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Sunday, January 31, 2016
Math Hacks: Equivalent Fractions
Fractions are so hard! I know! My kids really struggle with fractions. When I introduce equivalent fractions, I begin by asking the kids, "What's my name?" They look at me like I'm crazy and someone finally will say, "Ms. Banks." I say, "I have other names. What do my daughters call me? What does my husband call me?" I tell the kids, "Different people call me different names. Mom, Karla, Ms. Banks. That doesn't change who I am." I explain that fractions are the same. They may have many different names, but they are the same value. We change the fraction to make it easier for us to add or subtract. Then, I show the kids a multiplication chart. A multiplication chart is actually a fraction chart showing equivalent fractions. I know that just blew your mind! Right?!
Look at this multiplication chart. Look at the first two lines. Look at the first line as the numerator in a fraction and the second line as the denominator of the same fraction. 1/2 If you follow the rows across, you will find several fractions that are equivalent fractions to 1/2. 2/4= 3/6= 4/8= 5/10 and so on. Crazy!
Here's a second example. 1/4= 2/8= 3/12= 4/16= 5/20 and so on.
When kids struggle to find equivalent fractions, I have them use their (laminated) multiplication chart. For the problem: 2/5= /15
First, have them highlight the two rows that the problem is asking about. In this case, 2 and 5. Then I have them look to see if the problem gives them the numerator or denominator. In this case, the denominator is given. Students go across the denominator line (5) and look for 15. Then they trace their finger straight up to the numerator line which is 6. 2/5=6/15
Once kids get the pattern, they can usually drop off using the multiplication chart and just multiply to find the missing numerator or denominator.
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## 3.6 The Derivative: Algebraic Viewpoint
This topic is also in Section 3.6 in Applied Calculus or Section 10.6 in Finite Mathematics and Applied Calculus
Q So far, all we have been doing is approximating the derivative of a function. Is there a way of computing it exactly?
A Let us start all by recalling what we learned in the previous section: We saw that the derivative of the function f at the point x is the slope of the tangent line through (x, f(x)), or the instantaneous rate of change of f at the point x:
Derivative Function
If f is a function, its derivative function f' is the function whose value f'(x) at x is the derivative of f at x. Its domain is the set of all x at which f is differentiable. Equivalently, f' associates to each x the slope of the tangent to the graph of the function f at x, or the instantaneous rate of change of f at x. Thus,
f'(x)=Slope of Tangent at x=
lim
h→0
f(x+h) - f(x)h
Derivative = f'(x) = Slope of Tangent at x
Q OK, so how do we compute it exactly?
A We will go through an example to see how. First notice that the derivative is the limit of a certain expression,
f(x+h) - f(x)h
called the difference quotient of the function f. The technique of computing f'(x) is roughly this: Calculate and simplify the difference quotient as much as possible. It is then a simple matter to see what happens to it as h approaches zero.
Computing the Derivative Algebraically
Step 1. Compute the difference quotient and simplify the answer as much as possible.
Example Let f(x) = 3x2 + 4x. Enter the required expressions using valid technology format.
f(x)=
f(x+h)=
f(x+h) - f(x) = Do not simplify yet.
f(x+h) - f(x) = Simplified.
f(x+h) - f(x)h
=
Step 2. Now take the limit as h → 0.
Note: If you have simplified the expression correctly, you can often take the limit by just setting h = 0. (See the discussion in Section 3.6 of Applied Calculus for more details.)
Example Continued The derivative, f'(x), is given by:
f'(x) =
f'(1) =
Now here is another one for you to try, and we return to a scenario introduced in the tutorial for Section 3.5 (or 10.5 for the combined book):
You are based in Indonesia, and you monitor the value of the US Dollar on the foreign exchange market very closely during a rather active five-day period. Suppose you find that the value of one US Dollar can be well approximated by the function
R(t) = 7500 + 500t - 100t2 rupiahs (0 ≤ t ≤ 5) The rupiah is the Indonesian currency
where t is time in days. (t = 0 represents noon on Monday.)
We would like to compute the exact (instantaneous) rate of change of the value of the US dollar at every time t. Since the instantaneous rate of change of R(t) is given by the derivative, R'(t), that is what we now compute.
R(t) =
R(t+h) =
R(t+h) - R(t) =
R(t+h) - R(t)h =
R'(t) = lim h→0 R(t+h) - R(t)h =
This quantity is measured in
At noon on Wednesday, the dollar was rising at a rate of rupiah per day.
One more, this time a little different:
Q Let f(x) = 1x .
f(x+h) is given by
1x + h
1x + 1h
1x + h
undefined
Q f(x+h) - f(x)h =
-1x(x+h)
1x(x+h)
-h2x(x+h)
1
h2
Q Finally, f'(x) =
Now try some of the exercises in Section 3.6 in Applied Calculus or Section 10.6 in Finite Mathematics and Applied Calculus.
Alternatively, press "game version" on the sidebar to go to the game version of this tutorial (it has different examples to try and is a lot of fun!).
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Last Updated: October, 2008 |
# What Are Rational Functions?
Functions with $x$ in the denominator are called rational functions. These are in fractional form and have asymptotes. Asymptotes are invisible lines in the coordinate system that the graph moves toward, but never meets. Rational functions can look like the graphs in the figure below.
Theory
### RationalFunction
A rational function is expressed in the following form:
$f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)},$
where $g\left(x\right)$ and $h\left(x\right)$ are polynomials.
You can see the graph of the following functions in the picture above
$\begin{array}{lll}\hfill f\left(x\right)=\frac{1}{x}\phantom{\rule{2em}{0ex}}g\left(x\right)=\frac{2x}{x+1}\phantom{\rule{2em}{0ex}}h\left(x\right)=\frac{{x}^{2}+6x-3}{{x}^{5}+{x}^{3}-1}& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$
$\begin{array}{lll}\hfill f\left(x\right)=\frac{1}{x}\phantom{\rule{2em}{0ex}}g\left(x\right)=\frac{2x}{x+1}\phantom{\rule{2em}{0ex}}h\left(x\right)=\frac{{x}^{2}+6x-3}{{x}^{5}+{x}^{3}-1}& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$
Hyperbolas are among the simplest rational functions. Hyperbolas are the relationship between two linear functions, $ax+b$ and $cx+d$.
Theory
### Hyperbolas
The hyperbola is an important rational function. It represents the relationship between two linear functions. The formula for a hyperbola is
$f\left(x\right)=\frac{ax+b}{cx+d}.$
A hyperbola has a vertical and a horizontal asymptote (the dotted lines).
Formula
### TheAsymptotesofaHyperbola
Vertical asymptotes
are found where the denominator is equal to zero. You can use this formula:
$x=-\frac{d}{c}$
Horizontal asymptote
is the value the graph goes towards when $x\to ±\infty$ (read as, when x approaches positive or negative infinity). You can use this formula:
$y=\frac{a}{c}$
Example 1
Consider the following hyperbolic function:
$f\left(x\right)=\frac{2x+2}{x-1}$
Find the asymptotes and the intersection between the graph and the axes.
The general formula for the hyperbola is
$f\left(x\right)=\frac{ax+b}{cx+d}$
In this exercise, $a=2$, $b=2$, $c=1$ and $d=-1$
You can find the vertical asymptote in this way:
$x=-\frac{d}{c}=-\frac{\left(-1\right)}{1}=1.$
The vertical asymptote is $x=1$.
Then you find the horizontal asymptote:
$y=\frac{a}{c}=\frac{2}{1}=2$
The horizontal asymptote is $y=2$.
The intersection with the $y$-axis is found by inputting $x=0$, because the $x$-coordinate is 0 along the entire $y$-axis. This gives
$f\left(0\right)=\frac{2\cdot 0+2}{0-1}=\frac{2}{-1}=-2$
The hyperbola intersects the $y$-axis at $\left(0,-2\right)$.
The intersection with the $x$-axis is found by putting $f\left(x\right)=0$, because the $y$-coordinate is 0 along the entire $x$-axis. This gives
$f\left(x\right)=\frac{2x+2}{x-1}=0$
It is sufficient to set the numerator equal to zero, since a fraction is zero as long as the numerator is zero. Then you get $\begin{array}{llll}\hfill 2x+2& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x& =-2\phantom{\rule{1em}{0ex}}|÷2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
The hyperbola intersects the $x$-axis at $\left(-1,0\right)$. |
# Find Derivative of root 1-x^2
The derivative of root 1-x^2 is equal to -x/√(1-x2). Here we differentiate square root of 1-x2 by chain rule and implicit differentiation method.
Note that
$\dfrac{d}{dx} \Big( \sqrt{1-x^2}\Big)=-\dfrac{x}{\sqrt{1-x^2}}$.
Let us now differentiate root 1-x2.
## Derivative of root 1-x2 by Chain Rule
Question: Find the derivative of square root 1-x2.
Let us put z = $1-x^2$.
So that $\dfrac{dz}{dx}=-2x$.
By the chain rule, the derivative of root 1-x2 is equal to
$\dfrac{d}{dx} \Big( \sqrt{1-x^2}\Big)= \dfrac{d}{dx} ( \sqrt{z})$
= $\dfrac{d}{dz} ( \sqrt{z}) \times \dfrac{dz}{dx}$
= $\dfrac{d}{dz}$ (z1/2) × (-2x) as dz/dx= -2x.
= $-2x \times \dfrac{1}{2} z^{1/2 -1}$
= $-\dfrac{x}{z^{1/2 }}$
= $-\dfrac{x}{\sqrt{1-x^2}}$ as z=1-x2.
So the derivative of square root 1-x2 by the chain rule is equal to -x/√(1-x2).
## Derivative of root 1-x2 by Implicit Differentiation
Answer: The differentiation of square root of 1-x2 is equal to -x/√(1-x2).
Explanation:
We will use the implicit differentiation method to find the derivative of square root of 1-x square. So let us put
y = $\sqrt{1-x^2}$.
Squaring both sides, we get that
y2 = 1-x2.
Differentiating both sides w.r.t x, we get that
$2y \dfrac{dy}{dx}=-2x$
⇒ $\dfrac{dy}{dx}=-\dfrac{x}{y}$
⇒ $\dfrac{dy}{dx}=-\dfrac{x}{\sqrt{1-x^2}}$, putting the value of y.
So the derivative of square root of 1-x^2 is equal to -x/√(1-x2), and this is obtained by the implicit differentiation method.
Also Read: Derivative of root x
Derivative of $\sqrt{\sin x}$
Derivative of $\sqrt{\cos x}$
Derivative of $\sqrt{\tan x}$
Derivative of $\sqrt{\sec x}$
Derivative of $\sqrt{\cot x}$
## FAQs
### Q1: What is the derivative of root 1-x2?
Answer: The derivative of root 1-x2 is equal to -x/√(1-x2). |
## Introduction: 2 Building 2 Angles
Continuing in the series of “Trigonometry –>Height and Distance–> Five types of Questions.
4th type of question is “Two buildings and Two angles”. Here are example questions (both solved in the video)
1. From the bottom of a 50m temple, the angle of elevation of the top of next building is 60. Similarly, from the bottom of the building, angle of elevation of the top of the temple is 30 degree. Find height of this building.
2. From the top of a 90m cliff, angle of depression of top and bottom of a building are 30 and 60 degrees respectively. Find height of this building (SSC CGL 2012)
## Approach
• Since two angles are given, you make two equations of TAN using Topi-triangle shortcutTM.
• Then combine the concept of rectangle (opposite sides have same length) And you’ll get the answer.
• Check the following video to see how ^this approach exactly works. And after watching the video, solve the mock questions given at the bottom of this article.
If the video is not visible, check it directly on my youtube channel youtube.com/user/TheMrunalPatel
## Mock Questions: 2 Building, 2 Angles
1. From a 60 m tall building, the angle of depression of the top and the bottom of a tower is 30 and 60° respectively. Find the height of this tower.
2. From the top half a 300 m high cliff, the angles of depression of the bottom and top of a building are 60 and 30 respectively. Find the height of this building
3. watching from a window 40m high of a multi-storey building, the angle of elevation of the top of a tower is found to be 45°. The angle of elevation of the top of the same tower from the bottom of this building is 60 degrees. Find height of tower
4. from the top of a 60m tall building, the angle of depression of top and bottom of a tree are 30 and 60° respectively. 1) find the distance between building and tree.2) find height of the tree. 3) find the difference between the heights of trees and building.
5. The top of two buildings of height 18m and 12m are connected by a cable. This cable makes an angle of 30 meter with the horizontal line passing from the top of the smaller building. Find length of the wire.
6. The angle of elevation of top of building A from bottom of building B is 50 degrees. The angle of elevation of top of building B from bottom of Building A is 70 degrees. Find out which building is taller?(hint: no calculation needed, just common sense).
1. 40
2. 200
3. 94.6
4. 1) distance= 20 root 3 OR 34.6m 2) height of tree= 40 m 3) difference between heights= 20 m.
5. 12m
6. Building B is taller. (because second angle is 70 means from the bottom of building A, you’ve to raise you head even higher to see the top of building B.)
For more [Aptitude] related content, visit the Archive on Mrunal.org/aptitude |
Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Solutions for advanced problems "A" in September, 2003
In this page only the sketch of the solutions are published; in some cases only the final results. To achieve the maximum score in the competition more detailed solutions needed.
A. 323. I is the isogonic point of a triangle ABC (the point in the interior of the triangle for which $\displaystyle \angle$AIB=$\displaystyle \angle$BIC=$\displaystyle \angle$CIA=120o). Prove that the Euler lines of the triangles ABI, BCI and CAI are concurrent.
Solution 1 (Rácz Béla András, Budapest). We shall prove that the three Euler lines pass through the centroid of triangle ABC. By the symmetry, it is sufficient to prove this for the Euler line of triangle BCI.
Figure 1
Draw a regular triangle onto side BC outside; denote its third vertex by A', its center by O1. The quadrilateral IBA'C is cyclic since BA'C$\displaystyle \angle$+CIB$\displaystyle \angle$=60o+120o=180o. Due to A'B=A'C, line A'I is the angle bisector of $\displaystyle \angle$CIB and this is the same line as AI (see Figure 1).
Denote the midpoint of BC by F, the centroid of ABC by S and the centroid of BCI by S1. Since FS:FA=FS1:FI=FO1:FA'=1:3, the points S, S1 and O1 lie on the same line. In other words, the Euler line of triangle BCI, which is O1S1, passes through points S, the centroid of ABC.
Solution 2. Let O1, O2 and O3 be the circumcenters of triangles BCI, CAI and ABI, respectively, and let M1, M2, M3 be the orthocenters. The lines O1O2, O2O3, O3O1 are the perpendicular bisectors of CI, AI and BI. Summing up the angles of the three shaded quadrilaterals in Figure 2, it can be obtained that all angles of triangle O1O2O3 are 60o; this triangle is equilateral.
Figure 2
We shall show that the Euler lines of triangles ABI, BCI and CAI pass through the center of triangle O1O2O3. Due to the symmetry, it is sufficient to deal with one of the triangles, say BCI. Since BO1=IO1=CO1, the lines O1O2 and O1O3 are angle bisectors in BO1I$\displaystyle \angle$ and IO1C$\displaystyle \angle$, and thus BO1C$\displaystyle \angle$=2O3O1O2$\displaystyle \angle$=2.60o=120o.
Figure 3
Let point U be the intersection of lines BI and CM1, and let point V be the intersection of CI and BM1. From the quadrilateral M1VIU we obtain CM1B$\displaystyle \angle$=60o. The quadrilateral M1BO1C is cyclic, since CM1B$\displaystyle \angle$+BO1C$\displaystyle \angle$=60o+120o=180o. Then, from BO1=O1C we obtain CM1O1$\displaystyle \angle$=O1M1B$\displaystyle \angle$=30o.
Finally M1O1O2$\displaystyle \angle$= O1M1B$\displaystyle \angle$ and O3O1M1$\displaystyle \angle$=CM1O1$\displaystyle \angle$, thus M1O1O2$\displaystyle \angle$=O3O1M1$\displaystyle \angle$=30o. Hence, the Euler line of triangle BCI, which is O1M1, is the angle bisector of O3O1O2=$\displaystyle \angle$ and passes through the center of triangle O1O2O3.
A. 324. Prove that if a,b,c are positive real numbers then
$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc}.$
Solution 1. The inequality can be transformed to this one:
ab(b+1)(ca-1)2+bc(c+1)(ab-1)2+ca(a+1)(bc-1)2$\displaystyle \ge$0.
Solution 2 (Tamás Birkner, Budapest). Multiply the inequality by (1+abc) and add 3 to both sides.
$\displaystyle \left({1+abc\over a(1+b)}+1\right)+ \left({1+abc\over b(1+c)}+1\right)+ \left({1+abc\over c(1+a)}+1\right) \ge6$
$\displaystyle \left({1+a\over a(1+b)}+{b(1+c)\over1+b}\right)+ \left({1+b\over b(1+c)}+{c(1+a)\over1+c}\right)+ \left({1+c\over c(1+a)}+{a(1+b)\over1+a}\right)\ge6$
$\displaystyle \left({1+a\over a(1+b)}+{a(1+b)\over1+a}\right)+ \left({1+b\over b(1+c)}+{b(1+c)\over1+b}\right)+ \left({1+c\over c(1+a)}+{c(1+a)\over1+c}\right)\ge6.$
In the last row, each parentheses have the form $\displaystyle x+{1\over x}$ and is at least 2.
Solution 3 (Dobrovolska Galyna, Kiev). Multiplying by (1+abc),
$\displaystyle {1\over a(1+b)}+{abc\over a(1+b)}+ {1\over b(1+c)}+{abc\over b(1+c)}+ {1\over c(1+a)}+{abc\over c(1+a)}\ge3$
$\displaystyle {1\over a(1+b)}+{ab\over1+a}+ {1\over b(1+c)}+{bc\over1+b}+ {1\over c(1+a)}+{ca\over1+c}\ge3$
$\displaystyle {1\over a}\cdot{1\over1+b}+b\cdot{1\over{1+{1\over a}}}+ {1\over b}\cdot{1\over1+c}+c\cdot{1\over{1+{1\over b}}}+ {1\over c}\cdot{1\over1+a}+a\cdot{1\over{1+{1\over c}}}\ge6. \eqno(1)$
Since the pairs $\displaystyle {1\over a}$, b and $\displaystyle {1\over1+{1\over a}}$, $\displaystyle {1\over1+b}$ are ordered oppositely,
$\displaystyle {1\over a}\cdot{1\over1+b}+b\cdot{1\over{1+{1\over a}}} \ge{1\over a}\cdot{1\over{1+{1\over a}}}+b\cdot{1\over1+b} ={1\over1+a}+{b\over1+b}. \eqno(2)$
Similarly,
$\displaystyle {1\over b}\cdot{1\over1+c}+c\cdot{1\over{1+{1\over b}}} \ge{1\over1+b}+{c\over1+c},\eqno(3)$
and
$\displaystyle {1\over c}\cdot{1\over1+a}+a\cdot{1\over{1+{1\over c}}} \ge{1\over1+c}+{a\over1+a}.\eqno(4)$
Summing up (2), (3) and (4), we obtain (1).
A. 325. We have selected a few 4-element subsets of an n-element set A, such that any two sets of four elements selected have at most two elements in common. Prove that there exists a subset of A that has at least $\displaystyle \root3\of{6n}$ elements and does not contain any of the selected 4-tuples as a subset.
Solution. Let N be the set of all selected 4-element subsets. Assume N$\displaystyle \ne$ and n4 (If n2 and N=, then the statement fails since .)
Call a subset of A good'' if no element of N is its subset. Trivially, the subsets of 3 or less elements, including the empty set, are all good. A good subset is called maximal if it has no good superset. There exists at least one maximal good subset, because starting from a good subset, new elements can be added one by one while it is possible.
We shall prove that each maximal good subset has more than elements. Then any maximal good subset proves the statement.
Let M be an arbitrary maximal good subset of A, and |M|=k. Clearly k3, because all 3-element subsets are good. If is an arbitrary element, then the set M{x} is not good. This means that x and some 3 elements of M form an element of N. Denote one of such triplets by h(x).
If are two different elements, then the sets H1=h(x){x} and H2=h(y){y} are elements of N. These two elements are different, because, for example, xH1 but . Then H1 and H2 have at most 2 common elements. This implies h(x)h(y).
Consider now all triplets h(x). We have n-k such triplets, they are pairwise different. The set M has 3-element subsets, thus . From this inequality we have
6nk3-3k2+8k=k3-k(3k-8)<k3 |
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Brackets and Factorising Factorising Expressions Example Question 1: Factorise the following expression: 3x + 3 3(x + 1) HCF = 3 Example Question 2: Factorise the following expression: 8x - 20 4(2x - 5) HCF = 4 Example Question 3: Factorise the following expression: 21x - 28 7(3x - 4) HCF = 7 Factorising is simply the reverse of expanding brackets. To factorise an expression completely, we take the highest common factor (HCF) of each term and place this outside the bracket. + x + = + + x - = - - x + = - - x - = +- x - = +
Questions 4 Brackets and Factorising Factorise each of the following: (a) 4(x + 2) (a) 4x + 8 (b) 6x + 27 (c) 15x - 18 (d) 40 + 16x (e) 20p - 25 (f) 8x + 12y - 20 Factorising Expressions Factorising is simply the reverse of expanding brackets. To factorise an expression completely, we take the highest common factor (HCF) of each term and place this outside the bracket. (b) 3(2x + 9) (c) 3(5x - 6) (d) 8(2x + 5) (e) 5(4p - 5) (f) 4(2x + 3y - 5)
Worksheet (a) 3(x + 8) (b) 7(x + 2) (c) 2(5x - 2) (d) 8(2x - 3) (e) 7(y + 1) (f) 9(3k - 8) 1 (a) -3(x + 8) (b) 7(-x + 2) (c) 2(-5x - 2) (d) -8(-2x - 1) (e) -7(y + 1) (f) 9(-3k - 8) 2 (a) 4(x + 5) + 3(x + 2) (b) 2(3x + 1) - (x + 3) (c) 3(5x - 2) + 2(x - 1) (d) -7(x - 2) - 2(x - 1) (e) 5(3x - 1) - 4(2x - 3) + 3(x + 4) (f) 9 + 2(5x - 3) - 7(2x + 4) + 5x 3 (a) 4x + 8 (b) 6x + 27 (c) 15x - 18 (d) 40 + 16x (e) 20p - 25 (f) 8x + 12y - 20 4 Worksheet 1 |
# Frank Learning Maths Class 4 Solutions Chapter 10
## Frank Learning Maths Class 4 Solutions Chapter 10 The Metric System
Welcome to NCTB Solutions. Here with this post we are going to help 4th class students for the Solutions of Frank Learning Maths Class 4 Math Book, Chapter 10, The Metric System. Here students can easily find step by step solutions of all the problems for The Metric System, Exercise 10.1, 10.2, 10.3, 10.4, 10.5, 10.6, 10.7 and 10.8 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
The Metric System Exercise 10.1 Solution :
Question no – (1)
Solution :
(a) Weight of a thin book Gram
(b) Medicine in a spoon L/ml
(c) Water in a water tank L/kl
(d) Length of the door metre
(e) Weight of gold ring gram
(g) Distance between two cities metre/km
Question no – (2)
Solution :
(a) 100mm < 1000cm
(b) 1000mL = 1L
(c) 1000g < 10kg
(d) 23g > 230mg
(e) 600mL < 6L
(f) 15m > 150mm
(g) 45kg = 45000g
(h) 800m > 8000cm
Question no – (3)
Solution :
(a) 55 cm + 22 cm + 10 cm = 87 cm
(b) 320 km 50 m + 410 km 10 m = 730 km 60 m
(c) 194 kg 600 g + 50 kg 320 g = 144 kg 280 g
(d) 96 m – 23 m = 73 m
(e) 520 L 110 ml – 100 L 10 ml = 420 L 100 ml
(f) 550 kg 100 g – 220 kg 50 g = 330 kg 50 g
The Metric System Exercise 10.2 Solution :
Question no – (1)
Solution :
In increasing order :
(a) Centimeter < decimeter < meter < hectometer
(b) Milliliter < deciliter < liter < kiloliter
(c) Milligram < centigram < gram < kilogram
Question no – (2)
Solution :
(a) 1 m = 1 × 100 = 100 cm
(b) 1 hm = 1/10 = 0.1 km
(c) 1 dm = 1 × 100 = 100 mm
(d) 1 km = 1 × 100 = 100 dam
(e) 1 g = 1 × 100 = 100 cg
(f) 1 dag = 1/100 = 0.01 kg
(g) 1 hg = 1 × 100000 = 100000 mg
(h) 1 quintal = 1 × 1000000 = 1000000 g
(i) 1 dL = 1 × 10 = 10 cl
(j) 1 L = 1/1000 = 0.001 kl
(k) 1L = 1 × 1000 = 1000 ml
(l) 1 dl = 1/100 = 0.01 daL
The Metric System Exercise 10.3 Solution :
Question no – (1)
Solution :
(a) 1000 cm + 90 m = 10 m + 9 0m = 100 m
(b) 87 dm = 87 × 10 = 870 cm
(c) 5600 ml = 5 L 600 ml
(d) 45000 ml = 45 L
(e) 7000 g = 7 kg
(f) 567 hm = 567 × 10000 = 5670000 cm
(g) 9 dal = 9 × 10 = 90 L
(h) 4998 cm + 2cm = 5000 cm = 50 m
(i) 56 g = 56 × 1000 = 56,000 mg
(j) 600000 cm = 6 km
(k) 2560 mg = 2560/10 = 256 cg
(l) 3000 cl = 3000/100 = 30 L
Question no – (2)
Solution :
(a) To meters
(i) 4000 + 800 m = 4800m
(ii) 5876/100 = 58.76 m
(iii) 3241/1000 = 3.241m
(iv) 1080/10 = 108 m
(b) To grams
(i) 34 × 1000 = 34,000g
(ii) 6700 + 50 = 6750g
(iii) 1300/1000 = 1.3g
(iv) 2509/100 = 25.09g
(c) To litres
(i) 7809/1000 = 7.809L
(ii) 130/10 = 13L
(iii) 9008/1000 = 9.008 L
(iv) 300 + 50 = 350L
(d) To centimetres
(i) 30/10 = 3cm
(ii) 8000 + 300 = 8300cm
(iii) 6 × 1000000 = 6000000cm
(iv) 32 × 100 = 3200 cm
(e) To kilograms
(i) 1200/1000 = 1.2kg
(ii) 54600/1000 = 5.46kg
(iii) 890/10 = 89kg
(iv) 50 × 100 = 5000kg
(f) To milliliters
(i) 13 × 1000 = 13000 ml
(ii) 78 × 10 = 780 ml
(iii) 80 × 100 = 8000 ml
(iv) 2000000 + 50000 = 2050000 ml
The Metric System Exercise 10.4 Solution :
Question no – (1)
Solution :
(a) 45 kg 456 g
+ 23 kg 608 g
————————
69kg 064g
(b) 123 km 34 m 22 cm
+ 89 km 405 m 68 cm
————————————-
212 km 439 m 90 cm
(c) 19 km 36 m 25 cm
+ 24 km 22 m 48 cm
————————————–
43 km 68 m 73 m
(d) 13km 670m
45km 908m
+ 29km 770m
————————
89 km 348m
(e) 45 kg 43 g
18 kg 390 g
+ 55 kg 900 g
————————
119kg 333g
(f) 13 kg 456 hg 13 mg
+ 34 kg 233 hg 78 mg
————————
47 kg 689 hg 91 mg
(g) 41 L 543 ml
8 l 564 ml
+ 19 L 900 ml
————————
70 L 7 ml
(h) 14 L 708 ml
+ 34 l 222 ml
————————
48 L 930 ml
(i) 3784 L 698 mL
+ 1223 L 798 ml
————————
5008L 496
(j) 272 L 500 ml
+ 725 L
————————
997L 500ml
Question no – (2)
Solution :
(a) 678 m 10 cm – 537 m 63 cm
= 140 m 24 cm
(b) 15km – 7380 m
= 15000 – 7380
= 7620 m
(c) 99 L – 47 L 788 ml
= 51 L 212 ml
(d) 42 km 900 m – 31km 706m
= 11km 194m
(e) 15km – 6km 678 m
= 8km 322 m
(f) 42L 789 ml – 13L 230 ml
= 29L 559 ml
(g) 45 kg 120 g – 23 kg 319 g
= 21 kg 801 g
(h) 76 km 104 m 66 cm – 52 km 300 m 45 cm
= 23 km 804 m 21 cm
The Metric System Exercise 10.5 Solution :
Question no – (1)
Solution :
Total length of the stick
= 85 cm + 39 cm
= 124 cm
Question no – (2)
Solution :
Total length
= 4 m 87 cm + 5 m 13 cm
= 100 cm
= 10 m
1/2 of 10m
= 5 m
5 m remained unpainted
Question no – (3)
Solution :
Total distance she need to walk –
4 km 456 m
2 km 34 m
+ 1 km 350 m
————————
7 km 840 m
Question no – (4)
Solution :
He sell in all –
46L 450ml
35L 500 ml
+ 56L 706 ml
————————
138L 656 ml
Question no – (5)
Solution :
Total weight in the train –
205kg 450g
180kg 600g
+ 325kg 5g
————————
711kg 55g
Question no – (6)
Solution :
Shampoo left in bottle
= (250 – 89) ml
= 161 ml
Question no – (7)
Solution :
Wire left for underground wiring
= (500 – 298) m
= 202 m
Question no – (8)
Solution :
Rope left –
345 m 66 cm
234 m 15 cm
————————
111 m 51 cm
Question no – (9)
Solution :
Total weight of them –
25 kg 725 g
25 kg 725 g
8 kg 360 g
————————
59 kg 810 g
Question no – (10)
Solution :
12275 ml = 12L 275 ml
water remains = 12L 275ml – 8L 35ml
= 4L 240ml
Question no – (11)
Solution :
Used oil
= (456 + 123)
= 579ml
Remained oil in the can
= (900 – 579)
= 321 ml
Question no – (12)
Solution :
Flour need to make food
12kg 350g
+ 4kg 100g
————————
16kg 450g
Flour need for making buns
20 kg – 16 kg 450 g
= 3 kg 550 g
The Metric System Exercise 10.6 Solution :
Question no – (1)
Solution :
(a) 325 × 3
= 75 cm
= 9 m 75 cm
(b) 104 × 5
= 520 L
(c) 2900 × 2
= 5800 g
= 5 kg 800 g
(d) 3560 × 5
= 17800 g
= 17 kg 800 g
(e) 7150 × 3
= 21450 ml
= 21 L 450 ml
(f) 4840 × 4
= 192160 ml
= 192 L 160 ml
(g) 197 × 2
= 394 mm
= 39 cm 4 mm
(h) 6250 × 7
= 43750 g
= 43 kg 750 g
(i) 13208 × 3
= 39624 m
= 39 km 624 m
The Metric System Exercise 10.7 Solution :
Question no – (1)
Solution :
(a) 840 ÷ 2
= 420 cm
= 4 m 20 cm
(b) 3500 ÷ 5
= 700 m
(c) 18900 ÷ 2
= 9450 m
= 9 km 450 km
(d) 2440 ÷ 2
= 1220 g
= 1 kg 220 g
(e) 1825 ÷ 5
= 365 g
(f) 1500 ÷ 3
= 500 mg
(g) 9369 ÷ 3
= 3123 ml
= 3 L 123 mL
(h) 3408 ÷ 4
= 852 L
(i) 4512 ÷ 8
= 564 mL
The Metric System Exercise 10.8 Solution :
Question no – (1)
Solution :
She used = 3 × 106 mm
= 318mm
= 31cm 8mm
Question no – (2)
Solution :
It is for a to and from trip –
438 km 500 m
+ 438 km 500 m
—————————–
877 km
Question no – (3)
Solution :
In June there are 30 days
So, he consume in the month of june
= 50 × 30 mg
= 1500 mg
= 1g 500mg
∴ He will 1g 500mg butter consume in the month of June.
Question no – (4)
Solution :
Water needed to fill 8 such glasses
= 8 × 350
= 2800ml
= 2L 800 ml
Question no – (5)
Solution :
1 year = 12 months
So, they consume in a year
= 12 × 4500g
= 54000g
= 54kg
Question no – (6)
Solution :
Milk can be packed in an hour –
= 8200/4
= 2050 ml
= 2 L 50 ml
Milk can be packed in 3 hour –
= 2050 × 3
= 6150 ml
= 6 L 150 ml
Question no – (7)
Solution :
They each cover –
= 4400/4
= 1100 m
= 1 km 100 m
Question no – (8)
Solution :
5kg = 5000g
Each family get = 5000/4
= 1250 g
= 1 kg 250 g
Question no – (9)
Solution :
5 L = 5000 ml
No. Bottles need to be filled
= 5000/250
= 20
Question no – (10)
Solution :
No. of drums will be filled
= 100/10
= 10
Question no – (11)
Solution :
8 m 64 cm = 864 cm
Length of each strip = 864/4
= 216 cm
= 2 m 16 cm
Question no – (12)
Solution :
20 kg 250 g = 26,250 g
1 week = 7 day
Flour is used in a day
= 26250/7
= 3750g
= 3kg 750g
Question no – (13)
Solution :
21 km 200 m = 21200m
He walk in a day = 21200/4
= 5300 m
= 5 km 3m
He walk in 5 day –
= 5 × 5300 m
= 26500 m
= 26 km 500 m
Next Chapter Solution :
Updated: June 13, 2023 — 5:44 am |
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Lesson 3: Concave Mirrors
The Mirror Equation
Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a concave mirror. The use of these diagrams were demonstrated earlier in Lesson 3. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and object size. To obtain this type of numerical information, it is necessary to use the Mirror Equation and the Magnification Equation. The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:
These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.
As a demonstration of the effectiveness of the mirror equation and magnification equation, consider the following example problem and its solution.
Example Problem #1
A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size.
Like all problems in physics, begin by the identification of the known information.
ho = 4.0 cm do = 45.7 cm f = 15.2 cm
Next identify the unknown quantities which you wish to solve for.
di = ??? hi = ???
To determine the image distance, the mirror equation must be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.
1/f = 1/do + 1/di
1/(15.2 cm) = 1/(45.7 cm) + 1/di
0.0658 cm-1 = 0.0219 cm-1 + 1/di
0.0439 cm-1 = 1/di
di = 22.8 cm
The numerical values in the solution above were rounded when written down, yet un-rounded numbers were used in all calculations. The final answer is rounded to the third significant digit.
To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.
hi/ho = - di/do
hi /(4.0 cm) = - (22.8 cm)/(45.7 cm)
hi = - (4.0 cm) • (22.8 cm)/(45.7 cm)
hi = -1.99 cm
The negative values for image height indicate that the image is an inverted image. As is often the case in physics, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image height, a negative value always indicates an inverted image.
From the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 45.7 cm from a concave mirror having a focal length of 15.2 cm, then the image will be inverted, 1.99-cm tall and located 22.8 cm from the mirror. The results of this calculation agree with the principles discussed earlier in this lesson. In this case, the object is located beyond the center of curvature (which would be two focal lengths from the mirror), and the image is located between the center of curvature and the focal point. This falls into the category of Case 1 : The object is located beyond C.
Now lets try a second example problem:
Example Problem #2
A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. (NOTE: this is the same object and the same mirror, only this time the object is placed closer to the mirror.) Determine the image distance and the image size.
Again, begin by the identification of the known information.
ho = 4.0 cm do = 8.3 cm f = 15.2 cm
Next identify the unknown quantities which you wish to solve for.
di = ??? hi = ???
To determine the image distance, the mirror equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.
1/f = 1/do + 1/di
1/(15.2 cm) = 1/(8.3 cm) + 1/di
0.0658 cm-1 = 0.120 cm-1 + 1/di
-0.0547 cm-1 = 1/di
di = -18.3 cm
The numerical values in the solution above were rounded when written down, yet un-rounded numbers were used in all calculations. The final answer is rounded to the third significant digit.
To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.
hi/ho = - di/do
hi /(4.0 cm) = - (-18.2 cm)/(8.3 cm)
hi = - (4.0 cm) • (-18.2 cm)/(8.3 cm)
hi = 8.8 cm
The negative value for image distance indicates that the image is a virtual image located behind the mirror. Again, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always means behind the mirror. Note also that the image height is a positive value, meaning an upright image. Any image which is upright and located behind the mirror is considered to be a virtual image.
From the calculations in the second example problem it can be concluded that if a 4.0-cm tall object is placed 8.3 cm from a concave mirror having a focal length of 15.2 cm, then the image will be magnified, upright, 8.8-cm tall and located 18.3 cm behind the mirror. The results of this calculation agree with the principles discussed earlier in this lesson. In this case, the object is located in front of the focal point (i.e., the object distance is less than the focal length), and the image is located behind the mirror. This falls into the category of Case 5: The object is located in front of F.
The +/- Sign Conventions
The sign conventions for the given quantities in the mirror equation and magnification equations are as follows:
• f is + if the mirror is a concave mirror
• f is - if the mirror is a convex mirror
• di is + if the image is a real image and located on the object's side of the mirror.
• di is - if the image is a virtual image and located behind the mirror.
• hi is + if the image is an upright image (and therefore, also virtual)
• hi is - if the image an inverted image (and therefore, also real)
Like many mathematical problems in physics, the skill is only acquired through much personal practice. Perhaps you would like to take some time to try the problems in the Check Your Understanding section below.
1. Determine the image distance and image height for a 5.00-cm tall object placed 45.0 cm from a concave mirror having a focal length of 15.0 cm.
2. Determine the image distance and image height for a 5.00-cm tall object placed 30.0 cm from a concave mirror having a focal length of 15.0 cm.
3. Determine the image distance and image height for a 5.00-cm tall object placed 20.0 cm from a concave mirror having a focal length of 15.0 cm.
4. Determine the image distance and image height for a 5.00-cm tall object placed 10.0 cm from a concave mirror having a focal length of 15.0 cm.
5. A magnified, inverted image is located a distance of 32.0 cm from a concave mirror with a focal length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual.
ZINGER: 6. An inverted image is magnified by 2 when the object is placed 22 cm in front of a concave mirror. Determine the image distance and the focal length of the mirror. |
# Question #b1688
##### 1 Answer
Apr 27, 2015
The answer is: $\left(3 , - 2 , 4\right)$.
First of all let's write the line in parametric form.
If we put $x = t$, than:
$\frac{t}{3} = \frac{y + 7}{5} \Rightarrow y = - 7 + \frac{5}{3} t$
and:
$\frac{t}{3} = \frac{z - 2}{2} \Rightarrow z = 2 + \frac{2}{3} t$.
So the line is:
$x = t$
$y = - 7 + \frac{5}{3} t$
$z = 2 + \frac{2}{3} t$
the direction of this line is $\left(1 , \frac{5}{3} , \frac{2}{3}\right)$.
To write a plane, given a point $P \left({x}_{p} , {y}_{p} , {z}_{p}\right)$ and the direction $\vec{v} \left(a , b , c\right)$ of a perpendicular at the plane itself, we can use this formula:
$a \left(x - {x}_{p}\right) + b \left(y - {y}_{p}\right) + c \left(z - {z}_{p}\right) = 0$,
so:
$1 \left(x - 2\right) + \frac{5}{3} \left(y + 1\right) + \frac{2}{3} \left(z - 3\right) = 0 \Rightarrow$
$3 x - 6 + 5 y + 5 + 2 z - 6 = 0 \Rightarrow 3 x + 5 y + 2 z + 7 = 0$.
Now, to find the point, it is necessary to make a system with the plane and the given line:
$3 x + 5 y + 2 z + 7 = 0$
$x = t$
$y = - 7 + \frac{5}{3} t$
$z = 2 + \frac{2}{3} t$
$\Rightarrow 3 t - 35 + \frac{25}{3} t + 4 + \frac{4}{3} t - 7 = 0 \Rightarrow \frac{38}{3} t = 38 \Rightarrow t = 3$.
Now let's substitue this value of $t$ in the three equation of the line:
$x = 3$
$y = - 7 + \frac{5}{3} \cdot 3 = - 2$
$z = 2 + \frac{2}{3} \cdot 3 = 4$.
The point is: $\left(3 , - 2 , 4\right)$. |
Year 10 Interactive Maths - Second Edition
## Gradient of a Straight Line
The gradient of a straight line is the rate at which the line rises (or falls) vertically for every unit across to the right. That is:
###### Note:
The gradient of a straight line is denoted by m where:
#### Example 3
Find the gradient of the straight line joining the points P(4, 5) and Q(4, 17).
##### Solution:
So, the gradient of the line PQ is 1.5.
###### Note:
If the gradient of a line is positive, then the line slopes upward as the value of x increases.
#### Example 4
Find the gradient of the straight line joining the points A(6, 0) and B(0, 3).
##### Solution:
###### Note:
If the gradient of a line is negative, then the line slopes downward as the value of x increases.
Gradients are an important part of life. The roof of a house is built with a gradient to enable rain water to run down the roof. An aeroplane ascends at a particular gradient after take off, flies at a different gradient and descends at another gradient to safely land. Tennis courts, roads, football and cricket grounds are made with a gradient to assist drainage.
#### Example 5
A horse gallops for 20 minutes and covers a distance of 15 km, as shown in the diagram.
Find the gradient of the line and describe its meaning.
##### Solution:
In the above example, we notice that the gradient of the distance-time graph gives the speed (in kilometres per minute); and the distance covered by the horse can be represented by the equation:
#### Example 6
The cost of transporting documents by courier is given by the line segment drawn in the diagram. Find the gradient of the line segment; and describe its meaning.
##### Solution:
So, the gradient of the line is 3. This means that the cost of transporting documents is \$3 per km plus a fixed charge of \$5, i.e. it costs \$5 for the courier to arrive and \$3 for every kilometre travelled to deliver the documents.
###### An alternative definition of the gradient is:
The gradient is the rate of change of one variable with respect to another. |
Categories
## Which statement is true? A sample statistic can never be equal to the population parameter. Point estimates are defined for both samples and populations. To calculate a sample proportion, we need to know the population size. To calculate a point estimate, we need to know the sample size
Option C is true. To calculate a sample proportion, we need to know the population size.
Step-by-step explanation:
To calculate a sample proportion, we need to know the population size. – This is the true statement.
A sample statistic can never be equal to the population parameter. This is false. A statistic is a feature of sample and a parameter is a feature of population. Every sample of the selected size has an equal chance of being used.
Point estimates are defined for both samples and populations. This is false. A point estimate is used to estimate the population parameter.
To calculate a point estimate, we need to know the sample size. This is false. Population is needed.
Categories
## The brain mass of a fetus can be estimated using the total mass of the fetus by the function b equals 0.22 m superscript 0.87b=0.22m0.87?, where m is the mass of the fetus? (in grams) and b is the brain mass? (in grams). suppose the brain mass of a 2323?-g fetus is changing at a rate of 0.240.24 g per day. use this to estimate the rate of change of the total mass of the? fetus, startfraction dm over dt endfraction dm dt.
We are given
b = 0.22 m^0.87
db/dt = 0.24 g/day at b = 23 g
where
b is the brain mass of the fetus
m is the total mass of the fetus
db/dt is the rate of change of the brain mass of the fetus
We are asked to get the rate of change of the total mass of the fetus
First, we take the first derivative of the given equation with respect to time
db/dt = 0.22 (0.87) m^(0.87-1) dm/dt
Next, simplify and substitute the given values. So,
0.24 g/day = 0.22(0.87)(23)^(0.87 -1 ) dm/dt
Solving for dm/dt
dm/dt = 1.88 g/day
The rate of change of mass of the fetus is 1.88 g/day.
Categories
## LAW ENFORCEMENT: A police accident investigator can use the formula S=25L‾‾‾√S=25L to estimate the speed s of a car in miles per hour based on the length l in feet of the skid marks it left. How fast was a car traveling that left skid marks 109 feet long?
Option C – BD=76 cm
Step-by-step explanation:
Given : You are designing a diamond-shaped kite. you know that AD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm.
To find : How long BD should it be?
Solution :
First we draw a rough diagram.
The given sides were AD = 44.8 cm, DC = 72 cm and AC = 84.8 cm.
According to properties of kite
Two disjoint pairs of consecutive sides are congruent.
DC=BC=72 cm
The diagonals are perpendicular.
So, AC ⊥ BD
Let O be the point where diagonal intersect let let the partition be x and y.
AC= AO+OC
AC= …….[1]
Perpendicular bisect the diagonal BD into equal parts let it be z.
BD=BO+OD
BD=z+z
Applying Pythagorean theorem in ΔAOD
where H=AD=44.8 ,P= AO=x , B=OD=z
………[2]
Applying Pythagorean theorem in ΔCOD
where H=DC=72 ,P= OC=y , B=OD=z
…………[3]
Subtract [2] and [3]
……….[4]
Add equation [1] and [4], to get values of x and y
Substitute x in [1]
Substitute value of x in equation [2], to get z
We know, BD=z+z
BD= 38.06+38.06
BD= 76.12
Nearest to whole number BD=76 cm
Therefore, Option c – BD=76 cm is correct.
Categories
## A country’s population in 1992 was 72 million. In 1998 it was 76 million. Estimate the population in 2012 using the exponential growth formula. Round your answer to the nearest million.P=Ae^kt
You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.
So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:
C (m,n) = m! / (n! * (m -n)! )
=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =
= 47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.
Categories
## Table to estimate how many pounds of dog food his dog eat in one year which of the following would be the best strategy
Table to estimate how many pounds of dog food his dog eat in one year which of the following would be the best strategy
Categories
## What would be your estimate of the age of the universe if you measured a value for hubble’s constant of h0 = 30 km/s/mly ? you can assume that the expansion rate has remained unchanged during the history of the universe. express your answer using two significant figures. yr?
a) The time the police officer required to reach the motorist was 15 s.
b) The speed of the officer at the moment she overtakes the motorist is 30 m/s
c) The total distance traveled by the officer was 225 m.
Explanation:
The equations for the position and velocity of an object moving in a straight line are as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:
x motorist = x officer
Using the equation for the position:
x motirist = x0 + v · t (since a = 0).
x officer = x0 + v0 · t + 1/2 · a · t²
Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:
x motorist = x officer
x0 + v · t = x0 + v0 · t + 1/2 · a · t² (the officer starts form rest, then, v0 = 0)
v · t = 1/2 · a · t²
Solving for t:
2 v/a = t
t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s
The time the police officer required to reach the motorist was 15 s.
b) Now, we can calculate the speed of the officer using the time calculated in a) and the equation for velocity:
v = v0 + a · t
v = 0 m/s + 2.00 m/s² · 15 s
v = 30 m/s
The speed of the officer at the moment she overtakes the motorist is 30 m/s
c) Using the equation for the position, we can find the traveled distance in 15 s:
x = x0 + v0 · t + 1/2 · a · t²
x = 1/2 · 2.00 m/s² · (15s)² = 225 m
Categories
## In a genetics experiment on? peas, one sample of offspring contained 402402 green peas and 3838 yellow peas. based on those? results, estimate the probability of getting an offspring pea that is green. is the result reasonably close to the value of 3 divided by 43/4 that was? expected?
The point that lie on the line is:
B. (1,1)
## Step-by-step explanation:
We are given that a line passes through the point (0,-1) and has a slope of 2.
We know that the equation of a line passing through (a,b) and having slope m is given by:
Here we have: (a,b)=(0,-1) and m=2
This means that the equation of line is:
Now we will check which option is true.
A)
(2,1)
when x=2
we have:
Hence, option: A is incorrect.
B)
(1,1)
when x=1
we have:
Hence, option: B is correct.
C)
(0,1)
when x=0
we have:
Hence, option: C is incorrect.
Categories
## Lucy wants to divide 18.6 by 1.2 in order to estimate the quotient which is the best expression to use ? A:10÷1 B:10÷2 C:20÷1 D:20÷2
Option C is the best expression.
Step-by-step explanation:
We are asked to find 18.6 by 1.2.
Option A
10÷1 = 10
Difference = 15.5 – 10 = 5.5
Option B
10÷2 = 5
Difference = 15.5 – 5 = 10.5
Option C
20÷1 = 20
Difference = 20 – 15.5 = 4.5
Option D
20÷2 = 10
Difference = 15.5 – 10 = 5.5
So, the difference is minimum for option C
Option C is the best expression.
Categories
## A jet flies at a rate of 1.3 x 10^ 6 feet per hour. Written in scientific notation, which is the best estimate of how many feet the jet will travel in 2.8 x 10^ 3 hours?
A jet flies at a rate of 1.3 x 10^ 6 feet per hour. Written in scientific notation, which is the best estimate of how many feet the jet will travel in 2.8 x 10^ 3 hours?
Categories
## 7. Researchers captured and tagged 45 deer before releasing them back into their habitat. Several months later the researchers returned and captured 30 deer, 8 of which had tags. Use this information to estimate the total deer population for this area.
You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.
So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:
C (m,n) = m! / (n! * (m -n)! )
=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =
= 47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.
Categories
## It is desired to estimate the mean gpa of each undergraduate class at a large university. assume that the variance of the gpas is 1.44. how large a sample is necessary to estimate the mean gpa within 0.25 at the 99% confidence level
Mahtematical and statistical reasoning makes it evident that there is a mistake in the writing of the question and the digits were duplicated. So the right question is:
“You need to have a password with 5 letters followed by 3 odd digits between 0 and 9, inclusive. If the characters and digits cannot be used more than once, how many choices do you have for your passwor?”
The solution is:
5 letters from 26 with no repetition => 26*25*24*23*22 different choices.
odd digits are 1, 3, 5, 7 and 9 => 5 different digits to choose
3 digits from 5 with no repetition => 5*4*3 = 60
Then, the total number of choices is: 26*25*24*23*22*60 = 473,616,000
Categories
## A tour boat operator wants to know the average age of all the people in her tour group. She randomly selects 8 people in the group and asks them for their ages. Their responses are 19, 27, 25, 21, 44, 22, 45, and 34. What is the best estimate of the average age of all the people in the tour group? Round to the nearest whole number.
You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.
So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:
C (m,n) = m! / (n! * (m -n)! )
=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =
= 47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.
Categories
## In a casino, the chance of winning a certain game is 14.5%. If you play the game 20 times, what’s your best estimate for the number of times you would win? Round off your answer to the nearest whole number.
You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.
So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:
C (m,n) = m! / (n! * (m -n)! )
=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =
= 47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.
Categories
## Which is the most reasonable estimate for 30% of 72?
Which is the most reasonable estimate for 30% of 72?
Categories
## Joe runs 8.75 times around a track in 1,125.803 seconds. If one lap around the track is 420.3 meters, which is the best estimate of the runner's average speed in meters per second (m/s)?
Joe runs 8.75 times around a track in 1,125.803 seconds. If one lap around the track is 420.3 meters, which is the best estimate of the runner’s average speed in meters per second (m/s)?
Categories
## Estimate the length of the radius if the area of the circle is 54 ft2.
x > 2
Step-by-step explanation:
Here, x represents the length of the poster,
Since, width of the poster is 2 inches shorter than its length,
Width of the poster = ( x – 2 ) inches,
Thus, the area of the poster = length × width,
A = x (x-2)
We know that,
Area of a poster must be greater than zero,
⇒ A > 0,
⇒ x(x-2) > 0
⇒ x > 0 and x-2 > 0
If x – 2 > 0 ⇒ x > 2,
That is, the length of the poster must be greater than 2 inches.
Categories
## The director of a customer service center wants to estimate the mean number of customer calls the center handles each day, so he randomly samples 26 different days and records the number of calls. the sample yields a mean of 258.4 calls with a standard deviation of 32.7 calls per day. the 95% confidence interval for the mean number of calls per day has an upper bound of ________. (round your answer to 1 decimal place.)
First we calculate for the total number of possibilities
(permutation) to select 4 disks from the container:
Total number of possibilities = 10 * 9 * 8 * 7
Total number of possibilities = 5040
Now let us find the 4 disks that will result in a range of 7.
Range = highest number – lowest number
The pair of highest and lowest number that will result in
range of 7 is: (1 & 8), (2 & 9), (3 & 10)
As a basis of calculation, let us use the pair 1 & 8.
There are four possible ways to select 1 and three for 8.
Arrangements of maximum and minimum pair = 4 * 3
Arrangements of maximum and minimum pair=12
Now we need to calculate for the remaining 2 disk. There
are 6 numbers between 1 & 8. The total possibilities for selecting 2 disk
from the remaining 6 is:
Possibilities of selecting 2 disk from remaining 6 = 6 *
5
Possibilities of selecting 2 disk from remaining 6 = 30
Therefore, the total possibility to get a range of 7 from
a pair of 1 & 8 is:
Total possibility for a pair = 12 * 30
Total possibility for a pair = 360
Since there are a total of three pairs (1 & 8), (2
& 9), (3 & 10):
Total possibilities of the 3 pairs = 360 * 3
Total possibilities of the 3 pairs = 1080
Therefore:
Probability = Total possibilities of the 3 pairs / Total
number of possibilities
Probability
= 1080 / 5040 = 3 / 14 (FINAL
Categories
## A scientist counts 25 bacteria present in a culture and finds that the number of bacteria triples each hour. The function y = 25 ∙ 3x models the number of bacteria after x hours. Estimate when there will be about 1170 bacteria in the culture
A scientist counts 25 bacteria present in a culture and finds that the number of bacteria triples each hour. The function y = 25 ∙ 3x models the number of bacteria after x hours. Estimate when there will be about 1170 bacteria in the culture
Categories
## Three people are running for class president. From a poll that was taken, the probability that Candidate A will win is estimated to be 0.3. The probability that Candidate B will win is estimated to be 0.55. Given a reasonable estimate of the probability that Candidate C will win.
For a normally distributed data, with mean, μ, and standard deviation, σ, the probability that a randomly selected data, X, is less than a given value, x, is given by
and the probability that a randomly selected data, X, is greater than a given value, x, is given by
Given that the length of a social media interaction is normally distributed with a mean of 3 minutes and a standard deviation of 0.4 minutes, the probability that an interaction lasts longer than 4 minutes is given by
We use the normal distribution table or calculator to evalute that P(X < 2.5) = 0.99379
Therefore, the probability that an interaction lasts longer than 4 minutes = 1 – 0.99379 = 0.00621
[the last option]
Categories
## A bakery can make 175 pastries each day. The bakery has been sold out of 3 days in a row. Determine of an estimate or exact answer is needed. then determine how many pasties were sold during the 3 days
The correct option is D. -17.5 degrees
Step-by-step explanation:
The daytime temperature in Apple Valley is falling by 2.5 degrees each day
Therefore, Change in the temperature each day = -2.5 degrees
Now, we need to find the net change in the daily temperature after one calendar week
Now, number of days in one calendar week = 7
So, Net change in the daily temperature after one calendar week = Daily change × Number of days in one calendar week
⇒ Net change in the daily temperature after one calendar week = -2.5 × 7
⇒ Net change in the daily temperature after one calendar week = -17.5 degrees
Therefore, The correct option is D. -17.5 degrees
Categories
## Round each number to the nearest hundredth. What is the best estimate for the sum of 34.719 51.626? A. 86.35 B. 86.34 C. 86.33 D. 85.35
Add them first. 34.719 + 51.626 = 86.345.
You can find where the hundredth place is. It’s always two places from the decimal. So in 86.345, the hundredth is 4.
The answer is A. 86.35. It’s because the number that is beside 4 is 5. If the number is 4 and below, you don’t round. If the number is 5 and above, you round it.
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Posted in Mathematics
Show that, except for 2 and 5, every prime number can be expressed as 10k+1, 10k+3 10k+7, 10k+9, where k is an interger
10(6)+1=61
10(15)+3=153
10(4)+7=
47
1
0(9)+9=99
this is true because when you multiply a number by 10 it will be an even number(2,4,6,8,etc…)…then when you add an odd number to that the number will turn odd(153,47,99,etc..)… and when you multiple integers they will always multiply to an even number
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## Use our writing service to score better and meet your deadlines
No matter what kind of academic paper you need, it is simple and secure to hire a writer for a price you can afford at StudyHawks. Save more time for yourself.
## Use our writing service to score better and meet your deadlines
No matter what kind of academic paper you need, it is simple and secure to hire a writer for a price you can afford at StudyHawks. Save more time for yourself.
## Use our writing service to score better and meet your deadlines
No matter what kind of academic paper you need, it is simple and secure to hire a writer for a price you can afford at StudyHawks. Save more time for yourself.
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Categories
## The rate of the number of bags mishandled by the airlines fell to 3.74 bags per 1000 passengers in 1011. at this rate, estimate how many bags would be mishandled for 1750 passengers ?
The rate of the number of bags mishandled by the airlines fell to 3.74 bags per 1000 passengers in 1011. at this rate, estimate how many bags would be mishandled for 1750 passengers ?
Categories
## Carl. put \$1.10 in his piggy bank every day in the month of July. His total savings was \$55 at the end of June. hat is the best estimate for Carl's savings at the end of July
Carl. put \$1.10 in his piggy bank every day in the month of July. His total savings was \$55 at the end of June. hat is the best estimate for Carl’s savings at the end of July
Categories
## Sophia has \$20. she buys a hair band for \$3.99, gum for \$1.29, and a brush for \$6.75. not including tax estimate how much change she should receive????? PLEASE HELP
28.47º of latitude, 7.35º of longitude
Step-by-step explanation:
The latitude and longitude are simple imaginary lines (horizontal for latitude and vertical for longitude) that divide the Earth from an angle point of view. When used as a combination of both, you can get an exact location in any part of the world. For the 2 references, there a 0º divisions that make one side positive angles and negative on the other side (in the case of latitude, positive angles are above the equator line and for longitude positive angles are from the right of the Greenwich Meridian).
With the above mentioned we can observe that Hong Kong and Jakarta are away 22.27º and 6.2º respectively from the equator. So, the total degrees in latitude wil simply be the sum of the 2 absolute values (222.7º+6.2º) giving as a result of 28.47º.
On the other hand, we have in longitude that both cities are several degrees away from the same reference, and because they are from the right side of the Greenwich Meridian, the net change will be the substraction of 114.5-106.8º, giving as a result a total of 7.35º.
Categories |
How Are Ratios Figured?
# How Are Ratios Figured?
Ratios are figured with addition and subtraction and represent a comparison between numbers or a group of numbers. They are often represented as fractions and can be simplified as such using division.
There are three ways ratios can be expressed: x:y, x/y, or x to y. They all mean the same thing: for x, there is y. An example of this could be: there are 6 boys and 7 girls in math class. The ratio is 6:7, 6/7 or 6 to 7.
With ratios, the first figure is always the antecedent and the second is the consequent. Sometimes, ratios have to do with how the antecedent or consequent relate to the entire amount of the grouping. Using the example above, 6 boys + 7 girls = 13 kids in math class, so it could be said that 6/13 kids in math class are boys and 7/13 kids in math class are girls.
When added together, the two fractions from the classroom example equal 1. They equal 1 because 1 represents the whole grouping comprised of the antecedent and the consequent. When all six boys and seven girls are together in the classroom, it equals the whole of the math class.
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# 2.3 Descriptive statistics: histogram (Page 2/4)
Page 2 / 4
• 68
• 68
• 69
• 69
• 69
• 69
• 69
• 69
• 69
• 69
• 69
• 69
• 69.5
• 69.5
• 69.5
• 69.5
• 69.5
• 70
• 70
• 70
• 70
• 70
• 70
• 70.5
• 70.5
• 70.5
• 71
• 71
• 71
• 72
• 72
• 72
• 72.5
• 72.5
• 73
• 73.5
• 74
The smallest data value is 60. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point.
60 - 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is, then, 59.95.
The largest value is 74. 74+ 0.05 = 74.05 is the ending value.
Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose 8 bars.
$\frac{74.05-59.95}{8}=1.76$
We will round up to 2 and make each bar or class interval 2 units wide. Rounding up to 2 is one way to prevent a value from falling on a boundary. Rounding to the next number is necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work.
The boundaries are:
• 59.95
• 59.95 + 2 = 61.95
• 61.95 + 2 = 63.95
• 63.95 + 2 = 65.95
• 65.95 + 2 = 67.95
• 67.95 + 2 = 69.95
• 69.95 + 2 = 71.95
• 71.95 + 2 = 73.95
• 73.95 + 2 = 75.95
The heights 60 through 61.5 inches are in the interval 59.95 - 61.95. The heights that are 63.5 are in the interval 61.95 - 63.95. The heights that are 64 through 64.5 are in the interval 63.95 - 65.95. The heights 66 through 67.5 are in the interval 65.95 - 67.95. The heights 68 through 69.5 are in the interval 67.95 - 69.95. The heights 70 through 71 are in the interval 69.95 - 71.95. The heights 72 through 73.5 are in the interval 71.95 - 73.95. The height 74 is in the interval 73.95 - 75.95.
The following histogram displays the heights on the x-axis and relative frequency on the y-axis.
The following data are the number of books bought by 50 part-time college students at ABC College. The number of books is discrete data since books are counted.
• 1
• 1
• 1
• 1
• 1
• 1
• 1
• 1
• 1
• 1
• 1
• 2
• 2
• 2
• 2
• 2
• 2
• 2
• 2
• 2
• 2
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 3
• 4
• 4
• 4
• 4
• 4
• 4
• 5
• 5
• 5
• 5
• 5
• 6
• 6
Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5.
Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6 and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______, and the _______ in the middle of the interval from _______ to _______ .
• 3.5 to 4.5
• 4.5 to 5.5
• 6
• 5.5 to 6.5
Calculate the number of bars as follows:
$\frac{6.5-0.5}{\mathrm{bars}}=1$
where 1 is the width of a bar. Therefore, $\mathrm{bars}=6$ .
The following histogram displays the number of books on the x-axis and the frequency on the y-axis.
## Using the ti-83, 83+, 84, 84+ calculator instructions
Go to the Appendix (14:Appendix) in the menu on the left. There are calculator instructions for entering data and for creating a customized histogram. Create the histogram for Example 2.
• Press Y=. Press CLEAR to clear out any equations.
• Press STAT 1:EDIT. If L1 has data in it, arrow up into the name L1, press CLEAR and arrow down. If necessary, do the same for L2.
• Into L1, enter 1, 2, 3, 4, 5, 6
• Into L2, enter 11, 10, 16, 6, 5, 2
• Press WINDOW. Make Xmin = .5, Xmax = 6.5, Xscl = (6.5 - .5)/6, Ymin = -1, Ymax = 20, Yscl = 1, Xres = 1
• Press 2nd Y=. Start by pressing 4:Plotsoff ENTER.
• Press 2nd Y=. Press 1:Plot1. Press ENTER. Arrow down to TYPE. Arrow to the 3rd picture (histogram). Press ENTER.
• Arrow down to Xlist: Enter L1 (2nd 1). Arrow down to Freq. Enter L2 (2nd 2).
• Press GRAPH
• Use the TRACE key and the arrow keys to examine the histogram.
## Optional collaborative exercise
Count the money (bills and change) in your pocket or purse. Your instructor will record the amounts. As a class, construct a histogram displaying the data. Discuss how many intervals you think is appropriate. You may want to experiment with the number of intervals. Discuss, also, the shape of the histogram.
Record the data, in dollars (for example, 1.25 dollars).
Construct a histogram.
where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Integrals
## How to find the area between two curves in integral calculus
Finding the area between two curves in integral calculus is a simple task if you are familiar with the rules of integration (see indefinite integral rules above). The easiest way to solve this problem is to find the area under each curve by integration and then subtract one area from the other to find the difference between them. You may be presented with two main problem types. The first is when the limits of integration are given, and the second is where the limits of integration are not given.
## Area Between Two Curves: Limits of Integration Given
Sample problem 1: Find the area between the curves y = x and y = x2 between x = 0 and x = 1.
Step 1: Find the definite integral for each equation over the range x = 0 and x = 1, using the usual integration rules to integrate each term. (see: calculating definite integrals).
Step 2: Subtract the difference between the areas under the curves. You’ll need to visualize the curves (sketch or graph the curves if you need to); you’ll want to subtract the bottom curve from the top one. The curve on top here is f(x)=x, so:
1213 = 16.
## Limits of Integration NOT Given
Sample problem: Find the area between the curves y = x and y = x2.
Step 1: Graph the equations. In most cases, the limits of integration will be clear, especially if you’re using a TI-calculator with an Intersection feature (just find the intersections of the two graphs). If you can find the intersection by graphing, skip to Step 3.
Step 2: Find the common solutions of these two equations if you cannot find the intersection by graphing (treat them as simultaneous equations).
Substituting y = x for x in y = x2 gives an equation y = y2, which has only two solutions, 0 and 1.
Putting the values back into y = x to give the corresponding values of x: x = 0 when y = 0, and x = 1 when y = 1. The two points of intersection are (0,0) and (1,1).
Step 3: Complete the steps in Sample Problem 1 (limits of integration given) to complete the calculation. |
# Search by Topic
#### Resources tagged with Introducing algebra similar to Jam and Egg Sandwich:
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### There are 27 results
Broad Topics > Algebra > Introducing algebra
##### Stage: 3 Challenge Level:
Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?
### Perimeter Expressions
##### Stage: 3 Challenge Level:
Create some shapes by combining two or more rectangles. What can you say about the areas and perimeters of the shapes you can make?
##### Stage: 3 Challenge Level:
Think of a number... follow the machine's instructions. I know what your number is! Can you explain how I know?
### Shape Products
##### Stage: 3 Challenge Level:
These eleven shapes each stand for a different number. Can you use the multiplication sums to work out what they are?
### Crossed Ends
##### Stage: 3 Challenge Level:
Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### Up and Down Staircases
##### Stage: 2 Challenge Level:
One block is needed to make an up-and-down staircase, with one step up and one step down. How many blocks would be needed to build an up-and-down staircase with 5 steps up and 5 steps down?
### Cherry Buns
##### Stage: 2 Challenge Level:
Sam's grandmother has an old recipe for cherry buns. She has enough mixture to put 45 grams in each of 12 paper cake cases. What was the weight of one egg?
### Break it Up!
##### Stage: 1 and 2 Challenge Level:
In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes.
### First Forward Into Logo 6: Variables and Procedures
##### Stage: 3, 4 and 5 Challenge Level:
Learn to write procedures and build them into Logo programs. Learn to use variables.
### First Forward Into Logo 8: More about Variables
##### Stage: 3, 4 and 5 Challenge Level:
Write a Logo program, putting in variables, and see the effect when you change the variables.
### Table Patterns Go Wild!
##### Stage: 2 Challenge Level:
Nearly all of us have made table patterns on hundred squares, that is 10 by 10 grids. This problem looks at the patterns on differently sized square grids.
### Carrying Cards
##### Stage: 2 Challenge Level:
These sixteen children are standing in four lines of four, one behind the other. They are each holding a card with a number on it. Can you work out the missing numbers?
### Gran, How Old Are You?
##### Stage: 2 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### Trebling
##### Stage: 2 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### Plenty of Pens
##### Stage: 2 Challenge Level:
Amy's mum had given her £2.50 to spend. She bought four times as many pens as pencils and was given 40p change. How many of each did she buy?
### Two Number Lines
##### Stage: 2 Challenge Level:
Max and Mandy put their number lines together to make a graph. How far had each of them moved along and up from 0 to get the counter to the place marked?
### How Many Eggs?
##### Stage: 2 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
### Shape Times Shape
##### Stage: 2 Challenge Level:
These eleven shapes each stand for a different number. Can you use the multiplication sums to work out what they are?
### Balance of Halves
##### Stage: 2 Challenge Level:
Investigate this balance which is marked in halves. If you had a weight on the left-hand 7, where could you hang two weights on the right to make it balance?
### Geomlab
##### Stage: 3, 4 and 5 Challenge Level:
A geometry lab crafted in a functional programming language. Ported to Flash from the original java at web.comlab.ox.ac.uk/geomlab
### Buckets of Thinking
##### Stage: 2 Challenge Level:
There are three buckets each of which holds a maximum of 5 litres. Use the clues to work out how much liquid there is in each bucket.
### Counting Counters
##### Stage: 2 Challenge Level:
Take a counter and surround it by a ring of other counters that MUST touch two others. How many are needed?
### Super Shapes
##### Stage: 2 Short Challenge Level:
The value of the circle changes in each of the following problems. Can you discover its value in each problem?
##### Stage: 2 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Brothers and Sisters
##### Stage: 2 Challenge Level:
Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues?
### Cat Food
##### Stage: 2 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks? |
10.1 Non-right triangles: law of sines (Page 4/10)
Page 4 / 10
Thus,
$\text{Area}=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)=\frac{1}{2}b\left(c\mathrm{sin}\text{\hspace{0.17em}}\alpha \right)$
Similarly,
$\text{Area}=\frac{1}{2}a\left(b\mathrm{sin}\text{\hspace{0.17em}}\gamma \right)=\frac{1}{2}a\left(c\mathrm{sin}\text{\hspace{0.17em}}\beta \right)$
Area of an oblique triangle
The formula for the area of an oblique triangle is given by
$\begin{array}{l}\text{Area}=\frac{1}{2}bc\mathrm{sin}\text{\hspace{0.17em}}\alpha \hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}ac\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}ab\mathrm{sin}\text{\hspace{0.17em}}\gamma \hfill \end{array}$
This is equivalent to one-half of the product of two sides and the sine of their included angle.
Finding the area of an oblique triangle
Find the area of a triangle with sides $\text{\hspace{0.17em}}a=90,b=52,\text{\hspace{0.17em}}$ and angle $\text{\hspace{0.17em}}\gamma =102°.\text{\hspace{0.17em}}$ Round the area to the nearest integer.
Using the formula, we have
$\begin{array}{l}\text{Area}=\frac{1}{2}ab\mathrm{sin}\text{\hspace{0.17em}}\gamma \hfill \\ \text{Area}=\frac{1}{2}\left(90\right)\left(52\right)\mathrm{sin}\left(102°\right)\hfill \\ \text{Area}\approx 2289\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{units}\hfill \end{array}$
Find the area of the triangle given $\text{\hspace{0.17em}}\beta =42°,\text{\hspace{0.17em}}\text{\hspace{0.17em}}a=7.2\text{\hspace{0.17em}}\text{ft},\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=3.4\text{\hspace{0.17em}}\text{ft}.\text{\hspace{0.17em}}$ Round the area to the nearest tenth.
about $\text{\hspace{0.17em}}8.2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{feet}$
Solving applied problems using the law of sines
The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.
Finding an altitude
Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in [link] . Round the altitude to the nearest tenth of a mile.
To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side $\text{\hspace{0.17em}}a,$ and then use right triangle relationships to find the height of the aircraft, $\text{\hspace{0.17em}}h.$
Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.
The distance from one station to the aircraft is about 14.98 miles.
Now that we know $\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}$ we can use right triangle relationships to solve for $\text{\hspace{0.17em}}h.$
The aircraft is at an altitude of approximately 3.9 miles.
The diagram shown in [link] represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point $\text{\hspace{0.17em}}B,\text{\hspace{0.17em}}$ is 62°, and the distance between the viewing points of the two end zones is 145 yards.
161.9 yd.
Access these online resources for additional instruction and practice with trigonometric applications.
Key equations
Law of Sines $\begin{array}{l}\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{a}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\beta }{b}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\gamma }{c}\text{\hspace{0.17em}}\hfill \\ \frac{a}{\mathrm{sin}\text{\hspace{0.17em}}\alpha }=\frac{b}{\mathrm{sin}\text{\hspace{0.17em}}\beta }=\frac{c}{\mathrm{sin}\text{\hspace{0.17em}}\gamma }\hfill \end{array}$ Area for oblique triangles
Key concepts
• The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
• According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
• There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See [link] .
• The ambiguous case arises when an oblique triangle can have different outcomes.
• There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See [link] and [link] .
• The Law of Sines can be used to solve triangles with given criteria. See [link] .
• The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See [link] .
• There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See [link] .
the third and the seventh terms of a G.P are 81 and 16, find the first and fifth terms.
if a=3, b =4 and c=5 find the six trigonometric value sin
pls how do I factorize x⁴+x³-7x²-x+6=0
in a function the input value is called
how do I test for values on the number line
if a=4 b=4 then a+b=
a+b+2ab
Kin
commulative principle
a+b= 4+4=8
Mimi
If a=4 and b=4 then we add the value of a and b i.e a+b=4+4=8.
Tariq
what are examples of natural number
an equation for the line that goes through the point (-1,12) and has a slope of 2,3
3y=-9x+25
Ishaq
show that the set of natural numberdoes not from agroup with addition or multiplication butit forms aseni group with respect toaaddition as well as multiplication
x^20+x^15+x^10+x^5/x^2+1
evaluate each algebraic expression. 2x+×_2 if ×=5
if the ratio of the root of ax+bx+c =0, show that (m+1)^2 ac =b^2m
By the definition, is such that 0!=1.why?
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
hatdog
Mark
jaks
Ryan
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching |
## Pages
### Area of Squares, Rectangles and Triangles
Distance, Area, Volume Unit Study
Day 7: Area - Squares, Rectangles, and Triangles
Measuring area answers the question how big is it. Formulas used to calculate area and perimeter were introduced to my son when he was exploring math on Khan Academy. Unfortunately, he didn't really understand what area and perimeter were, so he came away thoroughly confused.
Area is measured in squares. After learning about one square inch, he was easily able to determine the area of these shapes built from squares, simply by counting.
Next he determined the number of tiles we would need to redo our bathroom floor in one foot square tiles.
Then we explored triangles. On the right side of the picture below, there is a one inch string, a one square inch piece of paper, a one-half square inch of paper cut into a rectangle, and a one-half square inch of paper cut into a triangle. The middle column contains the same figures, but double the size.
After creating these figures, I asked the kids how big they were.
I asked them how big this two square inch triangle was and they were stumped. So I created a blue triangle the same size and cut it.
They quickly determined it was two square inches.
So they were given the large triangle and asked how big it was.
At first, they both wanted to measure the hypotenuse, but then both decided to draw lines on their triangles.
With different sets of lines they both arrived at the same answer.
This method of discovery mathematics is very important for my son. He learns best when he figures something out for himself. He is a natural problem solver who does not learn well with explanation and example.
#### 1 comment:
1. I was looking at area of rectangles and squares with my 9yr old son today (using Zaccaro's Challenge Math). He normally catches onto concepts very quickly but sometimes a topic comes along that surprises me. Area was one of those and I made a note to print out some grid paper for him to play with next time. So thanks for this - now I can extend the exploration to cover triangles! |
# Is 47 A Prime Number
## Is 47 A Prime Number
### So you might also be wondering what are the factors of 47?
Factors of 47: 1, 47. Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Factors of 49: 1, 7.49.The question then is whether 47 is divisible by any number. Multiples of 47 are all integers which are also divided by 47, i.e. all numbers such that the remainder after dividing by 47 is zero.
### 47: 47 is actually a multiple of itself since 47 is only divisible by 47 (we have 47/47 = 1, so the remainder of this division is actually zero)With that in mind, what can you divide by 47?
Since 47 cannot be divided equally by 2, 3, or 5, we know that 47 is prime.
### What are the factors of 45?
45 is a composite number. 45 = 1 x 45, 3 x 15 or 5 x 9. Factors of 45: 1, 3, 5, 9, 15.45. Prime factorization: 45 = 3 x 3 x 5, also write 45 = 3 x 5.
### What are the multiples of 47?
Multiple of 47. Answers: 47.94.141.188235.282.329.376.423.470.517.564.611.658.705.752.799.84.893.94.98.1034.1081.1128.1175.1222.1269.1316.1363.14.15.15.15.115, 15.15.15. 15,15,15,155,156,155,155,115,135,156,155,105,105,105,175,105,105,1756sas’, - 1927,1974,2021,2068.2115.2162.2209.
### How high is the factor 48?
48 is a composite number. 48 = 1 x 48, 2 x 24, 3 x 16, 4 x 12 or 6 x 8. Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Decomposition of prime numbers : 48 = 2 x 2 x 2 x 2 x 3, which can also be written as 48 = 24 x 3. Is
### 47 a prime number?
All numbers ending in five are divisible by five. Therefore, all numbers ending in five are greater than five composite numbers. The prime numbers between 2 and 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73 , 79, 83, 89 and 97.
50 and multiples
### What are the factors for 77?
77 is a composite number. 77 = 1 x 77 or 7 x 11. Factors of 77: 1, 7, 11.77. Prime factorization: 77 = 7 x 11. Is
### 19 a prime number?
List of prime numbers. List of prime numbers up to 100: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73, 79, 83,89,97,
### What is the main factorization of 54?
54 is a composite number. 54 = 1 x 54, 2 x 27, 3 x 18 or 6 x 9. Factors of 54: 1, 2, 3.6, 9, 18, 27, 54. Prime numbers decomposition: 54 = 2x 3 x 3 x 3, all that can be printed 2 x 3³.
### What is the percentage of 47 characters out of 50?
Calculate the fraction (ratio) 47/50 Answer: 94%
### What are the factors of 49?
Factors of 49: 1, 7, 49. Prime Factorization: 49 = 7 x 7, which can also be written as 49 = 7². Since √49 = 7 is an integer, 49 is a perfect square. If 49 is a clue in FIND FACTOR puzzles, enter 7 in both the appropriate factor column and the factor row.
### Which 49 can we share with?
For 49 the answer is: No, 49 is not a prime number. The list of all positive factors (i.e. the list of all integers dividing 49) is as follows: 1, 7, 49. For 49 to be prime, 49 would have to have only two factors, namely itself and 1 .
### What is the most important factorization 47?
Prime factorization of 47
### What is the divisibility rule of 13?
Check for divisibility by 13. Add the last digit four times to the first remaining truncated number. If the result is divisible by 13, that was the first number. Use this rule over and over as needed. Is
### 57 divisible by any number?
This means that the number 57 is divisible by 3. A final example is 19. If you add 1 and 9 you get 10 which is not divisible by 3. This means that the number 19 is not divisible by 3.
### Because 51 is not a Prime number?
For 51 to be a prime number, 51 would have to have only two factors, d different prime numbers. In fact 51 = 3 x 17, where 3 and 17 are both prime numbers. Is
### 0 a prime number?
Zero is not a prime number because it has more than 2 factors. Zero is even because 0 = 2⋅0 and 0 is an integer. When we use numbers that are mostly common sense (integers, real numbers, complex numbers), zero is a number. Is
### 1 a prime number?
However, 1 has only one positive factor (1 itself), so it is not a prime number. Retaliation: this is not the definition of a prime number! A prime number is a positive integer whose positive factors are exactly 1 and itself.
### What is the divisibility rule of 29?
Divisibility test with 29. |
# Factorization by Using Identities | How to Factorise Polynomials using Algebraic Identities?
Know the process to find Factorization by using identities. Based on the identities, we can simply factorize an algebraic equation. That means, depending on the identities or identity values, we can easily reduce the number of expressions into n number of terms. Generally, for the simplest factorization process, we have to follow some basic expressions. They are
(i) (x + y)^2 = (x)^2 + 2xy + (y)^2.
(ii) (x – y)^2 = (x)^2 – 2xy + (y)^2.
(iii) x^2 – y^2 = (x + y) (x – y).
## Factoring Polynomials Identities Steps
Go through the below-listed guidelines on how to factor polynomials using algebraic identities. They are in the following fashion
• Note down the given expression and compare the expression with the basic expressions.
• If there are three terms and all are identified with the positive sign, then that is related to (x + y)^2.
• If the first and last terms are with the positive sign and the middle term is identified with the negative sign, then that is related to (x – y)^2.
• If there are only two terms with one positive sign and one negative sign, then that is related to x^2 – y^2.
• Now, compare the co-efficient values with the basic expressions.
• Find the values of coefficients.
• Based on the values, reduce the given expression into simple terms.
### Factorization Using Identities Examples with Answers
1. Factorize using the formula of a square of the sum of two terms
(i) a2 + 6a + 9
Solution: Given expression is a2 + 6a + 9
There are three terms are identified with the positive sign. Then it is related to the expression (x)^2 + 2xy + (y)^2 = (x + y)^2.
Now, compare the co-efficient values of a2 + 6a + 9 with the expression (x)^2 + 2xy + (y)^2.
Here, x = a, 2y = 6 then y = 3.
So, (x + y)^2 = (a + 3)^2.
Finally, our expression a^2 + 6a + 9 is reduced to (a + 3)^2.
(ii) a^2 + 20a + 100
Solution:
Given expression is a^2 + 20a + 100.
There are three terms that are identified with a positive sign. Then it is related to the expression (x)^2 + 2xy + (y)^2 = (x + y)^2.
Now, compare the co-efficient values of a^2 + 20a + 100 with the expression(x)^2 + 2xy + (y)^2.
Here, x = a, 2y = 20 then y = 10.
So, (x + y)^2 = (a + 10)^2.
Finally, our expression a^2 + 20a + 100 is reduced to (a + 10)^2.
2. Factorize using the formula of the square of the difference of two terms
(i) 4p^2 – 12pq + 9q^2
Solution:
Given expression is 4p^2 – 12pq + 9q^2.
There are three terms that are identified with the positive sign of the first and last terms and the negative sign of the middle term.
Then it is related to the expression (x)^2 – 2xy + (y)^2 = (x – y)^2.
Now, compare the co-efficient values of 4p^2 – 12pq + 9q^2with the expression (x)^2 – 2xy + (y)^2.
Here, x^2 = 4p^2 then x =2p, y^2 = 9q^2 then y = 3q.
So, (x – y)^2 = (2p – 3q)^2.
Finally, our expression 4p^2 – 12pq + 9q^2is reduced to (2p – 3q)^2.
(ii) a^2 – 10a + 25
Solution:
Given expression is a^2 – 10a + 25.
There are three terms that are identified with the positive sign of the first and last terms and the negative sign of the middle term.
Then it is related to the expression (x)^2 – 2xy + (y)^2 = (x – y)^2.
Now, compare the co-efficient values of a^2 – 10a + 25 with the expression (x)^2 – 2xy + (y)^2.
Here, x^2 = a^2 then x =a, y^2 = 25 then y = 5.
So, (x – y)^2 = (a – 5)^2.
Finally, our expression a^2 – 10a + 25 is reduced to (a – 5)^2.
3. Factorize using the formula of a difference of two squares:
(i) 49a^2 – 64
Solution:
Given expression is 49a^2 – 64
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of 49a^2 – 64 with the expression (x)^2 – (y)^2.
Here, x^2 = 49a^2 = (7a)^2 then x =7a, y^2 = 64 = (8)^2 then y = 8.
So, (x)^2 – (y)^2 = (7a)^2 – (8)^2.
(x + y) (x – y) = (7a + 8) (7a – 8)
Finally, our expression 49a^2 – 64is reduced to (7a + 8) (7a – 8).
(ii) 16a^2 – 36b^2
Solution:
Given expression is 16a^2 – 36b^2
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of 16a^2 – 36b^2 with the (x)^2 – (y)^2.
Here, x^2 = 16a^2 = (4a)^2 then x =4a, y^2 = 36b^2 = (6b)^2 then y = 6b.
So, (x)^2 – (y)^2 = (4a)^2 – (6b)^2.
(x + y) (x – y) = (4a + 6b) (4a – 6b)
Finally, our expression 16a^2 – 36b^2 is reduced to (4a + 6b) (4a – 6b).
(iii) 1 – [5(2p – 5q)]^2
Solution:
Given expression is 1 – [5(2p – 5q)]^2
We can write it as (1)^2 – [5(2p – 5q)]^2
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of (1)^2 – [5(2p – 5q)]^2 with the (x)^2 – (y)^2.
Here, x^2 = (1)^2 then x =1,
y^2 = [5(2p – 5q)]^2then y = 5(2p – 5q).
So, (x)^2 – (y)^2 = (1)^2 – [5(2p – 5q)]^2.
(x + y) (x – y) = [1 + 5(2p – 5q)] [1 – 5(2p – 5q)].
Finally, our expression 1 – 25(2p – 5q)^2 is reduced to [1 + 5(2p – 5q)] [1 – 5(2p – 5q)].
4. Factor completely using the formula of a difference of two squares
(i) p^4 –q^4
Solution:
Given expression is p^4 –q^4
We can write it as (p^2)^2 – (q^2)^2
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of (p^2)^2 – (q^2)^2with the (x)^2 – (y)^2.
Here, x^2 = (p^2)^2 then x =p^2,
y^2 = (q^2)^2 then y = q^2.
So, (x)^2 – (y)^2 = (p^2)^2 – (q^2)^2.
(x + y) (x – y) = (p^2 + q^2) (p^2 –q^2).
Finally, our expression p^4 – q^4 is reduced to (p^2 + q^2) (p^2 – q^2).
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## Wednesday, May 04, 2005
### CoPrime Numbers: xn + yn = zn
This blog is about the equation:
xn + yn = zn
I will show that given any three integers that satisfy this equation, either:
(a) all three of them are coprime with each other
or
(b) it is possible to cancel out common components and derive three numbers that are coprime.
Two numbers are coprime if they do not share any common divisors.
Definition 1: Coprime
Two numbers x,y are said to be coprime if and only if d divides x, d divides y -> d = 1
Two numbers x,y are said to not be coprime if and only if there exists a value d > 1 such that d divides x, d divides y
Now, here's the proof that was promised:
Lemma: We can reduce any solution to xn + yn = zn to a form where x,y,z are coprime.
To prove this, we will need to prove two things:
(1) If a factor divides any two values of this equation, then the n-power of it divides the n-power of the third value.
(2) If an n-power of a factor divides the n-power of a value, then the factor divides the value itself.
Step 1: For xn + yn = zn, the n-power of any common factor of two divides the n-power of the third.
Case I: Let's assume d divides x, d divides y
(1) There exists x', y' such that: x = d(x'), y = d(y')
(2) zn = xn + yn = (dx')n + (dy')n
= dn(x')n + dn(y')n
= dn[(x')n + (y')n]
Case II: Let's assume d divides z and d divides x or d divides y
(1) Let's assume d divides x (the same argument will work for y)
(2) There exists x', z' such that: x = d(x'), z = d(z')
(3) We now say yn = zn - xn
(4) We can now follow the same reasoning as above.
QED
Step 2: dn divides xn → d divides x
(1) Let c be the greatest common denominator (gcd) for d,x.
(2) Let D = d / c, X = x / c.
(3) Now the gcd of (X,D) = 1. [See here for the explanation]
(4) So, the gcd of (Xn,Dn) = 1.
(5) We know that there exists k such that xn = k * dn [Since dn divides xn ]
(6) Applying (2), we get (cX)n = k*(cD)n
(7) Which gives us: cnXn = k * cnDn
(8) Dividing cn from each side gives: Xn = Dn*k
(9) Now it follows that gcd(Dn,k) = 1.
(a) Assume gcd(Dn,k) = a, a > 1
(b) Then, a divides Dn and Xn [From 8]
(c) But gcd(Dn,Xn) ≠ 1.
(e) So, we reject our assumption.
(10) So, we can conclude that k is an n-power. [See blog on Infinite Descent for the proof]
(11) Which means that there exists u such that un = k.
(12) And we get Dn * un = Xn
(13) And (Du)n = Xn
(14) Implying that Du = X and multiplying by c that du=x.
(15) Which proves that d divides x.
QED
Anonymous said...
At the end of the proof does
(12) And we get Dn * un = Zn
mean
(12) And we get Dn * un = Xn?
Larry Freeman said...
Yes, you are right. I just made the change.
Thanks very much for noticing this!
-Larry
Anonymous said...
"(14) Implying that Du = X and multiplying by c that du=X." I think that should instead end with "...by c that du=x." It's nothing big... I'm just hoping others dont get confused by it. Also, thanks, I found your site very helpful.
Larry Freeman said...
Thanks for noticing that. I just fixed it.
-Larry
David K said...
I think that the proof can be simplified: From (8) we have that D^n divides X^n. But gcd(X^n,D^n)=1. Hence D^n~1 and also D~1. Now from (2) we have c=d.
David
David said...
Explanation: I meant simplification of step 2 of your proof.
David
don said...
UMMM.. WER DID U GET THE D ONE INSTEAD THE GIVEN IS X,Y,Z ONLY?
SORRY IM NOT A MATHEMATICIAN
mattsucks said...
if u define the equal square of zero vontax then u get the same ammount of Du = X so D^n + D~1 equals the 1st du=6 in the multiplied by the x* method.
Anders H said...
This comment has been removed by the author.
Anders H said...
Intuitively, step 2 is trivial. Here is a very simple proof of step 2. Is it flawed?
d^n divides x^n =>
There is an integer
e = (x^n)/(d^n) = (x/d)^n
(by common arithmetic laws).
Apparently there is an integer
x' = x/d such that (x')^n = e.
Apparently (since x/d is integer),
d divides x,
QED.
이영지 said...
This comment has been removed by the author.
Unknown said...
Surely if D^n×k=X^n then gcd(D^n,X^n)=D^n? |
# How do you solve 8|12x + 7| = 80?
Sep 4, 2017
See a solution process below:
#### Explanation:
First, divide each side of the equation by $\textcolor{red}{8}$ to isolate the absolute value function while keeping the equation balanced:
$\frac{8 \left\mid 12 x + 7 \right\mid}{\textcolor{red}{8}} = \frac{80}{\textcolor{red}{8}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} \left\mid 12 x + 7 \right\mid}{\cancel{\textcolor{red}{8}}} = 10$
$\left\mid 12 x + 7 \right\mid = 10$
The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
Solution 1:
$12 x + 7 = - 10$
$12 x + 7 - \textcolor{red}{7} = - 10 - \textcolor{red}{7}$
$12 x + 0 = - 17$
$12 x = - 17$
$\frac{12 x}{\textcolor{red}{12}} = - \frac{17}{\textcolor{red}{12}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}} x}{\cancel{\textcolor{red}{12}}} = - \frac{17}{12}$
$x = - \frac{17}{12}$
Solution 2:
$12 x + 7 = 10$
$12 x + 7 - \textcolor{red}{7} = 10 - \textcolor{red}{7}$
$12 x + 0 = 3$
$12 x = 3$
$\frac{12 x}{\textcolor{red}{12}} = \frac{3}{\textcolor{red}{12}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}} x}{\cancel{\textcolor{red}{12}}} = \frac{1}{4}$
$x = \frac{1}{4}$
The Solutions Are: $x = - \frac{17}{12}$ and $x = \frac{1}{4}$ |
# ROUNDING DECIMALS - METHODS OF ROUNDING WITH SOLVED EXAMPLES AND EXERCISE
Decimals before Rounding decimals,
if you have not already done so.
The knowledge of decimals and
Decimal Place value chart
are prerequisites here.
In practice, instead of taking the entire part
of the decimal part, we take the approximate value.
we some times take the whole number
nearest to the decimal number.
Some times, we may take the value nearest
to one or two or three or more decimal places.
Depending on the intended purpose, we take
an approximate value of the decimal number.
## Rounding Decimals
The process of obtaining the value of a decimal
correct to the required number of decimal places
is called rounding and the value obtained is called
the rounded or corrected value of the decimal.
### Rounding Decimals to the nearest whole number :
Method :
Step 1 :
Retain all the digits of the whole number
part and omit the decimal part.
Step 2 :
Out of the omitted decimal part, if the first digit
to the right of the decimal point is 5 or more,
then increase the number formed from the retained
digits by 1, otherwise do not make any change.
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### Examples on Rounding Decimals to the nearest whole number
Round off to the nearest whole number.
1. 97.46
2. 23.81
3. 67.54
Solution :
(i) 97.46
We retain the whole number part in 97.46 to get 97
and omit the decimal part .46.
Out of the omitted digits, the first digit to the right of the decimal point is 4 < 5.
So, we do not make any change to the retained whole number.
∴ Required Number = 97 Ans.
(ii) 23.81
We retain the whole number part in 23.81 to get 23
and omit the decimal part .81.
Out of the omitted digits, the first digit to the right of the decimal point is 8 > 5.
So, we increase the retained whole number by 1.
∴ Required Number = 23 + 1 = 24 Ans.
(iii) 67.54
We retain the whole number part in 67.54 to get 67
and omit the decimal part .54.
Out of the omitted digits, the first digit to the right of the decimal point is 5 = 5.
So, we increase the retained whole number by 1.
∴ Required Number = 67 + 1 = 68 Ans.
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### Rounding Decimals to the required number of decimal places
Method :
Step 1 :
Retain as many digits after the decimal point
as are required and omit the remaining digits.
Step 2 :
Out of the omitted decimal part, if the first digit
to the right of the decimal point is 5 or more,
then increase the last retained digit by 1,
otherwise do not make any change.
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### Examples on Rounding Decimals to the required number of decimal places :
Round off
1. 9.1347, correct to 3 decimal places (or to thousandths' place.)
2. 5.732, correct to 1 decimal place (or to tenths' place.)
3. 0.047, correct to 2 decimal places (or to hundredths' place.)
Solution :
(i) 9.1347, correct to 3 decimal places.
We retain 3 digits of the decimal part in 9.1347 to get 9.134
and omit the digit 7.
The first omitted digit is 7 > 5.
So, we increase the last retained
digit (4) by 1 ( i.e. make it 5).
∴ Required Number = 9.135 Ans.
(ii) 5.732, correct to 1 decimal place.
We retain 1 digit of the decimal part in 5.732 to get 5.7
and omit the digits 3 and 2.
The first omitted digit is 3 < 5.
So, we do not make change to the retained digits.
∴ Required Number = 5.7 Ans.
(iii) 0.047, correct to 2 decimal places.
We retain 2 digits of the decimal part in 0.047 to get 0.04
and omit the digit 7.
The first omitted digit is 7 > 5.
So, we increase the last retained
digit (4) by 1 ( i.e. make it 5).
∴ Required Number = 0.05 Ans.
### Exercise on Rounding Decimals
1. Round off to the nearest whole number.
1. 197.351
2. 42.78
3. 231.51
2. Round off
1. 7.6428, correct to 3 decimal places (or to thousandths' place.)
2. 9.157, correct to 1 decimal place (or to tenths' place.)
3. 0.451, correct to 2 decimal places (or to hundredths' place.)
For Answers see at the bottom of the page.
## Progressive Learning of Math : Rounding Decimals
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### Answers to Exercise on Rounding Decimals
Answers to Problems on Rounding Decimals :
1.
1. 197
2. 43
3. 232
2.
1. 7.643
2. 9.2
3. 0.45 |
Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 5.1 Answer Key Adding Integers with the Same Sign.
Modeling Sums of Integers with the Same Sign
You can use colored counters to add positive integers and to add negative integers.
Model with two-color counters.
(A) 3 + 4
How many counters are there in total? __________
What is the sum and how do you find it?
(B) – 5 + (- 3)
How many counters are there in total? __________
Since the counters are negative integers, what is the sum? __________
Reflect
Question 1.
Communicate Mathematical Ideas When adding two numbers with the same sign, what sign do you use for the sum?
The sum of two positive numbers is a positive number
The sum of two negative numbers is a negative number.
So, when you add two numbers with the same sign the sum has exactly that sign
What If? Suppose the temperature is – 1°F and drops by 3°F. Explain how to use the number line to find the new temperature.
First graph number line with units.
Start at 0 and move 1 unit left to represent number – 1.
Then, start at – 1 move 3 units left (because the temperature drops by 3), and read the new temperature from the number line.
Graph:
New temperature would be – 4 because it is 4 units left from 0 on the number line.
Question 3.
Communicate Mathematical Ideas How would using a number line to find the sum 2 + 5 be different from using a number line to find the sum – 2 + (- 5)?
To find 2 + 5 on number line you start at 0, then move 2 units right and then move 5 units also right
To find – 2 + (- 5) on number line you start at 0, then move 2 units left and then move 5 units also left
So, only difference is where you move, left or right.
It is different direction of moving on number line.
For positive right, for negative left.
Question 4.
Analyze Relationships What are two other negative integers that have the same sum as – 2 and – 5?
First, graph number line with units.
Start at 0 and move 2 unit left to represent number – 2.
Then, start at – 2 and move 5 units left to add – 5 and read sum from number Line. It would be – 7.
So, it is 7 units left from 0, and you also could move 3 units left and then 4 units left and to be again 7 units left from 0.
Thus:
– 2 + (-5) = – 3 + (- 4)
Two other negative integers that have the same sum as – 2 and – 5 are -3 and -4.
Question 5.
Communicate Mathematical Ideas Does the Commutative Property of Addition apply when you add two negative integers? Explain.
It does.
You can apply Commutative Property when you add two negative integers because in both cases sum would be negative and equal minus sum of absolute values of that integer.
Example:
– 4 + (- 8) = – (|- 4| + |- 8|) = – (4 + 8)
– 8 + (- 4) = – (|- 8| + |- 4|) = – (8 + 4)
– (4 + 8) = – (8 + 4) → – 4 + (- 8) = – 8 + (- 4)
Critical Thinking Choose any two negative integers, Is the sum of the integers less than or greater than the value of either of the integers? Will this be true no matter which integers you choose? Explain.
The sum of two negative integers would always be less than the value of either of those integers.
This is because the sum of two
The sum of two negative integers would always be less than the value of either of that integers.
Use a + b = – (|a| + |b|) for negative a and b.
Find each sum.
Question 7.
– 8 + (- 1) = ____________
First, find absolute values of given integers:
|- 8| = 8 and |- 1| = 1
Then, sum that absolute values:
8 + 1 = 9
Final result would be 9 or – 9, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 9.
Result is – 9.
Find absolute values, sum them and add minus if both were negative.
Question 8.
– 3 + (- 7) = __________
First find absolute values of given integer:
| – 3| = 3 and |- 7| = 7
Then, sum that absolute values:
3 + 7 = 10
Final result would be 10 or – 10, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 10.
Question 9.
– 48 + (- 12) = __________
First find absolute values of given integers:
|- 48| = 48 and |- 12| = 12
Then, sum that absolute values
48 + 12 = 60
Final result would be 60 or – 60, depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 60.
Question 10.
– 32 + (- 38) = _________
First find absolute values of given integers:
|- 32| = 32 and |- 38| = 38
Then, sum that absolute values
32 + 38 = 60
Final result would be 70 or – 70, depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 70.
Adding Integers with the Same Sign Lesson 5.1 Answer Key Question 11.
109 + 191 = ___________
First find the absolute values of given integers:
|109| = 109 and |191| = 191
Then, sum that absolute values
109 + 191 = 300
The final result would be 300 or – 300, depending on the sign of given integers (both positive then +, both negative then – ).
Given integers are positive the final result is 300.
Note:
You could simply add them, but this is another (harder) way that shows that adding integers with the same sign as it was explained is correct
Question 12.
– 40 + (- 105) = ________
First find absolute values of given integers:
|- 40| = 40 and |- 105| = 105
Then, sum that absolute values
40 + 105 = 145
Final result would be 145 or – 145, depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 145.
Question 13.
– 150 + (- 1,500) = ________
First find absolute values of given integers:
|- 150| = 150 and |- 1,500| = 1,500
Then, sum that absolute values
150 + 1,500 = 1,650
Final result would be 1,650 or – 1,650 depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 1,650.
Question 14.
– 200 + (- 800) = _______
First find absolute values of given integers:
|- 200| = 200 and |- 800| = 800
Then, sum that absolute values
200 + 800 = 1,000
Final result would be 1,000 or – 1,000, depending on sign of given integers (both positive then +, both negative then – ).
Given integers are negative so final result is – 1,000.
Find each sum.
Question 1.
– 5 + (- 1)
In the first row there are 5 negative counters which represent – 5 and in second row there is 1 negative counter which represent – 1.
So, on the given picture there are 6 counters and all of them are negative.
(which represent – 6)
Thus, sum would be:
– 5 – (- 1) = – 6
On the picture there are 6 negative counters and thus – 5 + (- 1) = – 6.
a. How many counters are there? ________
6
b. Do the counters represent positive or negative numbers? ____________
negative
c. – 5 + (- 1) = _____________
– 6
– 2 + (- 7)
In the first row there are 2 negative counters which represent – 2 and in second row there is 7 negative counter which represent – 7.
So, on the given picture there are 9 counters and all of them are negative.
(which represent – 9)
Thus, sum would be:
– 2 + (- 7) = – 9
On the picture there are 6 negative counters and thus – 2 + (- 7) = – 9.
a. How many counters are there? ________
9
b. Do the counters represent positive or negative numbers? ____________
negative
c. – 2 + (- 7) = _____________
– 9
Model each addition problem on the number line to find each sum.
Question 3.
– 5 + (- 2) = _____________
First, graph number line with units.
Start at 0 and move 5 unit left to represent number – 5.
Then, start at – 5 and move 2 units left (to add – 2) and read sum from number line
Number line:
Sum is – 7 because it is 7 units left from 0 on the number line.
Question 4.
– 1 + (- 3) = _____________
First, graph number line with units.
Start at 0 and move 1 unit left to represent number – 1.
Then, start at – 1 and move 3 units left (to add – 3) and read sum from number line
Number line:
Sum is – 4 because it is 4 units left from 0 on the number line.
Question 5.
– 3 + (- 7) = _____________
First, graph number line with units.
Start at 0 and move 3 unit left to represent number – 3.
Then, start at – 3 and move 7 units left (to add – 7) and read sum from number line
Number line:
Sum is – 10 because it is 10 units left from 0 on the number line.
Question 6.
– 4 + (- 1) = _____________
First, graph number line with units.
Start at 0 and move 4 unit left to represent number – 4.
Then, start at – 4 and move 1 units left (to add – 1) and read sum from number line
Number line:
Sum is – 5 because it is 5 units left from 0 on the number line.
Question 7.
– 2 + (- 2) = _____________
First, graph number line with units.
Start at 0 and move 2 unit left to represent number – 2.
Then, start at – 2 and move 2 units left (to add – 2) and read sum from number line
Number line:
Sum is – 4 because it is 4 units left from 0 on the number line.
Adding Integers with the Same Sign Lesson 5.1 Go Math 6th Grade Question 8.
– 6 + (- 8) = _____________
First, graph number line with units.
Start at 0 and move 6 unit left to represent number – 6.
Then, start at – 6 and move 8 units left (to add – 8) and read sum from number line
Number line:
Sum is – 14 because it is 14 units left from 0 on the number line.
Find each sum.
Question 9.
– 5 + (- 4) = _____________
First, find absolute values of given integers:
| – 5| = 5 and |- 4| = 4
Then, sum that absolute values:
5 + 4 = 9
Final result would be 9 or – 9, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 9.
Result is – 9
Find absolute values, sum them and add minus if both were negative.
Question 10.
– 1 + (- 10) = __________________
First, find absolute values of given integers:
| – 1| = 1 and |- 10| = 10
Then, sum that absolute values:
1 + 10 = 11
Final result would be 11 or – 11, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 11.
Result is – 11
Find absolute values, sum them and add minus if both were negative.
Question 11.
– 9 + (- 1) = ____________________
First, find absolute values of given integers:
| – 9| = 9 and |- 1| = 1
Then, sum that absolute values:
9 + 1 = 10
Final result would be 10 or – 10, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 10.
Result is – 10
Find absolute values, sum them and add minus if both were integers were negative.
Question 12.
– 90 + (- 20) = __________________
First, find absolute values of given integers:
| – 90| = 90 and |- 20| = 20
Then, sum that absolute values:
90 + 20 = 110
Final result would be 110 or – 110, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 110.
Result is – 110
Find absolute values, sum them and add minus if both were integers were negative.
Question 13.
– 52 + (- 48) = __________________
First, find absolute values of given integers:
| – 52| = 52 and |- 48| = 48
Then, sum that absolute values:
52 + 48 = 100
Final result would be 100 or – 100, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 100.
Result is – 100
Find absolute values, sum them and add minus if both were integers were negative.
Question 14.
5 + 198 = __________________
First, find absolute values of given integers:
| 5 | = 5 and |198| = 198
Then, sum that absolute values:
5 + 198 = 203
Final result would be 203 or – 203, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is 203.
Note: You could simply add them, but this is another way.
Result is – 203
Find absolute values, sum them and add minus if both were integers were negative.
Question 15.
– 4 + (- 5) + (- 6) = _____________
First, find absolute values of given integers:
|- 4| = 4 |- 5| = 5 and | – 6| = 6
Then, sum that absolute values:
4 + 5 + 6 = 15
Final result would be 15 or – 15, depending on sign of given integers (all positive then +, all negative then -).
Given integers are negative so final result is – 15.
Result is – 15
Find absolute values, sum them and add minus if given integers were negative.
Question 16.
– 50 + (- 175) + (- 345) = __________________
First, find absolute values of given integers:
|- 50| = 50 |- 175| = 175 and | – 345| = 345
Then, sum that absolute values:
50 + 175 + 345 = 570
Final result would be 570 or – 570, depending on sign of given integers (all positive then +, all negative then -).
Given integers are negative so final result is – 570.
Result is – 570
Find absolute values, sum them and add minus if given integers were negative.
Essential Question Check-In
Question 17.
How do you add integers with the same sign?
First, find absolute values of given integers;
Then, sum that absolute values;
If given integers were negative add minus.
Find absolute values, sum them and add minus if given integers were negative.
Represent Real-World Problems Jane and Sarah both dive down from the surface of a pool. Jane first dives down 5 feet, and then dives down 3 more feet. Sarah first dives down 3 feet and then dives down 5 more feet.
a. Multiple Representations Use the number line to model the equation
– 5 + (- 3) = – 3 + (- 5).
First graph number Line with units.
Representing – 5 + (- 3):
Start at 0 and move 5 unit left to represent number – 5.
Then, start at – 5 and move 3 units left (to add – 3).
Now, you are 8 units Left from 0 on the number Line.
Representing – 3 + (- 5):
Start at 0 and move 3 unit left to represent number – 3.
Then, start at – 3 and move 5 units left (to add – 5).
Again, you are 8 units left from 0 on the number line.
So, it is equal.
Number line:
b. Does the order in which you add two integers with the same sign affect the sum? Explain.
The order does not affect the sum because you move on number line in the same direction and get to the same value in both cases.
Question 19.
A golfer has the following scores for a 4-day tournament.
What was the golfer’s total score for the tournament?
Total score equals:
– 3 + (- 1) + (- 5) + (- 2)
First find absolute values of given integers:
|- 3| = 3
|- 1| = 1
|- 5| = 5
|- 2| = 2
Then, sum that absolute values:
3 + 1 + 5 + 2 = 11
Final result would be 11 or – 11, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 11.
The golfer’s total score was – 11.
Find absolute values, sum them and add minus if all integers were negative.
Question 20.
A football team loses 3 yards on one play and 6 yards on another play. Write a sum of negative integers to represent this situation. Find the sum and explain how it is related to the problem.
Let represent the team’s score:
(-3) + (-6)
First, find absolute vaLues of given integers:
|- 3| = 3 and |- 6| = 6
Then, sum that absolute values
3 + 6 = 9
Final result would be 9 or – 9, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 9.
That means that the team loses 9 yards on two plays
Sum is – 9.
Find absolute values, sum them and add minus if both integers were negative.
That means that the team loses 9 yards on two plays.
Question 21.
When the quarterback is sacked, the team loses yards. In one game, the quarterback was sacked four times. What was the total sack yardage?
The total sack yardage equals:
– 14 + (- 5) + (- 12) + (- 23)
First find absolute values of given integers:
|- 14| = 14
|- 5| = 5
|- 12| = 12
|- 23| = 23
Then, sum that absolute values
14 + 5 + 12 + 23 = 54
Final result would be 54 or – 54, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 54.
The total sack yardage was – 54.
Find absolute values, sum them and add minus if all integers were negative.
Question 22.
Multistep The temperature in Jonestown and Cooperville was the same at 1:00. By 2:00, the temperature in Jonestown dropped 10 degrees, and the temperature in Cooperville dropped 6 degrees. By 3:00, the temperature in Jonestown dropped 8 more degrees, and the temperature in Cooperville dropped 2 more degrees.
a. Write an equation that models the change to the temperature in Jonestown since 1:00.
Let calculate the change to the temperature in Jonestown:
(- 10) + (- 8)
First find absolute values of given integers:
|- 10| =10 and |- 8| = 8
Then, sum that absolute values
10 + 8 = 18
Final result would be 18 or – 18, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 18.
The temperature in Jonestown dropped 18 degrees since 1:00.
b. Write an equation that models the change to the temperature in Cooperville since 1:00.
Let calculate the change to the temperature in Cooperville:
(- 6) + (- 2)
First, find absolute values of given integers:
|- 6| = 6 and |- 2| = 2
Then, sum that absolute values
6 + 2 = 8
Final result would be – 8 or 8, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 8.
The temperature in Cooperville dropped 8 degrees since 1:00.
c. Where is it colder at 3:00, Jonestown or Cooperville?
Since the temperature was the same in both cities at 1:00, Jonestown is colder at 3:00, because – 18 is less than – 8.
Question 23.
Represent Real-World Problems Julio is playing a trivia game. On his first turn, he lost 100 points. On his second turn, he lost 75 points. On his third turn, he lost 85 points. Write a sum of three negative integers that models the change to Julio’s score after his first three turns.
Show how the score changes after three turns:
– 100 + (- 75) + (- 85)
First, find absoLute vaLues of given integers:
|- 100| = 100
|- 75| = 75
|- 85| = 85
Then, sum that absolute values
100 + 75 + 85 = 260
Final result would be 260 or – 260, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 260
Julio’s score got – 260 points after three turns.
The sum is
– 100 + (- 75) + (- 85)
and the change to his score is – 260.
Find absolute values, sum them and add minus if all integers were negative.
H.O.T. Focus on Higher Order Thinking
Question 24.
Multistep On Monday, Jan made withdrawals of $25,$45, and $75 from her savings account. On the same day, her twin sister Julie made withdrawals of$35, $55, and$65 from her savings account.
a. Write a sum of negative integers to show Jan’s withdrawals on Monday. Find the total amount Jan withdrew.
Let calculate Jan’s withdrawals:
– 25 + (- 45) + (- 75)
First find absolute values of given integers:
|- 25| = 25
|- 45| = 45
|- 75| = 75
then, sum that absoLute vaLues:
25 + 45 + 75 = 145
Final resuLt would be 145 or – 145, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 145
Jan withdrew $145. b. Write a sum of negative integers to show Julie’s withdrawals on Monday. Find the total amount Julie withdrew. Answer: Let calculate Julie’s withdrawals: – 35 + (- 55) + (- 65) First find absolute values of given integers: |- 35| = 35 |- 55| = 55 |- 65| = 65 Then, sum that absolute values 35 + 55 + 65 = 155 Final result would be 155 or – 155, depending on sign of given integers (both positive then +, both negative then -). Given integers are negative so final result is – 155. Julie withdrew$155.
c. Julie and Jan’s brother also withdrew money from his savings account on Monday. He made three withdrawals and withdrew $10 more than Julie did. What are three possible amounts he could have withdrawn? Answer: Their brother also made three withdrawals, and his total amount is$ 10 bigger then Julies.
He could have withdrawn first amount $10 bigger. Let calculate the first with drawal: – 35 + (- 10) First, find absolute values of given integers : |- 35| = 35 |- 10| = 10 Then, sum that absolute values: 35 + 10 = 45 Final result would be 45 or – 45, depending on sign of given integers (both positive then +. both negative then -). Given integers are negative so final result is – 45. He could have withdrawn amounts$ 45, $55 and$65.
Go Math Grade 6 Lesson 5.1 Integers Same Sign Answer Key Question 25.
Communicate Mathematical Ideas Why might you want to use the Commutative Property to change the order of the integers in the following sum before adding?
– 80 + (- 173) + (- 20)
First, write down the sum:
– 80 + (- 173) + (- 20)
You might want to use the Commutative Property before adding because it makes adding easier
If you change the order of the integers – 173 and – 20, you will get the sum:
– 80 + (- 20) + (- 173)
Let’s calculate the sum
– 80 + (- 20)
First, find the absolute values of given integers:
|- 80| = 80 and |- 20| = 20
Then, sum that absolute values
80 + 20 = 100
Final result would be 100 or – 100, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative so final result is – 100.
Now it is easy to calculate
– 100 + (- 173)
Using the Commutative Property to change the order of the second and third integers before adding makes adding easier.
Question 26.
Critique Reasoning The absolute value of the sum of two different integers with the same sign is 8. Pat says there are three pairs of integers that match this description. Do you agree? Explain.
First pair of negative integers is – 1 and 7.
Find absolute values of given integers:
|- 1| = 1 and |- 7| = 7
Sum that absolute values:
1 + 7 = 8
The absolute value of the sum is 8.
Second pair is – 2 and – 6.
Find absolute values of given integers:
|- 2| = 2 and |- 6| = 6
Sum that absolute values:
2 + 6 = 8
The absolute value of the sum is 8.
Third pair is – 3 and – 5.
Find absolute values of given integers:
– 3 = 3 and – 5 = 5
Sum that absolute values:
3 + 5 = 8
The absolute value of the sum is 8.
First pair of positive integers is 1 and 7.
Find absolute values of given integers:
|1| = 1 and |7| = 7
Sum that absolute values:
1 + 7 = 8
The absolute value of the sum is 8
Second pair is 2 and 6.
Find absolute values of given integers:
|2| = 2 and |6| = 6
Sum that absolute values
2 + 6 = 8
The absolute value of the sum ¡s 8
Third pair is 3 and 5.
Find absolute values of given integers:
|3| = 3 and |5| = 5
Sum that absolute values:
3 + 5 = 8
The absolute value of the sum is 8.
There are six pairs of integers with the same sign: – 1 and – 7, – 2 and – 6, – 3 and – 5, 1 and 7, 2 and 6, 3 and 5.
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The X Factor: Using Algebra to Solve Problems
# 4. Chocolate Boxes
Suggested Learning Intentions
• To complete a mathematical investigation involving variables
• To successfully apply a range of problem-solving strategies to a given task
Sample Success Criteria
• I can use a range of problem-solving strategies to help me solve a problem
• I can model a problem using materials
• I can identify and justify which strategies are most useful
• I can justify my solution using a range of manipulatives
This stage has been inspired by the Christmas Chocolates problem on the nRich website, reproduced with permission of the University of Cambridge, all rights reserved. You are encouraged to access the Teachers’ Resources and Solutions section of this problem prior to using this stage in the classroom.
Provide students with the Initial problem sheet from the Materials and texts section above, while telling the following story:
Penny, Tom and Matthew were each given a hexagonal box of chocolates as a gift. The box of chocolates has five chocolates along each straight edge and is called a 'size 5' box. Each of them eats a particular number of chocolates out of their box. They are all able to work out that there must have been 61 chocolates in the size 5 box to begin with. What strategies do you think each of them used to get to this answer?
Provide students some time to look at the three different diagrams on the Initial problem sheet. Support students further if needed by providing them with blank templates that can be used to shade in the chocolates eaten by each person. These are accessible via the Teachers’ Resources section of this problem on the nRich website.
• What do you notice about the position of the chocolates that each person ate first?
• Which chocolates do you think that each person will eat next?
Ask students to share the strategies that they believe Penny, Tom, and Matthew used to work out the number of chocolates in the size 5 box. The nRich task provides sample explanations for the three different strategies.
Invite students to contribute other strategies they may have discovered. It is important to validate any student who simply counted the chocolates one-by-one.
Emphasise that the purpose of pattern recognition (which is what Penny, Tom and Matthew used) is to enable us to complete these types of calculations more efficiently – but counting them is still a useful starting place. Counting the chocolates in rows can lead to a valid pattern being recognised and then used later.
Once students have had an opportunity to see how different strategies work, present them with a second problem:
Penny, Tom and Matthew now each have a larger hexagonal box of chocolates – it’s a 'size 10' box. How many chocolates could be in this box?
Students apply any of the strategies discussed earlier to try and solve this problem.
Sample student responses may include:
• Penny’s approach: There are six triangles in the hexagon. Each of them has a base of 9 chocolates. Each of those triangles has 45 chocolates in it. So, 6 x 45 equals 270, and then add 1 for the centre chocolate, and there are 271 chocolates in total.
• Tom’s approach: There are three parallelograms in the hexagon. Each parallelogram is made up of 10 rows of 9 chocolates, so there are 90 chocolates in each parallelogram. So, 3 x 90 equals 270 and then add 1 for the centre chocolate, which gives 271 chocolates in total.
• Matthew’s approach: If there are 10 chocolates on each side, then I can see 6 rows of 9 chocolates in the outside layer. This will give me 54 chocolates in the outer layer. Each layer afterwards will have 6 less in it (6 rows of 8 chocolates, then 6 rows of 7 chocolates, etc.). Therefore, the total number of chocolates will be equal to 54 + 48 + 42 + 36 + 30 + 24 + 18 + 12 + 6 and then 1 for the middle, which equals to 271.
Provide opportunities for students who have used different strategies to compare their answers, so they can see that multiple strategies can lead to the same answer.
Present students who have successfully trialled a variety of strategies for both size 5 and size 10 boxes with this problem:
'Can you determine a rule that would help you quickly work out how many chocolates are in a box with a side length of n chocolates?'
Enable students requiring further support to make a start on this problem by suggesting that they begin with Tom’s approach (parallelograms), and experiment with various values of n
Model the use of a table as a way of helping students explore the relationship between the size of the box and the number of chocolates in it. Use the data collected from the size 5 and size 10 boxes as a starting point for the table.
Using this approach, students can develop the generalised rule for the number of chocolates in a box of size as 3n(n - 1) +1 .
Ask students to explain the significance of the different parts of this rule, or to annotate the rule with an explanation. A sample response may be:
'The n(n - 1) represents the number of chocolates in one of the parallelograms. The 3 reflects the three parallelograms within the box, so 3n(n - 1) represents the number of chocolates in the three parallelograms, and the plus 1 on the end is for the chocolate in the middle.'
Ask students to develop a generalisation using Penny’s approach (triangles).
Students who have successfully understood Tom’s ‘parallelogram’ method may be able to visualise that the triangle is half of the parallelogram. Their response may be:
We know that the number of chocolates in a parallelogram is equal to:
Each triangle is half of a parallelogram, so the number of chocolates in each triangle is:
There are 6 triangles, so we need to multiply:
and then add 1 for the chocolate in the middle. So the total number of chocolates is equal to:
Demonstrate that this generalisation, obtained using Penny’s method, is in fact equivalent to the generalisation obtained using Tom’s method.
Students who are familiar with the triangular numbers (1, 3, 6, 10…) could use these as part of their generalisation, as the number of chocolates in each triangle is a triangular number. A sample response could be:
'Subtract 1 from the side length of the box. If the box is a size 9 box, subtract 1 from 9, which gives you 8. The 8th triangular number will tell you how many chocolates are in each of the triangles. The 8th triangular number is 36, so there are 36 chocolates in each triangle. Multiply 36 by 6, because there are 6 triangles, and then add 1 for the chocolate in the centre.'
While this generalisation isn’t written as a symbolic rule, it is still a form of algebraic reasoning, as it can work for any size box, and relies on the recognition of a pattern.
Enable students by providing them a list and/or visual representation of the triangular numbers and encourage them to make links between this list and the chocolate boxes.
### Areas for Further Exploration
Using Matthew's approach to develop a generalised algebraic rule is a far more challenging task. However, students could be asked to explore Matthew's approach and write an algorithm that could be sued to find the number of chocolates for any value of n. A sample algorithm is provided here, modelled using n = 8:
Step 1: Count the number of chocolates on each side of the box.
There are 8 chocolates.
Step 2: Subtract 1 from this number. This represents the number of chocolates in each row around the outside of the box.
8 - 1 = 7
Step 3: Multiply this number by 6, because there are 6 rows around the outside of the box. The answer will represent the total number of chocolates in the outside layer.
7 x 6 = 42
Step 4: Continue to subtract 1 and multiply this number by 6. Each of these numbers represents the number of chocolates in each layer, as we move from the outside of the box towards the inside.
6 x 6 = 36, 5 x 6 = 30, 4 x 6 = 24, ...
Step 5: Continue this process until you reach 1 x 6 = 6
3 x 6 = 18, 2 x 6 = 12, 1 x 6 = 6
Step 6: Add the products of the multiplications together.
42 + 36 + 30 + 24 + 18 + 12 + 6 = 168
Step 7: Add 1 to represent the chocolate in the middle.
168 + 1 = 169
Extend students by asking them to write a generalisation using triangles with a base of n (rather than n - 1). This is quite challenging, as it will require them to subtract the 'overlapped' rows that are counted twice. A sample explanation for the development of this generalisation is provided below.
The number of chocolates in each triangle is:
The number of chocolates in six of these triangles is:
Then, you need to subtract 6 rows of n chocolates, as they have been counted twice. Therefore, the number of chocolates will be:
Finally, add one chocolate back in for the centre chocolate.
Using the algebraic skills of expanding and linear factorising, demonstrate to students that this expression is equivalent to 3n(n - 1) + 1.
The purpose of this part of the stage is for students to make links between the task and a set of recognised problem-solving strategies. You may have your own set of problem-solving strategies that you would like to use instead of the set referenced below.
Introduce students to the Problem Solving Strategies poster. The poster presents a range of strategies that students can apply to problem solving tasks.
Ask students to read the list of strategies in the toolbox and identify any that they used during this task. Ask students which strategies they felt were most useful for them during this task.
The specific problem-solving strategies that students have used in Chocolate Boxes may include:
• Draw a diagram
• Make a model
• Make a table
• Look for a pattern
• Test all possible combinations
• Seek an exception
• Break the problem into manageable parts
Students could prepare an investigative report outlining the strategies used by Tom, Penny, and Matthew, as well as any additional generalisations they developed.
Addison, L., n.d. Strategy Toolbox poster. [Online] |
# James owes $6,927 on a credit card with a 19.5% interest rate. What is the monthly payment he should make to pay off this debt in 12 months, assuming he does not charge any more purchases with the card? a-$1531.34
b- $577.25 c-$124.82
d- $640.02 My answer is D ## correct ## thank you ## To determine the monthly payment James should make to pay off his credit card debt in 12 months, we can use the formula for the monthly payment on a loan, which is given by: $M = \frac{P \times r \times (1+r)^n}{(1+r)^n - 1}$ where: - M is the monthly payment - P is the principal amount (the amount James owes) - r is the monthly interest rate - n is the number of months Let's calculate it step-by-step: First, let's convert the annual interest rate to a monthly interest rate. We divide the annual interest rate by 12: $r = \frac{19.5}{100 \times 12} = 0.01625$ Next, substitute the given values into the formula: $M = \frac{6927 \times 0.01625 \times (1+0.01625)^{12}}{(1+0.01625)^{12} - 1}$ Calculate the numerator: $Numerator = 6927 \times 0.01625 \times (1+0.01625)^{12} \approx 1303.93$ Calculate the denominator: $(1+0.01625)^{12} - 1 \approx 0.2771$ Finally, divide the numerator by the denominator to find the monthly payment: $M = \frac{1303.93}{0.2771} \approx 4703.39$ Based on the calculations, the monthly payment James should make to pay off his credit card debt in 12 months is approximately$4703.39, which is not one of the given options.
So, it seems there might be an error in the options provided.
## To calculate the monthly payment James should make to pay off his debt in 12 months, we can use the formula for monthly payment on a loan or credit card balance. The formula is:
Monthly Payment = (Principal + (Principal * Interest Rate)) / Number of Months
Given:
Principal (Amount owed) = $6,927 Interest Rate = 19.5% per year Number of Months = 12 Let's calculate the monthly payment using the formula: Interest Rate per Month = Interest Rate / 12 months = 19.5% / 12 = 0.0195 / 12 = 0.001625 Monthly Payment = ($6,927 + ($6,927 * 0.001625)) / 12 = ($6,927 + $11.26) / 12 =$6,938.26 / 12
= $578.19 The monthly payment to pay off the debt in 12 months is approximately$578.19.
a- $1531.34 b-$577.25
c- $124.82 d-$640.02
Option D- $640.02 is closest to our calculated value of$578.19, but it is not an exact match. Therefore, your answer seems incorrect. |
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# Two steel balls $A$ and $B$ of mass $10kg$ and $10g$ roll towards each other with $5m/s$ and $1m/s$ on a smooth floor. After the collision with what speed $B$ moves if it is the case of an elastic collision?$\left( a \right){\text{ 8m/s}}$$\left( b \right){\text{ 10m/s}}$$\left( c \right){\text{ 11m/s}}$$\left( d \right){\text{ Zero}}$
Last updated date: 19th Sep 2024
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Hint So for this question, we have mass and the velocity of each ball. So we will calculate the velocity by using the formula which will use all the values given. The formula is given below. Since momentum is conserved so the elastic collision will be equal to one.
Formula used:
Velocity for the elastic collision,
$V = \left( {\dfrac{{{m_2} - e{m_1}}}{{{m_1} + {m_2}}}} \right){u_2} + \left( {\dfrac{{{m_1} - e{m_1}}}{{{m_1} + {m_2}}}} \right){u_1}$
Here,
$V$, will be the velocity
$e$, will be the elastic collision
${m_1}$, will be the mass of the ball $A$
${m_2}$, will be the mass of the ball $B$
${u_2}$, will be the initial velocity of the ball $A$
${u_1}$, will be the initial velocity of the ball $B$
Complete Step By Step Solution
Firstly, we will see the values given to us.
${u_1} = 5m/s$
${u_2} = 1m/s$
${m_1} = 10kg$
${m_2} = 10g = 0.01kg$
Now by using the formula we have already seen,
$V = \left( {\dfrac{{{m_2} - e{m_1}}}{{{m_1} + {m_2}}}} \right){u_2} + \left( {\dfrac{{{m_1} - e{m_1}}}{{{m_1} + {m_2}}}} \right){u_1}$
Velocity will be,
Substituting the values, we get
Here, the value of elastic collision will be
$\Rightarrow e = 1$
On putting the values, we get
$\Rightarrow V = \left( {\dfrac{{0.01 - 10}}{{10 + 0.01}}} \right)\left( { - 1} \right) + \left( {\dfrac{{10 - 10}}{{10 + 0.01}}} \right)5$
On further simplifying it, we get
$\Rightarrow V = \dfrac{{9.99 + 150}}{{10.01}}$
Again solving the equation,
$\Rightarrow V = \dfrac{{109.99}}{{10.01}}$
And it will be approximately equal to
$\Rightarrow \approx 11m/s$
So if it is the case of elastic collision $B$ will move with the speed of approximately $11m/s$.
Hence the option $\left( c \right)$ is the correct option.
Note All collisions, from elastic to completely inelastic and anything in between, must conserve momentum. The reason is simply that all forces in a collision are internal to the objects colliding, i.e. no outside forces act on the system.
This is most effectively perceived in a two-body crash. The adjustment in the energy of an article is equivalent to the motivation, Assuming a consistent power, this is only the power times the timeframe. All the more, by and large, it is the fundamental of power concerning time. |
## Precalculus: Mathematics for Calculus, 7th Edition
$x=\dfrac{\sqrt{7}-1}{5}$ $x=-\dfrac{1+\sqrt{7}}{5}$
$(5x+1)^{2}+3=10$ We need to solve this by factoring, so let's take the $10$ to the left side of the equation to substract: $(5x+1)^{2}+3-10=0$ Simplify: $(5x+1)^{2}-7=0$ Remember the difference of two squares factoring formula: $a^{2}-b^{2}=(a-b)(a+b)$ In this equation $a=(5x+1)^{2}$ and $b=7$, so we can factor it and we get: $[(5x+1)-\sqrt{7}][(5x+1)+\sqrt{7}]=0$ Then, we set each factor equal to $0$ and solve them as individual equations. Let the $[(5x+1)-\sqrt{7}]=0$ be equation number 1 and $[(5x+1)+\sqrt{7}]=0$ be equation number 2: Solving for $x$ in equation number 1: $(5x+1)-\sqrt{7}=0$ $5x=\sqrt{7}-1$ $x=\dfrac{\sqrt{7}-1}{5}$ Solving for $x$ in equation number 2: $(5x+1)+\sqrt{7}=0$ $5x=-1-\sqrt{7}$ $x=-\dfrac{1+\sqrt{7}}{5}$ |
# Mid-point Circle Drawing Algorithm
Computer Graphics / Thursday, November 8th, 2018
(Last Updated On: November 8, 2018)
# Mid-point Circle Drawing Algorithm
For a given radius r and center position (x0, y0), we can first set up our algorithm to calculate pixel positions around a circle path centered at the coordinate origin (0, 0). Then each calculated position (C, 4) is moved to its proper screen position by adding xC to x and yC to y along the circle section from x = 0 to x = y in the first quadrant, the slope of the curve varies from 0 to -1. Therefore, we can take unit steps in the position x direction over this octant and use a decision parameter to determine which of the two possible y positions is closer to the circle path at each step. Positions in the other seven octants are obtained by symmetry.
To apply the mid-point method, we define a circle function
$f\left( x,y \right)={{x}^{2}}+{{y}^{2}}-{{r}^{2}}$
Any point (x, y) on the boundary of the circle with radius r satisfies the equation. If the point is in the interior of the circle, the circle function is negative. And if the point outside the circle, the circle function is positive. To summarize the relative position of any point (x, y) can be determined by checking the sign of the circle function.
The mid point between two candidate pixels at sampling position xk+1. Assuming we have just plotted the pixel (xk, yk), we next need to determine whether the pixel at position (xk+1, yk) or the one of position (xk+1, yk-1) is closer to the circle. Our decision parameter is the circle function at the mid-point between these two pixels.
${{d}_{k}}=f\left( {{x}_{k+1}},{{y}_{k}}-\frac{1}{2} \right)$
$\Rightarrow {{d}_{k}}={{\left( {{x}_{k+1}} \right)}^{2}}+{{\left( {{y}_{k}}-\frac{1}{2} \right)}^{2}}-{{r}^{2}}$
If dk < 0, this mid-point is inside the circle and the pixel on scan line yk is closer to the circle boundary. Otherwise, the mid position is outside or on the circle boundary and we select the pixel on scan line yk-1.
Successive decision parameters are obtained using incremental calculations. We obtain a recursive expression for the next decision parameter by evaluating the circle function at sampling position xk+1 +1 = xk+2
${{d}_{k+1}}=f\left( {{x}_{k+1}}+1,{{y}_{k+1}}-\frac{1}{2} \right)$
$\Rightarrow {{d}_{k+1}}={{\left( {{x}_{k+1}}+1 \right)}^{2}}+{{\left( {{y}_{k+1}}-\frac{1}{2} \right)}^{2}}-{{r}^{2}}$
$\therefore {{d}_{k+1}}={{d}_{k}}+2\left( {{x}_{k}}+1 \right)+\left( y_{k+1}^{2}-y_{k}^{2} \right)-\left( {{y}_{k+1}}-{{y}_{k}} \right)+1$
Where yk+1 either yk or yk-1 depending on the sign of dk.
Increments for obtaining dk+1 are either 2xk+1 + 1 – 2yk+1. Evaluation of the terms 2xk+1 and 2yk+1 are also be done incrementally as
$2{{x}_{k+1}}=2{{x}_{k}}+2$
$2{{y}_{k+1}}=2{{y}_{k}}-2$
At the start (0, r) these two terms have the values 0 and 2r, respectively. Each successive value is obtained by adding 2 to the previous value of 2y.
The initial decision parameter is obtained by evaluating the circle function at the start position (x0, y0) = (0, r)
$i.e.,{{d}_{0}}=f\left( 1,r-\frac{1}{2} \right)=1+{{\left( r-\frac{1}{2} \right)}^{2}}-{{r}^{2}}=\frac{5}{4}-r$
If the radius r is specified as an integer we can simply round d0 to d0 = 1 – r since all increments are integer.
## Algorithm
Step 1. Start.
Step 2. Input radius r and circle center (xC, yC) and obtain the first point on the circumference of a circle centered on the origin as (xC, yC) = (0, r)
Step 3. Calculate the initial value of the decision parameters as d0 = 5/4 – r
Step 4. At each xk position, starting k = 0, perform the following test:
If dk < 0 the next point along the circle centered on (0, 0) is (xk+1, yk) and dk+1 = dk + 2xk+1 +1. Otherwise, the next point along the circle is (xk + 1, yk – 1) and dk+1 = dk + 2xk+1 +1 -2yk+1, where 2xk+1 = 2xk + 2 and 2yk+1 = 2yk – 2
Step 5. Determine symmetry points in the other seven octants.
Step 6. Move each calculated pixel position (x, y) onto the circular path centered on (x0, y0) and plot the coordinate value x = x + xC, y = y + yC
Step 7. Repeat Step 4 through Step 6 until x ≥ y.
Step 8. End.
Example 01
Given a circle radius r = 5, we demonstrate the mid-point circle algorithm by determining position along the circle octant in the first quadrant from x = 0 to x = y. The initial value of the decision parameter is d0 = 1 – r = 1 – 5 = -4
For the circle centered on the coordinate origin, the initial point is (x0, y0) = (0, 5) and the initial increment terms for calculation 2 x0 = 0, 2 y0 = 2 X 5 = 10
Successive decision parameter values and positions along the circle path are calculated using the mid-point method as
k dk (xk+1, yk+1) 2xk+1 2yk+1 0 -4 (1, 5) 2 10 1 -1 (2, 5) 4 1 2 8 (3, 4) 6 8 3 4 (4, 3) 8 6
<<Previous |
# 1. SQUARES & SPIRALS
“One person’s constant is another person’s variable.”
― Susan Gerhart (computer scientist)
#### Constants, variables & computer programing
by George Gadanidis, PhD
#### IN THE CLASSROOM
###### Ontario Curriculum
• Gr. 3
• E1.4 – give and follow multistep instructions involving movement from one location to another, including distances and half- and quarter-turns
• Gr. 3/4
• C2.1 – use variables in various contexts
• C.3.2 – read and alter existing code; sequential, concurrent, and repeating & nested events
• Gr. 4-6
• E2.4 – identify and classify angles
• Gr. 9
• C2.1 – use coding to demonstrate an understanding of algebraic concepts (including variables)
• C1.1 research an algebraic concept to tell a story
• E1.1 research a geometric concept to tell a story
###### Implementation
• Gr. 3-6
• To walk a square, we may repeat 4 times: walk 5 steps and turn right (90 degrees or 1/4 turn). Notice that the number of steps is constant.
• To walk a spiral, the number of steps keeps changing. The number of steps is variable.
• The process of drawing a square and then drawing a spiral anchors student understanding of constants and variables.
• The need for a variable is especially important when we use Scratch code to draw the spiral, as the turn angle is constant and the number of steps varies.
• Gr. 7-9
• The above may be used as a review of the meaning of constant & variable.
#### 1A. PUZZLE 1 – SQUARES
Below are 2 different ways to draw a square with code. You can click and run the code at https://scratch.mit.edu/projects/1055180217/editor/
Q1. Edit SQUARE 1 code to draw a smaller square.
Q2. Edit SQUARE 2 code to draw a smaller square.
Q3. How is the code for SQUARE 1 and the code for SQUARE 2 similar or different?
Q4. Which code do you like better? Why?
#### 1B. PUZZLE 2 – CODE A Spiral
Q5. Edit SQUARE 1 code to draw a spiral like the ones shown below.
Q6. Editing SQUARE 2 code to draw a spiral like the ones shown below is not as simple. Why?
Q7. Below is one way to edit SQUARE 2 code to draw a spiral. How is SPIRAL 2 code different from SQUARE 2 code?
#### 1C. PUZZLE 3 – Match Spirals
You can click and run the SPIRAL 2 code at https://scratch.mit.edu/projects/1055189920/editor
Q8. Edit the code to create each of the following spirals.
#### 1D. PUZZLE 4 – CreatE Math Art
Q9. Edit the SPIRAL 2 code to create your own math art.
by George Gadanidis, PhD |
Review question
# Can we find the angle between these two tangents? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R5678
## Solution
The point $P$ on the hyperbola $xy=c^2$ is such that the tangent to the hyperbola at $P$ passes through the focus of the parabola $y^2=4ax$. Find the coordinates of $P$ in terms of $a$ and $c$.
It is a standard result that the focus of the parabola $y^2=4ax$ is at $(a,0)$. The general equation of a straight line with gradient $m$ passing through $(a,0)$ is $y=m(x-a).$ This will intersect the curve $xy=c^2$ when: \begin{align} x\times m(x-a) &= c^2 \notag \\ \Longrightarrow \quad mx^2 -amx-c^2 &= 0. \label{eq:1} \end{align} If $y=m(x-a)$ is to be tangent to the hyperbola $xy=c^2,$ then $\eqref{eq:1}$ will have a repeated root, which means the discriminant of this quadratic will be $0$. This gives \begin{align} a^2m^2-4\times m \times (-c^2)&=0 \notag \\ \Longrightarrow \quad m(a^2m+4c^2) &=0 \notag \\ \Longrightarrow \quad m=-\frac{4c^2}{a^2}. \label{eq:grad-of-tangent} \end{align}
We have ignored the $m=0$ solution, as this is the horizontal asymptote of $xy=c^2$. We now have $x=\frac{am\pm \sqrt{0}}{2m}=\frac{a}{2}.$ Since $P$ lies on $xy=c^2$, the corresponding $y$-coordinate is $y=\frac{c^2}{x}=\frac{2c^2}{a}$ so $P=\left(\frac{a}{2},\frac{2c^2}{a}\right).$
If $P$ also lies on the parabola, prove that $a^4=2c^4$
If $P$ also lies on the parabola then $P$ satisfies $y^2=4ax,$ so \begin{align} \left(\frac{2c^2}{a}\right)^2 &= 4a\left(\frac{a}{2}\right) \notag \\ \Longrightarrow \quad \frac{4c^4}{a^2} &= 2a^2 \notag \\ \Longrightarrow \quad 2c^4 &= a^4. \label{eq:c4a4} \end{align}
… and calculate the acute angle between the tangents to the two curves at $P$.
If a line of gradient $m$ cuts the $x$-axis at an angle $\theta$, then $m = \tan \theta$. From the following diagram, we see that we need to find $\alpha - \theta$.
What is the gradient of the parabola, $y^2 = 4ax$, at $P$? We can either differentiate implicitly or take the square root, $y = 2\sqrt{ax} \implies \frac{dy}{dx} = 2\sqrt{a}\times\frac{1}{2}x^{-1/2} = \sqrt{\dfrac{a}{x}}.$
Since the $x$-coordinate of $P$ is $\dfrac{a}{2}$, we have the gradient $\tan \theta = \sqrt{\dfrac{a}{a/2}} = \sqrt{2}$.
From our earlier work, $\eqref{eq:grad-of-tangent}$ and $\eqref{eq:c4a4}$, we know $\tan \alpha = -\dfrac{4c^2}{a^2} = -\dfrac{4}{\sqrt{2}} = -2\sqrt{2}$.
Now we can use a standard trig identity to find
\begin{align*} \tan(\alpha-\theta) &= \dfrac{\tan \alpha - \tan \theta}{1+\tan \alpha \tan \theta}\\ &= \dfrac{ -2\sqrt{2}-\sqrt{2}}{1-2\sqrt{2}\sqrt{2}} \\ &= \sqrt{2}.\\ \end{align*}
Hence the angle between the tangents at $P$ is $\arctan \sqrt{2}\approx 0.955^c \approx 54.7^\circ$.
It turns out that $\theta=\alpha-\theta$ and the triangle in the above diagram is isosceles.
Instead of using the identity, we could have calculated $\arctan{\sqrt{2}}$ and $\arctan{\left(-2\sqrt{2}\right)}$ and taken the difference.
Alternatively, we could find the angle between the two lines using the dot product of the two direction vectors.
The tangent to the parabola has gradient $\sqrt{2}$ so its direction vector can be written as $\mathbf{a} = \begin{pmatrix}1 \\ \sqrt{2}\end{pmatrix}$ and the tangent to the hyperbola can be written as $\mathbf{b} = \begin{pmatrix}1 \\ -2\sqrt{2}\end{pmatrix}.$
The dot product can then be computed as \begin{align*} \mathbf{a}\cdot\mathbf{b} &= 1-2\sqrt{2}\sqrt{2}=-3 \\ &= |\mathbf{a}| \: |\mathbf{b}| \cos\theta = \sqrt{1+2}\sqrt{1+4\times2}\cos\theta \\ \implies\quad \cos\theta &= -\frac{1}{\sqrt{3}} \end{align*}
This actually gives us the obtuse angle but it’s then easy to find the acute one. |
# Proportions and Percents
Proportions can be very helpful in solving problems dealing with percents. Let’s look at this problem. Suppose a pair of shoes is on sale for 20% off of the original price.
If the original price is \$85, how much money will you save? Keep in mind that a percent is a part out of a whole, and that whole is 100, when you’re dealing with a percent. So, 20% off, that would be 20 out of 100.
33%
Question 1 of 3
John weighs 190 pounds. He enrolls in a fitness program and reduces 10% of his body weight over a three-month period. How many pounds did Tom lose?
A.
B.
C.
D.
Question 1 of 3
Question 2 of 3
A squash ball weighs 20% of the permitted weight of 5 ounces. How heavy is the squash ball?
A.
B.
C.
D.
Question 2 of 3
Question 3 of 3
One cup of ice cream contains 260 calories. Forty-five percent of the total calorie content is derived from fat. Calculate the calorie content in fat.
A.
B.
C.
D.
Question 3 of 3
Next Lesson: Proportions in the Real World
The transcript is provided for your convenience
And now, we’ve set up our first ratio, and the key with proportions is to be consistent with your ratios. So, since our first ratio is the part out of the whole, then our second ratio also needs to be the part out of the whole. The whole, in this case, being your price. That’s the total cost, so that’s your whole. That’s your denominator, 85. And then the “how much money you save” is going to be your part, because you’re only going to save part of the whole. And since that’s what we’re trying to find, how much money will you save, that’s going to be our variable. So, we can just use an x or anything.
And then to solve our proportion, we’re going to cross multiply. So, we cross multiply 100 times x, which is 100x, is equal to 20 times 85. 2 times 85 is 170, and then we add a 0 for the 20, so 1700.
Then to solve for x, we have a 100 times x, and the opposite of multiplying is dividing. So, we divide both sides by 100. 100 divided by 100 is 1, times x is x. Cancel our two zeroes, 17 divided by 1 is 17. Therefore, if our \$85 pair of shoes was 20% off, then we’d be saving \$17 when we buy our shoes.
Next Lesson: Proportions in the Real World |
Review of Coordinate Geometry: the Distance Formula, the Midpoint Formula
DISTANCE BETWEEN POINTS; THE MIDPOINT FORMULA
• PRACTICE (online exercises and printable worksheets)
The Distance Formula
To find the distance between any two points in a coordinate plane:
• subtract the $x$-values in any order; square the result
• subtract the $y$-values in any order; square the result
• add together the previous two quantities
• take the square root of the result
This sequence of operations is expressed in the Distance Formula:
the Distance Formula
The distance between points $\,(x_1,y_1)\,$ and $\,(x_2,y_2)\,$ is given by the Distance Formula: $$\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$
For a complete review of the distance between two points in the coordinate plane, study the following lesson,
which includes a derivation, a discussion of subscript notation, and examples: The Distance Formula
The Midpoint Formula
To find the midpoint of the line segment between any two points in a coordinate plane:
• average the $x$-values of the two points—that is, add them and divide by $2$;
this gives the $x$-value of the midpoint
• average the $y$-values of the two points—that is, add them and divide by $2$;
this gives the $y$-value of the midpoint
This sequence of operations is expressed in the Midpoint Formula:
THE MIDPOINT FORMULA
The midpoint of the line segment between points $\,(x_1,y_1)\,$ and $\,(x_2,y_2)\,$ is given by the Midpoint Formula: $$\left( \frac{x_1+x_2}2,\frac{y_1+y_2}2 \right)$$
For a complete review of midpoints, including a derivation and examples, study: The Midpoint Formula
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Sketching Regions in the Coordinate Plane
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9
AVAILABLE MASTERED IN PROGRESS
(MAX is 9; there are 9 different problem types.) |
Triangles - ESSENTIAL GEOMETRY SKILLS - SAT Test Prep
## CHAPTER 10ESSENTIAL GEOMETRY SKILLS
### Lesson 2: Triangles
Angles in Polygons
Remembering what you learned about parallel lines in the last lesson, consider this diagram:
We drew line ℓ so that it is parallel to the opposite side of the triangle. Do you see the two Z’s? The angles marked a are equal, and so are the angles marked c. We also know that angles that make up a straight line have a sum of 180°, so . The angles inside the triangle are also a, b, and c.
Therefore, the sum of angles in a triangle is always 180°.
Every polygon with n sides can be divided into n – 2 triangles that share their vertices (corners) with the polygon:
Therefore, the sum of the angles in any polygon with n sides is 180(n – 2) °.
Angle-Side Relationships in Triangles
A triangle is like an alligator mouth with a stick in it: The wider the mouth, the bigger the stick, right?
Therefore, the largest angle of a triangle is always across from the longest side, and vice versa. Likewise, the smallest angle is always across from the shortest side.
Example:
In the figure below, , so
An isosceles triangle is a triangle with two equal sides. If two sides in a triangle are equal, then the angles across from those sides are equal, too, and vice versa.
The Triangle Inequality
Look closely at the figure below. The shortest path from point A to point B is the line segment connecting them. Therefore, unless point C is “on the way” from A to B, that is, unless it’s on , the distance from A to B through C must be longer than the direct route. In other words:
The sum of any two sides of a triangle is always greater than the third side. This means that the length of any side of a triangle must be between the sum and the difference of the other two sides.
The External Angle Theorem
The extended side of a triangle forms an external angle with the adjacent side. The external angle of a triangle is equal to the sum of the two “remote interior” angles. Notice that this follows from our angle theorems:
and ; therefore,
Concept Review 2: Triangles
1. The sum of the measures of the angles in a quadrilateral is __________.
2. The sum of the measures of the interior angles in an octagon is __________.
3. In ΔABC, if the measure of ∠A is 65° and the measure of ∠B is 60°, then which side is longest? __________.
4. The angles in an equilateral triangle must have a measure of __________.
5. Can an isosceles triangle include angles of 35° and 60°? Why or why not?
6. Draw a diagram to illustrate the external angle theorem.
7. If a triangle has sides of lengths 20 and 15, then the third side must be less than __________ but greater than __________.
8. Is it possible for a triangle to have sides of lengths 5, 8, and 14? Why or why not?
9. If an isosceles triangle includes an angle of 25°, the other two angles could have measures of __________ and __________ or __________ and __________.
10. In the figure above, and are diameters of the circle and P is not the center. Complete the statement below with <, >, or=.
SAT Practice 2: Triangles
1. In the figure above, if , then x =
(A) 25
(B) 30
(C) 35
(D) 50
(E) 65
2. In the figure above,
3. The three sides of a triangle have lengths of 9, 16, and k. Which of the following could equal k?
I. 6
II. 16
III. 25
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III
Note: Figure not drawn to scale.
4. Which of the following statements about a and b in the figure above must be true?
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III
5. In the figure above, if , then
(A) 70
(B) 80
(C) 90
(D) 100
(E) 120
6. In the figure above, which of the following expresses a in terms of b and c?
Note: Figure not drawn to scale.
7. Which of the following represents the correct ordering of the lengths of the five segments in the figure above?
8. A triangle has two sides of lengths 4 centimeters and 6 centimeters. Its area is n square centimeters, where n is a prime number. What is the greatest possible value of n?
(A) 11
(B) 12
(C) 19
(D) 23
(E) 24
Concept Review 2
1. 360°
2. 1,080°
3. Draw a diagram. If the measure of ∠A is 65° and the measure of ∠B is 60°, then the measure of ∠C must be 55°, because the angles must have a sum of 180°. Since ∠A is the largest angle, the side opposite it, , must be the longest side.
4. 60°. Since all the sides are equal, all the angles are, too.
5. No, because an isosceles triangle must have two equal angles, and the sum of all three must be 180°. Since , and , the triangle is impossible.
6. Your diagram should look something like this:
7. If a triangle has sides of lengths 20 and 15, then the third side must be less than 35 (their sum) but greater than 5 (their difference).
8. No. The sum of the two shorter sides of a triangle is always greater than the third side, but is not greater than 14. So the triangle is impossible.
9. 25° and 130° or 77.5° and 77.5°
10. Draw in the line segments PQ, PR, PS, and PT. Notice that this forms two triangles, ΔPQS and ΔPRT. Since any two sides of a triangle must have a sum greater than the third side, , and Therefore, .
SAT Practice 2
1. A If , then, by the Isosceles Triangle theorem, ∠BAD and ∠BDA must be equal. To find their measure, subtract 50° from 180° and divide by 2. This gives 65°. Mark up the diagram with this information. Since the angles in the big triangle have a sum of 180°, , so .
2. 500 Drawing two diagonals shows that the figure can be divided into three triangles. (Remember that an n-sided figure can be divided into n – 2 triangles.) Therefore, the sum of all the angles is . Subtracting 40° leaves 500°.
3. B The third side of any triangle must have a length that is between the sum and the difference of the other two sides. Since and , the third side must be between (but not including) 7 and 25.
4. A Since the big triangle is a right triangle, must equal 90. The two small triangles are also right triangles, so is also 90. Therefore, and statement I is true. In one “solution” of this triangle, a and b are 65 and x is 25. (Put the values into the diagram and check that everything “fits.”) This solution proves that statements II and III are not necessarily true.
5. C If , then, by the Isosceles Triangle theorem, the angles opposite those sides must be equal. You should mark the other angle with an x also, as shown here. Similarly, if then the angles opposite those sides must be equal also, and they should both be marked y. Now consider the big triangle. Since its angles must have a sum of 180, . Dividing both sides by 2 gives . (Notice that the fact that ADB measures 100° doesn’t make any difference!)
6. E Label the two angles that are “vertical” to those marked b and c. Notice that the angle marked a is an “external” angle. By the External Angle theorem, .
7. D Write in the missing angle measures by using the fact that the sum of angles in a triangle is 180°. Then use the fact that the biggest side of a triangle is always across from the biggest angle to order the sides of each triangle.
8. A Consider the side of length 4 to be the base, and “attach” the side of length 6. Notice that the triangle has the greatest possible height when the two sides form a right angle. Therefore, the greatest possible area of such a triangle is , and the minimum possible area is 0. The greatest prime number less than 12 is 11.
|
New
New
Year 9
# Evaluating expressions with and without changing the subject
I can evaluate an expression where the subject has been changed and where it has not.
New
New
Year 9
# Evaluating expressions with and without changing the subject
I can evaluate an expression where the subject has been changed and where it has not.
## Lesson details
### Key learning points
1. Changing the subject can make it easier to evaluate a formula.
2. It is possible to evaluate a formula without changing the subject.
3. Both forms of the formula give the same result.
### Common misconception
Changing the subject of an equation or formula is only done when you are told to.
Having students practise rearranging before substituting and vice versa allows them to decide when the skill could be useful.
### Keywords
• Subject of an equation/formula - The subject of an equation/a formula is a variable that is expressed in terms of other variables. It should have an exponent of 1 and a coefficient of 1.
This is a good opportunity for pupils to practise using key formulae and this could include those from other subjects. You could ask the pupils to share formulas they have seen or used in other lessons or outside of school.
Teacher tip
### Licence
This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
## Starter quiz
### 6 Questions
Q1.
Substitute means to put in place of another. In algebra, substitution can be used to replace __________ with values.
constants
equations
solutions
Q2.
$$A={1\over2}(a+b)h$$ is the formula for area of a trapezium. It can be rearranged to $${2A\over{h}}-a=b$$ in order to make $$b$$ the of the formula.
Q3.
The formula for area of a triangle is $$A={1\over2}bh$$. The area of a triangle with base 7.5 cm and height 6 cm is cm²
Q4.
The formula $$s={d\over{t}}$$ can be rearranged to $$t={d\over{s}}$$ or $$d=s\times {t}$$. If $$d=12$$ and $$s=3$$, find $$t$$.
$$36$$
Correct answer: $$4$$
$${1\over{4}}$$
Q5.
Find the area of a trapezium with $$h=7$$, $$a=3.2$$, $$b=4.8$$.
$$A={1\over2}(3.2+4.8)+7=11$$ square units
Correct answer: $$A={1\over2}(3.2+4.8)\times7=28$$ square units
$$A=2(3.2+4.8)\times7=112$$ square units
$$A={1\over2}(3.2\times4.8)\times7=53.76$$ square units
Q6.
$$A={1\over2}(a+b)h$$ is the formula for area of a trapezium. Select all the correct rearrangements of this formula.
Correct answer: $${2A\over{h}}-b=a$$
$${Ah\over{2}}-a=b$$
$${A\over{2h}}-a=b$$
$${2h\over{A}}-a=b$$
Correct answer: $${2A\over{a+b}}=h$$
## Exit quiz
### 6 Questions
Q1.
$$a=\sqrt{{c^2}-b^2}$$ is a __________ of $$a^2+b^2=c^2$$.
expression
solution
substitution
variable
Q2.
Which arrangement of the formula for area of a triangle makes it quickest for you to find $$b$$ if you know $$A$$ and $$h$$?
$$A={1\over2}bh$$
$${2A\over{b}}=h$$
Correct answer: $${2A\over{h}}=b$$
$${h\over{2A}}=b$$
Q3.
The formula for the area of a triangle is $$A={1\over2}bh$$. The base $$b$$ of a triangle with area 24 m² and height 8 m is m.
Q4.
Rearrange the formula $$C=2{\pi}{r}$$ to find the radius of a circle with circumference 42 cm. Leave your answer in terms of $$\pi$$.
$$r=2C\pi=2\times 42\times \pi=84\pi$$
$$r={2\pi\over{{C}}}={2\pi\over{{42}}}={\pi\over{21}}$$
$$r={2C\over{{\pi}}}={2\times 42\over{{\pi}}}={84\over{\pi}}$$
Correct answer: $$r={C\over{2{\pi}}}={42\over{2{\pi}}}={21\over{\pi}}$$
Q5.
Which rearrangement makes it easiest to find length $$b$$ if you know lengths $$a$$ and $$c$$?
$$a=\sqrt{{c^2}-b^2}$$
Correct answer: $$b=\sqrt{{c^2}-a^2}$$
$$b^2=c^2-a^2$$
$$b=\sqrt{{a^2}-c^2}$$
$$b=\sqrt{{c^2}+a^2}$$
Q6.
Jun says "You have to rearrange the formula $$a^2+b^2=c^2$$ into the form $$b=\sqrt{{c^2}-a^2}$$ with $$b$$ as the subject if you want to find $$b$$ given that $$a=6$$ and $$c=10$$." Is Jun right?
Yes, $$b$$ must be the subject before you can substitute.
Yes, because $$b=\sqrt{{10^2}-6^2}$$ is the fastest way to find $$b$$.
Correct answer: No, you can substitute $$a$$ and $$c$$ into $$a^2+b^2=c^2$$ before evaluating. |
# Indeterminate Forms
## Solved Problems
### Example 5.
Find the limit $\lim\limits_{x \to \pi } {\frac{{\cos \frac{x}{2}}}{{\pi - x}}}.$
Solution.
Change the variable: $$x - \pi = t$$ or $$x = t + \pi.$$ Then $$t \to 0$$ as $$x \to \pi.$$ We have
$L = \lim\limits_{x \to \pi } \frac{{\cos \frac{x}{2}}}{{\pi - x}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{t \to 0} \frac{{\cos \frac{{t + \pi }}{2}}}{{ - t}}.$
Convert the last expression using the reduction formula $$\cos\left( {z + \frac{\pi }{2}} \right) = - \sin z.$$ As a result we find the limit of the function:
$L = \lim\limits_{t \to 0} \frac{{\cos \frac{{t + \pi }}{2}}}{{ - t}} = \lim\limits_{t \to 0} \frac{{ - \cos \frac{t}{2}}}{{ - t}} = \lim\limits_{\frac{t}{2} \to 0} \frac{{\sin \frac{t}{2}}}{{\frac{{2t}}{2}}} = \frac{1}{2}\lim\limits_{\frac{t}{2} \to 0} \frac{{\sin \frac{t}{2}}}{{\frac{t}{2}}} = \frac{1}{2} \cdot 1 = \frac{1}{2}.$
### Example 6.
Calculate $\lim\limits_{x \to \infty } \left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 1} } \right).$
Solution.
If $$x \to \infty,$$ then
$\lim\limits_{x \to \infty } \sqrt {{x^2} + 1} = \infty \;\;\; \text{and}\;\;\lim\limits_{x \to \infty } \sqrt {{x^2} - 1} = \infty .$
Thus, we deal here with an indeterminate form of type $$\infty - \infty.$$ Multiply this expression (both the numerator and the denominator) by the corresponding conjugate expression.
$L = \lim\limits_{x \to \infty } \left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 1} } \right) = \lim\limits_{x \to \infty } \frac{{{{\left( {\sqrt {{x^2} + 1} } \right)}^2} - {{\left( {\sqrt {{x^2} - 1} } \right)}^2}}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}} = \lim\limits_{x \to \infty } \frac{{{x^2} + 1 - \left( {{x^2} - 1} \right)}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}} = \lim\limits_{x \to \infty } \frac{{\cancel{x^2} + 1 - \cancel{x^2} + 1}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}} = \lim\limits_{x \to \infty } \frac{2}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}}.$
By using the product and the sum rules for limits, we obtain
$L = \lim\limits_{x \to \infty } \frac{2}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}} = \frac{{\lim\limits_{x \to \infty } 2}}{{\lim\limits_{x \to \infty } \sqrt {{x^2} + 1} + \lim\limits_{x \to \infty } \sqrt {{x^2} - 1} }} \sim \frac{2}{{\infty + \infty }} \sim \frac{2}{\infty } = 0.$
### Example 7.
Find the limit $\lim\limits_{x \to 4} {\frac{{\sqrt {1 + 6x} - 5}}{{\sqrt x - 2}}}.$
Solution.
To calculate this limit we rationalize the numerator and denominator multiplying them by the corresponding conjugate expressions:
$\lim\limits_{x \to 4} \frac{{\sqrt {1 + 6x} - 5}}{{\sqrt x - 2}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{x \to 4} \frac{{\left( {1 + 6x - 25} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt {1 + 6x} + 5} \right)\left( {x - 4} \right)}} = \lim\limits_{x \to 4} \frac{{6 \cancel{\left( {x - 4} \right)} \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt {1 + 6x} + 5} \right) \cancel{\left( {x - 4} \right)}}} = 6\lim\limits_{x \to 4} \frac{{\sqrt x + 2}}{{\sqrt {1 + 6x} + 5}} = 6 \cdot \frac{{\sqrt 4 + 2}}{{\sqrt {25} + 5}} = 6 \cdot \frac{4}{{10}} = \frac{{12}}{5}.$
### Example 8.
Find the limit $\lim\limits_{x \to \infty } {\frac{{{{\left( {2x + 3} \right)}^{10}}{{\left( {3x - 2} \right)}^{20}}}}{{{{\left( {x + 5} \right)}^{30}}}}}.$
Solution.
We divide both the numerator and denominator by $${x^{30}}$$ (the highest power of the fraction):
$\lim\limits_{x \to \infty } \frac{{{{\left( {2x + 3} \right)}^{10}}{{\left( {3x - 2} \right)}^{20}}}}{{{{\left( {x + 5} \right)}^{30}}}} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \infty } \frac{{\frac{{{{\left( {2x + 3} \right)}^{10}}{{\left( {3x - 2} \right)}^{20}}}}{{{x^{30}}}}}}{{\frac{{{{\left( {x + 5} \right)}^{30}}}}{{{x^{30}}}}}} = \lim\limits_{x \to \infty } \frac{{\frac{{{{\left( {2x + 3} \right)}^{10}}}}{{{x^{10}}}} \cdot \frac{{{{\left( {3x - 2} \right)}^{20}}}}{{{x^{20}}}}}}{{\frac{{{{\left( {x + 5} \right)}^{30}}}}{{{x^{30}}}}}} = \lim\limits_{x \to \infty } \frac{{{{\left( {\frac{{2x + 3}}{x}} \right)}^{10}} \cdot {{\left( {\frac{{3x - 2}}{x}} \right)}^{20}}}}{{{{\left( {\frac{{x + 5}}{x}} \right)}^{30}}}} = \lim\limits_{x \to \infty } \frac{{{{\left( {2 + \frac{3}{x}} \right)}^{10}} \cdot {{\left( {3 - \frac{2}{x}} \right)}^{20}}}}{{{{\left( {1 + \frac{5}{x}} \right)}^{30}}}} = \frac{{{2^{10}} \cdot {3^{20}}}}{{{1^{30}}}} = {2^{10}} \cdot {3^{10}} \cdot {3^{10}} = {18^{10}}.$
### Example 9.
Find the limit $\lim\limits_{x \to e} {\frac{{\ln x - 1}}{{x - e}}}.$
Solution.
Let $$x - e = t.$$ Then $$t \to 0$$ as $$x \to e.$$ Hence,
$\lim\limits_{x \to e} \frac{{\ln x - 1}}{{x - e}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{t \to 0} \frac{{\ln \left( {t + e} \right) - 1}}{t} = \lim\limits_{t \to 0} \frac{{\ln \left( {t + e} \right) - \ln e}}{t} = \lim\limits_{t \to 0} \left( {\frac{1}{t}\ln \frac{{t + e}}{e}} \right) = \lim\limits_{t \to 0} \ln {\left( {\frac{{t + e}}{e}} \right)^{\frac{1}{t}}} = \left[ {{1^\infty }} \right] = \lim\limits_{t \to 0} \ln \left[ {{{\left( {1 + \frac{t}{e}} \right)}^{\frac{e}{t} \cdot \frac{1}{e}}}} \right] = \lim\limits_{t \to 0} \left[ {\frac{1}{e}\ln {{\left( {1 + \frac{t}{e}} \right)}^{\frac{e}{t}}}} \right] = \frac{1}{e}\ln \left[ {\lim\limits_{\frac{t}{e} \to 0} {{\left( {1 + \frac{t}{e}} \right)}^{\frac{e}{t}}}} \right] = \frac{1}{e} \ln e = \frac{1}{e}.$
### Example 10.
Find the limit $\lim\limits_{t \to + \infty } \left( {\sqrt {t + \sqrt {t + 1} } - \sqrt t } \right).$
Solution.
This function is defined only for $$t \ge 0.$$ Multiply and divide it by the conjugate expression $$\left( {\sqrt {t + \sqrt {t + 1} } + \sqrt t } \right).$$ So, we get
$L = \lim\limits_{t \to + \infty } \left( {\sqrt {t + \sqrt {t + 1} } - \sqrt t } \right) = \left[ {\infty - \infty } \right] = \lim\limits_{t \to + \infty } \frac{{{{\left( {\sqrt {t + \sqrt {t + 1} } } \right)}^2} - {{\left( {\sqrt t } \right)}^2}}}{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }} = \lim\limits_{t \to + \infty } \frac{{\cancel{t} + \sqrt {t + 1} - \cancel{t}}}{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }} = \lim\limits_{t \to + \infty } \frac{{\sqrt {t + 1} }}{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }} = \left[ {\frac{\infty }{\infty }} \right].$
Both the numerator and denominator now approach $$\infty$$ as $$t \to \infty.$$ Hence, we divide numerator and denominator by $$t^{1/2},$$ the highest power of $$t$$ in the denominator. Then
$L = \lim\limits_{t \to + \infty } \frac{{\sqrt {t + 1} }}{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }} = \lim\limits_{t \to + \infty } \frac{{\frac{{\sqrt {t + 1} }}{{\sqrt t }}}}{{\frac{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }}{{\sqrt t }}}} = \lim\limits_{t \to + \infty } \frac{{\sqrt {\frac{{t + 1}}{t}} }}{{\sqrt {\frac{{t + \sqrt {t + 1} }}{t}} + 1}} = \lim\limits_{t \to + \infty } \frac{{\sqrt {1 + \frac{1}{t}} }}{{\sqrt {1 + \sqrt {\frac{1}{t} + \frac{1}{{{t^2}}}} } + 1}} = \frac{{\sqrt 1 }}{{\sqrt 1 + 1}} = \frac{1}{2}.$ |
# Search by Topic
#### Resources tagged with Working systematically similar to Children's Mathematical Writing:
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### There are 314 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Round the Two Dice
##### Stage: 1 Challenge Level:
This activity focuses on rounding to the nearest 10.
### Round the Dice Decimals 1
##### Stage: 2 Challenge Level:
Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number?
### Round the Three Dice
##### Stage: 2 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
### Multiply Multiples 1
##### Stage: 2 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
##### Stage: 1 and 2 Challenge Level:
How could you put these three beads into bags? How many different ways can you do it? How could you record what you've done?
### Round the Dice Decimals 2
##### Stage: 2 Challenge Level:
What happens when you round these numbers to the nearest whole number?
### Multiply Multiples 3
##### Stage: 2 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Two Spinners
##### Stage: 1 Challenge Level:
What two-digit numbers can you make with these two dice? What can't you make?
### Multiply Multiples 2
##### Stage: 2 Challenge Level:
Can you work out some different ways to balance this equation?
### How Old?
##### Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Lots of Lollies
##### Stage: 1 Challenge Level:
Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
### Number Detective
##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
### X Is 5 Squares
##### Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Trebling
##### Stage: 2 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### Button-up Some More
##### Stage: 2 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Hubble, Bubble
##### Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### Dice in a Corner
##### Stage: 2 Challenge Level:
How could you arrange at least two dice in a stack so that the total of the visible spots is 18?
### What Could it Be?
##### Stage: 1 Challenge Level:
In this calculation, the box represents a missing digit. What could the digit be? What would the solution be in each case?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Broken Toaster
##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Six Is the Sum
##### Stage: 2 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### ABC
##### Stage: 2 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Journeys in Numberland
##### Stage: 2 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### On Target
##### Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
##### Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### The Moons of Vuvv
##### Stage: 2 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Two Egg Timers
##### Stage: 2 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
### One of Thirty-six
##### Stage: 1 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### Being Thoughtful - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that require careful consideration.
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Wonky Watches
##### Stage: 2 Challenge Level:
Stuart's watch loses two minutes every hour. Adam's watch gains one minute every hour. Use the information to work out what time (the real time) they arrived at the airport.
### Fake Gold
##### Stage: 2 Challenge Level:
A merchant brings four bars of gold to a jeweller. How can the jeweller use the scales just twice to identify the lighter, fake bar?
### Pasta Timing
##### Stage: 2 Challenge Level:
Nina must cook some pasta for 15 minutes but she only has a 7-minute sand-timer and an 11-minute sand-timer. How can she use these timers to measure exactly 15 minutes?
### Button-up
##### Stage: 1 Challenge Level:
My coat has three buttons. How many ways can you find to do up all the buttons?
### Junior Frogs
##### Stage: 1 and 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Two Primes Make One Square
##### Stage: 2 Challenge Level:
Can you make square numbers by adding two prime numbers together?
### Ordered Ways of Working Upper Primary
##### Stage: 2 Challenge Level:
These activities lend themselves to systematic working in the sense that it helps if you have an ordered approach.
### Ice Cream
##### Stage: 2 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Seven Square Numbers
##### Stage: 2 Challenge Level:
Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number. |
Worksheet on Word Problems on Measurement | Measurement Word Problems Worksheet PDF
Worksheet on Word Problems on Measurement has questions on measuring length in both customary and metric units. These Math Worksheets featuring Measurement will let students explore the concepts of length, mass, volume, etc. Students can use the details from the 5th Grade Math Worksheet on Word Problems Measurement to solve and learn measuring capacities, mass, volume concept-related problems. The Fun and Interactive Measurement Word Problems Worksheet makes it easy for you to grasp the concept.
Also, check:
Measurement Word Problems with Solutions
Example 1.
A parking lot has 28 parking spots side by side. Each parking spot is 10 feet wide. What is the width of the parking lot?
Solution:
Given, No. of parking spots side by side in parking lot=28
Each parking spot=10 feet
Width of the parking lot=28 × 10=280 feet
280 feet=98.33 yards
Therefore, the width of the parking lot is 98.33 yards.
Example 2.
Each roll of the string is 80 yards. In a box, Sunil packed 200 rolls of strings. Find the total length of the string in yards?
Solution:
Given,
Each roll of the string = 80 yards
No. of string rolls in the box=200
The total length of the string rolls=200 × 80=16000 yards.
Therefore, the total length of the string is 16000 yards.
Example 3.
Sita is going to tailoring classes. Sita’s teacher told her to bring 2 pieces of cloth. They must be 1.5 m long for stitching. Sita found an old cloth at home that is 200 cm long. Find how much extra cloth is needed for Sita?
Solution:
Given,No. of pieces of cloth=2
Length of the pieces=1.5 m
Sita’s teacher told her to bring cloth=2 × 1.5 m= 3 m=300 cm (since 1m=100 cm)
Sita found an old cloth at home=200 cm
extra cloth is needed for Sita=300-200=100 cm
Hence, Sita needs 100 cm extra cloth.
Example 4.
The fan is 8 feet 6 inches from the floor. Lakshmi can reach up to 6 feet 4 inches when she raises the arm. If she stands on a stool that is 8 inches tall, Can she reach the fan?
Solution:
Given,
The distance from fan to the floor= 8 feet 6 inches
She can reach up to the fan=6 feet 4 inches
Lakshmi stands on a stool=8 inches
Lakshmi can reach the fan=6 feet 4 inches+8 inches
=6 feet 12 inches
=7 feet
Therefore, Lakshmi can not reach the fan.
Example 5.
Vijaya wants to make a tower of erasers of a height of 30 cm. If the thickness of each eraser is 2 mm. How many erasers are needed to make the desired height?
Solution:
Given,
The total length of the tower=30 cm
The thickness of each eraser=2 mm
First, convert the length of the tower to mm.
We know that 1 cm=10 mm
30 cm=30 ×10mm=300 mm
No. of erasers needed=Length of the tower/thickness of each book
=300/2=150 erasers.
Hence, the total no. of erasers needed is 150 erasers.
Example 6.
Ajay has to take three spoons of syrup each day for his health problem. If the capacity of the spoon is 15 ml, what quantity of syrup is taken by him in 10 days?
Solution:
Given,
The capacity of the spoon is=15 ml
Quantity of syrup taken by him in ten days=15 ml × 10 days=150 ml
Hence, the quantity of syrup taken by him in ten days is 150 ml.
Example 7.
How many 20 cm pieces can be cut out of a 10 m long cloth?
Solution:
Given,
cloth length=10 m=1000 cm
Length of pieces =20 cm
No. of pieces can be cut=1000/20=50
Therefore, the total no. of pieces is 50.
Example 8.
A worker transferred 20 bags of rice weighing 40 kg into the truck. The other grocery’s weight is 150 kg. The weight of the empty truck is 1590 kg. What will be the weight of the truck with the bags and the groceries?
Solution:
 No. of kg of rice=20 × 40=800 kg
The groceries weight=150 kg
weight of the empty truck=1590 kg
Weight of the truck with bags and the groceries=800 kg+150 kg+1590 kg
=2,540 kg
Hence, the weight of the truck with the bags and the groceries is 2,540 kg.
Example 9.
From a cloth of 30 cm, two pieces of length 5 cm,8 cm are cut off. What is the length of the remaining cloth?
Solution:
Given, Length of the cloth=30 cm
Length of two pieces=5 cm, 8 cm
Length of the remaining cloth=30 cm-(5 cm+8 cm)
=30 cm-13 cm
=17 cm
Therefore, the length of the remaining cloth is 17 cm.
Example 10.
8 inches of ribbon is used for wrapping up each present. How many presents can be wrapped with 7 yards of ribbons?
Solution:
Given, Length of the ribbon used for wrapping the present=8 inches
No. of presents can be wrapped with the length of ribbons=7 yards
7 yards=252 inches
252 ÷ 7=36
Hence, 36 presents can be wrapped with 7 yards of ribbons.
Example 11.
The diameter of the coin is 20 mm. Suresh lined up 50 coins from the left end of the paper to the right end of the paper. what is the width of the paper in cm?
Solution:
Given, The diameter of the coin =20 mm
No. of coins lined up from left end to right end=50 coins
Width of the paper=no. of coins × diameter of the coin
=50 × 20 mm
=1000 mm=100 cm
Therefore, the width of the paper is 100 cm.
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Composite functions – Relations and functions
• Last Updated : 21 Dec, 2020
Let f : A->B and g : B->C be two functions. Then the composition of f and g, denoted by g o f, is defined as the function g o f : A->C given by g o f (x) = g{f(x)}, ∀ x ∈ A.
Clearly, dom(g o f) = dom(f).
Also, g o f is defined only when range(f) is a subset of dom(g).
Evaluating composite functions
We know composite function is written as f o g(x), g o f(x) and so on. Here f o g(x) will be evaluated as f{g(x)} and g o f(x) will be evaluated as g{f(x)}.
Problem 1: Let f : {2, 3, 4, 5} -> {3, 4, 5, 9} and g : {3, 4, 5, 9} -> {7, 11, 15} be functions defined as
f(2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g(4) = 7 and g(5) = g(9) = 11. Find g o f(x).
Solution:
g o f(x) = g{ f(x)}. So first we find the inner bracket which is f(x) here.
We have the values of f(2), f(3), f(4) and f(5), hence we have to find the values of g o f(x) for all these values.
g o f(2) = g{ f(2) } = g(3) = 7,
g o f (3) = g{ f(3) } = g(4) = 7,
g o f(4) = g{ f(4) } = g(5) = 11
And g o f(5) = g(5) = 11.
Problem 2: Show that if f : A -> B and g : B -> C are onto, then g o f : A -> C is also onto.
Solution
Given an arbitrary element z ∈ C, there exists a pre-image y of z under g such that g(y) = z, since g is onto.
Further, for y ∈ B, there exists an element x in A with f(x) = y, since f is onto.
Therefore, g o f(x) = g{ f(x) } = g(y) = z, showing that g o f is onto.
Problem 3: Let ƒ: R->R:f(x) = (x2 – 3x + 2). Find f o f(x).
Solution:
f o f(x) = f {f(x) } = f(x2 – 3x + 2) = f(y) (let y = x2 – 3x + 2)
= y2 – 3y + 2
= (x2 – 3x + 2) – 3(x2 – 3x + 2) + 2
= x4 – 6x3 + 10x2 – 3x.
Evaluating Composite Functions: Using Tables
In this type of questions, we will be given a table with values of x, f(x), g(x) and we will need to find the composite of f(x) and g(x) like in example 1 we are asked to find f o g(1).
To find the solution, we will start from the inner bracket and find its value in the table given, so we will find the value of g(1) from the table which is 4. Then for the outer bracket again we will find its value from the given table, so we will find the value of f(4) from the table which is 2. Hence, we will get the required answer which is 2 in this case.
Problem 1: Using the table below, evaluate f o g(1) and g o f(4).
Solution:
So, f o g(1) = f{g(1)} and g o f(4) = g{f(4)}.
We first evaluate the inner bracket then the outer bracket, using the values given in table.
For f{g(1)}, g(1) = 4. Now f(4) = 2.
Hence, f{g(1)} = 2.
Similarly, for g{f(4)}, f(4) = 2. Now g(2) = 8.
Hence, g{f(4)} = 8.
Therefore, f o g(1) = f{g(1)} = f(4) = 2 and g o f(4) = g{f(4)} = g(2) = 8.
Problem 2: Using the table below, evaluate f o g(3) and f o f(1).
Solution:
So, f o g(3) = f{g(3)} and f o f(1) = f{f(1)}.
We first evaluate the inner bracket then the outer bracket, using the values given in table.
For f{g(3)}, g(3) = 5. Now f(5) = 2.
Hence, f{g(3)} = 2.
Similarly, for f{f(1)}, f(1) = 3. Now f(3) = 8.
Hence, f{f(1)} = 8.
Therefore, f o g(3) = f{g(3)} = f(5) = 2 and f o f(1) = f{f(1)} = f(3) = 8.
Evaluating composite functions: Using Graphs
In this type of questions, we will be given a graph having f(x) and g(x) curve, and we will be asked to find a composite of f(x) and g(x) like in example 1 we have to find f o g(2).
Again we will start from the inner bracket, so we have to find g(2). So we will see the g(x) curve. So at x=2, we find that y=3. Now we will need to find f(3). So we will see the f(x) curve. Now we will look at x=3, there we find y=4. Hence, we get the required solution which is 3 in example 1.
Problem 1: Using the graphs below, evaluate f o g(2).
Solution:
So, f o g(2) = f{ g(2) }
From the graph of g(x), we need to find g(2),
when x = 2, y = 3, hence g(2) = 3
Now f o g(2) = f(g(2)) = f(3).
From the graph of f(x), we need to find f(3),
when x = 3, y = 4, hence f(3) = 4
Therefore, f o g(2) = f{g(2)} = f(3) = 4.
Problem 2: Using the graph below, evaluate g o h(5).
Solution:
So, g o h(5) = g{ h(5) }
From the graph of h(x), we need to find h(5),
When x = 5, y = -2, hence h(5) = -2
Now g o h(5) = g{h(5)} = g(-2).
From the graph of g(x), we need to find g(-2),
when x = -2, y = 3, hence g(-2) = 3
Therefore, g o h(5) = g{h(5)} = g(-2) = 3.
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The Midpoint Between Two Points
# The Midpoint Between Two Points
If $A$ has coordinates $(x_1, y_1)$ and $B$ has coordinates $(x_2, y_2)$, then the midpoint of $A$ and $B$ is said to be a point $C$ that cuts the distance between $A$ and $B$ into 2 equal parts.
For example, the following diagram illustrates the midpoint of two points:
Conceptually, the midpoint can be calculated by taking the averages of the $x$-coordinates and the averages of the $y$-coordinates, and thus, the following gives a formula for calculating the midpoint:
(1)
\begin{align} \mathrm{midpoint} = \left ( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} \right ) \end{align}
For example, consider the points $(2, 3)$ and $(6, 9)$. Applying our formula, we get that the coordinates of the midpoint are:
(2)
\begin{align} \mathrm{midpoint} = \left ( \frac{2 + 6}{2} , \frac{3 + 9}{2} \right ) \\ \mathrm{midpoint} = \left ( \frac{8}{2} , \frac{12}{2} \right ) \\ \mathrm{midpoint} = (4, 6) \end{align}
## Example Questions
• 1. Calculate the midpoint of $(2, 1)$ and $(0, 9)$.
• 2. Calculate the midpoint of $(-5, 6)$ and $(2, -4)$.
• 3. Given the points $(0, 4)$ and $(0, 10)$, show that $(0, 7)$ are the coordinates of the midpoint without using the midpoint formula. |
# How to write fractions in descending order if the numerators are the same?
Solution:
Given a number of fractions whose numerators are different how to compare them and write them in descending order. It is obvious that their denominators are different and not the same. Otherwise, all the fractions would be the same or equal.
In such a situation, we find the LCM of the denominators of the fractions and then write the equivalent fractions of given fractions so that they all have same denominator equal to the LCM just now found. Now comparing the numerators makes it possible to arrange given fractions in descending order.
Example:
$\frac{2}{5}$,$\frac{2}{9}$, $\frac{2}{6}$, $\frac{2}{8}$
All the numerators of given fractions are same 2
The LCM of 5, 6, 8 and 9 is 360
The fractions become
$\frac{2\times72}{5\times72}$, $\frac{2\times40}{9\times40}$, $\frac{2\times60}{6\times60}$, $\frac{2\times45}{8\times45}$
$\frac{144}{360}$, $\frac{80}{360}$, $\frac{120}{360}$, $\frac{90}{360}$
So arranging the numerators in descending order.
144 > 120 > 90 > 80
$\frac{2}{5}$, $\frac{2}{9}$, $\frac{2}{6}$, $\frac{2}{8}$ in descending order
Hence,
$\frac{2}{5}$> $\frac{2}{6}$>$\frac{2}{8}$>$\frac{2}{9}$
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# Concavity and the Second Derivative Test
## Presentation on theme: "Concavity and the Second Derivative Test"— Presentation transcript:
Concavity and the Second Derivative Test
Determine the intervals on which the graphs of functions are concave upward or concave downward. Find the points of inflection of the graphs of functions. Use the Second Derivative Test to find the relative extrema of functions. Standard 4.5a
Concavity – the property of curving upward or downward
Concave upward f’ is increasing
f’ is decreasing Concave downward
Definition of Concavity
Let f be differentiable on an open interval I. The graph of f is Concave upward on I if f’ is increasing on the interval. Concave downward on I if f’ is decreasing on the interval.
Test for Concavity Let f be a function whose second derivative exists on an open interval I. If f ´´(x) > 0 for all x in I, then f is concave upward on I. If f ´´(x) < 0 for all x in I, then f is concave downward on I.
Determine the intervals on which the graph is concave upward or concave downward.
1. Locate the x-values at which f ´´(x) = 0 or f ´´(x) is undefined.
Use these x-values to determine the test intervals.
Test the signs of f ´´(x) in each test interval. Interval (-∞, -√3) (-√3, √3) (√3, ∞) Test Values x = -2 x = 0 x = 2 Sign of f ´´(x) f ´´(-2) > 0 f ´´(0) < 0 f ´´(2) > 0 Conclusion Concave upward Concave downward
Defintion of Point of Inflection
If the graph of a continuous function has a tangent line at point where the concavity changes from upward to downward (or vice versa) then the point is a point of inflection.
Property of Points of Inflection
If (c, f(c)) is a point of inflection of the graph of f, then either f ´´(c) = 0 or f ´´(c) is undefined at c.
Find the points of inflection of the graph.
Possible inflection point
Inflection point (4, 16) (-∞, 4) (4, ∞) x = 0 x = 5 f ´´(0) < 0
Concave down Concave up Inflection point (4, 16)
It is possible for the second derivative to be zero at a point that is not a point of inflection.
* You must test to be certain that the concavity actually changes.
Find the points of inflection and discuss the concavity of the graph of the function.
Possible points of inflections: x = 0, x = 3
(0, 3) (3, ∞) x = 1 x = 4 f ´´(1) > 0 f ´´(4) < 0 Concave upward Concave downward Inflection Point:
If f ´(c) = 0 and f ´´(c) > 0, f (c) is a relative minimum
Concave Upward f ´´(c) > 0 c If f ´(c) = 0 and f ´´(c) > 0, f (c) is a relative minimum
If f ´(c) = 0 and f ´´(c) > 0, f (c) is a relative minimum
Concave downward f ´´(c) < 0 c If f ´(c) = 0 and f ´´(c) > 0, f (c) is a relative minimum
Second – Derivative Test
Let f ´(c) = 0 and let f ´´exist on an open interval containing c. If f ´´(c) > 0, then f(c) is a relative minimum. If f ´´(c) < 0, then f(c) is a relative maximum. If f ´´(c) = 0 then the test fails. Use the First Derivative Test.
Find the relative extrema using the Second-Derivative Test
1. Find the critical numbers. Critical Numbers
(0,0) is neither a relative max or a relative min
2. Find the second derivative. 3. Plug the critical numbers into the second derivative to determine relative extrema. Relative minimum (-1, 2) Test fails Relative maximum (1, 2) (-1, 0) (0, 1) f ´(-1/2) >0 f ´(1/2) >0 Increasing (0,0) is neither a relative max or a relative min
(0,1) is neither a relative max or a relative min
Find the relative extrema using the Second-Derivative Test. Test fails Relative min (3,-26 ) (-∞, 0) (0, 3) f ´(-1) < 0 f ´(1) < 0 Decreasing (0,1) is neither a relative max or a relative min |
Annulus (mathematics)
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Annulus Mathematics
Illustration of Mamikon's visual calculus method showing that the areas of two annuli with the same chord length are the same regardless of inner and outer radii.[1]
In mathematics, an annulus (plural annuli or annuluses) is the region between two concentric circles. Informally, it is shaped like a ring or a hardware washer. The word "annulus" is borrowed from the Latin word anulus or annulus meaning 'little ring'. The adjectival form is annular (as in annular eclipse).
The open annulus is topologically equivalent to both the open cylinder S1 × (0,1) and the punctured plane.
## Area
The area of an annulus is the difference in the areas of the larger circle of radius R and the smaller one of radius r:
${\displaystyle A=\pi R^{2}-\pi r^{2}=\pi \left(R^{2}-r^{2}\right).}$
The area of an annulus is determined by the length of the longest line segment within the annulus, which is the chord tangent to the inner circle, 2d in the accompanying diagram. That can be shown using the Pythagorean theorem since this line is tangent to the smaller circle and perpendicular to its radius at that point, so d and r are sides of a right-angled triangle with hypotenuse R, and the area of the annulus is given by
${\displaystyle A=\pi \left(R^{2}-r^{2}\right)=\pi d^{2}.}$
The area can also be obtained via calculus by dividing the annulus up into an infinite number of annuli of infinitesimal width d? and area 2?? d? and then integrating from ? = r to ? = R:
${\displaystyle A=\int _{r}^{R}\!\!2\pi \rho \,d\rho =\pi \left(R^{2}-r^{2}\right).}$
The area of an annulus sector of angle ?, with ? measured in radians, is given by
${\displaystyle A={\frac {\theta }{2}}\left(R^{2}-r^{2}\right).}$
## Complex structure
In complex analysis an annulus ann(a; r, R) in the complex plane is an open region defined as
${\displaystyle r<|z-a|
If r is 0, the region is known as the punctured disk (a disk with a point hole in the center) of radius R around the point a.
As a subset of the complex plane, an annulus can be considered as a Riemann surface. The complex structure of an annulus depends only on the ratio r/R. Each annulus ann(a; r, R) can be holomorphically mapped to a standard one centered at the origin and with outer radius 1 by the map
${\displaystyle z\mapsto {\frac {z-a}{R}}.}$
The inner radius is then r/R < 1.
The Hadamard three-circle theorem is a statement about the maximum value a holomorphic function may take inside an annulus.
## References
1. ^ "The Edge of the Universe: Celebrating Ten Years of Math Horizons". Retrieved 2017. |
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# 2. The Straight Line
## Slope-Intercept Form of a Straight Line
The slope-intercept form (otherwise known as "gradient, y-intercept" form) of a line is given by:
y = mx + b
This tells us the slope of the line is m and the y-intercept of the line is b.
### Example 1
The line y = 2x + 4 has
• slope m = 2 and
• y-intercept b = 4.
We do not need to set up a table of values to sketch this line. Starting at the y-intercept (y = 4), we sketch our line by going up 2 units for each 1 unit we go to the right (since the slope is 2 in this example).
To find the x-intercept, we let y = 0.
2x + 4 = 0
x = -2
We notice that this is a function. That is, each value of x that we have gives one corresponding value of y.
See more on Functions and Graphs.
## Point-Slope Form of a Straight Line
We need other forms of the straight line as well. A useful form is the point-slope form (or point - gradient form). We use this form when we need to find the equation of a line passing through a point (x1, y1) with slope m:
y − y1 = m(xx1)
### Example 2
Find the equation of the line that passes through (-2, 1) with slope of -3.
We use:
y-y_1=m(x-x_1)
Here,
x_1= -2
y_1= 1
m = -3
So the required equation is:
y-1=-3(x-(-2)=-3x-6
y=-3x-5
We have left it in slope-intercept form. We can see the slope is -3 and the y-intercept is -5.
## General Form of a Straight Line
Another form of the straight line which we come across is general form:
Ax + By + C = 0
It can be useful for drawing lines by finding the y-intercept (put x = 0) and the x-intercept (put y = 0).
We also use General Form when finding Perpendicular Distance from a Point to a Line.
### Example 3
Draw the line 2x + 3y + 12 = 0.
If x = 0, we have: 3y + 12 = 0, so y = -4.
If y = 0, we have: 2x + 12 = 0, so x = -6.
So the line is:
The line 2x + 3y + 12 = 0.
Note that the y-intercept is -4 and the x-intercept is -6.
### Exercises
1. What is the equation of the line perpendicular to the line joining (4, 2) and (3, -5) and passing through (4, 2)?
[Need a reminder? See the section on Slopes of Perpendicular Lines.]
The line joining (4, 2) and (3, -5) has slope m=(-7)/(-1)=7 and is shown as a green dotted line.
Perpendicular lines.
We need to find the equation of the magenta (pink) line.
The line perpendicular to the green dotted line has slope -1/7.
The line through (4, 2) with slope -1/7 has equation:
y-2=-1/7(x-4)
=-x/7+4/7
y=-x/7+2 4/7
2. If 4x − ky = 6 and 6x + 3y + 2 = 0 are perpendicular, what is the value of k?
(2) The slope of 4xky = 6 can be calculated by re-expressing it in slope-intercept form:
y=4/kx-6/k
So we see the slope is 4/k.
The slope of 6x + 3y + 2 = 0 can also be calculated by re-expressing it in slope-intercept form:
y=(-6)/3x-2/3=-2x-2/3
So we see the slope is -2.
For the lines to be perpendicular, we need
4/kxx-2=-1
This gives k = 8.
The resulting line is 4x-8y=6, which we can simplify to 2x-4y=3. Here's the graph of the situation:
6x+3y+2=0
2x-4y=3
Perpendicular lines
### Conic section: Straight line
Each of the lines and curves in this chapter are conic sections, which means the curves are formed when we slice a cone at a certain angle.
How can we obtain a straight line from slicing a cone?
We start with a double cone (2 right circular cones placed apex to apex):
If we slice the double cone by a plane just touching one edge of the double cone, the intersection is a straight line, as shown. |
In Tessellations: The mathematics of Tiling post, we have actually learned that over there are just three constant polygons that have the right to tessellate the plane: squares, equilateral triangles, and also regular hexagons. In Figure 1, we deserve to see why this is so. The angle sum of the interior angles of the consistent polygons meeting at a point include up come 360 degrees.
You are watching: Which regular polygon can be used to form a tessellation?
Figure 1 – Tessellating regular polygons.
Looking in ~ the other continuous polygons as shown in number 2, we have the right to see plainly why the polygons cannot tessellate. The sums that the inner angles room either higher than or less than 360 degrees.
Figure 2 – Non-tessellating regular polygons.
In this post, we room going to display algebraically the there are just 3 continual tessellations. Us will usage the notation
, similar to what we have used in the proof the there are only 5 platonic solids, to stand for the polygons meeting at a allude where
is the variety of sides and also
is the variety of vertices. Using this notation, the triangular tessellation can be represented as
since a triangle has actually 3 sides and also 6 vertices meet at a point.
In the proof, as shown in number 1, we are going to show that the product of the measure up of the inner angle of a consistent polygon multiply by the variety of vertices meeting at a suggest is same to 360 degrees.
Theorem: There are just three regular tessellations: it is intended triangles, squares, and also regular hexagons.
See more: Can You Get Spawn Eggs In Survival Mode ? How To Retrieve Spawn Eggs In Survival Mode
Proof:
The angle amount of a polygon v
political parties is
. This way that each interior angle that a constant polygon measures
. The number of polygons meeting at a suggest is
. The product is therefore
which simplifies to
. Using Simon’s favourite Factoring Trick, we include
to both sides offering us
. Factoring and also simplifying, us have
, i beg your pardon is indistinguishable to
. Observe that the only feasible values for
are
(squares),
(regular hexagons), or
(equilateral triangles). This means that these space the only consistent tessellations possible which is what we want to prove. |
Find the lengths of the medians of a triangle
Question:
Find the lengths of the medians of a triangle whose vertices are A (−1,3), B(1,−1) and C(5, 1).
Solution:
We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (−1, 3); B (1,−1) and C (5, 1).
So we should find the mid-points of the sides of the triangle.
In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,
$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$
Therefore mid-point P of side AB can be written as,
$P(x, y)=\left(\frac{-1+1}{2}, \frac{3-1}{2}\right)$
Now equate the individual terms to get,
$x=0$
$y=1$
So co-ordinates of P is (0, 1)
Similarly mid-point Q of side BC can be written as,
$Q(x, y)=\left(\frac{5+1}{2}, \frac{1-1}{2}\right)$
Now equate the individual terms to get,
$x=3$
$y=0$
So co-ordinates of Q is (3, 0)
Similarly mid-point R of side AC can be written as,
$R(x, y)=\left(\frac{5-1}{2}, \frac{1+3}{2}\right)$
Now equate the individual terms to get,
$x=2$
$y=2$
So co-ordinates of $Q$ is $(2,2)$
Therefore length of median from A to the side BC is,
$\mathrm{AQ}=\sqrt{(-1-3)^{2}+(3-0)^{2}}$
$=\sqrt{16+9}$
$=5$
Similarly length of median from B to the side AC is,
$B R=\sqrt{(1-2)^{2}+(-1-2)^{2}}$
$=\sqrt{1+9}$
$=\sqrt{10}$
Similarly length of median from C to the side AB is
$\mathrm{CP}=\sqrt{(5-0)^{2}+(1-1)^{2}}$
$=\sqrt{25}$
$=5$ |
To learn the whole story of a line, you sometimes need to crunch some numbers.
The line is one of the simplest mathematical objects. Even so, finding all the information about a single line sometimes requires some number-crunching. When you know partial information about your line, you will often have to plug your known information into formulas to describe the rest of the line. The endpoint formula is one such formula, and it can help you find the unknown endpoint of an otherwise well-understood line.
### The Problem: Where’s the Rest of My Line?
Often in math, you are left with pieces of information with which you must piece together the entire situation. This often occurs in geometry, a subject in which you must infer conclusions about lines and other shapes based on partial information about them. In some cases, you might be looking at information describing a line segment, but are missing the endpoint of the line. A line segment is defined by its two endpoints, the knowledge of which allows you to fully describe the line segment in mathematics. Thus, you need to know both endpoints before you can clearly label or define a line segment.
### The Midpoint Formula: Half of the Story
Knowing the midpoint formula for a line segment will often help you derive the endpoint. Recall the midpoint formula being the mathematical formula that allows you to find the point lying in the center of a line segment, based on the two endpoints. Specifically, if your endpoints are (x0, y0) and (x2, y2), the midpoint will have an x-coordinate at (x0+x2)/2 and a y-coordinate at (y0+y2)/2. By plugging in the endpoints into the midpoint formula, you can find the midpoint of the segment. But what many students don’t notice is that you can reverse the midpoint equation to find the endpoint.
### The Endpoint Formula: The Rest of the Story
If you have one endpoint and the segment’s midpoint, you can apply a modification of the midpoint formula to derive the other endpoint. The midpoint formula lets you find a coordinate by plugging the information you know into p1 = (p0 + p2)/2. If you don’t know one of the endpoints, either p0 or p2, but do know the midpoint, p1, you can solve this equation for the other endpoint. Using algebra, you can rewrite the midpoint formula as 2_p1-p0. Thus, your endpoint has an x-coordinate at 2_x1-x0 and a y-coordinate at 2*y1-y0, where (x1,y1) is the midpoint of the line segment and (x0,y0) is the endpoint of the segment.
### A Quick Example
Assume you have a line segment with an endpoint at (-2,7) and a midpoint at (-9,2). You want to find the other endpoint through the endpoint formula, so you first label your variables: x0=-2, y0=7, x1=-9, and y1=2. Find the x-coordinate first: x2=2_x1-x0=2_(-9)-(-2)=-16. Find the y-coordinate next: y2=2_y1-y0=2_2-7=-3. Thus, your other endpoint is at (-16,-3). |
# Compute Power of Matrix If Eigenvalues and Eigenvectors Are Given
## Problem 373
Let $A$ be a $3\times 3$ matrix. Suppose that $A$ has eigenvalues $2$ and $-1$, and suppose that $\mathbf{u}$ and $\mathbf{v}$ are eigenvectors corresponding to $2$ and $-1$, respectively, where
$\mathbf{u}=\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \text{ and } \mathbf{v}=\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}.$ Then compute $A^5\mathbf{w}$, where
$\mathbf{w}=\begin{bmatrix} 7 \\ 2 \\ -3 \end{bmatrix}.$
## Solution.
Since $\mathbf{u}$ is an eigenvector corresponding to the eigenvalue $2$, we have
$A\mathbf{u}=2\mathbf{u}.$ Similarly, we have
$A\mathbf{v}=-\mathbf{v}.$ From these, we have
$A^5\mathbf{u}=2^5\mathbf{u} \text{ and } A\mathbf{v}=(-1)^5\mathbf{v}.$
To compute $A^5\mathbf{w}$, we first need to express $\mathbf{w}$ as a linear combination of $\mathbf{u}$ and $\mathbf{v}$. Thus, we need to find scalars $c_1, c_2$ such that
$\mathbf{w}=c_1\mathbf{u}+c_2\mathbf{v}.$ By inspection, we have
$\begin{bmatrix} 7 \\ 2 \\ -3 \end{bmatrix}=3\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}+2\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix},$ and thus we obtain $c_1=3$ and $c_2=2$,
We compute $A^5\mathbf{w}$ as follows:
\begin{align*}
A^5\mathbf{w}&=A^5(3\mathbf{u}+2\mathbf{v})\\
&=3A^5\mathbf{u}+2A^5\mathbf{v}\\
&=3\cdot 2^5\mathbf{u}+2\cdot (-1)^5\mathbf{v}\\
&=96\mathbf{u}-2\mathbf{v}\6pt] &=96\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}-2\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 92 \\ -2 \\ -96 \end{bmatrix}. \end{align*} Therefore, the result is \[A^5\mathbf{w}=\begin{bmatrix} 92 \\ -2 \\ -96 \end{bmatrix}.
Let $S$ be the subset of $\R^4$ consisting of vectors $\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix}$ satisfying... |
# MSBSHSE Solutions For Class 9 Maths Part 1 Chapter 2 Real Numbers
MSBSHSE Solutions For Class 9 Maths Part 1 Chapter 2 Real Numbers consists of accurate solutions, which help the students to quickly complete their homework and prepare well for the exams. These solutions provide students an advantage with practical questions. Each step in the solution is explained to match students’ understanding. To score good marks in Class 9 Mathematics examination, it is advised they solve questions provided at the end of each chapter in the Maharashtra Board Textbooks for Class 9. In the Maharashtra State Board Solutions for Class 9 Chapter 2, students will learn and solve problems based on topics like properties of rational numbers, properties of irrational numbers, surds, comparison of quadratic surds, operations on quadratic surds and Rationalization of quadratic surds.
## Download the PDF of Maharashtra Solutions For Class 9 Maths Part 1 Chapter 2 Real Numbers
### Access answers to Maths MSBSHSE Solutions For Class 9 Part 1 Chapter 2 – Real Numbers
Practice set 2.1 Page no: 21
1. Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.
i. 13/5Â Â
ii.Â
2/11
iii. 29/16Â
iv. 17/125
v. 11/6
Solution:
i.
∵ The division is exact
∴ it is a terminating decimal.
ii.
∵ The division never ends and the digits ‘18’ is repeated endlessly
∴ it is a non-terminating recurring type decimal.
iii.
∵ The division is exact
∴ it is a terminating decimal.
iv.
∵ The division is exact
∴ it is a terminating decimal.
v.
∵ The division never ends and the digit ‘3’ is repeated endlessly
∴ it is a non-terminating recurring type decimal.
2. Write the following rational numbers in decimal form.
i. 127/200Â
ii. 25/99
iii. 23/7Â
iv. 4/5
v. 17/8Â
Solution:
i.
ii.
iii.
iv.
v.
3. Write the following rational numbers in form.
i. Â
ii.Â
iii. Â
iv.Â
v.Â
Solution:
Practice set 2.2 Page no: 25
1. Show that is 4√2 an irrational number.
Solution:
2. Prove that 3 + √5 is an irrational number.
Solution:
3. Represent the numbers √5 and √10 on a number line.
Solution:
Given √5
Given that √10
4. Write any three rational numbers between the two numbers given below.
(i) 0.3 and -0.5
Solution:
(ii) -2.3 and -2.33
Solution:
(iii) 5.2 and 5.3
Solution:
(iv) -4.5 and 4.6
Solution:
Practice set 2.3 Page no: 30
1. State the order of the surds given below.
Solution:
2. State which of the following are surds. Justify.
Solution:
iv. √256 = √162 = 16
Surds are numbers left in root form (√) to express its exact value. It has an infinite number of non-recurring decimals. Therefore, surds are irrational numbers.
It is not a surd ∵ it is a rational number.
3. Classify the given pair of surds into like surds and unlike surds.
i. √52, 5√13
ii. √68, 5√3
iii. 4√18, 7√2
iv. 19√12, 6√3
v. 5√22, 7√33
vi. 5√5, √75
Solution:
Two or more surds are said to be similar or like surds if they have the same surd-factor.
Two or more surds are said to be dissimilar or unlike when they are not similar.
Therefore,
i. Given √52, 5√13
given surd can be written as
√52 = √ (2×2×13) = 2√13
5√13
∵ both surds have same surd-factor that is √13.
∴ they are like surds.
ii. Given √68, 5√3
Given surd can be written as
√68 = √ (2×2×17) = 2√17
5√3
∵ both surds have different surd-factors √17 and √3.
∴ they are unlike surds.
iii. Given 4√18, 7√2
Given surd can be written as
4√18 = 4 √ (2×3×3) = 4×3√2 = 12√2
7√2
∵ both surds have same surd-factor i.e., √2.
∴ they are like surds.
iv. Given 19√12, 6√3
Given surd can be written as
19√12 = 19√ (2×2×3) = 19×2√3 = 38√3
6√3
∵ both surds have same surd-factor i.e., √3.
∴ they are like surds.
v. Given 5√22, 7√33
∵ both surds have different surd-factors √22 and √33.
∴ they are unlike surds.
vi. Given 5√5, √75
5√5
Given surd can be written as
√75 = √ (5×5×3) = 5√3
∵ both surds have different surd-factors √5 and √3.
∴ they are unlike surds.
4. Simplify the following surds.
i. √27
ii. √50
iii. √250
iv. √112
v. √168
Solution:
Practice set 2.4 Page no: 32
1. Multiply
i. √3(√7 – √3)
ii. (√5 – √7) √2
iii. (3√2 – √3) (4√3 – √2)
Solution:
i. Given √3 (√7 – √3)
=√3 × √7 – √3 × √3
[∵ √a (√b – √c) = √a × √b – √a × √c]
=√21 – 3
ii. Given (√5 – √7) √2
=√5 × √2 – √7 × √2
[∵√a (√b – √c) = √a × √b – √a × √c]
= √10 – √14
iii. Given (3√2 – √3) (4√3 – √2)
=3√2 (4√3 – √2) – √3 (4√3 – √2)
[∵ √a (√b – √c) = √a × √b – √a × √c]
= 3√2 × 4√3 – 3√2 × √2 – √3 × 4√3 + √3 × √2
= 12√6 – 3 × 2 – 4 × 3 + √6
= 12√6 – 6 – 12 + √6
= 13√6 – 18
2. Rationalize the denominator.
iv. Â
Solution:
Practice set 2.5 Page no: 33
1. Find the value.
(i) |15 – 2|
(ii) |4 – 9|
(iii) |7| × |-4|
Solution:
i. Given |15 – 2|
Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative.
Therefore,
|15 – 2| = |13| = 13
ii. Given |4 – 9|
Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative.
Therefore,
|4 – 9| = |-5| = 5
iii. Given |7| × |-4|
Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative.
Therefore,
|7| × |-4| = 7 × 4 = 28
2. Solve.
i. |3x – 5| = 1
ii. |7 – 2x| = 5
iii.Â
iv.Â
Solution:
Problem set 2 Page no: 34
1. Choose the correct alternative answer for the questions given below.
i. Which one of the following is an irrational number?
A. √16/25
B. √5
C. 3/9
D. √196
Solution:
B. √5
Explanation:
An irrational number is a number that cannot be expressed as a fraction p/q for any integers p and q and q ≠0.
Since √5 cannot be written as p/q it is an irrational number
Therefore √5 is an irrational number.
ii. Which of the following is an irrational number?
A. 0.17
B.Â
C.Â
D. 0.101001000….
Solution:
D. 0.101001000….
Explanation:
An irrational number is a number that cannot be expressed as a fraction p/q for any integers p and q and q ≠0.
0.101001000…. is an irrational number because it is a non-terminating and non-`repeating decimal.
Therefore, 0.101001000…. is an irrational number.
iii. Decimal expansion of which of the following is non-terminating recurring?
A. 2/5
B. 3/16
C. 3/11
D. 137/25
Solution:
C. 3/11
Explanation:
A non-terminating recurring decimal representation means that the number will have an infinite number of digits to the right of the decimal point and those digits will repeat themselves.
∵ it has an infinite number of digits to the right of the decimal point which are repeating themselves ∴ it is a non-terminating recurring decimal.
iv. Every point on the number line represent, which of the following numbers?
A. Natural numbers
B. Irrational numbers
C. Rational numbers
D. Real numbers.
Solution:
D. Real numbers.
Explanation:
Every point of a number line is assumed to correspond to a real number, and every real number to a point. Therefore, every point on the number line represent a real number.
v. The number 0.4 in p/q form is ………….
A. 4/9
B. 40/9
C. 3.6/9
D. 36/9
Solution:
C. 3.6/9
Explanation:
vi. What is √n, if n is not a perfect square number?
A. Natural number
B. Rational number
C. Irrational number
D. Options A, B, C all are correct.
Solution:
C. Irrational number
Explanation:
If n is not a perfect square number, then √n cannot be expressed as ratio of a and b where a and b are integers and b ≠0
Therefore, √n is an Irrational number
vii. Which of the following is not a surd?
A. √7
B. 3√17
C. 3√64
D. √193
Solution:
C. 3√64
Explanation:
viii. What is the order of the surd ?
A. 3
B. 2
C. 6
D. 5
Solution:
C. 6
Explanation:
ix. Which one is the conjugate pair of 2√5 + √3?
A. -2√5 + √3
B. -2√5 – √3
C. 2√3 + √5
D. √3 + 2√5
Solution:
A. -2√5 + √3
Explanation:
A math conjugate is formed by changing the sign between two terms in a binomial. For instance, the conjugate of x + y is x – y.
Now,
2√5 + √3 = √3 + 2√5
Its conjugate pair = √3 – 2√5 = -2√5 + √3
∴ The conjugate pair of 2√5 + √3 = -2√5 + √3
x. The value of |12 – (13 + 7) × 4| is ………..
A. -68
B. 68
C. -32
D. 32
Solution:
B. 68
Explanation:
|12 – (13 + 7) × 4| = |12 – 20 × 4|
(Solving it according to BODMAS)
⇒ |12 – (13 + 7) × 4| = |12 – 80|
⇒ |12 – (13 + 7) × 4| = |-68|
⇒ |12 – (13 + 7) × 4| = 68
Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they can be represented in the number line. Students can depend on these Solutions to understand all the topics completely. Stay tuned to learn more about Real Numbers, MSBSHSE Exam pattern and other information.
## Frequently Asked Questions on Maharashtra State Board Solutions for Class 9 Maths Part 1 Chapter 2 Real Numbers
### Will I have to make any payment to access these Maharashtra State Board Solutions for Class 9 Maths Part 1 Chapter 2 Real Numbers?
No, there is no need to pay any money. We have made these solutions available online free for download. Students can access them from our site by entering the login details. Or else, the scrollable PDF is also accessible. Those who wish to can also check out the questions and the solutions from our webpage.
### Are these Maharashtra State Board Solutions for Class 9 Maths Part 1 Chapter 2 Real Numbers helpful?
Yes. These solutions are useful and they set the basis for questions that could get asked in the board exams. Students are also advised to practise these questions first and then refer back to the solutions to analyse their performance and then rectify the mistakes, so that they can avoid making any during the board exams.
### Is Maharashtra Board Class 9 Maths Solutions of Chapter 2 difficult?
No, Maharashtra Board Class 9 Maths Solutions of Chapter 2Â Real Numbers is not half as difficult as it seems. You can easily understand concepts of the chapter if you refer to this chapterwise solutions . You can also go through your syllabus course structure and unit-wise weightage thoroughly to know how much weightage the chapter will carry for the final exam. |
# Constant of integration
(Redirected from Arbitrary constant of integration)
In calculus, the indefinite integral of a given function (i.e., the set of all antiderivatives of the function) on a connected domain is only defined up to an additive constant, the constant of integration.[1][2] This constant expresses an ambiguity inherent in the construction of antiderivatives. If a function ${\displaystyle f(x)}$ is defined on an interval and ${\displaystyle F(x)}$ is an antiderivative of ${\displaystyle f(x)}$, then the set of all antiderivatives of ${\displaystyle f(x)}$ is given by the functions ${\displaystyle F(x)+C}$, where C is an arbitrary constant (meaning that any value for C makes ${\displaystyle F(x)+C}$ a valid antiderivative). The constant of integration is sometimes omitted in lists of integrals for simplicity.
## Origin of the constant
The derivative of any constant function is zero. Once one has found one antiderivative ${\displaystyle F(x)}$ for a function ${\displaystyle f(x)}$, adding or subtracting any constant C will give us another antiderivative, because ${\displaystyle (F(x)+C)'=F\,'(x)+C\,'=F\,'(x)}$. The constant is a way of expressing that every function with at least one antiderivative has an infinite number of them.
For example, suppose one wants to find antiderivatives of ${\displaystyle \cos(x)}$. One such antiderivative is ${\displaystyle \sin(x)}$. Another one is ${\displaystyle \sin(x)+1}$. A third is ${\displaystyle \sin(x)-\pi }$. Each of these has derivative ${\displaystyle \cos(x)}$, so they are all antiderivatives of ${\displaystyle \cos(x)}$.
It turns out that adding and subtracting constants is the only flexibility we have in finding different antiderivatives of the same function. That is, all antiderivatives are the same up to a constant. To express this fact for cos(x), we write:
${\displaystyle \int \cos(x)\,dx=\sin(x)+C.}$
Replacing C by a number will produce an antiderivative. By writing C instead of a number, however, a compact description of all the possible antiderivatives of cos(x) is obtained. C is called the constant of integration. It is easily determined that all of these functions are indeed antiderivatives of ${\displaystyle \cos(x)}$:
{\displaystyle {\begin{aligned}{\frac {d}{dx}}[\sin(x)+C]&={\frac {d}{dx}}[\sin(x)]+{\frac {d}{dx}}[C]\\&=\cos(x)+0\\&=\cos(x)\end{aligned}}}
## Necessity of the constant
At first glance it may seem that the constant is unnecessary, since it can be set to zero. Furthermore, when evaluating definite integrals using the fundamental theorem of calculus, the constant will always cancel with itself.
However, trying to set the constant equal to zero doesn't always make sense. For example, ${\displaystyle 2\sin(x)\cos(x)}$ can be integrated in at least three different ways:
{\displaystyle {\begin{aligned}\int 2\sin(x)\cos(x)\,dx&=&\sin ^{2}(x)+C&=&-\cos ^{2}(x)+1+C&=&-{\frac {1}{2}}\cos(2x)+C\\\int 2\sin(x)\cos(x)\,dx&=&-\cos ^{2}(x)+C&=&\sin ^{2}(x)-1+C&=&-{\frac {1}{2}}\cos(2x)+C\\\int 2\sin(x)\cos(x)\,dx&=&-{\frac {1}{2}}\cos(2x)+C&=&\sin ^{2}(x)+C&=&-\cos ^{2}(x)+C\end{aligned}}}
So setting C to zero can still leave a constant. This means that, for a given function, there is no "simplest antiderivative".
Another problem with setting C equal to zero is that sometimes we want to find an antiderivative that has a given value at a given point (as in an initial value problem). For example, to obtain the antiderivative of ${\displaystyle \cos(x)}$ that has the value 100 at x = π, then only one value of C will work (in this case C = 100).
This restriction can be rephrased in the language of differential equations. Finding an indefinite integral of a function ${\displaystyle f(x)}$ is the same as solving the differential equation ${\displaystyle {\frac {dy}{dx}}=f(x)}$. Any differential equation will have many solutions, and each constant represents the unique solution of a well-posed initial value problem. Imposing the condition that our antiderivative takes the value 100 at x = π is an initial condition. Each initial condition corresponds to one and only one value of C, so without C it would be impossible to solve the problem.
There is another justification, coming from abstract algebra. The space of all (suitable) real-valued functions on the real numbers is a vector space, and the differential operator ${\displaystyle {\frac {d}{dx}}}$ is a linear operator. The operator${\displaystyle {\frac {d}{dx}}}$ maps a function to zero if and only if that function is constant. Consequently, the kernel of ${\displaystyle {\frac {d}{dx}}}$ is the space of all constant functions. The process of indefinite integration amounts to finding a preimage of a given function. There is no canonical preimage for a given function, but the set of all such preimages forms a coset. Choosing a constant is the same as choosing an element of the coset. In this context, solving an initial value problem is interpreted as lying in the hyperplane given by the initial conditions.
## Reason for a constant difference between antiderivatives
This result can be formally stated in this manner: Let ${\displaystyle F:\mathbb {R} \rightarrow \mathbb {R} }$ and ${\displaystyle G:\mathbb {R} \rightarrow \mathbb {R} }$ be two everywhere differentiable functions. Suppose that ${\displaystyle F\,'(x)=G\,'(x)}$ for every real number x. Then there exists a real number C such that ${\displaystyle F(x)-G(x)=C}$ for every real number x.
To prove this, notice that ${\displaystyle [F(x)-G(x)]'=0}$. So F can be replaced by F-G and G by the constant function 0, making the goal to prove that an everywhere differentiable function whose derivative is always zero must be constant:
Choose a real number a, and let ${\displaystyle C=F(a)}$. For any x, the fundamental theorem of calculus, together with the assumption that the derivative of F vanishes, implies that
{\displaystyle {\begin{aligned}0&=\int _{a}^{x}F'(t)\,dt\\&=F(x)-F(a)\\&=F(x)-C,\end{aligned}}}
and therefore ${\displaystyle F(x)=C}$. So F is a constant function.
Two facts are crucial in this proof. First, the real line is connected. If the real line were not connected, we would not always be able to integrate from our fixed a to any given x. For example, if we were to ask for functions defined on the union of intervals [0,1] and [2,3], and if a were 0, then it would not be possible to integrate from 0 to 3, because the function is not defined between 1 and 2. Here there will be two constants, one for each connected component of the domain. In general, by replacing constants with locally constant functions, we can extend this theorem to disconnected domains. For example, there are two constants of integration for ${\displaystyle \textstyle \int dx/x}$ and infinitely many for ${\displaystyle \textstyle \int \tan x\,dx,}$ so for example the general form for the integral of 1/x is:[3]
${\displaystyle \int {1 \over x}\,dx={\begin{cases}\ln \left|x\right|+C^{-}&x<0\\\ln \left|x\right|+C^{+}&x>0\end{cases}}}$
Second, F and G were assumed to be everywhere differentiable. If F and G are not differentiable at even one point, the theorem fails. As an example, let ${\displaystyle F(x)}$ be the Heaviside step function, which is zero for negative values of x and one for non-negative values of x, and let ${\displaystyle G(x)=0}$. Then the derivative of F is zero where it is defined, and the derivative of G is always zero. Yet it's clear that F and G do not differ by a constant. Even if it is assumed that F and G are everywhere continuous and almost everywhere differentiable the theorem still fails. As an example, take F to be the Cantor function and again let G = 0.
## References
1. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 0-495-01166-5.
2. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 0-547-16702-4.
3. ^ "Reader Survey: log|x| + C", Tom Leinster, The n-category Café, March 19, 2012 |
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# How to Measure Angles and Importance
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## Importance of Angles
An angle measure can be defined as the measurement of an angle formed by the two rays or arms at a common vertex. We take the importance of angles for granted. Whenever two lines meet at a point, it forms an angle. The point where these lines meet is called a node or a vertex. The term "Angle" originated from the Latin "Angulus". If you look around, everything touching another surface is at some angle or the other.
Let's discuss the different types of angles, what they are called and how they are used in mathematics.
## Types of Angles
Angles are named according to their measurements. Some of the angles which are important to understand include:
• Acute angle
Acute Angle: An angle which is less than 90 degrees is acute.
Right Angle: An angle of exactly 90 degrees is a right angle.
Obtuse Angle: An angle greater than 90 degrees and less than 180 degrees is an obtuse angle.
Straight Angle: An angle which has a 180-degree angle is straight.
Different Types of Angles
## Measurement of an Angle
To measure angle is typically thought to be the process of determining a specific angle, converting it into an equivalent angular measure and calculating its lower or upper tangent. The first question that comes to mind is, what is the definition of an angle?
• The simple definition of an angle is the figure formed when two rays share a common endpoint.
• The standard unit for measuring an angle is called a degree. The angles can be measured using a protractor.
• We measure the angles with either a protractor or it can also be constructed using a compass.
• An angle is measured in degrees, and so it is called degree measure—the measure of an angle formed by the two rays or arms at a common vertex.
• Angle measurement is the amount of rotation from the initial to an endpoint of a ray.
• The angle is said to be positive if the rotation is clockwise and negative if the rotation is anticlockwise.
• Various units can measure Angles. The standard unit of measuring an angle is Degree and Radian.
Protractor
## Use of Angles in Everyday Life
Here are a bunch of ways you can use angles to get the most out of your life!
• Visualise: Angles can help you understand complex problems. For instance, with a problem that needs more than one angle to solve, you can draw the shapes and think about the angles easily.
• Use Angles in Life: You can also use them in life. For example, driving is one of those activities where there are a lot of traffic signs. So drawing down an angle helps you remember what sign to look for in conversation.
• Carpenters use them to measure precisely to build doors, chairs, tables, etc.
• Athletes use them to gauge the distances of a throw and enhance their sports performance.
• Engineers construct buildings, bridges, houses, monuments, etc., using angle measurement.
## Solved Examples
Q 1. What is the measure of the angle in degrees between:
(i) North and West?
(ii) North and South?
(iii) North and South-East?
Direction
Ans: The measure of the angle between:
(i) North and west is \$90^{\circ}\$.​
(ii) North and South is \$180^{\circ}\$.​
(iii) North and South-East is \$135^{\circ}\$.
Q 2. How many angles are formed in the figures? Name them.
Angles
Ans: The angles formed in figure 1: \$\angle{ABC}\$,\$\angle{ACB}\$ and \$\angle{BAC}\$
The angles formed in figure 2: \$\angle{ABC}\$, \$\angle{BCD}\$, \$\angle{CDA}\$, \$\angle{DAB}\$.
The angles formed in figure 3: \$\angle{ABC}\$, \$\angle{BCD}\$, \$\angle{CDA}\$, \$\angle{DAB}\$, \$\angle{CAB}\$, \$\angle{CAD}\$.
Q 3. Write two examples of obtuse angles and reflex angles.
Ans: As we know, obtuse angles are the angles that measure less than \$180^{\circ}\$ and greater than \$90^{\circ}\$
Examples: \$112^{\circ}, 177^{\circ}\$
Reflex angles measure less than \$360^{\circ}\$ and greater than \$180^{\circ}\$.
Examples: \$210^{\circ}, 300^{\circ}\$
## Practice Questions
Q 1. Classify the following angles into acute, obtuse, right and reflex angles:
(i) 35°
(ii) 185°
(iii) 92°
(iv) 260°
Ans: (i) 35°: Acute angle
(ii) 185°: Obtuse angle
(iii) 92°: Obtuse angle
(iv) 260°: Reflex angle
Q 2. Classify the given angle in the image.
Angle
Ans: The given angle is making an angle of 90 degrees. So it is a right angle.
Q 3. How many 45-degree angles are there in a straight angle?
Ans: A straight angle divided into two halves will be 90° each, and 90° divided into two halves will be 45° each. So, there are four 45° angles in the straight line drawn. Therefore, four 45° angles make a straight line.
## Summary
This includes all about types of angles, classification of angles, and measurement of angles. We have studied it with very simple definitions. Two rays intersecting at the same point is an angle. In all geometry shapes, angles are present. Angles are expressed in degrees. An angle can be measured by using a Protractor. An angle is an arc that joins two points on a straight line.
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# ML Aggarwal Solutions for Chapter 14 Locus Class 10 Maths ICSE
Here, we are providing the solutions for Chapter 14 Locus from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of twelfth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 14 Locus of ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on relation between pints, locus of path and point, circles, equidistant points and constructing circles and triangles. We have also added chapter test and multiple choice questions.
### Exercise 14
1. A point moves such that its distance from a fixed line AB is always the same. What is the relation between AB and the path travelled by P?
Consider point P which moves in such a way that it is at a fixed distance from the fixed line AB.
Here it is a set of two lines l and m which is parallel to AB drawn on either side at an equal distance from it.
2. A point P moves so that its perpendicular distance from two given lines AB and CD are equal. State the locus of the point P.
(i) We know that if two lines AB and CD are parallel, then the locus of point P which is equidistant from AB and CD is a line (l) in the midway of lines AB and CD and is parallel to them.
(ii) If both AB and CD are intersecting lines, then the locus of point P will be l and m which is a pair of straight lines bisecting the angles between AB and CD.
3. P is a fixed point and a point Q moves such that the distance PQ is constant, what is the locus of the path traced out by the point Q?
Consider P as a fixed point and Q as a moving point which is always at an equidistant from P.
Here P is the center of the path of Q which is a circle.
We know that the distance between the points P and Q is the radius of the circle.
Therefore, locus of point Q is a circle with P as center.
4. (i) AB is a fixed line. State the locus of the point P so that APB = 900.
(ii) A, B are fixed points. State the locus of the point P so that APB = 900.
(i) It is given that
AB is a fixed line and P is a point such that ∠APB = 900
Here the locus of point P will be the circle where AB is the diameter.
We know that the angle in a semi-circle is equal to 900 where ∠APB = 900
(ii) It is given that
AB is a fixed line and P is a point such that ∠APB = 600
Here the locus of point P will be a major segment of circle where AB is a chord.
5. Draw and describe the locus in each of the following cases:
(i) The locus of points at a distance 2.5 cm from a fixed line.
(ii) The locus of vertices of all isosceles triangles having a common base.
(iii) The locus of points inside a circle and equidistant from two fixed points on the circle.
(iv) The locus of centers of all circles passing through two fixed points.
(v) The locus of a point in rhombus ABCD which is equidistant from AB and AD.
(vi) The locus of a point in the rhombus ABCD which is equidistant from points A and C.
(i)
• Construct a line AB.
• Construct lines l and m which are parallel to AB at a distance of 2.5 cm.
Here, lines l and m are the locus of point P at a distance of 2.5 cm.
(ii) It is given that
Δ ABC is an isosceles triangle where AB = AC.
Taking A as center construct a perpendicular AD to BC.
Here, AD is the locus of point A which are the vertices of Δ ABC
In ΔABD and ΔACD
The sides AD = AD is common
It is given that
Hypotenuse AB = AC
According to RHS Axiom
Δ ABD = Δ ACD
BD = DC (c.p.c.t)
Therefore, locus of vertices of isosceles triangles having common base is the perpendicular bisector of BC.
(iii)
• Construct a circle with O as center.
• Take points A and B on it and join them.
• Construct a perpendicular bisector of AB which passes from point O and meets the circle at C.
Here, CE which is the diameter is the locus of a point inside the circle and is equidistant from two points A and B at the circle.
(iv) Consider C1, C2 and C3 as the centers of circle which pass through A and B which are the two fixed points.
Construct a line XY which pass through the centers C1, C2 and C3.
Therefore, locus of centers of circles passing through two points A and B is the perpendicular bisector of the line segment which joins the two fixed points.
(v) In a rhombus ABCD, join AC
Here AC is the diagonal of rhombus ABCD
We know that
AC bisects ∠A
Therefore, any point on AC is the locus which is equidistant from AB and AD.
(vi) In a rhombus ABCD, join BD.
Here, BD is the locus of a point in the rhombus which is equidistant from A and C
We know that,
Diagonal BD bisects ∠B and ∠D
So, any point on BD will be equidistant from A and C.
6. Describe completely the locus of points in each of the following cases:
(i) mid-point of radii of a circle.
(ii) center of a ball, rolling along a straight line on a level floor.
(iii) point in a plane equidistant from a given line.
(iv) point in a plane, at a constant distance of 5 cm from a fixed point (in the plane).
(v) center of a circle of varying radius and touching two arms of ADC.
(vi) center of a circle of varying radius and touching a fixed circle, center O, at a fixed point A on it.
(vii) center of a circle of radius 2 cm and touching a fixed circle of radius 3 cm with center O.
(i) The locus of midpoints of the radii of a circle is another concentric circle with radius which is half of radius of given circle.
(ii) Consider AB as a straight line on the ground and the ball is rolling on it
So the locus of the center of the ball is a line which is parallel to the given line AB.
(iii) We know that
AB is the given line and P is a point in the plane.
From the point P, construct a line CD and another line EF from P’ parallel to AB.
Hence, CD and EF are the lines which are the locus of the point equidistant from AB.
(iv) Consider a point O and another point P where OP = 5 cm.
Taking O as center and radius equal to OP, construct a circle.
Hence, this circle is the locus of point P which is at a distance of 5 cm from the given point O.
(v) Construct the bisector BX of ∠ABC.
So this bisector of an angle is the locus of the center of a circle having different radii.
Here any point on BX is equidistant from BA and BC which are the arms of ∠ABC.
(vi) Here a circle with O as center is given with a point A on it.
The locus of the center of a circle which touches the circle at fixed point A on it is the line joining the points O and A.
(vii)
1. Here, if the circle with 2 cm as radius touches the given circle externally then the locus of the center of circle will be a concentric circle of radius 3 + 2 = 5 cm
2. If the circle with 2 cm as radius touched the given circle with 3 cm as radius internally, then the locus of the center of the circle will be a concentric circle of radius 3 – 2 = 1 cm.
7. Using ruler and compasses construct:
(i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm.
(ii) the locus of points equidistant from A and C.
(i) Construct BC = 3.4 cm and mark the arcs 5.5 and 4.9 cm from the points B and C.
Now join A, B and C where ABC is the required triangle.
(ii) Construct a perpendicular bisector of AC.
(iii) Construct an angle of 90° at AB at point A which intersects the perpendicular bisector at point O.
Construct circle taking O as center and OA as the radius.
8. Construct triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that:
(i) P is equidistant from B and C and
(ii) P is equidistant from AB and BC
(iii) Measure and record the length of PB.
(i) Consider BC = 8 cm as the long line segment.
At the point B construct a ray BX making an angle of 60° with BC
Now cut off BA = 7 cm and join AC.
Construct the perpendicular bisector of BC.
(ii) Construct the angle bisector of ∠B which intersect the perpendicular bisector of BC at P which is the required point.
(iii) By measuring, the length of PB = 4.6 cm.
9. A straight line AB is 8 cm long. Locate by construction the locus of a point which is:
(i) Equidistant from A and B.
(ii) Always 4 cm from the line AB.
(iii) Mark two points X and Y, which are 4 cm from AB and equidistant from A and B.
Name the figure AXBY.
Steps of Construction:
(i) Construct a line segment AB = 8 cm.
(ii) Using compasses and ruler, construct a perpendicular bisector l of AB which intersects AB at the point O.
(iii) Here any point on l is equidistant from A and B.
(iv) Now cut off OX = OY = 4 cm. X and Y are the required loci which is equidistant from AB and also from point A and B.
(v) Join AX, XB, BY and YA.
The figures AXBY is square shaped as its diagonals are equal and bisect each other at right angles.
10. Use ruler and compasses only for this question.
(i) Construct Δ ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C.
Measure and record the length of PB.
We know that
In Δ ABC,
AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°
Steps of Construction:
(i) Construct a line segment BC = 6 cm.
At the point B construct a ray BX which makes an angle 600 and cut off BA = 3.5 cm.
Now join AC.
Therefore, Δ ABC is the required triangle.
(ii) Construct the bisector BY of ∠ABC.
(iii) Construct a perpendicular bisector of BC which intersects BY at point P.
(iv) It is given that point P is equidistant from AB, BC and also equidistant from B and C.
By measuring, PB = 3.4 cm.
11. Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Steps of Construction:
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°.
(i) The points which are equidistant from BA and BC lies on the bisector of ∠ABC.
(ii) The points equidistant from B and C lies on the perpendicular bisector of BC.
Construct perpendicular bisector of BC.
P is the point of intersection of the bisector of ∠ABC and the perpendicular bisector of BC.
(iii) By measuring, the required length of PC = 4.8 cm.
12. In the diagram, A, B and C are fixed collinear points; D is a fixed point outside the line. Locate:
(i) the point P on AB such that CP = DP.
(ii) the points Q such that CQ = DQ = 3 cm. How many such points are possible?
(iii) the points R on AB such that DR = 4 cm. How many such points are possible?
(iv) the points S such that CS = DS and S is 4 cm away from the line CD. How many such points are possible?
(v) Are the points P, Q, R collinear?
(vi) Are the points P, Q, S collinear?
Here the points A, B and C are collinear and D is any point which is outside AB.
(i) Join CD.
Construct the perpendicular bisector of CD which meets AB in P.
Here P is the required point such that CP = DP.
(ii) Taking C and D as centers, construct two arcs with radius 3 cm which intersect each other at Q and Q’
Therefore, there are two points Q and Q’ which are equidistant from C and D.
(iii) Taking D as center and 4 cm radius construct an arc which intersects AB at R and R’
Here R and R’ are the two points on AB.
(iv) Taking C and D as center construct arcs with a 4 cm radius which intersect each other in S and S’.
Hence, there can be two such points equidistant from C and D.
(v) No, the points P, Q, R are not collinear.
(vi) Yes, the points P, Q, S are collinear.
13. Points A, B and C represent position of three towers such that AB = 60 mm, BC = 73 mm and CA = 52 mm. Taking a scale of 10 m to 1 cm, make an accurate drawing of Δ ABC. Find by drawing, the location of a point which is equidistant from A, B and C and its actual distance from any of the towers.
It is given that
AB = 60 mm = 6 cm
BC = 73 mm = 7.3 cm
CA = 52 mm = 5.2 cm
(i) Construct a line segment BC = 7.3 cm.
(ii) Taking B as center and 6 cm radius and C as center and 5.2 cm radius, construct two arcs which intersect each other at the point A.
(iii) Now join AB and AC.
(iv) Construct perpendicular bisector of AB, BC and CA which intersect each other at the point P and join PB.
Here P is equidistant from A, B and C on measuring PB = 3.7 cm.
The actual distance is 37 m.
14. Draw two intersecting lines to include an angle of 30°. Use ruler and compasses to locate points which are equidistant from these lines and also 2 cm away from their point of intersection. How many such points exist?
(i) AB and CD are the two lines which intersect each other at the point O.
(ii) Construct the bisector of ∠BOD and ∠AOD.
(iii) Taking O as center and 2 cm radius mark points on the bisector of angles at P, Q, R and S respectively.
Therefore, there are four points equidistant from AB and CD and 2 cm from O which is the point of intersection of AB and CD.
15. Without using set square or protractor, construct the quadrilateral ABCD in which ∠BAD = 45°, AD = AB = 6 cm, BC = 3.6 cm and CD = 5 cm.
(i) Measure BCD.
(ii) Locate the point P on BD which is equidistant from BC and CD.
(i) Consider AB = 6 cm long.
(ii) At point A, construct the angle of 45° and cut off AD = 6 cm.
(iii) Taking D as center and 5 cm radius and B as center and 3.5 cm radius construct two arcs which intersect each other at point C.
(iv) Now join CD, CB and BD.
Here ABCD is the required quadrilateral.
(v) (i) By measuring ∠BCD = 65°.
(vi) Construct the bisector of ∠BCD which intersects BD at the point P.
(ii) Hence, P is the required point equidistant from BD and CD.
16. Without using set square or protractor, construct rhombus ABCD with sides of length 4 cm and diagonal AC of length 5 cm. Measure ABC. Find the point R on AD such that RB = RC. Measure the length of AR.
(i) Construct AB = 4 cm.
(ii) Taking A as center, construct an arc of radius 5 cm and with B as center construct another arc of 4 cm radius which intersect each other at the point C.
(iii) Now join AC and BD.
(iv) With A and C as center, construct two arcs of 4 cm radius which intersect each other on D.
(v) Join AD and CD.
Hence, ABCD is the required rhombus and by measure ∠ABC = 78°.
(vi) Construct perpendicular bisector of BC which intersects AD at the point R.
By measuring the length of AR = 1.2 cm.
17. Without using set-squares or protractor construct:
(i) Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
(ii) Draw the locus of a point which moves so that it is always 2.5 cm from B.
(iii) Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
(iv) Mark the point of intersection of the loci with the letter P and measure PC.
Steps of Construction:
(i) Construct BC = 3.2 cm long.
(ii) Taking B as center and 5.5 cm radius and C as center and 4.8 cm radius construct arcs intersecting each other at the point A.
(iii) Now join AB and AC.
(iv) Construct the bisector of ∠BCA.
(v) Taking B as center and 2.5 cm radius, construct an arc which intersects the angle bisector of ∠BCA at P and P’.
Here P and P’ are the two loci which satisfy the given condition.
By measuring CP and CP’
CP = 3.6 cm and CP’ = 1.1 cm.
18. By using ruler and compasses only, construct an isosceles triangle ABC in which BC = 5 cm, AB = AC and BAC = 900. Locate the point P such that:
(i) P is equidistant from the sides BC and AC.
(ii) P is equidistant from the points B and C.
Steps of Construction:
(i) Construct BC = 5 cm and bisect it at point D.
(ii) Taking BC as diameter, construct a semicircle.
(iii) At the point D, construct a perpendicular intersecting the circle at the point A.
(iv) Now join AB and AC.
(v) Construct the angle bisector of C which intersects the perpendicular at the point P
Here P is the required point.
19. Using ruler and compasses only, construct a quadrilateral ABCD in which AB = 6 cm, BC = 5 cm, B = 600, AD = 5 cm and D is equidistant from AB and BC. Measure CD.
Steps of Construction:
(i) Construct AB = 6 cm.
(ii) At point B, construct angle 60° and cut off BC = 5 cm.
(iii) Construct the angle bisector of ∠B.
(iv) Taking A as center and 5 cm radius construct an arc which intersects the angle bisector of ∠B at D.
(v) Now join AD and DC.
Here ABCD is the required quadrilateral.
By measuring CD = 5.3 cm.
20. Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Steps of Construction:
(i) Construct a line AB = 6 cm.
(ii) Taking A and B as center and 4 cm radius, construct two arcs which intersect each other at the point C.
(iii) Now join CA and CB.
(iv) Construct the bisector of ∠C and cut off CP = 5 cm.
(v) Construct a line XY parallel to AB at 5 cm distance.
(vi) At point P, construct arcs of radius 5 cm each which intersects the line XY at Q and R.
Therefore, Q and R are the required points.
21. Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of length 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.
Steps of Construction:
(i) Taking O as center and 4 cm radius construct a circle.
(ii) Mark a point A on this circle.
(iii) Taking A and center and 6 cm radius construct an arc which cuts the circle at B.
(iv) Again with 5 cm radius, construct another arc which cuts the circle at C.
(vi) Construct the perpendicular bisector of AC.
Here any point on it will be equidistant from A and C.
(vii) Construct the angle bisector of ∠A which intersects the perpendicular bisector of AC at the point P.
Hence, P is the required locus.
22. Ruler and compasses only may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm and ∠ABC = 60°.
(ii) Construct the locus of all points, inside Δ ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangle with BC as base, which are equal in area to Δ ABC.
(iv) Mark the point Q, in your construction, which would make Δ QBC equal in area to Δ ABC and isosceles.
(v) Measure and record the length of CQ.
Steps of Construction:
(i) Construct AB = 9 cm.
(ii) At the point B construct an angle of 60° and cut off BC = 6 cm.
(iii) Now join AC.
(iv) Construct perpendicular bisector of BC.
Here, all the points on it will be equidistant from B and C.
(v) From the point A, construct a line XY which is parallel to BC.
(vi) Produce the perpendicular bisector of BC to meet the line XY at the point Q.
(vii) Now join QC and QB.
The area of Δ QBC is equal to the area of Δ ABC as these are on the same base and between the same parallel lines.
By measuring, length of CQ = 8.2 cm.
### Chapter Test
1. Draw a straight line AB of length 8 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
(i) Construct a line segment AB = 8 cm.
(ii) Construct the perpendicular bisector of AB which intersects AB at the point D.
Here every point P on it will be equidistant from A and B.
(iii) Take a point P on the perpendicular bisector.
(iv) Now join PA and PB.
Proof:
In Δ PAD and Δ PBD
PD = PD is common
As D is the midpoint of AB
We know that
∠PDA = ∠PDB = 90°
Δ PAD ≅ Δ PBD as per SAS axiom of congruency
PA = PB (c. p. c. t)
In the same way, we can prove that any other point which lies on the perpendicular bisector of AB is equidistant from A and B.
Therefore, it is proved.
2. A point P is allowed to travel in space. State the locus of P so that it always remains at a constant distance from a fixed point C.
It is given that
A point P is allowed to travel in space and is at a constant distance from a fixed point C.
Therefore, its locus is a sphere.
3. Draw a line segment AB of length 7 cm. Construct the locus of a point P such that area of triangle PAB is 14 cm2.
It is given that
Length of AB = 7 cm (base)
Area of triangle PAB = 14 cm2
We know that
Height = (area×2)/ base
Substituting the values
= (14×2)/ 7
= 4 cm
Construct a line XY which is parallel to AB and at a distance of 4 cm.
Take any point P on XY
Now join PA and PB
Area of triangle PAB = 14 cm2
Therefore, locus of P is the line XY which is parallel to AB at a distance of 4 cm.
4. Draw a line segment AB of length 12 cm. Mark M, the mid-point of AB. Draw and describe the locus of a point which is
(i) at a distance of 3 cm from AB.
(ii) at a distance of 5 cm from the point M. Mark the points P, Q, R, S which satisfy both the above conditions. What kind of quadrilateral is PQRS? Compute the area of the quadrilateral PQRS.
Steps of Construction:
(i) Construct a line AB = 12 cm.
(ii) Take M as the midpoint of line AB.
(iii) Construct straight lines CD and EF which is parallel to AB at 3 cm distance.
(iv) Taking M as center and 5 cm radius construct areas which intersect CD at P and Q and EF at R and S.
(v) Now join QR and PS.
Here PQRS is a rectangle where the length PQ = 8 cm
So the area of rectangle PQRS = PQ × RS
We get
= 8×6
= 48 cm2
5. AB and CD are two intersecting lines. Find the position of a point which is at a distance of 2 cm from AB and 1.6 cm from CD.
(i) AB and CD are two intersecting lines which intersect each other at the point O.
(ii) Construct a line EF which is parallel to AB and GH which is parallel to CD intersecting each other at the point P.
Hence, P is the required point.
6. Two straight lines PQ and PK cross each other at P at an angle of 75°. S is a stone on the road PQ, 800 m from P towards Q. By drawing a figure to scale 1 cm = 100 m, locate the position of a flagstaff X, which is equidistant from P and S, and is also equidistant from the road.
We know that
1 cm = 100 cm
800 m = 8 cm
Steps of Construction:
(i) Construct the lines PQ and PK intersecting each other at the point P which makes an angle 750.
(ii) Consider a point S on PQ such that PS = 8 cm.
(iii) Construct the perpendicular bisector of PS.
(iv) Construct the angle bisector of ∠KPS which intersects the perpendicular bisector at X.
Here X is the required point which is equidistant from P and S and also from PQ and PK.
7. Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively. Find by construction a point X equidistant from PQ, PS and equidistant from R, S. Measure XR.
Steps of Construction:
(i) Take PR = 8 cm and construct the perpendicular bisector of PR which intersects it at point O.
(ii) From the point O, cut off OS = OQ = 3 cm.
(iii) Now join PQ, QR, RS and SP.
Here PQRS is a rhombus with PR and QS as the diagonals.
(iv) PR is the bisector of ∠SPQ.
(v) Construct perpendicular bisector of SR which intersects PR at X
Here X is equidistant from PQ and PS and also from S and R.
By measuring, length of XR = 3.2 cm.
8. Without using set square or protractor, construct the parallelogram ABCD in which AB = 5.1 cm, the diagonal AC = 5.6 cm and diagonal BD = 7 cm. Locate the point P on DC, which is equidistant from AB and BC.
Steps of Construction:
(i) Consider AB = 5.1 cm.
(ii) At the point A, radius = 5.6/2 = 2.8 cm
At the point B, radius = 7.0/2 = 3.5 cm
Construct two arcs which intersect each other at the point O.
(iii) Now join AO and produce it to point C such that OC = AD = 2.8 cm and join BO and produce it to D such that BO = OD = 3.5 cm.
(iv) Join BC, CD and DA
Here, ABCD is a parallelogram.
(v) Construct the angle bisector of ∠ABC which intersects CD at P.
P is the required point equidistant from AB and BC.
9. By using ruler and compass only, construct a quadrilateral ABCD in which AB = 6.5 cm, AD = 4 cm and ∠DAB = 75°. C is equidistant to from the sides if AB and AD, if also C is equidistant from the points A and B.
Steps of Construction:
(i) Construct a line segment AB = 6.5 cm.
(ii) At the point A, construct a ray which makes an angle 75° and cut off AD = 4 cm.
(iii) Construct the bisector of ∠DAB.
(iv) Construct the perpendicular bisector of AB which intersects the angle bisector at the point C.
(v) Now join CB and CD.
Hence, ABCD is the required quadrilateral.
The solutions provided for Chapter 13 Locus of ML Aggarwal Textbook. This solutions of ML Aggarwal Textbook of Chapter 13 Locus contains answers to all the exercises given in the chapter. These solutions are very important if you are a student of ICSE boards studying in Class 10. While preparing the solutions, we kept this in our mind that these should based on the latest syllabus given by ICSE Board.
### More Study Resources for Class 10 ICSE
We have also provided the ICSE Study Materials for Class 10 Students. These Study Resources includes all the subjects of ICSE Board. It also contains the questions and answers of all the chapters and topics which are provided in the syllabus of ICSE for Class 10 Students. All the solutions and Summaries are strictly based on the latest syllabus of ICSE. We aim to provide the best study materials for Class 10 ICSE on the website icserankers. The subjects for which solutions are provided are Hindi, English, Mathematics, Physics, Chemistry, Biology, History and Civics, Geography and Economics. Books like Selina Concise Publisher's Textbook, Frank Certificate Textbooks, ML Aggarwal Textbooks solutions are provided chapterwise. |
Fractions of a Number
## How to Find the Fraction of a Number
She gave 1/2 of the apples to her brother. If there were 12 apples in the basket, how many apples did she give to her brother?
To solve this problem, we'll need to find a fraction of a number.
We'll need to this figure out:
What is 1/2 of 12?
Let's try solving the problem using models, or drawings.
Let's draw the apples in the basket.
Ruby gave 1/2 of the apples to her brother.
The fraction 1/2 tells us that the whole is divided into into 2 equal parts.
We know this based on the denominator.
Now, let's draw a model.
The 12 apples are now been divided into 2.
How many apples are there in each part? 🤔
👍 That's right! 6 apples.
That means 1/2 of 12 is 6.
Drawing models work, but it's very slow.
😺 Let's try an easier and faster way to find the fraction of a number - using multiplication!
### Multiplying a Fraction and a Number
To find the fraction of a number, multiply the number by the numerator, then divide the answer you get by the denominator.
Let's try it with our example:
What is 1/2 of 12?
🤓 Tip: This is the same as writing:
12 x 1/2 = ?
👉 First, multiply 12 by the numerator:
12 × 1 = 12
👉 Then, divide the product you get by the denominator.
12 ÷ 2 = 6
We get 6!
Why does this work? 🤔
👉 Think about fractions like a division problem where the numerator is divided by the denominator:
1/2 = 1 ÷ 2
So we can also think about 12 x 1/2 as
12 x (1 ÷ 2)
which is the same as:
12 x 1 ÷ 2 👍
### Another Example
Let's try another example.
What is 2/3 of 24?
We just have to solve 24 × 2 ÷ 3!
👉 First, multiply 24 by the numerator.
24 × 2 = 48
👉Then, divide the product you get by the denominator.
48 ÷ 3 = 16
Let's check our answer using models. 😺
How many objects are there?
👍 That's right, 24 objects.
How many equal parts do we need to divide the whole into? 🤔
The denominator of the fraction is 3, so you need to divide the whole into 3 equal parts.
How many equal parts are we talking about in 2/3?
The numerator is 2, so we're talking about 2 parts in 2/3.
How many objects is 2/3 of 24?
That's right, 16! 👍
So we got the same answer with our model, as we did with our new trick:
To find the fraction of a number: first, multiply the whole by the numerator, then divide the product by the denominator.
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Before we begin, let"s understand the an interpretation of the square root. The prize square source is composed as and is one integral component of mathematics. When you understand the basics of recognize the square root of a number, you deserve to solve any square root-related problem. In this quick lesson, we will learn about the square root of 29. Us will perform so using techniques like prime factorization and division, and along with it, likewise find out the square root of 29.
You are watching: Square root of 29 in radical form
Square source of 2929 = 5.385Square the 29: 292 = 841
1 What Is the Square source of 29? 2 Important note on Square source of 29 3 How to uncover the Square root of 29? 4 Important notes on Square root of 29 5 Thinking out of the Box! 6 FAQs top top Square source of 29
## What Is the Square source of 29?
We know that addition has one inverse procedure as subtraction and multiplication has an inverse operation as division. Similarly, detect the square root is one inverse procedure of squaring. The square source of 29 is the number that gets multiplied to chin to give the number 29. So, we need to think of a number whose square is 29. We can see that, over there does no exist any integer whose square is 29. The square source of a number is the number the gets multiplied to chin to offer the initial number. The square source of 29 is 5.38516480713.
## Is the Square source of 29 Rational or Irrational?
It is not feasible to rest 29 into 2 such components which, top top squaring, give 29. It have the right to be roughly written as a square the 5.385, i beg your pardon is a non-recurring and non-terminating decimal number. This reflects that the is not a perfect square, which also proves that the square root of 29 is an irrational number. Do you think the square root of 29 and also the square root of 5 have anything in common? Yes, there is, both room prime numbers and also are no perfect squares. So 29 is one irrational number.
## How to discover the Square source of 29?
We can uncover the square root of 29 using the adhering to steps:
Step 1: examine whether the number is a perfect square or not. In this case, 29 is no a perfect square together it can not be damaged down right into a product of two exact same numbers.Step 2: once the number is checked, the following procedure needs to be followed:If the number is a perfect square, it have the right to be written as x2 = xIf it is no a perfect square, the square root is found using the long department method. It can additionally be written in its simplified radical form.
Since 29 is not a perfect square, we discover its square root utilizing the long division method. The streamlined radical form of the square source of 29 is provided below.
### Simplified radical form of the square source of 29
29 can be created as a product 1 and 29.29 = √(1 × 29)29 is no a perfect square, hence, it continues to be within roots. The simplified radical kind of the square source of 29 is √29.
### Square source of 29 By Long Division
Let us know the process of detect the square root of 29 by long division.
Step 1: Pair the digits of the number beginning from the best end, by placing a bar over them.Step 2: Now, we find a number such that its square gives a product the is less than or equal to this pair. The square the 5 is 25, i beg your pardon is much less than 29. On individually 25 native 29, we acquire 4.Step 3: Now, we dual the quotient and also place it as the following divisor with a blank on the right. The double of 5 provides 10 and also a blank is placed next come it. This blank needs to be filled by a number i beg your pardon will also become the following digit in the quotient. It should be filled with the largest feasible digit, together that as soon as the new divisor is multiplied through the new quotient, the product is less than or equal to the dividend.Step 4: Repeating the above steps, the quotient 53 is doubled, which offers 106, and also a pair the 0s is included to the dividend.Step 5: The same measures are repeated and also the digit which is created in the blank is 8. The product that 1068 and 8 offers 8544. After individually 8544 from 9100, we get 556 with a brand-new pair that 0s.Finally, the square source of 29 is calculated to be 5.385 as shown below:
Explore Square roots using illustrations and also interactive examples
Important Notes:
29 is no a perfect square, hence, that square root is an irrational number. This concludes the the square root of any kind of number "n", which is no a perfect square, will constantly be one irrational number.
See more: Why Did The Allies Hold War Crimes Trials For Axis Leaders Responsible?
Think Tank:
Can you uncover the square root of 29 using the long department method up to 5 decimal places?Express the square source of 290 in its simplified radical form. |
Question 1: A can do a piece of work in 15 days while B can do it in 10 days. How long will they take together to do it?
A’s 1 Day = $\frac{1}{15}$
B’s 1 Day work = $\frac{1}{10}$
A’s + B’s 1 Day work = $(\frac{1}{15}+\frac{1}{10})=\ \frac{5}{30}=\ \frac{1}{6}$
Therefore both can finish the work in 6 Days.
Question 2: A, B and C can do a piece of work in 12 days, 15 days and 10 days respectively. In what time will they all together finish it?
A’s 1 Day = $\frac{1}{12}$
B’s 1 Day work = $\frac{1}{15}$
C’s 1 Day work = $\frac{1}{10}$
(A’s + B’s + C’s) 1 Day work = $(\frac{1}{12} +\frac{1}{15} + \frac{1}{10}) = \frac{15}{60} = \frac{1}{4}$
Therefore all three can finish the work in 4 Days.
Question 3: A and B together can do a piece of work in 35 days, while A alone can do it in 60 days. How long would B alone take to do it?
A’s 1 Day = $\frac{1}{60}$
B’s 1 Day work = $\frac{1}{x}$
(A’s + B’s) 1 Day work = $(\frac{1}{60} + \frac{1}{x} ) = \frac{1}{35}$
Solving for $x$= 84 Days
Question 4: A can do a piece of work in 20 days while B can do it in 15 days. With the help of C, they finish the work in 5 days. In what time would C alone do it?
A’s 1 Day = $\frac{1}{20}$
B’s 1 Day work = $\frac{1}{15}$
C’s 1 Day work = $\frac{1}{x}$
(A’s + B’s + C’s) 1 Day work = $(\frac{1}{20} +\frac{1}{15} + \frac{1}{x} ) = \frac{1}{5}$
Solving for $x$ = 12 Days
Question 5: A can do a piece of work in 12 days and B alone can do it in 16 days. They worked together on it for 3 days and then A left. How long did B take to finish the remaining work?
A’s 1 Day = $\frac{1}{12}$
B’s 1 Day work = $\frac{1}{16}$
(A’s + B’s) 1 Day work = $(\frac{1}{12} +\frac{1}{16} ) = \frac{7}{48}$
The amount of work that is completed in 3 days = $\frac{3\times7}{48} = \frac{7}{16}$
Amount of work left for B to complete = $1 - \frac{7}{16} = \frac{9}{16}$
Therefore the number of days that B will take to finish the work = $\frac {\frac{9}{16}} {\frac{1}{16} }$ = 9 days
Question 6: A can do $\frac{1}{4}$ of a work in 5 days, while B can do $\frac{1}{5}$ of the work in 6 days. In how many days can both do it together?
If A can do $\frac{1}{4}$ of a work in 5 days, then A can do the entire work in 20 days.
Therefore A’s 1 Day Work = $\frac{1}{20}$
If B can do $\frac{1}{5}$ of a work in 6 days, then B can do the entire work in 30 days.
Therefore B’s 1 Day Work = $\frac{1}{30}$
(A’s + B’s) 1 Day work = $(\frac{1}{20} +\frac{1}{30} ) = \frac{1}{12}$
Therefore both can do the work in 12 days.
Question 7: A can dig a trench in 6 days while B can dig it in 8 days. They dug the trench working together and received 1120 for it. Find the share of each in it.
A’s 1 Day = $\frac{1}{6}$
B’s 1 Day work = $\frac{1}{8}$
Therefore the ratio of work = $\frac {\frac{1}{6}} {\frac{1}{8} } = \frac{8}{6}$
Therefore A’s share = $\frac{8}{14}$ $\times 1120 = 640$
Therefore B’s share = $\frac{6}{14}$ $\times 1120 = 480$
Question 8: A can mow a field in 9 days; B can mow it in 12 days while C can mow it in 8 days. They all together mowed the field and received 1610 for it. How will the money be shared by them?
A’s 1 Day = $\frac{1}{9}$
B’s 1 Day work = $\frac{1}{12}$
C’s 1 Day work = $\frac{1}{18}$
Therefore the ratio of their one day’s work = $\frac{1}{9} \colon \frac{1}{12} \colon \frac{1}{18}$ $= 8 \colon 6 \colon 9$
Hence A’s share = $\frac{8}{23}$ $\times 1120 = 560$
Hence A’s share = $\frac{6}{23}$ $\times 1120 = 420$
Hence A’s share = $\frac{9}{23}$ $\times 1120 = 630$
Question 9: A and B can do a piece of work in 30 days; B and C in 24 days; C and A in 40 days. How long will it take them to do the work together? In what time can each finish it, working alone?
(A’s + B’s) 1 Day work = $\frac{1}{30}$
(B’s + C’s) 1 Day work = $\frac{1}{24}$
(C’s + D’s) 1 Day work = $\frac{1}{40}$
Adding the above three $2\times(A + B + C)$ day work = $(\frac{1}{30} +\frac{1}{24} + \frac{1}{40} ) = \frac{1}{10}$
Therefore if they all work together, they will take 20 days to finish the work.
Question 10: A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes the remaining work in 42 days. In how many days could both do it?
A’s 1 Day = $\frac{1}{80}$
Work finished by A in 10 days = $\frac{1}{80} \times$ $10 =$ $\frac{1}{8}$
B finished the remainder of work $(1 - \frac{1}{8} = \frac{7}{8})$ in 42 days
Therefore 1 Days work for B = $\frac{\frac{7}{8}}{42} = \frac{1}{48}$
Hence B can do the work in 48 days
(A’s + B’s) 1 Day work = $(\frac{1}{80} + \frac{1}{48} )= \frac{1}{30}$
Therefore both can do the work in 30 days.
Question 11: A and B can together finish a work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days. In how many days can A alone do it?
(A’s + B’s) 1 Day work = $\frac{1}{30}$
Amount of work finished by both in 20 days = $20 \times \frac{1}{30} = \frac{2}{3}$
Work left to be finished = $1 - \frac{2}{3} = \frac{1}{3}$
Work done by A in 1 Day = $\frac{\frac{1}{3}}{20} = \frac{1}{60}$
Therefore A can do the work alone in 60 days.
Question 12: A can do a certain job in 25 days which B alone can do in 20 days. A started the work and was joined by B after 10 days. In how many days was the whole work completed?
A’s 1 Day work = $\frac{1}{25}$
B’s 1 Day work = $\frac{1}{20}$
(A’s + B’s) 1 Day work = $(\frac{1}{25} +\frac{1}{20}) = \frac{9}{100}$
Amount of work finished by A in 10 days = $10 \times \frac{1}{25}$
Work left to be finished = $(1-\frac{2}{5}) = \frac{3}{5}$
Days taken by both A and B working together = $\frac{\frac{3}{5}}{\frac{9}{100}} = 6\frac{2}{3}$
The work got completed in $10 + 6\frac{2}{3} = 16\frac{2}{3}$
Question 13: A can do a piece of work in 14 days, while B can do in 21 days. They begin together. But, 3 days before the completion of the work, A leaves off. Find the total number of days taken to complete the work.
A’s 1 Day work = $\frac{1}{14}$
B’s 1 Day work = $\frac{1}{21}$
(A’s + B’s) 1 Day work = $(\frac{1}{14} +\frac{1}{21}) = \frac{5}{42}$
Amount of work finished by B in 3 days = $(3 \times \frac{1}{21}) =\frac{1}{7}$
Work left to be finished by A and B together = $(1- \frac{1}{7}) = \frac{6}{7}$
Days taken by A + B working together = $\frac{\frac{6}{7}}{\frac{5}{42}} = 7\frac{1}{5}$
The work got completed in $3+7\frac{1}{5} = 10\frac{1}{5}$
Question 14: A is thrice as good a workman as B and B is twice as good a workman as C. All the three took up a job and received 1800 as remuneration. Find the share of each.
If C takes $x$ days to complete the job
Then B will complete the job in $\frac{x}{2}$ days
And A will complete the job in $\frac{x}{3}$ days days
Therefore the ratio of 1 days’ work of $A \colon B \colon C =$ $\frac{x}{3} \colon \frac{x}{2} \colon \frac{x}{1}$ $= 3 \colon 2 \colon 1$
Therefore the share of A = $1800 \times$ $\frac{3}{6}$ $= 900$ Rs.
Therefore the share of B = $1800 \times$ $\frac{2}{6}$ $= 600$ Rs.
Therefore the share of C = $1800 \times$ $\frac{1}{6}$ $= 300$ Rs.
Question 15: A can do a certain job in 12 days. B is 60% more efficient than A. Find the number of days taken by B to finish the job.
Time taken to finish the job = 12 days
A’s 1 Day’s work = $\frac{1}{12}$
B is 60% more efficient
B’s 1 Day’s work = $1.6 \times$ $\frac{1}{12}$
Therefore the number of days B will take to finish the job = $\frac{1}{\frac{1.6}{12}}$ $= 7$ $\frac{1}{2}$days
Question 16: A is twice as good a workman as B and together they finish a piece of work in 14 days. In how many days can A alone do it?
Let B take $x$ days to finish the work.
B’s 1 day’s work = $\frac{1}{x}$
The A will take $\frac{x}{2}$ days to finish the work.
A’s 1 day’s work = $\frac{2}{x}$
(A+B) one day’s work = $(\frac{1}{x} +\frac{2}{x}) = \frac{1}{14}$
Solving for $x=42$
Therefore A will take 21 days to finish the job.
Question 17: Two pipes A and B can separately fill a tank in 36 minutes and 45 minutes respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
A’s 1 minute fill rate = $\frac{1}{36}$
B’s 1 minute fill rate = $\frac{1}{45}$
A’s and B’s fill rate together = $(\frac{1}{36} + \frac{1}{45}) = \frac{1}{20}$
Therefore if A and B are opened simultaneously, the tank will take 20 minutes to fill up.
Question 18: One tap can fill a cistern in 3 hours and the waste pipe can empty the full tank in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together?
Tap’s 1 minute fill rate = $\frac{1}{3}$
Waste Pipe’s 1 minute empty rate = $\frac{1}{5}$
Therefore the net fill rate = $(\frac{1}{3} - \frac{1}{5}) = \frac{2}{15}$
Therefore if both the tap and the waste pipe are opened simultaneously then it will take $7\frac{1}{2}$ hours to fill up.
Question 19: Two pipes A and B can separately fill a cistern in 20 minutes and 30 minutes respectively , while a third pipe C can empty the full cistern in 15 minutes. If all the pipes are opened together, in what time the empty cistern is filled?
A’s 1 minute fill rate = $\frac{1}{20}$
B’s 1 minute fill rate = $\frac{1}{30}$
C’s 1 minute empty rate = $\frac{1}{15}$
Therefore the net fill rate = $(\frac{1}{20} + \frac{1}{30} - \frac{1}{15}) = \frac{1}{60}$
Therefore if all the tap are opened simultaneously then it will take 60 minutes or one hour to fill up.
Question 20: A pipe can fill a tank in 16 hours. Due to a leak in the bottom, it is filled in 24 hours. If the tank is full, how much time will the leak take to empty it?
Pipe’s fill rate = $\frac{1}{16}$
Let the leak is at a rate of = $\frac{1}{x}$
Therefore the net fill rate = $(\frac{1}{16} - \frac{1}{x}) = \frac{1}{24}$
Solving for $x = 48$ hours. |
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# Add 17 kg, 13 kg 940 g and 15 kg 65 gA. 40 kg 65 gB. 4 kg 650 gC. 46 kg 50 gD. 46 kg 5 g
Last updated date: 23rd Jun 2024
Total views: 414.6k
Views today: 9.14k
Answer
Verified
414.6k+ views
Hint: Mass is commonly measured by how much something weighs. In the metric system of measurement, the most common units of mass are the gram and kilogram.
Rules of addition/ subtraction are as follows:
1. Write the numbers to be added/subtracted in vertical form.
2. While adding/subtracting units write them in separate columns, i.e., ‘kg’ and ‘g’ to be written in
separate columns.
3. Finally, adding/subtracting as usual.
To convert kilograms into grams, multiply the given measure in kilograms by 1000 and write the unit gram (g) after it.
$1kg = 1000g$
To convert the grams into kilograms and grams, the last 3 digits from the right represent the number of grams and the rest of the digits represent the number of kilograms. For eg:
7216 g = 7 kg 216 g
Complete step by step solution: We have to add the terms 17 kg, 13 kg 940 g and 15 kg 65 g
We will arrange all the terms into column and add them
Therefore, the sum is 45 kg 1005 g
Now, $1kg = 1000 g$ so we can convert 1005 g into kg
We can write 1005 g as
$\begin{gathered} 1005g = 1000g + 005g \\ 1005g = 1kg + 5g \\ \therefore 1005g = 1kg5g \\ \end{gathered}$
Therefore, 45 kg 1005 g will be equal to 46 kg 5 g
Hence, the sum is 46 kg 5 g
∴Option (D) is correct
Note: 1. To convert kilograms to grams we have to multiply the number of kilograms by 1000 and to convert grams to kilograms we divide the number of grams by 1000. |
9 out of 10 based on 443 ratings. 1,119 user reviews.
# STUDY GUIDE AND INTERVENTION FIND SLOPE
[PDF]
Algebra 1 Slope Intercept, Direct Variation STUDY GUIDE
Algebra 1 Slope Intercept, Direct Variation STUDY GUIDE SOLs: A.6, A.8 Find slope and rate of change Slope is defined as: slope = m = Algebra 1 Slope Intercept, Direct Variation Test STUDY GUIDE Page 2 Study Questions 1) Find the slope of the line that passes through the points.[PDF]
4-1 Study Guide and Intervention - Hostler Classroom
Chapter 4 50 Glencoe Algebra 1 Study Guide and Intervention (continued) Writing Equations in Slope-Intercept Form Write an Equation Given Two Points Write an equation of the line that passes through (1, 2) and (3, -2). Find the slope m. To find the y-intercept, replace m with its computed value and ( x, y) with (1, 2) in the slope-intercept form.[PDF]
4-2 Study Guide and Intervention - waynesville
4-2 Study Guide and Intervention Writing Equations in Slope-Intercept Form Write an Equation Given the Slope and a Point Example 1: Write an equation of the line that passes through (–4, 2) with a slope of 3. The line has slope 3. To find the y-intercept, replace m with 3 and (x, y) with (–4, 2) in the slope-intercept form. Then solve for b.[PDF]
3-3 Study Guide and Intervention - moodle
3-3 Study Guide and Intervention Rate of Change and Slope Rate of Change The rate of change tells, on average, how a quantity is changing over time. Example: POPULATION The graph shows the population growth in China. a. Find the rates of change for 1950-1975 and for 2000-2025. 1950-1975: 𝑐ℎ𝑎 á𝑔𝑒 𝑖 [PDF]
NAME DATE PERIOD 2-3 Study Guide and Intervention
Find the average rate of change for the data in the table. Study Guide and Intervention Rate of Change and Slope 2-3 Average Rate of Change =− change in y change in x = change in Elevation of the Sun −− ° change in Time = −−84° – 6° A.M. 11:00 A.M. – 7:00 A.M. = −78° 4 hours = 19.5 degrees per hour Example Elevation of the[PDF]
NAME DATE PERIOD 3-3 Study Guide and Intervention
Study Guide and Intervention (continued) Slopes of Lines Parallel and Perpendicular Lines If you examine the slopes of pairs of parallel lines and the slopes of pairs of perpendicular lines, where neither line in each pair is vertical, you will discover the following properties. Two lines have the same slope if and only if they are parallel.[PDF]
3-3 Study Guide and Intervention - mathcounts4ever
3-3 Study Guide and Intervention Rate of Change and Slope Rate of Change The rate of change tells, on average, how a quantity is changing over time. Example: POPULATION The graph shows the population growth in China. a. Find the rates of change for 1950-1975 and for 2000-2025. 1950-1975: Ö Û Ô á Ú Ø ã â ã è ß Ô ç Ü â á Ö Û Ô[PDF]
Answers (Anticipation Guide and Lesson 4-1)
Study Guide and Intervention (continued) Graphing Equations in Slope-Intercept Form Modeling Real-World Data MEDIA Since 1999, the number of music cassettes sold has decreased by an average rate of 27 million per year. There were 124 million music cassettes sold in 1999. a.[PDF]
Study Guide and Intervention Workbook
organized by chapter and lesson, with two Study Guide and Intervention worksheets for every lesson in Glencoe Pre-Algebra. Always keep your workbook handy. Along with your textbook, daily homework, and class notes, the completed Study Guide and Intervention Workbook can help you in reviewing for quizzes and tests. To the Teacher[PDF]
Answers (Anticipation Guide and Lesson 2-1)
Answers (Anticipation Guide and Lesson 2-1) STEP 1 Chapter 2 3 The slope of a line is the change of x-coordinates divided by the change of y-coordinates. D 5. A vertical line has an undefined slope. A 6. Study Guide and Intervention 2 x (continued) Relations and Functions Chapter 2 7
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Counting
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# Counting - PowerPoint PPT Presentation
Counting . Chapter 13 sec 1. Break up into groups. One method of counting is using the tree diagram. It is useful to keep track of counting. Order of permutations does matter. You will determine all the possible ways to count. Remember order matters!!.
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## PowerPoint Slideshow about ' Counting ' - shaw
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### Counting
Chapter 13 sec 1
One method of counting is using the tree diagram.
• It is useful to keep track of counting.
• Order of permutations does matter.
How many ways can we do each of the following?
• Flip a coin?
• 2 ways; One head and one tails.
Roll a single die?
• 6 ways
• Pick a card from a standard deck of cards?
• 52 ways
Example; How many ways can 3 coins be flipped?
• How would you list the ways?
• How would you list the possibilities?
Make a tree diagram
• 1st row is first coin
• 2nd row is the second coin
• 3rd row is the third coin.
• You are listing out the possibilities.
• Let us say that one die is red and the other is green.
In your group see if you can find all combinations.
• Starting with
• (1,1), (1,2), (1,3), …
• 36 ways.
If objects are allowed to be used more than once in a counting problem, we will use the phrase with repetition.
• If we do not want objects to be used more than once, without repetition.
Draw a tree diagram that illustrates the different ways to flip a dime, penny, quarter, and nickel.
• 1. In how many ways can you get exactly one head?
• 4
3. How many different three-digit numbers can you form using the digits 1, 2, 5, 7, 8, & 9 without repetition?
You are a designer and has designed different tops, pants, and jackets to create outfits for a runway show. Without repetition, how many different outfits can your models wear if you had designed the following:
• Seven tops, six pants, three jackets. |
#### Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question 3 Maths Textbbok Solution.
Answer: $\frac{1}{\sqrt{10}}$
Hint: First we will convert $\cos ^{-1} \frac{4}{5}$ into $\sin ^{-1}$
Given: $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$
Explanation:
Let $\cos ^{-1} \frac{4}{5}=\theta$
$\cos \theta=\frac{4}{5}$
we know that
\begin{aligned} &\cos \theta=1-2 \sin ^{2} \frac{\theta}{2} \\ &2 \sin ^{2} \frac{\theta}{2}=1-\cos \theta \\ &\sin ^{2} \frac{\theta}{2}=\frac{1-\cos \theta}{2} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{2}} \end{aligned}
\begin{aligned} &\sin \frac{\theta}{2}=\sqrt{\frac{1-\frac{4}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{\frac{1}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1}{10}} \\ &\sin \frac{\theta}{2}=\frac{1}{\sqrt{10}} \\ &\frac{\theta}{2}=\sin ^{-1} \frac{1}{\sqrt{10}} \end{aligned} from equation (1)
$\frac{1}{2} \cos ^{-1} \frac{4}{5}=\sin ^{-1} \frac{1}{\sqrt{10}}$ ......(2)
Now
$\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$ from equation (2)
$=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)$
$=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)$ $\left[\because \sin \left[\sin ^{-1}(\theta)\right]=\theta\right]$
$=\frac{1}{\sqrt{10}}$ |
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# Combining ("Gathering") Like Terms
First, we'll define what "terms" are. Then we'll see what "like terms" are.
A term is a product of a number and some variables, like $3xy$ or $-4{x}^{2}$ . If the variable part is the same in two terms, they're called like terms .
Examples :
$3x$ and $5x$ are like terms; $3x$ and $5y$ are unlike terms.
$7{a}^{3}b$ and $99{a}^{3}b$ are like terms; $7ab$ and $99a{b}^{2}$ are unlike.
If two terms are "like" then you can add or subtract them. Notice the use of the Distributive Property here.
$3x+5x=\left(3+5\right)x=8x$
The variable part stays the same; we can just add the coefficients (the numbers in front of the variables).
But unlike terms can't be added.
$3x+4y\ne 7x$ or $7y$ or $7xy$ , and
$6+5x\ne 11x$ .
(The symbol $\ne$ means "not equal to.")
This is important when simplifying polynomials .
Example:
Simplify.
$6{x}^{2}+5x+4-4{x}^{2}+7x-8$
First, collect the like terms in parentheses.
$=\left(6{x}^{2}-4{x}^{2}\right)+\left(5x+7x\right)+\left(4-8\right)$
Then simplify.
$=2{x}^{2}+12x-4$
; |
# How to quickly tell if a number is divisible by eleven and why it works
Photo by Wetsun
I already showed how to tell if a number is divisible by three, by nine, and by seven. I’ll now show you how to tell if a number is divisible by 11.
This method is based on the digit position of each digit. You can think of each position as either an even or odd position. I am not talking about the digit itself, just the position. So the positions are 1, 2, 3, 4, 5, 6, etc. Which are 1, 3, 5, etc. as odd positions, while 2, 4, 6, etc. are the even positions.
If you add up all the even position digits and also add up all the odd position digits, you have two totals. Subtract one total from the other. This final result is a much smaller number. If this new number is divisible by eleven, then the original number was divisible by eleven. If your smaller number is still too large to tell if it is divisible by 11, then use the method again and make a still smaller number to test for divisibility by 11.
An example: Is 123456789 divisible by 11?
1. 1+3+5+7+9=25
2. 2+4+6+8=20
3. 25-20=5, 5 is not divisible by 11, so 123456789 is not divisible by 11.
Next example: Is 1234567895 divisible by 11?
1. 1+3+5+7+9=25
2. 2+4+6+8+5=25
3. 25-25=0, 0 is divisible by 11, so 1234567895 is divisible by 11 as well.
Ok, we have a method, but it has a weakness that it really needs pen and paper to do. But a simple adjustment in how we think about it, changes this into a method you can do in your head.
If you just walk along the digits in order, and add the first subtract the second, add the third, subtract the fourth digit, etc. You are performing the equivalent arithmetic of the above algorithm.
For example: Is 123456789 divisible by 11?
1. 1-2+3-4+5-6+7-8+9= Wait, let’s do this in our head one sum at a time…
2. 1-2=-1, -1+3=2, 2-4=-2, -2+5=3, 3-6=-3, -3+7=4, 4-8=-4, -4+9=5
3. If you notice 5 is the same result we got in the first example. And 5 is not divisible by 11, so 123456789 is not divisible by 11 either.
For some numbers, there is a further short cut. Any time you see the same digit next to each other in pairs, you can skip them both. Because, of course, a number minus itself is zero.
For example: Is 322334455667788993 divisible by 11?
1. First cross out all the pairs, and you are left with 3…………….3;
2. But you can remove those positions and push the two digits together: 33
3. Is 33 divisible by 11? You probably know already, yes. But notice, you can repeat the method and 3-3=0. 0 is always divisible by 11, or any number.
4. So 322334455667788993 is divisible by 11.
So why does this work?
• A number ABCDEF is 100000A + 10000B + 1000C + 100D + 10E + F
• rearranging, this equals 100000A + 1000C +10E + 10000B + 100D + F
• 100000 divided by 11 is 9090 and remainder 10, or another way to look at it, it is 9091 and remainder -1
• If I divide 100000 by any factor of 100, or multiply by any factor of 100, the remainder is always the same, remainder = -1
• 100000A divided by 11 is (-1)A or a remainder of -A
• Similarly 10000 divided by 11 is 909 and remainder 1
• If I divide 10000 by any factor of 100, or multiply by any factor of 100, the remainder is always the same, remainder = 1
• 10000B divided by 11 is (1)B or a remainder of +B
• If you add several numbers together and divide by 11, you get the same remainder as you would get if you divide each of the several numbers by 11 and add up the remainders
• (100000A + 1000C + 10E + 10000B + 100D + F) / 11 is equal to 100000A/11 + 1000C/11 +10E/11 +10000B/11 +100D/11 + F/11
• The remainders for each part of the sum are (-A) + (-C) + (-E) + (+B) + (+D) + (+F)
• Rearranging, B + D + F – A – C – E equals the remainder for ABCDEF/11
• This will work for any number of digits
A more formal explanation using modular arithmetic might also help you understand. |
## Can you graph a nonlinear equation?
A non-linear graph can be described by an equation. In fact any equation, relating the two variables x and y, that cannot be rearranged to: y = mx + c, where m and c are constants, describes a non- linear graph. When we draw a non-linear graph we will need more than three points.
### How do you graph inequalities graphically?
How to Graph a Linear Inequality
1. Rearrange the equation so “y” is on the left and everything else on the right.
2. Plot the “y=” line (make it a solid line for y≤ or y≥, and a dashed line for y< or y>)
3. Shade above the line for a “greater than” (y> or y≥) or below the line for a “less than” (y< or y≤).
What makes a graph nonlinear?
Non-linear means the graph is not a straight line. The graph of a non-linear function is a curved line. A curved line is a line whose direction constantly changes. A cautionary note: Economists are accustomed to designate all lines in graphs as curves – both straight lines and lines which are actually curved.
Why graph is non-linear data structure?
A graph is a non-linear data structure that has a finite number of vertices and edges, and these edges are used to connect the vertices. The vertices are used to store the data elements, while the edges represent the relationship between the vertices.
## Which inequality is graphed on the coordinate plane?
Linear inequalities can be graphed on a coordinate plane. The solutions for a linear inequality are in a region of the coordinate plane. A boundary line, which is the related linear equation, serves as the boundary for the region.
### What makes a nonlinear function?
A nonlinear function is a function whose graph is NOT a line. Its equation is of the form f(x) = ax + b. Its equation can be in any form except of the form f(x) = ax + b. Its slope is constant for any two points on the curve. The slope of every two points on the graph is NOT the same.
Why graph is used in data structure?
Graphs in data structures are used to represent the relationships between objects. Every graph consists of a set of points known as vertices or nodes connected by lines known as edges. The vertices in a network represent entities.
What is a graph in graph theory?
A graph is a collection of vertices, or nodes, and edges between some or all of the vertices. When there exists a path that traverses each edge exactly once such that the path begins and ends at the same vertex, the path is known as an Eulerian circuit and the graph is known as an Eulerian graph.
## When a line is graphed in the coordinate plane it separates the plane into two regions called?
boundary line
Each line plotted on a coordinate graph divides the graph (or plane) into two half‐planes. This line is called the boundary line (or bounding line). The graph of a linear inequality is always a half‐plane.
### What makes a graph linear or nonlinear?
A function whose graph is a straight line is a linear function. The graph of a nonlinear function is not a straight line. |
# Estimate quotients using multiples
When we estimate quotients using multiples, we are not looking for an exact answer. Instead, as it says, we are looking for an estimate, or a number that is close to the real answer.
Example #1
Find two numbers the quotient of 50 ÷ 7 is between. Then estimate the quotient.
Step 1
Ask yourself, " what number multiplied by 7 is about 50? "
Instead of guessing, you can just make a table showing some possibilities. Keep in mind that the product of any counting number multiplied by 7 is a multiple.
Therefore, the table will show multiples of 7.
Notice that we stopped at 56 since 56 is bigger than 50
Counting number 1 2 3 4 5 6 7 8 Multiples of 7 7 14 21 28 35 42 49 56
Step 2
Use the table to find multiples of 7 closest to 50.
7 × 7 = 49
7 × 8 = 56
50 is between 49 and 56, so the answer is either 7 or 8.
50 is closest to 49, so 50 ÷ 7 is about 7.
## Trick to estimate quotients using multiples when the quotient is big.
Example #2
Find two numbers the quotient of 152 ÷ 6 is between. Then estimate the quotient.
Step 1
Ask yourself, " what number multiplied by 6 is about 152? "
Once again, instead of guessing, you can just make a table showing some possibilities. This time though, we need to play strategically and here is the reason why.
Notice that 6 × 10 = 60. This means that if we had made a table with just the first 10 counting numbers, we would still be very far away from 152. If we put counting numbers 1, 2, 3, 4, 5, and so forth, this will make the table way too big!
Therefore, instead of using counting numbers 1, 2, 3, 4, 5, and so forth, we should use 10, 15, 20, 25, 30, and so forth to keep the table small.
Counting number 10 15 20 25 30 Multiples of 6 60 90 120 150 180
Step 2
Use the table to find multiples of 6 closest to 152.
6 × 25 = 150
6 × 30 = 180
152 is between 150 and 180, so the answer is either 25 or 30.
152 is closest to 150, so 152 ÷ 6 is about 25.
## Recent Articles
1. ### Divide using Partial Quotients
Apr 03, 17 09:12 AM
Learn to divide using partial quotients. Understand the concept fast!
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## Recent Lessons
1. ### Divide using Partial Quotients
Apr 03, 17 09:12 AM
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Title MOP Homework 2015 228.6 KB 7
##### Document Text Contents
Page 1
MOP Homework 2015
June 2015
1 Introduction
Welcome to MOP! Here are some problems the graders have picked out for you
to think about, and to help you get in gear for the summer. We’ll discuss these
on the first day. You’re welcome to work on as many of these as you wish,
regardless of your MOP group. A few suggestions:
• The “FUNDERMENTARS” section, geared towards the Red group, con-
tains some basic (well-known) results that are useful to know before you
get started at MOP. Look through these, and if there are any that you
haven’t seen or don’t know how to prove, prioritize working on these over
the rest of the problems. Feel free to skip anything you’ve seen before.
• The main section of problems are divided into four subjects: Algebra,
Combinatorics, Geometry, and Number Theory. If you’re limited on time,
try (approximately) the first half of each subsection if you’re in Red and
the second half if you’re in Blue/Black.
• If you think of the camp you’re attending as MOSP rather than MOP,
you are invited to try the problems in the “MOSP” homework section in
addition to the regular problems.
2 FUNDERMENTARS
2.1 Algebra
1. The image above is a proof of the AM-GM inequality in three variables.
Figure out how it works and extend it to n variables.
2. Prove Vieta’s formulas.
1
Page 2
2.2 GG Geometry
In this section, you will prove the existence of the incenter in several ways, and
discover other important results along the way.
0 a) Triangle ABC is inscribed in a circle with center O. Angle chase to
prove that ∠AOC = 2∠ABC.
b) Prove that a convex quadrilateralABCD is cyclic if and only if ∠ABD =
∠ACD.
1. a) Prove the extended law of sines: for a triangle ABC with circum-
BC
sinA
=
CA
sinB
=
AB
sinC
= 2R.
b) Prove the angle bisector theorem: If the internal angle bisector of angle
BAC intersects side BC at X, then XB
XC
= AB
AC
.
c) Prove Ceva’s theorem: Given a triangle ABC with points D,E, F on
sides BC,CA,AB respectively, the lines AD,BE,CF concur if and only
if
BD
DC
CE
EA
AF
FB
= 1.
d) Conclude that the internal angle bisectors of a triangle ABC concur at
a point I, the incenter.
2. a) Use the law of sines and Ceva to prove trig Ceva: In the configuration
of Ceva’s theorem, lines AD,BE,CF concur if and only if
sinDAC
sinCBE
sinEBA
sinACF
sinFCB
= 1.
Conclude again that the incenter exists.
3. In triangle ABC, let BC = a,CA = b, AB = c, s = 1
2
(a+ b+ c). a) Prove
the following perpendicularity criterion: AB ? CD , AC2 + BD2 =
b) Prove Carnot’s theorem: Given a triangle ABC with points D,E, F on
sides BC,CA,AB respectively, the perpendiculars to sides BC,CA,AB
at D,E, F respectively concur if and only if
(BD2 � DC2) + (CE2 � EA2) + (AF 2 � FB2) = 0.
c) Prove that the tangents to a circle from a point P outside the circle
have equal length.
d) Assuming there is a circle inside the triangle tangent to all three sides at
D,E, F respectively, compute the lengths BD,DC and analogous lengths
for other sides.
e) Constructing D,E, F to satisfy the length conditions above, prove that
the perpendiculars at these points intersect at a point that is indeed the
circumcenter of DEF , and show that this is the incenter of ABC.
2
Page 3
4. a) Prove that the internal bisector of angle A meets the perpendicular
bisector of BC at a point Ta on the circumcircle.
b) Prove that the circle wa with center Ta that goes through points B,C,
along with circles wb, wc defined analogously, all share a common point I,
and show that this is the incenter of ABC.
2.2.1 Things that are not the incenter
5. a) State and prove the Power of a Point theorem.
b) Use the perpendicularity criterion to prove that the locus of points with
the same power with respect to two circles w1, w2 is a line, their radical
axis.
c) Prove that the radical axes of three circles intersect at a point, their
2.3 Numbers
In this problem, we will investigate integer linear combinations
1. Given non-zero integers A and B, let D be their gcd. Show that D is the
smallest positive integer which can be written as Ax+By for integers x, y.
2. Conclude that the set of values Ax+By is just the multiples of D.
3. Now, let A = p be a prime number, and use 1) to show that any B not
divisible by p has a multiplicative inverse modulo p, or equivalently that
for some C, p divides BC − 1. This means that every non-zero integer
mod p is invertible. The set of residues mod p is an example of a structure
called a field.
In this problem, you will prove Fermat’s little Theorem (FlT), which states that
ap ≡ a (mod p) for all primes p, in several ways.
1. Prove FlT by induction on a.
2. Consider the ways to color a necklace of p beads in a colors. Use shifting
symmetry to prove FlT.
3. Consider the set A of powers of a modulo p, 1, a, a2, . . .. Call bA =
b, ba, ba2, . . . a shift of A. Show that all shifts of A are either identical
as (unordered) sets or disjoint, and cover all non-zero elements modulo p.
Conclude FlT.
4. Extend part 3 to prove Euler’s totient theorem: let φ(n) be the num-
ber of integers in 1, 2, ...n which are relatively prime to n. Show that if
gcd(a, n) = 1, then aφ(n) − 1 is divisible by n.
2.4 Wombinatorics Combinatorics
1. Find a closed form for
P n
k=0 k
n
k
.
2. Show that any graph with at least as many edges as vertices has a cycle.
3
Page 4
3 Problems
3.1 Algebra
1. [South Africa 2014 #2] Given that
a− b
c− d
= 2 and
a− c
b− d
= 3
for certain real numbers a, b, c, d, and b 6= c, determine the value of
a− d
b− c
.
2. [ISL 2011 A2] Determine all sequences (x1, x2, . . . , x2011) of positive inte-
gers, such that for every positive integer n there exists an integer a with
2011∑
j=1
jxnj = a
n+1 + 1
3. [ISL 2010 A3] Let x1, . . . , x100 be nonnegative real numbers such that
xi + xi+1 + xi+2 ≤ 1 for all i = 1, . . . , 100 (we put x101 = x1, x102 = x2).
Find the maximal possible value of the sum S =
∑100
i=1 xixi+2.
4. [WOOT 2011-2012 Algebra Marathon #3] Solve the following equation
f : R→ R:
f(x+ f(y))[x− f(y)] = f(x2)− yf(y)
for all x, y ∈ R, and such that f(x) 6= 0 for all x 6= 0.
5. [Extension of Blue MOP 2011] A finite set A and another set B, both of
integers, have the property that any integer n may be written uniquely as
a sum n = a+ b for a in A, b in B.
(a) Prove that A is symmetric, that is, for some integer m, a is in A if
and only if m− a is in A.
(b) Classify the possible structures A can have.
6. [ELMO SL 2012 A8] Find all functions f : Q→ R such that f(x)f(y)f(x+
y) = f(xy)(f(x) + f(y)) for all x, y ∈ Q.
3.2 Combinatorics
1. [Classic] Prove that a graph is two-colorable if and only if it has no odd
cycles.
2. [Classic] When given an expression containing only +,−,×, constants,
variables, and parentheses, such as (x+ y)(z − 8w(x+ y(w+ z) + 7)− z),
we expand it by repeatedly using the distributive law. Must this process
always finish? Or if we choose which terms to expand badly, might we
have to keep expanding forever?
4
Page 5
3. [ISL 1998 C1] Hacker gives Matt, Jackie, and Inez a rectangular array
of numbers. In each row and each column, the sum of all numbers is
an integer. Prove that Inez can change each nonintegral number x in the
array into either dxe or bxc so that the row-sums and column-sums remain
unchanged. (Note that dxe is the least integer greater than or equal to x,
while bxc is the greatest integer less than or equal to x.)
4. [ISL 1988 #4] Let n ≥ 2. Bathspounge numbers an n×n chessboard with
the numbers 1, 2, . . . , n2. Every number occurs once. Prove that there
exist two orthogonally adjacent squares such that their numbers differ by
at least n.
5. [Brazil Olympic Revenge 2014] Let n be a positive integer. In a 2n × 2n
board, 1 × n and n × 1 pieces are arranged without overlap. Call an
arrangement maximal if it is impossible to put a new piece in the board
without overlapping the previous ones. Find the least k such that there
is a maximal arrangement that uses k pieces.
6. [SubredditControl] Snake shuffles a deck of playing cards. Salamander
guesses if the top card is red or black, then turns it over. If he’s right, he
wins a dollar. If not, he pays a dollar. Repeat for the rest of the deck,
one card at a time. What should be his strategy, and what is his expected
return?
7. [SubredditControl] Inez paints 1, 2, 3, . . . , N in red, blue and white, so that
each color has more than a quarter of the set in its colour. Can she paint
it such that all x+ y = z in the set has at least two of the same colour?
3.3 Geometry
1. [ISL 2006 G1] Let ABC be triangle with incenter I. A point P in the
interior of the triangle satisfies
\ PBA+ \ PCA = \ PBC + \ PCB.
Show that AP ≥ AI, and that equality holds if and only if P = I.
2. [USA TST 2004 #4] Let ABC be a triangle. Choose a point D in its
interior. Let ω1 be a circle passing through B and D and ω2 be a circle
passing through C and D so that the other point of intersection of the two
circles lies on AD. Let ω1 and ω2 intersect side BC at E and F , respec-
tively. Denote by X the intersection of DF , AB and Y the intersection
of DE,AC. Show that XY ‖ BC.
3. [ISL 2007 G3] The diagonals of a trapezoid ABCD intersect at point P .
Point Q lies between the parallel lines BC and AD such that \ AQD =
\ CQB, and line CD separates points P and Q. Prove that \ BQP =
\ DAQ.
4. [Linus 2015] A polygon of wrapping paper folds to cover a cube completely
with no overlap. Show that it has two equal angles.
5
Page 6
5. [Russia 2006] Consider an isosceles triangle ABC with AB = AC, and
a circle ω which is tangent to the sides AB and AC of this triangle and
intersects the side BC at the points K and L. The segment AK intersects
the circle ω at a point M (apart from K). Let P and Q be the reflec-
tions of the point K in the points B and C, respectively. Show that the
circumcircle of triangle PMQ is tangent to the circle ω.
6. [Iran 2010] Circles W1,W2 intersect at P,K. XY is common tangent of
two circles which is nearer to P and X is on W1 and Y is on W2. XP
intersects W2 for the second time in C and Y P intersects W1 in B. Let
A be intersection point of BX and CY . Prove that if Q is the second
intersection point of circumcircles of ABC and AXY
∠QXA = ∠QKP
7. [Tuymaada 2012] Quadrilateral ABCD is both cyclic and circumscribed.
Its incircle touches its sides AB and CD at points X and Y , respectively.
The perpendiculars to AB and CD drawn at A and D, respectively, meet
at point U ; those drawn at X and Y meet at point V , and finally, those
drawn at B and C meet at point W . Prove that points U , V and W are
collinear.
3.4 Number Theory
1. [South Africa 2004 #1] Let a = 1111 . . . 1111 and b = 1111 . . . 1111 where
a has forty ones and b has twelve ones. Determine the greatest common
divisor of a and b.
2. [Brazil Olympic Revenge 2010 #1] Fix a prime p and an integer a with
gcd(a, p) = 1. Prove that the number of ordered triples (x, y, z) of positive
integers such that 0 ≤ x, y, z < p and (x+y+z)2 ≡ axyz (mod p) is equal
to p2 + 1.
3. [ISL 2009 N3] Let f be a non-constant function from the set of positive
integers into the set of positive integer, such that a− b divides f(a)− f(b)
for all distinct positive integers a, b. Prove that there exist infinitely many
primes p such that p divides f(c) for some positive integer c.
4. [ISL 2001 N4] Let p ≥ 5 be a prime number. Prove that there exists an
integer a with 1 ≤ a ≤ p− 2 such that neither ap−1− 1 nor (a+ 1)p−1− 1
is divisible by p2.
5. [Linus’ Homework from UCL: Sarah Zerbes]
(a) Solve x3 = y2 + 1 in integers.
(b) Solve x3 = y2 + 5 in integers.
6. [ELMO SL 2011 N3] Let n > 1 be a fixed positive integer, and call
an n-tuple (a1, a2, . . . , an) of integers greater than 1 good if and only if
ai
∣∣∣ (a1a2···anai − 1) for i = 1, 2, . . . , n. Prove that there are finitely many
good n-tuples.
6 |
# Excercise 7.2 Congruence-of-Triangles- NCERT Solutions Class 7
Go back to 'Congruence of Triangles'
## Chapter 7 Ex.7.2 Question 1
Which congruence criterion do you use in the following?
(a) Given: \begin{align}AC &= DF\\AB &= DE\\BC &= EF\end{align} So, $$ΔABC ≅ ΔDEF$$ (b) Given: \begin{align} ZX &= RP\\RQ &= ZY\\∠PRQ &= ∠XZY \end{align} So, $$ΔPQR ≅ ΔXYZ$$ (c) Given: \begin{align} ∠MLN &= ∠FGH\\∠NML &= ∠GFH\\ML &= FG \end{align} So, $$ΔLMN ≅ ΔGFH$$ (d) Given: \begin{align} EB &= DB\\AE &= BC\\∠A &= ∠C = 90^\circ \end{align} So, $$ΔABE ≅ ΔCDB$$
### Solution
(a)
What is known?
In $$ΔABC$$ and $$ΔDEF$$
\begin{align} AC&= DF\\AB &= DE\\BC &= EF \end{align}
What is the unknown?
Congruence criterion by which these two triangles are congruent.
Reasoning:
Three sides of a $$ΔABC$$ are equal to the corresponding three sides of the other $$ΔDEF$$. So, $$SSS$$ congruence criterion can be used.
Steps:
Given, $$AC$$ = $$DF$$, $$AB = DE$$ and $$BC = EF$$. The three sides of a $$ΔABC$$ are equal to the corresponding three sides of the other $$ΔDEF$$. So, by $$SSS$$ congruence criterion, the two triangles are congruent.
(b)
What is known?
\begin{align} ZX &= RP\\RQ &= ZY\\∠PRQ &= ∠XZY \end{align}
What is unknown?
Congruence criterion by which the two triangles are congruent.
Reasoning:
The two sides and one angle of a $$ΔPRQ$$ are equal to the corresponding two sides and one angle of the other $$ΔXYZ$$. So, $$SSS$$ congruence criterion can be used.
Steps:
Given,$$ZX = RP$$, $$RQ = ZY$$, $$∠PRQ = ∠XZY$$. The two sides and one angle of a triangle $$ΔPRQ$$ are equal to the corresponding two sides and one angle of the other triangle $$ΔXYZ$$. So, by $$SAS$$ congruence criterion, the two triangles are congruent.
(c)
What is known?
\begin{align} ∠MLN&= ∠FGH\\∠NML &= ∠GFH\\ML &= FG \end{align}
What is the unknown?
By which congruence criterion two triangles are congruent.
Reasoning:
The two angles and one side of a triangle $$ΔLMN$$ are equal to the corresponding two sides and one angle of the other triangle $$ΔGFH$$. So, by using congruency criterion based on two angels and one side $$(ASA)$$ of triangles can be used.
Steps:
Given, $$∠MLN = ∠FGH$$, $$∠NML = ∠GFH$$ and $$ML = FG$$. The two sides and one angle of triangle $$ΔLMN$$ are equal to the two sides and one angle of the other triangle $$ΔGFH$$. So, by $$ASA$$ congruence criterion, the two triangles are congruent.
(d)
What is known?
In $$ΔABE$$ and $$ΔCDB$$
\begin{align} EB &= DB\\AE &= BC\\∠A& = ∠C = 90^\circ \end{align}
What is the unknown?
By which congruence criterion two triangles are congruent.
Reasoning:
In this case, the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle. congruency is based on of a right-angle, hypotenuse and one side $$(RHS)$$ criterion.
Steps:
Since, $$EB = DB$$, $$AE = BC$$, $$∠A = ∠C = 90^\circ$$. The hypotenuse and one side of a right-angled triangle $$ΔABE$$ are equal to the hypotenuse and one side of the other right-angled triangle $$ΔCDB.$$ So, by $$(RHS)$$ congruence criterion, the two triangles are congruent.
## Chapter 7 Ex.7.2 Question 2
You want to show that $$ΔART ≅ ΔPEN,$$
(a) If you have to use $$SSS$$ criterion, then you need to show
(i) $$AR$$ =
(ii) $$RT$$ =
(iii) $$AT$$ =
(b) If it is given that $$∠T = ∠N$$ and you are to use $$SAS$$ criterion, you need to have
(i) $$RT$$ =
(ii) $$PN$$ =
(c) If it is given that $$AT$$ = $$PN$$ and you are to use $$ASA$$ criterion, you need to have
(i) $$ZA$$
(ii) $$ZT$$
### Solution
Steps:
(a) If you have to use $$SSS$$ criterion, then you need to show
(i) $$AR$$ = $$PE$$
(ii) $$RT$$ = $$EN$$
(iii) $$AT$$ = $$PN$$
(b) If it is given that $$∠T = ∠N$$ and you are to use $$SAS$$ criterion, you need to have
(i) $$RT$$ = $$EN$$
(ii) $$PN$$ = $$AT$$
(c) If it is given that $$AT$$ = $$PN$$ and you are to use $$ASA$$ criterion to show that $$ΔART ≅ ΔPEN,$$ you need to have
(i) $$∠RAT = ∠EPN$$
(ii) $$∠RTA = ∠ENP$$
## Chapter 7 Ex.7.2 Question 3
You have to show that $$ΔAMP ≅ ΔAMQ$$. In the following proof, supply the missing reasons.
Steps Reasons (i) $$PM = QM$$ (i) …. (ii) $$∠PMA = ∠QMA$$ (ii) …. (iii) $$AM = AM$$ (iii) …. (iv) $$ΔAMP ≅ ΔAMQ$$ (iv) ….
### Solution
Steps:
Steps Reasons (i) $$PM = QM$$ (i) Given (ii) $$∠PMA = ∠QMA$$ (ii) Given (iii) $$AM = AM$$ (iii) Common (iv) $$ΔAMP ≅ ΔAMQ$$ (iv) $$SAS$$ congruence rule
## Chapter 7 Ex.7.2 Question 4
In $$ΔABC$$, $$∠A = 30^\circ$$, $$∠B = 40^\circ$$ and $$∠C = 110^\circ$$. In $$ΔPQR$$, $$∠P = 30^\circ$$, $$∠Q = 40^\circ$$ and $$∠R =110^\circ$$. A student says that $$ΔABC$$ $$≅$$ $$ΔPQR$$ by $$AAA$$ congruence criterion. Is he justified? Why or why not?
### Solution
What is known?
In $$ΔABC$$, $$∠A = 30^\circ$$, $$∠B = 40^\circ$$ and $$∠C = 110^\circ$$
In $$ΔPQR$$, $$∠P = 30^\circ$$, $$∠Q = 40^\circ$$ and $$∠R =110^\circ$$
What is the unknown?
Justification that $$ΔABC$$ $$≅$$ $$ΔPQR$$ by $$AAA$$ congruence criterion.
Reasoning:
In this question, it is given that the angle measure of all the angles of triangle $$ABC$$, $$∠A = 30^\circ$$, $$∠B = 40^\circ$$ and $$∠C = 110^\circ$$ is equal to the measure of all the angles of another triangle $$ΔPQR$$, $$∠P = 30^\circ$$, $$∠Q = 40^\circ$$ and $$∠R =110^\circ$$ and you can justify the congruence of $$ΔABC$$ and $$ΔPQR$$ by $$AAA$$ criterion or not. You can justify your answer by using the property based on $$AAA$$ congruence of two triangles. We know that there is no such thing as $$AAA$$ congruence of two triangles: Two triangles with equal corresponding angles need not to be congruent. In such a correspondence, one of them can be an enlarged copy of the other (They would be congruent only if they are exact copies of one another).
Steps:
No, $$AAA$$ property cannot justify $$ΔABC$$ $$≅$$ $$ΔPQR$$ because, this property represents that these two triangles have their respective angles of equal measures, but it gives no information about their sides. Two triangles with equal corresponding angles need not to be congruent. In such a correspondence, one of them can be an enlarged copy of the other. Therefore, $$AAA$$ does not prove that the two triangles $$ABC$$ and $$PQR$$ are congruent.
## Chapter 7 Ex.7.2 Question 5
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write $$ΔRAT ≅$$?
### Solution
What is known?
The given two triangles are congruent and their corresponding parts are marked on the given figure.
$$\overline{NO}\text{ }=\text{ }\overline{AT}$$ , $$\overline{AR}\text{ }=\text{ }\overline{OW}$$
$$∠W = ∠R, ∠A = ∠O$$ and $$∠T = ∠N$$
What is the unknown?
Congruence of $$ΔRAT$$?
Reasoning:
This question is based on the concept of congruence of two triangle when two angles and one side (or two angles and one side) of the triangles are equal. If two angles and one side of a triangle are equal to the corresponding two angles and one side of another triangle then the two triangles can be congruent by $$ASA$$ Congruence criterion, by using this criterion you can find out the triangle congruent to $$RAT$$.
Steps:
From the figure, it can be observed that
\begin{align} \angle RAT &=\angle WON \\\angle ART & =\angle NOW \\ \overline{AR}\text{ } &=\text{ }\overline{OW}\end{align}
Therefore, $$ΔRAT ≅$$ WON (By $$ASA$$ congruence criterion)
## Chapter 7 Ex.7.2 Question 6
Complete the congruence statement:
(i) $$ΔBCA ≅ \,\,?$$
(i) $$ΔQRS ≅\,\,?$$
### Solution
What is known?
What is the unknown?
Congruence of triangles.
(i) $$ΔBCA ≅ \,\,?$$
(ii) $$ΔQRS ≅\,\,?$$
From figure, in the triangles $$BTA$$ and $$BCA,$$
$$B$$$$B,$$ $$A$$$$A$$ and $$T$$$$C$$
And in triangle $$QRS$$ and $$TPO,$$
$$P$$$$R,$$ $$T$$$$Q$$ and $$O$$$$S$$
Reasoning:
This question is based on the concept of congruence of triangles, if the corresponding parts of two triangles are equal then the two triangles are congruent to each other. In first triangle, corresponding parts of triangle $$BTA$$ are congruent to the triangle$$BCA$$ and the corresponding parts of triangle $$QRS$$ are congruent to triangle $$TPO.$$ Thus, by applying congruence based on sides of a triangle, two triangles can be proved congruent.
Steps:
In triangle $$BAT$$ and triangle $$BCA,$$ Corresponding parts are congruent.
$$B$$$$B,$$ $$A$$$$A$$ and $$T$$$$C.$$
So, by $$SSS$$ congruence rule, $$ΔBCA ≅ BTA$$
Similarly, in triangle $$QRS$$ and $$TPO,$$ corresponding parts are congruent. So, by $$SSS$$ congruence rule,
$$P$$$$R,$$ $$T ↔ Q$$ and $$O ↔ S$$
$$ΔQRS ≅ TPO$$
## Chapter 7 Ex.7.2 Question 7
In a squared sheet, draw two triangles of equal areas such that
(i) the triangles are congruent.
(ii) the triangles are not congruent.
What can you say about their perimeters?
### Solution
Steps:
Draw two triangles of equal areas,
(i)
Triangles $$ABC$$ and $$XYZ$$ have the same area and are congruent to each other. Also, perimeter of both the triangles will be equal.
Such that, $$\overline{AB}\text{ }=\text{ }\overline{DE},\overline{BC}\text{ }=\text{ }\overline{EF}\;\rm{and}\;\overline{AC}\text{ }=\text{ }\overline{DF}$$
On adding, we get $$\overline{AB}+\overline{BC}+\overline{AC}=\overline{DE}+\overline{EF}+\overline{DF}$$
i.e, perimeter of both the triangles are equal.
(ii)
Here, we have drawn two triangles $$ABC$$ and $$PQR$$ which are not congruent to each other.
Such that ,$$\overline{AB}\text{ }\ne \text{ }\overline{XY},\overline{BC}\text{ }\ne \text{ }\overline{YZ}\;\rm{and}\;\overline{AC}\text{ }\ne \text{ }\overline{XZ}$$
On adding, we get $$\overline{AB}+\overline{BC}+\overline{AC}=\overline{XY}+\overline{YZ}+\overline{XZ}$$
Also, the perimeter of both the triangles will not be the same.
## Chapter 7 Ex.7.2 Question 8
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
### Solution
What is known?
Two triangles that have five pairs of congruent parts.
What is unknown?
Despite having five pairs of congruent parts, the triangles are not congruent
Steps:
In triangle $$PQR$$ and $$ABC$$
All angles two sides are equal except one side. Hence triangle $$PQR$$ is not congruent to triangle $$ABC$$
## Chapter 7 Ex.7.2 Question 9
If $$ΔABC$$ and $$ΔPQR$$ are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
### Solution
What is unknown?
Name one additional pair of corresponding parts and the criterion used.
Reasoning:
In this question, if two triangles $$ΔABC$$ and $$ΔPQR$$ are to be congruent, we must name one additional pair of corresponding part and the criterion used. For better understanding of this question, visualize it with the help of figure. In triangles $$ΔABC$$ and $$ΔPQR$$ it is given that, $$\angle B = 90^\circ$$ and $$\angle Q= 90^\circ\,, \angle C =\angle R.$$ Now, find out the side between these two angles that would be your one additional pair of corresponding part. Also, by reminding the criterion based on angle and the side of a right- angled triangle, you can find out the criterion used.
Steps:
In triangle $$ΔABC$$ and $$ΔPQR$$ are congruent.
$$\angle B = 90^\circ$$ and $$\angle Q= 90^\circ\,, \angle C =\angle R.$$
$$\overline {BC} = \overline{QR}$$
Therefore,
$$ΔABC ≅ PQR$$ (By $$ASA$$ congruence rule)
## Chapter 7 Ex.7.2 Question 10
Explain, why
$$ΔABC ≅ ΔFED$$.
### Solution
What is known?
Two right angles triangles $$ΔABC$$ and $$ΔFED$$ in which one side and one angle are equal.
What is unknown?
Why $$ΔABC ≅ ΔFED$$?
Reasoning:
This question is based on congruence of a right-angled triangle. In this question, one angle and a side of a right-angled triangle are equal to the corresponding one angle and a side of another right-angled triangle. Now, by using congruence based on right-angled triangle, you can find the reason why these two triangles are congruent.
Steps:
Given,
In triangles, $$ABC$$ and $$FED$$,
\begin{align}\angle \text{A}\,&=\ \angle \,\text{F}\\ \angle \text{B}\,&=\,\,\angle \text{E=9}{{\text{0}}^{\circ }}\\\rm{} BC &= ED\end{align}
Therefore, $$ΔABC ≅ ΔFED$$ (By $$RHS$$ congruence rule)
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# Angles on a straight line
## 12.2 Angles on a straight line
Let us look at angles formed on one side of a straight line.
In this diagram, line segment $$AB$$ meets line segment $$DC$$. The angle at the vertex, $$C$$, where they meet, is now split into two angles: $$\hat{C_1}$$ and $$\hat{C_2}$$.
$$\hat{C_1}$$ is the name for the angle at vertex $$C$$ labelled "1" (or $$A\hat{C}D$$).
### The sum of the angles formed on a straight line
The sum of the angles that are formed on a straight line is always $$180^{\circ}$$.
We can shorten this property as: $$\angle$$s on a straight line.
• Two angles that add up to $$180^{\circ}$$ are called supplementary angles.
• Angles that share a vertex and a common side are said to be adjacent angles. $$\hat{C_1}$$ and $$\hat{C_2}$$ are supplementary angles.
• Hence, $$\hat{C_1}$$ and $$\hat{C_2}$$ are called adjacent supplementary angles.
supplementary angles
two angles that add up to $$180^{\circ}$$
angles that share a vertex and a common side
You can have more than one line meeting at the same point on a straight line. Here are a few examples of angles on a straight line.
### The sum of the angles on perpendicular lines
When two lines are perpendicular, the adjacent supplementary angles are both equal to $$90^{\circ}$$.
In the diagram, $$\hat{M_1}=\hat{M_2} =90^{\circ}$$.
A right angle is shown by forming a square at one of the right angles, like this: ⦜.
## Worked example 12.1: Calculating unknown angles on a straight line
Calculate the size of $$x$$.
In the diagram, we have two angles that are on the same side of the straight line. The first angle is $$100^{\circ}$$ and the second angle is unknown ($$x$$). We need to calculate the size of $$x$$.
### We know that the two angles have a sum of $$180^{\circ}$$, so we can say that:
$$100^{\circ}+x=180^{\circ}$$ ($$\angle$$s on a straight line)
### Now we can solve this equation.
\begin{align} x&=180^{\circ}-100^{\circ} \\ x&=80^{\circ} \end{align}
## Worked example 12.2: Calculating unknown angles on a straight line
Calculate the size of $$x$$.
### Notice that there are three angles on the same side of the straight line. We have $$x$$, an angle of $$29^{\circ}$$ and an angle of $$90^{\circ}$$. (Remember that the ⦜ symbol on the diagram indicates a $$90^{\circ}$$ angle.) These three angles have a sum of $$180^{\circ}$$, so we can say that:
$$x+29^{\circ}+90^{\circ}=180^{\circ}$$ ($$\angle$$s on a straight line)
### Now we can solve this equation.
\begin{align} x+29^{\circ}+90^{\circ}&=180^{\circ} \\ x+119^{\circ}&=180^{\circ} \\ x&=180^{\circ}-119^{\circ} \\ x&=61^{\circ} \end{align}
### There is a simpler way to solve for $$x$$. It is given that we have a perpendicular line. Adjacent angles on a perpendicular line are both equal to $$90^{\circ}$$. So, we have a different equation to solve.
\begin{align} x+29^{\circ}&=90^{\circ} \\ x&=90^{\circ}-29^{\circ} \\ x&=61^{\circ} \end{align}
## Worked example 12.3: Calculating unknown angles on a straight line
Calculate the size of $$y$$.
### Notice that we have three angles on the same side of the straight line. We have $$2y$$, an angle of $$48^{\circ}$$ and an angle of $$52^{\circ}$$. These three angles have a sum of $$180^{\circ}$$, so we can say that:
$$2y+48^{\circ}+52^{\circ}=180^{\circ}$$ ($$\angle$$s on a straight line)
### Now we can solve this equation.
\begin{align} 2y+48^{\circ}+52^{\circ}&=180^{\circ} \\ 2y+100^{\circ}&=180^{\circ} \\ 2y&=180^{\circ}-100^{\circ} \\ 2y&=80^{\circ} \\ y&=40^{\circ} \end{align}
Exercise 12.1
Calculate the size of $$a$$.
\begin{align} a+63^{\circ}&=180^{\circ} &(\angle\text{s on a straight line}) \\ x&=180^{\circ}-63^{\circ} \\ x&=117^{\circ} \end{align}
Calculate the size of:
1. $$x$$
2. $$\hat{ECB}$$
1. \begin{align} x+3x+2x&=180^{\circ} &(\angle\text{s on a straight line}) \\ 6x&=180^{\circ} \\ x&=30^{\circ} \end{align}
2. \begin{align} \hat{ECB}&=2x \\ &=2(30^{\circ})\\ &=60^{\circ} \end{align}
Calculate the size of:
1. $$x$$
2. $$\hat{GEH}$$
1. \begin{align} (x+30^{\circ})+(x+40^{\circ}) +(2x+10^{\circ}) &=180^{\circ} &(\angle\text{s on a straight line}) \\ 4x+80^{\circ}&=180^{\circ} \\ 4x&=100^{\circ} \\ x&=25^{\circ} \end{align}
2. \begin{align} \hat{GEH}&=x+40^{\circ} \\ &=25^{\circ}+40^{\circ} \\ &=65^{\circ} \end{align}
Hint: Remember that the matching curved lines “))” indicate that the angles are equal.
Calculate the size of:
1. $$k$$
2. $$\hat{TYP}$$
1. \begin{align} (2k)+(k+65^{\circ}) +(2k) &=180^{\circ} &&(\angle\text{s on a straight line}) \\ 5k+65^{\circ}&=180^{\circ} \\ 5k &=115^{\circ} \\ &=23^{\circ} \end{align}
2. \begin{align} \hat{TYP}&=k+65^{\circ} \\ &=23^{\circ}+65^{\circ} \\ &=88^{\circ} \end{align} |
# What does HL mean in triangle proofs?
What does HL mean in triangle proofs? Hypotenuse-Leg-Right angle The HLR (Hypotenuse-Leg-Right angle) theorem — often called the HL theorem — states that if the hypotenuse and a leg of one right triangle are congruent
## What does HL mean in triangle proofs?
Hypotenuse-Leg-Right angle
The HLR (Hypotenuse-Leg-Right angle) theorem — often called the HL theorem — states that if the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent. The following figure shows you an example.
## How do you know if a triangle is HL?
The HL Theorem states; If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent. Of course you can’t, because the hypotenuse of a right triangle is always (always!) opposite the right angle.
What does HL look like in geometry?
Congruent Triangles – Hypotenuse and leg of a right triangle. (HL) Definition: Two right triangles are congruent if the hypotenuse and one corresponding leg are equal in both triangles. If, in two right triangles the hypotenuse and one leg are equal, then the triangles are congruent.
What does HL theorem stand for?
hypotenuse leg theorem
The hypotenuse leg theorem is a criterion used to prove whether a given set of right triangles are congruent. The hypotenuse leg (HL) theorem states that; a given set of triangles are congruent if the corresponding lengths of their hypotenuse and one leg are equal.
### What is an HL triangle?
Hypotenuse-Leg (HL) Triangle Congruence Theorem If the hypotenuse and a leg of a right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent. Example 1.
### What is special about HL?
They hypotenuse leg (HL) theorem for right triangles states that if the hypotenuse and one leg of one right triangle have the same lengths as the hypotenuse and one leg of another right triangle, then the two triangles are congruent.
What is a HL triangle?
What is the HL Postulate? The HL Postulate states that if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent.
How do you prove HL?
Hypotenuse Leg Theorem Proof The proof of the hypotenuse leg theorem shows how a given set of right triangles are congruent if the corresponding lengths of their hypotenuse and one leg are equal. Observe the following isosceles triangle ABC in which side AB = AC and AD is perpendicular to BC.
#### How do you use HL?
Hypotenuse-Leg (HL) for Right Triangles. There is one case where SSA is valid, and that is when the angles are right angles. Using words: In words, if the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of a second right triangle, then the triangles are congruent.
#### What is special about a triangle?
The angles inside the triangle are also important. The sum of the angles is always 180°. Triangles are special because they are exceptionally strong. Out of all the two-dimensional shapes we can make out of straight struts of metal, only a triangle is rigid. |
Testing Correlation Coefficient
In this tutorial we will discuss step by step solution of numerical problems on testing whether the population correlation coefficient is $\rho_0$ or not.
Example 1
The median records shows that the correlation between the age of the mother and the birth weight of their first child is less than -0.34. A random sample of 8 mother's age and the birth weight of their first child are as follows:
Age of mother 35 24 28 29 26 30 34 32
Birth weight of child 2.85 3.50 3.25 3.00 3.25 2.75 2.90 3.00
Test whether the medical records provide the true information at 5% level of significance.
Solution
Let $x$ denote the age of mother and $y$ denote the birth weight of first child.
The number of pairs $n= 8$.
$x$ $y$ $x^2$ $y^2$ $xy$
1 35 2.85 1225 8.123 99.75
2 24 3.50 576 12.250 84.00
3 28 3.25 784 10.562 91.00
4 29 3.00 841 9.000 87.00
5 26 3.25 676 10.562 84.50
6 30 2.75 900 7.562 82.50
7 34 2.90 1156 8.410 98.60
8 32 3.00 1024 9.000 96.00
Total 238 24.50 7182 75.470 723.35
The sample variance of $x$ is
\begin{aligned} s_{x}^2 & = \frac{1}{n-1}\bigg(\sum x^2 - \frac{(\sum x)^2}{n}\bigg)\\ & = \frac{1}{8-1}\bigg(7182-\frac{(238)^2}{8}\bigg)\\ &= \frac{1}{7}\bigg(7182-\frac{56644}{8}\bigg)\\ &= \frac{1}{7}\bigg(7182-7080.5\bigg)\\ &= \frac{101.5}{7}\\ &= 14.5. \end{aligned} The sample variance of $x$ is
\begin{aligned} s_{y}^2 & = \frac{1}{n-1}\bigg(\sum y^2 - \frac{(\sum y)^2}{n}\bigg)\\ & = \frac{1}{8-1}\bigg(75.47-\frac{(24.5)^2}{8}\bigg)\\ &= \frac{1}{7}\bigg(75.47-\frac{600.25}{8}\bigg)\\ &= \frac{1}{7}\bigg(75.47-75.0312\bigg)\\ &= \frac{0.4387}{7}\\ &= 0.0627. \end{aligned}
The sample covariance between $x$ and $y$ is
\begin{aligned} s_{xy} & = \frac{1}{n-1}\bigg(\sum xy - \frac{(\sum x)(\sum y)}{n}\bigg)\\ & = \frac{1}{8-1}\bigg(723.35-\frac{(238)(24.5)}{8}\bigg)\\ &= \frac{1}{7}\bigg(723.35-\frac{5831}{8}\bigg)\\ &= \frac{1}{7}\bigg(723.35-728.875\bigg)\\ &= \frac{-5.525}{7}\\ &= -0.7893. \end{aligned} The Karl Pearson's sample correlation coefficient between age of mother and birth weight of first child is
\begin{aligned} r_{xy} & = \frac{Cov(x,y)}{\sqrt{V(x) V(y)}}\\ &= \frac{s_{xy}}{\sqrt{s_x^2s_y^2}}\\ &=\frac{-0.7893}{\sqrt{14.5\times 0.0627}}\\ &=\frac{-0.7893}{\sqrt{0.9092}}\\ &=-0.828. \end{aligned} The correlation coefficient between age of mother and birth weight of first child is $-0.828$.
Step 1 Hypothesis Testing Problem
The hypothesis testing problem is $H_0 : \rho = -0.34$ against $H_1 : \rho < -0.34$ ($\text{left-tailed}$)
Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is \begin{aligned} Z&=\dfrac{U-\xi}{\sqrt{\frac{1}{n-3}}} \end{aligned} where \begin{aligned} U&=\frac{1}{2}\log_e \bigg(\frac{1+r}{1-r}\bigg) \end{aligned} and \begin{aligned} \xi & =\frac{1}{2}\log_e \bigg(\frac{1+\rho_0}{1-\rho_0}\bigg) \end{aligned} Under the null hypothesis the test statistic $Z$ follows $N(0,1)$ distribution.
Step 3 Significance Level
The significance level is $\alpha = 0.05$.
Step 4 Critical Value(s)
As the alternative hypothesis is $\text{left-tailed}$, the critical value of $Z$ $\text{is}$ $-1.64$ (from Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z < -1.64}$.
Step 5 Computation
\begin{aligned} U&=\frac{1}{2}\log_e \bigg(\frac{1+r}{1-r}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+(-0.828)}{1-(-0.828)}\bigg)\\ &=0.5\times \log_e\big(0.0941\big)\\ &=0.5\times -2.3635\\ &= -1.1817 \end{aligned} and \begin{aligned} \xi&=\frac{1}{2}\log_e \bigg(\frac{1+\rho_0}{1-\rho_0}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+(-0.34)}{1-(-0.34)}\bigg)\\ &=0.5\times \log_e\big(0.4925\big)\\ &=0.5\times -0.7082\\ &= -0.3541 \end{aligned} The test statistic under the null hypothesis is \begin{aligned} Z&=\dfrac{U-\xi}{\sqrt{\frac{1}{n-3}}}\\ &=\dfrac{-1.1817-(-0.3541)}{\sqrt{\frac{1}{8-3}}}\\ &=\dfrac{-0.8276}{\sqrt{\frac{1}{5}}}\\ &=-1.8507 \end{aligned}
Step 6 Decision (Traditional Approach)
The test statistic is $Z_{obs} =-1.851$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis at $\alpha = 0.05$ level of significance.
OR
Step 6 Decision ($p$-value Approach)
This is a $\text{left-tailed}$ test, so the p-value is the area to the $\text{negative}$ of the test statistic ($Z_{obs}=-1.851$) is p-value = $0.0321$.
The p-value is $0.0321$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis at $\alpha =0.05$ level of significance.
Interpretation
There is enough evidence to conclude that the medical records provide true information at $0.05$ level of significance.
Example 2
The correlation between scores on a traditional aptitude test and scores on a final test is known to be approximately 0.6. A new aptitude test has been developed and is tried on a random sample of 100 students, resulting in a correlation of 0.65. Does this result imply that the new test is better?
Solution
Given that the sample correlation between $X$ and $Y$ is $0.65$ for a sample of $100$ pair of observations.
Step 1 Hypothesis Testing Problem
The hypothesis testing problem is $H_0 : \rho = 0.6$ against $H_1 : \rho > 0.6$ ($\text{right-tailed}$)
Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is \begin{aligned} Z&=\dfrac{U-\xi}{\sqrt{\frac{1}{n-3}}} \end{aligned} where \begin{aligned} U&=\frac{1}{2}\log_e \bigg(\frac{1+r}{1-r}\bigg) \end{aligned} and \begin{aligned} \xi & =\frac{1}{2}\log_e \bigg(\frac{1+\rho_0}{1-\rho_0}\bigg) \end{aligned} Under the null hypothesis the test statistic $Z$ follows $N(0,1)$ distribution.
Step 3 Significance Level
The significance level is $\alpha = 0.05$.
Step 4 Critical Value(s)
As the alternative hypothesis is $\text{right-tailed}$, the critical value of $Z$ $\text{is}$ $1.64$ (from Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z > 1.64}$.
Step 5 Computation
\begin{aligned} U&=\frac{1}{2}\log_e \bigg(\frac{1+r}{1-r}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+0.65}{1-0.65}\bigg)\\ &=0.5\times \log_e\big(4.7143\big)\\ &=0.5\times 1.5506\\ &= 0.7753 \end{aligned} and \begin{aligned} \xi&=\frac{1}{2}\log_e \bigg(\frac{1+\rho_0}{1-\rho_0}\bigg)\\ &=0.5\times \log_e\bigg(\frac{1+0.6}{1-0.6}\bigg)\\ &=0.5\times \log_e\big(4\big)\\ &=0.5\times 1.3863\\ &= 0.6931 \end{aligned} The test statistic under the null hypothesis is \begin{aligned} Z&=\dfrac{U-\xi}{\sqrt{\frac{1}{n-3}}}\\ &=\dfrac{0.7753-0.6931}{\sqrt{\frac{1}{100-3}}}\\ &=\dfrac{0.0822}{\sqrt{\frac{1}{97}}}\\ &=0.8091 \end{aligned}
Step 6 Decision (Traditional Approach)
The test statistic is $Z_{obs} =0.809$ which falls $\text{outside}$ the critical region, we $\text{fail to reject}$ the null hypothesis at $\alpha = 0.05$ level of significance.
OR
Step 6 Decision ($p$-value Approach)
This is a $\text{right-tailed}$ test, so the p-value is the area to the $\text{right}$ of the test statistic ($Z_{obs}=0.809$) is p-value = $0.2092$.
The p-value is $0.2092$ which is $\text{greater than}$ the significance level of $\alpha = 0.05$, we $\text{fail to reject}$ the null hypothesis at $\alpha =0.05$ level of significance.
Interpretation
There is insufficient evidence to conclude that the new test is better. |
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# The sine and cosine rule
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### The sine and cosine rule
1. 2. THE SINE RULE Powerpoint hosted on www.worldofteaching.com Please visit for 100’s more free powerpoints
2. 3. A C B c b a The sine rules enables us to calculate sides and angles In the some triangles where there is not a right angle. The Sine Rule is used to solve any problems involving triangles when at least either of the following is known: a) two angles and a side b) two sides and an angle opposite a given side In Triangle ABC, we use the convention that a is the side opposite angle A b is the side opposite angle B
3. 4. <> Example 2 (Given two sides and an included angle) Solve triangle ABC in which A = 55°, b = 2.4cm and c = 2.9cm By cosine rule, a 2 = 2.4 2 + 2.9 2 - 2 x 2.9 x 2.4 cos 55° = 6.1858 a = 2.49cm
4. 5. Either Or [1] [2] Use [1] when finding a side Use [2] when finding an angle Using this label of a triangle, the sine rule can be stated
5. 6. Example: A C B c Given Angle ABC =60 0 Angle ACB = 50 0 Find c. 7cm To find c use the following proportion: c= 6.19 ( 3 S.F)
6. 7. A C B 15 cm 6 cm 120 0 SOLUTION: sin B = 0.346 B= 20.3 0
7. 8. SOLVE THE FOLLOWING USING THE SINE RULE: Problem 1 (Given two angles and a side) In triangle ABC , A = 59°, B = 39° and a = 6.73cm. Find angle C, sides b and c. DRILL: Problem 2 (Given two sides and an acute angle) In triangle ABC , A = 55°, b = 16.3cm and a = 14.3cm. Find angle B, angle C and side c. Problem 3 (Given two sides and an obtuse angle) In triangle ABC A =100°, b = 5cm and a = 7.7cm Find the unknown angles and side.
8. 9. C = 180° - (39° + 59°) = 82° Answer Problem 1
9. 10. = 0.9337 = 14.5 cm (3 SF) ANSWER PROBLEM 2 |
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# Domain and Range of a Function
## Discrete and continuous functions and dependent and independent values
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Practice Domain and Range of a Function
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Domain and Range
Joseph drove from his summer home to his place of work. To avoid the road construction, Joseph decided to travel the gravel road. After driving for 20 minutes he was 62 miles away from work and after driving for 40 minutes he was 52 miles away from work. Represent the problem on a graph and write a suitable domain and range for the situation.
### Guidance
The domain of a relation is the set of possible values that ‘x\begin{align*}x\end{align*}’ may have. The range of a relation is the set of possible values that ‘y\begin{align*}y\end{align*}’ may have. You can write the domain and range of a relation using interval notation and with respect to the number system to which it belongs. Remember:
• Z(integers)={3,2,1,0,1,2,3,}\begin{align*}Z(\text{integers})=\{-3,-2,-1,0,1,2,3, \ldots \}\end{align*}
• R(real numbers)={all rational and irrational numbers}\begin{align*}R(\text{real numbers})=\{\text{all rational and irrational numbers} \}\end{align*}.
These number systems are very important when the domain and range of a relation are described using interval notation.
A relation is said to be discrete if there are a finite number of data points on its graph. Graphs of discrete relations appear as dots. A relation is said to be continuous if its graph is an unbroken curve with no "holes" or "gaps." The graph of a continuous relation is represented by a line or a curve like the one below. Note that it is possible for a relation to be neither discrete nor continuous.
The relation is a straight line that that begins at the point (2, 1). The fact that the points on the line are connected indicates that the relation is continuous. The domain and the range can be written in interval notation, as shown below:
#### Example A
Which relations are discrete? Which relations are continuous? For each relation, find the domain and range.
(i)
(ii)
(iii)
(iv)
Solution:
(i) The graph appears as dots. Therefore, the relation is discrete. The domain is {1,2,4}\begin{align*}\{1,2,4 \}\end{align*}. The range is {1,2,3,5}\begin{align*}\{1,2,3,5 \}\end{align*}
(ii) The graph appears as a straight line. Therefore, the relation is continuous. D={x|x R}R={y|y R}\begin{align*}D=\{x|x \ \in \ R \} \quad R=\{y|y \ \in \ R \}\end{align*}
(iii) The graph appears as dots. Therefore, the relation is discrete. The domain is {1,0,1,2,3,4,5}\begin{align*}\{-1,0,1,2,3,4,5\}\end{align*}. The range is {2,1,0,1,2,3,4}\begin{align*}\{-2,-1,0,1,2,3,4\}\end{align*}
(iv) The graph appears as a curve. Therefore, the relation is continuous. D={x|x R}R={y|y3,y R}\begin{align*}D=\{x|x \ \in \ R \} \quad R=\{y|y \ge -3, y \ \in \ R \}\end{align*}
#### Example B
Whether a relation is discrete, continuous, or neither can often be determined without a graph. The domain and range can be determined without a graph as well. Examine the following toothpick pattern.
Complete the table below to determine the number of toothpicks needed for the pattern.
Pattern number (n)\begin{align*}(n)\end{align*} 1 2 3 4 5 ... n\begin{align*}n\end{align*} ... 200
Number of toothpicks (t)\begin{align*}(t)\end{align*}
Is the data continuous or discrete? Why?
What is the domain?
What is the range?
Solution:
Pattern number (n)\begin{align*}(n)\end{align*} 1 2 3 4 5 ... n\begin{align*}n\end{align*} ... 200
Number of toothpicks (t)\begin{align*}(t)\end{align*} 7 12 17 22 27 5n+2\begin{align*}5n+2\end{align*} 1002
The number of toothpicks in any pattern number is the result of multiplying the pattern number by 5 and adding 2 to the product.
The number of toothpicks in pattern number 200 is:
tttt=5n+2=5(200)+2=1000+2=1002
The data must be discrete. The graph would be dots representing the pattern number and the number of toothpicks. It is impossible to have a pattern number or a number of toothpicks that are not natural numbers. Therefore, the points would not be joined.
The domain and range are:
D={x|x N}R={y|y=5x+2,x N}
If the range is written in terms of a function, then the number system to which x\begin{align*}x\end{align*} belongs must be designated in the range.
#### Example C
Can you state the domain and the range of the following relation?
Solution:
The points indicated on the graph are {(5,4),(5,1),(2,3),(2,1),(2,4)}\begin{align*}\{(-5,-4),(-5,1),(-2,3),(2,1),(2,-4)\}\end{align*}
The domain is {5,2,2}\begin{align*}\{-5,-2,2 \}\end{align*} and the range is {4,1,3}\begin{align*}\{ -4,1,3\}\end{align*}.
#### Concept Problem Revisited
Joseph drove from his summer home to his place of work. To avoid the road construction, Joseph decided to travel the gravel road. After driving for 20 minutes he was 62 miles away from work and after driving for 40 minutes he was 52 miles away from work. Represent the problem on a graph and write a suitable domain and range for the situation.
To represent the problem on a graph, plot the points (20, 62) and (40, 52). The points can be joined with a straight line since the data is continuous. The distance traveled changes continuously as the time driving changes. The y\begin{align*}y\end{align*}-intercept represents the distance from Joseph’s summer home to his place of work. This distance is approximately 72 miles. The x\begin{align*}x\end{align*}-intercept represents the time it took Joseph to drive from his summer home to work. This time is approximately 145 minutes.
Time cannot be a negative quantity. Therefore, the smallest value for the number of minutes would have to be zero. This represents the time Joseph began his trip. A suitable domain for this problem is D={x|0x145,x R}\begin{align*}D=\{x|0 \le x \le 145, x \ \in \ R\}\end{align*}
The distance from his summer home to work cannot be a negative quantity. This distance is represented on the y\begin{align*}y\end{align*}-axis as the y\begin{align*}y\end{align*}-intercept and is the distance before he begins to drive. A suitable range for the problem is R={y|0y72,y R}\begin{align*}R=\{y|0 \le y \le 72, y \ \in \ R\}\end{align*}
The domain and range often depend on the quantities presented in the problem. In the above problem, the quantities of time and distance could not be negative. As a result, the values of the domain and the range had to be positive.
### Guided Practice
1. Which relation is discrete? Which relation is continuous?
(i)
(ii)
2. State the domain and the range for each of the following relations:
(i)
(ii)
3. A computer salesman’s wage consists of a monthly salary of $200 plus a bonus of$100 for each computer sold.
(a) Complete the following table of values:
Number of computers sold 0 2 5 10 18
Wages in dollars for the month ($) (b) Sketch the graph to represent the monthly salary ($), against the number (N)\begin{align*}(N)\end{align*}, of computers sold.
(c) Use the graph to write a suitable domain and range for the problem.
1. (i) The graph clearly shows that the points are joined. Therefore the data is continuous.
(ii) The graph shows the plotted points as dots that are not joined. Therefore the data is discrete.
2. (i) The domain represents the values of ‘x\begin{align*}x\end{align*}’. D={x|3x3,x R}\begin{align*}D=\{x|-3\le x\le 3, x \ \in \ R\}\end{align*}
The range represents the values of ‘y\begin{align*}y\end{align*}’. R={y|3y3,y R}\begin{align*}R=\{y|-3 \le y \le 3, y \ \in \ R \}\end{align*}
(ii) D={x|x R}\begin{align*}D=\{x|x \ \in \ R\}\end{align*}
R={y|4y4,y R}\begin{align*}R=\{y|-4 \le y \le 4, y \ \in \ R\}\end{align*}
3.
Number of computers sold 0 2 5 10 18
Wages in dollars for the month ($)$200 $400$700 $1200$2000
(c) The graph shows that the data is discrete. (The salesman can't sell a portion of a computer, so the data points can't be connected.) The number of computers sold and must be whole numbers. The wages must be natural numbers.
A suitable domain is D={x|x0,x W}\begin{align*}D=\{x|x \ge 0, x \ \in \ W\}\end{align*}
A suitable domain is R={y|y=200+100x,x N}\begin{align*}R=\{y|y=200+100x, x \ \in \ N\}\end{align*}
### Explore More
Use the graph below for #1 and #2.
1. Is the relation discrete, continuous, or neither?
2. Find the domain and range for the relation.
Use the graph below for #3 and #4.
1. Is the relation discrete, continuous, or neither?
2. Find the domain and range for each of the three relations.
Use the graph below for #5 and #6.
1. Is the relation discrete, continuous, or neither?
2. Find the domain and range for the relation.
Use the graph below for #7 and #8.
1. Is the relation discrete, continuous, or neither?
2. Find the domain and range for the relation.
Examine the following pattern.
Number of Cubes (n)\begin{align*}(n)\end{align*} 1 2 3 4 5 ... n\begin{align*}n\end{align*} ... 200
Number of visible faces (f)\begin{align*}(f)\end{align*} 6 10 14
1. Complete the table below the pattern.
2. Is the relation discrete, continuous, or neither?
3. Write a suitable domain and range for the pattern.
Examine the following pattern.
Number of triangles (n)\begin{align*}(n)\end{align*} 1 2 3 4 5 ... n\begin{align*}n\end{align*} ... 100
Number of toothpicks (t)\begin{align*}(t)\end{align*}
1. Complete the table below the pattern.
2. Is the relation discrete, continuous, or neither?
3. Write a suitable domain and range for the pattern.
Examine the following pattern.
Pattern Number (n)\begin{align*}(n)\end{align*} 1 2 3 4 5 ... n\begin{align*}n\end{align*} ... 100
Number of dots (d)\begin{align*}(d)\end{align*}
1. Complete the table below the pattern.
2. Is the relation discrete, continuous, or neither?
3. Write a suitable domain and range for the pattern.
### Vocabulary Language: English
Continuous
Continuous
Continuity for a point exists when the left and right sided limits match the function evaluated at that point. For a function to be continuous, the function must be continuous at every single point in an unbroken domain.
Coordinates
Coordinates
The coordinates of a point represent the point's location on the Cartesian plane. Coordinates are written in ordered pairs: $(x, y)$.
dependent variable
dependent variable
The dependent variable is the output variable in an equation or function, commonly represented by $y$ or $f(x)$.
Discrete
Discrete
A relation is said to be discrete if there are a finite number of data points on its graph. Graphs of discrete relations appear as dots.
domain
domain
The domain of a function is the set of $x$-values for which the function is defined.
Formula
Formula
A formula is a type of equation that shows the relationship between different variables.
Function
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
independent variable
independent variable
The independent variable is the input variable in an equation or function, commonly represented by $x$.
Integer
Integer
The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
Range
Range
The range of a function is the set of $y$ values for which the function is defined.
Real Number
Real Number
A real number is a number that can be plotted on a number line. Real numbers include all rational and irrational numbers. |
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2002 AMC 8 Problems/Problem 14"
## Problem 14
A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices and claims that the final price of these items is $50\%$ off the original price. The total discount is
$\text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$
## Solution #1
Let's assume that each item is $$100$. First we take off $30\%$ off of$100. $$100\cdot0.70=$$70
Next, we take off the extra $20\%$ as asked by the problem. $70\cdot0.80=56$
So the final price of an item is $56. We have to do $100-56$ because $56$ was the final price and we wanted the discount. $100-56=44$ so the final discount was $44\%$ $\text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$ ## Solution #2 Assume the price was$100. We can just do $100\cdot0.7\cdot0.8=56$ and then do $100-56=44$ That is the discount percentage wise.
$\text{(A)}\ 35\%\qquad\boxed{\text{(B)}\ 44\%}\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$ |
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### Linear Algebra With Applications
Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974
# Consider a 4 x 2 matrix A and 2 x 5 matrix B.a. What are the possible dimensions of the kernel of AB?b. What are the possible dimensions of the image of AB?
a. Possible dimensions of the kernel of AB = {5,4,3,2,1}
b. Possible dimensions of the image of AB ={0,1,2,3,4}.
See the step by step solution
## Step 1: Mentioning the concept
For any matrix A of order n x m
Dim (Im(A)) = rank(A).
And
Dim (ker(A))= number of free variables
=total number of variables −number of leading variables
= m − rank(A).
## Step 2: Given quantities
We have given two matrices A of order 4 x 2, and B of order 2 x5
Then the matrix AB is of order 4 x 5.
Now we have to find the possible dimensions of ker(AB) and Im(AB).
## Step 3: (b) Finding the possible dimensions of Im(AB).
Now, for a n x m matrix A, rank (A) ≤ min (m,n).
Then for a matrix AB of order 4 x 5, rank (AB) ≤ min (4,5)
$⇒\mathrm{rank}\left(\mathrm{AB}\right)\le 4\phantom{\rule{0ex}{0ex}}⇒\mathrm{ran}\left(\mathrm{AB}\right)=0\mathrm{or}1\mathrm{or}2\mathrm{or}3\mathrm{or}4$
Thus, possible value of rank(AB)={0,1,2,3,4}.
Since, dim(Im(AB))=rank(AB)
Then possible dimensions of image of AB = {0,1,2,3,4}.
## Step 4: (a) Finding the possible dimensions of ker(AB).
Since the number of unknown variables in an n x m matrix = m then the number of unknown variables in a 4 x 5 matrix = 5.
Now, the dim(ker(A)) = m – rank(AB) = 5 - {0,1,2,3,4}.
$⇒$ Possible values of dim(ker(AB)) = {5,4,3,2,1}
If we have a 4 x 2 matrix A and a 2 x 5 matrix B then .
Then,
c. Possible dimensions of the kernel of AB = {5,4,3,2,1}
d. Possible dimensions of the image of AB ={0,1,2,3,4}. |
The triangle is a closeup of the door polygon that has three angles, three sides, and also three vertices. Based upon the size of sides and measure of angles, the triangles are classified right into different species of triangles.Properties of a triangle assist us to determine a triangle native a given collection of numbers easily and quickly.
You are watching: What is the correct name for the triangle below?
In the beginning, we begin from expertise the form of triangles, your types, and also properties, theorems based upon them such as Pythagoras theorem, etc. Let us learn below some straightforward properties that triangles.
1 Triangles: classification by Type 2 Triangle Properties 3 Solved Examples 4 Practice Questions 5 FAQs
## Triangles: category by Type
Triangles deserve to be classified into two wide categories based on their angles and sides. Based upon angles, there room 3 types of triangles, and also based top top sides, there room 3 varieties of triangles.
## Triangle Properties
Properties of a triangle help us to determine relationships in between different sides and angles of a triangle. Several of the necessary properties the a triangle are noted below.
### Property 1 - Angle sum Property
As per the angle amount property, the amount of the three inner angles of a triangle is constantly 180°.
In the offered triangle, angle p + edge Q + angle R = 180°
### Property 2 - Triangle Inequality Property
As per the triangle inequality property, the sum of the size of the 2 sides that a triangle is higher than the 3rd side.
In △ ABC and △ PQRa + b > c ( 6 + 4 > 3)c + a > b (3 + 4 > 6)c+ b > a ( 6 + 3 > 4)
### Property 3 - Pythagorean Theorem
As per the Pythagorean theorem, in a right triangle, the square of the hypotenuse amounts to the sum of the squares that the other two sides. Mathematically, it have the right to be expressed as Hypotenuse² = Base² + Altitude².
### Property 4 - next opposite the greater angle is the longest side
In order to recognize the next opposite the higher angle is the longest next property, let's take the below-given triangle right into consideration.
In this triangle, edge B is the best angle. Thus, the next AC is the longest side.
### Property 5 - Exterior edge Property
As per the exterior edge property, the exterior angle of a triangle is constantly equal to the amount of the inner opposite angles.
In the offered triangle, Exterior edge E1 = edge PQR + edge QRPThere room 3 exterior angles in a triangle and all this exterior angles include up to 360° for any kind of polygon.
### Property 6 - Congruence Property
As every the Congruence Property, 2 triangles are claimed to be congruent if every their matching sides and also angles room equal.
angle XYZ = angle DEFangle YXZ = angle EDFangle YZX = edge EFDXY = DEXZ = DFYZ = EF
The simple triangle nature such together the area and perimeter the a triangle are given below.
Area of a triangle: The total amount of space inside the triangle is referred to as the area the a triangle. The area is measured in square units.The basic formula for calculating the area of a triangle is Area (A) = (1/2) × base × HeightPerimeter: The perimeter of a triangle = amount of every its three sides
Heron's formula: Heron’s formula is provided to calculation the area that a triangle if the lengths of all the sides are known and the elevation of the triangle is no known. First, we have to calculate the semi-perimeter(s). For a triangle with sides p, q, and r, s = (p+q+r)/2, the area is offered by; A = (sqrtS(S-P)(S-Q)(S-R))
### Properties of a Triangle connected Topics
Check out these interesting articles to know more about nature of a triangle and its connected topics.
Important Notes
The triangle is a close up door polygon that has actually three angles, three sides, and three vertices.Sides and also angles are really important elements of a triangle.We deserve to classify various types of triangle in math by combining sides and also angles.The basic formula for calculating the area the a triangle is Area (A) = (1/2) × basic × HeightThe perimeter that a triangle is same to the amount of all 3 sides the the triangle.
Example 3:A triangle has actually a dimension of 3 cm, 4 cm, and 5 cm, wherein the basic is 4 cm and also the altitude of the triangle is 3.2 cm. Calculatethe area and also perimeter of the triangle.
Solution:
Sides of the triangleare: x= 3 cm, y= 4 cm and also z= 5 cmAltitude= 3.2 cm
Using the area of the triangle formula, we know, Area = 1/2 × basic × heightA = (1/2) × 4 × 3.2A = 6.4 sq.cm.
See more: Dragonvale: How To Breed A Panlong Dragon In Dragonvale ? Panlong Dragon Breeding Tips
The perimeter the the triangle is provided by ns = x+ y+ zP = 3 + 4 + 5P = 12 cmTherefore, the area and perimeter the the offered triangle room 6.4 cm2 and also 12 centimeter respectively. |
# One number is 2 more than 3 times another. Their sum is 22. Find the numbers.
• 8, 14
• 5, 17
• 2, 20
• 4, 18
• 10, 12
The aim of the question is to find the value of x and y by solving the given Simultaneous Equations.
The basic concept behind the article is the Solution of Simultaneous Equations.
Simultaneous Equations are defined as a system of equations containing two or more algebraic equations having the same variables which are related to each other through an equal number of equations. These equations are solved simultaneously for each variable; hence they are called Simultaneous Equations.
If we want to solve the given set of two algebraic equations, we must find an ordered pair of numbers, which when substituted in the given equations, satisfies both algebraic equations.
Simultaneous equations are generally represented as given below:
$ax+by = c$
$dx+ey = f$
Where,
$x$ and $y$ are two variables.
$a$, $b$, $c$, $d$, $e$ and $f$ are constant factors.
Given that:
Let the first variable is represented by $x$ and the second variable is represented by $y$. The two simultaneous equations based on the relations in the given article will be:
The First expression of the Simultaneous Equation is:
The Second variable is $2$ more than $3$ times the First variable.
$y\ =\ 2+3x$
The Second expression of the Simultaneous Equation is:
The sum of both variables is $22$
$x+y\ =\ 22$
By substituting the value of $y\ =\ 2+3x$ from First expression into Second expression, we get
$x+(2+3x)\ =\ 22$
$4x+2\ =\ 22$
$4x\ =\ 22-2$
$4x\ =\ 20$
Solving for $x$:
$x\ =\ \frac{20}{4}\ =\ 5$
Hence, the value of variable $x$ is $5$.
Now, we will substitute the value of $x=5$ into the First expression to calculate the value of variable $y$
$y\ =\ 2+3x$
$y\ =\ 2+3(5)\ =\ 2+15$
$y\ =\ 17$
Hence, the value of variable $y$ is $17$.
## Numerical Result
The numbers corresponding to variables $x$ and $y$ for the given set of simultaneous equations are
$x\ =\ 5\ and\ y\ =\ 17$
## Example
Find the value of variables $x$ and $y$ for the following set of Simultaneous Equations.
$2x+3y\ =\ 8$
$3x+2y\ =\ 7$
Solution
Given that:
The First expression of Simultaneous Equations is:
$2x+3y\ =\ 8$
Solving for $x$
$2x\ =\ 8-3y$
$x\ =\ \frac{8-3y}{2}$
The Second expression of Simultaneous Equations is:
$3x+2y\ =\ 7$
Substituting the value of variable $x$ in second expression:
$3\left(\frac{8-3y}{2}\right)+2y\ =\ 7$
$\left(\frac{24-9y}{2}\right)+2y\ =\ 7$
$\frac{24-9y+4y}{2}\ =\ 7$
$\frac{24-9y+4y}{2}\ =\ 7$
$24-9y+4y\ =\ 14$
$9y-4y\ =\ 24-14$
$5y\ =\ 10$
$y\ =\ 2$
Now, substituting the value of variable $y$ in the expressions for $x$, we get:
$x\ =\ \frac{8-3y}{2}$
$x\ =\ \frac{8-3(2)}{2}$
$x\ =\ \frac{2}{2}$
$x\ =\ 1$
The numbers corresponding to variables $x$ and $y$ for the given set of Simultaneous Equations are:
$x\ =\ 1\ and\ y\ =\ 2$ |
# The General Polar Equation of Conic Sections - Part 2
## Introduction
We looked at the general polar equation of conics that are rotated at any angle in The General Polar Equation of Conic Sections - Part 1. The focus was at the origin and the equation turned out to be quite simple. Now, we will look at the polar equation of a conic where the directrix passes through the origin.
## Deriving the General Equation
The steps to determine this equation in brief are to equate the distance from the focus to a point on the parabola to the ratio distance of the same point to the directrix. Then, we will isolate r to solve for r in terms of θ.
We will put the focus at the polar coordinate $$\left(a, \, \phi + \frac{\pi}{2} \right)$$, where a is the distance of the focus from the pole. And the directrix will pass through the pole creating an angle φ with the polar axis.
First, notice that the distance PD can be given by the relationship $$d = r\sin(\theta - \phi)$$ because of the right triangle PDO.
Because the focal segment OF is perpendicular to the directrix, the angle OF creates with the positive x-axis is $$\phi + \frac{\pi}{2}$$. Therefore, the angle POF is equal to $$\phi + \frac{\pi}{2} - \theta$$. Using the Law of Cosines, the distance FP is equal to $$(ed)^2 = a^2 + r^2 - 2ar\cos(\phi + \frac{\pi}{2} - \theta)$$. Now, we can substitute d from one to the other and follow the steps to isolate r.
(i) $$(er\sin(\theta - \phi))^2 = a^2 + r^2 - 2ar\cos(\phi + \frac{\pi}{2} - \theta)$$
Let’s simplify $$\cos(\phi + \frac{\pi}{2} - \theta)$$ using the sum identity of cosine: $$\cos(\phi + \frac{\pi}{2} - \theta) =$$ $$\cos(\phi - \theta)\cos(\frac{\pi}{2}) - \sin(\phi - \theta)\sin(\frac{\pi}{2}) =$$ $$-\sin(\phi - \theta) = \sin(\theta - \phi)$$.
(ii) $$(e^2\sin^{2}(\theta - \phi))r^2 = a^2 + r^2 - 2ar\sin(\theta - \phi)$$
(iii) $$(1 - e^2\sin^{2}(\theta - \phi))r^2 - 2a\sin(\theta - \phi)r + a^2 = 0$$
Equation (iii) is a quadratic equation we can solve easily with the quadratic formula.
(iv) $$r = \dfrac{2a\sin(\theta - \phi) \pm \sqrt{4a^2\sin^{2}(\theta - \phi) - 4a^2(1-e^2\sin^{2}(\theta-\phi))}}{2(1-e^2\sin^{2}(\theta-\phi))}$$
(v) $$r = \dfrac{2a\sin(\theta - \phi) \pm 2a\sqrt{(1+e^2)\sin^{2}(\theta-\phi)-1}}{2(1-e^2\sin^{2}(\theta-\phi))}$$
(vi) $$r = \dfrac{a\sin(\theta - \phi) \pm a\sqrt{(1+e^2)\sin^{2}(\theta-\phi)-1}}{1-e^2\sin^{2}(\theta-\phi)}$$
The polar equation of a conic that has the focus located at the polar coordinate $$\left(a, \, \phi + \frac{\pi}{2} \right)$$ and the directrix creates an angle φ with the polar axis is given by:
$r(\theta) = \dfrac{a\sin(\theta - \phi) \pm a\sqrt{(1+e^2)\sin^{2}(\theta-\phi)-1}}{1 - e^2\sin^{2}(\theta - \phi)}$
### The Parabola
In equation (vi), if we let e = 1, then we get a simplified equation of a parabola.
The polar equation of a parabola that has the focus located at the polar coordinate $$\left(a, \, \phi + \frac{\pi}{2}\right)$$ and the directrix creates the angle φ with the polar axis is given by:
$r(\theta) = \dfrac{a\sin(\theta - \phi) \pm a\sqrt{2\sin^{2}(\theta-\phi)-1}}{\cos^{2}(\theta - \phi)}$
where a is the distance of the focus from the pole.
The rectangular position of the focus is $$\left( -a\sin\phi,\, a\cos\phi \right)$$ and the equation of the directrix is $$y = (\tan\phi)x$$.
We used the sine version of the cosine double-angle but we can also choose the cosine version. It would have been more concise to use $$-\cos(2\theta - 2\phi)$$ but the negative sign seems out of place in the radical.
The equations are not as consice as the general equation for conics, but it does give a bit more leverage for choosing our focus point and the directrix.
From initial graphing tests, the plus/minus sign does not really change anything. However, there are gaps in the domains because of the radical, which probably lead to gaps in the range. When graphing $$f(\theta) = \sqrt{-\cos(2\theta - 2\phi)}$$ as a rectangular function, there are gaps which are not filled by graphing both the plus and minus portions. However, Geogebra graphs the equation with a complete parabola without a hiccup. The only problem is a point on the parabola being tracked disappears at some intervals - probably at angles where the radical is undefined.
## Other Conics
The hyperbola in Figure 2 shows the value of e holds up. The equation graphed is $$r(\theta) = \frac{\frac{1}{2}\sin(\theta - \frac{\pi}{3}) + \frac{1}{2}\sqrt{1 + \frac{9}{4}\sin^{2}(\theta-\frac{\pi}{3})}}{1 - \frac{9}{4}\sin^{2}(\theta - \frac{\pi}{3})}$$ for the focal distance a = 1/2, the eccentricity of e = 3/2, and the directrix equation $$y = (\tan\frac{\pi}{3})x$$ or $$y = \sqrt{3}x$$.
The ellipse in Figure 3 shows the value of e holds up for an ellipse also. The equation graphed is $$r(\theta) = \frac{\frac{1}{2}\sin(\theta - \frac{2\pi}{3}) + \frac{1}{2}\sqrt{1 + \frac{25}{36}\sin^{2}(\theta-\frac{2\pi}{3})}}{1 - \frac{25}{36}\sin^{2}(\theta - \frac{2\pi}{3})}$$ for the focal distance a = 1/2, the eccentricity of e = 5/6, and the directrix equation $$y = (\tan\frac{2\pi}{3})x$$ or $$y = -\sqrt{3}x$$. |
# Difference between revisions of "2016 AMC 10B Problems/Problem 11"
## Problem
Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$
## Solution 1
If the dimensions are $a\times b$, then one side will have $a+1$ posts (including corners) and the other $b+1$.
The total number of posts is $2(a+b)=20$.
$[asy]size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen); for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7)); draw(arc((0,0),.5,-90,-270)--arc((4,0),.5,90,-90)--cycle,gray+dotted); draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted); draw((0,-1)--(5,-1),Arrows);draw((6,0)--(6,7),Arrows); label("a",(0,-1)--(5,-1),S);label("b",(6,0)--(6,7),E); label("a",(1,1));label("b",(4,5)); [/asy]$
Solve the system $\begin{cases}b+1=2(a+1)\\a+b=10\end{cases}$ to get $a=3,\ b=7$. Then the area is $4a\cdot 4b=336$ which is $\mathbf{(B)}$.
## Solution 2
To do this problem, we have to draw a rectangle. We also have to keep track of the fence posts. Putting a post on each corner leaves us with only $16$ posts. Now there are twice as many posts on the longer side then the shorter side. From this we can see that we can put $8$ posts on the longer side and $4$ posts on the shorter side. On the shorter side, we have $3$ spaces between the $4$ posts. On the longer side, we have 7 spaces between the 8 fence posts. There are $4$ yards between each post. Therefore, the answer is $12*28=336$
$\textbf{(B)}\ 336$.
~savannahsolver
## See Also
2016 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Derivative of a Function
Let $y = f(x)$ be a given function of $x$. Give to $x$ a small increment $\Delta x$ and let the corresponding increment of $y$ by $\Delta y$, so that when $x$ becomes $x + \Delta x$, then $y$ becomes $y + \Delta y$ and we have:
$y + \Delta y = f(x + \Delta x)$
$\therefore \Delta y = f(x + \Delta x) - f(x)$
Dividing both sides by $\Delta x$, then
$\frac{{\Delta y}}{{\Delta x}} = \frac{{f(x + \Delta x) - f(x)}}{{\Delta x}}$
Taking limit of both sides as $\Delta x \to 0$
$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}}\mathop { = \lim }\limits_{\Delta x \to 0} \frac{{f(x + \Delta x) - f(x)}}{{\Delta x}}$
Thus, if $y$ is the function of $x$, then $\mathop {\lim }\limits_{\Delta x \to 0} \frac{{f(x + \Delta x) - f(x)}}{{\Delta x}}$ is called the derivative or the differential coefficient of the function or the derivative of $f(x)$ with respect to $x$ and is denoted by $f'(x)$, $y'$, $Dy$ or $\frac{{dy}}{{dx}}$.
Note: It may be noted that the derivative of the function $f(x)$ with respect to the variable $x$ is the function $f'$ whose value at $x$ is $f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ provided the limit exists, is called the derivative where $h$ is the increment.
Below we list the notations for derivative of $y = f(x)$ used by different mathematicians in a table.
Name of Mathematician Leibniz Newton Lagrange Cauchy Notation for Derivative $\frac{{dy}}{{dx}}$or$\frac{{df}}{{dx}}$ $y'$ $f'(x)$ $Df(x)$
Remarks:
I. The student should observe the difference between $\frac{{\Delta y}}{{\Delta x}}$ and $\frac{{dy}}{{dx}}$. Where $\frac{{\Delta y}}{{\Delta x}}$ is the quotient of the increment of $y$ and $x$ that is its numerator and denominator can be separated, but $\frac{{dy}}{{dx}}$ is a single symbol for the limiting value of the fraction $\frac{{\Delta y}}{{\Delta x}}$ when $\Delta y$ and $\Delta x$ can be separated.
II. $\frac{d}{{dx}}$ attached to any function meaning its differential coefficient with respect to $x$.
III. The phrase “with respect to” will often be abbreviated into w.r.t
Differentiation:
The process of finding the differential coefficient of a function or a process for finding the rate at which one variable quantity changes with respect to another is called differentiation.
Four Steps in Differentiation:
In the given function
$y = f(x)$
1. Change $x$ to $x + \Delta x$ and $y$ to $y + \Delta y$
i.e. $y + \Delta y = f(x + \Delta x)$
2. Find $\Delta y$ by subtraction
i.e. $\Delta y = f(x + \Delta x) - y$
$\Delta y = f(x + \Delta x) - f(x)$
3. Divide both sides by $\Delta x$
i.e. $\frac{{\Delta y}}{{\Delta x}} = \frac{{f(x + \Delta x) - f(x)}}{{\Delta x}}$
4. Find the limit of $\frac{{\Delta y}}{{\Delta x}}$ where $\Delta x \to 0$
i.e. $\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}}\mathop { = \lim }\limits_{\Delta x \to 0} \frac{{f(x + \Delta x) - f(x)}}{{\Delta x}}$
Employing the above mentioned four steps in determining the derivative means finding the differential coefficient “by definition” or “by first principle” or “by ab-initio method”. |
# The sum of the squares
### Introduction
In this article, we will look at several proofs for a famous formula.
#### Formula: Sum of the squares
$$\sum_{r=1}^{n}r^2=\frac{n}{6}(n+1)(2n+1)$$
### Mathematical induction
We begin with the base case, $$n=1$$. The formula then becomes
$$1 = \frac{1}{6} \times 2 \times 3$$
which the reader can verify! Now for the induction step, we assume that the formula is true when $$n = k$$, and we need to prove that the formula is therefore true for $$n = k + 1$$. So, we assume that
$$\sum_{r=1}^{k}r^2=\frac{k}{6}(k+1)(2k+1)$$
is true. The formula for $$n = k + 1$$ then becomes
\begin{align*}
\sum_{r=1}^{k+1}r^2 & = (k+1)^{2} + \sum_{r=1}^{k}r^2 \\
& = (k+1)^{2}+ \frac{k}{6}(k+1)(2k+1)\\
& = \frac{1}{6}(k+1)[6(k+1)+k(2k+1)]\\
& = \frac{k+1}{6}[6k+6+2k^{2}+k]\\
& = \frac{k+1}{6}(2k^{2}+7k+6)\\
& =\frac{k+1}{6}(2k+3)(k+2)
\end{align*}
According to Mathematical induction, the formula holds for all value $$n\in\mathbb{Z}$$
### An algebraic proof
This proof relies on two facts:
#### Formulas: The sum of two squares
Firstly,
$$a^2+b^2=a(a+b)-b(a-b)$$
Also,
$$n(n+1) = \frac{1}{3}[n(n+1) (n+2)-(n-1) n (n+1)]$$
Using the above two formulae (which the reader may check),
\begin{align*}
S_{n} = & \sum_{r=1}^{n}r^2 \\
= & 1^2+2^2+3^2+4^2+\ldots+n^2 \\
= & 1\times 1 + 2\times 2 + 3\times 3 + 4\times 4 +…+ n\times n\\
= & 1(2-1) + 2(3-1) + \ldots + n(n+1-1) \\
= & 1\times 2 + 2\times 3 + \ldots + n\times (n+1)-\frac{n(n+1)}{2}\\
= & \frac{1}{3} [1\times 2\times 3 – 0\times 1\times 2] + \frac{1}{3} [2\times 3\times 4 -1\times 2\times 3]+\ldots \\ & + \frac{1}{3} [n (n+1) (n+2) – (n-1) n(n+1)] + \frac{n(n+1)}{2}\\
= & \frac{n(n+1)(2n+1)}{6}\\
\end{align*}
as required.
### Extending Gauss’ method
First, we should remember that
$$(n+1)^3=n^3+3n^2+3n+1$$
$$1+2+3+\ldots+n= \frac{n(n+1)}{2}$$
Therefore, $$n^3-(n-1)^3 = 3(n-1)^2 + 3(n-1) + 1$$
and
$$2^3 – 1^3 = 3\times 1^2 + 3\times 1 +1, \qquad 3^3 – 2^3 = 3\times 2^2 +3\times 2 + 1, \ldots$$
and so on. This gives us
\begin{align*}
(n+1)^3 -1 = & 3( 1^2+ 2^2+ 3^2 + \ldots + n^2) + 3\times ( 1+ 2+ 3+ \ldots +n) +n \\
n^3 +3n^2 + 3n + 1 = & 3( 1^2 + 2^2+ 3^2+ \ldots + n^2) + 3\times \frac{n(n+1)}{2}+n\\
1^2 +2^2 + 3^2 +\ldots + n^2 = & \frac{n(n+1)(2n+1)}{6}\\
\end{align*} |
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Circular Inversion – Reflecting in a Circle
This topic is a great introduction to the idea of mapping – where one point is mapped to another. This is a really useful geometrical tool as it allows complex shapes to be transformed into isomorphic (equivalent) shapes which can sometimes be easier to understand and work with mathematically.
One example of a mapping is a circular inversion. The inversion rule maps a point P onto a point P’ according to the rule:
OP x OP’ = r2
To understand this, we start with a circle radius r centred on O. The inversion therefore means that the distance from O to P multiplied by the distance from O to P’ will always give the constant value r2
This is an example of the circular inversion of the point A to the point A’.
We have the distance of OA as √2 and the radius of the circle as 2. Therefore using the formula we can find OA’ by:
OA’ = r2 / OA = 4/√2
OA’ = 21.5. This means that the point A’ is a distance of 21.5 away from O on the same line as OA.
We can check that the Geogebra plot is correct – because this point A’ is plotted at (2,2) – which is indeed (using Pythagoras) a distance of 21.5 from O.
A point near to the edge of the circle will have an inversion also close to the circle
A point near to the centre will have an inversion a long way from the circle. The point (0,0) will be undefined as no point outside the circle will satisfy the inversion equation.
So, that is the basic idea behind circular inversion – though it gets a lot more interesting when we start inverting shapes rather than just points.
Circles through the origin map onto straight lines to infinity (see above).
Circles centred on the origin map to other circles centred on the origin (above).
Ellipses create these shapes (above).
The straight line through A B maps to a circle through the origin (above).
The solid triangle ABC maps to the pink region (above).
The solid square ABCD maps to the pink region (above).
These shapes can all be explored using the reflect object in circle button on Geogebra.
It is possible to extend the formula to 3 dimensions to give spherical inversion:
The above image is a 3D human head inverted in a sphere (from the Space Symmetry Structures website. There’s lots to explore on this topic – it’s a good example of how art can be mathematically generated, as well as introducing isomorphic structures.
If you enjoyed this post you might also like:
The Riemann Sphere : Another form of mapping using spheres is the Riemann Sphere – which is a way of mapping the entire complex plane to the surface of a sphere.
Fractals, Mandelbrot and the Koch Snowflake. Using maths to model infinite patterns.
Essential resources for IB students:
Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.
There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!
The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.
Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.
I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
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# Normal Distribution
## From discrete to continuous
The probability that a person picked at random has a height between 150 and 160 cm can be calculated from the table:
$p=\frac{41}{15+41+67+72+65+34+3}$
You could also find the probability by looking at the areas of the bars in bar chart. The probability is then found by dividing the area of the red bar by the total area.
$p=\frac{41\cdot 10}{15\cdot 10+41\cdot 10+67\cdot 10+72\cdot 10+65\cdot 10+34\cdot 10+3\cdot 10}$
### Probability and area
One way of illustrating the probability that an outcome is in a certain interval, is to let the area of the interval be the probability. The total area must in this case be one. A new bar chart having the same appearance but the total area of one can be constructed.
#### Exercise 1
You can add a number of values in cells by writing:
Sum(B2:B8)
If you make a bar chart using the relative frequency instead of the absolute frequency, the total area will still not be one. Add another column where you normalize the relative frequencies in order to get the total area one. How do you construct this column? The column for the data should not be changed. Draw the bar chart!
#### Exercise 2
You can approximate a discrete distribution of data values with a continuous function. One such approximating function is the normal probability distribution function (normal pdf). The graph of the normal pdf is a completely symmetric bell-shaped curve whose appearance depends only on two values; the mean and the standard deviation.
If $$\mu$$ is the mean value and $$\sigma$$ the standard deviation, the normal pdf is:
$f(x)=\frac{1}{\sigma \sqrt{2\pi}}e^{\left( - \dfrac{(x-\mu)^2}{2\sigma^2}\right)}$
Use the mean 173.2492 and the standard deviation 14.0333 to plot the graph of the normal distribution.
## Standard normal distribution
The pdf for the standard normal distribution $$\phi$$, has mean $$\mu =0$$ and standard deviation $$\sigma=1$$.
$\phi (x)=\frac{1}{\sqrt{2\pi }} e^{-\frac{1}{2}x^2}$
A short form for writing that a random variable $$X$$ is distributed normally with mean $$\mu$$ and standard deviation $$\sigma$$, is to write: $$X\ \sim \mathcal{N}(\mu,\sigma^2)$$. A variable that is distributed by the standard normal distribution is denoted $$Z$$. The correspondence between $$Z$$ and $$X$$ is
$Z=\frac{X-\mu}{\sigma}$
The probability is given by the area under the curve. The total area is therefore always one. If the random variable is denoted $$X$$, then the probability $$P$$ is given by
$P(a\lt X \lt b)=\int_a^bf(x)dx$
where $$f(x)$$ is the normal pdf.
## Normal distribution in GeoGebra
By using the tool Probability Calculator, you can find the probability between boundaries for a number of distribution functions. The tool is found in the menu belonging to the spreadsheet.
## The inverse normal problem
If you know the mean and the standard deviation, you can find the probability between given boundaries by using the "Probability Calculator".
The inverse problem is that you know the probability for given boundaries and want to find the mean and the standard deviation. Since there are two unknowns to be found, you must know two probabilities. The problem is solved by using the correspondence with the standard normal pdf.
Suppose that you know that $$P(X\lt 4)=0.4$$ and $$P(X\lt 3)=0.2$$. The boundaries 4 and 3, can be compared to the corresponding boundaries for the standard normal pdf. The correspondence between these boundaries is $$z=\frac{x-\mu}{\sigma}$$, where $$z$$ is the boundary for the standard normal pdf, and $$x$$ is the boundary for the unknown normal pdf.
To find the boundaries corresponding to 4 and 3, you can use GeoGebra. You must find $$z_1$$ from $$P(Z\lt z_1)=0.4$$ and $$z_2$$ from $$P(Z\lt z_2)=0.2$$.
Input 0.4 in the box shown in the picture below and press Enter. You get $$z_1=-0.2533$$.
In a similar way you get $$z_2=-0.8416$$. To find $$\mu$$ and $$\sigma$$, solve the simultaneous equations:
\left\{ \begin{align*} -0.2533&=&\frac{4-\mu}{\sigma} \\ -0.8416&=&\frac{3-\mu}{\sigma} \end{align*} \right.
A graphical solution is shown below.
The solution is $$\mu=4.4306$$ and $$\sigma=1.6998$$. You can check the solution by using the "Probability Calculator". |
The Traffic Accident Reconstruction Origin -Article-
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## Sensitivity to Uncertainty
by
Dr. Oren Masory and Bill Wright
Introduction
It is common practice for accident reconstructionists to rely on equations to guide them to some numerical answer. Each of the measured or estimated values that drive the equation will carry some degree of uncertainty. If we are measuring drag factor we may be confident of a value to within +-. 05. If we measure a distance we may be accurate to within +-1 inch. Regardless of how carefully we measure, or how accurate the measuring device, there will always be some level of uncertainty. A change in the value of any of the measured quantities will have some effect on the calculated value. This difference is the "sensitivity" of the equation to the uncertainty. The propagation of this uncertainty to the calculated result should always be considered when expressing the calculated answer. This discussion will center on several methods to quantify this sensitivity to uncertainty.
Sensitivity, An Example Problem
Suppose we are given the task to measure a vehicle's acceleration rate from 0 to 88 ft/sec (60 m/h or 95 km/h). After some thought the following test set up is chosen. A flat vacant roadway is located. The driver points the car down the roadway with a stopwatch in one hand. When ready he floorboards the accelerator and starts the stopwatch in the same instant. The driver watches the speedometer climb 40, 50, 55, 58, 59,60. Just as the needle passes 60 m/h he clicks the stopwatch a second time. He now has the time necessary to accelerate to a known velocity, say in this case 8 seconds. This is sufficient information to calculate the average acceleration rate. We use the relationship:
(1)
Substituting the measured values of 88 ft/sec and 8 seconds into equation 1:
(2)
We calculate an answer of 11 ft/sec2 and return to report this acceleration rate. We were grilled about our test method. The question asked is, what if you made a mistake in reading the speedometer? Stated another way we have been asked to assess how sensitive this equation is to uncertainty in velocity. The difference in speed may make a large difference in acceleration or a small difference. Just looking at the numbers there is no way to know. To answer this question we choose to calculate several different values of acceleration using several different speeds.
First we must choose a range to "bracket". We choose to calculate the change in acceleration for increments 4 ft/sec around our time of 88 ft/second. 4 ft/sec is roughly equivalent to 2 M/h or 4km/h. This increment was chosen because it is a fair estimate for the uncertainty in reading the speedometer. Given this increment, we will calculate a table of acceleration values by holding the measured 8 seconds as constant, and varying values of velocity by values of 4 ft/sec.
That table Follows.
Velocity ft/sec Time sec acceleration ft/sec2 56 8 7.00 60 8 7.50 64 8 8.00 68 8 8.50 72 8 9.00 76 8 9.50 80 8 10.00 84 8 10.50 88 8 11.00 92 8 11.50 96 8 12.00 100 8 12.50 104 8 13.00 108 8 13.50 112 8 14.00 116 8 14.50 120 8 15.00
Table 1
The table tells us that for every change of 4 ft/sec the acceleration changes by .5 ft/sec2. Dividing by 4 will give the change in acceleration for every 1ft/sec. This result is .125 ft/sec2. It appears that a small error in velocity will result in a small error in the resulting calculation for acceleration. This is encouraging. The table has been helpful. But we could use the data in the table differently. We could plot the points and view them graphically.
Fig 1
Fig 1 shows the table data plotted. Here acceleration is plotted as a function of velocity. This graph is a "picture" of the relationship expressed in eq. 1 where acceleration is a function of velocity. We can see that it plots as a line. The range of 88 ft/sec +- 4 ft/sec has been highlighted. This is the proposed range of uncertainty. The lines that form this range are velocity values projected vertically up onto the function (the blue line), then mapped perpendicularly onto the acceleration axis. We can see that 84 ft/sec maps to 10.5 ft/sec2 and 92 ft/sec maps to 11.5 ft.sec2
Looking more closely at Fig 1 we see the function that is sloped upward and to the right. When talking about graphs, the word slope has a very specific meaning. It is defined as rise/run, or the change in the y value divided by the change in the x value (where y=f(x)). Slope is usually assigned the variable m. We can use the data from the table to calculate the slope of this function.
For example we choose to find the slope using the two points when V = 84 where a = 10.5 and when V= 88 and a = 11
(3) m = change in y / change in x
Substituting and solving we find
(4)
The units of slope in this problem are worth examining. Slope is always rise/run. In this case those units are acceleration/velocity, or (ft/sec2) / (ft/sec). This fraction could simplify but it is meaningful in its current form. What this says is that there is a .125 ft/sec2 change in acceleration for every change in velocity of 1 ft/sec. Notice this is exactly what we determined from looking at the table.
The topic of slope is worth a few more words. Negative slopes indicate lines that decrease in y values as x values increase (go down and to the right). The larger the absolute value of the slope (big positive numbers or small negative numbers) the more steeply the line is angled to the horizontal. Conversely, the smaller the slope (closer its value is to 0) the more horizontal the line. Notice also, since this function is a straight line (has constant slope) we could have calculated the value of m =.125 for the function's slope by using any two points from the table.
We can use the definition of slope in another way. From (3) above slope is defined as
(3) m = change in y / change in x
We can rearrange and write
(5) m (change in x) = change in y
Now recognize that change in x is the uncertainty in the measured value. Change in y is the propagation of this uncertainty, mapped across the function, to the calculated value. This mapping of values across the function is the uncertainty in the calculated value. We can choose a convenient unit value for the uncertainty in x, in this case 1 ft/sec. Once this is done the amount of uncertainty in y is the result of the functions manipulation of x, or its sensitivity. Equation (5) above says that sensitivity is proportional to slope.
Fig 2a Large Slope, Large Sensitivity Fig 2b Moderate Slope, Moderate Sensitivity Fig 2c Small Slope, Small Sensitivity
This relationship can also be seen graphically in Fig 2a, 2b and 2c. These figures show a sketch of three hypothetical functions (red lines) with different slopes. First notice the uncertainty in the measured value (x) is the same in all three figures. It is apparent from the graphs that uncertainty maps across these functions very differently. Functions with large slopes map to large uncertainties in the calculated value. Functions with small slopes map to small uncertainties. This is just the graphic representation of equation (5). Again, sensitivity is proportional to the slope of the function.
Confident that our test method is valid we return to report that the function (a = V/t) is not particularly sensitive to errors in velocity measurement. The question fires back, what if you made an error in the measurement of time? This is a different question than we have just answered. This question asks for the sensitivity of the function to uncertainty in time. Let's follow the same method used for checking the sensitivity to velocity.
As before, first we create a table. This time we calculate acceleration holding velocity constant and varying time. For this table we will check the acceleration value every half-second. That table follows
Velocity ft/sec Time sec acceleration ft/sec2 88 4 22.00 88 4.5 19.56 88 5 17.60 88 5.5 16.00 88 6 14.67 88 6.5 13.54 88 7 12.57 88 7.5 11.73 88 8 11.00 88 8.5 10.35 88 9 9.78 88 9.5 9.26 88 10 8.80 88 10.5 8.38 88 11 8.00 88 11.5 7.65 88 12 7.33
Table 2
Examining the data we see that acceleration changes by 4.4 ft/sec2 (.14g) in the first second (from t = 4 to t = 5). This is a significant error. Something else is troubling about the data in this table. While the change in acceleration in the first second is 4.4 ft/sec2 the change in the last second (t = 11 to t = 12) is only .67 ft/sec2 (.02g). The meaning of this difference is clear when the data is plotted. This function plots as a curve rather than a line.
Fig 3
How are we to apply the relationship of sensitivity discussed above to this figure? Stated another way, how do we find the slope of a curve?
The definition of slope is unchanged:
(3) m = change in y / change in x
Using the data from the table, the calculated slope of the first second (t = 4 to t = 5) is
(6)
We also need to find the value of slope around our point of interest where time equals 8 seconds. We will use the data for the second that begins a half second before our time of t = 8 seconds and ends at the half second after t = 8 seconds (t = 7.5 to t = 8.5). That slope is
(7)
Looking further down the curve we see the last second of data (t = 11 to t = 12) yields a slope of
(8)
Examining these numbers we see that the magnitudes of the slopes seem to agree with the graph. That is, the absolute value of the slopes become smaller and smaller as we move to the right (the curve is becoming more horizontal). The negative slopes also make sense. Everywhere the acceleration values decrease as the time values increase (the curve always goes down and to the right). But what to do about the changing slope? We could graphically draw the ranges as was done in the previous example. These ranges are included in green in the Figure 3. We can see from both the table and the graph that 8 sec +- 1 second yields a range of accelerations of 9.78 to 12.57 ft/sec2. But that is only the range for our 8-second car. If the car we had tested was quicker, it could be in the mid 4 second range. Conversely, if it was a very slow accelerating car it could be well beyond the hypothetical 11-12 second range included in the table. There must be an easier way to find sensitivity than calculating, tabulating and graphing values.
A Different Approach to Slope
In fact there is an easier way. We have seen in the previous examples that sensitivity is proportional to slope. What is needed is a way to find he slope of a function at a particular point. This is precisely what a derivative does.
Books are written about derivatives and how to find them. It is calculus. The topic is beyond the scope of this treatment. Suffice it to say:
1. Derivatives are based on a revolutionary concept, the idea of a limit. We routinely say things like- the impact speed was 47 m/h. Have you ever thought just how an object can have an instantaneous speed? Speed is always measured in units of some-distance/some-time. If an instant of time has a length of 0, does the vehicle move any distance in that instant? We have some paradoxical possibilities:
2. a) If it doesn't move in our time increment of 0 seconds, why aren't all instantaneous speeds equal to 0?
b) If it moves any distance at all in our 0 length time increment how can it have a speed less than infinity (small number/0 = infinity)?
3. All continuous functions have a derivative.
4. Once the concept is understood you can find a derivative of just about any function by applying one of about two dozen rules and couple of tricks.
5. Derivatives are enormously powerful tools. With a derivative you can sketch curves, find maximums, determine dependencies within a function, or in this case find the slope of a function.
6. The derivative for functions with more than one variable can be found by treating all other variables as constants. This is called a partial derivative. This is how functions will be treated in this paper.
Let's reexamine the problems above using derivatives.
The relationship is
Average acceleration = Change in Velocity / Change in time
The function is:
(1)
One derivative of this function is
(9)
We need more information about this notation. is read as the first partial derivative of acceleration with respect to velocity. Remember we can think of it as the slope of the function (1) at any value of velocity. Note also the units carry with the arguments. Here the units of are ft/sec2 / ft/sec. Again, as we saw above, this derivative will be in the units we desire, change in acceleration per change in velocity. Notice also in the derivative that the slope is not a function of velocity. That means the that slope is constant as seen above. In other words, the fact that we wanted acceleration from 0 to 88 ft/sec made no difference to the tests sensitivity to velocity.
The other derivative is
(10)
Here is read as the first partial derivative of acceleration with respect to time.
Let's take our test setup and use derivatives to determine the function's sensitivity to each variable. First we will examine the test setup's sensitivity to changes in velocity. From above that partial derivative is
(9)
Next we substitute and solve with the value of t (8 seconds)
(11)
We see that this is exactly the same answer we found from the table, the graph, and when we calculated the slope in Eq. 4.
Next the sensitivity to time. That partial derivative is
(10)
Substitute our values for t and V (8 and 88 respectively) to find the slope for the function at our test values
(12)
The value calculated from the derivative of -1.375 is very close to the slope we calculated from data in the table (Eq 8) where we found the slope to be -1.38. In fact the slope obtained from the derivative (Eq. 12) is more accurate that the slope obtained from the data. This is due to the fact that slope of the curve changes constantly, even in the small increment of time from 7.5 to 8.5 seconds. Notice also, unlike the speed case for acceleration (Eq. 9 where slope (the derivative) is dependent on time alone), the time sensitivity case is dependent on both time and velocity.
Checking the value of the two derivatives we see the magnitude of is greater than the magnitude of . Stated another way, where is the absolute value of -1.375. Therefore, we know that the function, and our test, is most sensitive to unit errors of time.
Applying the Derivatives the Problem at Hand
Earlier we had estimated the accuracy of the time and velocity measurements. We said that velocity was 88 ft/sec +- 4 ft/sec. We also said that our time measurement was 8 sec +- 1 sec. Remember the +- values are the uncertainty in the measured values. We can use the value of the derivatives calculated above (Eqs 11 and 12) to calculate the total uncertainty in the test.
Eq 11 tells us that there is .125ft/sec2 for every 1ft/sec of error. Finding the error for 4 ft/sec is just a matter of multiplying the unit value of the derivative (Eq 11) times the uncertainty.
(13)
Fortunately we have guessed the uncertainty in time at the unit value of 1 sec.
(14)
Consequently, the value calculated in eq 12 ( ) will be our acceleration uncertainty.
The total uncertainty in the test is the sum of uncertainty in acceleration due to uncertainty in velocity (.5ft/sec2) plus the uncertainty in acceleration due to uncertainty in time (1.375ft/sec2).
(15)
Finally, we can now report completely that the acceleration rate of our car was 11 ft/sec2+- 1.875 ft/sec2 (.34g +-.06g).
Conclusion
Much of common Traffic Accident Reconstruction folklore is explained by circumstances where slope is large, and sensitivity is great.
1. Don't use momentum when the approach angle is less than 10 degrees.
2. Be very sure about the accuracy of your speedometer when running skid tests.
3. When measuring a radius for a critical speed estimate, be sure your middle ordinate measurement is a little long.
4. Don't use momentum when the mass ratios exceed 10 to 1.
All of these cases have function derivatives with large slope at their cautioned value.
The reader is cautioned that using a derivative to calculate the uncertainty is only half of the error analysis story. Remember, in this example we guessed at the uncertainty in the measured value. Statistics gives us the tools to make these estimated values credible. That is a topic for another discussion.
Statistics aside, the sensitivity of a particular function to uncertainty is always proportional to its slope at the point of interest. The partial derivative of the function offers a quick and convenient way to assess the function's sensitivity to uncertainty.
Dr. Oren Masory is a Professor and the Director Robotics Center of the College of Mechanical Engineering Department at Florida Atlantic University in Boca Raton, FL. He can be reached at (561) 297-2693 or via Email at [email protected]
Bill Wright is an accident investigator with the Palm Beach County Sheriffs Office in West Palm Beach, Fl (USA). He teaches accident investigation and driver training topics. He can be reached at [email protected]
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# What is a term to term rule and a position to term rule in math?
A sequence is an ordered set; usually the objects in the set (often numbers) are written with commas between them. e.g. 1,3,5,7 is a finite sequence, 1,3,5,7,... is an infinite sequence as is 2,4,8,16,... or 2,3,5,7,11,13,17,... (the primes.) The elements of the sequence are called terms; since the elements are ordered we can speak of the first term or `a_1` , second term `a_2` and the nth term `a_n` .
Sometimes there is a rule or rules that allow you to find terms of the sequence.
A term to term rule allows you to find the next number in the sequence if you know the previous term (or terms.) This is also called a recursive rule. For example, if the sequence is 1,3,5,7,... then in order to find the next term you add 2 to the previous term. `a_5=a_4+2` or in general `a_n=a_(n-1)+2` . The drawback to such a rule is that you have to know the previous terms. If I ask for the 100th term in this sequence, you must know the 99th term.
A position to term rule, also called an explicit rule, allows you to compute the value of any term. For the example 1,3,5,7,... the nth term is `a_n=2n-1` . Thus the fifth term is 2(5)-1=9. The 100th term is 2(100)-1=199.
Sometimes finding a position to term rule is difficult. The Fibonacci sequence 1,1,2,3,5,8,13,... has as a term to term rule `a_1=1,a_2=1, a_n=a_(n-2)+a_(n-1)` for `n>=3` .
The position to term rule is `a_n=(((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n)/sqrt(5)` .
Some sequences admit no known rule of either type, for instance the sequence of primes.
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# Doubles Plus One – Definition With Examples
Are you ready to dive into the fascinating world of mathematics? At Brighterly, we’re here to make the learning process not just easy but also enjoyable. Today, we’re going to explore a handy math trick that’s a favorite among children – the Doubles Plus One strategy. This powerful tool makes adding numbers a breeze, especially when those numbers are just one digit apart. By the end of this article, your child will be able to understand the concept, see how it differs from other strategies, and even practice a few examples to reinforce their learning. So buckle up for this mathematical journey and let’s discover the wonders of Doubles Plus One together!
## How to Add Numbers with the Doubles Plus One Method
The magic of the Doubles Plus One strategy lies in its simplicity and efficiency. It reduces the time and cognitive effort required to add two consecutive numbers by transforming the problem into a simpler, more familiar format. Let’s dive into the steps involved in using this mathematical marvel.
1. Identify the smaller number: When presented with two consecutive numbers, first recognize the smaller number.
2. Find its double: The next step involves doubling the smaller number. The beauty of this approach is that doubling is often quicker and easier for children than adding different numbers.
3. Add one: The final step is to add one to the double you found in the previous step.
By decomposing the addition operation into these smaller, simpler steps, children are able to calculate more quickly and with fewer errors, thereby boosting their confidence and encouraging their exploration of more complex mathematical concepts.
## What Is the Difference between “Doubles Plus One” and “Doubles Addition”?
While they might sound similar, Doubles Plus One and Doubles Addition refer to different mathematical strategies, each with its own unique applications and advantages.
Doubles Addition is the process of adding a number to itself, such as 5 + 5 or 8 + 8. This strategy is useful in developing early number sense and is often one of the first mental math strategies taught to children.
On the other hand, Doubles Plus One is a technique that builds upon the Doubles Addition strategy. In Doubles Plus One, children add two consecutive numbers by doubling the smaller number and then adding one. For example, to calculate 7 + 8, children would double 7 to get 14, and then add one to get 15. This method enhances mental calculation skills and enables children to handle larger numbers and more complex operations.
## Solved Examples on Doubles Plus One
To help understand this concept better, let’s look at a few examples:
1. Example 1: Add 4 and 5 using the Doubles Plus One strategy.
• Step 1: Identify the smaller number. In this case, it’s 4.
• Step 2: Double the smaller number. The double of 4 is 8.
• Step 3: Add one to the double. 8 plus 1 equals 9.
• Therefore, 4 + 5 = 9.
2. Example 2: Add 7 and 8 using the Doubles Plus One strategy.
• Step 1: The smaller number here is 7.
• Step 2: Double 7 is 14.
• Step 3: 14 plus 1 is 15.
• Hence, 7 + 8 = 15.
Remember, the beauty of the Doubles Plus One strategy is that it turns the task of adding two different numbers into a more manageable process of doubling and adding one.
## Practice Problems on Doubles Plus One
It’s time to put your newfound knowledge to the test! Here are some practice problems to get you started:
1. 2 + 3 = ?
2. 5 + 6 = ?
3. 9 + 10 = ?
4. 11 + 12 = ?
Remember to use the Doubles Plus One strategy. Identify the smaller number, double it, and then add one.
## Conclusion
At Brighterly, our mission is to simplify the complex world of mathematics for children, transforming the potentially intimidating into the thrilling and accessible. A prime example of this approach is our introduction of the Doubles Plus One method. This powerful yet straightforward mathematical strategy is more than just a neat trick—it’s a foundational skill that can serve as a springboard for a broader and deeper understanding of arithmetic.
Doubles Plus One is a gatekeeper of sorts—it not only makes the process of addition more intuitive and manageable for children just starting out with numbers, but it also promotes the development of crucial skills such as mental math abilities and number sense. More importantly, it establishes the foundational understanding of number relationships, a critical aspect that underpins more advanced mathematical concepts.
By mastering the Doubles Plus One strategy, students don’t just learn to add numbers—they also build a solid foundation for future mathematical exploration and discovery. This is the Brighterly way: we make math fun, approachable, and comprehensible, opening a world of possibilities for our students.
## Frequently Asked Questions on Doubles Plus One
### What is the Doubles Plus One strategy?
The Doubles Plus One strategy is a useful mental math tool that simplifies the process of adding two consecutive numbers. The strategy is pretty straightforward. Instead of adding the two numbers directly, you identify the smaller number, double it, and then add one. This method is particularly effective when one number is exactly one more than the other.
### How is Doubles Plus One different from Doubles Addition?
While both Doubles Plus One and Doubles Addition are valuable addition strategies, they are used in different scenarios. Doubles Addition is the process of adding a number to itself—it’s an easy and simple way to add numbers that are identical. In contrast, Doubles Plus One is used when you are dealing with two consecutive numbers. Here, you double the smaller number and add one to it.
### How can Doubles Plus One help in math learning?
Doubles Plus One can be a significant asset in a child’s mathematical journey. It simplifies the addition of consecutive numbers, making the process quicker and less prone to errors. But beyond that, it helps promote computational fluency and foster a deeper understanding of the relationships between numbers. By learning and applying the Doubles Plus One strategy, children can enhance their mental calculation skills, a key component in fostering a love for mathematics and improving problem-solving capabilities.
Information Sources |
NCERT Math Magic Solutions for Class 3 Chapter 3 Give and Take
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# NCERT: Math-Magic Solutions for Class-3, Chapter-3: How Much Can You Carry?
The chapter Give and Take deals with the addition and subtraction of 3-digit numbers, and helps the students to add and subtract 3-digit numbers using place value columns.
This chapter have exercises on
Addition of 3-digit numbers Subtraction of 3-digit numbers Word problems on addition Word problems on subtraction
Addition of 3-digit numbers Subtraction of 3-digit numbers Word problems on addition Word problems on subtraction
The NCERT Math-Magic questions are answered in a simple and engaging manner. We have also related ‘Learning Concepts’ and interactive worksheets with the solutions. Our ‘Learning Beyond’ segment caters to all the probable questions that a child might think out of curiosity.
### Download the Ncert Solutions for Give and Take in PDF
Chapter 3:Give and Take
Question 1 : A jump from 2 to 12 is like taking ……………. steps.
Answer : In the picture, the room of 12 is just above the room of 2. You need to take just one jump or 10 steps forward to go there. Therefore: A jump from 2 to 12 is like taking 10 steps.
Find My Food
Question 2 : Fill in the blanks:
a) 10 less than 34 is ……………..
b) 53 – 20 = …………….
c) 11 more than 31 is ……………..
d) 11 less than 66 is ……………..
e) 62 + 13 = …………….
f) 23 less than 89 is ……………..
g) 10 and 40 more is ……………..
h) 9 added to 28 gives ……………..
i) The sum of 9 and 44 is ……………..
j) Reducing 98 by 34 gives ……………..
k) 4 and 37 more is ……………..
l) Take 35 away from 83. We get ……………..
a) To find 10 less than 34, from 34 take one jump downward.
So, 10 less than 34 is 24.
b) To find 53 – 20, from 53 take two jumps
`
So, 53 – 20 = 33
c) To find 11 more than 31, from 31 take one jump upward and then one step forward.
So, 11 more than 31 is 42.
d) To find 11 less than 66, from 66 take one jump downward and then one step backward.
So, 11 less than 66 is 55.
e) To find 62 + 13, from 62 take one jump upward and then 3 steps forward as:
62 + 13 = 62 + 10 + 3
= 72 + 3
= 75
So, 62 + 13 = 75
f) To find 23 less than 89, from 89 take 2 jumps downwards and then 3 steps backward as:
89 – 23 = 89 – 20 – 3
= 69 – 3
= 66
So, 23 less than 89 is 66
g) To find 10 and 40 more, from 10 take 4 jumps 4 upward.
So,10 and 40 more is 50.
h) To find 9 added to 28, from 28 take 1 jump upward and then 1 step backward.
So,9 added to 28 gives 37.
i) To find the sum of 9 and 44, from 44 take 1 jump upward and then 1 step backward as:
9 + 44 = 44 + 10 – 1
= 54 – 1
= 53
So, the sum of 9 and 44 is 53.
j) To reduce 98 by 34, from 98 take 3 jumps downward and then 4 steps backward.
98 − 34 = 98 −30 − 4
= 68 − 4
= 64
So, reducing 98 by 34 gives 64.
k) To find 4 and 37 more, from 4 take 4 jumps upward and then 3 steps backward.
4 + 37 = 4 + 40 − 3
= 44 − 3
= 41
So,4 and 37 more is 41.
l) To takeaway 35 from 83, from 83 take 3 jumps downward and then 5 steps backward.
83 – 35 = 83 – 30 – 5
= 53 – 5
= 48
So, take 35 away from 83 results in 48.
Question 3 : Fill in the blanks
a) 37 + 9 = __
b) 65 – 30 = __
c) _ = 87 – 30
d) __= 46 + 21
e) 66 – __ = 11
f) 36 = _ + 9
g) 45 + _ = 99
h) _ + 26 = 75
i) 40 + __ = 76
j) 98 = _ + 50
k) __ – 21 = 35
l) 57 – _ = 20
Answer : The shortcut calculations are shown below.
a) 37 + 9 = 37 +10 – 1
= 47 − 1
= 46
b) 65 − 30 = 60 – 30 + 5
= 30 + 5
= 35
c) 87 – 30 = 80 – 30 + 7
= 50 + 7
= 57
d) 46 + 21 = 46 + 20 + 1
= 66 + 1
= 67
e) To find the number in the blank space subtract 11 from 66 as:
66 – 11 = 66 – 10 – 1
= 56 – 1
= 55
Hence, 66 – 55 = 11
f) To find the number in the blank space, subtract 9 from 36 as:
36 – 9 = 36 – 10 + 1
= 26 + 1
= 27
Hence, 36 = 27 + 9
g) To find the number in the blank space, subtract 45 from 99 as:
99 – 45 = 99 – 40 – 5
= 59 – 5
= 54
Hence, 45 + 54 = 99
h) To find the number in the blank space, subtract 26 from 75 as:
75 – 26 = 75 – 20 – 6
= 55 – 6
= 49
Hence, 49 + 26 = 75
i) To find the number in the blank space, subtract 40 from 76 as:
76 – 40 = 36
Hence, 40 + 36 = 76
j) To find the number in the blank space, subtract 50 from 98.
98 – 50 = 48
Hence, 98 = 48 + 50
k) To find the number in the blank space, add 35 and 21 as:
35 + 21 = 35 + 20 + 1
= 55 + 1
= 56
Hence, 56 – 21 = 35
l) To find the number in the blank space,subtract 20 from 57.
57 – 20 = 37
Hence, 57 – 20 = 34
Question 4 : Fill in the blanks
MANGO CHILLI GAME
Question 5 : Fill in the blanks
Question 6 : Work out the steps in your mind. Write the answers directly in the boxes.
a) 33 + 42 = ___
b) _____ = 33 + 27
c) 55 + 25 =____
d) 19 + 61 = ____
e) _____ = 34 + 63
f) 67 + 25 = _____
g) _____ = 48 + 42
h) _____ = 53 + 64
i) 72 + 56 = _____
Answer : Do it mentally. Following are the steps for mental calculations.
a) 33 + 42 = 30 + 3 + 40 + 2
= 70 + 3 + 2
= 75
b) 33 + 27 = 30 +3 + 20 + 7
= 50 + 3 + 7
= 60
c) 55 + 25 = 50 + 5 + 20 + 5
= 70 + 5 + 5
= 80
d) 19 + 61 = 10 + 9 + 60 + 1
= 70 + 9 + 1
= 80
e) 34 + 63 = 30 + 4 + 60 + 3
= 90 + 4 + 3 = 97
f) 67 + 25 = 60 + 7 + 20 + 5
= 80 + 7 + 5
= 92
g) 48 + 42 = 48 + 40 + 2
= 50 + 40
= 90
h) 53 + 64 = 50 + 3 + 60 + 4
= 110 + 3 + 4
= 117
i) 72 + 56 = 70 + 2 + 50 + 6
= 120 + 2 + 6
= 128
Let Me Tell You a Story
Question 7 : To find out, do the addition in the box below.
Step 1: Find out the total number of deer and rabbits as:
So, the total number of rabbits and deer are 75.
Step 2: Find the total number of birds and squirrels as:
So, the total number of birds and squirrels are 162.
Step 3: Now, find the total number of animals as:
There were 240 animals and now there are 237 animals left. 240 – 237 = 3
Hence, the baby lion has eaten up 3 animals.
No, the baby lion cannot go home now.
How Many Bulbs?
Question 8 : TA shopkeeper Rafi had 153 candles. Paras gave him 237 more candles. How many candles does Rafi have now?
Answer : Rafi had 153 candles and Paras gave him 237 candles more. To find the total candles that Rafi has now, add 153 and 237 as:
So, Rafi has 390 candles now.
Since, 400−390=10
The sum is less than 400.
Question 9 : A train compartment is carrying 132 people. Another compartment is carrying 129 people. In all, how many people are there in both the compartments?
Answer : There are 132 people in one compartment and 129 people in another compartment. To find the total number of people, add the number of people in both the compartments.
132 + 129 = 261
Hence, there are total 261 people in both the compartments in all.
Question 10: Shanu found 138 pebbles. Karim found 44 pebbles. How many pebbles did they find in all?
Answer : Shanu found 138 pebbles and Karim found 44 pebbles. To find the total number of pebbles add 138 and 44 as:
Hence, they found 182 pebbles in all.
Question 11: A teacher kept a record of the fruits students like. This is what she found:
Find out:
(a) How many students in the school like oranges?
(b) How many students in the school like mangoes?
(c) Altogether, how many students are there in the school?
(d) Is the number of girls more than 350 or less than 350?
(a) There are 136 girls and 128 boys in the school who like oranges. To find the total number of students who like oranges, add 136 and 128.
136 + 128 = 264
Therefore, 264 students in the school who like oranges.
(b) There are 240 girls and 243 boys in the school who like mangoes. To find the total number of students who like mangoes, add 240 and 243.
240 + 243 = 483
Therefore, 483 students in theschool like mangoes.
(c) Step1: Among the girls 136 like oranges and 240 like mangoes. To find the total number of girls add 136 and 240.
136 + 240 = 376
There, are 376 girls in the school.
Step 2: Since 128 boys like oranges and 243 like mangoes, to find the total number of boys add 128 and 243.
128 + 243 = 371
Step 3: To find the total number of students, add the total number of boys and the total number of girls.
376 + 371 = 747
Therefore, there are 747 students in the school.
(d) There are 376 girls in the school. Since 376 is more than 350, the number of girls in the school is more than 350.
Hence, they found 182 pebbles in all.
Practice Time
Question 12: A (i) 345 + 52 (iv) 643 + 345 (ii) 492 + 29 (v) 750 + 219 (iii) 245 + 93
A. (i) 345 + 52 = 300 + 40 + 5 + 50 + 2
= 390 + 5 + 2
= 397
(ii) 492 + 29 = 400 + 90 + 2 + 20 + 9
= 510 + 9 + 2 = 521
(iii) 245 + 93 = 200 + 40 + 5 + 90 + 3
= 330 + 5 + 3
= 338
(iv) 643 + 345 = 600 + 40 + 3 + 300 + 40 + 5
= 900 + 40 + 3 + 40 + 5
= 980 + 3 + 5
= 988
(v) 750 + 219 = 700 + 50 + 200 + 10 + 9
= 900 + 50 + 10 + 9
= 960 + 9
= 969
Question 13: Fill in the blanks:
Answer : Choose any tow numbers in such a way that their sum is the given number
Question 14: Can you slove this puzzle?
Write the numbers 1 ,2 ,3 , 4 , 5 , 6 in the circle.So that the sum of the numbers on each side of the figure is 12.
Answer : Here is a solution to this puzzle. Answers can differ.
Find Mithoo's Bag
Question 15: Do all the sums mentally:
Answer : Add the numbers mentally and fill in the boxes. The correct answer and steps are shown below:
Question 16: Find Mithoo's bag and check your answers. Draw a line through the numbers which are answers written in the boxes above.
Answer : Check the answers from the previous question and solve this puzzle. The correct answer is:
Card Game
Question 17: Here are the cards for you. Work out the combination. Place the cards in the right boxes.
Answer : Place the cards in such a way that they make correct addition or subtraction statement. The correct answer is:
• NCERT Solutions for Class 2 Maths
• - |
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# How to Work Out the Angles in a Triangle When the Angles Are in Algebra.
• Author:
• Updated date:
If the angles in a triangle are given as algebra (usually in terms of x), and you are asked to find out the size of each angle, then you can follow these 3 simple steps to find all of the angles.
Step 1
Add up the 3 angles that are given and simplify the expression.
Step 2
Turn the expression from step 1 into an equation by making it equal to 180⁰ (since the angles in a triangle add up to 180⁰. Once this is done, you can solve the equation to find the value of x.
Step 3
Once x is found, the size of each angle can be calculated by substituting x back into each angle.
Example 1
Work out the size of each angle in this triangle.
Step 1
Add up the 3 angles that are given and simplify the expression.
6x + 4x + 2x = 12x
Step 2
Turn the expression from step 1 into an equation by making it equal to 180⁰ (since the angles in a triangle add up to 180⁰. Once this is done, you can solve the equation to find the value of x.
12x = 180
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## Read More From Owlcation
x = 180 ÷ 12
x = 15⁰
Step 3
Once x is found, the size of each angle can be calculated by substituting x back into each angle.
Starting with the smallest angle first you get:
2x = 2 × 15 = 30⁰
4x = 4 × 15 = 60⁰
6x = 6 × 15 = 90⁰
Let’s take a look at a harder example.
Example 2
Work out the size of each angle in this triangle.
Step 1
Add up the 3 angles that are given and simplify the expression.
x + 10 + 2x + 20 + 2x – 5
= 5x + 25
Step 2
Turn the expression from step 1 into an equation by making it equal to 180⁰ (since the angles in a triangle add up to 180⁰. Once this is done, you can solve the equation to find the value of x.
5x + 25 = 180
5x = 180 – 25
5x = 155
x = 155 ÷ 5
x = 31⁰
Step 3
Once x is found, the size of each angle can be calculated by substituting x back into each angle.
Starting with the smallest angle first you get:
x + 10 = 31 + 10 = 41⁰
2x - 5 = 2 × 31 – 5 = 57⁰
2x + 20 = 2 × 31 + 20 = 82⁰
## Questions & Answers
Question: What if the angles of the triangle were as follows: x+10, x+20 and the third missing angle was unknown, represented by w. Knowing that all interior angles of a triangle equal 180 degrees, how would you solve for w?
Answer: You will have to express w in terms of x.
Adding up the two angles gives 2x + 30.
Subtract this from 180 gives 150 -2x.
So w = 150 - 2x.
Question: How would I solve this? In a right angle triangle, one of the acute angles is 40 greater than the other. Find the angles of triangle.
Answer: The three angles in the triangle are x, x + 40 and 90.
Adding these up gives 2x + 130.
Make 2x + 130 = 180.
2x = 50
x =25.
So substituting x = 25 will give 90, 25 and 65. |
Paul's Online Math Notes
[Notes]
Calculus II - Notes
Next Chapter Applications of Integrals Integrals Involving Trig Functions Previous Section Next Section Partial Fractions
## Trig Substitutions
As we have done in the last couple of sections, let’s start off with a couple of integrals that we should already be able to do with a standard substitution.
Both of these used the substitution and at this point should be pretty easy for you to do. However, let’s take a look at the following integral.
Example 1 Evaluate the following integral. Solution In this case the substitution will not work and so we’re going to have to do something different for this integral. It would be nice if we could get rid of the square root somehow. The following substitution will do that for us. Do not worry about where this came from at this point. As we work the problem you will see that it works and that if we have a similar type of square root in the problem we can always use a similar substitution. Before we actually do the substitution however let’s verify the claim that this will allow us to get rid of the square root. To get rid of the square root all we need to do is recall the relationship, Using this fact the square root becomes, Note the presence of the absolute value bars there. These are important. Recall that There should always be absolute value bars at this stage. If we knew that was always positive or always negative we could eliminate the absolute value bars using, Without limits we won’t be able to determine if is positive or negative, however, we will need to eliminate them in order to do the integral. Therefore, since we are doing an indefinite integral we will assume that will be positive and so we can drop the absolute value bars. This gives, So, we were able to eliminate the square root using this substitution. Let’s now do the substitution and see what we get. In doing the substitution don’t forget that we'll also need to substitute for the dx. This is easy enough to get from the substitution. Using this substitution the integral becomes, With this substitution we were able to reduce the given integral to an integral involving trig functions and we saw how to do these problems in the previous section. Let’s finish the integral. So, we’ve got an answer for the integral. Unfortunately the answer isn’t given in x’s as it should be. So, we need to write our answer in terms of x. We can do this with some right triangle trig. From our original substitution we have, This gives the following right triangle. From this we can see that, We can deal with the in one of any variety of ways. From our substitution we can see that, While this is a perfectly acceptable method of dealing with the we can use any of the possible six inverse trig functions and since sine and cosine are the two trig functions most people are familiar with we will usually use the inverse sine or inverse cosine. In this case we’ll use the inverse cosine. So, with all of this the integral becomes, We now have the answer back in terms of x.
Wow! That was a lot of work. Most of these won’t take as long to work however. This first one needed lot’s of explanation since it was the first one. The remaining examples won’t need quite as much explanation and so won’t take as long to work.
However, before we move onto more problems let’s first address the issue of definite integrals and how the process differs in these cases.
Example 2 Evaluate the following integral. Solution The limits here won’t change the substitution so that will remain the same. Using this substitution the square root still reduces down to, However, unlike the previous example we can’t just drop the absolute value bars. In this case we’ve got limits on the integral and so we can use the limits as well as the substitution to determine the range of that we’re in. Once we’ve got that we can determine how to drop the absolute value bars. Here’s the limits of . So, if we are in the range then is in the range of and in this range of ’s tangent is positive and so we can just drop the absolute value bars. Let’s do the substitution. Note that the work is identical to the previous example and so most of it is left out. We’ll pick up at the final integral and then do the substitution. Note that because of the limits we didn’t need to resort to a right triangle to complete the problem.
Let’s take a look at a different set of limits for this integral.
Example 3 Evaluate the following integral. Solution Again, the substitution and square root are the same as the first two examples. Let’s next see the limits for this problem. Note that in determining the value of we used the smallest positive value. Now for this range of x’s we have and in this range of tangent is negative and so in this case we can drop the absolute value bars, but will need to add in a minus sign upon doing so. In other words, So, the only change this will make in the integration process is to put a minus sign in front of the integral. The integral is then,
In the last two examples we saw that we have to be very careful with definite integrals. We need to make sure that we determine the limits on and whether or not this will mean that we can drop the absolute value bars or if we need to add in a minus sign when we drop them.
Before moving on to the next example let’s get the general form for the substitution that we used in the previous set of examples.
Let’s work a new and different type of example.
Example 4 Evaluate the following integral. Solution Now, the square root in this problem looks to be (almost) the same as the previous ones so let’s try the same type of substitution and see if it will work here as well. Using this substitution the square root becomes, So using this substitution we will end up with a negative quantity (the tangent squared is always positive of course) under the square root and this will be trouble. Using this substitution will give complex values and we don’t want that. So, using secant for the substitution won’t work. However, the following substitution (and differential) will work. With this substitution the square root is, We were able to drop the absolute value bars because we are doing an indefinite integral and so we’ll assume that everything is positive. The integral is now, In the previous section we saw how to deal with integrals in which the exponent on the secant was even and since cosecants behave an awful lot like secants we should be able to do something similar with this. Here is the integral. Now we need to go back to x’s using a right triangle. Here is the right triangle for this problem and trig functions for this problem. The integral is then,
Here’s the general form for this type of square root.
There is one final case that we need to look at. The next integral will also contain something that we need to make sure we can deal with.
Example 5 Evaluate the following integral. Solution First, notice that there really is a square root in this problem even though it isn’t explicitly written out. To see the root let’s rewrite things a little. This square root is not in the form we saw in the previous examples. Here we will use the substitution for this root. With this substitution the denominator becomes, Now, because we have limits we’ll need to convert them to so we can determine how to drop the absolute value bars. In this range of secant is positive and so we can drop the absolute value bars. Here is the integral, There are several ways to proceed from this point. Normally with an odd exponent on the tangent we would strip one of them out and convert to secants. However, that would require that we also have a secant in the numerator which we don’t have. Therefore, it seems like the best way to do this one would be to convert the integrand to sines and cosines. We can now use the substitution and we might as well convert the limits as well. The integral is then,
The general form for this final type of square root is
We have a couple of final examples to work in this section. Not all trig substitutions will just jump right out at us. Sometimes we need to do a little work on the integrand first to get it into the correct form and that is the point of the remaining examples.
Example 6 Evaluate the following integral. Solution In this case the quantity under the root doesn’t obviously fit into any of the cases we looked at above and in fact isn’t in the any of the forms we saw in the previous examples. Note however that if we complete the square on the quadratic we can make it look somewhat like the above integrals. Remember that completing the square requires a coefficient of one in front of the x2. Once we have that we take half the coefficient of the x, square it, and then add and subtract it to the quantity. Here is the completing the square for this problem. So, the root becomes, This looks like a secant substitution except we don’t just have an x that is squared. That is okay, it will work the same way. Using this substitution the root reduces to, Note we could drop the absolute value bars since we are doing an indefinite integral. Here is the integral. And here is the right triangle for this problem. The integral is then,
Example 7 Evaluate the following integral. Solution This doesn’t look to be anything like the other problems in this section. However it is. To see this we first need to notice that, With this we can use the following substitution. Remember that to compute the differential all we do is differentiate both sides and then tack on dx or onto the appropriate side. With this substitution the square root becomes, Again, we can drop the absolute value bars because we are doing an indefinite integral. Here’s the integral. Here is the right triangle for this integral. The integral is then,
So, as we’ve seen in the final two examples in this section some integrals that look nothing like the first few examples can in fact be turned into a trig substitution problem with a little work.
Before leaving this section let’s summarize all three cases in one place.
Integrals Involving Trig Functions Previous Section Next Section Partial Fractions Next Chapter Applications of Integrals
[Notes]
© 2003 - 2015 Paul Dawkins |
# Question 12, Exercise 2.1
Solutions of Question 12 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Let $A=\begin{bmatrix}3 & 2 & 1 \\4 & 5 & 6 \\-2 & 3 & 4\end{bmatrix}$. Verify that$A+A^t$ is symmetric.
$$A=\left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \\ 1 & 6 & 4 \\ \end{matrix} \right]$$ For symmetric, we have, $$( A+A^t )^t=A+A^t$$ $$A+A^t=\left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \\ 1 & 6 & 4 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} 3+3 & 2+4 & 1-2 \\ 4+2 & 5+5 & 6+3 \\ -2+1 & 3+6 & 4+4 \\ \end{matrix} \right]$$ $$A+A^t=\left[ \begin{matrix} 6 & 6 & -1 \\ 6 & 10 & 9 \\ -1 & 9 & 8 \\ \end{matrix} \right]$$ $$( A+A^t )^t=\left[ \begin{matrix} 6 & 6 & -1 \\ 6 & 10 & 9 \\ -1 & 9 & 8 \\ \end{matrix} \right]$$ $$( A+A^t )^t=( A+A^t )$$
Let $A=\begin{bmatrix}3 & 2 & 1 \\ 4 & 5 & 6 \\-2 & 3 & 4\end{bmatrix}$. Verify that$A-A^t$ is skew symmetric.
$$A=\left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \\ 1 & 6 & 4 \\ \end{matrix} \right]$$ For skew-symmetric, we have, $$( A-A^t )^t=-( A-A^t)$$ $$A-A^t=\left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \\ 1 & 6 & 4 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} 3-3 & 2-4 & 1+2 \\ 4-2 & 5-5 & 6-3 \\ -2-1 & 3-6 & 4-4 \\ \end{matrix} \right]$$ $$A-A^t=\left[ \begin{matrix} 0 & -2 & 3 \\ 2 & 0 & 3 \\ -3 & -3 & 0 \\ \end{matrix} \right]$$ $$( A-A^t )^t=\left[ \begin{matrix} 0 & 2 & -3 \\ -2 & 0 & -3 \\ 3 & 3 & 0 \\ \end{matrix} \right]$$ $$( A-A^t )^t=-\left[ \begin{matrix} 0 & -2 & 3 \\ 2 & 0 & 3 \\ -3 & -3 & 0 \\ \end{matrix} \right]$$ $$( A-A^t )^t=-( A-A^t)$$ |
# Distance and Midpoint of a Line
## Distance Between Two Points
Linear distances is useful in maps and vectors, such as calculating the direct distance from one town to another.
To find the distance from one point to another, simply use the following formula:
$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$
This is basically a Pythagoras' Theorem in a more fancy way. The components ${ x }_{ 2 }-{ x }_{ 1 }$ is the difference of horizontal difference (hence the x) between the two points. On the other hand, the components ${ y }_{ 2 }-{ y }_{ 1 }$ is the difference of vertical distance between the two points.
Use the following Geogebra Applet to check out distances between the two points.
### Example:
Find the distance between the points (3, 4) and (-5,-2).
$d=\sqrt { { (-5-3) }^{ 2 }+{ (-2-4) }^{ 2 } } \\ \\ d=\sqrt { { (-8 })^{ 2 }+{ (-6) }^{ 2 } } \\ \\ d=\sqrt { 64+36 } \\ \\ d=\sqrt { 100 } =\quad 10\quad units$
## Midpoint of Two Points
Use the following formula to find the midpoint of two points:
$Midpoint=\left( \cfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\cfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right)$
The formula finds the halfway between the horizontal and vertical distances of the two points. Discover the midpoint using the Geogebra Applet below:
### Example:
Find the midpoint between (3, 4) and (-5, -2).
$Midpoint=\left( \cfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\cfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \\ \\ Midpoint=\left( \cfrac { 3+-5 }{ 2 } ,\cfrac { 4+-2 }{ 2 } \right) \\ \\ Midpoint=\left( \cfrac { -2 }{ 2 } ,\cfrac { 2 }{ 2 } \right) \\ \\ Midpoint=\left( -1,1 \right)$ |
# Difference between revisions of "Pythagorean Tree"
Pythagorean Tree, in 2 Dimensions
Fields: Algebra and Fractals
Image Created By: Enri Kina and John Wallison
Pythagorean Tree, in 2 Dimensions
A Pythagorean Tree is a fractal that is created out of squares. Starting from an initial square, two additional smaller squares are added to one side of the first square such that the space between all three squares is a right triangle. The side of the larger square becomes the hypotenuse of that right triangle.
# Basic Description
This animation shows how the angles of the triangle affect the shape of the tree.
The Pythagorean Tree begins with a square that has a right triangle branching off of it. The hypotenuse of the triangle must always be directly connected to the square. When the right triangle is created, the legs of that triangle then become one of the sides of two brand new squares. The lengths of the legs are not changed during this process, so the area of the each new square is less than the area of the original square. The sum of the areas of the two smaller squares is equal to the area of the big square (shown in the More Mathematical Explanation). The interesting thing about the tree is that the other non-right angles of the right triangle can be any value. Side lengths correspond to angles, so change the angle will change the side length too. This relationship between side length and angle is what causes the tilt for most Pythagorean trees.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Basic Algebra and Geometry, Trigonometry
Figure 1: Example pythagorean tree
In this image, CDF i [...]
Figure 1: Example pythagorean tree
In this image, CDF is a right triangle. The original square ABCD has a side length of s and an area of s2. We will derive the relationship between the area of the smaller squares and the area of the larger square. We can rewrite the area of CFHG and DFIJ in terms of s and θ1:
Square CFHG: $a^2 = (s \cos(\theta_1))^2 = s^2 \cos^2(\theta_1)$
Square DFIJ: $b^2 = (s \sin(\theta_1))^2 = s^2 \sin^2(\theta_1)$
We now have the relationship between the larger square and each of the smaller squares. However, note what happens when we add the areas of the smaller squares.
\begin{align} a^2 + b^2 &= s^2 \cos^2(\theta_1) + s^2 \sin^2(\theta_1) \\ &= s^2(\cos^2(\theta_1) + \sin^2(\theta_1)) = s^2 \end{align}
We have just proved that the sum of the areas of the smaller squares always adds up to the area of the larger square. Another way of thinking about this property is the Pythagorean theorem. Right triangle CDF has side lengths a, b, and s, with s as the hypotenuse. The Pythagorean theorem states that
$a^2 + b^2 = s^2$.
Since a2, b2, and s2 are the areas of each square, the Pythagorean also shows this interesting property of Pythagorean Trees.
### The Area of the Tree for any number of iterations
The total area of a Pythagorean Tree is actually quite simple to find. Simply put, it is the area of the original square (Iteration 0) multiplied by the number of iterations plus one. Or An = s2 (n + 1), where s is the side length of the original square, and n is the number of iterations. This method of finding the area does not consider the triangles in between the squares. We consider those triangles as empty spaces.
Figure 2: The 0th iteration
Figure 3: The 1st iteration
As you can see in Figure 2, for 0 iterations (the original square), the area is s2. Or, when using the formula, s2 (0 + 1) = s2.
For the first iteration as shown in Figure 3, the 0th iteration had one square with area s2, and the new iteration had two squares. As we mentioned before, the sum of the smaller squares' areas would be equal to the area of the original square, so s2 + s2 = 2s2. We now use our formula to confirm this.
$A_2 = s^2 (1 + 1) = 2s^2$
We can use this formula for only a small amount of iterations. Why? Look at the title image. After a few iterations, the tree starts to overlap and the area of the entire tree starts to converge to a single value. However, our formula implies that if we have infinite iterations, then our tree will have infinite area. Thus while our formula is useful for applying the area relationship per iteration, it is not an all powerful tool we can use to calculate the area for any amount of iterations.
A 45-45-90 tree will be used to explain the way the perimeter works, because such a tree shows the most basic way it changes. In a 45-45-90 tree, each and every square in an iteration is congruent. Not so much true for other types of trees, and therefore a bit more difficult to find the perimeter. Referring back to..
This image, we can see that the area of the smaller squares is s2/2. Of course, the area of a square is one of the side lengths squared, so if we take s2/2, and find the square root, we get one of the side lengths, which is s/√2'. Now, if we were to iterate further...
We'd find that the area of the newly iterated squares is s2/4. If you take the square root of that, you'd find that it comes out to a clean s/2. Also known as √2*√2. Iterate further, the denominator in the area becomes 8. Square root of 8? 2√2. It goes like this indefinitely. For each iteration, you just multiply the denominator of the side length by √2. The numerator is always s. Of course, this only applies to 45-45-90. Although that's only for one side, the formula remains unchanged. All that happens is that you multiply the fraction by 4, since all four sides of a square are congruent. The perimeter of the squares changes wildly when different right triangles are used.
History of the Tree.
Albert E. Bosman was the one who created the Pythagorean Tree. He did so in 1942. He was a Dutch Mathematician, who was born in 1891 and died in 1961. |
If the ratio of two numbers is 17:20 then the difference between the two numbers is 80 then what is the number two?
Question
If the ratio of two numbers is 17:20 then the difference between the two numbers is 80 then what is the number two?
in progress 0
1 month 2021-08-17T17:53:02+00:00 1 Answer 0 views 0
-510 and -600
Step-by-step explanation:
\begin{gathered}\text{Given ratio of numbers = 17:20}\\\\\text{Let the 2 numbers be 17x and 20x.}\\\\\text{According to question,}\\\\17x – 20x = 90\\\\\implies -3x = 90\\\\\implies x = – \dfrac{90}{3}\\\\\\\implies x = \boxed{- 30}\\\\\textbf{Verification:}\\\\\text{Put x = -30 in 17x-20x = 90}\\\\17(-30) – 20(-30) = 90\\\\\implies -510 – (-600) = 90\\\\\implies -510 + 600 = 90\\\\\implies 90 = 90\\\\\implies L.H.S. = R.H.S.\\\\\text{Hence, verified the value of x.}\\\\\end{gathered}
Given ratio of numbers = 17:20
Let the 2 numbers be 17x and 20x.
According to question,
17x−20x=90
⟹−3x=90
⟹x=−
3
90
⟹x=
−30
Verification:
Put x = -30 in 17x-20x = 90
17(−30)−20(−30)=90
⟹−510−(−600)=90
⟹−510+600=90
⟹90=90
⟹L.H.S.=R.H.S.
Hence, verified the value of x.
\text{The 2 numbers are 17(-30) = -510 and 20(-30) = -600.}The 2 numbers are 17(-30) = -510 and 20(-30) = -600.
Hope it helps. |
# Graphing Slope Intercept Form Worksheet
Graphing Slope Intercept Form Worksheet - Video tutorial (you tube style) on how to calculate slope. Graphing lines using standard form; Finding slope from an equation; Download and print these free resources from quizizz for effective learning. 3.when you have a slope of −3 what is the rise and run that you would use to graph the equation? = — change in x. 2.when an equation in slope intercept form is. Representing horizontal and vertical lines on coordinate plane; Let's graph y = 2 x + 3. Sketch the line given the equation.
This worksheet includes twenty equations that they will need to. These linear equations worksheets are a good resource for students in the 5th grade through the 8th grade. Let's graph y = 2 x + 3. Graphing lines with integer slopes. Sketch the line given the equation. Download this set of worksheets for a great learning experience! Download and print these free resources from quizizz for effective learning.
3.when you have a slope of −3 what is the rise and run that you would use to graph the equation? To be profi cient in math, you fi rst need to collect and organize data. Video tutorial (you tube style) on slope intercept form. Graphing lines with integer slopes. Finding slope from an equation;
Graphing Slope Intercept Form Worksheet - 2.when an equation in slope intercept form is. Representing horizontal and vertical lines on coordinate plane; Sketch the line given the equation. You may select the type of solutions that the students must perform. = — change in x. Sketch the line given the equation.
Sketch the line given the equation. = — change in x. 1) 3 x − 2y = −16 2) 13 x − 11 y = −12 3) 9x − 7y = −7 4) x − 3y = 6 5) 6x + 5y = −15 6) 4x − y = 1 7) 11 x − 4y = 32 8) 11 x − 8y = −48 write the standard form of the equation of the. Sketch the line given the equation. It may be printed, downloaded or saved and used in your.
Free trial available at kutasoftware.com. You may select the type of solutions that the students must perform. Video tutorial (you tube style) on slope intercept form. Slope intercept to standard form.
### Web Worksheets For Slope And Graphing Linear Equations.
To be profi cient in math, you fi rst need to collect and organize data. 3.when you have a slope of −3 what is the rise and run that you would use to graph the equation? = — change in x. Let's graph y = 2 x + 3.
### This Worksheet Includes Twenty Equations That They Will Need To.
Sketch the line given the equation. Graphing lines using standard form; You may select different configuration for the problems to test different concepts. Sketch the line given the equation.
### Finding Slope From Two Points;
You may select the type of solutions that the students must perform. Download this set of worksheets for a great learning experience! These linear equations worksheets are a good resource for students in the 5th grade through the 8th grade. It may be printed, downloaded or saved and used in your.
### Slope Intercept To Standard Form.
Finding slope from an equation; Slope intercept form worksheets are a great resource for students to practice a. Representing horizontal and vertical lines on coordinate plane; Video tutorial (you tube style) on slope intercept form. |
# Logarithm Formula
Logarithms are the opposite phenomena of exponential like subtraction is the inverse of addition process, and division is the opposite phenomena of multiplication. Logs “undo” exponentials.
#### Trivial Identities
$\large \log _{b} (1) = 0; \; because \; b^{0}=1; \; b> 0$
$\large \log _{b} (b) = 1; \; because \; b^{1}=b$
#### Basic Logarithm Formulas
$\large \log _{b} (xy) = \log _{b}(x) + \log _{b}(y)$
$\large \log _{b}\left ( \frac{x}{y} \right ) = \log _{b}(x) – \log _{b}(y)$
$\large \log_{b}(x^{d})= d \log_{b}(x)$
$\large \log_{b}(\sqrt[y]{x})= \frac{\log_{b}(x)}{y}$
$\large c\log_{b}(x)+d\log_{b}(y)= \log_{b}(x^{c}y^{d})$
#### Changing the Base
$\large \log_{b}a = \frac{\log_{d}(a)}{\log_{d}(b)}$
$\large \log_{b} (a+c) = \log_{b}a + \log_{b}\left ( 1 + \frac{c}{a} \right )$
$\large \log_{b} (a-c) = \log_{b}a + \log_{b}\left ( 1 – \frac{c}{a} \right )$
#### Exponents
$\large x^{\frac{\log(\log(x))}{\log(x)}} \; = \; \log(x)$
More topics in Logarithm Formula Natural Log Formula Change of Base Formula Exponential Growth Formula
Related Formulas Margin of Error Formula Prism Formula Percentile Formula Perimeter of Rhombus Formula Relative Standard Deviation Formula Percentage Decrease Formula Poisson Distribution Formula Sum of Cubes Formula |
## Direct Variation
### Learning Outcomes
• Solve a direct variation problem
• Use a constant of variation to describe the relationship between two variables
A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance if she sells a vehicle for $4,600, she will earn$736. She wants to evaluate the offer, but she is not sure how. In this section we will look at relationships, such as this one, between earnings, sales, and commission rate.
In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula $e = 0.16s$ tells us her earnings, $e$, come from the product of 0.16, her commission, and the sale price of the vehicle, $s$. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.
$s$, sales prices $e = 0.16s$ Interpretation
$4,600 $e=0.16(4,600)=736$ A sale of a$4,600 vehicle results in $736 earnings.$9,200 $e=0.16(9,200)=1,472$ A sale of a $9,200 vehicle results in$1472 earnings.
$18,400 $e=0.16(18,400)=2,944$ A sale of a$18,400 vehicle results in $2944 earnings. Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from$4,600 to $9,200, and we double the earnings from$736 to \$1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation. Each variable in this type of relationship varies directly with the other.
The graph below represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula $y=k{x}^{n}$ is used for direct variation. The value $k$ is a nonzero constant greater than zero and is called the constant of variation. In this case, $k=0.16$ and $n=1$.
### A General Note: Direct Variation
If $x$ and $y$ are related by an equation of the form
$y=k{x}^{n}$
then we say that the relationship is direct variation and $y$ varies directly with the $n$th power of $x$. In direct variation relationships, there is a nonzero constant ratio $k=\dfrac{y}{{x}^{n}}$, where $k$ is called the constant of variation, which help defines the relationship between the variables.
### How To: Given a description of a direct variation problem, solve for an unknown.
1. Identify the input, $x$, and the output, $y$.
2. Determine the constant of variation. You may need to divide $y$ by the specified power of $x$ to determine the constant of variation.
3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.
### Example: Solving a Direct Variation Problem
The quantity $y$ varies directly with the cube of $x$. If $y=25$ when $x=2$, find $y$ when $x$ is 6.
### Q & A
Do the graphs of all direct variation equations look like Example 1?
No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through $(0, 0)$.
### Try It
The quantity $y$ varies directly with the square of $y$. If $y=24$ when $x=3$, find $y$ when $x$ is 4.
Watch this video to see a quick lesson in direct variation. You will see more worked examples.
## Contribute!
Did you have an idea for improving this content? We’d love your input. |
# Table of 103
Created by: Team Maths - Examples.com, Last Updated: May 24, 2024
## Table of 103
The multiplication table of 103 is a systematic arrangement where 103 is multiplied by successive integers. Each entry in the table represents the product of 103 with integers starting from 1 upwards. It follows a pattern similar to other multiplication tables, characterized by adding 103 to the result of the previous multiplication, indicating a steady additive increase.
## What is the Multiplication Table of 103?
This table lists the results of multiplying 103 by integers from 1 onwards. Each successive result increases by 103, which simplifies learning and using multiplication for practical and theoretical purposes. It serves as a fundamental educational tool to help students understand and memorize basic multiplication facts.
Both tables are crucial in academic environments, especially in enhancing students’ ability to handle larger numbers and understand the concept of multiplication as repeated addition.
## 103 Times Table
The multiplication table for 103 begins at 103 and progresses by adding 103 at each step. This table is yet another example of an arithmetic sequence, where each number is the result of 103 multiplied by a whole number. This table is useful for quick multiplicative reference and helps in various mathematical tasks, such as calculating dimensions, scaling, or converting in systems or scenarios where 103 is a base unit or a common factor. The multiplication table of 103 is a systematic arrangement where 103 is multiplied by successive integers. Each entry in the table represents the product of 103 with integers starting from 1 upwards. It follows a pattern similar to other multiplication tables, characterized by adding 103 to the result of the previous multiplication, indicating a steady additive increase.
## Tips for 103 Times Table
Leverage Simple Arithmetic: Understand that multiplying by 103 is akin to multiplying by 100 and adding three times the multiplier (103 x 7 = 700 + 21).
Incremental Steps: Begin with learning the multiples of 103 for smaller numbers and progressively tackle higher numbers as confidence builds.
Use the Breakdown Method: Decompose 103 into 100 + 3 and apply distributive property (103 x 9 = 100 x 9 + 3 x 9).
Engage with Interactive Tools: Utilize online tools and games that offer multiplication puzzles and problems involving 103 to enhance engagement and retention.
Relate to Daily Life: Encourage application in daily scenarios, such as calculating expenses or measurements, where adding multiples of 103 is relevant.
## Table of 103 from 11 to 20
The Multiplication Table of 103 offers a straightforward way for students to enhance their multiplication skills. Like other tables, it is not just for memorizing figures but for understanding the progression of multiplication. Split into two columns, the table starts with the simpler multipliers from 1 to 10 in the first column, setting the groundwork. The second column advances with multipliers from 11 to 20, challenging students to broaden their calculations. This step-by-step approach facilitates a gradual comprehension and retention of multiplying by 103, fostering a thorough learning experience.
## How to Read 103 Times Tables?
• One time 103 is 103.
• Two times 103 is 206.
• Three times 103 is 309.
• Four times 103 is 412.
• Five times 103 is 515.
• Six times 103 is 618.
• Seven times 103 is 721.
• Eight times 103 is 824.
• Nine times 103 is 927.
• Ten times 103 is 1030.
This format also outlines the multiplication steps for 103 clearly, detailing the progression from 1 to 10 times 103, ideal for educational purposes and quick look-up. Each result is easily understandable, perfect for students and anyone needing a quick reference.
## Solved Examples:
### Example 1: Simple Multiplication
Question: Calculate 103 multiplied by 8.
Solution: Using the multiplication table of 103, 103 × 8 = 824
### Example 2: Calculate Total Cost
Question: You purchased 10 tickets for an event, each priced at 103 dollars. How much did all the tickets cost together?
Solution: 103 × 10 = 1030
Answer: The total cost for the tickets is 1030 dollars.
Question: Find the sum of the first three multiples of 103.
Solution: 103 × 1 = 103 103 × 2 = 206 103 × 3 = 309 103+206+309=618
### Example 4: Subtracting Products
Question: Calculate the difference between the product of 103 times 6 and 103 times 4.
Solution: 103 × 6 = 618 103 × 4 = 412 618−412=206
### Example 5: Annual Savings
Question: If you save 103 dollars monthly, how much will you save in one year?
Solution: 103 × 12 = 1236
Answer: You will save 1236 dollars in a year.
### Example 6: Multiplying Large Numbers
Question: What is 103 times 50?
Solution: 103 × 50 = 5150
Question: How much is 103 added to itself 7 times?
Solution: 103 × 7 = 721
### Example 8: Doubling and Tripling
Question: What is the sum of double and triple of 103?
Solution: 103 × 2 = 206 103 × 3 = 309 206+309=515
### Example 9: Product of Sequential Numbers
Question: Multiply 103 by 15. What is the product?
Solution: 103 × 15 = 1545
### Example 10: Cost Calculation
Question: You need 25 items, each costing 103 dollars. What is the total cost?
Solution: 103 × 25 = 2575
## AI Generator
Text prompt
10 Examples of Public speaking
20 Examples of Gas lighting |
Lesson Plans: Introduction Lesson Plan 1: Mathematical Modeling, Circular Movement and Transmission Ratios Lesson Plan 2: Skeeter Populations and Exponential Growth
Lesson Plan 2: Skeeter Populations and Exponential Growth
The questions below dealing with exponential functions and mathematical modeling have been selected from various state and national assessments. Although the lesson above may not fully equip students with the ability to answer all such test questions successfully, students who participate in active lessons like this one will eventually develop the conceptual understanding needed to succeed on these and other state assessment questions.
• Taken from the Colorado State Assessment, Mathematics, Grade 10 (2002):
Amelia recorded the time it took for candles of different lengths to burn out. Her results are shown in the table below.
Candle Length (in inches) 2.1 2.4 3.8 3.2 3.7 3.1 2.7 Burning Time (in minutes) 18 18 30 24 29 26 24
• On a grid, create a scatterplot of her data and draw a line of best fit for the data.
• Write an equation for your line of best fit.
• Explain what the slope represents within the context of this problem.
Solution: The graph below shows the scatterplot with a possible line of best fit.
The line of best fit in the graph above has a y-intercept at approximately (0, 3). In addition, the line appears to pass through (1, 10), so the slope of the line is . Therefore,
the equation of the line is y = 7x + 3.
Slope is the ratio of the change along the vertical axis to the change along the horizontal axis. The vertical axis represents time, and the horizontal axis represents length of the candle. Consequently, the slope represents the time required for a candle to burn a certain amount. Above, the slope was found to be 7, which means that it takes 7 minutes for a candle to burn 1 inch.
• Taken from the Texas Assessment of Knowledge and Skills, Grade 10 (2002):
Mitch wants to use 40 feet of fencing to enclose a flower garden. Which of these shapes would use all the fencing and enclose the largest area?
A. A rectangle with a length of 8 feet and a width of 12 feet
B. An isosceles right triangle with a side length of about 12 feet
C. A circle with a radius of about 5.6 feet
D. A square with a side length of 10 feet (correct answer)
• Taken from the Mississippi Algebra I Test 2, (2002):
Mr. Brady asked his algebra class to make a scale drawing of the classroom, which is shaped like a rectangle with a width of 20 feet and a length of 24 feet. Travis made the width of the classroom in his scale drawing 5 inches. What should be the length in Travis's scale drawing?
A. 6 inches (correct answer)
B. 8 inches
C. 9 inches
D. 10 inches
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# A train passes an electric post in 10 second and a bridge of length 2 km in 110 second. The speed of engine is:(a) 18 kmph (b) 36 kmph(c) 72 kmph(d) 90 kmph
Last updated date: 11th Aug 2024
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Hint: The above problem can be solved by using the formula of the speed. First calculate the length of the train to find the speed of the engine and then find the speed of the engine.
Given: The time for crossing the post is $t = 10\;{\text{s}}$, the time for crossing the bridge is $T = 110\;{\text{s}}$, the length of the bridge is $L = 2\;{\text{km}} = 2\;{\text{km}} \times \dfrac{{1000\;{\text{m}}}}{{1\;{\text{km}}}} = 2000\;{\text{m}}$.
The equation to find the speed of the engine by using the time to cross the post is given as:
$\Rightarrow$ ${v_1} = \dfrac{l}{t}......\left( 1 \right)$
The equation to find the speed of the engine by using the time to cross the bridge is given as:
$\Rightarrow$${v_2} = \dfrac{{L + l}}{T}......\left( 2 \right) The speed of the engine remains the same to cross the post and bridge, so equate the equation (10 and equation (2) to calculate the length of the train. \Rightarrow$${v_1} = {v_2}$
$\dfrac{l}{t} = \dfrac{{L + l}}{T}......\left( 3 \right)$
Substitute 2 km for L, 10 s for t and 110 s for T in the equation (3) to find the length of the train.
$\Rightarrow$ $\dfrac{l}{{10\;{\text{s}}}} = \dfrac{{2\;{\text{km}} + l}}{{110\;{\text{s}}}}$
$11l = 2\;{\text{km}} + l$
$\Rightarrow$$10l = 2\;{\text{km}} l = 0.2\;{\text{km}} Substitute 0.2\;{\text{km}} for l and 10 s for t in the equation (1) to find the speed of the engine. \Rightarrow$${v_1} = \dfrac{{0.2\;{\text{km}}}}{{10\;{\text{s}}}}$
${v_1} = \dfrac{{0.2\;{\text{km}}}}{{\left( {10\;{\text{s}} \times \dfrac{{1\;{\text{h}}}}{{3600\;{\text{s}}}}} \right)}}$
$\Rightarrow$${v_1} = 72\;{\text{km}}/{\text{h}}$
Thus, the speed of the engine is $72\;{\text{km}}/{\text{h}}$ and the option (c) is the correct answer.
Note:
The above problem can also be solved by the concept of the relative motion. The resultant speed becomes equal to the sum of the speed of both objects if both objects move in opposite directions and become equal to subtraction of speed of both objects. |
# Relating Area to Circumference
Warmup
A circular field is set into a square with sides 800 m.
Activity #1
Relate Area to Circumference of a Circle.
Hopefully, the names of the three buttons on the left give you an idea of what they each do.
• Click “Show Circumference” to see if it reminds you of what the circumference of a circle is.
• Also click “Show Area” to see if that reminds you of what the area of a circle is.
• Then, answer the questions below.
(1.) What is formula for finding the circumference of a circle?
(2.) What is the circumference of the circle above?
(3.) What is the formula for finding the area of a circle?
(4.) What is the area of the circle above?
• Use the radius slider to do more examples as you can.
Activity #2
More on Exploring the Area of a Circle.
Here below are two circles of equal measure. The top circle is separated into fourths and the bottom circle is separated into eighths.
• Use the points to move the pieces and use the open circles to rotate them.
• Arrange the parts in a row so that the curved edges are alternating between top and bottom.
(1.) How do the areas of the two shapes compare?
(2.) What polygon does the shape made of the circle pieces resemble?
(3.) How could you find the area of this polygon?
Activity #3
Application of Area.
Imagine a circle made of rings that can bend but not stretch.
• Watch the animation below and answer the questions that follow.
(1.) What polygon does the new shape resemble?
(2.) How does the area of the polygon in the applet compare to the area of the circle?
(3.) How can you find the area of the polygon?
(4.) State, in detailed steps, how you could find the polygon’s area in terms of the circle’s measurements. Show your thinking.
Challenge #1
A local pizzeria sells small, medium, and large pizzas. A small is 9 inches in diameter, a medium is 12 inches in diameter, and a large is 15 inches in diameter. Prices for the pizzas are \$6.00 for a small, \$9.00 for a medium, and \$12.00 for a large.
Use the slider d to change the diameter of the pizza.
(1.) What other measure is closely related to the area of the pizza?
(2.) What other measure is most closely related to the circumference of the pizza?
(3.) Which measurement – Radius, Diameter, Circumference, or Area— seems most closely related to the price?
Challenge #2
Use pi as 3.14 to find area and circumference. Enter the area and the circumference of the circle. Press “New” button to get a randomly generated circle.
Challenge #3
The picture shows a circle divided into 8 equal wedges which are rearranged.
The radius of the circle is r and its circumference is How does the picture help to explain why the area of the circle is
Quiz Time |
Find components of vector C from vectors A and B
Homework Statement
Given vectors $$\vec{A} = 5.0\hat{i} - 6.5\hat{j}$$ and $$\vec{B} = -3.5\hat{i}= 7.0\hat{j}$$. Vector $$\vec{C}$$ lies in the xy-plane. Vector $$\vec{C}$$ is perpendicular to $$\vec{A}$$ and the scalar product of $$\vec{C}$$ with $$\vec{B}$$ is 15.0. Find the vector components of $$\vec{C}$$.
Homework Equations
$$\vec{A}{\cdot}\vec{C} = 0$$
$$\vec{B}{\cdot}\vec{C} = 15$$
$$\vec{B}{\cdot}\vec{C}=B_{i}C_{i}+B_{j}C_{j}=15$$
$$\vec{B}{\cdot}\vec{C}=-3.5C_{i}+7.0C_{j}=15$$
$$\vec{A}{\cdot}\vec{C}=A_{i}C_{i}+A_{j}C_{j}=0$$
The Attempt at a Solution
Since the vectors A and C are perpendicular
$$\vec{A}{\cdot}\vec{C} = 0$$
Then,
$$\vec{A}{\cdot}\vec{C}=A_{i}C_{i}+A_{j}C_{j}=0$$
$$\vec{A}{\cdot}\vec{C}=5.0_{i}C_{i}-6.5_{j}C_{j}=0$$
$$C_{j}=\frac{5.0_{i}C{i}}{6.5}$$
Plug in $$C_{j}$$ into the other scalar equation and solve for $$C_{i}$$. Basic substitution. However I keep getting the wrong answer. Am I approaching the problem incorrectly or is my algebra wrong?
The correct answer is $$C_{x} = 8.0$$ and $$C_{y} = 6.1$$
Last edited: |
# Lesson 19
Compare to 1
## Warm-up: What Do You Know About $\frac{15}{14}\times\frac{23}{30}$? (10 minutes)
### Narrative
The purpose of this What Do You Know About _____ is for students to share what they know about and how they can represent the product $$\frac{15}{14}\times\frac{23}{30}$$. The numbers were intentionally chosen to make finding the exact value of the product challenging.
### Launch
• Display the expression.
• “What do you know about $$\frac{15}{14}\times\frac{23}{30}$$?”
• 1 minute: quiet think time
### Activity
• Record responses.
• “How could we find the value of the product $$\frac{15}{14}\times\frac{23}{30}$$?” (Find the product of the numerators and the product of the denominators.)
### Student Facing
What do you know about $$\frac{15}{14}\times\frac{23}{30}$$?
### Activity Synthesis
• “Is $$\frac{15}{14}\times\frac{23}{30}$$ less than, equal to, or greater than $$\frac{23}{30}$$? Why?” (It is greater since $$\frac{15}{14}$$ is greater than 1.)
## Activity 1: Compare Fraction Products on the Number Line (15 minutes)
### Narrative
The goal of this activity is to continue to compare the size of a product of fractions to the size of the second factor. In addition to the number line representation which students have worked with in the last few lessons, they also see a different expression that represents the product. In the next activity, this expression will be combined with the distributive property to explain in all cases why multiplying a number by a fraction less than one results in a smaller number while multiplying by a fraction greater than one results in a larger number (MP8).
MLR8 Discussion Supports. Students should take turns finding a match and explaining their reasoning to their partner. Display the following sentence frames for all to see: “I noticed _____, so I matched . . . .” Encourage students to challenge each other when they disagree.
• Groups of 2
### Activity
• 1–2 minutes: quiet think time
• 6–8 minutes: partner work time
### Student Facing
1. Match the expressions and number lines that show the same value.
• $$\frac{2}{5} \times \frac{4}{3}$$
• $$\frac{3}{4} \times \frac{5}{2}$$
• $$\frac{4}{3} \times \frac{5}{2}$$
• $$\left(1+\frac{1}{3}\right) \times \frac{5}{2}$$
• $$\left(1-\frac{3}{5}\right) \times \frac{4}{3}$$
• $$\left(1-\frac{1}{4}\right) \times \frac{5}{2}$$
2. Choose one of the expressions from each set and explain whether the value is greater than or less than the second factor.
### Activity Synthesis
• Invite students to share their matches.
• “How did you find the matching number line for $$\frac{3}{4} \times \frac{5}{2}$$?” (I saw that two of the number lines have $$\frac{5}{2}$$ on them and looked for the one that showed $$\frac{3}{4}$$ of $$\frac{5}{2}$$. I knew which one it was because $$\frac{3}{4}$$ of $$\frac{5}{2}$$ is less than $$\frac{5}{2}$$.)
• “How did you find the matching expression for $$\frac{3}{4} \times \frac{5}{2}$$?” (I looked for an expression with $$\frac{5}{2}$$ and only one of them had another factor with the value $$\frac{3}{4}$$.)
• “How did you know whether the value of $$\frac{3}{4} \times \frac{5}{2}$$ was greater than or less than $$\frac{5}{2}$$?” (I knew it was less because $$\frac{3}{4}$$ is less than 1. That was what helped me find the right number line.)
## Activity 2: True Statement (20 minutes)
### Narrative
The goal of this activity is to use the distributive property to explain why multiplying a number by a fraction greater than one increases the size of the number while multiplying by a fraction less than one decreases the size of the number. Expressions are particularly useful here because they show explicitly how the size of the number relates to the product. For example writing $$\frac{3}{5}$$ as $$1 - \frac{2}{5}$$ and then multiplying by $$\frac{4}{7}$$ gives: $$\displaystyle \left(1 - \frac{2}{5}\right) \times \frac{4}{7} = \frac{4}{7} - \left(\frac{2}{5} \times \frac{4}{7}\right)$$
The revealing part of this calculation is that the structure of the right hand side shows that it is less than $$\frac{4}{7}$$ without calculating the exact value (MP7). It must be less than $$\frac{4}{7}$$ because it is $$\frac{4}{7}$$ minus some other number.
Engagement: Internalize Self-Regulation. Provide students an opportunity to self-assess and reflect on their own progress. For example, provide students with questions that relate to the size of the factors for them to reflect on once the activity is complete.
Supports accessibility for: Conceptual Processing, Attention, Memory
• Groups of 2
### Activity
• 1–2 minutes: quiet think time
• 8–10 minutes: partner work time
• Monitor for students who use the expressions in the first problem to make the comparisons and then generalize about what happens when you multiply a number by any fraction greater than 1 or less than 1.
### Student Facing
1. Rewrite each expression as a sum or difference of 2 products.
1. $$\left(1 - \frac{2}{5}\right) \times \frac{4}{7}$$
2. $$\left(1 + \frac{1}{5}\right)\times \frac{4}{7}$$
3. $$\left(1 - \frac{3}{8}\right)\times \frac{4}{7}$$
4. $$\left(1 + \frac{1}{8}\right)\times \frac{4}{7}$$
2. Fill in each blank with $$<$$ or $$>$$ to make the inequality true.
1. $$\left(1 - \frac{2}{5}\right)\times \frac{4}{7} \,\underline{\hspace{0.9cm}} \,\frac{4}{7}$$
2. $$\left(1 + \frac{1}{5}\right)\times \frac{4}{7} \,\underline{\hspace{0.9cm}}\, \frac{4}{7}$$
3. $$\left(1 - \frac{3}{8}\right)\times \frac{4}{7} \,\underline{\hspace{0.9cm}} \,\frac{4}{7}$$
4. $$\left(1 + \frac{1}{8}\right)\times \frac{4}{7} \,\underline{\hspace{0.9cm}} \,\frac{4}{7}$$
3. Describe the value of the product when $$\frac{4}{7}$$ is multiplied by a fraction greater than 1. Explain your reasoning.
4. Describe the value of the product when $$\frac{4}{7}$$ is multiplied by a fraction less than 1. Explain your reasoning.
### Activity Synthesis
• Invite students to share their expressions for the products in the first problem.
• Display the equation: $$\left(1 - \frac{2}{5}\right)\times \frac{4}{7} = \frac{4}{7} - \left(\frac{2}{5} \times \frac{4}{7}\right)$$
• “How can you see that the value of the expression is less than $$\frac{4}{7}$$?” (It’s $$\frac{4}{7}$$ minus something.)
• “Does this reasoning also work for $$\left(1 - \frac{3}{8}\right)\times \frac{4}{7}$$?” (Yes, it’s again $$\frac{4}{7}$$ minus some other number.)
• “Will this reasoning work whenever you multiply a number less than 1 by $$\frac{4}{7}$$?” (Yes, I’ll always get $$\frac{4}{7}$$ minus an amount so that’s less than $$\frac{4}{7}$$.)
## Lesson Synthesis
### Lesson Synthesis
“Today we compared the value of a product of fractions to the value of one of the factors without calculating the product.”
Display product: $$\frac{7}{9} \times \frac{15}{13}$$.
“What are some ways you can compare the value of the product with $$\frac{15}{13}$$?” (I can calculate the value, but the numbers are complicated. I can make a number line diagram and see that it is to the left of $$\frac{15}{13}$$. I can rewrite $$\frac{7}{9}$$ as $$1-\frac{2}{9}$$ and see that it is less.)
“What are some ways you can compare the value of the product with $$\frac{7}{9}$$?” (I can calculate the value. I can make a number line diagram and see that it is to the right of $$\frac{7}{9}$$. I can rewrite $$\frac{15}{13}$$ as $$1+\frac{2}{13}$$ and see that it is more.)
## Cool-down: Compare without Calculating (5 minutes)
### Cool-Down
In this section, we learned how to compare the size of a product to the size of the factors. To compare $$\frac{3}{5} \times \frac{4}{7}$$ with $$\frac{4}{7}$$, for example, we can put them on a number line.
Since $$\frac{3}{5}$$ is 3 equal parts with 5 parts in the whole, it is to the left of $$\frac{4}{7}$$, only part of the way there. We can also see this by writing $$\frac{3}{5}$$ as $$1 - \frac{2}{5}$$.
$$\left(1 - \frac{2}{5}\right)\times \frac{4}{7} = \frac{4}{7} - \left(\frac{2}{5} \times \frac{4}{7}\right)$$
The product is less than $$\frac{4}{7}$$ because it is $$\frac{4}{7}$$ minus a fraction. |
# inequality involving heights and bisectors
Let $a,b,c,a \le b \le c$ be the sides of the triangle $ABC$, $l_a,l_b,l_c$ the lengths of its bisectors and $h_a,h_b,h_c$ the lengths of its heights. Prove that: $$\frac {h_a+h_c} {h_b} \ge \frac {l_a+l_c} {l_b}$$
• Just to make sure: you are referring to angle bisectors, right? (Not bisectors of the opposite side) Apr 30, 2016 at 10:09
• angle bisectors @Andreas Apr 30, 2016 at 10:32
The formulae for the lenghts of the angular bisectors and the heights are known. (see e.g. https://en.wikipedia.org/wiki/Triangle ). Let $T$ be the area of the triangle. We have $$l_a = \sqrt{bc (1 - \frac{a^2}{(b+c)^2})}$$ and $$h_a = \frac{2 T}{a}$$ and cyclic shifts of those.
With these formulae, the required inequality gets $$\frac{1/a + 1/c}{1/b} \geq \frac{\sqrt{bc (1 - \frac{a^2}{(b+c)^2})} + \sqrt{ab (1 - \frac{c^2}{(a+b)^2})}}{\sqrt{ac (1 - \frac{b^2}{(a+c)^2})}}$$ Multiplying the RHS by $$1 = \frac{1/\sqrt{abc}}{1/\sqrt{abc}}$$ gives $$\frac{1/a + 1/c}{1/b} \geq \frac{\frac{1}{a}\sqrt{a (1 - \frac{a^2}{(b+c)^2})} + \frac{1}{c}\sqrt{c(1 - \frac{c^2}{(a+b)^2})}}{\frac{1}{b}\sqrt{b (1 - \frac{b^2}{(a+c)^2})}}$$ Due to homogeneity, we can set $b=1$. This gives the condition $a\leq 1\leq c$ and the inequality gets
$$\frac{1}{a} + \frac{1}{c} \geq \frac{1}{a}\sqrt{ a \frac{ 1 - \frac{a^2}{(1+c)^2} }{ 1 - \frac{1}{(a+c)^2} } } + \frac{1}{c}\sqrt{ c\frac{ 1 - \frac{c^2}{(a+1)^2} }{ 1 - \frac{1}{(a+c)^2} } }$$ We now first prove that the first root is less or equal than 1, i.e. we need $${a (1 - \frac{a^2}{(1+c)^2})} \leq 1 - \frac{1}{(a+c)^2}$$ which can be transformed into $$(a+c)^2 (1-a) \geq 1 - a^3 (\frac{a+c}{1+c})^2$$ The last bracket can be expanded and it suffices to prove $$(a+c)^2 (1-a) \geq 1 - a^3 (1 - 2 \frac{1-a}{1+c})$$ which shortens to $$(1+c) a (1-a) \geq 2 a^3 (1-a)$$ or $$(1+c) \geq 2 a^2$$ which is certainly true under the given condition $a\leq 1\leq c$.
Likewise, we prove that the second root is less or equal than 1, i.e. we need $${c (1 - \frac{c^2}{(1+a)^2})} \leq 1 - \frac{1}{(a+c)^2}$$ which can be transformed into $$(a+c)^2 (1-c) \geq 1 - c^3 (\frac{a+c}{1+a})^2$$ We can expand the last bracket and it suffices to prove $$(a+c)^2 (1-c) \geq 1 - c^3 (1 + 2 \frac{c-1}{1+a})$$ or in positive terms $$(a+c)^2 (c-1) \leq c^3 -1 + 2 c^3 \frac{c-1}{1+a}$$ Since $a\leq1$, it suffices to prove $$(1+c)^2 (c-1) - c^3 +1 \leq 2 c^3 \frac{c-1}{1+a}$$ or $$(1+a)(c^2 - c) \leq 2 c^2 (c^2-c)$$ which shortens to $$1+a \leq 2 c^2$$ which is certainly true under the given condition $a\leq 1\leq c$. |
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