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Krishna
1
Step 1: Take a look at the given equations
NOTE: \cosec \theta + \cot \theta = k ......................(1)
PROVE: \cos \theta = \frac{k^2 - 1}{k^2 + 2}
Step 2: Search for the appropriate formula among the trigonometric identities
IDENTITY: 1 + \cot^2 \theta = \cosec^2 \theta
We can write this as
\cosec^2 \theta - \cot^2 \theta = 1
Step 3: Use the formula of (a + b)(a - b) = a^2 - b^2 to simplify
EXAMPLE: \cosec^2 \theta - \cot^2 \theta = 1
(\cosec \theta + \cot \theta)(\cosec \theta - \cot \theta) = 1
We know that \cosec \theta + \cot \theta = k
k (\cosec \theta - \cot \theta) = 1
\cosec \theta - \cot \theta = \frac{1}{k} ..........................(2)
Step 4: Add equation (1) and (2)
EXAMPLE: \operatorname{cosec}\theta+\cot\theta+\operatorname{cosec}\theta-\cot\theta=k+\frac{1}{k}
2\cosec \theta = \frac{k^2 + 1}{k}.....................(3)
Step 4: Subtracting the equation (2) from (1)
EXAMPLE: \operatorname{cosec}\theta+\cot\theta-\operatorname{cosec}\theta+\cot\theta=k-\frac{1}{k}
2\cot \theta = \frac{k^2 - 1}{k}.............................(4)
Step 5: Dividing equation (4) by (3)
\frac{2\cot \theta}{2\cosec \theta} = \frac{ \frac{k^2 - 1}{k}}{ \frac{k^2 + 1}{k}}
\frac{\cos \theta}{\sin \theta} \sin \theta = \frac{k^2 - 1}{k^2 + 1}
\cos \theta = \frac{k^2 - 1}{k^2 + 1}
Hence proved |
# Height of a Building: Indirect Measurement Lesson | Grades 7-8
Looking for a fun way to help your kids practice and understand indirect measurement? This free indirect measurement lesson will get kids outside, applying their math skills to real life.
One of the biggest overarching math concepts in middle school is proportional relationships. We see proportional relationships in real life all the time, and it’s important that students can recognize them, but more than that apply proportions to solve real problems. There are a lot of applications, but one easy way to show students a proportional relationship is through indirect measurement. By creating similar triangles, we can create proportions to find a good estimate of things that are too tall or difficult to measure more precisely. In this lesson, Find the Height of a Building, students will apply their knowledge as they get outside and see it in action.
## What is Indirect Measurement?
Indirect measurement means finding an unknown measurement of an object that is difficult to measure directly, with a ruler or measuring stick.
When you cannot directly measure an object, you can use other measurements to indirectly estimate the measurement you’re looking for.
For example, you may not be able to easily measure the height of a tree in your back yard, the height of your house or the height of an office building downtown.
These things are significantly taller than you, so you can’t simply hold up a measuring tape to find the height.
You can, however, use a very clever application of similar triangles to estimate the heights of these objects. This is known as indirect measurement.
## Applying Similar Triangles to Indirect Measurement:
What does it mean for two triangles to be similar? Before we can begin our lesson on indirect measurement, we have to understand similar triangles.
Similar triangles have the same shape, but are different sizes. This means all corresponding angles are equal, but corresponding side lengths are not equal. This is what makes similar triangles different from congruent triangles.
Congruent triangles have equal corresponding angles and equal corresponding sides. So they are the same shape and the same size.
Although similar triangles do not have equal corresponding sides, they have an interesting relationship that we will put to work in our indirect measurement lesson: corresponding sides all have equal ratios. In other words:
Similar triangles have equal corresponding angles and all sides are of equal proportion.
Because we can create a proportion out of the corresponding sides of similar triangles, we can use that proportional relationship to find the measurements of known side lengths.
So how in the world does this relate to finding the height of a building?? Let’s find out.
## Finding the Height of a Building: An Indirect Measurement Lesson
To use indirect measurement, we will need two similar triangles.
The fun part of this lesson is that when you step outside on a sunny day, you (and other objects around you) cast a shadow on the ground.
The best part? At that moment, everything that is casting a shadow on the ground now has created a similar right triangle to you and your shadow.
### Materials Needed:
• Student handout for each student (grab it free in my shop!)
• Clipboard
• Measuring tape for each pair/small group
### Overview of the Lesson:
So the basic premise of this lesson is that you can use direct measurement to find the height and shadow length of an object that is easily measurable.
Then, you measure the shadow length of the object that is too difficult to measure directly.
Then you set up a proportion using the two similar triangles to solve for the unknown measurement (in this case, the height of a tall building).
Although the basic idea is pretty straightforward, the lesson includes a few more parts to help guide students and make sure they understand what they’re doing and why it works.
### Lesson Steps:
In step one, students will briefly review similar triangles and using proportions.
I recommend reviewing this problem in your classroom and discussing it together before you head outside.
In step two, students will use direct and indirect measurement to find the height of something small and measurable.
The goal of this step is to compare their measurements and begin to think about why precision matters and see how close their estimate is to the actual measurement.
This is also another opportunity to review measuring and creating proportions with similar triangles.
In step three, students will measure the shadow of their school building and once again create a proportion to estimate the height of the building.
In step four (which you can do back in your classroom), students answer some questions as they reflect on what they did and what they learned.
### Indirect Measurement Lesson Wrap-Up:
Once all students have completed the tasks, head back inside to compare and discuss.
When I used this lesson with my 8th grade students, we will in a brand new school building. Lucky for me, I knew the builder and I was able to reach out and get the actual, exact height of our new school building.
We were able to turn this into a little competition, with the group finding the closest estimate to the actual height earning a prize.
Although this is certainly not necessary, it definitely helps to be able to wrap up the lesson by revealing the actual height of the building.
This naturally leads into a discussion of why the estimates were different from each other and also different from the actual.
You can also find the average of all the estimates. Is this closer to the actual? Why?
Lastly, you can discuss other possible applications for indirect measurement.
And once you’re finished, if your students need additional practice with similar triangles, proportions and indirect measurement, you might try this similar triangles classroom scavenger hunt.
It includes 10 problems to post around your room. Students walk around solving them and finding the solutions until they have completed all 10.
It also includes two independent practice worksheets, which you can use for informal assessments or review.
If you enjoy this lesson, become a Math Geek Mama+ member and gain access to the entire library of engaging math lessons like this one, hundreds of math games and low-prep practice worksheets for grades 5-8!
Ready to get started? Click the link below to go to my shop and grab this free math lesson! |
Green’s Theorem and Area of Polygons « Stack Exchange Mathematics Blog
# Green’s Theorem and Area of Polygons
June 4, 2014 by . 9 comments
A common method used to find the area of a polygon is to break the polygon into smaller shapes of known area. For example, one can separate the polygon below into two triangles and a rectangle:
By breaking this composite shape into smaller ones, the area is at hand: \begin{align}A_1 &= bh = 5\cdot 2 = 10 \\ A_2 = A_3 &= \frac{bh}{2} = \frac{2\cdot 1}{2} = 1 \\ A_{total} &= A_1+A_2+A_3 = 12\end{align}
Unfortunately, this approach can be difficult for a person to use when they cannot physically (or mentally) see the polygon, such as when a polygon is given as a list of many vertices.
## Formula
Happily, there is a formula for the area of any simple polygon that only requires knowledge of the coordinates of each vertex. It is as follows: $$A = \sum_{k=0}^{n} \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2} \tag{1}$$ (Where $${n}$$ is the number of vertices, $${(x_k, y_k)}$$ is the $${k}$$-th point when labelled in a counter-clockwise manner, and $${(x_{n+1}, y_{n+1}) = (x_0, y_0)}$$; that is, the starting vertex is found both at the start and end of the list of vertices.)
It should be noted that the formula is not “symmetric” with respect to the signs of the $${x}$$ and $${y}$$ coordinates. This can be explained by considering the “negative areas” incurred when adding the signed areas of the triangles with vertices $${(0,0)-(x_k, y_k)-(x_{k+1}, y_{k+1})}$$.
In the next sections, I derive this formula using Green’s Theorem, show an example of its use, and provide some applications.
## Derivation
Green’s Theorem states that, for a “well-behaved” curve $${C}$$ forming the boundary of a region $${D}$$:
$$\displaystyle \oint_C P(x, y)\;\mathrm dx + Q(x, y)\;\mathrm dy = \iint_D \frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y}\;\mathrm dA \ \ \ \ \ (2)$$
(In this context, “well behaved” means, among other things, that $${C}$$ is piecewise smooth. For a full listing of the requirements for $${C}$$ and $${D}$$, please see here.)
Recalling that the area of $${D}$$ is equal to $${\iint_D \;\mathrm dA}$$, we can use Green’s Theorem to calculate area if we choose $${P}$$ and $${Q}$$ such that $${\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y}=1}$$. Clearly, choosing $${P(x, y) = 0}$$ and $${Q(x, y) = x}$$ satisfies this requirement.
Thus, letting $${A}$$ be the area of the region $${D}$$:
$$\displaystyle A = \oint_C x\;\mathrm dy \ \ \ \ \ (3)$$
Now, consider the polygon below, bordered by the piecewise-smooth curve $${C = C_0\cup C_1\cup\cdots\cup C_{n-1}\cup C_n}$$, where $${C_k}$$ starts at the point $${(x_k, y_k)}$$ and ends at the “next” point along the polygon’s edge when proceeding counter-clockwise. (N.B. If we proceed in a clockwise manner, we get the negative of the area of the polygon.)
Note: the dashed lines indicate that there could be any number of points between $${(x_2, y_2)}$$ and $${(x_{n-1}, y_{n-1})}$$; this is a complete polygon.
Because line integrals over piecewise-smooth curves are additive over length, we have that:
$$\displaystyle A = \oint_C x\;\mathrm dy = \int_{C_0} x\;\mathrm dy + \cdots + \int_{C_n} x\;\mathrm dy \ \ \ \ \ (4)$$
To compute the $${k}$$-th line line integral above, parametrize the segment from $${(x_k, y_k)}$$ to $${(x_{k+1}, y_{k+1})}$$:
$$\displaystyle C_k:\; \vec{r}=\left((x_{k+1} – x_k)t + x_k,\; (y_{k+1} – y_k)t + y_k\right),\quad 0\le t\le 1 \ \ \ \ \ (5)$$
Substituting this parametrization into the integral, we find: \begin{align}\int_{C_k} x\;\mathrm dy &= \int_0^1 \left((x_{k+1} – x_k)t + x_k\right)\left(y_{k+1} – y_k\right)\;\mathrm dt \\ &= \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2}\end{align}
Summing all of the $$C_k$$’s, we then find the total area: $$A = \sum_{k=0}^{n} \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2} \tag{6}$$
## Example
To demonstrate use of this formula, let us apply this to the shape at the beginning of the document. A copy of the image is found below, but this one is marked with the coordinates of the vertices.
Starting with the coordinate $${(0, 0)}$$ and proceeding counter-clockwise (again, if we go in the opposite direction, we get the negative of the correct area), we apply our formula:
\begin{align} A &= \sum_{k=0}^{n} \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2} \\ &= \frac{(1.5 + 0)(0 – 0)}{2} + \frac{(2.5 + 1.5)(-1 – 0)}{2} + \frac{(3.5 + 2.5)(0 – (-1))}{2} + \frac{(5+3.5)(0-0)}{2} \\ &\quad + \frac{(5+5)(2-0)}{2} + \frac{(3.5+5)(2-2)}{2} + \frac{(2.5+3.5)(3-2)}{2}\\ &\quad +\frac{(1.5+2.5)(2-3)}{2} + \frac{(0+1.5)(2-2)}{2} + \frac{(0+0)(0-2)}{2}\\ &= 0 + (-2) + 3 + 0 + 10 + 0 + 3 + (-2) + 0 + 0\\ &= 12 \end{align}
We arrive at the same answer we had above, which is good. (Obviously!)
## Possible Applications
Although this formula is an interesting application of Green’s Theorem in its own right, it is important to consider why it is useful. As can be seen above, this approach involves a lot of tedious arithmetic. Thus, its main benefit arises when applied in a computer program, when the tedium is no longer an issue.
From an engineering perspective, this formula could be used to determine the area (and, using similarly-derived formulas for the moments of a polygon about the $${x}$$ and $${y}$$ axes, the centroid) of component in a CAD program. Many computer programs estimate all shapes as polygons or connected line segments (rather than exact curves/circles), which means this formula can be directly applied to the existing representation in software.
In a contrived context, this formula can also be useful in collegiate programming competitions. (This was why I first took note of this formula.) Some problems in the “computational geometry” category may involve computing the area of specific polygons, such as the convex hull of a set of points. Knowing the derivation can make it easier to recall during a competition event.
As a final example, one can use this formula to approximate the area of a plot of land. Over a small region, surface area on a sphere can be approximated as that on a rectangle. Thus, an irregularly-shaped plot of land can be modelled as an polygon. Once coordinates are established for each vertex, the formula derived above can be applied to find its area.
## 9 Comments
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• Not only can this formula be used to find areas, centroids, etc., but it regularly is; I’ve used it myself for determining the “best” location to place a letter in an arbitrarily-shaped polygon.
• You say that there are $n$ vertices and your $k$ is from $0$ to $n$, so wouldn’t that involve $n+1$ vertices? Shouldn’t the first vertex be called $(x_1,y_1)$ and $k$ be from $1$ to $n$?
• Also, in your example, shouldn’t the last term in the sum be (0+0)(0-2)/2?
• apnorton says:
The first point and the last point are the same. That, is $(x_0, y_0) = (x_n, y_n)$, So, although there are $n$ vertices, there are $n+1$ terms in the summation.
You are correct regarding the typo in the last term of the summation, though–I’ll change that.
• Interested Reader says:
So, how does this generalize to 3 or 4 dimensional polyhedra?
• Norbert says:
There is no natural order on the set of vertices of $n$ dimensional polyhedra. I think there is no good answer here.
• apnorton says:
I considered this, but didn’t make much progress. I believe that some similar formula for volume could be deduced through an application of Stoke’s Theorem, but I did not work on that long enough to produce anything of note.
• The problem of computing the volume of a 3D polyhedron given a list of its vertices as input is underspecified. Consider the following example. Start with a cube (8 vertices). Add a ninth vertex somewhere inside the cube. Remove from the cube a pyramid shaped part that has one of the faces of the original cube as its base and the freshly added ninth vertex as its peak. Depending on the choice of the face you get six different non-convex polyhedra, possibly all with distinct volumes (with a generically placed ninth vertex).
To describe the polyhedron you need to give its faces. Vertices alone won’t be enough.
• The list of vertices of a polygon can also be seen as the set of oriented boundary elements (=edge segments) of the polygon. So the analogy with the 3D polyhedron would be to compute the volume based on the set of oriented boundary elements (=faces) of the 3D polyhedron. But I think one has to use Gauss’ theorem (sometimes also called Divergence theorem or Ostrogradsky’s theorem), not the Kelvin–Stokes theorem. |
# Converting the Mixed fractions into Improper fractions
A mixed fraction is basically an improper fraction but it is expressed in a special form for representing an integer part and a proper fraction at a time.
$Mixed \, fraction$ $\,=\,$ $(Integer \, part) \, \dfrac{Numerator}{Denominator}$
In some cases, a mixed fraction should be converted into an improper fraction. So, it is essential to learn how to change a mixed a fraction into an improper fraction mathematically.
### Steps
There are three simple steps to change any mixed fraction as an improper fraction.
1. Multiply the Integer part by the denominator of the fraction.
2. Add the product to the numerator of the proper fraction.
3. Display the sum as numerator with same denominator.
The three steps can be written as a formula, which helps us to change any mixed fraction as an improper fraction.
$Improper \, fraction$ $\,=\,$ $\dfrac{(Integer \, part) \times Denominator + Numerator}{Denominator}$
### Example
Convert the mixed fraction $7\dfrac{2}{5}$ as an improper fraction.
In this example, $7$ is an integer, $2$ is a numerator and $5$ is a denominator. Multiply the integer $7$ by the denominator $5$ and then add the product to the numerator $2$. The sum of them is written as numerator of the improper fraction but its denominator is same as the denominator of the fraction in the mixed fraction.
$Improper \, fraction$ $\,=\,$ $\dfrac{(7 \times 5)+2}{5}$
$\implies$ $Improper \, fraction$ $\,=\,$ $\dfrac{35+2}{5}$
$\,\,\, \therefore \,\,\,\,\,\,$ $Improper \, fraction$ $\,=\,$ $\dfrac{37}{5}$
Therefore, the mixed fraction $7\dfrac{2}{5}$ is converted as an improper fraction $\dfrac{37}{5}$ mathematically.
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NCERT Class 7 Maths Simple Equations
# NCERT Class 7 Maths Simple Equations
The chapter 4 begins with an introduction to Simple Equations by explaining the basic concept of setting up of an equation and recalling the concepts learnt earlier classes such as variables.Then the concept of equation and equality is discussed in detail.Now, the procedure to solve an equation and the steps involved in it is explained elaborately.Now, the reverse process i.e. to obtain the equation from the solution is explained.In the last section of the chapter, applications of simple equations to practical situations is dealt in detail.
## Chapter 4 Ex.4.1 Question 1
Complete the last column of the table.
S. No Equation Value Say, whether the equation is satisfied. (Yes/No) (i) $$x + 3 = 0$$ $$x= 3$$ (ii) $$x + 3 = 0$$ $$x= 0$$ (iii) $$x + 3 = 0$$ $$x= -3$$ (iv) $$x - 7 = 1$$ $$x= 7$$ (v) $$x - 7 = 1$$ $$x= 8$$ (vi) $$5x= 25$$ $$x= 0$$ (vii) $$5x= 25$$ $$x= 5$$ (viii) $$5x= 25$$ $$x= -5$$ (ix) \begin{align}\frac{m}{3} = 2\end{align} $$m= -6$$ (x) \begin{align}\frac{m}{3} = 2\end{align} $$m= 0$$ (xi) \begin{align}\frac{m}{3} = 2\end{align} $$m= 6$$
### Solution
What is Known?
Equation and the value of the variable.
What is unknown?
Whether the given value is a solution of the equation or not.
Reasoning:
Put the value of the variable in the given equation. If LHS is equal to the RHS then the equation is satisfied and if it is not that means this given value is not a solution of the equation.
Steps:
(i) \begin{align}x + 3 = 0,x = 3\end{align}
L.H.S $$=$$ $$x+ 3$$, R.H.S $$= 0$$
By putting, $$x= 3$$
L.H.S $$=3 + 3= 6 ≠$$R.H.S
Therefore, “No”, the equation is not satisfied.
(ii) \begin{align}x+ 3 = 0,x= 0\end{align}
L.H.S $$=$$ $$x+ 3$$, R.H.S$$= 0$$
By putting, $$x= 0;$$
L.H.S $$=0+3= 3 ≠$$R.H.S
Therefore, “No”, the equation is not satisfied.
(iii) \begin{align}x+ 3 = 0, x= -3\end{align}
L.H.S $$=$$ $$x+ 3$$, R.H.S $$= 0$$
By putting, $$x= -3$$
L.H.S $$= -3 + 3 = 0 =$$ R.H.S
Therefore, “Yes”, the equation is satisfied.
(iv) \begin{align}x– 7 = 1, x= 7\end{align}
L.H.S $$=$$ $$x\,– 7$$, R.H.S $$= 1$$
By putting, $$x= 7$$;
L.H.S $$= 7 \,– 7 = 0 ≠$$ R.H.S
Therefore, “No”, the equation is not satisfied.
(v) \begin{align}x– 7 = 1, x= 8\end{align}
L.H.S $$=$$ $$x– 7$$, R.H.S $$= 1$$
By putting, $$x= 8;$$
L.H.S $$= 8 \,– 7 = 1 =$$ R.H.S
Therefore, “Yes”, the equation is satisfied.
(vi) \begin{align}5x= 25, x= 0,\end{align}
L.H.S $$=$$ $$5x$$, R.H.S$$= 25$$
By putting, $$x= 0;$$
L.H.S $$=$$ $$5 \times 0=0 ≠$$R.H.S
Therefore, “No”, the equation is not satisfied.
(vii) \begin{align}5x= 25, x= 5,\end{align}
L.H.S $$= 5x$$, R.H.S $$= 25$$
By putting, $$x= 5;$$
L.H.S $$=$$ $$5 \times 5 = 25$$= R.H.S
Therefore, “Yes”, the equation is satisfied.
(viii) \begin{align}5x = 25, x= – 5,\end{align}
L.H.S $$=$$ $$5x$$, R.H.S $$= 25$$
By putting, $$x= – 5;$$
L.H.S $$=$$ $$5 \times \left( { - 5} \right) = - 25≠$$R.H.S
Therefore, “No”, the equation is not satisfied.
(ix) \begin{align}\frac{m}{3} = 2, m= – 6,\end{align}
L.H.S $$=$$ \begin{align}\frac{m}{3}\end{align}, R.H.S $$= 2$$
By putting, $$m$$$$= – 6;$$
L.H.S $$=$$ \begin{align}\frac{{ - 6}}{3}=−2≠\end{align} R.H.S
Therefore, “No”, the equation is not satisfied.
(x) \begin{align}\frac{m}{3} = 2, m=0\end{align}
L.H.S $$=$$ \begin{align}\frac{m}{3}\end{align}, R.H.S $$= 2$$
By putting, $$m= 0;$$
L.H.S $$=$$ \begin{align}\frac{0}{3}=0≠\end{align}R.H.S
Therefore, “No”, the equation is not satisfied.
(xi) \begin{align}\frac{m}{3} = 2 m= 6,\end{align}
L.H.S $$=$$ \begin{align}\frac{m}{3}\end{align}, R.H.S $$= 2$$
By putting, $$m$$ $$= 6;$$
L.H.S $$=$$ \begin{align}\frac{6}{3}= 2 =\end{align} R.H.S
Therefore, “Yes”, the equation is satisfied
S. No Equation Value Say, weather the equation is satisfied. (Yes/No) (i) $$x + 3 = 0$$ $$x= 3$$ No (ii) $$x + 3 = 0$$ $$x= 0$$ No (iii) $$x + 3 = 0$$ $$x= -3$$ Yes (iv) $$x - 7 = 1$$ $$x= 7$$ No (v) $$x - 7 = 1$$ $$x= 8$$ Yes (vi) $$5x= 25$$ $$x= 0$$ No (vii) $$5x= 25$$ $$x= 5$$ Yes (viii) $$5x= 25$$ $$x= -5$$ No (ix) \begin{align}\frac{m}{3} = 2\end{align} $$m= -6$$ No (x) \begin{align}\frac{m}{3} = 2\end{align} $$m= 0$$ No (xi) \begin{align}\frac{m}{3} = 2\end{align} $$m= 6$$ Yes
## Chapter 4 Ex.4.1 Question 2
Check whether the value given in the brackets is a solution to the given equation or not:
(a) $$n + 5 = 19\left( {n = 1} \right)$$
(b) $$7n + 5 = 19\left( {n = -2} \right)$$
(c) $$7n + 5 = 19\left( {n = 2} \right)$$
(d) $$4p-3 = 13\left( {p = 1} \right)$$
(e) $$4p-3 = 13\left( {p = -4} \right)$$
(f) $$4p-3 = 13\left( {p = 0} \right)$$
### Solution
What is unknown?
Whether the given value is a solution of the equation or not.
What is Known?
Equation and the value of the variable.
Reasoning:
Put the value of the given variable in the equation. If LHS is equal to the RHS then the equation is satisfied and if it is not that means this given value is not a solution of the equation
Steps:
a) Here, $$n+ 5$$ is L.H.S, $$19$$ is R.H.S and $$n= 1$$ (given)
L.H.S = $$n+ 5,$$
By putting, $$n= 1,$$
L.H.S $$= 1 + 5 = 6 ≠$$ R.H.S
L.H.S $$≠$$ R.H.S, so $$n= 1$$ is not a solution of the equation.
b) Here, $$7n+ 5$$ is L.H.S, $$19$$ is R.H.S and $$n= – 2$$ (given)
L.H.S = $$7n+ 5$$,
By putting, $$n= \,– 2$$,
L.H.S$$=7 \times (−2) + 5= −9 ≠$$ R.H.S
As, L.H.S $$≠$$ R.H.S, so $$n= \,– 2$$ is not a solution of the equation.
c) Here, $$7n+ 5$$ is L.H.S, $$19$$ is R.H.S and $$n= 2$$ (given)
L.H.S = $$7n+ 5$$,
By putting, $$n= 2$$,
L.H.S $$= 7 \times (2) + 5 = 19 =$$R.H.S
As, L.H.S = R.H.S, so $$n= 2$$ is a solution of the equation.
d) Here, $$4p\,– 3$$ is L.H.S, $$13$$ is R.H.S and $$p= 1$$ (given)
L.H.S = $$4p\,– 3,$$
By putting, $$p= 1$$,
L.H.S $$=4 \times (1)\, – 3 = 1 ≠$$ R.H.S
As, L.H.S $$≠$$ R.H.S, so $$p= 1$$ is not a solution of the equation.
e) Here, $$4p– 3$$ is L. H.S, $$13$$ is R.H.S and $$p= -4$$ (given)
L.H.S $$=4p– 3$$,
By putting, $$p= 1$$,
L.H.S $$=4 \times (-4) – 3 = -19 ≠$$ R.H.S
As, L.H.S $$≠$$ R.H.S, so $$p= 1$$ is not a solution.
f) Here, $$4p– 3$$ is L. H.S, $$13$$ is R.H.S and $$p= 0$$ (given data)
L.H.S $$=$$ $$4p\,– 3$$,
By putting, $$p= 0$$,
L.H.S $$=4 \times(0)–3=−3≠$$R.H.S
As, L.H.S$$≠$$R.H.S, so $$p= 0$$ is not a solution of the equation.
## Chapter 4 Ex.4.1 Question 3
Solve the equation given below by trial and error method:
(i) \begin{align}5p + 2 = 17\end{align}
(ii) \begin{align}3m-{\rm{1}}4 = 4\end{align}
### Solution
What is Known?
Equations
What is unknown?
Solution of the equation or the value of the variable.
Reasoning:
Put the different values of the variable in the given equation. If LHS is equal to the RHS then the equation is satisfied and if it is not that means this variable is not a solution of the equation.
Steps:
(i) $$5p + 2 = 17$$
$$5p + 2=$$ L.H.S
By putting, $$p= 0$$,
$$5 \times 0+2=2≠17$$
By putting, $$p=1$$,
$$5 \times (1)+2=7≠17$$
By putting, $$p= 2$$,
$$5 \times (2) + 2 = 12 ≠$$ R.H.S
By putting, $$p= 3$$,
$$5\times (3) + 2 = 17 =$$ R.H.S
Therefore, $$p= 3$$ is a solution of the equation.
(ii) $$3m-{\rm{1}}4 = 4$$
$$3m-{\rm{1}}4=$$ L.H.S
By putting, $$m= 5$$,
$$3 \times (5)–14=1≠6$$
By putting, $$m= 6$$,
$$3 \times (6)–14 =4 =$$R.H.S.
Therefore, $$m= 6$$ is a solution of the equation.
## Chapter 4 Ex.4.1 Question 4
Write equations for the following statements:
(i) The sum of numbers $$x$$ and is
(ii) subtracted from $$y$$ is
(iii) Ten times $$a$$ is
(iv) The number $$b$$ divided by gives
(v) Three fourth of $$t$$ is
(vi) Seven times $$m$$ plus gets you
(vii) One fourth of a number $$x$$ minus gives
(viii) If you take away from times $$y$$, you get
(ix) If you add to one third of , you get
### Solution
What is Known?
Statements of the equations.
What is unknown?
Equations for the given statement.
Reasoning:
This question is very simple. Read the statement carefully and frame the equation in steps.
Steps:
(i) $$x~+4=9$$
(ii) $$y-2 = 8$$
(iii) $$10a = 70$$
(iv) \begin{align}\frac{b}{5} = 6\end{align}
(v)\begin{align} \frac{3}{4}t = 15\end{align}
(vi) $$7m + 7 = 77$$
(vii) \begin{align}\frac{1}{4}x-4 = 4\end{align}
(viii) $$6y-6 = 60$$
(ix)\begin{align}\frac{1}{3}z + 3 = 30\end{align}
## Chapter 4 Ex.4.1 Question 5
Write the following equations in statement forms:
(i) $$p + 4 = 15$$
(ii) $$m-7=3$$
(iii) $$2m=7$$
(iv)\begin{align}\frac{m}{5} = 3\end{align}
(v) \begin{align}\frac{3m}{5}=6~\end{align}
(vi) $$3p+4=25~$$
(vii) $$4p-2=18~$$
(viii) \begin{align}\frac{p}{2} + 2{\rm{ = }}8\end{align}
### Solution
What is Known?
Equations
What is unknown?
Statements of the given equations.
Reasoning:
To solve this question, look at the equation carefully and decide the variable and numbers. Then write the suitable statement using the variables and numbers.
Steps:
(i) The sum of $$p$$ and $$4$$ is $$15.$$
(ii) 7 subtracted from $$m$$ is $$3.$$
(iii) Two times $$m$$ is $$7.$$
(iv) One-fifth of $$m$$ is $$3.$$
(v) Three-fifth of $$m$$ is $$6.$$
(vi) When $$4$$ is added to three times of a number $$p,$$ it gives $$25.$$
(vii) When $$2$$ is subtracted from four times of a number $$p$$, gives $$18.$$
(viii) When $$2$$ is added to half of $$p$$ gives $$8.$$
## Chapter 4 Ex.4.1 Question 6
Set up an equation in the following cases:
(i) Irfan says that he has $$7$$ marbles more than five times the marbles Parmit has. Irfan has $$37$$ marbles. (Take $$‘m’$$ to be the number of Parmit’s marbles.)
(ii) Laxmi’s father is $$49$$ years old. He is $$4$$ years older than three times Laxmi’s age. (Take Laxmi’s age to be $$y$$ years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus $$7.$$ The highest score is $$87.$$ (Take the lowest score to be $$l$$.)
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be $$b$$ in degrees. Remember that the sum of angles of a triangle is $$180$$ degrees).
### Solution
What is unknown?
Equations for the different statements.
What is Known?
Statements of the equations.
Reasoning:
Read the statement carefully and frame the equation.
Steps:
(i) Let permit has $$m$$ number of marbles
Number of marbles Irfan has $$=5m + 7$$
Total number of marbles Irfan has $$37$$
So, $$5m + 7=37$$
(ii) Let the age of Laxmi be $$y$$ years
Laxmi’s father is four years older than three times Laxmi’s age $$=3y + 4$$
Age of Laxmi’s father is $$49$$ years
According to question,
$$3y + 4= 49$$
(iii) Let the lowest marks obtained by the student be $$l$$
Highest marks obtained by the student be $$2l + 7$$
And the highest score is $$87$$
Therefore, According to question
$$2l + 7= 87$$
(iv) Let the base angle of a triangle be $$b$$
Vertex angle of the triangle $$= 2b$$
According to question,
\begin{align}b + b + 2b = 180^\circ\\ 4b = 180^\circ\end{align}
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# α, β and γ are the zeroes of cubic polynomial p(x) = ax3 + bx2 + cx + d, (a ≠ 0). The product of their zeroes [αβγ] is?
By Ritesh|Updated : November 14th, 2022
α, β and γ are the zeroes of cubic polynomial p(x) = ax3 + bx2 + cx + d, (a ≠ 0). The product of their zeroes [αβγ] is -d/a.
It is given that p(x) = ax3 + bx2 + cx + d, (a ≠ 0) …. (1)
If α, β, and γ are the zeroes of the cubic polynomial P(x).
Then, we can write as:
p(x) = (x - α) (x - β) (x - γ)
The above equation becomes
p(x) = x3 - (α + β + γ)x2 + (αβ + βγ + γα)x - αβγ …. (2)
Due to the similarity of equations (1) and (2).
The result of comparing the coefficients is:
a/1 = b/ -α - β - γ = c/αβ + βγ + γα = d/-αβγ
On solving, the above equation we get:
α + β + γ = -b/a = sum of roots
αβ + βγ + γα = c/a = sum of product of the roots
αβγ = -d/a = product of the roots
Hence, the product of zeroes is αβγ = -d/a.
### Polynomial
• The words Nominal, which means "terms," and Poly, which means "many," are combined to make the word polynomial.
• A polynomial is an equation made up of exponents, constants, and variables that are combined using mathematical operations including addition, subtraction, multiplication, and division (No division operation by a variable).
• Depending on how many terms are present, the expression is classified as a monomial, binomial, or trinomial.
Summary:
## α, β and γ are the zeroes of cubic polynomial p(x) = ax3 + bx2 + cx + d, (a ≠ 0). The product of their zeroes [αβγ] is?
α, β and γ are the zeroes of cubic polynomial p(x) = ax3 + bx2 + cx + d, (a ≠ 0). The product of their zeroes [αβγ] is -d/a. P(x) stands for the polynomial function, where x stands for the variable. |
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### Section 2.11 : Linear Inequalities
4. Solve the following inequality and give the solution in both inequality and interval notation.
$8 \le 3 - 5z < 12$
Show All Steps Hide All Steps
Hint : Solving double inequalities uses the same basic process as solving single inequalities. Just remember that what you do to one part you have to do to all parts of the inequality.
Start Solution
Just like with single inequalities solving these follow pretty much the same process as solving a linear equation. The only difference between this and a single inequality is that we now have three parts of the inequality and so we just need to remember that what we do to one part we need to do to all parts.
Also, recall that the main goal is to get the variable all by itself in the middle and all the numbers on the two outer parts of the inequality.
So, let’s start by subtracting 3 from all the parts. This gives,
$5 \le - 5z < 9$ Show Step 2
Finally, all we need to do is divide all three parts by -5 to get,
$- 1 \ge z > - \frac{9}{5}$
Don’t forget that because we were dividing everything by a negative number we needed to switch the direction of the inequalities.
So, the inequality form of the solution is $$\require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{9}{5} < z \le - 1}}$$ (we flipped the inequality around to get the smaller number on the left as that is a more “standard” form). The interval notation form of the solution is $$\require{bbox} \bbox[2pt,border:1px solid black]{{\left( { - \frac{9}{5}, - 1} \right]}}$$ .
For the interval notation form remember that the smaller number is always on the left (hence the reason for flipping the inequality form above!) and be careful with parenthesis and square brackets. We use parenthesis if we don’t include the number and square brackets if we do include the number. |
## Equivalence rules
Recall that two propositions are logically equivalent if and only if they entail each other. In other words, equivalent propositions have the same truth value in all possible circumstances: whenever one is true, so is the other; and whenever one is false, so is the other. The proposition P is equivalent to the proposition ~~P, for example. In fact, it is somewhat misleading to say that P and ~~P are two different propositions. They mean exactly the same thing; they are just different ways of representing the same proposition. If any two well-formed formulas (WFFs) are logically equivalent, they represent the same proposition.
An equivalence rule is a pair of equivalent proposition forms, with lowercase letters used as variables for which we can substitute any WFF (just as we did previously with inference rules). By memorizing a few simple equivalence rules, we can more easily recognize when two sentences mean the same thing—a useful skill in philosophy. Familiarity with equivalence rules is also necessary for constructing logical proofs, as we’ll see on the next page.
Here are six inference rules worth memorizing:
• Double negation (DN) says that a pair of tildes can be added or removed from any WFF:
x is equivalent to ~~x
• Commutation (Com) says that the two component propositions of a conjunction, disjunction, or biconditional can switch places:
(x • y) is equivalent to (y • x)
(x ∨ y) is equivalent to (y ∨ x)
(x ≡ y) is equivalent to (y ≡ x)
This rule is similar to the commutative property of addition and multiplication in mathematics: 1+2 = 2+1 and 2×3 = 3×2.
• Association (Assoc) allows us to rearrange the parentheses that associate the components of two conjunctions, two disjunctions, or two biconditionals:
(x • (y • z)) is equivalent to ((x • y) • z)
(x ∨ (y ∨ z)) is equivalent to ((x ∨ y) ∨ z)
(x ≡ (y ≡ z)) is equivalent to ((x ≡ y) ≡ z)
This rule is similar to the associative property of addition and multiplication: (1+2)+3 = 1+(2+3) and (1×2)×3 = 1×(2×3).
• De Morgan’s law (DM) says that a negated conjunction is equivalent to a disjunction with both components negated, and vice versa:
~(x • y) is equivalent to (~x ∨ ~y)
~(x ∨ y) is equivalent to (~x • ~y)
This rule is analogous in some ways to the distributive property of addition over multiplication. We can think of the ‘~’ (tilde) outside the parentheses as being “distributed” to each of the components inside the parentheses, in the same way we distribute a multiple: 2×(3+4) = 2×3 + 2×4. Alternatively, we can imagine “factoring out” the tilde from each component inside the parentheses, just as we factor out a multiple: 2×3 + 2×4 = 2×(3+4). The analogy isn’t perfect, though. When “distributing” or “factoring out” the tilde, we also have to change the ‘•’ to a ‘∨’ or vice versa, whereas we don’t change the ‘+’ when distributing or factoring a multiple in mathematics.
• Contraposition (Contra) says that the antecedent and the consequent of a conditional can switch places if we negate both of them:
(x ⊃ y) is equivalent to (~y ⊃ ~x)
• Implication (Imp) says that a conditional is equivalent to a disjunction in which the first disjunct is the negation of the conditional’s antecedent:
(x ⊃ y) is equivalent to (~x ∨ y)
Since logically equivalent WFFs represent the same proposition, they can be substituted for one another in any context, even when they appear as components of a larger WFF. Since P is equivalent to ~~P by the “double negation” rule, for example, (Q • P) is likewise equivalent to (Q • ~~P), by that same rule. Moreover, the substitution can go in either direction. For example, by De Morgan’s law, we can replace ~(A • B) with (~A ∨ ~B) and vice versa: we can replace (~A ∨ ~B) with ~(A • B). Here are a few more examples:
((P • Q) ≡ ~~R) is equivalent to ((P • Q) ≡ R) by double-negation. (A pair of tildes is removed from the right side of the biconditional by DN.)
(A ⊃ (B ∨ C)) is equivalent to (A ⊃ (C ∨ B)) by commutation. (The consequent of the conditional is replaced with an equivalent formula by Com.)
(~A ∨ (C ∨ ~D)) is equivalent to ((~A ∨ C) ∨ ~D) by association.
(~A • ~(B ⊃ C)) is equivalent to ~(A ∨ (B ⊃ C)) by De Morgan’s law. (A tilde is “factored out” from the two conjuncts and the ‘•’ is replaced with a ‘∨’.)
((~A • ~B) ⊃ C) is equivalent to (~(A ∨ B) ⊃ C) by De Morgan’s law. (The antecedent of the conditional is replaced with an equivalent formula by DM.)
(~(Q • R) ⊃ ~P) is equivalent to (P ⊃ (Q • R)) by contraposition.
(P • (Q ⊃ R)) is equivalent to (P • (~R ⊃ ~Q)) by contraposition. (The second conjunct is replaced with an equivalent formula by Contra.)
(P ⊃ (Q • R)) is equivalent to (~P ∨ (Q • R)) by implication.
(~(A ∨ B) ∨ C) is equivalent to ((A ∨ B) ⊃ C) by implication. |
# Algebra II Notes Unit Six: Polynomials Syllabus Objectives: 6.2 The student will simplify polynomial expressions.
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1 Algebra II Notes Unit Si: Polnomials Sllabus Objectives: 6. The student will simplif polnomial epressions. Review: Properties of Eponents (Allow students to come up with these on their own.) Let a and b be real numbers, and let m and n be integers. Product of Powers Propert Quotient of Powers Propert m n m n a a a m m a mn a 1 a or, a 0 n n nm a a a m Power of a Power Propert n a a mn ab a b Power of a Product Propert m m m m a a Power of a Quotient Propert, b 0 m b b m Negative Eponent Propert Zero Eponent Propert n m 1 a b a m or, a 0 a b a a 0 1, a 0 n Evaluating Numerical Epressions with Eponents E 1: Evaluate. Use the power of a power propert: Use the negative eponent propert: E : Evaluate. Use the power of a product propert: Use the negative eponent propert: Use the power of a quotient propert: 9 4 Page 1 of McDougal Littell
2 E : Evaluate 6 0. Algebra II Notes Unit Si: Polnomials Use the zero eponent propert: Use the quotient of a power propert: 6 Use the negative eponent propert: Note: We can use the quotient of a power propert to keep the eponent positive. Use the quotient of a power propert: Simplifing Algebraic Epressions E 1: Simplif the epression 8 1 and write with positive eponents. Use the power of a product propert: 1 Use the power of a power propert: Use the product of a power propert: Use the negative eponent propert: E : Simplif the epression 0 t v and write with positive eponents. Use the zero eponent propert: t 1 t 1 Use the negative eponent propert: t t t E : Simplif the epression 1 6 Use the product of a power propert: Use the power of a power propert: Use the quotient of a power propert: Simplif: 4 and write with positive eponents Page of McDougal Littell
3 Algebra II Notes Unit Si: Polnomials You Tr: Simplif the epression and write with positive eponents. Identif which properties of eponents ou used QOD: Which properties of eponents require ou to check that two or more bases are the same before appling the propert? Sample SAT Question(s): Taken from College Board online practice problems If m is a positive integer, which of the following is NOT equal to m? (A) (B) 4 m 4 m (C) m m (D) 4 m (E) Grid-In 16 m. For all positive integers a and b, let a b be defined b b a 1 ab a 1. What is the value of 4?. The positive integer n is not divisible b 7. The remainder when n is divided b are each equal to k. What is k? (A) 1 (B) (C) 4 (D) 6 (E) It cannot be determined from the information given. Page of McDougal Littell n is divided b 7 and the remainder when
4 Algebra II Notes Unit Si: Polnomials Sllabus Objective: 6.1 The student will graph a polnomial function with and without technolog. f a a... a a a, a 0 n n1 Polnomial Function: a function of the form n n1 1 0 Note: Eponents are whole numbers and coefficients are real numbers. n Leading Coefficient: a n Constant Term: a 0 Degree: n Note: Consider using the Fraer Model (Vocabular Concept Grid) Activit See resource pages. Standard Form of a Polnomial Function: The terms are written in descending order of the eponents Names of Polnomial Functions: This is kind of trick but a n is the name of the coefficient with the same degree. So, a n is the coefficient of the term that is the n th degree and a n 1 is the coefficient of the term that is degree n 1. Degree Tpe Standard Form 0 Constant f a0 1 Linear f a1 a0 Quadratic f a a a 1 0 Cubic f a a a a Quartic 4 f a a a a a Identifing Polnomial Functions f 8 a polnomial function? If es, write it in standard form. E 1: Is No. In order to be a polnomial function, all eponents must be whole numbers. f 8 a polnomial function? If es, write it in standard form. E : Is 4 Yes. All eponents are whole numbers and all coefficients are real numbers. f 8 Note: This is a quartic trinomial (degree = 4). Standard Form: 4 Evaluating Polnomial Functions Using Direct Substitution E 1: Find 4 f if f f So : f () 47 Page 4 of McDougal Littell
5 Algebra II Notes Unit Si: Polnomials Evaluating Polnomial Functions Using Snthetic Substitution E 1: Find 4 f if f 6 1 using snthetic substitution. Using the polnomial in standard form, write the coefficients in a row. Put the -value to the upper left Bring down the first coefficient, then multipl b the -value. multipl Add straight down the columns, and repeat The number in the bottom right is the value of So : f () 47 f. E : Find f if f 7 11 using snthetic substitution. This polnomial function is in standard form, however it is missing two terms. We can rewrite the f to fill in the missing terms. function as This also means that (, 17) is an ordered pair that would be a point on the graph. f 17 Graphing Polnomial Functions: To graph a polnomial function, make a table of values using snthetic substitution, plot the points, and determine the end behavior to draw the rest of the graph. End Behavior: the behavior of the graph as gets ver large (approaches positive infinit ) OR as gets ver small (or approaches negative infinit ). Notation: ( approaches positive infinit ) (The ver far right end of a graph). ( approaches negative infinit) (The ver far left end of a graph). Page of McDougal Littell
6 Algebra II Notes Unit Si: Polnomials Eploration Activit: Graph each function on the calculator. Determine the end behavior of f as approaches negative and positive infinit. Fill in the table and write our conclusion regarding the degree of the function and the end behavior. (Teacher Note: Answers are in red.) f Degree Sign of Leading Coefficient f f + + f + f + + f + 4 f f 4 + f + + f + 6 f f Page 6 of McDougal Littell
7 Conclusion: The graph of a polnomial function n n 1 f an an 1... a a1 a0 has the following end behavior. These patterns are ver predicatable. Algebra II Notes Unit Si: Polnomials Degree Lead Coefficient End Behavior Even Positive as, f as, f Even Negative as, f as, f Think of end behavior as what happens on either end of the graph. There can be a lot of curves, etc. in the middle, but polnomial functions either increase or decrease at the far ends (as, f( ) ). Odd Positive as, f as, f Odd Negative as, f as, f E 1: Graph the polnomial function 4 f 1 b hand. Check our graph on the graphing calculator. Step One: Make a table of values using snthetic substitution f Step Two: Determine end behavior using the degree and sign of the leading coefficient. The degree is even, and the leading coefficient is positive. So as, f as, f. Step Three: Graph the polnomial function Page 7 of McDougal Littell
8 Algebra II Notes Unit Si: Polnomials E : Graph the polnomial function f 4 b hand. Check our graph on the graphing calculator. Step One: Make a table of values using snthetic substitution f Wh didn t we use snthetic subsitution to find f(0)? Step Two: Determine end behavior using the degree and sign of the leading coefficient. The degree is odd and the leading coefficient is negative. So. Step Three: Graph the polnomial function. 1 as, f as, f You Tr: Graph the polnomial function f 1b hand. Check our graph on the graphing calculator. QOD: Which term of the polnomial function is most important when determining the end behavior of the function? Page 8 of McDougal Littell
9 Sample CCSD Common Eam Practice Question(s): Algebra II Notes Unit Si: Polnomials Which best represents the graph of the polnomial function? 4 Page 9 of McDougal Littell
10 Algebra II Notes Unit Si: Polnomials Sllabus Objective: 6. The student will simplif polnomial epressions. Adding Polnomials E 1: Add the polnomials Subtracting Polnomials. Vertical Method: Write each polnomial in standard form and line up like terms. Then add the like terms E 1: Subtract the polnomials To subtract, we will rewrite the problem as an addition problem b adding the opposite Horizontal Method: Combine each set of like terms Write the final answer in standard form Multipling Polnomials E 1: Find the product 4. Horizontal Method: Use the distributive propert b distributing each term of the first polnomial Combine like terms and write the answer in standard form Page 10 of McDougal Littell
11 Algebra II Notes Unit Si: Polnomials E : Multipl the polnomials Vertical Method: Use long multiplication E : Multipl the polnomials 14. Multipl the polnomials two at a time. Because the are binomials, we can use FOIL to multipl the first two Use the distributive propert Combine like terms and write in standard form Review: Special Products (Allow students to come up with these on their own.) Memorize these! Sum and Difference Product a bab a b Square of a Binomial Cube of a Binomial a b a abb a b a abb a b a a bab b a b a a bab b E 1: Simplif the epression. Using the cube of a binomial: Page 11 of McDougal Littell
12 Application Problems Algebra II Notes Unit Si: Polnomials E 1: Find a polnomial epression for the volume of a rectangular prism with sides, 4, and. Volume of a Rectangular Prism = Length Width Height FOIL: 4 1 Vertical Method: E : From 198 through 1996, the number of flu shots given in one cit can be modeled b A t t t t for adults and b C t t t t Write a model for the total number F of flu shots given in these ears for children, where t is the number of ears since 198. To find the total flu shots, we need to add the polnomials. Vertical Method: Solution: 4 11.t 8.t 194t 4190t t 106t 1t 1t t 97.67t 194t 40t 81 F t t t t You Tr: Find the product: 7 1 QOD: What is the advantage of the vertical method when adding, subtracting, or multipling polnomials? Sample CCSD Common Eam Practice Question(s):? Which polnomial represents the product of 8 A. B. C. D Page 1 of McDougal Littell
13 Algebra II Notes Unit Si: Polnomials Sllabus Objectives: 6. The student will solve polnomial equations b factoring and graphing. Review: Factoring Patterns Factoring a General Trinomial E 1: Factor the trinomial 1. ac Method: ac 4 Split the middle term: Factor b grouping: Factoring a Perfect Square Trinomial E 1: Factor the trinomial 6 9. Difference of Two Squares Use a abb a b. E 1: Factor Common Monomial Factor Use a b aba b E 1: Factor the trinomial completel. 7 Factor the GCF and the binomial square Since this is not completel factored, use a b a ba b Sum and Difference of Two Cubes 4 a b ab a abb a b ab a abb. Page 1 of McDougal Littell
14 E 1: Factor the binomial Algebra II Notes Unit Si: Polnomials 8 1. Use a b a ba ab b E : Factor the binomial 7. 6 Use a b a ba ab b. 4 9 Factoring b Grouping E 1: Factor the polnomial Group each pair of terms and factor the GCF. 9 Factor the common binomial factor. 9 Factor the remaining terms if possible. Review: Zero-Product Propert If ab 0, then a 0 or b 0. Solving Polnomial Equations b Factoring E 1: Solve the equation Step One: Set the equation equal to zero. Step Two: Factor the polnomial Step Three: Set each factor equal to zero and solve. Solutions:,,0,, , 0, Page 14 of McDougal Littell
15 Algebra II Notes Unit Si: Polnomials E : Find the real-number solutions of the equation 0. Step One: Set the equation equal to zero. Step Two: Factor the polnomial Step Three: Set each factor equal to zero and solve. Real-Number Solution: 0 0, i i Application Problem E 1: An optical compan is going to make a glass prism that has a volume of 1 cm. The height will be h cm, and the base will be a right triangle with legs of length h cm and h cm. What will be the height? 1 h h h Volume of a Prism = Area of the Base Height 1 To solve this equation for h, we must set it equal to zero. 1 h h h 1 h h h 1 h h h Before factoring, we can multipl both sides of the equation b to eliminate (clear) the fractions. 1 0 h h 6h 1 0 h h 1h 0 Factor b grouping. 0 h h 6 h h h 0 6 Solve b setting each factor equal to 0. The height of the prism will be cm. h 0 h cm h h 60 6 No real solution Page 1 of McDougal Littell
16 Algebra II Notes Unit Si: Polnomials You Tr: Solve the equation QOD: Give an eample of a binomial that can be factored either as the difference of two squares or as the difference of two cubes. Show the complete factorization of our binomial. Sample CCSD Common Eam Practice Question(s): 1. Which of the following represents the solution set of the polnomial equation A.,, i, i B.,,, C., i, i i, i D., i, i, 7 1 0? 4. What is the factored form of the polnomial A. 9 B. 9 C. 9 D. 9 7?. Which lists the set of all real zeros of the following polnomial function? A. B., f 4 1 C.,, D.,,1, Page 16 of McDougal Littell
17 Algebra II Notes Unit Si: Polnomials Sllabus Objective: 6.6 The student will divide polnomials and relate the result to the remainder theorem and the factor theorem. Dividing Polnomials Using Long Division On Your Own: Find the quotient of 1,6 and 4 using long division. On Your Own: Find the quotient of and 4 1. For each step of long division, we will divide the term with the highest power in the dividend b the first term of the divisor Remember to put a place for the missing term. 4 (add the opposite) (bring down the net term) (remainder) Eploration: Use the polnomial function f. Use long division to divide f. Then use snthetic substitution to find f. What do ou notice? b Remainder Theorem: If a polnomial is f divided b k, then the remainder is r f k. Dividing Polnomials Using Snthetic Division (Note: This can onl be used when the divisor is in the form k.) E 1: Divide the polnomial 7 6 b. Use snthetic substitution for k. The coefficients of the quotient and remainder appear in snthetic substitution. Quotient: R Note for graphing: This means that (, 7) is an ordered pair that is on the graph of the function. Page 17 of McDougal Littell
18 Factor Theorem: A polnomial Algebra II Notes Unit Si: Polnomials f has a factor k if and onl if f k 0. E 1: Factor f given that f 6 0. Because f 6 0, we know that 6 is a factor of f snthetic division to find the other factors E : One zero of f 6 f b the Factor Theorem. We will use f 9 1 is 7. Find the other zeros of the function. f 7 0, we know that 7 Because snthetic division to find the other factors. is a factor of f Note for graphing: This means that ( 6,0) is an ordered pair that is on the graph of the function. 6 is called a zero. It is also an intercept. b the Factor Theorem. We will use f 7 1 f Set each factor equal to zero and 4 You Tr: Use long division to find the quotient of You Tr: Given that t is a zero of the function. f t 4t 9t t 1, find the other zeros. QOD: If f is a polnomial that has a as a factor, what do ou know about the value of f a? Sample CCSD Common Eam Practice Question(s): What is 9 divided b? A. B. C. D Page 18 of McDougal Littell
19 Algebra II Notes Unit Si: Polnomials Sllabus Objectives: 6.7 The student will identif all possible rational zeros of a polnomial function b using the rational root theorem. 6.4 The student will find rational zeros of a polnomial. Using the Rational Zero Theorem Review: Rational zero is a rational number that produces a function value of 0. It can be visualized as f( ) 0 where is a rational number. On the graph it is an -intercept. The Rational Zero Theorem If f ( ) a n n... a1a0has integer coefficients, then ever rational zero of p factor of constant term a0 f has the following form: q factor of leading coefficient an The first important step is to list the possible rational zeros. After the are listed we can test them to determine if the are rational zeros. If the value of the possible rational zeros =0, the are called zeros. List the possible rational zeros: E 1: Find the possible rational zeros of f ( ) 7 1 Step 1: The leading coefficient is 1. 1 is the onl factor of 1. Step : The constant is 1. All of the factors of 1 are 1,,, 4, 6, 1. Step : List the possible factors ,,,,,and *If we tested for actual zeros using snthetic substitution from previous lessons we would find that and 4 are zeros. This also means that could not be a zero, 7 could not be a zero, 1 could not be a zero. E : Find the possible rational zeros of f ( ) 7 1 Step 1: The leading coefficient is. The factors of are 1 and. Step : The constant is -. All of the factors of - are, Step : List the possible factors -,,, 1 1 *We will not test for actual zeros for this eample. This also means that could not be a zero, 4 could not be a zero, could not be a zero. When the leading coefficient is not 1, the list of possible zeros can increase dramaticall. There are man tools that are used to find the rational zeros. We know some of those tools now and others will be introduced later in the class. Eamples that follow will demonstrate some of them. Page 19 of McDougal Littell
20 Algebra II Notes Unit Si: Polnomials f 7 18 E : Find all the real zeros of f 18 7 Step 1: Put the function in standard order. Step : List possible rational zeros (1,,,4,6,8,9,1,18,4,6,7) Step : Tr the possible zeros until ou find one From previous lessons, the function can be reduced to: Then factored: 4 f f 6 ( 4) Zeros: ( 4) Note: Finding rational zeros is also referred to as finding real zeros. Rational numbers are also real numbers. There is a distinction between listing possible rational zeros and finding rational (real) zeros. f E 4: Find all of the real zeros of Step 1: Notice that each term contains a common factor of. The problem can be factored to f ( 4 6) and since 0onl ( 4 6) can be =0. Step : List possible rational zeros (1,,,6) (Since the leading coefficient is now 1) Step : Tr the possible zeros until ou find one From previous lessons, the function can be reduced to: Then factored: f f ( 1) 0 0 ( 1) 0 Zeros: 1 Page 0 of McDougal Littell
21 E : Find all the real zeros of 4 Algebra II Notes Unit Si: Polnomials f 4 81 Step 1: Mabe this could be graphed first. Step : Look at the graph for reasonable choices 1 It appears the might be,, and Step : Check the chosen values using snthetic division. Start with -. Wh not ½?? Wh not 7? Is a root (zero). f ( )( ) The factored form so far is 4 Step 4: Repeat the steps above using a different reasonable choice. Tr Is a root (zero) Step : Repeat the steps above using a different reasonable choice. Tr Is a root (zero). Step 6: The function 7 is left to be factored. ( 9) has no real factors. 1 Solution: There are real zeros:,, and 1 Yes, all three work! And each time, the function (polnomial) is reduced b one degree. f You Tr: Find all real zeros of QOD: If the leading coefficient of a polnomial with integer coefficients is 1, what tpe of numbers must an possible rational zeros be? Page 1 of McDougal Littell
22 Sample CCSD Common Eam Practice Question(s): Algebra II Notes Unit Si: Polnomials Which lists the set of all real zeros of the following polnomial function? A. B., C.,, D.,,1, f 4 1 Page of McDougal Littell
23 Algebra II Notes Unit Si: Polnomials Sllabus Objective: 6. The student will use the Fundamental Theorem of Algebra to determine the number of zeros of a polnomial function. The Fundamental Theorem of Algebra If f is a polnomial of degree n where n0, then the equation f( ) 0 has at least one root in the set of comple numbers. Finding the number of solutions or zeros Review: Find the solutions of the following eamples. State how man solutions each has and classif each zero as rational, irrational, or comple (imaginar). E 1: 1 0 E : 9 0 E : 1 0 for the quadratic factor.) (Hint: Use the factorization for the difference of cubes, then use the quadratic formula Do ou notice a pattern with the degree of the polnomial and the number of solutions each has? E 4: How man different solutions are there to How do ou eplain this number? ? E : How man different solutions are there to Solution: 4, i 4i 16 0? Note: On the graph, the imaginar roots do not cross the ais. Note: 4, i 4i are comple conjugate pairs. 1i,1 i are comple conjugate pairs. The comple roots of polnomial functions with real coefficients alwas occur in comple conjugate pairs. Is this also the case for irrational zeros? Page of McDougal Littell
24 Algebra II Notes Unit Si: Polnomials Finding the zeros of a polnomial function This activit involves finding the rational zeros as learned in the previous section, then using other tools, such as the quadratic formula or technolog, to find the irrational or comple roots. E 1: Find all zeros of f 4 ( ) Using the rational root theorem and snthetic division, it can be shown that is a repeated root and and -1 are roots. The factored form looks like this: ( ) ( )( 1). The graph is shown. When a factor k is raised to an odd power, the graph crosses through the -ais. When a factor k is raised to an even power, the graph is tangent to the -ais. Solution: There are four real zeros is a repeated root and and -1 are roots. E : Find all zeros of f 4 ( ) 0 Using the rational root theorem and snthetic division, it can be shown that and - are roots. 4 Using the pattern of E it can be shown that f( ) 0 factors to ( )( )( ) Using the quadratic formula ields zeros of = i Solution: There are four zeros, and - and i. Two are real and two are comple conjugates Using Zeros to Write Polnomial Functions E 1: Write a polnomial function f of least degree that has real coefficients, a leading coefficient of 1, and, and - as zeros. Step 1: Write f ( ) in factored form: f( ) ( )( )( ) Step : Review - Multipl the polnomials two at a time. Because the are binomials, we can use FOIL to multipl the first two. f( ) ( 6)( ) f( ) 190 Solution: f ( ) 19 0 Page 4 of McDougal Littell
25 Algebra II Notes Unit Si: Polnomials E : Write a polnomial function f of least degree that has real coefficients, a leading coefficient of 1, and 1, -, and 1- i as zeros. Step 1: Since 1- i is a zero, so is 1+ i Step : Write f ( ) in factored form: f ( ) ( 1)( )( (1- i) ( (1+ i)) Step : Regroup: f ( ) ( 1)( ) ( 1)- i ( 1)+ i 10 Step 4: Epand, multipl polnomials, and combine like terms. f( ) ( ) ( 1) - i f ( ) ( ) ( 1 1 f ( ) ( )( ) f 4 ( ) Note: This graph onl has two intercepts. Wh? Using Technolog to Approimate Zeros Specific instructions should be given based on the calculator used. This section will provide onl general direction. E 1: 4 Approimate the real zeros of f( ) Use a graphing calculator to graph and calculate the zeros. You Tr: State the number of zeros of f 1 and find what the are. QOD: What is the conjugate of a comple number, and wh is it important when finding all of the zeros of a polnomial function? Page of McDougal Littell
26 Sample CCSD Common Eam Practice Question(s): Algebra II Notes Unit Si: Polnomials According to the Fundamental Theorem of Algebra, how man comple zeros does the polnomial f 1 have? A. B. C. 4 D. 4 Page 6 of McDougal Littell
27 Algebra II Notes Unit Si: Polnomials Sllabus Objective: 6.8 The student will analze graphs of polnomial functions to determine its characteristics. Analzing polnomial graphs Concept Summar n n 1 Let f an an 1... a a1a0 be a polnomial function. The following statements are equivalent: Zero: k is a zero of the function f. Factor: Solution: k is a factor of polnomial f(). k is a solution of the polnomial function f()=0. - Intercept: k is an -intercept of the graph of the polnomial function f. Using -Intercepts to Graph a Polnomial Function f 1 E 1: Graph the function Step 1: Plot the -intercepts. Since + and 1 are factors, and 1 are zeros (-intercepts) Note: + is raised to an odd power so the graph crosses the -ais at =. 1 is raised to an even power so the graph is tangent to the -ais at = 1. When a factor k is raised to an odd power, the graph crosses through the -ais. When a factor k is raised to an even power, the graph is tangent to the -ais. Step : Plot a few points between the -intercepts. f(0) ; f( 1) 4 Step : Determine the end behavior of the graph. Cubic function (odd degree) with positive leading coefficient as, f as, f Step 4: Sketch the graph Page 7 of McDougal Littell
28 Finding Turning Points Algebra II Notes Unit Si: Polnomials Turning points of polnomial functions: Another important characteristic of graphs of polnomial functions is that the have turning points corresponding to local maimum and minimum values. The coordinate of a turning point is a local maimum if the point is higher than all nearb points. The coordinate of a turning point is a local minimum if the point is lower than all nearb points. The graph of ever polnomial function of degree n has at most n 1 turning points. Moreover, if a polnomial has n distinct real zeros, then its graph has eactl n 1 turning points. E 1: Identif the zeros and turning points (estimate the zeros and turning points) Turning point (ma) zero zero Turning point (min) Turning point (ma) zero Turning point (min) Leading coefficient positive real zeros (including the double zero) {, 1, 1} turning points ( 1,4); (1,0) 1 local ma; 1 local min Leading coefficient positive 1 real zero, imaginar zeros {} turning points (0, ); (, 8) 1 local ma; 1 local min You Tr: Leading coefficient real zeros -10 Leading coefficient real zeros -10 Leading coefficient real zeros -10 Leading coefficient real zeros turning points turning points turning points turning points local ma; local min local ma; local min local ma; local min local ma; local min Page 8 of McDougal Littell
29 Algebra II Notes Unit Si: Polnomials E : Use a graphing calculator to graph and calculate the approimate local maimum(s) and local minimum(s) of f( ) ( )( )( ) Local maimum Coordinates:( 1, 6) Local ma is 6 at = 1 Local minimum You Tr: Graph the function use the calculator to find the turning points of the function. You Tr: Graph the function f 1 b hand. Check our graph on the graphing calculator and f 9 1 on the calculator and find the local etrema. QOD: What is the difference between local and absolute maima and minima? Sample CCSD Common Eam Practice Question(s): 1. Which describes the end behavior of the graph of 4 A. f B. f C. f 0 D. f f as? Page 9 of McDougal Littell
30 Algebra II Notes Unit Si: Polnomials. Use the graph of a polnomial function below. A. {} What are the zeros of the polnomial? B. { } C. {, 1, 4} D. {, 1, 4} Eploring Data and Statistics (Notes are not provided for this material) Modeling with Polnomial Functions Write a polnomial function whose intercepts are given. Finding and Using Finite Differences Properties of finite differences 1. If a polnomial function f() has degree n, then the n th -order differences of function values for equall spaced values are non-zero and constant.. Conversel, if the n th order differences of equall spaced data are non-zero and constant, then the data can be represented b a polnomial function of degree n. Polnomial Modeling with Technolog Graphing calculators make it eas to enter data, make a scatter plot, and calculate linear, quadratic, cubic, and quartic regressions. Page 0 of McDougal Littell
31 Unit Summar: Algebra II Notes Unit Si: Polnomials Polnomial equations provide some of the most classic problems in all of algebra. Finding zeros and etrema have man real-world applications. Real-life situations are modeled b writing equations based on data and using those equations to determine or estimate other data points (speed, volume, time, profits, patterns, etc). Page 1 of McDougal Littell
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Manipulating systems of equations
In this lesson, we will manipulate systems of equations in order to eliminate an unknown. We will look at cases where one equation can be altered to enable elimination by addition or subtraction.
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.3x+2y=5 AND 3x-2y=7. If I add these two equations together, which unknown will be eliminated?
1/5
Q2.3x+2y=5 AND 3x-2y=7. If I subtract the two equations, which unknown will be eliminated?
2/5
Q3.3x-5y=17 AND 2x-5y=10. Subtract the first equation from the second equation.
3/5
Q4.2x-5y=10 AND 3x-5y=17. Add the two equations together.
4/5
Q5.3x-5y=17 AND 7x-5y=20. How can I eliminate y from these simultaneous equations?
5/5
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.3x+2y=5 AND 3x-2y=7. If I add these two equations together, which unknown will be eliminated?
1/5
Q2.3x+2y=5 AND 3x-2y=7. If I subtract the two equations, which unknown will be eliminated?
2/5
Q3.3x-5y=17 AND 2x-5y=10. Subtract the first equation from the second equation.
3/5
Q4.2x-5y=10 AND 3x-5y=17. Add the two equations together.
4/5
Q5.3x-5y=17 AND 7x-5y=20. How can I eliminate y from these simultaneous equations?
5/5
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Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Manipulating systems of equations
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.x+y=6. Multiply this equation by 3.
1/5
Q2.x+y=3 AND 2x-y=5 This pair of equations has been transformed. Which of the following are equivalent to the original pair?
2/5
Q3.x+y=3 AND 2x-y=5 This pair of equations has been transformed. Which of the following are equivalent to the original pair?
3/5
Q4.2x+5y=9 AND 5x+4y=14. Solve the simultaneous equations.
4/5
Q5.10x+5y=5 AND 3x-5y=34. Solve the simultaneous equations.
5/5
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Manipulating systems of equations
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.x+y=6. Multiply this equation by 3.
1/5
Q2.x+y=3 AND 2x-y=5 This pair of equations has been transformed. Which of the following are equivalent to the original pair?
2/5
Q3.x+y=3 AND 2x-y=5 This pair of equations has been transformed. Which of the following are equivalent to the original pair?
3/5
Q4.2x+5y=9 AND 5x+4y=14. Solve the simultaneous equations.
4/5
Q5.10x+5y=5 AND 3x-5y=34. Solve the simultaneous equations.
5/5
Lesson summary: Manipulating systems of equations
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CLASS_11_MATHS_SOLUTIONS_NCERT
Class XI Chapter 14 –
Class XI Chapter 14 – Mathematical Reasoning Maths ______________________________________________________________________________ Exercise 14.4 Question 1: Rewrite the following statement with “ if-then †in five different ways conveying the same meaning. If a natural number is odd, then its square is also odd. Solution 1: The given statements can be written in five different ways as follows. (i)A natural number is odd implies that its square is odd. (ii)A natural number is odd only if its square is odd. (iii)For a natural number to be odd, it is necessary that its square is odd. (iv)For the square of a natural number to be odd, it is sufficient that the number is odd. (v)If the square of a natural number is not odd, then the natural number is not odd. Question 2: Write the contrapositive and converse of the following statements. (i) If x is a prime number, then x is odd. (ii) It the two lines are parallel, then they do not intersect in the same plane. (iii) Something is cold implies that it has low temperature. (iv) You cannot comprehend geometry if you do not know how to reason deductively. (v) x is an even number implies that x is divisible by 4 Solution 2: (i) The contrapositive is as follows. If a number x is not odd, then x is not a prime number. The converse is as follows. If a number x is odd, then it is a prime number. (ii) The contrapositive is as follows. If two lines intersect in the same plane, then they are not parallel. The converse is as follows. If two lines do not intersect in the same plane, then they are parallel. (iii) The contrapositive is as follows. If something does not have low temperature, then it is not cold. The converse is as follows. If something is at low temperature, then it is cold. (iv) The contrapositive is as follows. If you know how to reason deductively, then you can comprehend geometry. The converse is as follows. If you do not know how to reason deductively, then you cannot comprehend geometry. Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
Class XI Chapter 14 – Mathematical Reasoning Maths ______________________________________________________________________________ (v) The given statement can be written as follows. If x is an even number, then x is divisible by 4. The contrapositive is as follows. If x is not divisible by 4, then x is not an even number. The converse is as follows. If x is divisible by 4, then x is an even number. Question 3: Write each of the following statement in the form “ if-then â€. (i) You get a job implies that your credentials are good. (ii) The Banana trees will bloom if it stays warm for a month. (iii) A quadrilateral is a parallelogram if its diagonals bisect each other. (iv) To get A+ in the class, it is necessary that you do the exercises of the book. Solution 3: (i)If you get a job, then you credentials are good. (ii)If the Banana tree stays warm for a month, then it will bloom. (iii)If the diagonals of a quadrilateral bisect each other, then it is parallelogram. (iv)If you want to get an A + in the class, then you do all the exercise of the book. Question 4: Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other. (a) If you live in Delhi, then you have winter clothes. (i) If you do not have winter clothes, then you do not live in Delhi. (ii) If you have winter clothes, then you live in Delhi. (b) If a quadrilateral is a parallelogram, then its diagonals bisect each other. (i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram. (ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Solution 4: (a) (i) This is the contrapositive of the given statement (a). (ii) This is the converse of the given statement (a). (b) (i) This is the contrapositive of the given statement (b). (ii) This is the converse of the given statement (b). Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
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# Illustrative Mathematics Grade 7, Unit 8, Lesson 17: More about Sampling Variability
Learning Targets:
• I can use the means from many samples to judge how accurate an estimate for the population mean is.
• I know that as the sample size gets bigger, the sample mean is more likely to be close to the population mean.
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Illustrative Math
Grade 7
#### Lesson 17: More about Sampling Variability
Let’s compare samples from the same population.
Illustrative Math Unit 7.8, Lesson 17 (printable worksheets)
#### Lesson 17 Summary
The following diagram as the sample size gets bigger, the sample mean is more likely to be close to the population mean.
#### Lesson 17.1 Average Reactions
The other day, you worked with the reaction times of twelfth graders to see if they were fast enough to help out at the track meet. Look back at the sample you collected.
1. Calculate the mean reaction time for your sample.
2. Did you and your partner get the same sample mean? Explain why or why not.
#### Lesson 17.2 Reaction Population
Your teacher will display a blank dot plot.
1. Plot your sample mean from the previous activity on your teacher’s dot plot.
2. What do you notice about the distribution of the sample means from the class?
a. Where is the center?
b. Is there a lot of variability?
c. Is it approximately symmetric?
3. The population mean is 0.442 seconds. How does this value compare to the sample means from the class?
Pause here so your teacher can display a dot plot of the population of reaction times.
4. What do you notice about the distribution of the population?
a. Where is the center?
b. Is there a lot of variability?
c. Is it approximately symmetric?
5. Compare the two displayed dot plots.
6. Based on the distribution of sample means from the class, do you think the mean of a random sample of 20 items is likely to be:
a. within 0.01 seconds of the actual population mean?
b. within 0.1 seconds of the actual population mean?
Explain or show your reasoning.
#### Lesson 17.3 How Much Do You Trust the Answer?
The other day you worked with 2 different samples of viewers from each of 3 different television shows. Each sample included 10 viewers. Here are the mean ages for 100 different samples of viewers from each show.
1. For each show, use the dot plot to estimate the population mean.
a. Trivia the Game Show
b. Science Experiments YOU Can Do
c. Learning to Read
2. For each show, are most of the sample means within 1 year of your estimated population mean?
3. Suppose you take a new random sample of 10 viewers for each of the 3 shows. Which show do you expect to have the new sample mean closest to the population mean? Explain or show your reasoning.
#### Are you ready for more?
Market research shows that advertisements for retirement plans appeal to people between the ages of 40 and 55. Younger people are usually not interested and older people often already have a plan. Is it a good idea to advertise retirement plans during any of these three shows? Explain your reasoning.
#### Lesson 17 Practice Problems
1. One thousand baseball fans were asked how far they would be willing to travel to watch a professional baseball game. From this population, 100 different samples of size 40 were selected. Here is a dot plot showing the mean of each sample.
Based on the distribution of sample means, what do you think is a reasonable estimate for the mean of the population?
2. Last night, everyone at the school music concert wrote their age on a slip of paper and placed it in a box. Today, each of the students in a math class selected a random sample of size 10 from the box of papers. Here is a dot plot showing their sample means, rounded to the nearest year.
a. Does the number of dots on the dot plot tell you how many people were at the concert or how many students are in the math class?
b. The mean age for the population was 35 years. If Elena picks a new sample of size 10 from this population, should she expect her sample mean to be within 1 year of the population mean? Explain your reasoning.
c. What could Elena do to select a random sample that is more likely to have a sample mean within 1 year of the population mean?
3. A random sample of people were asked which hand they prefer to write with.
l means they prefer to use their left hand.
r means they prefer to use their right hand.
Based on this sample, estimate the proportion of the population that prefers to write with their left hand.
4. Andre would like to estimate the mean number of books the students at his school read over the summer break. He has a list of the names of all the students at the school, but he doesn’t have time to ask every student how many books they read.
What should Andre do to estimate the mean number of books?
5. A hockey team has a 75% chance of winning against the opposing team in each game of a playoff series. To win the series, the team must be the first to win 4 games.
a. Design a simulation for this event.
b. What counts as a successful outcome in your simulation?
c. Estimate the probability using your simulation.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 8 / 1/3 = 24/1 = 24
Spelled result in words is twenty-four.
### How do you solve fractions step by step?
1. Divide: 8 : 1/3 = 8/1 · 3/1 = 8 · 3/1 · 1 = 24/1 = 24
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 1/3 is 3/1) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators.
In words - eight divided by one third = twenty-four.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Product of two fractions
Product of two fractions is 9 3/5 . If one of the fraction is 9 3/7. Find the other fraction.
• Four people
Four people want waffles for breakfast. There are 6 waffles left. How can 6 waffles be shared equally among 4 people? How much does each person get? Draw a picture and write a division expression to model the problem.
• A baker
A baker has 5 1/4 pies in her shop. She cut the pies in pieces that are each 1/8 of a whole pie. How many pieces of pie does she have?
• Candy bars
Sheldon has 4 candy bars and wants to split them among his five friends. If each person gets the same amount, what part of the candy bar will each friend get? Show your work.
• Pizza 5
You have 2/4 of a pizza, and you want to share it equally between 2 people. How much pizza does each person get?
There are 12 slices of bread and each person gets 3/4 of a slice of bread. How many people get bread?
• Larry 2
Larry spends half of his workday teaching piano lessons. If he sees 6 students, each for the same amount of time, what fraction of his workday is spent with each student?
• One fourth
One fourth of an apple pie is left for 2 family members to share equally. What fraction of the original pie will each member get?
• A quotient
What is the quotient of 3/10 divided by 2/4 as a fraction?
• Mrs. Glover
Mrs. Glover is making brownies for the girls’ tennis team. She took 1/5 of the leftover brownies to school to give to her 3 friends. How much did each friend get?
• Fractions 4
How many 2/3s are in 6?
• Barbara 2
Barbara get 6 pizzas to divide equally among 4 people. How much of a pizza can each person have?
• Divide 13
Divide. Simplify your answer and write as an improper fraction or whole number. 14÷8/3 |
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10th grade math skills: Find out what you need to know for your student
In 10th grade, students will continue to deepen their understanding of complex math concepts.
Algebra
Overview
For high school students, math skills and understandings are organized not by grade level but by concept. In algebra, students work with creating and reading expressions, rational numbers and polynomials, and the conventions of algebraic notations. They apply these skills and understandings to solve real-world problems.
Understanding equations
Understand an equation as a mathematical statement that uses letters to represent unknown numbers (such as 2x-6y+z=14) and is a statement of equality between two expressions (“this equals that”). Explain each step in solving a simple equation, and construct a practical argument to justify a solution method. Graph these equations on coordinate axes with labels and scales.
Rewriting expressions
Identify ways to rewrite the structure of an expression.
Equation solutions
Understand that some equations have no solutions in a given number system, but have a solution in a larger system. For example, the solution of x + 1 = 0 is an integer, not a whole number; the solution of 2x + 1 = 0 is a rational number, not an integer; the solutions of x² – 2 = 0 are real numbers, not rational numbers; and the solutions of x² + 2 = 0 are complex numbers, not real numbers.
Polynomials
Add, subtract, and multiply polynomials (expressions with multiple terms, such as 5xy² + 2xy - 7). Understand the relationship between the zeros and the factors of polynomials.
Polynomial identities
Use polynomial identities to solve real world problems.
A rectangular garden has a length of x + 2 ft. and a width of x + 8 ft. What must x be in order for the garden to have an area of 91 sq. ft.?
One-variable equations
Create equations and inequalities in one variable, and use them to solve problems, including weighted averages, calculation of mortgage and interest rates, and rate of travel.
Example:
A plane takes off from Chicago O’Hare airport, heading east and traveling at 580 miles an hour. Another plane takes off from O’Hare at the same time, heading west and traveling at 530 miles an hour. The two planes will be 1000 miles apart in how many hours?
Graphs
Represent, interpret, and solve equations and inequalities on graphs, plotted in the coordinate plane, and using technology to graph the functions and make tables of values.
Geometry
Overview
For high school students, math skills and understandings are organized not by grade level but by concept. In geometry, students work primarily with plane, or Euclidean geometry (with and without coordinates). Students build on geometry concepts learned through 8th grade, using more precise definitions and develop careful proofs of theorems (statements that can be proved true).
Transformation
Understand geometric transformation (moving a shape so it is in a different position, but still has same size, area, angles, and lengths) – especially rigid motions: translations, rotations, reflections, and combinations of these – involving angles, circles, perpendicular lines, parallel lines, and line segments.
Geometric theorems
Understand and prove geometric theorems about lines and angles, triangles, parallelograms, and circles. For example, Pythagorean Theorem, Line Intersection Theorem, Exterior Angle Theorem.
Trigonometry
Understand trigonometry as a measurement of triangles (and circles, such as orbits). Apply trigonometry to general triangles. Define the sine, cosine, and tangent trigonometric ratios.
Algebraic reasoning
Understand and use algebraic reasoning to prove geometric theorems.
Volume formulas
Explain volume formulas and use them to solve problems.
Example:
What is the volume of a cylinder that is 10m high, and has a radius of 9m? (Use π = 3.14)
Real-life situations
Apply geometric concepts to model real-life situations.
• Use measures and properties of geometric shapes to describe objects – for example, model a tree trunk or a human torso as a cylinder.
• Apply concepts of density based on area and volume – for example, persons per square mile, BTUs per cubic foot.
• Design objects or structures to satisfy specific physical constraints or minimize cost.
Numbers
Overview
For high school students, math skills and understandings are organized not by grade level but by concept. In High School Math: Number and Quantity, students extend their understanding of number to imaginary numbers and complex numbers, and work with a variety of measurement units in modeling. Emphasis is on using numbers – in calculations, equations, and measurements – to solve real-world problems, including those that students themselves quantify and define.
Rational and irrational numbers
Understand and explain why:
• the sum of two rational numbers is rational (sum can be written as a fraction or decimal)
• the sum of a rational number and an irrational number is irrational (sum cannot be written as a fraction; written in decimal form, is non-repeating and unending)
Interpreting and converting units
Consistently choose and interpret units in formulas; scale drawings and figures in graphs, data displays and maps. Convert rates and measurements (grams to centigrams, inches to feet, meters to kilometers, miles to kilometers, square inches into square feet, etc.).
Real-world problems
Use measurement units in modeling to solve real-world problems – for example: acceleration, currency conversions, per capita income, safety statistics, disease incidence, batting averages, etc.)
Complex numbers
Understand that complex numbers are formed by real numbers and imaginary numbers – imaginary numbers that, when squared, give a negative result: i&2sup; = -1. Use the relation i&2sup; = -1 to add, subtract and multiply complex numbers.
Understanding vectors
Understand a vector as a quantity that has both magnitude (length) and direction. Add and subtract vectors.
Velocity
Solve problems involving velocity and other quantities represented by vectors.
Example:
• Drew leaves home for a morning walk. He goes 13.5 km south and 5.5 km west. What is his velocity relative to his brother, who is still asleep in bed at home?
• Jack is doing push-ups. Which requires smaller muscular force – if his hands are 0.25m apart, or his hands are 0.5m apart? |
# A n g l e s a n d T h e i r M e a s u r e. An angle is formed by joining the endpoints of two half-lines called rays. The side you measure from is called.
## Presentation on theme: "A n g l e s a n d T h e i r M e a s u r e. An angle is formed by joining the endpoints of two half-lines called rays. The side you measure from is called."— Presentation transcript:
A n g l e s a n d T h e i r M e a s u r e
An angle is formed by joining the endpoints of two half-lines called rays. The side you measure from is called the initial side. Initial Side The side you measure to is called the terminal side. Terminal Side This is a counterclockwise rotation. This is a clockwise rotation. Angles measured counterclockwise are given a positive sign and angles measured clockwise are given a negative sign. Positive Angle Negative Angle
It’s Greek To Me! It is customary to use small letters in the Greek alphabet to symbolize angle measurement. alpha betagamma theta phi delta
We can use a coordinate system with angles by putting the initial side along the positive x-axis with the vertex at the origin. positive Initial Side T e r m i n a l S i d e negative We say the angle lies in whatever quadrant the terminal side lies in. Quadrant I angle Quadrant II angle Quadrant IV angle If the terminal side is along an axis it is called a quadrantal angle.
We will be using two different units of measure when talking about angles: Degrees and Radians Let’s talk about degrees first. You are probably already somewhat familiar with degrees. If we start with the initial side and go all of the way around in a counterclockwise direction we have 360 degrees You are probably already familiar with a right angle that measures 1/4 of the way around or 90° = 90° If we went 1/4 of the way in a clockwise direction the angle would measure -90° = - 90° = 360°
= 45° What is the measure of this angle? You could measure in the positive direction = - 360° + 45° You could measure in the positive direction and go around another rotation which would be another 360 ° = 360° + 45° = 405° You could measure in the negative direction There are many ways to express the given angle. Whichever way you express it, it is still a Quadrant I angle since the terminal side is in Quadrant I. = - 315°
If the angle is not exactly to the next degree it can be expressed as a decimal (most common in math) or in degrees, minutes and seconds (common in surveying and some navigation). 1 degree = 60 minutes1 minute = 60 seconds = 25°48'30" degrees minutes seconds To convert to decimal form use conversion fractions. These are fractions where the numerator = denominator but two different units. Put unit on top you want to convert to and put unit on bottom you want to get rid of. Let's convert the seconds to minutes 30"= 0.5'
1 degree = 60 minutes1 minute = 60 seconds = 25°48'30" Now let's use another conversion fraction to get rid of minutes. 48.5'=.808° = 25°48.5'= 25.808°
initial side terminal side radius of circle is r r r arc length is also r r This angle measures 1 radian Given a circle of radius r with the vertex of an angle as the center of the circle, if the arc length formed by intercepting the circle with the sides of the angle is the same length as the radius r, the angle measures one radian. Another way to measure angles is using what is called radians.
Arc length s of a circle is found with the following formula: arc lengthradiusmeasure of angle IMPORTANT: ANGLE MEASURE MUST BE IN RADIANS TO USE FORMULA! Find the arc length if we have a circle with a radius of 3 meters and central angle of 0.52 radian. 3 = 0.52 arc length to find is in black s = r 3 0.52 = 1.56 m What if we have the measure of the angle in degrees? We can't use the formula until we convert to radians, but how? s = r
We need a conversion from degrees to radians. We could use a conversion fraction if we knew how many degrees equaled how many radians. Let's start with the arc length formula s = r If we look at one revolution around the circle, the arc length would be the circumference. Recall that circumference of a circle is 2 r 2 r = r cancel the r's This tells us that the radian measure all the way around is 2 . All the way around in degrees is 360°. 2 = 2 radians = 360°
Convert 30° to radians using a conversion fraction. 30° The fraction can be reduced by 2. This would be a simpler conversion fraction. 180° radians = 180° Can leave with or use button on your calculator for decimal. Convert /3 radians to degrees using a conversion fraction. = radians 0.52 = 60°
Area of a Sector of a Circle The formula for the area of a sector of a circle (shown in red here) is derived in your textbook. It is: r Again must be in RADIANS so if it is in degrees you must convert to radians to use the formula. Find the area of the sector if the radius is 3 feet and = 50° = 0.873 radians
A Sense of Angle Sizes See if you can guess the size of these angles first in degrees and then in radians. You will be working so much with these angles, you should know them in both degrees and radians.
Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.www.mathxtc.com Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au
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Home >> Numbers >> Order of Numbers >> Descending order of Numbers >>
Descending order of Numbers
Ascending order of Numbers Descending order of Numbers
Definition
By Descending Order, we mean the arrangement of whole numbers from the largest to smallest.
We can also say that in Descending Order, the largest whole number comes first and the smallest whole number comes in the last.
Following examples further illustrate Descending Order of Whole Numbers
Example 1 = Arrange the following series of whole numbers in descending order.
87, 57, 29 ,70, 15, 35, 90
Answer = In the above given series of whole number,
The Largest Whole Number is 90, so put it in First Place in the order.
Now, 87 is smaller than 90; but larger than rest of the whole numbers of the given series, so put it in Second Place in the order.
Similarly, put 70 in Third Place,
57 in Fourth Place,
35 in Fifth Place,
29 in Sixth Place
and 15 in The Last as it is The Smallest Whole Number of the given series.
Finally, The Descending Order of the given series of Whole Numbers = 90, 87, 70, 57, 35, 29, 15
Example 2 = Arrange the following series whole numbers in descending order.
50, 45, 71, 48, 79, 55
Answer = In the above given series of whole numbers,
The Largest Whole Number is 79, so put it in First Place in the order.
Now, 71 is smaller than 79; but larger than rest of the whole numbers of the given series, so put it in Second Place in the order.
Similarly, put 55 in Third Place,
50 in Fourth Place,
48 in Fifth Place,
and 45 in The Last as it is The Smallest Whole Number of the given series.
Finally, The Descending Order of the given series of Whole Numbers = 79, 71, 55, 50, 48, 45. |
# HCF and LCM
1) Factors & Multiples– if a number A divides another number B, exactly we say that A is a factor of B. In this case B is called a multiple of A.
2) Highest common factor or a greatest common measure (CGM) or greatest common divisor. The H.C.F of two or more than two numbers is the greatest number that divides each of them exactly.
Method- factorization- express each of the given number as the product of prime factors the product of least power of common prime factors gives H.C.F.
12 = 2×2×3
18 = 2×3×3 H.C.F 2×3 = 6
Division method- to find the H.C.F of two given number divide the larger number by smaller one. now divide the divisor by the reminder, repeat the process of dividing, the proceeding number by the reminder last obtained till zero is obtained as reminder. The last divisor is the required H.C.F same procedure for more than 2 number.
e.g- H.C.F of 513, 1134& 1215
1134 ) 1215 ( 1 H.C.F of 1134 & 1215 = 81
__1134_____
81 ) 1134 ( 14
____1134____
××
H.C.F of 513 & 31
H.C.F of 513 & 81 = 27
Least common multiple (LCM) – the least number which is exactly divisible by each one of the given number is called their L.C.M.
Method- factorization- resource each one of the given numbers into a product of prime factor then L.C.M is the product of highest power so fall the factors.
e.g. 72, 108 & 2100
Common division method (shortcut method)
Arrange the given number in a row in any order, divide by a number which divides exactly at least two of the given number and carry forwarded the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible. By the same number except the product of the divisor & the co divided numbers is the repaired L.C.M.
e.g- L.C.M of 16, 24, 36, 54
4) Product of two numbers = product of their H.C.F & L.C.M
Example
5) co- primes– the two numbers are said to be co-prime. If their H.C.F is 1
6) H.C.F & L.C.M of fractions
Example , , ,
L.C.M of 3, 9,27 81=81
H.C.F & L.C.M of decimal fraction– the given no make the same number of decimal places by annexing zero in some numbers if necessary considering these numbers without decimal point. Find H.C.F & L.C.M as the case maybe.
e.g. L.C.M & H.C.F of 0.63, 1.05 & 2.10
Without decimal- 63, 105, 210
HCF of 63, 105, 210 is 21
HCF of 0.63,0.105, 2.10 is 0.21
L.C.M of 63, 105, 210 is 630
L.C.M of 0.63, 1.05, 2.10 is 6.30
8) Comparison of Fraction– Find the L.C.M of denominator & convert them equivalent fraction with L.C.M as the denominator by multiples both the numerator & denominator.
Example
then
Typical Question:- Find the largest number which divided 62, 132, 237 to leave the same remainder in each case.
Reqd. number = H.C.F of (132-62), (237-132), & (237-62)
= H.C.F of 70,105 & 175 = 35 |
## Cross Multiplication Method Questions 1
In this page Cross Multiplication Method Questions 1 we are going to see solution for first problem of the worksheet cross multiplication method.
While solving a pair of linear equations in two unknowns x and y using elimination method,we have to think about how to eliminate any one of the variable in two equations to get value of one unknown and we have to plug it in the second to get the value of other variable. There is another method called cross multiplication which simplifies the procedure.
Solve the following system of equations using cross multiplication method
(i) 3 x + 4 y = 24, 20 x - 11 y = 47
Solution:
First we have to make the given equations in the form of a₁ x + b₁ y + c₁ = 0,a₂ x + b₂ y + c₂ = 0.
3 x + 4 y - 24 = 0 ----- (1)
20 x - 11 y - 47 = 0 ----- (2)
x/(-188-264) = y/(-480 -(-141)) = 1/(-33-80)
x/(-452) = y/(-480+141)) = 1/(-33-80)
x/(-452) = y/(-339) = 1/(-113)
x/(-452) = 1/(-113) y/(-339) = 1/(-113)
x = (-452)/(-113) y = (-339)/(-113)
x = 4 y = 3
Therefore solution is (4,3).
Verification:
Now let us apply the answer that we got in the first or second equation to check whether we got correct answer or not.
3 x + 4 y = 24
3(4) + 4(3) = 24
12 + 12 = 24
24 = 24 cross multiplication method question1 cross multiplication method question1
Formulate the following problems as a pair of equations,and hence find their solutions:
(i) One number is greater than thrice the other number by 2. If 4times the smaller number exceeds the greater by 5, find the numbers.
(ii) The ratio of income of two persons is 9:7 and the ratio of their expenditure is 4:3.If each of them manages to save \$2000 per month, find their monthly income.
(iii) A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.
(iv) Three chairs and two tables cost \$700 and five chairs and three tables cost \$1100.What is the total cost of 2 chairs and 3 tables.
(v) In a rectangle,if the length is increased and breadth is reduced each by 2 cm then the area is reduced by 28 cm². If the length is reduced by 1 cm and the breadth is increased by 2 cm,then the area increases by 33cm². Find the area of the rectangle.
(vi) A train traveled a certain distance at a uniform speed. If the train had been 6 km/hr faster,it would have taken 4 hours less than the scheduled time. If the train were slower by 6km/hr, then it would have taken 6 hours more than the scheduled time. Find the distance covered by the train. |
# Work-energy theorem: Definition, Equation, Derivation, Examples
We all know that in order to throw something at great speed, we must first perform some work on it. In such instances, the work-energy theorem, also known as the work-energy principle, can provide the relationship between the work done on the item and its speed.
This article explains the work-energy theorem in detail and includes some practice numerical examples for a clear understanding of the topic.
Contents:
## What is the Work-energy theorem?
The work-energy theorem states that the net work done by the forces on the object is equal to the change in kinetic energy of the object. It states the relationship between the net work done on an object and the change in the kinetic energy of an object.
This theorem obeys the law of energy conservation.
Let me explain it with the example of golf.
The golfer in the above figure tries to push the ball by applying force (F) on the ball for the distance of ‘x’.
The vertical component of the force (Fsinθ) is not responsible for doing work on the ball. To push the golf ball, the horizontal component of the force (Fcosθ) in the direction of the displacement does the work which is given by,
W = F.cosθ.x
During the interaction period between the golf stick and the ball, the work done by the golf stick gets transformed to raise the velocity of the golf ball. It means that the net work done onto the ball is responsible for the change in kinetic energy of the ball.
\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}= W
KE_{2}-KE_{1} = W
It means that the change in kinetic energy of the ball (ΔKE) is equal to the W_{\text{net}} on the ball.
## Work-energy theorem derivation:
Consider an object of mass ‘m’ moving at velocity V_{1} is subjected to a variable force ‘F’.
During the smaller time interval of ‘dt’, the object shows a smaller displacement dx. Therefore the small work done by force on the object is given by,
dw = F.dx
dw = m.a.dx \cdots \cdots [ \because F = ma]
dw = m.\frac{dv}{dt}.dx \cdots \cdots [ \because a = \frac{dv}{dt}]
dw = m\frac{dx}{dt}.dv
dw = m.v.dv \cdots[ \because \frac{dx}{dt} = v]
Integrate the equation to find the work done to change the velocity of the object from V_{1} to V_{2}.
W = \int dw = \int_{V_{1}}^{V_{2}}m.v.dv
W = m \int_{V_{1}}^{V_{2}}v.dv
W = m[\frac{v^{2}}{2}]_{V_{1}}^{V_{2}}
W = m[\frac{V_{2}^{2}}{2}-\frac{V_{1}^{2}}{2}]
W = \frac{mV_{2}^{2}}{2} – \frac{mV_{1}^{2}}{2}
W = KE_{2}-KE_{1}
W = \DeltaKE
Thus it proves that the work done by the force onto the object is equal to the change in kinetic energy of the object.
Alternate method:-
As shown in the above figure, the net force F acting on the object causes to change its velocity. Thus the work done by force on an object to move it from position 1 to 2 is given by,
W = F.S
According to one of the newtons equations of linear motion,
V_{2}^{2} =V_{1}^{2} + 2aS
∴ S = \frac{V_{2}^{2}-V_{1}^{2}}{2a}
Put this value of S in the equation of work.
W = F \times \frac{V_{2}^{2}-V_{1}^{2}}{2a}
W = m.a \times \frac{V_{2}^{2}-V_{1}^{2}}{2a} \cdots [ \because F = ma]
W = \frac{mV_{2}^{2}}{2}-\frac{mV_{1}^{2}}{2}
W = KE_{2}-KE_{1}
W = \Delta KE
Therefore it proves the work energy theorem.
## Work-energy theorem solved examples:
1] A bullet of a mass of 10 grams hits the wooden object at the velocity of 790 m/s and it penetrates the object up to 100 mm to stop. By assuming the force applied by the bullet is constant, find the average force applied by the bullet on the object.
Given:-
m = 10 g = 0.01 Kg
V_{1} = 790 m/s
V_{2} = 0 m/s
S = 100 mm = 0.1 m
Solution:-
By the work-energy theorem,
Work done by bullet = Change in kinetic energy of the bullet
-(F \times S) = KE_{2}-KE_{1}
- (F \times S) = \frac{mV_{2}^{2}}{2}-\frac{mV_{1}^{2}}{2}
– (F \times 0.1) = \frac{0.01 \times 0^{2}}{2}-\frac{0.01 \times 790^{2}}{2}
F = 31205 N
∴ F = 31.205 KN
This is the magnitude of average force applied by a bullet on the wooden object.
2] The children are pulling two blocks A & B from the rest that has a mass of 30 kg each. After pulling up to the distance of 3 m, the block gains a velocity of 1.5 m/s. If the coefficient of friction between block and ground is 0.3 then find the average uniform force applied by the children.
Given:-
m_{A} = m_{B} = 30 Kg
\text{Weight}, w_{A}=w_{B}=mg =30 \times 9.81=294.3 N
V_{1} = 0 m/s
V_{2} = 1.5 m/s
µ = 0.3
S = 3 m
Solution:-
Free body diagram:-
The net force acting on blocks is given by,
F_{\text{net}} = F-\muw_{B}-\muw_{A}
F_{\text{net}} = F-0.3(294.3)- 0.3(294.3)
F_{\text{net}} = (F-176.58)N
The total change in kinetic energy of the system is given by,
\DeltaKE = KE_{2}-KE_{1}
= \frac{1}{2}(m_{A}+m_{B})V_{2}^{2}-\frac{1}{2}(m_{A}+m_{B})V_{1}^{2}
= \frac{1}{2}(30+30)1.5^{2}-\frac{1}{2}(30+30)0^{2}
\DeltaKE = 67.5 N.m
Now as per the work-energy theorem,
W_{\text{net}} = \DeltaKE
F_{\text{net}} \times S = \DeltaKE
(F – 176.58) x 3 = 67.5
F = 199.08 N
It is the magnitude of the average uniform force applied by children.
## Application:
The work-energy theorem has application in analyzing the motion of a rigid object that is subjected to several motion-affecting forces.
## FAQs:
1. Why is the work-energy theorem important?
The work-energy theorem gives the relationship between the forces applied to the object and the motion of the object.
2. What is the formula of the work-energy theorem?
The work-energy theorem formula is as follows,
Change in kinetic energy = Work
3. When can you use the work-energy theorem?
The work-energy theorem can be used to find the unknown forces that are responsible for the motion or it can be used for the analysis of the motion of the object.
4. How to calculate speed using the work-energy theorem?
From the work-energy theorem, the initial or final speed of the object can be calculated as follows,
Net work = [(1/2)mV₂²] – [(1/2)mV₁²]
Where,
V₂ = Final speed
V₁ = Initial speed
5. How to find distance using the work-energy theorem?
From the work-energy theorem, the distance responsible for the work done can be calculated as follows,
F.d.cosθ = Change in kinetic energy
Where,
F = Net force
d = Distance
θ = Angle between the force and displacement vector
6. What does the work-kinetic energy theory say about the speed of a satellite?
According to the work-energy theorem, since the net work done on the satellite is zero, the change in kinetic energy of the satellite is also zero. That is, the satellite moves at a constant speed.
We included all of the work-energy theorem’s necessary information. I hope you grasped the topic completely. You can get it in PDF format by clicking the button below.
Related post:
Pratik is a Graduated Mechanical engineer. He enjoys sharing the engineering knowledge learned by him with people. |
#### Provide solution for RD Sharma maths class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 6
$B.\; \; -42$
Hint:
We are given the distance travelled x is a function of time, we can calculate velocity V and acceleration $a$ by
$V t=\frac{d x}{d t} \text { and } a(t)=\frac{d^{2} x}{d t^{2}}$
Given:
Here, we use the formula,
$x=t^{3}-12 t^{2}+6 t+8$
Solution:
Differentiating w.r.t time we get
$V(t)=\frac{d x}{d t}=3 t^{2}-24 t+6$
Again Differentiating w.r.t time we get
\begin{aligned} &a(t)=\frac{d^{2} x}{d t^{2}}=6 t-24\\ &a=0 \Rightarrow 6 t-24=0 \text { or } t=4 \text { units }\\ &\text { So, Velocity at } t=4\\ &V(t)=\frac{d x}{d t}=3 t^{2}-24 t+6\\ &V(4)=3 \times 4^{2}-24 \times 4+6\\ &V(4)=(-42) \end{aligned} |
# Calculus 2 : Applications in Physics
## Example Questions
← Previous 1 3 4 5 6
### Example Question #1 : Integral Applications
Determine the length of the following function between
Explanation:
In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:
where ds is given by the equation below:
We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:
Now we can plug this into the given equation to find ds:
Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:
### Example Question #2 : Integral Applications
Find the coordinates of the center of mass for the region bounded by the following two functions:
Explanation:
Before we can set up any sort of integral, we must find out where our two functions intersect, which tells of what the bounds of our region are. To do this, we set our functions equal to each other and solve for the x values at which they intersect:
Our next step is to use the formulas for the x and y coordinates of any region's center of mass, which are given below:
We know the bounds of our region, as well as the functions f(x) and g(x) that make up its upper and lower boundaries, respectively, so the only thing we have left to do is calculate the area of our region using the following integral:
Now that we know the area, bounds, and functions f(x) and g(x) that make up our region, we can simply plug these values into the formulas for the x and y coordinates of the center of mass for the region:
So by evaluating our integrals, we can see that the center of mass of the region bounded by our two functions is .
### Example Question #3 : Integral Applications
Calculate the surface area of the solid obtained by rotating
about the x-axis from to .
Explanation:
Before we can begin the problem, we must remember the formula for finding the surface area of a solid formed from rotation about the x-axis:
where ds is given by the following equation:
So in order to set up our integral for surface area, we must first find ds, which requires us to evaluate the derivative of our function:
In order to add the two terms in our radicand, we must find their common denominator. Because one of our terms is just 1, we can simplify the equation by writing 1 as (16-x^2)/(16-x^2), which allows us to add the two terms together and gives us:
Now we have our expression for ds in terms of x, and we can write the y in the formula for surface area by replacing it with its equation in terms of x given in the problem statement:
### Example Question #1 : Applications In Physics
In physics, the work done on an object is equal to the integral of the force on that object dotted with its displacent.
This looks like ( is work, is force, and is the infinitesimally small displacement vector). For a force whose direction is the line of motion, the equation becomes .
If the force on an object as a function of displacement is , what is the work as a function of displacement ? Assume and the force is in the direction of the object's motion.
Not enough information
Explanation:
, so .
Both the terms of the force are power terms in the form , which have the integral , so the integral of the force is .
We know
.
This means
.
### Example Question #5 : Integral Applications
. Find if . Assume is measured in meters and is measured in seconds.
Explanation:
By the fundamental theorem of calculus, we know that
So,
evaluated from to
### Example Question #181 : Integrals
The velocity of a rocket, in meters per second, seconds after it was launched is modeled by
.
What is the total distance travelled by the rocket during the first four seconds of its launch?
Explanation:
The distance traveled by an object over an interval of time is the total area under the velocity curved during the interval. Therefore, we need to evaluate the definite integral
Rewriting expressions written using radical notation using fractional exponents can make applying the power rule to find the antiderivative easier.
Evaluating this integral using the power rule,
, we find:
### Example Question #7 : Integral Applications
A force of
is applied to an object. How much work is done, in Joules, moving the object from to meters?
Explanation:
The work done moving an object is the intergral of force applied as a function of distance,
.
To find the work done moving the object from x=1 to x=4, evaluate the definite interval using x=1 and x=4 as the limits of integration. Recall that the antiderivative of a polynomial is found using the power rule,
### Example Question #8 : Integral Applications
The velocity of an moviing object at time is , and the position at time is . The acceleration of the object is modeled by the function for
Explanation:
The position function is the second antiderivative of acceleration. First, integrate the acceration function to find the velocity function. Since the acceleration function is a polynomial, the power rule is needed. Recall that the power rule is
.
.
Next, we can find the the value of the constant using the fact that .
, so . The velocity function is therefore .
Now integrate the velocity function to find the position function.
.
Use the fact that to find the constant
.
The velocity function is
### Example Question #1 : Initial Conditions
The temperature of an oven is increasing at a rate degrees Fahrenheit per miniute for minutes. The initial temperature of the oven is degrees Fahrenheit.
What is the temperture of the oven at ? Round your answer to the nearest tenth.
Explanation:
Integrating over an interval will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral
to find the change in temperature that occurs during the first five minutes.
A substitution is useful in this case. Let. We should also express the limits of integration in terms of . When , and when Making these substitutions leads to the integral
.
To evaluate this, you must know the antiderivative of an exponential function.
In general,
.
Therefore,
This tells us that the temperature rose by approximately degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at minutes is
degrees.
### Example Question #2 : Initial Conditions
Find the equation for the velocity of a particle if the acceleration of the particle is given by:
and the velocity at time of the particle is
Explanation:
In order to find the velocity function, we must integrate the accleration function:
We used the rule
to integrate.
Now, we use the initial condition for the velocity function to solve for C. We were told that
so we plug in zero into the velocity function and solve for C:
C is therefore 30.
Finally, we write out the velocity function, with the integer replacing C:
← Previous 1 3 4 5 6 |
# What Is The Slope In Standard Form?
## How do you find the slope and y intercept of a word problem?
In the equation of a straight line (when the equation is written as “y = mx + b”), the slope is the number “m” that is multiplied on the x, and “b” is the y-intercept (that is, the point where the line crosses the vertical y-axis).
This useful form of the line equation is sensibly named the “slope-intercept form”..
## What does C stand for in standard form?
Standard Form: the standard form of a line is in the form Ax + By = C where A is a positive integer, and B, and C are integers. Discussion. The standard form of a line is just another way of writing the equation of a line.
## How do I find the slope of the line?
The slope of a line characterizes the direction of a line. To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points.
## What is an example of point slope form?
Examples of Applying the Concept of Point-Slope Form of a Line. … The slope is given as m = 3 m = 3 m=3, and the point (2,5) has coordinates of x 1 = 2 {x_1} = 2 x1=2 and y 1 = 5 {y_1} = 5 y1=5. Now plug the known values into the slope-intercept form to get the final answer.
## What is standard form in math?
Standard form is a way of writing down very large or very small numbers easily. 103 = 1000, so 4 × 103 = 4000 . This idea can be used to write even larger numbers down easily in standard form. … Small numbers can also be written in standard form.
## How do you find the slope and y intercept?
The slope-intercept form of a line is: y=mx+b where m is the slope and b is the y-intercept. The y-intercept is always where the line intersects the y-axis, and will always appear as (0,b) in coordinate form.
## How do you find slope in standard form?
When we find slope in standard form, we take the A term, divide by B, then change the sign (or we just say -A/B. In this example, we have -3/-1, which is the same as 3/1, or 3.
## Is Standard Form point-slope form?
Remember, the point-slope formula is only one type of linear equation. … The standard form of an equation is Ax + By = C. In this kind of equation, x and y are variables and A, B, and C are integers.
## How do you find standard form?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.
## What does Standard Form tell you?
An equation written in standard form is yet another equation that forms a parabola when graphed. Each letter in the standard form equation tells us a piece of information about the parabola, just like the letters from the vertex form equation had. “a”, can also tell us the width of a parabola. …
## What is the rule for standard form?
The Standard Form for a linear equation in two variables, x and y, is usually given as Ax + By = C where, if at all possible, A, B, and C are integers, and A is non-negative, and, A, B, and C have no common factors other than 1.
## How do you find the vertex in standard form?
y=ax2+bx+c . In this equation, the vertex of the parabola is the point (h,k) . You can see how this relates to the standard equation by multiplying it out: y=a(x−h)(x−h)+ky=ax2−2ahx+ah2+k .
## What is the Y intercept in standard form?
The y-intercept is on the y-axis, where x = 0. Plug x = 0 into the equation and solve for y.
## What are the 3 slope formulas?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form. We review all three in this article. |
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# Presentation 20:
Presentation 20:. UNIT TRIANGLE AND RAFTER THEORY. Unit Triangle. Unit Triangle is found on the floor plans. It may be in many shapes. Unit Triangle vs. House. Doubling the Unit Triangle gives a shape that is similar to the building, but smaller. Triangle Names.
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## Presentation 20:
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### Presentation Transcript
1. Presentation 20: UNIT TRIANGLE AND RAFTER THEORY
2. Unit Triangle • Unit Triangle is found on the floor plans. • It may be in many shapes.
3. Unit Triangle vs. House • Doubling the Unit Triangle gives a shape that is similar to the building, but smaller.
4. Triangle Names • The sides of the triangles have names. Length Rise Run Length Rise Run Each side of the unit triangle has UNIT added to it.
5. Triangle Names • Thus the Unit Triangle Unit Length Unit Rise Unit Triangle Unit Run Building
6. 12′- 0″ Unit/House Relationship • How many units of run will fit into the building triangle? This building has six unit triangles. It depends on the size of the building.
7. Unit/House Relationship • How many units of rise will fit into the building triangle? In this building, six unit triangles.
8. Unit/House Relationship • How many units of length will fit into the building triangle? Again, six unit triangles.
9. 12′- 0″ Unit/House Relationship • There are equal numbers of unit triangles along the run, rise, and length. If this building is 12′ wide, then six unit triangles fit under the rafter.
10. Unit Triangle Theory • The base of the unit triangle is always 12″ for a common rafter. • The unit rise is given on the house plans. • If the unit rise is 6″, 13.42″ • Then the unit length is … 6″ Pythagorean Theorem 12″ = 1 foot = 1 unit of run or Rafter Tables
11. 20′- 0″ Example • If building is 20′ wide and unit rise is 6″, what is the rafter Total Rise and Line Length? 6″
12. 134 3/16″ 6″ 134 3/16″ 60″ 10′- 0″ = 10 units of run Rafter Calculation Answers • Total Rise = Unit Rise x RUN 6″ x 10 = 60″ • Line Length = Unit Length x RUN 13.42″ x 10 = 134.2″ = 134 3/16”
13. Conclusions 12″ • Unit run for a common rafter is ____. on the floor plans • The unit rise is found _____________. • Unit rise is converted to unit length by _______________________. using the Pythagorean Theorem Total Rise • Run x unit rise = _____________. Line Length • Run x unit length = _____________.
More Related |
# Find the following product:
Question:
Find the following product:
$-\frac{8}{27} x y z\left(\frac{3}{2} x y z^{2}-\frac{9}{4} x y^{2} z^{3}\right)$
Solution:
To find the product, we will use the distributive law in the following way:
$-\frac{8}{27} x y z\left(\frac{3}{2} x y z^{2}-\frac{9}{4} x y^{2} z^{3}\right)$
$=\left\{\left(-\frac{8}{27} x y z\right)\left(\frac{3}{2} x y z^{2}\right)\right\}-\left\{\left(-\frac{8}{27} x y z\right)\left(\frac{9}{4} x y^{2} z^{3}\right)\right\}$
$=\left\{\left(-\frac{8}{27} \times \frac{3}{2}\right)(x \times x) \times(y \times y) \times\left(z \times z^{2}\right)\right\}-\left\{\left(-\frac{8}{27} \times \frac{9}{4}\right)(x \times x) \times\left(y \times y^{2}\right) \times\left(z \times z^{3}\right)\right\}$
$=\left\{\left(-\frac{8}{27} \times \frac{3}{2}\right)\left(x^{1+1} y^{1+1} z^{1+2}\right)\right\}-\left\{\left(-\frac{8}{27} \times \frac{9}{4}\right)\left(x^{1+1} y^{1+2} z^{1+3}\right)\right\}$
$=-\frac{4}{9} x^{2} y^{2} z^{3}+\frac{2}{3} x^{2} y^{3} z^{4}$
Thus, the answer is $-\frac{4}{9} x^{2} y^{2} z^{3}+\frac{2}{3} x^{2} y^{3} z^{4}$. |
Math Blog Post- Parts of a Circle
We just started a chapter devoted to circles in Geometry, and I decided to show you what I’m learning.
A Chord’s endpoints lie in the circle.
A Secant goes straight through the circle.
A Tangent only touches the circle at one point.
The Point of the Tangent is where the tangent touches the circle.
Now, try to classify all of the lines in this problem:
Chords- AB, AC, AD
Secant- BC
Tangent- E
Point of the Tangent- E
Diameter- BC
Math Blog Post
In Geometry, we finished chapter about Spatial Reasoning. Spatial Reasoning is all about using formulas to find the lateral area, surface area, and volume of 3-D figures. First, we had to determine what figure could be made from a net, so I made a few of those problems for you to solve:
For the top one, the answer is a cylinder, because the two circles go on top of the rectangle to make a cylinder.
For to bottom one, the answer is a pentagonal pyramid, because the triangles go up to make a pyramid over the pentagon, which is the base.
Math Blog Post
One lesson that we learned in Geometry this week is estimating the area of irregular shapes. I made a problem below:
You have to count the squares and the half-squares to make an estimate.
Did you get it yet?
The answer is 37.
For this week’s Math Blog Post, I drew a picture of how you can solve the height of a shadow. It doesn’t have to be just used for shadows, however.
Let me explain:
Step 1: Convert the feet into inches by multiplying by 12. Fred is 5 feet, 2 inches, which is 62 inches. The building is 50 feet, 2 inches, which is 602 inches. The building’s shadow is 60 feet, which is 720 inches. Remember that we are solving for Fred’s shadow.
Step 2: Put the inches into a formula:
Fred’s height Building’s height
__________ = _____________
The above formula is put into effect in the picture above. When you multiply 62 (Fred’s height) by 720 (Building’s shadow), you get 44,640.
Step 3: Divide.
Divide 44,640 by 600, and you get 74.4. But that’s not the answer. You must divide 74.4 by 12 to put it into feet and inches. 74.4 divided by 12 is 6.2, which is rounded to 6 feet, 2.5 inches.
Math Blog Post- The Triangle Inequality Theorem
The Triangle Inequality Theorem goes as follows:
“The sum of any two side lengths of a triangle is greater than third side length.”
This means that A and B must be greater than C. If it isn’t greater, then the triangle’s sides don’t match.
For example: A is equal to 4, B is equal to 3, and C is equal to 7.
This WOULD NOT be a complete triangle because 4+3=7.
Let’s try another one: A is equal to 34, B is equal to 21, and C is equal to 50.
This WOULD be a complete triangle because 34+21=55, and 55 is more than 50.
Got it? Good!
Two Triangle Theorms- Math Blog Post
Today I will be explaining two geometry theroems- the Isosceles Triangle Theorm and the Equiangular Triangle Theorm.
Isosceles Triangle Theorm– This theorm states, “If two sides of a triangle are congruent, then the angles opposite it are congruent.” This theorm can be used the prove congruence in isosceles triangles.
Equiangular Triangle Theorm– This theorm says, “If a triangle is equilaterial, then it’s also equiangular.” It can be used for algebra problems within geometry. |
# SSAT Upper Level Math : How to find the part from the whole
## Example Questions
### Example Question #1 : How To Find The Part From The Whole
What is of ?
Explanation:
Multiply by to find the answer.
.
### Example Question #2 : How To Find The Part From The Whole
Susie won from the lottery. If she has to pay of the winnings to taxes and gives her parents of the remaining amount, how much does she have left over?
Explanation:
First, find out how much she has left after she pays her taxes.
Now, take away of the after-tax amount to find out how much she has left.
### Example Question #3 : How To Find The Part From The Whole
Kerry spends of his monthly salary on rent, on food, and on movies. If he spent on rent, how much did he spend on food?
Explanation:
First, find out how much Kerry makes per month. Let be his monthly salary. We know that he spends of his monthly salary on rent, and that he spends on rent. Knowing this, we can set up an equation and solve for , how much Kerry makes per month.
Now, we can find out how much he spends on food.
### Example Question #4 : How To Find The Part From The Whole
A class of students was polled. If of them claimed that they enjoy eating sushi, how many students enjoy eating sushi?
Explanation:
Multiply the total number of students by the fraction to find out how many of them enjoy eating sushi.
students in the class enjoy eating sushi.
### Example Question #5 : How To Find The Part From The Whole
A car is on sale for off. If the original price is , how much does the car cost when it is on sale?
Explanation:
Subtract the discount from the original price of the car.
The discounted car costs .
### Example Question #6 : How To Find The Part From The Whole
Jeremy bought worth of toothpaste. At the checkout, he found that he had a coupon that would take off the cost of the toothpaste. How much will Jeremy pay for the toothpaste if he uses the coupon?
Explanation:
Subtract the amount the coupon would take off from the total amount the toothpaste would cost.
If Jeremy uses the coupon, the toothpaste will cost .
### Example Question #7 : How To Find The Part From The Whole
The tax on the sale of a house worth was . What was the tax rate?
Explanation:
To find the tax rate, divide the tax amount by the total worth of the house.
Now, convert this decimal into a percentage.
The tax rate on this house sale is .
### Example Question #8 : How To Find The Part From The Whole
If Hannah needs to pay a tax on her yearly income of , how much does she need to pay for the tax?
Explanation:
Multiply by to find the amount Hannah needs to pay in taxes.
Hannah needs to pay in taxes.
### Example Question #9 : How To Find The Part From The Whole
A dress goes on sale for off. If the dress originally cost , how much does it cost while it is on sale?
Explanation:
Subtract the amount discounted from the original cost of the dress to find its sale price.
The dress cost on sale.
### Example Question #182 : Number Concepts And Operations
Initially, number of students were going to chip in to buy their teacher a present that cost dollars. If students decide not to pay, which of the following expressions represent how much each remaining student must now pay? |
Rs Aggarwal 2020 2021 Solutions for Class 9 Maths Chapter 7 Lines And Angles are provided here with simple step-by-step explanations. These solutions for Lines And Angles are extremely popular among Class 9 students for Maths Lines And Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 9 Maths Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.
#### Page No 198:
(i) Two rays OA and OB, with a common end-point O, form an angle AOB that is represented as $\angle AOB$.
(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.
(iii) An angle greater than $90°$ but less than $180°$ is called an obtuse angle.
(iv) An angle greater than $180°$ but less than $360°$ is called a reflex angle.
(v) Two angles are said to be complementary if the sum of their measures is $90°$.
(vi) Two angles are said to be supplementary if the sum of their measures is $180°$.
#### Page No 198:
Two angles whose sum is 90° are called complementary angles.
(i) Complement of 55° = 90° − 55° = 35°
(ii) Complement of $16°=\left(90-16\right)°$$=74°$
(iii) Complement of 90° = 90° − 90° = 0°
(iv)
#### Page No 198:
Two angles whose sum is 180° are called supplementary angles.
(i) Supplement of 42° = 180° − 42° = 138°
(ii) Supplement of 90° = 180° − 90° = 90°
(iii) Supplement of 124° = 180° − 124° = 56°
(iv)
Supplement of
#### Page No 198:
(i) Let the measure of the required angle be $x°$.
Then, in case of complementary angles:
$x+x=90°\phantom{\rule{0ex}{0ex}}⇒2x=90°\phantom{\rule{0ex}{0ex}}⇒x=45°$
Hence, measure of the angle that is equal to its complement is $45°$.
(ii) Let the measure of the required angle be $x°$.
Then, in case of supplementary angles:
$x+x=180°\phantom{\rule{0ex}{0ex}}⇒2x=180°\phantom{\rule{0ex}{0ex}}⇒x=90°$
Hence, measure of the angle that is equal to its supplement is $90°$.
#### Page No 198:
Let the measure of the required angle be $x°$.
Then, measure of its complement $=\left(90-x\right)°$.
Therefore,
$x-\left(90°-x\right)=36°\phantom{\rule{0ex}{0ex}}⇒2x=126°\phantom{\rule{0ex}{0ex}}⇒x=63°$
Hence, the measure of the required angle is $63°$.
#### Page No 198:
Let the measure of the angle be x°.
∴ Supplement of x° = 180° − x°
It is given that,
(180° − x°) − x° = 30°
⇒ 180° − 2x°= 30°
⇒ 2x° = 180° − 30° = 150°
x° = 75°
Thus, the measure of the angle is 75°.
#### Page No 198:
Let the measure of the required angle be $x$.
Then, measure of its complement $=\left(90°-x\right)$.
Therefore,
$x=\left(90°-x\right)4\phantom{\rule{0ex}{0ex}}⇒x=360°-4x\phantom{\rule{0ex}{0ex}}⇒5x=360°\phantom{\rule{0ex}{0ex}}⇒x=72°$
Hence, the measure of the required angle is $72°$.
#### Page No 198:
Let the measure of the required angle be $x$.
Then, measure of its supplement $=\left(180°-x\right)$.
Therefore,
$x=\left(180°-x\right)5\phantom{\rule{0ex}{0ex}}⇒x=900°-5x\phantom{\rule{0ex}{0ex}}⇒6x=900°\phantom{\rule{0ex}{0ex}}⇒x=150°$
Hence, the measure of the required angle is $150°$.
#### Page No 198:
Let the measure of the required angle be $x°$.
Then, measure of its complement $=\left(90-x\right)°$.
And, measure of its supplement$=\left(180-x\right)°$.
Therefore,
$\left(180-x\right)=4\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒180-x=360-4x\phantom{\rule{0ex}{0ex}}⇒3x=180\phantom{\rule{0ex}{0ex}}⇒x=60$
Hence, the measure of the required angle is $60°$.
#### Page No 198:
Let the measure of the required angle be $x°$.
Then, the measure of its complement $=\left(90-x\right)°$.
And the measure of its supplement$=\left(180-x\right)°$.
Therefore,
$\left(90-x\right)=\frac{1}{3}\left(180-x\right)\phantom{\rule{0ex}{0ex}}⇒3\left(90-x\right)=\left(180-x\right)\phantom{\rule{0ex}{0ex}}⇒270-3x=180-x\phantom{\rule{0ex}{0ex}}⇒2x=90\phantom{\rule{0ex}{0ex}}⇒x=45$
Hence, the measure of the required angle is $45°$.
#### Page No 198:
Let the two angles be 4x and 5x, respectively.
Then,
$4x+5x=90\phantom{\rule{0ex}{0ex}}⇒9x=90\phantom{\rule{0ex}{0ex}}⇒x=10°$
Hence, the two angles are .
#### Page No 198:
Two angles whose sum is 90° are called complementary angles.
It is given that the angles (2x – 5)° and (x – 10)° are the complementary angles.
∴ (2x – 5)° + (x – 10)° = 90°
⇒ 3x° – 15° = 90°
⇒ 3x° = 90° + 15° = 105°
⇒ x° = $\frac{105°}{3}$ = 35°
Thus, the value of x is 35.
#### Page No 206:
We know that the sum of angles in a linear pair is $180°$.
Therefore,
$\angle AOC+\angle BOC=180°\phantom{\rule{0ex}{0ex}}⇒62°+x°=180°\phantom{\rule{0ex}{0ex}}⇒x°=\left(180°-62°\right)\phantom{\rule{0ex}{0ex}}⇒x=118°\phantom{\rule{0ex}{0ex}}$
Hence, the value of x is $118°$.
#### Page No 206:
As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is $180°$.
Therefore,
$\angle \mathrm{AOC}+\angle \mathrm{COD}+\angle \mathrm{BOD}=180°\phantom{\rule{0ex}{0ex}}⇒\left(3x-7\right)°+55°+\left(x+20\right)°=180\phantom{\rule{0ex}{0ex}}⇒4x=112°\phantom{\rule{0ex}{0ex}}⇒x=28°$
Hence,
$\angle \mathrm{AOC}=3x-7$
$=3×28-7\phantom{\rule{0ex}{0ex}}=77°$
and $\angle \mathrm{BOD}=x+20$
$=28+20\phantom{\rule{0ex}{0ex}}=48°$
#### Page No 207:
AOB is a straight line. Therefore,
$\angle AOC+\angle COD+\angle BOD=180°\phantom{\rule{0ex}{0ex}}⇒\left(3x+7\right)°+\left(2x-19\right)°+x°=180°\phantom{\rule{0ex}{0ex}}⇒6x=192°\phantom{\rule{0ex}{0ex}}⇒x=32°$
Therefore,
#### Page No 207:
Let
XOY is a straight line. Therefore,
$\angle XOP+\angle POQ+\angle YOQ=180°\phantom{\rule{0ex}{0ex}}⇒5a+4a+6a=180°\phantom{\rule{0ex}{0ex}}⇒15a=180°\phantom{\rule{0ex}{0ex}}⇒a=12°$
Therefore,
#### Page No 207:
AOB will be a straight line if
$3x+20+4x-36=180°\phantom{\rule{0ex}{0ex}}⇒7x=196°\phantom{\rule{0ex}{0ex}}⇒x=28°$
Hence, x = 28 will make AOB a straight line.
#### Page No 207:
We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, $\angle AOC=\angle BOD=50°\phantom{\rule{0ex}{0ex}}$
Let $\angle AOD=\angle BOC=x°$
Also, we know that the sum of all angles around a point is $360°$.
Therefore,
$\angle AOC+\angle AOD+\angle BOD+\angle BOC=360°\phantom{\rule{0ex}{0ex}}⇒50+x+50+x=360°\phantom{\rule{0ex}{0ex}}⇒2x=260°\phantom{\rule{0ex}{0ex}}⇒x=130°$
Hence, $\angle AOD=\angle BOC=130°$
Therefore, .
#### Page No 207:
We know that if two lines intersect, then the vertically opposite angles are equal.
$\therefore \angle BOD=\angle AOC=90°\phantom{\rule{0ex}{0ex}}$
Hence, $t=90°$
Also,
$\angle DOF=\angle COE=50°$
Hence, $z=50°$
Since, AOB is a straight line, we have:
$\angle AOC+\angle COE+\angle BOE=180°\phantom{\rule{0ex}{0ex}}⇒90+50+y=180°\phantom{\rule{0ex}{0ex}}⇒140+y=180°\phantom{\rule{0ex}{0ex}}⇒y=40°$
Also,
$\angle BOE=\angle AOF=40°$
Hence, $x=40°$
#### Page No 207:
We know that if two lines intersect, then the vertically-opposite angles are equal.
Since, AOB is a straight line, we have:
$\angle AOE+\angle COE+\angle BOC=180°\phantom{\rule{0ex}{0ex}}⇒3x+5x+2x=180°\phantom{\rule{0ex}{0ex}}⇒10x=180°\phantom{\rule{0ex}{0ex}}⇒x=18°$
Therefore,
$\angle AOD=2×18°=36°\phantom{\rule{0ex}{0ex}}\angle COE=5×18°=90°\phantom{\rule{0ex}{0ex}}\angle AOE=3×18°=54°$
#### Page No 207:
Let the two adjacent angles be 5x and 4x, respectively.
Then,
$5x+4x=180°\phantom{\rule{0ex}{0ex}}⇒9x=180°\phantom{\rule{0ex}{0ex}}⇒x=20°$
Hence, the two angles are .
#### Page No 207:
We know that if two lines intersect, then the vertically-opposite angles are equal.
And let $\angle BOC=\angle AOD=x$
Also, we know that the sum of all angles around a point is $360°$
$\therefore \angle AOC+\angle BOD+\angle AOD+\angle BOC=360°\phantom{\rule{0ex}{0ex}}⇒90°+90°+x+x=360°\phantom{\rule{0ex}{0ex}}⇒2x=180°\phantom{\rule{0ex}{0ex}}⇒x=90°$
Hence, $\angle BOC=\angle AOD=90°$
$\therefore \angle AOC=\angle BOD=\angle BOC=\angle AOD=90°$
Hence, the measure of each of the remaining angles is 90o.
#### Page No 208:
We know that if two lines intersect, then the vertically-opposite angles are equal.
Let $\angle BOC=\angle AOD=x°$
Then,
$x+x=280\phantom{\rule{0ex}{0ex}}⇒2x=280\phantom{\rule{0ex}{0ex}}⇒x=140°\phantom{\rule{0ex}{0ex}}\therefore \angle BOC=\angle AOD=140°$
Also, let $\angle AOC=\angle BOD=y°$
We know that the sum of all angles around a point is $360°$.
$\therefore \angle AOC+\angle BOC+\angle BOD+\angle AOD=360°\phantom{\rule{0ex}{0ex}}⇒y+140+y+140=360°\phantom{\rule{0ex}{0ex}}⇒2y=80°\phantom{\rule{0ex}{0ex}}⇒y=40°$
Hence, $\angle AOC=\angle BOD=40°$
#### Page No 208:
Let ∠AOC = 5k and ∠AOD = 7k, where k is some constant.
Here, ∠AOC and ∠AOD form a linear pair.
∴ ∠AOC + ∠AOD = 180º
⇒ 5k + 7k = 180º
⇒ 12k = 180º
⇒ k = 15º
∴ ∠AOC = 5k = 5 × 15º = 75º
∠AOD = 7= 7 × 15º = 105º
Now, ∠BOD = ∠AOC = 75º (Vertically opposite angles)
∠BOC = ∠AOD = 105º (Vertically opposite angles)
#### Page No 208:
In the given figure,
∠AOC = ∠BOD = 40º (Vertically opposite angles)
∠BOF = ∠AOE = 35º (Vertically opposite angles)
Now, ∠EOC and ∠COF form a linear pair.
∴ ∠EOC + ∠COF = 180º
⇒ (∠AOE + ∠AOC) + ∠COF = 180º
⇒ 35º + 40º + ∠COF = 180º
⇒ 75º + ∠COF = 180º
⇒ ∠COF = 180º − 75º = 105º
Also, ∠DOE = ∠COF = 105º (Vertically opposite angles)
#### Page No 208:
Here, ∠AOC and ∠BOC form a linear pair.
∴ ∠AOC + ∠BOC = 180º
⇒ xº + 125º = 180º
⇒ xº = 180º − 125º = 55º
Now,
∠AOD = ∠BOC = 125º (Vertically opposite angles)
∴ yº = 125º
∠BOD = ∠AOC = 55º (Vertically opposite angles)
∴ zº = 55º
Thus, the respective values of xy and z are 55, 125 and 55.
#### Page No 208:
Let AB and CD be the two lines intersecting at a point O and let ray OE bisect $\angle AOC$. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let .
We know that vertically-opposite angles are equal.
But, $\angle 1=\angle 2$ [Since OE bisects $\angle AOC$ ]
$\therefore \angle 4=\angle 3$
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.
#### Page No 208:
Let AOB denote a straight line and let be the supplementary angles.
Then, we have:
Let .
Then, we have:
Therefore,
$\angle COE+\angle FOC=\frac{1}{2}x+\frac{1}{2}\left(180°-x\right)$
$=\frac{1}{2}\left(x+180°-x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(180°\right)\phantom{\rule{0ex}{0ex}}=90°$
#### Page No 223:
We have, $\angle 1=120°$. Then,
#### Page No 223:
In the given figure, ∠7 and ∠8 form a linear pair.
∴ ∠7 + ∠8 = 180º
⇒ 80º + ∠8 = 180º
⇒ ∠8 = 180º − 80º = 100º
Now,
∠6 = ∠8 = 100º (Vertically opposite angles)
∠5 = ∠7 = 80º (Vertically opposite angles)
It is given that, || m and is a transversal.
∴ ∠1 = ∠5 = 80º (Pair of corresponding angles)
∠2 = ∠6 = 100º (Pair of corresponding angles)
∠3 = ∠7 = 80º (Pair of corresponding angles)
∠4 = ∠8 = 100º (Pair of corresponding angles)
#### Page No 223:
Let ∠1 = 2k and ∠2 = 3k, where k is some constant.
Now, ∠1 and ∠2 form a linear pair.
∴ ∠1 + ∠2 = 180º
⇒ 2k + 3k = 180º
⇒ 5k = 180º
k = 36º
∴ ∠1 = 2k = 2 × 36º = 72º
∠2 = 3k = 3 × 36º = 108º
Now,
∠3 = ∠1 = 72º (Vertically opposite angles)
∠4 = ∠2 = 108º (Vertically opposite angles)
It is given that, || m and is a transversal.
∴ ∠5 = ∠1 = 72º (Pair of corresponding angles)
∠6 = ∠2 = 108º (Pair of corresponding angles)
∠7 = ∠1 = 72º (Pair of alternate exterior angles)
∠8 = ∠2 = 108º (Pair of alternate exterior angles)
#### Page No 223:
For the lines l and m to be parallel
#### Page No 224:
$BC\parallel ED$ and CD is the transversal.
Then,
$AB\parallel CD$ and BC is the transversal.
#### Page No 224:
$EF\parallel CD$ and CE is the transversal.
Then,
Again, $AB\parallel CD$ and BC is the transversal.
Then,
#### Page No 224:
$AB\parallel CD$ and let EF and EG be the transversals.
Now, $AB\parallel CD$ and EF is the transversal.
Then,
Also,
And,
We know that the sum of angles of a triangle is $180°$
#### Page No 224:
(i)
Draw $EF\parallel AB\parallel CD$.
Now, $AB\parallel EF$ and BE is the transversal.
Then,
Again, $EF\parallel CD$ and DE is the transversal.
Then,
(ii)
Draw $EO\parallel AB\parallel CD$.
Then, $\angle EOB+\angle EOD=x°$
Now, $EO\parallel AB$ and BO is the transversal.
Again, $EO\parallel CD$ and DO is the transversal.
Therefore,
(iii)
Draw $EF\parallel AB\parallel CD$.
Then, $\angle AEF+\angle CEF=x°$
Now, $EF\parallel AB$ and AE is the transversal.
Again, $EF\parallel CD$ and CE is the transversal.
Therefore,
#### Page No 225:
Draw $EF\parallel AB\parallel CD$.
$EF\parallel CD$ and CE is the transversal.
Then,
Again, $EF\parallel AB$ and AE is the transversal.
Then,
#### Page No 225:
Given, $AB\parallel PQ\phantom{\rule{0ex}{0ex}}$.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
We know that the sum of angles of a triangle is $180°$.
$\therefore \angle GEF+\angle EGF+\angle EFG=180\phantom{\rule{0ex}{0ex}}⇒85°+x+25°=180°\phantom{\rule{0ex}{0ex}}⇒110°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=70°$
And
#### Page No 225:
$AB\parallel CD$ and AC is the transversal.
Then,
And,
We know that the sum of the angles of a triangle is $180°$.
$\angle ECF+\angle CFE+\angle CEF=180°\phantom{\rule{0ex}{0ex}}⇒105°+30°+x=180°\phantom{\rule{0ex}{0ex}}⇒135°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=45°$
#### Page No 225:
$AB\parallel CD$ and PQ is the transversal.
Then,
And,
Also,
We know that the sum of angles of a triangle is $180°$.
$⇒\angle QGH+\angle GHQ+\angle GQH=180°\phantom{\rule{0ex}{0ex}}⇒95°+65°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=20°\phantom{\rule{0ex}{0ex}}\therefore x=20°$
#### Page No 226:
We know that the sum of the angles of a triangle is $180°$.
#### Page No 226:
Draw $EF\parallel AB\parallel CD$ through E.
Now, $EF\parallel AB$ and AE is the transversal.
Then,
Again, $EF\parallel CD$ and CE is the transversal.
Then,
#### Page No 226:
Draw $PFQ\parallel AB\parallel CD$.
Now, $PFQ\parallel AB$ and EF is the transversal.
Then,
Also, $PFQ\parallel CD$.
#### Page No 226:
In the given figure,
#### Page No 226:
It is given that, AB || CD and is a transversal.
∠BEF + ∠EFD = 180° .....(1) (Sum of the interior angles on the same side of a transversal is supplementary)
EG is the bisector of ∠BEF. (Given)
∴ ∠BEG = ∠GEF = $\frac{1}{2}$∠BEF
⇒ ∠BEF = 2∠GEF .....(2)
Also, FG is the bisector of ∠EFD. (Given)
∴ ∠EFG = ∠GFD = $\frac{1}{2}$∠EFD
⇒ ∠EFD = 2∠EFG .....(3)
From (1), (2) and (3), we have
2∠GEF + 2∠EFG = 180°
⇒ 2(
∠GEF + ∠EFG) = 180°
⇒
∠GEF + ∠EFG = 90° .....(4)
In ∆EFG,
∠GEF + ∠EFG + ∠EGF = 180° (Angle sum property)
⇒ 90° + ∠EGF = 180° [Using (4)]
⇒ ∠EGF = 180° − 90° = 90°
#### Page No 227:
It is given that, AB || CD and t is a transversal.
∴ ∠AEF = ∠EFD .....(1) (Pair of alternate interior angles)
EP is the bisectors of ∠AEF. (Given)
∴ ∠AEP = ∠FEP = $\frac{1}{2}$∠AEF
⇒ ∠AEF = 2∠FEP .....(2)
Also, FQ is the bisectors of ∠EFD.
∴ ∠EFQ = ∠QFD = $\frac{1}{2}$∠EFD
⇒ ∠EFD = 2∠EFQ .....(3)
From (1), (2) and (3), we have
2∠FEP = 2∠EFQ
⇒ ∠FEP = ∠EFQ
Thus, the lines EP and FQ are intersected by a transversal EF such that the pair of alternate interior angles formed are equal.
∴ EP || FQ (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)
#### Page No 227:
It is given that, BA || ED and BC || EF.
Construction: Extend DE such that it intersects BC at J. Also, extend FE such that it intersects AB at H.
Now, BA || JD and BC is a transversal.
∴ ∠ABC = ∠DJC .....(1) (Pair of corresponding angles)
Also, BC || HF and DJ is a transversal.
∴ ∠DJC = ∠DEF .....(2) (Pair of corresponding angles)
From (1) and (2), we have
∠ABC = ∠DEF
#### Page No 227:
It is given that, BA || ED and BC || EF.
Construction: Extend ED such that it intersects BC at G.
Now, BA || GE and BC is a transversal.
∴ ∠ABC = ∠EGC .....(1) (Pair of corresponding angles)
Also, BC || EF and EG is a transversal.
∴ ∠EGC + ∠GEF = 180° .....(2) (Interior angles on the same side of the transversal are supplementary)
From (1) and (2), we have
∠ABC + ∠GEF = 180°
Or ∠ABC + ∠DEF = 180°
#### Page No 227:
AP is normal to the plane mirror OA and BP is normal to the plane mirror OB.
It is given that the two plane mirrors are perpendicular to each other.
Therefore, BP || OA and AP || OB.
So, BP ⊥ AP (OA ⊥ OB)
⇒ ∠APB = 90° .....(1)
In ∆APB,
∠2 + ∠3 + ∠APB = 180° (Angle sum property)
∴ ∠2 + ∠3 + 90° = 180° [Using (1)]
⇒ ∠2 + ∠3 = 180° − 90° = 90°
⇒ 2∠2 + 2∠3 = 2 × 90° = 180° .....(2)
By law of reflection, we have
∠1 = ∠2 and ∠3 = ∠4 .....(3) (Angle of incidence = Angle of reflection)
From (2) and (3), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠BAC + ∠ABD = 180° (∠1 + ∠2 = ∠BAC and ∠3 + ∠4 = ∠ABD)
Thus, the lines CA and BD are intersected by a transversal AB such that the interior angles on the same side of the transversal are supplementary.
∴ CA || BD
#### Page No 227:
Here, ∠BAC = ∠ACD = 110°
Thus, lines AB aand CD are intersected by a transversal AC such that the pair of alternate angles are equal.
∴ AB || CD (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)
Thus, line AB is parallel to line CD.
Also, ∠ACD + ∠CDE = 110° + 80° = 190° ≠ 180°
If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.
Therefore, line AC is not parallel to line DE.
#### Page No 228:
Let the two lines m and n be respectively perpendicular to the two parallel lines p and q.
To prove: is parallel to n.
Proof: Since, m is perpendicular to p
$\therefore$
$\angle 1=90°$
Also, n is perpendicular to
$\therefore$
$\angle 3=90°$
Since p and are parallel and is a transversal line
$\therefore$ $\angle 2=\angle 1=90°$ [Corresponding angles]
Also, $\angle 2=\angle 3=90°$
We know that if two corresponding angles are equal then the two lines containing them must be parallel.
Therefore, the lines m and n are parallel to each other.
#### Page No 231:
Let ∆ABC be such that ∠A = ∠B + ∠C.
In ∆ABC,
∠A + ∠B + ∠C = 180º (Angle sum property)
⇒ ∠A + ∠A = 180º (∠A = ∠B + ∠C)
⇒ 2∠A = 180º
⇒ ∠A = 90º
Therefore, ∆ABC is a right triangle.
Thus, if one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right triangle.
Hence, the correct answer is option (d).
#### Page No 232:
Let the measure of each of the two equal interior opposite angles of the triangle be x.
In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.
∴ xx = 110°
⇒ 2x = 110°
⇒ x = 55°
Thus, the measure of each of these equal angles is 55°.
Hence, the correct answer is option (b).
#### Page No 232:
(a) acute-angled
Let the angles measure .
Then,
$3x+5x+7x=180°\phantom{\rule{0ex}{0ex}}⇒15x=180°\phantom{\rule{0ex}{0ex}}⇒x=12°$
Therefore, the angles are .
Hence, the triangle is acute-angled.
#### Page No 232:
Let ∆ABC be such that ∠A = 130°.
Here, BP is the bisector of ∠B and CP is the bisector of ∠C.
∴ ∠ABP = ∠PBC = $\frac{1}{2}$∠B .....(1)
Also, ∠ACP = ∠PCB = $\frac{1}{2}$∠C .....(2)
In ∆ABC,
∠A + ∠B + ∠C = 180° (Angle sum property)
⇒ 130° + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 180° − 130° = 50°
$\frac{1}{2}$∠B + $\frac{1}{2}$∠C = $\frac{1}{2}$ × 50° = 25°
⇒ ∠PBC + ∠PCB = 25° .....(3) [Using (1) and (2)]
In ∆PBC,
∠PBC + ∠PCB + ∠BPC = 180° (Angle sum property)
⇒ 25° + ∠BPC = 180° [Using (3)]
⇒ ∠BPC = 180° − 25° = 155°
Thus, if one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles is 155°.
Hence, the correct answer is option (d).
#### Page No 232:
It is given that, AOB is a straight line.
∴ 60º + (5xº + 3xº) = 180º (Linear pair)
⇒ 8xº = 180º − 60º = 120º
⇒ xº = 15º
Thus, the value of x is 15.
Hence, the correct answer is option (b).
#### Page No 232:
Suppose ∆ABC be such that ∠A : ∠B : ∠C = 2 : 3 : 4.
Let ∠A = 2k, ∠B = 3and ∠C = 4k, where k is some constant.
In ∆ABC,
∠A + ∠B + ∠C = 180º (Angle sum property)
⇒ 2k + 3k + 4k = 180º
⇒ 9k = 180º
⇒ k = 20º
∴ Measure of the largest angle = 4k = 4 × 20º = 80º
Hence, the correct answer is option (c).
#### Page No 232:
In the given figure, OA || CD.
Construction: Extend OA such that it intersects BC at E.
Now, OE || CD and BC is a transversal.
∴ ∠AEC = ∠BCD = 130° (Pair of corresponding angles)
Also, ∠OAB + ∠BAE = 180° (Linear pair)
∴ 110° + ∠BAE = 180°
⇒ ∠BAE = 180° − 110° = 70°
In ∆ABE,
∠AEC = ∠BAE + ∠ABE (In a triangle, exterior angle is equal to the sum of two opposite interior angles)
∴ 130° = 70° + x°
⇒ x° = 130° − 70° = 60°
Thus, the measure of angle ∠ABC is 60°.
Hence, the correct answer is option (c).
#### Page No 232:
(a) an acute angle
If two angles are complements of each other, that is, the sum of their measures is $90°$, then each angle is an acute angle.
#### Page No 232:
An angle which measures more than 180° but less than 360° is called a reflex angle.
Hence, the correct answer is option (d).
#### Page No 232:
(d) 75°
Let the measure of the required angle be $x°$.
Then, the measure of its complement will be $\left(90-x\right)°$.
$\therefore x=5\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒x=450-5x\phantom{\rule{0ex}{0ex}}⇒6x=450\phantom{\rule{0ex}{0ex}}⇒x=75$
#### Page No 233:
(b) 54°
Let the measure of the required angle be $x°$.
Then, the measure of its complement will be$\left(90-x\right)°$.
$\mathbf{\therefore }2x=3\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒2x=270-3x\phantom{\rule{0ex}{0ex}}⇒5x=270\phantom{\rule{0ex}{0ex}}⇒x=54$
(c) 80°
We have :
(b) 86°
We have :
#### Page No 233:
(c) 80°
We have :
$\therefore \angle AOC={\left[3×30-10\right]}^{°}\phantom{\rule{0ex}{0ex}}⇒\angle AOC=80°$
#### Page No 233:
(a) Through a given point, only one straight line can be drawn.
Clearly, statement (a) is false because we can draw infinitely many straight lines through a given point.
#### Page No 233:
(b) 30°
Let the measure of the required angle be $x°$
Then, the measure of its supplement will be ${\left(180-x\right)}^{°}$
$\therefore x=\frac{1}{5}\left(180°-x\right)\phantom{\rule{0ex}{0ex}}⇒5x=180°-x\phantom{\rule{0ex}{0ex}}⇒6x=180°\phantom{\rule{0ex}{0ex}}⇒x=30°$
(a) 60°
Let
Then, we have:
(c) 45°
We have :
(b) 115°
We have :
Now,
#### Page No 234:
(c) 36°
We know that angle of incidence = angle of reflection.
Then, let $\angle AQP=\angle BQR=x°$
Now,
#### Page No 234:
(c) 50°
Draw $EOF\parallel AB\parallel CD$.
Now, is the transversal.
Also,
is the transversal.
Now,
#### Page No 234:
(a) 130°
Draw $OE\parallel AB\parallel CD$
Now, is the transversal.
Also,
is the transversal.
#### Page No 234:
(c) 45°
is the transversal.
Side EC of triangle EFC is produced to D.
$\therefore \angle CEF+\angle EFC=\angle DCF\phantom{\rule{0ex}{0ex}}⇒\angle CEF+25°=80°\phantom{\rule{0ex}{0ex}}⇒\angle CEF=55°$
#### Page No 234:
(b) 126°
Let
Let the transversal intersect AB at P, CD at O and EF at Q.
Then, we have:
Also,
$\angle APO+\angle COP=180°\phantom{\rule{0ex}{0ex}}⇒x+54°=180°\phantom{\rule{0ex}{0ex}}⇒x=126°$
#### Page No 235:
(a) 50°
is the transversal.
Side QR of traingle PQR is produced to D.
$\therefore \angle PQR+\angle QPR=\angle PRD\phantom{\rule{0ex}{0ex}}⇒70°+\angle QPR=120°\phantom{\rule{0ex}{0ex}}⇒\angle QPR=50°$
#### Page No 235:
(c) 70°
is the transversal.
In $∆ABE$, we have:
#### Page No 235:
(c) 30°
In $∆OAB$, we have:
In $∆OCD$, we have: |
# Similarity and Distance Functions
Let’s go back to our posts on distance functions. The initial idea was to calculate the distance between two points or between one point and a set of points in a 2d space. Within the discussion, we often stated that one point is more “similar” to the set, than the other point. So, maybe what we are actually trying to calculate is the similarity between two entities (either being a point or a set). The problem is that similarity is a subjective matter, so we need an objective measure, and that’s why we have chosen distance.
In other words, we aim to calculate the similarity, and we do so indirectly through the distance function. The inverse relationship between these two functions is clear, as one increases the other should decrease and vice versa.
So we can easily use the inverse of the distance function, and name it as the similarity function:
$s(a,b) = \frac{1}{d(a,b)}$
where $s(a,b)$ is the similarity between points a and b. Let’s look at the behavior of the function and see what we have achieved:
The inverse similarity function.
The problem with this approach is related to the integral of this function. The integral goes to infinity if calculated for all the values the distance function can take. This might seem a bit out of the scope now, but later we observe that bounded integral is essential for having a well-formed probability density function.
A function which is usually used for that purpose is the exponential function. The function is defined as:
$s(a,b) = \frac{1}{e^{d^2(a,b)}} = e^{-d^2(a,b)}$
The exponential similarity function.
Which does not have the problem related to the inverse similarity function. Notice that the mentioned function is not the only function that can be used for this purpose. For example, another possibility is:
$s(a,b) = \frac{1}{1+d^2(a,b)}$
The polynomial similarity function.
Actually these two functions look very much like each other, but there are many more possibilities for similarity function which do not necessarily result in the same shape. We will see later that the choice of different similarity function will result in different probability density functions. |
How to Factor Expressions More Than Once
Sometimes you factor expressions more than once — and with different factoring techniques. To determine if an expression needs to be factored more than once, just take another look at the expression after you first factor it to see if you need to factor it again.
Usually, you can get to the correct answer regardless of which factoring method — and whatever order — you choose. If you have to factor an expression more than once, however, the first step should always be an attempt to factor out the greatest common factor.
Example:
1. Factor out the greatest common factor.
40 is the greatest common factor.
1. Determine all the ways you can multiply two numbers to get a.
In this case, a is 1.
The 1 can be written only as 1 × 1.
2. Determine all the ways you can multiply two numbers to get c.
In this case, c is 6.
The 6 can be written as 1 × 6 or 2 × 3.
3. Look at the sign of c and your lists from Steps 1 and 2 to see if you want a sum or difference.
If c is positive, find a value from your Step 1 list and another from your Step 2 list such that the sum of their product and the product of the two remaining numbers in those steps results in b.
If c is negative, find a value from your Step 1 list and another from your Step 2 list such that the difference of their product and the product of two remaining numbers from those steps results in b.
In this case, c is negative, so you want the difference of the products to be 1.
4. Choose a product from Step 1 and a product from Step 2 that result in the correct sum or difference determined in Step 3.
You need a combination that results in the difference of 1.
Using 1 × 1 and 2 × 3, multiply (1)(2) to get 2, and multiply (1)(3) to get 3. The difference of these products equals 1.
5. Arrange your choices in the binomials so the results are those you want.
(1x 2)(1x 3)
6. Place the signs to give the desired results.
The middle term, x, is negative, so you want the 3x, the outer terms, to be negative.
(x + 2)(x – 3)
7. FOIL the two binomials to check your work.
3. Put the result of both factoring methods together.
Put back into the answer the 40 that you factored out at the beginning. |
# 20 percent off twelve
How to calculate 20 percent-off \$twelve. How to figure out percentages off a price. Using this calculator you will find that the amount after the discount is \$9.6.
### Inputs
Original price: \$
Discount percentage: %
Discount: \$
Final Price: \$
### Details
Using this calculator you can find the discount value and the discounted price of an item. It is helpfull to answer questions like:
• 1) What is 20 percent (%) off \$twelve?
• 2) How much will you pay for an item where the original price before discount is \$ twelve when discounted 20 percent (%)? What is the final or sale price?
• 3) \$2.4 is what percent off \$12?
• See how to solve these questions just after the Percent-off Calculator (or Discount) below.
## How to work out discounts - Step by Step
To calculate discount it is ease by using the following equations:
• Amount Saved = Orig. Price x Discount % / 100 (a)
•
• Sale Price = Orig. Price - Amount Saved (b)
Now, let's solve the questions stated above:
### 1) What is 20 percent off \$12? Find the amount of discount.
Suppose you have a Kohls coupon of \$12 and you want to know how much you will save for an item if the discount is 20.
Solution:
Replacing the given values in formula (a) we have:
Amount Saved = Original Price x Discount in Percent / 100. So,
Amount Saved = 12 x 20 / 100
Amount Saved = 240 / 100
In other words, a 20% discount for a item with original price of \$12 is equal to \$2.4 (Amount Saved).
Note that to find the amount saved, just multiply it by the percentage and divide by 100.
### 2) How much to pay for an item of \$12 when discounted 20 percent (%)? What is item's sale price?
Suppose you have a L.L. Bean coupon of \$12 and you want to know the final or sale price if the discount is 20 percent.
Using the formula (b) and replacing the given values:
Sale Price = Original Price - Amount Saved. So,
Sale Price = 12 - 2.4
This means the cost of the item to you is \$9.6.
You will pay \$9.6 for a item with original price of \$12 when discounted 20%.
In this example, if you buy an item at \$12 with 20% discount, you will pay 12 - 2.4 = \$9.6.
### 3) 2.4 is what percent off \$12?
Using the formula (b) and replacing given values:
Amount Saved = Original Price x Discount in Percent /100. So,
2.4 = 12 x Discount in Percent / 100
2.4 / 12 = Discount in Percent /100
100 x 2.4 / 12 = Discount in Percent
240 / 12 = Discount in Percent, or
Discount in Percent = 20 (answer).
To find more examples, just choose one at the bottom of this page.
## Percent-off Table for 12
1 percent off of 12 = 11.88 2 percent off of 12 = 11.76 3 percent off of 12 = 11.64 4 percent off of 12 = 11.52 5 percent off of 12 = 11.4 6 percent off of 12 = 11.28 7 percent off of 12 = 11.16 1 percent off of 12 = 11.88 8 percent off of 12 = 11.04 9 percent off of 12 = 10.92 10 percent off of 12 = 10.8 12 percent off of 12 = 10.56 15 percent off of 12 = 10.2 20 percent off of 12 = 9.6 25 percent off of 12 = 9
## Percent-off Table for 12
30 percent off of 12 = 8.4 33 percent off of 12 = 8.04 35 percent off of 12 = 7.8 40 percent off of 12 = 7.2 45 percent off of 12 = 6.6 50 percent off of 12 = 6 50 percent off of 12 = 6 60 percent off of 12 = 4.8 65 percent off of 12 = 4.2 67 percent off of 12 = 3.96 70 percent off of 12 = 3.6 75 percent off of 12 = 3 80 percent off of 12 = 2.4 85 percent off of 12 = 1.8 90 percent off of 12 = 1.2 |
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# Sanya has a piece of land which is in the shape of rhombus. She wants her one daughter and one son to work on the land and produce different crops. She divided the land in two equal parts. If the perimeter of the land is 400m and one of the diagonals is 160 m, how much area each of them will get for their crops.
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Hint: In this question, we have to find the area of the triangle formed by the diagonals of the rhombus. As we know that all the sides of the rhombus are equal and the area of any triangle with the given sides can be found by using the Heron’s formula:
$\sqrt {s(s - a)(s - b)(s - c)}$ square unit.
[Where S is semi-perimeter and a, b, c are the lengths of the sides.]
We also know that semi-perimeter is
$s = \dfrac{{a + b + c}}{2}$
As the lengths of sides of the rhombus are equal, we can find the length of each side by the given perimeter
To find the length of side of a rhombus, we have to use the formula
i.e., $\dfrac{{perimeter}}{4}$
Therefore, the length of side of the rhombus is
$\dfrac{{400}}{4}$
$\Rightarrow 100$m
We also know that one diagonal of rhombus divides it in two equal parts and we get a triangle
Therefore, $a = 100,b = 100\& c = 160$
i.e., lengths of sides of triangle are $100m,100m{\text{ }}and{\text{ }}160m$
Find the semi-perimeter:
To find the area of the triangle with the sides $100$m,$100$m and $160$m, we can use the Heron’s formula
i.e., $\sqrt {s(s - a)(s - b)(s - c)}$ square unit.
${ \Rightarrow \sqrt {180(180 - 160)(180 - 100)(180 - 100} \\ \Rightarrow \sqrt {180(20)(80)(80)} \\ \Rightarrow \Delta = 4800 \\ }$[Where $\vartriangle$= Area of the triangle]
Hence, answer is $4800$ ${m^2}$
Note: Whenever we face such types of problems the key concept is to use Heron’s formula as stated in the solution. One can go wrong while calculating the square root. We must take care about the calculations. Alternative, we can also find the side of the rhombus by using area’s formula-
$\dfrac{1}{2} \times$product of diagonals. |
# Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 23
## Maharashtra State Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 23
Question 1.
Write three rational numbers that lie between the two given numbers.
i. $$\frac{2}{7}, \frac{6}{7}$$
ii. $$\frac{4}{5}, \frac{2}{3}$$
iii. $$-\frac{2}{3}, \frac{4}{5}$$
iv. $$\frac{7}{9},-\frac{5}{9}$$
v. $$\frac{-3}{4}, \frac{+5}{4}$$
vi. $$\frac{7}{8}, \frac{-5}{3}$$
vii. $$\frac{5}{7}, \frac{11}{7}$$
viii. $$0, \frac{-3}{4}$$
Solution:
i. $$\frac{2}{7}, \frac{6}{7}$$
The three numbers lying between $$\frac { 2 }{ 7 }$$ and $$\frac { 6 }{ 7 }$$ are $$\frac{3}{7}, \frac{4}{7}, \frac{5}{7}$$
ii. $$\frac{4}{5}, \frac{2}{3}$$
$$\frac{4}{5}=\frac{24}{30}, \frac{2}{3}=\frac{20}{30}$$
The three numbers between $$\frac { 4 }{ 5 }$$ and $$\frac { 2 }{ 3 }$$ are $$\frac{21}{30}, \frac{22}{30}, \frac{23}{30}$$
iii. $$-\frac{2}{3}, \frac{4}{5}$$
$$\frac{-2}{3}=\frac{-10}{15}, \frac{4}{5}=\frac{12}{15}$$
The three numbers between $$\frac { -2 }{ 3 }$$ and $$\frac { 4 }{ 5 }$$ are $$\frac{-9}{15}, \frac{-7}{15}, \frac{4}{15}$$
iv. $$\frac{7}{9},-\frac{5}{9}$$
The three numbers between $$\frac { 7 }{ 9 }$$ and $$\frac { -5 }{ 9 }$$ are $$\frac{6}{9}, 0, \frac{-4}{9}$$
v. $$\frac{-3}{4}, \frac{+5}{4}$$
The three numbers between $$\frac { -3 }{ 4 }$$ and $$\frac { +5 }{ 4 }$$ are $$\frac{-2}{4}, \frac{-1}{4}, \frac{3}{4}$$
vi. $$\frac{7}{8}, \frac{-5}{3}$$
$$\frac{7}{8}=\frac{21}{24}, \frac{-5}{3}=\frac{-40}{24}$$
The three numbers between $$\frac { 7 }{ 8 }$$ and $$\frac { -5 }{ 3 }$$ are $$\frac{17}{24}, \frac{11}{24}, \frac{-13}{24}$$
vii. $$\frac{5}{7}, \frac{11}{7}$$
The three numbers between $$\frac { 5 }{ 7 }$$ and $$\frac { 11 }{ 7 }$$ are $$\frac{6}{7}, \frac{8}{7}, \frac{9}{7}$$
viii. $$0, \frac{-3}{4}$$
The three numbers between 0 and $$\frac { -3 }{ 4 }$$ are $$\frac{-1}{8}, \frac{-2}{8}, \frac{-5}{8}$$
Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 23 Intext Questions and Activities
Question 1.
Answer the following questions: (Textbook pg. no. 36)
1. Write all the natural numbers between 2 and 9.
2. Write all the integers between -4, and 5.
3. Which rational numbers are there between $$\frac { 1 }{ 2 }$$ and $$\frac { 3 }{ 4 }$$ ?
Solution:
1. 3, 4, 5, 6, 7, 8
2. -3, -2, -1, 0, 1, 2, 3, 4
3. $$\frac{1}{2}=\frac{1 \times 2}{2 \times 2}=\frac{2}{4}=\frac{2 \times 10}{4 \times 10}=\frac{20}{40}$$
$$\frac{3}{4}=\frac{3 \times 10}{4 \times 10}=\frac{30}{40}$$
∴ The rational numbers between $$\frac { 1 }{ 2 }$$ and $$\frac { 3 }{ 4 }$$ are $$\frac{21}{40}, \frac{22}{40}, \frac{25}{40}, \frac{27}{40}$$ etc. |
# Bus Stop Method For Long Division Made Easy
Dec 12, 2023 By Vicky Gayle
Originally Published on Jul 07, 2020
info_i
It's remarkable how when you're faced with helping children with their maths homework, much more comes flooding back to you than what you first thought.
The 'bus stop' method is a tried and tested way of doing long division when you're asked to divide larger numbers by two or three-digit numbers, as well as when a number is being divided by a single digit, known as short division.
Bus stop division is simply another name for a step-by-step long division method and is suitable for Key Stage 2 children but is typically introduced in Year 5.
If you need more Maths help for your Key Stage 2 children, you can check out our classes on Kidadl TV.
### Why Is It Called The 'Bus Stop' Method?
The bracket you need to draw over the 'dividend', the number you're being asked to divide, resembles a bus stop. The long edge shelters the dividend while the short edge, drawn towards the metaphorical ground, separates the 'divisor', the number you're dividing by.
This method differs from short division as here you need to draw the bracket on top and work out the sum using a slightly more complex process.
### How To Do The Bus Stop Method With A 2-Digit Divisor?
QUESTION: What is 1,722 ÷ 15?
15│1722
Step A) 15 is too big to go into 1, so carry the 1 to the 7 to make 17. Cross out the big 1 and rewrite it smaller and closer to the 17 if that helps you visually.
Step B) 15 goes into 17 just once so write a 1 above the bracket over the 7. Now, what's the difference between 15 and 17? The answer to that is 2 so write a small 2 alongside the next number which makes 22.
Step C) How many times does 15 go into 22? Just once so write a 1 above the bracket over the 2. Now, the difference between 15 and 22 is 7, so write a small 7 next to the following number.
What we have now is 72, so we must calculate 15 ÷ 72 and use our times tables...
Step D) 15 x 4 = 60 and that's as close as we can get to 72 so now, write 4 above the 72. Now, we must work out the difference between 60 and 72, so do 72 – 60 = 12. But we've run out of numbers so where will we put the 12?
Immediately, add a decimal point after both the 114 and 1722, but after the decimal point below the bracket, put a 0.
Step E) Write a small 12 next to the 0 to make 120. Now, how many times does 15 go into 120? 15 x 8 = 120. Finally! Write 8 above the 120 and there you have it...
ANSWER: 1,722 ÷ 15 = 114.8!
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Written by Vicky Gayle
Vicky is an avid explorer living in Birmingham and is an auntie to four nieces and nephews. She believes in being a tourist in her own town and enjoys discovering new experiences. She has a Bachelor's degree in Journalism from the University of Central Lancashire. Vicky loves to share her finds and regularly sends event ideas to her friends. As many of her friends are parents, she is always on the lookout for child-friendly activities to recommend. Vicky is open-minded and willing to try most things at least once. |
# 12.3: Division of Polynomials
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
At the end of this lesson, students will be able to:
• Divide a polynomial by a monomial.
• Divide a polynomial by a binomial.
• Rewrite and graph rational functions.
## Vocabulary
Terms introduced in this lesson:
rational expression
numerator
denominator
common denominator
dividend
divisor
quotient
remainder
## Teaching Strategies and Tips
Students learned in chapter Factoring Polynomials how to add, subtract, and multiply polynomials. This lesson completes that discussion with dividing polynomials.
• Emphasize that the quotient of two polynomials forms a rational expression which is studied in its own right (rational functions).
Use Example 1 to demonstrate dividing a polynomial by a monomial.
• Remind students that each term in the numerator must be divided by the monomial in the denominator. See Example 2.
Use Example 3 to motivate long division of polynomials.
• To write the answer, remind students that:
\begin{align*}\frac{\text{dividend}}{\text{divisor}} = \text{quotient} + \frac{\text{remainder}}{\text{divisor}}\end{align*}
• To check an answer, have students use the equivalent form:
\begin{align*}\text{dividend} = (\text{divisor} \times \text{quotient}) + \text{remainder}\end{align*}
Have students rewrite for themselves the four cases for graphing rational functions preceding Example 5.
## Error Troubleshooting
General Tip: Students often incorrectly cancel a factor not common to all the terms.
• Example:
\begin{align*}\frac{\cancel{a}x+b}{\cancel{a}y} \neq \frac{x+b}{y}\end{align*}
• When students cancel the \begin{align*}a\end{align*} above, they violate order of operations. Remind students that the fraction sign is a grouping symbol (parentheses) and therefore the numerator and denominator must be simplified before dividing.
• Otherwise, if the numerator and denominator are completely factored, then the order of operations says to multiply or divide; therefore, canceling is justified.
• Have students write out the step preceding the canceling:
Example:
\begin{align*}\frac{ax+ab}{ay} = \frac{a(x+b)}{ay}\end{align*}
Then canceling is apparent:
\begin{align*}\frac{ax+ab}{ay} = \frac{a(x+b)}{ay} = \frac{\cancel{a}(x+b)}{\cancel{a}y} - \frac{x+b}{y}\end{align*}
• Other common cancelling errors are:
a. \begin{align*}\frac{\cancel{a}x+ab}{\cancel{a}y} \neq \frac{x+ab}{y}\end{align*} (forgetting to remove the canceled factor)
b. \begin{align*}\frac{\cancel{a}}{x+\cancel{a}} \neq \frac{1}{x}\end{align*}
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# Help Zone
### Student Question
Secondary II • 2mo.
how can I solve for x in inequalities
How can I solve for x in inequalities
Mathematics
## Explanations (1)
• Explanation from Alloprof
Explanation from Alloprof
This Explanation was submitted by a member of the Alloprof team.
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Team Alloprof • 2mo.
Hi!
To solve an inequality, you must always place like terms on one side of the inequality, and constants on the other side. Let's take an example to better understand.
We have the inequality:
$$4x - 6 < 2x + 10$$
Like terms are terms that have the same variables (the same unknowns), and these variables are assigned the same exponents. Our like terms here are $$4x$$ and $$2x$$, since they both contain the variable x which has an exponent 1.
Constants are terms that do not contain variables, here $$-6$$ and $$10$$.
Our goal will first be to place the two like terms on one side of the inequality, and the constants on the other side. To do this, we will start by moving one of the two like terms to the other side (it does not matter which one).
Let's move $$2x$$ to the left side of the inequality. Since the inverse of addition is subtraction, we'll need to subtract $$2x$$ from each side of the inequality, like this:
$$4x - 6 -2x< 2x + 10-2x$$
By subtracting it from each side, this allows us to eliminate it from the right side of the inequality:
$$4x - 6 -2x< 10$$
We thus moved the term $$2x$$ so that it is on the same side as $$4x$$.
Now, let's move on to constants. We will move the constant $$6$$ to the other side. Since the inverse operation of a subtraction is an addition, we will therefore add $$6$$ on each side:
$$4x - 6 -2x+6< 10+6$$
$$4x -2x< 10+6$$
We succeeded in placing our like terms on one side and our constants on the other! The next step will be to add the constants, and add the coefficients of like terms. Let's start with the constants. Since 10+6 gives 16, we have:
$$4x -2x< 16$$
To subtract like terms, you must subtract their coefficient, which is the number in front of the variable x.
$$(4-2)x< 16$$
$$2x< 16$$
Finally, the last step will be to eliminate the coefficient of the variable x, i.e. $$2$$, by performing the opposite operation of multiplication, i.e. division:
$$\frac{2x}{2} < \frac{16}{2}$$
$$x< 8$$
Here are some sheets on these concepts that might be useful to you:
I hope I was able to help you! Feel free to ask us more questions if you have any! :) |
# Basic Analysis I & II: Introduction to Real Analysis, Volumes I & II
## Section11.3Power series and analytic functions
Note: 2–3 lectures
### Subsection11.3.1Analytic functions
A (complex) power series is a series of the form
\begin{equation*} \sum_{n=0}^\infty c_n {(z-a)}^n \end{equation*}
for $$c_n, z, a \in \C\text{.}$$ We say the series converges if the series converges for some $$z \not= a\text{.}$$
Let $$U \subset \C$$ be an open set and $$f \colon U \to \C$$ a function. Suppose that for every $$a \in U$$ there exists a $$\rho > 0$$ and a power series convergent to the function
\begin{equation*} f(z) = \sum_{n=0}^\infty c_n {(z-a)}^n \end{equation*}
for all $$z \in B(a,\rho)\text{.}$$ Then we say $$f$$ is an analytic function. Similarly, given an interval $$(a,b) \subset \R\text{,}$$ we say that $$f \colon (a,b) \to \C$$ is analytic or perhaps real-analytic if for each point $$c \in (a,b)$$ there is a power series around $$c$$ that converges in some $$(c-\rho,c+\rho)$$ for some $$\rho > 0\text{.}$$ As we will sometimes talk about real and sometimes about complex power series, we will use $$z$$ to denote a complex number and $$x$$ a real number. We will always mention which case we are working with.
An analytic function has different expansions around different points. Moreover, convergence does not automatically happen on the entire domain of the function. For example, if $$\sabs{z} < 1\text{,}$$ then
\begin{equation*} \frac{1}{1-z} = \sum_{n=0}^\infty z^n . \end{equation*}
While the left-hand side exists on all of $$z \not= 1\text{,}$$ the right-hand side happens to converge only if $$\sabs{z} < 1\text{.}$$ See a graph of a small piece of $$\frac{1}{1-z}$$ in Figure 11.5. We cannot graph the function itself, we can only graph its real or imaginary parts for lack of dimensions in our universe.
### Subsection11.3.2Convergence of power series
We proved several results for power series of a real variable in Section 2.6. For the most part the convergence properties of power series deal with the series $$\sum_{k=0}^\infty \sabs{c_k} \, \sabs{z-a}^k$$ and so we have already proved many results about complex power series. In particular, we computed the so-called radius of convergence of a power series.
The number $$\rho$$ is the radius of convergence. See Figure 11.6. The radius of convergence gives a disc around $$a$$ where the series converges. A power series is convergent if $$\rho > 0\text{.}$$
#### Proof.
We use the real version of this proposition, Proposition 2.6.10. Let
\begin{equation*} R \coloneqq \limsup_{n\to\infty} \sqrt[n]{\sabs{c_n}} . \end{equation*}
If $$R = 0\text{,}$$ then $$\sum_{n=0}^\infty \sabs{c_n} \, \sabs{z-a}^n$$ converges for all $$z\text{.}$$ If $$R = \infty\text{,}$$ then $$\sum_{n=0}^\infty \sabs{c_n} \, \sabs{z-a}^n$$ converges only at $$z=a\text{.}$$ Otherwise, let $$\rho \coloneqq \nicefrac{1}{R}$$ and $$\sum_{n=0}^\infty \sabs{c_n} \, \sabs{z-a}^n$$ converges when $$\sabs{z-a} < \rho\text{,}$$ and diverges (in fact the terms of the series do not go to zero) when $$\sabs{z-a} > \rho\text{.}$$
To prove the Furthermore,’’ suppose $$0 < r < \rho$$ and $$z \in C(a,r)\text{.}$$ Then consider the partial sums
\begin{equation*} \abs{\sum_{n=0}^k c_n {(z-a)}^n} \leq \sum_{n=0}^k \sabs{c_n} \sabs{z-a}^n \leq \sum_{n=0}^k \sabs{c_n} r^n . \qedhere \end{equation*}
If $$\sum_{n=0}^\infty c_n {(z-a)}^n$$ converges for some $$z\text{,}$$ then
\begin{equation*} \sum_{n=0}^\infty c_n {(w-a)}^n \end{equation*}
converges absolutely whenever $$\sabs{w-a} < \sabs{z-a}\text{.}$$ Conversely, if the series diverges at $$z\text{,}$$ then it must diverge at $$w$$ whenever $$\sabs{w-a} > \sabs{z-a}\text{.}$$ Hence, to show that the radius of convergence is at least some number, we simply need to show convergence at some point by any method we know.
#### Example11.3.2.
We list some series we already know:
\begin{equation*} \begin{aligned} & & & \sum_{n=0}^\infty z^n & & \text{has radius of convergence } 1. & & \\ & & & \sum_{n=0}^\infty \frac{1}{n!} z^n & & \text{has radius of convergence } \infty. & & \\ & & & \sum_{n=0}^\infty n^n z^n & & \text{has radius of convergence } 0. & & \end{aligned} \end{equation*}
#### Example11.3.3.
Note the difference between $$\frac{1}{1-z}$$ and its power series. Let us expand $$\frac{1}{1-z}$$ as power series around a point $$a \not= 1\text{.}$$ Let $$c \coloneqq \frac{1}{1-a}\text{,}$$ then
\begin{equation*} \frac{1}{1-z} = \frac{c}{1-c(z-a)} = c \sum_{n=0}^\infty c^{n} {(z-a)}^n = \sum_{n=0}^\infty \left( \frac{1}{{(1-a)}^{n+1}} \right) {(z-a)}^n . \end{equation*}
The series $$\sum_{n=0}^\infty c^n {(z-a)}^n$$ converges if and only if the series on the right-hand side converges and
\begin{equation*} \limsup_{n\to\infty} \sqrt[n]{\sabs{c^n}} = \sabs{c} = \frac{1}{\sabs{1-a}} . \end{equation*}
The radius of convergence of the power series is $$\sabs{1-a}\text{,}$$ that is the distance from $$1$$ to $$a\text{.}$$ The function $$\frac{1}{1-z}$$ has a power series representation around every $$a\not= 1$$ and so is analytic in $$\C \setminus \{ 1 \}\text{.}$$ The domain of the function is bigger than the region of convergence of the power series representing the function at any point.
It turns out that if a function has a power series representation converging to the function on some ball, then it has a power series representation at every point in the ball. We will prove this result later.
### Subsection11.3.3Properties of analytic functions
#### Proof.
For $$z_0 \in B(a,\rho)\text{,}$$ pick $$r < \rho$$ such that $$z_0 \in B(a,r)\text{.}$$ On $$B(a,r)$$ the partial sums (which are continuous) converge uniformly, and so the limit $$f|_{B(a,r)}$$ is continuous. Any sequence converging to $$z_0$$ has some tail that is completely in the open ball $$B(a,r)\text{,}$$ hence $$f$$ is continuous at $$z_0\text{.}$$
In Corollary 6.2.13, we proved that we can differentiate real power series term by term. That is, we proved that if
\begin{equation*} f(x) \coloneqq \sum_{n=0}^\infty c_n {(x-a)}^n \end{equation*}
converges for real $$x$$ in an interval around $$a \in \R\text{,}$$ then we can differentiate term by term and obtain a series
\begin{equation*} f'(x) = \sum_{n=1}^\infty n c_n {(x-a)}^{n-1} = \sum_{n=0}^\infty (n+1)c_{n+1} {(x-a)}^{n} \end{equation*}
with the same radius of convergence. We only proved this theorem when $$c_n$$ is real, however, for complex $$c_n\text{,}$$ we write $$c_n = s_n + i t_n\text{,}$$ and as $$x$$ and $$a$$ are real
\begin{equation*} \sum_{n=0}^\infty c_n {(x-a)}^n = \sum_{n=0}^\infty s_n {(x-a)}^n + i \sum_{n=0}^\infty t_n {(x-a)}^n . \end{equation*}
We apply the theorem to the real and imaginary part.
By iterating this theorem, we find that an analytic function is infinitely differentiable:
\begin{equation*} f^{(\ell)}(x) = \sum_{n=\ell}^\infty n(n-1)\cdots(n-\ell+1)c_k {(x-a)}^{n-\ell} = \sum_{n=0}^\infty (n+\ell)(n+\ell-1)\cdots (n+1) c_{n+\ell} {(x-a)}^{n} . \end{equation*}
In particular,
$$f^{(\ell)}(a) = \ell! \, c_\ell .\tag{11.1}$$
The coefficients are uniquely determined by the derivatives of the function, and vice versa.
On the other hand, just because we have an infinitely differentiable function doesn’t mean that the numbers $$c_n$$ obtained by $$c_n = \frac{f^{(n)}(0)}{n!}$$ give a convergent power series. There is a theorem, which we will not prove, that given an arbitrary sequence $$\{ c_n \}_{n=1}^\infty\text{,}$$ there exists an infinitely differentiable function $$f$$ such that $$c_n = \frac{f^{(n)}(0)}{n!}\text{.}$$ Moreover, even if the obtained series converges, it may not converge to the function we started with. For an example, see Exercise 5.4.11: The function
\begin{equation*} f(x) \coloneqq \begin{cases} e^{-1/x} & \text{if } x > 0,\\ 0 & \text{if } x \leq 0, \end{cases} \end{equation*}
is infinitely differentiable, and all derivatives at the origin are zero. So its series at the origin would be just the zero series, and while that series converges, it does not converge to $$f$$ for $$x > 0\text{.}$$
We can apply an affine transformation $$z \mapsto z+a$$ that converts a power series at $$a$$ to a series at the origin. That is, if
\begin{equation*} f(z) = \sum_{n=0}^\infty c_n {(z-a)}^n, \qquad \text{we consider} \qquad f(z+a) = \sum_{n=0}^\infty c_n {z}^n. \end{equation*}
Therefore, it is usually sufficient to prove results about power series at the origin. From now on, we often assume $$a=0$$ for simplicity.
### Subsection11.3.4Power series as analytic functions
We need a theorem on swapping limits of series, that is, Fubini’s theorem for sums. For real series this was Exercise 2.6.15, but we have a slicker argument now.
#### Proof.
Let $$E$$ be the set $$\{ \nicefrac{1}{n} : n \in \N \} \cup \{ 0 \}\text{,}$$ and treat it as a metric space with the metric inherited from $$\R\text{.}$$ Define the sequence of functions $$f_k \colon E \to \C$$ by
\begin{equation*} f_k(\nicefrac{1}{n}) \coloneqq \sum_{m=1}^n a_{k,m} \qquad \text{and} \qquad f_k(0) \coloneqq \sum_{m=1}^\infty a_{k,m} . \end{equation*}
As the series converges, each $$f_k$$ is continuous at $$0$$ (since 0 is the only cluster point, they are continuous at every point of $$E\text{,}$$ but we don’t need that). For all $$x \in E\text{,}$$ we have
\begin{equation*} \sabs{f_k(x)} \leq \sum_{m=1}^\infty \sabs{a_{k,m}} . \end{equation*}
As $$\sum_k \sum_m \sabs{a_{k,m}}$$ converges (and does not depend on $$x$$), we know that
\begin{equation*} \sum_{k=1}^n f_k(x) \end{equation*}
converges uniformly on $$E\text{.}$$ Define
\begin{equation*} g(x) \coloneqq \sum_{k=1}^\infty f_k(x) , \end{equation*}
which is, therefore, a continuous function at $$0\text{.}$$ So
\begin{equation*} \begin{split} \sum_{k=1}^\infty \left( \sum_{m=1}^\infty a_{k,m} \right) & = \sum_{k=1}^\infty f_k(0) = g(0) = \lim_{n\to\infty} g(\nicefrac{1}{n}) \\ &= \lim_{n\to\infty}\sum_{k=1}^\infty f_k(\nicefrac{1}{n}) = \lim_{n\to\infty}\sum_{k=1}^\infty \sum_{m=1}^n a_{k,m} \\ &= \lim_{n\to\infty}\sum_{m=1}^n \sum_{k=1}^\infty a_{k,m} = \sum_{m=1}^\infty \left( \sum_{k=1}^\infty a_{k,m} \right) . \qedhere \end{split} \end{equation*}
Now we prove that once we have a series converging to a function in some interval, we can expand the function around every point.
The power series at $$a$$ could of course converge in a larger interval, but the one above is guaranteed. It is the largest symmetric interval about $$a$$ that fits in $$(-\rho,\rho)\text{.}$$
#### Proof.
Given $$a$$ and $$x$$ as in the theorem, write
\begin{equation*} \begin{split} f(x) &= \sum_{k=0}^\infty a_k {\bigl((x-a)+a\bigr)}^k \\ &= \sum_{k=0}^\infty a_k \sum_{m=0}^k \binom{k}{m} a^{k-m} {(x-a)}^m . \end{split} \end{equation*}
Define $$c_{k,m} \coloneqq a_k \binom{k}{m} a^{k-m}$$ if $$m \leq k$$ and $$0$$ if $$m > k\text{.}$$ Then
$$f(x) = \sum_{k=0}^\infty \, \sum_{m=0}^\infty c_{k,m} {(x-a)}^m .\tag{11.2}$$
Let us show that the double sum converges absolutely.
\begin{equation*} \begin{split} \sum_{k=0}^\infty \, \sum_{m=0}^\infty \abs{ c_{k,m} {(x-a)}^m} & = \sum_{k=0}^\infty \, \sum_{m=0}^k \abs{ a_k \binom{k}{m} a^{k-m} {(x-a)}^m } \\ & = \sum_{k=0}^\infty \sabs{a_k} \sum_{m=0}^k \binom{k}{m} \sabs{a}^{k-m} {\sabs{x-a}}^m \\ & = \sum_{k=0}^\infty \sabs{a_k} {\bigl(\sabs{x-a}+\sabs{a}\bigr)}^k , \end{split} \end{equation*}
and this series converges as long as $$(\sabs{x-a}+\sabs{a}) < \rho$$ or in other words if $$\sabs{x-a} < \rho-\sabs{a}\text{.}$$
Using Theorem 11.3.5, swap the order of summation in (11.2), and the following series converges when $$\sabs{x-a} < \rho-\sabs{a}\text{:}$$
\begin{equation*} f(x) = \sum_{k=0}^\infty \, \sum_{m=0}^\infty c_{k,m} {(x-a)}^m = \sum_{m=0}^\infty \left( \sum_{k=0}^\infty c_{k,m} \right) {(x-a)}^m . \end{equation*}
The formula in terms of derivatives at $$a$$ follows by differentiating the series to obtain (11.1).
Note that if a series converges for real $$x \in (a-\rho,a+\rho)$$ it also converges for all complex numbers in $$B(a,\rho)\text{.}$$ We have the following corollary, which says that functions defined by power series are analytic.
#### Proof.
Without loss of generality assume that $$a=0\text{.}$$ We can rotate to assume that $$b$$ is real, but since that is harder to picture, let us do it explicitly. Let $$\alpha \coloneqq \frac{\bar{b}}{\sabs{b}}\text{.}$$ Notice that
\begin{equation*} \abs{\nicefrac{1}{\alpha}} = \sabs{\alpha} = 1 . \end{equation*}
Therefore the series $$\sum_{k=0}^\infty c_k {(\nicefrac{z}{\alpha})}^k = \sum_{k=0}^\infty c_k \alpha^{-k} {z}^k$$ converges to $$f(\nicefrac{z}{\alpha})$$ in $$B(0,\rho)\text{.}$$ When $$z=x$$ is real we apply Theorem 11.3.6 at $$\sabs{b}$$ and get a series that converges to $$f(\nicefrac{z}{\alpha})$$ on $$B(\sabs{b},\rho-\sabs{b})\text{.}$$ That is, there is a convergent series
\begin{equation*} f(\nicefrac{z}{\alpha}) = \sum_{k=0}^\infty a_k {\bigl(z - \sabs{b}\bigr)}^k . \end{equation*}
Using $$\alpha b = \sabs{b}\text{,}$$ we find
\begin{equation*} f(z) = f(\nicefrac{\alpha z}{\alpha}) = \sum_{k=0}^\infty a_k {(\alpha z - \sabs{b})}^k = \sum_{k=0}^\infty a_k\alpha^k {\bigl(z - \nicefrac{\sabs{b}}{\alpha}\bigr)}^k = \sum_{k=0}^\infty a_k\alpha^k {(z - b)}^k , \end{equation*}
and this series converges for all $$z$$ such that $$\bigl\lvert \alpha z-\sabs{b}\bigr\rvert < \rho-\sabs{b}$$ or $$\sabs{z - b} < \rho-\sabs{b}\text{.}$$
We proved above that a convergent power series is an analytic function where it converges. We have also shown before that $$\frac{1}{1-z}$$ is analytic outside of $$z=1\text{.}$$
Note that just because a real analytic function is analytic on the entire real line it does not necessarily mean that it has a power series representation that converges everywhere. For example, the function
\begin{equation*} f(x) = \frac{1}{1+x^2} \end{equation*}
happens to be real analytic function on $$\R$$ (exercise). A power series around the origin converging to $$f$$ has a radius of convergence of exactly $$1\text{.}$$ Can you see why? (exercise)
### Subsection11.3.5Identity theorem for analytic functions
#### Proof.
By continuity we know $$f(0) = 0$$ so $$a_0 = 0\text{.}$$ Suppose there exists some nonzero $$a_k\text{.}$$ Let $$m$$ be the smallest $$m$$ such that $$a_m \not= 0\text{.}$$ Then
\begin{equation*} f(z) = \sum_{k=m}^\infty a_k z^k = z^m \sum_{k=m}^\infty a_k z^{k-m} = z^m \sum_{k=0}^\infty a_{k+m} z^{k} . \end{equation*}
Write $$g(z) = \sum_{k=0}^\infty a_{k+m} z^{k}$$ (this series converges in on the same set as $$f$$). $$g$$ is continuous and $$g(0) = a_m \not= 0\text{.}$$ Thus there exists some $$\delta > 0$$ such that $$g(z) \not= 0$$ for all $$z \in B(0,\delta)\text{.}$$ As $$f(z) = z^m g(z)\text{,}$$ the only point in $$B(0,\delta)$$ where $$f(z) = 0$$ is when $$z=0\text{,}$$ but this contradicts the assumption that $$f(z_n) = 0$$ for all $$n\text{.}$$
Recall that in a metric space $$X\text{,}$$ a cluster point (or sometimes limit point) of a set $$E$$ is a point $$p \in X$$ such that $$B(p,\epsilon) \setminus \{ p \}$$ contains points of $$E$$ for all $$\epsilon > 0\text{.}$$
In most common applications of this theorem $$E$$ is an open set or perhaps a curve.
#### Proof.
Without loss of generality suppose $$E$$ is the set of all points $$z \in U$$ such that $$g(z)=f(z)\text{.}$$ Note that $$E$$ must be closed as $$f$$ and $$g$$ are continuous.
Suppose $$E$$ has a cluster point. Without loss of generality assume that $$0$$ is this cluster point. Near $$0\text{,}$$ we have the expansions
\begin{equation*} f(z) = \sum_{k=0}^\infty a_k {z}^k \qquad \text{and} \qquad g(z) = \sum_{k=0}^\infty b_k {z}^k , \end{equation*}
which converge in some ball $$B(0,\rho)\text{.}$$ Therefore the series
\begin{equation*} 0 = f(z)-g(z) = \sum_{k=0}^\infty (a_k-b_k) z^k \end{equation*}
converges in $$B(0,\rho)\text{.}$$ As $$0$$ is a cluster point of $$E\text{,}$$ there is a sequence of nonzero points $$\{ z_n \}_{n=1}^\infty$$ such that $$f(z_n) -g(z_n) = 0\text{.}$$ Hence, by the lemma above $$a_k = b_k$$ for all $$k\text{.}$$ Therefore, $$B(0,\rho) \subset E\text{.}$$
Thus the set of cluster points of $$E$$ is open. The set of cluster points of $$E$$ is also closed: A limit of cluster points of $$E$$ is in $$E$$ as it is closed, and it is clearly a cluster point of $$E\text{.}$$ As $$U$$ is connected, the set of cluster points of $$E$$ is equal to $$U\text{,}$$ or in other words $$E = U\text{.}$$
By restricting our attention to real $$x\text{,}$$ we obtain the same theorem for connected open subsets of $$\R\text{,}$$ which are just open intervals.
### Subsection11.3.6Exercises
#### Exercise11.3.1.
Let
\begin{equation*} a_{k,m} \coloneqq \begin{cases} 1 & \text{if } k=m,\\ -2^{k-m} & \text{if } k<m,\\ 0 & \text{if } k>m. \end{cases} \end{equation*}
Compute (or show the limit doesn’t exist):
a) $$\displaystyle \sum_{m=1}^\infty \sabs{a_{k,m}}~$$ for all $$k\text{,}$$ b) $$\displaystyle \sum_{k=1}^\infty \sabs{a_{k,m}}~$$ for all $$m\text{,}$$ c) $$\displaystyle \sum_{k=1}^\infty \sum_{m=1}^\infty \sabs{a_{k,m}}\text{,}$$ d) $$\displaystyle \sum_{k=1}^\infty \sum_{m=1}^\infty a_{k,m}\text{,}$$ e) $$\displaystyle \sum_{m=1}^\infty \sum_{k=1}^\infty a_{k,m}\text{.}$$
Hint: Fubini for sums does not apply, in fact, answers to d) and e) are different.
#### Exercise11.3.2.
Let $$f(x) \coloneqq \frac{1}{1+x^2}\text{.}$$ Prove that
1. $$f$$ is analytic function on all of $$\R$$ by finding a power series for $$f$$ at every $$a \in \R\text{,}$$
2. the radius of convergence of the power series for $$f$$ at the origin is 1.
#### Exercise11.3.3.
Suppose $$f \colon \C \to \C$$ is analytic. Show that for each $$n\text{,}$$ there are at most finitely many zeros of $$f$$ in $$B(0,n)\text{,}$$ that is, $$f^{-1}(0) \cap B(0,n)$$ is finite for each $$n\text{.}$$
#### Exercise11.3.4.
Suppose $$U \subset \C$$ is open and connected, $$0 \in U\text{,}$$ and $$f \colon U \to \C$$ is analytic. Treating $$f$$ as a function of a real $$x$$ at the origin, suppose $$f^{(n)}(0) = 0$$ for all $$n\text{.}$$ Show that $$f(z) = 0$$ for all $$z \in U\text{.}$$
#### Exercise11.3.5.
Suppose $$U \subset \C$$ is open and connected, $$0 \in U\text{,}$$ and $$f \colon U \to \C$$ is analytic. For real $$x$$ and $$y\text{,}$$ let $$h(x) \coloneqq f(x)$$ and $$g(y) \coloneqq -i \, f(iy)\text{.}$$ Show that $$h$$ and $$g$$ are infinitely differentiable at the origin and $$h'(0) = g'(0)\text{.}$$
#### Exercise11.3.6.
Suppose a function $$f$$ is analytic in some neighborhood of the origin, and that there exists an $$M$$ such that $$\sabs{f^{(n)}(0)} \leq M$$ for all $$n\text{.}$$ Prove that the series of $$f$$ at the origin converges for all $$z \in \C\text{.}$$
#### Exercise11.3.7.
Suppose $$f(z) \coloneqq \sum_{n=0}^\infty c_n z^n$$ with a radius of convergence 1. Suppose $$f(0) = 0\text{,}$$ but $$f$$ is not the zero function. Show that there exists a $$k \in \N$$ and a convergent power series $$g(z) \coloneqq \sum_{n=0}^\infty d_n z^n$$ with radius of convergence 1 such that $$f(z) = z^k g(z)$$ for all $$z \in B(0,1)\text{,}$$ and $$g(0) \not= 0\text{.}$$
#### Exercise11.3.8.
Suppose $$U \subset \C$$ is open and connected. Suppose that $$f \colon U \to \C$$ is analytic, $$U \cap \R \not= \emptyset$$ and $$f(x) = 0$$ for all $$x \in U \cap \R\text{.}$$ Show that $$f(z) = 0$$ for all $$z \in U\text{.}$$
#### Exercise11.3.9.
For $$\alpha \in \C$$ and $$k=0,1,2,3\ldots\text{,}$$ define
\begin{equation*} \binom{\alpha}{k} \coloneqq \frac{\alpha(\alpha-1)\cdots(\alpha-k)}{k!} . \end{equation*}
1. Show that the series
\begin{equation*} f(z) \coloneqq \sum_{k=0}^\infty \binom{\alpha}{k} z^k \end{equation*}
converges whenever $$\abs{z} < 1\text{.}$$ In fact, prove that for $$\alpha = 0,1,2,3,\ldots$$ the radius of convergence is $$\infty\text{,}$$ and for all other $$\alpha$$ the radius of convergence is 1.
2. Show that for $$x \in \R\text{,}$$ $$\abs{x} < 1\text{,}$$ we have
\begin{equation*} (1+x) f'(x) = \alpha f(x) , \end{equation*}
meaning that $$f(x) = (1+x)^\alpha\text{.}$$
#### Exercise11.3.10.
Suppose $$f \colon \C \to \C$$ is analytic and suppose that for some open interval $$(a,b) \subset \R\text{,}$$ $$f$$ is real valued on $$(a,b)\text{.}$$ Show that $$f$$ is real-valued on $$\R\text{.}$$
#### Exercise11.3.11.
Let $$\D \coloneqq B(0,1)$$ be the unit disc. Suppose $$f \colon \D \to \C$$ is analytic with power series $$\sum_{n=0}^\infty c_n z^n\text{.}$$ Suppose $$\sabs{c_n} \leq 1$$ for all $$n\text{.}$$ Prove that for all $$z \in \D\text{,}$$ we have $$\sabs{f(z)} \leq \frac{1}{1-\sabs{z}}\text{.}$$
For a higher quality printout use the PDF versions: https://www.jirka.org/ra/realanal.pdf or https://www.jirka.org/ra/realanal2.pdf |
# (solved)Question 8 SSC-CGL 2018 June 4 Shift 1
An amount becomes 8,028 in 3 years at a fixed percentage interest rate and 12,042 in 6 years, when the interest is compounded annually. What is the actual amount?
(Rev. 18-Jun-2024)
## Categories | About Hoven's Blog
,
### Question 8SSC-CGL 2018 June 4 Shift 1
An amount becomes 8,028 in 3 years at a fixed percentage interest rate and 12,042 in 6 years, when the interest is compounded annually. What is the actual amount?
1. 5352
2. 5235
3. 5253
4. 5325
### Solution in Short
Periods of growth are 3 year each, so we can apply unitary method.
Observe that Re. 12042 is obtained from 8028 in three years.
Hence, Re. 8028 was obtained from:
$\displaystyle \frac{8028}{12042} \times 8028 = 5352 \text{ Rs. } \:\underline{Ans}$
### Solution in Detail
Use the CI formula on 8028, 3 years:
$\displaystyle 8028 = P \bigg(1 + r\bigg)^3 \text{ . . . (1)}$
Use the CI formula on 12042, 6 years:
$\displaystyle 12042 = P \bigg(1 + r\bigg)^6\text{ . . . (2)}$
Squaring (1), then dividing by (2):
$\displaystyle P = \frac{8028 \times 8028}{12042}$
Either calculate or conclude that P must be even. Hence (a) is the answer!
### Solution 3
In compound interest, percent growth is same for same periods of time.
$\displaystyle P \xrightarrow [\text{3 yrs}]{\times 1.5} 8028 \xrightarrow [\text{3 yrs}]{\times 1.5} 12042$
Hence P must be $\displaystyle \frac{2}{3} \times 8028= 5352$. Hence (a) is the answer! |
# What is the circle of Apollonius?
## What is the circle of Apollonius?
The circles of Apollonius of a triangle are three circles, each of which passes through one vertex of the triangle and maintains a constant ratio of distances to the other two. The isodynamic points and Lemoine line of a triangle can be solved using these circles of Apollonius.
How do you construct a circle tangent to three circles?
Construct the radical center of the three circles, and from that point, draw a line to each of the three poles. In this case, each line intersects its respective circle at two points. The six intersection points are points of tangency for two solution circles, with three tangent points on each solution.
What is the radical axis of two circles?
A radical axis of two circles is the locus of a point that moves in such a way that the tangent lines drawn from it to the two circles are of the same lengths. The radical axis of 2 circles is a line perpendicular to the line joining the centres. Consider two circles S1 and S2 with centres c1 and c2.
### Who created the Apollonian Gasket?
Apollonius
Apollonius discovered that there are two other non-intersecting circles, C4 and C5, which have the property that they are tangent to all three of the original circles – these are called Apollonian circles.
How do you construct a radical axis?
The construction is surprisingly simple: Draw any circle C(E,F) that intersects both A(B) and C(D) – one, say, in points G,H, the other in points I,J. Let K be the intersection of GH and IJ. The radical axis of A(B) and C(D) is the line through K perpendicular to the line of centers AC.
How do you construct a tangent to a circle without using the center?
Draw Any Circle. Take Any Point a on It and Construct Tangent at a Without Using the Centre of the Circle. – Geometry
1. Steps of Construction:
2. Step 1: Draw a circle with any radius and mark any point A on it.
3. Step 2: Draw chord AB and an inscribed ∠BCA.
#### How do you draw a tangent in engineering drawings?
Join A to the centre of the circle O. Bisect line AO so that point B is the mid-point of AO. With centre B, draw a semi-circle to intersect the given circle at point C. Line AC is the required tangent.
Is an Apollonian Gasket a fractal?
In mathematics, an Apollonian gasket or Apollonian net is a fractal generated by starting with a triple of circles, each tangent to the other two, and successively filling in more circles, each tangent to another three.
Is a circle a fractal?
The most iconic examples of fractals have bumps along their boundaries, and if you zoom in on any bump, it will be covered in bumps, etc etc. Both a circle and a line segment have Hausdorff dimension 1, so from this perspective it’s a very boring fractal. |
# What is the tangent side of a triangle?
## What is the tangent side of a triangle?
In any right triangle, the tangent of an angle is the length of the opposite side (O) divided by the length of the adjacent side (A). In a formula, it is written simply as ‘tan’.
What is SEC the opposite of?
The secant is the reciprocal of the cosine. It is the ratio of the hypotenuse to the side adjacent to a given angle in a right triangle.
What is the longest side of a right triangle?
hypotenuse
We define the side of the triangle opposite from the right angle to be the hypotenuse, h. It is the longest side of the three sides of the right triangle. The word “hypotenuse” comes from two Greek words meaning “to stretch”, since this is the longest side.
### What is a tangent in conversation?
A tangent is an entirely different topic or direction. The non-mathematical meaning of tangent comes from this sense of barely touching something: when a conversation heads off on a tangent, it’s hard to see how or why it came up.
What is sec theta equal to?
The reciprocal cosine function is secant: sec(theta)=1/cos(theta). The reciprocal sine function is cosecant, csc(theta)=1/sin(theta).
What is the tangent of a right triangle?
The tangent of a given angle in a right triangle is the ratio of the length of the opposite side to the length of the adjacent side. The tangent function is defined by the formula:
#### How to use the tangent function to find the adjacent?
Using the Tangent Function to Find the Adjacent. Using the Tangent Function to Find the Adjacent of a Right Triangle. The tangent function relates a given angle to the opposite side and adjacent side of a right triangle.
How many adjacent sides are there in a triangle?
every triangle will have 3 pairs of adjacent sides (each side will be an ‘adjacent’ side to 2 of the angles. each side of the triangle will be an opposite side to the angle that it doesn’t form a vertex with.
How is the tangent of an angle calculated?
Tangent (tan) function – Trigonometry. In a right triangle, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side. Try this Drag any vertex of the triangle and see how the tangent of A and C are calculated. |
# How do you solve (3x+3)(x+2)=2 by factoring?
Sep 13, 2015
$x = \frac{- 9 + \sqrt{33}}{6} , \frac{- 9 - \sqrt{33}}{6}$
#### Explanation:
$\left(3 x + 3\right) \left(x + 2\right) = 2$
Expand the left side by using the foil method.
$3 {x}^{2} + 6 x + 3 x + 6 = 2$
$3 {x}^{2} + 9 x + 6 = 2$
Subtract $2$ from both sides.
$3 {x}^{2} + 9 x + 4 = 0$
You now have a quadratic equation $a {x}^{2} + b x + c$, where $a = 3 , b = 9 , \mathmr{and} c = 4$.
Use the quadratic formula to solve for $x$.
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Substitute the values for a, b, and c into the equation.
$x = \frac{- 9 \pm \sqrt{{9}^{2} - 4 \cdot 3 \cdot 4}}{2 \cdot 3} =$
$x = \frac{- 9 \pm \sqrt{81 \cdot - 48}}{6} =$
$x = \frac{- 9 \pm \sqrt{33}}{6}$
Separate the equation into two separate equations to find both solutions for $x$.
$x = \frac{- 9 + \sqrt{33}}{6}$
$x = \frac{- 9 - \sqrt{33}}{6}$ |
# Solving Quadratic Equations by Extracting the Square Root
In the previous post, we have learned about quadratic equations or equations of the form $ax^2 + bx + c = 0$, where a is not equal to 0. In this equation, we want to find the value of x which we call the root or the solution to the equation.
There are three strategies in finding the root of the equation: by extracting the roots, by completing the square, and by the quadratic formula. In this example, we will discuss, how to find the root of the quadratic equation by extracting the root.
Just like in solving equations, if we want to find the value of x, we put all the numbers on one side, and all the x’s on one side. Since quadratic equations contain the term $x^2$, we can find the value of x by extracting the square roots. Below are five examples on how to do this.
Example 1: $2x^2 = 0$
Solution
Dividing both sides by 2, we have
$\frac{2x^2}{2} = \frac{0}{2}$.
This gives us
$x^2 = 0$.
Extracting the square root of both sides, we have
$x = 0$.
Therefore, the root $x = 0$.
Example 2: $x^2 - 36 = 0$
Solution
$x^2 - 36 = 0$.
Adding 36 to both side, we have
$x^2 - 36 + 36 = 0 + 36$.
$x^2 = 36$.
Extracting the square root of both sides, we have
$\sqrt{x^2} = \sqrt{36}$.
$x = \pm 6$.
In this example, x has two roots: x = 6 and x = -6.
Example 3: $x^2 + 81 = 0$
Solution
Subtracting 81 from both sides, we have
$x^2 + 81 - 81 = 0 - 81$
$x^2 = -81$
$\sqrt{x^2} = - 81$
$x = \sqrt{-81}$.
In this case, there is no number that when multiplied by itself is negative. For example, negative times negative is equal to positive, and positive times positive is equal to positive. Therefore, there is no real root. There is, however, what we call a complex root as shown in the video below.
Example 4: $5x^2 = 12$
Solution
$5x^2 = 12$
Dividing both sides by 5, we have
$x^2 = \frac{12}{5}$.
Extracting the square root of both sides, we have
$\sqrt{x^2} = \sqrt{\frac{12}{5}} = \frac{\sqrt{12}}{\sqrt{5}}$
$x = \frac{\sqrt{4}\sqrt{3}}{\sqrt{5}}$
$x = \frac{2 \sqrt{3}}{\sqrt{5}}$.
Example 5: $3x^2 = - 4$
$3x^2 = -4$
Dividing both sides by 3, we have
$x^2 = \frac{-4}{3}$.
Extracting the square root of both sides, we have
$\sqrt{x^2} = \sqrt{\frac{-4}{3}}$.
Again, the sign of the number inside the radical is negative, so there is no real root. To know how to compute for the complex root, watch the video below.
### 6 Responses
1. Hartford Denver Acapulco says:
THANK YOU VERY MUCH.. I am Hartford Denver M. Acapulco from Iloilo City, Job Hire of LGU-Iloilo City.
2. Hartford Denver Acapulco says:
Thanks again… Maraming salamat po sa inyong lahat… I took the Professional Civil Service Exam twice but still failed to pass. I am presently on a self-review, hoping to pass the SUB-PROFESSIONAL Paper-Pencil-Test this coming April. I am presently employed at the City Government of Iloilo as a JOB HIRE for 5 years and 6 months.
• Civil Service Reviewer says:
You’re welcome po. Sana makapasa kayo. 🙂
1. December 23, 2015
[…] the previous post, we have learned how to solve quadratic equations by extracting the roots. In this post, we are going to learn how to solve quadratic equations by […]
2. January 3, 2016
[…] the previous posts, we have learned how to solve quadratic equations by getting the extracting the square root, by factoring, and by quadratic formula. We continue this series by learning how to solve math word […]
3. January 23, 2016
[…] where , and are real numbers and . Depending on the form of the equation, you can solve for by extracting the quare root, factoring, or using the quadratic formula.This type of equation appears in various problems that […] |
Examples
Chapter 7 Class 11 Binomial Theorem
Serial order wise
### Transcript
Example 4 Using binomial theorem, prove that 6n – 5n always leaves remainder 1 when divided by 25. Writing 6n = (1 + 5)n (6)n = nC0 1n 50 + nC1 1n – 1 51 + nC2 1n – 2 52…... + nCn 1n – n 5n = nC0 50 + nC1 51 + nC2 52 + ………..…... + nCn 5n = 1 × 1 + 𝑛!/1!( 𝑛 −1)! 51 + 𝑛!/2!( 𝑛 −2)! 52 +………. + 1 × 5n = 1 + (𝑛(𝑛−1)!)/1( 𝑛−1)! 5 + (𝑛(𝑛 − 1)(𝑛 − 2)!)/2( 𝑛 −2)! 52 +………. + 5n = 1 + n(5) + (𝑛(𝑛 − 1) )/2 52 +………. + 5n We know that (a + b)n = nC0 an + nC1 an – 1 b 1 + nC2 an – 2 b2 + ….. + nCn bn Putting a = 1 ,b = 5 Thus, (6)n = 1 + 5n + (𝑛(𝑛 − 1) )/2 52………. + 5n 6n – 5n = 1 + (𝑛(𝑛 − 1) )/2 52………. + 5n 6n – 5n = 1 + 52 ((𝑛(𝑛 − 1) )/2 " ………. + 5n – 2" ) 6n – 5n = 1 + 25 ((𝑛(𝑛 − 1) )/2 " ………. + 5n – 2" ) 6n – 5n = 1 + 25k where k = ( (𝑛(𝑛 − 1) )/2 ………. + 5n – 2 ) The above equation is of the form Dividend = Divisor × Quotient + Remainder Thus, (6)n = 1 + 5n + (𝑛(𝑛 − 1) )/2 52………. + 5n 6n – 5n = 1 + (𝑛(𝑛 − 1) )/2 52………. + 5n 6n – 5n = 1 + 52 ((𝑛(𝑛 − 1) )/2 " ………. + 5n – 2" ) 6n – 5n = 1 + 25 ((𝑛(𝑛 − 1) )/2 " ………. + 5n – 2" ) 6n – 5n = 1 + 25k where k = ( (𝑛(𝑛 − 1) )/2 ………. + 5n – 2 ) The above equation is of the form Dividend = Divisor × Quotient + Remainder Thus, (6)n = 1 + 5n + (𝑛(𝑛 − 1) )/2 52………. + 5n 6n – 5n = 1 + (𝑛(𝑛 − 1) )/2 52………. + 5n 6n – 5n = 1 + 52 ((𝑛(𝑛 − 1) )/2 " ………. + 5n – 2" ) 6n – 5n = 1 + 25 ((𝑛(𝑛 − 1) )/2 " ………. + 5n – 2" ) 6n – 5n = 1 + 25k where k = ( (𝑛(𝑛 − 1) )/2 ………. + 5n – 2 ) The above equation is of the form Dividend = Divisor × Quotient + Remainder Putting Dividend = 6n – 5n , Divisor = 25 , Remainder = 1 6n – 5n = 25k + 1 Hence 6n – 5n always leave Remainder 1 when dividing by 25 Hence Proved Example 4 Using binomial theorem, prove that 6n – 5n always leaves remainder 1 when divided by 25. Writing 6n = (1 + 5)n (6)n = nC0 1n 50 + nC1 1n – 1 51 + nC2 1n – 2 52…... + nCn 1n – n 5n = nC0 50 + nC1 51 + nC2 52 + ………..…... + nCn 5n = 1 × 1 + 𝑛!/1!( 𝑛 −1)! 51 + 𝑛!/2!( 𝑛 −2)! 52 +………. + 1 × 5n = 1 + (𝑛(𝑛−1)!)/1( 𝑛−1)! 5 + (𝑛(𝑛 − 1)(𝑛 − 2)!)/2( 𝑛 −2)! 52 +………. + 5n = 1 + n(5) + (𝑛(𝑛 − 1) )/2 52 +………. + 5n We know that (a + b)n = nC0 an + nC1 an – 1 b 1 + nC2 an – 2 b2 + ….. + nCn bn Putting a = 1 ,b = 5 Thus, (6)n = 1 + 5n + (𝑛(𝑛 − 1) )/2 52………. + 5n 6n – 5n = 1 + (𝑛(𝑛 − 1) )/2 52………. + 5n 6n – 5n = 1 + 52 ((𝑛(𝑛 − 1) )/2 " ………. + 5n – 2" ) 6n – 5n = 1 + 25 ((𝑛(𝑛 − 1) )/2 " ………. + 5n – 2" ) 6n – 5n = 1 + 25k where k = ( (𝑛(𝑛 − 1) )/2 ………. + 5n – 2 ) The above equation is of the form Dividend = Divisor × Quotient + Remainder Putting Dividend = 6n – 5n , Divisor = 25 , Remainder = 1 6n – 5n = 25k + 1 Hence 6n – 5n always leave Remainder 1 when dividing by 25 Hence Proved |
# Unitary Method Online Test Aptitude Questions With Solutions, Online Test On Unitary Method
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The method in which you can find the value of a unit and the a required number of units.
Let's assume you go to the market to purchase 6 Mangoes. The shopkeeper tells you that he is selling 10 Mangoes for Rs 100. In this case, the Units are Mangoes and the Value is the cost of the Mangoes. While solving a problem using the unitary method, it is important to identify the units and values.
For simplification, it is always suggested to express the calculated quantities on the right-hand side and the known quantities on the left-hand side. In the given problem, we are informed about the amount of Mangoes, but the value of Mangoes is unknown. It is necessary to keep in mind that the concept of ratio and proportion are commonly applied to this problems that are used this method.
The best Strategy to enhance your knowledge in Unitary Method sharpen your skills by utilizing the Free Online Test platform, where you can acquire various practice questions and their solutions.
## Some Important Unitary Method Formulas and Questions
Question-1)
A Bus travelling at a speed of 140 kmph covers 420 km. How much time will it take to cover 280 km?
First, we need to find the time required to cover 420 km.
Speed = $\text"Distance"/\text"Time"$
140 = $420/T$
T = 3 hours
Applying the unitary method,
420 km = 3 hours
1 km = $3/420$ hour
280 km = $(3/420) × 280$ = 2 hours
Question-2)
If 6 identical machines can produce a total of 270 bottles per minute, what is the total number of bottles that 10 such machines could produce in 4 minutes, assuming they run at the same constant rate?
Anwer:
From question,
6 ${\text"machines can produce"} = 270$ bottles in 1 minute
then,$1 {\text"machine can produce"} = {270} / 6$ bottles in 1 minute
So,10 ${\text"machines can produce"} = {270 × 10 }/ 6$ bottles in 1 minute
Then, 10${\text"machines can produce"} = {270 × 10 × 4} / 6$ bottles in 4 minute
So, 10 ${\text"machines can produce"} = 45 × 10 × 4$ bottles in 4 minute
then,10 ${\text"machines can produce"}= 1800$ bottles in 4 minute
Question-3)
P finishes his work in 15 days while Q takes 10 days. How many days will the same work be done if they work together?
If P takes 15 days to finish his work then,
P’s One day of work = $1/15$
Similarly, Q’s One day of work = $1/10$
Now, total work is done by P and Q in 1 day = $1/15 + 1/10$
Taking LCM(15, 10), we have,
One day’s work of P and Q = ${(2+3)}/30$
One day’s work of (P + Q) = $1/6$
Thus, P and Q can finish the work in 6 days if they work together.
Question-4)
If a worker makes a toy every 2 hours, how many toys will he make if he works for 80 hours?
According to question,
worker makes 1 toy in every 2 hours,
then in 1 hour, worker makes $1/2$ toy
In, 80 hour worker can makes $(1/2 )$ × (80 hours) = 40 toys
then, the worker will make 40 toys in 80 hours
Question-5)
If the price of $1/4$ kg is Rs. 0.60, then find the cost of 200 grams?
We know,
1 kg = 1000 grams
⇒$1/4$ kg = 1000 × $1/4$ = 250 g
from question, the price of 250 g = 60 paise
then, the price of 200 g = $60 / {250} × 200$
= 48 paise
Online Test - 1 (Unitary Method) TAKE TEST Number of questions : 20 | Time : 30 minutes
Online Test - 2 (Unitary Method) TAKE TEST Number of questions : 20 | Time : 30 minutes |
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# KS3 Maths Lesson Plan – Explore How Proportions Work By Mixing Paints
Thinking proportionally helps to make sense of lots of different real-life situations
Proportionality is one of the hardest and most important ideas in lower secondary mathematics.
Students need to encounter different examples of situations in which proportions feature, and to gain experience at representing proportions visually in different ways.
In this lesson, students work on a task about pink paint, which is made from a mixture of red paint and white paint. How much red paint do you need to add to pink paint which is 60% red to turn it into pink paint which is 80% red?
The answer may be more than you expect! Students work on several problems relating to a situation of ‘sunk costs’, where paint has been mixed in the wrong proportions and has to be corrected.
They will make up their own problems for each other to solve and experiment with using different ways to represent the makeup of the different mixtures of pink paint. This lesson provides lots of opportunities to explore how proportions work and gain a better sense of what something like “60% pink” really means.
Generalisations are possible, and the lesson ends with students describing what they have found out, what problems they have created for each other, and what they have learned from all of this about proportionality.
### Starter activity
Q. Shifa makes pink paint by mixing red paint and white paint. “60% pink paint” means that 60% of the mixture is red paint and the other 40% is white paint. How much red and white paint does Shifa need to use to make 5 litres of 60% pink?
This starter question should assess how well students have understood the scenario and whether they can calculate 60% of something. They need to be able to work out that 60% of 5 litres is 3 litres, meaning that Shifa needs to mix 3 litres of red with 2 litres of white.
### Main activity
Q. Shifa makes 5 litres of 60% pink, but then realises that she meant to make 5 litres of 80% pink instead! Oh dear! What can she do to fix it? She wants to use the smallest possible amount of extra paint to fix her mistake.
Clearly, she could tip it all away and start all over again, but instead she wants to use some or all of the 60% pink paint that she has made.
If students have no idea what to do, you could ask:
Q. Does it need to get redder or whiter?
Students should realise that 80% pink is redder than 60% pink, so Shifa is going to need to add some red paint. (She can’t remove white paint, as it is all mixed in!) But how much? It should be clear that there is no point adding white paint – that is going in the wrong direction, and will just increase the amount of red paint she will need to add later – so Shifa just needs to add red paint.
This is quite a tricky problem. Give students some time to think about it and let them write down and talk about their ideas. You might need to emphasise that it is not possible to separate out the red and white paint from the pink mixture – it is all completely stirred in! If students are really stuck you could suggest that they start by working out how much red paint and white paint is contained in the paint that Shifa has, and in the paint that Shifa wants.
One way to think about it is to treat the white paint in the paint that Shifa has as fixed. We know that 5 litres of 60% pink paint contains 2 litres of white. So this white paint must make up 20% of the 80% pink paint, meaning that there must be another 8 litres of red. We know that 3 of these 8 litres of red paint are already in the 60% pink mixture, so this means that we need to add another 5 litres of red paint. This will give us 10 litres of 80% paint.
Students will probably be a bit surprised by the large amount of red paint that needs to be added to get from 60% to 80%. They will probably have imagined that just a litre or two would have been enough. One way to see what is going on is to make a sketch like this:
So, we have ended up with 10 litres of 80% paint, whereas we only wanted 5 litres. We had to use 5 litres of red paint to fix Shifa’s mistake, and that is the same amount of paint that we would have needed to use to make 5 litres of 80% paint just starting from red paint and white paint!
We could have just mixed 4 litres of red with 1 litre of white and got what we needed without using the 60% paint that Shifa made at all!
Q. Can we do better than this? What if we don’t use all of the 60% paint that Shifa made?
Give students some time to experiment with this. The idea is that we tip away some of the pink paint before we start, because that means that we don’t need to add as much red paint as before to turn it pink enough.
Since we ended up with twice as much 80% pink paint as we wanted, students might start by imagining tipping out half of the 60% paint, so that they have just 2.5 litres of 60% pink. This will contain 1 litre of white paint and 1.5 litres of red paint.
To make this into 80% pink paint, 4 litres of red is needed in total, meaning that a further 2.5 litres of red paint must be added. This way we end up with 5 litres of 80% pink paint, which is exactly what we need, but this time it only entailed using 2.5 more litres of red paint.
Q. What if we only wanted 4 litres of 80% pink, rather than 5 litres of it?
The same “half and half” approach will still work. We take 2 litres of the 60% pink and add 2 litres of red paint. The 2 litres of 60% pink contains 0.8 litres of white and the remaining 4 – 0.8 = 3.2 litres of paint is red. This means that we have 80% pink paint, as desired.
Q. Try working out how to get other quantities of 80% pink, and how to get 90% pink and 40% pink and so on.
Students could present their questions and solutions on poster paper, explaining for each example how they worked it out. They might want to represent portions of the paint as rectangles (as above) or in some other way. They might be able to generalise some of their methods and write a “recipe” for how to do problems like this.
### Discussion
You could conclude the lesson with a plenary in which the students talk about the questions that they posed and the solutions that they came up with.
Q. How did you work these out? What did you learn about how to solve problems like this? Which problems were easy and which were tricky? Why? What problems did you make up? How did you solve them? Did anything surprise you?
It is unlikely that students will generalise this far, but it may be helpful for the teacher to have a complete sense of what is going on:
Suppose you start with a litres of p% paint and you want b litres of q% paint, and suppose that p<q, which means that we are going to need to add some red paint.
If we decide to use k litres of our starting p% paint (where ka), this will give us $\inline&space;\dpi{100}&space;\bg_white&space;\large&space;\frac{k(100-p)}{100}$ litres of white paint, which, to minimise wasted paint, needs to be all of the white paint in the q% paint that we want to end up with.
This means that $\inline&space;\dpi{100}&space;\bg_white&space;\large&space;\frac{k(100-p)}{100}=\frac{b(100-q)}{100}$, which means that we need to start by using only $\inline&space;\dpi{120}&space;\bg_white&space;\large&space;k=\frac{b(100-q)}{100-p}$ litres of our starting paint.
Doing this gets us $\inline&space;\dpi{120}&space;\bg_white&space;\large&space;\frac{bp(100-q)}{100-p}$ litres of red paint, and we need $\inline&space;\dpi{120}&space;\bg_white&space;\large&space;\frac{bq}{100}$ litres of red paint.
So the amount of red paint that we have to add is $\inline&space;\dpi{120}&space;\bg_white&space;\large&space;\frac{bq}{100}-\frac{bp(100-q)}{100-p}$, which simplifies to $\inline&space;\dpi{120}&space;\bg_white&space;\large&space;\frac{b(q-p)}{100-p}$. (Note that the restriction that k < a – ie we cannot use more of the starting paint than there is! – sets a limit on how large b can be: $\inline&space;\dpi{120}&space;\bg_white&space;\large&space;b<\frac{a(100-p)}{100-q}$.)
In the problem in the main activity, p% was 60% and q% was 80%, and because $\inline&space;\dpi{120}&space;\bg_white&space;\large&space;\frac{80-60}{100-60}$ was ½ we always needed to add red paint equal to half of the amount of paint we wanted. But for other values of p% and q% this will not necessarily be the case. So there is much to explore here that will be challenging for students to explain.
A nice related problem, with a video introduction, is available here at blog.mrmeyer.com.
### Stretch them further
There should be plenty here to stretch all students, but anyone in need of a greater challenge could suppose that the red paint is a bit more expensive than the white paint. This complicates the model in interesting ways!
Colin Foster is an Assistant Professor in mathematics education in the School of Education at the University of Nottingham. He has written many books and articles for mathematics teachers (see foster77.co.uk).
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# Integration by Partial Fractions – Integrals
• Difficulty Level : Medium
• Last Updated : 03 Jan, 2021
If f(x) and g(x) are polynomial functions such function. that g(x) ≠ 0 then f(x)/g(x) is called Rational Functions. If degree f(x) < degree g(x) then f(x)/g(x) is called a proper rational function. If degree f(x) < degree g(x) then f(x)/g(x) is called an improper rational function. If f(x)/g(x) is an improper rational function then by dividing f(x) by g(x), we can express f(x)/g(x) as the sum of a polynomial and a proper rational function.
### Partial Fractions
Any proper rational function p(x)/q(x) can be expressed as the sum of rational functions, each having the simplest factor q(x), each such fraction is known as a partial function and the process of obtaining them is called the decomposition or resolving of the given function into partial fractions.
### Integration by Partial Fractions
For example lets say we want to evaluate ∫[p(x)/q(x)] dx where p(x)/q(x) is in a proper rational fraction. In cases like these, we can write the integrand as in a form of the sum of simpler rational functions by using partial fraction decomposition after that integration can be carried out easily. Here The values of A, B, C, etc. can be obtained as per the question.
### Examples
Example 1: Evaluate ∫(x – 1)/(x + 1)(x – 2) dx?
Solution:
Let (x – 1)/(x + 1)(x – 2) = A/(x + 1) + B/(x – 2)
Then, (x – 1) = A(x – 2) + B(x + 1) ………………(i)
Putting x = -1 in (i), we get A = 2/3
Putting x = 2 in (i), we get B = 1/3
Therefore,
(x – 1)/(x + 1)(x – 2) = 2/3(x + 1) + 1/3(x – 2)
=> ∫(x – 1)/(x + 1)(x – 2) = 2/3∫dx/(x + 1) + 1/3∫dx/(x – 2)
= 2/3l
x + 1 | + 1/3 log | x – 2 | + C
Example 2: Evaluate ∫dx/x{6(log x)2 + 7log x + 2}?
Solution:
Putting log x = t and 1/x dx = dt, we get
I = ∫dx/x{6(log x)2 + 7log x + 2} = ∫dt/(6t2 + 7t + 2) = ∫dt/(2t + 1)(3t + 2)
Let 1/(2t + 1)(3t + 2) = A/(2t + 1) + B/(3t + 2)
Then, 1 ≡ A(3t + 2) + B(2t + 1) …………………….(i)
Putting t = -1/2 in (i), we get A = 2
Putting t = -2/3 in (i), we get B = -3
Therefore, 1/(2t + 1)(3t + 2) = 2/(2t + 1) + (-3)/(3t +
p>
=> I = ∫dt/(2t + 1)(3t + 2)
= ∫2dt/(2t + 1) – ∫3dt/(3t – 2)
= log | 2t + 1 | – log | 3t + 2 |
= log | 2t + 1 |/log | 3t + 2 | + C
= log | 2 log x + 1 | / log | 3 log x + 2 | + C
Example 3: Evaluate ∫dx/(x3 + x2 + x + 1)?
Solution:
We have 1/(x3 + x2 + x + 1) = 1/x2(x + 1) + (x + 1) = 1/(x + 1)(x2 + 1)
Let 1/(x + 1)(x2 + 1) = A/(x + 1) + Bx + C/(x2 + 1) ……………………(i)
=> 1 ≡ A(x2 + 1) + (Bx + C) (x + 1)
Putting x
on both sides of (i), we get A = 1/2.
Comparing coefficients of x2 on the both sides of (i), we get
A + B = 0 => B = -A = -1/2
Comparing coefficients of x on the both sides of (i), we get
B + C = 0 => C = -B = 1/2
Therefore, 1/(x + 1) (x2 + 1) = 1/2(x + 1) + (-1/2x + 1/2)/(x2 + 1)
Therefore, ∫1/(x + 1) (x2 + 1) = ∫dx/(x + 1) (x2 + 1)
= 1/2∫dx/(x + 1) – 1/2∫x/(x2 + 1)dx + 1/2∫dx/(x2 + 1)
= 1/2∫dx/(x + 1) – 1/4∫2x/(x2 + 1)dx + 1/2∫dx/(x2 + 1)
= 1/2 log | x + 1 | – 1/4 log | x2 + 1 | + 1/2 tan-1x + C
Example 4: Evaluate ∫x2/(x2
p> + 2)(x2 + 3)dx?
Solution:
Let x2/(x2 + 2) (x2 + 3) = y/(y + 2)(y + 3) where x2 = y.
Let y/(y + 2) (y + 3) = A/(y + 2) + B/(y + 3)
=> y ≡ A(y + 3) + B/(y + 2) ………………(i)
Putting y = -2 on the both sides of (i), we get A = -2.
Putting y = -3 on the both sides of (i), we get B = 3.
Therefore, y/(y + 2) (y + 3) = -2/(y + 2) + 3/(y + 3)
=> x2/(x2 + 2) (x2 + 3) = -2/(x2 + 2) + 3/(x2 + 3)
=> ∫x2/(x2 + 2) (x2 + 3) = -2∫dx/(x2 + 2) + 3∫dx/(x2 + 3)
= -2/√2tan-1(x/
) + 3/√3tan-1(x/√3) + C
= -√2tan-1(x/√2) + √3tan-1(x/√3) + C
Example 5: Evaluate ∫dx/x(x4 + 1)?
Solution:
We have
I = ∫dx/x(x4 + 1) = ∫x3/x4 (x4 + 1) dx [multiplying numerator and denominator by x3].
Putting x4 = t and 4x3dx = dt, we get
I = 1/4∫dt/t(t + 1)
= 1/4∫{1/t – 1/(t + 1)}dt [by partial fraction]
= 1/4∫1/t dt – 1/4∫1/(t + 1)dt
= 1/4 log | t | – 1/4 log | t + 1 | + C
= 1/4 log | x4 | – 1/4 log | x4 + 1 | + C
= (1/4 * 4) log | x | – 1/4 log | x4 + 1 | + C
= log | x | – 1/4 log | x4 + 1 | + C
My Personal Notes arrow_drop_up |
In Sam and Jill's garden there are two sorts of ladybirds. There are red Seven-Spot ladybirds with $7$ black spots and shiny black Four-Spot ladybirds with $4$ red spots.
Sam and Jill looked at a leaf with three ladybirds on it.
"One Seven-Spot ladybird," said Sam, "and two Four-Spot ones."
"That's $15$ spots altogether!" laughed Jill.
"I wonder if we could find ladybirds whose spots add to other numbers. I know how to do $16$."
"And $14$ is easy too," added Sam.
How would you make $16$ and $14$ spots with the Seven-Spot and Four-Spot ladybirds?
What other numbers can you make with adding $4$s and $7$s?
Can you get lots of numbers from say $4$ to $35$?
Are there some numbers you can't get?
### Why do this problem?
Ladybirds in the Garden is a good problem for children to practise addition, subtraction and possibly multiplication. It is a useful context in which to encourage learners to "have a go" and "play" with numbers, but then you can focus on having a system to find all the possible totals and giving reasons why some can't be made.
### Possible approach
It would be good to have the two pictures of the ladybirds in the problem displayed for the children to see, either printed off or on screen. You could get them started simply by asking how many spots there are altogether on the two, then how many on the picture of the three. You might say that you're picturing some ladybirds on a leaf whose spots add to $16$ and ask children in pairs to think about which ladybirds they are.
Set the class off on the task of finding other possible totals, saying very little more. Wander round the group as they work, looking for different ways of recording and different ways of working. After 5/10 minutes, draw attention to some of these. Some might be writing number sentences, others might be drawing pictures. Some might be starting with a total, others might be starting from a combination of ladybirds. It is important to stress that all of these are valid and it depends what each pair finds most helpful. At this stage, remind the children that you'd like them to be sure they have found all the possible totals. How will they know? Invite them to share their ideas which could lead into a discussion of having a system, for example working in numerical order through the totals, or by using only four-spot ladybirds first, then only seven-spot, then both.
You could bring their ideas together by asking them to write each different total, and how they made it, on a strip of paper. Gather these strips on the board or a wall and ask the children to arrange them in numerical order. Draw out a list of numbers which haven't been made and ask all pairs to check that they're sure they are impossible. The important point here is for pupils to try to explain why these totals can't be made. Ask pairs to convince each other and then invite some explanations for all to hear.
### Key questions
Which totals have you found so far? Which ladybirds made these totals?
What totals could we make if there are no four-spot ladybirds?
What totals could we make if there are no seven-spot ladybirds?
What totals could we made if there are both four-spot and seven-spot ladybirds?
How do you know you've made all the totals that are possible?
Why can't we make these totals?
### Possible extension
The article, Opening Out , suggests possible routes for further investigation in general, but this problem is also mentioned specifically. Children could be asked what combination of spots would be needed to be able to generate all the numbers under $30$. Would it be possible to do this without a one-spot ladybird? If we couldn't use a one-spot ladybird, what totals would be possible?
### Possible support
Some children might find it useful to have a sheet with consecutive numbers from $1$ to $29$ written on so that they can fill in ways of making each total as they go along. Calculators might be available, or other apparatus to support the arithmetic, for example a hundred square, numberline, Cuisenaire rods or counters/cubes. This document contains multiple copies of each ladybird which could be cut into cards. To make the task even more accessible, you could choose and create ladybirds with $2$ or $3$ spots instead. |
# The function f is defined as f(x) = x/(x-1), how do you find f(f(x))?
Apr 8, 2017
Substitute f(x) for every x and then simplify.
#### Explanation:
Given: $f \left(x\right) = \frac{x}{x - 1}$
Substitute f(x) for every x
$f \left(f \left(x\right)\right) = \frac{\frac{x}{x - 1}}{\left(\frac{x}{x - 1}\right) - 1}$
Multiply numerator and denominator by 1 in the form of $\frac{x - 1}{x - 1}$
$f \left(f \left(x\right)\right) = \frac{\frac{x}{x - 1}}{\left(\frac{x}{x - 1}\right) - 1} \frac{x - 1}{x - 1}$
$f \left(f \left(x\right)\right) = \frac{x}{x - x + 1}$
$f \left(f \left(x\right)\right) = \frac{x}{1}$
$f \left(f \left(x\right)\right) = x$
This means that $f \left(x\right) = \frac{x}{x - 1}$ is its own inverse. |
# The Constant Difference Concept
The Constant Difference Concept is derived from the
Comparison Concept . This concept is applicable when the problems deal with an equal quantity being transferred in or transferred out of the two variables concerned. This leaves the two variables with an equal increase or decrease in value. The unique feature in this concept lies in the fact that after the transfer in or transfer out of quantities, the difference between the two variables remains unchanged, hence the name "Constant Difference" Concept.
To illustrate this concept, take a look at the following problems.
### (A) Equal Amount Transferred into 2 Variables
Question: Ken had 14 pens and Ben had 2 pens. When they received an equal number of pens from their teacher, the ratio of Ken's pens to Ben's pens became 3:1. How many pens did each of them receive from their teacher?
Answer: For this question, we will work backwards. It is always easier to start drawing models where a multiples-relationship exist, i.e., a variable is a multiple of another variable. In this example, Ken is 3 times of Ben after the transfer in.
Step 1: Since the ratio of Ken's pens to Ben's pens after the transfer in is 3:1, we draw 3 boxes to represent the number of units that Ken had and 1 box to represent the number of units that Ben had.
Step 2: Since an equal amount was transferred in, we mark out an equal amount from both Ken's and Ben's model bars to show this amount.
Step 3: Next, we label the models with the number of pens they each had at first.
Step 4: After all information have been put into the model, we can then mark out all the known parts and try to make all the unknown parts equal.
Step 5: From Ken's bar, we can see that,
2 units + 2 pens + 2 pens + 2 pens ----------> 14 pens
2 units + 6 pens ----------> 14 pens
2 units ----------> 14 pens - 6 pens = 8 pens
1 unit ----------> 8 pens / 2 units = 4 pens
Thus, they reach receive 4 pens from their teacher.
### (B) Equal Amount Transferred out of 2 Variables
Question: Chloe had 18 stickers and Jane had 6 stickers. When they both gave away an equal number of stickers, Chloe had 4 times as many stickers as Jane. How many stickers did they each gave away?
Answer: For this question, we will again work backwards. It is always easier to start drawing models where a multiples-relationship exist, i.e., a variable is a multiple of another variable. In this example, Chloe is 4 times of Jane after the transfer out.
Step 1: Since the ratio of Chloe's pens to Jane's pens after the transfer out is 4:1, we draw 4 boxes to represent the number of units that Chloe had and 1 box to represent the number of units that Jane had.
Step 2: Since an equal amount was transferred out, we add an equal amount back to represent the part that each of them gave away.
Step 3: Next, we label the models with the number of stickers they each had at first.
Step 4: Rearrange the units in Chloe's bar.
From the model, we can see that after we shift the last unit of Chloe's bar forward, the remaining 3 unknown but equal units add up to 12.
3 units ----------> 12 stickers
1 unit -----------> 12 stickers / 3 = 4 stickers
Step 5: Putting the value of 4 stickers per unit back into the model, we get:
Looking at Jane's model,
6 stickers - 4 stickers = 2 stickers
Therefore, they each gave away 2 stickers.
### (C) Age Difference
Question: Matthew is 29 years old and his son is 5 yrs old now. In how many years will Matthew be thrice as old as his son?
Answer: In such questions, do bear in mind that the age difference between any 2 persons will always remain a constant.
Step 1: It is easier to draw the "After" model first and work backwards to the "Before" model. First, we draw 3 boxes to represent Matthew's age and 1 box to represent the son's age after their age increases.
Step 2: Since an equal amount was added to their ages, we mark out an equal box on both their model bars to represent the number of years they have grown from their present age.
Step 3: Next, we label the models with their present age.
Step 4: We then divide the model to reflect the known and unknown units.
From the model,
2 units + 5 years + 5 years + 5 years ----------> 29 years
2 units + 15 years ----------> 29 years
2 units ----------> 29 years - 15 years
2 units ----------> 14 years
1 unit ----------> 14 years / 2 = 7 years
Therefore, Matthew will be thrice as old as his son in 7 years' time.
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# Trigonometric Functions of a General Angle
So far, we dealt with right triangles and considered the trigonometric functions of an acute angle. In this section, we extend the definition of the trigonometric functions to any angles.
### Unit Circle and Angles
To find trig functions of an arbitrary angle, it is convenient to use a unit circle. A unit circle is a circle with a radius of one, centered at the origin of the Cartesian plane.
An angle is said to be in standard position if its vertex is located at the origin and the initial side lies on the positive $$x-$$axis. If the rotation is counterclockwise, the angle has a positive measure. Respectively, if the rotation is clockwise, the angle is negative.
The $$x-$$ and $$y-$$ axes divide the coordinate plane into $$4$$ quarters called quadrants. The unit circle has $$360^\circ.$$ Hence, each quadrant is equal to $$90^\circ.$$ Angles in the $$1\text{st}$$ quadrant range from $$0^\circ$$ to $$90^\circ.$$ The $$2\text{nd}$$ quadrant angles range from $$90^\circ$$ to $$180^\circ,$$ and so on.
Angles that have a measure multiple of $$90^\circ$$ do not belong to a quadrant. Their terminal sides lie on the $$x-$$ or $$y-$$axis. Such angles are called quadrantal.
The reference angle of an angle $$\alpha$$ is a positive acute angle formed by the terminal side of $$\alpha$$ and the $$x-$$axis. For example, if an angle $$\alpha$$ lies in the $$1\text{st}$$ quadrant, its reference angle is equal to $$\alpha$$ itself. For an angle $$\alpha$$ in the $$2\text{nd}$$ quadrant, the reference angle is equal to $$180^\circ – \alpha.$$
### Unit Circle Definitions of Sine and Cosine
Let $$\alpha$$ be any angle in standard position in the unit circle. Suppose that the terminal side of the angle intersects the circle at a point $$M\left( {x,y} \right).$$
The sine of the angle $$\alpha$$ is defined by
Similarly, the cosine function is defined on the coordinate plane by the formula
These expressions conform the definition of sine and cosine in a right triangle. Indeed, the triangle $$OAM$$ in Figure $$1$$ is a right triangle. In this triangle, the radius $$r = OM$$ is the hypotenuse, and the $$y-$$and $$x-$$coordinates form, respectively, the opposite and adjacent legs.
Since we consider a unit circle, the distance $$r$$ is equal to $$1$$ for any point $$\left( {x,y} \right)$$ of the circle:
$r = \sqrt {{x^2} + {y^2}} = 1.$
Therefore, the sine and cosine of an angle $$\alpha$$ can be simply defined as the coordinates $$y$$ and $$x$$ of the point $$M\left( {x,y} \right)$$ determined by the angle $$\alpha:$$
${\sin \alpha = y ,\;\;}\kern0pt{\cos \alpha = x.}$
When the angle $$\alpha$$ is measured in radians, we can consider any real values of $$\alpha.$$ So, the unit circle definition of sine and cosine is more general than in a right triangle.
### Unit Circle Definitions of Tangent and Cotangent
Consider again the unit circle and an angle $$\alpha$$ in standard position. The angle $$\alpha$$ determines the location of a point $$M\left( {x,y} \right)$$ on the circle.
The tangent of $$\alpha$$ on the coordinate plane is defined as
The tangent function is expressed in terms of sine and cosine in the form
${\tan \alpha = \frac{y}{x} }={ \frac{{\frac{y}{r}}}{{\frac{x}{r}}} }={ \frac{{\sin \alpha }}{{\cos \alpha }}.}$
As you can see, the tangent function is defined at the points where $$cos \alpha = x = 0.$$
On the unit circle in Figure $$3,$$ the value of the tangent function is displayed by the vertical segment line $$BK$$ passing through the point $$B\left( {1,0} \right).$$ Indeed, the triangles $$OAM$$ and $$OBK$$ are similar. Therefore,
$\frac{{AM}}{{OA}} = \frac{{BK}}{{OB}}.$
Here $$AM= OS = y = \sin \alpha,$$ $$OA = x = \cos \alpha,$$ $$OB = 1.$$ Hence,
${BK = \frac{{AM}}{{OA}} \times OB }={ \frac{{\sin \alpha }}{{\cos \alpha }} \times 1 }={ \tan \alpha .}$
The cotangent of angle $$\alpha$$ is given by
We can also express the cotangent function as the ratio of cosine and sine:
${\cot \alpha = \frac{x}{y} }={ \frac{{\frac{x}{r}}}{{\frac{y}{r}}} }={ \frac{{\cos \alpha }}{{\sin \alpha }}.}$
The cotangent function is not defined at the points where $$\sin \alpha = y = 0.$$
In Figure $$3$$ above, the cotangent of $$\alpha$$ is numerically equal to the length of the horizontal segment line $$PL.$$ Since the triangles $$OSM$$ and $$OPL$$ are similar, we have
$\frac{{SM}}{{OS}} = \frac{{PL}}{{OP}}.$
Then
${PL = \frac{{SM}}{{OS}} \times OP }={ \frac{{\cos \alpha }}{{\sin \alpha }} \times 1 }={ \cot \alpha .}$
### Unit Circle Definitions of Secant and Cosecant
The secant is the reciprocal of cosine. Therefore it is defined by the formula
In the unit circle below, the secant function of $$\alpha$$ is represented by the segment line $$OK.$$ This follows from the similarity of triangles $$OAM$$ and $$OBK.$$ For these triangles,
$\frac{{OM}}{{OA}} = \frac{{OK}}{{OB}}.$
Hence,
${OK = \frac{{OM}}{{OA}} \times OB }={ \frac{1}{{\cos \alpha }} \times 1 }={ \sec \alpha .}$
The cosecant is the reciprocal of sine. So we have
The value of the cosecant of $$\alpha$$ in Figure $$4$$ is equal to the length of the segment line $$OL.$$ Indeed, since $$\triangle OSM \sim \triangle OPL,$$ we can write
$\frac{{OM}}{{OS}} = \frac{{OL}}{{OP}}.$
Therefore,
${OL = \frac{{OM}}{{OS}} \times OP }={ \frac{1}{{\sin \alpha }} \times 1 }={ \csc \alpha .}$
### Trigonometric Functions of Special Angles
There are certain common angles that are more frequently used in trigonometry. The following table contains the values of trigonometric functions for such angles.
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
Find the value of the expression $\sin \frac{\pi }{4}\cos \frac{\pi }{6}\tan \frac{\pi }{3}.$
### Example 2
Find the value of the expression $\cot \frac{\pi }{6}\sec \frac{\pi }{3}\csc \frac{\pi }{4}\tan \frac{\pi }{6}.$
### Example 3
Calculate the values of the six trigonometric functions of $$\alpha = \large{\frac{\pi }{6}}\normalsize.$$
### Example 4
Calculate the values of the six trigonometric functions of $$\alpha = \large{\frac{\pi }{4}}\normalsize.$$
### Example 5
The terminal side of an angle $$\alpha$$ in standard position contains the point $$P\left( { – 2,3} \right).$$ Find the six trigonometric functions of the angle $$\alpha.$$
### Example 6
The terminal side of an angle $$\beta$$ in standard position contains the point $$Q\left( { – 8,-6} \right).$$ Find the six trigonometric functions of the angle $$\beta.$$
### Example 7
Calculate the sum of the series $1 + \sin \frac{\pi }{6} + {\sin ^2}\frac{\pi }{6} + {\sin ^3}\frac{\pi }{6} + \ldots$
### Example 8
Calculate the sum of the series $1 – \cos \frac{\pi }{4} + {\cos ^2}\frac{\pi }{4} – {\cos ^3}\frac{\pi }{4} + \ldots$
### Example 1.
Find the value of the expression $\sin \frac{\pi }{4}\cos \frac{\pi }{6}\tan \frac{\pi }{3}.$
Solution.
This expression contains trig functions of special angles. The values of these functions are given in the table above:
${\sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{2},\;\;}\kern0pt{\cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2},\;\;}\kern0pt{\tan \frac{\pi }{3} = \sqrt 3 .}$
Hence
${\sin \frac{\pi }{4}\cos \frac{\pi }{6}\tan \frac{\pi }{3} }={ \frac{{\sqrt 2 }}{2} \times \frac{{\sqrt 3 }}{2} \times \sqrt 3 }={ \frac{{3\sqrt 2 }}{4}.}$
### Example 2.
Find the value of the expression $\cot \frac{\pi }{6}\sec \frac{\pi }{3}\csc \frac{\pi }{4}\tan \frac{\pi }{6}.$
Solution.
Using the table above, we find that
${\cot \frac{\pi }{6} = \sqrt 3 ,\;\;}\kern0pt{\sec \frac{\pi }{3} = 2,\;\;}\kern0pt{\csc \frac{\pi }{4} = \sqrt 2 ,\;\;}\kern0pt{\tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}.}$
Substitute these values into our expression:
${\cot \frac{\pi }{6}\sec \frac{\pi }{3}\csc \frac{\pi }{4}\tan \frac{\pi }{6} }={ \sqrt 3 \times 2 \times \sqrt 2 \times \frac{1}{{\sqrt 3 }} }={ 2\sqrt 2 .}$
### Example 3.
Calculate the values of the six trigonometric functions of $$\alpha = \large{\frac{\pi }{6}}\normalsize.$$
Solution.
The right triangle $$OAM$$ is a special $$30\text{-}60\text{-}90$$ triangle in which the hypotenuse is twice the length of the shorter leg. Therefore, we have
${AM = \frac{{OM}}{2},}\;\; \Rightarrow {\sin \frac{\pi }{6} = \frac{r}{2} = \frac{1}{2}.}$
The cosine of $$\alpha = \large{\frac{\pi }{6}}\normalsize$$ can be found by the Pythagorean trig identity:
${\cos \frac{\pi }{6} }={ \sqrt {1 – {{\sin }^2}\left( {\frac{\pi }{6}} \right)} }={ \sqrt {1 – \left( {\frac{1}{2}} \right)} }={ \sqrt {1 – \frac{1}{4}} }={ \sqrt {\frac{3}{4}} }={ \frac{{\sqrt 3 }}{2}.}$
The other trigonometric functions of $$\alpha = \large{\frac{\pi }{6}}\normalsize$$ are given by
${\tan \frac{\pi }{6} = \frac{{\sin \frac{\pi }{6}}}{{\cos \frac{\pi }{6}}} }={ \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} }={ \frac{1}{{\sqrt 3 }};}$
${\cot \frac{\pi }{6} = \frac{{\cos \frac{\pi }{6}}}{{\sin \frac{\pi }{6}}} }={ \frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{1}{2}}} }={ \sqrt 3 ;}$
${\sec \frac{\pi }{6} = \frac{1}{{\cos \frac{\pi }{6}}} }={ \frac{1}{{\frac{{\sqrt 3 }}{2}}} }={ \frac{2}{{\sqrt 3 }};}$
${\csc \frac{\pi }{6} = \frac{1}{{\sin \frac{\pi }{6}}} }={ \frac{1}{{\frac{1}{2}}} }={ 2.}$
The table above with the values of trig functions for special angles is composed on the basis of these calculations.
### Example 4.
Calculate the values of the six trigonometric functions of $$\alpha = \large{\frac{\pi }{4}}\normalsize.$$
Solution.
We deal here with a $$45\text{-}45\text{-}90$$ triangle. This is a right isosceles triangle, so it has equal legs. Suppose the length of a leg be $$x.$$ By the Pythagorean theorem,
${{x^2} + {x^2} = {r^2},}\;\; \Rightarrow {{x^2} = \frac{{{r^2}}}{2},}\;\; \Rightarrow {x = \frac{r}{{\sqrt 2 }} }={ \frac{{r\sqrt 2 }}{2}.}$
Since $$r = 1,$$ we have
${x = \sin \frac{\pi }{4} }={ \cos \frac{\pi }{4} }={ \frac{{\sqrt 2 }}{2}.}$
Compute the values of the other trig functions:
$\require{cancel}{\tan \frac{\pi }{4} = \frac{{\sin \frac{\pi }{4}}}{{\cos \frac{\pi }{4}}} }={ \frac{\cancel{\frac{{\sqrt 2 }}{2}}}{\cancel{\frac{{\sqrt 2 }}{2}}} }={ 1;}$
$\require{cancel}{\cot \frac{\pi }{4} = \frac{{\cos \frac{\pi }{4}}}{{\sin \frac{\pi }{4}}} }={ \frac{\cancel{\frac{{\sqrt 2 }}{2}}}{\cancel{\frac{{\sqrt 2 }}{2}}} }={ 1;}$
${\sec \frac{\pi }{4} = \frac{1}{{\cos \frac{\pi }{4}}} }={ \frac{1}{{\frac{{\sqrt 2 }}{2}}} }={ \frac{2}{{\sqrt 2 }} }={ \sqrt 2 ;}$
${\csc \frac{\pi }{4} = \frac{1}{{\sin \frac{\pi }{4}}} }={ \frac{1}{{\frac{{\sqrt 2 }}{2}}} }={ \frac{2}{{\sqrt 2 }} }={ \sqrt 2 .}$
### Example 5.
The terminal side of an angle $$\alpha$$ in standard position contains the point $$P\left( { – 2,3} \right).$$ Find the six trigonometric functions of the angle $$\alpha.$$
Solution.
Here $$x = -2,$$ $$y = 3.$$ The distance of the point $$P\left( { – 2,3} \right)$$ from the origin is equal to
${r = \sqrt {{x^2} + {y^2}} }={ \sqrt {{{\left( { – 2} \right)}^2} + {3^2}} }={ \sqrt {4 + 9} }={ \sqrt {13} .}$
The trig functions of the angle $$\alpha$$ are given by
${\sin \alpha = \frac{y}{r} }={ \frac{{ – 2}}{{\sqrt {13} }} }={ – \frac{{2\sqrt {13} }}{{13}};}$
${\cos \alpha = \frac{x}{r} }={ \frac{{3}}{{\sqrt {13} }} }={\frac{{3\sqrt {13} }}{{13}};}$
${\tan \alpha = \frac{y}{x} }={ \frac{3}{{ – 2}} }={ – \frac{3}{2};}$
${\cot \alpha = \frac{x}{y} }={ \frac{-2}{{ 3}} }={ – \frac{2}{3};}$
${\sec \alpha = \frac{r}{x} }={ \frac{{\sqrt {13} }}{{ – 2}} }={ – \frac{{\sqrt {13} }}{2};}$
${\csc \alpha = \frac{r}{y} }={ \frac{{\sqrt {13} }}{{3}}. }$
### Example 6.
The terminal side of an angle $$\beta$$ in standard position contains the point $$Q\left( { – 8,-6} \right).$$ Find the six trigonometric functions of the angle $$\beta.$$
Solution.
Determine the distance $$r$$ from the origin to the point $$Q\left( { – 8,-6} \right):$$
${r = \sqrt {{x^2} + {y^2}} }={ \sqrt {{{\left( { – 8} \right)}^2} + {{\left( { – 6} \right)}^2}} }={ \sqrt {64 + 36} }={ \sqrt {100} }={ 10.}$
Calculate the values of the trig functions:
${\sin \beta = \frac{y}{r} }={ \frac{{ – 6}}{{10}} }={ – \frac{3}{5};}$
${\cos \beta = \frac{x}{r} }={ \frac{{ – 8}}{{10}} }={ – \frac{4}{5};}$
${\tan \beta = \frac{y}{x} }={ \frac{{ – 6}}{{-8}} }={ \frac{3}{4};}$
${\cot \beta = \frac{x}{y} }={ \frac{{ – 8}}{{- 6}} }={ \frac{4}{3};}$
${\sec \beta = \frac{r}{x} }={ \frac{{ 10}}{{-8}} }={ – \frac{5}{4};}$
${\csc \beta = \frac{r}{y} }={ \frac{{ 10}}{{-6}} }={ – \frac{5}{3}.}$
### Example 7.
Calculate the sum of the series $1 + \sin \frac{\pi }{6} + {\sin ^2}\frac{\pi }{6} + {\sin ^3}\frac{\pi }{6} + \ldots$
Solution.
Since $$\sin \large{\frac{\pi }{6}}\normalsize = \large{\frac{1}{2}}\normalsize,$$ we can write this expression in the form:
$1 + \frac{1}{2} + {\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^3} + \ldots$
We have here an infinite geometric series with the initial term $$a_1 =1$$ and the common ratio $$q = \large{\frac{1}{2}}\normalsize.$$ The sum of the geometric series is given by
${S = \frac{{{a_1}}}{{1 – q}} }={ \frac{1}{{1 – \frac{1}{2}}} }={ \frac{1}{{\frac{1}{2}}} }={ 2.}$
### Example 8.
Calculate the sum of the series $1 – \cos \frac{\pi }{4} + {\cos ^2}\frac{\pi }{4} – {\cos ^3}\frac{\pi }{4} + \ldots$
Solution.
We have here an infinite geometric series with the initial term $$a_1 = 1$$ and the negative common ratio $$q = – \cos \large{\frac{\pi }{4}}\normalsize = – \large{\frac{{\sqrt 2 }}{2}}\normalsize.$$ Determine the sum of the series:
${S = \frac{{{a_1}}}{{1 – q}} }={ \frac{1}{{1 – \left( { – \frac{{\sqrt 2 }}{2}} \right)}} }={ \frac{1}{{1 + \frac{1}{{\sqrt 2 }}}} }={ \frac{1}{{\frac{{\sqrt 2 + 1}}{{\sqrt 2 }}}} }={ \frac{{\sqrt 2 }}{{\sqrt 2 + 1}}.}$ |
# Square number
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In mathematics, a square number or perfect square is an integer that is the square of an integer;[1] in other words, it is the product of some integer with itself. For example, 9 is a square number, since it can be written as 3 × 3.
The usual notation for the formula for the square of a number n is not the product n × n, but the equivalent exponentiation n2, usually pronounced as "n squared". The name square number comes from the name of the shape; see below.
Square numbers are non-negative. Another way of saying that a (non-negative) number is a square number, is that its square roots are again integers. For example, 9 = ±3, so 9 is a square number.
A positive integer that has no perfect square divisors except 1 is called square-free.
For a non-negative integer n, the nth square number is n2, with 02 = 0 being the 0-th one. The concept of square can be extended to some other number systems. If rational numbers are included, then a square is the ratio of two square integers, and, conversely, the ratio of two square integers is a square (e.g., 4/9 = (2/3)2).
Starting with 1, there are $\lfloor \sqrt{m} \rfloor$ square numbers up to and including m, where the expression $\lfloor x \rfloor$ represents the floor of the number x.
## Examples
The squares (sequence A000290 in OEIS) smaller than 602 are:
02 = 0
12 = 1
22 = 4
32 = 9
42 = 16
52 = 25
62 = 36
72 = 49
82 = 64
92 = 81
102 = 100
112 = 121
122 = 144
132 = 169
142 = 196
152 = 225
162 = 256
172 = 289
182 = 324
192 = 361
202 = 400
212 = 441
222 = 484
232 = 529
242 = 576
252 = 625
262 = 676
272 = 729
282 = 784
292 = 841
302 = 900
312 = 961
322 = 1024
332 = 1089
342 = 1156
352 = 1225
362 = 1296
372 = 1369
382 = 1444
392 = 1521
402 = 1600
412 = 1681
422 = 1764
432 = 1849
442 = 1936
452 = 2025
462 = 2116
472 = 2209
482 = 2304
492 = 2401
502 = 2500
512 = 2601
522 = 2704
532 = 2809
542 = 2916
552 = 3025
562 = 3136
572 = 3249
582 = 3364
592 = 3481
The difference between any perfect square and its predecessor is given by the identity $n ^ 2 - (n - 1) ^ 2 = 2 n - 1$. Equivalently, it is possible to count up square numbers by adding together the last square, the last square's root, and the current root, that is, $n ^ 2 = (n - 1) ^ 2 + (n - 1) + n$.
## Properties
The number m is a square number if and only if one can compose a square of m equal (lesser) squares:
m = 12 = 1 m = 22 = 4 m = 32 = 9 m = 42 = 16 m = 52 = 25 Note: White gaps between squares serve only to improve visual perception. There must be no gaps between actual squares.
The unit of area is defined as the area of unit square (1 × 1). Hence, a square with side length n has area n2.
The expression for the nth square number is n2. This is also equal to the sum of the first n odd numbers as can be seen in the above pictures, where a square results from the previous one by adding an odd number of points (shown in magenta). The formula follows:
$n^2 = \sum_{k=1}^n(2k-1).$
So for example, 52 = 25 = 1 + 3 + 5 + 7 + 9.
There are several recursive methods for computing square numbers. For example, the nth square number can be computed from the previous square by $n^2 = (n-1)^2 + (n-1) + n = (n-1)^2 + (2n-1)$. Alternatively, the nth square number can be calculated from the previous two by doubling the (n − 1)-th square, subtracting the (n − 2)-th square number, and adding 2, because n2 = 2(n − 1)2 − (n − 2)2 + 2. For example,
2 × 52 − 42 + 2 = 2 × 25 − 16 + 2 = 50 − 16 + 2 = 36 = 62.
A square number is also the sum of two consecutive triangular numbers. The sum of two consecutive square numbers is a centered square number. Every odd square is also a centered octagonal number.
Another property of a square number is that it has an odd number of positive divisors, while other natural numbers have an even number of positive divisors. An integer root is the only divisor that pairs up with itself to yield the square number, while other divisors come in pairs.
Lagrange's four-square theorem states that any positive integer can be written as the sum of four or fewer perfect squares. Three squares are not sufficient for numbers of the form 4k(8m + 7). A positive integer can be represented as a sum of two squares precisely if its prime factorization contains no odd powers of primes of the form 4k + 3. This is generalized by Waring's problem.
A square number can end only with digits 0, 1, 4, 6, 9, or 25 in base 10, as follows:
1. If the last digit of a number is 0, its square ends in an even number of 0s (so at least 00) and the digits preceding the ending 0s must also form a square.
2. If the last digit of a number is 1 or 9, its square ends in 1 and the number formed by its preceding digits must be divisible by four.
3. If the last digit of a number is 2 or 8, its square ends in 4 and the preceding digit must be even.
4. If the last digit of a number is 3 or 7, its square ends in 9 and the number formed by its preceding digits must be divisible by four.
5. If the last digit of a number is 4 or 6, its square ends in 6 and the preceding digit must be odd.
6. If the last digit of a number is 5, its square ends in 25 and the preceding digits must be 0, 2, 06, or 56.
In base 16, a square number can end only with 0, 1, 4 or 9 and
• in case 0, only 0, 1, 4, 9 can precede it,
• in case 4, only even numbers can precede it.
In general, if a prime p divides a square number m then the square of p must also divide m; if p fails to divide mp, then m is definitely not square. Repeating the divisions of the previous sentence, one concludes that every prime must divide a given perfect square an even number of times (including possibly 0 times). Thus, the number m is a square number if and only if, in its canonical representation, all exponents are even.
Squarity testing can be used as alternative way in factorization of large numbers. Instead of testing for divisibility, test for squarity: for given m and some number k, if k2m is the square of an integer n then kn divides m. (This is an application of the factorization of a difference of two squares.) For example, 1002 − 9991 is the square of 3, so consequently 100 − 3 divides 9991. This test is deterministic for odd divisors in the range from kn to k + n where k covers some range of natural numbers k m.
A square number cannot be a perfect number.
The sum of the series of power numbers
$\sum_{n=0}^N n^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + \cdots + N^2$
can also be represented by the formula
$\frac{N(N+1)(2N+1)}{6}.$
The first terms of this series (the square pyramidal numbers) are:
0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201... (sequence A000330 in OEIS).
All fourth powers, sixth powers, eighth powers and so on are perfect squares.
## Special cases
• If the number is of the form m5 where m represents the preceding digits, its square is n25 where n = m × (m + 1) and represents digits before 25. For example the square of 65 can be calculated by n = 6 × (6 + 1) = 42 which makes the square equal to 4225.
• If the number is of the form m0 where m represents the preceding digits, its square is n00 where n = m2. For example the square of 70 is 4900.
• If the number has two digits and is of the form 5m where m represents the units digit, its square is AABB where AA = 25 + m and BB = m2. Example: To calculate the square of 57, 25 + 7 = 32 and 72 = 49, which means 572 = 3249.
• If the number ends in 5, its square will end in 5; similarly for ending in 25, 625, 0625, 90625, ... 8212890625, etc. If the number ends in 6, its square will end in 6, similarly for ending in 76, 376, 9376, 09376, ... 1787109376. For example, the square of 55376 is 3066501376, both ending in 376. (The numbers 5, 6, 25, 76, etc. are called automorphic numbers. They are sequence A003226 in the OEIS.)
## Odd and even square numbers
Squares of even numbers are even (and in fact divisible by 4), since (2n)2 = 4n2.
Squares of odd numbers are odd, since (2n + 1)2 = 4(n2 + n) + 1.
It follows that square roots of even square numbers are even, and square roots of odd square numbers are odd.
As all even square numbers are divisible by 4, the even numbers of the form 4n + 2 are not square numbers.
As all odd square numbers are of the form 4n + 1, the odd numbers of the form 4n + 3 are not square numbers.
Squares of odd numbers are of the form 8n + 1, since (2n + 1)2 = 4n(n + 1) + 1 and n(n + 1) is an even number.
Every odd perfect square is a centered octagonal number. The difference between any two odd perfect squares is a multiple of 8. The difference between 1 and any higher odd perfect square always is eight times a triangular number, while the difference between 9 and any higher odd perfect square is eight times a triangular number minus eight. Since all triangular numbers have an odd factor, but no two values of 2n differ by an amount containing an odd factor, the only perfect square of the form 2n - 1 is 1, and the only perfect square of the form 2n + 1 is 9.
## Notes
1. ^ Some authors also call squares of rational numbers perfect squares. |
# Sum of squares of natural numbers
In this section, we will discuss the formulas of the sum of the squares of natural numbers. These formulas are very useful in various competitive exams.
#### Sum of squares of first n natural numbers:
The sum of squares of consecutive natural numbers is determined by the formula below:
Proof: Let $S$ denote the desired sum. That is, $S=1^2+2^2+\cdots+n^2.$ We will use the following fact:
$n^3-(n-1)^3=3n^2-3n+1$ $\cdots$ (I)
This is obtained by applying the formula $a^3-b^3$$=(a-b)(a^2+ab+b^2)$
Using (I) we get the following relations:
$1^3-0^3=3.1^2-3.1+1$
$2^3-1^3=3.2^2-3.2+1$
$3^3-2^3=3.2^2-3.3+1$
$\quad \vdots \quad \quad \quad \vdots$
$n^3-(n-1)^3=3n^2-3n+1$
——————————————-
$n^3-0^3$ $=3(1^2+2^2+\cdots +n^2)$ $-3(1+2+\cdots +n)+$ $(1+1+\cdots$ till n terms)
$\Rightarrow n^3=3S-3 \frac{n(n+1)}{2}+n$, since the sum of the first n natural numbers is n(n+1)/2.
$\Rightarrow 3S=3 \frac{n(n+1)}{2}+n^3-n$
$\Rightarrow 3S=3 \frac{n(n+1)}{2}$ $+n(n-1)(n+1)$
$\Rightarrow 3S=n(n+1)[\frac{3}{2}$ $+(n-1)]$
$\Rightarrow 3S=n(n+1)[\frac{3+2n-2}{2}]$
$\Rightarrow 3S=n(n+1)[\frac{2n+1}{2}]$
$\Rightarrow S=\frac{n(n+1)(2n+1)}{6}$
Remark: Note that the above sum $1^2+2^2+\cdots +n^2$ can be written as $\sum_{k=1}^n k^2$
SOLVED EXAMPLES
Problem 1: Find the sum of the squares of the first 100 natural numbers.
Solution:
We need to find the sum $1^2+2^2+\cdots+100^2$
By the above formula, we know that the sum of the squares of first n natural numbers is $\frac{n(n+1)(2n+1)}{6}$
So we have the sum =
$1^2+2^2+\cdots+100^2$ $=\frac{n(n+1)(2n+1)}{6}$ where $n=100$
$=\frac{100(100+1)(2.100+1)}{6}$
$=\frac{100 \times 101 \times 201}{6}$
$=50 \times 101 \times 67$
$=338350$
#### Sum of squares of first n even natural numbers:
Proof: Note that
$2^2+4^2+6^2+\cdots+(2n)^2$
$=\sum_{k=1}^n(2k)^2$
$=\sum_{k=1}^n4k^2$
$=4\sum_{k=1}^nk^2$
$=4 \times \dfrac{n(n+1)(2n+1)}{6}$ as the sum of squares of first n natural numbers is n(n+1)(2n+1)/6.
$=\dfrac{2n(n+1)(2n+1)}{3}$
#### FAQs on Sum of Squares
Q1: What is sum of squares?
Answer: The sum of squares formula is generally referred by the sum of squares of first n natural numbers. It means 12 + 22 + 32 + … + n2 = Σ n2. The formula of sum of squares is as follows: 12 + 22 + 32 + … + n2 = [n(n+1)(2n+1)] / 6.
Q2: What is the formula for the sum of squares of even natural numbers?
Answer: The sum of squares of even natural numbers formula is Σ(2n)2 = 22 + 42 + 62 + … + (2n)2 = [2n(n + 1)(2n + 1)] / 3.
Q3: Write down the sum of squares of odd natural numbers formula.
Answer: The formula of the sum of squares of odd natural numbers is Σ(2n-1)2 = 12 + 32 + 52 + … + (2n-1)2 = [n(2n+1)(2n-1)] / 3.
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How to calculate definite integrals of a function.
This topic is part of the TCS FREE high school mathematics 'How-to Library', and will help you to calculate definite integrals of a function. (See the index page for a list of all available topics in the library.) To make best use of this topic, you need to download the Maths Helper Plus software. Click here for instructions.
### Theory:
Consider the problem of finding the area ‘A’ between a curved line: y = f(x) and the ‘x’ axis, between x = a and x = b:
We can approximate the area by slicing it into rectangles and adding up the areas of the rectangles. In the diagram below, there are three rectangles of equal width. The height of each rectangle equals the height of the function graph at its midpoint, and the width of each rectangle = (b-a)/3:
The total area under the curve from x = a to x = b can be approximated by: A = A1 + A2 + A3.
Notice that the area A1 is an overestimate of the area under the curve, A2 is about right, and A3 is too small. Thus this method may be accurate for some types of curves, but not for others. It depends on their shape.
One way to improve the accuracy for any shaped curve is to increase the number of rectangles. If there are ‘n’ rectangles, then as ‘n’ increases, the width of each rectangle decreases.
For very thin rectangles, areas at the top that don’t fit the curve become insignificant, then the sum of the rectangles becomes very close to the area under the curved graph line.
Let be the area under the graph: y = f(x) from x = a to x = b. We can calculate the area by summing up the areas of the rectangular slices. If there are ‘n’ slices, then the total area is approximated by:
The true area is called the 'definite integral' of the function, and can be written as the limit of this sum as ‘n’ approaches infinity, like this:
The 'Method' section below shows you how to use Maths Helper Plus to calculate the definite integral of a function by adding rectangles under its graph. This can be very accurate because the computer can add many thousands of tiny rectangles in a fraction of a second.
### Method:
To explain how to use Maths Helper Plus to calculate a definite integral, we will use the function:
y = 0.02x³ + 0.05x² - 0.3x + 2
We will calculate the definite integral from x = -3 to x = 4.
If you have just launched the software then you already have an empty document, otherwise hold down ‘Ctrl’ while you briefly press the ‘N’ key.
#### Step 2 Graph your function
1. Press the F3 key to activate the 'input box' for typing (see below):
2. Type the function into the input box:
3. Press ENTER to complete the entry
For more help on entering functions, see the 'Easy Start' tutorial in the Maths Helper Plus 'help'. To access the tutorial, click 'Help' on the Maths Helper Plus menu bar, then select: 'Tutorial...'
If the graph of the function does not appear, you need to adjust the graph scale.
First reduce the graph scale. Press the F10 key enough times until the main parts of the graph are visible.
To enlarge the graph, hold down 'Ctrl' while you press F10.
For more help on setting graph scales, click 'Help' on the Maths Helper Plus menu bar, then select 'Index'. Click the button at the top left corner of the help screen. Click 'Training', then 'Essential skills', then 'Graph scale magic'.
#### Step 3 Display the options dialog box and set options
Carefully point to the plotted function graph line with the mouse and double click the mouse. This will display the function options dialog box. Click to select the 'Integrals' tab at the top of the dialog box:
1. Click on the 'Limits of integration' edit box and type the lower and upper boundaries for the integral in square brackets like this: [a,b]
For the example, we type: [-3,4]
2. Click on the 'Number of intervals' edit box, and type these values of 'n':
2,4,10,50,100,200,300,500
The area under the graph between 'a' and 'b' will be divided up into 'n' slices to calculate the total area. Each number in this list will be used for 'n'. Higher values of 'n' will usually give a more accurate result, so by comparing results for increasing value of 'n', we can decide how accurate our answer is.
3. Select 'Shade areas on graph' to shade the area being calculated.
4. There are several different methods of calculating area under a curve. These can be selected by clicking the required options under 'Methods of Integration'. The method described in 'Theory' above is Rectangular (mid-point). the 'Trapezoidal rule' reduces errors by joining points on the graph with straight lines, while 'Simpson's rule' fits parabolas between points on the graph. You should select several methods for comparison.
5. Make sure 'Calculation mode' is set to 'definite integral', unless you are calculating the true area between the graph and the 'x' axis. The true area is the same as the definite integral, unless part of the area lies under the 'x' axis. For definite integrals, areas below the 'x' axis are negative, and subtract from the total answer.
Click OK to finish.
#### Step 4 Read the integral value from the table
The text view displays the table of calculated integrals. This is how the table looks for the example integral:
Definite integral from x = -3 to x = 4 n Simpson Trapezoidal Midpoint 2 15.3417 16.485 14.77 4 15.3417 15.6275 15.1988 10 15.3417 15.3874 15.3188 20 15.3417 15.3531 15.3359 50 15.3417 15.3435 15.3408 100 15.3417 15.3421 15.3414 200 15.3417 15.3418 15.3416 300 15.3417 15.3417 15.3416 500 15.3417 15.3417 15.3417
Simpson's rule was fastest to achieve four decimal place accuracy.
For the trapezoidal rule, four decimal place accuracy is achieved at n=300. This is proved because the last digit (7) does not change for n = 500.
For the midpoint method, n = 500 is required to achieve four decimal place accuracy.
NOTE: We could add another higher 'n' value to the list, say n = 700, to check out the accuracy of the last digit for the midpoint method, but since we already know this answer is correct from the other two columns, we don't need to do so. |
# Lesson Explainer: Proportional Equations Mathematics • 8th Grade
In this explainer, we will learn how to write an equation to describe the proportional relationship of data in a table or a graph.
Let us recall what it means when two quantities are in a directly proportional relationship.
### Directly Proportional Relationship
Two quantities and are directly proportional or in direct proportion when, from one situation to another, both quantities have been multiplied (or divided) by the same number. It follows that the ratios of quantity to quantity are in all situations equivalent.
Mathematically, this means that if, in a first situation, quantity is and quantity is and, in another situation, quantity is and quantity is , then and when is directly proportional to .
So, two proportional quantities are always in the same ratio. Remember that when the quantities are different in nature (expressed with a different unit), we talk about rates.
Let us take an example of the price paid for different numbers of milk bottles, shown in the table below.
Number of Bottles Price (\$) 2 3 4 β― π₯ 2.4 3.6 4.8 β― π¦
We can check that the ratio of the price to the number of bottles is the same for the three pairs of values given, namely, that
Hence, we can say that the price is proportional to the number of bottles, and the constant of proportionality is \$1.20 per bottle. The constant of proportionality (or unit rate) here is the unit price. We have seen that the ratio of the two quantities in each pair is equal to the constant of proportionality, here the unit price. It means that for any number of bottles (greater than zero), the price is such that
Multiplying both sides of this equation by , we find that it is equivalent to
This equation is just saying that the price of a given number of bottles is given by multiplying the unit price by this number of bottles. This relationship is often indicated on a proportionality table or a double-line diagram.
Remember that the constant of proportionality or unit rate of the proportional relationship between quantities and is the value of quantity when quantity is equal to 1. In other words, it is the result of carrying out the division of quantity by quantity for any pair of values we know. It is expressed with the compound unit βunit of quantity per unit of quantity ,β where βperβ means βfor each.β
Looking at the units is very helpful to make sure we have set up the equation correctly. Indeed, as we can see with our example, the units are used consistently:
On the right-hand side of the equation, we have a division of the unit βnumber of bottlesβ by βnumber of bottles,β which leads to a number without any unit (the units βcancel outβ). In terms of unit, we then get
This is consistent.
Let us summarize what we have just found out.
### How To: Describing a Proportional Relationship with an Equation
A directly proportional relationship between two quantities and is described mathematically with an equation of the form where is the constant of proportionality, or unit rate, of this relationship. The unit for is a so-called compound unit, written βunit for /unit for β (the slash is read βper,β which means βfor eachβ).
### Example 1: Writing an Equation to Describe a Proportional Relationship
In a pasta recipe that serves 4 people, it says to use 440 g of spaghetti.
• How much spaghetti should be used for 2 people? How much should be used for people?
• Write an equation in the form for the quantity of spaghetti needed for people.
If 440 g of spaghetti serves 4 people, then we will need half as much to serve 2 people, that is, 220 g.
By doing this, we have assumed that every person eats the same quantity of spaghetti or that we can use an average value of the quantity of spaghetti needed per person.
To find out the quantity needed for people, we need to find the value needed for one person and then multiply this value by the number of people . We have the value for 4 people, 440 g, so the value for one person is 440 g divided by 4:
The quantity that should be used for people is then
If we call the quantity of spaghetti needed for people, we have
We recognize that the 110 in the equation is the unit rate: it is 110 g/person. When it is multiplied by a number of people, we get a quantity in g.
### Example 2: Identifying a Unit Rate from an Equation
The amount of meat required to feed a captive lion is given by the equation , where is the weight of the meat in kilograms needed to feed a lion for days. What is the unit rate of this proportional relationship?
We are given the equation , which describes the proportional relationship between the weight of meet needed to feed a lion () and the number of days he can then be fed with this amount of meat. Recall that the unit rate is the multiplying constant in the general equation . By comparing with , we see that the constant is 9. Now, we need to think about the unit of . If we look at our equation in terms of units, it gives
By dividing both sides of the βunit equationβ by the unit βnumber of days,β we find that
We see that is expressed as kilogram per day. Hence, the unit rate of this proportional relationship is 9 kg/day.
### Example 3: Identifying a Unit Rate from an Equation
The quantity of paint needed to cover a certain wall area is given by , where is expressed in litres and in square metres. The quantity of paint is thus proportional to the wall area. What is the unit rate of this proportional relationship?
Recall that the unit rate is the multiplying constant in the general equation , which describes the proportional relationship between and , and is expressed in βunit for /unit for .β Here, we have .
Hence, we see that the unit rate is 0.1; however, we need to find its unit. For this, we need to look at the units of and . The quantity of paint is expressed in litres and the area in square metres. Therefore, the unit of is litres per square metre.
Our answer is that the unit rate is 0.1 L/m2.
### Example 4: Writing an Equation to Describe a Proportional Relationship from a Table
The table shows how many pages of a book Seif has read at different times.
Time (Minutes) Number of Pages 12 28 36 48 60 9 21 27 36 45
• Is Seif reading at a constant speed? Why?
1. No, because the number of pages read is not proportional to the reading time.
2. Yes, because the number of pages read is proportional to the reading time.
• What is the constant of proportionality (or unit rate)? What does it represent?
1. 9 pages, the first number of pages
2. 45 pages, the last number of pages
3. 60 pages per minute, the total reading time
4. 0.75 pages per minute, the time needed to read a page
5. 0.75 pages per minute, the reading speed
• Write an equation in the form for the number of pages read in minutes.
If Seif is reading at a constant speed, then the number of pages he reads is proportional to the reading time. This means that this number of pages is then given by the time multiplied by a certain constant, which is called constant of proportionality or unit rate.
To decide using the values given in the table whether Seif is reading at a constant speed, we need to check if the values given in each column are in the same ratio. The first ratio is 9 pages in 12 minutes. This can be reduced to 3 pages read every 4 minutes. The second ratio is 21 pages read in 28 minutes. As , we need to check if . This is true, so or
We can check in the same way that all the pairs of values given in the table are in the same ratio.
12=4Γ3 28=4Γ7 36=4Γ9 48=4Γ12 60=4Γ15 9=3Γ3 21=3Γ7 27=3Γ9 36=3Γ12 45=3Γ15
The ratio of pages read to time in minutes is constant; we can say that the number of pages read is proportional to the reading time. Therefore, the reading speed is constant since it is the proportionality constant or unit rate of this relationship.
So, the answer is βyes,β because the number of pages read is proportional to the reading time.
We have just seen that the ratio of pages read to time in minutes is 3 pages for 4 minutes. It can be written as . This ratio is the unit rate of this proportional relationship. It is the constant that is multiplied by the time in to give the read. We can express it as a decimal by dividing 3 by 4.
We find it is 0.75 /.
Observe that the unit is consistent: we have divided 3 by , so we get 0.75 (which is 3 divided by 4) divided by , which is written / (and is read β per β).
The reading speed is the unit rate, it represents the number of pages read in a time unit, here a minute.
We see that, for any time expressed in minutes, the number of pages read is given by multiplying this time by the unit rate, 0.75 pages/minute. The corresponding equation is
### Example 5: Writing an Equation to Describe a Proportional Relationship
Yara paints one garden chair in . Write an equation for the number of chairs that she could paint in hours.
If Yara works at a constant rate, then the number of chairs she can paint is proportional to the time she spends painting. To write the corresponding equation, we need to find the unit rate of this proportional relationship, that is, the number of chairs she paints in one hour. We know that she paints one chair in 12 minutes. As an hour is made up of 60 minutes, 12 minutes is of an hour.
Using a double-line diagram, we visualize that the number of chairs that Yara paints in one hour is given by multiplying 1 (the number of chairs she paints in 12 minutes) by .
And we find that (as 12 minutes is indeed one-fifth of an hour). The unit rate is therefore 5 chairs per hour and is nothing else than the ratio of the number of chairs to be painted to the time needed to paint them. Using the pair of values we are given, we find it is indeed
Now that we have found the unit rate, it is easy to write an equation for the number of chairs she can paint in hours since it is given by multiplying any time expressed in hours by the unit rate expressed in βnumber of chairs per hour.β Hence, we have
Note that a proportional relationship can be taken in the reverse way, that is, by swapping the two quantities. In our last example, this would be working out the time needed to paint a given number of chairs. The unit rate would then be the time needed to paint one chair. We know it is 12 minutes, which, expressed in hours, is of an hour. Hence, the unit rate is of an hour per chair. And the corresponding equation linking and is .
Both equations are actually one and the same equation. One gets from simply by dividing by the unit rate 5 chairs/hour.
Let us have a quick look at the units in both equations here. In the first one, we have and in the second one
### Key Points
• Two quantities and are directly proportional or in direct proportion when, from one situation to another, both quantities have been multiplied (or divided) by the same number.
• This means that the ratio of to is constant: , where is a constant called the constant of proportionality, or unit rate, of the proportional relationship between and
• A directly proportional relationship between two quantities and is described mathematically with an equation of the form , where is the constant of proportionality, or unit rate, of this relationship.
• The constant of proportionality has a compound unit: unit of per unit of , for instance, kilometres per hour, dollars per kilogram, minutes per page, and so on. |
# Let f(x) = x^2and g(x) = 5x. Find the composite. what is (f ∘ g)(x)?
Apr 5, 2018
See below.
#### Explanation:
To find the composite function, all we need to do is to replace the $x$ in the initial expression, with that of the second.
To further explain, we are given these two functions:
$f \left(x\right) = \textcolor{b l u e}{{x}^{2}}$
$g \left(x\right) = \textcolor{red}{5 x}$
We are asked to find the composite function of (f ∘ g)(x).
So, with the above explanation, we should have this:
(f ∘ g)(x)=(color(red)(5x))color(blue)(^2)
Since the $^ 2$ is affecting the entirety of $5 x$, it will distribute to both, making it:
(f ∘ g)(x)=5^2x^2
=> (f ∘ g)(x)=25x^2 |
# Convert Polar to Rectangular
In this post, I am going to describe the theory that allows you to convert polar to rectangular coordinates. This follows up my previous post that describes graphing polar coordinates, where I introduced the concept of this new method of graphing. However, the relationships that I will show you here will hopefully allow you to see the connection between polar and Cartesian coordinate systems, which will make them easier for you to work with!
To quickly refresh what I explained last time, start at the origin (or pole) of your graph and extend the polar axis line out to the right. This is your reference line that will help you describe the location of any other point. Now, pick a point P somewhere, and this is described by its distance (radius), r, from the origin (if you were to connect it to there) and the angle, ɵ, by which that line has rotated away from the polar axis. Whereas points in the rectangular coordinate system are described as P (x, y), points in the polar coordinate system are described as P (r, ɵ).
Now, let's take a closer look at the relationship between P (x, y) and P (r, ɵ). This will use a little bit of trigonometry, so review it in my posts about sine, cosine, and tangent if you need to brush up!
To do this, let's superimpose the two coordinate systems, meaning that we will assume that the origins of each are in the same place, and the polar axis is the same as the positive x-axis. You should be able to see, with the help of the following handy figure, that our point P can be described by P (x, y) and P (r, ɵ). You can also see that, by combining the two coordinate systems, we have formed a triangle which has sides of lengths x and y and hypotenuse r. This triangle will be the basis that allows us to convert polar to rectangular coordinates.
Now, if we apply our rules and identities of trigonometry, the relationships of the triangle sides and angles will connect the polar coordinate system and rectangular coordinate system. And with these relationships, you will be able to convert from polar to rectangular, and also back again to convert rectangular to polar.
The relationships are quite simply the basic trigonometry identities that you already know:
sin ɵ = opposite / hypotenuse
cos ɵ = adjacent / hypotenuse
tan ɵ = opposite / adjacent
Now, if we substitute in the names for our sides, and rearrange to have polar terms on one side and Cartesian terms on the other, we arrive at the following relationships:
sin ɵ = y / r -----> y = r sin ɵ
cos ɵ = x / r -----> x = r cos ɵ
tan ɵ = y / x
Furthermore, we can apply the Theorem of Pythagoras to give us another useful relationship:
r2 = x2 + y2
And there you have it! Easily derived connections between polar and rectangular coordinate systems. You will find as you work through you studies that sometimes some expressions may be easier to work with in one coordinate system or the other. Keep this in mind, especially when you begin to work with polar equations. Using these quick methods to convert polar to rectangular coordinates may help you to get through your problems a lot faster (and easier)!
In my next post, I'll show you some more things about polar coordinates!
#### 1 comment:
1. well describe about the conversion form polar to rectangular and I think this conversion is very important in coordinate question. |
Order of Operations in Algebra How to Reorder Operations with the Commutative Property Permutations When Order Matters
# How to Apply the Associative Property
The associative property comes in handy when you work with algebraic expressions. Use the associative property to change the grouping in an algebraic expression to make the work tidier or more convenient. Just keep in mind that you can use the associative property with addition and multiplication operations, but not subtraction or division, except in a few special cases.
Think about what the word associate means. When you associate with someone, you’re close to the person, or you form a group with the person. Say that Anika, Becky, and Cora associate. Whether Anika drives over to pick up Becky and the two of them go to Cora’s and pick her up, or Cora is at Becky’s house and Anika picks up both of them at the same time, the same result occurs — the same people are in the car at the end.
Check out how the associative property works in the following examples:
• Addition: a + (b + c) = (a + b) + c
4 + (5 + 8) = 4 + 13 = 17, and (4 + 5) + 8 = 9 + 8 = 17
You can group the numbers however you want to and still reach the same result, 17.
• Multiplication: a × (b × c) = (a × b) × c
3 × (2 × 5) = 3 × 10 = 30, and (3 × 2) × 5 = 6 × 5 = 30
This example shows you two options for grouping the numbers — but the result, 30, is the same regardless of how you group the numbers.
• Subtraction: a – (bc) ≠ (ab) – c (except in a few special cases)
13 – (8 – 2) = 13 – 6 = 7, but (13 – 8) – 2 = 5 – 2 = 3
Now you can see how subtraction doesn't follow the associative property. Regrouping the numbers resulted in two different answers.
• Division: a ÷ (b ÷ c) ≠ (a ÷ b) ÷ c (except in a few special cases)
48 ÷ (16 ÷ 2) = 48 ÷ 8 = 6, but (48 ÷ 16) ÷ 2 = 3 ÷ 2 = 1.5
This example illustrates how division doesn't follow the associative property. Regrouping the numbers resulted in two different answers.
You can always find a few cases where the property works even though it isn’t supposed to. For instance, in the subtraction problem 5 – (4 – 0) = (5 – 4) – 0 the property seems to work. Also, in the division problem
6 ÷ (3 ÷ 1) = (6 ÷ 3) ÷ 1, it seems to work. |
# Some Mathematical Symbols
## Multiplication
There are three commonly used means of indicating multiplication
• The symbol "x", e.g., 5 x 6 = 30. Note that this symbol is generally avoided in algebraic equations because of the common use of "x" to indicate an unknown quantity.
• The symbol "*", e.g., 5 * 8 = 40. The use of the asterisk to indicate multiplication is commonly used in spreadsheets (e.g., Excel) and in algerbraic expressions.
• Or simply a number next to a parenthetic expression, e.g., 5(6+2) = 40
## Division
There are three commonly used ways to indicate division.
• "/", e.g., 40/5 = 8
• "÷", e.g., 30 ÷ 5 = 6
• Division can also be indicated by placing one quantity (the numerator) over another quantity (the denominator) as shown below.
44/123 = 0.3577
## Equals (=) & Doesn't Equal (≠)
2+3 = 5
2+3 4
(Read as "doesn't equal" or "is not equal to."
## Less than (<) and greater than (>)
• The symbol < means less than. For example,
7 < 8
200 < 300
• The symbol > means greater than. For example,
6 > 4
3000 > 2750
• The symbol means less than or equal to.
• The symbol means greater than or equal to.
## Approximately Equal
• The symbol means approximately equal to.
# The Order of Math Operations
When you are given a mathematical expression or an equation, the order in which mathematical operations are performed is very important. The rules for this are quite simple. Consider the following example:
2 + (7+3) * 32 + 4* (3-1) + 10
At first this may look daunting, but it is really quite simple. The rules are: summarized in the table below.
Order of Operations Solve within Parentheses and Brackets from the inside out Compute Exponents Perform Multiplication and Division in the order they appear. Perform Addition and Subtraction in the order they appear..
So, for the example above your would:
1. Solve within parentheses
2. Compute exponents
3. Perform multiplication and division |
# Tag Info
5
If $x$ and $y$ are identical, then $x^2+x=x(x+1)$ has to be a square where $x^2<x(x+1)<(x+1)^2$. If $x$ and $y$ are different, then wlog $y<x$ and then $x^2<x^2+y<x(x+1)<(x+1)^2$. There may be a problem here ... .
4
If $Ax=u$ has no solution, it is also true that $Ax = -u$ has no solution. So we are saying your $v=-u$ Next, $Ax = u + v = u +(-u) = u-u=0$ always has a solution
3
The first equation is $xb+ya = ab$, while the second one is $a^3y - b^3x = xy(b^2-a^2).$ Let's rearrange the second one as $ay(a^2+ax) = bx(b^2+by).$ Using the first one, we get $a(a+x)(ab-xb) = b(b+y)(ab-ya)$. Thus, $(a+x)(a-x) = (b+y)(b-y)$, which implies the thesis.
3
Rewrite the two given equations as $$ab-2cd=3 \quad\text{and}\quad ac+bd=1,$$ and note for later use that $a\ne0$ and $b\ne0$ and that at most one $c$ and $d$ can equal $0$, and also that at least one variable is even. Squaring both sides of both equations and rearranging yields $$a^2b^2+4c^2d^2=9+4abcd \quad\text{and}\quad a^2c^2+b^2d^2=1-abcd,$$ and ...
2
If $x^2+y=m^2\Rightarrow m\geq x+1\Rightarrow y\geq 2x+1$ . This shows that if $y^2+x$ is a square then $x\geq 2y+1$. Combine the last two inequalities to see that $y\geq2x+1\geq 2(2y+1)+1=4y+3$ which is of course impossible.
1
Luca's solution works, and is likely how they intended this to be solved. I'd like to point out that OP's approach of "substitute the first equation into the second" can be forced through to completion. We have $y = \frac{b(a-x) } { x }$ from the first equation. Substituting into the second equation, we get $\frac{a^3}{x} - \frac{ab^2}{a-x} = b^... 1 You consider the function$\|x(t)\|_2$. If you want to show that it is increasing, then it's enough to show that$\|x(t)\|_2^2$is increasing, since$\|x(t)\|_2$is non-negative and the square-root with this co-domain is increasing. However,$\|x(t)\|_2^2 = x(t)^T x(t)$. This is a product of differentiable functions of$t$, hence differentiable. We must use ... 1 Welcome to MSE! I think some of the confusion regarding the responses to this question come from the (admittedly subtle) distinction between a closed form solution to a recurrence and an extension of a recurrence (to all of$\mathbb{C}$, say). If you're able to solve a recurrence, so that$T(n) = f(n)$where$f$is some "nice" function, then we get ... 1 You claim that The only continuous function F that satisfies the above system of equations is, as we probably all know, $$F(n) = \frac{\varphi^n - \psi^{n}}{\sqrt{5}},$$ This claim in false even if$\,F\,$is restricted to real functions. Given any integer$\,k,\,$define the confinuous complex function $$F_k(x) := e^{2\pi i k x}\frac{\varphi^x - \psi^{... 1 Let me use an example with t=6.$$\frac {x^5}{\prod_{n=1}^5 \left(1-\frac{x}{a_n}\right)}=\sum_{n=1}^5 \frac {c_n}{1-\frac{x}{a_n}}$$Cross multiply to have$$x^5=\Bigg[\sum_{n=1}^5 \frac{c_n}{1-\frac{x}{a_n}}\Bigg]\,\times\, \Bigg[\prod_{n=1}^5\frac{1}{1-\frac{x}{a_n}}\Bigg]$$Now, take the$\color{red}{\text{limit}}$of the rhs when$x\to a_1\$. You will ...
Only top voted, non community-wiki answers of a minimum length are eligible |
# 5-1& 5-2 Prime Factors and Greatest Common Factors Vocab: Prime Numbers(5-1) A whole number with exactly 2 factors (1 and itself) ex 17: 1,17 1 is not.
## Presentation on theme: "5-1& 5-2 Prime Factors and Greatest Common Factors Vocab: Prime Numbers(5-1) A whole number with exactly 2 factors (1 and itself) ex 17: 1,17 1 is not."— Presentation transcript:
5-1& 5-2 Prime Factors and Greatest Common Factors Vocab: Prime Numbers(5-1) A whole number with exactly 2 factors (1 and itself) ex 17: 1,17 1 is not Prime!! Composite Numbers(5-1) A whole number with more than 2 factors ex 8: 1,2,4,8 Prime Factorization: Write a composite number as a product of only primes. 18 = 2x3x3 Greatest Common factor: The largest factor 2 or more numbers share (will divide into both numbers evenly)
To find the GCF using lists of factors Write all the factors of the following numbers 20 : 24: 20:1, 2, 4, 5, 10, 20 24:1, 2, 3, 4, 6, 8, 12, 24 5 10 20 124124 3 6 8 24 12 Factors of 20 Factors of 24 The greatest common factor(GCF) is 4
Find the GCF of 6 and 12 6:1,2,3,6 12:1,2,3,4,6,12 1 2 3 4 6 12 The GCF is 6
Find GCF with Prime Factors Find the Prime factors of 45 and 36 36 = 6 x 6 = 2x3x2x3 ←all primes =2x2x3x3←rewrite in order 45 = 9 X 5 = 3x3x5←all primes (in order) So 36 = 2x2x3x3 45 = 3x3x5 3x 3 is on both lists so the GCF is 3x3 =9
Find the GCF of 30 a 2 and 18 a 3 b 30 a 2 = 2x3x5xaxa 18a 3 b = 2x3x3xaxaxaxb 2x3xaxa is on both lists so the GCF is 2x3xaxa=6a 2
Period 6 & 7 5-1/199/33-35,50,51 5-2/205-206/26-33,39-44,47 Periods 1, 3, & 8 5-2/205-206/9-19 o,39-44,46,47
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## Sunday, December 11, 2011
### Danilyn's Scribe Post
Page 148. Question 4.
(1) binder= \$4.99******
(2) binders= \$9.98 (\$4.99 multiplied by 2)
(1) math set= \$3.99
(1) backpack= \$19.99
total= \$33.96
5% (GST) + 7% (PST) =
12%
112÷100= 1.12%(you could rotate it sideways to get the fraction version, but for this scribe I prefer using decimal sentences) Percentage multiplied by total 1.12% multiplied by \$33.96= 38.0352 (round it up)
38.0352 = \$38.04
-:.;_;.:-
-:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:-
Page 149. Ques
tion 7.
Calculate 10% of 100 to find the first year (100%÷10=10%)
What you do to the o
ne side you always have to do to the other side, if you are dividing the
100(%) with 10, you would have to divide 100(caribou) by 10 also. Since it said increasing, you would have to add both of your tens (percentage and caribou) to your totals (100 on both percentage and caribou)
So far: 110 caribou = 110% (First Year)
Next, calculate 20% of your new whole (new whole (whole= 100%))= first year. first year= 110 caribou). This is another way to sh
ow my explanation, yet this way might be more space taking on some explanations****
100 % = 110
then divide by 10
10% = 11
then multiply by 2
20% = 22
Now, we move onto a)
To find the population after the second year (meaning, the total caribou you have calculated on both first and second year) You would need to add the first year total (which was 110 caribou) to the 20% (which was 22 caribou) increase we had on the
second year.
110 + 22 =
132 After the second year, the total population for the caribou is now 132.
Moving onto
b)
To explain why there wasn't a 30% increase over those 2 years is because both of the years don't have the same whole. During the second year, we have switched the whole(the whole=100%) (100- during first year) into (110- after first
year).
-:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:-
Page 150. Question 7.
a) 0.4% (decimal= explosion) =
b) 12% =
c) 115% (more than "100") =
-:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:--:.;_;.:-
Here is a video on finding percents.
Anyone kind enough to give me a link to a game? Please?
#### 1 comment:
1. Great job Danilyn! Usually I would tell people to use some colour in their font and bold the most important words we needed to see, but as you could see, you already did everything I expected! For different questions and answers, you used some sort of pattern to separate them, and I found it very cool because at first I didn't know what that was, but I finally understood what you did afterwards. I liked how you showed your work for both pictures and the typed work because it makes it extra understandable. For page 150, number 7 a), I really liked how you used the "zoom in" trick Mr. Harbeck taught us and also coloured in the grid with a coloured pencil instead of highlighting or just shading it in with a regular pencil. You did everything really great and impressing, but the one thing I would just have to say is when I clicked your video, there was a tiny error opening it. Other than that, Danilyn Sanchez did a terrific job. Keep it up! (= |
LCM the 12 and also 15 is the smallest number amongst all common multiples that 12 and also 15. The first couple of multiples that 12 and also 15 space (12, 24, 36, 48, 60, 72, 84, . . . ) and also (15, 30, 45, 60, . . . ) respectively. There are 3 typically used techniques to find LCM that 12 and also 15 - by department method, by element factorization, and also by listing multiples.
You are watching: Find the least common multiple of 15 and 12.
1 LCM of 12 and also 15 2 List that Methods 3 Solved Examples 4 FAQs
Answer: LCM the 12 and 15 is 60.
Explanation:
The LCM of two non-zero integers, x(12) and also y(15), is the smallest optimistic integer m(60) that is divisible by both x(12) and y(15) without any kind of remainder.
The approaches to find the LCM of 12 and 15 are explained below.
By Listing MultiplesBy prime Factorization MethodBy division Method
### LCM that 12 and also 15 through Listing Multiples
To calculate the LCM of 12 and 15 through listing out the typical multiples, we deserve to follow the given listed below steps:
Step 1: list a couple of multiples that 12 (12, 24, 36, 48, 60, 72, 84, . . . ) and 15 (15, 30, 45, 60, . . . . )Step 2: The common multiples indigenous the multiples of 12 and also 15 room 60, 120, . . .Step 3: The smallest usual multiple the 12 and also 15 is 60.
∴ The least common multiple the 12 and 15 = 60.
### LCM of 12 and 15 by element Factorization
Prime administrate of 12 and 15 is (2 × 2 × 3) = 22 × 31 and also (3 × 5) = 31 × 51 respectively. LCM that 12 and also 15 deserve to be obtained by multiplying prime determinants raised to their respective highest power, i.e. 22 × 31 × 51 = 60.Hence, the LCM the 12 and also 15 by prime factorization is 60.
### LCM of 12 and also 15 by department Method
To calculate the LCM that 12 and 15 by the division method, we will certainly divide the numbers(12, 15) by your prime determinants (preferably common). The product of this divisors gives the LCM that 12 and 15.
Step 3: proceed the actions until just 1s space left in the last row.
The LCM that 12 and 15 is the product of all prime numbers on the left, i.e. LCM(12, 15) by division method = 2 × 2 × 3 × 5 = 60.
## FAQs ~ above LCM that 12 and 15
### What is the LCM the 12 and 15?
The LCM the 12 and 15 is 60. To find the least typical multiple (LCM) the 12 and also 15, we require to find the multiples that 12 and also 15 (multiples the 12 = 12, 24, 36, 48 . . . . 60; multiples the 15 = 15, 30, 45, 60) and choose the the smallest multiple that is specifically divisible through 12 and also 15, i.e., 60.
### Which of the following is the LCM of 12 and also 15? 36, 60, 35, 11
The worth of LCM the 12, 15 is the smallest typical multiple the 12 and also 15. The number satisfying the given condition is 60.
### If the LCM of 15 and also 12 is 60, find its GCF.
LCM(15, 12) × GCF(15, 12) = 15 × 12Since the LCM the 15 and 12 = 60⇒ 60 × GCF(15, 12) = 180Therefore, the greatest common factor = 180/60 = 3.
### How to uncover the LCM that 12 and 15 by prime Factorization?
To find the LCM of 12 and also 15 using prime factorization, us will uncover the prime factors, (12 = 2 × 2 × 3) and also (15 = 3 × 5). LCM that 12 and also 15 is the product that prime determinants raised to their respective greatest exponent amongst the number 12 and also 15.⇒ LCM of 12, 15 = 22 × 31 × 51 = 60.
See more: When Did Guy Penrod Heart Attack, Who Did Taranda Greene Married
### What is the Relation in between GCF and also LCM that 12, 15?
The complying with equation have the right to be provided to express the relation between GCF and LCM that 12 and also 15, i.e. GCF × LCM = 12 × 15. |
# NCERT Class 9 Maths Formula CBSE Board Sample Problems Part 7 (For CBSE, ICSE, IAS, NET, NRA 2022)
Doorsteptutor material for CBSE/Class-9 is prepared by world's top subject experts: get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-9.
## 7. Triangles
S. no Terms Descriptions 1 Congruence Two Geometric figure are said to be congruence if they are exactly same size and shape Symbol used is Two angles are congruent if they are equalTwo circle are congruent if they have equal radiiTwo squares are congruent if the sides are equal 2 Triangle Congruence Two triangles are congruent if three sides and three angles of one triangle is congruent to the corresponding sides andangles of the otherTriangle CongruenceCorresponding sides are equal AB = DE, BC = EF, AC = DFCorresponding angles are equal We write this asThe above six equalities are between the corresponding parts of the two congruent triangles. In short form this is called C. P. C. TWe should keep the letters in correct order on both sides 3 Inequalities in Triangles 1) In a triangle angle opposite to longer side is larger2) In a triangle side opposite to larger angle is larger3) The sum of any two sides of the triangle is greater than the third sideIn triangle ABC
### Different Criterion for Congruence of the Triangles
N Criterion Description Figures and expression 1 Side angle Side (SAS) congruence Two triangles are congruent if the two sides and included angles of one triangle is equal to the two sides and included angleIt is an axiom as it cannot be proved so it is an accepted truthASS and SSA type two triangles may not be congruent always Side Angle Side (SAS) CongruenceIf following conditionAB = DE, BC = EFThen 2 Angle side Two triangles are congruent if the angle (ASA) two angles and included side of congruence one triangle is equal to the corresponding angles and sideIt is a theorem and can be proved If following conditionThen 3 Angle angle side (AAS) congruent Two triangles are if the any two pair of angles and any congruence side of one triangle is equal to the corresponding angles and sideIt is a theorem and can be proved Side-Side-Side (SSS) CongruenceIf following conditionBC = EFThen 4 Side-Side-Side (SSS) congruence Two triangles are congruent if the three sides of one triangle is equal A to the three sides of the another Side-Side-Side (SSS) CongruenceIf following conditionBC = E F, AB = DE, DF = ACThen 5 Right angle hypotenuse-side (RHS) congruence Two right triangles are congruent if A the hypotenuse and a side of the one triangle are equal to corresponding hypotenuse and side of the another Right Angle Hypotenuse-Side (RHS) CongruenceIf following conditionAC = DF, BC = EFThen
### Some Important Points on Triangles
Point of intersection of the three altitude of the triangle Orthocenter Triangle whose all sides are equal and all angles are equal to 60° Equilateral A line Segment joining the corner of the triangle to the midpoint of the opposite side of the triangle Median A line Segment from the corner of the triangle and perpendicular to the opposite side of the triangle Altitude A triangle whose two sides are equal Isosceles Centroid Point of intersection of the three median of the triangle is called the centroid of the triangle In center All the angle bisector of the triangle passes through same point Circumcenter The perpendicular bisector of the sides of the triangles passes through same point Scalene triangle Triangle having no equal angles and no equal sides Right Triangle Right triangle has one angle equal to Obtuse Triangle One angle is obtuse angle while other two are acute angles Acute Triangle All the angles are acute
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# Illustrative Mathematics Grade 8, Unit 6, Lesson 6: The Slope of a Fitted Line
Learning Targets:
• I can use the slope of a line fit to data in a scatter plot to say how the variables are connected in real-world situations.
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Illustrative Math
#### Lesson 6: The Slope of a Fitted Line
Let’s look at how changing one variable changes another.
Illustrative Math Unit 8.6, Lesson 6 (printable worksheets)
#### Lesson 6 Summary
The following diagram shows how to use the slope of a line fit to data in a scatter plot to say how the variables are connected in real-world situations.
#### Lesson 6.1 Estimating Slope
Estimate the slope of the line.
#### Lesson 6.2 Describing Linear Associations
For each scatter plot, decide if there is an association between the two variables, and describe the situation using one of these sentences:
• For these data, as __________ increases, ___________ tends to increase.
• For these data, as __________ increases, ___________ tends to decrease.
• For these data, ___________ and ___________ do not appear to be related.
#### Lesson 6.3 Interpreting Slopes
For each of the situations, a linear model for some data is shown.
1. What is the slope of the line in the scatter plot for each situation?
2. What is the meaning of the slope in that situation?
#### Are you ready for more?
The scatter plot shows the weight and fuel efficiency data used in an earlier lesson along with a linear model represented by the equation y = -0.0114x + 41.3021
1. What is the value of the slope and what does it mean in this context?
Slope = -0.0114
For every 1 kg increase of weight there is a decrease of 0.0114 in fuel efficiency.
2. What does the other number in the equation represent on the graph? What does it mean in context?
The graph will intersect the y-axis at 41.3021. In this context, when the weight is 0, then fuel efficiency is 41.3021.
3. Use the equation to predict the fuel efficiency of a car that weighs 100 kilograms.
When x = 100 then y = -0.0114(100) + 41.031 = 29.631
4. Use the equation to predict the weight of a car that has a fuel efficiency of 22 mpg.
5. Which of these two predictions probably fits reality better? Explain.
The second prediction fits reality better.
That is because cars are heavier than 100kg.
#### Lesson 6.4 Positive or Negative?
1. For each of the scatter plots, decide whether it makes sense to fit a linear model to the data. If it does, would the graph of the model have a positive slope, a negative slope, or a slope of zero?
2. Which of these scatter plots show evidence of a positive association between the variables? Of a negative association? Which do not appear to show an association?
#### Lesson 6 Practice Problems
1. Which of these statements is true about the data in the scatter plot?
2. Here is a scatter plot that compares hits to at bats for players on a baseball team.
Describe the relationship between the number of at bats and the number of hits using the data in the scatter plot.
3. The linear model for some butterfly data is given by the equation . Which of the following best describes the slope of the model?
4. Nonstop, one-way flight times from O’Hare Airport in Chicago and prices of a one-way ticket are shown in the scatter plot.
5. Solve:
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Implied Domain
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In general, the implied domain of a function is the set of all real numbers for which the function is defined [1]. If we’re given a function and the domain isn’t specified, we assume that the domain is the largest possible subset of real numbers: (-∞, ∞).
However, there are some notable exceptions [2].
1. The radicand of a square root function (the expression contained within the square root) must be zero or greater.
2. The denominator in a rational function cannot equal zero.
3. Logarithmic functions must have positive real numbers in the domain, because we cannot take the logarithm of negative numbers.
has an implied domain of all reals if n is an even numbered natural number (i.e. the counting numbers 2, 4, 6,….). However, if n is an odd numbered natural number, we can’t take even roots of negative numbers so the implied domain is [0, ∞).
## Implied Domain: Examples
Example #1: (Radical Function) Find the implied domain of f(x) = √(x – 2).
Step 1: Set the radicand equal to or greater than zero:
(x – 2) ≥ 0.
Step 2: Solve for x:
x ≥ 2.
This tells us that the implied domain is the set of all real numbers greater than or equal to 2. In interval notation, we can write that as [2, ∞).
Example #2 (Rational Function): Find the implied domain of the following function:
Step 1: Solve the denominator for x.
• x + 4 ≠ 0
• x ≠ -4.
This gives a domain of all real numbers except for -4. In interval notation, the domain is (-∞ -4) ∪ (-4, ∞).
Example #3: (Polynomial Function): What is the implied domain of f(x) = x2 + 5x?
Solution: All polynomial functions have an implied domain of all real numbers. We can write that in interval notation as (-∞, ∞). Or, we can say the domain is ℝ.
## A Practical Example
Sometimes you may be asked to find an implied domain for a word problem instead of an algebraic equation. For example, this precalculus worksheet asks you for the implied domain for a function concerning the monthly cost in dollars of a phone call. The input, “x” is the number of minutes used. As there are 43,200 minutes in a month, the longest possible phone call is 43,200 minutes and the shortest call is 0 minutes, giving an implied domain of [0, 43200].
## References
[1] Functions. Retrieved October 31, 2021 from: https://people.richland.edu/james/lecture/m116/functions/functions.html
[2] Handout. Retrieved October 31, 2021 from: https://somerset.kctcs.edu/current-students/media/student-support-services/workshops/implied-domain-handout.docx
CITE THIS AS:
Stephanie Glen. "Implied Domain" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/implied-domain/
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Find the value of ${\text{k}}$, if the points ${\text{A}}\left( {8,1} \right),{\text{ B}}\left( {3, - 4} \right),{\text{ and C}}\left( {2,k} \right)$ are collinear.
Last updated date: 20th Jun 2024
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Hint: - If three points are given as A,B,C then they will be collinear if the slopes of the line segment between 2 points are equal. I.e. slope of AB = slope of BC.
Given points are
${\text{A}}\left( {8,1} \right),{\text{ B}}\left( {3, - 4} \right),{\text{ and C}}\left( {2,k} \right)$
Now we know two points are collinear if their slopes are equal
Therefore slope of AB $=$ Slope of BC
Collinearity of points: - Collinear points always lie on the same line.
Now we know
Slope between two points ${\text{ = }}\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)$
Consider $A\left( {8,1} \right) \equiv \left( {{x_1},{y_1}} \right),{\text{ }}B\left( {3, - 4} \right) \equiv \left( {{x_2},{y_2}} \right),{\text{ }}C\left( {2,k} \right) \equiv \left( {{x_3},{y_3}} \right)$
Therefore slope of AB${\text{ = }}\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right){\text{ = }}\dfrac{{ - 4 - 1}}{{3 - 8}} = \dfrac{{ - 5}}{{ - 5}} = 1$
Therefore slope of BC ${\text{ = }}\left( {\dfrac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}}} \right){\text{ = }}\dfrac{{k - \left( { - 4} \right)}}{{2 - 3}} = \dfrac{{k + 4}}{{ - 1}} = - k - 4$
Points are collinear
Therefore slope of AB $=$ Slope of BC
$\begin{gathered} \Rightarrow 1 = - k - 4 \\ \Rightarrow k = - 1 - 4 = - 5 \\ \end{gathered}$
So, $k = - 5$ is the required answer.
Note: - If three points are collinear then the area formed by these points should be zero because collinear points always lie on the same line so the area formed by these points is zero. We can also use this property to find the collinearity of the points and the condition is $\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0$, with the help of this we can easily calculate the collinearity of the given points. |
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F Distribution and One-Way ANOVA: Facts About the F Distribution
Summary: This module states the factors associated with F Distributions and provides students with some examples to help further understand the concept. Students will be given the opportunity to see F Distributions in action through participation in an optional classroom exercise.
Note: You are viewing an old version of this document. The latest version is available here.
1. The curve is not symmetrical but skewed to the right.
2. There is a different curve for each set of dfs dfs .
3. The F statistic is greater than or equal to zero.
4. As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal.
5. Other uses for the F distribution include comparing two variances and Two-Way Analysis of Variance. Comparing two variances is discussed at the end of the chapter. Two-Way Analysis is mentioned for your information only.
Example 1
One-Way ANOVA: Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown below:
Table 1
Sorority 1 Sorority 2 Sorority 3 Sorority 4
2.17 2.63 2.63 3.79
1.85 1.77 3.78 3.45
2.83 3.25 4.00 3.08
1.69 1.86 2.55 2.26
3.33 2.21 2.45 3.18
Problem 1
Using a significance level of 1%, is there a difference in grade means among the sororities?
Solution
Let μ 1 μ 1 , μ 2 μ 2 , μ 3 μ 3 , μ 4 μ 4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each size 5.
Note:
This is an example of a balanced design, since each factor (i.e. Sorority) has the same number of observations.
H o : μ 1 = μ 2 = μ 3 = μ 4 H o : μ 1 = μ 2 = μ 3 = μ 4
H a H a : Not all of the means μ 1 , μ 2 , μ 3 , μ 4 μ 1 , μ 2 , μ 3 , μ 4 are equal.
Distribution for the test: F 3 , 16 F 3 , 16
where k = 4 groupsk=4 groups and n = 20 samples in total n=20 samples in total
df(num) = k - 1 = 4 - 1 = 3 df(num)=k-1=4-1=3
df(denom) = n - k = 20 - 4 = 16 df(denom)=n-k=20-4=16
Calculate the test statistic: F = 2.23 F=2.23
Graph:
Probability statement: p-value = P ( F > 2.23 ) = 0.1241 p-value=P(F>2.23)=0.1241
Compare α α and the p-value p-value: α = 0.01 p-value = 0.1241 α=0.01p-value=0.1241 α < p-value α<p-value
Make a decision: Since α < p-value α<p-value, you cannot reject H o H o .
Conclusion: There is not sufficient evidence to conclude that there is a difference among the grade means for the sororities.
TI-83+ or TI 84: Put the data into lists L1, L2, L3, and L4. Press STAT and arrow over to TESTS. Arrow down to F:ANOVA. Press ENTER and Enter (L1,L2,L3,L4). The F statistic is 2.2303 and the p-value p-value is 0.1241. df(numerator) = 3df(numerator) = 3 (under "Factor") and df(denominator) = 16df(denominator) = 16 (under Error).
Example 2
A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew 5 plants. At the end of the growing period, each plant was measured, producing the following data (in inches):
Table 2
Tommy's Plants Tara's Plants Nick's Plants
24 25 23
21 31 27
23 23 22
30 20 30
23 28 20
Problem 1
Does it appear that the three media in which the bean plants were grown produce the same average height? Test at a 3% level of significance.
Solution
This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants so we will use the formula F' = n s x_ 2 s 2 pooled F' = n s x_ 2 s 2 pooled .
First, calculate the sample mean and sample variance of each group.
Table 3
Tommy's Plants Tara's Plants Nick's Plants
Sample Mean 24.2 25.4 24.4
Sample Variance 11.7 18.3 16.3
Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 = s x¯ 2s x 2
Then MS between = n s x¯ 2 = ( 5 ) ( 0.413 ) MS between =ns x 2=(5)(0.413) where n=5n=5 is the sample size (number of plants each child grew).
Calculate the average of the three sample variances (Calculate the average of 11.7, 18.3, and 16.3). Average of the sample variances = 15.433 = s 2 pooled = s 2 pooled size 12{ $$s rSub { size 8{ ital "pooled"} }$$ rSup { size 8{2} } ={}} {}
Then MS within = s 2 pooled = 15.433 MS within = s 2 pooled size 12{ $$s rSub { size 8{ ital "pooled"} }$$ rSup { size 8{2} } ={}} {}=15.433.
The F F statistic (or F F ratio) is F = MS between MS within = n s x_ 2 s 2 pooled = ( 5 ) ( 0.413 ) 15.433 = 0.134 F= MS between MS within = n s x_ 2 s 2 pooled = ( 5 ) ( 0.413 ) 15.433 =0.134
The dfs for the numerator = the number of groups - 1 = 3 - 1 = 2 the number of groups-1=3-1=2
The dfs for the denominator = the total number of samples - the number of groups = 15 - 3 = 12 the total number of samples-the number of groups=15-3=12
The distribution for the test is F 2 , 12 F 2 , 12 and the F statistic is F = 0.134 F=0.134
The p-value is P ( F > 0.134 ) = 0.8759 P(F>0.134)=0.8759.
Decision: Since α = 0.03 α=0.03 and the p-value = 0.8759 p-value=0.8759, do not reject H o H o . (Why?)
Conclusion: With a 3% the level of significance, from the sample data, the evidence is not sufficient to conclude that the average heights of the bean plants are different.
(This experiment was actually done by three classmates of the son of one of the authors.)
Another fourth grader also grew bean plants but this time in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32.
Problem 2
Do an ANOVA test on the 4 groups. You may use your calculator or computer to perform the test. Are the heights of the bean plants different? Use a solution sheet.
Solution
• FF = 0.9496
• p-valuep-value = 0.4402
From the sample data, the evidence is not sufficient to conclude that the average heights of the bean plants are different.
Optional Classroom Activity
From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1% level of significance. Use one of the solution sheets at the end of the chapter (after the homework).
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## How to Find the Area of Squares, Rectangles, and Parallelograms
### What is the Parallelogram’s Area?
A parallelogram’s area is the space bounded by the parallelogram in a provided 2-dimension space. As stated earlier, a parallelogram is a particular kind of quadrilateral having $$4$$ sides, as well as the pair of opposite sides, are parallel. With a parallelogram, the opposite sides are an equal length and the opposite angles are also an equal measure. Because the rectangle and the parallelogram have properties that are similar, the rectangular’ s area equals the parallelogram’s area.
### Ways to Calculate a Parallelogram’s Area
Step one: Note down this formula: $$A \ = \ bh$$. $$A$$ stands for the area, while $$b$$ stands for the parallelogram’s length, and $$h$$ stands for the parallelogram’s height.
Step two: Find the parallelogram’s base. The base equals the length of the bottom side of a parallelogram.
Step three: Find the parallelogram’s height. The height is the length a perpendicular line has to go from the bottom to the top of a parallelogram.
Step four: Multiply the height and the base together.
### What is the Square’s Area?
Squares are closed 2-dimensional shapes having $$4$$ equal sides as well as $$4$$ equal angles. The square’s $$4$$ sides create the $$4$$ angles at the vertices. The total of the full length of the sides of a square is the perimeter, as well as the full space the square occupies, is the square’s area. It’s a quadrilateral with these properties.
• Its opposite sides are parallel.
• All $$4$$sides are the same.
• All the angles are $$90°$$.
The formula to determine the area of a square when the sides are provided is:
Area of a square $$= \ Side \times Side \ = \ s^2$$
A square’s area may additionally be calculated via the assistance of the square’s diagonal. The formula utilized for finding a square’s area if the diagonal is provided is:
Area of a square using diagonals $$= \ \frac{Diagonal^2}{2}$$
### What is the Rectangle’s area?
The area of a rectangle is calculated via the sides. Usually, a rectangle’s area equals the product of its length and width. The rectangle’s perimeter, though, equals the total of all its $$4$$ sides. So, the rectangle’s area is described as the area encompassed by the perimeter. Rectangles are quadrilaterals whose opposite sides are equal and parallel to one another. Because rectangles have $$4$$ sides, there are also $$4$$ angles. All the rectangle’s angles equal $$90$$ degrees; so, all the rectangle’s angles are right angles.
### Formula for Determining the Area of a Rectangle
A rectangle’s area is determined in units via multiplying the width (or breadth) by the rectangle’s length.
Area of a Rectangle $$A \ = \ l \times b$$
Follow these steps to determine the area:
• Step one: Write down the length and width dimensions using provided data
• Step two: Multiply the values of the length and width
• Step three: Write down the solution in square units
### Exercises for Area of Squares, Rectangles, and Parallelograms
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
1) $$\ \color{red}{77 \ m}$$
2) $$\ \color{red}{100 \ cm^2}$$
3) $$\ \color{red}{143 \ cm^2}$$
4) $$\ \color{red}{168 \ cm^2}$$
5) $$\ \color{red}{437 \ cm^2}$$
6) $$\ \color{red}{80 \ m^2}$$
7) $$\ \color{red}{9 \ cm^2}$$
8) $$\ \color{red}{16 \ cm^2}$$
9) $$\ \color{red}{121 \ cm^2}$$
10) $$\ \color{red}{460 \ cm^2}$$
## Area of Squares, Rectangles, and Parallelograms Practice Quiz
### ISEE Upper Level Math for Beginners
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### AFOQT Math Practice Workbook
$25.99$14.99
### 5 Full-Length STAAR Grade 7 Math Practice Tests
$16.99$11.99
### CHSPE Math Full Study Guide
$25.99$13.99 |
# Examples on How to Multiply Fractions
Lesson Objective
In this lesson, we will see more examples on how to multiply fractions. Also, we will learn how to use common factor to make the multiplication easier.
The basics on multiplying fractions have already been explained in the previous lesson.
This lesson will show more examples on multiplying two fractions that are:
1. 2 proper fractions
2. 1 proper and 1 improper fraction
3. 2 mixed fractions.
Also, we will learn how to use common factor to simplify the multiplication when possible.
The study tips and math video below will explain more.
# Study Tips
Tip #1 - The basics on how to multiply fractions
The lesson on multiplying fractions have already explained the basics needed. Please go through it if you are not sure.
Tip #2 - Simplify the multiplication using Common Factor
It is possible to simplify the multiplication if there is one or more common factor in the fractions. For example, let's multiply:
1. Multiplying the fraction gives:
2. Notice that 3 and 9 have the common factor, 3. Therefore we can divide these two numbers with 3.
3. This gives 1 and 3 respectively:
4. Since we unable to simplify it further, we can continue by doing the multiplication. This gives:
# Math Video
Lesson Video
If the above player doesn't work, try this direct link.
Math Video Transcript
# Practice Questions & More
Multiple Choice Questions (MCQ)
Now, let's try some MCQ questions to understand this lesson better.
You can start by going through the series of questions on How to Multiply Fractions or pick your choice of question below.
1. Question 1 on simplify and multiply two fractions
2. Question 2 on simplify and multiply two fractions |
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Sect. 1.5: Probability Distributions for Large N: (Continuous Distributions)
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Sect. 1.5: Probability Distributions for Large N (Continuous Probability Distributions). We’ve found that, for the one-dimensional Random Walk Problem , the probability distribution is the Binomial Distribution : W N (n 1 ) = [N!/(n 1 !n 2 !)]p n1 q n2 Here, q = 1 – p, n 2 = N - n 1. - PowerPoint PPT Presentation
### Transcript of Sect. 1.5: Probability Distributions for Large N (Continuous Probability Distributions)
Sect. 1.5: Probability Distributions for Large N: (Continuous Distributions)
For the 1 Dimensional Random Walk ProblemWe’ve found: The Probability Distribution is Binomial:
WN(n1) = [N!/(n1!n2!)]pn1qn2
Mean number of steps to the right: <n1> = NpDispersion in n1: <(Δn1)2> = Npq
Relative Width:(Δ*n1)/<n1> = (q½)(pN)½
for N increasing, mean value increases N, & relative width decreases (N)-½
N = 20 p = q = ½
q = 1 – p n2 = N - n1
• Imagine N getting larger & larger. Based on what we just said, the relative width of WN(n1) gets smaller & smaller & the mean value <n1> gets larger & larger.
• If N is VERY, VERY large, we can treat W(n1) as a
continuous function of a continuous variable n1. • For N large, it’s convenient to look at the natural log
ln[W(n1)] of W(n1), rather than the function itself.
• Do a Taylor’s series expansion of ln[W(n1)] about value of n1 where W(n1) has a maximum.
• Detailed math (in text) shows that this value of n1 is it’s average value <n1> = Np.
• It also shows that the width is equal to the value of the width <(Δn1)2> = Npq.
• For N VERY, VERY large, treat W(n1) as a continuous function of n1. For N large, look at ln[W(n1)], rather than the function itself.
• Do a Taylor’s series expansion of ln[W(n1)] about the n1 for W(n1) = its maximum. Detailed math shows that this
value of n1 is it’s mean <n1> = Np. It also shows that the width is equal to <(Δn1)2> = Npq.
• For ln[W(n1)], use Stirling’s Approximation (Appendix A-6) for logs of large factorials.
Stirling’s Approximation• If n is a large integer, the natural log of it’s factorial is
approximately:
ln[n!] ≈ n[ln(n) – 1]
• In this large N, large n1 limit, the Binomial
Distribution W(n1) becomes (shown in detail in the text):
W(n1) = Ŵexp[-(n1 - <n1>)2/(2<(Δn1)2>)] Here, Ŵ = [2π <(Δn1)2>]-½
• This is called the Gaussian Distribution or the Normal Distribution. We’ve found that <n1> = Np, <(Δn1)2> = Npq.
• The reasoning which led to this for large N & continuous n1
limit started with the Binomial Distribution. But this is a very general result. Starting with ANY discrete probability distribution & taking the limit of LARGE N, will result in the Gaussian or Normal Distribution. This is called
The Central Limit Theoremor The Law of Large Numbers.
• One of the most important results of probability theory is
The Central Limit Theorem:• The distribution of any random
phenomenon tends to be Gaussian or Normal if we average it over a large number of independent repetitions.
• This theorem allows us to analyze and predict the results of chance phenomena when we average over many observations.
Related to the Central Limit Theorem is
The Law of Large Numbers:• As a random phenomenon is repeated a
large number of times, the proportion of trials on which each outcome occurs gets closer and closer to the probability of that outcome, and
• The mean of the observed values gets closer and closer to the mean of a Gaussian Distribution which describes the data.
Sect. 1.6: The Gaussian Probability Distribution• In the limit of a large number of steps in the random
walk, N (>>1), the Binomial Distribution becomes a
Gaussian Distribution: W(n1) = [2π<(Δn1)2>]-½exp[-(n1 - <n1>)2/(2<(Δn1)2>)]
<n1> = Np, <(Δn1)2> = Npq
• Recall that n1 = ½(N + m), where the displacement
x = mℓ & that <m> = N(p – q). We can use this to convert to the probability distribution for displacement m, in the large N limit (after algebra):
P(m) = [2π<(Δm)2>]-½exp[-(m - <m>)2/(2<(Δm)2>)]
<m> = N(p – q), <(Δm)2> = 4Npq
P(m) = [2πNpq]-½exp[-(m – N{p – q})2/(8Npq)]
We can express this in terms of x = mℓ. As N >> 1, x can betreated as continuous. In this case, |P(m+2) – P(m)| << P(m)& discrete values of P(m) getcloser & closer together.
• Now, ask: What is the probability that, afterN steps, the particle is in the range x to x + dx?• Let the probability distribution for this ≡ P(x).• Then, we have: P(x)dx = (½)P(m)(dx/ℓ). • The range dx contains (½)(dx/ℓ) possible values
of m, since the smallest possible dx is dx = 2ℓ.
• After some math, we obtain the standard form of the
Gaussian (Normal) DistributionP(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]
μ ≡ N(p – q)ℓ ≡ mean value of xσ ≡ 2ℓ(Npq)-½ ≡ width of the distribution
NOTE: The generality of the arguments
we’ve used is such that a
Gaussian Distribution occurs in the limit oflarge numbers for all discrete distributions!
P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]μ ≡ N(p – q)ℓ σ ≡ 2ℓ(Npq)-½
• Note: To deal with Gaussian distributions, you need to get used to doing integrals with them! Many are tabulated!!
• Is P(x) properly normalized? That is, doesP(x)dx = 1? (limits - < x < )
P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]dx = (2π)-½σ-1exp[-y2/2σ2]dy (y = x – μ)
= (2π)-½σ-1 [(2π)½σ] (from a table)
P(x)dx = 1
P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]μ ≡ N(p – q)ℓ σ ≡ 2ℓ(Npq)-½
• Compute the mean value of x (<x>):
<x> = xP(x)dx = (limits - < x < )
xP(x)dx = (2π)-½σ-1xexp[-(x – μ)2/2σ2]dx= (2π)-½σ-(y + μ)exp[-y2/2σ2]dy (y = x – μ)
= (2π)-½σ-1yexp[-y2/2σ2]dy + μ exp[-y2/2σ2]dyyexp[-y2/2σ2]dy = 0 (odd function times even function)
exp[-y2/2σ2]dy = [(2π)½σ] (from a table)
<x> = μ ≡ N(p – q)ℓ
P(x)dx = (2π)-½σ-1exp[-(x – μ)2/2σ2]
μ ≡ N(p – q)ℓ σ ≡ 2ℓ(Npq)-½
• Compute the dispersion in x (<(Δx)2>)<(Δx)2> = <(x – μ)2> = (x – μ)2P(x)dx (limits - < x < )
<(Δx)2> = xP(x)dx = (2π)-½σ-1xexp[-(x – μ)2/2σ2]dx
= (2π)-½σ-1y2exp[-y2/2σ2]dy (y = x – μ)
= (2π)-½σ-1(½)(π)½σ(2σ2)1.5 (from a table)
<(Δx)2> = σ2 = 4Npqℓ2
0 2 4 6 8 10
x
0.00
0.05
0.10
0.15
0.20
0.25
fxComparison of Binomial &
Gaussian Distributions
Dots = BinomialCurve = Gaussian
with the same mean μ & the same width σ
Some Well-known & Potentially Useful Properties of Gaussians
Gaussian Width = 2σ
P(x) =
Areas Under Portions of a Gaussian Distribution
Two Graphs with the Same Informationin Different Forms
Again, Two Forms ofthe Same Information
Areas Under Portions of a Gaussian Distribution
Sect. 1.7: Probability Distributions Involving Several Variables: Discrete or Continuous
• Consider a statistical description of a situation with more than one random variable:
Example, 2 variables, u, v• The possible values of u are: u1,u2,u3,…uM
• The possible values of v are: v1,v2,v3,…vM
P(ui,vj) ≡ Probability that u = ui, & v = vj
SIMULTANEOUSLY• We must have:
∑i = 1 M ∑j = 1 N P(ui,vj) = 1
P(ui,vj) ≡ Probability that u = ui, & v = vj
SIMULTANEOUSLY∑i = 1 M ∑j = 1 N P(ui,vj) = 1
• Let Pu(ui) ≡ Probability that u = ui independent of the value v = vj
So, Pu(ui) ≡ ∑j = 1 N P(ui,vj)
• Similarly, let Pv(vj) ≡ Probability that
v = vj independent of value u = ui So, Pv(vj) ≡ ∑i = 1 M P(ui,vj)
• Of course, it must also be true that∑i = 1 M Pu(ui) = 1 & ∑j = 1 N Pv(vj) = 1
In the special case that u & v are
Statistically Independent
or Uncorrelated:
Then & only then can we write:
P(ui,vj) ≡ Pu(ui)Pv(vj)
A General Discussion of Mean Values• If F(u,v) = any function of u,v, it’s mean value is:
<F(u,v)> ≡ ∑i = 1 M ∑j = 1 N P(ui,vj)F(ui,vj) • If F(u,v) & G(u,v) are any 2 functions of u, v, we
can easily show:
<F(u,v) + G(u,v)> = <F(u,v)> + <G(u,v)> • If f(u) is any function of u & g(v) is any function
of v, we can easily show:
<f(u)g(v)> ≠ <f(u)><g(v)> • The only case when the inequality becomes an
equality is if u & v are statistically independent.
Sect. 1.8: Comments on Continuous Probability Distributions
• Everything we’ve discussed for discrete distributions generalizes to continuous distributions in obvious ways.
• Let u ≡ a continuous random variable in the range:
a1 ≤ u ≤ a2 • The probability of finding u in the range u to u + du
≡ P(u) ≡ P(u)du P(u) ≡ Probability Density
of the distribution function • Normalization: P(u)du = 1 (limits a1 ≤ u ≤ a2)
• Mean values: <F(u)> ≡ F(u)P(u)du.
• Consider two continuous random variables:u ≡ continuous random variable in range: a1 ≤ u ≤ a2
v ≡ continuous random variable in range: b1 ≤ v ≤ b2
• The probability of finding u in the range u to u + du AND v in the range v to v + dv is
P(u,v) ≡ P(u,v)dudv P(u,v) ≡ Probability Density function
• Normalization:
P(u,v)dudv = 1(limits a1 ≤ u ≤ a2, b1 ≤ v ≤ b2)
• Mean values:
<G(u,v)> ≡ G(u,v)P(u,v)dudv
Functions of Random VariablesAn important, often occurring problem is:
• Consider a random variable u. • Suppose φ(u) ≡ any continuous function of u.
Question• If P(u)du ≡ Probability of finding u in the range
u to u + du, what is the probability W(φ)dφ of finding φ in the range φ to φ + dφ?
• Answer using essentially the “Chain Rule” of differentiation, but take the absolute value to make sure that probability W ≥ 0:
W(φ)dφ ≡ P(u)|du/dφ|dφCaution!!
φ(u) may not be a single valued function of u!
• Equally Likely The probability of finding θ between θ & θ + dθ is: P(θ)dθ ≡ (dθ/2π)
Question• What is the probability W(Bx)dBx that the x component
of B lies between Bx & Bx + dBx?
• Clearly, we must have –B ≤ Bx ≤ B. Also, each value of dBx corresponds to 2 possible values of dθ.
Also, dBx = |Bsinθ|dθ
• So, we have:W(Bx)dBx = 2P(θ)|dθ/dBx|dBx = (π)-1dBx/|Bsinθ|
Note also that: |sinθ| = [1 – cos2θ]½ = [1 – (Bx)2/B2]½ so finally,
W(Bx)dBx = (π)-1dBx[1 – (Bx)2/B2]-½, –B ≤ Bx ≤ B = 0, otherwise
• W not only has a maximum at Bx = B, it diverges there! It has a minimum at Bx = 0.
So, it looks like
• W diverges at Bx = B, but it can be shown that it’s integral is finite. So that W(Bx) is a properly normalized probability:
W(Bx)dBx= 1 (limits: –B ≤ Bx ≤ B) |
Update all PDFs
# Joining two midpoints of sides of a triangle
Alignments to Content Standards: G-SRT.B.4
Suppose $ABC$ is a triangle. Let $M$ be the midpoint of $\overline{AB}$ and $\ell$ the line through $M$ parallel to $\overleftrightarrow{AC}$:
1. Show that angle $CAB$ is congruent to angle $PMB$ and that angle $BPM$ is congruent to angle $BCA$. Conclude that triangle $MBP$ is similar to triangle $ABC$.
2. Use part (a) to show that $P$ is the midpoint of $\overline{BC}$.
## IM Commentary
This task is closely related to very important material about similarity and ratios in geometry. In this case, the point $M$ selected on the triangle $ABC$ is the midpoint of $\overline{AB}$. More generally, consider the picture below:
Here $F$ is any point on $\overline{AB}$ (other than one of the endpoints) and $G$ is the point on $\overline{BC}$ so that $\overleftrightarrow{FG}$ parallel to line $\overleftrightarrow{AC}$. Then we have the relationship $$\frac{|FA|}{|AB|} = \frac{|GC|}{|BC|}.$$ The converse of this result holds as well: if $F$ is a point on $\overline{AB}$ and $G$ is a point on $\overline{BC}$ so that the above equality holds then then $\overleftrightarrow{FG}$ is parallel to $\overleftrightarrow{AC}$.
The argument presented here for midpoints works also in the more general setting mentioned in the above paragraph. Essential to this argument is the use of the AAA criterion for triangle similarity, G-SRT.3. The task can be used for assessment provided students are aware that they may use the AAA result in part (b) of this problem.
## Solution
1. We require two auxiliary points on line $\ell$: a point $D$ so that $M$ is between $P$ and $D$ and a point $E$ so that $P$ is between $M$ and $E$ as in the picture below
Lines $\ell$ and $\overleftrightarrow{AC}$ are parallel and $\overleftrightarrow{AB}$ is a transverse for these parallel lines. Therefore angle $CAB$ is congruent to angle $DMA$ since these are alternate interior angles for a transverse meeting parallel lines. Angle $DMA$ is congruent to angle $PMB$ by the vertical angle theorem. So angle $CAB$ is congruent to angle $PMB$.
The same argument will show that angles $BPM$ and $BCA$ are congruent: angle $BCA$ is congruent to angle $EPC$ (alternate interior angles) and angle $EPC$ is congruent to angle $BPM$ (vertical angles) establishing that angles $BCA$ and $BPM$ are congruent.
We have $m(\angle ABC) = m(\angle MBP)$ since these are the same angles. Therefore the triangles $MBP$ and $ABC$ share three congruent angles and they are similar triangles. Alternatively, we could stop after producing two congruent angles and use the AA criterion for similarity of triangles.
2. Since triangles $MBP$ and $ABC$ are similar it follows that $$\frac{|PB|}{|CB|} = \frac{|BM|}{|BA|}.$$ Since $M$ is the midpoint of $\overline{BA}$, we have $\frac{|BM|}{|BA|} = \frac{1}{2}$ and so $\frac{|PB|}{|CB|} = \frac{1}{2}$ which means that $P$ is the midpoint of segment $\overline{CB}$. |
# Ch. - 6 Application of Derivatives
## Chapter 6 Ex.6.1 Question 1
Find the rate of change of the area of a circle with respect to its radius $$r$$ when
a. $$r = 3\;cm$$
b. $$r = 4\;cm$$
### Solution
We know that the area of a circle, $$A=\pi {{r}^{2}}$$
Therefore, the rate of change of the area with respect to its radius is given by
\begin{align} \frac{dA}{dr}&=\frac{d}{dr}\left( \pi {{r}^{2}} \right) \\ & =2\pi r \end{align}
a. When $$r=3\;cm$$
Then,
\begin{align} \frac{dA}{dr}&=2\pi \left( 3 \right) \\ & =6\pi \\ \end{align}
Thus, the area is changing at the rate of $$6\pi$$.
b. When $$r=4\;cm$$
Then,
\begin{align} \frac{dA}{dr}&=2\pi \left( 4 \right) \\ & =8\pi \end{align}
Thus, the area is changing at the rate of $$8\pi$$.
## Chapter 6 Ex.6.1 Question 2
The volume of a cube is increasing at the rate of $$8{\text{ cm}^{3}}/s$$. How fast is the surface area increasing when the length of its edge is $$12\, \rm{cm}$$?
### Solution
Let the side length, volume and surface area respectively be equal to $$x$$, $$V$$ and $$S$$ .
Hence, $$V = {x^3}$$ and $$S=6{{x}^{2}}$$
We have,
\begin{align} \frac{dV}{dt}=8\;{{cm}^{3}}/s \end{align}
Therefore,
\begin{align} \frac{dV}{dt}&=\frac{d}{dt}\left( {{x}^{3}} \right) \\ 8 &=\frac{d}{dx}\left( {{x}^{3}} \right)\frac{dx}{dt} \\ 8 &=3{{x}^{2}}\frac{dx}{dt} \\ \frac{dx}{dt}&=\frac{8}{3{{x}^{2}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) \end{align}
Now,
\begin{align} \frac{dS}{dt}&=\frac{d}{dt}\left( 6{{x}^{2}} \right) \\ & =\frac{d}{dx}\left( 6{{x}^{2}} \right)\frac{dx}{dt} \\ & =12x\frac{dx}{dt} \\ & =12x\left( \frac{8}{3{{x}^{2}}} \right)\ \ \ \ \ \ \ \ \ \ \left[ \text{from }\left( 1 \right) \right] \\ & =\frac{32}{x} \end{align}
So, when $$x=12\, \rm{cm}$$
Then,
\begin{align} \frac{dS}{dt}&=\frac{32}{12}{\text{cm}^{2}}/s \\ & =\frac{8}{3}{\text{cm}^{2}}/s \end{align}
## Chapter 6 Ex.6.1 Question 3
The radius of a circle is increasing uniformly at the rate of $$3 \,\rm{cm}/s$$. Find the rate at which the area of the circle is increasing when the radius is $$10\, \rm{cm}/s$$.
### Solution
We know that $$A=\pi {{r}^{2}}$$
Now,
\begin{align} \frac{dA}{dt}&=\frac{d}{dr}\left( \pi {{r}^{2}} \right) \\ & =2\pi r\frac{dr}{dt} \end{align}
We have,
$\frac{dr}{dt}=3 \,\rm{cm}/s$
Hence,
\begin{align} \frac{dA}{dt}& =2\pi r\left( 3 \right) \\ & =6\pi r \end{align}
So, when $$r=10\, \rm{cm}/s$$
Then,
\begin{align} \frac{dA}{dt} &=6\pi \left( 10 \right) \\ & =60\pi \,\rm{{cm}^{2}}/s \end{align}
## Chapter 6 Ex.6.1 Question 4
An edge of a variable cube is increasing at the rate of $$3\, \rm{cm}/s$$ . How fast is the volume of the cube increasing when the edge is $$10\, \rm{cm}$$ long?
### Solution
Let the length and the volume of the cube respectively be $$x$$ and $$V$$.
Hence, $$V = {x^3}$$
Now,
\begin{align} \frac{dV}{dt}&=\frac{d}{dt}\left( {{x}^{3}} \right) \\ & =\frac{d}{dx}\left( {{x}^{3}} \right)\frac{dx}{dt} \\ & =3{{x}^{2}}\frac{dx}{dt} \end{align}
We have,
\begin{align} \frac{dx}{dt}=3 \,\rm{cm}/s \end{align}
Hence,
\begin{align} \frac{dV}{dt} &=3{{x}^{2}}\left( 3 \right) \\ & =9{{x}^{2}} \end{align}
So, when $$x=10\,cm$$
Then,
\begin{align} \frac{dV}{dt}&=9{{\left( 10 \right)}^{2}} \\ & =900\,{{cm}^{3}}/s \end{align}
## Chapter 6 Ex.6.1 Question 5
A stone is dropped into a quiet lake and waves move in circles at the speed of $$5\, \rm{cm}/s$$. At the instant when the radius of the circular wave is $$8 \,\rm{cm}$$, how fast is the encoding area is increasing?
### Solution
We know that $$A=\pi {{r}^{2}}$$
Now,
\begin{align} \frac{dA}{dt}&=\frac{d}{dt}\left( \pi {{r}^{2}} \right) \\ & =2\pi r\frac{dr}{dt} \end{align}
We have,
\begin{align} \frac{dr}{dt}=5\;cm/s \end{align}
Hence,
\begin{align} \frac{dA}{dt}&=2\pi r\left( 5 \right) \\ & =10\pi r \end{align}
So, when $$r=8\;cm$$
Then,
\begin{align} \frac{dA}{dt}&=10\pi \left( 8 \right) \\ & =80\pi \;{{cm}^{2}}/s \end{align}
## Chapter 6 Ex.6.1 Question 6
The radius of a circle is increasing at the rate of $$0.7\,\rm{cm}/s$$. What is the rate of increase of its circumference?
### Solution
We know that $$C=2\pi r$$
Now,
\begin{align} \frac{dC}{dt}&=\frac{d}{dt}\left( 2\pi r \right) \\ & =\frac{d}{dr}\left( 2\pi r \right)\frac{dr}{dt} \\ & =2\pi \frac{dr}{dt} \end{align}
We have, \begin{align} \frac{dr}{dt}=0.7\pi \,\rm{cm/s}\end{align}
Hence,
\begin{align}\frac{dC}{dt}&=2\pi \left( 0.7 \right) \\ & =1.4\pi \,\rm{cm/s} \end{align}
## Chapter 6 Ex.6.1 Question 7
The length $$x$$ of a rectangle is decreasing at the rate of $$5 \text{ cm/minute}$$ and the width $$y$$ is increasing at the rate of $$4\text{ cm/minute}$$. When $$x=8\,\rm{cm}$$ and $$y=6\,\rm{cm}$$, find the rate of change of
(a) Perimeter of the rectangle
(b) Area of the rectangle
### Solution
It is given that \begin{align} \frac{dx}{dt}=-5\text{ cm/minute}, \frac{{dy}}{{dt}} = 4\text{ cm/minute},\end{align} $$x=8\,\rm{cm}$$ and $$y=6\,\rm{cm}$$
(a) Perimeter of a rectangle is given by $$P=2\left( x+y \right)$$
Therefore,
\begin{align} \frac{dP}{dt}&=2\left( \frac{dx}{dt}+\frac{dy}{dt} \right) \\ & =2\left( -5 + 4 \right) \\& = -2 \text{ cm/minute} \end{align}
(b) The Area of a rectangle is given by $$A=xy$$
Therefore,
\begin{align}\frac{d A}{d t} &=\frac{d x}{d t} y+x \frac{d y}{d t} \\&= - 5y + 4x\end{align}
When $$x = 8\,\rm{cm}$$ and $$y=6\,\rm{cm}$$
Then,
\begin{align} \frac{dA}{dt}&=\left( -5\times 6+4\times 8 \right)\,{\rm{cm}^{2}}\text{/minute} \\& = 2\,{\rm{cm}^{2}}\text{/minute} \end{align}
## Chapter 6 Ex.6.1 Question 8
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is $$15cm$$.
### Solution
We know that $$V=\frac{4}{3}\pi {{r}^{3}}$$
Hence,
\begin{align} \frac{dV}{dt}&=\frac{dV}{dt}\left( \frac{4}{3}\pi {{r}^{3}} \right) \\ & =\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{3}} \right)\frac{dr}{dt} \\ & =4\pi {{r}^{2}}\frac{dr}{dt} \end{align}
We have,
$$\frac{dV}{dt}=900c{{m}^{2}}/s$$
Therefore,
\begin{align} 4\pi {{r}^{2}}\frac{dr}{dt}&=900 \\ \frac{dr}{dt}&=\frac{900}{4\pi {{r}^{2}}} \\ & =\frac{225}{\pi {{r}^{2}}} \end{align}
When radius, $$r=15cm$$
Then,
\begin{align} \frac{dr}{dt}&=\frac{225}{\pi {{\left( 15 \right)}^{2}}} \\ & =\frac{1}{\pi }cm/s \end{align}
## Chapter 6 Ex.6.1 Question 9
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is $$10\,\rm{cm}$$.
### Solution
We know that $$V=\frac{4}{3}\pi {{r}^{2}}$$
Therefore,
\begin{align} \frac{dV}{dt}&=\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{2}} \right) \\ & =\frac{4}{3}\pi \left( 3{{r}^{2}} \right) \\ & =4\pi {{r}^{2}} \end{align}
When radius, $$r = 10\,\rm{cm}$$
Then,
\begin{align} \frac{dV}{dt}&=4\pi {{\left( 10 \right)}^{2}} \\ & =400\pi \end{align}
Thus, the volume of the balloon is increasing at the rate of $$400\pi\, {\rm{cm}^{3}}/s$$.
## Chapter 6 Ex.6.1 Question 10
A ladder is $$5\,\rm{m}$$ long is leaning against the wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $$2\, \rm{cm/s}$$. How fast is its height on the wall decreasing when the foot of the ladder is $$4\,\rm{m}$$ away from the wall?
### Solution
Let the height of the wall at which the ladder is touching it be $$ym$$ and the distance of its foot from the wall on the ground be $$x\,\rm{m}$$.
Hence,
\begin{align} & {{x}^{2}}+{{y}^{2}}={{5}^{2}} \\ & \Rightarrow {{y}^{2}}=25-{{x}^{2}} \\ & \Rightarrow\; y=\sqrt{25-{{x}^{2}}} \\ \end{align}
Therefore,
\begin{align} \frac{dy}{dt}&=\frac{d}{dt}\left( \sqrt{25-{{x}^{2}}} \right) \\ & =\frac{d}{dx}\left( \sqrt{25-{{x}^{2}}} \right)\frac{dx}{dt} \\ & =\frac{-x}{\left( \sqrt{25-{{x}^{2}}} \right)}\frac{dx}{dt} \end{align}
We have,
$$\frac{dx}{dt}=2\,\rm{cm/s}$$
Thus,
$$\frac{dy}{dt}=\frac{-2x}{\sqrt{25-{{x}^{2}}}}$$
When $$x=4\,\rm{cm}$$
Then,
\begin{align} \frac{dy}{dt}&=\frac{-2\times 4}{\sqrt{25-16}} \\ & =-\frac{8}{3}\,\rm{cm/s} \end{align}
## Chapter 6 Ex.6.1 Question 11
A particle is moving along the curve $$6y = {x^3} + 2$$. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
### Solution
The equation of the curve is $$6y={{x}^{3}}+2$$
Differentiating with respect to time, we have
\begin{align} & \Rightarrow 6\frac{dy}{dt}=3{{x}^{2}}\frac{dx}{dt} \\ & \Rightarrow 2\frac{dy}{dt}={{x}^{2}}\frac{dx}{dt} \\ \end{align}
According to the question, $$\left( \frac{dy}{dt}=8\frac{dx}{dt} \right)$$
Hence,
\begin{align} & \Rightarrow 2\left( 8\frac{dx}{dt} \right)={{x}^{2}}\frac{dx}{dt} \\ & \Rightarrow 16\frac{dx}{dt}={{x}^{2}}\frac{dx}{dt} \\ & \Rightarrow \left( {{x}^{2}}-16 \right)\frac{dx}{dt}=0 \\ & \Rightarrow {{x}^{2}}=16 \\ & \Rightarrow x=\pm 4 \\ \end{align}
When $$x=4$$
Then,
\begin{align} y&=\frac{{{4}^{3}}+2}{6} \\ & =\frac{66}{6} \\ & =11 \end{align}
When $$x = - 4$$
Then,
\begin{align} y&=\frac{\left( -{{4}^{3}} \right)+2}{6} \\ & =-\frac{62}{6} \\ & =-\frac{31}{3} \end{align}
Thus, the points on the curve are $$\left( 4,11 \right)$$ and $$\left( -4,\frac{-31}{3} \right)$$.
## Chapter 6 Ex.6.1 Question 12
The radius of an air bubble is increasing at the rate of $$\frac{1}{2}\,\rm{cm/s}$$. At which rate is the volume of the bubble increasing when the radius is $$1\,\rm{cm}$$?
### Solution
Assuming that the air bubble is a sphere, $$V=\frac{4}{3}\pi {{r}^{3}}$$
Therefore,
\begin{align} \frac{dV}{dt}&=\frac{d}{dt}\left( \frac{4}{3}\pi {{r}^{3}} \right)\frac{dr}{dt} \\ & =4\pi {{r}^{2}}\frac{dr}{dt} \end{align}
We have,
$$\frac{{dr}}{{dt}} = \frac{1}{2}\,\rm{cm/s}$$
When $$r=1\,\rm{cm}$$
Then,
\begin{align} \frac{dV}{dt}&=4\pi {{\left( 1 \right)}^{2}}\left( \frac{1}{2} \right) \\ & =2\pi {\rm{cm}^{3}}/s \end{align}
## Chapter 6 Ex.6.1 Question 13
A balloon, which always remains spherical, has a variable diameter $$\frac{3}{2}\left( 2x+1 \right)$$. Find the rate of change of its volume with respect to $$x$$.
### Solution
We know that $$V=\frac{4}{3}\pi {{r}^{3}}$$
It is given that diameter, $$d=\frac{3}{2}\left( 2x+1 \right)$$
Hence, $$r = \frac{3}{4}\left( {2x + 1} \right)$$
Therefore,
\begin{align} V&=\frac{4}{3}\pi {{\left( \frac{3}{4} \right)}^{3}}{{\left( 2x+1 \right)}^{3}} \\ & =\frac{9}{16}\pi {{\left( 2x+1 \right)}^{3}} \end{align}
Thus,
\begin{align} \frac{dV}{dx}&=\frac{9}{16}\pi \frac{d}{dx}{{\left( 2x+1 \right)}^{3}} \\ & =\frac{27}{8}\pi {{\left( 2x+1 \right)}^{3}} \end{align}
## Chapter 6 Ex.6.1 Question 14
Sand is pouring from a pipe at the rate of $$12{\rm{cm}^{3}}/s$$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when height is $$4\,\rm{cm}$$?
### Solution
We know that $$V=\frac{1}{3}\pi {{r}^{2}}h$$
It is given that, $$h=\frac{1}{6}r$$
Hence, $$r=6h$$
Therefore,
\begin{align} V&=\frac{1}{3}\pi {{\left( 6h \right)}^{2}}h \\ & =12\pi {{h}^{3}} \end{align}
Thus,
\begin{align} \frac{dV}{dt}&=12\pi \frac{d}{dt}\left( {{h}^{3}} \right)\frac{dh}{dt} \\ & =12\pi \left( 3{{h}^{2}} \right)\frac{dh}{dt} \\ & =36\pi {{h}^{2}}\frac{dh}{dt} \end{align}
We have,
$$\frac{dV}{dt}=12\,{\rm{cm}^{2}}/s$$
When $$h=4\,\rm{cm}$$
Then,
\begin{align} 12&=36\pi {{\left( 4 \right)}^{2}}\frac{dh}{dt} \\ & \frac{dh}{dt}=\frac{12}{36\pi \left( 16 \right)} \\ & =\frac{1}{48\pi }\,\rm{cm/s} \end{align}
## Chapter 6 Ex.6.1 Question 15
The total cost $$C\left( x \right)$$ in Rupees associated with the production of $$x$$ units of an item is given by $$C\left( x \right)=0.007{{x}^{3}}-0.003{{x}^{2}}+15x+4000$$. Find the marginal cost when 17 units are produced.
### Solution
Marginal cost (MC) is the rate of change of the total cost with respect to the output.
Therefore,
\begin{align} MC&=\frac{dC}{dx}=0.007\left( 3{{x}^{2}} \right)-0.003\left( 2x \right)+15 \\ & =0.021{{x}^{2}}-0.006x+15 \end{align}
When $$x=17$$
Then,
\begin{align} MC&=0.021{{\left( 17 \right)}^{2}}-0.006\left( 17 \right)+15 \\ & =0.021\left( 289 \right)-0.006\left( 17 \right)+15 \\ & =6.069-0.102+15 \\ & =20.967 \end{align}
So, when $$17$$ units are produced, the marginal cost is ₹ 20.967.
## Chapter 6 Ex.6.1 Question 16
The total revenue in Rupees received from the sale of $$x$$ units of a product given by $$R\left( x \right)=13{{x}^{2}}+26x+15$$. Find the marginal revenue when $$x=7$$.
### Solution
Marginal revenue (MR) is the rate of change of the total revenue with respect to the number of units sold.
Therefore,
\begin{align} MR&=\frac{dR}{dx}=13\left( 2x \right)+26 \\ & =26x+26 \end{align}
When, $$x=7$$
Then,
\begin{align} MR&=26\left( 7 \right)+26 \\ & =182+26 \\ & =208 \end{align}
Thus, the marginal revenue is $$₹$$ $$208$$.
## Chapter 6 Ex.6.1 Question 17
The rate of change of the area of a circle with respect to its radius $$r$$ is $$r=6\, \rm{cm}$$
(A) $$10\pi$$
(B) $$12\pi$$
(C) $$8\pi$$
(D) $$11\pi$$
### Solution
We know that $$A=\pi {{r}^{2}}$$
Therefore,
\begin{align} \frac{dA}{dr}&=\frac{d}{dr}\left( \pi {{r}^{2}} \right) \\ & =2\pi r \end{align}
When $$r=6cm$$
Then,
\begin{align} \frac{dA}{dr}&=2\pi \times 6 \\ & =12\pi c{{m}^{2}}/s \end{align}
Thus, the rate of change of the area of the circle is $$12\pi c{{m}^{2}}/s$$.
Hence, the correct option is B.
## Chapter 6 Ex.6.1 Question 18
The total revenue is Rupees received from the sale of $$x$$ units of a product is given by $$R\left( x \right)=3{{x}^{2}}+36x+5$$. The marginal revenue, when $$x=15$$ is
(A) 116
(B) 96
(C) 90
(D) 126
### Solution
Marginal revenue (MR) is the rate of change of the total revenue with respect to the number of units sold.
Therefore,
\begin{align} MR&=\frac{dR}{dx}=3\left( 2x \right)+36 \\ & =6x+36 \end{align}
When, $$x=15$$
Then,
\begin{align} MR&=6\left( 15 \right)+36 \\ & =90+36 \\ & =126 \end{align}
Thus, the marginal revenue is ₹ 126.
Hence, the correct option is D.
## Miscellaneous Exercise
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# Question Video: Computing Area of Parallelogram Using Matrices Mathematics • 10th Grade
Use determinants to calculate the area of the parallelogram with vertices (1, 1), (−4, 5), (−2, 8), and (3, 4).
02:27
### Video Transcript
Use determinants to calculate the area of the parallelogram with vertices one, one; negative four, five; negative two, eight; and three, four.
So, what I’ve done here is that I’ve drawn a sketch. And this is a sketch of our parallelogram. So, how are we gonna use determinants to calculate the area? So, let’s consider how we’d actually work out the area of the parallelogram? Well, we’d multiply its width by its length.
So, in our diagram, we can represent these using vectors. So, our width is going to be two, three. So, it’s the vector two, three. And that’s because we’ve gone from negative four to negative two. So, that’s two in the positive direction. And then, we’ve also gone from five to eight. So, that’s up by three in the positive direction. And then, our vector for our length would be five, negative four. So, we’re gonna use these two vectors to determine the area of our parallelogram.
So, we’ve got the vectors two, three; five, negative four. And what we’re gonna do is we’re gonna put them together to form a two-by-two matrix where the columns are these two vectors. And it’s the determinant of this matrix that we’re gonna form that’s gonna find the area cause we can say that the area of the parallelogram is equal to the modulus of the determinant of the matrix two, five, three, negative four. So, now we’re gonna use the method we have to find the determinant of a two-by-two matrix.
And that is that if we have a matrix 𝑎, 𝑏, 𝑐, 𝑑, then the determinant of this matrix is equal to 𝑎𝑑 minus 𝑏𝑐. So therefore, we can say our area is going to be equal to the modulus, or absolute value, of two multiplied by negative four minus five multiplied by three, which is gonna be the absolute value, or modulus, of negative eight minus 15, which is gonna give us the result, which is the modulus, or absolute value, of negative 23.
Remembering, if we wanted to find the absolute value, then we disregard the negative result. Which makes sense in this situation because an area won’t be a negative value. So therefore, we can say that if we’ve got a parallelogram with vertices one, one; negative four, five; negative two, eight; and three, four, the area is going to be equal to 23. |
# Time and Distance - Aptitude Test Tricks, Shortcuts & Formulas
Quantitative aptitude problems on time and distance are sure questions in any competitive exam like GRE, GMAT, CAT, XAT, MAT, Bank exams or any other campus recruitment tests for that matter. These problems are often lengthy and test takers often skip them, thinking these questions will consume most amount of time. This is a mistake.
Time and distance aptitude questions are easy to score area if you know the very basic formula that you learnt in high school. All shortcuts to solve time and distance problems can be easily derived and learnt. This article provides Tips and Tricks to Solve "Time and Distance" Aptitude Problems with Important Formulas, Shortcuts, Core Concepts to crack Placement Test and Competitive exams.
Do not simply byheart the shortcuts or the tips and tricks. You should actually take time and learn how one arrives at these shortcuts, try it out yourself, then solve as many problems you can. By this you will automatically use the shortcut when you solve questions later on. We have compiled time and distance concepts and formulas for you in this section. We have also given how we derive each of these formulas. They should surely help you crack any aptitude test on time and distance.
To enhance your knowledge and skills to solve Time and Distance aptitude test problems, go through the tutorial on Time and Distance.
## Basic Concepts of Time and Distance
Most of the aptitude questions on time and distance can be solved if you know the basic correlation between speed, time and distance which you have learnt in your high school class.
• Relation between time, distance and speed: Speed is distance covered by a moving object in unit time.
Speed = "Distance Covered" / "Time taken"
• Ratio of the varying components when other is constant: Consider 2 objects A and B having speed S_a, S_b. Let the distance travelled by them are D_a and D_b respectively and time taken to cover these distances be T_a and T_b respectively.
Let's see the relation between time, distance and speed when one of them is kept constant
1. When speed is constant distance covered by the object is directly proportional to the time taken.
ie; If S_a = S_b, then D_a/D_b = T_a/T_b
2. When time is constant speed is directly proportional to the distance travelled.
ie; If T_a = T_b, then S_a/S_b = D_a/D_b
3. When distance is constant speed is inversely proportional to the time taken ie if speed increases then time taken to cover the distance decreases.
ie; If D_a = D_b, then S_a/S_b = T_b/T_a
• We know that when distance travelled is constant, speed of the object is inversely proportional to time taken.
1. If the speeds given are in Harmonic progression or HP then the corresponding time taken will be in Arithmetic progression or AP
2. If the speeds given are in AP then the corresponding time taken is in HP
• Unit conversion:While answering multiple choice time and dinstance problems in quantitative aptitude test, double check the units of values given. It could be in m/s or km/h. You can use the following formula to convert from one unit to other
1. x km/hr = x5/18 m/s
2. x m/s = x18/5 km/hr
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## Average Speed
Average speed is always equal to total distance travelled to total time taken to travel that distance.
"Average speed" = "Total distance" / "Total time"
• Distance Constant
If distance travelled for each part of the journey, ie d_1 = d_2 = d_3 = ... = d_n = d, then average speed of the object is Harmonic Mean of speeds.
Let each distance be covered with speeds s_1, s_2, ... s_n in t_1, t_2, ... t_n times respectively.
Then t_1 = d/s_1, t_2 = d/s_2, … t_n=d/s_n
"Average Speed" = (d + d + d +… "ntimes") / ((d/s_1) + (d/s_2) + (d/s_3) + ... (d/s_n))
"Average Speed" = n/((1/s_1) + (1/s_2) + … + (1/s_n))
• Time Constant
If time taken to travel each part of the journey, ie t_1 = t_2 = t_3 = … t_n = t, then average speed of the object is Arithmetic Mean of speeds
Let distance of parts of the journey be d_1, d_2, d_3, ... d_n and let them be covered with speed s_1, s_2, s_3, ... s_n respectively.
Then d_1 = s_1t, d_2 = s_2t, d_3 = s_3t, ... d_n = s_nt
"Average Speed" = (s_1t + s_2t + ... + s_nt) / (t + t + ... + "ntimes")
"Average Speed" = (s_1 + s_2 + s_3 + ... + s_n) / n"
## Relative Speed
• If two objects are moving in same direction with speeds a and b then their relative speed is |a - b|
• If two objects are moving is opposite direction with speeds a and b then their relative speed is (a + b)
## Important shortcuts to solve time and distance problems quickly
Using the shortcuts provided below, you can solve the aptitude problems on time and distance quickly
1. Given a person covers a distance with speed a km/hr and further covers same distance with speed b km/hr, then the average speed of the person is:
"Average speed" = "Total distance travelled" / "Total time taken"
Let the distance covered be d km.
Given d km be covered with speed a km/hr in time t_1 hour => t_1 = d/a
Given next d km be covered with speed b km/hr in time t_2 hour => t_2 = d/b
"Average speed" = 2d/(d/a + d/b)
"Average speed" = (2ab)/(a+b)
Shortcut: As discussed in "Basic Concepts" section, average speed is the HM (Harmonic Mean) of speeds a & b
2. Given a person covers a certain distance d km with speed a km/hr and returns back to the starting point with speed b km/hr.
• If the total time taken for the whole journey is given as T hours, then to find d:
We know "average speed" = "total distance travelled" / "total time taken"
Also "average speed" = (2ab) / (a+b)
=> (2ab)/(a+b) = 2d/T
d = T (ab/(a+b)) km
• If the difference between the individual time taken are given that is, if distance d is covered in t_1 hours with speed a km/hr and same distance is covered with speed b km/hr in t_2 hours, then to find d:
"Difference between individual time" = t_1 – t_2 (if t_1 > t_2)
Also t_1 = d/a and t_2 = d/b
So t_1 – t_2 = d/a – d/b
d = (t_1 – t_2) ((ab)/(b – a)) km
3. If a person covers p^(th) part of a distance at x km/hr, q^(th) part of the distance at y km/hr, r^(th) part of the distance at zkm/hr, then average speed is
"Average speed" = 1 / ((p/x) + (q/y) + (r/z))
4. Two persons A and B start at the same time from two points P and Q at the same time towards each other. They meet at a point R and A takes t_a time to reach Q and B takes t_b time to reach B. If speed of A and B are S_a and S_b respectively.
• Then S_a/S_b is: Let PQ = d and also let PR = l => RQ = d – l
Time taken by A to cover PR is same as time taken by B to cover QR.
We know that when time is constant, speed is directly proportional to distance covered.
So, S_a/S_b = (PR)/(QR) = l / (d – l)
Also, B takes t_b time to cover PR => PR = S_bt_b => l = S_bt_b
A takes t_a time to cover RQ => RQ = S_at_a => d – l = S_at_a
Substituting these values in above equation, we get S_a/S_b = (S_bt_b) / (S_at_a)
=> S_a/S_b = sqrt(t_b / t_a)
• Then time taken by A and B to meet at point R is:
We know t = (PR) / S_a
From previous analysis we also know PR = S_bt_b and S_a/S_b = sqrt(t_b / t_a)
So t = (S_bt_b )/S_a = t_b sqrt(t_b / t_a)
Thus t = sqrt(t_at_b)
• Both equations are valid even if A and B start at 2 different times from P and Q towards each other where A takes t_a time to reach R and B takes t_b time to reach R. After meeting at R they take the same time t to reach Q and P respectively
5. Two persons A and B start at the same time from two points P and Q at the same time towards each other with speeds S_1 and S_2 respectively. They reach their respective destinations and reverse their directions. They continue this to and fro motion. If S_1 > S_2 and S_1 < 2S_2 and D is the initial distance separating them, then,
• "Total distance covered till nth meeting" = (2n-1)D
• "Time taken by them to meet for the nth time" = ((2n-1)D) / (S_1 + S_2)
"Time taken by the persons to meet first time" = ("Distance travelled by A " + " Distance travelled by B") / ("Speed of A " + " Speed of B")
= D/ (S_1 + S_2)
After meeting, they continue to Q and P respectively. When they reach their destinations, they have together covered 2D distance.
Then they reverse directions. By the time they meet for second time, they will have covered 3D distance. Total distance covered by A and B for their 3rd meeting is 5D.
With this logic, "by nth meeting, they will have covered a total distance" = (2n-1)D.
"Time taken by them to meet for the nth time" = ((2n -1) D )/ (S_1 + S_2)
• Point of meeting when they meet for the nth time
1. Distance covered by A till nth meeting = Speed of A * Time taken by A till nth meeting = S1((2n -1)D)/(S1 + S2)
2. Divide distance obtained in step 1 by 2D, if value > 2D.
3. Remainder obtained in step 2 will give you the distance of meeting point from P
When you learn time and distance, you should keep in mind to cover following sections:
• Time and distance
• Relative speed and average speed
• Trains
• Boats and streams
• Races
• Clocks
All these are just different variations of that one simple formula speed = distance / time.
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# Lesson 1
The Areas of Squares and Their Side Lengths
### Problem 1
Find the area of each square. Each grid square represents 1 square unit.
### Problem 2
Find the length of a side of a square if its area is:
1. 81 square inches
2. $$\frac{4}{25}$$ cm2
3. 0.49 square units
4. $$m^2$$ square units
### Problem 3
Find the area of a square if its side length is:
1. 3 inches
2. 7 units
3. 100 cm
4. 40 inches
5. $$x$$ units
### Problem 4
Evaluate $$(3.1 \times 10^4) \boldcdot (2 \times 10^6)$$. Choose the correct answer:
A:
$$5.1 \times 10^{10}$$
B:
$$5.1 \times 10^{24}$$
C:
$$6.2 \times 10^{10}$$
D:
$$6.2 \times 10^{24}$$
### Solution
(From Unit 7, Lesson 13.)
### Problem 5
Noah reads the problem, “Evaluate each expression, giving the answer in scientific notation.” The first problem part is: $$5.4 \times 10^5 + 2.3 \times 10^4$$.
Noah says, “I can rewrite $$5.4 \times 10^5$$ as $$54 \times 10^4$$. Now I can add the numbers: $$54 \times 10^4 + 2.3 \times 10^4 = 56.3 \times 10^4$$.”
Do you agree with Noah’s solution to the problem? Explain your reasoning.
### Solution
(From Unit 7, Lesson 14.)
### Problem 6
Select all the expressions that are equivalent to $$3^8$$.
A:
$$(3^2)^4$$
B:
$$8^3$$
C:
$$3 \boldcdot 3 \boldcdot 3 \boldcdot 3 \boldcdot 3 \boldcdot 3 \boldcdot 3 \boldcdot 3$$
D:
$$(3^4)^2$$
E:
$$\frac{3^6}{3^{\text-2}}$$
F:
$$3^6 \boldcdot 10^2$$ |
# Dividing Unit Fractions
Vocabulary:
Unit Fraction: A fraction with a 1 in the numerator such as 1/3, 1/4, 1/8. All fractions are built on unit fractions. For example 2/3 is 1/3 + 1/3 and is made from 2 unit fractions.
This week we will change focus a little bit. Last week we were dividing whole numbers into unit fractions. This week we will take a unit fraction and break it into same size pieces.
We can start by asking What do I have? and What do I do with it?
Today we have 1/4 and we are partitioning, or dividing, our amount into 3 same size pieces.
Start by showing what you have: 1/4
Once you show what you have, 1/4 of your whole. You can break your 1/4 into 3 same size pieces.
Breaking your 1/4 into 3 same size pieces
Your model now needs to show how much of your whole each new piece is. You also need to break up the rest of your fourths into 3 same size pieces.
Each one of your new pieces is 1/12 of the whole
This model can now be represented with the equation 1/3 ÷ 4 = 1/12
# Creating Your Own Word Problem
Tonight you will be creating a word problem that matches an expression. We can start with the dividend, or the first number in a division expression. The dividend is the whole, or what you have. The second number is the divisor. This is what you are dividing your whole by.
When you are creating your word problem start with what you have. When creating these sort of problems it is easiest to use things that can cut into smaller pieces. Pieces of yarn, rope and ribbon are common. Making necklaces from yarn, bracelets from rope, and wrapping presents with ribbon are examples. Here is an example of a word problem that the students have seen.
The models that you have been creating all week can be used to represent your situations. Each fraction would be, in the example above, one present being wrapped since 1/3 of a meter of ribbon is used to wrap each present.
# Rules that Expire
Vocabulary:
As students explore math they begin to make rules, and they expect their work to obey these rules. As students progress they bump into math where their rules does not work anymore. This is an opportunity to explore how to alter the rule to accommodate this new information.
A rule that expires this week is that division always makes your dividend smaller. Students will discover that 5 ÷ 1/3 = 15. This quotient will initially be a bit disorienting as it breaks a rule they have created.
Students can make sense of this by exploring what 5 ÷ 1/3 is asking them to do. This expression is asking them to find out how many 1/3s there are in 5 wholes. This connects to the work you did with them last week converting fractions. An example of a model you may see is below:
In this example 5 was broken into thirds, and then the thirds were counted.
At this point we are not yet teaching an algorithm, but we can ask students to start looking for patterns they see. The algorithm will be further explored in coming posts.
# Multiplying Mixed Numbers-Converting Improper Fractions to Mixed Numbers
When students are multiplying mixed numbers they generally get an improper fraction as their product. This improper fraction will need to be converted to a mixed number number. Strategies involve making as many wholes as you can, and then adding the left over unit fractions.
Below is an example of ways that you may see your child thinking about his work. Even if the work your child is doing does not look exactly like this, you will see aspects of this example in your child’s work. The 3, 6, 9 on the bottom are keeping track of how many thirds your have. Notice that the 12 is crossed out. This step shows that 12 thirds is more than we have, so we have 3 wholes and now need to find how many more thirds we have left over.
# Homework
Adults often tell me that they want to help their children with math, but aren’t sure how to. The math seems to look so much different from when we were kids. Students will begin bringing home homework that will allow you to better understand how students are thinking about math on a daily basis.
The best way to understand how math is being done is to spend some time exploring the math. Every Monday-Thursday students will bring home some math for you to both work on together. You will take turns showing your strategies to each other. It is important to talk about what you are doing. Ask questions about each others’ work, point out what you like about what each other did. Be transparent, when your child does something you are not familiar with tell them that. Ask them to explain what they did. This will help to empower your child to become the teacher and help build their math confidence.
Check in here to see examples of work being done and to get updates on information and strategies for the work. Please leave comments if you have questions or if there is math that you would like to see demonstrated.
Have fun and be creative and brave in your explorations!!
# Multiplying Mixed Numbers and Fractions- Changing Mixed Numbers to Improper Fractions
## Vocabulary
Mixed Number: A fraction with a whole number and a fraction.
Improper Fraction: A fraction with a value greater than 1.
Before students are able to multiply mixed numbers they need a strategy to distribute the values of both numbers to each other. An efficient strategy is to place the value of the whole number in the numerator. This will give us an improper fraction. When doing this work with your mathematician have them first make a visual model and explain it. After they can explain the process then they can use an algorithm. An example of a model students may use and an algorithm are below.
# Math Explorations
Adults have been telling me for years that they are having difficulty helping their children with math. Students are thinking about math in much different ways than most of the adults did when we were in school. This website is designed to introduce adults to the mathematical concepts being taught in schools throughout the country. You can use the resources to first explore and do some of the math yourself, then to share in your exploration with your little mathematicians at home. |
Question Video: Finding the Distance between Parallel Lines Mathematics
Find, to the nearest hundredth, the distance between the parallel lines π₯ = 1 + 3π‘, π¦ = 7 + 2π‘, π§ = 4 + 5π‘ and π₯ = 3 β 3π‘, π¦ = 6 β 2π‘, π§ = 4 β 5π‘.
04:33
Video Transcript
Find, to the nearest hundredth, the distance between the parallel lines π₯ equals one plus three π‘, π¦ equals seven plus two π‘, π§ equals four plus five π‘ and π₯ equals three minus three π‘, π¦ equals six minus two π‘, π§ equals four minus five π‘.
Okay, so here we have these two parallel lines, and weβll call the first line line one and the second line two. We want to solve for the perpendicular distance between these lines. Weβll call that π. To do this, there are three bits of information weβll need to find out. First, weβll need to know the coordinates of a point on the first line; weβll call that π one. Weβll also need to know a point on line two; weβll call that π two. And lastly, weβll need to know the components of a vector thatβs parallel to both these lines. Weβll call this vector π¬. Once we know all this, weβll be able to use this expression to compute the distance between our two parallel lines.
We see this equation involves a vector π¬, thatβs the vector thatβs parallel to our lines, and a second vector weβve called π one π two. On our sketch, thatβs a vector that looks like this. It goes from point one on line one to point two on line two. Letβs start by figuring out a point on line one, in other words, the coordinates of a point π one. Weβll do this by looking at the equation of line one, which we see is given to us in parametric form. Written this way, we have separate equations for the π₯-, π¦-, and π§-coordinates of every point on this line, and itβs possible to convert this lineβs equation from parametric form to whatβs called vector form. This involves combining all three equations into one, where π« is a vector with components π₯, π¦, and π§.
This form of the lineβs equation begins with the vector from the origin of our coordinate frame to the point one, seven, four. This point lies on line one, and then it moves out along this vector three, two, five multiplied by the scale factor π‘. We can say then that the point with coordinates one, seven, four lies along line πΏ one. So weβll call this point π one. Furthermore, this vector three, two, five is parallel to our line, and therefore we can call this the vector π¬.
Now that we know a point π one and a vector π¬, letβs clear a bit of space and start looking at the given equation for line two. Our goal in doing this is to discover a point π two that lies along this line. Just like for line one, line two is given to us in parametric form. That means we can write this line in vector form as a vector to the point three, six, four plus π‘ times another vector parallel to line two. Since the point three, six, four is on πΏ two, we can call this π two.
And so, now we have all the information we need to begin calculating the distance π. The first thing weβll do is solve for the components of this vector π one π two. We find these components by subtracting the coordinates of point π one from those of π two. With those values substituted in, we find that the vector π one π two has components three minus one or two, six minus seven or negative one, and four minus four, zero. Now that we know this vector, next, weβll take the cross product of this vector with π¬. That cross product is given by the determinant of this three-by-three matrix. In the top row, we have the π’, π£, and π€ unit vectors and then the π₯-, π¦-, and π§-components of π one, π two, and π¬, respectively.
The magnitude of the π’-component is given by the determinant of this two-by-two matrix. Negative one times five minus zero times two is negative five. And then the π£-component is negative the determinant of this matrix, which is two times five minus zero times three or 10. And lastly, thereβs the π€-component of this cross product equal to the determinant of this matrix. Two times two minus negative one times three is positive seven. This, then, is our overall cross product, which we can write in vector form with components negative five, negative 10, seven.
Okay, weβre ready now to calculate π by computing the magnitude of π one π two cross π¬ and dividing that by the magnitude of π¬. The magnitude of our cross product equals the square root of negative five squared plus negative 10 squared plus seven squared, while the magnitude of π¬ equals the square root of three squared plus two squared plus five squared. Entering this whole expression on our calculator, to the nearest hundredth, our answer is 2.14. This is the minimum distance between these two parallel lines. |
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Have you ever honored someone with a gift? Take a look at this dilemma.
Mr. Samuels, the custodian, has been working overtime getting ready for the Olympics at Montgomery Middle School. To help the students, he has been creating long jump fields, bringing in tons of sand and offering to help out after school and on weekends.
“I think that we should do something nice for Mr. Samuels,” Crystal said at lunch.
“I agree. But what?” Kenneth responded.
“How about giving him one of our prism awards? I could wrap it up and we could present it to him at the Olympics,” Crystal suggested.
Kenneth, Marcy and Dylan all agreed. So that week, Crystal found a box for the prism award and began to wrap it. Because of the size of the award, she needed a good sized box.
How much wrapping paper will she need?
Surface area is the topic for this Concept, and Crystal’s wrapping paper dilemma is all about surface area. By the end of this Concept, you will know how much wrapping paper she will need to cover this box.
Guidance
A three–dimensional or solid figure has length, width and depth.
This Concept focuses on prisms and surface area.
What is a prism?
A prism is a three–dimensional figure with two parallel congruent polygons as bases. The side faces of a prism are rectangular in shape.
One of the things that we can measure when working with three–dimensional figures is called surface area . Surface area is the total of the areas of each face of a solid figure. Imagine you could wrap one of the figures above in wrapping paper, like a present.
The amount of wrapping paper needed to cover the figure represents its surface area. To find the surface area, we must be able to calculate the area of each face and then add these areas together.
There are several different ways to calculate surface area.
One way is to use a net .
A net is a two-dimensional diagram of a three-dimensional figure. Imagine you could unfold a box so that it is completely flat. You would have something that looks like this.
If we folded this up, you could see that it would form a cube. A cube is made up of faces that are squares. If we wanted to figure out the surface area or measurement of the outer covering of this cube, then we could find the area of each surface of the cube and then add the products together.
We could also look at a net of a rectangular prism.
A rectangular prism is made up of rectangles. To find the surface area of a prism, we would need to calculate the area of each of the faces and then add them together.
Let’s begin by calculating the area of a rectangular prism.
Now that we have all the information we need, we can calculate the area of each face and then add their areas together.
$& \mathbf{bottom \ face} && \mathbf{top \ face} && \mathbf{long \ side} && \mathbf{long \ side} && \mathbf{short \ side} && \mathbf{short \ side}\\& A=lw && A=lw && A=lh && A=lh && A=wh && A=wh\\& 12 \times 7 \quad + && 12 \times 7 \quad + && 12 \times 3 \quad + && 12 \times 3 \quad + && 7 \times 3 \quad + && 7 \times 3\\& 84 \qquad \ \ + && 84 \qquad \ \ + && 36 \qquad \ \ + && 36 \qquad \ \ + && 21 \qquad \ + && 21 \qquad \ \ = \ 282 \ in.^2$
We found the area of each rectangular face and then added all of these areas together. The total surface area of the rectangular prism is 282 square inches. Using a net helped us to locate all of the faces and find the measurements of each side.
Nets let us see each face so that we can calculate their area and then add them together. However, we can also use a formula to represent the faces as we find their area. The formula gives us a nice short cut that we can use for any kind of prism, no matter what shape its base is. Take a look at the formula below.
$SA=Ph+2B$
Let’s look at the first part of the formula. $P$ represents the perimeter of the base, and $h$ represents the height of the prism. By multiplying the perimeter and height, we are finding the area of all of the side faces at once. This will be very useful if the prism that we are working with isn’t just a cube or a rectangular prism.
The second part of the formula represents the area of the top and bottom faces. $B$ represents the area of one base, which we find using whichever area formula is appropriate for the shape of the base. Then we multiply it by 2 to show the area of the top and bottom faces at once. Let’s give it a try to see how this works.
Find the surface area of this figure using a formula.
We have all the measurements we need. Let’s find the perimeter of the base first. It is a rectangle, so we add the lengths and widths: $21 + 21 + 14 + 14 = 70$ . We can put this number in for $P$ in the formula. The height, we can see, is 5 centimeters.
Now let’s solve for $B$ , the area of the base. The base of this prism is a rectangle, so we use the formula $A = lw$ to find its area.
$B &= lw\\B &=21 \times 14\\B &=294 \ cm^2$
Now we have all of the information we need to fill in the formula. Let’s put it in and solve for $SA$ , surface area.
$SA &=Ph + 2B\\SA &= 70 (5) + 2 (294)\\SA &= 350 + 588\\SA &= 938 \ cm^2$
This rectangular prism has a surface area of 938 square centimeters.
Write this formula for finding the surface area of a prism down in your notebook.
Example A
True or false. The surface area includes the inside of a prism.
Solution: False. The surface area is the measurement of the outer covering of a prism.
Example B
True or false. A net shows all three dimensions of a prism.
Solution: True.
Example C
True or false. You know a figure is a prism because the faces are rectangles.
Solution: True
Now let's go back to the dilemma from the beginning of the Concept.
First of all, what kind of solid figure is this? All of the faces are rectangles, including the base, so it is a rectangular prism. The picture clearly shows us what its length, width, and height are, so let’s use the formula for finding the surface area of prisms.
What is the perimeter of the base?
$12 + 12 + 9 + 9 = 42 \ inches$ .
We’ll put this in for $P$ .
We also need to find the area of the base, $B$ . This base is a rectangle, so we use the formula $B = lw$ .
$B &=lw \\B &=12 (9)\\B &=108 \ in.^2$
Now we have all of the measurements to put in for the appropriate variables in the formula.
$SA &=Ph + 2B\\SA &=42(6)+2(108)\\SA &=252 + 216\\SA &=468 \ in.^2$
Crystal will need 468 square inches of wrapping paper in order to cover the present.
Vocabulary
Three Dimensional Figures
solid figures that have length, width and height.
Prisms
three – dimensional figures with parallel, congruent polygons as bases and rectangular side faces.
Surface Area
the measurement of the outer covering on a solid figure.
Net
the pattern of a solid figure-what a solid figure would look like if it were drawn out as a pattern.
Guided Practice
Here is one for you to try on your own.
What is the surface area of the figure below?
Solution
Let’s look at the base first to find its perimeter. The triangle has two sides of 5 inches and one that is 8 inches: $5 + 5 + 8 = 18 \ inches$ . This will be $P$ in the formula. The height of the prism is 15 inches. Be careful not to confuse the height of the prism with the height of the triangular base!
To find $B$ , we need to use the area formula for triangles: $A =\frac{1}{2} bh$ . The base of the triangle is 8 inches, and the height is 3 inches.
$A &=\frac{1}{2} bh\\A &= \frac{1}{2} (8) (3)\\A &= 4 (3)\\A &= 12 \ in.^2$
The area of the triangular base is 12 square inches, so we put this in for $B$ in the formula. Let’s put all of the values in and solve.
$SA &= Ph + 2B\\SA &= 18 (15) + 2 (12)\\SA &= 270 + 24\\SA &= 294 \ in.^2$
The surface area of this triangular prism is 294 square inches.
Practice
Directions: Look at each figure and then answer the following questions about each.
1. What is the name of the figure pictured above?
2. What is the surface area of this figure?
3. What is the shape of it's base?
4. What is the height of this figure?
5. What is the area of this figure's base?
1. What is the name of this figure?
2. What is the surface area of this figure?
3. What is the shape of it's base?
4. What is the height of this figure?
5. What is the area of this figure's base?
1. What is the name of this figure?
2. What is the shape of the base
3. How many bases does this figure have?
4. How many side faces are there?
5. What is the surface area of this figure?
Vocabulary Language: English
Net
Net
A net is a diagram that shows a “flattened” view of a solid. In a net, each face and base is shown with all of its dimensions. A net can also serve as a pattern to build a three-dimensional solid.
Prism
Prism
A prism is a three-dimensional object with two congruent parallel bases that are polygons.
Surface Area
Surface Area
Surface area is the total area of all of the surfaces of a three-dimensional object.
Three – Dimensional
Three – Dimensional
A figure drawn in three dimensions is drawn using length, width and height or depth.
Explore More
Sign in to explore more, including practice questions and solutions for Surface Area of Prisms. |
two math problems and a math discussion
Please show all your work if needed for the math questions –
1) In order to help pay for college, the grandparents of a child invest \$1700 in a bond that pays 4% interest compounded quarterly. How much money will there be in 7 years?
A) \$2296.69
B) \$2225.94
C) \$2346.31
D) \$2246.19
2) A house sells for \$495,000 and a 35% down payment is made. A 30-year mortgage at 7.5% was obtained.
(i) Find the down payment.
(ii) Find the amount of the mortgage.
(iii) Find the monthly payment.
(iv) Find the total interest paid
Math discussion –
• Identify the types of errors that you make most often and what strategy you will use to minimize making this type of error in the future.
• State what you have learned by analyzing your mistakes.
Why do we make mistakes? The old adage that “nobody is perfect†is certainly true but we can LEARN from our mistakes. Start to analyze your mistakes. In this analysis consider these categories for your mistakes.
1. Conceptual Error: The error is due to a misunderstanding of a concept. Name the concept, at least to yourself. For example, the concept might be finding the absolute value of a number. The process of naming the concept will help you to find additional information about that concept and strengthen your understanding.
2. Calculation Error: In your judgment, you understood the concept or concepts involved but you made a miscalculation. Name the type of calculation error. For example, a type of calculation error might be that you applied the wrong rule for adding signed numbers or that you rounded the number incorrectly.
3. Transcription Error: The error resulted from miscopying of the problem or miscopying a portion of the problem from one step to another. Identify the transcription error. For example, you may have copied the wrong problem or copied part of one problem and part of another. Another common transcription error is to leave out a number or sign as you copy from one line to the next.
4. Interpretation Error. We answer the wrong question or do not answer all the questions.We often find ourselves anticipating what a problem is asking us to do without carefully reading the instructions. For example, you might be asked to give the new amount in a percent increase problem instead of just the amount of increase. |
# AP Statistics Curriculum 2007 Hypothesis S Mean
(Difference between revisions)
Revision as of 18:42, 21 November 2008 (view source)IvoDinov (Talk | contribs) (→Normal Process with Known Variance: fixed a typo with definition of Z_o)← Older edit Current revision as of 04:36, 7 June 2013 (view source)IvoDinov (Talk | contribs) m (→Example) (5 intermediate revisions not shown) Line 3: Line 3: === Testing a Claim about a Mean: Small Samples=== === Testing a Claim about a Mean: Small Samples=== - The [[AP_Statistics_Curriculum_2007_Hypothesis_L_Mean | previous section discussed inference on the population mean for large smaples]]. Now, we show how to do hypothesis testing about the mean for small sample-sizes. + The [[AP_Statistics_Curriculum_2007_Hypothesis_L_Mean | previous section discussed inference on the population mean for large smaples]]. Now, we show how to do hypothesis testing of the mean for small sample-sizes. ===[[AP_Statistics_Curriculum_2007_Estim_L_Mean | Background]]=== ===[[AP_Statistics_Curriculum_2007_Estim_L_Mean | Background]]=== Line 16: Line 16: * Alternative Research Hypotheses: * Alternative Research Hypotheses: ** One sided (uni-directional): $H_1: \mu >\mu_o$, or $H_o: \mu<\mu_o$ ** One sided (uni-directional): $H_1: \mu >\mu_o$, or $H_o: \mu<\mu_o$ - ** Double sided: $H_1: \mu \not= \mu_o$ + ** Double sided: $$H_1: \mu \not= \mu_o$$ ====Normal Process with Known Variance==== ====Normal Process with Known Variance==== Line 24: Line 24: ====(Approximately) Normal Process with Unknown Variance==== ====(Approximately) Normal Process with Unknown Variance==== * If the population is approximately Normally distributed and we do not know the population variance, then the [http://en.wikipedia.org/wiki/Hypothesis_testing#Common_test_statistics Test statistics] is: * If the population is approximately Normally distributed and we do not know the population variance, then the [http://en.wikipedia.org/wiki/Hypothesis_testing#Common_test_statistics Test statistics] is: - : $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {\overline{x} - \mu_o \over {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}})} \sim T_{(df=n-1)}$. + : $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {\overline{x} - \mu_o \over {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}} \sim T_{(df=n-1)}$. ===Example=== ===Example=== - Let's use again the small-sample example of the [[AP_Statistics_Curriculum_2007_Estim_L_Mean | ''number of sentences per advertisement'']], where we measure of readability for magazine advertisements. A random sample of the number of sentences found in 10 magazine advertisements is listed below. Suppose we want to test at $\alpha=0.01$ a null hypothesis: $H_o: \mu=12$ against a one-sided research alternative hypothesis: $H_1: \mu > 12$. Recall that we had the following [[AP_Statistics_Curriculum_2007_Estim_S_Mean#Example |sample statistics: '''sample-mean=22.1, sample-variance=737.88 and sample-SD=27.16390579''']] for these data. + Let's use again the small-sample example of the [[AP_Statistics_Curriculum_2007_Estim_L_Mean | ''number of sentences per advertisement'']], where we measure the readability for magazine advertisements. A random sample of the number of sentences found in 10 magazine advertisements is listed below. Suppose we want to test at $\alpha=0.01$ a null hypothesis: $H_o: \mu=12$ against a one-sided research alternative hypothesis: $H_1: \mu > 12$. Recall that we had the following [[AP_Statistics_Curriculum_2007_Estim_S_Mean#Example |sample statistics: '''sample-mean=22.1, sample-variance=737.88 and sample-SD=27.16390579''']] for these data.
{| class="wikitable" style="text-align:center; width:75%" border="1" {| class="wikitable" style="text-align:center; width:75%" border="1" Line 36: Line 36: As the population variance is not given, we have to use the [[AP_Statistics_Curriculum_2007_StudentsT |T-statistics]]: $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} \sim T(df=9)$ As the population variance is not given, we have to use the [[AP_Statistics_Curriculum_2007_StudentsT |T-statistics]]: $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} \sim T(df=9)$ - : $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {22.1 - 12 \over {{1\over \sqrt{10}} \sqrt{\sum_{i=1}^{10}{(x_i-22.1)^2\over 9}}})}=1.176$. + : $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {22.1 - 12 \over {{1\over \sqrt{10}} \sqrt{\sum_{i=1}^{10}{(x_i-22.1)^2\over 9}}}}=1.176$. - : $p-value=P(T_{(df=9)}>T_o=1.176)=0.134882$ for this (one-sided) test. Therefore, we '''can not reject''' the null hypothesis at $\alpha=0.01$! The left white area at the tails of the T(df=9) distribution depict graphically the probability of interest, which represents the strenght of the evidence (in the data) against the Null hypothesis. In this case, this area is 0.134882, which is larger than the initially set [[AP_Statistics_Curriculum_2007_Hypothesis_Basics | Type I]] error $\alpha = 0.01$ and we can not reject the null hypothesis. + : $p-value=P(T_{(df=9)}>T_o=1.176)=0.134882$ for this (one-sided) test. Therefore, we '''cannot reject''' the null hypothesis at $\alpha=0.01$! The left white area at the tails of the T(df=9) distribution depicts graphically the probability of interest, which represents the strength of the evidence (in the data) against the Null hypothesis. In this case, this area is 0.134882, which is larger than the initially set [[AP_Statistics_Curriculum_2007_Hypothesis_Basics | Type I]] error $\alpha = 0.01$ and we cannot reject the null hypothesis.
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig4.jpg|600px]]
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig4.jpg|600px]]
Line 48: Line 48: ====Cavendish Mean Density of the Earth==== ====Cavendish Mean Density of the Earth==== - A number of famous early experiments of measuring physical constants have later been shown to be biased. In the 1700's [http://en.wikipedia.org/wiki/Henry_Cavendish Henry Cavendish] measured the [http://www.jstor.org/stable/pdfplus/107617.pdf mean density of the Earth]. Formulate and test null and research hypotheses about these data regarding the now know exact mean-density value = 5.517. These sample statistics may be helpful + A number of famous early experiments of measuring physical constants have later been shown to be biased. In the 1700's [http://en.wikipedia.org/wiki/Henry_Cavendish Henry Cavendish] measured the [http://www.jstor.org/stable/pdfplus/107617.pdf mean density of the Earth]. Formulate and test null and research hypotheses about these data regarding the now known exact mean-density value = 5.517. These sample statistics may be helpful : n = 23, sample mean = 5.483, sample SD = 0.1904 : n = 23, sample mean = 5.483, sample SD = 0.1904
Line 73: Line 73: * Hypothesis (Significance) testing: Only one possible value for the parameter, called the hypothesized value, is tested. We determine the strength of the evidence (confidence) provided by the data against the proposition that the hypothesized value is the true value. * Hypothesis (Significance) testing: Only one possible value for the parameter, called the hypothesized value, is tested. We determine the strength of the evidence (confidence) provided by the data against the proposition that the hypothesized value is the true value. + + ===[[EBook_Problems_Hypothesis_S_Mean|Problems]]===
## General Advance-Placement (AP) Statistics Curriculum - Testing a Claim about a Mean: Small Samples
### Testing a Claim about a Mean: Small Samples
The previous section discussed inference on the population mean for large smaples. Now, we show how to do hypothesis testing of the mean for small sample-sizes.
### Background
• Recall that for a random sample {$X_1, X_2, X_3, \cdots , X_n$} of the process, the population mean may be estimated by the sample average, $\overline{X_n}={1\over n}\sum_{i=1}^n{X_i}$.
• For a given small α (e.g., 0.1, 0.05, 0.025, 0.01, 0.001, etc.), the (1 − α)100% Confidence interval for the mean is constructed by
$CI(\alpha): \overline{x} \pm t_{df=n-1,{\alpha\over 2}} {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}$
and $t_{df=n-1, {\alpha\over 2}}$ is the critical value for a T-distribution of df=(sample size - 1) at ${\alpha\over 2}$.
### Hypothesis Testing about a Mean: Small Samples
• Null Hypothesis: Ho:μ = μo (e.g., 0)
• Alternative Research Hypotheses:
• One sided (uni-directional): H1:μ > μo, or Ho:μ < μo
• Double sided: $$H_1: \mu \not= \mu_o$$
#### Normal Process with Known Variance
• If the population is Normally distributed and we know the population variance, then the Test statistics is:
$Z_o = {\overline{x} - \mu_o \over {\sigma \over \sqrt{n}}} \sim N(0,1)$.
#### (Approximately) Normal Process with Unknown Variance
• If the population is approximately Normally distributed and we do not know the population variance, then the Test statistics is:
$T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {\overline{x} - \mu_o \over {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}} \sim T_{(df=n-1)}$.
### Example
Let's use again the small-sample example of the number of sentences per advertisement, where we measure the readability for magazine advertisements. A random sample of the number of sentences found in 10 magazine advertisements is listed below. Suppose we want to test at α = 0.01 a null hypothesis: Ho:μ = 12 against a one-sided research alternative hypothesis: H1:μ > 12. Recall that we had the following sample statistics: sample-mean=22.1, sample-variance=737.88 and sample-SD=27.16390579 for these data.
16 9 14 11 17 12 99 18 13 12
As the population variance is not given, we have to use the T-statistics: $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} \sim T(df=9)$
$T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {22.1 - 12 \over {{1\over \sqrt{10}} \sqrt{\sum_{i=1}^{10}{(x_i-22.1)^2\over 9}}}}=1.176$.
pvalue = P(T(df = 9) > To = 1.176) = 0.134882 for this (one-sided) test. Therefore, we cannot reject the null hypothesis at α = 0.01! The left white area at the tails of the T(df=9) distribution depicts graphically the probability of interest, which represents the strength of the evidence (in the data) against the Null hypothesis. In this case, this area is 0.134882, which is larger than the initially set Type I error α = 0.01 and we cannot reject the null hypothesis.
### Examples
#### Cavendish Mean Density of the Earth
A number of famous early experiments of measuring physical constants have later been shown to be biased. In the 1700's Henry Cavendish measured the mean density of the Earth. Formulate and test null and research hypotheses about these data regarding the now known exact mean-density value = 5.517. These sample statistics may be helpful
n = 23, sample mean = 5.483, sample SD = 0.1904
5.36 5.29 5.58 5.65 5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.1 5.27 5.39 5.42 5.47 5.63 5.34 5.46 5.3 5.75 5.68 5.85
### Hypothesis Testing Summary
Important parts of Hypothesis test conclusions:
• Decision (significance or no significance)
• Parameter of interest
• Variable of interest
• Population under study
• (optional but preferred) P-value
### Parallels between Hypothesis Testing and Confidence Intervals
These are different methods for coping with the uncertainty about the true value of a parameter caused by the sampling variation in estimates.
• Confidence intervals: A fixed level of confidence is chosen. We determine a range of possible values for the parameter that are consistent with the data (at the chosen confidence level).
• Hypothesis (Significance) testing: Only one possible value for the parameter, called the hypothesized value, is tested. We determine the strength of the evidence (confidence) provided by the data against the proposition that the hypothesized value is the true value. |
$$\require{cancel}$$
Any reader of this book will know that the solutions to the quadratic equation
$ax^2 + bx + c = 0 \label{1.3.1}$
are
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \label{1.3.2}$
and will have no difficulty in finding that the solutions to
$2.9x^2 - 4.7x + 1.7 = 0 \nonumber$
are
$x = 1.0758 \text{ or } 0.5449. \nonumber$
We are now going to look, largely for fun, at two alternative iterative numerical methods of solving a quadratic equation. One of them will turn out not to be very good, but the second will turn out to be sufficiently good to merit our serious attention.
In the first method, we re-write the quadratic equation in the form
$x = \frac{-\left(ax^2 + c \right)}{b} \nonumber$
We guess a value for one of the solutions, put the guess in the right hand side, and hence calculate a new value for $$x$$. We continue iterating like this until the solution converges.
For example, let us guess that a solution to the equation $$2.9x^2 − 4.7x + 1.7 = 0$$ is $$x = 0.55$$. Successive iterations produce the values
\begin{array}{c c}
0.54835 & 0.54501 \\
0.54723 & 0.54498 \\
0.54648 & 0.54496 \\
0.54597 & 0.54495 \\
0.54562 & 0.54494 \\
0.54539 & 0.54493 \\
0.54524 & 0.54493 \\
0.54513 & 0.54494 \\
0.54506 & 0.54492 \\
\nonumber
\end{array}
We did eventually arrive at the correct answer, but it was very slow indeed even though our first guess was so close to the correct answer that we would not have been likely to make such a good first guess accidentally.
Let us try to obtain the second solution, and we shall try a first guess of 1.10, which again is such a good first guess that we would not be likely to arrive at it accidentally. Successive iterations result in
\begin{array}{c}
1.10830 \\
1.11960 \\
1.13515 \\
\nonumber
\end{array}
and we are getting further and further from the correct answer!
Let us try a better first guess of 1.05. This time, successive iterations result in
\begin{array}{c}
1.04197 \\
1.03160 \\
1.01834 \\
\nonumber
\end{array}
Again, we are getting further and further from the solution.
No more need be said to convince the reader that this is not a good method, so let us try something a little different.
$ax^2 + bx = -c \label{1.3.3}$
Add $$ax^2$$ to each side:
$2ax^2 + bx = ax^2 - c \label{1.3.4}$
or $(2ax + b)x = ax^2 - c \label{1.3.5}$
Solve for $$x$$: $x = \frac{ax^2-c}{2ax+b} \label{1.3.6}$
This is just the original equation written in a slightly rearranged form. Now let us make a guess for $$x$$, and iterate as before. This time, however, instead of making a guess so good that we are unlikely to have stumbled upon it, let us make a very stupid first guess, for example $$x = 0$$. Successive iterations then proceed as follows.
\begin{array}{c}
0.00000 \\
0.36170 \\
0.51751 \\
0.54261 \\
0.54491 \\
0.54492 \\
\nonumber
\end{array}
and the solution converged rapidly in spite of the exceptional stupidity of our first guess. The reader should now try another very stupid first guess to try to arrive at the second solution. I tried $$x = 100$$, which is very stupid indeed, but I found convergence to the solution $$1.0758$$ after just a few iterations.
Even although we already know how to solve a quadratic equation, there is something intriguing about this. What was the motivation for adding $$ax^2$$ to each side of the equation, and why did the resulting minor rearrangement lead to rapid convergence from a stupid first guess, whereas a simple direct iteration either converged extremely slowly from an impossibly good first guess or did not converge at all? |
# Elementary algebra -variables
In mathematics, variables are symbols or letters used to represent unknown or changing quantities in mathematical expressions, equations, and formulas. They allow us to generalize mathematical relationships and describe patterns and rules without specifying specific numerical values. Variables serve as placeholders for numbers or other mathematical entities, and their values can vary depending on the context.
Here are some key points to understand about variables in mathematics:
1. Representation of Unknowns: In many mathematical problems, certain quantities are unknown, and we use variables to represent these unknown values. For example, if we want to find the area of a rectangle but do not know the length and width, we can use variables like “l” and “w” to represent these unknown dimensions.
2. Flexibility: Variables provide flexibility and generality in mathematical expressions. They allow us to work with formulas that can be applied to various scenarios by substituting different values for the variables.
3. Algebraic Manipulations: Variables are fundamental in algebra, where they are used to form algebraic expressions and equations. By manipulating these expressions and equations, we can solve for the values of the variables and find specific solutions to problems.
4. Independent and Dependent Variables:
• Independent Variable: The independent variable is the variable that is not affected by other variables in the equation or function. It is the input variable, and its value is chosen independently. In other words, the value of the independent variable is not dependent on any other variable in the system. In mathematical equations and functions, it is often represented by the symbol “x.”
• Example 1: In the equation y = 2x + 3, “x” is the independent variable. The value of “x” can be chosen arbitrarily, and the corresponding value of “y” will be determined based on the equation.
• Example 2: In the function f(x) = x^2, “x” is the independent variable. Whatever value we assign to “x,” the function will calculate the square of that value.
• Dependent Variable: The dependent variable is the variable whose value depends on the value of the independent variable. It is the output variable, and its value is determined based on the value of the independent variable. In mathematical equations and functions, it is often represented by the symbol “y.”
• Example 1: In the equation y = 2x + 3, “y” is the dependent variable. Its value is determined by the value of “x” according to the equation.
• Example 2: In the function f(x) = x^2, “f(x)” (or simply “y”) is the dependent variable. Its value is the square of the value of the independent variable “x.”
• In mathematical relationships, the independent variable is typically plotted on the horizontal axis (x-axis), while the dependent variable is plotted on the vertical axis (y-axis) in a Cartesian coordinate system.
5. Consistency Across Equations: In mathematical equations, a variable typically retains the same symbol throughout the equation or mathematical expression. This consistency allows us to understand the relationship between different variables and perform operations accordingly.
6. Contextual Interpretation: The meaning of a variable depends on the context in which it is used. For instance, if “t” represents time in one equation and temperature in another, the interpretation of “t” will be different in each case.
7. Constants: While variables represent changing quantities, constants are fixed values that do not change within a specific context. Constants are also used in mathematical expressions but are not represented by variables. For example, in the equation A = πr^2, π is a constant, and “r” is the variable representing the radius.
Variables are a fundamental concept in mathematics, enabling us to represent and analyze relationships between quantities, solve problems, and model real-world situations. They play a crucial role in various mathematical branches, including algebra, calculus, statistics, and many others. |
# How Do You Find The Surface Area Of A Pyramid
## Understanding the Surface Area of a Pyramid
When it comes to understanding the surface area of a pyramid, it’s essential to consider both the lateral surface area and the total surface area. The lateral surface area of a regular pyramid is the sum of the areas of its lateral faces, while the total surface area includes the lateral faces and the base. To calculate the lateral surface area of a regular pyramid, the general formula L.S.A. = 1/2pl is used, where p represents the perimeter of the base and l represents the slant height.
## Calculating the Lateral Surface Area
For example, if we have a regular pyramid with a triangular base where each edge of the base measures 8 inches and the slant height is 5 inches, we can calculate the perimeter of the base by summing the sides, which gives us p = 3(8) = 24 inches. Using the formula, we find that the lateral surface area is 1/2(24)(5) = 60 square inches.
## Determining the Total Surface Area
On the other hand, to find the total surface area of a regular pyramid, the general formula T.S.A. = 1/2pl + B is used, where p represents the perimeter of the base, l represents the slant height, and B represents the area of the base.
## Example of Total Surface Area Calculation
For instance, if we have a regular pyramid with a square base where each edge of the base measures 16 inches, the slant height of a side is 17 inches, and the altitude is 15 inches, we can calculate the perimeter of the base by multiplying the number of sides for a square base, giving us p = 4(16) = 64 inches. The area of the base is found by squaring the length of one side, resulting in B = 16^2 = 256 square inches. By using the formula, the total surface area is determined to be 1/2(64)(17) + 256 = 688 square inches.
## FAQs about Finding the Surface Area of a Pyramid
### 1. What is the lateral surface area of a pyramid?
The lateral surface area of a pyramid is the sum of the areas of its lateral faces. It can be calculated using the formula L.S.A. = 1/2pl, where p represents the perimeter of the base and l represents the slant height.
### 2. How is the total surface area of a pyramid calculated?
The total surface area of a pyramid is calculated by adding the lateral surface area to the area of the base. The formula T.S.A. = 1/2pl + B is used, where p represents the perimeter of the base, l represents the slant height, and B represents the area of the base.
### 3. What are the steps to find the lateral surface area of a pyramid?
To find the lateral surface area of a pyramid, first calculate the perimeter of the base and the slant height. Then, use the formula L.S.A. = 1/2pl to find the lateral surface area.
### 4. Can the lateral surface area of a pyramid be greater than the total surface area?
No, the lateral surface area of a pyramid cannot be greater than the total surface area. The total surface area includes the lateral faces and the base, so it will always be greater than or equal to the lateral surface area.
### 5. Why is it important to calculate the surface area of a pyramid?
Calculating the surface area of a pyramid is important in various real-world applications such as architecture, engineering, and construction. It helps in determining the amount of material needed to construct or cover the pyramid.
By understanding the formulas and methods for finding the surface area of a pyramid, one can effectively apply this knowledge in practical scenarios and mathematical problems.
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# M09#12
Author Message
Intern
Joined: 14 Jun 2013
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### Show Tags
14 Jun 2013, 06:56
is it not as simple as 0.9/1.2?
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Joined: 25 Jul 2012
Posts: 31
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GMAT Date: 08-03-2013
WE: Engineering (Computer Software)
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### Show Tags
15 Jun 2013, 03:36
santas wrote:
is it not as simple as 0.9/1.2?
No. We shouldn't consider like that. We cannot simply add 0.1 for every cylinder. Actually we need to do add 10% of the sum.
Consider Power of one cylinder engine is 1. For every cylinder increase, power increases by 10%. So then
2 Cylinder Engine power = 1+(10% of 1) = 1.1
3 Cylinder Engine power = 1.1+(10% of 1.1) = 1.21
4 Cylinder Engine power = 1.21+(10% of 1.21) = 1.33
5 Cylinder Engine power = 1.33+(10% of 1.33) = 1.46
6 Cylinder Engine power = 1.46+(10% of 1.46) = 1.61
7 Cylinder Engine power = 1.61+(10% of 1.61) = 1.77
8 Cylinder Engine power = 1.77+(10% of 1.77) = 1.95
9 Cylinder Engine power = 1.95+(10% of 1.95) = 2.15
10 Cylinder Engine power = 2.15+(10% of 2.15) = 2.37
11 Cylinder Engine power = 2.37+(10% of 2.37) = 2.61
12 Cylinder Engine power = 2.61+(10% of 2.61) = 2.87
So $$\frac{2.15}{2.87}=0.75$$
Hope this helps !
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### Show Tags
15 Jun 2013, 03:38
very simple...
each time a cylinder is added it increases by 1.1
So the required ans is
= (1.1)^9/(1.1)^12
= 1/(1.1)^3
= 1000/11*11*11
= 1000/1331(lol more simpler)
= 0.75 ...which is D
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WE: Supply Chain Management (Other)
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### Show Tags
16 Jun 2013, 05:09
Ratio= (1/1.333)=3/4=0.75
_________________
Cheers!
+1 Kudos if you like my post!
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### Show Tags
19 Jun 2013, 01:33
1
KUDOS
Power of 9 cylinder = X
Power of 10 cylinder = 1.1X
Power of 11 cylinder = 1.1^2 X
Power of 12 cylinder = 1.1^3 X
==> Power of 9 cylinder / Power of 12 cylinder = X / 1.1^3X = 1/1.33
Tip:
when you see 0.333 or 1.3333 or 2.3333 ==> Think about "3 times of them" because 0.3333 x 3 = 1; 1.333 x 3 = 4
We time both numerator and denominator by 3
==> [1 x 3] / [1.33 x 3] = 3/4 = 0.75
Hope it helps.
_________________
Please +1 KUDO if my post helps. Thank you.
"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."
Chris Bangle - Former BMW Chief of Design.
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### Show Tags
23 Jun 2013, 18:06
This problem would use the percentage increase or percentage decrease formula. As we know the increase in one cylinder means 10% increase, hence an increase from 9 to 12 cylinders is (1.1)^3 times. To get that, an increase by 10% is
P(1+i%) or
P(1+0.1)
= P(1.1)
Doing that 3 times
= P(1.1)(1.1)(1.1)
Take a simple value of P, or 9 cylinder power as 100 = Percentage increase from 9 to 12 = 100 (1.1)^3
As we need the ratio here:
= 100/ (100(1.1)^3) = approx. .75
Hope this helps!
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### Show Tags
26 Jun 2013, 14:49
I used 10 as my base and eventually got to 10/13.31.
But how would you put this in decimal form and get .75 without a calculator?
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### Show Tags
28 Jun 2013, 20:27
Quote:
If the power of an engine grows by 10% when the number of its cylinders is increased by one, which of the following is closest to the ratio of the power of a 9-cylinder engine to that of a 12-cylinder engine?
(A) 0.69
(B) 0.71
(C) 0.72
(D) 0.75
(E) 0.78
this worked best for me
Assume that 9 cylinders has a power output of 100%, then...
9 cylinders: 100
10 cylinders: 100 + 10 = 110
11 cylinders: 110 + 11 = 121
12 cylinders: 121 + 12 = 133
Thus, the ratio of 9 cylinders : 12 cylinders is...
100/133 = 0.75
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### Show Tags
02 Jun 2014, 21:50
If the power of an engine grows by 10% when the number of its cylinders is increased by one, which of the following is closest to the ratio of the power of a 9-cylinder engine to that of a 12-cylinder engine?
(A) 0.69
(B) 0.71
(C) 0.72
(D) 0.75
(E) 0.78
The power of engine decreases by 9.09% when the number of its cylinders is decreased by one.
Hence, power of engine decreases by (9+9-0.81)% or 17.37% when the number of its cylinders is decreased by two
Likewise, power power of engine decreases by (17.37+9.09-0.15)% or 24.89% when the number of cylinders is decreased by three.
Let the power of 12-cylinder engine be 100.
So, the ratio of power of a 9 cylinder engine to that of 12 cylinder engine is (100-24.89)/100=0.75 (approximated)
The question can be solved quickly by approximating 9.09 as 9.
Re: M09#12 [#permalink] 02 Jun 2014, 21:50
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# M09#12
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# Lesson: Properties of Fractions - Part II
## Comment on Properties of Fractions - Part II
### Dear Brent,
Dear Brent,
in the fifth problem I am having troubles with the expression's simplification. More specifically, where does that 9 come from? Which property of fractions has been used?
One approach is to simplify the expression.
(27x + 23y)/(3x + 2y) = (27x + 18y + 5y)/(3x + 2y)
= (27x + 18y)/(3x + 2y) + (5y)/(3x + 2y)
= 9 + (5y)/(3x + 2y)
Thank you very much!
I'm using the RULE that says: (a + b)/c = a/c + b/c
So, in my solution, I first rewrote 23y as (18y + 5y)
This allowed me to take 27x + 18y + 5y and add brackets as follows to get the equivalent form: (27x + 18y) + 5y
So, we get: (27x + 23y)/(3x + 2y) = [27x + 18y + 5y]/(3x + 2y)
= [(27x + 18y) + 5y]/(3x + 2y)
= (27x + 18y)/(3x + 2y) + 5y/(3x + 2y) {I applied the RULE at the top}
= 9 + [5y/(3x + 2y)]
How did I get 9?
Well 27x + 18y = 9(3x + 2y)
So, (27x + 18y)/(3x + 2y) = (9)(3x + 2y)/(3x + 2y)
= 9 {since (3x + 2y)/(3x + 2y) = 1}
Does that help?
### Thanks for explaining. I too
Thanks for explaining. I too got confused by 9.
### Now let's take a closer look
Now let's take a closer look at (5y)/(3x + 2y)
Notice that (5y)/(3y + 2y) = 5y/5y = 1 [since the numerator and denominator are EQUAL]
However, since we're told that y < x, we know that 3y + 2y < 3y + 2x
This means that (5y)/(3x + 2y) < 1, [since the numerator is LESS THAN the denominator]
If (5y)/(3x + 2y) < 1, then we can conclude that 9 + (5y)/(3x + 2y) < 10
I did not understand this part of the solution. Can you please explain?
If (5y)/(3x + 2y) < 1, then we can add 9 to both sides of the inequality to get:
(5y)/(3x + 2y) + 9 < 10
Does that help?
Cheers,
Brent
### I just assumed x to be 3 and
I just assumed x to be 3 and y to be 2. When i do this, it makes sense. But is that how you concluded? By assuming values of x and y?
### x = 3 and y = 2 is ONE pair
x = 3 and y = 2 is ONE pair of values that satisfy the given restriction that 0 < y < x
However, there are infinitely many possible pairs of values for x and y that satisfy the given restriction.
### On the second reinforcement
On the third reinforcement activity for this video the question is as follows "If m is the product of all the integers from 2 to 11, inclusive, and n is the product of all the integers from 4 to 11, inclusive, what is the value of n/m?" - when coming to the final answer to the question, how is the numerator 1? Shouldn't it be x^2/2(3)? Please let me know! Thank you!
n = (4)(5).....(10)(11)
m = (2)(3)(4)(5).....(10)(11)
Notice that every value in m is also in n
So, all of m's values will pair with some of n's values to get many 1's
That is: 4/4 = 1, 5/5 = 1, 6/6 = 1, . . . 10/10 = 1 and 11/11 = 1
All that remains, is (2)(3) in the denominator.
Another approach is to first notice that m = (2)(3)(n)
So, n/m = n/(2)(3)(n) = 1/(2)(3) = 1/6
Does that help?
Cheers,
Brent
### Hi there,
Hi there,
I am struggling with the second problem in the practice list.
When using the property (x+y)/z=x/z+y/z
If b/(a+b) = 7/12, then we can also say that (a + b)/b = 12/7
What I don't understand is why we would add 1 next.
the explanation suggest:
Simplify: a/b + 1 = 12/7
Subtract 1 from both sides to get: a/b = 12/7 - 1
Rewrite 1 as follows: a/b = 12/7 - 7/7
Evaluate: a/b = 5/7
We know that: (a + b)/b = 12/7
Applying the fraction property, we get: a/b + b/b = 12/7
Since b/b = 1, we can write: a/b + 1 = 12/7
Does that help?
Cheers,
Brent
### understood thank you!
understood thank you!
### https://gre.myprepclub.com/forum
https://gre.myprepclub.com/forum/compare-for-x-2155.html
Hi Brent, I've gone over this problem a few times, but I seem to not understand it. Could you please explain this one again.
Also, for comparison questions, are there certain criteria of a question that would be able to tell me that plugging in numbers could solve the question easily?
I posted a 3rd solution here: https://gre.myprepclub.com/forum/compare-for-x-2155.html#p28345
Please let me know if that helps.
Plugging in values has its drawbacks. The biggest problem is that, unless we're able to show conflicting outcomes (meaning the correct answer is D), we can never be 100% certain of the correct answer.
That said, I'd say that plugging in numbers is a great approach when you're not sure how to proceed with a question.
Cheers,
Brent
### Thank you so much. That
Thank you so much. That helped Brent!
### https://gre.myprepclub.com/forum
https://gre.myprepclub.com/forum/qotd-21-if-bc-0-and-3b-2c-18-then-which-of-the-2748.html
Can you give me the answer break down for this problem?
### Hey Brent, quick question on
Hey Brent, quick question on this one. https://gre.myprepclub.com/forum/if-0-x-y-then-which-of-the-following-must-be-true-3579.html
y>x (which is still a true statement)
You're absolutely right. Good catch - thanks!
I have edited my solution accordingly.
Cheers,
Brent
### https://gre.myprepclub.com/forum
https://gre.myprepclub.com/forum/if-0-y-x-then-which-of-the-following-3202.html
How come we are able to conclude that 3x+2y > 3y+2y? And therefore, the value of 3x+2y will be less than 1?
Thanks Bretnt for the help! |
#### Need Help?
Get in touch with us
Sep 23, 2022
## Key Concepts
• Add numbers using place value.
• Use properties of operation to find the sum.
• Solve few model questions related to addition by using place
### Introduction
Example 1:
3 – 2 = 1.
Find 2 + ____ = 3
Solution:
3 – 2 = 1.
So, 2 + 1 = 3.
## Use addition fact to subtract
Example 2:
Find 5 – 3.
To find 5 – 3, you can think:
The missing number is the same in both equations.
3 + 2 = 5
You also know the subtraction fact.
5 – 3 = 2
Example 3:
Find 15 – 4.
To find 15 – 4, you can think:
The missing number is the same in both equations.
4 + 11 = 15
You also know the subtraction fact.
15 – 4 = 11
### Make a 10 to Subtract
Example 4:
Find how many more balls are needed to be added to make a ten?
Solution:
Add 2 more balls to make a ten.
Example 5:
Find 12 – 9
Solution:
10 + 2 = 12
Example 6
Find 12 – 9
Solution:
10 – 7 = 3
## Exercise
6 – 4 =?
4 + ____ = 6
So, 6 – 4 = ____
9 – 3 =?
3 + ____ = 9
So, 9 – 3 = ____
12 – 4 =?
4 + ____ = 12
So, 12 – 4 = ____
17 – 7 = ____
7 + ____ = 17
• Complete the addition fact and subtract,
14 – 8 = ____
8 + ____ = 14
• Make a ten to subtract by using counters and your workmate.
11
– 4
______
• Make a ten to subtract by using counters and your workmate.
16
– 9
______
• Chen had 13 animal stickers. He gave 5 of the stickers away. How many animal stickers does Chen have now?
• Angie bought 13 strawberries. She ate 8 of the strawberries. How many strawberries does Angie have now?
### What we have learned
• Use addition fact and subtract two numbers.
• Use the ten frames to make a 10 and subtract two numbers.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
#### System of Linear Inequalities and Equations
Introduction: Systems of Linear Inequalities: A system of linear inequalities is a set of two or more linear inequalities in the same variables. The following example illustrates this, y < x + 2…………..Inequality 1 y ≥ 2x − 1…………Inequality 2 Solution of a System of Linear Inequalities: A solution of a system of linear inequalities […] |
# How do you simplify (-2-i)^2?
Dec 14, 2016
${\left(- 2 - i\right)}^{2} = 3 + 4 i$
#### Explanation:
In general
$\textcolor{w h i t e}{\text{XXX}} {\left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right)}^{2} = {\textcolor{red}{a}}^{2} + 2 \textcolor{red}{a} \textcolor{b l u e}{b} + {\textcolor{b l u e}{b}}^{2}$
${\left(- 2 - i\right)}^{2} = {\left(\textcolor{red}{\text{(-2))+color(blue)(} \left(- i\right)}\right)}^{2}$
Substituting color(red)a=color(red)(""(-2)) and color(blue)b=color(blue)(""(-i))
We have
${\textcolor{w h i t e}{\text{XXX")color(red)(""(-2))^2+2 * color(red)(""(-2)) * color(blue)(""(-i))+color(blue)(} \left(- i\right)}}^{2}$
$\textcolor{w h i t e}{\text{XXX}} = 4 + 4 i - 1$
$\textcolor{w h i t e}{\text{XXX}} = 3 + 4 i$ |
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2011 AMC 12A Problems/Problem 12"
## Problem
A power boat and a raft both left dock $A$ on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock $B$ downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock $A.$ How many hours did it take the power boat to go from $A$ to $B$?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 3.5 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 4.5 \qquad \textbf{(E)}\ 5$
## Solution
### Solution 1
Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of $0$. In this case, when the powerboat travels from $A$ to $B$, the raft remains at $A$. Thus the trip from $A$ to $B$ takes the same time as the trip from $B$ to the raft. Since these times are equal and sum to $9$ hours, the trip from $A$ to $B$ must take half this time, or $4.5$ hours. The answer is thus $\boxed{\textbf{D}}$.
### Solution 2
What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as $b$ and the speed of the raft at $r$.
$[asy] size(8cm,8cm); pair A, B; A=(-3,6); B=(3,6.2); draw(A--B); label("A",A,S); label("B",B,S); pair A, B; arrow((-2.5,4),dir(180),blue); arrow((-2.5,4.2),dir(180),red); A=(-3,5); B=(3,5); draw(A--B); label("A",A,S); label("B",B,S); pair A, B; A=(-3,6); B=(3,6); draw(A--B); label("A",A,S); label("B",B,S);[/asy]$ |
# 2009 AMC 12A Problems/Problem 21
## Problem
Let $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are complex numbers. Suppose that
$$p(2009 + 9002\pi i) = p(2009) = p(9002) = 0$$
What is the number of nonreal zeros of $x^{12} + ax^8 + bx^4 + c$?
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12$
## Solution
From the three zeroes, we have $p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)$.
Then $p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002)=x^{12}+ax^8+bx^4+c$.
Let's do each factor case by case:
• $x^4 - (2009 + 9002\pi i) = 0$: Clearly, all the fourth roots are going to be complex.
• $x^4 - 2009 = 0$: The real roots are $\pm \sqrt [4]{2009}$, and there are two complex roots.
• $x^4 - 9002 = 0$: The real roots are $\pm \sqrt [4]{9002}$, and there are two complex roots.
So the answer is $4 + 2 + 2 = 8\ \mathbf{(C)}$.
## Alternative Thinking
Consider the graph of $x^4$. It is similar to a parabola, but with a wider "base". First examine $x^4-2009$ and $x^4-9002$. Clearly they are just being translated down some large amount. This will result in the $x$-axis being crossed twice, indicating $2$ real zeroes. From the Fundamental Theorem of Algebra we know that a polynomial must have exactly as many roots as its highest degree, so we are left with $4-2$ or $2$ nonreal roots for each of the graphs. For the graph of $x^4-(2009+9002\pi i)$, it's not even possible to graph it on the Cartesian plane, so all $4$ roots will be nonreal. This is $2+2+4 = 8$ total nonreal roots $\Rightarrow \boxed{\text{C}}$. |
# Question Video: Solving Using Combination Mathematics
Evaluate 23πΆ19.
01:49
### Video Transcript
Evaluate 23 πΆ 19.
In order to answer this question, we begin by recalling the formula for combinations. ππΆπ is equal to π factorial divided by π minus π factorial multiplied by π factorial. In this question, π is equal to 23 and π is equal to 19. 23 πΆ 19 is therefore equal to 23 factorial divided by 23 minus 19 factorial multiplied by 19 factorial. The denominator simplifies to four factorial multiplied by 19 factorial. We can rewrite the numerator as 23 multiplied by 22 multiplied by 21 multiplied by 20 multiplied by 19 factorial. And by dividing the numerator and denominator by 19 factorial, we have 23 multiplied by 22 multiplied by 21 multiplied by 20 divided by four factorial.
Writing four factorial as four multiplied by three multiplied by two multiplied by one, we can reduce the factors as shown. This leaves us with 23 multiplied by 11 multiplied by seven multiplied by five, which is 8,855. 23 πΆ 19 is equal to 8,855. |
# 1 10 equivalent fractions
How to find equivalent fractions?
1. 20 is equivalent to 1 10 because 1 x 2 = 2 and 10 x 2 = 20
2. 30 is equivalent to 1 10 because 1 x 3 = 3 and 10 x 3 = 30
3. 40 is equivalent to 1 10 because 1 x 4 = 4 and 10 x 4 = 40
## How do you write an equivalent fraction?
Summary:
• You can make equivalent fractions by multiplying or dividing both top and bottom by the same amount.
• You only multiply or divide, never add or subtract, to get an equivalent fraction.
• Only divide when the top and bottom stay as whole numbers.
## How to make an equivalent fraction?
It is possible by these methods:
• Method 1: Make the Denominators the same
• Method 2: Cross Multiply
• Method 3: Convert to decimals
## What are the rules of equivalent fractions?
Equivalent Fractions Rule. A rule stating that if the numerator and denominator of a fraction are multiplied by the same nonzero number, the result is a fraction that is equivalent to the original fraction. This rule can be represented as: a//b = (n * a)//(n * b).
## How to solve equivalent fractions?
Methods to Determine Equivalent Fractions
• Method 1: By making the denominators the same, we can evaluate if two fractions are equivalent. For example, find if 2/3 and 6/9 are equivalent.
• Method 2: Since, both the values are equal, therefore, 1/2 and 3/6 are equivalent fractions.
• Method 3: If two fractions are given, we can simply find their decimals to check if they are equivalent fractions.
## What is the fraction of 1 10?
0.11/10 = 110 = 0.1 Spelled result in words is one tenth.
## How do you find fractions equivalent?
To find the equivalent fractions for any given fraction, multiply the numerator and the denominator by the same number. For example, to find an equivalent fraction of 3/4, multiply the numerator 3 and the denominator 4 by the same number, say, 2. Thus, 6/8 is an equivalent fraction of 3/4.
## What is 1/10 as a percentage?
Convert 1/10 to Percentage by Converting to Decimal We can see that this gives us the exact same answer as the first method: 1/10 as a percentage is 10%.
## What are the 10 fractions?
Example ValuesPercentDecimalFraction10%0.11/1012½%0.1251/820%0.21/525%0.251/412 more rows
## What is equivalent calculator?
Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds.
## What is 1/4 equal to as a fraction?
2/8Answer: The fractions equivalent to 1/4 are 2/8, 3/12, 4/16, etc. Equivalent fractions have the same value in their reduced form.
## How do you find 1/10 of a number?
To find one tenth of a number, simply divide it by 10 .
## What is the ratio of 1 to 10?
A ratio of 1 to 10 can be written as 1 to 10, 1:10, or 1/10. Furthermore, 1 and 10 can be the quantity or measurement of anything, such as students, fruit, weights, heights, speed and so on. A ratio of 1 to 10 simply means that for every 1 of something, there are 10 of something else, with a total of 11.
## What is the decimal of 1 10?
As you can see, in one quick calculation, we’ve converted the fraction 110 into it’s decimal expression, 0.1.
## What is a base 10 fraction?
A decimal fraction (base 10 fraction) is a fraction whose denominator is a power of ten. Specifically, we would say the denominator is a multiple of 10: 10, 100, 1000, 10,000,…
## What is 1/3 equivalent to as a fraction?
Fractions equivalent to 1/3: 2/6, 3/9, 4/12, 5/15 and so on …
## How do you make 10 into a fraction?
10% as a fraction is expressed as 1/10.
## How do you find equivalent fractions for kids?
0:245:15Equivalent Fractions | Math for 3rd Grade | Kids Academy – YouTubeYouTubeStart of suggested clipEnd of suggested clipIn so for example an equivalent fraction to 1 4 could be 2 8. to show this in a picture instead ofMoreIn so for example an equivalent fraction to 1 4 could be 2 8. to show this in a picture instead of cutting our fraction into four pieces we’ll cut our fraction into eight pieces.
## What is the equivalent fraction of 2 by 5?
Equivalent fractions of 2/5 : 4/10 , 6/15 , 8/20 , 10/
## What is the fraction 3/4 equivalent to?
Decimal and Fraction Conversion ChartFractionEquivalent Fractions1/32/63/92/34/66/91/42/83/123/46/89/1223 more rows
## How do you know if two fractions are equivalent?
It’s called the cross multiply rule. The rule is shown below: This formula says that if the numerator of one fraction times the denominator of the other fraction equals the denominator of the first fraction times the numerator of the second fraction, then the fractions are equivalent.
## How to find equivalent fractions?
To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 1 10) by the same natural number, ie, multiply by 2, 3, 4, 5, 6
## Is 1 10 a fraction?
Important: 1 10 looks like a fraction, but it is actually an improper fraction.
## Can you convert fractions to decimals?
This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.
## How to make a fraction equivalent?
Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction , and you’ll create an equivalent fraction.
## What are Equivalent Fractions?
Equivalent fractions are fractions with different numbers representing the same part of a whole. They have different numerators and denominators, but their fractional values are the same.
## What is half of a fraction?
For example, think about the fraction 1/2. It means half of something. You can also say that 6/12 is half, and that 50/100 is half. They represent the same part of the whole. These equivalent fractions contain different numbers but they mean the same thing: 1/2 = 6/12 = 50/100
## How to find equivalent fractions?
To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 10 1) by the same natural number, ie, multiply by 2, 3, 4, 5, 6
## Is 10 1 a fraction?
Important: 10 1 looks like a fraction but it is actually the whole number 10.
## Can you convert fractions to decimals?
This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.
## What is equivalent fraction?
Equivalent fractions are fractions that, although they have different digits in the numerator and denominator, represent the same value. For example, the fractions , , and all represent the same part of a whole: 1/2. This can be depicted using fraction bars:
## What happens when unlike fractions are converted into equivalent fractions?
If they are unlike fractions, then the unlike fractions must be converted into equivalent fractions that share the same denominator in order to be added or subtracted.
## What happens when you multiply a fraction by 1?
Since multiplying any number by 1 results in the same number, multiplying any fraction by a fractional representation of the number 1 results in an equivalent fraction.
## What is the least common multiple of 2 and 5?
1. : The least common multiple of 2 and 5 is 10 , so we will convert the above fractions to equivalent fractions that have a denominator of 10 in order to add them. 2. : The least common multiple of 7 and 21 is 21, so we only need to convert the first fraction to an equivalent fraction.
## Is each fraction bar the same?
Even though each of the fraction bars is broken up into a different number of parts, each of the fraction bars represents the same fraction, albeit with different numerators and denominators. Any given fraction has an infinite number of equivalent fractions.
## Can you divide to find equivalent fractions?
Depending on the starting fraction, it is also possible to divide to find equivalent fractions . For example, dividing the numerator and denominator of by 4 would give us , another equivalent fraction of .
## What are Equivalent Fractions?
Equivalent fractions state that two or more than two fractions are said to be equal if both results are the same fraction after simplification. Let us say, a/b and c/d are two fractions, after the simplification of these fractions, both result in equivalent fractions, say e/f, then they are equal to each other.
## Why are fractions the same?
These fractions are actually the same because when we multiply or divide both the numerator and the denominator by the same number, the value of the fraction actually doesn’t change. Therefore, equivalent fractions, when reduced to their simplified value, will all be the same. For example, consider the fraction 1/5.
## What happens when two fractions are given?
If two fractions are given, we can simply find their decimals to check if they are equivalent fractions.
## How to make the denominator equal to 9?
Multiply 2/3 by 3/3 to make the denominator equal to 9.
## Is a fraction equivalent to a decimal?
Since, both the fractions results in the same decimal, thus they are equivalent.
## Do denominators have a common multiple?
The answer to this question is that, as the numerator and denominator are not co-prime numbers, therefore they have a common multiple, which on division gives exactly the same value.
## Is 10/12 a fraction?
But, ⅚ and 10/12 are equivalent fractions because of 10/12 = ⅚.
## Why are fractions the same?
Why are they the same? Because when you multiply or divide both the top and bottom by the same number, the fraction keeps it’s value.
## Can you multiply to get an equivalent fraction?
You only multiply or divide, never add or subtract, to get an equivalent fraction.
## What is Meant by the Equivalent Fractions?
In mathematics, the fraction defines the parts of the whole. It is the ratio of two numbers. The fraction is of the form a/b. For example, ⅝ is a fraction. Here, the top number 5 is called the numerator, and the bottom number 8 is called the denominator. The fractions can be classified into different types, namely proper fraction, improper fraction, mixed fraction, like fraction, unlike fractions, equivalent fractions. Let us consider two fractions, 15/20 and 9/12. These fractions are said to be an equivalent fraction because when the fractions are simplified, the resultant fraction will be the same. If 15/20 is simplified, we get ¾. Similarly, if 9/12 is simplified, we get ¾. Hence, it is called equivalent fraction
## What are the different types of fractions?
The fractions can be classified into different types, namely proper fraction, improper fraction, mixed fraction, like fraction, unlike fractions, equivalent fractions. Let us consider two fractions, 15/20 and 9/12. These fractions are said to be an equivalent fraction because when the fractions are simplified, the resultant fraction will be … |
# Conditional Probability
## P(B|A) = P(A and B)/ P(A)
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Intersection of Compound Events
There is a classic example of probability studies involving a coin flip. Everyone knows that the probability of getting heads on a single flip is 50%, which means that every time you flip it, there is also a 50% probability of getting tails. The question is, if you have flipped a coin 99 times and got heads every time, what is the probability of getting heads the next time?
Credit: Expert Infantry
Source: https://www.flickr.com/photos/expertinfantry/5458168907/
After this lesson on the intersection of compound events, we’ll return to this question and see how it does (and doesn’t!) fit with the concept.
### Intersection of Compound Events
It should make sense intuitively that the more specific or restricted you make the details of an event, the less probable it becomes for that event to occur. The concept of calculating the total probability of multiple events strung together is the same idea.
If I flip a coin once there are only two possible outcomes:
If I flip the coin twice, there are four possibilities:
We know there are a total of four possible outcomes from two coin flips: , , , and , and only one of them: , results in the outcome we want to calculate. Using the simple probability formula, we get:
#### Calculating Probability
1. What is the probability of flipping a coin four times and getting tails all four times?
Create a table listing all of the possible outcomes:
Now we can look at the bottom row and see that there are a total of 16 possibilities, only one of which is four tails in a row. The probability, therefore, is:
2. What is the probability of rolling two even numbers in a row on a standard six-sided die?
Create a table listing all possible outcomes:
Reducing to:
3. What is the probability of spinning two 2’s in a row OR two 4’s in a row on a spinner with the numbers 1-4?
Create a table listing all possible outcomes, and highlight the favorable ones:
Out of a total of 16 possible outcomes, only 2 fit our description, which gives us:
#### Earlier Problem Revisited
The question is, if you have flipped a coin 99 times and got heads every time, what is the probability of getting heads the next time?
This is a very common example of something called the gambler’s fallacy. It is not a good example of calculating the intersection of compound events because of the way it is worded. The question as written is essentially asking about a single flip of the coin, which is always , because a coin has no memory.
From the standpoint of an example of what we have been studying in this chapter, the more useful, and dramatically more difficult question would be:
What is the probability of flipping a coin 100 times and getting heads every time?
If you want to know the probability of flipping 100 heads in a row, you could either draw a really long chart of all of the possibilities (like the one in Example A, but much longer), or you could use the multiplication rule that we will be learning in the next lesson. Check it out!
### Example 1
#### Example 1
What is the probability of pulling 1 red marble, replacing it, then pulling another red marble out of a bag containing 4 red and 2 white marbles?
Make a chart:
The four sets of four red “’s” represent the favorable outcomes out of the total of 36, therefore
#### Example 2
What is the probability of a spinner landing on “2” and then a “3”, or “6” if there are 6 equally spaced points on the spinner?
Make a chart:
The red numbers 3 and 6 represent the two favorable outcomes out of 36 total, therefore
#### Example 3
What is the probability of pulling a red and then a black card at random from a standard deck (replacing the first card after drawing)?
There are 26 black and 26 red cards in the deck, so the probability on the first pull is On the second pull, we again have a 50% chance of favorable outcome, but that 50% only applies to the half of the first pulls that were favorable. Therefore:
#### Example 4
What probability of picking a red and then a green marble from a bag with 5 red and 1 green marbles in it (replacing the first marble after the draw)?
Make a chart: Of the 36 possible outcomes, only 5 fit the description of red the first time, and green the second time (noted by the red “’s”. Therefore
#### Example 5
What is the probability of shaking the hand of a student wearing red and then a student wearing blue if you randomly shake the hands of two people in a row in a room containing 3 students in blue and 2 in red?
Make a chart:
So, out of the 25 possible handshake possibilities, 6 of them fit the requirements of red first, then blue:
### Review
Questions 1-6: Suppose you have an opaque bag filled with 4 red and 3 green balls. Assume that each time a ball is pulled from the bag, it is random, and the ball is replaced before another pull.
1. Create a chart of all possible outcomes of an experiment consisting of pulling one ball from the bag at random, noting the color and replacing it, then pulling another.
2. How many possible outcomes are there?
3. What is the probability of randomly pulling a red ball from the bag, returning it, and pulling agreen ball on your second pull?
4. What is the probability of randomly pulling a red ball both times?
5. What is the probability of pulling a green ball both times?
6. Is the probability of pulling a red followed by a green different than pulling a green followed by a red?
Questions 7 – 12: Suppose you have two standard dice, one red and one blue.
7. Construct a probability distribution table or diagram for an experiment consisting of one roll of the red die followed by one roll of the blue one.
8. How many possible outcomes are there?
9. Is there an apparent mathematical relationship between the number of sides on the dice and the number of possible outcomes?
10. What is the probability of rolling a 2 on the red die and a 1, 3, or 5 on the blue one?
11. What is the probability of rolling an even number on the red die and an odd on the blue one?
12. Do the probabilities of a particular outcome change based on which die is rolled first? Why or why not?
Questions 13 – 16: Suppose you have a spinner with 5 equally-spaced color sections: red, blue, green, yellow, and orange.
13. Construct a probability distribution detailing the possible outcomes of three consecutive spins. You may wish to use only the first letter, or a single color-coded hash mark, to represent each possibility, as there will be many of them.
14. How many possible outcomes are there?
15. Is there an apparent mathematical relationship between the number of sections on the spinner, the number of spins, and the number of possible outcomes? If so, what is the relationship?
16. What is the probability of spinning red, then green, and then orange?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
conditional probability The probability of a particular dependent event given the outcome of the event on which it occurs.
conditional probability formula The conditional probability formula is P(A/B) = P(AUB)/P(B)
Dependent Events In probability situations, dependent events are events where one outcome impacts the probability of the other.
Favorable Outcome A favorable outcome is the outcome that you are looking for in an experiment.
Independent Events Two events are independent if the occurrence of one event does not impact the probability of the other event.
Multiplication Rule States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).
Mutually Exclusive Events Mutually exclusive events have no common outcomes.
Sample Space In a probability experiment, the sample space is the set of all the possible outcomes of the experiment. |
### Theory:
Often we meet fractions with denominators of $$10, 100, 1000$$, etc.
For example, $$1$$ g $$=$$ $\frac{1}{1000}$kg, $$1$$ mm $$=$$ $\frac{1}{10}$ cm, $$4$$ cm $$3$$ mm $$=$$ $4\frac{3}{10}$ cm, etc.
Numbers with denominators of $$10, 100, 1000$$, etc., agreed to write down without a denominator.
First, write the integer part, and then the numerator of the fractional part. The whole number part is separated from the fractional part by a point.
For example, instead of $4\frac{3}{10}$, we write $$4.3$$ (we read: "$$4$$ integers and $$3$$ tenths").
Instead $5\frac{19}{100}$ we write $$5.19$$ (we read: "$$5$$ as a whole and $$19$$ hundredths").
Any number whose denominator of the fractional part is expressed as one with one or more zeros can be represented as a decimal fraction. If the fraction is correct, then the digit $$0$$ is written before the decimal point.
For example, instead of $\frac{21}{100}$, we write $$0.21$$ (we read: "$$0$$ integers and $$21$$ hundredths").
Important!
After the decimal point, the numerator of the fractional part should have as many digits as there are zeros in the denominator.
Therefore, for example, the number $1\frac{3}{100}$ should be written like this: $$1.03$$ (read: "$$1$$ whole and $$3$$ hundredths"). |
# Convert thousandths between simplified fractions and decimals
Lesson
When we rename fractions as tenths, hundredths or thousandths, we are also able to use our knowledge of place value to then write our fraction as a decimal.
Remember how we converted fractions to decimals, using tenths and hundredths? This time, we'll use a similar approach, but look at thousandths.
## Place value
Place value is important when we express numbers in decimals. Let's say we want to compare $0.353$0.353 and $0.6$0.6. We may think that because $353$353 is larger than $6$6, that $0.353$0.353 is larger than $0.6$0.6, but we can see when we use a place value table, this is not the case.
Place value
Units . Tenths Hundredths Thousandths
0 . 3 5 3
0 . 6 0 0
$0.6$0.6 is actually $6$6 tenths, where $0.353$0.353 only has $3$3 tenths, so is smaller. Similarly, if we consider them both as fractions of a thousand, then $0.6$0.6 is $\frac{600}{1000}$6001000, which is more than $\frac{353}{1000}$3531000.
## Changing our fractions to thousandths
The first thing we need to do is rename our fraction so that it is expressed in thousandths if it's not already in thousandths. This video shows you how to convert your fraction to thousandths, allowing you to then easily change it to decimals. By the way, renaming our fractions to thousandths is not changing the value of our number at all, we're just finding an equivalent fraction.
Remember!
To convert fractions to decimals, you can rename them to tenths, hundredths or thousandths.
#### Examples
##### Question 1
Write the fraction $\frac{11}{200}$11200 as a decimal.
##### QUESTION 2
Write the decimal $0.872$0.872 as a fraction in simplest form.
##### QUESTION 3
Write the number represented by $98$98 thousandths:
1. As a simplified fraction.
2. As a decimal.
### Outcomes
#### NA4-6
Know the relative size and place value structure of positive and negative integers and decimals to three places. |
# The Shapes of Numbers — The Curve
Earlier in our Shapes of Numbers block, we explored triangle numbers, square numbers, and primes. We looked at the different ways they related to each other by exploring their sums and differences. Next up — curves — a particular curve — the parabola. I think I only mention the word once though, at the end. Really our work is with doubles with an introduction to exponents. The beginning is a review of doubles that we’ve done before from my Addition and Subtraction curriculum.
We start out with the one white cube from our Cuisenaire rods, and just keep doubling. From here, we determine what each line column of rods were:
White – 1
Red – 1 x 2 = 2
Purple – 1 x 2 x 2 = 4
Brown – 1 x 2 x 2 x 2 = 8
Each time we were building the doubles, we are showing the multiplication with the red rods (not pictured,) and it was getting cumbersome, and we were running out of red rods. Enter Gattegno’s crosses for review. We talk about how we could use these crosses to represent the area model of multiplication, and each time we added another red rod, making a tower of red rods, it represented doubling the entire number.
Orange and dark green – 1 x 2 x 2 x 2 x 2 = 16
3 orange and a red – 1 x 2 x 2 x 2 x 2 x 2 = 32
Anyone else tired of writing ‘x 2’? Yea, me too. Let’s come up with a shortcut notation. This ‘2’ will be our base, and a superscript will represent how many times we are multiplying the 2. So how many 2’s have we multiplied now? 5, so 25. That works! We’ll call that superscript an ‘exponent.’ That simplifies things.
Oh, but wait, our towers keep falling. Geez! How many red rods can you stack and maintain stability?!? I know! Let’s come up with another system. What do you notice about our towers? The number of 2’s we are multiplying correspond with the height of a rod. Let’s make an L, with the red rod being the base, and the exponent being represented by how tall the second rod is (Gattegno). Whew! That’s easier.
So now we have
26 = 64
27 = 128
28 = 256
29 = 512
210 = 1,024
Now let’s go back and compare a regular staircase (1, 2, 3, 4, 5….) to our doubles (1, 2, 4, 8, 16, 32.) What do you notice? What shape does the staircase have as it rises?
Place a dot at the upper left corner of each rod, move the rods over, and connect the dots. What do you have? A line! Now, place a dot on the upper left corner of each rod for the doubles sequence. Move the rods, and connect the dots. What shape do you have? A curve. This curve is called a parabola. |
# Quick Answer: What Is Slope Of A Straight Line?
## What does a positive slope look like?
Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises.
We will learn in other sections that “price” and “quantity supplied” have a positive relationship; that is, firms will supply more when the price is higher..
## What is the equation for a zero slope?
The equation for a zero slope line is one where the X value may vary but the Y value will always be constant. An equation for a zero slope line will be y = b, where the line’s slope is 0 (m = 0).
## What does a zero slope look like?
‘ When the ‘rise’ is zero, then the line is horizontal, or flat, and the slope of the line is zero. Put simply, a zero slope is perfectly flat in the horizontal direction. The equation of a line with zero slope will not have an x in it. It will look like ‘y = something.
## Does a straight line on a graph have a slope?
A straight line has one slope. That is, whether we measure the slope by moving from point A to point B, or by moving from point C to point D, the number we obtain will be the same.
## Is the slope of a straight line zero?
A line that goes straight across (Horizontal) has a Gradient of zero.
## What is the slope of a straight vertical line?
The slope of a line is change in Y / change in X. The slope of a horizontal line = 0, not undefined. The slope of a vertical line = undefined.
## What does slope look like?
The slope equals the rise divided by the run: . You can determine the slope of a line from its graph by looking at the rise and run. One characteristic of a line is that its slope is constant all the way along it. So, you can choose any 2 points along the graph of the line to figure out the slope.
## What does vertical line look like?
A vertical line is a line, parallel to y-axis and goes straight, up and down, in a coordinate plane. Whereas the horizontal line is parallel to x-axis and goes straight, left and right.
## Which axis is Y =- 1 parallel to?
Thus, if P(x, y) is any point on AB, then x = a. Hence, the equation of a straight line parallel to y-axis at a distance a from it is x = a. The equation of y-axis is x = 0, since, y-axis is a parallel to itself at a distance 0 from it.
## Is zero slope a function?
The slope of a line can be positive, negative, zero, or undefined. … because division by zero is an undefined operation. Vertical lines are symbolically represented by the equation, x = a where a is the x-intercept. Vertical lines are not functions; they do not pass the vertical line test at the point x = a.
## What do you mean by slope of a straight line?
Definition: The slope of a line is a number that measures its “steepness”, usually denoted by the letter m. It is the change in y for a unit change in x along the line. … The slope of a line (also called the gradient of a line) is a number that describes how “steep” it is. In the figure above press ‘reset’.
## How do you find the slope of a straight line?
The slope will be the same for a straight line no matter which two points you pick as you know. All you need to do is to calculate the difference in the y coordinates of the 2 points and divide that by the difference of the x coordinates of the points(rise over run). That will give you the slope. |
NCERT Solutions for Class 7 Maths Exercise 7.2 Chapter 7 Congruence of Triangles
NCERT Solutions for Class 7 Maths Exercise 7.2 Chapter 7 Congruence of Triangles in simple PDF are given here. Criteria for congruence of triangles and congruence among right-angled triangles are the two topics covered in this exercise of NCERT Solutions for Class 7. Students of Class 7 are suggested to solve NCERT Solutions for Class 7 Maths Chapter 7 Congruence to strengthen the fundamentals and be able to solve questions that are usually asked in the examination.
Download the PDF of NCERT Solutions For Class 7 Maths Chapter 7 Congruence of Triangles – Exercise 7.2
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Access Another Exercise of NCERT Solutions For Class 7 Chapter 7 – Congruence of Triangles
Exercise 7.1 Solutions
Access Answers to NCERT Class 7 Maths Chapter 7 – Congruence of Triangles Exercise 7.2
1. Which congruence criterion do you use in the following?
(a) Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
Solution:-
By SSS congruence property:- Two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.
ΔABC ≅ ΔDEF
(b) Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
Solution:-
By SAS congruence property:- Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.
ΔACB ≅ ΔDEF
(c) Given: ∠MLN = ∠FGH
∠NML = ∠GFH
∠ML = ∠FG
So, ΔLMN ≅ ΔGFH
Solution:-
By ASA congruence property:- Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.
ΔLMN ≅ ΔGFH
(d) Given: EB = DB
AE = BC
∠A = ∠C = 90o
So, ΔABE ≅ ΔACD
Solution:-
By RHS congruence property:- Two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and one side of the second.
ΔABE ≅ ΔACD
2. You want to show that ΔART ≅ ΔPEN,
(a) If you have to use SSS criterion, then you need to show
(i) AR = (ii) RT = (iii) AT =
Solution:-
We know that,
SSS criterion is defined as, two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.
∴ (i) AR = PE
(ii) RT = EN
(iii) AT = PN
(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
(i) RT = and (ii) PN =
Solution:-
We know that,
SAS criterion is defined as, two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.
∴ (i) RT = EN
(ii) PN = AT
(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ? (ii) ?
Solution:-
We know that,
ASA criterion is defined as, two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.
Then,
(i) ∠ATR = ∠PNE
(ii) ∠RAT = ∠EPN
3. You have to show that ΔAMP ≅ ΔAMQ.
In the following proof, supply the missing reasons.
Steps Reasons (i) PM = QM (i) … (ii) ∠PMA = ∠QMA (ii) … (iii) AM = AM (iii) … (iv) ΔAMP ≅ ΔAMQ (iv) …
Solution:-
Steps Reasons (i) PM = QM (i) From the given figure (ii) ∠PMA = ∠QMA (ii) From the given figure (iii) AM = AM (iii) Common side for the both triangles (iv) ΔAMP ≅ ΔAMQ (iv) By SAS congruence property:- Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.
4. In ΔABC, ∠A = 30o, ∠B = 40o and ∠C = 110o
In ΔPQR, ∠P = 30o, ∠Q = 40o and ∠R = 110o
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or Why not?
Solution:-
No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be enlarged copy of the other.
5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ΔRAT ≅ ?
Solution:-
From the given figure,
We may observe that,
∠TRA = ∠OWN
∠TAR = ∠NOW
∠ATR = ∠ONW
Hence, ΔRAT ≅ ΔWON
6. Complete the congruence statement:
ΔBCA ≅ ΔQRS ≅
Solution:-
First consider the ΔBCA and ΔBTA
From the figure, it is given that,
BT = BC
Then,
BA is common side for the ΔBCA and ΔBTA
Hence, ΔBCA ≅ ΔBTA
Similarly,
Consider the ΔQRS and ΔTPQ
From the figure, it is given that
PT = QR
TQ = QS
PQ = RS
Hence, ΔQRS ≅ ΔTPQ
7. In a squared sheet, draw two triangles of equal areas such that
(i) The triangles are congruent.
(ii) The triangles are not congruent.
What can you say about their perimeters?
Solution:-
(ii)
Â
In the above figure, ΔABC and ΔDEF have equal areas.
And also, ΔABC ≅ ΔDEF
So, we can say that perimeters of ΔABC and ΔDEF are equal.
(ii)
In the above figure, ΔLMN and ΔOPQ
ΔLMN is not congruent to ΔOPQ
So, we can also say that their perimeters are not same.
8. Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Solution:-
Let us draw triangles LMN and FGH.
In the above figure, all angles of two triangles are equal. But, out of three sides only two sides are equal.
Hence, ΔLMN is not congruent to ΔFGH.
9. If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
Solution:-
By observing the given figure, we can say that
∠ABC = ∠PQR
∠BCA = ∠PRQ
The other additional pair of corresponding part is BC = QR
∴ ΔABC ≅ ΔPQR
10. Explain, why ΔABC ≅ ΔFED
Solution:-
From the figure, it is given that,
∠ABC = ∠DEF = 90o
∠BAC = ∠DFE
BC = DE
By ASA congruence property, two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.
ΔABC ≅ ΔFED
2 Comments
1. This is a very useful app
2. Mahima bisht
This is very helpful app for students |
# Question #64b09
Nov 12, 2016
$f \left(- 4\right) = 4 {e}^{\frac{1}{8}}$ is the maximum value of $f$ on $\left[- 4 , 16\right]$
and
$f \left(16\right) = - 16 {e}^{2}$ is the minimum value of $f$ on $\left[- 4 , 16\right]$
#### Explanation:
Given a continuous function $f \left(x\right)$ a closed interval $\left[a , b\right]$, any extrema will occur either at an endpoint or a critical point, that is, a point where $f ' \left(x\right) = 0$ or a point at which $f ' \left(x\right)$ does not exist.
In our case, we have $f \left(- x\right) = x {e}^{{\left(- x\right)}^{2} / 128}$, meaning $f \left(x\right) = - x {e}^{{x}^{2} / 128}$.
Differentiating, we can use the product rule and the chain rule to get
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} - x {e}^{{x}^{2} / 128}$
$= - x \left(\frac{d}{\mathrm{dx}} {e}^{{x}^{2} / 128}\right) + {e}^{{x}^{2} / 128} \left(\frac{d}{\mathrm{dx}} - x\right)$
$= - x \left[{e}^{{x}^{2} / 128} \left(\frac{d}{\mathrm{dx}} {x}^{2} / 128\right)\right] + {e}^{{x}^{2} / 128} \left(- 1\right)$
$= - x {e}^{{x}^{2} / 128} \left(\frac{x}{64}\right) - {e}^{{x}^{2} / 128}$
$= - {e}^{{x}^{2} / 128} \left({x}^{2} / 64 + 1\right)$
As this is defined for all real numbers, we only need look for points at which $f ' \left(x\right) = 0$. Doing so, we get
$- {e}^{{x}^{2} / 128} \left({x}^{2} / 64 + 1\right) = 0$
As ${e}^{{x}^{2} / 128} > 0$ for all real numbers, ${x}^{2} / 64 + 1 > {x}^{2} / 64 \ge 0$ for all real numbers, and the product of two nonzero values is nonzero, we know that there are no real solutions to the above. Thus, our function has no critical points.
As our function has no critical points on $\left[- 4 , 16\right]$ (or anywhere, for that matter), its extrema must occur at the endpoints. Evaluating, we get
$f \left(- 4\right) = 4 {e}^{{\left(- 4\right)}^{2} / 128} = 4 {e}^{\frac{1}{8}}$
$f \left(16\right) = - 16 {e}^{{16}^{2} / 128} = - 16 {e}^{2}$
So the above are the extrema of the function on $\left[- 4 , 16\right]$, giving us our answer:
$f \left(- 4\right) = 4 {e}^{\frac{1}{8}}$ is the maximum value of $f$ on $\left[- 4 , 16\right]$
and
$f \left(16\right) = - 16 {e}^{2}$ is the minimum value of $f$ on $\left[- 4 , 16\right]$ |
Here showcasing some of the trigonometry math problems
## Trigonometry math problem 1
A man 30 meters away from a tree and looking at the top of the tree at an angle of 30 degrees then find the height of the tree (Height of the man is 1.5 meters)
## Solution to problem 1
We can picture this math problem like this
Here OA = 1.5 meters is the height of the man and XY is the height of the tree, We need to find the value of XY to solve this math problem
AB is parallel to the ground then ∠BAY = 30° (Man looking at 30 degrees) and AB = 30 meters (distance from the tree)
From triangle BAY, Let BY = x then
tan ∠BAY = BY/AB
⇒ tan 30° = x/30
⇒ 1/√3 = x/30
so, x = 10√3 meters
⇒ XY = 10√3 +BX = 10√3 + 1.5
Height of the tree = 10√3 + 1.5 = meters
## Trigonometry math problem 2
ABC is a triangle with sides 4 cm & 5 cm and ∠ABC = 60°, then find the radius of the circumcircle
## Solution to problem 2
Apply cosine rule in triangle ABC
BC2 = AB2 + AC2 – 2 × AB × AC × cos A
⇒ BC2 = 52 + 42 – 2 × 5 × 4 × cos 60°
⇒ BC2 = 25 + 16 – 2 × 5 × 4 × 1/2 = 41 – 20
so, BC2 = 21
⇒ BC = 21 cm
Apply sine rule in triangle ABC then
D ( Diameter) = BC/sin A
⇒ D = 21/sin 60 = 21/(3/2)
⇒ D = 27 cm
then, Radius of the circumcircle = 7 cm
## Trigonometry math problem 3
If sin A = 3/2 then find the value of cos A
## Solution to problem 3
We can solve this problem with help of sin2 A + cos2 A = 1 identity
sin2 A + cos2 A = 1
⇒ (√3/2)2 + cos2 A = 1
⇒ 3/4 + cos2 A = 1
so, cos2 A = 1 – 3/4
⇒ cos2 A = 1/4
then, cos A = 1/2
## Trigonometry math problem 4
ABC is a triangle with sides 4 cm & 6 cm and ∠ABC = 30°, then find the area of the triangle
## Solution to problem 4
Area of the triangle = ½ × AB × AC × sin A
⇒ Area of the triangle = ½ × 6 × 4 × sin 30°
⇒ Area of the triangle = ½ × 6 × 4 × ½
so, the area of the triangle = 6 cm2 |
Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
# Mark the Correct Alternative in of the Following: If Y = Sin ( X + 9 ) Cos X Then D Y D X at X = 0 is - Mathematics
MCQ
Mark the correct alternative in of the following:
If $y = \frac{\sin\left( x + 9 \right)}{\cos x}$ then $\frac{dy}{dx}$ at x = 0 is
• cos 9
• sin 9
• 0
• 1
#### Solution
$y = \frac{\sin\left( x + 9 \right)}{\cos x}$
Differentiating both sides with respect to x, we get
$\frac{dy}{dx} = \frac{\cos x \times \frac{d}{dx}\sin\left( x + 9 \right) - \sin\left( x + 9 \right) \times \frac{d}{dx}\cos x}{\cos^2 x} \left( \text{ Quotient rule } \right)$
$= \frac{\cos x \times \cos\left( x + 9 \right) - \sin\left( x + 9 \right) \times \left( - \sin x \right)}{\cos^2 x}$
$= \frac{\cos\left( x + 9 \right)\cos x + \sin\left( x + 9 \right)\sin x}{\cos^2 x}$
$= \frac{\cos\left( x + 9 - x \right)}{\cos^2 x}$
$= \frac{\cos9}{\cos^2 x}$
Putting x = 0, we get
$\left( \frac{dy}{dx} \right)_{x = 0} = \frac{\cos9}{\cos^2 0} = \cos9$
Thus, $\frac{dy}{dx}$ at x = 0 is cos 9.
Hence, the correct answer is option (a).
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Q 10 | Page 48 |
variance calculator with steps
Let ¯ X = 1 n ∑ Xi be the sample mean and s2 = 1 n − 1(∑ni = 1x2i − (∑ xi)2 n) be the sample variance. It will have a high variance if the numbers are very widely distributed. These two categories are numbers on the left of the mean and numbers on the right of the mean if your data is visualized in a number line. Need some help? You can calculate anything on Calculators.tech. $$\bar{x} = \dfrac{\sum x}{n} = \dfrac{42 + 48 + 30 + 36 + 46 + 53 + 62}{7} = \dfrac{317}{7} = 45.28$$. Find the mean value of the sample taken from the shop by adding all values dividing it by the total number of days. This means that the mean deviation is always zero, so that nothing tells how the results are distributed. You did not waste the time, though, because the standard deviation is the square root of the variance. A data sample is a collection of data from a population in statistics. The squared deviations cannot amount to zero and do not show any variability in the data. This means that the uncertainty or risk is often represented as SD rather than variances because the former is understood more easily. Let's use the formula for the population variance given above. You can find variance and standard deviation for your statistics problems and assignments on just one click. Calculator with step by step explanations to find standard deviation, variance, skewness and kurtosis. Step 3- Calculate inventory variance using the formula. Through analyzing the total numbers of apples sold in a store, we track the random results for seven days. The symbol μ is the arithmetic mean when analyzing a population. • Take a square of each result from the previous step. You probably know what it means in our daily routines. In this step, we will evaluate the expression $$\sum (x_i - \bar{x})^2$$, which is the numerator in the formula for sample variance. What is Meant by the Variance? Let C = 1 − α be the confidence coefficient. In statistics, the variance of a random variable is the average value of the squared distance from the average. For example, the narrow bell curve has a small variance in the normal distribution, and the wide bell curve has a large variance. Making all the deviations positive will ensure that summing up will not result in zero. First, write the sample data set that you have with you. For each data point in your sample, now you have the value $$(x_i - \bar{x}) 2$$. After entering the values in the input box, click the "Calculate" button to get the result. Standard Deviation Calculator with Step by Step Solution Step 3 – Add up All The Values in the First Column. Step 3: Finally, the variance for the given set of data will be displayed in the output field. The sample variation is denoted by s2 and is used to determine how different a sample is from the mean value. We can have an average voltage or current value for electronics. You will see the result for four values as soon as you click the button. However, remember that a sample is only a larger population estimate. you can contact us anytime. You can also use our Median, Mode and Range calculators. You can also use the population variance calculator above to calculate variance for your set of data. Samplel variance calculator uses the following formula to calculate the Variance(σ2). The variance is the square of standard deviation and denoted by the σ2 which is a Greek letter “sigma.”. It can be calculated by taking the average of the squared difference from the mean. Let's calculate the sample variance by using an example. In Mathematics, the variance is a measure which can be calculated from the set of data. Mean in general is the central value of a data set. The average and the mean are mathematically exactly the same: you add every value into a set and divide it by the total number of those items in the data set. If you are a student, you can use this tool to understand and solve the complex and lengthy variance problems. A More Operational Form. If you look at a set of 20 results and see only values of 8, 9, and 10 in the results, it is intuitively obvious that the average is about 9.
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