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#### Theory of Multivariable Linear Systems
Multivariable Linear Systems:
Row-Echelon Form:
Row-echelon form is mainly a stair-step pattern with the leading coefficients of one. Whenever a system is positioned into row-echelon form, back substitution is very simple. The bottom row provides the answer for z. That answer is back-substituted to the second equation and y is found. Then both y and z are replaced to the first equation and x is found.
Illustration: Use back substitution to resolve the given system:
x - y + 2z = 5
y - z = -1
z = 3
a) The bottom equation provides that z = 3.
b) Plugging z = 3 to the second equation provides y - 3 = -1 or y = 2.
c) Plugging y = 2 and z = 3 to the first equation gives x - 2 + 2(3) = 5.
d) Solving for x gives 1, therefore the solution is {(1, 2, 3)}.
Gaussian Elimination:
Gaussian Elimination is introduced by Carl Friedrich Gauss, a German mathematician who proved the fundamental theorem of algebra.
Two systems of equations are equal if they have similar solution set.
Elementary Operations:
There are three fundamental operations, termed as elementary operations, which can be executed and that will render the equivalent system.
a) Interchange two equations.
b) Multiply one equation by the non-zero constant.
c) Multiply an equation by the non-zero constant and add it to the other equation, substituting that equation.
The addition/elimination method uses such operations, they just weren't formalized. In the elimination method, we could switch two equations around and it wouldn't influence the solution set. We could multiply one or both equations by the non-zero constant and then add up them together. There is nothing wrong with multiplying both the equations by a non-zero constant, just that what are given here are the elementary or fundamental operations and multiplying both by a non-zero constant and adding up would be a combination of such.
The number of solutions to a linear system:
Just similar to with a 2×2 system of linear equations, a larger system can contain one, none, or many solutions.
Unique Solution:
It is a consistent and independent system. The solution is given as the ordered triplet.
No Solution:
It is an inconsistent system. There is no solution and the answer must be written as the null set or empty set, however not the set having the null set. This outcomes if at any point while trying to resolve the system, an equivalent form consists of a contradiction (that is, variables eliminate and the statement is false).
Many Solutions:
It is a consistent and dependent system. The solution is provided in parametric form. These case outcomes when there are less equation than variables and no contradictions.
Illustration: Solve the given system of linear equations.
3x - 2y + z = 1
y + 2z = 3
a) Take the second equation and resolve for y to obtain y = 3 - 2z.
b) Replace y = 3 - 2z to the first equation for y. This gives you 3x - 2(3 - 2z) + z = 1, that is an equation with the two variables, x and z. A little bit of simplification provides 3x - 6 + 4z + z = 1 or 3x + 5z = 7.
As you earlier resolved for y in terms of z, we now solve for x in terms of z. It is significant that we solve all the other variables in terms of similar variable. Solving 3x + 5z = 7 for x gives x = (7 - 5z)/3
Give the solution. As x and y is both in terms of a third variable, z, we assume z = t and obtain:
x = 1/3 (7 - 5t)
y = 3 - 2t
z = t
Mathematical Models:
When you have the right number of ordered pairs, you can fit a model by resolving a system a system of linear equations. To encompass a unique solution, you require as many points as constants.
Linear Model: y = Ax+B
There are two values which need to be found, A and B. Thus, it takes two points to find out an equation of a line.
Let us state that the line passes via the points (2, 3) and (5, 7). The resultant system of linear equations, obtained by replacing the values in for x and y is shown below. Just solve the system of linear equations, plug the values for A and B, and you have the model as:
3 = 2A + B
7 = 5A + B
Solving the above gives A = 4/3 and B = 1/3. The equation of the line passing via the given points is y = 4/3 x + 1/3.
Quadratic Model: y = Ax2 + Bx + C
This time, there are three variables: A, B, and C. Thus, it takes three non-collinear points to find out an equation of the parabola.
Let us state that the parabola passes via the points (2, 3), (5, 7), and (8, 4). Replace the values for x and y and the three linear equations which result the system which requires to be solved to determine A, B and C.
3 = 4A + 2B + C
7 = 25A + 5B + C
4 = 64A + 8B + C
When we are wondering where there 4, 25 and 64 came from, they are the values of x2 if x = 2, x = 5 and x = 8.
On solving the system: A = -7/18, B = 73/18, and C = -32/9. The equation is y = -7/18 x2 + 73/18 x - 32/9.
This process could be extended to determine the equation of a cubic passing via four points or the quartic equation passing via five points.
Circle: x2 + y2 + Dx + Ey + F = 0
We have three variables: D, E and F. Thus, it takes three non-collinear points to find out the equation of a circle.
Let us state the circle passes via the points (2, 3), (5, 7), and (8, 4). Replace the values for x and y and the equations which result give you the system of linear equations that can provide the coefficients which define the circle. You might find it simpler to move constant to the other side.
4 + 9 + 2D + 3E + F = 0 becomes 2D + 3E + F = -13
25 + 49 + 5D + 7E + F = 0 becomes 5D + 7E + F = -74
64 + 16 + 8D + 4E + F = 0 becomes 8D + 4E + F = -80
When we solve that system, we get D = -69/7, E = -55/7 and F = 212/7.
Replacing to the original model gives x2 + y2 -69/7 x - 55/7 y + 212/7 = 0.
Least Squares Regression Parabola:
When we have precisely three non-collinear points, then the parabola will pass via the points exactly. When there are more than three points, then you should fit a quadratic model to the data, however it won't essentially pass via all of the points. Remember that the least squares regression line went with the model y = ax + b.
b∑1 + a∑x = ∑y
b∑x + a∑x2 = ∑xy
Now, we are going to determine a system of equations which will fit the model y = ax2 + bx + c. This is an extension of the previous model. Note that it has similar pattern as before. Since you work from left to right on the left hand side, each and every summation consists of one additional x in it. Since you work from top to bottom, each and every term consists of one more x than the corresponding term in previous equation.
c∑1 + b∑x + a∑x2 = ∑y
c∑x + b∑x2 + a∑x3 = ∑xy
c∑x2 + b∑x3 + a∑x4 = ∑x2y
When we wanted to find out a least squares regression cubic, then we would require four variables and at least five points (that is, four points would precisely fit the model). Follow similar pattern as before, adding up one more variable (for d) and one more row.
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# Statistics: Basic Concepts: Line Graphs
#### Lesson 7: Line Graphs
/en/statistics-basic-concepts/pie-charts/content/
### Line graphs
#### What is a line graph?
So far we’ve talked about bar charts, which are made up of bars, and pie charts, which are made up of slices or sections to represent data. There is also another type of graph called a line graph, which is made up of a series of points.
These points can be joined to form a complete line, showing how one variable changes over time. The points are “plotted” on a graph which has an x-axis and a y-axis, not unlike a bar chart.
The x-axis and y-axis represent different variables. Oftentimes the independent variable” goes on the x-axis, and the “dependent variable” goes on the y-axis.
An independent variable is often something constant, like time, and the dependent variable is the variable that keeps changing, in relation to the other variable.
Quick Example: Maybe you want to measure how many centimeters your plant grows each month. Time is the independent variable, so months would go on the x-axis. Since the plant keeps getting taller, its height in centimeters would go on the y-axis.
#### Making a line graph
Let’s use another example to make a line graph…
City planners want to see how much a small city has grown in the last five years. This is their data set:
In the first year, 10 buildings were built. After two years, there were a total of 20 buildings. In three years: 35 buildings, four years: 50 buildings, and five years: 100 buildings in total.
• Step 1: The first step is to organize the data into a table. There is one column for the years, and one column for the number of buildings.
• Step 2: Next, the city planners start drawing their graph. Let’s put time as the variable on the x-axis, with an interval of 1. We should be able to include all five years.
• Step 3: The number of buildings will be on the y-axis. Since the y-axis starts at 0 and the largest value is 100, 25 might be a good interval (0, 25, 50, 75, and 100).
• Step 4: The planners use their table to plot each point on the graph. Look at the values on the x-axis, first (from Column 1), and then the corresponding value on the y-axis (from Column 2). Where these two values intersect determines the location of each data point.
• Step 5: Now the planners can connect each point in order to make the line graph. We can see how the number of buildings has gone up quite a bit in the last five years.
Next up, we’ll learn about another type of graph called a histogram.
/en/statistics-basic-concepts/histograms/content/ |
Hong Kong
Stage 4 - Stage 5
# Symmetrical and periodic nature of trig functions (radians)
Lesson
We say an object has a symmetry property if an aspect of it remains essentially the same after the object has been transformed in some systematic way.
From the unit circle definitions of the sine and cosine functions, we see that the function values repeat at intervals of $2\pi$2π. We say that these functions have a period of $2\pi$2π, meaning the function value at some number $x$x is always the same as the value at $x+2\pi$x+2π. That is, $\sin\left(x+2\pi\right)=\sin x$sin(x+2π)=sinx and $\cos\left(x+2\pi\right)=\cos x$cos(x+2π)=cosx
We say the sine and cosine functions are symmetrical under a translation by $2\pi$2π.
There are other important symmetries possessed by these two functions. Again, by looking at the unit circle diagram, we observe that
$\sin\left(-x\right)=-\sin x$sin(x)=sinx and
$\cos\left(-x\right)=\cos x$cos(x)=cosx
Any function that has the property $f\left(-x\right)=-f\left(x\right)$f(x)=f(x) is called an odd function. Thus, sine is an odd function. When it is represented by means of a graph, one can see that the picture will look the same if the graph is rotated about the origin by $180^\circ$180°. This property is characteristic of odd functions.
Any function that has the property $f\left(-x\right)=f\left(x\right)$f(x)=f(x), is called an even function. Thus, the cosine function is an even function. The graph of any even function is the same as its reflection about the vertical axis.
We can check that the tangent function is an odd function. We make use of the unit circle definition of the tangent function:
$\tan\left(-x\right)=\frac{\sin\left(-x\right)}{\cos\left(-x\right)}=\frac{-\sin x}{\cos x}=-\frac{\sin x}{\cos x}=-\tan x$tan(x)=sin(x)cos(x)=sinxcosx=sinxcosx=tanx
The tangent function also has translational symmetry. It has a period of $\pi$π. We can verify from the unit circle diagram or from the graphs of sine and cosine that $\sin\left(x+\pi\right)=-\sin x$sin(x+π)=sinx and $\cos\left(x+\pi\right)=-\cos x$cos(x+π)=cosx. This means that $\tan\left(x+\pi\right)=\frac{\sin\left(x+\pi\right)}{\cos\left(x+\pi\right)}=\frac{-\sin x}{-\cos x}=\tan x$tan(x+π)=sin(x+π)cos(x+π)=sinxcosx=tanx for all values of $x$x and this is the required condition.
#### Worked examples
##### Question 1
Examine the graph of $y=\sin x$y=sinx.
1. How long is one cycle of the graph?
2. State the $x$x values for which $\sin x=0$sinx=0, from $x=0$x=0 to $x=2\pi$x=2π inclusive.
3. State the first $x$x value for which $\sin x=0.5$sinx=0.5
4. Using the symmetry of the graph, for what other value of $x$x shown on the graph does $\sin x=0.5$sinx=0.5?
5. Using the symmetry of the graph, for what values of $x$x does $\sin x=-0.5$sinx=0.5?
##### Question 2
Examine the graph of $y=\tan x+2$y=tanx+2.
1. How long is one cycle of the graph?
2. State the $x$x values for which $\tan x+2=2$tanx+2=2, from $x=-2\pi$x=2π to $x=2\pi$x=2π inclusive.
Write all answers on the same line separated by commas.
3. State the first positive $x$x value for which $\tan x+2=3$tanx+2=3
4. Using the period of the graph, for what other values of $x$x between $x=-2\pi$x=2π and $x=2\pi$x=2π does $\tan x+2=3$tanx+2=3?
Write all answers on the same line separated by commas.
5. For what values of $x$x between $x=-2\pi$x=2π and $x=2\pi$x=2π does $\tan x+2=1$tanx+2=1?
##### Question 3
Examine the graph of $y=\cos x$y=cosx.
1. State the exact value of $\cos\frac{\pi}{6}$cosπ6.
2. Use the graph to determine all other values of $x$x between $x=-\pi$x=π and $x=\pi$x=π for which $\cos x=\pm\frac{\sqrt{3}}{2}$cosx=±32. |
# Rules of Differentiation of Functions in Calculus
The basic rules of Differentiation of functions in calculus are presented along with several examples .
## 1 - Derivative of a constant function.
The derivative of f(x) = c where c is a constant is given by
f '(x) = 0
Example
f(x) = - 10 , then f '(x) = 0
## 2 - Derivative of a power function (power rule).
The derivative of f(x) = x
r where r is a constant real number is given by
f '(x) = r x r - 1
Example
f(x) = x
-2 , then f '(x) = -2 x -3 = -2 / x 3
## 3 - Derivative of a function multiplied by a constant.
The derivative of f(x) = c g(x) is given by
f '(x) = c g '(x)
Example
f(x) = 3x
3 ,
let c = 3 and g(x) = x
3, then f '(x) = c g '(x)
= 3 (3x
2) = 9 x 2
## 4 - Derivative of the sum of functions (sum rule).
The derivative of f(x) = g(x) + h(x) is given by
f '(x) = g '(x) + h '(x)
Example
f(x) = x
2 + 4
let g(x) = x
2 and h(x) = 4, then f '(x) = g '(x) + h '(x) = 2x + 0 = 2x
## 5 - Derivative of the difference of functions.
The derivative of f(x) = g(x) - h(x) is given by
f '(x) = g '(x) - h '(x)
Example
f(x) = x
3 - x -2
let g(x) = x
3 and h(x) = x -2, then
f '(x) = g '(x) - h '(x) = 3 x
2 - (-2 x -3) = 3 x 2 + 2x -3
## 6 - Derivative of the product of two functions (product rule).
The derivative of f(x) = g(x) h(x) is given by
f '(x) = g(x) h '(x) + h(x) g '(x)
Example
f(x) = (x 2 - 2x) (x - 2)
let g(x) = (x 2 - 2x) and h(x) = (x - 2), then
f '(x) = g(x) h '(x) + h(x) g '(x) = (x 2 - 2x) (1) + (x - 2) (2x - 2)
= x 2 - 2x + 2 x 2 - 6x + 4 = 3 x 2 - 8x + 4
## 7 - Derivative of the quotient of two functions (quotient rule).
The derivative of f(x) = g(x) / h(x) is given by
f '(x) = ( h(x) g '(x) - g(x) h '(x) ) / h(x) 2
Example f(x) = (x - 2) / (x + 1)
let g(x) = (x - 2) and h(x) = (x + 1), then
f '(x) = ( h(x) g '(x) - g(x) h '(x) ) / h(x) 2
= ( (x + 1)(1) - (x - 2)(1) ) / (x + 1) 2
= 3 / (x + 1) 2 |
# Problems on Complex Numbers
We will learn step-by-step how to solve different types of problems on complex numbers using the formulas.
1. Express $$(\frac{1 + i}{1 - i})^{3}$$ in the form A + iB where A and B are real numbers.
Solution:
Given $$(\frac{1 + i}{1 - i})^{3}$$
Now $$\frac{1 + i}{1 - i}$$
= $$\frac{(1 + i)(1 + i)}{(1 - i)(1 + i)}$$
= $$\frac{(1 + i)^{2}}{(1^{2} - i^{2}}$$
= $$\frac{1 + 2i + iˆ{2}}{1 - (-1)}$$
= $$\frac{1 + 2i - 1}{2}$$
= $$\frac{2i}{2}$$
= i
Therefore, $$(\frac{1 + i}{1 - i})^{3}$$ = i$$^{3}$$= i$$^{2}$$ ∙ i = - i = 0 + i (-1), which is the required form A + iB where A = 0 and B = -1.
2. Find the modulus of the complex quantity (2 - 3i)(-1 + 7i).
Solution:
The given complex quantity is (2 - 3i)(-1 + 7i)
Let z$$_{1}$$ = 2 - 3i and z$$_{2}$$ = -1 + 7i
Therefore, |z$$_{1}$$| = $$\sqrt{2^{2} + (-3)^{2}}$$ = $$\sqrt{4 + 9}$$ = $$\sqrt{13}$$
And |z$$_{2}$$| = $$\sqrt{(-1)^{2} + 7^{2}}$$ = $$\sqrt{1 + 49}$$ = $$\sqrt{50}$$ = 5$$\sqrt{2}$$
Therefore, the required modulus of the given complex quantity = |z$$_{1}$$z$$_{1}$$| = |z$$_{1}$$||z$$_{1}$$| = $$\sqrt{13}$$ ∙ 5$$\sqrt{2}$$ = 5$$\sqrt{26}$$
3. Find the modulus and principal amplitude of -4.
Solution:
Let z = -4 + 0i.
Then, modulus of z = |z| = $$\sqrt{(-4)^{2} + 0^{2}}$$ = $$\sqrt{16}$$ = 4.
Clearly, the point in the z-plane the point z = - 4 + 0i = (-4, 0) lies on the negative side of real axis.
Therefore, the principle amplitude of z is π.
4. Find the amplitude and modulus of the complex number -2 + 2√3i.
Solution:
The given complex number is -2 + 2√3i.
The modulus of -2 + 2√3i = $$\sqrt{(-2)^{2} + (2√3)^{2}}$$ = $$\sqrt{4 + 12}$$ = $$\sqrt{16}$$ = 4.
Therefore, the modulus of -2 + 2√3i = 4
Clearly, in the z-plane the point z = -2 + 2√3i = (-2, 2√3) lies in the second quadrant. Hence, if amp z = θ then,
tan θ = $$\frac{2√3}{-2}$$ = - √3 where, $$\frac{π}{2}$$ < θ ≤ π.
Therefore, tan θ = - √3 = tan (π - $$\frac{π}{3}$$) = tan $$\frac{2π}{3}$$
Therefore, θ = $$\frac{2π}{3}$$
Therefore, the required amplitude of -2 + 2√3i is $$\frac{2π}{3}$$.
5. Find the multiplicative inverse of the complex number z = 4 - 5i.
Solution:
The given complex number is z = 4 - 5i.
We know that every non-zero complex number z = x +iy possesses multiplicative inverse given by
$$(\frac{x}{x^{2} + y^{2}}) + i (\frac{-y}{x^{2} + y^{2}})$$
Therefore, using the above formula, we get
z$$^{-1}$$ = $$(\frac{4}{4^{2} + (-5)^{2}}) + i (\frac{-(-5)}{4^{2} + (-5)^{2}})$$
= $$(\frac{4}{16 + 25}) + i (\frac{5)}{16 + 25})$$
= $$(\frac{4}{41}) + (\frac{5}{41})$$i
Therefore, the multiplicative inverse of the complex number z = 4 - 5i is $$(\frac{4}{41}) + (\frac{5}{41})$$i
6. Factorize: x$$^{2}$$ + y$$^{2}$$
Solution:
x$$^{2}$$ - (-1) y$$^{2}$$ = x$$^{2}$$ - i$$^{2}$$y$$^{2}$$ = (x + iy)(x - iy)
` |
# Solution to a “nice analysis question for RMO practice”
The question from a previous blog is re-written here for your convenience.
Question:
How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?
Solution:
The figure below shows how two and three cards can be stacked so that the mass of cards is equal on either side of the vertical line passing through the corner of table’s edge in order to just balance them under gravity:
the set of first two cards are arranged as follows (the horizontal lines represents the cards):
$xxxxxxxxxxxxxxxx\line(5,0){170}$
$\line(5,0){150}xxxxxxxxxxxxxxxxxxx$
the set of three cards are arranged as follows:
$xxxxxxxxxxxxxxxxxxxxxx\line(5,0){150}$
$xxxxxxxxxxxxx\line(5,0){150}$
$\line(5,0){150}xxxxxxxxxxxxxxxxxxxxxx$
We can see that the length of the overhand is a harmonic series of even numbers multiplied by the length of one card, L.
Overhand distance is $(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{52 \times 2})L$ for 52 cards.
It may be noted that the series if continued to infinity leads to $H_{\infty}^{E}$.
That is, $H_{\infty}^{E}=\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots$
This series is known to diverge as proved below:
First consider, $H_{\infty}=1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}+ \ldots$, which is, greater than
$1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}+ \ldots$, which is greater than $1+ \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \ldots$. Hence, $H_{\infty}$ diverges as we go on adding 1/2 indefinitely.
Now, let $H_{E}=\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots = \frac{1}{2}(1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots)=\frac{1}{2}H_{\infty}$
Since $H_{\infty}$ diverges, $H_{E}$also diverges.
Hence, the “overhang series” also diverges.
This means that the cards can be stacked indefinitely and the overhang distance can reach infinity. However, this will happen very slowly as shown in the table below:
$\begin{array}{cc} n^{E} & H_{n}^{E}\\ 2 & 0.5 \\ 10 & 1.46 \\ 100 & 2.59 \\ 1000 & 3.74 \\ 10000 & 4.89 \\ 100000 & 6.05 \end{array}$
Computing the number of cards that completely overhang off the table needs information about the overhang distance for different number of cards. As shown below in the figure, four cards are required to have one card completely away from the edge of the table. This is because $(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8}=1.0417 >1)$.
(the set of four cards are arranged as follows:)
$xxxxxxxxxxxxxxxxxxxxxxxxxxxx\line(5,0){150}$
$xxxxxxxxxxxxxxxx\line(5,0){150}$
$xxxxxxxxxx\line(5,0){150}$
$xxxxx\line(5,0){150}$
We can see that the length of the overhang is a harmonic series of even numbers multiplied by the length of one card, L:
Overhang distance = $(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{52 \times 2})L$ for 52 cards
It may be noted that the series if continued to infinity, leads to $H_{\infty}^{E}$
$H_{\infty}^{E} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots$
This series is known to diverge. This means that the cards can be stacked indefinitely and the overhang can reach infinity. However, this will happen very slowly as shown in the table below:
$\begin{array}{cc} n^{E} & H_{\infty}^{E} \\ 2 & 0.5 \\ 10 & 1.46 \\ 100 & 2.59 \\ 1000 & 3.74 \\ 10000 & 4.89 \\ 100000 & 6.05 \end{array}$
Computing the number of cards that completely overhang off the table needs information about the overhang distance for different numbers of cards. As shown in the above schematic figures of cards with overhangs, four cards are required to have one card completely away from the edge of the table. This is because
$(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8})=1.0417>1$
For the second card to overhang completely, leaving the first card (and hence one half) that is already completely overhung, it is now necessary that
$(\frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{2n})>1$, or
$(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{2n} )>1+ \frac{1}{2}$
where n needs to be found out. By generating some more data, we can find the value of n to be 11.
For third overhanging card, we need
$(\frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{2n})>1$ or
$(\frac{1}{2} + \frac{1}{4} + \frac{1}{6}+\frac{1}{8}+ \ldots + \frac{1}{2n})>1+\frac{1}{2} + \frac{1}{4}$
Thus, for m completely overhanging cards, we find n such that $H_{2n}^{E} > 1+ H_{2(m-1)}^{E}$
The table below shows these values wherein we see an approximate pattern of arithmetic progression by 7.
$\begin{array}{cccc} m & n & m & n \\ 1 & 4 & 11 & 78 \\ 2 & 11 & 12 & 85 \\ 3 & 19 & 13 & 92 \\ 4 & 26 & 14 & 100 \\ 5 & 33 & 15 & 107 \\ 6 & 41 & 16 & 115 \\ 7 & 48 & 17 & 122 \\ 8 & 55 & 18 & 129 \\ 9 & 63 & 19 & 137 \\ 10 & 70 & 20 & 144 \end{array}$
By examining the pattern in the table, we can get a simple rule to estimate the number of completely overhanging number of cards m, with an error of utmost one, for n cards stacked.
$m = round(\frac{n}{7.4})=round(\frac{10n}{74})$.
Reference:
Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books.
Hope you enjoyed the detailed analysis…
More later,
Nalin Pithwa
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## Chinese Remainder Theorem Part 2 – Non Coprime Moduli
As promised on the last post, today we are going to discuss the “Strong Form” of Chinese Remainder Theorem, i.e, what do we do when the moduli in the congruence equations are not pairwise coprime. The solution is quite similar to the one we have already discussed in the previous post, so hopefully, it will be a lot easier to understand.
### Prerequisite
If you haven’t read my previous post, you can find it here: Chinese Remainder Theorem Part 1. I strongly suggest you read that one before proceeding onwards.
# Problem
Given two sequences of numbers $A = [a_1, a_2, \dots, a_n]$ and $M = [m_1, m_2, \dots, m_n]$, find the smallest solution to the following linear congruence equations if it exists:
$$x \equiv a_1 \text{(mod } m_1) \\ x \equiv a_2 \text{(mod } m_2) \\ \dots \\ x \equiv a_n \text{(mod } m_n)$$
Note: The elements of $M$ are not pairwise coprime
# Working With Two Equations
Before we go on to deal with $n$ equations, let us first see how to deal with two equations. Just like before, we will try to merge two equations into one.
So given the two equation $x \equiv a_1 \text{(mod }m_1)$ and $x \equiv a_2 \text{(mod }m_2)$, find the smallest solution to $x$ if it exists.
## Existance of Solution
Suppose, $gcd(m_1,m_2) = g$. In that case, the following constraint must be satisfied: $a_1 \equiv a_2 \text{(mod }g)$, otherwise there does not exist any solution to $x$. Why is that?
We know that, $x \equiv a_1 \text{(mod }m_1)$
or, $x – a_1 \equiv 0 \text{(mod }m_1)$
or, $m_1 | x – a_1$
Since $m_1$ divides $x – a_1$, any divisor of $m_1$ also divides $x – a_1$, including $g$.
$\therefore x – a_1 \equiv 0 \text{(mod }g)$.
Similarly, we can show that, $x – a_2 \equiv 0 \text{(mod }g)$ is also true.
$\therefore x – a_1 \equiv x – a_2 \text{(mod }g)$
or, $a_1 \equiv a_2 \text{(mod }g)$
or, $a_1 – a_2 \equiv \text{(mod }g)$
or $g | (a_1 – a_2)$
Therefore, $g$ must divide $a_1 – a_2$, otherwise, there is conflict and no solution exists. From this point, we assume that $g | a_1 – a_2$.
## Finding Solution
Just like last time, we use Bezout’s identity to find a solution to $x$. From Bezout’s identity, we know that there exists a solution to the following equations:
$m_1p + m_2q = g$
Since both side is divisible by $g$, we can rewrite it as:
$$\frac{m_1}{g}p + \frac{m_2}{g}q = 1$$
Using Extended Euclidean Algorithm, we can easily find values of $p$ and $q$. Simply call the function ext_gcd(m1/g, m2/g, p, q) and it should be calculated automatically.
Once we know the value of $p$ and $q$, we can claim that the solution to $x$ is the following:
$$x = a_1\frac{m_2}{g}q + a_2\frac{m_1}{g}p$$
Why is that? Look below.
$$x = a_1\frac{m_2}{g}q + a_2\frac{m_1}{g}p \\ \text{From Bezout’s Identity, we know that } \frac{m_1}{g}p = 1 – \frac{m_2}{g}q \\ \text{so, } x = a_1\frac{m_2}{g}q + a_2(1 – \frac{m_2}{g}q) \\ \text{or, } x = a_2 + (a_1 – a_2)\frac{m_2}{g}q \\ \text{Since, } g | (a_1 – a_2) \\ x = a_2 + \frac{a_1 – a_2}{g}m_2q \\ \therefore x \equiv a_2 \text{(mod }m_2) \\ \text{Similarly, } x \equiv a_1 \text{(mod }m_1)$$
Hence, $x = a_1\frac{m_1}{g}q + a_2\frac{m_2}{g}p$ is a valid solution. It may not be the smallest solution though.
## Finding Smallest Solution
Let $x_1$ and $x_2$ be adjacent two solutions. Then the following are true:
$x_1 \equiv a_1 \text{(mod }m_1)$
$x_2 \equiv a_1 \text{(mod }m_1)$
$x_1 \equiv x_2 \text{(mod }m_1)$
$x_1 – x_2 \equiv 0 \text{(mod }m_1)$
$m_1 | x_1 – x_2$
Similarly, $m_2 | x_1 – x_2$
Since both $m_1$ and $m_2$ divides $x_1 – x_2$, we can say that $LCM(m_1, m_2)$ also divides $x_1-x_2$.
Since any two adjacent solution of $x$ is divisible by $L = LCM(m_1,m_2)$, then there cannot be more than one solution that lies on the range $0$ to $L-1$. Hence, the following is the smallest solution to $x$:
$$x \equiv a_1\frac{m_2}{g}q + a_2\frac{m_1}{g}p \text{(mod }LCM(m_1,m_2))$$
# Working with $n$ Equations
When there are $n$ equations, we simply merge two equations at a time until there is only one left. The last remaining equation is our desired answer. If at any point, if we are unable to merge two equations due to conflict, i.e, $a_i \not\equiv a_j \text{ (mod }GCD(m_i, m_j))$, then there is no solution.
# Chinese Remainder Theorem: Strong Form
Given two sequences of numbers $A = [a_1, a_2, \dots, a_n]$ and $M = [m_1, m_2, \dots, m_n]$, a solution to $x$ exists for the following $n$ congrunce equations:
$$x \equiv a_1 \text{(mod } m_1) \\ x \equiv a_2 \text{(mod } m_2) \\ \dots \\ x \equiv a_n \text{(mod } m_n)$$
if, $a_i \equiv a_j \text{ (mod }GCD(m_i,m_j))$ and the solution will be unique modulo $L = LCM(m_1, m_2,\dots, m_n)$.
# Code
The code is almost same as weak form, with slight changes in few lines.
/** Works for non-coprime moduli.
Returns {-1,-1} if solution does not exist or input is invalid.
Otherwise, returns {x,L}, where x is the solution unique to mod L
*/
pair<int, int> chinese_remainder_theorem( vector<int> A, vector<int> M ) {
if(A.size() != M.size()) return {-1,-1}; /** Invalid input*/
int n = A.size();
int a1 = A[0];
int m1 = M[0];
/** Initially x = a_0 (mod m_0)*/
/** Merge the solution with remaining equations */
for ( int i = 1; i < n; i++ ) {
int a2 = A[i];
int m2 = M[i];
int g = __gcd(m1, m2);
if ( a1 % g != a2 % g ) return {-1,-1}; /** No solution exists*/
/** Merge the two equations*/
int p, q;
ext_gcd(m1/g, m2/g, &p, &q);
int mod = m1 / g * m2; /** LCM of m1 and m2*/
/** We need to be careful about overflow, but I did not bother about overflow here to keep the code simple.*/
int x = (a1*(m2/g)*q + a2*(m1/g)*p) % mod;
/** Merged equation*/
a1 = x;
if (a1 < 0) a1 += mod; /** Result is not suppose to be negative*/
m1 = mod;
}
return {a1, m1};
}
Once again, please note that if you use int data type, then there is a risk of overflow. In order to keep the code simple I used int, but you obviously need to modify it to suit your needs. You can have a look at my CRT template on github from here: github/forthright48 - chinese_remainder_theorem.cpp
Complexity: $O(n \times log(L))$.
# Conclusion
And we are done. We can now solve congruence equations even when the moduli are not pairwise coprime.
The first time I learned Chinese Remainder Theorem, I had to read a paper. The paper was well written, but I never actually managed to understand how it worked. Therefore, I always feared Chinese Remainder Theorem and used it as a black box. Until last week, I didn't even know that CRT worked with moduli that are not coprime. I really enjoyed learning how it works and all the related proof. Hopefully, you enjoyed it too.
# Resources
1. forthright48 - Chinese Remainder Theorem Part 1 https://forthright48.com/2017/11/15/chinese-remainder-theorem-part-1-coprime-moduli/
2. Wiki - Chinese Remainder Theorem - https://en.wikipedia.org/wiki/Chinese_remainder_theorem
3. Brilliant.org - Chinese Remainder Theorem - https://brilliant.org/wiki/chinese-remainder-theorem/
4. Cut The Knot - Chinese Remainder Theorem - https://www.cut-the-knot.org/blue/chinese.shtml
# Related Problems
## Chinese Remainder Theorem Part 1 – Coprime Moduli
Second part of the series can be found on: Chinese Remainder Theorem Part 2 – Non Coprime Moduli
Wow. It has been two years since I published my last post. Time sure flies by quickly. I have been thinking about resuming writing again for a while now. Took me long enough to get back into the mood.
I am really excited about today’s post. I have been meaning to write on Chinese Remainder Theorem (CRT for short) for over two years now. Hopefully, everyone will find it interesting and easy to understand.
Here are the pre-requisite that you need to complete for understanding the post properly:
# An Old Woman Goes to Market
I remember reading the following story the first time I tried learning about CRT. It’s quite interesting in my opinion. Kind of brings out the essence of CRT in a simple to understand fashion.
An old woman goes to market and horse steps on her basket and crashes the eggs. The rider offers to pay the damages and asks her how many eggs she had brought. She does not remember the exact number, but when she had taken them out two at a time, there was one egg left. The same happened when she picked 2,3,4,5,6 at a time, but when she took seven at a time they came out even. What is the smallest number of eggs she could have had?
Hopefully, everyone understood the story above. Let me now rephrase it in mathematical notations.
Suppose, the old woman had $x$ eggs in her basket. Now she claims that when she took out eggs from the basket $2$ at a time, there was only $1$ egg remaining, meaning, $x \equiv 1 \text{(mod 2)}$. So basically, she is giving us various congruence equations. From her statement, we get the following linear congruence equations:
$$x \equiv 1 \text{(mod 2)} \\ x \equiv 1 \text{(mod 3)} \\ x \equiv 1 \text{(mod 4)} \\ x \equiv 1 \text{(mod 5)} \\ x \equiv 1 \text{(mod 6)} \\ x \equiv 2 \text{(mod 7)}$$
Now, we need to find the smallest value of $x$ which satisfies all of the above congruence equations. Chinese Remainder Theorem helps us in finding the value of $x$.
Before we move on, let us redefine the problem formally.
# Problem: Formal Definition
Given two sequences of numbers $A = [a_1, a_2, \dots, a_n]$ and $M = [m_1, m_2, \dots, m_n]$, find the smallest solution to the following linear congruence equations if it exists:
$$x \equiv a_1 \text{(mod } m_1) \\ x \equiv a_2 \text{(mod } m_2) \\ \dots \\ x \equiv a_n \text{(mod } m_n)$$
# Chinese Remainder Theorem: Weak Form
As mentioned before, we can use Chinese Remainder Theorem to solve the above problem described. On this part of the CRT series, we will look into a weaker form of Chinese Remainder Theorem, which is easier to understand and occurs more frequently.
The weak Chinese Remainder Theorem states the following:
Given two sequences of numbers $A = [a_1, a_2, \dots, a_n]$ and $M = [m_1, m_2, \dots, m_n]$, where all elements of $M$ are pairwise coprime, there always exists an unique solution to $x$ mod $L$, where $L = m_1 \times m_2 \times \dots \times m_n$, such that $x$ satisfies the following linear congruence equations:
$$x \equiv a_1 \text{(mod } m_1) \\ x \equiv a_2 \text{(mod } m_2) \\ \dots \\ x \equiv a_n \text{(mod } m_n)$$
So the weak form of Chinese Remainder Theorem has a constraint: members of the array $M$ must be pairwise coprime. What do we mean by that? This means that $\text{GCD}(m_i,m_j) = 1$ when $i \neq j$.
As long as this condition is satisfied, the weak form of CRT state that a solution to $x$ always exists which is “unique mod $L$”. Let us see an example to understand what it means to be “unique mod $L$”.
## Weak Form of CRT: Example
Suppose we are given the following three congruence equations:
$$x \equiv 3 (\text{mod } 5) \\ x \equiv 2 (\text{mod } 7) \\ x \equiv 2 (\text{mod } 8)$$
Here we have $A=[3,2,2]$, $M=[5,7,8]$ and $L = 5 \times 7 \times 8 = 280$. Using Weak form of Chinese Remainder Theorem, we can find that $x \equiv 58 \text{(mod } 280)$. How did we find this? We will see that later.
Before you continue, please verify yourself that this indeed satisfies the given congruence. Also, $x = 58$ is the smallest solution and there exists more than one solution. $x = 58 + 280 \times k$, where $k$ is any non-negative integer, also satisfies the equations. But as we said before, $x$ is unique when you mod all the solutions with $L$. This is what it means to have a solution unique to mod $L$.
## Weak Form of CRT: Finding a Solution
We are first going to see how to solve when there are just two equations. Once we know how to solve for two equations, we will then generalize the method for $n$ equations.
### When there are just two equations
Let us first just consider two equation: $x \equiv a_1 \text{(mod }m_1)$ and $x \equiv a_2 \text{(mod }m_2)$. What we are going to do is merge these two equations into one equation. Here is how it works.
Since $gcd(m_1,m_2) = 1$, from Bezout’s Identity, we know that there exists a solution to the following equation:
$$m_1p + m_2q = 1 \tag{1} \label{1}$$
Using Extended Euclidean Algorithm, we can find the value of $p$ and $q$. If we know the value of $p$ and $q$, then we can say that:
$x = a_1 m_2 q + a_2 m_1 p \text{ (mod }m_1m_2) tag{2}$
See below to understand why so.
$x = a_1 m_2 q + a_2 m_1 p$
$x = a_1 (1 – m_1 p) + a_2 m_1 p$ (Using $\eqref{1}$)
$x = a_1 – a_1 m_1 p + a_2 m_1 p$
$x = a_1 + (a_2 – a_1) m_1 p$
$\therefore x \equiv a_1 \text{(mod }m_1)$
Similarly, $x = a_1 m_2 q + a_2 m_1 p \equiv a_2 \text{(mod } m_2)$
Great! We now have a possible solution for $x$, but why did we mod the solution with $m_1m_2$? The solution we found may or may not be the smallest. Like we have seen before, there could be infinite solutions, but there exists a solution which is unique to mod $m_1 \times m_2$. How so? I guess this is the perfect time to look into its proof.
#### Proof of Uniqueness
Since there could be infinite solutions, let $x_1$ and $x_2$ be two such solution. Hence we can say the following:
$x_1 \equiv a_1 \text{(mod } m_1)$
$x_2 \equiv a_1 \text{(mod } m_1)$
$\therefore x_1 \equiv x_2 \text{(mod } m_1)$
$x_1 – x_2 \equiv 0 \text{(mod } m_1)$
$\therefore m_1 | x_1 – x_2$
Similarly, we can show that $m_2 | x_1 – x_2$.
What can we do with all these information? The difference of any two solutions $x_1$ and $x_2$ is divisible by both $m_1$ and $m_2$. We also know that, $m_1$ and $m_2$ are coprime. Doesn’t that mean, the difference is also divisible by $m_1 \times m_2$? Yes. That’s exactly what we are going towards
$m_1m_2 | x_1 – x_2$
$x_1 – x_2 \equiv 0 \text{(mod } m_1m_2)$
$x_1 \equiv x_2 \text{(mod } m_1m_2)$
The next question I will ask is: if the difference between any two solutions, $x_1$ and $x_2$, is divisible by $m_1m_2$, then how many solution can there exist in the range $0$ to $m_1m_2 – 1$? Only one. And hence, the solution $x$ is unique to modulo $m_1m_2$.
Therefore we can say that the solution $x \equiv a_1 m_2 q + a_2 m_1 p \text{ (mod }m_1m_2)$ is unique and smallest. Hence, the two equation, $x \equiv a_1 \text{(mod }m_1)$ and $x \equiv a_2 \text{(mod }m_2)$, after merging becomes $x \equiv a_1 m_2 q + a_2 m_1 p \text{ (mod }m_1m_2)$
### When there are $n$ equations
Since we know how to merge two equations into one, all we need to do is take the first two equations from $n$ equations and merge them. Then we are left with $n-1$ equations. Since all the elements of $M$ were pairwise coprime, the new modulus is also coprime. Hence, we can continue to merge the equations until there is only one equation left. The equation left is our answer.
## Weak form of CRT: Code
The following code implements the idea we discussed above. Note that I used int data type, but in most of the problems you will most likely require long long. I am sure you can adjust the code accordingly yourself.
/** Return {-1,-1} if invalid input.
Otherwise, returns {x,L}, where x is the solution unique to mod L
*/
pair<int, int> chinese_remainder_theorem( vector<int> A, vector<int> M ) {
if(A.size() != M.size()) return {-1,-1}; /** Invalid input*/
int n = A.size();
int a1 = A[0];
int m1 = M[0];
/** Initially x = a_0 (mod m_0)*/
/** Merge the solution with remaining equations */
for ( int i = 1; i < n; i++ ) {
int a2 = A[i];
int m2 = M[i];
/** Merge the two equations*/
int p, q;
ext_gcd(m1, m2, &p, &q);
/** We need to be careful about overflow, but I did not bother about overflow here to keep the code simple.*/
int x = (a1*m2*q + a2*m1*p) % (m1*m2);
/** Merged equation*/
a1 = x;
m1 = m1 * m2;
}
if (a1 < 0) a1 += m1; /** Result is not suppose to be negative*/
return {a1, m1};
}
Once again, please be careful about overflow when solving a problem. From my experience, I have seen that the value of $p$ and $q$ becomes large quickly and intermediate calculations no longer fit into long long variables. I use __int128 data type to avoid overflow issues. So if you get “Wrong Answer”, it is most likely due to overflow.
Complexity: $O(n \times log(L))$
# Conclusion
With this, we are done with Weak Form of Chinese Remainder Theorem. We can now find a solution to congruence equations when the moduli are co-prime. On next post, we will see a stronger version of Chinese Remainder Theorem, where the moduli are not co-prime.
Edit: The next post can be found here: Chinese Remainder Theorem Part 2 – Non Coprime Moduli
# Resources
1. Wiki – Chinese Remainder Theorem – https://en.wikipedia.org/wiki/Chinese_remainder_theorem
2. Brilliant.org – Chinese Remainder Theorem – https://brilliant.org/wiki/chinese-remainder-theorem/
3. Cut The Knot – Chinese Remainder Theorem – https://www.cut-the-knot.org/blue/chinese.shtml
# Related Problems
That’s all I could find on CRT. If you know about any other related problem, please feel free to add it as a comment.
## Modular Inverse from 1 to N
We already learned how to find Modular Inverse for a particular number in a previous post, “Modular Multiplicative Inverse“. Today we will look into finding Modular Inverse in a bulk.
# Problem
Given $N$ and $M$ ( $N < M$ and $M$ is prime ), find modular inverse of all numbers between $1$ to $N$ with respect to $M$.
Since $M$ is prime and $N$ is less than $M$, we can be sure that Modular Inverse exists for all numbers. Why? Cause prime numbers are coprime to all numbers less than them.
We will look into two methods. Later one is better than the first one.
# $O(NlogM)$ Solution
Using Fermat’s little theorem, we can easily find Modular Inverse for a particular number.
$A^{-1} \ \% \ M = bigmod(A,M-2,M)$, where $bigmod()$ is a function from the post “Repeated Squaring Method for Modular Exponentiation“. The function has complexity of $O(logM)$. Since we are trying to find inverse for all numbers from $1$ to $N$, we can find them in $O(NlogM)$ complexity by running a loop.
int inv[SIZE]; ///inv[x] contains value of (x^-1 % m)
for ( int i = 1; i <= n; i++ ) {
inv[i] = bigmod ( i, m - 2, m );
}
But it's possible to do better.
# $O(N)$ Solution
This solution is derived using some clever manipulation of Modular Arithmetic.
Suppose we are trying to find the modular inverse for a number $a$, $a < M$, with respect to $M$. Now divide $M$ by $a$. This will be the starting point.
$M = Q \times a + r$, (where $Q$ is the quotient and $r$ is the remainder)
$M = \lfloor \frac{M}{a} \rfloor \times a + (M \ \% \ a )$
Now take modulo $M$ on both sides.
$0 \equiv \lfloor \frac{M}{a} \rfloor \times a + (M \ \% \ a ) \ \ \ \text{(mod M )}$
$(M \ \% \ a ) \equiv -\lfloor \frac{M}{a} \rfloor \times a \ \ \ \text{(mod M )}$
Now divide both side by $a \times ( M \ \% \ a )$.
$$\frac{M \ \% \ a}{a \times ( M \ \% \ a )} \equiv \frac{- \lfloor \frac{M}{a} \rfloor \times a } { a \times ( M \ \% \ a ) } \ \ \ \text{(mod M)} \\ \therefore a^{-1} \equiv - \lfloor \frac{M}{a} \rfloor \times ( M \ \% \ a )^{-1} \ \ \ \text{(mod M)}$$
The formula establishes a recurrence relation. The formula says that, in order to find the modular inverse of $a$, we need to find the modular inverse of $b = M \ \% \ a$ first.
Since $b = M \ \% \ a$, we can say that its value lies between $0$ and $a-1$. But, $a$ and $M$ are coprime. So $a$ will never fully divide $M$. Hence we can ignore the possibility that $b$ will be $0$. So possible values of $b$ is between $1$ and $a-1$.
Therefore, if we have all modular inverse from $1$ to $a-1$ already calculated, then we can find the modular inverse of $a$ in $O(1)$.
# Code
We can now formulate our code.
int inv[SIZE];
inv[1] = 1;
for ( int i = 2; i <= n; i++ ) {
inv[i] = (-(m/i) * inv[m%i] ) % m;
inv[i] = inv[i] + m;
}
In line $2$, we set the base case. Modular inverse of $1$ is always $1$. Then we start calculating inverse from $2$ to $N$. When $i=2$, all modular inverse from $1$ to $i-1=1$ is already calculated in array inv[]. So we can calculate it in $O(1)$ using the formula above at line $4$.
At line $5$, we make sure the modular inverse is non-negative.
Next, when $i=3$, all modular inverse from $1$ to $i-1=2$ is already calculated. This is process is repeated until we reach $N$.
Since we calculated each inverse in $O(1)$, the complexity of this code is $O(N)$.
# Conclusion
I saw this code first time on CodeChef forum. I didn't know how it worked back then. I added it to my notebook and have been using it since then. Recently, while searching over the net for resources on Pollard Rho's algorithm, I stumbled on an article from Come On Code On which had the explanation. Thanks, fR0DDY, I have been looking for the proof.
# Reference
1. forthright48 - Modular Multiplicative Inverse
2. forthright48 - Repeated Squaring Method for Modular Exponentiation
3. Come On Code On - Modular Multiplicative Inverse |
Saltar al contenido principal
# 7.2: El teorema del límite central para las medias muestrales (promedios)
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Supongamos que$$X$$ es una variable aleatoria con una distribución que puede ser conocida o desconocida (puede ser cualquier distribución). Usando un subíndice que coincida con la variable aleatoria, supongamos:
1. $$\mu_{x} =$$la media de$$X$$
2. $$\sigma_{x} =$$la desviación estándar de$$X$$
Si dibujas muestras aleatorias de tamaño$$n$$, entonces a medida que$$n$$ aumenta, la variable aleatoria$$\bar{X}$$ que consiste en medias de muestra, tiende a distribuirse normalmente y
$\bar{X} \sim N \left(\mu_{x}, \dfrac{\sigma_{x}}{\sqrt{n}}\right).$
El teorema del límite central para las medias muestrales dice que si sigues dibujando muestras cada vez más grandes (como rodar uno, dos, cinco y finalmente, diez dados) y calculando sus medias, las medias muestrales forman su propia distribución normal (la distribución muestral). La distribución normal tiene la misma media que la distribución original y una varianza que equivale a la varianza original dividida por, el tamaño de la muestra. La variable$$n$$ es el número de valores que se promedian juntos, no el número de veces que se realiza el experimento.
Para decirlo de manera más formal, si dibujas muestras aleatorias de tamaño$$n$$, la distribución de la variable aleatoria$$\bar{X}$$, que consiste en medias muestrales, se denomina distribución muestral de la media. La distribución muestral de la media se aproxima a una distribución normal a medida$$n$$ que aumenta el tamaño de la muestra.
La variable aleatoria$$\bar{X}$$ tiene una$$z$$ puntuación diferente asociada a la misma de la variable aleatoria$$X$$. La media$$\bar{x}$$ es el valor de$$\bar{X}$$ en una muestra.
$z = \dfrac{\bar{x}-\mu_{x}}{\left(\dfrac{\sigma_{x}}{\sqrt{n}}\right)}$
• $$\mu_{x}$$es el promedio de ambos$$X$$ y$$\bar{X}$$.
• $$\sigma \bar{x} = \dfrac{\sigma_{x}}{\sqrt{n}} =$$desviación estándar de$$\bar{X}$$ y se llama el error estándar de la media.
2 ° DISTR./
2:normalcdf
$$\text{normalcdf} \left(\text{lower value of the area, upper value of the area, mean}, \dfrac{\text{standard deviation}}{\sqrt{\text{sample size}}}\right)$$
donde:
• media es la media de la distribución original
• desviación estándar es la desviación estándar de la distribución original
• tamaño de la muestra$$= n$$
Ejemplo$$\PageIndex{1}$$
Una distribución desconocida tiene una media de 90 y una desviación estándar de 15. Las muestras de tamaño$$n = 25$$ se extraen aleatoriamente de la población.
1. Encuentra la probabilidad de que la media muestral esté entre 85 y 92.
2. Encuentra el valor que está dos desviaciones estándar por encima del valor esperado, 90, de la media de la muestra.
Contestar
a.
Dejar$$X =$$ un valor de la población desconocida original. La pregunta de probabilidad le pide encontrar una probabilidad para la media de la muestra.
Dejar$$\bar{X} =$$ la media de una muestra de tamaño 25. Desde$$\mu_{x} = 90, \sigma_{x} = 15$$, y$$n = 25$$,
$\bar{X} \sim N(90, \dfrac{15}{\sqrt{25}}). \nonumber$
Encuentra$$P(85 < x < 92)$$. Dibuja una gráfica.
$P(85 < x < 92) = 0.6997 \nonumber$
La probabilidad de que la media muestral esté entre 85 y 92 es de 0.6997.
normalcdf (valor inferior, valor superior, media, error estándar de la media)
La lista de parámetros se abrevia (valor inferior, valor superior,$$\mu$$,$$\dfrac{\sigma}{\sqrt{n}}$$)
normalcdf$$(85,92,90,\dfrac{15}{\sqrt{25}}) = 0.6997$$
b.
Para encontrar el valor que es dos desviaciones estándar por encima del valor esperado 90, utilice la fórmula:
\begin{align*} \text{value} &= \mu_{x} + (\#\text{ofTSDEVs})\left(\dfrac{\sigma_{x}}{\sqrt{n}}\right) \\[5pt] &= 90 + 2 \left(\dfrac{15}{\sqrt{25}}\right) = 96 \end{align*}
El valor que está dos desviaciones estándar por encima del valor esperado es de 96.
El error estándar de la media es
$\dfrac{\sigma_{x}}{\sqrt{n}} = \dfrac{15}{\sqrt{25}} = 3. \nonumber$
Recordemos que el error estándar de la media es una descripción de qué tan lejos (en promedio) estará la media muestral de la media poblacional en muestras aleatorias simples repetidas de tamaño$$n$$.
Ejercicio$$\PageIndex{1}$$
Una distribución desconocida tiene una media de 45 y una desviación estándar de ocho. Muestras de tamaño$$n$$ = 30 se extraen aleatoriamente de la población. Encuentra la probabilidad de que la media muestral esté entre 42 y 50.
Contestar
$$P(42 < \bar{x} < 50) = \left(42, 50, 45, \dfrac{8}{\sqrt{30}}\right) = 0.9797$$
Ejemplo$$\PageIndex{2}$$
El tiempo, en horas, que tarda un grupo de personas “más de 40" en jugar un partido de fútbol se distribuye normalmente con una media de dos horas y una desviación estándar de 0.5 horas. Se extrae aleatoriamente una muestra de tamaño$$n = 50$$ de la población. Encuentra la probabilidad de que la media de la muestra esté entre 1.8 horas y 2.3 horas.
Contestar
Que$$X =$$ el tiempo, en horas, se necesita para jugar un partido de futbol.
La pregunta de probabilidad te pide encontrar una probabilidad para el tiempo medio de la muestra, en horas, se necesita para jugar un partido de fútbol.
Que$$\bar{X} =$$ el tiempo medio, en horas, se necesita para jugar un partido de futbol.
Si$$\mu_{x} =$$ _________,$$\sigma_{x} =$$ __________, y$$n =$$ ___________, entonces$$X \sim N$$ (______, ______) por el teorema del límite central para las medias.
$$\mu_{x} = 2, \sigma_{x} = 0.5, n = 50$$, y$$X \sim N \left(2, \dfrac{0.5}{\sqrt{50}}\right)$$
Encuentra$$P(1.8 < \bar{x} < 2.3)$$. Dibuja una gráfica.
$$P(1.8 < \bar{x} < 2.3) = 0.9977$$
normalcdf$$\left(1.8,2.3,2,\dfrac{.5}{\sqrt{50}}\right) = 0.9977$$
La probabilidad de que el tiempo medio esté entre 1.8 horas y 2.3 horas es 0.9977.
Ejercicio$$\PageIndex{2}$$
El tiempo empleado en el SAT para un grupo de estudiantes se distribuye normalmente con una media de 2.5 horas y una desviación estándar de 0.25 horas. Se extrae aleatoriamente$$n = 60$$ un tamaño de muestra de la población. Encuentra la probabilidad de que la media de la muestra esté entre dos horas y tres horas.
Contestar
$P(2 < \bar{x} < 3) = \text{normalcdf}\left(2, 3, 2.5, \dfrac{0.25}{\sqrt{60}}\right) = 1 \nonumber$
Para encontrar percentiles para medias en la calculadora, siga estos pasos.
• 2 º Distr
• 3:InvNorm
$$k = \text{invNorm} \left(\text{area to the left of} k, \text{mean}, \dfrac{\text{standard deviation}}{\sqrt{sample size}}\right)$$
donde:
• $$k$$= el$$k$$ percentil th
• media es la media de la distribución original
• desviación estándar es la desviación estándar de la distribución original
• tamaño de la muestra =$$n$$
Ejemplo$$\PageIndex{3}$$
En un estudio reciente reportado el 29 de octubre de 2012 en el Blog Flurry, la edad media de los usuarios de tabletas es de 34 años. Supongamos que la desviación estándar es de 15 años. Toma una muestra de tamaño$$n = 100$$.
1. ¿Cuáles son la media y la desviación estándar para las edades medias de la muestra de los usuarios de tabletas?
2. ¿Qué aspecto tiene la distribución?
3. Encontrar la probabilidad de que la edad media de la muestra sea mayor a 30 años (la edad media reportada de los usuarios de tabletas en este estudio en particular).
4. Encontrar el percentil 95 para la edad media de la muestra (a un decimal).
Contestar
1. Dado que la media muestral tiende a apuntar a la media poblacional, tenemos$$\mu_{x} = \mu = 34$$. La desviación estándar de la muestra viene dada por:$\sigma_{x} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{15}{\sqrt{100}} = \dfrac{15}{10} = 1.5 \nonumber$
2. El teorema del límite central establece que para tamaños de muestra grandes ($$n$$), la distribución muestral será aproximadamente normal.
3. La probabilidad de que la edad media de la muestra sea mayor a 30 viene dada por:$P(Χ > 30) = \text{normalcdf}(30,E99,34,1.5) = 0.9962 \nonumber$
4. Let$$k$$ = el percentil 95. $k = \text{invNorm}\left(0.95, 34, \dfrac{15}{\sqrt{100}}\right) = 36.5 \nonumber$
Ejercicio$$\PageIndex{3}$$
En un artículo en Flurry Blog, se identifica una brecha de marketing de juegos para hombres entre las edades de 30 y 40 años. Estás investigando un juego de inicio dirigido al grupo demográfico de 35 años. Tu idea es desarrollar un juego de estrategia que pueda ser jugado por hombres desde finales de los 20 hasta finales de los 30. Con base en los datos del artículo, la investigación de la industria muestra que el jugador de estrategia promedio tiene 28 años con una desviación estándar de 4.8 años. Tomas una muestra de 100 jugadores seleccionados al azar. Si tu mercado objetivo es de 29 a 35 años, ¿deberías continuar con tu estrategia de desarrollo?
Contestar
Es necesario determinar la probabilidad de que los hombres cuya edad media esté entre 29 y 35 años de edad quieran jugar un juego de estrategia.
$P(29 < \bar{x} < 35) = \text{normalcdf} \left(29, 35, 28,\dfrac{4.8}{\sqrt{100}}\right) = 0.0186$
Puedes concluir que hay aproximadamente un 1.9% de probabilidad de que tu juego sea jugado por hombres cuya edad media esté entre 29 y 35 años.
Ejemplo$$\PageIndex{4}$$
El número medio de minutos para la interacción de la aplicación por parte de un usuario de tableta es de 8.2 minutos. Supongamos que la desviación estándar es de un minuto. Toma una muestra de 60.
1. ¿Cuáles son la media y la desviación estándar para la muestra del número medio de interacción de la aplicación por parte de un usuario de tableta?
2. ¿Cuál es el error estándar de la media?
3. Encuentre el percentil 90 para el tiempo medio de muestra para la interacción de la aplicación para un usuario de tableta. Interpretar este valor en una oración completa.
4. Encuentra la probabilidad de que la media de la muestra esté entre ocho minutos y 8.5 minutos.
Contestar
1. $$\mu = \mu = 8.2 \sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{1}{\sqrt{60}} = 0.13$$
2. Esto nos permite calcular la probabilidad de medias muestrales de una distancia particular de la media, en muestras repetidas de tamaño 60.
3. Let$$k$$ = el percentil 90
$$k = \text{invNorm}\left(0.90, 8.2, \dfrac{1}{\sqrt{60}}\right) = 8.37$$. Estos valores indican que el 90 por ciento del tiempo promedio de interacción de la aplicación para los usuarios de la mesa es inferior a 8.37 minutos.
4. $$P(8 < \bar{x} < 8.5) = \text{normalcdf}\left(8, 8.5, 8.2, \dfrac{1}{\sqrt{60}}\right) = 0.9293$$
Ejercicio$$\PageIndex{4}$$
Las latas de una bebida de cola afirman contener 16 onzas. Se miden las cantidades en una muestra y las estadísticas son$$n = 34$$,$$\bar{x} = 16.01$$ onzas. Si las latas se llenan para que$$\mu = 16.00$$ onzas (como etiquetadas) y$$\sigma = 0.143$$ onzas, encuentre la probabilidad de que una muestra de 34 latas tenga una cantidad promedio mayor a 16.01 onzas. ¿Los resultados sugieren que las latas se llenan con una cantidad mayor a 16 onzas?
Contestar
Tenemos$$P(\bar{x} > 16.01) = \text{normalcdf} \left(16.01,E99,16, \dfrac{0.143}{\sqrt{34}}\right) = 0.3417$$. Dado que existe una probabilidad de 34.17% de que el peso promedio de la muestra sea mayor a 16.01 onzas, debemos ser escépticos sobre el volumen reclamado por la compañía. Si soy consumidor, debería alegrarme de que probablemente esté recibiendo cola gratis. Si soy el fabricante, necesito determinar si mis procesos de embotellado están fuera de los límites aceptables.
## Resumen
En una población cuya distribución puede ser conocida o desconocida, si el tamaño ($$n$$) de las muestras es suficientemente grande, la distribución de las medias muestrales será aproximadamente normal. La media de las medias de la muestra será igual a la media poblacional. La desviación estándar de la distribución de las medias muestrales, denominada error estándar de la media, es igual a la desviación estándar poblacional dividida por la raíz cuadrada del tamaño muestral ($$n$$).
## Revisión de Fórmula
• El teorema del límite central para las medias de la muestra:$\bar{X} \sim N\left(\mu_{x}, \dfrac{\sigma_{x}}{\sqrt{n}}\right) \nonumber$
• La Media$$\bar{X}: \sigma_{x}$$
• Teorema del límite central para las medias muestrales z-score y error estándar de la media:$z = \dfrac{\bar{x}-\mu_{x}}{\left(\dfrac{\sigma_{x}}{\sqrt{n}}\right)} \nonumber$
• Error estándar de la media (desviación estándar ($$\bar{X}$$)):$\dfrac{\sigma_{x}}{\sqrt{n}} \nonumber$
## Glosario
Promedio
un número que describe la tendencia central de los datos; hay una serie de promedios especializados, incluyendo la media aritmética, la media ponderada, la mediana, el modo y la media geométrica.
Teorema de Límite Central
Dada una variable aleatoria (RV) con media conocida$$\mu$$ y desviación estándar conocida,$$\sigma$$, estamos muestreando con tamaño$$n$$, y estamos interesados en dos nuevas RV: la media de la muestra,$$\bar{X}$$, y la suma de la muestra,$$\sum X$$. Si el tamaño ($$n$$) de la muestra es suficientemente grande, entonces$$\bar{X} \sim N\left(\mu, \dfrac{\sigma}{\sqrt{n}}\right)$$ y$$\sum X \sim N(n\mu, (\sqrt{n})(\sigma))$$. Si el tamaño ($$n$$) de la muestra es suficientemente grande, entonces la distribución de las medias muestrales y la distribución de las sumas muestrales se aproximarán a distribuciones normales independientemente de la forma de la población. La media de las medias de la muestra será igual a la media de la población, y la media de las sumas de la muestra será igual a$$n$$ veces la media de la población. La desviación estándar de la distribución de las medias muestrales$$\dfrac{\sigma}{\sqrt{n}}$$,, se denomina error estándar de la media.
Distribución Normal
una variable aleatoria continua (RV) con pdf$$f(x) = \dfrac{1}{\sigma \sqrt{2 \pi}}e^{\dfrac{-(x-\mu)^{2}}{2 \sigma^{2}}}$$, donde$$\mu$$ es la media de la distribución y$$\sigma$$ es la desviación estándar; notación:$$X \sim N(\mu, \sigma)$$. Si$$\mu = 0$$ y$$\sigma = 1$$, el RV se denomina distribución normal estándar.
Error estándar de la media
la desviación estándar de la distribución de las medias de la muestra, o$$\dfrac{\sigma}{\sqrt{n}}$$.
## Referencias
1. Baran, Daya. “El 20 por ciento de los estadounidenses nunca han usado el correo electrónico”. WebGuild, 2010. Disponible en línea en www.webguild.org/20080519/20-... ver-usado-correo electrónico (consultado el 17 de mayo de 2013).
2. Datos de The Flurry Blog, 2013. Disponible en línea en blog.flurry.com (consultado el 17 de mayo de 2013).
3. Datos del Departamento de Agricultura de los Estados Unidos.
This page titled 7.2: El teorema del límite central para las medias muestrales (promedios) is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. |
# How do you divide (x^4+x^2-3x-3)/(5x-2)?
Jan 19, 2016
Long divide the coefficients of the polynomials to find:
$\frac{{x}^{4} + {x}^{2} - 3 x - 3}{5 x - 2}$
$= \frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625} - \frac{2509}{625 \left(5 x - 2\right)}$
#### Explanation:
I like to divide polynomials by long dividing their coefficients, including $0$'s for any missing powers of $x$ ...
This is similar to long division of numbers.
Write the dividend $1 , 0 , 1 , - 3 , - 3$ under the bar and the divisor $5 , - 2$ to the left of the bar.
Write each term of the quotient in turn, chosen to match the leading term of the running remainder when multiplied by the divisor.
So we write $\textcolor{b l u e}{\frac{1}{5}}$ which multiplied by $5 , - 2$ results in $1 , - \frac{2}{5}$, matching the leading term $1$ of the dividend.
Subtract the product from the dividend and bring down the next term of the dividend alongside the result as the running remainder.
Choose the next term $\textcolor{b l u e}{\frac{2}{25}}$ to match the leading term of this running remainder and subtract the product, etc.
Repeat until the running remainder is shorter than the divisor and there is no remaining term to bring down from the dividend. This is the final remainder.
In this particular example we find that the quotient polynomial is:
$\frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625}$
with remainder $- \frac{2509}{625}$
That is:
$\frac{{x}^{4} + {x}^{2} - 3 x - 3}{5 x - 2}$
$= \frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625} - \frac{2509}{625 \left(5 x - 2\right)}$
or if you prefer:
${x}^{4} + {x}^{2} - 3 x - 3$
$= \left(5 x - 2\right) \left(\frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625}\right) - \frac{2509}{625}$ |
Assignment 6
Construction of a triangle given its three medians
Lucia Zapata ~Claudette Tucker
Given three line segments j, k and m. If these are the medians of a triangle, construct the triangle.
We now that the three medians of a triangle meet in a point at 2/3 from each vertex.
Now we need to take 2/3 of each median to construct a triangle.
Next, prolong segment BF 1/2 times BF until P, prolong EF a length equal to EF until A. Through F trace a parallel to BE until that parallel meets the prolongation of AP
Now construct the segments AB, BC and CA. Now ABC is our triangle constructed with the three medians.
We can wonder what is the relationships between the area of a triangle and the area of a triangle constructed with its medians. You can explore here and make some conjectures. We found that the areas of those triangles have a ratio of 3/4.
We can observe the following relationships between areas:
triangle PIR = 1/6 triangle ABC
triangle FPR = 1/12 triangle ABC
triangle BFR = 1/8 of ABC
Besides, triangle PBI = 1/2 ABC
Therefore, area of BPI is 1/6+1/12+1/8 of ABC area. So, area of BPI is 3/4 of ABC area. |
# Is f(x)=x^2lnx increasing or decreasing at x=1?
Jan 17, 2016
Increasing.
#### Explanation:
The derivative of a function can be used to determine if a function is increasing or decreasing at a point.
• If $f ' \left(1\right) < 0$, then $f \left(x\right)$ is decreasing at $x = 1$.
• If $f ' \left(1\right) > 0$, then $f \left(x\right)$ is increasing at $x = 1$.
First, find $f ' \left(x\right)$ through the product rule.
$f ' \left(x\right) = \ln x \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] + {x}^{2} \frac{d}{\mathrm{dx}} \left[\ln x\right]$
$f ' \left(x\right) = \ln x \cdot \left(2 x\right) + {x}^{2} \left(\frac{1}{x}\right)$
$f ' \left(x\right) = 2 x \ln x + x$
Now, find $f ' \left(1\right)$.
$f ' \left(1\right) = 2 \left(1\right) \ln \left(1\right) + 1$
$f ' \left(1\right) = 2 \left(0\right) + 1$
$f ' \left(1\right) = 1$
Since $f ' \left(1\right) > 0$, the function is increasing at $x = 1$.
We can check a graph of $f \left(x\right)$:
graph{x^2lnx [-0.221, 1.6745, -0.353, 0.595]} |
# How do you simplify and make the decimal 0.225 into a fraction?
Oct 14, 2015
9/40
#### Explanation:
First convert 0.225 into a whole number, so you have to multiply it with 1000.
Now divide it by the 1000, so 225/1000.
Now simplify that. They are both divisible by 25.
225/25=9
1000/25=40
Hence, 9/40 is the simplest form. And 9/40=0.225
Oct 14, 2015
0.225 = 9/40
#### Explanation:
First, there is an answer that is evident from the beginning. Take the literal decimal, .225, and determine what fraction it represents.
There are three digits in the decimal 0.225, so the last digit is the 1000ths place. This shows that 0.225 = 225/1000.
The fraction 225/1000 is not in its simplest form. To reduce the fraction, find the Greatest Common Divisor (GCD) / Greatest Common Factor (GCF). (*) We can reduce this fraction to its lowest terms by dividing the numerator and denominator by 25 because it is the GCD/GCF.
So now, divide 225 and 1000 by 25... |
# Year Six Mathematics Worksheets
Math is an essential subject that teaches children problem-solving skills and helps them develop critical thinking. Understanding the basics of math at a young age sets the foundation for future success in this subject and in life. One of the important concepts in mathematics is the calculation of surface area.
Surface area is a measure of the total area that surrounds a three-dimensional object. It helps us understand the amount of material needed to cover a particular object, such as a box or a sphere. Surface area is calculated by adding up the area of each of the object’s faces or surfaces.
There are different formulas for calculating the surface area of different 3-dimensional objects, including:
1. Cube: The surface area of a cube is calculated by adding up the area of all six of its faces. The formula for the surface area of a cube is 6 x (side length)^2.
2. Rectangular Prism: The surface area of a rectangular prism is calculated by adding up the area of all six of its faces. The formula for the surface area of a rectangular prism is 2lw + 2lh + 2wh, where l is the length, w is the width, and h is the height.
3. Sphere: The surface area of a sphere is calculated by adding up the area of all of its curved surfaces. The formula for the surface area of a sphere is 4πr^2, where r is the radius of the sphere.
4. Cylinder: The surface area of a cylinder is calculated by adding up the area of its two circular ends and the area of its curved side. The formula for the surface area of a cylinder is 2πr^2 + 2πrh, where r is the radius and h is the height of the cylinder.
It’s important for kids to understand the concept of surface area and how to calculate it so that they can use this knowledge in real-life situations. For example, when building a shelter, you need to know the surface area of the walls so that you know how much material is needed to cover them.
Math worksheets for kids can be a great tool for practicing the calculation of surface area. These worksheets provide kids with a variety of problems to solve and allow them to check their answers and get feedback on their work. This can help them understand the concept better and improve their skills.
In conclusion, surface area is an important concept in mathematics that helps us understand the amount of material needed to cover a three-dimensional object. Understanding this concept can be useful in real-life situations and math worksheets for kids are a great tool for practicing the calculation of surface area.
# Year Six Math Worksheet for Kids – Surface Area
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## Applied Machine Learning & Data Science Projects and Coding Recipes for Beginners
A list of FREE programming examples together with eTutorials & eBooks @ SETScholars
# Projects and Coding Recipes, eTutorials and eBooks: The best All-in-One resources for Data Analyst, Data Scientist, Machine Learning Engineer and Software Developer
Topics included: Classification, Clustering, Regression, Forecasting, Algorithms, Data Structures, Data Analytics & Data Science, Deep Learning, Machine Learning, Programming Languages and Software Tools & Packages.
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`Disclaimer: The information and code presented within this recipe/tutorial is only for educational and coaching purposes for beginners and developers. Anyone can practice and apply the recipe/tutorial presented here, but the reader is taking full responsibility for his/her actions. The author (content curator) of this recipe (code / program) has made every effort to ensure the accuracy of the information was correct at time of publication. The author (content curator) does not assume and hereby disclaims any liability to any party for any loss, damage, or disruption caused by errors or omissions, whether such errors or omissions result from accident, negligence, or any other cause. The information presented here could also be found in public knowledge domains.` |
# FRACTION WORD PROBLEMS 16
In this page fraction word problems 16 we are going to see practice problem in the topic fraction and you can find the solution for this problem with detailed steps and explanation. fraction word problems 16
## What are the points to be noticed when we add or subtract two or more fractions?
(i) Before adding or subtracting any two or three fractions first we need to consider the denominator of the fractions.
(ii) If the denominators are same,we can put only one denominator and we can add or subtract the numerator.
(iii) If we have different denominators, we should take L.C.M of the fraction and we need change them as same.Then we can add or subtract.
(iv) For multiplying two fractions we do not have to consider the denominators.We should multiply numerator with numerator and denominator with denominator.
## Fraction Word Problems 16
Mary used 1 ¾ bars of chocolate to make a cake. She used 1 ⅜ bars of chocolate to make a pudding.How many bars of chocolate did she use altogether?
Basic idea:
Mary is using 1 ¾ bars of chocolate to prepare a cake and
she is using 1 ⅜ bars of chocolate to make a pudding.We need to find how many bars of chocolate she had used altogether.To solve this problem we need to add two quantity of chocolates used for making cake and pudding.
Solution :
Quantity of chocolate she had used for cake = 1 ¾ bars
Quantity of chocolate she had used for pudding = 1 ⅜ bars
Number of bars of chocolate bars = 1 ¾ + 1
[Now we have to convert this mixed fraction as improper fraction]
= (4+3)/4 + (8+3)/8
= (7/4) + (11/8)
The denominators of the two fractions are not same.So we have to take L.C.M to make them same.
L.C.M = 8
= (7/4) x (2/2) + (11/8)
= (14/8) + (11/8)
= (14+11)/8
= 25 /8
Now we have to convert this improper fraction as mixed fraction.For that we have to divide 25 by 8.so that we will get 3 as quotient 8 as divisor and 1 as remainder.
So the final answer is 3 ⅛ fraction word problems 16
(1) A fruit merchant bought mangoes in bulk. He sold
⅝ of the mangoes. 1/16 of the mangoes were spoiled. 300 mangoes remained with him. How many mangoes did he buy? Solution
(2) A family requires 2 ½ liters of milk per day.How much milk would family require in a month of 31 days?
Solution
(3) A ream of paper weighs 12 ½ kg What is the weight per quire, if 20 quire make one ream? Solution
(4) It was Richard's birthday.He distributed 6 kg of sweets among her friends.If he gave ⅛ kg of sweet to each.How many friends are there? Solution
(5) 6 students went on a picnic.One student agreed to bear half of the expenses. The remaining 5 students shared the remaining expenses equally. What fraction of the expenses does each of 5 students pay?Solution
(6) I have 2 ½ times money that David has. If i have \$100,how much money does David have? Solution
(7) In a basket there are two kinds of sweet packets. There are 7 packets of the first kind each weighing 1 ¼ kg and 9 packets of the second kind each weighing ¾ kg . What is the total weight of the sweets in the basket?
(8) How many half-liter bottles can be filled from a can containing 37 ½ liter of milk? Solution
(9) A gentleman bought 200 liter of milk for a function. ⅘ of it was used for preparing candies. ¾ of the remaining milk was used for preparing coffee. How much of the milk remained. Solution
(10) Two third of a tank can be filled in 18 minutes. How many minutes will it require to fill the whole tank?
Solution
(11) A clerk works 4 ½ hours in the morning and 2 ⅘ hours in the afternoon. Find how many hours does she work each day? Solution
(12) Mary had 4 ⅘ m of lace. She used 3m of a lace for a hand kerchief. How much lace did she have left?
Solution
(13) An artist took 1 ½ hours to paint one picture. He took 1 ⅙ hours to paint another picture. How much time did he take to paint both pictures? Solution
(14) John had 5 ½ tins of paint. After painting a hall, he had 2 ¾ tins of paint left. How many tins of paint did he use?Solution
(15) A brick of weights 2 ½ kg .A stone is 1 kg lighter than the brick. Find the weight of the stone required.
Solution
(16) Mary used 1 ¾ bars of chocolate to make a cake. She used 1 ⅜ bars of chocolate to make a pudding.How many bars of chocolate did she use altogether? Solution
(17) A pen is 13 ⅝ cm long. A pencil is 9 ½ cm long. Find the difference in length. Solution
(18) Jennifer made 4 cakes for a class party. The pupils ate 2 ½ cakes. How much cakes was left over?
Solution
(19) Peter spends 3 ½ hours for doing his home work,3 hours for reading novels and 1 ¾ hours for playing games. How much time does he spend in all? Solution
(20) Peter bought 1 ⅛ m of iron wire and 2 ⅔ m of copper wire. How much wire did he buy altogether? Solution |
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     Â
# Slope Intercept to Standard Form
For converting Slope intercept to standard form, we use following steps -
Step 1: As we all know that, slope intercept form equation in mathematics is y = mx + K, where ‘m’ is a Slope of given line and ‘K’ is a constant. So, first of all we calculate slope ‘m’ and constant ‘K’ form slope intercepts form.
Step 2 : After evaluation of slope intercepts form, now we shift all variables to one side and constants to other side, like we have a slope intercepts form y = 7x + 11, then
y = 7x + 11,
=> y – 7x = 11 or 7x – y = -11,
These are two possible standard forms of a given line equation.
Now we take an example to understand the process of converting slope intercept to standard form -
Example: Convert the given slope intercepts form into standard form -
y = (4 / 3) x + 12?
Solution: We use following steps to evaluate standard form from Slope Intercept Form-
Step 1: First of all, we analyze given slope intercept form-
y = (4 / 3)x + 12,
Here slope = m = (4 / 3) and constant K = 12,
Step 2: Now we shift variables on one side and constants on other side -
y = (4 / 3) x + 12,
=> y – (4 / 3) x = 12 or (4 / 3) x – y = -12,
=> 3y – 4x = 36 or 4x – 3y = -36
So, these are two possible standard forms are generated from slope intercept form.
Therefore these two steps are useful for converting slope intercepts form to standard form. |
# Video: Calculating the Force Applied That Brings a Moving Object to Rest over a Distance
A bullet has a mass of 2.60 g and moves horizontally at a speed of 335 m/s as it collides with a stack of eight pine boards, each 0.750′ thick. The bullet decelerates as it penetrates the boards. And the bullet comes to rest just as it has moved the full distance through all eight boards. Find the average force exerted by the boards on the bullet. Assume that the motion of the boards due to the bullet’s impact is negligible.
03:23
### Video Transcript
A bullet has a mass of 2.60 grams and moves horizontally at a speed of 335 meters per second as it collides with a stack of eight pine boards, each 0.750 inches thick. The bullet decelerates as it penetrates the boards. And the bullet comes to rest just as it has moved the full distance through all eight boards. Find the average force exerted by the boards on the bullet. Assume that the motion of the boards due to the bullet’s impact is negligible.
Looking to solve here for the average force exerted by the boards on a bullet, we can call that force 𝐹. And we’ll start on our solution by drawing a diagram of this situation. We have in this scenario a bullet with an initial speed 𝑣 sub 𝑖 of 335 meters per second encountering a stack of eight pine boards. Each of the boards has a thickness 𝑇 of 0.750 inches. And we’re told that by the time the bullet makes it all the way through the last board, it comes to a stop.
Knowing all this, we want to solve for the average force 𝐹 that the boards exert on the bullet as it decelerates. Seeking to solve for average force may remind us of Newton’s second law of motion, which says that the net force on an object equals its mass times its acceleration 𝑎. Using this law, we want to solve for 𝐹. And we’re given 𝑚 in the problem statement. But we don’t yet know the acceleration of the bullet as it comes to a stop. However, if we assume that acceleration 𝑎 is constant, then that means the kinematic equations apply for describing the motion of the decelerating bullet.
Looking over these four equations of motion, we see that the second one helps us solve for what we want to know — acceleration — in terms of values we’re given. Written in terms of the variables for our particular situation, we can say that the final speed of the bullet squared is equal to its initial speed squared plus two times its acceleration times the distance it travels eight times the thickness of a board.
We know the bullet ends up at rest. So 𝑣 sub 𝑓 is equal to zero. So when we rearrange to solve for 𝑎, it equals negative 𝑣 sub 𝑖 squared all over two times eight 𝑇. We can substitute this expression for acceleration in for 𝑎 in our equation for force. And we understand that even though the minus sign implies that the bullet is decelerating, which is true. Since we want to solve for the average force the boards exert on the bullet, which will be positive, we’ll change that to a plus sign.
Looking at this expression, we were given the mass of the bullet 𝑚 in the problem statement as well as the initial speed of the bullet 𝑣 sub 𝑖. We’re also told the thickness 𝑇 of the boards. But that thickness is currently expressed in inches and we like to convert it to meters. 0.750 inches is approximately 1.905 centimeters. When we plug our values into this expression, we’re careful to use a mass in units of kilograms and a distance of the thickness of each board in units of meters. Entering this expression on our calculator, it comes out to 960 newtons. To two significant figures, that’s the average force that the boards exert on the bullet. |
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# Chapter 9: Correlation and Regression: Solutions
9.1 Correlation
In this section, we aim to answer the question: Is there a relationship between A and B?
Is there a relationship between the number of employee training hours and the number of
on-the-job accidents? Is there a relationship between the number of hours a person sleeps
and their reaction time? Is there a relationship between the number of hours a student
spends studying for a calculus test and the student’s score on that calculus test?
Definition: a correlation is a relationship between two variables.
Typically, we take x to be the independent variable. We take y to be the dependent
variable. Data is represented by a collection of ordered pairs (x, y).
Mathematically, the strength and direction of a linear relationship between two variables is
represented by the correlation coefficient. Suppose that there are n ordered pairs (x, y)
that make up a sample from a population. The correlation coefficient r is given by:
P P P
(xy) − ( x) ( y)
n
r=q P P 2q P 2
n x − ( x) n y − ( y)2
2
P
## This will always be a number between -1 and 1 (inclusive).
• If r is close to 1, we say that the variables are positively correlated. This means there
is likely a strong linear relationship between the two variables, with a positive slope.
• If r is close to -1, we say that the variables are negatively correlated. This means there
is likely a strong linear relationship between the two variables, with a negative slope.
• If r is close to 0, we say that the variables are not correlated. This means that there
is likely no linear relationship between the two variables, however, the variables may
still be related in some other way.
1
(Image from Laerd Statistics) The correlation coefficient of the population is denoted by ρ
– and is usually unknown.
Example 1: The time x in years that an employee spent at a company and the employee’s
hourly pay, y, for 5 employees are listed in the table below. Calculate and interpret the
correlation coefficient r. Include a plot of the data in your discussion.
x y x2 y2 xy
5 25 25 625 125
3 20 9 400 60
4 21 16 441 84
10 35 100 1225 350
P 15 P 38 P 225
2
P 1444
2
P 570
x = 37 y = 139 x = 375 y = 4135 xy = 1189
## Hint: Calculate the numerator:
X X X
n (xy) − x y = 5 · 1189 − 37 · 139 = 802
## Then calculate the denominator:
r X 2 r X X 2 q q
X
n 2
x − x n 2
y − y = 5 · 375 − (37) 5 · 4135 − (139)2
2
√ √
= 506 1354 ≈ 827.72
802
Now, divide to get r ≈ ≈ 0.97.
827.72
Interpret this result: There is a strong positive correlation between the number of years and
employee has worked and the employee’s salary, since r is very close to 1.
2
Example 2: The table below shows the number of absences, x, in a Calculus course and the
final exam grade, y, for 7 students. Find the correlation coefficient and interpret your result.
x 1 0 2 6 4 3 3
y 95 90 90 55 70 80 85
You may use the facts that (double check this for practice)
X X X X X
x = 19, y = 565, x2 = 75, y 2 = 46, 775, xy = 1, 380.
## Calculate the numerator:
X X X
n (xy) − x y = 7 · 1380 − 19 · 565 = −1075
## Then calculate the denominator:
r X 2 r X X 2 q q
X
n 2
x − x n 2
y − y = 7 · 75 − (19) 7 · 46775 − (565)2
2
√ √
= 164 8200 ≈ 1159.66
−1075
Now, divide to get r ≈ ≈ −0.93.
1159.66
Interpret this result: There is a strong negative correlation between the number of absences
and the final exam grade, since r is very close to −1. Thus, as the number of absences
increases, the final exam grade tends to decrease.
3
Example 3: The table below shows the height, x, in inches and the pulse rate, y, per minute,
for 9 people. Find the correlation coefficient and interpret your result.
x 68 72 65 70 62 75 78 64 68
y 90 85 88 100 105 98 70 65 72
You may use the facts that (double check this for practice)
X X X X X
x = 622, y = 773, x2 = 43, 206, y 2 = 68, 007, xy = 53, 336.
## Calculate the numerator:
X X X
n (xy) − x y = 9 · 53336 − 622 · 773 = −782
## Then calculate the denominator:
r X 2 r X X 2 q q
X
n 2
x − x n 2
y − y = 9 · 43206 − (622) 9 · 68007 − (773)2
2
√ √
= 1970 14534 ≈ 5350.89
−782
Now, divide to get r ≈ ≈ −0.15.
5350.89
Interpret this result: There appears to be an extremely weak, if any, correlation between
height and pulse rate, since r is close to 0.
Example 4: The table below shows the number of absences, x, in a Calculus course and the
final exam grade, y, for 7 students. Find the correlation coefficient and interpret your result.
x 1 0 2 6 4 3 3
y 85 80 70 55 90 90 95
There are 7 ordered pairs (x, y), so n = 7. Calculate the needed sums:
x y x2 y2 xy
1 85 1 7225 85
0 80 0 6400 0
2 70 4 4900 140
6 55 36 3025 330
4 90 16 8100 360
3 90 9 8100 270
P 3 P 95 P 29 P 29025 P 285
x = 19 y = 565 x = 75 y = 46775 xy = 1470
4
Calculate the numerator:
X X X
n (xy) − x y = 7 · 1470 − 19 · 565 = −445
## Then calculate the denominator:
r X 2 r X X 2 q q
X
n 2
x − x n 2
y − y = 7 · 75 − (19) 7 · 46775 − (565)2
2
√ √
= 164 8200 ≈ 1159.66
−445
Now, divide to get r ≈ ≈ −0.38.
1159.66
Interpret this result: There is a weak negative correlation between the study time and final
exam grade, since r is closer to 0 than it is to −1. (Compare this problem with Example 2).
Interpreting the Correlation Between Two Variables:
Suppose that you find a strong positive or negative correlation between two variables. Is
there a cause-and-effect relationship between these variables?
## • There could be a reverse cause-and-effect relationship: that is, y causes x.
• There could be a third (or fourth? or more?) variable that leads to the relationship
between x and y.
## • The “relationship” between x and y may just be a coincidence.
5
9.2 Linear Regression
If there is a “significant” linear correlation between two variables, the next step is to find the
equation of a line that “best” fits the data. Such an equation can be used for prediction: given
a new x-value, this equation can predict the y-value that is consistent with the information
known about the data. This predicted y-value will be denoted by ŷ. The line represented
by such an equation is called the linear regression line.
The equation for a line is
ŷ = mx + b,
where m is the slope of the line and b is the y-intercept (the y-value for which x is 0).
In general, the regression line, will not pass through each data point. For each data point,
there is an error: the difference between the y-value from the data and the y-value on the
line, ŷ. By definition, this linear regression line is such that the sum of the squares of the
errors is the least possible. It turns out, given a set of data, there is only one such line. The
slope m and y-intercept b are given by
P P P P P
n xy − ( x) ( y) y x
m= b= −m
n (x2 ) − ( x)2 n n
P P
Examples: Find the equation of the regression line for each of the two examples and two
practice problems in section 9.1.
Example 1:
First, find the slope m. Start by determining the numerator:
X X X
n xy − x y 5 · 1189 − 37 · 139 = 802
## Next, find the denominator:
X X 2
n 2
(x ) − x = 5 · 375 − (37)2 = 506
802
Divide to obtain m = ≈ 1.58
506
P P
y x 139 37
Now, find the y-intercept: b = −m ≈ − 1.58 · ≈ 16.11
n n 5 5
Therefore, the equation of the regression line is ŷ = 1.58x + 16.11
Additional Questions: Use the equations to (Ex 1) predict the hourly pay rate of an employee
who has worked for 20 years, and (Ex 2) predict the test score for a student with 5 absences.
6
For an employee who has worked 20 years, x = 20. Plug this into the equation for the
regression line: ŷ = 1.58 · 20 + 16.11 = 47.71 is the predicted salary, based on the data.
Example 2:
First, find the slope m. Start by determining the numerator:
X X X
n xy − x y = 7 · 1380 − 19 · 565 = −1075
## Next, find the denominator:
X X 2
n (x2 ) − x = 7 · 775 − (19)2 = 164
−1075
Divide to obtain m = ≈ −6.55
164
P P
y x 565 19
Now, find the y-intercept: b = −m ≈ − (−6.55) · = 98.49
n n 7 7
Therefore, the equation of the regression line is ŷ = −6.55x + 98.49
Additional Questions: Use the equations to (Ex 1) predict the hourly pay rate of an employee
who has worked for 20 years, and (Ex 2) predict the test score for a student with 5 absences.
For a student with 5 absences, x = 5. Plug this into the equation for the regression line:
ŷ = −6.55 · 5 + 98.49 = 65.74 is the predicted score, based on the data.
Example 3:
First, find the slope m. Start by determining the numerator:
X X X
n xy − x y = 9 · 53336 − 622 · 773 = −782
## Next, find the denominator:
X X 2
n 2
(x ) − x = 9 · 43206 − (622)2 = 1970
−782
Divide to obtain m = ≈ −0.40
1970
P P
y x 773 622
Now, find the y-intercept: b = −m ≈ − (−0.40) · = 113.53
n n 9 9
Therefore, the equation of the regression line is ŷ = −0.40x + 113.53. Even though we found
an equation, recall that the correlation between x and y in this example was weak. Thus,
this regression line many not work very well for the data.
7
Example 4:
First, find the slope m. Start by determining the numerator:
X X X
n xy − x y = 7 · 1470 − 19 · 565 = −445
## Next, find the denominator:
X X 2
n (x2 ) − x = 7 · 75 − (19)2 = 164
−445
Divide to obtain m = ≈ −2.71
164
P P
y x 565 19
Now, find the y-intercept: b = −m ≈ − (−2.71) · = 88.07
n n 7 7
Therefore, the equation of the regression line is ŷ = −2.71x + 88.07. Even though we found
an equation, recall that the correlation between x and y in this example was weak. Thus,
this regression line many not work very well for the data. For example, for a student with
x = 0 absences, plugging in, we find that the grade predicted by the regression line is 88. |
Theory:
Construct a triangle $$ABC$$ in which $$BC = 8 \ cm$$, $$\angle B = 60^\circ$$ and $$AB + AC = 12 \ cm$$.
Step 1: Draw a line segment $$BC = 8 \ cm$$.
Step 2: Make $$\angle XBC = 60^\circ$$.
Step 3: Given that, $$AB + AC = 12 \ cm$$. Let's take $$12 \ cm$$ as the radius, with $$B$$ as centre, draw an arc that cuts $$BX$$ at $$D$$. Then, join $$CD$$.
Step 4: Now, draw a perpendicular bisector $$EF$$ of line segment $$CD$$. Mark a point $$A$$, where perpendicular bisector intersects $$BD$$.
Step 5: Join $$AC$$. Thus, $$ABC$$ is the required triangle.
JUSTIFICATION:
The point where $$EF$$ cuts $$CD$$ is marked as $$G$$.
In triangles $$AGD$$ and $$AGC$$:
$$DG = GC$$ [Since $$EF$$ bisects $$CD$$ by construction]
$$\angle AGD = \angle AGC$$ [Since $$EF$$ perpendicular $$CD$$ by construction]
$$AG = AG$$ [Common side]
Therefore, $$\Delta AGD \cong \Delta AGC$$ [by $$SAS$$ congruence rule]
$$\Rightarrow AD = AC$$ [by CPCT] - - - - - - (I)
Now, $$AB = BD - AD$$
$$AB = BD - AC$$ [Using (I)]
$$AB + AC = BD = 12 \ cm$$
Therefore, our construction is justified. |
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# A tower is $5\sqrt 3$ meter high. Find the angle of elevation of its top from a point 5 meter away from its foot.
Last updated date: 11th Aug 2024
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Hint- Here, we will be making diagram according to the problem statement and then we will use the formula for tangent trigonometric function i.e, $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$ in order to evaluate the value for the angle of elevation (i.e., $\theta$).
Given, height of the tower AB = $5\sqrt 3$ meter
Let us suppose that the angle of elevation of the top of the tower (i.e., point A) from point C is $\theta$.
As we know that in any right angled triangle, $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}{\text{ }} \to {\text{(1)}}$
$\tan \theta = \dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \dfrac{{5\sqrt 3 }}{5} \\ \Rightarrow \tan \theta = \sqrt 3 {\text{ }} \to {\text{(2)}} \\$
Also we know that $\tan {60^0} = \sqrt 3 {\text{ }} \to {\text{(3)}}$
i.e., $\tan \theta = \tan {60^0} \\ \Rightarrow \theta = {60^0} \\$
Therefore, the required angle of elevation of the top of the tower from a point 5 meter away from the foot of the tower is ${60^0}$.
Note- In any right angled triangle, the hypotenuse is the side opposite to ${90^0}$ (in this case the right angle is at B and the side opposite to vertex B is AC), the perpendicular is the side opposite to the considered angle $\theta$ (in this case the perpendicular is AB) and the base is the remaining side (in this case base is BC). |
### I can use multiple strategies to divide whole numbers of 4
```I can use multiple strategies
to divide whole numbers of 4digit dividends with 1-digit
divisors with remainders.
4.M.NBT.06
Vocabulary
• Divisor – the number you divide by
• Dividend – the amount that you want to
divide up
• Quotient – the answer to a division
problem
Dividend ÷ Divisor = Quotient
Grouping Strategy
• How many groups of 4 balls can we
make with 28 balls?
• 28 balls ÷ 4 balls in each group
1
2
3 4
5
6
7
• We can make 7 groups with 4 balls in
each group.
Grouping Strategy Practice
• How many groups of 5 balls can we
make with 30 balls?
Grouping Strategy Practice
• 30 balls ÷ 5 balls in a group
1
2
3
4
5
6
• We can make 6 groups with 5 balls in
each group.
Distributive (Place Value)
Strategy
• 395 ÷ 5
Step 1: (300 + 90 + 5) ÷ 5
Step 2:
(300 ÷ 5) + (90 ÷ 5) + (5 ÷ 5)
Step 3 & 4: 60 + 18 + 1 = 79
Step 1: Write the
numbers in expanded
notation
Step 2: Divide each
number in the dividend
(first number) by the
divisor (second number)
Step 3: Divide each
section
Distributive Strategy Practice
420 ÷ 5
Step 1: (400 + 20 + 0) ÷ 5
Step 2:
(400 ÷ 5) + (20 ÷ 5) + (0 ÷ 5)
Step 3 & 4: 80 + 4 + 0 = 84
Step 1: Write the
numbers in expanded
notation
Step 2: Divide each
number in the dividend
(first number) by the
divisor (second number)
Step 3: Divide each
section
Long Division
• If Donna sets up 139 chairs into equal
rows of 6 chairs. How many rows will
there be?
• 139 (dividend) ÷ 6 (divisor)
6 cannot go into 1 so put a x
x23
over the 1
6 139
6 can go into 13 two times so put
a 2 over the 3
-12
Subtract 12 from 13 write the 1
under the 3 -2 and bring down
19
the 9
-18
6 can go into 19 three times so
put a 3 over the 9
1
Subtract 18 from 19 and write
the 1 under the 9-8
Remainder
Donna can set up 23 rows of chairs with 6 in each
row and will have 1 chair left over.
Long Division Practice
• If Debby sets up 279 chairs into equal
rows of 5 chairs. How many rows will
there be?
5 cannot go into 2 so put a x
over the 2
5 can go into 27 five times so
put a 5 over the 7
Subtract 25 from 27 write
the 2 under the 7-5 and bring
down the 9
279 (dividend) ÷ 5 (divisor)
x55
5 279
-25
29
-25
4
Remainder
5 can go into 29 five times so Debby will have 55 rows with 5
put a 5 over the 9
chairs in each row and 4 chairs left
over.
Subtract 25 from 29 and
write the 4 under the 9-5
Rectangular Array
• 1,946 ÷ 7
• 1,946 ÷ 7
200
70
+ 8
278
**Using long
division would
help with filling
out the array**
x278
7 1946
-14
54
-49
56
-56
0
(No Remainder)
Rectangular Array Practice
• 3,336 ÷ 8
Rectangular Array Practice
• 3,336 ÷ 8
400
10
+ 7
417
**Using long division
would help with filling
out the array**
x417
8 3336
-32
13
-8
56
-56
0
(No Remainder)
Practice 1
• How many groups of 7 stars can we
make with 28 stars?
• 28 stars ÷ 7 stars in each group
1
2
3
4
• We can have 4 groups with 7 stars in
each group.
Practice 2
• Use the Distributive (Place Value)
Strategy to solve:
985 ÷ 5
985 ÷ 5
Step 1: Write the numbers in
expanded notation
(900 + 80 + 5) ÷ 5
Step 2: Divide each number
in the dividend (first number)
by the divisor (second
(900 ÷ 5) + (80 ÷ 5) + (5 ÷ 5) number)
180 + 16 + 1 = 197
Step 3: Divide each section
Step 4: add up the quotients
Practice 3
• Solve using Long Division:
If James sets up 259 chairs into equal
rows of 6 chairs. How many rows will
there be?
6 cannot go into 1 so put a x
over the 1
6 can go into 13 two times so
put a 2 over the 3
Subtract 12 from 13 write the
1 under the 3 -2 and bring
down the 9
6 can go into 19 three times
so put a 3 over the 9
Subtract 18 from 19 and
write the 1 under the 9-8
259 ÷ 6
x43
6 259
-24
19
-18
1
Remainder
James can set up 43 rows of 6
chairs with 1 chair left over.
Practice 4
• Use rectangular array to solve:
3,534 ÷ 6
3,534 ÷ 6
**Long division will help with
filling out the array**
x589
6 3534
-30
53
-48
54
-54
0
(No Remainder) |
# Unit 8 Polygons And Quadrilaterals Test Answer Key
Last Modified: Published: 2023/04
## Introduction
If you are a student who has recently taken a Unit 8 Polygons and Quadrilaterals test, then you might be eagerly waiting for the answer key. An answer key is a document that contains the correct answers to all the questions asked in the test. It helps students to evaluate their performance and identify their strengths and weaknesses. In this article, we will provide you with the answer key for the Unit 8 Polygons and Quadrilaterals test.
## What are Polygons and Quadrilaterals?
Polygons and quadrilaterals are two-dimensional figures that are commonly studied in geometry. A polygon is a closed figure with three or more straight sides, while a quadrilateral is a polygon with four sides. Some examples of polygons include triangles, rectangles, pentagons, and hexagons. Some examples of quadrilaterals include squares, rectangles, rhombuses, and trapezoids.
## Test Overview
The Unit 8 Polygons and Quadrilaterals test is designed to evaluate your understanding of various concepts related to polygons and quadrilaterals. The test might include questions on topics such as identifying polygons and quadrilaterals, calculating angles and sides, and finding the perimeter and area of polygons and quadrilaterals.
## Answer Key
Here is the answer key for the Unit 8 Polygons and Quadrilaterals test: 1. B 2. D 3. A 4. C 5. B 6. A 7. D 8. C 9. B 10. A
## Explanation of Answers
1. The sum of the interior angles of a triangle is 180 degrees. 2. A quadrilateral is a polygon with four sides. 3. A rectangle has four right angles. 4. The opposite sides of a parallelogram are parallel and congruent. 5. The area of a triangle is calculated using the formula 1/2 * base * height. 6. A square is a special type of rectangle where all the sides are equal. 7. A rhombus has opposite sides parallel and congruent, and all four sides are equal. 8. A trapezoid is a quadrilateral with one pair of parallel sides. 9. The perimeter of a polygon is the sum of the lengths of all its sides. 10. The area of a rectangle is calculated using the formula length * width.
## Conclusion
In conclusion, the Unit 8 Polygons and Quadrilaterals test is an important assessment tool for students studying geometry. The answer key provided in this article can help you evaluate your performance and identify areas where you need to improve. We hope this article has been helpful to you. Good luck with your studies! |
# Algebra Homework Help: How to Factor Polynomials
If there is a constant in the math universe, it is the letter X. In algebra, X is the source of many a mental math meltdown. “Finding ” is the instruction in every single problem. Learn here to factor polynomials.
But X can be fun! We can play with it, move it around, multiply it, divide it, and cancel it out. X really lets us see the relationship between numbers and how math theorems work.
Totally Awesome, I know!
But I know the only reason you’re here is to learn about polynomials and what to do with them.
Factoring polynomials is a crucial step to using the quadratic formula. Factoring polynomials is easier, though, and faster, too.
Let’s look at some X factors.
3x, x, -x2, 5x2, 5x, 7x, 20, -4
The first thing we do is put them in descending order of x.
-x2, 5x2, 3x, x, 5x, 7x, 20, -4
We can simplify them by adding like-factors together: all of the x2, then all the x, then the numbers without x
4x2 + 16x + 16
The next step is to look for common factors in all three groups. For example, these groups all have even numbered coefficients, so we know that 2 is a factor. Even more significant, we can see that all these integers are divisible by 4!
Take out the four from all of them.
4(x2 + 4x + 4)
We’re almost there!
Let’s look at (x2 + 4x + 4) . Can we break this down anymore?
The answer is yes! With the FOIL system!
FOIL stands for “first, outside, inside, last”. This is a cute way of saying that we got this expression by multiplying two parts separate factors together. Each of these factors is called a Binomial. (This is because each factor has two parts: an unknown and a number).
Here we are:
(a + b) X (c + d) = (x2 + 4x + 4) .
Basically, we end up with: ac+ ad + bc + bd. When we add them all together, we get (x2 + 4x + 4).
In this case, bd = 4, the last part of our expression, while ac = x2, the first part of our expression. The other two parts, ad+ bc, need to equal 4x.
We know that ac = x2, so A and C must be x. So that’s easy.
The hard part is figuring out ad and bc. But to solve that, we can look to bd. What are factors that, when multiplied together, give us 4? 4 breaks down to 1, 2, and 4. That means B or D must be 1, 2, or 4.
*****Let’s try 1 and 4. Plug them in and see what happens
(x + 1)(x + 4)
Use FOIL to multiply them together: First, Outside, Inside, Last.
We get x2 + 4x + x + 4
Add them together: x2 + 5x + 4 ≠ x2 + 4x + 4
Yikes! That didn’t work.****
Then it must be 2. This actually makes sense, since 2 + 2 = 4 or 2x + 2x = 4x!
Let’s double check
(x +2)(x + 2) → x2 + 2x + 2x + 4 → x2 + 4x + 4 Bravo!
We can even make this simpler. Since (x + 2) is the same, we can just square the binomial (x + 2)2
So the final answer is 4(x + 2)2
Here’s a rough one: 3x3 + 15x2 – 36x
First thing is to take out the greatest common factor: 3x Check it and see!
3x(x2 + 5x – 12)
Now, we can FOIL the larger expression.
****Here’s a trick: look at the coefficients in the middle (5) and end (12). You know that two factors multiplied together will bring you the end number, but they must be added/subtracted to form the middle number 5.
What are the multiplication factors of 12? 1 and 12, 2 and 6, and 3 and 4. None of these sets of factors, when added or subtracted, give you 5.
If you don’t believe me, try finding the binomials through trial and error. You’ll be at it forever!
Rats! We’ve been “foiled!” We can’t break this down further!
3x(x2 + 5x – 12)
One more: -2x2 + 17x – 36
First of all, are there any common factors? Unfortunately, no. One number is odd, so 2 is not a factor in all three. The last number has no x, so that can’t be a factor, either.
We’ll just proceed to the FOIL, then.
This is a bit hard, since we have THREE GROUPS of numbers to deal with: 2, 13, 36. But we can still do this! Just take your time.
What are the factors of 36 that can add up to 36?
36 is broken down to: 1 and 36, 2 and 18, 3 and 12, 4, and 9, and 6 and 6.
None of these really add up to 17, but we have to consider the 2. There are two factors that make up 2: 1 and 2. Can we multiply any of the factors that make up 36 by 1 and 2, and then add them together to get 17?
YES!
Look at factors 4 and 9. They match up with 1 and 2. Multiply and add the sums:
1 x 9 = 9
2 x 4 = 8
9 + 8 = 17!!! Hooray!
Since the 2 and the 4 are multiplied together, they must be in separate factors: (2x, 9)(x, 4)
Now to determine which ones are positive and negative.
This might sound harder than it is.
We know that two positives multiplied together form a positive. (5 x 4) = 20
Two negatives multiplied together form a positive. (-7 x -9) = 63
A positive and negative multiplied together form a negative. (-8 x 2) = -16
Go back to the original equation and our factors:
-2x2 + 17x – 36
(2x, 9)(x, 4)
We know the first and last coefficients are negative. That means one integer in each binomial must be negative and one must be positive so that, when multiplied, we get negative numbers. But we also know that the middle coefficient is positive.
To get the middle coefficient we need to make sure they are positive. That’s the only way they can ADD up to a positive 17. We need a 9x and an 8x. We can multiply two positives together or two negatives together to get positives. |
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# What Is The Prime Factorization Of 132
What Is The Prime Factorization Of 132. Is 132 a perfect number? 132 is divisible by 2, 132/2 = 66. 132 is not a prime number. In number theory, the prime factors of a positive integer are the prime numbers that divide that integer exactly.
11 is a prime number. The factorization or decomposition of 132 = 2 2 •3•11. 5 rows in this lesson, we will find the factors of 132, prime factors of 132, and factors of 132 in.
## 33 ÷ 11 = 3.
No, 135 is not a prime number. It is the list of the integer's prime factors. 132 ÷ 2 = 66 now we repeat this action until the result equals 1:
## Gcf Of 121 And 132 Is 11.Method Of Prime Factorization:prime Factorization.
The prime factorization of 132 = 2 2 •3•11. No, 135 is not a prime number. Is 135 a prime number? 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
### 11 Is A Prime Number.
Find factors of 132 in pairs pair factors of a number are any two numbers which, which on multiplying together, give that.
### Kesimpulan dari What Is The Prime Factorization Of 132.
Prime factors of 132 are 2 x 2 x 3 x 11. As a simple example, below is the prime factorization of 820 using trial division: 66 is divisible by 2, 66/2 = 33. |
Differentiate the following functions with respect to x :
Question:
Differentiate the following functions with respect to $x$ :
$x^{\sin x}$
Solution:
Let $y=x^{\sin x}$
Taking log both the sides:
$\log y=\log \left(x^{\sin x}\right)$
$\log y=\sin x \log x\left\{\log x^{a}=\operatorname{alog} x\right\}$
Differentiating with respect to $\mathrm{x}$ :
$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}(\sin \mathrm{x} \log \mathrm{x})}{\mathrm{dx}}$
$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\sin \mathrm{x} \times \frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}+\log \mathrm{x} \times \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}$
$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\sin x \times \frac{1}{x} \frac{d x}{d x}+\log x(\cos x)$
$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}} \& \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}\right\}$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{\sin x}{x}+\log x \cos x$
$\Rightarrow \frac{d y}{d x}=y\left(\frac{\sin x}{x}+\log x \cos x\right)$
Put the value of $y=x^{\sin x}$ :
$\Rightarrow \frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)$ |
# LAW OF COSINES AND SINES WORKSHEET
## About "Law of Cosines and Sines Worksheet"
Law of Cosines and Sines Worksheet :
Here we are going to see some practice questions on law of cosines and sines.
## Law of Cosines and Sines Worksheet - Practice questions
(1) In a triangle ABC, if sin A/sin C = sin(A − B)/sin(B − C), prove that a2, b2, c2 are in Arithmetic Progression.
Solution
(2) The angles of a triangle ABC, are in Arithmetic Progression and if b : c = √3 : √2, find ∠A.
Solution
(3) In a triangle ABC, if cos C = sin A / 2 sin B, show that the triangle is isosceles. Solution
(4) In a triangle ABC, prove that sin B/sinC = (c − a cosB)/(b − a cosC) Solution
(5) In a triangle ABC, prove that a cosA + b cosB + c cosC = 2a sinB sinC. Solution
(6) In a triangle ABC, ∠A = 60°. Prove that b + c = 2a cos (B − C)/2 Solution
In a triangle ABC, prove the following
(i) a sin (A/2 + B) = (b + c) sin A/2 Solution
(ii) a(cos B + cos C) = 2(b + c) sin2 A/2 Solution
(iii) (a2 − c2) / b2 = sin(A − C) / sin(A + C) Solution
(iv) a sin(B − C)/(b2 − c2) = b sin(C − A)/c2 − a2 = c sin(A − B)/(a2 − b2
(v) (a + b)/(a − b) = tan (A + B)/2 cot (A − B)/2 Solution
(8) In a triangle ABC, prove that (a2 − b2 + c2) tanB = (a2 + b2 − c2) tanC. Solution
(9) An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park. Solution
(10) A rope of length 12 m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed Solution
(11) Derive Projection formula from (i) Law of sines, (ii) Law of cosines. Solution
After having gone through the stuff given above, we hope that the students would have understood, "Law of Cosines and Sines Worksheet"
Apart from the stuff given in "Law of Cosines and Sines Worksheet", if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
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### Linear Algebra with Applications
Waves have a lot of nice properties that make them easy to analyze.
So when a scientist or engineer needs to analyze a signal that's not exactly a wave, they often approximate it as a wave.
It turns out that vector spaces show us the "best" possible way to do this!
In this quiz, we'll sketch the basic ideas and fill the details in later.
Note on Notation:
$(f+g)(t)$ means "evaluate $f +g$ at $t$," not "multiply $f + g$ by $t.$"
# Why Vector Spaces?
Any signal of interest to a scientist is a real-valued function $f$ of time $t.$ Let's call the set of all such functions $F.$
Waves live in $F,$ too, but they're special since they repeat. Put all the waves $\big($i.e. functions obeying $w(t+1) = w(t) \big)$ in a subset called $W.$
We can add functions and scalar-multiply with the following rules: $(f+g)(t) = f(t) + g(t) \hspace{0.5cm} \text{and} \hspace{0.5cm} (c f)(t) = c f(t).$
Which of the following statements is true?
Note: For reference, you can find the definition of a vector space here.
# Why Vector Spaces?
Both $F$ and $W$ are vector spaces, so we can rephrase our original problem of approximating a signal by a wave this way:
What vector in the subspace $W$ is as close to $|f\rangle \in F$ as possible?
We call $W$ a subspace because it is a space that's also a subset.
There's a way of solving this problem based solely on vector ideas. That means we can develop a strategy for solving it in, say, the plane (which is easier to visualize) and carry what we learn there over to the space $F!$
# Why Vector Spaces?
In the picture below, the point plays the role of the signal, the line plays the role of $W,$ and the Cartesian plane corresponds to $F.$
Before we can take the analogy too seriously, we have to prove that the Cartesian plane is, in fact, a vector space.
The plane is made up of all real number pairs $(x_{1},x_{2}),$ so we can define addition and scalar multiplication naturally as \begin{aligned} (x_{1} , x_{2}) + (y_{1} , y_{2}) & = (x_{1} +y_{1}, x_{2} + y_{2}) \\ c (x_{1},x_{2}) & = (c x_{1} , c x_{2} ). \end{aligned} With these definitions, is the Cartesian plane a vector space?
# Why Vector Spaces?
With $(x_{1} , x_{2} ) + (y_{1},y_{2})= (x_{1}+y_{1}, x_{2} + y_{2})$ and $c(x_{1},x_{2}) = (c x_{1}, c x_{2} ),$ the Cartesian plane is a vector space.
As a set, the line in the picture below consists of all pairs $(x_{1}, m x_{1} )$ where $m$ is the slope.
It inherits addition and scalar multiplication from the plane, but to be a vector subspace, it has to be closed under these operations. This means
• the sum of any two points on the line is another point on the line;
• scaling a point on the line produces another point on the line.
Select the correct statement from the list of options.
# Why Vector Spaces?
So far, our analogy works quite well: both the Cartesian plane $($playing the role of $F)$ and the line $($playing the role of $W)$ are vector spaces.
Now, we need to decide how to tell if the point $(a,b)$ (representing the signal) is "close" to a point $(x_{1} , m x_{1} )$ on the line.
The natural way to do this is to use the distance between $(a,b)$ and $(x_{1} , m x_{1} ),$ which is the length of the line segment connecting them.
The interactive below lets you pick a point on the line with slope $m =0.5.$ It then computes the distance between it and $(a,b) = (2,3).$
(We choose these numbers just to be concrete!)
Is there a smallest possible distance?
# Why Vector Spaces?
With a bit more work, we can actually show that the line segment minimizing the distance is perpendicular to the original line.
So here's the strategy we should import to the signal problem:
• Define distances between vectors in $F$ and $W. \\\\ \text{} \\\\$
• Minimize the distance from $W$ to the signal $|f \rangle \in F.$
It'll take some work, but in later chapters we'll show there's a purely vector-based way to do these things!
# Why Vector Spaces?
We can't fill in the details here, but we can certainly peek ahead and see the result of applying this strategy to a slightly simpler problem.
Suppose instead that $F$'s functions are defined on the smaller interval $[0,1],$ and $W$ is built up from the waves $\big | \sin\left( 2 \pi t \right) \big\rangle , \big | \sin\left( 4 \pi t \right) \big\rangle , \big | \sin\left(6 \pi t \right) \big\rangle , \ldots, \big | \sin\left( 2 \pi n t \right) \big\rangle,$where $n$ is a positive integer.
The interactive below shows the result of approximating a step function by a combination of these sine waves:
The more waves we use, the better, but even with just a few waves the signal approximation is pretty good.
The scheme we outlined in this quiz applies to many real-world problems from physics and statistics, as we'll see.
This is just one example of how useful the vector space concept can be.
# Why Vector Spaces?
×
Problem Loading...
Note Loading...
Set Loading... |
## Linear Algebra and Its Applications, Exercise 1.5.10
Exercise 1.5.10. (a) Both Lc = b and Ux = c take approximately $n^2/2$ multiplication-substraction steps to solve. Explain why.
(b) Assume A is a 60 by 60 coefficient matrix. How many steps are required to use elimination to solve ten systems involving A?
Answer: (a) L is lower triangular, so we have
$b = Lc \Rightarrow \begin{bmatrix} 1&&& \\ l_{21}&1&& \\ \vdots&\vdots&\ddots& \\ l_{n1}&l_{n2}&\cdots&1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix}$
The first entry $c_1$ can be found in a single operation, while solving for $c_2$ takes two operations, solving for $c_3$ three operations, and so on until solving for $c_n$ takes n operations.
The total number of operations is then
$1 + 2 + 3 + \cdots + n = n(n+1)/2 \approx n^2/2$
We have a similar situation with Ux = c, since U is upper triangular:
$Ux = c \Rightarrow \begin{bmatrix} u_{11}&u_{12}&\cdots&u_{1n} \\ &u_{22}&\cdots&u_{2n} \\ &&\ddots&\vdots \\ &&&u_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}$
Here solving for $x_n$ takes one step, solving for $x_{n-1}$ two steps, and so on until solving for $x_1$ takes n steps. Again the total number of steps is $n(n+1)/2 \approx n^2/2$.
(b) As discussed on p. 15, doing elimination on an arbitrary n by n matrix takes approximately $n^3/3$ steps for large n. As a result of performing elimination on A we obtain its two factors L and U. Given the particular system Ax = b we can use Lc = b to solve for c in approximately $n^2/2$ steps, and can then use Ux = c to solve for x, also in approximately $n^2/2$ steps, for a total of $n^2$ steps in all.
For ten systems with the same 60 by 60 matrix the total number of steps is then approximately
$n^3/3 + 10n^2 = 60^3/3 + 10 \cdot 60^2= (60 \cdot 60^2)/3 + 10 \cdot 60^2$
$= 20 \cdot 60^2 + 10 \cdot 60^2 = 30 \cdot 60^2 = 30 \cdot 3,600 = 108,000$
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra. Bookmark the permalink. |
# List of Notes - Category: Fractions
## What is Fraction
So, what is fraction?
Suppose you divided a cake into 10 pieces and eat 3 pieces. How much of cake do you eat?
We write it as 3/10.
3 is a number of pieces you eat and 10 is total number of pieces.
## Proper Fractions
Proper Fraction is a fraction in which numerator is less than denominator.
Examples of proper fractions are 1/5, 7/19, 127/128.
Note, that 5/5, 8/7, 125/3 are not proper fractions.
We saw example with cake, when talked about fractions. This is example of proper fractions.
## Improper Fractions
Improper Fraction is a fraction in which numerator is greater or equal than denominator. In other words improper fractions are fractions that are not proper.
Examples of improper fractions are 5/5, 8/7, 125/3.
## Mixed Numbers/Fractions
Mixed number (or mixed fraction) is a number that consists of integer and proper fraction.
In other words mixed numbers are "mix" of integer and proper fraction.
We write it in the following way: 4 1/5. Here, 4 is integer and 1/5 is proper fraction.
## Mixed Numbers on a Number Line
So, where are mixed numbers located on a number line?
For example, where is 2 4/5 located? We see that this number consists of integer 2 and proper fraction 4/5.
This means that it is greater than 2 (because there is 2 and proper fraction), but it is less than 2+1=3 because proper fraction is less than 1.
## Equivalent Fractions
Here we will learn about equivalent fractions.
Suppose you have a cake. You decide to divide it into 3 pieces and eat one piece. In this case you eat 1/3 (1 out of 3) of the cake.
But you see that the piece is too big, so you divide each piece into two pieces, so there are 3*2=6 pieces now.
## Reducing Fractions
Let's learn how to reduce fractions.
We already learned about equivalent fractions. There are many fractions that are equivalent to the given one, but there is one special fraction.
Fraction is irreducible if numerator and denominator have no common factors.
## Adding Fractions with Like Denominators
To add fractions with like denominators we just add numerators (just like we add integers) and denominator leave the same: color(red)(a/c+b/c=(a+b)/c).
It is possible that you need to reduce fraction after finding sum.
## Subtracting Fractions with Like Denominators
To subtract fractions with like denominators we just subtract numerators (just like we subtract integers) and denominator leave the same: color(red)(a/c-b/c=(a-b)/c).
It is possible that you need to reduce fraction after finding difference.
## Adding Fractions with Unlike Denominators
It is a bit harder to add fractions with unlike denominators than with like denominators.
We saw that it is very simple to add fractions with like denominators.
But how to transform fractions that have different denominators into fractions that have same denominators? In fact, very easy. We use equivalence of fractions for this.
## Subtracting Fractions with Unlike Denominators
It is a bit harder to subtract fractions with unlike denominators than with like denominators.
We saw that it is very simple to subtract fractions with like denominators.
But how to transform fractions that have different denominators into fractions that have same denominators? In fact, very easy. We use equivalence of fractions for this.
## Converting Mixed Numbers to Improper Fractions
Let's see how to convert mixed numbers into improper fractions.
Actually, we convert them almost in the same way as we add fractions with unlike denominators.
Recall that every integer m can be expressed as fraction m/1.
## Converting Improper Fractions to Mixed Numbers
Converting improper fractions to mixed numbers is inverse of converting mixed number to improper fractions.
Suppose you want to convert 3 4/5 to improper fraction.
We already know that 3 4/5=3/1+4/5=(3*5)/5+4/5=(3*5+4)/5=19/9.
## Adding Fractions with Whole Numbers
Adding fractions with whole numbers is essentially the same as converting mixed number to improper fraction (just remember how to add integers correctly).
Indeed, suppose we want to add whole number m and fraction n/q.
## Subtracting Fractions with Whole Numbers
Subtracting fractions with whole numbers doesn't differ much from adding fractions with whole numbers (just remember how to subtract integers correctly).
Indeed, suppose we want to subtract whole number m from fraction n/q.
## Adding Mixed Numbers
Adding mixed numbers is quite easy.
We know that mixed number consists of integer part and fractional part.
To add mixed numbers three steps are needed:
1. Convert each mixed number to improper fraction.
2. Add improper fractions (using addition of fractions with unlike denominators)
3. Convert improper fraction to mixed number if needed (and if possible).
Example 1. Find 1 3/5+2 4/9.
## Subtracting Mixed Numbers
Subtracting mixed numbers is quite easy.
We know that mixed number consists of integer part and fractional part.
To subtract mixed numbers three steps are needed:
1. Convert each mixed number to improper fraction.
2. Subtract improper fractions (using subtraction of fractions with unlike denominators)
3. Convert improper fraction to mixed number if needed (and if possible).
Example 1. Find 1 3/5-2 4/9.
## Comparing Fractions
To compare two fractions we first need to make same denominators using equivalence of fractions. After this denominators are already equal, so we compare numerators just like we compared integers.
If we compare mixed numbers then we compare integer parts. If integer parts are equal then we need to compare fractional parts.
## Multiplying Fractions
To multiply fractions multiply separately numerators and separately denominators: color(green)(a/b*c/d=(ac)/(bd)).
After this you, possibly, need to reduce fraction.
Note! Rules for determining sign of the result are same as when multiplying integers.
## Multiplying Mixed Numbers
To multiply mixed numbers
1. convert mixed numbers to improper fractions
2. multiply fractions
3. reduce improper fraction (if possible) and convert to mixed number (if needed).
Note! Rules for determining sign of the result are same as when multiplying integers.
## Dividing Fractions by Whole Number
To divide fraction by whole number multiply denominator of the fraction by whole number, i.e. fraction n/q divided by the whole number m is color(green)(n/q-:m=(n/q)/m=n/(qm)).
After this you, possibly, need to reduce fraction.
## Dividing Fractions
To divide fraction by a fraction multiply numerator of the first fraction by the denominator of the second fraction and denominator of the first fraction by the numerator of the second fraction, i.e. color(green)(a/b-:c/d=(a/b)/(c/d)=(ad)/(bc)).
## Dividing Mixed Numbers
To divide mixed numbers, convert them to improper fractions and then divide fractions.
After this you, possibly, need to reduce fraction.
Also, you can convert improper fraction back to mixed number.
## Reciprocals
Reciprocal of the fraction is fraction that is turned "upside down", i.e. reciprocal of the fraction color(green)(a/b) is color(red)(b/a).
There is very nice fact about reciprocals.
Fact. Product of fraction and its reciprocal always equals 1.
## Negative Exponents
Let's learn about negative integer exponents.
When we talked about exponents and integers, we assumed that exponent is positive integer.
But what if we want to raise number to negative integer exponent?
## Rational Numbers
Rational numbers are integers plus fractions.
In other words number is rational, if it can be written as fraction p/q (we know that every integer m can be written as fraction m/1).
So, -4, -10, 0, 2, 23, -4/5, -3 2/5, -15/7, 2/3, 9/4, 5 10/11 are all rational numbers. |
Criteria for a Subgroup to be Normal
Criteria for a Subgroup to be Normal
Recall from the Normal Subgroups page that if $(G, \cdot)$ is a group and $(H, \cdot)$ is a subgroup then $(H, \cdot)$ is said to be a normal subgroup of $G$ if $gH = Hg$ for all $g \in G$, that is, the left and right cosets of $H$ with representative $g$ are equal for all $g \in G$.
We will now look at some criteria for when a subgroup of a group will be normal.
Theorem 1: Let $G$ be a group and let $H$ be a subgroup of $G$. The following statements are equivalent: a) $H$ is a normal subgroup of $G$. b) For all $g \in G$ we have that $gHg^{-1} \subseteq H$. c) For all $g \in G$ we have that $gHg^{-1} = H$, i.e., $N_G(H) = \{ g \in G : gHg^{-1} = H\} = G$.
• Proof of $a) \Rightarrow c)$ Suppose that $H$ is a normal subgroup of $G$. Then $gH = Hg$ for all $g \in G$. Fix $g \in G$. Let $z \in gH$. Then $z = gh'$ for some $h \in H$. Since $gH = Hg$ we have that $z \in Hg$ and so $z = h''g$ for some $h'' \in H$. Thus $gh' = h''g$. So $gh'g^{-1} = h''$. Since $h'$ and $h''$ are arbitrary elements of $H$, we have that $gHg^{-1} = H$.
• Let $h' \in H$. Then $h'g \in Hg$. Since $gH = Hg$ we have that $h'g \in gH$. So there exists an $h'' \in H$ such that $h'g = gh''$. So$h' = gh''g^{-1} \in gHg^{-1}$ which shows that $H \subseteq gHg^{-1}$.
• Now let $h' \in gHg^{-1}$. Then there exists an $h'' \in H$ such that $h' = gh''g^{-1}$. So $h'g = gh'' \in gH$. Since $gH = Hg$ we have that $h'g \in Hg$. So $h' \in H$. This shows that $H \supseteq gHg^{-1}$.
• Thus, for all $g \in G$ we have that $gHg^{-1} = H$.
• Proof of $c) \Rightarrow b)$ Trivial.
• Proof of $b) \Rightarrow a)$ Suppose that for all $g \in G$ we have that $gHg^{-1} \subseteq H$. Then $gH \subseteq Hg$. Now since $g \in G$ we have that $g^{-1} \in G$ and by hypothesis, $g^{-1}Hg \subseteq H$. Therefore $Hg \subseteq gH$. So $gH = Hg$ for all $g \in G$, i.e., $H$ is a normal subgroup of $G$. $\blacksquare$ |
# ONE FOR THE AGES
Date of the Problem
January 22, 2024
Cara was born on January 1, 2010, and her mother, Sydney, was born on January 1, 1982. In what year will Sydney’s age be twice Cara’s age?
If we let Cara’s age be C, then Sydney’s age is S = C + 28, since Cara was born 28 years after Sydney. We are interested in determining when Sydney’s age is twice Cara’s age, in other words, when S = 2C. Substituting, we have C + 28 = 2C. Solving for C we get C = 28. Therefore, Sydney’s age will be twice Cara’s age in the year 2010 + 28 = 2038.
In 2018, Cara’s brother, Nile, celebrated a birthday on January 4th. Cara’s age at that time was 4/5 Nile’s age at that time. How old was Nile’s mom, Sydney, when he was born?
Cara’s age in 2018 was 2018 – 2010 = 8 years old. We are told that Cara’s age in 2018 was 4/5 Nile’s age in 2018, N. That means (4/5)N = 8. Solving for N, we see that Nile’s age in 2018 was N = (5/4)8 = 10 years old. Sydney’s age in 2018, S, was 2018 – 1982 = 36 years. Therefore, ten years prior, when Nile was born, Sydney’s age was 36 – 10 = 26 years old.
The sum of the ages of Cara, Nile and Sydney each year forms an arithmetic progression. The sum of their ages in 2018 was 54. In what year will the sum of their ages be 78?
From the previous problem, we know that Cara’s, Nile’s and Sydney’s ages in 2018 were 8, 10 and 36, respectively. The sum of these ages is 8 + 10 + 36 = 54. The following year, the sum of their ages was 9 + 11 + 37 = 57. The following year, the sum of their ages was 10 + 12 + 38 = 60. Notice that the common difference in the arithmetic progression is +3, since each year the ages of Cara, Nile and Sydney each increase by 1. To determine in how many years the sum of the ages will be 78, we can find how many times 3 is added to 54 to get to 78. In other words, we can solve the equation 54 + 3x = 78 → 3x = 24 → x = 8. So, the sum of the ages of Cara, Nile and Sydney will be 78 in the year 2018 + 8 = 2026.
Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Math topic
CCSS (Common Core State Standard)
Difficulty |
# How do you solve x² - x - 20 = 0?
Jul 13, 2015
Find two numbers whose product is $20$ and whose difference is $1$. The pair $5$, $4$ works.
Hence ${x}^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right)$.
So solutions are $x = 5$ or $x = - 4$.
#### Explanation:
$\left(x - a\right) \left(x + b\right) = {x}^{2} - \left(a - b\right) x - a b$
Matching this against ${x}^{2} - x - 20$ we see that if we can find $a$ and $b$ such that $a - b = 1$ and $a b = 20$, then we can factor the quadratic into two linear terms.
The pair $a = 5$, $b = 4$ works, giving us
${x}^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right)$
This will be $0$ when $\left(x - 5\right) = 0$ or $\left(x + 4\right) = 0$, which is when $x = 5$ or $x = - 4$. |
1. ## Problem solving
A farmer wishes to make a rectangular pen using an existing section of straight fence and 36 m of relocatable fencing materials. What is the largest possible area of the pen?
2. Is the existing section however long you need it to be?
Any rectangle must have a length L and a width W. For the perimeter to be $\displaystyle 32+x$, $\displaystyle 2L + 2W$ must be equal to $\displaystyle 32+x$.
We know that by the definition of a rectangle, the section of pre-existing fence will be equal in length to either the length or the width. So, we set $\displaystyle x$equal to $\displaystyle L$.
This gives us $\displaystyle 2L + 2W = 36+L$, or $\displaystyle L + 2W = 36$.
Also, we know that the maximum area of any rectangle is actually a rectangle with four equal sides. This is also known as a square.
With that knowledge, when looking for the maximum area we can say that $\displaystyle L = W$.
If $\displaystyle L = W$, and $\displaystyle L + 2W = 36$, then $\displaystyle W + 2W = 36$.
After that, it's just basic algebra.
$\displaystyle 3W = 36$
$\displaystyle W = 36/3$
We know the width, and since all the sides are equal, we can square it to find the area.
$\displaystyle (36/3)^2 = 144 m^2$
3. Originally Posted by Mr Rayon
A farmer wishes to make a rectangular pen using an existing section of straight fence and 36 m of relocatable fencing materials. What is the largest possible area of the pen?
Perimeter must equal 36m: $\displaystyle 36=2(w+l)$
We want the largest area: $\displaystyle A=wl$
Solve the first equal for either length or width and plug it into the second equation. Calculus will give you the value for the largest area, but without knowledge of that math, you will need to use some intuition/trial and error...
$\displaystyle 18=w+l$
$\displaystyle A=(18-w)w$
$\displaystyle A=18w-w^2$
$\displaystyle w=9; l=9$
$\displaystyle A=81$
4. Originally Posted by colby2152
Perimeter must equal 36m: $\displaystyle 36=2(w+l)$
We want the largest area: $\displaystyle A=wl$
Solve the first equal for either length or width and plug it into the second equation. Calculus will give you the value for the largest area, but without knowledge of that math, you will need to use some intuition/trial and error...
$\displaystyle 18=w+l$
$\displaystyle A=(18-w)w$
$\displaystyle A=18w-w^2$
$\displaystyle w=9; l=9$
$\displaystyle A=81$
I am sorry but with all due respect, that is just not right. Your response ignores the fact that there is another section of straight fence. |
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This states that when taking the reciprocal of a number, the exponent becomes the opposite sign. So, to solve a fraction with a negative exponent, you simply take the reciprocal of the base number and then change the sign of the exponent.
Once I have a good understanding of the problem, I start brainstorming potential solutions. I try to come up with as many possible solutions as I can, no matter how crazy they might sound at first. After I have a good list of potential solutions, I start to narrow them
In this case, we can subtract 4 from both sides of the equation to get: y - 4 = 2x Then, we need to divide both sides of the equation by 2 to get: y/2 = x Finally, we can multiply both sides of the equation by 2 to get: y = 2x
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# Partial Products: A Comprehensive Guide
## Introduction
Partial products are a useful tool for solving multiplication problems. By breaking down numbers into smaller parts, it makes it easier to solve more complex equations. In this guide, we’ll explore what partial products are, how they work, and how they can be used to teach kids about math.
## Explaining Partial Products: A Guide
First, let’s take a look at what partial products are and how they can help solve multiplication problems.
### What Are Partial Products?
Partial products are a way of breaking down a multiplication problem into smaller parts in order to make it easier to solve. This method is particularly helpful when dealing with large numbers or multiple factors. Rather than trying to calculate the entire equation in one go, partial products allow you to break it down into smaller chunks that are easier to comprehend.
### How Partial Products Help Solve Multiplication Problems
Partial products can be used to simplify multiplication problems by breaking them down into manageable parts. For example, if you were trying to multiply 24 and 12, rather than trying to solve the entire equation at once, you could break it down into two parts: (24 x 10) + (24 x 2). This makes it easier to understand and calculate the answer.
## Teaching Kids About Partial Products
Partial products can also be used to teach kids about multiplication. By introducing the concept early on, it can help make the subject easier to understand and learn. Here’s how you can incorporate partial products into everyday math lessons.
### Introducing Partial Products
When teaching kids about partial products, it’s important to start off by explaining what they are and why they’re useful. You can do this by providing examples of how they can be used to simplify multiplication problems. Once they understand the concept, you can move on to more complex examples.
### Incorporating Partial Products into Everyday Math Lessons
Once your students have a basic understanding of partial products, you can start incorporating them into everyday math lessons. For example, when teaching multiplication, you can use partial products to break up larger numbers into smaller, more manageable chunks. This will make it easier for kids to understand and solve the problem.
## Breaking Down Partial Products for Beginners
If you’re just starting out with partial products, here are some tips for getting started.
### Examples of Partial Products
A great way to get familiar with partial products is to look at some real-world examples. This will help you understand how they work and how they can be used to solve multiplication problems. Here are some examples of partial products:
• 6 x 7 = (6 x 5) + (6 x 2)
• 15 x 12 = (15 x 10) + (15 x 2)
• 20 x 25 = (20 x 20) + (20 x 5)
### Step-by-Step Instructions for Calculating Partial Products
Once you’ve become familiar with the concept of partial products, you can start calculating them yourself. Here are the steps you should follow:
1. Identify the factors you need to multiply.
2. Break up each factor into its individual digits.
3. Calculate the partial product for each digit.
4. Add up all the partial products to get the final answer.
For example, if you were trying to calculate 24 x 12, you would first break up the factors into their individual digits: 24 = 2 x 10 + 4 x 1, and 12 = 1 x 10 + 2 x 1. Then, you would calculate the partial products for each digit: (2 x 10) + (4 x 1) + (1 x 10) + (2 x 1). Finally, you would add up the partial products to get the final answer: 24 x 12 = 288.
## Using Partial Products to Simplify Multiplication
Partial products can also be used to simplify complex multiplication problems. Here’s how:
### Identifying Factors to Break Up Multiplication Problems
The first step is to identify the factors that you need to multiply. For example, if you were trying to calculate 24 x 12 x 8, you would need to break up the factors into two-digit numbers: 24 = 2 x 10 + 4 x 1, 12 = 1 x 10 + 2 x 1, and 8 = 0 x 10 + 8 x 1.
### Utilizing Partial Products to Simplify Complex Multiplication Problems
Once you’ve identified the factors, you can then use partial products to simplify the equation. For our example, the equation would look like this: (2 x 10 x 0 x 10) + (2 x 10 x 8 x 1) + (4 x 1 x 0 x 10) + (4 x 1 x 8 x 1). Once you’ve broken down the equation into its constituent parts, it becomes much easier to solve.
## Conclusion
Partial products are a powerful tool for simplifying complex multiplication problems. By breaking down factors into smaller parts, it makes it easier to understand and calculate the answer. Additionally, partial products can be used to teach kids about math, as it helps make the subject more accessible. So if you’re looking for a way to simplify multiplication problems or teach kids about math, partial products may be the solution you’ve been looking for.
### Summary of Benefits of Partial Products
Partial products provide many benefits, including:
• Simplifying complex multiplication problems
• Making math easier to understand and learn
• Helping kids master basic multiplication skills
### Final Thoughts on Partial Products
Partial products are a great tool for solving multiplication problems and teaching kids about math. With a little practice, anyone can learn how to use partial products to simplify complex equations and make learning math easier. |
# Solve the inequality 16^(x-1)>2^(2x+2)
hala718 | Certified Educator
Given the inequality:
16^(x-1) > 2^2x+2
First we will write: 16 = 2^4.
==> 2^4^(x-1) > 2^2x+2
We know that x^a^b = x^ab
==> 2^4(x-1) > 2^(2x+2)
==> 2^(4x-4) > 2^(2x+2)
Now that the bases are equal, then the powers should the inequality
==> 4x -4 >2x + 2
We will combine like terms.
==> 4x-2x > 2 + 4
==> 2x > 6
Divide by 2.
==> x > 3
x belongs to the interval ( 3, inf)
justaguide | Certified Educator
We have to solve the inequality 16^(x-1)>2^(2x+2)
16^(x-1)>2^(2x+2)
=> 2^4*(x - 1) > 2^(2x + 2)
as the base is the same and positive we can write
4*(x - 1) > (2x + 2)
=> 4x - 4 > 2x +2
=> 4x - 2x > 2 + 4
=> 2x > 6
=> x > 6/2
=> x > 3
Therefore x > 3.
giorgiana1976 | Student
We'll write both bases as power of 2:
2^4(x-1)>2^(2x+2)
Since the bases are bigger than unit value, the function is increasing and we'll get:
4(x-1)> 2x+2
We'll divide by 2:
2(x- 1) > x + 1
We'll remove the brackets:
2x - 2 > x + 1
We'll subtract x both sides:
2x - x - 2 > 1
x - 2 > 1 |
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# Write and Solve Equations with Rational Numbers
Sep 14, 2022
### Key Concepts
• Solve multiplication equations with decimals.
• Solve addition, subtraction, and division equations with decimals.
## 4.5.2 Write and Solve Equations with Rational Numbers
### Solving equations with decimals:
You can solve equations with decimals the same way that you solve equations with rational numbers: using inverse relationships and properties of equality to isolate the variable.
For example, let us see an additional equation.
### 4.5.2.1 Solve multiplication equations with decimals
Example 1:
Keith bought all the four T-shirts for \$47.8. He paid the same amount for each T-shirt. Write and solve an equation to find m, the cost of each T-shirt.
Solution:
Use a bar diagram to show how the quantities are related and to write an equation.
Example 2:
Alex bought a bag of 8 apples for \$3.60. Write and solve an equation to find how much Alex paid for each apple.
Solution:
Use a bar diagram to show how the quantities are related and to write an equation.
### 4.5.2.2 Solve addition, subtraction, and division equations with decimals
Example 3:
Diane spent \$12.50 for a new notebook and a compass. The notebook cost is 6.35. Write and solve an equation to find c, the cost of the compass.
Solution:
Example 4:
Use inverse relationship to solve the following subtraction equation.
y – 8.4 = 3.5
Solution:
Example 5:
Use inverse relationship to solve the following division equation.
m ÷ 4.5 = 50
Solution:
## Exercise:
1. In the human body, 1.5 L of blood are supplied to the liver every minute. Solve the following equation to find the number of liters per hour. h+ 60 = 1.5
2. The length of the small intestine is 640 cm, or 51.2 times the length of the pancreas. Solve the following equation to find the length of the pancreas. 51.2p= 640
3. Aidan says that he solved the equation x3.5 – 7.2 by adding 3.5 to the left side of the equation. Explain whether Aidan is correct
4. Solve the following equation. 13.27 = t- 24.45
5. Solve the following equation. r+ 5.5 = 18.2
6. Solve the following equation. 1.8x= 40.14
7. Solve the following equation. 17.3 + v= 22.32
8. Solve the following equation. w – 3.2 = 5.6
9. The area, A, of a triangle is 15.3 square centimeters. Its base, A is 4.5 centimeters. The formula for finding the area of a triangle is A = 1/2 b/z. Write and solve an equation to find the height, 1), of the triangle.
10. In one study, the number of women classified as supertasters was 2.25 times the number of men classified as supertasters. Suppose 72 women were classified as supertasters. Write an equation that represents the number of men, m, who were classified as supertasters. Then solve the equation. How many men were classified as supertasters?
### What have we learned:
• Solve multiplication equations with decimals by using inverse relationships and properties of equality.
• Solve addition, subtraction, and division equations with decimals by using inverse relationships and properties of equality.
#### 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 […] |
# Solution to puzzle 73: Unobtuse triangle
A triangle has internal angles A, B, and C, none of which exceeds 90°. Show that
• sin A + sin B + sin C > 2
• cos A + cos B + cos C > 1
• tan (A/2) + tan (B/2) + tan (C/2) < 2
## sin A + sin B + sin C > 2
Consider the graph of y = sin x, below, and the line segment joining the origin to the point (/2,1).
Note that the second derivative, y'' = −sin x, is negative for 0 < x < /2, and so that section of the graph is concave.
The equation of the line segment is y = (2/) · x. Note that the line segment intersects the concave curve at x = 0 and x = /2.
Hence, for 0 < x /2, we have sin x (2/) · x, with equality only for x = /2.
Since 0 < A, B, C /2, with equality in at most one case, we have: sin A + sin B + sin C > (2/) · (A + B + C).
Since A + B + C = , it follows that sin A + sin B + sin C > 2.
By judicious choice of line segment we can similarly establish the other two results.
## cos A + cos B + cos C > 1
Consider the graph of y = cos x, below, and the line segment from (0,1) to (/2,0).
The second derivative, y'' = −cos x, is negative for 0 < x < /2, and so that section of the graph is concave.
The equation of the line segment is y = 1 − (2/) · x.
Hence, for 0 < x /2, we have cos x 1 − (2/) · x, with equality only for x = /2.
Then cos A + cos B + cos C > 3 − (2/) · (A + B + C).
Therefore cos A + cos B + cos C > 1.
## tan (A/2) + tan (B/2) + tan (C/2) < 2
Finally, consider the graph of y = tan x, and the line segment from the origin to (/4,1).
The second derivative, y'' = 2 tan x sec2 x, is positive for 0 < x < /4, and so that section of the graph is convex.
The equation of the line segment is y = (4/) · x.
Hence, for 0 < x /4, we have tan x (4/) · x, with equality only for x = /4.
Then tan (A/2) + tan (B/2) + tan (C/2) < (4/) · (A/2 + B/2 + C/2).
Therefore tan (A/2) + tan (B/2) + tan (C/2) < 2.
## Remarks
A real-valued function is said to be convex on an interval I if, and only if
f(ta + (1 − t)b) tf(a) + (1 − t)f(b)
for all a, b in I and 0 t 1. It can be shown that if f''(x) 0 for all x in I, then f is convex on I. A real-valued function is said to be concave on an interval I if, and only if, −f is convex on I. |
# How do you differentiate f(x)=ln(3x^2) using the chain rule?
Dec 28, 2015
Step by step explanation and working is given below.
#### Explanation:
$f \left(x\right) = \ln \left(3 {x}^{2}\right)$
For chain rule, first break the problem into smaller links and find their derivatives. The final answer would be the product of all the derivatives in the link.
$y = \ln \left(u\right)$
$u = 3 {x}^{2}$
The differentiation using chain rule would be
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
$y = \ln \left(u\right)$
Differentiate with respect to $u$
$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$
$u = 3 {x}^{2}$
Differentiate with respect to $x$
$\frac{\mathrm{du}}{\mathrm{dx}} = 3 \frac{d \left({x}^{2}\right)}{\mathrm{dx}}$
$\frac{\mathrm{du}}{\mathrm{dx}} = 3 \cdot 2 x$
$\frac{\mathrm{du}}{\mathrm{dx}} = 6 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \cdot 6 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x}{u}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x}{3 {x}^{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x}$ Final answer
Note: The above process might look big, this was given in such a form to help you understand step by step working. With practice, you can do it quickly and with less steps. |
# 3.4a Definite Integral of f(x) from x=a to x=b
3.4a Definite Integral of f(x) from x=a to x=b
Example:
Evaluate each of the following.
$\begin{array}{l}\text{(a)}{\int }_{-1}^{0}\left(3{x}^{2}-2x+5\right)dx\\ \text{(b)}{\int }_{0}^{2}{\left(2x+1\right)}^{3}dx\end{array}$
Solution:
$\begin{array}{l}\text{(a)}{\int }_{-1}^{0}\left(3{x}^{2}-2x+5\right)dx\\ ={\left[\frac{3{x}^{3}}{3}-\frac{2{x}^{2}}{2}+5x\right]}_{-1}^{0}\\ ={\left[{x}^{3}-{x}^{2}+5x\right]}_{-1}^{0}\\ =0-\left[{\left(-1\right)}^{3}-{\left(-1\right)}^{2}+5\left(-1\right)\right]\\ =0-\left(-1-1-5\right)\\ =7\\ \\ \\ \text{(b)}{\int }_{0}^{2}{\left(2x+1\right)}^{3}dx\\ ={\left[\frac{{\left(2x+1\right)}^{4}}{4\left(2\right)}\right]}_{0}^{2}\\ ={\left[\frac{{\left(2x+1\right)}^{4}}{8}\right]}_{0}^{2}\\ =\left[\frac{{\left(2\left(2\right)+1\right)}^{4}}{8}\right]-\left[\frac{{\left(2\left(0\right)+1\right)}^{4}}{8}\right]\\ =\frac{625}{8}-\frac{1}{8}\\ =78\end{array}$ |
LOGARITHMS
Objectives:
• To learn what a logarithm does and how to use it.
• Evaluate a logarithmic expression.
Related topics: 1-4, 7, 9, 10, 13, 25.
Recall 1: You may have forgotten the basic rules for manipulating exponents. Here are some of those rules.
1.1 There is a particular number, denoted e, which is used extensively in science and mathematics. Its value is approximately 2.718 and may be found as the "e" button on most calculators. Computers generally expect you to enter the number e as Exp(1). Therefore, if you wish to enter the number e^4 on a computer, you would type Exp(4). In the computer algebra system Mathematica, you would type Exp[4]. Note the capital letter and the square brackets around the 4.
1.2 To enter mathematical operations of multiplication and exponentiation on a computer, use the * for multiplication and ^ for exponentiation. Therefore,
3*4 = 12 and 3^4 = 81.
1.3 You may wish to think of the exponentials a^x and e^x as functions of x. You may enter e^x in Mathematica as Exp[x].
The rules for Exp[x] are then expressed as
Exp[x+y] = Exp[x] *Exp[y]
Exp[x]^y = Exp[xy]
Exp[x-y] = Exp[x]/Exp[y]
Exp[0] = 1
Recall 2: If a is different from 0, then a^x is different from 0 for all x.
Recall 3: When two functions are connected by the properties y = f(x) and x = g(y), we say that f(x) is the inverse function of g(y) and g(y) is the inverse function of f(x). Of course this also means that g(f(x)) = x and f(g(y)) = y.
Recall 4: If you plot the graphs of a function and its inverse you will find that they are mirror images of each other when the imaginary mirror is placed along the diagonal line y = x. See illustration below.
The illustration below shows the functions Exp[x] and its inverse.
Introduction 1 (Logarithms):
The inverse function of a^x is called the logarithm to the base a and is denoted
The inverse function of e^x is called the natural logarithm of x and is often denoted Ln(x).
The graphs of e^x and Ln(x) are displayed in the illustration above. To enter a logarithm to the base a in Mathematica type Log[a,x]. If you type Log[x], Mathematica will interpret the base as e. Most calculators have buttons labeled "log" and "ln" for Log10(x) and Ln(x), respectively.
Recall 5 As you would expect, the rules for manipulating logarithms look very much like the reverse of those for manipulating exponents. They are:
Rules of Logarithms Let a be a positive number such that a does not equal 1, let n be a real number, and let u and v be positive real numbers.
Important Note: The domain of the logarithm function is the set of all positive numbers. Therefore, Log(x) makes no sense when x is a negative number.
Recall 6 Since the logarithm and exponential functions are inverses of each other, we have the following relationships:
Recall 7 One of the many great uses of the logarithm function is to reduce exponents to products. So, for example, if you are ever faced with an equation like 4 = 2^(x-3) and you need to know what x is, all you have to do is take the logarithm to the base 2 of both sides of the equation to get
Log2(4) = Log2[2^(x-3)]
and you can proceed by knowing that the left side is equal to 2 and the right side is equal to x - 3. Then all you have to do is solve the equation 2 = x - 3 to get x = 5.
Recall 8 You know that a^b = a^c if and only if b = c. Therefore you may solve equations such as
because you would know that x^2 + 6 = -x, and that would mean
x^2 + x + 6 = 0, or
(x + 3)(x + 2) = 0.
This means that x = -3 or x = -2.
Recall 9 You know that Logab = Logac if and only if b = c. Therefore you may solve equations of the form
Log2(x^2 - 6) = Log2(x) by what we just said in the last paragraph.
From Recall 8, you know that
From this it follows that x^2 - 6 = x and therefore x = 3 or x = -2. But x cannot equal -2 because when we check it we find that we have to make sense of Log (-2). The logarithm function does not have negative numbers in its domain. (See the important note above.) So the only solution is x = 3.
Recall 10 If you know the logarithm to the base "a" of a number then you know it to any other base by using the conversion formula below:
Calculators often only have two kinds of log buttons, log and ln. If you wish to find Log2(x) then all you have to do is find the quotient ln(x)/ln(2). If you are simply converting from common logarithms (base 10) to natural logarithms (base e) or back, you could use the conversion factors given by
Ln(x) = 2.30259 Lg(x), or Lg(x) = .43429 ln(x).
Examples
Example 1.
Example 2.
Example 3. The expression
can be written as a single logarithm
Example 4. To calculate Log5(1/3) you may have to calculate the ratio -Ln(3)/Ln(5)
Example 5. To solve an equation Log2(x - 4) = 1 - Log2(x - 3) you would collect the logarithms to one side to get Log2(x - 4) + Log2(x - 3) = 1, then group the logarithms together to get Log2((x - 4)(x - 3)) = 1. Now recall that if Loga(y) = 1, then y = a. Therefore (x - 4)(x-3)=2, which gives x = 2 or x = 5. But x = 2 cannot be a solution because Log2(x - 4) is the logarithm of a negative number when x = 2 and there is no such thing as the logarithm of a negative number. So the only solution is x = 5.
Example 6. What if you want to find x when Log2(x - 4) + Log2(x - 3) = 2, then collect the logarithms to get Log2((x - 4)(x - 3)) = 2. Now use the fact that a^b = a^c if and only if b = c to see that
Therefore, (x - 4)(x - 3) = 4. If you multiply out the left side you get x^2 - 7x - 12 = 4. The equation then becomes x^2 - 7x - 16 = 0. This equation does not factor; so we have to use the quadratic formula to get the two solutions.
But one of the solutions is less than 3 and therefore is not in the domain of Log2(x - 3). So we have only one valid solution.
EXERCISES
Note: log(x) is a shorthand notation for the logarithm to the base 10 of x.
1. Find
2. If Log4(x) = 5/2, what is x?
3 Simplify Log5(5)
4 Simplify Exp[ln7].
5 Write Log3(x + 2) + 2Log3x - Log32 as a single logarithm
6 Use natural logarithms to evaluate Log527
7 If Loga(8/27) = 3, what is a?
8 Solve the equation 9^x = 12 for x.
9 Solve the equation 7^(2x-1) = 7^(x+2) for x.
10. Solve the equation 4^(2x-1) = 3^(x+2) for x.
11 Solve the equation exp(x^2)=100 for x.
12 Solve the equation 5 = 2(3 - e^x) for x.
13 Solve the equation Logx(x + 6) - Logx(x + 2) = Logxx for x.
14 Solve the equation Log3(3x^2)^2 - 1 = 3 for x.
15 If y = Log2(x^2 + 6), write x as a function of y.
16 Solve the equation (Logx)^2 = Log(x^2) for x.
1. 6
2. x = 32
3. 1
4. 7
5. Log3[x^2(x+2)/2]
6. Ln(27)/Ln(5)
7. a = 2/3
8. Log912
9. x = 3
10.. x = 2 Log43 + 1
11. x = ±sqrt(Ln(100)
12. x = -Ln(2)
13. x = 2
14. x = sqrt(3)
15. x = ± sqrt(2)^y - 6
16. x = 1 or x = 100 |
New Zealand
Level 8 - NCEA Level 3
# Constructing Discrete Probability Distribution
Lesson
When we are required to construct our own discrete probability distribution, there are a number of things that this could mean.
• Read and interpret an experiment and represent the probability distribution as a table
• Read and interpret an experiment and represent the probability distribution as a function
In this section where we're mainly dealing with general discrete random variables, we'll mostly be constructing tables to represent our distributions.
## How to construct a discrete probability distribution
Step One: Define exactly what your random variable $X$X represents.
Note: You can use any capital letter for your random variable, but most often we use $X$X
Step Two: Consider how many possible outcomes $X$X can take.
Step Three: Devise a method for calculating the probabilities for each outcome.
Note: The main methods you will use here is either constructing a sample space (the most common being a tree diagram or a table) or using counting techniques (combinations)
Step Four: Calculate your probabilities and represent them in a table.
You have now created an individual probability distribution for your random variable!
#### Worked Examples
##### Example 1:
A fair standard dice is thrown and the number of dots on the uppermost face is noted.
Let $X$X be the number of dots on the uppermost face.
(a) Construct the probability distribution for $X$X.
Think: In this example the random variable has already been defined for us. All we need to do is determine the number of possible outcomes and then work out their associated probabilities.
Do: The outcomes for $X$X are $1,2,3,4,5$1,2,3,4,5 and $6$6 since we're talking about rolling a standard dice. We also know the probabilities for each are $\frac{1}{6}$16 since each outcome is equally likely.
Now we construct a table of values.
$x$x $1$1 $2$2 $3$3 $4$4 $5$5 $6$6
$P(X=x)$P(X=x) $\frac{1}{6}$16 $\frac{1}{6}$16 $\frac{1}{6}$16 $\frac{1}{6}$16 $\frac{1}{6}$16 $\frac{1}{6}$16
And there we have our probability distribution.
(b) State the type of distribution which $X$X represents.
Think: Our discrete distributions can either be uniform or non - uniform. Later on you'll learn about specific types of non-uniform discrete random variables.
Do: Since all outcomes have the same probability, this distribution is a uniform distribution.
##### QUESTION 1
A random variable $X$X can take any of the values $0$0, $1$1, $2$2 or $3$3. The following facts about the distribution of $X$X are known.
• $P\left(X=0\right)=\frac{1}{2}P\left(X=1\right)$P(X=0)=12P(X=1)
• $P\left(X=1\right)=4P\left(X=2\right)$P(X=1)=4P(X=2)
• $P\left(X=2\right)=\frac{1}{5}P\left(X=3\right)$P(X=2)=15P(X=3)
1. Use these facts to complete the probability distribution table for $X$X. Give each probability in simplest form.
$x$x $P\left(X=x\right)$P(X=x) $0$0 $1$1 $2$2 $3$3 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
##### QUESTION 2
A pencil case contains $9$9 blue pens and $5$5 green pens. $4$4 pens are drawn randomly from the pencil case, one at a time, each being replaced before the next one is drawn.
1. What is the probability of drawing one blue pen from the pencil case?
2. What is the probability of drawing three blue pen from the pencil case?
3. Let $X$X be the number of blue pens drawn. Complete the probability distribution table.
$x$x $P\left(X=x\right)$P(X=x) $0$0 $1$1 $2$2 $3$3 $4$4 $\frac{625}{38416}$62538416 $\editable{}$ $\frac{6075}{19208}$607519208 $\editable{}$ $\editable{}$
### Outcomes
#### S8-4
Investigate situations that involve elements of chance: A calculating probabilities of independent, combined, and conditional events B calculating and interpreting expected values and standard deviations of discrete random variables C applying distributions such as the Poisson, binomial, and normal
#### 91586
Apply probability distributions in solving problems |
# Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle
## Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle
Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle
### The Circle Exercise 29A – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
Use the figure given below to fill in the blanks :
(i) R is the …… of the circle.
(ii) Diameter of a circle is …… .
(iii) Tangent to a circle is … .
(iv) EF is a …… of the circle.
(v) …… is a chord of the circle.
(vi) Diameter = 2 x …..
(vii) ……. is a radius of the circle.
(viii) If the length of RS is 5 cm, the length of PQ = ……
(ix) If PQ is 8 cm long, the length of RS =…..
(x) AB is a ….. of the circle
Solution:
Question 2.
Draw a circle of radius 4.2 cm. Mark its centre as O. Take a point A on the circumference of the circle. Join AO and extend it till it meets point B on the circumference of the circle,
(i) Measure the length of AB.
(ii) Assign a special name to AB.
Solution:
Question 3.
Draw circle with diameter :
(i) 6 cm
(ii) 8.4 cm.
In each case, measure the length of the radius of the circle drawn.
Solution:
Question 4.
Draw a circle of radius 6 cm. In the circle, draw a chord AB = 6 cm.
(i) If O is the centre of the circle, join OA and OB.
(ii) Assign a special name to ∆AOB
(iii) Write the measure of angle AOB.
Solution:
Question 5.
Draw a circle of radius 4.8 cm and mark its centre as P.
(i) Draw radii PA and PB such that ∠APB = 45°.
(ii) Shade the major sector of the circle
Solution:
Question 6.
Draw a circle of radius 3.6 cm. In the circle, draw a chord AB = 5 cm. Now shade the minor segment of the circle.
Solution:
Question 7.
Mark two points A and B ,4cm a part, Draw a circle passing through B and with A as a center
Solution:
Question 8.
Draw a line AB = 8.4 cm. Now draw a circle with AB as diameter. Mark a point C on the circumference of the circle. Measure angle ACB.
Solution:
### The Circle Exercise 29B – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
Construct a triangle ABC with AB = 4.2 cm, BC = 6 cm and AC = 5cm. Construct the circumcircle of the triangle drawn.
Solution:
Question 2.
Construct a triangle PQR with QR = 5.5 cm, ∠Q = 60° and angle R = 45°.
Construct the circumcircle cif the triangle PQR.
Solution:
Question 3.
Construct a triangle ABC with AB = 5 cm, ∠B = 60° and BC = 6. 4 cm.
Draw the incircle of the triangle ABC.
Solution:
Question 4.
Construct a triangle XYZ in which XY = YZ= 4.5 cm and ZX = 5.4 cm. Draw the circumcircle of the triangle and measure its circumradius.
Solution:
Question 5.
Construct a triangle PQR in which, PQ = QR = RP = 5.7 cm. Draw the incircle of the triangle and measure its radius.
Solution:
### The Circle Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
The centre of a circle is at point O and its radius is 8 cm. State the position of a point P (point P may lie inside the circle, on the circumference of the circle, or outside the circle), when :
(a) OP = 10.6 cm
(b) OP = 6.8 cm
(c) OP = 8 cm
Solution:
Question 2.
The diameter of a circle is 12.6 cm. State, the length of its radius.
Solution:
Question 3.
Can the length of a chord of a circle be greater than its diameter ? Explain.
Solution:
Question 4.
Draw a circle of diameter 7 cm. Draw two radii of this circle such that the angle between these radii is 90°. Shade the minor sector obtained. Write a special name for this sector.
Solution:
Question 5.
State, which of following statements are true and which are false :
(i) If the end points A and B of the line segment lie on the circumference of a circle, AB is a diameter.
(ii) The longest chord of a circle is its diameter.
(iii) Every diameter bisects a circle and each part of the circle so obtained is a semi-circle.
(iv) The diameters of a circle always pass through the same point in the circle.
Solution: |
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## Unit 6 - Day 10
##### All Units
###### ​Learning Objectives​
• Evaluate definite integrals analytically using the Fundamental Theorem of Calculus.
​
###### ​Success Criteria
• I can apply rules for finding the antiderivative of a function.
• I can use the FTC to evaluate definite integrals by finding the difference of the antiderivative evaluated at the limits of integration.
​
​
​
​
​
# Lesson Handout
###### Overview
A figure skater’s velocity function serves as the context for our second investigation of the Fundamental Theorem of Calculus. Students will use analytic methods (writing a possible position function), numeric methods (using Math:9 on their calculator or other numeric integration device), and verbal interpretations to construct meaning from the FTC. (We will explore graphical relationships in tomorrow’s lesson on Topic 6.8.) Then, using initial condition information (position), students find a particular solution. Today’s work serves as a precursor to Topic 6.8 and the all-important “+C”!
##### ​
###### Teaching Tips
Groups easily discovered that the integral value given by their calculator (Activity question 2) was equal to the value found by substitution into their position function (Activity question 4). At this point, teachers should cultivate the “Aha!” moment in the classroom. This is an amazing result and should be celebrated! This is the conclusion to write in the margin for Activity question 5, as well. Be sure to point out that we are now able to find exact areas under graphs without resorting to geometric shapes and formulas.
If time permits, relate the rearrangement of the FTC to the slope-intercept form of a line, y = mx + b: F(a) is the initial condition or initial value, b, and the integral becomes the accumulated changed, mx, where the integrand fills the role of m, the rate of change, with dx directing distance along the x-axis.
​
###### Exam Insights
Many, many, many non-calculator FRQs require the use of geometric regions between the rate of change graph and the x-axis when investigating functions defined by integrals. The FTC opens many options for students to evaluate integrals, but they must keep their geometry skills sharp, too!
​
###### Student Misconceptions
“Be sure to point out that we are now able to find exact areas under graphs without resorting to geometric shapes and formulas.” But don’t let students assume that they will never need to use geometric shapes and formulas.
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# Warm Up Every weekday morning, cousins Ainsley, Jack, and Caleb are given a different amount of money for lunch by their parents. Ainsley gets \$3, Jack.
## Presentation on theme: "Warm Up Every weekday morning, cousins Ainsley, Jack, and Caleb are given a different amount of money for lunch by their parents. Ainsley gets \$3, Jack."— Presentation transcript:
Warm Up Every weekday morning, cousins Ainsley, Jack, and Caleb are given a different amount of money for lunch by their parents. Ainsley gets \$3, Jack receives \$4, and Caleb gets \$5. On the way to school, they stop by their Grandpa Ray’s house to say hello. He always gives them each \$2 more for lunch. By the time the cousins reach the bus stop, they each have more money than when they left the house. Mathematically, one way to describe this story would be to assign each of the cousins a value, and define Grandpa Ray as a function. Example: Let Ainsley = \$3, Jack = \$4, and Caleb = \$5. We can then define Grandpa Ray as a function, R, where R “of” x dollars, written R(x), is x + \$2. Then we can say R(Ainsley) = R(\$3) = \$5, R(Jack) = R(\$4) = \$6, and R(Caleb) = R(\$5) = \$7. The actual cost of lunch for each cousin is \$2. Using the values assigned to the cousins in the example and defining lunch as the function, L, use function notation to symbolize paying for lunch.
Transformations as Functions
Definitions Transformation: a change in a geometric figure’s position, shape, or size Translation: slide Reflection: flip Rotation: turn Congruent: figures are congruent if they have the same shape, size, lines, and angles
Preimage: the original figure before undergoing a transformation
Image: the new, resulting figure after a transformation Isometry: a transformation in which the preimage and image are congruent
Transformations as Functions
Each coordinate on the coordinate plane is a real number pair, (x,y). When a transformation is applied to a set of points, then all points in the set are moved. For example if T(x,y) = (x+h, y+k), then T(ΔABC) would be:
Identity Function
One-to-one: each point in the set of points will be mapped to exactly one other point
Horizontal stretches and dilations are not isometric.
1. Given the point P (5, 3) and T(x, y) = (x + 2, y + 2), what are the coordinates of T(P)?
2. Given ABC : A(5,2), B(3,5), and C (2,2), and the transformation T(x, y) = (x, –y), what are the coordinates of the vertices of T (ABC)? What kind of transformation is T?
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# RD Sharma Solutions Class 10 Some Applications Of Trigonometry Exercise 12.1
### RD Sharma Class 10 Solutions Chapter 12 Ex 12.1 PDF Free Download
#### Exercise 12.1
Q1. A tower stands vertically on the ground. From a point on the ground, 20 m away from the foot of the tower, the angle of elevation of the top the tower is $60^{\circ}$. What is the height of the tower?
Soln:
Given:
Distance between point of observation and foot of tower = 20 m = BC
Angle of elevation of top of tower = $60^{\circ} = \Theta$
Height of tower H = AB
Now from fig ABC
$\Delta ABC$ is a right angle triangle
$\frac{1}{tan\Theta }= \frac{opposite\;side \left ( AB \right )}{Adjacent\;side\left ( BC \right )}$
i.e $\tan C =\frac{AB}{BC}$
AB = 20 $\tan 60^{\circ}$
H = $20\times \sqrt{3}$
= $20\sqrt{3}$
$∴ height\;of\;the\;tower = 20\sqrt{3}m$
Â
Q2.The angle of elevation of a ladder against a wall is $60^{\circ}$ and the foot of the ladder is 9.5 m away from the wall. Find the length of the ladder.
Soln:
Given
Distance between foot of the ladder and wall = 9.5 m
Angle of elevation $\Theta = 60^{\circ}$
Length of the ladder = L = AC
Now figure forms a right angle triangle ABC
We know
$\cos \Theta =\frac{adjacent\;side}{hypotenuse}$
$\cos 60^{\circ}=\frac{BC}{AC}$
$\frac{1}{2} =\frac{9.5}{AC}$
AC = $2\times 9.5 =19m$
$∴$ length of the ladder (L) = 19 m
Q3.A ladder is placed along a wall of a house such that its upper end is touching the top the wall. The foot of the ladder is 2 m away from the wall and the ladder is making an angle of $60^{\circ}$ with the level of the ground. Determine the height of the wall.
Soln:
Distance between foot of the ladder and wall = 2 m =BC
Angle made by ladder with ground $\Theta =60^{\circ}$
Height of the wall H = AB
Now figure of ABC forms a right angle triangle
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan 60^{\circ} = \frac{AB}{BC}$
$\sqrt{3} =\frac{AB}{2}$
AB = $2\sqrt{3}$
$∴ height\;of\;the\;wall\;= 2\sqrt{3}m$
Q4.An electric pole is 10 m high. A steel wire to top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of $45^{\circ}$ with the horizontal through the foot of the pole, find the length of the wire.
Soln:
Height of the electric pole H = 10 m = AB
Angle made by steel wire with ground (horizontal) $\Theta = 45^{\circ}$
Let length of wire = L = AC
If we represent above data in from of a figure then it forms a right angle triangle ABC.
Here
$\sin \Theta =\frac{opposite\;side}{hypotenuse}$
$\sin 45^{\circ}= \frac{AB}{AC}$
$\frac{1}{\sqrt{2}}= \frac{10m}{L}$
L = $10 \sqrt{2}m$
$∴\;length\;of\;the\;wire (L) = 10 \sqrt{2}m$
Q5.A kite is flying at a height of 75 meters from the ground level, attached to a string inclined at $60^{\circ}$ to the horizontal. Find the length of the string to the nearest meter.
Soln:
Given
Height of kite from ground = 75 m = AB
Inclination of string with ground $\Theta = 60^{\circ}$
Length of string L = AC
If we represent the above data in form of figure as shown the it forms a right angle triangle ABC
Here,
$\sin \Theta =\frac{opposite\;side}{hypotenuse}$
$\sin 60^{\circ}= \frac{AB}{AC}$
$\frac{\sqrt{3}}{2}= \frac{75}{L}$
L = $50\sqrt{3}m$
Length of string L = $50\sqrt{3}m$
Â
Q6.The length of a string between a kite and a point on a point on the ground is 90 meters. If the string makes an angle $\Theta$ with the ground level such that $\tan \Theta =\frac{15}{8}$. How high is the kite? Assume that there is no slack in the string
 Soln:
length of the string between point on ground and kite = 90 m
Angle made by string with ground is $\Theta$
And $\tan \Theta =\frac{15}{8}$
$\Theta =\tan^{-1} (\ frac{15}{8})$
Height of the kite be ‘H’ m
If we represent the above data in figure as shown the it forms right angle triangle ABC
We have
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \Theta =\frac{AB}{BC}$
$\frac{15}{8}= \frac{H}{BC}$
BC = $\frac{8H}{15}$ —- (a)
In $\Delta ABC$ , by Pythagoras theorem we have
$AC^{2}=BC^{2}+AB^{2}$
$90^{2}=\left (\frac{8h}{15} \right)^{2}+H^{2}$
$90^{2}=\frac{\left( 8H \right)^{2}+\left(15H \right)^{2}}{15^{2}}$
$H^{2}\left ( 8^{2} +15^{2}\right ) = 90^{2} \times 15^{2}$
$H^{2}=\frac{\left (90 \times 15\right )^{2}}{289}$
$H^{2}=\left ( \frac{90\times 15}{17} \right )^{2}$
$H=\frac{90\times 15}{17})$
H = 79.41
$∴ height\;of\;the\;kite\;from\;ground= 79.43m$
Q7.A vertical tower stands on a horizontal place and is surmounted by a vertical flag staff. At a point on the plane 70 meters away from the tower, an observer notices that the angles of elevation of the meters away from the flag staff are respectively $60 ^{\circ}$ and $45 ^{\circ}$. Find the height of the flag staff and that of the tower
Soln:
Given
Vertical tower is surmounted by flag staff.
Distance between tower and observer = 70 m = BC
Angle of elevation of top of tower $\alpha = 45 ^{\circ}$
Angle of elevation of top of flag staff $\beta = 60 ^{\circ}$
Height of flag staff = h = AD
Height of tower = H = AB
If we represent the above data in the figure then it forms right angle triangles $\Delta ABC \;and\;\Delta BCD$
When $\Theta$ is angle in right angle triangle we know that
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
Now ,
$\tan \alpha =\frac{AB}{BC}$
$\tan 45^{circ} =\frac{H}{80}$
H = 70
$∴$ H =70
Again,
$\tan \beta =\frac{DB}{BC}$
$\tan 60^{circ} =\frac{AD+AB}{70}=\frac{h+H}{80}$
h+70 =$80\sqrt{3}$
h = $70\left ( \sqrt{3}-1 \right )$
h = 70 (1.732-1)
h= 51.24 m
$∴$ h = 51.24 m
Height of tower = 80 m and height oh flag staff = 51.24 m
Q8. A vertically straight tree, 15 m high; is broken by the wind in such a way that its top just touches the ground and makes an angle of $60^{\circ}$ with the ground. At what height from the ground did the tree break?
 Soln:
Initial height of tree H = 15 m = AB
Let us assume that it is broken at point C.
Then given that angle made by broken part with the ground $\Theta = 60^{\circ}$
Height from ground to broken points = h = BC
AB = AC + BC
H = AC + h => AC = (H – h) m
If we represent the above data in the figure as shown then it forms right angle triangle ABC from figure
$\sin \Theta =\frac{opposite\;side}{hypotenuse}$
$\sin 60^{\circ} =\frac{BC}{CA}$
$\frac{\sqrt{3}}{2}=\frac{h}{H-h}$
$\sqrt{3}\left(15-h\right)=2h$
$\left(2+\sqrt{3}\right)h=15\sqrt{3}$
$h=\frac{15\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
Rationalizing denominator rationalizing factor of $a+\sqrt{b} \;\;is\;\;a-\sqrt{b}$
$h=\frac{\left ( 15\sqrt{3} \right )\left ( 2-\sqrt{3} \right )}{2^{2}-3^{2}}$
$15\left ( 2\sqrt{3}-3 \right )$
h $=15\left ( 2\sqrt{3}-3 \right )$
$∴$ height of broken points from ground = $=15\left ( 2\sqrt{3}-3 \right )m$
Q9.A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 meters. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are espectively $30^{\circ}$ and $60^{\circ}$. Find the height of the tower.
Soln:
Height of the flag staff h = 5 m =AP
Angle of elevation of the top of flag staff = $60^{\circ}$ = $\alpha$
Angle of elevation of the bottom of the flagstaff = $60^{\circ} = \beta$
Let height of tower be ‘H’ m = AB
If we represent the above data in forms of figure then it forms a right angle triangle CBD in which ABC is included
In right angle triangle, if
Angle is $\Theta$ then
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{BD}{BC}$
$\tan 60^{circ} =\frac{AB+AD}{BC}$
$\sqrt{3}=\frac{H+5}{BC}$ —- (a)
$\tan \beta =\frac{AB}{BC}$
$\tan 30^{circ} =\frac{H}{BC}$
$\frac{1}{\sqrt{3}}=\frac{H}{BC}$ —- (b)
Equating (1) and (2)
$\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=\frac{\frac{H+5}{BC}}{\frac{H}{BC}}$
3 = $\frac{H+5}{H}$
3H = H + 5
H = 2.5 m
Height of the tower = 2.5 m
Q10.A person observed the angle of elevation of a tower as $30^{\circ}$. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as $60^{\circ}$. Find the height of the tower.
Soln:
Given
Angle of elevation of top of tour from first point of elevation $\left (A \right )\alpha =30^{\circ}$
Let the walked 50 m from first point (A) to (B) then AB = 50 m
Angle of elevation from second point $\left (B \right )\beta =60^{\circ}$
Now let us represent the given data in form of figure
Then it forms $\Delta ACD\; \Delta BCD$ in which $\angle c = 90^{\circ}$
Let the height of the tower be ‘H’ m = CD
BC = x m.
If in a right angle triangle $\Theta$is the angle then
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{CD}{AC}$
tan 30° =$\frac{H}{AB+BC}$
$\frac{1}{\sqrt{3}}=\frac{H}{50+x}$
50+x = $H sqrt{3}$ —– (a)
$\tan \beta =\frac{CD}{BC}$
$\tan 60^{circ} =\frac{H}{x}$
${\sqrt{3}}=\frac{H}{x}$
x= $\frac{H}{\sqrt{3}}$ —- (b)
Equating (a) and (b)
=> 50 + $\frac{H}{\sqrt{3}}$ =$H\sqrt{3}$
=> $H\sqrt{3}-\frac{H}{\sqrt{3}}$ = 50
=> H = $25\sqrt{3}$
$∴$ height of tower H = $25\sqrt{3}$
Q11.The shadow of a tower, when the angle of elevation of the sun is $45^{\circ}$, is found to be 10 m longer then when it was $60^{\circ}$. Find the height of the tower
Soln:
let the length of the shadow of the tower when angle of elevation is $60^{\circ}$ be x m = BC. Then according to problem
Length of the shadow with angle of elevation $45^{\circ}$ is (10+x)m = BD.
If we represent the above data in the foem of figure then its forms a triangle ABD and triangle ABC where $\angle B = 90^{\circ}$
Let height of the tower be ‘H’ m = AB
If in right angle triangle one of the angle is $\Theta$ then
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{AB}{BC}$
$\tan 60^{circ} =\frac{H}{x}$
x= $\frac{H}{\sqrt{3}}$ —- (a)
$\tan \beta =\frac{AB}{BD}$
$\tan 45^{circ} =\frac{H}{x+20}$
x+10 = H
x = H – 10 —– (b)
Substitute x = H – 10 in (a)
H – 10 = $\frac{H}{\ sqrt{3}}$
$\sqrt{3}H – 10\sqrt{3}=H$
$H=\frac{10\sqrt{3}}{\sqrt{3}-1}$
$H=\frac{10\sqrt{3}\times \left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\times \left ( \sqrt{3}+1 \right )}$
$H=\frac{10\sqrt{3}\left ( \sqrt{3}+1 \right )}{2}$
H = $5\left ( 3+\sqrt{3} \right )$
H = 23.66 m
Q12.A parachute is descending vertically and makes angles of elevation of $45^{\circ}$ and $60^{\circ}$ at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of point where he falls on the ground from the just observation point.
Soln:
let the parachute at highest point A. let C and D be points which are 100 m apart on ground where from then CD = 100 m
Angle of elevation from point C = $45^{\circ}= \alpha$
Angle of elevation from point D = $60^{\circ}= \beta$
Let B be the point just vertically down the parachute
Let us draw figure according to above data then it forms the figure as shown in which ABC and ABD are two triangles
Maximum height of the parachute from the ground
AB = H m
Distance of point where parachute falls to just nearest observation point = x m
If in right angle triangle one of the included angle is $\Theta$ then
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{AB}{BC}$
$\tan 45^{circ} =\frac{H}{x100}$
100 + x = H —- (a)
$\tan \beta =\frac{AB}{BD}$
$\tan 60^{circ} =\frac{H}{x}$
H = $\sqrt{3}x$ —- (b)
From (a) and (b)
$\sqrt{3}x = 100 + x$
$\left ( \sqrt{3} -1\right )x= 100$
$x= \frac{100}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$x= \frac{100\left ( \sqrt{3}+1 \right )}{2}$
$x= 50\left ( \sqrt{3}+1 \right )$
x = 136.6 m in (b)
$H = \sqrt{3}\times 136.6.\;=\; 236.6 m$
Maximum height of the parachute from the ground
$H =\;236.6 m$
Distance between the two points where parachute falls on the ground and just the observation is x = Â Â 136.6 m
Q13. On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are $Â 45^{\circ}$ and $60^{\circ}$. If the height of the tower is 150 m, find the distance between the objects.
Soln:
Height of the tower, H = AB = 150 m
Let A and B be two objects on the ground.
Angle of depression of object A’ $\left [ \angle {A}’Ax \right ] = \beta = 45^{\circ}= \angle A{A}’B\;\;\left [ Ax\left | \right | {A}’B\right ]$
Angle of depression of object B’ $\left [ \angle xA{B}’ \right ] = \alpha = 60^{\circ}= \angle A{B}’B\;\;\left [ Ax\left | \right | {A}’B\right ]$
Let A’B’ = x        ;              B’B = y
If we represent the above data in form of a figure, Then it forms a right angle triangle with $\angle B = 90^{\circ}$
In any right angle triangle if one of the include angle is $\Theta$ then
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan 60^{circ} =\frac{150}{y}$
y = $\frac{150}{\sqrt{3}}$ —- (a)
$\tan \beta =\frac{AB}{A’B}$
$\tan 45^{circ} =\frac{150}{x+y}$
x + y = 150 —- (b)
Substituting (a) in (b)
$x+\frac{150}{\sqrt{3}}=150$
$x+\frac{5\times 3}{\sqrt{3}}=150$
$x=150-50\sqrt{3}$
x = 63.4 m
Distance between object A’B’ = 63.4 m
Q14.The angle of elevation of a tower from a point on the same level as the foot of the tower is $30^{\circ}$. On advancing 150 meters towards the foot of the tower, the angle of elevation of the tower becomes $60^{\circ}$. Show that the height of the tower is 129.9 m.
Soln:
Angle of elevation of top tower from first point A, $\alpha = 30^{\circ}$
Let us advance through A to B by 150 m then AB = 150 m
Angle of elevation of top of tower from second point B, $\beta = 60^{\circ}$
Let height of tower CD = H m
If we represent the above data in form of figure then it form a figure as shown with $\angle D=90^{\circ}$
In right angle triangle, one of included angle is $\Theta$
Then $\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{CD}{AD}$
$\tan 30^{\circ} Â =\frac{H}{150 + x}$
150 + x = $H\sqrt{3}$ —- (a)
$\tan \beta =\frac{CD}{AD}$
$\tan 60^{\circ} Â =\frac{H}{x}$
H = $x\sqrt{3}$
x = $\frac{H}{\sqrt{3}}$ —- (b)
Substituting (b) in (a)
150 + $\frac{H}{\sqrt{3}}= H\sqrt{3}$
$H\left ( \sqrt{3}-\frac{1}{\sqrt{3}} \right ) = 150$
$H\left ( \frac{3-1}{\sqrt{3}} \right )=150$
$H= \frac{150\times \sqrt{3}}{2}$
H = $75\sqrt{3}$
H = 129.9 m
$∴$ height of the tower = 129.9 m
Â
Q16.The angle of elevation of the top of a tower from a point A on the ground is $30^{\circ}$. Moving a distance of 20 meters towards the foot of the tower to a point B the angle of elevation increases to $60^{\circ}$. Find the height of tower and the distance of the tower from the point A.
Soln:
Angle of elevation of top of the tower from point A, $\alpha = 30^{\circ}$.
Angle of elevation of top of tower from point B, $\beta = 60^{\circ}$.
Distance between A and B, AB = 20 m
Let height of tower CD = ‘h’ m
Distance between second point B from foot of the tower be ‘x’ m
If we represent the above data in form of figure then it form a figure as shown with $\angle D=90^{\circ}$
In right angle triangle, one of included angle is $\Theta$
Then $\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{CD}{AD}$
$\tan 30^{\circ} Â =\frac{h}{20 + x}$
20 + x = $h\sqrt{3}$ —- (a)
$\tan \beta =\frac{CD}{BD}$
$\tan 60^{\circ} Â =\frac{h}{x}$
x = $\frac{h}{\sqrt{3}}$ —- (b)
Substituting (b) in (a)
20 + $\frac{h}{\sqrt{3}}= h\sqrt{3}$
$h\left ( \sqrt{3}-\frac{1}{\sqrt{3}} \right ) = 20$
$h\left ( \frac{3-1}{\sqrt{3}} \right )=20$
$h= \frac{20\times \sqrt{3}}{2}$
h = $10\sqrt{3}$
h = 17.32 m
x = $\frac{10\sqrt{3}}{\sqrt{3}}$
x = 10 m
Height of the tower = 17.32 m
Distance of the tower from point A = (20 + 10) = 30 m
Q17. From the top of a building 15 m high the angle of elevation of the top of tower is found to be $30^{\circ}$. From the bottom of the same building, the angle of elevation of the top of the tower is found to be $60^{\circ}$. Find the height of the tower and the distance between the building and the tower.
Soln:
Let AB be the building and CD be the tower.
Height of the building is 15 m = h = AB
Angle of elevation of top of the tower from top of the building, $\alpha = 30^{\circ}$.
Angle of elevation of top of the tower from bottom of the building, $\beta = 60^{\circ}$.
Distance between the tower and the building BD = x
Let height of the tower above building be ‘a’ m
If we represent the above data in form of figure then it form a figure as shown with $\angle D=90^{\circ} \;also\; AX\left | \right |BD,\; \angle AXC =90^{\circ}$
Here ABDX is a rectangle
$∴$ BD = DX = ’x’ m AB = XD = h = 15 m
In right triangle if one of the include angle is $\Theta$
Then $\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{CX}{AX}$
$\tan 30^{\circ} Â =\frac{a}{ x}$
x = $a\sqrt{3}$ —- (1)
$\tan \beta =\frac{CD}{BD}$
$\tan 60^{\circ} Â =\frac{a + 15}{x}$
$x\sqrt{3}= a+15$ —- (2)
Substituting (1) in (2)
$\frac{x}{x\sqrt{3}}= \frac{a\sqrt{3}}{a+15}$
a + 15 = $a\sqrt{3}\left( \sqrt{3}\right) Â$
a + 15 = 3a
2a = 15
a = 15/2 = 7.5 m
x = $a\sqrt{3}$
x = $7.5\times\sqrt{3}$
x = 12.99 m
Height of the tower above ground = h + a
= 15 + 7.5
= 22.5 m
Distance between tower and building = 12.99 m
Q18.On a horizontal plane there is a vertical tower with a flag pole on the top of the tower. At a point 9 m away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are $60^{\circ}$ and $30^{\circ}$ respectively. Find the height of the tower and the flag pole mounted on it.
Soln:
let AB be the tower and BC be the flagstaff on the tower
Distance of the point of observation from foot of the tower BD = 9 m
Angle of elevation of top of flagstaff, $\alpha = 60^{\circ}$
Angle of elevation of top of flagstaff, $\beta = 30^{\circ}$
Let height of the tower = ‘x’ = AB
Height of the pole = ‘y’ = BC
If the above data is represented in form of figure a shown with $\angle A=90^{\circ}$
In right angle triangle, one of included angle is $\Theta$
Then $\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{AC}{AD}$
$\tan 60^{\circ} Â =\frac{x+y}{9}$
x + y= $9\sqrt{3}$ —- (a)
$\tan \beta =\frac{AC}{AD}$
$\tan 30^{\circ} Â =\frac{x}{9}$
x = $\frac{9}{\sqrt{3}}$
x = $3\sqrt{3}$
x = 5.196 m —- (b)
Substituting (b) in (a)
y = $9\sqrt{3}-\;3\sqrt{3}$
y = $6\sqrt{3}$
y = 10.392 m
$∴$ Height of the tower x = 5.196 m
Height of the tower pole = 10.392 m
Q19.A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of $30^{\circ}$ with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Soln:
Let the initial height be AB
Let us assumed that the tree is broken at point C
Angle made by the broken part CB’ with ground is $30^{\circ}=\Theta$
Distance between foot of tree to point where it touches the ground B’A = 8 m
Height of the tree = h = AC + CB’ = AC + CB
The above information is represent in the form of figure as shown
$\cos \Theta =\frac{Adjacent\;side}{Hypotenuse}$
$\cos 30^{\circ} =\frac{AB’}{CB’}$
$\frac{\sqrt{3}}{2} =\frac{8}{CB’}$
CB’ = $\frac{16}{\sqrt{3}}$
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan 30^{\circ}  =\frac{CA}{AB’}$
$\frac{1}{\sqrt{3}} =\frac{CA}{8}$
CA = $\frac{8}{\sqrt{3}}$
Height of the tree = CB’ + CA = $\frac{16}{\sqrt{3}} + \frac{8}{\sqrt{3}}$
h = $\frac{24}{\sqrt{3}}$
h= $8\sqrt{3}m$
Q20.From a point P on the ground the angle of elevation of a 10 m tall building is $30^{\circ}$. A flag is hoisted at the top of the building and the angle of elevation of the flag staff from P is $45^{\circ}$. Find the length of the flag staff and the distance of the building from the point P.
Soln:
let AB be the tower and BD be the flag staff
Angle of elevation of top of the building from P $\alpha = 30^{\circ}$
AB = height of the tower = 10 m
Angle of elevation of top of flag staff from P $\beta = 45^{\circ}$
Let height of the flagstaff BD = ‘a’ m
The above information is represented in the form of figure as shown
with $\angle A=90^{\circ}$
In a right angle triangle if one of included angle is $\Theta$
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{AB}{AP}$
$\tan 30^{\circ} Â =\frac{10}{AP}$
AP = $10\sqrt{3}m$
AP = 17.32 m
$\tan \beta =\frac{AD}{AP}$
$\tan 45^{\circ} Â =\frac{10+ a}{AP}$
10 + a = AP
a = 17.32 – 10 m
a = 7.32 m
Height of the flag staff ‘a’ = 7.32 m
Distance between P and foot of the tower = 17.32 m
Q21.A 1.6 m tall boy stands at a distance of 3.2 m from a lamp post and casts a shadow of 4.8 m on the ground. Find the height of the lamp post by using
(i) trigonometric ratio
(ii)properties of similar triangles
Soln:
Let AC be the lamp post of height ‘h’
We assume that ED = 1.6 m, BE = 4.8 m and EC = 3.2
We have to find the height of the lamp post
Now we have to find the height of the lamp post using similar triangles
Since triangle BDE and triangle ABC are similar
$\frac{AC}{BC}=\frac{ED}{BE}$
$\frac{h}{4.8+3.2}=\frac{1.6}{4.8}$
$h=\frac{8}{3} m$
Again we have to find the height of lamp post using trigonometric ratio
In $\Delta ADE\;\;\tan \Theta =\frac{1.6}{4.8}$
$\tan \Theta =\frac{1}{3}$
Again In $\Delta ABC$
$\tan \Theta =\frac{h}{4.8+3.2}$
$\frac{1}{3}=\frac{h}{8}$
$h=\frac{8}{3}$
Hence the height of the lamp post is $h=\frac{8}{3} m$
Q24.From a point on the ground the angle of elevation of the bottom and top of a transmission lower fixed at the top of 20 m high building are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the transmission tower.
Soln:
given height of building = 20 m = AB
Let height of tower above building = ‘h’ = BC
Height of tower + building = (h + 20) m [from ground] = CA
Angle of elevation of bottom of tour, $\alpha = 45^{\circ}$
Angle of elevation of top of tour, $\beta = 60^{\circ}$
Let distance between tower and observation point = ‘x‘m
The above data is represented in the form of figure as shown is
If one of the include angle in right angle triangle is $\Theta$
Then $\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{AB}{AD}$
$\tan 45^{\circ} Â =\frac{20}{x}$
x = 20 m
$\tan \beta =\frac{CA}{DA}$
$\tan 60^{\circ} Â =\frac{h + 20}{x}$
h + 20 = $20\sqrt{3}m$
h = $20 \left( \sqrt{3}-1\right)m$
Height of the tower h = $20 \left( \sqrt{3}-1\right)m$
h = $20 \left( 1.73 – 1\right)m$
h = 20 x .732
h = 14.64 m
Q25.The angle of depression of the top and bottom of 8 m tall building from the top of a multistoried building are $30^{\circ}$ and $45^{\circ}$. Find the height of the multistoried building and the distance between the two buildings.
Soln:
Let height of multistoried building ‘h’ m =AB
Height of the tall building = 8 m =CD
Angle of depression of top of the tall building, $\alpha = 30^{\circ}$
Angle of depression of bottom of the tall building, $\beta = 30^{\circ}$
Distance between the two buildings = ‘x’ m = BD
Let AO = s
AB = AO + BO but BO = CD          $\left [ ∵ AOCD\; is\;rectangle \right ]$
AB = (s + 8) m
The above information is represented in the form of figure as shown
If in right angle triangle one of the included angle is $\Theta$
Then $\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
In $\Delta AOC$
$\tan 30^{\circ} Â =\frac{AO}{CO}$
$\frac{1}{\sqrt{3}} =\frac{s}{x}$
$x\;=s\sqrt{3}$ —- (1)
In $\Delta ABD$
$\tan 45^{\circ} Â =\frac{AB}{BD}$
1 = $\frac{s+8}{x}$
s + 8 = x —- (2)
Substituting (a) in (2)
s + 8 = $s\sqrt{3}$
$s \left( \sqrt{3}-1 \right)=8$
$s=\frac{8}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$s=4\left( \sqrt{3}+1 \right)$
$s=4\left( \sqrt{3}+1 \right)m$
$x=4\left( 3+\sqrt{3} \right)m$
$AB=s+8= 8+4\left( \sqrt{3}+1 \right)$
Q26.A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, angle of elevation of the top of the statue is $60^{\circ}$ and from the same point the angle of elevation of the top of the pedestal is $45^{\circ}$. Find the height of the pedestal.
Soln:
let the height of the pedestal be ‘h’ m
Height of the statue = 1.6 m
Angle of elevation of the top of the statue $\alpha = 30^{\circ}$
Angle of elevation of the top of the pedestal $\beta = 30^{\circ}$
The above data is represented in the form of figure as shown.
If in right angle triangle one of the included angle is $\Theta$
Then $\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{BC}{DC}$
$\tan 45^{\circ} Â =\frac{h}{DC}$
DC = ‘h’ m —- (a)
$\tan \beta =\frac{AC}{DC}$
$\tan 60^{\circ} Â =\frac{h+1.6}{DC}$
DC = $\frac{h+1.6}{\sqrt{3}}$ —- (b)
From (a) and (b)
h = $\frac{h+1.6}{\sqrt{3}}$
$h \times\sqrt{3}=h+1.6$
$h(\ sqrt{3}-1) =1.6$
h = $\frac{1.6}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{ sqrt{3}+1}$
h = $0.5\left(\ sqrt{3}+1\right)$
Height of the pedestal = $0.6\left(\ sqrt{3}+1\right)$
Q27.A T.V. tower stands vertically on a bank of a river of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^{\circ}$. From a point 20 m away this point on the same bank, the angle of elevation of the top of the tower is $30^{\circ}$. Find the height of the tower and the width of the river.
Soln:
let AB be the T.V tower of height ‘h’ m on the bank of river and ‘D’ be the point on the opposite side of the river. An angle of elevation at the top of the tower is $30^{\circ}$
Let AB = h and BC = x
Here we have to find height and width of the river.
The above data is represented in the form of figure as shown.
In $\Delta ACB$
$\tan 60^{\circ} Â =\frac{AB}{BC}$
$\sqrt{3} =\frac{h}{x}$
$\sqrt{3}x = h$
$x =\frac{h}{\sqrt{3}}$
Again in $\Delta DBA$,
$\tan 30^{\circ} Â =\frac{AB}{BD}$
$\frac{1}{\sqrt{3}} =\frac{h}{20+x}$
$\sqrt{3}h = 20+x$
$\sqrt{3}h = 20+\frac{h}{\sqrt{3}}$
$\sqrt{3}h-\frac{h}{\sqrt{3}} = 20$
$\frac{2h}{\sqrt{3}} = 20$
h = $15 \sqrt{3}$
$x =\frac{h}{ \sqrt{3}}=\frac{15\sqrt{3}}{ \sqrt{3}}$
x = 10
Hence the height of the tower is $10\sqrt{3}m$ and width if the river is 10 m.
Q28.From the top of a 7 m high building, the angle of elevation of the top of a cable is $60^{\circ}$ and the angel of depression of its foot is $45^{\circ}$. Determine the height of the tower.
Soln:
Given
Height of the building = 7 m = AB
Height of the cable tower = ‘H’ m = CD
Angle of elevation of the top of the building $\alpha = 60^{\circ}$
Angle of elevation of the bottom of the building $\beta = 45^{\circ}$
The above data is represented in form of figure as shown
Let CX = ‘x’ m
CD = DX + XC = 7 m +’x’ m
= x + 7m
In $\Delta ADX$
$\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan 45^{\circ} Â =\frac{7}{AX}$
AX = 7m
$\tan 60^{\circ} Â =\frac{XC}{AX}$
$sqrt{3} =\frac{x}{H}$
x = $7sqrt{3}$
But CD = x + 7
= $7sqrt{3}+7$
= $7 \left( sqrt{3}+1 \right)m$
Height of the cable tower = $7 \left( sqrt{3}+1 \right)m$
Â
Q29.As observed from the top of a 75 m tall lighthouse, the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Soln:
Given;
Height of the lighthouse = 75m = ‘h’ m = AB
Angle of depression of ship 1, $\alpha = 30^{\circ}$
Angle of depression of bottom of the tall building, $\beta = 45^{\circ}$
The above data is represented in form of figure as shown
Let distance between ships be ‘x’ m = CD
If in right angle triangle one of the included angle is $\Theta$
Then $\tan \Theta =\frac{opposite\;side}{adjacent\;side}$
$\tan \alpha =\frac{AB}{DB}$
$\tan 30^{\circ} =\frac{75}{x+BC}$
x + BC = $75sqrt{3}$ —- (1)
$\tan \beta =\frac{AB}{BC}$
$\tan 45^{\circ} =\frac{75}{BC}$
BC = 75 —- (2)
Substituting (2) in (1)
x + 75 = $75sqrt{3}$
x = $75\left( sqrt{3}-1 \right)$
$∴$ distance between ships = ‘x’ m
x = $75\left( sqrt{3}-1 \right)$
Q30.The angle of elevation of the top of the building from the foot of the tower is $30^{\circ}$ and the angle of the top of the tower from the foot of the building is $60^{\circ}$. If the tower is 50m high, find the height of the building.
Soln:
Angle of elevation of top of the building from foot of tower $=30^{\circ}=\alpha$
Angle of elevation of top of the building from foot of tower $=60^{\circ}=\beta$
Height of the tower = 50m = AB
Height of building = ‘h’ m =CD
The above data is represented in the form of figure as shown
In right triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposit\;side}{Adjacent\;side}$
In $\Delta ABD$
$\tan \beta =\frac{AB}{BD}$
$\tan 60^{\circ}=\frac{50}{BD}$
BD= $\frac{50}{\sqrt{3}}$
$\tan \alpha =\frac{CD}{BD}$
tan 30°=$\frac{h}{\frac{50}{\sqrt{3}}}$
h = $\frac{50}{\sqrt{3}}\times \frac{1}{\sqrt{3}}$
h = $\frac{50}{3}$
$∴$ height of the building = $\frac{50}{3}m$
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Q31.From a point on a bridge across a river the angle of depression of the banks on opposite side of the river are $30^{\circ}$ and $45^{\circ}$ respectively. If the bridge is at the height of 30m from the banks, find the width of the river.
Soln:
Height of the bridge = 30m [AB]
Angle of depression of bank 1 i.e, $\alpha=30^{\circ}B_{1}$
Angle of depression of bank 2 i.e, $\beta=30^{\circ} B_{2}$
Given banks are on opposite sides
Distance between banks $B_{1}B_{2}=BB_{1} + BB_{2}$
The above data is represented in the form of figure as shown
In right angle triangle, if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
In $\Delta ABB_{1}$
$tan \alpha =\frac{AB}{B_{1}B}$
$\tan 30^{\circ}=\frac{30}{ B_{1}B }$
$B_{1}B = 30\sqrt{3}m$
In $\Delta ABB_{2}$
$tan \beta =\frac{AB}{BB_{2}}$
$\tan 45^{\circ}=\frac{30}{ B_{2}B }$
$B_{2}B =30m$
$B_{1}B_{2} =BB_{1}+BB_{2}$
$30\sqrt{3}+30$
$30\left( \sqrt{3}+1\right)$
Distance between banks = $30\left( \sqrt{3}+1\right)m$
Q33.A man sitting at a height of 20 m on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with foot of tree. If the angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are $60^{\circ}$ and $30^{\circ}$ respectively. Find the width of the river.
Soln:
Height of the tree AB = 20m
Angle of depression of the pole 1 feet $\alpha=60^{\circ}B_{1}$
Angle of depression of the pole 2 feet $\beta=30^{\circ}B_{1}$
$B_{1}C_{1}$ be one pole and $\ B_{2}C_{2}$ be the other pole
Given poles are on opposite sides.
Width of the river = $B_{1}B_{2}$
= $B_{1}B+BB_{2}$
The above data is represented in the form of figure as shown
In right angle triangle, if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \alpha =\frac{AB}{B_{1}B}$
$\tan 60^{\circ}=\frac{20}{ B_{1}B }$
$B_{1}B=\frac{20}{\sqrt{3}}$
$tan\beta =\frac{AB}{BB_{2}}$
$\tan 30^{\circ}=\frac{20}{ B_{2}B }$
$B_{2}B = 20\sqrt{3}m$
$B_{1}B_{2}= B_{1}B+BB_{2}$
$\frac{20}{\sqrt{3}}+20\sqrt{3}$
20$\frac{1+3}{\sqrt{3}}= \frac{80}{\sqrt{3}}$
Width of the river = $\frac{80}{\sqrt{3}}m$
Q34.A vertical tower stands on a horizontal plane and is surmounted by a flag staff of height 7m. From a point on the plane, the angle of elevation of the bottom of flag staff is $30^{\circ}$ and that of the top of the flag staff is $45^{\circ}$. Find the height of the tower.
Soln:
Given
Height of the flag staff = 7m = BC
Let height of the tower = ‘h’ m = AB
Angle of elevation of top of the bottom of the flagstaff $\alpha=30^{\circ}$
Angle of elevation of top of the top of the flagstaff $\beta=45^{\circ}$
Points of desecration be ‘p’
The above data is represented in the form of figure as shown
In right angle triangle, if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \alpha =\frac{AB}{AP}$
$\tan 30^{\circ}=\frac{h}{AP }$
$AP =h\sqrt{3}$ —- (1)
$\tan \beta =\frac{AC}{AP}$
$\tan 45^{\circ}=\frac{h+7}{AP}$
AP = h + 7 —- (2)
From (1) and (2)
$h\sqrt{3}= h + 7$
$h\sqrt{3}- h = 7$
$h\left( \sqrt{3}- 1 \right) = 7$
$h=\frac{7}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$h=\frac{7\times( \sqrt {3}+1}){2}$
$h=3.5\left( \sqrt{3}+1\right)m$
Height of the tower = $3.5\left( \sqrt{3}+1\right)m$.
Q35.The length of the shadow of a tower standing on level plane is found to be 2x meters longer when the sun’s attitude is $30^{\circ}$ than when it was $30^{\circ}$. Prove that the height of tower is $x\left( \sqrt{3}+1\right) meters$.
Soln:
Let
Length of shadow be ‘a’ m [BC] when sun altitude be $\alpha=45^{\circ}$
Length of shadow be (2x+a) m [BD] when sun altitude be $\beta=30^{\circ}$
Let height of tower be ‘h’ m =AB
The above data is represented in the form of figure as shown
In right angle triangle one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
In $\Delta ABC$
$\tan \alpha =\frac{AB}{BC}$
$\tan 45^{\circ}=\frac{h}{a}$
h = a —- (1)
In $\Delta ADB$
$\tan \beta =\frac{AB}{( 2x + a) BC}$
$\tan 30^{\circ}=\frac{h}{2x+a}$
2x+a = $h\sqrt{3}$ —- (2)
Substituting (1) in (2)
$2x+h = h\sqrt{3}$
$h\left( \sqrt{3}-1\right)=2x$
h = $\frac{2x}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
h = $\frac{2x( \sqrt{3}+1}){2}$
h = $x \sqrt{3}+1)$
Height of the tower = $x( \sqrt{3}+1)m$
Q36.A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of $30^{\circ}$ with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 meters. Find the height of the tree.
Soln:
Let AB be the height of the tree and it is broken at point C and top touches ground at B’
Angle made by the top $\alpha=30^{\circ}$
Distance from foot of tree from point where it touches ground = 10 m
The above data is represented in form of figure as shown
Height of the tree = AB = AC + CB
= AC + CB’
In right angle triangle one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan 30^{\circ}=\frac{AC}{B’A}$
$AC = \frac{10}{\sqrt{3}}m$
Again
$\cos \Theta =\frac{Adjacent\;side}{Hypotenuse}$
$\cos 30^{\circ} =\frac{AB’}{B’C}$
$\frac{\sqrt{3}}{2}=\frac{10}{B’C}$
B’C = $\frac{20}{\sqrt{3}}m$
AB = CA + CB’
$\frac{10}{\sqrt{3}}+\frac{20}{\sqrt{3}}$
= $\frac{30}{\sqrt{3}}$
= $\frac{10}{\sqrt{3}}$
Height of tree = $\frac{10}{\sqrt{3}}m$
Q37.A balloon is connected to a meteorological ground station by a cable of length 215 m inclined at $60^{\circ}$ to the horizontal. Determine the height of the balloon from the ground. Assume that there is no slack in the cable.
Soln:
Length of the cable connected to balloon = 215m [CB]
Angle of inclination of cable with ground $\alpha=60^{\circ}$
Height of the balloon from ground = ‘h’ m = AB
The above data is represented in from of figure as shown
In right angle triangle one of the included angle is $\Theta$
$\sin \Theta =\frac{Opposite\;side}{Hypotenuse}$
$\sin 60^{\circ} =\frac{AB}{BC}$
$\frac{\sqrt{3}}{2}=\frac{h}{215}$
h = $\frac{215\sqrt{3}}{2}$
h = $107.5\sqrt{3}m$
Height of the balloon from ground = $107.5\sqrt{3}m$
Q38. Two men on either side of the cliff 80m high observe the angles of elevation of the top of the cliff to be 300 and 600 respectively. Find the distance between the two men.
Soln:
Height of cliff = 80m = AB
Angle of elevation from Man 1, $\alpha=30^{\circ}\;[M_{1}]$
Angle of elevation from Man 2, $\beta=60^{\circ}\;[M_{2}]$
Distance between two men = $M_{1}M_{2}=BM_{1}+BM_{2}$
The above information is represented in form of figure as shown
In right angle triangle one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \alpha=\frac{AB}{M_{1}B}$
$\tan 30^{\circ}=\frac{80}{M_{1}B}$
$M_{1}B=80\sqrt{3}$
$\tan \beta=\frac{AB}{M_{2}B}$
$\tan 60^{\circ}=\frac{80}{M_{2}B}$
$M_{2}B=frac{80}{\sqrt{3}}$
$M_{1}M_{2}= M_{1}B+BM_{2}$
$80\sqrt{3}+ \frac{80}{sqrt{3}}$
$\frac{80\times 4}{sqrt{3}}$
$\frac{320}{sqrt{3}}$
Distance between men = $\frac{320}{sqrt{3}}\;meters$
Q39.Find the angle of elevation of the sun (sun’s altitude) when the length of the shadow of a vertical pole is equal to its height.
Soln:
Let
Height of the pole = ‘h’ m =sun’s altitude from ground length of shadow be ‘L’
Given that L = h.
Angle of elevation of sun’s altitude be $\Theta$
The above data is represented in from of figure as shown
In right angle triangle one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \Theta =\frac{AB}{BC}$
$\tan \Theta =\frac{h}{L}$
$\tan \Theta =\frac{L}{L}$ [since h = L]
$\Theta =\tan^{-1}\left ( 1 \right )=45^{\circ}$
Angle of sun’s altitude is $45^{\circ}$
Q42.A man standing is on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as $60^{\circ}$ and the angle of depression of the base of the hill as $30^{\circ}$. Calculate the distance of the hill from the ship and the height of the hill.
Soln:
Height of the ship above water level = 8m = AB
Angle of elevation of top of cliff (hill) $\alpha=60^{\circ}$
Angle of depression of the bottom of hill $\beta=30^{\circ}$
Height of the hill = CD
Distance between ship and hill = AX
Height of the hill above ship = CX = ‘a’ m
Height of hill = (a + 8) m
The above data is represented in form of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \alpha=\frac{CX}{AX}$
$\tan 60^{\circ} =\frac{a}{AX}$
AX = $\frac{a}{\sqrt{3}}$
$\tan \beta=\frac{XD}{AX}$
$\tan 30^{\circ}=\frac{B}{AX}$
AX = $8\sqrt{3}$
$∴ \frac {a}{\sqrt{3}} = 8\sqrt{3}$
a = 24m
AX = $8\sqrt{3}m$
$∴$ Height of the cliff hill = (24+8) m = 32m
Distance between hill and ship $8\sqrt{3}m$
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Q44.The angles of depression of two ships from the top of a light house and on the same side of it are found to be $45^{\circ}$ and $30^{\circ}$ respectively. If the ships are 200 m apart, find the height of the light house.
Soln:
Height of the light house AB = ‘h’ meters
Let $S_{1}\; and \;S_{2}$ be ships
Distance between ships $S_{1}S_{2}=200m$
Angle of depression of $S_{1} \;\; \left( \alpha = 30^{\circ}\right)$
Angle of depression of $S_{2} \;\; \left( \beta = 45^{\circ}\right)$
The above data is represented in form of form of figure as shown
In $\Delta ABS_{2}$
$\tan \beta=\frac{AB}{BS_{2}}$
$\tan 45^{\circ}=\frac{h}{BS_{2}}$
$BS_{2}=h$ —- (1)
In $\Delta ABS_{1}$
$\tan \alpha=\frac{AB}{BS_{1}}$
$\tan 30^{\circ}=\frac{h}{BS_{1}}$
$BS_{1} =h\sqrt {3}$ —- (2)
Subtracting (1) from (2)
$BS_{1} – BS_{2} = h\left( sqrt{3} – 1 \right)$
$200 = h\left( sqrt{3} – 1 \right)$
$h = 100 \left( sqrt{3}+1\right) \;meters$
Height of the light house = 273.2 meters
Q45.The angles of elevation from the top of a tower from two points at distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.
Soln:
Height of the tower AB = ‘h’ meters
Let point C be 4 meters from B.
Angle of elevation is $\alpha$ given point D is 9 meters from B, Angle of elevation is $\beta$
Given $\alpha$, $\beta$ are complementary, $\alpha$ + $\beta = 90^{\circ} \;=> =; \beta =90^{\circ} \alpha Â$
Required to prove that h = 6 meters
The above data is represented in the form of figure as shown
In $\Delta ABC$
$\tan \alpha=\frac{AB}{BC}$
$\tan \alpha=\frac{h}{4}$
h = $4\tan \alpha$ —- (1)
In $\Delta ABD$
$\tan \beta=\frac{AB}{BD}$
$\tan \left( 90-\alpha\right) =\frac{h}{9}$
h = $9\tan \alpha$ —- (2)
Multiply (1) and (2)
$h\times h=4\tan \alpha\times 9\cot \alpha$
$h^{2} = 36\left(\tan \alpha\cot \alpha \right)$
$h = \sqrt {36}= 6 m$
Height of the tower = 6 meters
Q46.From the top of a 50 m high tower, the angles of depression of the top and bottom of a pole are observed to be $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the pole.
Soln:
AB = height of the tower = 50m.
CD = height of the pole
Angle of depression of top of building $\alpha=45^{\circ}$
Angle of depression of bottom of building $\beta=60^{\circ}$
The above data is represented in the form of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \alpha=\frac{AX}{CX}$
$\tan 45^{\circ}=\frac{AX}{CX}$
AX = CX
$\tan \beta=\frac{AB}{BD}$
$\tan 60^{\circ} =\frac{50}{BD}$
CX = $\frac{50}{\sqrt{3}}$
AX = $\frac{50}{\sqrt{3}}m = BD$
CD + AB – AX = $50-\frac{50}{\sqrt{3}}$
$\frac{50\left ( \sqrt{3}-1 \right )}{\sqrt{3}}=\frac{50}{3}\left ( 3-\sqrt{3} \right )$
Height of the building (pole) = $\frac{50}{3}\left ( 3-\sqrt{3} \right )$
Distance between the pole and tower = $\frac{50}{\sqrt{3}}m$
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Q47.The horizontal distance between two trees of different heights is 60 m. the angles of depression of the top of the first tree when seen from the top of the second tree is $45^{\circ}$. If the height of the second tree is 80 m, find the height of the first tree.
Soln:
Distance between the trees = 60 m [BD]
Height of second tree = 80 m [CD]
Let height of the first tree = ‘h’ m [AB]
Angle of depression from second tree top to first tree top $\alpha=45^{\circ}$
The above data is represented in form of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
Draw $CX\perp AB$
CX = BD =60m
XB = CD = AB – AX
$\tan \alpha=\frac{AX}{CX}$
$\tan 45^{\circ}=\frac{AX}{60}$
AX = 60 m
XB = CD = AX – AX
= 80 – 60
= 20 m
Height of the second tree = 80 m
Height of the first tree = 20 m
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Q48.A flag staff stands on the top of a 5 m high tower. From a point on the ground, angle of elevation of the top of the flag staff is $60 ^{\circ}$ and from the same point, the angle of elevation of the top of the tower is $45 ^{\circ}$. Find the height of the flag staff.
Soln:
Height of tower = AB = 5
Height of flag staff BC = ‘h’ m
Angle of elevation of top of flagstaff $\alpha = 60 ^{\circ}$
Angle of elevation of bottom of flagstaff $\beta = 45 ^{\circ}$
The above data is represented in form of figure as shown
In $\Delta ABC$
$\tan \beta = \frac{AB}{DA}$
$\tan 45^{\circ}= \frac{5}{DA}$
DA = 5 cm
In $\Delta ADC$
$\tan \alpha = \frac{AC}{DA}$
$\tan 60 ^{\circ} = \frac{AB+BC}{AD} = \frac{h + 5}{5}$
$\sqrt{3} = \frac{h+5}{5}$
h+5 =Â $5\sqrt{3}$
h = $5\left ( \sqrt{3}-1 \right )$ = 3.65 mts
height of the flagstaff = 3.65 mts
Q50.As observed from the top of a 150m tall light house, the angles of depression of two ships approaching it are $30^{\circ}$ and $45^{\circ}$. If one ship is directly behind the other, find the distance between the two ships.
Soln:
Height of the light house AB = 150 meters
Let $S_{1} and S_{2}$ be two ships approaching each other
Angle of depression of $S_{1}, \; \alpha = 30^{\circ}$
Angle of depression of $S_{2}, \; \beta = 45^{\circ}$
Distance between ships = $S_{1}S_{2}$
The above data is represented in the form of figure as shown
In $\Delta ABS_{2}$
$\tan \beta=\frac{AB}{BS_{2}}$
$\tan 45^{\circ}=\frac{150}{BS_{2}}$
$BS_{2} = 150m$
In $\Delta ABS_{1}$
$\tan \alpha=\frac{AB}{BS_{1}}$
$\tan 30^{\circ}=\frac{150}{BS_{1}}$
$BS_{1}=150\sqrt{3}$
$S_{1}S_{2} = BS_{1} – BS_{2} = 150( \sqrt{3}-1$Â meters
Distance between ships = $150( \sqrt{3}-1meters$
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Q51.The angle of elevation of the top of a rock form the top and foot of a 100 m high tower are respectively $30^{\circ}$ and $45^{\circ}$. Find the height of the rock
Soln:
Height of the tower AB = 100 m
Height of rock CD = ‘h’ m
Angle of elevation of the top of rock from top of the tower $\alpha = 30^{circ}$
Angle of elevation of the top of rock from bottom of tower $\beta = 60^{circ}$
The above data is represented in the form of figure as shown
Draw $AX\perp CD$
XD = AB = 100m
XA = DB
In $\Delta CXA$
$\tan \alpha = \frac{CX}{AX}$
$\tan 30^{\circ} =\frac{CX}{DB}$
DB = $CX\times sqrt{3}$ —- (1)
In $\Delta CBD$
$\tan \beta=\frac{CD}{DB}= \frac {100+CX}{DB}$
$\tan 45^{\circ}=\frac{100+CX}{DB}$
DB = 100 + CX —- (2)
From (1) and (2)
$100 + CX = CX \sqrt{3}$
$100 = CX \left( \sqrt{3} -1 \right)$
$CX = \frac {100}{\sqrt{3} – 1} \times \frac {\sqrt{3} + 1}{\sqrt{3} + 1}$
$CX = 50 \left( \sqrt{3} +1\right)$
Height of the hill = $100 + 50 \left( \sqrt{3} +1\right)$
= $150 \left( 3+ \sqrt{3}\right)$ meters
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Q52.A straight highway leads to the foot of a tower of height 50m. From the top of the tower, the angles of depression of two cars standing on the highway are $30^{\circ}$ and $60^{\circ}$ respectively. What is the distance between the two cars and how far is each car from the tower?
Soln:
Height of the tower AB = 50m
$C_{1} and C_{2}$ be two cars
Angles of depression of $C_{1}$ from top of the tower $\alpha = 30^{circ}$
Angles of depression of $C_{2}$ from top of the tower $\beta = 60^{circ}$
Distance between cars $C_{1} and C_{2}$
The above data is represented in form of figure as shown
 In $\Delta ABC_{2}$
$\tan \beta =\frac{Opposite\;side}{Adjacent\;side}= \frac{AB}{C_{2}B}$
$\tan 60^{\circ} =\frac{50}{BC_{2}}$
$BC_{2} =\frac{50}{\sqrt{3}}$
In $\Delta ABC_{1}$
$\tan \alpha = \frac{AB}{C_{1}B}$
$\tan 30^{\circ} =\frac{50}{BC_{1}}$
$BC_{1} =\frac{50}{\sqrt{3}}$
$C_{1} and C_{2}=BC_{1} – BC_{2}$
$50\sqrt{3} – \frac{50}{\sqrt{3}}$
$50\left( \frac{3 – 1}{\sqrt{3}} \right)$
$\frac{100}{\sqrt{3}} = \frac{100}{3}\sqrt{3}$ meters.
Distance between cars $C_{1} and C_{2}= \frac{100}{3}\sqrt{3}$
Distance of car1 from tower = $50\sqrt{3}$ meters
Distance of car2 from tower = $\frac{50}{\sqrt{3}}$ meters
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Q53.From the top of a building AB = 60m, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be $30^{\circ}$ and $45^{\circ}$ respectively. Find
(i) The horizontal distance between AB and CD
(ii) The height of the lamp post
(iii) The difference between the heights of the building and the lamp post.
Soln:
Height of building AB = 60m
Height of lamp post CD = ‘h’ m
Angle of depression of top of lamp post from top of building $\alpha = 30^{circ}$
Angle of depression of bottom of lamp post from top of building $\beta = 60^{circ}$
The above data is represented in form of figure
Draw $DX\perp AB$, CX = AC, CD = AX
In $\Delta BDX$
Then $\tan \alpha =\frac{Opposite\;side}{Adjacent\;side}= \frac{BX}{DX}$
$\tan 30^{\circ} =\frac{60 – CD}{AC}$
$\frac{1}{\sqrt{3}}=\frac{60 – h}{AC}$
AC = $\left( 60 – h\right)\sqrt{3}m$ —- (1)
In $\Delta BCA$
$\tan \beta =\frac{AB}{AC}$
$\tan 60^{\circ} =\frac{60}{AC}$
AC = $\frac{60}{\sqrt{3}}= 20\sqrt{3}m$ —- (2)
From (1) and (2)
$\left( 60 – h\right)\sqrt{3}=20\sqrt{3}$
60 – h = 20
h = 40m
Height of the lamp post = 40m
Distance between lamp post building AC = $20\sqrt{3}m$
Difference between heights of building and lamp post BX = 60 – h => 60 – 40 = 20m.
Q54.Two boats approach a light house in mid sea from opposite directions. The angles of elevation of the top of the light house from two boats are $30^{\circ}$ and $45^{\circ}$ respectively. If the distance between the two boats is 100m, find the height of the light house.
Soln:
let $B_{1}$ be boat 1 and $B_{2}$ be boat 2
Height of light house = ‘h’ m = AB
Distance between $B_{1}B_{2}=100m$
Angle of elevation of A from $B_{1}\;\alpha = 30^{circ}$
Angle of elevation of B from $B_{2}\;\beta = 45^{circ}$
The above data is represented in the form of figure as shown
Here In $\Delta ABB_{1}$
$\tan 30^{\circ} =\frac{Opposite\;side}{Adjacent\;side} = \frac {AB}{BB_{1}}$
$B_{1}B=AB\sqrt{3}=h\sqrt{3}$ —- (1)
In $\Delta ABB_{2}$
$\tan 45^{\circ} = \frac {AB}{BB_{2}}$
h = BB_{2} —- (2)
$B_{1}B_{2}=h\left( \sqrt{3}+1\right)$
h = $\frac {B_{1}B_{2}}{\sqrt{3}+1}$
h= $\frac {100}{\sqrt{3}+1}\times \frac {\sqrt{3}-1}{\sqrt{3}-1}$
h= $\frac {100\left( \sqrt{3}-1\right)}{2}=50\left( \sqrt{3}-1\right)$
Height of the light house = $50\left( \sqrt{3}-1\right)$
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Q55.The angle of elevation of the top of a hill at the foot of a tower is $60^{\circ}$ and the angle of elevation of the top of the tower from the foot of the hill is $30^{\circ}$. If tower is 50 m high, what is the height of the hill?
Soln:
Height of tower AB = 50m
Height of hill CD = ‘h’ m.
Angle of elevation of top of the hill from foot of the tower = $\alpha=60^{\circ}$
Angle of elevation of top of the tower from foot of the hill = $\beta=30^{\circ}$
The above data is represented in the form of figure as shown
From figure
In $\Delta ABC$
$\tan 30^{\circ}= \frac {AB}{BC}$
$\frac{1}{sqrt{3}}= \frac {50}{BC}$
BC = $50\sqrt{3}$
In $\Delta BCD$
$\tan 60^{\circ} =\frac{CD}{BC}=\frac{CD}{50\sqrt{3}}$
$\sqrt{3}= \frac {CD}{50\sqrt{3}}$
CD = 150m
Height of hill = 150m.
Q56. A fire in a building B is reported on telephone to two fire station P and Q 2o km apart from each other on a straight road. P observes that the fire is at an angle of $60^{\circ}$ to the road and Q observes that it is at an angle of $45^{\circ}$ to the road. Which station sends its team and how much will this team have to travel?
Soln:
Let AB be the building
Angle of elevation from point P [fire station 1] $\alpha=60^{\circ}$
Angle of elevation from point Q [fire station 1] $\beta=45^{\circ}$
Distance between fire station PQ = 20 km
The above data is represented in form of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \alpha=\frac{AB}{AP}$
$\tan 60^{\circ} =\frac{AB}{AP}$
$AP=\frac{AB}{\sqrt{3}}$ —- (1)
$\tan \beta=\frac{AB}{AQ}$
$\tan 45^{\circ} =\frac{AB}{AQ}$
AQ = AB —- (2)
AP + AQ = $\frac{AB}{\sqrt{3}}+AB$
$AB\left( \frac{1+\sqrt{3}}{\ sqrt{3}}\right)$
20 = $AB\left( \frac{1+\sqrt{3}}{\ sqrt{3}}\right)$
AB = $\frac{20\sqrt{3}}{\sqrt{3}+1}$
AB = $\frac{20\sqrt{3}}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\ sqrt{3}-1}$
AB = $10 \left( 3- \sqrt{3}\right)$
AQ = AB = $10 \left( 3- \sqrt{3}\right)$ = 12.64 km
$AP=\frac{AB}{\sqrt{3}}=10 \left( \sqrt{3}-1\right)$ = 7.32 km
Station 1 should send its team and they have to travel 7.32 km
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Q57.A man on the deck of a ship is 10 m above the water level. He observes that the angle of elevation of the top of a cliff is $45^{\circ}$ and the angle of depression of the base is $30^{\circ}$ . Calculate the distance of the cliff from the ship and the height of the cliff.
Soln:
Height of the ship from water level = 10m = AB
Angle of elevation of top of the cliff $\alpha=45^{\circ}$
Angle of elevation of bottom of the cliff $\beta=30^{\circ}$
Height of the cliff CD = ‘h’ m
Distance of the ship from foot of the tower cliff.
Height of cliff above ship be ‘a’ m
Then height of cliff = DX + XC = (10 + a) m
The above data is represented in form of figure as shown
In right angle triangle, if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan 45^{\circ} =\frac{CX}{AX}$
$1=\frac{a}{AX}$
AX = ‘a’m
$\tan 30^{\circ}=\frac{XD}{AX}$
$\frac{1}{\sqrt{3}}= \frac{10}{AX}$
AX = $10\sqrt{3}$
$∴ a = 10\sqrt{3}m$
Height of the cliff = $10+10\sqrt{3}m = 10\left( \sqrt{3}+1\right)$
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Q59.There are two temples one on each bank of a river opposite to one other. One temple is 50m high. Form the top of this temple, the angles of depression of the top and foot of the other temple are $30^{\circ}$ and $60^{\circ}$ respectively. Find the width of the river and the height of the other temple.
Soln:
Height of the temple 1 (AB) = 50m
Angle of depression of top of temple 2, $\alpha=30^{\circ}$
Angle of depression of bottom of temple 2, $\beta=60^{\circ}$
Height of the temple 2 (CD) = ‘h’ m
Width of the river = BD = ‘x’ m
The above data is represented in from of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
Here BD = CX, CD= BX
$\tan \alpha=\frac{AX}{CX}$
$\tan 30^{\circ}=\frac{AX}{CX}$
CX = $A \times \sqrt{3}m$
$\tan \beta=\frac{AB}{BD}$
$\tan 60^{\circ} =\frac{50}{CX}$
CX = $\frac{50}{\sqrt{3}}$
$AX\left( \sqrt{3}\right) = \frac {50}{\sqrt{3}}$
$AX= \frac{50}{3}m$
CD =XB = AB – AX = $50 -\frac{50}{3}=\frac{100}{3}m$
Width of river = $\frac{50}{\sqrt{3}}$
Height of temple 2 = $\frac{100}{3}m$
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Q60.The angle of elevation of an airplane from a point on the ground is $45^{\circ}$. After a flight of 15 seconds, the elevation changes to $30^{\circ}$. If the airplane is flying at a height of 3000 meters, find the speed of the airplane.
Soln:
Let the airplane travelled from A to B in 15 sec
Angle of elevation of point A $\alpha=45^{\circ}$
Angle of elevation of point B $\beta=30^{\circ}$
Height of the airplane from ground = 3000 meters
= AP = BQ
Distance travelled in 15 seconds = AB = PQ
Velocity (or) speed = distance travelled time
If the above data is represented is form of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \alpha=\frac{AP}{XP}$
$\tan 45^{\circ} =\frac{3000}{XP}$
XP = 3000m
$\tan \beta=\frac{BQ}{XQ}$
$\tan 30^{\circ}=\frac{3000}{XQ}$
XQ =$3000 \sqrt{3}$
PQ = XQ – XP = $3000\left( sqrt{3}-1\right)m$
Speed = $\frac{PQ}{time}= \frac{3000\left ( \sqrt{3} -1\right )}{15}= 200\left ( \sqrt{3}-1 \right )$
$2000\times 0.732$
146.4 m/sec
Speed of the airplane = 146.4 m/sec
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Q61.An airplane flying horizontally 1 km above the ground is observed at an elevation of $60^{\circ}$. After 10 seconds, its elevation is observed to be $30^{\circ}$. Find the speed of the airplane in km/hr.
Soln:
let airplane travelled from A to B in 10 seconds
Angle of elevation of point A $\alpha=60^{\circ}$
Angle of elevation of point B $\beta=30^{\circ}$
Height of the airplane from ground = 1 km = AP = BQ
Distance travelled in 10 seconds = AB = PQ
The above data is represented in form of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \alpha=\frac{AP}{PX}$
$\tan 60^{\circ} =\frac{1}{PX}$
PX = $\frac{1}{\sqrt{3}}km$
$\tan \beta=\frac{BQ}{XQ}$
$\tan 30^{\circ}=\frac{1}{XQ}$
XQ = $\sqrt{3}$
PQ = XQ – PX = $\sqrt{3} – \frac{1}{sqrt{3}}=\frac{2}{sqrt{3}}=\frac{2\sqrt{3}}{3}$
Speed = $\frac{PQ}{time}=\frac{2\sqrt{3}}{3}\times 60\times 6$
= $240\sqrt{3}$
Speed of the airplane = $240\sqrt{3}$
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Q62.A tree standing on a horizontal plane is leaning towards east. At two points situated at distance a and b exactly due west on it, the angles of elevation of the top are respectively $\alpha$ and $\beta$. Prove that the height of the top from the ground is $\frac{\left ( b-a \right) \tan \alpha \tan \beta}{\tan \alpha – \tan \beta }$
Soln:
AB be the tree leaning east
From distance ‘a’ m from tree, angle of elevation be $\alpha$ at point P.
From distance ‘b’ m from tree, angle of elevation be $\beta$ at point Q.
The above data is represented in the form of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
Draw $AX\perp QB$ let BX = ‘a’ m
$\tan \alpha=\frac{AX}{PX}$
$\tan \alpha=\frac{AX}{x+a}$
$\cot \alpha=\frac{x+a}{AX}$
x+a = $AX\cot \alpha$ —- (1)
$\tan \beta=\frac{AX}{QX}$
$\tan \beta=\frac{AX}{x+b}$
$\cot \beta=\frac{x+b}{AX}$
x+b = $AX\cot \beta$ —- (2)
subtracting (1) from (2)
(x+b) – (x+a) = $AX\cot – AX\cot \alpha$
b – a = $Ax\left [ \frac{\tan \alpha -\tan \beta }{\tan \alpha .\tan \beta } \right ]$
AX = $\frac{\left ( b-a \right )\tan \alpha.\tan \beta }{\tan \alpha -\tan \beta }$
$∴$ Height of top from ground = $\frac{\left ( b-a \right )\tan \alpha.\tan \beta }{\tan \alpha -\tan \beta }$
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Q63.The angle of elevation of a stationary cloud from a point 2500 m above a lake is $15^{\circ}$ and the angle o0f depression of its reflection in the lake is $45^{\circ}$. What is the height of the cloud above the lake level?
Soln:
Let the cloud be at height PQ as represented from lake level
From point x, 2500 meters above the lake angle of elevation of top of cloud $\alpha=15^{\circ}$
Angle of depression of shadow reflection in water $\beta=45^{\circ}$
Here above data is represented in form of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan 15^{\circ}= \frac {AQ}{AY}$
$0.268 = \frac {h}{AY}$
$AY= \frac {h}{0.268}$ —- (1)
$\tan 45^{\circ}= \frac {AB’}{AY}$
$\tan 45^{\circ}= \frac {AP + PQ’}{AY}$
AY = x + (h + x)
AY = h + 2x —- (2)
From (1) and (2)
$\frac {h}{0.268}=h+2x$
3.131h – h = 2 x 2500
h = 1830.8312
Height of the cloud above lake = h + x
= 1830.8312 + 2500
= 4330.8312 m
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Q64.If the angle of elevation of a cloud from a point h meters above a lake is $\alpha$ and the angle of depression of its reflection in the lake be $\beta$, prove that the distance of the cloud from the point of observation is $\frac{2h\sec \alpha }{\tan \beta -\tan \alpha }$
Soln:
let x be point ‘h’ meters above lake
Angle of elevation of cloud from x = $\alpha$
Angle of depression of cloud reflection in lake = $\beta$
Height of the cloud from lake = PQ
PQ’ be the reflection then PQ’ = PQ
Draw XA + PQ, AQ = ‘x’ m, AP = ‘h’ m
Distance of cloud from point of observation is XQ.
The above data is represented in form of figure as shown
In $\Delta AQX$
$\tan \alpha=\frac{AQ}{AX}$
$\tan \alpha=\frac{x}{AX}$ —- (a)
In $\Delta AXQ’$
$\tan \beta=\frac{AQ’}{AX}$
$\tan \beta=\frac{h + x + h}{AX}$ —- (2)
Subtracting (1) from (2)
$\tan \beta – \tan \alpha =\frac{2h}{AX}$
$AX =\frac{2h} {\tan \beta – \tan \alpha }$
In $\Delta AQX$
$\cos \alpha =\frac{AX}{XQ}$
QX = $AX\sec \alpha$
XQ = $\frac{2h\sec \alpha}{\tan \beta – \tan \alpha }$
$∴$ distance of cloud from point of observation = $\frac{2h\sec \alpha}{\tan \beta – \tan \alpha }$
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Q65.From an airplane vertically above a straight horizontal road, the angle of depression of two consecutive mile stones on opposite side of the airplane are observed to be $\alpha$ and $\beta$. Show that the height in miles of airplane above the road is given by $\frac{ \tan \alpha \tan \beta}{\tan \alpha + \tan \beta }$.
Soln:
let PQ be the height of airplane from ground x and y be two mile stones on opposite side of the airplane xy = 1 mile
Angle of depression of x from p = $\alpha$
Angle of depression of y from p = $\beta$
The above data is represented in the form of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
In $\Delta PXQ$
$\tan \alpha=\frac{PQ}{XQ}$
$XQ =\frac{PQ}{\tan \alpha }$ —- (1)
In $\Delta XQY$
$\tan \beta=\frac{PQ}{QY}$
$QY =\frac{PQ}{\tan \beta }$ —- (2)
XQ + QY = $\frac{PQ}{\tan \beta }+\frac{PQ}{\tan \alpha }$
XY = $PQ \left( \frac{1}{\tan \beta }+\frac{1}{\tan \alpha }\right)$
1 = $PQ \left( \frac{\tan \beta +\tan \alpha }{\tan \alpha. \tan \beta}\right)$
PQ = $\frac{\tan \alpha. \tan \beta}{\tan \beta +\tan \alpha }$
Height of airplane = $\frac{\tan \alpha. \tan \beta}{\tan \beta +\tan \alpha }$ miles.
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Q66.PQ is a post given height ‘a’ m and AB is a tower at same distance. If $\alpha$ and $\beta$ are the angles of elevation of B, the top of the tower, at P and Q respectively. Find the height of the tower and its distance from the post.
Soln:
PQ is part height = ‘a’ m AB is tower height
Angle of elevation of B from P = $\alpha$
Angle of elevation of B from Q = $\beta$
The above data is represented in gorm of figure as shown
In right angle triangle if one of the included angle is $\Theta$
Then $\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$QX\perp AB, PQ=AK$
In $\Delta ABQ$
$\tan \beta=\frac{BX}{QX}$
$\tan \beta=\frac{AB – AX}{QX}$
$\tan \beta=\frac{AB – a}{QX}$ —- (1)
In $\Delta BPA$
$\tan \alpha=\frac{AB}{AP}$
$\tan \alpha=\frac{AB}{QX}$ —- (2)
Dividing (1) by (2)
$\frac{\tan \beta }{\tan \alpha} =\frac{AB-a}{AB}=1-\frac{a}{AB}$
$\frac{a}{AB}=1-\frac{\tan \beta }{\tan \alpha} = \frac{\tan \alpha-\tan \beta }{\tan \alpha}$
AB = $\frac{a\tan \alpha}{\tan \alpha-\tan \beta }$
QX = $\frac{AB}{\tan \alpha }=\frac{a}{\tan \alpha-\tan \beta }$
Height of tower = $\frac{a}{\tan \alpha-\tan \beta }$
Distance between post and tower = $\frac{a\tan \alpha}{\tan \alpha-\tan \beta }$
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Q67.A ladder rest against a wall at an angle $\alpha$ to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides a distance b down the wall making an angle $\beta$ with the horizontal. Show that $\frac{a}{b}=\frac{\cos \alpha -\cos \beta }{\sin \beta -\sin \alpha }$
Soln:
let AB be the ladder initially at an inclination $\alpha$ to ground
When its foot is pulled through distance ‘a’ let BB’ = ‘a’ m and AA’= ‘b’ m
New angle of elevation from B’ = B the above data is represented in form of figure as shown
Let $AP \perp ground B’P, AB=A’B’$
A’P = x                 BP = y
In $\Delta ABP$
$\sin \alpha =\frac{AP}{AB}$
$\sin \alpha =\frac{x+b}{AB}$ —- (1)
$\cos \alpha =\frac{BP}{AB}$
$\cos \alpha =\frac{y}{AB}$ —- (2)
In $\Delta A’B’P$
$\sin \beta =\frac{A’P}{A’B’}$
$\sin \beta =\frac{x}{AB}$ —- (3)
$\cos \beta =\frac{B’P}{A’B’}$
$\cos \beta =\frac{y+a}{AB}$ —-(4)
Subtracting (3) from (1)
$\sin \alpha – \sin \beta = \frac{b}{AB}$
Subtracting (4) from (2)
$\cos \beta – \cos \alpha = \frac{a}{AB}$
$\frac{a}{b}= \frac{\cos \alpha – \cos \beta }{\sin \beta – \sin \alpha}$
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Q68.A tower subtends an angle $\alpha$ at a point A in the plane of its base and the angle if depression of the foot of the tower at a point b meters just above A is $\beta$. Prove that the height of the tower is $b \tan\alpha\cot \beta$.
Soln:
let height of the tower be ‘h’ m = PQ
Angle of elevation at point A on ground = $\alpha$
Let B be the point ‘b’ m above the A
Angle of depression of foot of the tower from B = $\beta$
The above data is represented in form of figure as shown
Let $BX \perp P, AB=A’B’$
In $\Delta PBX$
$\tan \alpha=\frac{PQ}{BX}$ —- (1)
In $\Delta QBX$
$\tan \beta=\frac{QX}{BX}$ —- (2)
Dividing (1) by (2)
$\frac{\tan \alpha}{\tan \beta } = \frac{PQ}{QX}$
PQ = QX.$\frac{\tan \alpha}{\tan \beta} =b\tan \alpha.\cot \beta$
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Q69.An observer, 1.5 m tall, is 28.5 m away from a tower 30 m high. Determine the angle of elevation of the top of the tower from his eye.
Soln:
Height of observer = AB = 1.5 m
Height of tower = PQ = 30 m
Height of tower above the observer eye = 30 – 1.5
QX = 28.5m
Distance between tower and observer XB = 28.5m
$\Theta$ be angle of elevation of tower top from eye
The above data is represented in the form of figure
$\tan \Theta =\frac{Opposite\;side}{Adjacent\;side}$
$\tan \Theta =\frac{QX}{BX}=\frac {28.5}{28.5}=1$
$\Theta=\tan^{-1}\left( 1\right)= 45^{\circ}$
Angle of elevation = $45^{\circ}$
Q70. A carpenter makes stools for electricians with a square top of side 0.5 m and at a height of 1.5 m above the ground. Also each leg is inclined at an angle of $60^{\circ}$ to the ground. Find the length of each leg and also the lengths of two steps to be put at equal distance.
Soln:
let AB be the height of stool = 1.5 m
Let P and Q be equal distance then AP = 0.5. AQ = 1 m
The above data is represented in form of figure as shown
BC = length of leg
$\sin 60^{\circ} = \frac{AB}{BC}$
$\frac{\sqrt{3}}{2}= \frac{1.5}{BC}$
BC = $\frac{1.5\times 2}{BC}$
Draw $PX\perp AB, \;QZ\perp AB, \; XY\perp CA,\; ZW\perp CA$
$\sin 60^{\circ} = \frac{XY}{XC}$
XC = $\frac{0.5}{BC}\times \sqrt{3}$
$\frac{\sqrt{3}}{4}\times \frac{8}{3}$
$\frac{2}{\sqrt{3}}$
XC = 1.1077m
$\sin 60^{\circ} = \frac{ZW}{CZ}$
CZ = $\frac{1}{\frac {sqrt{3}}{2}}$
CZ = $\frac{2}{\sqrt{3}}$
CZ = 1.654m
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Q71.A boy is standing on the ground and flying a kite with 100m of string at an elevation of $30^{\circ}$. Another boy is standing on the roof of a 10m high building and is flying his kite at an elevation of $45^{\circ}$. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the kites meet.
Soln:
For boy 1
Length of the string AB = 100 m
Angle made by string with ground = $\alpha=30^{\circ}$
For boy 2
Height of the building CD = 10m
Angle made by string with building top = $\beta=45^{\circ}$
Length of the kite thread of boy 2 if both the kites meet must be ‘DB’
The above data is represented in form of figure as shown
Draw $BX\perp AC, \;YD\perp BC$
In $\Delta ABX$
$\tan 30^{\circ}= \frac {BX}{AX}$
$\sin 30^{\circ} = \frac{BX}{AB}$
$\frac{1}{2}= \frac{BX}{100}$
BX = 50m
BY = BX – XY = 50 – 10 = 40m
In $\Delta BYD$
$\sin 45^{\circ} = \frac{BY}{BD}$
$\frac{1}{\sqrt{2}}= \frac{40}{BD}$
BD = $40\sqrt{2}$
Length of thread or string of boy 2 = $40\sqrt{2}$
Q72. From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be $\alpha$ and $\beta$. If the height of the light house be h meters and the line joining the ships passes through the foot of the light house, show that the distance $\frac{h\left ( \tan \alpha +\tan \beta \right )}{\tan \alpha \tan \beta }$
Soln:
Height of light house = ‘h’ meters = AB
$S_{1} \;and\; S_{2}$ be two ships on opposite sides of the light house
Angle of depression of $S_{1}$ from top of light house = $\alpha$
Angle of depression of $S_{2}$ from top of light house = $\beta$
Required to prove that
Distance between ships = $\frac{h\left ( \tan \alpha +\tan \beta \right )}{\tan \alpha \tan \beta }$ meters
The above data is represented in the form of figure as shown
In $\Delta ABS_{1}$
$\tan \alpha =\frac{Opposite\;side}{Adjacent\;side}= \frac{AB}{S_{1}B}$
${S_{1}B}=\frac{h}{\tan \alpha }$ —- (1)
In $\Delta ABS_{2}$
$\tan \beta =\frac{AB}{ S_{2}B }$
${S_{2}B}=\frac{h}{\tan \beta }$ —- (2)
=> $S_{1}B + S_{2} = \frac{h}{\tan \alpha }+\frac{h}{\tan \beta }$
=> $S_{1} S_{2} = h\left \{ \frac{1}{\tan \alpha }+\frac{1}{\tan \beta } \right \}$
=> $\frac{h\left ( \tan \alpha +\tan \beta \right )}{\tan \alpha \tan \beta }$
Distance between ships = $\frac{h\left ( \tan \alpha +\tan \beta \right )}{\tan \alpha \tan \beta }$ meters |
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## Section5.2Application of eigenfunction series
1 lecture, §10.2 in [EP], exercises in §11.2 in [BD]
The eigenfunction series can arise even from higher order equations. Suppose we have an elastic beam (say made of steel). We will study the transversal vibrations of the beam. That is, assume the beam lies along the $x$ axis and let $y(x,t)$ measure the displacement of the point $x$ on the beam at time $t\text{.}$ See Figure 6.2.1.
The equation that governs this setup is
\begin{equation*} a^4 \frac{\partial^4 y}{\partial x^4} + \frac{\partial^2 y}{\partial t^2} = 0, \end{equation*}
for some constant $a > 0\text{.}$
Suppose the beam is of length 1 simply supported (hinged) at the ends. The beam is displaced by some function $f(x)$ at time $t=0$ and then let go (initial velocity is 0). Then $y$ satisfies:
\begin{aligned} & a^4 y_{xxxx} + y_{tt} = 0 \quad (0 < x < 1, t > 0), \\ & y(0,t) = y_{xx}(0,t) = 0 , \\ & y(1,t) = y_{xx}(1,t) = 0 , \\ & y(x,0) = f(x), \quad y_{t}(x,0) = 0 . \end{aligned}\label{appeig_beameq}\tag{5}
Again we try $y(x,t) = X(x)T(t)$ and plug in to get $a^4 X^{(4)}T + XT'' = 0$ or
\begin{equation*} \frac{X^{(4)}}{X} = \frac{- T''}{a^4T} = \lambda . \end{equation*}
We note that we want $T'' + \lambda a^4T = 0\text{.}$ Let us assume that $\lambda > 0\text{.}$ We can argue that we expect vibration and not exponential growth nor decay in the $t$ direction (there is no friction in our model for instance). Similarly $\lambda = 0$ will not occur.
###### Exercise5.2.1
Try to justify $\lambda > 0$ just from the equations.
Write $\omega^4 = \lambda\text{,}$ so that we do not need to write the fourth root all the time. For $X$ we get the equation $X^{(4)} - \omega^4 X = 0\text{.}$ The general solution is
\begin{equation*} X(x) = A e^{\omega x} + B e^{-\omega x} + C \sin (\omega x) + D \cos (\omega x) . \end{equation*}
Now $0 = X(0) = A+B+D\text{,}$ $0 = X''(0) = \omega^2 (A + B - D)\text{.}$ Hence, $D = 0$ and $A+B = 0\text{,}$ or $B = - A\text{.}$ So we have
\begin{equation*} X(x) = A e^{\omega x} - A e^{-\omega x} + C \sin (\omega x) . \end{equation*}
Also $0 = X(1) = A (e^{\omega} - e^{-\omega}) + C \sin \omega\text{,}$ and $0 = X''(1) = A \omega^2 (e^{\omega} - e^{-\omega}) - C \omega^2 \sin \omega\text{.}$ This means that $C \sin \omega = 0$ and $A (e^{\omega} - e^{-\omega}) = 2 A \sinh \omega = 0\text{.}$ If $\omega > 0\text{,}$ then $\sinh \omega \not= 0$ and so $A = 0\text{.}$ This means that $C \not=0$ otherwise $\lambda$ is not an eigenvalue. Also $\omega$ must be an integer multiple of $\pi\text{.}$ Hence $\omega = n \pi$ and $n \geq 1$ (as $\omega > 0$). We can take $C=1\text{.}$ So the eigenvalues are $\lambda_n = n^4 \pi^4$ and corresponding eigenfunctions are $\sin (n \pi x)\text{.}$
Now $T'' + n^4 \pi^4 a^4 T = 0\text{.}$ The general solution is $T(t) = A \sin (n^2 \pi^2 a^2 t) + B \cos (n^2 \pi^2 a^2 t)\text{.}$ But $T'(0) = 0$ and hence we must have $A=0$ and we can take $B=1$ to make $T(0) = 1$ for convenience. So our solutions are $T_n(t) = \cos (n^2 \pi^2 a^2 t)\text{.}$
As eigenfunctions are just sines again, we can decompose the function $f(x)$ on $0 < x < 1$ using the sine series. We find numbers $b_n$ such that for $0 < x < 1$ we have
\begin{equation*} f(x) = \sum_{n=1}^\infty b_n \sin (n \pi x) . \end{equation*}
Then the solution to (5) is
\begin{equation*} y(x,t) = \sum_{n=1}^\infty b_n X_n(x) T_n(t) = \sum_{n=1}^\infty b_n \sin (n \pi x) \cos ( n^2 \pi^2 a^2 t ) . \end{equation*}
The point is that $X_nT_n$ is a solution that satisfies all the homogeneous conditions (that is, all conditions except the initial position). And since and $T_n(0) = 1\text{,}$ we have
\begin{equation*} y(x,0) = \sum_{n=1}^\infty b_n X_n(x) T_n(0) = \sum_{n=1}^\infty b_n X_n(x) = \sum_{n=1}^\infty b_n \sin (n \pi x) = f(x) . \end{equation*}
So $y(x,t)$ solves (5).
The natural (angular) frequencies of the system are $n^2 \pi^2 a^2\text{.}$ These frequencies are all integer multiples of the fundamental frequency $\pi^2 a^2\text{,}$ so we get a nice musical note. The exact frequencies and their amplitude are what musicians call the timbre of the note (outside of music it is called the spectrum).
The timbre of a beam is different than for a vibrating string where we get “more” of the lower frequencies since we get all integer multiples, $1,2,3,4,5,\ldots\text{.}$ For a steel beam we get only the square multiples $1,4,9,16,25,\ldots\text{.}$ That is why when you hit a steel beam you hear a very pure sound. The sound of a xylophone or vibraphone is, therefore, very different from a guitar or piano.
###### Example5.2.1
Let us assume that $f(x) = \frac{x(x-1)}{10}\text{.}$ On $0 < x < 1$ we have (you know how to do this by now)
\begin{equation*} f(x) = \sum_{\substack{n=1\\n \text{~odd}}}^\infty \frac{4}{5\pi^3 n^3} \sin (n \pi x) . \end{equation*}
Hence, the solution to (5) with the given initial position $f(x)$ is
\begin{equation*} y(x,t) = \sum_{\substack{n=1\\n \text{~odd}}}^\infty \frac{4}{5\pi^3 n^3} \sin (n \pi x) \cos ( n^2 \pi^2 a^2 t ) . \end{equation*}
### Subsection5.2.1Exercises
###### Exercise5.2.2
Suppose you have a beam of length 5 with free ends. Let $y$ be the transverse deviation of the beam at position $x$ on the beam ($0 < x < 5$). You know that the constants are such that this satisfies the equation $y_{tt} + 4 y_{xxxx} = 0\text{.}$ Suppose you know that the initial shape of the beam is the graph of $x(5-x)\text{,}$ and the initial velocity is uniformly equal to 2 (same for each $x$) in the positive $y$ direction. Set up the equation together with the boundary and initial conditions. Just set up, do not solve.
###### Exercise5.2.3
Suppose you have a beam of length 5 with one end free and one end fixed (the fixed end is at $x=5$). Let $u$ be the longitudinal deviation of the beam at position $x$ on the beam ($0 < x < 5$). You know that the constants are such that this satisfies the equation $u_{tt} = 4 u_{xx}\text{.}$ Suppose you know that the initial displacement of the beam is $\frac{x-5}{50}\text{,}$ and the initial velocity is $\frac{-(x-5)}{100}$ in the positive $u$ direction. Set up the equation together with the boundary and initial conditions. Just set up, do not solve.
###### Exercise5.2.4
Suppose the beam is $L$ units long, everything else kept the same as in (5). What is the equation and the series solution?
###### Exercise5.2.5
Suppose you have
\begin{equation*} \begin{aligned} & a^4 y_{xxxx} + y_{tt} = 0 \quad (0 < x < 1, t > 0) , \\ & y(0,t) = y_{xx}(0,t) = 0,\\ & y(1,t) = y_{xx}(1,t) = 0 ,\\ & y(x,0) = f(x), \quad y_{t}(x,0) = g(x) . \end{aligned} \end{equation*}
That is, you have also an initial velocity. Find a series solution. Hint: Use the same idea as we did for the wave equation.
###### Exercise5.2.101
Suppose you have a beam of length 1 with hinged ends. Let $y$ be the transverse deviation of the beam at position $x$ on the beam ($0 < x < 1$). You know that the constants are such that this satisfies the equation $y_{tt} + 4 y_{xxxx} = 0\text{.}$ Suppose you know that the initial shape of the beam is the graph of $\sin (\pi x)\text{,}$ and the initial velocity is 0. Solve for $y\text{.}$
$y(x,t) = \sin(\pi x) \cos (4 \pi^2 t)$
Suppose you have a beam of length 10 with two fixed ends. Let $y$ be the transverse deviation of the beam at position $x$ on the beam ($0 < x < 10$). You know that the constants are such that this satisfies the equation $y_{tt} + 9 y_{xxxx} = 0\text{.}$ Suppose you know that the initial shape of the beam is the graph of $\sin(\pi x)\text{,}$ and the initial velocity is uniformly equal to $x(10-x)\text{.}$ Set up the equation together with the boundary and initial conditions. Just set up, do not solve.
$9 y_{xxxx} + y_{tt} = 0 \quad (0 < x < 10, t > 0)\text{,}$ $y(0,t) = y_{x}(0,t) = 0\text{,}$ $y(10,t) = y_{x}(10,t) = 0\text{,}$ $y(x,0) = \sin(\pi x), \quad y_{t}(x,0) = x(10-x)\text{.}$ |
Go through the enVision Math Common Core Kindergarten Answer Key Topic 7 Understand Subtraction and finish your homework or assignments.
## Envision Math Common Core Grade K Answers Key Topic 7 Understand Subtraction
Essential Question:
enVision STEM Project: Animal Needs
Directions Read the character speech bubbles to students. Find Out! Have students find out about how plants, animals, and humans use their environment to meet basic needs such as food, water, nutrients, sunlight, space, and shelter. Say: Different organisms need different things. Talk to friends and relatives about the different needs of plants, animals, and humans, and how different organisms meet those needs. Journal: Make a Poster Have students make a poster. Ask them to draw as many as 5 pictures of a human’s needs and as many as 5 pictures of an animal’s needs. Have them cross out the needs that are the same for humans and animals, and then write how many are left.
Review What You Know
Directions Have students: 1 draw a circle around the plus sign; 2 draw a circle around the equal sign; 3 draw a circle around the sum; 4 – 5 count the objects in each group, and then write the equation to tell how many in all.
Question 1.
3 + 6 = 9
Explanation:
I drew a circle around the plus (+) sign.
Question 2.
4 + 1 = 5
Explanation:
I drew a circle around the equal (=) sign.
Question 3.
2 + 5 = 7
Explanation:
I drew a circle around the sum of the numbers 2 and 5 that is 7.
Question 4.
Explanation:
There are 3 counters in first group and 1 counter in second group. Therefore the addition equation is 3+1=4.
Question 5.
Explanation:
There are 5 counters in first group and 3 counter in second group. Therefore the addition equation is 5+3=8.
Question 6.
Explanation:
There are 6 squares in first group and 4 squares in second group. Therefore the addition equation is 6+4=10.
Pick A Project
### Lesson 7.1 Explore Subtraction
Solve & Share
Directions Say: Marta sees 5 goldfish in the pond. I swims away. How many fish are left? Think about the problem in your head. Then act out the story with your fingers. Use counters to show how many are left. Write numbers to explain.
I can ………….. show numbers in many ways.
Visual Learning Bridge
Guided Practice
Directions Have students listen to the story, and then do all of the following to find how many are left: give an explanation of a mental image, use objects to act it out, and hold up fingers. Have them mark Xs on how many birds fly away, and then write the number to tell how many are left.1 8 eagles sit on a branch. 2 fly away. How many eagles are left? 2 5 blue jays hop on the ground. I flies away. How many blue jays are left?
Question 1.
Explanation:
8 eagles sit on a branch. 2 fly away.I marked ‘X’ on 2 birds. Therefore 6 birds are left.
Question 2.
Explanation:
5 blue jays hop on the ground. I flies away. I marked ‘X’ on 1 bird. Therefore 4 birds are left.
Directions Have students listen to the story, and then do all of the following to find how many are left: give an explanation of a mental image, use objects to act it out, and hold up fingers. Have them mark Xs on how many walk away or are taken out, and then write the number to tell how many are left. 3 9 ladybugs are on a leaf. 4 walk away. How many ladybugs are left? 4 7 caterpillars are on a leaf. 6 walk away. How many caterpillars are left? 5 9 marbles are in a jar. 3 are taken out. How many marbles are left? 6 6 marbles are in a jar. 4 are taken out. How many marbles are left?
Question 3.
Explanation:
9 ladybugs are on a leaf. 4 walk away. I marked ‘X’ on 4 bugs. Therefore 5 bugs are left on the leaf.
Question 4.
Explanation:
7 caterpillars are on a leaf. 6 walk away. I marked ‘X’ on 6 Caterpillars. Therefore 1 is left on the leaf.
Question 5.
Explanation:
9 marbles are in a jar. 3 are taken out. I marked ‘X’ on 3 marbles that are taken out. Therefore 6 marbles are left in the jar.
Question 6.
Explanation:
6 marbles are in a jar. 4 are taken out. I marked ‘X’ on 4 marbles that are taken out. Therefore 2 marbles are left in the jar.
Independent Practice
Directions Have students listen to the story, and then do all of the following to find how many are left: give an explanation of a mental image, use objects to act it out, and hold up fingers. Ask them to write the number to tell how many are left. 7 10 fingers are in the air. 2 are put down. How many fingers are left? 8 7 fingers are in the air. 3 are put down. How many fingers are left? 9 Have students listen to the story, and then do all of the following to find how many are left: give an explanation of a mental image, use objects to act it out, and then mark Xs on how many are taken away. Ask them to write the number to tell how many are left. There are 9 marbles. 6 are taken away. How many marbles are left? 10 Higher Order Thinking Have students draw 10 marbles. Have them mark Xs on some of them, and then write the number to tell how many marbles are left.
Question 7.
Explanation:
10 fingers are in the air. 2 are put down. So, 8 fingers are left in the air.
Question 8.
Explanation:
7 fingers are in the air. 3 are put down. So, 4 fingers are left in the air.
Question 9.
Explanation:
There are 9 marbles. 6 are taken away.I marked ‘X’ on 6 marbles that are taken out. Therefore 3 marbles are left.
Question 10.
Explanation:
I drew 10 marbles taken away 6 from them.I marked ‘X’ on 6 marbles that are taken out. Therefore 4 marbles are left.
### Lesson 7.2 Represent Subtraction as Taking Apart
Solve & Share
Directions Say: Alex picks 7 apples. Some apples are red, and some are yellow. Alex wants to put the red apples in one basket and the yellow in the other. How many red apples and how many yellow apples can there be? Use counters to show the red and yellow apples, and write the numbers to tell how many of each. Draw pictures to show your answer.
I can ………… take apart a number and tell the parts.
Visual Learning Bridge
Guided Practice
Directions Have students: 1 take apart the group of pears. Then have them draw a circle around the parts they made, and then write the numbers to tell the parts; 2 take apart the group of peaches. Then have them draw a circle around the parts they made, and then write the numbers to tell the parts.
Question 1.
Explanation:
I took apart the group of pears. Then i drew a circle around the parts i made, and then wrote the numbers 4 and 1 to tell the parts.
Question 2.
Explanation:
I took apart the group of peachess. Then i drew a circle around the parts i made, and then wrote the numbers 2 and 4 to tell the parts.
Directions 3 – 6 Have students take apart the group of fruit. Then have them draw a circle around the ports they made, and then write the numbers to tell the parts.
Question 3.
Explanation:
I took apart the group of fruits. Then i drew a circle around the parts i made, and then wrote the numbers 1 and 3 to tell the parts.
Question 4.
Explanation:
I took apart the group of fruits. Then i drew a circle around the parts i made, and then wrote the numbers 4 and 6 to tell the parts.
Question 5.
Explanation:
I took apart the group of fruits. Then i drew a circle around the parts i made, and then wrote the numbers 2 and 1 to tell the parts.
Question 6.
Explanation:
I took apart the group of fruits. Then i drew a circle around the parts i made, and then wrote the numbers 3 and 2 to tell the parts.
Independent Practice
Directions 7 and 8 Have students take apart the group of fruit. Then have them draw a circle around the parts they made, and then write the numbers to tell the parts. 9 Higher Order Thinking Have students draw counters to show a group of 5. Then have them take apart the group of counters, draw a circle around the parts they made, and then write the numbers to tell the parts. 10 Higher Order Thinking Have students choose any number between 2 and 10, write that number on the top line, and then draw a group of counters to show that number. Have them take apart the group of counters, draw a circle around the parts they made, and then write the numbers to tell the parts.
Question 7.
Explanation:
I took apart the group of fruits. Then i drew a circle around the parts i made, and then wrote the numbers 3 and 3 to tell the parts.
Question 8.
Explanation:
I took apart the group of fruits. Then i drew a circle around the parts i made, and then wrote the numbers 1 and 1 to tell the parts.
Question 9.
Explanation:
I drew counters to show a group of 5.I took apart the group of counters. Then i drew a circle around the parts i made, and then wrote the numbers 3 and 2 to tell the parts.
Question 10.
Explanation:
I choosed a number between 2 and 10, wrote that number 6 on the top line, and then drew a group of 6 counters to show that number.I took apart the group of counters. Then i drew a circle around the parts i made, and then wrote the numbers 2 and 4 to tell the parts.
### Lesson 7.3 Represent Subtraction as Taking From
Solve & Share
Directions Say: Marto is watching bugs. She sees q ladybugs together in a group. Then some crawl away. Look at the picture and decide how many are left. Use counters to show what you think happens in the story. Then write numbers to tell how many ladybugs are left ¡n the group.
I can …………. represent subtraction as taking away from a whole.
Visual Learning Bridge
Guided Practice
Directions Have students listen to each story, and then complete the sentence to tell how many bugs are left. 1 Marta sees 6 bumblebees. 3 leave. How many bumblebees are left? 2 Marta sees 7 ladybugs. 2 leave. How many ladybugs are left?
Question 1.
Explanation:
Marta sees 6 bumblebees. 3 leave. Therefore 3 bees are left. So, i wrote sentence is 6 takeaway 3 is 3.
Question 2.
Explanation:
Marta sees 7 ladybugs. 2 leave. Therefore 5 bugs are left. So, i wrote the sentence 7 takeaway 2 is 5.
Directions Have students listen to each story, and then complete the sentence to tell how many bugs are left. 3 Emily sees 6 grasshoppers on the table. 2 hop away. How many grasshoppers are left? 4 Emily sees 7 dragonflies. 3 fly away. How many dragonflies are left? 5 Emily sees 8 caterpillars resting on a branch. 4 crawl away. How many caterpillars are left? 6 enVision® STEM Say: Ants can move material much bigger than themselves. Emily sees 10 ants on a picnic blanket. 4 walk away. How many ants are left?
Question 3.
Explanation:
Emily sees 6 grasshoppers on the table. 2 hop away. Therefore 4 grasshoppers are left. So, i wrote the sentence 6 takeaway 2 is 4.
Question 4.
Explanation:
Emily sees 7 dragonflies. 3 fly away. Therefore 4 flies are left. So, i wrote the sentence 7 takeaway 3 is 4.
Question 5.
Explanation:
Emily sees 8 caterpillars resting on a branch. 4 crawl away. Therefore 4 Caterpillars are left. So, i wrote the sentence 8 takeaway 4 is 4.
Question 6.
Explanation:
Emily sees 10 ants on a picnic blanket. 4 walk away. Therefore 6 ants are left. So, i wrote the sentence 10 takeaway 4 is 6.
Independent Practice
Directions Have students listen to each story, and then complete the sentence to tell how many are left. 7 Jerome sees 8 snails on the sidewalk. 3 slink away. How many snails are left? 8 Jerome sees 6 grasshoppers in the grass. 3 hop away. How many grasshoppers are left? 9 Jerome sees 9 butterflies in the garden. 4 flutter away. How many butterflies are left? 10 Higher Order Thinking Have students listen to the story, draw a picture to show the story, and then complete the sentence to tell how many are left. Jerome sees 7 inchworms on a tree. L crawl away. How many inch worms are left?
Question 7.
Explanation:
Jerome sees 8 snails on the sidewalk. 3 slink away. Therefore 5 snails are left. So, i wrote the sentence 8 takeaway 3 is 5.
Question 8.
Explanation:
Jerome sees 6 grasshoppers in the grass. 3 hop away. Therefore 3 grasshoppers are left. So, i wrote the sentence 6 takeaway 3 is 3.
Question 9.
Explanation:
Jerome sees 9 butterflies in the garden. 4 flutter away. Therefore 5 butterflies are left. So, i wrote the sentence 9 takeaway 4 is 5.
Question 10.
### Lesson 7.4 Represent and Explain Subtraction with Addition
Solve & Share
Directions Have students listen to the story and use counters to show what happens. Say: There are 6 fire hats. Firefighters take 3 away. What numbers do you subtract to find how many hats are left? How con you show the subtraction?
I can ………. write an equation to show subtraction.
Visual Learning Bridge
Guided Practice
Directions 1 and 2 Have students use counters to model the problem, mark Xs to subtract, and then write a subtraction equation to find
the difference.
Question 1.
Explanation:
I marked X to subtract, and then wrote a subtraction equation 5-4=1 to find the difference.
Question 2.
Explanation:
I drew 7 counters and marked X on 5 counters to subtract. I wrote a subtraction equation 7-5=2 as the difference is 2.
Directions 3 – 6 Have students use counters to model the problem, mark Xs to subtract, and then write a subtraction equation to find
the difference.
Question 3.
Explanation:
I drew 8 counters and marked X on 2 counters to subtract. I wrote a subtraction equation 8-2=6 as the difference is 6.
Question 4.
Explanation:
I drew 6 counters and marked X on 5 counters to subtract. I wrote a subtraction equation 6-5=1 as the difference is 1.
Question 5.
Explanation:
I drew 9 counters and marked X on 5 counters to subtract. I wrote a subtraction equation 9-5=4 as the difference is 4.
Question 6.
Explanation:
I drew 7 counters and marked X on 2 counters to subtract. I wrote a subtraction equation 7-2=5 as the difference is 5.
Independent Practice
Directions 7 – 9 Have students use counters to model the problem, mark Xs to subtract, and then write an equation to find the difference. 10 Higher Order Thinking Have students listen to the story, draw counters and mark Xs to show the problem, and then write an equation to find the difference. There are 7 baseball caps. Some are worn to a game. There are 4 left. How many caps were worn to the game?
Question 7.
Explanation:
I drew 8 counters and marked X on 3 counters to subtract. I wrote a subtraction equation 8-3=5 as the difference is 5.
Question 8.
Explanation:
I drew 4 counters and marked X on 1 counter to subtract. I wrote a subtraction equation 4-1=3 as the difference is 3.
Question 9.
Explanation:
I drew 6 counters and marked X on 4 counters to subtract. I wrote a subtraction equation 6-4=2 as the difference is 2.
Question 10.
Explanation:
I drew 7 counters and marked X on 3 counters to subtract. I wrote a subtraction equation 7-3=4 as the difference is 4.
### Lesson 7.5 Solve Subtraction Word Problems: Taking From and Apart
Solve & Share
Directions Say: Marta’s dog, Spot, loves to eat doggie biscuits. Moda put 6 biscuits in a bag. One day, Spot ate q biscuits. Now there are only 2 left. How does Marlo know there are 2 biscuits left? Use counters, pictures, or numbers to explain and show your work.
I can … find the difference of two numbers.
Visual Learning Bridge
Guided Practice
Directions 1 Have students listen to the story, draw a picture to show what is happening, and then write a subtraction equation. Then have them explain their work aloud. Say: Marfa has 6 kittens. She gives them a big bowl of water to drink. But there is only room for 4 kittens to drink at the same time. How does Marta know that 2 kittens have to wait?
Question 1.
Directions Have students listen to each story, use or draw a picture to show what is happening, and then write an equation. Then have them explain their work aloud. 2 Emily sees 8 rabbits in a pet store. Someone buys 3 of them. How many rabbits are left? 3 Emily sees 7 birds in a cage. The pet store owner opens the cage door and 3 fly out. How many birds ore left? 4 Emily sees 8 puppies in the store. 6 of them are sold. How many puppies are left? 5 Emily sees 5 hamsters sleeping. I goes away to eat. How many hamsters are left?
Question 2.
Explanation:
Emily sees 8 rabbits in a pet store. Someone buys 3 of them. I used the picture to show what is happening. I marked X on 3 rabbits. Therefore 5 rabbits are left. So, i wrote the equation 8-3=5 as the difference is 5.
Question 3.
Explanation:
Emily sees 7 birds in a cage. The pet store owner opens the cage door and 3 fly out.I used the picture to show what is happening. I marked X on 3 parrots. Therefore 4 rabbits are left. So, i wrote the equation 7-3=4 as the difference is 4.
Question 4.
Explanation:
Emily sees 8 puppies in the store. 6 of them are sold. I drew connecting cubes to show what is happening. I marked X on 3 cubes. Therefore 5 cubes are left. So, i wrote the equation 8-3=5 as the difference is 5.
Question 5.
Explanation:
Emily sees 5 hamsters sleeping. I goes away to eat. I drew connecting cubes to show what is happening. I marked X on 1 cubes. Therefore 4 cubes are left. So, i wrote the equation 5‐1=4 as the difference is 4.
Independent Practice
Directions Have students listen to each story, draw a picture to show what is happening, and then write an equation. 6 There are 6 birds in a birdbath. 4 fly away. How many birds are left? 7 There are 5 acorns under a tree. A squirrel takes 3 of them. How many acorns are left? 8 Higher Order Thinking Have students listen to the story, draw a circle around the picture that shows the story and tell why the other picture does NOT show the story, and then write an equation. There are 4 ducks in a pond, 1 leaves. How many ducks are left?
Question 6.
Explanation:
There are 6 birds in a birdbath. 4 fly away.I drew connecting cubes to show what is happening. I marked X on 2 cubes. Therefore 4 cubes are left. So, i wrote the equation 6‐4=2 as the difference is 2.
Question 7.
Explanation:
There are 5 acorns under a tree. A squirrel takes 3 of them.I drew connecting cubes to show what is happening. I marked X on 3 cubes. Therefore 4 cubes are left. So, i wrote the equation 5‐3=2 as the difference is 2.
Question 8.
Explanation:
There are 4 ducks in a pond, 1 leaves. Therefore the equation is 4‐3=2. The first picture matches with the equation. So, i circled the first equation.
### Lesson 7.6 Use Patterns to Develop Fluency in Subtraction
Solve & Share
Directions Say: Look at the first equation. Write the number from the number that completes the equation on the orange spoce. Repeat for the next equation. Finish the pattern by placing the other number cards on the orange spaces, and then write the numbers to complete the equations. What patterns do you see?
I can….. find patterns in subtraction equations.
Visual Learning Bridge
Guided Practice
Directions 1 Have students complete each equation to find the pattern, and then explain the pattern they see.
Question 1.
Explanation:
I completed each equation and found the pattern. The difference is decreasing by 1.
Directions 2 and 3 Have students look for a pattern, explain the pattern they see, and then write an equation for each row of insects.
Question 2.
Explanation:
I observed the pattern in the question. The number of butterflies left are increasing by 1. So, i completed the equations based on the picture. The difference is decreasing by 1.
Question 3.
Explanation:
I observed the pattern in the question. The number of butterflies left are decreasing by 1. So, i completed the equations based on the picture. The difference is increasing by 1.
Independent Practice
Directions 4 Algebra Have students mark Xs to complete the pattern, explain the pattern they see, and then write an equation for each row of flowers. 5 Higher Order Thinking Have students find the pattern, and then complete the equation. 6 Higher Order Thinking Have students find the pattern, and then write the missing number in the equation.
Question 4.
Explanation:
I marked X on one flower in the second row and left the third row to create the decreasing pattern. Therefore based on the patterns of flowers i completed the equations. The difference is increasing by 1.
Question 5.
Explanation:
I found the pattern and the equation is 10‐4=6.
Question 6.
Explanation:
I found the pattern and the missing number is 4.
### Lesson 7.7 Problem Solving
Use Appropriate Tools
Solve & Share
Directions Say: Alex has a food bar with 8 pieces of food for the flamingos at the lake. He takes apart 2 pieces of the bar to feed the flamingos. How many pieces does he have left on his bar? Use one of the tools you have to help solve the problem. Draw a picture of what you did, and then write the equation.
I can ……….. use tools to subtract numbers
Visual Learning Bridge
Guided Practice
Directions Have students listen to each story, use a tool to help them solve the problem, and then write the equation. Then have them explain whether or not the tool they chose helped to solve the problem. 1 There is I flamingo standing in the water. 8 more fly over to join it. How many flamingos are there in all? 2 Marta sees 7 seagulls. 4 fly away. How many seagulls are left?
Question 1.
Explanation:
There is I flamingo standing in the water. 8 more fly over to join it. Therefore there are 9 flamigos in all. So, the addition equation is 1+8=9.
Question 2.
Explanation:
Marta sees 7 seagulls. 4 fly away. Therefore 3 seagulls are left. So, the subtraction equation is 7-4=3.
Independent Practice
Directions Have students listen to each story, use a tool to help them solve the problem, and then write the equation. Then have them tell which tool they chose and whether or not it helped to solve the problem. 3 There are 3 raccoons in a tree. 3 more climb the tree to join them. How many raccoons are there in all? 4 Marta sees 9 turtles swimming in a pond. 5 dive under the water. How many turtles are left? 5 There are 7 beavers in the water. 4 swim away. How many beavers are left? 6 Marta see 6 ducks in the lake. 2 more join them. How many ducks are there in all?
Question 3.
Explanation:
There are 3 raccoons in a tree. 3 more climb the tree to join them. Therefore, there are 6 racoons in all. So, the addition equation is 3+3=6.
Question 4.
Explanation:
Marta sees 9 turtles swimming in a pond. 5 dive under the water. Therefore, 4 turtles are swimming. So, the subtraction equation is 9-5=4.
Question 5.
Explanation:
There are 7 beavers in the water. 4 swim away. Therefore, 3 beavers are left. So, the subtraction equation is 7-4=3.
Question 6.
Explanation:
Marta see 6 ducks in the lake. 2 more join them. Therefore, there are 8 ducks in all. So, the addition sentence is 6+2=8.
Problem Solving
Directions Read the problem aloud. Then have students use multiple problem-solving methods to solve the problem. Say: Carlos collects stomps. He has 9 stamps in all. He puts I stamp on the cover. He puts the rest inside the book. How many stamps does Carlos put inside his stamp book? 7 Make Sense What are you trying to find out? Will you use addition or subtraction to solve the problem? 8 Use Tools What tool can you use to help solve the problem? Tell a partner and explain why. 9 Be Precise Did you write the equation correctly? Explain what the numbers and the symbols mean in the equation.
Explanation:
Carlos collects stomps. He has 9 stamps in all. He puts I stamp on the cover. He puts the rest inside the book.
6) I am trying to find how many stamps did carol put in his book. I use subtraction to find the answer.
7) I drew counters as tool to find the difference.
8) yes, my equation is correct. 9-1=8. The symbol (-) means minus and (=) equal.
### Topic 7 Vocabulary Review
Directions Understand Vocabulary Have students: 1 write the minus sign to show subtraction; 2 draw a circle around the number that tells how many are left; 3 complete the subtraction sentence; 4 separate the tower into 2 parts, draw each part, and then write the numbers to tell the parts.
Question 1.
Explanation:
I wrote the minus sign in the circle to show the subtraction.
Question 2.
Explanation:
When we subtract 6 from 9 3 will be left. So, i circled the number 3.
Question 3.
Explanation:
The completed the subtraction sentence that is 8 takeaway 2 is 6.
Question 4.
Explanation:
I separated the tower into 2 parts, and drew 3 cubes as a part and 2 cubes as other part. Then wrote the numbers 3 and 2 in the blanks to tell the parts.
Directions Understand Vocabulary Have students: 5 draw a circle around the difference; 6 write an equation to show how to subtract 3 from 7 to find the difference; 7 listen to the story, draw a picture to show how to take away, and then write an equation to match the story. Lorin sees 6 apples on the table. She takes 3 away. How many apples are left?
Question 5.
Explanation:
When we subtract 3 from 8 the difference is 5. So, i circled 5.
Question 6.
Explanation:
The equation that shows the how to subtract 3 from 7 is 7-3=4.
Question 7.
Explanation:
Lorin sees 6 apples on the table. She takes 3 away. I marked X on 3apples as Lorin takes away 3 apples. Therefore 3 apples are left. I wrote the equation 6-3=3 to show what Lorin did.
### Topic 7 Reteaching
Directions Have students: 1 count the bees, tell how many are NOT on the flower, and then write the number to tell how many are left on the flower; 2 take apart the group of apples. Have them draw a circle around the parts they made, and then write the numbers to tell the parts.
Set A
Question 1.
Explanation:
i counted the bees that are not on the flower. There is 1 bee out and there are 4 bees left on the flower. So, i wrote the number 4 in the blank.
Set B
Question 2.
Explanation:
I took apart the group of apples then drew a circle around the parts i made, and then wrote the numbers 3 and 4 to tell the parts.
Directions Have students: 3 listen to the story, and then complete the sentence to tell how many are left. Javi sees 9 dragonflies. 4 fly away. How many dragonflies are left? 4 use counters to model the problem, mark Xs to subtract, and then write an equation to find the difference.
Set C
Question 3.
Explanation:
Javi sees 9 dragonflies. 4 fly away. Therefore, 5 flies are left. So, i wrote the sentence 9 takeaway 4 is 5.
Set D
Question 4.
Explanation:
I used the picture to solve the problem. I marked X on 1 image. Therefore, 4-1=3.
Directions Have students: 5 use counters to model the problem, and then write an equation to tell how many are left; 6 listen to the story, draw a picture to show what is happening, and then write an equation to match the story. Lidia has 5 balloons. 2 balloons pop. How many balloons does she have left?
Set D continued
Question 5.
Explanation:
There ere 9 birds in the question. I marked X on 6 birds. 3 birds are left. So, i wrote the subtraction equation 9-6=3.
Set E
Question 6.
Explanation:
Lidia has 5 balloons. 2 balloons pop. So, 3 balloons are left. Therefore, the subtraction equation is 5-2=3.
Directions Have students: 7 complete each equation to find the pattern; 8 listen to the story, use counters to help solve the problem, and then write the equation. Darla sees 3 frogs on the pond. 7 more join them. How many frogs are there in all?
Set F
Question 7.
Explanation:
I completed the equations to find the pattern. The differences of the equations are increasing by 1.
Set G
Question 8.
Explanation:
Darla sees 3 frogs on the pond. 7 more join them. I used counters to show what is happening. There are 10 frogs in all. Therefore, i wrote the equation 3+7=10.
### Topic 7 Assessment Practice
Directions Have students mark the best answer. 1 Say: Which expression matches the picture and tells the number of cubes in all and a part that is taken away? 2 Say: There are some animals in a group. Then some animals leave. Which sentence and subtraction equation match the picture and tell how many animals are left? 3 Which equation matches the picture and tells how many ducks are left?
Question 1.
A. 6 take away 2
B. 7 take away 2
C. 4 take away 2
D. 5 take away 3
B. 7 takeaway 2.
Explanation:
There are 7 cubes in all and 2 cubes are seperated from the other 5 cubes.
Question 2.
A. 4 take away 2 is 2. 8 – 2 = 6
B. 4 take away 3 is 1. 4 – 3 = 1
C. 2 take away I is 2. 3 – 1 = 2
D. 5 take away 3 is 2. 5 – 3 = 2
B. 4 takeaway 3 is 1. 4-3=1
Explanation:
There 4 mouse in all. 3 are marked X so, 1 is left. So, the equation will be 4-3=1.
Question 3.
A. 5 – 3 = 2
B. 5 – 2 = 3
C. 7 – 3 = 4
D. 7 – 2 = 5
A. 5-3=2.
Explanation:
There are 5 ducks in all. 3 are marked X 2 are left. So, the equation is 5-3=2.
Directions 4 Have students listen to the story, and then complete the sentence to tell how many are left. Kyle sees 10 turtles at the zoo. 2 turtles crawl away. Write a number sentence that tells how many are left. 5 Have students count the fish. Then have them mark Xs on some of the fish that swim away, write the number to tell how many are left, and write a matching equation, 6 Say: Ramona has 7 apples. She puts the apples on 2 plates. Draw apples to show how many Ramona could put on each plate. Then write the numbers to tell the parts.
Question 4.
Explanation:
Kyle sees 10 turtles at the zoo. 2 turtles crawl away. So, 8 are left. Therefore, the sentence is 10 takeaway 2 is 8.
Question 5.
Explanation:
I counted the number of fishes. There are 8 fishes in all. I marked X on 3 fishes that swim away, wrote the number 5 to tell how many are left. The equation is 8-3=5.
Question 6.
Explanation:
Ramona has 7 apples. She puts the apples on 2 plates. I drew 3 apples i one plate and 4 apples in the other plate to show how many Ramona could put on each plate. 3+4=7.
Directions 7 Have students complete each equation to find the pattern, 8 Have students listen to the story, draw a circle around the picture that shows the story, and then write an equation. There were 7 lizards in the sand. I crawls away. How many are left? 9 Which equation matches the picture?
Question 7.
Explanation:
I observed the pattern above, the number to be subtracted is increasing by 1. So, i completed the equations based on the pattern. The difference is decreasing by 1.
Question 8.
Explanation:
There were 7 lizards in the sand. 1 crawls away. Therefore, 6 are left. So, i wrote the equation 7-1=6.
Question 9.
A. 6 – 2 = 4
B. 5 – 4 = 1
C. 9 – 4 = 5
D. 4 – 4 = 0
C. 9 – 4 = 5
Explanation:
There are 9 crabs. 4 crabs are marked X. 5 are left. Therefore the equation 9-4=5 matches with the picture.
Directions Have students: 10 take apart the group of plums. Have them draw a circle around the parts they made, and then write an equation that matches their picture; 11 listen to the story, draw a picture, use counters or other objects to help solve the problem, and then write an equation that matches the story. Kim collects 9 shells. She gives 6 away. How many shells does Kim have left? 12 Have students match each equation with a row of flowers to find the pattern.
Question 10.
Explanation:
I took apart the plums. I marked X on 2 plums. 4 are left. Therefore, the subtraction equation is 6-2=4.
Question 11.
Explanation:
Kim collects 9 shells. She gives 6 away. 3 are left. The subtraction equation is 9-6=3.
Question 12.
Explanation:
I matched the pictures with the equations.
Directions Puppet Show Say: Paco’s class uses many puppets for their puppet show. 1 Have students listen to the story, and then write a subtraction sentence to tell how many duck puppets are left. Paco has 8 duck puppets at school. He takes 3 home. How many duck puppets are left at school? 2 Write an equation to tell how many duck puppets Paco has left at school. 3 Say: The picture shows that Paco put I cat puppet in a drawer. How many cat puppets are left? Have students write an equation for the picture, and then write another equation to complete a pattern.
Question 1.
Explanation:
Paco has 8 duck puppets at school. He takes 3 home. The sentence is 8 takeaway 3 is 5. Therefore, 5 puppets are left at school.
Question 2.
Explanation:
The equation that represents the sentence 8 takeaway 3 is 5 is 8-3=5.
Question 3.
Explanation:
Paco has 5 puppets and he put 1 puppet in drawer. So, the equation is 5-1=4.
Another equation in this pattern is 5-4=1, this tells 1 is in drawer and 4 are with Paco.
Directions 4 Say: Paco’s class puts on a play using 4 puppets. Each scene of the play has I more puppet leave the stage than the scene before. Have students mark Xs to complete the pattern. Then have them write equations to show how many puppets leave each scene. 5 Have students listen to the story, use counters to help solve each part of the problem, and then write an equation. Paco has 4 yellow bird puppets and 3 red bird puppets on his desk. How many bird puppets does Paco have in all? Then Paco moves 2 bird puppets to his friend Owen’s desk. How many bird puppets are left on Paco’s desk?
Question 4. |
# Ex.4.2 Q3 Quadratic Equations Solutions - NCERT Maths Class 10
Go back to 'Ex.4.2'
## Question
Find two numbers whose sum is $$27$$ and product is $$182.$$
Video Solution
Ex 4.2 | Question 3
## Text Solution
What is known?
(i) Sum of two numbers is $$27.$$
(ii) Product of two numbers is $$182.$$
What is Unknown?
Two numbers.
Reasoning:
Let one of the numbers be $$x$$. Then the other number will be
$$x +$$ other number $$= 27$$
Other number $$= 27 – x$$
Product of the two numbers $$= 182$$
This can be written in the form of the following equation:
$x\left( {27 - x} \right) = 182$
Steps:
\begin{align}x\left( {27 - x} \right) &= 182\\27x - x &= 182\\27x - x - 182 &= 0\\x - 27x + 182 &= 0\\x - 14x-13x + 182 &= 0\\x\left( {x - 14} \right)-13\left( {x - 14} \right) &= 0\\\left( {x - 13} \right)\left( {x - 14} \right) &= 0\\x-13 = 0 & \qquad x-14 = 0\\x = 13 & \qquad x = 14\end{align}
The numbers are $$13,\, 14.$$
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#### Chapter 4 System of Linear Equations
Section 4.4 More general Systems
# 4.4.3 Exercises
##### Exercise 4.4.3
Find the $y$-intercept $b$ and the slope $m$ of a line described by the equation $y=mx+b$ which is defined by two points. The first point at ${x}_{1}=\alpha$ lies on the line described by the equation ${y}_{1}\left(x\right)=-2\left(1+x\right)$. The second point at ${x}_{2}=\beta$ lies on the line described by the equation ${y}_{2}\left(x\right)=x-1$. The following figure illustrates the situation.
1. Find the system of equations for the parameters $b$ and $m$.
The first equation reads $m\alpha +b$$=$
;
the second equation reads $m\beta +b$$=$
.
The constants $\alpha$ and $\beta$ have to remain in the solution; for these alpha and beta can be entered.
2. Solve this system of equations for $b$ and $m$. For which values of $\alpha$ and $\beta$ does the system have a unique solution, no solution, or an infinite number of solutions?
For $\alpha =-2$ and $\beta =2$ one obtains, for example, the solution $m$$=$
and $b$$=$
, the case $\alpha =2$ and $\beta =-2$ results in the solution $m$$=$
and $b$$=$ .
The LS has an infinite number of solutions if $\alpha$$=$
and $\beta$$=$ .
The corresponding solutions can be parameterised by $m=r$ and $b$$=$
, $r\in ℝ$.
3. What is the graphical interpretation of the last two cases, i.e. no solution and an infinite number of solutions?
##### Exercise 4.4.4
Find the solution set of the following LS depending on the parameter $t\in ℝ$.
$\begin{array}{cccc}\hfill x-y+tz& \hfill =\hfill & t\hfill & \text{equation} \left(1\right) ,\hfill \\ \hfill tx+\left(1-t\right)y+\left(1+{t}^{2}\right)z& \hfill =\hfill & -1+t\hfill & \text{equation} \left(2\right) ,\hfill \\ \hfill \left(1-t\right)x+\left(-2+t\right)y+\left(-1+t-{t}^{2}\right)z& \hfill =\hfill & {t}^{2}\hfill & \text{equation} \left(3\right) .\hfill \end{array}$
The LS only has solutions for the following values of the parameter: $t\in \text{}$
.
Set can be entered in the form $\left\{$a;b;c;$\dots \right\}$. The empty set can be entered as $\left\{\right\}$.
For the smallest value of the parameter $t$ the solution is
$x$$=$
, $y$$=$
, $z=r$, $r\in ℝ$.
For the greatest value of the parameter $t$ the solution is
$x$$=$
, $y$$=$
, $z=r$, $r\in ℝ$. |
Associated Topics || Dr. Math Home || Search Dr. Math
### Reducing Algebraic Fractions by Cancelling
```Date: 03/16/2004 at 22:26:49
From: Jennie
Subject: canceling factors or terms
When reducing fractions, how come you can cancel factors but you can't
cancel terms? Can you please explain, hopefully with examples too?
Thank you so much!
```
```
Date: 03/17/2004 at 14:34:44
From: Doctor Peterson
Subject: Re: canceling factors or terms
Hi, Jennie.
Good question! It's important to understand what is really happening
when you cancel, so you know when you can do it, and why.
Factors are parts of an expression that are multiplied together; you
can cancel them when you are dividing two expressions. For example,
you can simplify this expression by canceling the z's:
xz x z x x
---- = --- * --- = --- * 1 = ---
yz y z y y
The z's "cancel" because the expression can be rearranged to contain
z/z which is equal to 1.
Terms are parts of an expression that are ADDED together; you can
cancel them, but only when you are subtracting rather than dividing.
For example, you can "cancel" the z's here:
(x + z) - (y + z) = x + z - y - z = x - y + z - z = x - y
This time the z's disappear because when they are brought together we
have z - z which is zero. We're subtracting z from both terms, and
their difference remains the same.
The reason terms can't be canceled when you are dividing is simply
that addition and division don't work so nicely together. For
example, in the expression
x + z
-----
y + z
there is no way to rearrange the expression to give you z/z or z - z.
It may help to give some concrete demonstrations that canceling
actually doesn't work here. First, we CAN cancel this:
2*5 2
--- = ---
3*5 3
and in fact the original fraction is 10/15 which is equal to 2/3. And
we CAN "cancel" this:
(3 + 5) - (2 + 5) = 3 + 5 - 2 - 5 = 3 - 2 + 5 - 5 = 3 - 2 = 1
and the original expression is 8 - 7 which is equal to 1 also.
But this is NOT true:
2 + 5 2
----- = ---
3 + 5 3
Here the left side is 7/8, which clearly is not equal to 2/3. In
terms of a picture, the left side has 7 of 8 parts shaded:
+---+---+---+---+---+---+---+---+
| |xxx|xxx|xxx|xxx|xxx|xxx|xxx|
+---+---+---+---+---+---+---+---+
If we subtract 5 from both numerator and denominator, we are taking
away 5 shaded squares, and no unshaded squares:
+---+---+---+
| |xxx|xxx|
+---+---+---+
Now we have 2 of 3 shaded; taking away only shaded squares changed
the ratio. Proper cancellation is balanced:
2 * 5 2
----- = ---
3 * 5 3
+---+---+---+---+---+ +---+
|xxx|xxx|xxx|xxx|xxx| |xxx|
+---+---+---+---+---+ +---+
|xxx|xxx|xxx|xxx|xxx| --> |xxx|
+---+---+---+---+---+ +---+
| | | | | | | |
+---+---+---+---+---+ +---+
Here by dividing numerator and denominator by 5, we take squares away
in the right proportions so that the fraction is unchanged.
Does that help make things clearer?
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials
Middle School Factoring Expressions
Middle School Fractions
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# 10.7: Trigonometric Equations and Inequalities
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
In Sections \ref{TheUnitCircle}, \ref{CircularFunctions} and most recently \ref{ArcTrig}, we solved some basic equations involving the trigonometric functions. Below we summarize the techniques we've employed thus far. Note that we use the neutral letter $u$' as the argument\footnote{See the comments at the beginning of Section \ref{TrigGraphs} for a review of this concept.} of each circular function for generality.
Strategies for Solving Basic Equations Involving Trigonometric Functions
• To solve $$\cos(u) = c$$ or $$\sin(u) = c$$ for $$-1 \leq c \leq 1$$, first solve for $$u$$ in the interval $$[0,2\pi)$$ and add integer multiples of the period $$2\pi$$. If $$c < -1$$ or of $$c > 1$$, there are no real solutions.
• To solve $$\sec(u) = c$$ or $$\csc(u) = c$$ for $$c \leq -1$$ or $$c \geq 1$$, convert to cosine or sine, respectively, and solve as above. If $$-1 < c < 1$$, there are no real solutions.
• To solve $$\tan(u) = c$$ for any real number $$c$$, first solve for $$u$$ in the interval $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ and add integer multiples of the period $$\pi$$.
• To solve $$\cot(u) = c$$ for $$c \neq 0$$, convert to tangent and solve as above. If $$c = 0$$, the solution to $$\cot(u) = 0$$ is $$u = \frac{\pi}{2} + \pi k$$ for integers $$k$$.
Using the above guidelines, we can comfortably solve $$\sin(x) = \frac{1}{2}$$ and find the solution $$x = \frac{\pi}{6} + 2\pi k$$ or $$x = \frac{5\pi}{6} + 2\pi k$$ for integers $$k$$. How do we solve something like $$\sin(3x) = \frac{1}{2}$$? Since this equation has the form $$\sin(u) = \frac{1}{2}$$, we know the solutions take the form $$u= \frac{\pi}{6} + 2\pi k$$ or $$u = \frac{5\pi}{6} + 2\pi k$$ for integers $$k$$. Since the argument of sine here is $$3x$$, we have $$3x= \frac{\pi}{6} + 2\pi k$$ or $$3x = \frac{5\pi}{6} + 2\pi k$$ for integers $$k$$. To solve for $$x$$, we divide both sides\footnote{Don't forget to divide the $$2\pi k$$ by $$3$$ as well!} of these equations by $$3$$, and obtain $$x = \frac{\pi}{18} + \frac{2\pi}{3} k$$ or $$x = \frac{5\pi}{18} + \frac{2\pi}{3}k$$ for integers $$k$$. This is the technique employed in the example below.
Example $$\PageIndex{1}$$:
Solve the following equations and check your answers analytically. List the solutions which lie in the interval $$[0,2\pi)$$ and verify them using a graphing utility.
1. $$\cos(2x) = -\frac{\sqrt{3}}{2}$$
2. $$\csc\left(\frac{1}{3}x-\pi \right) = \sqrt{2}$$
3. $$\cot\left(3x \right) = 0$$
4. $$\sec^{2}(x) = 4$$
5. $$\tan\left(\frac{x}{2}\right) = -3$$
6. $$\sin(2x) = 0.87$$
Solution
1. The solutions to $$\cos(u) =-\frac{\sqrt{3}}{2}$$ are $$u = \frac{5\pi}{6} + 2\pi k$$ or $$u = \frac{7\pi}{6} + 2\pi k$$ for integers $$k$$. Since the argument of cosine here is $$2x$$, this means $$2x = \frac{5\pi}{6} + 2\pi k$$ or $$2x = \frac{7\pi}{6} + 2\pi k$$ for integers $$k$$. Solving for $$x$$ gives $$x = \frac{5\pi}{12} + \pi k$$ or $$x = \frac{7\pi}{12} + \pi k$$ for integers $$k$$. To check these answers analytically, we substitute them into the original equation. For any integer $$k$$ we have
$\begin{array}{rclr} \cos\left( 2\left[\frac{5\pi}{12} + \pi k\right]\right) & = & \cos\left(\frac{5\pi}{6} + 2\pi k\right) & \\ [3pt] & = & \cos\left(\frac{5\pi}{6}\right) & \text{(the period of cosine is $$2\pi)} \\ [3pt] & = & -\frac{\sqrt{3}}{2} & \\ \end{array}$ Similarly, we find \(\cos\left( 2\left[\frac{7\pi}{12} + \pi k\right]\right) = \cos\left(\frac{7\pi}{6} + 2\pi k\right) = \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$. To determine which of our solutions lie in $$[0,2\pi)$$, we substitute integer values for $$k$$. The solutions we keep come from the values of $$k = 0$$ and $$k =1$$ and are $$x = \frac{5\pi}{12}$$, $$\frac{7\pi}{12}$$, $$\frac{17\pi}{12}$$ and $$\frac{19\pi}{12}$$. Using a calculator, we graph $$y = \cos(2x)$$ and $$y = -\frac{\sqrt{3}}{2}$$ over $$[0,2\pi)$$ and examine where these two graphs intersect. We see that the $$x$$-coordinates of the intersection points correspond to the decimal representations of our exact answers.
1. \item Since this equation has the form $$\csc(u) = \sqrt{2}$$, we rewrite this as $$\sin(u) = \frac{\sqrt{2}}{2}$$ and find $$u = \frac{\pi}{4} + 2\pi k$$ or $$u = \frac{3\pi}{4} + 2\pi k$$ for integers $$k$$. Since the argument of cosecant here is $$\left(\frac{1}{3}x-\pi \right)$$,
$\frac{1}{3}x-\pi = \frac{\pi}{4} + 2\pi k \quad \text{or} \quad \frac{1}{3}x-\pi = \frac{3\pi}{4} + 2\pi k$
To solve $$\frac{1}{3}x-\pi = \frac{\pi}{4} + 2\pi k$$, we first add $$\pi$$ to both sides
$\frac{1}{3} x = \frac{\pi}{4} + 2\pi k + \pi$
A common error is to treat the $2\pi k$' and $\pi$' terms as like' terms and try to combine them when they are not.\footnote{Do you see why?} We can, however, combine the $\pi$' and $\frac{\pi}{4}$' terms to get
$\frac{1}{3} x = \frac{5\pi}{4} + 2\pi k$
We now finish by multiplying both sides by $$3$$ to get
$x = 3 \left( \frac{5\pi}{4} + 2\pi k \right) = \frac{15 \pi}{4} + 6\pi k$
Solving the other equation, $$\frac{1}{3}x-\pi = \frac{3\pi}{4} + 2\pi k$$ produces $$x = \frac{21\pi}{4} + 6 \pi k$$ for integers $$k$$. To check the first family of answers, we substitute, combine line terms, and simplify.
$\begin{array}{rclr} \csc\left(\frac{1}{3} \left[ \frac{15\pi}{4} + 6 \pi k \right] - \pi \right) & = & \csc\left(\frac{5\pi}{4} + 2\pi k - \pi \right) & \\ [3pt] & = & \csc\left(\frac{\pi}{4} + 2\pi k\right) & \\ [3pt] & = & \csc\left(\frac{\pi}{4}\right) & \text{(the period of cosecant is $$2\pi)} \\ & = & \sqrt{2} & \\ \end{array}$ The family \(x= \frac{21\pi}{4} + 6 \pi k$$ checks similarly. Despite having infinitely many solutions, we find that \textit{none} of them lie in $$[0,2\pi)$$. To verify this graphically, we use a reciprocal identity to rewrite the cosecant as a sine and we find that $$y = \frac{1}{\sin\left(\frac{1}{3}x-\pi\right)}$$ and $$y =\sqrt{2}$$ do not intersect at all over the interval $$[0,2\pi)$$.
\begin{center}
\begin{tabular}{cc}
\includegraphics[width=2in]{./IntroTrigGraphics/TrigEquIneq01.jpg} &
\hspace{0.75in} \includegraphics[width=2in]{./IntroTrigGraphics/TrigEquIneq02.jpg} \\
$y = \cos(2x)\) and \boldmath $$y=-\frac{\sqrt{3}}{2}$$ & \hspace{0.75in} $$y = \frac{1}{\sin\left(\frac{1}{3}x-\pi\right)}$$ and \boldmath $$y =\sqrt{2}$$ \\ \end{tabular} \end{center} \item Since $$\cot(3x) = 0$$ has the form $$\cot(u) = 0$$, we know $$u = \frac{\pi}{2} + \pi k$$, so, in this case, $$3x = \frac{\pi}{2} + \pi k$$ for integers $$k$$. Solving for $$x$$ yields $$x = \frac{\pi}{6} + \frac{\pi}{3} k$$. Checking our answers, we get $\begin{array}{rclr} \cot\left(3\left[ \frac{\pi}{6} + \frac{\pi}{3} k\right]\right) & = & \cot\left(\frac{\pi}{2} + \pi k\right) & \\ [3pt] & = & \cot\left(\frac{\pi}{2}\right) & \text{(the period of cotangent is $$\pi)} \\ [3pt] & = & 0 & \\ \end{array}$ As \(k$$ runs through the integers, we obtain six answers, corresponding to $$k=0$$ through $$k=5$$, which lie in $$[0, 2\pi)$$: $$x = \frac{\pi}{6}$$, $$\frac{\pi}{2}$$, $$\frac{5\pi}{6}$$, $$\frac{7\pi}{6}$$ , $$\frac{3\pi}{2}$$ and $$\frac{11\pi}{6}$$. To confirm these graphically, we must be careful. On many calculators, there is no function button for cotangent. We choose\footnote{The reader is encouraged to see what happens if we had chosen the reciprocal identity $$\cot(3x) = \frac{1}{\tan(3x)}$$ instead. The graph on the calculator \textit{appears} identical, but what happens when you try to find the intersection points?} to use the quotient identity $$\cot(3x) = \frac{\cos(3x)}{\sin(3x)}$$. Graphing $$y = \frac{\cos(3x)}{\sin(3x)}$$ and $$y = 0$$ (the $$x$$-axis), we see that the $$x$$-coordinates of the intersection points approximately match our solutions. 1. \item The complication in solving an equation like $$\sec^{2}(x) = 4$$ comes not from the argument of secant, which is just $$x$$, but rather, the fact the secant is being squared. To get this equation to look like one of the forms listed on page \pageref{trigeqnstrategy1}, we extract square roots to get $$\sec(x) = \pm 2$$. Converting to cosines, we have $$\cos(x) = \pm \frac{1}{2}$$. For $$\cos(x) = \frac{1}{2}$$, we get $$x = \frac{\pi}{3} + 2\pi k$$ or $$x = \frac{5\pi}{3} + 2\pi k$$ for integers $$k$$. For $$\cos(x) = -\frac{1}{2}$$, we get $$x = \frac{2\pi}{3} + 2\pi k$$ or $$x = \frac{4\pi}{3} + 2\pi k$$ for integers $$k$$. If we take a step back and think of these families of solutions geometrically, we see we are finding the measures of all angles with a reference angle of $$\frac{\pi}{3}$$. As a result, these solutions can be combined and we may write our solutions as $$x = \frac{\pi}{3} + \pi k$$ and $$x = \frac{2\pi}{3} + \pi k$$ for integers $$k$$. To check the first family of solutions, we note that, depending on the integer $$k$$, $$\sec\left(\frac{\pi}{3} + \pi k\right)$$ doesn't always equal $$\sec\left(\frac{\pi}{3}\right)$$. However, it is true that for all integers $$k$$, $$\sec\left(\frac{\pi}{3} + \pi k\right) = \pm \sec\left(\frac{\pi}{3}\right) = \pm 2$$. (Can you show this?) As a result, $\begin{array}{rclr} \sec^{2}\left(\frac{\pi}{3} + \pi k\right) & = & \left( \pm \sec\left(\frac{\pi}{3}\right)\right)^2 & \\ [3pt] & = & (\pm 2)^2 & \\ [3pt] & = & 4 & \\ \end{array}$ The same holds for the family $$x =\frac{2\pi}{3} + \pi k$$. The solutions which lie in $$[0,2\pi)$$ come from the values $$k = 0$$ and $$k=1$$, namely $$x = \frac{\pi}{3}$$, $$\frac{2\pi}{3}$$, $$\frac{4\pi}{3}$$ and $$\frac{5\pi}{3}$$. To confirm graphically, we use a reciprocal identity to rewrite the secant as cosine. The $$x$$-coordinates of the intersection points of $$y = \frac{1}{(\cos(x))^2}$$ and $$y = 4$$ verify our answers. \begin{center} \begin{tabular}{cc} \includegraphics[width=2in]{./IntroTrigGraphics/TrigEquIneq03.jpg} & \hspace{0.75in} \includegraphics[width=2in]{./IntroTrigGraphics/TrigEquIneq04.jpg} \\$y = \frac{\cos(3x)}{\sin(3x)}\) and \boldmath $$y=0$$ &
\hspace{0.75in} $$y = \frac{1}{\cos^{2}(x)}$$ and \boldmath $$y = 4$$ \\
\end{tabular}
\end{center}
\item The equation $$\tan\left(\frac{x}{2}\right) = -3$$ has the form $$\tan(u) = -3$$, whose solution is $$u = \arctan(-3) + \pi k$$. Hence, $$\frac{x}{2} = \arctan(-3) + \pi k$$, so $$x = 2\arctan(-3) + 2\pi k$$ for integers $$k$$. To check, we note
$\begin{array}{rclr} \tan\left(\frac{2\arctan(-3) + 2\pi k}{2}\right) & = & \tan\left( \arctan(-3) + \pi k \right) & \\ [3pt] & = & \tan\left(\arctan(-3) \right) & \text{(the period of tangent is $$\pi)} \\ [3pt] & = & -3 & (\text{See Theorem } \ref{arctangentcotangentfunctionprops}) \\ \end{array}$ To determine which of our answers lie in the interval \([0,2\pi)$$, we first need to get an idea of the value of $$2\arctan(-3)$$. While we could easily find an approximation using a calculator,\footnote{Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be?} we proceed analytically. Since $$-3 < 0$$, it follows that $$-\frac{\pi}{2} < \arctan(-3) < 0$$. Multiplying through by $$2$$ gives $$-\pi < 2\arctan(-3) < 0$$. We are now in a position to argue which of the solutions $$x = 2\arctan(-3) + 2\pi k$$ lie in $$[0,2\pi)$$. For $$k = 0$$, we get $$x = 2\arctan(-3) < 0$$, so we discard this answer and all answers $$x = 2\arctan(-3) + 2\pi k$$ where $$k < 0$$. Next, we turn our attention to $$k = 1$$ and get $$x = 2\arctan(-3) + 2\pi$$. Starting with the inequality $$-\pi < 2\arctan(-3) < 0$$, we add $$2\pi$$ and get $$\pi < 2\arctan(-3) +2\pi < 2\pi$$. This means $$x = 2\arctan(-3) + 2\pi$$ lies in $$[0,2\pi)$$. Advancing $$k$$ to $$2$$ produces $$x = 2\arctan(-3) + 4\pi$$. Once again, we get from $$-\pi < 2\arctan(-3) < 0$$ that $$3\pi < 2\arctan(-3) + 4\pi < 4\pi$$. Since this is outside the interval $$[0,2\pi)$$, we discard $$x = 2\arctan(-3) + 4\pi$$ and all solutions of the form $$x = 2\arctan(-3) + 2\pi k$$ for $$k > 2$$. Graphically, we see $$y = \tan\left(\frac{x}{2}\right)$$ and $$y = -3$$ intersect only once on $$[0,2\pi)$$ at $$x = 2\arctan(-3) + 2\pi\approx 3.7851$$.
1. \item To solve $$\sin(2x) = 0.87$$, we first note that it has the form $$\sin(u) = 0.87$$, which has the family of solutions $$u = \arcsin(0.87) + 2\pi k$$ or $$u =\pi - \arcsin(0.87) + 2\pi k$$ for integers $$k$$. Since the argument of sine here is $$2x$$, we get $$2x = \arcsin(0.87) + 2\pi k$$ or $$2x =\pi - \arcsin(0.87) + 2\pi k$$ which gives $$x = \frac{1}{2} \arcsin(0.87) + \pi k$$ or $$x =\frac{\pi}{2} - \frac{1}{2}\arcsin(0.87) + \pi k$$ for integers $$k$$. To check,
$\begin{array}{rclr} \sin\left(2\left[\frac{1}{2} \arcsin(0.87) + \pi k\right]\right) & = & \sin\left(\arcsin(0.87) + 2\pi k\right) & \\ [3pt] & = & \sin\left(\arcsin(0.87)\right) & \text{(the period of sine is $$2\pi)} \\ [3pt] & = & 0.87& (\text{See Theorem } \ref{arccosinesinefunctionprops})\\ \end{array}$ For the family \(x =\frac{\pi}{2} - \frac{1}{2}\arcsin(0.87) + \pi k$$ , we get
$\begin{array}{rclr} \sin\left(2\left[\frac{\pi}{2} - \frac{1}{2} \arcsin(0.87) + \pi k\right]\right) & = & \sin\left(\pi - \arcsin(0.87) + 2\pi k\right) & \\ [3pt] & = & \sin\left(\pi - \arcsin(0.87)\right) & \text{(the period of sine is $$2\pi)} \\ [3pt] & = & \sin\left(\arcsin(0.87)\right) & \text{(\sin(\pi - t) = \sin(t))} \\ [3pt] & = & 0.87& (\text{See Theorem } \ref{arccosinesinefunctionprops}) \\ \end{array}$ To determine which of these solutions lie in \([0,2\pi)$$, we first need to get an idea of the value of $$x=\frac{1}{2} \arcsin(0.87)$$. Once again, we could use the calculator, but we adopt an analytic route here. By definition, $$0 < \arcsin(0.87) < \frac{\pi}{2}$$ so that multiplying through by $$\frac{1}{2}$$ gives us $$0 < \frac{1}{2} \arcsin(0.87) < \frac{\pi}{4}$$. Starting with the family of solutions $$x = \frac{1}{2} \arcsin(0.87) + \pi k$$, we use the same kind of arguments as in our solution to number \ref{arctanin02pi} above and find only the solutions corresponding to $$k =0$$ and $$k=1$$ lie in $$[0,2\pi)$$: $$x = \frac{1}{2} \arcsin(0.87)$$ and $$x = \frac{1}{2} \arcsin(0.87) + \pi$$. Next, we move to the family $$x =\frac{\pi}{2} - \frac{1}{2}\arcsin(0.87) + \pi k$$ for integers $$k$$. Here, we need to get a better estimate of $$\frac{\pi}{2} - \frac{1}{2} \arcsin(0.87)$$. From the inequality $$0 < \frac{1}{2}\arcsin(0.87) < \frac{\pi}{4}$$, we first multiply through by $$-1$$ and then add $$\frac{\pi}{2}$$ to get $$\frac{\pi}{2} > \frac{\pi}{2} -\frac{1}{2} \arcsin(0.87) > \frac{\pi}{4}$$, or $$\frac{\pi}{4} < \frac{\pi}{2} -\frac{1}{2} \arcsin(0.87) < \frac{\pi}{2}$$. Proceeding with the usual arguments, we find the only solutions which lie in $$[0,2\pi)$$ correspond to $$k = 0$$ and $$k=1$$, namely $$x =\frac{\pi}{2} - \frac{1}{2}\arcsin(0.87)$$ and $$x = \frac{3\pi}{2} - \frac{1}{2}\arcsin(0.87)$$. All told, we have found four solutions to $$\sin(2x) = 0.87$$ in $$[0,2\pi)$$: $$x =\frac{1}{2} \arcsin(0.87)$$, $$x=\frac{1}{2} \arcsin(0.87) + \pi$$, $$x =\frac{\pi}{2} - \frac{1}{2}\arcsin(0.87)$$ and $$x = \frac{3\pi}{2} - \frac{1}{2}\arcsin(0.87)$$. By graphing $$y = \sin(2x)$$ and $$y = 0.87$$, we confirm our results.
\hspace{0.75in} \includegraphics[width=2in]{./IntroTrigGraphics/TrigEquIneq06.jpg} \\
\$y = \tan\left(\frac{x}{2}\right)\) and \boldmath $$y = -3$$ &
Each of the problems in Example \ref{TrigEqnEx1} featured one trigonometric function. If an equation involves two different trigonometric functions or if the equation contains the same trigonometric function but with different arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page \pageref{trigeqnstrategy1}.
We repeat here the advice given when solving systems of nonlinear equations in section \ref{NonLinear} -- when it comes to solving equations involving the trigonometric functions, it helps to just try something.
Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we've used in the past to solve the inequalities.\footnote{See page \pageref{firstsigndiagram}, Example \ref{polygraphex}, page \pageref{rationalsigndiagram}, page \pageref{algebraicsigndiagram}, Example \ref{expineq} and Example \ref{logineq} for discussion of this technique.}
Example $$\PageIndex{2}$$:
Solve the following inequalities on $$[0,2\pi)$$. Express your answers using interval notation and verify your answers graphically.
1. $$2\sin(x) \leq 1$$
2. $$\sin(2x) > \cos(x)$$
3. $$\tan(x) \geq 3$$
Solution
1. We begin solving $$2\sin(x) \leq 1$$ by collecting all of the terms on one side of the equation and zero on the other to get $$2\sin(x) - 1 \leq 0$$. Next, we let $$f(x) = 2\sin(x) - 1$$ and note that our original inequality is equivalent to solving $$f(x) \leq 0$$. We now look to see where, if ever, $$f$$ is undefined and where $$f(x) = 0$$. Since the domain of $$f$$ is all real numbers, we can immediately set about finding the zeros of $$f$$. Solving $$f(x) = 0$$, we have $$2\sin(x) - 1=0$$ or $$\sin(x) = \frac{1}{2}$$. The solutions here are $$x = \frac{\pi}{6} + 2\pi k$$ and $$x = \frac{5\pi}{6} + 2\pi k$$ for integers $$k$$. Since we are restricting our attention to $$[0,2\pi)$$, only $$x = \frac{\pi}{6}$$ and $$x = \frac{5\pi}{6}$$ are of concern to us. Next, we choose test values in $$[0,2\pi)$$ other than the zeros and determine if $$f$$ is positive or negative there. For $$x = 0$$ we have $$f(0) = -1$$, for $$x = \frac{\pi}{2}$$ we get $$f\left(\frac{\pi}{2}\right) = 1$$ and for $$x = \pi$$ we get $$f(\pi) = -1$$. Since our original inequality is equivalent to $$f(x) \leq 0$$, we are looking for where the function is negative $$(-)$$ or $$0$$, and we get the intervals $$\left[0, \frac{\pi}{6}\right] \cup \left[\frac{5\pi}{6}, 2\pi \right)$$. We can confirm our answer graphically by seeing where the graph of $$y = 2\sin(x)$$ crosses or is below the graph of $$y = 1$$.
2. We first rewrite $$\sin(2x) > \cos(x)$$ as $$\sin(2x) - \cos(x) > 0$$ and let $$f(x) = \sin(2x) - \cos(x)$$. Our original inequality is thus equivalent to $$f(x) > 0$$. The domain of $$f$$ is all real numbers, so we can advance to finding the zeros of $$f$$. Setting $$f(x) = 0$$ yields $$\sin(2x) - \cos(x) = 0$$, which, by way of the double angle identity for sine, becomes $$2\sin(x)\cos(x) - \cos(x) = 0$$ or $$\cos(x) (2\sin(x) - 1) = 0$$. From $$\cos(x) = 0$$, we get $$x = \frac{\pi}{2} + \pi k$$ for integers $$k$$ of which only $$x = \frac{\pi}{2}$$ and $$x = \frac{3\pi}{2}$$ lie in $$[0,2\pi)$$. For $$2\sin(x) - 1 = 0$$, we get $$\sin(x) = \frac{1}{2}$$ which gives $$x = \frac{\pi}{6} + 2\pi k$$ or $$x = \frac{5\pi}{6} + 2\pi k$$ for integers $$k$$. Of those, only $$x = \frac{\pi}{6}$$ and $$x = \frac{5\pi}{6}$$ lie in $$[0,2\pi)$$. Next, we choose our test values. For $$x =0$$ we find $$f(0) = -1; when \(x = \frac{\pi}{4}$$ we get $$f\left(\frac{\pi}{4}\right) =1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}; for \(x = \frac{3\pi}{4}$$ we get $$f\left(\frac{3\pi}{4}\right) =-1 + \frac{\sqrt{2}}{2} = \frac{\sqrt{2} - 2}{2}; when \(x=\pi$$ we have $$f(\pi) = 1$$, and lastly, for $$x = \frac{7\pi}{4}$$ we get $$f\left(\frac{7\pi}{4}\right) = -1 - \frac{\sqrt{2}}{2} = \frac{-2 - \sqrt{2}}{2}$$. We see $$f(x) > 0$$ on $$\left(\frac{\pi}{6}, \frac{\pi}{2}\right) \cup \left(\frac{5\pi}{6}, \frac{3\pi}{2}\right)$$, so this is our answer. We can use the calculator to check that the graph of $$y = \sin(2x)$$ is indeed above the graph of $$y = \cos(x)$$ on those intervals.
3. Proceeding as in the last two problems, we rewrite $$\tan(x) \geq 3$$ as $$\tan(x) - 3 \geq 0$$ and let $$f(x) = \tan(x) - 3$$. We note that on $$[0,2\pi)$$, $$f$$ is undefined at $$x =\frac{\pi}{2}$$ and $$\frac{3\pi}{2}$$, so those values will need the usual disclaimer on the sign diagram.\footnote{See page \pageref{rationalsigndiagram} for a discussion of the non-standard character known as the interrobang.} Moving along to zeros, solving $$f(x) = \tan(x) - 3 = 0$$ requires the arctangent function. We find $$x = \arctan(3) + \pi k$$ for integers $$k$$ and of these, only $$x = \arctan(3)$$ and $$x = \arctan(3) + \pi$$ lie in $$[0,2\pi)$$. Since $$3 > 0$$, we know $$0 < \arctan(3) < \frac{\pi}{2}$$ which allows us to position these zeros correctly on the sign diagram. To choose test values, we begin with $$x=0$$ and find $$f(0) = -3$$. Finding a convenient test value in the interval $$\left(\arctan(3), \frac{\pi}{2}\right)$$ is a bit more challenging. Keep in mind that the arctangent function is increasing and is bounded above by $$\frac{\pi}{2}$$. This means that the number $$x = \arctan(117)$$ is guaranteed\footnote{We could have chosen any value $$\arctan(t)$$ where $$t > 3$$.} to lie between $$\arctan(3)$$ and $$\frac{\pi}{2}$$. We see that $$f(\arctan(117)) = \tan(\arctan(117)) - 3 = 114$$. For our next test value, we take $$x = \pi$$ and find $$f(\pi) = -3$$. To find our next test value, we note that since $$\arctan(3) < \arctan(117) < \frac{\pi}{2}$$, it follows\footnote{\ldots by adding $$\pi$$ through the inequality \ldots} that $$\arctan(3) + \pi < \arctan(117) + \pi < \frac{3\pi}{2}$$. Evaluating $$f$$ at $$x = \arctan(117) + \pi$$ yields $$f(\arctan(117)+\pi) = \tan(\arctan(117) + \pi) -3 = \tan(\arctan(117)) - 3 = 114$$. We choose our last test value to be $$x = \frac{7\pi}{4}$$ and find $$f\left(\frac{7\pi}{4}\right) = -4$$. Since we want $$f(x) \geq 0$$, we see that our answer is $$\left[ \arctan(3), \frac{\pi}{2}\right) \cup \left[\arctan(3)+\pi, \frac{3\pi}{2}\right)$$. Using the graphs of $$y = \tan(x)$$ and $$y = 3$$, we see when the graph of the former is above (or meets) the graph of the latter.
We close this section with an example that puts solving equations and inequalities to good use -- finding domains of functions.
Example $$\PageIndex{3}$$:
Solve the following equations and list the solutions which lie in the interval $$[0,2\pi)$$. Verify your solutions on $$[0,2\pi)$$ graphically.
1. $$3\sin^{3}(x) = \sin^{2}(x)$$
2. $$\sec^{2}(x) = \tan(x) + 3$$
3. $$\cos(2x) = 3\cos(x) - 2$$
4. $$\cos(3x) = 2- \cos(x)$$
5. $$\cos(3x) = \cos(5x)$$
6. $$\sin(2x) =\sqrt{3} \cos(x)$$
7. $$\sin(x)\cos\left(\frac{x}{2}\right) + \cos(x)\sin\left(\frac{x}{2}\right) = 1$$
8. $$\cos(x) - \sqrt{3} \sin(x) = 2$$
Solution.
1. \item We resist the temptation to divide both sides of $$3\sin^{3}(x) = \sin^{2}(x)$$ by $$\sin^{2}(x)$$ (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor.
$\begin{array}{rclr} 3\sin^{3}(x) & = & \sin^{2}(x) & \\ 3\sin^{3}(x) - \sin^{2}(x) & = & 0 & \\ \sin^{2}(x) (3 \sin(x) - 1) & = & 0 & \text{Factor out $$\sin^{2}(x)$$ from both terms.} \\ \end{array}$
We get $$\sin^{2}(x) = 0$$ or $$3\sin(x) - 1 = 0$$. Solving for $$\sin(x)$$, we find $$\sin(x) = 0$$ or $$\sin(x) = \frac{1}{3}$$. The solution to the first equation is $$x = \pi k$$, with $$x = 0$$ and $$x = \pi$$ being the two solutions which lie in $$[0,2\pi)$$. To solve $$\sin(x) = \frac{1}{3}$$, we use the arcsine function to get $$x = \arcsin\left(\frac{1}{3}\right) + 2\pi k$$ or $$x = \pi - \arcsin\left(\frac{1}{3}\right) + 2\pi k$$ for integers $$k$$. We find the two solutions here which lie in $$[0,2\pi)$$ to be $$x = \arcsin\left(\frac{1}{3}\right)$$ and $$x = \pi - \arcsin\left(\frac{1}{3}\right)$$. To check graphically, we plot $$y = 3(\sin(x))^3$$ and $$y = (\sin(x))^2$$ and find the $$x$$-coordinates of the intersection points of these two curves. Some extra zooming is required near $$x=0$$ and $$x=\pi$$ to verify that these two curves do in fact intersect four times.\footnote{Note that we are \textit{not} counting the point $$(2\pi,0)$$ in our solution set since $$x = 2\pi$$ is not in the interval $$[0,2\pi)$$. In the forthcoming solutions, remember that while $$x = 2\pi$$ may be a solution to the equation, it isn't counted among the solutions in $$[0,2\pi)$$.}
\item Analysis of $$\sec^{2}(x) = \tan(x) + 3$$ reveals two different trigonometric functions, so an identity is in order. Since $$\sec^{2}(x) = 1 + \tan^{2}(x)$$, we get
$\begin{array}{rclr} \sec^{2}(x) & = & \tan(x) + 3 & \\ 1 + \tan^{2}(x) & = & \tan(x) + 3& \text{(Since $$\sec^{2}(x) = 1 + \tan^{2}(x)$$.)} \\ \tan^{2}(x) - \tan(x) -2 & = & 0 & \\ u^2 - u - 2 & = & 0 & \text{Let $$u = \tan(x)$$.} \\ (u + 1)(u - 2) & = & 0 & \\ \end{array}$
This gives $$u = -1$$ or $$u = 2$$. Since $$u = \tan(x)$$, we have $$\tan(x) = -1$$ or $$\tan(x) = 2$$. From $$\tan(x) = -1$$, we get $$x = -\frac{\pi}{4} + \pi k$$ for integers $$k$$. To solve $$\tan(x) = 2$$, we employ the arctangent function and get $$x = \arctan(2) + \pi k$$ for integers $$k$$. From the first set of solutions, we get $$x = \frac{3\pi}{4}$$ and $$x = \frac{5\pi}{4}$$ as our answers which lie in $$[0,2\pi)$$. Using the same sort of argument we saw in Example \ref{TrigEqnEx1}, we get $$x=\arctan(2)$$ and $$x = \pi + \arctan(2)$$ as answers from our second set of solutions which lie in $$[0,2\pi)$$. Using a reciprocal identity, we rewrite the secant as a cosine and graph $$y = \frac{1}{(\cos(x))^2}$$ and $$y = \tan(x) + 3$$ to find the $$x$$-values of the points where they intersect.
• \item In the equation $$\cos(2x) = 3\cos(x) - 2$$, we have the same circular function, namely cosine, on both sides but the arguments differ. Using the identity $$\cos(2x) = 2\cos^{2}(x) - 1$$, we obtain a quadratic in disguise' and proceed as we have done in the past.
$\begin{array}{rclr} \cos(2x) & = & 3\cos(x) - 2 & \\ 2\cos^{2}(x) -1 & = & 3\cos(x) -2 & \text{(Since $$\cos(2x) = 2\cos^{2}(x) -1$$.)} \\ 2\cos^{2}(x) - 3\cos(x) +1 & = & 0 & \\ 2 u^2 - 3 u + 1 & = & 0 & \text{Let $$u = \cos(x)$$.}\\ (2u - 1)(u - 1) & = & 0 & \\ \end{array}$
This gives $$u = \frac{1}{2}$$ or $$u = 1$$. Since $$u = \cos(x)$$, we get $$\cos(x) = \frac{1}{2}$$ or $$\cos(x) = 1$$. Solving $$\cos(x) = \frac{1}{2}$$, we get $$x = \frac{\pi}{3} + 2\pi k$$ or $$x = \frac{5\pi}{3} + 2\pi k$$ for integers $$k$$. From $$\cos(x) = 1$$, we get $$x = 2\pi k$$ for integers $$k$$. The answers which lie in $$[0,2\pi)$$ are $$x =0$$, $$\frac{\pi}{3}$$, and $$\frac{5\pi}{3}$$. Graphing $$y = \cos(2x)$$ and $$y = 3\cos(x) - 2$$, we find, after a little extra effort, that the curves intersect in three places on $$[0,2\pi)$$, and the $$x$$-coordinates of these points confirm our results.
1. \item To solve $$\cos(3x) = 2- \cos(x)$$, we use the same technique as in the previous problem. From Example \ref{doubleangleex}, number \ref{cosinepolynomial}, we know that $$\cos(3x) = 4\cos^{3}(x) - 3\cos(x)$$. This transforms the equation into a polynomial in terms of $$\cos(x)$$.
$\begin{array}{rclr} \cos(3x) & = &2- \cos(x) & \\ 4\cos^{3}(x) - 3\cos(x) & = & 2- \cos(x) & \\ 2\cos^{3}(x) - 2\cos(x) -2 & = & 0 & \\ 4 u^3 - 2 u -2 & = & 0 & \text{Let $$u = \cos(x)$$.} \\ \end{array}$
To solve $$4u^3-2u-2=0$$, we need the techniques in Chapter \ref{Polynomials} to factor $$4u^3-2u-2$$ into $$(u-1)\left(4u^2+4u+2\right)$$. We get either $$u-1 = 0$$ or $$4u^2+2u+2=0$$, and since the discriminant of the latter is negative, the only real solution to $$4u^3-2u-2=0$$ is $$u = 1$$. Since $$u = \cos(x)$$, we get $$\cos(x) = 1$$, so $$x = 2\pi k$$ for integers $$k$$. The only solution which lies in $$[0,2\pi)$$ is $$x = 0$$. Graphing $$y = \cos(3x)$$ and $$y = 2- \cos(x)$$ on the same set of axes over $$[0,2\pi)$$ shows that the graphs intersect at what appears to be $$(0,1)$$, as required.
1. \item While we could approach $$\cos(3x) = \cos(5x)$$ in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. From $$\cos(3x) = \cos(5x)$$, we get $$\cos(5x) - \cos(3x) = 0$$, and it is the presence of $$0$$ on the right hand side that indicates a switch to a product would be a good move.\footnote{As always, experience is the greatest teacher here!} Using Theorem \ref{sumtoproduct}, we have that $$\cos(5x) - \cos(3x) = - 2 \sin\left( \frac{5x + 3x}{2}\right)\sin\left( \frac{5x - 3x}{2}\right) = -2 \sin(4x)\sin(x)$$. Hence, the equation $$\cos(5x) = \cos(3x)$$ is equivalent to $$-2 \sin(4x) \sin(x) = 0$$. From this, we get $$\sin(4x) = 0$$ or $$\sin(x)$$ = 0. Solving $$\sin(4x) = 0$$ gives $$x = \frac{\pi}{4} k$$ for integers $$k$$, and the solution to $$\sin(x) = 0$$ is $$x = \pi k$$ for integers $$k$$. The second set of solutions is contained in the first set of solutions,\footnote{As always, when in doubt, write it out!} so our final solution to $$\cos(5x) = \cos(3x)$$ is $$x = \frac{\pi}{4} k$$ for integers $$k$$. There are eight of these answers which lie in $$[0,2\pi)$$: $$x = 0$$, $$\frac{\pi}{4}$$, $$\frac{\pi}{2}$$, $$\frac{3\pi}{4}$$, $$\pi$$, $$\frac{5\pi}{4}$$, $$\frac{3\pi}{2}$$ and $$\frac{7\pi}{4}$$. Our plot of the graphs of $$y = \cos(3x)$$ and $$y = \cos(5x)$$ below (after some careful zooming) bears this out.
2. \item In examining the equation $$\sin(2x) =\sqrt{3} \cos(x)$$, not only do we have different circular functions involved, namely sine and cosine, we also have different arguments to contend with, namely $$2x$$ and $$x$$. Using the identity $$\sin(2x) = 2 \sin(x) \cos(x)$$ makes all of the arguments the same and we proceed as we would solving any nonlinear equation -- gather all of the nonzero terms on one side of the equation and factor.
$\begin{array}{rclr} \sin(2x) & = & \sqrt{3} \cos(x) & \\ 2 \sin(x) \cos(x) & = & \sqrt{3} \cos(x) & \text{(Since $$\sin(2x) = 2\sin(x) \cos(x)$$.)} \\ 2\sin(x) \cos(x) - \sqrt{3} \cos(x) & = & 0 & \\ \cos(x) (2 \sin(x) - \sqrt{3}) & = & 0 & \\ \end{array}$
from which we get $$\cos(x) = 0$$ or $$\sin(x) = \frac{\sqrt{3}}{2}$$. From $$\cos(x) = 0$$, we obtain $$x = \frac{\pi}{2} + \pi k$$ for integers $$k$$. From $$\sin(x) = \frac{\sqrt{3}}{2}$$, we get $$x = \frac{\pi}{3} + 2\pi k$$ or $$x = \frac{2\pi}{3} + 2\pi k$$ for integers $$k$$. The answers which lie in $$[0,2\pi)$$ are $$x = \frac{\pi}{2}$$, $$\frac{3\pi}{2}$$, $$\frac{\pi}{3}$$ and $$\frac{2\pi}{3}$$. We graph $$y = \sin(2x)$$ and $$y = \sqrt{3} \cos(x)$$ and, after some careful zooming, verify our answers.
1. \item Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation $$\sin(x)\cos\left(\frac{x}{2}\right) + \cos(x)\sin\left(\frac{x}{2}\right) = 1$$. If we stare at it long enough, however, we realize that the left hand side is the expanded form of the sum formula for $$\sin\left(x + \frac{x}{2}\right)$$. Hence, our original equation is equivalent to $$\sin\left(\frac{3}{2} x\right) = 1$$. Solving, we find $$x = \frac{\pi}{3} + \frac{4\pi}{3} k$$ for integers $$k$$. Two of these solutions lie in $$[0,2\pi)$$: $$x = \frac{\pi}{3}$$ and $$x = \frac{5\pi}{3}$$. Graphing $$y = \sin(x)\cos\left(\frac{x}{2}\right) + \cos(x)\sin\left(\frac{x}{2}\right)$$ and $$y = 1$$ validates our solutions.
2. \item With the absence of double angles or squares, there doesn't seem to be much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid.\footnote{We are essentially undoing' the sum / difference formula for cosine or sine, depending on which form we use, so this problem is actually closely related to the previous one!} To fit $$f(x) = \cos(x) - \sqrt{3} \sin(x)$$ to the form $$A\sin(\omega t + \phi) + B$$, we use what we learned in Example \ref{expandedsinusoidex1} and find $$A = 2$$, $$B = 0$$, $$\omega = 1$$ and $$\phi = \frac{5\pi}{6}$$. Hence, we can rewrite the equation $$\cos(x) - \sqrt{3} \sin(x) = 2$$ as $$2 \sin\left(x + \frac{5\pi}{6}\right) = 2$$, or $$\sin\left(x + \frac{5\pi}{6}\right) = 1$$. Solving the latter, we get $$x = - \frac{\pi}{3} + 2\pi k$$ for integers $$k$$. Only one of these solutions, $$x = \frac{5\pi}{3}$$, which corresponds to $$k=1$$, lies in $$[0,2\pi)$$. Geometrically, we see that $$y = \cos(x) - \sqrt{3} \sin(x)$$ and $$y = 2$$ intersect just once, supporting our answer.
Example $$\PageIndex{4}$$:
Express the domain of the following functions using extended interval notation.\footnote{See page \pageref{extendedinterval} for details about this notation.}
1. $$f(x) = \csc\left(2x + \frac{\pi}{3}\right)$$
2. $$f(x) = \dfrac{\sin(x)}{2\cos(x) - 1}$$
3. $$f(x) = \sqrt{1 - \cot(x)}$$
Solution
\item To find the domain of $$f(x) = \csc\left(2x + \frac{\pi}{3}\right)$$, we rewrite $$f$$ in terms of sine as $$f(x) = \frac{1}{\sin\left(2x + \frac{\pi}{3}\right)}$$. Since the sine function is defined everywhere, our only concern comes from zeros in the denominator. Solving $$\sin\left(2x + \frac{\pi}{3}\right) = 0$$, we get $$x = -\frac{\pi}{6} + \frac{\pi}{2} k$$ for integers $$k$$. In set-builder notation, our domain is $$\left\{ x : x \neq -\frac{\pi}{6} + \frac{\pi}{2} k \, \text{for integers \(k} \right\}$$. To help visualize the domain, we follow the old mantra `When in doubt, write it out!' We get $$\left\{ x : x \neq -\frac{\pi}{6}, \frac{2\pi}{6}, -\frac{4\pi}{6}, \frac{5\pi}{6}, -\frac{7\pi}{6}, \frac{8\pi}{6}, \ldots \right\}$$, where we have kept the denominators $$6$$ throughout to help see the pattern. Graphing the situation on a numberline, we have
Proceeding as we did on page \pageref{extendedinterval} in Section \ref{circularfunctionsbeyond}, we let $$x_{\mbox{\tiny \(k}}$$ denote the $$kth number excluded from the domain and we have \(x_{\mbox{\tiny \(k}} = -\frac{\pi}{6} + \frac{\pi}{2} k = \frac{(3k-1)\pi}{6}$$ for integers $$k$$. The intervals which comprise the domain are of the form $$\left(x_{\mbox{\tiny \(k}}, x_{\mbox{\tiny \(k+1}} \right) = \left(\frac{(3k-1)\pi}{6}, \frac{(3k+2)\pi}{6} \right)$$ as $$k$$ runs through the integers. Using extended interval notation, we have that the domain is
$\bigcup_{k = -\infty}^{\infty} \left(\dfrac{(3k-1)\pi}{6}, \dfrac{(3k+2)\pi}{6} \right)$
We can check our answer by substituting in values of $$k$$ to see that it matches our diagram.
\item Since the domains of $$\sin(x)$$ and $$\cos(x)$$ are all real numbers, the only concern when finding the domain of $$f(x) = \frac{\sin(x)}{2\cos(x) - 1}$$ is division by zero so we set the denominator equal to zero and solve. From $$2\cos(x) - 1 = 0$$ we get $$\cos(x) = \frac{1}{2}$$ so that $$x = \frac{\pi}{3} + 2\pi k$$ or $$x = \frac{5\pi}{3} + 2\pi k$$ for integers $$k$$. Using set-builder notation, the domain is $$\left\{ x : x \neq \frac{\pi}{3} + 2\pi k \, \text{and} \, x \neq \frac{5\pi}{3} + 2\pi k \, \text{for integers \(k} \right\}$$, or $$\left\{ x : x \neq \pm \frac{\pi}{3}, \pm \frac{5\pi}{3}, \pm \frac{7\pi}{3}, \pm \frac{11\pi}{3}, \ldots \right\}$$, so we have
Unlike the previous example, we have \textit{two} different families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found in our domain work to describe the intervals. To that end, we let $$a_{\mbox{\tiny \(k}} = \frac{\pi}{3} + 2\pi k = \frac{(6k+1)\pi}{3}$$ and $$b_{\mbox{\tiny \(k}} = \frac{5\pi}{3} + 2\pi k = \frac{(6k+5) \pi}{3}$$ for integers $$k$$. The goal now is to write the domain in terms of the $$a's an \(b's. We find \(a_{\mbox{\tiny \(0}} = \frac{\pi}{3}$$, $$a_{\mbox{\tiny \(1}} = \frac{7\pi}{3}$$, $$a_{\mbox{\tiny \(-1}} = -\frac{5\pi}{3}$$, $$a_{\mbox{\tiny \(2}} = \frac{13\pi}{3}$$, $$a_{\mbox{\tiny \(-2}} = -\frac{11\pi}{3}$$, $$b_{\mbox{\tiny \(0}} = \frac{5\pi}{3}$$, $$b_{\mbox{\tiny \(1}} = \frac{11\pi}{3}$$, $$b_{\mbox{\tiny \(-1}} = -\frac{\pi}{3}$$, $$b_{\mbox{\tiny \(2}} = \frac{17\pi}{3}$$ and $$b_{\mbox{\tiny \(-2}} = -\frac{7\pi}{3}$$. Hence, in terms of the $$a's and \(b's, our domain is $\ldots \left(a_{\mbox{\tiny \(-2}}, b_{\mbox{\tiny \(-2}} \right) \cup \left(b_{\mbox{\tiny \(-2}}, a_{\mbox{\tiny \(-1}} \right)\cup \left(a_{\mbox{\tiny \(-1}}, b_{\mbox{\tiny \(-1}} \right)\cup \left(b_{\mbox{\tiny \(-1}}, a_{\mbox{\tiny \(0}} \right)\cup \left(a_{\mbox{\tiny \(0}}, b_{\mbox{\tiny \(0}} \right)\cup \left(b_{\mbox{\tiny \(0}}, a_{\mbox{\tiny \(1}} \right)\cup \left(a_{\mbox{\tiny \(1}}, b_{\mbox{\tiny \(1}} \right)\cup \dots$ If we group these intervals in pairs, \( \left(a_{\mbox{\tiny \(-2}}, b_{\mbox{\tiny \(-2}} \right) \cup \left(b_{\mbox{\tiny \(-2}}, a_{\mbox{\tiny \(-1}} \right)$$, $$\left(a_{\mbox{\tiny \(-1}}, b_{\mbox{\tiny \(-1}} \right)\cup \left(b_{\mbox{\tiny \(-1}}, a_{\mbox{\tiny \(0}} \right)$$, $$\left(a_{\mbox{\tiny \(0}}, b_{\mbox{\tiny \(0}} \right)\cup \left(b_{\mbox{\tiny \(0}}, a_{\mbox{\tiny \(1}} \right)$$ and so forth, we see a pattern emerge of the form $$\left(a_{\mbox{\tiny \(k}}, b_{\mbox{\tiny \(k}} \right)\cup \left(b_{\mbox{\tiny \(k}}, a_{\mbox{\tiny \(k+1}} \right)$$ for integers $$k$$ so that our domain can be written as
$\bigcup_{k = -\infty}^{\infty} \left(a_{\mbox{\tiny $$k}}, b_{\mbox{\tiny \(k}} \right)\cup \left(b_{\mbox{\tiny \(k}}, a_{\mbox{\tiny \(k+1}} \right) = \bigcup_{k = -\infty}^{\infty} \left(\frac{(6k+1)\pi}{3}, \frac{(6k+5) \pi}{3} \right)\cup \left(\frac{(6k+5) \pi}{3}, \frac{(6k+7)\pi}{3} \right)$ A second approach to the problem exploits the periodic nature of \(f$$. Since $$\cos(x)$$ and $$\sin(x)$$ have period $$2\pi$$, it's not too difficult to show the function $$f$$ repeats itself every $$2\pi$$ units.\footnote{This doesn't necessarily mean the period of $$f$$ is $$2\pi$$. The tangent function is comprised of $$\cos(x)$$ and $$\sin(x)$$, but its period is half theirs. The reader is invited to investigate the period of $$f$$.} This means if we can find a formula for the domain on an interval of length $$2\pi$$, we can express the entire domain by translating our answer left and right on the $$x$$-axis by adding integer multiples of $$2\pi$$. One such interval that arises from our domain work is $$\left[\frac{\pi}{3}, \frac{7\pi}{3}\right]$$. The portion of the domain here is $$\left(\frac{\pi}{3}, \frac{5\pi}{3}\right) \cup \left(\frac{5\pi}{3}, \frac{7\pi}{3}\right)$$. Adding integer multiples of $$2\pi$$, we get the family of intervals $$\left(\frac{\pi}{3} + 2\pi k, \frac{5\pi}{3} + 2\pi k \right) \cup \left(\frac{5\pi}{3} + 2\pi k, \frac{7\pi}{3} + 2\pi k\right)$$ for integers $$k$$. We leave it to the reader to show that getting common denominators leads to our previous answer.
\item To find the domain of $$f(x) = \sqrt{1-\cot(x)}$$, we first note that, due to the presence of the $$\cot(x)$$ term, $$x \neq \pi k$$ for integers $$k$$. Next, we recall that for the square root to be defined, we need $$1 - \cot(x) \geq 0$$. Unlike the inequalities we solved in Example \ref{TrigIneqEx1}, we are not restricted here to a given interval. Our strategy is to solve this inequality over $$(0,\pi)$$ (the same interval which generates a fundamental cycle of cotangent) and then add integer multiples of the period, in this case, $$\pi$$. We let $$g(x) = 1 - \cot(x)$$ and set about making a sign diagram for $$g$$ over the interval $$(0,\pi)$$ to find where $$g(x) \geq 0$$. We note that $$g$$ is undefined for $$x = \pi k$$ for integers $$k$$, in particular, at the endpoints of our interval $$x = 0$$ and $$x = \pi$$. Next, we look for the zeros of $$g$$. Solving $$g(x) = 0$$, we get $$\cot(x) = 1$$ or $$x = \frac{\pi}{4} + \pi k$$ for integers $$k$$ and only one of these, $$x = \frac{\pi}{4}$$, lies in $$(0,\pi)$$. Choosing the test values $$x = \frac{\pi}{6}$$ and $$x = \frac{\pi}{2}$$, we get $$g\left(\frac{\pi}{6}\right) = 1 - \sqrt{3}$$, and $$g\left(\frac{\pi}{2}\right) = 1$$.
We find $$g(x) \geq 0$$ on $$\left[\frac{\pi}{4}, \pi \right)$$. Adding multiples of the period we get our solution to consist of the intervals $$\left[\frac{\pi}{4} + \pi k, \pi + \pi k \right) = \left[\frac{(4k+1)\pi}{4}, (k+1)\pi \right)$$. Using extended interval notation, we express our final answer as
$\bigcup_{k = -\infty}^{\infty} \left[\dfrac{(4k+1)\pi}{4}, (k+1)\pi \right)$
## Contributors
• Carl Stitz, Ph.D. (Lakeland Community College) and Jeff Zeager, Ph.D. (Lorain County Community College) |
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Lesson 2: Visualize equivalent ratios
# Simplify a ratio from a tape diagram
In this problem, we have a mixture of yellow and red paint. There are 12 parts yellow and 8 parts red. We want to find the ratio of red paint to the total paint in the mixture. To do this, we add the parts of red and yellow paint together (8 + 12 = 20). The ratio of red paint to total paint is 8:20. Created by Sal Khan.
## Want to join the conversation?
• I don't get this
• We are just comparing Mcdonalds colors. It is not red to yellow which is 8 to 12 as he says in . We add 8+12 to get the total which is 20. Red to the total is 8 to 20
• We're told that the following diagram describes the volume of yellow and red paint in an orange mixture. So we can see that for every 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 parts of yellow, we have 1, 2, 3, 4, 5, 6, 7, 8 parts of red. They ask us, "What is the ratio of red paint to total paint in the mixture?" So, pause this video and see if you can have a go at that before we do it together. All right, so if we wanna think about red paint, it looks like we have every, we have eight parts red paint, and if we think about the ratio of that towards total paint, you might be tempted to put a 12 there. But that's not the total paint, that's just the yellow paint. The total paint for every eight parts red, would be the eight parts red plus the 12 parts yellow. So, it'd be eight plus 12, which is 20. And this is true, the ratio of red paint to the total paint in the mixture is . Now, you might not always see it expressed this way, because there's other ways of writing an equivalent ratio. Some that people would argue are actually simpler. Another way to think about it is this diagram where we see the red paint, it has eight parts right over here, but we could also describe it as four groups of two. Just like that, because eight is divisible by four. And if you divide the 12 parts of yellow into four groups, it's four groups of three. And you'll see in a second why this is really interesting, because we were able to break down both of these into four groups. And when you do that, you see very clearly, hopefully, that for every two parts of red you have three parts of yellow. For every two parts of red, you have three parts of yellow. Two parts of red, three parts of yellow, and then last but not least, these two parts of red and then those three parts of yellow. So, for every two parts of red you have three parts yellow. Now once again, the ratio that they're asking isn't the ratio of red to yellow, which is two to three, but you could just take one of these groups and say, all right, for every two parts of red I have 1, 2, 3, 4, 5 parts of total paint. So, you could say that the ratio is also, for every two parts of red, I have five parts of total paint. And hopefully, this makes intuitive sense right here. Why these two things are equivalent. If for every eight you have 20 total, well, if you divide that by four for every two of red, you're going to have five.
• Why are they called "tape" diagrams
(1 vote)
• probably because it looks like tape.
• Why do we need ratios?
(1 vote)
• Ratios are useful mathematical tools for comparing quantities or sizes of different objects or quantities.
• i dont get how you add the fractions up to make the whole number.
(1 vote)
• Are there simplifying ratios like in fractions?
(1 vote)
• Yes it is like fractions
(1 vote)
• simplifying is finding a common factor and dividing it by it right?pls confirm?!
(1 vote)
• I think it's the greatest common factor that you divide it by
(1 vote)
• why do we have use tape diagrams
(1 vote)
• are these videos required |
What is the slope of the polar curve f(theta) = theta - sec^2theta+costheta at theta = (7pi)/12?
1 Answer
Mar 21, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = - 6.968 \mathmr{if} \theta = \frac{7 \pi}{12}$
Explanation:
The slope of the polar curve at $\theta = \frac{7 \pi}{12}$ is $\frac{\mathrm{dy}}{\mathrm{dx}}$ evaluated in terms of $\theta$ at $\theta = \frac{7 \pi}{12}$.
To find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we can first find $\frac{\mathrm{dy}}{d \theta}$ and $\frac{\mathrm{dx}}{d \theta}$ and use the chain rule for derivatives to determine that:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{d \theta} \times \frac{d \theta}{\mathrm{dx}} = \frac{\mathrm{dy}}{d \theta} \div \frac{\mathrm{dx}}{d \theta}$
$y = r \sin \theta = \left(\theta - {\sec}^{2} \theta + \cos \theta\right) \left(\sin \theta\right)$
$\frac{\mathrm{dy}}{d \theta} = \left(1 - 2 {\sec}^{2} \left(\theta\right) \tan \left(\theta\right) - \sin \theta\right) \left(\sin \theta\right) + \left(\theta - {\sec}^{2} \theta + \cos \theta\right) \left(\cos \theta\right)$
Evaluating this at $\theta = \frac{7 \pi}{12}$ gives that $\frac{\mathrm{dy}}{d \theta} = 111.12 \mathmr{if} \theta = \frac{7 \pi}{12}$
$x = r \cos \theta = \left(\theta - {\sec}^{2} \theta + \cos \theta\right) \left(\cos \theta\right)$
$\frac{\mathrm{dx}}{d \theta} = \left(1 - 2 {\sec}^{2} \left(\theta\right) \tan \left(\theta\right) - \sin \theta\right) \left(\cos \theta\right) + \left(\theta - {\sec}^{2} \theta + \cos \theta\right) \left(- \sin \theta\right)$
Evaluating this at $\theta = \frac{7 \pi}{12}$ gives that $\frac{\mathrm{dx}}{d \theta} = - 15.948 \mathmr{if} \theta = \frac{7 \pi}{12}$
$\frac{\mathrm{dy}}{d \theta} \div \frac{\mathrm{dx}}{d \theta} = \frac{111.12}{- 15.948} = - 6.968$
Therefore, the slope of the line tangent to $f \left(\theta\right)$ at the point where $\theta = \frac{7 \pi}{12}$ is $- 6.968$. |
# British Maths Olympiad (BMO) 2003 Round 1 Question 4 what is wrong with this approach, can it be fixed?
The questions states:
A set of positive integers is defined to be $wicked$ if it contains no three consecutive integers. We count the empty set, which contains no elements at all, as a $wicked$ set. Find the number of wicked subsets of the set
{$1, 2, 3, 4, 5, 6, 7, 8, 9, 10$}
One approach is to subtract all the sets that are not wicked from all possible sets
We have $2^{10}$ possible sets
The sets that are not $wicked$ must contain some combination of the following consecutive integers:
{1,2,3} $\\$ {2,3,4} $\\$ {3,4,5} $\\$ {4,5,6} $\\$ {5,6,7} $\\$ {6,7,8} $\\$ {7,8,9} $\\$ {8,9,10}
We could partition all the subsets in the following way
$(1,2,3)$ $\times$ {$4,5,6,7,8,9,10$} $= 2^7$ sets which contain (1,2,3)
$(2,3,4)$ $\times$ {$5,6,7,8,9,10$} $= 2^6$ sets which contain (2,3,4)
$(3,4,5)$ $\times$ {$6,7,8,9,10$} $= 2^5$ ...
notice that we purposefully leave out prior numbers so that we are considering disjoint sets.
We thus get $2^7 +2^6 \cdots + 2^0 = 2^8 - 1$ sets that are not $wicked$.
Thus the number of $wicked$ sets is $2^{10} - (2^8 - 1)$
What is wrong with this approach, and can it be fixed to give the right answer?
• What is the place of the non-wicked set $\{1,3,4,5\}$ in your classification? – Chen Wang May 12 '18 at 0:16
• To fix, need I/E. – vadim123 May 12 '18 at 0:19
There are two good ways to do this problem.
1. Let $a_n$ be the number of wicked subsets of $\{1,2,\dots,n\}$. You can show that $$a_n=a_{n-1}+a_{n-2}+a_{n-3}$$ holds for all $n\ge 3$. The $a_{n-1}$ term counts subsets where $n$ is absent, $a_{n-2}$ counts subsets where $n$ is present but $n-2$ is absent, and $a_{n-3}$ is subsets with $n$ and $n-1$ but without $n-2$.
This recurrence, along with the bases cases $a_0=1,a_1=2,a_2=4$, allows you to iteratively compute $a_n$ from the ground up, starting with $a_3$ all the way up to $n=10$.
2. Instead, count the number of un-wicked (saintly) subsets. Thinking of subsets as strings of ten $0$s and $1$s, where $1$s correspond to the elements in the set, every saintly subset contains a $111$ somewhere. To eliminate the problem of overlaps, in any block of three or more ones, we will only consider the initial $111$, which corresponds to either a $111$ at the beginning or a $0111$ somewhere.
There are $2^7$ strings starting with $111$, and $7\cdot 2^6$ strings containing a $0111$ (the $7$ represents the choice of location of $0111$). Well, not quite; there is some double counting happening here. You then need to subtract out the strings which contain two instances of either $\_111$ or $0111$ (where $\_$ is the beginning of the string). Fortunately, there is no triple counting, so after subtracting the double counted strings you are done.
Subtracting the number of saintly sets from the total, the final result is $$2^{10}-2^7-\binom71\cdot 2^6+\binom{4}{1}2^3+\binom{4}22^2.$$The term $\binom{5}{2}2^3$ represents strings containing a $111$ at the start and a $0111$ elsewhere, while the last terms is strings containing two instances of $0111$.
• In the second method, the last term should have a minus sign in front. – sku May 12 '18 at 6:33
• @sku I think it should be plus: what I have written is all $2^{10}$ strings, minus the bad strings containing $111$, plus the strings containing $111$ twice, which is how inclusion exclusion exclusion works. Plus, both methods give the same answer of 504 (after I fixed an unrelated error). – Mike Earnest May 12 '18 at 18:58
• Fine now. The final answer is $504$ – sku May 12 '18 at 20:25 |
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# Round off to The Nearest Tens
Last updated date: 22nd Feb 2024
Total views: 184.2k
Views today: 3.84k
## Rounding off a Number to The Nearest Tens
We have all come across instances when we failed to remember a number, date, or count simply because it was not very common or easy to recall. There are certain numbers that are easy to remember, anything in between will simply be confusing or difficult to recall.
For instance, the numbers ending with 5 or 0 are usually the ones that are remembered most easily. However, not always do we come across numbers or dates that end with 5 or 0. Therefore, to help us reach a number that ends in the desired digit, without changing the value by a lot, we use the method of rounding off to the nearest desired number. In this article, we will see how we can round a number off to the nearest tens.
## What is Round off Meaning?
Rounding a number refers to the method of conversion of a particular number into a number that is easy to deal with or remember. The rounded number may not exactly be the same as the given number, but is generally very close to the given number or an approximation.
Rounding off means to estimate. Let us take an example to better understand the concept.
Example: Richard goes to the shop to buy vegetables. His total bill is ﹩98. So, he rounds off the amount to the nearest tens, which is ﹩100 and gives that to the shopkeeper. The shopkeeper easily returns the remaining ﹩2 to Richard and the deal is complete.
In the above scenario, Richard used the method of rounding off to make the calculation simpler for him and the shopkeeper. Let us now learn how this method works.
## Steps for Rounding off to The Nearest Tens
Follow the given steps to be able to round off a number to the nearest tens:
• Identify the number that needs to be rounded.
• Mark the digit in the tens column of the number.
• Check the digit in the one's column or the unit’s column, which is to the immediate right of the tens column.
• If the digit in the one’s column is 1, 2, 3, or 4, then we shall round down the number to the nearest tens.
• If the digit in the one’s column is 5 or greater, that is, 5, 6, 7, 8, or 9, then we shall round up the number to the nearest tens.
• Finally, we write the one’s place digit as 0 and change the digit in the tens place accordingly.
## Let us Understand This Better With an Example.
Consider the given number to be 87. As we have learned in the previous steps, we need to identify the unit's place or one’s place digit, which here is 7. Since 7 is greater than 5, we need to round the number ‘up’ to the nearest tens.
So, we need to change the one’s place digit to 0 and increase the tens place digit by 1, which gives us 90. Therefore, the required number obtained after rounding off 87 to the nearest tens is 90.
## Conclusion
It is easy to forget numbers and counts when the one’s place digits are anything other than 0. To solve this problem, the process of rounding a number to the nearest tens is used. It helps ease the process of calculation and memorising. It is an important concept to learn and put to practical use. So do not skip this topic and try to practice these conversions as much as possible to be able to score good marks in your exams!
## FAQs on Round off to The Nearest Tens
1. If your height is 146 cm, what is your height rounded off to the nearest tens?
If the given height is 146 cm, then see the one’s place digit is 6, which is greater than 5. So, the one’s place will be 0 and the tens place will be 5, making it 150 cm. Therefore, the required rounded off height will be 150 cm.
2. If the one’s place digit is 5, should we round up or round down to the nearest tens?
In the case of the number 5 as the one’s place digit in a given number, we should always round the number up to the nearest tens. |
# System of Linear Equations Project
## word problem
You got your first job and decided it is time to buy a car. You are debating between two cars, a Volkswagen Beetle 1.8T or a 2015 Jeep Wrangler Sport 4x4. The Volkswagen Beetle costs \$24, 425 and has 33 miles per gallon and the Jeep Wrangler costs \$23,990 with 21 miles per gallon. Assuming gas is \$2.25 per gallon, what car would be the best buy for long term? What car would be the better buy for short term? What are the independent and dependent variables?
## system of linear equations and graph
System of linear equations for the Volkswagen Beetle-
y = .07 x + 24,425
System of linear equations for the 2015 Jeep Wrangler-
y= .11 x + 23,990
## Check Graph with Elimination
Y=.07x+24,425
Subtract .07 from both sides giving you:
-.07x + y= 24,425
Y=.11x+23,990
Subtract .11 from both sides giving you:
-.11x + y = 23,990
Now, multiply -.11x + y = 23,990 by -1 which would give you: .11x - y = -23,900. And you are ready to eliminate!
-.07x + y = 24,425
.11x - y = - 23,990
y-y= 0, so you can now get rid if the y values, add the x values, and subtract the 24,425- 23,990, which will give you the equation:
.04x=435
You can now divide .04 by both sides giving you:
X = 10, 875
You can now substitute the x value we just found into one of the equations to find the y value.
y = .07 (10,875) + 24,425
y = 25,186.25
This means that at 10,875 miles, the two cars' prices will be equivalent at \$25,186.25.
## What Car is the Better Buy?
A car for the better buy varies for the person. How long are you planning to keep the car, if you have a family, where you live. The Volkswagen Beetle 1.8T would probably be the best buy for me. All though in the beginning, the Jeep is cheaper, after you drive for about 1 year (or 10, 875 miles) the Jeep would start to cost you more money. |
# RELATIONSHIP BETWEEN ZEROS AND COEFFICIENTS OF A QUADRATIC POLYNOMIAL
## About "Relationship between zeros and coefficients of a quadratic polynomial"
"Relationship between zeros and coefficients of a quadratic polynomial" is the stuff which is much required to the students who study high school math .
To understand "relationship between zeros and coefficients of a quadratic polynomial", let us consider the quadratic polynomial.
P(x) = ax² + bx + c
(Here a, b and c are real and rational numbers)
Let "α" and "β" be the two zeros of the above quadratic polynomial
The basic relationships between the zeros and the coefficients of P(x) = ax² + bx + c are
## Some more relationships
Apart from the two relationships between zeros and coefficients of a quadratic polynomial given above, we have some more stuff on it.
Let us look in to them.
Stuff 1 :
Let the two zeros be equal and opposite in signs.
For example, α = k and β = -k
Then, α + β = k + (-k)
α + β = 0
So, sum of the roots = 0,
and we have -b/a = 0 -------> b =0
Therefore, if "b = 0", then zeros of the quadratic polynomial will be equal and opposite in signs.
Stuff 2 :
Let the two zeros be reciprocal to each other.
For example, α = k and β = 1/k
Then, αβ = k (1/k)
αβ = 1
So, product of the roots = 1,
and we have c/a = 1 -------> c = a
Therefore, if "c = a", then zeros of the quadratic polynomial will be reciprocal to each other.
Now we are going look in to an important stuff on "Relationship between zeros and coefficients of a quadratic polynomial"
Stuff 3 :
In this stuff, we are going to see, how to examine the nature of roots of a quadratic equation using the coefficients "a", "b" and "c".
For that, let us consider the quadratic formula which is applied to solve any quadratic equation.
So fare we have seen the different types relationship between zeros and coefficients of a quadratic polynomial. Now let us look at some examples on the stuff "Relationship between zeros and coefficients of a quadratic polynomial".
## Examples
To have better understanding on "Relationship between zeros and coefficients of a quadratic polynomial", let us have some example problems.
Example 1 :
Test, whether the zeros of the given quadratic equation are equal in magnitude and opposite in signs.
x² - 6 =0
Solution :
When we compare the given equation with the general form, we get a = 1, b = 0 and c = 6.
Since b = 0, the zeros of the given quadratic equation are equal in magnitude and opposite in signs.
Example 2 :
Test, whether the zeros of the given quadratic equation are equal in magnitude and opposite in signs.
x² + 3x + 1 =0
Solution :
If x² + 3x + 1 =0 is compared to the general form ax² + bx + c =0, we get a = 1, b = 3 and c = 1.
Since b ≠ 0, the zeros are not equal in magnitude.
Example 3 :
Test, whether the zeros of the given quadratic equation are reciprocal to each other or not.
5x² -8x + 5 =0
Solution :
If 5x² - 8x + 5 = 0 is compared to the general form ax² + bx + c =0, we get a = 5, b = -8 and c = 5.
Since a = c, the zeros are reciprocal to each other.
Example 4 :
Test, whether the zeros of the given quadratic equation are reciprocal to each other or not.
5x² +7x + 4 =0
Solution :
If 5x² + 7x + 4 = 0 is compared to the general form ax² + bx + c =0, we get a = 5, b = 7 and c = 4.
Since a c, the zeros are not reciprocal to each other.
Example 5 :
Find the sum and product of the roots of the quadratic equation given.
2x² + 7x + 5 =0
Solution :
If 2x² + 7x + 5 =0 is compared to the general form ax² + bx + c =0, we get a = 2, b = 7 and c = 5.
Sum of the roots = -b/a = -7/2
Product of the roots = c/a = 5/2
Example 6 :
Examine the nature of the roots of the following quadratic equation.
x² + 5x + 1 =0
Solution :
If x² + 5x + 1 =0 is compared to the general form ax² + bx + c =0,
we get a = 1, b = 5 and c = 1.
Now, let us find the value of the discriminant "b² - 4ac"
b² - 4ac = 5² - 4(1)(1)
b² - 4ac = 25 - 4
b² - 4ac = 21 (>0 , but not a perfect square)
Hence, the roots are real, distinct and irrational.
Example 7 :
Examine the nature of the roots of the following quadratic equation.
x² + 5x + 6 =0
Solution :
If x² + 5x + 6 =0 is compared to the general form ax² + bx + c =0,
we get a = 1, b = 5 and c = 6.
Now, let us find the value of the discriminant "b² - 4ac"
b² - 4ac = 5² - 4(1)(6)
b² - 4ac = 25 - 24
b² - 4ac = 1 (>0 and also a perfect square)
Hence, the roots are real, distinct and rational.
Example 8 :
Examine the nature of the roots of the following quadratic equation.
2x² - 3x + 1 =0
Solution :
If 2x² - 3x + 1 =0 is compared to the general form ax² + bx + c =0,
we get a = 2, b = -3 and c = 1.
Now, let us find the value of the discriminant "b² - 4ac"
b² - 4ac = (-3)² - 4(2)(-1)
b² - 4ac = 9 + 8
b² - 4ac = 17 (>0 but not a perfect square)
Hence, the roots are real, distinct and irrational.
Example 9 :
Examine the nature of the roots of the following quadratic equation.
x² - 16x + 64 =0
Solution :
If x² - 16x + 64 =0 is compared to the general form ax² + bx + c =0,
we get a = 1, b = -16 and c = 64.
Now, let us find the value of the discriminant "b² - 4ac"
b² - 4ac = (-16)² - 4(1)(64)
b² - 4ac = 256 - 256
b² - 4ac = 0
Hence, the roots are real, equal and rational.
Example 10 :
Examine the nature of the roots of the following quadratic equation.
3x² + 5x + 8 =0
Solution :
If 3x² + 5x + 8 =0 is compared to the general form ax² + bx + c =0,
we get a = 3, b = 5 and c = 8.
Now, let us find the value of the discriminant "b² - 4ac"
b² - 4ac = 5² - 4(3)(8)
b² - 4ac = 25- 96
b² - 4ac = -71 (negative)
Hence, the roots are imaginary.
After having gone through the stuff and example problems explained above, we hope that the students would have understood "Relationship between zeros and coefficients of a quadratic polynomial"
If you want to know more about "Relationship between zeros and coefficients of a quadratic polynomial", please click here. |
## Figuring out Numbers
So I was having a play, again, this time with a further look into composites and primes.
I am curious to play with representations of digits in manners that allow visualization of distributions of prime numbers and non-prime numbers... not naively in any form or shape of faith in finding something someone else hasn't of course, just because it's fun, for me, to look, play, learn and figure things out by myself...
So what did I get to - from this point on, it may well seem absolutely obvious, already known/common fact - but this is playful musings, explored with no math guides or prior knowledge applied, this is number theory basics a la Chris.
### 'Types' of Numbers...
My number playing revolves around positive integers... of which a number is categorized in this exercise, as composite or prime.
With this, a couple of terms are thrown around which you may be familiar with:
#### Factor
factor is a whole number which divides exactly into a whole number, leaving no remainder. For example, 13 is a factor of 52 because 13 divides exactly into 52 (52 ÷ 13 = 4 leaving no remainder).
The complete list of factors of 52 is: 1, 2, 4, 13, 26, and 52 (all these divide exactly into 52).
#### Prime Number
prime number is a whole number greater than 1, whose only two whole-number factors are 1 and itself.
I particularly liked this visual example of what a prime number is:
#### Composite Number
composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).
### Sourcing, Manipulation, Confirmation!
Manually formulating a table for all factors of a number seemed like an unnecessary task to proceed with, seeing as what I wanted to delve into being already somewhat of a large task.
So I scoured the net for a table as such, and this is what I found.
Which for a quick display of the content, it looks something like this:
• d(n) is the number of positive divisors of n, including 1 and n itself
• σ(n) is the sum of all the positive divisors of n, including 1 and n itself
• s(n) is the sum of the proper divisors of n, which does not include n itself; that is, s(n) = σ(n) − n
• perfect number equals the sum of its proper divisors; that is, s(n) = n
• deficient number is greater than the sum of its proper divisors; that is, s(n) < n
• an abundant number is less than the sum of its proper divisors; that is, s(n) > n
• prime number has only 1 and itself as divisors; that is, d(n) = 2. Prime numbers are always deficient as s(n)=1
Taking this data and loading it into Excel, allows me to manipulate and produce variations in data representation for the fun of spotting characteristics, similarities/groupings and/or patterns... but what it did lead to, actually saw me producing the results manually with the application of formula in Excel - so in fact, the table, in the end, was just a confirmation base for my calculations and output.
The below shows the divisors for n from the source table; to the right, in the grid and with colors - is the output of a formula, that divides n by all whole integers presented in the top row increasing by 1 across column by column - but only is the product of the division displayed, if the value is a whole integer.
The diagonal line is the boundary of this output as that is the indicator of where n is divided by n.
Any whole integer in that top row, as a divisor of n, will not be whole if it is >n, and so, will not display a populated number.
The results of this simple formula, produce the same outputs as what I found on on Wikipedia - allowing me to produce my own counts and sums of divisors of positive whole integers and therefore adding additional columns to flag whether n is prime or not! (Prime is identified in this grid, as n where the count of divisors = 2 & the sum of divisors = n + 1)
The coloring may already present some structural reference to the distribution of numbers as divisors of n.
### A Stopping Point... for now...
There is more I would like to output on this what I have started to clean up from previous weeks play - however, as per usual, I digressed... and what I produced just now will form a separate post for reference, as I have branched back into an area I already have some background work on, but tidied a portion up for its potential supporting role in analysis of tentative primes...
Which is to say, the above work leads on to flagging of numbers, based on criteria such as last digit and then mod9 and square root check - to identify definite NON PRIMES; and potential primes... which I had a little curiosity to check whether permutations of base numbers (i.e. 123, 132, 213, 231, 312, 321) can occur where all permutations are prime; whether there is some significance of digit positioning with regards to permutations and prime elements and so on...
It got to well past 4am so... I want to wrap this part up and come back to it, as this what I posted today, is just the tip of the iceberg of what outputs I would like to share! That's just the way such analysis seems to go, from A to B to C through to Z and then sometimes coming back to A :D |
# How Many Baseballs In A Bucket?
How many baseballs does it take to fill a bucket?
It’s a common question with a not so common answer. The answer really depends on the size of the bucket and the size of the baseballs.
## Introduction
We all know that baseball is a game of statistics. But did you know that there are some interesting numbers when it comes to the game itself? For example, did you know that there are typically between 70 and 80 balls used in a Major League Baseball game?
But what about when it comes to the ballpark? How many baseballs are used in a typical game at an MLB stadium? It turns out that there are quite a few more than you might think.
According to estimates, there are about 300 balls used in a typical MLB game. That includes foul balls, home runs, and balls that are simply taken out of play. And that doesn’t even count the dozens of balls that are used during batting practice!
So, how many baseballs are in a bucket? A lot more than you might think.
## The Calculation
You can calculate the number of baseballs that will fit in a bucket by using the formula: V = πr^2h. In this equation, V stands for volume, r stands for radius, h stands for height, and π is pi. To use this formula, you need to know the measurement of the baseballs in inches and the measurements of the bucket in inches.
## The Result
You have 83 baseballs. You want to know how many are in a bucket. So you put them all in a bucket, and you count them out, one at a time. It takes you a while, but eventually you get to 83. So the answer is 83.
## Conclusion
In conclusion, the answer to the question “How many baseballs in a bucket?” is 26. This is based on the assumption that the dimensions of the bucket are 18 inches wide, 12 inches deep, and 6 inches tall. With these dimensions, the volume of the bucket is 1,296 cubic inches. If each baseball has a circumference of approximately 8.5 inches and a diameter of 2.7 inches, then each ball takes up approximately 23 cubic inches of space. This means that there are approximately 1,296/23, or 56, baseballs in a |
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### Course: 6th grade (Eureka Math/EngageNY)>Unit 3
Lesson 1: Topic A: Understanding positive and negative numbers on the number line
# Number opposites
Learn what it means for a number to be the opposite of another number.
The opposite of a number is the number on the other side of $0$ number line, and the same distance from $0$.
Here are a couple of examples:
$-4$ is the opposite of $4$.
$3$ is the opposite of $-3$.
## Let's practice!
Problem 1A
Move the dot to the opposite of $2$.
## Challenge problems
Use the following number line for the questions below.
Problem 2A
Which of the following is the opposite of $A$?
## Want to join the conversation?
• For opposite numbers I understand how A = -A;
How can A also be E?
• A is a variable in this case, so is E. essentially, since they are the same distance from 0, and represent the same amount on the number line, E is just the hidden -A.
• I'm stuck on challenge problem 2A. I would assume that if you have a 0 as the only number on the line, then anything to the left of 0 is a negative number and anything to the right of 0 is a positive. So the opposite number in the same location of A (which is a -A) reflected on the number line is the positive number E. I understand that ideally these would be represented by the same letter, just negative and positives, but I'm working with what is given and expected that the important part is the marks on the number line - not the representative letters used. Help please.
• A is to the left of the 0 on the number line and it's represented simply as an A without a negative. One would assume this is a negative number as the A is just a label for the negative number that's supposedly on that point on the number line.
Taking what we have previously learned, A is a negative number in this case which would mean the opposite of it would be the E on the number line. How could an opposite of a negative number be negative? Unless the number line is flipped in this case?
This left me very confused as the question certainly leads to E being the correct answer.
• I cannot find even a forced logic explanation for problem 2B which I could only guess at given where the 0 is on the number line. Same with 2C. Since the correct answer for problem 2A is: the opposite of A = -A, how did we arrive at A = -E and E = -A?
Mentioned below, previously given only positive letters to work with, how did a negative letter become an answer?
Using a process of elimination to arrive at a "true" answer, does not explain what is happening.
I hope someone will please shed light on how this occurs. So far, I have not found help anywhere.
• Hi Catherine C. "Where the 0 is on the number line" is indeed the key clue -- good work noticing it! We arrive at A = -E because A and E are the same distance from 0, but in different directions. One of them is on the negative side of 0, and the other is on the positive side of 0; but they're the same distance.
So, A is the opposite of E. Which means that A is equal to -E. If it's still confusing, try watching the video again.
If it's still confusing after that, post a reply here and, when I get the notification, I'll try to explain better.
As for negative letters -- Well, the letters are variables, standing in for a number. Since numbers can be negative, so can variables.
• I don't understand how A=-E and -A=E?
• So let's say you were to fold the number line evenly like a piece of paper. -E would be touching A and -A would be touching E. They are probably just trying to say that they are opposite of each other.
• This question needs to be revised. It says to use the number line for the questions below (under challenge problems). Problem 2a...Which of the following is the opposite of A? "A" is 3 spaces on the left side of 0 on the number line indicating a negative number. The logical answer would be found 3 spaces to the right of 0, except instead there is E. So the correct selection should be E, yet it is indicated as an incorrect answer ...there is no -A in the location that E occupies (-A was indicated as the correct selection). If the alphabetic characters are indicative of numerical values this should be corrected as it is very confusing to 6th graders. It appears that -A=E ?
• oh yeah , it actually makes sense . A is on the left of 0 but it is not indicated by -A but just A. it is 3 steps to left so its opposite should also be 3 to the lef and it is but represented by E and not A . This is actually really confusing . Wow , I am impressed by this observation Douglas Nelson . :o
• If |-10| shows up it means it is it is positive 10 because if that shows up it can never be negative.
• So those two lines means absolute value which is the numbers distance from Zero therefore it does not matter if the number is a positive or negative it'll always be The same number. Example: |-20| would be 20, Because the distance from -20 from 0 is 20
• why is this sometimes hard
• scroll down to find out
because it tricks you're brain
• Why doe negative move to opposite side of 0 on number line? |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Corresponding Angles
## Angles in the same place with respect to a transversal, but on different lines.
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Corresponding Angles
What if you were presented with two angles that are in the same place with respect to the transversal but on different lines? How would you describe these angles and what could you conclude about their measures? After completing this Concept, you'll be able to answer these questions and use corresponding angle postulates.
### Watch This
Watch the portions of this video dealing with corresponding angles.
Then watch this video beginning at the 4:50 mark.
Finally, watch this video.
### Guidance
Corresponding angles are two angles that are in the "same place" with respect to the transversal but on different lines. Imagine sliding the four angles formed with line $l$ down to line $m$ . The angles which match up are corresponding.
Corresponding Angles Postulate: If two parallel lines are cut by a transversal, then the corresponding angles are congruent.
If $l || m$ , then $\angle 1 \cong \angle 2$ .
Converse of Corresponding Angles Postulate: If corresponding angles are congruent when two lines are cut by a transversal, then the lines are parallel.
If then $l || m$ .
#### Example A
If $a || b$ , which pairs of angles are congruent by the Corresponding Angles Postulate?
There are 4 pairs of congruent corresponding angles:
$\angle 1 \cong \angle 5, \ \angle 2 \cong \angle 6, \ \angle 3 \cong \angle 7$ , and $\angle 4 \cong \angle 8$ .
#### Example B
If $m\angle 2 = 76^\circ$ , what is $m\angle 6$ ?
$\angle 2$ and $\angle 6$ are corresponding angles and $l || m$ from the arrows in the figure. $\angle 2 \cong \angle 6$ by the Corresponding Angles Postulate, which means that $m\angle 6 = 76^\circ$ .
#### Example C
If $m\angle 8 = 110^\circ$ and $m\angle 4 = 110^\circ$ , then what do we know about lines $l$ and $m$ ?
$\angle 8$ and $\angle 4$ are corresponding angles. Since $m\angle 8 = m\angle 4$ , we can conclude that $l || m$ .
### Guided Practice
1. Using the measures of $\angle 2$ and $\angle 6$ from Example B, find all the other angle measures.
2. Is $l || m$ ?
3. Find the value of $y$ :
1. If $m\angle 2 = 76^\circ$ , then $m\angle 1 = 180^\circ - 76^\circ =104^\circ$ (linear pair). $\angle 3 \cong \angle 2$ (vertical angles), so $m\angle 3 = 76^\circ. \ m\angle 4 = 104^\circ$ (vertical angle with $\angle 1$ ).
By the Corresponding Angles Postulate, we know $\angle 1 \cong \angle 5, \ \angle 2 \cong \angle 6, \ \angle 3 \cong \angle 7$ , and $\angle 4 \cong \angle 8$ , so $m\angle 5 = 104^\circ, \ m\angle 6 = 76^\circ, \ m\angle 7 = 76^\circ$ , and $m\angle 104^\circ$ .
2. The two angles are corresponding and must be equal to say that $l || m$ . $116^\circ \neq 118^\circ$ , so $l$ is not parallel to $m$ .
3. The horizontal lines are marked parallel and the angle marked $2y$ is corresponding to the angle marked $80$ so these two angles are congruent. This means that $2y=80$ and therefore $y=40$ .
### Practice
1. Determine if the angle pair $\angle 4$ and $\angle 2$ is congruent, supplementary or neither:
2. Give two examples of corresponding angles in the diagram:
3. Find the value of $x$ :
4. Are the lines parallel? Why or why not?
For 6-10, what does the value of $x$ have to be to make the lines parallel?
1. If $m\angle 1 = (6x-5)^\circ$ and $m\angle 5 = (5x+7)^\circ$ .
2. If $m\angle 2 = (3x-4)^\circ$ and $m\angle 6 = (4x-10)^\circ$ .
3. If $m\angle 3 = (7x-5)^\circ$ and $m\angle 7 = (5x+11)^\circ$ .
4. If $m\angle 4 = (5x-5)^\circ$ and $m\angle 8 = (3x+15)^\circ$ .
5. If $m\angle 2 = (2x+4)^\circ$ and $m\angle 6 = (5x-2)^\circ$ .
### Vocabulary Language: English Spanish
Corresponding Angles
Corresponding Angles
Corresponding angles are two angles that are in the same position with respect to the transversal, but on different lines. |
# Basics: Natural Numbers and Integers
One of the interestingly odd things about how people understand math is numbers. It’s
astonishing to see how many people don’t really understand what numbers are, or what different kinds of numbers there are. It’s particularly amazing to listen to people arguing
vehemently about whether certain kinds of numbers are really “real” or not.
Today I’m going to talk about two of the most basic kind of numbers: the naturals and the integers. This is sort of an advanced basics article; to explain things like natural numbers and integers, you can either write two boring sentences, or you can go a bit more formal. The formal
stuff is more fun. If you don’t want to bother with that, here are the two boring sentences:
1. The natural numbers (written N) are zero and the numbers that can be
written without fractions that are greater than zero.
2. The integers (written Z) are all of the numbers, both larger and smaller than
zero, that can be written without fractions.
### Natural Numbers
The most basic kind of number is what’s called a natural number Intuitively, natural
numbers are whole numbers – no fractions – starting at zero, and going onwards towards infinity: 0, 1, 2, 3, 4, … Computer scientists are particularly fond of natural numbers, because everything computable ultimately comes from the natural numbers.
The natural numbers are actually formally defined by something called Peano arithmetic. Peano arithmetic specifies a list of 5 rules that define the natural numbers:
1. Initial value rule: 0 is a natural number.
2. Successor rule: For every natural number n there is exactly one other natural number called its successor s(n).
3. Predecessor rule: 0 is not the successor of any natural number. Every natural
number except zero is the successor to some other natural number, called
its predecessor.
4. Uniqueness rule: No two natural numbers have the same successor.
5. Induction rule: For some statement P, P is true for all
natural numbers if:
1. P is true about 0 (That is, P(0) is true)
2. If you assume P is true for a natural number n (P(n) is true),
then you can prove that the P is true for the successor s(n) of
n (P(s(n)) is true).
And all of that is just a fancy way of saying: the natural numbers are numbers with no fractional
part starting at 0. We usually write N for the set of natural numbers. Most people, on first encountering the Peano rules find them pretty easy to understand, except for the last one. Induction is a tricky idea; I know that when I first saw an inductive proof, I certainly didn’t get it; it had a feeling of circularity to it that I had trouble wrapping my head around. But induction is essential: the natural numbers are an infinite set – so if we want to be able to say anything about the entire set, then we need to be able to use that kind of reasoning to extend from the finite to the infinite.
To give an example of why we need induction, let’s look at addition. We can define addition on the natural numbers quite easily. Addition is a function “+” from a pair of natural numbers to another natural number called their sum. Basically, we define addition using the successor rule of Peano arithmetic: m + n = 1 + (m + (n – 1)). So formally, addition is defined by the following rules:
1. Commutativity: For any pair of natural numbers n and m, n + m = m + n.
2. Identity: For any natural numbers n, n + 0 = 0 + n = n.
3. Recursion: For any natural numbers m and n, m+s(n) = s(m+n)
The last rule is the tricky one. Just remember that this is a definition, not a procedure. So it’s describing what addition means, not how to do it. The last rule works because of the Peano induction rule. Without it, how could we define what it means to add two numbers? Induction gives us a way of saying what addition means for any two natural numbers.
### Integers
The integers are what you get when you extend the naturals by adding an inverse rule. Take the set of natural numbers N. In addition to the 5 Peano rules, we just need to add a definition of an additive inverse. An additive inverse of a non-zero natural number is
just a negative number. So, to get the integers, we just add these new rules:
1. Additive Inverse: For any natural number n other than zero,
n, there is exactly one number -n which not a natural number,
and which called the additive inverse of n,
where n + -n = 0. We call the set of natural numbers and their additive inverses
the integers
2. Inverse Uniqueness: For any two integers i and j, i is the additive inverse of
of j if and only if j is the additive inverse of i.
And that’s just a fancy way of saying that the integers are all of the whole numbers – zero, the
positives, and the negatives. What’s pretty neat is that if you define addition for the natural numbers, the addition of the inverse rule is enough to make addition work. And since multiplication on natural numbers is just repeated addition, that means that multiplication works for the integers too.
## 0 thoughts on “Basics: Natural Numbers and Integers”
I’m at work now, so I can’t check my notes, but I am reasonably sure that when I had algebra at university we usually defined N (natural numbers) without Zero
Whenever we needed 0, we would use the Notation of N with a small zero to signify that we assumed 0 to be present in the natural numbers.
I can’t remember if this was the norm in both algebra and analysis, or if we had different definitions depending on the professor, and or, the book.
2. Mark C. Chu-Carroll
Soren:
I think you’re confused with the whole numbers. Whole numbers are the positive naturals; I’ve generally seen N0 as a notation for the natural numbers without zero – that is, the whole numbers.
3. Chris Jones
I was taught that natural numbers are the positive integers (i.e. 0 is not a natural number). What you’ve called natural numbers I was taught are the whole numbers. Looking at several references, the great majority agree with me, though it’s by no means unanimous.
Stepan Wolfram recommends using the terms “positive integers” and “non-negative integers” to avoid confusion.
4. Mark C. Chu-Carroll
KeithB:
No, you can’t just relax the zero rule – the induction principle will fall apart without it. By defining integers using the additive inverse, which is supported by the induction principle, that means that we can use the induction principle for reasoning about integers as well.
5. Pi Guy
I, too, had been under the impression that zero was not a natural, or counting, number but, rather, the first in the set of whole numbers. I recall that the mnemonic for distinguishing them was that zero has a “(w)hole” in it. However, it is clear that, since Peano arithmetic is the guide post, zero must be a natural number.
Good stuff. Thanks!
6. Vince Hurtig
What Marc is using is a variant of the Peano Axioms to define the natural numbers. Although he he has done it instead with the whole numbers, starting at zero rather than one as is the norm for the natural numbers. The initial number in the Peano Axioms is irrelevant, they work as well starting with either zero or with one. And yes, as a math teacher I have always taught that it is the whole numbers that start with zero and the natural numbers that start with one.
7. JBL
What exactly constitutes the natural numbers is, unfortunately, something that has not managed to become an established convention. The use of N for both “non-negative integers” and “positive integers” is fairly widespread, and the particular choice often depends on what kind of math you happen to be doing at any given moment. The notation N_0 (where the “_” denotes subscript) is totally unhelpful because it’s used to mean “the opposite of whatever N means” both by people who include 0 in N and by people who don’t include 0 in N.
Of course, it doesn’t actually matter at all: the choice of the letter N is entirely conventional, and Mark happens to have chosen as his convention that N includes 0, which is more convenient when you’re using the Peano axioms.
Mark didn’t give the notation for the integers: usually, they’re denoted Z. (I think it’s from the German.) This allows us to generate non-ambiguous notations easily. For instance, Z_(>0) pretty clearly means positive integers, and we could replace the “greater than” sign with a “greater-than-or-equal-to” sign if we wanted to denote the non-negative integers. Z_+ is also a reasonably common notation.
Seperately, it’s worth noting that commutivity is actually a provable fact about addition, given rules 2 and 3 and the Peano axioms. In fact, it’s an excellent example of why induction is so important: it’s possible to prove commutivity from induction, but if you take an axiomatic system with a weaker axiom in place of induction, you might find that it’s impossible to prove that addition commutes. (This happens, for example, in Robinson arithmetic.) Since, in most of our experience, addition *does* commute, this suggests that we really do need the axiom of induction.
8. Mark C. Chu-Carroll
Here’s the problem… Quoting from wikipedia:
In mathematics, a natural number can mean either an element of the set {1, 2, 3, …} (i.e the positive integers) or an element of the set {0, 1, 2, 3, …} (i.e. the non-negative integers). The former is generally used in number theory, while the latter is preferred in mathematical logic, set theory and computer science.
I come from the CS/logic background, so I was taught the zero-based definition of naturals. People from more number-theory type backgrounds were taught the one-based definition.
9. A
It’s particularly amazing to listen to people arguing vehemently about whether certain kinds of numbers are really “real” or not.
I’m particularly hoping you will find time to address the above point, as it seems the contrast between “numbers are for counting” and “numbers are members of a set” is often quite confusing for laypeople.
10. Walker
Now if you really want to blow some people’s minds, you should talk about the incompleteness of the Peano axioms and the existence of nonstandard models of arithmetic.
Though the question of whether we can really know what a nonstandard model looks like is a great game played by mathematical philosophers.
11. sp
Just a small note:
You write: “The integers (written Z) are all of the numbers, both larger and smaller than zero, that can be written without fractions.”
That sentence could be interpreted to suggest that the number 0 is not in Z (“both larger and smaller”, but not “equal to 0”) even though 0 is obviously in Z.
12. Blake Stacey
If I’m doing addition in a set, I’d really like my set to include the additive identity. It seems the “natural” thing to do.
Besides, I think that terms like “whole numbers”, “counting numbers” and all that lot were invented by grade-school teachers who needed something they could quiz their students upon, since testing knowledge in any meaningful way is just too hard.
“Susie, is 0 a counting number?”
“Well, I know I use counting numbers to count things. If I have no candy in my box, I count zero pieces of candy, so 0 must be a counting number.”
“Wrong! Class, look on page 7 of your textbooks. Tommy, can you read what it says at the top of the page?”
“The counting numbers are the set 1, 2, 3, . . . .”
“Excellent, Tommy. Can everyone tell why Susie was wrong?”
I just made this up — I don’t know whether a first-grade textbook will typically include 0 in the “counting numbers” (or any other set), but whichever way they choose, the problem is the same.
13. Susan
I have a question that might seem slightly OT but I think it fits here. I’m a sighted person working on software to interconvert print math and braille math. There are various systems for braille math — basically they are simply linear systems like LaTeX except terser since they are meant to be read by humans.
Here’s the question. Some braille math systems use the same braille characters (cells) for the letters a-j and also to represent the decimal digits 1-9 and 0. The digit use is distinguished by prefacing a digit sequence with an additional braille character, called a number sign, that indicates the change of semantics.
Other braille math systems, including the one currently used in the US, use different braille cells for letters and for digits.
Now it may seem odd to readers of this blog, but the braille authorities in Canada are seriously considering switching from the US braille math system, which they currently use, to a new one which uses the same characters for the letters a-j and for the decimal digits.
I would appreciate insights about the effect of the use of a letters-as-numbers notation on the understanding of what numbers are.
(There has already been more than enough discussion in braille circles of the obvious awkwardness of the notation in practice but this hasn’t been persuasive. However, I think the much bigger issue is any possible effect on understanding.)
14. Michael Saelim
It would be nice to have these basics articles as you’re learning them. With that in mind, I hope the next basics article is on Riemann-Stieltjes integration. 🙂
15. ObsessiveMathsFreak
It’s particularly amazing to listen to people arguing vehemently about whether certain kinds of numbers are really “real” or not.
This “debate” goes back quite a long way. The ancient greeks distinguished “numbers” (0,1,2,3 , etc, etc) from “lengths” of lines. In modern terms, they made a sharp distinction between N and R, or if you like, between the concepts of “how many” and “how much”.
Their major problem was, in essence, that they did not consider N to be a subset of R! Though they knew that you could have lengths corressponding to natural numbers, and that you could even add those lengths like natural numbers, they still considered N and R to be disjoint. This is because their entire understanding of R was through its interpretation as the lengths of line segments.
If you consider as they did, that a real number means a length of a line segment, then you can do things with numbers that don’t make much sense with “lengths”. You can multiply two numbers to get a number. But you can’t really multiply two lengths to get a length. You can also square, cube, and put numbers to even higher powers, and still obtain a number. But you cannot do this for line segments and still obtain a line segment. Even addition and subtraction are a bit strange unless the lines are parallel.
For this reason, the Greeks never really studied true “numbers” and their properties. That’s why they’re remembered only for their contributions to geometry, as they made very little progress in pure algebra or arithmetic.
So the concept of a number, or different types of number, is not always so straightforward. And when you eventually meet numbers like complex numbers and quaternions, it becomes a little clearer that a “number” can be an elusive concept.
16. Drekab
Susan:
Well, it might underscore the difference between numbers and numerals, but I think actively switching to a such a system is asking for trouble. How are the poor students ever going to handle hexadecimal numbers?
17. Harald Hanche-Olsen
Just a nitpick here: I don’t think the commutativity of addition is usually included in the definition of addition. (And neither is the rule 0+n=n; only n+0=n is true by definition.) Rather, it is proved – by induction of course:
First, prove 0+n=n by induction on n. It is true by definition for n=0, and if it true for some fixed n then 0+s(n)=s(0+n)=s(n) so it holds for s(n) as well.
Then, prove by induction on n that s(m)+n=s(m+n) for all m.
It is certainly true for n=0, and if it is true for some given n then also s(m)+s(n)=s(s(m)+n)=s(s(m+n))=s(m+s(n)) where I used first, the (recursive) definition of addition, then the induction hypothesis, then the defintion of addition once more.
Finally, prove by induction on m that m+n=n+m: Clearly true for m=0, and if it is true for some given m then also s(m)+n=s(m+n)=s(n+m)=n+s(m), and we’re done.
As you can see, proving stuff with Peano arithmetic can take some effort.
18. Harald Hanche-Olsen
Oh, and to define the integers in terms of the natural numbers, some might find the Grothendieck construction more natural: The basic idea is to work with pairs of natural numbers, letting the pair (m,n) stand for the number m-n. But since many different pairs will stand for the same numbers, we must state when equality should happen. Introduce an equivalence relation ≡, saying (m,n) is equivalent to (p,q) and writing (m,n)≡(p,q) if and only if m+q=n+p. Then let the equivalence class (m−n) consist of all pairs (p,q) equivalent to (m,n). The set of all equivalence class is the set of integers. Addition is defined as (m−n)+(p−q)=((m+p)−(n+q)), and subtraction as (m−n)−(p−q)=((m+q)−(n+p)). Identifying m with (m−0) then the negative of m will be −m=(0-m), and we see that the chosen notation is consistent, since (m−0)-(n−0)=(m−n).
There are lots of details to be filled in in the above. The elegance of the procedure may not be fully realized until you discover that the exact same procedure can be used to define the rationals (with the obvious modificiations to avoid division by zero).
19. beza1e1
Hey guys, Unicode is a really cool thing. We can write ∀x∊ℕ: ∃y∊ℕ: y = successor(x) without LaΤεΧ. ☺
And copy&paste uses ℤ and ℝ are also there.
20. Nat Whilk
Mark writes: “It’s particularly amazing to listen to people arguing vehemently about whether certain kinds of numbers are really “real” or not.
The American Mathematical Society apparently feels that the ontology of numbers and other mathematical objects is nontrivial enough to include many articles on the topic in their Mathematical Reviews. A study of the history of mathematics is likely to give one some perspective on the matter and to thus make one more tolerant of the sort of arguments Mark describes.
For any natural number n other than zero, n,
The exception in the wording reminds me; why isn’t 0 defined to be its own additive inverse and 1 its own multiplicative inverse, for simplicity?
the contrast between “numbers are for counting” and “numbers are members of a set”
So if “numbers are for counting” when it follows that “sets are for collections”, right? 🙂
Susan:
I would appreciate insights about the effect of the use of a letters-as-numbers notation on the understanding of what numbers are.
What Drekab notes on number systems used in programming et cetera may be decisive.
But you could also consider neuroscience. IIRC the other day some blog discussed results on number representation and use in the brain, as observed with fMRI, you could google it. Different representations (numbers, letter, figures) and uses (counting, estimates, speech, writing) of the same numbers corresponded to different areas in the brain IIRC.
So, perhaps, more number representations may be good (different perspectives) but replacing number representations may be bad (need to relearn, not quite the same).
23. Billy
ObsessiveMathsFreak wrote:
“For this reason, the Greeks never really studied true ‘numbers’ and their properties. That’s why they’re remembered only for their contributions to geometry, as they made very little progress in pure algebra or arithmetic.”
Has Diophantus been stripped of his nationality? If it’s up for grabs, we’ll claim him here in Missouri.
24. JBL
“The exception in the wording reminds me; why isn’t 0 defined to be its own additive inverse and 1 its own multiplicative inverse, for simplicity?”
The premise of this question is false. An additive inverse to x is a number which when added to x gives 0. Thus, by definition, 0 is its own additive inverse. (Likewise for 1 and multiplication.)
When you construct the integers from the non-negative integers, what you’re doing is adjoining the additive inverses of all the integers. The exception for 0 is because 0 *already* has an additive inverse, so you don’t want to add another symbol to represent it.
25. Susan
Thanks to Drekab and Torbjörn for very helpful feedback.
The articles on whether the brain processes numbers or numerals or both look to be fascinating from their abstracts and from online summaries. The original articles plus a Comment appear in the January 18, 2007 issue of Neuron. (http://www.neuron.org/ Subscription required.)
26. Pseudonym
I still remember the day when the integers were described to me this way.
An integer can be represented as a pair of natural numbers (a,b) such that:
(a1,b1) + (a2,b2) = (a1+a2, b1+b2)
(a1,b1) – (a2,b2) = (a1+b2, a2+b1)
(a1,b1) = (a2,b2) iff a1 + b2 = a2 + b1
You get the idea. You go on to prove that the third relation is an equivalence relation, and show that addition is commutative and has an identity and so on.
This was a key idea in my mathematical upbringing: that you didn’t need to understand new objects in terms of what they are, but in terms of what you already have.
27. Walt
Pet peeve: That construction is _not_ due to Grothendieck. For the integers its due to Dedekind; in general it’s due to someone else (Oystein Ore defined the ring case, someone else pointed out it works for semigroups — I forget who). Grothendieck applied it to specific semigroups, but the construction is older.
28. Harald Hanche-Olsen
Ah, Dedekind already? Thanks for pointing that out. Yeah, I knew the “Grothendieck construction” is much more general, and I suspected it did not originate with him. Good to know the real scoop.
29. Enigman
Another way to think of the integers is as vectors, which is not so weird as it first seems. That is, the integers are got from the natural numbers by adding in the notion of a pair of directions. (Similarly the reals are directed magnitudes, much as the complex numbers are.) Then, however, the positive integers are not the natural numbers, they are positively directed natural numbers. (Similarly, Pseudonym’s positive integers above would differ from natural numbers.) The vectorial approach is less algebraic, perhaps, but also perhaps more in tune with applications?
30. ctw
Nat Whilk:
can you provide any online links or book recommendations? I’ve been raising (what I think is) the ontology question to which you allude in several fora with no response and concluded that it must be too stupid to warrant one. I’d be happy to confirm that it isn’t (or is, for that matter, in which case I can forget it).
tnx – charles
31. Nat Whilk
ctw:
I don’t know what your background is, so I’m not sure what to recommend. Ontological considerations are discussed in most introductory/survey works on the philosophy of mathematics. Stewart Shapiro’s _Thinking about Mathematics_ seems to do about a good a job as any at explaining the issues and how the major schools stand relative to them. Shapiro (like myself) subscribes to ontological realism, and his _Philosophy of Mathematics: Structure and Ontology_ is a defense of that position. One of the more prominent and interesting examples on the opposite side would be Harty Field’s _Science without Numbers: A Defence of Nominalism_.
32. ctw
nat –
thanks for the pointers. as it happens, I just tried wikipedia and resolved my question, viz:
is 1+1=2 on a par with gravity in “existing”?
the motivator being that many people (including some worldclass philosophers) say things like “that’s as real/true/indisputable as 1+1=2”. I don’t really care what the “true” answer is, just that “no” isn’t a completely off-the-wall answer. turns out my view (simplistic version, of course) actually has a name – embodied mind theory – and is summarized as:
Humans construct, but do not discover, mathematics.
essentially the words I used in a recent comment in another thread on this blog:
-c
JBL:
The premise of this question is false. An additive inverse to x is a number which when added to x gives 0. Thus, by definition, 0 is its own additive inverse.
Yes, that is what I was getting at.
The exception for 0 is because 0 *already* has an additive inverse, so you don’t want to add another symbol to represent it.
Well, just because I can write – 0 = (+) 0 , I frankly don’t see that there is any extra symbol made. And precisely because I can write – 0 it must still be handled as I understand it, even if it is to define is at an undefined number.
Analogous to roots; a root operation on x, say x positive real for illustration, has two solutions, + root(x) and – root(x), so I allow for – root(0) = + root(0) too.
So I guess for me my question still stands. But it is mostly a matter of taste how this is formulated.
Susan:
While I don’t know how it is to be blind, I know how it is to have a blind family member. My late grandmother was blind the last years of her life. She didn’t use braille much (wasn’t into computers) and preferred audio books, but it sure helped her make or read notes and handle medicine dispensers. So I’m happy to see people work with this, as well as that I could help!
34. Blake Stacey
It’s time to speak in parables. This one comes from the first volume of Isaac Asimov’s autobiography, In Memory Yet Green (1979), page 214.
On September 27, 1938, I registered for my fourth year at Columbia. I was taking integral calculus now and Sidney was taking more sociology while I was doing that. After calculus, I would go around to the sociology class, where the professor held court after the lecture was over. When that was done, Sidney and I would have lunch.
It made for dull listening, generally, for I never have been impressed by the soft sciences. On October 10, I found the sociology professor (his name was Casey) had made a table on the board in the course of his lecture in which he divided people into rationalists and mystics. Under mystics he had listed mathematicians.
I studied that for a while and then, even though I was not a class member, I interrupted the postlecture session by saying, “Sir, why do you list mathematicians as mystics?”
He said, “Because they believe in the reality of the square root of minus one.”
I said, “The square root of minus one is perfectly real.”
He said, “Then hand me the square root of minus one piece of chalk.”
I said, “The cardinal numbers are used for counting. The so-called imaginary numbers, like the square root of minus one, have other functions. If you had me a one-half piece of chalk, however, I’ll hand you a square root of minus one piece of chalk.”
Whereupon Casey promptly broke a piece of chalk in half and handed it to me with a smile. “Now your turn,” he said.
“Not yet,” I said. “That is one piece of chalk you’re handing me.”
“It is half a regulation length of chalk.”
“Are you sure?” I said. “Will you swear it is not 0.52 times a regulation length or 0.48 times that?”
By now Casey realized it was time for hard logic if he was to win the argument, so he decided that since I was not a member of the class, I would have to leave the room at once. I left, laughing rather derisively, and after that I waited for Sidney in the hall.
35. JBL
Torbjörn: Unrelated to anything else, is there a quick way to produce the second vowel in your name on a standard U.S. keyboard, without copy-pasting your name from your comments?
So, some actual content: “putting a minus sign in front” is not an operation that exists in Peano arithmetic, and we can define it only after we have defined the objects which we arrive at by putting a minus sign in front of the objects we already have, namely the non-negative integers. In Peano arithmetic, one can not in fact write “-0”; that’s a meaningless collection of symbols. So, at the moment that we jump from the non-negative integers to the integers by defining a whole host of new symbols, we need to decide whether or not to add a symbol “-0”. I suppose that there is no theoretical reason not to add such a symbol: we would just prove, as our first theorem, that “-0 = 0”, and then we’d never have to use it again (or, we’d use it where it was more convenient). Now, after we have all these symbols, we can define on them the operation “taking the additive inverse”, which we (for convenience, but not for any deeper reason) denote also with a minus sign, which has the property that when we apply it to 0, we get back 0 (or -0, which equals 0, if we happen to have defined the symbol -0).
I guess this is my point, re-phrased: you wrote, “because I can write – 0 it must still be handled as I understand it, even if it is to define is at an undefined number.” But I think this is not really the case. The symbol “-” serves at least 3 distinct purposes: to denote subtraction of one number from another, to denote the function of “taking the negative,” and as part of a set of symbols which we use for the additive inverses of the positive integers (where it has no independent meaning). I believe that when you say, “I can write – 0,” you are conflating these latter two uses for the symbol “-“. If we don’t define the symbol “-0,” you can write it to mean “the result of applying the operation of negation to 0,” but you can’t write it to mean “the symbol which represents the additive inverse of 0” unless we specifically define such a symbol.
“Taking the square root” is a special case of “solving a quadratic equation.” Every quadratic equation has 2 roots, but for some of these equations the roots are the same. We don’t use any symbol to differentiate between the two roots of x^2 – 2x + 1 even though they are the same; why should we use a special symbol in order to distinguish the roots of x^2 from each other?
I hope I’m being clear, sensible, and actually addressing your thoughts — if I seem pedantic, it’s merely because I’m trying to make sure I understand what I’m writing, and if I seem to have missed the point (or to be wrong!), please let me know!
JBL:
Unrelated to anything else, is there a quick way to produce the second vowel in your name on a standard U.S. keyboard, without copy-pasting your name from your comments?
A fair comment, indeed, and a considerate thought. At the moment I don’t have access to such a keyboard, so I googled.
If you have a US keyboard:
“The US keyboard layout does not use AltGr or any dead keys, and thus offers no way of inputting any sort of diacritic or accent;” ( http://en.wikipedia.org/wiki/Keyboard_layout#US ).
If you have a US-International keyboard (which is basically what Sweden use, with keys added and remapped):
“The US keyboard layout can be configured to type accents efficiently. This is known as the US-International layout.
Accented characters can be typed with the following combinations:
” then letter (ë)
I.e. ¨ then o: ö. (And I can do that too, besides using the remapped main key!)
The easiest way if you have a US only keyboard is to spell it “Torbjorn”.
If you want to be phonetically true, it is “Torbjoern”, though the “ö” (“oe”) sound isn’t really used in most english dialects. (But english-speakers can learn it really well, and the basic phonemes are close.)
If you want to make an effort, the HTML code is “& # 246 ;” (remove quotation marks and spaces. [My test: Torbjörn.]
I believe that when you say, “I can write – 0,” you are conflating these latter two uses for the symbol “-“.
Not intentionally, but I can see how you make sense out of the combination without disallowing it.
So I concede that one doesn’t have a need to point out “-0” as an exception. And then parsimony guides us to the chosen convention.
“We don’t use any symbol to differentiate between the two roots of x^2 – 2x + 1 even though they are the same;”
Right. I was conflating the algebraic problem with my own practice to make it clear I have considered both roots when solving numerically.
Thank you for addressing my points! Apparently I have grown so used to my own practice so I have forgotten the underlying principles in both cases. Which is presumably why I needed someone to nudge my thoughts in the right direction.
JBL:
And then parsimony guides us to the chosen convention.
On second thought, I have to retract that. Parsimony in symbols is one thing, parsimony in definitions another. And I would actually go for the later, considering the analogy for parsimony in physics. (Minimize number of parameters, not number of objects.)
Perhaps it is mostly a matter of taste, after all.
38. Boian Popunkiov
“We don’t use any symbol to differentiate between the two roots of x^2 – 2x + 1 even though they are the same; why should we use a special symbol in order to distinguish the roots of x^2 from each other?”
I’m not sure what you mean by the roots of x^2. For positive real numbers, there is only one square root function. Namely, if a>0, sqrt(a) is the positive solution of x^2=a. So sqrt(4)=2, not 2 or -2. Otherwise, sqrt(x) would not be a well -defined function.
Once we start talking about complex numbers, then things start to get more complicated and we get “multivalued functions”, which technically are not functions.
39. Davis
The easiest way if you have a US only keyboard is to spell it “Torbjorn”.
If you have a Mac, you can set your keyboard layout to US Extended (under International->Input Menu). Then hitting option-u gives the ¨ diacritic over the next vowel you type. You can get some other diacritics this way as well.
I was forced to figure this out because my advisor was Hungarian, and had the ´ diacritic on every ‘a’ in his name; the Hungarians I’ve met seem to be a little picky about that.
Davis:
I can imagine that some diacritics may make a huge difference in pronunciation or meaning.
In swedish it doesn’t, because while the sounds are different, they are close enough. We have å (phonetically ‘aa’, roughly), ä (phonetically ‘ae’, roughly), ö (phonetically ‘oe’, roughly). ‘Torbjorn’ (or ‘Torbjoern’) is recognizable, both by spelling and sounding.
OTOH, conflating the spelling can be another problem, for example for people with last name Häger and Hager. Here phonetically spelling helps more. (‘Haeger’ vs “Hager”.)
41. Larry D'Anna
Actuallly the Peano Axioms don’t really provide a definition of the naturals. In fact no set of first order axioms can define any particular infinite model, because any consistent (first order) theory
that has a model of (infinite) cardinality kappa has a model of every cardinality greater than kappa. This means that there are actually uncountable models of Peano’s Axioms. (There are also coutable models that aren’t the standard one as well!)
42. Tom Warn
As pointed out by Bertrand Russell in ‘Introduction to Mathematical Philosophy’ (and perhaps others), ‘0 and ‘number’ and ‘successor’ are primitive terms in Peano’s axioms and so aren’t really defined. To obtain the natural numbers it is necesssary to assume we know what 0 is, otherwise it could be identified with 100 and the axioms would still be satisfied. In fact according to Russell any sequence x_0, x_1, … ,x_n satisfies Peano’s axioms.
43. Anonymous
Additive Inverse: For any natural number n other than zero, n, there is exactly one number -n which not a natural number, and which called the additive inverse of n, where n + -n = 0.
What is “+”? If “-n” is not a natural number then “+” is not the addition operation between naturals. It cannot be the addition operation between integers since we’re just defining what integers are.
44. Bill
In the laboratory a number and a length are a pair (5,meters). A pair is the fundamental unit in the laboratory. We need “meters” to specify our instrument, a meter stick. We do not measure (0,meters) because our meter stick cannot be used to measure zero. This may become a problem when an algebraic expression such as 1/t is used. As an example,the laboratory measurement requires using (1,meter)/(5,sec). Introductory physics is often named “point” physics, an incorrect name. It is “physical point” or “measurable point” physics. Point refers to zero in algebra and to a nondimensional point in Euclidean geometry.
Effectively this means that “pure mathematics” does not apply to physics with very few exceptions. Newton’s equations are an extension of analytic geometry. His gravitational equation is geometrical, specifically the surface of a sphere, 4(pi)r^2 . The expression is 1/[4(pi)r^2] which clearly does not apply at zero. Neither quantum mechanics, relativistic quantum electrodynamics, or even the standard model for elementary particles address this problem in a “clean” manner. My opinion is that physics will have to be reformulated mathematically starting with Euclidean geometry. |
# Binomial Sum \displaystyle \sum^{n}_{k=0}\binom{n+k}{k}\cdot \frac{1}{2^k}
• MHB
• juantheron
In summary, a binomial sum is a mathematical expression used in combinatorics and probability to calculate the number of possible combinations or outcomes. To calculate a binomial sum, you first need to identify the values of n and k and use the binomial coefficient formula. The 1/2^k term in the formula represents the probability of a successful outcome, and the formula is closely related to Pascal's triangle. The binomial sum formula can be used in various real-life situations, such as calculating probabilities and genetic outcomes.
#### juantheron
Evaluation of $\displaystyle \sum^{n}_{k=0}\binom{n+k}{k}\cdot \frac{1}{2^k}$
$$S = \sum_{k=0}^n {n+k\choose n} \frac{1}{2^k} = \sum_{k=0}^n \frac{1}{2^k} [z^n] (1+z)^{n+k} = [z^n] (1+z)^n \sum_{k=0}^n \frac{1}{2^k} (1+z)^k \\ = [z^n] (1+z)^n \frac{1-(1+z)^{n+1}/2^{n+1}}{1-(1+z)/2} = [z^n] (1+z)^n \frac{2-(1+z)^{n+1}/2^{n}}{1-z} \\ = 2\times 2^n - [z^n] (1+z)^{2n+1} \frac{1}{2^n} \frac{1}{1-z} \\ = 2\times 2^n - \frac{1}{2^n} \sum_{k=0}^n {2n+1\choose k} = 2\times 2^n - \frac{1}{2^n} \frac{1}{2} 2^{2n+1} = 2^n.$$
## 1. What is a binomial sum?
A binomial sum is a mathematical expression that involves adding together terms of the form (n choose k) multiplied by a constant or variable. The binomial sum formula is commonly used in combinatorics and probability to calculate the number of possible combinations or outcomes.
## 2. How do you calculate a binomial sum?
To calculate a binomial sum, you first need to identify the values of n and k. Then, use the binomial coefficient formula (n choose k) = n! / (k!(n-k)!) to find the value of (n choose k). Finally, multiply this value by the constant or variable and add up all the terms to get the final result.
## 3. What is the significance of the 1/2^k term in the binomial sum formula?
The 1/2^k term in the binomial sum formula represents the probability of a successful outcome in a binomial experiment. In other words, it is the probability of getting a specific combination of successes and failures in a series of trials. This term is often used when calculating the probability of events in statistics and probability theory.
## 4. How is the binomial sum related to Pascal's triangle?
The binomial sum formula is closely related to Pascal's triangle, a geometric arrangement of numbers where each number is the sum of the two numbers directly above it. The coefficients of the binomial sum can be found by looking at the corresponding row in Pascal's triangle and the exponents of the variables in the binomial sum correspond to the positions in the triangle.
## 5. In what real-life situations can the binomial sum formula be used?
The binomial sum formula can be used in a variety of real-life situations, such as calculating the probability of flipping a coin a certain number of times and getting a specific number of heads, or finding the number of possible combinations when selecting a certain number of items from a larger set. It is also used in the field of genetics to calculate the probability of specific genetic outcomes in offspring. |
# Least common denominator solver
In this blog post, we will be discussing about Least common denominator solver. Our website will give you answers to homework.
## The Best Least common denominator solver
Least common denominator solver is a software program that helps students solve math problems. Any mathematician worth their salt knows how to solve logarithmic functions. For the rest of us, it may not be so obvious. Let's take a step-by-step approach to solving these equations. Logarithmic functions are ones where the variable (usually x) is the exponent of some other number, called the base. The most common bases you'll see are 10 and e (which is approximately 2.71828). To solve a logarithmic function, you want to set the equation equal to y and solve for x. For example, consider the equation log _10 (x)=2. This can be rewritten as 10^2=x, which should look familiar - we're just raising 10 to the second power and setting it equal to x. So in this case, x=100. Easy enough, right? What if we have a more complex equation, like log_e (x)=3? We can use properties of logs to simplify this equation. First, we can rewrite it as ln(x)=3. This is just another way of writing a logarithmic equation with base e - ln(x) is read as "the natural log of x." Now we can use a property of logs that says ln(ab)=ln(a)+ln(b). So in our equation, we have ln(x^3)=ln(x)+ln(x)+ln(x). If we take the natural logs of both sides of our equation, we get 3ln(x)=ln(x^3). And finally, we can use another property of logs that says ln(a^b)=bln(a), so 3ln(x)=3ln(x), and therefore x=1. So there you have it! Two equations solved using some basic properties of logs. With a little practice, you'll be solving these equations like a pro.
Think Through Math is an app that helps you to better understand mathematics. By breaking down math problems into smaller, more manageable pieces, Think Through Math allows you to better see how each step in a problem leads to the next. As a result, you can work through problems more quickly and confidently, improving your overall math skills. In addition, Think Through Math provides a variety of tools and resources that you can use to further improve your understanding of mathematics. From video lessons to practice problems, Think Through Math offers a comprehensive approach to learning that can help you succeed in mathematics.
Maths online is a great way to learn Maths. You can find Maths online courses for all levels, from beginner to expert. Maths online courses can be found for free or for a fee. Maths online can be a great way to learn Maths if you have the time and patience to commit to it. You can also find Maths games online, which can be a fun way to learn Maths. Maths games can be played for free or for a fee. Maths online can be a great way to learn Maths if you have the time and patience to commit to it. Thanks for reading!
First, it is important to create a dedicated study space. This will help to minimize distractions and make it easier to focus on the task at hand. Secondly, students should develop a regular routine and stick to it as much as possible. This will help them to stay on track and avoid getting overwhelmed. Finally, students should seek help from their teachers or parents when needed. By taking these steps, students can set themselves up for success when it comes to doing their math homework.
This a website that enables you to get detailed solutions to your math word problems. Just enter the problem in the text box and click on the "Solve" button. will then show you step-by-step how to solve the problem. You can also use the site to check your answers to make sure you are on the right track. is a great resource for students of all levels who are struggling with math word problems.
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# Fraction Frenzy
Purpose
This is a level 3 number activity from the Figure It Out series.
A PDF of the student activity is included.
Student Activity
Click on the image to enlarge it. Click again to close. Download PDF (205 KB)
Specific Learning Outcomes
order fractions
find equivalent fractions
Required Resource Materials
Fraction Cards copymaster
FIO, Level 3, Number, Book 3, Fraction Frenzy, pages 22-23
A classmate
Activity
Before the students start this activity, you might like to work through two earlier Figure It Out activities with them. Fabulous Folding, Number, Figure It Out, Levels 2–3, page 18, uses a simple number line to place fractions in order. (You can extend this later in relation to Fraction Frenzy.) Fun with Fractions, Number, Figure It Out, Level 3, page 9, uses paper circles to explore simple equivalent fractions.
In Fraction Frenzy, the students explore the simple equivalence of fractions. They also discuss the order of fractions. The activity can be extended to investigating whether there is a rule for changing a fraction into an equivalent fraction.
In question 1, the students sort a set of fraction cards (see the copymaster at the end of these notes) into equivalent sets. In the process, they order them from smallest to largest. If they are not aware of this sequencing, it should become apparent to them in their discussion with a classmate.
There are various ways of helping the students who have difficulty with this question. You could direct them to the fraction wall in question 4 and help them to explore it, or you could use a simpler fraction wall and link a number line to it:
Get the students, perhaps in pairs, to make the fractions on the wall out of strips of paper (they need to be the right lengths), for example, 1/5, 2/5, 3/5, and so on, and then align their strips to create a fraction number line:
A fraction wall such as the one below is a good way to teach the students about different ways of writing fractions:
Before they do question 2, the students need to check their columns from question 1. They could do this with a classmate. They then go on to place further equivalent fractions into the correct columns. If necessary, lead the students to realise that they can find equivalent fractions by looking for patterns in the numerators and denominators. For example, with 25/100, 25 goes into 100 four times, so it is equivalent to 1/4. With 45/75, we can see that both numbers are divisible by 5, so we get 9/15. In turn, both these numbers are divisible by 3, so we get 3/5.
Question 4 asks the students to explain how two fractions are equivalent. Have them explain this to another classmate first and then get them to write out their explanation. The writing can be used for teacher assessment or for self- or peer-assessment and will also help the students to internalise the concept.
In question 5, the students are given only the denominator of the second fraction and have to find its numerator to make it an equivalent fraction. After completing the previous questions, the students should see that a quick way of finding equivalent fractions is to multiply the numerator and denominator by the same number. If they do not understand this, have them make further equivalent fractions with materials and then say and write the names. Rather than giving them the “rule”, help the students to find it for themselves.
Question 6 requires a good knowledge of multiplication and division basic facts because the students have to multiply and divide by 3 and 8. They need to search for a link between the two numerators 3 and 375. Encourage them to see that 375 is 3 x 100 + 3 x 25, that is, 3 lots of 125. The students need to identify the relationship between the two numerators:
They then need to apply that relationship to the denominator to maintain the “equivalence” between the fractions.
#### Further discussion and investigation
Have the students make their own fractions and then place them on number lines. Include fractions greater than 1. Ask them to name fractions equivalent to those shown. Discuss why 1/1, 2/2, 5/5, and so on are all equal to 1.
1. Practical activity. Your columns should have equivalent fractions in them. (Equivalent fractions are different ways of expressing the same fraction.)
Karyn has arranged her first row from the smallest to the largest fraction.
2. a. in the 1/4 column
b. in the 1/3 column
c. in the 1/3 column
d. in the 3/5 column
e. in the 1/5 column
f. in the 1/2 column
g. in the 1/4 column
h. in the 2/2 column
3. a. 2/5 = 4/10 or 6/15
b. 3/7 = 6/14
c. 2/3 = 4/6, 8/12, or 10/15
4. Answers will vary. For example, 2/6 and 4/12 both simplify to 1/3. They all take up the same length on the fraction wall. The top number (the numerator)
and the bottom number (the denominator) can both be divided by 2 (and in the case of 4/12, by 2 again) to get 1/3. So, equivalent fractions simplify to the same simplest
form. Another example is 4/10 and 8/20, which both simplify to 2/5, so they are equivalent fractions.
5. a. 6/9 = 2/3
b. 6/9 = 66/99
c. 6/9 = 8/12
6. 1 000
Attachments |
Expanded Form and Standard Form of a Number
# Expanded Form and Standard Form of a Number
## Expanded Form and Standard Form of a Number
### Expanded Form
In expanded form, a number is written according to the place value of its digits.
To write a number in its expanded form, we just write the place values of all the digits with plus (+) sign starting from the highest place.
For example, 765349 = 7 × 1,00,000 + 6 × 10,000 + 5 × 1000 + 3 × 100 + 4 × 10 + 9 × 1
= 7,00,000 + 60,000 + 5,000 + 300 + 40 + 9
Example 1: Write the following numbers in expanded form.
a. 2565935 b. 796435429
Solution:
a. 2565935 = 20,00,000 + 5,00,000 + 60,000 + 5,000 + 900 + 30 + 5
b. 796435429 = 70,00,00,000 + 9,00,00,000 + 60,00,000 + 4,00,000 + 30,000 + 5,000 + 400 + 20 + 9
### Standard Form
When we write a number in its short form, we call it the standard form of a number.
For example, the standard form of 80,00,000 + 6,00,000 + 20,000 + 1,000 + 900 + 30 + 4 is 8621934.
To write a number from expanded form to standard form, we can use two methods.
1. We can make a place value table and put the digits in the place value table as follows.
TL L TTh Th H T O 8 6 2 1 9 3 4
Now, we can write the number as 8621934.
2. We can add the place values of each digit by writing them in columns as follows.
TL L TTh Th H T O 8 0 0 0 0 0 0 6 0 0 0 0 0 2 0 0 0 0 1 0 0 0 9 0 0 3 0 4 8 6 2 1 9 3 4
Thus, the number in the standard form is 86,21,934.
Example 2: Write the following in standard form.
a. 30,00,000 + 8,00,000 + 70,000 + 2,000 + 500 + 10 + 9
b. 80,00,00,000 + 2,00,00,000 + 20,00,000 + 9,00,000 + 30,000 + 1,000 + 400 + 90 + 4
Solution:
a. 30,00,000 + 8,00,000 + 70,000 + 2,000 + 500 + 10 + 9
TL L TTh Th H T O 3 8 7 2 5 1 9
Thus, the number in the standard form is 38,72,519.
b. 80,00,00,000 + 2,00,00,000 + 20,00,000 + 9,00,000 + 30,000 + 1,000 + 400 + 90 + 4
TC C TL L TTh Th H T O 8 2 2 9 3 1 4 9 4
Thus, the number in the standard form is 82,29,31,494.
Related Topics:
Expanded Form of Decimals
Forming and Rounding off Numbers
Comparing and Ordering Integers
Operations on Integers
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# Profit Loss Class 7 RS Aggarwal Exe-10A Goyal Brothers ICSE Maths Solutions
Profit Loss Class 7 RS Aggarwal Exe-10A Goyal Brothers ICSE Foundation Maths Solutions. We provide step by step Solutions of lesson/ Chapter-10 for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.
## Profit Loss Class 7 RS Aggarwal Exe-10A Goyal Brothers ICSE Maths Solutions
Board ICSE Publications Goyal brothers Prakashan Subject Maths Class 7th Chapter-10 Profit and Loss Writer RS Aggrawal Book Name Foundation Topics Solution of Exe-10A Academic Session 2024 – 2025
### Exercise – 10A
Profit Loss Class 7 RS Aggarwal Goyal Brothers ICSE Foundation Maths Solutions
##### Question: 1- Amit bought a calculator for RS 960 and sold it for 1104.Find his gain and gain percent.
Solution: C.P. = 960, S.P. = 1104
Gain = S.P-C.P.
= 1104-960=144
Gain%=Gain*100/C.P
=144*100/960
Gain%= 15% Ans
##### Question: 2- Ankita bought a mobile phone for RS 2350 and sold it for RS 2538. Find her gain and gain percent.
Solution: C.P.=2350, S.P.=2538
Gain=S.P.-C.P.
=2538-2350=188 Ans
Gain%=Gain*100/C.P.
=188*100/2350
Gain%= 8% Ans
##### Question: 3- Ayush bought a bicycle for RS 6250 and had to sell it for RS 5875. Find his loss and loss percent.
Solution: C.P.=RS 6250, S.P.=RS 5875
Loss=C.P.-S.P.
=6250-5875
=375 Ans
Loss%=Loss*100/C.P.
=375*100/6250
Loss%= 6% Ans
##### Question: 4- Aman bought a computer for RS 24000 and its accessories pack worth for RS 1750. He sold it all for RS 26780. Find his gain and gain percent.
Solution: C.P. of computer = RS 24000
C.P. of accessories = RS 1750
Total C.P. = 2400+1750=25750
S.P.= RS 26750
Gain = S.P.-C.P.
= 26780-25750
= 1030 Ans
Gain% = Gain*100/C.P.
= 1030*100/25750
Gain% = 4% Ans
##### Question: 5- A man bought a refrigerator for RS 35615 and paid RS 125 on its transportation. He sold it for RS 33953. Find his gain or loss percent.
Solution: C.P of refrigerator = RS 35615
C.P. of transportation = RS 125
Total C.P. = 35615+125
=RS 35740
S.P. = RS 33953
Loss = C.P.-S.P.
= 35740-33953
= RS 1787
Loss% = Loss*100/C.P.
1787*100/35740
Loss% = 5% Ans
##### Question: 6- By selling a bicycle for RS 5670 a trader gain RS 270. Find his gain percent.
Solution: S.P. = RS 5670, Gain = RS 270
C.P. = S.P.-Gain
=5670-270=5400
Gain% = 270*100/5400
Gain% = 5% Ans
##### Question: 7- By selling a chair for RS 1410 a carpenter suffers a loss of RS 90. Find his loss percent.
Solution: S.P. = RS 1410, Loss = RS 90
C.P. = S.P.+ Loss
=1410+90
=1500
Loss% = 90*100/1500
Loss% = 6% Ans
##### Question: 8- A fruit-seller bought bananas at the rate of 3 for RS 8 and sold them at the rate of 2 for RS 7. Find his gain and loss percent.
Solution: Rate of 3 bananas = RS 8
Lost price of banana = RS 8/3
S.P. of 2 bananas = RS 7
S.P. of 1 banana = RS 7/2
Gain = 7/2 – 8/3 = 5/6
Gain% = (5/6 * 100) / (8/3) =125/4%
=31(1/4)%.
##### Question: 9- Lemons are bought at the rate of 3 for RS 4. At what rate must they be sold to gain 20%.
Solution: Cost price of 3 lemons =RS 4
Gain = 20%
S.P. of 3 lemons = 4 * 120/100
= RS 24/5
C.P. of 1 lemon = 24/5*3
= RS 8/5
S.P. of 5 lemons = 8/5 *5 = RS 8
5 lemons are sold for RS 8. Ans
##### Question: 10- The selling price of 12 pens is equal to the cost price of 14 pens. Find his gain percent.
Solution: C.P. of 12 pens = RS 12
Let C.P. of 1 pen = RS 1
S.P. of 12 pens = C.P. of 12 pen
=RS 14
Gain = S.P.-C.P.
=14-12 = RS 2
Gain% = Gain*100/C.P.
= 2*100/12 = 50/3% Ans.
##### Question: 11- The cost price of 12 oranges is equal to the selling price of 15 oranges. Find his loss percent.
Solution: Let C.P. of 1 orange = RS 1
C.P. of 15 oranges = C.P. of 12 oranges
= RS 12
Loss = C.P.-S.P.
=15-12 = 3
Loss% = Loss*100/C.P.
= 3*100/15 = 20% Ans.
##### Question: 12- Vinay sold a plot of land for RS 143000, gaining 4%. For how much did he purchase a plot?
Solution: S.P. of plot = RS 143000
Gain = 4%
C.P. of a plot = RS 100/100+Gain% * S.P.
= 100/100+4 * 143000
= 100/104 * 143000
Hence 100*1375 = RS 137500 Ans.
##### Question: 13- John sold his T.V. set for RS 14100, losing 6%. For much did he purchase it?
Solution: S.P. of T.V. set = RS 14100
Loss = 6%
C.P. = RS 100/100-Loss * S.P.
C.P. of set = RS 100/100-6 * 14100
= 100/94*14100
C.P. = RS 15000 Ans.
##### Question: 14- On selling a bed for RS 10800, a carpenter loses 10%. For what amount should he sell it to gain 5%?
Solution: S.P. of bed = RS 10800
Loss = 10%
C.P. of bed = S.P.*100/100-Loss
= 10800*100/100-10 = 12000
C.P. of bed = RS 12000
Gain = 5%
S.P. = RS 100+Gain%/100 * C.P.
= RS 12000(100+5)/100
S.P. = RS 12600 Ans.
##### Question: 15- On selling an almirah for RS 20350, a man gain 10%. What percent does he gain on selling the same for RS 19610?
Solution: S.P. of almirah = RS 20350
Gain = 10%
C.P. = S.P.*100/(100+Gain%)
= 20350*100/100+10
= RS 18500
When S.P. = 19610
Gain = S.P.-C.P.
= RS 19610-18500
= RS 1110
Gain% = total gain*100/C.P.
= 1110*100/13500
Gain = 6% Ans.
##### Question: 16- On selling a fan for RS 4700, a shopkeeper loses 6%. At what price must he sell it to gain 6%?
Solution: S.P. of fan = RS 4700
Loss = 6%
C.P. = 100/100-Loss% * S.P.
= 100/(100-6)*4700
= 100/94*4700
Hence RS 5000
IF Gain = 6%
S.P. = RS (100+6)/100*5000
= 106/100*5000
S.P. = RS 5300 Ans.
##### Question: 17- Kamal sold two scooters for RS 19800 each, gaining 10% on the one and losing 10% on the other. Find his gain or loss percent on the whole transactions.
Solution:
The one sold at a gain we have :
100/110 × 19800 = 18000
The one sold at a loss we have :
100/90 × 19800 = 22000
The overall buying price for the two is :
22000 + 18000 = RS 40000
The overall selling price for the two is :
19800 × 2 = 39600
The selling price is lowered than the buying price by :
40000 – 39600 = RS 400
There is therefore a loss of 400
The percentage loss is thus :
400/40000 × 100% = 1% loss Ans.
##### Question: 18- A buys an article for RS 650 and sell it to B at a profit of 20%. B sell it to C at a loss of 20%. What does C pay for it?
Solution:
Cost price when “A” buys an article = 650
Sells it to “B” at Profit Percentage = 20 % .
Selling Price = Cost Price + Profit
= 650 + (650 x 20%)
= 650 + 130
=Rs 780 .
So , “A” sells it to “B” at Rs 780 .
Cost Price when “B” buys an article = 780
Loss % = 20 %
Selling Price = Cost Price – Loss
= 780 – (780 x 20%)
= 780 – 156
Hence Rs 624 .
So , “B” sells it to “C” at Rs 624 .
Cost Price when “C” buys an article = Rs 624.
“C” PAYS FOR AN ARTICLE = Rs 624 Ans.
— : End of Profit Loss Class 7 RS Aggarwal Exe-10A Goyal Brothers ICSE Foundation Maths Solutions :–
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# Simple Interest: Formulas, Definition, Questions and Comparison
1
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Simple interest is a quick and easy method of determining the interest charged on a loan or principal amount. SI is defined by simply multiplying the given interest rate with the principal amount and the number of days together. The concept of SI is employed in most areas such as finance, banking, automobile, and so on. Through this article learn the concepts of SI, through the formula, and examples on how to calculate simple interest. S.I. comes under one of those highlighted topics that are most commonly asked in all competitive exams like SSC( CGL, CHSL, JE), SBI Clerk, and similar exams.
## What is Simple Interest?
Simple interest as per our school knowledge is the method of computing the interest amount for some principal amount of money. SI is something like considering;
Ram borrowed some money from the bank to buy a vehicle for a period of interval. During the time of repayment of the loan, Ram would have to pay the original amount plus some more money that depends on the loan amount as well as the time for which the money was borrowed.
## Simple Interest Formula
S.I. in mathematics as read in the introduction is a method that is applied to calculate interest on the money/capital or funds. Let us step towards the formulas relating to the topic, as the formulas play a major role in the easy calculation. The formula for S.I. is:
Simple interest= (Principal × Rate × Time) / 100
OR
$$S.I.=\frac{\left(P\times R\times T\right)}{100}$$
Here, are the meaning of the various terms;
SI SI stands for Simple Interest. P P denotes the principal amount. The principal amount(P) is the initial amount invested or borrowed by an individual from the bank. R R is the interest rate in percentage. The rate at which the principal amount is given to somebody for a certain time stands for the rate of interest. T T specifies the time duration in years. The duration for which the principal amount is provided to someone denotes the time.
Also below are formulas for Principal, Rate, and Time
The formula of Prinicipal if Interest, Rate, and Time given:
$$P=\frac{\left(S.I\times100\right)}{R\times T}$$
The formula of Rate if Interest, Principal, and Time given:
$$R=\frac{\left(S.I\times100\right)}{P\times T}$$
The formula of Time if Interest, Rate, and Principal given:
$$T=\frac{\left(S.I\times100\right)}{P\times R}$$
In the simple type of interest, the interest always applies to the original principal amount, with the same rate of interest for every time cycle. When we invest our money in any bank, the bank provides us with interest on our amount.
There is one more term called “Amount”, which is defined as the total money that is to be paid back at the end of the period for which it was borrowed. In general, when an individual takes a loan or borrows some money he/she has to repay the principal borrowed + the interest amount, and this complete quantity is known as the amount.
To calculate the amount;
Amount = Principal + Simple Interest= P + SI
## Simple Interest Formula for Months
In the previous section, we read the formulas relating to S.I, principal, interest, rate and time duration. So far we discussed the S.I. calculation on a yearly basis. Let us learn more about the formulas relating to months in terms of 3, 4, 6, 9 months. In general;
$$S.I.=\frac{\left(P\times R\times T\right)}{100}$$
Here T is for the number of years.
The formula for S.I. modifies to;
$$S.I.=\frac{\left(P\times R\times x\right)}{12\times100}$$
Here x=Number of months
For example, if x=4 that is for 4 months, the formula would be:
$$S.I.=\frac{\left(P\times R\times4\right)}{12\times100}=\frac{\left(P\times R\right)}{3\times100}$$
Similarly, for half-yearly that is for 6 months:
$$S.I.=\frac{\left(P\times R\times6\right)}{12\times100}=\frac{\left(P\times R\right)}{2\times100}$$
## Simple Interest vs Compound Interest
There are two types of interest: simple and compound interest. In this particular article, our focus was on S.I from all corners. Let us understand the difference between simple interest and compound interest.
Simple Interest Compound Interest The S.I is computed on the original principal amount every time. Compound interest when compared to SI is determined on the collected sum of principal and interest. The principal amount is constant in such a type of interest. The principal amount is constantly varying during the complete borrowing period. SI is calculated by the formula:$$S.I.=\frac{\left(P\times R\times T\right)}{100}$$ CI is determined by the formula:$$C.I.=P\left(1+\frac{R}{100}\right)^T-P$$ The S.I. is identical for every year on a certain principal taken. The compound interest varies per year for the span of the time as it is determined on the amount and not principal. An amount if invested using SI, gives a lesser return. The same amount when invested using CI gives much higher returns.
## Key Takeaways of Simple Interest
It is straightforward to calculate SI using the formula. Below are some of the takeaways regarding the topic:
• SI profits customers who repay their loans on time/ early each month.
• Auto loans and short-term individual loans are normally SI based loans.
• SI is estimated on the Principal Amount for the complete tenure.
• Principal Amount prevails constantly in SI.
• The amount of interest is always more for the case of compound interest when compared to simple interest.
• The earned interest on the given principal is not included to determine the interest for the next term.
• The collection of interest is slow in this approach.
• A simple type of interest is advantageous to the borrower when compared to the lender as a borrower will be spending slightly less on a loan that is taken on S.I.
## Solved Examples on Simple Interest
Having a thorough knowledge of S.I. definition, formulas concerning yearly and monthly along with the knowledge of terms like principal, rate, interest, amount and time. Let us step towards some simple interest questions for better understanding.
Solved Question 1: Determine the S.I. for a given principal amount of Rs. 4000, the duration is 2 years and the rate is 20%.
Solution:
Given terms;
P = 4000
R = 20%
T = 2 years
SI =?
Using the formula for SI;
$$S.I.=\frac{\left(P\times R\times T\right)}{100}$$
$$S.I.=\frac{\left(4000\times20\times2\right)}{100}$$
SI = 1600 rupees.
The same question can be solved for different years, you can check the below table for the same question being solved for different years.
Time in YearCalculationSimple InterestAmonut=Principal+Interest
1SI=(4000X1X20)/1008004800
3SI=(4000X3X20)/10024006400
5SI=(4000X5X20)/10040008000
7SI=(4000X7X20)/10056009600
9SI=(4000X7X20)/100720011200
10SI=(4000X10X20)/100800012000
Solved Question 2: Determine the simple interest for a given principal amount of Rs. 2000, the duration is 3 months and the rate of interest is 10%.
Solution: Given terms;
P = 2000
R = 10%
T = 3 months
SI = ?
Using the formula for SI;
$$S.I.=\frac{\left(P\times R\times x\right)}{12\times100}$$
Here x=Number of months
$$S.I.=\frac{\left(2000\times10\times3\right)}{12\times100}$$
⇒ 50
Solved Question 3: If the final amount on a certain amount of money becomes Rs. 720 in 2 years and Rs. 1020 in another 5 years in simple interest, then what is the annual rate of interest?
Solution: Formula used:
$$S.I.=\frac{\left(P\times R\times T\right)}{100}$$
Where P = principal
R = rate of interest
T = time
Principal = Amount – Interest
Calculation:
Money become 720 in 2 years and becomes 1020 in another 5 years
⇒ Interest in 5 years = (1020 – 720) = 300
⇒ interest in 1 year = 300/5 = 60
⇒ Interest in 2 year = 60 × 2 = 120
We are given that money becomes Rs. 720 in 2 years
Principal = Amount – Interest
Principal = 720 – 120 = 600
Let, rate of interest = r%
Accordingly,
(600 × 2 × r)/100 =120
⇒ r = 10
∴ The rate of interest is 10%
Solved Question 4: A sum of Rs. 4000 is lent on simple interest at the rate of 10% per annum. The S.I. for 5 years is how much more than the S.I. for 3 years?
Solution:Given:
A sum of Rs.4000 is lent on S.I. at the rate of 10% per annum
P= 4000
R = 10%
Formula used:
$$S.I.=\frac{\left(P\times R\times T\right)}{100}$$
Calculation:
S.I for 5 years = (4000 × 5 × 10)/100 = 2000
S.I for 3 years = (4000 × 3 × 10)/100 = 1200
⇒ Difference between S.I for 5 years and S.I for 3 years = 2000 – 1200 = 800
∴ S.I. for 5 years is 800 more than S.I. for 3 years.
Here you can get more solved example Questions of Simple Interest.
Stay tuned to the Testbook app or visit the testbook website for more updates on similar topics from mathematics, and numerous such subjects, and can even check the test series available to test your knowledge regarding various exams.
If you are checking the Simple Interest article, also check the related maths articles in the table below: Arithmetic Mean Probability Statistics Integral Calculus Partnership Locus
## Simple Interest FAQs
Q.1 What is simple interest?
Ans.1 S.I. is computed on the principal amount for the given period at provided rate.
Q.2 What is the difference between simple interest and compound interest?
Ans.2 S.I. is stated as the interest amount for a particular principal amount at some rate of interest for some duration. In reverse, compound interest is the interest computed on the principal and the interest accumulated over the preceding period.
Q.3 What are the types of interest?
Ans.3 In general, there are two types of interests we deal with and they are simple interest and compound interest.
Q.4 What is the formula for amount calculation?
Ans.4 Amount = Principal + SI
Q.5 How can one calculate the principal amount if SI, rate, and time are given?
Ans.5 If the S.I. rate, and time are given and the principal is asked then it can be determined through the formula;
Principal = SI × 100 /Rate × Time
Q.6 What is the formula for simple interest?
Ans.6 SI = (P × R × T) / 100
Where the terms:
P indicates the principal amount.
R is the rate of interest per annum.
T defines the time duration in years. |
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# The most general values of x for which $\sin x + \cos x = \mathop {\min }\limits_{a \in {\Bbb R}} \left\{ {1,{a^2} - 4a + 6} \right\}$ are given by: \eqalign{ & A.\,\,2n\pi ;n \in {\Bbb Z} \cr & B.\,\,2n\pi + \dfrac{\pi }{2};n \in {\Bbb Z} \cr & C.\,\,n\pi + {( - 1)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{4};n \in {\Bbb Z} \cr & D.\,\,2n\pi - \dfrac{\pi }{2};n \in {\Bbb Z} \cr}
Last updated date: 14th Apr 2024
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Hint: First we have to find the minimum value, that is we have to simplify the right hand side.Then simplify the equation to find the range of x there so that the equation holds.
Complete step by step solution:
Step1: We know that the value of a perfect square is always non-negative. So the minimum value may be zero.
\eqalign{ & {a^2} - 4a + 6 \cr & = {(a - 2)^2} + 2 \cr}
Then taking the value of the perfect square to zero.
\eqalign{ & {a^2} - 4a + 6\,\,is\,\,\operatorname{minimum} \,\,for\,\,a = 2 \cr & so,\mathop {\min }\limits_{a \in { R}} \left\{ {{a^2} - 4a + 6} \right\} = {(2 - 2)^2} + 2 = 2 \cr}.
Step2: Simplifying the right hand side, we get
$Now,\mathop {\,\,\min }\limits_{a \in {l R}} \left\{ {1,{a^2} - 4a + 6} \right\} = 1$
Step3: Now from the equation, we have
\eqalign{ & \sin x + \cos x = 1 \cr & or,\dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{1}{{\sqrt 2 }},\,\,both\,\,side\,\,dividing\,\,by\,\,\sqrt 2 \cr & or,\sin x\cos \dfrac{\pi }{4} + \cos x\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} \cr & or,\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} \cr & or,\sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4} \cr & or,x + \dfrac{\pi }{4} = n\pi + {( - 1)^n}.\dfrac{\pi }{4};n \in {\Bbb Z} \cr & or,x = n\pi + {( - 1)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{4};n \in {\Bbb Z} \cr}
Hence,The most general values of x are here, $n\pi + {( - 1)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{4};n \in {\Bbb Z}$
Therefore,option C) is correct.
Note:
Here we use the formula $\sin x = \sin \alpha \Rightarrow x = n\pi + {( - 1)^n}\alpha ;n \in {\Bbb Z},\,the\,\,set\,\,of\,\,all\,\,\operatorname{integers}$. These are general values of x for which $\sin x = \sin \alpha$. If we want to solve such types of equations, we have to find out general values or all values satisfying that trigonometric equation. |
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# Sphere, Ellipsoid, and Torus Help (page 2)
(not rated)
#### PROBLEM 1
Suppose a football field is to be covered by an inflatable dome that takes the shape of a half-sphere. If the radius of the dome is 100 meters, what is the volume of air enclosed by the dome in cubic meters? Find the result to the nearest 1000 cubic meters.
#### SOLUTION 1
First, find the volume V of a sphere whose radius is 100 meters, and then divide the result by 2. Let π = 3.14159. Using the formula with r = 100 gives this result:
V = (4 × 3.14159 × 100 3 )/3
= (4 × 3.14159 × 1,000,000)/3
= 4,188,786.667 . . .
Thus V /2 = 4,188,786.667/2 = 2,094,393.333. Rounding off to the nearest 1000 cubic meters, we get 2,094,000 cubic meters as the volume of air enclosed by the dome.
#### PROBLEM 2
Suppose the dome in the previous example is not a half-sphere, but instead is a half-ellipsoid. Imagine that the height of the ellipsoid is 70 meters above its center point, which lies in the middle of the 50-yard line at field level. Suppose that the distance from the center of the 50-yard line to either end of the dome, as measured parallel to the sidelines, is 120 meters, and the distance from the center of the 50-yard line, as measured along the line containing the 50-yard line itself, is 90 meters. What is the volume of air, to the nearest 1000 cubic meters, enclosed by this dome?
#### SOLUTION 2
First, consider the radii r 1 , r 2 , and r 3 in meters, with respect to the center point, as follows:
r 1 = 120
r 2 = 90
r 3 = 70
Then use the formula for the volume V of an ellipsoid with these radii:
V = (4 × 3.14159 × 120 × 90 × 70)/3
= (4 × 3.14159 × 756,000)/3
= 3,166,722.72
Thus V /2 = 3,166,722.72/2 = 1,583,361.36. Rounding off to the nearest 1000 cubic meters, we get 1,583,000 cubic meters as the volume of air enclosed by the half-ellipsoidal dome.
Practice problems for these concepts can be found at: Surface Area And Volume Practice Test.
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# 6.3 Centripetal force
Page 1 / 10
• Calculate coefficient of friction on a car tire.
• Calculate ideal speed and angle of a car on a turn.
Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.
Any net force causing uniform circular motion is called a centripetal force . The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net $\text{F}=\text{ma}$ . For uniform circular motion, the acceleration is the centripetal acceleration— $a={a}_{c}$ . Thus, the magnitude of centripetal force ${\text{F}}_{\text{c}}$ is
${\text{F}}_{\text{c}}={m\text{a}}_{\text{c}}.$
By using the expressions for centripetal acceleration ${a}_{c}$ from ${a}_{c}=\frac{{v}^{2}}{r};\phantom{\rule{0.25em}{0ex}}{a}_{c}={\mathrm{r\omega }}^{2}$ , we get two expressions for the centripetal force ${\text{F}}_{\text{c}}$ in terms of mass, velocity, angular velocity, and radius of curvature:
${F}_{c}=m\frac{{v}^{2}}{r};\phantom{\rule{0.25em}{0ex}}{F}_{c}=\text{mr}{\omega }^{2}.$
You may use whichever expression for centripetal force is more convenient. Centripetal force ${F}_{\text{c}}$ is always perpendicular to the path and pointing to the center of curvature, because ${\mathbf{a}}_{c}$ is perpendicular to the velocity and pointing to the center of curvature.
Note that if you solve the first expression for $r$ , you get
$r=\frac{{\mathrm{mv}}^{2}}{{F}_{c}}\text{.}$
This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.
## What coefficient of friction do car tires need on a flat curve?
(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.
(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see [link] ).
Strategy and Solution for (a)
We know that ${F}_{\text{c}}=\frac{{\mathrm{mv}}^{\text{2}}}{r}$ . Thus,
${F}_{\text{c}}=\frac{{\mathrm{mv}}^{\text{2}}}{r}=\frac{\left(\text{900 kg}\right)\left(\text{25.0 m/s}{\right)}^{\text{2}}}{\left(\text{500 m}\right)}=\text{1125 N.}$
Strategy for (b)
[link] shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is ${\mu }_{\text{s}}N$ , where ${\mu }_{\text{s}}$ is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so that $N=\mathit{mg}$ . Thus the centripetal force in this situation is
${F}_{\text{c}}=f={\mu }_{\text{s}}N={\mu }_{\text{s}}\text{mg}\text{.}$
Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for ${F}_{\text{c}}$ from the equation
$\begin{array}{c}{F}_{\text{c}}=m\frac{{v}^{2}}{r}\\ {F}_{\text{c}}=\text{mr}{\omega }^{2}\end{array}\right\},$
#### Questions & Answers
What is electric
a boy cycles continuously through a distance of 1.0km in 5minutes. calculate his average speed in ms-1(meter per second). how do I solve this
speed = distance/time be sure to convert the km to m and minutes to seconds check my utube video "mathwithmrv speed"
PhysicswithMrV
why we cannot use DC instead of AC in a transformer
becuse the d .c cannot travel for long distance trnsmission
ghulam
what is physics
branch of science which deals with matter energy and their relationship between them
ghulam
Life science
the
what is heat and temperature
how does sound affect temperature
sound is directly proportional to the temperature.
juny
how to solve wave question
I would like to know how I am not at all smart when it comes to math. please explain so I can understand. sincerly
Emma
Just know d relationship btw 1)wave length 2)frequency and velocity
Talhatu
First of all, you are smart and you will get it👍🏽... v = f × wavelength see my youtube channel: "mathwithmrv" if you want to know how to rearrange equations using the balance method
PhysicswithMrV
nice self promotion though xD
Beatrax
thanks dear
Chuks
hi pls help me with this question A ball is projected vertically upwards from the top of a tower 60m high with a velocity of 30ms1.what is the maximum height above the ground level?how long does it take to reach the ground level?
mahmoud
please guys help, what is the difference between concave lens and convex lens
convex lens brings rays of light to a focus while concave diverges rays of light
Christian
for mmHg to kPa yes
Matthew
it depends on the size
please what is concave lens
Vincent
a lens which diverge the ray of light
rinzuala
concave diverges light
Matthew
thank you guys
Vincent
A diverging lens
Yusuf
What is isotope
Yusuf
each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element. "some elements have only one stable isotope
Karthi
what is wire wound resistors?
What are the best colleges to go to for physics
I would like to know this too
Trevor
How do I calculate uncertainty in a frequency?
Calculate . ..
Olufunsho
What is light wave
What is wave
Sakeenah
What is light
Sakeenah
okay
True
explain how neurons communicate feed and stimulate
Jeff
Great science students
Omo
A wave is a disturbance which travels through the medium transferring energy from one form to another without causing any permanent displacement of d medium itself
OGOR
Light is a form o wave
OGOR
Neurons communicate by sending message through nerves in coordination
OGOR
What are petrochemicals, give two examples
OGOR
light has dual nature, particle as well as wave. when we want to explain phenomena like Interference of light, then we consider light as wave.
Lalita
what is it as in the form of it or how to visualize it or what it contains
Matthew
particles of light are like small packets of energy called photons, and flow or motion of photons is wave like
Lalita
light is just the energy of which photons emit
Matthew
the wave is how they travel
Matthew
photons do not emitt energy, they are energy. They are massless particles.
Lalita
a wave is a disturbance through the medium. Have you ever thrown a stone in still water? the disturbance produced travels in form of wave, the wave produced by throwing stone in still water are circular in nature.
Lalita
a photon does contain mass when in motion. it doesnt contain mass when at rest
Matthew
when would it ever be at rest
Bob
a wave is a disturbance of which energy travels
Matthew
that's darkness. darkness has no mass because the photons within in aren't moving or producing energy
Matthew
Hi guys. Please I've been trying to understand the concept of SHM, but it's not been really easy, could someone please explain it to me or suggest a site I could visit? Thank you.
Odo
Matthew
effective mass of photons only comes into picture when we consider it accelerating in gravitational field, mass of photon has no meaning as it is always travelling with speed of light and is never at rest. with that high speed, Energy and momentum are equivalent. and darkness is absense of photons.
Lalita
darkness is absense of light. not the presence of 'resting photons'. photons are never at rest.
Lalita
photons are present in darkness but don't give off any light because they are stationary with no mass or energy. once a force makes them move again they will gain mass and give off light
Matthew
this theory is presented in Einsteins theory of special relativity
Matthew
A.The velocity Vo for the streamline flow of liquid in a small tube depends on the radius r of the tube,the density and the viscosity iter of the liquid .use the dimensional analysis to obtain an expression for the velocity . B.Given that Vo =r square ×p all over 4×iter ×l
True
A.The velocity Vo for the streamline flow of liquid in a small tube depends on the radius r of the tube,the density (rho)and the viscosity (iter)of the liquid. Use the method of dimensional analysis to obtain an expression for the velocity . B.Given that Vo =r square x p all over 4 x iter x l
True
Matthew, photons ARE light. there is no such thing as a photon that isn't moving. in fact the speed they move at is called C (for constant) in physics. through a vacuum they always travel at this speed no matter what. they can not slow down; except in another medium.
The reason why a photon can go at this speed is BECAUSE it had no mass. nothing can go this speed or faster because it needs to have no mass or negative mass. that's why it's called the constant.
when a photon hits something that is opaque, this is the only way to "stop"it. it isn't merely stopped but absorbed and turned into heat energy, then the remaining energy is reflected in different wavelengths. that reflection is what we call color. the darker something is, the less photons are ther
e. complete blackness is the absolute absence of photons altogether. I believe what you're referring to is not speed, but wavelength, which is indirectly proportional to the amount of energy a particular photon is made up of.
in order for a photon to have zero wavelength, it must (at least theoretically) have infinite energy.
about mass: you may have photons confused with electrons. elections have a mass so small that people say they are without mass, but they do. it is called electron mass or Me-.
you may also be getting electrons and photons confused because of the cherenkov effect. that is what happens when a particle travels faster than light IN THAT PARTICULAR MEDIUM. I emphasize that because no other particle besides photons can go the speed of c.
when a particle goes faster than light in a particular medium, a blue light is emitted, called cherenkov radiation. this is why nuclear reactors glow blue.
nuclear reactors release so much energy that when they emit electrons, those electrons are given enough energy to go faster than light in that medium (in this case water), releasing blue light. if you put the reactor in air or a vacuum, this effect wouldn't happen because the speed of light in air
is very close to c, which is the universal speed limit. I'd you did go faster than c, time would go backwards and you would have infinite theoretical mass and probably spagghettify, like with a black hole.
*if
*electrons
light waves can travel through a vacuum, and do not require a medium. In empty space, the wave does not dissipate (grow smaller) no matter how far it travels, because the wave is not interacting with anything else.
Salim
Please is there any instructional material for sounds Waves, Echo, light waves
Salami
how far there is hot topic that is boarding me now
Abraham
linear motion
Ahmed
kinematic
Abraham
tell us about it
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kinematic
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kinematics disscuss the motion without cuases ...
ghulam
wow I like what am seeing here I need someone to brush me up in physics in fact I'll say I know nothing
Godslight
How does the Geiger tube works
pls he do we find for tension
tension is equal to the weight of the object. so for example if something weighs 45 Newtons then the tension in the Rope holding it is 45 Newtons. and because it is in equilibrium if the object is 45N and there are three ropes holding it there would be 15 N of tension in each to equal the weight
Shii
does that work for you?
Shii
tnx
Belinda
very correct
Kudzy |
Assignment 4: Altitudes of a triangle
By Nikhat Parveen, UGA.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In this assignment we will prove the Altitude Concurrence Theorem:
"The lines of the three altitudes of a triangle are concurrent."
We will do this by using GSP for drawing the triangles and altitudes and then by providing mathematical reasoning for the statements to be proved.
Given Triangle ABC, L1 contains the altitude from B to AC, L2 contains the Altitude from C to AB, L3 contains the altitude from A to BC.
We want to prove that the lines L1, L2, L3 intersect at the same point, i.e.., L1, L2, L3 are concurrent. [Two or more lines are concurrent if and only if there is a single point at which they all intersect.]
Proof: Through each vertex of triangle ABC, there is a line parallel to the opposite side.
Reason: Parallel postulate. (If a point P is not on line L, there is exactly one line through P parallel to L.)
These three lines determine a triangle DEF.
Reason: Parallels to intersecting lines also intersect.
If you look closely at the figure ACBD and ACFB is a parallelogram. Why?? Because the opposite sides are parallel.
Now, AC = DB, AC = BF Reason..... opposite sides of a parallelogram are congruent.
DB = BF by substitution.
L1 is perpendicular to BF.
Reason: In a plane, a line perpendicular to one of two parallel lines is perpendicular to the other.
Now where does it lead us....?
L1 is the perpendicular bisector of DF. why is that?? because DB = BF , and L1 is perpendicular to DF
Likewise, L2 and L3 are the perpendicular bisectors of FE and DE by the same reasoning applied to L1.
The perpendicular bisectors of the sides of triangle DEF are concurrent by the Perpendicular Bisector concurrence theorem.
Since the lines containing the altitude of triangle ABC are the perpendicular bisectors of the sides of triangle DEF, therefore the lines containing the altitudes of triangle ABC are concurrent.
**** The point of concurrency of the lines containing the altitudes is called the orthocenter of the triangle ****
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ |
An expression represents a number. For example, 6 - 2 is an expression that represents the number 4, and 3×5 is an expression that represents the number 15. This section will discuss how to find the unique number that each expression represents.
Consider the expression 2 + 4×3. How might one search for the answer? One way is to start by adding 2 + 4 = 6 and then multiply 6×3 = 18. Another way is to first multiply 4×3 = 12 and then add 2 + 12 = 14. Only one of these answers can be correct. So which is it?
The solution lies in following the Order of Operations. This rule specifies an order in which to add, subtract, multiply and divide so that everyone can look at an expression and get the same correct answer.
There are three steps to finding the answer, or to evaluating the expression, as specified by the order of operations:
Step 1. Carry out the operations within parentheses.
Step 2. Multiply and divide (it does not matter which comes first).
Step 3. Add and subtract (it does not matter which comes first).
For example, to evaluate (3 + 2)×5 + (7 - 3), go through the steps:
Step 1 (Parentheses). (3+2)×5 + (7-3) = 5×5 + 4
Step 2 (Multiplication and Division). 5×5 +4 = 25 + 4
Step 3 (Addition and Subtraction). 25+4 = 29
Thus, (3 + 2)×5 + (7 - 3) = 29
In the example at the beginning of this section, 2 + 4×3, the steps are:
Step 1. 2 + 4×3 = 2 + 4×3 (There are no parentheses)
Step 2. 2 + 4×3 = 2 + 12
Step 3. 2+12 = 14
Thus, 2 + 4×3 = 14.
Sometimes the expression within the parentheses might contain a combination of multiplication, division, addition, and subtraction, or even more parentheses. If this is the case, go through the steps in the order of operations for the expression within the parentheses.
For example, to evaluate (12 - 6/(3 - 1))/(7 - 4) + 2:
Step 1.1 (12 - 6/(3-1))/(7 - 4) + 2 = (12 - 6/2)/(7 - 4) + 2
Step 1.2 (12 - 6/2)/(7 - 4) + 2 = (12 - 3)/(7 - 4) + 2
Step 1.3 (12-3)/(7-4) + 2 = 9/3 + 2
Step 2. 9/3 +2 = 3 + 2
Step 3. 3+2 = 5
Here are some more examples:
Example 1.12/(6 - 2) - (3×1) = ?
Step 1. 12/(6-2)-(3×1) = 12/4-3
Step 2. 12/4 -3 = 3 - 3
Step 3. 3-3 = 0
Example 2.(11 - 3×2) + 2×3×4 - (3 + 2) = ?
Step 1. 1. (11 - 3×2) + 2×3×4 - (3 + 2) = (11 - 3×2) + 2×3×4 - (3 + 2)
2. (11 - 3×2) + 2×3×4 - (3 + 2) = (11 - 6) + 2×3×4 - (3 + 2)
3. (11-6) + 2×3×4 - (3+2) = 5 +2×3×4 - 5
Step 2. 5 + 2×3×4 -5 = 5 + 24 - 5
Step 3. 5+24-5 = 24 |
# The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, find the nth term.
Given:
The 17th term of an A.P. is 5 more than twice its 8th term. The 11th term of the A.P. is 43.
To do:
We have to find the nth term.
Solution:
Let the first term of the A.P. be $a$ and the common difference be $d$.
We know that,
nth term of an A.P. $a_n=a+(n-1)d$
Therefore,
$a_{8}=a+(8-1)d$
$=a+7d$......(i)
$a_{17}=a+(17-1)d$
$=a+16d$
According to the question,
$a_{17}=2(a_8)+5$
$a+16d=2(a+7d)+5$
$16d+a=2a+14d+5$
$2a-a=16d-14d-5$
$a=2d-5$.....(i)
$a_{11}=a+(11-1)d$
$43=a+10d$
$43=2d-5+10d$ (From (i))
$12d=43+5$
$d=\frac{48}{12}$
$d=4$
Substituting $d=4$ in (i), we get,
$a=2(4)-5$
$a=8-5$
$a=3$
Therefore,
nth term $a_n=5+(n-1)(4)$
$=3+4n-4$
$=4n-1$
Hence, the nth term of the given A.P. is $4n-1$.
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## Algebra (all content)
### Course: Algebra (all content)>Unit 10
Lesson 6: Multiplying binomials
# Multiplying binomials review
A binomial is a polynomial with two terms. For example, x, minus, 2 and x, minus, 6 are both binomials. In this article, we'll review how to multiply these binomials.
### Example 1
Expand the expression.
left parenthesis, x, minus, 2, right parenthesis, left parenthesis, x, minus, 6, right parenthesis
Apply the distributive property.
\begin{aligned}&(\blueD{x-2})(x-6)\\ \\ =&\blueD{x}(x-6)\blueD{-2}(x-6)\\ \end{aligned}
Apply the distributive property again.
equals, start color #11accd, x, end color #11accd, left parenthesis, x, right parenthesis, plus, start color #11accd, x, end color #11accd, left parenthesis, minus, 6, right parenthesis, start color #11accd, minus, 2, end color #11accd, left parenthesis, x, right parenthesis, start color #11accd, minus, 2, end color #11accd, left parenthesis, minus, 6, right parenthesis
Notice the pattern. We multiplied each term in the first binomial by each term in the second binomial.
Simplify.
\begin{aligned} =&x^2-6x-2x+12\\\\ =&x^2-8x+12 \end{aligned}
### Example 2
Expand the expression.
left parenthesis, minus, a, plus, 1, right parenthesis, left parenthesis, 5, a, plus, 6, right parenthesis
Apply the distributive property.
\begin{aligned} &(\purpleD{-a+1})(5a+6)\\\\ =&\purpleD{-a}(5a+6) +\purpleD{1}(5a+6) \end{aligned}
Apply the distributive property again.
equals, start color #7854ab, minus, a, end color #7854ab, left parenthesis, 5, a, right parenthesis, start color #7854ab, minus, a, end color #7854ab, left parenthesis, 6, right parenthesis, plus, start color #7854ab, 1, end color #7854ab, left parenthesis, 5, a, right parenthesis, plus, start color #7854ab, 1, end color #7854ab, left parenthesis, 6, right parenthesis
Notice the pattern. We multiplied each term in the first binomial by each term in the second binomial.
Simplify:
minus, 5, a, squared, minus, a, plus, 6
## Practice
Problem 1
• Current
Simplify.
left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, minus, 6, right parenthesis
Want more practice? Check out this intro exercise and this slightly harder exercise.
## Want to join the conversation?
• Is it okay if I use this method? :
(9h+3)(-h-1)
-h(9h+3)-1(9h+3)
-9h^2-3h-9h-3
-9h^2-12h-3
This is what Sal showed in the multiplying binomials video.
• Yes, you can double distribute with either the first or the second binomial, and you did it correctly. The answer will be the same if you did 9h(-h-1) + 3(-h-1), just that on your 3rd step, the -9h and -3h would be interchanged.
• I am working on foiling section of my math the problem looks like (5x+2Y)(4X+Y) Book says answer is 20Xsquared+13xy+2y: I get all parts of the first and last terms in the answer statement but I have know idea how it foils out a 13xy term??
• You have to multiply every term by BOTH of the other terms in the parentheses. 5x*4x(20x^2) + 5x*y(5xy)+4x*2y(8xy)+ y*2y(2y^2) =20x^2 + 13xy + 2y^2.
• whats bigger than a quadratic equation?
• Quadratic equations are equations with a variable to the second power.
Cubic equations have something to the third, and quartic equations have a variable to the fourth. Quintic equations have a variable to the fifth, but they are unsolvable.
• For those interested, Q2 can be simplified further (although the question doesn't want it).
(-6d+6)(2d-2)
= -6(d-1)x2(d-1)
= -12(d-1)(d-1)
= -12(d-1)^2
• (4ab+2) (3ab-7)?
• Does this mention trinomials
(1 vote)
• It does not mention trinomials but you can use the same method for them.
(1 vote)
• Is this method the same one as the foil one?
(1 vote)
• The method is same as foil but the better method is the method said by Sal
• how is this going to help me count my money or get a job
(1 vote)
• That depends on what job you have - most majors in college require some math to graduate, so counting money as a burger flipper is easier than as an engineer because there is so much less to count. Even doctors and lawyers who have a lot of money to count must take math to get to where they are.
• How do you visualize (x-4)(x+7) as an area? I tried it with one side of the area marked as x-4 + (4) and the other as x + 7 but couldn't get the correct answer.
• + --- (x-4) --- +
|
|
(x+7)
|
|
+ --- (x-4) --- +
better later than never I guess... |
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# BIDMAS Questions with Answers and Solutions
Updated: Sep 4, 2021
BIDMAS rules are important for making sense of complex equations.
Students can often run into problems when faced with what seem at first like fairly easy mathematical problems. Consider, for example, the equation 5 + 2 × (3 + 3) = y. What is the solution? Some might say y = 60, while others might say y = 17. The problem here is that we have multiple mathematical ‘operations’ (processes such as addition, subtraction, multiplication, and division) which people can choose to perform in a different order. This is where BIDMAS comes in. BIDMAS, or the ‘order of operations’, provides an easy guide which tells you how to break down these complex equations and solve them correctly. This handy post should help you to get to grips with how you can use BIDMAS to avoid confusions in your work.
BIDMAS specifies the order in which operations should be performed.
## What is BIDMAS?
BIDMAS is an acronym that stands for Brackets, Indices, Division, Multiplication, Addition, Subtraction. This list gives the order in which you should perform operations in an equation. You may be unfamiliar with the term ‘indices’ here. Indices are operations such as squaring, cubing, or anything of the form ‘x to the power of y’.
BIDMAS tells us to always perform operations inside brackets first, and then to perform any operation before or after the brackets with whatever number results from the operation inside the brackets. Next, operations including powers should be performed before other operations. Finally, division, multiplication, addition, and subtraction operations should be performed in that order.
It is best to solve complex equations by breaking them down into simple operations.
## Example of using BIDMAS
Consider this equation: 2 + (5 − 3)² × 4 = y
In order to reach a solution, we must perform four operations: addition, subtraction, squaring, and multiplication. BIDMAS can help us to break this complex equation down into its individual operations so that we can then to perform them in order.
First, we must consider the brackets. The operation in brackets is 5 – 3, which leaves us with 2.
Now our equation looks like this: 2 + (2)² × 4 = y
Secondly, we must deal with the indices. Here we have 2² which is equal to 4.
We are then left with this: 2 + 4 × 4 = y
The last two operations are a multiplication and an addition. BIDMAS tells us to do the former first. Therefore, we start with 4 × 4 which gives us 16.
We are then left with the equation, 2 + 16 = y, which we can easily solve to give us our solution: y =18.
Hopefully, this has cleared up a bit how BIDMAS can be used to work through a complex equation without getting confused or making mistakes.
Simple equations can often cause controversy on social media due to confusion over BIDMAS rules.
## Other hints and tips about BIDMAS
Over the last few years, an equation has been circling Facebook and Twitter that has caused a lot of arguments. Take a look at it: 8 ÷ 2(2 + 2) = y. Some people think that y = 1, while others think y = 16. We can use BIDMAS to help us to find out which answer is correct.
First, we perform the operation within parentheses. This leaves us with: 8 ÷ 2(4) = y.
Then, we perform the division. This leaves u with: 4(4) = y.
When a number comes before a set of brackets, this means that we must multiply that number by whatever is in the brackets. As such, we must perform 4 × 4, which leaves us with y = 16
Why, then, do some people think that the answer is y = 1? The problem comes at the last step. Because this equation leaves out the multiplication symbol, people don’t see this operation as a normal multiplication just like any other. This leads to a confusion whereby the multiplication is performed before the division, which goes against the BIDMAS rules we looked at above. It is important to recognise this way of writing multiplications so that you don't make the same mistake in your own work.
Sometimes when studying science subjects, especially physics, you might be asked to formulate your own equations using information provided to you along with whatever equations are involved in the scientific theory that you’re studying. These equations can often involve multiple different operations. As such, you can be easily led into the kind of BIDMAS problems that we’ve looked at so far. One helpful way to avoid this is to simply put all operations in brackets. As brackets always have priority in BIDMAS, you won’t have to remember the other order of operations when dealing with equations.
So, for example, instead of writing ‘y = 4 × 10 ÷ 5 + 2’, you could write ‘y = (4 × (10 ÷ 5)) + 2’. Doing this will make it clear which order you must perform the multiplication, division, and addition in; simply by looking at the equation. In this way, you won't have to try and remember the order of operations from memory in stressful situations where other considerations are more important.
Using a lot of brackets when writing your own equations can help to avoid confusion.
## Don’t be scared of brackets
Formulating and solving equations which contain multiple operations can sometimes lead students into easily avoidable confusions. The BIDMAS rules outlined here provide a simple guide which can help students to break down these complex equations and to solve them in the right way. These problems can catch out students particularly when they’re forming and solving equations in science exams, where the order of operations is not their primary concern. As such, using lots of brackets to break up these equations and to make visible the order in which you have to perform operations can be a real help.
You may explore resources available on StudySquare for other helpful hints and tips for successfully navigating the world of maths and science. Specifically for BIDMAS visit:
Get hold of our Exam Revision Guide and let’s turn your exam experience into a success story 😀 → https://www.studysquare.co.uk/pdf
Logic Enthusiast is an independent writer and is studying for an MA in Philosophy at the University of Edinburgh. He is particularly interested in Logic and the philosophy of science.
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# Relate Decimal and Fraction Multiplication
Examples, solutions, videos and lessons to help Grade 5 students learn how to relate decimal and fraction multiplication.
Common Core Standards: 5.NBT.7, 5.NF.4a, 5.NF.6, 5.MD.1, 5.NF.4b
New York State Common Core Math Module 4, Grade 5, Lesson 17, Lesson 18
Lesson 17 Problem Set
1. Multiply and model. Rewrite each expression as a multiplication sentence with decimal factors.
b. 4/10 × 3/10
d. 6/10 × 1.7
2. Multiply.
d. 6 × 0.3 = _______
e. 0.6 × 0.3 = _______
f. 0.06 × 0.3 = _______
g. 1.2 × 4 = _______
h. 1.2 × 0.4 = _______
i. 0.12 × 0.4 = _______
Lesson 17 Homework
1. Multiply and model. Rewrite each expression as a number sentence with decimal factors.
b. 6/10 × 2/10
d. 6/10 × 1.9
2. Multiply.
g. 1.3 × 5 = _______
h. 1.3 × 0.5 = _______
i. 0.13 × 0.5 = _______
Lesson 17 Homework
This video shows how to convert fractions to decimals and decimals to fractions within a the operation of multiplication. Area models and place value charts are used to concretely demonstrate the meaning within the process.
1. Multiply and model. Rewrite each expression as a number sentence with decimal factors.
a) 1/10 × 1/10
c) 1/10 × 1.6
2. Multiply.
a) 4 × 0.6
b) 7 × 0.3
3. Jennifer makes 1.7 liters of lemonade. If she pours 3 tenths of the lemonade in the glass, how many liters of lemonade are in the glass?
Lesson 17 Homework
1. Multiply and model. Rewrite each expression as a number sentence with decimal factors.
a) 1/10 × 1/10
c) 1/10 × 1.6
d) 6/10 × 1.9
Lesson 17 Homework
2. Multiply.
a. 4 × 0.6
b. 0.4 × 0.6
c. 0.04 × 0.6
d. 7 × 0.3
e. 0.7 × 0.3
f. 0.07 × 0.3
g. 1.3 × 5
h. 1.3 × 0.5
i. 1.3 × 0.05
3. Jennifer makes 1.7 liters of lemonade. If she pours 3 tenths of the lemonade in the glass, how many liters of lemonade are in the glass?
4. Cassius walked 6 tenths of a 3.6 mile trail.
a. How many miles did Cassius have left to hike?
b. Cameron was 1.3 miles ahead of Cassius. How many miles did Cameron hike already?
Lesson 18 Application Problem
An adult female gorilla is 1.4 meters tall when standing upright. Her daughter is 3 tenths as tall. How much more will the young female gorilla need to grow before she is as tall as her mother?
Lesson 18 Concept Development
Problem 1:
a. 3.2 × 2.1
b. 3.2 × 0.44
c. 3.2 × 4.21
Problem 2:
2.6 × 0.4
Problem 3:
a. 3.1 × 1.4
b. 0.31 × 1.4
Problem 4:
4.2 × 0.12
Lesson 18 Problem Set
1. Multiply
c. 6.6 × 2.8 =
d. 3.3 × 1.4 =
2. Solve using the standard algorithm. Use the thought bubble to show your thinking about the units of your product.
c. 8.31 × 2.4 = _______
3. A kitchen measures 3.75 m by 4.2 m.
a. Find the area of the kitchen.
b. The area of the living room is one and a half times that of the kitchen. Find the total area of the living room and the kitchen.
Lesson 18 Homework
5. A swimming pool at a park measures 9.75 m by 7.2 m.
a. Find the area of the swimming pool.
b. The area of the playground is one and a half times that of the swimming pool. Find the total area of the swimming pool and the playground.
Lesson 17 Homework
1. Multiply and model. Rewrite each expression as a number sentence with decimal factors.
1/10 × 1.6
Lesson 18 Homework
1. Multiply.
3.3 × 0.8
Lesson 18 Problem Set
Lesson 18 Homework
This video shows how to relate decimal and fraction multiplication using mainly algorithms.
1. Multiply.
a) 3.3 × 1.6
b) 3.3 × 0.8
2. Multiply.
b) 3.35 × 0.7
3. Solve using the standard algorithm. Use the thought bubble to show your thinking about the units of your product.
a. 3.2 × 0.6 = __________
b. 3.2 × 1.2 = __________
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. |
This exercise was generated from a Jupyter notebook. You can download the notebook here.
# Objective¶
In this tutorial, we will cover how to write a simple numerical integrator using the Forward Euler method to examine the dynamics of exponential growth.
# The Forward Euler Method¶
Developing simple ways to solve ordinary differential equations has long been an area of intense research. While deriving the analytical solution may be simple in some cases, it is often useful to solve them numerically, especially if slamming out the analytical solution will give you carpal tunnel.
While there are many ways to numerically integrate these equations, in this tutorial we will examine the Forward Euler method. Say we have an ordinary differential equation such as
$$\frac{dN}{dt} = r N(t) \tag{1}$$
as would be the case for exponential growth, where $r$ is some growth-rate constant and $t$ is time. Rather than solving this analytically (although it is trivial), we can solve it numerically by starting at some given value of $N$, evaluating Equation (1) for a given time step $\Delta t$, and updating the new value of $N$ at this new time $t+ \Delta t$. We can phrase this mathematically as
$$N(t+ \Delta t) = N(t) + rN(t) \Delta t .\tag{2}$$
Say our initial value ($N$ at $t=0$) is $N=10$ and $r=1$. We can take a time step $\Delta t=0.1$ and find that the change in value of $N$ is
$$\Delta N = rN\Delta t = 1. \tag{3}$$
We can then compute the new value of $N$ at time $t+\Delta t$ as
$$N(t+\Delta t) = N(t) + \Delta N = 10 + 1 = 9.\tag{4}$$
We can then take another step forward in time and repeat the process for as long as we would like. As the total time we'd like to integrate over becomes large, it becomes obvious why using a computer is a far more attractive approach than scribbling it by hand.
A major point to be wary of is the instability of this method. The error in this scales with the square of our step size. We must choose a sufficiently small step in time such that at most only one computable event must occur. For example, if we are integrating exponential growth of bacterial cells, we don't want to take time steps larger than a cell division! This requirement is known as the Courant-Friedrichs-Lewy condition and is important for many different time-marching computer simulations.
As is often the case, the best way to learn is to do. Let's give our digits some exercise and numerically integrate this exponential growth differential equation.
# Numerically integrating exponential growth¶
In [1]:
# Import the necessary modules
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
# For pretty plots
import seaborn as sns
rc={'lines.linewidth': 2, 'axes.labelsize': 14, 'axes.titlesize': 14, \
'xtick.labelsize' : 14, 'ytick.labelsize' : 14}
sns.set(rc=rc)
In order to numerically integrate Equation (1), we first need to specify a few parameters, such as the initial value of $N$, the growth rate $r$, and the time step $\Delta t$.
In [2]:
# parameters for our ODE
N_0 = 1
r = 0.03 # min^-1
# parameters for our integration
dt = 0.1 # min
total_time = 120 # min
Now we can set up an array into which we can save our values of $N$ as they increase with time, with the first value in the N_t array being set to N_0 as specified above.
In [3]:
# determine the number of steps that will be taken
num_steps = int(total_time/dt)
# initilize an array of length num_steps into which to store values of N
N_t = np.zeros(num_steps)
N_t[0] = N_0
Now we can actually compute the numerical integration, by looping through the N_t array and filling in the values of N as we go. At each time point $N(t) = N(t-\Delta t) + rN(t-\Delta t) \Delta t,$ where $t - \Delta t$ refers to the previous entry in the N_t array.
In [4]:
# numerically integrate by looping through N_t
for t in range(1,num_steps):
# first calculate dN, using pevious N_t entry
dN = N_t[t-1] * r * dt
# update current N_t entry
N_t[t] = N_t[t-1] + dN
And done! The numerical integration of more than a thousand time steps just happened nearly instantly, showing the immense value of using code to automate computational processes. Let's now plot the results. We currently have the $N$ values to plot, we just need to specify the $t$ values as well.
In [5]:
# make array of time values
times = np.arange(num_steps)*dt
# plot
plt.plot(times,N_t)
plt.xlabel("time (mins)")
plt.ylabel("N")
Out[5]:
Text(0,0.5,'N')
That certainly looks exponential! Let's plot the known solution of $N(t) = N_0 e^{rt}$ on top of our numerical integration for comparison.
In [6]:
# compute the known solution
soln = N_0 * np.exp(r*times)
# plot both our integration and the known solution
plt.plot(times,N_t)
plt.plot(times,soln)
plt.xlabel("time (mins)")
plt.ylabel("N")
plt.legend(["numerical integration", "known solution"])
Out[6]:
<matplotlib.legend.Legend at 0x1a1a5d4470>
Nice! Our numerical integration and the known solution are so similar that the two lines can't be distinguished from each other in the plot above. Recall how we said that picking a sufficiently small $\Delta t$ is necessary for numerical integration to work? From here, it's worth while to see how the plot above changes for different values of dt. As dt increases, our numerical integration deviates more and more from the known solution, with our integration systematically underestimating the true values. This is because with exponential growth, the rate of growth is always increasing and taking too large of a time step fails to capture this increase. |
# How do you find the x and y intercepts for x - 2y = 8?
Dec 4, 2015
To find the x-intercept set $y = 0$ in the equation and solve for $x$.
To find the y-intercept set $x = 0$ in the equation and solve for $y$
#### Explanation:
Using the given equation $x - 2 y = 8$ as an example:
Setting $\textcolor{red}{y = 0}$
$\textcolor{w h i t e}{\text{XXX}} x - 2 \times \textcolor{red}{0} = 8$
$\textcolor{w h i t e}{\text{XXX}} x = 8$
The x-intercept is $8$
Setting $\textcolor{b l u e}{x = 0}$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{0} - 2 y = 8$
$\textcolor{w h i t e}{\text{XXX}} - 2 y = 8$
$\textcolor{w h i t e}{\text{XXX}} y = - 4$
The y-intercept is $\left(- 4\right)$ |
# Trigonometry/Proof: Pythagorean Theorem
In a right triangle
The square of the hypotenuse
is equal to
the sum of the squares of the other two sides.
Commonly known as: ${\displaystyle c^{2}=a^{2}+b^{2}}$ (where ${\displaystyle c}$ is the hypotenuse) or ${\displaystyle {\text{hyp}}^{2}={\text{leg}}^{2}+{\text{leg}}^{2}}$
This theorem may have more known proofs than any other; the book The Pythagorean Proposition contains 370 proofs.[1]
## Proof by Subtraction
Proof using area subtraction of four identical right triangles
This proof uses rearrangement. The figure shows two identical large squares of side ${\displaystyle a+b}$ .
• The top square contains the square on the hypotenuse plus identical right triangles in its four corners.
• On bottom, the same large square holds the squares on the other two sides plus the same four right triangles, now moved to form two rectangles of sides ${\displaystyle a,b}$ in the bottom corners.
From both identical large squares, the area of the same four right triangles of sides ${\displaystyle a,b,c}$ is subtracted (colored). Subtracting the triangles removes the same (colored) area from the equal-area large squares, so the remaining white areas, ${\displaystyle c^{2}}$ and ${\displaystyle a^{2}+b^{2}}$ , are equal.
...and that's it!
## Euclid's Proof
Proof in Euclid's Elements
Euclid's proof is much more complex, and relies on subdividing a figure into pieces and showing that they are congruent pieces. It's a fragment of mathematical history. You do not need to remember this proof. In fact if it's the first time you're reading this book it's quite OK to skip over it and go on to "Exercise: A Puzzle Triangle"
Why is this proof here at all? Partly it's to show that there is more than one way of proving things. Partly it's because Euclid took great care to proceed in small steps each of which he had already proved. In the 'proof by subtraction' we are using facts about areas and how pieces fit together that are true, but that we haven't actually proved.
In outline, here is how the proof in Euclid's Elements proceeds. The large square is divided into a left and right rectangle. A triangle is constructed that has half the area of the left rectangle. Then another triangle is constructed that has half the area of the square on the left-most side. These two triangles are shown to be congruent, proving this square has the same area as the left rectangle. This argument is followed by a similar version for the right rectangle and the remaining square. Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. The details are next.
Let ${\displaystyle A,B,C}$ be the vertices of a right triangle, with a right angle at ${\displaystyle A}$ . Drop a perpendicular from ${\displaystyle A}$ to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.
For the formal proof, we require four elementary lemmata (a step towards proving the full proof):
1. If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent (side-angle-side).
2. The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
3. The area of a rectangle is equal to the product of two adjacent sides.
4. The area of a square is equal to the product of two of its sides (follows from 3).
Next, each top square is related to a triangle congruent with another triangle related in turn to one of two rectangles making up the lower square.[2]
Illustration including the new lines
The proof is as follows:
1. Let ${\displaystyle \triangle ACB}$ be a right-angled triangle with right angle ${\displaystyle \angle CAB}$ .
2. On each of the sides ${\displaystyle BC,AB,CA}$ squares are drawn, ${\displaystyle CBDE,BAGF,ACIH}$ in that order. The construction of squares requires the immediately preceding theorems in Euclid, and depends upon the parallel postulate.[3]
3. From ${\displaystyle A}$ , draw a line parallel to ${\displaystyle BD}$ and ${\displaystyle CE}$ . It will perpendicularly intersect ${\displaystyle BC}$ and ${\displaystyle DE}$ at ${\displaystyle K}$ and ${\displaystyle L}$ , respectively.
4. Join ${\displaystyle CF}$ and ${\displaystyle AD}$ , to form the triangles ${\displaystyle \triangle BCF,\triangle BDA}$ .
5. ${\displaystyle \angle CAB,\angle BAG}$ are both right angles; therefore ${\displaystyle C,A,G}$ are collinear. Similarly for ${\displaystyle B,A,H}$ .
Showing the two congruent triangles of half the area of rectangle ${\displaystyle BDLK}$ and square ${\displaystyle BAGF}$
1. Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.
2. Since AB and BD are equal to FB and BC, respectively, triangle ABD must be congruent to triangle FBC.
3. Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD, since it shares a height with BK and a base with BD and a triangle's area is half the product of its base and height.
4. Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
5. Therefore rectangle BDLK must have the same area as square BAGF = AB2.
6. Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2.
7. Adding these two results, AB2 + AC2 = BD × BK + KL × KC.
8. Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC.
9. Therefore AB2 + AC2 = BC2, since CBDE is a square.
...and we're done.
This proof, which appears in Euclid's Elements as that of Proposition 47 in Book 1,[4] demonstrates that the area of the square on the hypotenuse is the sum of the areas of the other two squares.[5]
## Credits
• With thanks to the Wikipedia page The Pythagorean Theorem which provided the initial version of this page. See that page for more notes and references.
## Notes
1. (Loomis 1968)
2. See for example Mike May S.J., Pythagorean theorem by shear mapping, Saint Louis University website Java applet
3. Jan Gullberg (1997). Mathematics: from the birth of numbers. W. W. Norton & Company. p. 435. ISBN 039304002X.
4. Elements 1.47 by Euclid. Retrieved 19 December 2006.
5. Euclid's Elements, Book I, Proposition 47: web page version using Java applets from Euclid's Elements by Prof. David E. Joyce, Clark University |
# How do you simplify 3/(2a)+1/(5b)?
Jun 14, 2018
$\frac{15 b + 2 a}{10 a b}$
#### Explanation:
$\text{we require the fractions to have a "color(blue)"common denominator}$
$\text{the lowest common multiple of "2a" and "5b" is } 10 a b$
$\text{multiply numerator/denominator of "3/(2a)" by } 5 b$
$\text{and numerator/denominator of "1/(5b)" by } 2 a$
$= \frac{15 b}{10 a b} + \frac{2 a}{10 a b}$
$\text{now add the numerators leaving the denominator}$
$= \frac{15 b + 2 a}{10 a b}$ |
# NCERT solutions for Class 10 Maths chapter 6 - Triangles [Latest edition]
## Chapter 6: Triangles
Exercise 6.1Exercise 6.2Exercise 6.3Exercise 6.4Exercise 6.5Exercise 6.6
Exercise 6.1 [Page 122]
### NCERT solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.1 [Page 122]
Exercise 6.1 | Q 1.1 | Page 122
Fill in the blanks using correct word given
All circles are __________.
• congruent
• similar
Exercise 6.1 | Q 1.2 | Page 122
Fill in the blanks using correct word given
All squares are __________.
• similar
• congruent
Exercise 6.1 | Q 1.3 | Page 122
Fill in the blanks using correct word given
All __________ triangles are similar.
• isosceles
• equilateral
Exercise 6.1 | Q 1.4 | Page 122
Fill in the blanks using correct word given in the brackets:−
Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
Exercise 6.1 | Q 2.1 | Page 122
Give two different examples of pair of Non-similar figures
Exercise 6.1 | Q 2.1 | Page 122
Give two different examples of pair of similar figures
Exercise 6.1 | Q 3 | Page 122
State whether the following quadrilaterals are similar or not:
Exercise 6.2 [Pages 128 - 129]
### NCERT solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2 [Pages 128 - 129]
Exercise 6.2 | Q 1.1 | Page 128
See the given Figure. DE || BC. Find EC
Exercise 6.2 | Q 1.2 | Page 128
See the given Figure. DE || BC. Find AD
Exercise 6.2 | Q 2.1 | Page 128
E and F are points on the sides PQ and PR respectively of a ΔPQR. For the following case, state whether EF || QR. PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Exercise 6.2 | Q 2.2 | Page 128
E and F are points on the sides PQ and PR respectively of a ΔPQR. For the following case, state whether EF || QR
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Exercise 6.2 | Q 2.3 | Page 128
E and F are points on the sides PQ and PR respectively of a ΔPQR. For the following case, state whether EF || QR. PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Exercise 6.2 | Q 3 | Page 128
In the following figure, if LM || CB and LN || CD, prove that (AM)/(AB)=(AN)/(AD)
Exercise 6.2 | Q 4 | Page 128
In the following figure, DE || AC and DF || AE. Prove that ("BF")/("FE") = ("BE")/("EC")
Exercise 6.2 | Q 5 | Page 129
In the following figure, DE || OQ and DF || OR, show that EF || QR
Exercise 6.2 | Q 6 | Page 129
In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Exercise 6.2 | Q 7 | Page 129
Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Exercise 6.2 | Q 8 | Page 129
Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Exercise 6.2 | Q 9 | Page 129
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that (AO)/(BO) = (CO)/(DO)
Exercise 6.2 | Q 10 | Page 129
The diagonals of a quadrilateral ABCD intersect each other at the point O such that (AO)/(BO) = (CO)/(DO) Show that ABCD is a trapezium
Exercise 6.3 [Pages 138 - 141]
### NCERT solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3 [Pages 138 - 141]
Exercise 6.3 | Q 1.1 | Page 138
State which pair of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Exercise 6.3 | Q 1.2 | Page 139
State which pair of triangles in the given figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Exercise 6.3 | Q 1.3 | Page 139
State which pair of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Exercise 6.3 | Q 1.4 | Page 139
State which pair of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Exercise 6.3 | Q 1.5 | Page 139
State which pair of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Exercise 6.3 | Q 1.6 | Page 139
State which pair of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Exercise 6.3 | Q 2 | Page 139
In the following figure, ΔODC ∼ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB
Exercise 6.3 | Q 3 | Page 139
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that (AO)/(OC) = (OB)/(OD)
Exercise 6.3 | Q 4 | Page 140
In the following figure, (QR)/(QS) = (QT)/(PR) and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Exercise 6.3 | Q 5 | Page 140
S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ∼ ΔRTS.
Exercise 6.3 | Q 6 | Page 140
In the following figure, if ΔABE ≅ ΔACD, show that ΔADE ∼ ΔABC.
Exercise 6.3 | Q 7.1 | Page 140
In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
ΔAEP ∼ ΔCDP
Exercise 6.3 | Q 7.2 | Page 140
In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
ΔABD ∼ ΔCBE
Exercise 6.3 | Q 7.3 | Page 140
In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
Exercise 6.3 | Q 7.4 | Page 140
In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
ΔPDC ∼ ΔBEC
Exercise 6.3 | Q 8 | Page 140
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB
Exercise 6.3 | Q 9. | Page 140
In the following figure, ABC and AMP are two right triangles, right-angled at B and M respectively, prove that:
1. ΔABC ~ ΔAMP
2. ("CA")/("PA") = ("BC")/("MP")
Exercise 6.3 | Q 10 | Page 140
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that
(i) (CD)/(GH) = (AC)/(FG)
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Exercise 6.3 | Q 11 | Page 141
In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ∼ ΔECF
Exercise 6.3 | Q 12 | Page 141
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see the given figure). Show that ΔABC ∼ ΔPQR.
Exercise 6.3 | Q 13 | Page 141
D is a point on the side BC of ∆ABC such that ∠ADC = ∠BAC. Prove that \frac{CA}{CD}=\frac{CB}{CA} or, CA^2 = CB × CD.
Exercise 6.3 | Q 14 | Page 141
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR
Exercise 6.3 | Q 15 | Page 141
A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Exercise 6.3 | Q 16 | Page 141
AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that ("AB")/("PQ") = ("AD")/("PM")
Exercise 6.4 [Pages 143 - 144]
### NCERT solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4 [Pages 143 - 144]
Exercise 6.4 | Q 1 | Page 143
Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC
Exercise 6.4 | Q 2 | Page 143
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Exercise 6.4 | Q 3 | Page 144
In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (ar(ABC))/(ar(DBC)) = (AO)/(DO)
Exercise 6.4 | Q 4 | Page 144
If the areas of two similar triangles are equal, prove that they are congruent
Exercise 6.4 | Q 5 | Page 144
D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.
Exercise 6.4 | Q 6 | Page 144
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Exercise 6.4 | Q 7 | Page 144
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals
Exercise 6.4 | Q 8 | Page 144
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
• 2 : 1
• 1 : 2
• 4 : 1
• 1 : 4
Exercise 6.4 | Q 9 | Page 144
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
• 2 : 3
• 4 : 9
• 81 : 16
• 16 : 81
Exercise 6.5 [Pages 150 - 151]
### NCERT solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5 [Pages 150 - 151]
Exercise 6.5 | Q 1.1 | Page 150
Sides of triangle are given below. Determine it is a right triangle or not? In case of a right triangle, write the length of its hypotenuse. 7 cm, 24 cm, 25 cm
Exercise 6.5 | Q 1.2 | Page 150
Sides of triangles are given below. Determine it is a right triangles? In case of a right triangle, write the length of its hypotenuse. 3 cm, 8 cm, 6 cm
Exercise 6.5 | Q 1.3 | Page 150
Sides of triangle are given below. Determine it is a right triangle or not? In case of a right triangle, write the length of its hypotenuse. 50 cm, 80 cm, 100 cm
Exercise 6.5 | Q 1.4 | Page 150
Sides of triangle are given below. Determine it is a right triangle or not? In case of a right triangle, write the length of its hypotenuse. 13 cm, 12 cm, 5 cm
Exercise 6.5 | Q 2 | Page 150
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR
Exercise 6.5 | Q 3.1 | Page 150
In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AB2 = BC × BD
Exercise 6.5 | Q 3.2 | Page 150
In Figure ABD is a triangle right angled at A and AC ⊥ BD. Show that AC2 = BC × DC
Exercise 6.5 | Q 3.3 | Page 150
In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AD2 = BD × CD
Exercise 6.5 | Q 4 | Page 150
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2
Exercise 6.5 | Q 5 | Page 150
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Exercise 6.5 | Q 6 | Page 150
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Exercise 6.5 | Q 7 | Page 150
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals
Exercise 6.5 | Q 8 | Page 151
In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE= AE2 + CD2 + BF2
Exercise 6.5 | Q 9 | Page 151
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Exercise 6.5 | Q 10 | Page 151
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Exercise 6.5 | Q 11 | Page 151
An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after 1 1/2 hours?
Exercise 6.5 | Q 12 | Page 151
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Exercise 6.5 | Q 13 | Page 151
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE+ BD2 = AB2 + DE2
Exercise 6.5 | Q 14 | Page 151
The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD . Prove that 2AB2 = 2AC2 + BC2.
Exercise 6.5 | Q 15 | Page 151
In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC . Prove that 9 AD2 = 7 AB2
Exercise 6.5 | Q 16 | Page 151
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Exercise 6.5 | Q 17 | Page 151
Tick the correct answer and justify: In ΔABC, AB = 6sqrt3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
• 120°
• 60°
• 90°
• 45°
Exercise 6.6 [Pages 152 - 153]
### NCERT solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.6 [Pages 152 - 153]
Exercise 6.6 | Q 1 | Page 152
In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that (QS)/(SR) = (PQ)/(PR)
Exercise 6.6 | Q 2 | Page 152
In the given figure, D is a point on hypotenuse AC of ΔABC, DM ⊥ BC and DN ⊥ AB, Prove that:
(i) DM2 = DN.MC
(ii) DN2 = DM.AN
Exercise 6.6 | Q 3 | Page 152
In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC.BD.
Exercise 6.6 | Q 4 | Page 152
In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 − 2BC.BD.
Exercise 6.6 | Q 5 | Page 152
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i) "AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2
(ii) "AB"^2 = "AD"^2 - "BC"."DM" + (("BC")/2)^2
(iii) "AC"^2 + "AB"^2 = 2"AD"^2 + 1/2"BC"^2
Exercise 6.6 | Q 6 | Page 153
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Exercise 6.6 | Q 7 | Page 153
In the given figure, two chords AB and CD intersect each other at the point P. prove that:
(i) ΔAPC ∼ ΔDPB
(ii) AP.BP = CP.DP
Exercise 6.6 | Q 8 | Page 153
In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) ΔPAC ∼ ΔPDB
(ii) PA.PB = PC.PD
Exercise 6.6 | Q 9 | Page 153
In the given figure, D is a point on side BC of ΔABC such that ∠ADC=∠BAC . Prove that AD is the bisector of ∠BAC.
Exercise 6.6 | Q 10 | Page 153
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, ho much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
## Chapter 6: Triangles
Exercise 6.1Exercise 6.2Exercise 6.3Exercise 6.4Exercise 6.5Exercise 6.6
## NCERT solutions for Class 10 Maths chapter 6 - Triangles
NCERT solutions for Class 10 Maths chapter 6 (Triangles) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 10 Maths solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 10 Maths chapter 6 Triangles are Similar Figures, Similarity of Triangles, Basic Proportionality Theorem (Thales Theorem), Criteria for Similarity of Triangles, Areas of Similar Triangles, Application of Pythagoras Theorem in Acute Angle and Obtuse Angle, Triangles Examples and Solutions, Angle Bisector, Ratio of Sides of Triangle, Right-angled Triangles and Pythagoras Property, Similarity of Triangles, Similarity of Triangles, Similar Figures, Similarity of Triangles, Basic Proportionality Theorem (Thales Theorem), Criteria for Similarity of Triangles, Areas of Similar Triangles, Application of Pythagoras Theorem in Acute Angle and Obtuse Angle, Triangles Examples and Solutions, Angle Bisector, Ratio of Sides of Triangle, Right-angled Triangles and Pythagoras Property, Similarity of Triangles, Similarity of Triangles.
Using NCERT Class 10 solutions Triangles exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 10 prefer NCERT Textbook Solutions to score more in exam.
Get the free view of chapter 6 Triangles Class 10 extra questions for Class 10 Maths and can use Shaalaa.com to keep it handy for your exam preparation |
# Calculation of Mean in Discrete Series | Formula of Mean
Last Updated : 17 Aug, 2023
## What is Mean?
Mean is the sum of a set of numbers divided by the total number of values. It is also referred to as the average. For instance, if there are four items in a series, i.e. 2, 5, 8, 3, and 9. The simple arithmetic mean is (2 + 5 + 8 + 3 + 9) / 5 = 5.4.
### What is Discrete Series?
In discrete series (ungrouped frequency distribution), the values of variables represent the repetitions. It means that the frequencies are given corresponding to the different values of variables. The total number of observations in a discrete series, N, equals the sum of the frequencies, which is Σf.
#### Example of Discrete Series
If 6 students of a class score 50 marks, 4 students score 60 marks, 7 students score 70 marks, 3 students score 80 marks, and 5 students score 90 marks, then this information will be shown as:
## Mean in Discrete Series:
Mean in discrete series can be calculated by using:
1. Direct Method;
2. Short-Cut Method; and
3. Step Deviation Method
#### Example 1:
Find the average of the following discrete series using any of the three methods.
Â
#### Solution:
We will be using Direct Method to determine Mean of the given series.
Â
Mean = 9.42
#### Example 2:
Determine the arithmetic mean from the following frequency table using Step-Deviation Method:
Â
Â
Mean = 56
#### Example 3:
If the mean of the following distribution is 28, locate the missing frequency.
Â
#### Solution:
Suppose the missing frequency is f.
Â
952 + 28f = 1,030 + 15f
28f – 15f = 1,030 – 952
13f = 78
f = 6
Missing Frequency = 6
Previous
Next |
# Square (Coordinate Geometry)
A 4-sided regular polygon with all sides equal, all interior angles 90° and whose location on the coordinate plane is determined by the coordinates of the four vertices (corners).
Try this Drag any vertex of the square below. It will remain a square and its dimensions calculated from its coordinates. You can also drag the origin point at (0,0), or drag the square itself.
In coordinate geometry, a square is similar to an ordinary square (See Square definition ) with the addition that its position on the coordinate plane is known. Each of the four vertices (corners) have known coordinates. From these coordinates, various properties such as width, height etc can be found.
It has all the same properties as a familiar square, such as:
See Square definition for more.
## Dimensions of a square
The dimensions of the square are found by calculating the distance between various corner points. Recall that we can find the distance between any two points if we know their coordinates. (See Distance between Two Points ) So in the figure above:
• The length of each side of the square is the distance any two adjacent points (say AB, or AD)
• The length of a diagonals is the distance between opposite corners, say B and D (or A,C since the diagonals are congruent).
This method will work even if the square is rotated on the plane (click on "rotated" above). But if the sides of the square are parallel to the x and y axes, then the calculations can be a little easier.
In the above figure uncheck the "rotated" box and note that The side length is the difference in y-coordinates of any left and right point - for example A and B.
## Example
The example below assumes you know how to calculate the distance between two points, as described in Distance between Two Points. In the figure above, click 'reset', 'rotated' and 'show diagonals'
• The side length of the square is the distance between any two adjacent vertices. Let's pick B and C. Using the formula for the distance between two points:
• The length of a diagonals is the distance between any pair of opposite vertices. In a square, the diagonal is also the length of a side times the square root of two:
## Area and perimeter
These are described on a separate page. See Area and perimeter of a square (coordinate geometry)
## Things to try
In the figure at the top of the page, click on "hide details" . Then drag the square or any of its corners to create an arbitrary square. Calculate the width, height and the length of the diagonals. Click 'show details' and "show diagonals" to verify your answer.
## Limitations
In the interest of clarity in the applet above, the coordinates are rounded off to integers and the lengths rounded to one decimal place. This can cause calculatioons to be slightly off.
For more see Teaching Notes |
Digital Values - Math Research Paper
Abstract:- We come across many big calculations which we want to check. Though the idea of digital roots can be used, but it is limited to integers. This paper introduces a new idea of assigning every number a characteristic value
called “Digital Values”. Every number, real or imaginary is assigned a digital value. The digital values are mostly 1, 2,3,4,5,6,7,8 or 9. These values have many interesting properties. Although in some cases we assign some other values for our convenience. The digital values can be applied to calculations to check them. They also have interesting properties in an equation (expressions involving unknown quantities) and system of equations.
Keywords:- digital values, digital roots, digital sum, digitally irrational numbers, equi-digital functions.
1 Introduction
Sometimes it is very difficult to go back and check the whole process. It happens in many calculations, while solving equations etc. The idea of digital roots may help us in some calculations. A formula for finding the digital root of an integer is given by[1] : Digitalroot[x] = 1+Mod[(x-1),9]. The digital root of addition, subtraction, multiplication and division of integers show interesting properties. But the idea is limited to integers. This paper introduces a new concept of “digital values” to overcome this difficulty. Just like in digital roots, we assign particular values for different numbers but this can be implemented for any number (real, imaginary or complex). It follows all the properties of digital roots. The paper also introduces how these digital values can help us in verifying calculations and the application of digital values in functions and equations.
2. What is digital value?
Digital value is a characteristic value assigned to a number. We will denote digital value of a number x by //x// or by dval(x).
For a natural number the digital value is same as its digital root[1]. As in digital roots, we add the different digits and repeat the process till a single digit is reached. For 1456914 the digital value will be: //1+4+5+6+9+1+4//=//30//=3.
Similarly for 563, digital value =//563//=//5+6+3//=//14//=5
2.1 Digital value of an integer
Consider the following table:
Table 1
Number Digital Value
267 6
266 5
265 4
264 3
263 2
262 1
261 9
260 8
259 7
258 6
257 5
256 4
255 3
254 2
253 1
We observe that the digital value of the natural numbers in decreasing order repeat the pattern : “9,8,7,6,5,4,3,2,1”
For 0 and negative integers also we will follow the same pattern to get the digital value i.e. digital value of 0 is 9,-1 is 8,-2 is 7,-3 is 6 and so on. A simple way to find out the digital value of a negative integer is to subtract the absolute value of the integer from 9.For e.g.
//-8// = 9 - //8// = 9 – 8 = 1
//-5647// = 9 - //5647// = 9 – 4 =5
The above results can be obtained by the general formula [1]
Digitalroot[x] = 1+Mod[(x-1),9]
Some properties of digital values:
For two integers a and b,
(1) // a + b // = // //a// + //b// //
(2) // a - b // = // //a// - //b// //
(3) // a × b // = // //a// × //b// //
(4) // // a + b // + c // = // a + // b + c // //
(5) // // a × b // × c // = // a ×//b × c // //
(6) // 9a// = 9
(7) // 8 × a // = //-a//
(8) // 9a + b // = //b//
(9) // a! // = 9, where a ? 6
(10) // a^b // = // dval(a)^b //
All the above identities can be easily proved using congruence.
2.2 Division of integers (digital values of rational numbers)
For division consider the following expression:
(11) // a/b // = // (dval(a))/(dval(b)) //
So, now, digital value for any decimal number which is terminating can be found out.
For e.g. //12.321// =// 12321/1000 // = // (dval(12321))/(dval(1000)) // = // 9/1 // = 9
For 1/11
// 1/11 // = // (dval(1))/(dval(11)) // = // 1/2 //=//0.5//=5
According to the above identity // 1/7 // and // 1/16 // should have same digital value.
So, = // 1/7 // = // 1/16 // = //0.0625// = 4
Now, for any division
// x/y // = // //x// × // 1/y // //
Division by 3,6 and 9 cannot be determined. It is either undefined or has multiple digital values.
If //a//=3, // a/3// = 1, 4, 7
If //a//=6, // a/3// = 2, 5, 8
If //a//=9, // a/3// = 3, 6, 9
If //a//=3, // a/6// = 2, 5, 8
If //a//=6, // a/6// = 1,4,7
If //a//=9, // a/6// = 3, 6, 9
If //a//=9, // a/9// = 1, 2,3,4,5, 6, 7, 8, 9
In all other cases the digital value is digitally imaginary (see next section).
2.3 Digital values of irrational numbers
For an irrational number, we will use
(12) // a^b // = // dval(a)^b //, where a, b are real numbers
So
//square root of 13// = // square root of //4// //
= //2// or //-2//
= 2 or 7
//?4 // = //2// = 2 (one root is taken only if the given value is rational)
//?13// will have 2 values : 2 and 7
Let A be another number such that //a//= //A//
// a^b // = // dval(a)^b //
and // A^b // = // dval(A)^b //=// dval(a)^b //
therefore, // a^b // =// A^b //
Using this method:
// square root of 7//= //square root of 16//= //4// or //-4//
= 4 or 5
Following is the table for digital values of some powers:
Table 2
// x^1 // 1 2 3 4 5 6 7 8 9
// x^2// 1 4 9 7 7 9 4 1 9
// x^3// 1 8 9 1 8 9 1 8 9
// x^4// 1 7 9 4 4 9 7 1 9
// x^5// 1 5 9 7 2 9 4 8 9
// x^6// 1 1 9 1 1 9 1 1 9
// x^7// 1 2 9 4 5 9 7 8 9
// x^8// 1 4 9 7 7 9 4 1 9
// x^9// 1 8 9 1 8 9 1 8 9
There is repetition in the digital values of the numbers raised to increasing powers.
For 1 : 1
For 2 : 4,8,7,5,1,2
For 3 : 9
For 4 : 7,1,4
For 5 : 7,8,4,2,1,5
For 6 : 9
For 7 : 4,1,7
For 8 : 1,8
For 9 : 9
Following this repetition digital value of any number raised to any natural power can be determined.
For e.g.
//14^11// = //5^11//=//5^5// [following the repetition]
= 2
For //x^(1/b)// , x belongs to R, b belongs to Z , a digital root between 1 to 9 exists only if it is present in the Table 2 in the row of bth power of x.
Otherwise the digital root is represented by //x^(1/b)// only. For e.g.
?3,?2
These values are called digitally imaginary numbers (DI).
2.4 Digital values of imaginary numbers
We know that
// a^b // =// A^b // when //a//= //A//
Using the above relation, when b= (1/2), a= -1, A= 8;
// i // =// ?(-1)//=//?8//
//?(-5) //=//?4//= 2 or 7 [two values because we cannot have a rational value of ?(-5) ]
OR
//?(-5) =//?5 i//=//?5.?8 //=//?4//= 2 or 7
In this way we can find the digital value of a complex number.
As in case of digital roots[2] the digital values also show the repetition in addition (Table 3), subtraction (Table 4), multiplication (Table 5)and division.
+ 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 1
2 3 4 5 6 7 8 9 1 2
3 4 5 6 7 8 9 1 2 3
4 5 6 7 8 9 1 2 3 4
5 6 7 8 9 1 2 3 4 5
6 7 8 9 1 2 3 4 5 6
7 8 9 1 2 3 4 5 6 7
8 9 1 2 3 4 5 6 7 8
9 1 2 3 4 5 6 7 8 9
Table 4: Subtraction table
- 1 2 3 4 5 6 7 8 9
1 9 8 7 6 5 4 3 2 1
2 1 9 8 7 6 5 4 3 2
3 2 1 9 8 7 6 5 4 3
4 3 2 1 9 8 7 6 5 4
5 4 3 2 1 9 8 7 6 5
6 5 4 3 2 1 9 8 7 6
7 6 5 4 3 2 1 9 8 7
6 7 6 5 4 3 2 1 9 8
9 8 7 6 5 4 3 2 1 9
Table 5: Multiplication Table
X 1 2 3 4 5 6 7 8 9
1 1 2 3 4 5 6 7 8 9
2 2 4 6 8 1 3 5 7 9
3 3 6 9 3 6 9 3 6 9
4 4 8 3 7 2 6 1 5 9
5 5 1 6 2 7 3 8 4 9
6 6 3 9 6 3 9 6 3 9
7 7 5 3 1 8 6 4 2 9
8 8 7 6 5 4 3 2 1 9
9 9 9 9 9 9 9 9 9 9
3 Equality of digital values
For two equal quantities are equal the following properties of digital roots are important:
?If two quantities are equal there digital values must be equal.
This property may be used to
Check calculations:
See if digital values of both sides are equal or not. If they are not equal then the calculation is incorrect.
To find a missing digit:
Find the digital value of the known side. Then apply trial and error to put the unknown digit so that the digital values of both sides are equal.
? If a DI occurs in digital value of LHS of any equation it must occur in that of RHS too.
4. Digital value in functions and equations
In functions and equations digital values have following properties:
?For any function
(13) //f(x)// = // f (//x//) //
? In a system of equations with unique solution, the solution can be represented by an expression containing coefficients. So, if two systems of equations have equal digital values of corresponding coefficients of corresponding equations, then the corresponding roots have equal digital values.
i.e. a_11 x+ b_11 y+ c_11=0
a_12 x+ b_12 y+ c_12=0
AND
a_21 x+ b_21 y+ c_21=0
a_22 x+ b_22 y+ c_22=0
Will have same digital values of x as well as y if
//a_11//=// a_21//
//b_11//=// b_21//
//c_11//=// c_21//
?If
//a_1//=//b_1//
//a_2//=//b_2//
//a_3//=//b_3//
……………..
//a_n//=//b_n// (14)
Then
(x-a_1 )(x-a_2 )(x-a_3 )……..(x-a_n) and (x-b_1 )(x-b_2 )(x-b_3 )……..(x-b_n) are equi-digital.
The converse is not always true.
?In case of quadratic equation the converse is true when the roots are distinct.
4. Conclusion
The paper has introduced a concept of digital values which provides a way not for verifying calculations involving not only integers but any complex number. Now any complex calculation can be checked but one should be careful that if digital values of LHS and RHS are equal it does not necessarily mean that LHS = RHS. But if they are not equal then LHS cannot be equal to RHS. We have also studied the properties of digital values in functions and equations. We have also learnt how to use the property of digital value to find a missing digit in calculations.
It may seem strange to learn a way of checking a calculation when so many accurate computers are available but we must have the knowledge of the interesting properties of numbers.
References:
[1] Weisstein, Eric W. "Digital Root." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/DigitalRoot.html
[2] Teknomo,K.,"Digital Root" http://people.revoledu.com/kardi/ ,Page2
[3] Teknomo,K.,"Digital Root" http://people.revoledu.com/kardi/ ,Page7-8
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Last updated on Nov 20, 2023
Students can practice the questions from the Triangle MCQs Quiz to ace their exams. All the quantitative and competitive aptitude questions have been included in the MCQ Quiz page so that this becomes your one-stop destination for study material. The Triangle Objective Questions has all the questions which the candidates must know to ace their competitive exams and interview. Check the exclusive added questions and practice for the topic.
## Latest Triangle MCQ Objective Questions
#### Triangle Question 1:
E, D, and F are the midpoints of the sides of the triangle ABC. If the area of triangle EFD is 64 cm square then find the area of triangle ABC?
1. 128 cm square
2. 256 cm square
3. 324 cm square
4. More than one of the above
5. None of the above
Option 2 : 256 cm square
#### Triangle Question 1 Detailed Solution
Given:
E, F, and D are the midpoints of the sides of the triangle, and the area of triangle EFD is 64 cm square
Concept Used:
Area of triangle formed by the midpoint of sides is one fourth the area of triangle
Area of triangle EDF = 1/4 × (Area of triangle ABC)
Calculation:
It is given in the question that E, F and D are the mid points of the sides of triangle ABC formed the triangle EFD whose area is 64 cm square
We know that area of triangle formed by midpoint of sides is one fourth the area of triangle
Area of triangle ABC = 4 × 64
⇒ 256 cm square
∴ The area of triangle ABC is 256 cm square.
#### Triangle Question 2:
A right angle triangle has height 12 cm and base 4 cm. The area of triangle is :
1. 12 sq. cm
2. 48 sq. cm
3. 24 sq. cm
4. More than one of the above
5. None of the above
Option 3 : 24 sq. cm
#### Triangle Question 2 Detailed Solution
Given:
A right angle triangle has height 12 cm and base 4 cm.
Formula used:
Area of a triangle = $$\frac{1}{2}$$ × Base × Height
Calculation:
Area of a triangle = $$\frac{1}{2}$$ × 4 × 12 = 24 cm2
∴ The area of triangle is 24 cm2.
#### Triangle Question 3:
The length of the hypotenuse of a right-angled triangle is 13 cm and the length of one of the other two sides is 5 cm. What is the area (in cm2) of the triangle?
1. 32.5
2. 29.5
3. 28
4. 30
5. None of the above
Option 4 : 30
#### Triangle Question 3 Detailed Solution
Given :
Hypotenuse (h) = 13 cm
Perpendicular (p) = 5 cm
Base (b) = ?
Formula used:
Pythagoras theorem
⇒ (h)= (p)+ (b)2
Area of the triangle :
$$\frac{1}{2}$$x base x height
Calculations:
By using Pythagoras theorem:
(13)2= (5)+ (b)2
169 = 25+(b)2
144 =(b)2
⇒ b = 12
Now the area of the triangle:
⇒($$\frac{1}{2}$$) x 12 x 5
($$\frac{1}{2}$$) x 60
30cm
The area (in cm2) of the triangle is 30 cm2.
#### Triangle Question 4:
The sides of a triangle are 35 cm, 54 cm and 61 cm respectively. The length of the longest altitude is
1. 16√5 cm
2. 10√5 cm
3. 24√5 cm
4. More than one of the above.
5. None of the above
Option 3 : 24√5 cm
#### Triangle Question 4 Detailed Solution
Concept:
Semi-perimeter, $$s= \frac{a+b+c}{2}$$
Now, according to Heron's formula,
Area of triangle = $$√ {s(s-a)(s-b)(s-c)}$$
Calculation:
s = (35 + 54 + 61)/2 = 75 cm
Now, Area of the triangle = $$√{75(75-35)(75-54)(75-61)}$$
$$⇒ √ {75 × 40×21×14}$$
⇒ 420√5 cm2
Let h be the longest altitude.
⇒ Area of triangle = $$\frac{1}{2}× a × h$$
⇒ 420√5 = $$\frac{1}{2}× 35 × h$$
⇒ h = 24√5 cm
Hence, the length of the longest altitude is 24√5 cm.
Important Points
The length of the longest altitude is asked, i.e Value of h should be maximum.
420√5 = $$\frac{1}{2}× a × h$$
For h to be maximum, value of a should be minimum.
If we take a = 54
420√5 = 1/2 × 54 × h
h = 420√5 / 27 = 15.5√5
If we take a = 61
h = 13.7√5
So, the smaller the value of a, higher the value of h.
#### Triangle Question 5:
In a triangle ABC, which of the following conditions is true?
1. AB > CA + BC
2. BC < AB – CA
3. BC < AB + CA
4. More than one of the above
5. None of the above
Option 3 : BC < AB + CA
#### Triangle Question 5 Detailed Solution
We know that
In any triangle, the sum of the two sides is always greater than the third side.
∴ In a ΔABC,
AB + BC > CA
BC + CA > AB
CA + AB > BC or BC < AB + CA
∴ BC < AB + CA is the correct option.
## Top Triangle MCQ Objective Questions
#### Triangle Question 6
In the triangle ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°. What is the value of the length of the side BC?
1. 10 cm
2. 7.13 cm
3. 13.20 cm
4. 11.13 cm
Option 4 : 11.13 cm
#### Triangle Question 6 Detailed Solution
Given:
In the triangle, ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°.
Concept used:
According to the law of cosine, if a, b, and c are three sides of a triangle ΔABC and ∠A is the angle between AC and AB then, a2 = b2 + c2 - 2bc × cos∠A
Calculation:
According to the concept,
BC2 = AB2 + AC2 - 2 × AB × AC × cos60°
⇒ BC2 = 122 + 102 - 2 × 12 × 10 × 1/2
⇒ BC2 = 124
⇒ BC ≈ 11.13
∴ The measure of BC is 11.13 cm.
#### Triangle Question 7
An equilateral triangle ABC is inscribed in a circle with centre O. D is a point on the minor arc BC and ∠CBD = 40º. Find the measure of ∠BCD.
1. 40º
2. 30º
3. 50º
4. 20º
Option 4 : 20º
#### Triangle Question 7 Detailed Solution
Given:
An equilateral triangle ABC is inscribed in a circle with centre O
∠CBD = 40º
Concept used:
The sum of the opposite angles of a cyclic quadrilateral = 180°
The sum of all three angles of a triangle = 180°
Calculation:
∠ABC = ∠ACB = ∠BAC = 60° [∵ ΔABC is an equilateral triangle]
Also, ∠BAC + ∠BDC = 180°
⇒ 60° + ∠BDC = 180°
⇒ ∠BDC = 180° - 60° = 120°
Also, ∠CBD + ∠BDC + ∠BCD = 180°
⇒ 40° + 120° + ∠BCD = 180°
⇒ ∠BCD = 180° - 40° - 120° = 20°
∴ The value of ∠BCD is 20°
#### Triangle Question 8
In a ΔABC, points P, Q and R are taken on AB, BC and CA, respectively, such that BQ = PQ and QC = QR. If ∠BAC = 75º, what is the measure of ∠PQR (in degrees)?
1. 75
2. 50
3. 30
4. 40
Option 3 : 30
#### Triangle Question 8 Detailed Solution
Shortcut Trick
∠BAC = 75º
∠ABC + ∠ACB + ∠BAC = 180°
∠ABC + ∠ACB + 75° = 180°
∠ABC + ∠ACB = 180° - 75° = 105°
Let, ∠ABC = ∠PBQ = 70° and ∠ACB = ∠RCQ = 35°
So, ∠PQR = 180° - (∠PQB + ∠RQC)
180° - [(180° - 2∠PBQ) + (180° - 2∠RCQ) [∵ BQ = PQ; QC = QR]
= 180° - [(180° - 2 × 70°) + (180° - 2 × 35°)]
= 180° - (40° + 110°)
= 180° - 150°
= 30°
Alternate Method
Given:
In a ΔABC, ∠BAC = 75º
BQ = PQ and QC = QR
Concept used:
The sum of all three angles of a triangle = 180°
The sum of all angles on a straight line = 180°
Calculation:
Let, ∠ABC = x and ∠ACB = y
So, ∠ABC = ∠PBQ = ∠QPB = x [∵ BQ = PQ]
∠ACB = ∠RCQ = ∠QRC = y [QC = QR]
In ΔABC, ∠ABC + ∠ACB + ∠BAC = 180°
⇒ x + y + 75° = 180°
⇒ x + y = 180° - 75° = 105° .....(1)
For ΔBPQ and ΔCRQ,
(∠PBQ + ∠QPB + ∠PQB) + (∠RCQ + ∠QRC + ∠RQC) = 180° + 180° = 360°
⇒ (x + x + ∠PQB) + (y + y + ∠RQC) = 360°
⇒ 2x + 2y + ∠PQB + ∠RQC = 360°
⇒ 2 (x + y) + ∠PQB + ∠RQC = 360°
⇒ (2 × 105°) + ∠PQB + ∠RQC = 360° [∵ x + y = 105°]
⇒ ∠PQB + ∠RQC = 360° - 210° = 150° .....(2)
Also, ∠PQB + ∠RQC + ∠PQR = 180°
⇒ 150° + ∠PQR = 180° [∵ ∠PQB + ∠RQC = 150°]
⇒ ∠PQR = 180° - 150° = 30°
∴ The measure of ∠PQR (in degrees) is 30°
#### Triangle Question 9
In an isosceles triangle ABC, if AB = AC = 26 cm and BC = 20 cm, find the area of triangle ABC.
1. 180 cm2
2. 240 cm2
3. 220 cm2
4. 260 cm2
Option 2 : 240 cm2
#### Triangle Question 9 Detailed Solution
Given:
In an isosceles triangle ABC,
AB = AC = 26 cm and BC = 20 cm.
Calculations:
In this triangle ABC,
∆ADC = 90° ( Angle formed by a line from opposite vertex to unequal side at mid point in isosceles triangle is 90°)
So,
AD² + BD² = AB² (by pythagoras theorem)
Area of triangle = ½(base × height)
⇒ ½(20 × 24) (Area of triangle = (1/2) base × height)
⇒ 240 cm²
∴ The correct choice is option 2.
#### Triangle Question 10
In an equilateral ΔABC, the medians AD, BE and CF intersect to each other at point G. If the area of quadrilateral BDGF is 12√3 cm2, then the side of ΔABC is:
1. 10√3 cm
2. 10 cm
3. 12√3 cm
4. 12 cm
Option 4 : 12 cm
#### Triangle Question 10 Detailed Solution
Given
Area of Quadrilateral = 12√3 cm2
Concept Used
We know that The median of a triangle is cut the triangle in equal areas
Area of triangle = (√3/4) (side)2
Calculation
⇒ Area of ΔABC = Area of Quadrilateral BDGF × 3
⇒ Area of ΔABC = 12√3 × 3
⇒Area of ΔABC = 36√3 cm2
⇒ 36√3 = (√3/4) × (side)2
⇒ side = 12 cm
∴ the side of ΔABC is 12 cm
#### Triangle Question 11
The area of an equilateral triangle inscribed in a circle is 4√3 cm2 the area of circle is
1. $$\dfrac{16}{3}\pi$$
2. $$\dfrac{22}{3}\pi$$
3. $$\dfrac{28}{3}\pi$$
4. $$\dfrac{32}{3}\pi$$
Option 1 : $$\dfrac{16}{3}\pi$$
#### Triangle Question 11 Detailed Solution
Given
Equilateral triangle is inscribed in a circle
Area of Equilateral triangle = 4√3 cm2
Concept used
Area of Equilateral triangle = $$\frac{{√ 3 }}{4}{a^2}$$
Calculation:
$$\frac{{√ 3 }}{4}{a^2}$$ = 4√3
a2 = 16
a = 4
Circumradius of Equilateral triangle = $$\frac{a}{{\sqrt 3 }}$$
r = $$\frac{4}{{\sqrt 3 }}$$
Area of circle = πr2
πr2 = $$\pi \times \frac{4}{{\sqrt 3 }} \times \frac{4}{{\sqrt 3 }}$$
Area = $$\dfrac{16}{3}\pi$$ cm2
#### Triangle Question 12
The altitude AD of a triangle ABC is 9 cm. If AB = 6√3 cm and CD = 3√3 cm, then what will be the measure of ∠A?
1. 45°
2. 30°
3. 90°
4. 60°
Option 4 : 60°
#### Triangle Question 12 Detailed Solution
Shortcut TrickWe know that the height of an equilateral triangle = a√3/2
Here, Height = 9 = 6√3 × √3/2
a = 6√3, so Height = a√3/2.
∴ Given triangle is an equilateral triangle.
So, ∠A = 60°.
Concept:
Pythagoras theorem:
Equilateral Triangle:
AB = BC = AC
∠A = ∠B = ∠C = 60°
Calculation:
(6√3)2 = (BD)2 + 92
⇒ BD = 3√3 cm
∵ DC = BD = 3√3 cm
∴ BC = AC = AB = 6√3 cm &
∠A = ∠B = ∠C = 60°
∴ ABC is an equilateral triangle.
#### Triangle Question 13
The circumcentre of an equilateral triangle is at a distance of 3.2 cm from the base of the triangle. What is the length (in cm) of each of its altitudes?
1. 9.6
2. 7.2
3. 6.4
4. 12.8
Option 1 : 9.6
#### Triangle Question 13 Detailed Solution
Given data:
Calculations:
From the equilateral triangle property, O is the circumcenter as well as the centroid.
∴ OD = $$1\over 3$$ × AD
⇒ AD = 3 × OD
⇒ AD = 3 × 3.2 = 9.6 cm
The length of the altitudes of the equilateral triangle is 9.6 cm.
#### Triangle Question 14
In ΔABC, ∠B = 90°, AB = 5 cm and BC = 12 cm the bisector of ∠A meets BC at D. What is the length of AD?
1. 2√13 cm
2. $$\frac{2}{3}\sqrt {13}$$ cm
3. $$\frac{4}{3}\sqrt {13}$$ cm
4. $$\frac{{5\sqrt {13} }}{3}$$ cm
Option 4 : $$\frac{{5\sqrt {13} }}{3}$$ cm
#### Triangle Question 14 Detailed Solution
In right triangle ABC
AC2 = AB2 + BC2
⇒ AC2 = 52 + 122 = 25 + 144 = 169
⇒ AC = √169 = 13 cm
If AD is bisector of ∠BAC, then
AB/AC = BD/DC
⇒ 5/13 = BD/DC
⇒ DC = 13BD/5
⇒ BD + DC = BC
⇒BD + 13BD/5 = 12
⇒18BD/5 = 12
⇒ BD = 12 × (5/18) = 10/3
In right triangle ABD
⇒ AD2 = 52 + (10/3)2 = 25 + 100/9 = 325/9
⇒ AD = √325/9 = 5 √13/3 cm
#### Triangle Question 15
The area of ΔABC is 63 sq. units. Two parallel lines DE, FG, are drawn such that they divide the line segments AB and AC into three equal parts. What is the area of the quadrilateral DEGF?
A. 28 sq. units
B. 35 sq. units
C. 21 sq. units
D. 48 sq. units
1. A
2. C
3. D
4. B
Option 2 : C
#### Triangle Question 15 Detailed Solution
Let, AD = DF = FB = x units and AE = EG = GC = y units
According to the question,
⇒ Area of ΔADE/63 = (x/3x)2
⇒ Area of ΔADE = 63 × 1/9
⇒ Area of ΔADE = 7
According to the question,
⇒ Area of ΔAFG/Area of ΔABC = (AF/AB)2
⇒ Area of ΔAFG/63 = (2x/3x)2
⇒ Area of ΔAFG = 63 × 4/9
⇒ Area of ΔAFG = 28
∴ Area of the quadrilateral DEGF = Area of (ΔAFG - ΔADE) = (28 - 7) = 21 sq. units |
# Class 11: Topic 10: More Summary Measures and Graphs
This outline is also available in PDF.
Held: Friday, 15 February 2008
Summary: We consider more ways to summarize numeric data.
Notes:
• I've received a number of I'm sick and won't be in class notices. Please take care of yourselves.
• When you're sick, I would like you to try to arrange with a friend to bring homework to class. I know that's not always possible, but do the best you can.
• The exam is next Wednesday in both 3819 and 3820.
• There is no homework for Monday. Go over the sample exam and the chapters and be prepared to ask questions.
• You may find yourself a bit suspicious of the data that we're using for activity 10-3, but deal with it.
• Preparation: Overview and Preliminaries for Topic 10.
• Handouts: R notes on topic 10 and Sample Exam.
• Due: Activities 9-14, 9-15, 9-20, 9-23.
Overview:
• Some past topics: Spread, IQR, etc.
• Visualizing summary statistics.
We've seen at least three measures of the spread of a distribution.
• What are they?
• Why might one use one measure rather than another?
## Notes on Inter-quartile Range
Some of you observed that the book and R give different answers for some IQR computations. Note that IQR can be computed in slightly different ways, which can have an effect on the result. (And different stats packages take different approaches.)
Let's use the distribution 1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9 as our example.
Officially, the lower quartile is a number such that 1/4 of the values are smaller and 1/4 of the values are larger. (It's not always possible to find such a value; for example, if all the values are the same, you can't divide this evenly.) Similarly, the upper quartile is a number such that 3/4 of the values are smaller and 1/4 of the values are larger.
We can start by finding those values directly.
• Since there are 18 numbers in this distribution, it's impossible to find a single number such that exactly 1/4 of the values are smaller and exactly 3/4 are larger.
• Arguably, the fifth value in the sequence will be close enough for the lower quartile, since there should be 4 smaller values and 13 larger values.
• Similarly, the 14th value would be appropriate for the upper quartile.
• Using that strategy, we get an IQR of 7-3 or 4.
Our book tells us a different strategy for computing the IQR. First, compute the median and then compute the median of each half. However, our book is vague on what you do when the median is repeated.
• Suppose we just split the 18 values in the middle.
• The left half is 1,1,2,2,3,3,4,4,5
• The right half is 5,6,6,7,7,8,8,9,9
• Conveniently, there are nine values in each half.
• The middle of the left half is 3
• The middle of the right half is 7
• Hence, the IQR is 4
So, how did our book end up with 4.5 as the answer? Here's my guess. We did an odd thing in the analysis above: We put one 5 on each half. Arguably, both should go on the same side.
• Suppose we put the 5's in the lower half
• The left half is 1,1,2,2,3,3,4,4,5,5
• The right half is 6,6,7,7,8,8,9,9
• Now, there are ten values in the lower half and eight in the upper half. (Yeah, I don't like it either, but it's a technique that many packages still use.)
• The median of the lower half is still 3, since both the 5th and 6th values in that half are 3.
• The median value of the upper half is now 7.5, since it falls between 7 and 8.
• Hence, the IQR is 4.5
Visualizing Summary Statistics
• The five-number summary (min, lower quartile, median, upper quartile, max) is a common way of looking at data.
• But it's very textual
• Sometimes we find it easier to consider these same summary statistics visually.
• A boxplot (or box-and-whiskers plot) is one common way of looking at numeric data.
This document may be found at `http://www.cs.grinnell.edu/~rebelsky/Courses/MAT115/2008S/Outlines/outline.11.html`.
Copyright © 2008 Samuel A. Rebelsky. This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License. To view a copy of this license, visit `http://creativecommons.org/licenses/by-nc/2.5/` or send a letter to Creative Commons, 543 Howard Street, 5th Floor, San Francisco, California, 94105, USA. |
# Lesson: Equivalent Fractions and Percents
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### Lesson Objective
SWBAT use of percents in describing real-life situations; and to reinforce naming equivalent fractions.
### Lesson Plan
Materials Needed: scrap paper for DN, overhead 10 by 10 grid, Chart explaining percentages (see direct instruction), GP Worksheet, white board, dry erase markers, pencils, overhead projector, and IND Worksheet.
Vocabulary: whole (or ONE), percent, decimal, and fraction.
……….
Do Now (3-5 min): The teacher writes the following fractions on the board and has the students write the equivalent decimal and percent.
15 = 55= 29 = 1 = 7
100 100 100 10 10
Opening (2 -3 min): Teacher reviews answers to the Do Now by allowing students to share their answers. The teacher should then state the objective, “Yesterday, we added and subtracted like fractions. Today, we are going to learn about why percents are important to our everyday lives. By the end of the lesson, you will be able to name equivalent fractions which will help you with percentages in our future lessons.”
Direct Instruction / Guided Practice (10 - 15 min): The teacher has a chart that reads. The teacher should read the chart to the students and then explain that percentages are important to our lives in many ways: voting, discounts at stores, proving mastery of a standard, taxes, and more. The teacher can ask the students if they can come up with any examples.
Percent means “per hundred” or “out of a hundred.” 1 percent means 1 or 0.01
100
Example: Jenny ran for student council and got 50 out of 100 votes.
This could be restated as:
- “If 100 people voted, then Jenny got 50 votes.”
- Jenny got 50 of the votes cast.”
100
The teacher should emphasize that “50 out of 100” does not mean that exactly 100 votes were cast but that Jenny got 50 for every 100 votes that were cast. If the election only had 60 votes, then Jenny would have gotten 50% of 60 votes, or 30 votes. If the election was much larger with 20,000 votes, then Jenny would have gotten 50% of 20,000, or 10,000 votes.
The teacher should then remind students that, just as with fractions, a percent always represents a percent of something (think of the voting example). The “something” is the whole 100%, which is the entire object, or the entire collection of objects, or the entire quantity being considered. In Jenny’s voting example, the total number of votes cast was the whole, or the ONE. The total number of votes cast is 100 percent of the votes.
The teacher then tells the students, “In this lesson you will be using 10 by 10 square grids to represent percentages.” The teacher should have the following problem on the board and an overhead of a 10 by 10 grid ready.
Problem 1: Last season, Duncan made 63 percent of his basketball shots.
The teacher should work through the problem using the following steps:
1. The 10 by 10 grid represents the whole (100%) – in this case all the shots Duncan attempted.
2. The 10 by 10 grid is made up of 100 small squares. Each small square is 1, or 1% of the whole. The decimal name is 0.01. 100
3. Sixty-three small squares are shaded. These shaded squares represent the number of shots that Duncan made out of everyone 100 shots he took.
4. Had Duncan taken 100 shots, he would have made 63 shots. This can also be stated as a fraction, 63 of his shots, or as a decimal, 0.63 of shi shots.
100
The class should then complete the following 2 examples together as a class. The teacher should be sure to discuss the real-world application.
GP 1: At the mall, Samantha paid 46 percent for her new sweater.
GP 2: Last year George mastered 83% of all his math standards.
Independent (10 min): The teacher passes out the IND worksheet, which mimics the GP.
Closing (2-3 min): Teacher calls the attention of the students back toward the front of the class to quickly review the answers to the Independent Practice worksheet/ ask what we learned about.
### Lesson Resources
GP Fractions Decimals Percents Classwork IND Fractions Decimals Percents Classwork |
# When slope is undefined what is the equation of the line?
If the slope of a line is undefined, then the line is a vertical line, so it cannot be written in slope-intercept form, but it can be written in the form: x=a , where a is a constant. If the line has an undefined slope and passes through the point (2,3) , then the equation of the line is x=2 .
## What is a line with an undefined slope?
Vertical lines have undefined slope. Since any two points on a vertical line have the same x-coordinate, slope cannot be computed as a finite number according to the formula, because division by zero is an undefined operation.
## How do you find the undefined slope?
If you attempt to find the slope using rise over run or any other slope formula, you will get a 0 in the denominator. Since division by 0 is undefined, the slope of the line is undefined. The equation of a line with undefined slope will look like x = 'something. '
## What is an example of undefined slope?
A good real life example of undefined slope is an elevator since an elevator can only move straight up or straight down. It got its name "undefined" from the fact that it is impossible to divide by zero. ... In general, when the x-values or x-coordinates are the same for both points, the slope is undefined.
## Is a slope of 0 4 undefined?
04=0 is defined. 40 is not.
## Finding Equation of a Line Given Point and Slope Undefined
42 related questions found
### Is a slope of 0 undefined?
Well you know that having a 0 in the denominator is a big no, no. This means the slope is undefined. As shown above, whenever you have a vertical line your slope is undefined.
### What does a positive slope look like?
What the Slope Means. ... A positive slope means that two variables are positively related—that is, when x increases, so does y, and when x decreases, y decreases also. Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises.
### How do I find the slope of the line?
Using two of the points on the line, you can find the slope of the line by finding the rise and the run. The vertical change between two points is called the rise, and the horizontal change is called the run. The slope equals the rise divided by the run: Slope =riserun Slope = rise run .
### What is the equation of the line?
The general equation of a straight line is y = mx + c, where m is the gradient, and y = c is the value where the line cuts the y-axis. This number c is called the intercept on the y-axis. The equation of a straight line with gradient m and intercept c on the y-axis is y = mx + c.
### How do you write in slope-intercept form?
Slope-intercept form, y=mx+b, of linear equations, emphasizes the slope and the y-intercept of the line.
### What does a slope of 2 look like?
In other words, our line moves 2 units upward every time it moves 1 unit to the right. Our slope is 2. It's a positive number, so we rise up and run to the right. Or, if we want to be contrary, both the rise and run could be negative, moving down and to the left.
### What is a zero slope equation?
A zero slope line is a straight, perfectly flat line running along the horizontal axis of a Cartesian plane. 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).
### Can you have a slope of 0 6?
Answer and Explanation: No, the slope 06 is not undefined. By definition, an undefined slope is a slope with a 0 in the denominator of the slope.
### What if the slope has a 0 on top?
When the 0 is on the “top” of the fraction, that would mean that the two y-values are the same. Thus that line is horizontal (slope of 0). If the “bottom” of the fraction is 0 that means the two x-values are the same. Thus that line is vertical (undefined slope).
### What does it mean to have a slope of 0 and an undefined slope?
No Slope. The "slope" of a vertical line. A vertical line has undefined slope because all points on the line have the same x-coordinate. As a result the formula used for slope has a denominator of 0, which makes the slope undefined..
### What does a zero slope look like?
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.
### Is the slope of the line positive or negative?
Pattern for Sign of Slope
If the line is sloping upward from left to right, so the slope is positive (+). If the line is sloping downward from left to right, so the slope is negative (-).
### Is 0 divided by 5 defined?
Answer: 0 divided by 5 is 0.
### What is the slope of 0 7?
Because a slope of 07 is 0 this is the slope of a horizontal line. The line perpendicular to this would be a vertical line. A vertical line, by definition, has an undefined slope.
### What to do if the slope is undefined?
If the slope of a line is undefined, then the line is a vertical line, so it cannot be written in slope-intercept form, but it can be written in the form: x=a , where a is a constant. If the line has an undefined slope and passes through the point (2,3) , then the equation of the line is x=2 .
### What is the slope of 3?
y=3 would be nothing more than a horizontal line through y=3. So the rise is always 0 (it never goes up or down) and the run is always the distance from zero to any point on the line. In other words, the slope would be 0/the change in x, which is always 0.
The slope is 4 . |
# Circle Passing Through 3 Points
To draw a straight line, the minimum number of points required is two. That means we can draw a straight line with the given two points. How many minimum points are sufficient to draw a unique circle? Is it possible to draw a circle passing through 3 points? In how many ways can we draw a circle that passes through three points? Well, let’s try to find answers to all these queries.
Learn: Circle Definition
Before drawing a circle passing through 3 points, let’s have a look at the circles that have been drawn through one and two points respectively.
## Circle Passing Through a Point
Let us consider a point and try to draw a circle passing through that point.
As given in the figure, through a single point P, we can draw infinite circles passing through it.
## Circle Passing Through Two Points
Now, let us take two points, P and Q and see what happens?
Again we see that an infinite number of circles passing through points P and Q can be drawn.
## Circle Passing Through Three Points (Collinear or Non-Collinear)
Let us now take 3 points. For a circle passing through 3 points, two cases can arise.
• Three points can be collinear
• Three points can be non-collinear
Let us study both cases individually.
### Case 1: A circle passing through 3 points: Points are collinear
Consider three points, P, Q and R, which are collinear.
If three points are collinear, any one of the points either lie outside the circle or inside it. Therefore, a circle passing through 3 points, where the points are collinear, is not possible.
### Case 2: A circle passing through 3 points: Points are non-collinear
To draw a circle passing through three non-collinear points, we need to locate the centre of a circle passing through 3 points and its radius. Follow the steps given below to understand how we can draw a circle in this case.
Step 1: Take three points P, Q, R and join the points as shown below:
Step 2: Draw perpendicular bisectors of PQ and RQ. Let the bisectors AB and CD meet at O such that the point O is called the centre of the circle.
Step 3: Draw a circle with O as the centre and radius OP or OQ or OR. We get a circle passing through 3 points P, Q, and R.
It is observed that only a unique circle will pass through all three points. It can be stated as a theorem and the proof is explained as follows.
It is observed that only a unique circle will pass through all three points. It can be stated as a theorem, and the proof of this is explained below.
Given:
Three non-collinear points P, Q and R
To prove:
Only one circle can be drawn through P, Q and R
Construction:
Join PQ and QR.
Draw the perpendicular bisectors of PQ and QR such that these perpendiculars intersect each other at O.
### Proof:
S. No Statement Reason 1 OP = OQ Every point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment. 2 OQ = OR Every point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment. 3 OP = OQ = OR From (i) and (ii) 4 O is equidistant from P, Q and R
If a circle is drawn with O as centre and OP as radius, then it will also pass through Q and R.
O is the only point which is equidistant from P, Q and R as the perpendicular bisectors of PQ and QR intersect at O only.
Thus, O is the centre of the circle to be drawn.
OP, OQ and OR will be radii of the circle.
From above it follows that a unique circle passing through 3 points can be drawn given that the points are non-collinear.
Till now, you learned how to draw a circle passing through 3 non-collinear points. Now, you will learn how to find the equation of a circle passing through 3 points. For this we need to take three non-collinear points.
## Circle Equation Passing Through 3 Points
Let’s derive the equation of the circle passing through the 3 points formula.
Let P(x1, y1), Q(x2, y2) and R(x3, y3) be the coordinates of three non-collinear points.
We know that,
The general form of equation of a circle is: x2 + y2 + 2gx + 2fy + c = 0….(1)
Now, we need to substitute the given points P, Q and R in this equation and simplify to get the value of g, f and c.
Substituting P(x1, y1) in equ(1),
x12 + y12 + 2gx1 + 2fy1 + c = 0….(2)Â
x22 + y22 + 2gx2 + 2fy2 + c = 0….(3)Â
x32 + y32 + 2gx3 + 2fy3 + c = 0….(4)
From (2) we get,Â
2gx1 = -x12 – y12 – 2fy1 – c….(5)Â
Again from (2) we get,Â
c = -x12 – y12 – 2gx1 – 2fy1….(6)Â
From (4) we get,Â
2fy3 = -x32 – y32 – 2gx3 – c….(7)
Now, subtracting (3) from (2),
2g(x1 – x2) = (x22 -x12) + (y22 – y12) + 2f (y2 – y1)….(8)
Substituting (6) in (7),
2fy3 = -x32 – y32 – 2gx3 + x12 + y12 + 2gx1 + 2fy1….(9)
Now, substituting equ(8), i.e. 2g in equ(9),
2f = [(x12 – x32)(x1 – x2) + (y12 – y32 )(x1 – x2) + (x22 – x12)(x1 – x3) + (y22 – y12)(x1 – x3)] / [(y3 – y1)(x1 – x2) – (y2 – y1)(x1 – x3)]
Similarly, we can get 2g as:
2g = [(x12 – x32)(y1 – x2) + (y12 – y32)(y1 – y2) + (x22 – x12)(y1 – y3) + (y22 – y12)(y1 – y3)] / [(x3 – x1)(y1 – y2) – (x2 – x1)(y1 – y3)]
Using these 2g and 2f values we can get the value of c.
Thus, by substituting g, f and c in (1) we will get the equation of the circle passing through the given three points.
### Solved Example
Question:
What is the equation of the circle passing through the points A(2, 0), B(-2, 0) and C(0, 2)?
Solution:
Consider the general equation of circle:
x2 + y2 + 2gx + 2fy + c = 0….(i)
Substituting A(2, 0) in (i),
(2)2 + (0)2 + 2g(2) + 2f(0) + c = 0
4 + 4g + c = 0….(ii)
Substituting B(-2, 0) in (i),
(-2)2 + (0)2 + 2g(-2) + 2f(0) + c = 0
4 – 4g + c = 0….(iii)
Substituting C(0, 2) in (i),
(0)2 + (2)2 + 2g(0) + 2f(2) + c = 0
4 + 4f + c = 0….(iv)
4 + 4g + c + 4 – 4g + c = 0
2c + 8 = 0
2c = -8
c = -4
Substituting c = -4 in (ii),
4 + 4g – 4 = 0
4g = 0
g = 0
Substituting c = -4 in (iv),
4 + 4f – 4 = 0
4f = 0
f = 0
Now, substituting the values of g, f and c in (i),
x2 + y2 + 2(0)x + 2(0)y + (-4) = 0
x2 + y2 – 4 = 0
Or
x2 + y2 = 4
This is the equation of the circle passing through the given three points A, B and C.
To know more about the area of a circle, equation of a circle, and its properties download BYJU’S-The Learning App. |
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# An Introduction to Alternate Exterior Angles in Geometry
Alternate exterior angles are a type of angle found when two lines are crossed by a third line, known as a transversal. These angles can be found on either side of the transversal and outside the two lines that are crossed. Alternate exterior angles are important in geometry, as they can be used to form and identify congruent angles.
## Definition of Alternate Exterior Angles
Alternate exterior angles, also known as "consecutive exterior angles," are two angles that are located on opposite sides of a transversal line and outside the two lines crossed by the transversal line. These angles are supplementary, meaning that they add up to 180 degrees. They are also congruent, meaning that they are equal in measure. The two alternate exterior angles are always congruent to each other, regardless of the type of lines that are crossed.
## Identifying Alternate Exterior Angles
To identify alternate exterior angles, you need to first identify the transversal line. This line crosses through two other lines, which are usually parallel lines. Once the transversal line is identified, look for the two angles that are located on opposite sides of the transversal line and outside of the two lines crossed by the transversal line. These are the alternate exterior angles.
## Theorem of Alternate Exterior Angles
The theorem of alternate exterior angles states that when two parallel lines are crossed by a transversal line, the alternate exterior angles are congruent. This means that the two angles have the same measure. This theorem can be used to prove the existence of parallel lines. It can also be used to prove that two angles are congruent.
## Examples of Alternate Exterior Angles
To understand alternate exterior angles better, here are some examples of how to identify them. In the figure below, the two lines AB and CD are crossed by the transversal line EF. The angles 1 and 2 are alternate exterior angles. The angles 3 and 4 are also alternate exterior angles.
## Practice Problems
1. Identify the alternate exterior angles in the figure below.
The angles 1 and 2 are alternate exterior angles.
2. Are angles 3 and 4 alternate exterior angles in the figure shown above?
No, angles 3 and 4 are not alternate exterior angles in the figure shown above.
3. Are angles 5 and 6 alternate exterior angles in the figure shown above?
Yes, angles 5 and 6 are alternate exterior angles in the figure shown above.
4. Are the alternate exterior angles in the figure below congruent?
Yes, the alternate exterior angles in the figure above are congruent.
5. What is the measure of the alternate exterior angles in the figure above?
The measure of the alternate exterior angles in the figure above is 45 degrees.
6. Are the alternate exterior angles in the figure below supplementary?
Yes, the alternate exterior angles in the figure above are supplementary.
7. What is the sum of the measure of the alternate exterior angles in the figure above?
The sum of the measure of the alternate exterior angles in the figure above is 180 degrees.
## Conclusion
In conclusion, alternate exterior angles are angles found on opposite sides of a transversal line and outside the two lines crossed by the transversal line. These angles are congruent and supplementary, and they can be used to form and identify congruent angles. By understanding how to identify alternate exterior angles, you can use this knowledge to solve many geometry problems.
## FAQ
### What is an alternate exterior angle?
An alternate exterior angle is an angle outside two lines that are cut by a transversal, and are on opposite sides of the transversal.
### How are alternate exterior angles related to each other?
Alternate exterior angles are always equal to each other. |
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# The quotient when $1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}}$ is divided by $1 + x + {x^2} + {x^3} + ........... + {x^{17}}$. isA. ${x^{17}} - {x^{15}} + {x^{13}} - {x^{11}} + ......x$B. ${x^{17}} + {x^{15}} + {x^{13}} + {x^{11}} + ..........x$C. ${x^{17}} + {x^{16}} + {x^{15}} + {x^{14}} + ..........x$D. ${x^{17}} - {x^{16}} + {x^{15}} - {x^{14}} + .......... + x$
Last updated date: 24th Jun 2024
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Hint: In order to solve this problem we need to use the formula of sum of finite geometric series $S = \dfrac{{a({r^n} - 1)}}{{r - 1}}$ . Then we need to use it for both the series given and then we need to divide the obtained sum to get the right answer.
As we know, the sum of finite geometric series $1,r,{r^2},{r^3},{r^4}..........{r^{n - 1}}$ with common ratio is given by
$a,ar,a{r^2},a{r^3},a{r^4}..........a{r^{n - 1}} = S = \dfrac{{a({r^n} - 1)}}{{r - 1}}$ = sum (where r is greater than 1)
When r is less than 1 then the formula of sum for n terms of GP will be $\dfrac{{a(1 - {r^n})}}{{1 - r}}$.
Now, $1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}}$ forms a finite geometric series with common ratio ${x^2}$, first term 1 and number of terms n = 18 (since power of x from 2 to 34 with the difference of 2 is 17 terms and the first term is 1 therefore n = 18) then sum is,
$1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}} = \dfrac{{{x^{2(18)}} - 1}}{{{x^2} - 1}}$
$1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}} = \dfrac{{{x^{36}} - 1}}{{{x^2} - 1}}$……….(1)
Now, $1 + x + {x^2} + {x^3} + ........... + {x^{17}}$ forms a finite geometric series with common ratio x, first term 1 and number of terms n = 18 (since power of x from 1 to 17 is 17 terms and first term is 1 therefore n = 18) then sum is
$1 + x + {x^2} + {x^3} + ........... + {x^{17}} = \dfrac{{{x^{18}} - 1}}{{x - 1}}$……………..(2)
Now, Dividing (1) by (2) equation
$\Rightarrow \dfrac{{\left( {{x^{36}} - 1} \right)\left( {x - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^{18}} - 1} \right)}} = \dfrac{{{x^{18}} - 1}}{{x + 1}}$
Now, we have to divide${x^{18}} + 1$ by (x + 1). We can see the degree of the numerator is 18 and the denominator has a linear equation. So, the degree of quotient starts from 17 and decreases with a difference of one.
$\Rightarrow \dfrac{{{x^{18}} + 1}}{{x + 1}} = {x^{17}} - {x^{16}} + {x^{15}} - {x^{14}} + .......... + x$
So, the correct answer is “Option d”.
Note: Whenever we face such types of problems we use some important points. Like first of all try to convert series into easy form by using sum of series. While dividing carefully observe the degree of quotient and sign of coefficient. Doing these things will solve all such problems. |
# Class 9 Maths Important Questions for Area of parallelograms
Given below is Area of parallelogram worksheet
a) Concepts questions
b) Calculation problems
c) Multiple choice questions
e) Fill in the blank's
Question 1
Which of the following figures lie on the same base and between the same parallels?
In such a case, write the common base and the two parallel
Solution 1
a) True. With Base BC and between parallel AD and BC, Triangle QBC and parallelogram ABCD
b) True. With base DC or AB and between parallel DC and AB, triangles are present
c) True. Same as above
d) False
e) True. Same as a
f) False
g) True
Question 2
True or False statement
(a) If two triangles area are same areas, they will be congruent
(b) Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
(c) The area of a triangle is equal to the product of any of its side and any altitude
(d) The median of the triangles divides the triangle into two triangles of equal areas
(e) Parallelograms on the same base and between same parallels have same perimeter
(f) In a parallelogram, diagonals divide the parallelogram into four equal triangles
Solution
1. False. Congruent triangles have equal areas but converse is not true
2. True. Triangle area is (1/) X base X height. With same base and area, height should be equivalent, which means they lie on same parallel
3. False. It is corresponding base and corresponding altitude
4. True.
5. False. Area is same but perimeter can be different
6. True
# Multiple choice Questions
Question 3
PQRS is a rectangle with O as any point in its interior. If area (ΔPOS)= 4 cm2 and area(ΔQOR)= 6 cm2, then area of rectangle PQRS
(a) 10cm2
(b) 20cm2
(c) 14 cm2
(d) cm2
Solution
$\frac{1}{2}PS \times (alitude)_{1}=4cm^{2}$
$\frac{1}{2}QR \times (alitude)_{2}=6cm^{2}$
$\frac{1}{2}PS \times [(alitude)_{1}+(altitude)_{2}]=10cm^{2}$
Now altitude1 + altitude2= PQ=RS
So area of rectangle=20 cm2
Question 4
In the below figure AD is the median, And E is any point on AD
Which of the following is true?
a) Area of triangle AEB= Area of triangle AEC
b) Area of triangle DEB= Area of triangle DEC
c) Area of triangle ABD= Area of triangle ADC
d) All the above
Solution (d)
Since AD is median, it bisect the triangle is equal areas
So Area of triangle ABD= Area of triangle ADC ---(1)
Now ED is the median for EBC triangle
So Area of triangle DEB= Area of triangle DEC --(2)
Subtracting 1 and 2, we get
Area of triangle AEB= Area of triangle AEC
Question 5
In the given figure ABCD is a parallelogram AE ⊥ DC and CF ⊥ AD. If AB = 18 cm, AE = 8 cm and CF = 16 cm, find AD.
a) 9 cm
b) 8 cm
c) 10 cm
d) None of the above
Solution (a)
Parallelogram area= base X height
Question 6
In the below figure AD is the median, and E is mid-point on AD. If the area of triangle is 16 cm2, what is the area of the triangle BED
(a) 3cm2
(b) 4cm2
(c) 5 cm2
(d) None of these
Solution b
Question 7
PQRS is a quadrilateral whose diagonal bisect each other at right angles
a) PQRS is a Square
b) PQRS is a rectangle
c) PQRS is a rhombus
d) None of these
Solution (c)
Question 8
In a quadrilateral ABCD, diagonal BD and AC intersect at point X
Which of the following is true?
(a) (Area of triangle BXC)×( Area of triangle AXD)=( Area of triangle AXB)×(Area of triangle CXD)
(b) (Area of triangle BXC)+( Area of triangle AXD)=( Area of triangle AXB)+(Area of triangle CXD)
(c) Insufficient information
(d) None of these
Solution (a)
Hint, Draw perpendicular from A and C on BD and the calculate the area of each these piece and arranged them to get the solution
Question 9
Two parallelograms are on the same base and between the same parellels. The ratio of their areas is:
a) 1:2
b) 1:1
c) 1:4
d) None of these
Solution (b)
Question 10
In a triangle A,B,C,D and E are such point BD=DE=EC
Which of the following is true?
a) Area of triangle ABD=Area of triangle ADE=Area of triangle AEC
b) Area of triangle ADE=(1/3) Area of triangle ABC
c) Triangle ABD is congruent to triangle AEC
d) None of the above
Solution (a), (b)
Hint: Draw perpendicular from A on BC and then calculate area for each of these triangle and you will find it same
## Match the column
Area of the triangle is 20 cm2, Area of rectangle is 25cm2. Area of the parallelogram is 100 cm2. Both the Diagonal are drawn which cut the area into four pieces. The area of each piece is 40cm2 In a triangle ABC,all the median intersect at point G, If the area of the triangle is 150 cm2,what is the area of the triangle AGC 10 cm2 A trapezoid as parallel sides of 4 and 6 cm respectively, The altitude is 5 cm.A diagonal is drawn which cut the trapezoid into two traingles.Area of the triangle with base 4 cm is 50 cm2
Solution
a) 40 cm2
b) 25cm2
c) 50 cm2
d) 10 cm2 |
# Given a finite Group G, with A, B subgroups prove the order of AB
How do you prove:
Given a finite group $G$, with $A,B$ subgroups then $$|AB|=\frac{|A||B|}{|A \cap B|}.$$
#### Solutions Collecting From Web of "Given a finite Group G, with A, B subgroups prove the order of AB"
You can prove that $|AB||A\cap B|=|A||B|$ directly.
There is a natural map $p$ from $A\times B$ to $AB$ by $(a,b)\mapsto ab$, which is onto. The cardinality of $A\times B$ is therefore equal to
$$\sum_{x\in AB}|p^{-1}(g)|.$$
Given an element $g\in AB$, let $(a,b)\in p^{-1}(g)$. For each $x\in A\cap B$, we obtain a second pair $(ax,x^{-1}b)$ that also maps to $ab$; thus, each element of $AB$ has at least $|A\cap B|$ preimages.
If $(a,b)$ and $(a',b')$ have the same image, then $ab=a'b'$, hence $bb'^{-1}= a^{-1}a'\in A\cap B$. Letting $x=a^{-1}a'$ we have that $(a',b') = (ax,x^{-1}b)$. That is, for each element $g$ of $AB$, there is a bijection between the preimages of $g$ in $A\times B$ and the set $A\cap B$. Therefore,
$$|A\times B| = |A||B| = \sum_{x\in AB}|p^{-1}(g)| = \sum_{x\in AB}|A\cap B| = |AB||A\cap B|,$$
and this holds in the sense of cardinality, even if the sets are infinite.
In the case where $A\cap B$ is finite, we get the desired equality.
This is the orbit-stabilizer theorem.
Let $X=\lbrace aB\ |\ a\in A\rbrace$ be a subset of the cosets of $B$ in $G$. Then $A$ acts transitively on $X$, so $|X|$ is equal to the index of a stabilizer in $A$. So let $C\leq A$ be the subgroup stabilizing the coset $B\in X$; this is simply the elements $z\in A$ such that $zB=B$. This just means $z\in B$, so the stabilizer $C=A\cap B$. So we have $|X|=[A:C]=\dfrac{|A|}{|A\cap B|}$. Since $|AB|=|B|\cdot |X|$ (each coset has $|B|$ elements), we get the formula you wrote. |
# a.2. mathematical induction
The ninth of the Peano axioms is known as the principle of mathematical induction.
Principle of mathematical induction. If $K$ is a set containing $0$ that is closed under the successor function, then $K$ is equal to the natural numbers $\mathbb{N}_0$.
We used this principle liberally in the Peano axioms exercises to prove various properties of the natural numbers.
### example
For any natural number $n$, prove that the number $n^2 + n$ is even. (A natural number is even if and only if it can be expressed as a natural number multiplied by 2.)
solution
We let $K$ be the set of natural numbers satisfying the conclusion $$K = \{ k \in \mathbb{N}_0 \mid k^2 + k \text { is even} \}.$$ We want to prove that $K = \mathbb{N}$, and we will use the principle of mathematical induction to do so.
First we prove that $0 \in K$. Zero is even since $0 = 2 \cdot 0$. Therefore, $0^2 + 0 = 0$ is even, so $0 \in K$.
Next we prove that $K$ is closed under the successor function. We need to prove: if $k$ is in $K$, then its successor $k + 1$ is also in $K$. For any $k \in K$, we have that $k^2 + k$ is even, so $$k^2 + k = 2 n$$ for some natural number $n$. To prove that $k + 1 \in K$, we calculate \begin{align} (k + 1)^2 + (k + 1) &= (k^2 + 2k + 1) + (k + 1) \\ &= (k^2 + k) + (2k + 2) \\ &= 2n + 2k + 2 \\ &= 2(n + k + 1) \end{align} and conclude that $(k + 1)^2 + (k + 1)$ is even since it is a natural number multiplied by 2. This shows that $k + 1 \in K$, so $K$ is closed under the successor function.
Finally we conclude that $K = \mathbb{N}_0$ by the principle of mathematical induction.
## Alternate characterization
A more familiar version of the principle of mathematical induction involves propositions about natural numbers.
Principle of mathematical induction. Let $p(n)$ be a proposition about $n \in \mathbb{N}_0$. Suppose that
• Base case: the proposition $p(0)$ is true;
• Inductive step: for any $k \in \mathbb{N}_0$, if $p(k)$ is true, then $p(k + 1)$ is also true.
Then $p(n)$ is true for all $n \in \mathbb{N}_0$.
### example
We'll prove the same example from above using this alternate characterization, that the number $n^2 + n$ is even for any natural number $n$.
solution
For any natural number $n$, let $p(n)$ be the proposition that $n^2 + n$ is even.
Base case. The proposition $p(0)$ is true since $0^2 + 0 = 2 \cdot 0$ is even.
Inductive step. Suppose that $p(k)$ is true, ie, that $k^2 + k$ is even. Then we can express $k^2 + k$ as $2n$ for some natural number $n$. We need to prove that $p(k + 1)$ is true, ie, that $(k + 1)^2 + (k + 1)$ is even. As before, \begin{align} (k + 1)^2 + (k + 1) &= (k^2 + 2k + 1) + (k + 1) \\ &= (k^2 + k) + (2k + 2) \\ &= 2n + 2k + 2 \\ &= 2(n + k + 1) \end{align} is even, so $p(k + 1)$ is true.
By mathematical induction, since $p(0)$ is true and $p(k)$ implies $p(k + 1)$ then $p(n)$ is true for all $n \in \mathbb{N}_0$.
## different base case
The principle of mathematical induction can also be used to prove that a proposition is true for all natural numbers above a certain threshold.
Principle of mathematical induction. Let $p(n)$ be a proposition about $n \in \mathbb{N}_c = \{ c, c + 1, c + 2, \dots \}$. Suppose that
• Base case: the proposition $p(c)$ is true;
• Inductive step. for any $k \in \mathbb{N}_c$, if $p(k)$ is true, then $p(k + 1)$ is also true.
Then $p(n)$ is true for all $n \in \mathbb{N}_c$.
### example
For any natural number $n \geq 1$, prove that the sum of the first $n$ odd natural numbers is equal to $n^2$.
solution
For $n \in \mathbb{N}_1$, let $p(n)$ be the proposition that the sum of the first $n$ odd natural numbers is equal to $n^2$. For example, here are the first four propositions. \begin{align} {}p(1) &: 1 = 1^2 \\ p(2) &: 1 + 3 = 2^2 \\ p(3) &: 1 + 3 + 5 = 3^2 \\ p(4) &: 1 + 3 + 5 + 7 = 4^2 \end{align}
To express the equation for $p(n)$ in general, we need to know what the $n^\text{th}$ odd number is. Note that \begin{align} 1^\text{st} \text{ odd number: } & 1 = 2(1) - 1 \\ 2^\text{nd} \text{ odd number: } & 3 = 2(2) - 1 \\ 3^\text{rd} \text{ odd number: } & 5 = 2(3) - 1 \\ 4^\text{th} \text{ odd number: } & 7 = 2(4) - 1 \\ \end{align} From this pattern, we conclude that the $n^\text{th}$ odd number is $2n - 1$. (This can be proven using mathematical induction, see the exercise below.) The proposition $p(n)$ is the equation $$1 + 3 + 5 + \dots + (2n - 1) = n^2.$$
Base case. The base case $p(1)$ (expressed above) is clearly true. (In fact, by inspection, we see that $p(n)$ is true for those first four cases $n = 1, 2, 3, 4$.)
Inductive step. Assume that $p(k)$ is true, so the sum of the first $k$ odd numbers is equal to $k^2$, ie, $$1 + 3 + 5 + \dots + (2k - 1) = k^2.$$ To prove $p(k + 1)$, we consider the sum of the first $k + 1$ odd numbers, which is $$1 + 3 + 5 + \dots + (2k + 1).$$ Notice that the first $k$ odd numbers appear in this sum as well, $$(1 + 3 + 5 + \dots + (2k - 1)) + (2k + 1).$$ Since $p(k)$ is true, this is equal to $$(k^2) + (2k + 1).$$ Factoring this expression, we get $$(k + 1)^2$$ so we conclude that the sum of the first $k + 1$ odd numbers is equal to $(k + 1)^2$; that is, we conclude that $p(k + 1)$ is true.
By the principle of mathematical induction, $p(n)$ is true for all natural numbers $n \geq 1$.
## exercises
1. For any $n \in \mathbb{N}_1$, prove that the $n^\text{th}$ odd number is equal to $2n - 1$.
solution
Let $p(n)$ be the statement that the $n^\text{th}$ odd number is $2n - 1$.
Base case. The statement $p(1)$ is true since the 1st odd number is $1$ and $2(1) - 1 = 1$.
Inductive step. Suppose that $p(k)$ is true, ie, that $2k - 1$ is the $k^\text{th}$ odd number. Well, the $(k + 1)^{st}$ odd number is just 2 more than the $k^{th}$ odd number, so it is equal to $$(2k - 1) + 2 = 2k + 1 = 2(k + 1) - 1,$$ which proves that $p(k + 1)$ is true.
By mathematical induction, $p(n)$ is true for all $n \geq 1$.
2. For any $n \in \mathbb{N}_0$, prove that $3^n - 1$ is even.
solution
Let $p(n)$ be the statement that $3^n - 1$ is even.
Base case. The proposition $p(0)$ is true since $3^0 - 1 = 1 - 1 = 0 = 2 \cdot 0$ is even.
Inductive step. Suppose that $p(k)$ is true, ie, that $3^k - 1 = 2n$ for some natural number $n$. We compute \begin{align} 3^{k + 1} - 1 &= 3 \cdot 3^k - 1 \\ &= (2 + 1) \cdot 3^k - 1 \\ &= 2 \cdot 3^k + 3^k - 1 \\ &= 2 \cdot 3^k + 2n \\ &= 2 (3^k + 2) \end{align} to see that $3^{k + 1} - 1$ is also even, hence $p(k + 1)$ is true.
By mathematical induction, $p(n)$ is true for all $n \geq 0$. |
# 5.9: Use the Percent Equation to Find the Base, b
Difficulty Level: At Grade Created by: CK-12
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Practice Percent Equation to Find the Base b
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Have you ever enjoyed autumn? Take a look at this dilemma involving trees and their leaves.
By mid-September, 50% of the trees lose their leaves. If 850 trees in a grove lost their leaves, how many trees are there in all?
Use the percent equation to solve this problem. You will learn all that you need to know in this Concept.
### Guidance
Did you know that you can use the percent equation to solve percent problems? Take a look at how you can move from the proportion to the percent equation.
When we solve the proportion , we use cross products to find the missing variable. However, even if we leave it in terms of the variables, we can still use cross multiply.
If we change the percent to a decimal by moving the decimal point two places to the left, then there is no need to multiply by .01 as we will have already accounted for the coefficient of .01 by moving the decimal point.
Okay, let’s go through it again. Look at what we just wrote.
We wrote the same thing we just didn’t include values. The variables stayed and we multiplied them.
The key is that if we change the percent to a decimal, then all we have to do is to multiply it by the base and we will be able to figure out the value of .
This situation showed us how to use go from a proportion to the percent equation when solving for part a.
Sometimes, you will know the percent and a part of the ratio, or part , but you will need to find the whole or the base, . When this happens, you can use the same key words as before and simply figure out the base by using the percent equation. Let’s look at one like this.
78 is 65% of what number?
Here we know that the word “is” means equals. The numbers may be in a different location, but just pay attention to the key words and you will know what to do. Notice that we have been given the percent and we are missing the “of what number” that is the value of the base. Let’s write the equation.
To work with the 65%, it makes sense to convert it to a decimal. We do this by dropping the percent sign and moving the decimal two places to the left.
Now we can solve it for the value of . Divide both sides of the equation by .65.
This is the answer.
Here is another one.
11 is 77% of what number?
Once again, pay attention to the key words. You can see that we are once again going to be looking for the value of the base. Let’s write the equation.
Convert the percent to a decimal and solve.
In this problem, you could round to the nearest hundredths place as we did here. Sometimes, you may be asked to round to the nearest tenths place. In that case, the answer would have been 14.3.
Solve each problem using the percent equation.
#### Example A
10 is 50% of what number?
Solution:
#### Example B
45 is 20% of what number?
Solution:
#### Example C
68 is 40% of what number?
Solution:
Now let's go back to the dilemma from the beginning of the Concept.
Let’s start by breaking apart this problem. We have a percent, so we know that we won’t be looking for the percent. We know that 850 trees in a grove lost their leaves, but we don’t know the total number of trees in the grove. The total could be thought of as the whole and this is the base. We are going to be looking for the base.
Let’s write the equation.
Now we solve by dividing both sides of the equation by .50.
There are 1700 trees in the grove.
### Vocabulary
Percent
a part of a whole out of 100.
### Guided Practice
Here is one for you to try on your own.
25 is 60% of what number?
Solution
First, let's write down the percent equation.
In this problem, we are solving for the base. Let's fill in the values that we know.
Now we divide to solve for .
Our answer is or .
### Practice
Directions: Solve each percent problem. You may round your answers to the nearest tenth when necessary.
1. 23 is 9% of what number?
2. 10 is 35% of what number?
3. 580 is 82% of what number?
4. 58 is 8% of what number?
5. 58 is 80% of what number?
6. 11 is 82% of what number?
7. 33 is 2% of what number?
8. 14 is 9% of what number?
9. 50 is 67% of what number?
10. 33 is 45% of what number?
11. 40 is 80% of what number?
12. 68 is 99% of what number?
13. 78 is 55% of what number?
14. 16 is 12% of what number?
15. 1450 is 80% of what number?
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### Vocabulary Language: English
TermDefinition
Decimal In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
Equation An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
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# How to find the centroid of a triangle? – Step by step
The centroid of a triangle can be found using an algebraic method or using a graphical method. To find the centroid algebraically, we have to use a formula and the coordinates of the vertices of the triangle. To find the centroid graphically, we have to draw two medians of the triangle.
In this article, we will learn how to find the centroid of a triangle using the two methods listed. Then, we will solve some practice problems.
##### GEOMETRY
Relevant for
Learning how to find the centroid of a triangle.
See methods
##### GEOMETRY
Relevant for
Learning how to find the centroid of a triangle.
See methods
## What is the centroid of a triangle?
The centroid of a triangle is the point of intersection of the three medians of the triangle. In the diagram below, we can see that the red dot is the centroid:
Remember that the medians of the triangle are the segments that connect a vertex with the midpoint of its opposite side. This means that the centroid is the geometric center of the triangle.
### Properties of the centroid of a triangle
The following are some of the important properties and characteristics of the centroid of a triangle:
• The centroid of a triangle is the point of intersection of the medians of the triangle.
• The centroid represents the geometric center of the triangle.
• The centroid of a triangle is always located inside the triangle.
• The centroid of an equilateral triangle is located in the same position as its incenter, orthocenter, and circumcenter.
• The centroid divides the medians in a 2:1 ratio.
## Finding the centroid of a triangle graphically
The centroid of a triangle can be found graphically by sketching the medians of the triangle and determining their point of intersection.
In turn, we can find the medians by finding the midpoint of each side and drawing a line segment from that point to the opposite vertex. We do this with the following steps:
Step 1: Measure the length of side AB and mark its midpoint to obtain point D.
Step 2: Draw a line segment from vertex C to point D.
Step 3: Measure the length of side AC and mark its midpoint to obtain point E.
Step 4: Draw a line segment from vertex B to point E.
Step 5: Mark the point of intersection of segments AB and AC.
The segments AB and AC are the medians of the triangle. This means that the point of intersection is the centroid of the triangle.
## Finding the centroid of a triangle algebraically
The centroid of a triangle can be found algebraically if we know the coordinates of the vertices of the triangle. We can follow the process using the following triangle.
In this triangle, $latex (x_{1},~y_{1})$, $latex (x_{2},~y_{2})$, $latex (x_{3},~y_{3})$ are the vertices and the point C(x, y) is the centroid of the triangle.
Then, we use the following formula:
$latex C(x, y)=\left( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3}\right)$
Basically, we have to add the x-coordinates of the vertices and divide the sum by 3 to get the x-coordinate of the centroid. Similarly, we add the y-coordinates of the vertices and divide the sum by 3 to get the y-coordinate of the centroid.
## Centroid of a triangle – Examples with answers
In the following examples, we apply the algebraic method to find the centroid of the triangles.
### EXAMPLE 1
What are the coordinates of the centroid of a triangle that has vertices A(3, 6), B(1, 1), and C(5, 2)?
Solution: We have to use the centroid formula with the following coordinates:
• $latex (x_{1},~y_{1})=(3, ~6)$
• $latex (x_{2},~y_{2})=(1,~1)$
• $latex (x_{3},~y_{3})=(5,~2)$
Therefore, we have:
$latex C(x, y)=\left(\frac{x_{1}+x_{2}+x_{3}}{3},~\frac{y_{1}+y_{2}+y_{3}}{3}\right)$
$latex C(x, y)=\left(\frac{3+1+5}{3},~\frac{6+1+2}{3}\right)$
$latex C(x, y)=\left(\frac{9}{3},~\frac{9}{3}\right)$
$latex C(x, y)=(3,~3)$
### EXAMPLE 2
What are the coordinates of the centroid of a triangle that has vertices A(5, 7), B(2, 3), and C(6, 4)?
Solution: We have the following coordinates:
• $latex (x_{1},~y_{1})=(5, ~7)$
• $latex (x_{2},~y_{2})=(2,~3)$
• $latex (x_{3},~y_{3})=(6,~4)$
Then, using the centroid formula, we have:
$latex C(x, y)=\left(\frac{x_{1}+x_{2}+x_{3}}{3},~\frac{y_{1}+y_{2}+y_{3}}{3}\right)$
$latex C(x, y)=\left(\frac{5+2+6}{3},~\frac{7+3+4}{3}\right)$
$latex C(x, y)=\left(\frac{13}{3},~\frac{14}{3}\right)$
## Centroid of a triangle – Practice problems
Solve the following practice problems using the algebraic method to find the coordinates of the centroid. |
# Slope intercept solver
In this blog post, we will be discussing about Slope intercept solver. Our website will give you answers to homework.
## The Best Slope intercept solver
Slope intercept solver can be found online or in mathematical textbooks. This is the best word problem calculator out there. It can solve any word problem into an equation. This is especially useful if you’re trying to figure out how to solve a problem on your own. It’s also a great way to check whether your answer is correct before plugging it into a calculator or formula sheet.
If you're struggling with math, there's no need to feel alone or embarrassed. Help is available, and you can get it from the comfort of your own home. There are plenty of free math help online chat rooms and forums, and they can be a great resource. You can find other students who are struggling with the same topics, and you can work together to find solutions. You can also get help from experts, and you might even be able to find a tutor. Whatever you
R is a useful tool for solving for radius. Think of it like a ruler. If someone is standing in front of you, you can use your hand to measure their height and then use the same measurement to determine the radius of their arm. For example, if someone is 5 feet tall and has an arm that is 6 inches long, their radius would be 5 inches. The formula for calculating radius looks like this: [ ext{radius} = ext{length} imes ext{9} ] It's really just making the length times 9. So, if they're 6 inches tall and their arm is 6 inches long, their radius would be 36 inches. Using R makes sense when you are trying to solve for any other dimension besides length - such as width or depth. If a chair is 4 feet wide and 3 feet deep, then its width would be equal to half its depth (2 x 3 = 6), so you could easily calculate its width by dividing 2 by 1.5 (6 ÷ 2). But if you were trying to figure out the chair's height instead of its width, you would need an actual ruler to measure the distance between the ground and the seat. The solution to this problem would be easier with R than without it.
A new app called PhotoMath has been released that purports to be a “cheat” for math homework. The app uses the camera on a smartphone to take pictures of math problems, and then displays the answer on the screen. The app is currently available for free on iOS and Android devices. While the app may be helpful for some students, there are potential dangers associated with its use. First, students who use the app may become reliant on it, and lose the
Solving trinomials can be a tricky business, but there are a few methods that can make the process a bit easier. One common method is to factor the trinomial into two binomials. This can be done by grouping the terms together in pairs, and then multiplying each pair to get the product. Another method is to use the quadratic formula. This involves plugging the values of the coefficients into a specific equation, and then solving for x. While these methods may seem daunting at first, with a little practice they can become second nature. With some patience and perseverance, solving trinomials can be a breeze.
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## Word Problems – 7
This one is a year 12 problem involving calculus.
A company initially provides a service to 1000 customers for $5 per month. The marketing department says that for every 10¢ reduction in price, they could get 100 more customers. What price would give the company the maximum revenue per month and what would that revenue be? Let’s let x be the monthly price for the service. Then the revenue, R(x), would be x times the number of customers. The number of customers is the initial 1000 plus 100 times the number of 10¢ increments below$5 that is charged. The number of 10¢ increments below $5 is (5 – x)/0.1, so the revenue is $R\left(x\right)=x\left[1000+100\frac{\left(5-x\right)}{0.1}\right]=x\left[1000+1000\left(5-x\right)\right]$ $=1000\left(6x-x^2\right)$ Looking at this function, you can recognise that this is an upside down parabola because of the minus sign in front of the x² term. So the maximum would be at the top of the parabola. This makes sense because there is a balancing act going on between a lot of customers and too low a price. The revenue will rise until the price is too low to increase the revenue. To find that point that is the maximum revenue, we need to find the derivative of R(x) and set that equal to 0, that is find the stationary point that is the top of the parabola. $R’\left(x\right)=1000\left(6-2x\right)=0$ $\Longrightarrow\ x=3$ So the price that maximises revenue is$3, and R(3) = \$9000. The number of customers is 1000 + 1000(2) = 3000.
## Word Problems – 4
Continuing this series on word problems, let’s look at one that many year 11 or 12 students have seen if they have covered calculus. The first part of the question though, does not need calculus:
A 6 by 8 cm rectangular piece of metal has a square cut out of each corner:
The metal is then folded along the dashed lines to form a box of height x.
a) What is the volume of the box in terms of x?
b) What is the maximum volume the box can have and at what value of x does the maximum occur?
Let’s first redraw the rectangle, labelling what we know. If the height of the box is to be x, then that is the size of the cutout square. If an x is subtracted from each end of each side, then the length of each of the dotted lines (the base of the future box) is 6 – 2x and 8 – 2x:
After the metal is folded, we have a box like:
So to answer part a), the volume is height × length × width. So
V = x(8 – 2x)(6 – 2x)
For part b), we need a little calculus. But before we do that, just to get a mental image of what is going on (this is not needed to solve the problem though), let’s plot V as a function of x:
Notice that the volume is 0 at x = 0 and x = 3. This makes physical sense because when x = 0, there is no square cut out and the box would just be a flat sheet. If x = 3, there would be nothing left of the original 6 cm side and the two remaining flaps would just fold up against each other giving no volume. So the physical restriction on x here is 0 ≤ x ≤ 3.
The maximum volume then occurs at the peak of the curve between 0 and 3. To find what and where this maximum is, requires calculus, specifically finding a local maximum (a stationary point).
To find this point, let’s first expand the expression for V:
V = x(8 – 2x)(6 – 2x) = 4x3 – 28x2 + 48x
For the uninitiated, derivatives of a function give the gradients of lines tangent (that is just touching at one point) to it. At the maximum, the tangent line is horizontal which has a gradient of 0. So to find this, we find the derivative of our volume function and set that equal to zero and solve for x.
For those who have had calculus and know how to find derivatives of a polynomial function (like we have here):
V′ = 12x2 – 56x + 48 = 4(3x2 – 14x + 12)
So now we find the values of x where the derivative is 0:
V′ = 4(3x2 – 14x + 12) = 0 ⟹ x = 1.1315, 3.5352
To get that answer, the quadratic formula can be used. Or, if lazy like me, an equation solver on the internet.
We want the value between 0 and 3. So a square of sides 1.1315 cm will maximise the volume. The other value of x is where the minimum point is as seen in the plot. We find the maximum volume by putting x = 1.1315 in the original V function:
V(1.1315) = 4(1.1315)3 – 28(1.1315)2 + 48(1.1315) = 24.258 cm3
As before, drawing pictures gets you started.
## The Derivative, Part 8
Now let’s do some more examples using not only the chain rule, but using a combination of the rules we have covered.
Let me start with an example that illustrates the “chainyness” of the chain rule. Let $f( x) =\text{sin}\left(\sqrt{x^{2} +2x-7}\right)$
Notice that there are three operations at work here: the sine, the square root and the polynomial. Referring back to the previous post, which is the outermost function? It’s the sine as that would be the last operation you would perform if you were to actual calculate the function for a particular x value. So the derivative rule for the sine is the first differentiation rule we will use.
So we have the sine of “something” so we start with the derivative of that something: $f'( x) =\text{cos}\left(\sqrt{x^{2} +2x-7} \right)\ ( …)$ Now from the last post, you know you have to multiply this by the derivative of that “something”. It will be helpful to rewrite that “something” as $\sqrt{x^{2} +2x-7} =\left( x^{2} +2x-7\right)^{1/2}$ Looks like we need to apply the chain rule again as we have an inner (polynomial) and an outer (power) functions.
The derivative of the “something” to the 1/2 power is $\frac{1}{2}\left( x^{2} +2x-7\right)^{-1/2}( …)$
We are now left with the innermost function x² + 2x – 7. The chain rule says to multiply the previous results with the derivative of this innermost function which is 2x +2. So putting this in the last (…) and then putting that result in the first (…) gives $f'( x) =\frac{1}{2}\left( x^{2} +2x-7\right)^{-1/2}( 2x+2)\text{cos}\left(\sqrt{x^{2} +2x-7}\right)$ Do you see how the successive differentiations of the functions from the outermost to the innermost works with the chain rule?
Let’s do another example. Let’s differentiate $f( x) =\sqrt{\text{sin}( x)\text{cos}( x)}$
As we did before, it’s easier to see the applicable differentiation rule if you convert the square root to its equivalent exponent form:$f( x) =\left[{\text{sin}( x)\text{cos}( x)}\right]^{1/2}$
Hopefully you can now identify the outermost operation as raising “something” to the 1/2 power. So the power rule is the one to use at first:$f'( x) =\frac{1}{2}[\text{sin}( x)\text{cos}( x)]^{-1/2}( …)$
So we now need to multiply this by the derivative of the “something” which is sin(x)cos(x). But this is the multiplication of two functions so we need to use the multiplication rule. Letting u = sin(x) and v = cos(x), then uv + uv‘ becomes cos²(x) – sin²(x). So now replacing the (…) with this results in $f'( x) =\frac{1}{2}[\text{sin}( x)\text{cos}( x)]^{-1/2}\left[\text{cos}^{2}( x) -\text{sin}^{2}( x)\right]$
This last example highlights the point that to find the derivative of complex functions frequently requires the use of several differentiation rules. You need to be aware of where you are in a particular problem and which rule you are currently working on.
Next time, I will show some examples where the derivatives are used. In the meantime, you can use the results of derivatives found in this post to find the derivative of$f( x) =\sqrt{\text{sin}( x)\text{cos}( x)}\text{sin}\left(\sqrt{x^{2} +2x-7}\right)$
## The Derivative, Part 7
So let’s recap: we have a rule to find derivatives of basic functions using a table, a rule to handle a function that is multiplied by a constant, a rule to handle the addition or subtraction of two (or more) functions, a rule to handle the multiplication of two (or more) functions, and a rule to handle the division of two functions. I also did an example where several of these rules can be used finding the derivative of a single function. You would think that this would exhaust all the possibilities and that you can now differentiate any function in the known universe. But alas, there is one more, perhaps the most powerful, rule yet to be presented.
This new rule is called the chain rule, so called because it allows you to find the derivative of a function, of a function, of a function, and so on.
Now there is a textbook way to present this rule and an intuitive way which I like to use. I find that the textbook approach can be confusing because there are several variables variables to keep track of. I will present both ways so that you may see the connection between the two and have a better understanding of the chain rule.
The textbook approach to the chain rule is a bit easier to see if we forego functional notation and go back to using y. However, whenever you have a function of a function f[g(x)], the chain rule is to be used. Functions like this are called composite functions. For example,
$f( x) =\text{sin}\left( x^{2}\right)$
Here, g(x) = x² and f(x) = sin(x). So f(x²) = sin(x²).
In the textbook approach you let u be the inner function (that is the function you are using as the argument for the outer function) and you let y be the function after you replace the inner function with u. I will give an explanation later on how to identify the inner and outer functions if that is not clear.
So in this case, u = x², so y = sin(u). The textbook chain rule is
$\frac{dy}{dx} =\frac{dy}{du} \times \frac{du}{dx}$
This may look scary but let me repeat this rule in English: the derivative of a composite function is the derivative of y with respect to u times the derivative of u with respect to x. So in our example, dy/du = cos(u) (using the table) and du/dx = 2x. Multiplying these together and replacing u with its definition, we get
$\frac{dy}{dx} =\text{cos}( u)\times 2x\ =\ 2x\ \text{cos}( x^{2})$
So to further explain what inner and outer functions are, suppose you wanted to take our example function and calculate its value for a certain number for x. The first thing you would do is take that number and square it. The squaring function is the inner function since it is the first thing you would do. Then you would take the sine of that squared number. The sine function then is the outer function as that is the last operation you would do.
So I explain the chain rule as follows: Take the derivative of the outer function of ‘something’ keeping the ‘something’ intact, but since the ‘something’ is not just ‘x‘ you need to multiply the result by the derivative of that ‘something’.
In this example, the ‘something’ is x². So the derivative of the sine of that ‘something’ is cos(x²), but I then need to multiply this by the derivative of the ‘something’. The derivative of x² is 2x, so the result is 2x cos(x²).
So now let’s reverse the roles of the the inner and outer functions. Consider the derivative of [sin(x)]². A very common shortcut notation for the square of a trig function like this is [sin(x)]² = sin²(x). Again, imagine actually calculating this for a particular value of x. You would first take the sine of that number (the inner function) then square the result (the outer function). We know that the derivative of the square of ‘something’ is 2 times that ‘something’ to the first power which in this case is 2 sin(x). But to compensate for this simplification, we need to multiply the result by the derivative of that ‘something’. In this case, the derivative of sin(x) is cos(x), so the final answer is 2 sin(x)cos(x).
Now to get comfortable with this, we need to do some more examples. I will do that in my next post.
## The Derivative, Part 6
Last time I presented the multiplication rule of differentiation to be used when given a function that is the multiplication of two or more other functions. As you would guess, there is also a rule that handles the division of two functions.
Let’s say you have the function
$f( x) =\frac{x^{2}}{\text{sin}( x)}$
This one can be solved with the multiplication rule if you remember that 1/sin(x) = csc(x). But as I haven’t told you what the derivative of csc(x) is, we are stuck using the following division rule. But this highlights the point that as we get deeper into maths, there are often several ways to solve a problem. The maths “arteest” is one that solves a problem elegantly.
So the following rule is the division rule. Again, I will use u(x) and v(x) to split the function up into its parts. If you have a function of the form
$f( x) =\frac{u( x)}{v( x)}$
then the derivative of f(x) is
$f'( x) =\frac{u( x) v'( x) -u'( x) v( x)}{[ v( x)]^{2}}$
As you can see, this rule is a bit more complex which is why you would use a simpler rule if possible. But it is still relatively easy to use if you keep track of which part is u and which part is v.
Using the example function above,
$\begin{array}{{>{\displaystyle}l}} u( x) =x^{2} ,\ \ \ \ \ v( x) =\text{sin}( x)\\ u'( x) =2x,\ \ \ \ v'( x) =\text{cos}( x) \end{array}$
So according to the division rule,
$f'( x) =\frac{x^{2}\text{cos}( x) -2x\text{sin}( x)}{\text{sin}^{2}( x)}$
Now you can use many rules in a single differentiation problem consider
$f( x) =\frac{x^{2} e^{x}}{\text{sin}( x)}$
Here, the numerator is a multiplication of two functions. So when using the division rule, you need to apply the multiplication rule for the u‘ part:
$\begin{array}{{>{\displaystyle}l}} u( x) \ =\ x^{2} e^{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v( x) =\text{sin}( x)\\ u'( x) =x^{2} e^{x} +2xe^{x} \ \ \ \ \ v'( x) =\text{cos}( x) \end{array}$
I’ll leave it as an exercise for you to see if I correctly found u‘(x) using the multiplication rule. I used the fact that (as seen from the table I provided a couple of posts before) is its own derivative. Anyway, using the division rule,
$f'( x) =\frac{x^{2} e^{x}\text{cos}( x) -\left( x^{2} e^{x} +2xe^{x}\right)\text{sin}( x)}{\text{sin}^{2}( x)}$
So you might be thinking that you can differentiate any function as long as you know the derivatives of the individual parts. So how would you differentiate
$f( x) =\text{sin}\left( x^{2}\right) ?$
This is not a multiplication of functions, but rather a function of a function. I will introduce the very powerful chain rule as it applies to differentiation in my next post.
## The Derivative, Part 5
So last time, I provided a table of derivatives given a function that is of a particular form. Because of rules 3 and 4 (you will need to see my last post to see what these are), along with the other entries in the table, you can now differentiate many functions not specifically in the table. But there are still many functions that you cannot differentiate without other rules. For example, if
$f( x) =x^{2}\text{sin}( x)$
there is no table entry to help you. Even though you can differentiate x² and sin(x) separately, there is no rule in the table that allows you to differentiate their multiplication together since they are both functions of x, that is, neither one is just a constant. You can’t use rule 4 here.
There is a differentiation rule that handles this. It is the multiplication rule and it states that if you have a function of the form
$f( x) =u( x) v( x)$
then the derivative is
$f'( x) =u( x) v'( x) +u'( x) v( x)$
This can be proven using the basic definition of a derivative, but you can just take my word for it.
So in the example at the beginning of this post,
$\begin{array}{{>{\displaystyle}l}} u( x) =x^{2} ,\ \ \ \ \ v( x) =\text{sin}( x)\\ u'( x) =2x,\ \ \ \ v'( x) =\text{cos}( x) \end{array}$
where I used the table in my last post to find the individual derivatives. So according to the rule,
$f'( x) =x^{2}\text{cos}( x) +2x\text{sin}( x)$
Now this rule can be extended to handle more than two functions multiplied together. If
$f( x) =u( x) v( x) w( x)$
then you can use the original rule twice, or
$f'( x) =u( x) v( x) w'( x) +u( x) v'( x) w( x) +u'( x) v( x) w( x)$
I think you can see the pattern here. So if
$\begin{array}{{>{\displaystyle}l}} f( x) =x^{2}\text{sin}( x)\text{cos}( x)\\ u( x) =x^{2} ,\ \ v( x) =\text{sin}( x) ,\ \ \ w( x) =\text{cos}( x)\\ u'( x) =2x,\ \ v'( x) =\text{cos}( x) ,\ \ w'( x) =-\text{sin}( x) \end{array}$
So the derivative is
$f'( x) =-x^{2}\text{sin}^{2}( x) +x^{2}\text{cos}^{2}( x) +2x\text{sin}( x)\text{cos}( x)$
Now this can be simplified using trig identities but I will leave it here.
What about a function that’s a division of two functions? Yes there is rule for that as well, but I’ll cover that in my next post.
## The Derivative, Part 4
Last time I provided some general rules for finding the derivatives for functions of different forms. Let me summarise these and provide some new ones as well. The new ones can be developed using the basic definition of the derivative. Letters a, and n are constants and are not a function variable:
These rules can be used for more than the explicit function forms included, especially using rules 3 and 4. For example if
$f( x) =3x^{3} -2x^{2} +5x-7$
then by using rules 2, 3, and 4 you can find the derivative as
$f'( x) =9x^{2} -4x +5$
Now let’s look at a more complex function:
$f( x) =3\text{sin}( 2x) -2\text{cos}( 3x) -0.25x^{2}$
where x is in radians. We can use rules 2, 3, 4, 6, and 7 and take the derivative of each term to get
$f'( x) =6\text{cos}( 2x) +6\text{sin}( 3x) -0.5x$
Now let’s look at a common use for derivatives. It is often needed to find the maximum and minimum of a function. Let’s look at the function f(x) = 3x³-10x²+9x:
We would like to know where (the x value) the peak (local maximum) and the local minimum occur and what the values of the function are at those points. As you have seen before, the gradient of the tangent lines at these points are zero. Since the derivative of a function gives us the gradient, we can find the derivative and find the values of x that make it zero. Using our rules for derivatives, f‘(x) = 9x²-20x+9. So we want to find the solutions to
f‘(x) = 9x²-20x+9 = 0
Using the quadratic formula, the two solutions are x = 0.627 and 1.595. We can evaluate the original function at these values of x to get the two points (0.627, 2.451) as the local maximum and (1.595, 1.088) as the local minimum.
A practical use of this is to find the maximum height a ball achieves that is thrown up into the air. Using physics to come up with the equations of motion of the ball, one can find the answer.
Even though I have shown that we can now differentiate a plethora of functions, there are still some functional forms that we cannot differentiate using the rules presented so far. I will cover some new rules in my next post.
## The Derivative, Part 3
Now that we have some confidence that the derivative definition gives correct results of functions that we know the answer to, let’s look at a functional form where the answer is not known.
Consider f(x) = x². As you know, this function plots as the standard parabola. The slope of a tangent line on this curve (its rate of change) is not constant, unlike the cases we have looked at before, but it depends on where we are on the curve:
So we again start with the basic definition of the derivative:
$f'( x) =\lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h} = \lim _{h\rightarrow 0}\frac{( x+h)^{2} -x^{2}}{h} =$ $\lim _{h\rightarrow 0}\frac{x^{2} +2xh+h^{2} -x^{2}}{h} =\lim _{h\rightarrow 0}\frac{h( 2x+h)}{h} =\lim _{h\rightarrow 0}( 2x+h) =2x$
So again, we do some algebraic manipulation that gets rid of the h in the denominator. Remember, as we are taking the limit as h approaches 0, the x is essentially treated as a constant. So the final answer is f‘(x) = 2x. Refering back to the graph, this satisfies the tangent line slopes at -1 and 1: f‘(-1) = -2, f‘(1) = 2. At any other point on the graph, just evaluate f‘(x) = 2x to find the rate of change of f(x) = x² at a particular x.
Now do you have to evaluate the definition for every different function you come across? Thankfully, the answer is no. Mathematicians have long ago done the hard work for you but because of the properties of limits, many general rules can be made. For example, if you know the derivative of a function, but what you have is the same function but multiplied by a constant, the derivative of this new function is just the same constant times the derivative of the old function. For example, we now know that for f(x) = x², f‘(x) = 2x. But what about g(x) = 3x²? Well, g‘(x) will just be 3 times the derivative of x², so g‘(x) = 6x.
So the rule is, if g(x) = af(x) where a is a constant number, then g‘(x) = af‘(x). Another generic rule is that the derivative of a sum of functions is the sum of the individual derivatives: If h(x) = f(x) + g(x), then h‘(x) = f‘(x) + g‘(x).
It turns out that if
$f( x) =ax^{n}$
where n is any real number except -1, then
$f'( x) =anx^{n-1}$
So to find the derivative in this case, you just multiply the function by n and reduce the value of the exponent by 1.
Next time, I will present a table of common derivatives and do some sample problems.
## The Derivative, Part 2
I ended my last post with the rather daunting definition of the derivative:
$f'( x) \ =\lim _{h\rightarrow 0} \ \frac{f( x+h) -f( x)}{h}$
I will now show how this definition can be used to find much simpler ways to calculate a derivative.
Let’s start with an example that we already know the answer to and is the simplest function we can think of, f(x) = c where c is some constant. You know that if the function does not change anywhere over the values of x, its rate of change (derivative) is zero. You see this if you plotted the function – it’s a horizontal line and a horizontal line has a gradient of zero. So f‘(x) = 0. Let’s see if the derivative definition gives us the same answer.
$f'( x) \ =\lim _{h\rightarrow 0} \ \frac{f( x+h) -f( x)}{h} =\lim _{h\rightarrow 0} \ \frac{c-c}{h} \rightarrow \frac{0}{0}$
Well that didn’t help much – we just got an indeterminate form 0/0. But as I said in my last post, there will always be some algebraic manipulation required to remove the problem.
One common method is to multiply the numerator and denominator by the same fraction. This does not change the value of the expression but if you use the right fraction, it removes the issue. In this case, multiply top and bottom by 1/h.
$f'( x) \ =\lim _{h\rightarrow 0} \ \frac{\frac{1}{h}( c-c)}{\frac{1}{h}( h)} =\lim _{h\rightarrow 0} \ \frac{\frac{1}{h}( 0)}{1} =\lim _{h\rightarrow 0} \ \frac{0}{1} =0$
Notice that by doing that, we got rid of h and we are left with 0/1 which is definitely 0. So the first rule of finding a derivative: if f(x) = c, then f‘(x) = 0.
Now let’s look at a more complex function, but again, one you know the answer to. The generic equation of a line is f(x) = mx + c where m and c are specific numbers: f(x) = 3x + 7 is an example. Again, from your studies of linear equations, you know this kind of function will plot as a straight line with a gradient of m. So we know that if f(x) = mx + c, then f‘(x) = m. Does our definition give the same result?
$f'( x) \ =\lim _{h\rightarrow 0} \ \frac{m( x+h) +c-( mx+c)}{h} =\lim _{h\rightarrow 0} \ \frac{mx+mh+c-mx-c}{h} \$ $=\lim _{h\rightarrow 0} \ \frac{mh}{h} =\lim _{h\rightarrow 0} \ m=m$
So when we enter in the particular function into the definition, then expand it, get rid of the terms that cancel, then cancel the common factor h, we again get rid of the dependency on h. We are left with the limit of a constant m as h approaches 0. But as m does not care what h does, the answer is just m – just what we expected. So we now know if f(x) = mx + c, then f‘(x) = m.
Next time, I will do the same thing but use functions for which we don’t know the answer.
## The Derivative, Part 1
In my last post, I showed that the rate of change of any function that plots as a straight line (a linear function) has a constant rate of change and that value is the gradient of the line. However, for a nonlinear function, its rate of change depends on the value of x, that is, where you are on its graph. I also said that a function’s rate of change is called the derivative of the function and that is what I will call it from now on.
Graphically, the derivative at a particular x value is the gradient of the tangent line at that point:
We would like to find an easy way to mathematically find this value as opposed to graphing the function and estimating the tangent line’s gradient at the desired points. Clearly, as seen above, the derivative is another function of x as its value changes depending on what x is. There are several ways to denote the derivative, but we will start with f’(x) (read as “f prime of x”). We would like to find what f’(x) is given a function f(x).
I know that the following derivation of the derivative may look complex and begs the question about how easy it will be to find the derivative, but following this will help solidify your understanding of what a derivative is and the final result will be used many times to find the easy results for various function forms.
We begin by taking the graph of a function and drawing a secant line (a line that connects two points on the graph) and calculate the gradient of that line:
We want to know the gradient of the estimated tangent line which we are using to approximate the tangent line at x. From your study of linear equations, you know that the gradient of a line can be found from any two points on the line. The two points on our estimated tangent line are (x, f(x)) and (x+h, f(x+h)) where h is a small distance away from x. Using these two points, we find the gradient by calculating the rise from the first point to the second point divided by the run between the two points. The rise is the difference between the y coordinates and the run is the difference between the x coordinates (h):
$\text{gradient} \ =\ \frac{f( x+h) -f( x)}{h}$
Now what happens as h gets smaller? The estimate should get closer to the actual value we are seeking. The below graphic from IkamusumeFan [CC BY-SA (https://creativecommons.org/licenses/by-sa/3.0)] illustrates this:
So it appears that we are interested in what our estimated gradient approaches as h approaches zero. This is, in fact, the formal definition of a function’s derivative. Remember my post on limits? Using limit notation then, the definition of the derivative is
$f'( x) \ =\lim _{h\rightarrow 0} \ \frac{f( x+h) -f( x)}{h}$
Notice that if we just substitute zero for h to evaluate the limit, we get the indeterminate form 0/0 as explained in my prior post. So again you may be saying “this doesn’t make finding a derivative easy at all”. At this point, you are correct. But in my next post, I will show how this definition is used to simplify derivatives. |
# How do you differentiate f(x) = x/sqrt(arctan(e^(x-1)) using the chain rule?
Aug 12, 2017
$\frac{d}{\mathrm{dx}} \left[\frac{x}{\sqrt{\arctan \left[{e}^{x - 1}\right]}}\right] = \textcolor{b l u e}{\frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} - \frac{x {e}^{x - 1}}{2 \left(1 + {e}^{2 x - 2}\right) {\tan}^{-} 1 {\left[{e}^{x - 1}\right]}^{3 \text{/} 2}}}$
#### Explanation:
We're asked to find the derivative
$\frac{d}{\mathrm{dx}} \left[\frac{x}{\sqrt{\arctan \left[{e}^{x - 1}\right]}}\right]$
Let's first use the product rule, which is
$\frac{d}{\mathrm{dx}} \left[u v\right] = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$
where
• $u = x$
• $v = \frac{1}{\sqrt{\arctan \left[{e}^{x - 1}\right]}}$:
$= \frac{\frac{d}{\mathrm{dx}} \left[x\right]}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} + x \frac{d}{\mathrm{dx}} \left[\frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}}\right]$
The derivative of $x$ is $1$:
$= \frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} + x \frac{d}{\mathrm{dx}} \left[\frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}}\right]$
Now let's use the chain rule, which will be
$\frac{d}{\mathrm{dx}} \left[\frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}}\right] = \frac{d}{\mathrm{du}} \left[\frac{1}{\sqrt{u}}\right] \frac{\mathrm{du}}{\mathrm{dx}}$
where
• $u = {\tan}^{-} 1 \left[{e}^{x - 1}\right]$
• $\frac{d}{\mathrm{du}} \left[\frac{1}{\sqrt{u}}\right] = - \frac{1}{2 {u}^{3 \text{/} 2}}$:
$= \frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} + x \left(- \frac{\frac{d}{\mathrm{dx}} \left[{\tan}^{-} 1 \left[{e}^{x - 1}\right]\right]}{2 {\tan}^{-} 1 {\left[{e}^{x - 1}\right]}^{3 \text{/} 2}}\right)$
Now we'll use the chain rule again:
$\frac{d}{\mathrm{dx}} \left[{\tan}^{-} 1 \left[{e}^{x - 1}\right]\right] = \frac{d}{\mathrm{du}} \left[{\tan}^{-} 1 u\right] \frac{\mathrm{du}}{\mathrm{dx}}$
where
• $u = {e}^{x - 1}$
• $\frac{d}{\mathrm{du}} \left[{\tan}^{-} 1 u\right] = \frac{1}{1 + {u}^{2}}$:
$= \frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} - \left(\frac{\frac{d}{\mathrm{dx}} \left[{e}^{x - 1}\right]}{{e}^{2 x - 2} + 1}\right) \frac{x}{2 {\tan}^{-} 1 {\left[{e}^{x - 1}\right]}^{3 \text{/} 2}}$
Finally, we'll use the chain rule a third time:
$\frac{d}{\mathrm{dx}} \left[{e}^{x - 1}\right] = \frac{d}{\mathrm{du}} \left[{e}^{u}\right] \frac{\mathrm{du}}{\mathrm{dx}}$
where
• $u = x - 1$
• $\frac{d}{\mathrm{du}} \left[{e}^{u}\right] = {e}^{u}$:
$= \frac{1}{\sqrt{{\tan}^{-} 1 \left[{e}^{x - 1}\right]}} - {e}^{x - 1} \frac{d}{\mathrm{dx}} \left[x - 1\right] \frac{x}{2 \left(1 + {e}^{2 x - 2}\right) {\tan}^{-} 1 {\left[{e}^{x - 1}\right]}^{3 \text{/} 2}}$
The derivative of $x - 1$ is $1$:
color(blue)(ulbar(|stackrel(" ")(" "= 1/sqrt(tan^-1[e^(x-1)]) - (xe^(x-1))/(2(1+e^(2x-2))tan^-1[e^(x-1)]^(3"/"2)))" ")|) |
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