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#### Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 25 maths textbook solution
$f(x) \text { is increasing on } (-\frac{\pi }{3},\frac{\pi }{3})$
Given:
$f(x)=-\frac{x}{2}+sin\: x$
To prove:
$\text { We have to determine whether } f(x) \text { is increasing or decreasing on } (-\frac{\pi }{3},\frac{\pi }{3})$
Hint:
For f(x) to be increasing we must have f ’(x) > 0 and f ‘(x) < 0 for decreasing.
Solution:
We have
$f(x)=-\frac{x}{2}+sin\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(-\frac{x}{2}+\sin x\right) \\ &\Rightarrow f^{\prime}(x)=-\frac{1}{2}+\cos x \\ &\text { Now, } x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right) \\ &\Rightarrow-\frac{\pi}{3}
\begin{aligned} &\Rightarrow \cos \left(-\frac{\pi}{3}\right)<\cos x<\cos \frac{\pi}{3} \\ &\Rightarrow \frac{1}{2}<\cos x,\left[\therefore \cos \left(-\frac{\pi}{3}\right)=\cos \frac{\pi}{3}=\frac{1}{2}\right] \\ &\Rightarrow-\frac{1}{2}+\cos x>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
$\text { Hence } f(x) \text { is increasing function for x } \in \: (-\frac{\pi }{3},\frac{\pi }{3})$
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Cosine + sine - math problems
Number of problems found: 99
• Angles by cosine law
Calculate the size of the angles of the triangle ABC, if it is given by: a = 3 cm; b = 5 cm; c = 7 cm (use the sine and cosine theorem).
• Sin cos tan
In triangle ABC, right-angled at B. Sides/AB/=7cm, /BC/=5cm, /AC/=8.6cm. Find to two decimal places. A. Sine C B. Cosine C C. Tangent C.
• Right triangle trigonometrics
Calculate the size of the remaining sides and angles of a right triangle ABC if it is given: b = 10 cm; c = 20 cm; angle alpha = 60° and the angle beta = 30° (use the Pythagorean theorem and functions sine, cosine, tangent, cotangent)
• Q-Exam
If tg α = 9.6, Calculating sin α, cos α, cotg α .
• Cotangent
If the angle α is acute, and cotan α = 1/3. Determine the value of sin α, cos α, tg α.
• Trigonometry
If you know that cos(γ) = sin (806°), what is the angle γ?
• Trigonometric functions
In the right triangle is: ? Find the value of s and c: ? ?
• Pentagon
Calculate the area of a regular pentagon, which diagonal is u=16.
• Right
Determine angles of the right triangle with the hypotenuse c and legs a, b, if: ?
• One side
One side is 36 long with a 15° incline. What is the height at the end of that side?
The angle of a straight road is approximately 12 degrees. Determine the percentage of this road.
• The cone
The lateral surface area of the cone is 4 cm2, the area of the base of the cone is 2 cm2. Determine the angle in degrees (deviation) of the cone sine and the cone base plane. (Cone side is the segment joining the vertex cone with any point of the base c
• ABCD
AC= 40cm , angle DAB=38 , angle DCB=58 , angle DBC=90 , DB is perpendicular on AC , find BD and AD
• Sphere in cone
A sphere of radius 3 cm describes a cone with minimum volume. Determine cone dimensions.
• Chord - TS
The radius of circle k measures 68 cm. Arc GH = 47 cm. What is TS?
• The aspect ratio
The aspect ratio of the rectangular triangle is 13: 12: 5. Calculate the internal angles of the triangle.
• Triangle
Calculate the area of the triangle ABC if b = c = 17 cm, R = 19 cm (R is the circumradius).
• Right triangle
A right triangle ABC is given, c is a hypotenuse. Find the length of the sides a, b, the angle beta if c = 5 and angle alfa = A = 35 degrees.
• Regular 5-gon
Calculate the area of the regular pentagon with side 7 cm.
• Isosceles triangle 10
In an isosceles triangle, the equal sides are 2/3 of the length of the base. Determine the measure of the base angles.
Do you have an interesting mathematical word problem that you can't solve it? Submit a math problem, and we can try to solve it.
We will send a solution to your e-mail address. Solved examples are also published here. Please enter the e-mail correctly and check whether you don't have a full mailbox.
Please do not submit problems from current active competitions such as Mathematical Olympiad, correspondence seminars etc...
Most natural application of trigonometry and trigonometric functions is a calculation of the triangles. Common and less common calculations of different types of triangles offers our triangle calculator. Word trigonometry comes from Greek and literally means triangle calculation. Cosine - math problems. Sine - math problems.
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# Find dydx, if y = xxx - Mathematics and Statistics
Sum
Find ("d"y)/("d"x), if y = x^(x^x)
#### Solution
y = x^(x^x)
Taking logarithm of both sides, we get
log y = log x^(x^x)
∴ log y = xx log x
Differentiating both sides w.r.t. x, we get
"d"/("d"x)(log y) = x^x*"d"/("d"x)(log x) + logx*"d"/("d"x)(x^x)
∴ 1/y*("d"y)/("d"x) = x^x*1/x + logx*"d"/("d"x)(x^x) ......(i)
Let u = xx
Taking logarithm of both sides, we get
log u = log xx
∴ log u = x log x
Differentiating both sides w.r.t. x, we get
"d"/("d"x)(log "u") = x*"d"/("d"x)(log x) + logx*"d"/("d"x)(x)
∴ 1/"u"*"du"/("d"x) = x*1/x + logx*1
∴ 1/"u"*"du"/("d"x) = 1 + log x
∴ "du"/("d"x) = u(1 + log x)
∴ "d"/("d"x)(x^x) = xx(1 + log x) ......(ii)
Substituting (ii) in (i), we get
1/y*("d"y)/("d"x) = x^x*1/x + logx*x^x(1 + log x)
∴ ("d"y)/("d"x) = yx^x[1/x + logx(1 + logx)]
∴ ("d"y)/("d"x) = x^(x^x)*x^x[1/x + logx(1 + logx)]
Concept: The Concept of Derivative - Derivatives of Logarithmic Functions
Is there an error in this question or solution?
Chapter 1.3: Differentiation - Q.5
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Paul's Online Notes
Home / Calculus I / Extras / Area and Volume Formulas
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### Section 7-6 : Area and Volume Formulas
In this section we will derive the formulas used to get the area between two curves and the volume of a solid of revolution.
#### Area Between Two Curves
We will start with the formula for determining the area between $$y = f\left( x \right)$$ and $$y = g\left( x \right)$$ on the interval $$\left[ {a,b} \right]$$. We will also assume that $$f\left( x \right) \ge g\left( x \right)$$ on $$\left[ {a,b} \right]$$.
We will now proceed much as we did when we looked that the Area Problem in the Integrals Chapter. We will first divide up the interval into $$n$$ equal subintervals each with length,
$\Delta x = \frac{{b - a}}{n}$
Next, pick a point in each subinterval, $$x_i^*$$, and we can then use rectangles on each interval as follows.
The height of each of these rectangles is given by,
$f\left( {x_i^*} \right) - g\left( {x_i^*} \right)$
and the area of each rectangle is then,
$\left( {f\left( {x_i^*} \right) - g\left( {x_i^*} \right)} \right)\Delta x$
So, the area between the two curves is then approximated by,
$A \approx \sum\limits_{i = 1}^n {\left( {f\left( {x_i^*} \right) - g\left( {x_i^*} \right)} \right)\Delta x}$
The exact area is,
$A = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {f\left( {x_i^*} \right) - g\left( {x_i^*} \right)} \right)\Delta x}$
Now, recalling the definition of the definite integral this is nothing more than,
$A = \int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}}$
The formula above will work provided the two functions are in the form $$y = f\left( x \right)$$ and $$y = g\left( x \right)$$. However, not all functions are in that form.
Sometimes we will be forced to work with functions in the form between $$x = f\left( y \right)$$ and $$x = g\left( y \right)$$ on the interval $$\left[ {c,d} \right]$$ (an interval of $$y$$ values…). When this happens, the derivation is identical. First we will start by assuming that $$f\left( y \right) \ge g\left( y \right)$$ on $$\left[ {c,d} \right]$$. We can then divide up the interval into equal subintervals and build rectangles on each of these intervals. Here is a sketch of this situation.
Following the work from above, we will arrive at the following for the area,
$A = \int_{{\,c}}^{{\,d}}{{f\left( y \right) - g\left( y \right)\,dy}}$
So, regardless of the form that the functions are in we use basically the same formula.
#### Volumes for Solid of Revolution
Before deriving the formula for this we should probably first define just what a solid of revolution is. To get a solid of revolution we start out with a function, $$y = f\left( x \right)$$, on an interval $$\left[ {a,b} \right]$$.
We then rotate this curve about a given axis to get the surface of the solid of revolution. For purposes of this derivation let’s rotate the curve about the $$x$$-axis. Doing this gives the following three dimensional region.
We want to determine the volume of the interior of this object. To do this we will proceed much as we did for the area between two curves case. We will first divide up the interval into $$n$$ subintervals of width,
$\Delta x = \frac{{b - a}}{n}$
We will then choose a point from each subinterval, $$x_i^*$$.
Now, in the area between two curves case we approximated the area using rectangles on each subinterval. For volumes we will use disks on each subinterval to approximate the area. The area of the face of each disk is given by $$A\left( {x_i^*} \right)$$ and the volume of each disk is
${V_i} = A\left( {x_i^*} \right)\Delta x$
Here is a sketch of this,
The volume of the region can then be approximated by,
$V \approx \sum\limits_{i = 1}^n {A\left( {x_i^*} \right)\Delta x}$
The exact volume is then,
\begin{align*}V & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {A\left( {x_i^*} \right)\Delta x} \\ & = \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\end{align*}
So, in this case the volume will be the integral of the cross-sectional area at any $$x$$, $$A\left( x \right)$$. Note as well that, in this case, the cross-sectional area is a circle and we could go farther and get a formula for that as well. However, the formula above is more general and will work for any way of getting a cross section so we will leave it like it is.
In the sections where we actually use this formula we will also see that there are ways of generating the cross section that will actually give a cross-sectional area that is a function of $$y$$ instead of $$x$$. In these cases the formula will be,
$V = \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}},\,\,\,\hspace{0.25in}c \le y \le d$
In this case we looked at rotating a curve about the $$x$$-axis, however, we could have just as easily rotated the curve about the $$y$$-axis. In fact, we could rotate the curve about any vertical or horizontal axis and in all of these, case we can use one or both of the following formulas.
$V = \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}V = \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}}$
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# Greta uses 3 ounces of pasta to make 3/4 of a serving of pasta. How many ounces of pasta are there per serving?
May 3, 2018
4 ounces, and here's why:
#### Explanation:
We're given a ratio of ounces to servings; 3 ounces of pasta is 3/4 of a serving. If we write this out as an algebra equation:
$\frac{3}{4} S = 3$
Where $S$ is the serving size. Solving for $S$, we'll multiply both sides by the reciprocal of $S$'s coefficient:
$\cancel{\frac{3}{4}} S \times \textcolor{red}{\cancel{\frac{4}{3}}} = \cancel{3} \times \textcolor{red}{\frac{4}{\cancel{3}}}$
$\textcolor{g r e e n}{S = 4 \Leftarrow a n s}$
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## Tuesday, December 29, 2020
### Completing the square
For all quadratic expressions, that is, of the form $ax^2 + bx + c$, which we'll call canonical forms, there exists an identical expression of the form $a(x + p)^2 + q$, which we'll call completed square forms. For example, $3x^2 + 12x + 27 \equiv 3(x + 2)^2 + 15$. Among other reasons, completed square forms are convient for finding the root of the quadratic's graph because it is easier to manipulate algebraically. Below is the method for converting a quadratic expression in canonical form into an identical quadratic in completed square form.
The trick to completing the square is to see what would happen if, given $x^2 + bx + c$, you assume that the completed square form is $(x + \frac{b}{2})^2$ and correct it. If we have an $a$ in front of $x^2$, we can factor it out of the expression so that we have $a(x^2 + \frac{b}{a}x + \frac{c}{a})$ and then complete the square of what's inside the brackets. Let's use $3x^2 + 12x + 27$ as an example:
$3x^2 + 12x + 27 \equiv 3(x^2 + 4x + 9)$
Focus on $x^2 + 4x + 9$
Assume that $x^2 + 4x + 9 \equiv (x + 2)^2$
Expanding the brackets:
$(x + 2)^2 \equiv x^2 + 2 \times 2 \times x + 2^2 \equiv x^2 + 4x + 4$
We can see that dividing $b$ (that is, 4) by 2 is useful because that term gets doubled after expanding the brackets. We can further see that instead of 9, the last term is 4. So we need to correct for this. Fortunately, we can introduce the term $q$ to our completed square form, which can be used to correct it. We correct it by adding 5 (that is, $9 - 4$) after the squared brackets, like this:
$(x + 2)^2 + 5$
Now, when we expand the brackets, we get:
$(x + 2)^2 + 5 \equiv (x^2 + 2 \times 2 \times x + 2^2) + 5 \equiv x^2 + 4x + 9$
which is exactly what we want. Therefore,
$3x^2 + 12x + 27 \equiv 3(x^2 + 4x + 9) \equiv 3((x + 2)^2 + 5) \equiv 3(x + 2)^2 + 15$
Let's try this with the general case:
$ax^2 + bx + c \equiv a(x^2 + \frac{b}{a}x + \frac{c}{a})$
Focus on $x^2 + \frac{b}{a}x + \frac{c}{a}$
Assume that $x^2 + \frac{b}{a}x + \frac{c}{a} \equiv (x + \frac{b}{2a})^2$
$(x + \frac{b}{2a})^2 \equiv x^2 + 2 \times \frac{b}{2a} \times x + (\frac{b}{2a})^2 \equiv x^2 + \frac{b}{a}x + (\frac{b}{2a})^2$
Add $\frac{c}{a} - (\frac{b}{2a})^2$ to the completed square expression.
$(x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2$
Therefore,
\begin{align*} ax^2 + bx + c &\equiv a(x^2 + \frac{b}{a}x + \frac{c}{a}) \\ &\equiv a((x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2) \\ &\equiv a(x + \frac{b}{2a})^2 + a\frac{c}{a} - a(\frac{b}{2a})^2 \\ &\equiv a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a}) \end{align*}
You can use the last line to convert the expression directly.
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Find The Equation Of A Line Perpendicular To Pdf Through E NCRT Class 11 Mathematics Solutions Chapter 10 Straight
Ixl is easy online learning designed for busy parents. Enter the line equation and the coordinates of a point by which the line passes through, in the input field.
[4] find the acute angle between the lines and. Y = m 2 x + c 2. What is meant by the perpendicular line?
Vector Equation Of A Line Through Two Points Pdf
This is the slope for the perpendicular line.
X = 4y + 7
To find the equation of a line, we need to know the slope and a point that passes through the line. Let m 1 be the slope of the given line and m 2 be the slope of a line perpendicular to the given line. Wnte down the two parallel lines. To solve this type of problem, 1) identify the slope of the equation given.
Example 71 find the parametric and symmetric equations of the line through p 1 (1;2;3) and p 2 (2;4;1).
Want to see the full answer? 2i using equation 1.10, we get 8 <: Check out a sample q&a here. Write down the two perpendicular lines.
Given that lines 8 and 7 are perpendicular, find the coordinates of the point where they intersect.
Writing an equation of a perpendicular line write an equation of the line passing through the point (2, 3) that is perpendicular to the line 2x + y = 2. [4] (i) find a vector perpendicular to both lines. Flip the slope and change the sign). The line 8 has equation %−2!−49=0 the line 7 has equation 9%−!+9=0 9 is a constant.
Up to 10% cash back explanation:
Download free worksheet(pdf) on finding the equation of a line perpendicular to another line and through a given point, includes model problems worked out step by step, and video tutorial. Show that the lines and are skew. X /2 + y = 5. Check out a sample q&a here.
Want to see the full answer?
Once we know this, we can use the equation where m is the slope of the line, and is a point on the line. Here are the equations of five straight lines. Solving equations involving parallel and perpendicular lines worksheet find the slope of a line that is parallel and the slope of a line that is perpendicular to each line whose equation is given. The equation of a line passing through the point (x 1, y 1) and having the slope ‘m’ is given as:
Since the line goes through p 1 and p 2, the vector!
The line 2x + y = 2, or y = −2x + 2, has a slope of −2. P 1p 2 = h1;2; Use the slopes of perpendicular lines theorem. Finally, the perpendicular line equation will be displayed in the output field.
P 1p 2 will be parallel to the line.!
Ad we're here to support your family! Up to 24% cash back so, the slope of the line perpendicular to lv undefined and hence it is a vertical line. (ii) hence determine an equation of the line that is perpendicular to both and and intersects both lines. Want to see the full answer?
Asked to find a line perpendicular to the given equation and that your parallel line goes through the point (x,y).
X= 1 + t y= 2 + 2t z= 3 2t if we solve for t, we get x. Challenge find the equations of the lines that contain the sides of zlwkyhuwlfhv x (±2, 0), y(1, 3), and z(3, ±1). The equations of the lines and are. Now click the button “calculate perpendicular line” to get the equation.
(total for question 10 is 1 mark) line.
Find the acute angle between the planes with equations and [10] c. 2) find the negative reciprocal (i.e. In mathematics, the perpendicular lines are the lines in which they. For perpendicular lines, the slopes are negative reciprocals of each other.
So, equation of a vertical line through the point ( ±1, 3) is x = ±1.
Solution step 1 find the slope m of the perpendicular line. Show that line 5%−!+2=0 is perpendicular to line 2%+10!−4=0. Want to see the full answer? Y = 4 x + 2 2.
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Chapter 7 Alligator Math
Chapter 7 Alligator Math
Chomp Chomp to Find the Bigger Number
Who knew that an alligator could be so helpful in math??? Well since we have learned how to count and model all kinds of numbers it is time to compare them and the alligator is going to be our helper in this process. This chapter will focus on finding out which number is bigger or smaller; greater than or less than.
VOCABULARY
> is greater than
< is less than
is greater than - more in quantity or amount
is less than - fewer in quantity or amount
When comparing numbers we are trying to tell if one number is more or less than another number. In class we use the wonderful and loved by all example of chocolate. Would you rather have 15 pieces of chocolate or 25 pieces of chocolate? Well now we are kicking it up a notch… we are going to compare the numbers to see how many more and how many less using the bigger numbers!!
· An example of this is:
7 is 1 less than 8.
8 is 1 more than 7
70 is 10 less than 80.
80 is 10 more than 70.
Counting on or counting back as we did when we were adding and subtracting numbers a few chapters ago, will help students develop these concepts. It is also helpful to have some models or counters to help. Again using dimes and pennies and other small objects can help. The 100s chart is also very helpful for students to see the numbers and realize that charts are helpful.
The key understandings here are:
· Numbers get bigger by 1 when adding to the ONES and get bigger by 10 when adding to the TENS.
· When comparing numbers like 56 and 75 look at the number in the tens to see which is bigger or smaller. We also will start using the < and > signs to show less than and greater than. We use an alligator to demonstrate which is greater than and less than. The alligator eats the bigger number. The open part of the mouth makes the shape of the open triangle. So the alligator that looks like this:
would be the one that is the greater than.
The alligator that looks like this:
would be the one that is less than. Better watch out for that crazy alligator!! Munch! Munch!!
Activities for Chapter 7
1. Alligator Card Game
Materials: a deck of cards, copies of alligators to show the greater than and less than signs
Procedures:
1. Each player draws a card. Each player decides which
is greater faster by pulling out the greater alligor card
first. Whoever gets it right gets to keep the card. The
player with the most cards at the end wins.
2. Repeat play by looking for the lesser card.
2. Yummy Numbers
Materials: favorite small candy or raisins, cups
Procedures:
1. Spill the treat on the table and each player takes a handful of the treat and puts it in the cup.
2. Each side guesses if they have more or less by putting out an alligator greater than or less than card.
3. Swap cups and count how many are in each. Whoever has the most gets to eat some of the others!
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One-Step Equations: Multiplication Problems with Fractions
SOLVING ONE STEP EQUATIONS
In this video we will be looking at how to solve equations which involve multiplying by a fraction, using the Multiplicative Inverse Property.
One-Step Equations: Multiplication Problems with Fractions
We have looked at multiplying and dividing one-step equations with integers,
Now lets see what happens when fractions are involved.
Lets start by looking at the problem: $\frac{-5x}{7}=35$
In this problem we are dividing by $7$, so we can get rid of that by multiplying each side by $7$, since as long as we do the same thing to both sides of the equation the equation will remain equal.
$7*\frac{-5x}{7}=35*7$
Giving us $-5x =245$
Now we can divide each side of the equation by $-5$
Leaving us with $x = -45$
Or we could solve the problem in one step. Just like before we will be using the multiplicative inverse property: which says that any number times its reciprocal equals 1.
The multiplicative inverse of $\frac{-5}{7}$ is $\frac{7x}{-5}$
So if we multiply both sides of the equation by $\frac{7x}{-5}$ on the left we apply the multiplicative inverse property and get 1x and on the left hand side of the equation we can simplify the problem before multiplying and we would get
So $x = -45$
A little confused?
You may want to look at:
Hey!
This might be fun to look at next:
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# SSC CPO Quantitative Aptitude Quiz 12
5 Steps - 3 Clicks
# SSC CPO Quantitative Aptitude Quiz 12
### Introduction
What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.
A candidate with quantitative aptitude knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an important measure for a prospective business executive’s abilities.
The article SSC CPO Quantitative Aptitude Quiz 12 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. The article SSC CPO Quantitative Aptitude Quiz 12 will assist the students to know the expected questions from Quantitative Aptitude.
### Quiz
.1. If “*” is called “+”, “/” is called “*”, “-” is called “/”, “+” is called “-”. 40/20 – 5 * 10 + 5 = ?
A. 170
B. 160
C. 150
D. 165
Answer: Option D
Explanation:
Given : $$\frac{40}{20}$$ – 5 * 10 + 5 = ?
Substituting the coded symbols for mathematical operations, we get,
40 * $$\frac{20}{5 }$$ +10 – 5 = ?
40 * 4 + 10 – 5 = ?
160 + 10 – 5 = ?
170 – 5 = 165
2. If Suresh distributes his pens in the ratio of 1/2 : 1/4 : 1/5 : 1/7 between his four friends A, B, C and D, then find the total number of pens Suresh should have?
A. 153
B. 150
C. 100
D. 125
Answer: Option A
Explanation:
Here, A : B : C : D = $$\frac{1}{2}$$ : $$\frac{1}{4 }$$ : $$\frac{1}{5 }$$ : $$\frac{1}{7 }$$
1) L.C.M of 2, 4, 5, 7 is 140
2) Find the number of pens each friend received ——— (To find no. of pens each friend has, multiply the ratio with the L.C.M. calculated)
A = ($$\frac{1}{2}$$) x 140 = 70
B = ($$\frac{1}{4 }$$) x 140 = 35
C = ($$\frac{1}{5 }$$) x 140 = 28
D = ($$\frac{1}{7 }$$) x 140 = 20
3) Total number of pens = (70 x + 35 x + 28 x + 20 x) = 153 x
Minimum number of pens (x) = 1
Therefore, total number of pens = 153 pens
3. When folded into two equal halves a rectangular sheet had a perimeter of 48cm for each part folded along one set of sides and the same is 66cm when folded along the other set of sides. Find the area of the sheet.
A. 1584
B. 1120
C. 792
D. 1320
Answer: Option B
Explanation:
Let the sheet be folded along its breadth and its perimeter = 48cm
Therefore, ($$\frac{L}{2 }$$ + b) = 48 …… (i)
Now, let the sheet be folded along its length, and the perimeter = 66cm
(l + $$\frac{b}{2 }$$)= 66 …… (ii)
Solving (i) and (ii), we get,
l = 56cm, b = 20cm
Area = l*b
Area = 1120 c$${m}^{2 }$$
4. A clock was set right at 2 p.m. It gains 5 seconds in 3 minutes, and it indicates 8.30 a.m. the next morning, then the true time is:
A. 8.00 a.m.
B. 7.45 a.m.
C. 8.15 a.m.
D. 7.30 a.m.
Answer: Option A
Explanation:
Time elapsed from 2 p.m. to 8.30 a.m. = 18 hours 30 minutes = 1110 minutes.
Now, 3 minutes and 5 seconds of the given clock is 3 minutes of the correct clock.
Therefore, 1110 minutes of this clock is $$\frac{(1110 * 3)}{(37/12) }$$ minutes of the correct clock.
= 1080 minutes = 18 hours of the correct clock
Hence, the correct time is 8 a.m.
5. Find the number of times the hour hand and the minute hand of a clock are at right angle in a day.
A. 48
B. 24
C. 22
D. 44
Answer: Option D
Explanation:
In 12 hours, the hands of a clock are at right angle 22 times.
Hence in a day they are at right angle 44 times.
1. The reflex angle between the hands of a clock at 9.35 is:
A. 270°
B. 77.5°
C. 282.5°
D. 300°
Answer: Option C
Explanation:
Angle traced by the hour hand in 12 hours = 360°
Angle traced by the hour hand in 9 hours 35 minutes = $$\frac{(360 * 115) }{(12 * 12) }$$= 287.5°
Angle traced by the minute hand in 60 minutes = 360°
Angle traced by the minute hand in 35 minutes = 210°
Therefore, the angle between the hour hand and the minute hand at 9.35
= (287.5 – 210) = 77.5°
Reflex angle = 360 – 77.5 = 282.5°
2. Find the angle between the hour hand and the minute hand of a clock when the time is 5.45.
A. 97.5°
B. 90°
C. 100°
D. 95°
Answer: Option A
Explanation:
Angle traced by the hour hand in 12 hours = 360°
Angle traced by the hour hand in 5 hours 45 minutes = $$\frac{(360 * 23) }{(12 *4) }$$= 172.5°
Angle traced by the minute hand in 60 minutes = 360°
Angle traced by the minute hand in 45 minutes = 270°
Therefore, the angle between the hour hand and the minute hand at 5.45
= (270 – 172.5) = 97.5°
3. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then some of the number is:
A. 28
B. 32
C. 40
D. 64
Answer: Option C
Explanation:
Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8.
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
4. Two trains moving in same direction run at a speed of 60 km/hr and 40 km/hr respectively. If a man sitting in slow train is passed by the fast train in 10 seconds, then what is the length of the faster train?
A. 53.2 m
B. 55.6 m
C. 150 m
D. 200 m
Answer: Option B
Explanation:
Given: Speed of slow train = 60 km/hr, speed of fast train = 40 km/hr
Here both the trains move in same direction. Hence their relative speed is obtained by subtracting the individual speeds of trains.
Relative speed = 60 – 40 = 20 km/hr
1) Convert km/hr into m/s
20 x 5 = 100 = 5.56 m/s
18 18
2) Distance (Length of faster train) = Speed x Time
Length of faster train = 5.56 x 10 m = 55.6 m
5. Find the largest 4 digit number which is divisible by 88.
A. 8844
B. 9999
C. 9944
D. 9930
Answer: Option C
Explanation:
We know that the largest 4 digit number is 9999.
Simply divide 9999 by 88.
After dividing 9999 by 88 we get, 55 as remainder.
The number is said to be completely divisible, only if the remainder is zero.
Hence, we can find the required answer by subtracting the remainder obtained from the 4 digit number.
Therefore, required number = 9999 – 55 = 9944
1. The remainder is 29, when a number is divided 56. If the same number is divided by 8, then what is the remainder?
A. 3
B. 4
C. 7
D. 5
Answer: Option D
Explanation:
Dividend = [(Divisor × Quotient)] + Remainder
It is given that, the remainder is 29, when a number (dividend) is divided 56(divisor).
Dividend and quotient are unknown, hence assume dividend as X and quotient as Y.
Therefore,
X = 56 × Y + 29
56 is completely divisible by 8, but 29 is not completely divisible and we get remainder as 5, which is the required answer.
OR
X = 56 × Y + 29
= (8 × 7Y) + (8 × 3) + 5
5 is the required remainder
2. If 3 spiders make 3 webs in 3 days, then 1 spider will make 1 web in how many days?
A. 1 day
B. 1.5 days
C. 3 days
D. 6 days
Answer: Option C
Explanation:
Indirect proportion: Less spiders (↓) More days (↑)
Direct proportion: Less webs (↓) Less days (↓)
We are given, that 3 spiders make 3 webs in 3 days. This means that 3 spiders make a web in each day. 3 days for 3 webs. Hence, work done by each spider is $$\frac{1 }{3 }$$.
Now a complete web by a single spider can be done by working for 1/3rd each day.
1st day spider will work for 1/3rd
2nd day = 1/3rd
3rd day = 1/3rd
Therefore one complete web can be made in = $$\frac{1 }{3 }$$ + $$\frac{1 }{3 }$$ +$$\frac{1 }{3 }$$ = $$\frac{3 }{3 }$$ = 1 complete web
This means a spider took 3 days to complete 1 complete web.
3. What is the investment made if one invests in 15% stock at 50 and earns Rs.2000?
A. 8000
B. 7000
C. 5000
D. 6000
Answer: Option C
Explanation:
To earn Rs.15, investment = Rs.50.
Hence, to earn Rs.1500, investment = $$\frac{(1500*50) }{15 }$$
= Rs.5000
4. A company pays 12.5% dividend to its investors. If an investor buys Rs.50 shares and gets 25% on investment, at what price did the investor buy the shares?
A. 6.25
B. 25
C. 50
D. 12.5
Answer: Option B
Explanation:
Dividend on 1 share = $$\frac{(12.5 * 50) }{100 }$$ = Rs.6.25
Rs.25 is income on an investment of Rs.100
Rs.6.25 is income on an investment of Rs. $$\frac{(6.25 * 100) }{25 }$$ = Rs.25
5. Find the number of shares that can be bought for Rs.8200 if the market value is Rs.20 each with brokerage being 2.5%
450
B. 500
C. 400
D. 410
Answer: Option C
Explanation:
Let the initial investments of A and B be 3x and 5x.
A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5.
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# Exponential Properties Involving Quotients
## Subtract exponents to divide exponents by other exponents
%
Progress
Practice Exponential Properties Involving Quotients
Progress
%
Recognize and Apply the Power of a Quotient Property
Have you ever been in a science laboratory? Take a look at this dilemma.
One of the places that the students were able to visit when they went downtown was a laboratory at the city college. Downtown, the city college had some of its classrooms and one of the classrooms was a laboratory.
“This is a good friend of mine Professor Smith,” Mr. Travis said introducing the students to a woman with blonde hair and a wide smile.
“Welcome,” Professor Smith said. “Are you enjoying your trip downtown?”
Many students responded yes and then were drawn over to one of the laboratory tables where a lot of work was taking place.
“What is happening here?” Sam asked.
“Well, I started with a very small sample of cobalt. I actually had 10 grams of it and I took a third of a third of a third of a third of it,” She explained.
The students began figuring the math out in their heads.
Can you figure it out? How many grams did the sample end up being? By the end of this Concept, you will be able to solve this dilemma.
### Guidance
This may sound confusing, but in math, we can rewrite this as $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}$ or $\left(\frac{1}{2} \right)^4$ . We can use exponents with fractions or quotients, too. In order to answer the question above, we would multiply the numerators and denominators across, like this: $\frac{1 \cdot 1 \cdot 1 \cdot 1}{2 \cdot 2 \cdot 2 \cdot 2}=\frac{1}{16}$ . Half of a half of a half of a half is one sixteenth. Once again, we have repeating multiplication of the same number which we could write more easily as $\frac{1^4}{2^4}=\frac{1}{16}$ .
The Power of a Quotient Property says that for any nonzero numbers $a$ and $b$ and any integer $n$ :
$\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$
Here is one for you to try.
$\left(\frac{5}{3}\right)^4=\frac{5^4}{3^4}=\frac{625}{81}$
You can see in this situation that we have simplified the expression by figuring out what five to the fourth is and what three to the fourth is. The next step in this problem would be to divide.
Take a look at this one.
$\left(\frac{3k}{2j}\right)^4=\frac{(3k)^4}{(2j)^4}=\frac{(3k)(3k)(3k)(3k)}{(2j)(2j)(2j)(2j)}=\frac{81k^4}{16j^4}$
This problem has different variables, so this is as far as we can take this problem.
Simply each quotient.
#### Example A
$\left(\frac{4}{5}\right)^3$
Solution: $\frac{64}{125} = .512$
#### Example B
$\left(\frac{2a}{3b}\right)^2$
Solution: $\frac{4a^2}{9b^2}$
#### Example C
$\left(\frac{a}{5b}\right)^3$
Solution: $\frac{a^3}{125b^3}$
Now let's go back to the dilemma from the beginning of the Concept.
To figure out the number of grams in the sample, we must use what we have learned about monomials and powers.
Professor Smith started off with 10 grams.
Then she took a third of a third of a third of a third of it. That is $\frac{1}{3}$ to the fourth power.
Here is how we can set up the problem.
$10 \left(\frac{1}{3}\right)^4=10 \left(\frac{1^4}{3^4}\right)=10 \left (\frac{1}{81}\right)=\frac{10}{81} \ grams$
We can convert that into a decimal by dividing the numerator by the denominator.
.12 grams is our answer as a decimal.
### Guided Practice
Here is one for you to try on your own.
Simplify the following quotient.
$\left(\frac{-4x}{3y}\right)^3$
Solution
First, let's work with the numerator.
$(-4x)^3 = -64x^3$
Now let's work with the denominator.
$(3y)^3 = 27y^3$
$\frac{-64x^3}{27y^3}$
### Explore More
Directions: Simplify.
1. $\left(\frac{2}{3}\right)^4$
2. $\left(\frac{1}{3}\right)^3$
3. $\left(\frac{7}{8}\right)^2$
4. $\left(\frac{2}{5}\right)^4$
5. $\left(\frac{7k}{-2m}\right)^3$
6. $\left(\frac{3x}{-2y}\right)^3$
7. $\left(\frac{4x}{-3y}\right)^4$
8. $\left(\frac{5y}{-2z}\right)^5$
9. $\left(\frac{-2y}{4z}\right)^4$
10. $\left(\frac{4xy}{-2z^5}\right)^5$
11. $\left(\frac{12x^2y^4}{-6z^3}\right)^2$
12. $\left(\frac{7x^2y}{-2z^3}\right)^3$
13. $\left(\frac{2x^3y^2}{-2z^3}\right)^3$
14. $\left(\frac{x^{11}}{y^9}\right)^5$
15. $\left(\frac{-5x^3}{3h^2 j^8}\right)^5$
### Vocabulary Language: English
Base
Base
When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent.
Coefficient
Coefficient
A coefficient is the number in front of a variable.
Expanded Form
Expanded Form
Expanded form refers to a base and an exponent written as repeated multiplication.
Exponent
Exponent
Exponents are used to describe the number of times that a term is multiplied by itself.
Monomial
Monomial
A monomial is an expression made up of only one term.
Power
Power
The "power" refers to the value of the exponent. For example, $3^4$ is "three to the fourth power".
Power of a Product Property
Power of a Product Property
The power of a product property states that $(ab)^m = a^m b^m$.
Power of a Quotient Property
Power of a Quotient Property
The power of a quotient property states that $\left( \frac{a}{b} \right)^m = \frac{a^m}{b^m}$.
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
|
# An explanation of fast power and matrix fast Power algorithm for integer
Source: Internet
Author: User
In case the link fails to lose such a good blog, so copy one to prevent loss.
Explanation of Matrix Fast Power Foundation
1. Basic Knowledge Reserve Chapter
The related operations of matrices are then learned in linear algebra.
1.1 Definition of the matrix:
? N-order matrices (n-order matrix): a matrix with the same number of rows m as the number of columns N, as shown in a 4-4 Phalanx: *
Row matrix (Row vector): Only one row of the matrix is a row matrix:
?
Column matrix (column vector): Only one column of the matrix is a column matrix:
?
Matrix A and matrix B, matrix A and the number of columns and Matrix B are the same, then the matrix A, B is the same type matrix.
?
Unit matrix: In the multiplication of matrices, there is a matrix that plays a special role, like 1 of the Multiplication of numbers, which is called the unit matrix. It is a square, with 1 elements on the diagonal from the upper-left corner to the lower-right corner (called the main diagonal). All of them are 0. As shown is a 3-order unit matrix.
···
1.2 Related operations of the matrix:
Matrix multiplication
2. Matrix Fast Power Introduction Chapter
Integer Fast power:
To elicit the fast power of the Matrix, and to illustrate the benefits of a fast power algorithm, we can first find the power of integers.
If you want to calculate x^8 now: Then xx xx x x x x according to theusual idea, one by one, multiply the multiplication operation 7 times.
(X
x)(xx)(xx)(xx)
In this method, the multiplication is x^2 first, then the x^2 is performed three times, so the multiplication is performed 4 times. have been less than seven times. So in order to quickly calculate the power of an integer, it will consider this combination of thought.
Now consider how the calculations should be made faster. Next, calculate the integer fast power. For example: x^19 the second party.
The binary of 19 is: 1 0 0 1 1.
by (X^m)(x^n) = x^ (m+n)
Then x^19 = (x^16)
(x^2) * (x^1)
So how to solve the fast power. Take a look at the following code:
Solves the value of the x^n.
Integer fast power, calculate x^n
``Int Quickpow (int X, int N) {int res = x; int ans = 1; while (N) {if (N&< Span class= "DV" >1) {ans = ans * res;} res = res*res; N = N>>1;} return ans;} ``
So let's see if the following code is right:
For x^19:
The 19 binary is: 1 0 0 1 1
Initial:
``1; res = x;``
Then 10011 the last one is 1, so it is odd.
`` ans = res*ans = x; res = res*res = x^2;``
Then move right one bit, 1 0 0 1
Then 1001 the last one is 1, so it's odd.
`` ans = res*ans = x*(x^2) = x^3 res = res*res = x^2*x^2 = x^4``
Then move right one bit, 1 0 0
The last digit is 0, so the current number is even.
`` res = res*res = x^4*x^4 = x^8``
Then move right one bit, 1 0
The last one is 0, and the current number is even.
`` res = res*res =x^8*x^8= x^16``
Then move right one, 1.
The last one is 1, the current number is odd.
`` ans = ans*res = (x^3)*（x^16） = x^19 res = res*res = x^32``
You can see that res = X^m,m is always relative to the weights at the binary position. When bits is 0 o'clock, we only let Resres make the power exponent 2. Corresponds to the next bits weight, when bits is 1 o'clock, ans = ans*res. is multiplied by the X power of the multiplication.
2. Matrix Fast Power Algorithm Chapter
Looking at the fast power of an integer number, we now formally introduce the Matrix fast power algorithm. If there is now a n*n square a. The so-called Square is the number of rows and the number of columns equal to the matrix, first give a number m, let the calculation matrix A's M power, a^m. It is not necessary to delve into the meaning of this matrix. The above code can be converted to.
The above is just a simple calculation of the power of the Matrix, we will feel very abstract, because the above matrix does not have a specific meaning,
Now we illustrate the significance of the fast power of matrix in practical application:
Take the most common Fibonacci sequence as an example: it is well known that the Fibonacci number recursion formula is:
F[n] = f[n-1] + f[n-2]. By F[0]=0,f[1]=1, you can recursively push back all the numbers.
In the past, we would often use a for loop, which is the most straightforward algorithm.
POJ 3070, let the Fibonacci sequence, its n is up to 1 billion.
Limitations of direct recursion:
(1) The Fibonacci number n up to 1 billion that you're pushing. The test time is only 1 seconds, and the for loop causes a time-out with recursion formula recursion.
(2) It is impossible to make random access to a table, first to get the Fibonacci sequence to 1 billion, and then to go for random access. The topic does not give so much memory, the array is not open to 1 billion.
So it can be written with a fast power to the matrix.
Observe f[n] = f[n-1]+f[n-2] The nth phase is recursive by the n-1 and the n-2.
In the same vein, item n+1 is recursive by items N and n-1.
It can therefore be represented by a matrix:
Then, know f[n-1], f[n-2] then multiply the left matrix, you can get the equality matrix, the first position
That is the requirement of f[n].
An explanation of fast power and matrix fast Power algorithm for integer
Related Keywords:
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# Chain, Product and Quotient Rule: Summary
by Batool Akmal
(1)
My Notes
• Required.
Learning Material 2
• PDF
DLM Quotient Rule, Chain Rule and Product Rule Calculus Akmal.pdf
• PDF
Report mistake
Transcript
00:01 So there we have it, the three rules, and examples and their proofs.
00:06 So hopefully we're fully convinced and ready to try some of these questions but before we do this let?s just look at a recap and go over all the rules before.
00:15 Remember the chain rule, this is the rule that you use when you have function inside of a function.
00:23 The first thing that you need to do is spot that function, so recognize that this is a function within a function and then you apply the chain rule.
00:32 We discussed the full definition of the chain rule but also looked at the faster way of doing it which is the way that we have given here which says that you can differentiate the outside function as a whole first and then multiply it with the differential of the inside function.
00:49 So if you're dealing with the function that looks like f of g of x, where g of x is the inside function, you differentiate f as a whole first and then you differentiate g which is the inside function by itself.
01:02 We then looked at the product rule. Now this is when you have two functions that are multiplying with each other so once again you need to observe and look at your question.
01:13 You need to look at what kind of function it is, whether it?s a function inside of have a function or whether it?s two functions multiplying or whether it?s both.
01:21 You recognize that and then you can differentiate it using the product rule.
01:26 So the product rule states that you leave one of the functions as it is, multiply it with the differential of another, plus leave the second function as it is and differentiate it with the derivative or the differential of the first.
01:41 So you could read that as vdudx plus udvdx where you done them separately, and then you bring them together into this rule.
01:50 The last type of function we looked at is when you have two functions dividing each other.
01:57 So you have a function at the top and a function at the bottom.
01:59 In order to do that you can use the quotient rule which is a little bit different to the product rule because it takes into account the function at the bottom, so you always call the top function u and the bottom function v.
02:13 It?s really important that you stick to this because if you mix it up this quotient rule wouldn?t be valid for any other combination. So to use the quotient rule you leave v as it is which is the function at the bottom multiply it with the differential of u and then subtract the u as it is, multiply it with the differential of v and divide it by v squared.
02:35 Let?s look at some quick examples now before I let you, have you go at the exercise lecture.
02:42 So we're going to try and use all of these three rules, the faster more efficient way of using them and see if we can find the gradients of the following.
The lecture Chain, Product and Quotient Rule: Summary by Batool Akmal is from the course Quotient Rule, Chain Rule and Product Rule.
### Included Quiz Questions
1. f(g(x))' = f'(g(x)) g'(x)
2. f(g(x))' = f(g(x)) g(x)
3. f(g(x))' = f'(x)g(x) + f(x)g'(x)
1. f(x)g(x) = g(x)f'(x) + f(x)g'(x)
2. f(x)g(x) = g(x)f(x) + f(x)g(x)
3. f(x)g(x) = g'(x)f'(x) + f'(x)g'(x)
4. f(x)g(x) = f(x)f'(x) + g(x)g'(x)
1. u / v = ( v u' - u v' ) / v^2
2. u / v = ( v u' - u v' )
3. u / v = ( v u' + u v' ) / v^2
4. u / v = ( v u' - u v' ) / u^2
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Class 9 NCERT Math Solution
TOPICS
Unit-15(Examples)
Example-1 :- A coin is tossed 1000 times with the following frequencies:
Head : 455, Tail : 545
Compute the probability for each event.
Solution :-
``` Since the coin is tossed = 1000 times,
The total number of trials = 1000.
Let H = Event of outcomes Head, T = Event of outcomes Tail
Then, the number of times H happens, i.e., the number of times a head come up, is 455.
So, P(H) = (no. of Heads)/(Total no. of trials)
P(H) = 455/1000
P(H) = 0.455
The number of times T happens, i.e., the number of times a tail come up, is 545.
Similarly,
P(T) = (no. of Tails)/(Total no. of trials)
P(T) = 545/1000
P(T) = 0.545
```
Example-2 :- Two coins are tossed simultaneously 500 times, and we get
Find the probability of occurrence of each of these events.
Solution :-
``` Let us denote the events of getting two heads, one head and no head by E₁, E₂ and E₃, respectively. So,
Total outcomes = 105 + 275 + 120 = 500
P(E₁) = 105/500 = 0.21
P(E₂) = 275/500 = 0.55
P(E₃) = 120/500 = 0.24
```
Example-3 :- A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table : Find the probability of getting each outcome.
Solution :-
``` Let Eᵢ denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6.
Total outcomes = 179 + 150 + 157 + 149 + 175 + 190 = 1000.
P(E₁) = 179/1000 = 0.179
P(E₂) = 150/1000 = 0.150
P(E₃) = 157/1000 = 0.157
P(E₄) = 149/1000 = 0.149
P(E₅) = 175/1000 = 0.175
P(E6) = 190/1000 = 0.190
```
Example-4 :- On one page of a telephone directory, there were 200 telephone numbers. The frequency distribution of their unit place digit (for example, in the number 25828573, the unit place digit is 3) is given in Table: Without looking at the page, the pencil is placed on one of these numbers, i.e., the number is chosen at random. What is the probability that the digit in its unit place is 6?
Solution :-
``` Let E be the event that the digit 6 being in the unit place.
No. of frequency of 6 = 14
Total no. of telephone numbers = 200
P(E) = 14/200
P(E) = 0.07
```
Example-5 :- The record of a weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times.
(i) What is the probability that on a given day it was correct?
(ii) What is the probability that it was not correct on a given day?
Solution :-
``` The total number of days for which the record is available = 250
(i) Number of days when the forecast was correct = 175
P(the forecast was correct on a given day) = 175/250 = 0.7
(ii) Number of days when the forecast was not correct = 250 - 175 = 75
P(the forecast was not correct on a given day) = 75/250 = 0.3
```
Example-6 :- A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results of 1000 cases. If you buy a tyre of this company, what is the probability that :
(i) it will need to be replaced before it has covered 4000 km?
(ii) it will last more than 9000 km?
(iii) it will need to be replaced after it has covered somewhere between 4000 km and 14000 km?
Solution :-
```(i) The total number of trials = 1000. The frequency of a tyre that needs to be replaced before it covers 4000 km is 20.
So, P(tyre to be replaced before it covers 4000 km) = 20/1000 = 0.02
(ii) The frequency of a tyre that will last more than 9000 km is 325 + 445 = 770
So, P(tyre will last more than 9000 km) = 770/1000 = 0.77
(iii) The frequency of a tyre that requires replacement between 4000 km and 14000 km is 210 + 325 = 535.
So, P(tyre requiring replacement between 4000 km and 14000 km) = 535/1000 = 0.535
```
Example-7 :- The percentage of marks obtained by a student in the monthly unit tests are given below: Based on this data, find the probability that the student gets more than 70% marks in a unit test.
Solution :-
``` The total number of unit tests held is 5.
The number of unit tests in which the student obtained more than 70% marks is 3.
So, P(scoring more than 70% marks) = 3/5 = 0.6
```
Example-8 :- An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained are given in the following table: Find the probabilities of the following events for a driver chosen at random from the city:
(i) being 18-29 years of age and having exactly 3 accidents in one year.
(ii) being 30-50 years of age and having one or more accidents in a year.
(iii) having no accidents in one year.
Solution :-
``` Total number of drivers = 2000.
(i) The number of drivers who are 18-29 years old and have exactly 3 accidents in one year is 61.
So, P (driver is 18-29 years old with exactly 3 accidents) = 61/2000 = 0.0305
(ii) The number of drivers 30-50 years of age and having one or more accidents in one year = 125 + 60 + 22 + 18 = 225 .
So, P(driver is 30-50 years of age and having one or more accidents) = = 225/2000 = 0.1125
(iii) The number of drivers having no accidents in one year = 440 + 505 + 360 = 1305.
Therefore, P(drivers with no accident) = 1305/2000 = 0.653
```
Example-9 :- Consider the frequency distribution table, which gives the weights of 38 students of a class.
(i) Find the probability that the weight of a student in the class lies in the interval 46-50 kg.
(ii) Give two events in this context, one having probability 0 and the other having probability 1.
Solution :-
```(i) The total number of students is 38, and the number of students with weight in the interval 46 - 50 kg is 3.
So, P(weight of a student is in the interval 46 - 50 kg) = 3/38 = 0.079
(ii) For instance, consider the event that a student weighs 30 kg.
Since no student has this weight, the probability of occurrence of this event is 0.
Similarly, P (a student weighing more than 30 kg) = 38/38 = 1
```
Example-10 :- Fifty seeds were selected at random from each of 5 bags of seeds, and were kept under standardised conditions favourable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follows: What is the probability of germination of
(i) more than 40 seeds in a bag?
(ii) 49 seeds in a bag?
(iii) more that 35 seeds in a bag?
Solution :-
``` Total number of bags is 5.
(i) Number of bags in which more than 40 seeds germinated out of 50 seeds is 3.
P(germination of more than 40 seeds in a bag) = 3/5 = 0.6
(ii) Number of bags in which 49 seeds germinated = 0.
P(germination of 49 seeds in a bag) = 0/5 = 0
(iii) Number of bags in which more than 35 seeds germinated = 5.
So, the required probability = 5/5 = 1
```
CLASSES
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# NCERT Exemplar solutions for Mathematics Class 9 chapter 9 - Areas of Parallelograms & Triangles [Latest edition]
## Solutions for Chapter 9: Areas of Parallelograms & Triangles
Below listed, you can find solutions for Chapter 9 of CBSE NCERT Exemplar for Mathematics Class 9.
Exercise 9.1Exercise 9.2Exercise 9.3Exercise 9.4
Exercise 9.1 [Pages 85 - 87]
### NCERT Exemplar solutions for Mathematics Class 9 Chapter 9 Areas of Parallelograms & Triangles Exercise 9.1 [Pages 85 - 87]
#### Choose the correct alternative:
Exercise 9.1 | Q 1 | Page 85
The median of a triangle divides it into two ______.
• Triangles of equal area
• Congruent triangles
• right triangles
• Isosceles triangles
Exercise 9.1 | Q 2 | Page 85
In which of the following figures, you find two polygons on the same base and between the same parallels?
Exercise 9.1 | Q 3 | Page 86
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is ______.
• a rhombus of area 24 cm2
• a rectangle of area 24 cm2
• a square of area 26 cm2
• a trapezium of area 14 cm2
Exercise 9.1 | Q 4 | Page 86
In figure, the area of parallelogram ABCD is ______.
• AB × BM
• BC × BN
• DC × DL
Exercise 9.1 | Q 5 | Page 86
In figure if parallelogram ABCD and rectangle ABEF are of equal area, then ______.
• Perimeter of ABCD = Perimeter of ABEM
• Perimeter of ABCD < Perimeter of ABEM
• Perimeter of ABCD > Perimeter of ABEM
• Perimeter of ABCD = 1/2 (Perimeter of ABEM)
Exercise 9.1 | Q 6 | Page 87
The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to ______.
• 1/2 ar (ABC)
• 1/3 ar (ABC)
• 1/4 ar (ABC)
• ar (ABC)
Exercise 9.1 | Q 7 | Page 87
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is ______.
• 1:2
• 1:1
• 2:1
• 3:1
Exercise 9.1 | Q 8 | Page 87
ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD ______.
• Is a rectangle
• Is always a rhombus
• Is a parallelogram
• Need not be any of (A), (B) or (C)
Exercise 9.1 | Q 9 | Page 87
If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is ______.
• 1:3
• 1:2
• 3:1
• 1:4
Exercise 9.1 | Q 10 | Page 87
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (figure). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is ______.
• a : b
• (3a + b) : (a + 3b)
• (a + 3b) : (3a + b)
• (2a + b) : (3a + b)
Exercise 9.2 [Page 88]
### NCERT Exemplar solutions for Mathematics Class 9 Chapter 9 Areas of Parallelograms & Triangles Exercise 9.2 [Page 88]
#### State whether the following statement is True or False:
Exercise 9.2 | Q 1 | Page 88
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
• True
• False
Exercise 9.2 | Q 2 | Page 88
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm2.
• True
• False
Exercise 9.2 | Q 3 | Page 88
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ ASR = 90 cm2.
• True
• False
Exercise 9.2 | Q 4 | Page 88
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = 1/4 ar (ABC).
• True
• False
Exercise 9.2 | Q 5 | Page 88
In figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = 1/2 ar(EFGD).
• True
• False
Exercise 9.3 [Pages 89 - 92]
### NCERT Exemplar solutions for Mathematics Class 9 Chapter 9 Areas of Parallelograms & Triangles Exercise 9.3 [Pages 89 - 92]
Exercise 9.3 | Q 1 | Page 89
In figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).
Exercise 9.3 | Q 2 | Page 90
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Prove that ar (LZY) = ar (MZYX)
Exercise 9.3 | Q 3.(i) | Page 90
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
Exercise 9.3 | Q 3.(ii) | Page 90
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
Exercise 9.3 | Q 3.(iii) | Page 90
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
Exercise 9.3 | Q 4 | Page 90
In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (figure), then prove that ar (BPQ) = 1/2 ar(∆ABC).
Exercise 9.3 | Q 5 | Page 90
ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (figure), prove that ar (AER) = ar (AFR)
Exercise 9.3 | Q 6 | Page 91
O is any point on the diagonal PR of a parallelogram PQRS (figure). Prove that ar (PSO) = ar (PQO).
Exercise 9.3 | Q 7 | Page 91
ABCD is a parallelogram in which BC is produced to E such that CE = BC (figure). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.
Exercise 9.3 | Q 8 | Page 91
In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (figure). Prove that ar (ABCD) = ar (APQD)
Exercise 9.3 | Q 9 | Page 92
If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (figure).
Exercise 9.4 [Pages 94 - 96]
### NCERT Exemplar solutions for Mathematics Class 9 Chapter 9 Areas of Parallelograms & Triangles Exercise 9.4 [Pages 94 - 96]
Exercise 9.4 | Q 1 | Page 94
A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)
Exercise 9.4 | Q 2 | Page 94
The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.
Exercise 9.4 | Q 3 | Page 94
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
Exercise 9.4 | Q 4 | Page 95
In figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).
Exercise 9.4 | Q 5 | Page 95
ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = 7/9 ar (XYBA
Exercise 9.4 | Q 6 | Page 95
In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)
Exercise 9.4 | Q 7 | Page 95
In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)
Exercise 9.4 | Q 8 | Page 96
If the medians of a ∆ ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = 1/3 ar (ABC)
Exercise 9.4 | Q 9 | Page 96
In figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ).
Exercise 9.4 | Q 10 | Page 96
In figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD).
## Solutions for Chapter 9: Areas of Parallelograms & Triangles
Exercise 9.1Exercise 9.2Exercise 9.3Exercise 9.4
## NCERT Exemplar solutions for Mathematics Class 9 chapter 9 - Areas of Parallelograms & Triangles
Shaalaa.com has the CBSE Mathematics Mathematics Class 9 CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT Exemplar solutions for Mathematics Mathematics Class 9 CBSE 9 (Areas of Parallelograms & Triangles) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.
Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT Exemplar textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.
Concepts covered in Mathematics Class 9 chapter 9 Areas of Parallelograms & Triangles are Corollary: Triangles on the same base and between the same parallels are equal in area., Corollary: A rectangle and a parallelogram on the same base and between the same parallels are equal in area., Theorem: Parallelograms on the Same Base and Between the Same Parallels., Concept of Area.
Using NCERT Exemplar Mathematics Class 9 solutions Areas of Parallelograms & Triangles exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Exemplar Solutions are essential questions that can be asked in the final exam. Maximum CBSE Mathematics Class 9 students prefer NCERT Exemplar Textbook Solutions to score more in exams.
Get the free view of Chapter 9, Areas of Parallelograms & Triangles Mathematics Class 9 additional questions for Mathematics Mathematics Class 9 CBSE, and you can use Shaalaa.com to keep it handy for your exam preparation.
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# How do you differentiate f(x)= (2x+1)(4-x^2)(1+x^2) using the product rule?
Jul 21, 2016
$d f \left(x\right) = \left[2 \left(4 - {x}^{2}\right) \left(1 + {x}^{2}\right) - 2 x \left(2 x + 1\right) \left(1 + {x}^{2}\right) + 2 x \left(2 x + 1\right) \left(4 - {x}^{2}\right)\right] d x$
#### Explanation:
$y = a \cdot b \cdot c \text{ ; } {y}^{'} = a ' \cdot b \cdot c + {b}^{'} \cdot a . c + {c}^{'} \cdot a \cdot b$
$f \left(x\right) = \left(2 x + 1\right) \left(4 - {x}^{2}\right) \left(1 + {x}^{2}\right)$
$\frac{d f \left(x\right)}{d x} = 2 \left(4 - {x}^{2}\right) \left(1 + {x}^{2}\right) - 2 x \left(2 x + 1\right) \left(1 + {x}^{2}\right) + 2 x \left(2 x + 1\right) \left(4 - {x}^{2}\right)$
$d f \left(x\right) = \left[2 \left(4 - {x}^{2}\right) \left(1 + {x}^{2}\right) - 2 x \left(2 x + 1\right) \left(1 + {x}^{2}\right) + 2 x \left(2 x + 1\right) \left(4 - {x}^{2}\right)\right] d x$
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# Math Snap
## Which of the following is equivalent to the expression $1,000(16)^{\left(\frac{1}{4}\right)(t+2)}$ ? $16,000^{\left(\frac{1}{4}\right)(t+2)}$ $4000(2)^{t}$ $4000^{t+2}$ $2000(4)^{t}$
#### STEP 1
Assumptions 1. We are given the expression $1,000(16)^{\left(\frac{1}{4}\right)(t+2)}$. 2. We need to find an equivalent expression among the given options. 3. The properties of exponents will be used to simplify the expression.
#### STEP 2
First, let's simplify the base of the exponent by recognizing that $16$ is a power of $2$. $16 = 2^4$
#### STEP 3
Now, substitute $16$ with $2^4$ in the original expression. $1,000(16)^{\left(\frac{1}{4}\right)(t+2)} = 1,000(2^4)^{\left(\frac{1}{4}\right)(t+2)}$
#### STEP 4
Using the property of exponents that states $(a^m)^n = a^{mn}$, we can simplify the expression further. $1,000(2^4)^{\left(\frac{1}{4}\right)(t+2)} = 1,000 \cdot 2^{4 \cdot \left(\frac{1}{4}\right)(t+2)}$
#### STEP 5
Multiply the exponents inside the power of $2$. $1,000 \cdot 2^{4 \cdot \left(\frac{1}{4}\right)(t+2)} = 1,000 \cdot 2^{(t+2)}$
#### STEP 6
Now, let's distribute the exponent over the sum $(t+2)$. $1,000 \cdot 2^{(t+2)} = 1,000 \cdot 2^t \cdot 2^2$
#### STEP 7
Calculate $2^2$. $2^2 = 4$
#### STEP 8
Substitute $2^2$ with $4$ in the expression. $1,000 \cdot 2^t \cdot 2^2 = 1,000 \cdot 2^t \cdot 4$
#### STEP 9
Now, multiply $1,000$ by $4$. $1,000 \cdot 2^t \cdot 4 = 4,000 \cdot 2^t$
#### STEP 10
We have simplified the original expression to $4,000 \cdot 2^t$. Let's compare this with the given options. The equivalent expression to $1,000(16)^{\left(\frac{1}{4}\right)(t+2)}$ is $4,000 \cdot 2^t$.
##### SOLUTION
Match the simplified expression to the given options. The correct option is $4000(2)^{t}$. Karan needs to pay \$803 each month.
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# Section 4 - derivatives
```Section 3-1 : The Definition Of The Derivative
In the first section of the Limits chapter we saw that the computation of the
slope of a tangent line, the instantaneous rate of change of a function, and
the instantaneous velocity of an object at x=ax=a all required us to compute
the following limit.
limx→af(x)−f(a)x−alimx→af(x)−f(a)x−a
We also saw that with a small change of notation this limit could also be
written as,
limh→0f(a+h)−f(a)h(1)(1)limh→0f(a+h)−f(a)h
This is such an important limit and it arises in so many places that we give
it a name. We call it a derivative. Here is the official definition of the
derivative.
Defintion of the Derivative
The derivative of f(x)f(x) with respect to x is the function f′(x)f′(x) and is
defined as,f′(x)=limh→0f(x+h)−f(x)h(2)(2)f′(x)=limh→0f(x+h)−f(x)h
Note that we replaced all the a’s in (1)(1) with x’s to acknowledge the fact
that the derivative is really a function as well. We often “read” f′(x)f′(x) as
“f prime of x”.
Let’s compute a couple of derivatives using the definition.
Example 1 Find the derivative of the following function using the definition
of the derivative.f(x)=2x2−16x+35f(x)=2x2−16x+35
Hide Solution
So, all we really need to do is to plug this function into the definition of the
derivative, (2)(2), and do some algebra. While, admittedly, the algebra will
get somewhat unpleasant at times, but it’s just algebra so don’t get excited
about the fact that we’re now computing derivatives.
First plug the function into the definition of the derivative.
f′(x)=limh→0f(x+h)−f(x)h=limh→02(x+h)2−16(x+h)+35−(2x2−16x+35)hf′(x)=lim
h→0f(x+h)−f(x)h=limh→02(x+h)2−16(x+h)+35−(2x2−16x+35)h
Be careful and make sure that you properly deal with parenthesis when
doing the subtracting.
Now, we know from the previous chapter that we can’t just plug
in h=0h=0 since this will give us a division by zero error. So, we are going
to have to do some work. In this case that means multiplying everything out
and distributing the minus sign through on the second term. Doing this
gives,
f′(x)=limh→02x2+4xh+2h2−16x−16h+35−2x2+16x−35h=limh→04xh+2h2−16hhf′(
x)=limh→02x2+4xh+2h2−16x−16h+35−2x2+16x−35h=limh→04xh+2
h2−16hh
Notice that every term in the numerator that didn’t have an h in it canceled
out and we can now factor an h out of the numerator which will cancel
against the h in the denominator. After that we can compute the limit.
f′(x)=limh→0h(4x+2h−16)h=limh→04x+2h−16=4x−16f′(x)=limh→0h(4x+2h−
16)h=limh→04x+2h−16=4x−16
So, the derivative is,
f′(x)=4x−16
```
Polynomials
21 Cards
Abstract algebra
19 Cards
Algebraists
68 Cards
Group theory
34 Cards
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# 4.4: Random Numbers
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Given the same inputs, most computer programs generate the same outputs every time, so they are said to be deterministic. Determinism is usually a good thing, since we expect the same calculation to yield the same result. For some applications, though, we want the computer to be unpredictable. Games are an obvious example, but there are more.
Making a program truly nondeterministic turns out to be not so easy, but there are ways to make it at least seem nondeterministic. One of them is to use algorithms that generate pseudorandom numbers. Pseudorandom numbers are not truly random because they are generated by a deterministic computation, but just by looking at the numbers it is all but impossible to distinguish them from random.
The random module provides functions that generate pseudorandom numbers (which I will simply call "random" from here on).
The function random returns a random float between 0.0 and 1.0 (including 0.0 but not 1.0). Each time you call random, you get the next number in a long series. To see a sample, run this loop:
import random
for i in range(10):
x = random.random()
print(x)
This program produces the following list of 10 random numbers between 0.0 and up to but not including 1.0.
0.11132867921152356
0.5950949227890241
0.04820265884996877
0.841003109276478
0.997914947094958
0.04842330803368111
0.7416295948208405
0.510535245390327
0.27447040171978143
0.028511805472785867
Exercise 1: Run the program on your system and see what numbers you get. Run the program more than once and see what numbers you get.
##### Code 4.4.1 (Python)
import random
for i in range(10):
x = random.random()
print(x)
The random function is only one of many functions that handle random numbers. The function randint takes the parameters low and high, and returns an integer between low and high (including both).
>>> random.randint(5, 10)
5
>>> random.randint(5, 10)
9
To choose an element from a sequence at random, you can use choice:
>>> t = [1, 2, 3]
>>> random.choice(t)
2
>>> random.choice(t)
3
The random module also provides functions to generate random values from continuous distributions including Gaussian, exponential, gamma, and a few more.
This page titled 4.4: Random Numbers is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Chuck Severance.
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# ps3sol[1] - 1/111 – 1 = 8.41 To find the FV of the first...
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Problem Set #3 – Solutions 1. The simple interest per year is: \$5,000 × .06 = \$300 So after 10 years you will have: \$300 × 10 = \$3,000 in interest. The total balance will be \$5,000 + 3,000 = \$8,000 With compound interest we use the future value formula: FV = PV(1 + r ) t FV = \$5,000(1.06) 10 = \$8,954.24 The difference is: \$8,954.24 – 8,000 = \$954.24 9. To answer this question, we can use either the FV or the PV formula. Both will give the same answer since they are the inverse of each other. We will use the FV formula, that is: FV = PV(1 + r ) t Solving for t , we get: t = ln(FV / PV) / ln(1 + r ) t = ln (\$170,000 / \$40,000) / ln 1.062 = 24.05 years 13. To answer this question, we can use either the FV or the PV formula. We will use the FV formula, that is: FV = PV(1 + r ) t Solving for r , we get: r = (FV / PV) 1 / t – 1 r = (\$1,170,000 / \$150)
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Unformatted text preview: 1/111 – 1 = 8.41% To find the FV of the first prize, we use: FV = PV(1 + r ) t FV = \$1,170,000(1.0841) 34 = \$18,212,056.26 14. To find the PV of a lump sum, we use: PV = FV / (1 + r) t PV = \$485,000 / (1.2590) 67 = \$0.10 19. We need to find the FV of a lump sum. However, the money will only be invested for six years, so the number of periods is six. FV = PV(1 + r ) t FV = \$25,000(1.079) 6 = \$35,451.97 20. To answer this question, we will use the FV formula, that is: FV = PV(1 + r ) t Solving for t , we get: t = ln(FV / PV) / ln(1 + r ) t = ln(\$100,000 / \$10,000) / ln(1.11) = 22.06 So, the money must be invested for 22.06 years. However, you will not receive the money for another two years. From now, you’ll wait: 2 years + 22.06 years = 24.06 years...
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# Exponential Functions PowerPoint PPT Presentation
Exponential Functions. The exponential function f with base b is defined by f (x) = b x or y = b x Where b is a positive constant other than and x is any real number. /. Base is 2. Base is 10. Base is 3. Definition of the Exponential Function.
Exponential Functions
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### Exponential Functions
The exponential function f with base b is defined by
f (x) = bx or y = bx
Where b is a positive constant other than and x is any real number.
/
Base is 2.
Base is 10.
Base is 3.
## Definition of the Exponential Function
Here are some examples of exponential functions.
f (x) = 2xg(x) = 10xh(x) = 3x+1
f (x) = 13.49(0.967)x – 1
This is the given function.
f (31) = 13.49(0.967)31 – 1
Substitute 31 for x.
## Text Example
The exponential function f (x) = 13.49(0.967)x – 1 describes the number of O-rings expected to fail, when the temperature is x°F. On the morning the Challenger was launched, the temperature was 31°F, colder than any previous experience. Find the number of O-rings expected to fail at this temperature.
SolutionBecause the temperature was 31°F, substitute 31 for x and evaluate the function at 31.
f (31) = 13.49(0.967)31 – 1=3.77
f (x) = bx
0 < b < 1
f (x) = bx
b > 1
## Characteristics of Exponential Functions
• The domain of f (x) = bx consists of all real numbers. The range of f (x) = bx consists of all positive real numbers.
• The graphs of all exponential functions pass through the point (0, 1) because f (0) = b0 = 1.
• If b > 1, f (x) = bx has a graph that goes up to the right and is an increasing function.
• If 0 < b < 1, f (x) = bx has a graph that goes down to the right and is a decreasing function.
• f (x) = bx is a one-to-one function and has an inverse that is a function.
• The graph of f (x) = bx approaches but does not cross the x-axis. The x-axis is a horizontal asymptote.
Transformation
Equation
Description
Horizontal translation
g(x) = bx+c
• Shifts the graph of f (x) = bx to the left c units if c > 0.
• Shifts the graph of f (x) = bx to the right c units if c < 0.
Vertical stretching or shrinking
g(x) = c bx
• Multiplying y-coordintates of f (x) = bx by c,
• Stretches the graph of f (x) = bx if c > 1.
• Shrinks the graph of f (x) = bx if 0 < c < 1.
Reflecting
g(x) = -bx
g(x) = b-x
• Reflects the graph of f (x) = bx about the x-axis.
• Reflects the graph of f (x) = bx about the y-axis.
Vertical translation
g(x) = -bx + c
• Shifts the graph of f (x) = bx upward c units if c > 0.
• Shifts the graph of f (x) = bxdownward c units if c < 0.
g(x) = 3x+1
f (x) = 3x
(-1, 1)
(0, 1)
1
2
3
4
5
6
-5
-4
-3
-2
-1
## Text Example
Use the graph of f (x) = 3x to obtain the graph of g(x) = 3 x+1.
SolutionExamine the table below. Note that the function g(x) = 3x+1 has the general form g(x) = bx+c, where c = 1. Because c > 0, we graph g(x) = 3 x+1 by shifting the graph of f (x) = 3x one unit to the left. We construct a table showing some of the coordinates for f and g to build their graphs.
### Problems
Sketch a graph using transformation of the following:
1.
2.
3.
Recall the order of shifting: horizontal, reflection (horz., vert.), vertical.
f (x) = 3x
f (x) = ex
4
(1, 3)
f (x) = 2x
3
(1, e)
2
(1, 2)
(0, 1)
1
## The Natural Base e
An irrational number, symbolized by the letter e, appears as the base in many applied exponential functions. This irrational number is approximately equal to 2.72. More accurately,
The number e is called the natural base. The function f (x) = ex is called the natural exponential function.
-1
### Formulas for Compound Interest
After t years, the balance, A, in an account with principal P and annual interest rate r (in decimal form) is given by the following formulas:
• For n compoundings per year:
• For continuous compounding: A = Pert.
more
more
## Example:Choosing Between Investments
You want to invest \$8000 for 6 years, and you have a choice between two accounts. The first pays 7% per year, compounded monthly. The second pays 6.85% per year, compounded continuously. Which is the better investment?
SolutionThe better investment is the one with the greater balance in the account after 6 years. Let’s begin with the account with monthly compounding. We use the compound interest model with P = 8000,
r = 7% = 0.07, n = 12 (monthly compounding, means 12 compoundings per year), and t = 6.
The balance in this account after 6 years is \$12,160.84.
## Example:Choosing Between Investments
You want to invest \$8000 for 6 years, and you have a choice between two accounts. The first pays 7% per year, compounded monthly. The second pays 6.85% per year, compounded continuously. Which is the better investment?
SolutionFor the second investment option, we use the model for continuous compounding with P = 8000, r = 6.85% = 0.0685, and t = 6.
The balance in this account after 6 years is \$12,066.60, slightly less than the previous amount. Thus, the better investment is the 7% monthly compounding option.
Use A= Pert to solve the following problem: Find the accumulated value of an investment of \$2000 for 8 years at an interest rate of 7% if the money is compounded continuously
Solution:
A= PertA = 2000e(.07)(8)A = 2000 e(.56)A = 2000 * 1.75A = \$3500
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# 2014 UNC) Math Contest II Problems
Twenty-second Annual UNC Math Contest Final Round January 25, 2014 Three hours; no electronic devices. Show your work and justify your answers. Clearer presentations will earn higher rank. We hope you enjoy thinking about these problems, but you are not expected to do them all.
You may write answers in terms of the Fibonacci numbers $F_n$.
The Fibonacci numbers are $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, F_6 = 8, \ldots$
They are defined by the equations $F_1 = F_2 = 1$ and, for $n \ge 3, F_n = F_{n-1} + F_{n-2}.$
## Problem 1
The Duchess had a child on May 1st every two years until she had five children. This year the youngest is $1$ and the ages of the children are $1, 3, 5, 7$, and $9$. Alice notices that the sum of the ages is a perfect square: $1 + 3 + 5 + 7 + 9 = 25$. How old will the youngest be the next time the sum of the ages of the five children is a perfect square, and what is that perfect square?
## Problem 2
Define the Cheshire Cat function $\fbox{:)}$ by
\begin{align*} \fbox{:)}(x) &= -x \quad \text {if } x \text{ is even and} \\ \fbox{:)}(x) &= x \quad \text{ if }x \text{ is odd} \end{align*}
Find the sum $\fbox{:)}(1) + \fbox{:)}(2) + \fbox{:)}(3) + \fbox{:)}(4) + . . .+ \fbox{:)}(289)$
## Problem 3
Find $x$ and $y$ if $\frac{1}{1+\frac{1}{x}}=2$ and $\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{y}}}}=2$
## Problem 4
On the first slate, the Queen’s jurors write the number $1$. On the second slate they write the numbers $2$ and $3$. On the third slate the jurors write $4, 5$, and $6$, and so on, writing $n$ integers on the $n$th slate.
(a) What is the largest number they write on the $20$th slate?
(b) What is the sum of the numbers written on the $20$th slate?
(c) What is the sum of the numbers written on the $n$th slate?
## Problem 5
(a) The White Rabbit has a square garden with sides of length one meter. He builds a square cucumber frame in the center by connecting each corner of the garden to the midpoint of a far side of the garden, going clockwise, as shown in the diagram. What is the area of the region that is enclosed in the inner square frame?
(b) Suppose that the White Rabbit builds his square cucumber frame by connecting each corner of the garden to a point a distance $x$ from the next corner, going clockwise, as shown in the diagram. Now what is the area of the region that is enclosed in the inner square frame?
## Problem 6
(a) Alice falls down a rabbit hole and finds herself in a circular room with five doors of five different sizes evenly spaced around the circumference. Alice tries keys in some or all of the doors. She must leave no pair of adjacent doors untried. How many different sets of doors left untried does Alice have to choose from? For example, Alice might try doors $1$, $2$, and $4$ and leave doors $3$ and $5$ untried. There are no adjacent doors in the set of untried doors. Note: doors $1$ and $5$ are adjacent.
(b) Suppose the circular room in which Alice finds herself has nine doors of nine different sizes evenly spaced around the circumference. Again, she is to try keys in some or all of the doors and must leave no pair of adjacent doors untried. Now how many different sets of doors left untried does Alice have to choose from?
## Problem 7
The Caterpillar owns five different matched pairs of socks. He keeps the ten socks jumbled in random order inside a silk sack. Dressing in the dark, he selects socks, choosing randomly without replacement. If the two socks he puts on his first pair of feet are a mismatched pair and the two socks he puts on his second pair of feet are a mismatched pair, then what is the probability that the pair he selects for his third set of feet is a mismatched pair?
## Problem 8
In the Queen’s croquet, a game begins with the ball at the bottom wicket. All players hit the same ball. Each player hits the ball from the place the previous player has left it. When the ball is hit from the bottom wicket, it has a $50$% chance of going to the top wicket and a $50$% chance of staying at the bottom wicket. When hit from the top wicket, it has a $50$% chance of hitting the goal post and a 50% chance of returning to the bottom wicket.
(a) If Alice makes the first hit and alternates hits with the Queen, what is the probability that Alice is the first player to hit the goal post with the ball?
(b) Suppose Alice, the King, and the Queen take turns hitting the ball, with Alice playing first. Now what is the probability that Alice is the first player to hit the goal post with the ball?
## Problem 9
In the Queen’s croquet, as described in Problem $8$, what is the probability that the ball hits the goal post the $n$th time the ball is hit?
## Problem 10
The March Hare invites $11$ guests to a tea party. He randomly assigns to each guest either tea or cake, but no guest receives both. The guests know that the March Hare always does this, but they never know which guests will receive tea and which will receive cake. The guests decide to play a game. Each one tries to guess who of all $11$ guests will get cake and who will get tea. If one guest has more correct guesses than all the others, that guest wins. When several guests tie for the most correct guesses, then the Dormouse selects one to be the winner by selecting at random one of the guessers who has tied.
(a) All the guests make their guesses at random, perhaps by tossing a coin. What is the probability that Tweedledee, the last guest to arrive, is the winner?
(b) Tweedledum is the first guest to arrive. What is the probability that one or the other of Tweedledee and Tweedledum is the winner?
(c) Suppose that instead of guessing randomly, Tweedledee always makes the guess opposite to Tweedledum’s guess. If Tweedledum guesses that a guest will have tea, then Tweedledee will guess cake. If all the other guests have guessed randomly, what is the probability that one or the other of Tweedledee and Tweedledum is the winner? Your answer should be an explicit number, but partial credit may be given for reasonable formulae.
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# Factorial of a Number using Loop in C++
## Factorial of a Number using Loop in C++
In this article, I am going to discuss the Program to Print Factorial of a Number using Loop in C++ with Examples. Please read our previous articles, where we discussed the Sum of N natural numbers using loop in C++ with Examples.
##### Factorial of a Number:
Let us understand the procedure then the flowchart and then the program. If we have a number ‘n = 6’, integer number, it should not be a decimal number. Factorial of that number means the product of first ‘n’ natural numbers that are
That is factorial. So, 6! = 720. Here, we are multiplying numbers i.e. 1 * 2 * … up to that number which we want factorial. Multiplication is a repeating step, when it is repeating then we have to do it by using a loop. So how it is repeated? To understand that we have data below in the form of a table. Let us explain these step by step.
1. First, we multiply ‘1’ with ‘1’, because there is no other number, so the result is ‘1’. If we multiply with ‘0’ then the result of the factorial will be zero.
2. Now, multiply ‘2’ with the previous result that is ‘1 * 2 = 2’.
3. Now, again multiply ‘3’ to the result of previous multiplication ‘2 * 3 = 6’.
4. Multiply ‘4’ with the previous result that is ‘6 * 4 = 24’.
5. Multiply ‘5’ with the result of the previous addition that is ’24 * 5 = 120’.
6. Multiply ‘6’ to the previous result that is ’120 * 6 = 720’.
The result of ‘6!’ is 720. In this way, we can calculate the factorial of any number. So ‘i’ is multiplied with the value in every step. And the result is stored in that value. So let us call this value ‘fact’. And in each step, we are multiplying this ‘fact’ value with ‘i’ as ‘fact = fact * i’. So, we are modifying ‘fact’ in each and every step. So initially the value of ‘fact’ is ‘1’. Now let us look at the flowchart:
##### Factorial of a Number Flowchart:
Let us study that flow chart. First, we take a number from the user. Then we should initialize the counter to ‘1’ as well as our ‘fact’ variable to ‘1’, we will not initialize ‘fact’ to ‘0’ as the result will be 0. Both ‘i’ and ‘fact’ will be modified in every iteration of the loop.
Then up to where we want to find the factorial, let’s take that number be ‘n’. So, we will check the condition as ‘if (i <= n)’. If our counter is equal to the given number then the loop will be exit.
If the condition will true then modify the ‘fact’ variable as ‘fact = fact * i’ and increment the ‘i’ variable by 1. After this step, control goes back to the condition and check if the condition is true then again modify the ‘sum’ and ‘i’ variables and it will continue until the condition become false. If the condition will false, then simply stop the loop. Now let us look at the program.
##### Program to find Factorial of a number using loop in C++:
```#include <iostream>
using namespace std;
int main()
{
int n, fact = 1;
cout << "Enter number:" << endl;
cin >> n;
cout << endl;
for (int i = 1; i <= n; i++)
{
fact *= i;
}
cout << "Factorial of " << n << " is " << fact << endl;
return 0;
}
```
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## Geometry: Common Core (15th Edition)
$x = 3$
We know that the three interior angles of a triangle equal $180^{\circ}$. Let us set up the equation to add the three angles together. One angle measures $20x + 10$, the second measures $x - 20$, and the third measures $x + 25$: $(20x + 10) + (30x - 2) + (7x + 1) = 180$ $20x + 10 + 30x - 2 + 7x + 1 = 180$ Group like terms on the left side of the equation: $(20x + 30x + 7x) + (10 - 2 + 1) = 180$ Combine like terms: $57x + 9 = 180$ Subtract $9$ from each side of the equation to isolate the $x$ term: $57x + 9 - 9 = 180 - 9$ Subtract to simplify: $57x = 171$ Divide both sides by $57$ to solve for $x$: $x = 3$
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Scire ad trascendere - Learn to transcend.
Topics covered:
• Relations
• Indirect proof
## Relations
• Cartesian product
The Cartesian product of $A$ and $B$ (denoted by $A \times B$) is
Note: $A$ and $B$ must be sets!
If $\lvert A\rvert =m$, $\lvert B \rvert =n$, then $\lvert A \times B \rvert = m \times n$. $\newcommand{\rrel}{\mathrel{R}}$
If $A = \{1, 2, 3\}$, $B = \{4, 5, 6\}$, what is $A \times B$?
%
• What are relations?
A relation $R$ is a set of ordered pairs.
• Where is a relation defined?
• A relation $R$ is said to be on $A$ if $R \subseteq A \times A$.
• A relation $R$ is said to be from $A$ to $B$ if $R \subseteq A \times B$.
If $R$ is on $A = \{2, 5\}$, what elements can be present in $R$?
\begin{aligned} R \subseteq A \times A = \{ (2, 2), (2, 5), (5, 2), (5, 5) \}. \end{aligned}
If $R_1$, $R_2$ are on $A = \{2, 5\}$, give an example of $R_1 \subseteq R_2$.
%
If $a \mathrel{R_1} b$, $R_1 \subseteq R_2$, is $a \mathrel{R_2} b$ true?
Yes. Because $a \mathrel{R_1} b \rightarrow (a, b) \in R_1$. Given the fact that $R_1 \subseteq R_2$, we have $(a, b) \in R_2$.
## Indirect proof
### Proof by contrapositive
If we look at the truth table of $a \rightarrow b$ and $(\neg b) \rightarrow \neg a$, it is not difficult to find out that they are logically equivalent. Therefore, proving $a \rightarrow b$ is equivalent to proving $(\neg b) \rightarrow \neg a$.
T T F F T T
T F F T F F
F T T F T T
F F T T T T
Write the contrapositive of “if $a$ and $b$, then $c$ or $d$”.
The proposition can be written as $(a \land b) \rightarrow (c \lor d)$. The contrapositive is
In plain English, it means “if not $c$ and not $d$, then not $a$ or not $b$”.
Write the contrapositive of “if $n \in \mathbb{Z}$ is a even number, then $\exists k \in \mathbb{Z}$, $n = 2k$”.
The proposition can be written as $n~\mbox{is even}\rightarrow \exists k \in \mathbb{Z}, n = 2k$. The contrapositive is
In plain English, it means “If for all integer $k$, $n$ is not equal to $2k$, then $n$ is not even.”
We can observe that $a \rightarrow b$ is logically equivalent to $(a \land \neg b) \rightarrow \mathrm{False}$. Therefore, to prove $a \rightarrow b$ is True, we can assume both $a$ and $\neg b$ are True, and verify that this will imply something False.
T T F F T T
T F T T F F
F T F F T T
F F T F T T
Proof by contradiction: if $x$ is a real number, then $x^2$ is not negative.
Suppose $x$ is a real number, but $x^2$ is negative. If $x \in R$, we know that either $% $, $x = 0$ or $x > 0$.
1. When $% $, $x^2 > 0$.
2. When $x = 0$, $x^2 = 0$.
3. When $x > 0$, $x^2 > 0$.
Therefore, such real number $x$ does not exist. $\Rightarrow\Leftarrow$
Remarks. The $\Rightarrow\Leftarrow$ notation means: thus we have reached a conclusion. Therefore the supposition must be False, which means the original proposition must be True.
Proof by contradiction: if $a$ and $b$ are real numbers and $ab = 0$, then $a = 0$ or $b = 0$.
Suppose $a$ and $b$ are real numbers and $ab = 0$, but $a \neq 0$ and $b \neq 0$. It is obvious that $ab \neq 0$. $\Rightarrow\Leftarrow$
The technique is very useful in the proof of some very tricky statements, for example:
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# Solve Systems of Equations by Elimination
## Presentation on theme: "Solve Systems of Equations by Elimination"— Presentation transcript:
Solve Systems of Equations by Elimination
Methods to Solve Systems of Equations:
Graphing (y = mx + b) Substitution Graphing with x- and y- intercepts. Remember, to Solve a System of Equations, you are finding where the lines INTERSECT.
How to solve systems of equations by Elimination:
This method is used when the equations are in standard form: Ax + By = C Set up the equations so the variables line up. 2x + y = 4 x – y = 2 Notice that the x’s & y’s line up on top of each other. Notice that the “y” terms in both equations are opposites of each other. If we add this system together, the “y” terms will cancel out and we can solve for “x”.
How to solve systems of equations by Elimination:
2x + y = x – y = 2 3x = 6 x = 2 Substitute the “x” value into either equation and solve for “y.” 2x + y = 4 2(2) + y = 4 4 + y = 4 y = 0 Add the equations vertically. Get “x” alone. Divide. This is the first part of the ordered pair for your solution. (2, ___) The solution to the system is (2, 0).
Solve the system of equations using the elimination method.
x + 3y = 2 -x + 2y = 3 5y = 5 y = 1 Substitute the “y” value into either equation and solve for “x.” x + 3y = 2 x + 3(1) = 2 x + 3 = 2 x = -1 Add the equations vertically. Get “y” alone. Divide. This is part of the ordered pair for your solution. (___, 1) -1 The solution to the system is (-1, 1).
Solve the system of equations using the elimination method.
2x – y = 2 4x + 3y = 24 In these equations, neither the “x” or the “y” will cancel out, we will have to do an extra step first. THINK of what you can multiply the first equation by so that one of the variables could cancel out. If we multiply the top equation by “3” then the “y”s will cancel out. (3)2x – (3)y = (3)2 6x – 3y = 6 4x + 3y = 24 10x = 30 x = 3 2x – y = 2 2(3) – y = 2 6 – y = 2 -y = -4 y = 4 Now we can add the equations vertically. Divide by “-1” in order to get “y” alone. Get “x” alone. Divide. (3, __) 4 The solution to the system is (3, 4).
The solution to the system is (5, -4).
Solve the system of equations using the elimination method. 4x + 3y = 8 x - 2y = 13 In these equations, neither the “x” or the “y” will cancel out, we will have to do an extra step first. THINK of what you can multiply one of the equations by so that one of the variables could cancel out. If we multiply the bottom equation by “-4” then the “x”s will cancel out. (-4)x – (-4)2y = (-4)13 4x + 3y = x + 8y = -52 11y = -44 y = -4 4x + 3y = 8 4x + 3(-4)= 8 4x – 12 = 8 4x = 20 x = 5 Now we can add the equations vertically. Get “y” alone. Divide. (__, -4) 5 The solution to the system is (5, -4).
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# Suppose T_4(x) = 7-3(x-2)+7(x-2)^2-6(x-2)^2+8(x-2)^4 is the degree 4 Taylor polynomial centered at x=2 for some function f, how do you estimate the value of f'(1.9)?
##### 1 Answer
Aug 25, 2017
Depending on the final details in your approximation method, you could either say $f ' \left(1.9\right) \approx - 4.612$ (more accurate) or $f ' \left(1.9\right) \approx - 4.4$ (less accurate).
#### Explanation:
We have
$f \left(x\right) \approx {T}_{4} \left(x\right) = 7 - 3 \left(x - 2\right) + 7 {\left(x - 2\right)}^{2} - 6 {\left(x - 3\right)}^{3} + 8 {\left(x - 2\right)}^{4}$ for $x \approx 2$.
Therefore,
$f ' \left(x\right) \approx {T}_{4} ' \left(x\right) = - 3 + 14 \left(x - 2\right) - 18 {\left(x - 2\right)}^{2} + 32 {\left(x - 2\right)}^{3}$ for $x \approx 2$.
To get the more accurate approximation, plug $x = 1.9$ into this last equation:
$f ' \left(1.9\right) \approx {T}_{4} ' \left(1.9\right) = - 3 + 14 \cdot \left(- .1\right) - 18 \cdot {\left(- .1\right)}^{2} + 32 \cdot {\left(- .1\right)}^{3} \approx - 4.612$.
To get a less accurate approximation, we can get a linear approximation to ${T}_{4} ' \left(x\right)$ itself as ${T}_{4} ' \left(x\right) \approx - 3 + 14 \left(x - 2\right)$. Then $f ' \left(1.9\right) \approx {T}_{4} ' \left(1.9\right) \approx - 3 + 14 \cdot \left(- .1\right) = - 4.4$.
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Categories:
## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2
Ex 11.2 Class 7 Maths Question 1.
Find the area of each of the following parallelograms:
Solution:
(a) Area of the parallelogram
= base × altitude = 7 cm × 4 cm
= 28 cm2
(b) Area of the parallelogram
= base × altitude = 5 cm × 3 cm
= 15 cm2
(c) Area of the parallelogram
= base × altitude = 2.5 cm × 3.5 cm
= 8.75 cm2
(d) Area of the parallelogram
= base × altitude = 5 cm × 4.8 cm
= 24.0 cm2
(e) Area of the parallelogram
= base × altitude = 2 cm × 4.4 cm
= 8.8 cm2
Ex 11.2 Class 7 Maths Question 2.
Find the area of each of the following triangles:
Solution:
Area of the triangle = (frac{1}{2}) × b × h
= (frac{1}{2}) × 4 cm × 3 cm
= 6m2
(b) Area of the triangle = (frac{1}{2}) × b × h
= (frac{1}{2}) × 5 cm × 3.2 cm
= 8.0 cm2
(c) Area of the triangle = (frac{1}{2}) × b × l
= (frac{1}{2}) × 3 cm × 4 cm
= 6 cm2
(d) Area of the triangle = (frac{1}{2}) × b × h
= (frac{1}{2}) × 3 cm × 2 cm
= 3 cm2
Ex 11.2 Class 7 Maths Question 3.
Find the missing values:
S.No. Base Height Area of the parallelogram (a) 20 cm 246 cm2 (6) 15 cm 154.5 cm2 (c) 8.4 cm 48.72 cm2 (d) 15.6 16.38 cm2
Solution:
(a) Area of the parallelogram =b × h
246 = 20 × h
(b) Area of the parallelogram = b × h
154.5 = b × 15
(c) Area of the parallelogram = b × h
48.72 = b × 8.4
(d) Area of the parallelogram = b × h
16.38 = 15.6 × h
Ex 11.2 Class 7 Maths Question 4.
Find the missing values:
Base Height Area of the triangle 15 cm — 87 cm2 — 31.4 mm 1256 mm2 22 cm — 170.5 cm2
Solution:
(i) Area of the triangle = (frac{1}{2}) × b × h
So, the height =11.6 cm
(ii) Area of the triangle = (frac{1}{2}) × b × h
So, the required base = 80 mm.
(iii) Area of the triangle = (frac{1}{2}) × b × h
So, the required height = 15.5 cm
Ex 11.2 Class 7 Maths Question 5.
PQRS is a parallelogram. QM is the height of Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm
Solution:
(a) Area of the parallelogram PQRS
= SR × QM (∵ Area = Base × Height)
= 12 cm × 7.6 cm
= 91.2 cm2
(b) Area of the parallelogram PQRS
Ex 11.2 Class 7 Maths Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Solution:
Area of the parallelogram ABCD
= AB × DL
⇒ 1470 cm2 = 35 cm × DL
⇒ (frac{1470}{35}) DL
∴ DL = 42 cm
Area of the parallelogram ABCD = AD × BM
1470 cm2 = 49 cm × BM
⇒ (frac{1470}{49}) = 30 cm
∴ BM = 30 cm
Hence, BM = 30 cm and DL = 42 cm
Ex 11.2 Class 7 Maths Question 7.
∆ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ∆ABC. Also find the length of AD.
Solution:
Area of right triangle ABC
Ex 11.2 Class 7 Maths Question 8.
∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?
Solution:
Area of ∆ABC = (frac{1}{2}) × base × height
<!– –>
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# Help me accelerate linear recurrence relation!
## Background
A linear recurrence relation is a description of a sequence, defined as one or more initial terms and a linear formula on last $$\k\$$ terms to calculate the next term. (For the sake of simplicity, we only consider homogeneous relations, i.e. the ones without a constant term in the formula.)
A formal definition of a linear recurrence relation looks like this, where $$\y_n\$$ is the desired sequence (1-based, so it is defined over $$\n\ge 1\$$) and $$\x_i\$$'s and $$\a_i\$$'s are constants:
$$y_n = \begin{cases} x_n, & 1\le n\le k \\ a_1y_{n-1}+a_2y_{n-2}+\cdots+a_ky_{n-k}, & k
In this challenge, we will accelerate this sequence by converting it to a matrix form, so that the $$\n\$$-th term can be found by repeated squaring of the matrix in $$\O(\log n)\$$ steps, followed by inner product with the vector of initial terms.
For example, consider the famous Fibonacci sequence: its recurrence relation is $$\y_n=y_{n-1} + y_{n-2}\$$ with $$\k=2\$$, and let's use the initial values $$\x_1=x_2=1\$$. The recurrence relation can be converted to a matrix form:
$$\begin{bmatrix} y_{n-1} \\ y_{n} \end{bmatrix} = \begin{bmatrix} y_{n-1} \\ y_{n-1}+y_{n-2} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} y_{n-2} \\ y_{n-1} \end{bmatrix}$$
So multiplying the matrix once advances the sequence by one term. Since this holds for any $$\n\$$, it can be extended all the way until we reach the initial terms:
$$\begin{bmatrix} y_{n-1} \\ y_{n} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} y_{n-2} \\ y_{n-1} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}^2\begin{bmatrix} y_{n-3} \\ y_{n-2} \end{bmatrix} \\ = \cdots = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}^{n-2}\begin{bmatrix} y_{1} \\ y_{2} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}^{n-2}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
In general, one way to construct such a matrix is the following:
$$\begin{bmatrix} y_{n-k+1} \\ y_{n-k+2} \\ \vdots \\ y_{n-1} \\ y_{n} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ & \vdots & & & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ a_k & a_{k-1} & a_{k-2} & \cdots & a_1 \end{bmatrix}\begin{bmatrix} y_{n-k} \\ y_{n-k+1} \\ \vdots \\ y_{n-2} \\ y_{n-1} \end{bmatrix}$$
Note that, if you reverse the vectors and the matrix in every dimension, the equation still holds, retaining the property of "advancing a term by matmul-ing once". (Actually any permutation will work, given that the rows and columns of the matrix are permuted in the same way.)
## Challenge
Given the list of coefficients $$\a_1,\cdots,a_k\$$, construct a matrix that represents the recurrence relation (so that its powers can be used to accelerate the computation of $$\n\$$-th term of the sequence).
You can take the coefficients in reverse order, and you can optionally take the value $$\k\$$ as a separate input. $$\k\$$ (the number of terms) is at least 1.
Standard rules apply. The shortest code in bytes wins.
## Test cases
In all cases, any other matrix that can be formed by permuting rows and columns in the same way is also valid.
Input
[1,1]
Output
[[0, 1],
[1, 1]]
Input
[5]
Output
[[5]]
Input
[3, -1, 19]
Output
[[0, 1, 0],
[0, 0, 1],
[19, -1, 3]]
or reversed in both dimensions:
[[3, -1, 19],
[1, 0, 0],
[0, 1, 0]]
or cycled once in both dimensions:
[[3, 19, -1],
[0, 0, 1],
[1, 0, 0]]
etc.
# MATL, 8 7 bytes
-1 byte thanks to @LuisMendo
Xy4LY)i
Takes the coefficients in reverse order
Try it online!
## Explanation
Xy4LY)i
Xy : Create an identity matrix of size equal to input
4LY) : Remove the first row
i : Insert input onto the stack
# J, 10 8 bytes
Returns the matrix reversed in both dimensions.
,}:@=@/:
Try it online!
### How it works
,}:@=@/: input: 3 _1 19
/: indices that sort: 1 0 2
(just to get k different numbers)
=@ self-classify: 1 0 0
0 1 0
0 0 1
}:@ drop last row: 1 0 0
0 1 0
, prepend input: 3 _1 19
1 0 0
0 1 0
# JavaScript (ES6), 36 bytes
a=>a.map((_,i)=>i?a.map(_=>+!--i):a)
Try it online!
Returns:
$$\begin{bmatrix} a_1 & a_2 & a_3 & \cdots & a_{k-1} & a_k \\ 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \end{bmatrix}$$
# Io, 56 bytes
method(a,a map(i,v,if(i<1,a,a map(I,v,if(I==i-1,1,0)))))
Try it online!
## Explanation
method(a, ) // Input an array.
a map(i,v, ) // Map. i = index, v = value
if(i<1, ) // If the indice is 0,
a, // Return the inputted list
a map(I,v, ) // Otherwise, map: (I is the current index)
if(I==i-1, ) // If I == i-1,
1, // Return 1,
0 // Otherwise 0
# APL (Dyalog Unicode), 10 bytes
⊢⍪¯1↓⍋∘.=⍋
Try it online!
Tacit function taking the list of coefficients on the right.
## Explanation
⊢⍪¯1↓⍋∘.=⍋
⍋ ⍋ ⍝ Grade up to obtain a list of k distinct values
∘.= ⍝ Outer product with operation equals (identity matrix)
¯1↓ ⍝ Drop the last row
⊢⍪ ⍝ Prepend the list of coefficients
# Python 2, 46 bytes
lambda l,k:[l]+zip(*[iter(([1]+[0]*k)*~-k)]*k)
Try it online!
Takes input as a tuple l and number of terms k, and outputs with both rows and columns reversed.
The idea is to use the zip/iter trick to create an identity-like matrix by splitting a repeating list into chunks. The is similar to my solution to construct the identity matrix but changed to have one fewer row by changing the inner multiplier k to k-1 (written ~-k).
• Cool, a nice trick of which I was unaware! Commented Jul 31, 2020 at 11:51
# Charcoal, 12 bytes
ＩＥθ⎇κＥθ⁼⊖κμθ
Try it online! Link is to verbose version of code. Produces the "reversed in both directions" output. Works by replacing the first row of a shifted identity matrix with the input. Explanation:
Ｅθ Map over input list
⎇κ If this is not the first row then
Ｅθ Map over input list
⁼⊖κμ Generate a shifted identity matrix
θ Otherwise replace the first row with the input
Ｉ Cast to string for implicit print
• @xash Actually I was thinking of a different permutation, but either way I've probably misunderstood the question. I'll rewrite it to use the reversed in both directions option.
– Neil
Commented Jul 30, 2020 at 12:48
# R, 34 bytes
function(r,k)rbind(diag(k)[-1,],r)
Try it online!
Takes the length as well; the TIO link has a k=length(r) argument so you can just input the recurrence relation.
# Python 3, 60 58 bytes
-2 bytes thanks to @JonathanAllan
lambda a,k:[map(i.__eq__,range(k))for i in range(1,k)]+[a]
Try it online!
Takes the coefficients in reverse order
• -1 byte: lambda a,k,r=range(k):[[i==j for j in r]for i in r[1:]]+[a] Commented Jul 30, 2020 at 19:05
• @ManfP - that's invalid, k is not defined at the point it's being used by range (TIO). Commented Jul 31, 2020 at 1:12
• lambda a,k:[map(i.__eq__,range(k))for i in range(1,k)]+[a] is 58. Returns a list of iterables, and False==0 and True==1. Commented Jul 31, 2020 at 1:25
• @JonathanAllan oh sorry you are totally right, the perils of testing locally with shadowing variable names... Commented Jul 31, 2020 at 10:08
# 05AB1E, 7 bytes
āDδQ\)
Outputs reversed in both dimensions.
Explanation:
ā # Push a list in the range [1,length] (without popping the implicit input-list)
D # Duplicate it
δ # Apply double-vectorized:
Q # Check if it's equal
# (this results in an L by L matrix filled with 0s, with a top-left to
# bottom-right diagonal of 1s; where L is the length of the input-list)
# Pop and push all rows of this matrix separated to the stack
\ # Discard the last row
) # And wrap all list on the stack into a list
# (after which the matrix is output implicitly as result)
# Jelly, 8 bytes
W;J⁼þṖ$$ A monadic Link accepting a list which yields a list of lists in the reversed rows & columns permutation. Try it online! ### How? W;J⁼þṖ$$ - Link: list A e.g. [5,2,5,4]
W - wrap (A) in a list [[5,2,5,4]]
$- last two links as a monad - f(A): J - range of length (A) [1,2,3,4]$ - last two links as a monad - f(J):
Ṗ - pop [1,2,3]
þ - (J) outer product (that) with:
⁼ - equals? [[1,0,0,0],[0,1,0,0],[0,0,1,0]]
; - (W) concatenate (that) [[5,2,5,4],[1,0,0,0],[0,1,0,0],[0,0,1,0]]
# C (gcc), 90 89 80 bytes
Saved 9 bytes thanks to ceilingcat!!!
i;j;f(a,k)int*a;{for(i=k;i--;puts(""))for(j=k;j--;)printf("%d ",i?i-1==j:a[j]);}
Try it online!
Inputs an array of coefficients (in forward order) along with its length.
Prints a matrix that represents the recurrence relation.
# Google Sheets, 52
Closing Parens discounted.
• Input cells is Row 1, starting in column B.
• A2 - =COUNTA(1:1). Rules say that we can take this as input too, so I have discounted this as well. (Our "k")
• A3 - =ArrayFormula(IFERROR(0^MOD(SEQUENCE(A2-1,A2)-1,A2+1)))
The output matrix starts in B1.
## How it works
1. Since this is a spreadsheet, the input cells give us free output too. As long is it's the first row and we end up with a square set of cells, we're good. If this didn't count, we'd have to do this with Column 1 instead to use TRANSPOSE() to copy the input. (Because it's smaller than ArrayFormula())
2. Cache the number of columns in A2
3. Generate a k-1 x k matrix using SEQUENCE. Values are MOD number of columns + 1. (Diagonals are 0, otherwise something else).
4. Since 0^0 is 1 in Sheets, that means this effectively is a Boolean NOT() converted to an integer.
5. IFERROR handles input size of 1. (Output a Blank)
# Pari/GP, 29 bytes
a->matcompanion(x^#a-Pol(a))~
Try it online!
|
# 2.1 Use a General Strategy to Solve Linear Equations
Topics covered in this section are:
## 2.1.1 Solve Linear Equations Using a General Strategy
Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that makes it a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!
### SOLUTION OF AN EQUATION
solution of an equation is a value of a variable that makes a true statement when substituted into the equation.
To determine whether a number is a solution to an equation, we substitute the value for the variable in the equation. If the resulting equation is a true statement, then the number is a solution of the equation.
### HOW TO: Determine Whether a Number is a Solution to an Equation.
1. Substitute the number for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. Determine whether the resulting equation is true.
• If it is true, the number is a solution.
• If it is not true, the number is not a solution.
#### Example 1
Determine whether the values are solutions to the equation: $5y+3=10y-4$.
• $y=\frac{3}{5}$
• $y=\frac{7}{5}$
Solution
Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.
Part 1.
Since $y=\frac{3}{5}$ does not result in a true equation, $y=\frac{3}{5}$ is not a solution to the equation $5y+3=10y-4$.
Part 2.
Since $y=\frac{7}{5}$ results in a true equation, $y=\frac{7}{5}$ is a solution to the equation $5y+3=10y-4$.
There are many types of equations that we will learn to solve. In this section we will focus on a linear equation.
### LINEAR EQUATION
linear equation is an equation in one variable that can be written, where $a$ and $b$ are real numbers and $a≠0$, as:
$ax+b=0$
To solve a linear equation it is a good idea to have an overall strategy that can be used to solve any linear equation. In the next example, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.
#### Example 2
Solve $7(n-3)-8= -15$.
Solution
These steps are summarized in the General Strategy for Solving Linear Equations below.
### HOW TO: Solve linear equations using a general strategy.
1. Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.
2. Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality.
3. Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality.
4. Make the coefficient of the variable term equal to 1.
Use the Multiplication or Division Property of Equality.
State the solution to the equation.
5. Check the solution.
Substitute the solution into the original equation to make sure the result is a true statement.
#### Example 3
Solve: $\frac{2}{3}(3m-6)=5-m$.
Solution
We can solve equations by getting all the variable terms to either side of the equal sign. By collecting the variable terms on the side where the coefficient of the variable is larger, we avoid working with some negatives. This will be a good strategy when we solve inequalities later in this chapter. It also helps us prevent errors with negatives.
#### Example 4
Solve: $4(x-1)-2=5(2x+3)+6$.
Solution
#### Example 5
Solve: $10[3-8(2s-5)]=15(40-5s)$.
Solution
## 2.1.2 Classify Equations
Whether or not an equation is true depends on the value of the variable. The equation $7x+8=-13$ is true when we replace the variable, $x$, with the value $-3$, but not true when we replace $x$ with any other value. An equation like this is called a conditional equation. All the equations we have solved so far are conditional equations.
### CONDITIONAL EQUATION
An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.
Now let’s consider the equation $7y+14=7(y+2)$. Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for $y$.
Solve:
This means that the equation $7y+14=7(y+2)$ is true for any value of $y$. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable is called an identity.
### IDENTITY
An equation that is true for any value of the variable is called an identity.
The solution of an identity is all real numbers.
What happens when we solve the equation $-8z=-8z+9$?
Solving the equation $-8z=-8z+9$ led to the false statement $0≠9$. The equation $-8z=-8z+9$ will not be true for any value of $z$. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.
An equation that is false for all values of the variable is called a contradiction.
The next few examples will ask us to classify an equation as conditional, an identity, or as a contradiction.
#### Example 6
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $6(2n-1)+3=2n-8+5(2n+1)$.
Solution
#### Example 7
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $8+3(a-4)=0$.
Solution
#### Example 8
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $5m+3(9+3m)=2(7m-11)$.
Solution
We summarize the methods for classifying equations in the table.
## 2.1.3 Solve Equations with Fraction or Decimal Coefficients
We could use the General Strategy to solve the next example. This method would work fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.
We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator(LCD) of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but without fractions. This process is called clearing the equation of fractions.
To clear an equation of decimals, we think of all the decimals in their fraction form and then find the LCD of those denominators.
#### Example 9
Solve: $\frac{1}{12}x + \frac{5}{6} = \frac{3}{4}$.
Solution
Notice in the previous example, once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve. We then used the General Strategy for Solving Linear Equations.
### HOW TO: Solve Equations with Fraction or Decimal Coefficients.
1. Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
3. Solve using the General Strategy for Solving Linear Equations.
#### Example 10
Solve: $5=\frac{1}{2}y + \frac{2}{3}y – \frac{3}{4}y$.
Solution
We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.
In the next example, we’ll distribute before we clear the fractions.
#### Example 11
Solve: $\frac{1}{2}(y-5)=\frac{1}{4}(y-1)$.
Solution
An alternate way to solve this equation is to clear the fractions without distributing first. If you multiply the factors correctly, this method will be easier.
When you multiply both sides of an equation by the LCD of the fractions, make sure you multiply each term by the LCD—even if it does not contain a fraction.
#### Example 12
Solve: $\frac{4q+3}{2}+6 = \frac{3q+5}{4}$.
Solution
Some equations have decimals in them. This kind of equation may occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, $0.7 = \frac{7}{10}$ and $0.29= \frac{29}{100}$. So, with an equation with decimals, we can use the same method we used to clear fractions—multiply both sides of the equation by the least common denominator.
The next example uses an equation that is typical of the ones we will see in the money applications in a later section. Notice that we will clear all decimals by multiplying by the LCD of their fraction form.
#### Example 13
Solve: $0.25x+0.05(x+3)=2.85$.
Solution
Look at the decimals and think of the equivalent fractions:
$0.25=\frac{25}{100}$, $0.05=\frac{5}{100}$, $2.85=2\frac{85}{100}$.
Notice, the LCD is $100$. By multiplying by the LCD we will clear the decimals from the equation.
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# Class 7 – Ch- 2 Fractions And Decimals
## Exercise 2.1
1. Solve:
(i) 2 – (3/5)
Solution:-
For subtraction of two unlike fractions, first change them to the like fractions.
LCM of 1, 5 = 5
Now, let us change each of the given fraction into an equivalent fraction having 5 as the denominator.
= [(2/1) × (5/5)] = (10/5)
= [(3/5) × (1/1)] = (3/5)
Now,
= (10/5) – (3/5)
= [(10 – 3)/5]
= (7/5)
(ii) 4 + (7/8)
Solution:-
For addition of two unlike fractions, first change them to the like fractions.
LCM of 1, 8 = 8
Now, let us change each of the given fraction into an equivalent fraction having 8 as the denominator.
= [(4/1) × (8/8)] = (32/8)
= [(7/8) × (1/1)] = (7/8)
Now,
= (32/8) + (7/8)
= [(32 + 7)/8]
= (39/8)
=
(iii) (3/5) + (2/7)
Solution:-
For addition of two unlike fractions, first change them to the like fractions.
LCM of 5, 7 = 35
Now, let us change each of the given fraction into an equivalent fraction having 35 as the denominator.
= [(3/5) × (7/7)] = (21/35)
= [(2/7) × (5/5)] = (10/35)
Now,
= (21/35) + (10/35)
= [(21 + 10)/35]
= (31/35)
(iv) (9/11) – (4/15)
Solution:-
For subtraction of two unlike fractions, first change them to the like fractions.
LCM of 11, 15 = 165
Now, let us change each of the given fraction into an equivalent fraction having 165 as the denominator.
= [(9/11) × (15/15)] = (135/165)
= [(4/15) × (11/11)] = (44/165)
Now,
= (135/165) – (44/165)
= [(135 – 44)/165]
= (91/165)
(v) (7/10) + (2/5) + (3/2)
Solution:-
For addition of two unlike fractions, first change them to the like fractions.
LCM of 10, 5, 2 = 10
Now, let us change each of the given fraction into an equivalent fraction having 35 as the denominator.
= [(7/10) × (1/1)] = (7/10)
= [(2/5) × (2/2)] = (4/10)
= [(3/2) × (5/5)] = (15/10)
Now,
= (7/10) + (4/10) + (15/10)
= [(7 + 4 + 15)/10]
= (26/10)
= (13/5)
=
(vi)
Solution:-
First convert mixed fraction into improper fraction,
== 8/3
= 3 ½ = 7/2
For addition of two unlike fractions, first change them to the like fractions.
LCM of 3, 2 = 6
Now, let us change each of the given fraction into an equivalent fraction having 6 as the denominator.
= [(8/3) × (2/2)] = (16/6)
= [(7/2) × (3/3)] = (21/6)
Now,
= (16/6) + (21/6)
= [(16 + 21)/6]
= (37/6)
=
(vii) -
Solution:-
First convert mixed fraction into improper fraction,
= 8 ½ = 17/2
== 29/8
For Subtraction of two unlike fractions, first change them to the like fractions.
LCM of 2, 8 = 8
Now, let us change each of the given fraction into an equivalent fraction having 35 as the denominator.
= [(17/2) × (4/4)] = (68/8)
= [(29/8) × (1/1)] = (29/8)
Now,
= (68/8) – (29/8)
= [(68 – 29)/8]
= (39/8)
=
2. Arrange the following in descending order:
(i) 2/9, 2/3, 8/21
Solution:-
LCM of 9, 3, 21 = 63
Now, let us change each of the given fraction into an equivalent fraction having 63 as the denominator.
[(2/9) × (7/7)] = (14/63)[(2/3) × (21/21)] = (42/63)[(8/21) × (3/3)] = (24/63)
Clearly,
(42/63) > (24/63) > (14/63)
Hence,
(2/3) > (8/21) > (2/9)
Hence, the given fractions in descending order are (2/3), (8/21), (2/9)
(ii) 1/5, 3/7, 7/10
Solution:-
LCM of 5, 7, 10 = 70
Now, let us change each of the given fraction into an equivalent fraction having 70 as the denominator.
[(1/5) × (14/14)] = (14/70)[(3/7) × (10/10)] = (30/70)[(7/10) × (7/7)] = (49/70)
Clearly,
(49/70) > (30/70) > (14/70)
Hence,
(7/10) > (3/7) > (1/5)
Hence, the given fractions in descending order are (7/10), (3/7), (1/5)
3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?
4/11 9/11 2/11 3/11 5/11 7/11 8/11 1/11 6/11
Solution:-
4. A rectangular sheet of paper is 12 ½ cm long and 10 2/3 cm wide. Find its perimeter.
Solution:-
From the question, it is given that,
Length = 12 ½ cm = 25/2 cm
10 2/3 cm = 32/3 cm
We know that,
Perimeter of the rectangle = 2 × (length + breadth)
= 2 × [(25/2) + (32/3)]
= 2 × {[(25 × 3) + (32 × 2)]/6}
= 2 × [(75 + 64)/6]
= 2 × [139/6]
= 139/3 cm
Hence, the perimeter of the sheet of paper is
46 1/3cm
5. Find the perimeters of (i) Triangle ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
7. Ritu ate (3/5) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?
Solution:-
From the question, it is given that,
Part of apple eaten by Ritu is = (3/5)
Part of apple eaten by Somu is = 1 – Part of apple eaten by Ritu
= 1 – (3/5)
The LCM of 1, 5 = 5
Now, let us change each of the given fraction into an equivalent fraction having 10 as the denominator.
= [(1/1) × (5/5)] – [(3/5) × (1/1)]
= (5/5) – (3/5)
= (5 – 3)/5
= 2/5
∴ Part of apple eaten by Somu is (2/5)
So, (3/5) > (2/5) hence, Ritu ate larger size of apple.
Now, the difference between the 32 shares = (3/5) – (2/5)
= (3 – 2)/5
= 1/5
Thus, Ritu’s share is larger than share of Somu by (1/5)
8. Michael finished colouring a picture in (7/12) hour. Vaibhav finished colouring the same picture in (3/4) hour. Who worked longer? By what fraction was it longer?
Solution:-
From the question, it is given that,
Time taken by the Michael to colour the picture is = (7/12)
Time taken by the Vaibhav to colour the picture is = (3/4)
The LCM of 12, 4 = 12
Now, let us change each of the given fraction into an equivalent fraction having 12 as the denominator.
(7/12) = (7/12) × (1/1) = 7/12
(3/4) = (3/4) × (3/3) = 9/12
Clearly, (7/12) < (9/12)
Hence, (7/12) < (3/4)
Thus, Vaibhav worked for longer time.
So, Vaibhav worked longer time by = (3/4) – (7/12)
= (9/12) – (7/12)
= (9 – 7)/12
= (2/12)
= (1/6) of an hour.
## Exercise 2.2
1. Which of the drawings (a) to (d) show:
(i) 2 × (1/5) (ii) 2 × ½ (iii) 3 × (2/3) (iv) 3 × ¼
Solution:-
(i) 2 × (1/5) represents the addition of 2 figures, each represents 1 shaded part out of the given 5 equal parts.
∴ 2 × (1/5) is represented by fig (d).
(ii) 2 × ½ represents the addition of 2 figures, each represents 1 shaded part out of the given 2 equal parts.
∴ 2 × ½ is represented by fig (b).
(iii) 3 × (2/3) represents the addition of 3 figures, each represents 2 shaded part out of the given 3 equal parts.
∴ 3 × (2/3) is represented by fig (a).
(iii) 3 × ¼ represents the addition of 3 figures, each represents 1 shaded part out of the given 4 equal parts.
∴ 3 × ¼ is represented by fig (c).
2. Some pictures (a) to (c) are given below. Tell which of them show:
(i) 3 × (1/5) = (3/5) (ii) 2 × (1/3) = (2/3) (iii) 3 × (3/4) = 2 ¼
Solution:-
(i) 3 × (1/5) represents the addition of 3 figures, each represents 1 shaded part out of the given 5 equal parts and (3/5) represents 3 shaded parts out of 5 equal parts.
∴ 3 × (1/5) = (3/5) is represented by fig (c).
(ii) 2 × (1/3) represents the addition of 2 figures, each represents 1 shaded part out of the given 3 equal parts and (2/3) represents 2 shaded parts out of 3 equal parts.
∴ 2 × (1/3) = (2/3) is represented by fig (a).
(iii) 3 × (3/4) represents the addition of 3 figures, each represents 3 shaded part out of the given 4 equal parts and 2 ¼ represents 2 fully and 1 figure having 1 part as shaded out of 4 equal parts.
∴ 3 × (3/4) = 2 ¼ is represented by fig (b).
3. Multiply and reduce to lowest form and convert into a mixed fraction:
(i) 7 × (3/5)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (7/1) × (3/5)
= (7 × 3)/ (1 × 5)
= (21/5)
=
(ii) 4 × (1/3)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (4/1) × (1/3)
= (4 × 1)/ (1 × 3)
= (4/3)
=1 1/2
iii) 2 × (6/7)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/1) × (6/7)
= (2 × 6)/ (1 × 7)
= (12/7)
=1 5/7
(iv) 5 × (2/9)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/1) × (2/9)
= (5 × 2)/ (1 × 9)
= (10/9)
=1 1/9
(v) (2/3) × 4
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/3) × (4/1)
= (2 × 4)/ (3 × 1)
= (8/3)
=2 2/3
(vi) (5/2) × 6
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/2) × (6/1)
= (5 × 6)/ (2 × 1)
= (30/2)
= 15
(vii) 11 × (4/7)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (11/1) × (4/7)
= (11 × 4)/ (1 × 7)
= (44/7)
=6 2/7
(viii) 20 × (4/5)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (20/1) × (4/5)
= (20 × 4)/ (1 × 5)
= (80/5)
= 16
(ix) 13 × (1/3)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (13/1) × (1/3)
= (13 × 1)/ (1 × 3)
= (13/3)
=4 1/2
(x) 15 × (3/5)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (15/1) × (3/5)
= (15 × 3)/ (1 × 5)
= (45/5)
= 9
(i) $frac{1}{2}$ of the circles in box (a)
(ii) $frac{2}{3}$ of the circles in box (b)
(iii) $frac{3}{5}$ of the circles in box (c)
5. Find:
6. Multiply and express as a mixed fraction:
(a) 3 × 5 1/5
Solution:-
First convert the given mixed fraction into improper fraction.
== 26/5
Now,
= 3 × (26/5)
= 78/5
=15 3/5
(b) 5 × 6 ¾
Solution:-
First convert the given mixed fraction into improper fraction.
= 6 ¾ = 27/4
Now,
= 5 × (27/4)
= 135/4
= 33 ¾
(c) 7 × 2 ¼
Solution:-
First convert the given mixed fraction into improper fraction.
= 2 ¼ = 9/4
Now,
= 7 × (9/4)
= 63/4
= 15 ¾
(d) 4 ×
Solution:-
First convert the given mixed fraction into improper fraction.
== 19/3
Now,
= 4 × (19/3)
= 76/3
=
(e) 3 ¼ × 6
Solution:-
First convert the given mixed fraction into improper fraction.
= 3 ¼ = 13/4
Now,
= (13/4) × 6
= (13/2) × 3
= 39/2
= 19 ½
(f) × 8
Solution:-
First convert the given mixed fraction into improper fraction.
== 17/5
Now,
= (17/5) × 8
= 136/5
=
7. Find:
(a) ½ of (i) 2 ¾ (ii)
Solution:-
(i) 2 ¾
First convert the given mixed fraction into improper fraction.
= 2 ¾ = 11/4
Now,
= ½ × 11/4
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= ½ × (11/4)
= (1 × 11)/ (2 × 4)
= (11/8)
=
(ii)
First convert the given mixed fraction into improper fraction.
== 38/9
Now,
= ½ × (38/9)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= ½ × (38/9)
= (1 × 38)/ (2 × 9)
= (38/18)
= 19/9
=
(b) 5/8 of (i) (ii)
Solution:-
(i)
First convert the given mixed fraction into improper fraction.
== 23/6
Now,
= (5/8) × (23/6)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/8) × (23/6)
= (5 × 23)/ (8 × 6)
= (115/48)
=
(ii)
First convert the given mixed fraction into improper fraction.
== 29/3
Now,
= (5/8) × (29/3)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/8) × (29/3)
= (5 × 29)/ (8 × 3)
= (145/24)
=
8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 liters water. Vidya consumed 2/5 of the water. Pratap consumed the remaining water.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
Solution:-
(i) From the question, it is given that,
Amount of water in the water bottle = 5 liters
Amount of water consumed by Vidya = 2/5 of 5 liters
= (2/5) × 5
= 2 liters
So, the total amount of water drank by Vidya is 2 liters
(ii) From the question, it is given that,
Amount of water in the water bottle = 5 liters
Then,
Amount of water consumed by Pratap = (1 – water consumed by Vidya)
= (1 – (2/5))
= (5-2)/5
= 3/5
∴ Total amount of water consumed by Pratap = 3/5 of 5 liters
= (3/5) × 5
= 3 liters
So, the total amount of water drank by Pratap is 3 liters
## Exercise 2.3
Question 1.
Find:
Solution:
Question 2.
Multiply and reduce to lowest form (if possible):
Solution:
Question 3.
Multiply the following fractions:
Solution:
Question 4.
Which is greater:
(i) (2/7) of (3/4) or (3/5) of (5/8)
Solution:-
We have,
= (2/7) × (3/4) and (3/5) × (5/8)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/7) × (3/4)
= (2 × 3)/ (7 × 4)
= (1 × 3)/ (7 × 2)
= (3/14) … [i]
And,
= (3/5) × (5/8)
= (3 × 5)/ (5 × 8)
= (3 × 1)/ (1 × 8)
= (3/8) … [ii]
Now, convert [i] and [ii] into like fractions,
LCM of 14 and 8 is 56
Now, let us change each of the given fraction into an equivalent fraction having 56 as the denominator.
[(3/14) × (4/4)] = (12/56)[(3/8) × (7/7)] = (21/56)
Clearly,
(12/56) < (21/56)
Hence,
(3/14) < (3/8)
(ii) (1/2) of (6/7) or (2/3) of (3/7)
Solution:-
We have,
= (1/2) × (6/7) and (2/3) × (3/7)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (1/2) × (6/7)
= (1 × 6)/ (2 × 7)
= (1 × 3)/ (1 × 7)
= (3/7) … [i]
And,
= (2/3) × (3/7)
= (2 × 3)/ (3 × 7)
= (2 × 1)/ (1 × 7)
= (2/7) … [ii]
By comparing [i] and [ii],
Clearly,
(3/7) > (2/7)
Question 5.
Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is ¾ m. Find the distance between the first and the last sapling.
Solution:-
From the question, it is given that,
The distance between two adjacent saplings = ¾ m
Number of saplings planted by Saili in a row = 4
Then, number of gap in saplings = ¾ × 4
= 3
∴The distance between the first and the last saplings = 3 × ¾
= (9/4) m
= 2 ¼ m
Hence, the distance between the first and the last saplings is 2 ¼ m.
Question 6.
Lipika reads a book for 1 ¾ hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution:-
From the question, it is given that,
Lipika reads the book for = 1 ¾ hours every day = 7/4 hours
Number of days she took to read the entire book = 6 days
∴Total number of hours required by her to complete the book = (7/4) × 6
= (7/2) × 3
= 21/2
= 10 ½ hours
Hence, the total number of hours required by her to complete the book is 10 ½ hours.
Question 7.
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 ¾ litres of petrol.
Solution:-
From the question, it is given that,
The total number of distance travelled by a car in 1 liter of petrol = 16 km
Then,
Total quantity of petrol = 2 ¾ liter = 11/4 liters
Total number of distance travelled by car in 11/4 liters of petrol = (11/4) × 16
= 11 × 4
= 44 km
∴Total number of distance travelled by car in 11/4 liters of petrol is 44 km.
Question 8.
(a) (i) provide the number in the box [ ], such that (2/3) × [ ] = (10/30)
Solution:-
Let the required number be x,
Then,
= (2/3) × (x) = (10/30)
By cross multiplication,
= x = (10/30) × (3/2)
= x = (10 × 3) / (30 × 2)
= x = (5 × 1) / (10 × 1)
= x = 5/10
∴The required number in the box is (5/20)
(ii) The simplest form of the number obtained in [ ] is
Solution:-
The number in the box is 5/10
Then,
The simplest form of 5/10 is ½
(b) (i) provide the number in the box [ ], such that (3/5) × [ ] = (24/75)
Solution:-
Let the required number be x,
Then,
= (3/5) × (x) = (24/75)
By cross multiplication,
= x = (24/75) × (5/3)
= x = (24 × 5) / (75 × 3)
= x = (8 × 1) / (15 × 1)
= x = 8/15
∴The required number in the box is (8/15)
(ii) The simplest form of the number obtained in [ ] is
Solution:-
The number in the box is 8/15
Then,
The simplest form of 8/15 is 8/15
## Exercise 2.4
1. Find:
(i) 12 ÷ 3/4
Solution:-
We have,
= 12 × reciprocal of ¾
= 12 × (4/3)
= 4 × 4
= 16
(ii) 14 ÷ (5/6)
Solution:-
We have,
= 14 × reciprocal of (5/6)
= 14 × (6/5)
= 84/5
(iii) 8 ÷ (7/3)
Solution:-
We have,
= 8 × reciprocal of (7/3)
= 8 × (3/7)
(iv) 4 ÷ (8/3)
Solution:-
We have,
= 4 × reciprocal of (8/3)
= 4 × (3/8)
= 1 × (3/2)
= 3/2
(v) 3 ÷
Solution:-
While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction
We have,
== 7/3
Then,
= 3 ÷ (7/3)
= 3 × reciprocal of (7/3)
= 3 × (3/7)
= 9/7
(vi) 5 ÷
Solution:-
While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction
We have,
== 25/7
Then,
= 5 ÷ (25/7)
= 5 × reciprocal of (25/7)
= 5 × (7/25)
= 1 × (7/5)
= 7/5
2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
Solution:-
(i) Reciprocal of $frac{3}{7}=frac{7}{3}$, which is improper fraction.
(ii) Reciprocal of $frac{5}{8}=frac{8}{5}$, which is improper fraction.
(iii) Reciprocal of $frac{9}{7}=frac{7}{9}$, which is proper fraction.
(iv) Reciprocal of $frac{6}{5}=frac{5}{6}$, which is proper fraction.
(vi) Reciprocal of $frac{12}{7}=frac{7}{12}$, which is proper fraction.
(vi) Reciprocal of $frac{1}{8}=8$, which is whole number.
(vii) Reciprocal of $frac{1}{11}=11$, which is whole number.
3. Find:
(i) (7/3) ÷ 2
Solution:-
We have,
= (7/3) × reciprocal of 2
= (7/3) × (1/2)
= (7 × 1) / (3 × 2)
= 7/6
=
(ii) (4/9) ÷ 5
Solution:-
We have,
= (4/9) × reciprocal of 5
= (4/9) × (1/5)
= (4 × 1) / (9 × 5)
= 4/45
(iii) (6/13) ÷ 7
Solution:-
We have,
= (6/13) × reciprocal of 7
= (6/13) × (1/7)
= (6 × 1) / (13 × 7)
= 6/91
(iv) ÷ 3
Solution:-
First covert the mixed fraction into improper fraction.
We have,
== 13/3
Then,
= (13/3) × reciprocal of 3
= (13/3) × (1/3)
= (13 × 1) / (3 × 3)
= 13/9
(v) 3 ½ ÷ 4
Solution:-
First covert the mixed fraction into improper fraction.
We have,
= 3 ½ = 7/2
Then,
= (7/2) × reciprocal of 4
= (7/2) × (1/4)
= (7 × 1) / (2 × 4)
= 7/8
(vi) ÷ 7
Solution:-
First covert the mixed fraction into improper fraction.
We have,
== 31/7
Then,
= (31/7) × reciprocal of 7
= (31/7) × (1/7)
= (31 × 1) / (7 × 7)
= 31/49
4. Find:
(i) (2/5) ÷ (½)
Solution:-
We have,
= (2/5) × reciprocal of ½
= (2/5) × (2/1)
= (2 × 2) / (5 × 1)
= 4/5
(ii) (4/9) ÷ (2/3)
Solution:-
We have,
= (4/9) × reciprocal of (2/3)
= (4/9) × (3/2)
= (4 × 3) / (9 × 2)
= (2 × 1) / (3 × 1)
= 2/3
(iii) (3/7) ÷ (8/7)
Solution:-
We have,
= (3/7) × reciprocal of (8/7)
= (3/7) × (7/8)
= (3 × 7) / (7 × 8)
= (3 × 1) / (1 × 8)
= 3/8
(iv) ÷ (3/5)
Solution:-
First covert the mixed fraction into improper fraction.
We have,
== 7/3
Then,
= (7/3) × reciprocal of (3/5)
= (7/3) × (5/3)
= (7 × 5) / (3 × 3)
= 35/9
(v) 3 ½ ÷ (8/3)
Solution:-
First covert the mixed fraction into improper fraction.
We have,
= 3 ½ = 7/2
Then,
= (7/2) × reciprocal of (8/3)
= (7/2) × (3/8)
= (7 × 3) / (2 × 8)
= 21/16
(vi) (2/5) ÷ 1 ½
Solution:-
First covert the mixed fraction into improper fraction.
We have,
= 1 ½ = 3/2
Then,
= (2/5) × reciprocal of (3/2)
= (2/5) × (2/3)
= (2 × 2) / (5 × 3)
= 4/15
(vii) ÷
Solution:-
First covert the mixed fraction into improper fraction.
We have,
== 16/5
== 5/3
Then,
= (16/5) × reciprocal of (5/3)
= (16/5) × (3/5)
= (16 × 3) / (5 × 5)
= 48/25
(viii) ÷
Solution:-
First covert the mixed fraction into improper fraction.
We have,
== 11/5
== 6/5
Then,
= (11/5) × reciprocal of (6/5)
= (11/5) × (5/6)
= (11 × 5) / (5 × 6)
= (11 × 1) / (1 × 6)
= 11/6
## Exercise 2.5
1. Which is greater?
(i) 0.5 or 0.05
Solution:-
By comparing whole number, 0 = 0
By comparing the tenths place digit, 5 > 0
∴ 0.5 > 0.05
(ii) 0.7 or 0.5
Solution:-
By comparing whole number, 0 = 0
By comparing the tenths place digit, 7 > 5
∴ 0.7 > 0.5
(iii) 7 or 0.7
Solution:-
By comparing whole number, 7 > 0
∴ 7 > 0.7
(iv) 1.37 or 1.49
Solution:-
By comparing whole number, 1 = 1
By comparing the tenths place digit, 3 < 4
∴ 1.37 < 1.49
(v) 2.03 or 2.30
Solution:-
By comparing whole number, 2 = 2
By comparing the tenths place digit, 0 < 3
∴ 2.03 < 2.30
(vi) 0.8 or 0.88
Solution:-
By comparing whole number, 0 = 0
By comparing the tenths place digit, 8 = 8
By comparing the hundredths place digit, 0 < 8
∴ 0.8 < 0.88
2. Express as rupees as decimals:
(i) 7 paise
Solution:-
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 7 paise = Rs. (7/100)
= Rs. 0.07
(ii) 7 rupees 7 paise
Solution:-
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 7 rupees 7 paise = Rs. 7 + Rs. (7/100)
= Rs. 7 + Rs. 0.07
= Rs. 7.07
(iii) 77 rupees 77 paise
Solution:-
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 77 rupees 77 paise = Rs. 77 + Rs. (77/100)
= Rs. 77 + Rs. 0.77
= Rs. 77.77
(iv) 50 paise
Solution:-
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 50 paise = Rs. (50/100)
= Rs. 0.50
(v) 235 paise
Solution:-
We know that,
= Rs. 1 = 100 paise
= 1 paise = Rs. (1/100)
∴ 235 paise = Rs. (235/100)
= Rs. 2.35
3. (i) Express 5 cm in meter and kilometer
Solution:-
We know that,
= 1 meter = 100 cm
Then,
= 1 cm = (1/100) m
= 5 cm = (5/100)
= 0.05 m
Now,
= 1 km = 1000 m
Then,
= 1 m = (1/1000) km
= 0.05 m = (0.05/1000)
= 0. 00005 km
(i) Express 35 mm in cm, m and km
Solution:-
We know that,
= 1 cm = 10 mm
Then,
= 1 mm = (1/10) cm
= 35 mm = (35/10) cm
= 3.5 cm
And,
= 1 meter = 100 cm
Then,
= 1 cm = (1/100) m
= 3.5 cm = (3.5/100) m
= (35/1000) m
= 0.035 m
Now,
= 1 km = 1000 m
Then,
= 1 m = (1/1000) km
= 0.035 m = (0.035/1000)
= 0. 000035 km
4. Express in kg:
(i) 200 g
Solution:-
We know that,
= 1 kg = 1000 g
Then,
= 1 g = (1/1000) kg
= 200 g = (200/1000) kg
= (2/10)
= 0.2 kg
(ii) 3470 g
Solution:-
We know that,
= 1 kg = 1000 g
Then,
= 1 g = (1/1000) kg
= 3470 g = (3470/1000) kg
= (3470/100)
= 3.470 kg
(ii) 4 kg 8 g
Solution:-
We know that,
= 1 kg = 1000 g
Then,
= 1 g = (1/1000) kg
= 4 kg 8 g = 4 kg + (8/1000) kg
= 4 kg + 0.008
= 4.008 kg
5. Write the following decimal numbers in the expanded form:
(i) 20.03
Solution:-
We have,
20.03 = (2 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))
(ii) 2.03
Solution:-
We have,
2.03 = (2 × 1) + (0 × (1/10)) + (3 × (1/100))
(iii) 200.03
Solution:-
We have,
200.03 = (2 × 100) + (0 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))
(iv) 2.034
Solution:-
We have,
2.034 = (2 × 1) + (0 × (1/10)) + (3 × (1/100)) + (4 × (1/1000))
6. Write the place value of 2 in the following decimal numbers:
(i) 2.56
Solution:-
From the question, we observe that,
The place value of 2 in 2.56 is ones
(ii) 21.37
Solution:-
From the question, we observe that,
The place value of 2 in 21.37 is tens
(iii) 10.25
Solution:-
From the question, we observe that,
The place value of 2 in 10.25 is tenths.
(iv) 9.42
Solution:-
From the question, we observe that,
The place value of 2 in 9.42 is hundredth.
(v) 63.352
Solution:-
From the question, we observe that,
The place value of 2 in 63.352 is thousandth.
7. Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?
Solution:-
From the question, it is given that,
Distance travelled by Dinesh = AB + BC
= 7.5 + 12.7
= 20.2 km
∴Dinesh travelled 20.2 km
Distance travelled by Ayub = AD + DC
= 9.3 + 11.8
= 21.1 km
∴Ayub travelled 21.1km
Clearly, Ayub travelled more distance by = (21.1 – 20.2)
= 0.9 km
∴Ayub travelled 0.9 km more than Dinesh.
8. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Solution:-
From the question, it is given that,
Fruits bought by Shyama = 5 kg 300 g
= 5 kg + (300/1000) kg
= 5 kg + 0.3 kg
= 5.3 kg
Fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g
= (4 + (800/1000)) + (4 + (150/1000))
= (4 + 0.8) kg + (4 + .150) kg
= 4.8 kg + 4.150kg
= 8.950 kg
So, Sarala bought more fruits.
9. How much less is 28 km than 42.6 km?
Solution:-
Now, we have to find the difference of 42.6 km and 28 km
42.6
-28.0
14.6
∴ 14.6 km less is 28 km than 42.6 km.
## Exercise 2.6
Find:
(i) 0.2 × 6
Solution:-
We have,
= (2/10) × 6
= (12/10)
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
Then,
= 1.2
(ii) 8 × 4.6
Solution:-
We have,
= (8) × (46/10)
= (368/10)
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
Then,
= 36.8
(iii) 2.71 × 5
Solution:-
We have,
= (271/100) × 5
= (1355/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 13.55
(iv) 20.1 × 4
Solution:-
We have,
= (201/10) × 4
= (804/10)
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
Then,
= 80.4
(v) 0.05 × 7
Solution:-
We have,
= (5/100) × 7
= (35/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 0.35
(vi) 211.02 × 4
Solution:-
We have,
= (21102/100) × 4
= (84408/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 844.08
(vii) 2 × 0.86
Solution:-
We have,
= (2) × (86/100)
= (172/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 1.72
2. Find the area of rectangle whose length is 5.7cm and breadth is 3 cm.
Solution:-
From the question, it is given that,
Length of the rectangle = 5.7 cm
Breadth of the rectangle = 3 cm
Then,
Area of the rectangle = length × Breadth
= 5.7 × 3
= 17.1 cm2
3. Find:
(i) 1.3 × 10
Solution:-
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
We have,
= 1.3 × 10 = 13
(ii) 36.8 × 10
Solution:-
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
We have,
= 36.8 × 10 = 368
(iii) 153.7 × 10
Solution:-
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
We have,
= 153.7 × 10 = 1537
(iv) 168.07 × 10
Solution:-
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
We have,
= 168.07 × 10 = 1680.7
(v) 31.1 × 100
Solution:-
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
We have,
= 31.1 × 100 = 3110
(vi) 156.1 × 100
Solution:-
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
We have,
= 156.1 × 100 = 15610
(vii) 3.62 × 100
Solution:-
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
We have,
= 3.62 × 100 = 362
(viii) 43.07 × 100
Solution:-
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
We have,
= 43.07 × 100 = 4307
(ix) 0.5 × 10
Solution:-
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
We have,
= 0.5 × 10 = 5
(x) 0.08 × 10
Solution:-
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
We have,
= 0.08 × 10 = 0.8
(xi) 0.9 × 100
Solution:-
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
We have,
= 0.9 × 100 = 90
(xii) 0.03 × 1000
Solution:-
On multiplying a decimal by 1000, the decimal point is shifted to the right by three places.
We have,
= 0.03 × 1000 = 30
4. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Solution:-
From the question, it is given that,
Distance covered by two-wheeler in 1L of petrol = 55.3 km
Then,
Distance covered by two wheeler in 10L of petrol = (10 × 55.3)
= 553 km
∴Two-wheeler covers a distance in 10L of petrol is 553 km.
5. Find:
(i) 2.5 × 0.3
Solution:-
We have,
= (25/10) × (3/10)
= (75/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 0.75
(ii) 0.1 × 51.7
Solution:-
We have,
= (1/10) × (517/10)
= (517/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 5.17
(iii) 0.2 × 316.8
Solution:-
We have,
= (2/10) × (3168/10)
= (6336/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 63.36
(iv) 1.3 × 3.1
Solution:-
We have,
= (13/10) × (31/10)
= (403/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 4.03
(v) 0.5 × 0.05
Solution:-
We have,
= (5/10) × (5/100)
= (25/1000)
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
Then,
= 0.025
(vi) 11.2 × 0.15
Solution:-
We have,
= (112/10) × (15/100)
= (1680/1000)
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
Then,
= 1.680
(vii) 1.07 × 0.02
Solution:-
We have,
= (107/100) × (2/100)
= (214/10000)
On dividing a decimal by 10000, the decimal point is shifted to the left by four places.
Then,
= 0.0214
(viii) 10.05 × 1.05
Solution:-
We have,
= (1005/100) × (105/100)
= (105525/10000)
On dividing a decimal by 10000, the decimal point is shifted to the left by four places.
Then,
= 10.5525
(ix) 101.01 × 0.01
Solution:-
We have,
= (10101/100) × (1/100)
= (10101/10000)
On dividing a decimal by 10000, the decimal point is shifted to the left by four places.
Then,
= 1.0101
(x) 100.01 × 1.1
Solution:-
We have,
= (10001/100) × (11/10)
= (110011/1000)
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
Then,
= 110.011
## Exercise 2.7
1. Find:
(i) 0.4 ÷ 2
Solution:-
We have,
= (4/10) ÷ 2
Then,
= (4/10) × (1/2)
= (2/10) × (1/1)
= (2/10)
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
Then,
= 0.2
(ii) 0.35 ÷ 5
Solution:-
We have,
= (35/100) ÷ 5
Then,
= (35/100) × (1/5)
= (7/100) × (1/1)
= (7/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 0.07
(iii) 2.48 ÷ 4
Solution:-
We have,
= (248/100) ÷ 4
Then,
= (248/100) × (1/4)
= (62/100) × (1/1)
= (62/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 0.62
(iv) 65.4 ÷ 6
Solution:-
We have,
= (654/10) ÷ 6
Then,
= (654/10) × (1/6)
= (109/10) × (1/1)
= (109/10)
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
Then,
= 10.9
(v) 651.2 ÷ 4
Solution:-
We have,
= (6512/10) ÷ 4
Then,
= (6512/10) × (1/4)
= (1628/10) × (1/1)
= (1628/10)
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
Then,
= 162.8
(vi) 14.49 ÷ 7
Solution:-
We have,
= (1449/100) ÷ 7
Then,
= (1449/100) × (1/7)
= (207/100) × (1/1)
= (207/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 2.07
(vii) 3.96 ÷ 4
Solution:-
We have,
= (396/100) ÷ 4
Then,
= (396/100) × (1/4)
= (99/100) × (1/1)
= (99/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 0.99
(viii) 0.80 ÷ 5
Solution:-
We have,
= (80/100) ÷ 5
Then,
= (80/100) × (1/5)
= (16/100) × (1/1)
= (16/100)
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then,
= 0.16
2. Find:
(i) 4.8 ÷ 10
Solution:-
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
We have,
= 4.8 ÷ 10
= (4.8/10)
= 0.48
(ii) 52.5 ÷ 10
Solution:-
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
We have,
= 52.5 ÷ 10
= (52.5/10)
= 5.25
(iii) 0.7 ÷ 10
Solution:-
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
We have,
= 0.7 ÷ 10
= (0.7/10)
= 0.07
(iv) 33.1 ÷ 10
Solution:-
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
We have,
= 33.1 ÷ 10
= (33.1/10)
= 3.31
(v) 272.23 ÷ 10
Solution:-
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
We have,
= 272.23 ÷ 10
= (272.23/10)
= 27.223
(vi) 0.56 ÷ 10
Solution:-
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
We have,
= 0.56 ÷ 10
= (0.56/10)
= 0.056
(vii) 3.97 ÷10
Solution:-
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
We have,
= 3.97 ÷ 10
= (3.97/10)
= 0.397
3. Find:
(i) 2.7 ÷ 100
Solution:-
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
We have,
= 2.7 ÷ 100
= (2.7/100)
= 0.027
(ii) 0.3 ÷ 100
Solution:-
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
We have,
= 0.3 ÷ 100
= (0.3/100)
= 0.003
(iii) 0.78 ÷ 100
Solution:-
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
We have,
= 0.78 ÷ 100
= (0.78/100)
= 0.0078
(iv) 432.6 ÷ 100
Solution:-
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
We have,
= 432.6 ÷ 100
= (432.6/100)
= 4.326
(v) 23.6 ÷100
Solution:-
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
We have,
= 23.6 ÷ 100
= (23.6/100)
= 0.236
(vi) 98.53 ÷ 100
Solution:-
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
We have,
= 98.53 ÷ 100
= (98.53/100)
= 0.9853
4. Find:
(i) 7.9 ÷ 1000
Solution:-
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
We have,
= 7.9 ÷ 1000
= (7.9/1000)
= 0.0079
(ii) 26.3 ÷ 1000
Solution:-
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
We have,
= 26.3 ÷ 1000
= (26.3/1000)
= 0.0263
(iii) 38.53 ÷ 1000
Solution:-
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
We have,
= 38.53 ÷ 1000
= (38.53/1000)
= 0.03853
(iv) 128.9 ÷ 1000
Solution:-
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
We have,
= 128.9 ÷ 1000
= (128.9/1000)
= 0.1289
(v) 0.5 ÷ 1000
Solution:-
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
We have,
= 0.5 ÷ 1000
= (0.5/1000)
= 0.0005
5. Find:
(i) 7 ÷ 3.5
Solution:-
We have,
= 7 ÷ (35/10)
= 7 × (10/35)
= 1 × (10/5)
= 2
(ii) 36 ÷ 0.2
Solution:-
We have,
= 36 ÷ (2/10)
= 36 × (10/2)
= 18 × 10
= 180
(iii) 3.25 ÷ 0.5
Solution:-
We have,
= (325/100) ÷ (5/10)
= (325/100) × (10/5)
= (325 × 10)/ (100 × 5)
= (65 × 1)/ (10 × 1)
= 65/10
= 6.5
(iv) 30.94 ÷ 0.7
Solution:-
We have,
= (3094/100) ÷ (7/10)
= (3094/100) × (10/7)
= (3094 × 10)/ (100 × 7)
= (442 × 1)/ (10 × 1)
= 442/10
= 44.2
(v) 0.5 ÷ 0.25
Solution:-
We have,
= (5/10) ÷ (25/100)
= (5/10) × (100/25)
= (5 × 100)/ (10 × 25)
= (1 × 10)/ (1 × 5)
= 10/5
= 2
(vi) 7.75 ÷ 0.25
Solution:-
We have,
= (775/100) ÷ (25/100)
= (775/100) × (100/25)
= (775 × 100)/ (100 × 25)
= (155 × 1)/ (1 × 5)
= (31 × 1)/ (1 × 1)
= 31
(vii) 76.5 ÷ 0.15
Solution:-
We have,
= (765/10) ÷ (15/100)
= (765/10) × (100/15)
= (765 × 100)/ (10 × 15)
= (51 × 10)/ (1 × 1)
= 510
(viii) 37.8 ÷ 1.4
Solution:-
We have,
= (378/10) ÷ (14/10)
= (378/10) × (10/14)
= (378 × 10)/ (10 × 14)
= (27 × 1)/ (1 × 1)
= 27
(ix) 2.73 ÷ 1.3
Solution:-
We have,
= (273/100) ÷ (13/10)
= (273/100) × (10/13)
= (273 × 10)/ (100 × 13)
= (21 × 1)/ (10 × 1)
= 21/10
= 2.1
6. A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?
Solution:-
From the question, it is given that,
Total distance covered by vehicle in 2.4 litres of petrol = 43.2 km
Then,
Distance covered in 1 litre of petrol = 43.2 ÷ 2.4
= (432/10) ÷ (24/10)
= (432/10) × (10/24)
= (432 × 10)/ (10 × 24)
= (36 × 1)/ (1 × 2)
= (18 × 1)/ (1 × 1)
= 18 km
∴Total distance covered in 1 liter of petrol is 18 km.
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# Conversions of Length, Mass, Capacity in Metric Units
## Determine equivalent units of metric measurement for length, mass and capacity.
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Convert Metric Units of Measurement
### Themetric systemof measurement is the primary measurement system in many countries; it contains units such as meters, kilometers and liters.
You can remember the conversions by learning the prefixes: Milli-means thousandth, centi-means hundredth, and kilo-means thousand. So a millimeter is one-thousandth of a meter, and a kilometer is one thousand meters.
Now that you have reviewed these units of measurement, we can look at converting among the different units of measurement. Just like we used proportions when we converted among customary units of measurement, we can use proportions and ratios here too.
How do we use proportions to convert among metric units of measure?
First, set up the proportion in the same way you used to find actual measurements from scale drawings. Use the conversion factor as the first ratio, and the known and unknown units in the second ratio.
How many centimeters are in 5 meters?
First, set up a proportion.
The conversion factor is the number of centimeters in 1 meter. We can look at the chart above and see that there are 100 centimeters in 1 meters. That is our first ratio: \begin{align*}\frac{100 \ centimeters}{1 \ meter}\end{align*}.
Now write the second ratio.
The known unit is 5 meters. The unknown unit is \begin{align*}x\end{align*} centimeters. Make sure that the second ratio follows the form of the first ratio: centimeters over meters.
\begin{align*}\frac{100 \ centimeters}{1 \ meters} = \frac{x \ centimeters}{5 \ meters}\end{align*}
Now cross-multiply to solve for \begin{align*}x\end{align*}.
\begin{align*}(1)x &= 100(5)\\ x &= 500\end{align*}
There are 500 centimeters in 5 meters.
Henry is making a recipe for lemonade that uses 2 liters of water. If he makes 3 batches of the recipe, how many milliliters of water will he need?
First find the total number of liters he needs.
If there are 2 liters in one batch, and he is making 3 batches, then he will need \begin{align*}2 \times 3 = 6 \ liters\end{align*}.
Next, set up a proportion.
The conversion factor is the number of milliliters in a liter.
\begin{align*}\frac{1000 \ milliliters}{1 \ liter}\end{align*}
Now write the second ratio, making sure it follows the form of the first ratio.
\begin{align*}\frac{1000 \ milliliters}{1 \ liter} = \frac{x \ milliliters}{6 \ liters}\end{align*}
Cross-multiply to solve for \begin{align*}x\end{align*}.
\begin{align*}(1)x &= 1000(6)\\ x &= 6000\end{align*}
He will need 6000 milliliters of water.
Convert each measurement.
4500 ml = ____ Liters
5.5 grams = ____milligrams
40 mm = ____centimeters
### Vocabulary
Metric System
a system of measurement commonly used outside of the United States. It contains units such as meters, milliliters and grams.
### Guided Practice
Kyle is going to be traveling with his family over the winter holidays. He wants to figure out how many kilometers it is from his home in Cincinatti to his grandparents home in Chicago. Which unit of measurement should Kyle use?
First, let’s think about the correct unit of measurement for Kyle to use.
If Kyle is measuring a far distance, he needs a measure of length. We know that the metric units for measuring length are millimeters, centimeters, meters and kilometers. Kyle is measuring the distance between two cities. It makes the most sense for him to use the largest unit for measuring length, and that is kilometers.
Kyle would use kilometers to measure the distance.
### Practice
Directions: Solve each problem.
1. 3 km = _____ m
2. 2000 m = _____ km
3. 5.5 km = _____ m
4. 2500 m = _____ km
5. 12000 m = _____ km
6. 500 cm = _____ m
7. 6000 cm = _____ m
8. 4 m = _____ cm
9. 11 m = _____ cm
10. 50 mm = _____ cm
11. 3 cm = _____ mm
12. 15 cm = _____ mm
13. 2000 g = _____ kg
14. 35000 g = _____ kg
15. 7 kg = _____ g
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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# 10/1 Decimal Conversion
10/1 as a decimal is a significant conversion that one should know. It is crucial in understanding mathematical concepts and solving equations related to ratios, proportions, and percentages. Converting 10/1 to a decimal is a simple calculation that offers accuracy and simplicity in mathematical operations. In this article, we will guide you through the process of converting 10/1 to a decimal.
Contents
## Understanding Decimal Notation
Decimal notation is a representation of a number in the form of a fraction, using base 10 and a decimal point to separate the whole number and fractional part. This notation uses digits 0-9 and allows for efficient calculation, as it functions with other mathematical concepts like addition, subtraction, multiplication, and division. It is an essential part of mathematics and is commonly used in everyday life.
## Method #1: Long Division
If you want to convert 10/1 to a decimal, you can use long division. Here’s how it works:
10 | 1 | 10 | 1.0
When dividing 10 by 1, the quotient is 10, so the decimal starts with 10. Then, add a decimal point after 10 to get 10.0. This is the final decimal form of 10/1.
## Method #2: Dividing the Numerator by the Denominator
Converting 10/1 to a decimal can also be done by dividing the numerator (10) by the denominator (1). The process is simple, just divide 10 by 1, which is equal to 10. Therefore, 10/1 is equivalent to 10 as a decimal.
This method is easy and straightforward, especially for fractions with a denominator of 1. Just divide the numerator by the denominator to get the decimal equivalent. However, for fractions with larger denominators, it might be more efficient to use long division, which shows the steps in solving the problem.
Remember that decimal notation is just another way of representing a number using a base of 10, with digits ranging from 0 to 9 separated by a decimal point. Converting a fraction to a decimal is simply finding its decimal equivalent. So, whether you use division or long division, the important thing is to understand the process and be able to use it for any given fraction.
## Method #3: Converting Mixed Numbers to Improper Fractions
Converting mixed numbers to improper fractions is a crucial step in converting fractions to decimals. A mixed number is a combination of a whole number and a fraction. To convert a mixed number to an improper fraction, follow these steps:
1. Multiply the denominator (bottom number) of the fraction by the whole number.
2. Add the product from step 1 to the numerator (top number) of the fraction.
3. Write the sum from step 2 as the numerator of the improper fraction, and keep the same denominator.
For example, let’s convert the mixed number 10 1/100 to an improper fraction:
1. Multiply the denominator 100 by the whole number 10: 100 x 10 = 1000
2. Add the product 1000 to the numerator 1: 1000 + 1 = 1001
3. Write the sum 1001 as the numerator of the improper fraction: 1001/100
Now we can use the division method to convert 1001/100 to a decimal: divide the numerator 1001 by the denominator 100, which results in 10.01. Therefore, 10 1/100 represented as a decimal is 10.01.
## 10/1 as a Decimal: Calculation Examples
If you are wondering how to convert 10/1 to a decimal, it is essential to understand decimal notation. Decimal notation is a way of representing a number as a fraction with the base as 10 and a decimal point. The digits in decimal notation range from 0-9 and are written in two parts, a whole number and a fractional part that is separated by a dot called the decimal point.
The easiest way to convert a fraction to a decimal is by using the division method. You can simply divide the numerator by the denominator to get a decimal equivalent. To convert 10/1 to a decimal, we just need to divide 10 by 1. This will give the result of 10.00 as 10/1 is already in whole number format.
If you want to convert other fractions to decimals, you need to divide the numerator by the denominator. For example, to convert 77/80 to a decimal, divide 77 by 80 which gives 0.9625. Similarly, to convert 12/16 to a decimal, divide 12 by 16 resulting in 0.75.
Converting mixed numbers to improper fractions is necessary before converting them to decimals. You can convert a mixed number to an improper fraction by multiplying the denominator with the whole number and then adding the numerator to the product. For instance, to convert 3 1/4 to an improper fraction, we need to multiply the denominator 4 and the whole number 3, and add the numerator 1 to the product, resulting in 13/4. Once you have the improper fraction, you can proceed to convert it to a decimal by dividing the numerator by the denominator.
By following these simple calculation methods, you can convert any fraction into a decimal with ease.
## Decimal Equivalents Chart
The decimal equivalent chart is a useful tool for converting fractions to decimals. It provides a visual representation of common fractions and their corresponding decimal equivalents, making it easier for individuals to perform mathematical calculations.
For example, the fraction 1/2 is equivalent to 0.5 in decimal form, while the fraction 3/4 is equivalent to 0.75. To convert a fraction to a decimal, one must simply divide the numerator by the denominator.
The table below shows some of the most common fractions and their decimal equivalents:
Fraction Decimal Equivalent 1/8 0.125 1/4 0.25 1/2 0.5 3/4 0.75 1 1.0
With the use of a decimal equivalent chart, individuals can easily convert fractions to decimals and perform basic mathematical operations.
## Common Mistakes to Avoid
When converting 10/1 to a decimal, there are some common mistakes that people make. Avoid these by ensuring that you understand the process of converting fractions to decimals correctly.
One common mistake is misplacing the decimal and getting the incorrect answer. When converting fractions to decimals, the decimal point should always sit in the same place as it did in the fraction – just transferred to the new number.
Another mistake is simply getting the math wrong. Be careful when adding and subtracting, and especially when multiplying or dividing to get the answer you need.
Finally, ensure that you keep track of units throughout the process. Units can make a big difference when converting fractions to decimals, so be sure to label clearly as you go.
## Conclusion
Understanding decimal notation is crucial in everyday life where higher levels of precision are required. Knowing how to convert a fraction such as 10/1 to a decimal is important and can be done easily by dividing the numerator by the denominator. This knowledge can be beneficial in various circumstances, from solving mathematical problems to calculating measurements, conversion rates, and many more. Therefore, it is essential to master decimal notation and apply it accurately in different situations.
## References
Math is Fun: Decimal Definition
Sciencing: How to Calculate Fraction to Decimal
ThoughtCo: Worksheets for Converting Fractions to Decimals
Being a web developer, writer, and blogger for five years, Jade has a keen interest in writing about programming, coding, and web development.
Posts created 491
## Calculate 2/5 of 35: Quick and Easy Guide
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# How do you write the equation of a line in slope intercept, point slope and standard form given (3, -2) and (-2, 3)?
Jun 11, 2018
$\text{1. Point - Slope form of equation is } y + 2 = - \left(x - 3\right)$
$\text{2. Slope - intercept form is } y = - x + 1$
$3. \text{Standard form is } x = y = 1$
#### Explanation:
$\left({x}_{1} , {y}_{1}\right) \left(3 , - 2\right) , \left({x}_{2} , {y}_{2}\right) = \left(- 2 , 3\right)$
$\text{Slope } m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{3 + 2}{- 2 - 3} = - 1$
$\text{1. Point - Slope form of equation is }$
$y - {y}_{1} = m \left(x - {x}_{1}\right) = y + 2 = - 1 \cdot \left(x - 3\right)$
$y + 2 = - \left(x - 3\right)$
$\text{2. Slope - intercept form is }$
$y = m x + c = - x + 1$
$3. \text{Standard form is } x = y = 1$
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# How do you simplify (x-3)(2x-5)-(x+1)(x-6)?
Oct 21, 2017
${x}^{2} - 6 x + 21$
#### Explanation:
You can expend the expression.
$\left(x - 3\right) \left(2 x - 5\right) - \left(x + 1\right) \left(x - 6\right)$
$= \left(x\right) \left(2 x\right) + \left(x\right) \left(- 5\right) - 3 \left(2 x\right) - 3 \left(- 5\right) - \left[\left(x\right) \left(x\right) + \left(x\right) \left(- 6\right) + 1 \left(x\right) + 1 \left(- 6\right)\right]$
$= 2 {x}^{2} - 5 x - 6 x + 15 - \left({x}^{2} - 6 x + x - 6\right)$
$= 2 {x}^{2} - 11 x + 15 - {x}^{2} + 5 x + 6$
$= {x}^{2} - 6 x + 21$
Oct 21, 2017
$\left(x - 3\right) \left(2 x - 5\right) - \left(x + 1\right) \left(x - 6\right) = {x}^{2} - 6 x + 21$
#### Explanation:
(x-3)(2x-5) - (x+1)(x-6)
Use the distributive law to expand both equations:
(a+b)*(c+d)= (ac + ad + bc + bd)
=$\left(\left(x \cdot 2 x\right) + \left(x \cdot - 5\right) + \left(- 3 \cdot 2 x\right) + \left(- 3 \cdot - 5\right)\right) - \left(\left(x \cdot x\right) + \left(x \cdot - 6\right) + \left(1 \cdot x\right) + \left(1 \cdot - 6\right)\right)$
=$\left(2 {x}^{2} - 5 x - 6 x + 15\right) - \left({x}^{2} - 6 x + x - 6\right)$
=$\left(2 {x}^{2} - 11 x + 15\right) - \left({x}^{2} - 5 x - 6\right)$
=$2 {x}^{2} - 11 x + 15 - {x}^{2} + 5 x + 6$
=${x}^{2} - 6 x + 21$
Therefore:
$\left(x - 3\right) \left(2 x - 5\right) - \left(x + 1\right) \left(x - 6\right) = {x}^{2} - 6 x + 21$
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Parallelogram facts for kids
Kids Encyclopedia Facts
A parallelogram is a polygon with four sides (a quadrilateral). It has two pairs of parallel sides (sides which never meet) and four edges. The opposite sides of a parallelogram have the same length (they are equally long). The word "parallelogram" comes from the Greek word "parallelogrammon" (bounded by parallel lines). Rectangles, rhombuses, and squares are all parallelograms.
As shown in the picture on the right, because triangles ABE and CDE are congruent (have the same shape and size),
$AE = CE$
$BE = DE.$
In all Parallelogram's opposite angles are equal to each other. Angles which are not opposite in the Parallelogram will add up to 180 degrees.
Characterizations
A simple (non self-intersecting) quadrilateral is a parallelogram if and only if any one of the following statements is true:
• Two pairs of opposite sides are equal in length
• Two pairs of opposite angles are equal in measure
• The diagonals bisect each other
• One pair of opposite sides are parallel and equal in length
• Each diagonal divides the quadrilateral into two congruent triangles
• The sum of the squares of the sides equals the sum of the squares of the diagonals. (This is the parallelogram law)
• It has rotational symmetry of order 2
• It has two lines of symmetry
Properties
1. Opposite sides of parallelogram are parallel.
2. Any line through the midpoint of a parallelogram bisects the area.
this is a parallelogram
a good fact paralleogram parall in it A parallelogram is a polygon with four sides (a quadrilateral). It has two pairs of parallel sides (sides which never meet) and four edges. The opposite sides of a parallelogram have the same length (they are equally long). The word "parallelogram" comes from the Greek word "parallelogrammon" (bounded by parallel lines).[1] Rectangles, rhombuses, and squares are all parallelograms.
As shown in the picture on the right, because triangles ABE and CDE are congruent (have the same shape and size),
{\displaystyle AE=CE} {\displaystyle AE=CE} {\displaystyle BE=DE.} {\displaystyle BE=DE.} In all Parallelogram's opposite angles are equal to each other. Angles which are not opposite in the Parallelogram will add up to 180 degrees.
Contents 1 Characterizations 2 Properties 3 References 4 Other websites Characterizations A simple (non self-intersecting) quadrilateral is a parallelogram if and only if any one of the following statements is true:[2][3]
Two pairs of opposite sides are equal in length Two pairs of opposite angles are equal in measure The diagonals bisect each other One pair of opposite sides are parallel and equal in length Adjacent angles are supplementary Each diagonal divides the quadrilateral into two congruent triangles The sum of the squares of the sides equals the sum of the squares of the diagonals. (This is the parallelogram law) It has rotational symmetry of order 2 It has two lines of symmetry Properties Opposite sides of parallelogram are parallel. Any line through the midpoint of a parallelogram bisects the area. Parallelograms are quadrilaterals. References "Online Etymology Dictionary". etymonline.com. Retrieved 10 January 2011. Owen Byer, Felix Lazebnik and Deirdre Smeltzer, Methods for Euclidean Geometry, Mathematical Association of America, 2010, pp. 51-52. Zalman Usiskin and Jennifer Griffin, "The Classification of Quadrilaterals. A Study of Definition", Information Age Publishing, 2008, p. 22. Other websites Wikimedia Commons has media related to Parallelograms. Parallelogram and Rhombus - Animated course (Construction, Circumference, Area) Interactive Parallelogram --sides, angles and slope
This short article about mathematics can be made longer. You can help Wikipedia by adding to it.
this is a parallelogram a good fact paralleogram parall in it
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## How To Multiply Faster Than A Calculator
I’m sorry to tell you that you’ve been duped! I didn’t want to be the one to inform you that nobody can multiply faster than a calculator. It just isn’t possible. I’ve misled you. However, those of you who did not take my title literally, I do have good news.
Consider the following scenario. The process of using a calculator is not nearly as fast as it is portrayed to be. First, you need to get it out and turn it on. Then, you need to enter in the numbers. Finally, it will display the answer for you. Think about it. How long does it really take to do all this? I claim that I can do this task faster in my head, and that I can teach you to do the same. I’ll teach you how to be lazy, save time, and avoid the hassle of getting out the calculator all at the same time!
There are two methods that I would like to share with you: Round And Adjust, and Multiply In Parts. Along with these two techniques, I’ll share with you a couple of tricks that I apply to simplify both of these methods.
Trick #1: Ignore the trailing zeros
What this means is that if you have a number such as 250, pretend it is just 25. Once you do the multiplication, you just add the zero back to the end of your answer. This works for any number of zeros. 250000, 250, and 25000000 should all be treated the same as 25. Just remember to add the respective amount of zeros to the end of your answer. However, you cannot simplify a problem when the zeros are in the middle. You can only use this for trailing zeros (zeros at the far right side). For example, 205 must be thought of as 205. This cannot be reduced to anything smaller. On the contrary, 2050 can be treated as 205, because there was one trailing zero. This helps quite a bit.
Trick #2: Ignore the decimals
Now for the main course, my methods!
I’d like to explain this method by using an example. Let’s say you want to multiply 98 by 52, and try and do it quickly in your head. The answer is 5096, and I did that in around 8-10 seconds – much faster than getting out a calculator. So, what did I do? I rounded, did the multiplication, and then adjusted my result. I rounded 98 up to 100. Then, it is extremely easy to multiply 100 by 52: 5200. Then, I think about how much I rounded by. I rounded up by 2. 100 x 52 is 2 x 52 more than 98 x 52. This means, that 98 x 5equals 100 x 52 minus 2 x 52. Therefore, I find a quick answer by multiplying 100 by 52, then I subtracted 2 times 52 from that. 2 times 52 is 104. 5200 – 104 is 5096. Simple as that.
Now, what if you round down? What is 72 times 45? 3240. In my head, this was the process that I followed. 72 is close to 70, so let’s start with that. 70 times 45 is like 7 times 45. 7 times 45 is 315. Since I turned 70 into 7, I’ve got to add a zero to the end. 3150. I rounded down by 2, this means that I now have to add this 2 back in. To do so I multiply 2 by 45, and add this to my answer. 2 times 45 is 90, and 3150 + 90 is 3240. I know that took a little while to read, but it goes much faster in your head once you understand the method.
Why does this work?
98 x 52 = (100 – 2) x 52 = (100 x 52) – (2 x 52) = 5200 – 104 = 5096.
See? This is the algebra that allows for this method to be used. The same proof can be applied to rounding down. Basically, you look to round one of the numbers to a multiple of ten. It needs to be somewhat close to that number already, in order for this method to really be useful. Remember that you can only round ONE of the numbers. Do not round both numbers. Only one. Try it out =)
Multiply In Parts
What the heck does this one mean? Well, actually, it is very similar to my first method, however it is a bit more universal. It works because of the same logic, but it does not require rounding or adjusting. It is simply breaking down a number piece by piece. What do I mean by that? I’ll show you an example, using algebra to solve it.
1234 x 32 = (1000 + 200 + 30 + 4) x 32 = (1000 x 32) + (200 x 32) + (30 x 32) + (4 x 32)
= 32000 + 6400 + 960 + 128 = 39488
Could you have done that in your head? I think you can =) You need to break up the larger number into it’s parts. For example, think of 345 as a 3, 4, and 5 separately. Multiply them individually, and then add the results. Remember that 3 is really 300, so you must add two zeros to that result, and add one zero to the result from the 4. This is the same way we do multiplication when we do it on paper. The difference is the perspective in which you are thinking about it. You are multiplying each number individually, and then adding the results. It’s not all that hard one you get used to it. 64 x 55. 6 x 55 is 330, and 4 times 55 is 220. The answer is simply 3300 + 220. The extra zero on the 3300 was inserted because the 6 is really a 60. The answer is just 3520!
What do you think? I promise that doing multiplication in your head is much faster than doing it on paper, and can even be faster than using a calculator. If you have a calculator out already, then by all means use it. However, if it is out of the way to go get it, you should try to do it in your head! I’ll do a couple of examples using all the tricks I showed you above.
31 x 4.5?
31 x 4.5 is like 30 x 45, which is like 3 times 45. 135. 1350. When you add back in the 1 that we rounded down from, this becomes 1395. There was 1 digit to the right of the decimal. The final answer is 139.5. This is exactly what I thought in my head. I know it takes a while to read, however it is much faster when I’m thinking it. Try it out! It is really easy to make up an example for yourself.
2.40 x 314?
2.40 x 314 is like 24 x 314. 24 x 3 is 72. 24 x 1 is 24. 24 x 4 is 96. 7200 + 240 + 96 = 7536. When you add back the zero and decimal, you get 753.60.
To be honest, I don’t add all those large numbers in my head all at once. I keep a running tally. This helps when the numbers get much, much larger. What does that mean? In the example above, I say 24 x 3 is 72, which is 7200. I repeat that number in my head. 7200. 24 x 1 is 24, which is really 240. 7440. I then repeat that in my head, so that I remember it better. 7440. I continue this, until I get to the last number, 24 x 4 which is 96. 7536. Then I add in my zeros and decimal, and get the final result.
The most common use for this is shopping. Do you ever go to a store and see that the price is 30% off? Is that really a good deal? What if it is 30% off, but then take an additional 20% off? Doesn’t that get a bit more complicated? Of course, cell phones can act as a calculator on the go, but is it worth the time to get it out and type in the numbers? I think not.
You go into a store and see a pair of shoes you’d like to purchase. The price for them is 49.99, and they are onsale for 30% off. However, they are having a special 1 day only sale, that allows you to take an additional 20% off the original sale price.
So, you need to take 30% off of the original price. Then, take 20% off of that.
First, sales need to be thought of in an opposite way of thinking. If something is 30% off, that means you are paying 70%. If you multiply the price by 0.7, this will give you the sale price. So, find 49.99 x 0.7.
49.99 x 0.7 is like 5 x 7, or 35. With the zeros, 35000. Subtract 1 x 7, from rounding. 34993. Now, add back in the decimal. 34.993. This is the new price! 34.993, or just 34.99! Now, you can take 20% off of that price. So, we will multiply by 0.8, because you are really paying 80% of the price when you take off 20%.
34.99 x 0.8 is like 35 x 8, 280. 28000. Subtract 1 x 8. 28992. The answer is 28.992, with the decimal. The final cost of the shoes will be 28.99! That is a great deal =)
How else could we have done this problem? Well, you are taking off 30%, and then another 20%. This is the same as paying 70%, and then paying 80% of that. We could just do 0.7 x 0.8 = 0.56. Multiply this by 49.99, and you will get the same answer. Just another way to think about it =)
I know I’m really geeky and all, but when I’m bored I multiply things in my head sometimes. It keeps my mind sharp, I believe. I also like to do some calculus in my head. I work a part-time job after school, and this helps pass the time when things are slow. I know it seems useless to do this sort of thing randomly, however when the time comes when you need to use it, you can do it much, much faster. I hope this has been somewhat helpful!
If there is anything you would like to be explained further, please comment below or email me at admin@calculustricks.com. Thanks!
## Calculus I: The Chain Rule
The chain rule is used extremely often, and is a concept you can’t do without.
If it’s so important, why? What is it, exactly? Before I get to the teaching, I need to warn you. This is one of the most common things that people forget to do. The chain rule is needed every single time you take a derivative. I’ll show you why I haven’t mentioned it until now, but from now on, you need to consider the effects of the chain rule EVERY time you take a derivative. Up until now, you haven’t been able to take the derivative of anything complicated. I’ve only given you things that are fairly simple. I’m sorry to tell you that things can get much, much harder! Here are a couple of examples, to show you what I mean.
Before now, you wouldn’t be able to take the derivative of the functions above. They are too complex, and require the chain rule. With some problems, the effects of the chain rule are negligible, and those are the functions we’ve been working with thus far.
What’s the chain rule?
First, I’ll give you the formal definition. I know it is going to not make much sense, but I’ll thoroughly explain it afterward.
If you don’t understand that, don’t worry. Nobody really expects you to at first. So, what the heck does that mean? Simply, this means when you have one function inside of another. Here’s an example:
The inside function is , and the outside function is . One function is inside of the other. That’s the best way to think about a complicated function. You should think of it in parts, like I just showed. Now, how do we take the derivative of this? To begin, you pretend that the inside of the function is just a simple x, and you don’t do anything with it. You leave it alone completely. For this part, pretend that we are taking the derivative of . It doesn’t matter what the “something” is. Just leave it alone, and take the derivative as if the “something” is just an “x”.
That is exactly what I just told you to do, however the problem isn’t complete. Next, you must multiply that answer by the derivative of the “something”. In the first part, we completely ignore what the “something” is. Now, you look at ONLY the “something”, and take the derivative of that. This is going to be our final answer, unless we can simplify something.
As you can see, I didn’t leave the word “something” in the middle of my answer. I just left it alone completely in the first part. I didn’t touch it. I just rewrote it. I’m being deliberately redundant here. I want to say this as many ways as possible, so that you completely understand what I am saying. This is extremely important, and I would venture to guess that it is the most common mistake made in calculus. Everyone messes up the chain rule at some point.
This is an extremely fundamental topic in calculus. This rule should be applied to every single derivative that you are taking. Sometimes, it will not affect your answer, but most of the time it will. If you forget to use the chain rule, you will get the answer wrong! This is a rule. This is how you take derivatives. It is as important as the order of operations. It is needed. This is how you take a derivative properly. You need to practice this thoroughly!
Before I get to another example for the chain rule, I want to show you why the chain rule was not needed for all the derivatives we’ve been taking until now. The reason is simply that every time we take the derivative, the inside function has just been an “x”. For example,
The outside function is “sin”, and the inside is just “x”. What happens if we use the chain rule here?
The only difference from what we used to do is when we multiply the derivative of the inside “something”. The “something” has always been just an “x”. The derivative of x is just 1, and when you multiply by 1 nothing changes! The chain rule, the part where we multiply by the derivative of the inside, does absolutely nothing. It is completely negligible. That is what we’ve never needed it! However, now that you know about the chain rule, you should always use it. Even when it isn’t going to change the answer. You need to make it a habit, or you will forget to do it at a time where you really need to. It’s inevitable that you will mess this up a few times, but hopefully not much more than that.
Remember this:
To take a derivative, you look at the most “outside” function. In most cases, it will be an exponent on parenthesis. You pretend that everything on the inside is just an x. It could be absolutely anything, and it doesn’t matter. Take the derivative as usual, but leave the inside completely untouched. Just copy it from one line to the next. After you do that, you must multiply that answer by the derivative of the inside stuff. The same stuff you previously skipped over and ignored. That is the most simple way to describe and understand the chain rule.
Want to try another example?
The most outside part is the “sin”, so start there. Take the derivative, and then move to the inside part.
See how that works?
One more basic problem, before I show you some hard ones?
I have two things to admit to you. First of all, you don’t have to actually write “something”, like I’ve been doing in every problem. That is just the way I suggest you think about these problems. Visualize that step as you are using the chain rule, but don’t write it down. Second, you really don’t need to show the step where I write the symbol for the derivative. You can just take the derivative. If you look in this last example, you should skip the entire first step. You can go directly to the second step that I’ve shown. I’m just showing extra so that you understand it a little bit better.
I have two more examples for you. These are a bit more challenging. If you can do this type of problem on your own, you will be able to do any chain rule problem. That said, I suggest you try doing the following two examples yourself. Don’t look at my solution, and work it out on your own. Only scroll further to see my answer when you get stuck. That is the best way for you to learn! (I promise)
These can get pretty complicated, but I hope you could follow me! This answer is technically correct, however many teachers will want you to go further than this. When you have to use both the product rule and chain rule, often times you can factor and simplify things a great deal. I’ll show you what to do with this answer.
See how much nicer that looks? I think so at least, and many of your teachers will require you to do this simplification! Be on alert whenever you are using the product rule to take the derivative of something complicated.
For my last example, I will give you my favorite challenge problem. Whenever I tutor calculus I, I always ask this type of function. Like I said before, try to do it without my help. If you can, that is extremely impressive! This is like the test of all tests. If you can figure this one out, consider yourself an expert of the chain rule! I’m going to break it down step by step, and explain why it needs to be done this way.
The way you should always start taking a derivative is by asking yourself, “What is the most outside part?” This is where we always begin. If you remember, you need to ignore the part on the inside, and take the derivative of only the outermost part of the function. Work your way from the outside to the inside. So, what’s the answer? What is the outermost part of this function? What should we start taking the derivative of?
To be honest, I’m being sneaky. Most of you would say, “cosine of course!” That is exactly what I expect you to say, however that is wrong! Let me rewrite the problem. If you don’t know what I am talking about, you will in just a second.
The reason this may have tricked you, is that I placed the exponent right next to the “cos”. This is a very common way to write exponents on trigonometric functions, but it always leads to confusion! If you get a problem with a trig function that has a power on it, rewrite it like I just did. Be careful though, the parenthesis around the “cos” are extremely important. Otherwise, the exponent isn’t in the correct place, and you will confuse yourself even further.
Now, I’ll ask you one more time. What is the most outside part? Of course, you can now see the most outside part is the exponent. For the first part, you should ignore everything inside the parenthesis with the exponent on it. I’m going to show you the full solution now, and then give a brief explanation afterward.
There was a chain rule needed inside of a chain rule! Isn’t that really neat? Basically, you work your way in from the most outside piece of the function. Then you take one step inside. Then another step inside. Until you finally reach the innermost basic function that was trapped inside everything else. As you go along, you need to respect the chain rule, and multiply by the derivative of the respective next inside piece of the function. You multiply by that derivative. Then you need to determine if you will need to use the chain rule on the that derivative as well. You just keep multiplying them!
Do you see how I simplified my answer? Make sure you bring all of the coefficients to the very front of the expression. Thats is very important for style points, and you might get points off on a test if you don’t do that!
That was my last example. I really like teaching that one. These problems can get pretty complicated, and I suggest you practice the chain rule. It needs to be second nature, and a habit for every time you take a derivative. Always write it! This is the most common mistake in calculus. Don’t forget the chain rule! Remind yourself before you even begin a problem.
Good Luck!
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# Permutations and Combinations Questions
FACTS AND FORMULAE FOR PERMUTATIONS AND COMBINATIONS QUESTIONS
1. Factorial Notation: Let n be a positive integer. Then, factorial n, denoted n! is defined as: n!=n(n - 1)(n - 2) ... 3.2.1.
Examples : We define 0! = 1.
4! = (4 x 3 x 2 x 1) = 24.
5! = (5 x 4 x 3 x 2 x 1) = 120.
2. Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.
Ex1 : All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
Ex2 : All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)
Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:
$P_{r}^{n}=n\left(n-1\right)\left(n-2\right)....\left(n-r+1\right)=\frac{n!}{\left(n-r\right)!}$
Ex : (i) $P_{2}^{6}=\left(6×5\right)=30$ (ii) $P_{3}^{7}=\left(7×6×5\right)=210$
Cor. number of all permutations of n things, taken all at a time = n!.
Important Result: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,
such that $\left({p}_{1}+{p}_{2}+...+{p}_{r}\right)=n$
Then, number of permutations of these n objects is :
3. Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.
Ex.1 : Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
Note that AB and BA represent the same selection.
Ex.2 : All the combinations formed by a, b, c taking ab, bc, ca.
Ex.3 : The only combination that can be formed of three letters a, b, c taken all at a time is abc.
Ex.4 : Various groups of 2 out of four persons A, B, C, D are : AB, AC, AD, BC, BD, CD.
Ex.5 : Note that ab ba are two different permutations but they represent the same combination.
Number of Combinations: The number of all combinations of n things, taken r at a time is:
Note : (i) (ii)$C_{r}^{n}=C_{\left(n-r\right)}^{n}$
Examples : (i) $C_{4}^{11}=\frac{11×10×9×8}{4×3×2×1}=330$ (ii)$C_{13}^{16}=C_{\left(16-13\right)}^{16}=C_{3}^{16}=560$
Q:
A box contains 5 green, 4 yellow and 3 white marbles. Threemarbles are drawn at random. What is the probability thatthey are not of the same colour ?
A) 40/44 B) 44/41 C) 41/44 D) 40/39
Explanation:
n(E) = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15n(S) = 12C3 = 220
8 1592
Q:
From a pack of 52 cards, 3 cards are drawn together atrandom, What is the probability of both the cards are king?
A) 1/5225 B) 1/5525 C) 5525 D) 1/525
Explanation:
n(S) = 52C3 = 132600/6 = 22100
n(E) = 4C3 = 24/6 = 4
6 1493
Q:
If it is possible to make a meaningful word with the first, the seventh, the ninth and the tenth letters of the word RECREATIONAL, using each letter only once, which of the following will be the third letter of the word? If more than one such word can be formed, give ‘X’ as the answer. If no such word can be formed, give ‘Z’ as the answer.
A) T B) X C) N D) R
Explanation:
The first, the seventh, the ninth and the tenth letters of the word RECREATIONAL are R, T, O and N respectively. Meaningful word from these letters is only TORN. The third letter of the word is ‘R’.
6 1450
Q:
If each of the vowels in the word 'MEAT' is kept unchanged and each of the consonants is replaced by the previous letter in the English alphabet, how many four-lettered meaningful words can be formed with the new letters, using each letter only once in each word?
A) 3 B) 4 C) 1 D) 2
Explanation:
16 994
Q:
In a bag containing red, green and pink tokens, the ratio of red to green tokens was 5 : 12 while the ratio of pink to red tokens was 7 : 15. What was the ratio of green to pink tokens?
A) 25 : 28 B) 36 : 7 C) 8 :25 D) 12 : 7
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# High School Math : How to solve one-step equations with fractions in pre-algebra
## Example Questions
### Example Question #1 : How To Solve One Step Equations With Fractions In Pre Algebra
Solve for .
Explanation:
Perform the same operation on both sides of the equation.
It will be easier to write the right side of the equation as a fraction.
Now, we add two-fifths to both sides of the equation.
### Example Question #2 : How To Solve One Step Equations With Fractions In Pre Algebra
is % of what number?
Explanation:
To find the number of which is %, use this equation with % expressed as a fraction coefficient:
To solve this equation, multiply both sides of the equation by the reciprocal of the fraction on the left side, then reduce the result to simplest terms.
### Example Question #1 : How To Solve One Step Equations With Fractions In Pre Algebra
Solve the equation for .
Explanation:
Multiply both sides of the equation by .
We can check our answer by plugging it back into the equation.
We know that our answer works.
### Example Question #4 : How To Solve One Step Equations With Fractions In Pre Algebra
Solve for if
Explanation:
To solve for we must get all of the numbers on the other side of the equation of .
To do this in a problem where a number is being added to , we must subtract the number from both sides of the equation.
In this case the number is so we subtract from each side of the equation to make it look like this
To subtract fractions we must first ensure that we have the same denominator which is the bottom part of the fraction.
To do this we must find the least common multiple of the denominators.
The least common multiple is the smallest number that multiples of both of the denominators multiply to.
In this case the LCM is
We then multiply the numerator and denominator of by to get the same denominator because anything divided by itself is one so the fractions maintain their same value as the numbers change into the format we need to determine the answer.
To multiply a fraction you multiply the top number of the first fraction by the top number of the second fraction and the bottom number of the first fraction by the bottom number of the second fraction so it would look like this
After doing this we then subtract the first numerator (top part of the fraction) from the second numerator and place the result over the new denominator
### Example Question #5 : How To Solve One Step Equations With Fractions In Pre Algebra
A student is 5'4" tall. What is their height in cm?
Explanation:
First, we need to convert the student's height to inches. There are 12 inches in one foot and the student is 5 feet and 4 inches tall. We need to covert feet to inches, and add 4.
So, the student is 64 inches tall. Now we need to convert to centimeters.
The key to this kind of analysis is to make sure the units cancel correctly, leaving you with the units that you need.
### Example Question #4 : How To Solve One Step Equations With Fractions In Pre Algebra
Solve for .
Explanation:
Multiply both sides by 3 to isolate .
### Example Question #7 : How To Solve One Step Equations With Fractions In Pre Algebra
Solve for the value of .
Explanation:
We need to isolate the variable. Multiply both sides by .
### Example Question #8 : How To Solve One Step Equations With Fractions In Pre Algebra
Solve for .
Explanation:
To solve , we need to isolate . That means we need to multiply both sides by .
### Example Question #5 : How To Solve One Step Equations With Fractions In Pre Algebra
Solve for .
Explanation:
To solve , we need to isolate . That means we need to divide by .
Remember, dividing by a fraction is the same as multiplying by the reciprocal. The reciprocal of is .
From here you can either plug this into your calculator or work in pieces:
What is ?
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Integer Addition Unit 5 > Lesson 4 of 11
Problem: George owes his friend Jeanne \$3. If he borrows another \$6, how much will he owe her altogether? Solution: This problem is quite simple: just add \$3 and \$6 and the result is \$9.
The problem above can be solved using addition of integers. Owing \$3 can be represented by -3 and owing \$6 can be represented by -6. The problem becomes: -3 + -6 = -9
Look at the number line below. If we start at 0, and move 3 to the left, we land on -3. If we then move another 6 to the left, we end up at -9.
Rule: The sum of two negative integers is a negative integer.
Example 1: Find the sum of each pair of integers. You may draw a number line to help you solve this problem.
Adding Negative Integers Integers Sum -2 + -9 = -11 -5 + -8 = -13 -13 + -7 = -20
Do not confuse the sign of the integer with the operation being performed. Remember that: -2 + -9 = -11 is read as Negative 2 plus negative 9 equals negative 11.
Rule: The sum of two positive integers is a positive integer.
Example 2: Find the sum of each pair of integers. You may draw a number line to help you solve this problem.
Adding Positive Integers Integers Sum +2 + +9 = +11 +17 + +5 = +22 +29 + +16 = +45
Do not confuse the sign of the integer with the operation being performed. Remember that: +29 + +16 = +45 is read as Positive 29 plus positive 16 equals positive 45.
So far we have added integers with like signs (either both negative or both positive). What happens when we add integers with unlike signs? How do we add a positive and a negative integer, or a negative and a positive integer?
Procedure: To add a positive and a negative integer (or a negative and a positive integer), follow these steps:
1 Find the absolute value of each integer. 2 Subtract the smaller number from the larger number you get in Step 1. 3 The result from Step 2 takes the sign of the integer with the greater absolute value.
We will use the above procedure to add integers with unlike signs in Examples 3 through 7. Refer to the number line to help you visualize the process in each example. We will use money as an alternative method for adding integers.
Example 3: Find the sum of +7 and -4. Step 1: |+7| = 7 and |-4| = 4 Step 2: 7 - 4 = 3 Step 3: The number 3 will take a positive sign since +7 is farther from zero than -4. Solution 1: +7 + -4 = +3 Solution 2: If you start with \$7 and you owe \$4, then you end up with \$3.
Example 4: Find the sum of -9 and +5. Step 1: |-9| = 9 and |+5| = 5 Step 2: 9 - 5 = 4 Step 3: The number 4 will take a negative sign since -9 is farther from 0 than +5. Solution 1: -9 + +5 = -4 Solution 2: If you owe \$9 and you are paid \$5, then you are still short \$4.
Example 5: Find the sum of +6 and -7. Step 1: |+6| = 6 and |-7| = 7 Step 2: 7 - 6 = 1 Step 3: The number 1 will take a negative sign since -7 is farther from 0 than +6. Solution 1: +6 + -7 = -1 Solution 2: If you start with \$6 and you owe \$7, then you are still short \$1.
Example 6: Find the sum of -6 and +7. Step 1: |-6| = 6 and |+7| = 7 Step 2: 7 - 6 = 1 Step 3: The number 1 will take a positive sign since +7 is farther from 0 than -6. Solution 1: -6 + +7 = +1 Solution 2: If you owe \$6 and you are paid \$7, then you end up with \$1.
Example 7: Find the sum of +9 and -9. Step 1: |+9| = 9 and |-9| = 9 Step 2: 9 - 9 = 0 Step 3: The integer 0 has no sign. Solution 1: +9 + -9 = 0 Solution 2: If you start with \$9 and you owe \$9, then you end up with \$0.
In Example 7 you will notice that the integers +9 and -9 are opposites. Look at the problems below. Do you see a pattern?
-100 + +100 = 0 +349 + -349 = 0 -798 + +798 = 0
Rule: The sum of any integer and its opposite is equal to zero.
Summary: Adding two positive integers always yields a positive sum; adding two negative integers always yields a negative sum. To find the sum of a positive and a negative integer, take the absolute value of each integer and then subtract these values. The result takes the sign of the integer with the larger absolute value. The sum of any integer and its opposite is equal to zero.
### Exercises
Directions: Read each question below. Click once in an ANSWER BOX and type in your answer; then click ENTER. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR.
1 -6 + -9 = ? ANSWER BOX: RESULTS BOX:
2 +7 + +11 = ? ANSWER BOX: RESULTS BOX:
3 +5 + -3 = ? ANSWER BOX: RESULTS BOX:
4 -2 + +7 = ? ANSWER BOX: RESULTS BOX:
5 +9 + -12 = ? ANSWER BOX: RESULTS BOX:
This lesson is by Gisele Glosser. You can find me on Google.
Lessons on Integers Introduction Absolute Value Compare and Order Integer Addition Integer Subtraction Integer Multiplication Integer Division Operations With Integers Practice Exercises Challenge Exercises Solutions
Related Activities Integers and Science Integer Football Game Integer Word Search Puzzle Pre-Algebra Worksheet Generator
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# 1. - PowerPoint PPT Presentation
1. Review: What are Factors?. 1. Factors come in pairs and divide a bigger number evenly without a remainder Examples: Factors of 12: 1, 12 2, 6 3, 4 Factors of 30: 1, 30 2, 15 3, 10 5, 6 We can use factor trees, always check 1 through 9 until you get repeats.
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Presentation Transcript
1.
Review: What are Factors?
1.
• Factorscome in pairs and divide a bigger number evenly without a remainder
• Examples:
• Factors of 12:
• 1, 12
• 2, 6
• 3, 4
• Factors of 30:
• 1, 30
• 2, 15
• 3, 10
• 5, 6
• We can use factor trees, always check 1 through 9 until you get repeats
30
15
2
5
3
6
2.
How do I Reduce Fractions to Lowest Terms?
2.
• A fraction is a proportion that expresses a part over a whole
• The top number is the numerator
• The bottom number is the denominator
• To convert fractions to equivalent decimals and percents, we first need to reduce fractions to their lowest terms
• To reduce fractions, divide the numerator and denominator by common factors until you cannot divide evenly
2.
How do I Reduce Fractions to Lowest Terms?
2.
• I DO
÷ 5 =
÷ 12 =
÷ 5 =
÷ 12 =
• WE DO
÷ 7 =
÷ 3 =
÷ 7 =
÷ 3 =
• YOU DO
÷ 4 =
÷ 6 =
÷ 4 =
÷ 6 =
3.
How do I Convert Fractions to Decimals
3.
• Reduce a fraction to its lowest terms
• Check your list of common fractions
• Memorize the common fractions and their decimal/percent equivalents
• WE DO
÷ 3 =
= 2.60
÷ 3 =
• YOU DO
÷ 7 =
= 0.875
÷ 7 =
4.
How do I Convert Decimals to Fractions
4.
• YOUR BRAIN IS A FRACTION BUTTON!
• Check your list of common fractions
• Memorize the common fractions and their decimal/percent equivalents
• Or, express as a fraction over a multiple of 10:
• 8 tenths = 0.8 =
• 65 hundredths = 0.65 =
• 143 hundredths = 1.43 =
• Then, simplify the fraction to its lowest terms
4.
How do I Convert Decimals to Fractions
4.
• I DO
0.4 =
• WE DO
0.85 =
• YOU DO
0.75 =
5.
How do I Convert Percents to Decimals and Decimals to Percents
5.
• Percent means “out of 100”
• 50% means 50 hundredths
• 50% means 50 out of 100
• To convert decimal to percent
• Move the decimal two places to the right
• To convert percent to decimal
• Move the decimal two places to the left
Decimals to Percent
WE DOYOU DO
0.40 = 0.35 =
0.875 = 1.42 =
0.005 = 0.625 =
Percent to Decimal
WE DOYOU DO
55% = 32.5% =
37.5% = 80% =
132% = 250% =
40%
87.5%
0.5%
0.55
0.375
1.32
35%
142%
62.5%
0.325
0.80
2.50
6.
Ordering Numbers on a Number Line
6.
• Convert all fractions and percents to decimals
• Put in order in decreasing order of place value
• Hundreds, then tens, then ones
• Tenths, then hundredths, then thousandths
• Cross out numbers as you put them in order
0.75 0.375 0.50 0.60 0.625
0.375 0.50 0.60 0.625 0.75
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Q:
# A student wants to know how far above the ground the top of a leaning flagpole is. At high noon, when the sun is almost directly overhead, the shadow cast by the pole is 6 ft long. The student holds a plumb bob with a string 3 ft long up to the flagpole and determines that the point of the plumb bob touches the ground 14 in. from the base of the flagpole. How far above the ground is the top of the pole?
Accepted Solution
A:
Answer:15.43 ftStep-by-step explanation:To solve this question, one should use the concept of similar triangles.The pole's shadow and the distance to the ground (x) form a similar triangle to the distance of the plumb bob from the base and the length of the plumb bob, respectively. Note that we do not need to know the measurements of the third side of the triangles to solve the problem. Therefore, the distance of the top of the pole to the ground is:$$14 in = 1.1667 ft$$$$1.1667x=6*3\\x = 15.43$$The distance of the top of the pole to the ground is 15.43 ft.
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The vertex of a parabola can tell us the extreme value (local maximum or minimum) of a parabola, along with the line of symmetry of the curve. However, we still need to know how to find the vertex in a variety of situations.
So, how do you find the vertex of a parabola? To find the vertex of a parabola, you can use the graph (find the maximum/minimum of the curve), use two points (using a parabola’s symmetry), or use the corresponding quadratic equation. You can find the vertex of a parabola from a quadratic equation in vertex form, factored form, or standard form.
Of course, given any three points on a parabola, we can always find the equation of the corresponding quadratic and use it to find the vertex.
In this article, we’ll talk about how to find the vertex of a parabola. We’ll also show some examples to make the concepts clear.
Let’s get started.
## How To Find The Vertex Of A Parabola
There are a few ways you can find the vertex of a parabola:
• From an equation: if you have a quadratic equation in vertex form, factored form, or standard form, you can use it to find the vertex of the corresponding parabola.
• From two points (symmetry): if you have two points on a horizontal line that are an equal distance from the vertex of a parabola, you can use symmetry to find the vertex.
• From a graph: if you have the graph of a quadratic, you can find the local maximum (top of the curve) or local minimum (bottom of the curve) to find the vertex of the parabola.
Let’s start off with the first method: finding the vertex from an equation.
How To Find The Vertex Of A Parabola From An Equation
The easiest way to find the vertex of a parabola from an equation is to convert it to vertex form or factored form. However, we can convert any form of a quadratic equation into any other form.
The procedure is slightly different in each case, so let’s go through each one in turn.
How To Find The Vertex Of A Parabola In Vertex Form
To find the vertex of a parabola in vertex form, look at the constants h and k in the corresponding quadratic equation:
• y = a(x – h)2 + k
This form is easiest to find the vertex from, since all we need to do is read the coordinates from the equation.
The vertex will be the point (h, k).
Example: How To Find The Vertex Of A Parabola From An Equation In Vertex Form
Let’s say we have the following quadratic equation in vertex form:
• y = 2(x – 5)2 + 7
In this case, h = 5 and k = 7. This means that the vertex of the corresponding parabola is (h, k) = (5, 7).
We can verify this with the graph below.
How To Find The Vertex Of A Parabola In Factored Form
Find the vertex of a parabola in factored form is a little more involved, but it is still not too difficult. The steps are as follows:
• 1.) Find the two zeros (roots), r and s, of the quadratic from the factored form. These value comes from the factored form y = a(x – r)(x – s).
• 2.) Take the average of r and s to get h = (r + s) / 2 (h is the x-coordinate of the vertex).
• 3.) Substitute x = h into the quadratic factored form to find y. This will always give us a y-coordinate of k = -a(r – s)2 / 4
• 4.) The vertex is the point (h, k) = ((r + s) / 2, -a(r – s)2 / 4)
Let’s look at an example.
Example: How To Find The Vertex Of A Parabola From An Equation In Factored Form
Let’s say we have the following quadratic equation in factored form:
• y = 4(x – 2)(x – 8)
Step 1: In this case, r = 2 and s = 8 are the two zeros (roots) of this quadratic equation.
Step 2: The average of r = 2 and s = 8 is (2 + 8) / 2 = 5. So, h = 5 is the x-coordinate of the vertex.
Step 3: We substitute x = 5 into the quadratic equation to get 4(5 – 2)(5 – 8) = 4(3)(-3) = -36. So, k = -36 is the y-coordinate of the vertex.
Step 4: The vertex of the parabola is the point (h, k) = (5, -36).
We can verify this with the graph below.
Note: we can also convert the quadratic equation to vertex form and then read the coordinates as in the last example:
• y = 4(x – 2)(x – 8)
• y = 4(x2 – 8x – 2x + 16) [used FOIL]
• y = 4(x2 – 10x + 16) [combine like terms]
• y = 4(x2 – 10x + 25 – 25 + 16) [complete the square for x2 + 10x by adding 25]
• y = 4((x – 5)2 – 9) [factor x2 – 10x + 25 as a perfect square trinomial (x – 5)2]
• y = 4(x – 5)2 – 36
We get the same answer for the vertex: (5, -36).
How To Find The Vertex Of A Parabola In Standard Form
To find the vertex of a parabola in standard form, we can convert to either vertex form or factored form and then follow the steps in the previous examples.
We also have the option of using the shortcut formula for the vertex of a parabola in standard form. If the quadratic has the equation
• y = ax2 + bx + c
then the vertex has an x-coordinate of –b/2a. We can then substitute x = -b/2a into the quadratic equation to find the value of y.
This gives us a y-coordinate of c – (b2 / 4a) for the vertex of the parabola.
So, the coordinates of the vertex are (-b / 2a, c – (b2 / 4a)).
Example: How To Find The Vertex Of A Parabola From An Equation In Standard Form
Let’s say we have the following quadratic equation in standard form:
• y = 2×2 – 12x + 16
Let’s convert it to factored form and find the vertex that way:
• y = 2(x2 – 6x + 8)
• y = 2(x – 2)(x – 4)
So, r = 2 and s = 4. Taking the average gives us (2 + 4) / 2 = 3.
The x-coordinate of the vertex is 3. If we substitute this into the quadratic, we can find y:
• y = 2(x – 2)(x – 4)
• y = 2(3 – 2)(3 – 4)
• y = 2(1)(-1)
• y = -2
The y-coordinate of the vertex is -2. So, the vertex of the parabola is (3, -2).
We can also find this with our formulas that involve a, b, and c (the coefficients of the quadratic equation in standard form).
First, the x-coordinate of the vertex:
• x = -b / 2a
• x = -(-12) / 2(2)
• x = 12 / 4
• x = 3
Now, the y-coordinate of the vertex:
• y = c – (b2 / 4a)
• y = 16 – ((-12)2 / 4(2))
• y = 16 – (144 / 8)
• y = 16 – 18
• y = -2
So, the vertex of the parabola is (3, -2), just as we found before. You can also see this in the graph below.
How To Find The Vertex Of A Parabola From Two Points (Symmetry)
If we have two points on a horizontal line that are equidistant from the vertex, then we can find the x-coordinate.
This is due to the symmetry of a parabola about the line x = h (where h is the x-coordinate of the vertex).
Since a parabola has symmetry, a horizontal line y = d will intersect the parabola at two points that are an equal distance from the vertex (or from the line of symmetry).
We can use this to our advantage to take the average of these two x-coordinates to find the x-coordinate of the vertex.
Example: How To Find The Vertex Of A Parabola From Two Points (Symmetry)
Let’s say we have two points (3, 7) and (9, 7) that lie on a parabola. Since the y-coordinates are the same (both are y = 7), we know that these two points lie on the same horizontal line (y = 7).
This means that the parabola will intersect this horizontal line at two points that are the same distance from the line of symmetry (the horizontal line x = h, which goes through the vertex).
Taking the average of the two x-values, we get (3 + 9) / 2 = 6. So, the x-coordinate of the vertex is 6.
We would need more information to find the y-coordinate of the vertex (without more information, we don’t even know if the parabola is concave or convex!)
How To Find The Vertex Of A Parabola From A Graph
To find the vertex of a parabola from a graph, all we need to do is take a look at where the curve has a local minimum (for a convex/concave up curve) or local maximum (for a concave/concave down curve).
This point (the local minimum or maximum) is the vertex of the parabola.
Example: How To Find The Vertex Of A Parabola From A Graph
Given the graph of the parabola y = 3×2 – 12x + 9 (pictured below):
We can see by visual inspection that the vertex is at (2, -3), since that point is the local minimum of the curve.
## Conclusion
Now you know how to find the vertex of a parabola from an equation, from two points on a horizontal line, or from a graph.
You can also learn more about how to change the shape of a parabola in my article here, or about whether a parabola is a function in my article here.
You can learn about the focus of a parabola and what it means here.
You can learn how to find the domain and range of a parabola here.
You might also want to read my article on common questions about ellipses (a parabola is a conic section, just like an ellipse).
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09
Feb 2021
## Triangle Inequality Rule
By: Rich Zwelling, Apex GMAT Instructor
Date: 9th February, 2021
One of the less-common but still need-to-know rules tested on the GMAT is the “triangle inequality” rule, which allows you to draw conclusions about the length of the third side of a triangle given information about the lengths of the other two sides.
Often times, this rule is presented in two parts, but I find it is easiest to condense it into one, simple part that concerns a sum and a difference. Here’s what I mean, and we’ll use a SCENARIO:
Suppose we have a triangle that has two sides of length 3 and 5:
What can we say about the length of the third side? Of course, we can’t nail down a single definitive value for that length, but we can actually put a limit on its range. That range is simply the difference and the sum of the lengths of the other two sides, non-inclusive.
So, in this case, since the difference between the lengths of the other two sides is 2, and their sum is 8, we can say for sure that the third side of this triangle must have a length between 2 and 8, non-inclusive. [Algebraically, this reads as (5-3) < x < (5+3) OR 2 < x < 8.]
If you’d like to see that put into words:
**The length of any side of a triangle must be shorter than the sum of the other two side lengths and longer than the difference of the other two side lengths.**
It’s important to note that this works for any triangle. But why did we say non-inclusive? Well, let’s look at what would happen if we included the 8 in the above example. Imagine a “triangle” with lengths 3, 5, and 8. Can you see the problem? (Think about it before reading the next paragraph.)
Imagine a twig of length 3 inches and another of length 5 inches. How would you form a geometric figure of length 8 inches? You’d simply join the two twigs in a straight line to form a longer, single twig of 8 inches. It would be impossible to form a triangle with a side of 8 inches with the original two twigs.
If you wanted to form a triangle with the twigs of 3 and 5, you’d have to “break” the longer twig of 8 inches and bend the two twigs at an angle for an opportunity to have a third side, guaranteed to be shorter than 8 inches:
The same logic would hold for the other end of the range (we couldn’t have a triangle of 3, 5, and 2, as the only way to form a length of 5 from lengths of 2 and 3 would be to form a longer line segment of 5.)
Now that we’ve covered the basics, let’s dive into a few problems, starting with this Official Guide problem:
If k is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths 2, 7, and k?
(A) one
(B) two
(C) three
(D) four
(E) five
As usual with the GMAT, it’s one thing to know the rule, but it’s another when you’re presented with a carefully worded question that tests your ability to pay close attention to detail. First, we are told that two of the lengths of the triangle are 2 and 7. What does that mean for the third side, given the triangle inequality rule? We know the third side must have a length between 5 (the difference between the two sides) and 9 (the sum of the two sides).
Here, you can actually use the answer choices to your advantage, at least to eliminate some answers. Notice that k is specified as an integer. How many integers do we know now are possible? Well, if k must be between 5 and 9 (and remember, it’s non-inclusive), the only options possibly available to us are 6, 7, and 8. That means a maximum of three possible values of k, thus eliminating answers D and E.
Since the GMAT is a time-intensive test, you might have to end up guessing now and then, so if you can strategically eliminate answers, it increases your chances of guessing correctly.
Now for this problem, there’s another condition given, namely that 2 < k < 7. We already determined that k must be 6, 7, or 8. However, of those numbers, only 6 fits in the given range 2 < k < 7. This means that 6 is the only legal value that fits for k. The correct answer is A.
#### Note:
It’s important to emphasize that the eliminate answers strategy is not a mandate. We’re simply presenting it as an option that works here because it is useful on many GMAT problems and should be explored and practiced as often as possible.
Check out the following links for our other articles on triangles and their properties:
Posted on
26
Jan 2021
## Isosceles Triangles and Data Sufficiency
By: Rich Zwelling, Apex GMAT Instructor
Date: 21st January, 2021
Although we’ve already discussed isosceles triangles a bit during our discussion of 45-45-90 (i.e. isosceles right) triangles, it’s worth discussing some other contexts in which you may see isosceles triangles on the GMAT, specifically on Data Sufficiency problems.
As we discussed before, an isosceles triangle is any triangle that features two equal sides and thus two equal opposite angles:
That’s an easy enough definition to remember, but how does the GMAT turn this into more challenging problems? For that, let’s take a look at the following Official Guide problem. Try to solve before reading the explanation below the problem:
In the figure above, what is the value of x + y ?
(1) x = 70
(2) ABC and ADC are both isosceles triangles
#### Explanation
In this case, it’s straightforward enough to determine that each statement alone will be insufficient. Statement (1) gives us a definitive value for x, but no information about y, thus we cannot answer the question (the value of x+y). And although Statement (2) labels each triangle in the diagram as isosceles, we have no way of knowing the specific angles involved nor their relationships.
However, as with many Data Sufficiency problems, especially those involving Geometry, things can get thorny when we have to combine the statements. The two statements look very complimentary, and that could lead us to prematurely conclude the answer is C (i.e. the two statements are sufficient when combined). But we must do a thorough check.
#### Reframing the question
Remember that at any point during a Data Sufficiency problem — beginning, middle, or end — you can reframe the question for simplicity. The question asks for the value of x+y. But now that we are combining the statements, we already know that x=70. In terms of sufficiency, then, what information do we need? The only thing missing is a definitive value of y. The question now might as well be “What is the value of y?”
Now, here’s where the GMAT thinking really comes into play. It’s one thing to understand what an isosceles triangle is. It’s quite another to judge what a diagram of an isosceles triangle does or does not tell you and what you can or cannot extrapolate from it.
One of my personal favorite things about Geometry Data Sufficiency problems is that they tend to be very intuitive visually. You can often answer them by manipulating figures.
We know that triangle ADC is isosceles, but is that enough to give us definitive measurements? Visually, which of these does it look like?
Without any numerical evaluations, we can see that we can’t get a definitive measure for the angle at D, which in this case is our y. So even when we combine the statements, we cannot get an answer to our question. The correct answer is E
Here’s another case of a tricky Data Sufficiency problem involving isosceles triangles:
In isosceles triangle RST, what is the measure of angle R?
• The measure of angle T is 100 degrees
• The measure of angle S is 40 degrees
Again, give the problem a shot before reading the answer and explanation.
#### Explanation
This is one for which you can draw a diagram, but it’s not necessary. The trick here is to remember another key property of triangles, namely that all angles in the triangle must sum to 180 degrees.
Since the triangle is isosceles, and since each statement gives you only one angle of three, the temptation can be to say that each statement is insufficient on its own. This is certainly the case for Statement (2), because the 40-degree angle could be one of a pair (in which case we would have a 40-40-100 triangle) or the 40-degree angle could be the odd angle out (in which case we would have a 40-70-70 triangle).
Because the problem asks for the value of R, and since R could be 40, 70, or 100 depending on the situations outlined above, Statement (2) is INSUFFICIENT.
However, there’s a catch when evaluating Statement (1). Notice that angle T is an obtuse angle, meaning it is greater than 90 degrees. Is it possible that there are two 100-degree angles in a triangle? This would produce a total of 200 degrees, which would exceed the 180-degree total for any triangle. As such, the only possibility is that the 100 degree angle is the odd angle out, and the other two angles are equal acute angles (specifically, we have a 40-40-100 triangle).
Now we know R must be 40 degrees. Statement (1) is sufficient, and the correct answer is A.
But notice how the GMAT sets the statements up to bait you into thinking that you must combine the two statements to figure out the value of angle R.
Now that we’ve finished talking about the basic triangle types, we can move on to talking about what happens when triangles are used within different shapes. In the meantime, here are links to our other triangle articles:
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Q)
# For the curve $$y = 4x^3 – 2x^5$$, find all the points at which the tangent passes through the origin.
$\begin{array}{1 1} (A)\;(0,0),(1,-2),(-1,2) \\ (B)\;(0,0),(-1,-2),(1,2)\\ (C)\;(0,0),(1,2),(-1,2) \\ (D)\;(0,0),(1,-2) \end{array}$
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A)
Toolbox:
• Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
• Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1) Step 1: Let (x_1,y_1) be the required point on the given curve y=4x^3-2x^5 \Rightarrow y_1=4x_1^3-2x_1^5-----(1) Differentiating equ(1) w.r.t x we get, \large\frac{dy}{dx}$$=12x^2-10x^4$
Therefore $\large\frac{dy}{dx}_{(x_1,y_1)}$$=12x_1^2-10x_1^4$
Step 2:
Equation of the tangent is $y-y_1=(12x_1^2-10x_1^4)(x-x_1)$
This passes through the origin hence $0-y_1=(12x_1^2+10x_1^4)(0-x_1)$
$\Rightarrow y_1=12x_1^3-10x_1^5$------(2)
Subtract equ(2) from equ(1)
$0=-8x_1^3+8x_1^5$
$\Rightarrow 8x_1^3(x_1^2-1)=0$
$\Rightarrow x_1=0$ or $x_1=\pm 1$
Step 3:
When $x_1=0$ from equ(2) $y=0$
When $x_1=1$ from equ (2)
$y_1=4(1)^2-2(1)^5$
$\;\;\;\;=4-2$
$\;\;\;\;=2$
When $x_1=-1$ from equ(2)
$y_1=4(-1)^3-2(-1)^5$
$\;\;\;\;=-4-(-2)$
$\;\;\;\;=-2$
Step 4:
Hence the required points are $(0,0),(1,2)$ and $(-1,-2)$
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Q: Why is the integral/antiderivative the area under a function?
Physicist: If you’ve taken calculus, then at some point you learned that to find the area under a function (generally written ) you need to find the anti-derivative of that function. The most natural response to these types of theorems is “wait… what?… why?”.
This theorem is so important and widely used that it’s called the “fundamental theorem of calculus”, and it ties together the integral (area under a function) with the antiderivative (opposite of the derivative) so tightly that the two words are essentially interchangeable. However, there are some mathematicians who may take issue with mixing up the two terms.
It comes back (in a roundabout way) to the fact that the derivative of a function is the slope of that function or the “rate of change”. In what follows “f” is a function, and “F” is its anti-derivative (that is: F’ = f).
Intuitively: Say you’ve got a function f(x), and the area under f(x) (up to some value x) is given by A(x).
Then the statement “the area, A, is given by the anti-derivative of f” is equivalent to “the derivative of A is given by f”.
In other words, the rate at which the area increases (as you slide x to the right) is given by the height, f(x).
For example, if the height of the function were 3, then, for a moment, the area under the function is increasing by 3 for every 1 unit of distance you slide to the right. Keep in mind that the function can move up and down as much as it wants. As far as the function “knows”, at any particular moment it may as well be constant (dotted line in picture above).
So if the height of the function (which is just the function) is the rate at which the area changes, then f is the derivative of the area: A’=f. But that’s exactly the same as saying that the area is the anti-derivative of the function.
Mathematically: There’s a theorem called the mean value theorem that states that if you have a “smooth” function with no sudden bends or kinks, then over any interval the derivative will be equal to the average slope at least once. This needs a picture:
More precisely, if you have a function on the interval [A,B], then there’s a point c between A and B such that . You can just as easily write this as or (since F’ =f).
So if you drive 60 miles in one hour, then at some instant you must have been driving at exactly 60 mph, even though for almost the entire trip you may have been traveling much faster or much slower than 60 mph.
Keep that stuff in the back of your mind for a moment, and ponder instead how to go about approximating the area under a function.
You can divide up the area between x=A and x=B under a function by putting a mess of rectangles under it. Divide up the interval [A,B] by picking a string of points x 0 , x 1 , x 2 , …, x N , and use these as the left and right sides of your rectangles (and set x 0 =A and x N =B).
The point, c i , that you pick in between each x i-1 and x i is unimportant. To get the exact area you let N, the total number of rectangles, go flying off to infinity, and you’ll find that the highest value of f and the lowest value of f in each tiny interval gets squeezed together.
So, why not choose a value of c i so that in each rectangle you can say ?
Holy crap! The area under the function (the integral) is given by the antiderivative! Again, this approximation becomes an equality as the number of rectangles becomes infinite.
As an aside (for those of you who really wanted to read an entire post about integrals), integrals are surprisingly robust. That is to say, if your function has a kink in it (the way |x| has a kink at zero, for example) then you can’t find a derivative at that kink, but integrals don’t have that problem. If there’s a kink or even a discontinuity; no problem!
You can just put the edge of a rectangle at the problem point, and then ignore it. In fact, think of (almost) any function in your head… You can take the integral of that. It may have an infinite value, or something awful like that, but you can still take the integral.
To make a function that can’t be integrated you have to make it infinitely messed up. Mathematicians live for this sort of thing. There is almost nothing in the world they enjoy more than coming up with ways to break each other’s theories. One of the classic examples is the function
Over any interval you pick, f still jumps around infinitely often, so the whole “things will get better as the number of rectangles increases” thing can never get off the ground. There are fixes to this, but they come boiling and howling up out of the ever-darker, stygian abyss that is measure theory.
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Who Wants Pizza? - Part 5
The last time that Ms. Martinez ordered pizza, there was 2/3 of a pizza left. Bobby came in and ate 1/2 of what was left. The other children were mad that Bobby ate 1/2 of it. Bobby said, "I only ate 2 pieces." Was he right? If there were12 pieces in a whole pizza, how many pieces did he eat? What fraction of a whole pizza did he eat?
Let's Explore
Here are the 12 original pieces of the pizza.
The children left 2/3 of it.
121 x 23 = 8
Bobby ate 1/2 of the 8 pieces left.
81 x 12 = 4
So Bobby ate 4 pieces of pizza.
On the next page we will investigate the fraction of the pizza that Bobby ate.
Something for You to Do
Directions:Use colored markers and the square grid paper to draw the fractions as we have above and multiply the following.
1 1/4 x 2/3 of 12 pieces
(a. 2) (b. 1) (c. 12) (d. 8) (e. 4)
2 4/5 x 1/4 of 20 pieces
(a. 2) (b. 1) (c. 12) (d. 8) (e. 4)
3 2/3 x 3/4 of 24 pieces
(a. 2) (b. 1) (c. 12) (d. 8) (e. 4)
4 4/5 x 1/2 of 20 pieces
(a. 2) (b. 1) (c. 12) (d. 8) (e. 4)
5 1/2 x 1/2 of 4 pieces
(a. 2) (b. 1) (c. 12) (d. 8) (e. 4)
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Check it out!
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# Check it out! - PowerPoint PPT Presentation
Check it out!. A top fuel dragster (a car built for drag racing) can travel
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Check it out!
4.3.1: Interpreting Slope and y-intercept
mile in 4 seconds. The dragster’s distance over time is graphed on the next slide. The graph assumes a constant speed. Use the graph to complete problems 1 and 2. Then use what you know about slope-intercept form to answer the remaining questions.
4.3.1: Interpreting Slope and y-intercept
Write the algebraic equation of the line.
What is the slope of a line with the equation y = –x + 7?
What is the y-intercept of a line with the equation
y= 3x – 2?
4.3.1: Interpreting Slope and y-intercept
The slope is or . To calculate the slope, find any two points on the line.
The graph shows that (0, 0) and (16, 1) are both points on the line. The formula to find the slope between two points (x1, y1) and (x2, y2) is . Substitute (0, 0) and (16, 1) into the formula to find the slope.
4.3.1: Interpreting Slope and y-intercept
The slope between the two points (0, 0) and (16, 1) is 16.
The y-intercept is the point at which the graph crosses the y-axis. The graph shows that the y-intercept is 0, or (0, 0).
4.3.1: Interpreting Slope and y-intercept
Write the algebraic equation of the line.
The equation of a line can be written in the form y= mx + b, where m is the slope of the line and b is the y-intercept. The equation of the line is y= 16x + 0 or y = 16x.
4.3.1: Interpreting Slope and y-intercept
What is the slope of a line with the equation
y= –x + 7?
If the equation of a line is in the form y = mx + b, m is the slope of the line.
The slope of the line y = –x + 7 is –1.
4.3.1: Interpreting Slope and y-intercept
What is the y-intercept of a line with the equation
y= 3x – 2?
If the equation of a line is in the form y = mx + b, then b is the y-intercept.
The y-intercept of y = 3x – 2 is –2, or the point (0, –2).
4.3.1: Interpreting Slope and y-intercept
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Five Number Summary
Quartiles or medians and the minimum and maximum values.
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Practice Five Number Summary
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Five Number Summary
When given a long list of numbers, it is useful to summarize the data. One way to summarize the data is to give the lowest number, the highest number and the middle number. In addition to these three numbers it is also useful to give the median of the lower half of the data and the median of the upper half of the data. These five numbers give a very concise summary of the data.
What is the five number summary of the following data?
0, 0, 1, 2, 63, 61, 27, 13
The Five Number Summary
Suppose you have ordered data with m\begin{align*}m\end{align*} observations. The rank of each observation is shown by its index.The rank of an observation is the number of observations that are less than or equal to the value of that observation.
y1y2y3ym\begin{align*}y_1 \le y_2 \le y_3 \le \cdots \le y_m\end{align*}
In data sets that are large enough, you can divide the numbers into four parts called quartiles. The quartiles of interest are the first quartile, Q1\begin{align*}Q1\end{align*}, the second quartile, Q2\begin{align*}Q2\end{align*}, and the third quartile Q3\begin{align*}Q3\end{align*}. The second quartile, Q2\begin{align*}Q2\end{align*}, is defined to be the median of the data. The first quartile, Q1\begin{align*}Q1\end{align*}, is defined to be the median of the lower half of the data. The third quartile, Q3\begin{align*}Q3\end{align*}, is similarly defined to be the median of the upper half of the data.
These three numbers in addition to the minimum and maximum values are the five number summary. Note that there are variations of the five number summary that you can study in a statistics course.
Take the following data:
2, 7, 17, 19, 25, 26, 26, 32
There are 8 observations total in this set of data.
• Lowest value (minimum) : 2
• Q1:7+172=12\begin{align*}Q1:\frac{7+17}{2}=12\end{align*} (Note that this is the median of the first half of the data - 2, 7, 17, 19)
• Q2:19+252=22\begin{align*}Q2:\frac{19+25}{2}=22\end{align*} (Note that this is the median of the full set of data)
• Q3:26\begin{align*}Q3: 26\end{align*} (Note that this is the median of the second half of the data - 25, 26, 26, 32)
• Upper value (maximum) : 32
The 5 number summary is 2, 12, 22, 26, 32.
Examples
Example 1
Earlier you were asked to compute the five number summary for 0, 0, 1, 2, 63, 61, 27, 13. It helps to order the data.
0, 0, 1, 2, 13, 27, 61, 63
• Since there are 8 observations, the median is the average of the 4th\begin{align*}4^{th}\end{align*} and 5th\begin{align*}5^{th}\end{align*} observations: 2+132=7.5\begin{align*}\frac{2+13}{2}=7.5\end{align*}
• The lowest observation is 0.
• The highest observation is 63.
• The middle of the lower half is 0+12=0.5\begin{align*}\frac{0+1}{2}=0.5\end{align*}
• The middle of the upper half is 27+612=44\begin{align*}\frac{27+61}{2}=44\end{align*}
The five number summary is 0, 0.5, 7.5, 44, 63
Example 2
Create a set of data that meets the following five number summary:
{2, 5, 9, 18, 20}
Suppose there are 8 data points. The lowest point must be 2 and the highest point must be 20. The middle two points must average to be 9 so they could be 8 and 10. The second and third points must average to be 5 so they could be 4 and 6. The sixth and seventh points need to average to be 18 so they could be 18 and 18. Here is one possible answer:
2, 4, 6, 8, 10, 18, 18, 20
Example 3
Compute the five number summary for the following data:
1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 8, 9, 10, 15
There are 20 observations.
• Lower : 1
• Q1:2+32=2.5\begin{align*}Q1 : \frac{2+3}{2}=2.5\end{align*}
• Q2:4+52=4.5\begin{align*}Q2 : \frac{4+5}{2}=4.5\end{align*}
• Q3:6+72=6.5\begin{align*}Q3 : \frac{6+7}{2}=6.5\end{align*}
• Upper : 15
Example 4
Compute the five number summary for the following data:
4, 8, 11, 11, 12, 14, 16, 20, 21, 25
There are 10 observations total in this set of data.
• Lowest value (minimum) : 4
• Q1:11\begin{align*}Q1 : 11\end{align*} (Note that this is the median of the first half of the data - 4, 8, 11, 11, 12)
• Q2:12+142=13\begin{align*}Q2 : \frac{12+14}{2}=13\end{align*} (Note that this is the median of the full set of data)
• Q3:20\begin{align*}Q3 : 20\end{align*} (Note that this is the median of the second half of the data - 14, 16, 20, 21, 25)
• Upper value (maximum) : 25
The five number summary is 4, 11, 13, 20, 25.
Example 5
Compute the five number summary for the following data:
3, 7, 10, 14, 19, 19, 23, 27, 29
There are 9 observations total. To calculate Q1\begin{align*}Q1\end{align*} and Q3\begin{align*}Q3\end{align*}, you should include the median in both the lower half and upper half calculations.
• Lowest value (minimum) : 3
• Q1:10\begin{align*}Q1 : 10\end{align*} (this is the median of 3, 7, 10, 14, 19)
• Q2:19\begin{align*}Q2 : 19\end{align*}
• Q3:23\begin{align*}Q3 : 23\end{align*} (this is the median of 19, 19, 23, 27, 29)
• Upper value (maximum) : 29
The five number summary is 3, 10, 19, 23, 29.
Review
Compute the five number summary for each of the following sets of data:
1. 0.16, 0.08, 0.27, 0.20, 0.22, 0.32, 0.25, 0.18, 0.28, 0.27
2. 77, 79, 80, 86, 87, 87, 94, 99
3. 79, 53, 82, 91, 87, 98, 80, 93
4. 91, 85, 76, 86, 96, 51, 68, 92, 85, 72, 66, 88, 93, 82, 84
5. 335, 233, 185, 392, 235, 518, 281, 208, 318
6. 38, 33, 41, 37, 54, 39, 38, 71, 49, 48, 42, 38
7. 3, 7, 8, 5, 12, 14, 21, 13, 18
8. 6, 22, 11, 25, 16, 26, 28, 37, 37, 38, 33, 40, 34, 39, 23, 11, 48, 49, 8, 26, 18, 17, 27, 14
9. 9, 10, 12, 13, 10, 14, 8, 10, 12, 6, 8, 11, 12, 12, 9, 11, 10, 15, 10, 8, 8, 12, 10, 14, 10, 9, 7, 5, 11, 15, 8, 9, 17, 12, 12, 13, 7, 14, 6, 17, 11, 15, 10, 13, 9, 7, 12, 13, 10, 12
10. 49, 57, 53, 54, 49, 67, 51, 57, 56, 59, 57, 50, 49, 52, 53, 50, 58
11. 18, 20, 24, 21, 5, 23, 19, 22
12. 900, 840, 880, 880, 800, 860, 720, 720, 620, 860, 970, 950, 890, 810, 810, 820, 800, 770, 850, 740, 900, 1070, 930, 850, 950, 980, 980, 880, 960, 940, 960, 940, 880, 800, 850, 880, 760, 740, 750, 760, 890, 840, 780, 810, 760, 810, 790, 810, 820, 850
13. 13, 15, 19, 14, 26, 17, 12, 42, 18
14. 25, 33, 55, 32, 17, 19, 15, 18, 21
15. 149, 123, 126, 122, 129, 120
To see the Review answers, open this PDF file and look for section 15.3.
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Color Highlighted Text Notes
Vocabulary Language: English
first quartile
The first quartile, also known as $Q_1$, is the median of the lower half of the data.
five number summary
The five number summary of a set of data is the minimum, first quartile, second quartile, third quartile, and maximum.
Lower quartile
The lower quartile, also known as $Q_1$, is the median of the lower half of the data.
Maximum
The largest number in a data set.
Median
The median of a data set is the middle value of an organized data set.
Minimum
The minimum is the smallest value in a data set.
Quartile
A quartile is each of four equal groups that a data set can be divided into.
rank
The rank of an observation is the number of observations that are less than or equal to the value of that observation.
second quartile
The second quartile, also known as $Q_2$, is the median of the data.
third quartile
The third quartile, also known as $Q_3$, is the median of the upper half of the data.
Upper Quartile
The upper quartile, also known as $Q_3$, is the median of the upper half of the data.
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Simplification & Approximation Tips & Tricks
The Quantitative Aptitude section of the Bank exam consists of questions such as Simplification, Number Series, Permutation & Combination, Quadratic Equation, Data Interpretation, Data Analysis and other Miscellaneous questions.
There are almost 10 to 12 questions in the paper which deal with topics such as Percentage and Average, ratio & proportions, partnerships, Profit & loss, calculations of simple & compound interests etc.
Here is a short study-guide to help you crack questions on “Simplification and Approximation“,
As we all know that simplification is most widely asked topic in almost every banking exam. So let us try to understand what is actually meant by word ‘Simplification’.
Simplification means to find out a final answer for a complex calculation.
Simplification questions are asked to check the ability of a student to deal with numbers which can be in one of the following two types.
• Sometimes, a calculation is given and one of the numbers is missing from the calculation. To find out the missing number, we have to approximate the given numbers or do the basic operations.
• Sometimes all the numbers are given with some operations between them & we have to simplify the calculation.
Rules related to Simplification
Rule-(I) Replace ‘of’ by ‘Multiplication’ & ‘/’ by ‘Division’
Explanation: Whenever we find ‘of’ in a simplification problem, we can replace that by ‘multiplication(*)’. Similiarly ‘/’ can be replaced by ‘÷’.
Example: Find ¼ of 20
Solution: (¼) x 20 = 20÷4 = 5
Rule-(II) Always keep in mind the “BODMAS” rule. These operations have priorities in the same order as mentioned.
Explanation: Whenever we have more than one operation in the given calculation, we have to do the operations according to the priority specified by ‘BODMAS’
• B-Bracket
• O-Of (means multiplication)
• D-Division
• M-Multiplication
• S-Subtraction
Example: Simplify: (2+3)*30
Solution: In this question, we have two things-Bracket & Multiplication. According to the BODMAS rule, we have to solve bracket first and not multiplication. So now coming to bracket, we have only one operation-Addition, so we will do addition.
(2+3)*30 = 5*30
Now we have only one operation to do – Multiplication
5*30 = 150
Example: Simplify: (2+5) of 80
Solution: In this question, we have three things – bracket, addition & of. Replacing ‘of’ by ‘multiplication’.
(2+5) of 30 = (2+5)*80
Now we have three things – bracket, addition & Multiplication. According to the BODMAS rule, we have to solve bracket first and not multiplication. So now coming to bracket, we have only one operation-Addition, we will do addition.
(2+5)*80 = 7*80
Now we will do multiplication.
7*80 = 560
Rule-(III) Multiplication & Division have same priority(Do that operation first which is on left).
Explanation: Though division has more priority than multiplication according to ‘BODMAS’ but we can perform multiplication first.
Example: 8*30/15
8*30 ÷ 15
Solution: In this question, we have two things – Multiplication & Division. So we can perform any operation first as they have same priority.
Doing Multiplication first:
240 ÷ 15
16
Doing division first:
8*2
16
Rule-(IV) Addition & Subtraction have same priority.
Explanation: Though addition has more priority than division according to ‘BODMAS’ but we can perform any of the two operations first.
Example: 30+40-15
Solution: In this question, we have two things – Addition & Subtraction. So we can perform any operation first as they have same priority.
70 – 15
55
Doing Subtraction first:
30 + 25
55
Rule-(V) Don’t hesitate in rounding the numbers to nearest integers.
Explanation: Most of the times the numbers are given in such a way that you can round them quickly and get the answer (Rounding should be done or not, It can be realised by looking at the given options).
Example: (324.5*15)/(5.01*24.98)
Solution: (325*15)/(5*25)
=13*3
=39
Some previous year questions asked in banking exams from simplification
Now let us see some of the previous year questions asked from ‘Simplification’ & try to apply the rules learnt so far.
Q.1) Simplify: 127.001 * 7.998 + 6.05 * 4.001
1. 1000
2. 1020
3. 1040
4. 1080
5. None of these
Solution: Using the rounding concept
127 * 8 + 6 * 4
Using the BODMAS rule
1016 + 24
1040 (Option 3)
Q.2) What will come at place of ?: 9876 ÷ 24.96 + 215.005 – ? = 309.99
1. 270
2. 280
3. 290
4. 300
5. 310
Solution: Using the rounding concept
9875 ÷ 25 + 215 – ? = 310
Using the BODMAS rule
395 + 215 – ? = 310
610 – ? = 310
? = 300 (Option 4)
Q.3) What will come at place of a: (128 ÷ 16 x a – 7*2)/(72-8*6+a2) = 1
1. 1
2. 5
3. 9
4. 13
5. 17
Solution: Using the BODMAS rule
(8*a – 14)/(49-48+a2) = 1
(8*a – 14)/(1 + a2) = 1
8a – 14 = 1 + a2
a2 – 8a + 15 = 0
a=3 or 5 (Option 2)
Q.4) What will come at place of ?: 85.147 + 34.192*6.2 + ? = 802.293
1. 400
2. 450
3. 550
4. 600
5. 500
Solution: Using the rounding concept
85 + 35*6 + ? = 803
Using the BODMAS rule
85 + 210 + ? = 803
295 + ? = 803
? = 508 [approx. = 500] (Option 5)
Q.5) What will come at place of ?: 3/8 of 168*15 ÷ 5 + ? = 549 ÷ 9 + 235
1. 189
2. 107
3. 174
4. 296
5. None of these
Solution: Using the BODMAS rule
(3*168÷ 8)*15 ÷ 5 + ? = 549 ÷ 9 + 235
(504÷ 8)*3 + ? = 61 + 235
63*3 + ? = 296
189 + ? = 296
? = 107 (Option 2)
Key points to remember related to Simplification:
• Replace ‘of’ by ‘Multiplication’.
• Replace ‘/’ by ‘Division’.
• Always do the operations in priority according to ‘BODMAS’.
• Division & Multiplication have same priority (Start from left).
• Addition & Subtraction have same priority.
• Rounding can be done to simplify problems.
• When the given options are very close then rounding doesn’t help much.
• Always look at the options before doing simplification that can help in elimination of options.
Basic Rules of Simplification
BODMAS Rule
It defines the correct sequence in which operations are to be performed in a given mathematical expression to find the correct value. This means that to simplify an expression, the following order must be followed –
B = Bracket,
O = Order (Powers, Square Roots, etc.)
D = Division
M = Multiplication
S = Subtraction
1. Hence, to solve simplification questions correctly, you must apply the operations of brackets first. Further, in solving for brackets, the order – (), {} and [] – should be stricly followed.
2.
Next you should evaluate exponents (for instance powers, roots etc.)
3. Next, you should perform division and multiplication, working from left to right. (division and multiplication rank equally and are done left to right).
4. Finally, you should perform addition and subtraction, working from left to right. (addition and subtraction rank equally and are done left to right).
EXAMPLE 1: Solve 12 + 22 ÷ 11 × (18 ÷ 3)^2 – 10
= 12 + 22 ÷ 11 × 6^2 – 10 (Brackets first)
= 12 + 22 ÷ 11 × 36 – 10 (Exponents)
= 12 + 2 × 36 – 10 = 12 + 72 – 10 (Division and multiplication, left to right)
= 84 – 10 = 74 (Addition and Subtraction, left to right)
EXAMPLE 2: Solve 4 + 10 – 3 × 6 / 3 + 4
= 4 + 10 – 18/3 + 4 = 4 + 10 – 6 + 4 (Division and multiplication, left to right)
= 14 – 6 + 4 = 8 + 4 = 12 (Addition and Subtraction, left to right)
To Solve Modulus of a Real Number
The Modulus (or the absolute value) of x is always either positive or zero, but never negative. For any real number x, the absolute value or modulus of x is denoted by |x| and is defined as
|x|x {if ≥ 0} and x {if 0}
EXAMPLE 1: Solve |8|
|8| = |-8| = 8
Tips to Crack Approximation
Conversion of decimal numbers to nearest number
To solve such questions, first convert the decimal to nearest value. Then simplify the given equation using the new values that you have obtained.
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EXAMPLE 1: Solve 4433.764 – 2211.993 – 1133.667 + 3377.442
Here,
4433.764 = 4434
2211.993 = 2212
1133.667 = 1134
3377.442 = 3377
Now simplify, 4434 – 2212 – 1134 + 3377 = 4466
EXAMPLE 2: Solve 530 x 20.3% + 225 x 16.8%
Here, 20.3% becomes 20% and 16.8% becomes 17%
Now, simplify 530 x 20% + 225 x 17%
= 106 + 38.25 = 144.25
Approximation of Square Roots
(1) To simplify a square root, you can follow these steps:
(2) Factor the number inside the square root sign.
(3) If a factor appears twice, cross out both and write the factor one time to the left of the square root sign. If the factor appears three times, cross out two of the factors and write the factor outside the sign, and leave the third factor inside the sign. Note: If a factor appears 4, 6, 8, etc. times, this counts as 2, 3, and 4 pairs, respectively.
(4) Multiply the numbers outside the sign.
(5) Multiply the numbers left inside the sign.
(6) To simplify the square root of a fraction, simplify the numerator and simplify the denominator.
NOTE: Check that the outside number squared times the inside number should equal the original number inside the square root.
Thanks
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# 数学代写|微积分代写Calculus代写|MAST10006
## 数学代写|微积分代写Calculus代写|Vertical tangents
There is yet one more feature on a graph that can indicate nondifferentiability.
Example 6 Let $f(x)=x^{1 / 3}$. Find $f^{\prime}(0)$.
Solution We use the derivative formula with $k=0$ :
\begin{aligned} f^{\prime}(0) &=\frac{f(0+\alpha)-f(0)}{\alpha}=\frac{\alpha^{1 / 3}-0^{1 / 3}}{\alpha} \ &=\frac{\alpha^{1 / 3}}{\alpha}=\frac{1}{\alpha^{2 / 3}}=A^{2 / 3} \ & \doteq \infty . \end{aligned}
We are not stuck at $A^{2 / 3}$ because of the even numerator in the exponent; $A^{2 / 3}=\left(A^2\right)^{1 / 3}$, and $A^2$ is positive. We can render $\infty$. Because we rendered $\infty$ rather than a real number, $f^{\prime}(0)$ does not exist.
Since $f^{\prime}(0)=\infty$, it seems the slope of the tangent line at $x=0$ is positive infinite. This is true, as verified by the graph; see figure 6 .
The procedure for determining the equation of the tangent line to the curve $y=x^{1 / 3}$ at $x=0$ is not the same as before. Because the equation of a vertical line is of the form $x=k$, we cannot use the pointslope equation of the line. Instead, we simply use the $x$-coordinate at which we are finding the derivative; the equation of the tangent line is $x=0$.
If the derivative renders an infinite value, the tangent line is vertical. Figure 7 shows four of the possibilities.The two graphs on the right contain a cusp, where there is a corner with a vertical tangent line. The cusp is a point on the graph; the function is defined there. Because of the manner in which calculators and computer software calculate and display graphs of functions, it is sometimes difficult to distinguish cusps from vertical asymptotes using only a graph, but calculus can help you determine the difference!
## 数学代写|微积分代写Calculus代写|Proofs of theorems
The local linearity formula is part of the local linearity theorem.
Theorem 2 LOCAL LINEARITY THEOREM Let $f$ be differentiable at $x=k$ and let $\alpha$ be an infinitesimal.
(1) If not both $f(k)=0$ and $f^{\prime}(k)=0$, then $f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k)$.
(2) If $f(k)=0=f^{\prime}(k)$, then $f(k+\alpha)$ is zero or is on a lower level than $\alpha$.
Proof. (1) Case 1: $f^{\prime}(k) \neq 0$.
Since $f$ is differentiable at $k, \frac{f(k+\alpha)-f(k)}{\alpha} \doteq f^{\prime}(k)$. Since $f^{\prime}(k) \neq 0$ is real, $\frac{f(k+\alpha)-f(k)}{\alpha} \approx f^{\prime}(k)$, and in fact $\frac{f(k+\alpha)-f(k)}{\alpha}=f^{\prime}(k)+\beta$, where $\beta$ is zero or infinitesimal. Then, $f(k+\alpha)-f(k)=\alpha f^{\prime}(k)+\alpha \beta$, and $f(k+\alpha)=f(k)+\alpha f^{\prime}(k)+\alpha \beta$. Because $\alpha \beta$ is either zero or is on a lower level than $\alpha f^{\prime}(k)$, this term can be discarded and $f(k+\alpha) \approx$ $f(k)+\alpha f^{\prime}(k)$
(1) Case 2: $f(k) \neq 0, f^{\prime}(k)=0$.
Then, since $\frac{f(k+\alpha)-f(k)}{\alpha} \doteq f^{\prime}(k)=0, \frac{f(k+a)-f(k)}{\alpha}=\beta$, where $\beta$ is zero or infinitesimal. Then, $f(k+\alpha)=f(k)+\alpha \beta \approx f(k)=f(k)+\alpha f^{\prime}(k)$.
(2) Now suppose $f^{\prime}(k)=f(k)=0$. Then, $\frac{f(k+a)-f(k)}{\alpha}=\frac{f(k+\alpha)}{\alpha} \doteq 0$, so $\frac{f(k+\alpha)}{\alpha}$ is zero or infinitesimal. Thus, either $f(k+\alpha)=0$ or $f(k+\alpha)$ is on a lower level than $\alpha$.
It remains to furnish a complete proof of theorem 1. Let $f$ be a function that is differentiable at $x=k$. We wish to prove that $f$ is continuous at $x=k$.
Proof. (theorem 1) Case (1): $f(k) \neq 0$. By the local linearity theorem part $(1), f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k) \approx f(k)$. By definition, $f$ is continuous at $x=k$.
Case (2): $f(k)=0, f^{\prime}(k) \neq 0$. By the local linearity theorem part (1), $f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k)=\alpha f^{\prime}(k) \doteq 0=f(k)$. By definition, $f$ is continuous at $x=k$.
Case (3): $f(k)=0, f^{\prime}(k)=0$. By the local linearity theorem part (2), $f(k+\alpha) \doteq 0=f(k)$. By definition, $f$ is continuous at $x=k$.
Formal proofs that involve the local linearity formula $f(k+\alpha) \approx$ $f(k)+\alpha f^{\prime}(k)$ nearly always require the use of both parts of the local linearity theorem. In this text, results are usually derived using the local linearity formula only and the rest of the details are left unstated.
# 微积分代考
## 数学代写|微积分代写Calculus代写|Vertical tangents
$$f^{\prime}(0)=\frac{f(0+\alpha)-f(0)}{\alpha}=\frac{\alpha^{1 / 3}-0^{1 / 3}}{\alpha} \quad=\frac{\alpha^{1 / 3}}{\alpha}=\frac{1}{\alpha^{2 / 3}}=A^{2 / 3} \doteq \infty$$
## 数学代写|微积分代写Calculus代写|Proofs of theorems
(1) 如果不是两者 $f(k)=0$ 和 $f^{\prime}(k)=0$ ，然后 $f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k)$.
(2) 如果 $f(k)=0=f^{\prime}(k)$ ，然后 $f(k+\alpha)$ 为雴或低于 $\alpha$.
$(-)$ 案例二: $f(k) \neq 0, f^{\prime}(k)=0$.
(2) 现在假设 $f^{\prime}(k)=f(k)=0$. 然后， $\frac{f(k+a)-f(k)}{\alpha}=\frac{f(k+\alpha)}{\alpha} \doteq 0$ ，所以 $\frac{f(k+\alpha)}{\alpha}$ 为零或无穷小。因此， 无论是 $f(k+\alpha)=0$ 或者 $f(k+\alpha)$ 低于 $\alpha$.
$f(k+\alpha) \approx f(k)+\alpha f^{\prime}(k)=\alpha f^{\prime}(k) \doteq 0=f(k)$. 根据定义, $f$ 是连续的 $x=k$.
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## Engage NY Eureka Math 4th Grade Module 5 Lesson 40 Answer Key
### Eureka Math Grade 4 Module 5 Lesson 40 Problem Set Answer Key
Question 1.
The chart to the right shows the height of some football players.
Player Height (in feet) A 6$$\frac{1}{4}$$ B 5$$\frac{7}{8}$$ C 6$$\frac{1}{2}$$ D 6$$\frac{1}{4}$$ E 6$$\frac{2}{8}$$ F 5$$\frac{7}{8}$$ G 6$$\frac{1}{8}$$ H 6$$\frac{5}{8}$$ I 5$$\frac{6}{8}$$ J 6$$\frac{1}{8}$$
A: 6(1/4) = 6.2
Explanation:
In the above-given question,
given that,
The height of the Football A.
6(1/4) = 6 x 4.
6 x 4 = 24.
24 + 1/4 = 25/4.
25/4 = 6.2.
B: 5(7/8) = 5.8.
Explanation:
In the above-given question,
given that,
The height of the Football B.
5(7/8) = 5 x 8.
5 x 8 = 40.
40 + 7/8 = 47/8.
47/8 = 5.8.
C: 6(1/2) = 6.5.
Explanation:
In the above-given question,
given that,
The height of the Football A.
6(1/2) = 6 x 2.
6 x 2 = 12.
12 + 1/2 = 13/2.
13/2 = 6.5.
D: 6(1/4) = 6.2
Explanation:
In the above-given question,
given that,
The height of the Football D.
6(1/4) = 6 x 4.
6 x 4 = 24.
24 + 1/4 = 25/4.
25/4 = 6.2.
E: 6(2/8) = 6.2
Explanation:
In the above-given question,
given that,
The height of the Football A.
6(2/8) = 6 x 8.
6 x 8 = 48.
48 + 2/8 = 50/8.
50/8 = 6.2.
F: 5(7/8) = 5.8
Explanation:
In the above-given question,
given that,
The height of the Football F.
5(7/8) = 5 x 8.
5 x 8 = 40.
40 + 7/8 = 47/8.
47/8 = 5.8.
G: 6(1/8) = 6.1
Explanation:
In the above-given question,
given that,
The height of the Football G.
6(1/8) = 6 x 8.
6 x 8 = 48.
48 + 1/8 = 49/8.
49/8 = 6.1.
H: 6(5/8) = 6.6.
Explanation:
In the above-given question,
given that,
The height of the Football H.
6(5/8) = 6 x 8.
6 x 8 = 48.
48 + 5/8 = 53/8.
53/8 = 6.6.
I: 5(6/8) = 5.7
Explanation:
In the above-given question,
given that,
The height of the Football I.
5(6/8) = 5 x 8.
5 x 8 = 40.
40 + 6/8 = 46/8.
46/8 = 5.7.
J: 6(1/8) = 6.1
Explanation:
In the above-given question,
given that,
The height of the Football J.
6(1/8) = 6 x 8.
6 x 8 = 48.
48 + 1/8 = 49/8.
49/8 = 6.1.
a. Use the data to create a line plot at the bottom of this page and to answer the questions below.
The points on the co-ordinate plane increase.
Explanation:
In the above-given question,
given that,
the players on the y-axis.
the height of the football on the x-axis.
the points on the graph increases.
b. What is the difference in height of the tallest and shortest players?
The difference in height of the tallest and shortest players = 0.9.
Explanation:
In the above-given question,
given that,
the height of the tallest player = 6.6,
the height of the shortest player = 5.7.
6.6 – 5.7 = 0.9.
so the difference in height of the tallest and shortest players = 0.9.
c. Player I and Player B have a combined height that is 1$$\frac{1}{8}$$ feet taller than a school bus. What is the height of a school bus?
The height of the school bus = 10.375.
Explanation:
In the above-given question,
given that,
player I and player B have a combined height that is 1(1/8) feet.
1(1/8) = 8 + 1/8.
9/8 = 1.125.
Player I = 5.7.
Player H = 5.8.
5.7 + 5.8 = 11.5.
11.5 – 1.125 = 10.375.
so the height of the school bus = 10.375 ft.
Question 2.
One of the players on the team is now 4 times as tall as he was at birth, when he measured 1$$\frac{5}{8}$$ feet. Who is the player?
The Player = C.
Explanation:
In the above-given question,
given that,
One of the players on the team is now 4 times as tall as he was at birth.
1(5/8) = 1 x 8 = 8.
8 + 5/8 = 13/8.
13/8 = 1.6.
1.6 x 4 = 2.4.
so player C has the 6.4 ft height.
Question 3.
Six of the players on the team weigh over 300 pounds. Doctors recommend that players of this weight drink at least 3$$\frac{3}{4}$$ quarts of water each day. At least how much water should be consumed per day by all 6 players?
The water should be consumed per day by all 6 players = 22.2 quarts.
Explanation:
In the above-given question,
given that,
Six of the players on the team weigh over 300 pounds.
3(3/4) quarts.
3 x 4 = 12.
12 + 3/4 = 15/4.
15/4 = 3.7.
3.7 x 6 = 22.2 quarts.
Question 4.
Nine of the players on the team weigh about 200 pounds. Doctors recommend that people of this weight each eat about 3$$\frac{7}{10}$$ grams of carbohydrates per pound each day. About how many combined grams of carbohydrates should these 9 players eat per pound each day?
The combined grams of carbohydrates should these 9 players eat per pound each day = 33.3 ft.
Explanation:
In the above-given question,
given that,
Nine of the players on the team weigh about 200 pounds.
3(7/10) = 3 x 10.
30 + 7/10 = 37/10.
37/10 = 3.7.
3.7 x 9 = 33.3 ft.
### Eureka Math Grade 4 Module 5 Lesson 40 Exit Ticket Answer Key
Coach Taylor asked his team to record the distance they ran during practice. The distances are listed in the table.
Team Members Distance (in miles) Alec 1$$\frac{3}{4}$$ Henry 1$$\frac{1}{2}$$ Charles 2$$\frac{1}{8}$$ Steve 1$$\frac{3}{4}$$ Pitch 2$$\frac{2}{4}$$ Raj 1$$\frac{6}{8}$$ Pam 2$$\frac{1}{2}$$ Tony 1$$\frac{3}{8}$$
Alec: 1(3/4) = 1.7miles.
Explanation:
In the above-given question,
given that,
The distance of Team members in miles.
1(3/4) = 1 x 4.
1 x 4 = 4.
4 + 3/4 = 7/4.
25/4 = 1.75.
Henry: 1(1/2) = 1.5miles.
Explanation:
In the above-given question,
given that,
The distance of Team members in miles.
1(1/2) = 1 x 2.
1 x 2 = 2.
2 + 1/2 = 3/2.
3/2 = 1.5.
Charles: 2(1/8) = 2.1miles.
Explanation:
In the above-given question,
given that,
The distance of Team members in miles.
2(1/8) = 2 x 8.
2 x 8 = 16.
16 + 1/8 = 17/8.
17/8 = 2.1.
Steve: 1(3/4) = 1.7miles.
Explanation:
In the above-given question,
given that,
The distance of Team members in miles.
1(3/4) = 1 x 4.
1 x 4 = 4.
4 + 3/4 = 7/4.
25/4 = 1.75.
Pitch: 2(2/4) = 1miles.
Explanation:
In the above-given question,
given that,
The distance of Team members in miles.
2(2/4) = 2 x 4.
2 x 4 = 8.
2 + 2/4 = 4/4.
4/4 = 1.
Raj: 1(6/8) = 1.7miles.
Explanation:
In the above-given question,
given that,
The distance of Team members in miles.
1(6/8) = 1 x 8.
1 x 8 = 8.
8 + 6/8 = 14/8.
14/8 = 1.75.
Pam: 2(1/2) = 2.5miles.
Explanation:
In the above-given question,
given that,
The distance of Team members in miles.
2(1/2) = 2 x 2.
2 x 2 = 4.
4 + 1/2 = 5/2.
5/2 = 2.5.
Tony: 1(3/8) = 1.37miles.
Explanation:
In the above-given question,
given that,
The distance of Team members in miles.
1(3/8) = 1 x 8.
1 x 8 = 8.
8 + 3/8 = 11/8.
11/8 = 1.37.
Question 1.
Use the table to locate the incorrect data on the line plot.
Circle any incorrect points.
Mark any missing points.
There are no incorrect points.
Explanation:
In the above-given question,
given that,
Question 2.
Of the team members who ran 1$$\frac{6}{8}$$ miles, how many miles did those team members run combined?
The team members run combined = 5.25 miles.
Explanation:
In the above-given question,
given that,
Of the team members who ran 1(6/8) miles.
1 x 6/8 = 8 + 6/8.
14/8 = 1.75 miles.
1.75 + 1.75 + 1.75 = 5.25 miles.
### Eureka Math Grade 4 Module 5 Lesson 40 Homework Answer Key
The chart to the right shows the total monthly rainfall for a city.
Month Rainfall (in inches) January 2$$\frac{2}{8}$$ February 1$$\frac{3}{8}$$ March 2$$\frac{3}{8}$$ April 2$$\frac{5}{8}$$ May 4$$\frac{1}{4}$$ June 2$$\frac{1}{4}$$ July 3$$\frac{7}{8}$$ August 3$$\frac{1}{4}$$ September 1$$\frac{5}{8}$$ October 3$$\frac{2}{8}$$ November 1$$\frac{3}{4}$$ December 1$$\frac{5}{8}$$
January:2(2/8) = 2.2 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
2(2/8) = 2 x 8.
16 + 2/8 = 18/8.
18/8 = 2.25 in.
Febraury:1(3/8) = 1.3 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
1(3/8) = 1 x 8.
8 + 3/8 = 11/8.
11/8 = 1.3 in.
March:2(3/8) = 2.3 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
2(3/8) = 2 x 8.
16 + 3/8 = 19/8.
19/8 = 2.3 in.
April:2(5/8) = 2.6 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
2(5/8) = 2 x 8.
16 + 5/8 = 21/8.
21/8 = 2.6 in.
May:4(1/4) = 4.2 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
4(1/4) = 4 x 4.
16 + 1/4 = 17/4.
17/4 = 4.25 in.
June:2(1/4) = 2.2 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
2(1/4) = 2 x 4.
8 + 1/4 = 9/4.
9/4 = 2.25 in.
July:3(7/8) = 3.8 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
3(7/8) = 3 x 8.
24 + 7/8 = 31/8.
31/8 = 3.8 in.
August:3(1/4) = 3.2 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
3(1/4) = 3 x 4.
12 + 1/4 = 13/4.
13/4 = 3.25 in.
September:1(5/8) = 1.6 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
1(5/8) = 1 x 8.
8 + 5/8 = 13/8.
13/8 = 1.6 in.
October:3(2/8) = 3.2 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
3(2/8) = 3 x 8.
24 + 2/8 = 26/8.
26/8 = 3.25 in.
November:1(3/4) = 1.7 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
1(3/4) = 1 x 4.
4 + 3/4 = 7/4.
7/4 = 1.75 in.
December:1(5/8) = 1.6 in.
Explanation:
In the above-given question,
given that,
the monthly rainfall for a city.
1(5/8) = 1 x 8.
8 + 5/8 = 13/8.
13/8 = 1.625 in.
Question 1.
Use the data to create a line plot at the bottom of this page and to answer the following questions.
The points on the co-ordinate plane increase.
Explanation:
In the above-given question,
given that,
the months on the y-axis.
the rainfall on the x-axis.
the points on the graph increases.
Question 2.
What is the difference in rainfall from the wettest and driest months?
The difference in rainfall from the wettest and driest months = 2.95 in.
Explanation:
In the above-given question,
given that,
the rainfall from the wettest months = 4.25 in.
the rainfall from the driest months = 1.3 in.
4.25 – 1.3 = 2.95 in.
Question 3.
How much more rain fell in May than in April?
The more rain fell in May than April = 1.65 in.
Explanation:
In the above-given question,
given that,
the rain fell in may = 4.25 in.
the rain fell in April = 2.6 in.
4.25 – 2.6 = 1.65in.
so the rain fell in may than April = 1.65 in.
Question 4.
What is the combined rainfall amount for the summer months of June, July, and August?
The combined rainfall amount for the summer months of June, July, and August = 9.2 in.
Explanation:
In the above-given question,
given that,
the rainfall in June = 2.2 in.
the rainfall in July = 3.8 in.
the rainfall in August = 3.2 in.
2.2 + 3.8 + 3.2 = 9.2 in.
so the combined amount for the summer month of June, July, and August = 9.2 in.
Question 5.
How much more rain fell in the summer months than the combined rainfall for the last 4 months of the year?
The combined rainfall for the last 4 months of the year = 8.1 in.
Explanation:
In the above-given question,
given that,
the much more rain fell in the summer months = 9.2 in.
September = 1.6 in.
October = 3.2 in.
November = 1.75 in.
December = 1.625 in.
1.6 + 3.2 + 1.75 + 1.625 = 8.1 in.
Question 6.
In which months did it rain twice as much as it rained in December?
The two months it rains twice as much as it rained in December = August and October.
Explanation:
In the above-given question,
given that,
the rainfall in December = 1.625 in.
1.625 + 1.625 = 3.25 in
August = 3.2 in.
October = 3.2 in.
Question 7.
Each inch of rain can produce ten times that many inches of snow. If all of the rainfall in January was in the form of snow, how many inches of snow fell in January?
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# What Is A Unit Fraction?
What Is A Unit Fraction:
A unit fraction is a number that can be expressed as a proper fraction with one whole, and the denominator is 1.
e.g:- ½ = 2/4, ¼ = 4/9, ⅛ = 8/9, etc.
The definition of a unit fraction tells us that it can be expressed in terms of its numerator divided by its denominator.
### In other words, we have:
In mathematics, the symbol for “or” is … so this equation may also be written as:
### as shown below:
A formula is an expression that provides detailed information about a particular problem or situation. An excellent example of this is the following famous geometry recipe related to circles: Area of Circle = π × (radius)2. This equation tells us how to work out the area of a process if we know its radius.
e.g.:- If the circumference of a circular plate is 18 cm, what is its radius?
To find the radius, plugin for “C” and solve for “r.” The answer will be 6 cm.
Since unit fractions are ratios between whole numbers, they may also be expressed as proportions. We have already seen this with equivalence proofs so far, so it won’t be necessary to re-visit them again here. It should also be noted that the following two equations are equivalent:
i) A/B = 1/X
ii) B = X/A
i) A/B = 1/X
ii) B = X/A
### For example, let’s have a look at the following question:
If ½ is of ¼, what is the value of x?
To solve this problem, first, write down both of the equations. The second equation can be written as two separate equations where “x” appears on each side of the equal sign. This gives us:
1/4 = x/½ and ½ = × ¼
Now solve for x by multiplying both sides of the equation by four, so we get 4x=1 which means x=1/. Now recall how fractional multiplication works, i.e., multiply both top and bottom by the denominator, so the new fractions are all in lowest terms.
This gives us 1/8 = x, the final answer to the question.
One way of thinking about a unit fraction is a “building block” for other fractions. In other words, we can use unit fractions to create more complex fractions by multiplying (or dividing) them together. For example, let’s take the fraction ¾ and see how we can make new fractions from it.
We can create the fraction ¾ by multiplying 3 unit fractions together, i.e., 1/3, 2/3, and 3/3.
Similarly, we can create the fraction ¼ by dividing 3 unit fractions together, i.e., 1/3, 2/3, and 3/3.
In general, we can create a fraction of n in unit fractions by multiplying together the first n natural numbers, i.e., 1, 2, 3, 4… etc. The product of these unit fractions will be the fraction of n.
### One example is to show that 1⁄6 = 6⁄1x:
(1/6) * (6/1) = (1*6)/(6*1) = (6)/(1) = 6 – Therefore we know that 1⁄6= 6⁄1x and vice versa.
The next example is to show that 8⁄9 + 7⁄9 = 15⁄9: –
8/9 + 7/9 = (8+7)/9 = 15/9 – Therefore we know that 8⁄9 + 7⁄9 = 15⁄9.
Lastly, let’s show that 1² + 2² = 3²: –
### We can use the same method as the previous example to show that 1² + 2² = 3². In this case, we have:
1/1 + 2/1 = (1+2)/1 = 3/1 = 3 – Therefore we know that 1² + 2² = 3².
Now that we understand what a unit is and how unit fractions can be used to create other bits, we will explore the concept of a unitary operation.
A unitary operation is any process that reduces a fraction or complex number down to its simplest form by multiplying or dividing both the top and bottom of the fraction or number by the same multiplier, divisor, or factor. In this case, it means reducing all denominators in a fraction down to 1 to simplify them. For example, 4⁄9 = 4/9 but 9 = 1.
Similarly, 3² can also be reduced because 3*3 = 9 and 2² can also be reduced because 2*2 = 4, while 1² cannot be reduced since 1*1=1.
For two numbers to be in a unitary relationship, one must represent the inverse of the other – this is to say that their product will equal 1. We can determine whether two fractions are in a unitary relationship by multiplying them and seeing if their product equals 1. If it is, then they are unitarily related.
### For example:
2/3 × 3/2 = (2*3)/(3*2) = 6/6 = 1
– This means that 2⁄3 and 3⁄2 are in a unitary relationship since 1 represents the multiplicative inverse of either fraction (i.e., x * 1 =1).
Now let’s look at some examples involving negative numbers.
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# Horizontal and Vertical Line Graphs
## Line y = # is parallel to x-axis and line x = # is parallel to y-axis
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Practice Horizontal and Vertical Line Graphs
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Horizontal and Vertical Line Graphs
What if you were given the graph of a vertical or horizontal line? How could you write the equation of this line? After completing this Concept, you'll be able to write horizontal and vertical linear equations and graph them in the coordinate plane.
### Guidance
How do you graph equations of horizontal and vertical lines? See how in the example below.
#### Example A
“Mad-cabs” have an unusual offer going on. They are charging \$7.50 for a taxi ride of any length within the city limits. Graph the function that relates the cost of hiring the taxi \begin{align*}(y)\end{align*} to the length of the journey in miles \begin{align*}(x)\end{align*}.
To proceed, the first thing we need is an equation. You can see from the problem that the cost of a journey doesn’t depend on the length of the journey. It should come as no surprise that the equation then, does not have \begin{align*}x\end{align*} in it. Since any value of \begin{align*}x\end{align*} results in the same value of \begin{align*}y (7.5)\end{align*}, the value you choose for \begin{align*}x\end{align*} doesn’t matter, so it isn’t included in the equation. Here is the equation:
\begin{align*}y = 7.5\end{align*}
The graph of this function is shown below. You can see that it’s simply a horizontal line.
Any time you see an equation of the form “\begin{align*}y =\end{align*} constant,” the graph is a horizontal line that intercepts the \begin{align*}y-\end{align*}axis at the value of the constant.
Similarly, when you see an equation of the form \begin{align*}x =\end{align*} constant, then the graph is a vertical line that intercepts the \begin{align*}x-\end{align*}axis at the value of the constant. (Notice that that kind of equation is a relation, and not a function, because each \begin{align*}x-\end{align*}value (there’s only one in this case) corresponds to many (actually an infinite number) \begin{align*}y-\end{align*}values.)
#### Example B
Plot the following graphs.
(a) \begin{align*}y = 4\end{align*}
(b) \begin{align*}y = -4\end{align*}
(c) \begin{align*}x = 4\end{align*}
(d) \begin{align*}x = -4\end{align*}
(a) \begin{align*}y = 4\end{align*} is a horizontal line that crosses the \begin{align*}y-\end{align*}axis at 4.
(b) \begin{align*}y = -4\end{align*} is a horizontal line that crosses the \begin{align*}y-\end{align*}axis at −4.
(c) \begin{align*}x = 4\end{align*} is a vertical line that crosses the \begin{align*}x-\end{align*}axis at 4.
(d) \begin{align*}x = -4\end{align*} is a vertical line that crosses the \begin{align*}x-\end{align*}axis at −4.
#### Example C
Find an equation for the \begin{align*}x-\end{align*}axis and the \begin{align*}y-\end{align*}axis.
Look at the axes on any of the graphs from previous examples. We have already said that they intersect at the origin (the point where \begin{align*}x = 0\end{align*} and \begin{align*}y = 0\end{align*}). The following definition could easily work for each axis.
\begin{align*}x-\end{align*}axis: A horizontal line crossing the \begin{align*}y-\end{align*}axis at zero.
\begin{align*}y-\end{align*}axis: A vertical line crossing the \begin{align*}x-\end{align*}axis at zero.
So using example 3 as our guide, we could define the \begin{align*}x-\end{align*}axis as the line \begin{align*}y = 0\end{align*} and the \begin{align*}y-\end{align*}axis as the line \begin{align*}x = 0\end{align*}.
Watch this video for help with the Examples above.
### Vocabulary
• Horizontal lines are defined by the equation \begin{align*}y=\end{align*} constant and vertical lines are defined by the equation \begin{align*}x= \end{align*} constant.
• Be aware that although we graph the function as a line to make it easier to interpret, the function may actually be discrete.
### Guided Practice
Write the equation of the horizontal line that is 3 units below the x-axis.
Solution:
The horizontal line that is 3 units below the x-axis will intercept the y-axis at \begin{align*}y=-3\end{align*}. No matter what the value of x, the y value of the line will always be -3. This means that the equations for the line is \begin{align*}y=-3\end{align*}.
### Practice
1. Write the equations for the five lines (\begin{align*}A\end{align*} through \begin{align*}E\end{align*}) plotted in the graph below.
For 2-10, use the graph above to determine at what points the following lines intersect.
1. \begin{align*}A\end{align*} and \begin{align*}E\end{align*}
2. \begin{align*}A\end{align*} and \begin{align*}D\end{align*}
3. \begin{align*}C\end{align*} and \begin{align*}D\end{align*}
4. \begin{align*}B\end{align*} and the \begin{align*}y-\end{align*}axis
5. \begin{align*}E\end{align*} and the \begin{align*}x-\end{align*}axis
6. \begin{align*}C\end{align*} and the line \begin{align*}y = x\end{align*}
7. \begin{align*}E\end{align*} and the line \begin{align*}y = \frac {1} {2} x\end{align*}
8. \begin{align*}A\end{align*} and the line \begin{align*}y = x + 3\end{align*}
9. \begin{align*}B\end{align*} and the line \begin{align*}y=-2x\end{align*}
### Vocabulary Language: English
Horizontally
Horizontally
Horizontally means written across in rows.
Vertically
Vertically
Vertically means written up and down in columns.
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# Difference between revisions of "2013 AMC 12B Problems/Problem 17"
## Problem
Let $a,b,$ and $c$ be real numbers such that
$$a+b+c=2, \text{ and}$$ $$a^2+b^2+c^2=12$$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3}$
## Solution 1
$a+b= 2-c$. Now, by Cauchy-Schwarz, we have that $(a^2+b^2) \ge \frac{(2-c)^2}{2}$. Therefore, we have that $\frac{(2-c)^2}{2}+c^2 \le 12$. We then find the roots of $c$ that satisfy equality and find the difference of the roots. This gives the answer, $\boxed{\textbf{(D)} \ \frac{16}{3}}$.
## Solution 2
This is similar to the first solution but is far more intuitive. From the given, we have $$a + b = 2 - c$$ $$a^2 + b^2 = 12 - c^2$$ This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have $$2\,(a^2 + b^2) \geq (a + b)^2$$ Substitution of the above results and some algebra yields $$3c^2 - 4c - 20 \leq 0$$ This quadratic inequality is easily solved, and it is seen that equality holds for $c = -2$ and $c = \frac{10}{3}$.
The difference between these two values is $\boxed{\textbf{(D)} \ \frac{16}{3}}$.
## Solution 3
(no Cauchy-Schwarz)
From the first equation, we know that $c=2-a-b$. We substitute this into the second equation to find that $$a^2+b^2+(2-a-b)^2=12.$$ This simplifies to $2a^2+2b^2-4a-4b+2ab=8$, which we can write as the quadratic $a^2+(b-2)a+(b^2-2b-4)=0$. We wish to find real values for $a$ and $b$ that satisfy this equation. Therefore, the discriminant is nonnegative. Hence, $$(b-2)^2-4(b^2-2b-4)\ge0,$$ or $-3b^2+4b+20\ge 0$. This factors as $-(3b-10)(b+2)\ge 0$. Therefore, $-2\le b\le \frac{10}{3}$, and by symmetry this must be true for $a$ and $c$ as well.
Now $a=b=2$ and $c=-2$ satisfy both equations, so we see that $c=-2$ must be the minimum possible value of $c$. Also, $c=\frac{10}{3}$ and $a=b=-\frac{2}{3}$ satisfy both equations, so we see that $c=\frac{10}{3}$ is the maximum possible value of $c$. The difference between these is $\frac{10}{3}-(-2)=\frac{16}{3}$, or $\boxed{\textbf{(D)}}$.
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# Michigan - Grade 1 - Math - Numbers and Operations in Base Ten - Comparing Numbers - 1.NBT.3
### Description
Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.
• State - Michigan
• Standard ID - 1.NBT.3
• Subjects - Math Common Core
### Keywords
• Math
• Numbers and Operations in Base Ten
## More Michigan Topics
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
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# Mathematical Induction
Mathematical Induction
Mathematical Induction – Meaning
In general, the word Induction means the generalisation from particular cases or facts. In Mathematics, there are certain results or statements that are formulated in terms of n, where n is a positive integer. We may prove such general cases from Initial Base step to next one through the Mathematical Induction.
Mathematical Reasoning
Mathematical Reasoning is the process of finding the proof for a certain mathematical statement by using logic and deductions. Inductive and deductive reasoning are two fundamental forms of reasoning for mathematicians. The formal theorems and proofs that we rely on today all began with these two types of reasoning.
• Mathematical Deduction : Deduction is drawing a conclusion from something known or assumed, normally used in almost every step in a mathematical argument.
• Mathematical Induction : Mathematical induction is a particular type of mathematical argument. It is most often used to prove general statements about Positive Integers.
To prove the formula, we may verify the statement for as many positive integral values of n as we like, but this process will not prove the formula for all values of n. We need is some kind of chain effect which will have the effect that once the formula is proved for a particular positive integer, the formula will automatically follow for the next positive integer, and the next indefinitely. Such chain effect is produced by Mathematical Induction.
Mathematical Induction Principles
Suppose there is a given statement P(n) involving the natural number n such that :
(i) The statement is true for n = 1, i.e., P (1) is true, and
(ii) If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P(k) implies the truth of P (k + 1). Then, P(n) is true for all natural numbers n.
Here, Property (i) is simply a statement of fact. When a statement is true for all n ≥4, we start from n = 4 and verify the result for n = 4, i.e., P(4).
Property (ii) is a conditional property. It does not assert that the given statement is true for n = k, but that if it is true for n = k, then it is also true for n = k +1. So, to prove that the property holds, prove the conditional proposition:
If the statement is true for n = k, then it is also true for n = k + 1.
This is referred to as the inductive step. The assumption that the given statement is true for n = k in this inductive step is called the inductive hypothesis.
Mathematical Induction – Example 1
Mathematical Induction – Example 2
Mathematical Induction – Example 3
Mathematical Induction – Example 4
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Math Train
# Number Friends
In this lesson, children will learn to count and recognize the number of objects in a set, match numbers to quantities and practice simple addition
30-45 min.
PreK-K
## Connect
• Gather the children around the train set and show them two train cars with different quantities of bricks loaded onto
them (e.g., one with six green bricks and the number six brick, and another with four light green bricks and the number four brick).
• Place the cars side by side and ask the children to compare.
- Which train car has the most bricks?
- Which train car has the least bricks?
• Explain that sometimes people want to know how many items there are altogether and that we can figure this out by adding.
• Encourage the children to count the total number of bricks on both train cars. Point out that the two quantities (i.e., six and four) added together make ten, and
so we can call six and four best friends of ten!
## Tip
If making sums of ten is too difficult, use smaller numbers.
## Construct
• Explain that several sets of numbers are “best friends of ten.”
• Tell the children they will be working with a partner to create “best friends of ten” train cars.
• Ask each pair of children to find two number bricks that add u p to ten. You may use the in-box card as an example.
• Ask each child to place a number brick on a train car and find the corresponding
number of bricks in the same color. If they need help, show them the inspiration photos, which they can duplicate.
• Once the children have finished building, have them count the bricks on each train car.
Then ask them to count their pair’s bricks altogether--which make ten!
## Contemplate
• Facilitate a discussion about “best friends of ten.”
- How many sets of “best friends of ten” there are?
- What other numbers add up to ten?
• Write out the sets as children say them. If time allows, encourage the children to
build each set of “best friends” and share with the group.
## Continue
• Challenge children to create “best friends of twelve” train cars.
• The combinations which can be built using the set include:
- Two and ten
- Three and nine
- Four and eight
- Five and seven
• Look at the in-box cards and use the ideas shown on the cards to help the children find the bricks needed to add up to twelve.
## Did you notice?
Observing the following skills can help you monitor whether the children are developing the necessary competencies in math.
• Counting using number names, and beginning to recognize the number of objects in a set
• Comparing two or more objects
• Exploring simple operations, such as adding
## Teacher Support
Children will:
• Count and recognize the number of objects in a set
• Match numbers to quantities
• Compare quantities
For up to 4 children
The Mathematics guidelines from the National Association for the Education of Young Children (NAEYC) and HeadStart have been used to develop the Math Train lessons. Please refer to the learning grid for an overview of the learning values referenced throughout this Teacher Guide.
The learning goals listed at the end of each lesson can be used to determine whether or not each child is developing the relevant early math skills. These bullet points target specific skills or pieces of information that are practiced or presented during each lesson.
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# Graphing y = nx2 Lesson 5.4.1.
## Presentation on theme: "Graphing y = nx2 Lesson 5.4.1."— Presentation transcript:
Graphing y = nx2 Lesson 5.4.1
Graphing y = nx2 5.4.1 California Standards: What it means for you:
Lesson 5.4.1 Graphing y = nx2 California Standards: Algebra and Functions 3.1 Graph functions of the form y = nx2 and y = nx3 and use in solving problems. Mathematical Reasoning 2.3 Estimate unknown quantities graphically and solve for them by using logical reasoning and arithmetic and algebraic techniques. Mathematical Reasoning 2.5 Use a variety of methods, such as words, numbers, symbols, charts, graphs, tables, diagrams, and models, to explain mathematical reasoning. What it means for you: You’ll learn how to plot graphs of equations with squared variables in them. Key words: parabola vertex
Lesson 5.4.1 Graphing y = nx2 Think about the monomial x2. You can put any number in place of x and work out the result — different values of x give different results. The results you get form a pattern. And the best way to see the pattern is on a graph.
Graphing y = nx2 5.4.1 The Graph of y = x2 is a Parabola
Lesson 5.4.1 Graphing y = nx2 The Graph of y = x2 is a Parabola You can find out what the graph of y = x2 looks like by plotting points. Examples 1 and 2 show you how to plot a graph of y = x2 for positive and negative values of x.
Lesson 5.4.1 Graphing y = nx2 Example 1 Plot the graph of y = x2 for values of x between 0 and 6. Solution The best thing to do first is to make a table for the integer values of x like the one below. y 3 2 1 x 4 5 6 y (= x2) 9 16 25 36 2 4 6 1 3 5 10 20 30 40 Then you can plot points on a set of axes using the x- and y-values as coordinates, and join the points with a smooth curve. The curve passes through all the values that fit the equation between the integer points. x Solution follows…
Lesson 5.4.1 Graphing y = nx2 Example 2 Plot the graph of y = x2 for values of x between –6 and 6. Solution The table of values looks like this: –3 –2 –1 x –4 –5 –6 y (= x2) 9 4 1 16 25 36 y 2 4 6 10 20 30 40 –2 –4 –6 And the curve looks like this. This kind of curve is called a parabola. x Solution follows…
Graphing y = nx2 5.4.1 Guided Practice
Lesson 5.4.1 Graphing y = nx2 Guided Practice 1. Which of the following points are on the graph of y = x2? (1, 1), (–1, 1), (–2, –4), (2, 4), (3, 9), (–4, –16), (5, 25), (6, –36) In Exercises 2–5, calculate the y-coordinate of the point on the graph of y = x2 whose x-coordinate is shown. –10 4. – (1, 1), (–1, 1), (2, 4), (3, 9), and (5, 25) 36 100 1 3 1 9 6.25 Solution follows…
Graphing y = nx2 5.4.1 Guided Practice
Lesson 5.4.1 Graphing y = nx2 Guided Practice In Exercises 6–9, calculate the two possible x-coordinates of the points on the graph of y = x2 whose y-coordinate is shown. 4 and –4 5 and –5 7 and –7 and – Solution follows…
Graphing y = nx2 5.4.1 You Can Use a Graph to Solve an Equation
Lesson 5.4.1 Graphing y = nx2 You Can Use a Graph to Solve an Equation Graphs can be useful if you need to solve an equation. Using them means you don’t have to do any tricky calculations — and they often show you how many solutions the equation has. The downside is that it can be impossible to get an exact answer by reading off a graph.
Lesson 5.4.1 Graphing y = nx2 Example 3 Using the graph of y = x2 in Example 2, solve x2 = 12. Solution Since the graph shows y = x2, you need to find where y = Then you can find the corresponding value (or values) of x. y 2 4 6 10 20 30 40 –2 –4 –6 There are two different values of x that correspond to y = 12, at approximately x = 3.5 and x = –3.5. y = 12 x x = –3.5 x = 3.5 Solution continues… Solution follows…
Lesson 5.4.1 Graphing y = nx2 Example 3 Using the graph of y = x2 in Example 2, solve x2 = 12. Solution (continued) There are two different values of x that correspond to y = 12 because 12 has two square roots — a positive one (3.5) and a negative one (–3.5). Or you can look at it another way, and say that the numbers 3.5 and –3.5 can both be squared to give 12 (approximately).
Graphing y = nx2 5.4.1 Guided Practice
Lesson 5.4.1 Graphing y = nx2 Guided Practice Use the graph of y = x2 shown below to solve the equations in Exercises 10–13. 10. x2 = 16 11. x2 = 25 12. x2 = 10 13. x2 = 30 x = 4 or x = –4 y 2 4 6 10 20 30 40 –2 –4 –6 x = 5 or x = –5 x = –5.5 x = 5.5 x = –5 x = 5 x = 3.2 or x = –3.2 (approximately) x = –4 x = 4 x = –3.2 x = 3.2 x x = 5.5 or x = –5.5 (approximately) Solution follows…
Graphing y = nx2 5.4.1 The Graph of y = nx2 is Also a Parabola
Lesson 5.4.1 Graphing y = nx2 The Graph of y = nx2 is Also a Parabola The graph of y = x2 is y = nx2 where n = 1. It has the U shape of a parabola. Other values of n give graphs that look very similar.
Lesson 5.4.1 Graphing y = nx2 Example 4 Plot the graphs of the following equations for values of x between –5 and 5. a) y = 2x2 b) y = 3x2 c) y = 4x2 d) y = x2 1 2 Solution All these equations are of the form y = nx2, for different values of n (2 then 3 then 4 then ). 1 2 The best place to start is with a table of values, just like before. Solution continues… Solution follows…
Lesson 5.4.1 Graphing y = nx2 Example 4 Plot the graphs of the following equations for values of x between –5 and 5. a) y = 2x2 b) y = 3x2 c) y = 4x2 d) y = x2 1 2 Solution (continued) x 2x2 1 and –1 2 and –2 3 and –3 4 and –4 5 and –5 2 8 18 32 50 3x2 3 12 27 48 75 4x2 4 16 36 64 100 ½ x2 0.5 4.5 12.5 The table on the right shows values for parts a)–d). You then need to plot the y-values in each colored column against the x‑values in the first column. Solution continues…
Graphing y = nx2 5.4.1 Solution (continued) Lesson Example 4 y 4 –2 –4
Increasing values of n Decreasing values of n (n = 4) (n = 3) (n = 2) (n = ½) 4 –2 –4 2 3 1 –1 –3 20 40 60 80 100 5 –5 y = 4x2 Solution (continued) x 1 and –1 2 and –2 3 and –3 4 and –4 5 and –5 4x2 4 16 36 64 100 x 2x2 1 and –1 2 and –2 3 and –3 4 and –4 5 and –5 2 8 18 32 50 x 1 and –1 2 and –2 3 and –3 4 and –4 5 and –5 3x2 3 12 27 48 75 x 1 and –1 2 and –2 3 and –3 4 and –4 5 and –5 ½ x2 0.5 2 4.5 8 12.5 y = 3x2 y = 2x2 y = x2 1 2 x
Lesson 5.4.1 Graphing y = nx2 Notice how all the graphs are “u-shaped” parabolas. And all the graphs have their vertex (the lowest point) at the same place, the origin. 4 –2 –4 2 3 1 –1 –3 20 40 60 80 100 5 –5 y = 4x2 y = 3x2 y = 2x2 y = x2 y x In fact, this is a general rule — if n is positive, the graph of y = nx2 will always be a “u-shaped” parabola with its vertex at the origin.
Lesson 5.4.1 Graphing y = nx2 Also, the greater the value of n, the steeper the parabola will be. 4 –2 –4 2 3 1 –1 –3 20 40 60 80 100 5 –5 y = 4x2 y = 3x2 y = 2x2 y = x2 y x In Example 4, the graph of y = 4x2 had the steepest parabola, while the graph of y = ½ x2 was the least steep.
Graphing y = nx2 5.4.1 Guided Practice
Lesson 5.4.1 Graphing y = nx2 Guided Practice For Exercises 14–17, draw on the same axes the graph of each of the given equations. 14. y = 5x2 15. y = x2 16. y = 10x2 17. y = x2 1 4 10 y 2 4 10 20 30 40 –2 –4 50 60 70 y = 10x2 y = 5x2 y = x2 1 4 y = x2 1 10 x Solution follows…
Graphing y = nx2 5.4.1 Guided Practice
Lesson 5.4.1 Graphing y = nx2 Guided Practice y In Exercises 18–23, use the graphs from Example 4 to solve the given equations. 18. 2x2 = x2 = 25 20. 4x2 = x2 = 10 22. 3x2 = x2 = 42 2 4 10 20 30 40 –2 –4 50 60 70 y = 4x2 y = 3x2 y = 2x2 y = x2 1 x » 3.2 or x » –3.2 x » 2.9 or x » –2.9 1 2 x » 1.9 or x » –1.9 x » 4.5 or x » –4.5 x » 4.8 or x » –4.8 x » 4.6 or x » –4.6 x Solution follows…
Graphing y = nx2 5.4.1 Independent Practice 1 3
Lesson 5.4.1 Graphing y = nx2 Independent Practice Using a table of values, plot the graphs of the equations in Exercises 1–3 for values of x between –4 and 4. 1. y = 1.5x2 2. y = 5x2 3. y = x2 y 4 –2 –4 2 3 1 –1 –3 20 40 60 80 100 y = 5x2 1 3 y = 1.5x2 y = x2 1 3 x Solution follows…
Graphing y = nx2 5.4.1 Independent Practice 2 3
Lesson 5.4.1 Graphing y = nx2 Independent Practice On the same set of axes as you used for Exercises 1–3, sketch the approximate graphs of the equations in Exercises 4–6. 4. y = 2.5x2 5. y = 6x2 6. y = x2 2 3 y 4 –2 –4 2 3 1 –1 –3 20 40 60 80 100 y = 1.5x2 y = 5x2 y = x2 y = 6x2 y = 2.5x2 y = x2 2 3 x Solution follows…
Graphing y = nx2 5.4.1 Independent Practice
Lesson 5.4.1 Graphing y = nx2 Independent Practice 7. If s is the length of a square’s sides, then a formula for its area, A, is A = s2. Plot a graph of A against s, for values of s up to 10. 10 20 40 60 80 100 A s –10 Solution follows…
Graphing y = nx2 5.4.1 Independent Practice 2 3 8 5
Lesson 5.4.1 Graphing y = nx2 Independent Practice 8. On a graph of y = x2, what is the y-coordinate when x = 103? For Exercises 9–12, find the y-coordinate of the point on the graph of y = x2 for each given value of x. 9. x = 10– x = 10–4 11. x = x = 106 10–2 10–8 2 3 8 5 4 9 64 25 Solution follows…
Graphing y = nx2 5.4.1 Independent Practice
Lesson 5.4.1 Graphing y = nx2 Independent Practice For Exercises 13–15, find the x-coordinates of the point on the y = x2 graph for each given value of y. 13. y = 102 14. y = 10–6 15. y = 28 10 and –10 10–3 and –10–3 24 and –24 Solution follows…
Lesson 5.4.1 Graphing y = nx2 Round Up In this Lesson you’ve looked at graphs of the form y = nx2, where n is positive. The basic message is that these graphs are all u-shaped. And the greater the value of n, the narrower and steeper the parabola is. Remember that, because in the next Lesson you’re going to look at graphs of the same form where n is negative.
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## The Best Quick check math answers
Quick check math answers can support pupils to understand the material and improve their grades. Online math graphing calculators are a great tool for visual learners. They can help you see patterns and relationships that might be difficult to spot on paper. They can also be a great way to check your work. Most online graphing calculators are free to use, and they’re easy to find with a simple Google search. Whether you’re working on a school project or just trying to better understand a concept, an online math graph can be a valuable resource.
In this case, we are looking for the distance travelled by the second train when it overtakes the first. We can rearrange the formula to solve for T: T = D/R. We know that the second train is travelling at 70 mph, so R = 70. We also know that the distance between the two trains when they meet will be the same as the distance travelled by the first train in one hour, which we can calculate by multiplying 60 by 1 hour (60 x 1 = 60). So, plugging these values into our equation gives us: T = 60/70. This simplifies to 0.857 hours, or 51.4 minutes. So, after 51 minutes of travel, the second train will overtake the first.
A complex number can be represented on a complex plane, which is similar to a coordinate plane. The real part of the complex number is represented on the x-axis, and the imaginary part is represented on the y-axis. One way to solve for a complex number is to use the quadratic equation. This equation can be used to find the roots of any quadratic equation. In order to use this equation, you must first convert the complex number into its rectangular form. This can be done by using the following formula: z = x + yi. Once the complex number is in rectangular form, you can then use the quadratic equation to find its roots. Another way to solve for a complex number is to use De Moivre's theorem. This theorem states that if z = x + yi is a complex number, then its nth roots are given by: z1/n = x1/n(cos (2π/n) + i sin (2π/n)). This theorem can be used to find both the real and imaginary parts of a complex number. There are many other methods that can be used to solve for a complex number, but these two are some of the most commonly used.
Solving problems can be difficult, but by breaking the problem down into smaller steps, it becomes much easier to handle. When you take the time to solves step by step, you are better able to understand the problem and find the best solution. In addition, by solving problems step by step, you can avoid making mistakes that can make the problem worse. So next time you are faced with a difficult problem, remember to Solve step by step!
Solving a system of equations by graphing is a visual way to find the point of intersection for two linear equations. To do this, first plot the two equations on a coordinate plane. Then, use a straightedge to draw a line through the points of intersection. The point where the line intersects the x-axis is the solution to the system of equations. This method can be used to solve systems of two or more equations. However, it is important to note that not all systems of equations will have a unique solution. In some cases, the lines may be parallel and will not intersect. In other cases, the lines may intersect at more than one point. When this happens, the system of equations is said to be inconsistent and has no solution.
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## Cell Division
How quickly does a chick embryo grow? Remember that the way any embryo develops is by repeated cell division. It starts off as a single cell which spilts into two. Each of those two cells then splits into two, and so on. We have illustrated this schematically with a square box representing the initial egg cell
Move your cursor into the box to `fertilize the egg' and start the cell divison procedure. The box divides into 2 regions, then 4, then 8, etc. The number of regions grows in the same way that the number of cells in a real embryo does. At the left, we graph the number of cells against the number of times cell division has occurred. Notice how quickly the number of cells grows!
## Exponential Functions
We can very efficiently describe the way the number of cells grows if we use some mathematical notation. Let's denote the number of times division has occured by the letter t, and let's denote the total number of cells by N(t). Thus N(1) denotes the number of cells after 1 division; so N(1)=2. Similarly, N(2)=2x2=4, N(3)=2x2x2=8, etc. A more compact way to write 2x2x2 is as 23; 2x2x2x2 is 24 and so on. With this convenient notation, we can write
N(1) = 21, N(2) = 22, N(3) = 23, N(4) = 24.
Following this pattern, we can write down a formula for N(t), namely:
N(t)=2t
In mathematical language, N(t) is an exponential function of t.
In the graph next to the cell division simulation, we have plotted N(t) for t=1,2,3,.... We can join up the dots to get a smooth curve, from which we can read off values for 2t corresponding to any value for t, whole number or not. (Later we will see a more precise way to define 2t for any value of t.)
In the exponential function N(t)=2t, the number 2 is called the base of the exponential. We can change this base and get other functions of t, all of which are called exponentials. For example, using base 3, we get the exponential function 3t.
The graphs of exponential functions are shown in this demonstration. You can change the base of the exponential (denoted by b) by using the slider at the top.
Notice that the graphs extend into the region where t is negative. This doesn't make much sense if t represents the number of times that a chick egg has divided. Mathematically, however, we can make sense of it as follows. For the exponential with base b, i.e. for the function bt, we use the rule that
b-t = 1 / bt
This rule tells us how to compute bt for negative values of t. With this rule it is still true that bt+1 = b bt.
If you see this message, your browser doesn't support Java, or Java isn't enabled. If it worked, you'd see something like this:
If you experiment with a few different choices for the base, you will notice that the graphs of all the exponential functions have a few things in common:
1. The graphs all pass through the same point when t=0
2. Going off to one direction (left or right), the value of the function (the height of the curve) grows very rapidly, but going off to the other direction, the value decays slowly towards zero, getting ever closer to the horizontal axis but never actually reaching it.
Notice that for some choices of base b, the value increases when we move to the left and decays to the right, while for other choices of b, the value increases to the right and decays to the left. See if you can figure out what determines which alternative applies to a given exponential function.
3. There is a third, more subtle, property shared by all the exponential functions. To understand what this is, we need to look at the straight lines tangent to the graphs. In particular, we need to look at the directions of these tangent lines, as measured by their slopes.
Select a point on the graph by clicking on it. The demonstration shows the tangent line to the graph at the selected point. The three numbers shown at the bottom of the demonstration are
• the value of the exponential, N = bt, at that point,
• the slope (denoted m) of the tangent line at that point, and
• the ratio m/bt
If you see this message, your browser doesn't support Java, or Java isn't enabled. If it worked, you'd see something like this:
Notice that along a given curve, while the first two numbers change, the third one remains the same no matter which point you select! This is the third special property of exponentials:
The slope of the tangent line is proportional to the value of the exponential at the point of tangency
The ratio (the third number displayed in the applet) is called the constant of proportionality.
## The special base
If you explore this property of exponentials in more detail, you will notice that this constant of proportionality changes depending on the base of the exponential. For instance, for the exponential to base 2 the constant is .693147, while for the exponential to base 3 the constant is 1.0986.
This prompts the following question:
• Is there some base for which the constant of proportionality is exactly 1?
The answer to this question is YES! This base, for which the exponential function has slope equal to its value, is traditionally denoted by the letter e. This number e has many wonderful properties. You can explore the graph of the exponential function et, and verify that it has slope equal to value by clicking the e button in the demonstration above.
Continue to the number e or go back to the the main EggMath page.
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# Terminating Decimal Definition
Those decimals that have a finite, limited, or known number of digits after the decimal point are called Terminating Decimals. For Example, 3.14, 2.578, and 1.987 are terminating decimals because all of these have limited numbers after the Decimal Point.
Such decimals are used to represent a whole number and a fraction collectively in a single term. In a terminating decimal, the left side of the decimal point gives the whole number while the right side shows the fractional part. For example, in 3.14, “3” represents the whole number while “14” shows the decimal fraction.
Note:
Read about Whole Numbers and a Fraction here.
## What is the opposite of Terminating a Decimal?
In Mathematics, Decimals are divided into two major types. One of which is terminating decimals about which we are reading here. In contrast, the other type is called Non-terminating decimals. These decimals have infinite or unknown numbers after the decimal point.
Such numbers are normally represented by putting “Three Dots” after the last number on the right side of the decimal point. For example, 3.3333333… is a non-terminating decimal because we don’t know the ending point of this decimal.
Note:
Read about Non-terminating Decimals here.
### How to convert a fraction into a termination decimal?
The only way to convert a fraction into a terminating decimal is by division. We only have to divide the numerator by the denominator of the fraction to get this decimal. While dividing, we can get the following three types of numbers.
• Whole number
• Terminating Decimal
• Non-terminating Decimal
It means that we can get the answer of a fraction in any of these formats. So, there is no restriction that we will always get a terminating decimal by dividing a fraction.
Note:
Read about Division, Numerator, and Denominator here.
### How to identify a terminating decimal?
As mentioned above, a terminating decimal has a limited number of digits after the decimal point. But we can also identify these numbers even without solving the fraction. To do so, we only have to check the denominator of the given fraction.
If the denominator can be written in the power of 2 and 5, it means that the fraction will give a terminating decimal as the answer. It is the simplest and easiest way to identify a decimal of this type without division.
### Fun facts about terminating decimals
• All terminating decimals are also rational numbers.
• The left side of the terminating decimal gives the whole part of the fraction.
• All fractions with a denominator that can be written in the form of an exponent of 2 and 5 are terminating decimals.
• A non-terminating decimal can be converted to a terminating decimal with the Round-Off technique.
### FAQ's
What does terminating mean?
Terminating means to end. In Mathematics, it is used for those decimals that are going to an end after some points.
How to check the termination of a fraction without division?
You can check it just by looking at the denominator of the fraction. If the denominator can be written as an exponent of 2 and 5, the fraction will give a terminating decimal.
Is Pi a terminating decimal?
No, Pi is not a terminating decimal because the numbers after the decimal point are now known.
Are terminating and repeating decimals the same?
No, terminating decimals are those that have limited numbers after the decimal point. On the other hand, repeating decimals is a type of non-terminating decimal that has a specific pattern repeating itself after the decimal point.
Related Definitions
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# Finding the Volume and Surface Area of a Cube
### Learning Outcomes
• Find the volume and surface area of a cube
A cube is a rectangular solid whose length, width, and height are equal. See Volume and Surface Area of a Cube, below. Substituting, s for the length, width and height into the formulas for volume and surface area of a rectangular solid, we get:
$\begin{array}{ccccc}V=LWH\qquad & & & & S=2LH+2LW+2WH\qquad \\ V=s\cdot s\cdot s\qquad & & & & S=2s\cdot s+2s\cdot s+2s\cdot s\qquad \\ V={s}^{3}\qquad & & & & S=2{s}^{2}+2{s}^{2}+2{s}^{2}\qquad \\ & & & & S=6{s}^{2}\qquad \end{array}$
So for a cube, the formulas for volume and surface area are
$V={s}^{3}$
and
$S=6{s}^{2}$
.
### Volume and Surface Area of a Cube
For any cube with sides of length
$s$
,
### example
A cube is
$2.5$
inches on each side. Find its 1. volume and 2. surface area.
Solution
Step 1 is the same for both 1. and 2., so we will show it just once.
Step 1. Read the problem. Draw the figure and label it with the given information.
1. Step 2. Identify what you are looking for. the volume of the cube Step 3. Name. Choose a variable to represent it. let V = volume Step 4. Translate. Write the appropriate formula. $V={s}^{3}$ Step 5. Solve. Substitute and solve. $V={\left(2.5\right)}^{3}$ $V=15.625$ Step 6. Check: Check your work. Step 7. Answer the question. The volume is $15.625$ cubic inches.
2. Step 2. Identify what you are looking for. the surface area of the cube Step 3. Name. Choose a variable to represent it. let S = surface area Step 4. Translate. Write the appropriate formula. $S=6{s}^{2}$ Step 5. Solve. Substitute and solve. $S=6\cdot {\left(2.5\right)}^{2}$ $S=37.5$ Step 6. Check: The check is left to you. Step 7. Answer the question. The surface area is $37.5$ square inches.
### example
$2$
inches on each side. Find its 1. volume and 2. surface area.
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# Common denominator of 2/3+10/15?
Then teach the underlying concepts
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#### Explanation
Explain in detail...
#### Explanation:
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Feb 14, 2017
$\frac{4}{3}$ or $1 \frac{1}{3}$
#### Explanation:
$\frac{2}{3} + \frac{10}{15}$
$\therefore \frac{10 + 10}{15}$
$\therefore {\cancel{20}}^{4} / {\cancel{15}}^{3}$
$\therefore \frac{4}{3}$ or$1 \frac{1}{3}$
2nd option:
$\therefore \frac{2}{3} + {\cancel{10}}^{2} / {\cancel{15}}^{3}$
$\therefore \frac{2}{3} + \frac{2}{3}$
$\frac{4}{3}$ or $1 \frac{1}{3}$
Then teach the underlying concepts
Don't copy without citing sources
preview
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#### Explanation
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#### Explanation:
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Feb 13, 2017
#### Explanation:
The denominator of $\frac{2}{3}$ is $3$
and denominator of $\frac{10}{15}$ is $15$
We can add two fractions only if their denominators are same or common.
As a fraction $\frac{2}{3}$ does not change when numerator and denominator are multiplied by same number,
hence to find common denominator, we will have to multiply numerator and denominator of each fraction are by a number (albeit different numbers), so that denominators become same.
This is done by identifying Least Common Multiple (LCM) of the two denominators.
Here LCM of $3$ and $15$ is $15$ and hence we can convert both denominators tp $15$, but in second fraction, it is already so.
Hence multiplying numerator and denominator of first by $5$, we get $\frac{2 \times 5}{3 \times 5} = \frac{10}{15}$
Hence, $\frac{2}{3} + \frac{10}{15} = \frac{10}{15} + \frac{10}{15} = \frac{20}{15}$.
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# What is 44 factorial ?
Steps to calculate factorial of 44
## How to calculate the factorial of 44
To find 44 factorial, or 44!, simply use the formula that multiplies the number 44 by all positive whole numbers less than it.
Let's look at how to calculate the Factorial of Forty-four :
44! is exactly : 2658271574788448768043625811014615890319638528000000000
Factorial of 44 can be calculated as:
44! = 44 x 43 x 42 x 41 x ... x 3 x 2 x 1
The number of trailing zeros in 44! is 10
The number of digits in 44 factorial is 55.
## What Is Factorial?
A factorial is showed by an integer and an exclamation tag. In arithmetic, factorial is a multiplication operation of natural numbers .
It multiplies the number by every standard number that is less than it .
Symbolically, it is listed as "!".
The function is used, among other things, to get the "n" way elements can be arranged .
## Factorial Formula
To find the factorial of any given number, exchange the exact value for n in the given equation :
n! = n × (n-1) × (n-2) × (n-3) × ….× 3 × 2 × 1
The expansion of the formula provides numbers to be replicated with each other to receive the factorial of the number.
We can also figure out a factorial from the prior one. The factorial of any number is that number multiplied the factorial of (that number minus 1).
So the rule is : n! = n × (n−1)!
Example :
44! Factorial = 44 x 43 x 42 x 41 x ... x 3 x 2 x 1 = 44 × 43! = 2658271574788448768043625811014615890319638528000000000
## What are Factorials Used For?
The best use of factorial is in Combinations and Permutations.
Example : Determine how to arrange letters without repeating?
There one way for 1 letter "a":
2 ways for two letters "ab": ab, ba.
There are 6 ways for 3 letters "abc": abc acb cab bac bca.
There are 24 ways for 1234 of the letters "abcd"
## Frequently Asked Questions on Factorial
### Can we have factorials for negative numbers ?
Negative number factorials are undefined
### What Is 0!
Zero factorial or Factorial of 0 is simple, and its value is corresponding to 1. So, 0! = 1.
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# 4th Class Mathematics Fractions and Decimals Comparison of Fraction
## Comparison of Fraction
Category : 4th Class
### Comparison of Fraction
Comparison of Like Fractions
Let$\frac{p}{q}$and$\frac{r}{q}$are like fractions.
If p is greater than $q,\frac{p}{q}>\frac{r}{q}$
Compare between$\frac{\mathbf{7}}{\mathbf{9}}$and$\frac{5}{\mathbf{9}}$.Which is greater?
Solution:
$7>5$ $\frac{7}{9}>\frac{5}{9}$
Comparison of Fractions Having Same Numerator
If the two fractions have same numerator, the fraction which has smaller denominator is greater. Like$\frac{P}{Q}$is greater than$\frac{P}{R}$if$\text{Q}<\text{R}$.
Find the greatest fraction out of the given fractions:
$\frac{18}{23},\frac{18}{17},\frac{18}{19},\frac{18}{20},\frac{18}{12}$
Solution:
$\frac{18}{12}$is the greatest fraction among the given fractions. As it has smallest denominator.
Comparison of Unlike Fractions
Compare between$\frac{7}{13}$and$\frac{6}{9}$
Step 1: Convert the fractions into like fractions.
$\frac{7\times 9}{13\times 9}=\frac{63}{117}$
And$\frac{6\times 13}{9\times 13}=\frac{78}{117}$
Step 2: The fraction having greater numerator is greater.
$\because$$78>63$
$\therefore$$\frac{78}{117}>\frac{63}{117}$or$\frac{6}{9}>\frac{7}{13}$
Compare between$\frac{21}{22}$and$\frac{22}{23}$, which is greater?
Solution: $\frac{21}{22}=\frac{21\times 23}{22\times 23}=\frac{483}{506}$
$\frac{22}{23}=\frac{22\times 22}{23\times 22}=\frac{484}{506}$
$\because$$484>483$
$\therefore$$\frac{484}{506}>\frac{483}{506}$or$\frac{22}{23}>\frac{21}{22}$
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# A sum of money is to be divided among A,Band C in the ratio 2:3:5.The smallest share amounts to \$600.Calculate the total sum of money to be shared.
## To find the total sum of money to be shared, we need to determine the value of the smallest share in the given ratio.
Let's assume the smallest share is represented by the value "x".
According to the given ratio, the share of A:B:C is 2:3:5.
Since the smallest share amounts to \$600, we can set up the following equation:
2x = \$600
To solve for x, we divide both sides of the equation by 2:
x = \$600 / 2 = \$300
Now that we know the value of x, we can find the total sum of money to be shared by summing up the individual shares of A, B, and C.
A's share = 2x = 2 * \$300 = \$600
B's share = 3x = 3 * \$300 = \$900
C's share = 5x = 5 * \$300 = \$1500
Total sum of money to be shared = A's share + B's share + C's share
= \$600 + \$900 + \$1500
= \$3000
Therefore, the total sum of money to be shared is \$3000.
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2/10 s = 600
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4 Dynamics: Force and Newton’s Laws of Motion
# 28 4.7 Further Applications of Newton’s Laws of Motion
### Summary
• Apply problem-solving techniques to solve for quantities in more complex systems of forces.
• Integrate concepts from kinematics to solve problems using Newton’s laws of motion.
There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills.
### Example 1: Drag Force on a Barge
Suppose two tugboats push on a barge at different angles, as shown in Figure 1. The first tugboat exerts a force ofin the x-direction, and the second tugboat exerts a force ofin the y-direction.
If the mass of the barge isand its acceleration is observed to bein the direction shown, what is the drag force of the water on the barge resisting the motion? (Note: drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object.)
Strategy
The directions and magnitudes of acceleration and the applied forces are given in Figure 1(a). We will define the total force of the tugboats on the barge asso that:
Since the barge is flat bottomed, the drag of the waterwill be in the direction opposite toas shown in the free-body diagram in Figure 1(b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Our strategy is to find the magnitude and direction of the net applied forceand then apply Newton’s second law to solve for the drag force
Solution
Sinceandare perpendicular, the magnitude and direction ofare easily found. First, the resultant magnitude is given by the Pythagorean theorem:
The angle is given by
which we know, because of Newton’s first law, is the same direction as the acceleration.is in the opposite direction ofsince it acts to slow down the acceleration. Therefore, the net external force is in the same direction asbut its magnitude is slightly less thanThe problem is now one-dimensional. From Figure 1(b), we can see that
But Newton’s second law states that
Thus,
This can be solved for the magnitude of the drag force of the waterin terms of known quantities:
Substituting known values gives
The direction ofhas already been determined to be in the direction opposite toor at an angle ofsouth of west.
Discussion
The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, whereis less than 1/600th of the weight of the ship.
In the earlier example of a tightrope walker we noted that the tensions in wires supporting a mass were equal only because the angles on either side were equal. Consider the following example, where the angles are not equal; slightly more trigonometry is involved.
### Example 2: Different Tensions at Different Angles
Consider the traffic light (mass 15.0 kg) suspended from two wires as shown in Figure 2. Find the tension in each wire, neglecting the masses of the wires.
Strategy
The system of interest is the traffic light, and its free-body diagram is shown in Figure 2(c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in part (d) of the figure. There are two unknowns in this problem (and), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.
Solution
First consider the horizontal or x-axis:
Thus, as you might expect,
This gives us the following relationship betweenand
Thus,
Note thatandare not equal in this case, because the angles on either side are not equal. It is reasonable thatends up being greater thanbecause it is exerted more vertically than
Now consider the force components along the vertical or y-axis:
This implies
Substituting the expressions for the vertical components gives
There are two unknowns in this equation, but substituting the expression forin terms ofreduces this to one equation with one unknown:
which yields
Solving this last equation gives the magnitude ofto be
Finally, the magnitude ofis determined using the relationship between them,= 1.225found above. Thus we obtain
Discussion
Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker).
The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed: will the scale still read more than your weight at rest? Consider the following example.
### Example 3: What Does the Bathroom Scale Read in an Elevator?
Figure 3 shows a 75.0-kg man (weight of about 165 lb) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of , and (b) if the elevator moves upward at a constant speed of 1 m/s.
Strategy
If the scale is accurate, its reading will equalthe magnitude of the force the person exerts downward on it. Figure 3(a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn as in Figure 3(b). Analysis of the free-body diagram using Newton’s laws can produce answers to both parts (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weightand the upward force of the scaleAccording to Newton’s third law andare equal in magnitude and opposite in direction, so that we need to findin order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,
From the free-body diagram we see thatso that
Solving forgives an equation with only one unknown:
or, because simply
No assumptions were made about the acceleration, and so this solution should be valid for a variety of accelerations in addition to the ones in this exercise.
Solution for (a)
In this part of the problem,so that
yielding
Discussion for (a)
This is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:
So, the scale reading in the elevator is greater than his 735-N (165 lb) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators.
Solution for (b)
Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero becauseand
Thus,
Now
which gives
Discussion for (b)
The scale reading is 735 N, which equals the person’s weight. This will be the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.
The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward,is negative, and the scale reading is less than the weight of the person, until a constant downward velocity is reached, at which time the scale reading again becomes equal to the person’s weight. If the elevator is in free-fall and accelerating downward atthen the scale reading will be zero and the person will appear to be weightless.
# Integrating Concepts: Newton’s Laws of Motion and Kinematics
Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters. When approaching problems that involve various types of forces, acceleration, velocity, and/or position, use the following steps to approach the problem:
Problem-Solving Strategy
Step 1. Identify which physical principles are involved. Listing the givens and the quantities to be calculated will allow you to identify the principles involved.
Step 2. Solve the problem using strategies outlined in the text. If these are available for the specific topic, you should refer to them. You should also refer to the sections of the text that deal with a particular topic. The following worked example illustrates how these strategies are applied to an integrated concept problem.
### Example 4: What Force Must a Soccer Player Exert to Reach Top Speed?
A soccer player starts from rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What was his average acceleration? (b) What average force did he exert backward on the ground to achieve this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.
Strategy
1. To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters in which they are found. Part (a) of this example considers acceleration along a straight line. This is a topic of kinematics. Part (b) deals with force, a topic of dynamics found in this chapter.
2. The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so forth.
Solution for (a)
We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity isWe are given the elapsed time, and soThe unknown is acceleration, which can be found from its definition:
[latex size =”2″]\boldsymbol{\frac{\Delta{v}}{\Delta{t}}}[/latex].
Substituting the known values yields
Discussion for (a)
This is an attainable acceleration for an athlete in good condition.
Solution for (b)
Here we are asked to find the average force the player exerts backward to achieve this forward acceleration. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes his acceleration. Since we now know the player’s acceleration and are given his mass, we can use Newton’s second law to find the force exerted. That is,
Substituting the known values ofandgives
Discussion for (b)
This is about 50 pounds, a reasonable average force.
This worked example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These strategies are found throughout the text, and many worked examples show how to use them for single topics. You will find these techniques for integrated concept problems useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles.
# Summary
• Newton’s laws of motion can be applied in numerous situations to solve problems of motion.
• Some problems will contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whetheror
• The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating, the normal force will be less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force will always be less than the full weight of the object.
• Some problems will contain various physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics in order to solve these problems of motion.
### Conceptual Questions
1: To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward atWhy will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?
2: A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward ride. Can this really happen without the person being tied to the floor of the elevator? Explain your answer.
### Problems & Exercises
1: A flea jumps by exerting a force ofstraight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force ofon the flea. Find the direction and magnitude of the acceleration of the flea if its mass isDo not neglect the gravitational force.
2: Two muscles in the back of the leg pull upward on the Achilles tendon, as shown in Figure 4. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?
3: A 76.0-kg person is being pulled away from a burning building as shown in Figure 5. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.
4: Integrated Concepts A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow him if he was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)
5: Integrated Concepts When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?
6: Integrated Concepts A large rocket has a mass ofat takeoff, and its engines produce a thrust of(a) Find its initial acceleration if it takes off vertically. (b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust? (c) In reality, the mass of a rocket decreases significantly as its fuel is consumed. Describe qualitatively how this affects the acceleration and time for this motion.
7: Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.
8: Integrated Concepts A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.
9: Integrated Concepts Repeat Exercise 8 for a shell fired at an anglefrom the vertical.
10: Integrated Concepts An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate offor 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate offor 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?
11: Unreasonable Results (a) What is the final velocity of a car originally traveling at 50.0 km/h that decelerates at a rate offor 50.0 s? (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?
12: Unreasonable Results A 75.0-kg man stands on a bathroom scale in an elevator that accelerates from rest to 30.0 m/s in 2.00 s. (a) Calculate the scale reading in newtons and compare it with his weight. (The scale exerts an upward force on him equal to its reading.) (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?
### Solutions
Problems & Exercises
1:
3:
5:
(a)
(b) 2.97 m
7:
(a)
(b)
(c)
9:
(a) 47.1 m/s
(b)
(c)The average force is 252 times the shell’s weight.
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## Engage NY Eureka Math 2nd Grade Module 1 Lesson 4 Answer Key
### Eureka Math Grade 2 Module 1 Lesson 4 Problem Set Answer Key
Solve.
Question 1.
9 + 3 = _12___
9 + 3= 12,
Explanation:
Adding 9 to 3 we will get 12.
Question 2.
9 + 5 = _14___
9 + 5 = 14,
Explanation:
Adding 9 to 5 we will get 14.
Question 3.
8 + 4 = __12___
8 + 4 = 12,
Explanation:
Adding 8 to 4 we will get 12.
Question 4.
8 + 7 = __15___
8 + 7= 15,
Explanation:
Adding 8 to 7 we will get 15.
Question 5.
7 + 5 = __12___
7 + 5 = 12,
Explanation:
Adding 5 to 7 we will get 12.
Question 6.
7 + 6 = ___13__
7 + 6 = 13,
Explanation:
Adding 7 to 6 we will get 13.
Question 7.
8 + 8 = __16___
8 + 8 = 16,
Explanation:
Adding 8 to 8 we will get 16.
Question 8.
9 + 8 = __17___
9 + 8 = 17,
Explanation:
Adding 9 to 8 we will get 17.
Solve.
Question 9.
10 + __2__ = 12
9 + __3__ = 12
10 + 2 = 12,
9 + 3= 12,
Explanation:
Adding 10 to 2 we will get 12
Adding 9 to 3 we will get 12.
Question 10.
10 + _3___ = 13
9 + __4__ = 13
10 + 3 = 13,
9 + 4 = 13,
Explanation:
Adding 10 to 3 we will get 13,
Adding 9 to 4 we will get 13.
Question 11.
10 + __4__ = 14
8 + __6__ = 14
10 + 4 = 14,
8 + 6 = 14,
Explanation:
Adding 10 to 4 we will get 14.
Adding 8 to 6 we will get 14.
Question 12.
10 + _6___ = 16
7 + __9__ = 16
10 + 6 = 16,
7 + 9 = 16,
Explanation:
Adding 10 to 6 we will get 16,
Adding 7 to 9 we will get 16.
Question 13.
Lisa has 2 blue beads and 9 purple beads. How many beads does Lisa have in all?
Lisa has __11___ beads in all.
Lisa has 11 beads in all,
Explanation:
So in all lisa has 9 + 2 = 11 beads.
Question 14.
Ben had 8 pencils and bought 5 more.
How many pencils does Ben have altogether?
________13_______
Ben have altogether 13 pencils,
Explanation:
Given Ben had 8 pencils and bought 5 more so in total
Ben have altogether 13 pencils.
### Eureka Math Grade 2 Module 1 Lesson 4 Exit Ticket Answer Key
Solve.
Question 1.
9 + 6 = _15___
9 + 6 = 15,
Explanation:
Adding 6 to 9 we will get 15.
Question 2.
8 + 5 = _13__
8 + 5 = 13,
Explanation:
Adding 5 to 8 we will get 13.
### Eureka Math Grade 2 Module 1 Lesson 4 Homework Answer Key
Solve.
Question 1.
8 + 4 = 12,
8 + 2 + 2 = 12,
Explanation:
Adding 4 to 8 we will get 12,
adding 2 to 2 we get 4 and adding 4 to 8
we will get 12.
Question 2.
9 + 7 = __16__
9 + 7 = 16,
Explanation:
Adding 7 to 9 we will get 16.
Question 3.
9 + 3 = __12___
9 + 3 = 12,
Explanation:
Adding 3 to 9 we will get 12.
Question 4.
8 + 6 = __14___
8 + 6 = 14,
Explanation:
Adding 6 to 8 we will get 14.
Question 5.
7 + 6 = __13___
7 + 6 = 13,
Explanation:
Adding 6 to 7 we will get 13.
Question 6.
7 + 8 = __15___
7 + 8 = 15,
Explanation:
Adding 8 to 7 we will get 15.
Question 7.
8 + 8 = __16___
8 + 8 = 16,
Explanation:
Adding 8 to 8 we will get 16.
Question 8.
8 + 9 = __17___
8 + 9 = 17,
Explanation:
Adding 9 to 8 we will get 17.
Question 9.
Solve and match.
Explanation:
Solved and matched as shown above,
as 10 + 2 = 12 and 3 + 9 = 12 both results
are same 12 so matched,
10 + _____ = 13 means 13 -10 =3 and 7 + 6 = 13,
So 10 + 3 = 13 to 7 + 6 = 13 matched them,
10 + _____ = 17 means 17 – 10 = 7 and 9 + 8 = 17,
So 10 + 7 = 17 matched to 9 + 8 = 17,
10 + 5 = 15 means 15 – 10 = 5 and 9 + 6 = 15,
So 10 + 5 = 15 matched to 9 + 6 = 15,
4 + ____= 14 means 14 – 4 = 10 and 6 + 8 = 14,
So 4 + 10 = 14 matched to 6 + 8 = 14.
Question 10.
Ronnie uses 5 brown bricks and 8 red bricks to build a fort.
How many bricks does Ronnie use in all?
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Find the HCF by successive division method: 567, 621, 675.
Last updated date: 18th Jun 2024
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Hint: We will first understand the concept of successive division method which is given as if a quotient of the dividend is taken and this is used as the dividend in the next division, such division is called “successive division”. We will solve this in two phases i.e. first we will find HCF of 567, 621. On getting an answer, we will find HCF of that answer, 675. On getting an answer to this, we will find the HCF of 567, 621, 675.
Here, we will first understand the concept of successive division method.
If a quotient of the dividend is taken and this is used as the dividend in the next division, such division is called “successive division”. This process can continue up to any number of steps until the quotient in the first division is taken and dividend in the second division; the quotient in the second division is taken as the dividend in the third division; the quotient in the third division is taken as the dividend in the fourth division and so on.
Now, we solve this in two phases. First, we will take 567, 621.
So, now we will divide 621 by 567. We get as
567\overset{1}{\overline{\left){\begin{align} & 621 \\ & 567 \\ & \overline{\text{ }54} \\ \end{align}}\right.}}
Now, we will take divisor 567 as dividend and remainder 54 as divisor. On solving, we get as
54\overset{10}{\overline{\left){\begin{align} & 567 \\ & 540 \\ & \overline{\text{ 27}} \\ \end{align}}\right.}}
Again, we will take divisor 54 as dividend and remainder as divisor. We will get as
27\overset{2}{\overline{\left){\begin{align} & 54 \\ & 54 \\ & \overline{\text{ 0}} \\ \end{align}}\right.}}
Now, we got remainder 0 so, we will consider the divisor to be HCF of 567, 621 i.e. 27.
Now, we will find HCF of HCF $\left( 567,621 \right)$ , 675 i.e. 27, 675.
Using the same concept done above, we will solve.
27\overset{24}{\overline{\left){\begin{align} & 675 \\ & 54\downarrow \\ & \overline{\text{135}} \\ & 132 \\ & \overline{\text{ }3} \\ \end{align}}\right.}}
Now, we will take divisor 27 as dividend and remainder 3 to be divisor. We get as
3\overset{9}{\overline{\left){\begin{align} & 27 \\ & 27 \\ & \overline{\text{ }0} \\ \end{align}}\right.}}
Thus, we can now say that divisor 3 is HCF of 567, 621, 675.
Thus, $HCF\left( 567,621,675 \right)$ is 3.
Note: Do not assume a successive division method to be the same as normal division we do in prime factorization. Though the answer will be correct, the method of solving will be wrong. So, students should be clear with the methods of finding the highest common factor and then solve as per asked in question.
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# 7.4: AA Similarity
Difficulty Level: At Grade Created by: CK-12
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What if you were given a pair of triangles and the angle measures for two of their angles? How could you use this information to determine if the two triangles are similar? After completing this Concept, you'll be able to use AA Similarity to decide if two triangles are similar.
### Watch This
Watch this video beginning at the 2:09 mark.
### Guidance
The Third Angle Theorem states if two angles are congruent to two angles in another triangle, the third angles are congruent too. Because a triangle has 180\begin{align*}180^\circ\end{align*}, the third angle in any triangle is 180\begin{align*}180^\circ\end{align*} minus the other two angle measures. Let’s investigate what happens when two different triangles have the same angle measures.
##### Investigation: Constructing Similar Triangles
Tools Needed: pencil, paper, protractor, ruler
1. Draw a 45\begin{align*}45^\circ\end{align*} angle. Extend the horizontal side and then draw a 60\begin{align*}60^\circ\end{align*} angle on the other side of this side. Extend the other side of the 45\begin{align*}45^\circ\end{align*} angle and the 60\begin{align*}60^\circ\end{align*} angle so that they intersect to form a triangle. What is the measure of the third angle? Measure the length of each side.
2. Repeat Step 1 and make the horizontal side between the 45\begin{align*}45^\circ\end{align*} and 60\begin{align*}60^\circ\end{align*} angle at least 1 inch longer than in Step 1. This will make the entire triangle larger. Find the measure of the third angle and measure the length of each side.
3. Find the ratio of the sides. Put the sides opposite the 45\begin{align*}45^\circ\end{align*} angles over each other, the sides opposite the 60\begin{align*}60^\circ\end{align*} angles over each other, and the sides opposite the third angles over each other. What happens?
AA Similarity Postulate: If two angles in one triangle are congruent to two angles in another triangle, the two triangles are similar.
The AA Similarity Postulate is a shortcut for showing that two triangles are similar. If you know that two angles in one triangle are congruent to two angles in another, which is now enough information to show that the two triangles are similar. Then, you can use the similarity to find the lengths of the sides.
#### Example A
Determine if the following two triangles are similar. If so, write the similarity statement.
Find the measure of the third angle in each triangle. mG=48\begin{align*}m \angle G = 48^\circ\end{align*} and mM=30\begin{align*}m \angle M = 30^\circ\end{align*} by the Triangle Sum Theorem. Therefore, all three angles are congruent, so the two triangles are similar. FEGMLN\begin{align*}\triangle FEG \sim \triangle MLN\end{align*}.
#### Example B
Determine if the following two triangles are similar. If so, write the similarity statement.
mC=39\begin{align*}m \angle C = 39^\circ\end{align*} and mF=59\begin{align*}m \angle F = 59^\circ\end{align*}. The angles are not equal, ABC\begin{align*}\triangle ABC\end{align*} and DEF\begin{align*}\triangle DEF\end{align*} are not similar.
#### Example C
Are the following triangles similar? If so, write the similarity statement.
Because AE¯¯¯¯¯¯¯¯ || CD¯¯¯¯¯¯¯¯,AD\begin{align*}\overline{AE} \ || \ \overline{CD}, \angle A \cong \angle D\end{align*} and CE\begin{align*}\angle C \cong \angle E\end{align*} by the Alternate Interior Angles Theorem. Therefore, by the AA Similarity Postulate, ABEDBC\begin{align*}\triangle ABE \sim \triangle DBC\end{align*}.
Watch this video for help with the Examples above.
### Vocabulary
Two triangles are similar if all their corresponding angles are congruent (exactly the same) and their corresponding sides are proportional (in the same ratio).
### Guided Practice
Are the following triangles similar? If so, write a similarity statement.
1.
2.
3.
1. Yes, DGEFGDFDE\begin{align*}\triangle DGE \sim \triangle FGD \sim \triangle FDE\end{align*}.
2. Yes, HLIHMJ\begin{align*}\triangle HLI \sim \triangle HMJ\end{align*}.
3. No, though MNQONP\begin{align*}\angle MNQ \cong \angle ONP\end{align*} because they are vertical angles, we need to have two pairs of congruent angles in order to be able to say that the triangles are similar.
### Practice
Use the diagram to complete each statement.
1. SAM\begin{align*}\triangle SAM \sim \triangle \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
2. SA?=SM?=?RI\begin{align*}\frac{SA}{?}=\frac{SM}{?}=\frac{?}{RI}\end{align*}
3. SM=\begin{align*}SM = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
4. TR=\begin{align*}TR = \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
5. 9?=?8\begin{align*}\frac{9}{?}=\frac{?}{8}\end{align*}
Answer questions 6-9 about trapezoid ABCD\begin{align*}ABCD\end{align*}.
1. Name two similar triangles. How do you know they are similar?
2. Write a true proportion.
3. Name two other triangles that might not be similar.
4. If AB=10,AE=7,\begin{align*}AB = 10, AE = 7,\end{align*} and DC=22\begin{align*}DC = 22\end{align*}, find AC\begin{align*}AC\end{align*}. Be careful!
5. Writing How many angles need to be congruent to show that two triangles are similar? Why?
6. Writing How do congruent triangles and similar triangles differ? How are they the same?
Use the triangles below for questions 12-15.
AB=20,DE=15,\begin{align*}AB = 20, DE = 15,\end{align*} and BC=k\begin{align*}BC = k\end{align*}.
1. Are the two triangles similar? How do you know?
2. Write an expression for FE\begin{align*}FE\end{align*} in terms of k\begin{align*}k\end{align*}.
3. If FE=12\begin{align*}FE = 12\end{align*}, what is k\begin{align*}k\end{align*}?
4. Fill in the blanks: If an acute angle of a _______ triangle is congruent to an acute angle in another ________ triangle, then the two triangles are _______.
Use the diagram below to answer questions 16-20.
1. Draw the three separate triangles in the diagram.
2. Explain why GDEDFEGFD\begin{align*}\triangle GDE \sim \triangle DFE \sim \triangle GFD\end{align*}.
Complete the following proportionality statements.
1. GFDF=?FE\begin{align*}\frac{GF}{DF}=\frac{?}{FE}\end{align*}
2. GFGD=?GE\begin{align*}\frac{GF}{GD}=\frac{?}{GE}\end{align*}
3. GEDE=DE?\begin{align*}\frac{GE}{DE}=\frac{DE}{?}\end{align*}
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
AA Similarity Postulate
If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar.
Dilation
To reduce or enlarge a figure according to a scale factor is a dilation.
Triangle Sum Theorem
The Triangle Sum Theorem states that the three interior angles of any triangle add up to 180 degrees.
Rigid Transformation
A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure.
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Home » Using the answer in (b)(i), solve the system of equations
# Using the answer in (b)(i), solve the system of equations
(a) Differentiate (frac{x^{2} + 1}{(x + 1)^{2}}) with respect to x.
(b)(i) Evaluate (begin{vmatrix} 1 & 2 & -1 \ 2 & 3 & -1 \ -1 & 1 & 3 end{vmatrix}).
(ii) Using the answer in (b)(i), solve the system of equations.
(x + 2y – z = 4)
(2x + 3y – z = 2)
(-x + y + 3z = -1).
##### Explanation
(a) Let (y = frac{x^{2} + 1}{(x + 1)^{2}})
Let (u = x^{2} + 1 ; v = (x + 1)^{2})
(frac{mathrm d u}{mathrm d x} = 2x ; frac{mathrm d v}{mathrm d x} = 2(x + 1))
Using the quotient rule,
(frac{mathrm d y}{mathrm d x} = frac{v frac{mathrm d u}{mathrm d x} – u frac{mathrm d v}{mathrm d x}}{v^{2}})
= (frac{(x + 1)^{2} (2x) – (x^{2} + 1)(2(x + 1))}{((x + 1)^{2})^{2}})
= (frac{(x + 1)[(x + 1)(2x) – (x^{2} + 1)(2)}{(x + 1)^{4}})
= (frac{2x^{2} + 2x – 2x^{2} – 2}{(x + 1)^{3}})
= (frac{2x – 2}{(x + 1)^{3}})
(b)(i) (begin{vmatrix} 1 & 2 & -1 \ 2 & 3 & -1 \ -1 & 1 & 3 end{vmatrix})
= (1(9 + 1) – 2(6 – 1) – 1(2 + 3))
= (10 – 10 – 5 = -5)
(ii) (x + 2y – z = 4)
(2x + 3y – z = 2)
(-x + y + 3z = -1)
In matrix form, this is
(begin{pmatrix} 1 & 2 & -1 \ 2 & 3 & -1 \ -1 & 1 & 3 end{pmatrix} begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 4 \ 2 \ -1 end{pmatrix})
Cofactors :
(A_{11} = +(3 times 3 + 1) = 10)
(A_{12} = -(2 times 3 – 1) = -5)
(A_{13} = +(2 times 1 + 3) = 6)
(A_{21} = -(2 times 3 + 1) = -7)
(A_{22} = +(1 times 3 – 1) = 2)
(A_{23} = -(1 times 1 + 1 times 2) = -3)
(A_{31} = +(2 times -1 + 3) = 1)
(A_{32} = -(-1 + 2) = -1)
(A_{33} = +(1 times 3 – 2 times 2) = -1)
(C = begin{pmatrix} 10 & -5 & 5 \ -7 & 2 & -3 \ 1 & -1 & -1 end{pmatrix})
adj A = (C^{T} = begin{pmatrix} 10 & -7 & 1 \ -5 & 2 & -1 \ 5 & -3 & -1 end{pmatrix})
= (frac{1}{-5} begin{pmatrix} 10 & -7 & 1 \ -5 & 2 & -1 \ 5 & -3 & -1 end{pmatrix})
= (begin{pmatrix} -2 & frac{7}{5} & frac{-1}{5} \ 1 & frac{-2}{5} & frac{1}{5} \ -1 & frac{3}{5} & frac{1}{5} end{pmatrix})
(therefore x = A^{-1} . b )
= (begin{pmatrix} -2 & frac{7}{5} & frac{-1}{5} \ 1 & frac{-2}{5} & frac{1}{5} \ -1 & frac{3}{5} & frac{1}{5} end{pmatrix} begin{pmatrix} 4 \ 2 \ -1 end{pmatrix})
= (begin{pmatrix} – 8 + frac{14}{5} + frac{1}{5} \ 4 – frac{4}{5} + frac{1}{5} \ -4 + frac{6}{5} – frac{1}{5} end{pmatrix})
= (begin{pmatrix} -5 \ 3 \ -3 end{pmatrix})
x = -5 ; y = 3 ; z = -3.
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# Continuity And Differentiability
Continuity and Differentiability is one of the most important topics which help students to understand the concepts like, continuity at a point, continuity on an interval, derivative of functions and many more. However, continuity and Differentiability of functional parameters are very difficult. Let us take an example to make this simpler:
Consider the function,
$$\begin{array}{l}\left\{\begin{matrix} x+3 & if\ x \leq 0\\ x & if\ x>0 \end{matrix}\right.\end{array}$$
For any point on the Real number line, this function is defined.Â
It can be seen that the value of the function x = 0 changes suddenly. Following the concepts of limits, we can say that;
Right-hand limit ≠Left-hand limit.
It implies that this function is not continuous at x=0.
In simple words, we can say that a function is continuous at a point if we are able to graph it without lifting the pen.
## Definition of Continuity
In Mathematically, A function is said to be continuous at a point x = a, if
$$\begin{array}{l}\lim_{x\rightarrow a}\end{array}$$
f(x) Exists, and
$$\begin{array}{l}\lim_{x\rightarrow a}\end{array}$$
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Continuity in open interval (a, b)
f(x) will be continuous in the open interval (a,b) if at any point in the given interval the function is continuous.
Continuity in closed interval [a, b]
A function f(x) is said to be continuous in the closed interval [a,b] if it satisfies the following three conditions.
1) f(x) is be continuous in the open interval (a, b)
2) f(x) is continuous at the point a from right i.e.
$$\begin{array}{l}\lim\limits_{x \to a}f(x)\end{array}$$
=f(a)
3) f(x) is continuous at the point b from left i.e.
$$\begin{array}{l}\lim\limits_{x \to b}f(x)\end{array}$$
=f(b)
Lets Work Out- Example: Check whether the function $$\begin{array}{l}\frac{4x^{2}-1}{2x-1}\end{array}$$ is continuous or not? Solution: At x=1/2, the value of denominator is 0. So the function is discontinuous at x = 1/2.
## Definition of Differentiability
f(x) is said to be differentiable at the point x = a if the derivative f ‘(a) exists at every point in its domain. It is given by
f'(a)=$$\begin{array}{l}\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}\end{array}$$
For a function to be differentiable at any point x=a in its domain, it must be continuous at that particular point but vice-versa is not always true.
The derivatives of the basic trigonometric functions are;
1. $$\begin{array}{l}\frac{d}{dx}\end{array}$$
(sinx) = cosx
2. $$\begin{array}{l}\frac{d}{dx}\end{array}$$
(cosx) = −sinx
3. $$\begin{array}{l}\frac{d}{dx}\end{array}$$
(tanx) = sec2x
4. $$\begin{array}{l}\frac{d}{dx}\end{array}$$
(cotx) = –cosec2x
5. $$\begin{array}{l}\frac{d}{dx}\end{array}$$
(secx) = secx tanx
6. $$\begin{array}{l}\frac{d}{dx}\end{array}$$
(cosecx) = −cosecx cotx
Lets Work Out- Example: Consider the function $$\begin{array}{l}f(x)=(2x-3)^{\frac{1}{5}}\end{array}$$.Discuss its continuity and differentiability at $$\begin{array}{l}x= \frac{3}{2}\end{array}$$. Solution: For checking the continuity, we need to check the left hand and right-hand limits and the value of the function at a point x=a. L.H.L. = $$\begin{array}{l}\lim\limits_{x \to a^{-}}f(x)= \lim_{x \to \frac{3}{2}}(2x-3)^{\frac{1}{5}}\end{array}$$ $$\begin{array}{l}= \left ( 2 \times \frac{3}{2} -3 \right )^{\frac{1}{5}}\end{array}$$ = 0 R.H.L. = $$\begin{array}{l}\lim\limits_{x \to a^{+}}f(x)= \lim_{x \to \frac{3}{2}}(2x-3)^{\frac{1}{5}}\end{array}$$ $$\begin{array}{l}= \left ( 2 \times \frac{3}{2} -3\right )^{\frac{1}{5}} \end{array}$$ = 0 L.H.L = R.H.L = f(a) = 0. Thus the function is continuous at about the point $$\begin{array}{l}x= \frac{3}{2}\end{array}$$. Now to check differentiability at the given point, we know $$\begin{array}{l}f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}\end{array}$$ = $$\begin{array}{l}\lim\limits_{h \to 0}\frac{f(\frac{3}{2}+h)-f(\frac{3}{2})}{h}\end{array}$$ = $$\begin{array}{l}\lim\limits_{h \to 0}\frac{\left ( [2(\frac{3}{2})+h]-3 \right )^{\frac{1}{5}}- \left ( 2(\frac{3}{2})-3 \right )^{\frac{1}{5}}}{h}\end{array}$$ = $$\begin{array}{l}\lim\limits_{h \to 0}\frac{\left ( 3 + 2h -3 \right )^{\frac{1}{5}}- \left ( 3 -3 \right )^{\frac{1}{5}}}{h}\end{array}$$ = $$\begin{array}{l}\lim\limits_{h \to 0}\frac{\left ( 2h \right )^{\frac{1}{5}}- 0}{h}\end{array}$$ = $$\begin{array}{l}\lim\limits_{h \to 0}\frac{ 2^\frac{1}{5}}{h^\frac{4}{5}} = \infty\end{array}$$ Thus f is not differentiable at $$\begin{array}{l}x= \frac{3}{2}\end{array}$$. We see that even though the function is continuous but it is not differentiable.
Learn more about limits and continuity and other topics with BYJU’S and make learning more fun.Â
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# What Are The Prime Factors Of 18
What are the prime factors of 18? Answer: 2, 3
The number 18 has 2 prime factors. Primes can only have two factors(1 and itself) and only be divisible by those two factors. Any figure where this rule applies can be called a prime factor. The biggest prime factor of 18 is 3. The smallest prime factor of 18 is 2.
## What Is The Factor Tree Of 18
How to use a factor tree to find the prime factors of 18? A factor tree is a diagram that organizes the factoring process.
• First step is to find two numbers that when multiplied together equal the number we start with.
18
↙ ↘
2 × 9
• Second step is to check the multiplication(of the first step) for numbers that are not primes.
9
↙ ↘
3 × 3
We found 2 prime factors(2, 3) using the factor tree of 18. Now let us explain the process to solving factor trees in more detail. Our goal is to find all prime factors of a given whole number. In each step of our factor tree diagram for 18 we always checked both multiplication numbers if they were primes or not. If one or both of the integers are not prime numbers then this means that we will have to make diagrams for them too. This process continues until only prime numbers are left.
Remember that often a factor tree for the same integer can be solved in more than one correct way! An example of this is the figure 12 where 2*6=12 and 4*3=12. The primes of a factor tree for 12 are the same regardles if we start the factor tree with 2*6 or 4*3.
## How To Verify If Prime Factors Of 18 Are Correct Answers
To know if we got the correct prime factors of 18 we have to get the prime factorization of 18 which is 2 * 3 * 3. Because when you multiply the primes of the prime factorization the answer has to be equal with 18.
After having checked the prime factorization we can now safely say that we got all prime factors.
## General Mathematical Properties Of Number 18
18 is a composite number. 18 is a composite number, because it has more divisors than 1 and itself. This is an even number. 18 is an even number, because it can be divided by 2 without leaving a comma spot. This also means that 18 is not an odd number. When we simplify Sin 18 degrees we get the value of sin(18)=-0.75098724677168. Simplify Cos 18 degrees. The value of cos(18)=0.66031670824408. Simplify Tan 18 degrees. Value of tan(18)=-1.1373137123377. When converting 18 in binary you get 10010. Converting decimal 18 in hexadecimal is 12. The square root of 18=4.2426406871193. The cube root of 18=2.6207413942089. Square root of √18 simplified is 3√2. All radicals are now simplified and in their simplest form. Cube root of ∛18 simplified is 18. The simplified radicand no longer has any more cubed factors.
## Determine Prime Factors Of Numbers Smaller Than 18
Learn how to calculate primes of smaller numbers like:
## Determine Prime Factors Of Numbers Bigger Than 18
Learn how to calculate primes of bigger numbers such as:
## Single Digit Properties For Number 18 Explained
• Integer 1 properties: 1 is an odd figure. In set theory, the 1 is constructed starting from the empty set obtaining {∅}, whose cardinality is precisely 1. It is the neutral element of multiplication and division in the sets of natural, integer, rational and real numbers. The first and second digit of the Fibonacci sequence(before 2). Second to the succession of Lucas(after 2). First element of all the successions of figured numbers. One is a part of the Tetranacci Succession. 1 is a number of: Catalan, Dudeney, Kaprekar, Wedderburn-Etherington. It is strictly non-palindrome, integer-free, first suitable digit, first issue of Ulam and the first centered square. The first term of the succession of Mian-Chowla. Complete Harshad, which is a number of Harshad in any expressed base. 1 is the first highly totest integer and also the only odd number that is not non-tottering.
• Integer 8 properties: Eight is even and a cube of 2. It is a composite, with the following 4 divisors:1, 2, 4, 8. Since the total of the divisors(excluding itself) is 7<8, it is a defective number. The sixth of the Fibonacci sequence, after 5 and before 13. It is the quantity of the twin primes 3 and 5. The first octagonal value. 8 is a Ulam, centered heptagonal and Leyland number. All amounts are divisible by 8 if and only if the result formed by its last three digits is. A refactorizable, being divisible by the count of its divisors. At the same time a highly totter and highly cototent quantity. It is the fourth term of the succession of Mian-Chowla. Any odd greater than or equal to 3, elevated to the square, to which subtract is subtracted 1 is divisible by 8 (example: 7²=49 49-1=48 divisible by 8). The sum of two squares, 8=2²+2². The sum of the digits of its cube: 8³=512, 5+1+2=8. The first 4-digit binary:1000. Part of the Pythagorean triples (6, 8, 10), (8, 15, 17). Eight is a repeated number in the positional numbering system based on 3 (22) and on the base 7 (11).
## Finding All Prime Factors Of A Number
We found that 18 has 2 primes. The prime factors of 18 are 2, 3. We arrived to this answer by using the factor tree. However we could have also used upside down division to get the factorization primes. There are more that one method to factorize a integer.
## List of divisibility rules for finding prime factors faster
Rule 1: If the last digit of a figure is 0, 2, 4, 6 or 8 then it is an even number. All even numbers are divisible by 2.
Rule 2: If the sum of digits of a integer is divisible by 3 then the figure is also divisible by 3 and 3 is a prime factor(example: the digits of 12 are 1 and 2 so 1+2=3 and 3 is divisible by 3, meaning that 12 is divisible by 3). The same logic also works for 9.
Rule 3: If the last two digits of a number are 00 then this integer is divisible by 4(example: we know that 124=100+24 and 100 has two zeros in the end making it divisible with 4. We also know that 4 is divisible with 24). In order to use this rule to it's fullest it is best to know multiples of 4.
Rule 4: If the last digit of a number is 0 or 5 then 5 it is divisible by 5.
Rule 5: All integers that are divisible by both 2 and 3 are also divisible by 6. This is logical because 2*3=6.
## What Are Prime Factors Of A Number?
All numbers that are only divisible by one and itself are called prime factors in mathematics. A prime factor is a figure that has only two factors(one and itself).
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# How do you graph the equation y-8=-x?
Mar 30, 2018
$\text{see explanation}$
#### Explanation:
$\text{one way is to find the intercepts, that is where the}$
$\text{graph crosses the x and y axes}$
• " let x = 0, in the equation for y-intercept"
• " let y = 0, in the equation for x-intercept"
$x = 0 \Rightarrow y - 8 = 0 \Rightarrow y = 8 \leftarrow \textcolor{red}{\text{y-intercept}}$
$y = 0 \Rightarrow 0 - 8 = - x \Rightarrow x = 8 \leftarrow \textcolor{red}{\text{x-intercept}}$
$\text{plot the points "(0,8)" and } \left(8 , 0\right)$
$\text{draw a straight line through them for graph}$
graph{-x+8 [-20, 20, -10, 10]}
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# Types of Number Systems
Today in this article we will discuss about the types of number systems. The description of each number system will be given here.
Anything that refers to the measurement of things is known as number. The system of counting or expressing a number is called number system.
That means number system is the method of expressing or writing a number and performing various calculations like addition, subtraction, multiplication, division etc with these numbers.
The fundamental symbols to write any number of a system is called digit. So, digit is the smallest symbol to make a number. For example, 123 is a number which consists of 3 digits 1, 2 and 3.
In this guide, we will discuss broadly about different types of number systems and after that I will try to explain the conversion of number from one system to another. Now, let’s be familiar with different types of number systems.
## Types of Number Systems
Every number system that invented still now are categorized in two vast types of number systems.
Then the last one is sub divided into many categories from which I will discuss about most uses 4 types of number systems here. See the list bellow;
1. Non-positional number system
2. Positional number system
1. Binary Number System
2. Octal Number System
3. Decimal Number System
### 1. Non-positional number system
Non-positional number system is the oldest method of counting a number. The digits have no their positional value in this number system.
The total value of a number is determined by individual value of every digit. It is very complex to perform mathematical calculation in this number system.
For this, there is no available use of this number system today although in ancient times it has been used.
Roman number system is the existing example of non-positional number system. In roman number system value of x is 10 and for this xxx is equivalent to 30. Here, x has no positional value.
### 2. Positional number system
Now a days we are using positional number system to perform our daily mathematical calculations.
In this types of number systems every digit has a positional value and this positional value depends on the method of writing this number.
To know the actual value of a number in positional number system we have to know the value of every digit, base of the writing method and position of every digit.
For example, in decimal number we use 10 digits from 0 to 9 and the positional value depends on the position of these digits.
In positional number system every number is divided into two part of integer and fraction by a radix (.) point.
Suppose a decimal number 135.265
Here, 135 is the integer portion, (.) is the radix point and 265 is the fraction part.
### Types of positional number system
There are mainly 4 types of positional number system we use still today. Other types have no practically used now a days.
So, we should learn only this 4 types. You will get the idea of other types of number by learning these systems.
• Binary Number System
• Octal Number System
• Decimal Number System
#### Binary number system
There are only two digits to write a number in binary number system. These are 0 and 1. For this, the base of binary number system is 2.
So, the number system in which only two digits are used to write any number is called binary number system.
The shape of a number is largest compare to other number system as there are only two digits to express a number in binary number system.
Every electronic device including computer use this number system to perform mathematical calculation. Every digit is called bit (binary digit) in binary number system.
For example, (1011001)2 is a binary number which consists of 7 bits. We can easily convert a binary number to octal, decimal or hexadecimal number which we will discuss later.
#### Octal number system
Meaning of octal is eight. In octal number system total eight digits can use to express a number and they are 0, 1, 2, 3, 4, 5, 6 and 7.
For this, the base of octal number system is 8. So, the number system in which total 8 digits are used to write any number is called octal number system.
It needs minimum two digits to write a number greater than 7 in octal number system. Consider, (127)8 is an octal number which consists of 3 digits 1, 2 and 7.
#### Decimal number system
Decimal number system is the most popular number system which we use in our normal life. In decimal number system we use 10 digits to write any number which are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
For this, the base of decimal number system is 10. So, the number system in which we use total 10 digits to express any number is called decimal number system.
For example, (99)10 is a decimal number which consists of 2 digits 9 and 9. We can easily convert a decimal number to equivalent binary, octal or hexadecimal number.
In hexadecimal number system we can use total 16 digits to write a number. These are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A (10), B (11), C (12), D (13), E (14) and F (15).
For this, the base of hexadecimal number system is 16. So, the number system in which total 16 digits can be used to write any number is called hexadecimal number system.
For example, (7D)16 is a hexadecimal number which consists of 2 digits 7 and D. Learn how to convert a hexadecimal number to its equivalent binary, octal or decimal number.
### Comparison among binary, octal, decimal and hex
Here is a simple comparison among four main number systems. You can get a clear concept about all the main number systems seeing this table bellow;
Now, think about some different types which may 4 based, 5 based or 6, 7 9 etc. All you have to think that there will be the same number of digits as their base. You will learn details in our next several posts.
I hope, now you have no doubt about different types of number systems which we have discussed here in this post. This can be a very useful post for you we believe.
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Probability distribution
Chapter 13 Class 12 Probability (Term 2)
Concept wise
### Transcript
Ex 13.4, 9 The random variable X has a probability distribution P(X) of the following form, where k is some number : P(X) = ๐, ๐๐ ๐ฅ=0๏ทฎ2๐, ๐๐ ๐ฅ=1๏ทฎ3๐, ๐๐ ๐ฅ=2๏ทฎ0, ๐๐กโ๐๐๐ค๐๐ ๐๏ทฏ๏ทฏ (a) Determine the value of k. Making in tabular format Since X is a random variable , its Sum of Probabilities is equal to 1 0๏ทฎ4๏ทฎ๐(๐)๏ทฏ = 1 P(x = 0) + P(x = 1) + P(x = 2) = 1 k + 2k + 3k = 1 6k = 1 k = ๐๏ทฎ๐๏ทฏ Ex 13.4, 9 (b) Find P (X < 2), P (X โค 2), P(X โฅ 2). Our probability distribution table is P(X < 2) = P(X = 0) + P(X = 1) = k + 2k = 3k = 3 ร 1๏ทฎ6๏ทฏ = ๐๏ทฎ๐๏ทฏ P(X โค 2) = P(X = 0) + P(X = 1) + P(X = 2) = k + 2k + 3k = 6k = 6 ร 1๏ทฎ6๏ทฏ = 1 P(X โฅ 2) = P(X = 2) + P(x = 3) + P(X = 4) + โฆโฆ = 3k + 0 + 0 + โฆโฆโฆ. = 3k = 3 ร 1๏ทฎ6๏ทฏ = ๐๏ทฎ๐๏ทฏ
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# Vedic Math Puzzle: Only 1% Of Genius Can Find The Product of The Two Digits Below With Vedic Sutra in 19 Seconds. Hurry Up!
## Vedic Math Puzzle: Vedic math involves learning mental calculation methods, shortcuts, and patterns to perform arithmetic operations, solve complex equations, and make calculations easier. Try this one!
Jun 8, 2023, 18:26 IST
Vedic Math Puzzle: Vedic math puzzles are mathematical problems or equations that use methods from the ancient Indian system of Vedic mathematics to solve them more quickly and effectively.
Vedic math puzzles are vital because they enhance general mathematical ability and cultivate a respect for Vedic mathematics while encouraging critical thinking, problem-solving abilities, and mental agility. Your task in this equation is to use a Vedic Math formula to calculate the product of the two units.
Look for the equation below:
Do You Think You Will Be Able To Solve?
Don’t Say It’s Not Your Cup Of Tea!
What is Vedic Math?
Vedic math multiplication is a collection of methods that were derived from classical Indian mathematics that provide quick and different ways to multiply integers. These techniques help to speed up and simplify calculations while simplifying challenging multiplication problems.
Are You Skilled Enough To Solve Basic Multiplication Using Vedic Formula in 27 Seconds?
Hey Genius…
You are Losing…
The benefits of solving Vedic math equations are numerous. Math becomes more interesting and approachable as a result of these improvements in mental calculation abilities, speed and precision of mathematical operations, logical thinking and problem-solving skills, and alternate approaches to hard computations.
Tick…
Tock…
Tick…
I am sure you have a digit. Just tally your steps and answers with the below picture.
Understand the steps here:
Step 1: 6*2 = 12 (write 2, carry 1)
Step 2: 6*3 + 2*5 = 28 + 1 = 29 (write 9, carry 2)
Step 3: 6*1 + 3*5 + 2*4 + 2 (carried) = 31 (write 1, carry 3)
Step 4: 1*5 + 3*4 = 17 + 3 (carried) = 20 (write 0, carry 2)
Step 5: 1*4 = 4 + 2 = 6 (write 6)
Therefore, the correct answer to this is 60192.
This three-digit multiplication equation is solved using the Sutra (formula) of Urdhva-Tiryagbyham. It is a Vedic math technique that involves vertical and crosswise multiplication, enabling efficient multiplication of multi-digit numbers by breaking them down into simpler computations.
You did a great job. Scroll back to JagranJosh to discover more about Vedic Maths and its various Sutras.
Only A Record-Holding Genius Can Find The Odd Digit in 3 Seconds. Try Your Skills!
Get here current GK and GK quiz questions in English and Hindi for India, World, Sports and Competitive exam preparation. Download the Jagran Josh Current Affairs App.
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# Project Euler 285
Problem 285 of Project Euler was released yesterday; it requires some geometry and probability sense.
The problem goes like this:
Albert chooses two real numbers a and b randomly between 0 and 1, and a positive integer k.
He plays a game with the following formula:
$k' = \sqrt{(ka+1)^2 + (kb+1)^2}$
The result k’ is rounded to the nearest integer. If k’ rounded is equal to k, then Albert gets k points. Otherwise, he gets nothing.
Find the expected value of the total score if Albert plays this game with $k = 1, 2, \cdots , 10^5$.
For example (from the problem statement), let k=6, a=0.2, and b=0.85.
Here the expression $\sqrt{(ka+1)^2 + (kb+1)^2}$ gives the number 6.484. When rounded to the nearest integer, it becomes 6, which is equal to k. Therefore, Albert gets 6 points.
Yup – another expected value problem. But the concept of this problem is simple:
For any value of k, there are certain values of a and b where Albert will get points. Otherwise, he does not get points. Here the gray blob is the area where he does get points, and the white area is where he does not.
The red square is the sample space of the random variables a and b, since $0 and $0.
The probability of Albert getting the points is the fraction of the square that’s gray.
The area of the red square is always 1. The area of the gray area varies with k. Thus the idea is to calculate the area of the gray area for each value of k from 1 to $10^5$.
Now we can get to the details of the problem.
Notice that if $round(k') = k$, then $k-0.5 < k' < k+0.5$ (the difference between less-equal and less is irrelevant in this problem):
In a similar way, the equation of the larger circle is $(a+\frac{1}{k})^2 + (b+\frac{1}{k})^2 < (\frac{k+0.5}{k})^2$.
These are equations of circles. A circle has the equation $(x-a)^2 + (y-b)^2 = r^2$ where $(a,b)$ is the center and r is the radius.
Thus both the smaller and the larger circles have a center at $(-\frac{1}{k},-\frac{1}{k})$. The radii of the circles are different: the radius of the smaller circle is $\frac{k-0.5}{k}$ and the radius of the larger circle $\frac{k+0.5}{k}$.
Plotting the inequalities:
We can see now that this is a calculus problem! Indeed, the shaded area is the area of the larger circle inside the boxarea of the smaller circle inside the box.
Initially I tried integrating the circles directly, but I ran into problems while integrating and I was unable to correctly integrate the equation. Instead, I integrated an equivalent problem:
It’s obvious that the areas do not change with this transformation.
So now for the smaller circle the equation is $a^2 + (b+\frac{1}{k})^2 > (\frac{k-0.5}{k})^2$. Solving for b, we get:
$b > \sqrt{(\frac{k-0.5}{k})^2-a^2} - \frac{1}{k}$
.. And this equation for the larger circle:
$b < \sqrt{(\frac{k+0.5}{k})^2-a^2} - \frac{1}{k}$
The variable of integration here is a, and k is only a constant. Therefore we can replace the radius with r and get this equation to integrate:
$b = \sqrt{r^2-a^2} - \frac{1}{k}$
With the help of an integration table, this can be integrated fairly easily.
There’s one additional problem – computing the integrating ‘limits’ of the two circles. Let’s call them $L_S$ and $L_L$ for small and large respectively.
Both of them can be computed using the pythagorean theorem.
The formula for $L_S$ (it should be obvious by now what $L_L$ should be):
$L_S = \sqrt{(\frac{k-0.5}{k})^2-\frac{1}{k^2}}$
Now we can get this really big formula for computing the area for any integer k:
$\begin{array}{l} \int_{\frac{1}{k}}^{\sqrt{(\frac{k+0.5}{k})^2-\frac{1}{k^2}}} (\sqrt{(\frac{k+0.5}{k})^2-a^2}-\frac{1}{k}) da \\ - \int_{\frac{1}{k}}^{\sqrt{(\frac{k-0.5}{k})^2-\frac{1}{k^2}}} (\sqrt{(\frac{k-0.5}{k})^2-a^2}-\frac{1}{k}) da \end{array}$
Also, fortunately for me, it is impossible for the upper bound to cross the circle at the top or right, I mean, we never have something like this –
Anyways, here’s a Haskell implementation of the algorithm:
integral a r k = 0.5 * (a * sqrt (r^2 - a^2) + r^2 * atan (a/sqrt (r^2-a^2))) - a/k
area k = let area_1 1 = 0
area_1 k = integral (sqrt (((k-0.5)/k)^2 - (1/k)^2)) ((k-0.5)/k) k - integral (1/k) ((k-0.5)/k) k
area_2 k = integral (sqrt (((k+0.5)/k)^2 - (1/k)^2)) ((k+0.5)/k) k - integral (1/k) ((k+0.5)/k) k
in area_2 k - area_1 k
main = print . sum . zipWith (*) [1..] \$ map area [1..10^5]
The code runs in about 3 seconds.
There seems to be an edge case where k=1 (probably an asymptote) so I have to specially make it be zero.
This problem was released 10 pm last night and I was tired, lol, but I still managed to solve it and get into the top 20.
## One thought on “Project Euler 285”
1. Cody says:
Albert has far too much time on his hands.
Like
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# 6 is what percent of 612 - step by step solution
## Simple and best practice solution for 6 is what percent of 612. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the calculator fields your own values, and You will get the solution.
To get the solution, we are looking for, we need to point out what we know.
1. We assume, that the number 612 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 100% equals 612, so we can write it down as 100%=612.
4. We know, that x% equals 6 of the output value, so we can write it down as x%=6.
5. Now we have two simple equations:
1) 100%=612
2) x%=6
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
100%/x%=612/6
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.
7. Solution for 6 is what percent of 612
100%/x%=612/6
(100/x)*x=(612/6)*x - we multiply both sides of the equation by x
100=102*x - we divide both sides of the equation by (102) to get x
100/102=x
0.=x
x=0.
now we have:
6 is 0.% of 612
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# 11- 3D GEOMETRY
Here we are going through the ncert solutions for class 12 maths chapter 11– 3D geometry . so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily.
# EXERCISE 11.1
#### Question 1:
If a line makes angles 90°, 135°, 45° with xy and z-axes respectively, find its direction cosines.
Let direction cosines of the line be lm, and n.
Therefore, the direction cosines of the line are
#### Question 2:
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Let the direction cosines of the line make an angle α with each of the coordinate axes.
l = cos αm = cos αn = cos α
Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are
#### Question 3:
If a line has the direction ratios −18, 12, −4, then what are its direction cosines?
If a line has direction ratios of −18, 12, and −4, then its direction cosines are
Thus, the direction cosines are.
#### Question 4:
Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.
The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7).
It is known that the direction ratios of line joining the points, (x1y1z1) and (x2y2z2), are given by, x2 − x1y2 − y1, and z2 − z1.
The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3.
The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6.
It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are proportional.
Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B, and C are collinear.
#### Question 5:
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2)
The vertices of ΔABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).
The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.
Therefore, the direction cosines of AB are
The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4.
Therefore, the direction cosines of BC are
−217√−317√−217√-217, -317, -217
The direction ratios of CA are 3−(−5), 5−(−5) and −4−(−2) i.e. 8, 10 and -2.
Therefore the direction cosines of CA are
8(8)2 + (10)2 + (−2)210(8)2 + (10)2 + (−2)2−2(8)2 + (10)2 + (−2)28242√10242√−2242√442√542√−142√882 + 102 + -22, 1082 + 102 + -22, -282 + 102 + -228242, 10242, -2242442, 542, -142
#### Question 1:
Show that the three lines with direction cosines
are mutually perpendicular.
Two lines with direction cosines, l1m1n1 and l2m2n2, are perpendicular to each other, if l1l2 + m1m2 + n1n2 = 0
(i) For the lines with direction cosines, and , we obtain
Therefore, the lines are perpendicular.
(ii) For the lines with direction cosines, and , we obtain
Therefore, the lines are perpendicular.
(iii) For the lines with direction cosines, and , we obtain
Therefore, the lines are perpendicular.
Thus, all the lines are mutually perpendicular.
#### Question 2:
Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6).
The direction ratios, a1b1c1, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4.
The direction ratios, a2b2c2, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4.
AB and CD will be perpendicular to each other, if a1a2 + b1b2c1c2 = 0
a1a2 + b1b2c1c2 = 2 × 3 + 5 × 2 + (− 4) × 4
= 6 + 10 − 16
= 0
Therefore, AB and CD are perpendicular to each other.
#### Question 3:
Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).
Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5).
The directions ratios, a1b1c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4.
The direction ratios, a2b2c2, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4.
AB will be parallel to CD, if
Thus, AB is parallel to CD.
#### Question 4:
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector.
It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is
It is known that the line which passes through point A and parallel to is given by is a constant.
This is the required equation of the line.
#### Question 5:
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector and is in the direction .
It is given that the line passes through the point with position vector
It is known that a line through a point with position vector and parallel to is given by the equation,
This is the required equation of the line in vector form.
Eliminating λ, we obtain the Cartesian form equation as
This is the required equation of the given line in Cartesian form.
#### Question 6:
Find the Cartesian equation of the line which passes through the point
(−2, 4, −5) and parallel to the line given by
It is given that the line passes through the point (−2, 4, −5) and is parallel to
The direction ratios of the line, are 3, 5, and 6.
The required line is parallel to
Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0
It is known that the equation of the line through the point (x1y1z1) and with direction ratios, abc, is given by
Therefore the equation of the required line is
#### Question 7:
The Cartesian equation of a line is . Write its vector form.
The Cartesian equation of the line is
The given line passes through the point (5, −4, 6). The position vector of this point is
Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of vector,
It is known that the line through position vector and in the direction of the vector is given by the equation,
This is the required equation of the given line in vector form.
#### Question 8:
Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
The required line passes through the origin. Therefore, its position vector is given by,
The direction ratios of the line through origin and (5, −2, 3) are
(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3
The line is parallel to the vector given by the equation,
The equation of the line in vector form through a point with position vector and parallel to is,
The equation of the line through the point (x1y1z1) and direction ratios abc is given by,
Therefore, the equation of the required line in the Cartesian form is
Question 9:
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.
Since PQ passes through P (3, −2, −5), its position vector is given by,
The direction ratios of PQ are given by,
(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11
The equation of the vector in the direction of PQ is
The equation of PQ in vector form is given by,
The equation of PQ in Cartesian form is
i.e.,
#### Question 10:
Find the angle between the following pairs of lines:
(i)
(ii) and
(i) Let Q be the angle between the given lines.
The angle between the given pairs of lines is given by,
The given lines are parallel to the vectors, and , respectively.
(ii) The given lines are parallel to the vectors, and , respectively.
#### Question 11:
Find the angle between the following pairs of lines:
(i)
(ii)
• Let and be the vectors parallel to the pair of lines, respectively.
and
The angle, Q, between the given pair of lines is given by the relation,
(ii) Let be the vectors parallel to the given pair of lines, and , respectively.
If Q is the angle between the given pair of lines, then
#### Question 12:
Find the values of p so the line and
are at right angles.
The given equations can be written in the standard form as
and
The direction ratios of the lines are −3,, 2 and respectively.
Two lines with direction ratios, a1b1c1 and a2b2c2, are perpendicular to each other, if a1a2 + b1 b2 + c1c2 = 0
Thus, the value of p is .
#### Question 13:
Show that the lines and are perpendicular to each other.
The equations of the given lines areand
The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.
Two lines with direction ratios, a1b1c1 and a2b2c2, are perpendicular to each other, if a1a2 + b1 b2 + c1c2 = 0
7 × 1 + (−5) × 2 + 1 × 3
= 7 − 10 + 3
= 0
Therefore, the given lines are perpendicular to each other.
#### Question 14:
Find the shortest distance between the lines
The equations of the given lines are
It is known that the shortest distance between the lines, and , is given by,
d = ∣∣∣∣(b1→×b2).(a2→−a1)∣∣∣b1→×b2→∣∣∣∣∣∣∣d = b1→×b2→.a2→-a1→b1→×b2→
Comparing the given equations, we obtain
Substituting all the values in equation (1), we obtain
Therefore, the shortest distance between the two lines is units.
#### Question 15:
Find the shortest distance between the lines and
The given lines are and
It is known that the shortest distance between the two lines, is given by,
Comparing the given equations, we obtain
Substituting all the values in equation (1), we obtain
Since distance is always non-negative, the distance between the given lines is units.
#### Question 16:
Find the shortest distance between the lines whose vector equations are
The given lines are and
It is known that the shortest distance between the lines, and , is given by,
Comparing the given equations with and , we obtain
Substituting all the values in equation (1), we obtain
Therefore, the shortest distance between the two given lines is units.
#### Question 17:
Find the shortest distance between the lines whose vector equations are
The given lines are
r→=(s+1)iˆ+(2s−1)jˆ−(2s+1)kˆ⇒r→=(iˆ−jˆ−kˆ)+s(iˆ+2jˆ−2kˆ) ...(2)r→=(s+1)i^+(2s-1)j^-(2s+1)k^r→=(i^-j^-k^)+s(i^+2j^-2k^) ...(2)
It is known that the shortest distance between the lines, and , is given by,
For the given equations,
Substituting all the values in equation (3), we obtain
Therefore, the shortest distance between the lines isunits.
#### Question 1:
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a)z = 2 (b)
(c)
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Chapter 3 Class 10 Pair of Linear Equations in Two Variables
Class 10
Important Questions for Exam - Class 10
### Transcript
Example 8 The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs.2000 per month, find their monthly incomes. Given that Ratio of income of two persons is 9 : 7 Let Income of 1st person be 9x & Income for 2nd person be 7x Similarly, Ratio of expenditures of two persons is 4 : 3 Let Expenditure of 1st person be 4y & Expenditure of 2nd person be 3y Now, Both save Rs 2000 per month We know that, Income – Expenditure = Savings For 1st person 9x – 4y = 2000 Similarly for 2nd person 7x – 3y = 2000 Now, our equations are 9x – 4y = 2000 …(1) 7x – 3y = 2000 …(2) We will solve them by elimination We know that 12 is the multiple of 3 and 4 Multiplying (1) with 3 3 (9x – 4y) = 3(2000) 27x – 12y = 6000 Multiplying (2) with 4 4(7x – 3y) = 4(2000) 28x – 12y = 8000 Now we use elimination with equation (3) & (4) –x = –2000 x = 2000 Putting value of x in (1) 9x – 4y = 2000 9(2000) – 4y = 2000 18000 – 4y = 2000 –4y = 2000 – 18000 –4y = –16000 y = 16000/4 y = 4000 Hence, x = 2000, y = 4000 is the solution of the equation. Hence, Monthly income of 1st person = 9x = 9 × 2000 = Rs 18,000 Monthly income of 2nd person = 7x = 7 × 2000 = Rs 14,000
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## Algebra: A Combined Approach (4th Edition)
Published by Pearson
# Chapter 5 - Section 5.4 - Adding and Subtracting Polynomials - Exercise Set: 12
#### Answer
$-\dfrac{1}{7}n^{2}+\dfrac{5}{12}m-\dfrac{7}{20}$
#### Work Step by Step
$\Big(-\dfrac{4}{7}n^{2}+\dfrac{5}{6}m-\dfrac{1}{20}\Big)+\Big(\dfrac{3}{7}n^{2}-\dfrac{5}{12}m-\dfrac{3}{10}\Big)$ Remove both parentheses: $-\dfrac{4}{7}n^{2}+\dfrac{5}{6}m-\dfrac{1}{20}+\dfrac{3}{7}n^{2}-\dfrac{5}{12}m-\dfrac{3}{10}=...$ Simplify by combining like terms: $...=\Big(\dfrac{3}{7}-\dfrac{4}{7}\Big)n^{2}+\Big(\dfrac{5}{6}-\dfrac{5}{12}\Big)m-\Big(\dfrac{1}{20}+\dfrac{3}{10}\Big)=...$ $...=-\dfrac{1}{7}n^{2}+\dfrac{60-30}{72}m-\dfrac{10+60}{200}=...$ $...=-\dfrac{1}{7}n^{2}+\dfrac{30}{72}m-\dfrac{70}{200}=...$ $...=-\dfrac{1}{7}n^{2}+\dfrac{5}{12}m-\dfrac{7}{20}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Boats and Streams
1. Downstream/Upstream:
In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream.
2. If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then:
Speed downstream = (u + v) km/hr.
Speed upstream = (u - v) km/hr.
3. If the speed downstream is a km/hr and the speed upstream is b km/hr, then:
Speed in still water = 1 (a + b) km/hr. 2
Rate of stream = 1 (a - b) km/hr. 2
## TIP on cracking Aptitude Questions on Boats and Streams
Tip: Understand that the water current can help (when downstream) or hinder (when upstream) the boat
If Sb, Sc, Su and Sd be the speeds of the boat in still water, speed of the current in the stream, speed of the boat while running upstream and speed of the boat while running downstream respectively, then
Sb + Sc = Sd and Sb – Sc = Su
Question: A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?
Solution:
Let the distance travelled be ‘d’ meters
Sb – Sc = d/ 528 meters/ min [When travelling upstream]
Sb + Sc = d/240 meters/ min [When travelling downstream]
Solving the equations above, we get
Sb: Ss = 8: 3
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# A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 50% of this population prefers the color green. If 14 buyers are randomly selected, what is the probability that exactly 12 buyers would prefer green
The probability that exactly 12 buyers would prefer green
=0.00555
Step-by-step explanation:
We are given that
p=50%=50/100=0.50
n=14
We have to find the probability that exactly 12 buyers would prefer green.
q=1-p
q=1-0.50=0.50
Using binomial distribution formula
Hence, the probability that exactly 12 buyers would prefer green
=0.00555
## Related Questions
For this case we have the following equation:
y = 4 (x-2) ^ 2-1
Rewriting we have:
y = 4 (x ^ 2-4x + 4) ^ 2-1
Multiplying the common factor 2 we have:
y = 4x ^ 2-16x + 16-1
Adding the constant term we have:
y = 4x ^ 2-16x + 15
y = 4x ^ 2-16x + 15
option D
According to a poll, 76% of California adults (385 out of 506 surveyed) feel that education is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education is one of the top issues facing California. Find the error bound. (Round your answer to three decimal places.)
The error bound is 3.125%.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence interval , we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
For this problem, we have that:
A sample of 506 California adults.. This means that .
76% of California adults (385 out of 506 surveyed) feel that education is one of the top issues facing California. This means that
We wish to construct a 90% confidence interval
So , z is the value of Z that has a pvalue of , so .
The lower limit of this interval is:
The upper limit of this interval is:
The error bound of the confidence interval is the division by 2 of the subtraction of the upper limit by the lower limit. So:
The error bound is 3.125%.
The smaller of two consecutive numbers is x + 3. Find the sum of the two numbers.
x+3. is the first num
x +4 is the second num
the sum is 2x +7
2x + 7
Difference is 1
Step-by-step explanation:
Consecutive numbers are numbers that increase with a steady difference.
For example; 1, 2, 3 all increase with a difference of 1 (2 = 1 + 1; 3 = 2 + 1) and so on.
If the smaller of two consecutive numbers is x + 3, that means the other number which should be bigger is x + 4
Therefore the sum of both numbers is (x + 3) + (x + 4) = 2x + 7.
Their difference is (x + 4) - (x + 3) = 1
And subtracting their sum from x + 8 gives (x + 8) - (2x + 7) = x + 8 - 2x - 7 = -x + 1.
a discount voucher offering 15% off is used to pay a bill. after using the voucher the bill is reduced to 36.72. how much was the bill before applying the voucher discount
We can use the is/of=p/100 method for this problem. Since it's 15% off, this would mean that the bill is 85% of what it was initially. Plug the values into the is/of=p/100 formula.
36.72/x = 85/100
Solve for x.
x=36.72/0.85
x=43.2
So, the bill was $43.20 before applying the voucher discount of 15% off. The answer is 244.80 You can calculate this by taking 36.72 and dividing it by 0.15 (15%). You would do this because it is the opposite of multiplying an original number by 0.15 (15%) to get 36.72. Hope It helps! Please vote Brainliest! In a survey of 269 college students, it is found that69 like brussels sprouts, 90 like broccoli, 59 like cauliflower, 28 like both Brussels sprouts and broccoli, 20 like both Brussels sprouts and cauliflower, 24 like both broccoli and cauliflower, and 10 of the students like all three vegetables. a) How many of the 269 college students do not like any of these three vegetables? b) How many like broccoli only? c) How many like broccoli AND cauliflower but not Brussels sprouts? d) How many like neither Brussels sprouts nor cauliflower? ### Answers Answer: a) 83, b) 28, c) 14, d) 28. Step-by-step explanation: Since we have given that n(B) = 69 n(Br)=90 n(C)=59 n(B∩Br)=28 n(B∩C)=20 n(Br∩C)=24 n(B∩Br∩C)=10 a) How many of the 269 college students do not like any of these three vegetables? n(B∪Br∪C)=n(B)+n(Br)+n(C)-n(B∩Br)-n(B∩C)-n(Br∩C)+n(B∩Br∩C) n(B∪Br∪C)= So, n(B∪Br∪C)'=269-n(B∪Br∪C)=269-156=83 b) How many like broccoli only? n(only Br)=n(Br) -(n(B∩Br)+n(Br∩C)+n(B∩Br∩C)) n(only Br)= c) How many like broccoli AND cauliflower but not Brussels sprouts? n(Br∩C-B)=n(Br∩C)-n(B∩Br∩C) n(Br∩C-B)= d) How many like neither Brussels sprouts nor cauliflower? n(B'∪C')=n(only Br)= 28 Hence, a) 83, b) 28, c) 14, d) 28. Jimmy purchased a government bond which has maturity value of$2500 after 9 months at 11 % simple interest. How much should he pay for this bond?
Answer: he should pay $23095 for this bond. Step-by-step explanation: We would apply the formula for determining simple interest which is expressed as I = PRT/100 Where I represents interest paid on the bond purchased. P represents the principal or amount of bond purchased. R represents interest rate T represents the duration of the bond in years. From the given information, R = 11% T = 9 months = 9/12 = 0.75 I = 25000 - P Therefore 25000 - P = (P × 11 × 0.75)/100 25000 - P = 8.25P/100 = 0.0825P P + 0.0825P = 25000 1.0825P = 25000 P = 25000/1.0825 P =$23095
Answer:$2706.25 Step-by-step explanation: Principal(t)=$2500
Time(t)=9months=9/12=0.75year
Rate(r)=11%
Simple interest(si)=?
Si=(pxrxt)/100
Si=(2500x11x0.75)/100
Si=20625/100
Si=$206.25 Total amount=p + si Total amount=2500+206.25 Total amount=$2706.25
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# For which values of $p$ are events $A$ and $B$ independent?
Flip a coin three times. Assume the probability of tails is $p$ and that successive flips are independent. Let $A$ be the event that we have exactly one tails among the first two coin flips and $B$ the event that we have exactly one tails among the last two coin flips. For which values of $p$ are events $A$ and $B$ independent?
I know that $P(AB)=(1/4)=(1/2)*(1/2)=P(A)P(B)$ when $p=(1/2)$ for heads and tails. Does changing the value of $p$ make a set of unequally likely outcomes, leading to the fact you can't just count the number of events?
Given $p$ we compute: $$P(A)=2p(1-p)=P(B)$$
Now, let's compute $P(A|B)$:
Given $B$ we know that the last two tosses were $HT$ or $TH$, with equal probability. If they were $HT$ then $A$ would require that the first toss was a $T$, a probability $p$ event. If they were $TH$ then $A$ would require that the first toss was an $H$, a probability $1-p$ event. Thus $$P(A|B)=\frac 12\,p+\frac 12 (1-p)= \frac 12$$.
Independence requires that $P(A|B)=P(A)$ and it is easily seen that this entails $p=\frac 12$.
• So you are just generalizing the events A and B for any given p, then simply using the definition of independence, thanks. Commented Sep 30, 2015 at 23:42
By definition, $A$ and $B$ are independent when $P(A|B)=A$ and $P(B|A)=B$.
There are $2\cdot2=4$ ways we have exactly one tails among the first two coin flips, out of a total $\frac18$ ways to flip the coins, which gives us $A=\frac48=\frac12$. By symmetry, we know that $B=\frac12$ as well. Thus, in order for $P(A|B)=A$ and $P(B|A)=B$, $p$ must be equal to $\boxed{\frac12}$.
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# How do you simplify (2v)^2*2v^2 and write it using only positive exponents?
Feb 8, 2017
See the entire simplification process below:
#### Explanation:
First, we will use these two rules for exponents to simplify the term on the left of this expression:
$a = {a}^{\textcolor{red}{1}}$
${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$
${\left(2 v\right)}^{2} \cdot 2 {v}^{2} \to {\left({2}^{\textcolor{red}{1}} {v}^{\textcolor{red}{1}}\right)}^{\textcolor{b l u e}{2}} \cdot 2 {v}^{2} \to {2}^{\textcolor{red}{1} \times \textcolor{b l u e}{2}} {v}^{\textcolor{red}{1} \times \textcolor{b l u e}{2}} \cdot 2 {v}^{2} \to {2}^{2} {v}^{2} \cdot 2 {v}^{2}$
We can now use these two rule for exponents to complete the simplification:
$a = {a}^{\textcolor{red}{1}}$
${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
${2}^{\textcolor{red}{2}} {v}^{\textcolor{red}{2}} \cdot {2}^{\textcolor{b l u e}{1}} {v}^{\textcolor{b l u e}{2}} \to {2}^{\textcolor{red}{2} + \textcolor{b l u e}{1}} {v}^{\textcolor{red}{2} + \textcolor{b l u e}{2}} \to {2}^{3} {v}^{4} \to 8 {v}^{4}$
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# Week 13
## Topics
This is the second of three weeks devoted to elementary probability and statistics. This week, we will cover the second half of chapter 8.
• Section 8-3 : Conditional Probability, Intersection, and Independence
• Section 8-4 : Bayes' Formula
• Section 8-5 : Random Variable, Probability Distribution, and Expected Value
## Overview of the Sections
Section 8-3 deals with conditional probability. The conditional probability P(A|B) (pronounced "P of A given B") is the probability that event A will happen if it is known that event B will happen. The conditional probability may or may not be the same as P(A).
Example: If you roll one die, the probability that you will get an even number is 1/2. If you know that the number is greater than 3, the probability of rolling an even number changes to 2/3 (there are three possible outcomes left [4,5, and 6], and two of them are even). This is the probability of rolling an even number given that the number is greater than 3.
You can calculate a conditional probability either with the Conditional Probobility formula (1) in Section 8.3, or by drawing a tree diagram or Venn diagram and reading off the probabilities.
Two events A and B are independent if P(A|B) is the same as P(A). This also implies that P(B|A) is the same as P(B). Whether B happens or not has no influence on A, and vice versa.
Section 8-4 deals with Bayes' formula. This formula lets you go back and forth between P(A|B) and P(B|A). The book explains the formula, but doesn't do much with it. By the time you have drawn a tree diagram to go along with a problem, you can read off what you need without using the formula.
Section 8-5 is very important if you plan to take a statistics course later. It introduces some basic concepts you will see over and over again later.
A random variable is a mapping that assigns a number to each elementary outcome in a sample space. This produces a new sample space consisting of numbers, with a probability distribution inherited from the original. The book uses a capital P for the probability distribution on the original space, and a small p for the probability distribution on the new space. The probability of each outcome in the new sample space is obtained by adding up the probabilities of all outcomes in the original sample space that get mapped into that outcome.
Example: You roll two dice. The original sample space has 36 elements (1,1),...,(6,6). Let X be the sum of the two numbers. That is a random variable. The new sample space consists of the numbers from 2 to 12. The original outcomes (1,4), (2,3), (3,2) and (4,1) are the ones that get mapped into 5, so p(X=5) = P(1,4) + P(2,3) + P(3,2) + P(4,1) = 4/36.
Example: Let Y be the larger of the two numbers you rolled. That is a different random variable on the same original sample space. This one has a new sample space of {1,...,6}, and p(Y=5) = P(1,5)+P(5,1)+ P(2,5)+P(5,2)+P(3,5)+P(5,3)+P(4,5)+P(5,4)+P(5,5)=9/36.
So what's the point? The point is that the new sample space consists of numbers, and with numbers you can do calculations. If you draw a card from a deck, your sample space consists of 52 playing cards. How would you calculate an average, for example? With random variables, your sample space consists of numbers, and now you can add them, average them, whatever. Doing calculations with random variables is what probabilists and statisticians do most of the time.
When you take a statistics course, that is mostly what you will be doing, too, but that is a later course. In Math 150 there is only one calculation we do with random variables: computing the expected value or expectation, written E(X). This is the average outcome. The formula is in the definition of Expected Value of a Random Variable in Section 8.5.
For example, if the experiment consists of playing a casino game, and the random variable X is the amount of money you win or lose, the expected value is the amount of money you win per play on the average. Most likely, this will be a negative amount, otherwise the casino would go broke.
## Assignments
Read the textbook and do the homework assignment HW 10
Last Updated: Wednesday, August 5, 2015
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# 2007 AMC 12A Problems/Problem 11
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The following problem is from both the 2007 AMC 12A #11 and 2007 AMC 10A #22, so both problems redirect to this page.
## Problem
A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$?
$\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43$
## Solution
A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \cdot 37k$, so $S$ must be divisible by $37\ \mathrm{(D)}$. To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{(D)}}$.
## Solution 2 (Using the provided sequence/Guessing)
Since a prime number will always divide $S$, the prime number must work on any sequence that follows the stated rule. Since the solution kindly provides us with the terms $247,475,756,824$,all we have to do is finish the sequence.
One way of finishing the sequence is by adding $568,682$ (before 824) to the sequence and adding them all together. After this, we just divide the result by each of the answer choices. Of the five, $37$ is the largest prime number that divides this sequence, so the answer is ${\boxed{(D)}}$. ~Banspeedrun
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# How to Write a Polynomial in Factored Form
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Writing a polynomial in factored form is something that you can do as X = 2X squared - 12X + 22. Write a polynomial in factored form with help from an experienced mathematics educator in this free video clip.
Part of the Video Series: Math Lessons
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## Video Transcript
Hello, my name is Walter Unglaub. And this is how to write a polynomial in factored form. So, if we're given a polynomial, in this case, a quadratic polynomial. We have F of X is equal to two-X squared minus 12-X plus 22. The first step in terms of factoring a polynomial, is to find a common factor that exists in all of the terms. In this case, we can easily identify that common factor as two. Since two goes evenly into itself, 12 and 22. So, what we do is we multiply our function, our polynomial by one. But one here is in terms of two divided by two. And on the left hand side, that simply becomes F of X times one. And on the right hand side, I leave the two in the numerator, untouched and I divide the coefficients of each of the terms in the polynomial by this denominator. So, I have two-X squared divided two minus 12-x divided two, plus 22 divided two. This simplifies because this first term, you just get one in the coefficient. For the second term, you get a six and finally, for this constant right here, you get 11. So, we have two times X squared minus six-X plus 11. The next step is to complete the square. Since this is a quadratic relationship, we can identify the form X minus A squared plus B. And we want to rewrite this portion of our worked out example. So, to get A, what we do is, we take the coefficient of the linear term and we divide it by two. So, we have negative six divided two, which gives us negative three. So, we see that A is going to be equal to three in this case. So, we have F of X is equal to two times X minus three squared plus a constant B. In order to determine this constant B, we need to subtract from this constant, the square of this number. So, if we square negative three, we get nine. And so 11 minus nine is equal to two. And this is our B. So, finally we have a factored form for the original polynomial and it's written. Which is the common factor for all terms in the original expression, times X minus three squared plus two. Which could be rewritten as two-X minus three squared plus four. My name is Walter Unglaub. And this is how to write a polynomial in a factored form.
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# Difference between revisions of "009A Sample Midterm 2, Problem 2"
The function ${\displaystyle f(x)=3x^{7}-8x+2}$ is a polynomial and therefore continuous everywhere.
(a) State the Intermediate Value Theorem.
(b) Use the Intermediate Value Theorem to show that ${\displaystyle f(x)}$ has a zero in the interval ${\displaystyle [0,1].}$
Foundations:
What is a zero of the function ${\displaystyle f(x)?}$
A zero is a value ${\displaystyle c}$ such that ${\displaystyle f(c)=0.}$
Solution:
(a)
Intermediate Value Theorem
If ${\displaystyle f(x)}$ is continuous on a closed interval ${\displaystyle [a,b]}$
and ${\displaystyle c}$ is any number between ${\displaystyle f(a)}$ and ${\displaystyle f(b),}$
then there is at least one number ${\displaystyle x}$ in the closed interval such that ${\displaystyle f(x)=c.}$
(b)
Step 1:
First, ${\displaystyle f(x)}$ is continuous on the interval ${\displaystyle [0,1]}$ since ${\displaystyle f(x)}$ is continuous everywhere.
Also,
${\displaystyle f(0)=2}$
and
${\displaystyle f(1)=3-8+2=-3.}$.
Step 2:
Since ${\displaystyle y=0}$ is between ${\displaystyle f(0)=2}$ and ${\displaystyle f(1)=-3,}$
the Intermediate Value Theorem tells us that there is at least one number ${\displaystyle x}$
such that ${\displaystyle f(x)=0.}$
This means that ${\displaystyle f(x)}$ has a zero in the interval ${\displaystyle [0,1].}$
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# Chain Rule Integration – Definition and Examples
In this article, we’ll dive deep into the world of ‘Chain Rule Integration,’ illuminating its importance, mechanics, and the various techniques that hinge on its use.
Whether you’re a calculus newbie or an established mathematician looking for a refresher, this comprehensive guide aims to demystify the complexities and equip you with the essential tools to effectively wield the chain rule in your mathematical arsenal.
## Definition of Chain Rule Integration
In calculus, the chain rule is predominantly mentioned in the context of differentiation. However, the chain rule can also be applicable to integration via the method known as u-substitution or the substitution method, which is essentially the reverse process of the chain rule in differentiation.
In a nutshell, the chain rule integration, or the substitution method, allows us to transform a complicated integral into a simpler one by substituting a part of the integral (often the inner function of a composite function) with a new variable. This rule is usually expressed as follows:
If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then:
∫f(g(x)) * g'(x) dx = ∫f(u) du
Where g'(x) is the derivative of g with respect to x.
By using the substitution method, we can integrate complex expressions that would be difficult or impossible to integrate directly. It is particularly useful when dealing with composite functions, where one function is nested within another.
Figure-1.
## Properties
The method of integration by substitution, which is essentially the application of the chain rule Integration, relies on the following properties:
### Function Substitution
The central idea behind the chain rule for integration is to substitute a part of the integral (usually the inner function of a composite function) with a new variable. This technique simplifies complex integrals and makes it possible to integrate functions that might otherwise be difficult or impossible to handle.
For example, if we have a function of the form f(g(x)) where g(x) = u, then the integral becomes ∫f(u) * g'(x) dx.
### The Differential dx
In the substitution method, we replace a part of the function and its differential with the new variable and its differential. So if u = g(x), then the differential du is equal to g'(x) dx. This replacement allows us to express the original integral entirely in terms of the new variable.
### Inverse Substitution
After performing the integration in terms of the new variable, we substitute the original function back into the result to express it in terms of the original variable. This is because the original integral was in terms of x, so the result must also be in terms of x.
### Definite Integrals
For definite integrals, we must also change the limits of integration. If the original limits of integration were a and b in terms of x, then the new limits will be g(a) and g(b) in terms of u.
### Nested Functions
The chain rule integration is especially useful when dealing with nested or composite functions. This is where we have a function within a function, like sin(x^2) or (e^x)^2. These are usually prime candidates for the substitution method.
### Integration Techniques
Sometimes, the chain rule integration is combined with other integration techniques, like integration by parts or partial fractions. This can make it possible to solve even more complex integrals.
### Choice of Substitution
The effectiveness of the substitution method often depends on the choice of substitution. It’s often an art as much as a science, and with practice, it becomes easier to identify the best substitution to simplify an integral.
These properties together form the basis of the chain rule integration, a powerful tool for solving a wide range of integrals in calculus.
## Ralevent Formulas
In the context of chain rule integration or integration by substitution, there are a few key formulas and principles to keep in mind:
### Basic Formula
The fundamental concept of chain rule integration, or integration by substitution, is represented by the formula:
∫f(g(x)) * g'(x) dx = ∫f(u) du
Where u = g(x). The right-hand side of the formula shows the simplified integral after the substitution.
### Differential Substitution
An important aspect of substitution is translating the dx term to the new variable. If u = g(x), then the differential of u (du) is calculated as:
du/dx = g'(x)
Which can be rearranged as:
du = g'(x) dx
This du replaces the g'(x) dx in the integral, which assists in transforming the integral into terms of u.
### Definite Integrals
In the case of definite integrals, when the limits of the original integral are a and b (with respect to x), after the substitution, the new limits of integration are u(a) and u(b), where u = g(x). So the definite integral becomes:
∫f(g(x)) * g'(x) dx; from a to b = ∫f(u) du; from g(a) to g(b)
### Inverse Substitution
After performing the integration, we typically need to substitute back to express the answer in terms of the original variable. If the integral has been calculated to be F(u), where F'(u) = f(u), then the integral in terms of x is given by:
∫f(g(x)) * g'(x) dx = F(u) + C = F(g(x)) + C
Where C is the constant of integration. The result is in the same variable as the original problem.
## Computation Process of Chain Rule Integration
The process of evaluating or computing the chain rule integration, commonly known as u-substitution, is typically as follows:
### Choose Your Substitution (u)
Identify a function within the integral that will simplify the integral when substituted out. Often, this will be a function within another function or part of a function whose derivative is also in the integral.
### Compute du/dx and dx
Differentiate the function you chose for u with respect to x to get du/dx. Then, solve for dx by rearranging the equation: dx = du / (du/dx).
### Make the Substitution
Substitute u into the function and dx into the integral.
### Solve the Integral
Now that you have a simpler integral, solve it as you would any other integral. This may require additional techniques, such as integration by parts or trigonometric substitution, for more complex integrals.
### Back-Substitute
Once the integral is solved, substitute the original function back in for u to express the solution in terms of the original variable.
### Include Constant of Integration
Don’t forget to include the constant of integration, typically denoted as “C,” at the end of your solution. This accounts for the fact that the derivative of a constant is zero and thus doesn’t show up in the original integral.
### For Definite Integrals
If you’re solving a definite integral, you’ll need to change the limits of integration to match your u-substitution. When you substitute back in step 5, replace u with its original expression in terms of x, but leave the limits of integration in their substituted form. Then, simply evaluate the antiderivative at the upper limit and subtract the antiderivative evaluated at the lower limit.
This method can dramatically simplify integrals, especially when dealing with composite functions or functions that contain their own derivatives. Practice and familiarity with different types of integrals can help you quickly identify the best function to choose for u in the substitution method.
## Exercise
### Example 1
Integrate ∫x * sin() dx
Figure-2.
### Solution
Let u = . Then du/dx = 2x, so dx = du/(2x).
Substitute u and dx into the integral:
∫(1/2) * sin(u) du = -(1/2) * cos(u) + C
= -(1/2) * cos() + C
### Example 2
Evaluate ∫(2x) / (1 + ) dx
### Solution
Let u = 1 + . Then du/dx = 2x, so dx = du/(2x).
Substitute u and dx into the integral:
∫1/u du = ln|u| + C
= ln|1 +| + C
### Example 3
evaluate ∫$e^(3x)$ dx
Figure-3.
### Solution
Let u = 3 * x. Then du/dx = 3, so dx = du/3.
Substitute u and dx into the integral:
∫(1/3) * $e^u$ du = (1/3) * $e^u$ + C
= (1/3) * $e^{(3x)}$ + C
Example 4:
Evaluate ∫ cos(√(x)) dx
### Solution
Let u = √(x), so x = u². Then dx/du = 2u, so dx = 2u du.
Substitute u and dx into the integral:
∫2u * cos(u) du
This is an integration by parts problem. If we let dv = cos(u) du, v = sin(u), and w = 2u, dw = 2 du. Then the integral becomes:
∫w dv = w * v – ∫v dw
∫w dv = 2u * sin(u) – ∫2sin(u) du
∫w dv = 2u * sin(u) + 2cos(u) + C
Substitute u back in for (x):
= 2√(x) * sin(√(x)) + 2cos(√(x)) + C
Example 5:
Evaluate * (x⁴ + 1) dx
### Solution
Let u = x⁴ + 1. Then du/dx = 4, so dx = du/(4).
Substitute u and dx into the integral:
∫(1/4) *(u) du = (1/4) * (2/3) * $u^{(3/2)}$ + C
= (1/6) * $(x⁴ + 1)^{(3/2)}$ + C
Example 6:
Evaluate ∫ln(x)/x dx
### Solution
Let u = ln(x). Then du/dx = 1/x, so dx = x du.
Substitute u and dx into the integral:
∫u du = (1/2) * u² + C
= (1/2) * (ln(x))² + C
Example 7:
Evaluate ∫sin³(x) cos(x) dx
Figure-4.
### Solution
Let u = sin(x). Then du/dx = cos(x), so dx = du/cos(x).
Substitute u and dx into the integral:
∫u³ du = (1/4) * u + C
= (1/4) * sin(x) + C
## Applications
The chain rule integration, or integration by substitution, is a fundamental concept in calculus with a variety of applications in many fields. Here are a few notable examples:
### Physics
In physics, chain rule integration is used extensively, especially in the fields of kinematics, quantum mechanics, and electromagnetism. It helps in solving complex integrals that describe physical phenomena like the movement of particles, the interaction of electromagnetic fields, or the behavior of waves.
### Engineering
Engineers frequently use calculus, including chain rule integration, to model and solve real-world problems. It can be used to determine the stress and strain distribution in materials, solve electrical circuit problems, calculate fluid flow in pipes, and much more.
### Economics and Finance
In economics and finance, integration plays a crucial role in solving problems related to cost, revenue, and profit functions. It can also be used in modeling economic growth, calculating the present value of cash flows, and optimizing resource allocation, among other applications.
### Biology
In biology and related fields, chain rule integration is used in modeling population dynamics (like the growth and decay of populations over time), chemical kinetics (like the rate of reactions), and the spread of diseases, to name a few.
### Computer Science
Chain rule integration is applied in computer graphics to model continuous change and to render curves and surfaces. It’s also used in machine learning algorithms, particularly in backpropagation in neural networks, which requires both the chain rule for differentiation and its integration counterpart.
### Statistics and Probability
In statistics and probability theory, chain rule integration is used to calculate the expected value, variance, and other statistical parameters. It’s also used in the formulation of continuous probability distributions.
All images were created with GeoGebra and MATLAB.
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# BASIC DC THEORY
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### Basic DC Circuit Calculations
Each type of DC circuit contains certain characteristics that determine the way its voltage and current behave. To begin analysis of the voltages and currents at each part of a circuit, an understanding of these characteristics is necessary.
4.5 Resistivity 5.1 Batteries 5.2 DC Generator 5.4 Rectifiers 5.5 Forward Bias 5.6 Reverse Bias 7.2 Polarities
### Series Resistance
The total resistance in a series circuit is equal to the sum of all the parts of that circuit, as shown in equation (2-3).
Example:
A series circuit has a 60Ω, a 100Ω, and a 150Ω resistor in series (Figure 1). What is the total resistance of the circuit?
Solution:
Figure 1: Resistance in a Series Circuit
The total voltage across a series circuit is equal to the sum of the voltages across each resistor in the circuit (Figure 2) as shown in equation (2-4).
Figure 2: Voltage Drops in a Series Circuit
Ohms law may now be applied to the entire series circuit or to individual component parts of the circuit. When used on individual component parts, the voltage across that part is equal to the current times the resistance of that part. For the circuit shown in Figure 3, the voltage can be determined as shown below.
Figure 3: Voltage Total in a Series Circuit
To find the total voltage across a series circuit, multiply the current by the total resistance as shown in equation (2-5).
Example 1: A series circuit has a 50Ω, a 75Ω, and a 100Ω resistor in series (Figure 4). Find the voltage necessary to produce a current of 0.5 amps.
Figure 4: Example 1 Series Circuit
Solution:
Step 1: Find circuit current. As we already know, current is the same throughout a series circuit, which is already given as 0.5 amps.
Step 2: Find RT.
Step 3: Find VT. Use Ohms law.
Example 2: A 120 V battery is connected in series with three resistors: 40Ω, 60Ω, and 100Ω (Figure 5). Find the voltage across each resistor.
Figure 5: Example 2 Series Circuit
Solution:
Step 1: Find total resistance.
Step 2: Find circuit current (I).
Step 3: Find the voltage across each component.
The voltages of V1, V2, and V3 in Example 2 are known as "voltage drops" or "IR drops." Their effect is to reduce the available voltage to be applied across the other circuit components. The sum of the voltage drops in any series circuit is always equal to the applied voltage. We can verify our answer in Example 2 by using equation (2-4).
### Parallel Currents
The sum of the currents flowing through each branch of a parallel circuit is equal to the total current flow in the circuit. Using Ohms Law, the branch current for a three branch circuit equals the applied voltage divided by the resistance as shown in equations (2-6), (2-7), and (2-8).
Example 1: Two resistors, each drawing 3A, and a third resistor, drawing 2A, are connected in parallel across a 115 volt source (Figure 6). What is total current?
Figure 6: Example 1 Parallel Circuit
Example 2: Two branches, R1 and R2, are across a 120 V power source. The total current flow is 30 A (Figure 7). Branch R1 takes 22 amps. What is the current flow in Branch R2?
Figure 7: Example 2 Parallel Circuit
Example 3: A parallel circuit consists of R1 = 15Ω, R2 = 20Ω and R3 = 10Ω, with an applied voltage of 120 V (Figure 8). What current will flow through each branch?
Figure 8: Example 3 Parallel Circuit
### Resistance in Parallel
Total resistance in a parallel circuit can be found by applying Ohms Law. Divide the voltage across the parallel resistance by the total line current as shown in equation (2-9).
Example: Find the total resistance of the circuit shown in Figure 8 if the line voltage is 120 V and total current is 26A.
The total load connected to a 120 V source is the same as the single "equivalent resistance" of 4.62Ω connected across the source (Figure 9). Equivalent resistance is the total resistance a combination of loads present to a circuit.
Figure 9 Equivalent Resistance in a Parallel Circuit
The total resistance in a parallel circuit can also be found by using the equation (2-10).
Example 1: Find the total resistance of a 4Ω, an 8Ω, and a 16Ωresistor in parallel (Figure 10).
Figure 10 Total Resistance in a Parallel Circuit
Solution:
Note: Whenever resistors are in parallel, the total resistance is always smaller than any single branch.
Example 2: Now add a fourth resistance of 4Ω in parallel to the circuit in Figure 11. What is the new total resistance of the circuit?
Solution:
### Simplified Formulas
Total resistance of equal resistors in a parallel circuit is equal to the resistance of one resistor divided by the number of resistors.
Example: Five lamps, each with a resistance of 40Ω, are connected in parallel. Find total resistance.
When any two resistors are unequal in a parallel circuit, it is easier to calculate RT by multiplying the two resistances and then dividing the product by the sum, as shown in equation (2-11). As shown in equation (2-11), this is valid when there are only two resistors in parallel.
Example: Find the total resistance of a parallel circuit which has one 12Ω and one 4Ω resistor.
In certain cases involving two resistors in parallel, it is useful to find an unknown resistor, Rx, to obtain a certain RT. To find the appropriate formula, we start with equation (2-10) and let the known resistor be R and the unknown resistor be Rx.
Example: What value of resistance must be added, in parallel, with an 8Ω resistor to provide a total resistance of 6Ω (Figure 11)?
Figure 11 Example Parallel Circuit
Solution:
### Voltage Divider
A voltage divider, or network, is used when it is necessary to obtain different values of voltage from a single energy source. A simple voltage divider is shown in Figure 12. In this circuit, 24 volts is applied to three resistors in series. The total resistance limits the current through the circuit to one ampere. Individual voltages are found as follows using equation (2-12).
Figure 12 Voltage Divider
### Current Division
Sometimes it is necessary to find the individual branch currents in a parallel circuit when only resistance and total current are known. When only two branches are involved, the current in one branch will be some fraction of IT. The resistance in each circuit can be used to divide the total current into fractional currents in each branch. This process is known as current division.
Note that the equation for each branch current has the opposite R in the numerator. This is because each branch current is inversely proportional to the branch resistance.
Example: Find branch current for I1 and I2 for the circuit shown in Figure 13.
Figure 13 Current Division Example Circuit
Solution:
Since I1 and IT were known, we could have also simply subtracted I1 from IT to find I2:
### DC Circuit Analysis
All of the rules governing DC circuits that have been discussed so far can now be applied to analyze complex DC circuits. To apply these rules effectively, loop equations, node equations, and equivalent resistances must be used.
### Loop Equations
As we have already learned, Kirchhoff's Laws provide a practical means to solve for unknowns in a circuit. Kirchhoff's current law states that at any junction point in a circuit, the current arriving is equal to the current leaving. In a series circuit, the current is the same at all points in that circuit. In parallel circuits, the total current is equal to the sum of the currents in each branch. Kirchhoff's voltage law states that the sum of all potential differences in a closed loop equals zero.
Using Kirchhoff's laws, it is possible to take a circuit with two loops and several power sources (Figure 14) and determine loop equations, solve loop currents, and solve individual element currents.
Figure 14 Example Circuit for Loop Equations
The first step is to draw an assumed direction of current flow (Figure 15). It does not matter whether the direction is correct. If it is wrong, the resulting value for current will be negative.
Figure 15 Assumed Direction of Current Flow
Second, mark the polarity of voltage across each component (Figure 16). It is necessary to choose a direction for current through the center leg, but it is not necessary to put in a new variable. It is simply I2 - I1.
Figure 16 Making Polarity
Third, apply Kirchhoff's voltage law to loops one and two by picking a point in each loop and writing a loop equation of the voltage drops around the loop; then set the equation equal to zero.
Figure 17 Applying Voltage Law to Loop 1
Figure 17 shows Loop 1.
From Point A to Point B, there is an increase in voltage of 8 volts. From Point C to Point D, there is an increase in voltage of 200 (I2 - I1). From Point D to Point E, there is a decrease in voltage of 10 volts. From Point E to Point A, there is a voltage decrease of 50I1 volts. The result in equation form is illustrated in equation (2-16).
Using the same procedure for Loop 2 of Figure 16, the resulting equation is shown in equation (2-18).
Fourth, solve equations (2-17) and (2-18) simultaneously. First, rearrange and combine like terms in the equation for Loop 1.
Divide both sides by two.
Rearrange and combine like terms in the Loop 2 equation.
Multiplying the Loop 1 equation by 3, and add it to the Loop 2 equation.
Solving for I1:
Solving for I2 using the Loop 1 equation:
The current flow through R1 (50Ω) is I1. The current flow through R2(100Ω) is I2, and through R3(200Ω) is I2 - I1:
Fifth, apply Ohms Law to obtain the voltage drops across Resistors R1, R2, and R3:
Sixth, check the calculations by applying Kirchhoff's Laws:
Check 1: Apply Kirchhoff's voltage law to the larger outer loop (Figure 18).
Figure 18 Applying Voltage Laws to Outer Loop
The sum of the voltage drops around the loop is essentially zero. (Not exactly zero due to rounding off.)
Therefore, the solution checks.
Check 2: Use Kirchhoff's current law at one of the junctions (Figure 19).
Figure 19 Applying Current Law to Junction
The sum of the currents out of the junction is:
The current into the junction is 345.8 ma. he current into the junction is equal to the current out of the junction. Therefore, the solution checks.
### Node Equations
Kirchhoff's current law, as previously stated, says that at any junction point in a circuit the current arriving is equal to the current leaving. Let us consider five currents entering and leaving a junction shown as P (Figure 20). This junction is also considered a node.
Assume that all currents entering the node are positive, and all currents that leave the node are negative. Therefore, I1, I3, and I4 are positive, and I2 and I5 are negative. Kirchhoff's Law also states that the sum of all the currents meeting at the node is zero. For Figure 20, Equation (2-19) represents this law mathematically.
Figure 20 Node Point
By solving node equations, we can calculate the unknown node voltages. To each node in a circuit we will assign a letter or number. In Figure 21, A, B, C, and N are nodes, and N and C are principal nodes. Principal nodes are those nodes with three or more connections. Node C will be our selected reference node. VAC is the voltage between Nodes A and C; VBC is the voltage between Nodes B and C; and VNC is the voltage between Nodes N and C. We have already determined that all node voltages have a reference node; therefore, we can substitute VA for VAC, VB for VBC, and VN for VNC.
Figure 21 Circuit for Node Analysis
Assume that loop currents I1 and I2 leave Node N, and that I3 enters Node N (Figure 21).
From Kirchhoffs current law:
Using Ohms Law and solving for the current through each resistor we obtain the following.
Substitute these equations for I1, I2, and I3 into Kirchhoffs current equation (2-20) and yields the following.
The circuit shown in Figure 22 can be solved for voltages and currents by using the node-voltage analysis.
Figure 22 Node-Voltage Analysis
First, assume direction of current flow shown. Mark nodes A, B, C, and N, and mark the Analysis polarity across each resistor.
Second, using Kirchhoff's current law at Node N, solve for VN.
Clear the fraction so that we have a common denominator:
Third, find all voltage drops and currents.
The negative value for V3 shows that the current flow through R3 is opposite that which was assumed and that the polarity across R3 is reversed.
The negative value for I3 shows that the current flow through R3 is opposite that which was assumed.
### Series-Parallel Circuit Analysis
When solving for voltage, current, and resistance in a series-parallel circuit, follow the rules that apply to the series part of the circuit, and follow the rules that apply to the parallel part of the circuit. Solving these circuits can be simplified by reducing the circuit to a single equivalent resistance circuit, and redrawing the circuit in simplified form. The circuit is then called an equivalent circuit (Figure 23).
Figure 23 Redrawn Circuit Example
The easiest way to solve these types of circuits is to do it in steps.
Step 1: Find the equivalent resistance of the parallel branch:
Step 2: Find the resistance of the equivalent series circuit:
Step 3: Find total current (IT):
Step 4: Find I2 and I3. The voltage across R1 and R2 is equal to the applied voltage (V), minus the voltage drop across R1.
Then, I2 and I3 are calculated.
### Y and Delta Network Calculation
Because of its shape, the network shown in Figure 24 is called a T (tee) or Y (wye) network. These are different names for the same network.
Figure 24 T or Y Network
The network shown in Figure 25 is called π (pi) or Δ(delta) because the shapes resemble Greek letters π and Ω. These are different names for the same network.
Figure 25 π or Δ Network
In order to analyze the circuits, it may be helpful to convert Y to Δ, or Δ to Y, to simplify the solution. The formulas that will be used for these conversions are derived from Kirchhoff's laws. The resistances in these networks are shown in a three-terminal network. After we use the conversion formulas, one network is equivalent to the other because they have equivalent resistances across any one pair of terminals (Figure 26).
Rule 1: The resistance of any branch of a Y network is equal to the product of the two adjacent sides of a Δ network, divided by the sum of the three Δ resistances.
Figure 26 Y - Δ Equivalent
Rule 2: The resistance of any side of a Δ network is equal to the sum of the Y network resistance, multiplied in pairs, divided by the opposite branch of the Y network.
Let us consider a bridge circuit (Figure 27).
Figure 27 Bridge Circuit
Find RT at terminals a and d.
Step 1: Convert the Y network (b-e, e-c, e-d) to the equivalent Δ network.
• Using Rule 2:
Step 2: Now, we can redraw the Y circuit as a Δ circuit and reconnect it to the original circuit (Figure 28):
Figure 28 Y- Δ Redrawn Circuit
Step 3: Reduce and simplify the circuit. Note that the 20Ω and 60 Ω branches are in parallel in Figure 28. Refer to Figures 28 and 29 for redrawing the circuit in each step below.
Figure 29 Steps to Simplify Redrawn Circuit
### DC Circuit Faults
Faults within a DC circuit will cause various effects, depending upon the nature of the fault. An understanding of the effects of these faults is necessary to fully understand DC circuit operation.
### Open Circuit Series
A circuit must have a "complete" path for current flow, that is, from the negative side to the positive side of a power source. A series circuit has only one path for current to flow. If this path is broken, no current flows, and the circuit becomes an open circuit (Figure 30).
Figure 30 Open Series Circuit
Circuits can be opened deliberately, such as by the use of a switch, or they may be opened by a defect, such as a broken wire or a burned-out resistor.
Since no current flows in an open series circuit, there are no voltage drops across the loads. No power is consumed by the loads, and total power consumed by the circuit is zero.
A parallel circuit has more than one path for current to flow. If one of the paths is opened, current will continue to flow as long as a complete path is provided by one or more of the remaining paths. It does not mean that you cannot stop current flow through a parallel circuit by opening it at one point; it means that the behavior of a parallel circuit depends on where the opening occurs (Figure 31).
Figure 31 Open Parallel Circuit-Total
If a parallel circuit is opened at a point where only a branch current flows, then only that branch is open, and current continues to flow in the rest of the circuit (Figure 32).
Figure 32 Open Parallel Circuit-Branch
### Short Circuit Series
In a DC circuit, the only current limit is the circuit resistance. If there is no resistance in a circuit, or if the resistance suddenly becomes zero, a very large current will flow. This condition of very low resistance and high current flow is known as a "short circuit" (Figure 33).
Figure 33 Shorted DC Circuit
A short circuit is said to exist if the circuit resistance is so low that current increases to a point where damage can occur to circuit components. With an increase in circuit current flow, the terminal voltage of the energy source will decrease. This occurs due to the internal resistance of the energy source causing an increased voltage drop within the energy source. The increased current flow resulting from a short circuit can damage power sources, burn insulation, and start fires. Fuses are provided in circuits to protect against short circuits.
### Short Circuit Parallel
When a parallel circuit becomes short circuited, the same effect occurs as in a series circuit: there is a sudden and very large increase in circuit current (Figure 34).
Figure 34 Shorted DC Circuit
Parallel circuits are more likely than series circuits to develop damaging short circuits. This is because each load is connected directly across the power source. If any of the load becomes shorted, the resistance between the power source terminals is practically zero. If a series load becomes shorted, the resistance of the other loads keeps the circuit resistance from dropping to zero.
### DC Circuit Terminology
Before operations with DC circuits can be studied, an understanding of the types of circuits and common circuit terminology associated with circuits is essential.
### Schematic Diagram
Schematic diagrams are the standard means by which we communicate information in electrical and electronics circuits. On schematic diagrams, the component parts are represented by graphic symbols. Because graphic symbols are small, it is possible to have diagrams in a compact form. The symbols and associated lines show how circuit components are connected and the relationship of those components with one another.
As an example, let us look at a schematic diagram of a two-transistor radio circuit (Figure 35). This diagram, from left to right, shows the components in the order they are used to convert radio waves into sound energy. By using this diagram it is possible to trace the operation of the circuit from beginning to end. Due to this important feature of schematic diagrams, they are widely used in construction, maintenance, and servicing of all types of electronic circuits.
Figure 35 Schematic Diagram
### One-Line Diagram
The one-line, or single-line, diagram shows the components of a circuit by means of single lines and the appropriate graphic symbols. One-line diagrams show two or more conductors that are connected between components in the actual circuit. The one-line diagram shows all pertinent information about the sequence of the circuit, but does not give as much detail as a schematic diagram. Normally, the one-line diagram is used to show highly complex systems without showing the actual physical connections between components and individual conductors.
As an example, Figure 36 shows a typical one-line diagram of an electrical substation.
Figure 36 One-Line Diagram
### Block Diagram
A block diagram is used to show the relationship between component groups, or stages in a circuit. In block form, it shows the path through a circuit from input to output (Figure 37). The blocks are drawn in the form of squares or rectangles connected by single lines with arrowheads at the terminal end, showing the direction of the signal path from input to output. Normally, the necessary information to describe the stages of components is contained in the blocks.
Figure 37 Block Diagram
### Wiring Diagram
A wiring diagram is a very simple way to show wiring connections in an easy-to-follow manner. These types of diagrams are normally found with home appliances and automobile electrical systems (Figure 38). Wiring diagrams show the component parts in pictorial form, and the components are identified by name. Most wiring diagrams also show the relative location of component parts and color coding of conductors or leads.
Figure 38 Wiring Diagram
### Resistivity
Resistivity is defined as the measure of the resistance a material imposes on current flow. The resistance of a given length of conductor depends upon the resistivity of that material, the length of the conductor, and the cross-sectional area of the conductor, according to Equation (2-1).
The resistivity ρ(rho) allows different materials to be compared for resistance, according to their nature, without regard to length or area. The higher the value of ρ, the higher the resistance.
Table 1 gives resistivity values for metals having the standard wire size of one foot in length and a cross-sectional area of 1 cm.
### Temperature Coefficient of Resistance
Temperature coefficient of resistance, α(alpha), is defined as the amount of change of the resistance of a material for a given change in temperature. A positive value of α indicates that R increases with temperature; a negative value of α indicates R decreases; and zero α indicates that R is constant. Typical values are listed in Table 2.
For a given material, α may vary with temperature; therefore, charts are often used to describe how resistance of a material varies with temperature.
An increase in resistance can be approximated from equation (2-2).
### Electric Circuit
Each electrical circuit has at least four basic parts: (1) a source of electromotive force, (2) conductors, (3) load or loads, and (4) some means of control. In Figure 39, the source of EMF is the battery; the conductors are wires which connect the various component parts; the resistor is the load; and a switch is used as the circuit control device.
Figure 39 Closed Circuit
A closed circuit (Figure 39) is an uninterrupted, or unbroken, path for current from the source (EMF), through the load, and back to the source.
An open circuit, or incomplete circuit, (Figure 40) exists if a break in the circuit occurs; this prevents a complete path for current flow.
Figure 40 Open Circuit
A short circuit is a circuit which offers very little resistance to current flow and can cause dangerously high current flow through a circuit (Figure 41). Short circuits are usually caused by an inadvertent connection between two points in a circuit which offers little or no resistance to current flow. Shorting resistor R in Figure 41 will probably cause the fuse to blow.
Figure 41 Short Circuit
### Series Circuit
A series circuit is a circuit where there is only one path for current flow. In a series circuit (Figure 42), the current will be the same throughout the circuit. This means that the current flow through R1 is the same as the current flow through R2 and R3.
Figure 42 Series Circuit
### Parallel Circuit
Parallel circuits are those circuits which have two or more components connected across the same voltage source (Figure 43). Resistors R1, R2, and R3 are in parallel with each other and the source. Each parallel path is a branch with its own individual current. When the current leaves the source V, part I1 of IT will flow through R1; part I2 will flow through R2; and part I3 will flow through R3. Current through each branch can be different; however, voltage throughout the circuit will be equal.
Figure 43 Series Circuit
### Equivalent Resistance
In a parallel circuit, the total resistance of the resistors in parallel is referred to as equivalent resistance. This can be described as the total circuit resistance as seen by the voltage source. In all cases, the equivalent resistance will be less than any of the individual parallel circuit resistors. Using Ohms Law, equivalent resistance (REQ) can be found by dividing the source voltage (V) by the total circuit current (IT), as shown in Figure 43.
### DC Sources
When most people think of DC, they usually think of batteries. In addition to batteries, however, there are other devices that produce DC which are frequently used in modern technology.
### Batteries
A battery consists of two or more chemical cells connected in series. The combination of materials within a battery is used for the purpose of converting chemical energy into electrical energy. To understand how a battery works, we must first discuss the chemical cell.
The chemical cell is composed of two electrodes made of different types of metal or metallic compounds which are immersed in an electrolyte solution. The chemical actions which result are complicated, and they vary with the type of material used in cell construction. Some knowledge of the basic action of a simple cell will be helpful in understanding the operation of a chemical cell in general.
In the cell, electrolyte ionizes to produce positive and negative ions (Figure 44, Part A). Simultaneously, chemical action causes the atoms within one of the electrodes to ionize.
Figure 44 Basic Chemical Battery
Due to this action, electrons are deposited on the electrode, and positive ions from the electrode pass into the electrolyte solution (Part B). This causes a negative charge on the electrode and leaves a positive charge in the area near the electrode (Part C).
The positive ions, which were produced by ionization of the electrolyte, are repelled to the other electrode. At this electrode, these ions will combine with the electrons. Because this action causes removal of electrons from the electrode, it becomes positively charged.
### DC Generator
A simple DC generator consists of an armature coil with a single turn of wire. The armature coil cuts across the magnetic field to produce a voltage output. As long as a complete path is present, current will flow through the circuit in the direction shown by the arrows in Figure 45. In this coil position, commutator segment 1 contacts with brush 1, while commutator segment 2 is in contact with brush 2.
Rotating the armature one-half turn in the clockwise direction causes the contacts between the commutator segments to be reversed. Now segment 1 is contacted by brush 2, and segment 2 is in contact with brush 1.
Figure 45 Basic DC Generator
Due to this commutator action, that side of the armature coil which is in contact with either of the brushes is always cutting the magnetic field in the same direction. Brushes 1 and 2 have a constant polarity, and pulsating DC is delivered to the load circuit.
### Thermocouples
A thermocouple is a device used to convert heat energy into a voltage output. The thermocouple consists of two different types of metal joined at a junction (Figure 46).
Figure 46 Production of a DC Voltage Using a Thermocouple
As the junction is heated, the electrons in one of the metals gain enough energy to become free electrons. The free electrons will then migrate across the junction and into the other metal. This displacement of electrons produces a voltage across the terminals of the thermocouple. The combinations used in the makeup of a thermocouple include: iron and constantan; copper and constantan; antimony and bismuth; and chromel and alumel.
Thermocouples are normally used to measure temperature. The voltage produced causes a current to flow through a meter, which is calibrated to indicate temperature.
### Rectifiers
Most electrical power generating stations produce alternating current. The major reason for generating AC is that it can be transferred over long distances with fewer losses than DC; however, many of the devices which are used today operate only, or more efficiently, with DC. For example, transistors, electron tubes, and certain electronic control devices require DC for operation. If we are to operate these devices from ordinary AC outlet receptacles, they must be equipped with rectifier units to convert AC to DC. In order to accomplish this conversion, we use diodes in rectifier circuits. The purpose of a rectifier circuit is to convert AC power to DC.
The most common type of solid state diode rectifier is made of silicon. The diode acts as a gate, which allows current to pass in one direction and blocks current in the other direction. The polarity of the applied voltage determines if the diode will conduct. The two polarities are known as forward bias and reverse bias.
### Forward Bias
A diode is forward biased when the positive terminal of a voltage source is connected to its anode, and the negative terminal is connected to the cathode (Figure 47A). The power source's positive side will tend to repel the holes in the p-type material toward the p-n junction by the negative side. A hole is a vacancy in the electron structure of a material. Holes behave as positive charges. As the holes and the electrons reach the p-n junction, some of them break through it (Figure 47B). Holes combine with electrons in the n-type material, and electrons combine with holes in the p-type material.
Figure 47 Forward-Biased Diode
When a hole combines with an electron, or an electron combines with a hole near the p-n junction, an electron from an electron-pair bond in the p-type material breaks its bond and enters the positive side of the source. Simultaneously, an electron from the negative side of the source enters the n-type material (Figure 47C). This produces a flow of electrons in the circuit.
### Reverse Bias
Reverse biasing occurs when the diodes anode is connected to the negative side of the source, and the cathode is connected to the positive side of the source (Figure 48A). Holes within the p-type material are attracted toward the negative terminal, and the electrons in the n-type material are attracted to the positive terminal (Figure 48B). This prevents the combination of electrons and holes near the p-n junction, and therefore causes a high resistance to current flow. This resistance prevents current flow through the circuit.
Figure 48 Reverse-Biased Diode
### Half-Wave Rectifier Circuit
When a diode is connected to a source of alternating voltage, it will be alternately forward-biased, and then reverse-biased, during each cycle of the AC sine-wave. When a single diode is used in a rectifier circuit, current will flow through the circuit only during one-half of the input voltage cycle (Figure 50). For this reason, this rectifier circuit is called a half-wave rectifier. The output of a half-wave rectifier circuit is pulsating DC.
Figure 50 Half-Wave Rectifier
### Full-Wave Rectifier Circuit
A full-wave rectifier circuit is a circuit that rectifies the entire cycle of the AC sine-wave. A basic full-wave rectifier uses two diodes. The action of these diodes during each half cycle is shown in Figure 51.
Figure 51 Full-Wave Rectifier
Another type of full-wave rectifier circuit is the full-wave bridge rectifier. This circuit utilizes four diodes. These diodes actions during each half cycle of the applied AC input voltage are shown in Figure 52. The output of this circuit then becomes a pulsating DC, with all of the waves of the input AC being transferred. The output looks identical to that obtained from a full-wave rectifier (Figure 52).
Figure 52 Full-Wave Rectifier
### Kirchhoff's Laws
Kirchhoff's two laws reveal a unique relationship between current, voltage, and resistance in electrical circuits that is vital to performing and understanding electrical circuit analysis.
In all of the circuits examined so far, Ohm's Law described the relationship between current, voltage, and resistance. These circuits have been relatively simple in nature. Many circuits are extremely complex and cannot be solved with Ohm's Law. These circuits have many power sources and branches which would make the use of Ohm's Law impractical or impossible.
Through experimentation in 1857 the German physicist Gustav Kirchhoff developed methods to solve complex circuits. Kirchhoff developed two conclusions, known today as Kirchhoff's Laws.
Law 1: The sum of the voltage drops around a closed loop is equal to the sum of the voltage sources of that loop (Kirchhoff's Voltage Law).
Law 2: The current arriving at any junction point in a circuit is equal to the current leaving that junction (Kirchhoff's Current Law).
Kirchhoff's two laws may seem obvious based on what we already know about circuit theory. Even though they may seem very simple, they are powerful tools in solving complex and difficult circuits.
Kirchhoff's laws can be related to conservation of energy and charge if we look at a circuit with one load and source. Since all of the power provided from the source is consumed by the load, energy and charge are conserved. Since voltage and current can be related to energy and charge, then Kirchhoff's laws are only restating the laws governing energy and charge conservation.
The mathematics involved becomes more difficult as the circuits become more complex. Therefore, the discussion here will be limited to solving only relatively simple circuits.
### Kirchhoff's Voltage Law
Kirchhoff's first law is also known as his "voltage law." The voltage law gives the relationship between the "voltage drops" around any closed loop in a circuit, and the voltage sources in that loop. The total of these two quantities is always equal. In equation form:
where the symbol Σ (the Greek letter sigma) means "the sum of."
Kirchhoff's voltage law can be applied only to closed loops (Figure 53). A closed loop must meet two conditions:
1. It must have one or more voltage sources.
2. It must have a complete path for current flow from any point, around the loop, and back to that point.
Figure 53 Closed Loop
You will remember that in a simple series circuit, the sum of the voltage drops around the circuit is equal to the applied voltage. Actually, this is Kirchhoff's voltage law applied to the simplest case, that is, where there is only one loop and one voltage source.
### Applying Kirchhoff's Voltage Law
For a simple series circuit, Kirchhoff's voltage law corresponds to Ohm's Law. To find the current in a circuit (Figure 54) by using Kirchhoff's voltage law, use equation (2-15).
Figure 54 Using Kirchhoff's Voltage Law to find Current with one Source
In the problem above, the direction of current flow was known before solving the problem. When there is more than one voltage source, the direction of current flow may or may not be known. In such a case, a direction of current flow must be assumed in the beginning of the problem. All the sources that would aid the current in the assumed direction of current flow are then positive, and all that would oppose current flow are negative. If the assumed direction is correct, the answer will be positive. The answer would be negative if the direction assumed was wrong. In any case, the correct magnitude will be attained.
For example, what is the current flow in Figure 55? Assume that the current is flowing in the direction shown.
Figure 55 Using Kirchhoff's Voltage Law to find Current with Multiple Battery Sources
The result is negative. The current is actually 0.5 ampere in the opposite direction to that of the assumed direction.
### Kirchhoff's Current Law
Kirchhoff's second law is called his current law and states: "At any junction point in a circuit, the current arriving is equal to the current leaving." Thus, if 15 amperes of current arrives at a junction that has two paths leading away from it, 15 amperes will divide among the two branches, but a total of 15 amperes must leave the junction. We are already familiar with Kirchhoff's current law from parallel circuits, that is, the sum of the branch currents is equal to the total current entering the branches, as well as the total current leaving the branches (Figure 56).
Figure 56 Illustration of Kirchhoff's Current Law
In equation form, Kirchhoff's current law may be expressed:
Normally, Kirchhoff's current law is not used by itself, but with the voltage law, in solving a problem.
Example: Find I2 in the circuit shown in Figure 57 using Kirchhoffs voltage and current laws.
Figure 57 Using the Current Law
Solution:
First, apply Kirchhoff's voltage law to both loops.
Since Kirchhoff's current law states Itotal = I1 + I
### Voltage Polarity and Current Direction
Before introducing the laws associated with complex DC circuit analysis, the importance of voltage polarity and current direction must be understood. This section will introduce the polarities and current direction associated with DC circuits.
### Conventional and Electron Flow
The direction of electron flow is from a point of negative potential to a point of positive potential. The direction of positive charges, or holes, is in the opposite direction of electron flow. This flow of positive charges is known as conventional flow. All of the electrical effects of electron flow from negative to positive, or from a high potential to a lower potential, are the same as those that would be created by flow of positive charges in the opposite direction; therefore, it is important to realize that both conventions are in use, and they are essentially equivalent. In this manual, the electron flow convention is used.
### Polarities
All voltages and currents have polarity as well as magnitude. In a series circuit, there is only one current, and its polarity is from the negative battery terminal through the rest of the circuit to the positive battery terminal. Voltage drops across loads also have polarities. The easiest way to find these polarities is to use the direction of the electron current as a basis. Then, where the electron current enters the load, the voltage is negative (Figure 58). This holds true regardless of the number or type of loads in the circuit. The drop across the load is opposite to that of the source. The voltage drops oppose the source voltage and reduce it for the other loads. This is because each load uses energy, leaving less energy for other loads.
Figure 58 Voltage Polarities
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Factors the 320 room the perform of confident and an unfavorable whole numbers that can be separated evenly into 320. An adverse factors the 320 room just components with a an adverse sign. There space 14 factors of 320 in all. Among the 14 factors, 320 itself is the best factor. In this lesson, we will certainly discuss an unfavorable and positive factors of 320, element pairs with the aid of exciting solved examples.
You are watching: What are the factors of 320
Factors of 320: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 160 and also 320Factors that -320: -1, -2, -4, -5, -8, -10, -16, -20, -32, -40, -64, -80, -160 and also -320Prime administrate of 320: 26 × 5
1 What space the factors of 320? 2 How to Calculate components of 320? 3 Factors the 320 by element Factorization 4 Factors of 320 in Pairs 5 Tips and Tricks 6 Challenging Questions 7 FAQs on components of 320
Factors the 320 room all the number which as soon as multiplied give the value 320. 320 is a composite number and also it has actually a full of 14 components i.e. 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 160 and also 320. Explore factors of various other numbers with illustrations and also interactive examples:
Step 2: find the 2 numbers whose product offers 320.As we recognize 320 is a product that 10 and also 32.10 is no a prime number. Hence, 10 = 2 × 532 can additionally be factored as 2 × 16.Thus we attain the determinants 1, 2, 4, 5, 8, 10, and 16 and also their pair components 20,32, 40, 64, 80,160, and 320 the make 320.Step 3: Thus the factors of 320 space 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 160 and 320.
Prime factorization is finding which element numbers multiply with each other to do the initial number.
Step 1: 320 can be factored together a product that 16 and also 20.Step 2: We watch the factors, whether each among them is element or not. 16 is no a prime number and also can be factored together 4 × 4. 4 is a composite number and also it have the right to be factored together 2 × 2. 20 is no a element number and can be factored as 2 × 10. 10 is a composite number and it deserve to be factored together 2 × 5. 2 and also 5 room prime, and also hence we can stop factoring.Step 3: as per the criterion, the element factorization the 320 can also be created as 320 = 26 × 5
Let"s check out the other possibilities because that finding the factors of 320 through the help of the following variable trees.
Method 1:
Method 2:
Pair components of the number 320 space the two totality numbers that multiply to acquire the original number. Pair factors might be either optimistic or an unfavorable but no a portion or a decimal number.
Positive factors of 320: (1, 320), (2, 160), (4, 80), (5, 64), (8, 40), (10, 32), (16, 20 ), (20, 16), (32, 10), (40, 8), (64, 5), (80, 4), (160, 2), (320, 1)Negative determinants of 320: (-1, -320), (-2, -160), (-4, -80), (-5, -64), (-8, -40), (-10, -32), (-16, -20 ), (-20, -16), (-32, -10), (-40, -8), (-64, -5), (-80, -4), (-160,-2), (-320, -1)
1 is the smallest factor of every number. Hence, 1 is a aspect of 320320 is an even number, 2 as one of their factors.320 is a number the ends in 0, for this reason 1, 2, 5, and 10 space its factors.
Suzy has to travel 320 mile to with her home from she uncle"s home. There room restaurants at every 12 miles on the way. How countless times can she protect against in in between to have actually her meals?How would certainly Isabella \$320 in 6 parts such that each component gets an same amount?
Example 1: Johnathan owns a bakery. He obtained an order of 320 cookie to be baked. If that bakes 5 cookies in one hour, just how long will it take because that him to bake every the cookies?
Solution:
Johnathan can bake 5 cookie in one hour. The quantity of time essential for Johnathan to bake 320 cookies will certainly be offered by 320 ÷ 5.This provides Johnathan can bake 320 cookie in 64 hours.
Example 2: Josephine"s teacher offered her 2 numbers, 320 and 203 to uncover the typical factors amongst both that them. Assist her uncover the typical factors in both the 320 and 203.
See more: P A Cell Is Like A School Analogy To School By Andi Hur, Cell Analogy Project Ideas
Solution:
Josephine knows factors of 320 space 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 160 and also 320. 203 have the right to be factored together 203 = 7 × 29. For this reason the determinants of 203 space 1, 7, 29 and 203.
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# Basic Geometry : How to find an angle of a line
## Example Questions
← Previous 1 3 4 5 6
### Example Question #1 : How To Find An Angle Of A Line
Examine the diagram. Which of these conditions does not prove that ?
and
Any of these statements can be used to prove that .
Explanation:
If and , then , since two lines parallel to the same line are parallel to each other.
If , then , since two same-side interior angles formed by transversal are supplementary.
If , then , since two alternate interior angles formed by transversal are congruent.
However, regardless of whether and are parallel; they are vertical angles, and by the Vertical Angles Theorem, they must be congruent.
### Example Question #1 : How To Find An Angle Of A Line
An isosceles triangle has an interior angle that measures . What are the measures of its other two angles?
This triangle cannot exist.
Explanation:
By the Isosceles Triangle Theorem, two interior angles must be congruent. However, since a triangle cannot have two obtuse interior angles, the two missing angles must be the ones that are congruent. Since the total angle measure of a triangle is , each of the missing angles measures .
### Example Question #3 : How To Find An Angle Of A Line
How would you classify the following angle?
Acute
Scalene
Obtuse
Straight
Right
Obtuse
Explanation:
Obtuse angles are greater than .
Scalene is a designation for triangles that have one angle greater than , but this figure is not a triangle.
Acute angles are less than , right angles are , and straight angles are .
Therefore this angle is obtuse.
### Example Question #3 : How To Find An Angle Of A Line
What is the measure of ?
Explanation:
When two parallel lines are crossed by a third line (called the transversal), the measure of the angles follows a specific pattern. The pairs of angles inside the two lines and on opposite sides are called alternate interior angles. Alternate interior angles, such as and , have the same degree measure. Therefore, the measure of is .
### Example Question #2 : How To Find An Angle Of A Line
Mark is training for cross country and comes across a new hill to run on. After Mark runs meters, he's at a height of meters. What is the hill's angle of depression when he's at an altitude of meters?
The same as the angle of inclination
Cannot be determined
Explanation:
Upon reading the question, we're left with this spatial image of Mark in our heads. After adding in the given information, the image becomes more like
The hill Mark is running on can be seen in terms of a right triangle. This problem quickly becomes one that is asking for a mystery angle given that the two legs of the triangle are given. In order to solve for the angle of depression, we have to call upon the principles of the tangent function. Tan, Sin, or Cos are normally used when there is an angle present and the goal is to calculate one of the sides of the triangle. In this case, the circumstances are reversed.
Remember back to "SOH CAH TOA." In this problem, no information is given about the hypotenuse and nor are we trying to calculate the hypotenuse. Therefore, we are left with "TOA." If we were to check, this would work out because the angle at Mark's feet has the information for the opposite side and adjacent side.
Because there's no angle given, we must use the principles behind the tan function while using a fraction composed of the given sides. This problem will be solved using arctan (sometimes denoted as ).
### Example Question #1 : Geometry
Two angles are supplementary and have a ratio of 1:4. What is the size of the smaller angle?
Explanation:
Since the angles are supplementary, their sum is 180 degrees. Because they are in a ratio of 1:4, the following expression could be written:
### Example Question #1 : Intersecting Lines And Angles
AB and CD are two parrellel lines intersected by line EF. If the measure of angle 1 is , what is the measure of angle 2?
Explanation:
The angles are equal. When two parallel lines are intersected by a transversal, the corresponding angles have the same measure.
### Example Question #3 : How To Find An Angle Of A Line
Lines A and B in the diagram below are parallel. The triangle at the bottom of the figure is an isosceles triangle.
What is the degree measure of angle ?
Explanation:
Since A and B are parallel, and the triangle is isosceles, we can use the supplementary rule for the two angles, and which will sum up to . Setting up an algebraic equation for this, we get . Solving for , we get . With this, we can get either (for the smaller angle) or (for the larger angle - must then use supplementary rule again for inner smaller angle). Either way, we find that the inner angles at the top are 80 degrees each. Since the sum of the angles within a triangle must equal 180, we can set up the equation as
degrees.
### Example Question #1 : Geometry
Figure not drawn to scale.
In the figure above, APB forms a straight line. If the measure of angle APC is eighty-one degrees larger than the measure of angle DPB, and the measures of angles CPD and DPB are equal, then what is the measure, in degrees, of angle CPB?
50
33
114
40
66
66
Explanation:
Let x equal the measure of angle DPB. Because the measure of angle APC is eighty-one degrees larger than the measure of DPB, we can represent this angle's measure as x + 81. Also, because the measure of angle CPD is equal to the measure of angle DPB, we can represent the measure of CPD as x.
Since APB is a straight line, the sum of the measures of angles DPB, APC, and CPD must all equal 180; therefore, we can write the following equation to find x:
x + (x + 81) + x = 180
Simplify by collecting the x terms.
3x + 81 = 180
Subtract 81 from both sides.
3x = 99
Divide by 3.
x = 33.
This means that the measures of angles DPB and CPD are both equal to 33 degrees. The original question asks us to find the measure of angle CPB, which is equal to the sum of the measures of angles DPB and CPD.
measure of CPB = 33 + 33 = 66.
### Example Question #2 : How To Find An Angle Of A Line
One-half of the measure of the supplement of angle ABC is equal to the twice the measure of angle ABC. What is the measure, in degrees, of the complement of angle ABC?
72
54
18
90
36
54
Explanation:
Let x equal the measure of angle ABC, let y equal the measure of the supplement of angle ABC, and let z equal the measure of the complement of angle ABC.
Because x and y are supplements, the sum of their measures must equal 180. In other words, x + y = 180.
We are told that one-half of the measure of the supplement is equal to twice the measure of ABC. We could write this equation as follows:
(1/2)y = 2x.
Because x + y = 180, we can solve for y in terms of x by subtracting x from both sides. In other words, y = 180 – x. Next, we can substitute this value into the equation (1/2)y = 2x and then solve for x.
(1/2)(180-x) = 2x.
Multiply both sides by 2 to get rid of the fraction.
(180 – x) = 4x.
180 = 5x.
Divide both sides by 5.
x = 36.
The measure of angle ABC is 36 degrees. However, the original question asks us to find the measure of the complement of ABC, which we denoted previously as z. Because the sum of the measure of an angle and the measure of its complement equals 90, we can write the following equation:
x + z = 90.
Now, we can substitute 36 as the value of x and then solve for z.
36 + z = 90.
Subtract 36 from both sides.
z = 54.
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10th Maths Chapter 2. Numbers And Sequences Exercise 2.2
10th Standard Maths Chapter 2 Exercise 2.2 Numbers And Sequences Guide Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here
1. For what values of natural number n, 4n can end with the digit 6?
Solution:
4n = (2 × 2)n = 2n × 2n
2 is a factor of 4n.
So, 4n is always even and end with 4 and 6.
When n is an even number say 2, 4, 6, 8 then 4n can end with the digit 6.
Example:
42 = 16
43 = 64
44 = 256
45 = 1,024
46 = 4,096
47 = 16,384
48 = 65, 536
49 = 262,144
2. If m, n are natural numbers, for what values of m, does 2n × 5m ends in 5?
2n is always even for any values of n.
[Example. 22 = 4, 23 = 8, 24 = 16 etc]
5m is always odd and it ends with 5.
[Example.
52 = 25,
53 = 125,
54 = 625 etc]
But 2n × 5m is always even and end in 0.
[Example. 23 × 53 = 8 × 125 = 1000
= 22 × 52
= 4 × 25
= 100]
∴ 2n × 5m cannot end with the digit 5 for any values of m.
3. Find the H.C.F. of 252525 and 363636.
Solution:
To find the H.C.F. of 252525 and 363636
Using Euclid’s Division algorithm
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0.
∴ Again by division algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0.
∴ Again by division algorithm.
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0.
∴ Again by division algorithm
30303 = 20202 × 1 + 10101
The remainder 10101 ≠ 0.
∴ Again using division algorithm
20202 = 10101 × 2 + 0
The remainder is 0.
∴ 10101 is the H.C.F. of 363636 and 252525.
4.If 13824 = 2a × 3b then find a and b.
Solution:
If 13824 = 2a × 3b
Using the prime factorisation tree
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 29 × 33 = 2a × 3b
∴ a = 9, b = 3.
5.If p1x1 × p2x2 × p3x3 × p4x4 = 113400 where p1, p2, p3, p4 are primes in ascending order and x1, x2, x3, x4 are integers, find the value of P1, P2, P3, P4 and x1, x2, x3, x4.
Solution:
If p1x1 × p2x2 × p3x3 × p4x4 = 113400
p1, p2, p3, P4 are primes in ascending order, x1, x2, x3, x4 are integers.
using Prime factorisation tree.
113400 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 7
= 23 × 34 × 52 × 7
= p1x1 × p2x2 × p3x3 × p4x4
∴ p1= 2, p2 = 3, p3 = 5, p4 = 7, x1 = 3, x2 = 4, x3 = 2, x4 = 1.
6. Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of arithmetic.
Solution:
408 and 170.
408 = 23 × 31 × 171
170 = 21 × 51 × 171
∴ H.C.F. = 21 × 171 = 34.
To find L.C.M, we list all prime factors of 408 and 170, and their greatest exponents as follows.
∴ L.C.M. = 23 × 31 × 51 × 171
= 2040.
7.Find the greatest number consisting of 6 digits which is exactly divisible by 24, 15, 36?
Solution:
To find L.C.M of 24, 15, 36
24 = 23 × 3
15 = 3 × 5
36 = 22 × 32
∴ L.C.M = 23 × 32 × 51
= 8 × 9 × 5
= 360
If a number has to be exactly divisible by 24, 15, and 36, then it has to be divisible by 360. The greatest 6 digit number is 999999.
Common multiplies of 24, 15, 36 with 6 digits are 103680, 116640, 115520, …933120, 999720 with six digits.
∴ The greatest number consisting 6 digits which is exactly divisible by 24, 15, 36 is 999720.
8. What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves the remainder 7 in each case?
Find the L.C.M of 35, 56, and 91
35 – 5 × 7 56
56 = 2 × 2 × 2 × 7
91 = 7 × 13
L.C.M = 23 × 5 × 7 × 13
= 3640
Since it leaves the remainder 7
The required number = 3640 + 7
= 3647
The smallest number is = 3647
9. Find the least number that is divisible by the first ten natural numbers.
Solution:
The least number that is divisible by the first ten natural numbers is 2520.
Hint:
1,2, 3,4, 5, 6, 7, 8,9,10
The least multiple of 2 & 4 is 8
The least multiple of 3 is 9
The least multiple of 7 is 7
The least multiple of 5 is 5
∴ 5 × 7 × 9 × 8 = 2520.
L.C.M is 8 × 9 × 7 × 5
= 40 × 63
= 2520
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Mental Math to Solve Proportions
## Solve proportions by comparing numerators and denominators of equal fractions.
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Practice Mental Math to Solve Proportions
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Mental Math to Solve Proportions
Chase, Marc and Kris are all working at the supermarket in the stock room. After school, they work to stock the shelves at the local supermarket with all kinds of cans. In fact, they have gotten quite fast at it and they love to have contests to see who is the fastest. Often the one who loses has to treat the others to ice cream after work. It takes Chase 15 minutes to stock three shelves of canned goods, and it takes Marc 45 minutes to stock nine shelves.
“I am definitely faster,” Chase tells Marc one afternoon.
“I don’t think so, I think we are both the same.” Marc disagrees.
Chase and Marc continue to argue. Who is correct? Their friend Kris stocks shelves at the same rate as Chase does. If he stocks 12 shelves at this rate, how many minutes does it take him?
Previously we worked on how to identify a proportion. In this Concept, you will learn how to use mental math to solve proportions. Solving proportions using mental math is exactly what Marc and Chase need to help them with their argument.
### Guidance
In an earlier Concept on unit rates and rates, you practiced figuring out a rate or a unit rate. Essentially, there was a missing part of the rate that you had to figure out.
Here we have a unit rate of twenty-five campers for one tent. Then we have another rate that says that we have 75 campers and we are trying to figure out how many tents are needed for the 75 campers.
These two ratios are equal and they form a proportion.
How can we figure out the missing number of tents?
We can use equal ratios to do this, or we can simply examine the problem and use mental math. Sometimes it makes more sense to simply figure out an answer in your head!
Often we can use mental math to quickly figure out the missing part of a proportion. We call this “solving a proportion.”
We can think, “Four is half of eight, what is half of sixteen?” Our answer is eight.
Practice using mental math to solve the following proportions.
#### Example A
\begin{align*}\frac{1}{4} = \frac{x}{16}\end{align*}
Solution: 4
#### Example B
\begin{align*}\frac{3}{9} = \frac{x}{18}\end{align*}
Solution: 6
#### Example C
\begin{align*}\frac{5}{15} = \frac{1}{x}\end{align*}
Solution: 3
Now let's go back and help Marc and Chase to figure out their argument.
Chase, Marc and Kris are all working at the supermarket as stock guys. After school, they work to stock the shelves at the local supermarket with all kinds of cans. In fact, they have gotten quite fast at it and they love to have contests to see who is the fastest. Often the one who loses has to treat the ice cream after work. It takes Chase 15 minutes to stock three shelves of canned goods, and it takes Marc 45 minutes to stock nine shelves.
“I am definitely faster,” Chase tells Marc one afternoon.
“I don’t think so, I think we are both the same.” Marc disagrees.
Chase and Marc continue to argue. Who is correct? Their friend Kris stocks shelves at the same rate as Chase does. If he stocks 12 shelves at this rate, how many minutes does it take him?
The first problem is to figure out which boy is faster at stocking shelves. To do this, we need to write two ratios and see if they form a proportion. If they form a proportion, then Marc is correct.
We can look at the relationship between the numerators and the denominators to see that the ratios are the same. Three is one - fifth of fifteen and nine is one - fifth of forty - five. Using mental math has helped us to solve this dilemma.
The ratios form a proportion, so Marc is correct. The boys both work at the same pace.
Now Kris works at the same rate as Chase (Marc too). He stocked 12 shelves. Given the rate, how long did it take him? Once again, we need to write two ratios to form a proportion. The unknown will be the time that it takes Kris.
If we look at this proportion and ask ourselves, “What times 3 equals twelve?” The answer is four. We can use that to form an equal ratio.
15 \begin{align*}\times\end{align*} 4 \begin{align*}=\end{align*} 60.
It takes Kris 60 minutes to stock his twelve shelves. All three boys are equal in their rate of speed. They decide to take turns buying the ice cream.
### Vocabulary
Proportion
two equal ratios.
Ratio
a comparison of two quantities can be written in fraction form, with a colon or with the word “to”.
Cross Products
to multiply the diagonals of each ratio of a proportion.
### Guided Practice
Here is one for you to try on your own.
\begin{align*}\frac{2}{3} = \frac{x}{33}\end{align*}
To figure this out, you can look at the relationship between the denominators and the numerators. Then using mental math, you can solve for the missing value.
\begin{align*} 3 \times 11 = 33\end{align*}
Therefore, we multiply the given numerator by 11 to find the unknown numerator.
### Practice
Directions: Use mental math to solve the unknown part of each proportion.
1. \begin{align*}\frac{1}{2} = \frac{x}{8}\end{align*}
2. \begin{align*}\frac{1}{2} = \frac{5}{x}\end{align*}
3. \begin{align*}\frac{1}{3} = \frac{4}{x}\end{align*}
4. \begin{align*}\frac{2}{3} = \frac{x}{6}\end{align*}
5. \begin{align*}\frac{1}{2} = \frac{x}{16}\end{align*}
6. \begin{align*}\frac{5}{6} = \frac{x}{12}\end{align*}
7. \begin{align*}\frac{14}{16} = \frac{x}{8}\end{align*}
8. \begin{align*}\frac{1}{2} = \frac{x}{18}\end{align*}
9. \begin{align*}\frac{1}{4} = \frac{x}{20}\end{align*}
10. \begin{align*}\frac{1}{4} = \frac{x}{24}\end{align*}
11. \begin{align*}\frac{1}{4} = \frac{x}{40}\end{align*}
12. \begin{align*}\frac{2}{4} = \frac{x}{40}\end{align*}
13. \begin{align*}\frac{25}{50} = \frac{2}{x}\end{align*}
14. \begin{align*}\frac{4}{12} = \frac{x}{48}\end{align*}
15. \begin{align*}\frac{6}{7} = \frac{36}{x}\end{align*}
### Vocabulary Language: English
Cross Products
Cross Products
To simplify a proportion using cross products, multiply the diagonals of each ratio.
Proportion
Proportion
A proportion is an equation that shows two equivalent ratios.
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# Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
Question:
Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
$4 x^{2}-4 a^{2} x+\left(a^{4}-b^{4}\right)=0$
Solution:
The given equation is $4 x^{2}-4 a^{2} x+\left(a^{4}-b^{4}\right)=0$.
Comparing it with $A x^{2}+B x+C=0$, we get
$A=4, B=-4 a^{2}$ and $C=a^{4}-b^{4}$
$\therefore$ Discriminant, $D=B^{2}-4 A C=\left(-4 a^{2}\right)^{2}-4 \times 4 \times\left(a^{4}-b^{4}\right)=16 a^{4}-16 a^{4}+16 b^{4}=16 b^{4}>0$
So, the given equation has real roots.
Now, $\sqrt{D}=\sqrt{16 b^{4}}=4 b^{2}$
$\therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-\left(-4 a^{2}\right)+4 b^{2}}{2 \times 4}=\frac{4\left(a^{2}+b^{2}\right)}{8}=\frac{a^{2}+b^{2}}{2}$
$\beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-\left(-4 a^{2}\right)-4 b^{2}}{2 \times 4}=\frac{4\left(a^{2}-b^{2}\right)}{8}=\frac{a^{2}-b^{2}}{2}$
Hence, $\frac{1}{2}\left(a^{2}+b^{2}\right)$ and $\frac{1}{2}\left(a^{2}-b^{2}\right)$ are the roots of the given equation.
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# Multiplying polynomials
Multiplying polynomials with the following two good examples will help you master this topic once and for all. That is all you will need. It could be very useful though to review the multiplication of binomials
Example #1:
Multiply 4x3 + 2x + 5 by 3x4 + x + 6
(4x3 + 2x + 5) × (3x4 + x + 6)
Important concept
You must know what a term is when multiplying polynomials. It is because the goal is to multiply each term of the polynomial on the left by each term of the polynomial on the right and then adding the whole thing!
We will show the result of each multiplication or whatever will be added together in bold
The polynomial on the left is 4x3 + 2x + 5
Look at it carefully. Each term is separated by an addition sign.
So the first term is 4x3
The second term is 2x
The third term is 5
The polynomial on the right is 3x4 + x + 6
Again, each term is separated by an addition sign.
So the first term is 3x4
The second term is x
The third term is 6
Now multiply the first term of the polynomial on the left that is 4x3 by each term of the polynomial on the right and these are 3x4, x, and 6.
Let's do it!
4x3 × 3x4 = 4 × 3 × x3 × x4 = 12x 3 + 4 = 12x7
4x3 × x = 4 × x3 × x = 4 × x3 × x1 = 4x 3 + 1 = 4x4
4x3 × 6 = 4 × 6x3 = 24x3
Next, multiply the second term of the polynomial on the left that is 2x by each term of the polynomial on the right and these are 3x4, x, and 6.
2x × 3x4 = 2 × 3 × x × x4 = 2 × 3 × x1 × x4 = 6x 1 + 4 = 6x5
2x × x = 2 × x × x = 2 × x1 × x1 = 2x 1 + 1 = 2x2
2x × 6 = 2 × 6x = 12x
Finally, multiply the third term of the polynomial on the left that is 5 by each term of the polynomial on the right and these are 3x4, x, and 6.
5 × 3x4 = 15x4
5 × x = 5x
5 × 6 = 30
Adding the result in bold together, we get:
12x7 + 4x4 + 24x3 + 6x5 + 2x2 + 12x + 15x4 + 5x + 30
Combine like terms
12x7 + (4x4 + 15x4) + 24x3 + 6x5 + 2x2 + (12x + 5x) + 30
12x7 + 19x4 + 24x3 + 6x5 + 2x2 + 17x + 30
Example #2:
Multiply 4x3 − 2x + 5 by 3x4 + x − 6
Almost the same problem. We just incorporated a couple of subtraction signs. My teaching experience has taught me that when multiplying polynomials, it is best to say to students to replace minus with + -
Then, do the exact same thing you did in example #1
(4x3 − 2x + 5) × (3x4 + x − 6) = (4x3 + -2x + 5) × (3x4 + x + -6)
4x3 × 3x4 = 12x7
4x3 × x = 4x4
4x3 × -6 = 4 × -6x3 = -24x3
Notice this time that the second term of the polynomial on the left has a negative next to it! same thing for the third term of the polynomial on the right
Pay careful attention to this when multiplying polynomials!
-2x × 3x4 = -2 × 3 × x × x4 = -2 × 3 × x1 × x4 = -6x 1 + 4 = -6x5
-2x × x = -2 × x × x = -2 × x1 × x1 = -2x 1 + 1 = -2x2
-2x × -6 = -2 × -6x = 12x
5 × 3x4 = 15x4
5 × x = 5x
5 × -6 = -30
Adding the result in bold together, we get:
12x7 + 4x4 + -24x3 + -6x5 + -2x2 + 12x + 15x4 + 5x + -30
Combine like terms
12x7 + (4x4 + 15x4) + -24x3 + -6x5 + -2x2 + (12x + 5x) + -30
12x7 + 19x4 + -24x3 + -6x5 + -2x2 + 17x + -30
Multiplying polynomials should be a breeze if you really understand the two examples above.
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# Instantaneous Rates of Change
## Derivative is the slope of the tangent line at a point.
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Instantaneous Rate of Change
When you first learned about slope you learned the mnemonic device “rise over run” to help you remember that to calculate the slope between two points you use the following formula:
In Calculus, you learn that for curved functions, it makes more sense to discuss the slope at one precise point rather than between two points. The slope at one point is called the slope of the tangent line and the slope between two separate points is called a secant line.
Consider a car driving down the highway and think about its speed. You are probably thinking about speed in terms of going a given distance in a given amount of time. The units could be miles per hour or feet per second, but the units always have time in the denominator. What happens when you consider the instantaneous speed of the car at one instant of time? Wouldn’t the denominator be zero?
#### Watch This
http://www.youtube.com/watch?v=7CvLzpzGhJI Brightstorm: Definition of a Derivative
#### Guidance
The slope at a point \begin{align*}P\end{align*} (also called the slope of the tangent line) can be approximated by the slope of secant lines as the “run” of each secant line approaches zero.
Because you are interested in the slope as the “run” approaches zero, this is a limit question. One of the main reasons that you study limits in calculus is so that you can determine the slope of a curve at a point (the slope of a tangent line).
Example A
Estimate the slope of the following function at -3, -2, -1, 0, 1, 2, 3. Organize the slopes in a table.
Solution: By mentally drawing a tangent line at the following \begin{align*}x\end{align*} values you can estimate the following slopes.
\begin{align*}x\end{align*} slope -3 0 -2 0 -1 -1 0 -1 1 2 2 0 3 0
If you graph these points you will produce a graph of what’s known as the derivative of the original function.
Example B
Estimate the slope of the function\begin{align*}f(x)=\sqrt{x}\end{align*} at the point \begin{align*}(1,1)\end{align*} by calculating 4 successively close secant lines.
Solution: Calculate the slope between \begin{align*}(1,1)\end{align*} and 4 other points on the curve:
• The slope of the line between \begin{align*}\left(5,\sqrt{5}\right)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_1=\frac{\sqrt{5}-1}{5-1} \approx 0.309\end{align*}
• The slope of the line between \begin{align*}(4,\ 2)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_2=\frac{2-1}{4-1}\approx 0.333\end{align*}
• The slope of the line between \begin{align*}\left(3,\sqrt{3}\right)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_3=\frac{\sqrt{3}-1}{3-1} \approx 0.366\end{align*}
• The slope of the line between \begin{align*}\left(2,\sqrt{2}\right)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_4=\frac{\sqrt{2}-1}{2-1} \approx 0.414\end{align*}
If you had to guess what the slope was at the point \begin{align*}(1,1)\end{align*} what would you guess the slope to be?
Example C
Evaluate the following limit and explain its connection with Example B.
\begin{align*}\lim \limits_{x \to 1} \left(\frac{\sqrt{x}-1}{x-1}\right)\end{align*}
Solution: Notice that the pattern in the previous problem is leading up to \begin{align*}\frac{\sqrt{1}-1}{1-1}\end{align*} . Unfortunately, this cannot be computed directly because there is a zero in the denominator. Luckily, you know how to evaluate using limits.
The slope of the function \begin{align*}f(x)=\sqrt{x}\end{align*} at the point \begin{align*}(1,1)\end{align*} is exactly \begin{align*}m=\frac{1}{2}\end{align*}.
Concept Problem Revisited
If you write the ratio of distance to time and use limit notation to allow time to go to zero you do seem to get a zero in the denominator.
\begin{align*}\lim \limits_{time \to 0}\left(\frac{distance}{time}\right)\end{align*}
The great thing about limits is that you have learned techniques for finding a limit even when the denominator goes to zero. Instantaneous speed for a car essentially means the number that the speedometer reads at that precise moment in time. You are no longer restricted to finding slope from two separate points.
#### Vocabulary
A tangent line to a function at a given point is the straight line that just touches the curve at that point. The slope of the tangent line is the same as the slope of the function at that point.
A secant line is a line that passes through two distinct points on a function.
A derivative is a function of the slopes of the original function.
#### Guided Practice
1. Sketch a complete cycle of a sine graph. Estimate the slopes at \begin{align*}0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi\end{align*}.
2. Logan travels by bike at 20 mph for 3 hours. Then she gets in a car and drives 60 mph for 2 hours. Sketch both the distance vs. time graph and the rate vs. time graph.
3. Approximate the slope of \begin{align*}y=x^3\end{align*} at \begin{align*}(1,1)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?
1.
\begin{align*}x\end{align*} Slope 0 1 \begin{align*}\frac{\pi}{2}\end{align*} 0 \begin{align*}\pi\end{align*} -1 \begin{align*}\frac{3\pi}{2}\end{align*} 0 \begin{align*}2\pi\end{align*} 1
You should notice that these are the exact values of cosine evaluated at those points.
2. Distance vs. Time:
Rate vs. Time: (this is the graph of the derivative of the original function shown above)
3.
• The slope of the line between \begin{align*}(0.7,0.7^3)\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_1=\frac{0.7^3-1}{0.7-1} \approx 2.19\end{align*}
• The slope of the line between \begin{align*}(0.8,0.8^3 )\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_2=\frac{0.8^3-1}{0.8-1} \approx 2.44\end{align*}
• The slope of the line between \begin{align*}(0.9,0.9^3 )\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_3=\frac{0.9^3-1}{0.9-1} \approx 2.71\end{align*}
• The slope of the line between \begin{align*}(0.95,0.95^3 )\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_1=\frac{0.95^3-1}{0.95-1} \approx 2.8525\end{align*}
• The slope of the line between \begin{align*} (0.975,0.975^3 )\end{align*} and \begin{align*}(1,1)\end{align*} is: \begin{align*}m_1=\frac{0.975^3-1}{0.975-1} \approx 2.925625\end{align*}
The slope at \begin{align*}(1,1)\end{align*} will be slightly greater than the estimates because of the way the slope curves. The slope at \begin{align*}(1,1)\end{align*} appears to be about 3.
#### Practice
1. Approximate the slope of \begin{align*}y=x^2\end{align*} at \begin{align*}(1,1)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?
2. Evaluate the following limit and explain how it confirms your answer to #1.
\begin{align*}\lim \limits_{x \to 1} \left(\frac{x^2-1}{x-1}\right)\end{align*}
3. Approximate the slope of \begin{align*}y=3x^2+1\end{align*} at \begin{align*} (1,4)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?
4. Evaluate the following limit and explain how it confirms your answer to #3.
\begin{align*}\lim \limits_{x \to 1} \left(\frac{3x^2+1-4}{x-1}\right)\end{align*}
5. Approximate the slope of \begin{align*}y=x^3-2\end{align*} at \begin{align*}(1,-1)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?
6. Evaluate the following limit and explain how it confirms your answer to #5.
\begin{align*}\lim \limits_{x \to 1}\left(\frac{x^3-2-(-1)}{x-1}\right)\end{align*}
7. Approximate the slope of \begin{align*}y=2x^3-1\end{align*} at \begin{align*}(1,1)\end{align*} by using secant lines from the left. Will the actual slope be greater or less than the estimates?
8. What limit could you evaluate to confirm your answer to #7?
9. Sketch a complete cycle of a cosine graph. Estimate the slopes at \begin{align*}0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi\end{align*}.
10. How do the slopes found in the previous question relate to the sine function? What function do you think is the derivative of the cosine function?
11. Sketch the line \begin{align*}y=2x+1\end{align*}. What is the slope at each point on this line? What is the derivative of this function?
12. Logan travels by bike at 30 mph for 2 hours. Then she gets in a car and drives 65 mph for 3 hours. Sketch both the distance vs. time graph and the rate vs. time graph.
13. Explain what a tangent line is and how it relates to derivatives.
14. Why is finding the slope of a tangent line for a point on a function the same as the instantaneous rate of change at that point?
15. What do limits have to do with finding the slopes of tangent lines?
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 14.8.
### Vocabulary Language: English
Average rate of change
Average rate of change
The average rate of change of a function is the change in $y$ coordinates of a function, divided by the change in $x$ coordinates.
instantaneous rate of change
instantaneous rate of change
The instantaneous rate of change of a curve at a given point is the slope of the line tangent to the curve at that point.
limit
limit
A limit is the value that the output of a function approaches as the input of the function approaches a given value.
secant line
secant line
A secant line is a line that joins two points on a curve.
Slope
Slope
Slope is a measure of the steepness of a line. A line can have positive, negative, zero (horizontal), or undefined (vertical) slope. The slope of a line can be found by calculating “rise over run” or “the change in the $y$ over the change in the $x$.” The symbol for slope is $m$
Tangent line
Tangent line
A tangent line is a line that "just touches" a curve at a single point and no others.
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