive
Catherine L.
Complementary angles sum to 90 degrees. So the complementary angle can be modeled by the equation x + 14 = 90 Therefore, by subtracting 14 we get x = 76 Supplementary angles sum to 180 degrees. So the complementary angle can be modeled by the equation x + 14 = 180 Therefore, by subtracting 14 we get x = 166
### Subject:Calculus
TutorMe
Question:
Find the critical points of the following function: f(x) = 8x^3 + 81x^2 - 42x - 8
Inactive
Catherine L.
1. Take the derivative using the power rule. (Power Rule: If f(x) = x^n, then f'(x) = n * x^(n-1) f'(x) = 24x^2 + 162x - 42 2. Factor f'(x) 6(4x^2 + 27x - 7) 6(x+7)(4x-1) 3. Set f'(x) equal to zero 6(x+7)(4x-1) = 0 x = -7 x = 1/4
### Subject:Algebra
TutorMe
Question:
The ratio of two numbers is 5 to 1. The sum is 18. What are the two numbers?
Inactive
Catherine L.
The key words in this problem are "ratio" and "sum". This let's us know how to set up the equations. "Ratio" means that we need to set up a division problem. Let the two unknown numbers be 'x' and 'y'. We can set up a division of x / y and set it equal to 5 / 1. So the overall equation is x/y = 5/1. We can rearrange this to solve for x. Multiply each side by y x = 5y "Sum" means that we need to set up an addition problem. So the equation we write it x + y = 18. Since we solved for x in the first equation, we can replace it in the addition equation so we get only one variable. 5y + y = 18 Next, combine like terms: 6y = 18 Divide by 6: y = 3 y = 3 is one answer. To solve for x you can plug y into either of the original equations. Let's choose the addition equation: x + y = 18 x + 3 = 18 Subtract x from each side: x = 15
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# Second Kirchhoff law or mesh law!
Knowing how to analyze the various types of electrical circuits, using different methods is essential for the professional of the Electrical World! Thinking to help professionals, students and lovers of electricity to better understand the methods and concepts for analysis of electrical circuits, we created this article explaining in detail what is Kirchhoff’s second law and where we apply Kirchhoff’s second law, in addition to explaining step by step how to use Kirchhoff’s second law. Come on!
Kirchhoff’s laws are generally applied to more complex electrical circuits, such as circuits with more than one source and several components, whether they are in series or in parallel. Created and developed by the German physicist Gustav Robert Kirchhoff, Kirchhoff’s laws have fundamental concepts for the analysis of electrical circuits, from the simplest circuits to the most complex circuits.
To better understand what Kirchhoff’s second law is, it is important to understand what meshes are in electrical circuits. We can define that the grid is a closed path of electrical conductors, where the circulation of the electric current is the same in all points of this grid. Disregarding the association of resistors, the image below shows a parallel series circuit that has three loops, respectively indicated.
Three loops in an electrical circuit.
## Kirchhoff’s Second Law – Definition
Kirchhoff’s second law, which is also known as mesh law or Kirchhoff’s law for voltages (LKT) basically shows how electrical voltage is distributed across the elements of electrical circuits.
The second Kirchhoff’s Law states that when traversing a loop in a certain direction, starting and reaching the same point, the algebraic sum of the DDP is zero, that is, the sum of the stresses in a loop must be equal to zero.
To facilitate understanding, in the example below we have a series circuit , where the sum of all these voltages must be zero, or the sum of all voltage drops on the resistors must equal the voltage of the source.
Definitions of the second Kirchhoff law.
## Kirchhoff’s Second Law – Application
Before applying the second law of Kirchhoff in the circuit of the previous image, it is important to highlight that it is necessary to arbitrate the direction that the electric current will travel in the circuit and establish the polarity of all the components of this same circuit.
### Determining the polarity of components
The first step to start the analysis of the circuit is to determine the direction of the electric current in the loop, so that the direction of the electric current is arbitrated equally for all the loops of the circuit to be analyzed, however when determining the direction of the electric current try do this in the most coherent way possible, as in this example, which was chosen to perform the analysis clockwise, as the current tends to flow from the highest electrical potential to the lowest potential.
Determining the direction of the current in the circuit.
Performing the analysis of this circuit is relatively simple, but the great tip for determining what we will add or subtract in the expression of the mesh analysis is to make the following considerations:
If the direction of the electrical current that was arbitrated is the same direction as the direction of the electrical current, considering the source of greatest value, we must consider the signal of the resistors as positive in the equation.
Actual and arbitrary sense of the current are the same.
However, if the direction of the electric current that was arbitrated is different from the direction of the electric current, considering the source with the highest value, we must consider the signal of the resistors in the load as negative in the equation.
Actual and arbitrary direction of the electric current are distinct.
When analyzing the electrical circuit and following the flow of electrical current that was arbitrated, we must consider the signal that the arrow is entering the source, that is, if the arrow is coming from the most positive (positive) end of the source, we will consider adding this source in the equation, however if the arrow is entering the less positive (negative) end we must subtract the source in the equation, as we can see in the image below.
Consider the source signal as negative in the equation.
To better understand mesh analysis we will determine the electrical voltage over the resistor R5 of the circuit shown in the image below and check if the value found will be correct, come on!
As seen previously, the first step is to determine the direction of the current arbitrarily, so we will make two examples, arbitrating the direction of the electric current in a clockwise and counterclockwise direction.
### Application – Clockwise
In this example we will arbitrate the direction of the electric current to clockwise, which is the same direction as the electric current, so we will consider the signal of the voltage drops on the resistors as positive in the equation.
When analyzing the circuit, we will have 4V of R1, plus 6V of R2, plus 3V of R3, plus 5V of R4, plus the voltage of R5, which we don’t know. Following the flow of the electric current that was arbitrated, the arrow is entering the less positive (negative) end of the source, so that the source in the equation has the negative sign and this whole sum must be equal to zero, as we can see in the image below.
Loop analysis step by step of the circuit.
Now to know the value of the electrical voltage on resistor R5 we must isolate it in the equation, passing all other values to the other side of the equality, exchanging their respective signals. Solving the equation, the voltage drop on resistor R5 is 2V.
Step by step solving the equation.
To verify that it is correct, just do the same analysis as we did previously, but replacing R5 with 2V and adding all the grid voltages. When performing the calculation we have that the result is equal to zero, as Kirchhoff’s second law states.
The sum of the voltage drops in the circuit is equal to zero.
### Application – Counterclockwise
We will arbitrate the direction of the electric current to the counterclockwise direction, which in this case is the opposite direction to that of the electric current, so we will consider the sign of the voltage drops on the resistors as negative in the equation.
When analyzing the circuit, we will have the voltage drop on resistor R5, a voltage that we do not know, minus the voltage drops on resistors R4, R3, R2 and R1, with voltages -5V, -3V, -6V and -4V respectively.
Following the flow of the electric current that was arbitrated, the arrow is entering the most positive end of the source, so that the source in the equation has the positive sign and this whole sum must be equal to zero.
Step-by-step resolution of the equation counterclockwise.
In order to know the value of the electrical voltage on resistor R5, we must isolate it in the equation, passing all other values to the other side of the equality, exchanging their respective signals, as was done in the previous resolution. Solving the equation, the voltage drop on resistor R5 is -2V.
To verify that it is correct, just do the same analysis as we did previously, but replacing R5 with -2V and adding all the grid voltages. When performing the calculation we have that the result is equal to zero, as Kirchhoff’s second law states.
The sum of the stresses in a mesh must equal zero.
To facilitate understanding of Kirchhoff’s second law, the video below the Mundo da Elétrica channel shows a little more about the mesh law.
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# Algorithmic concepts
By Afshine Amidi and Shervine Amidi
## Overview
### Algorithm
Given a problem, an algorithm $\mathcal{A}$ is a set of well-defined instructions that runs in a finite amount of time and space. It receives an input $I$ and returns an output $O$ that satisfies the constraints of the problem.
As an example, a problem can be to check whether a number is even. In order to do that, an algorithm could be to check whether the number is divisible by 2.
### Iteration
An iterative algorithm is an algorithm that runs through a sequence of actions. It is characterized by either a for or a while loop.
Suppose we want to return the sum of all of the elements of a given list. An example of an iterative algorithm would be to sequentially add each element of the list to a variable, and return its final value.
### Recursion
A recursive algorithm uses a function that calls itself. It is composed of the following components:
• Base case: This is the set of inputs for which the outputs are known.
• Recursive formula: The answer of the current step is based on function calls relying on previous steps, eventually using the base case answer.
A classic problem is to compute the power of a number $x^n$ without explicitly using the power operation. In order to do that, a recursive solution could rely on the following cases:
$\begin{cases} \: x^0=1&\textrm{ is known}\\ \: x^n=x^{\frac{n}{2}}\times x^{\frac{n}{2}}&\textrm{when }n\in\mathbb{N}^*\textrm{ is even}\\ \: x^n=x\times x^{\frac{n-1}{2}}\times x^{\frac{n-1}{2}}&\textrm{when }n\in\mathbb{N}^*\textrm{ is odd} \end{cases}$
### Call stack
In a recursive algorithm, the space used by function calls $c_i$ is called the stack space.
### Stack overflow
The problem of stack overflow occurs when a recursive algorithm uses more stack space than the maximum allowed $N$.
A solution to circumvent this bottleneck is to convert the code from being recursive to being iterative so that it relies on memory space, which is typically bigger than stack space.
### Memoization
Memoization is an optimization technique aimed at speeding up the runtime by storing results of expensive function calls and returning the cache when the same result is needed.
## Types of algorithms
### Brute-force
A brute-force approach aims at listing all the possible output candidates of a problem and checking whether any of them satisfies the constraints. It is generally the least efficient way of solving a problem.
To illustrate this technique, let's consider the following problem: given a sorted array $A$, we want to return all pairs of elements that sum up to a given number.
• A brute-force approach would try all possible pairs $(a_i, a_j)$ and return those that sum up to that number. This method produces the desired result but not in minimal time.
• A non-brute-force approach could use the fact that the array is sorted and scan the array using the two-pointer technique.
### Backtracking
A backtracking algorithm recursively generates potential solutions and prunes those that do not satisfy the problem constraints. It can be seen as a version of brute-force that discards invalid candidates as soon as possible.
As an example, the $N$-Queens problem aims at finding a configuration of $N$ queens on a $N\times N$ chessboard where no two queens attack each other. A backtracking approach would consist of placing queens one at a time and prune solution branches that involve queens attacking each other.
### Greedy
A greedy algorithm makes choices that are seen as optimal at every given step. However, it is important to note that the resulting solution may not be optimal globally. This technique often leads to relatively low-complexity algorithms that work reasonably well within the constraints of the problem.
To illustrate this concept, let's consider the problem of finding the longest path from a given starting point in a weighted graph. A greedy approach constructs the final path by iteratively selecting the next edge that has the highest weight. The resulting solution may miss a longer path that has large edge weights "hidden" behind a low-weighted edge.
### Divide and conquer
A divide and conquer (D&C) algorithm computes the final result of a problem by recursively dividing it into independent subproblems:
• Divide: The problem is divided into several independent subproblems.
• Conquer: Each subproblem is solved independently.
• Combine: The result of each subproblem is combined to find the final answer.
Algorithms following the D&C principle include sorting algorithms such as merge sort and quick sort.
### Dynamic Programming
Dynamic programming (DP) is a method of problem resolution that relies on finding answers to overlapping subproblems.
A common example of problem resolution using DP is the computation of Fibonacci numbers.
There are two main approaches:
Top-down This approach finds the target value by recursively computing previous values.
• Step 1: Try computing the desired value $F_n$ and notice that it is based on previous values.
• Step 2: Try computing the previous values, which themselves rely on earlier values, some of which may have already been computed. In this case, we can use memoization to avoid duplicate operations.
• Step 3: Use the newly computed values to deduce $F_n$.
Bottom-up This approach starts from already-known results and iteratively computes succeeding values until it reaches the target value.
• Step 1: Compute $F_0$, $F_1$, $F_2$, etc. in a predetermined way. These values are typically stored in an array.
• Step 2: Deduce $F_n$.
In summary, the two main ways of solving a problem with DP are:
Top-downBottom-up
Recursive approachIterative approach
Remark
A key difference between DP and D&C strategies is that DP relies on overlapping subproblems whereas D&C bases itself on independent subproblems.
## Complexity
### Definition
The concept of complexity is used to quantify the efficiency of an algorithm. There are two types of complexities:
• Time: How many operations are made?
• Space: How much extra space is needed?
Both measures are usually given as a function of the input size $n$, although other parameters can also be used.
### Notations
The complexity $f$ of an algorithm can be described using a known function $g$ with the notations described in the table below:
NotationDefinitionMeaningIllustration
$f=o(g)$
"little oh of $g$"
$\forall\epsilon>0, \exists n_0, \forall n \geqslant n_0$
$\vert f(n)\vert\leqslant\epsilon\vert g(n)\vert$
Negligible compared to $g$
$f(n) \underset{n\rightarrow+\infty}{\ll} g(n)$
$f=\mathcal{O}(g)$
"big oh of $g$"
$\exists c>0, \exists n_0, \forall n\geqslant n_0$
$\vert f(n)\vert \leqslant c\vert g(n)\vert$
Upper-bounded by $g$
$f(n)\underset{n\rightarrow+\infty}{\leqslant}g(n)$
$f=\Omega(g)$
"omega of $g$"
$\exists c>0, \exists n_0, \forall n\geqslant n_0$
$\vert f(n)\vert \geqslant c\vert g(n)\vert$
Lower-bounded by $g$
$f(n)\underset{n\rightarrow+\infty}{\geqslant}g(n)$
$f=\Theta(g)$
"theta of $g$"
$\exists c_1, c_2>0, \exists n_0, \forall n\geqslant n_0$
$\vert f(n)\vert \geqslant c_1\vert g(n)\vert$
$\vert f(n)\vert \leqslant c_2\vert g(n)\vert$
Similar to $g$
$f(n)\underset{n\rightarrow+\infty}{\sim}g(n)$
Remark
The big oh notation is frequently used to describe the time and space complexity of a given algorithm.
### Orders of magnitude
The table below highlights the main kinds of runtime complexities $T(n)$ as a function of the input size $n$:
ComplexityExample of application
O(1)
"constant"
Hash table lookup: Accessing the value of a given key does not depend on the size of the hash table.
O(log(n))
"logarithmic"
Binary search: The search space is divided by 2 at each iteration.
O(n)
"linear"
Linear search: All elements of the input are visited.
O(n log(n))
"linearithmic"
Merge sort: The array is broken down by two at every step and each step requires all elements to be checked.
O(n**2)
Bubble sort: Each pair of elements of the input array is checked.
O(2**n)
"exponential"
0/1 knapsack problem: The naive approach would consist of trying out every combination of items in the final set.
O(n!)
"factorial"
Traveling salesman problem: The naive approach consists of trying all possible permutations of cities to visit.
The following graph shows the difference in their evolutions:
As a rule of thumb, we have:
$\boxed{\mathcal{O}(1) < \mathcal{O}(\log(n)) < \mathcal{O}(n) < \mathcal{O}(n\log(n)) < \mathcal{O}(n^2) < \mathcal{O}(2^n) < \mathcal{O}(n!)}$
### Master theorem
The master theorem gives an explicit solution of a runtime $T(n)$ that satisfies a recursive relationship of the form below:
$\boxed{T(n)=aT\left(\frac{n}{b}\right)+\Theta(n^d)}$
The solution depends on the relationship between $a\in\mathbb{N}^*$, $b\in\mathbb{N}^*\backslash\{1\}$ and $d\geqslant0$:
CaseSolution $T(n)$Example of application
$d < \log_b(a)$$\Theta(n^{\log_b(a)})$Binary tree traversal: After a node is visited, we move on
to its left and right subtrees.
$\displaystyle T(n)=2T\left(\frac{n}{2}\right)\quad\longrightarrow\quad T(n)=\Theta(n)$
$d = \log_b(a)$$\Theta(n^{d}\log(n))$Binary search: The search space is divided by 2 at
each pass.
$\displaystyle T(n)=T\left(\frac{n}{2}\right)\quad\longrightarrow\quad T(n)=\Theta(\log(n))$
$d > \log_b(a)$$\Theta(n^{d})$$\displaystyle T(n)=4T\left(\frac{n}{2}\right)+\Theta(n^3)\quad\longrightarrow\quad T(n)=\Theta(n^3)$
### Problem complexity
Problems can be divided into classes that quantify how hard it is to solve them and to verify whether a proposed solution works. The table below presents two well-known classes:
Problem classDescriptionExample
P
polynomial time
The problem can be solved in polynomial time.Sorting: An array of n elements can be sorted in polynomial time with selection sort.
NP
nondeterministic polynomial time
A solution to the problem can be verified inpolynomial time.Traveling salesman problem: Given a path of n cities and a target value, we can determine whether the lengthof the path is lower or equal than the target value in O(n) time.
We note the following:
• $\textrm{P}\subseteq \textrm{NP}$: If it is possible to find a solution in polynomial time, then it is possible to verify a candidate in polynomial time by computing the solution directly.
• $\textrm{P}\stackrel{?}{\supseteq}\textrm{NP}$: It is unclear whether being able to verify a solution in polynomial time implies that we can solve the problem in polynomial time. The general consensus is that $\textrm{P}\not\supseteq\textrm{NP}$ (hence $\textrm{P}\neq\textrm{NP}$) but this has not been formally proven yet.
Any problem in NP can be reduced in polynomial time to the following set of problems:
• NP-hard problems, that are at least as hard as NP problems
• NP-complete problems, that are NP-hard problems in NP
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Review question
# If $x_2 = 1/(1 - x_1), x_3 = 1/(1 - x_2)$, can we show $x_1x_2x_3+1=0$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R5995
## Solution
It is given that $\begin{equation*} x_2 = \frac{1}{1 - x_1}, \quad x_3 = \frac{1}{1 - x_2}, \quad x_4 = \frac{1}{1 - x_3}, \quad\text{where x_1 \ne 0 or 1.} \end{equation*}$
Prove that (i) $x_1x_2x_3 + 1 = 0$, (ii) $x_4 = x_1$.
As $x_1$ neither equals $0$ nor $1$, it follows that $x_2$, $x_3$, and $x_4$ likewise neither equal $0$ nor $1$.
We can write \begin{align*} x_1x_2x_3 = \dfrac{x_1}{1 - x_1} \dfrac{1}{1 - x_2} &= \dfrac{x_1}{1 - x_1} \dfrac{1}{1 - \dfrac{1}{1 - x_1}} \\ &= \dfrac{x_1}{1 - x_1} \dfrac{1 - x_1}{(1 - x_1) - 1} \\ &= \dfrac{x_1}{1 - x_1} \dfrac{1 - x_1}{-x_1} \\ &= -1. \end{align*}
from which the result immediately follows.
For the second part, note that what we have just proved for $x_1 x_2 x_3$ applies similarly to $x_2 x_3 x_4$. So we also have $x_2 x_3 x_4 = -1$.
Then
$x_2 x_3 x_4 = x_1 x_2 x_3,$
so dividing both sides by $x_2 x_3$ yields
$x_4 = x_1.$
If $f(x) = \dfrac{1}{1 - x}$, then we have shown $f^3(x) = x$.
We might wonder, if $f(x) = \dfrac{p(x)}{q(x)}$, where $p$ and $q$ are polynomials with rational coefficients, and $f^n(x) = x$ for (almost ) all $x$, what are the possible values for $n$?
These beautiful periodic creatures are called Lyness cycles, after Robert Cranston Lyness.
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# What Is Applied Combinatorics?
Applied Combinatorics is an open-source textbook for a course covering the fundamental enumeration techniques (permutations, combinations, subsets, pigeon hole principle), recursion and mathematical induction, more advanced enumeration techniques (inclusion-exclusion, generating functions, recurrence relations, Polyá
What is combinatorics used for?
Combinatorics is used frequently in computer science to obtain formulas and estimates in the analysis of algorithms. A mathematician who studies combinatorics is called a combinatorialist .
why is combinatorics so hard? In short, combinatorics is difficult because there is no easy, ready-made algorithm for counting things fast. You need to identify patterns/regularities offered by the particular problem at hand, and exploit them in a clever way to break down the big counting problem into smaller counting problems.
### what are combinatorics in mathematics?
Combinatorics is the branch of mathematics studying the enumeration, combination, and permutation of sets of elements and the mathematical relations that characterize their properties. Mathematicians sometimes use the term “combinatorics” to refer to a larger subset of discrete mathematics that includes graph theory.
Is graph theory part of combinatorics?
One of the oldest and most accessible parts of combinatorics is graph theory, which also has numerous natural connections to other areas. Combinatorics is used frequently in computer science to obtain estimates on the number of elements of certain sets.
### How do you use combinatorics?
The basic rules of combinatorics one must remember are: The Rule of Product: The product rule states that if there are number of ways to choose one element from and number of ways to choose one element from , then there will be X × Y number of ways to choose two elements, one from and one from . The Rule of Sum:
### What does n choose k mean?
N choose K is called so because there are (n/k) number of ways to choose k elements, irrespective of their order from a set of n elements. To calculate the number of happening of an event, N choose K tool is used. N is the sum of data and K is the number that we chose from the sum of data.
### Is combinatorics useful for machine learning?
Of course, there are computations done in machine learning applications that have combinatoric elements, but that’s true of virtually any broad area of computation. Combinatorics are certainly not central to machine learning algorithms or proofs.
### How many combinations of 4 numbers are there?
For each choice of the first two digits you have 10 choices for the third digit. Thus you have 10x10x10 = 1000 choices for the first three digits. Finally you have 10 choices for the fourth digit and thus there are 10x10x10x10 = 10 000 possible 4 digit combinations from 0-9.
### How do permutations work?
A permutation is an arrangement of items or events in which order is important. Permutations help us find the total number of ways that items can be chosen when order does matter. To find the factorial of a number, multiply all of the positive integers equal to or less than that number. For example, 7!
### How do you find probability using combinatorics?
Combinatorics and Probability Here is an example: Four children, called A, B, C and D, sit randomly on four chairs. What is the probability that A sits on the first chair? P(A sits on the first chair) = number of outcomes where A sits on the first chairtotal number of possible outcomes = 624 = 14.
### Is Combinatoric a number theory?
“Combinatorial number theory”, in very loose terms, can be described as an area of mathematics which is a cross between combinatorics and number theory. In recent years, more and more mathematical olympiad style problems related to the area have also appeared.
### How do you calculate combinations?
Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula nCr = n! / r! * (n – r)!, where n represents the total number of items, and r represents the number of items being chosen at a time.
### Who Discovered number theory?
Euclid of Alexandria
### What is permutation formula?
The number of permutations of n objects taken r at a time is determined by the following formula: P(n,r)=n! (n−r)! Example. A code have 4 digits in a specific order, the digits are between 0-9.
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# How do I find the vertical and horizontal asymptotes of the function f(x)=(3x-1)/(x+4)?
Apr 3, 2018
See below.
#### Explanation:
$f \left(x\right) = \frac{3 x - 1}{x + 4}$
Vertical asymptotes occur where the function is undefined.
This happens when $x = - 4$
So the line $\textcolor{b l u e}{x = - 4}$ is a vertical asymptote.
We next examine the end behaviour as $x \to - \pm \infty$
Starting with:
$\frac{3 x - 1}{x + 4}$
Divide by $x$:
$\frac{\frac{3 x}{x} - \frac{1}{x}}{\frac{x}{x} + \frac{4}{x}}$
Cancelling:
$\frac{\frac{3 \cancel{x}}{\cancel{x}} - \frac{1}{x}}{\cancel{\frac{x}{x}} + \frac{4}{x}} = \frac{3 - \frac{1}{x}}{1 + \frac{4}{x}}$
as $x \to \infty$ , $\frac{3 - \frac{1}{x}}{1 + \frac{4}{x}} \to \frac{3 - 0}{1 + 0} = 3$
as $x \to - \infty$ , $\frac{3 - \frac{1}{x}}{1 + \frac{4}{x}} \to \frac{3 - 0}{1 + 0} = 3$
This show that the line $\textcolor{b l u e}{y = 3}$ is a horizontal asymptote.
These findings are confirmed by the graph of $f \left(x\right) = \frac{3 x - 1}{x + 4}$:
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# 2013 AIME I Problems/Problem 12
## Problem 12
Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$. A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$, side $\overline{CD}$ lies on $\overline{QR}$, and one of the remaining vertices lies on $\overline{RP}$. There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$, where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$.
## Solution
First, find that $\angle R = 45^\circ$. Draw $ABCDEF$. Now draw $\bigtriangleup PQR$ around $ABCDEF$ such that $Q$ is adjacent to $C$ and $D$. The height of $ABCDEF$ is $\sqrt{3}$, so the length of base $QR$ is $2+\sqrt{3}$. Let the equation of $\overline{RP}$ be $y = x$. Then, the equation of $\overline{PQ}$ is $y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3$. Solving the two equations gives $y = x = \frac{\sqrt{3} + 3}{2}$. The area of $\bigtriangleup PQR$ is $\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}$. $a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}$
## Cartesian Variation Solution
Also use a coordinate system, but upon drawing the diagram call $Q$ the origin and $QP$ be on the x-axis. It is easy to see that $F$ is the vertex on $RP$. After labeling coordinates (noting additionally that $QBC$ is an equilateral triangle), we see that the area is $QP$ times $0.5$ times the ordinate of $R$. Draw a perpendicular of $F$, call it $H$, and note that $QP = 1 + \sqrt{3}$ after using the trig functions for $75$ degrees.
Now, get the lines for $QR$ and $RP$: $y=\sqrt{3}x$ and $y=-(2+\sqrt{3})x + (5+\sqrt{3})$, whereupon we get the ordinate of $R$ to be $\frac{3+2\sqrt{3}}{2}$, and the area is $\frac{5\sqrt{3} + 9}{4}$, so our answer is $\boxed{021}$.
|
# Volume of Rectangular Prisms with Fractions
A rectangular prism is a 3-dimensional solid shape having 12 edges and 6 faces. This shape is given the name of a rectangular prism because of its rectangular base. It spreads in x, y, and z all three dimensions having length, width, and height.
Because of the involvement of these dimensions, it has a particular volume that can be measured quickly. Sometimes, the measure of one side is given in terms of the other side.
For example, the width of a rectangular prism is given as 2/3 of its length. In this regard, the measure of its one side is given in the fractional term. It makes the process to calculate the volume of rectangular prisms difficult.
But you can still solve it easily by following a few steps. This blog will highlight the steps that you have to take for finding the volume of a rectangular prism with a fraction.
## What is the volume of a rectangular prism?
The volume of a rectangular prism is the total space it covers in all dimensions. It means that it is the space covered by its length, width, and height in the x, y, and z axes respectively. This particular measurement shows the space surrounded by the prism’s sides
As it involves three sides, the unit representing its volume will be the volume of the units in which measurements have been done. For example, if the measures of length, height, and width are given in meters, the unit of the volume will be “meter3 or m3.
## Find the Volume of Rectangular Prism
Calculating the volume of a rectangular prism is not a complex task if you are familiar with basic mathematical operations. You only have to use a rectangular prism volume formula for its calculation which involves the multiplication of its measures.
Here is the formula to follow for this calculation:
Volume of a Rectangular Prism = Length x Width x Height (cubic units)
In simple words, you only have to multiply the measure of length, width, and height to estimate the rectangular prism volume.
## Use of fractions in a rectangular prism volume
It is common to get the measure of one side of a rectangular prism as the fraction of the other side. For example, if the width is half of its length, it will be represented as 1/2 of the length.
Also, it might be possible that you are getting fractional measurements for its sides. You must have to solve them first to find the volume.
No doubt, you can directly insert those values in the above formula to find the volume. But it may be tricky as you have to deal with division and multiplication side by side. So, it is good to solve the fraction first and then find the volume of a rectangular prism.
## How to calculate the volume of a rectangular prism?
To explain the above process of finding the volume of a rectangular prism, we have solved an example here.
example1:
Find the volume of a rectangular prism if its measures are, length = 3/4 m, width = 5/9 m, and height = 2/5 m.
Solution:
We can either solve these fractions first to find their decimals or put them directly in the above formula. Let’s first solve the example directly without decimals conversion:
Volume of a rectangular prism = 3/4 × 5/9 × 2/5 m3
Volume of a rectangular prism = 1/6 m3
It is the volume of a rectangular prism in fractional format. We can also convert it to decimal format by dividing 1 by 6. So,
Volume of a rectangular prism = 0.167 m3
In the second method, we have to convert given fractions to decimals first and then multiply all of them to find the volume.
## Conclusion
In the above blog, we have shared a comprehensive guide about the calculation of the volume of a rectangular prism with fractions. It might be possible that you are ready to do so manually and complete your work.
But if you are still unable to get the solution, you can use a volume of a rectangular prism calculator. You can get the final answer just by inserting the measurements you have for length, width, and height.
### FAQ
How do I find volume with fractions?
You can find the volume with fractions by converting fractions to decimals or multiplying all fractional terms given for length, width, and height.
How can you find the volume of a rectangular prism with fraction edge lengths?
We can find the volume of a rectangular prism just by multiplying its length, width, and height. If the measurements are given in fractions, you can whether multiply them or first convert them to decimals.
What fraction of the rectangular prism volume is the pyramid volume?
The volume of a pyramid will be 1/3 of the total volume of the rectangular prism.
Related Blogs
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# What is the imaginary unit i ? What is a complex number? Identify the imaginary and real parts of a complex number. Practice - see textbook p.103 #7 –
## Presentation on theme: "What is the imaginary unit i ? What is a complex number? Identify the imaginary and real parts of a complex number. Practice - see textbook p.103 #7 –"— Presentation transcript:
What is the imaginary unit i ? What is a complex number? Identify the imaginary and real parts of a complex number. Practice - see textbook p.103 #7 – 42 1.3 Complex Numbers http://www.youtube.com/watch?v=NeTRNpBI17I
Add and Subtract Complex Numbers Practice - see textbook p.103 #43 - 50 Add the real parts and the imaginary parts, just like combining like terms. Be sure to distribute the negative sign when subtracting! http://www.youtube.com/watch?v=eyy_SR1JfyQ&feature=youtu.be
http://www.youtube.com/watch?v=O9xQaIi0NX0 Multiplying easy complex numbers * Remember i 2 = -1
Multiplying complex numbers Practice - see textbook p.103 #51- 68 http://www.youtube.com/watch?v=Fmr3o2zkwLM&feature=youtu.be
Complex numbers and their conjugate The conjugate of a complex number is simply the same complex number with the sign changed on the imaginary part. When you multiply a complex number by the conjugate, the product simplifies to a real number with no imaginary part. Complex conjugates are needed for division of complex numbers. http://www.youtube.com/watch?v=VbOzEkvMb_A&feature=youtu.be
Dividing complex numbers To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator and simplify. Practice - see textbook p.103 #69 – 80 http://www.youtube.com/watch?v=XBJjbJAwM1c
Practice - see textbook p.103 #85 - 96 Simplifying powers of i http://www.youtube.com/watch?v=sfP6SmEYHRw
What you should know: Add and Subtract complex numbers Imaginary part Real part Complex number Imaginary unit Divide complex numbers Multiply complex numbers Simplify radicals with negative radicands Standard form Quotient Complex conjugate Numerator Denominator Product Powers of i 1.3 Complex Numbers
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You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 6 Go to the latest version.
9.5: Angle Classification
Difficulty Level: At Grade Created by: CK-12
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Have you ever had to get something approved?
Marc and Isaac finished their drawing on time and presented it to the city council. The city council loved their ideas, but did not agree to rebuild the skateboard park in the park. Marc and Isaac were feeling very defeated when they left the meeting.
When they got back to Isaac’s house, there was a message from Principal Fuller at their school. It seems the school has decided to move the soccer field to a bigger space across the street and the boys can submit the design for the new skateboard park right at the school.
“This is great!” Marc says when he hears the news. “Now we can ride before and after school.”
“Yes, but we have to redo our design,” Isaac says. “Let’s get to work.”
The space of the soccer field has been designated by the white lines on the map. To complete their design, the boys will need to count all of the different angles on the map. This way they can figure out where the ramps are going to go and where the rails will also go.
Do you know how to figure out which angles are which? How many right angles can you count? How many 180 degree angles? Use the information in this Concept to answer these questions.
Guidance
In the last few Concepts you learned to identify different geometric figures and one of them was the angle. Remember that an angle is formed when two rays connect at a single endpoint. You can measure an angle in degrees. When measuring an angle, you are measuring the space between the two rays.
The arc represents the space of the angle that you are measuring. This angle is very small as you can see by the size of the arc.
This arc is very large. The space between the rays is large so the angle is large too.
We can classify or organize angles according to the size of the angle. Because we measure them in degrees, the angle is classified according to the number of degrees that it has.
What types of angles are there?
1. Right Angle–the first type of angle is a right angle. It is an angle that is easy to recognize because it forms a corner that is straight. Often with a right angle you will see a little box in the corner too.
The corner of this building forms a right angle. You can see buildings like this all the time in the real world. It is an example of a right angle.
2. Acute Angle-an acute angle is an angle that is less than 90 degrees. Here is a picture of an acute angle.
Here is a picture of an acute angle. You can see that it has been labeled 45\begin{align*}45^\circ\end{align*} to show that it is less than 90 degrees. An acute angle is smaller than a right angle.
3. Straight Angle-a straight angle is the same as a straight line. A straight line is equal to 180\begin{align*}180^\circ\end{align*}. The angle of a straight line stretches from one side of the line to the other side as indicated by the arc in this picture.
This bike path shows a very straight line in real life.
4. Obtuse angle-an obtuse angle is an angle that is greater than 90 degrees but less than 180 degrees. Here is a picture of an obtuse angle.
This corner forms an obtuse angle. Even if we made a sharp corner from the rounded one, it would still be greater than 90 degrees, but not a straight line, so it is less than 180 degrees.
Now it is time to try a few on your own. Identify the following angles as acute, obtuse, right or straight.
Solution: Acute
Solution: Right
Example C
Solution: Obtuse
Now it is time to reconsider this map and figure out where the different types of angle are located. Be sure to reread the problem!!
Marc and Isaac finished their drawing on time and presented it to the city council. The city council loved their ideas, but did not agree to rebuild the skateboard park in the park. Marc and Isaac were feeling very defeated when they left the meeting.
When they got back to Isaac’s house, there was a message from Principal Fuller at their school. It seems the school has decided to move the soccer field to a bigger space across the street and the boys can submit the design for the new skateboard park right at the school.
“This is great!” Marc says when he hears the news. “Now we can ride before and after school.”
“Yes, but we have to redo our design,” Isaac says. “Let’s get to work.”
The space of the soccer field has been designated by the white lines on the map. To complete their design, the boys will need to count all of the different angles on the map. This way they can figure out where the ramps are going to go and where the rails will also go.
Do you know how to figure out which angles are which? How many right angles can you count? How many 180 degree angles?
By drawing arrows on the school map, we can see where all of the right angles are located. There are eight right angles located on the outside border of the plan for the soccer field.
These are the angles of the skate park.
Vocabulary
Here are the vocabulary words in this Concept.
Acute angle
an angle less than 90 degrees.
Right angle
an angle equal to 90 degrees.
Obtuse angle
an angle greater than 90 degrees but less than 180 degrees.
Straight angle
a straight line equal to 180 degrees
Guided Practice
Here is one for you to try on your own. Look at this picture.
Would you describe the panes of the window as acute, right or obtuse?
If you look at the panes, you can see that they form corners. These "corners" are a big hint as to what kind of angles are present. The panes make up right angles. Right angles are 90 degrees.
Video Review
Here is a video for review.
Practice
Directions: Classify each angle as acute, right, obtuse or straight.
1.
2.
3.
4.
5. An angle measuring 88\begin{align*}88^\circ\end{align*}
6. An angle measuring 90\begin{align*}90^\circ\end{align*}
7. An angle measuring 180\begin{align*}180^\circ\end{align*}
8. An angle measuring 105\begin{align*}105^\circ\end{align*}
9. An angle measuring 118\begin{align*}118^\circ\end{align*}
10. An angle measuring 5\begin{align*}5^\circ\end{align*}
11. An angle measuring 17\begin{align*}17^\circ\end{align*}
12. An angle measuring 135\begin{align*}135^\circ\end{align*}
13. An angle measuring 75\begin{align*}75^\circ\end{align*}
14. An angle measuring 5\begin{align*}5^\circ\end{align*}
15. An angle measuring 15\begin{align*}15^\circ\end{align*}
Vocabulary Language: English
Acute Angle
Acute Angle
An acute angle is an angle with a measure of less than 90 degrees.
Obtuse angle
Obtuse angle
An obtuse angle is an angle greater than 90 degrees but less than 180 degrees.
Perpendicular
Perpendicular
Perpendicular lines are lines that intersect at a $90^{\circ}$ angle. The product of the slopes of two perpendicular lines is -1.
Right Angle
Right Angle
A right angle is an angle equal to 90 degrees.
Straight angle
Straight angle
A straight angle is a straight line equal to $180^{\circ}$.
Oct 29, 2012
Jul 08, 2015
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# Exploring horizontal and vertical lines
In this lesson, we will solve problems involving horizontal and vertical lines.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.Fill in the gap: The equation y = 3 describes a line through every point with y-coordinate ..............
1/8
Q2.Fill in the gap: The equation .......... describes a line through every point with x-coordinate 4.
2/8
Q3.Fill in the gap: Every point on a line has a 𝑦-coordinate of 2. The equation of the line is ...........
3/8
Q4.Fill in the gap: Every point on a line has a 𝑥-coordinate of −3. The equation of the line is ........
4/8
Q5.What is the equation for line C?
5/8
Q6.What is the equation for line A?
6/8
Q7.What is the equation for line D?
7/8
Q8.What is the equation for line E?
8/8
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.Fill in the gap: The equation y = 3 describes a line through every point with y-coordinate ..............
1/8
Q2.Fill in the gap: The equation .......... describes a line through every point with x-coordinate 4.
2/8
Q3.Fill in the gap: Every point on a line has a 𝑦-coordinate of 2. The equation of the line is ...........
3/8
Q4.Fill in the gap: Every point on a line has a 𝑥-coordinate of −3. The equation of the line is ........
4/8
Q5.What is the equation for line C?
5/8
Q6.What is the equation for line A?
6/8
Q7.What is the equation for line D?
7/8
Q8.What is the equation for line E?
8/8
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Exploring horizontal and vertical lines
Exit Quiz
Q1.What is the midpoint of the line segment shown below?
1/6
Q2.What is the name of quadrilateral X?
2/6
Q3.What is the name of quadrilateral Z?
3/6
Q4.What is the equation of the line of symmetry for the triangle shown below?
4/6
Q5.What is the equation of one of the lines of symmetry for the following rhombus?
5/6
Q6.What is the equation of the line of symmetry for the kite shown below?
6/6
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Exploring horizontal and vertical lines
Exit Quiz
Q1.What is the midpoint of the line segment shown below?
1/6
Q2.What is the name of quadrilateral X?
2/6
Q3.What is the name of quadrilateral Z?
3/6
Q4.What is the equation of the line of symmetry for the triangle shown below?
4/6
Q5.What is the equation of one of the lines of symmetry for the following rhombus?
5/6
Q6.What is the equation of the line of symmetry for the kite shown below?
6/6
# Lesson summary: Exploring horizontal and vertical lines
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Climb stairs
On the spot:
Chair yoga
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# Problem Statement
Let ‘A’ be a R x C Matrix. Elements in ‘A’ are filled from 1 to R*C in row-major order. ‘B’ is also a R x C Matrix where the elements are filled in column-major order.
Find number of cells which satisfy the property Ai,j == Bi,j.
# Solution
Consider an element ( i, j ). In Matrix A, the element is located at
Posold = i * C + j.
In Matrix B (Transposed Matrix of A), the same element will be located at
Posnew = j * R + i
To satisfy the property Ai,j == Bi,j. Posold and Posnew should be equal.
i * C + j == j * R + i
The equation can be simplified as i * (C-1) – j*(R-1) == 0
i = ( (R-1) / (C-1) ) * j
Clearly ‘i’ should be an integer. The eq. is of the form i = ( a / b ) * j. For ‘i’ to be integer, ‘j’ should be a multiple of ‘b’.
a = ( R – 1 ) / GCD( R-1,C-1) —————————————————————–> 1
b = ( C-1 ) / GCD( R-1,C-1 ). —————————————————————–> 2
‘j’ is the column index. Its values ranges from 0 to C-1. So, the maximum value ‘j’ can take is C-1. Therefore , the no. of unique possible values of ‘j’ are j / b . In other words, number of multiples of ‘b’ in the range 0 to C – 1.
From 2, j / b = ( ( C – 1 ) * GCD ( R-1, C-1 ) ) / ( C – 1 ) which is simply GCD ( R-1, C-1 ).
Therefore, No. of Elements which satisfy the property is GCD(R-1,C-1) + 1.
Note: +1 is the first element in the Matrix.
Boundary Conditions
• M == 1, No. of Elements which satisfy the property is N
• N == 1, No. of Elements which satisfy the property is M.
Code:
```#include<iostream>
using namespace std;
int gcd(int a,int b)
{
if(b == 0)
return a;
return gcd(b,a%b);
}
int main()
{
std::ios_base::sync_with_stdio(false);
int t;
cin>>t;
while(t--)
{
int m,n,g;
cin>>m>>n;
if( m < n )
std::swap(m,n);
g = gcd(m-1,n-1);
if( m == 1 || n == 1 )
cout<<std::max(m,n)<<endl;
else
cout<<g+1<<endl;
}
}
```
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# PSEB 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4
Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.4 Textbook Exercise Questions and Answers.
## PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.4
1. Find H.C.F. of the following numbers by prime factorisation:
Question (i)
30, 42
Solution:
First we write the prime factorization of the given numbers
We find that 2 occurs two times and 3 occurs two times as common factors.
∴ HCF of 30 and 42 = 2 × 3 = 6
Question (ii)
135,225
Solution:
First we write the prime factorization of the given number
We find that 3 occurs two times and 5 occurs once as common factors
∴ HCF of 135 and 225 = 3 x 3 x 5 = 45
Question (iii)
180,192
Solution:
First we write the prime factorisation of the given numbers
We find that 2 occurs twice and 3 occurs once as common factors
HCF of 180 and 192
= 2 × 2 × 3 = 12
Question (iv)
49,91,175
Solution:
First we write the prime factorization of the given numbers
We find that 7 occurs once as a common factor.
∴ HCF of 49, 91 and 175 = 7
Question (v)
144, 252, 630.
Solution:
First we write the prime factorisation of the given numbers
We find that 2 occurs once and 3 occurs twice as common factors.
∴ HCF of 144, 252 and 630
= 2 × 3 × 3 = 18
2. Find H.C.F. of the following numbers using division method:
Question (i)
170, 238
Solution:
Given numbers are 170 and 238
Hence, H.C.F. of 170 and 238 = 34
Question (ii)
54, 144
Solution:
Given numbers are 54 and 144
Hence, H.C.F. of 54 and 144 = 18
Question (iii)
72, 88
Solution:
Given numbers are 72 and 88.
Hence, H.C.F. of 72 and 88 = 8
Question (iv)
96, 240, 336
Solution:
Given numbers are 96, 240 and 336 Consider any two numbers say 96 and 240
∴ H.C.F. of 96 and 240 = 48
Now, we find H.C.F. of 48 and 336
∴ H.C.F. of 48 and 336 = 48
Hence, H.C.F. of 96, 240 and 336 = 48
Question (v)
120, 156, 192.
Solution:
Given numbers are 120, 156 and 192 Consider any two numbers say 120 and 156
∴ H.C.F. of 12 and 192 = 12
Hence, H.C.F. of 120, 156 and 192 = 12
3. What is the H.C.F. of two prime numbers?
Solution:
H.C.F. of two prime numbers = 1.
4. What is the H.C.F. of two consecutive even numbers?
Solution:
The H.C.F. of two consecutive even numbers = 2.
5. What is the H.C.F. of two consecutive natural numbers?
Solution:
H.C.F. of two consecutive natural numbers = 1.
6. What is the H.C.F. of two consecutive odd numbers?
Solution:
H.C.F. of two conseutive odd numbers = 1.
7. Find the greatest number which divides 245 and 1029, leaving a remainder 5 in each case.
Solution:
Given that, required number when divides 245 and 1029, the remainder is 5 in each case.
⇒ 245 – 5 = 240 and 1029 – 5 = 1024 are completely divisible by the required number.
⇒ Required number is the highest common factor of 240 and 1024. Since it is given that required number is the greatest number.
∴ Required number is the H.C.F. 240 and 1024.
Hence, required number (H.C.F.) of 240 and 1024 = 16
8. Find the greatest number that can divide 782 and 460 leaving remainder 2 and 5 respectively.
Solution:
Required greatest number = H.C.F. of (782 – 2) and (460 – 5)
= H.C.F. of 780 and 455 = 65
Hence required greatest number = 65
9. Find the greatest number that will divide 398,437 and 540 leaving remainders 7,12 and 13 respectively.
Solution:
Required greatest number = H.C.F. of (398 – 7), (437 – 12) and (540 – 13)
= H.C.F. of 391, 425 and 527
∴ 391 = 17 × 23
425 = 5 × 5 × 17
and 527 = 17 × 31
∴ H.C.F. = 17
Hence, required greatest number = 17
10. Two different containers contain 529 litres and 667 litres of milk respectively. Find the maximum capacity of container which can measure the milk of both containers in exact number of times.
Solution:
We have to find, maximum capacity of a container which measure both conainers in exact number of times.
⇒ We required the maximum number which divides 529 and 667
⇒ Required number = H.C.F. of 529 and 667 = 23
Hence required capacity of container = 23 litres
11. There are 136 apples, 170 mangoes and 255 oranges. These are to be packed in boxes containing the same number of fruits. Find the greatest number of fruits possible in each box.
Solution:
We have to find the greatest number of fruits in each box ,
So, we required greatest numbers which divides 136, 170 and 255
∴ Required greatest number of fruits possible in each box
= H.C.F. of 136, 170 and 255
Now take any two numbers, say 136 and 170
H.C.F. of 136 and 170 = 34
Now find H.C.F. of 34 and 255
∴ H.C.F. of 34 and 255 = 17
H.C.F. of 136, 170 and 255 = 17
∴ Hence the greatest number of fruits possible in each box = 17
12. Three pieces of timber 54 m, 36 m and 24 m long, have to be divided into planks of the same length. What is the greatest possible length of each plank?
Solution:
We have to find the greatest possible length of each plank.
So, we required the maximum number which divides 54 m, 36 m and 24 m.
∴ Required length of each plank = H.C.F. of 54 m, 36 m and 24 m
Now, take any two numbers, say 54 and 36
H.C.F. of 54 and 36 = 18
Now find the H.C.F. of 18 and 24
H.C.F. 18 and 24 = 6
H.C.F. 54, 36 and 24 = 6
Hence, the greatest length of each plank = 6m
13. A room Measures 4.8 m and 5.04 m. Find the size of the largest square tile that can be used to tile the floor without cutting any tile.
Solution:
We have to find the size of largest square tile that can be used to the floor without cutting any tile.
∴ Required size of tile = H.C.F. of 4.8 and 5.04 m
= H.C.F. of 480 cm and 504 cm [1 m – 100 cm]
∴ H.C.F. of 480 cm and 504 cm = 24 cm
Hence size of each square tile = 24 cm
14. Reduce each of the following fractions to lowest forms:
Question (i)
$$\frac {85}{102}$$
Solution:
In order to reduce given fraction to the lowest terms,
We divide numerator and denominator by their H.C.F.
Now we find H.C.F. of 85 and 102 Clearly H.C.F. of 85 and 102 = 17
Question (ii)
$$\frac {52}{130}$$
Solution:
We find H.C.F. of 52 and 130
Clearly H.C.F. of 52 and 130 = 26
Question (iii)
$$\frac {289}{391}$$
Solution:
We find H.C.F. of 289 and 391
Clearly, H.C.F. of 289 and 391 = 17
|
# What is 143/92 as a decimal?
## Solution and how to convert 143 / 92 into a decimal
143 / 92 = 1.554
Convert 143/92 to 1.554 decimal form by understanding when to use each form of the number. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). Now, let's solve for how we convert 143/92 into a decimal.
## 143/92 is 143 divided by 92
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 143 divided by 92. We use this as our equation: numerator(143) / denominator (92) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. This is how we look at our fraction as an equation:
### Numerator: 143
• Numerators sit at the top of the fraction, representing the parts of the whole. 143 is one of the largest two-digit numbers you'll have to convert. 143 is an odd number so it might be harder to convert without a calculator. Values closer to one-hundred make converting to fractions more complex. Now let's explore the denominator of the fraction.
### Denominator: 92
• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 92 is one of the largest two-digit numbers to deal with. And it is nice having an even denominator like 92. It simplifies some equations for us. Overall, two-digit denominators are no problem with long division. So without a calculator, let's convert 143/92 from a fraction to a decimal.
## Converting 143/92 to 1.554
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 92 \enclose{longdiv}{ 143 }$$
To solve, we will use left-to-right long division. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Solve for how many whole groups you can divide 92 into 143
$$\require{enclose} 00.1 \\ 92 \enclose{longdiv}{ 143.0 }$$
We can now pull 92 whole groups from the equation. Multiply by the left of our equation (92) to get the first number in our solution.
### Step 3: Subtract the remainder
$$\require{enclose} 00.1 \\ 92 \enclose{longdiv}{ 143.0 } \\ \underline{ 92 \phantom{00} } \\ 1338 \phantom{0}$$
If your remainder is zero, that's it! If you have a remainder over 92, go back. Your solution will need a bit of adjustment. If you have a number less than 92, continue!
### Step 4: Repeat step 3 until you have no remainder
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. And the same is true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Here are examples of when we should use each.
### When you should convert 143/92 into a decimal
Contracts - Almost all contracts leverage decimal format. If a worker is logging hours, they will log 1.155 hours, not 1 and 143/92 hours. Percentage format is also used in contracts as well.
### When to convert 1.554 to 143/92 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 143/92 = 1.554 what would it be as a percentage?
• What is 1 + 143/92 in decimal form?
• What is 1 - 143/92 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 1.554 + 1/2?
|
# 12 percent of 25
## What's 12 percent of 25?
Answer: 12 percent of 25 it is 3
(Three)
3 is 12% of 25
## 12 percent of 25 calculation explanation
In order to calculate 12% of 25 let's write it as fractional equation.
We have 25 = 100% and X = 12%. So our fraction will look like:
25
/
X
=
100%
/
12%
Now we can solve our fraction by writing it as an equation:
X = (25 × 12) ÷ 100
=
X = 300 ÷ 100
=
X = 3
Therefore, 12% of 25 is 3
Another way to solve our problem is to find the value of 1% of the number and then multiply it by the number of percent (12). To find 1% of a number 25 you need to divide it by 100:
X = (25 ÷ 100) × 12
=
X = 0.25 × 12
=
X = 3
So we got the same result again: 3
PercentValue
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
11% of 252.75
3
3.25
3.5
3.75
4
4.25
4.5
4.75
5
21% of 255.25
22% of 255.5
5.75
6
6.25
6.5
27% of 256.75
7
29% of 257.25
7.5
## About "Percent of Number" Calculator
This calculator will help you to calculate percent of a given number. For example, it can help you find out what's 12 percent of 25? (The answer is: 3). Enter the percent (e.g. '12') and the number (e.g. '25'). Then hit the 'Calculate' button.
## FAQ
### What's 12 percent of 25?
12 percent of 25 it is 3
|
# Question 564a5
Jun 13, 2015
The time of flight is $t = \text{10 s}$.
The horizontal component of its displacement is ${d}_{x} = \text{1500 m}$.
#### Explanation:
Here's what's going on after the object is dropped from the plane.
There are two distinct directions for which you must describe the motion of the object.
Horizontally, the initial velocity of the object will be equal to that of the plane. Moreover, since no horizontal forces act upon the object, its horizontal motion takes place at constant speed.
This means that the horizontal component of its displacment can be written as
${d}_{x} = {v}_{0 x} \cdot t$, where
${v}_{0 x}$ - the horizontal component of the initial velocity;
$t$ - the total time of the fall.
Vertically, falls under the influence of gravity. This means that its vertical motion will be accelerated, the actual acceleration being equal to the gravitational acceleration, $g$.
The object starts at the height at which the plance cruises, and its initial velocity has a Zero vertical component. Moreover, it drops down in the same total time $t$. This means that you can write
${d}_{y} = {\underbrace{{v}_{0 y}}}_{\textcolor{b l u e}{\text{=0}}} \cdot t + \frac{1}{2} g {t}^{2}$
${d}_{y} = \frac{1}{2} g {t}^{2}$
Plug in your data and solve for $t$
$500 \frac{\cancel{{\text{m") = 1/2 * 10cancel("m")/"s"^2 * t^2 => t^2 = (500"s}}^{2}}}{5}$
$t = \sqrt{100 \text{s"^2) = color(green)("10 s}}$
The object drops for a total time of 10 seconds. This means that the horizontal component of its displacement will be
d_x = 150"m"/cancel("s") * 10cancel("s") = color(green)("1500 m")#
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# Help me solve a math problem
Here, we will show you how to work with Help me solve a math problem. Our website can solve math problems for you.
## The Best Help me solve a math problem
This Help me solve a math problem helps to quickly and easily solve any math problems. There are several apps available on the market today that can do everything from helping kids learn basic math facts to helping them practice their multiplication tables. There are also apps that can help kids learn how to solve general-purpose math equations and even ones that can help kids practice using fractions and decimals. Regardless of what type of word problem solver you choose, one thing is certain: you'll be able to get your child started on the right track if you take the time to learn how each app works.
A right triangle is a triangle with two right angles. By definition, it has one leg that's longer than the other. A right triangle has three sides. A right triangle has three sides: the hypotenuse (the longest side) and two shorter sides. These are called legs. The legs are always equal in length. They have equal lengths to each other and to the hypotenuse. The hypotenuse is the longest side of a right triangle and is therefore the opposite side from the one with the highest angle. It is also called the altimeter or longer leg. Right triangles always have an altimeter (the longest side). It is opposite to the hypotenuse and is also called the longer leg or hypotenuse. The other two sides of a right triangle are called legs or short sides. These are always equal in length to each other and to the longer leg of the triangle, which is called the hypotenuse. The sum of any two angles in a right triangle must be 180 degrees, because this is one full turn in any direction around a vertical line from vertex to vertex of an angle-triangle intersection. An angle-triangle intersection occurs when two lines that intersect at a common point meet back together at another point on their way down from both vertexes to that point where they intersected at first!
The substitution method is a step-by-step process used to solve systems of linear equations with two variables. First, one of the equations is solved for one of the variables. Then, the value of the variable is substituted into the other equation. The resulting equation is solved for the other variable. Finally, the values of both variables are substituted into the original equation to check the solution.
Trig equations are a type of equation that involves three numbers. They can be used to solve both simple and complex problems. For example, the trig equation 4x + 5 = 14 is used to solve the problem: "If x is equal to 4, then how much is 5?" To do this, you would subtract 5 from 14 and divide the answer by 2. The result is 9. This means that when x equals 4, 5 must be equal to 9. To solve this problem, you would plug in the value of 4 into the trig equation and solve for x. To solve a trig equation, you will usually need to carry out some calculations and follow some steps. Here's a step-by-step guide to solving trig equations: 1) Set up the equation. Start by writing down all the numbers in your problem in order from least to greatest. Put a plus sign (+) in front of each number except for one big number on top that represents your unknown number (the one you're trying to find). Write a corresponding minus sign (-) in front of this big number to represent the solution number (the one you want). For example, if you have 4x + 5 = 14 (shown above), your equation would look like this: -4 + 5 = 14 so your unknown number is -4 and your solution number is 14. 2)
## Instant help with all types of math
This app helps A TON when you don’t have time to do lots of work yourself. It is very efficient and explains to you, in a very neat fashion by the way, how to solve it. I'd recommend this calculating app to anyone who needs help on their math.
### Zaynab Hernandez
I suck at math but this help explains the problem to you and gives you different ways on how to do the problem I highly recommend it’s worth the ill cheap fee they charge the free version is good to. Yah, it helps if your struggle in math it helps me find a way that works for me that some teachers won't show you.
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# Copyright©amberpasillas2010. A mixed number has a part that is a whole number and a part that is a fraction. = 1 3 4.
## Presentation on theme: "Copyright©amberpasillas2010. A mixed number has a part that is a whole number and a part that is a fraction. = 1 3 4."— Presentation transcript:
A mixed number has a part that is a whole number and a part that is a fraction. = 1 3 4
copyright©amberpasillas2010 1 3 1 6 + Shortcut to find the LCD. Will 3 divide into 6? Use 6 as your LCD. 6 6 10 3
copyright©amberpasillas2010 1 3 1 6 + Now find the equivalent fractions for 1/3 & 1/6. 6 6 x 1 = x 2 = Ask what do you multiply 3 by to get 6 and what do you multiply 6 by to get 6. 10 3
copyright©amberpasillas2010 1 3 1 6 + You are writing equivalent fractions, so fill in your Giant One. 6 6 x 1 = x 2 = x 1 = x 2 = 10 3
copyright©amberpasillas2010 1 3 1 6 + 6 6 x 1 = x 2 = x 1 = x 2 = 10 3 Now multiply across. 2 1
copyright©amberpasillas2010 1 3 1 6 + 6 6 x 1 = x 2 = x 1 = x 2 = 10 3 2 1 Add numerators. 3 6
copyright©amberpasillas2010 1 3 1 6 + 6 6 x 1 = x 2 = x 1 = x 2 = 10 3 2 1 3 6 Add whole numbers. 13
copyright©amberpasillas2010 4 5 1 3 + Find the least common denominator for 5 and 3 15 11 2
copyright©amberpasillas2010 4 5 1 3 + Ask yourself what you multiply the denominator by to get 15. 15 x 5 = x 3 = 11 2
copyright©amberpasillas2010 4 5 1 3 + Fill in your giant one. 15 x 5 = x 3 = x 5 = x 5 = 2 11
copyright©amberpasillas2010 4 5 1 3 + 15 x 5 = x 3 = x 5 = x 3 = 2 11 Multiply straight across. 12 5
copyright©amberpasillas2010 4 5 1 3 + 15 x 5 = x 3 = x 5 = x 3 = 2 11 12 5 Add the Numerators. Denominator stays the same. 17 15
copyright©amberpasillas2010 4 5 1 3 + 15 x 5 = x 3 = x 5 = x 3 = 2 11 12 5 17 15 15 ) 17 1 15 2 2 IMPROPER
copyright©amberpasillas2010 4 5 1 3 + 15 x 5 = x 3 = x 5 = x 3 = 2 11 12 5 17 15 15 ) 17 1 15 2 2 IMPROPER Add Your 1 to Your Whole Numbers 1
copyright©amberpasillas2010 4 5 1 3 + 15 x 5 = x 3 = x 5 = x 3 = 2 11 12 5 17 15 15 ) 17 1 15 2 2 IMPROPER Put remainder fraction by answer 1 2 15 14
copyright©amberpasillas2010 5 6 5 8 + 24 x 3= x 4 = x 3= x 4 = 15 20 35 24 3 24 ) 35 1 11 24 1 28 11 24 IMPROPER
copyright©amberpasillas2010 3 5 1 3 - 15 x 5= x 3 = x 5= x 3 = 5 9 4 15 8 2 6
copyright©amberpasillas2010 3 4 1 3 - 12 x 4= x 3 = x 4= x 3 = 4 9 5 12 5 3 2
|
Surface Area of Rectangle
Before finding the surface area of a rectangle, let’s recall what a rectangle is. It would be easy to understand the concept better if we know the basics of it first.
Definition of a Rectangle
A rectangle is a two dimensional figure with opposite sides equal and parallel to each other. Al so all the four angles of the rectangle of a right angle i.e 90 degrees.
Formula for area of a rectangle
The surface area of a rectangle can be found by the formula
Surface area of a Rectangle = L × B, where L is the length of the Rectangle and B is the breadth of the rectangle.
The unit for the surface area of a rectangle is cm2 or m2.
Examples of Surface area of a Rectangle
Problem: Find the area of a rectangle whose length is 6 cm and breadth is 3 cm?
Solution: The surface area of a rectangle is found by using the formula l × b.
The length , l in this question is 6 cm and breadth, b is 3 cm.
A = l ×b = 6cm × 3cm = 18 cm2
Problem: Find the area of a rectangle whose length is 16 cm and breadth is 5 cm?
Solution: The surface area of a rectangle is found by using the formula l × b.
The length , l in this question is 16 cm and breadth, b is 5 cm.
A = l ×b = 16 cm × 5 cm = 80 cm2
Problem: Find the length of a rectangle whose area is 27cm2 and breadth is 3 cm?
Solution: The surface area of a rectangle is found by using the formula l × b.
The length , l in this question is to be found out and breadth , b is 3 cm.
A = l × b
27 cm2 = l × 3 cm
l = 27 / 3 = 9 cm
So, the length of the rectangle with an area of 27 cm2 is 9 cm.
Register at BYJU’S to learn more about all mathematical shapes and formulas.
Explore more on Rectangle
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## Introduction
In this lesson, you will learn how to solve 2-step equations using reverse operations. This is a great review for our previous posts on equations because we will be handling addition, subtraction, multiplication, and division in this lesson.
## Solving 2-step Equations
5x + 13 = 48
The above equation is a 2-step equation. In order to solve this, we need to reverse the operation applied so that we can find the value of x.
First let's look at the +13. 5x was added to 13 to get 48. We need to use the reverse of adding, which is subtracting, to find the value of x. So, 13-13 equals 0 and 48-13 equals 35.
Now we are left with
5x = 35
Now, we have to divide because the reverse of multiplication is division:
5/5 equals 1 or 1x which is the same thing as x.
35/5 equals 7.
That means the value of x=7!
We can check this by solving the equation:
5 (7)+ 13 = 48
According to the Order of Operations, multiplication comes before addition:
35 + 13 = 48
So our answer, or solution, is correct!
5x + 13 = 48
x = 7
Sheet 1 Solving 2-step Equations
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## Intermediate Algebra (12th Edition)
$p^{-3} \left( 3+2p \right)$
$\bf{\text{Solution Outline:}}$ Factor the variable with the lesser exponent in the given expression, $3p^{-3}+2p^{-2} .$ Then, divide the given expression and the variable with the lesser exponent. $\bf{\text{Solution Details:}}$ Factoring $p^{-3}$ (the variable with the lesser exponent), the expression above is equivalent to \begin{array}{l}\require{cancel} p^{-3} \left( \dfrac{3p^{-3}}{p^{-3}}+\dfrac{2p^{-2}}{p^{-3}} \right) .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} p^{-3} \left( 3p^{-3-(-3)}+2p^{-2-(-3)} \right) \\\\= p^{-3} \left( 3p^{-3+3}+2p^{-2+3} \right) \\\\= p^{-3} \left( 3p^{0}+2p^{1} \right) \\\\= p^{-3} \left( 3(1)+2p \right) \\\\= p^{-3} \left( 3+2p \right) .\end{array}
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# What are the critical values of f(x)=x/sqrt(x^2+2)-(x-1)^2?
Nov 21, 2017
$x \cong 1.1629$
#### Explanation:
The critical values of a function can be found by looking at its derivative. If the derivative is either 0 or undefined, this is a critical value.
So, we first start by computing $f ' \left(x\right)$:
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 2}} - {\left(x - 1\right)}^{2}\right)$
Let's split this into two derivatives:
$= \frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 2}}\right) - \frac{d}{\mathrm{dx}} \left({\left(x - 1\right)}^{2}\right)$
We start by computing the first expression:
$\frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 2}}\right)$
By the quotient rule, this is equal to
$= \frac{\frac{\mathrm{dx}}{\mathrm{dx}} \sqrt{{x}^{2} + 2} - x \frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + 2}\right)}{\sqrt{{x}^{2} + 2}} ^ 2$
$= \frac{\sqrt{{x}^{2} + 2} - x \frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + 2}\right)}{{x}^{2} + 2}$
We now have to solve for the derivative of $\sqrt{{x}^{2} + 2}$. If we let $u = {x}^{2} + 2$, then we can get to this expression using the chain rule:
$= \frac{d}{\mathrm{du}} \left(\sqrt{u}\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 2\right)$
$= \frac{2 x}{2 \sqrt{u}}$
Now let's substitute back in for $u$:
$= \frac{2 x}{2 \sqrt{{x}^{2} + 2}}$
So now we know that
$\frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 2}}\right) = \frac{\sqrt{{x}^{2} + 2} - x \left(\frac{2 x}{2 \sqrt{{x}^{2} + 2}}\right)}{{x}^{2} + 2}$
$= \frac{\sqrt{{x}^{2} + 2} - \frac{{x}^{2}}{\sqrt{{x}^{2} + 2}}}{{x}^{2} + 2} = \frac{\frac{\sqrt{{x}^{2} + 2} \sqrt{{x}^{2} + 2}}{\sqrt{{x}^{2} + 2}} - {x}^{2} / \left(\sqrt{{x}^{2} + 2}\right)}{{x}^{2} + 2}$
$= \frac{\frac{{x}^{2} + 2 - {x}^{2}}{\sqrt{{x}^{2} + 2}}}{{x}^{2} + 2} = \frac{2}{\left({x}^{2} + 2\right) \sqrt{{x}^{2} + 2}}$
And there we have the first part of the derivative. Now for the second:
$\frac{d}{\mathrm{dx}} \left({\left(x - 1\right)}^{2}\right)$
We can let $u = x - 1$
$= \frac{d}{\mathrm{du}} \left({u}^{2}\right) \frac{d}{\mathrm{dx}} \left(x - 1\right)$
$= 2 u$
Substituting back in, we get:
$= 2 \left(x - 1\right) = 2 x - 2$
Now we know the value of $f ' \left(x\right)$:
$f ' \left(x\right) = \frac{2}{\left({x}^{2} + 2\right) \sqrt{{x}^{2} + 2}} - 2 x + 2$
To find the critical points, we set this expression equal to 0 and solve for x.
$\frac{2}{\left({x}^{2} + 2\right) \sqrt{{x}^{2} + 2}} - 2 x + 2 = 0$
Now, this equation is a very hard one to solve, and I'm not the one to explain how to solve it. But I can tell you that the only real solution is around $1.1629$
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top of page
# What is a Simplified Fraction?
Some fractions can be simplified.
For example, the fraction 2/4 can be simplified to 1/2.
# Simplifying Fractions
A simplified fraction is a fraction that is already in its simplest form. If you can’t find an equivalent fraction with a smaller numerator and denominator then you’ve got a simplified fraction.
General Information
BASIC CONCEPTS
• Any number multiplied or divided by 1 doesn’t change its value.
• Similarly, dividing the numerator and the denominator by the same number won’t change the value of the fraction.
• The simplest form of a fraction has no common factors in the numerator and denominator.
• Factors are numbers that are multiplied to give a product.
What Fraction of Fish can be Shaded Blue
• 6 blue fish
• 12 total fish
• 6/12 of the fish are blue
• We can simplify this fraction further because 6 and 12 both have common factors.
1/2 is the simplest form of 6/12
How to Simplify Fractions
A fraction can be simplified by dividing both the numerator and denominator by a common factor.
6/8 = 6/8 ÷ 2/2 = 3/4
Let’s look at the fraction :
• The numerator and denominator are both multiples of 2.
• If we divide the numerator and denominator by 2, we’ll get an equivalent fraction.
• 6/8 can be simplified to 3/4
How to Simplify Fractions
• If you divide the numerator and denominator of a fraction by the same number, you get an equivalent fraction. In the previous example we divided both 6 and 8 by 2.
• 6 and 8 are both divisible by 2. This means 6 and 8 can be divided by 2 without a remainder.
• Since 6 and 8 are both divisible by 2, we can also say that 2 is a factor of 6 and 8.
• 2 as a factor of 6 can be written as: 2 x 3 = 6
• 2 as a factor of 8 can be written as: 4 x 2 = 8
Therefore 2 is a factor of both 6 and 8. This is also why we could divide the numerator, 6, and the denominator, 8, by 2 to get 3/4.
Finding Factors
Let’s review factors!
• Factors are numbers that multiply together to make a larger number.
Example: What are the factors of 10?
Example Factors
Find the Simplest Form of the Fraction
We can see above that 2 and 4 are both divisible by 2 because
2÷2 = 1
4÷2 = 2
Since 2 and 4 are both divisible by 2, it means that 2 is a factor of these numbers.
Divide the numerator and denominator both by 2 to get the simplest fraction 1/2.
Find the Simplest Form of the Fraction
We can see above that 4 and 10 are both divisible by 2 because
4÷2 = 2
10÷2 = 5
Since 4 and 10 are both divisible by 2, it means that 2 is a factor of these numbers.
Divide the numerator and denominator both by 2 to get the simplest fraction 2/5 .
##
Notice: The value of the fraction doesn’t change.
Dividing the numerator and denominator by the same number is the same as dividing by 1.
Find the Simplest Form of the fraction
4/16
• First, find the common factors of the numerator and denominator.
Factors of 4: 1 , 2 , 4
(1 × 4 = 4; 2 × 2 = 4; 4 × 1 = 4)
Factors of 16: 1 , 2 , 4 , 8 , 16
(1 × 16 = 16; 2 × 8 = 16; 4 × 4 = 16;
8 × 2 = 16; 16 × 1 = 16)
Find the Simplest Form of the fraction
• Greatest Common Factor (GCF)
• GCF is the greatest number that is both a factor for the numerator and denominator.
• In the previous example, it is 4.
• Divide the numerator and denominator by the GCF.
4 ÷ 4 = 1
16 ÷ 4 = 4
• So, the simplest form of 4/16 is 1/4 .
Find the Simplest Form of the fraction
• Greatest Common Factor (GCF)
• GCF is the greatest number that is both a factor for the numerator and denominator.
• In the previous example, it is 4.
• Divide the numerator and denominator by the GCF.
4 ÷ 4 = 1
16 ÷ 4 = 4
• So, the simplest form of 4/16 is 1/4 .
### Find the Simplest Form of the Fraction
Common factors:
16 : 1 , 2 , 4 , 8 , 16
84 : 1 , 2 , 4 , 6 , 7, 12 , 14 , 21 , 42 , 84
• Once past the maximum value of the numerator, there won’t be a common factor.
• Greatest Common Factor = 4
16/84 ÷ 4/4 = 4/21
Find the Simplest Form of the Fraction
####
Alex has six toy cars. Three of the cars are blue. How can you write a fraction for the blue cars in simplest form?
Fraction 3/6 Blue cars (numerator)
Total cars (denominator)
• Common factors
3 : 1, 3
6 : 1,2, 3, 6
3/6 ÷ 3/3 = 1/2
## Why Do We Find the Common Denominator?
• In order to add, subtract or compare two fractions, we need to first find a common denominator.
Example: The fractions 1/3 and 1/2 have different denominators.
• The fraction 1/3 has a denominator of 3.
• The fraction 1/2 has a denominator of 2.
• The way to find a common denominator is to find a pair of fractions that are equivalent to the fractions 1/3 and 1/2.
How to Find Common Denominators
Here’s how we’ll find a common denominator of
1/3 and 1/2 :
• We want to multiply the numerator and denominator of 1/2 by a different number to make a new fraction that is equivalent to 1/2.
• We want to multiply the numerator and denominator of 1/3 by a different number to make a new fraction that is equivalent to 1/3.
• We’ll choose the number for each case such that the two new fractions have the same denominator.
1/3 1/2
Let’s Think: What number is a multiple of both 2 and 3?
• The number 6 is the smallest multiple of both 2 and 3.
• We’ll make a common denominator of 6.
• Multiply the numerator and denominator by 2 to make an equivalent fraction with a denominator of 6.
1/3 = 1/3 X 2/2 = 2/6
Multiply the numerator and denominator of 1/2 by 3 to make an equivalent fraction with a denominator of 6.
1/2 = 1/2 X 3/3 = 3/6
• The fraction 2/6 is equivalent to the original fraction 1/3.
• The fraction 3/6 is equivalent to the original fraction 1/2.
• Since 2/6 and 3/6 now have the same denominator, we have successfully found a common denominator and we can add, subtract, or compare these fractions!
We have also found the lowest common denominator because 6 is the smallest multiple of the original denominators (3 and 2).
The examples in this chapter show you how to find the lowest common denominator. We’ll practice the skill in this chapter, and then learn how to apply it in the next chapters when we compare, add, and subtract fractions.
In the previous example, we used the smallest multiple of both denominators as the least common denominator.
What Are Multiples?
• The multiples of a number are the products that you can make from multiplication. They are whole numbers.
###
Let’s review the first ten multiples of numbers 2 through 10:
###
• The least common multiple (LCM) is the lowest shared multiple between two numbers.
• To find the common denominator, we’ll use the LCM.
###
Consider the fractions 1/4and 1/3. What is the lowest common denominator of these two fractions?
Step 1:
Find the LCM: The least common multiple of 3 and 4 is the number 12.
Step 2:
Multiply the numerator and denominator of 1/4 by 3 (because 4 x 3 = 12) to make an equivalent fraction with a denominator of 12.
1/4 = 1/4 X 3/3 = 3/12
Step 3:
Multiply the numerator and denominator of 1/3 by 4 (because 3 x 4 = 12) to make an equivalent fraction with a denominator of 12.
1/3 = 1/3 X 4/4 = 4/12
Step 4:
3/12 is an equivalent fraction of 1/4
1/3 is an equivalent fraction of 4/12
The lowest common denominator of 1/3 and 1/4 is 12.
###
Consider the fractions 3/8 and 2/3. What is the lowest common denominator of these two fractions?
Step 1:
Find the LCM: The least common multiple of 8 and 3 is 24.
Step 2:
Multiply the numerator and denominator of 3/8 by 3 (because 8 x 3 = 24) to make an equivalent fraction with a denominator of 24.
3/8 = 3/8 X 3/3 + 9/24
Step 3:
Multiply the numerator and denominator of 2/3 by 8 (because 3 x 8 = 24) to make an equivalent fraction with a denominator of 24.
2/3 = 2/3 X 8/8 = 16/24
Step 4:
9/24 is an equivalent fraction of 3/8
16/24 is an equivalent fraction of 2/3
The lowest common denominator of 3/8 and 2/3 is 24.
## Video: Least Common Multiples
Why did the fraction feel down?
Because it couldn't find its common denominator! 😄📊
###
We need to understand equivalent fractions before we can compare fractions. Let’s take a look at equivalent fractions.
Look at the fractions below. Do you notice any similarities?
2/3 = 4/6 = 8/12 = 16/24
• These are equivalent fractions because they have the same value.
• The simplest form of all these fractions is 2/3.
###
2/3 = 4/6 = 8/12 = 16/24
• To find equivalent fractions multiply the numerator and denominator by the same non-zero whole number.
Note: Multiplying the numerator and the denominator by the same non-zero whole number will change that fraction into an equivalent fraction, but it will not change the value.
Which fraction is larger 1/2 or 3/4?
We can use what we learned in the previous chapter about common denominators to compare fractions.
Step 1:
Let’s find the common denominator for 1/2 and 3/4 .
• We find the common denominator by finding a pair of fractions that are equivalent to the fractions 1/2 and 3/4 ..
• Multiply the numerator and the denominator of the fraction by the same number to get an equivalent fraction.
• We’ll choose the numbers in each case so that the two new fractions (which are equivalent to 1/2 and 3/4 .) have the same denominator.
• The denominators of these fractions are 2 and 4. What number is a multiple of both 2 and 4?
The number 4 is the smallest multiple of both 2 and 4. Therefore, we’ll make a common denominator of 4.
Step 2:
• Multiply the numerator and denominator of 1/2 by 2 to make an equivalent fraction with a denominator of 4.
Step 3:
• Multiply the numerator and denominator of 3/4 by 1 to make an equivalent fraction with a denominator of 4. Multiplying by 1 doesn’t change the fraction so we still have 3/4
Now, the two fractions 2/4 and 3/4 have the same denominator.
Step 4:
We can compare the fractions now because they have the same denominator.
• When we compare fractions with the same denominator, we look at each numerator.
• In this case, the numerator 3 is larger than the numerator 2.
3/4 is greater than 2/4
###
Instead of saying greater than, less than, or equal to, we can use symbols to compare fractions.
Greater Than
We use > for greater than.
Less Than
We use < for less than.
Equal To
We use = for equal to.
3/4 > 2/4
3/4 is greater than 2/4
2/3 x 2/2 = 4/6
4/6 x 2/2 = 8/12
4/6 x 2/2 = 8/12
8/12 x 2/2 = 16/24
Equivalent Fractions Examples
## Video: Equivalent Fractions
When you learned how to compare fractions in the previous chapter, you were well on your way to adding and subtracting fractions!
Remember how we first made the denominators the same. Then we compared the numerators to tell which fraction was bigger.
Similarly, we can add or subtract fractions by first finding a common denominator, using the strategy in Section 3. Then we simply add or subtract the numerators of the two fractions.
Step 1:
• Find the common denominator of 3 and 4. We’ll use the method we used in Chapters 3 and 4.
• To find the common denominator, we need to find equivalent fractions for the original fractions
The equivalent fractions we make from 2/3 and 1/4
• need to have the same denominator.
• We know that 12 is a multiple of both 3 and 4, so we can make 12 the common denominator.
Step 2:
• Multiply the numerator and denominator of 2/3 by 4 (because 4 x 3 = 12) to make an equivalent fraction with a denominator of 12.
• Now we have a denominator of 12, which is what we want.
Step 3:
• Multiply the numerator and denominator of 1/4 by 3 (because 4 x 3 = 12).
• Now, we have a denominator of 12, which is what we want.
Step 4:
• Now we can simply add the numerators of these two fractions.
• Since 3/12 is an equivalent fraction of 1/4 is an equivalent fraction of 2/3, we can say that
Step 1:
• Find the common denominator of 2 and 4. We’ll use the method we used in Chapters 3 and 4.
• To find the common denominator, we need to find equivalent fractions for the original fractions 3/4 and 1/2
• The equivalent fractions we make from 3/4 and 1/2 need to have the same denominator.
• We know that 4 is a multiple of 2 and 4, so we can make 4 the common denominator.
Step 2:
• Multiply the numerator and denominator of 3/4 by 1 (because 4 x 1 = 4) to make an equivalent fraction with a denominator of 4.
• We can keep 3/4 the same because it already has a denominator of 4.
Step 3:
• Multiply the numerator and denominator of 1/2 by 2 (because 2 x 2 = 4).
• Now, we have a denominator of 4.
Step 4:
We’re ready to subtract the fractions.
• Now, we can simply subtract the numerators of these two fractions.
• Since 2/4 is an equivalent fraction of 1/2 we know that
## 3/4 – 1/2 = 1/4
• What are the Steps in Simplifying Fractions?
Follow the steps mentioned below to reduce a fraction to its simplest form: Find the highest common factor of the numerator and denominator. Divide the numerator and denominator by the highest common factor. The fraction so obtained is in the simplest form.
• What does Simplifying Fractions Mean?
Simplifying fractions mean reducing the fraction in its lowest form. It helps us to do calculations involving fractions much easily. For an instance, it is easier to add 1/2 and 1/2 as compared to 2/4 + 4/8.
• What is the Rule for Simplifying Fractions?
The rule of simplifying fractions is to cancel out the common factors in the numerator and the denominator of the given fraction. In other words, we have to make sure that the numerator and denominator should be co-prime numbers.
• How do you Simplify Large Fractions?
How to Teach Simplifying Fractions? Simplifying fractions usually come in grade 5 or 6. To teach simplifying fractions, follow the points given below: Allow learners to work on hands-on activities including rectangular or circular fraction models to arrive at an understanding that 2/4 is the same as 1/2. Use real-life examples of simplifying fractions. Use simplifying fractions worksheets.
• What is the Easiest way to Simplifying Fractions?
One of the quickest ways to reduce a fraction to its simplest form is to divide the numerator and denominator of the fraction by their highest common factor.
• How to Explain Simplifying Fractions?
A fraction is said to be in the simplest form when there is no common factor of numerator and denominator other than 1. For example, 11/23 is a simplified fraction as 11 and 23 do not have any common factors.
• How to Convert an Improper Fraction to its Simplified Form?
Divide the numerator by denominator to obtain quotient and remainder. Then, the mixed fraction or the simplified fraction can be written as Quotient=Remainder/Divisor.
• How are Fractions Reduced to their Simplified Form?
To reduce a fraction into its simplest form, divide the numerator and denominator by their highest common factor.
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# Factors of 87 | Prime Factorization of 87 | Factor Tree of 87
Written by Prerit Jain
Updated on: 15 Feb 2023
### Factors of 87 | Prime Factorization of 87 | Factor Tree of 87
Factors of 87: Factors of a number are the numbers that can divide the given number perfectly with zero as the remainder and also without any decimal points in the quotient. The factors of a number can be labeled as prime numbers and composite numbers based on the count of the factors of that number.
Composite numbers have more than two numbers as their factors and prime numbers have two or less than that as the factors. Further on reading this article, you can understand the factors of 87 and various methods to find the factors of 87.
## What Are the Factors of 87?
The numbers that can divide the number 87 by zero remain, without any decimal points in the quotient are known as factors of 87. The number 87 is a composite number as it has more than two factors and they are 1, 3, 29, and 87.
## What Are the Factors of 87 in Pairs?
The pair factors of 87 are the numbers that, on multiplying with each other, give 87 as the value. Pair factors of a number can be positive and negative.
### Positive Pair Factors of 87
The positive pair factors of 87 are (1, 87) and (3, 29).
### Negative Pair Factors of 87
The negative pair factors of 87 are (-1, -87) and (-3, -29).
## What Is the Basic Prime Factorization Method of 87?
The Basic prime factorization method helps to find the prime factors of 87. The following are steps given below:
• Step 1: For this purpose, let’s divide the number 87 by its smallest prime number 3.
• Step 2: And the remainder obtained is 29. The number 29 is divisible by itself only.
• Step 3: Now, the remainder is 1. Therefore, the prime factors of 87 are 3 and 29.
The image of the prime factorization of 87 is given below:
## How to Find Factors of 87 Through Division Method?
The division method helps to find the factors of a number by dividing the given number by the smallest prime number and again the obtained remainder is divided by the smallest number that can divide it. This process is continued till the remainder becomes 1. The steps involved in this process are given below:
• 87 ÷ 1= 87
• 87 ÷ 3 = 29
• 87÷ 29 = 3
• 87 ÷ 87 = 1
Therefore, the factors of 87 are 1, 3, 29, and 87.
## What Is the Factor Tree of 87?
The factor tree method is really a good way to represent the factors of 87. As it shows the prime numbers and the composite numbers that can divide the given number. The steps involved in the factor tree method are given below:
• Step 1: Let’s place the given number 87 at the top of the factor tree method and place the first pair factors of 87 in its branches which are 3 and 29.
• Step 2: The numbers 3 and 29 are not further divisible. Therefore, the prime factors of 87 are 3 and 29.
Therefore, the prime factors of 87 are 3 and 29.
## Solved Examples
What are the pair factors of 87?
The pair factors of 87 are (1, 87) and (3, 29).
What is the sum of the factors of 87?
The sum of the factors of 87 is 120.
What are the factors of 87?
The factors of 87 are 1, 3, 29, and 87.
What is the sum of the prime factors of 87?
The sum of the prime factors of 87 is 32.
What are the prime factors of 87?
The prime factors of 87 are 3 and 29.
How is 87 composite?
Yes, 87 is a composite number as it has more than two factors.
How many divisors does 87 have?
The factors of 87 itself are the divisors of 87 and they are 1, 3, 29, and 87.
What is the HCF of 87?
The HCF of 87 is 29.
What is the square of 87?
The square of 87 is 7569.
Written by
Prerit Jain
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# How Many Numbers up to 500 are Divisible by 23
If you’ve ever pondered over mathematical questions and found yourself wondering about the divisibility of numbers, you’re not alone. One intriguing question that often comes up is, “how many numbers up to 500 are divisible by 23?” In this informative article, we will dive into this mathematical puzzle, providing you with detailed explanations, expert insights, and answers to frequently asked questions.
Mathematics has a unique way of intriguing our minds. It’s not just about numbers; it’s about patterns, logic, and solving puzzles. One such puzzle that piques our curiosity is understanding how many numbers up to 500 can be divided evenly by 23. This might sound like a daunting task, but fear not; we’re here to break it down for you.
How Many Students Appeared for JELET 2023
## Exploring Divisibility
Divisibility Rules
Before we dive into the numbers, let’s quickly recap some basic rules of divisibility. These rules will help us determine if a number is divisible by 23 or not.
• Divisible by 2: Any number that ends in 0, 2, 4, 6, or 8 is divisible by 2.
• Divisible by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
• Divisible by 5: Numbers ending in 0 or 5 are divisible by 5.
• Divisible by 23: To check for divisibility by 23, we’ll need to perform some calculations.
### The Magic of 23
The number 23 has some interesting properties. It’s a prime number, which means it has only two distinct positive divisors: 1 and 23. This characteristic simplifies our task. We need to find multiples of 23 within the range of 1 to 500.
## How Many numbers up to 500 are divisible by 23
Calculating how many numbers up to 500 are divisible by 23 is a straightforward task. We’ll divide 500 by 23 and see how many times it goes evenly.
500 รท 23 = 21.74 (approximately)
So, 23 goes into 500 approximately 21 times without leaving a remainder. That means there are 21 numbers up to 500 that are divisible by 23.
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# How to Solve Motion Problems Part 4
In Part 1, Part 2, and Part 3 of the How to Solve Motion Problems Series, we have learned how to solve problems involving objects moving in the same direction as well as those which move toward each other. In this post, we are going to learn about objects which move on opposite directions. The method in solving this problem is very similar to the method used in Part 3 of this series.
We now solve the sixth problem in this series.
Problem 6
Two jet planes left Naria Airport at 9:00 am and travels in opposite directions. One jet travels at an average speed of 450 kilometers per hour and the other jet travels at an average speed of 550 kilometers per hour. By what time will the two jets be 2500 kilometers apart?
Solution 1
Just like in the previous parts, we can solve this problem using a table. As we can see, in the first hour, the jets will be 1000 kilometers apart. After three hours they will be 3000 kilometers apart.
The question, however, is the time when they are 2500 kilometers. From the table above, since after each hour, the distance traveled is 1000 kilometers, then 2500 kilometers will require 2 and a half hours. Now, 2 and a half hours after 9 am is 11:30 am.
Solution 2
We name the jet planes A and B as shown below. Plane A travels at 450 kilometers per hour and plane B at 550 kilometers per hour. The time traveled by both planes, we let $x$ and since they both start and the same time, they will have the same time (duration) traveled. The distance d is the product of the rate and the time.
Note: The time of departure so (9:00 am) is irrelevant at first in solving the problem. It can only be used after the answer is obtained.
Now that we have the table, let us examine the figure below. In the question, how many hours will the two planes be 2500 kilometers apart. Since the planes are traveling in opposite directions, the word “apart” in the problem means the distance traveled by the first plane (450x) and the distance traveled by the second plane (550x). Therefore, we can form the equation
d traveled by Plane A + d traveled by plane B = 2500
If we substitute the values in the table in the preceding equation, then we have
$450x + 550x = 2500$
$1000x = 2500$
$x = 2.5$ hours
Meaning, if the planes are 2500 kilometers apart, two and a half hours would have past. Therefore, 2 and a half hours from the time of the departure which is 9:00 am is 11:30 am.
Therefore the correct answer is 11:30 am. This confirms the answer in Solution 1.
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# polynomials
### 2 個解答
• 6 年前
最愛解答
An algebraic expression of the form a + bx + cx^2 + dx^3 + ………….. in which a, b, c, d are constants (numbers) and x is a variable, is called a polynomial in x.
The numbers a, b, c, d…….. are also called coefficients (therefore constants as they are numbers that do not change like variable x)
現在將 term with the highest power 寫先,
For quadratic expression, ax^2 + bx + c
For cubic expression, ax^3 + bx^2 + cx + d
For quartic expression, ax^4 + bx^3 + cx^2 + dx + e
When the variable x is raised to an index p to give x^p, the result is a power of x
Power refers to the number of times x is multiplied. For example, x^4 is the fourth power of x, x is raised to the power of 4..
Say a cubic expression, 2x^3 + 5x^2 + 6x + 7
The power of the first term is 3 (2x^3)
The power of the second term is 2 (5x^2)
The power of the third term is 1 (6x^1)
The power of the fourth term is 0 (7x^0), x^0 = 1,
The degree of a polynomial is the highest power of the variable.
For quadratic expression, ax^2 + bx + c , the degree is 2
For cubic expression, ax^3 + bx^2 + cx + d the degree is 3
For quartic expression, ax^4 + bx^3 + cx^2 + dx + e, the degree is 4
In secondary school, we don’t do polynomial of several variables.
Degree of a polynomial in two variables:
The degree of a polynomial which contains two or more variables is the greatest of the different sums of the powers of the variables in the various terms of the polynomial
Examples of degree of a polynomial in two or more variables:
In 2x^2y + 3xy + 4, the sum of the powers of variables x and y in:
2x^2y is (2 + 1) = 3,
3xy is (1 +1) = 2,
0 in 4.
Among the various sums of the powers of the variables in the terms, 3 is the greatest and therefore the degree of the above polynomial is taken as 3.
2014-07-02 04:03:20 補充：
Correction:
In secondary school, we don’t do polynomial equation of several variables, especially solving it.
• 6 年前
Power is the highest index of the polynomial.
Eg. X^3y^2+ x^1y^2 , power =3 since the largest index is 3.
Degree is the largest sum of powers of a term in polynomials, that is in the above example
First term x^3y^2 has degree 3+2=5
Second term 1+2= 3
5 >3
Degree is 5
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Volume of Cylinders, Cones & Spheres | Math for Kids Grade 6, 7, 8
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HOW DO YOU FIND THE VOLUME OF CYLINDERS, CONES & SPHERES?
Volume is the space inside a 3D solid. Volume is measured in cubic units. Formulas are rules that are written using mathematical symbols that relate quantities. You have already used volume formulas to find the volume of rectangular prisms and cubes. You can also use volume formulas to find the volumes of cones, cylinders, and spheres.
To better understand how to find the volume of cylinders, cones, and sphere…
HOW DO YOU FIND THE VOLUME OF CYLINDERS, CONES & SPHERES?. Volume is the space inside a 3D solid. Volume is measured in cubic units. Formulas are rules that are written using mathematical symbols that relate quantities. You have already used volume formulas to find the volume of rectangular prisms and cubes. You can also use volume formulas to find the volumes of cones, cylinders, and spheres. To better understand how to find the volume of cylinders, cones, and sphere…
## LET’S BREAK IT DOWN!
### Review Volume
You can find the volume of a rectangular prism by seeing how many cubic units fit inside it. The number of unit cubes that fit inside tells you the volume. You can also use a formula. The volume of a rectangular prism is length times width times height. If the length is 3 units, the width is 2 units, and the height is 4 units, the volume is 3 × 2 × 4 = 24 cubic units. Try this one yourself: A rectangular prism has length 5 units, width 6 units, and height 2 units. What is the volume of the rectangular prism?
Review Volume You can find the volume of a rectangular prism by seeing how many cubic units fit inside it. The number of unit cubes that fit inside tells you the volume. You can also use a formula. The volume of a rectangular prism is length times width times height. If the length is 3 units, the width is 2 units, and the height is 4 units, the volume is 3 × 2 × 4 = 24 cubic units. Try this one yourself: A rectangular prism has length 5 units, width 6 units, and height 2 units. What is the volume of the rectangular prism?
### Volume of a Cylinder
How can you find the volume of a cake in the shape of a cylinder? Think about this shape by comparing it to a rectangular prism. The formula for the volume of a rectangular prism is length times width times height. The length times the width is the area of the base. So, another way to think of this formula is the area of the base times the height. You can think of the volume of a cylinder in the same way: the area of the base times the height. The base is a circle. A circle has area πr2. So, the formula for the volume of a cylinder is πr2h. Amari’s cake is a cylinder with diameter 44 cm and height 50 cm. What is the volume of the cake? The radius is half the diameter, so the radius is 22 cm. Use the formula. 3.14 × 222 × 50 = 75,988 cubic centimeters. Remember that when you use 3.14 to approximate pi, your answer is an approximation. Try this one yourself: A cylinder has height 20 cm and diameter 16 cm. What is the volume of the cylinder?
Volume of a Cylinder How can you find the volume of a cake in the shape of a cylinder? Think about this shape by comparing it to a rectangular prism. The formula for the volume of a rectangular prism is length times width times height. The length times the width is the area of the base. So, another way to think of this formula is the area of the base times the height. You can think of the volume of a cylinder in the same way: the area of the base times the height. The base is a circle. A circle has area πr2. So, the formula for the volume of a cylinder is πr2h. Amari’s cake is a cylinder with diameter 44 cm and height 50 cm. What is the volume of the cake? The radius is half the diameter, so the radius is 22 cm. Use the formula. 3.14 × 222 × 50 = 75,988 cubic centimeters. Remember that when you use 3.14 to approximate pi, your answer is an approximation. Try this one yourself: A cylinder has height 20 cm and diameter 16 cm. What is the volume of the cylinder?
### Volume of a Cone
How can you find the volume of a plastic cup in the shape of a cone? Compare the cone to a cylinder with the same diameter and same height. If you fill the cone with water and pour it into the cylinder, it takes exactly 3 cones full of water to fill the cylinder. The volume of the cone is one third the volume of the cylinder. Remember, the formula for the volume of a cylinder is πr2h. So, the formula for the volume of a cone is [ggfrac]1/3[/ggfrac]πr2h. Emily has a cone cup. The height is 10 cm and the diameter is 6 cm. What is the volume of the cup? The radius is half the diameter, or 3 cm. Substitute the numbers into the volume formula. The volume of the cone cup is approximately 94.2 cubic centimeters. Try this one yourself: A cone has height 14 cm and diameter 12 cm. What is the volume of the cone?
Volume of a Cone How can you find the volume of a plastic cup in the shape of a cone? Compare the cone to a cylinder with the same diameter and same height. If you fill the cone with water and pour it into the cylinder, it takes exactly 3 cones full of water to fill the cylinder. The volume of the cone is one third the volume of the cylinder. Remember, the formula for the volume of a cylinder is πr2h. So, the formula for the volume of a cone is [ggfrac]1/3[/ggfrac]πr2h. Emily has a cone cup. The height is 10 cm and the diameter is 6 cm. What is the volume of the cup? The radius is half the diameter, or 3 cm. Substitute the numbers into the volume formula. The volume of the cone cup is approximately 94.2 cubic centimeters. Try this one yourself: A cone has height 14 cm and diameter 12 cm. What is the volume of the cone?
### Volume of a Sphere
How can you find the volume of a sphere? Compare the sphere to a cone with the same radius and the same height. If you fill the cone with water and pour it into the sphere, it takes exactly 2 cones full of water to fill the sphere. The volume of the sphere is twice the volume of the cone. Remember, the formula for the volume of a cone was [ggfrac]1/3[/ggfrac]πr2h. So, [ggfrac]1/3[/ggfrac]πr2h + [ggfrac]1/3[/ggfrac]πr2h is the volume of a sphere. You can make this formula simpler. The height of the sphere, h, is equal to the diameter, or 2r. Now you have [ggfrac]1/3[/ggfrac]πr2(2r) + [ggfrac]1/3[/ggfrac]πr2(2r). You can simplify this expression. Since you can multiply in any order, move the 2s to the front. Remember, r2 times r is r3. Now you have [ggfrac]2/3[/ggfrac]πr3 + [ggfrac]2/3[/ggfrac]πr3. Add these parts together: [ggfrac]4/3[/ggfrac]πr3. So, the formula for the volume of a sphere in its simplest form is [ggfrac]4/3[/ggfrac]πr3. You could use another version of the formula, but this version is simplest. Use this formula to solve a problem. A fishbowl in the shape of a sphere has diameter 20 cm. What is the volume of the fishbowl? The radius of the fishbowl is half the diameter, so it is 10 cm. Substitute that number into the formula. The volume of the fishbowl is approximately 4,187 cm3 Try this one yourself: A sphere has diameter 18 cm. What is the volume of the sphere?
Volume of a Sphere How can you find the volume of a sphere? Compare the sphere to a cone with the same radius and the same height. If you fill the cone with water and pour it into the sphere, it takes exactly 2 cones full of water to fill the sphere. The volume of the sphere is twice the volume of the cone. Remember, the formula for the volume of a cone was [ggfrac]1/3[/ggfrac]πr2h. So, [ggfrac]1/3[/ggfrac]πr2h + [ggfrac]1/3[/ggfrac]πr2h is the volume of a sphere. You can make this formula simpler. The height of the sphere, h, is equal to the diameter, or 2r. Now you have [ggfrac]1/3[/ggfrac]πr2(2r) + [ggfrac]1/3[/ggfrac]πr2(2r). You can simplify this expression. Since you can multiply in any order, move the 2s to the front. Remember, r2 times r is r3. Now you have [ggfrac]2/3[/ggfrac]πr3 + [ggfrac]2/3[/ggfrac]πr3. Add these parts together: [ggfrac]4/3[/ggfrac]πr3. So, the formula for the volume of a sphere in its simplest form is [ggfrac]4/3[/ggfrac]πr3. You could use another version of the formula, but this version is simplest. Use this formula to solve a problem. A fishbowl in the shape of a sphere has diameter 20 cm. What is the volume of the fishbowl? The radius of the fishbowl is half the diameter, so it is 10 cm. Substitute that number into the formula. The volume of the fishbowl is approximately 4,187 cm3 Try this one yourself: A sphere has diameter 18 cm. What is the volume of the sphere?
### Relationship Between Cones, Cylinders, and Spheres
A restaurant serves smoothies in containers in three different shapes: cones, cylinders, and spheres. All three containers have the same radius and height. All of them are the same price. Which shape container should you order if you want to get the greatest amount of smoothie? Remember, 1 cylinder has the same volume as 3 cones. 1 sphere has the same volume as 2 cones. The cylinder container gives you the greatest amount of smoothie. Try this one yourself: A cylinder and a cone have the same radius. They hold the same amount of water. How many times as tall as the cylinder is the cone?
Relationship Between Cones, Cylinders, and Spheres A restaurant serves smoothies in containers in three different shapes: cones, cylinders, and spheres. All three containers have the same radius and height. All of them are the same price. Which shape container should you order if you want to get the greatest amount of smoothie? Remember, 1 cylinder has the same volume as 3 cones. 1 sphere has the same volume as 2 cones. The cylinder container gives you the greatest amount of smoothie. Try this one yourself: A cylinder and a cone have the same radius. They hold the same amount of water. How many times as tall as the cylinder is the cone?
## VOLUME OF CYLINDERS, CONES & SPHERES VOCABULARY
Volume
The amount of space inside a 3D object.
Cone
A 3D shape with one circular base.
Cylinder
A 3D shape with two circular bases.
Sphere
A 3D shape in the shape of a ball.
Pi (π)
The ratio of the circumference of a circle to its diameter. This is always the same number: approximately 3.14. The actual number is an irrational number (a decimal that never ends or repeats).
The measurement from the center of a circle or sphere to its edge. The radius is half the diameter.
The distance straight across a circle or sphere. The diameter always passes through the center point of the circle.
## VOLUME OF CYLINDERS, CONES & SPHERES DISCUSSION QUESTIONS
### What is the formula for the volume of a cylinder? What does each part of the formula mean?
V = πr2h. r is the radius of the circular base of the cylinder and h is the height of the cylinder (the distance between the two circular ends).
### What is the formula for the volume of a cone? What does each part of the formula mean?
V = [ggfrac]1/3[/ggfrac]πr2h. r is the radius of the circle at the base and h is the distance between the center point of the circular base and the point at the top of the cone.
### What is the formula for the volume of a sphere? What does each part of the formula mean?
V = [ggfrac]4/3[/ggfrac]πr3. If I cut the sphere exactly in half I could see a circle on the face. r is the radius of the circle at the very center of the sphere.
### [Draw a cylinder with radius 6 cm and height 10 cm.] What is the volume of this shape?
360π cm3 or about 1,130.4 cm3.
### Suppose you have a cone and a cylinder that have the same height and the same radius. Which has a greater volume?
The cylinder.
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# Matrix-vector multiplication
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Matrix-vector multiplication is an operation between a matrix and a vector that produces a new vector. In this post, I’ll define matrix vector multiplication as well as three angles from which to view this concept. The third angle entails viewing matrices as functions between vector spaces
## Introduction
In a previous post, we discussed three ways one can view a matrix: as a table of values, as a list of vectors, and finally, as a function. It is the third way of viewing matrices that really gives matrices their power. Here, we’ll introduce an operation between matrices and vectors, called matrix-vector multiplication, which will enable us to use matrices as functions.
Matrix-vector multiplication is an operation between a matrix and a vector that produces a new vector. Notably, matrix-vector multiplication is only defined between a matrix and a vector where the length of the vector equals the number of columns of the matrix. It is defined as follows:
Definition 1 (Matrix-vector multiplication): Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$ and vector $\boldsymbol{x} \in \mathbb{R}^n$ the matrix-vector multiplication of $\boldsymbol{A}$ and $\boldsymbol{x}$ is defined as
$$\boldsymbol{A}\boldsymbol{x} := x_1\boldsymbol{a}_{*,1} + x_2\boldsymbol{a}_{*,2} + \dots + x_n\boldsymbol{a}_{*,n}$$
where $\boldsymbol{a}_{*,i}$ is the $i$th column vector of $\boldsymbol{A}$.
Like most mathematical concepts, matrix-vector multiplication can be viewed from multiple angles, at various levels of abstraction. These views come in handy when we attempt to conceptualize the various ways in which we utilize matrix-vector multiplication to model real-world problems. Below are three ways that I find useful for conceptualizing matrix-vector multiplication ordered from least to most abstract:
1. As a “row-wise”, vector-generating process: Matrix-vector multiplication defines a process for creating a new vector using an existing vector where each element of the new vector is “generated” by taking a weighted sum of each row of the matrix using the elements of a vector as coefficients
2. As taking a linear combination of the columns of a matrix: Matrix-vector multiplication is the process of taking a linear combination of the column-space of a matrix using the elements of a vector as the coefficients
3. As evaluating a function between vector spaces: Matrix-vector multiplication allows a matrix to define a mapping between two vector spaces.
I find all three of the perspectives useful. The first two perspectives provide a way of understanding the mechanism of matrix-vector multiplication whereas the third perspective provides the essence of matrix-vector multiplication. It is this third perspective of matrix-vector multiplication that enables us to view matrices as functions, as we discussed in the previous post.
## Matrix-vector multiplication as a “row-wise”, vector-generating process
A useful way for viewing the mechanism of matrix-vector multiplication between a matrix $\boldsymbol{A}$ and a vector $\boldsymbol{x}$ is to see it as a sort of “process” (or even as a computer program) that constructs each element of the output vector in an iterative fashion where we iterate over each row of $A$. Specifically, for each row $i$ of $\boldsymbol{A}$, we take $\boldsymbol{x}$ and compute the dot product between $\boldsymbol{x}$ and the $i$th row of the matrix thereby producing the $i$th element of the output vector (See Theorem 1 in the Appendix to this post):
$\boldsymbol{Ax} = \begin{bmatrix} \boldsymbol{a}_{1,*} \cdot \boldsymbol{x} \\ \boldsymbol{a}_{2,*} \cdot \boldsymbol{x} \\ \vdots \\ \boldsymbol{a}_{m,*} \cdot \boldsymbol{x} \\ \end{bmatrix}$
where $\boldsymbol{a}_{i,*}$ is the $i$th row-vector in $\boldsymbol{A}$. This process is illustrated schematically in the figure below:
## Matrix-vector multiplication as taking a linear combination of the columns of a matrix
Matrix-vector multiplication between a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$ and vector $\boldsymbol{x} \in \mathbb{R}^m$ can be understood as taking a linear combination of the column vectors of $\boldsymbol{A}$ using the elements of $\boldsymbol{x}$ as the coefficients. This is illustrated schematically below:
We can also view this geometrically:
In Panel A, we depict two column vectors of some matrix $\boldsymbol{A} \in \mathbb{R}^{3,2}$. In Panel B, we take a linear combination of the two column vectors of $\boldsymbol{A}$ according to the elements of some vector $\boldsymbol{x}$ thus producing the vector $\boldsymbol{Ax}$ (black vector) as shown in Panel C.
## Matrix-vector multiplication as evaluating a function between vector spaces
If we hold a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$ as fixed, this matrix maps vectors in $\mathbb{R}^n$ to vectors in $\mathbb{R}^m$. Making this more explicit, we can define a function $T : \mathbb{R}^n \rightarrow \mathbb{R}^m$ as: $T(\boldsymbol{x}) := \boldsymbol{Ax}$ where $T$ uses the matrix $\boldsymbol{A}$ to performing the mapping.
This is illustrated schematically below:
In fact, as we show in a later post, such a matrix-defined function is a linear function. Even more importantly, any linear function between finite-dimensional vector spaces is uniquely defined by some matrix.
## Appendix
Theorem 1: Given a matrix $\boldsymbol{A} \in \mathbb{R}^{m \times n}$ and vector $\boldsymbol{x} \in \mathbb{R}^n$, it holds that
$$\boldsymbol{Ax} = \begin{bmatrix}\boldsymbol{a}_{1,*} \cdot \boldsymbol{x} \\ \boldsymbol{a}_{2,*} \cdot \boldsymbol{x} \\ \vdots \\ \boldsymbol{a}_{m,*} \cdot \boldsymbol{x} \\ \end{bmatrix}$$
where $\boldsymbol{a}_{i,*}$ is the $i$th row-vector in $\boldsymbol{A}$.
Proof:
\begin{align*} \boldsymbol{A}\boldsymbol{x} &:= x_1\boldsymbol{a}_{*,1} + x_2\boldsymbol{a}_{*,2} + \dots + x_n\boldsymbol{a}_{*,n} \\ &= x_1 \begin{bmatrix}a_{1,1} \\ a_{2,1} \\ \vdots \\ a_{m, 1} \end{bmatrix} + x_2\begin{bmatrix}a_{1,2} \\ a_{2,2} \\ \vdots \\ a_{m, 2} \end{bmatrix} + \dots + x_n\begin{bmatrix}a_{1,n} \\ a_{2,n} \\ \vdots \\ a_{m, n} \end{bmatrix} \\ &= \begin{bmatrix}x_1a_{1,1} \\ x_1a_{2,1} \\ \vdots \\ x_1a_{m, 1} \end{bmatrix} + \begin{bmatrix}x_2a_{1,2} \\ x_2a_{2,2} \\ \vdots \\ x_2a_{m, 2} \end{bmatrix} + \dots + \begin{bmatrix}x_na_{1,n} \\ x_na_{2,n} \\ \vdots \\ x_na_{m, n} \end{bmatrix} \\ &= \begin{bmatrix} x_1a_{1,1} + x_2a_{1,2} + \dots + x_na_{1,n} \\ x_1a_{2,1} + x_2a_{2,2} + \dots + x_na_{2,n} \\ \vdots \\ x_1a_{m,1} + x_2a_{m,2} + \dots + x_na_{m,n} \\ \end{bmatrix} \\ &= \begin{bmatrix} \sum_{i=1}^n a_{1,i}x_i \\ \sum_{i=1}^n a_{2,i}x_i \\ \vdots \\ \sum_{i=1}^n a_{m,i}x_i \\ \end{bmatrix} \\ &= \begin{bmatrix}\boldsymbol{a}_{1,*} \cdot \boldsymbol{x} \\ \boldsymbol{a}_{2,*} \cdot \boldsymbol{x} \\ \vdots \\ \boldsymbol{a}_{m,*} \cdot \boldsymbol{x} \\ \end{bmatrix}\end{align*}
$\square$
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## Evaluate and estimate numerical square and cube roots
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Suppose that a shoemaker has determined that the optimal weight in ounces of a pair of running shoes is $\sqrt[4]{20000}$ . How many ounces would this be? Is there a way that you could rewrite this expression to make it easier to grasp? In this Concept, you'll learn how to simplify radical expressions like this one so that you can write them in multiple ways.
### Guidance
Radicals are the roots of values. In fact, the word radical comes from the Latin word “radix,” meaning “root.” You are most comfortable with the square root symbol $\sqrt{x}$ ; however, there are many more radical symbols.
A radical is a mathematical expression involving a root by means of a radical sign.
$\sqrt[3]{y}=x && \text{because} \ x^3=y && \sqrt[3]{27}=3, \ because \ 3^3=27\\\sqrt[4]{y}=x && \text{because} \ x^4=y && \sqrt[4]{16}=2 \ because \ 2^4=16\\\sqrt[n]{y}=x && \text{because} \ x^n=y &&$
Some roots do not have real values; in this case, they are called undefined .
Even roots of negative numbers are undefined .
$\sqrt[n]{x}$ is undefined when $n$ is an even whole number and $x<0$ .
#### Example A
• $\sqrt[3]{64}$
• $\sqrt[4]{-81}$
Solution:
$\sqrt[3]{64} = 4$ because $4^3=64$
$\sqrt[4]{-81}$ is undefined because $n$ is an even whole number and $-81<0$ .
In a previous Concept, you learned how to evaluate rational exponents:
$a^{\frac{x}{y}} \ where \ x=power \ and \ y=root$
This can be written in radical notation using the following property.
Rational Exponent Property: For integer values of $x$ and whole values of $y$ :
$a^{\frac{x}{y}}= \sqrt[y]{a^x}$
#### Example B
Rewrite $x^{\frac{5}{6}}$ using radical notation.
Solution:
This is correctly read as the sixth root of $x$ to the fifth power. Writing in radical notation, $x^{\frac{5}{6}}=\sqrt[6]{x^5}$ , where $x^5>0$ .
You can also simplify other radicals, like cube roots and fourth roots.
#### Example C
Simplify $\sqrt[3]{135}$ .
Solution:
Begin by finding the prime factorization of 135. This is easily done by using a factor tree.
$&\sqrt[3]{135}= \sqrt[3]{3 \cdot 3 \cdot 3 \cdot 5} = \sqrt[3]{3^3} \cdot \sqrt[3]{5}\\& 3 \sqrt[3]{5}$
### Guided Practice
Evaluate $\sqrt[4]{4^2}$ .
Solution: This is read, “The fourth root of four to the second power.”
$4^2=16$
The fourth root of 16 is 2; therefore,
$\sqrt[4]{4^2}=2$
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Radical Expressions with Higher Roots (8:46)
1. For which values of $n$ is $\sqrt[n]{-16}$ undefined?
1. $\sqrt{169}$
2. $\sqrt[4]{81}$
3. $\sqrt[3]{-125}$
4. $\sqrt[5]{1024}$
Write each expression as a rational exponent.
1. $\sqrt[3]{14}$
2. $\sqrt[4]{zw}$
3. $\sqrt{a}$
4. $\sqrt[9]{y^3}$
Write the following expressions in simplest radical form.
1. $\sqrt{24}$
2. $\sqrt{300}$
3. $\sqrt[5]{96}$
4. $\sqrt{\frac{240}{567}}$
5. $\sqrt[3]{500}$
6. $\sqrt[6]{64x^8}$
### Vocabulary Language: English Spanish
A mathematical expression involving a root by means of a radical sign. The word radical comes from the Latin word radix, meaning root.
Rational Exponent Property
Rational Exponent Property
For integer values of $x$ and whole values of $y$: $a^{\frac{x}{y}}= \sqrt[y]{a^x}$
A radical expression is an expression with numbers, operations and radicals in it.
Rationalize the denominator
Rationalize the denominator
To rationalize the denominator means to rewrite the fraction so that the denominator no longer contains a radical.
Variable Expression
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
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# What Is The Square Root Of 14?
• Square root √14 cannot be reduced, because it already is in its simplest form.
• All radicals are now simplified. The radicand no longer has any square factors.
## Determine The Square Root Of 14?
• The square root of fourteen √14 = 3.7416573867739
## How To Calculate Square Roots
• In mathematics, a square root of a number a is a number y such that y² = a, in other words, a number y whose square (the result of multiplying the number by itself, or y * y) is a. For example, 4 and -4 are square roots of 16 because 4² = (-4)² = 16.
• Every non-negative real number a has a unique non-negative square root, called the principal square root, which is denoted by √a, where √ is called the radical sign or radix. For example, the principal square root of 9 is 3, denoted √9 = 3, because 32 = 3 ^ 3 = 9 and 3 is non-negative. The term whose root is being considered is known as the radicand. The radicand is the number or expression underneath the radical sign, in this example 9.
• The justification for taking out the square root of any number is this theorem to help simplify √a*b = √a * √b. The square root of a number is equal to the number of the square roots of each factor.
## Mathematical Information About Numbers 1 4
• About Number 1. The number 1 is not a prime number, but a divider for every natural number. It is often taken as the smallest natural number (however, some authors include the natural numbers from zero). Your prime factorization is the empty product with 0 factors, which is defined as having a value of 1. The one is often referred to as one of the five most important constants of analysis (besides 0, p, e, and i). Number one is also used in other meanings in mathematics, such as a neutral element for multiplication in a ring, called the identity element. In these systems, other rules can apply, so does 1 + 1 different meanings and can give different results. With 1 are in linear algebra and vectors and one Einsmatrizen whose elements are all equal to the identity element, and refers to the identity map.
• About Number 4. Four is linear. It is the first composite number and thus the first non-prime number after one. The peculiarity of the four is that both 2 + 2 = 4 and 2 * 2 = 4 and thus 2^2 = 4. Four points make the plane of a square, an area with four sides. It is the simplest figure that can be deformed while keeping it's side lengths, such as the rectangle to parallelogram. Space let's us arrange equidistantly a maximum of four points. These then form a tetrahedron (tetrahedron), a body with four identical triangular faces. Another feature of the four is the impossibility of an algebraic equation of higher degree than four square roots using simple arithmetic and basic operations dissolve.
## What is a square root?
A square root of a number is a number that, when it is multiplied by itself (squared), gives the first number again. For example, 2 is the square root of 4, because 2x2=4. Only numbers bigger than or equal to zero have real square roots. A number bigger than zero has two square roots: one is positive (bigger than zero) and the other is negative (smaller than zero). For example, 4 has two square roots: 2 and -2. The only square root of zero is zero. A whole number with a square root that is also a whole number is called a perfect square. The square root radical is simplified or in its simplest form only when the radicand has no square factors left. A radical is also in simplest form when the radicand is not a fraction.
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# Convert 343 as a power of 7
Grade:12th pass
## 1 Answers
Pawan Prajapati
askIITians Faculty 60787 Points
2 years ago
Here, we have to use the concept of the factorization. Factorization is the process in which a number is written in the forms of its small factors which on multiplication give the original number. We will factorize the given number and then we will write it in the exponent form. Complete step-by-step answer: Given number is 343. Now we will factorize the given number into its small factors. Factors are the smallest numbers with which the given number is divisible and their multiplication will give the original number. Therefore, the number 343 in terms of its factors is written as 343=7×7×7 Now we will write it in the exponent form or power form. An exponent shows how many times a number is being multiplied to itself. In our case it is three as number 7 is being multiplied with itself three times. Therefore, we get ⇒343=73 Hence, 343 is equal to 73 or we can write it as 7 raised to the power 3. Note: Here we found out the exponent of 7 is 3, therefore we can say 343 is the cube of 7. A Cube of a number is defined as the number which is being multiplied by itself three times. We should not get confused between cubes with the square. The square of a number is defined as the number which is being multiplied by itself two times. Square of a number =a2 Cube of a number =a3 We generally pronounce it as a number raised to the power two or three according to the condition.
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Systems of Linear Equations and Gaussian Elimination
What is a System of Linear Equations?
A linear equation in n unknowns x1, x2, … , xn is an equation of the form:
a1 x1 + a2 x2 + … + an xn = b
where a1, a2, … , an and b are constants.
The name linear comes from the fact that such an equation in two unknowns or variables represents a straight line. A set of such equations is called a system. An example of a system of three linear equations in the three unknowns x, y and z is:
Some lessons to learn from graphing 2 equations in 2 unknowns
The graphical method is not very useful as a computational tool but it is useful for visualizing concepts such as the uniqueness of the solution or the meaning of inconsistent or redundant systems. Consider the following system of two linear equations in two unknowns:
In this method we simply draw graphs of the equations as we have done to the right. Notice that the graph of each equation is a straight line. (This is the hallmark of a linear system. There are no curves, only straight lines.)
Any point on one line is a solution of one equation and any point on the other line is a solution of the other equation. The point where the lines cross {x=3.6,y=0.4} is the solution that satisfies both equations simultaneously. Notice that the solution is unique. This is because the lines are straight and there is only one point where they can cross.
A system of linear equations with a unique solution is the "normal" situation. However it is possible to have a system of equations with no solution. Such a system of equations is called inconsistent. It is often the result of an inaccurate or incorrect analysis of the physical system being described by the system of equations.
Consider the following system of two linear equations in two unknowns:
This system of equations is inconsistent because there is no way that x+y can equal 2 and 4 at the same time. The figure to the right shows that the graph of this system consists of two parallel lines that never cross. Thus there is no solution.
It is also possible to have a system of equations with an infinite number of solutions. Such a system of equations is called redundant. It is often the result of an incomplete analysis of the physical system.
Consider the following system of equations:
This system is redundant because the second equation is equivalent to the first one. The graph consists of two lines that lie on top of one another. They 'cross' at an infinite number of points, so there are an infinite number of solutions.
To summarize, a linear system with 2 unknowns must have at least 2 equations to get a unique solution. Having 1 equation is not enough, because 1 equation in 2 unknowns is represented by an entire line. Having 2 equations is exactly enough, as long as they are not redundant or inconsistent. Having 3 (or more) equations is too many. The third equation must be either redundant or inconsistent.
These ideas can be generalized to linear systems of equations with more unknowns:
A linear equation in n variables represents a "hyperplane" in an n dimensional space. A linear system of equations with n unknowns must have at least n equations to get a unique solution. Having any fewer is not enough; the solution will not be unique. Having n equations is exactly enough, as long as they are not redundant or inconsistent. Having any more than n equations is too many; the system will be either redundant or inconsistent.
The Augmented Matrix
We will represent a system of equations by a rectangular array of numbers called an augmented matrix. Here is the augmented matrix for the above example:
Some Terminology:
• The entries in the augmented matrix are called elements.
• Rows run across the matrix.
• Columns run down the matrix.
• The diagonal of the matrix is the set of elements that starts at the top, left corner and runs diagonally down and to the right. The diagonal of the above matrix consists of the numbers 4, 1 and 2.
• Any elements in position a are said to lie above the diagonal, and any in position b are below the diagonal:
Keep in mind the following:
• the i-th row of the augmented matrix represents the i-th equation
• the j-th column (to the left of the vertical line) represents the (coefficients of the) j-th variable or unknown
• the vertical line represents the equal signs
Elementary Row Operations
Recall that an equation such as:
7(x−4)=14,
may be solved for x by applying the following operations:
• dividing both sides of the equation by the same value, namely 7, to yield x−4=2,
• then adding the same quantity to both sides, namely 4, to yield x=6.
The solution is x=6, as can be verified by substituting it back into the original equation and finding the identity 14=14.
Similarly, the solution of a system of equations is any set of values of all of the variables that satisfies all of the equations simultaneously. For example the system:
has the solution:
{x = 7, y = 5, z = 3}.
This can be verified by substituting these values into all three of the equations and producing three identities.
A system of equations can be solved by generalizing the two operations described above and observing that the solution of a system of equations is not changed by:
• dividing an both sides of an equation by a constant, or
• subtracting a multiple of one equation from another equation.
These same operations can be applied to the rows of an augmented matrix, since each row just represents an equation. They are then called Elementary Row Operations.
The Elementary Row Operations (E.R.O.'s) are:
• E.R.O.#1: Choose a row of the augmented matrix and divide (every element of) the row by a constant.
Example:
The notation means to divide the first row of the augmented matrix by 2 to produce the new augmented matrix.
• E.R.O.#2: Choose any row of the augmented matrix and subtract a multiple of any other row from it (element by element).
Example:
The notation means to take row 2 and subtract 3 times row 1 from it to produce the new augmented matrix.
We will apply the E.R.O.'s in a certain sequence (the Gaussian elimination method, described below) to transform the augmented matrix into triangular echelon form. In this form the augmented matrix has 1's on the diagonal, 0's below the diagonal and any numbers above the diagonal. For example, the augmented matrix:
transformed into the triangular echelon form is:
This new augmented matrix represents the system of equations:
It is solved by back-substitution. Substituting z = 3 from the third equation into the second equation gives y = 5, and substituting z = 3 and y = 5 into the first equation gives x = 7. Thus the complete solution is:
{x = 7, y = 5, z = 3}.
Gaussian Elimination
In the Gaussian Elimination Method, Elementary Row Operations (E.R.O.'s) are applied in a specific order to transform an augmented matrix into triangular echelon form as efficiently as possible.
This is the essence of the method: Given a system of m equations in n variables or unknowns, pick the first equation and subtract suitable multiples of it from the remaining m-1 equations. In each case choose the multiple so that the subtraction cancels or eliminates the same variable, say x1. The result is that the remaining m-1 equations contain only n-1 unknowns (x1 no longer appears).
Now set aside the first equation and repeat the above process with the remaining m-1 equations in n-1 unknowns.
Continue repeating the process. Each cycle reduces the number of variables and the number of equations. The process stops when either:
• There remains one equation in one variable. In that case, there is a unique solution and back-substitution is used to find the values of the other variables.
• There remain variables but no equations. In that case there is no unique solution.
• There remain equations but no variables (ie. the lowest row(s) of the augmented matrix contain only zeros on the left side of the vertical line). This indicates that either the system of equations is inconsistent or redundant. In the case of inconsistency the information contained in the equations is contradictory. In the case of redundancy, there may still be a unique solution and back-substitution can be used to find the values of the other variables.
Examples of all these possibilities are given below.
Algorithm for Gaussian Elimination:
Transform the columns of the augmented matrix, one at a time, into triangular echelon form. The column presently being transformed is called the pivot column. Proceed from left to right, letting the pivot column be the first column, then the second column, etc. and finally the last column before the vertical line. For each pivot column, do the following two steps before moving on to the next pivot column:
1. Locate the diagonal element in the pivot column. This element is called the pivot. The row containing the pivot is called the pivot row. Divide every element in the pivot row by the pivot (ie. use E.R.O. #1) to get a new pivot row with a 1 in the pivot position.
2. Get a 0 in each position below the pivot position by subtracting a suitable multiple of the pivot row from each of the rows below it (ie. by using E.R.O. #2).
Upon completion of this procedure the augmented matrix will be in triangular echelon form and may be solved by back-substitution.
Example: Use Gaussian elimination to solve the system of equations:
Solution: Perform this sequence of E.R.O.'s on the augmented matrix. Set the pivot column to column 1. Get a 1 in the diagonal position (underlined):
Next, get 0's below the pivot (underlined):
Now, let pivot column = second column. First, get a 1 in the diagonal position:
Next, get a 0 in the position below the pivot:
Now, let pivot column = third column. Get a 1 in the diagonal position:
This matrix, which is now in triangular echelon form, represents:
It is solved by back-substitution. Substituting z = 3 from the third equation into the second equation gives y = 5, and substituting z = 3 and y = 5 into the first equation gives x = 7. Thus the complete solution is:
{x = 7, y = 5, z = 3}.
Example of Gaussian Elimination Applied to a Redundant System of Linear Equations
Use Gaussian elimination to put this system of equations into triangular echelon form and solve it if possible:
Solution: Perform this sequence of E.R.O.'s on the augmented matrix. Set the pivot column to column 1. There is already a 1 in the pivot position, so proceed to get 0's below the pivot:
Now, set the pivot column to the second column. First, get a 1 in the diagonal position:
Next, get a 0 in the position below the pivot:
Now, set the pivot column to the third column. The first thing to do is to get a 1 in the diagonal position, but there is no way to do this. In fact this matrix is already in triangular echelon form and represents:
This system of equations can't be solved by back-substitution because we have no value for z. The last equation merely states that 0=0. There is no unique solution because z can take on any value.
In general, one or more rows of zeros at the bottom of an augmented matrix that has been put into triangular echelon form indicates a redundant system of equations.
Example of Gaussian Elimination Applied to an Inconsistent System of Linear Equations
Use Gaussian elimination to put this system of equations into triangular echelon form and solve it if possible:
Solution: Perform this sequence of E.R.O.'s on the augmented matrix. Set the pivot column to column 1. There is already a 1 in the pivot position, so proceed to get 0's below the pivot:
Now, set the pivot column to the second column. There is already a 1 in the pivot position, so proceed to get 0's below the pivot:
Now, set the pivot column to the third column. The first thing to do is to get a 1 in the diagonal position, but there is no way to do this. In fact this matrix is already in triangular echelon form and represents:
This system of equations is inconsistent and has no solution. The last equation states a contradiction, namely 0 = −50.
In general, an augmented matrix which has been put into triangular echelon form and which contains one or more bottom rows consisting of all zeros to the left of the vertical line and a non-zero number to the right indicates an inconsistent system of equations with no solution.
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Several examples with detailed solutions and exercises with answers on the addition of fractions, are presented.
A fraction represents a part of a whole. Take a whole divide into b equal parts and take a parts; this is represented by the fraction $$\dfrac{a}{b}$$. a is called the numerator and b is called the denominator and must be non zero.
Examples of fractions
$$\dfrac{2}{3}$$ , $$\dfrac{3}{4}$$ , and $$\dfrac{7}{2}$$ are examples of fractions.
Examples of Mixed Numbers
A combination of a whole number and a fraction is called a mixed number.
$$2\dfrac{1}{3}$$ , $$5\dfrac{3}{5}$$ , and $$1\dfrac{7}{2}$$ are examples of mixed numbers.
NOTES:
1) Any fraction with denominator equal to 1 is equal to its numerator.
Examples of Fractions with Denominator Equal to 1
$$\dfrac{3}{1} = 3$$ , $$\dfrac{10}{1} = 10$$
2) Fraction with the same denominator are called like fractions.
Examples of like Fractions
$$\dfrac{2}{5}$$ , $$\dfrac{ - 3}{5}$$ , $$\dfrac{21}{5}$$ are like fractions
3) Fraction with different denominators are called unlike fractions.
Examples of unlike Fractions
$$\dfrac{2}{6}$$ , $$\dfrac{ - 3}{7}$$ , $$\dfrac{21}{2}$$ are unlike fractions
4) Any fraction with denominator equal to zero is undefined.
Examples of Fractions with Denominator Equal to Zero
$$\dfrac{2}{0}$$ is undefined because it deniminator is equal to zero..
A fraction calculator that helps you develop further the skills of how to reduce, add and multiply fractions with steps is included.
## Adding Fractions: Examples with Solutions
### 1. Add fractions with same denominators (like fractions)
Example 1
$$\dfrac{2}{3} + \dfrac{4}{3}$$
Solution to Example 1
To add fractions with the same denominator, you add the numerators, keep the same denominator
$$\dfrac{2}{3} + \dfrac{4}{3} = \dfrac{2+4}{3} = \dfrac{6}{3}$$
3 is a common factor to the numerator 6 and the denominator 3, hence we reduce (or simplify) the fraction by dividing the numerator and denominator by the common factor 3 as follows:
$$= \dfrac{6\div3}{3\div3} = \dfrac{2}{1} = 2$$
Example 2
$$\dfrac{12}{14} + \dfrac{34}{14}$$
Solution to Example 2
The two fractions have the same denominator, hence we add the numerators and keep the common denominator as follows:
$$\dfrac{12}{14} + \dfrac{34}{14} = \dfrac{12+34}{14} = \dfrac{46}{14}$$
divide numerator and denominator by $$2$$ and simplify
$$= \dfrac{46\div 2}{14 \div 2} = \dfrac{23}{7}$$
### 2. Add fractions with different denominators (unlike fractions)
Example 3
$$\dfrac{2}{9} + \dfrac{4}{6}$$
Solution to Example 3
We first need to find the lowest common multiple (LCM) of the two denominators 9 and 6 by factoring into prime factors.
$$9 = 3 \times 3$$
$$6 = 2 \times 3$$
The LCM of 9 and 6 $$= 3 \times 3 \times 2 = 18$$
We next convert the two given fractions so that they have common denominator equal to the LCM = 18. The denominator of the fraction $$\dfrac{2}{9}$$ is 9 and we need to multiply numerator and denominator by 2 in order to change the denominator to 18
$$\dfrac{2}{9} = \dfrac{2 \times 2}{9 \times 2} = \dfrac{4}{18}$$
The denominator of the fraction $$\dfrac{4}{6}$$ is 6 and we need to multiply numerator and denominator by 3 in order to change the denominator to 18
$$\dfrac{4}{6} = \dfrac{4 \times 3}{6 \times 3} = \dfrac{12}{18}$$
Now that we have converted the two fractions so that they have common denominator, we can easily add them as follows
$$\dfrac{2}{9} + \dfrac{4}{6} = \dfrac{4}{18} + \dfrac{12}{18} = \dfrac{16}{18}$$
The numerator 16 and denominator 18 have a common factor 2, hence we divide numrator and denominator by 2 to reduce (simplify) the fraction as follows:
$$= \dfrac{16\div2}{18\div2} = \dfrac{8}{9}$$
Example 4
Add, simplify and express the final answer as a fraction and as a mixed number.
$$\dfrac{23}{15} + \dfrac{27}{55}$$
Solution to Example 4
Find the LCM of the denominators 15 and 55
$$15 = 3 \times 5$$
$$55 = 5 \times 11$$
The LCM of 5 and 55 $$= 3 \times 5 \times 11 = 165$$
We convert the given fractions so the they have common denominator equal to the LCM = 165. The denominator of the fraction $$\dfrac{23}{15}$$ is 15 and both numerator and denominator need to be multiplied by 11 in order to change the denominator to 165
$$\dfrac{23}{15} = \dfrac{23\times11}{15\times11} = \dfrac{253}{165}$$
The second fraction $$\dfrac{27}{55}$$ is converted to one with the denominator equal to 165 by multiplying numerator and denominator by 3 in order to change the denominator to 165
$$\dfrac{27}{55} = \dfrac{27\times3}{55\times3} = \dfrac{81}{165}$$
$$\dfrac{23}{15} + \dfrac{27}{55} = \dfrac{253}{165} + \dfrac{81}{165} = \dfrac{334}{165}$$
Because the numerator is larger than the denominator, it is possible to write the above fraction as a mixed number using the long division as follows
$$= \dfrac{334}{165} = 2 + \dfrac{4}{165} = 2 \dfrac{4}{165}$$
2 is the quotient and 4 is the remainder of the divison 334 by 165.
Example 5
$$2 + \dfrac{1}{5}$$
Solution to Example 5
We first convert 2 into a fraction
$$2 = \dfrac{2}{1}$$
The common denominator will be 5
$$2 = \dfrac{2}{1} = \dfrac{10}{5}$$
$$2 + \dfrac{1}{5} = \dfrac{10}{5} + \dfrac{1}{5} = \dfrac{11}{5}$$
Example 6
Add the mixed numbers and simplify
$$4\dfrac{5}{6} + 5\dfrac{7}{8}$$
Solution to Example 6
We first rewrite the two fractions included in the mixed numbers to the same denominator. The LCM of 6 and 8 is 24. Hence
$$\dfrac{5}{6} = \dfrac{20}{24}$$ and $$\dfrac{7}{8} = \dfrac{21}{24}$$
We now substitute the fractions by those with equal denominator
$$4\dfrac{5}{6} + 5\dfrac{7}{8} = 4\dfrac{20}{24} + 5\dfrac{21}{24}$$
We now add the whole numbers and the fractions
$$= (4 + 5) + (\dfrac{20}{24} + \dfrac{21}{24}) = 9 + \dfrac{41}{24}$$
We now rewrite the improper fraction $$\dfrac{41}{24}$$ as a mixed number.
$$\dfrac{41}{24} = 1\dfrac{17}{24}$$
Substitute $$\dfrac{41}{24}$$ by $$1\dfrac{17}{24}$$ in the expression $$9 + \dfrac{41}{24}$$ to obtain the final answer.
$$= 9 + 1\dfrac{17}{24} = 9 + 1 + \dfrac{17}{24} = 10\dfrac{17}{24}$$
Example 7
$$\dfrac{5}{-6} + \dfrac{7}{6}$$
Solution to Example 7
We first change the fraction $$\dfrac{5}{-6}$$ so that its denominator is positive.
$$\dfrac{5}{-6} = \dfrac{-5}{6}$$
$$\dfrac{5}{-6} + \dfrac{7}{6} = \dfrac{-5}{6} + \dfrac{7}{6} = \dfrac{-5+7}{6} = \dfrac{2}{6}$$
$$\dfrac{1}{3}$$
## Execises
Add, simplify and express the answer as a proper fraction or a mixed number.
1. $$\dfrac{3}{7} + \dfrac{6}{7}$$
2. $$\dfrac{2}{3} + \dfrac{5}{7}$$
3. $$4\dfrac{23}{30} + 1\dfrac{8}{25}$$
4. $$5 - \dfrac{2}{6}$$
5. $$\dfrac{11}{8} + \dfrac{3}{-8}$$
1. $$1\dfrac{2}{7}$$
2. $$1\dfrac{8}{21}$$
3. $$6\dfrac{13}{150}$$
4. $$4\dfrac{2}{3}$$
5. $$1$$
|
### Card Counting: Simple Math Part 2
Yet, at the same time, both possible draws from an un-shuffled deck are dependent on each other. This is known as joint probability and, because of it, math nerds set the probability of drawing two cards in succession differently than they would for drawing two of them separately. For situations where the individual propositions are NOT mutually exclusive but still dependent, the probability of one event happening, say drawing a King and an Ace in succession, is the probability of drawing a King, times the probability of drawing the Ace after the King is subtracted from the deck. So, the math works out something like this:
The probability of drawing a King, if there are four Kings in a 52-card deck is 4:52, or – simplified by dividing both numbers by four – 1:12 (i.e., for every 12 cards you draw in a well shuffled deck, one of them is bound to be a King).
The probability of drawing an Ace, if there are four Aces in a 52-card deck and one of the cards (the King) has been removed is 4:51.
The probability of drawing the King, then drawing the Ace right after it is 1:12 x 4:52 or:
Taken to an extreme, it is easy to see how this concept plays out in card counting. If, for instance, every 2,3,4,5,6,7,8 and 9 has been removed, the probability of drawing an Ace is 4:20 or 1:5, and the probability of drawing a ten-point card, like a King, after that Ace has been removed is 16:19. What’s more, the probability of doing both and landing yourself a cool, soft natural is 16:95, which when you divide 16 by 95 to determine the percentage, comes out to about a 17 percent likelihood. Now compare that to the 1:153, which is the equivalent of a 0.65 percent likelihood.
That in a nutshell is what card counting literally is: using the laws of probability to get one over on the house. But as for how counting happens, that’s another story altogether.
Essentially, counting works according to count systems. The number of decks isn’t exactly important because the number of trump cards remains proportionate to the whole shoe. Determining the number of decks though is key, and when we’re talking upwards of five or more, even those ritzy MIT kids can’t do all the math in their heads. What generations of them have done, however, is figure out short-hand methods for recognizing when a deck or shoe is and is not loaded, as well as the best ways to insert themselves into a table when the pickins are ripe.
The most basic of these systems, and the one featured in “21,” is known as “Hi-Lo” and was pioneered by the granddaddy of all counters, Ed Thorp. In this system, you assign one positive point to every 2, 3, 4, 5 and 6 you see, no points to 7-9 and one negative point to each 10, J, Q, K and A. Long story short, when the count is high, “winner-winner-chicken-dinner.” When it’s low, “loser-loser….” Eh, whatever rhymes with loser.
Needless to say, over the past 50-odd years, “single-level” counts have become obsolete, and counting gurus have devised other plans to keep one step ahead of casinos’ G-men. The following are the most commonly attempted counts (though chances are, if it’s something you’re able to find on a gambling e-zine like GP, casinos already know how to spot it):
Card Strategy 2 3 4 5 6 7 8 9 10 J Q K A Wizard Ace/Five 0 0 0 1 0 0 0 0 0 0 0 0 −1 KO 1 1 1 1 1 1 0 0 −1 −1 −1 −1 −1 Hi-Lo 1 1 1 1 1 0 0 0 −1 −1 −1 −1 −1 Hi-Opt I 0 1 1 1 1 0 0 0 −1 −1 −1 −1 0 Hi-Opt II 1 1 2 2 1 1 0 0 −2 −2 −2 −2 0 Zen Count 1 1 2 2 2 1 0 0 −2 −2 −2 −2 −1 Omega II 1 1 2 2 2 1 0 −1 −2 −2 −2 −2 0
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# Three baskets with numbered balls
Three baskets and some balls are given to you. You are supposed to put these balls into these baskets by writing some positive numbers on the balls. But:
• The numbers you are supposed to write on them have to be positive integer number,
• In each basket, the numbered balls have to be distinct. (That also means you can put the same number in different baskets actually.)
• The sum of the numbers on the balls in all baskets has to be at most $31$.
• You do not have to put balls in every basket.
What is the maximum value of the multiplication of the numbers written on the balls?
• The rule: "You may not put any ball in any basket" is confusing, can you clarify this? Oct 11, 2017 at 19:43
• To restate the problem: you are to maximize the product of a series of integers such that: (1) all integers are positive; (2) their sum is <= 31; (3) no integer may appear in the series more than 3 times. Oct 12, 2017 at 0:06
To make the product of the ball values as large as possible one needs to split up the 'budget' of $31$ into as many pieces as possible greater than $1$. Why $> 1$? Because balls with $1$ on them contribute to the sum but don't increase the product. Why as many as possible? Because $2 \cdot N \ge 2 + N$ (for $N \ge 2$), i.e. the total product will increase more by an additional factor of $2$ than by one of the factors being increased by $2$.
Edit: With regard to ffao's comment, I should add that there is exactly 1 case where the maximum number of factors is not the optimum, namely $2 \cdot 2 \cdot 2 < 3 \cdot 3$. In the final solution one will see that further 'finetuning' by replacing three $2$s by two $3$s is not possible, because there are already three $3$s (one for each basket) used.
So let's start with filling $2$s into each basket. The product is $8$ so far and the sum is $6$. Continuing with $3$s in each basket we have a product of $216$ and a sum of $15$. Next would be 3 $4$s, resulting in a product of $13824$ and a sum of $27$. But as the rest 'budget' to fill up to $31$ is just $4$, this can not be used to write another number on a ball. Instead one needs to use this to increase the existing numbers.
This increase should be spread among as many numbers as possible because $\left( N_1 + 1 \right) \left( N_2 + 1 \right) > N_1 \left( N_2 + 2 \right)$ for $N_1$, $N_2 \ge 2$. So let's first promote two $4$s to $5$s. Then we have a rest budget of $2$ from which we can promote a $2$ to $4$.
In total we then have in the baskets:
1. basket: $2, 3, 4, 5$
2. basket: $2, 3, 4, 5$
3. basket: $3$
Sum is exactly $31$ and product is $43200$.
• "As many as possible" needs more justification. For example, 3*3 > 2*2*2.
– ffao
Oct 11, 2017 at 21:29
• @ffao Thank you for pointing that out. I have added this to my answer. Oct 11, 2017 at 21:54
• The sum of the baskets shown is 36. Did you mean:$$(1): 3, 4, 5\\(2): 2, 3, 5\\(3): 2, 3, 4$$ That matches your description and has the indicated sum and product. Oct 11, 2017 at 23:07
• @PaulSinclair You're right. One $5$ was too much. Oct 12, 2017 at 6:18
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# Equal Chords and their Distances from Center
Last updated date: 17th May 2024
Total views: 109.5k
Views today: 3.09k
## Equal Chords and their Distance from Center: Preface
Well, we see many round objects in daily life like coins, clocks, wheels, bangles, and many more. In this article, we will learn about the equal chords theorem i.e. the equal chords of a circle are equidistant from the center of the circle. And then will learn its converse too. After that, we discussed the theorem regarding the intersection of equal chords. At last, we will learn the diameter is the largest chord of the circle and we will solve examples to understand the concepts more easily.
## Perpendicular Bisector of a chord
Perpendicular Bisector of the Chord
Here, two points are joined to form a line segment which we call as Chord of the circle. The longest chord of a circle is called the diameter. When a line crosses the line at ${90^ \circ }$(perpendicular) and cuts in half (bisector) is called the perpendicular bisector of the chord. Also, this perpendicular bisector of the chord passes through the center of the circle. To represent a perpendicular line is ‘$\bot$’. Thus, the perpendicular drawn from the center of a circle to a chord bisects the chord.
## Chords and their Distance from the Center
As we know that there are infinite numbers of points on a line segment. We take a circle with center O having chord AB as shown below:
Chord and their Distance from the Center
Next, to find the distance between the chord and the center O, if we join the infinite points on the line segment AB with O then we will get infinite line segments of different lengths. As we can see, line ON is perpendicular to AB and it has the shortest length i.e. $ON \bot AB$. This means the shortest length ON is the distance between O and AB. Thus we can say that the length of a perpendicular from a point to a line is the distance between the point from the line.
## Equal Chords and their Distance from the Center Theorem
Statement: Equal chords of a circle (or of congruent circles) are equidistant from the center (or centers).
Given: We have a circle with center O. AB and CD are chords that are equal i.e. AB = CD. Also, OX and OY are perpendiculars to AB and CD respectively
To prove: OX = OY
Equal Chords and their Distance from the Center Theorem
Proof:
Draw OA and OC as shown in the figure.
As we know, perpendicular from the center to the chord bisects the chord.
Since, $OX \bot AB$
$\Rightarrow AX = BX = \frac{{AB}}{2}$ -------- (1)
Similarly for $OY \bot CD$
$\Rightarrow CY = DY = \frac{{CD}}{2}$ ---------- (2)
Next, given that AB = CD
$\Rightarrow \frac{{AB}}{2} = \frac{{CD}}{2}$
Using equations (1) and (2), we will get,
$\Rightarrow AX = CY$ -------- (3)
In $\Delta AOX$ and $\Delta COY$,
As both are right-angled triangles i.e. ${90^ \circ }$
$\Rightarrow \angle OXA = \angle OYC$
$\Rightarrow OA = OC$ (Both are the radii of the circle)
$\Rightarrow AX = CY$ (From (1))
So, RHS rule we can say that, $\Delta AOX \cong \Delta COY$
$\Rightarrow OX = OY$ (From CPCT rule)
Hence it is proved that equal chords of a circle are equidistant from the center.
## The Converse of this Above Theorem
Statement: Chords equidistant from the center of a circle are equal in length.
Given: We have a circle with center O. Also, AB and CD are chords of the circle where OX and OY are the distances from the center i.e. $OX \bot AB$ ($AX = BX = \frac{{AB}}{2}$) and $OY \bot CD$ ($CY = DY = \frac{{CD}}{2}$) respectively. And, OX = OY.
To prove: The chords are equal i.e. AB = CD
Converse of Equal Chords and their Distance from the Center Theorem
Proof:
Draw OA and OC as shown in the figure.
In $\Delta AOX$ and $\Delta COY$
As both are right-angled triangles i.e. ${90^ \circ }$
$\Rightarrow \angle OXA = \angle OYC$
$\Rightarrow OA = OC$ (Both are the radii of the circle)
$\Rightarrow OX = OY$ (Given)
So, from RHS rule we can say that, $\Delta AOX \cong \Delta COY$
$\Rightarrow AX = CY$ (From CPCT rule) ------- (1)
Next, perpendiculars from the center to the chord bisect the chord of the circle.
So, for the chord AB:
$\therefore$ $OX \bot AB$
$\Rightarrow$ X bisects AB
$\Rightarrow AX = BX = \frac{{AB}}{2}$ ------ (2)
And, for the chord CD:
$\therefore$$OY \bot CD$
$\Rightarrow$ Y bisects CD
$\Rightarrow CY = DY = \frac{{CD}}{2}$ ------ (3)
Lastly, from (1), we have:
$\therefore AX = CY$
$\Rightarrow \frac{{AB}}{2} = \frac{{CD}}{2}$ (From (2) and (3))
$\Rightarrow AB = CD$
Hence it is proven.
## The Intersection of Equal Chords
Statement: If two equal chords of a circle intersect within the circle. Prove that, the segments of one chord are equal to the corresponding segments of another chord.
Given: AB = CD and both the chords intersect at point E in the circle.
To prove: AE = ED and CE = BE
The Intersection of Equal Chords
Proof:
First, join points O and E.
Then, draw a line OM from the center to the chord AB.
$\Rightarrow OM \bot AB$
$\Rightarrow$M bisects AB, which means M is the midpoint on AB.
$\Rightarrow AM = MB$ ---------------- (1)
Similarly, draw a line ON from the center to the chord CD.
$\Rightarrow ON \bot CD$
$\Rightarrow$N bisects CD, which means N is the midpoint on CD.
$\Rightarrow CN = ND$ ----------------- (2)
Now, we will see the proof:
As we have given AB = CD
$\Rightarrow \frac{1}{2}AB = \frac{1}{2}CD$
$\Rightarrow AM = ND$
$\Rightarrow MB = CN$
$\Rightarrow MB = AM = ND = CN$ ------------------- (3)
In $\Delta OME$ and $\Delta ONE$
$\Rightarrow \angle OME = \angle ONE = {90^ \circ }$ ( By construction)
$\Rightarrow OE = OE$ (Common side)
As the distance from the center is equal if the chords (AB=CD) are equal, we have
$\Rightarrow OM = ON$
$\therefore \Delta OME \cong \Delta ONE$ (RHS rule)
$\Rightarrow ME = EN$ ( CPCT rule) ----- (4)
Next, from (1) and (3), we will get,
$\Rightarrow AM = ND$
Add (4) on both sides, we will get,
$\Rightarrow AM + ME = EN + ND$
$\Rightarrow AE = DE$ ---------- (5)
And, subtract (5) from AB=CD
$\Rightarrow AB - AE = CD - DE$
$\Rightarrow BE = CE$
Hence, it is proven.
## Interesting Facts
• The radius of the circle bisects the chord at ${90^ \circ }$.
• When two radii join the two ends of a chord, they form an isosceles triangle.
• The converse of intersecting equal chords: Two intersecting chords of a circle make equal angles with the diameter that passes through their point of intersection then the chords are equal.
• When a line is drawn through the center of a circle that bisects the chord is perpendicular to the chord.
## Important Questions
1. A circular road of radius 20m. Three girls A, S and D are sitting at equal distances on its boundary, each having a toy telephone in their hands to talk to each other. Find the length of the string of each phone.
Solution:
We are given that, OS = 20m and AS = SD = DA =?
From this, we can say that $\Delta ASD$ is an equilateral triangle.
$\Rightarrow OS = OA = OD = 20$
As AM is the median and altitude of $\Delta ASD$
$\Rightarrow$M is the midpoint of SD --------- (1)
And point O is the centroid.
$\therefore$O divides AM in 2:1
$\Rightarrow \frac{{OA}}{{OM}} = \frac{2}{1}$
$\Rightarrow \frac{{20}}{{OM}} = \frac{2}{1}$
$\Rightarrow OM = 10m$
Next, from this we can see that $\Delta OMS = {90^ \circ }$
$\Rightarrow O{S^2} = O{M^2} + S{M^2}$
$\Rightarrow {(20)^2} = {(10)^2} + S{M^2}$
$\Rightarrow S{M^2} = 300$
$\Rightarrow SM = 10\sqrt 3 m$
From (1) we can say that,
$\Rightarrow SM = DM = 10\sqrt 3 m$
$\Rightarrow SD = 2SM = 2 \times 10\sqrt 3$
$\Rightarrow SD = 20\sqrt 3$m
Thus, we have $AS = SD = AD = 20\sqrt 3$m.
2. In this circle with center O, the radius is 8 cm and XY=PQ=12 cm. Find the lengths of OS and OT.
Solution:
According to the given information, we will get,
A Perpendicular on the Chord
Given: OY=OQ=8 cm ------- (1)
And, XY = 12 cm
According to the figure,
$\Rightarrow XS + SY = SY$
Also, the perpendicular drawn from the center to the chord bisects the chord
$\Rightarrow XS = SY$
Next, $SY + SY = XY$
$\Rightarrow SY = \frac{1}{2} \times XY$
$\Rightarrow SY = \frac{1}{2} \times 12$
$\Rightarrow SY = 6$ cm ----- (2)
In $\Delta OSY$, using Pythagoras Theorem,
$\Rightarrow O{Y^2} = S{Y^2} + O{S^2}$
$\Rightarrow O{S^2} = O{Y^2} - S{Y^2}$
$\Rightarrow O{S^2} = {8^2} - {6^2}$
$\Rightarrow O{S^2} = 28$
$\Rightarrow OS = 2\sqrt 7$cm -------- (3)
According to the question, XY=PQ
$\Rightarrow OS = OT$ ( Equal chords of a circle are equidistant from the center)
$\Rightarrow OT = 2\sqrt 7$ cm -------- (4)
Thus, the length of OS and OT is $2\sqrt 7$ cm.
3. If a line intersects two concentric circles (circles with the same center) with center O at A, B, C and D. Prove that AB = CD.
Solution:
We have taken point E which joins from the center. According to the given information, we get the following figure:
Two concentric circles
As $OE \bot AD$
$\Rightarrow AE = ED$ ----- (1)
And, $OE \bot BC$
$\Rightarrow BE = EC$ ----- (2)
From (1) – (2), we will get,
$\Rightarrow AE - BE = ED - EC$
$\Rightarrow AB = CD$
Thus, it is proved.
## Conclusion
The article summarizes the theorem of equal chords and their distances from the center i.e. equal chords are equidistant from the center and then it's converse too. Then we learned the theorem of the intersection of equal chords too. Also, this article has solved examples to understand the concepts easily and see that the largest chord of the circle is known as the diameter.
## Practice Questions
1. We have two circles of radii 5 cm and 3 cm which intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.
A. 3 cm
B. 6 cm
C. 4 cm
2. Equal chords of a circle make equal angles at the center of a circle. Is this statement true?
A. Yes
B. No
C. Can’t determined
1) B
2) A
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## FAQs on Equal Chords and their Distances from Center
1. Why is the diameter the longest chord of a circle?
A circle has an infinite number of chords. When the chords move closer to the center, then the length of the chords increases. This means the chord nearer to the center of a circle is the longest chord. So the chord which is away from the center is the least. Thus, the longest chord of a circle is called as diameter which passes through the center of a circle.
2. Does a circle has an infinite number of equal chords?
A circle has only a finite number of equal chords.
3. Can we say that every chord is the radius of the circle?
We know that the line which connects the point from the center of the circle is called the radius. And, any two points on the circumference join to form the chord. Thus, we can say that a radius is not the chord of the circle.
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# Understanding Fractions
Second
Time Needed
60 minutes
Materials Needed
Common area
Whiteboard
Document camera
Pencil
Scrap paper
Student books
Fraction manipulatives
Vocabulary
Equal
Unequal
Whole
Fraction
One-half
One-third
One-fourth
Introduction:
This lesson is part of a unit on fractions. It is the first
lesson following the introduction to what fractions are.
This lesson first has students make equal parts,
introduces notation to describe these equal parts as part
of a whole, has students name fractional parts, and
finally use models drawings to show a whole in different
ways.
Background:
The background knowledge needed consists of knowing
being introduced to the idea of half in the first grade
curriculum. It also consists of the previous days
instruction to fractions, which consisted mostly of
dealing with equal parts.
Standard(s):
CCSS.MATH.CONTENT.2.G.A.2
Partition a rectangle into rows and columns of same-size
squares and count to find the total number of them.
CCSS.MATH.CONTENT.2.G.A.3
Partition circles and rectangles into two, three, or four equal shares, describe
the shares using the words halves, thirds, half of, a third of, etc., and
describe the whole as two halves, three thirds, four fourths. Recognize that
equal shares of identical wholes need not have the same shape.
Guiding Question:
How does one represent fractions in different ways?
I Can Statement(s):
I can understand and use halves, thirds, and fourths as equal parts of a
whole.
Focus Lesson:
A brief focus lesson will focus on introducing the vocabulary of the lesson.
[Equalsame as; unequaldifferent or not the same; wholeall of, entire;
fractionpart of a whole; one-half1/2; one-third1/3; one-fourth1/4].
Important to note that the language used is equal parts of the whole.
## First partwrite I can statement and vocabulary definitions on board.
Second partwill focus on equal vs. unequal. Draw three circles and a
triangle. Divide the triangle into fourths, one of circles into fourths, one circle
in half, and the last in unequal portions. Talk about which ones are equal and
which ones arent.
Third partReview notation. Draw rectangle split up into a whole, half,
thirds, and fourths and label the pieces.
Fourth partintroduce notation of adding like-fractions i.e. 1/3+1/3+1/3=1
whole.
Guided Practice:
Students will copy the shape the teacher
makes on the board and make lines in
portions the teacher indicates. First example is 1/3
Should look like this. Have students fill in 1/3, write one-third, and 1/3 under
each portion and write the sentence 1/3+1/3+1/3=3/3=1 whole.
Have students follow suit with 1/2 and 1/4. Have students complete this with
triangles, squares, and circles.
IF TIME (need five minutes)
Introduce the idea of 2/3, 2/4, and 3/4
IF TIME (need 15 minutes so would need to start by 1:15)
Bring back to circle area and demonstrate how to play a quick game.
With a partner I will make a shape and draw in either a fourth, third, or
half of that shape. Partner will need to label what the portions are and
give me the number sentence to go along.
Independent Practice:
As an evaluative tool, all students will complete independent practice at the
end of the lesson. This will be modeled at 1:33 to give the students from 1:35
to 1:45 to complete the two pages from their student book (p. 81-82). This
will be entirely independent though Mr. Schmidt will walk around helping
explain directions. Students will raise their hands to indicate that they have
completed the pages. Mr. Schmidt will do a quick check of the work and
release those who have completed them correct to read quietly while others
are finishing.
Differentiation:
Some of the lower students (Thomas, Kahli, and Chase) will work with Mrs.
Tollas in a small group given her availability.
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# How do you find all critical numbers for the function f(x)= (x + 2)^3*(x -1)^4?
Aug 13, 2017
Alternatively, we can get the same result by using logarithmic differentiation.
#### Explanation:
Remember the basic concept of the logarithm: it is the opposite of exponentiation. That is to say, if:
${x}^{y} = z$
Then we can use a logarithm to "cancel out" the exponent, bringing it to the front like so:
$y \ln x = \ln z$
Now, we don't have to use the natural logarithm ($\ln x$) all the time - we could use, say, ${\log}_{2} \left(x\right)$ - but mathematicians think the natural log is cool (and it's extremely useful when differentiating, as we'll see in a moment).
Okay, so what's the point? Well, according to the rules of logs, if we take the log of both sides:
$\ln \left(f \left(x\right)\right) = \ln \left({\left(x + 2\right)}^{3} {\left(x - 1\right)}^{4}\right)$
We can split the multiplication into addition:
$\ln \left(f \left(x\right)\right) = \ln {\left(x + 2\right)}^{3} + \ln {\left(x - 1\right)}^{4}$
And remember, exponents come down to the front:
$\ln \left(f \left(x\right)\right) = 3 \ln \left(x + 2\right) + 4 \ln \left(x - 1\right)$
Now we can take the derivative, making sure to use the chain rule on the left side:
$\frac{f ' \left(x\right)}{f} \left(x\right) = \frac{3}{x + 2} + \frac{4}{x - 1} \to$ because the derivative of $\ln x$ is $\frac{1}{x}$
Now just multiply both sides by $f \left(x\right)$:
$f ' \left(x\right) = \left(\frac{3}{x + 2} + \frac{4}{x - 1}\right) f \left(x\right)$
And since $f \left(x\right) = {\left(x + 2\right)}^{3} {\left(x - 1\right)}^{4}$:
$f ' \left(x\right) = \left(\frac{3}{x + 2} + \frac{4}{x - 1}\right) \left({\left(x + 2\right)}^{3} {\left(x - 1\right)}^{4}\right)$
Finally, set $f ' \left(x\right) = 0$ and solve:
$0 = \left(\frac{3}{x + 2} + \frac{4}{x - 1}\right) \left({\left(x + 2\right)}^{3} {\left(x - 1\right)}^{4}\right)$
We have three solutions:
$0 = \frac{3}{x + 2} + \frac{4}{x - 1}$ and
$0 = {\left(x + 2\right)}^{3}$ and
$0 = {\left(x - 1\right)}^{4}$
The last two clearly mean that $x = - 2$ and $x = 1$ are solutions. For the first, we set up the equation:
$- \frac{3}{x + 2} = \frac{4}{x - 1}$
$\to - 3 \left(x - 1\right) = 4 \left(x + 2\right)$
Which is a simple equation that yields $x = - \frac{5}{7}$.
So there you have it: same answer, different method. Use which one you feel most comfortable with - that's the beauty of math.
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How do you solve a right triangle with side B = 46° and side c = 30 feet?
Jan 9, 2016
First, draw a diagram to represent the situation.
Explanation:
As you can see, we are looking for two sides and one angle. Let's do the simplest task first, which is to find angle A.
A= 180˚ - 90˚ - 46˚
A = 44˚
Now we can use the primary trigonometric ratios to find sides a and b.
Starting with b:
$\frac{b}{30}$ = $\sin \frac{46}{1}$
Solve for b -->b = 21.58
Finishing with a:
$\frac{a}{30}$ = $\cos \frac{46}{1}$
a = 20.84
So, A = 44˚, a = 20.84 feet and b = 21.58 feet
Hopefully this helps. Below I have included a few exercises for you to practice yourself with should you choose to practise more.
1. Solve the following triangles:
a) ∆ABC where A = 56˚ and c = 24 inches
b) ∆ABC where b = 35 cm and c = 27 cm.
1. A photographer is standing 34 meters from a skyscraper, and the angle of elevation between him [the photographer] and the top of the building is 65˚. Find the height of the building rounded to two decimals
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Percents: Overview
Real-estate salesman Z is selling a house at a 20% discount from its retail price. Real-estate salesman X vows to match this price and then offers an additional 10% discount. Real-estate salesman Y decides to average the prices of salesmen Z and X and then offer an additional 25% discount. Salesman Y's final price is what fraction of salesman X's final price?
Correct. [[snippet]] Z offers a 20% discount on the house, so he's offering it for $80. X matches the offer of$80, then offers another 10% discount. Careful here—X's additional discount is calculated from his matching offer of $80, not from the original! Thus, X offers an additional $$10\% \times 80 = 8$$ discount, and his final price is$72. Along comes Y, averages X's and Z's price, and then offers a 25% discount on the new price. The average of 80 and 72 is >$$\frac{80+72}{2} = \frac{152}{2} = 76$$. We then calculate that 25% of 76 is >$$\frac{25}{100} \times 76 = \frac{1}{4}\times 76 = 19$$. Thus, Y's final price is $$76-19 = 57$$. Place Y's price over X's price to find the fraction: $$\frac{57}{72}$$. Reduce by 3 to get to $$\frac{19}{24}$$.
Incorrect. [[snippet]]
Incorrect. [[snippet]]
Incorrect. [[snippet]] This is the fraction of X's from the original retail price. However, what did the question ask?
Incorrect. [[snippet]]
$$\frac{72}{80}$$
$$\frac{72}{100}$$
$$\frac{57}{80}$$
$$\frac{19}{24}$$
$$\frac{76}{100}$$
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How do you divide (b^3+2b^2-15b+49)div(b+6) using synthetic division?
Jul 29, 2018
$\left({b}^{3} + 2 {b}^{2} - 15 b + 49\right) = \left(b + 6\right) \left({b}^{2} - 4 b + 9\right) + \left(- 5\right)$
Explanation:
$\left({b}^{3} + 2 {b}^{2} - 15 b + 49\right) \div \left(b + 6\right)$
Using synthetic division :
We have , $p \left(b\right) = {b}^{3} + 2 {b}^{2} - 15 b + 49 \mathmr{and} \text{divisor : } b = - 6$
We take ,coefficients of $p \left(b\right) \to 1 , 2 , - 15 , 49$
$- 6 |$ $1 \textcolor{w h i t e}{\ldots \ldots . .} 2 \textcolor{w h i t e}{\ldots \ldots .} - 15 \textcolor{w h i t e}{\ldots \ldots .} 49$
$\underline{\textcolor{w h i t e}{\ldots .}} |$ ul(0color(white)( .....)-6color(white)(..........)24color(white)(...)-54
color(white)(......)1color(white)(......)-4color(white)(........)color(white)(..)9color(white)(.....)color(violet)(ul|-5|
We can see that , quotient polynomial :
$q \left(b\right) = {b}^{2} - 4 b + 9 \mathmr{and} \text{the Remainder} = - 5$
Hence ,
$\left({b}^{3} + 2 {b}^{2} - 15 b + 49\right) = \left(b + 6\right) \left({b}^{2} - 4 b + 9\right) + \left(- 5\right)$
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# How do I find the limit as $x$ approaches negative infinity of $\ln x$?
Last updated date: 19th Jul 2024
Total views: 384k
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Hint: In order to find the solution to this question, we will first convert the term in limits form and then find the limits. We are using limits because infinity is not a number as $x$ approaches infinity.
From the question, we can see that we have been asked to find the value of $\ln x$ when $x$ approaches negative infinity.
So, to start with the solution, we will first convert the given statement into a mathematical expression. Therefore, we can convert into limits since infinity is not a number.
Therefore, we get:
$\displaystyle \lim_{x \to -\infty }\left( \ln \left( x \right) \right)$
As we can see above, x approaches to negative infinity, therefore the answer is undefined because $-\infty$ is not in the domain of $\ln \left( x \right)$, the limit does not exist.
$\ln \left( -\infty \right)$ is undefined.
If the scenario is:
x approaches infinity that is positive infinity, then:
$\displaystyle \lim_{x \to \infty }\ln \left( x \right)=\infty$
That is the limit of the natural logarithm of $x$ when $x$ approaches positive infinity is infinity.
We can understand with the help of the following graph:
As when $x$ approaches minus infinity:
In this case, the natural logarithm of minus infinity is undefined for real numbers, since the natural logarithm function is undefined for negative numbers:
Therefore, we get:
$\displaystyle \lim_{x \to \infty }\ln \left( x \right)$ is undefined.
We can understand with the help of the following graph:
Therefore, we can conclude at this point:
$\ln \left( \infty \right)=\infty$
And
$\ln \left( x \right)$ is undefined.
Note:
In a hurry, we might end up reading the question wrong and then getting the wrong answer. So, we have to be very careful about that. Also, we can solve this question from the graph of ln x, but sometimes we get confused about the graph, so we have used the conventional method to find the answer.
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# What is the maximum flow speed of water for laminar flow in a pipe having a diameter $5cm$ given that coefficient of viscosity of water is ${10^{ - 3}}Pa\sec .$
Last updated date: 15th Aug 2024
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Hint:To solve this problem,we will apply Reynold's number formula for the laminar flow. Reynold’s number is dimensionless number as it is the ratio of Inertial force to viscous force. Reynold’s number is denoted by ${N_{\operatorname{Re} }}$. Mathematically, the Reynold’s number is calculated as,
${N_{\operatorname{Re} }} = \dfrac{{\rho vd}}{\mu }$ ……(1)
Where, ${N_{\operatorname{Re} }}$ is Reynolds’s number and its numerical value for the laminar flow is 2000.
$\rho$ Is the density of the fluid
$v$ is the average flow speed or average velocity of the flow
$d$ is the diameter of the pipe
$\mu$ is the coefficient of viscosity or dynamic viscosity
Complete step by step solution:
Given that:
Diameter of the pipe is, $d = 5cm$
The coefficient of viscosity is, $\mu = {10^{ - 3}}Pa\sec .$
The density of water is, $\rho = 1000\dfrac{{kg}}{{m{}^3}}$
Now put all the given values in equation (1)
${N_{\operatorname{Re} }} = \dfrac{{\rho vd}}{\mu } \\ v = \dfrac{{{N_{\operatorname{Re} }}\mu }}{{\rho d}} \\ v = \dfrac{{2000 \times {{10}^{ - 3}}}}{{1000 \times 0.05}} \\ v = 4\dfrac{{\text{m}}}{{\text{s}}} \\$
As we know, for laminar flow the maximum flow speed is twice the average flow speed.
Thus the maximum flow speed of water for laminar flow in a pipe is calculated as,
${v_{\max }} = 2 \times v \\ = 2 \times 4 \\ = 8\dfrac{{\text{m}}}{{\text{s}}} \\$
Therefore, the maximum flow speed of water for laminar flow in a pipe is ${v_{\max }} = 8\dfrac{{\text{m}}}{{\text{s}}}$.
Note: Laminar flow is also known as stream line flow because the fluid particles travel in a regular, smooth and in straight line path. On the other side, if the fluid particles travel in irregular or zig-zag ways then the flow is known as turbulent flow.
If the value of Reynolds’s number is equal to or less than 2000 then the flow is considered as laminar flow. If the value of Reynolds’s number is more than 4000 then the flow is considered as turbulent flow.
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# How do you simplify 5t − 3(1 + t) ?
Apr 5, 2018
$2 t - 3$
Here's how I did it:
#### Explanation:
$5 t - 3 \left(1 + t\right)$
The first thing we want to do is distribute (multiply) the $- 3$ to everything in the parenthesis:
$- 3 \cdot 1 = - 3$
$- 3 \cdot t = - 3 t$
And when we combine them we get $- 3 - 3 t$. Now the expression is:
$5 t - 3 - 3 t$
Now we do $5 t - 3 t$:
$2 t - 3$ (final simplified answer)
Apr 5, 2018
Your answer is $2 t - 3$
#### Explanation:
Firstly open up the bracket $- 3 \left(1 + t\right)$ this is equal to $- 3 - 3 t$. I am sure you noticed the sign change this is because minus times plus equals minus. Your equation know is $5 t - 3 - 3 t$ which is then $5 t - 3 t - 3$. The simplification is $2 t - 3$. Hope this helps
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# Fraction Explained
A fraction (from Latin: fractus, "broken") represents a part of a whole or, more generally, any number of equal parts. When spoken in everyday English, a fraction describes how many parts of a certain size there are, for example, one-half, eight-fifths, three-quarters. A common, vulgar, or simple fraction (examples:
\tfrac{1}{2}
and
\tfrac{17}{3}
) consists of a numerator, displayed above a line (or before a slash like), and a non-zero denominator, displayed below (or after) that line. Numerators and denominators are also used in fractions that are not common, including compound fractions, complex fractions, and mixed numerals.
In positive common fractions, the numerator and denominator are natural numbers. The numerator represents a number of equal parts, and the denominator indicates how many of those parts make up a unit or a whole. The denominator cannot be zero, because zero parts can never make up a whole. For example, in the fraction, the numerator 3 indicates that the fraction represents 3 equal parts, and the denominator 4 indicates that 4 parts make up a whole. The picture to the right illustrates of a cake.
A common fraction is a numeral which represents a rational number. That same number can also be represented as a decimal, a percent, or with a negative exponent. For example, 0.01, 1%, and 10−2 are all equal to the fraction 1/100. An integer can be thought of as having an implicit denominator of one (for example, 7 equals 7/1).
Other uses for fractions are to represent ratios and division.[1] Thus the fraction can also be used to represent the ratio 3:4 (the ratio of the part to the whole), and the division (three divided by four). The non-zero denominator rule, which applies when representing a division as a fraction, is an example of the rule that division by zero is undefined.
We can also write negative fractions, which represent the opposite of a positive fraction. For example, if represents a half-dollar profit, then − represents a half-dollar loss. Because of the rules of division of signed numbers (which states in part that negative divided by positive is negative), −, and all represent the same fraction negative one-half. And because a negative divided by a negative produces a positive, represents positive one-half.
In mathematics the set of all numbers that can be expressed in the form, where a and b are integers and b is not zero, is called the set of rational numbers and is represented by the symbol Q, which stands for quotient. A number is a rational number precisely when it can be written in that form (i.e., as a common fraction). However, the word fraction can also be used to describe mathematical expressions that are not rational numbers. Examples of these usages include algebraic fractions (quotients of algebraic expressions), and expressions that contain irrational numbers, such as $\frac$ (see square root of 2) and (see proof that π is irrational).
## Vocabulary
In a fraction, the number of equal parts being described is the numerator (from Latin: numerātor, "counter" or "numberer"), and the type or variety of the parts is the denominator (from Latin: dēnōminātor, "thing that names or designates").[2] [3] As an example, the fraction amounts to eight parts, each of which is of the type named "fifth". In terms of division, the numerator corresponds to the dividend, and the denominator corresponds to the divisor.
Informally, the numerator and denominator may be distinguished by placement alone, but in formal contexts they are usually separated by a fraction bar. The fraction bar may be horizontal (as in), oblique (as in 2/5), or diagonal (as in). These marks are respectively known as the horizontal bar; the virgule, slash (US), or stroke (UK); and the fraction bar, solidus,[4] or fraction slash. In typography, fractions stacked vertically are also known as "en" or "nut fractions", and diagonal ones as "em" or "mutton fractions", based on whether a fraction with a single-digit numerator and denominator occupies the proportion of a narrow en square, or a wider em square.[5] In traditional typefounding, a piece of type bearing a complete fraction (e.g.) was known as a "case fraction", while those representing only part of fraction were called "piece fractions".
The denominators of English fractions are generally expressed as ordinal numbers, in the plural if the numerator is not 1. (For example, and are both read as a number of "fifths".) Exceptions include the denominator 2, which is always read "half" or "halves", the denominator 4, which may be alternatively expressed as "quarter"/"quarters" or as "fourth"/"fourths", and the denominator 100, which may be alternatively expressed as "hundredth"/"hundredths" or "percent".
When the denominator is 1, it may be expressed in terms of "wholes" but is more commonly ignored, with the numerator read out as a whole number. For example, may be described as "three wholes", or simply as "three". When the numerator is 1, it may be omitted (as in "a tenth" or "each quarter").
The entire fraction may be expressed as a single composition, in which case it is hyphenated, or as a number of fractions with a numerator of one, in which case they are not. (For example, "two-fifths" is the fraction and "two fifths" is the same fraction understood as 2 instances of .) Fractions should always be hyphenated when used as adjectives. Alternatively, a fraction may be described by reading it out as the numerator "over" the denominator, with the denominator expressed as a cardinal number. (For example, may also be expressed as "three over one".) The term "over" is used even in the case of solidus fractions, where the numbers are placed left and right of a slash mark. (For example, 1/2 may be read "one-half", "one half", or "one over two".) Fractions with large denominators that are not powers of ten are often rendered in this fashion (e.g., as "one over one hundred seventeen"), while those with denominators divisible by ten are typically read in the normal ordinal fashion (e.g., as "six-millionths", "six millionths", or "six one-millionths").
## Forms of fractions
### Simple, common, or vulgar fractions
A simple fraction (also known as a common fraction or vulgar fraction, where vulgar is Latin for "common") is a rational number written as a/b or
\tfrac{a}{b}
, where a and b are both integers. As with other fractions, the denominator (b) cannot be zero. Examples include
\tfrac{1}{2}
,
-\tfrac{8}{5}
,
\tfrac{-8}{5}
, and
\tfrac{8}{-5}
. The term was originally used to distinguish this type of fraction from the sexagesimal fraction used in astronomy.[6]
Common fractions can be positive or negative, and they can be proper or improper (see below). Compound fractions, complex fractions, mixed numerals, and decimals (see below) are not common fractions; though, unless irrational, they can be evaluated to a common fraction.
• A unit fraction is a common fraction with a numerator of 1 (e.g.,
\tfrac{1}{7}
). Unit fractions can also be expressed using negative exponents, as in 2−1, which represents 1/2, and 2−2, which represents 1/(22) or 1/4.
\tfrac{1}{8}=\tfrac{1}{23}
.
In Unicode, precomposed fraction characters are in the Number Forms block.
### Proper and improper fractions
Common fractions can be classified as either proper or improper. When the numerator and the denominator are both positive, the fraction is called proper if the numerator is less than the denominator, and improper otherwise.[7] The concept of an "improper fraction" is a late development, with the terminology deriving from the fact that "fraction" means "a piece", so a proper fraction must be less than 1.[6] This was explained in the 17th century textbook The Ground of Arts.[8] [9]
In general, a common fraction is said to be a proper fraction, if the absolute value of the fraction is strictly less than one—that is, if the fraction is greater than −1 and less than 1.[10] [11] It is said to be an improper fraction, or sometimes top-heavy fraction,[12] if the absolute value of the fraction is greater than or equal to 1. Examples of proper fractions are 2/3, −3/4, and 4/9, whereas examples of improper fractions are 9/4, −4/3, and 3/3.
### Reciprocals and the "invisible denominator"
The reciprocal of a fraction is another fraction with the numerator and denominator exchanged. The reciprocal of
\tfrac{3}{7}
, for instance, is
\tfrac{7}{3}
. The product of a fraction and its reciprocal is 1, hence the reciprocal is the multiplicative inverse of a fraction. The reciprocal of a proper fraction is improper, and the reciprocal of an improper fraction not equal to 1 (that is, numerator and denominator are not equal) is a proper fraction.
When the numerator and denominator of a fraction are equal (for example,
\tfrac{7}{7}
), its value is 1, and the fraction therefore is improper. Its reciprocal is identical and hence also equal to 1 and improper.
Any integer can be written as a fraction with the number one as denominator. For example, 17 can be written as
\tfrac{17}{1}
, where 1 is sometimes referred to as the invisible denominator. Therefore, every fraction or integer, except for zero, has a reciprocal. For example. the reciprocal of 17 is
\tfrac{1}{17}
.
### Ratios
A ratio is a relationship between two or more numbers that can be sometimes expressed as a fraction. Typically, a number of items are grouped and compared in a ratio, specifying numerically the relationship between each group. Ratios are expressed as "group 1 to group 2 ... to group n". For example, if a car lot had 12 vehicles, of which
• 2 are white,
• 6 are red, and
• 4 are yellow,
then the ratio of red to white to yellow cars is 6 to 2 to 4. The ratio of yellow cars to white cars is 4 to 2 and may be expressed as 4:2 or 2:1.
A ratio is often converted to a fraction when it is expressed as a ratio to the whole. In the above example, the ratio of yellow cars to all the cars on the lot is 4:12 or 1:3. We can convert these ratios to a fraction, and say that of the cars or of the cars in the lot are yellow. Therefore, if a person randomly chose one car on the lot, then there is a one in three chance or probability that it would be yellow.
### Decimal fractions and percentages
A decimal fraction is a fraction whose denominator is not given explicitly, but is understood to be an integer power of ten. Decimal fractions are commonly expressed using decimal notation in which the implied denominator is determined by the number of digits to the right of a decimal separator, the appearance of which (e.g., a period, an interpunct (·), a comma) depends on the locale (for examples, see decimal separator). Thus, for 0.75 the numerator is 75 and the implied denominator is 10 to the second power, namely, 100, because there are two digits to the right of the decimal separator. In decimal numbers greater than 1 (such as 3.75), the fractional part of the number is expressed by the digits to the right of the decimal (with a value of 0.75 in this case). 3.75 can be written either as an improper fraction, 375/100, or as a mixed number,
3\tfrac{75}{100}
.
Decimal fractions can also be expressed using scientific notation with negative exponents, such as, which represents 0.0000006023. The represents a denominator of . Dividing by moves the decimal point 7 places to the left.
Decimal fractions with infinitely many digits to the right of the decimal separator represent an infinite series. For example, = 0.333... represents the infinite series 3/10 + 3/100 + 3/1000 + ....
Another kind of fraction is the percentage (from Latin: per centum, meaning "per hundred", represented by the symbol %), in which the implied denominator is always 100. Thus, 51% means 51/100. Percentages greater than 100 or less than zero are treated in the same way, e.g. 311% equals 311/100, and −27% equals −27/100.
The related concept of permille or parts per thousand (ppt) has an implied denominator of 1000, while the more general parts-per notation, as in 75 parts per million (ppm), means that the proportion is 75/1,000,000.
Whether common fractions or decimal fractions are used is often a matter of taste and context. Common fractions are used most often when the denominator is relatively small. By mental calculation, it is easier to multiply 16 by 3/16 than to do the same calculation using the fraction's decimal equivalent (0.1875). And it is more accurate to multiply 15 by 1/3, for example, than it is to multiply 15 by any decimal approximation of one third. Monetary values are commonly expressed as decimal fractions with denominator 100, i.e., with two decimals, for example \$3.75. However, as noted above, in pre-decimal British currency, shillings and pence were often given the form (but not the meaning) of a fraction, as, for example, "3/6" (read "three and six") meaning 3 shillings and 6 pence, and having no relationship to the fraction 3/6.
### Mixed numbers
A mixed numeral (also called a mixed fraction or mixed number) is a traditional denotation of the sum of a non-zero integer and a proper fraction (having the same sign). It is used primarily in measurement:
2\tfrac{3}{16}
inches, for example. Scientific measurements almost invariably use decimal notation rather than mixed numbers. The sum can be implied without the use of a visible operator such as the appropriate "+". For example, in referring to two entire cakes and three quarters of another cake, the numerals denoting the integer part and the fractional part of the cakes can be written next to each other as
2\tfrac{3}{4}
2+\tfrac{3}{4}.
Negative mixed numerals, as in
-2\tfrac{3}{4}
, are treated like
\scriptstyle-\left(2+
3 4
\right).
Any such sum of a whole plus a part can be converted to an improper fraction by applying the rules of adding unlike quantities.
This tradition is, formally, in conflict with the notation in algebra where adjacent symbols, without an explicit infix operator, denote a product. In the expression
2x
, the "understood" operation is multiplication. If is replaced by, for example, the fraction
\tfrac{3}{4}
, the "understood" multiplication needs to be replaced by explicit multiplication, to avoid the appearance of a mixed number.
When multiplication is intended,
2\tfrac{b}{c}
may be written as
2
b c
,
or
2 x
b c
,
or
2\left(
b c
\right),\ldots
An improper fraction can be converted to a mixed number as follows:
1. Using Euclidean division (division with remainder), divide the numerator by the denominator. In the example,
\tfrac{11}{4}
, divide 11 by 4. 11 ÷ 4 = 2 remainder 3.
1. The quotient (without the remainder) becomes the whole number part of the mixed number. The remainder becomes the numerator of the fractional part. In the example, 2 is the whole number part and 3 is the numerator of the fractional part.
2. The new denominator is the same as the denominator of the improper fraction. In the example, it is 4. Thus,
\tfrac{11}{4}=2\tfrac{3}{4}
.
### Historical notions
#### Egyptian fraction
An Egyptian fraction is the sum of distinct positive unit fractions, for example
\tfrac{1}{2}+\tfrac{1}{3}
. This definition derives from the fact that the ancient Egyptians expressed all fractions except
\tfrac{1}{2}
,
\tfrac{2}{3}
and
\tfrac{3}{4}
in this manner. Every positive rational number can be expanded as an Egyptian fraction. For example,
\tfrac{5}{7}
can be written as
\tfrac{1}{2}+\tfrac{1}{6}+\tfrac{1}{21}.
Any positive rational number can be written as a sum of unit fractions in infinitely many ways. Two ways to write
\tfrac{13}{17}
are
\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{68}
and
\tfrac{1}{3}+\tfrac{1}{4}+\tfrac{1}{6}+\tfrac{1}{68}
.
#### Complex and compound fractions
Fraction should not be confused with Complex numbers.
In a complex fraction, either the numerator, or the denominator, or both, is a fraction or a mixed number,[13] [14] corresponding to division of fractions. For example,
\tfrac{1 2
} and
12\tfrac{3 4
} are complex fractions. To reduce a complex fraction to a simple fraction, treat the longest fraction line as representing division. For example:
\tfrac{1 2
}=\tfrac\times\tfrac=\tfrac
12\tfrac{3 4
} = 12\tfrac \cdot \tfrac = \tfrac \cdot \tfrac = \tfrac \cdot \tfrac = \tfrac
\tfrac{3 2
}5=\tfrac\times\tfrac=\tfrac
8 \tfrac{1
{3}}=8 x \tfrac{3}{1}=24.
If, in a complex fraction, there is no unique way to tell which fraction lines takes precedence, then this expression is improperly formed, because of ambiguity. So 5/10/20/40 is not a valid mathematical expression, because of multiple possible interpretations, e.g. as
5/(10/(20/40))=
5 10/\tfrac{20
{40}}=
1 4
or as
(5/10)/(20/40)=
\tfrac{5 10
} = 1
A compound fraction is a fraction of a fraction, or any number of fractions connected with the word of, corresponding to multiplication of fractions. To reduce a compound fraction to a simple fraction, just carry out the multiplication (see the section on multiplication). For example,
\tfrac{3}{4}
of
\tfrac{5}{7}
is a compound fraction, corresponding to
\tfrac{3}{4} x \tfrac{5}{7}=\tfrac{15}{28}
. The terms compound fraction and complex fraction are closely related and sometimes one is used as a synonym for the other. (For example, the compound fraction
\tfrac{3}{4} x \tfrac{5}{7}
is equivalent to the complex fraction
\tfrac{3/4}{7/5}
.)
Nevertheless, "complex fraction" and "compound fraction" may both be considered outdated[15] and now used in no well-defined manner, partly even taken synonymously for each other[16] or for mixed numerals.[17] They have lost their meaning as technical terms and the attributes "complex" and "compound" tend to be used in their every day meaning of "consisting of parts".
## Arithmetic with fractions
Like whole numbers, fractions obey the commutative, associative, and distributive laws, and the rule against division by zero.
### Equivalent fractions
Multiplying the numerator and denominator of a fraction by the same (non-zero) number results in a fraction that is equivalent to the original fraction. This is true because for any non-zero number
n
, the fraction
\tfrac{n}{n}
equals
1
. Therefore, multiplying by
\tfrac{n}{n}
is the same as multiplying by one, and any number multiplied by one has the same value as the original number. By way of an example, start with the fraction
\tfrac{1}{2}
. When the numerator and denominator are both multiplied by 2, the result is
\tfrac{2}{4}
, which has the same value (0.5) as
\tfrac{1}{2}
. To picture this visually, imagine cutting a cake into four pieces; two of the pieces together (
\tfrac{2}{4}
) make up half the cake (
\tfrac{1}{2}
).
#### Simplifying (reducing) fractions
Dividing the numerator and denominator of a fraction by the same non-zero number yields an equivalent fraction: if the numerator and the denominator of a fraction are both divisible by a number (called a factor) greater than 1, then the fraction can be reduced to an equivalent fraction with a smaller numerator and a smaller denominator. For example, if both the numerator and the denominator of the fraction
\tfrac{a}{b}
are divisible by
c,
then they can be written as
a=cd
and
b=ce,
and the fraction becomes
\tfrac{cd}{ce}
, which can be reduced by dividing both the numerator and denominator by
c
to give the reduced fraction
\tfrac{d}{e}.
If one takes for the greatest common divisor of the numerator and the denominator, one gets the equivalent fraction whose numerator and denominator have the lowest absolute values. One says that the fraction has been reduced to its lowest terms.
If the numerator and the denominator do not share any factor greater than 1, the fraction is already reduced to its lowest terms, and it is said to be irreducible, reduced, or in simplest terms. For example,
\tfrac{3}{9}
is not in lowest terms because both 3 and 9 can be exactly divided by 3. In contrast,
\tfrac{3}{8}
is in lowest terms—the only positive integer that goes into both 3 and 8 evenly is 1.
Using these rules, we can show that
\tfrac{5}{10}=\tfrac{1}{2}=\tfrac{10}{20}=\tfrac{50}{100}
, for example.
As another example, since the greatest common divisor of 63 and 462 is 21, the fraction
\tfrac{63}{462}
can be reduced to lowest terms by dividing the numerator and denominator by 21:
\tfrac{63}{462}=\tfrac{63 ÷ 21}{462 ÷ 21}=\tfrac{3}{22}
The Euclidean algorithm gives a method for finding the greatest common divisor of any two integers.
### Comparing fractions
Comparing fractions with the same positive denominator yields the same result as comparing the numerators:
\tfrac{3}{4}>\tfrac{2}{4}
because, and the equal denominators
4
are positive.
If the equal denominators are negative, then the opposite result of comparing the numerators holds for the fractions:
\tfrac{3}{-4}<\tfrac{2}{-4}because\tfrac{a}{-b}=\tfrac{-a}{b}and-3<-2.
If two positive fractions have the same numerator, then the fraction with the smaller denominator is the larger number. When a whole is divided into equal pieces, if fewer equal pieces are needed to make up the whole, then each piece must be larger. When two positive fractions have the same numerator, they represent the same number of parts, but in the fraction with the smaller denominator, the parts are larger.
One way to compare fractions with different numerators and denominators is to find a common denominator. To compare
\tfrac{a}{b}
and
\tfrac{c}{d}
, these are converted to
and
\tfrac{bc}{bd}
(where the dot signifies multiplication and is an alternative symbol to ×). Then bd is a common denominator and the numerators ad and bc can be compared. It is not necessary to determine the value of the common denominator to compare fractions – one can just compare ad and bc, without evaluating bd, e.g., comparing
\tfrac{2}{3}
?
\tfrac{1}{2}
gives
\tfrac{4}{6}>\tfrac{3}{6}
.
For the more laborious question
\tfrac{5}{18}
?
\tfrac{4}{17},
multiply top and bottom of each fraction by the denominator of the other fraction, to get a common denominator, yielding
\tfrac{5 x 17}{18 x 17}
?
\tfrac{18 x 4}{18 x 17}
. It is not necessary to calculate
18 x 17
– only the numerators need to be compared. Since 5×17 (= 85) is greater than 4×18 (= 72), the result of comparing is
\tfrac{5}{18}>\tfrac{4}{17}
.
Because every negative number, including negative fractions, is less than zero, and every positive number, including positive fractions, is greater than zero, it follows that any negative fraction is less than any positive fraction. This allows, together with the above rules, to compare all possible fractions.
The first rule of addition is that only like quantities can be added; for example, various quantities of quarters. Unlike quantities, such as adding thirds to quarters, must first be converted to like quantities as described below: Imagine a pocket containing two quarters, and another pocket containing three quarters; in total, there are five quarters. Since four quarters is equivalent to one (dollar), this can be represented as follows:
\tfrac24+\tfrac34=\tfrac54=1\tfrac14
.
1.
For adding quarters to thirds, both types of fraction are converted to twelfths, thus:
14 +
13= 1 x 3 4 x 3
+
1 x 4 = 3 x 4
3{12} +
4{12}= 7{12}.
Consider adding the following two quantities:
35+ 23
First, convert
\tfrac35
into fifteenths by multiplying both the numerator and denominator by three:
\tfrac35 x \tfrac33=\tfrac9{15}
. Since
\tfrac33
equals 1, multiplication by
\tfrac33
does not change the value of the fraction.
Second, convert
\tfrac23
into fifteenths by multiplying both the numerator and denominator by five:
\tfrac23 x \tfrac55=\tfrac{10}{15}
.
Now it can be seen that:
35+ 23
is equivalent to:
9{15}+ 10 15
=1
= 19 15
4{15}
This method can be expressed algebraically:
a b
+
c d
=
This algebraic method always works, thereby guaranteeing that the sum of simple fractions is always again a simple fraction. However, if the single denominators contain a common factor, a smaller denominator than the product of these can be used. For example, when adding
\tfrac{3}{4}
and
\tfrac{5}{6}
the single denominators have a common factor
2,
and therefore, instead of the denominator 24 (4 × 6), the halved denominator 12 may be used, not only reducing the denominator in the result, but also the factors in the numerator.
\begin{align}
34+ 56 &=
+
3 ⋅ 6 4 ⋅ 6
4 ⋅ 5 = 4 ⋅ 6
18 24
+
20 &= 24
19 \\ &= 12
3 ⋅ 3 + 4 ⋅ 3
2 ⋅ 5 = 2 ⋅ 6
9 12
+
10 &= 12
19 12
\end{align}
The smallest possible denominator is given by the least common multiple of the single denominators, which results from dividing the rote multiple by all common factors of the single denominators. This is called the least common denominator.
### Subtraction
The process for subtracting fractions is, in essence, the same as that of adding them: find a common denominator, and change each fraction to an equivalent fraction with the chosen common denominator. The resulting fraction will have that denominator, and its numerator will be the result of subtracting the numerators of the original fractions. For instance,
\tfrac23-\tfrac12=\tfrac46-\tfrac36=\tfrac16
### Multiplication
#### Multiplying a fraction by another fraction
To multiply fractions, multiply the numerators and multiply the denominators. Thus:
2 3
x
3 4
=
6 12
To explain the process, consider one third of one quarter. Using the example of a cake, if three small slices of equal size make up a quarter, and four quarters make up a whole, twelve of these small, equal slices make up a whole. Therefore, a third of a quarter is a twelfth. Now consider the numerators. The first fraction, two thirds, is twice as large as one third. Since one third of a quarter is one twelfth, two thirds of a quarter is two twelfth. The second fraction, three quarters, is three times as large as one quarter, so two thirds of three quarters is three times as large as two thirds of one quarter. Thus two thirds times three quarters is six twelfths.
A short cut for multiplying fractions is called "cancellation". Effectively the answer is reduced to lowest terms during multiplication. For example:
2 3
x
3 4
=
\cancel{2 ~1
} \times \frac = \frac \times \frac = \frac
A two is a common factor in both the numerator of the left fraction and the denominator of the right and is divided out of both. Three is a common factor of the left denominator and right numerator and is divided out of both.
#### Multiplying a fraction by a whole number
Since a whole number can be rewritten as itself divided by 1, normal fraction multiplication rules can still apply.
6 x \tfrac{3}{4}=\tfrac{6}{1} x \tfrac{3}{4}=\tfrac{18}{4}
This method works because the fraction 6/1 means six equal parts, each one of which is a whole.
#### Multiplying mixed numbers
When multiplying mixed numbers, it is considered preferable to convert the mixed number into an improper fraction.[18] For example:
3 x 2
3 4
=3 x \left(
8 4
+
3 4
\right)=3 x
11 4
=
33 4
=8
1 4
In other words,
2\tfrac{3}{4}
is the same as
\tfrac{8}{4}+\tfrac{3}{4}
, making 11 quarters in total (because 2 cakes, each split into quarters makes 8 quarters total) and 33 quarters is
8\tfrac{1}{4}
, since 8 cakes, each made of quarters, is 32 quarters in total.
### Division
To divide a fraction by a whole number, you may either divide the numerator by the number, if it goes evenly into the numerator, or multiply the denominator by the number. For example,
\tfrac{10}{3} ÷ 5
equals
\tfrac{2}{3}
and also equals
\tfrac{10}{35}=\tfrac{10}{15}
, which reduces to
\tfrac{2}{3}
. To divide a number by a fraction, multiply that number by the reciprocal of that fraction. Thus,
\tfrac{1}{2} ÷ \tfrac{3}{4}=\tfrac{1}{2} x \tfrac{4}{3}=\tfrac{14}{23}=\tfrac{2}{3}
.
### Converting between decimals and fractions
To change a common fraction to a decimal, do a long division of the decimal representations of the numerator by the denominator (this is idiomatically also phrased as "divide the denominator into the numerator"), and round the answer to the desired accuracy. For example, to change to a decimal, divide by (" into "), to obtain . To change to a decimal, divide by (" into "), and stop when the desired accuracy is obtained, e.g., at decimals with . The fraction can be written exactly with two decimal digits, while the fraction cannot be written exactly as a decimal with a finite number of digits. To change a decimal to a fraction, write in the denominator a followed by as many zeroes as there are digits to the right of the decimal point, and write in the numerator all the digits of the original decimal, just omitting the decimal point. Thus
12.3456=\tfrac{123456}{10000}.
#### Converting repeating decimals to fractions
See also: Repeating decimal. Decimal numbers, while arguably more useful to work with when performing calculations, sometimes lack the precision that common fractions have. Sometimes an infinite repeating decimal is required to reach the same precision. Thus, it is often useful to convert repeating decimals into fractions.
A conventional way to indicate a repeating decimal is to place a bar (known as a vinculum) over the digits that repeat, for example = 0.789789789... For repeating patterns that begin immediately after the decimal point, the result of the conversion is the fraction with the pattern as a numerator, and the same number of nines as a denominator. For example:
= 5/9
= 62/99
= 264/999
= 6291/9999If leading zeros precede the pattern, the nines are suffixed by the same number of trailing zeros:
= 5/90
= 392/999000
= 12/9900If a non-repeating set of decimals precede the pattern (such as), one may write the number as the sum of the non-repeating and repeating parts, respectively:
0.1523 + Then, convert both parts to fractions, and add them using the methods described above:
1523 / 10000 + 987 / 9990000 = 1522464 / 9990000
Alternatively, algebra can be used, such as below:
1. Let x = the repeating decimal:
x =
1. Multiply both sides by the power of 10 just great enough (in this case 104) to move the decimal point just before the repeating part of the decimal number:
10,000x =
1. Multiply both sides by the power of 10 (in this case 103) that is the same as the number of places that repeat:
10,000,000x =
1. Subtract the two equations from each other (if a = b and c = d, then ac = bd):
10,000,000x − 10,000x = −
1. Continue the subtraction operation to clear the repeating decimal:
9,990,000x = 1,523,987 − 1,523
9,990,000x = 1,522,464
1. Divide both sides by 9,990,000 to represent x as a fraction
x =
## Fractions in abstract mathematics
In addition to being of great practical importance, fractions are also studied by mathematicians, who check that the rules for fractions given above are consistent and reliable. Mathematicians define a fraction as an ordered pair
(a,b)
of integers
a
and
b\ne0,
for which the operations addition, subtraction, multiplication, and division are defined as follows:[19]
(a,b)(c,d)=(ac,bd)
These definitions agree in every case with the definitions given above; only the notation is different. Alternatively, instead of defining subtraction and division as operations, the "inverse" fractions with respect to addition and multiplication might be defined as:
Furthermore, the relation, specified as
is an equivalence relation of fractions. Each fraction from one equivalence class may be considered as a representative for the whole class, and each whole class may be considered as one abstract fraction. This equivalence is preserved by the above defined operations, i.e., the results of operating on fractions are independent of the selection of representatives from their equivalence class. Formally, for addition of fractions
(a,b)\sim(a',b')
and
(c,d)\sim(c',d')
imply
((a,b)+(c,d))\sim((a',b')+(c',d'))
and similarly for the other operations.
In the case of fractions of integers, the fractions with and coprime and are often taken as uniquely determined representatives for their equivalent fractions, which are considered to be the same rational number. This way the fractions of integers make up the field of the rational numbers.
More generally, a and b may be elements of any integral domain R, in which case a fraction is an element of the field of fractions of R. For example, polynomials in one indeterminate, with coefficients from some integral domain D, are themselves an integral domain, call it P. So for a and b elements of P, the generated field of fractions is the field of rational fractions (also known as the field of rational functions).
## Algebraic fractions
See main article: Algebraic fraction. An algebraic fraction is the indicated quotient of two algebraic expressions. As with fractions of integers, the denominator of an algebraic fraction cannot be zero. Two examples of algebraic fractions are
3x x2+2x-3
and
\sqrt{x+2
}. Algebraic fractions are subject to the same field properties as arithmetic fractions.
If the numerator and the denominator are polynomials, as in
3x x2+2x-3
, the algebraic fraction is called a rational fraction (or rational expression). An irrational fraction is one that is not rational, as, for example, one that contains the variable under a fractional exponent or root, as in
\sqrt{x+2
}.
The terminology used to describe algebraic fractions is similar to that used for ordinary fractions. For example, an algebraic fraction is in lowest terms if the only factors common to the numerator and the denominator are 1 and −1. An algebraic fraction whose numerator or denominator, or both, contain a fraction, such as
1+\tfrac{1 x
}, is called a complex fraction.
The field of rational numbers is the field of fractions of the integers, while the integers themselves are not a field but rather an integral domain. Similarly, the rational fractions with coefficients in a field form the field of fractions of polynomials with coefficient in that field. Considering the rational fractions with real coefficients, radical expressions representing numbers, such as
style\sqrt{2}/2,
are also rational fractions, as are a transcendental numbers such as $\pi/2,$ since all of
\sqrt{2},\pi,
and
2
are real numbers, and thus considered as coefficients. These same numbers, however, are not rational fractions with integer coefficients.
The term partial fraction is used when decomposing rational fractions into sums of simpler fractions. For example, the rational fraction
2x x2-1
can be decomposed as the sum of two fractions:
1 x+1
+
1 x-1
.
This is useful for the computation of antiderivatives of rational functions (see partial fraction decomposition for more).
See main article: Nth root and Rationalization (mathematics). A fraction may also contain radicals in the numerator or the denominator. If the denominator contains radicals, it can be helpful to rationalize it (compare Simplified form of a radical expression), especially if further operations, such as adding or comparing that fraction to another, are to be carried out. It is also more convenient if division is to be done manually. When the denominator is a monomial square root, it can be rationalized by multiplying both the top and the bottom of the fraction by the denominator:
3 \sqrt{7
} = \frac \cdot \frac = \frac
The process of rationalization of binomial denominators involves multiplying the top and the bottom of a fraction by the conjugate of the denominator so that the denominator becomes a rational number. For example:
3 3-2\sqrt{5
} = \frac \cdot \frac = \frac = \frac = - \frac
3 3+2\sqrt{5
} = \frac \cdot \frac = \frac = \frac = - \frac
Even if this process results in the numerator being irrational, like in the examples above, the process may still facilitate subsequent manipulations by reducing the number of irrationals one has to work with in the denominator.
## Typographical variations
In computer displays and typography, simple fractions are sometimes printed as a single character, e.g. ½ (one half). See the article on Number Forms for information on doing this in Unicode.
Scientific publishing distinguishes four ways to set fractions, together with guidelines on use:[20]
• Special fractions: fractions that are presented as a single character with a slanted bar, with roughly the same height and width as other characters in the text. Generally used for simple fractions, such as: ½, ⅓, ⅔, ¼, and ¾. Since the numerals are smaller, legibility can be an issue, especially for small-sized fonts. These are not used in modern mathematical notation, but in other contexts.
• Case fractions: similar to special fractions, these are rendered as a single typographical character, but with a horizontal bar, thus making them upright. An example would be
\tfrac{1}{2}
, but rendered with the same height as other characters. Some sources include all rendering of fractions as case fractions if they take only one typographical space, regardless of the direction of the bar.[21]
• Shilling or solidus fractions: 1/2, so called because this notation was used for pre-decimal British currency (£sd), as in "2/6" for a half crown, meaning two shillings and six pence. While the notation "two shillings and six pence" did not represent a fraction, the forward slash is now used in fractions, especially for fractions inline with prose (rather than displayed), to avoid uneven lines. It is also used for fractions within fractions (complex fractions) or within exponents to increase legibility. Fractions written this way, also known as piece fractions,[22] are written all on one typographical line, but take 3 or more typographical spaces.
• Built-up fractions:
1 2
. This notation uses two or more lines of ordinary text and results in a variation in spacing between lines when included within other text. While large and legible, these can be disruptive, particularly for simple fractions or within complex fractions.
## History
The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on.[23] The Egyptians used Egyptian fractions BC. About 4000 years ago, Egyptians divided with fractions using slightly different methods. They used least common multiples with unit fractions. Their methods gave the same answer as modern methods.[24] The Egyptians also had a different notation for dyadic fractions in the Akhmim Wooden Tablet and several Rhind Mathematical Papyrus problems.
The Greeks used unit fractions and (later) continued fractions. Followers of the Greek philosopher Pythagoras ( BC) discovered that the square root of two cannot be expressed as a fraction of integers. (This is commonly though probably erroneously ascribed to Hippasus of Metapontum, who is said to have been executed for revealing this fact.) In Jain mathematicians in India wrote the "Sthananga Sutra", which contains work on the theory of numbers, arithmetical operations, and operations with fractions.
A modern expression of fractions known as bhinnarasi seems to have originated in India in the work of Aryabhatta, Brahmagupta, and Bhaskara .[25] Their works form fractions by placing the numerators (Sanskrit: amsa) over the denominators (Sanskrit: cheda), but without a bar between them.[25] In Sanskrit literature, fractions were always expressed as an addition to or subtraction from an integer. The integer was written on one line and the fraction in its two parts on the next line. If the fraction was marked by a small circle or cross, it is subtracted from the integer; if no such sign appears, it is understood to be added. For example, Bhaskara I writes:[26]
६ १ २
१ १ १
४ ५ ९which is the equivalent of
6 1 2
1 1 −1
4 5 9and would be written in modern notation as 6, 1, and 2 − (i.e., 1).
The horizontal fraction bar is first attested in the work of Al-Hassār,[25] a Muslim mathematician from Fez, Morocco, who specialized in Islamic inheritance jurisprudence. In his discussion he writes: "for example, if you are told to write three-fifths and a third of a fifth, write thus, [27] The same fractional notation—with the fraction given before the integer[25] —appears soon after in the work of Leonardo Fibonacci in the 13th century.
In discussing the origins of decimal fractions, Dirk Jan Struik states:[28]
The introduction of decimal fractions as a common computational practice can be dated back to the Flemish pamphlet De Thiende, published at Leyden in 1585, together with a French translation, La Disme, by the Flemish mathematician Simon Stevin (1548–1620), then settled in the Northern Netherlands. It is true that decimal fractions were used by the Chinese many centuries before Stevin and that the Persian astronomer Al-Kāshī used both decimal and sexagesimal fractions with great ease in his Key to arithmetic (Samarkand, early fifteenth century).[29]
While the Persian mathematician Jamshīd al-Kāshī claimed to have discovered decimal fractions himself in the 15th century, J. Lennart Berggren notes that he was mistaken, as decimal fractions were first used five centuries before him by the Baghdadi mathematician Abu'l-Hasan al-Uqlidisi as early as the 10th century.[30]
## In formal education
### Pedagogical tools
In primary schools, fractions have been demonstrated through Cuisenaire rods, Fraction Bars, fraction strips, fraction circles, paper (for folding or cutting), pattern blocks, pie-shaped pieces, plastic rectangles, grid paper, dot paper, geoboards, counters and computer software.
### Documents for teachers
Several states in the United States have adopted learning trajectories from the Common Core State Standards Initiative's guidelines for mathematics education. Aside from sequencing the learning of fractions and operations with fractions, the document provides the following definition of a fraction: "A number expressible in the form 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.)"[31] The document itself also refers to negative fractions.
• Encyclopedia: The Online Encyclopaedia of Mathematics . Fraction, arithmetical .
• Encyclopedia: Encyclopædia Britannica . Fraction .
• Encyclopedia: Citizendium . Fraction (mathematics) .
• Encyclopedia: PlanetMath . Fraction . 29 September 2019 . live . https://web.archive.org/web/20191025060020/https://planetmath.org/fraction . 25 October 2019 .
## Notes and References
1. H. Wu, "The Mis-Education of Mathematics Teachers", Notices of the American Mathematical Society, Volume 58, Issue 03 (March 2011), p. 374. .
2. Book: Schwartzman, Steven . The Words of Mathematics: An Etymological Dictionary of Mathematical Terms Used in English . registration. Mathematical Association of America . 1994. 978-0-88385-511-9 .
3. Web site: Fractions . 2020-08-27 . www.mathsisfun.com.
4. Web site: Weisstein. Eric W.. Fraction. 2020-08-27. mathworld.wolfram.com. en.
5. Book: Ambrose, Gavin . Paul Harris . 1 . . 74 . The Fundamentals of Typography . 2nd . AVA Publishing . Lausanne . 2006 . 978-2-940411-76-4 . 2016-02-20 . https://web.archive.org/web/20160304022742/https://books.google.co.jp/books?id=IW9MAQAAQBAJ&printsec=frontcover . 2016-03-04 . live . .
6. Book: David E. Smith. History of Mathematics. 1 June 1958. Courier Corporation. 978-0-486-20430-7. 219.
7. Web site: World Wide Words: Vulgar fractions . World Wide Words . 2014-10-30 . https://web.archive.org/web/20141030183347/http://www.worldwidewords.org/qa/qa-vul1.htm . 2014-10-30 . live.
8. Book: Jack Williams . Robert Recorde: Tudor Polymath, Expositor and Practitioner of Computation . 19 November 2011 . Springer Science & Business Media . 978-0-85729-862-1 . 87–.
9. Book: Record, Robert . Record's Arithmetick: Or, the Ground of Arts: Teaching the Perfect Work and Practise of Arithmetick ... Made by Mr. Robert Record ... Afterward Augmented by Mr. John Dee. And Since Enlarged with a Third Part of Rules of Practise ... By John Mellis. And Now Diligently Perused, Corrected ... and Enlarged ; with an Appendix of Figurative Numbers ... with Tables of Board and Timber Measure ... the First Calculated by R. C. But Corrected, and the Latter ... Calculated by Ro. Hartwell ... . 1654 . James Flesher, and are to be sold by Edward Dod . 266–.
10. Web site: Math Forum – Ask Dr. Math: Can Negative Fractions Also Be Proper or Improper? . Laurel . 31 March 2004 . 2014-10-30 . https://web.archive.org/web/20141109010850/http://mathforum.org/library/drmath/view/65128.html . 9 November 2014 . live.
11. Web site: New England Compact Math Resources . 2011-12-31 . https://web.archive.org/web/20120415053421/http://www.necompact.org/ea/gle_support/Math/resources_number/prop_fraction.htm . 2012-04-15 . dead.
12. Book: Greer . A. . New comprehensive mathematics for 'O' level . 1986 . Thornes . Cheltenham . 978-0-85950-159-0 . 5 . 2nd ed., reprinted . 2014-07-29 . https://web.archive.org/web/20190119204758/https://books.google.com/books?id=wX2dxeDahAwC&pg=PA5 . 2019-01-19 . live.
13. Book: Trotter, James. A complete system of arithmetic. 65. 1853.
14. Book: Barlow, Peter. A new mathematical and philosophical dictionary. 1814.
15. complex fraction . Collins English Dictionary . https://web.archive.org/web/20171201182513/https://www.collinsdictionary.com/dictionary/english/complex-fraction . 2017-12-01 . 29 August 2022.
16. Complex fraction definition and meaning . Collins English Dictionary . 2018-03-09 . 2018-03-13 . https://web.archive.org/web/20171201182513/https://www.collinsdictionary.com/dictionary/english/complex-fraction . 2017-12-01 . live .
17. Web site: Compound Fractions . Sosmath.com . 1996-02-05 . 2018-03-13 . https://web.archive.org/web/20180314105714/http://www.sosmath.com/algebra/fraction/frac5/frac5.html . 2018-03-14 . live .
18. Book: Schoenborn . Barry . Simkins . Bradley . 2010 . Technical Math For Dummies . 8. Fun with Fractions . . 120 . Hoboken . en . 978-0-470-59874-0 . 719886424 . https://archive.org/details/technical-math-for-dummies_202007/page/120 . 28 September 2020.
19. Web site: Fraction . Encyclopedia of Mathematics . 2012-04-06 . 2012-08-15 . https://web.archive.org/web/20141021043927/http://www.encyclopediaofmath.org/index.php/Fraction . 2014-10-21 . live .
20. Putting Fractions in Their Place . Leslie Blackwell . Galen . . March 2004 . 111 . 238–242 . 3 . 10.2307/4145131 . 4145131 . 2010-01-27 . https://web.archive.org/web/20110713044149/http://www.integretechpub.com/research/papers/monthly238-242.pdf . 2011-07-13 . live .
|
# How do you solve the triangle given A= 30 degrees, b= 8.9 mi c= 6.0 mi?
May 2, 2018
Here's the situation
Let's put this in a table:
color(white)(..)A n g l e color(white)(. . . . ?) | color(white)(..) L e n g t h
color(white)(..) A = 30^o color(white)(..?) | color(white)(..) a =?
color(white)(..) B = ? color(white)(30^o)color(white)(..) | color(white)(..) b =8.9
color(white)(..) C = ? color(white)(30^o)color(white)(..) | color(white)(..) c =6.0
Law of Cosines
${a}^{2} = {b}^{2} + {c}^{2} - 2 \times b \times c \times \cos \left(A\right)$
$a = \sqrt{{b}^{2} + {c}^{2} - 2 \times b \times c \times \cos \left(A\right)}$
$a = \sqrt{{8.9}^{2} + {6}^{2} - 2 \times 8.9 \times 6 \times \cos \left(30\right)}$
$a = \sqrt{79.21 + 36 - 92.5}$
$a = 4.77$
color(white)(..)A n g l e color(white)(. . . . ?) | color(white)(..) L e n g t h
color(white)(..) A = 30^o color(white)(..?) | color(white)(..) a = 4.77
color(white)(..) B = ? color(white)(30^o)color(white)(..) | color(white)(..) b =8.9
color(white)(..) C = ? color(white)(30^o)color(white)(..) | color(white)(..) c =6.0
Law of Sine
$\sin \frac{A}{a} = \sin \frac{C}{c}$
${\sin}^{- 1} \left(c \times \sin \frac{A}{a}\right) = C$
${\sin}^{- 1} \left(6.0 \times \sin \frac{30}{4.77}\right) = C = {39}^{o}$
color(white)(..)A n g l e color(white)(. . . . ?) | color(white)(..) L e n g t h
color(white)(..) A = 30^o color(white)(..?) | color(white)(..) a = 4.77
color(white)(..) B = ? color(white)(30^o)color(white)(..) | color(white)(..) b =8.9
color(white)(..) C = 39^o color(white)(..?) | color(white)(..) c =6.0
Since the angle of a triangle must add up to ${180}^{o}$, we already know the angle of $B$:
$180 - 30 - 39 = 111$
color(white)(..)A n g l e color(white)(. . . . ?1) | color(white)(..) L e n g t h
color(white)(..) A = 30^o color(white)(..?1) | color(white)(..) a = 4.77
color(white)(..) B = 111^o color(white)(. .?) | color(white)(..) b =8.9
color(white)(..) C = 39^o color(white)(. .? 1) | color(white)(..) c =6.0
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# Integration by Substitution
## Solved Problems
### Example 9.
Compute the integral $\int {\frac{{xdx}}{{1 + {x^4}}}}.$
Solution.
We can try the substitution $$u = {x^2}.$$ Then
$du = 2xdx,\;\; \Rightarrow xdx = \frac{{du}}{2}.$
Hence, the integral is equal to
$\int {\frac{{xdx}}{{1 + {x^4}}}} = \int {\frac{{\frac{{du}}{2}}}{{1 + {u^2}}}} = \frac{1}{2}\int {\frac{{du}}{{1 + {u^2}}}} = \frac{1}{2}\arctan u + C = \frac{1}{2}\arctan {x^2} + C.$
### Example 10.
Evaluate the integral $\int {\frac{{xdx}}{{{x^4} + 2{x^2} + 1}}}.$
Solution.
If we complete the square in the denominator:
${x^4} + 2{x^2} + 1 = {\left( {{x^2} + 1} \right)^2} = {\left( {1 + {x^2}} \right)^2},$
we can use the substitution $$u = 1 + {x^2}.$$ Write the differential $$du:$$
$du = d\left( {1 + {x^2}} \right) = 2xdx.$
We see that the numerator can be expressed in terms of $$u:$$
$xdx = \frac{{du}}{2}.$
Hence, the integral is given by
$\int {\frac{{xdx}}{{{x^4} + 2{x^2} + 1}}} = \int {\frac{{xdx}}{{{{\left( {1 + {x^2}} \right)}^2}}}} = \int {\frac{{\frac{{du}}{2}}}{{{u^2}}}} = \frac{1}{2}\int {\frac{{du}}{{{u^2}}}} = \frac{1}{2}\int {{u^{ - 2}}du} = \frac{1}{2} \cdot \frac{{{u^{ - 1}}}}{{\left( { - 1} \right)}} + C = - \frac{1}{{2u}} + C = - \frac{1}{{2\left( {1 + {x^2}} \right)}} + C.$
### Example 11.
Calculate the integral $\int {{2^x}{e^x}dx}.$
Solution.
We rewrite the integral in the following way:
$\int {{2^x}{e^x}dx} = \int {{{\left( {2e} \right)}^x}dx} .$
Denoting $$2e = a$$ (this is not a change of variable, since $$x$$ still remains the independent variable), we get the table integral:
$\int {{{\left( {2e} \right)}^x}dx} = \int {{a^x}dx} = \frac{{{a^x}}}{{\ln a}} + C = \frac{{{{\left( {2e} \right)}^x}}}{{\ln \left( {2e} \right)}} + C = \frac{{{2^x}{e^x}}}{{\ln 2 + \ln e}} + C = \frac{{{2^x}{e^x}}}{{\ln 2 + 1}} + C.$
### Example 12.
Find the integral $\int {x{e^{ - {x^2}}}dx}.$
Solution.
Using the substitution $$u = - {x^2},$$ we have
$du = d\left( { - {x^2}} \right) = - 2xdx.$
Note that
$xdx = - \frac{{du}}{2},$
so we can rewrite the integral in terms of the variable $$u$$ and solve it:
$\int {x{e^{ - {x^2}}}dx} = \int {{e^u}\left( { - \frac{{du}}{2}} \right)} = - \frac{1}{2}\int {{e^u}du} = - \frac{1}{2}{e^u} + C = - \frac{{{e^{ - {x^2}}}}}{2} + C.$
### Example 13.
Evaluate the integral $\int {\frac{{\sin x}}{{1 - \cos x}} dx}.$
Solution.
We make the substitution $$u = 1 - \cos x.$$ Hence
$du = - \left( { - \sin x} \right)dx = \sin xdx.$
This gives
$\int {\frac{{\sin x}}{{1 - \cos x}}dx} = \int {\frac{{du}}{u}} = \ln \left| u \right| + C = \ln \left| {1 - \cos x} \right| + C.$
### Example 14.
Evaluate the integral $\int {x\sqrt {x + 1} dx}.$
Solution.
To get rid of the square root, we make the substitution $$u = \sqrt {x + 1} .$$ Then
${u^2} = x + 1,\;\; \Rightarrow x = {u^2} - 1,\;\; \Rightarrow du = 2udu.$
The integral becomes
$\int {x\sqrt {x + 1} dx} = \int {\left( {{u^2} - 1} \right)u \cdot 2udu} = 2\int {\left( {{u^2} - 1} \right){u^2}du} = 2\int {\left( {{u^4} - {u^2}} \right)du} = 2\int {{u^4}du} - 2\int {{u^2}du} = 2 \cdot \frac{{{u^5}}}{5} - 2 \cdot \frac{{{u^3}}}{3} + C = \frac{2}{5}{\left( {x + 1} \right)^{\frac{5}{2}}} - \frac{2}{3}{\left( {x + 1} \right)^{\frac{3}{2}}} + C.$
### Example 15.
Calculate the integral $\int {\cot \left( {3x + 5} \right)dx}.$
Solution.
We can write the integral as
$\int {\cot \left( {3x + 5} \right)dx} = \int {\frac{{\cos \left( {3x + 5} \right)}}{{\sin\left( {3x + 5} \right)}}dx} .$
Changing the variable
$u = \sin\left( {3x + 5} \right),\;\;\;du = 3\cos \left( {3x + 5} \right)dx,\;\; \Rightarrow \cos \left( {3x + 5} \right)dx = \frac{{du}}{3},$
$\int {\cot \left( {3x + 5} \right)dx} = \int {\frac{{\cos \left( {3x + 5} \right)}}{{\sin\left( {3x + 5} \right)}}dx} = \int {\frac{{\frac{{du}}{3}}}{u}} = \frac{1}{3}\int {\frac{{du}}{u}} = \frac{1}{3}\ln \left| u \right| + C = \frac{1}{3}\ln \left| {\sin\left( {3x + 5} \right)} \right| + C.$
### Example 16.
Find the integral $\int {{\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}} dx}.$
Solution.
We make the following substitution:
$u = 1 + {\cos ^2}x,\;\; \Rightarrow du = {\left( {1 + {{\cos }^2}x} \right)^\prime }dx = 2\cos x \cdot \left( { - \sin x} \right)dx = - \sin 2xdx.$
Hence,
$\int {\frac{{\sin 2x}}{{\sqrt {1 + {{\cos }^2}x} }}dx} = \int {\frac{{\left( { - du} \right)}}{{\sqrt u }}} = - 2\int {\frac{{du}}{{2\sqrt u }}} = - 2\sqrt u + C = - 2\sqrt {1 + {{\cos }^2}x} + C.$
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# What Is The Probability Of Getting 53 Fridays?
## What is the probability of getting 53 Fridays or 53 Saturdays in a leap year?
Let E be the event that the leap year has 53 Fridays or 53 Saturdays.
Now, we will find the required probability using the formula P(E)=n(E)n(S) .
Hence, the required probability is 37..
## How often are there 53 Fridays in a year?
17.75%Exactly 17.75% of all years of the Gregorian calendar have 53 Fridays each. Years that have 53 Fridays are all years that begin on a Friday plus leap years that begin on a Thursday. Every 400-year cycle of the Gregorian calendar includes 58 years begin on a Friday and 13 leap years that begin on a Thursday.
## What is the probability of getting 53?
Answer: (2) In 365 days, Number of weeks = 52 weeks and 1 day is remaining. 1 remaining day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. Total of 7 outcomes, the favourable outcome is 1. ∴ probability of getting 53 Sundays = 1 / 7.
## What is the probability of 52 Sundays in a leap year?
In a leap year, we have 366 days. So, we have 52 weeks and 2 days. Out of these, 7 pairs of combinations, only 2 pairs have Sunday, and the other 5 pairs do not have Sundays. Therefore, the probability that a leap year will have only 52 Sundays is 5/7.
## When a die is thrown what is the probability of getting?
If it’s not strictly between 1 and 6, then any dice roll will give you a 1, 2, 3, 4, 5, or 6. Therefore, your probability is 6/6=1. A fair ordinary dice is thrown once.
## What is the probability of getting 53 Fridays in a non leap year?
1/7 is the probability of getting 53 fridays in a non-leap year.
## Are there 53 Fridays in 2021?
There are exactly 53 Fridays in the year 2021. Most years have 365 days, but a leap year has 366 days.
## How many Fridays are there in 2020 a year?
52 FridaysThere are exactly 52 Fridays in the year 2020. Most years have 365 days, but a leap year has 366 days. That adds up to 52 weeks (where each week is exactly 7 days) PLUS 1 or 2 additional days.
## How many work days were there in 2021?
261 working daysThere are a total of 261 working days in the 2021 calendar year.
## What is the probability of getting 53 Tuesday in a year is?
Hence the probability of getting 53 Tuesdays in an ordinary year is =71.
## What is the probability that a leap year has 52 Mondays?
0.710.71 is probability for 52 Mondays in a leap year.
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Home » Math Theory » Graphs and Charts » All Four Quadrants
# All Four Quadrants
## Introduction
You plot ordered pairs on the Cartesian plane, also known as the x-y plane, which is a two-line graph. Four distinct quadrants are created by the two intersecting lines of the Cartesian plane. A quadrant is a region bounded by the intersection of the x – axis and the y – axis.
We will learn more about the four quadrants of a coordinate system in this article, as well as how to plot and locate points in each quadrant.
## The Four Quadrants
The rectangle coordinate system is split into four equal parts by the intersection of the x – axis and the y – axis. The x – axis is the horizontal number line, while the y – axis is the vertical number line.
As shown below, the coordinate system has four regions representing one-quarter of the entire coordinate system. Each quadrant has its properties and is frequently labeled using Roman numerals.
The movement of the label from Quadrant I to IV follows an anti-clockwise rotation starting from the upper right quadrant to the bottom right quadrant.
Quadrant I, or the first quadrant, is the upper right quadrant. Both the x – axis and y – axis have positive numbers in Quadrant I.
Quadrant II, or the second quadrant, is the upper left quadrant. In Quadrant II, the x – axis has negative numbers while the y – axis has positive numbers.
Quadrant III, or the third quadrant, is the bottom left quadrant. Both the x – axis and y – axis have negative numbers in Quadrant III.
Quadrant IV, or the fourth quadrant, is the bottom right quadrant. In Quadrant IV, the x – axis has positive numbers while the y – axis has negative numbers.
## Points on the Quadrants
The location of a point has something to do with its vertical distance from the x – axis and its horizontal distance from the y – axis. The intersection of the two axes, the x- and y – axis, is known as the origin, which is point ( 0, 0 ).
Let us say, for example, in the illustration below, we have point M, and we want to identify its location in the coordinate system. To find the location of point M, let us draw two straight lines starting from this point. One line is horizontal towards the y – axis, while the other is vertical towards the x – axis.
The location of point M is 3 units on the x – axis while 4 units on the y – axis. Therefore, the location of point M using an ordered pair is ( 3, 4 ), or we may also say M ( 3, 4 ).
The points on the coordinate system are always written in the form ( x, y ), in which x is referred to as the abscissa, and y is called the ordinate.
As for another example, in the figure below, let us locate point K in the coordinate system. To find the location of point K, let us draw two straight lines starting from this point. One line is horizontal towards the y – axis, while the other is vertical towards the x – axis.
The location of point K is 2 units on the x – axis while 1 unit on the y – axis. Therefore, the location of point K using an ordered pair is ( 2, -1 ), or we may also say K ( 2, -1 ).
## Signs of the Coordinates in Quadrants
There are four different quadrants, depending on the sign of the coordinates. To identify which quadrant the point is in, you must know the ordinate and abscissa signs of an ordered pair.
The signs of the coordinates for each quadrant
are displayed in the table below.
## Plotting Points on the Coordinate System
In plotting points on the coordinate system, you must be mindful of the signs of the x-coordinate and the y-coordinate. Remember also that the intersection of the two axes, x and y, is the origin ( 0, 0 ).
Let us say, for example, we want to plot point Z with coordinates ( -2, 3 ). Since the x-coordinate is negative 2, let us move 2 units to the left starting from the origin.
Since the y-coordinate is positive 3, we will move 3 units upward. As shown below, point Z is in Quadrant II.
As for another example, let us plot point H with coordinate ( 3, -3 ). Since the x-coordinate is positive, we will move 3 units to the right starting from the origin. Then, move 3 units down since the y-coordinate is negative 3. The figure below shows that H ( 3, -3 ) is in Quadrant IV.
## Examples
Example 1
Determine the quadrant in which each point is located.
( a ) ( 5, 8 )
( b ) ( -9, 12 )
( c ) ( 20, -16 )
( d ) ( -3, -10 )
( e ) ( -4, 11 )
( f ) ( 12, -5 )
Solution
( a ) ( 5, 8 )
Since the x-coordinate and the y-coordinate are both positive, ( 5, 8 ) is in Quadrant I.
( b ) ( -9, 12 )
Since the x-coordinate is negative and the y-coordinate is positive, ( -9, 12 ) is in Quadrant II.
( c ) ( 20, -16 )
Since the x-coordinate is positive and the y-coordinate is negative, ( 20, -16 ) is in Quadrant IV.
( d ) ( -3, -10 )
Both the x-coordinate and the y-coordinate are negative; hence, ( -3, -10 ) is in Quadrant III.
( e ) ( -4, 11 )
Since the x-coordinate is negative and the y-coordinate is positive, ( -4, 11 ) is in Quadrant II.
( f ) ( 12, -5 )
Since the x-coordinate is positive and the y-coordinate is negative, ( 12, -5 ) is in Quadrant IV.
Example 2
Plot the following points in a coordinate system and note which quadrant each one is in.
( a ) A ( 1, 5 )
( b ) B ( 2, -3 )
( c ) C ( -4, -1 )
( d ) D ( -1, 3 )
( e ) E ( 5, 4 )
Solution
The graph below shows the location of each point.
( a ) A ( 1, 5 )
Since both the x-coordinate and the y-coordinate are positive, point A is in Quadrant I.
( b ) B ( 2, -3 )
Since the x-coordinate is positive while the y-coordinate is negative, point B is in Quadrant IV.
( c ) C ( -4, -1 )
Both the x-coordinate and the y-coordinate are negative. Hence, point C is in Quadrant III.
( d ) D ( -1, 3 )
Since the x-coordinate is negative and the y – coordinate is positive, point D is in Quadrant II.
( e ) E ( 5, 4 )
Since both the x-coordinate and the y-coordinate are positive, point E is in Quadrant I.
Example 3
Which points below lie in the third quadrant or QIII?
L ( -3, 9 )
M ( -4, -8 )
N ( 9, -8 )
O ( -10, 25 )
P ( -12, -17 )
Q ( 4, 15 )
Solution
The points in the third quadrant or QIII must have both negative x- and y-coordinates. Therefore, points M ( -4, -8 ) and P ( -12, -17) are in QIII.
On the other hand, points L ( -3, 9 ) and O ( -10, 25 ) are in Quadrant II, point N ( 9, -8 ) is in Quadrant IV, and point Q ( 4, 15 ) is in Quadrant I.
Example 4
Where does each point lie in the coordinate system’s four quadrants?
( a ) P ( 2, 0 )
( b ) Q ( 0, -4 )
( c ) R ( -6, 0 )
( d ) S ( 0, 7 )
( e ) S ( 5, 0 )
Solution
( a ) P ( 2, 0 )
Starting from the origin, point P with coordinates ( 2, 0 ) is 2 units to the right. It does not lie in any of the four quadrants. Point P ( 2, 0 ) is along the positive x – axis.
( b ) Q ( 0, -4 )
Starting from the origin, point Q with coordinates ( 0, -4 ) is 4 units downward. It does not lie in any of the four quadrants. Point Q ( 0, -4 ) is along the negative y – axis.
( c ) R ( -6, 0 )
Starting from the origin, point R with coordinates ( -6, 0 ) is 6 units to the left. It does not lie in any of the four quadrants. Point R ( -6, 0 ) is along the negative x – axis.
( d ) S ( 0, 7 )
Starting from the origin, point S with coordinates ( 0, 7 ) is 7 units upward. It does not lie in any of the four quadrants. Point S ( 0, 7 ) is along the positive y – axis.
( e ) P ( 5, 0 )
Starting from the origin, point P with coordinates ( 2, 0 ) is 5 units to the right. It does not lie in any of the four quadrants. Point P ( 5, 0 ) is along the positive x – axis.
Example 5
Give 5 examples of points in each quadrant in a coordinate system.
Solution
In Quadrant I, both x – coordinate and y – coordinate must be positive; hence, the following are some examples: ( 1, 2 ). ( 4, 8 ), ( 16, 78 ), ( 3, 7 ), and ( 100, 1000 ).
In Quadrant II, the x – coordinate is negative while the y – coordinate is positive; thus, some examples are ( -8, 9 ). ( -6, 10 ), ( -21, 18 ), ( -3, 15 ), and ( -25, 100 ).
In Quadrant III, both x – coordinate and y – coordinate must be negative; hence, the following examples are ( -6, -21 ). ( -13, -5 ), ( -2, -1 ), ( -4, -18 ), and ( -100, -8 ).
In Quadrant IV, the x – coordinate is positive while the y – coordinate is negative; thus, some examples are ( 9, -8 ). ( 21, -6 ), ( 4, -1 ), ( 11, -12 ), and ( 19, -5 ).
Example 6
In which quadrant does ( x, 3 ) lie?
Solution
Since ( x, 3 ) does not indicate a definite x – coordinate, it means that x is a collection of positive or negative numbers. If we try to plot it in the coordinate system, it forms a horizontal at y = 3.
Therefore, we can say that the quadrants are QI and QII.
## Trigonometric Functions in Quadrants
A trigonometric function’s sign is determined by the signs of the x and y – coordinates of the points on the angle’s terminal side. You can determine the signs of all the trigonometric functions by determining which quadrant an angle’s terminal side is located.
All six trigonometric functions are positive in the first quadrant or Quadrant I since both x- and y – coordinates are positive.
In the second quadrant or Quadrant II, since y – coordinate is positive and x – coordinate is negative, only sine and cosecant are positive.
In the third quadrant or Quadrant III, since both x- coordinate and y – coordinate are negative, only tangent and cotangent are positive.
In the fourth quadrant or QIV, since x – coordinate is positive and y – coordinate is negative, only cosine and cosecant are positive.
The chart below illustrates the trigonometric functions that are positive in each quadrant.
## Summary
The rectangle coordinate system is split into four equal parts by the intersection of the x – axis and the y – axis. The x – axis is the horizontal number line, while the y – axis is the vertical number line.
Each one-quarter region of the entire coordinate system is a quadrant. The quadrants are called the first quadrant ( QI ), second quadrant ( QII ), third quadrant ( QIII ), and fourth quadrant ( QIV ).
The signs of the coordinates for each quadrant are displayed in the table below.
Angles in Each Quadrant
The angles in the first quadrant or QI lie between 0° and 90°.
The angles in the second quadrant or QII lie between 90° and 180°.
The angles in the third quadrant or QIII lie between 180° and 270°.
The angles in the fourth quadrant or QIV lie between 270° and 360°.
The signs of the six trigonometric functions in each quadrant are displayed in the table below.
## Frequently Asked Questions on All Four Quadrants ( FAQs )
### What is meant by quadrants in a coordinate system?
The rectangle coordinate system is split into four equal parts by the intersection of the x – axis and the y – axis. The x – axis is the horizontal number line, while the y – axis is the vertical number line. Each one-quarter region of the entire coordinate system is a quadrant. The quadrants are called the first quadrant ( QI ), second quadrant ( QII ), third quadrant ( QIII ), and fourth quadrant ( QIV ).
The image below shows the four quadrants in a coordinate system.
### What is a coordinate system?
Any point can be located using a Cartesian coordinate system or coordinate system, and that point can be displayed as an ordered pair (x, y). The x and y are known as the coordinates of the ordered pair. The origin, indicated by the letter “O,” is the place where the two number lines intersect. The horizontal number line is known as the x – axis, while the vertical number line is known as the y – axis.
### Where is the origin in the coordinate system?
The coordinates coordinate system’s origin is the intersection of the x – and y – axis. It is point ( 0, 0 ) as an ordered pair.
### How do you identify the quadrant of a point?
A coordinate system has four quadrants. Each quadrant has its properties and is frequently labeled using Roman numerals.
Quadrant I, or the first quadrant, is the upper right quadrant. Both the x – axis and y – axis have positive numbers in Quadrant I.
Quadrant II, or the second quadrant, is the upper left quadrant. In Quadrant II, the x – axis has negative numbers while the y – axis has positive numbers.
Quadrant III, or the third quadrant, is the bottom left quadrant. Both the x – axis and y – axis have negative numbers in Quadrant III.
Quadrant IV, or the fourth quadrant, is the bottom right quadrant. In Quadrant IV, the x – axis has positive numbers while the y – axis has negative numbers.
The signs of the coordinates for each quadrant are displayed in the table below.
### What are the measures of the angles in each quadrant?
The coordinate system is split into four quadrants, I, II, III, and IV. If we are referring to angles in standard position, they have different measures depending on the location of the terminal side in the coordinate system.
As shown below, 0°, 90°, 180°, 270°, and 360° lie along the axes, and they are considered as quadrantal angles. Note that the labels below are for angles in standard position with positive rotations.
The angles in the first quadrant or QI lie between 0° and 90°.
The angles in the second quadrant or QII lie between 90° and 180°.
The angles in the third quadrant or QIII lie between 180° and 270°.
The angles in the fourth quadrant or QIV lie between 270° and 360°.
### What are the sign conventions for the coordinates in quadrants?
The signs of the coordinates for each quadrant are displayed in the table below.
### What are the trigonometric values in each quadrant?
All six trigonometric functions are positive in the first quadrant or Quadrant I since both x- and y – coordinates are positive.
In the second quadrant or Quadrant II, since y – coordinate is positive and x – coordinate is negative, only sine and cosecant are positive.
In the third quadrant or Quadrant III, since both x- coordinate and y – coordinate are negative, only tangent and cotangent are positive.
In the fourth quadrant or QIV, since x – coordinate is positive and y – coordinate is negative, only cosine and cosecant are positive.
The signs of the six trigonometric functions in each quadrant are displayed in the table below.
### What are all four quadrants in order?
The movement of the label from Quadrant I to IV follows an anti-clockwise rotation starting from the upper right quadrant to the bottom right quadrant.
Quadrant I, or the first quadrant, is the upper right quadrant. Both the x – axis and y – axis have positive numbers in Quadrant I.
Quadrant II, or the second quadrant, is the upper left quadrant. In Quadrant II, the x – axis has negative numbers while the y – axis has positive numbers.
Quadrant III, or the third quadrant, is the bottom left quadrant. Both the x – axis and y – axis have negative numbers in Quadrant III.
Quadrant IV, or the fourth quadrant, is the bottom right quadrant. In Quadrant IV, the x – axis has positive numbers while the y – axis has negative numbers.
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These worksheets will give your students a review in working with the rules of division.
#### What Are the Operation Rules of Division? With Respect to Division - Basics - We use the method of division as the inverse of multiplication. In division, we split a larger value into the smaller one and we call the little value is quotient. For example; we split the set of 8 boxes into 4 equal parts then, you will have 4 sets of 2 boxes. In mathematical expression; 8/4 = 2. Operation rules of divisions with negative numbers. In arithmetic, signs rules are same for all and similarly in division too such as; When we divide two positive number, the answer will be a positive number. When we divide one +ve and one -ve number, the answer will be negative one. By dividing two -ve values, we get positive number. In division, you will always have a positive sign when the dividend and the divisor contain the same sign. Understand this by mathematical expression mentioned below; 8 / (- 2) = - 4 and (- 8) / 2 = - 4. But, ( - 8 ) / ( - 2 ) = 4.
This section looks at how integer rules work with fractions and algebraic fractions that are undergoing a quotient operation. In many case, cross multiplication will be your best friend of all time. You will find the reciprocal of the second fraction and then just multiply. The key is to remember that you then multiply the numerators together and then the denominators together. These worksheets cover working with the rules of division. Questions cover solving fractions, converting between division and multiplication of fractions, positive and negative numbers, and more.
# Integer Operation Rules with Division Worksheets
## Lesson
This goes over the very basics to remind students where they are with this skill. Solve the numerator first: (a - a = 0) Now, substitute the value of the numerator in the above expression Here, we substitute the value of (a - a)/p>
## Integer Operations Rules with Division Lesson and Practice
Students will practice working with the rules of division in a series of operations. A sample problem is solved and two practice problems are provided.
## Worksheet
Students will simplify each expression that is presented over the series of ten problems that are provided.
## Practice
Change to multiplication; then simplify to find the quotient of these mixed values.
## Skill Drill
Students will translate the sentences into mathematical expressions. Eight problems are provided.
## Warm Up
Students demonstrate skill with working with the rules of division in an operation based setting. Three problems are provided.
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# What is 19/84 as a decimal?
## Solution and how to convert 19 / 84 into a decimal
19 / 84 = 0.226
Fraction conversions explained:
• 19 divided by 84
• Numerator: 19
• Denominator: 84
• Decimal: 0.226
• Percentage: 0.226%
19/84 converted into 0.226 begins with understanding long division and which variation brings more clarity to a situation. Both represent numbers between integers, in some cases defining portions of whole numbers But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. Once we've decided the best way to represent the number, we can dive into how to convert 19/84 into 0.226
19 / 84 as a percentage 19 / 84 as a fraction 19 / 84 as a decimal
0.226% - Convert percentages 19 / 84 19 / 84 = 0.226
## 19/84 is 19 divided by 84
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. To solve the equation, we must divide the numerator (19) by the denominator (84). Here's how you set your equation:
### Numerator: 19
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. 19 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. Values closer to one-hundred make converting to fractions more complex. Let's take a look at the denominator of our fraction.
### Denominator: 84
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. 84 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. Overall, two-digit denominators are no problem with long division. Now let's dive into how we convert into decimal format.
## How to convert 19/84 to 0.226
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 84 \enclose{longdiv}{ 19 }$$
We will be using the left-to-right method of calculation. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 84 \enclose{longdiv}{ 19.0 }$$
Because 84 into 19 will equal less than one, we can’t divide less than a whole number. So that means we must add a decimal point and extend our equation with a zero. Now 84 will be able to divide into 190.
### Step 3: Solve for how many whole groups you can divide 84 into 190
$$\require{enclose} 00.2 \\ 84 \enclose{longdiv}{ 19.0 }$$
We can now pull 168 whole groups from the equation. Multiply by the left of our equation (84) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.2 \\ 84 \enclose{longdiv}{ 19.0 } \\ \underline{ 168 \phantom{00} } \\ 22 \phantom{0}$$
If your remainder is zero, that's it! If you still have a remainder, 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.
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 19/84 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. This is also true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 19/84 into a decimal
Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.22 per hour and not$20 and 19/84.
### When to convert 0.226 to 19/84 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 19/84 = 0.226 what would it be as a percentage?
• What is 1 + 19/84 in decimal form?
• What is 1 - 19/84 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.226 + 1/2?
### Convert more fractions to decimals
From 19 Numerator From 84 Denominator What is 19/74 as a decimal? What is 9/84 as a decimal? What is 19/75 as a decimal? What is 10/84 as a decimal? What is 19/76 as a decimal? What is 11/84 as a decimal? What is 19/77 as a decimal? What is 12/84 as a decimal? What is 19/78 as a decimal? What is 13/84 as a decimal? What is 19/79 as a decimal? What is 14/84 as a decimal? What is 19/80 as a decimal? What is 15/84 as a decimal? What is 19/81 as a decimal? What is 16/84 as a decimal? What is 19/82 as a decimal? What is 17/84 as a decimal? What is 19/83 as a decimal? What is 18/84 as a decimal? What is 19/84 as a decimal? What is 19/84 as a decimal? What is 19/85 as a decimal? What is 20/84 as a decimal? What is 19/86 as a decimal? What is 21/84 as a decimal? What is 19/87 as a decimal? What is 22/84 as a decimal? What is 19/88 as a decimal? What is 23/84 as a decimal? What is 19/89 as a decimal? What is 24/84 as a decimal? What is 19/90 as a decimal? What is 25/84 as a decimal? What is 19/91 as a decimal? What is 26/84 as a decimal? What is 19/92 as a decimal? What is 27/84 as a decimal? What is 19/93 as a decimal? What is 28/84 as a decimal? What is 19/94 as a decimal? What is 29/84 as a decimal?
### Convert similar fractions to percentages
From 19 Numerator From 84 Denominator 20/84 as a percentage 19/85 as a percentage 21/84 as a percentage 19/86 as a percentage 22/84 as a percentage 19/87 as a percentage 23/84 as a percentage 19/88 as a percentage 24/84 as a percentage 19/89 as a percentage 25/84 as a percentage 19/90 as a percentage 26/84 as a percentage 19/91 as a percentage 27/84 as a percentage 19/92 as a percentage 28/84 as a percentage 19/93 as a percentage 29/84 as a percentage 19/94 as a percentage
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# What is the standard form of y= (x-2)(3x+2)(-x+8)?
Dec 10, 2017
$- 3 {x}^{3} + 28 {x}^{2} - 28 x - 32$
#### Explanation:
$\text{expand the first 2 factors using FOIL then multiply the}$
$\text{result by the third factor}$
$\left(x - 2\right) \left(3 x + 2\right) \leftarrow \textcolor{b l u e}{\text{first 2 factors}}$
$= 3 {x}^{2} + 2 x - 6 x - 4$
$= 3 {x}^{2} - 4 x - 4$
$\Rightarrow \left(3 {x}^{2} - 4 x - 4\right) \left(- x + 8\right) \leftarrow \textcolor{b l u e}{\text{multiply by third factor}}$
$= - 3 {x}^{3} + 24 {x}^{2} + 4 {x}^{2} - 32 x + 4 x - 32$
$= - 3 {x}^{3} + 28 {x}^{2} - 28 x - 32 \leftarrow \textcolor{b l u e}{\text{in standard form}}$
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# A rectangular piece is 20 m long and 15 m wide.
Question:
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.
Solution:
It is given that, the quadrants of radius r have been cut from the four corners of a rectangular piece is of length and width.
We have to find the area of remaining part.
We know that,
Area of rectangle $=I \times w$
$=20 \times 15$
$=300 \mathrm{~m}^{2}$
Area of quadrant $=\frac{1}{4} \pi r^{2}$
$=\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5$
$=9.625 \mathrm{~m}^{2}$
Now,
Area of remaining part $=$ Area of rectangle $-4 \times$ Area of quadrant
$=300-4 \times 9.625$
$=300-38.5$
$=261.5 \mathrm{~m}^{2}$
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# A Simple Definition of Probability Theory
Imagine that a street performer invites you to play a game: you need to guess the total of the 2 rolled dice to win a prize. It’s impossible to predict which total from 2 to 12 will be on the dice. But if you say 7, you’ll have a better chance of winning than if you say 2. This is what probability theory is about.
At 5-Minute Crafts, we would like to explain, in simple terms, what probability theory is.
### What is probability theory?
Let’s make the situation with the dice simpler by using an ordinary coin. If you flip it, you’ll either get heads or tails. Try flipping the coin 100 times and you’ll see that each side of the coin comes up about the same number of times. It means the probability of flipping either heads or tails is 50%.
If you roll the dice, you’ll get a total from 2 to 12 (according to the numbers on the sides of the dice). The total number rolled on the dice will be a random variable. In other words, a random variable is a numerical expression of the result of a random event.
If you flip a coin, you get either heads or tails. These are 2 random events. Each of them may or may not happen. An event is a basic element of probability theory.
Probability theory is the branch of mathematics that studies patterns associated with random phenomena: random events, random variables, their properties, and operations on them.
### What’s sample space?
Let’s take 2 coins. If you flip each of them, you’ll get heads or tails, 1 of 2 possible events. And how many possible events can you get if you flip 2 coins at the same time?
Let’s take a look at the picture above. The letter T stands for when you get tails, and the letter H stands for the event when you get heads. Let’s say that when you flip the coins for the first time, you get heads on one coin and tails on the other. When you flip the coins for the second time, each coin can show either heads or tails, which will eventually give you 4 possible combinations based on the outcome of 2 flips. The thing that you see in the picture is called a tree diagram. It depicts all possible independent events in the given series.
So the combinations of HH, HT, TH, and TT represent all the different outcomes of our series and are called a sample space, which is denoted by the label S.
### How to find probability with a formula
The probability that an event will or will not happen in a particular series can be calculated using the formula, P(A) = m/n, where “P” is the probability of the event, “A” is the event, “n” is the total number of all possible outcomes of this series, and “m” is the number of outcomes due to which event “A” can happen.
Let’s take candies of different colors as an example. There are 51 candies in the package, 6 of which are blue. The probability that you will get a blue candy from this package (event A in this case) is: P(A) = m/n. It means 6/51, or a little more than 11% (instead of using a fraction, we can write the probability as a percentage).
• If all candies in the package are blue, then the probability that you will get a blue candy will be 51/51 = 1.
• If you want to find the probability of getting a purple candy from this package, you will get 0/51 = 0.
Based on the above, we now know that probability has a few properties:
• If the probability of an event is 1, then it will 100% happen.
• If the probability of an event is 0, then it will never happen.
• Between 0 and 1, there is a positive number that shows the probability of a particular random event. If the probability of an event is 0.3, this means that it will happen with a probability of 30%.
Taking this into account, we can conclude that the probability of any event is 0 ≤ P(A) ≤ 1.
### How to solve easy probability practice problems
Problem 1: There are 15 apples in the basket — 10 of them are green and 5 are red. What is the probability of getting a yellow apple from the basket?
Solution: Let’s use the formula, P(A) = m/n. In this case, n = 15 and m = 0. Therefore, the probability of getting a yellow apple out of the basket is P(A) = 0/15 = 0. In other words, it’s impossible.
Problem 2: There are 36 cards in the deck. You take one card out of it. What is the probability that this is a card of spades?
Solution: Let’s use the formula, P(A) = m/n. In this case, n = 36, and m = 9 (that’s how many cards of spades are in the entire deck). This means that the probability of the desired outcome is equal to P(A) = 9/36 = 0.25, or 25% probability that you will get a card of spades.
### Addition and multiplication theorems of probability
Imagine that you have 3 piles in front of you, and each has 10 pieces of a construction toy. The first pile has 8 pieces for building a house, the second pile has 7 pieces, and the third one has 9 pieces. All other pieces are decorative elements and parts of mini-figures. You pick one random piece from each pile. What is the probability that all 3 pieces you’ve picked will be building pieces?
Let’s specify the conditions: If you take a building piece from one pile, this doesn’t mean that you will or will not get a similar piece from the other 2. In other words, these events are independent of each other (the probability of the occurrence of any of them doesn’t depend on the other events).
So, we can consider 3 independent events. Let’s find the probability of getting a building piece for each pile:
• First pile: 8/10, or 0.8, or 80%
• Second pile: 7/10, or 0.7, or 70%
• Third pile: 9/10, or 0.9, or 90%
To find the probability of simultaneous occurrence of these 3 independent events, we need to multiply them: 0.8 × 0.7 × 0.9 = 0.504. So the probability that all 3 pieces will be building pieces is 0.504, or more than 50%.
We solved this problem using the multiplication theorem of probability, which looks like this:
P(A × B) = P(A) × P(B|A)
The formula means that the probability of the product of 2 events is equal to the product of the probability of one of them, and the conditional probability of the other is calculated under the condition that the first event occurred (the part of the formula written as P (B|A) denotes the probability of event B, provided that event A happened). Wherein:
• Event A is independent of event B if the probability of event A doesn’t depend on whether event B occurred or not.
• Event A is dependent on event B if the probability of event A changes depending on whether event B occurred or not.
If we are dealing with independent events, then the formula looks like this:
P(A × B) = P(A) × P(B)
Problem: There are balls in a bowl — 2 of them are orange and 3 are blue. You take 2 balls out of the bowl simultaneously. What is the probability that both balls are orange?
Solution: Let’s consider the simultaneous extraction of 2 orange balls to be “event A.” It will be the product of 2 events: A = A1 × A2, where:
A1 is an orange ball taken out of the bowl during the first attempt.
A2 is an orange ball taken out of the bowl during the second attempt.
P(A1) = 2/5
P(A2) = (2-1) / (5-1) = 1/4
P(A) = P(A1 × A2) = 2/5 × 1/4 = 0.1. Therefore, the probability that both balls will be orange is only 0.1, or 10%.
The sum of 2 events A and B is event A + B, at which either event A, event B, or both events will occur simultaneously. If the events are incompatible (meaning the occurrence of one event excludes the occurrence of the other within the same series), then either event A or event B may occur.
By flipping a coin, we get 2 opposite and mutually exclusive events: you may get either heads or tails. When 2 events are opposite to each other, they are indicated by the same letters, but a dash or an apostrophe is added above one of them. For example, it would be A and Ā or A and A’.
Imagine that you have 9 pears in a bag of fruit — 4 of them are green, 3 are red, and 2 are yellow. You take one of them at random. What is the probability that it is not yellow?
To answer the question, you need to find the probability that you will get a green pear (A) and a red one (B), and then add them together (A + B) to find the probability that you will get one or the other (but not yellow).
P(A) = 4/9
P(B) = 3/9 = 1/3
These 2 events are incompatible. Let’s add their probabilities together:
P(A+B) = P(A) + P(B) = 4/9 + 1/3 = 7/9.
So the probability that the pear you’ve taken is not yellow is 7/9, or more than 77%.
To solve this problem, we used the formula, P (A + B) = P (A) + P (B), from the probability addition theorem, meaning, “The probability of occurrence of 1 of 2 incompatible events is equal to the sum of the probabilities of these events.”
If we are dealing with compatible events (meaning the occurrence of one event doesn’t exclude the occurrence of the other one within the same test), the probability of their sum is calculated by the formula: P(A + B) = P(A) + P(B) − P(A×B).
These formulas can be used to solve simple probability problems.
### Bonus: what “a chance of rain” in a weather forecast really means
Imagine that tomorrow’s weather forecast says there is a 30% chance of rain. This figure seems insignificant. So the following day, you go for a walk around the city without an umbrella and end up getting soaked. How could this happen if they said that there was only a 30% chance of rain?
If the chance of rain mentioned in the weather forecast for your city is 30%, this indicates a 100% chance of rain, but it will only happen in 30% of the area. If you move around the city on this day, then you will increase your chances of getting wet.
5-Minute Crafts/Science/A Simple Definition of Probability Theory
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# How do you solve x/12-4=20?
Mar 5, 2018
See a solution process below:
#### Explanation:
First, add $\textcolor{red}{4}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:
$\frac{x}{12} - 4 + \textcolor{red}{4} = 20 + \textcolor{red}{4}$
$\frac{x}{12} - 0 = 24$
$\frac{x}{12} = 24$
Now, multiply each side of the equation by $\textcolor{red}{12}$ to solve for $x$ while keeping the equation balanced:
$\textcolor{red}{12} \times \frac{x}{12} = \textcolor{red}{12} \times 24$
$\cancel{\textcolor{red}{12}} \times \frac{x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}}} = 288$
$x = 288$
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+0
# Help!
0
244
3
+1028
Suppose a is directly proportional to b, but inversely proportional to c. If a = 2 when b = 5 and c = 9, then what is c when b =3?
Aug 23, 2018
#1
+391
+4
You can first find the second $$a$$, by setting up an equation and solving it. We can make the equation $$\frac{2}{5} = \frac{a}{3}$$, which we can solve using cross-multiplication. Then, we get $$6 = 5a$$, which gives us $$a = \frac{6}{5}$$. So if $$a = \frac{6}{5}$$, and $$a$$ is inversely proportional to $$c$$, than we can see that $$\frac{6}{5} \times c$$ must equal $$18$$, because $$2 \times 9 = 18$$$$\frac{18}{\frac{6}{5}} = 18 \times \frac{5}{6} = 15$$. So $$c = 15$$.
- Daisy
Aug 23, 2018
#2
+1
2 = 5k
k=2/5 constant of direct proportionality. But:............................(1)
2 =k/9
k=18 constant of inverse proportionality. But when:..................(2)
b =3, from (1) above:
a =2/5 x 3=6/5 = 1.2. Using (2) above for the new value of a:
1.2 =18 / c
c = 18 / 1.2
c =15 when b = 3
Aug 24, 2018
edited by Guest Aug 24, 2018
#3
0
Look at this problem in a much simpler and intuitive way as follows:
When "b" = 5 goes to "b"= 3, that is a direct decrease of 3/5 =0.6, when c=9
And since "b" is directly proportional to "a" and inversely proportional to "c", it follows that:
New "c" = 9 x 1/0.6 =15
Aug 24, 2018
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# Illustrative Mathematics Grade 7, Unit 8, Lesson 19: Comparing Populations With Friends
Learning Targets:
• I can decide what information I need to know to be able to compare two populations based on a sample from each.
Related Pages
Illustrative Math
#### Lesson 19: Comparing Populations With Friends
Let’s ask important questions to compare groups.
Illustrative Math Unit 7.8, Lesson 19 (printable worksheets)
#### Lesson 19 Summary
When using samples to comparing two populations, there are a lot of factors to consider.
• Are the samples representative of their populations? If the sample is biased, then it may not have the same center and variability as the population.
• Which characteristic of the populations makes sense to compare—the mean, the median, or a proportion?
• How variable is the data? If the data is very spread out, it can be more difficult to make conclusions with certainty.
Knowing the correct questions to ask when trying to compare groups is important to correctly interpret the results.
#### Lesson 19.1 Features of Graphic Representations
Dot plots, histograms, and box plots are different ways to represent a data set graphically. Which of those displays would be the easiest to use to find each feature of the data?
1. the mean
2. the median
3. the mean absolute deviation
4. the interquartile range
5. the symmetry
#### Lesson 19.2 Info Gap: Comparing Populations
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the problem card:
3. Explain to your partner how you are using the information to solve the problem.
If your teacher gives you the data card:
2. Ask your partner, “What specific information do you need?” Wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your partner!)
3. Before telling your partner the information, ask “Why do you need that information?”
4. After your partner solves the problem, ask them to explain their reasoning, and listen to their explanation.
#### Are you ready for more?
Is there a meaningful difference between top sports performance in two different decades? Choose a variable from your favorite sport (for example, home runs in baseball, kills in volleyball, aces in tennis, saves in soccer, etc.) and compare the leaders for each year of two different decades. Is the performance in one decade meaningfully different from the other?
#### Lesson 19.3 Comparing to Known Characteristics
1. A college graduate is considering two different companies to apply to for a job. Acme Corp lists this sample of salaries on their website:
What typical salary would Summit Systems need to have to be meaningfully different from Acme Corp? Explain your reasoning.
2. A factory manager is wondering whether they should upgrade their equipment. The manager keeps track of how many faulty products are created each day for a week.
The new equipment guarantees an average of 4 or fewer faulty products per day. Is there a meaningful difference between the new and old equipment? Explain your reasoning.
#### Lesson 19 Practice Problems
1. An agent at an advertising agency asks a random sample of people how many episodes of a TV show they watch each day. The results are shown in the dot plot.
The agency currently advertises on a different show, but wants to change to this one as long as the typical number of episodes is not meaningfully less.
a. What measure of center and measure of variation would the agent need to find for their current show to determine if there is a meaningful difference? Explain your reasoning.
b. What are the values for these same characteristics for the data in the dot plot?
c. What numbers for these characteristics would be meaningfully different if the measure of variability for the current show is similar? Explain your reasoning.
2. Jada wants to know if there is a meaningful difference in the mean number of friends on social media for teens and adults. She looks at the friend count for the 10 most popular of her friends and the friend count for 10 of her parents’ friends. She then computes the mean and MAD of each sample and determines there is a meaningful difference. Jada’s dad later tells her he thinks she has not come to the right conclusion. Jada checks her calculations and everything is right. Do you agree with her dad? Explain your reasoning.
3. The mean weight for a sample of a certain kind of ring made from platinum is 8.21 grams. The mean weight for a sample of a certain kind of ring made from gold is 8.61 grams. Is there a meaningful difference in the weights of the two types of rings? Explain your reasoning
4. The lengths in feet of a random sample of 20 male and 20 female humpback whales were measured and used to create the box plot.
Estimate the median lengths of male and female humpback whales based on these samples.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# 9.4 Series and their notations (Page 3/18)
Page 3 / 18
A man earns $100 in the first week of June. Each week, he earns$12.50 more than the previous week. After 12 weeks, how much has he earned?
$2,025 ## Using the formula for geometric series Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series . Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio , $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ We can write the sum of the first $n$ terms of a geometric series as ${S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n–1}{a}_{1}.$ Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ terms of a geometric series. We will begin by multiplying both sides of the equation by $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ $r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}$ Next, we subtract this equation from the original equation. Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for ${S}_{n},$ divide both sides by $\left(1-r\right).$ ## Formula for the sum of the first n Terms of a geometric series A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ terms of a geometric sequence is represented as Given a geometric series, find the sum of the first n terms. 1. Identify $\text{\hspace{0.17em}}{a}_{1},\text{\hspace{0.17em}}r,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}n.$ 2. Substitute values for $\text{\hspace{0.17em}}{a}_{1},\text{\hspace{0.17em}}r,$ and $n$ into the formula ${S}_{n}=\frac{{a}_{1}\left(1–{r}^{n}\right)}{1–r}.$ 3. Simplify to find ${S}_{n}.$ ## Finding the first n Terms of a geometric series Use the formula to find the indicated partial sum of each geometric series. 1. ${S}_{11}$ for the series 2. $\underset{6}{\overset{k=1}{{\sum }^{\text{}}}}3\cdot {2}^{k}$ 1. ${a}_{1}=8,$ and we are given that $n=11.$ We can find $r$ by dividing the second term of the series by the first. $r=\frac{-4}{8}=-\frac{1}{2}$ Substitute values for into the formula and simplify. $\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{11}=\frac{8\left(1-{\left(-\frac{1}{2}\right)}^{11}\right)}{1-\left(-\frac{1}{2}\right)}\approx 5.336\hfill \end{array}$ 2. Find ${a}_{1}$ by substituting $k=1$ into the given explicit formula. ${a}_{1}=3\cdot {2}^{1}=6$ We can see from the given explicit formula that $r=2.$ The upper limit of summation is 6, so $n=6.$ Substitute values for ${a}_{1},\text{\hspace{0.17em}}r,$ and $n$ into the formula, and simplify. $\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{6}=\frac{6\left(1-{2}^{6}\right)}{1-2}=378\hfill \end{array}$ Use the formula to find the indicated partial sum of each geometric series. ${S}_{20}$ for the series $\approx 2,000.00$ $\sum _{k=1}^{8}{3}^{k}$ 9,840 ## Solving an application problem with a geometric series At a new job, an employee’s starting salary is$26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.
The problem can be represented by a geometric series with ${a}_{1}=26,750\text{;}\text{\hspace{0.17em}}$ $n=5\text{;}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r=1.016.$ Substitute values for $\text{\hspace{0.17em}}{a}_{1}\text{,}\text{\hspace{0.17em}}$ $r\text{,}$ and $n$ into the formula and simplify to find the total amount earned at the end of 5 years.
$\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{5}=\frac{26\text{,}750\left(1-{1.016}^{5}\right)}{1-1.016}\approx 138\text{,}099.03\hfill \end{array}$
He will have earned a total of $138,099.03 by the end of 5 years. At a new job, an employee’s starting salary is$32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?
$275,513.31 ## Using the formula for the sum of an infinite geometric series Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first $n$ terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is $2+4+6+8+...$ #### Questions & Answers An investment account was opened with an initial deposit of$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
|
### Multiplication of a Polynomial by a Monomial
To multiply a polynomial by a monomial, use the distributive property: multiply each term of the polynomial by the monomial. This involves multiplying coefficients and adding exponents of the appropriate variables.
Example 1: 3y2(12y3 -6y2 + 5y - 1) =?
= 3y2(12y3) + (3y2)(- 6y2) + (3y2)(5y) + (3y2)(- 1)
= (3)(12)y2+3 + (3)(- 6)y2+2 + (3)(5)y2+1 + (3)(- 1)y2
= 36y5 -18y4 +15y3 -3y2
Example 2: -4x3y(- 2y2 + xy - x + 9) =?
= - 4x3y(- 2y2) + (- 4x3y)(xy) + (- 4x3y)(- x) + (- 4x3y)(9)
= (- 4)(- 2)x3y1+2 + (- 4)x3+1y1+1 + (- 4)(- 1)x3+1y + (- 4)(9)x3y
= 8x3y3 -4x4y2 +4x4y - 36x3y
### Multiplication of Binomials
To multiply a binomial by a binomial--(a + b)(c + d ), where a, b, c, and d are terms--use the distributive property twice. First, treat the second binomial as a single term and distribute over the first binomial:
(a + b)(c + d )= a(c + d )+ b(c + d )
Next, use the distributive property over the second binomial:
a(c + d )+ b(c + d )= ac + ad + bc + bd
At this point, there should be 4 terms in the answer -- every combination of a term of the first binomial and a term of the second binomial. Simplify the answer by combining like terms.
We can use the word FOIL to remember how to multiply two binomials (a + b)(c + d ):
• Multiply their First terms. (ac)
• Multiply their Outside terms. (ad )
• Multiply their Iinside terms. (bc)
• Multiply their Last terms. (bd )
• Finally, add the results together: ac + ad + bc + bd. Combine like terms.
Remember to include negative signs as part of their respective terms in the multiplication.
Example 1.(xy + 6)(x + 2y) =?
= (xy)(x) + (xy)(2y) + (6)(x) + (6)(2y)
= x2y + 2xy2 + 6x + 12y
Example 2.(3x2 +7)(4 - x2) =?
= (3x2)(4) + (3x2)(- x2) + (7)(4) + (7)(- x2)
= 12x2 -3x4 +28 - 7x2
= - 3x4 + (12 - 7)x2 + 28
= - 3x4 +5x2 + 28
Example 3: (y - x)(- 4y - 3x) =?
= (y)(- 4y) + (y)(- 3x) + (- x)(- 4y) + (- x)(- 3x)
= - 4y2 -3xy + 4xy + 3x2
= 3x2 + (- 3 + 4)xy - 4y2
= 3x2 + xy - 4y2
### Multiplication of Polynomials
The strategy for multiplying two polynomials in general is similar to multiplying two binomials. First, treat the second polynomial as a single term, and distribute over the first term:
(a + b + c)(d + e + f )= a(d + e + f )+ b(d + e + f )+ c(d + e + f )
Next, distribute over the second polynomial:
a(d + e + f )+ b(d + e + f )+ c(d + e + f )= ad + ae + af + bd + be + bf + cd + ce + cf
At this point, the number of terms in the answer should be the number in the first polynomial times the number in the second polynomial--every combination of a term of the first polynomial and a term of the second polynomial. Since there are 3 terms in each polynomial in this example there should be 3(3) = 9 terms in our answer so far. If the first polynomial had 4 terms and the second had 5, there would be 4(5) = 20 terms in the answer so far.
Finally, since the the terms in such a product of polynomials are often highly redundant (many have the same variables and exponents), it is important to combine like terms.
Example 1: (x2 -2)(3x2 - 3x + 7) =?
= x2(3x2 -3x + 7) - 2(3x2 - 3x + 7)
= x2(3x2) + x2(- 3x) + x2(7) - 2(3x2) - 2(- 3x) - 2(7) (6 terms)
= 3x4 -3x3 +7x2 -6x2 + 6x - 14
= 3x4 -3x3 + (7 - 6)x2 + 6x - 14
= 3x4 -3x3 + x2 + 6x - 14
Example 2: (x2 + x + 3)(2x2 - 3x + 1) =?
= x2(2x2 -3x + 1) + x(2x2 -3x + 1) + 3(2x2 - 3x + 1)
= x2(2x2) + x2(- 3x) + x2(1) + x(2x2) + x(- 3x) + x(1) + 3(2x2) + 3(- 3x) + 3(1) (9 terms)
= 2x4 -3x3 + x2 +2x3 -3x2 + x + 6x2 - 9x + 3
= 2x4 + (- 3 + 2)x3 + (1 - 3 + 6)x2 + (1 - 9)x + 3
= 2x4 - x3 +4x2 - 8x + 3
Note: To check your answer, pick a value for the variable and evaluate both the original expression and your answer--they should be the same.
|
Education.com
# Geometry Practice Problems for GED Math (page 2)
By LearningExpress Editors
LearningExpress, LLC
Updated on Jan 17, 2011
1. d. Angles 2, 4, 6, and 7 are alternating (vertical) angles. Therefore, their measures are equal. Angles 7 and 8 are supplementary. Therefore, angles 2 and 8 are also supplementary. 12x + 10 + 7x – 1 = 180, 19x + 9 = 180, 19x = 171, x = 9. Because x = 9, the measure of angle 2 is 12(9) + 10 = 108 + 10 = 118 degrees.
2. e. Angles 5 and 6 are supplementary. Therefore, the sum of their measures is 180 degrees. If the measure of angle 6 is x, then the measure of angle 5 is 5x. 5x + x = 180, 6x = 180, x = 30. The measure of angle 6 is 30 degrees, and the measure of angle 5 is 5(30) = 150 degrees.
3. c. Angles 4 and 7 are alternating angles. Therefore, their measures are equal: 6x + 20 = 10x – 40, 4x = 60, x = 15. Because x = 15, the measure of angles 4 and 7 is 6(15) + 20 = 90 + 20 = 110. Notice that replacing x with 15 in the measure of angle 7 also yields 110: 10(15) – 40 = 150 – 40 = 110. Because angles 6 and 7 are vertical angles, the measure of angle 6 is also 110 degrees.
4. a. If angle 3 measures 90 degrees, then angles 1, 6, and 7 must also measure 90 degrees, because they are alternating angles. Angles 3 and 4 are supplementary, because these angles form a line. Therefore, the measure of angle 4 is equal to 180 – 90 = 90 degrees. Angles 3 and 4 are congruent and supplementary. Because angles 2, 4, 5, and 8 are alternating angles, they are all congruent to each other. Every numbered angle measures 90 degrees. Therefore, every numbered angle is congruent and supplementary to every other numbered angle. Angles 5 and 7 are in fact adjacent, because they share a common vertex and a common ray. However, angles 1 and 2 are not complementary—their measures add to 180 degrees, not 90 degrees.
5. b. Angles 2 and 6 are alternating angles. Therefore, their measures are equal. 8x + 10 = x2 – 38, x2 – 8x – 48 = 0. Factor x2 – 8x – 48 and set each factor equal to 0. x2 – 8x – 48 = (x + 4)(x – 12), x + 4 = 0, x = –4. x – 12 = 0, x = 12. An angle cannot have a negative measure, so the –4 value of x must be discarded. If x = 12, then the measure of angle 2 is 8(12) + 10 = 106 degrees. Notice that replacing x with 12 in the measure of angle 6 also yields 106: (12)2 – 38 = 144 – 38 = 106. Angles 6 and 8 are supplementary, so the measure of angle 8 is equal to 180 – 106 = 74 degrees.
6. e. If x is the width of the rectangle, then 2x – 4 is the length of the rectangle. Because opposite sides of a rectangle are congruent, the perimeter of the rectangle is equal to 2x – 4 + x + 2x – 4 + x = 6x – 8.
7. c. One face of a cube is a square. The area of a square is equal to the length of one side of the square multiplied by itself. Therefore, the length of a side of this square (and edge of the cube) is equal to or units. Because every edge of a cube is equal in length and the volume of a cube is equal to e3, where e is the length of an edge (or lwh, l is the length of the cube, w is the width of the cube, and h is the height of the cube, which in this case, are all units), the volume of the cube is equal to cubic units.
8. d. The circumference of a circle is equal to 2πr. Because the radius of a circle is half the diameter of a circle, 2r is equal to the diameter, d, of a circle. Therefore, the circumference of a circle is equal to πd. If the diameter is doubled, the circumference becomes π2d, or two times its original size.
9. b. If the base of the triangle is b, then the height of the triangle is . The area of a triangle is equal to . Therefore, the area of this triangle is equal to .
10. e. Use the Pythagorean theorem: (x – 3)2 + (x + 4)2 = (2x – 3)2, x2 – 6x + 9 + x2 + 8x + 16 = 4x2 – 12x + 9, 2x2 + 2x + 25 = 4x2 – 12x + 9, 2x2 – 14x – 16 = 0, x2 – 7x – 8 = 0, (x – 8)(x + 1) = 0, x = 8. Disregard the negative value of x, because a side of a triangle cannot be negative. Because x = 8, the length of the hypotenuse is 2(8) – 3 = 16 – 3 = 13.
11. b. The slope of a line is the difference between the y values of two points divided by the difference between the x values of those two points: (4 – 6) ÷ 7 – (–3) = .
12. b. The midpoint of a line segment is equal to the average of the x values of the endpoints and the average of the y values of the endpoints:
13. c. To find the distance between two points, use the distance formula:
1. a. Perpendicular lines cross at right angles. Therefore, angle AOC is 90 degrees. Because angles 1 and 2 combine to form angle AOC, the sum of the measures of angles 1 and 2 must be 90 degrees. Therefore, they are complementary angles.
2. d. AE is perpendicular to OC, so angles AOC and EOC are both 90 degrees. Because angles 1 and 2 combine to form angle AOC and angles 3 and 4 combine to form angle EOC, these sums must both equal 90 degrees. Therefore, angle 1 + angle 2 = angle 3 + angle 4. Angles 1, 2, 3, and 7 form a line, as do angles 4, 5, and 6. Therefore, the measures of angles 1, 2, 3, and 7 add to 180 degrees, as do the measures of angles 4, 5, and 6. In the same way, angles 2, 3, 4, and 5 form a line, so the sum of the measures of those angles is also 180 degrees. Angles GOF and BOD are vertical angles. Therefore, their measures are equal. Angle 6 is angle GOF and angles 2 and 3 combine to form angle BOD, so the measure of angle 6 is equal to the sum of the measures of angles 2 and 3. However, the sum of angle 1 and angle 7 is not equal to the sum of angle 2 and angle 3. In fact, the sum of angles 2 and 3 is supplementary to the sum of angles 7 and 1, because angle 6 is supplementary to the sum of angles 7 and 1, and the sum of angles 2 and 3 is equal to the measure of angle 6.
3. c. Because AE is perpendicular to OC, angle EOC is a right angle (measuring 90 degrees). Angles 3 and 4 combine to form angle EOC; therefore, their sum is equal to 90 degrees: 2x + 2 + 5x – 10 = 90, 7x – 8 = 90, 7x = 98, x = 14. Angle 4 is equal to 5(14) – 10 = 70 – 10 = 60 degrees. Because angles 4 and 7 are vertical angles, their measures are equal, and angle 7, too, measures 60 degrees.
4. b. Because AE is perpendicular to OC, angle EOC is a right angle (measuring 90 degrees). Angles 3 and 4 combine to form angle EOC; therefore, their sum is equal to 90 degrees: 90 – 57 = 33. Angle 3 measures 33 degrees. Angle AOC is also a right angle, with angles 1 and 2 combining to form that angle. Therefore, the measure of angle 2 is equal to: 90 – 62 = 28 degrees. Angle 6 and angle BOD are vertical angles; their measures are equal. Because angles 2 and 3 combine to form angle BOD, the measure of BOD (and angle 6) is equal to: 33 + 28 = 61 degrees.
5. b. Because AE is perpendicular to OC, angle EOC is a right angle (measuring 90 degrees). Angles 3 and 4 combine to form angle EOC; therefore, their sum is equal to 90 degrees: 5x + 3 + 15x + 7 = 90, 20x + 10 = 90, 20x = 80, x = 4. Therefore, the measure of angle 4 is equal to: 15(4) + 7 = 67 degrees. Angles 4, 5, and 6 form a line; therefore, the sum of their measures is 180 degrees. If x is the sum of angles 5 and 6, then 67 + x = 180, and x = 113 degrees.
6. a. If the square and the rectangle share a side, then the width of the rectangle is equal to the length of the square. If the length of the square is x, then x + x + x + x = 2, 4x = 2, x = . Because the width of the rectangle is equal to the length of the square and the length of the rectangle is 4 times the length of the square, the width of the rectangle is and the length is 4() = 2. Therefore, the perimeter of the rectangle is equal to: + 2 + + 2 = 5 units.
7. c. The area of a rectangle is equal to its length times its width. Therefore, the width of the rectangle is equal to its area divided by its length: . x2 + 7x + 10 can be factored into (x + 2)(x + 5). Cancel the (x + 2) terms from the numerator and denominator of the fraction. The width of the rectangle is x + 5 units.
8. e. The volume of a rectangular solid is equal to lwh, where l is the length of the solid, w is the width of the solid, and h is the height of the solid. If x represents the width (and therefore, the height as well), then the length of the solid is equal to 2(x + x), or 2(2x) = 4x. Therefore, (4x)(x)(x) = 108, 4x3 = 108, x3 = 27, and x = 3. If the width and height of the solid are each 3 inches, then the length of the solid is 2(3 + 3) = 2(6) = 12 inches.
9. d. Use the Pythagorean theorem: 82 + x2 = (8)2, 64 + x2 = 320, x2 = 256, x = 16 (x cannot equal –16 because the side of a triangle cannot be negative).
10. b. The measures of the angles of a triangle add to 180 degrees. Therefore, 3x + 4x + 5x = 180, 12x = 180, and x = 15.
11. e. To find the distance between two points, use the distance formula:
12. c. The midpoint of a line segment is equal to the average of the x values of the endpoints and the average of the y values of the endpoints:
13. c. The slope of a line is the difference between the y values of two points divided by the difference between the x values of those two points: [–5 – (–5)] ÷ –5 – 5 = = 0.
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# Math Help: Understanding Reciprocals
You'll use reciprocals when you learn to divide fractions by whole numbers and other fractions. This skill is usually taught in 5th grade. Keep reading for help!
## Reciprocals in Math
### Reciprocals of Whole Numbers
A number's reciprocal is defined as one over that number (1/n). For example, the reciprocal of three is 1/3, and the reciprocal of 17 is 1/17.
Another way to think of reciprocals is that they are 'upside-down' fractions. Whole numbers, like 3, 17 and 345, can all be written as fractions by putting them over one. For instance, 2 = 2/1, 17 = 17/1 and 345 = 345/1. The reciprocal of each number is the reverse of each fraction. The reciprocal of two (2/1) is 1/2, and the reciprocal of 345 (345/1) is 1/345.
### Reciprocals of Fractions
Similarly, the reciprocal of a fraction is just that fraction reversed. For example, the reciprocal of 4/7 is 7/4, and the reciprocal of 13/8 is 8/13.
### Division with Reciprocals
Reciprocals are mainly used when you're dividing by a fraction. To divide by fractions, you need to find the reciprocal of the fraction you're dividing by and then multiply the fractions. For example, if you're dividing five by 2/3, you'd write five as a fraction (5/1) and multiply it by the reciprocal of 2/3, which is 3/2. Since 5/1 x 3/2 = 15/2, that's your answer.
### Reciprocal Practice Problems
Solve each of the following problems, and then check your answers using the key below. Don't forget to use a reciprocal for the division problems.
1. What is the reciprocal of 11?
2. Find the reciprocal of 7/8.
3. What is 1/6 the reciprocal of?
4. What fraction is 15/4 the reciprocal of?
6. Find the quotient of 1/10 ÷ 2.
1. Since the reciprocal of a number (n) is 1/n, the reciprocal of 11 is 1/11.
2. To find the reciprocal of 7/8, you can simply switch the numerator with the denominator to get 8/7.
3. Since 1/n is the reciprocal of a whole number, 1/6 is the reciprocal of six or 6/1.
4. The fraction 15/4 is the reciprocal of 4/15.
5. To solve 2/7 ÷ 1/7, start by finding the reciprocal of 1/7, which is 7/1. Then, multiply 2/7 x 7/1 to get 14/7. You can simplify 14/7 to two.
6. To find the quotient of 1/10 ÷ 2, you'll need to write two as a fraction (2/1) and then find its reciprocal (1/2). Next, multiply 1/10 x 1/2 to get 1/20.
Did you find this useful? If so, please let others know!
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# Syllogism II – Syllogism Tricks to solve Syllogism Questions for Bank Exams
In this post, we will discuss few syllogism tricks to solve syllogism questions in bank exams such as SBI PO, IBPS PO and SSC CGL Exam.
Syllogism questions are word based problems from the reasoning section of SBI PO, IBPS PO and SSC CGL Exams. There are generally 5-6 syllogism questions that are asked every time in bank exams. Syllogism Questions are the easiest type of questions that can be solved within few seconds by using syllogism tricks.
In our previous post, we have discussed what is a syllogism and few elements of syllogism questions for bank exams. In this post, we will discuss few Syllogism tricks to solve syllogism questions for bank exams.
Various types of Venn diagram to solve Syllogism Questions
1. All A’s are B’s.
In this type of Venn diagram, The smaller circle represents set A and the larger circle represents set B. The smaller circle A which is within the circle B, we can say that all- A’s are B. This is the basic diagram which we use to represent the statement- all A’s are B.
This is the second type of Venn diagram which is used to represent the statement all A’s are B. In this Venn diagram the circle A represents the set A. Now we can use the same circle to represent the set B. In this type, we can draw the conclusion that all -A’s are B.
The conclusion that can be taken from the above statement is that some- B’s are A.
2. No A is B.
To represent this statement, we need to take a circle A and a circle B which are not related in any way or there is no overlapping of the circles. There is no other way in which we can represent the statement- No A is B.
The conclusion that can be taken from the above statement is- No B is A.
3. Some A’s are B
This is the basic diagram which can be drawn from the statement some A’s are B. We draw two circles A and B which overlap each other at a certain point, from this diagram we can conclude that some A’s are B. The overlapping section represent the statement some A’s are B.
There are other possibilities in which we can show that some A’s are B.
We draw a circle A and then draw a smaller circle B inside A. From this we can conclude that some A’s are B.
We generally assume the statement some A’s are B's as all A’s are B.
The conclusion that can be taken from the above statement is that some B’s are A.
4. Some A’s are not B.
The basic diagram which we use to represent this statement, we draw a circle A and then draw a circle B overlapping the circle A. The part o f the circle which doesn’t overlap represents the statement some A’s are not B.
There are more possibilities in which you can represent the statement some A’s are not B.
We draw a circle A and then draw a smaller circle B inside A. The part which doesn’t come inside the circle B, represents the statement- Some A’s are not B.
Some A’s are not B can be assumed as all the A’s are not B. So, we draw circle A and circle B away from each other as such that they don’t overlap each other.
We cannot draw any conclusion from the above statement.
Complimentary pair of statement
It is a set of two statements in which if one statement fails then the other statement is true.
When we toss a coin there are two possible outcomes - Head or Tail. So if the coin doesn’t show head then it will definitely show tail. Similarly, if the coin doesn't show tails then it will definitely show head. Therefore, this is how a complimentary statement works.
There are two categories of complimentary statements in syllogisms.
1.No A is B; Some A's are B's
For one statement to be true the other statement should be false. To prove that Some A's are B, one of the A should move out and mix be B to make the other statement true. When some A's are B fails, No A is B becomes true.
2. All A’s are B's; Some A’s are not B.
For one statement to be false, the other statement should be true that is why- All A's are B's, and Some A's are B's, forms a complimentary form of a statement.
There are two types of syllogism questions that are asked in IBPS PO and SSC CGL Exams
Either the questions are asked on conclusions or possibilities. Let's define both the terms.
Conclusion A conclusion should be definitely true in all the cases.
Example: A is the wife of B = B is the husband of A.
In the above example, we can make out that if A is the wife of B then definitely B is the husband A. This is called as conclusion- drawing definite facts from the given statement.
Possibility - A possibility is true only one case. There are no definite conclusions that can be drawn from the given statement. There is always a possibility of only one statement to be true.
Example: A is the son of B = B is the Parent of A
There are two possibilities that can be drawn from the above statement.
The parent can be a mother or a father. We can not predict whether B is Female or Male, so there is a possibility of any one of the statements to be true.
Do write in the comments section how this blog helped you solve Syllogism Questions.
Stay tuned for more on Syllogism Questions.
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# 2012 AIME II Problems/Problem 8
## Problem 8
The complex numbers $z$ and $w$ satisfy the system $$z + \frac{20i}w = 5+i$$ $$w+\frac{12i}z = -4+10i$$ Find the smallest possible value of $\vert zw\vert^2$.
## Solution
Multiplying the two equations together gives us $$zw + 32i - \frac{240}{zw} = -30 + 46i$$ and multiplying by $zw$ then gives us a quadratic in $zw$: $$(zw)^2 + (30-14i)zw - 240 =0.$$ Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \pm \sqrt{(15-7i)^2 + 240}$ = $6+2i,$ $12i - 36.$ The smallest possible value of $\vert zw\vert^2$ is then obviously $6^2 + 2^2 = \boxed{040}$.
### Note
A key thing to note here is that $|zw|^2=|z|^2\cdot|w|^2,$ which can be proved as follows:
Proof: Using the values for $z$ and $w$ that we used above, we get:
\begin{align*} |zw|^2&=|(ac-bd)+i(bc+ad)|^2\\ &=(ac-bd)^2+(bc+ad)^2\\ &=a^2c^2+b^2d^2+b^2c^2+a^2d^2-2abcd+2abc\\ &=a^2c^2+b^2d^2+b^2c^2+a^2d^2 \end{align*} Also, $|z|^2=a^2+b^2$ and $|w|^2=c^2+d^2$. Therefore: $$|z|^2\cdot|w|^2=(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2$$ and our proof is complete.
Now, also note that we found $\sqrt{416-210i}$ by letting $416-210i=(a-bi)^2$ and solving for $a$ and $b$ by considering real and imaginary parts. Then, we substitute that into $a-bi$ which is the value of $\sqrt{416-210i}$ and continue from there.
mathboy282
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# Square Root Numbers and Other Roots
For numbers, a root is another number that produces that whole number when raised to a certain power/exponent.
With the number 8 for example:
2 is the 3rd root of 8, => 23 = 2 × 2 × 2 = 8
The 3rd root is commonly referred to as the cube root.
However, the most common root that most will encounter is likely to be the square root of numbers.
The square root is the 2nd root of a number.
So it is a number that multiplies with itself once.
4 is the ;2nd root of 16, => 42 = 4 × 4 = 16
5 is the 2nd root of 25, => 52 = 5 × 5 = 25
## Square Root of Numbers and other Roots Notation
The symbol for the square root of numbers is the radical symbol .
Numbers featuring this symbol are referred to as radical terms, but can sometimes just be called radicals for short.
As seen earlier in the page, it's the case that:
√16 = 4 , √25 = 5
But negative numbers can also be square roots too:
-4 × -4 = 16
So it's actually more accurate to say that 4 is a square root of 16, rather than the square root of 16, as -4 is also a root too.
√16+4
## Cube Root & Other Roots
The notation for the cube root is similar to the square root, with the cube root being the 3rd root, it is denoted by 3.
The 4th root is denoted by 4, the 5th root by 5, and so on.
Calculators today usually have the [ x ] button that can assist in finding specific roots of numbers quickly.
To establish 4√256 .
Type 4, followed by the [ x ] button, then type 256, which gives the answer 4.
[ 4 × 4 × 4 × 4 = 256 ]
Examples
(1.1)
3√512 = 8
(1.2)
5√243 = 3
## Even and Odd Roots, General Case
We can generalize the case of root of a number with the same exponent as the root, depending on whether the base number is positive or negative.
It can be helpful to know these facts, for more advanced situations when required to simplify a radical containing variables.
Firstly some examples of the square root of a squared number.
√22 = √4 = 2
√52 = √25 = 5
Looking at these two examples, it could be reasonable to assume that it's always the case that:
√a2 = a
But this is not quite the case. For example with -3.
-3 × -3 = 9
Thus: -32 = √9 = 3
So to ensure the answer is always positive for an even number root of the same even exponent, the absolute value of the base number has to be taken.
√a2 = |a| , 4√a4 = |a| etc.
So the general case is n√an = |a| when "n" is even.
With the same odd exponent and root however, things are simpler and we don't have to take the exact value of the base number.
-23 = -2 × -2 × -2 = -8
Thus: 3-23 = -2
In fact the general case is n√an = a when "n" is odd.
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# Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Additional Questions
## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Additional Questions
Question 1.
Solve the following system of linear equations in three variables. x + y + z = 6; 2x + 3y + 4z = 20;
3x + 2y + 5z = 22
Solution:
x + y + z = 6 ………….. (1)
2x + 3y + 4z = 20 ………… (2)
3x + 2y + 5z = 22 …………(3)
Sub. z = 3 in (5) ⇒ y – 2(3) = -4
y = 2
Sub. y = 2, z = 3 in (1), we get
x + 2 + 3 = 6
x = 1
x = 1, y = 2, z = 3
Question 2.
Using quadratic formula solve the following equations.
(i) p2x2 + (p2 – q2) x – q2 = 0
(ii) 9x2 – 9 (a + b)x + (2a2 + 5ab + 2b2) = 0
Solution:
(i) p2x2 + (p2 – q2)x – q2 = 0
Comparing this with ax2 + bx + c = 0, we have
a = p2
b = p2 – q2
c = -q2
Δ = b2 – 4ac
= (p2 – q2)-4 × p2 × -q2
= (p2 – q2)2 + 4p2 q2
= (p2 + q2)2 > 0
So, the given equation has real roots given by
$$\alpha=\frac{-b-\sqrt{\Delta}}{2 a}=\frac{-\left(p^{2}-q^{2}\right)+\left(p^{2}+q^{2}\right)}{2 p^{2}}=\frac{q^{2}}{p^{2}}$$
$$\beta=\frac{-b-\sqrt{\Delta}}{2 a}=\frac{-\left(p^{2}-q^{2}\right)-\left(p^{2}+q^{2}\right)}{2 p^{2}}$$
= -1
(ii) 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0
Comparing this with ax2 + bx + c = 0.
a =9
b = -9 (a + b)
c = (2a2 + 5 ab + 2b2)
Δ = B2 – 4AC
⇒ 81 (a + b)2 – 36(2a2 + 5ab + 2b2)
⇒ 9a2 + 9b2 – 18ab
⇒ 9(a – b)2 > 0
∴ the roots are real and given by
Question 3.
Find the HCF of x3 + x2 + x + 1 and x4 – 1.
x3 + x2 + x + 1 = x2 (x + 1) + 1 (x + 1)
= (x + 1) (x2 + 1)
x4-1 = (x2)2 – 1
= (x2 + 1) (x2– 1)
= (x2 + 1) (x + 1) (x – 1)
H.C.F. = (x2 + 1)(x + 1)
Question 4.
Prove that the equation x2(a2 + b2) + 2x(ac + bd) + (c2 + d2) = 0 has no real root if ad ≠ bc
Solution:
Δ = b2 – 4ac
⇒ 4(ac + bd)2 – 4(a2 + b2)(c2 + d2)
⇒ 4[(ac + bd)2 – (a2 + b2)(c2 + d2)]
⇒ 4(a2c2 + b2d2 + 2acbd – a2c2b2c2 – a2d2 – b2d2]
⇒ 4[2acbd – a2d2 – b2c2]
⇒ 4 [a2c2 + b2c2 – 2adbc]
∴ ad – bc > 0
⇒ (ad – bc)2 > 0
⇒ -4(ad – bc)2 < 0 ⇒ Δ < 0
Hence the given equation has no real roots.
Question 5.
Find the L.C.M of 2(x3 + x2 – x – 1) and 3(x3 + 3x2 – x – 3)
2[x3 + x2 – x – 1] = 2[x2(x+ 1)- 1 (x + 1)]
= 2(x + 1) (x2 – 1)
= 2(x + 1) (x + 1) (x – 1)
= 2(x + 1)2 (x – 1)
3[x3 + 3x2 – x – 3] = 3[x2(x + 3) -1 (x + 3)]
= 3[(x + 3)(x2 – 1)]
= 3(x + 3)(x + 1) (x – 1)
L.C.M. = 6(x + 1)2(x – 1) (x + 3)
Question 6.
A two digit number is such that the product of its digits is 12. When 36 is added to the number the digits interchange their places. Find the number.
Solution:
Let the ten’s digit of the number be x. It is given that the product of the digits is 12.
Unit’s digit = $$\frac{12}{x}$$
Number = 10x + $$\frac{12}{x}$$
If 36 is added to the number the digits interchange their places.
x = -6, 2.
But a number can never be (-ve). So, x = 2.
The number is 10 × 2 + $$\frac{12}{2}$$ = 26
Question 7.
Seven years ago, Vanin’s age was five times the square of swati’s age. Three years hence Swati’s age will be two fifth of Varun’s age. Find their present ages.
Solution:
Seven years ago, let Swathi’s age be x years.
Seven years ago, let Varun’s age was 5x2 years.
Swathi’s present age = x + 7 years
Varun’s present age = (5x2 + 7) years
3 years hence, we have Swathi’s age = x + 7 + 3 years
= x + 10 years
Varun’s age = 5x2 + 7 + 3 years
= 5x2 + 10 years
It is given that 3 years hence Swathi’s age will be $$\frac{2}{5}$$ of Varun’s age.
∴ x + 10 = $$\frac{2}{5}$$ (5x2 + 10)
⇒ x + 10 = 2x2 + 4
⇒ 2x2 – x – 6 = 0
⇒ 2x(x – 2) + 3(x – 2) = 0
⇒ (2x + 3)(x – 2) = 0
⇒ x – 2 = 0
⇒ x = 2 (∵ 2x + 3 ≠ 0 as x > 0)
Hence Swathi’s present age = (2 + 7) years
= 9 years
Varun’s present age = (5 × 22 + 7) years = 27 years
Question 8.
A chess board contains 64 equal squares and the area of each square is 6.25 cm2. A border round the board is 2 cm wide find its side.
Solution:
Let the length of the side of the chess board be x cm. Then,
Area of 64 squares = (x – 4)2
(x – 4)2 = 64 × 6.25
⇒ x2 – 8x+ 16 = 400
⇒ x2 – 8x-384 = 0
⇒ x2 – 24x + 16x – 384 = 0
⇒ (x – 24)(x + 16) = 0
⇒ x = 24 cm.
Question 9.
Find two consecutive natural numbers whose product is 20.
Solution:
Let a natural number be x.
The next number = x + 1
x (x + 1) = 20
x2 + x – 20 = 0
(x + 5)(x – 4) = 0
x = -5, 4
∴ x = 4 (∵ x ≠ -5, x is natural number)
The next number = 4 + 1 = 5
Two consecutive numbers are 4, 5
Question 10.
A two digit number is such that the product of its digits is 18, when 63 is subtracted from the number, the digits interchange their places. Find the number.
Solution:
Let the tens digit be x. Then the units digits = $$\frac{18}{x}$$
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Breaking News
# 20 Percent Of 9000
20 Percent Of 9000. Use this calculator to find percentages. How much is 20 percent of 90000? Always use this formula to find a percentage: = 20/100 x 9000, = 1800 20% of 9000 equals to 1800 where, 20 is the relative quantity in each 100, 9000 is the reference or base quantity, 1800 is 20 percent of.
Now we can solve our fraction by writing it as an equation: To find 1% of a. Use this calculator to find percentages.
## What is 20 percent of 90,000?
Working out 20% of 9000. Replacing the given values in formula (a) we have: Inicial value = 9000, percentage = 20.
## 20 Percent Of 9000 Equals To:
Amount saved = original price x discount in percent / 100. = 20/100 x 9000, = 1800 20% of 9000 equals to 1800 where, 20 is the relative quantity in each 100, 9000 is the reference or base quantity, 1800 is 20 percent of. To find 20 percent of a number, multiply the number by 0.2. 1) what is 20% of 9000?
### Calculate The Percentage Of A Number.
Therefore, 20% of 9000 is 1800.
### Kesimpulan dari 20 Percent Of 9000.
In this instance, 0.2 x 45000 = 9000. Amount saved = 180000 / 100. 20 x 9000 = 100 x part, or.
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# Solve the differential equation (dy)/(dx)=(x+2y+3)/(2x+y+3) ?
Nov 29, 2017
See below.
#### Explanation:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x + 2 y + 3}{2 x + y + 3} \Rightarrow \left(2 x + y + 3\right) \mathrm{dy} = \left(x + 2 y + 3\right) \mathrm{dx}$
Now we will do a series of variable transformations to get an amenable format in the differential equation
1) $\left\{\begin{matrix}x + 2 y = u \\ 2 x + y = v\end{matrix}\right. \Rightarrow \left\{\begin{matrix}\mathrm{dx} = \frac{1}{3} \left(- \mathrm{du} + 2 \mathrm{dv}\right) \\ \mathrm{dy} = \frac{1}{3} \left(2 \mathrm{du} - \mathrm{dv}\right)\end{matrix}\right.$
$\left(v + 3\right) \left(2 \mathrm{du} - \mathrm{dv}\right) = \left(u + 3\right) \left(- \mathrm{du} + 2 \mathrm{dv}\right)$
Now making the transformation
2) $\left\{\begin{matrix}U = u + 3 \\ V = v + 3\end{matrix}\right. \Rightarrow \left\{\begin{matrix}\mathrm{dU} = \mathrm{du} \\ \mathrm{dV} = \mathrm{dv}\end{matrix}\right.$
$V \left(2 \mathrm{dU} - \mathrm{dV}\right) = U \left(- \mathrm{dU} + 2 \mathrm{dV}\right)$ or
$2 V \mathrm{dU} - V \mathrm{dV} = - U \mathrm{dU} + 2 U \mathrm{dV}$
Now making
3) $\left\{\begin{matrix}\eta = {U}^{2} \\ \xi = {V}^{2}\end{matrix}\right. \Rightarrow \left\{\begin{matrix}\frac{1}{2} \frac{d \eta}{\sqrt{\eta}} = \mathrm{dU} \\ \frac{1}{2} \frac{d \xi}{\sqrt{\xi}} = \mathrm{dV}\end{matrix}\right.$
$2 \sqrt{\frac{\xi}{\eta}} d \eta - d \xi = - d \eta + 2 \sqrt{\frac{\eta}{\xi}} d \xi$ or
$\left(2 \sqrt{\frac{\xi}{\eta}} + 1\right) d \eta = \left(2 \sqrt{\frac{\eta}{\xi}} + 1\right) d \xi$
now introducing
4) $\eta = \lambda \xi \Rightarrow d \eta = \lambda d \xi + \xi d \lambda$
$\left(\frac{2}{\sqrt{\lambda}} + 1\right) \left(\lambda d \xi + \xi d \lambda\right) = \left(2 \sqrt{\lambda} + 1\right) d \xi$
or grouping variables
$\frac{d \xi}{\xi} + f \frac{\lambda}{\lambda f \left(\lambda\right) - 1} d \lambda = 0$
with $f \left(\lambda\right) = \frac{\frac{2}{\sqrt{\lambda}} + 1}{2 \sqrt{\lambda} + 1}$
After integration we obtain
$\log \xi + 3 \log \left(1 - \sqrt{\lambda}\right) - \log \left(1 + \sqrt{\lambda}\right) = {C}_{0}$ or
$\xi = \frac{{C}_{1} \left(1 + \sqrt{\lambda}\right)}{1 - \sqrt{\lambda}} ^ 3$ or
${V}^{2} = \frac{{C}_{1} \left(1 + \frac{U}{V}\right)}{1 - \frac{U}{V}} ^ 3$ and finally
${\left(2 x + y + 3\right)}^{2} = \frac{{C}_{1} \left(1 + \frac{x + 2 y + 3}{2 x + y + 3}\right)}{1 - \frac{x + 2 y + 3}{2 x + y + 3}} ^ 3$
An implicit form solution.
This can be reduced to
${\left(x - y\right)}^{3} = {C}_{2} \left(x + y + 2\right)$
Nov 29, 2017
${\left(x - y\right)}^{3} / \left(x + y + 2\right) = C$
#### Explanation:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x + 2 y + 3}{2 x + y + 3}$
$\left(2 x + y + 3\right) \cdot \mathrm{dy} = \left(x + 2 y + 3\right) \cdot \mathrm{dx}$
Now I solved $2 x + y + 3 = 0$ and $x + 2 y + 3 = 0$ equation system. From them, $x = y = - 1$
Hence I used $x = m - 1$, $y = p - 1$, $\mathrm{dx} = \mathrm{dm}$ and $\mathrm{dy} = \mathrm{dp}$ substitution, this differential equation became
$\left(2 m + p\right) \cdot \mathrm{dp} = \left(m + 2 p\right) \cdot \mathrm{dm}$
I used $m = p \cdot z$ and $\mathrm{dm} = p \mathrm{dz} + z \mathrm{dp}$ transformation, it became
$\left(2 p z + p\right) \cdot \mathrm{dp} = \left(p z + 2 p\right) \left(p \mathrm{dz} + z \mathrm{dp}\right) = 0$
$\left(2 z + 1\right) \cdot \mathrm{dp} = \left(z + 2\right) \left(p \mathrm{dz} + z \mathrm{dp}\right) = 0$
$\left(2 z + 1\right) \cdot \mathrm{dp} = p \left(z + 2\right) \cdot \mathrm{dz} + \left({z}^{2} + 2 z\right) \cdot \mathrm{dp}$
$\left({z}^{2} - 1\right) \cdot \mathrm{dp} + p \left(z + 2\right) \cdot \mathrm{dz} = 0$
$\frac{\mathrm{dp}}{p} + \frac{\left(z + 2\right) \cdot \mathrm{dz}}{{z}^{2} - 1} = 0$
$\frac{2 \mathrm{dp}}{p} + \frac{\left(2 z + 4\right) \cdot \mathrm{dz}}{\left(z + 1\right) \left(z - 1\right)} = 0$
$\frac{2 \mathrm{dp}}{p} + \frac{3 \mathrm{dz}}{z - 1} - \frac{\mathrm{dz}}{z + 1} = 0$
$2 L n p + 3 L n \left(z - 1\right) - L n \left(z + 1\right) = L n C$
$L n \left[{p}^{2} \cdot {\left(z - 1\right)}^{3} / \left(z + 1\right)\right] = L n C$
${p}^{2} \cdot {\left(z - 1\right)}^{3} / \left(z + 1\right) = C$
After using $m = p \cdot z$ and $z = \frac{m}{p}$ inverse transformation,
${p}^{2} \cdot {\left(\frac{m}{p} - 1\right)}^{3} / \left(\frac{m}{p} + 1\right) = C$
${\left(m - p\right)}^{3} / \left(m + p\right) = C$
After using $x = m - 1$, $y = p - 1$, $m = x + 1$ and $p = y + 1$ inverse transformation, I found
${\left(x - y\right)}^{3} / \left(x + y + 2\right) = C$
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How to Multiply Decimals
All you need to know about multiplying decimals by each other, whole numbers, and exponents. Boost your skills and confidence with this step-by-step guide.
Author
Tess Loucka
Published
November 7, 2023
How to Multiply Decimals
All you need to know about multiplying decimals by each other, whole numbers, and exponents. Boost your skills and confidence with this step-by-step guide.
Author
Tess Loucka
Published
Nov 1, 2023
How to Multiply Decimals
All you need to know about multiplying decimals by each other, whole numbers, and exponents. Boost your skills and confidence with this step-by-step guide. them to real problems
Author
Tess Loucka
Published
Nov 7, 2023
Key takeaways
• Knowing how to multiply decimals is a crucial skill that you’ll use throughout your daily life.
• To multiply decimals, convert them into whole numbers first. Multiply the whole numbers together, then place the decimal points back so the final answer has the same number of decimal spaces as the two factors.
• When multiplying decimals by 10, 100, 1,000, or other exponents, move the decimal point to the right the same number of spaces as there are zeros.
Multiplying decimals is a simple math calculation that you’ll be using for the rest of your life. Calculating money, time, weight, length, and distance are all likely to involve decimals, and knowing how to multiply decimals confidently will save you plenty of trouble and time in the future.
For example, you multiply decimals when calculating how much gas you can afford at a rate of \$3.67 per gallon, or how far you can walk at a rate of 2.5 miles per hour.
Decimal multiplication is easy once you know the trick!
How to Multiply Decimals Step by Step
The easiest way to multiply decimals is to convert them to whole numbers first, multiply them, then convert them back to decimals.
To explain how to multiply with decimals using this technique, let’s look at a few methods with examples.
Multiplying decimals by whole numbers
To start, let’s learn how to multiply decimals by whole numbers.
Take 0.15 x 4 as an example.
Step 1: To solve this problem, let’s turn the decimal number into a whole number by moving the decimal point to the right two places so that it falls after the digits.
In this example, 0.15 becomes 15 when we move the decimal point over 2 spaces.
Step 2: Now, multiply 15 x 4 to get 60.
Step 3: Next, convert this number back into a decimal number. Since we moved the decimal point over two spaces to the right in Step 1, we now have to move the decimal point over two spaces to the left.
The final answer is 0.15 x 4 = 0.60 or 0.6
Multiplying two decimals
Now let’s take a look at how we can multiply two decimal numbers together.
We will use 0.15 x 0.4 as our example.
Step 1: First, turn both numbers into whole numbers by moving their decimal points to the right so they fall after the digits.
Step 2: Next, we multiply our new factors, 15 and 4, to get 60.
Step 3: Lastly, convert the whole numbers back into decimals. We moved the decimal point over two spaces to the right for our first number, 0.15, and one space to the right for our second number, 0.4. That’s three spaces in total.
So, we have to move the decimal space back to the left by three spaces to get our final answer.
0.15 x 0.4 = 0.060 or 0.06
Multiplying decimals by 10, 100, and 1,000
To multiply a decimal by 10, 100, or 1,000, all we have to do is move the decimal point over to the right the same number of spaces as there are zeros.
Let’s look at an example:
11.7 x 10 = 117
There is only one zero in 10, so we move the decimal point over to the right by one to get our answer.
But how would we multiply this number by 100?
11.7 x 100 = 1,170
In this case, we have to move the decimal point over to the right by two spaces because there are two zeros in 100. That means we’ll have to move the decimal point a space past the 7. We can fill in that space with a 0.
11.7 x 1,000 is 11,700
Move the decimal point to the right by three spaces since there are three zeros in 1,000. The empty spaces can be replaced with zeros.
For more practice multiplying with decimals, download a math practice app or use a math website that offers practice problems, answers, and explanations to all your questions. The more often you practice multiplying decimals, the more confident you’ll be.
Practice problems
Click on the boxes below to see the answers!
Turn both numbers into whole numbers. Solve for 145 x 13 to get 1,885. Next, add the decimal points back in. You moved the decimal point to the right two spaces for 1.45 and two spaces for 0.13 to turn them into whole numbers, so move the decimal point to the left by 4 spaces for the final answer. That means 1,885 turns into 0.1885.
Write out your equation for the problem, 1.25 x 12. Turn 1.25 into a whole number. Solve for 125 x 12 to get 1,500. Then, add the decimal points back in by counting two spaces to the left. So, 1,500 becomes 15.00 or 15. Use a dollar sign since the problem refers to money.
When multiplying by 10, 100, or 1,000, move the decimal point to the right the same number of spaces as there are zeros. Since 100 has two zeros, move the decimal point to the right by two. So, 1.4 becomes 140.
Turn the decimal number into a whole number. Multiply the whole numbers together, then add the decimal point back in to get your final answer.
Turn both numbers into whole numbers and multiply. Put the decimal points back in by counting the decimal spaces you got rid of when converting them into whole numbers. There should be the same number of decimal spaces in the final answer.
Turn them into whole numbers, multiply them, and convert them back to decimals.
Three steps. Step 1: Make the decimal numbers whole numbers. Step 2: Multiply the whole numbers. Step 3: Add the decimal points back in.
Lesson credits
Tess Loucka
Tess Loucka discovered her passion for writing in high school and has not stopped writing since. Combined with her love of numbers, she became a math and English tutor, focusing on middle- and high-school-level topics. Since graduating from Hunter College, her goal has been to use her writing to spread knowledge and the joy of learning to readers of all ages.
Tess Loucka
Tess Loucka discovered her passion for writing in high school and has not stopped writing since. Combined with her love of numbers, she became a math and English tutor, focusing on middle- and high-school-level topics. Since graduating from Hunter College, her goal has been to use her writing to spread knowledge and the joy of learning to readers of all ages.
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# McGraw Hill My Math Grade 5 Chapter 12 Lesson 7 Answer Key Three Dimensional Figures
All the solutions provided in McGraw Hill My Math Grade 5 Answer Key PDF Chapter 12 Lesson 7 Three Dimensional Figures will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 5 Answer Key Chapter 12 Lesson 7 Three Dimensional Figures
A three-dimensional figure has length, width, and height.
Math in My World
Describe the faces, edges, and vertices of the figure outlined on the luggage bag. Then identify the shape of the figure.
faces: The figure has ______________ faces. Each face appears to be a rectangle.
edges: There are _____________ edges. The opposite edges are parallel and congruent.
vertices: The figure has _____________ vertices.
Prisms are three-dimensional figures. A prism has at least three faces that are rectangles. The top and bottom faces, called the bases, are congruent parallel polygons.
The figure above is a rectangular prism. In a rectangular prism, the bases are congruent rectangles. A rectangular prism has six rectangular faces, twelve edges, and eight vertices.
The shape of the bag is a rectangular prism.
– The figure has 6 faces. Each face appears to be a rectangle.
– There are 12 edges.
– The figure has 8 vertices.
Talk Math
Describe the differences between a triangular prism and a rectangular prism.
– A triangular prism has two bases and four faces, while a rectangular prism has one base and two faces.
Guided Practice
Question 1.
Describe the faces, edges, and vertices of the three-dimensional figure. Then identity it.
faces:
This figure has _____________ faces. The ___________ bases are congruent and parallel. The other faces are _____________.
edges:
There are _____________ edges. The edges that form the vertical sides of the rectangles are parallel and _____________.
vertices:
This figure has ____________ vertices.
The figure is a _____________.
The above-given shape:
– The figure has 5 faces. The triangular bases are congruent and parallel. The other faces are rectangles.
– There are 9 edges. The edges that form the vertical sides of the rectangles are parallel and congruent.
– The figure has 6 vertices.
– From the above information, we can say that the figure is a triangular prism.
Independent Practice
Describe the faces, edges, and vertices of each three-dimensional figure. Then identify it.
Question 2.
From the above-given figure count the faces, edges, and vertices.
– The figure has 6 faces. It is one of the simplest shapes in three-dimensional space. All six faces of a figure are squares, a two-dimensional shape.
– The figure has 12 edges.
– The figure has 8 vertices.
– All the faces are shaped as a square hence the length, breadth, and height are the same.
– The opposite planes or faces in a figure are parallel to each other.
– The opposite edges in a figure are parallel to each other.
– Each of the faces in a figure meets the other four faces.
– Each of the vertices in a figure meets the three faces and three edges.
– From the above information, we can say that the above-given figure is the cube.
Question 3.
– A figure has 5 faces, 9 edges, and 6 vertices.
– It is a polyhedron with 3 rectangular faces and 2 triangular faces.
– The two triangular bases are congruent to each other.
– From the above-given information, we can say that the figure is a triangular prism.
Question 4.
Explanation:
– It has 5 faces
– It has 6 edges.
– The vertices of the triangular prism are the vertices of the two triangular bases connected by lines that form rectangles. Its edges include 6 edges of two triangular bases (3 + 3) and 3 sides that join the bases.
Question 5.
– It has 6 faces
– It has 12 edges.
– It has 8 vertices.
– It has 3 dimensions which are length, width, and height.
– The opposite faces of a given figure are congruent.
– From the above information, we can say that the figure is a rectangular prism.
Question 6.
– It has 6 faces
– It has 12 edges.
– It has 8 vertices.
– It has 3 dimensions which are length, width, and height.
– The opposite faces of a given figure are congruent.
– From the above information, we can say that the figure is a rectangular prism.
Question 7.
– A figure has 5 faces, 9 edges, and 6 vertices.
– It is a polyhedron with 3 rectangular faces and 2 triangular faces.
– The two triangular bases are congruent to each other.
– From the above-given information, we can say that the figure is a triangular prism.
Problem Solving
Question 8.
Mathematical PRACTICE Identify Structure The Metropolitan Correction Center in Chicago is in the shape of a triangular prism. Circle the two-dimensional figures that make up the faces of the prism.
– A triangular prism has 5 faces, 9 edges, and 6 vertices.
– It is a polyhedron with 3 rectangular faces and 2 triangular faces.
– The two triangular bases are congruent to each other.
– Any cross-section of a triangular prism is in the shape of a triangle.
Question 9.
Describe the number of vertices and edges in an unopened cereal box. Identify the shape of the box.
Actually, the cereal box shape is like a rectangular prism.
For a rectangular prism:
– The number of vertices is 8
– The number of edges is 12.
HOT Problems
Question 10.
Mathematical PRACTICE Model Math What figure is formed if only the height of a cube is Increased? Draw the figure to support your answer.
The figure formed is a rectangular prism.
Question 11.
Building on the Essential Question Why is it important to know the different properties of three-dimensional figures?
It is important to know the different properties of three-dimensional figures because they preoccupy space and are applied in our day-to-day life.
### McGraw Hill My Math Grade 5 Chapter 12 Lesson 7 My Homework Answer Key
Practice
Describe the faces, edges, and vertices of each three-dimensional figure. Then identify it.
Question 1.
– It has 6 faces
– It has 12 edges.
– It has 8 vertices.
– It has 3 dimensions which are length, width, and height.
– The opposite faces of a given figure are congruent.
– From the above information, we can say that the figure is a rectangular prism.
Question 2.
From the above-given figure count the faces, edges, and vertices.
– The figure has 6 faces. Â A figure is one of the simplest shapes in three-dimensional space. All six faces of a figure are squares, a two-dimensional shape.
– The figure has 12 edges.
– The figure has 8 vertices.
– All the faces are shaped as a square hence the length, breadth, and height are the same.
– The opposite planes or faces in a figure are parallel to each other.
– The opposite edges in a figure are parallel to each other.
– Each of the faces in a figure meets the other four faces.
– Each of the vertices in a figure meets the three faces and three edges.
– From the above information, we can say that the above-given figure is the cube.
Problem Solving
Question 3.
Rhett made a simple drawing of his house. It is a three-dimensional figure with four faces that are rectangular and two that are square. What kind of figure is it?
The four faces are rectangular and two are square a rectangular prism.
Therefore, the figure is a rectangular prism.
Question 4.
A toy box has 6 faces that are squares. There are 12 edges and 8 vertices. Identify the shape of the toy box.
The above-given:
the number of faces that are squares = 6
The number of edges = 12
The number of vertices = 8
From the above-given information, we can say that the toy box is a cube.
Question 5.
Mathematical PRACTICE Make Sense of Problems Gabriel is playing a board game. When it is his turn, he tosses a three-dimensional figure that has 6 square faces. What kind of figure is it? How many edges and vertices does it have?
– The figure has 6 faces. Â A figure is one of the simplest shapes in three-dimensional space. All six faces of a figure are squares, a two-dimensional shape.
– The figure has 12 edges.
– The figure has 8 vertices.
– From the above information, we can say that the above-given figure is the cube.
Vocabulary Check
Fill in the blank with the correct term or number to complete the sentence.
Question 6.
A vertex is a point where ___________ or more edges meet.
Vertex is a point on a polygon where the sides or edges of the object meet or where two rays or line segments meet.
Test Practice
Question 7.
Which statement is true about the three-dimensional figure that most closely represents the slice of pie?
(A) The figure has 4 vertices.
(B) The figure has 6 vertices.
(C) The figure has 8 vertices.
(D) The figure has 9 vertices.
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A R I T H M E T I C
Lesson 32
# PRIME NUMBERS andPRIME FACTORIZATION
A natural number is a collection of indivisible ones.
1, 2, 3, 4, and so on. Although we often represent a natural number by a line,
the student should keep in mind that a natural number is not like a line.
A prime number is a special kind of natural number. In order to define a prime number, we must first define a proper divisor of a number.
By a "number" in what follows, we will mean a natural number.
1. What does it mean to say that a smaller number is a proper divisor of a larger number? It means that the larger number can be composed -- made up -- of the smaller number.
2 is a proper divisor of 10, because 10 can be composed -- made up -- of 2's. 5 and 1 are also proper divisors of 10, because 10 can be made up
of 5's and 1's. As for 10 itself, we are not interested that 10 is equal to one 10; we want to know what other numbers will compose it.
Nevertheless, 10 is called a divisor of itself, but not a proper divisor.
The proper divisors do not include the number itself.
It is important to note that 1 is a proper divisor of every natural number (except itself), because every natural number is made up of 1's. That is what a natural number is
5 = 1 + 1 + 1 + 1 + 1.
Problem 1.
a) Which numbers are the divisors of 12?
Do the problem yourself first!
1, 2, 3, 4, 6, 12.
a) Which are the proper divisors of 12?
1, 2, 3, 4, 6.
b) 16 can be composed of which other numbers?
1, 2, 4, 8.
Those are the proper divisors of 16.
c) 17 can be composed of which other numbers.
1.
1 is its only proper divisor.
d) Write all the proper divisors of 29.
1.
2. What is a prime number? A prime number is a number whose only proper divisor is 1.
Thus a prime number can be made up only of 1's. 17 and 29 are prime numbers.
Again, 1 is a proper divisor of every number (except itself), but a prime number has 1 as its only proper divisor.
Problem 2. Is 1 itself a prime number?
No. 1 has no proper divisors. 1 cannot be made up of other numbers.
1 in this case is the measure. It cannot be measured.
Problem 3. Write the first ten prime numbers.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
With the exception of 2, then, -- which is the only even prime -- a prime number is a kind of odd number.
3. What is a number called if it is not a prime? A composite number.
6 is a composite number, because it can be composed of other numbers besides 1.
1 itself is neither prime nor composite.
Square roots
When we multiply a number by itself, we say that we have "squared" the number. Thus the square of 5 is 5 × 5 = 25. We then say that 5 is the square root of 25. The square numbers are the numbers we get by squaring a number: 1, 4, 9, 16, 25, and so on.
50, for example, is not a square number, therefore it does not have an exact square root. Its square root is closest to 7, however, because
7 × 7 = 49.
Problem 4. The square root of 175 is closest to what number?
13. 13 × 13= 169.
Divisors
We can always find the divisors of a number in pairs. One member of the pair will be less than the square root of the number, and the other will be more. (If the number is a square number, then its square root will be its own partner.)
For example, here are the pairs of divisors of 24:
1 and 24. (Because 1 × 24 = 24.)
2 and 12. (Because 2 × 12 = 24.)
3 and 8. (Because 3 × 8 = 24.)
4 and 6. (Because 4 × 6 = 24.)
Each number on the left is less than the square root of 24, and each number on the right is more.
The point is:
When we look for divisors of a number,
it is necessary to look only up to its square root.
Because for each divisor we find less than the square root, we will have found its partner, which is more.
Problem 5. If we are looking for the divisors of 157, up to what number must we look?
12. Because 13 is more than the square root of 157.
In this Lesson we will be looking for the prime divisors of a number, and to that we now turn.
Prime divisors
Every number except 1 is either prime or composite. And if a number is composite, it will have a prime divisor.
(Euclid, VII. 31)
28 is composite. It has a prime divisor 2 and a prime divisor 7.
Example 1. What are the prime divisors of 63?
Answer. Here again are the first few prime numbers:
2, 3, 5, 7, 11, 13, 17.
We must test whether 2 is a divisor, or 3, or 5, and so on. But we need test only up to 7, because 11 is more than the square root of 63.
Is 2 a divisor of 63? No, it is not. How do we know? Because 63 is not an even number. And how do we know that? Because even numbers end in 0, 2, 4, 6, or 8.
Next, is 3 a divisor of 63? There is a test for divisibility by 3, and it is as follows:
A number is divisible by 3 if the sum of the digits is divisible by 3.
The sum of the digits of 63 is 6 + 3 = 9. 9 is divisible by 3. That tells us that 63 is divisible by 3. 3 is a prime divisor of 63.
Next, is 63 divisible by 5? There is a simple test for divisibility by 5, namely, the number ends in either 0 or 5. 63 does not end in 0 or 5. Therefore, 63 is not divisible by 5.
Finally, is 63 divisible by 7? Yes, it is: 63 = 7 × 9.
63, then, has two prime divisors: 3 and 7.
Example 2. What are the prime divisors of 78?
Answer. 2 is a prime divisor, and
78 = 2 × 39.
Now, 39 is composite. Therefore it has a prime divisor:
39 = 3 × 13.
And since 39 is made up of 13's, and 78 is made up of 39's, then 78 is made up of 13's. That is, 13 is a divisor of 78.
That is called the transitive property of " is made up of" or, equivalently, "is a divisor of;" namely:
If a is a divisor of b, and b is a divisor of c, then a is a divisor of c.
78 therefore has three prime divisors: 2, 3, and 13.
Problem 6. What is the smallest prime that is a divisor of the following?
a) 231. 3. The sum of the digits is divisible by 3.
b) 3,165. 3.
c) 3,265. 5.
d) 91. 7.
e) 121. 11.
Prime factorization
When numbers are multiplied, they are called factors. Since
30 = 2 × 15,
we say that 2 and 15 are factors of 30. Now, 2 is a prime factor -- but 15 is not. However, 15 = 3 × 5. Therefore we can express 30 as a product of prime factors only:
30 = 2 × 3 × 5.
"2 × 3 × 5" is called the prime factorization of 30. And it is unique. That is, apart from the order of the factors:
Every composite number can be uniquely factored
as a product of prime numbers only.
Note: We could have found those same factors by factoring 30 in any way. For example,
30 = 5 × 6.
And since 6 = 2 × 3,
30 = 5 × 2 × 3.
Apart from the order, we have found the same prime factors.
Example 3. Find the prime factorization of 102.
Solution. 2 is obviously a prime factor. Its partner will be half of 102
-- 51 (Lesson 16):
102 = 2 × 51.
Because the sum of the digits is divisible by 3, we know that 51 has a prime factor 3. Now, 3 times what is equal to 51? On mentally decomposing 51 into 30 + 21,
51 3 = 30 + 21 3 = 10 + 7 = 17.
51 = 3 × 17.
And 17 is a prime number. Therefore,
102 = 2 × 3 × 17.
That is the prime factorization of 102.
Problem 7. Which of these numbers is prime and which is composite? If a number is composite, write its prime factorization.
a) 29. Prime.
b) 50. 2 × 5 × 5.
c) 73. Prime.
d) 32. 2 × 2 × 2 × 2 × 2.
e) 60. 2 × 2 × 3 × 5.
f) 135. 3 × 3 × 3 × 5.
g) 137. Prime.
h) 143. 11 × 13.
i) 169. 13 × 13.
j) 360. 2 × 2 × 2 × 3 × 3 × 5.
k) 450. 2 × 5 × 5 × 3 × 3.
Square factors
We sometimes want to know whether a number has square factors. 50 for example has the square factor 25. 50 = 25 × 2. But for a less familiar number, such as 60, we can discover whether or not it has square factors by writing its prime factorization.
We could proceed as follows:
60 = 2 × 30 = 2 × 2 × 15 = 2 × 2 × 3 × 5.
When a prime appears twice, that product is a square number. 60, then, has one square factor, namely 2 × 2 = 4. 60 = 4 × 15.
Example 4. Does 180 have any square factors?
Solution. 180 = 2 × 90 = 2 × 2 × 45 = 2 × 2 × 9 × 5 = 2 × 2 × 3 × 3 × 5
180, then, has two square factors: 2 × 2 = 4, and 3 × 3 = 9.
But 4 × 9 is itself a square number -- 36. For,
A product of square numbers is itself a square number.
2 × 2 × 3 × 3 = 2 × 3 × 2 × 3 = 6 × 6.
Problem 8. Find the square factors of each number by writing its prime factorization.
a) 112 = 2 × 2 × 2 × 2 × 7 = 16 × 7.
b) 450 = 3 × 3 × 5 × 5 × 2 = 3 × 5 × 3 × 5 = 225 × 2.
c) 153 = 3 × 51 = 3 × 3 × 17 = 9 × 17.
d) 294 = 2 × 147 = 2 × 3 × 49 = 49 × 6.
e) 1225 = 25 × 49. 1225 is itself a square number. It is the square of 5 × 7 = 35.
Is there a last prime?
As the numbers get larger, the greater the possibility that they will have a divisor, so that there might in fact be a last prime.
Now, if there were a last prime, then we could imagine a list that contains every prime up to and including the last one. We will now prove that there is no such list, which is to say, There is no last prime.
Here is the theorem:
Given any list of prime numbers, there will always be a prime number
that is not on the list.
Let the following, then, be any list of prime numbers:
2, 3, 5, 7, 11, . . . , P.
Now construct the number N which will be the product of every prime on that list:
N = 2 × 3 × 5 × 7 × 11 × . . . × P.
Every prime on the list is thus a divisor of N.
N + 1 = (2 × 3 × 5 × 7 × 11 × . . . × P) + 1.
Now, N + 1 a number that is not on the list, because it is greater than every number on the list. And N + 1 is either prime or composite. If it is prime, then we have found a prime that is not on the list, and the theorem is proved.
If N + 1 is composite, then it has a prime factor p. But p is not one of the primes on the list For if it were, then p would be a divisor of both N + 1 and N. But that would imply that p divides their difference (Lesson 11), namely 1 -- which is absurd.
Therefore, if N + 1 is composite, then there is a prime p that is a divisor of N + 1 but not a divisor of N, which is to say, p is not a prime on the list.
Therefore, given any list of primes, there will always be a prime that is not on the list. Which is what we wanted to prove.
*
A modern enunciation of this theorem is: The number of primes is infinite. Euclid thus teaches us what we mean, or rather what we should mean, when we say that a collection of numbers is "infinite." It means: "No list of them is complete; there will always be more." That meaning of "infinite" refers to something that exists and that we could witness, namely an actual list. It does not refer to something that we cannot witness, namely a list that has no end.
Thus if we say that the number of primes is infinite, we mean they are potentially infinite, not actually infinite.
*
A famous problem in mathematics is the twin prime conjecture. It states that there are infinitely many pairs of primes that differ only by 2. For example, 5 and 7, 17 and 19, 41 and 43. The conjecture has never been proved.
What about prime triples, which are three primes that differ only by 2? For example, 3, 5, 7. What do you think? Are there many -- perhaps even an infinite number -- of such triples? Or is 3, 5, 7 the only one?
3, 5, 7 is the only such triple. Because in every sequence of three odd numbers, at least one of them is a multiple of 3. For if the first is a multiple of 3, then the proposition is proved; for example, 21, 23, 25. If the first is 1 more than a multiple of 3, then, on adding 2, the next will be a multiple of 3; for example, 25, 27, 29. Finally, if the first is 1 less than a multiple of 3, then the next will be 1 more, and the third will be a multiple of 3; for example, 35, 37, 39. In the sequence, 3, 5, 7, 3 is the only multiple of 3 that is a prime.
Next Lesson:
Greatest common divisor. Lowest common multiple.
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### 3.1 Introduction
You know that the paper is a model for a plane surface. When you join a number of points without lifting a pencil from the paper (and without retracing any portion of the drawing other than single points), you get a plane curve.
Try to recall different varieties of curves you have seen in the earlier classes.
Match the following: (Caution! A figure may match to more than one type).
Figure Type
(1) (a) Simple closed curve
(2) (b) A closed curve that is not simple
(3) (c) Simple curve that is not closed
(4) (d) Not a simple curve
### 3.2 Polygons
A simple closed curve made up of only line segments is called a polygon.
Curves that are polygons
Curves that are not polygons
Try to give a few more examples and non-examples for a polygon.
Draw a rough figure of a polygon and identify its sides and vertices.
##### 3.2.1 Classification of polygons
We classify polygons according to the number of sides (or vertices) they have.
##### 3.2.2 Diagonals
A diagonal is a line segment connecting two non-consecutive vertices of a polygon (Fig 3.1).
Fig 3.1
Can you name the diagonals in each of the above figures? (Fig 3.1)
Is a diagonal? What about ?
You already know what we mean by interior and exterior of a closed curve (Fig 3.2).
Interior Exterior
Fig 3.2
The interior has a boundary. Does the exterior have a boundary? Discuss with your friends.
##### 3.2.3 Convex and concave polygons
Here are some convex polygons and some concave polygons. (Fig 3.3)
Convex polygons .......................Concave polygons
Fig 3.3
Can you find how these types of polygons differ from one another? Polygons that are convex have no portions of their diagonals in their exteriors or any line segment joining any two different points, in the interior of the polygon, lies wholly in the interior of it . Is this true with concave polygons? Study the figures given. Then try to describe in your own words what we mean by a convex polygon and what we mean by a concave polygon. Give two rough sketches of each kind.
In our work in this class, we will be dealing with convex polygons only.
##### 3.2.4 Regular and irregular polygons
A regular polygon is both ‘equiangular’ and ‘equilateral’. For example, a square has sides of equal length and angles of equal measure. Hence it is a regular polygon. A rectangle is equiangular but not equilateral. Is a rectangle a regular polygon? Is an equilateral triangle a regular polygon? Why?
Regular polygons
Polygons that are not regular
[Note: Use of or indicates segments of equal length].
In the previous classes, have you come across any quadrilateral that is equilateral but not equiangular? Recall the quadrilateral shapes you saw in earlier classes – Rectangle, Square, Rhombus etc.
Is there a triangle that is equilateral but not equiangular?
##### 3.2.5 Angle sum property
Do you remember the angle-sum property of a triangle? The sum of the measures of the three angles of a triangle is 180°. Recall the methods by which we tried to visualise this fact. We now extend these ideas to a quadrilateral.
## Do This
1. Take any quadrilateral, say ABCD (Fig 3.4). Divide it into two triangles, by drawing a diagonal. You get six angles 1, 2, 3, 4, 5 and 6.
Fig 3.4
Use the angle-sum property of a triangle and argue how the sum of the measures of A, B, C and D amounts to 180° + 180° = 360°.
(i) (ii)
Fig 3.5
For doing this you may have to turn and match appropriate corners so that they fit.
2. Take four congruent card-board copies of any quadrilateral ABCD, with angles as shown [Fig 3.5 (i)]. Arrange the copies as shown in the figure, where angles 1, 2, 3, 4 meet at a point [Fig 3.5 (ii)].
What can you say about the sum of the angles 1, 2, 3 and 4?
[Note: We denote the angles by 1, 2, 3, etc., and their respective measures by m1, m2, m3, etc.]
The sum of the measures of the four angles of a quadrilateral is___________.
You may arrive at this result in several other ways also.
3. As before consider quadrilateral ABCD (Fig 3.6). Let P be any point in its interior. Join P to vertices A, B, C and D. In the figure, consider PAB. From this we see x = 180° – m2 – m3; similarly from PBC, y = 180° – m4 – m5, from PCD,
z = 180º – m6 – m7 and from PDA, w = 180º – m8
m1. Use this to find the total measure m1 + m2 + ...
+ m8, does it help you to arrive at the result? Remember
x + y + z + w = 360°.
Fig 3.6
Fig 3.7
4. These quadrilaterals were convex. What would happen if the quadrilateral is not convex? Consider quadrilateral ABCD. Split it into two triangles and find the sum of the interior angles (Fig 3.7).
### Exercise 3.1
1. Given here are some figures.
(1) (2) (3) (4)
(5) (6) (7) (8)
Classify each of them on the basis of the following.
(a) Simple curve (b) Simple closed curve (c) Polygon
(d) Convex polygon (e) Concave polygon
2. How many diagonals does each of the following have?
(a) A convex quadrilateral (b) A regular hexagon (c) A triangle
3. What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)
4. Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)
What can you say about the angle sum of a convex polygon with number of sides?
(a) 7
(b) 8
(c) 10
(d) n
5. What is a regular polygon?
State the name of a regular polygon of
(i) 3 sides (ii) 4 sides (iii) 6 sides
6. Find the angle measure x in the following figures.
(a)
(b)
(c)
(d)
7.
(a) Find x + y + z
(b) Find x + y + z + w
##### 3.3 Sum of the Measures of the Exterior Angles of a Polygon
On many occasions a knowledge of exterior angles may throw light on the nature of interior angles and sides.
## DO THIS
Draw a polygon on the floor, using a piece of chalk.
(In the figure, a pentagon ABCDE is shown) (Fig 3.8).
Fig 3.8
We want to know the total measure of angles, i.e,
m1 + m2 + m3 + m4 + m5. Start at A. Walk along . On reaching B, you need to turn through an angle of m1, to walk along . When you reach at C, you need to turn through an angle of m2 to walk along . You continue to move in this manner, until you return to side AB. You would have in fact made one complete turn.
Therefore, m1 + m2 + m3 + m4 + m5 = 360°
This is true whatever be the number of sides of the polygon.
Therefore, the sum of the measures of the external angles of any polygon is 360°
Example 1: Find measure x in Fig 3.9.
Solution: x + 90° + 50° + 110° = 360° (Why?)
x + 250° = 360°
x = 110°
Fig 3.9
## Try These
Take a regular hexagon Fig 3.10.
Fig 3.10
1. What is the sum of the measures of its exterior angles x, y, z, p, q, r?
2. Is x = y = z = p = q = r? Why?
3. What is the measure of each?
(i) exterior angle (ii) interior angle
4. Repeat this activity for the cases of
(i) a regular octagon (ii) a regular 20-gon
Example 2: Find the number of sides of a regular polygon whose each exterior angle has a measure of 45°.
Solution: Total measure of all exterior angles = 360°
Measure of each exterior angle = 45°
Therefore, the number of exterior angles = = 8
The polygon has 8 sides.
### Exercise 3.2
1. Find x in the following figures.
(a) (b)
2. Find the measure of each exterior angle of a regular polygon of
(i) 9 sides (ii) 15 sides
3. How many sides does a regular polygon have if the measure of an exterior angle is 24°?
4. How many sides does a regular polygon have if each of its interior angles
is 165°?
5. (a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
(b) Can it be an interior angle of a regular polygon? Why?
6. (a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Based on the nature of the sides or angles of a quadrilateral, it gets special names.
##### 3.4.1 Trapezium
Trapezium is a quadrilateral with a pair of parallel sides.
These are trapeziums These are not trapeziums
Study the above figures and discuss with your friends why some of them are trapeziums while some are not. (Note: The arrow marks indicate parallel lines).
## Do This
1. Take identical cut-outs of congruent triangles of sides 3 cm, 4 cm, 5 cm. Arrange them as shown (Fig 3.11).
Fig 3.11
You get a trapezium. (Check it!) Which are the parallel sides here? Should the
non-parallel sides be equal?
You can get two more trapeziums using the same set of triangles. Find them out and discuss their shapes.
2. Take four set-squares from your and your friend’s instrument boxes. Use different numbers of them to place side-by-side and obtain different trapeziums.
If the non-parallel sides of a trapezium are of equal length, we call it an isosceles trapezium. Did you get an isoceles trapezium in any of your investigations given above?
##### 3.4.2 Kite
Kite is a special type of a quadrilateral. The sides with the same markings in each figure are equal. For example AB = AD and BC = CD.
These are kites These are not kites
Study these figures and try to describe what a kite is. Observe that
(i) A kite has 4 sides (It is a quadrilateral).
(ii) There are exactly two distinct consecutive pairs of sides of equal length.
Check whether a square is a kite.
## DO THIS
Take a thick white sheet.
Fold the paper once.
Draw two line segments of different lengths as shown in Fig 3.12.
Fig 3.12
Cut along the line segments and open up.
Show that ∆ABC and ∆ADC are congruent . What do we infer from this?
You have the shape of a kite (Fig 3.13).
Has the kite any line symmetry?
Fold both the diagonals of the kite. Use the set-square to check if they cut at right angles. Are the diagonals equal in length?
Fig 3.13
Verify (by paper-folding or measurement) if the diagonals bisect each other.
By folding an angle of the kite on its opposite, check for angles of equal measure.
Observe the diagonal folds; do they indicate any diagonal being an angle bisector?
Share your findings with others and list them. A summary of these results are given elsewhere in the chapter for your reference.
3.4.3 Parallelogram
A parallelogram is a quadrilateral. As the name suggests, it has something to do with parallel lines.
These are parallelograms These are not parallelograms
Study these figures and try to describe in your own words what we mean by a parallelogram. Share your observations with your friends.
Check whether a rectangle is also a parallelogram.
## Do This
Take two different rectangular cardboard strips of different widths (Fig 3.14).
Strip 1 Strip 2
Fig 3.14
Place one strip horizontally and draw lines along its edge as drawn in the figure (Fig 3.15).
Fig 3.15
Now place the other strip in a slant position over the lines drawn and use this to draw two more lines as shown (Fig 3.16).
These four lines enclose a quadrilateral. This is made up of two pairs of parallel lines (Fig 3.17).
It is a parallelogram.
A parallelogram is a quadrilateral whose opposite sides are parallel.
Fig 3.16
Fig 3.17
3.4.4 Elements of a parallelogram
There are four sides and four angles in a parallelogram. Some of these are equal. There are some terms associated with these elements that you need to remember.
Fig 3.18
Given a parallelogram ABCD (Fig 3.18).
and , are opposite sides. and form another pair of opposite sides.
A and C are a pair of opposite angles; another pair of opposite angles would be B and D.
and are adjacent sides. This means, one of the sides starts where the other ends. Are and adjacent sides too? Try to find two more pairs of adjacent sides.
A and B are adjacent angles. They are at the ends of the same side. B and C are also adjacent. Identify other pairs of adjacent angles of the parallelogram.
## Do This
Take cut-outs of two identical parallelograms, say ABCD and ABCD (Fig 3.19).
Fig 3.19
Here is same as except for the name. Similarly the other corresponding sides are equal too.
Place over . Do they coincide? What can you now say about the lengths and ?
Similarly examine the lengths and . What do you find?
You may also arrive at this result by measuring and
Property: The opposite sides of a parallelogram are of equal length.
## TRY THESE
Take two identical set squares with angles 30° – 60° – 90° and place them adjacently to form a parallelogram as shown in Fig 3.20. Does this help you to verify the above property?
Fig 3.20
You can further strengthen this idea through a logical argument also.
Fig 3.21
Consider a parallelogram ABCD (Fig 3.21). Draw
any one diagonal, say
. Looking at the angles,
1 = 2 and 3 = 4 (Why?)
Since in triangles ABC and ADC, 1 = 2, 3 = 4 and is common, so, by ASA congruency condition,
ABC CDA (How is ASA used here?)
This gives AB = DC and BC = AD.
Example 3: Find the perimeter of the parallelogram PQRS (Fig 3.22).
Fig 3.22
Solution: In a parallelogram, the opposite sides have same length.
Therefore, PQ = SR = 12 cm and QR = PS = 7 cm
So, Perimeter = PQ + QR + RS + SP
= 12 cm + 7 cm + 12 cm + 7 cm = 38 cm
##### 3.4.5 Angles of a parallelogram
We studied a property of parallelograms concerning the (opposite) sides. What can we say about the angles?
## Do This
Let ABCD be a parallelogram (Fig 3.23). Copy it on a tracing sheet. Name this copy as ABCD. Place ABCDon ABCD. Pin them together at the point where the diagonals meet. Rotate the transparent sheet by 180°. The parallelograms still concide; but you now find A lying exactly on C and vice-versa; similarly B lies on D and vice-versa.
Fig 3.23
Does this tell you anything about the measures of the angles A and C? Examine the same for angles B and D. State your findings.
Property: The opposite angles of a parallelogram are of equal measure.
## Try These
Take two identical 30° – 60° – 90° set-squares and form a parallelogram as before. Does the figure obtained help you to confirm the above property?
You can further justify this idea through logical arguments.
If and are the diagonals of the parallelogram, (Fig 3.24) you find that
Fig 3.24
1 =2 and 3 = 4 (Why?)
Studying ABC and ADC (Fig 3.25) separately, will help you to see that by ASA congruency condition,
ABC CDA (How?)
Fig 3.25
This shows that B and D have same measure. In the same way you can get
m
A = m C.
Alternatively, 1 = 2 and 3 =4, we have, mA =1+4 =2+C mC
Example 4: In Fig 3.26, BEST is a parallelogram. Find the values x, y and z.
Solution: S is opposite to B.
Fig 3.26
So, x = 100° (opposite angles property)
y = 100° (measure of angle corresponding to x)
z = 80° (since y, z is a linear pair)
We now turn our attention to adjacent angles of a parallelogram.
In parallelogram ABCD, (Fig 3.27).
A and D are supplementary since
and with transversal , these two angles are interior opposite.
Fig 3.27
A and B are also supplementary. Can you say ‘why’?
and is a transversal, making A and B interior opposite.
Identify two more pairs of supplementary angles from the figure.
Property: The adjacent angles in a parallelogram are supplementary.
Example 5: In a parallelogram RING, (Fig 3.28) if mR = 70°, find all the other angles.
Fig 3.28
Solution: Given mR = 70°
Then mN = 70°
because R and N are opposite angles of a parallelogram.
Since R and I are supplementary,
mI = 180° – 70° = 110°
Also, mG = 110° since G is opposite to I
Thus, mR = mN = 70° and mI = mG = 110°
## THINK,DISCUSS AND WRITE
After showing mR = mN = 70°, can you find mI and mG by any other method?
3.4.6 Diagonals of a parallelogram
The diagonals of a parallelogram, in general, are not of equal length. (Did you check this in your earlier activity?) However, the diagonals of a parallelogram have an interesting property.
Fig 3.29
## Do This
Take a cut-out of a parallelogram, say,
ABCD (Fig 3.29). Let its diagonals and meet at O.
Find the mid point of by a fold, placing C on A. Is the mid-point same as O?
Does this show that diagonal bisects the diagonal at the point O? Discuss it with your friends. Repeat the activity to find where the mid point of could lie.
Property: The diagonals of a parallelogram bisect each other (at the point of their intersection, of course!)
To argue and justify this property is not very difficult. From Fig 3.30, applying ASA criterion, it is easy to see that
Fig 3.30
AOB COD (How is ASA used here?)
This gives AO = CO and BO = DO
Example 6: In Fig 3.31 HELP is a parallelogram. (Lengths are in cms). Given that
OE = 4 and HL is 5 more than PE? Find OH.
Fig 3.31
Solution : If OE = 4 then OP also is 4 (Why?)
So PE = 8, (Why?)
Therefore HL = 8 + 5 = 13
Hence OH = = 6.5 (cms)
### Exercise 3.3
1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD = ...... (ii) DCB = ......
(iii) OC = ...... (iv) m DAB + m CDA = ......
2. Consider the following parallelograms. Find the values of the unknowns x, y, z.
(i) (ii)
(iii) (iv) (v)
3. Can a quadrilateral ABCD be a parallelogram if
(i) D + B = 180°? (ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) A = 70° and C = 65°?
4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
5. The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
8. The following figures GUNS and RUNS are parallelograms.
Find x and y. (Lengths are in cm)
(i)
(ii)
9.
In the above figure both RISK and CLUE are parallelograms. Find the value of x.
10. Explain how this figure is a trapezium. Which of its two sides are parallel? (Fig 3.32)
Fig 3.32 Fig 3.33
11. Find mC in Fig 3.33 if .
Fig 3.34
12. Find the measure of P and S if in Fig 3.34.
(If you find mR, is there more than one method to find mP?)
### 3.5 Some Special Parallelograms
##### 3.5.1 Rhombus
We obtain a Rhombus (which, you will see, is a parallelogram) as a special case of kite (which is not a a parallelogram).
## Do This
Recall the paper-cut kite you made earlier.
Kite-cut Rhombus-cut
When you cut along ABC and opened up, you got a kite. Here lengths AB and BC were different. If you draw AB = BC, then the kite you obtain is called a rhombus.
Note that the sides of rhombus are all of same length; this is not the case with the kite.
A rhombus is a quadrilateral with sides of equal length.
Kite Rhombus
Since the opposite sides of a rhombus have the same length, it is also a parallelogram. So, a rhombus has all the properties of a parallelogram and also that of a kite. Try to list them out. You can then verify your list with the check list summarised in the book elsewhere.
The most useful property of a rhombus is that of its diagonals.
Property: The diagonals of a rhombus are perpendicular bisectors of one another.
## DO THIS
Take a copy of rhombus. By paper-folding verify if the point of intersection is the mid-point of each diagonal. You may also check if they intersect at right angles, using the corner of a set-square.
Here is an outline justifying this property using logical steps.
ABCD is a rhombus (Fig 3.35). Therefore it is a parallelogram too.
Fig 3.35
Since diagonals bisect each other, OA = OC and OB = OD.
We have to show that mAOD = mCOD = 90°
It can be seen that by SSS congruency criterion
AOD COD
Therefore, m AOD = m COD
Since AOD and COD are a linear pair,
m AOD = m COD = 90°
Since AO = CO (Why?)
OD = OD
Example 7:
RICE is a rhombus (Fig 3.36). Find x, y, z. Justify your findings.
Fig 3.36
Solution:
x = OE y = OR z = side of the rhombus
= OI (diagonals bisect) = OC (diagonals bisect) = 13 (all sides are equal )
= 5 = 12
##### 3.5.2 A rectangle
A rectangle is a parallelogram with equal angles (Fig 3.37).
Fig 3.37
What is the full meaning of this definition? Discuss with your friends.
If the rectangle is to be equiangular, what could be the measure of each angle?
Let the measure of each angle be x°.
Then 4x° = 360° (Why)?
Therefore, x° = 90°
Thus each angle of a rectangle is a right angle.
So, a rectangle is a parallelogram in which every angle is a right angle.
Being a parallelogram, the rectangle has opposite sides of equal length and its diagonals bisect each other.
In a parallelogram, the diagonals can be of different lengths. (Check this); but surprisingly the rectangle (being a special case) has diagonals of equal length.
Property: The diagonals of a rectangle are of equal length.
Fig 3.38
Fig 3.39
Fig 3.40
This is easy to justify. If ABCD is a rectangle (Fig 3.38), then looking at triangles ABC and ABD separately [(Fig 3.39) and (Fig 3.40) respectively], we have
ABC ABD
This is because AB = AB (Common)
m A = m B = 90° (Why?)
The congruency follows by SAS criterion.
Fig 3.41
Thus AC = BD
and in a rectangle the diagonals, besides being equal in length bisect each other (Why?)
Example 8: RENT is a rectangle (Fig 3.41). Its diagonals meet at O. Find x, if
OR = 2
x + 4 and OT = 3x + 1.
Solution: is half of the diagonal ,
is half of the diagonal .
Diagonals are equal here. (Why?)
So, their halves are also equal.
Therefore 3x + 1 = 2x + 4
or x = 3
BELT is a square, BE = EL = LT = TB
B, E, L, T are right angles.
BL = ET and .
OB = OL and OE = OT.
##### 3.5.3 A square
A square is a rectangle with equal sides.
This means a square has all the properties of a rectangle with an additional requirement that all the sides have equal length.
The square, like the rectangle, has diagonals of equal length.
In a rectangle, there is no requirement for the diagonals to be perpendicular to one another, (Check this).
In a square the diagonals.
(i) bisect one another (square being a parallelogram)
(ii) are of equal length (square being a rectangle) and
(iii) are perpendicular to one another.
Hence, we get the following property.
Property: The diagonals of a square are perpendicular bisectors of each other.
## Do This
Take a square sheet, say PQRS (Fig 3.42).
Fig 3.42
Fold along both the diagonals. Are their mid-points the same?
Check if the angle at O is 90° by using a set-square.
This verifies the property stated above.
We can justify this also by arguing logically:
ABCD is a square whose diagonals meet at O (Fig 3.43).
Fig 3.43
OA = OC (Since the square is a parallelogram)
By SSS congruency condition, we now see that
AOD ≅ ∆ COD (How?)
Therefore, mAOD = mCOD
These angles being a linear pair, each is right angle.
### Exercise 3.4
1. State whether True or False.
(a) All rectangles are squares
(b) All rhombuses are parallelograms
(c) All squares are rhombuses and also rectangles
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
2. Identify all the quadrilaterals that have.
(a) four sides of equal length (b) four right angles
3. Explain how a square is.
(i) a quadrilateral (ii) a parallelogram (iii) a rhombus (iv) a rectangle
4. Name the quadrilaterals whose diagonals.
(i) bisect each other (ii) are perpendicular bisectors of each other (iii) are equal
5. Explain why a rectangle is a convex quadrilateral.
6. ABC is a right-angled triangle and O is the mid point of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).
## THINK, DISCUSS AND WRITE
1. A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular?
2. A square was defined as a rectangle with all sides equal. Can we define it as rhombus with equal angles? Explore this idea.
3. Can a trapezium have all angles equal? Can it have all sides equal? Explain.
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# Solving Triangles
Field: Geometry
Image Created By: Orion Pictures
In the 1991 film Shadows and Fog, the eerie shadow of a larger-than-life figure appears against the wall as the shady figure lurks around the corner. How tall is the ominous character really? Filmmakers use the geometry of shadows and triangles to make this special effect.
The shadow problem is a standard type of problem for teaching trigonometry and the geometry of triangles. In the standard shadow problem, several elements of a triangle will be given. The process by which the rest of the elements are found is referred to as solving a triangle.
# Basic Description
A triangle has six total elements: three sides and three angles. Sides are valued by length, and angles are valued by degree or radian measure. According to postulates for congruent triangles, given three elements, other elements can always be determined as long as at least one side length is given. Math problems that involve solving triangles, like shadow problems, typically provide certain information about just a few of the elements of a triangle, so that a variety of methods can be used to solve the triangle.
Shadow problems normally have a particular format. Some light source, often the sun, shines down at a given angle of elevation. The angle of elevation is the smallest—always acute—numerical angle measure that can be measured by swinging from the horizon from which the light source shines. Assuming that the horizon is parallel to the surface on which the light is shining, the angle of elevation is always equal to the angle of depression. The angle of depression is the angle at which the light shines down, compared to the angle of elevation which is the angle at which someone or something must look up to see the light source. Knowing the angle of elevation or depression can be helpful because trigonometry can be used to relate angle and side lengths.
In the typical shadow problem, the light shines down on an object or person of a given height. It casts a shadow on the ground below, so that the farthest tip of the shadow makes a direct line with the tallest point of the person or object and the light source. The line that directly connects the tip of the shadow and the tallest point of the object that casts the shadow can be viewed as the hypotenuse of a triangle. The length from the tip of the shadow to the point on the surface where the object stands can be viewed as the first leg, or base, of the triangle, and the height of the object can be viewed as the second leg of the triangle. In the most simple shadow problems, the triangle is a right triangle because the object stands perpendicular to the ground.
In the picture below, the sun casts a shadow on the man. The length of the shadow is the base of the triangle, the height of the man is the height of the triangle, and the length from the tip of the shadow to top of the man's head is the hypotenuse. The resulting triangle is a right triangle.
In another version of the shadow problem, the light source shines from the same surface on which the object or person stands. In this case the shadow is projected onto some wall or vertical surface, which is typically perpendicular to the first surface. In this situation, the line that connects the light source, the top of the object and the tip of the shadow on the wall is the hypotenuse. The height of the triangle is the length of the shadow on the wall, and the distance from the light source to the base of the wall can be viewed as the other leg other leg of the triangle. The picture below diagrams this type of shadow problem, and this page's main picture is an example of one of these types of shadows.
More difficult shadow problems will often involve a surface that is not level, like a hill. The person standing on the hill does not stand perpendicular to the surface of the ground, so the resulting triangle is not a right triangle. Other shadow problems may fix the light source, like a street lamp, at a given height. This scenario creates a set of two similar triangles.
Ultimately, a shadow problem asks you to solve a triangle given only a few elements of the possible six total. In the case of some shadow problems, like the one that involves two similar triangles, information about one triangle may be given and the question may ask to find elements of another.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Trigonometry, Geometry
Shadows are useful in the set-up of a triangle pro [...]
Shadows are useful in the set-up of a triangle problems because of the way light works. A shadow is cast when light cannot shine through a solid surface. Light shines in a linear fashion, that is to say it does not bend. Light waves travel forward in the same direction in which the light was shined.
In addition to the linear fashion in which light shines, light has certain angular properties. When light shines on an object that reflects light, it reflects back at the same angle at which it shined. Say a light shines onto a mirror. The angle between the beam of light and the wall that the mirror is the angle of approach. The angle from the wall at which the light reflects off of the mirror is the angle of departure. The angle of approach is equal to the angle of departure.
Light behaves the same way a cue ball does when it is bounced off of the wall of a pool table at a certain angle. Just like the way that the cue ball bounces off the wall, light reflects off of the mirror at exactly the same angle at which it shines. The beam of light has the same properties as the cue ball in this case: the angle of departure is the same as the angle of approach. This property will help with certain types of triangle problems, particularly those that involve mirrors.
Shadow problems are just one type of problem that involves solving triangles. There are numerous other formats and set ups for unsolved triangle problems. Most of these problems are formatted as word problems; they set up a triangle problem in terms of some real life scenario.
There are, however, many problems that simply provide numbers that represent angles and side lengths. In this type of problem, angles are denoted with capital letters, ${A, B, C,...}$, and the sides are denoted by lower-case letters,${a,b,c,...}$, where $a$ is the side opposite the angle $A$.
## Ways to Solve Triangles
In all cases, a triangle problem will only give a few elements of a triangle and will ask to find one or more of the lengths or angle measures that is not given. There are numerous formulas, methods, and operations that can help to solve a triangle depending on the information given in the problem.
The first step in any triangle problem is drawing a diagram. A picture can help to show which elements of the triangle are given and which elements are adjacent or opposite one another. By knowing where the elements are in relation to one another, we can use the trigonometric functions to relate angle and side lengths.
There are many techniques which can be implemented in solving triangles:
• Trigonometry: The basic trigonometric functions relate side lengths to angles. By substituting the appropriate values into the formulas for sine, cosine, or tangent, trigonometry can help to solve for a particular side length or angle measure of a right triangle. This is useful when given a side length and an angle measure.
• Inverse Trigonometry: Provided two side lengths, the inverse trig functions use the ratio of the two lengths and output an angle measure in right triangle trigonometry. Inverse trig is particularly useful in finding an angle measure when two side lengths are given in a right triangle.
• Special Right Triangles: Special right triangles are right triangles whose side lengths produce a particular ratio in trigonometry. A 30°− 60°− 90° triangle has a hypotenuse that is twice as long as one of its legs. A 45°− 45°− 90° is called an isosceles right triangle since both of its legs are the same length. These special cases can help to quicken the process of solving triangles.
• Pythagorean Theorem: The Pythagorean Theorem relates the squares of all three side lengths to one another in right triangles. This is useful when a triangle problem provides two side lengths and a third is needed.
$a^{2}+b^{2} = c^{2}$
• Pythagorean Triples: A Pythagorean triple is a set of three positive integers that satisfy the Pythagorean Theorem. The set {3,4,5} is one of the most commonly seen triples. Given a right triangle with legs of length 3 and 4, for example, the hypotenuse is known to be 5 by Pythagorean triples.
• Law of Cosines: The law of cosines is a generalization of the Pythagorean Theorem which can be used for solving non-right triangles. The law of cosines relates the squares of the side lengths to the cosine of one of the angle measures. This is particularly useful given a SAS configuration, or when three side lengths are known and no angles for non-right triangles.
$c^{2} = a^{2} + b^{2} - 2ab \cos C$
• Law of Sines: The law of sines is a formula that relates the sine of a given angle to the length of its opposite side. The law of sines is useful in any configuration when an angle measure and the length of its opposite side are given. It is also useful given an ASA configuration, and often the ASS configuration . The ASS configuration is known as the ambiguous case since it does not always provide one definite solution to the triangle.
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
When solving a triangle, one side length must always be given in the problem. Given an AAA configuration, there is no way to prove congruency. According to postulates for congruent triangles, the AAA configuration proves similarity in triangles, but there is no way to find the side lengths of a triangle. Knowing just angle measures is not helpful in solving triangles.
## Example Triangle Problems
Example 1: Using Trigonometry
A damsel in distress awaits her rescue from the tallest tower of the castle. A brave knight is on the way. He can see the castle in the distance and starts to plan his rescue, but he needs to know the height of the tower so he can plan properly. The knight sits on his horse 500 feet away from the castle. He uses his handy protractor to find the measure of the angle at which he looks up to see the princess in the tower, which is 15°. Sitting on the horse, the knight's eye level is 8 feet above the ground. What is the height of the tower?
We can use tangent to solve this problem. For a more in depth look at tangent, see Basic Trigonometric Functions.
Use the definition of tangent.
$\tan =\frac{\text{opposite}}{\text{adjacent}}$
Plug in the angle and the known side length.
$\tan 15^\circ =\frac{x \text{ft}}{500 \text{ft}}$
Clearing the fraction gives us
$\tan 15^\circ (500) =x$
Simplify for
$(.26795)(500) =x$
Round to get
$134 \text{ft} \approx x$
But this is only the height of the triangle and not the height of the tower. We need to add 8 ft to account for the height between the ground and the knight's eye-level which served as the base of the triangle.
$134 \text{ft} + 8 \text{ft} = h$
simplifying gives us
$142 \text{ft} = h$
The tower is approximately 142 feet tall.
Example 2: Using Law of Sines
A man stands 100 feet above the sea on top of a cliff. The captain of a white-sailed ship looks up at a 45° angle to see the man, and the captain of a black-sailed ship looks up at a 30° angle to see him. How far apart are the two ships?
To solve this problem, we can use the law of sines to solve for the bases of the two triangles since we have an AAS configuration with a known right angle. To find the distance between the two ships, we can take the difference in length between the bases of the two triangles.
First, we need to find the third angle for both of the triangles. Then we can use the law of sines.
For the black-sailed ship, $180^\circ - 90^\circ - 30^\circ =60^\circ$ Let the distance between this ship and the cliff be denoted by $b$. By the law of sines, $\frac{100}{\sin 30^\circ} = \frac{b}{\sin 60^\circ}$ Clear the fractions to get, $100(\sin 60^\circ) = b(\sin 30^\circ)$ Compute the sines of the angle to give us $100\frac{\sqrt{3}}{2} = b\frac{1}{2}$ Simplify for $100(\sqrt{3}) = b$ YOU CAN'T SEE THIS! OOOOOOOOOOOOOOOOOOOOOOHHH SPOOOOOKY! For the white-sailed ship, $180^\circ - 90^\circ - 45^\circ = 45^\circ$ Let the distance between this ship and the cliff be denoted by $a$. By the law of sines, $\frac{100}{\sin 45^\circ} = \frac{a}{\sin 45^\circ}$ Clear the fractions to get, $100(\sin 45^\circ) = b(\sin 45^\circ)$ Compute the sines of the angle to give us $100\frac{\sqrt{2}}{2} = b\frac{\sqrt{2}}{2}$ Simplify for $a = 100 \text{ft}$
Multiply and round for
$b =173 \text{ft}$
The distance between the two ships, $x$, is the positive difference between the lengths of the bases of the triangle.
$b-a=x$
$173-100 = 73 \text{ft}$
The ships are about 73 feet apart from one another.
Example 3: Using Multiple Methods
At the park one afternoon, a tree casts a shadow on the lawn. A man stands at the edge of the shadow and wants to know the angle at which the sun shines down on the tree. If the tree is 51 feet tall and if he stands 68 feet away from the tree, what is the angle of elevation?
There are several ways to solve this problem. The following solution uses a combination of the methods described above.
First, we can use Pythagorean Theorem to find the length of the hypotenuse of the triangle, from the tip of the shadow to the top of the tree.
$a^{2}+b^{2} = c^{2}$
Substitute the length of the legs of the triangle for $a, b$
$51^{2}+68^{2} = c^{2}$
Simplifying gives us
$2601+4624 = c^{2}$
$7225 = c^{2}$
Take the square root of both sides for
$\sqrt{7225} = c$
$85 = c$
Next, we can use the law of cosines to find the measure of the angle of elevation.
$a^{2}=b^{2}+c^{2} - 2bc \cos A$
Plugging in the appropriate values gives us
$51^{2}=68^{2}+85^{2} - 2(85)(68) \cos A$
Computing the squares gives us
$2601= 4624+7225 - 11560 \cos A$
Simplify for
$2601= 11849 - 11560 \cos A$
Subtract $11849$ from both sides for
$-9248= -11560 \cos A$
Simplify to get
$.8 = \cos A$
Use inverse trigonometry to find the angle of elevation.
$A = 37^\circ$
# Why It's Interesting
Shadow Problems are one of the most common types of problems used in teaching trigonometry. A shadow problem sets up a scenario that is simple, visual, and easy to remember. Shadow problems are commonly used and highly applicable.
Shadows, while an effective paradigm in a word problem, can even be useful in real life applications. In this section, we can use real life examples of using shadows and triangles to calculate heights and distances.
## Example: Sizing Up Swarthmore
The Clothier Bell Tower is the tallest building on Swarthmore College's campus, yet few people know exactly how tall the tower stands. We can use shadows to determine the height of the tower. Here's how:
Step 1) Mark the shadow of the of the tower. Make sure to mark the time of day. The sun is at different heights throughout the day. The shadows are longest earlier in the morning and later in the afternoon. At around midday, the shadows aren't very long, so it might be harder to find a good shadow. When we marked the shadow of the bell tower, it was around 3:40 pm in mid-June.
Step 2) After marking the shadow, we can measure the distance from our mark to the bottom of the tower. This length will serve as the base of our triangle. In this case, the length of the shadow was 111 feet.
Step 3) Measure the angle of the sun at that time of day.
Use a yardstick to make a smaller, more manageable triangle. Because the sun shines down at the same angle as it does on the bell tower, the small triangle and the bell tower's triangle are similar and therefore have the same trigonometric ratios.
• Stand the yardstick so it's perpendicular to the ground so that it forms a right angle. The sun will cast a shadow. Mark the end of the shadow with a piece of chalk.
• Measure the length of the shadow. This will be considered the length of the base of the triangle.
Draw a diagram of the triangle made by connecting the top of the yardstick to the marked tip of the shadow. Use inverse trigonometry to determine the angle of elevation. $\tan X = \frac{36 \text{in}}{27 \text{in}}$ $\arctan \frac{36}{27} = X$ $\arctan \frac{4}{3} = X$ $X = 53^\circ$ Step 4) Now, we can use trigonometry to solve the triangle for the height of the bell tower. $\tan 53^\circ = \frac{h}{111 \text{ft}}$ Clearing the fractions, $111 (\tan 53^\circ) = h$ Plugging in the value of $\tan 53^\circ$ gives us $111 \frac{4}{3} = h$ Simplify for $148 \text{ft} = h$ According to our calculations, the height of the Clothier Bell Tower is 148 feet.
## History: Eratosthenes and the Earth
In ancient Greece, mathematician Eratosthenes made a name for himself in the history books by calculating the circumference of the Earth by using shadows. Many other mathematicians had attempted the problem before, but Eratosthenes was the first one to actually have any success. His rate of error was less than 2%.
Eratosthenes used shadows to calculate the distance around the Earth. As an astronomer, he determined the time of the summer solstice when the sun would be directly over the town of Syene in Egypt (now Aswan). On this day, with the sun directly above, there were no shadows, but in Alexandria, which is about 500 miles north of Syene, Eratosthenes saw shadows. He calculated based on the length of the shadow that the angle at which the sun hit the Earth was 7 °. He used this calculation, along with his knowledge of geometry, to determine the circumference of the Earth.
# References
All of the images on this page, unless otherwise stated on their own image page, were made or photographed by the author Richard Scott, Swarthmore College.
The information on Eratosthenes can be cited to http://www.math.twsu.edu/history/men/eratosthenes.html.
The main image and details about it were found at http://www.imdb.com/title/tt0105378/.
Some of the ideas for problems/pictures on this page are based from ideas or concepts in the Interactive Mathematics Program Textbooks by Fendel, Resek, Alper and Fraser.
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## A Multiplication Based Logic Puzzle
### When is 690 a Palindrome? In Base 16 and Base 29.
• 690 is a composite number.
• Prime factorization: 690 = 2 x 3 x 5 x 23
• The exponents in the prime factorization are 1, 1, 1, and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 x 2 = 16. Therefore 690 has exactly 16 factors.
• Factors of 690: 1, 2, 3, 5, 6, 10, 15, 23, 30, 46, 69, 115, 138, 230, 345, 690
• Factor pairs: 690 = 1 x 690, 2 x 345, 3 x 230, 5 x 138, 6 x 115, 10 x 69, 15 x 46, or 23 x 30
• 690 has no square factors that allow its square root to be simplified. √690 ≈ 26.267851
Here is today’s factoring puzzle:
Print the puzzles or type the solution on this excel file: 10 Factors 2015-11-23
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Here is a little more about the number 690:
690 is the sum of the six prime numbers from 103 to 131. Do you know what all of those prime numbers are?
690 is also the hypotenuse of Pythagorean triple 414-552-690. What is the greatest common factor of those three numbers?
In BASE 10 we use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. For bases smaller than 10 we use only the digits less than that base. For bases greater than 10, we might use A to represent 10, B to represent 11, and so forth all the way to Z representing 35 in BASE 36.
690 is a palindrome in two bases that require us to use letters of the alphabet to represent it:
• 2B2 in BASE 16; note that 2(256) + 11(16) + 2(1) = 690
• NN in BASE 29; note that 23(29) + 23(1) = 690. (N is the 14th letter of the alphabet and 14 + 9 = 23)
NN looks like it is divisible by 11, but remember that 11 base 29 is the same as 30 in base 10.
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# The End 2016 Mathematics A To Z: Yang Hui’s Triangle
Today’s is another request from gaurish and another I’m glad to have as it let me learn things too. That’s a particularly fun kind of essay to have here.
## Yang Hui’s Triangle.
It’s a triangle. Not because we’re interested in triangles, but because it’s a particularly good way to organize what we’re doing and show why we do that. We’re making an arrangement of numbers. First we need cells to put the numbers in.
Start with a single cell in what’ll be the top middle of the triangle. It spreads out in rows beneath that. The rows are staggered. The second row has two cells, each one-half width to the side of the starting one. The third row has three cells, each one-half width to the sides of the row above, so that its center cell is directly under the original one. The fourth row has four cells, two of which are exactly underneath the cells of the second row. The fifth row has five cells, three of them directly underneath the third row’s cells. And so on. You know the pattern. It’s the one that pins in a plinko board take. Just trimmed down to a triangle. Make as many rows as you find interesting. You can always add more later.
In the top cell goes the number ‘1’. There’s also a ‘1’ in the leftmost cell of each row, and a ‘1’ in the rightmost cell of each row.
What of interior cells? The number for those we work out by looking to the row above. Take the cells to the immediate left and right of it. Add the values of those together. So for example the center cell in the third row will be ‘1’ plus ‘1’, commonly regarded as ‘2’. In the third row the leftmost cell is ‘1’; it always is. The next cell over will be ‘1’ plus ‘2’, from the row above. That’s ‘3’. The cell next to that will be ‘2’ plus ‘1’, a subtly different ‘3’. And the last cell in the row is ‘1’ because it always is. In the fourth row we get, starting from the left, ‘1’, ‘4’, ‘6’, ‘4’, and ‘1’. And so on.
It’s a neat little arithmetic project. It has useful application beyond the joy of making something neat. Many neat little arithmetic projects don’t have that. But the numbers in each row give us binomial coefficients, which we often want to know. That is, if we wanted to work out (a + b) to, say, the third power, we would know what it looks like from looking at the fourth row of Yanghui’s Triangle. It will be $1\cdot a^4 + 4\cdot a^3 \cdot b^1 + 6\cdot a^2\cdot b^2 + 4\cdot a^1\cdot b^3 + 1\cdot b^4$. This turns up in polynomials all the time.
Look at diagonals. By diagonal here I mean a line parallel to the line of ‘1’s. Left side or right side; it doesn’t matter. Yang Hui’s triangle is bilaterally symmetric around its center. The first diagonal under the edges is a bit boring but familiar enough: 1-2-3-4-5-6-7-et cetera. The second diagonal is more curious: 1-3-6-10-15-21-28 and so on. You’ve seen those numbers before. They’re called the triangular numbers. They’re the number of dots you need to make a uniformly spaced, staggered-row triangle. Doodle a bit and you’ll see. Or play with coins or pool balls.
The third diagonal looks more arbitrary yet: 1-4-10-20-35-56-84 and on. But these are something too. They’re the tetrahedronal numbers. They’re the number of things you need to make a tetrahedron. Try it out with a couple of balls. Oranges if you’re bored at the grocer’s. Four, ten, twenty, these make a nice stack. The fourth diagonal is a bunch of numbers I never paid attention to before. 1-5-15-35-70-126-210 and so on. This is — well. We just did tetrahedrons, the triangular arrangement of three-dimensional balls. Before that we did triangles, the triangular arrangement of two-dimensional discs. Do you want to put in a guess what these “pentatope numbers” are about? Sure, but you hardly need to. If we’ve got a bunch of four-dimensional hyperspheres and want to stack them in a neat triangular pile we need one, or five, or fifteen, or so on to make the pile come out neat. You can guess what might be in the fifth diagonal. I don’t want to think too hard about making triangular heaps of five-dimensional hyperspheres.
There’s more stuff lurking in here, waiting to be decoded. Add the numbers of, say, row four up and you get two raised to the third power. Add the numbers of row ten up and you get two raised to the ninth power. You see the pattern. Add everything in, say, the top five rows together and you get the fifth Mersenne number, two raised to the fifth power (32) minus one (31, when we’re done). Add everything in the top ten rows together and you get the tenth Mersenne number, two raised to the tenth power (1024) minus one (1023).
Or add together things on “shallow diagonals”. Start from a ‘1’ on the outer edge. I’m going to suppose you started on the left edge, but remember symmetry; it’ll be fine if you go from the right instead. Add to that ‘1’ the number you get by moving one cell to the right and going up-and-right. And then again, go one cell to the right and then one cell up-and-right. And again and again, until you run out of cells. You get the Fibonacci sequence, 1-1-2-3-5-8-13-21-and so on.
We can even make an astounding picture from this. Take the cells of Yang Hui’s triangle. Color them in. One shade if the cell has an odd number, another if the cell has an even number. It will create a pattern we know as the Sierpiński Triangle. (Wacław Sierpiński is proving to be the surprise special guest star in many of this A To Z sequence’s essays.) That’s the fractal of a triangle subdivided into four triangles with the center one knocked out, and the remaining triangles them subdivided into four triangles with the center knocked out, and on and on.
By now I imagine even my most skeptical readers agree this is an interesting, useful mathematical construct. Also that they’re wondering why I haven’t said the name “Blaise Pascal”. The Western mathematical tradition knows of this from Pascal’s work, particularly his 1653 Traité du triangle arithmétique. But mathematicians like to say their work is universal, and independent of the mere human beings who find it. Constructions like this triangle give support to this. Yang lived in China, in the 12th century. I imagine it possible Pascal had hard of his work or been influenced by it, by some chain, but I know of no evidence that he did.
And even if he had, there are other apparently independent inventions. The Avanti Indian astronomer-mathematician-astrologer Varāhamihira described the addition rule which makes the triangle work in commentaries written around the year 500. Omar Khayyám, who keeps appearing in the history of science and mathematics, wrote about the triangle in his 1070 Treatise on Demonstration of Problems of Algebra. Again so far as I am aware there’s not a direct link between any of these discoveries. They are things different people in different traditions found because the tools — arithmetic and aesthetically-pleasing orders of things — were ready for them.
Yang Hui wrote about his triangle in the 1261 book Xiangjie Jiuzhang Suanfa. In it he credits the use of the triangle (for finding roots) as invented around 1100 by mathematician Jia Xian. This reminds us that it is not merely mathematical discoveries that are found by many peoples at many times and places. So is Boyer’s Law, discovered by Hubert Kennedy.
## Author: Joseph Nebus
I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.
## 10 thoughts on “The End 2016 Mathematics A To Z: Yang Hui’s Triangle”
1. This is first time that I have read an article about Pascal triangle without a picture of it in front of me and could still imagine it in my mind. :)
Like
1. Thank you; I’m glad you like it. I did spend a good bit of time before writing the essay thinking about why it is a triangle that we use for this figure, and that helped me think about how things are organized and why. (The one thing I didn’t get into was identifying the top row, the single cell, as row zero. Computers may index things starting from zero and there may be fair reasons to do it, but that is always going to be a weird choice for humans.)
Liked by 1 person
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# Basic Math & Pre-Algebra Cheat Sheet
### Groups of Numbers
• Natural or counting numbers: 1, 2, 3, 4,...
• Whole numbers: 0, 1, 2, 3,...
• Integers: ...-3, -2, -1, 0, 1, 2, 3,
• Rational numbers: Integers and fractions.
• Irrational numbers: Cannot be written as fractions: or π.
• Prime numbers: Divisible only by 1 and itself: 2, 3, 5, 7, 11, 13, . . . . (0 and 1 are not prime or composite.)
Composite numbers: Divisible by more than just 1: 4, 6, 8, 9, 10, 12, . . . .
### Properties of Addition and Multiplication
• Closure: All answers fall into original set.
• Commutative: Order does not make any difference: a + b = b + a, ab = ba.
• Associative: Grouping does not make any difference: (a + b) + c = a + (b + c), (ab)c = a(bc).
• Identity: 0 for addition, 1 for multiplication.
• Inverse: Negative for addition, reciprocal for multiplication.
### Order of Operations
1. Work within parentheses ( ), brackets [ ], and braces { } from innermost and work outward.
2. Simplify exponents and roots working from left to right.
3. Do multiplication and division, whichever comes first left to right.
4. Do addition and subtraction, whichever comes first left to right.
### Rounding Off
1. Underline the place value to which you're rounding off.
2. Look to the immediate right (one place) of your underlined place value.
3. Identify the number (the one to the right).
If it is 5 or higher, round your underlined place value up 1 and change all the other numbers to its right to zeros. If less than 5, leave your underlined place value as it is and change all the other numbers to the right to zeros.
### Decimals
• To add or subtract decimals, simply line up the decimal points and then add or subtract as usual.
• To multiply decimals, just multiply as usual and then count the total number of digits above the line that are to the right of all decimal points. Place the decimal point in your answer so that there are the same number of digits to the right of the decimal point as there are above the line.
• To divide decimals, if the number you're dividing by has a decimal, move the decimal to the right as many places as possible and then move it under the division sign just as many places (add zeros if necessary). Move the decimal up to your answer.
### Fractions
To add or subtract fractions, you must have a common denominator.
• If two fractions have a common denominator (like fractions), you simply add or subtract the numerator and keep the same denominator. (For example, 1/5 + 2/5 = 3/5.)
• If two fractions do not have a common denominator (unlike fractions), find a lowest common denominator (LCD), change each of the fractions to equivalent fractions with the new denominator, and then add or subtract the numerators and keep the same denominator. (For example, 1/2 + 1/3 = 3/6 + 2/6 = 5/6
• When subtracting mixed numbers, you may have to "borrow" from the whole number. When you borrow 1 from the whole number, the 1 must be changed to a fraction.
• To multiply fractions, simply multiply the numerators and then multiply the denominators. (For example, 2/3 × 1/5 = 2/15.) Reduce to lowest terms if necessary.
• To divide fractions, invert the second fraction and then multiply. (For example, 1/5 ÷ 1/4 = 1/5 × 4/1 = 4/5.)
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Home >> CBSE XII >> Math >> Matrices
# Find $A^2 -5A + 6I$ , if $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$
Toolbox:
• Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B:
• $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
• If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
• $I=\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0& 0 & 1\end{bmatrix}$
Consider $A^2-5A+6I$.
$A^2=A\times A\Rightarrow \begin{bmatrix}2 & 0& 1\\2 & 1 & 3\\1& -1 & 0\end{bmatrix}\begin{bmatrix}2 & 0& 1\\2 & 1 & 3\\1& -1 & 0\end{bmatrix}$
$\Rightarrow A^2 = \begin{bmatrix}2\times 2+0\times 2+1\times 1 & 2\times 0+0\times 1+1\times (-1)& 2\times 1+0\times 3+1\times 01\\2\times 2+1\times 2+3\times 1 & 2\times 0+1\times 1+3\times (-1) & 2\times 1+1\times 3+3\times 0\\1\times 2+(-1)\times 2+0\times 1& 1\times 0+(-1)\times 1+0\times (-1) & 1\times 1+(-1)\times 3+0\times 0\end{bmatrix}$
$A^2 = \begin{bmatrix}5 & -1& 2\\9 & -2 & 5\\0& -1 & -2\end{bmatrix}$
$-5A = (-1)(5A) = (-1)(5)\begin{bmatrix}2 & 0& 1\\2 & 1 & 3\\1& -1 & 0\end{bmatrix}$
$-5A = \begin{bmatrix}-10 & 0& -5\\-10 & -5 & -15\\-5& 5 & 0\end{bmatrix}$
Since we know that $I=\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0& 0 & 1\end{bmatrix}$, $6I = 6\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0& 0 & 1\end{bmatrix}$
$6I = \begin{bmatrix}6 & 0& 0\\0 & 6 & 0\\0& 0 & 6\end{bmatrix}$
Therefore, $A^2-5A+6I=\begin{bmatrix}5 & -1& 2\\9 & -2 & 5\\0& -1 & -2\end{bmatrix}+\begin{bmatrix}-10 & 0& -5\\-10 & -5 & -15\\-5& 5 & 0\end{bmatrix}+\begin{bmatrix}6 & 0& 0\\0 & 6 & 0\\0& 0 & 6\end{bmatrix}$:
$\Rightarrow A^2-5A+6I = \begin{bmatrix}5-10+6 & -1+0+0& 2-5+0\\9-10+0 & -2-5+6 & 5-15+0\\0-5+0& -1+5+0 & -2+0+6\end{bmatrix}$
$\Rightarrow A^2-5A+6I = \begin{bmatrix}1 & -1& 3\\-1 & -1 & -10\\-5& 4 & 4\end{bmatrix}$
edited Feb 27, 2013
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## Pages
Showing posts with label factoring. Show all posts
Showing posts with label factoring. Show all posts
## Sunday, November 4, 2012
### General Guidelines for Factoring Polynomials
Now that we have learned techniques for factoring 4, 3, and 2-term polynomials, we are ready to practice by mixing up the problems. The challenge is to first identify the type of factoring problem then decide which method to apply. The basic guidelines for factoring follow:
1. Look to factor out any GCF.
2. Four-Term Polynomials - Factor by grouping.
3. Trinomials - Factor using the "guess and check" method.
4. Binomials - Use the the special products in this order:
Sum and Difference of Squares
Sum and Difference of Cubes
* If a binomial is both a difference of squares and cubes, then to obtain a more complete factorization, factor it as a difference of squares first.
* Not all polynomials factor. In this case, beginning algebra students may write, "does not factor - DNF."
Factor.
Tip: Make some note cards to aid in helping memorize the formulas for the special products. Look for factors to factor further - sometimes factoring once is not enough.
Factor.
Take some time to understand the difference between the last two solved problems. Notice that x^6 - y^6 is both a difference of squares and a difference of cubes at the same time. Here we chose to apply the difference of squares formula first. On the other hand, for x^6 + y^6 we chose to apply the sum of cubes formula first because it does not factor as a sum of squares.
Factor.
Video Examples on YouTube: Factor the following polynomials.
### Factoring Trinomials of the Form x^2 + bx + c
In this section we will factor trinomials - polynomials with three terms. Students find this difficult at first. However, with much practice factoring trinomials becomes routine. If a trinomial factors, then it will factor into the product of two binomials.
Factor:
Step 1: Factor the first term: x^2 = x*x.
Step 2: Factor the last term. Choose factors that add or subtract to obtain the middle term.
Step 3: Determine the signs by adding or subtracting the product of the inner and outer terms.
Step 4: Check by multiplying.
Rather than trying all possible combinations of the factors that make up the last term spend some time looking at the factors before starting step two. Look for combinations that will produce the middle term. Here is the thought process in choosing 3 and 4 in step two above:
"Can I add or subtract 1 and 12 to obtain 7?" – NO
"Can I add or subtract 2 and 6 to obtain 7?" – NO
"Can I add or subtract 3 and 4 to obtain 7?" – YES, because +3 + 4 = +7
Factor the trinomials.
This process used for factoring trinomials is sometimes called guess and check or trial and error. The biggest problem occurs when the signs are improperly chosen. With this in mind, you should take care to check your results by multiplying. Also, since multiplication is commutative order does not matter, in other words
If the trinomial has a GCF you should factor that out first. Also, you should factor in such a way as to ensure a resulting trinomial with a positive leading coefficient.
Factor the trinomials.
Take care to perform the check. Most of the problems that you will encounter factor nicely but be sure to watch out for something like this . The middle term works but the last term does not:
because the sign of the last term is incorrect
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# Lesson
Lesson. Vectors Review. . R. head. tail. Scalars vs Vectors. Scalars have magnitude only Distance, speed, time, mass Vectors have both magnitude and direction displacement, velocity, acceleration. . x. A. Direction of Vectors.
## Lesson
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### Presentation Transcript
1. Lesson Vectors Review
2. R head tail Scalars vs Vectors • Scalars have magnitude only • Distance, speed, time, mass • Vectors have both magnitude and direction • displacement, velocity, acceleration
3. x A Direction of Vectors • The direction of a vector is represented by the direction in which the ray points. • This is typically given by an angle.
4. If vector A represents a displacement of three miles to the north… A B Then vector B, which is twice as long, would represent a displacement of six miles to the north! Magnitude of Vectors • The magnitude of a vector is the size of whatever the vector represents. • The magnitude is represented by the length of the vector. • Symbolically, the magnitude is often represented as │A │
5. Equal Vectors • Equal vectors have the same length and direction, and represent the same quantity (such as force or velocity).
6. A -A Inverse Vectors • Inverse vectors have the same length, but opposite direction.
7. B A R Graphical Addition of Vectors • Vectors are added graphically together head-to-tail. • The sum is called the resultant. • The inverse of the sum is called the equilibrant A + B = R
8. Component Addition of Vectors • Resolve each vector into its x- and y-components. Ax = Acos Ay = Asin Bx = Bcos By = Bsin etc. • Add the x-components together to get Rx and the y-components to get Ry. • Use the Pythagorean Theorem to get the magnitude of the resultant. • Use the inverse tangent function to get the angle.
9. Sample problem: Add together the following graphically and by component, giving the magnitude and direction of the resultant and the equilibrant. • Vector A: 300 m @ 60o • Vector B: 450 m @ 100o • Vector C: 120 m @ -120o
10. Lesson Unit Vectors
11. a Consider Three Dimensions Polar Angle z Azimuthal Angle az q ay y f ax xy Projection x
12. Unit Vectors • Unit vectors are quantities that specify direction only. They have a magnitude of exactly one, and typically point in the x, y, or z directions.
13. Unit Vectors z k j i y x
14. Unit Vectors • Instead of using magnitudes and directions, vectors can be represented by their components combined with their unit vectors. • Example: displacement of 30 meters in the +x direction added to a displacement of 60 meters in the –y direction added to a displacement of 40 meters in the +z direction yields a displacement of:
15. Adding Vectors Using Unit Vectors • Simply add all the i components together, all the j components together, and all the k components together.
16. Sample problem: Consider two vectors, A = 3.00 i + 7.50 j and B = -5.20 i + 2.40 j. Calculate C where C = A + B.
17. Sample problem: You move 10 meters north and 6 meters east. You then climb a 3 meter platform, and move 1 meter west on the platform. What is your displacement vector? (Assume East is in the +x direction).
18. Suppose I need to convert unit vectors to a magnitude and direction? • Given the vector
19. Sample problem: You move 10 meters north and 6 meters east. You then climb a 3 meter platform, and move 1 meter west on the platform. How far are you from your starting point?
20. Lesson Position, Velocity, and Acceleration Vectors in Multiple Dimensions
21. x: position x: displacement v: velocity a: acceleration r: position r: displacement v: velocity a: acceleration In Unit Vector Notation • r = x i + y j + z k • r = x i + y j + z k • v = vxi + vyj + vzk • a = axi + ayj + azk 1 Dimension 2 or 3 Dimensions
22. Sample problem: The position of a particle is given byr = (80 + 2t)i – 40j - 5t2k. Derive the velocity and acceleration vectors for this particle. What does motion “look like”?
23. Sample problem: A position function has the form r = x i + y j with x = t3 – 6 and y = 5t - 3. a) Determine the velocity and acceleration functions. b) Determine the velocity and speed at 2 seconds.
24. Miscellaneous • Let’s look at some video analysis. • Let’s look at a documentary. • Homework questions?
25. Lesson Multi-Dimensional Motion with Constant (or Uniform) Acceleration
26. Sample Problem: A baseball outfielder throws a long ball. The components of the position are x = (30 t) m and y = (10 t – 4.9t2) m a) Write vector expressions for the ball’s position, velocity, and acceleration as functions of time. Use unit vector notation! b) Write vector expressions for the ball’s position, velocity, and acceleration at 2.0 seconds.
27. Sample problem: A particle undergoing constant acceleration changes from a velocity of 4i – 3j to a velocity of 5i + j in 4.0 seconds. What is the acceleration of the particle during this time period? What is its displacement during this time period?
28. g g g g g Trajectory of Projectile • This shows the parabolic trajectory of a projectile fired over level ground. • Acceleration points down at 9.8 m/s2 for the entire trajectory.
29. Trajectory of Projectile • The velocity can be resolved into components all along its path. Horizontal velocity remains constant; vertical velocity is accelerated. vx vx vy vy vx vy vx vy vx
30. y y x x t t Position graphs for 2-D projectiles. Assume projectile fired over level ground.
31. Velocity graphs for 2-D projectiles. Assume projectile fired over level ground. Vy Vx t t
32. Acceleration graphs for 2-D projectiles. Assume projectile fired over level ground. ay ax t t
33. Vo,y = Vo sin Vo,x = Vo cos Remember…To work projectile problems… • …resolve the initial velocity into components. Vo
34. Sample problem: A soccer player kicks a ball at 15 m/s at an angle of 35o above the horizontal over level ground. How far horizontally will the ball travel until it strikes the ground?
35. Sample problem: A cannon is fired at a 15o angle above the horizontal from the top of a 120 m high cliff. How long will it take the cannonball to strike the plane below the cliff? How far from the base of the cliff will it strike?
36. Lesson Monkey Gun Experiment – shooting on an angle
37. Lesson A day of derivations
38. Sample problem: derive the trajectory equation.
39. Sample problem: Derive the range equation for a projectile fired over level ground.
40. Sample problem: Show that maximum range is obtained for a firing angle of 45o.
41. Will the projectile always hit the target presuming it has enough range? The target will begin to fall as soon as the projectile leaves the gun.
42. Punt-Pass-Kick Pre-lab • Purpose: Using only a stopwatch, a football field, and a meter stick, determine the launch velocity of sports projectiles that you punt, pass, or kick. • Theory: Use horizontal (unaccelerated) motion to determine Vx, and vertical (accelerated) motion to determine Vy. Ignore air resistance. • Data: Prepare your lab book to collect xi, xf, yo, and Dt measurements for each sports projectile. Analyze the data fully for at least three trials. • Make sure you dress comfortably tomorrow!
43. Lesson Punt-pass-kick lab
44. Lesson Review of Uniform Circular Motion Radial and Tangential Acceleration
45. Uniform Circular Motion • Occurs when an object moves in a circle without changing speed. • Despite the constant speed, the object’s velocity vector is continually changing; therefore, the object must be accelerating. • The acceleration vector is pointed toward the center of the circle in which the object is moving, and is referred to as centripetal acceleration.
46. v v a a a a v v Vectors inUniform Circular Motion a = v2 / r
47. Sample Problem The Moon revolves around the Earth every 27.3 days. The radius of the orbit is 382,000,000 m. What is the magnitude and direction of the acceleration of the Moon relative to Earth?
48. Sample problem: Space Shuttle astronauts typically experience accelerations of 1.4 g during takeoff. What is the rotation rate, in rps, required to give an astronaut a centripetal acceleration equal to this in a simulator moving in a 10.0 m circle?
49. Tangential acceleration • Sometimes the speed of an object in circular motion is not constant (in other words, it’s not uniform circular motion). • An acceleration component may be tangent to the path, aligned with the velocity. This is called tangential acceleration. It causes speeding up or slowing down. • The centripetal acceleration component causes the object to continue to turn as the tangential component causes the speed to change. The centripetal component is sometimes called the radial acceleration, since it lies along the radius.
More Related
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## What is the standard procedure for dividing?
All you do is divide the first number of your given dividend by the first digit of the divisor. Then write the result of that division in the space of sed quotient. Now you want to multiply the digit of the quotient by the divisor, finally write the result beneath the dividend and subtract it.
## How do you estimate using compatible numbers?
Compatible numbers are pairs of numbers that are easy to add, subtract, multiply, or divide mentally. When using estimation to approximate a calculation, replace actual numbers with compatible numbers. The numbers 500 and 300 are compatible for addition, since the sum of 800 can be easily calculated mentally.
How do you estimate quotients?
To estimate the quotient, we first round off the divisor and the dividend to the nearest tens, hundreds, or thousands and then divide the rounded numbers. In a division sum, when the divisor is made up of 2 digits or more than 2 digits, it helps if we first estimate the quotient and then try to find the actual number.
### How do you divide multi digit numbers?
Divide the first number of the dividend (or the two first numbers if the previous step took another digit) by the first digit of the divisor. Write the result of this division in the space of the quotient. Multiply the digit of the quotient by the divisor, write the result beneath the dividend and subtract it.
### What do you do with the divisor when you estimate the quotient?
When to teach two digit divisor in math?
This is a complete lesson with examples and exercises about two-digit divisor in long division, meant for initial teaching in 5th grade. The first exercises have grids to complete the division, and space for students to write the multiplication table of the divisor in the margin.
## Do you write the divisor before you divide?
Then there are conversion problems between inches/feet and ounces/pounds, because those are solved with division. Often, it is helpful to write the multiplication table of the divisor before you divide. Example 1. The division is by 16. Here is the multiplication table of 16:
## How do you do long division with two digits?
Multiply the digit of the quotient by the divisor, write the result beneath the dividend and subtract it. If you cannot, because the dividend is smaller, you will have to choose a smaller number in the quotient until it can subtract.
Which is an example of a two digit divisor?
Example 2. We are dividing by 32. Here is the multiplication table of 32: 32 goes into 47 once. 32 goes into 150 four times. 32 goes into 224 seven times. Notice there is a remainder. 1. Divide. First write a multiplication table for the divisor. Check each answer by multiplying. 2. Divide.
What is the standard procedure for dividing? All you do is divide the first number of your given dividend by the first digit of the divisor. Then write the result of that division in the space of sed quotient. Now you want to multiply the digit of the quotient by the divisor, finally write the…
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## What is the Cartesian Coordinate System?
The Cartesian coordinate system is a system of mapping points in space to sets of numerical coordinates. It is named after the French mathematician and philosopher René.Descartes, who developed it in the 17th century..
The Cartesian coordinate system is widely used in mathematics and science as a way to represent and analyze geometric shapes, functions, and physical phenomena.
In the Cartesian coordinate system, a point in space is represented by an ordered pair of numbers (x, y). The x-coordinate indicates the point’s distance from the y-axis, and the y-coordinate indicates the point’s distance from the x-axis. The point where the x-axis and y-axis intersect is called the origin and is assigned the coordinates (0, 0).
To find the coordinates of a point in the Cartesian plane, you can use the following steps:
Draw a horizontal line (the x-axis) and a vertical line (the y-axis) intersecting at the origin.
Locate the point on the plane and draw a line segment from the point to the y-axis. The length of this line segment is the point’s x-coordinate.
Draw a line segment from the point to the x-axis. The length of this line segment is the point’s y-coordinate.
READ MORE: Applications Of Angular Velocity
The Cartesian coordinate system is also useful for graphing functions. A function is a rule that assigns to each element in a set (called the domain) exactly one element in another set (called the range). In the Cartesian plane, the domain of a function is usually represented on the x-axis, and the range is represented on the y-axis. To graph a function, you can plot points on the plane corresponding to different values of the function’s inputs (x-values) and outputs (y-values). Connecting these points with a smooth curve gives you the graph of the function.
## Slope
The slope of a line in the Cartesian plane is a measure of how steeply the line rises or falls. It is calculated by taking the difference between the y-coordinates of two points on the line and dividing it by the difference between the x-coordinates of those same points. The y-intercept of a line is the point where the line crosses the y-axis. It is the y-coordinate of the point at which the line intersects the y-axis when the x-coordinate is 0.
READ MORE: Application and Uses of the Bomb Calorimeter
## Applications
The Cartesian coordinate system has many practical applications. In geometry, it is used to represent and analyze shapes and figures. In physics and engineering, it is used to model the motion of objects and to solve problems involving forces, velocities, and acceleration. The Cartesian coordinate system is also used in computer graphics, where it is used to represent the position and orientation of objects in 3D space.
In addition to these basic concepts, there are also more advanced topics in the Cartesian coordinate system. For example, complex numbers are a type of number that can be represented in the complex plane, which is a variant of the Cartesian plane. Polar coordinates are another way of representing points in the Cartesian plane, using the distance from the origin and the angle from a fixed reference direction.
## FAQs:
### How do I find the distance between two points in the Cartesian plane?
To find the distance between two points (x1, y1) and (x2, y2) in the Cartesian plane, you can use the following formula:
d = sqrt((x2-x1)^2 + (y2-y1)^2)
READ MORE: Laboratory and Industrial Uses of Multimeter and Voltmeter
This formula uses the Pythagorean theorem to calculate the distance between the two points.
### Can the Cartesian coordinate system be extended to three dimensions?
Yes, the Cartesian coordinate system can be extended to three dimensions by adding a third coordinate, called the z-coordinate. This allows you to represent points in 3D space and graph functions with three variables.
### How do I convert from polar coordinates to Cartesian coordinates?
To convert from polar coordinates (r, theta) to Cartesian coordinates (x, y), you can use the following formulas:
x = rcos(theta)
y = rsin(theta)
Theta is the angle from the positive x-axis, and r is the distance from the origin.
### Is the Cartesian coordinate system the only way to represent points in space?
No, there are other coordinate systems that can be used to represent points in space. For example, the polar coordinate system represents points in space using the distance from the origin and the angle from a fixed reference direction, while the spherical coordinate system represents points using the distance from the origin, the polar angle, and the azimuthal angle.
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# 2007 AIME I Problems/Problem 12
## Problem
In isosceles triangle $\triangle ABC$, $A$ is located at the origin and $B$ is located at (20,0). Point $C$ is in the first quadrant with $\displaystyle AC = BC$ and angle $BAC = 75^{\circ}$. If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$-axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$, where $\displaysytle p,q,r,s$ (Error compiling LaTeX. ! Undefined control sequence.) are integers. Find $\frac{p-q+r-s}2$.
## Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Call the vertices of the new triangle $AB'C'$ ($A$, the origin, is a vertex of both triangles). $B'C'$ and $AB$ intersect at a single point, $D$. $BC$ intersect at two points; the one with the higher y-coordinate will be $E$, and the other $F$. The intersection of the two triangles is a quadrilateral $ADEF$. Notice that we can find this area by subtracting $[\triangle ADB'] - [\triangle EFB']$.
Since $\displaystyle \angle B'AC'$ and $\displaystyle \angle BAC$ both have measures $75^{\circ}$, both of their complements are $15^{\circ}$, and $\angle DAC' = 90 - 2(15) = 60^{\circ}$. We know that $C'B'A = 75^{\circ}$, and since the angles of a triangle add up to $180^{\circ}$, we find that $ADB' = 180 - 60 - 75 = 45^{\circ}$.
So $ADB'$ is a $45 - 60 - 75 \triangle$. It can be solved by drawing an altitude splitting the $75^{\circ}$ angle into $30^{\circ}$ and $45^{\circ}$ angles – this forms a $\displaystyle 30-60-90$ right triangle and a $\displaystyle 45-45-90$ isosceles right triangle. Since we know that $DB' = 20$, the base of the $\displaystyle 30-60-90$ triangle is $10$, the height is $10\sqrt{3}$, and the base of the $\displaystyle 45-45-90$ is $10\sqrt{3}$. Thus, the total area of $[\triangle ADB'] = \frac{1}{2}(10\sqrt{3})(10\sqrt{3} + 10) = 150 + 50\sqrt{3}$.
Now, we need to find $[\triangle EFB']$, which is a $\displaystyle 15-75-90$ right triangle. We can find its base by subtracting $AF$ from $20$. $\triangle AFB$ is also a $\displaystyle 15-75-90$ triangle, so we find that $AF = 20\sin 75 = 20 \sin (30 + 45) = 20\frac{\sqrt{2} + \sqrt{6}}4 = 5\sqrt{2} + 5\sqrt{6}$. $FB' = 20 - AF = 20 - 5\sqrt{2} - 5\sqrt{6}$.
To solve $[\triangle EFB']$, note that $[\triangle EFB'] = \frac{1}{2} FB' \cdot EF = \frac{1}{2} FB' \cdot (\tan 75 FB')$. Through algebra, we can calculate $(FB')^2 \cdot \tan 75$:
$\frac{1}{2}\tan 75 \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2 == See also == {{AIME box|year=2007|n=I|num-b=11|num-a=13}}$ (Error compiling LaTeX. ! Missing \$ inserted.)
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# Created by the a.m.r.a.h.y group. first topic: number problems Of all the word problems, the number problems are the easiest to translate into equations.
## Presentation on theme: "Created by the a.m.r.a.h.y group. first topic: number problems Of all the word problems, the number problems are the easiest to translate into equations."— Presentation transcript:
Created by the a.m.r.a.h.y group
first topic: number problems Of all the word problems, the number problems are the easiest to translate into equations since the relationships among the numbers are directly stated in the problem.
ILLUSTRATIVE EXAMPLE One number is two more than thrice another. Their sum is 30. Find these numbers. Solution: READ: Reading the problem thoroughly, we know two things about the numbers. a a) their sizes : one of them is two more than thrice the other and b b) their sum: the sum is 30. REPRESENT: If we represent the numbers using the first sentence we have: l let x= first number t then: 3x+2= other number
R RELATE : The relationship between the number x and 3x+2 and the other number,30, gives as an equation. E EQUATE: X X+3X+2= 30 S SOLVE: In solving the equation we have: x+3x+2=30 4x+2=30 4x=28 x=7 Therefore, the first number is 7 and the other number is 3x+2=3(7)+2=23 Answer :7 &23 Prove: a)T heir sum is 30: 7+23=30 b) 2 3 is two more than thrice 7: 3(7)+2=21+2=23
Exercises 1.) The sum of two numbers is 29 and their difference is 6. find the numbers. 2.) the smaller of two numbers is thrice the larger. The larger number is eight more than the smaller one. Find the numbers.
Solution: 1.) x-(29-x)=5 x-29+x=5 2x-29=5 2x=34 x=17 Other number is 29-x=29-17=12 2.) x=3x+8 3x+8=x 2x=-8 x=-4 3x=-12 The answers are: -4 & -12
SECOND TOPIC: ODD, EVEN, AND CONSECUTIVE INTEGERS The word consecutive means following in order without interruption. Even numbers are whole numbers divisible by 2 while whole numbers which are not divisible by 2 are odd numbers. Consecutive even integers are even numbers in uninterrupted order which is the same as when you count by twos such as 2,4,6,8,&10. Consecutive odd integers are odd numbers in uninterrupted order, such as 7,9,11,13,& 15.
Illustrative example: 1.) The sum of three consecutive integers is 90. Find the integers. solution: let x= first number then x+1= next consecutive integer. and x+2= third consecutive integer. their sum is 90 manipulating the equation we have: x+(x+1)+(x+2)= 90 x+x+1+x+2=90
3x+3=90 3x+3=90 3x= 87 3x= 87 x=29 1 st integer x=29 1 st integer x+x1= 2 nd integer x+x1= 2 nd integer x+2= 3 rd integer x+2= 3 rd integer answer: the consecutive integers are 29,30&31 answer: the consecutive integers are 29,30&31 Prove: 29+30+31=90
exercises: 1.F ind two consecutive integer such that if we triple the first and double the second, the sum is 77. 2. the sides of a triangle are consecutive integers. If the perimeter of this triangle is 192 cm, find the length of each sides.
solution: 1.3 (x)+2(x+1)=77 3x+2x+2=77 5x+2= 75 x=15 1st integer x+1= 16 2nd integer Answer: the consecutive integers are 15 & 16 Proof: triple 15: 45 double 16:32 total:77 2. x+x+1+x+2=192 3x+3= 192 3x= 189 x=63 cm x+1= 64 cm x+2= 65 cm answer: the lengths of the sides of a triangle are 63,64,65 cm.
The delineation of a sphere on a flat plane is similar to painting. For just as the painters seek to imitate objects exactly...geometricians and astronomers delineate on a flat plane solid objects such as octahedrons and cubes and all spherical bodies, like the stars, the heavens, and the earth.
If all the arts aspire to the condition of music, all the sciences aspire to the condition of mathematics.
Let no one ignorant of mathematics enter here.
As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.
No, it is a very interesting number, it is the smallest number expressible as a sum of two cubes in two different ways.
There is no "royal road" to geometry.
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# ISEE Lower Level Math : How to add
## Example Questions
← Previous 1 3 4 5 6
### Example Question #1 : How To Add
Thomas rides the bus to school each morning. If the bus picks Thomas up at a quarter before eight, and it takes half an hour to get to school, what time will Thomas arrive at school?
a quarter after 8
8 o'clock
half past 9
half past 8
a quarter before 9
a quarter after 8
Explanation:
The phrase 'a quarter before' indicates that the time is 15 minutes before the hour.
The bus picks Thomas up at
"a quarter before 8"
which translates to 15 minutes before 8 o'clock, or
7:45 AM.
Because it takes half an hour, or
30 minutes
to get to school, Thomas will arrive at school 30 minutes after 7:45 AM.
7:45 AM + 30 minutes =
8:15 AM
'a quarter after' indicates that the time is 15 minutes after the hour, so 8:15 can be read as
a quarter after 8.
Note that there are 60 minutes in one hour.
### Example Question #1 : Addition And Subtraction With Money
Approximately how much will John need to pay if a pair of shoes cost $42.64, a pair of pants costs$37.99, and a shirt costs $21.34. Possible Answers:$100
$120$80
$110$90
$100 Explanation: The key word, "approximately," indicates that we must round each of these values. Let's round to the nearest tens place because each of our answer choices is rounded to the tens place. Shoes cost$42.64.
$42.64 rounded (down) to the nearest tens place is$40.
Pants cost $37.99.$37.99 rounded (up) to the nearest tens place is $40. A shirt costs$21.34.
$21.34 rounded (down) to the nearest tens place is$20.
Let's add up our estimated prices.
$40 +$40 + $20 =$100.
### Example Question #2 : How To Add
Evaluate:
Explanation:
Add the first two numbers as follows:
Now add this sum to the third number as follows:
### Example Question #3 : How To Add
Evaluate:
Explanation:
Add the first two numbers as follows:
Now add this sum to the third number as follows:
### Example Question #4 : How To Add
Add. Show as a mixed fraction.
Explanation:
When adding fractions, you must first make sure that the denominators are equal. If they are not equal, then you must make them equal by finding the LCD (lowest common denominator). In this case the LCD is 15. Now set up the new equation:
Then add the numerators to get:
Show as a mixed fraction:
### Example Question #6 : Addition And Subtraction With Money
Approximately how much will Sophie need to buy a shirt at $12.99, a loaf of bread at$3.99, and a television at $43.76? Possible Answers:$60
$100$80
$61 Correct answer:$61
Explanation:
Since we're looking for an approximate value, we need to round our numbers up. Adding the three items gives a total of $60.74, so at least$61 is needed - thus $61 is the correct answer. ### Example Question #1 : Addition And Subtraction With Money Laura is going bowling with her friends this weekend and she is using some of the money that she has saved up from allowance and her birthday. She will have to pay a flat fee of$8 to rent her shoes and $10.50 per game she bowls. If she has planned to bowl 4 games and she is only taking$50 with her to the bowling alley will she have enough money? If yes, how much money will she have when she leaves the bowling alley?
Yes,
No
Yes,
Yes, None.
Yes, None.
Explanation:
The flat fee for the shoes will only be paid once since the shoes will be used for all four games. The cost per game will be paid each time one game finishes and a new one is started.
To calculate her cost add the fee for the shoes and the fee for each game played. That would be $8 (shoes) +$10.50 (game 1) + $10.50 (game 2) +$10.50 (game 3) + $10.50 (game 4) =$50 (total cost).
Laura had \$50 with her so there was enough money and nothing left over when she left the bowling alley.
### Example Question #8 : Addition And Subtraction With Money
Danielle buys four items costing , , , and . What is the estimated total of Danielle's items?
Between and
Between and
Between and
Between and
Between and
Explanation:
To solve this problem, one should round either to the tenths place, or to the nearest dollar. can be rounded to , to , to , and to . When one adds up the estimated costs, it yields about .
### Example Question #11 : Money And Time
Sam buys four items, costing , , and . What is the estimated cost of Sam's items?
Between and dollars
Between and dollars
Between and dollars
Between and dollars
Between and dollars
Explanation:
In this problem, the costs of the items can be rounded to either the nearest dollar, or the nearest ten cents.
and can be rounded to and , respectively.
can be rounded to , and can be rounded to .
When the new, estimated costs of the items are added, the sum is , which is between and dollars.
### Example Question #5 : How To Add
Which of the following is a valid approximation of the problem
?
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# GRE Subject Test: Math : Product Rule
## Example Questions
### Example Question #1 : Product Rule
Derive:
Explanation:
This problem requires the product rule. The derivative using the product rule is as follows:
Let and .
Their derivatives are:
and
Substitute the functions into the product rule formula and simplify.
### Example Question #2 : Product Rule
Given and , find .
Explanation:
Recall the chain rule from calculus:
So we will want to begin by finding the first derivative of each of our functions:
Next, use the chain rule formula:
Expand everything to get
### Example Question #1 : Product Rule
Find given and .
Explanation:
Recall the product rule for differentiaion.
So we need to find the first derivative of each of our functions:
Recall that is a strange one:
Next, use the formula from up above.
Expand and simplify.
Rewrite in standard form and factor out an .
### Example Question #2 : Product Rule
What is the derivative of: ?
Explanation:
Step 1: Define and :
Step 2: Find and .
Step 3: Define Product Rule:
Product Rule=.
Step 4: Substitute the functions for their places in the product rule formula
Step 5: Expand:
Step 6: Combine like terms:
### Example Question #3 : Product Rule
Find the derivative, with respect to , of the following equation:
Explanation:
1) Starting Equation:
2) Simplifying:
3) Take the derivative, using the power rule.
Notes:
1) Easiest way to take the derivative is to simplify the equation first. In doing so, you should see that this is NOT an application for the chain rule. Although two variables are multiplied together, they are the same variable. The chain rule will give you the wrong answer.
2) Exponent Math... multiplied factors means you should add the exponents
3) Standard Power Rule. Bring the exponent down, multiplying it into the coefficient. Subtract 1 from the exponent. Constants go to 0.
### Example Question #6 : Product Rule
Find the derivative of .
None of the Above
None of the Above
Explanation:
Step 1: We need to define the product rule. The product rule says is defined as the derivative of f(x) multiplied by g(x). In the question, the first parentheses is f(x) and the second parentheses is g(x).
Step 2: We will first calculate and . To find the derivative of any term, we do one the following rules:
1) All terms with exponents (positive and negative) of the original equation are dropped down and multiplied by the coefficient of that term that you are working on. The exponent that gets written after taking the derivative is 1 less than (can also be thought of as (x-1, where x is the exponent that was dropped).
2) The derivative of a term ax, , is just the value , which is the coefficient of the term.
3) The derivative of any constant term, that is any term that does not have a variable next to it, is always .
Step 3: We will take derivative of first.
. Let us take the derivative of each and every term and then add everything back together. We denote (') as derivative.
. We are using rule 1 here (listed above). The two from the exponent dropped down and was multiplied by the coefficient of that term. The exponent is 1 less than the exponent that was dropped down, which is why you see in the exponent.
. We use rule 2 that was listed above. The derivative of this term is just the coefficient of that term, in this case, 3.
. We use rule 3. Since the derivative is , we won't write it in the final equation for the derivative of .
Let's put everything together:
Step 4: We will take derivative of g(x).
So, .
Step 5: Now that we have found the derivatives, let's substitute all the equations into the formula for product rule.
Step 6: Let's find .
In the equation above, . We will need to distribute and expand this multiplication.
When we expand, we get . Let's simplify that expansion.
We will get: .
Step 7: Let's find , which is defined as .
We will expand and simplify.
When we expand, we get: .
If we simplify, we get .
Step 8: Add the two products together and simplify.
If we add and simplify, we get:
. This is the final answer to the expansion of the product rule.
### Example Question #4 : Product Rule
Find derivative of:
Explanation:
Step 1: Define the two functions...
Step 2: Find the derivative of each function:
Step 3: Define the Product Rule Formula...
Step 4: Plug in the functions:
Step 5: Expand and Simplify:
The derivative of the product of and is .
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MAT 156 LAB 2Topic 1: DerivativesWe first define the function NiMvLSUiZkc2IyUieEcqJEYnIiIk and the expression f3=NiMqJEkieEc2IiIiIw==. Maple treats functions and expressions differently.f:= x->x^3; f3 := x^2;In these examples f is a function and f3 is an expression. To find the derivative of f we can use either of the following methodsD(f); diff(f(x),x); Notice that the first method produced a function and the second just an expression. To find the derivative of f3 we must use diff. There is also a way to find second derivative. In these examples I am assignming names to the derivative and the second derivative.df3 := diff(f3,x); d2f3:= diff(f3,x\$2); Now we can come back to the function g(x)=NiMsKCokSSJ4RzYiIiIlIiIiKiYiIidGKCokRiUiIiRGKCEiIkYoRig= we studied in Lab 1 and find its minimum. Recall that the points where g'(x)=0 are called critical pointsand we search among these points for (local) maxima and (local) minima. Compute the derivative of g(x) and solve the equation g'(x)=0.Now find the minimum value. It defers from what we saw in the graph only at the third and fourth decimal digit!We also see that the graph of g(x) changes concavity. The points where the graphs changes from concave upwards to concavedownwards (or vice versa) are called inflection points. They are found among the points where g''(x)=0. Find the inflection points for g(x).D(D(g));plot(g(x), x=-1..4);Example: Graph the function h(x)= NiMsKCokSSJ4RzYiIiIjIiIiKiYiI0NGKEYlRihGKCIjOkYo. How do you know that the initial graph is not good enough?Write a complete statement.This cannot be the whole story, since h(x) is a quadratic polynomial, so its graph should be a parabola. To find the peak of the parabola (called vertex), we notice that it is the minimum and we identify it by finding the critical point h'(x)=0. Plot the parabola to show its vertex.Topic 2: Velocity and Distance travelled.In this rest of this lab we will explore a velocity function and the distance we travelled. NiMtJSQ8LD5HNiktJSQ8fGdyPkc2JC0lJVRpbWVHNiMlJHNlY0ctJSlWZWxvY2l0eUc2IyomJSNmdEciIiJGLCEiIi1GJzYkIiIhIiM/LUYnNiRGMiIjSS1GJzYkIiIjIiNRLUYnNiQiIiQiI1ctRic2JCIiJSIjWy1GJzYkIiImIiNdNiMtJSdSVEFCTEVHNiUiKCEpXCpHLSUnTUFUUklYRzYjNyk3JC0lJVRpbWVHNiMlJHNlY0ctJSlWZWxvY2l0eUc2IyomJSNmdEciIiJGLyEiIjckIiIhIiM/NyRGNSIjSTckIiIjIiNRNyQiIiQiI1c3JCIiJSIjWzckIiImIiNdJSdNYXRyaXhHIt ends up that the table has been created using the following complicated function.f := t->-22108.111351860603123*t^12+267716.97873114933992*t^7-28550.588718383953421*t^4-.69089471139707261503e-1*t^18-593.25933891402415755*t^2+88214.018798090967877*t^5-182944.24328666353298*t^6+5693.6391073166057753*t^3-285572.24388626189800*t^8+227046.45357947547549*t^9-136597.49693854806554*t^10+62764.645490312705181*t^11+1.5172197744989322043*t^17+5957.3207937003999701*t^13-1217.1081971634308471*t^14+185.14666942246783552*t^15-20.307901075143421783*t^16+.14466922546287623311e-2*t^19+33.706872407075593812*t+20;We do not care about the formula above.Topic 3: Plotting left-hand-sums.Maple allows us to see graphically left-hand sums using the command leftbox(function, variable= lower limit .. upper limit, number of intervals), as seen below. Notice that we must use in the function the same variable as in variable. First we introduce the package for students:with(student): Digits:=20:leftbox(f(t), t=0..5, 5);The command Digits:=20 asks Maple to work with 20 decimal digits. This is a choice that I found to be good for this lab. It does not have to be the case always.For future use we will give names to our plots. This can be done the same way as any other expression. However, we do not want to seethe result immediately, so we will use colon, rather than semicolon.lhs5:=leftbox(f(t), t=0..5, 5):It ends up that we do not get a good approximation of the distance we travel. To improve our results we use data for our velocity every half second, rather than every second. This gave us the table: NiMtJSQ8LD5HNigtJSQ8fGdyPkc2JCIiISIjPy1GJzYkLSUmRmxvYXRHNiQiIiYhIiIiI0UtRic2JCIiIiIjSS1GJzYkLUYuNiQiIzpGMSIjTi1GJzYkIiIjIiNRLUYnNiQtRi42JCIjREYxIiNVNiMtJSdSVEFCTEVHNiUiKSc+QjUiLSUnTUFUUklYRzYjNyg3JCIiISIjPzckJCIiJiEiIiIjRTckIiIiIiNJNyQkIiM6RjEiI043JCIiIyIjUTckJCIjREYxIiNVJSdNYXRyaXhH<<3 | 44> , <3.5 | 46> , <4 | 48> , <4.5 | 49>, <5| 50>>;NiMtJSdSVEFCTEVHNiUiKSl5I1s2LSUnTUFUUklYRzYjNyc3JCIiJCIjVzckJCIjTiEiIiIjWTckIiIlIiNbNyQkIiNYRjEiI1w3JCIiJiIjXSUnTWF0cml4Rw==Write a command that produces graphically the left-hand-sum with 10 intervals.Write a command that gives a name to the previous plot. Create plots (and names) for the left-hand sum with 20, 40, 80 and 160 subintervals. You should be able to see how closer they fit under the curve and that the area under the curve seems to be the limit of the left-hand sums as the number of subintervals increases to infinity.Topic 4: Plotting right-hand sums.In Maple there is a command also for seeing graphically the right-hand sums. We use rightbox( function, variable= lower limit .. upper limit, number of subintervals). See the example:rightbox(f(t), t=0..5, 5);We can also name the plot. An appropriate name would be rhs5.rhs5:=rightbox(f(t), t=0..5, 5):Write commands to see graphically the right-hand sums with 10, 20, 40 , 80, 160 subintervals. Write commands that name these plots.Write your comments on what you notice for the right-hand sums, e.g. are they underestimates, or overestimates and why? Do they increase or decrease when the number of subintervals increase? Where do they converge? Topic 5: Comparing left-hand sums with right-hand sums.Now we would like to see the left-hand sums and the right-hand sums simultaneously. To do this we first load the plots package:with(plots):Warning, the name changecoords has been redefined Let us first display the left-hand sum and the right-hand sum with 5 subintervals. The command is display(name_1, name_2);display(lhs5, rhs5);Write commands that show you at the same time the left-hand sums and the right-hand sums with the same number of subintervals for NiMlIm5H=10, 20, 40, 80. What do you notice about the error in the approximation?Now we will work with the function g(x)=1/x and the integral NiMtJSRpbnRHNiQqJiIiIkYnJSJ4RyEiIi9GKDtGJyIiIw== .Graph the left-hand sums and right-hand sums with n=2, 4, 8, 16, 32, 64 subintervals.
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How Do You Calculate the Mark up of a Product?
# How Do You Calculate the Mark up of a Product?
To find the markup of goods, the formula to use is selling price of product = product cost + markup. An example of this type of problem is when a retailer buys an item at \$6 (cost) and sells it at \$7.20. Using the formula and this information, students can calculate the markup and the markup percent rate.
To make a profit, store owners use a selling price that is based on the original cost of an item and a markup percent rate. These types of problems entail finding the percent change relative to the original product cost of items.
1. Use the formula to calculate the markup
2. Substitute the values for the selling price and cost of product into the formula selling price of product = product cost + markup to get \$7.20 = \$6 + markup. Subtract \$6 from both sides to find the markup as \$1.20.
3. Find the markup percent rate
4. Calculate the markup percent rate by knowing that the markup of \$1.20 is a percent Y of the original product cost, or 1.20 = Y x 6. Find Y by dividing both sides of the equation by 6 to get the answer as 0.2. Convert 0.2 to a percentage by multiplying it by 100 percent to arrive at the answer of 20 percent as the markup percent rate.
If the store owner wants a higher selling price than \$7.20, then he needs to use a markup percent rate that is higher than 20 percent for the given cost of \$6.
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# How do you solve 8(2f - 2) = 7(3f + 2) ?
Apr 20, 2017
The final answer is $- 6 = f$.
#### Explanation:
$8 \left(2 f - 2\right) = 7 \left(3 f + 2\right)$
$\left(8 \cdot 2 f\right) + \left(8 \cdot - 2\right) = \left(7 \cdot 3 f\right) + \left(7 \cdot 2\right) \to$ Expand both sides by using the Distributive Property.
$16 f - 16 = 21 f + 14$
$\cancel{\textcolor{red}{16 f}} - 16 \cancel{\textcolor{red}{- 16 f}} = \cancel{21 f \textcolor{red}{- 16 f}} 5 f + 14 \to$ Subtract $16 f$ from both sides.
$\cancel{- 16 \textcolor{b l u e}{- 14}} - 30 = 5 f \cancel{\textcolor{b l u e}{+ 14}} \cancel{\textcolor{b l u e}{- 14}} \to$ Subtract $14$ from both sides.
$\cancel{- 30 \textcolor{g r e e n}{/ 5}} - 6 = \cancel{\textcolor{g r e e n}{5}} f \cancel{\textcolor{g r e e n}{/ 5}} \to$ Divide both sides by 5.
$- 6 = f \to$ Final answer!
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# Total Internal Reflection Problems and Solutions
On this page, all aspects of the total internal reflection phenomenon are explored in the form of problem-solving.
For reference, total internal reflection occurs when the incident light is in a medium with a higher index of refraction.
In this situation, if the angle of incidence is greater than the critical angle, then the refracted ray totally internally reflected back into the original medium.
## Total Internal Reflection Problems
Problem (1): An unknown glass has an index of refraction of $n=1.5$. For a beam of light originating in the glass, at what angles does the light $100\%$ reflected back into the glass? (The index of refraction of air is $n_{air}=1.00$).
Solution: When light moves through a medium with an index of refraction $n_1$ and strikes a boundary of a region with an index of refraction $n_2$ such that $n_1>n_2$, then total internal reflection can occur if the incident angle equals or is greater than a critical angle $\theta_c$ given by the following formula $\sin\theta_c=\frac{n_2}{n_1}$ If the incident ray equals the critical angle, then the refracted ray exit along the boundary at the angle of $90^\circ$.
For incident angles greater than the critical angle, $\theta_1 >\theta_c$, the refracted ray vanishes at the boundary, and $100\%$ is reflected back into the original medium. This process is called total internal reflection.
In this problem, we are told that light is initially in a medium with a higher index of refraction and wants to enter a medium with a lower index of refraction. So, the necessary condition for occurring total internal reflection is satisfied.
Thus, we must first find the critical angle whose magnitude is obtained as below $\sin\theta_c=\frac{n_{air}}{n_{glass}}=\frac{1}{1.5}$ Taking the inverse of the sine of both sides, gives $\theta_c=\sin^{-1}\left(\frac{1}{1.5}\right)=41.8^\circ$ Hence, if the angle of incidence in the glass exceeds about $42^\circ$, then all incident rays completely return back into the same medium.
Problem (2): A ray of light traveling through glass with an index of refraction $n=1.5$ strikes the interface of the glass-water. Let the index of refraction of water be $n=1.33$. At what angle, the ray does not enter the water?
Solution: We want no ray to exist in the water. We know that when a ray of light strikes a boundary of two different media, part of it is reflected back into the same medium and part of the ray enters the second medium (or refracted).
In this question, we are asked to find an angle at which the refracted ray is removed. This occurs when the refracted angle in the second medium (here, water) becomes greater than or equal to $90^\circ$. Applying Snell's law of refraction, we have \begin{align*} n_1 \sin\theta_1&=n_2 \sin\theta_2\\\\ (1.5)\sin\theta_1&=(1.33)\sin 90^\circ\\\\ \Rightarrow \quad \sin\theta_2&=\frac{1.33}{1.5}\sin 90^\circ\\\\&=0.887\end{align*} Next, take the inverse of the sine to find the angle whose sine is $1.128$ $\theta_2=\sin^{-1}(0.887)=62.5^\circ$ This incident angle at which the refracted angle becomes $90^\circ$ is called the critical angle. Any ray, in the glass, striking the boundary with an angle that exceeds 40 degrees, totally returns back into the same original medium. This phenomenon is also called total internal reflection.
Problem (3): Find the critical angle for a ray of light traveling from Ethyl alcohol to Benzene. Ethyl alcohol has an index of refraction $n=1.361$ and Benzene has $n=1.501$.
Solution: There occurs no totally internal reflection because there is no critical angle for this configuration. In this scenario, light is traveling from a rare medium (low $n$) to a dense medium (with high $n$) and this is contrary to the criteria for the formation of a total internal reflection.
Thus, total internal reflection occurs when light traveling in a medium strikes the boundary of a medium whose index of refraction is higher than the original medium.
Problem (4): The speed of sound in a room filled with air is $343\,{\rm m/s}$. The walls of this room are made of concrete, in which the speed of sound is $1850\,{\rm m/s}$.
(a) Find the critical angle for total internal reflection of sound waves striking the air-concrete interface.
(b) If we want the sound to undergo total internal reflection, in which medium must it be traveling?
Solution: To find the critical angle for a total internal reflection problem, we must use the following equation $\theta_c=\sin^{-1}\left(\frac{n_{rare}}{n_{dense}}\right)$ But in this question, neither of indices of refraction are given.
Recall that Snell's law of refraction is also written in terms of velocities instead of indices of refraction. Thus, the alternate form of Snell's law is given by the following equation $\frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2}$ where the subscripts $1$ and $2$ refer to the incident and refracted rays.
(a) We know that the critical angle is the angle at which the refracted ray becomes $90^\circ$. So, setting $\theta_2=90^\circ$ in the above equation, and solving for $\theta_1=\theta_{cri}$, we get \begin{align*} \sin\theta_{cri}&=\sin\theta_2 \left(\frac{v_1}{v_2}\right)\\\\&=\sin 90^\circ \left(\frac{343}{1850}\right)\\\\&=0.186 \end{align*} Now, take the inverse of sine from both sides to find the angle whose sine is $0.186$ $\theta_{cri}=\sin^{-1}(0.186)=10.7^\circ$ Hence, if the sound waves strike the wall at angles greater than about $11^\circ$, then all is reflected back into the wall without any waves passing through the wall.
(b) As you know, total internal reflection occurs when mechanical waves (sound) or electromagnetic waves (light) are incidents on a boundary of a medium having a lower index of refraction than the original medium.
We can modify the above statement to accommodate the velocities in both mediums where the indices of refraction are not known. In this case, the total internal reflection occurs when the waves strike the boundary of a medium, in which the speed of the waves is greater than the original medium.
In this problem, the total internal reflection can occur when sound waves travel from the air and strike the wall at an angle greater than the critical angle found above, i.e. $\theta>10.7^\circ$.
Problem (5): The boundary of the two mediums is shown in the figure below. The angle of incidence with the horizontal is $30^\circ$. The incident ray refracted parallel to the interface. Find the index of refraction of the second medium.
Solution: We are told in the problem that the refracted ray is parallel to the boundary of two mediums. We know that this situation always occurs when the incident ray is in a dense medium and strikes the boundary of a less dense medium.
So, the second medium must have a lower index of refraction than the original medium. Note that the incident angle is measured with a line perpendicular to the boundary. Here, the incident angle is $\theta_1=90^\circ-30^\circ=60^\circ$.
Setting the refracted angle as $\theta_2=90^\circ$ in Snell's law of refraction, and solving for the index of refraction $n_2$, we will have \begin{align*} n_1 \sin\theta_1&=n_2\sin\theta_2\\\\\sqrt{3}\sin 60^\circ&=n_2 \sin 90^\circ\\\\ \sqrt{3}\left(\frac{\sqrt{3}}2\right)&=n_2\\\\\Rightarrow n_2 &=\frac 32 \end{align*} As expected, the second medium must be less dense than the first medium.
Problem (6): A beam of light moves from air into an unknown glass. The angle of incidence is $30^\circ$, and the angle of refraction is $22^\circ$. For a beam of light originating from the glass, find the critical angle.
Solution: Critical angle is defined when a ray travels in a dense medium and strikes a boundary with a lower index of refraction. Its magnitude is calculated as $\theta_c=\sin^{-1}\left(\frac{n_2}{n_1}\right)$ where always $n_1>n_2$.
When a beam of light strikes the interface of the two media at the critical angle, the refracted ray exits parallel to the boundary. In this case, the refracted angle is $90^\circ$.
Here, we want to find the critical angle when the light is in the glass and enters the air. The index of refraction of the glass is unknown.
From the first part of the question, we can find the glass's index of refraction by applying Snell's law of refraction. \begin{align*} n_1 \sin\theta_1&=n_2 \sin\theta_2\\\\ (1)\sin 30^\circ&=n\sin 22^\circ\\\\ \Rightarrow \quad n&=\frac{\sin 30^\circ}{\sin 22^\circ}(1)\\\\&=\frac{0.5}{0.375}\\\\&=1.33 \end{align*} By having the index of refraction of the glass, the critical angle is found for this total internal reflection problem from the following equation $\theta_c=\sin^{-1}\left(\frac{n_2}{n_1}\right)=\sin^{-1}\left(\frac{1.00}{1.33}\right)=48.8^\circ$ Thus, if light originated from the water incident on the boundary of air at an angle greater than about $49^\circ$, then light is totally returned back into the water and does not enter the air.
Problem (7): The critical angle of a specific glass in the air is $\theta_c=37.8^\circ$. Now suppose that we put it into a tank of water. What will change its critical angle?
Solution: In the first case, the boundary is composed of glass-air. Using critical angle formula, we have $\sin\theta_c=\frac{n_{air}}{n_{glass}}$ In the second case, the glass surrounded by a water. So, the interface is glass-water. Again, its critical angle is found as $\sin\theta'_c=\frac{n_{water}}{n_{glass}}$ The index of refraction of the glass is unknown. Dividing the second relation by the first one, gives $\frac{\sin\theta'_c}{\sin\theta_c}=\frac{n_{water}}{n_{air}}$ Setting $n_{water}=1.33$ and solving for the unknown angle $\theta'_c$, we get \begin{align*} \sin\theta'_c&=\sin\theta_c \left(\frac{n_{water}}{n_{air}}\right)\\\\&=\sin(37.8^\circ)\frac{1.33}{1.00}\\\\&=0.816\end{align*} Take the inverse sine of both sides to find the angle whose sine is $0.816$ $\theta'_c=\sin^{-1}(0.816)=54.7^\circ$
Problem (8): A light bulb is placed 5 meters below a swimming pool. What is the diameter of the circle of light formed on the surface seen directly from above?
Solution: We can assume the light bulb to be a point source. Rays emit in all directions from that point source and strike the water-air interface. Some rays are incident on the boundary at less than the critical angle, refracted, and enter the air.
The rays have an incident angle greater than the critical angle, do not enter the air, and $100\%$ reflect back into the water.
These completely reflected rays are responsible for the formation of a circle of light on the surface of the water.
The diameter of the circle of light is the distance between the two points at which the rays reach the surface at the critical angle. First, using critical angle formula, we find it as below $\theta_c=\sin^{-1}\left(\frac{1.00}{1.33}\right)=48.7^\circ$ As you can see from the geometry, and using the definition of sine function, the radius of the circle of light is found as $\sin\theta_c=\frac{r}{H} \Rightarrow r=H\sin\theta_c$ where $H$ is the pool's depth. So, the radius or diameter of the circle $D=2r$ is computed as below $D=2H\sin\theta_c=2(5)(\sin 48.7^\circ)=7.5\,{\rm m}$
Problem (9): A beam of light is incident on the surface $AB$ of a prism surrounded by air at the critical angle, as shown in the figure. Find the angle of incidence, $\theta_1$.
Solution: we are told that the ray is incident on the surface $BC$ at the critical angle. Thus, the refracted ray must be parallel to $BC$.
To find the critical angle in a total internal reflection problem, using Snell's law, we arrive at the following formula $\sin\theta_c=\frac{n_{rare}}{n_{dense}}$ Here, the prism is surrounded by air, so $n_{rare}=1.00$. The index of refraction of the prism, $n_p$, is not given.
Thus, the critical angle at the boundary of $BC$ is found as below $\sin 42^\circ=\frac{1}{n_p} \quad (I)$ Now, we must apply Snell's law at the entering point into the prism for the incident ray, but the angle of refraction at that point is not known.
From geometry of the figure, we have $\beta=90^\circ-42^\circ=48^\circ$. On the other, recall that the sum of angles of a triangle equals $180^\circ$.
Thus, the angle $\alpha$ is determined as below \begin{gather*} \alpha+\beta+60^\circ=180^\circ \\ \alpha+48^\circ+60^\circ=180^\circ\\\Rightarrow \quad \alpha=72^\circ\end{gather*} Now that we have angle $\alpha$, the refracted ray at the entering surface $AB$ is simply found as below $\theta_2+\alpha=90^\circ \Rightarrow \theta_2=18^\circ$ Applying Snell's law at this surface $AB$, we get \begin{align*} n_1 \sin\theta_1&=n_2 \sin\theta_2 \\\\ (1)\sin\theta_1 &=n_p \sin 18^\circ \quad (II) \end{align*} From equation $(I)$, the index of refraction of prism is $n_p=\frac{1}{\sin 42^\circ}$. Substituting this into equation $(II)$, we can get the incident angle as below \begin{align*} \sin\theta_1 &=\frac{\sin 18^\circ}{\sin 42^\circ}\\\\&=0.462 \end{align*} By taking the inverse of sine from both sides, we can find the angle whose sine is $0.461$ $\theta_1=\sin^{-1}(0.461)=27.5^\circ$
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Basic Math Review Key Words and Symbols (continued) (continued) Rates, , Proportions, The following words and symbols are used for the and Percents MULTIPLYING AND DIVIDING WITH NEGATIVES operations listed. # Equivalent fractions are found by multiplying the numerator -a b =-ab and denominator of the by the same . In the RATES AND RATIOS previous example, - # - = A is a comparison of two quantities with different units. Important Properties Sum, total, increase, plus a b ab # # - 2 2 4 8 1 1 3 3 For example, a car that travels 110 miles in 2 hours is mov- a = a = = and = = . addend addend = sum # # ing at a rate of 110 miles/2 hours or 55 mph. NATURAL NUMBERS PROPERTIES OF ADDITION -b b 3 3 4 12 4 4 3 12 a A is a comparison of two quantities with the same {1, 2, 3, 4, 5, …} Property of Zero: a + 0 = a -a , b =- Difference, decrease, minus b MULTIPLYING AND DIVIDING FRACTIONS units. For example, a with 23 students has a WHOLE NUMBERS Inverse Property: a + -a = 0 When multiplying and dividing fractions, a common student–teacher ratio of 23:1 or23 . 1 2 minuend subtrahend = difference Some examples: 1 denominator is not needed. To multiply, take the {0, 1, 2, 3, 4, …} + = + - # =- : a b b a 3 5 15 of the numerators and the product of the denominators: PROPORTIONS Product, of, -7 -6 = 42 # INTEGERS : a + b + = a + b + c 1 21 2 a # c a c ac A proportion is a statement in which two ratios or rates are 1 2 1 2 * # - - = = = {…, 3, 2, 1, 0, 1, 2, …} a b, a b, a b , ab 24 8 3 # equal. PROPERTIES OF MULTIPLICATION 1 21 2 1 2>1 2 b d b d bd factor factor = product 18 36 An example of a proportion is the following statement: The Number # or 18 To divide fractions, invert the second fraction and then Property of Zero: a 0 = 0 2 36 30 dollars is to 5 hours as 60 dollars is to 10 hours. # 2 multiply the numerators and denominators: Identity Property of One: a 1 = a, when a Z 0. , per, divided by a c a d ad This is written –5–5 – 4–4 –3–3 –2–2 –1–1 0 1 2 3 4 5 # 1 , = = # = Z a b d b c bc \$30 = \$60 Negative integers Positive integers Inverse Property: a 1, when a 0. a b a b ba . a # # b > Fractions 5 hr 10 hr Commutative Property: a b = b a Some examples: Zero A typical proportion problem will have one unknown # # # # 3 2 6 Associative Property: a b c = a b c dividend = quotient LEAST COMMON # = quantity, such as 1 2 1 2 5 7 35 The LCM of a of numbers is the smallest number that is a 1 mile x miles RATIONAL NUMBERS PROPERTIES OF DIVISION = . multiple of all the given numbers. 5 1 5 2 10 5 20 min 60 min All numbers that can be written in the form a b , where a 0 , = # = = Z > Property of Zero: = 0, when a Z 0. For example, the LCM of 5 and 6 is 30, since 5 and 6 have no 12 2 12 1 12 6 and b are integers andb 0 . a We can solve this by cross multiplying as shown: 1st: Parentheses factors in common. IRRATIONAL NUMBERS a = # Property of One: = 1, when a Z 0. Simplify any expressions inside parentheses. REDUCING FRACTIONS 20x 60 1 Real numbers that cannot be written as the quotient of two a GREATEST COMMON FACTOR 2nd: Exponents To reduce a fraction, divide both the numerator and denom- 60 integers but can be represented on the . a # x = = 3. Identity Property of One: = a 1 The GCF of a set of numbers is the largest number that can inator by common factors. In the last example, 20 1 Work out any exponents. be evenly divided into each of the given numbers. 10 10 , 2 5 REAL NUMBERS 3rd: Multiplication and Division = = . So, it takes 60 minutes to walk 3 miles. For example, the GCF of 24 and 27 is 3, since both 24 and 12 12 , 2 6 Include all numbers that can be represented on the number Solve all multiplication and division, working from 27 are divisible by 3, but they are not both divisible by any line, that is, all rational and irrational numbers. PERCENTS left to right. numbers larger than 3. 4th: Addition and Subtraction MIXED NUMBERS A percent is the number of parts out of 100. To write a per- cent as a fraction, divide by 100 and drop the percent . Real Numbers The absolute value of a number is always ≥ 0. These are done last, from left to right. FRACTIONS A mixed number has two parts: a whole number part and a ƒ ƒ 3 _4 2 If a 7 0, a = a . fractional part. An example of a mixed number is 5 . This For example, Rational Numbers 23, 22.4, 21 5 , 0, 0.6, 1, etc. For example, Fractions are another way to express division. The top num- 8 Irrational 6 ƒ - ƒ = 57 Numbers If a 0, a a . - # + - , 2 ber of a fraction is called the numerator, and the bottom really represents = p 23, 22, 21, 0, 1, 2, 3, p 15 2 3 30 3 3 3 57% . 25VN3, Integers ƒ - ƒ = ƒ ƒ = # 1 2 number is called the denominator. 5 + , 100 For example, 5 5 and 5 5 . In each case, the = 15 - 2 3 + 27 , 9 VN2, p, etc. answer is positive. 8 Whole Numbers 0, 1, 2, 3, p = 15 - 6 + 3 ADDING AND SUBTRACTING FRACTIONS To write a fraction as a percent, first check to see if the which can be written as denominator is 100. If it is not, write the fraction as an = Natural Numbers 1, 2, 3, p 12. Fractions must have the same denominator before they can 40 3 43 equivalent fraction with 100 in the denominator. Then the + = . be added or subtracted. 8 8 8 numerator becomes the percent. For example, a b a + b 4 80 + = , with d Z 0 . Similarly, an improper fraction can be written as a mixed = = 80%. Integers d d d 5 100 PRIME NUMBERS number. For example, a b a - b - = Z 20 2 To find a percent of a quantity, multiply the percent by the A is a number greater than 1 that has only ADDING AND SUBTRACTING WITH NEGATIVES , with d 0. can be written as 6 , ISBN-13: 978-0-321-39476-7 d d d 3 3 quantity. itself and 1 as factors. - - = - + - ISBN-10: 0-321-39476-3 a b a b If the fractions have different denominators, rewrite them as Some examples: 1 2 1 2 since 20 divided by 3 equals 6 with a of 2. For example, 30% of 5 is 90000 -a + b = b - a equivalent fractions with a common denominator. Then add 2, 3, and 7 are prime numbers. 30 # 150 3 a - -b = a + b or subtract the numerators, keeping the denominators the 5 = = . COMPOSITE NUMBERS 1 2 same. For example, 100 100 2 Some examples: A composite number is a number that is not prime. For 2 1 8 3 11 -3 - 17 = -3 + -17 =-20 + = + = example, 8 is a composite number since 1 2 1 2 . # # -19 + 4 = 4 - 19 =-15 3 4 12 12 12 8 = 2 2 2 = 23. 9 780321 394767 more➤ more➤
1 2 3 NEWCOLORs_basic_math_rev 3/31/08 3:52 PM Page 1
Basic Math Review Key Words and Symbols Integers (continued) Fractions (continued) Rates, Ratios, Proportions, The following words and symbols are used for the and Percents MULTIPLYING AND DIVIDING WITH NEGATIVES operations listed. # Equivalent fractions are found by multiplying the numerator -a b =-ab and denominator of the fraction by the same number. In the RATES AND RATIOS Addition previous example, - # - = A rate is a comparison of two quantities with different units. Numbers Important Properties Sum, total, increase, plus a b ab # # - 2 2 4 8 1 1 3 3 For example, a car that travels 110 miles in 2 hours is mov- a = a = = and = = . addend addend = sum # # ing at a rate of 110 miles/2 hours or 55 mph. NATURAL NUMBERS PROPERTIES OF ADDITION -b b 3 3 4 12 4 4 3 12 Subtraction a A ratio is a comparison of two quantities with the same {1, 2, 3, 4, 5, …} Identity Property of Zero: a + 0 = a -a , b =- Difference, decrease, minus b MULTIPLYING AND DIVIDING FRACTIONS units. For example, a class with 23 students has a WHOLE NUMBERS Inverse Property: a + -a = 0 When multiplying and dividing fractions, a common student–teacher ratio of 23:1 or23 . 1 2 minuend subtrahend = difference Some examples: 1 denominator is not needed. To multiply, take the product {0, 1, 2, 3, 4, …} + = + Multiplication - # =- Commutative Property: a b b a 3 5 15 of the numerators and the product of the denominators: PROPORTIONS Product, of, times -7 -6 = 42 # INTEGERS Associative Property: a + b + c = a + b + c 1 21 2 a # c a c ac A proportion is a statement in which two ratios or rates are 1 2 1 2 * # - - = = = {…, 3, 2, 1, 0, 1, 2, …} a b, a b, a b , ab 24 8 3 # equal. PROPERTIES OF MULTIPLICATION 1 21 2 1 2>1 2 b d b d bd factor factor = product 18 36 An example of a proportion is the following statement: The Number Line # or 18 To divide fractions, invert the second fraction and then Property of Zero: a 0 = 0 2 36 30 dollars is to 5 hours as 60 dollars is to 10 hours. # Division 2 multiply the numerators and denominators: Identity Property of One: a 1 = a, when a Z 0. Quotient, per, divided by a c a d ad This is written –5–5 – 4–4 –3–3 –2–2 –1–1 0 1 2 3 4 5 # 1 , = = # = Z a b d b c bc \$30 = \$60 Negative integers Positive integers Inverse Property: a 1, when a 0. a b a b ba . a # # b > Fractions 5 hr 10 hr Commutative Property: a b = b a Some examples: Zero A typical proportion problem will have one unknown # # # # 3 2 6 Associative Property: a b c = a b c dividend divisor = quotient # = quantity, such as 1 2 1 2 5 7 35 The LCM of a set of numbers is the smallest number that is a 1 mile x miles RATIONAL NUMBERS PROPERTIES OF DIVISION = . multiple of all the given numbers. 5 1 5 2 10 5 20 min 60 min All numbers that can be written in the form a b , where a 0 , = # = = Z > Property of Zero: = 0, when a Z 0. Order of Operations For example, the LCM of 5 and 6 is 30, since 5 and 6 have no 12 2 12 1 12 6 and b are integers andb 0 . a We can solve this equation by cross multiplying as shown: 1st: Parentheses factors in common. IRRATIONAL NUMBERS a = # Property of One: = 1, when a Z 0. Simplify any expressions inside parentheses. REDUCING FRACTIONS 20x 60 1 Real numbers that cannot be written as the quotient of two a GREATEST COMMON FACTOR 2nd: Exponents To reduce a fraction, divide both the numerator and denom- 60 integers but can be represented on the number line. a # x = = 3. Identity Property of One: = a 1 The GCF of a set of numbers is the largest number that can inator by common factors. In the last example, 20 1 Work out any exponents. be evenly divided into each of the given numbers. 10 10 , 2 5 REAL NUMBERS 3rd: Multiplication and Division = = . So, it takes 60 minutes to walk 3 miles. For example, the GCF of 24 and 27 is 3, since both 24 and 12 12 , 2 6 Include all numbers that can be represented on the number Solve all multiplication and division, working from 27 are divisible by 3, but they are not both divisible by any line, that is, all rational and irrational numbers. PERCENTS Absolute Value left to right. numbers larger than 3. 4th: Addition and Subtraction MIXED NUMBERS A percent is the number of parts out of 100. To write a per- cent as a fraction, divide by 100 and drop the . Real Numbers The absolute value of a number is always ≥ 0. These are done last, from left to right. FRACTIONS A mixed number has two parts: a whole number part and a ƒ ƒ 3 _4 2 If a 7 0, a = a . fractional part. An example of a mixed number is 5 . This For example, Rational Numbers 23, 22.4, 21 5 , 0, 0.6, 1, etc. For example, Fractions are another way to express division. The top num- 8 Irrational 6 ƒ - ƒ = 57 Numbers If a 0, a a . - # + - , 2 ber of a fraction is called the numerator, and the bottom really represents = p 23, 22, 21, 0, 1, 2, 3, p 15 2 3 30 3 3 3 57% . 25VN3, Integers ƒ - ƒ = ƒ ƒ = # 1 2 number is called the denominator. 5 + , 100 For example, 5 5 and 5 5 . In each case, the = 15 - 2 3 + 27 , 9 VN2, p, etc. answer is positive. 8 Whole Numbers 0, 1, 2, 3, p = 15 - 6 + 3 ADDING AND SUBTRACTING FRACTIONS To write a fraction as a percent, first check to see if the which can be written as denominator is 100. If it is not, write the fraction as an = Natural Numbers 1, 2, 3, p 12. Fractions must have the same denominator before they can 40 3 43 equivalent fraction with 100 in the denominator. Then the + = . be added or subtracted. 8 8 8 numerator becomes the percent. For example, a b a + b 4 80 + = , with d Z 0 . Similarly, an improper fraction can be written as a mixed = = 80%. Integers d d d 5 100 PRIME NUMBERS number. For example, a b a - b - = Z 20 2 To find a percent of a quantity, multiply the percent by the A prime number is a number greater than 1 that has only ADDING AND SUBTRACTING WITH NEGATIVES , with d 0. can be written as 6 , ISBN-13: 978-0-321-39476-7 d d d 3 3 quantity. itself and 1 as factors. - - = - + - ISBN-10: 0-321-39476-3 a b a b If the fractions have different denominators, rewrite them as Some examples: 1 2 1 2 since 20 divided by 3 equals 6 with a remainder of 2. For example, 30% of 5 is 90000 -a + b = b - a equivalent fractions with a common denominator. Then add 2, 3, and 7 are prime numbers. 30 # 150 3 a - -b = a + b or subtract the numerators, keeping the denominators the 5 = = . COMPOSITE NUMBERS 1 2 same. For example, 100 100 2 Some examples: A composite number is a number that is not prime. For 2 1 8 3 11 -3 - 17 = -3 + -17 =-20 + = + = example, 8 is a composite number since 1 2 1 2 . # # -19 + 4 = 4 - 19 =-15 3 4 12 12 12 8 = 2 2 2 = 23. 9 780321 394767 more➤ more➤
1 2 3 NEWCOLORs_basic_math_rev 3/31/08 3:52 PM Page 1
Basic Math Review Key Words and Symbols Integers (continued) Fractions (continued) Rates, Ratios, Proportions, The following words and symbols are used for the and Percents MULTIPLYING AND DIVIDING WITH NEGATIVES operations listed. # Equivalent fractions are found by multiplying the numerator -a b =-ab and denominator of the fraction by the same number. In the RATES AND RATIOS Addition previous example, - # - = A rate is a comparison of two quantities with different units. Numbers Important Properties Sum, total, increase, plus a b ab # # - 2 2 4 8 1 1 3 3 For example, a car that travels 110 miles in 2 hours is mov- a = a = = and = = . addend addend = sum # # ing at a rate of 110 miles/2 hours or 55 mph. NATURAL NUMBERS PROPERTIES OF ADDITION -b b 3 3 4 12 4 4 3 12 Subtraction a A ratio is a comparison of two quantities with the same {1, 2, 3, 4, 5, …} Identity Property of Zero: a + 0 = a -a , b =- Difference, decrease, minus b MULTIPLYING AND DIVIDING FRACTIONS units. For example, a class with 23 students has a WHOLE NUMBERS Inverse Property: a + -a = 0 When multiplying and dividing fractions, a common student–teacher ratio of 23:1 or23 . 1 2 minuend subtrahend = difference Some examples: 1 denominator is not needed. To multiply, take the product {0, 1, 2, 3, 4, …} + = + Multiplication - # =- Commutative Property: a b b a 3 5 15 of the numerators and the product of the denominators: PROPORTIONS Product, of, times -7 -6 = 42 # INTEGERS Associative Property: a + b + c = a + b + c 1 21 2 a # c a c ac A proportion is a statement in which two ratios or rates are 1 2 1 2 * # - - = = = {…, 3, 2, 1, 0, 1, 2, …} a b, a b, a b , ab 24 8 3 # equal. PROPERTIES OF MULTIPLICATION 1 21 2 1 2>1 2 b d b d bd factor factor = product 18 36 An example of a proportion is the following statement: The Number Line # or 18 To divide fractions, invert the second fraction and then Property of Zero: a 0 = 0 2 36 30 dollars is to 5 hours as 60 dollars is to 10 hours. # Division 2 multiply the numerators and denominators: Identity Property of One: a 1 = a, when a Z 0. Quotient, per, divided by a c a d ad This is written –5–5 – 4–4 –3–3 –2–2 –1–1 0 1 2 3 4 5 # 1 , = = # = Z a b d b c bc \$30 = \$60 Negative integers Positive integers Inverse Property: a 1, when a 0. a b a b ba . a # # b > Fractions 5 hr 10 hr Commutative Property: a b = b a Some examples: Zero A typical proportion problem will have one unknown # # # # 3 2 6 Associative Property: a b c = a b c dividend divisor = quotient LEAST COMMON MULTIPLE # = quantity, such as 1 2 1 2 5 7 35 The LCM of a set of numbers is the smallest number that is a 1 mile x miles RATIONAL NUMBERS PROPERTIES OF DIVISION = . multiple of all the given numbers. 5 1 5 2 10 5 20 min 60 min All numbers that can be written in the form a b , where a 0 , = # = = Z > Property of Zero: = 0, when a Z 0. Order of Operations For example, the LCM of 5 and 6 is 30, since 5 and 6 have no 12 2 12 1 12 6 and b are integers andb 0 . a We can solve this equation by cross multiplying as shown: 1st: Parentheses factors in common. IRRATIONAL NUMBERS a = # Property of One: = 1, when a Z 0. Simplify any expressions inside parentheses. REDUCING FRACTIONS 20x 60 1 Real numbers that cannot be written as the quotient of two a GREATEST COMMON FACTOR 2nd: Exponents To reduce a fraction, divide both the numerator and denom- 60 integers but can be represented on the number line. a # x = = 3. Identity Property of One: = a 1 The GCF of a set of numbers is the largest number that can inator by common factors. In the last example, 20 1 Work out any exponents. be evenly divided into each of the given numbers. 10 10 , 2 5 REAL NUMBERS 3rd: Multiplication and Division = = . So, it takes 60 minutes to walk 3 miles. For example, the GCF of 24 and 27 is 3, since both 24 and 12 12 , 2 6 Include all numbers that can be represented on the number Solve all multiplication and division, working from 27 are divisible by 3, but they are not both divisible by any line, that is, all rational and irrational numbers. PERCENTS Absolute Value left to right. numbers larger than 3. 4th: Addition and Subtraction MIXED NUMBERS A percent is the number of parts out of 100. To write a per- cent as a fraction, divide by 100 and drop the percent sign. Real Numbers The absolute value of a number is always ≥ 0. These are done last, from left to right. FRACTIONS A mixed number has two parts: a whole number part and a ƒ ƒ 3 _4 2 If a 7 0, a = a . fractional part. An example of a mixed number is 5 . This For example, Rational Numbers 23, 22.4, 21 5 , 0, 0.6, 1, etc. For example, Fractions are another way to express division. The top num- 8 Irrational 6 ƒ - ƒ = 57 Numbers If a 0, a a . - # + - , 2 ber of a fraction is called the numerator, and the bottom really represents = p 23, 22, 21, 0, 1, 2, 3, p 15 2 3 30 3 3 3 57% . 25VN3, Integers ƒ - ƒ = ƒ ƒ = # 1 2 number is called the denominator. 5 + , 100 For example, 5 5 and 5 5 . In each case, the = 15 - 2 3 + 27 , 9 VN2, p, etc. answer is positive. 8 Whole Numbers 0, 1, 2, 3, p = 15 - 6 + 3 ADDING AND SUBTRACTING FRACTIONS To write a fraction as a percent, first check to see if the which can be written as denominator is 100. If it is not, write the fraction as an = Natural Numbers 1, 2, 3, p 12. Fractions must have the same denominator before they can 40 3 43 equivalent fraction with 100 in the denominator. Then the + = . be added or subtracted. 8 8 8 numerator becomes the percent. For example, a b a + b 4 80 + = , with d Z 0 . Similarly, an improper fraction can be written as a mixed = = 80%. Integers d d d 5 100 PRIME NUMBERS number. For example, a b a - b - = Z 20 2 To find a percent of a quantity, multiply the percent by the A prime number is a number greater than 1 that has only ADDING AND SUBTRACTING WITH NEGATIVES , with d 0. can be written as 6 , ISBN-13: 978-0-321-39476-7 d d d 3 3 quantity. itself and 1 as factors. - - = - + - ISBN-10: 0-321-39476-3 a b a b If the fractions have different denominators, rewrite them as Some examples: 1 2 1 2 since 20 divided by 3 equals 6 with a remainder of 2. For example, 30% of 5 is 90000 -a + b = b - a equivalent fractions with a common denominator. Then add 2, 3, and 7 are prime numbers. 30 # 150 3 a - -b = a + b or subtract the numerators, keeping the denominators the 5 = = . COMPOSITE NUMBERS 1 2 same. For example, 100 100 2 Some examples: A composite number is a number that is not prime. For 2 1 8 3 11 -3 - 17 = -3 + -17 =-20 + = + = example, 8 is a composite number since 1 2 1 2 . # # -19 + 4 = 4 - 19 =-15 3 4 12 12 12 8 = 2 2 2 = 23. 9 780321 394767 more➤ more➤
1 2 3 NEWCOLORs_basic_math_rev 3/31/08 3:52 PM Page 2
Basic Math Review
Decimal Numbers Percents to and (continued) Geometry (continued) Decimals to Percents U.S. Units The perimeter of a geometric figure is the around it The numbers after the represent fractions with MULTIPLYING AND DIVIDING IN SCIENTIFIC NOTATION To change a number from a percent to a decimal, divide by or the sum of the of its sides. denominators that are powers of 10. The decimal point sep- To multiply or divide numbers in scientific notation, we can in. = oz = ounce In any right , if a and b are the lengths of the legs arates the whole number part from the fractional part. 100 and drop the percent sign: The perimeter of a rectangle is 2 times the plus 2 change the order and grouping, so that we multiply or divide ft = c = cup times the width: and c is the length of the hypotenuse, then 9 58% = 58/100 = 0.58. first the decimal parts and then the powers of 10. For example, 2 + 2 = 2 For example, 0.9 represents 10 . min = minute mi = mile a b c . * -3 # * 8 To change a number from a decimal to a percent, multiply 3.7 10 2.5 10 sec = second hr = hour L 1 2# 1 - 2 Place Value Chart by 100 and add the percent sign: = 3.7 * 2.5 10 3 * 108 1 2 1 2 gal = gallon lb = W 0.73 = .73* 100 = 73%. = * 5 c 9.25 10 . yd = yard qt = quart a pt = pint T = ton P = 2L + 2W
tens ones tenths illionths millions b thousandshundreds m Statistics The perimeter of a is 4 times the length of a side: hundredths ten millions thousandthsn thousandths Simple Interest Metric Units ten thousands hundred millions hundred thousands te hundred thousandths There are several ways to study a list of data. 9 327604985326894 Given the principal (amount of money to be borrowed or mm = millimeter s invested), interest rate, and length of , the amount of Mean, or average, is the sum of all the data values divided by cm = centimeter = # 2 Whole numbers Decimals interest can be found using the the number of values. s : A p r # # km = kilometer = # = # # I = p r t Median is the number that separates the list of data into two : C p d 2 p r m = meter equal parts. To find the median, list the data in order from = where d is the , r is the radius, or half the diameter, ADDING AND SUBTRACTING DECIMAL NUMBERS where I = interest dollar amount P 4s 1 2 smallest to largest. If the number of data is odd, the median is mL = milliliter p 22 To add or subtract decimal numbers, line up the numbers so = and is approximately 3.14 or 7 . p principal the middle number. If the number of data is even, the median Area is always expressed in square units, since it is two- that the decimal points are aligned. Then add or subtract as cL = centiliter = is the average of the two middle numbers. dimensional. usual, keeping the decimal point in the same place. r rate of interest L = liter = is the number in the list that occurs the most fre- The formula for area of a rectangle is For example, 23 - 0.37 = 23.00 t time period. kL = kiloliter quently. There can be more than one mode. = # d For example, find the amount of simple interest on a \$3800 mg = milligram A L W. 0.37 For example, consider the following list of test scores: loan at an annual rate of 5.5% for 5 years: 22.63 cg = centigram The formula for area of a square is = {87, 56, 69, 87, 93, 82} # r p \$3800 g = gram A = s s orA = s2 . MULTIPLYING AND DIVIDING DECIMAL NUMBERS = = To find the mean, first add: r 5.5% 0.055 kg = kilogram = + + + + + = The area of a triangle is one-half the product of the height To multiply decimal numbers, multiply them as though they t 5 years 87 56 69 87 93 82 474. and : A has an of 360 degrees. were whole numbers. The number of decimal places in the I = 3800 0.055 5 = 1045. Then divide by 6: U.S. AND METRIC CONVERSIONS product is the sum of the number of decimal places in the 1 21 21 2 A straight line has an angle of 180 degrees. factors. For example, 3.72 * 4.5 is 474 U.S. The amount of interest is \$1045. = 79. 2 decimal places 6 12 in. = 1 ft 3 ft = 1 yd h The mean score is 79. 1760 yd = 1 mi 5280 ft = 1 mi Algebraic Terms 3.72 1 decimal place 4.5 Scientific Notation To find the median, first list the data in order: 2 c = 1 pt 1 c = 8 oz b : A variable is a letter that represents a number 56, 69, 82, 87, 87, 93. 4 qt = 1 gal 2 pt = 1 qt because the number is unknown or because it can change. 16.740 Scientific notation is a convenient way to express very large 1 # A = b h For example, the number of days until your vacation 3 decimal places or very small numbers. A number in this form is written as Since there is an even number of data, we take the average 2000 lb = 1 T 16 oz = 1 lb 2 * n … ƒ ƒ 6 changes every day, so it could be represented by a a 10 , where 1 a 10 and n is an . For of 82 and 87: * 5 - * -4 variable, x. To divide decimal numbers, first make sure the divisor is a example, 3.62 10 and 1.2 10 are expressed + Metric The sum of all three in any triangle always equals 82 87 = 169 = : A constant is a that does not change. For whole number. If it is not, move the decimal place to the right in scientific notation. 84.5. 180 degrees. 2 2 1000 mm = 1 m 100 cm = 1 m example, the number of days in the week, 7, does not (multiply by 10, 100, and so on) to make it a whole number. To change a number from scientific notation to a number 1000 m = 1 km 100 cL = 1 L change, so it is a constant. Then move the decimal point the same number of places in without exponents, look at the power of ten. If that number is The median score is 84.5. x the dividend. positive, move the decimal point to the right. If it is negative, The mode is 87, since this number appears twice and each 1000 mL = 1 L 100 cg = 1 g z : An consists of constants, move the decimal point to the left. The number tells you how 1000 mg = 1 g 1000 g = 1 kg y variables, numerals and at least one . For example, For example, of the other numbers appears only once. + many places to move the decimal point. x 7 is an expression. 0.42 , 1.2 = 4.2 , 12 0.001 m = 1 mm 0.01 m = 1 cm + + = x° y° z° 180° Equation: An equation is basically a mathematical sentence 0.35 For example, 0.001 g = 1 mg 0.01 g = 1 cg * 3 = Distance Formula A right triangle is a triangle with a 90° (right) angle. The indicating that two expressions are equal. For example, 124.20 . 3.97 10 3970. 0.001 L = 1 mL 0.01 L = 1 cL + = hypotenuse of a right triangle is the side opposite the right x 7 18 is an equation. To change a number to scientific notation, move the deci- Given the rate at which you are traveling and the length of The decimal point in the answer is placed directly above the angle. Solution: A number that makes an equation true is a mal point so it is to the right of the first nonzero digit. If the time you will be traveling, the distance can be found by new decimal point in the dividend. solution to that equation. For example, in using the above decimal point is moved n places to the left and this makes using the formula equation, x + 7 = 18, we know that the statement is true = # the number smaller, n is positive; otherwise, n is negative. If d r t if x = 11. the decimal point is not moved, n is 0. where d = distance hypotenuse - For example, 0.0000216 = 2.16 * 10 5. r = rate t = time more➤ 90°
more➤ 4 5 6 NEWCOLORs_basic_math_rev 3/31/08 3:52 PM Page 2
Basic Math Review
Decimal Numbers Percents to Decimals and Scientific Notation (continued) Measurements Geometry Geometry (continued) Decimals to Percents U.S. Measurement Units The perimeter of a geometric figure is the distance around it The numbers after the decimal point represent fractions with MULTIPLYING AND DIVIDING IN SCIENTIFIC NOTATION PYTHAGOREAN THEOREM To change a number from a percent to a decimal, divide by or the sum of the lengths of its sides. denominators that are powers of 10. The decimal point sep- To multiply or divide numbers in scientific notation, we can in. = inch oz = ounce In any right triangle, if a and b are the lengths of the legs arates the whole number part from the fractional part. 100 and drop the percent sign: The perimeter of a rectangle is 2 times the length plus 2 change the order and grouping, so that we multiply or divide ft = foot c = cup times the width: and c is the length of the hypotenuse, then 9 58% = 58/100 = 0.58. first the decimal parts and then the powers of 10. For example, 2 + 2 = 2 For example, 0.9 represents 10 . min = minute mi = mile a b c . * -3 # * 8 To change a number from a decimal to a percent, multiply 3.7 10 2.5 10 sec = second hr = hour L 1 2# 1 - 2 Place Value Chart by 100 and add the percent sign: = 3.7 * 2.5 10 3 * 108 1 2 1 2 gal = gallon lb = pound W 0.73 = .73* 100 = 73%. = * 5 c 9.25 10 . yd = yard qt = quart a pt = pint T = ton P = 2L + 2W
tens ones billions tenths illionths millions b thousandshundreds m Statistics The perimeter of a square is 4 times the length of a side: hundredths ten millions thousandthsn thousandths Simple Interest Metric Units ten thousands hundred millions hundred thousands te hundred thousandths There are several ways to study a list of data. 9 327604985326894 Given the principal (amount of money to be borrowed or mm = millimeter s CIRCLES invested), interest rate, and length of time, the amount of Mean, or average, is the sum of all the data values divided by cm = centimeter = # 2 Whole numbers Decimals interest can be found using the formula the number of values. s Area: A p r # # km = kilometer = # = # # I = p r t Median is the number that separates the list of data into two Circumference: C p d 2 p r m = meter equal parts. To find the median, list the data in order from = where d is the diameter, r is the radius, or half the diameter, ADDING AND SUBTRACTING DECIMAL NUMBERS where I = interest dollar amount P 4s 1 2 smallest to largest. If the number of data is odd, the median is mL = milliliter p 22 To add or subtract decimal numbers, line up the numbers so = and is approximately 3.14 or 7 . p principal the middle number. If the number of data is even, the median Area is always expressed in square units, since it is two- that the decimal points are aligned. Then add or subtract as cL = centiliter = is the average of the two middle numbers. dimensional. usual, keeping the decimal point in the same place. r percentage rate of interest L = liter = Mode is the number in the list that occurs the most fre- The formula for area of a rectangle is For example, 23 - 0.37 = 23.00 t time period. kL = kiloliter quently. There can be more than one mode. = # d For example, find the amount of simple interest on a \$3800 mg = milligram A L W. 0.37 For example, consider the following list of test scores: loan at an annual rate of 5.5% for 5 years: 22.63 cg = centigram The formula for area of a square is = {87, 56, 69, 87, 93, 82} # r p \$3800 g = gram A = s s orA = s2 . MULTIPLYING AND DIVIDING DECIMAL NUMBERS = = To find the mean, first add: r 5.5% 0.055 kg = kilogram = + + + + + = The area of a triangle is one-half the product of the height To multiply decimal numbers, multiply them as though they t 5 years 87 56 69 87 93 82 474. and base: A circle has an angle of 360 degrees. were whole numbers. The number of decimal places in the I = 3800 0.055 5 = 1045. Then divide by 6: U.S. AND METRIC CONVERSIONS product is the sum of the number of decimal places in the 1 21 21 2 A straight line has an angle of 180 degrees. factors. For example, 3.72 * 4.5 is 474 U.S. The amount of interest is \$1045. = 79. 2 decimal places 6 12 in. = 1 ft 3 ft = 1 yd h The mean score is 79. 1760 yd = 1 mi 5280 ft = 1 mi Algebraic Terms 3.72 1 decimal place 4.5 Scientific Notation To find the median, first list the data in order: 2 c = 1 pt 1 c = 8 oz b Variable: A variable is a letter that represents a number 56, 69, 82, 87, 87, 93. 4 qt = 1 gal 2 pt = 1 qt because the number is unknown or because it can change. 16.740 Scientific notation is a convenient way to express very large 1 # A = b h For example, the number of days until your vacation 3 decimal places or very small numbers. A number in this form is written as Since there is an even number of data, we take the average 2000 lb = 1 T 16 oz = 1 lb 2 * n … ƒ ƒ 6 changes every day, so it could be represented by a a 10 , where 1 a 10 and n is an integer. For of 82 and 87: * 5 - * -4 variable, x. To divide decimal numbers, first make sure the divisor is a example, 3.62 10 and 1.2 10 are expressed + Metric The sum of all three angles in any triangle always equals 82 87 = 169 = Constant: A constant is a term that does not change. For whole number. If it is not, move the decimal place to the right in scientific notation. 84.5. 180 degrees. 2 2 1000 mm = 1 m 100 cm = 1 m example, the number of days in the week, 7, does not (multiply by 10, 100, and so on) to make it a whole number. To change a number from scientific notation to a number 1000 m = 1 km 100 cL = 1 L change, so it is a constant. Then move the decimal point the same number of places in without exponents, look at the power of ten. If that number is The median score is 84.5. x the dividend. positive, move the decimal point to the right. If it is negative, The mode is 87, since this number appears twice and each 1000 mL = 1 L 100 cg = 1 g z Expression: An algebraic expression consists of constants, move the decimal point to the left. The number tells you how 1000 mg = 1 g 1000 g = 1 kg y variables, numerals and at least one operation. For example, For example, of the other numbers appears only once. + many places to move the decimal point. x 7 is an expression. 0.42 , 1.2 = 4.2 , 12 0.001 m = 1 mm 0.01 m = 1 cm + + = x° y° z° 180° Equation: An equation is basically a mathematical sentence 0.35 For example, 0.001 g = 1 mg 0.01 g = 1 cg * 3 = Distance Formula A right triangle is a triangle with a 90° (right) angle. The indicating that two expressions are equal. For example, 124.20 . 3.97 10 3970. 0.001 L = 1 mL 0.01 L = 1 cL + = hypotenuse of a right triangle is the side opposite the right x 7 18 is an equation. To change a number to scientific notation, move the deci- Given the rate at which you are traveling and the length of The decimal point in the answer is placed directly above the angle. Solution: A number that makes an equation true is a mal point so it is to the right of the first nonzero digit. If the time you will be traveling, the distance can be found by new decimal point in the dividend. solution to that equation. For example, in using the above decimal point is moved n places to the left and this makes using the formula equation, x + 7 = 18, we know that the statement is true = # the number smaller, n is positive; otherwise, n is negative. If d r t if x = 11. the decimal point is not moved, n is 0. where d = distance hypotenuse - For example, 0.0000216 = 2.16 * 10 5. r = rate t = time more➤ 90°
more➤ 4 5 6 NEWCOLORs_basic_math_rev 3/31/08 3:52 PM Page 2
Basic Math Review
Decimal Numbers Percents to Decimals and Scientific Notation (continued) Measurements Geometry Geometry (continued) Decimals to Percents U.S. Measurement Units The perimeter of a geometric figure is the distance around it The numbers after the decimal point represent fractions with MULTIPLYING AND DIVIDING IN SCIENTIFIC NOTATION PYTHAGOREAN THEOREM To change a number from a percent to a decimal, divide by or the sum of the lengths of its sides. denominators that are powers of 10. The decimal point sep- To multiply or divide numbers in scientific notation, we can in. = inch oz = ounce In any right triangle, if a and b are the lengths of the legs arates the whole number part from the fractional part. 100 and drop the percent sign: The perimeter of a rectangle is 2 times the length plus 2 change the order and grouping, so that we multiply or divide ft = foot c = cup times the width: and c is the length of the hypotenuse, then 9 58% = 58/100 = 0.58. first the decimal parts and then the powers of 10. For example, 2 + 2 = 2 For example, 0.9 represents 10 . min = minute mi = mile a b c . * -3 # * 8 To change a number from a decimal to a percent, multiply 3.7 10 2.5 10 sec = second hr = hour L 1 2# 1 - 2 Place Value Chart by 100 and add the percent sign: = 3.7 * 2.5 10 3 * 108 1 2 1 2 gal = gallon lb = pound W 0.73 = .73* 100 = 73%. = * 5 c 9.25 10 . yd = yard qt = quart a pt = pint T = ton P = 2L + 2W
tens ones billions tenths illionths millions b thousandshundreds m Statistics The perimeter of a square is 4 times the length of a side: hundredths ten millions thousandthsn thousandths Simple Interest Metric Units ten thousands hundred millions hundred thousands te hundred thousandths There are several ways to study a list of data. 9 327604985326894 Given the principal (amount of money to be borrowed or mm = millimeter s CIRCLES invested), interest rate, and length of time, the amount of Mean, or average, is the sum of all the data values divided by cm = centimeter = # 2 Whole numbers Decimals interest can be found using the formula the number of values. s Area: A p r # # km = kilometer = # = # # I = p r t Median is the number that separates the list of data into two Circumference: C p d 2 p r m = meter equal parts. To find the median, list the data in order from = where d is the diameter, r is the radius, or half the diameter, ADDING AND SUBTRACTING DECIMAL NUMBERS where I = interest dollar amount P 4s 1 2 smallest to largest. If the number of data is odd, the median is mL = milliliter p 22 To add or subtract decimal numbers, line up the numbers so = and is approximately 3.14 or 7 . p principal the middle number. If the number of data is even, the median Area is always expressed in square units, since it is two- that the decimal points are aligned. Then add or subtract as cL = centiliter = is the average of the two middle numbers. dimensional. usual, keeping the decimal point in the same place. r percentage rate of interest L = liter = Mode is the number in the list that occurs the most fre- The formula for area of a rectangle is For example, 23 - 0.37 = 23.00 t time period. kL = kiloliter quently. There can be more than one mode. = # d For example, find the amount of simple interest on a \$3800 mg = milligram A L W. 0.37 For example, consider the following list of test scores: loan at an annual rate of 5.5% for 5 years: 22.63 cg = centigram The formula for area of a square is = {87, 56, 69, 87, 93, 82} # r p \$3800 g = gram A = s s orA = s2 . MULTIPLYING AND DIVIDING DECIMAL NUMBERS = = To find the mean, first add: r 5.5% 0.055 kg = kilogram = + + + + + = The area of a triangle is one-half the product of the height To multiply decimal numbers, multiply them as though they t 5 years 87 56 69 87 93 82 474. and base: A circle has an angle of 360 degrees. were whole numbers. The number of decimal places in the I = 3800 0.055 5 = 1045. Then divide by 6: U.S. AND METRIC CONVERSIONS product is the sum of the number of decimal places in the 1 21 21 2 A straight line has an angle of 180 degrees. factors. For example, 3.72 * 4.5 is 474 U.S. The amount of interest is \$1045. = 79. 2 decimal places 6 12 in. = 1 ft 3 ft = 1 yd h The mean score is 79. 1760 yd = 1 mi 5280 ft = 1 mi Algebraic Terms 3.72 1 decimal place 4.5 Scientific Notation To find the median, first list the data in order: 2 c = 1 pt 1 c = 8 oz b Variable: A variable is a letter that represents a number 56, 69, 82, 87, 87, 93. 4 qt = 1 gal 2 pt = 1 qt because the number is unknown or because it can change. 16.740 Scientific notation is a convenient way to express very large 1 # A = b h For example, the number of days until your vacation 3 decimal places or very small numbers. A number in this form is written as Since there is an even number of data, we take the average 2000 lb = 1 T 16 oz = 1 lb 2 * n … ƒ ƒ 6 changes every day, so it could be represented by a a 10 , where 1 a 10 and n is an integer. For of 82 and 87: * 5 - * -4 variable, x. To divide decimal numbers, first make sure the divisor is a example, 3.62 10 and 1.2 10 are expressed + Metric The sum of all three angles in any triangle always equals 82 87 = 169 = Constant: A constant is a term that does not change. For whole number. If it is not, move the decimal place to the right in scientific notation. 84.5. 180 degrees. 2 2 1000 mm = 1 m 100 cm = 1 m example, the number of days in the week, 7, does not (multiply by 10, 100, and so on) to make it a whole number. To change a number from scientific notation to a number 1000 m = 1 km 100 cL = 1 L change, so it is a constant. Then move the decimal point the same number of places in without exponents, look at the power of ten. If that number is The median score is 84.5. x the dividend. positive, move the decimal point to the right. If it is negative, The mode is 87, since this number appears twice and each 1000 mL = 1 L 100 cg = 1 g z Expression: An algebraic expression consists of constants, move the decimal point to the left. The number tells you how 1000 mg = 1 g 1000 g = 1 kg y variables, numerals and at least one operation. For example, For example, of the other numbers appears only once. + many places to move the decimal point. x 7 is an expression. 0.42 , 1.2 = 4.2 , 12 0.001 m = 1 mm 0.01 m = 1 cm + + = x° y° z° 180° Equation: An equation is basically a mathematical sentence 0.35 For example, 0.001 g = 1 mg 0.01 g = 1 cg * 3 = Distance Formula A right triangle is a triangle with a 90° (right) angle. The indicating that two expressions are equal. For example, 124.20 . 3.97 10 3970. 0.001 L = 1 mL 0.01 L = 1 cL + = hypotenuse of a right triangle is the side opposite the right x 7 18 is an equation. To change a number to scientific notation, move the deci- Given the rate at which you are traveling and the length of The decimal point in the answer is placed directly above the angle. Solution: A number that makes an equation true is a mal point so it is to the right of the first nonzero digit. If the time you will be traveling, the distance can be found by new decimal point in the dividend. solution to that equation. For example, in using the above decimal point is moved n places to the left and this makes using the formula equation, x + 7 = 18, we know that the statement is true = # the number smaller, n is positive; otherwise, n is negative. If d r t if x = 11. the decimal point is not moved, n is 0. where d = distance hypotenuse - For example, 0.0000216 = 2.16 * 10 5. r = rate t = time more➤ 90°
more➤ 4 5 6
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# How do you divide ( 3i+8) / (i +4 ) in trigonometric form?
May 21, 2016
$\frac{6 i - 4}{- 3 i - 5} = \sqrt{\frac{73}{17}} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} \left(\frac{4}{35}\right)$
#### Explanation:
Let us first write $\left(3 i + 8\right)$ and $\left(i + 4\right)$ in trigonometric form.
$a + i b$ can be written in trigonometric form $r {e}^{i \theta} = r \cos \theta + i r \sin \theta = r \left(\cos \theta + i \sin \theta\right)$,
where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\tan \theta = \frac{b}{a}$ or $\theta = \arctan \left(\frac{b}{a}\right)$
Hence $3 i + 8 = \left(8 + 3 i\right) = \sqrt{{8}^{2} + {3}^{2}} \left[\cos \alpha + i \sin \alpha\right]$ or
$\sqrt{73} {e}^{i \alpha}$, where $\tan \alpha = \frac{3}{8}$ and
$i + 4 = \left(4 + i\right) = \sqrt{{4}^{2} + {1}^{2}} \left[\cos \beta + i \sin \beta\right]$ or
$\sqrt{17} {e}^{i \beta}$, where $\tan \beta = \frac{1}{4}$
Hence $\frac{3 i + 8}{i + 4} = \frac{\sqrt{73} {e}^{i \alpha}}{\sqrt{17} {e}^{i \beta}} = \sqrt{\frac{73}{17}} {e}^{i \left(\alpha - \beta\right)} = \sqrt{\frac{73}{17}} \left(\cos \left(\alpha - \beta\right) + i \sin \left(\alpha - \beta\right)\right)$
Now, $\tan \left(\alpha - \beta\right) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$
= $\frac{\frac{3}{8} - \frac{1}{4}}{1 + \frac{3}{8} \cdot \frac{1}{4}} = \frac{\frac{1}{8}}{1 + \frac{3}{32}} = \frac{1}{8} \cdot \frac{32}{35} = \frac{4}{35}$
Hence $\frac{6 i - 4}{- 3 i - 5} = \sqrt{\frac{73}{17}} \left(\cos \rho + i \sin \rho\right)$ where $\rho = {\tan}^{- 1} \left(\frac{4}{35}\right)$
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\require{cancel}\newcommand{\nicefrac}[2]{{{}^{#1}}\!/\!{{}_{#2}}} \newcommand{\unitfrac}[3][\!\!]{#1 \,\, {{}^{#2}}\!/\!{{}_{#3}}} \newcommand{\unit}[2][\!\!]{#1 \,\, #2} \newcommand{\noalign}[1]{} \newcommand{\qed}{\qquad \Box} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}
## Section3.5Two dimensional systems and their vector fields
1 lecture, part of §6.2 in [EP], parts of §7.5 and §7.6 in [BD]
Let us take a moment to talk about constant coefficient linear homogeneous systems in the plane. Much intuition can be obtained by studying this simple case. Suppose we have a $2 \times 2$ matrix $P$ and the system
$$\begin{bmatrix} x \\ y \end{bmatrix} ' = P \begin{bmatrix} x \\ y \end{bmatrix} .\label{pln_eq}\tag{2}$$
The system is autonomous (compare this section to Section 1.6) and so we can draw a vector field (see end of Section 3.1). We will be able to visually tell what the vector field looks like and how the solutions behave, once we find the eigenvalues and eigenvectors of the matrix $P\text{.}$ For this section, we assume that $P$ has two eigenvalues and two corresponding eigenvectors.
Case 1. Suppose that the eigenvalues of $P$ are real and positive. We find two corresponding eigenvectors and plot them in the plane. For example, take the matrix $\left[ \begin{smallmatrix} 1 & 1 \\ 0 & 2 \end{smallmatrix} \right]\text{.}$ The eigenvalues are 1 and 2 and corresponding eigenvectors are $\left[ \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \right]\text{.}$ See Figure 4.5.1.
Now suppose that $x$ and $y$ are on the line determined by an eigenvector $\vec{v}$ for an eigenvalue $\lambda\text{.}$ That is, $\left[ \begin{smallmatrix} x \\ y \end{smallmatrix} \right] = a \vec{v}$ for some scalar $a\text{.}$ Then
\begin{equation*} \begin{bmatrix} x \\ y \end{bmatrix} ' = P \begin{bmatrix} x \\ y \end{bmatrix} = P ( a \vec{v} ) = a ( P \vec{v} ) = a \lambda \vec{v} . \end{equation*}
The derivative is a multiple of $\vec{v}$ and hence points along the line determined by $\vec{v}\text{.}$ As $\lambda > 0\text{,}$ the derivative points in the direction of $\vec{v}$ when $a$ is positive and in the opposite direction when $a$ is negative. Let us draw the lines determined by the eigenvectors, and let us draw arrows on the lines to indicate the directions. See Figure 4.5.2.
We fill in the rest of the arrows for the vector field and we also draw a few solutions. See Figure 4.5.3. The picture looks like a source with arrows coming out from the origin. Hence we call this type of picture a source or sometimes an unstable node.
Case 2. Suppose both eigenvalues are negative. For example, take the negation of the matrix in case 1, $\left[ \begin{smallmatrix} -1 & -1 \\ 0 & -2 \end{smallmatrix} \right]\text{.}$ The eigenvalues are $-1$ and $-2$ and corresponding eigenvectors are the same, $\left[ \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \right]\text{.}$ The calculation and the picture are almost the same. The only difference is that the eigenvalues are negative and hence all arrows are reversed. We get the picture in Figure 4.5.4. We call this kind of picture a sink or sometimes a stable node.
Case 3. Suppose one eigenvalue is positive and one is negative. For example the matrix $\left[ \begin{smallmatrix} 1 & 1 \\ 0 & -2 \end{smallmatrix} \right]\text{.}$ The eigenvalues are 1 and $-2$ and corresponding eigenvectors are $\left[ \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} 1 \\ -3 \end{smallmatrix} \right]\text{.}$ We reverse the arrows on one line (corresponding to the negative eigenvalue) and we obtain the picture in Figure 4.5.5. We call this picture a saddle point.
For the next three cases we will assume the eigenvalues are complex. In this case the eigenvectors are also complex and we cannot just plot them in the plane.
Case 4. Suppose the eigenvalues are purely imaginary. That is, suppose the eigenvalues are $\pm ib\text{.}$ For example, let $P = \left[ \begin{smallmatrix} 0 & 1 \\ -4 & 0 \end{smallmatrix} \right]\text{.}$ The eigenvalues turn out to be $\pm 2i$ and eigenvectors are $\left[ \begin{smallmatrix} 1 \\ 2i \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} 1 \\ -2i \end{smallmatrix} \right]\text{.}$ Consider the eigenvalue $2i$ and its eigenvector $\left[ \begin{smallmatrix} 1 \\ 2i \end{smallmatrix} \right]\text{.}$ The real and imaginary parts of $\vec{v} e^{i2t}$ are
\begin{equation*} \begin{aligned} \operatorname{Re} \begin{bmatrix} 1 \\ 2i \end{bmatrix} e^{i2t} & = \begin{bmatrix} \cos (2t) \\ -2 \sin (2t) \end{bmatrix} , \\ \operatorname{Im} \begin{bmatrix} 1 \\ 2i \end{bmatrix} e^{i2t} & = \begin{bmatrix} \sin (2t) \\ 2 \cos (2t) \end{bmatrix} . \end{aligned} \end{equation*}
We can take any linear combination of them to get other solutions, which one we take depends on the initial conditions. Now note that the real part is a parametric equation for an ellipse. Same with the imaginary part and in fact any linear combination of the two. This is what happens in general when the eigenvalues are purely imaginary. So when the eigenvalues are purely imaginary, we get ellipses for the solutions. This type of picture is sometimes called a center. See Figure 4.5.6.
Case 5. Now suppose the complex eigenvalues have a positive real part. That is, suppose the eigenvalues are $a \pm ib$ for some $a > 0\text{.}$ For example, let $P = \left[ \begin{smallmatrix} 1 & 1 \\ -4 & 1 \end{smallmatrix} \right]\text{.}$ The eigenvalues turn out to be $1\pm 2i$ and eigenvectors are $\left[ \begin{smallmatrix} 1 \\ 2i \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} 1 \\ -2i \end{smallmatrix} \right]\text{.}$ We take $1 + 2i$ and its eigenvector $\left[ \begin{smallmatrix} 1 \\ 2i \end{smallmatrix} \right]$ and find the real and imaginary parts of $\vec{v} e^{(1+2i)t}$ are
\begin{equation*} \begin{aligned} \operatorname{Re} \begin{bmatrix} 1 \\ 2i \end{bmatrix} e^{(1+2i)t} & = e^t \begin{bmatrix} \cos (2t) \\ -2 \sin (2t) \end{bmatrix} , \\ \operatorname{Im} \begin{bmatrix} 1 \\ 2i \end{bmatrix} e^{(1+2i)t} & = e^t \begin{bmatrix} \sin (2t) \\ 2 \cos (2t) \end{bmatrix} . \end{aligned} \end{equation*}
Note the $e^t$ in front of the solutions. This means that the solutions grow in magnitude while spinning around the origin. Hence we get a spiral source. See Figure 4.5.7.
Case 6. Finally suppose the complex eigenvalues have a negative real part. That is, suppose the eigenvalues are $-a \pm ib$ for some $a > 0\text{.}$ For example, let $P = \left[ \begin{smallmatrix} -1 & -1 \\ 4 & -1 \end{smallmatrix} \right]\text{.}$ The eigenvalues turn out to be $-1\pm 2i$ and eigenvectors are $\left[ \begin{smallmatrix} 1 \\ -2i \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} 1 \\ 2i \end{smallmatrix} \right]\text{.}$ We take $-1 - 2i$ and its eigenvector $\left[ \begin{smallmatrix} 1 \\ 2i \end{smallmatrix} \right]$ and find the real and imaginary parts of $\vec{v} e^{(-1-2i)t}$ are
\begin{equation*} \begin{aligned} \operatorname{Re} \begin{bmatrix} 1 \\ 2i \end{bmatrix} e^{(-1-2i)t} & = e^{-t} \begin{bmatrix} \cos (2t) \\ 2 \sin (2t) \end{bmatrix} , \\ \operatorname{Im} \begin{bmatrix} 1 \\ 2i \end{bmatrix} e^{(-1-2i)t} & = e^{-t} \begin{bmatrix} -\sin (2t) \\ 2 \cos (2t) \end{bmatrix} . \end{aligned} \end{equation*}
Note the $e^{-t}$ in front of the solutions. This means that the solutions shrink in magnitude while spinning around the origin. Hence we get a spiral sink. See Figure 4.5.8.
We summarize the behavior of linear homogeneous two dimensional systems in Table 4.5.9.
### Subsection3.5.1Exercises
###### Exercise3.5.1
Take the equation $m x'' + c x' + kx = 0\text{,}$ with $m > 0\text{,}$ $c \geq 0\text{,}$ $k > 0$ for the mass-spring system. a) Convert this to a system of first order equations. b) Classify for what $m, c, k$ do you get which behavior. c) Can you explain from physical intuition why you do not get all the different kinds of behavior here?
###### Exercise3.5.2
What happens in the case when $P = \left[ \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right]\text{?}$ In this case the eigenvalue is repeated and there is only one independent eigenvector. What picture does this look like?
###### Exercise3.5.3
What happens in the case when $P = \left[ \begin{smallmatrix} 1 & 1 \\ 1 & 1 \end{smallmatrix} \right]\text{?}$ Does this look like any of the pictures we have drawn?
###### Exercise3.5.4
Which behaviors are possible if $P$ is diagonal, that is $P = \left[ \begin{smallmatrix} a & 0 \\ 0 & b \end{smallmatrix} \right]\text{?}$ You can assume that $a$ and $b$ are not zero.
###### Exercise3.5.101
Describe the behavior of the following systems without solving:
a) $x' = x + y\text{,}$ $y' = x-y\text{.}$
b) $x_1' = x_1 + x_2\text{,}$ $x_2' = 2 x_2\text{.}$
c) $x_1' = -2x_2\text{,}$ $x_2' = 2 x_1\text{.}$
d) $x' = x + 3y\text{,}$ $y' = -2x-4y\text{.}$
e) $x' = x - 4y\text{,}$ $y' = -4x+y\text{.}$
a) Two eigenvalues: $\pm \sqrt{2}$ so the behavior is a saddle. b) Two eigenvalues: $1$ and $2\text{,}$ so the behavior is a source. c) Two eigenvalues: $\pm 2i\text{,}$ so the behavior is a center (ellipses). d) Two eigenvalues: $-1$ and $-2\text{,}$ so the behavior is a sink. e) Two eigenvalues: $5$ and $-3\text{,}$ so the behavior is a saddle.
###### Exercise3.5.102
Suppose that $\vec{x}\,' = A \vec{x}$ where $A$ is a 2 by 2 matrix with eigenvalues $2\pm i\text{.}$ Describe the behavior.
Take $\left[ \begin{smallmatrix} x \\ y \end{smallmatrix}\right] ' = \left[ \begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\right] \left[ \begin{smallmatrix} x \\ y \end{smallmatrix}\right]\text{.}$ Draw the vector field and describe the behavior. Is it one of the behaviors that we have seen before?
The solution will not move anywhere if $y = 0\text{.}$ When $y$ is positive, then the solution moves (with constant speed) in the positive $x$ direction. When $y$ is negative, then the solution moves (with constant speed) in the negative $x$ direction. It is not one of the behaviors we have seen.
Note that the matrix has a double eigenvalue 0 and the general solution is $x = C_1 t + C_2$ and $y = C_1\text{,}$ which agrees with the above description.
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Why the factor and remainder theorem work
So there I was, merrily teaching the factor and remainder theorems, and my student asked me one of my favourite questions: "I accept that the method works, but why does it?"
(I like that kind of question because it makes me think on my feet in class, and that makes me feel alive!)
Let's get everyone up to speed, shall we?
The factor theorem says that $(x-a)$ is a factor of a polynomial $p(x)$ if and only if $p(a)=0$. That means, if a question says "show that $x-7$ is a factor of $8x^3 -58x^2 - x + 105$", all you need to do is work out what happens to that expression when you stick a 7 in. You get $8 \times 343 - 58 \times 49 - 7 + 105$, which is $2744 - 2842 - 7 + 105 = 0$. Boom, it's a factor.
The remainder theorem is a close cousin of the factor theorem, and says that when you divide $p(x)$ by $(x-a)$, the remainder you get is $p(a)$. Notice that this fits perfectly well with the factor theorem: if the remainder when you divide by something is zero, what you divided by is a factor!
What is a remainder?
Let's think about dividing numbers, which are much less weird things to divide by. Suppose I need to split 39 items between seven people. I can go one-for-you, one-for-you, and so on - at least until I hand out the 35th item. Everyone now has five items, and there are four left over.
That means, I can write $39 = 7 \times 5 + 4$. The five items everyone got is the quotient, and the four left over is the remainder.
Any positive integer divided by any other can be expressed this way: the quotient is how many times the second number goes completely into the first, and the remainder is what's left over. More formally, $a \div b$ gives an answer of $q$ with remainder $r$ if and only if $a = b \times q + r$, with $0 \le r \lt b$.
The same goes for polynomials... with a few wrinkles
If I want to divide $p(x)$ by $(x-a)$ -- and frankly, who doesn't? -- I'm working with polynomials rather than just numbers.1
My quotient $q(x)$ and my remainder $r(x)$ will both be polynomials; the degree of $q$ will be one smaller than the degree of $p$ in this context (for example, if $p$ is a quartic, $q$ will be a cubic); $r$, meanwhile, will be of degree 0 - which is to say, a constant.
Putting it together, $\frac{p(x)}{x-a}$ gives a quotient $q(x)$ and remainder $r$ if and only if $p(x) = (x-a) q(x) + r$, where $r$ is a constant.
Do you see where this is going?
Now, if you put $x=a$ into this equation, you get $p(a) = (a-a) q(a) + r$. The first term on the right-hand side is clearly 0, so $p(a) = r$.
If you stick $a$ into your polynomial, you get the remainder out. And, as a bonus, if there's no remainder, you've found a factor.
I think that's pleasing, don't you?
Colin
Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.
1. 'Just numbers' are a subset of the polynomials, by the way, but polynomial division and integer division are different things. Roll with it. []
4 comments on “Why the factor and remainder theorem work”
• J. Hall
Thank you. This answers my question exactly.
• Jack
Dude, thank you … I’ve been searching for this answer for a long time.now I can die in peace
• Ben
But why does it work? You’ve shown how we arrive at the remainder theorem using the division algorithm. But why does dividing a polynomial by (x-a) give us the same value as a remainder, as when we substitute ‘a’ for ‘x’ in that polynomial? What is the going on when we substitute ‘a’ for ‘x’ that relates to dividing by (x-a)?
• Colin
I shall have to have a think about this and see if I can clear it up!
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