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## What is Pascal’s Law?
Needed Supplies: Pressure Gauge
Instructions: Read the following about hydraulic force multiplication and how it relates to
the PFPD you previously worked on. This particular lesson will build upon your
knowledge of forces and area by relating them to the definition of pressure.
Theoretical Background
o Pascal’s Law
o For a particular position within a fluid at rest, the pressure is the same in
all directions.
o P=F/A (Equation 1.4)
P is pressure (in psi)
F is force (lbf)
A is area (in2)
o Hydraulic Cylinders
o As you learned in “What is Fluid Flow?”, each side of the piston has its
own area.
o The force that the rod can exert depends upon which side of the piston
has pressure acting upon it.
o When the cylinder extends, the system fluid pushes against an area equal
to the area of the cylinder (Ac).
The force that the cylinder rod exerts is:
F = PAc (Equation 1.5)
o When the cylinder retracts, the system fluid pushes against an area equal
to the cylinder area minus the area of the rod (Ac – Ar).
o The force that the cylinder rod exerts is:
F = P (Ac – Ar) (Equation 1.6)
Experiment
## 1. Remove cotter pin on cylinder that is attached to the bucket
2. Force produced during extension
a. Attach pressure gauge at appropriate port of cylinder
b. Move the valve that extends the top cylinder, and record the pressure at
the gauge
c. P = ___________ psi
d. Ac = .44 in2
e. F = PAc = (_______psi)(_______in2)
f. F = _______lbf
3. Force produced during retraction
a. Attach pressure gauge at appropriate port
b. Move the valve that retracts the top cylinder, and record the pressure at
the gauge
c. P = _________ psi
d. Ac = .44 in2
e. Ar = .049 in2
f. F = P (Ac – Ar) = (_________ psi)(_________ in2)
g. F = ________ lbf
Discussion
o Given the same pressure, which side of the piston can produce more force? Why
is this? (Refer to equations in theoretical background section)
o What would happen to the forces that the excavator cylinders produce if the
system pressure were decreased? (In other words, what if there was a leak in the
system?) |
# Change of variables
(Redirected from Substitution of variables)
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In mathematics, a change of variables is a basic technique used to simplify problems in which the original variables are replaced with functions of other variables. The intent is that when expressed in new variables, the problem may become simpler, or equivalent to a better understood problem.
Change of variables is an operation that is related with substitution. However these are different operations, as it can be seen when considering differentiation (chain rule) or integration (integration by substitution).
A very simple example of a useful variable change can be seen in the problem of finding the roots of the sixth order polynomial:
${\displaystyle x^{6}-9x^{3}+8=0.\,}$
Sixth order polynomial equations are generally impossible to solve in terms of radicals (see Abel–Ruffini theorem). This particular equation, however, may be written
${\displaystyle (x^{3})^{2}-9(x^{3})+8=0}$
(this is a simple case of a polynomial decomposition). Thus the equation may be simplified by defining a new variable x3 = u. Substituting x by ${\displaystyle {\sqrt[{3}]{u}}}$ into the polynomial gives
${\displaystyle u^{2}-9u+8=0,}$
which is just a quadratic equation with solutions:
${\displaystyle u=1\quad {\text{and}}\quad u=8.}$
The solutions in terms of the original variable are obtained by substituting x3 back in for u:
${\displaystyle x^{3}=1\quad {\text{and}}\quad x^{3}=8.}$
Then, assuming that x is real,
${\displaystyle x=(1)^{1/3}=1\quad {\text{and}}\quad x=(8)^{1/3}=2.}$
## Simple example
Consider the system of equations
${\displaystyle xy+x+y=71}$
${\displaystyle x^{2}y+xy^{2}=880}$
where ${\displaystyle x}$ and ${\displaystyle y}$ are positive integers with ${\displaystyle x>y}$. (Source: 1991 AIME)
Solving this normally is not very difficult, but it may get a little tedious. However, we can rewrite the second equation as ${\displaystyle xy(x+y)=880}$. Making the substitution ${\displaystyle s=x+y,t=xy}$ reduces the system to ${\displaystyle s+t=71,st=880.}$ Solving this gives ${\displaystyle (s,t)=(16,55)}$ or ${\displaystyle (s,t)=(55,16).}$ Back-substituting the first ordered pair gives us ${\displaystyle x+y=16,xy=55}$, which easily gives the solution ${\displaystyle (x,y)=(11,5).}$ Back-substituting the second ordered pair gives us ${\displaystyle x+y=55,xy=16}$, which gives no solutions. Hence the solution that solves the system is ${\displaystyle (x,y)=(11,5)}$.
## Formal introduction
Let ${\displaystyle A}$, ${\displaystyle B}$ be smooth manifolds and let ${\displaystyle \Phi :A\rightarrow B}$ be a ${\displaystyle C^{r}}$-diffeomorphism between them, that is: ${\displaystyle \Phi }$ is a ${\displaystyle r}$ times continuously differentiable, bijective map from ${\displaystyle A}$ to ${\displaystyle B}$ with ${\displaystyle r}$ times continuously differentiable inverse from ${\displaystyle B}$ to ${\displaystyle A}$. Here ${\displaystyle r}$ may be any natural number (or zero), ${\displaystyle \infty }$ (smooth) or ${\displaystyle \omega }$ (analytic).
The map ${\displaystyle \Phi }$ is called a regular coordinate transformation or regular variable substitution, where regular refers to the ${\displaystyle C^{r}}$-ness of ${\displaystyle \Phi }$. Usually one will write ${\displaystyle x=\Phi (y)}$ to indicate the replacement of the variable ${\displaystyle x}$ by the variable ${\displaystyle y}$ by substituting the value of ${\displaystyle \Phi }$ in ${\displaystyle y}$ for every occurrence of ${\displaystyle x}$.
## Other examples
### Coordinate transformation
Some systems can be more easily solved when switching to polar coordinates. Consider for example the equation
${\displaystyle U(x,y):=(x^{2}+y^{2}){\sqrt {1-{\frac {x^{2}}{x^{2}+y^{2}}}}}=0.}$
This may be a potential energy function for some physical problem. If one does not immediately see a solution, one might try the substitution
${\displaystyle \displaystyle (x,y)=\Phi (r,\theta )}$ given by ${\displaystyle \displaystyle \Phi (r,\theta )=(r\cos(\theta ),r\sin(\theta ))}$.
Note that if ${\displaystyle \theta }$ runs outside a ${\displaystyle 2\pi }$-length interval, for example, ${\displaystyle [0,2\pi ]}$, the map ${\displaystyle \Phi }$ is no longer bijective. Therefore ${\displaystyle \Phi }$ should be limited to, for example ${\displaystyle (0,\infty ]\times [0,2\pi )}$. Notice how ${\displaystyle r=0}$ is excluded, for ${\displaystyle \Phi }$ is not bijective in the origin (${\displaystyle \theta }$ can take any value, the point will be mapped to (0, 0)). Then, replacing all occurrences of the original variables by the new expressions prescribed by ${\displaystyle \Phi }$ and using the identity ${\displaystyle \sin ^{2}x+\cos ^{2}x=1}$, we get
${\displaystyle V(r,\theta )=r^{2}{\sqrt {1-{\frac {r^{2}\cos ^{2}\theta }{r^{2}}}}}=r^{2}{\sqrt {1-\cos ^{2}\theta }}=r^{2}\left|\sin \theta \right|}$.
Now the solutions can be readily found: ${\displaystyle \sin(\theta )=0}$, so ${\displaystyle \theta =0}$ or ${\displaystyle \theta =\pi }$. Applying the inverse of ${\displaystyle \Phi }$ shows that this is equivalent to ${\displaystyle y=0}$ while ${\displaystyle x\not =0}$. Indeed we see that for ${\displaystyle y=0}$ the function vanishes, except for the origin.
Note that, had we allowed ${\displaystyle r=0}$, the origin would also have been a solution, though it is not a solution to the original problem. Here the bijectivity of ${\displaystyle \Phi }$ is crucial. Note also that the function is always positive (for ${\displaystyle x,y\in \mathbb {R} }$), hence the absolute values.
### Differentiation
Main article: Chain rule
The chain rule is used to simplify complicated differentiation. For example, to calculate the derivative
${\displaystyle {\frac {d}{dx}}\left(\sin(x^{2})\right)\,}$
the variable x may be changed by introducing x2 = u. Then, by the chain rule:
${\displaystyle {\frac {d}{dx}}={\frac {d}{du}}{\frac {du}{dx}}={\frac {d}{dx}}\left(u\right){\frac {d}{du}}={\frac {d}{dx}}\left(x^{2}\right){\frac {d}{du}}=2x{\frac {d}{du}}\,}$
so that
${\displaystyle {\frac {d}{dx}}\left(\sin(x^{2})\right)=2x{\frac {d}{du}}\left(\sin(u)\right)=2x\cos(x^{2})\,}$
where in the very last step u has been replaced with x2.
### Integration
Difficult integrals may often be evaluated by changing variables; this is enabled by the substitution rule and is analogous to the use of the chain rule above. Difficult integrals may also be solved by simplifying the integral using a change of variables given by the corresponding Jacobian matrix and determinant. Using the Jacobian determinant and the corresponding change of variable that it gives is the basis of coordinate systems such as polar, cylindrical, and spherical coordinate systems.
### Differential equations
Variable changes for differentiation and integration are taught in elementary calculus and the steps are rarely carried out in full.
The very broad use of variable changes is apparent when considering differential equations, where the independent variables may be changed using the chain rule or the dependent variables are changed resulting in some differentiation to be carried out. Exotic changes, such as the mingling of dependent and independent variables in point and contact transformations, can be very complicated but allow much freedom.
Very often, a general form for a change is substituted into a problem and parameters picked along the way to best simplify the problem.
### Scaling and shifting
Probably the simplest change is the scaling and shifting of variables, that is replacing them with new variables that are "stretched" and "moved" by constant amounts. This is very common in practical applications to get physical parameters out of problems. For an nth order derivative, the change simply results in
${\displaystyle {\frac {d^{n}y}{dx^{n}}}={\frac {y_{\text{scale}}}{x_{\text{scale}}^{n}}}{\frac {d^{n}{\hat {y}}}{d{\hat {x}}^{n}}}}$
where
${\displaystyle x={\hat {x}}x_{\text{scale}}+x_{\text{shift}}}$
${\displaystyle y={\hat {y}}y_{\text{scale}}+y_{\text{shift}}.}$
This may be shown readily through the chain rule and linearity of differentiation. This change is very common in practical applications to get physical parameters out of problems, for example, the boundary value problem
${\displaystyle \mu {\frac {d^{2}u}{dy^{2}}}={\frac {dp}{dx}}\quad ;\quad u(0)=u(L)=0}$
describes parallel fluid flow between flat solid walls separated by a distance δ; µ is the viscosity and ${\displaystyle dp/dx}$ the pressure gradient, both constants. By scaling the variables the problem becomes
${\displaystyle {\frac {d^{2}{\hat {u}}}{d{\hat {y}}^{2}}}=1\quad ;\quad {\hat {u}}(0)={\hat {u}}(1)=0}$
where
${\displaystyle y={\hat {y}}L\qquad {\text{and}}\qquad u={\hat {u}}{\frac {L^{2}}{\mu }}{\frac {dp}{dx}}.}$
Scaling is useful for many reasons. It simplifies analysis both by reducing the number of parameters and by simply making the problem neater. Proper scaling may normalize variables, that is make them have a sensible unitless range such as 0 to 1. Finally, if a problem mandates numeric solution, the fewer the parameters the fewer the number of computations.
### Momentum vs. velocity
Consider a system of equations
${\displaystyle m{\dot {v}}=-{\frac {\partial H}{\partial x}}}$
${\displaystyle m{\dot {x}}={\frac {\partial H}{\partial v}}}$
for a given function ${\displaystyle H(x,v)}$. The mass can be eliminated by the (trivial) substitution ${\displaystyle \Phi (p)=1/m\cdot v}$. Clearly this is a bijective map from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$. Under the substitution ${\displaystyle v=\Phi (p)}$ the system becomes
${\displaystyle {\dot {p}}=-{\frac {\partial H}{\partial x}}}$
${\displaystyle {\dot {x}}={\frac {\partial H}{\partial p}}}$
### Lagrangian mechanics
Main article: Lagrangian mechanics
Given a force field ${\displaystyle \phi (t,x,v)}$, Newton's equations of motion are
${\displaystyle m{\ddot {x}}=\phi (t,x,v)}$.
Lagrange examined how these equations of motion change under an arbitrary substitution of variables ${\displaystyle x=\Psi (t,y)}$, ${\displaystyle v={\frac {\partial \Psi (t,y)}{\partial t}}+{\frac {\partial \Psi (t,y)}{\partial y}}\cdot w}$.
He found that the equations
${\displaystyle {\frac {\partial {L}}{\partial y}}={\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial {L}}{\partial {w}}}}$
are equivalent to Newton's equations for the function ${\displaystyle L=T-V}$, where T is the kinetic, and V the potential energy.
In fact, when the substitution is chosen well (exploiting for example symmetries and constraints of the system) these equations are much easier to solve than Newton's equations in Cartesian coordinates. |
# Double Angle Identities Calculator
Created by Davide Borchia
Reviewed by Komal Rafay
Last updated: Jan 18, 2024
Knowing how to calculate the double angle identities will help you with countless problems after your first encounter with trigonometry. From high school to university, and even after, this handy tool will give you the help you need, refreshing or calculating these critical mathematical formulas.
With this tool, you will learn the following:
• What are double angle trig identities;
• How to calculate the double angle identities in trigonometry;
and much more, as examples and related tools!
## What are trigonometric double angle identities?
Double angle identities are a class of trigonometric identities (that is, an equality that relates two mathematical formulas, being valid for all the values in a given range) which equate the value of a trigonometric function for twice the value of an angle to an algebraic combination of other trigonometric functions applied to the value of the angle.
Double angle identities allow you to calculate the value of functions such as $\sin(2\alpha)$, $\cos(4\beta)$, and so on. This class of identities is a particular case of the compound angle identities, which allow you to calculate the trigonometric functions of sums of angles.
## How do I calculate the double angle identities?
In this section, we will learn how to calculate the double angle identities for the three fundamental trigonometric functions (sine, cosine, and tangent). Let's see them one by one!
#### Calculate the double angle identity for the sine
The double angle identity for the sine is the first one we will meet in our journey. In mathematical terms, we can use the following formula:
$\scriptsize \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)$
Using other trigonometric identities, we can write the formula above in different ways.
• Using the relationship between sine and cosine (the Pythagorean identity), $\sin^2(\alpha)+\cos^2(\alpha) = 1$, and the formula for the square of a polynomial ($(a+b)^2 = a^2+b^2+2ab$), we can find another way to write the right-hand side of the identity:
$\scriptsize\quad\begin{split} \left(\sin(\alpha)+\cos(\alpha)\right)^2&=\sin^2(\alpha)+\cos^2(\alpha)\\ &+2\sin(\alpha)\cos(\alpha)\\ 2\sin(\alpha)\cos(\alpha)&=\left(\sin(\alpha)+\cos(\alpha)\right)^2\\ &-\sin^2(\alpha)-\cos^2(\alpha)\\ 2\sin(\alpha)\cos(\alpha)&=\left(\sin(\alpha)+\cos(\alpha)\right)^2-1 \end{split}$
• If we also use the definition of the tangent as the ratio between sine and cosine, we can find a third way to express $\sin(2\alpha)$:
$\scriptsize\quad\begin{split} 2\sin(\alpha)\cos(\alpha)&=2\left(\tan(\alpha)\cos(\alpha)\right)\cos(\alpha)\\ &=2\tan(\alpha)\cos^2(\alpha)\\[1em] &=\frac{2\tan(\alpha)}{\frac{1}{\cos^2(\alpha)}}\\[1em] &=\frac{2\tan(\alpha)}{\frac{\cos^2(\alpha)+\sin^2(\alpha)}{\cos^2(\alpha)}}\\[1em] &=\frac{2\tan(\alpha)}{1+\tan^2(\alpha)}\\[1em] \end{split}$
#### Calculate the double angle identity for the cosine
If the double angle identity for the sine uses the product of sine and cosine, the one for the cosine uses the difference of these functions:
$\scriptsize \cos(2\alpha) =\cos^2(\alpha) - \sin^2(\alpha)$
Again, we can find other ways to express this identity. The first one uses the Pythagorean identity:
$\scriptsize\begin{split} \cos(2\alpha) &=\cos^2(\alpha) - \sin^2(\alpha)\\ &=\cos^2(\alpha) - \left(1-\cos^2(\alpha)\right)\\ &=2\cos^2(\alpha)-1\\ &=1-2\sin^2(\alpha) \end{split}$
Introducing the tangent once again gives us the second form of the double angle identity for the cosine:
$\scriptsize\begin{split} \cos(2\alpha) &=2\cos^2(\alpha) -1\\ &=\frac{1}{1+\tan^2(\alpha)}-1\\[1em] &=\frac{1-\tan^2(\alpha)}{1+\tan^2(\alpha)} \end{split}$
#### Double angle trig identity for the tangent
To calculate the double angle formula for the tangent, we can use the ratio between the results found previously for the sine and cosine. There is a single neat way to express this identity:
$\scriptsize \tan(2\alpha) = \frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$
If our trig double angle identities calculator was helpful, we suggest you visit our other tools related to the topic:
## FAQ
### How do I find the double angle identity for the sine?
To find the double angle trig identity for the sine, follow these easy steps:
1. Start with the compound angle formula for the sine: sin(α + ß) = sin(α)cos(ß) + sin(ß)cos(α).
2. Substitute the angle ß with α,
3. The result is the following formula: sin(α + α) = sin(2α) = sin(α)cos(α) + sin(α)cos(α) = 2sin(α)cos(α).
### What are the double angle identities in trigonometry?
The double angle trig identities are a set of trigonometric identities that allow you to compute the values of the trigonometric functions of angles in the form 2α when the value of sin(α), cos(α), or tan(α) is known. Here are the identities for the three fundamental trigonometric functions:
• sin(2α) = 2sin(α)cos(α);
• cos(2α) = cos²(α) - sin²(α); and
• tan(2α) = 2tan(α)/(1 - tan²(α)).
### How do I find the cosine of 120 degrees?
To find the cosine of 120 degrees, you can use the double angle identity of trigonometry for the cosine. Use the following formula:
cos(2α) = cos²(α) - 1
If you know the cosine for this one:
cos(2α) = 1 + sin²(α)
If you know the sine.
Choosing 2α = 120°, and knowing that sin(60) = sqrt(3)/2, write:
cos(120°) = 1 - sin²(60°) = 1 - 3/4 = 2/4 = 1/2
Davide Borchia
Angle θ
deg
Step by step solution?
Double sine
Double cosine
Double tangent
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Lesson 14
Working with Pyramids
• Let’s use the pyramid volume formula to solve problems.
14.1: Volume Matching
Here is a pyramid.
Which, if either, of these solids has the same volume as the pyramid?
14.2: Practice with Pyramids
1. Calculate the volume of each solid. Round your answers to the nearest tenth if necessary.
2. A particular cone has radius $$r$$ and height $$h$$.
1. Write an expression for the volume of this cone in terms of $$r$$ and $$h$$.
2. What is the height of a cone whose volume is $$16\pi$$ cubic units and whose radius is 3 units?
3. What is the radius of a cone whose volume is $$16\pi$$ cubic units and whose height is 3 units?
The Pyramid of Giza is 455 feet tall. The base is square with a 756-foot side length. How many Olympic-size swimming pool volumes of water can fit inside the Pyramid of Giza? Explain or show your reasoning.
14.3: An Icy Pyramid
A caterer is making an ice sculpture in the shape of a pyramid for a party. The caterer wants to use 12 liters of water, which is about 720 cubic inches. The sculpture must be 15 inches tall. The caterer needs to decide how large to make the base, which can be any shape.
1. Draw and label the dimensions of a base that would work.
2. Find a second base that satisfies the baker’s requirements. You may use the applet to help, if you choose.
Directions for using the applet:
• Draw your base on the grid with the Polygon tool.
• Change to the 3D Graphics View by clicking on button.
• Click in the 3D window to switch back to the 3D menu.
• Select the Extrude to Pyramid tool and click on your polygon.
• When the dialog box opens, input the height.
• Use the Volume tool to verify your calculations and your figure.
• Refresh the page and repeat the steps with another base that works.
Summary
We can work backward from a given volume to find possible dimensions for a cone or pyramid.
Suppose we want to find dimensions for a cone so it has a volume of $$900\pi$$ cubic inches. Start by substituting the volume into the pyramid volume formula to get $$900\pi = \frac13 Bh$$. The base of a cone is a circle, so we can write $$900\pi = \frac13 \pi r^2 h$$. Multiply both sides of the equation by 3 and divide both sides by $$\pi$$ to get $$2,\!700 = r^2 h$$.
Now consider different possible values for $$r$$ and $$h$$. If we can find a perfect square that divides evenly into 2,700, we can set the square root of that number to be the radius. The number 25 is a perfect square and divdes into 2,700, so choose $$r=5$$. Now $$2,\!700 = 25h$$. This tells us that if the pyramid’s radius is 5 inches, its height is 108 inches because $$2,\!700 \div 25 = 108$$.
These aren’t the only possible values. Suppose we set the radius to be 20 inches. Substitute this into the original equation and rearrange to find the value of $$h$$.
$$900\pi = \frac13 \pi (20)^2 h$$
$$900\pi = \frac13 \pi \boldcdot 400 h$$
$$2,\!700 = 400 h$$
$$6.75=h$$
A height of 6.75 inches together with a radius of 20 inches gives the cone a volume of $$900\pi$$ cubic inches.
Glossary Entries
• apex
The single point on a cone or pyramid that is the furthest from the base. For a pyramid, the apex is where all the triangular faces meet. |
## Simplifying Complex Expressions I
### Learning Outcomes
• Simplify complex expressions using a combination of exponent rules
• Simplify quotients that require a combination of the properties of exponents
All the exponent properties we developed earlier in this chapter with whole number exponents apply to integer exponents, too. We restate them here for reference as we will be using them here to simplify various exponential expressions.
### Summary of Exponent Properties
If $a,b$ are real numbers and $m,n$ are integers, then
$\begin{array}{cccc}\mathbf{\text{Product Property}}\hfill & & & {a}^{m}\cdot {a}^{n}={a}^{m+n}\hfill \\ \mathbf{\text{Power Property}}\hfill & & & {\left({a}^{m}\right)}^{n}={a}^{m\cdot n}\hfill \\ \mathbf{\text{Product to a Power Property}}\hfill & & & {\left(ab\right)}^{m}={a}^{m}{b}^{m}\hfill \\ \mathbf{\text{Quotient Property}}\hfill & & & {\Large\frac{{a}^{m}}{{a}^{n}}}={a}^{m-n},a\ne 0\hfill \\ \mathbf{\text{Zero Exponent Property}}\hfill & & & {a}^{0}=1,a\ne 0\hfill \\ \mathbf{\text{Quotient to a Power Property}}\hfill & & & {\left({\Large\frac{a}{b}}\right)}^{m}={\Large\frac{{a}^{m}}{{b}^{m}}},b\ne 0\hfill \\ \mathbf{\text{Definition of Negative Exponent}}\hfill & & & {a}^{-n}={\Large\frac{1}{{a}^{n}}}\hfill \end{array}$
### Expressions with negative exponents
The following examples involve simplifying expressions with negative exponents.
### example
Simplify:
1. ${x}^{-4}\cdot {x}^{6}$
2. ${y}^{-6}\cdot {y}^{4}$
3. ${z}^{-5}\cdot {z}^{-3}$
Solution
1. ${x}^{-4}\cdot {x}^{6}$ Use the Product Property, ${a}^{m}\cdot {a}^{n}={a}^{m+n}$. ${x}^{-4+6}$ Simplify. ${x}^{2}$
2. ${y}^{-6}\cdot {y}^{4}$ The bases are the same, so add the exponents. ${y}^{-6+4}$ Simplify. ${y}^{-2}$ Use the definition of a negative exponent, ${a}^{-n}={\Large\frac{1}{{a}^{n}}}$. ${\Large\frac{1}{{y}^{2}}}$
3. ${z}^{-5}\cdot {z}^{-3}$ The bases are the same, so add the exponents. ${z}^{-5 - 3}$ Simplify. ${z}^{-8}$ Use the definition of a negative exponent, ${a}^{-n}={\Large\frac{1}{{a}^{n}}}$. ${\Large\frac{1}{{z}^{8}}}$
### try it
In the next two examples, we’ll start by using the Commutative Property to group the same variables together. This makes it easier to identify the like bases before using the Product Property of Exponents.
### example
Simplify: $\left({m}^{4}{n}^{-3}\right)\left({m}^{-5}{n}^{-2}\right)$
### try it
If we multipy two expressions with numerical coefficients, we multiply the coefficients together.
### example
Simplify: $\left(2{x}^{-6}{y}^{8}\right)\left(-5{x}^{5}{y}^{-3}\right)$
### try it
In the next two examples, we’ll use the Power Property and the Product to a Power Property to simplify expressions with negative exponents.
### example
Simplify: ${\left({k}^{3}\right)}^{-2}$.
### example
Simplify: ${\left(5{x}^{-3}\right)}^{2}$
### try it
In the following video we show another example of how to simplify a product that contains negative exponents.
The following examples involve solving exponential expressions with quotients.
### example
Simplify: ${\Large\frac{{\left({x}^{2}\right)}^{3}}{{x}^{5}}}$.
Solution
${\Large\frac{{\left({x}^{2}\right)}^{3}}{{x}^{5}}}$ Multiply the exponents in the numerator, using the Power Property. ${\Large\frac{{x}^{6}}{{x}^{5}}}$ Subtract the exponents. $x$
### example
Simplify: ${\Large\frac{{m}^{8}}{{\left({m}^{2}\right)}^{4}}}$
### example
Simplify: ${\left({\Large\frac{{x}^{7}}{{x}^{3}}}\right)}^{2}$
### example
Simplify: ${\left({\Large\frac{{p}^{2}}{{q}^{5}}}\right)}^{3}$
### example
Simplify: ${\Large{\left(\frac{2{x}^{3}}{3y}\right)}}^{4}$
### example
Simplify: ${\Large\frac{{\left({y}^{2}\right)}^{3}{\left({y}^{2}\right)}^{4}}{{\left({y}^{5}\right)}^{4}}}$
### try it
For more similar examples, watch the following video.
To conclude this section, we will simplify quotient expressions with a negative exponent.
### example
Simplify: ${\Large\frac{{r}^{5}}{{r}^{-4}}}$.
### try it
In the next video we share more examples of simplifying a quotient with negative exponents. |
Note: This is the second article of a planned 5-part series. The first article: Adventures with Generating Functions I: An Infinite Sum.
Second Adventure: How Many Ways Are There to Make a Million Dollars?
There are $88{,}265{,}881{,}340{,}710{,}786{,}348{,}934{,}950{,}201{,}250{,}975{,}072{,}332{,}541{,}120{,}001$ ways to make one million dollars using any combination of pennies, nickels, dimes, quarters, half-dollars, one-dollar bills, five-dollar bills, ten-dollar bills, twenty-dollar bills, fifty-dollar bills, and hundred-dollar bills.
Let's start by looking at a simpler problem, the number of ways to make one dollar using any combination of coins. We have an unlimited supply of currency in each denomination and the order of the units used is unimportant. For example, one way to make a dollar is $50+25+10+10+1+1+1+1+1$, and this is considered the same as $1+1+1+1+1+10+10+25+50$.
George Pólya showed how to solve this problem in his book How to Solve It. Let $p$, $n$, $d$, $q$, $h$, and $w$ stand for the number of pennies, nickels, dimes, quarters, half-dollars, and whole dollars, respectively. For each denomination $x$ and positive integer $k$, let $x_k$ be the number of ways to make the amount $k$ using no denomination higher than $x$. The following recurrence relations hold: \begin{align*} p_k&=1\\ n_k&=p_{k}+n_{k-5}\\ d_k&=n_{k}+d_{k-10}\\ q_k&=d_{k}+q_{k-25}\\ h_k&=q_{k}+h_{k-50}\\ w_k&=h_{k}+w_{k-100}, \end{align*} where we set $x_k=0$ for $k<0$. It's easy to complete the following table by hand and find that the number of ways to make one dollar is $w_{100}=293$ (this includes using a dollar itself). \begin{array}{c|cccccc} k & p_k & n_k & d_k & q_k & h_k & w_k\\ \hline 0 & 1 & 1 & 1 & 1 & 1 & 1\\ 5 & 1 & 2 & 2 & 2 & 2 & 2\\ 10 & 1 & 3 & 4 & 4 & 4 & 4\\ 15 & 1 & 4 & 6 & 6 & 6 & 6\\ 20 & 1 & 5 & 9 & 9 & 9 & 9\\ 25 & 1 & 6 & 12 & 13 & 13 & 13\\ 30 & 1 & 7 & 16 & 18 & 18 & 18\\ 35 & 1 & 8 & 20 & 24 & 24 & 24\\ 40 & 1 & 9 & 25 & 31 & 31 & 31\\ 45 & 1 & 10 & 30 & 39 & 39 & 39\\ 50 & 1 & 11 & 36 & 49 & 50 & 50\\ 55 & 1 & 12 & 42 & 60 & 62 & 62\\ 60 & 1 & 13 & 49 & 73 & 77 & 77\\ 65 & 1 & 14 & 56 & 87 & 93 & 93\\ 70 & 1 & 15 & 64 & 103 & 112 & 112\\ 75 & 1 & 16 & 72 & 121 & 134 & 134\\ 80 & 1 & 17 & 81 & 141 & 159 & 159\\ 85 & 1 & 18 & 90 & 163 & 187 & 187\\ 90 & 1 & 19 & 100 & 187 & 218 & 218\\ 95 & 1 & 20 & 110 & 213 & 252 & 252\\ 100 & 1 & 21 & 121 & 242 & 292 & 293 \end{array}
We can also solve the problem using generating functions.
The generating function of the sequence $\{a_k\}$, where $a_k$ is the number of ways to make the amount $k$ using pennies, nickels, dimes, quarters, half-dollars, and whole dollars, is $\sum_{k=0}^{\infty}a_k x^k = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots,$ where the coefficient of $x^k$ in this infinite polynomial'' is $a_k$. The generating function has a representation as a rational function, namely, $\frac{1}{(1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})(1-x^{100})}.$ To see how this works, consider this function as a product of geometric series: $(1+x+x^2+x^3+\cdots)(1+x^5+x^{2 \cdot 5}+ x^{3 \cdot 5}+\cdots)\ldots(1+x^{100}+x^{2 \cdot 100}+ x^{3 \cdot 100}+\cdots).$ Taking a single monomial from each sum in parentheses corresponds to one way of making an amount equal to the sum of the exponents in the chosen monomials. For example, choosing the sixth term in the first group, the third term in the second group, the second term in the third group, and the first term in each of the other groups results in the monomial $x^{5 \cdot 1 + 2 \cdot 5 + 1 \cdot 10}$, which corresponds to one way of making 25 cents, namely, five pennies, two nickels, and one dime.
A computer algebra system tells us that the coefficient of $x^{100}$ in our rational generating function is 293. Actually (as shown in Ivan Niven's book The Mathematics of Choice), we only need to compute the coefficient of $x^{100}$ in the polynomial product \begin{align*} &(1+x+x^2+\cdots+x^{100})(1+x^5+x^{10}+\cdots+x^{100})(1+x^{10}+x^{20}+\cdots+x^{100})\\ \cdot&(1+x^{25}+x^{50}+x^{75}+x^{100})(1+x^{50}+x^{100})(1+x^{100}), \end{align*} and we can do this by hand.
We can get more information from the generating function. The denominator of the rational function is a polynomial of degree 191; therefore $\{w_k\}$ satisfies a linear recurrence relation with constant coefficients of order 191. Also, we can modify the generating function to show all the relevant combinations: $\frac{1}{(1-px)(1-nx^5)(1-dx^{10})(1-qx^{25})(1-hx^{50})(1-wx^{100})}.$ The coefficient of, say, $x^{100}$, includes terms $p^{100}$, $p^{95}n$, and $p^{90}n^2$, etc., indicating that $100$ can be made with $100$ pennies, $95$ pennies and a nickel, $90$ pennies and two nickels, etc.
Similarly, we can find the number of ways to make 100 dollars (10000 cents) by using the generating function $\frac{1}{(1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})(1-x^{100}) (1-x^{500})(1-x^{1000})(1-x^{2000})(1-x^{5000})(1-x^{10000})}.$ The coefficient of $x^{10000}$ in the expansion is $1{,}424{,}039{,}612{,}939$.
Now we will find the number of ways to make a million dollars with the denominations mentioned at the beginning of this discussion. We use generating functions in a modified way as indicated in the book Concrete Mathematics, by Ronald L. Graham, Donald E. Knuth, and Oren Patashnik.
The number of pennies used to make a whole dollar amount must be a multiple of $5$, so let's think of $5$ cents as the smallest unit of currency. Let $c_k$ be the number of ways to make $5k$ cents using the given denominations. The generating function for $\{c_k\}$ is $\frac{1}{(1-x)(1-x)(1-x^{2})(1-x^{5})(1-x^{10})(1-x^{20}) (1-x^{100})(1-x^{200})(1-x^{400})(1-x^{1000})(1-x^{2000})}.$
Since 1 million is $20{,}000{,}000$ of our basic units, we must find $c_{20,000,000}$. To do this, we express each factor in the denominator as $(1-x^{2000})$ with a compensating factor (a geometric series) in the numerator. The new numerator is \begin{align*} &(1+x+\cdots+x^{1999})^2(1 + x^2 + x^4 + \cdots + x^{1998})(1+x^5+\cdots+x^{1995}) (1+x^{10}+\cdots+x^{1990})\\ & \cdot (1+x^{20}+\cdots+x^{1980})(1+x^{100}+\cdots+x^{1900})(1+x^{200}+\cdots+x^{1800})(1+x^{400}+\cdots+x^{1600})(1+x^{1000}), \end{align*} and a computer algebra system can multiply this quickly. The new denominator is $(1-x^{2000})^{11}$, and we can expand its reciprocal as a binomial series: $(1-x^{2000})^{-11} = \sum _{j=0}^{\infty} \binom{j+10}{10}x^{2000j}.$
We complete our calculation by multiplying this binomial series with contributing terms in the numerator to obtain the coefficient of $x^{20,000,000}$. In the numerator, the only powers of $x$ that matter are multiples of $2000$. If $\alpha_i$ is the coefficient of $x^i$, then the relevant coefficients are \begin{align*} \alpha_0 &= 1\\ \alpha_{2000} &= 1424039612928 \\ \alpha_{4000} &= 212561825179035\\ \alpha_{6000} &= 3224717280609587 \\ \alpha_{8000} &= 11601166434205649\\ \alpha_{10000} &= 12519790995056639\\ \alpha_{12000} &= 4102067385934937\\ \alpha_{14000} &= 334900882733305 \\ \alpha_{16000} &=3371148659578.\\ \alpha_{18000} &= 8008341. \end{align*} Hence \begin{align*} c_{20,000,000}&=\sum_{j=0}^9 \alpha_{2000j} \binom{10000+10-j}{10}\\ &=88265881340710786348934950201250975072332541120001\\ &\approx 9 \times 10^{49}. \end{align*}
The next article in the series: Adventures with Generating Functions III: Integer Triangles.
References
1. R. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science: Second Edition, Addison-Wesley Professional, 1994.
2. I. Niven , Mathematics of Choice: How to Count Without Counting, Mathematical Association of America, 1975.
3. G. Pólya , How to Solve It: A New Aspect of Mathematical Method, Princeton University Press , 2004. |
# Hexadecimal to Decimal Conversion (with Fractional Part)
A number can be converted from hexadecimal to decimal by following the steps outlined in this post, using the corresponding formula, and referring to the accompanying example. But before we get started, let's take a moment to discuss the meaning of these two numbers.
• Hexadecimal number: The hexadecimal number system has a base of 16, therefore it allows 16 digits that can be used to create or form a hexadecimal number. For example: 23F, 43, and 23D4F0, etc.
In other words, a number that consists of 16 digits, which are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F, is considered to be a hexadecimal number. The characters A-F refer to 10–15. That is, A refers to 10, B refers to 11, and so on.
• Decimal number: On the other hand, the decimal number system has a base of 10. Therefore, it has a total of 10 digits, which are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, which can be used to form a decimal number. For example: 230, 409, and 23022, etc.
## Hexadecimal to Decimal Conversion Steps
It is necessary to adhere to the rules that are outlined in the following lists in order to convert any hexadecimal number into the decimal system:
1. Write down the given hexadecimal number.
2. Write down the weight for different positions.
3. To get the product, multiply each digit of the given hexadecimal number by the corresponding weight.
4. Now add all the products to get the decimal equivalent.
Now, let's look at an example and see how we can better understand it by applying the rules that were just discussed.
Now that we have a number written in hexadecimal, let's say it's 1D7F; we need to translate that number into its representation in decimal.
That is, (1D7F)16 = ( ? )10.
The table given below shows you how to apply the first three rules given above to convert hexadecimal to decimal:
1 D or 13 7 F or 15 Given Number 163 162 161 160 Weight
After concentrating on the preceding table, apply rules 3 and 4, and we have
```= 1*163 + 13*162 + 7*161 + 15*160
= 4076 + 3328 + 112 + 15
= 7551```
which is the decimal equivalent of the given hexadecimal number, that is, 1D7F. Therefore, (1D7F)16 = (7551)10.
## Hexadecimal to Decimal Conversion Formula
Let's focus on the area that has been provided for us below, which explains how the hexadecimal number system can be converted into the decimal system. Consider the following scenario, in which the user has provided the hexadecimal input 5A9:
```5 A 9
162 161 160
(5*162) (A*161) (9*160)
= (5*162) + (10*161) + (9*160)
= 1280 + 160+9
= 1449```
As you can see from the above box:
• First, you have to write every hexadecimal digit.
• Then write 16 as the base and increment its exponent with 0, 1, 2, and so on from right to left below each and every hexadecimal digit.
• Write the multiplication of each hexadecimal digit with the corresponding powers of 16.
• Finally, add all the values as shown in the above box.
Therefore, when converting 5A9 into decimal, we get 1449. Therefore, (5A9)16 = (1449)10.
The rules presented above are only relevant for hexadecimal input that does not include a decimal point. Let's move on to the next step and learn how to convert numbers written in hexadecimal that contain fractional components into the corresponding value written in decimal.
## Hexadecimal (containing a decimal point) to Decimal Conversion
Here is another box that shows the conversion of hexadecimal (with a decimal point) to decimal. Let's suppose that the user has supplied 5A9.63 as hexadecimal input.
```5 A 9 6 3
162 161 160 16-1 162
(5*162) (10*161) (9*160) (6*16-1) (3*16-2)
= (5*162) + (10*161) + (9*160) + (6*16-1) + (3*16-2)
= 1280 + 160 + 9 + 0.375 + 0.01172
= 1449.3867```
#### Programs Created on Hexadecimal to Decimal Conversion
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# John tossed a fair coin 3 times. What is the probability that the outc
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John tossed a fair coin 3 times. What is the probability that the outcome was “tails” exactly twice?
A. 1/8
B. 1/4
C. 3/8
D. 1/2
E. 9/10
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25 Sep 2016, 12:49
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Bunuel wrote:
John tossed a fair coin 3 times. What is the probability that the outcome was “tails” exactly twice?
A. 1/8
B. 1/4
C. 3/8
D. 1/2
E. 9/10
Total number of outcomes = 2*2*2 = 8
Number of favorable outcomes = Number of ways two places (for tails) can be chosen out of 3 slots = 3C2 = 3
So probability = 3/8
Alternative
Probability of getting tail in single toss = 1/2
Probability of getting head in single toss = 1/2
Probability of getting First Tail = 1/2
Probability of getting Second tail (Such that first tail has occurred, this incidentally is also the probability when first was head and second is tail) = 1/2 * 1/2 = 1/4
Probability of getting Third Head ( such that first was tail and second was tail) = 1/2 * 1/2 * 12 = 1/8
As you can notice the probability will be 1/8 whatever the sequence of Heads and tails.
We can arrange 2 Tails and one head in 3 ways (3C2 ways OR (3P3)/2! OR enumeration, whichever method you prefer to get here)
Since the favorable even can happen in either of these 3 ways, so Probability = 3 * 1/8 = 3/8
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Re: John tossed a fair coin 3 times. What is the probability that the outc [#permalink]
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25 Sep 2016, 21:43
Total outcomes : 2^3 = 8
HHH
HHT
HTH
THH
TTH
THT
HTT
TTT
Exactly 2 tails are 3/8
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Cheers,
Shri
-------------------------------
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Re: John tossed a fair coin 3 times. What is the probability that the outc [#permalink]
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26 Sep 2016, 03:58
Bunuel wrote:
John tossed a fair coin 3 times. What is the probability that the outcome was “tails” exactly twice?
There are three possible scenarios here:
1) "Tails" on toss 1 and 2, and "Heads" on toss 3:
OR (1/2)*(1/2)*(1/2)=1/8
2) "Tails" on toss 1 and 3, and "Heads" on toss 2:
OR (1/2)*(1/2)*(1/2)=1/8
3) "Tails" on toss 3 and 2, and "Heads" on toss 1:
OR (1/2)*(1/2)*(1/2)=1/8
That's a total probability of (1/8)+(1/8)+(1/8)=3/8
Alternately, you can count: Only possible situation are: HTT, THT, TTH of the possible 8 outcomes. Hence 3/8.
Hope it helps
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Re: John tossed a fair coin 3 times. What is the probability that the outc [#permalink]
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18 Oct 2018, 18:35
Bunuel wrote:
John tossed a fair coin 3 times. What is the probability that the outcome was “tails” exactly twice?
A. 1/8
B. 1/4
C. 3/8
D. 1/2
E. 9/10
We need to determine the following:
P(TTH) = (1/2)^3 = 1/8
Since TTH can be arranged in 3!/2! = 3 ways, the probability is 3 x (1/8) = 3/8.
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Re: John tossed a fair coin 3 times. What is the probability that the outc [#permalink] 18 Oct 2018, 18:35
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# Making Subtraction Efficient – Just Think Addition!
Students who are efficient with the count-on, use-doubles, and make-ten strategies for addition have all they need to be proficient with subtraction number facts!
Subtraction is the inverse, or opposite action, of addition. Both operations involve a part-part-total structure. In the example below, 1 and 3 are two parts that make up the number 4.
## Parts of Subtraction
With addition, the parts are known, but not the total; with subtraction, the total and one of the parts are known, but not the other part. Because of this relationship between the two operations, using addition is the most effective thinking strategy for helping students learn the basic subtraction facts. Watch this ORIGO One video to learn more!
## Fact Families
Addition facts and subtraction facts that involve the same parts and total form fact families. Clusters of subtraction facts are named according to the strategy used for the related addition facts. For example, 8 – 3 = 5 is part of the count-on subtraction fact cluster because its related addition fact 5 + 3 = 8 involves counting on three. Students need to understand the connections between the number facts within a family and the related strategy.
## Subtraction Games: Teach Subtraction Using Addition
“Think addition to subtract” is one of the most effective strategies for subtracting mentally. This game reinforces the connection between addition and subtraction. The students are encouraged to use their knowledge of addition to make a true subtraction number sentence.
Materials needed:
• Take or Tally game board (access this subtraction game board file here)
• 2 blank cubes, marked as follows: (Be sure to underline the 9s and 6s!)
• Write 1, 2, 3, 1, 2, 3 on one cube
• Write 4-9 on the other.
Game Directions: The aim is to complete twelve true number sentences.
• The first player rolls the two number cubes.
• The player then writes the two numbers in one of the number sentences on his or her game board. The completed number sentence must be true.
Example: Sue rolls 4 and 3. She completes the number sentence 7 – 4 = 3.
• If a true number sentence cannot be made, the player makes a tally in the space provided at the bottom of his or her game board.
• The first player to complete twelve number sentences before making a total of ten tallies is the winner.
Strategy Tip: Ask, How did you know to place your numbers in that sentence?
Extending the game:
• Change the rules so two players share the one game board. Each player has his or her own column of number sentences to complete. The remaining rules can stay unchanged.
• Use the Take or Tally Again game board for greater numbers. Make a new number cube by writing numerals 6-11 and replace the cube with 4-9 from the original game.
Encourage children to apply the think-addition strategy for subtraction to numbers beyond the basic facts. For example, the strategy can be extended to solve 106 – 89. This is specifically highlighted in the ORIGO One video referenced earlier in this blog! |
## Simple and also ideal practice solution for 60% of 800. Check just how straightforward it is, and learn it for the future. Our solution is easy, and simple to understand, so don`t hesitate to use it as a solution of your homework.
If it"s not what You are trying to find form in the calculator areas your very own values, and also You will gain the solution.
You are watching: What is 60 of 800
To obtain the solution, we are trying to find, we have to point out what we know. 1. We assume, that the number 800 is 100% - because it"s the output value of the task. 2. We assume, that x is the worth we are searching for. 3. If 800 is 100%, so we can compose it dvery own as 800=100%. 4. We know, that x is 60% of the output value, so we have the right to compose it down as x=60%. 5. Now we have actually two straightforward equations:1) 800=100%2) x=60%where left sides of both of them have actually the very same systems, and both best sides have actually the very same devices, so we deserve to execute somepoint choose that:800/x=100%/60%6.
See more: What Channel Is Vh1 On Dish Network, Vh1 (Indian Tv Channel)
Now we just need to solve the easy equation, and we will certainly gain the solution we are looking for.7. Systems for what is 60% of 800800/x=100/60(800/x)*x=(100/60)*x - we multiply both sides of the equation by x800=1.66666666667*x - we divide both sides of the equation by (1.66666666667) to gain x800/1.66666666667=x 480=x x=480now we have: 60% of 800=480
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# Interpreting the Sign of the First Derivative (Increasing and Decreasing Functions)
Derivatives #12: Intro to Graphing Derivatives
Tip: This isn't the place to ask a question because the teacher can't reply.
## Key Questions
• A function is monotonically increasing if it is always increasing.
From a graph point of view, this means that moving to the right in the graph, the function will only increase.
In more mathemtical notation:
$\left(\forall x , y \in \mathbb{R}\right) \left(x > y\right) \implies \left(f \left(x\right) > f \left(y\right)\right)$
• If the first derivative at a given point $c$, denoted $f ' \left(c\right)$, is positive, then the function $f \left(x\right)$ is increasing at that point.
Remember that the derivative represents the rate of change of the function $f \left(x\right)$. If the first derivative is positive then the rate at which the function increases with $x$ is also positive, meaning that as we make $x$ larger, we make $f \left(x\right)$ larger too.
This is most obvious in the case of a simple function like $f \left(x\right) = 2 x$. We know that $f ' \left(x\right) = 2$ at every point on the real line: the rate of change is positive. Hence, we expect $f \left(x\right)$ to increase with $x$.
Here are 2 more examples:
The red function is $f \left(x\right)$ from E.S's example, the blue function is $g \left(x\right) = {x}^{2}$ and the green function is $h \left(x\right) = {x}^{3}$. The derivatives for the new functions are: $g ' \left(x\right) = 2 x$ and $h ' \left(x\right) = 3 {x}^{2}$.
You can see the the red line is always increasing.
The blue curve is decreasing when $x < 0$ and increasing when $x > 0$. This matches the derivative $g ' \left(x\right) = 2 x$; we can see that $g '$ is negative when $x < 0$ and positive when $x > 0$.
The green curve is increasing everywhere except $x = 0$. It's hard to tell from looking at the graph because it's pretty flat. However, examining the derivative $h ' \left(x\right) = 3 {x}^{2}$, we can see that ${x}^{2}$ is always positive except at $x = 0$.
So you can determine intervals of increase by looking at the graph although it may be difficult at some points. Or you can more accurately determine the intervals algebraically using the first derivative.
• It can be done by looking at the sign of the first derivative.
Remember: if the first derivative of a function is positive, then the function itself is increasing.
For $f \left(x\right) = \sin \left(x\right)$ the first derivative equals
$f ' \left(x\right) = \cos \left(x\right)$
So the question becomes: when is $\cos \left(x\right) > 0$?
The solution is pretty straight forward if you know your trigonometry:
$- \frac{\pi}{2} < x < \frac{\pi}{2} \mod 2 \pi$
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# Chapter 9: Arithmetic Progressions Exercise 9.3
### Question: 1
Find:
(i) 10th tent of the AP 1, 4, 7, 10....
(ii) 18th term of the AP √2, 3√2, 5√2, …….
(iii) nth term of the AP 13, 8, 3, -2, ..........
(iv) 10th term of the AP -40, -15, 10, 35, .............
(v) 8th term of the AP 11, 104, 91, 78, ...............
(vi) 11th tenor of the AP 10.0, 10.5, 11.0, 11.2, ..............
(vii) 9th term of the AP 3/4, 5/4, 7/4 + 9/4, ...........
### Solution:
Given A.P. is 1, 4, 7, 10, ..........
First term (a) = 1
Common difference (d) = second than first term
= 4 - 1 = 3.
nth term in an A.p = a + (n - 1)d
10th term in an 1 + (10 - 1)3
= 1 + 9.3
= 1 + 27
= 28
(ii) Given A.P. is
√2, 3√2, 5√2, …….
Fiat term (a) = √2
Common difference = Second term – First term
= 3√2 - √2
d = 2√2
nth term in an A. P. = a + (n - 1)d
18th term of A. P. = √2 + (18 - 1)2√2
= √2 + 17.2√2
= √2 (1+34)
= 35√2
∴ 18th term of A. P. is 35√2
(iii) Given A. P. is
13, 8, 3, - 2, ............
First term (a) = 13
Common difference (d) = Second term first term
= 8 - 13 = – 5
nth term of an A.P. an = a +(n - 1)d
= 13 + (n - 1) - 5
= 13 - 5n + 5
an = 18 - 5n
(iv) Given A. P. is
- 40, -15, 10, 35, ..........
First term (a) = -40
Common difference (d) = Second term - fast term
= -15 - (- 40)
= 40 - 15
= 25
nth term of an A.P. an = a + (n - 1)d
10th term of A. P. a10 = -40 + (10 - 1)25
= – 40 + 9.25
= – 40 + 225
= 185
(v) Given sequence is 117, 104, 91, 78, .............
First learn can = 117
Common difference (d) = Second term - first term
= 104 - 117
= – 13
nth term = a + (n - 1)d
8th term = a + (8 - 1)d
= 117 + 7(-13)
= 117 - 91
= 26
(vi) Given A. P is
10.0, 10.5, 11.0, 11.5,
First term (a) = 10.0
Common difference (d) = Second term - first term
= 10.5 - 10.0 = 0.5
nth term; an = a + (n - 1)d
11th term a11 = 10.0 + (11 - 1)0.5
= 10.0 + 10 x 0.5
= 10.0 + 5
=15.0
(vii) Given A. P is
3/4, 5/4, 7/4, 9/4, ............
First term (a) = 3/4
Common difference (d) = Second term - first term
= 5/4 - 3/4
= 2/4
nth term an = a + (n - 1)d
9th term a9 = a + (9 - 1)d
### Question: 2
(i) Which term of the AP 3, 8, 13, .... is 248?
(ii) Which term of the AP 84, 80, 76, ... is 0?
(iii) Which term of the AP 4. 9, 14, .... is 254?
(iv) Which term of the AP 21. 42, 63, 84, ... is 420?
(v) Which term of the AP 121, 117. 113, ... is its first negative term?
### Solution:
(i) Given A.P. is 3, 8, 13, ...........
First term (a) = 3
Common difference (d) = Second term - first term
= 8 - 3
= 5
nth term (an) = a + (n - 1)d
Given nth term an = 248
248 = 3+(n - 1).5
248 = -2 + 5n
5n = 250
n =250/5 = 50
50th term is 248.
(ii) Given A. P is 84, 80, 76, ............
First term (a) = 84
Common difference (d) = a2 - a
= 80 - 84
= – 4
nth term (an) = a +(n - 1)d
Given nth term is 0
0 = 84 + (n - 1) - 4
84 = +4(n - 1)
n - 1 = 84/4 = 21
n = 21 + 1 = 22
22nd term is 0.
(iii) Given A. P 4, 9, 14, ............
Fiat term (a) = 4
Common difference (d) = a2 - a
= 9 - 4
= 5
nth term (an) = a + (n - 1)d
Given nth term is 254
4 + (n - 1)5 = 254
(n - 1)∙5 = 250
n - 1 = 250/5 = 50
n = 51
∴ 51th term is 254.
(iv) Given A. P
21, 42, 63, 84, .........
a = 21, d = a2 - a
= 42 - 21
= 21
nth term (an) = a +(n - 1)d
Given nth term = 420
21 + (n - 1)21 = 420
(n - 1)21 = 399
n - 1 = 399/21 = 19
n = 20
∴ 20th term is 420.
(v) Given A.P is 121, 117, 113, ...........
Fiat term (a) = 121
Common difference (d) = 117 - 121
= - 4
nth term (a) = a + (n - 1)d
Given nth term is negative i.e., an < 0
121 + (n - 1) - 4 < 0
121 + 4 - 4n < 0
125 - 4n < 0
4n > 125
n > 125/4
n > 31.25
The integer which comes after 31.25 is 32.
∴ 32nd term is first negative term
### Course Features
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# The Cosine Rule. How to Work out an Angle Inside any Triangle Given 3 Side Lengths.
Updated on February 25, 2019
When you are given all of the 3 sides of a triangle then any angle inside the triangle can be found by applying the cosine rule (the angle version is easiest to apply). The formula for this is:
CosA = (b² + c² - a²)/(2bc)
A is the angle that you are finding and a is the side that is opposite the angle.
b and c are the lengths of the other 2 sides of the triangle. It doesn’t matter which way around you put b and c as you will get the same answer either way.
You can use a² = b² + c² -2bcCosA but this is harder to apply to questions that involve working out an angle.
Let’s take a look at a couple of questions that involve working out an angle using the cosine rule.
Example 1
Work out the missing angle in this triangle.
Since you given all 3 sides of the triangle and you need to work out an angle then the problem can be solved using the cosine rule.
First of all label up your triangle:
Since you are working out the angle call this big A.
Therefore, the side opposite this angle you are finding is little a.
The two other sides can be called b and c.
All you need to do know is apply the Cosine rule:
Substitute A = ?, a = 82, b = 46 and c = 117 into the formula above:
CosA = (b² + c² - a²)/(2bc)
Cos? = (82² + 46² - 117²)/(2×46×117)
Cos? = 9081/10764 (all you need to do next is cos inverse this fraction to convert it to an angle)
? = 32.5
Example 2
Work out the missing angle in this triangle.
Since you given all 3 sides of the triangle and you need to work out an angle then the problem can be solved using the cosine rule.
First of all label up your triangle:
Since you are working out the angle call this big A.
Therefore, the side opposite this angle you are finding is little a.
The two other sides can be called b and c.
All you need to do know is apply the Cosine rule:
Substitute A = ?, a = 117, b = 70 and c = 79 into the formula above:
CosA = (b² + c² - a²)/(2bc)
Cos? = (70² + 79² - 117²)/(2×70×79)
Cos? = -2548/11060 (all you need to do next is cos inverse this fraction to convert it to an angle)
? = 103.3⁰ (rounded to 1 decimal)
As you can see using this version of the cosine rule is actually quite easy to work out an angle. Just take care labelling up your triangle and subbing your values into the cosine rule formula. Also make sure that you have cos inversed your final answer.
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# Pythagoras Theorem is a2 + b2 = c2. 'a' being the shortest side, 'b' being the middle side and 'c' being the longest side (hypotenuse) of a right angled triangle.
Extracts from this document...
Introduction
Pythagoras Theorem is a2 + b2 = c2. ‘a’ being the shortest side, ‘b’ being the middle side and ‘c’ being the longest side (hypotenuse) of a right angled triangle.
The numbers 3, 4 and 5 satisfy this condition
32 + 42 = 52
because 32 = 3 x 3 = 9
42 = 4 x 4 = 16
52 = 5 x 5 = 25
and so 32 + 42 = 9 + 16 = 25 = 52
The numbers 5, 12, 13 and 7, 24, 25 also work for this theorem
52 + 122 = 132
because 52 = 5 x 5 = 25
122 = 12 x 12 = 144
132 = 13 x 13 = 169
and so 52 + 122 = 25 + 144 = 169 = 132
72 + 242 = 252
because 72 = 7 x 7 = 49
242 = 24 x 24 = 576
252 = 25 x 25 = 625
and so 72 + 242 = 49 + 576 = 625 = 252
3 , 4, 5
Perimeter = 3 + 4 + 5 = 12
Area = ½ x 3 x 4 = 6
5, 12, 13
Perimeter = 5 + 12 + 13 = 30
Area = ½ x 5 x 12 = 30
7, 24, 25
Perimeter = 7 + 24 + 25 = 56
Area = ½ x 7 x 24 = 84
From the first three terms I have noticed the following: -
• ‘a’ increases by +2 each term
• ‘a’ is equal to the term number times 2 then add 1
• the last digit of ‘b’ is in a pattern 4, 2, 4
• the last digit of ‘c’ is in a pattern 5, 3, 5
• the square root of (‘b’ + ‘c’) = ‘a’
• ‘c’ is always +1 to ‘b’
• ‘b’ increases by +4 each term
• (‘a’ x ‘n’) + n = ‘b’
Middle
90
180
5
11
60
61
132
330
I have worked out formulas for
1. How to get ‘a’ from ‘n’
2. How to get ‘b’ from ‘n’
3. How to get ‘c’ from ‘n’
4. How to get the perimeter from ‘n’
5. How to get the area from ‘n’
My formulas are
1. 2n + 1
2. 2n2 + 2n
3. 2n2 + 2n + 1
4. 4n2 + 6n + 2
5. 2n3 + 3n2 + n
To get these formulas I did the following
1. Take side ‘a’ for the first five terms 3, 5, 7, 9, 11. From these numbers you can see that the formula is 2n + 1 because these are consecutive odd numbers (2n + 1 is the general formula for consecutive odd numbers) You may be able to see the formula if you draw a graph
1. From looking at my table of results, I noticed that ‘an + n = b’. So I took my formula for ‘a’ (2n + 1) multiplied it by ‘n’ to get ‘2n2 + n’. I then added my other ‘n’ to get ‘2n2 + 2n’. This is a parabola as you can see from the equation and also the graph
1. Side ‘c’ is just the formula for side ‘b’ +1
2. The perimeter = a + b + c.
Conclusion
4n2 + 6n + 2
4(n +3)2 – 9 + 2
4(n +3)2 – 7
4(n +3)2 = 7
(n +3)2 = 1.75
n + 3 = 1.322875656
n + 3 = -1.322875656
n = -1.677124344
n = -4.322875656
## Arithmatic Progression
I would like to know whether or not the Pythagorean triple 3,4,5 is the basis of all triples just some of them.
To find this out I have been to the library and looked at some A-level textbooks and learnt ‘Arithmatic Progression’
3, 4, 5 is a Pythagorean triple
The pattern is plus one
If a = 3 and d = difference (which is +1) then
3 = a
4 = a + d
5 = a +2d
a, a +d, a + 2d
Therefore if you incorporate this into Pythagoras theorem
a2 + (a + d)2 = (a + 2d)2
a2 + (a + d)(a + d) = (a + 2d)2
a2 + a2 + ad + ad + d2 = (a + 2d)2
2a2 + 2ad + d2 = (a + 2d)2
2a2 + 2ad + d2 = (a + 2d)(a + 2d)
If you equate these equations to 0 you get
a2 – 3d2 – 2ad = 0
Change a to x
x2 – 3d2 – 2dx = 0
Factorise this equation to get
(x + d)(x – 3d)
Therefore
x = -d
x = 3d
x = -d is impossible as you cannot have a negative dimension
a, a+d, a + 2d
Is the same as
3d, 4d, 5d
This tells us that the only Pythagorean triples are 3, 4, 5 or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc.
This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.
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# Related GCSE Fencing Problem essays
1. ## Proving a2 + b2 = c2 Using Odd Numbers
Perimeter Evens Perimeter = a + b + c a = 2n + 4 b = n2 + 4n +
2. ## Beyond Pythagoras
cm 24 cm 25 cm 56 cm 84 cm� 9 cm 40 cm 41 cm 90 cm 180 cm� 11 cm 60 cm 61 cm 132 cm 330 cm� 13 cm 84 cm 85 cm 182 cm 546 cm� 15 cm 112 cm 113 cm 240 cm 840 cm� 17
1. ## Beyond pythagoras - First Number is odd.
= area 2 Multiply this out to get 4n3 + 6n2 + 2n = area 2 Then divide 4n3 + 6n2 + 2n by 2 to get 2n3 + 3n2+ n Summary of Formulas I have found out: 1. Use 2n + 1 to get 'a' from 'n' 2.
2. ## Beyond Pythagoras
To get from 5 to 13, I have to add 8; then from 13 to 25, I have to add 12. So moving further I have seen that the number increases by 4 each time, indicating the difference of 4 as we move further.
1. ## Beyond Pythagoras
This means that the formula is 2n. Middle side (b) I can tell just by looking at the numbers in this column that the numbers are the squares of the triangle numbers minus 1. This means that the formula for b is n2 - 1 Longest side (c)
2. ## Beyond Pythagoras.
This tells us that the only term where the perimeter = area is term number 2 Therefore when f(x) = 2n3 - n2 - 5n - 2 is divided by n - 2 there is no remainder and a quotient 2n2 + 3n + 1.
1. ## Beyond Pythagoras.
I will use T1, that way I can find the un-multiplied value of "b". an� + bn + c = 4 2(1�) + b(1) + 0 = 4 2 + b + 0 = 4 -2 -2 b = 4-2 b = 2 This shows that "a" = 2 "b"
2. ## Beyond Pythagoras.
this formula to find the nth term for side 'b' Bellow are the rules that I will use to find a, b and c 2a = 2nd diff 3a + b = 1st diff a + b + c = 1st term 2a = 2nd diff = 2a = 4
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# How to Find the Domain of a Function
19-03-2021, 02:50
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The domain of a function is the set of numbers that can go into a given function. In other words, it is the set of x-values that you can put into any given equation. The set of possible y-values is called the range. If you want to know how to find the domain of a function in a variety of situations, just follow these steps.
### Learning the Basics
1. Learn the definition of the domain. The domain is defined as the set of input values for which the function produces an output value. In other words, the domain is the full set of x-values that can be plugged into a function to produce a y-value.
2. Learn how to find the domain of a variety of functions. The type of function will determine the best method for finding a domain. Here are the basics that you need to know about each type of function, which will be explained in the next section:
• A polynomial function without radicals or variables in the denominator. For this type of function, the domain is all real numbers.
• A function with a fraction with a variable in the denominator. To find the domain of this type of function, set the bottom equal to zero and exclude the x value you find when you solve the equation.
• A function with a variable inside a radical sign. To find the domain of this type of function, just set the terms inside the radical sign to >0 and solve to find the values that would work for x.
• A function using the natural log (ln). Just set the terms in the parentheses to >0 and solve.
• A graph. Check out the graph to see which values work for x.
• A relation. This will be a list of x and y coordinates. Your domain will simply be a list of x coordinates.
3. Correctly state the domain. The proper notation for the domain is easy to learn, but it is important that you write it correctly to express the correct answer and get full points on assignments and tests. Here are a few things you need to know about writing the domain of a function:
• The format for expressing the domain is an open bracket/parenthesis, followed by the 2 endpoints of the domain separated by a comma, followed by a closed bracket/parenthesis.
• For example, [-1,5). This means that the domain goes from -1 to 5.
• Use brackets such as [ and ] to indicate that a number is included in the domain.
• So in the example, [-1,5), the domain includes -1.
• Use parentheses such as ( and ) to indicate that a number is not included in the domain.
• So in the example, [-1,5), 5 is not included in the domain. The domain stops arbitrarily short of 5, i.e. 4.999…
• Use “U” (meaning "union") to connect parts of the domain that are separated by a gap.'
• For example, [-1,5) U (5,10]. This means that the domain goes from -1 to 10, inclusive, but that there is a gap in the domain at 5. This could be the result of, for example, a function with “x - 5” in the denominator.
• You can use as many "U" symbols as necessary if the domain has multiple gaps in it.
• Use infinity and negative infinity signs to express that the domain goes on infinitely in either direction.
• Always use ( ), not [ ], with infinity symbols.
• Keep in mind that this notation may be different depending on where you live.
• The rules outlined above apply to the UK and USA.
• Some regions use arrows instead of infinity signs to express that the domain goes on infinitely in either direction.
• Usage of brackets varies wildly across regions. For example, Belgium uses reverse square brackets instead of round ones.
### Finding the Domain of a Function with a Fraction
1. Write the problem. Let's say you're working with the following problem:
• f(x) = 2x/(x - 4)
2. Set the denominator equal to zero for fractions with a variable in the denominator. When finding the domain of a fractional function, you must exclude all the x-values that make the denominator equal to zero, because you can never divide by zero. So, write the denominator as an equation and set it equal to 0. Here's how you do it:
• f(x) = 2x/(x - 4)
• x - 4 = 0
• (x - 2 )(x + 2) = 0
• x ≠ (2, - 2)
3. State the domain. Here's how you do it:
• x = all real numbers except 2 and -2
### Finding the Domain of a Function with a Square Root
1. Write the problem. Let's say you're working with the following problem: Y =√(x-7)
2. Set the terms inside the radicand to be greater than or equal to 0. You cannot take the square root of a negative number, though you can take the square root of 0. So, set the terms inside the radicand to be greater than or equal to 0. Note that this applies not just to square roots, but to all even-numbered roots. It does not, however, apply to odd-numbered roots, because it is perfectly fine to have negatives under odd roots. Here's how:
• x-7 ≧ 0
3. Isolate the variable. Now, to isolate x on the left side of the equation, just add 7 to both sides, so you're left with the following:
• x ≧ 7
4. State the domain correctly. Here is how you would write it:
• D = [7,∞)
5. Find the domain of a function with a square root when there are multiple solutions. Let's say you're working with the following function: Y = 1/√( ̅x -4). When you factor the denominator and set it equal to zero, you'll get x ≠ (2, - 2). Here's where you go from there:
• Now, check the area below -2 (by plugging in -3, for example), to see if the numbers below -2 can be plugged into the denominator to yield a number higher than 0. They do.
• (-3) - 4 = 5
• Now, check the area between -2 and 2. Pick 0, for example.
• 0 - 4 = -4, so you know the numbers between -2 and 2 don't work.
• Now try a number above 2, such as +3.
• 3 - 4 = 5, so the numbers over 2 do work.
• Write the domain when you're done. Here is how you would write the domain:
• D = (-∞, -2) U (2, ∞)
### Finding the Domain of a Function Using a Natural Log
1. Write the problem. Let's say you're working with this one:
• f(x) = ln(x-8)
2. Set the terms inside the parentheses to greater than zero. The natural log has to be a positive number, so set the terms inside the parentheses to greater than zero to make it so. Here's what you do:
• x - 8 > 0
3. Solve. Just isolate the variable x by adding 8 to both sides. Here's how:
• x - 8 + 8 > 0 + 8
• x > 8
4. State the domain. Show that the domain for this equation is equal to all numbers greater than 8 until infinity. Here's how:
• D = (8,∞)
### Finding the Domain of a Function Using a Graph
1. Look at the graph.
2. Check out the x-values that are included in the graph. This may be easier said than done, but here are some tips:
• A line. If you see a line on the graph that extends to infinity, then all versions of x will be covered eventually, so the domain is equal to all real numbers.
• A normal parabola. If you see a parabola that is facing upwards or downwards, then yes, the domain will be all real numbers, because all numbers on the x-axis will eventually be covered.
• A sideways parabola. Now, if you have a parabola with a vertex at (4,0) which extends infinitely to the right, then your domain is D = [4,∞)
3. State the domain. Just state the domain based on the type of graph you're working with. If you're uncertain and know the equation of the line, plug the x-coordinates back into the function to check.
### Finding the Domain of a Function Using a Relation
1. Write down the relation. A relation is simply a set of ordered pairs. Let's say you're working with the following coordinates: {(1, 3), (2, 4), (5, 7)}
2. Write down the x coordinates. They are: 1, 2, 5.
3. State the domain. D = {1, 2, 5}
4. Make sure the relation is a function. For a relation to be a function, every time you put in one numerical x coordinate, you should get the same y coordinate. So, if you put in 3 for x, you should always get 6 for y, and so on. The following relation is not a function because you get two different values of "y" for each value of "x": {(1, 4),(3, 5),(1, 5)} is not a function because X coordinate (1) has two different corresponding (4) and (5).
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# What is the surface area of the solid created by revolving f(x) =2x+5 , x in [1,2] around the x axis?
Jan 5, 2017
We need to calculate the area of te curved surface and the two end circles.
#### Explanation:
The end circles have radii $f \left(1\right) = 7$ and $f \left(2\right) = 9$ respectively and so have areas $49 \pi$ and $81 \pi$.
The are of the curved surface may be calculated by considering an infinitesimal ring of radius $f \left(x\right)$ between the values $x$ and $x + \mathrm{dx}$, which will have thickness $\mathrm{dl} = \sqrt{{\mathrm{df}}^{2} \left(x\right) + {\mathrm{dx}}^{2}}$ and therefore $A r e a = 2 \pi f \left(x\right) \mathrm{dl}$. The total curved area will then be $\setminus {\int}_{1}^{2} 2 \pi f \left(x\right) \mathrm{dl}$
Now, $\mathrm{dl} = \sqrt{{\mathrm{df}}^{2} \left(x\right) + {\mathrm{dx}}^{2}} = \mathrm{dx} \sqrt{{\left(\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}\right)}^{2} + 1} = \mathrm{dx} \sqrt{{2}^{2} + 1}$
$= \mathrm{dx} \sqrt{5}$.
So we have to evaluate the integral
$\setminus {\int}_{1}^{2} \mathrm{dx} 2 \pi f \left(x\right) \sqrt{5} = 2 \sqrt{5} \pi {\left[2 {x}^{2} / 2 + 5 x\right]}_{1}^{2} = 2 \sqrt{5} \pi \left[14 - 6\right]$
$= 16 \sqrt{5} \pi$.
The total area is then $= \left(16 \sqrt{5} + 49 + 81\right) \pi = \left(16 \sqrt{5} + 130\right) \pi$. |
What are the 7 laws of exponents?
There are seven exponent rules, or laws of exponents, that your students need to learn….Exponent rules
• Product of powers rule.
• Quotient of powers rule.
• Power of a power rule.
• Power of a product rule.
• Power of a quotient rule.
• Zero power rule.
What is the 5 Laws of exponent?
Laws of Exponent at a Glance
Law of Reciprocal a−n=1/an
Product Law of Exponents xm.xn=xm+n
Power of Product Rule (a×b)n=an×bn
Quotient Law of Exponents xm÷xn=xm-n
Power Law of Exponents (xm)n=xmn
How do you simplify laws of exponents?
This leads to another rule for exponents—the Power Rule for Exponents. To simplify a power of a power, you multiply the exponents, keeping the base the same. For example, (23)5 = 215. For any positive number x and integers a and b: (xa)b= xa· b.
What are the 3 laws of exponents?
Rule 1: To multiply identical bases, add the exponents. Rule 2: To divide identical bases, subtract the exponents. Rule 3: When there are two or more exponents and only one base, multiply the exponents.
What are the 8 laws of exponents?
Laws of Exponents
• Multiplying Powers with same Base.
• Dividing Powers with the same Base.
• Power of a Power.
• Multiplying Powers with the same Exponents.
• Negative Exponents.
• Power with Exponent Zero.
• Fractional Exponent.
What are the three laws of exponents?
What is the golden rule in laws of exponents?
The mathematical golden rule states that, for any fraction, both numerator and denominator may be multiplied by the same number without changing the fraction’s value.
How do you solve exponents without a calculator?
So, for example, this is how you would solve 6^3 without a calculator, from start to finish. Write: 6 6 6, because the base number is 6 and the exponent is 3. Then write: 6 x 6 x 6, to place multiplication signs between each of the base numbers. After that, multiply out the first multiplication sign, or 6 x 6 = 36. |
## Lesson: Functions and Graphing Introducing the Concept
Your students may have graphed on a coordinate plane before, but it may not be a skill they have mastered. You will need to review how to graph points, the names of the four quadrants, and terms such as coordinate plane, ordered pair, x-coordinate, y-coordinate, and origin.
Materials: graph paper for students, two worksheets shown below, graph paper for the overhead, a ruler
Preparation: Prepare the worksheets shown below with some points graphed on one and some figures (for which students will find the areas) drawn on the other. Use Learning Tool 8 in the Learning Tools Folder.
Prerequisite Skills and Background: Students should be able to perform the four basic operations, and they should be able to find the area of a triangle and square by using formulas.
• Say: Today we are going to work on graphing points in the coordinate plane. (Place the overhead graph paper on the overhead and draw in the axes as students watch. Label the x-axis and y-axis.) I am now drawing two perpendicular lines on my graph paper. These are called the x-axis and y-axis. Do the same on your sheet of graph paper using your ruler. Label your axes as I have done.
• Ask: Does anyone know what we call the point where the two perpendicular lines intersect? (the origin) The origin is the ordered pair (0, 0) because it is at zero units on the x-axis and zero units on the y-axis. Points to the right of the y-axis have positive x-values, points to the left of the y-axis have negative x-values, and points on the y-axis have an x-value of zero. (Label the x-axis.) Similarly, points above the x-axis have positive y-values, points below the x-axis have negative y-values and points on the x-axis have a y-value of zero.
(Label the y-axis.)
• Say: The 2 in the ordered pair (2, 4) is the x-coordinate for the point and the 4 is the y-coordinate for the point. What would we label a point that is 3 units to the right of the origin and 5 units below the x-axis? (3, -5)
Mark this point on the overhead and have students do the same on their graph paper.
• Say: Now, on your graph paper, graph the following ordered pairs and label the points.
(-4, 6) (-3, -3) (4, 0) (0, -5) (-3, 2.5)
While students are graphing and labeling those points, walk around the room to check students' progress. When they have finished, have volunteers come up and do the same on the overhead.
Distribute the worksheet shown below.
• Say: Use the worksheet I have just given you and write the ordered pairs for the 6 points on it.
Walk around the room to check students' progress. After students have finished, have volunteers come up and do the same on the overhead.
• Ask: If we go from point C on the graph to point F, how many units do we move?(6) When we have two points like C and F above, we can find how far apart they are by subtracting their x-coordinates. Point C is at (-4, -4) and point F is at (2, -4). Therefore, the distance between the points is 2 − (-4), or 6 units. In general, we can find the distance between two points that lie on a line parallel to the x-axis by subtracting their x-coordinates.
• Say: On your worksheet, plot point G at coordinates (-2, -1).
• Ask: How do you think we could find the distance between points G and D? (We could count them on the graph, or we could subtract their y-coordinates.) Let's do both. If we subtract the y-coordinates, we get 6 − (-1), or 7, units.
Have students count to check that the answer is 7.
Distribute the second worksheet, which has a triangle and a rectangle drawn on it as shown below.
• Say: Look at rectangle WXYZ and write the ordered pairs.
Give students some time to label the points, and then ask a volunteer to read off the coordinates for each point.
• Ask: Who can find the length of side WX without counting? (4 units) How did you figure that out? (subtracted the x-coordinates: 6 − 2 = 4) Find the length of XY the same way. (10 units)
Have a volunteer explain how to do it.
• Ask: How do we find the area of rectangles? (Multiply the length times the width.) Good. So we can find the area of rectangle WXYZ by multiplying: 10 x 4 = 40 square units.
Have students check the answer by counting the square units.
• Ask: Look at triangle ABC. What is the length of side AC? (3 units) What do we call AC? (base) What is the length of side BC? (6 units) What do we call BC? (height) Who remembers the formula for the area of a triangle?
(A = bh)
• Say: Using the formula, find the area of the triangle. (9 units)
When students have finished, ask a volunteer to show the work on the board. |
# Arithmetic progression problem
• Jan 8th 2010, 06:02 PM
mastermin346
Arithmetic progression problem
How many terms at least of the arithmetic progression 1,4,7,10,.............are needed to give a sum greater than 590,starting from the first term of the arithmetic progression?
• Jan 8th 2010, 06:29 PM
skeeter
Quote:
Originally Posted by mastermin346
How many terms at least of the arithmetic progression 1,4,7,10,.............are needed to give a sum greater than 590,starting from the first term of the arithmetic progression?
$S_n = \frac{n}{2}(t_1 + t_n) > 590
$
where $t_n = t_1 + d(n-1)$
solve for the least value of $n$ that makes the sum greater than 590
• Jan 8th 2010, 09:41 PM
mastermin346
Quote:
Originally Posted by skeeter
$S_n = \frac{n}{2}(t_1 + t_n) > 590
$
where $t_n = t_1 + d(n-1)$
solve for the least value of $n$ that makes the sum greater than 590
how?? i dont know how to do..
• Jan 8th 2010, 10:14 PM
Prove It
Do as you have been instructed.
$S_n = \frac{n}{2}(t_1 + t_n)$
$= \frac{n}{2}[t_1 + t_1 + (n - 1)d]$
$= \frac{n}{2}[2t_1 + (n - 1)d]$
You should be able to see that $t_1 = 1$ and $d = 3$
So $S_n = \frac{n}{2}[2(1) + 3(n - 1)]$
$= \frac{n}{2}(2 + 3n - 3)$
$= \frac{n}{2}(3n - 1)$
$= \frac{3n^2}{2} - \frac{n}{2}$.
You know that $S_n > 590$
So $\frac{3n^2}{2} - \frac{n}{2} > 590$
Solve for $n$.
Choose the first whole value of $n$ that is greater than what you found...
• Jan 8th 2010, 11:00 PM
mastermin346
Quote:
Originally Posted by Prove It
Do as you have been instructed.
$S_n = \frac{n}{2}(t_1 + t_n)$
$= \frac{n}{2}[t_1 + t_1 + (n - 1)d]$
$= \frac{n}{2}[2t_1 + (n - 1)d]$
You should be able to see that $t_1 = 1$ and $d = 3$
So $S_n = \frac{n}{2}[2(1) + 3(n - 1)]$
$= \frac{n}{2}(2 + 3n - 3)$
$= \frac{n}{2}(3n - 1)$
$= \frac{3n^2}{2} - \frac{n}{2}$.
You know that $S_n > 590$
So $\frac{3n^2}{2} - \frac{n}{2} > 590$
Solve for $n$.
Choose the first whole value of $n$ that is greater than what you found...
what do you mean by "Choose the first whole value of http://www.mathhelpforum.com/math-he...b31363a1-1.gif that is greater than what you found"???
• Jan 8th 2010, 11:23 PM
mr fantastic
Quote:
Originally Posted by mastermin346
what do you mean by "Choose the first whole value of http://www.mathhelpforum.com/math-he...b31363a1-1.gif that is greater than what you found"???
You have been given almost the entire solution. Your job is to read and process what you've been told and then finish off the problem.
Solve $\frac{3n^2}{2} - \frac{n}{2} = 590 \Rightarrow \frac{3n^2}{2} - \frac{n}{2} - 590 = 0$ for n. Choose the value of n that is positive and satisfies $\frac{3n^2}{2} - \frac{n}{2} - 590 \geq 0$. |
# Kaplan GMAT Sample Problem: Prime Factors
- Aug 12, 06:00 AM Comments [0]
Dealing with prime numbers and prime factors is an essential GMAT skill. Technically it is a skill we learned as early as elementary school/early school age, however just because it was learned in childhood does not mean that these questions are necessarily easy for GMAT test-takers. First of all, dealing with prime factors may not be something we do on a daily basis, so you’ll want to be sure you practice during your studies. Secondly, sometimes the wording or steps along the way can be challenging on the GMAT, not to mention the time limit. Here is a fairly straightforward question, though it may take some time to work through.
Problem:
What is the smallest positive integer that is a multiple of 18, 20, 24, 25 and 30?
(A) 360
(B) 900
(C) 1,800
(D) 2,400
(E) 3,600
Solution:
While taking each answer choice, starting with the smallest, and dividing by each of the five numbers listed until you find one that divides evenly by all of them will get you to the right answer, you can avoid this time consuming process by employing prime factorization.
First, find the prime factors of the five numbers included in the question. These are as follows:
18 = 3 x 3 x 2
20 = 2 x 2 x 5
24 = 2 x 2 x 2 x 3
25 = 5 x 5
30 = 2 x 3 x 5
Second, we need to put these prime factors together in order to find our answer. We want to find the smallest number that is a multiple of all of our given numbers, so let’s start with the smallest one. The smallest multiple of 18 is 18, which equals 3 x 3 x 2. Next, our number must be a multiple of 20, which equals 2 x 2 x 5. Thus, we need 3 x 3 x 2 x 2 x 5 for our number to be a multiple of both 18 and 20, as this number includes all of the prime factors of both 18 and 20. Our third number is 24, which equals 2 x 2 x 2 x 3. However, the number we are building already includes two 2’s and a 3. Therefore, we only need to include one more 2 in order to make our number a multiple of 24. Now we are at 3 x 3 x 2 x 2 x 5 x 2. For our number to be a multiple of 25, we must add in another 5, giving us 3 x 3 x 2 x 2 x 5 x 2 x 5. Finally, to be divisible by 30 our number must include 2 x 3 x 5. However, as 2 x 3 x 5 is can already be found, we do not need to add in any additional numbers.
Lastly, to reach the answer, we calculate 3 x 3 x 2 x 5 x 2 x 5 in the following manner:
3 x 3 x 2 x 2 x 5 x 2 x 5 =
18 x 10 x 10 =
18 x 100 =
1800, which is answer choice (C).
~Bret Ruber |
Intro text, can be displayed through an additional field
## What is 32/5 As A Decimal?
When dealing with fractions, it is sometimes necessary to convert them into decimal form for various mathematical calculations. One such fraction is 32/5, which we will be exploring in this article. In decimal form, 32/5 is expressed as 6.4.
### Understanding Fractions and Decimals
Fractions are numerical quantities that represent a part of a whole. They are expressed in the form of a numerator and a denominator, separated by a slash (/). The numerator represents the number of parts we have, while the denominator represents the total number of equal parts in the whole.
Decimals, on the other hand, are a way of expressing fractions or parts of a whole in a different format. They use a decimal point to separate the whole number from the fractional part. For example, 1.5 represents one whole unit and a half, while 0.25 represents one-fourth of a whole.
### Converting 32/5 to a Decimal
To convert the fraction 32/5 into a decimal, we need to divide the numerator (32) by the denominator (5). Let's see how it's done:
32 ÷ 5 = 6.4
Therefore, 32/5 as a decimal is 6.4.
### Why Convert Fractions to Decimals?
Converting fractions to decimals can be useful in various situations, such as:
• Performing mathematical operations: It is often easier to perform addition, subtraction, multiplication, and division with decimals rather than fractions.
• Comparing quantities: Decimals provide a more straightforward way to compare quantities, especially when dealing with different denominators.
• Working with measurements: Many measurements, such as inches and centimeters, are often expressed in decimal form.
### FAQs
#### Q: What is the process of converting a fraction to a decimal?
A: To convert a fraction to a decimal, divide the numerator by the denominator.
#### Q: Is there a shortcut to convert fractions to decimals?
A: Yes, some fractions can be converted to decimals using mental math. For example, fractions with denominators of 10 or its multiples can be easily converted by shifting the decimal point.
#### Q: Can all fractions be converted to decimals?
A: Yes, all fractions can be converted to decimals. However, some fractions result in recurring decimals that go on forever, such as 1/3 (0.33333...).
### Conclusion
Converting fractions to decimals is a fundamental skill in mathematics. In the case of 32/5, we found that it can be expressed as the decimal number 6.4. Understanding the relationship between fractions and decimals allows us to perform calculations and comparisons more efficiently. Whether you are working on a math problem or dealing with real-life measurements, converting fractions to decimals is a valuable tool to have in your mathematical toolkit.
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# Finding Ratio Amounts in Proportions
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### Finding Ratio Amounts in Proportions
1. 1. Images Source: Google Images
2. 2. A strawberry milkshake is made in the ratio of 2 partsstrawberry ice cream to 7 parts of milk.If we fill our blender with 3 scoops of ice cream,then how many scoops of milk shouldwe add to the blender? Ice Cream : Milk 2 : 7 3 : m Image Source: Google ImagesWhere m = scoops of Milk (Continued next slide)
3. 3. We can solve this question using a method called“Cross Multiplying” like this: Ice Cream : Milk 2 : 7 3 : m 2xm=3x7 Image Source: Google Images 2m = 21 2 2 m = 10.5 scoops of milk
4. 4. - Write the Ratios out Vertically- Make sure there are three numbers present and one unknown letter variable- Cross Multiply the Items- Divide both sides by the number which is in front of the letter variable- Write the simplified answer
5. 5. a : 15 = 2 : 5 Find the value of “a” 2 : 5 Write the ratios out vertically, and then Cross Multiply a : 152 x 15 = 5 x a (Note writing 2 x 15 = a x 5 is also okay) 30 = 5a “5” is the number in front of the letter variable and so divide 30 = 5a both sides of the equals sign 5 5 by 5 and simplify. 6=a a=6
6. 6. 4 : 11 = 80 : k Find the value of “k” 4 : 11 Write the ratios out vertically, and then Cross Multiply80 : k4 x k = 11 x 80 (Note writing k x 4 = 80 x 11 is also okay) 4k = 880 “4” is the number in front of the letter variable and so divide 4k = 880 both sides of the equals sign 4 4 by 4 and simplify. k = 220
7. 7. The instructions for mixing two stroke lawn mower fuelsay to mix 1 part Oil to 24 parts Petrol or Gasoline fuel.If we have a 200ml of Oil, then how much Petrol dowe need to add to make two stroke lawn mower fuel ?Note that the usual method for makingthe fuel is to first pour the oil into a 5litre fuel can, and then add the requiredamount of petrol in stages, giving the cana good shake, with the lid on, to mix thefuel thoroughly at the end of each stage.(Continued next slide) Image Source: http://www.hobuk.co.uk
8. 8. Oil : Petrol = 1 : 24 = 200 : P where “P” = Petrol Amount 1 : 24 Write the ratios out vertically, and then Cross Multiply 200 : P 1 x P = 24 x 200 1P = 4800 “1” is the number in front of the letter variable and so divide 1P = 4800 both sides of the equals sign 1 1 by 1 and simplify, or do 1P = PP = 4800ml or 4.8 litres (So we just fill the 5 litre can with Petrol)
9. 9. Famous Australian artist Ken Done has created the following designsfor a new Australian flag.The ratio for the flag size is Length : Width = 5 : 3Tiffany is going to make one of these flags for her Graphic DesignProject, and plans for the length of the flag to be 90 cm long.What width should she make the flag ?(Continued next slide) Image Source: http://www.ausflag.com.au
10. 10. Length : Width = 5 : 3 = 90 : w where “w” = Width of Flag 5 : 3 Write the ratios out vertically, and then Cross Multiply 90 : w 5 x w = 3 x 90 5w = 270 “5” is the number in front of the letter variable and so divide 5w = 270 both sides of the equals sign 5 5 by 5 and simplify. w = 54 cm |
## 4.2 Degree Centrality
How should we define the idea of centrality? We might imagine that someone “central” to the network is someone who holds some sort of important, advantaged position in the network. The problem is, that there are lots of specific ways we can think of centrality, and thus different ways of mathematically defining the concept.
The first way of defining centrality is simply as a measure of how many alters an ego shares ties with. This simply takes a nodes degree as introduced in Chapter 2, and begins to consider this measure as a reflection of centrality. The logic is that those with more alters, compared to those with fewer, hold a more prominent place in the network.
Equation 1 presents how degree centrality is calculated. Although it might seem a simple task to just add up the number of connections of each node, that is essentially what the below mathematical equation is doing! Mathematical notation plays an important role in expressing network measures in succint formats.
Degree Centrality $$$C_D(j) = \sum_{j= 1}^{n}A_{ij}$$$ {\tag{0.2}}
How to Read Equations for Matrix Operations
Depending on your background in math, you may or may not already know how to interpret Equation (???)(eq:degcenteq). Essentially, the number at the bottom of the sigma is where to start. The number at the top of the Sigma symbol ($$\Sigma$$) is where to end. The equation to the left is the operation to be performed. Thus, one reads the Equation (???)(eq:degcenteq) as, starting at row $$j=1$$, and ending at the last possible row $$n$$ ($$n$$ is simply the total number of rows in the matrix, $$n$$ means to go to the final value in the matrix), add up all possible values of the cells designated by the row $$i$$ and column $$j$$ combination in matrix A. Thus, to calculate the degree centrality of each $$i = a, b, c$$ in the below matrix, each of the following calculations would be performed.
a b c
a - 1 0
b 1 - 1
c 0 1 -
$$C_D(a)=aa+ab+ac=1$$
$$C_D(b)=ba+bb+bc=2$$
$$C_D(c)=ca+cb+cb=1$$
In the same way if we had the formula:
\begin{align*} C_D(i) = \sum_{j = 1}^{n}A_{ij} \end{align*}
Then it would be telling us to sum values of each column $$j$$ down each row $$i$$:
$$C_D(a)=aa+ba+ca=1$$
$$C_D(b)=ab+bb+cb=2$$
$$C_D(c)=ac+bc+cc=1$$
The sigma notation is useful for summarizing this repetitive process in a simple, condensed form.
The same equation is used to calculate the out-degree centrality for an asymmetric matrix is the same as the equation for degree centrality in a symmetric matrix. However, the general equation to calculate in-degree centrality is reversed, adding down the columns j rather than adding across the columns i.
$$$C_I(i) = \sum_{i = 1}^{n}A_{ij}$$$
Remember the example of a directed graph and corresponding asymmetric matrix from the previous chapters? We can use it to understand in- and out-degree centrality.
Figure 0.18: A directed graph.
$\begin{array}{ccccccccc} & A & B & C & D & E & F & G \\ A & - & 1 & 0 & 0 & 0 & 1 & 0 \\ B & 1 & - & 0 & 1 & 0 & 0 & 0 \\ C & 0 & 1 & - & 0 & 0 & 0 & 0 \\ D & 0 & 1 & 0 & - & 0 & 0 & 0 \\ E & 0 & 0 & 1 & 1 & - & 0 & 0 \\ F & 1 & 0 & 0 & 0 & 0 & - & 0 \\ G & 0 & 0 & 0 & 1 & 0 & 1 & - \\ \end{array}$
Starting with calculating out-degree, we can look at the matrix. Looking across row A (i), we can see that there are two nodes in the columns (j), which A directs ties to. Thus, the out-degree of node A is two. Likewise, looking down column A (j), we see that A receives ties from two nodes for an in-degree of two. Looking then at the graph, we can confirm that A sends out two ties and receives two ties as well. Doing the same procedure for node G, we would see that adding across the matrix computes that node G sends out two ties, yet adding down the column, G receives no ties, just as a look at the graph would confirm! |
## 17Calculus Precalculus - Equations of Lines
### Functions
Functions
Polynomials
Rational Functions
Matrices
Systems
Trigonometry
Complex Numbers
### Practice
Calculus 1 Practice
Calculus 2 Practice
Practice Exams
### Articles
Line Equations
slope
$$\displaystyle{ m = \frac{y_2 - y_1}{x_2 - x_1} }$$
point-slope form
$$y-y_1 = m(x-x_1)$$
slope-intercept form
$$y=mx+b$$
general form
$$Ax + By = C$$
Before we get down to the details of lines, let's watch this video to remind us of the slope of a line.
### Prof Leonard - Introduction to the Slope of a Line [12min-29secs]
video by Prof Leonard
In order to determine an equation of a line, you need only two pieces of information, the slope and one point on the line. If you are given two points and no slope, use the slope equation $$\displaystyle{ m = \frac{y_2 - y_1}{x_2 - x_1} }$$ to find the slope and then use one of the points (either one, its doesn't matter, you will get the same answer no matter which point you choose).
Once you have the slope, usually called m, and one point, $$(x_1,y_1)$$, you can find the equation of the line using two possible equations.
1. The one that is the easiest and requires the least amount of algebra is the point-slope form, $$y-y_1 = m(x-x_1)$$. Notice that this is just the slope equation in a little different form.
### Prof Leonard - Using the Point-Slope Equation of a Line [24min-6secs]
video by Prof Leonard
2. The second, which works too, is to use the slope-intercept form, $$y=mx+b$$ and solve for $$b$$. This seems to be the preferred method of most students.
For your final answer, most instructors prefer the slope-intercept form $$y=mx+b$$. This is the required form for answers on this site. However, you need to check with your instructor to see what they expect. Here is a video on why we need the slope-intercept form.
### Prof Leonard - Why We Need Slope-Intercept Form [23min-55secs]
video by Prof Leonard
Slope-Intercept Form of a Line
Adjust the sliders for the slope m and the y-intercept b to see the plot and the equation change.
As we said above, your goal is to get the slope and one point on the line in order to find the equation of the line. Start with these problems to get some practice on this.
Instructions - - Unless otherwise instructed, find the equation of the line, in slope-intercept form, with the given slope passing through the given point.
Find the equation of the line, in slope-intercept form, with slope $$-2/3$$ passing through the point $$(-4, 6)$$.
Problem Statement
Find the equation of the line, in slope-intercept form, with slope $$-2/3$$ passing through the point $$(-4, 6)$$.
$$y = -2x/3 +10/3$$
Problem Statement
Find the equation of the line, in slope-intercept form, with slope $$-2/3$$ passing through the point $$(-4, 6)$$.
Solution
### 2710 video
$$y = -2x/3 +10/3$$
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Find the equation of the line, in slope-intercept form, with slope $$-2/3$$ passing through the point $$(1, -1)$$.
Problem Statement
Find the equation of the line, in slope-intercept form, with slope $$-2/3$$ passing through the point $$(1, -1)$$.
$$y = -2x/3 - 1/3$$
Problem Statement
Find the equation of the line, in slope-intercept form, with slope $$-2/3$$ passing through the point $$(1, -1)$$.
Solution
We are given the slope $$m = -2/3$$ and a point on the line $$(1, -1)$$. Let's use the point-slope form of the equation of the line, $$y-y_1 = m(x-x_1)$$ $$y-(-1) = (-2/3)( x- 1)$$ $$y + 1 = -2x/3 + 2/3$$ $$y = -2x/3 + 2/3 - 1$$ Final Answer: $$y = -2x/3 - 1/3$$ Here is how we would use the slope-intercept form to solve this problem. $$y = mx + b \to -1 = (-2/3)(1) + b$$ $$b = -1 + 2/3 = -1/3$$ Final Answer: $$y = -2x/3 - 1/3$$
$$y = -2x/3 - 1/3$$
Log in to rate this practice problem and to see it's current rating.
Find the equation of the line, in slope-intercept form, with slope $$2$$ passing through the point $$(-1, -6)$$.
Problem Statement
Find the equation of the line, in slope-intercept form, with slope $$2$$ passing through the point $$(-1, -6)$$.
$$y = 2x - 4$$
Problem Statement
Find the equation of the line, in slope-intercept form, with slope $$2$$ passing through the point $$(-1, -6)$$.
Solution
### 2711 video
$$y = 2x - 4$$
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Find the equation of the line, in slope-intercept form, with slope $$3$$ passing through the point $$(-2, 3)$$.
Problem Statement
Find the equation of the line, in slope-intercept form, with slope $$3$$ passing through the point $$(-2, 3)$$.
$$y = 3x + 9$$
Problem Statement
Find the equation of the line, in slope-intercept form, with slope $$3$$ passing through the point $$(-2, 3)$$.
Solution
We are given the slope $$m = 3$$ and a point on the line at $$(-2, 3)$$. Let's use the point-slope form $$y-y_1 = m(x - x_1)$$. $$y-3 = 3( x - (-2))$$ $$y - 3 = 3x + 6$$ $$y = 3x + 6 + 3$$ Final Answer: $$y = 3x + 9$$ If we used the slope-intercept form to solve this problem, we would have this. $$y = mx + b \to 3 = 3(-2) + b$$ $$b = 3 + 6 \to b = 9$$ Final Answer: $$y = 3x + 9$$
$$y = 3x + 9$$
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Even though your goal is to have the slope and one point on the line, you will not usually be given this exact information in order to find the equation of the line. Sometimes you are given two points. From that information you need to determine the slope and then use one of the points to get the equation of the line. These problems will give you practice with this.
Instructions - - Unless otherwise instructed, find the equation of the line that passes through the two given points. Give your answer in slope-intercept form.
Find the equation of the line in slope-intercept form that passes through $$(-3,7)$$ and $$(5,-1)$$.
Problem Statement
Find the equation of the line in slope-intercept form that passes through $$(-3,7)$$ and $$(5,-1)$$.
$$y=-x+4$$
Problem Statement
Find the equation of the line in slope-intercept form that passes through $$(-3,7)$$ and $$(5,-1)$$.
Solution
### 2712 video
$$y=-x+4$$
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Find the equation of the line in slope-intercept form that passes through the points $$(3,-2)$$ and $$(2,-1)$$.
Problem Statement
Find the equation of the line in slope-intercept form that passes through the points $$(3,-2)$$ and $$(2,-1)$$.
$$y=-x+1$$
Problem Statement
Find the equation of the line in slope-intercept form that passes through the points $$(3,-2)$$ and $$(2,-1)$$.
Solution
### 2713 video
$$y=-x+1$$
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Find the equation of the line in slope-intercept form that contains $$(5,0)$$ and $$(-4,3)$$.
Problem Statement
Find the equation of the line in slope-intercept form that contains $$(5,0)$$ and $$(-4,3)$$.
$$y = -x/3 + 5/3$$
Problem Statement
Find the equation of the line in slope-intercept form that contains $$(5,0)$$ and $$(-4,3)$$.
Solution
### 2714 video
$$y = -x/3 + 5/3$$
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Determine the equation of the line passing through the points $$(1, 1)$$ and $$(5, -1)$$, giving your answer in slope-intercept form.
Problem Statement
Determine the equation of the line passing through the points $$(1, 1)$$ and $$(5, -1)$$, giving your answer in slope-intercept form.
$$y = -x/2 + 3/2$$
Problem Statement
Determine the equation of the line passing through the points $$(1, 1)$$ and $$(5, -1)$$, giving your answer in slope-intercept form.
Solution
### 2715 video
$$y = -x/2 + 3/2$$
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Determine the equation of the line passing through the points $$(-1, 1)$$ and $$(1, 7)$$, giving your answer in standard form $$Ax + By = C$$.
Problem Statement
Determine the equation of the line passing through the points $$(-1, 1)$$ and $$(1, 7)$$, giving your answer in standard form $$Ax + By = C$$.
$$3x - y = -4$$
Problem Statement
Determine the equation of the line passing through the points $$(-1, 1)$$ and $$(1, 7)$$, giving your answer in standard form $$Ax + By = C$$.
Solution
### 2716 video
$$3x - y = -4$$
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Determine the equation of the line passing through $$(-2, -3)$$ and $$(4, -2)$$. Give your answer in slope-intercept form.
Problem Statement
Determine the equation of the line passing through $$(-2, -3)$$ and $$(4, -2)$$. Give your answer in slope-intercept form.
$$y = x/6 - 8/3$$
Problem Statement
Determine the equation of the line passing through $$(-2, -3)$$ and $$(4, -2)$$. Give your answer in slope-intercept form.
Solution
### 2717 video
$$y = x/6 - 8/3$$
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Find the equation of the line that goes through $$(-3, 5)$$ and $$(2, 8)$$. Use the point-slope form $$y - y_1 = m(x - x_1)$$ and the second point. Give your answer in slope-intercept form.
Problem Statement
Find the equation of the line that goes through $$(-3, 5)$$ and $$(2, 8)$$. Use the point-slope form $$y - y_1 = m(x - x_1)$$ and the second point. Give your answer in slope-intercept form.
$$y = 3x/5+ 34/5$$
Problem Statement
Find the equation of the line that goes through $$(-3, 5)$$ and $$(2, 8)$$. Use the point-slope form $$y - y_1 = m(x - x_1)$$ and the second point. Give your answer in slope-intercept form.
Solution
### 2718 video
$$y = 3x/5+ 34/5$$
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Determine the equation of the line with an x-intercept of $$3$$ and a y-intercept of $$-4$$. Give your answer in slope-intercept form.
Problem Statement
Determine the equation of the line with an x-intercept of $$3$$ and a y-intercept of $$-4$$. Give your answer in slope-intercept form.
$$y = 4x/3 - 4$$
Problem Statement
Determine the equation of the line with an x-intercept of $$3$$ and a y-intercept of $$-4$$. Give your answer in slope-intercept form.
Solution
### 2719 video
$$y = 4x/3 - 4$$
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Okay, before going on, here is great video on what we are going to discuss next, parallel and perpendicular lines.
### Finding Equations of Parallel and Perpendicular Lines
Parallel Lines
When two lines are parallel, their slopes are equal. So, for two lines in slope intercept form $$y = mx + b_1$$ and $$y=mx + b_2$$, the slopes $$m$$ are the same but $$b_1 \neq b_2$$ (unless of course they are the same line). One of the nice things about the slope intercept form, $$y = mx + b$$ is that you can pull the slope $$m$$ and the y-intercept $$b$$ directly from the equation without having to do any work. For the two parallel lines, the slopes are the same but they cross the y-axis at different points.
Instructions - - Unless otherwise instructed, find the equation of the line in slope-intercept parallel to the given line going through the given point.
Find the equation of the line in slope-intercept parallel to $$3x-y/4 = 2$$ going through the point $$(1/6,-8)$$.
Problem Statement
Find the equation of the line in slope-intercept parallel to $$3x-y/4 = 2$$ going through the point $$(1/6,-8)$$.
$$y=12x-10$$
Problem Statement
Find the equation of the line in slope-intercept parallel to $$3x-y/4 = 2$$ going through the point $$(1/6,-8)$$.
Solution
### 2720 video
$$y=12x-10$$
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Write the equation of the line that is parallel to $$x - 2y = 6$$ and passes through the point $$(-4, 3)$$, giving your answer in slope-intercept form.
Problem Statement
Write the equation of the line that is parallel to $$x - 2y = 6$$ and passes through the point $$(-4, 3)$$, giving your answer in slope-intercept form.
$$y = x/2 + 5$$
Problem Statement
Write the equation of the line that is parallel to $$x - 2y = 6$$ and passes through the point $$(-4, 3)$$, giving your answer in slope-intercept form.
Solution
### 2721 video
$$y = x/2 + 5$$
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Determine the equation of the line that is parallel to $$y = -5x + 4$$ and passes through $$(-1, 4)$$, giving your answer in slope-intercept form.
Problem Statement
Determine the equation of the line that is parallel to $$y = -5x + 4$$ and passes through $$(-1, 4)$$, giving your answer in slope-intercept form.
$$y = -5x - 1$$
Problem Statement
Determine the equation of the line that is parallel to $$y = -5x + 4$$ and passes through $$(-1, 4)$$, giving your answer in slope-intercept form.
Solution
### 2722 video
$$y = -5x - 1$$
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Determine the equation of the line that is parallel to $$3x+2y = 12$$ and passes through $$(2, -5)$$, giving your answer in slope-intercept form.
Problem Statement
Determine the equation of the line that is parallel to $$3x+2y = 12$$ and passes through $$(2, -5)$$, giving your answer in slope-intercept form.
$$y = -3x/2 + 6$$
Problem Statement
Determine the equation of the line that is parallel to $$3x+2y = 12$$ and passes through $$(2, -5)$$, giving your answer in slope-intercept form.
Solution
### 2723 video
$$y = -3x/2 + 6$$
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Find the equation of the line parallel to the line $$y = 2x + 1$$ that passes through the point $$(1,5)$$, giving your answer in slope-intercept form.
Problem Statement
Find the equation of the line parallel to the line $$y = 2x + 1$$ that passes through the point $$(1,5)$$, giving your answer in slope-intercept form.
$$y = 2x + 3$$
Problem Statement
Find the equation of the line parallel to the line $$y = 2x + 1$$ that passes through the point $$(1,5)$$, giving your answer in slope-intercept form.
Solution
### 2724 video
$$y = 2x + 3$$
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Find the equation of the line parallel to the line $$y = 3x - 4$$ that passes through the point $$(6, 4)$$, giving your answer in slope-intercept form.
Problem Statement
Find the equation of the line parallel to the line $$y = 3x - 4$$ that passes through the point $$(6, 4)$$, giving your answer in slope-intercept form.
$$y = 3x - 14$$
Problem Statement
Find the equation of the line parallel to the line $$y = 3x - 4$$ that passes through the point $$(6, 4)$$, giving your answer in slope-intercept form.
Solution
### 2725 video
$$y = 3x - 14$$
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Find the equation of the line parallel to the line $$y = 3x/2 + 3$$ that passes through the point $$(4, -3)$$, giving your answer in slope-intercept form.
Problem Statement
Find the equation of the line parallel to the line $$y = 3x/2 + 3$$ that passes through the point $$(4, -3)$$, giving your answer in slope-intercept form.
$$y = 3x/2 - 9$$
Problem Statement
Find the equation of the line parallel to the line $$y = 3x/2 + 3$$ that passes through the point $$(4, -3)$$, giving your answer in slope-intercept form.
Solution
### 2726 video
$$y = 3x/2 - 9$$
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Find the equation of the line in slope-intercept parallel to $$x+7y=4$$ through the point $$(7,9)$$.
Problem Statement
Find the equation of the line in slope-intercept parallel to $$x+7y=4$$ through the point $$(7,9)$$.
$$y=-x/7+10$$
Problem Statement
Find the equation of the line in slope-intercept parallel to $$x+7y=4$$ through the point $$(7,9)$$.
Solution
### 2727 video
$$y=-x/7+10$$
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Perpendicular Lines
In the case of perpendicular lines, if the equation of one line is $$y = mx + b_1$$, the equation of the other is $$y=-x/m+b_2$$ So the slope of the second line is said to be the negative reciprocal of the first line.
Okay, time for the remaining practice problems.
Instructions - - Unless otherwise instructed, Find the equation of the line, in slope-intercept form, perpendicular to the given line going through the given point.
Find the equation of the line, in slope-intercept form, perpendicular to $$9x+12y=2$$ containing the point $$(15,7)$$.
Problem Statement
Find the equation of the line, in slope-intercept form, perpendicular to $$9x+12y=2$$ containing the point $$(15,7)$$.
$$y=4x/3 - 13$$
Problem Statement
Find the equation of the line, in slope-intercept form, perpendicular to $$9x+12y=2$$ containing the point $$(15,7)$$.
Solution
### 2728 video
$$y=4x/3 - 13$$
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Find the equation of the line, in slope-intercept form, perpendicular to $$10x+2y=9$$ going through the point $$(20,3)$$.
Problem Statement
Find the equation of the line, in slope-intercept form, perpendicular to $$10x+2y=9$$ going through the point $$(20,3)$$.
$$y = x/5 - 1$$
Problem Statement
Find the equation of the line, in slope-intercept form, perpendicular to $$10x+2y=9$$ going through the point $$(20,3)$$.
Solution
### 2729 video
$$y = x/5 - 1$$
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Determine the equation of the line that is perpendicular to $$x - 4y = 15$$ and passes through $$(3, 1)$$, giving your answer in slope-intercept form.
Problem Statement
Determine the equation of the line that is perpendicular to $$x - 4y = 15$$ and passes through $$(3, 1)$$, giving your answer in slope-intercept form.
$$y = -4x + 13$$
Problem Statement
Determine the equation of the line that is perpendicular to $$x - 4y = 15$$ and passes through $$(3, 1)$$, giving your answer in slope-intercept form.
Solution
### 2730 video
$$y = -4x + 13$$
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Write the equation of the line that contains the point $$(3, 5)$$ and is perpendicular to the line $$y = -3x + 2$$, giving your answer in slope-intercept form.
Problem Statement
Write the equation of the line that contains the point $$(3, 5)$$ and is perpendicular to the line $$y = -3x + 2$$, giving your answer in slope-intercept form.
$$y = x/3 + 4$$
Problem Statement
Write the equation of the line that contains the point $$(3, 5)$$ and is perpendicular to the line $$y = -3x + 2$$, giving your answer in slope-intercept form.
Solution
### 2731 video
$$y = x/3 + 4$$
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Determine the equation of the line that is perpendicular to $$y = 2x + 1$$ and passes through $$(-4, 5)$$, giving your answer in slope-intercept form.
Problem Statement
Determine the equation of the line that is perpendicular to $$y = 2x + 1$$ and passes through $$(-4, 5)$$, giving your answer in slope-intercept form.
$$y = -x/2 + 3$$
Problem Statement
Determine the equation of the line that is perpendicular to $$y = 2x + 1$$ and passes through $$(-4, 5)$$, giving your answer in slope-intercept form.
Solution
### 2732 video
$$y = -x/2 + 3$$
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Find the equation of the line perpendicular to the line $$y = 5x - 3$$ that passes through the point $$(2, -1)$$, giving your answer in slope-intercept form.
Problem Statement
Find the equation of the line perpendicular to the line $$y = 5x - 3$$ that passes through the point $$(2, -1)$$, giving your answer in slope-intercept form.
$$y = -x/5 - 3/5$$
Problem Statement
Find the equation of the line perpendicular to the line $$y = 5x - 3$$ that passes through the point $$(2, -1)$$, giving your answer in slope-intercept form.
Solution
### 2733 video
$$y = -x/5 - 3/5$$
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Find the equation of the line perpendicular to the line $$y = 3x/4 - 1$$ that passes through the point $$(8, -3)$$, giving your answer in slope-intercept form.
Problem Statement
Find the equation of the line perpendicular to the line $$y = 3x/4 - 1$$ that passes through the point $$(8, -3)$$, giving your answer in slope-intercept form.
$$y = -4x/3 + 23/3$$
Problem Statement
Find the equation of the line perpendicular to the line $$y = 3x/4 - 1$$ that passes through the point $$(8, -3)$$, giving your answer in slope-intercept form.
Solution
### 2734 video
$$y = -4x/3 + 23/3$$
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Find the equation of the line perpendicular to the line $$y = -2x/3 + 4$$ that passes through the point $$(6, -2)$$, giving your answer in slope-intercept form.
Problem Statement
Find the equation of the line perpendicular to the line $$y = -2x/3 + 4$$ that passes through the point $$(6, -2)$$, giving your answer in slope-intercept form.
$$y = 3x/2 - 11$$
Problem Statement
Find the equation of the line perpendicular to the line $$y = -2x/3 + 4$$ that passes through the point $$(6, -2)$$, giving your answer in slope-intercept form.
Solution
### 2735 video
$$y = 3x/2 - 11$$
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Find the equation of the line in slope-intercept perpendicular to $$y=2x+1$$ through the point $$(2,0)$$.
Problem Statement
Find the equation of the line in slope-intercept perpendicular to $$y=2x+1$$ through the point $$(2,0)$$.
$$y = -x/2 + 1$$
Problem Statement
Find the equation of the line in slope-intercept perpendicular to $$y=2x+1$$ through the point $$(2,0)$$.
Solution
### 2736 video
$$y = -x/2 + 1$$
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Find the equation of the line in slope-intercept perpendicular to $$y=-9x+5$$ through the point $$(3,9)$$.
Problem Statement
Find the equation of the line in slope-intercept perpendicular to $$y=-9x+5$$ through the point $$(3,9)$$.
$$y = x/9 + 26/3$$
Problem Statement
Find the equation of the line in slope-intercept perpendicular to $$y=-9x+5$$ through the point $$(3,9)$$.
Solution
### 2737 video
$$y = x/9 + 26/3$$
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Really UNDERSTAND Precalculus
### Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities
$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$
$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$
$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$
$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$
Set 2 - squared identities
$$\sin^2t + \cos^2t = 1$$
$$1 + \tan^2t = \sec^2t$$
$$1 + \cot^2t = \csc^2t$$
Set 3 - double-angle formulas
$$\sin(2t) = 2\sin(t)\cos(t)$$
$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$
Set 4 - half-angle formulas
$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$
$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$
Trig Derivatives
$$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$
Inverse Trig Derivatives
$$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$
Trig Integrals
$$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$
### Topics Listed Alphabetically
Single Variable Calculus
Multi-Variable Calculus
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Precalculus
Engineering
Circuits
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Home > Standard Error > Standard Error Of Mean Formula
# Standard Error Of Mean Formula
## Contents
To estimate the standard error of a student t-distribution it is sufficient to use the sample standard deviation "s" instead of σ, and we could use this value to calculate confidence For a value that is sampled with an unbiased normally distributed error, the above depicts the proportion of samples that would fall between 0, 1, 2, and 3 standard deviations above So if I were to take 9.3-- so let me do this case. Here, we're going to do a 25 at a time and then average them. Check This Out
Now, if it's 29, don't panic -- 30 is not a magic number, it's just a general rule of thumb. (The population standard deviation must be known either way.) Here's an So just for fun, I'll just mess with this distribution a little bit. This lesson shows how to compute the standard error, based on sample data. Well, Sal, you just gave a formula.
## Standard Error Of Mean Formula
Edwards Deming. So it's going to be a very low standard deviation. Notice that the population standard deviation of 4.72 years for age at first marriage is about half the standard deviation of 9.27 years for the runners.
1. To estimate the standard error of a student t-distribution it is sufficient to use the sample standard deviation "s" instead of σ, and we could use this value to calculate confidence
2. The graph below shows the distribution of the sample means for 20,000 samples, where each sample is of size n=16.
3. It can only be calculated if the mean is a non-zero value.
4. For example, when we take random samples of women's heights, while any individual height will vary by as much as 12 inches (a woman who is 5'10 and one who is
5. Of the 2000 voters, 1040 (52%) state that they will vote for candidate A.
6. And to make it so you don't get confused between that and that, let me say the variance.
7. So it equals-- n is 100-- so it equals one fifth.
8. Notice that the population standard deviation of 4.72 years for age at first marriage is about half the standard deviation of 9.27 years for the runners.
9. The variance is just the standard deviation squared.
This is more squeezed together. v t e Statistics Outline Index Descriptive statistics Continuous data Center Mean arithmetic geometric harmonic Median Mode Dispersion Variance Standard deviation Coefficient of variation Percentile Range Interquartile range Shape Moments But anyway, hopefully this makes everything clear. Standard Error Formula Statistics This chart can be expanded to other confidence percentages as well.
If people are interested in managing an existing finite population that will not change over time, then it is necessary to adjust for the population size; this is called an enumerative Standard Error Formula Excel For the runners, the population mean age is 33.87, and the population standard deviation is 9.27. So this is equal to 9.3 divided by 5. Statistic Standard Error Sample mean, x SEx = s / sqrt( n ) Sample proportion, p SEp = sqrt [ p(1 - p) / n ] Difference between means, x1 -
For any random sample from a population, the sample mean will very rarely be equal to the population mean. Standard Error Of Estimate Formula The next graph shows the sampling distribution of the mean (the distribution of the 20,000 sample means) superimposed on the distribution of ages for the 9,732 women. If the population standard deviation is finite, the standard error of the mean of the sample will tend to zero with increasing sample size, because the estimate of the population mean The notation for standard error can be any one of SE, SEM (for standard error of measurement or mean), or SE.
## Standard Error Formula Excel
So as you can see, what we got experimentally was almost exactly-- and this is after 10,000 trials-- of what you would expect. A practical result: Decreasing the uncertainty in a mean value estimate by a factor of two requires acquiring four times as many observations in the sample. Standard Error Of Mean Formula The sample mean x ¯ {\displaystyle {\bar {x}}} = 37.25 is greater than the true population mean μ {\displaystyle \mu } = 33.88 years. Standard Error Of Proportion Scenario 1.
So we got in this case 1.86. his comment is here T-distributions are slightly different from Gaussian, and vary depending on the size of the sample. Correction for correlation in the sample Expected error in the mean of A for a sample of n data points with sample bias coefficient ρ. So here, just visually, you can tell just when n was larger, the standard deviation here is smaller. Standard Error Of The Mean Definition
Contents 1 Introduction to the standard error 1.1 Standard error of the mean (SEM) 1.1.1 Sampling from a distribution with a large standard deviation 1.1.2 Sampling from a distribution with a If σ is not known, the standard error is estimated using the formula s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} where s is the sample The standard error can be computed from a knowledge of sample attributes - sample size and sample statistics. http://touchnerds.com/standard-error/standard-error-formula.html Next, consider all possible samples of 16 runners from the population of 9,732 runners.
Assumptions and usage Further information: Confidence interval If its sampling distribution is normally distributed, the sample mean, its standard error, and the quantiles of the normal distribution can be used to Standard Error Vs Standard Deviation The population standard deviation, will be given in the problem. You just take the variance divided by n.
## Blackwell Publishing. 81 (1): 75–81.
doi:10.2307/2340569. I. n is the size (number of observations) of the sample. Standard Error Regression In fact, data organizations often set reliability standards that their data must reach before publication.
They may be used to calculate confidence intervals. The graph below shows the distribution of the sample means for 20,000 samples, where each sample is of size n=16. The standard error is important because it is used to compute other measures, like confidence intervals and margins of error. navigate here This is equal to the mean.
When a statistical characteristic that's being measured (such as income, IQ, price, height, quantity, or weight) is numerical, most people want to estimate the mean (average) value for the population. In this scenario, the 2000 voters are a sample from all the actual voters. Look in the last row where the confidence levels are located, and find the confidence level of 95%; this marks the column you need. You take a random sample of 10 fingerlings and determine that the average length is 7.5 inches and the sample standard deviation is 2.3 inches. |
+0
0
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+15
Let x and y be nonnegative real numbers. If x^2+3y^2=18, then find the maximum value of xy.
Feb 16, 2024
#2
+2
As we need to find the max value of a product we can use AM-GM Inequality. the inequality states that for any real numbers $$x_1, x_2, \ldots, x_n \geq 0$$,
$$\LARGE\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}$$
with equality if and only if $$x_1 = x_2 = \cdots = x_n$$
Thus using this we write:
$$\Large\frac{x^2+3y^2}{2}\ge\sqrt[2]{x^2\cdot3y^2}$$
Plugging in values and solving-->
$$\sqrt3xy\le9\\ \boxed{xy\le3\sqrt3}$$
Feb 16, 2024
#1
+53
+1
I guess draw the ellipse
Feb 16, 2024
#2
+2
As we need to find the max value of a product we can use AM-GM Inequality. the inequality states that for any real numbers $$x_1, x_2, \ldots, x_n \geq 0$$,
$$\LARGE\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}$$
with equality if and only if $$x_1 = x_2 = \cdots = x_n$$
Thus using this we write:
$$\Large\frac{x^2+3y^2}{2}\ge\sqrt[2]{x^2\cdot3y^2}$$
Plugging in values and solving-->
$$\sqrt3xy\le9\\ \boxed{xy\le3\sqrt3}$$ |
Live Chat
# Matrix Exponentiation
## 1. What is Matrix Exponentiation
The matrix exponential plays an important role in the solution of systems of ordinary differential equations. It is a faster method that can be used to calculate the nth element of a series defined by a recurrence relation. This is one of the most used techniques in competitive programming.
For example,
In Fibonacci series problems, where we have to find the value of f(n) which is n’th Fibonacci number. When the value of n is sufficiently small, f(n) can be found by simple recursion F(n)=f(n-1)+f(n-2)
But, in the cases where the value of n is large, and the problem says
```{`
given 0< n <1000000000
find f(n) % 99999
`}```
The simple recursion method cannot be used also the dynamic programming will fail.
## 2. Matrix Exponential Algorithm
Now, let us see how matrix exponentiation can help us to represent recurrence relations.
### 2.1 RECURSIVE RELATIONS
The Fibonacci series is a sequence of numbers in which the first number is 0, the second number is 1 and all subsequent numbers are determined using the formula: f(n) = f(n-1) + f(n-2)
Such equation, in which one term of a sequence is defined using the previous terms i.e. value of the next term is a function of the previous term, is called a Recurrence relation.
A linear recurrence equation of degree k or order k is a recurrence equation which is in the format
```{`
X(n)=A1x(n−1) + A2x(n−1) + A3x(n−1) + …Akx(n) − kx(n)
= A1x(n−1) + A2x(n−1) + A3x(n−1) + …Akx(n−k)
`}```
(An is a constant and Ak≠0Ak≠0) on a sequence of numbers as a first-degree polynomial.
We can take the sequence defined by the recurrence relation a(n)= a(n−1) + 2, and a(0) = 1.
```{`
• The terms of the sequence are 1, 3, 5, 7, 9, ….
• A solution is a(n)=2n−1.
or
f(n)=f(n-3) + f(n-2) + f(n-1) which is an example of Tribonacci series
or
f(n)=3*f(n-1) + 7*f(n-2) an arbitrary example
are all recurrence relations.
`}```
### 2.2 REPRESENTING RECURSIVE RELATIONS USING MATRIX EXPONENTIAL
We can represent a recursive relation using a matrix M which can take us to the desired state from a set of known states.
Given, p states of a recurrence relation. Now, let us find the (p+1)th state of the recursive relation.
Let R be a p x p matrix, We have to create a matrix A:[p x 1] matrix from the known states of the recurrence relation given previously.
i.e. R is a p x p matrix & A is a p x 1 matrix
After building matrix A, let us get a matrix B:[p x 1] which will represent the set of next states
i.e. R x A = B
Hence, once matrix R is designed, then matrix B can be found easily and the recurrence relation can be represented using the matrix designed as above.
## 3. Matrix Exponential Implementation
Matrix Exponential can be implemented by using the following function in our code.
1. Multiplying two matrices of appropriate size
2. Creating an identity matrix In;
3. Raising a matrix to r-th power using fast exponentiation.
The matrix should be stored as well. The easiest approach is to store matrices as 2D-arrays:
```{`
int matrix1[30] [30]
int matrix2[30] [30]
`}```
Now, let us implement Matrix Exponential using some recurrence relation.
Given, a recurrence relation, as follows
```{`
f(n) = f(n-1) + f(n-2)
So, f(n+1) = f(n) + f(n-1)
`}```
The exponential matrix can be used to represent the recurrence relation f(n) i.e. value of f(n+1) when value of f(n) and f(n-1) are given.
Matrix A and B can be represented as,
Now , we have to design matrix M ( 2X2) such that,
M X A = B
Now, let us revise matrix multiplications
The product C of two matrices A and B is defined as
Cik = (Aij) (Bjk)
where j is summed over for all possible values of i and k and the notation above uses the Einstein summation convention. The implied summation over repeated indices without the presence of an explicit sum sign is called Einstein summation, and is commonly used in both matrix and tensor analysis. Therefore, in order for matrix multiplication to be defined, the dimensions of the matrices must satisfy
(n x m) (m x p)=(n x p)
Where
The matrix A contains f(n) and f(n-1) and matrix B contains f(n+1) and f(n). The first element of matrix B is f(n+1) which can be computed as f(n) + f(n-1).
Since,
M ( a, b) X A (f(n)) = B(f(n+1))
The value of 1st row of M can be computed as [1 1].
Similarly, the value of second row of matrix M can be computed which is [1 0]
Since,
In this way, the value of matrix M of size 2x2 could be found.
## 4. Matrix Exponential Code in C++
Below is the C++ implementation of the Matrix Exponentiation.
AIM:
To find value of f(n) where f(n) is defined as
f(n) = f(n-1) + f(n-2) + f(n-3),
Where n >= 3 and f(0)=0, f(1)=1, f(2)=1
```{`
AIM:
To find value of f(n) where f(n) is defined as
f(n) = f(n-1) + f(n-2) + f(n-3),
Where n >= 3 and f(0)=0, f(1)=1, f(2)=1
PROGRAM:
#include
using namespace std;
// Function to multiply two matrices a and b.
void multiply(int a[3][3], int b[3][3])
{
// Matrix to store elements at the time of matrix multiplication
int mul[3][3];
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
mul[i][j] = 0;
for (int k = 0; k < 3; k++)
mul[i][j] += a[i][k]*b[k][j];
}
}
// storing the value of multiplication in matrix a
for (int i=0; i<3; i++)
for (int j=0; j<3; j++)
a[i][j] = mul[i][j];
}
// Function to compute F raise to power n-2.
int power(int F[3][3], int n)
{
int M[3][3] = {1,1,1}, {1,0,0}, {0,1,0};
if (n==1)
return F[0][0] + F[0][1];
power(F, n/2);
multiply(F, F);
if (n%2 != 0)
multiply(F, M);
return F[0][0] + F[0][1] ;
}
// Function to find nth term of series
int findNthTerm(int n)
{
int F[3][3] = {1,1,1}, {1,0,0}, {0,1,0} ;
return power(F, n-2);
}
// Main function
int main()
{
int n = 5;
cout << "F(5) is " << findNthTerm(n);
return 0;
}
OUTPUT:
F(5) is 7
TIME COMPLEXITY:
O(log n)
`}```
## 5. Applications of Matrix Exponential
5.1 Finding the nth element of Fibonacci Series
Fibonacci numbers can be defined using f(n) as follows:
1. F0 = F1 = 1;
2. Fi = Fi - 1 + Fi - 2 for i ≥ 2.
We want to find FN modulo 1000000007, where N can be up to 1018.
Pseudo code:
```{`
function fibonacci_exponentiation(N):
if N <= 1:
return 1
initial = (1, 1)
exp = matrix_power_with_modulo(M, N - 1, 1000000007) // assuming we’ve defined M
return (initial * exp)[1][2] modulo 1000000007
`}```
### 5.2 Finding Nth element of Linear Recurrent Sequence
The sequence A satisfies two properties:
1. Ai = c1 * Ai- 1 + c2 * Ai- 2 + … + ck * Ai - k for i ≥ k (c1, c2 …, ck are given integers);
2. A0 = a0, A1 = a1, …, Ak- 1 = ak - 1 (a0, a1, …, ak - 1 are given integers). We need to find AN modulo 1000000007, when N is up to 1018 and k up to 50.
Pseudo code:
```{`
dp[1][a] = dp[1][b] = … = dp[1][z] = 1
for n = 2..L:
for last_letter = a..z:
for next_letter = a..z:
if banned[last_letter][next_letter] != 0:
dp[n + 1][next_letter] += dp[n][last_letter]
`}```
### 5.3 Finding the sum of Fibonacci numbers up to N
Let’s go back to Fibonacci numbers:
1. F0 = F1 = 1;
2. Fi = Fi - 1 + Fi - 2 for i ≥ 2.
Introduce a new sequence: Pi = F0 + F1 + … + Fi (i.e. sum of first i Fibonacci numbers). The problem is: find PN modulo 1000000007 for N up to 1018.
Pseudo code:
To find sum of first N numbers of any recurrent linear sequence Ai,
introduce the “sum sequence” Pi = A0 + A1 + … + Ai.
Put this sequence into the initial vector.
Then find the matrix that gives you Pi + 1, based on Pi and some A’s.
### 5.4 Computing multiple linear recurrent sequences at once
Given integer n ≥ 106, find the number of pairs of integers (x, y) such that:
1. 0 ≤ x < 2n;
2. 0 ≤ y < 2n;
3. Bitwise OR of x and y is not equal to x;
4. Bitwise OR of x and y is not equal to y.
Pseudo code:
let dpi be the answer for n = i.
Then it can be shown that:
dpi = dpi - 1 * 4 + gi - 1 * 2 for i > 1,
dp1 = 0,
where sequence g satisfies:
gi = gi - 1 * 3 + 2i - 1 for i > 1,
g1 = 1.
### 5.5 Solving dynamic programming problems with fixed linear transitions
Count the number of strings of length L with lowercase English letters, if some pairs of letters can not appear consequently in those strings. L is up to 107, there are up to 100 inputs.
Pseudo code:
Let dp[n][last_letter] be the number of valid strings of length n, which have last letter equal to last_letter. Then we can recalculate dp in order of increasing n:
```{`
dp[1][a] = dp[1][b] = … = dp[1][z] = 1
for n = 2..L:
for last_letter = a..z:
for next_letter = a..z:
if banned[last_letter][next_letter] != 0:
dp[n + 1][next_letter] += dp[n][last_letter]
`}```
### BIBLIOGRAPHY
[1] Mike Koltsov, August,2016. Matrix exponentiation. https://www.hackerearth.com/practice/notes/matrix-exponentiation-1/ |
# Understanding the Reciprocal of 5
Understanding the Reciprocal of 5
When it comes to learning mathematics, there are certain terms that can be confusing for many students. One such term is the reciprocal. A reciprocal is essentially a fraction where the numerator and denominator are flipped. For example, the reciprocal of 2 is 1/2 because the numerator and denominator have been flipped.
In this article, we will take a closer look at the reciprocal of 5 and understand its importance in mathematics.
The reciprocal of 5 is simply 1/5. This means that if we were to flip the numerator and denominator of 5, we would get the fraction 1/5. The reciprocal of any number is important because it helps us to perform mathematical operations more easily.
For example, let’s say we wanted to multiply 5 by its reciprocal, which is 1/5. We would simply multiply the numerators (5 and 1) and the denominators (1 and 5) separately as shown below:
5 x 1 = 5
1 x 5 = 5
So, the answer to 5 multiplied by its reciprocal is 1. This may seem like a pointless exercise, but it actually gives us an important insight into fractions.
Fractions represent a part of a whole. For example, if we have a pizza and we want to eat half of it, we would say that we are eating 1/2 of the pizza. Similarly, if we have a rectangle and we shade in one-third of it, we would say that one-third of the rectangle is shaded.
However, sometimes we need to perform operations on fractions. For example, we may need to add two fractions together or subtract one fraction from another. In these cases, we need to have a common denominator.
The denominator is the bottom number in a fraction and tells us how many parts the whole has been divided into. For example, in the fraction 2/5, the denominator is 5, which means the whole has been divided into 5 parts.
To add or subtract fractions, we need to have a common denominator. This means that both fractions have the same denominator, so we can simply add or subtract the numerators.
For example, let’s say we wanted to add 1/5 and 3/5. We first need to find a common denominator, which in this case is 5. To convert 1/5 into a fraction with a denominator of 5, we need to multiply both the numerator and denominator by 1:
1 x 1 = 1
5 x 1 = 5
So, 1/5 becomes 1/5. Similarly, to convert 3/5 into a fraction with a denominator of 5, we need to multiply both the numerator and denominator by 1:
3 x 1 = 3
5 x 1 = 5
So, 3/5 becomes 3/5. Now that both fractions have a common denominator of 5, we can simply add the numerators:
1/5 + 3/5 = (1+3)/5 = 4/5
This process of finding a common denominator can be time-consuming, especially when dealing with large numbers. However, if we know the reciprocal of the denominator, we can perform the operation more easily.
For example, let’s say we wanted to add 2/7 and 3/8. We could find a common denominator by multiplying 7 and 8 together to get 56. However, this would be a time-consuming process.
Instead, we can simply find the reciprocal of each denominator and multiply the denominators together. The reciprocal of 7 is 1/7 and the reciprocal of 8 is 1/8. Multiplying these two reciprocals together gives us 1/56, which is the common denominator:
1/7 x 1/8 = 1/56
Now that we have a common denominator, we can convert both fractions to have a denominator of 56:
2/7 x 8/8 = 16/56
3/8 x 7/7 = 21/56
Now we can add the numerators:
16/56 + 21/56 = 37/56 |
# 11.6 - Negative Binomial Examples
11.6 - Negative Binomial Examples
## Example 11-2
An oil company conducts a geological study that indicates that an exploratory oil well should have a 20% chance of striking oil. What is the probability that the first strike comes on the third well drilled?
#### Solution
To find the requested probability, we need to find $$P(X=3$$. Note that $$X$$is technically a geometric random variable, since we are only looking for one success. Since a geometric random variable is just a special case of a negative binomial random variable, we'll try finding the probability using the negative binomial p.m.f. In this case, $$p=0.20, 1-p=0.80, r=1, x=3$$, and here's what the calculation looks like:
$$P(X=3)=\dbinom{3-1}{1-1}(1-p)^{3-1}p^1=(1-p)^2 p=0.80^2\times 0.20=0.128$$
It is at the second equal sign that you can see how the general negative binomial problem reduces to a geometric random variable problem. In any case, there is about a 13% chance thathe first strike comes on the third well drilled.
What is the probability that the third strike comes on the seventh well drilled?
#### Solution
To find the requested probability, we need to find $$P(X=7$$, which can be readily found using the p.m.f. of a negative binomial random variable with $$p=0.20, 1-p=0.80, x=7, r=3$$:
$$P(X=7)=\dbinom{7-1}{3-1}(1-p)^{7-3}p^3=\dbinom{6}{2}0.80^4\times 0.20^3=0.049$$
That is, there is about a 5% chance that the third strike comes on the seventh well drilled.
What is the mean and variance of the number of wells that must be drilled if the oil company wants to set up three producing wells?
#### Solution
The mean number of wells is:
$$\mu=E(X)=\dfrac{r}{p}=\dfrac{3}{0.20}=15$$
with a variance of:
$$\sigma^2=Var(x)=\dfrac{r(1-p)}{p^2}=\dfrac{3(0.80)}{0.20^2}=60$$
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility |
# It x and y vary inversely, fill in the following blanks:
Question:
It x and y vary inversely, fill in the following blanks:
(i)
(ii)
(iii)
Solution:
(i) Since $x$ and $y$ vary inversely, we have :
$x y=k$
For $x=16$ and $y=6$, we have :
$16 \times 6=k$
$\Rightarrow k=96$
For $x=12$ and $k=96$, we have :
$x y=k$
$\Rightarrow 12 y=96$
$\Rightarrow y=\frac{96}{12}$
$=8$
For $y=4$ and $k=96$, we have :
$x y=k$
$\Rightarrow 4 x=96$
$\Rightarrow x=\frac{96}{4}$
$=24$
For $x=8$ and $k=96$, we have :
$x y=k$
$\Rightarrow 8 y=96$
$\Rightarrow y=\frac{96}{8}$
$=12$
For $y=0.25$ and $k=96$, we have :
$x y=k$
$\Rightarrow 0.25 x=96$
$\Rightarrow x=\frac{96}{0.25}$
$=384$
(ii) Since $x$ and $y$ vary inversely, we have:
$x y=k$
For $x=16$ and $y=4$, we have:
$16 \times 4=k$
$\Rightarrow k=64$
For $x=32$ and $k=64$, we have :
$x y=k$
$\Rightarrow 32 y=64$
$\Rightarrow y=\frac{64}{32}$
$=2$
For $x=8$ and $k=64$
$x y=k$
$\Rightarrow 8 y=64$
$\Rightarrow y=\frac{64}{8}$
$=8$
(iii) Since $x$ and $y$ vary inversely, we have :
$x y=k$
For $x=9$ and $y=27$
$9 \times 27=k$
$\Rightarrow k=243$
For $y=9$ and $k=243$, we have :
$x y=k$
$\Rightarrow 9 x=243$
$\Rightarrow y=\frac{243}{9}$
$=27$
For $x=81$ and $k=243$, we have :
$x y=k$
$\Rightarrow 81 y=243$
$\Rightarrow y=\frac{243}{81}$
$=3$ |
# Logaritma
## Sifat logaritma 1
###### Hubungan Eksponen dan Logaritma
Hubungan eksponen dan logaritma
$$^a\log b = c \rightarrow a^c = b$$
$$a > 0, \: b > 0$$ dan $$a \neq 1$$
Sifat Logaritma 1
$$^a \log c^p = p \:.\: ^a \log c$$
$$^{a^q} \log c = \frac{1}{q} \:.\: ^a \log c$$
$$^{a^q} \log c^p = \frac{p}{q} \:.\: ^a \log c$$
Contoh 01
Tentukan nilai x dariĀ $$^2 \log x = 3$$
\begin{equation*} \begin{split} & ^2 \log x = 3 \\\\ & x = 2^3 \\\\ & \bbox[5px, border: 2px solid magenta] {x = 8} \end{split} \end{equation*}
Contoh 02
\begin{equation*} \begin{split} & ^7\log \frac{1}{49} \\\\ & ^7\log 7^{-2} \\\\ & -2 \:.\: ^7\log 7 \\\\ & \bbox[5px, border: 2px solid magenta] {-2} \end{split} \end{equation*}
Contoh 03
\begin{equation*} \begin{split} & ^8\log 2 \\\\ & ^{\large{2^3}}\log 2 \\\\ & \frac{1}{3}\:.\: ^2 \log 2 \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{1}{3}} \end{split} \end{equation*}
Contoh 04
\begin{equation*} \begin{split} & ^{25}\log \frac{1}{625} \\\\ & ^{\large{5^2}}\log 5^{-4} \\\\ & \frac{-4}{2} \: .\: ^5 \log 5 \\\\ & \bbox[5px, border: 2px solid magenta] {-2} \end{split} \end{equation*}
##### SOAL LATIHAN
--- Buka halaman ini --- |
RATIO AND PROPORTION
Graphing Ratios
(Student Copy)
### Activity 7: Graphing Ratios
• completed copy of Activity 6
• calculator
• metric ruler
• graph paper (4 squares to the inch preferred)
Part A
Ratios can be written in a variety of forms. The ratio of 2 to 3 can be written using a colon (2:3) or a fraction (2/3). In order to answer the following questions, look back at the table in Part B of Activity 6 for the number of teeth on the front sprocket of a bicycle compared to the number of teeth on the rear sprocket.
Write the gear ratios in Activity 6 (Part B) in simplest form using a colon and using a fraction.
To determine the speed of the bicycle in each combination of gears, you divided the number of teeth in the rear sprocket into the number of teeth in the front sprocket. Explain, in writing, why we did not give these answers in colon or fraction form.
Part B
In manufacturing, it is important to keep track of the number of defective parts or products being produced. Rather than simply counting the number of defective parts, company officials usually consider the number of defects per hundred, thousand, or million parts.
Suppose that the Lightning Electric Company was making complex light switches and had a defect rate of 25 parts per hundred. Why would the figure 25 parts per hundred be considered a ratio?
Fill in the table below to show the number of parts produced in the top row and the number of defective parts in the bottom row. The top row should have the entries 100, 200, 300, 400, and 500.
Parts produced Defective parts
What number must you multiply the bottom row by to get the top row? (If you are not sure, take a guess and check it with your calculator.)
What number must you multiply the top row by to get the bottom row?
Part C
Draw an x-y graph on your graph paper for the data in the table; each line on the graph paper should represent 25 units. Label the x-axis "Number of Parts" and the y-axis "Number of Defects". Plot the points listed in your table and then connect them with a line.
Next, find a point on the line that is between two points. Carefully, draw a horizontal line (to the left) from your new point to the y-axis and measure it in millimeters. Record the distance below. Also, draw a vertical line (down) from your new point to the x-axis. Measure and record.
Horizontal distance:
Vertical distance:
What number must you multiply the horizontal distance by to get the vertical distance?
Explain how the numbers you just found relate to the top and bottom rows of the table.
Do you think that choosing a different point on the line and repeating the measurements would result in a different ratio of the horizontal distance to the vertical distance? Why or why not?
### Activity 8: Using Ratio to Approximate
• standard-units (English) ruler
• graph paper (4 squares to the inch preferred)
• calculator
• Reference Copy
Part A
Ö2 = _______
Draw squares with sides of 3 inches, 4 inches, 6 inches and 8 inches on graph paper. Draw and measure (to the nearest quarter of an inch) the length of one diagonal (a diagonal connects two opposite corners) for each square. In the space below, make a table showing the length of the side and the length of the diagonal for each square.
Length of the side (inches) 3 4 6 8 Length of the diagonal (inches) Divide the diagonal by side
Now, complete the table by dividing the length of each diagonal by the length of the side. Do you think the ratio will be the same for any square? Write your answer below.
Part B
Before people had sophisticated measure instruments, distances had to be approximated. Distances across a river or a canyon could be particularly troublesome. Even so, precise measurements were often desired. How would you judge the amount of rope needed to construct a bridge made of rope and wood that needed to be built over a rushing river if you didn't have any sophisticated tools? You will soon find out.
Refer to the Reference Sheet. Let's say that the picture represents the place where a bridge is to be built. You have noticed two prominent landmarks—a tree and a rock—and have marked two spots directly across the river from them. Next you measured the distance between the two marked spots and found that they were 300 feet apart.
To make the picture as realistic as possible, you first drew the river with the tree and the Marked Spot A. You located the rock in your picture at 28 degrees, the same angle that the rock is located from Marked Spot A. Finally, you completed with picture as shown.
The two marked spots and the rock form a triangle. Draw the triangle on the picture. The triangle on the page is similar to a triangle formed by the actual tree and the two marked spots. Your two triangles are similar, which means that the ratio of corresponding sides will be the same number. You decide to measure the distances on the picture and use that ratio to find the same ratio that works with a side of 300 feet.
Use your ruler to measure to the nearest 1/8 inch (rounding up for mid-distances): the distance from A to B (scaled-distance #1), and from B to the closest edge of the rock (scaled-distance #2). Next, divide scaled-distance #1 by scaled-distance #2 to get the ratio of length 1 to length 2. Decide how you can use this information to find the approximate real-life distance across the river.
1. distance from A to B is inches (scaled-distance #1)
2. distance from B to the rock is inches (scaled-distance #2)
3. ratio of scaled-distance #2 to scaled-distance #1
4. actual distance across the river is feet
Summary Questions
1. If your speedometer needle is pointing to 55, what ratio is indicated? What kind of ratio is it (same units or different units)? How would you state this ratio formally? How would you write it?
2. Most car speedometers have both English (miles per hour) and metric (kilometers per hour) units. How could you make a table to show the ratio of miles per hour to kilometers per hour? |
Fractions are used in everyday life A fraction is a part of a whole. It is a number between 0 and 1. You might see that you have a quarter of a tank of gas, or your child only ate half of their green vegetables at dinner. Whatever the case may be, knowing fractions and the relationship between fractions, improper fractions and mixed numbers are crutial.
What are Fractions?
Fractions are a way of conveying a relationship of numbers. The top part of the fraction is called the numerator; the bottom part is called the denominator. The numerator is the number of parts you are interested in and the denominator is the number of all the parts together.
For example, it’s pizza night at your house and you eat two slices of pizza and your child eats one slice of pizza. Now there are 5 out of 8 slices of pizza left. We can express that like this:
What is an Improper Fraction?
What we just saw is called a proper fraction. Now is there such a thing as an improper fraction? Yes. For a fraction to be “proper” it must be a part of a whole, meaning it is less than 1 whole thing (like a pizza or tank of gas) but greater than 0 – or, in other words, greater than nothing. So an improper fraction does not follow one of these rules. Well, for what we are talking about here, we cannot be less than 0, so the only way for a fraction to be improper is if it is greater than 1.
Let’s go back to pizza night. It turns out that there are actually 2 pizza pies, instead of 1. If we want to know how many slices there are left after the 3 that have been eaten, then we simply need to count. We already know that there are 8 slices in a pie so the denominator stays the same. Our numerator is no longer 5 but 13, because there are 13 slices of pizza left. This looks like this:
What is a Mixed Number?
Have you ever heard someone say “I have 13 8ths of a pizza left?” Probably not. They would probably say “I have 1 and 5 8ths of a pizza left.” This way of expressing fractions is called a mixed number. A mixed number is any whole number together with a fraction. Mixed numbers cannot be considered whole numbers because part of the number is a fraction. To write how much pizza is left as a mixed number, we would write:
Converting between Mixed Numbers and Improper Fractions
At some point your child will have to convert between mixed numbers and improper fractions. To do so is quite simple.
If we had for a mixed number and we need to make this an improper fraction, we simply multiply the denominator and the whole number to get As you can see, there are 17 shaded boxes but there are 2 whole rows filled and 3 boxes filled in the last row.
If we had the improper fraction , to make this into a mixed number we divide the numerator by the denominator. Most of the time there will be a remainder. This remainder becomes the new numerator while the quotient, the whole number answer to 32 divided by 5, is the whole number of the mixed number. as a mixed number isOnce again, there are 32 boxes shaded but also the first 6 rows are completely shaded and there are 2 boxes left over in the last row.
Fractions is something that every child should have some familiarity with. Some children know only fractions as "a half of a cookie" or "a slice of pizza". As the child gets older and as their studies continue fractions will always be there in same shape or form. Being familiar with fractions at an early age and being comfortable with them throughout a child's time at school will be a huge boost in confidence and aptitude. As you going about your every day life, point out some fractions to your child. They are everywhere! |
# 3.5 Transformation of functions (Page 3/21)
Page 3 / 21
Given a tabular function, create a new row to represent a horizontal shift.
1. Identify the input row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each input cell.
## Shifting a tabular function horizontally
A function $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is given in [link] . Create a table for the function $\text{\hspace{0.17em}}g\left(x\right)=f\left(x-3\right).$
$x$ 2 4 6 8 $f\left(x\right)$ 1 3 7 11
The formula $\text{\hspace{0.17em}}g\left(x\right)=f\left(x-3\right)\text{\hspace{0.17em}}$ tells us that the output values of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ are the same as the output value of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ when the input value is 3 less than the original value. For example, we know that $\text{\hspace{0.17em}}f\left(2\right)=1.\text{\hspace{0.17em}}$ To get the same output from the function $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ we will need an input value that is 3 larger . We input a value that is 3 larger for $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ because the function takes 3 away before evaluating the function $\text{\hspace{0.17em}}f.$
$\begin{array}{ccc}\hfill g\left(5\right)& =& f\left(5-3\right)\hfill \\ & =& f\left(2\right)\hfill \\ & =& 1\hfill \end{array}$
We continue with the other values to create [link] .
$x$ 5 7 9 11 $x-3$ 2 4 6 8 $f\left(x–3\right)$ 1 3 7 11 $g\left(x\right)$ 1 3 7 11
The result is that the function $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ has been shifted to the right by 3. Notice the output values for $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ remain the same as the output values for $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ but the corresponding input values, $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ have shifted to the right by 3. Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11.
## Identifying a horizontal shift of a toolkit function
[link] represents a transformation of the toolkit function $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}.\text{\hspace{0.17em}}$ Relate this new function $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ and then find a formula for $\text{\hspace{0.17em}}g\left(x\right).$
Notice that the graph is identical in shape to the $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}\text{\hspace{0.17em}}$ function, but the x- values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so
$g\left(x\right)=f\left(x-2\right)$
Notice how we must input the value $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ to get the output value $\text{\hspace{0.17em}}y=0;\text{\hspace{0.17em}}$ the x -values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ function to write a formula for $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ by evaluating $\text{\hspace{0.17em}}f\left(x-2\right).$
$\begin{array}{ccc}\hfill f\left(x\right)& =& {x}^{2}\hfill \\ \hfill g\left(x\right)& =& f\left(x-2\right)\hfill \\ \hfill g\left(x\right)& =& f\left(x-2\right)={\left(x-2\right)}^{2}\hfill \end{array}$
## Interpreting horizontal versus vertical shifts
The function $\text{\hspace{0.17em}}G\left(m\right)\text{\hspace{0.17em}}$ gives the number of gallons of gas required to drive $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ miles. Interpret $\text{\hspace{0.17em}}G\left(m\right)+10\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}G\left(m+10\right).$
$G\left(m\right)+10\text{\hspace{0.17em}}$ can be interpreted as adding 10 to the output, gallons. This is the gas required to drive $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ miles, plus another 10 gallons of gas. The graph would indicate a vertical shift.
$G\left(m+10\right)\text{\hspace{0.17em}}$ can be interpreted as adding 10 to the input, miles. So this is the number of gallons of gas required to drive 10 miles more than $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ miles. The graph would indicate a horizontal shift.
Given the function $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{x},\text{\hspace{0.17em}}$ graph the original function $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ and the transformation $\text{\hspace{0.17em}}g\left(x\right)=f\left(x+2\right)\text{\hspace{0.17em}}$ on the same axes. Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units?
The graphs of $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.
## Combining vertical and horizontal shifts
Now that we have two transformations, we can combine them. Vertical shifts are outside changes that affect the output ( y -) values and shift the function up or down. Horizontal shifts are inside changes that affect the input ( x -) values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and left or right.
#### Questions & Answers
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael |
# A class has 50 women and 82 men. If a student is picked at random, what's the probability of picking a woman?
$\frac{50}{132} = \frac{25}{66}$
#### Explanation:
There is a class of 50 women and 82 men, so 132 students all told. If we pick one person at random, what is the probability that person will be a woman?
Probability is a ratio, with the numerator being the number of ways we can satisfy a condition (or set of conditions) and the denominator being the number of ways we can get results.
In this case, there are 50 picks that will satisfy the condition of the pick being a woman and 132 total picks, so we have:
$\frac{50}{132}$
which can be reduced down:
$\frac{50}{132} = \frac{25}{66}$ |
# Factor by using a quadratic pattern
Learn how to factor a higher-degree polynomial by using a quadratic pattern with these carefully chosen examples.
Example #1
Factor x4 + 7x2 + 6 by using a quadratic pattern
Step 1
Write x4 + 7x2 + 6 in the pattern of a quadratic expression so you can factor it like one by making a temporary substitution of variables.
Let y = x2 and substitute y for x2
x4 + 7x2 + 6 = (x2)2 + 7(x2) + 6
x4 + 7x2 + 6 = (y)2 + 7(y) + 6
Step 2
Factor y2 + 7y + 6
y2 + 7y + 6 = (y + ___ )(y + ___ )
To fill in the blank above, look for factors of 6 that will add up to 7.
6 × 1 = 6 and 6 + 1 = 7.
Fill in the blank in the expression above with 1 and 6.
y2 + 7y + 6 = (y + 1 )(y + 6)
Step 3
Substitute back to the original variable
(y + 1 )(y + 6) = (x2 + 1)(x2 + 6)
Example #2
Factor x4 - 4x2 - 45 by using a quadratic pattern
Step 1
Write x4 - 4x2 - 45 in the pattern of a quadratic expression so you can factor it like one by making a temporary substitution of variables.
Let y = x2 and substitute y for x2
x4 - 4x2 - 45 = (x2)2 - 4(x2) - 45
x4 - 4x2 - 45 = (y)2 - 4(y) - 45
Step 2
Factor y2 - 4y - 45
y2 - 4y - 45 = (y + ___ )(y + ___ )
To fill in the blank above, look for factors of -45 that will add up to -4.
-9 × 5 = 45 and -9 + 5 = -4.
Fill in the blank in the expression above with -9 and 5.
y2 - 4y - 45 = (y - 9)(y + 5)
Step 3
Substitute back to the original variable
(y - 9)(y + 5) = (x2 - 9)(x2 + 5)
Factor completely
(y - 9)(y + 5) = (x2 - 9)(x2 + 5) = (x - 3)(x + 3)(x2 + 5)
## Recent Articles
Jan 26, 23 11:44 AM
Learn about the quadratic formula, the discriminant, important definitions related to the formula, and applications.
2. ### Area Formula - List of Important Formulas
Jan 25, 23 05:54 AM
What is the area formula for a two-dimensional figure? Here is a list of the ones that you must know! |
# Midpoint solver
Math can be a challenging subject for many students. But there is help available in the form of Midpoint solver. Keep reading to learn more!
## The Best Midpoint solver
In this blog post, we discuss how Midpoint solver can help students learn Algebra. As a log book keeper, you will be tracking logs from the beginning of your business to the end. You need to be able to keep track of where these logs are and what is going on with them. If you don’t have a good log book keeping system, this can be very difficult. The best way to solve this problem is to use a good log book or diary system. This will allow you to organize your logs and keep track of what is happening with them at all times. This can make it much easier to solve problems, as you can keep track of all of your logs in one place. With this information, you can be able to see exactly what is happening with your logs at any given time. This will help you to solve any issues quickly, as you will know exactly what is going on when there are any problems.
When calculating a circle’s radius, you need to take into account both the radius of the circle’s circumference and the radius of its diameter. You can use this formula to solve for either or both: With these formulas, all you have to do is find the radius of each side in relation to the other one. You should also remember that the radius increases as your circle gets larger. If a circle has a radius of 1 unit, then its radius will double (or triple) as it grows from 1 unit in size. Once you know how much bigger a circle is than another one, you can calculate its diameter. Divide the first circle’s circumference by the second one’s diameter and multiply by pi to get the answer.
The system of equations is the mathematical representation of a set of related equations. It is an ordered list of equations with and without solutions. The solution of a system of equations is the set of values that satisfies all the given equations. To solve system of equations, first we need to identify all the variables involved in the given system. Then we need to add all unknowns and solve for them individually. Once all unknowns are known, we can add all knowns and solve for them individually. This way, we get a single solution from a set of individual solutions. We use algebra to find a solution or to solve a system of linear equations or inequalities. Algebra is used to simplify, manipulate and evaluate expressions and questions involving variables. Algebra is also used for solving more complicated problems such as quadratic equations, polynomial equations, rational expressions, exponential expressions etc. Algebra can be used to solve systems with several variables or when there are different types of questions (such as multiple choice, fill-in-the-blank). There are various methods one can use to solve system of linear equations like substitution method, elimination method and combination method etc. In this article, we will discuss several approaches on solving systems of linear equation i.e substitution method etc.
It should be easy for you to learn and use. It should also provide you with lots of practice problems and other resources so that you can start building the kind of skills that will make it easier for you to learn algebra. You can find a good study program by doing some research and talking to friends who have used it before. Good luck!
Math problems may be difficult for some, but there is an easier way to do them. You can use a camera to help you solve math problems. You can take lots of pictures of the problem and then look at the pictures together with your child. You can also record the sounds of numbers being tapped out on a table or keyboard. When you have a recording of the problem sounds, you can play it back and ask your child to describe the problem in detail. This will help your child understand what they are doing more clearly and it may also help them solve the problem quicker than they otherwise would. When taking photos, make sure that you pose the subject correctly so that the camera focuses correctly. Also make sure that the background is not blurry or distracting. One of the best ways to teach kids math is through play.
Amazing I found that when I used the app for the first time it showed me how to use the app and I'd recommend to anyone struggling on a problem in math So amazing and so beneficial. This app as helped me a lot. However, the questions are, it will solve it for u. |
# Understanding the t-Test in Linear Regression
Linear regression is used to quantify the relationship between a predictor variable and a response variable.
Whenever we perform linear regression, we want to know if there is a statistically significant relationship between the predictor variable and the response variable.
We test for significance by performing a t-test for the regression slope. We use the following null and alternative hypothesis for this t-test:
• H0: β1 = 0 (the slope is equal to zero)
• HA: β1 ≠ 0 (the slope is not equal to zero)
We then calculate the test statistic as follows:
t = b / SEb
where:
• b: coefficient estimate
• SEb: standard error of the coefficient estimate
If the p-value that corresponds to t is less than some threshold (e.g. α = .05) then we reject the null hypothesis and conclude that there is a statistically significant relationship between the predictor variable and the response variable.
The following example shows how to perform a t-test for a linear regression model in practice.
### Example: Performing a t-Test for Linear Regression
Suppose a professor wants to analyze the relationship between hours studied and exam score received for 40 of his students.
He performs simple linear regression using hours studied as the predictor variable and exam score received as the response variable.
The following table shows the results of the regression model:
To determine if hours studied has a statistically significant relationship with final exam score, we can perform a t-test.
We use the following null and alternative hypothesis for this t-test:
• H0: β1 = 0 (the slope for hours studied is equal to zero)
• HA: β1 ≠ 0 (the slope for hours studied is not equal to zero)
We then calculate the test statistic as follows:
• t = b / SEb
• t = 1.117 / 1.025
• t = 1.089
The p-value that corresponds to t = 1.089 with df = n-2 = 40 – 2 = 38 is 0.283.
Note that we can also use the T Score to P Value Calculator to calculate this p-value:
Since this p-value is not less than .05, we fail to reject the null hypothesis.
This means that hours studied does not have a statistically significant relationship between final exam score. |
Question
# A patient in a hospital is given soup daily in a cylindrical bowl of diameter $7cm$. If the bowl is filled with soup to a height of $4cm$. How much soup hospital has to prepare daily to serve $250$ patients?
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Solution
## Step 1: Calculate the volume of one soap:The diameter of the bowl is $D=7cm$.The radius of the bowl is $r=\frac{D}{2}$.$⇒r=\frac{7}{2}cm$.The height of the soap is $h=4cm$.The volume of a cylinder can be given by, $\mathbf{V}\mathbf{=}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{h}$.Thus the volume of the cylindrical soap is $V=\frac{22}{7}{\left(\frac{7}{2}\right)}^{2}\left(4\right)c{m}^{3}\mathbf{\left[}\mathbf{\pi }\mathbf{=}\frac{\mathbf{22}}{\mathbf{7}}\mathbf{\right]}$$⇒V=\left(\frac{22}{7}×\frac{49}{4}×4\right)c{m}^{3}\phantom{\rule{0ex}{0ex}}⇒V=22×7c{m}^{3}\phantom{\rule{0ex}{0ex}}⇒V=154c{m}^{3}$Therefore, the volume of one soap is $154c{m}^{3}$.Step 2: Calculate the total volume of the soup required dailyThe volume of soup for one patient is $154c{m}^{3}$.The total number of patients is $250$.The total volume of soup required daily is $\left(154×250\right)c{m}^{3}=38500c{m}^{3}$.Hence, the total volume of soup required daily is $38500c{m}^{3}$.
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# According to the Chronicles of Contrived Statistics (March, 2016), the
According to the Chronicles of Contrived Statistics (March, 2016), the probability that Elon Musk will offer you a free trip into space in the next month is 49%. The probability that scientists from Area 51 will start selling pet aliens next month is 46%. Finally, the probability that either Elon Musk will offer you a free trip to space or that scientists will begin sales of pet aliens next month is 31%. What is the probability that both Elon Musk will offer you a free trip into space and scientists from Area 51 will start selling pet aliens in the next month?
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Step 1
The probability that EM will offer you a free trip into space in the next month is 49%. Let this be P(A) where A is the event.
The probability that scientists from Area 51 will start selling pet aliens next month is 46%. Let this be P(B) where B is the event.
The probability that either EM will offer you a free trip to space or that scientists will begin sales of pet aliens next month is 31%. This is equal to the probability $$\displaystyle{P}{\left({A}\cap{B}\right)}$$
The probability that both EM will offer you a free trip into space and scientists from Area 51 will start selling pet aliens in the next month=? This is equivalent to $$\displaystyle{P}{\left({A}\cup{B}\right)}$$
Step 2
Now, by the formula,
$$\displaystyle{P}{\left({A}\cup{B}\right)}={P}{\left({A}\right)}+{P}{\left({B}\right)}-{P}{\left({A}\cap{B}\right)}$$
$$\displaystyle={0.49}+{0.46}-{0.31}$$
$$\displaystyle={0.64}$$
The answer is 64%. |
# Arranging Numbers
Arranging numbers in ascending order and descending order.
We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order.
Suppose for example, 81, 97, 123, 137 and 201 are arranged in ascending order.
Vice-versa while arranging numbers from the largest number to the smallest number then the numbers are arranged in descending order.
Suppose for example, 187, 121, 117, 103 and 99 are arranged in descending order.
Examples on arranging numbers in ascending order:
1. Write the following numbers is ascending order:
42734; 5358; 42876; 52287.
Solution:
Count the digits in each number.
5358 Is the smallest number as it has only 4 digits.
Line up the number accordingly to place value.
Begin comparing from the left.
5358 ← smallest number
42734
42876
7 < 8
52287 ← Largest number
The ascending order is 5358; 42734; 42876; 52287
2. Arranging numbers in ascending order:
3679; 3542; 3797; 3545
Solution:
The digit in the hundreds place in each number is 3.
On comparing the hundreds place; 3679; 3542; 3797; 3545
We find: 3797 to be the greatest and 3679 to be smaller.
On comparing the tens place in the two remaining numbers we find both
numbers to be the same. 3542; 3545
On comparing the ones place, we find 3545 > 3542
So, the ascending order is 3542 < 3545 < 3679 < 3797
Example on arranging numbers in descending order:
Write in descending order:
32593; 60537; 28524; 57198
Solution:
Compare digits according to place value.
Descending order means arranging numbers from the largest number to the smallest number;
60537 > 57198 > 32593 > 28524
Related Concept
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## Documents
TRANSCRIPT
CONFIDENTIAL 1
by Factoringby Factoring
CONFIDENTIAL 2
Warm UpWarm Up
Solve each equation by graphing the related function.
1) x2 - 49 = 0
2) x2 = x + 12
3) - x2 + 8x = 15
CONFIDENTIAL 3
You have solved quadratic equations bygraphing. Another method used to solvequadratic equations is to factor and use
the Zero Product Property.
Zero Product PropertyZero Product Property
Notice that when writing a quadratic function as its related quadratic equation,
you replace y with 0. So y = 0.
y = ax2 + bx + c
0 = ax2 + bx + c
ax2 + bx + c = 0
CONFIDENTIAL 4
One way to solve a quadratic equation in standard form is to graph the related function and find the x-values where y = 0. In other words, find the zeros of the related function. Recall that a quadratic function
may have two, one, or no zeros.
Using the Zero Product PropertyUsing the Zero Product Property
WORDS NUMBERS ALGEBRA
If the product of two quantities equals zero,
at least one of thequantities equals zero.
3 (0) = 0
0(4) = 0
If ab = 0,
then a = 0 or b = 0.
CONFIDENTIAL 5
Solving Quadratic Equations by GraphingSolving Quadratic Equations by GraphingUse the Zero Product Property to solve each
A) (x - 3)(x + 7) = 0
x - 3 = 0 or x + 7 = 0
x= 3 or x = -7
Use the Zero Product Property.
Solve each equation.
Check
(x - 3)(x + 7) = 0
(3 - 3)(3 + 7) 0
(0)(10) 0
0 0
(-7 - 3)(x + 7) = 0
(-7 - 3)(-7 + 7) 0
(10)(0) 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are 3 and -7.
CONFIDENTIAL 6
B) (x)(x - 5) = 0
x = 0 or x - 5 = 0
x= 5
Use the Zero Product Property.
Solve each equation.
Check
(x)(x - 5) = 0
(0)(0 - 5) 0
(0)(-5) 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are 0 and 5.
(x)(x - 5) = 0
(5)(5 - 5) 0
(5)(0) 0
0 0
CONFIDENTIAL 7
Now you try!
Use the Zero Product Property to solve each equation. Check your answer.
1a. (x)(x + 4) = 0
1b. (x + 4)(x - 3) = 0
CONFIDENTIAL 8
If a quadratic equation is written in standard form, a x 2 + bx + c = 0,
then to solve the equation, you may need to factor before using the
Zero Product Property.
CONFIDENTIAL 9
A) x2 + 7x + 10 = 0
(x + 5) (x + 2) = 0
x + 5 = 0 or x + 2 = 0 Use the Zero Product Property.
Solve each equation.
Check
x2 + 7x + 10 = 0
(-5)2 + 7(-5) + 10 0
25 - 35 + 10 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are -5 and -2.
x = -5 or x = -2
Factor the trinomial.
x2 + 7x + 10 = 0
(-2)2 + 7(-2) + 10 0
4 - 14 + 10 0
0 0
CONFIDENTIAL 10
B) x2 + 2x = 8
-8 -8
Use the Zero Product Property.
Solve each equation.
The solutions are -4 and 2.
x = -4 or x = 2
Factor the trinomial.
x2 + 2x = 8
x2 + 2x – 8 = 0
The equation must be written in standard form. So subtract 8 from both sides.
(x + 4) (x - 2) = 0
x + 4 = 0 or x - 2 = 0
CONFIDENTIAL 11
Check: Graph the related quadratic function. The zeros of the related function should be the
same as the solutions from factoring.
The graph of y = x2 + 2x - 8 shows two zeros appear to be -
4 and 2, the same as the solutions from factoring.
CONFIDENTIAL 12
C) x2 + 2x + 1 = 0
Use the Zero Product Property.
Solve each equation.
Both factors result in the same solution, so there is one solution, -1.
x = -1 or x = -1
Factor the trinomial.(x + 1) (x + 1) = 0
x + 1 = 0 or x + 1 = 0
CONFIDENTIAL 13
Check: Graph the related quadratic function. The zeros of the related function should be the
same as the solutions from factoring.
The graph of y = x2 + 2x + 1 shows that one zero appears
to be -1, the same as the solution from factoring.
CONFIDENTIAL 14
D) -2x2 = 18 - 12x
-2 (x - 3) (x - 3) = 0
-2 ≠ 0 or x - 3 = 0 Use the Zero Product Property.
Solve each equation.
Check
The only solution is 3.
x = 3
Factor the trinomial.
-2x2 = 18 - 12x
-2(3)2 18 - 12(3)
-18 18 - 36
0 0
Write the equation in standard form.-2x2 + 12x – 18 = 0
-2( x2 - 6x + 9) = 0 Factor out the GCF, -2.
Substitute 3 into the original equation.
CONFIDENTIAL 15
Now you try!
2a. x2 - 6x + 9 = 0
2b. x2 + 4x = 5
CONFIDENTIAL 16
Sports ApplicationSports ApplicationThe height of a diver above the water during a dive can be modeled by h = -16t2 + 8t + 48, where h is
height in feet and t is time in seconds. Find the time it takes for the diver to reach the water.
h = -16t2 + 8t + 48
0 = -16t2 + 8t + 48
Use the Zero Product Property.
Solve each equation.
Factor the trinomial.
The diver reaches the water when h = 0.
Factor out the GCF, -8.0 = -8(2t2 - t - 6)
0 = -8(2t + 3) (t -2)
-8 ≠ 0, 2t + 3 = 0 or t - 2 = 0
2t = -3 or t = 2
t = -3 2
Since time cannot be negative, (-3/2 ) does not make sense in this situation.
CONFIDENTIAL 17
It takes the diver 2 seconds to reach the water.
Check
0 = -16 t 2 + 8t + 48
0 -16(2)2 + 8(2) + 48
0 -64 + 16 + 48
0 0
Substitute 3 into the original equation.
CONFIDENTIAL 18
Now you try!
3.) The equation for the height above the water for another diver can be modeled by h = -16t2 + 8t + 24. Find the time it takes this diver
to reach the water.
CONFIDENTIAL 19
BREAK
CONFIDENTIAL 21
Assessment
1) (x + 2) (x - 8) = 0
Use the Zero Product Property to solve each equation. Check your answer.
2) (x - 6) (x - 5) = 0
3) (x + 7) (x + 9) = 0
4) (x) (x - 1) = 0
CONFIDENTIAL 22
6) 3x2 - 4x + 1 = 0
5) 30x = -9x2 - 25
8) x2 - 8x - 9 = 0
7) x2 + 4x - 12 = 0
CONFIDENTIAL 23
9) A group of friends tries to keep a beanbag from touching the ground without using their hands. Once the beanbag has been kicked, its height can be modeled by h = -16t2 + 14t + 2, where h is the height in feet above the ground and t is the time in seconds. Find the time it
takes the beanbag to reach the ground.
CONFIDENTIAL 24
You have solved quadratic equations bygraphing. Another method used to solvequadratic equations is to factor and use
the Zero Product Property.
Zero Product PropertyZero Product Property
Notice that when writing a quadratic function as its related quadratic equation,
you replace y with 0. So y = 0.
y = ax2 + bx + c
0 = ax2 + bx + c
ax2 + bx + c = 0
Let’s review
CONFIDENTIAL 25
One way to solve a quadratic equation in standard form is to graph the related function and find the x-values where y = 0. In other words, find the zeros of the related function. Recall that a quadratic function
may have two, one, or no zeros.
Using the Zero Product PropertyUsing the Zero Product Property
WORDS NUMBERS ALGEBRA
If the product of two quantities equals zero,
at least one of thequantities equals zero.
3 (0) = 0
0(4) = 0
If ab = 0,
then a = 0 or b = 0.
CONFIDENTIAL 26
Solving Quadratic Equations by GraphingSolving Quadratic Equations by GraphingUse the Zero Product Property to solve each
A) (x - 3)(x + 7) = 0
x - 3 = 0 or x + 7 = 0
x= 3 or x = -7
Use the Zero Product Property.
Solve each equation.
Check
(x - 3)(x + 7) = 0
(3 - 3)(3 + 7) 0
(0)(10) 0
0 0
(-7 - 3)(x + 7) = 0
(-7 - 3)(-7 + 7) 0
(10)(0) 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are 3 and -7.
CONFIDENTIAL 27
A) x2 + 7x + 10 = 0
(x + 5) (x + 2) = 0
x + 5 = 0 or x + 2 = 0 Use the Zero Product Property.
Solve each equation.
Check
x2 + 7x + 10 = 0
(-5)2 + 7(-5) + 10 0
25 - 35 + 10 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are -5 and -2.
x = -5 or x = -2
Factor the trinomial.
x2 + 7x + 10 = 0
(-2)2 + 7(-2) + 10 0
4 - 14 + 10 0
0 0
CONFIDENTIAL 28
B) x2 + 2x = 8
-8 -8
Use the Zero Product Property.
Solve each equation.
The solutions are -4 and 2.
x = -4 or x = 2
Factor the trinomial.
x2 + 2x = 8
x2 + 2x – 8 = 0
The equation must be written in standard form. So subtract 8 from both sides.
(x + 4) (x - 2) = 0
x + 4 = 0 or x - 2 = 0
CONFIDENTIAL 29
Check: Graph the related quadratic function. The zeros of the related function should be the
same as the solutions from factoring.
The graph of y = x2 + 2x - 8 shows two zeros appear to be -
4 and 2, the same as the solutions from factoring.
CONFIDENTIAL 30
Sports ApplicationSports ApplicationThe height of a diver above the water during a dive can be modeled by h = -16t2 + 8t + 48, where h is
height in feet and t is time in seconds. Find the time it takes for the diver to reach the water.
h = -16t2 + 8t + 48
0 = -16t2 + 8t + 48
Use the Zero Product Property.
Solve each equation.
Factor the trinomial.
The diver reaches the water when h = 0.
Factor out the GCF, -8.0 = -8(2t2 - t - 6)
0 = -8(2t + 3) (t -2)
-8 ≠ 0, 2t + 3 = 0 or t - 2 = 0
2t = -3 or t = 2
t = -3 2
Since time cannot be negative, (-3/2 ) does not make sense in this situation.
CONFIDENTIAL 31
It takes the diver 2 seconds to reach the water.
Check
0 = -16 t 2 + 8t + 48
0 -16(2)2 + 8(2) + 48
0 -64 + 16 + 48
0 0
Substitute 3 into the original equation.
CONFIDENTIAL 32
You did a great job You did a great job today!today! |
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NCERT Solutions for Class 7 Maths, Chapter 1 - Integers
The NCERT Class 7 Syllabus is related to class 6 and class 8 to a great extent. This is because Class 7 mathematics syllabus is the continuation of class 6 syllabus and it, in turn, becomes the basis for class 8 learning. The NCERT Class 7 Mathematics Chapter 1 – Integers, is an extension of the Integers chapter that you have already studied in class 6. It acts as a revision of the topics learnt and helps the students understand the concept of Integers better. In this chapter, one will learn more about Integers, their properties and operations. The chapter deals with all the basic mathematical operations of Integers and the rules applicable while performing such operations. To gain more insight on Class 7 Maths Chapter 1, let’s look at the topics and sub-topics of this chapter:
1.2 Recall
1.3 Properties of Addition and Subtraction of Integers
1.4 Multiplication of Integers
1.5 Properties of Multiplication of Integers
1.6 Division of Integers
1.7 Properties of Division of Integers
In Class 7 Maths Chapter 1, you will learn -
Section 1.2 - Representing integers on a number line helps to learn the other operation of Integers. The integers are placed on the number line based on its positive or negative signs and the operations are done to find the solution. This is the simplest way of solving problems related to integers.
Section 1.3 - The different properties of addition and subtraction of integers need to be understood for doing the operations. There are properties like Closure Property for Addition and Subtraction, Commutative Property, Associative Property, Additive Identity of Integers. These properties form the basis of solving problems on integers.
Section 1.4 - This section deals with the multiplication of Integers and the rules associated with it. The multiplication of a Positive Number and a Negative Number, the solution can be found using the number line. Multiplication of two negative integers result in a Positive Integer; such properties need to be studied to understand the multiplication of Integers well. It also highlights the discoveries of the Swiss mathematician Leonhard Euler, who, in his book, was the first to attempt to prove (-1) x (-1)= +1.
Class 7 Integers chapter gives one a great exposure to the properties of integers. It also gives one a clear understanding of all the rules and different steps to be followed to derive the right solution for any problem related to integers. It gives one a practical exposure by a wide range of exercises, which help one to logically apply all the properties and find solutions to problems related to integers.
Section 1.5 - Just like the properties of addition, there are properties of multiplication of integers. In this one learns, the different properties of multiplication like Closure Property, Commutative Property, Multiplicity by Zero, Multiplicative Identity, Associative Property, Distributive Property. It consists of exercises to make multiplication easy.
Section 1.6 - This section deals with the division of integers. It gives a detailed explanation regarding the steps to be followed for the division of integers.
Section 1.7 - This section provides a deeper understanding of properties of the division of integers. The rules are applicable when dividing an integer by 1 or 0. These properties make it easier to solve any problem related to the division of integers.
To know more about the NCERT solutions Class 7 Maths, Chapter 1 – Integers, explore the exercises below:
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Question
Mon July 04, 2011 By: Sanya Wason
# determine algebrically the vertices of the triangle formed by the lines 3x-y=3,2x-3y=2,x+2y=8
Mon July 04, 2011
3x-y=3 -----(i)
2x-3y=2 -----(ii)
x+2y=8 -----(iii)
The vertices will be the points of intersection of the three equations, taking two at a time. Corresponding to the three pairs of equations, we get the three vertices.
Consider (i) and (ii) :
Multiplying (i) by (-3) and adding to (ii), we get
- 9x + 3y + 2x - 3y = -9 + 2
=> x = 1
(i) => y = 3x - 3 = 3.1 - 3 = 0
1st vertex is (1, 0)
Consider (ii) and (iii) :
Multiplying (iii) by (-2) and adding to (ii), we get,
2x - 3y - 2x - 4y = 2 - 16
=> y = 2
(iii) => x = 8 - 2y = 8 - 2.2 = 4
2nd vertex is ( 4, 2)
Consider (i) and (iii) :
Multiplying (i) by 2 and adding to (iii), we get
6x - 2y + x + 2y = 6 + 8
=> x = 2
(i) => y = 3x - 3 = 3.2 - 3 = 3
3rd vertex is ( 2,3)
Hence the three vertices are (1,0) , (4,2) & ( 2,3)
Related Questions
Tue September 13, 2016
# Determine algebraically the vertices of the triangle formed by the lines represented by the equation: 5x-y=5 x+2y=1 6x+y=17
Sat August 13, 2016
|
# Week 5 Dq 1math 117
Topics: Integer, Addition, Real number Pages: 1 (320 words) Published: August 27, 2010
Review section 10.2 (p. 692) of your text. Describe two laws of exponents and provide an example illustrating each law. Explain how to simplify your expression. How do the laws work with rational exponents? Provide the class with a third expression to simplify that includes rational (fractional) exponents.
The two laws of exponents are
For any real number a and any rational exponents m and n:
1. In multiplying, we can add exponents if the bases are the same. 2. In dividing, we can subtract exponents if the bases are the same. 3. To raise a power to a power, we can multiply the exponents. 4. To raise a product to a power, we can raise each factor to the power. 5. To raise a quotient to a power, we can raise both the numerator and the denominator to the power.
1. Convert radical expressions to exponential expressions.
2. Use arithmetic and the laws of exponents to simplify.
3. Convert back to radical notation when appropriate.
Important: This procedure works only when all expressions under radicals are nonnegative since rational exponents are not defined otherwise. With this assumption, no absolute-value signs will be needed.
Here are some example of it. EXAMPLES Use the laws of exponents to simplify. 16. 3 1/5 _ 33/5 _ 31/5_3/5 _ 34/5 Adding exponents
17. 71/4 / 71/2 _ 71/4_1/2 _ 71/4_2/4 _ 7_1/4 _1 71/4 Subtracting exponents 18. _7.22/3_3/4 _ 7.22/3 _ 3/4 _ 7.26/12 _ 7.21/2 Multiplying exponents 19. _ a_1/6b1/5 _b1/5a1/6 _a_1/3b2/5_1/2 _ a_1/3 _ 1/2 _ b2/5 _ ½ Raising a product to a power and multiplying exponents The rational exponents can be used to simplify some radical expressions. Example: _3 5 _ _2 |
# How do you write the standard form of the equation of the circle with the given the center (7,-3); tangent to the x-axis?
Feb 2, 2016
${\left(x - 7\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {3}^{2}$ (explicit standard circle format)
or ${\left(x - 7\right)}^{2} + {\left(y + 3\right)}^{2} = 9$ (implicit standard circle format)
or ${x}^{2} + {y}^{2} - 14 x + 6 y + 49$ (standard polynomial format for an equation that happens to be a circle).
#### Explanation:
If the circle has a center at $\left(7 , - 3\right)$ and is tangent to the X-axis
then it has a radius of $r = 3$
The explicit standard circle equation for a circle with center at $\left({x}_{c} , {y}_{c}\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {x}_{c}\right)}^{2} + {\left(x - {y}_{c}\right)}^{2} = {r}^{2}$
which given the first version above ${\left(x - 7\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {3}^{3}$
Some (but not all) teachers like to see this expanded to get rid of the double minus signs and express the squared radius as a single entity:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 7\right)}^{2} + {\left(y + 3\right)}^{3} = 9$
and still others are looking for the standard form of a general polynomial which is completely expanded with the right side as a series of terms in decreasing degree and the left side $= 0$
(The third form above). |
Name: ___________________Date:___________________
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### MEAP Preparation - Grade 7 Mathematics2.35 Linear Equations - 2
Examples: 2x + 8 = 14 Then x = ? 2x + 8 = 14 subtracting 8 from both sides 2x + 8 - 8 = 14 - 8 2x = 6 Dividing both sides by 2 2x/2 = 6/2 x = 3 6y - 15 = 2y + 9 Then y = ? 6y - 15 = 2y + 9 subtracting 2y from both sides 6y - 15 - 2y = 2y + 9 - 2y 4y - 15 = 9 Adding 15 both sides 4y - 15 + 15 = 9 + 15 4y = 24 dividing both sides by 4 4y/4 = 24/4 y = 6 4(s + 1) = -2(4 - s) distributive property Note: Negative number multiplied by a negative number is positive. -2 x -s = 2s 4s + 4 = -8 + 2s subtracting 2s both sides 4s + 4 - 2s= -8 + 2s - 2s 2s + 4 = -8 subtracting 4 both sides 2s + 4 - 4 = -8 - 4 2s = - 12 dividing both sides by 2 2s/2 = - 12/2 s = -6 2/x = 3/6 x/2 = 6/3 x/2 = 2 x = 4 t/6 + 2 = t/3 This equation contains fractions so multiply both sides by the LCD of 6 and 3 that is 6 6(t/6 + 2) = 6(t/3) t + 12 = 2t subtracting t from both sides t + 12 - t = 2t - t 12 = t Therefore, t = 12 Directions: Solve for the variables. Also write at least 10 examples of your own.
Name: ___________________Date:___________________
### MEAP Preparation - Grade 7 Mathematics2.35 Linear Equations - 2
Q 1: 5(y-1) - 2(y+6) = 2y + 12 then y = ?292527 Q 2: 2(x+5) - 3(x-6) = 201078 Q 3: 5x/6 = 50 then x = ?705060 Q 4: (-x/2) = -10212220 Q 5: (-x/3) = 24 then x = ?-21-54-72 Q 6: 1.5x = 30 then x = ?202224 Q 7: (-2x) = 20 then x = ?1020-10 Q 8: 2x/0.2 = - 0.8 then x=?-0.080.08-0.8 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Proofs and Use of the Pythagorean Theorem
## Identify triples and calculate missing sides
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Practice Proofs and Use of the Pythagorean Theorem
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Derive and Use the Pythagorean Theorem
Have you ever painted something taller than you are? Take a look at this dilemma.
While Mr. Kennedy’s class was working on the garden, the students in Ms. Richardson’s class decided to paint the equipment shed where all of the sports equipment was kept.
“That thing hasn’t been painted in decades,” Karen said in the first meeting.
“I agree, it does look awful,” Cameron added.
“Well, I don’t know about decades, but it does need to be painted, so that is what we are going to do. Now to work on this project, we will need to choose a ladder that will reach up high enough. How can we figure this out?” Ms. Richardson asked.
The class was silent.
“I have an idea,” Veria said smiling. “It has to do with triangles. We need to figure out the height of the ladder, compared to the height of the building.”
“Yes, but don’t forget that the ladder reaches out from the building, not up against it. So we have to consider that measurement too,” Aran chimed in.
“What can we use to figure that out?” Karen asked.
Once again, the group was silent.
A mathematical formula involving the Pythagorean theorem is needed to solve this problem. The students have their work cut out for them. So do you. Pay attention to this Concept so that you can explain the formula that they will need and why they will need it.
### Guidance
You have probably already studied many different types of triangles.
Acute triangles have angles that are all less than 90\begin{align*}90^{\circ}\end{align*}.
Obtuse triangles have one angle that is between 90\begin{align*}90^{\circ}\end{align*} and 180\begin{align*}180^{\circ}\end{align*}.
Right triangles have one angle that measures exactly 90\begin{align*}90^{\circ}\end{align*}—in other words, it has one right angle.
This Concept focuses entirely on properties specific to right triangles. While all of the equations and strategies you are about to learn are helpful, they apply only to right triangles – they will not work with acute or obtuse triangles.
To begin, let’s look at the parts of a right triangle.
The legs are the two sides of the triangle that are labeled a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}. The hypotenuse is the longest side of a right triangle and it is labeled c\begin{align*}c\end{align*}. There is a special relationship between the legs of a right triangle and the hypotenuse of a right triangle.
One of the special characteristics of right triangles is described by the Pythagorean Theorem, thought to have been developed around 500 B.C.E. It states that the squared value of the hypotenuse will equal the sum of the squares of the two legs. In the triangle above, the sum of the squares of the legs is a2+b2\begin{align*}a^2 + b^2\end{align*} and the square of the hypotenuse is c2\begin{align*}c^2\end{align*}. So, the Pythagorean theorem is commonly represented as a2+b2=c2\begin{align*}a^2 + b^2 = c^2\end{align*} where a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are the legs of the right triangle and c\begin{align*}c\end{align*} is the hypotenuse.
The Pythagorean Theorem is known as a2+b2=c2\begin{align*}a^2 + b^2 = c^2\end{align*}.
You may be asking yourself why that is the case. Well, we can think about the Pythagorean Theorem in terms of a square. We know that there is a relationship between a square and a right triangle. We can divide a square with a diagonal and because a square has four right angles, the diagonal will divide the square into two right triangles. Now because a right triangle comes from the square, the sides will also be related to the square. This is where the Pythagorean Theorem comes from.
Let’s look at one.
Use the measures of the triangle below to test the Pythagorean theorem.
The legs of the triangle above are 3 inches and 4 inches. The hypotenuse is 5 inches. So, a=3\begin{align*}a = 3\end{align*}, b=4\begin{align*}b = 4\end{align*}, and c=5\begin{align*}c = 5\end{align*}. We can test the formula to see if this is true.
a2+b232+42(3×3)+(4×4)9+1625=c2=52=(5×5)=25=25
Since both sides of the equation equal 25, the equation is true. Therefore, the Pythagorean theorem worked on this right triangle.
This combination of numbers (3, 4, 5) is referred to as a Pythagorean triple. In other words, these three numbers work together to make the Pythagorean Theorem true.
Now that you have learned how to derive and execute the Pythagorean Theorem, there are many different ways to apply it. Any time you have two out of three sides in a right triangle, you can find the third using the equation a2+b2=c2\begin{align*}a^2 + b^2 = c^2\end{align*}, where a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are the lengths of the legs of the triangle and c\begin{align*}c\end{align*} is the length of the hypotenuse.
When applying the Pythagorean Theorem, be sure to use exponents and square roots accurately.
What is the length of b\begin{align*}b\end{align*} in the triangle below?
Use the Pythagorean Theorem to identify the length of the missing leg, b\begin{align*}b\end{align*}. Be sure to simplify the exponents and roots carefully. Also remember to use inverse operations to solve the equation properly.
a2+b2=c2,\begin{align*}a^2 + b^2 = c^2,\end{align*} where a=6\begin{align*}a = 6\end{align*} and c=10\begin{align*}c = 10\end{align*}
62+b236+b236+b236b2b2b=102=100=10036=64=64=8
The length of the missing side is 8 inches.
You already know about the Pythagorean triple 3:4:5. Notice that this triangle is proportional to that ratio. If you divide the lengths of the triangle in the example by two, you find the same proportion—3:4:5. Whenever you find a Pythagorean triple, you can apply those ratios with greater factors as well. So, 6, 8, 10 is another Pythagorean triple.
Notice that as long as you use the Pythagorean Theorem you can figure out the missing length of any of the three sides of a right triangle.
Find the missing side length for each right triangle.
#### Example A
9,12\begin{align*}9,12 \end{align*}
Solution: 15\begin{align*}15\end{align*}
#### Example B
15,20,\begin{align*}15,20, \end{align*}
Solution: 25\begin{align*}25\end{align*}
#### Example C
\begin{align*}21,28, \end{align*}
Solution: \begin{align*}35\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
The students will need to use the Pythagorean Theorem to figure out this problem.
Why? They will need to use the Pythagorean Theorem because the ladder against the shed forms a right triangle with the ground. The shed and the distance that the ladder is placed from the shed form the sides \begin{align*}a\end{align*} and \begin{align*}b\end{align*} of the right triangle. The ladder is the \begin{align*}c\end{align*} side of the triangle.
Look at the diagram below.
### Vocabulary
Right Triangle
one angle is equal to \begin{align*}90^{\circ}\end{align*}.
Legs
the two shorter sides of a right triangle.
Hypotenuse
the longest side of a right triangle.
Pythagorean Theorem
\begin{align*}a^2 + b^2 = c^2\end{align*}
Pythagorean Triple
values that work perfectly in the Pythagorean Theorem. The ratio always simplifies to 3:4:5.
### Guided Practice
Here is one for you to try on your own.
Find the length of the missing side in the triangle below.
Solution
Use the Pythagorean Theorem to identify the length of the missing hypotenuse. Be sure to simplify the exponents and roots carefully. Also remember to use inverse operations to solve the equation properly.
\begin{align*}a^2 + b^2 = c^2,\end{align*} where \begin{align*}a = 5\end{align*} and \begin{align*}b = 12\end{align*}
The length of the missing side is 13 centimeters.
### Practice
Directions: Use the Pythagorean Theorem to find the missing dimensions of right triangles.
1. \begin{align*}a=3,b=4,c=?\end{align*}
2. \begin{align*}a=6,b=8,c=?\end{align*}
3. \begin{align*}a=9,b=12,c= ?\end{align*}
4. \begin{align*}a=27,b=36,c= ?\end{align*}
5. \begin{align*}a=15,b=20,c= ?\end{align*}
6. \begin{align*}a=18,b=24,c= ?\end{align*}
7. \begin{align*}a= ?,b=16,c= 20\end{align*}
8. \begin{align*}a= ?,b=28,c=35\end{align*}
9. \begin{align*}a=30,b= ?,c=50\end{align*}
10. \begin{align*}a=33,b= ?,c=55\end{align*}
11. \begin{align*}a=1.5,b= ?,c=2.5\end{align*}
12. \begin{align*}a=36,b= ?,c=60\end{align*}
Directions:Answer the following questions True or False.
13. The Pythagorean Theorem will work for any triangle.
14. The longest side of a right triangle is called the hypotenuse.
15. A Pythagorean Triple can only be found in a right triangle.
### Vocabulary Language: English
Hypotenuse
Hypotenuse
The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.
Legs of a Right Triangle
Legs of a Right Triangle
The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.
Pythagorean Theorem
Pythagorean Theorem
The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.
Pythagorean Triple
Pythagorean Triple
A Pythagorean Triple is a set of three whole numbers $a,b$ and $c$ that satisfy the Pythagorean Theorem, $a^2 + b^2 = c^2$.
Right Triangle
Right Triangle
A right triangle is a triangle with one 90 degree angle. |
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If the weight of the shark is $200kg$ and weight of one child is $15kg$, then how much more does the shark weigh than 12 children put together?(A) 20(B) 18(C) 22(D) 24
Last updated date: 17th Jul 2024
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Hint: We are given the weight of a shark and the weight of one child. We are asked in the question to find the difference in the weight of a shark and the weight of 12 children put together. So, we are given the weight of one child, which is, $15kg$ and so the weight of 12 such children will be the product of 12 and weight of one child and so we will have the total weight of the child. We will then subtract the weight of the shark and the weight of 12 children. Hence, we will have the required difference.
Complete step-by-step solution:
According to the given question, we are given the weight of the shark, which is, $200kg$ and the weight of one child, which is, $15kg$. We are then asked to find the difference between the weight of the shark and the weight of 12 children.
We are given,
Weight of one child = $15kg$
Weight of 12 children = $12\times 15$
$\Rightarrow$ Weight of 12 children = $180kg$
Now, we have the given weight of the shark = $200kg$
The difference in the weight of the shark and 12 children together, we have,
$\Rightarrow 200-180$
$\Rightarrow 20kg$
Therefore, the weight of the shark is $20kg$ more than the weight of 12 children together.
Note: The weights of the shark and of the children should not get intermixed. Also, while finding the weight of 12 children, we are assuming that all the children taken into consideration are of $15kg$ each. While subtracting the weight of children from the weight of shark, make sure that the calculation is done correctly. |
# Should You Shoot Free Throws Underhand?
A FiveThirtyEight Riddler puzzle.
mathematics
Riddler
Published
March 11, 2016
## Problem
Hark! The NCAA Tournament starts next week, and the “granny shot” has reappeared as a free throw technique. Its proponents claim that it improves accuracy because there are fewer moving parts — the elbows and wrists are held more stable, for example, and the move is symmetric because one’s arms are, more or less, equal length. Let’s find out how effective the granny shot really is.
Consider the following simplified model of free throws. Imagine the rim to be a circle (which we’ll call $$C$$) that has a radius of $$1$$, and is centered at the origin (the point $$(0,0)$$). Let $$V$$ be a random point in the plane, with coordinates $$X$$ and $$Y$$, and where $$X$$ and $$Y$$ are independent normal random variables, with means equal to zero and each having equal variance — think of this as the point where your free throw winds up, in the rim’s plane. If $$V$$ is in the circle, your shot goes in. Finally, suppose that the variance is chosen such that the probability that $$V$$ is in $$C$$ is exactly $$75$$ percent (roughly the NBA free-throw average).
But suppose you switch it up, and go granny-style, which in this universe eliminates any possible left-right error in your free throws. What’s the probability you make your shot now? (Put another way, calculate the probability that $$\|Y\| < 1$$.)
## Solution
$$R \sim \mathrm {Rayleigh} (\sigma )$$ is Rayleigh distributed if $$R={\sqrt {X^{2}+Y^{2}}}$$, where $$X \sim N(0,\sigma ^{2})$$ and $$Y\sim N(0,\sigma ^{2})$$ are independent normal random variables.
From the above, it is clear that $$V$$ follows the Rayleigh distribution.
The CDF of $$V$$ is given by $$1 - e^{-x^2/\sigma^2}$$.
The fact that the shot goes in $$75\%$$ of the time is equivalent to
$\begin{equation*} \mathbb{P}[V \leq 1] = 1 - e^{-1/2\sigma^2} = 0.75 \end{equation*}$
Therefore, $$\sigma^2 =0.36$$ and $$Y \sim N(0, 0.36)$$.
The probability that $$\|Y\|<1$$ is given by $$\mathbb{P}[-1 \leq Y \leq 1] = 0.904419$$. |
# Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expressions And Identities – Exercise 6.4 | Set 1
• Last Updated : 07 Apr, 2021
### Question 1. Find the product 2a3(3a + 5b)
Solution:
Using Distributive law,
2a3 (3a + 5b) = 2a3 × 3a + 2a3 × 5b
= 6a3+1 + 10a3b = 6a4 + 10a3b
Hence, the product is 6a4 + 10a3b
### Question 2. Find the product -11a(3a + 2b)
Solution:
Using Distributive law,
-11a (3a + 2b) = -11a × 3a + (-11a) × 2b
= -33a1+1 – 22ab = -33a2 – 22ab
Hence, the product is -33a2 – 22ab
### Question 3. Find the product -5a(7a – 2b)
Solution:
Using Distributive law,
-5a (7a – 2b) = -5a × 7a – (-5a) × 2b
= -35a1+1 + 10ab = -35a2 + 10ab
Hence, the product is -35a2 + 10ab
### Question 4. Find the product -11y2(3y + 7)
Solution:
Using Distributive law,
-11y2 (3y + 7) = -11y2 × 3y + (-11y2) × 7
= -33y2+1 – 77y2 = -33y3 – 77y2
Hence, the product is -33y3 – 77y2
### Question 5. Find the product of 6x/5(x3+y3)
Solution:
Using Distributive law,
6x/5 (x3+y3) = 6x/5 × x3 + 6x/5 × y3
= 6x3+1/5 + 6xy3/5 = 6x4/5 + 6xy3/5
Hence, the product is 6x4/5 + 6xy3/5
### Question 6. Find the product of xy(x3 – y3)
Solution:
Using Distributive law,
xy (x3-y3) = xy × x3 – xy × y3
= x3+1y – xy3+1 = x4y – xy4
Hence, the product is x4y – xy4
### Question 7. Find the product of 0.1y (0.1x5 + 0.1y)
Solution:
Using Distributive law,
0.1y (0.1x5 + 0.1y) = 0.1y × 0.1x5 + 0.1y × 0.1y
= 0.01x5y + 0.01y2
Hence, the product is 0.01x5y + 0.01y2
### Question 8. Find the product of (-7ab2c/4 – 6a2c2/25) (-50a2b2c2)
Solution:
Using Distributive law,
(-7ab2c/4 – 6a2c2/25) (-50a2b2c2) = (-50a2b2c2) × (-7ab2c/4) – (-50a2b2c2) × (6a2c2/25)
= 175a2+1b2+2c2+1/2 + 12a2+2b2c2+2
= 175a3b4c3/2 + 12a4b2c4
Hence, the product is 175a3b4c3/2 + 12a4b2c4
### Question 9. Find the product of -8xyz/27 (3xyz2/2 – 9xy2z3/4)
Solution:
Using Distributive law,
-8xyz/27 (3xyz2/2 – 9xy2z3/4) = (-8xyz/27) × (3xyz2/2) – (-8xyz/27) × (9xy2z3/4)
= -4x1+1y1+1z1+2/9 + 2x1+1y1+2z1+3/3
= -4x2y2z3/9 + 2x2y3z4/3
Hence, the product is -4x2y2z3/9 + 2x2y3z4/3
### Question 10. Find the product of (-4xyz/27) [9x2yz/2 – 3xyz2/4]
Solution:
Using Distributive law,
(-4xyz/27) [9x2yz/2 – 3xyz2/4] = (-4xyz/27) × (9x2yz/2) + (-4xyz/27) × (3xyz2/4)
= -2x1+2y1+1z1+1/3 + 1x1+1y1+1z1+2/9
= -2x3y2z2/3 + 1x2y2z3/9
Hence, the product is 2x3y2z2/3 + 1x2y2z3/9
### Chapter 6 Algebraic Expression and Identities – Exercise 6.4 | Set 2
My Personal Notes arrow_drop_up |
Undefined slope
Undefined slope is so often misunderstood and many students find it confusing that I thought it would be a good thing to dedicate a special lesson to explain it.
As the lesson about "what is slope" shows, the 4 different types of slopes are positive slope, negative slope, zero slope and undefined slope.
Basically, a slope that is undefined looks like the vertical lines on the coordinate system you see in the graph below.
When a slope is undefined all you do is moving straight up or straight down only. You are not moving horizontally at all. In other words, the run or horizontal change is zero. The slope is therefore at its steepest.
Real life examples 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.
• The walls in your bedroom or living room could also have a slope that is undefined since this is how these walls are usually built. A spirit level is used during construction to ensure the walls are as "vertical" as possible.
It got its name "undefined" from the fact that it is impossible to divide by zero. This can happen when the denominator is zero.
Recall that 6/2 equal 3 because 3 × 2 = 6
However, it is impossible to do 8/0 because there exists no number you can multiply 0 by to get 8. We say that this division is undefined.
Using the slope formula to show that each of the vertical lines above has an undefined slope
Now, let us try to get the slope for the lines above. Recall that the slope formula is
m = (y1 − y2)/(x1 − x2)
For the line on the left, the points are (-3, 4) and (-3, 1)
Let x1 = -3 , y1 = 4 and x2 = -3 and y2 = 1
m = (y1 − y2)/(x1 − x2)
m = (4 − 1)/(-3 − -3)
m = (4 − 1)/(-3 + 3)
m = 3/0
The denominator is zero. As a result, there exists no number you can multiply 0 by to get 3. Therefore, we say that the slope is undefined.
For the line in the middle, the points are (1, 3) and (1, 9)
Let x1 = 1 , y1 = 9 and x2 = 1 and y2 = 3
m = (y1 − y2)/(x1 − x2)
m = (9 − 3)/(1 − 1)
m = 6/0
The denominator is zero. As a result, there exists no number you can multiply 0 by to get 6. Therefore, we say that the slope is undefined.
For the line on the right, the points are (3, 7) and (3, 2)
Let x1 = 3 , y1 = 7 and x2 = 3 and y2 = 2
m = (y1 − y2)/(x1 − x2)
m = (7 − 2)/(3 − 3)
m = 5/0
The denominator is zero. As a result, there exists no number you can multiply 0 by to get 5. Therefore, we say that the slope is undefined.
Notice that for the line in the middle, the x-values are the same for both points.
x1 = x2 = 1
This is also the case for all the lines above.
In general, when the x-values or x-coordinates are the same for all points on a line, the slope is undefined. If you can make this observation, there is no need to compute the slope.
Undefined slope equation
Is it possible to write a line with an undefined slope in slope-intercept form or point-slope form?
Since the slope m cannot be found, this is not possible. However, you can still write an equation for the line.
The equation of a line with an undefined slope is x = x-coordinate. Notice that the y-coordinates can equal to any real number.
Example
Write an equation of the line with undefined slope and passing through (-2, 4), (-2,0), and (-2, -4)
Since the x-coordinate is -2, the equation of the line is x = -2
Take the undefined slope quiz below to see if you really understood this lesson.
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Recommended |
This is the fifth post in an article series about MIT's Linear Algebra course. In this post I will review lecture five that finally introduces real linear algebra topics such as vector spaces their subspaces and spaces from matrices. But before it does that it closes the topics that were started in the previous lecture on permutations, transposes and symmetric matrices.
Here is a list of the previous posts in this article series:
## Lecture 5: Vector Spaces and Subspaces
Lecture starts with reminding some facts about permutation matrices. Remember from the previous lecture that permutation matrices P execute row exchanges and they are identity matrices with reordered rows.
Let's count how many permutation matrices are there for an nxn matrix.
For a matrix of size 1x1, there is just one permutation matrix - the identity matrix.
For a matrix of size 2x2 there are two permutation matrices - the identity matrix and the identity matrix with rows exchanged.
For a matrix of size 3x3 we may have the rows of the identity matrix rearranged in 6 ways - {1,2,3}, {1,3,2}, {2,1,3}, {2,3,1}, {3,1,2}, {3,2,1}.
For a matrix of size 4x4 the number of ways to reorder the rows is the same as the number of ways to rearrange numbers {1,2,3,4}. This is the simplest possible combinatorics problem. The answer is 4! = 24 ways.
In general, for an nxn matrix, there are n! permutation matrices.
Another key fact to remember about permutation matrices is that their inverse P-1 is their transpose PT. Or algebraically PT·P = I.
The lecture proceeds to transpose matrices. The transpose of a matrix exchanges its columns with rows. Another way to think about it that it flips the matrix over its main diagonal. Transpose of matrix A is denoted by AT.
Here is an example of transpose of a 3-by-3 matrix. I color coded the columns to better see how they get exchanged:
A matrix does not have to be square for its transpose to exist. Here is another example of transpose of a 3-by-2 matrix:
In algebraic notation transpose is expressed as (AT)ij = Aji, which says that an element aij at position ij get transposed into the position ji.
Here are the rules for matrix transposition:
• The transpose of A + B is (A + B)T = AT + BT.
• The transpose of A·B is (A·B)T = BT·AT.
• The transpose of A·B·C is (A·B·C)T = CT·BT·AT.
• The transpose of A-1 is (A-1)T = (AT)-1.
Next the lecture continues with symmetric matrices. A symmetric matrix has its transpose equal to itself, i.e., AT = A. It means that we can flip the matrix along the diagonal (transpose it) but it won't change.
Here is an example of a symmetric matrix. Notice that the elements on opposite sides of the diagonal are equal:
Now check this out. If you have a matrix R that is not symmetric and you multiply it with its transpose RT as R·RT, you get a symmetric matrix! Here is an example:
Are you wondering why it's true? The proof is really simple. Remember that matrix is symmetric if its transpose is equal to itself. Now what's the transpose of the product R·RT? It's (R·RT)T = (RT)T·RT = R·RT - it's the same product, which means that R·RT is always symmetric.
Here is another cool fact - the inverse of a symmetric matrix (if it exists) is also symmetric. Here is the proof. Suppose A is symmetric, then the transpose of A-1 is (A-1)T = (AT)-1. But AT = A, therefore (AT)-1 = A-1.
At this point lecture finally reaches the fundamental topic of linear algebra - vector spaces. As usual, it introduces the topic by examples.
Example 1: Vector space R2 - all 2-dimensional vectors. Some of the vectors in this space are (3, 2), (0, 0), (?, e) and infinitely many others. These are all the vectors with two components and they represent the xy plane.
Example 2: Vector space R3 - all vectors with 3 components (all 3-dimensional vectors).
Example 3: Vector space Rn - all vectors with n components (all n-dimensional vectors).
What makes these vectors vector spaces is that they are closed under multiplication by a scalar and addition, i.e., vector space must be closed under linear combination of vectors. What I mean by that is if you take two vectors and add them together or multiply them by a scalar they are still in the same space.
For example, take a vector (1,2,3) in R3. If we multiply it by any number ?, it's still in R3 because ?·(1,2,3) = (?, 2?, 3?). Similarly, if we take any two vectors (a, b, c) and (d, e, f) and add them together, the result is (a+d, b+e, f+c) and it's still in R3.
There are actually 8 axioms that the vectors must satisfy for them to make a space, but they are not listed in this lecture.
Here is an example of not-a-vector-space. It's 1/4 of R2 (the 1st quadrant). The green vectors are in the 1st quadrant but the red one is not:
An example of not-a-vector-space.
This is not a vector space because the green vectors in the space are not closed under multiplication by a scalar. If we take the vector (3,1) and multiply it by -1 we get the red vector (-3, -1) but it's not in the 1st quadrant, therefore it's not a vector space.
Next, Gilbert Strang introduces subspaces of vector spaces.
For example, any line in R2 that goes through the origin (0, 0) is a subspace of R2. Why? Because if we take any vector on the line and multiply it by a scalar, it's still on the line. And if we take any two vectors on the line and add them together, they are also still on the line. The requirement for a subspace is that the vectors in it do not go outside when added together or multiplied by a number.
Here is a visualization. The blue line is a subspace of R2 because the red vectors on it can't go outside of line:
An example of subspace of R2.
And example of not-a-subspace of R2 is any line that does not go through the origin. If we take any vector on the line and multiply it by 0, we get the zero vector, but it's not on the line. Also if we take two vectors and add them together, they are not on the line. Here is a visualization:
An example of not-a-subspace of R2.
Why not list all the subspaces of R2. They are:
• the R2 itself,
• any line through the origin (0, 0),
• the zero vector (0, 0).
And all the subspaces of R3 are:
• the R3 itself,
• any line through the origin (0, 0, 0),
• any plane through the origin (0, 0, 0),
• the zero vector.
The last 10 minutes of the lecture are spent on column spaces of matrices.
The column space of a matrix is made out of all the linear combinations of its columns. For example, given this matrix:
The column space C(A) is the set of all vectors {?·(1,2,4) + ?·(3,3,1)}. In fact, this column space is a subspace of R3 and it forms a plane through the origin.
More about column spaces in the next lecture.
You're welcome to watch the video lecture five:
Topics covered in lecture five:
• [01:30] Permutations.
• [03:00] A=LU elimination without row exchanges.
• [03:50] How Matlab does A=LU elimination.
• [04:50] PA=LU elimination with row exchanges
• [06:40] Permutation matrices.
• [07:25] How many permutation matrices are there?
• [08:30] Permutation matrix properties.
• [10:30] Transpose matrices.
• [11:50] General formula for transposes: (AT)ij = Aji.
• [13:06] Symmetric matrices.
• [13:30] Example of a symmetric matrix.
• [15:15] R·RT is always symmetric.
• [18:23] Why is R·RT symmetric?
• [20:50] Vector spaces.
• [22:05] Examples of vector spaces.
• [22:55] Real vector space R2.
• [23:20] Picture of R2 - xy plane.
• [26:50] Vector space R3.
• [28:00] Vector space Rn.
• [30:00] Example of not a vector space.
• [32:00] Subspaces of vector spaces.
• [33:00] A vector space inside R2.
• [34:35] A line in R2 that is subspace.
• [34:50] A line in R2 that is not a subspace.
• [36:30] All subspaces of R2.
• [39:30] All subspaces of R3.
• [40:20] Subspaces of matrices.
• [41:00] Column spaces of matrices C(A).
• [44:10] Example of column space of matrix with columns in R3.
Here are my notes of lecture five:
My notes of linear algebra lecture 5 on vector spaces and subspaces.
The next post is going to be more about column spaces and null spaces of matrices. |
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# Fractions with like Denominators
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### Fractions with like Denominators
1. 1. Adding and Subtracting Fractions with like Denominators You know that the bottom number of a fraction tells how may parts each whole is divided into. In this picture each circle is divided into 4 parts so the bottom number for this fractions is 4. 4 We use or shade 5 parts so the top number of this fraction is 5. The picture shows the fraction 5 . 4 In a fraction the bottom number has a special name. The bottom number in a fraction is called the denominator. The denominator or the bottom number in a fraction tells how many parts each whole is divided into.
2. 2. What are the denominators in these fractions? <ul><li>Two </li></ul><ul><li>Six </li></ul><ul><li>Eight </li></ul><ul><li>Five </li></ul><ul><li>Three </li></ul><ul><li>Remember the bottom number in a fraction is called the denominator. </li></ul>1 2 4 6 7 8 7 5 2 3
3. 3. <ul><li>1 4 5 </li></ul><ul><li>3 3 3 </li></ul><ul><li>Fractions can be added and subtracted without using pictures. </li></ul><ul><li>Here’s a problem. </li></ul><ul><li>5 3 </li></ul><ul><li>4 4 </li></ul><ul><li>When you add and subtract fractions you do not work on the top and the bottom the same way. </li></ul>You have learned to add fractions using pictures. + = + = + =
4. 4. <ul><li>When you add and subtract fractions you COPY the denominator, then you work on the top. Remember, you copy the denominator and then you work on the top. </li></ul><ul><li>5 3 </li></ul><ul><li>4 4 </li></ul><ul><li>Look at this problem. What is the denominator? </li></ul><ul><li>Yes, it’s 4. </li></ul><ul><li>What do you do with the denominator? </li></ul><ul><li>Right, you copy it in the answer so the denominator in the answer is 4. </li></ul><ul><li>Now we can add the numbers on the top. What do we get when we add 5 + 3? </li></ul><ul><li>Correct, 5 + 3 = 8. </li></ul><ul><li>We put the 8 on top in the answer so 5 + 3 = 8 . </li></ul><ul><li> 4 4 4 </li></ul>+ = 4
5. 5. <ul><li>7 2 </li></ul><ul><li>5 </li></ul><ul><li>Here’s a different problem. First look at the sign. We are subtracting in this problem. </li></ul><ul><li>Next look at the denominators. What do we do with the denominator? </li></ul><ul><li>Yes, we copy it in the answer. </li></ul><ul><li>On the top the sign tells us to subtract. 7 – 2, what’s the answer? </li></ul><ul><li>Right, 7 – 2 = 5. We put the 5 on top. </li></ul><ul><li>5 </li></ul><ul><li>5 </li></ul><ul><li>7 2 5 </li></ul><ul><li>5 5 5 </li></ul>= 5 - = -
6. 6. Let’s try another problem. <ul><li>4 3 </li></ul><ul><li>2 </li></ul><ul><li>First you copy the denominator then you work the top. </li></ul><ul><li>What is the denominator? </li></ul><ul><li>Yes, it’s 2. </li></ul><ul><li>Now work on the top. What is 4 – 3? </li></ul><ul><li>Right, it’s 1. </li></ul><ul><li>So 4 3 1 </li></ul><ul><li>2 2 2 </li></ul>- = 2 - =
7. 7. Here’s a new problem. <ul><li>2 3 </li></ul><ul><li>4 </li></ul><ul><li>What do we do with the denominator? </li></ul><ul><li>Yes, we copy the 4. </li></ul><ul><li>4 </li></ul><ul><li>What do we get on top? </li></ul><ul><li>Good Job! 2 + 3 = 5 . </li></ul><ul><li>2 3 5 </li></ul><ul><li>4 4 4 </li></ul>+ = + =
8. 8. Your Turn <ul><li>Copy these problems on your paper and write the answer. </li></ul><ul><li>1. 8 2 </li></ul><ul><li>5 5 </li></ul><ul><li>2. 5 2 </li></ul><ul><li>3 3 </li></ul><ul><li>3. 8 4 </li></ul><ul><li>6 6 </li></ul><ul><li>4. 10 3 </li></ul><ul><li> 4 4 </li></ul><ul><li>5. 3 2 </li></ul><ul><li>4 4 </li></ul>= = = = = + + - - - Check your work 6 5 7 3 4 6 7 4 5 4
9. 9. <ul><li>If the fraction you are working does not have the same denominator you cannot work the problem the way it is written. </li></ul><ul><li>You can’t work this problem the way it is written because the denominators are not the same. </li></ul><ul><li>6 1 </li></ul><ul><li>4 3 </li></ul>= -
10. 10. Some of these problems can be worked the way they are written but some cannot. Look at the denominators to decide, then write the numbers of the problems that can be worked the way they are written. <ul><li>1. 7 2 4. 4 4 </li></ul><ul><li>3 4 5 6 </li></ul><ul><li>2. 8 2 5. 8 2 </li></ul><ul><li>3 5 5 5 </li></ul><ul><li>3. 6 1 6. 2 5 </li></ul><ul><li>2 2 3 4 </li></ul>- - - + + + = = = = = =
11. 11. Check your work. You can work problems 3 and 5 the way they are written. Remember, if you can’t copy the denominator of the fraction you are adding or subtracting you cannot work the problem the way it is written. <ul><li>1. 7 2 4. 4 4 </li></ul><ul><li> 3 4 5 6 </li></ul><ul><li>2. 8 2 5. </li></ul><ul><li> 3 5 </li></ul><ul><li>3. 6. 2 5 </li></ul><ul><li> 3 4 </li></ul>- - - + + + = = = = = = 6 2 1 2 8 5 2 5
12. 12. Look at these new problems. Write the numbers of the problems that can be worked the way they are written. <ul><li>1. 6 5 4. 2 1 </li></ul><ul><li>8 8 3 3 </li></ul><ul><li>2. 7 2 5. 6 3 </li></ul><ul><li>6 4 4 2 </li></ul><ul><li>3. 1 4 6. 4 3 </li></ul><ul><li>2 3 6 6 </li></ul>- - - + + + = = = = = =
13. 13. Check your work. You can work problems 1, 4, and 6. Now on your paper work the problems that can be worked the way they are written. <ul><li>6 5 </li></ul><ul><li>8 8 </li></ul><ul><li>2 1 </li></ul><ul><li>3 3 </li></ul><ul><li>4 3 </li></ul><ul><li>6 6 </li></ul>1 . 6 5 1 8 8 8 4 . 2 1 1 3 3 3 6 . 4 3 7 6 6 6 - - + = = = = = = + - -
14. 14. <ul><li>1 </li></ul><ul><li>3 </li></ul><ul><li>What do you do in the denominator? </li></ul><ul><li>Yes, you copy the denominator. </li></ul><ul><li>3 </li></ul><ul><li>What do you get on top? </li></ul><ul><li>That’s right, 2. 1+1=2. </li></ul><ul><li>1 1 2 </li></ul><ul><li>3 3 </li></ul>+ + = = Let’s try one more.
15. 15. <ul><li>New problem. </li></ul><ul><li>1 1 </li></ul><ul><li>3 3 </li></ul><ul><li>Do we work in the denominator? </li></ul><ul><li>No. What do we do in the denominator? </li></ul><ul><li>Correct, we copy the denominator. </li></ul><ul><li>3 </li></ul><ul><li>Do we work on top? </li></ul><ul><li>Yes. </li></ul><ul><li>What do we get on top? </li></ul><ul><li>Yes, 0. </li></ul>- =
16. 16. Think about the picture of this problem when you work it. <ul><li>You have 1 and you take away 1 . How many thirds do you have left? </li></ul><ul><li> 3 3 </li></ul><ul><li>That’s Correct. You have 0 thirds left. You have no thirds or 0 . </li></ul><ul><li> 3 </li></ul><ul><li>1 1 0 </li></ul><ul><li>3 3 3 </li></ul><ul><li>When you see a problem that is addition or subtraction first look at the denominators. If they are the same you can work the problem the way it is written. If the denominators are not the same you cannot work the problem the way it is written. </li></ul>- = 1 1 3 3 - =
17. 17. You have learned to work problems this way. <ul><li>3 1 4 or 7 5 2 </li></ul><ul><li>4 4 4 8 8 8 </li></ul><ul><li>You can also work fraction addition and subtraction problems when they are written this way. </li></ul><ul><li>5 8 </li></ul><ul><li>8 3 </li></ul><ul><li>2 2 </li></ul><ul><li>8 3 </li></ul><ul><li>7 6 </li></ul><ul><li>8 3 </li></ul><ul><li>Just like before check to see if you can work the problem the way it is written, copy the denominator, then add or subtract. </li></ul>+ - + - = =
18. 18. Your Turn <ul><li>Copy each problem you can work the way it is written, then work the problem. </li></ul><ul><li>1. 2. 3. 4. 5. </li></ul><ul><li>6 8 4 5 2 </li></ul><ul><li>8 4 5 3 6 </li></ul><ul><li>+ 1 - 3 + 2 - 6 - 1 </li></ul><ul><li>8 4 6 3 2 </li></ul>
19. 19. Check your work. Problems 3 and 5 cannot be worked the way they are written. Work problems 1, 2, and 4. <ul><li>1. 2. 3. 4. 5. </li></ul><ul><li>6 8 4 5 2 </li></ul><ul><li>8 4 5 3 6 </li></ul><ul><li>+ 1 - 3 + 2 - 2 - 1 </li></ul><ul><li>8 4 6 3 2 </li></ul><ul><li>Remember which ever way the problem is written to check the denominators. If the denominators are the same copy the denominator in the answer and add or subtract the top. </li></ul>7 8 5 4 3 3 |
# Divide by 10, 100 and 1000 Divisors | How to Divide Whole Numbers by 10, 100 and 1000?
The division is one the basic arithmetic operation which we have to learn in the primary level. Dividing by 10, 100, and 1000 Divisors means a number is divided into 10 or 100 or 1000 equal parts. For this, you need to learn the multiples of 10 which helps you to solve the problems quickly. Go to the below section to understand the concept of Divide by 10, 100, and 1000 Divisors.
Also, Refer:
### How to Divide by 10, 100 and 1000 Divisors?
The method to Divide by 10, 100, and 1000 Divisors is given below. Hence go through the below steps and practice the questions related to Dividing by 10, 100, and 1000 Divisors. Examples for Divide by 10, 100, and 1000 Divisors with answers are given in the coming section.
• When we divide a number by 10 then the very first 2-digit from the extreme right of the dividend will be divided by 10.
• Check whether the divisor 10 is greater than the 2 digit number.
• Then multiply 10 with the suitable digit.
• Then write the quotient and remainder.
• Write the next digit beside the remainder
• Again repeat the same process till you get the remainder as 0 or less than 10.
• The same process will be applied for 100 and 1000 divisors
### Division of Decimal by 10, 100, and 1000 Divisors
The division of decimals numbers is similar to the concept of fractions. When we divide the decimal number by 10, 100, or 1000 we need to shift the point to the left. If the divisor is 10 you have to shift the point one time to the left. If the divisor is 100 you have to shift the point two times to the left same with 1000.
Example:
25 ÷ 10 = 2.5
25 ÷ 100 = 0.25
25 ÷ 1000 = 0.025
### Dividing by 10 100 1000 Examples
Example: 1
A dealer bought 10 tables for $38000. What will be the cost of 1 table? Solution: Given, A dealer bought 10 tables for$38000.
Divide 38000 by 10 and find the cost of 1 table.
38000 ÷ 10 = 3800
The quotient is 3800 and the remainder is 0.
Example 2.
Divide the given numbers by 100 and write the quotient and remainder.
i. 28800 ÷ 100
ii. 10000 ÷ 100
iii. 1128 ÷ 100
Solution:
i.
It is a 3 digit divisor. Consider 3 digits in the dividend
288 > 100
Now multiply with 100
100 × 2 = 200
Write the result below the dividend and quotient to the right.
288 – 200 = 88
Write the remainder below and add 0 to the right it becomes 880
Again multiply with 100.
100 × 8 = 800
Subtract 880 and 800 we get 80.
Write the remainder below and add 0 to the right it becomes 800
Again multiply with 100.
100 × 8 = 800
Subtract 800 and 800 we get 0.
Thus the remainder is 0.
ii.
It is a 3 digit divisor. Consider 3 digits in the dividend
100 = 100
Now multiply with 100
100 × 1 = 100
Write the result below the dividend and quotient to the right.
100 – 100 = 0
Write the remainder below and add 0 to the right it becomes 00 and write 0 near the quotient.
Thus the remainder is 0.
iii.
It is a 3 digit divisor. Consider 3 digits in the dividend
112 > 100
Now multiply with 100
100 × 1 = 100
Write the result below the dividend and quotient to the right.
112 – 100 = 12
Write the remainder below and add 8 to the right it becomes 128.
Again multiply with 100.
100 × 1 = 100
Subtract 128 and 100 we get 28.
Thus the remainder is 28.
Example 3.
Find the quotient and remainder for the following numbers.
i. 34000 ÷ 1000
ii. 3720 ÷ 1000
iii. 1500 ÷ 1000
Solution:
i.
It is a 4 digit divisor. Consider 3 digits in the dividend
3400 > 1000
Now multiply with 1000
1000 × 3 = 3000
3400 – 3000 = 400
400 is the remainder and 3 is the quotient
Add 0 to the right side of the remainder it becomes 4000
1000 × 4 = 4000
Thus the remainder is 0 and the quotient is 34.
ii.
It is a 4 digit divisor. Consider 3 digits in the dividend
3720 > 1000
Now multiply with 1000
1000 × 3 = 3000
3720 – 3000 = 720
720 < 1000
Thus the remainder is 720 and the quotient is 3.
iii.
It is a 4 digit divisor. Consider 3 digits in the dividend
1500 > 1000
Now multiply with 1000
1000 × 1 = 1000
1500- 1000 = 500
500 < 1000
Thus the remainder is 500 and the quotient is 1.
### FAQs on Divide by 10, 100, and 1000 Divisors
1. What happens when you divide a 2 digit number by 10 100 or 1000?
When we divide by 100, the 2-digits in the ones and the tens place form the remainder while the remaining digits form quotient.
2. How do you divide a number by 1000?
To divide a number by 1000, move all of its digits to thousands place value columns to the right.
3. What is the rule for dividing by 10?
Move the decimal point one place to the left. Place value is the value of a digit based on its location in the number.
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If you find any mistakes, please make a comment! Thank you.
## Solution to Linear Algebra Hoffman & Kunze Chapter 7.1
#### Exercise 7.1.1
Let $T$ be a linear operator on $F^2$. Prove that any non-zero vector which is not a characteristic vector for $T$ is a cyclic vector for $T$. Hence, prove that either $T$ has a cyclic vector or $T$ is a scalar multiple of the identity operator.
Solution: Let $v\in F^2$ be nonzero such that $v$ is not a characteristic vector for $T$. Then $v,Tv$ is linearly independent. Hence $v,Tv$ forms a basis of $F^2$. Namely, $v,Tv$ spans the whole space $F^2$. So $v$ is a cyclic vector for $T$.
Suppose $T$ does not have a cyclic vector, we only need to show that $T$ is a scalar multiple of the identity operator. By the first part, any non-zero vector is a characteristic vector for $T$. We take a basis $v_1,v_2$ of $F^2$. Then $Tv_1=av_1$ and $Tv_2=bv_2$. We show that $a=b$. Since $v_1+v_2$ is non-zero and is a characteristic vector for $T$, there exists $c$ such that$T(v_1+v_2)=c(v_1+v_2).$Hence we have$av_1+bv_2=cv_1+cv_2.$Since $v_1$ and $v_2$ are linearly independent, we must have $a=b=c$. Hence $Tv_1=av_1$ and $Tv_2=av_2$. It is clear that $T$ is a scalar multiple of the identity operator.
#### Exercise 7.1.2
Solution: Since $T$ is diagonalizable, the minimal polynomial of $T$ is $(x-2)(x+1)$ but the characteristic polynomial is $(x-2)^2(x+1)$. By Theorem 2 and Corollary in page 230, if $T$ has a cyclic vector then the minimal polynomial and characteristic polynomial are the same. This is impossible. Hence $T$ has no cyclic vectors.
The $T$-cyclic subspace generated by $(1,-1,3)$ is $\{(a,-a,b)|a,b\in F\}$.
#### Exercise 7.1.3
Note that we have$T(1,0,0)=(1,-1,0),\quad T^2(1,0,0)=(1-i,-3,-1).$Hence $(1,0,0)$, $T(1,0,0)$, $T^2(1,0,0)$ are linearly independent. Thus $(1,0,0)$ is a cyclic vector and the T-annihilator of $(1,0,0)$ is exactly the characteristic polynomial of $T$ (please compute it by yourself).
Note that $T(1,0,i)=(1,0,i)$, hence the T-annihilator of $(1,0,i)$ is exactly $x-1$.
#### Exercise 7.1.4
Prove that if $T^2$ has a cyclic vector, then $T$ has a cyclic vector. Is the converse true?
Solution: It is clear that $Z(\alpha,T^2)\subset Z(\alpha,T)$. Let $\alpha$ be a cyclic vector of $T^2$, then we have$V=Z(\alpha,T^2)\subset Z(\alpha,T).$Hence $Z(\alpha,T)=V$, namely $\alpha$ is a cyclic vector of $T$.
The converse is false. Let $T$ be the operator on $F^2$ corresponding to the matrix $\begin{pmatrix} 0 &1\\ 0&0\end{pmatrix}$. Then $T$ is cyclic by Exercise 7.1.1. But $T^2=0$ is not cyclic.
#### Exercise 7.1.5
Solution: Recall that from Exercise 6.8.15, the characteristic polynomial of $N$ is $x^n$ and hence $N^n=0$. Since $N^{n-1}\alpha\ne 0$, the $T$-annihilator of $T$ is $x^n$. Hence by Theorem 1 (i) $\Z(\alpha,N)=V$. Thus $\alpha$ is a cyclic vector of $N$. The matrix has the form of (7-2) in page 229 where all $c_i$ are zero.
#### Exercise 7.1.6
Give a direct proof that if $A$ is the companion matrix of the monic polynomial $p$, then $T$ is the characteristic polynomial for $A$.
Solution: Expand $\det(xI-A)$ from the last column.
#### Exercise 7.1.7
Let $V$ be an $n$-dimensional vector space, and let $T$ be a linear operator on $V$. Suppose that $T$ is diagonalizable.
(a) If $T$ has a cyclic vector, show that
(b) If $T$ has $n$ distinct characteristic values, and if $\{\alpha_1,\dots,\alpha_n\}$ is a basis of characteristic vectors for $T$. Show that $\alpha=\alpha_1+\cdots+\alpha_n$ is a cyclic vector for $T$.
Solution:
(a) Since $T$ is diagonalizable, let $p$ be the minimal polynomial of $T$,$p(x)=(x-z_1)\cdots(x-z_k),$where $z_1,\dots,z_k$ are all distinct characteristic values for $T$, see Theorem 6 in page 204. Since $T$ has a cyclic vector. Hence $p$ is also the characteristic polynomial for $T$, which has degree $n$. Therefore $k=n$. That is $T$ has $n$ distinct characteristic values.
(b) Let $z_i$ be the characteristic value for $T$ corresponding to $\alpha_i$. Then $T^k\alpha_i=z_i^k\alpha_i$. Thus for any polynomial $g$, we have$g(T)\alpha=\sum_{i=1}^n g(z_i)\alpha_i.$Since $z_i$ are all distinct, we can choose$g_i(x)=\frac{(x-z_1)\cdots(x-z_{i-1})(x-z_{i+1})\cdots(x-z_n)}{(z_i-z_1)\cdots(z_i-z_{i-1})(z_i-z_{i+1})\cdots(z_i-z_n)}.$Recall from Section 4.3, we have $g_i(z_j)=\delta_{ij}$. Therefore$g_i(T)\alpha=\alpha_i.$Hence $Z(\alpha,T)$ contains a basis $\{\alpha_1,\dots,\alpha_n\}$ of $V$. So $\alpha$ is a cyclic vector for $T$.
#### Exercise 7.1.8
Let $T$ be a linear operator on the finite-dimensional vector space $V$. Suppose $T$ has a cyclic vector. Prove that if $U$ is any linear operator which commutes with $T$, then $U$ is a polynomial in $T$.
Solution: Let $\alpha$ be a cyclic vector of $V$ and $\dim V=n$. Then $\alpha,T\alpha,\cdots,T^{n-1}\alpha$ is a basis of $V$. Hence we have$U\alpha=\sum_{i=0}^{n-1}a_iT^i\alpha,$for some $a_i\in F$. Since $U$ commutes with $T$, we have\begin{align*}U(T\alpha)&=TU\alpha=T\sum_{i=0}^{n-1}a_iT^i\alpha\\&=\sum_{i=0}^{n-1}a_{i+1}T^{i+1}\alpha=\sum_{i=0}^{n-1}a_iT^i(T\alpha).\end{align*}Similarly, we have$U(T^j\alpha)=\sum_{i=0}^{n-1}a_iT^i(T^j\alpha),$for all $j>0$. Since $\alpha,T\alpha,\cdots,T^{n-1}\alpha$ is a basis of $V$, we conclude that$U=\sum_{i=0}^{n-1}a_iT^i$ is a polynomial in $T$. |
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# Divide the following: 1 Kg 825 g by 5
Last updated date: 06th Aug 2024
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Hint: We have to find the divided value of 1 Kg 825 g by 5. Here the value of 1 Kg 825 g will be the numerator and the value of 5 will be the denominator of the fraction part. Using the GCD we find the simplified form.
We need to find the division of 1 Kg 825 g by 5.
In this case we first convert the unit of the 1 Kg 825 g to a single unit. We know that the relation between kilogram and gram is that 1 kilogram is equal to 1000 gram.
Therefore, 1 Kg 825 g will be equal to $1000+825=1825$ gram.
We now assume the value 1825 as the dividend and the value of 5 as the divisor.
The value of the division is $\dfrac{1825}{5}$.
We need to find the simplified form of the proper fraction $\dfrac{1825}{5}$.
Simplified form is achieved when the G.C.D of the denominator and the numerator is 1.
For any fraction $\dfrac{p}{q}$, we first find the G.C.D of the denominator and the numerator. If it’s 1 then it’s already in its simplified form and if the G.C.D of the denominator and the numerator is any other number d then we need to divide the denominator and the numerator with d and get the simplified fraction form as $\dfrac{{}^{p}/{}_{d}}{{}^{q}/{}_{d}}$.
For our given fraction $\dfrac{1825}{5}$, the G.C.D of the denominator and the numerator is 5.
Now we divide both the denominator and the numerator with 5 and get $\dfrac{{}^{1825}/{}_{5}}{{}^{5}/{}_{5}}=\dfrac{365}{1}=365$.
The divided value of 1 Kg 825 g by 5 is 365 grams.
Note: we can also convert 1 Kg 825 g into kilogram form where 1825 grams is equal to $1.825$ kilogram. We now divide $1.825$ by 5 and we get $\dfrac{1.825}{5}=0.365$ kilogram. We can easily state that $0.365$ kilogram is equal to 365 grams. |
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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples.Objectives of this module: be able to add numbers with like signs and unlike signs, understand addition with zero.
## Overview
• Addition of Numbers with Like Signs
• Addition of Numbers with Unlike Signs
## Addition of numbers with like signs
Let us add the two positive numbers 2 and 3. We perform this addition on the number line as follows.
We begin at 0, the origin.
Since 2 is positive, we move 2 units to the right.
Since 3 is positive, we move 3 more units to the right.
We are now located at 5.
Thus, $2+3=5$ .
Summarizing, we have
$\left(2\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{units}\right)+\left(3\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{units}\right)=\left(5\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{units}\right)$
Now let us add the two negative numbers $-2$ and $-3$ . We perform this addition on the number line as follows.
We begin at 0, the origin.
Since $-2$ is negative, we move 2 units to the left.
Since $-3$ is negative, we move 3 more units to the left.
We are now located at $-5$ .
Thus, $\left(-2\right)+\left(-3\right)=-5$ .
Summarizing, we have
$\left(2\text{\hspace{0.17em}}\text{negative}\text{\hspace{0.17em}}\text{units}\right)+\left(3\text{\hspace{0.17em}}\text{negative}\text{\hspace{0.17em}}\text{units}\right)=\left(5\text{\hspace{0.17em}}\text{negative}\text{\hspace{0.17em}}\text{units}\right)$
These two examples suggest that
$\begin{array}{l}\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)\\ \left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)\end{array}$
## Adding numbers with the same sign
To add two real numbers that have the same sign, add the absolute values of the numbers and associate the common sign with the sum.
## Sample set a
Find the sums.
$3+7$
$\begin{array}{l}\begin{array}{cc}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Add}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\end{array}\\ \begin{array}{cc}\begin{array}{l}|3|=3\\ |7|=7\end{array}\right\}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3+7=10\end{array}\\ \begin{array}{cc}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{The}\text{\hspace{0.17em}}\text{common}\text{\hspace{0.17em}}\text{sign}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{"+}\text{."}\end{array}\end{array}$
$3+7=+10\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3+7=10$
$\left(-4\right)+\left(-9\right)$
$\begin{array}{l}\begin{array}{cc}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Add}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\end{array}\\ \begin{array}{cc}\begin{array}{l}|-4|=4\\ |-9|=9\end{array}\right\}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4+9=13\end{array}\\ \begin{array}{cc}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{The}\text{\hspace{0.17em}}\text{common}\text{\hspace{0.17em}}\text{sign}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{"}-\text{."}\end{array}\end{array}$
$\left(-4\right)+\left(-9\right)=-13$
## Practice set a
Find the sums.
$8+6$
14
$41+11$
52
$\left(-4\right)+\left(-8\right)$
$-12$
$\left(-36\right)+\left(-9\right)$
$-45$
$-14+\left(-20\right)$
$-34$
$-\frac{2}{3}+\left(-\frac{5}{3}\right)$
$-\frac{7}{3}$
$-2.8+\left(-4.6\right)$
$-7.4$
Notice that
$\begin{array}{l}\left(\text{0}\right)\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\left(\text{a}\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{number}\right)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(\text{that}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\text{}\text{positive}\text{\hspace{0.17em}}\text{number}\right)\\ \left(\text{0}\right)\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\left(\text{a}\text{\hspace{0.17em}}\text{negative}\text{\hspace{0.17em}}\text{number}\right)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(\text{that}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\text{}\text{negative}\text{\hspace{0.17em}}\text{number)}\end{array}$
## The additive identity is 0
Since adding 0 to a real number leaves that number unchanged, 0 is called the additive identity .
## Addition of numbers with unlike signs
Now let us perform the addition $2+\left(-6\right)$ . These two numbers have unlike signs. This type of addition can also be illustrated using the number line.
We begin at 0, the origin.
Since 2 is positive, we move 2 units to the right.
Since $-6$ is negative, we move, from the 2, 6 units to the left.
We are now located at $-4$ .
A rule for adding two numbers that have unlike signs is suggested by noting that if the signs are disregarded, 4 can be obtained from 2 and 6 by subtracting 2 from 6. But 2 and 6 are precisely the absolute values of 2 and $-6$ . Also, notice that the sign of the number with the larger absolute value is negative and that the sign of the resulting sum is negative.
## Adding numbers with unlike signs
To add two real numbers that have unlike signs, subtract the smaller absolute value from the larger absolute value and associate the sign of the number with the larger absolute value with this difference.
## Sample set b
Find the following sums.
$7+\left(-2\right)$
$\begin{array}{cc}\underset{\begin{array}{l}\text{Larger}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{value}\text{.}\\ \text{Sign}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}"+".\end{array}}{\underbrace{|7|=7}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& \underset{\text{Smaller}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{value}\text{.}}{\underbrace{|-2|=2}}\end{array}$
$\begin{array}{ll}\text{Subtract}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values:}\hfill & 7-2=5.\hfill \\ \text{Attach}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{proper}\text{\hspace{0.17em}}\text{sign:}\hfill & "+".\hfill \end{array}$
$7+\left(-2\right)=+5\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7+\left(-2\right)=5$
$3+\left(-11\right)$
$\begin{array}{cc}\underset{\text{Smaller}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{value}\text{.}}{\underbrace{|3|=3}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& \underset{\begin{array}{l}\text{Larger}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{value}\text{.}\\ \text{Sign}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}"-".\end{array}}{\underbrace{|-11|=11}}\end{array}$
$\begin{array}{ll}\text{Subtract}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values:}\hfill & 11-3=8.\hfill \\ \text{Attach}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{proper}\text{\hspace{0.17em}}\text{sign:}\hfill & "-".\hfill \end{array}$
$3+\left(-11\right)=-8$
The morning temperature on a winter's day in Lake Tahoe was $-12$ degrees. The afternoon temperature was 25 degrees warmer. What was the afternoon temperature?
We need to find $-12+25$ .
$\begin{array}{cc}\underset{\text{Smaller}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{value}\text{.}}{\underbrace{|-12|=12}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& \underset{\begin{array}{l}\text{Larger}\text{}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{value}\text{.}\\ \text{Sign}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{"+"}\text{.}\end{array}}{\underbrace{|25|=25}}\end{array}$
$\begin{array}{ll}\text{Subtract}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values:}\hfill & 25-12=13.\hfill \\ \text{Attach}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{proper}\text{\hspace{0.17em}}\text{sign:}\hfill & "+".\hfill \end{array}$
$-12+25=13$
Thus, the afternoon temperature is 13 degrees.
Add $-147+84$ . Display Reads
$\begin{array}{l}\begin{array}{ccc}\text{Type}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}147& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}147\end{array}\\ \begin{array}{ccc}\text{Press}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{||}\hline +/-\\ \hline\end{array}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{147}\end{array}\\ \begin{array}{ccc}\text{Press}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{||}\hline +\\ \hline\end{array}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{147}\end{array}\\ \begin{array}{ccc}\text{Type}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}84& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}84\end{array}\\ \begin{array}{ccc}\text{Press}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{||}\hline =\\ \hline\end{array}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-63\end{array}\end{array}$
## Practice set b
Find the sums.
$4+\left(-3\right)$
1
$-3+5$
2
$15+\left(-18\right)$
$-3$
$0+\left(-6\right)$
$-6$
$-26+12$
$-14$
$35+\left(-78\right)$
$-43$
$15+\left(-10\right)$
5
$1.5+\left(-2\right)$
$-0.5$
$-8+0$
$-8$
$0+\left(0.57\right)$
$0.57$
$-879+454$
$-425$
$-1345.6+\left(-6648.1\right)$
$-7993.7$
## Exercises
Find the sums for the the following problems.
$4+12$
16
$8+6$
$6+2$
8
$7+9$
$\left(-3\right)+\left(-12\right)$
$-15$
$\left(-6\right)+\left(-20\right)$
$\left(-4\right)+\left(-8\right)$
$-12$
$\left(-11\right)+\left(-8\right)$
$\left(-16\right)+\left(-8\right)$
$-24$
$\left(-2\right)+\left(-15\right)$
$14+\left(-3\right)$
11
$21+\left(-4\right)$
$14+\left(-6\right)$
8
$18+\left(-2\right)$
$10+\left(-8\right)$
2
$40+\left(-31\right)$
$\left(-3\right)+\left(-12\right)$
$-15$
$\left(-6\right)+\left(-20\right)$
$10+\left(-2\right)$
8
$8+\left(-15\right)$
$-2+\left(-6\right)$
$-8$
$-11+\left(-14\right)$
$-9+\left(-6\right)$
$-15$
$-1+\left(-1\right)$
$-16+\left(-9\right)$
$-25$
$-22+\left(-1\right)$
$0+\left(-12\right)$
$-12$
$0+\left(-4\right)$
$0+\left(24\right)$
24
$-6+1+\left(-7\right)$
$-5+\left(-12\right)+\left(-4\right)$
$-21$
$-5+5$
$-7+7$
0
$-14+14$
$4+\left(-4\right)$
0
$9+\left(-9\right)$
$84+\left(-61\right)$
23
$13+\left(-56\right)$
$452+\left(-124\right)$
328
$636+\left(-989\right)$
$1811+\left(-935\right)$
876
$-373+\left(-14\right)$
$-1221+\left(-44\right)$
$-1265$
$-47.03+\left(-22.71\right)$
$-1.998+\left(-4.086\right)$
$-6.084$
$\left[\left(-3\right)+\left(-4\right)\right]+\left[\left(-6\right)+\left(-1\right)\right]$
$\left[\left(-2\right)+\left(-8\right)\right]+\left[\left(-3\right)+\left(-7\right)\right]$
$-20$
$\left[\left(-3\right)+\left(-8\right)\right]+\left[\left(-6\right)+\left(-12\right)\right]$
$\left[\left(-8\right)+\left(-6\right)\right]+\left[\left(-2\right)+\left(-1\right)\right]$
$-17$
$\left[4+\left(-12\right)\right]+\left[12+\left(-3\right)\right]$
$\left[5+\left(-16\right)\right]+\left[4+\left(-11\right)\right]$
$-18$
$\left[2+\left(-4\right)\right]+\left[17+\left(-19\right)\right]$
$\left[10+\left(-6\right)\right]+\left[12+\left(-2\right)\right]$
14
$9+\left[\left(-4\right)+7\right]$
$14+\left[\left(-3\right)+5\right]$
16
$\left[2+\left(-7\right)\right]+\left(-11\right)$
$\left[14+\left(-8\right)\right]+\left(-2\right)$
4
In order for a small business to break even on a project, it must have sales of $21,000$ . If the amount of sales was $15,000$ , how much money did this company fall short?
Suppose a person has $56.00$ in his checking account. He deposits $100.00$ into his checking account by using the automatic teller machine. He then writes a check for $84.50$ . If an error causes the deposit not to be listed into this person's account, what is this person's checking balance?
$-28.50$
A person borrows $7.00$ on Monday and then $12.00$ on Tuesday. How much has this person borrowed?
A person borrows $11.00$ on Monday and then pays back $8.00$ on Tuesday. How much does this person owe?
$3.00$
## Exercises for review
( [link] ) Simplify $\frac{4\left({7}^{2}-6\cdot {2}^{3}\right)}{{2}^{2}}$ .
( [link] ) Simplify $\frac{35{a}^{6}{b}^{2}{c}^{5}}{7{b}^{2}{c}^{4}}$ .
$5{a}^{6}c$
( [link] ) Simplify ${\left(\frac{12{a}^{8}{b}^{5}}{4{a}^{5}{b}^{2}}\right)}^{3}$ .
( [link] ) Determine the value of $|-8|$ .
8
( [link] ) Determine the value of $\left(|2|+{|4|}^{2}\right)+{|-5|}^{2}$ .
where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Got questions? Join the online conversation and get instant answers! |
# Find all function $f: \mathbb{Q} \to \mathbb{Q}$ such that $f(x+f(x)+2y)=2x+2f(f(y))$ [closed]
Find all function $f: \mathbb{Q} \to \mathbb{Q}$ such that $f(x+f(x)+2y)=2x+2f(f(y))$ for all $x,y \in \mathbb{Q}$.
• Even if this is an interestng problem, you should say something on what you tried to solve it. – Crostul Jul 7 '15 at 14:46
• This is not the sort of site where you can just cut-and-paste a problem and expect a useful response. What have you tried so far? Where does the problem come from? – anomaly Jul 7 '15 at 14:46
First set $x = y = 0$. This gives us that $f(f(0)) = 2f(f(0))$ and so $f(f(0)) = 0$.
Now set $y=0$. This gives us $$f(x + f(x)) = 2x + 2f(f(0)) = 2x$$
Now replace $x$ with $x + f(x)$ in this equation to get
$$f(3x + f(x)) = f(x + f(x) + 2x) = f(x + f(x) + f(x + f(x))) = 2x + 2f(x)$$
But by setting $x=y$ in the original equation, we have that $$f(3x + f(x)) = 2x + 2f(f(x))$$
and so $$f(f(x)) = f(x)$$ for all $x$.
This immediately gives us that $f(0)=f(f(0))=0$. Now set $x=0$ in the original equation to obtain $$f(2y) = 2f(f(y)) = 2f(y)$$
We then get that $$2f(x) = f(2x) = f(f(x + f(x))) = f(x + f(x)) = 2x$$
and so
$$f(x) = x$$
for all $x \in \mathbb{Q}$. |
# Long Division Can Be EasyPage 1
Slide 1
LONG DIVISION
CAN BE EASY!
Slide 2
## Long Division
L.O - To know how to use DMSB to work out long division.
R.T-
(Remember To)
Divide
Multiply
Subtract
Bring Down
Slide 3
## DMSB
What is DMSB? You wonder?
This stands for the order of long division
Divide
Multiply
Subtract
Bring Down
Slide 4
Long Division
Take a look at this division problem:
3 )74
Looks scary, huh? Well, when you use DMSB It can be simple. Move to the next slide to get started!
This is the dividend. The dividend is the number we break into groups.
This is the divisor. The divisor is the number of groups we are dividing into or the number we are dividing by.
Slide 5
## Divide
First, we want to DIVIDE.
Ask yourself: “How many times can 3 go into 7?”
3 )74
Slide 6
DIVIDE
You want to find the closest number to 7 without going over!
In this case it’s 2 times because 3 x 2 is 6
3 times is too high since 3 x 3 = 9
Slide 7
DIVIDE
Now, place your 2 above the 7, because 3 goes into 7 two times.
2
3 )74
You’ve just completed the divide. Now, we’ll move onto multiply.
The 2 becomes the first part of your quotient. The quotient is the answer in division.
Slide 8
## Multiply
Next, you want to multiply 3 times 2:
2
3 )74
3 x 2= 6 6
We do this because we said that 3 can go into 7 two times, but now we have to find out how close to 7 we can get. So we multiply!
Slide 9
MULTIPLY
That was a really simple step!
Now, we want to move on to subtract!
Slide 10
## Subtract
Now that we know how many times 3 goes into 7, we have to subtract to see the difference between the numbers:
2
3 )74
-6
1
Good! We know 7 – 6 = 1. Leave your 1 in it’s spot, because we’re going to use it for the bring down!
Slide 11
## Bring down
When you bring down, you’re moving the second part of your dividend into your workspace:
Go to page:
1 2 3 |
# What Is 16/33 as a Decimal + Solution With Free Steps
The fraction 16/33 as a decimal is equal to 0.484.
The p/q form, where p and q are referred to as the Numerator and Denominator, can be used to represent fractions. The division is one of the most challenging mathematical operations because it is required when working with fractions. But if we employ the approach that will be covered later, we can simplify it.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 16/33. The Long Division method is shown in figure 1 below:
Figure 1
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 16
Divisor = 33
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 16 $\div$ 33
This is when we go through the Long Division solution to our problem.
## 16/33 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 16 and 33, we can see how 16 is Smaller than 33, and to solve this division, we require that 16 be Bigger than 33.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 16, which after getting multiplied by 10 becomes 160.
We take this 160 and divide it by 33; this can be done as follows:
 160 $\div$ 33 $\approx$ 4
Where:
33 x 4 = 132
This will lead to the generation of a Remainder equal to 160 – 132 = 28. Now this means we have to repeat the process by Converting the 28 into 280 and solving for that:
280 $\div$ 33 $\approx$ 8Â
Where:
33 x 8 = 264
This, therefore, produces another Remainder which is equal to 280 – 264 = 16. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 160.
160 $\div$ 33 $\approx$ 4Â
Where:
33 x 4 = 160
Finally, we have a Quotient generated after combining the three pieces of it as 0.484=z, with a Remainder equal to 28.
Images/mathematical drawings are created with GeoGebra. |
# Growing Patterns Concept. The figures as shown below are made up of identical tiles. Fig. 1 Fig. 2 Fig. 3 (a) Following the pattern, how many tiles is.
## Presentation on theme: "Growing Patterns Concept. The figures as shown below are made up of identical tiles. Fig. 1 Fig. 2 Fig. 3 (a) Following the pattern, how many tiles is."— Presentation transcript:
Growing Patterns Concept
The figures as shown below are made up of identical tiles. Fig. 1 Fig. 2 Fig. 3 (a) Following the pattern, how many tiles is Fig. 10 made up of ? (b) Which figure will you form if you are required to use as many of the 300 tiles given to you as possible ? Example
Solution (a)The figures as shown follow this pattern : 4, 7 ( 4 + 3 ), 10 ( 7 + 3 ),.. No. of tiles in Fig. 2 = 4 + 1 x 3 No. of tiles in Fig. 3 = 4 + 2 x 3 Therefore, the number of tiles in Fig. 10 = 4 + 9 x 3 = 31 (b) Note that number of tiles in Fig. 100 is 4 + 99 x 3 = 301 ( which just exceeds 300 ). Therefore, the figure that can be formed using as many of the 300 tiles given as possible is Fig.99
Question 1 How many different ways can you arrange 7 chairs, each of a different colour, in a row? Step 1 Start with 2 chairs, A and B. ABBA There are 2 ways.
Step 2 Add 1 more chair, C. Find the number of ways to arrange 3 chairs. There are 6 ways. ABCBCACAB BACCBAACB
Step 3 Add 1 more chair, D. Find the number of ways to arrange 4 chairs. There are 24 ways. ABCBACDBACBDABDC
Look at the numbers below. Can you see a pattern? Step 1: Number of ways = 1 x 2 = 2 Step 2: Number of ways = 1 x 2 x 3 = 6 Step 3: Number of ways = 1 x 2 x 3 x 4 = 24 Now calculate the number of different ways to arrange 7 chairs in a row. There are ________ ways to arrange 7 chairs in a row.
Question 2 Figure Number 123 Number of Circles 13 (1 + 2) 4 (1 + 2 + 3) Fig 1 Fig 2Fig 3
a) Calculate the number of circles in Fig 7. There are _______ circles in Fig 7. b) Calculate the number of circles in Fig 88. There are _______ circles in Fig 88. c) Which figure has 276 circles? Figure _______ has 276 circles.
Question 3 Figure Number 123 Number of Circles Fig 1 Fig 2Fig 3
a) Calculate the number of circles in Fig 9. There are _______ circles in Fig 9. b) Calculate the number of circles in Fig 50. There are _______ circles in Fig 50. c) Which figure has 130 circles? Figure _______ has 130 circles.
Question 4 Figure Number123 Number of sides48 Perimeter (cm)816 Area (cm 2 )412 Fig 1 Fig 2Fig 3 The following figures are made up of squares. Each side of the square is 2 cm.
a) Calculate the perimeter of Fig 6. The perimeter of Fig 6 is ______cm. b) Calculate the area of Fig 12. The area of Fig 12 is _______ cm 2. c) Which figure has an area of 840 cm 2 ? Figure _______ has an area of cm 2.
Question 5 How many squares are there altogether in the following figure? Hint: There are more than 4 squares. DimensionNumber of Squares 1 x 14 2 x 21 Total5
How many squares are there altogether in the figure below? Hint: There are more than 9 squares. DimensionNumber of Squares 1 x 19 2 x 24 3 X 31 Total14
Can you see a pattern? For a 2 by 2 square, the total number of squares is 1 x 1 + 2 x 2 = 5 For a 3 by 3 square, the total number of squares is 1 x 1 + 2 x 2 + 3 x 3 = 14
How many squares are there altogether in the figure below? Hint: There are more than 25 squares. There are _______ squares altogether in the figure.
Thats all for today… Thank you. A.L.2007
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# List of trigonometric identities
Cosines and sines around the unit circle
In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are identities potentially involving angles but also involving side lengths or other lengths of a triangle.
These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non-trigonometric functions: a common technique involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.
## Notation
### Angles
Signs of trigonometric functions in each quadrant. The mnemonic "All Science Teachers (are) Crazy" lists the basic functions ('All', sin, tan, cos) which are positive from quadrants I to IV.[1] This is a variation on the mnemonic "All Students Take Calculus".
This article uses Greek letters such as alpha (α), beta (β), gamma (γ), and theta (θ) to represent angles. Several different units of angle measure are widely used, including degree, radian, and gradian (gons):
1 full circle (turn) = 360 degree = 2π radian = 400 gon.
If not specifically annotated by (°) for degree or (${\displaystyle ^{\mathrm {g} }}$ ) for gradian, all values for angles in this article are assumed to be given in radian.
The following table shows for some common angles their conversions and the values of the basic trigonometric functions:
Conversions of common angles
${\displaystyle 0}$ ${\displaystyle 0^{\circ }}$ ${\displaystyle 0}$ ${\displaystyle 0^{\mathrm {g} }}$ ${\displaystyle 0}$ ${\displaystyle 1}$ ${\displaystyle 0}$
${\displaystyle {\dfrac {1}{12}}}$ ${\displaystyle 30^{\circ }}$ ${\displaystyle {\dfrac {\pi }{6}}}$ ${\displaystyle 33{\dfrac {1}{3}}^{\mathrm {g} }}$ ${\displaystyle {\dfrac {1}{2}}}$ ${\displaystyle {\dfrac {\sqrt {3}}{2}}}$ ${\displaystyle {\dfrac {\sqrt {3}}{3}}}$
${\displaystyle {\dfrac {1}{8}}}$ ${\displaystyle 45^{\circ }}$ ${\displaystyle {\dfrac {\pi }{4}}}$ ${\displaystyle 50^{\mathrm {g} }}$ ${\displaystyle {\dfrac {\sqrt {2}}{2}}}$ ${\displaystyle {\dfrac {\sqrt {2}}{2}}}$ ${\displaystyle 1}$
${\displaystyle {\dfrac {1}{6}}}$ ${\displaystyle 60^{\circ }}$ ${\displaystyle {\dfrac {\pi }{3}}}$ ${\displaystyle 66{\dfrac {2}{3}}^{\mathrm {g} }}$ ${\displaystyle {\dfrac {\sqrt {3}}{2}}}$ ${\displaystyle {\dfrac {1}{2}}}$ ${\displaystyle {\sqrt {3}}}$
${\displaystyle {\dfrac {1}{4}}}$ ${\displaystyle 90^{\circ }}$ ${\displaystyle {\dfrac {\pi }{2}}}$ ${\displaystyle 100^{\mathrm {g} }}$ ${\displaystyle 1}$ ${\displaystyle 0}$ Undefined
${\displaystyle {\dfrac {1}{3}}}$ ${\displaystyle 120^{\circ }}$ ${\displaystyle {\dfrac {2\pi }{3}}}$ ${\displaystyle 133{\dfrac {1}{3}}^{\mathrm {g} }}$ ${\displaystyle {\dfrac {\sqrt {3}}{2}}}$ ${\displaystyle -{\dfrac {1}{2}}}$ ${\displaystyle -{\sqrt {3}}}$
${\displaystyle {\dfrac {3}{8}}}$ ${\displaystyle 135^{\circ }}$ ${\displaystyle {\dfrac {3\pi }{4}}}$ ${\displaystyle 150^{\mathrm {g} }}$ ${\displaystyle {\dfrac {\sqrt {2}}{2}}}$ ${\displaystyle -{\dfrac {\sqrt {2}}{2}}}$ ${\displaystyle -1}$
${\displaystyle {\dfrac {5}{12}}}$ ${\displaystyle 150^{\circ }}$ ${\displaystyle {\dfrac {5\pi }{6}}}$ ${\displaystyle 166{\dfrac {2}{3}}^{\mathrm {g} }}$ ${\displaystyle {\dfrac {1}{2}}}$ ${\displaystyle -{\dfrac {\sqrt {3}}{2}}}$ ${\displaystyle -{\dfrac {\sqrt {3}}{3}}}$
${\displaystyle {\dfrac {1}{2}}}$ ${\displaystyle 180^{\circ }}$ ${\displaystyle \pi }$ ${\displaystyle 200^{\mathrm {g} }}$ ${\displaystyle 0}$ ${\displaystyle -1}$ ${\displaystyle 0}$
${\displaystyle {\dfrac {7}{12}}}$ ${\displaystyle 210^{\circ }}$ ${\displaystyle {\dfrac {7\pi }{6}}}$ ${\displaystyle 233{\dfrac {1}{3}}^{\mathrm {g} }}$ ${\displaystyle -{\dfrac {1}{2}}}$ ${\displaystyle -{\dfrac {\sqrt {3}}{2}}}$ ${\displaystyle {\dfrac {\sqrt {3}}{3}}}$
${\displaystyle {\dfrac {5}{8}}}$ ${\displaystyle 225^{\circ }}$ ${\displaystyle {\dfrac {5\pi }{4}}}$ ${\displaystyle 250^{\mathrm {g} }}$ ${\displaystyle -{\dfrac {\sqrt {2}}{2}}}$ ${\displaystyle -{\dfrac {\sqrt {2}}{2}}}$ ${\displaystyle 1}$
${\displaystyle {\dfrac {2}{3}}}$ ${\displaystyle 240^{\circ }}$ ${\displaystyle {\dfrac {4\pi }{3}}}$ ${\displaystyle 266{\dfrac {2}{3}}^{\mathrm {g} }}$ ${\displaystyle -{\dfrac {\sqrt {3}}{2}}}$ ${\displaystyle -{\dfrac {1}{2}}}$ ${\displaystyle {\sqrt {3}}}$
${\displaystyle {\dfrac {3}{4}}}$ ${\displaystyle 270^{\circ }}$ ${\displaystyle {\dfrac {3\pi }{2}}}$ ${\displaystyle 300^{\mathrm {g} }}$ ${\displaystyle -1}$ ${\displaystyle 0}$ Undefined
${\displaystyle {\dfrac {5}{6}}}$ ${\displaystyle 300^{\circ }}$ ${\displaystyle {\dfrac {5\pi }{3}}}$ ${\displaystyle 333{\dfrac {1}{3}}^{\mathrm {g} }}$ ${\displaystyle -{\dfrac {\sqrt {3}}{2}}}$ ${\displaystyle {\dfrac {1}{2}}}$ ${\displaystyle -{\sqrt {3}}}$
${\displaystyle {\dfrac {7}{8}}}$ ${\displaystyle 315^{\circ }}$ ${\displaystyle {\dfrac {7\pi }{4}}}$ ${\displaystyle 350^{\mathrm {g} }}$ ${\displaystyle -{\dfrac {\sqrt {2}}{2}}}$ ${\displaystyle {\dfrac {\sqrt {2}}{2}}}$ ${\displaystyle -1}$
${\displaystyle {\dfrac {11}{12}}}$ ${\displaystyle 330^{\circ }}$ ${\displaystyle {\dfrac {11\pi }{6}}}$ ${\displaystyle 366{\dfrac {2}{3}}^{\mathrm {g} }}$ ${\displaystyle -{\dfrac {1}{2}}}$ ${\displaystyle {\dfrac {\sqrt {3}}{2}}}$ ${\displaystyle -{\dfrac {\sqrt {3}}{3}}}$
${\displaystyle 1}$ ${\displaystyle 360^{\circ }}$ ${\displaystyle 2\pi }$ ${\displaystyle 400^{\mathrm {g} }}$ ${\displaystyle 0}$ ${\displaystyle 1}$ ${\displaystyle 0}$
Results for other angles can be found at Trigonometric constants expressed in real radicals. Per Niven's theorem, ${\displaystyle (0,\;30,\;90,\;150,\;180,\;210,\;270,\;330,\;360)}$ are the only rational numbers that, taken in degrees, result in a rational sine-value for the corresponding angle within the first turn, which may account for their popularity in examples.[2][3] The analogous condition for the unit radian requires that the argument divided by π is rational, and yields the solutions 0, π/6, π/2, 5π/6, π, 7π/6, 3π/2, 11π/6(, 2π).
### Trigonometric functions
The functions sine, cosine and tangent of an angle are sometimes referred to as the primary or basic trigonometric functions. Their usual abbreviations are sin(θ), cos(θ) and tan(θ), respectively, where θ denotes the angle. The parentheses around the argument of the functions are often omitted, e.g., sin θ and cos θ, if an interpretation is unambiguously possible.
The sine of an angle is defined, in the context of a right triangle, as the ratio of the length of the side that is opposite to the angle divided by the length of the longest side of the triangle (the hypotenuse).
${\displaystyle \sin \theta ={\frac {\text{opposite}}{\text{hypotenuse}}}.}$
The cosine of an angle in this context is the ratio of the length of the side that is adjacent to the angle divided by the length of the hypotenuse.
${\displaystyle \cos \theta ={\frac {\text{adjacent}}{\text{hypotenuse}}}.}$
The tangent of an angle in this context is the ratio of the length of the side that is opposite to the angle divided by the length of the side that is adjacent to the angle. This is the same as the ratio of the sine to the cosine of this angle, as can be seen by substituting the definitions of sin and cos from above:
${\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}={\frac {\text{opposite}}{\text{adjacent}}}.}$
The remaining trigonometric functions secant (sec), cosecant (csc), and cotangent (cot) are defined as the reciprocal functions of cosine, sine, and tangent, respectively. Rarely, these are called the secondary trigonometric functions:
${\displaystyle \sec \theta ={\frac {1}{\cos \theta }},\quad \csc \theta ={\frac {1}{\sin \theta }},\quad \cot \theta ={\frac {1}{\tan \theta }}={\frac {\cos \theta }{\sin \theta }}.}$
These definitions are sometimes referred to as ratio identities.
## Inverse functions
The inverse trigonometric functions are partial inverse functions for the trigonometric functions. For example, the inverse function for the sine, known as the inverse sine (sin−1) or arcsine (arcsin or asin), satisfies
${\displaystyle \sin(\arcsin x)=x\quad {\text{for}}\quad |x|\leq 1}$
and
${\displaystyle \arcsin(\sin x)=x\quad {\text{for}}\quad |x|\leq {\frac {\pi }{2}}.}$
Function Inverse sin cos tan sec csc cot arcsin arccos arctan arcsec arccsc arccot
## Pythagorean identities
In trigonometry, the basic relationship between the sine and the cosine is given by the Pythagorean identity:
${\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1,}$
where sin2 θ means (sin(θ))2 and cos2 θ means (cos(θ))2.
This can be viewed as a version of the Pythagorean theorem, and follows from the equation x2 + y2 = 1 for the unit circle. This equation can be solved for either the sine or the cosine:
{\displaystyle {\begin{aligned}\sin \theta &=\pm {\sqrt {1-\cos ^{2}\theta }},\\\cos \theta &=\pm {\sqrt {1-\sin ^{2}\theta }}.\end{aligned}}}
where the sign depends on the quadrant of θ.
Dividing this identity by either sin2 θ or cos2 θ yields the other two Pythagorean identities:
${\displaystyle 1+\tan ^{2}\theta =\sec ^{2}\theta \quad {\text{and}}\quad 1+\cot ^{2}\theta =\csc ^{2}\theta .}$
Using these identities together with the ratio identities, it is possible to express any trigonometric function in terms of any other (up to a plus or minus sign):
Each trigonometric function in terms of each of the other five.[4]
in terms of ${\displaystyle \sin \theta }$ ${\displaystyle \cos \theta }$ ${\displaystyle \tan \theta }$ ${\displaystyle \csc \theta }$ ${\displaystyle \sec \theta }$ ${\displaystyle \cot \theta }$
${\displaystyle \sin \theta =}$ ${\displaystyle \sin \theta }$ ${\displaystyle \pm {\sqrt {1-\cos ^{2}\theta }}}$ ${\displaystyle \pm {\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}}$ ${\displaystyle {\frac {1}{\csc \theta }}}$ ${\displaystyle \pm {\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}}$ ${\displaystyle \pm {\frac {1}{\sqrt {1+\cot ^{2}\theta }}}}$
${\displaystyle \cos \theta =}$ ${\displaystyle \pm {\sqrt {1-\sin ^{2}\theta }}}$ ${\displaystyle \cos \theta }$ ${\displaystyle \pm {\frac {1}{\sqrt {1+\tan ^{2}\theta }}}}$ ${\displaystyle \pm {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}}$ ${\displaystyle {\frac {1}{\sec \theta }}}$ ${\displaystyle \pm {\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}}$
${\displaystyle \tan \theta =}$ ${\displaystyle \pm {\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}}$ ${\displaystyle \pm {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}}$ ${\displaystyle \tan \theta }$ ${\displaystyle \pm {\frac {1}{\sqrt {\csc ^{2}\theta -1}}}}$ ${\displaystyle \pm {\sqrt {\sec ^{2}\theta -1}}}$ ${\displaystyle {\frac {1}{\cot \theta }}}$
${\displaystyle \csc \theta =}$ ${\displaystyle {\frac {1}{\sin \theta }}}$ ${\displaystyle \pm {\frac {1}{\sqrt {1-\cos ^{2}\theta }}}}$ ${\displaystyle \pm {\frac {\sqrt {1+\tan ^{2}\theta }}{\tan \theta }}}$ ${\displaystyle \csc \theta }$ ${\displaystyle \pm {\frac {\sec \theta }{\sqrt {\sec ^{2}\theta -1}}}}$ ${\displaystyle \pm {\sqrt {1+\cot ^{2}\theta }}}$
${\displaystyle \sec \theta =}$ ${\displaystyle \pm {\frac {1}{\sqrt {1-\sin ^{2}\theta }}}}$
${\displaystyle {\frac {1}{\cos \theta }}}$ ${\displaystyle \pm {\sqrt {1+\tan ^{2}\theta }}}$ ${\displaystyle \pm {\frac {\csc \theta }{\sqrt {\csc ^{2}\theta -1}}}}$ ${\displaystyle \sec \theta }$ ${\displaystyle \pm {\frac {\sqrt {1+\cot ^{2}\theta }}{\cot \theta }}}$
${\displaystyle \cot \theta =}$ ${\displaystyle \pm {\frac {\sqrt {1-\sin ^{2}\theta }}{\sin \theta }}}$ ${\displaystyle \pm {\frac {\cos \theta }{\sqrt {1-\cos ^{2}\theta }}}}$ ${\displaystyle {\frac {1}{\tan \theta }}}$ ${\displaystyle \pm {\sqrt {\csc ^{2}\theta -1}}}$ ${\displaystyle \pm {\frac {1}{\sqrt {\sec ^{2}\theta -1}}}}$ ${\displaystyle \cot \theta }$
## Historical shorthands
All of the trigonometric functions of an angle θ can be constructed geometrically in terms of a unit circle centered at O. Many of these terms are no longer in common use, however this diagram is not exhaustive.
The versine, coversine, haversine, and exsecant were used in navigation. For example, the haversine formula was used to calculate the distance between two points on a sphere. They are rarely used today.
Name Abbreviation Value[5][6]
versed sine, versine ${\displaystyle \operatorname {versin} \theta }$
${\displaystyle \operatorname {vers} \theta }$
${\displaystyle \operatorname {ver} \theta }$
${\displaystyle 1-\cos \theta }$
versed cosine, vercosine ${\displaystyle \operatorname {vercosin} \theta }$
${\displaystyle \operatorname {vercos} \theta }$
${\displaystyle \operatorname {vcs} \theta }$
${\displaystyle 1+\cos \theta }$
coversed sine, coversine ${\displaystyle \operatorname {coversin} \theta }$
${\displaystyle \operatorname {covers} \theta }$
${\displaystyle \operatorname {cvs} \theta }$
${\displaystyle 1-\sin \theta }$
coversed cosine, covercosine ${\displaystyle \operatorname {covercosin} \theta }$
${\displaystyle \operatorname {covercos} \theta }$
${\displaystyle \operatorname {cvc} \theta }$
${\displaystyle 1+\sin \theta }$
half versed sine, haversine ${\displaystyle \operatorname {haversin} \theta }$
${\displaystyle \operatorname {hav} \theta }$
${\displaystyle \operatorname {sem} \theta }$
${\displaystyle {\frac {1-\cos \theta }{2}}}$
half versed cosine, havercosine ${\displaystyle \operatorname {havercosin} \theta }$
${\displaystyle \operatorname {havercos} \theta }$
${\displaystyle \operatorname {hvc} \theta }$
${\displaystyle {\frac {1+\cos \theta }{2}}}$
half coversed sine, hacoversine
cohaversine
${\displaystyle \operatorname {hacoversin} \theta }$
${\displaystyle \operatorname {hacovers} \theta }$
${\displaystyle \operatorname {hcv} \theta }$
${\displaystyle {\frac {1-\sin \theta }{2}}}$
half coversed cosine, hacovercosine
cohavercosine
${\displaystyle \operatorname {hacovercosin} \theta }$
${\displaystyle \operatorname {hacovercos} \theta }$
${\displaystyle \operatorname {hcc} \theta }$
${\displaystyle {\frac {1+\sin \theta }{2}}}$
exterior secant, exsecant ${\displaystyle \operatorname {exsec} \theta }$
${\displaystyle \operatorname {exs} \theta }$
${\displaystyle \sec \theta -1}$
exterior cosecant, excosecant ${\displaystyle \operatorname {excosec} \theta }$
${\displaystyle \operatorname {excsc} \theta }$
${\displaystyle \operatorname {exc} \theta }$
${\displaystyle \csc \theta -1}$
chord ${\displaystyle \operatorname {crd} \theta }$ ${\displaystyle 2\sin {\frac {\theta }{2}}}$
## Reflections, shifts, and periodicity
Reflecting θ in α=0 (α=π)
By examining the unit circle, the following properties of the trigonometric functions can be established.
### Reflections
When the direction of a Euclidean vector is represented by an angle ${\displaystyle \theta }$ , this is the angle determined by the free vector (starting at the origin) and the positive x-unit vector. The same concept may also be applied to lines in a Euclidean space, where the angle is that determined by a parallel to the given line through the origin and the positive x-axis. If a line (vector) with direction ${\displaystyle \theta }$ is reflected about a line with direction ${\displaystyle \alpha ,}$ then the direction angle ${\displaystyle \theta '}$ of this reflected line (vector) has the value
${\displaystyle \theta '=2\alpha -\theta .}$
The values of the trigonometric functions of these angles ${\displaystyle \theta ,\;\theta '}$ for specific angles ${\displaystyle \alpha }$ satisfy simple identities: either they are equal, or have opposite signs, or employ the complementary trigonometric function. These are also known as reduction formulas.[7]
θ reflected in α = 0[8]
odd/even identities
θ reflected in α = π/4 θ reflected in α = π/2 θ reflected in α = π
compare to α = 0
${\displaystyle \sin(-\theta )=-\sin \theta }$ ${\displaystyle \sin \left({\tfrac {\pi }{2}}-\theta \right)=\cos \theta }$ ${\displaystyle \sin(\pi -\theta )=+\sin \theta }$ ${\displaystyle \sin(2\pi -\theta )=-\sin(\theta )=\sin(-\theta )}$
${\displaystyle \cos(-\theta )=+\cos \theta }$ ${\displaystyle \cos \left({\tfrac {\pi }{2}}-\theta \right)=\sin \theta }$ ${\displaystyle \cos(\pi -\theta )=-\cos \theta }$ ${\displaystyle \cos(2\pi -\theta )=+\cos(\theta )=\cos(-\theta )}$
${\displaystyle \tan(-\theta )=-\tan \theta }$ ${\displaystyle \tan \left({\tfrac {\pi }{2}}-\theta \right)=\cot \theta }$ ${\displaystyle \tan(\pi -\theta )=-\tan \theta }$ ${\displaystyle \tan(2\pi -\theta )=-\tan(\theta )=\tan(-\theta )}$
${\displaystyle \csc(-\theta )=-\csc \theta }$ ${\displaystyle \csc \left({\tfrac {\pi }{2}}-\theta \right)=\sec \theta }$ ${\displaystyle \csc(\pi -\theta )=+\csc \theta }$ ${\displaystyle \csc(2\pi -\theta )=-\csc(\theta )=\csc(-\theta )}$
${\displaystyle \sec(-\theta )=+\sec \theta }$ ${\displaystyle \sec \left({\tfrac {\pi }{2}}-\theta \right)=\csc \theta }$ ${\displaystyle \sec(\pi -\theta )=-\sec \theta }$ ${\displaystyle \sec(2\pi -\theta )=+\sec(\theta )=\sec(-\theta )}$
${\displaystyle \cot(-\theta )=-\cot \theta }$ ${\displaystyle \cot \left({\tfrac {\pi }{2}}-\theta \right)=\tan \theta }$ ${\displaystyle \cot(\pi -\theta )=-\cot \theta }$ ${\displaystyle \cot(2\pi -\theta )=-\cot(\theta )=\cot(-\theta )}$
### Shifts and periodicity
By shifting the arguments of trigonometric functions by certain angles, it is sometimes possible that changing the sign or applying complementary trigonometric functions can express particular results more simply. Some examples of shifts are shown below in the table.
• A full turn, or 360°, or 2π radian leaves the unit circle fixed and is the smallest interval for which the trigonometric functions sin, cos, sec, and csc repeat their values, and is thus their period. Shifting arguments of any periodic function by any integer multiple of a full period preserves the function value of the unshifted argument.
• A half turn, or 180°, or π radian is the period of tan(x) = sin(x)/cos(x) and cot(x) = cos(x)/sin(x), as can be seen from these definitions and the period of the defining trigonometric functions. So shifting the arguments of tan(x) and cot(x) by any multiple of π, does not change their function values.
For the functions sin, cos, sec, and csc with period 2π half a turn is half of their period. For this shift they change the sign of their values, as can be seen from the unit circle again. This new value repeats after any additional shift of 2π, so all together they change the sign for a shift by any odd multiple of π, i.e., by (2k + 1)⋅π, with k an arbitrary integer. Any even multiple of π is of course just a full period, and a backward shift by half a period is the same as a backward shift by one full period plus one shift forward by half a period.
• A quarter turn, or 90°, or π/2 radian is a half period shift for tan(x) and cot(x) with period π (180°), and yields the function value of applying the complementary function to the unshifted argument. By the argument above this also holds for a shift by any odd multiple (2k + 1)⋅π/2 of the half period.
For the four other trigonometric functions a quarter turn also represents a quarter period. A shift by an arbitrary multiple of a quarter period, that is not covered by a multiple of half periods, can be decomposed in an integer multiple of periods, plus or minus one quarter period. The terms expressing these multiples are (4k ± 1)⋅π/2. The forward/backward shifts by one quarter period are reflected in the table below. Again, these shifts yield function values, employing the respective complementary function applied to the unshifted argument.
Shifting the arguments of tan(x) and cot(x) by their quarter period (π/4) does not yield such simple results.
Shift by one quarter period Shift by one half period[9] Shift by full periods[10] Period
${\displaystyle \sin(\theta \pm {\tfrac {\pi }{2}})=\pm \cos \theta }$ ${\displaystyle \sin(\theta +\pi )=-\sin \theta }$ ${\displaystyle \sin(\theta +k\cdot 2\pi )=+\sin \theta }$ ${\displaystyle 2\pi }$
${\displaystyle \cos(\theta \pm {\tfrac {\pi }{2}})=\mp \sin \theta }$ ${\displaystyle \cos(\theta +\pi )=-\cos \theta }$ ${\displaystyle \cos(\theta +k\cdot 2\pi )=+\cos \theta }$ ${\displaystyle 2\pi }$
${\displaystyle \tan(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\tan \theta \pm 1}{1\mp \tan \theta }}}$ ${\displaystyle \tan(\theta +{\tfrac {\pi }{2}})=-\cot \theta }$ ${\displaystyle \tan(\theta +k\cdot \pi )=+\tan \theta }$ ${\displaystyle \pi }$
${\displaystyle \csc(\theta \pm {\tfrac {\pi }{2}})=\pm \sec \theta }$ ${\displaystyle \csc(\theta +\pi )=-\csc \theta }$ ${\displaystyle \csc(\theta +k\cdot 2\pi )=+\csc \theta }$ ${\displaystyle 2\pi }$
${\displaystyle \sec(\theta \pm {\tfrac {\pi }{2}})=\mp \csc \theta }$ ${\displaystyle \sec(\theta +\pi )=-\sec \theta }$ ${\displaystyle \sec(\theta +k\cdot 2\pi )=+\sec \theta }$ ${\displaystyle 2\pi }$
${\displaystyle \cot(\theta \pm {\tfrac {\pi }{4}})={\tfrac {\cot \theta \pm 1}{1\mp \cot \theta }}}$ ${\displaystyle \cot(\theta +{\tfrac {\pi }{2}})=-\tan \theta }$ ${\displaystyle \cot(\theta +k\cdot \pi )=+\cot \theta }$ ${\displaystyle \pi }$
## Angle sum and difference identities
Illustration of angle addition formulae for the sine and cosine. Emphasized segment is of unit length.
These are also known as the angle addition and subtraction theorems (or formulae). The identities can be derived by combining right triangles such as in the adjacent diagram, or by considering the invariance of the length of a chord on a unit circle given a particular central angle. The most intuitive derivation uses rotation matrices (see below).
Illustration of the angle addition formula for the tangent. Emphasized segments are of unit length.
For acute angles α and β, whose sum is non-obtuse, a concise diagram (shown) illustrates the angle sum formulae for sine and cosine: The bold segment labeled "1" has unit length and serves as the hypotenuse of a right triangle with angle β; the opposite and adjacent legs for this angle have respective lengths sin β and cos β. The cos β leg is itself the hypotenuse of a right triangle with angle α; that triangle's legs, therefore, have lengths given by sin α and cos α, multiplied by cos β. The sin β leg, as hypotenuse of another right triangle with angle α, likewise leads to segments of length cos α sin β and sin α sin β. Now, we observe that the "1" segment is also the hypotenuse of a right triangle with angle α + β; the leg opposite this angle necessarily has length sin(α + β), while the leg adjacent has length cos(α + β). Consequently, as the opposing sides of the diagram's outer rectangle are equal, we deduce
{\displaystyle {\begin{aligned}\sin(\alpha +\beta )&=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\\cos(\alpha +\beta )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \end{aligned}}}
Relocating one of the named angles yields a variant of the diagram that demonstrates the angle difference formulae for sine and cosine.[11] (The diagram admits further variants to accommodate angles and sums greater than a right angle.) Dividing all elements of the diagram by cos α cos β provides yet another variant (shown) illustrating the angle sum formula for tangent.
Illustration of the angle addition formula for the cotangent. Top right segment is of unit length.
Sine ${\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta }$ [12][13] ${\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta }$ [13][14] ${\displaystyle \tan(\alpha \pm \beta )={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}$ [13][15] ${\displaystyle \csc(\alpha \pm \beta )={\frac {\sec \alpha \sec \beta \csc \alpha \csc \beta }{\sec \alpha \csc \beta \pm \csc \alpha \sec \beta }}}$ [16] ${\displaystyle \sec(\alpha \pm \beta )={\frac {\sec \alpha \sec \beta \csc \alpha \csc \beta }{\csc \alpha \csc \beta \mp \sec \alpha \sec \beta }}}$ [16] ${\displaystyle \cot(\alpha \pm \beta )={\frac {\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha }}}$ [13][17] ${\displaystyle \arcsin x\pm \arcsin y=\arcsin \left(x{\sqrt {1-y^{2}}}\pm y{\sqrt {1-x^{2}}}\right)}$ [18] ${\displaystyle \arccos x\pm \arccos y=\arccos \left(xy\mp {\sqrt {\left(1-x^{2}\right)\left(1-y^{2}\right)}}\right)}$ [19] ${\displaystyle \arctan x\pm \arctan y=\arctan \left({\frac {x\pm y}{1\mp xy}}\right)}$ [20] ${\displaystyle \operatorname {arccot} x\pm \operatorname {arccot} y=\operatorname {arccot} \left({\frac {xy\mp 1}{y\pm x}}\right)}$
### Matrix form
The sum and difference formulae for sine and cosine follow from the fact that a rotation of the plane by angle α, following a rotation by β, is equal to a rotation by α+β. In terms of rotation matrices:
{\displaystyle {\begin{aligned}&{}\quad \left({\begin{array}{rr}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{array}}\right)\left({\begin{array}{rr}\cos \beta &-\sin \beta \\\sin \beta &\cos \beta \end{array}}\right)\\[12pt]&=\left({\begin{array}{rr}\cos \alpha \cos \beta -\sin \alpha \sin \beta &-\cos \alpha \sin \beta -\sin \alpha \cos \beta \\\sin \alpha \cos \beta +\cos \alpha \sin \beta &-\sin \alpha \sin \beta +\cos \alpha \cos \beta \end{array}}\right)\\[12pt]&=\left({\begin{array}{rr}\cos(\alpha +\beta )&-\sin(\alpha +\beta )\\\sin(\alpha +\beta )&\cos(\alpha +\beta )\end{array}}\right).\end{aligned}}}
The matrix inverse for a rotation is the rotation with the negative of the angle
${\displaystyle \left({\begin{array}{rr}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{array}}\right)^{-1}=\left({\begin{array}{rr}\cos(-\alpha )&-\sin(-\alpha )\\\sin(-\alpha )&\cos(-\alpha )\end{array}}\right)=\left({\begin{array}{rr}\cos \alpha &\sin \alpha \\-\sin \alpha &\cos \alpha \end{array}}\right)\,,}$
which is also the matrix transpose.
These formulae show that these matrices form a representation of the rotation group in the plane (technically, the special orthogonal group SO(2)), since the composition law is fulfilled and inverses exist. Furthermore, matrix multiplication of the rotation matrix for an angle α with a column vector will rotate the column vector counterclockwise by the angle α.
Since multiplication by a complex number of unit length rotates the complex plane by the argument of the number, the above multiplication of rotation matrices is equivalent to a multiplication of complex numbers:
${\displaystyle {\begin{array}{rcl}(\cos \alpha +i\sin \alpha )(\cos \beta +i\sin \beta )&=&(\cos \alpha \cos \beta -\sin \alpha \sin \beta )+i(\cos \alpha \sin \beta +\sin \alpha \cos \beta )\\&=&\cos(\alpha {+}\beta )+i\sin(\alpha {+}\beta ).\end{array}}}$
In terms of Euler's formula, this simply says ${\displaystyle e^{i\alpha }e^{i\beta }=e^{i(\alpha +\beta )}}$ , showing that ${\displaystyle \theta \ \mapsto \ e^{i\theta }=\cos \theta +i\sin \theta }$ is a one-dimensional complex representation of ${\displaystyle \mathrm {SO} (2)}$ .
### Sines and cosines of sums of infinitely many angles
When the series ${\displaystyle \sum _{i=1}^{\infty }\theta _{i}}$ converges absolutely then
${\displaystyle \sin \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{{\text{odd}}\ k\geq 1}(-1)^{\frac {k-1}{2}}\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}$
${\displaystyle \cos \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{{\text{even}}\ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)\,.}$
Because the series ${\displaystyle \sum _{i=1}^{\infty }\theta _{i}}$ converges absolutely, it is necessarily the case that ${\displaystyle \lim _{i\rightarrow \infty }\theta _{i}=0}$ , ${\displaystyle \lim _{i\rightarrow \infty }\sin \,\theta _{i}=0}$ , and ${\displaystyle \lim _{i\rightarrow \infty }\cos \theta _{i}=1}$ . In particular, in these two identities an asymmetry appears that is not seen in the case of sums of finitely many angles: in each product, there are only finitely many sine factors but there are cofinitely many cosine factors. Terms with infinitely many sine factors would necessarily be equal to zero.
When only finitely many of the angles θi are nonzero then only finitely many of the terms on the right side are nonzero because all but finitely many sine factors vanish. Furthermore, in each term all but finitely many of the cosine factors are unity.
### Tangents and cotangents of sums
Let ek (for k = 0, 1, 2, 3, ...) be the kth-degree elementary symmetric polynomial in the variables
${\displaystyle x_{i}=\tan \theta _{i}}$
for i = 0, 1, 2, 3, ..., i.e.,
{\displaystyle {\begin{aligned}e_{0}&=1\\[6pt]e_{1}&=\sum _{i}x_{i}&&=\sum _{i}\tan \theta _{i}\\[6pt]e_{2}&=\sum _{i
Then
{\displaystyle {\begin{aligned}\tan \left(\sum _{i}\theta _{i}\right)&={\frac {\sin \left(\sum _{i}\theta _{i}\right)/\prod _{i}\cos \theta _{i}}{\cos \left(\sum _{i}\theta _{i}\right)/\prod _{i}\cos \theta _{i}}}\\&={\frac {\sum _{{\text{odd}}\ k\geq 1}(-1)^{\frac {k-1}{2}}\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}{\sum _{{\text{even}}\ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}}={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }}\\\cot \left(\sum _{i}\theta _{i}\right)&={\frac {e_{0}-e_{2}+e_{4}-\cdots }{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}
using the sine and cosine sum formulae above.
The number of terms on the right side depends on the number of terms on the left side.
For example:
{\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2})&={\frac {e_{1}}{e_{0}-e_{2}}}={\frac {x_{1}+x_{2}}{1\ -\ x_{1}x_{2}}}={\frac {\tan \theta _{1}+\tan \theta _{2}}{1\ -\ \tan \theta _{1}\tan \theta _{2}}},\\[8pt]\tan(\theta _{1}+\theta _{2}+\theta _{3})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\[8pt]\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\[8pt]&={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}
and so on. The case of only finitely many terms can be proved by mathematical induction.[21]
### Secants and cosecants of sums
{\displaystyle {\begin{aligned}\sec \left(\sum _{i}\theta _{i}\right)&={\frac {\prod _{i}\sec \theta _{i}}{e_{0}-e_{2}+e_{4}-\cdots }}\\[8pt]\csc \left(\sum _{i}\theta _{i}\right)&={\frac {\prod _{i}\sec \theta _{i}}{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}
where ek is the kth-degree elementary symmetric polynomial in the n variables xi = tan θi, i = 1, ..., n, and the number of terms in the denominator and the number of factors in the product in the numerator depend on the number of terms in the sum on the left.[22] The case of only finitely many terms can be proved by mathematical induction on the number of such terms.
For example,
{\displaystyle {\begin{aligned}\sec(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{1-\tan \alpha \tan \beta -\tan \alpha \tan \gamma -\tan \beta \tan \gamma }}\\[8pt]\csc(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{\tan \alpha +\tan \beta +\tan \gamma -\tan \alpha \tan \beta \tan \gamma }}.\end{aligned}}}
## Multiple-angle formulae
Tn is the nth Chebyshev polynomial ${\displaystyle \cos(n\theta )=T_{n}(\cos \theta )}$ [23] ${\displaystyle \cos(n\theta )+i\sin(n\theta )=(\cos \theta +i\sin \theta )^{n}}$ [24]
### Double-angle, triple-angle, and half-angle formulae
#### Double-angle formulae
Formulae for twice an angle.[25]
${\displaystyle \sin(2\theta )=2\sin \theta \cos \theta ={\frac {2\tan \theta }{1+\tan ^{2}\theta }}}$
${\displaystyle \cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta =2\cos ^{2}\theta -1=1-2\sin ^{2}\theta ={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}}$
${\displaystyle \tan(2\theta )={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}$
${\displaystyle \cot(2\theta )={\frac {\cot ^{2}\theta -1}{2\cot \theta }}}$
${\displaystyle \sec(2\theta )={\frac {\sec ^{2}\theta }{2-\sec ^{2}\theta }}}$
${\displaystyle \csc(2\theta )={\frac {\sec \theta \csc \theta }{2}}}$
#### Triple-angle formulae
Formulae for triple angles.[25]
${\displaystyle \sin(3\theta )=3\sin \theta -4\sin ^{3}\theta =4\sin \theta \sin({\frac {\pi }{3}}-\theta )\sin({\frac {\pi }{3}}+\theta )}$
${\displaystyle \cos(3\theta )=4\cos ^{3}\theta -3\cos \theta =4\cos \theta \cos({\frac {\pi }{3}}-\theta )\cos({\frac {\pi }{3}}+\theta )}$
${\displaystyle \tan(3\theta )={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}=\tan \theta \tan({\frac {\pi }{3}}-\theta )\tan({\frac {\pi }{3}}+\theta )}$
${\displaystyle \cot(3\theta )={\frac {3\cot \theta -\cot ^{3}\theta }{1-3\cot ^{2}\theta }}}$
${\displaystyle \sec(3\theta )={\frac {\sec ^{3}\theta }{4-3\sec ^{2}\theta }}}$
${\displaystyle \csc(3\theta )={\frac {\csc ^{3}\theta }{3\csc ^{2}\theta -4}}}$
#### Half-angle formulae
{\displaystyle {\begin{aligned}&\sin {\frac {\theta }{2}}=\operatorname {sgn} \left(2\pi -\theta +4\pi \left\lfloor {\frac {\theta }{4\pi }}\right\rfloor \right){\sqrt {\frac {1-\cos \theta }{2}}}\\&\qquad {\text{where }}\operatorname {sgn} x=\pm 1{\text{ according to whether }}x{\text{ is positive or negative.}}\end{aligned}}}
${\displaystyle \sin ^{2}{\frac {\theta }{2}}={\frac {1-\cos \theta }{2}}}$
${\displaystyle \cos {\frac {\theta }{2}}=\operatorname {sgn} \left(\pi +\theta +4\pi \left\lfloor {\frac {\pi -\theta }{4\pi }}\right\rfloor \right){\sqrt {\frac {1+\cos \theta }{2}}}}$
${\displaystyle \cos ^{2}{\frac {\theta }{2}}={\frac {1+\cos \theta }{2}}}$
{\displaystyle {\begin{aligned}\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta =\pm \,{\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}={\frac {\sin \theta }{1+\cos \theta }}\\&={\frac {1-\cos \theta }{\sin \theta }}={\frac {-1\pm {\sqrt {1+\tan ^{2}\theta }}}{\tan \theta }}={\frac {\tan \theta }{1+\sec {\theta }}}\end{aligned}}}
${\displaystyle \cot {\frac {\theta }{2}}=\csc \theta +\cot \theta =\pm \,{\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}={\frac {\sin \theta }{1-\cos \theta }}={\frac {1+\cos \theta }{\sin \theta }}}$
Also
${\displaystyle \tan {\frac {\eta +\theta }{2}}={\frac {\sin \eta +\sin \theta }{\cos \eta +\cos \theta }}}$
${\displaystyle \tan \left({\frac {\theta }{2}}+{\frac {\pi }{4}}\right)=\sec \theta +\tan \theta }$
${\displaystyle {\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}={\frac {|1-\tan {\frac {\theta }{2}}|}{|1+\tan {\frac {\theta }{2}}|}}}$
#### Table
These can be shown by using either the sum and difference identities or the multiple-angle formulae.
Sine Cosine Tangent Cotangent
Double-angle formulae[28][29] {\displaystyle {\begin{aligned}\sin(2\theta )&=2\sin \theta \cos \theta \ \\&={\frac {2\tan \theta }{1+\tan ^{2}\theta }}\end{aligned}}} {\displaystyle {\begin{aligned}\cos(2\theta )&=\cos ^{2}\theta -\sin ^{2}\theta \\&=2\cos ^{2}\theta -1\\&=1-2\sin ^{2}\theta \\&={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}\end{aligned}}} ${\displaystyle \tan(2\theta )={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}$ ${\displaystyle \cot(2\theta )={\frac {\cot ^{2}\theta -1}{2\cot \theta }}}$
Triple-angle formulae[23][30] {\displaystyle {\begin{aligned}\sin(3\theta )\!&=\!-\sin ^{3}\theta \!+\!3\cos ^{2}\theta \sin \theta \\&=-4\sin ^{3}\theta +3\sin \theta \end{aligned}}} {\displaystyle {\begin{aligned}\cos(3\theta )\!&=\!\cos ^{3}\theta \!-\!3\sin ^{2}\theta \cos \theta \\&=4\cos ^{3}\theta -3\cos \theta \end{aligned}}} ${\displaystyle \tan(3\theta )={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}}$ ${\displaystyle \cot(3\theta )\!=\!{\frac {3\cot \theta \!-\!\cot ^{3}\theta }{1\!-\!3\cot ^{2}\theta }}}$
Half-angle formulae[26][27] {\displaystyle {\begin{aligned}&\sin {\frac {\theta }{2}}=\operatorname {sgn}(A)\,{\sqrt {\frac {1\!-\!\cos \theta }{2}}}\\\\&{\text{where}}\,A=2\pi -\theta +4\pi \left\lfloor {\frac {\theta }{4\pi }}\right\rfloor \\\\&\left({\text{or}}\,\,\sin ^{2}{\frac {\theta }{2}}={\frac {1-\cos \theta }{2}}\right)\end{aligned}}} {\displaystyle {\begin{aligned}&\cos {\frac {\theta }{2}}=\operatorname {sgn}(B)\,{\sqrt {\frac {1+\cos \theta }{2}}}\\\\&{\text{where}}\,B=\pi +\theta +4\pi \left\lfloor {\frac {\pi -\theta }{4\pi }}\right\rfloor \\\\&\left(\mathrm {or} \,\,\cos ^{2}{\frac {\theta }{2}}={\frac {1+\cos \theta }{2}}\right)\end{aligned}}} {\displaystyle {\begin{aligned}\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta \\&=\pm \,{\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}\\[8pt]&={\frac {\sin \theta }{1+\cos \theta }}\\[8pt]&={\frac {1-\cos \theta }{\sin \theta }}\\[10pt]\tan {\frac {\eta +\theta }{2}}\!&={\frac {\sin \eta +\sin \theta }{\cos \eta +\cos \theta }}\\[8pt]\tan \left(\!{\frac {\theta }{2}}\!+\!{\frac {\pi }{4}}\!\right)\!&=\!\sec \theta \!+\!\tan \theta \\[8pt]{\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}&={\frac {|1-\tan {\frac {\theta }{2}}|}{|1+\tan {\frac {\theta }{2}}|}}\\[8pt]\tan {\frac {\theta }{2}}\!&=\!{\frac {\tan \theta }{1\!+\!{\sqrt {1\!+\!\tan ^{2}\theta }}}}\\&{\text{for}}\quad \theta \in \left(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right)\end{aligned}}} {\displaystyle {\begin{aligned}\cot {\frac {\theta }{2}}&=\csc \theta +\cot \theta \\&=\pm \,{\sqrt {\frac {1\!+\!\cos \theta }{1\!-\!\cos \theta }}}\\[8pt]&={\frac {\sin \theta }{1\!-\!\cos \theta }}\\[8pt]&={\frac {1\!+\!\cos \theta }{\sin \theta }}\end{aligned}}}
The fact that the triple-angle formula for sine and cosine only involves powers of a single function allows one to relate the geometric problem of a compass and straightedge construction of angle trisection to the algebraic problem of solving a cubic equation, which allows one to prove that trisection is in general impossible using the given tools, by field theory.
A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x3 − 3x + d = 0, where x is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle. However, the discriminant of this equation is positive, so this equation has three real roots (of which only one is the solution for the cosine of the one-third angle). None of these solutions is reducible to a real algebraic expression, as they use intermediate complex numbers under the cube roots.
### Sine, cosine, and tangent of multiple angles
For specific multiples, these follow from the angle addition formulae, while the general formula was given by 16th-century French mathematician François Viète.
{\displaystyle {\begin{aligned}\sin(n\theta )&=\sum _{k{\text{ odd}}}(-1)^{\frac {k-1}{2}}{n \choose k}\cos ^{n-k}\theta \sin ^{k}\theta ,\\\cos(n\theta )&=\sum _{k{\text{ even}}}(-1)^{\frac {k}{2}}{n \choose k}\cos ^{n-k}\theta \sin ^{k}\theta \,,\end{aligned}}}
for nonnegative values of k up through n.
In each of these two equations, the first parenthesized term is a binomial coefficient, and the final trigonometric function equals one or minus one or zero so that half the entries in each of the sums are removed. The ratio of these formulae gives
${\displaystyle \tan(n\theta )={\frac {\sum _{k{\text{ odd}}}(-1)^{\frac {k-1}{2}}{n \choose k}\tan ^{k}\theta }{\sum _{k{\text{ even}}}(-1)^{\frac {k}{2}}{n \choose k}\tan ^{k}\theta }}\,.}$
### Chebyshev method
The Chebyshev method is a recursive algorithm for finding the nth multiple angle formula knowing the (n − 1)th and (n − 2)th values.[31]
cos(nx) can be computed from cos((n − 1)x), cos((n − 2)x), and cos(x) with
cos(nx) = 2 · cos x · cos((n − 1)x) − cos((n − 2)x).
This can be proved by adding together the formulae
cos((n − 1)x + x) = cos((n − 1)x) cos x − sin((n − 1)x) sin x
cos((n − 1)xx) = cos((n − 1)x) cos x + sin((n − 1)x) sin x.
It follows by induction that cos(nx) is a polynomial of cos x, the so-called Chebyshev polynomial of the first kind, see Chebyshev polynomials#Trigonometric definition.
Similarly, sin(nx) can be computed from sin((n − 1)x), sin((n − 2)x), and cos(x) with
sin(nx) = 2 · cos x · sin((n − 1)x) − sin((n − 2)x).
This can be proved by adding formulae for sin((n − 1)x + x) and sin((n − 1)xx).
Serving a purpose similar to that of the Chebyshev method, for the tangent we can write:
${\displaystyle \tan(nx)={\frac {\tan((n-1)x)+\tan x}{1-\tan((n-1)x)\tan x}}\,.}$
### Tangent of an average
${\displaystyle \tan \left({\frac {\alpha +\beta }{2}}\right)={\frac {\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }}=-\,{\frac {\cos \alpha -\cos \beta }{\sin \alpha -\sin \beta }}}$
Setting either α or β to 0 gives the usual tangent half-angle formulae.
### Viète's infinite product
${\displaystyle \cos {\frac {\theta }{2}}\cdot \cos {\frac {\theta }{4}}\cdot \cos {\frac {\theta }{8}}\cdots =\prod _{n=1}^{\infty }\cos {\frac {\theta }{2^{n}}}={\frac {\sin \theta }{\theta }}=\operatorname {sinc} \theta .}$
(Refer to sinc function.)
## Power-reduction formulae
Obtained by solving the second and third versions of the cosine double-angle formula.
Sine Cosine Other
${\displaystyle \sin ^{2}\theta ={\frac {1-\cos(2\theta )}{2}}}$ ${\displaystyle \cos ^{2}\theta ={\frac {1+\cos(2\theta )}{2}}}$ ${\displaystyle \sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos(4\theta )}{8}}}$
${\displaystyle \sin ^{3}\theta ={\frac {3\sin \theta -\sin(3\theta )}{4}}}$ ${\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos(3\theta )}{4}}}$ ${\displaystyle \sin ^{3}\theta \cos ^{3}\theta ={\frac {3\sin(2\theta )-\sin(6\theta )}{32}}}$
${\displaystyle \sin ^{4}\theta ={\frac {3-4\cos(2\theta )+\cos(4\theta )}{8}}}$ ${\displaystyle \cos ^{4}\theta ={\frac {3+4\cos(2\theta )+\cos(4\theta )}{8}}}$ ${\displaystyle \sin ^{4}\theta \cos ^{4}\theta ={\frac {3-4\cos(4\theta )+\cos(8\theta )}{128}}}$
${\displaystyle \sin ^{5}\theta ={\frac {10\sin \theta -5\sin(3\theta )+\sin(5\theta )}{16}}}$ ${\displaystyle \cos ^{5}\theta ={\frac {10\cos \theta +5\cos(3\theta )+\cos(5\theta )}{16}}}$ ${\displaystyle \sin ^{5}\theta \cos ^{5}\theta ={\frac {10\sin(2\theta )-5\sin(6\theta )+\sin(10\theta )}{512}}}$
and in general terms of powers of sin θ or cos θ the following is true, and can be deduced using De Moivre's formula, Euler's formula and the binomial theorem[citation needed].
Cosine Sine
${\displaystyle {\text{if }}n{\text{ is odd}}}$ ${\displaystyle \cos ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}{\binom {n}{k}}\cos {{\big (}(n-2k)\theta {\big )}}}$ ${\displaystyle \sin ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}(-1)^{\left({\frac {n-1}{2}}-k\right)}{\binom {n}{k}}\sin {{\big (}(n-2k)\theta {\big )}}}$
${\displaystyle {\text{if }}n{\text{ is even}}}$ ${\displaystyle \cos ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}{\binom {n}{k}}\cos {{\big (}(n-2k)\theta {\big )}}}$ ${\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{\left({\frac {n}{2}}-k\right)}{\binom {n}{k}}\cos {{\big (}(n-2k)\theta {\big )}}}$
## Product-to-sum and sum-to-product identities
The product-to-sum identities or prosthaphaeresis formulae can be proven by expanding their right-hand sides using the angle addition theorems. See amplitude modulation for an application of the product-to-sum formulae, and beat (acoustics) and phase detector for applications of the sum-to-product formulae.
Product-to-sum[32]
${\displaystyle 2\cos \theta \cos \varphi ={\cos(\theta -\varphi )+\cos(\theta +\varphi )}}$
${\displaystyle 2\sin \theta \sin \varphi ={\cos(\theta -\varphi )-\cos(\theta +\varphi )}}$
${\displaystyle 2\sin \theta \cos \varphi ={\sin(\theta +\varphi )+\sin(\theta -\varphi )}}$
${\displaystyle 2\cos \theta \sin \varphi ={\sin(\theta +\varphi )-\sin(\theta -\varphi )}}$
${\displaystyle \tan \theta \tan \varphi ={\frac {\cos(\theta -\varphi )-\cos(\theta +\varphi )}{\cos(\theta -\varphi )+\cos(\theta +\varphi )}}}$
{\displaystyle {\begin{aligned}\prod _{k=1}^{n}\cos \theta _{k}&={\frac {1}{2^{n}}}\sum _{e\in S}\cos(e_{1}\theta _{1}+\cdots +e_{n}\theta _{n})\\[6pt]&{\text{where }}S=\{1,-1\}^{n}\end{aligned}}}
Sum-to-product[33]
${\displaystyle \sin \theta \pm \sin \varphi =2\sin \left({\frac {\theta \pm \varphi }{2}}\right)\cos \left({\frac {\theta \mp \varphi }{2}}\right)}$
${\displaystyle \cos \theta +\cos \varphi =2\cos \left({\frac {\theta +\varphi }{2}}\right)\cos \left({\frac {\theta -\varphi }{2}}\right)}$
${\displaystyle \cos \theta -\cos \varphi =-2\sin \left({\frac {\theta +\varphi }{2}}\right)\sin \left({\frac {\theta -\varphi }{2}}\right)}$
### Other related identities
• ${\displaystyle \sec ^{2}x+\csc ^{2}x=\sec ^{2}x\csc ^{2}x.}$ [34]
• If x + y + z = π (half circle), then
${\displaystyle \sin(2x)+\sin(2y)+\sin(2z)=4\sin x\sin y\sin z.}$
• Triple tangent identity: If x + y + z = π (half circle), then
${\displaystyle \tan x+\tan y+\tan z=\tan x\tan y\tan z.}$
In particular, the formula holds when x, y, and z are the three angles of any triangle.
(If any of x, y, z is a right angle, one should take both sides to be . This is neither +∞ nor −∞; for present purposes it makes sense to add just one point at infinity to the real line, that is approached by tan θ as tan θ either increases through positive values or decreases through negative values. This is a one-point compactification of the real line.)
• Triple cotangent identity: If x + y + z = π/2 (right angle or quarter circle), then
${\displaystyle \cot x+\cot y+\cot z=\cot x\cot y\cot z.}$
### Hermite's cotangent identity
Charles Hermite demonstrated the following identity.[35] Suppose a1, ..., an are complex numbers, no two of which differ by an integer multiple of π. Let
${\displaystyle A_{n,k}=\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq k\end{smallmatrix}}\cot(a_{k}-a_{j})}$
(in particular, A1,1, being an empty product, is 1). Then
${\displaystyle \cot(z-a_{1})\cdots \cot(z-a_{n})=\cos {\frac {n\pi }{2}}+\sum _{k=1}^{n}A_{n,k}\cot(z-a_{k}).}$
The simplest non-trivial example is the case n = 2:
${\displaystyle \cot(z-a_{1})\cot(z-a_{2})=-1+\cot(a_{1}-a_{2})\cot(z-a_{1})+\cot(a_{2}-a_{1})\cot(z-a_{2}).}$
### Ptolemy's theorem
Ptolemy's theorem can be expressed in the language of modern trigonometry as:
If w + x + y + z = π, then:
{\displaystyle {\begin{aligned}\sin(w+x)\sin(x+y)&=\sin(x+y)\sin(y+z)&{\text{(trivial)}}\\&=\sin(y+z)\sin(z+w)&{\text{(trivial)}}\\&=\sin(z+w)\sin(w+x)&{\text{(trivial)}}\\&=\sin w\sin y+\sin x\sin z.&{\text{(significant)}}\end{aligned}}}
(The first three equalities are trivial rearrangements; the fourth is the substance of this identity.)
### Finite products of trigonometric functions
For coprime integers n, m
${\displaystyle \prod _{k=1}^{n}\left(2a+2\cos \left({\frac {2\pi km}{n}}+x\right)\right)=2\left(T_{n}(a)+{(-1)}^{n+m}\cos(nx)\right)}$
where Tn is the Chebyshev polynomial.
The following relationship holds for the sine function
${\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{n}}\right)={\frac {n}{2^{n-1}}}.}$
More generally [36]
${\displaystyle \sin(nx)=2^{n-1}\prod _{k=0}^{n-1}\sin \left(x+{\frac {k\pi }{n}}\right).}$
## Linear combinations
For some purposes it is important to know that any linear combination of sine waves of the same period or frequency but different phase shifts is also a sine wave with the same period or frequency, but a different phase shift. This is useful in sinusoid data fitting, because the measured or observed data are linearly related to the a and b unknowns of the in-phase and quadrature components basis below, resulting in a simpler Jacobian, compared to that of c and φ.
### Sine and cosine
The linear combination, or harmonic addition, of sine and cosine waves is equivalent to a single sine wave with a phase shift and scaled amplitude,[37][38]
${\displaystyle a\cos x+b\sin x=c\cos(x+\varphi )}$
where c and φ are defined as so:
${\displaystyle c=\operatorname {sgn}(a){\sqrt {a^{2}+b^{2}}},}$
${\displaystyle \varphi =\operatorname {arctan} \left(-{\frac {b}{a}}\right).}$
### Arbitrary phase shift
More generally, for arbitrary phase shifts, we have
${\displaystyle a\sin(x+\theta _{a})+b\sin(x+\theta _{b})=c\sin(x+\varphi )}$
where c and φ satisfy:
${\displaystyle c^{2}=a^{2}+b^{2}+2ab\cos \left(\theta _{a}-\theta _{b}\right),}$
${\displaystyle \tan \varphi ={\frac {a\sin \theta _{a}+b\sin \theta _{b}}{a\cos \theta _{a}+b\cos \theta _{b}}}.}$
### More than two sinusoids
${\displaystyle \sum _{i}a_{i}\sin(x+\theta _{i})=a\sin(x+\theta ),}$
where
${\displaystyle a^{2}=\sum _{i,j}a_{i}a_{j}\cos(\theta _{i}-\theta _{j})}$
and
${\displaystyle \tan \theta ={\frac {\sum _{i}a_{i}\sin \theta _{i}}{\sum _{i}a_{i}\cos \theta _{i}}}.}$
## Lagrange's trigonometric identities
These identities, named after Joseph Louis Lagrange, are:[39][40]
{\displaystyle {\begin{aligned}\sum _{n=1}^{N}\sin(n\theta )&={\frac {1}{2}}\cot {\frac {\theta }{2}}-{\frac {\cos \left(\left(N+{\frac {1}{2}}\right)\theta \right)}{2\sin \left({\frac {\theta }{2}}\right)}}\\[5pt]\sum _{n=1}^{N}\cos(n\theta )&=-{\frac {1}{2}}+{\frac {\sin \left(\left(N+{\frac {1}{2}}\right)\theta \right)}{2\sin \left({\frac {\theta }{2}}\right)}}\end{aligned}}}
A related function is the following function of x, called the Dirichlet kernel.
${\displaystyle 1+2\cos x+2\cos(2x)+2\cos(3x)+\cdots +2\cos(nx)={\frac {\sin \left(\left(n+{\frac {1}{2}}\right)x\right)}{\sin \left({\frac {x}{2}}\right)}}.}$
see proof.
## Other sums of trigonometric functions
Sum of sines and cosines with arguments in arithmetic progression:[41] if α ≠ 0, then
{\displaystyle {\begin{aligned}&\sin \varphi +\sin(\varphi +\alpha )+\sin(\varphi +2\alpha )+\cdots \\[8pt]&{}\qquad \qquad \cdots +\sin(\varphi +n\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\cdot \sin \left(\varphi +{\frac {n\alpha }{2}}\right)}{\sin {\frac {\alpha }{2}}}}\quad {\text{and}}\\[10pt]&\cos \varphi +\cos(\varphi +\alpha )+\cos(\varphi +2\alpha )+\cdots \\[8pt]&{}\qquad \qquad \cdots +\cos(\varphi +n\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\cdot \cos \left(\varphi +{\frac {n\alpha }{2}}\right)}{\sin {\frac {\alpha }{2}}}}.\end{aligned}}}
${\displaystyle \sec x\pm \tan x=\tan \left({\frac {\pi }{4}}\pm {\frac {x}{2}}\right).}$
The above identity is sometimes convenient to know when thinking about the Gudermannian function, which relates the circular and hyperbolic trigonometric functions without resorting to complex numbers.
If x, y, and z are the three angles of any triangle, i.e. if x + y + z = π, then
${\displaystyle \cot x\cot y+\cot y\cot z+\cot z\cot x=1.}$
## Certain linear fractional transformations
If f(x) is given by the linear fractional transformation
${\displaystyle f(x)={\frac {(\cos \alpha )x-\sin \alpha }{(\sin \alpha )x+\cos \alpha }},}$
and similarly
${\displaystyle g(x)={\frac {(\cos \beta )x-\sin \beta }{(\sin \beta )x+\cos \beta }},}$
then
${\displaystyle f{\big (}g(x){\big )}=g{\big (}f(x){\big )}={\frac {{\big (}\cos(\alpha +\beta ){\big )}x-\sin(\alpha +\beta )}{{\big (}\sin(\alpha +\beta ){\big )}x+\cos(\alpha +\beta )}}.}$
More tersely stated, if for all α we let fα be what we called f above, then
${\displaystyle f_{\alpha }\circ f_{\beta }=f_{\alpha +\beta }.}$
If x is the slope of a line, then f(x) is the slope of its rotation through an angle of α.
## Inverse trigonometric functions
{\displaystyle {\begin{aligned}\arcsin x+\arccos x&={\dfrac {\pi }{2}}\\\arctan x+\operatorname {arccot} x&={\dfrac {\pi }{2}}\\\arctan x+\arctan {\dfrac {1}{x}}&={\begin{cases}{\dfrac {\pi }{2}},&{\text{if }}x>0\\-{\dfrac {\pi }{2}},&{\text{if }}x<0\end{cases}}\end{aligned}}}
${\displaystyle \arctan {\frac {1}{x}}=\arctan {\frac {1}{x+y}}+\arctan {\frac {y}{x^{2}+xy+1}}}$ [42]
### Compositions of trig and inverse trig functions
{\displaystyle {\begin{aligned}\sin(\arccos x)&={\sqrt {1-x^{2}}}&\tan(\arcsin x)&={\frac {x}{\sqrt {1-x^{2}}}}\\\sin(\arctan x)&={\frac {x}{\sqrt {1+x^{2}}}}&\tan(\arccos x)&={\frac {\sqrt {1-x^{2}}}{x}}\\\cos(\arctan x)&={\frac {1}{\sqrt {1+x^{2}}}}&\cot(\arcsin x)&={\frac {\sqrt {1-x^{2}}}{x}}\\\cos(\arcsin x)&={\sqrt {1-x^{2}}}&\cot(\arccos x)&={\frac {x}{\sqrt {1-x^{2}}}}\end{aligned}}}
## Relation to the complex exponential function
With the unit imaginary number i satisfying i2 = −1,
${\displaystyle e^{ix}=\cos x+i\sin x}$ [43] (Euler's formula),
${\displaystyle e^{-ix}=\cos(-x)+i\sin(-x)=\cos x-i\sin x}$
${\displaystyle e^{i\pi }+1=0}$ (Euler's identity),
${\displaystyle e^{2\pi i}=1}$
${\displaystyle \cos x={\frac {e^{ix}+e^{-ix}}{2}}}$ [44]
${\displaystyle \sin x={\frac {e^{ix}-e^{-ix}}{2i}}}$ [45]
${\displaystyle \tan x={\frac {\sin x}{\cos x}}={\frac {e^{ix}-e^{-ix}}{i({e^{ix}+e^{-ix}})}}\,.}$
These formulae are useful for proving many other trigonometric identities. For example, that ei(θ+φ) = e e means that
cos(θ+φ) + i sin(θ+φ) = (cos θ + i sin θ) (cos φ + i sin φ) = (cos θ cos φ − sin θ sin φ) + i (cos θ sin φ + sin θ cos φ).
That the real part of the left hand side equals the real part of the right hand side is an angle addition formula for cosine. The equality of the imaginary parts gives an angle addition formula for sine.
## Infinite product formulae
For applications to special functions, the following infinite product formulae for trigonometric functions are useful:[46][47]
{\displaystyle {\begin{aligned}\sin x&=x\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)\\\sinh x&=x\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)\end{aligned}}\ \,{\begin{aligned}\cos x&=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}\left(n-{\frac {1}{2}}\right)^{2}}}\right)\\\cosh x&=\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}\left(n-{\frac {1}{2}}\right)^{2}}}\right)\end{aligned}}}
## Identities without variables
In terms of the arctangent function we have[42]
${\displaystyle \arctan {\frac {1}{2}}=\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}.}$
The curious identity known as Morrie's law,
${\displaystyle \cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 80^{\circ }={\frac {1}{8}},}$
is a special case of an identity that contains one variable:
${\displaystyle \prod _{j=0}^{k-1}\cos(2^{j}x)={\frac {\sin(2^{k}x)}{2^{k}\sin x}}.}$
The same cosine identity in radians is
${\displaystyle \cos {\frac {\pi }{9}}\cos {\frac {2\pi }{9}}\cos {\frac {4\pi }{9}}={\frac {1}{8}}.}$
Similarly,
${\displaystyle \sin 20^{\circ }\cdot \sin 40^{\circ }\cdot \sin 80^{\circ }={\frac {\sqrt {3}}{8}}}$
is a special case of an identity with the case x = 20:
${\displaystyle \sin x\cdot \sin(60^{\circ }-x)\cdot \sin(60^{\circ }+x)={\frac {\sin 3x}{4}}.}$
For the case x = 15,
${\displaystyle \sin 15^{\circ }\cdot \sin 45^{\circ }\cdot \sin 75^{\circ }={\frac {\sqrt {2}}{8}},}$
${\displaystyle \sin 15^{\circ }\cdot \sin 75^{\circ }={\frac {1}{4}}.}$
For the case x = 10,
${\displaystyle \sin 10^{\circ }\cdot \sin 50^{\circ }\cdot \sin 70^{\circ }={\frac {1}{8}}.}$
The same cosine identity is
${\displaystyle \cos x\cdot \cos(60^{\circ }-x)\cdot \cos(60^{\circ }+x)={\frac {\cos 3x}{4}}.}$
Similarly,
${\displaystyle \cos 10^{\circ }\cdot \cos 50^{\circ }\cdot \cos 70^{\circ }={\frac {\sqrt {3}}{8}},}$
${\displaystyle \cos 15^{\circ }\cdot \cos 45^{\circ }\cdot \cos 75^{\circ }={\frac {\sqrt {2}}{8}},}$
${\displaystyle \cos 15^{\circ }\cdot \cos 75^{\circ }={\frac {1}{4}}.}$
Similarly,
${\displaystyle \tan 50^{\circ }\cdot \tan 60^{\circ }\cdot \tan 70^{\circ }=\tan 80^{\circ },}$
${\displaystyle \tan 40^{\circ }\cdot \tan 30^{\circ }\cdot \tan 20^{\circ }=\tan 10^{\circ }.}$
The following is perhaps not as readily generalized to an identity containing variables (but see explanation below):
${\displaystyle \cos 24^{\circ }+\cos 48^{\circ }+\cos 96^{\circ }+\cos 168^{\circ }={\frac {1}{2}}.}$
Degree measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators:
{\displaystyle {\begin{aligned}&\cos {\frac {2\pi }{21}}+\cos \left(2\cdot {\frac {2\pi }{21}}\right)+\cos \left(4\cdot {\frac {2\pi }{21}}\right)\\[10pt]&{}\qquad {}+\cos \left(5\cdot {\frac {2\pi }{21}}\right)+\cos \left(8\cdot {\frac {2\pi }{21}}\right)+\cos \left(10\cdot {\frac {2\pi }{21}}\right)={\frac {1}{2}}.\end{aligned}}}
The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: they are those integers less than 21/2 that are relatively prime to (or have no prime factors in common with) 21. The last several examples are corollaries of a basic fact about the irreducible cyclotomic polynomials: the cosines are the real parts of the zeroes of those polynomials; the sum of the zeroes is the Möbius function evaluated at (in the very last case above) 21; only half of the zeroes are present above. The two identities preceding this last one arise in the same fashion with 21 replaced by 10 and 15, respectively.
Other cosine identities include:[48]
${\displaystyle 2\cos {\frac {\pi }{3}}=1,}$
${\displaystyle 2\cos {\frac {\pi }{5}}\times 2\cos {\frac {2\pi }{5}}=1,}$
${\displaystyle 2\cos {\frac {\pi }{7}}\times 2\cos {\frac {2\pi }{7}}\times 2\cos {\frac {3\pi }{7}}=1,}$
and so forth for all odd numbers, and hence
${\displaystyle \cos {\frac {\pi }{3}}+\cos {\frac {\pi }{5}}\times \cos {\frac {2\pi }{5}}+\cos {\frac {\pi }{7}}\times \cos {\frac {2\pi }{7}}\times \cos {\frac {3\pi }{7}}+\dots =1.}$
Many of those curious identities stem from more general facts like the following:[49]
${\displaystyle \prod _{k=1}^{n-1}\sin {\frac {k\pi }{n}}={\frac {n}{2^{n-1}}}}$
and
${\displaystyle \prod _{k=1}^{n-1}\cos {\frac {k\pi }{n}}={\frac {\sin {\frac {\pi n}{2}}}{2^{n-1}}}}$
Combining these gives us
${\displaystyle \prod _{k=1}^{n-1}\tan {\frac {k\pi }{n}}={\frac {n}{\sin {\frac {\pi n}{2}}}}}$
If n is an odd number (n = 2m + 1) we can make use of the symmetries to get
${\displaystyle \prod _{k=1}^{m}\tan {\frac {k\pi }{2m+1}}={\sqrt {2m+1}}}$
The transfer function of the Butterworth low pass filter can be expressed in terms of polynomial and poles. By setting the frequency as the cutoff frequency, the following identity can be proved:
${\displaystyle \prod _{k=1}^{n}\sin {\frac {\left(2k-1\right)\pi }{4n}}=\prod _{k=1}^{n}\cos {\frac {\left(2k-1\right)\pi }{4n}}={\frac {\sqrt {2}}{2^{n}}}}$
### Computing π
An efficient way to compute π is based on the following identity without variables, due to Machin:
${\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}$
or, alternatively, by using an identity of Leonhard Euler:
${\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}}$
or by using Pythagorean triples:
${\displaystyle \pi =\arccos {\frac {4}{5}}+\arccos {\frac {5}{13}}+\arccos {\frac {16}{65}}=\arcsin {\frac {3}{5}}+\arcsin {\frac {12}{13}}+\arcsin {\frac {63}{65}}.}$
Others include
${\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}};}$ [50][42]
${\displaystyle \pi =\arctan 1+\arctan 2+\arctan 3.}$ [50]
${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}.}$ [42]
Generally, for numbers t1, ..., tn−1 ∈ (−1, 1) for which θn = ∑n−1
k=1
arctan tk ∈ (π/4, 3π/4)
, let tn = tan(π/2 − θn) = cot θn. This last expression can be computed directly using the formula for the cotangent of a sum of angles whose tangents are t1, ..., tn−1 and its value will be in (−1, 1). In particular, the computed tn will be rational whenever all the t1, ..., tn−1 values are rational. With these values,
{\displaystyle {\begin{aligned}{\frac {\pi }{2}}&=\sum _{k=1}^{n}\arctan(t_{k})\\\pi &=\sum _{k=1}^{n}\operatorname {sign} (t_{k})\arccos \left({\frac {1-t_{k}^{2}}{1+t_{k}^{2}}}\right)\\\pi &=\sum _{k=1}^{n}\arcsin \left({\frac {2t_{k}}{1+t_{k}^{2}}}\right)\\\pi &=\sum _{k=1}^{n}\arctan \left({\frac {2t_{k}}{1-t_{k}^{2}}}\right)\,,\end{aligned}}}
where in all but the first expression, we have used tangent half-angle formulae. The first two formulae work even if one or more of the tk values is not within (−1, 1). Note that when t = p/q is rational then the (2t, 1 − t2, 1 + t2) values in the above formulae are proportional to the Pythagorean triple (2pq, q2p2, q2 + p2).
For example, for n = 3 terms,
${\displaystyle {\frac {\pi }{2}}=\arctan \left({\frac {a}{b}}\right)+\arctan \left({\frac {c}{d}}\right)+\arctan \left({\frac {bd-ac}{ad+bc}}\right)}$
for any a, b, c, d > 0.
### A useful mnemonic for certain values of sines and cosines
For certain simple angles, the sines and cosines take the form n/2 for 0 ≤ n ≤ 4, which makes them easy to remember. |
# PROBLEMS ON TRAINS (PART-II)
## PROBLEMS ON TRAINS (PART-II)
#### QUERY 11
A train got accidental after the journey of 50 km. Therefore its speed become 3/4th, so it took 35 more minutes to complete its destination. If accident was taken place before 24 km, then time taken to complete destination was 60 minutes more. What is the speed of the train?
A) 19.2 km/h
B) 19 km/h
C) 20 km/h
D) 20.6 km/h
MAHA GUPTA
Let the speed of the train = x km/h
and the total distance = d km
CASE-I
The trains covers the whole distance in two ways; first 50 km at its normal speed and the rest i.e. (d – 50) km at ¾ of its normal speed i.e. 3x/4 km/h.
Therefore, time taken for the whole journey = 50x + d – 503x/4
= 50x + (d – 50) × 43x
But it equals to 35 minutes more than the usual time
Hence, 50x + (d – 50) × 43x = dx + 7/12 (35 minutes = 35/60 hour = 7/12 hour)
=> 150+4d–2003x = 12d + 7x12x
=> 4d – 7x = 200 —–(i)
CASE-II
Now the accident takes 24 km before, means the trains covers first 50 – 24 = 26 km at its normal speed and the rest (d – 26) at ¾ of its normal speed i.e. 3x/4 km/h
Therefore, time taken for the whole journey = 26x + d–263x/4
= 26x + (d – 26) × 43x
But it equals to 60 minutes i.e. 1 hour more than the usual time
Hence, 26x + (d – 26) × 43x = dx + 1
=> 78+4d–1043x = d+xx
=> d – 3x = 26 —-(ii)
Solving (i) & (ii)
x = 96/5 = 19.2 km/h (option ‘A’)
SHORT
SUMER SINGH CHOUHAN
The difference in time in the two cases = (60 – 35) minutes = 25/60 hr = 5/12 hr
Obviously this difference in time is only because of that 24 km as all other conditions are same.
In the first case the trains runs this 24 km at its original speed and in the second case it runs this 24 km at 75% i.e. 3/4 of its original speed;
Now let the original speed of the train = x km/hr
We know that TIME = DISTANCE/SPEED
Hence, Time taken to run 24 km in the first case therefore = 24/x hr
And time taken to run this 24 km in the second case = 243x/4 = 24 × 43x hr i.e. 32/x hr
Therefore Time taken in the second case – Time taken in the first case = Difference in time
= 32/x – 24/x = 5/12
=> x = 19.2 km/h (option ‘A’)
#### QUERY 12
A train travelling got accidental 50 km after the journey and therefore proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late. Had the accident occurred 24 km further, it would have reached the destination only 15 minutes late. Normal speed of the train?
A) 19.2 km/h
B) 19 km/h
C) 20 km/h
D) 24 km/h
SUMER SINGH CHOUHAN
The difference in time in the two cases = (35 – 15) minutes = 20/60 hr = 1/3 hr
Obviously this difference in time is only because of that 24 km as all other conditions are same.
Obviously, in the first case the trains runs this 24 km at 3/4 of its original speed speed and in the second case it runs this 24 km at its original speed;
Now let the original speed of the train = x km/hr
We know that TIME = DISTANCE/SPEED
Hence, Time taken to run 24 km in the first case therefore = 243x/4 = 24 × 43x hr i.e. 32/x hr
And time taken to run this 24 km in the second case = 24/x hr
Therefore, Time taken in the first case – Time taken in the second case = Difference in time
=> 32/x – 24/x = 1/3
=> x = 24 km/h (option ‘D’)
#### QUERY 12 (A)
A train meets with an accident after travelling 30 km, after which it moves with 4/5 of its original speed and arrives at destination 45 minutes late. Had the accident happened 18 km further on, it would have reached 9 minutes before. Find the distance of the journey and original speed.
A) 120 km, 25km/h
B) 125 km, 25 km/h
C) 130km, 30km/h
D) 120 km, 30km/h
MAHA GUPTA
Let the speed of the train = x km/h
and the total distance = d km
CASE-I
The trains covers the whole distance in two ways; first 30 km at its normal speed and the rest i.e. (d – 30) km at 4/5 of its normal speed i.e. 4x/5 km/h.
Therefore, time taken for the whole journey = 30x + d–304x/5
= 30x + (d – 30) × 54x
But it equals to 45 minutes more than the usual time
Hence, 30x + (d – 30) × 54x = dx + 34 (45 minutes = 45/60 hour = 3/4 hour)
=> 120+5d–1504x = 4d+3x4x
=> d – 3x = 30 —–(i)
CASE-II
Now the accident takes 18 km further, means the trains covers first 30 + 18 = 48 km at its normal speed and the rest (d – 48) at 4/5 of its normal speed i.e. 4x/5 km/h
Therefore, time taken for the whole journey = 48x + d–484x/5
= 48x + (d – 48) × 54x
But it equals to 9 minutes before the time taken earlier, means 45 – 9 = 36 minutes later than the usual time Hence, 48x + (d – 48) × 54x = dx + 35 (36 minutes = 36/60 hour = 3/5 hour)
=> 192+5d–2404x = 5d+3x5x
=> 5d – 12x = 240 —-(ii)
Solving (i) & (ii)
d = 120, x = 30 km/h (option ‘D’)
#### QUERY 13
A train met with an accident 3 hours after starting, which detains it for an hour, after which it proceeds at 75% of its original speed and it arrives the destination 4 hours late. Had the accident taken place 150 km further along the railway line, the train would have arrived only 3½ hours late. The original speed of the train is?
A) 50 km/hr
B) 75 km/hr
C) 60 km/hr
D) 100 km/hr
SUMER SINGH CHOUHAN
The difference in time in the two cases = 4 – 3½ = 1/2 hr
Obviously this difference in time is only because of that 150 km as all other conditions are same.
In the first case the trains runs this 150 km at 75% i.e. 3/4 of its original speed; and in the second case it runs this 150 km at its original speed.
Now let the original speed of the train = x km/hr
We know that TIME = DISTANCE/SPEED
Hence, Time taken to run 150 km in the first case therefore = 1503x/4 = 150 × 43x hr i.e. 200/x hr
And time taken to run this 150 km in the second case = 150/x
Therefore Time taken in first case – Time taken in the second case = Difference in time
= 200x150x = 1/2
=> x = 100 (option ‘D’)
#### QUERY 14
Two trains, one from station A to station B, and the other from B to A start simultaneously. After meeting, the trains reach their destination after 9 hours and 16 hours respectively. The ratio of their speeds is?
A) 2:3
B) 4:3
C) 6:7
D) 9:16
MAHA GUPTA
Suppose both the trains meet at R
And let first train’s speed be x km/h and the second train’s speed be y km/h
Then AR = 9x kilometers and BR = 16y kilometers
Now the time taken by the first train = 16yx —-Distance covered by the second train/Speed of the first train
and the time taken by the other train = 9xy —-Distance covered by the first train/Speed of the second train
It’s obvious that these two times are equal
Therefore 16y= 9xy
=> 16y² = 9x²
=> = 16/9
=> xy = 4/3
=> x : y = 4 : 3 (option ‘B’)
#### QUERY 15
Two guns were fired from the same place at the difference of 26 minutes. A person sitting in a train coming towards the same direction hears the second sound after 25 minutes. If speed of the sound is 350 m/s, find the speed of the train.
A) 14 m/s
B) 60 m/s
C) 80 m/s
D) 40 m/s
MAHA GUPTA
Let the speed of the train be x m/m
Obviously, the distance traveled by train in 25 minutes = distance traveled by sound in 26 minutes – 25 minutes i.e. 1 minute
Therefore according to above; x*25 = 350*1
or x = 350/25 = 14 (option ‘A’)
#### QUERY 16
Two trains A and B are 110 km apart on a straight line. One train starts at 8 am and travels toward the train B at 40 km/h. Another train starts at 10 am and travels towards the train A at 50 km/h. At what time will they meet?
A) 10:30 am
B) 11:00 am
C) 10:20 am
D) 10:40 am
MAHA GUPTA
Train A travels alone for 2 hours
So distance travelled by it in 2 hours at 40 km/h = 2*40 = 80 km
Thus the remaining distance = 110 – 80 = 30 km
Obviously this distance of 30 km has to be travelled by both the trains running together in the opposite direction
Here you can do the sum by finding the relative speed of both the trains
The relative speed here = 40 + 50 = 90 km/h
Time taken to meet = DISTANCE/SPEED = 30/90 hours = 20 minutes
So the time at which both the trains will meet = 10:20 am (option ‘C’)
#### QUERY 17
Two trains A and B start from Howrah and Patna towards Patna and Howrah respectively at the same time. After passing each other they take 4h 48m and 3h 20m to reach Patna & Howrah respectively. If the train from Howrah is moving at 45 km/hr, then the speed of the other train is?
A) 45 km/h
B) 54 km/h
C) 40 km/h
D) 60 km/h
MAHA GUPTA
Suppose both the trains meet at R and let the second train’s speed be y km/h
Then AR = (24/5)45 = 216 kilometers and BR = (10/3)y kilometers [4 hours 48 minutes = 24/5 hours; and 3 hours 20 minutes = 10/3 hours]
Now time taken by the first train = [(10/3)y]/45 —-Distance covered by the second train/speed of the first train
and time taken by the other train = 216/y —-Distance covered by the first train/speed of the second train
It’s obvious that these two times are equal
Therefore [(10/3)y]/45 = 216/y
=> 2y/27 = 216/y
=> y = (108*27)/y
=> y² = 108*27
=> y² = 2*2*27*27
=> y = 54
Hence the speed of the second train = 54 km/h (option ‘B’)
#### QUERY 18
Without stoppages a train travels a certain distance with an average speed of 60 km/h and with stoppage it covers the same distance with the average speed of 40 km/h. On an average how many minutes per hour does the train stops during the journey?
A) 20 minutes
B) 18 minutes
C) 25 minutes
D) 15 minutes
MAHA GUPTA
Due to stoppages, the train covers 20 km/h less
Therefore time taken to cover that 20 km in minutes = (20/60)60 = 20 minutes (option ‘A’)
#### QUERY 19
Two trains start running from the same place on parallel lines in the same direction at speed of 45 km/h and 40 km/h respectively. Find the distance between them after 45 minutes.
A) 10/3 km
B) 15/4 km
C) 12/5 km
D) 15/7 km
MAHA GUPTA
Distance covered by the first train in 60 minutes = 45 km
Therefore distance covered by it in 45 minutes = (45/60)45 = 135/4 km
Now distance covered by the other train in 60 minutes = 40 km
Therefore distance covered by it in 45 minutes = (40/60)45 = 30 km
Hence the distance between them = 135/4 – 30 = 15/4 km (answer)
TRICK
To find the distance between them, find their relative speed and do
Now their relative speed = 45 – 40 = 5 km/h
Difference between them in 60 minutes = 5 km
Hence the distance between them in 45 minutes = (5/60)45 = 15/4 km (option ‘B’)
#### QUERY 20
A train crosses a 80 meter long platform in 20 seconds and a man standing on the platform crosses it in 12 seconds. Find the length of the train.
A) 100 meter
B) 110 meter
C) 120 meter
D) 200 meter
MAHA GUPTA
To remember:
To cross a pole or a standing man, a train is required to cover distance equal to its own length, and to cross a platform or any other stationary object of some length it requires to cover distance equal to the sum of its own length and the length of the platform or the object.
Now, let the length of the train = x meters
We know that SPEED = DISTANCE/TIME
Therefore the speed of train in case when it crosses the man = x/12 seconds per meter
And the speed in case when it crosses the platform = (x + 80)/20
As the speed in both the cases must be equal, therefore
x/12 = (x + 80)/20
=> x = 120
So the length of the train = 120 meters (option ‘C’)
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# Find the Number of Small Painted Cubes
### Problem Statement
A cube is painted with some colour on all faces. Now, we cut it into 1000 small cubes of equal size. How many small cubes are not painted?
### Approach 1
One basic approach is to count the number of painted small cubes and subtract it from the total number of small cubes. Note: While solving such questions, the most important part is to visualise the cube.
If we visualise the large cube, all smaller cubes will have at least one face facing inside. So none of the smaller cubes will have all faces painted. On the other side, the maximum number of faces of the larger cube that intersect at a point = 3 (at the corners). So, the smaller cubes can have a maximum of 3 faces painted. In other words, there are three types of painted small cubes.
#### Cubes painted on 3 sides
These smaller cubes are located at the corners of the large cube. The number of smaller cubes with 3 faces painted = The number of corners in the larger cube = 8.
#### Cubes painted on 2 sides
To find the number of smaller cubes with only 2 faces painted, we need to consider the cubes where 2 faces of the larger cube meet, i.e. the edges. So these smaller cubes are positioned on the edges of the large cube.
Remember, each edge includes the smaller cubes present at the corners as well, which are painted on 3 sides. So we need to remove those 2 cubes from the number of cubes on each edge. So the number of smaller cubes with 2 faces painted at each edge = n — 2 (Here n is the length of each side of the cube).
Overall, there will be 12 such edges on the larger cube. So the number of smaller cubes with 2 faces painted = 12*(n — 2) = 12 * 8 = 96. (Here n = 10).
#### Cubes painted on 1 side
These smaller cubes are located at the face of the larger cube, excluding the cubes at corners and edges. At each face, the number of such cubes = (n — 2) * (n — 2). There are 6 faces of larger cubes, so the total number of smaller cubes with one face painted = 6 * (n — 2) * (n — 2) = 6 * 8* 8 = 384.
From the above analysis, total number of painted cubes = 8 + 12*(n — 2) + 6 * (n — 2) * (n — 2) = 8 + 96 + 384 = 488. The number of cubes not painted = 1000 – 488 = 512.
### Approach 2
In n x n x n cube, if we remove the outer layer of all 1 x 1 x 1 painted small cubes, then the dimensions of the hidden cube (not painted) will be (n — 2) x (n — 2) x (n — 2). So number of small 1 x 1 x 1 cubes not painted = (n — 2)³ = 8³ = 512.
### Critical idea to think!
Suppose we have a cuboid of dimension a*b*c painted on all sides which is cut into smaller cubes of dimension 1*1*1. Then:
• Number of cubes with 0 sides painted = (a — 2) (b — 2) (c — 2)
• Number of cubes with 1 sides painted = 2[(a — 2) (b — 2) + (b — 2)(c — 2) + (a — 2)(c — 2) ]
• Number of cubes with 2 sides painted = 4(a + b + c — 6)
• Number of cubes with 3 sides painted = 8
Enjoy learning, Enjoy mathematics!
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Module 23 - Sequences and Series
Introduction | Lesson 1 | Lesson 2 | Lesson 3 | Self-Test
Lesson 23.2: Series and Sequences of Partial Sums
This lesson explores series and partial sums of infinite series. Series are used in many applications including integration, approximation, and the solution of differential equations. These applications arise in many disciplines, especially physics and chemistry.
Defining a Series
A series, which is not a list of terms like a sequence, is the sum of the terms in a sequence. If the series has a finite number of terms, it is a simple matter to find the sum of the series by adding the terms. However, when the series has an infinite number of terms the summation is more complicated and the series may or may not have a finite sum.
Defining Partial Sums
A partial sum of an infinite series is the sum of a finite number of consecutive terms beginning with the first term. When working with infinite series, it is often helpful to examine the behavior of the partial sums.
Suppose an infinite sequence is defined by . The terms of the sequence are
.
The first four partial sums of the associated infinite series are computed below, where sk represents the sum of the first k terms of the sequence.
Each of the results shown above is a partial sum of the series which is associated with the sequence .
Defining the Sequence of Partial Sums of a Series
The partial sums of a series form a new sequence, which is denoted as {s1, s2, s3, s4,...}. For the series given above, the sequence of partial sums is .
If the sequence of partial sums for an infinite series converges to a limit L, then the sum of the series is said to be L and the series is convergent. Otherwise, the infinite series does not have a sum and it is divergent.
Finding Partial Sums of Series
The TI-83's sum( function and the Sequence Graphing mode are useful tools in understanding the sequence of partial sums of series.
Using the sum( Function
Partial sums can be computed with the sum function and may be used to help explore whether or not an infinite series converges. If it is convergent, the partial sums can also help estimate the sum of the series.
The sum( Function The sum( function returns the sum of the terms of any list. The syntax is sum(list). If the seq( command is being used to generate the list, then the sum of the list can be found with the following syntax sum(seq (expression, var, min, max, step))
Investigate whether converges.
• Open the LIST MATH menu by pressing [LIST] .
"5: sum(" is the fifth item in this menu.
• Select "5:sum(" to paste the "sum(" command to the Home screen.
Beginning with Zero
Explore the partial sums of the series . The first term in this series corresponds to
k = 0, and that term is the first partial sum as well. The second partial sum is found by adding the first two terms, corresponding to k = 0 and k = 1. In general, the nth partial sum is found by adding the terms corresponding to k = 0, 1, ..., (n - 1).
• Complete the command sum(seq((2^K + 5)/3^K, K , 0 , 4)).
Recall seq( is in the LIST OPS menu.
This result is called the fifth partial sum because the first five terms corresponding to k = 0, 1, 2, 3, 4 were added.
23.2.1 Approximate the tenth partial sum of the infinite series .
Graphing the Sequence of Partial Sums
The graph of the sequence of partial sums for the infinite series can be created by defining the sequence of partial sums in the Y= Editor in Sequence Graph mode.
• Define u(n) = sum(seq((2^K+5)/3^K, K, 0, n)) in the Y= Editor.
• Recall that n is the independent variable in Sequence Graphing mode and it can be entered using the key.
• Use the following window values:
nMin = 0 Xmin = -1 Ymin = -1 nMax = 10 Xmax = 11 Ymax = 15 PlotStart = 1 Xscl = 1 Yscl = 1 PlotStep = 1
• Graph the sequence of partial sums by pressing
Each point of this graph represents a partial sum.
• Trace to the tenth partial sum (n = 9).
This should be the same value for the tenth partial sum that you computed earlier on the Home Screen.
23.2.2 The graph of the sequence of partial sums appears to level off. What does this suggest?
Creating a Table of Values for the Partial Sums
Make a table to show the partial sums.
• Open the Table Setup dialog box by pressing [TBLSET].
• Set TblStart = 0 and Tbl = 1.
• Open the table by pressing [TABLE]
You can compute partial sums of your choosing by returning to the Table Setup dialog box and selecting the Ask mode.
• Open the Table Setup dialog box by pressing [TBLSET].
• On the line labeled "Indpnt" highlight "Ask."
• Enter 9 for n in the first row and press .
• Move the cursor to the next row and enter 15 for n.
• Move the cursor to the next row and enter 25 for n.
• Move the cursor to the next row and enter 50 for n.
The table provides further evidence that the series converges.
It is impossible to determine for certain whether or not a series converges and/or to determine the sum of a series by exploring partial sums. Many divergent infinite series may appear to converge and have a sum when studied using the methods discussed so far in this lesson.
Tests for Convergence
Both the graph and table for the series of partial sums give some evidence a series converges. However, analytic methods must be used to be certain that a series converges. You should consult a calculus text for descriptions of tests for convergence and divergence for infinite series.
23.2.3 The ratio test is an analytic test for convergence and it shows that the series being discussed, , converges. To what value do you think the series converges? |
# Finding the roots of a function applied twice
I saw this problem a few days ago and still haven't cracked it. Therefore, I thought I might ask you all for help.
The graph of $$f(x)$$ has four roots on the interval $$[-5,5]$$. How many different roots does $$f(f(x))$$ have? Here is a picture of the graph of $$f(x)$$:
We stumbled across this problem a week ago during math club and we were unable to solve it; our teacher didn't know the answer, either. Any help would thus be greatly appreciated!
• $x$ is a root of $f\circ f$ if $f(x)$ is a root of $f$. so: $f(x)$ should be $-4$, $-2$, $2$ or $4$ Mar 15, 2021 at 19:25
The roots of $$f(x)$$ are $$x=-4,-2,2,4$$, according to your graph, so for $$f(f(x)) = 0$$, we have $$f(x) = -4,-2,2,4$$ because $$f(x)$$ took the place of $$x$$. So, we need to examine when $$f(x) = -4,-2,2,4$$.
From the graph we see that $$f(x)$$ is never $$-4$$
From the graph we see that $$f(x)=-2$$ at $$2$$ values of $$x$$.
From the graph we see that $$f(x) = 2$$ at $$4$$ values of $$x$$
From the graph we see that $$f(x) = 4$$ at $$2$$ values of $$x$$
Thus, that makes $$8$$ values of $$x$$ total that $$f(f(x)) = 0$$, so it has $$8$$ roots.
Notice that $$f(f(x)) = 0$$ if and only if $$f(x) \in S=\{-4,-2,2,4\} ,$$ since $$S$$ is the set of the roots of $$f$$.
So you should find how many times $$f(x) = y$$, with $$y$$ being a value in $$S$$. I suggest to draw four horizontal lines, respectively on $$y=-4$$ (which lies outside the graphics), $$y=-2$$, $$y=2 and$$y=4$. You can notice that the line $$y=-4$$ has zero intersections with the graph of the function $$f$$; the line $$y=-2$$ has 2 intersections; the line $$y=2$$ has 4 intersections and the line $$y=4$$ has other 2 intersections$.
The equation $$f(f(x)) = 0$$ has thus 8 solutions. |
# How do you use Heron's formula to determine the area of a triangle with sides of that are 25, 28, and 22 units in length?
Mar 10, 2016
Area of triangle is $262.72$ units
#### Explanation:
Heron's formula gives the area of a triangle with sides $a$, $b$ and $c$ as $\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ where $s = \frac{1}{2} \left(a + b + c\right)$.
Hence to determine the area of a triangle with sides of $25 , 28$ and $22$ units, first we find $s$, which is given by $s = \frac{1}{2} \left(25 + 28 + 22\right) = 37.5$.
Hence area of triangle is
$\sqrt{37.5 \times \left(37.5 - 25\right) \times \left(37.5 - 28\right) \times \left(37.5 - 22\right)}$ or
$\sqrt{37.5 \times 12.5 \times 9.5 \times 15.5}$ or $\sqrt{69 , 023.4375}$ or $262.72$ units
May 1, 2018
$16 {A}^{2} = \left(25 + 28 + 22\right) \left(- 25 + 28 + 22\right) \left(25 - 28 + 22\right) \left(25 + 28 - 22\right)$$= \left(75\right) \left(25\right) \left(19\right) \left(31\right)$
$A = \setminus \sqrt{\frac{3 \left(25\right) \left(25\right) \left(19\right) \left(31\right)}{16}} = \frac{25}{4} \sqrt{1767}$
#### Explanation:
Heron's formula is usually among the poorest choices. Here are some modern forms:
For a triangle with sides $a , b , c$ and area $A$,
$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2}$
$= {\left({a}^{2} + {b}^{2} + {c}^{2}\right)}^{2} - 2 \left({a}^{4} + {b}^{4} + {c}^{4}\right)$
$= \left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right)$
$= 16 s \left(s - a\right) \left(s - b\right) \left(s - c\right)$ where $s = \frac{1}{2} \left(a + b + c\right)$
The last is of course Heron.
Usually given integer sides the factored form is fast and leads to the least messy exact answer:
$16 {A}^{2} = \left(25 + 28 + 22\right) \left(- 25 + 28 + 22\right) \left(25 - 28 + 22\right) \left(25 + 28 - 22\right)$
$= \left(75\right) \left(25\right) \left(19\right) \left(31\right)$
$A = \setminus \sqrt{\frac{3 \left(25\right) \left(25\right) \left(19\right) \left(31\right)}{16}} = \frac{25}{4} \setminus \sqrt{3 \left(19\right) \left(31\right)} = \frac{25}{4} \sqrt{1767}$
If we're given 2d coordinates, the Shoelace Theorem is the quickest way to the area. If we're given 3 or more dimensional coordinates, the first two forms, which rely only on squared lengths, are best.
The first form is great for deriving the special cases as well:
$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2}$
Equilateral triangle : $a = b = c$
$16 {A}^{2} = 4 {a}^{4} - {a}^{4} = 3 {a}^{4} \quad$ or $\quad A = \setminus \frac{\sqrt{3}}{4} {a}^{2}$
Isosceles triangle: $a = c ,$ base $b .$
$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {b}^{4} = {b}^{2} \left(4 {a}^{2} - {b}^{2}\right) \mathmr{and} A = \frac{b}{4} \setminus \sqrt{4 {a}^{2} - {b}^{2}}$
Right Triangle ${c}^{2} = {a}^{2} + {b}^{2}$
$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2} = 4 {a}^{2} {b}^{2} - 0$ or $A = \frac{1}{2} a b$
Cool. I'm not sure why they don't teach it in school. |
# Ten Frames Part 2: Addition and subtraction
by C. Elkins, OK Math and Reading Lady
Last week’s focus was on using ten frames to help with students’ number sense and conceptual development of number bonds for amounts 1-10. This post will feature ways to use ten frames to enhance students’ understanding of addition and subtraction. Look for freebies and a video!
There are many addition and subtraction strategies to help students memorize the basic facts such as these below. The ten frame is a very good tool for students of all grade levels to make these strategies more concrete and visual. I will focus on some of these today.
• add or take away 1 (or 2)
• doubles, near doubles
• facts of 10
• make a ten
• add or sub. 10
• add or sub. 9
• add or sub. tens and ones
Doubles and near doubles (doubles +1, -1, +2, or -2): If the doubles are memorized, then problems near doubles can be solved strategically.
• Show a doubles fact on a single ten frame (for up to 5 + 5). Use a double ten-frame template for 6 + 6 and beyond.
• With the same doubles fact showing, show a near doubles problem. This should help students see that the answer is just one or two more or less.
• Repeat with other examples.
• Help student identify what a doubles + 1 more (or less) problem looks like. They often have a misconception there should be a 1 in the problem. Make sure they can explain where the “1” does come from. Examples: 7 + 8, 10+11, 24+25, 15 +16, etc.
• For subtraction, start with the doubles problem showing and turn over the 2-color counters or remove them.
Facts of 10: These are important to grasp for higher level addition / subtraction problems as well as rounding concepts. Continue reading
# Ten Frames Part 1: Number Sense
by C. Elkins, OK Math and Reading Lady
The focus in this post will be an introduction to ten frames and ways they can help your students gain number sense. Then stay tuned because ten frames can also be a great tool for addition, subtraction, multiplication, and division.
Subitizing: This is the ability to recognize an amount without physically counting. Looking at the picture of red counters: If the top row is full, does the student automatically know there are 5? Doing a Number Talk is a great way to practice subitizing using a ten frame:
• Use your own or pre-made dot cards. Flash the card for 1-2 seconds. Observe students. Are any of them trying to point and count? Or do they seem to know right away? Here’s a great video I recommend: KG Number Talk with ten frames
• Tell students to put their thumb in front of their chest (quietly) to signal they know how many there are.
• Ask a few students to name the amount.
• Then ask this very important question, “How did you know?”
• For the top picture you might hope a child says, “I knew there were 5 because when the top row is full, there are 5.”
• For the bottom picture, you might hope for these types of responses: “I saw 4 (making a square) and 1 more.” or “I saw 3 and 2 more.” or “I pictured the 2 at the bottom moving up to the top row and filling it up, which is 5.”
The idea is to keep building on this.
• What if I showed 4 in the top row? Can the student rationalize that it was almost 5? Do they see 2 and 2?
• What if I showed 5 in the top row and 1 in the bottom row? Can the student think “5 and 1 more is 6?”
Here are some resources you might like to help with subitizing using ten frames.
Number Bonds: Using ten frames to illustrate number bonds assists students with composing and decomposing numbers. Students then see that a number can be more than a counted amount or a digit on a jersey or phone number. Here is an example of number bonds for 6:
• 6 is 5 and 1 (or 1 and 5).
• 6 is 4 and 2 (or 2 and 4).
• 6 is 6 and 0 (or 0 and 6).
• 6 is 3 and 3.
Teaching strategies for number bonds using ten frames: Continue reading
# All About 10: “Make a 10” and “Adding Up”
by C. Elkins, OK Math and Reading Lady
Last time I focused on some basics about learning the number bonds (combinations) of 10 as well as adding 10 to any number. Today I want to show the benefits of making a 10 when adding numbers with sums greater than 10 (such as 8 + 5). Then I’ll show how to help students add up to apply that to addition and subtraction of larger numbers. I’ll model this using concrete and pictorial representations (which are both important before starting abstract forms).
Using a 10 Frame:
A ten frame is an excellent manipulative for students to experience ways to “Make a 10.” I am attaching a couple of videos I like to illustrate the point.
Let’s say the task is to add 8 + 5:
• Model this process with your students using 2 ten frames.
• Put 8 counters on one ten frame. (I love using 2-color counters.)
• Put 5 counters (in another color) on the second ten frame.
• Determine how many counters to move from one ten frame to the other to “make a 10.” In this example, I moved 2 to join the 8 to make a 10. That left 3 on the second ten frame. 10 + 3 = 13 (and 8 + 5 = 13).
The example below shows the same problem, but this time move 5 from the first ten frame to the second ten frame to “make a 10.” That left 3 on the first ten frame. 3 + 10 = 13 (and 8 + 5 = 13). Continue reading
# All About 10: Fluency with addition and subtraction facts
by C. Elkins, OK Math and Reading Lady
I’m sure everyone would agree that learning the addition / subtraction facts associated with the number 10 are very important. Or maybe you are thinking, aren’t they all important? Why single out 10? My feelings are that of all the basic facts, being fluent with 10 and the combinations that make 10 enable the user to apply more mental math strategies, especially when adding and subtracting larger numbers. Here are a few of my favorite activities to promote ten-ness! Check out the card trick videos below – great way to get kids attention, practice math, and give them something to practice at home. Continue reading |
# Radha made a picture of an airplane with colored paper as shown in Figure. Find the total area of the paper used.
To Find: Total area of the paper used.
Given: Picture of the airplane
Concept Used:
If sides of a triangle are a, b, and c, then area of a triangle is given by:
Where s = semiperimeter of the triangle
Area of Rectangle = length × breadth
Area of trapezium = 1/2 × height × sum of parallel sides
Area of right-angled triangle = 1/2 base × height
Diagram:
Explanation:
As we can see the airplane is divided into 5 parts. Let us calculate the area of each part.
Part I.
This is a triangle with sides,
a = 5 cm,
b = 5 cm, and
c = 1 cm
Part II.
Now, this is a rectangle with,
Length = 1 cm
Area of rectangle = 1 × 6.5 cm2 = 6.5 cm2
Part III.
Diagram:
In Δ AED, applying Pythagoras theorem, we get,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
(1)2 = (1/2)2 + (height)2
This part is a trapezium with,
Sum of parallel sides = (1 + 2) cm = 3 cm
Part IV.
This is a right-angled triangle with,
Base = 1.5 cm
Height = 6 cm
Area = (1/2 × 1.5 × 6) cm2
Area = 4.5 cm2
Part V.
This is a right-angled triangle with,
Base = 1.5 cm
Height = 6 cm
Area = (1/2 × 1.5 × 6) cm2
Area = 4.5 cm2
Total Area = Area I + Area II + Area III + Area IV + Area V
Let us put the value of √11 = 3.31 and √3 = 1.73, we have now,
Total Area = 0.75 × 3.31 + 6.5 + 0.75 × 1.73 + 4.5 + 4.5
Total Area = 2.49 + 6.5 + 1.3 + 9
Total Area = 19.29 cm2 = 19.3 cm2
Hence, total area of paper needed is 19.3 cm2.
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# Y-Wing Strategy for Sudoku
I recently got a new Sudoku app that produces really hard Sudoku's, which can't be solved using the standard strategies. So I had to learn a few new ones. One of these strategies is the Y-Wing Strategy. It is ranked under "Tough Strategies", but it is actually not that hard.
# Example
For this strategy only 4 cells are important. Therefore I ignored all other cells in the images.
We look at all candidates for each cell. In the following example we have a cell with the candidates 3 7 (that means we have already rejected the candidates 1 2 4 5 6 8 9, for instance because there is a 1 in the same row, a 2 in the same 3x3 box, ...), a cell with the candidates 6 7, a cell with the candidates 3 6 and a cell with the candidates 2 6. The Y-Wing strategy will suggest, that the 6 can be removed from the candidates of the downright cell, leaving only a 2 as candidate, which you can fill in. So we found a correct number and are one step closer in solving the full Sudoku.
Why can the 6 be removed?
## Explanation
Let us assume that the 6 is the correct number for the downright cell. Now there is a 6 in this column, therefore we can remove the 6 from the candidates of the top right cell, leaving only a 7, which we can fill in. The same happens with the down left cell. We can remove the 6 and fill in the 3. Now if we look at the top left cell we get a contradiction. Because now there's already a 7 in the same row and a 3 in the same column, so we can remove the 7 and the 3 of the candidates, leaving no candidates at all. Which is clearly not possible. Therefore the 6 cannot be the correct number of the downright cell.
More precisely: If we have 4 cells with the candidates [A B] [A C] [C D] [B C] (in this order or circular rotated) and the cells are connected (via same row, same column or same 3x3 box) in a circle (Cell 1 is connected to Cell 2, which is connected to Cell 3, which is connected to Cell 4, which is connected to Cell 1), than you can remove C from the [C D] cell. It is crucial, that the three cells [A B], [A C] and [B C] only contain two candidates each. Differently the cell [C D], which may contain more or less (D can be zero, one or even more candidates).
Notice that I explicitly said they can be connected either way. In the next example you can see the strategy applied again. But this time the 4 cells doesn't form a rectangle. The down left and downright cells are simply connected, because they are in the same 3x3 box. Y-Wing says, that we can remove the 1 as candidate of the top left cell. This time there are still 2 candidates in this cell left, so we didn't actually found a new correct number. But nevertheless the removal of the 1 can open doors to different strategies.
## Challenge Specifications
You will receive 4 lists as input, representing the candidates of the cells. The four cells are connected like a circle (Cell 1 is connected to Cell 2, which is connected to Cell 3, which is connected to Cell 4, which is connected to Cell 1). You can assume that each list is sorted in ascending order.
Your job is to remove one candidate (by applying the Y-Wing strategy) and returning the resulting candidates lists in the same order. If you can't apply the strategy, just return the same candidates lists.
If there are two possible solutions (you could remove A of cell B or remove C of cell D), than return only one solution. It doesn't matter which one.
The input can be in any native list or array format. You could also use a list of lists or something similar. You can receive the input via STDIN, command-line argument, prompt or function argument and return the output via return value or simply by printing to STDOUT.
This is code-golf. The shortest code (in bytes) wins.
## Test Cases
[3 7] [6 7] [2 6] [3 6] => [3 7] [6 7] [2] [3 6] # Example 1
[6 7] [2 6] [3 6] [3 7] => [6 7] [2] [3 6] [3 7] # Example 1, different order
[2 6] [3 6] [3 7] [6 7] => [2] [3 6] [3 7] [6 7] # Example 1, different order
[3 6] [3 7] [6 7] [2 6] => [3 6] [3 7] [6 7] [2] # Example 1, different order
[1 2 8] [1 8] [8 9] [1 9] => [2 8] [1 8] [8 9] [1 9] # Example 2
[3 8] [4 8] [3 4 8] [3 4] => [3 8] [4 8] [3 8] [3 4]
[1 3 6 7 8] [3 8] [3 4] [4 8] => [1 3 6 7] [3 8] [3 4] [4 8]
[7 8] [7 8] [4 7] [4 8] => [7 8] [8] [4 7] [4 8] or [7] [7 8] [4 7] [4 8]
[4 7] [7 8] [4 8] [4] => [4 7] [7 8] [4 8] [] # Fictional example
[3 7] [2 6] [6 7] [3 6] => [3 7] [2 6] [6 7] [3 6] # Y-Wing can't be applied here
[4 7] [2 7 8] [4 8] [1 4] => [4 7] [2 7 8] [4 8] [1 4] # -||-
• Can multiple sets in a single input be exactly same ? – Optimizer Apr 25 '15 at 17:46
• @Optimizer Yes, for instance in the 8th test case, 7 8 is are the candidates for the first and the second cell. The Y-Wing strategy can still be applied. – Jakube Apr 25 '15 at 17:48
• @Jakube ah okay, didn't see that. – Optimizer Apr 25 '15 at 17:52
• If more than 1 solutions are possible, can I output any one of them ? – Optimizer Apr 25 '15 at 18:32
• Yes, I clarified it in the question. – Jakube Apr 25 '15 at 19:02
# CJam, 90 bytes
Ugghhh, this is too long because of the constraint that the other 3 cells should have only 2 candidates.
l~_:_(a+2/::&_{,}$2>:&:Y;{:PY&Y{P1<}?~}%:X,3>1${,}$W=_,2>\Y&,1?*{X:_(+2/{~:I=}#)_2$=I-t}&p
This expects input as a list of list in CJam format. For ex.:
[[2 6] [3 6] [3 7] [6 7]]
gives output in a CJam list of list format:
[[2] [3 6] [3 7] [6 7]]
Will add explanation once I am done golfing..
# Mathematica, 124 110 bytes
Cases[e@n_:>n]/@(Plus@@e/@#&/@#/.NestList[RotateLeft/@#&,{x:a_+b_,y:b_+c_,z:c_+a_,w:a_+_.}->{x,y,z,w-a+1},3])&
Examples:
In[1]:= yWing = Cases[e@n_:>n]/@(Plus@@e/@#&/@#/.NestList[RotateLeft/@#&,{x:a_+b_,y:b_+c_,z:c_+a_,w:a_+_.}->{x,y,z,w-a+1},3])& ;
In[2]:= yWing[{{3, 7}, {6, 7}, {2, 6}, {3, 6}}]
Out[2]= {{3, 7}, {6, 7}, {2}, {3, 6}}
In[3]:= yWing[{{4, 7}, {7, 8}, {4, 8}, {4}}]
Out[3]= {{4, 7}, {7, 8}, {4, 8}, {}} |
An Isosceles Right Triangle is a right triangle that consists of two equal length legs. Since the two legs of the right triangle are equal in length, the corresponding angles would also be congruent. Thus, in an isosceles right triangle, two legs and the two acute angles are congruent.
Can a right triangle be scalene and isosceles? can a right triangle be isosceles.
Can a right triangle also be an isosceles triangle Why?
Explanation: To be a right triangle, one of the angles has to be 90 degrees. This means that the two remaining angles have to be 90 degrees when summed up. … That means that we can only have a right triangle, that also is a isosceles triangle, when the degrees are 45, 45 and 90.
Can a triangle be an isosceles and right triangle at the same time?
It is possible for a triangle to be a right-angled triangle and an isosceles triangle at the same time. In this case the angles would be 90°, 45° and 45°.
What triangle can never have a right angle?
Since a right-angled triangle has one right angle, the other two angles are acute. Therefore, an obtuse-angled triangle can never have a right angle; and vice versa. The side opposite the obtuse angle in the triangle is the longest.
How do you prove a right isosceles triangle?
1. In a triangle ABC , AC is hypotenuse, AB is altitude and BC is base. AB is equal to BC.
2. Angle ABC is 90 degree.
3. Angle BAC and angle BCA are 45 degrees.
4. Hence triangle ABC is a right isosceles triangle.
Can a triangle have two right angles?
There can never be two or more right angles in a triangle. A triangle has three sides, and the interior angles add up to 180 degrees.
Can a right angled triangle be isosceles and scalene?
Note: It is possible for a right triangle to also be scalene or isosceles. An obtuse triangle has one angle measuring more than 90º but less than 180º (an obtuse angle). It is not possible to draw a triangle with more than one obtuse angle. Note: It is possible for an obtuse triangle to also be scalene or isosceles.
Which of these triangle Cannot be isosceles?
A triangle that is not isosceles (having three unequal sides) is called scalene.
How many degrees is an isosceles triangle?
Isosceles Triangles : Example Question #7 Explanation: Every triangle has 180 degrees. An isosceles triangle has one vertex angle and two congruent base angles. Thus the vertex angle is 38 and the base angle is 71 and their sum is 109.
What kind of triangle has a right angle?
A right triangle (American English) or right-angled triangle (British), or more formally an orthogonal triangle (Ancient Greek: ὀρθόςγωνία, lit. ‘upright angle’), is a triangle in which one angle is a right angle (that is, a 90-degree angle).
What does an isosceles triangle look like?
An isosceles triangle has two equal sides (or three, technically) and two equal angles (or three, technically). The equal sides are called legs, and the third side is the base. … The angle between the two legs is called the vertex angle. The above figure shows two isosceles triangles.
Are all isosceles right triangles similar?
Yes, two right isosceles triangles are always similar. … A right triangle always has one right angle with measure 90°. If a right triangle is an isosceles triangle, then the two sides that have equal length are opposite the non-right angles in the triangle.
Can a spherical triangle have two right angles?
Because of the fact that the sum of the three interior angles of a triangle must be 180 degrees, a triangle could not have two right angles.
Can a right triangle be equilateral?
In an equilateral triangle, all the sides are equal. If we use the longest side theorem which say in the triangle the longest side is across the largest angle. Since all the sides are equal then the angles must be equal too. So we can’t have an Right angled equilateral triangle.
Are the base angles of an isosceles triangle complementary?
Base angles of an isosceles triangle are complementary. Base angles of an isosceles triangle can be equal to the vertex angle. Base angles of an isosceles triangle are acute.
Can a triangle be both acute and isosceles?
A triangle whose only two sides are equal is called an isosceles triangle. An isosceles triangle also has two equal angels. A triangle whose all angels are greater than 0∘ and less than 90∘ , i.e, all angels are acute is called an acute triangle. Given triangle has an angle of 36∘ and is both isosceles and acute.
Can an isosceles triangle be equilateral?
An equilateral triangle is a triangle whose sides are all equal. It is a specific kind of isosceles triangle whose base is equal to each leg, and whose vertex angle is equal to its base angles. … Every equilateral triangle is also an isosceles triangle, so any two sides that are equal have equal opposite angles.
What do you call the side opposite the right angle in a right triangle?
The hypotenuse of a right triangle is always the side opposite the right angle. It is the longest side in a right triangle.
Do isosceles triangles add up to 180?
Parts of an Isosceles Triangle The two equal sides of the isosceles triangle are legs and the third side is the base. The angle between the equal sides is called the vertex angle. All of the angles should equal 180 degrees when added together.
Which angles are the same in an isosceles triangle?
The angles situated opposite to the equal sides of an isosceles triangle are always equal. All the three angles situated within the isosceles triangle are acute, which signifies that the angles are less than 90°.
What is isosceles right angled?
An Isosceles triangle is a triangle in which at least two sides are equal. … Isosceles Right Triangle has one of the angles exactly 90 degrees and two sides which is equal to each other. Since the two sides are equal which makes the corresponding angle congruent.
Why is a right triangle called a right triangle?
Right Triangle. A right triangle is a triangle with one right angle (one angle equal to 90°). The side opposite the right angle (the longest side) is called the hypotenuse. The remaining two sides (the sides that intersect to determine the right angle) are called the legs of the right triangle.
What are the properties of a right isosceles triangle?
An isosceles right triangle has the characteristic of both the isosceles and the right triangles. It has two equal sides, two equal angles, and one right angle. (The right angle cannot be one of the equal angles or the sum of the angles would exceed 180°.)
Is 7cm 24cm and 25cm forms a right angle triangle?
The given sides suit the Pythagoras theorem as the sum of squares of two sides is equal to the square of the largest side. The triangle formed by sides 7cm, 24cm, 25cm is a right triangle having hypotenuse 25cm.
Are all isosceles triangles?
No, all isosceles triangles are not similar. An isosceles triangle is a triangle with two sides of equal length.
Why isosceles triangle is not similar?
All isosceles triangles are not similar for a couple of reasons. The length of the two equal sides can stay the same but the measure of the angle between the two equal side will change, as will the base and the base angles.
Can spherical triangles have three right angles?
A spherical triangle can have three angles of size pi/2 or a right angle. In this case, the sides of the spherical triangle are right angles as well. A triangle on a plane, a two dimensional space, can only have at most one right angle.
Which triangle exists only in spherical geometry?
Since spherical geometry violates the parallel postulate, there exists no such triangle on the surface of a sphere. The sum of the angles of a triangle on a sphere is 180°(1 + 4f), where f is the fraction of the sphere’s surface that is enclosed by the triangle. For any positive value of f, this exceeds 180°.
Why does spherical geometry exist?
Spherical geometry is important in navigation, because the shortest distance between two points on a sphere is the path along a great circle. Riemannian Postulate: Given a line and a point not on the line, every line passing though the point intersects the line. (There are no parallel lines). |
# linear congruence by euclidean algorithm
• Nov 10th 2012, 09:07 AM
jethrotulL
linear congruence by euclidean algorithm
Hey!
I have a question about solutions of linear congruences with Euclidean algorithm. I'd be very happy if you can help.
9x=1 (mod 16)
What I know is gcd of 9 and 16 is equal to 1 and it divides 1. Thus exists a unique solution.
But I cannot figure out how to find x=9 by Euclidean way.
If someone helps or just tries, thanks in advance :)
• Nov 10th 2012, 11:13 AM
Soroban
Re: linear congruence by euclidean algorithm
Hello, jethrotulL!
Even if you don't know (or understand) the Euclidean algorithm,
. . you can still solve it.
Quote:
Solve by the Euclidean algorithm: .$\displaystyle 9x\:\equiv\:1\text{ (mod 16) }$
We have: .$\displaystyle 9x \:=\:16a + 1\,\text{ for some integer }a.$
Solve for $\displaystyle x\!:\;x \:=\:\frac{16a+1}{9} \:=\:a + \frac{7a+1}{9}$ .[1]
Since $\displaystyle x$ is an integer, $\displaystyle 7a+1$ must be a multiple of 9.
. . That is: .$\displaystyle 7a+1 \:=\:9b\,\text{ for some integer }b.$
Solve for $\displaystyle a\!:\;a \:=\:\frac{9b-1}{7} \:=\:b + \frac{2b-1}{7}$ .[2]
Since $\displaystyle a$ is an integer, $\displaystyle 2b-1$ must be a multiple of 7.
. . The first time this happens is $\displaystyle b = 4$
Substitute into [2]: .$\displaystyle a \:=\:4 + \frac{2(4)-1}{7} \quad\Rightarrow\quad a\,=\,5$
Substitute into [1]: .$\displaystyle x \:=\:5 + \frac{7(5)+1}{9} \quad\Rightarrow\quad \boxed{x \,=\,9}$
• Nov 10th 2012, 11:25 AM
jethrotulL
Re: linear congruence by euclidean algorithm
Thank you, but I need to solve it by euclidean algorithm. Can you also solve by this way please? |
# Factoring a fourth degree polynomial
How would I go about factoring $$x^4-x^2-12=0$$
I'm supposed to find the roots of this without the use of calculators (well we do have TI's but some don't so I suppose they're assuming we can solve this manually). But I can't see how I'd factor this simply?
• As an intermediate step, write it as a polynomial in $y = x^2$. Commented Apr 13, 2014 at 18:41
• Sorry, I don't understand what you mean? Do you mean I write it as: $$x^2 = x^4-12$$? Commented Apr 13, 2014 at 18:42
• No, $y^2-y-12$. Commented Apr 13, 2014 at 18:44
Let $t=x^2$, then you have $t^2-t-12=0$ which is a lot easier to factor. Once you're done that, resubstitute $x^2$ back in.
Write $y=x^2$, giving $y^2-y-12=0=(x^2+3)(x^2-4)$. I'll leave the steps in the factorization of the quadratic $y^2-y-12=0$ to you.
• Ah right, that's how you did it. I thought I had to pull $12$ down by a factor of $x^2$ as well for some reason. Commented Apr 13, 2014 at 18:43
• @user3200098 No, doing that would be redundant and wrong.
– user122283
Commented Apr 13, 2014 at 18:44
Change the variable so you have $y=x^2$ and so solve for $$y^2 - y -12 =0$$
You will find $y = -3$ or $y = 4$, then solve $x^2 = -3$ and $x^2 = 4$. You will have multiple roots which are $$x_1 = -2$$
$$x_2 = 2$$ $$x_3 = -i\sqrt{3}$$ $$x_4 = i\sqrt{3}$$
You will end up with
$$x^4 - x^2 -12 = (x-2)(x+2)(x-i\sqrt{3})(x+i\sqrt{3})$$
Forget $x_3$ and $x_4$ if you are in $\mathbb{R}$ :
$$x^4 - x^2 -12 = (x-2)(x+2)(x^2+3)$$
Let $t=x^2$
Now we have
$t^2-t-12=0$
$\Delta=1+48=49$
$t_1=-3$, $t_2=4$
go back to x
$x_1=\sqrt-3$, $x_2=-\sqrt-3$, $x_3=2$, $x_4=-2$
If you had complex numbers in school, you can easily compute that $x_1=i\sqrt3$ and $x_2=-i\sqrt3$
If not, just write that there are no real roots of negative numbers.
Note that $x^4 - x^2 -12 = (x^2 - 4)(x^2 + 3)$; this factorization was guesswork on my part, based on the observation that $(x^2 - \mu_1)(x^2 - \mu_2) = x^4 - (\mu_1 + \mu_2)x^2 + \mu_1 \mu_2$; so I looked for integers $\mu_1$, $\mu_2$ such that $\mu_1 + \mu_2 = 1$ and $\mu_1 \mu_2 = -12$; it was easy to make such a guess. But this factorization could have been done systematically by setting $y = x^2$, as several others have suggested, and then realizing the roots of the quadratic $y^2 - y -12 = 0$ are $4, -3$; this of course can be done with aid of the quadratic formula. Once we have $x^4 - x^2 -12 = (x^2 - 4)(x^2 + 3)$, we can factor further by using $x^2 - 4 = (x + 2)(x - 2)$; thus $x^4 - x^2 -12 = (x + 2)(x - 2)(x^2 + 3)$; we can't go further over the reals since $x^2 + 3$ has no real zeroes.
That's how I'd do it, in any event.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
This "quartic" equation is actually a quadratic in $x^2$. Notice that $$x^4 - x^2 - 12 \equiv \left(x^2\right)^2 - \left(x^2\right) - 12$$ If we re-name $x^2$ as $u$ then we have $$x^4 - x^2 - 12 = 0 \iff u^2 - u - 12 = 0$$ We are able to factorise $u^2-u-12$. Indeed $u^2-u-12 \equiv (u+3)(u-4)$. It follows that $$u^2-u-12=0 \iff u=-3,\, 4$$ Since $u$ is used in place of $x^2$ we have $x^2=-3, \, 4$. It follows that $$x^4-x^2-12=0 \iff x = \pm\mathrm{i}\sqrt{3}, \, \pm 2$$ If you are working over the real numbers then $x= \pm 2$ are the only valid solutions.
$$x^4-x^2-12=0\,\,\text{can be written as: }\color{blue}{(x^2)}^2-\color{blue}{(x^2)}-12=0.$$ which with a change of variable is a quadratic equation, that can be solved using the quadratic formula to get: $$x^2=\dfrac{1\pm\sqrt{1+4\cdot12}}{2}.$$ And so, you will get $2$ roots, and recall that: If $\alpha$ is a root of $\mathrm P(x)$, then there exist one polynomial $\mathrm Q(x)$ such that $\mathrm P(x)=(x-\alpha)\cdot\mathrm Q(x)$ and so you'll have to use polynomial long division.
After doing all of those steps, you will find that: $$x^4-x^2-12=(x-2 ) (2 + x) (3 + x^2).$$ |
# APTITUDE – Problems on SIMPLE INTEREST
Important Facts and Formulae:
Principal or Sum:- The money borrowed or lent out for a certain period is called Principal or the Sum.
Interest:- Extra money paid for using others money is called Interest.
Simple Interest:- If the interest on a sum borrowed for a certain period is reckoned uniformly,then it is called Simple Interest.
Formulae:
Principal = P
Rate = R% per annum
Time = T years. Then,
(i)Simple Interest(S.I)= (P*T*R)/100
(ii) Principal(P) = (100*S.I)/(R*T)
Rate(R) = (100*S.I)/(P*T)
Time(T) = (100*S.I)/(P*R)
Simple Problems
1.Find S.I on Rs68000 at 16 2/3% per annum for 9months.
Sol:- P=68000
R=50/3% p.a
T=9/12 years=4/3 years
S.I=(P*R*T)/100
=(68000*(50/3)*(3/4)*(1/100))
=Rs 8500
Note:If months are given we have to converted into
years by dividing 12 ie., no.of months/12=years
2.Find S.I on Rs3000 at 18% per annum for the period from 4th Feb to 18th April 1995
Sol:- Time=(24+31+18)days
=73 days
=73/365=1/5 years
P= Rs 3000
R= 18% p.a
S.I = (P*R*T)/100
=(3000*18*1/5*1/100)
=Rs 108
Remark:- The day on which money is deposited is not
counted while the day on which money is withdrawn is
counted.
3. In how many years will a sum of money becomes triple at 10% per annum.
Sol:- Let principal =P
S.I = 2P
S.I = (P*T*R)/100
2P = (P*T*10)/100
T = 20 years
Note:
(1) Total amount = Principal + S.I
(2) If sum of money becomes double means Total amount
or Sum
= Principal + S.I
= P + P = 2P
Medium Problems
1.A sum at Simple interest at 13 1/2% per annum amounts to Rs 2502.50 after 4 years.Find the sum.
Sol:- Let Sum be x. then,
S.I = (P*T*R)/100
= ((x*4*27)/(100*2))
= 27x/100
Amount = (x+(27x)/100)
= 77x/50
77x/50 = 2502.50
x = (2502.50*50)/77
= 1625
Sum = 1625
2. A some of money becomes double of itself in 4 years in 12 years it will become how many times at the same rate.
Sol:- 4 yrs – – – – – – – – – P
12 yrs – – – – – – – – – ?
(12/4)* P =3P
Amount or Sum = P+3P = 4 times
3. A Sum was put at S.I at a certain rate for 3 years. Had it been put at 2% higher rate ,it would have fetched Rs 360 more .Find the Sum.
Sol:- Let Sum =P
original rate = R
T = 3 years
If 2% is more than the original rate ,it would have
fetched 360 more ie., R+2
(P*(R+2)*3/100) – (P*R*3)/100 = 360
3PR+ 6P-3PR = 36000
6P = 36000
P = 6000
Sum = 6000.
4.Rs 800 amounts to Rs 920 in 3yrs at S.I.If the interest rate is increased by 3%, it would amount to how much?
Sol:- S.I = 920 – 800 = 120
Rate = (100*120)/(800*3) = 5%
New Rate = 5 + 3 = 8%
Principal = 800
Time = 3 yrs
S.I = (800*8*3)/100 = 192
New Amount = 800 + 192
= 992
5. Prabhat took a certain amount as a loan from bank at the rate of 8% p.a S.I and gave the same amount to Ashish as a loan at the rate of 12% p.a . If at the end of 12 yrs, he made a profit of Rs. 320 in the deal,What was the original amount?
Sol:- Let the original amount be Rs x.
T = 12
R1 = 8%
R2 = 12%
Profit = 320
P = x
(P*T*R2)/100 – (P*T*R1)/100 =320
(x*12*12)/100 – (x*8*12)/100 = 320
x = 2000/3
x = Rs.666.67
6. Simple Interest on a certail sum at a certain rate is 9/16 of the sum . if the number representing rate percent and time in years be equal ,then the rate is.
Sol:- Let Sum = x .Then,
S.I = 9x/16
Let time = n years & rate = n%
n = 100 * 9x/16 * 1/x * 1/n
n * n = 900/16
n = 30/4 = 7 1/2%
Complex Problems
1. A certain sum of money amounts t 1680 in 3yrs & it becomes 1920 in 7 yrs .What is the sum.
Sol:- 3 yrs – – – – – – – – – – – – – 1680
7 yrs – – – – – – – – – – – – – 1920
then, 4 yrs – – – – – – – – – – – – – 240
1 yr – – – – – – – – – – – – – ?
(1/4) * 240 = 60
S.I in 3 yrs = 3*60 = 18012
Sum = Amount – S.I
= 1680 – 180
= 1500
we get the same amount if we take S.I in 7 yrs
I.e., 7*60 =420
Sum = Amount – S.I
= 1920 – 420
= 1500
2. A Person takes a loan of Rs 200 at 5% simple Interest. He returns Rs.100 at the end of 1 yr. In order to clear his dues at the end of 2yrs ,he would pay:
Sol:- Amount to be paid
= Rs(100 + (200*5*1)/100 + (100*5*1)/100)
= Rs 115
3. A Man borrowed Rs 24000 from two money lenders.For one loan, he paid 15% per annum and for other 18% per annum. At the end of one year,he paid Rs.4050.How much did he
borrowed at each rate?
Sol:- Let the Sum at 15% be Rs.x
& then at 18% be Rs (24000-x)
P1 = x R1 = 15
P2 = (24000-x) R2 = 18
At the end of ine year T = 1
(P1*T*R1)/100 + (P2*T*R2)/100 = 4050
(x*1*15)/100 + ((24000-x)*1*18)/100 = 4050
15x + 432000 – 18x = 405000
x = 9000
Money borrowed at 15% = 9000
Money borrowed at 18% = (24000 – 9000)
= 15000
4.What annual instalment will discharge a debt of Rs. 1092 due in 3 years at 12% Simple Interest ?
Sol:- Let each instalment be Rs x
(x + (x * 12 * 1)/100) + (x + (x * 12 * 2)/100) + x = 1092
28x/25 + 31x/25 + x =1092
(28x +31x + 25x) = (1092 * 25)
84x = 1092 * 25
x = (1092*25)/84 = 325
Each instalement = 325
5.If x,y,z are three sums of money such that y is the simple interest on x,z is the simple interest on y for the same time and at the same rate of interest ,then we have:
Sol:- y is simple interest on x, means
y = (x*R*T)/100
RT = 100y/x
z is simple interest on y,
z = (y*R*T)/100
RT = 100z/y
100y/x = 100z/y
y * y = xz
6.A Sum of Rs.1550 was lent partly at 5% and partly at 5% and partly at 8% p.a Simple interest .The total interest received after 3 years was Rs.300.The ratio of the money lent at 5% to that lent at 8% is:
Sol:- Let the Sum at 5% be Rs x
at 8% be Rs(1550-x)
(x*5*3)/100 + ((1500-x)*8*3)/100 = 300
15x + 1500 * 24 – 24x = 30000
x = 800
Money at 5%/ Money at 8% = 800/(1550 – 800)
= 800/750 = 16/15
7. A Man invests a certain sum of money at 6% p.a Simple interest and another sum at 7% p.a Simple interest. His income from interest after 2 years was Rs 354 .one fourth of the first sum is equal to one fifth of the second sum.The total sum invested was:
Sol:- Let the sums be x & y
R1 = 6 R2 = 7
T = 2
(P1*R1*T)/100 + (P2*R2*T)/100 = 354
(x * 6 * 2)/100 + (y * 7 * 2)/100 = 354
6x + 7y = 17700 ———(1)
also one fourth of the first sum is equal to one
fifth of the second sum
x/4 = y/5 => 5x – 4y = 0 —— (2)
By solving 1 & 2 we get,
x = 1200 y = 1500
Total sum = 1200 +1500
= 2700
8. Rs 2189 are divided into three parts such that their amounts after 1,2& 3 years respectively may be equal, the rate of S.I being 4% p.a in all cases. The Smallest part is:
Sol:- Let these parts be x,y and[2189-(x+y)] then,
(x*1*4)/100 = (y*2*4)/100 = (2189-(x+y))*3*4/100
4x/100 = 8y/100
x = 2y
By substituting values
(2y*1*4)/100 = (2189-3y)*3*4/100
44y = 2189 *12
y = 597
Smallest Part = 597
9. A man invested 3/3 of his capital at 7% , 1/4 at 8% and the remainder at 10%.If his annual income is Rs.561. The capital is:
Sol:- Let the capital be Rs.x
Then, (x/3 * 7/100 * 1) + ( x/4 * 8/100 * 1)
+ (5x/12 * 10/100 * 1) = 561
7x/300 + x/50 + x/24 = 561
51x = 561 * 600
x = 6600 |
StATS: A simple model of nonlinear growth (October 1, 2003)
Part of the challenge in nonlinear regression is to choose the correct form of the nonlinear relationship. Part of this is intuition, part of this is understanding some basic mathematics, and part of it is just trial and error.
A very simple growth model
The simplest model of growth uses the exponential function. If you are trying to describe growth that eventually levels off, then the formula
is a good place to start. You can generate this formula using differential equations that assume that the rate of growth is proportional to the amount of growth left until full maturity.
Here's a graph of this formula.
Notice that when x equals 1, about two-thirds of the growth has already occurred and that when x equals 3, about 95% of the growth has occurred.
Allowing for changes in the growth rate
We can change the rate of growth by modifying the equation to
Here's what the graph looks like for three values of b.
Notice that a larger value of b implies faster growth. As a rough rule of thumb, about 2/3 of the growth will occur at x=1/b and about 95% of the growth will occur by x=3/b. For example, when b=0.5, about two-thirds of the growth has occurred by x=2 and 95% has occurred by x=6 (which is off the scale of this graph).
Allowing for different starting and ending values
We need to make two more modifications to the formula, though. First, we need to account for a full grown size that is usually a value different from 1. Second, we need to account for a starting size that is usually larger than 0. These requirements modify the equation to
With this equation, the full-grown size is represented by c, and the initial size at x=0 is represented by c-a. Here's an example of this graph with c=0.9, a=0.7, and b=2.
You can change the values of c, a, and b to illustrate a variety of different growth patterns. For example, here are three growth curves that have different initial sizes, but which all reach the same full grown size.
Here are three growth curves that start at the same initial size but which reach different full grown sizes.
An example using real data
Here's some data on the growth of a child who was born prematurely.
When you ask SPSS to fit a nonlinear regression model (select ANALYZE | REGRESSION | NONLINEAR), your dialog box will look like this:
Notice that I specified the nonlinear model as "C-A*exp(-B*age)" and used upper case for the values C, A, and B. This makes it easier to distinguish between these parameters and the variables.
Now you have to provide SPSS with some starting values. They don't have to be too accurate, but they need to be in the right ball park. Click on the PARAMETERS button to provide starting values.
If you look at the graph the we drew for the data, it looks like it might level off around 120, so that's a good starting value for C. The initial size is around 30. Since C-A represents the initial size, a value of 90 would be a good starting value for A. Finally, it looks like more than half of the growth occurs in the first 24 months. So a good starting value for B would be 1/24, which I rounded to 0.04.
After you have added all your starting values, click on the continue button. Here is what the dialog box looks like now.
If you want to compute predicted values and residuals, click on the SAVE button. Otherwise, click on OK to run your nonlinear regression model. Here is a portion of the output that SPSS produced.
Our model shows an initial size of about 30 cm, a full-grown size of about 107 cm and a growth rate value of 0.037. Also notice how wide the confidence intervals are for each of the parameters. This is not too surprising, since our sample size is so small.
Be careful with this data. The full grown size represents an extrapolation beyond the range of the data (most children continue to grow quite a bit beyond 72 months). In fact, it looks like the estimate of full grown size is a serious underestimate (107 cm is about 42 inches). Perhaps this underestimate is also telling me that my choice for a nonlinear model is not a good one.
Here is the original data (red circles) and the values predicted by the nonlinear regression model (blue line).
There is a good level of agreement between the growth model and the data.
This page was written by Steve Simon while working at Children's Mercy Hospital. Although I do not hold the copyright for this material, I am reproducing it here as a service, as it is no longer available on the Children's Mercy Hospital website. Need more information? I have a page with general help resources. You can also browse for pages similar to this one at Category: Nonlinear regression. |
# RD Sharma Solutions for Class 11 Chapter 18 - Binomial Theorem
An algebraic expression containing two terms is called a binomial expression. The RD Sharma Class 11 Maths solutions are developed by our experienced faculty team at BYJU’S, which is created for the purpose of clarifying student’s doubts, as per their convenience. These solutions also provide guidance to students for solving problems confidently, which in turn, helps in improving their problem-solving skills, that is very important from the examination point of view. For students wishing to learn the right steps of solving such problems, the RD Sharma Class 11 Maths Solutions is made available in the pdf format, students can easily download the pdf from the links given below.
Chapter 18 – Binomial Theorem contains two exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.
• Binomial theorem for positive integral index.
• Some important conclusions from the binomial theorem.
• General term and middle terms in a binomial expansion.
## Download the pdf of RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem
### Access answers to RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem
EXERCISE 18.1 PAGE NO: 18.11
1. Using binomial theorem, write down the expressions of the following:
Solution:
(i) (2x + 3y) 5
Let us solve the given expression:
(2x + 3y) 5 = 5C0 (2x)5 (3y)0 + 5C1 (2x)4 (3y)1 + 5C2 (2x)3 (3y)2 + 5C3 (2x)2 (3y)3 + 5C4 (2x)1 (3y)4 + 5C5 (2x)0 (3y)5
= 32x5 + 5 (16x4) (3y) + 10 (8x3) (9y)2 + 10 (4x)2 (27y)3 + 5 (2x) (81y4) + 243y5
= 32x5 + 240x4y + 720x3y2 + 1080x2y3 + 810xy4 + 243y5
(ii) (2x – 3y) 4
Let us solve the given expression:
(2x – 3y) 4 = 4C0 (2x)4 (3y)04C1 (2x)3 (3y)1 + 4C2 (2x)2 (3y)24C3 (2x)1 (3y)3 + 4C4 (2x)0 (3y)4
= 16x4 – 4 (8x3) (3y) + 6 (4x2) (9y2) – 4 (2x) (27y3) + 81y4
= 16x4 – 96x3y + 216x2y2 – 216xy3 + 81y4
(iv) (1 – 3x) 7
Let us solve the given expression:
(1 – 3x) 7 = 7C0 (3x)07C1 (3x)1 + 7C2 (3x)27C3 (3x)3 + 7C4 (3x)4 – 7C5 (3x)5 + 7C6 (3x)67C7 (3x)7
= 1 – 7 (3x) + 21 (9x)2 – 35 (27x3) + 35 (81x4) – 21 (243x5) + 7 (729x6) – 2187(x7)
= 1 – 21x + 189x2 – 945x3 + 2835x4 – 5103x5 + 5103x6 – 2187x7
(viii) (1 + 2x – 3x2)5
Let us solve the given expression:
Let us consider (1 + 2x) and 3x2 as two different entities and apply the binomial theorem.
(1 + 2x – 3x2)5 = 5C0 (1 + 2x)5 (3x2)05C1 (1 + 2x)4 (3x2)1 + 5C2 (1 + 2x)3 (3x2)25C3 (1 + 2x)2 (3x2)3 + 5C4 (1 + 2x)1 (3x2)45C5 (1 + 2x)0 (3x2)5
= (1 + 2x)5 – 5(1 + 2x)4 (3x2) + 10 (1 + 2x)3 (9x4) – 10 (1 + 2x)2 (27x6) + 5 (1 + 2x) (81x8) – 243x10
= 5C0 (2x)0 + 5C1 (2x)1 + 5C2 (2x)2 + 5C3 (2x)3 + 5C4 (2x)4 + 5C5 (2x)5 – 15x2 [4C0 (2x)0 + 4C1 (2x)1 + 4C2 (2x)2 + 4C3 (2x)3 + 4C4 (2x)4] + 90x4 [1 + 8x3 + 6x + 12x2] – 270x6(1 + 4x2 + 4x) + 405x8 + 810x9 – 243x10
= 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5 – 15x2 – 120x3 – 3604 – 480x5 – 240x6 + 90x4 + 720x7 + 540x5 + 1080x6 – 270x6 – 1080x8 – 1080x7 + 405x8 + 810x9 – 243x10
= 1 + 10x + 25x2 – 40x3 – 190x4 + 92x5 + 570x6 – 360x7 – 675x8 + 810x9 – 243x10
(x) (1 – 2x + 3x2)3
Let us solve the given expression:
2. Evaluate the following:
Solution:
Let us solve the given expression:
= 2 [5C0 (2√x)0 + 5C2 (2√x)2 + 5C4 (2√x)4]
= 2 [1 + 10 (4x) + 5 (16x2)]
= 2 [1 + 40x + 80x2]
Let us solve the given expression:
= 2 [6C0 (√2)6 + 6C2 (√2)4 + 6C4 (√2)2 + 6C6 (√2)0]
= 2 [8 + 15 (4) + 15 (2) + 1]
= 2 [99]
= 198
Let us solve the given expression:
= 2 [5C1 (34) (√2)1 + 5C3 (32) (√2)3 + 5C5 (30) (√2)5]
= 2 [5 (81) (√2) + 10 (9) (2√2) + 4√2]
= 2√2 (405 + 180 + 4)
= 1178√2
Let us solve the given expression:
= 2 [7C0 (27) (√3)0 + 7C2 (25) (√3)2 + 7C4 (23) (√3)4 + 7C6 (21) (√3)6]
= 2 [128 + 21 (32)(3) + 35(8)(9) + 7(2)(27)]
= 2 [128 + 2016 + 2520 + 378]
= 2 [5042]
= 10084
Let us solve the given expression:
= 2 [5C1 (√3)4 + 5C3 (√3)2 + 5C5 (√3)0]
= 2 [5 (9) + 10 (3) + 1]
= 2 [76]
= 152
Let us solve the given expression:
= (1 – 0.01)5 + (1 + 0.01)5
= 2 [5C0 (0.01)0 + 5C2 (0.01)2 + 5C4 (0.01)4]
= 2 [1 + 10 (0.0001) + 5 (0.00000001)]
= 2 [1.00100005]
= 2.0020001
Let us solve the given expression:
= 2 [6C1 (√3)5 (√2)1 + 6C3 (√3)3 (√2)3 + 6C5 (√3)1 (√2)5]
= 2 [6 (9√3) (√2) + 20 (3√3) (2√2) + 6 (√3) (4√2)]
= 2 [√6 (54 + 120 + 24)]
= 396 √6
= 2 [a8 + 6a6 – 6a4 + a4 + 1 – 2a2]
= 2a8 + 12a6 – 10a4 – 4a2 + 2
3. Find (a + b) 4 – (a – b) 4. Hence, evaluate (√3 + √2)4 – (√3 – √2)4.
Solution:
Firstly, let us solve the given expression:
(a + b) 4 – (a – b) 4
The above expression can be expressed as,
(a + b) 4 – (a – b) 4 = 2 [4C1 a3b1 + 4C3 a1b3]
= 2 [4a3b + 4ab3]
= 8 (a3b + ab3)
Now,
Let us evaluate the expression:
(√3 + √2)4 – (√3 -√2)4
So consider, a = √3 and b = √2 we get,
(√3 + √2)4 – (√3 -√2)4 = 8 (a3b + ab3)
= 8 [(√3)3 (√2) + (√3) (√2)3]
= 8 [(3√6) + (2√6)]
= 8 (5√6)
= 40√6
4. Find (x + 1) 6 + (x – 1) 6. Hence, or otherwise evaluate (√2 + 1)6 + (√2 – 1)6.
Solution:
Firstly, let us solve the given expression:
(x + 1) 6 + (x – 1) 6
The above expression can be expressed as,
(x + 1) 6 + (x – 1) 6 = 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6 x0]
= 2 [x6 + 15x4 + 15x2 + 1]
Now,
Let us evaluate the expression:
(√2 + 1)6 + (√2 – 1)6
So consider, x = √2 then we get,
(√2 + 1)6 + (√2 – 1)6 = 2 [x6 + 15x4 + 15x2 + 1]
= 2 [(√2)6 + 15 (√2)4 + 15 (√2)2 + 1]
= 2 [8 + 15 (4) + 15 (2) + 1]
= 2 [8 + 60 + 30 + 1]
= 198
5. Using binomial theorem evaluate each of the following:
(i) (96)3
(ii) (102)5
(iii) (101)4
(iv) (98)5
Solution:
(i) (96)3
We have,
(96)3
Let us express the given expression as two different entities and apply the binomial theorem.
(96)3 = (100 – 4)3
= 3C0 (100)3 (4)03C1 (100)2 (4)1 + 3C2 (100)1 (4)23C3 (100)0 (4)3
= 1000000 – 120000 + 4800 – 64
= 884736
(ii) (102)5
We have,
(102)5
Let us express the given expression as two different entities and apply the binomial theorem.
(102)5 = (100 + 2)5
= 5C0 (100)5 (2)0 + 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 + 5C3 (100)2 (2)3 + 5C4 (100)1 (2)4 + 5C5 (100)0 (2)5
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032
(iii) (101)4
We have,
(101)4
Let us express the given expression as two different entities and apply the binomial theorem.
(101)4 = (100 + 1)4
= 4C0 (100)4 + 4C1 (100)3 + 4C2 (100)2 + 4C3 (100)1 + 4C4 (100)0
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401
(iv) (98)5
We have,
(98)5
Let us express the given expression as two different entities and apply the binomial theorem.
(98)5 = (100 – 2)5
= 5C0 (100)5 (2)05C1 (100)4 (2)1 + 5C2 (100)3 (2)25C3 (100)2 (2)3 + 5C4 (100)1 (2)45C5 (100)0 (2)5
= 10000000000 – 1000000000 + 40000000 – 800000 + 8000 – 32
= 9039207968
6. Using binomial theorem, prove that 23n – 7n – 1 is divisible by 49, where n ∈ N.
Solution:
Given:
23n – 7n – 1
So, 23n – 7n – 1 = 8n – 7n – 1
Now,
8n – 7n – 1
8n = 7n + 1
= (1 + 7) n
= nC0 + nC1 (7)1 + nC2 (7)2 + nC3 (7)3 + nC4 (7)2 + nC5 (7)1 + … + nCn (7) n
8n = 1 + 7n + 49 [nC2 + nC3 (71) + nC4 (72) + … + nCn (7) n-2]
8n – 1 – 7n = 49 (integer)
So now,
8n – 1 – 7n is divisible by 49
Or
23n – 1 – 7n is divisible by 49.
Hence proved.
EXERCISE 18.2 PAGE NO: 18.37
1. Find the 11th term from the beginning and the 11th term from the end in the expansion of (2x – 1/x2)25.
Solution:
Given:
(2x – 1/x2)25
The given expression contains 26 terms.
So, the 11th term from the end is the (26 − 11 + 1) th term from the beginning.
In other words, the 11th term from the end is the 16th term from the beginning.
Then,
T16 = T15+1 = 25C15 (2x)25-15 (-1/x2)15
= 25C15 (210) (x)10 (-1/x30)
= – 25C15 (210 / x20)
Now we shall find the 11th term from the beginning.
T11 = T10+1 = 25C10 (2x)25-10 (-1/x2)10
= 25C10 (215) (x)15 (1/x20)
= 25C10 (215 / x5)
2. Find the 7th term in the expansion of (3x2 – 1/x3)10.
Solution:
Given:
(3x2 – 1/x3)10
Let us consider the 7th term as T7
So,
T7 = T6+1
= 10C6 (3x2)10-6 (-1/x3)6
= 10C6 (3)4 (x)8 (1/x18)
= [10×9×8×7×81] / [4×3×2×x10]
= 17010 / x10
∴ The 7th term of the expression (3x2 – 1/x3)10 is 17010 / x10.
3. Find the 5th term in the expansion of (3x – 1/x2)10.
Solution:
Given:
(3x – 1/x2)10
The 5th term from the end is the (11 – 5 + 1)th, is., 7th term from the beginning.
So,
T7 = T6+1
= 10C6 (3x)10-6 (-1/x2)6
= 10C6 (3)4 (x)4 (1/x12)
= [10×9×8×7×81] / [4×3×2×x8]
= 17010 / x8
∴ The 5th term of the expression (3x – 1/x2)10 is 17010 / x8.
4. Find the 8th term in the expansion of (x3/2 y1/2 – x1/2 y3/2)10.
Solution:
Given:
(x3/2 y1/2 – x1/2 y3/2)10
Let us consider the 8th term as T8
So,
T8 = T7+1
= 10C7 (x3/2 y1/2)10-7 (-x1/2 y3/2)7
= -[10×9×8]/[3×2] x9/2 y3/2 (x7/2 y21/2)
= -120 x8y12
∴ The 8th term of the expression (x3/2 y1/2 – x1/2 y3/2)10 is -120 x8y12.
5. Find the 7th term in the expansion of (4x/5 + 5/2x) 8.
Solution:
Given:
(4x/5 + 5/2x) 8
Let us consider the 7th term as T7
So,
T7 = T6+1
∴ The 7th term of the expression (4x/5 + 5/2x) 8 is 4375/x4.
6. Find the 4th term from the beginning and 4th term from the end in the expansion of (x + 2/x) 9.
Solution:
Given:
(x + 2/x) 9
Let Tr+1 be the 4th term from the end.
Then, Tr+1 is (10 − 4 + 1)th, i.e., 7th, term from the beginning.
7. Find the 4th term from the end in the expansion of (4x/5 – 5/2x) 9.
Solution:
Given:
(4x/5 – 5/2x) 9
Let Tr+1 be the 4th term from the end of the given expression.
Then, Tr+1 is (10 − 4 + 1)th term, i.e., 7th term, from the beginning.
T7 = T6+1
∴ The 4th term from the end is 10500/x3.
8. Find the 7th term from the end in the expansion of (2x2 – 3/2x) 8.
Solution:
Given:
(2x2 – 3/2x) 8
Let Tr+1 be the 4th term from the end of the given expression.
Then, Tr+1 is (9 − 7 + 1)th term, i.e., 3rd term, from the beginning.
T3 = T2+1
∴ The 7th term from the end is 4032 x10.
9. Find the coefficient of:
(i)  x10 in the expansion of (2x2 – 1/x)20
(ii) x7 in the expansion of (x – 1/x2)40
(iii) x-15 in the expansion of (3x2 – a/3x3)10
(iv) x9 in the expansion of (x2 – 1/3x)9
(v) xm in the expansion of (x + 1/x)n
(vi) x in the expansion of (1 – 2x3 + 3x5) (1 + 1/x)8
(vii) a5b7 in the expansion of (a – 2b)12
(viii) x in the expansion of (1 – 3x + 7x2) (1 – x)16
Solution:
(i)  x10 in the expansion of (2x2 – 1/x)20
Given:
(2x2 – 1/x)20
If  x10 occurs in the (r + 1)th term in the given expression.
Then, we have:
Tr+1Â = nCr xn-r ar
(ii) x7 in the expansion of (x – 1/x2)40
Given:
(x – 1/x2)40
If x7 occurs at the (r + 1) th term in the given expression.
Then, we have:
Tr+1Â = nCr xn-r ar
40 − 3r =7
3r = 40 – 7
3r = 33
r = 33/3
= 11
(iii) x-15 in the expansion of (3x2 – a/3x3)10
Given:
(3x2 – a/3x3)10
If x−15 occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1Â = nCr xn-r ar
(iv) x9 in the expansion of (x2 – 1/3x)9
Given:
(x2 – 1/3x)9
If x9 occurs at the (r + 1)th term in the above expression.
Then, we have:
Tr+1Â = nCr xn-r ar
For this term to contain x9, we must have:
18 − 3r = 9
3r = 18 – 9
3r = 9
r = 9/3
= 3
(v) xm in the expansion of (x + 1/x)n
Given:
(x + 1/x)n
If xm occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1Â = nCr xn-r ar
(vi) x in the expansion of (1 – 2x3 + 3x5) (1 + 1/x)8
Given:
(1 – 2x3 + 3x5) (1 + 1/x)8
If x occurs at the (r + 1)th term in the given expression.
Then, we have:
(1 – 2x3 + 3x5) (1 + 1/x)8 = (1 – 2x3 + 3x5) (8C0 + 8C1 (1/x) + 8C2 (1/x)2 + 8C3 (1/x)3 + 8C4 (1/x)4 + 8C5 (1/x)5 + 8C6 (1/x)6 + 8C7 (1/x)7 + 8C8 (1/x)8)
So, ‘x’ occurs in the above expression at -2x3.8C2 (1/x2) + 3x5.8C4 (1/x4)
∴ Coefficient of x = -2 (8!/(2!6!)) + 3 (8!/(4! 4!))
= -56 + 210
= 154
(vii) a5b7 in the expansion of (a – 2b)12
Given:
(a – 2b)12
If a5b7 occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1Â = nCr xn-r ar
(viii) x in the expansion of (1 – 3x + 7x2) (1 – x)16
Given:
(1 – 3x + 7x2) (1 – x)16
If x occurs at the (r + 1)th term in the given expression.
Then, we have:
(1 – 3x + 7x2) (1 – x)16 = (1 – 3x + 7x2) (16C0 + 16C1 (-x) + 16C2 (-x)2 + 16C3 (-x)3 + 16C4 (-x)4 + 16C5 (-x)5 + 16C6 (-x)6 + 16C7 (-x)7 + 16C8 (-x)8 + 16C9 (-x)9 + 16C10 (-x)10 + 16C11 (-x)11 + 16C12 (-x)12 + 16C13 (-x)13 + 16C14 (-x)14 + 16C15 (-x)15 + 16C16 (-x)16)
So, ‘x’ occurs in the above expression at 16C1 (-x) – 3x16C0
∴ Coefficient of x = -(16!/(1! 15!)) – 3(16!/(0! 16!))
= -16 – 3
= -19
10. Which term in the expansion of contains x and y to one and the same power.
Solution:
Let us consider Tr+1Â th term in the given expansion contains x and y to one and the same power.
Then we have,
Tr+1 = nCr xn-r ar
11. Does the expansion of (2x2 – 1/x) contain any term involving x9?
Solution:
Given:
(2x2 – 1/x)
If x9 occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to contain x9, we must have
40 – 3r = 9
3r = 40 – 9
3r = 31
r = 31/3
It is not possible, since r is not an integer.
Hence, there is no term with x9 in the given expansion.
12. Show that the expansion of (x2 + 1/x)12 does not contain any term involving x-1.
Solution:
Given:
(x2 + 1/x)12
If x-1 occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to contain x-1, we must have
24 – 3r = -1
3r = 24 + 1
3r = 25
r = 25/3
It is not possible, since r is not an integer.
Hence, there is no term with x-1 in the given expansion.
13. Find the middle term in the expansion of:
(i) (2/3x – 3/2x)20
(ii) (a/x + bx)12
(iii) (x2 – 2/x)10
(iv) (x/a – a/x)10
Solution:
(i) (2/3x – 3/2x)20
We have,
(2/3x – 3/2x)20 where, n = 20 (even number)
So the middle term is (n/2 + 1) = (20/2 + 1) = (10 + 1) = 11. ie., 11th term
Now,
T11 = T10+1
= 20C10 (2/3x)20-10 (3/2x)10
= 20C10 210/310 × 310/210 x10-10
= 20C10
Hence, the middle term is 20C10.
(ii) (a/x + bx)12
We have,
(a/x + bx)12 where, n = 12 (even number)
So the middle term is (n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. ie., 7th term
Now,
T7 = T6+1
= 924 a6b6
Hence, the middle term is 924 a6b6.
(iii) (x2 – 2/x)10
We have,
(x2 – 2/x)10 where, n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle term is -8064x5.
(iv) (x/a – a/x)10
We have,
(x/a – a/x) 10 where, n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle term is -252.
14. Find the middle terms in the expansion of:
(i) (3x – x3/6)9
(ii) (2x2 – 1/x)7
(iii) (3x – 2/x2)15
(iv) (x4 – 1/x3)11
Solution:
(i) (3x – x3/6)9
We have,
(3x – x3/6)9 where, n = 9 (odd number)
So the middle terms are ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) = ((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and 6th.
Now,
T5 = T4+1
Hence, the middle term are 189/8 x17 and -21/16 x19.
(ii) (2x2 – 1/x)7
We have,
(2x2 – 1/x)7 where, n = 7 (odd number)
So the middle terms are ((n+1)/2) = ((7+1)/2) = 8/2 = 4 and
((n+1)/2 + 1) = ((7+1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5
The terms are 4th and 5th.
Now,
Hence, the middle term are -560x5 and 280x2.
(iii) (3x – 2/x2)15
We have,
(3x – 2/x2)15 where, n = 15 (odd number)
So the middle terms are ((n+1)/2) = ((15+1)/2) = 16/2 = 8 and
((n+1)/2 + 1) = ((15+1)/2 + 1) = (16/2 + 1) = (8 + 1) = 9
The terms are 8th and 9th.
Now,
Hence, the middle term are (-6435×38×27)/x6 and (6435×37×28)/x9.
(iv) (x4 – 1/x3)11
We have,
(x4 – 1/x3)11
where, n = 11 (odd number)
So the middle terms are ((n+1)/2) = ((11+1)/2) = 12/2 = 6 and
((n+1)/2 + 1) = ((11+1)/2 + 1) = (12/2 + 1) = (6 + 1) = 7
The terms are 6th and 7th.
Now,
T7 = T6+1
Hence, the middle term are -462x9 and 462x2.
15. Find the middle terms in the expansion of:
(i) (x – 1/x)10
(ii) (1 – 2x + x2)n
(iii) (1 + 3x + 3x2 + x3)2n
(iv) (2x – x2/4)9
(v) (x – 1/x)2n+1
(vi) (x/3 + 9y)10
(vii) (3 – x3/6)7
(viii) (2ax – b/x2)12
(ix) (p/x + x/p)9
(x) (x/a – a/x)10
Solution:
(i) (x – 1/x)10
We have,
(x – 1/x)10 where, n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle term is -252.
(ii) (1 – 2x + x2)n
We have,
(1 – 2x + x2)n = (1 – x)2n where, n is an even number.
So the middle term is (2n/2 + 1) = (n + 1)th term.
Now,
Tn = Tn+1
= 2nCn (-1)n (x)n
= (2n)!/(n!)2 (-1)n xn
Hence, the middle term is (2n)!/(n!)2 (-1)n xn.
(iii) (1 + 3x + 3x2 + x3)2n
We have,
(1 + 3x + 3x2 + x3)2n = (1 + x)6n where, n is an even number.
So the middle term is (n/2 + 1) = (6n/2 + 1) = (3n + 1)th term.
Now,
T2n = T3n+1
= 6nC3n x3n
= (6n)!/(3n!)2 x3n
Hence, the middle term is (6n)!/(3n!)2 x3n.
(iv) (2x – x2/4)9
We have,
(2x – x2/4)9 where, n = 9 (odd number)
So the middle terms are ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) = ((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and 6th.
Now,
T5 = T4+1
And,
T6 = T5+1
Hence, the middle term is 63/4 x13 and -63/32 x14.
(v) (x – 1/x)2n+1
We have,
(x – 1/x)2n+1 where, n = (2n + 1) is an (odd number)
So the middle terms are ((n+1)/2) = ((2n+1+1)/2) = (2n+2)/2 = (n + 1) and
((n+1)/2 + 1) = ((2n+1+1)/2 + 1) = ((2n+2)/2 + 1) = (n + 1 + 1) = (n + 2)
The terms are (n + 1)th and (n + 2)th.
Now,
Tn = Tn+1
And,
Tn+2 = Tn+1+1
Hence, the middle term is (-1)n.2n+1Cn x and (-1)n+1.2n+1Cn (1/x).
(vi) (x/3 + 9y)10
We have,
(x/3 + 9y)10 where, n = 10 is an even number.
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. i.e., 6th term.
Now,
T6 = T5+1
Hence, the middle term is 61236x5y5.
(vii) (3 – x3/6)7
We have,
(3 – x3/6)7 where, n = 7 (odd number).
So the middle terms are ((n+1)/2) = ((7+1)/2) = 8/2 = 4 and
((n+1)/2 + 1) = ((7+1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5
The terms are 4th and 5th.
Now,
T4 = T3+1
= 7C3 (3)7-3 (-x3/6)3
= -105/8 x9
And,
T5 = T4+1
= 9C4 (3)9-4 (-x3/6)4
Hence, the middle terms are -105/8 x9 and 35/48 x12.
(viii) (2ax – b/x2)12
We have,
(2ax – b/x2)12 where, n = 12 is an even number.
So the middle term is (n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. i.e., 7th term.
Now,
T7 = T6+1
Hence, the middle term is (59136a6b6)/x6.
(ix) (p/x + x/p)9
We have,
(p/x + x/p)9 where, n = 9 (odd number).
So the middle terms are ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) = ((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and 6th.
Now,
T5 = T4+1
And,
T6 = T5+1
= 9C5 (p/x)9-5 (x/p)5
Hence, the middle terms are 126p/x and 126x/p.
(x) (x/a – a/x)10
We have,
(x/a – a/x) 10 where, n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle term is -252.
16. Find the term independent of x in the expansion of the following expressions:
(i) (3/2 x2 – 1/3x)9
(ii) (2x + 1/3x2)9
(iii) (2x2 – 3/x3)25
(iv) (3x – 2/x2)15
(v) ((√x/3) + √3/2x2)10
(vi) (x – 1/x2)3n
(vii) (1/2 x1/3 + x-1/5)8
(viii) (1 + x + 2x3) (3/2x2 – 3/3x)9
(ix) (∛x + 1/2∛x)18, x > 0
(x) (3/2x2 – 1/3x)6
Solution:
(i) (3/2 x2 – 1/3x)9
Given:
(3/2 x2 – 1/3x)9
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of x, we must have
18 – 3r = 0
3r = 18
r = 18/3
= 6
So, the required term is 7th term.
We have,
T7 = T6+1
= 9C6 × (39-12)/(29-6)
= (9×8×7)/(3×2) × 3-3 × 2-3
= 7/18
Hence, the term independent of x is 7/18.
(ii) (2x + 1/3x2)9
Given:
(2x + 1/3x2)9
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of x, we must have
9 – 3r = 0
3r = 9
r = 9/3
= 3
So, the required term is 4th term.
We have,
T4 = T3+1
= 9C3 × (26)/(33)
= 9C3 × 64/27
Hence, the term independent of x is 9C3 × 64/27.
(iii) (2x2 – 3/x3)25
Given:
(2x2 – 3/x3)25
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
= 25Cr (2x2)25-r (-3/x3)r
= (-1)r 25Cr × 225-r × 3r x50-2r-3r
For this term to be independent of x, we must have
50 – 5r = 0
5r = 50
r = 50/5
= 10
So, the required term is 11th term.
We have,
T11 = T10+1
= (-1)10 25C10 × 225-10 × 310
= 25C10 (215 × 310)
Hence, the term independent of x is 25C10 (215 × 310).
(iv) (3x – 2/x2)15
Given:
(3x – 2/x2)15
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
= 15Cr (3x)15-r (-2/x2)r
= (-1)r 15Cr × 315-r × 2r x15-r-2r
For this term to be independent of x, we must have
15 – 3r = 0
3r = 15
r = 15/3
= 5
So, the required term is 6th term.
We have,
T6 = T5+1
= (-1)5 15C5 × 315-5 × 25
= -3003 × 310 × 25
Hence, the term independent of x is -3003 × 310 × 25.
(v) ((√x/3) + √3/2x2)10
Given:
((√x/3) + √3/2x2)10
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of x, we must have
(10-r)/2 – 2r = 0
10 – 5r = 0
5r = 10
r = 10/5
= 2
So, the required term is 3rd term.
We have,
T3 = T2+1
Hence, the term independent of x is 5/4.
(vi) (x – 1/x2)3n
Given:
(x – 1/x2)3n
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
= 3nCr x3n-r (-1/x2)r
= (-1)r 3nCr x3n-r-2r
For this term to be independent of x, we must have
3n – 3r = 0
r = n
So, the required term is (n+1)th term.
We have,
(-1)n 3nCn
Hence, the term independent of x is (-1)n 3nCn
(vii) (1/2 x1/3 + x-1/5)8
Given:
(1/2 x1/3 + x-1/5)8
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of x, we must have
(8-r)/3 – r/5 = 0
(40 – 5r – 3r)/15 = 0
40 – 5r – 3r = 0
40 – 8r = 0
8r = 40
r = 40/8
= 5
So, the required term is 6th term.
We have,
T6 = T5+1
= 8C5 × 1/(28-5)
= (8×7×6)/(3×2×8)
= 7
Hence, the term independent of x is 7.
(viii) (1 + x + 2x3) (3/2x2 – 3/3x)9
Given:
(1 + x + 2x3) (3/2x2 – 3/3x)9
If (r + 1)th term in the given expression is independent of x.
Then, we have:
(1 + x + 2x3) (3/2x2 – 3/3x)9 =
= 7/18 – 2/27
= (189 – 36)/486
= 153/486 (divide by 9)
= 17/54
Hence, the term independent of x is 17/54.
(ix) (∛x + 1/2∛x)18, x > 0
Given:
(∛x + 1/2∛x)18, x > 0
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of r, we must have
(18-r)/3 – r/3 = 0
(18 – r – r)/3 = 0
18 – 2r = 0
2r = 18
r = 18/2
= 9
So, the required term is 10th term.
We have,
T10 = T9+1
= 18C9 × 1/29
Hence, the term independent of x is 18C9 × 1/29.
(x) (3/2x2 – 1/3x)6
Given:
(3/2x2 – 1/3x)6
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of r, we must have
12 – 3r = 0
3r = 12
r = 12/3
= 4
So, the required term is 5th term.
We have,
T5 = T4+1
Hence, the term independent of x is 5/12.
17. If the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)18 are equal, find r.
Solution:
Given:
(1 + x)18
We know, the coefficient of the r term in the expansion of (1 + x)n is nCr-1
So, the coefficients of the (2r + 4) and (r – 2) terms in the given expansion are 18C2r+4-1 and 18Cr-2-1
For these coefficients to be equal, we must have
18C2r+4-1 = 18Cr-2-1
18C2r+3 = 18Cr-3
2r + 3 = r – 3 (or) 2r + 3 + r – 3 = 18 [Since, nCr = nCs => r = s (or) r + s = n]
2r – r = -3 – 3 (or) 3r = 18 – 3 + 3
r = -6 (or) 3r = 18
r = -6 (or) r = 18/3
r = -6 (or) r = 6
∴ r = 6 [since, r should be a positive integer.]
18. If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion of (1 + x)43 are equal, find r.
Solution:
Given:
(1 +Â x)43
We know, the coefficient of the r term in the expansion of (1 + x)n is nCr-1
So, the coefficients of the (2r + 1) and (r + 2) terms in the given expansion are 43C2r+1-1 and 43Cr+2-1
For these coefficients to be equal, we must have
43C2r+1-1 = 43Cr+2-1
43C2r = 43Cr+1
2r = r + 1 (or) 2r + r + 1 = 43 [Since, nCr = nCs => r = s (or) r + s = n]
2r – r = 1 (or) 3r + 1 = 43
r = 1 (or) 3r = 43 – 1
r = 1 (or) 3r = 42
r = 1 (or) r = 42/3
r = 1 (or) r = 14
∴ r = 14 [since, value ‘1’ gives the same term]
19. Prove that the coefficient of (r + 1)th term in the expansion of (1 + x)n + 1 is equal to the sum of the coefficients of rth and (r + 1)th terms in the expansion of (1 + x)n.
Solution:
We know, the coefficients of (r + 1)th term in (1 + x)n+1 is n+1Cr
So, sum of the coefficients of the rth and (r + 1)th terms in (1 + x)n is
(1 + x)n = nCr-1 + nCr
= n+1Cr [since, nCr+1 + nCr = n+1Cr+1]
Hence proved.
### Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem
Exercise 18.1 Solutions
Exercise 18.2 Solutions |
Derivatives The Easy Way
Constant Rule and Power Rule
We have seen the following derivatives:
1. If f(x) = c, then f '(x) = 0
2. If f(x) = x, then f '(x) = 1
3. If f(x) = x2, then f '(x) = 2x
4. If f(x) = x3, then f '(x) = 3x2
5. If f(x) = x4, then f '(x) = 4x3
This leads us the guess the following theorem.
Theorem d xn = nxn-1 dx
Proof:
We have
Applications
Example
Find the derivatives of the following functions:
1. f(x) = 4x3 - 2x100
2. f(x) = 3x5 + 4x8 - x + 2
3. f(x) = (x3 - 2)2
Solution
We use our new derivative rules to find
1. 12x2 - 200x99
2. 15x3+32x7-1
3. First we FOIL to get
[x6 - 4x3 + 4] '
Now use the derivative rule for powers
6x5 - 12x2
Example:
Find the equation to the tangent line to
y = 3x3 - x + 4
at the point (1,6)
Solution:
y' = 9x2 - 1
at x = 1 this is 8. Using the point-slope equation for the line gives
y - 6 = 8(x - 1)
or
y = 8x - 2
Example:
Find the points where the tangent line to
y = x3 - 3x- 24x + 3
is horizontal.
Solution:
We find
y' = 3x2 - 6x - 24
The tangent line will be horizontal when its slope is zero, that is, the derivative is zero. Setting the derivative equal to zero gives:
3x2 - 6x - 24 = 0
or
x2 - 2x - 8 = 0
or
(x - 4)(x + 2) = 0
so that
x = 4 or x = -2
Derivative of f(x) = sin(x)
d sin(x) = cos(x) dx
Proof:
d/dx cos(x)
d cos x = -sin x dx
Back to Math 105 Home Page |
# Measurement of Angles Examples
An angle can be defined as the rotation from the initial point to an endpoint of a ray. Angle measurement is the amount of rotation from the initial to an endpoint of a ray. The angle is said to be a positive angle if the rotation is clockwise and negative if the rotation is anticlockwise. Angles can be measured by various units.
## How to Measure an Angle?
The different systems for angle measurement are as follows:
• Measurement Of Angle – Degree Measure
A degree is defined as a complete rotation in either clockwise or anticlockwise direction, where the beginning and the ending point is the same. The rotation is divided into 360 units.
It is said to be 10 if the rotation from the initial and ending side is [1 / 360]th of the rotation. The degree is divided into hours, minutes and seconds. One degree is 60 minutes and one minute is 60 seconds.
• Measurement Of Angle – Radian Measure
Consider a circle of radius one unit. Let the arc of the circle be one unit. The measure of the angle is 1 radian if the arc subtends at the centre, provided the radius and arc lengths are equal.
The arc length of a circle with radius unity is equal to the angle in radian.
• Measurement Of Angle – Grade Measure
A grade can be defined as a right angle divided into a hundred equal parts. Further, each grade is divided into a hundred minutes and each minute into a hundred seconds.
## Measurement Of Angles Formula
The following formulae can be used in the measurement of angles.
Degree Measure
$1^0=[\frac{1}{360}]^{th}of \ a \ complete \ rotation$
$\theta =\frac{l}{r}$ where l is the arc length and r is the radius of the circle
It is known that a complete rotation in degrees is 3600 and 1 complete rotation = 2π radians in radian measure.
Assuming, π = 3.14159
$1 \ radian= \frac{180^0}{\pi} \\ 1 \ degree = \frac{\pi}{180}$
Let D be the number of degrees, R be the number of radians and G be the number of grades in an angle $\theta , \text then \ \frac{D}{90}=\frac{G}{100}=\frac{2R}{\pi }$
This is the required relationship between the three systems of angle measurement.
Therefore, one radian $\cos A.\cos 2A.\cos {{2}^{2}}A.\cos {{2}^{3}}A…….\cos {{2}^{n-1}}A=\frac{\sin {{2}^{n}}A}{{{2}^{n}}\sin A},\,\text{if }A=n\pi \text \ radians ={{180}^{o}}\\ i.e., 1 \text \ radian =57{}^\circ 1{7}’44.{{8}’}’\approx {{57}^{o}}1{7}’4{{5}’}’.$
## Measurement Of Angles Examples
Example 1: Find the missing angles in the diagram and justify your answer.
Solution:
a = 105° because a = e (corresponding angles)
b = 75° because b = 180° – a (supplementary angles)
c = 105° because a = c (opposite angles)
d = 75° because d = b (opposite angles)
e = 105° because e = 105° (opposite angles).
f = 75° because f = 180° – e (supplementary angles)
h = 75° because h = 180° – 105° (supplementary angles)
Example 2: Find the following missing angles.
Solution:
Find angle a:
a = 35° because 80° + 65° + a = 180° (supplementary angles)
a = 180° – 145°
or a = 35°
Find angle b:
b = 50° because 35° + 95° + b = 180° (sum of angles in a triangle)
b = 180° – 130°
or b = 60°
Find for c:
c = 85° because c + 95° = 180° (supplementary angles)
or c = 180° – 95° = 85°
Find for d:
d = 30° because 65° + 85° + c = 180° (sum of angles in a triangle)
d = 180° – 150° = 30°
Example 3: Find the unknown angles from the figure below.
Solution:
The triangle given is an isosceles triangle. The two base angles will be equal.
Hence a = 350
The two angles of the triangle are known. Adding the third angle makes 1800.
35° + 35° + b = 180°
b = 180° – 70°
b = 110°
The two angles of a quadrilateral are known. All four angles sum to 3600.
2c + 110° + 120° = 360°
2c = 360° – 230°
2c = 130°
c = 65°
Example 4: In the diagram given below, the lines l₁ and l₂ are parallel and line T is transversal. Find ∠GDE.
Solution:
Step 1 :
In the above diagram, m∠GDE and m∠ADE are the angles on the straight line l₁.
So, we have
m∠GDE + m∠ADE = 180° —– (1)
Step 2 :
In the above diagram, m∠ADE and m∠BEF are corresponding angles and corresponding angles are always congruent.
So, we have
Step 3 :
In (1) replace m∠ADE by m∠BEF
(1) —–> m∠GDE + m∠BEF = 180°
Step 4 :
From the diagram given above, we have m∠GDE = 4x and m∠BEF = 6x. So, replace m∠GDE by 4x and m∠BEF by 6x.
4x + 6x = 180°
Combine like terms.
10x = 180°
Divide both sides by 10.
10x/10 = 180°/10
Simplify.
x = 18°
Step 5 :
Plug x = 18° in m∠GDE = 4x
m∠GDE = 4 · 18°
m∠GDE = 72°
Example 5: The measures of two angles of a triangle are $\text{6}0{}^\circ \text{ 53}’\text{ 51}”$ and $\text{51}{}^\circ \text{ 22}’\text{ 5}0”$ respectively. The measure of third angle in radian is____.
(e) None of these
Solution:
The sum of two angles $=[(\text{62}”\text{ 53}’\text{ 51}”)+\text{51}{}^\circ \text{22}’\text{ 5}0”]=\text{114}{}^\circ \text{ 16}’\text{ 41}”$
Third angle of triangle $=\text{18}0{}^\circ -(\text{14}0{}^\circ \text{ 16}’\text{ 41}”) =\text{65}{}^\circ \text{ 43}’\text{ 19}”\\ ={{\left( 65+\frac{43}{60}+\frac{19}{3600} \right)}^{0}}=\frac{234000+2580+19}{3600}={{\left( \frac{236599}{3600} \right)}^{0}}={{(65.722)}^{0}}$
We know that
${{1}^{o}}=\frac{\pi }{180} \text \ radian \text{(65}.\text{722}){}^\circ =\left( \frac{\pi }{180}\times 65.722 \right) \text \ radian = 1.15 \text \ radian\\$
Example 6: Find the ratio of radii of two circles, if arc of equal length subtend angles $\text{3}0{}^\circ$ and $\text{45}{}^\circ$ at their centre.
(a) 2:3
(b) 4:3
(c) 3:2
(d) 3 :4
(e) None of these
Solution:
Suppose radii of circles be ${{r}_{1}} \text \ and \ {{r}_{2}}$ then ${{\theta }_{1}}=\frac{l}{{{r}_{1}}} \text \ and \ {{\theta }_{2}}=\frac{l}{{{r}_{2}}}$ where
${{\theta }_{1}}={{30}^{o}},{{\theta }_{2}}={{45}^{o}}\\ \frac{{{\theta }_{1}}}{{{\theta }_{2}}}=\frac{\frac{l}{{{r}_{1}}}}{\frac{l}{{{r}_{2}}}}\\ \Rightarrow \frac{{{\theta }_{1}}}{{{\theta }_{2}}}=\frac{{{r}_{2}}}{{{r}_{1}}} \Rightarrow \frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{\theta }_{2}}}{{{\theta }_{1}}}\Rightarrow \frac{{{r}_{1}}}{{{r}_{2}}}=\frac{45}{30} \Rightarrow \frac{{{r}_{1}}}{{{r}_{2}}}=\frac{3}{2}\\ {{r}_{1}}:{{r}_{2}}=3:2\\$
Example 7: The perimeter of a sector of a circle of area 64 π sq. cm is 56 cm, then area of sector is
(A) 140 sq.cm
(B) 150 sq.cm
(C) 160 sq.cm
(D) 170 sq.cm
Solution:
A = 64 π
$\pi r^{2}$ = 64π
∴ r = 8 cm
Since, perimeter of sector = 2r + S
∴ S + r + r = 56
∴ S + 8 + 8 = 56
∴ S = 40
Since, S = r $\Theta$
∴ 40 = 8 × $\Theta$
∴ θ = 5c
∴ Area of sector = $\frac{1}{2}r^{2} \Theta = \frac{1}{2}[8^{2} *5]=160$ sq.cm. |
# How To Find Number Of Subsets
Subset is part of one of the mathematical concepts called Sets. A set is a collection of objects or elements, grouped in curly braces, such as {a, b, c, d}. If a set A is the set of even numbers and the set B consists of {2,4,6}, then B is said to be a subset of A, denoted FIRE, and A is a subset of B Learn Subset And Superset to understand the difference. The elements of the set can be anything such as a group of real numbers, variables, constants, integers, etc. It also includes an empty set. Let us discuss the subset here with its types and examples.Table of contents:
• Definition
• Types
• Set of real numbers
• The appropriate subset notation
• Formula
• Subsets and Appropriate Subsets
• Subset does not match
• Characteristic
• Solved examples
## What is a subset in Mathematics?
Contents
Set A is said to be a subset of set B if all the elements of set A are also in set B. In other words, set A is contained within set B. Example: If set A is contained within set B, for example A has {X, Y} and set B has {X, Y, Z} then A is a subset of B because the elements of A are also in set B.
### Subset icons
In set theory, a subset is denoted by the symbol ⊆ and is read as ‘is a subset of’. Using this notation, we can represent the subsets as follows: A ⊆ B; means that Set A is a subset of Set B. Note: A subset can be equal to that set. That is, a subset can contain all the elements contained in that set.
### All subsets of a set
The subsets of any set include all possible sets including its elements and the empty set. Let us understand with the help of an example. Example: Find all subsets of set A = {1,2,3,4} Solution: Given, A = {1,2,3,4} Subset = {} {1 }, {2} , {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2, 3}, {2,3,4}, {1,3,4}, {1,2,4} {1,2,3,4}
## Types of subsets
The subsets are classified as
• Set of real numbers
• Subset does not match
The appropriate subset is the one that contains some elements of the original set while the unsuitable subset contains all the elements of the original set along with the empty set. For example, if set A = {2, 4, 6} then Number of subsets: {2}, {4}, {6}, {2,4}, {4,6}, {2, 6}, {2,4,6} and Φ or {} .Proper Subsets: { }, {2}, {4}, {6}, {2,4}, {4,6}, {2,6 } Inappropriate subset: {2,4,6} Read more: how to restart pokemon ultra moonThere is no specific recipe to find subsets, we have to list them all instead them to distinguish between subsets and nonconforming sets. Set theory notation was developed by mathematicians to describe sets of objects.
## What is the appropriate subset?
Set A is considered to be an appropriate subset of Set B if Set B contains at least one element that is not present in Set A. Example: If Set A has elements of {12, 24} and set B has elements {12, 24, 36} then set A is a suitable subset of B because 36 is not in set A.
### The appropriate subset notation
A suitable subset is denoted ⊂ and read as ‘is a suitable subset of’. Using this notation, we can express a suitable subset for set A and set B as; A B
### Proper Subset Formula
If we have to choose n number of elements from a set containing N number of elements then it can be done in number of ways NCn So the number of possible subsets containing n number of elements from a set containing N number elements are equal to NCn.
### How many subsets and appropriate subsets does a set have?
If a set has “n” elements, then the number of subsets of the given set is 2n and the number of appropriate subsets of the given subset is 2n-1. Consider an example, If the set A has elements, A = {a, b}, then the appropriate subset of the given subset is {}, {a} and {b}. Here, the number of elements in the set is 2 We know that the formula for the number of subsets is 2n – 1. = 22 – 1 = 4 – 1 = 3 Therefore, the appropriate number of subsets of the given set is 3 ({}, {a}, {b}).
## What is a non-conforming subset?
A subset containing all the elements of the original set is called a non-conforming subset. It is denoted by ⊆.For example: Let P = {2,4,6} Then the subsets of P are; {}, {2}, {4}, {6}, {2,4}, {4,6}, {2,6} and {2,4,6}. Where, {}, {2}, {4}, {6}, {2,4}, {4,6}, {2,6} are the appropriate subsets and {2 , 4,6} are subsets do not match. Therefore, we can write {2,4,6} ⊆ P.Note: The drum set Is one its own mismatched subset (because it is equal to itself) but it is an appropriate subset of any other set.
### The source
The the source is said to be the set of all subsets. It is represented by P(A). If A is a set with elements {a, b}. Then the set of powers of A will be; P(A) = {∅, {a}, {b}, {a, b}} To learn more briefly, click on the article link on power sets.
## Attribute of subset
Read more: how to reheat olive garden bread | Top Q&A Some important properties of a subset are:
• Every set is considered to be a subset of the given set itself. Means X ⊂ X or Y ⊂ Y, etc
• It can be said that an empty set is considered as a subset of all sets.
• X is a subset of Y. That means X is contained in Y
• If the set X is a subset of the set Y, we can say that Y is a subset of X .
• Set subsets and subsets
• Union of Ministries
• Universal set
## Subset example problem
Example 1: How many three-element subsets can be formed from that set?S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}Solution: Number of elements in a set = 10Number of elements in a subset = 3 Therefore, the number of subsets that can contain 3 elements = 10C3Therefore, the number of possible subsets containing 3 elements from the set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is 120. Example 2: Give any two real-life examples on the subset.Solution: We can find many examples of subsets in daily life like: |
# To estimate: The value of the function when x approaches zero by graphing the function f ( x ) .
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 2.3, Problem 25E
(a)
To determine
## To estimate: The value of the function when x approaches zero by graphing the function f(x).
Expert Solution
The estimated value of the function when x approaches zero f(x) approaches 0.6667 or 23.
### Explanation of Solution
Given:
The graph of the function f(x)=x1+3x1.
Draw the graph of the function f(x)=x1+3x1 by using the graphing calculator as shown below in Figure 1.
From the graph, as x=0, then f(x) is not defined. But x approaches 0, then f(x) goes to 23.
That is, limx0x1+3x123.
Thus, the estimated value of limx0x1+3x1 is 23_.
(b)
To determine
### To guess: The value of the limit by using the table of values of f(x) for x close to 0.
Expert Solution
The value of the limit by using the table of values of f(x) for x close to 0 is 23_.
### Explanation of Solution
Calculation:
Make the table of values of f(x) for x close to 0.
x 1+3x−1 f(x)=x1+3x−1 −0.001 1+3(−0.001)−1 ≈−0.0015011267 −0.001−0.0015011267≈0.666 166 3 −0.000 1 1+3(−0.0001)−1 ≈−0.000150011 −0.0001−0.000150011≈0.666 6167 −0.000 01 1+3(−0.00001)−1 ≈−0.00001500011 −0.00001−0.00001500011≈0.666 6617 −0.000 001 1+3(−0.000001)−1 ≈−0.000001500001125 −0.000001−0.0000015000011≈0.666 6662 0.000 001 1+3(0.000001)−1 ≈0.000001499998875 0.0000010.000001499998875≈0.666 6672 0.000 01 1+3(0.00001)−1 ≈0.0000149998875 0.000010.0000149998875≈0.666 6717 0.000 1 1+3(0.0001)−1 ≈0.00014998875 0.00010.00014998875≈0.666 7167 0.001 1+3(0.001)−1 ≈0.001498877 0.0010.00014998875≈0.667 1663
From the table, as x gets more close to 0, the value of f(x) approaches 0.66667 or 23_.
That is, limx0x1+3x123.
Thus, the limit appears to be 23_.
(c)
To determine
Expert Solution
### Explanation of Solution
Given:
The limit of the function as x approaches 0 is f(x)=x1+3x1.
Limit Laws:
Suppose that c is a constant and the limits limxaf(x) and limxag(x) exist. Then
Limit law 1: limxa[f(x)+g(x)]=limxaf(x)+limxag(x)
Limit law 2: limxa[f(x)g(x)]=limxaf(x)limxag(x)
Limit law 3: limxa[cf(x)]=climxaf(x)
Limit law 4: limxa[f(x)g(x)]=limxaf(x)limxag(x)
Limit law 5: limxaf(x)g(x)=limxaf(x)limxag(x) if limxag(x)0
Limit law 6: limxa[f(x)]n=[limxaf(x)]n where n is a positive integer
Limit law 7: limxac=c
Limit law 8: limxax=a
Limit law 9: limxaxn=an where n is a positive integer.
Limit law 10: limxaxn=an where n is a positive integer, if n is even, assume that a>0.
Limit law 11: limxaf(x)n=limxaf(x)n where n is a positive integer, if n is even, assume that limxaf(x)>0.
Note 1:
The Quotient rule is not applicable directly for the function f(x) because the limit of the denominator is zero.
limx0(1+3x1)=limx0(1+3x)limx01 (by limit law 2)=(limx0(1+3x))limx01 (by limit law 11)=limx01+limx0(3x)1 (by limit law 1 and 7)
=1+3limx0(x)1 (by limit law 3)=1+3(0)1 (by limit law 8)=11=0
Note 2:
The limit may be infinite or it may be some finite value when both numerator and denominator approach 0.”
Calculation:
By note 3, take the limit x approaches 0 but x0.
Simplify f(x) by using elementary algebra, f(x)=x1+3x1.
Take the conjugate of the denominator and multiply and divide of f(x).
f(x)=x1+3x1×1+3x+11+3x+1=x(1+3x+1)(1+3x1)(1+3x+1)
Use the difference of square formula,
f(x)=x(1+3x+1)((1+3x)2(1)2)=x(1+3x+1)(1+3x1)=x(1+3x+1)3x
Since the limit x approaches 0 but not equal to 0, cancel the common term x0 of both numerator and denominator,
f(x)=(1+3x+1)3=13(1+3x+1)
Use fact 1, f(x)=13(1+3x+1) and x0, then
limx0(x1+3x1)=limx0(13(1+3x+1)).
Use the limit laws to obtain the limit of the function as below:
limx0(13(1+3x+1))=13limx0(1+3x+1) (by limit law 3)=13(limx0(1+3x)+limx01) (by limit law 1)=13((limx0(1+3x))+limx01) (by limit law 11)
=13((limx01+3limx0x)+limx01) (by limit law 1)=13((1+3(0))+(1)) (by limit law 7, 8)=13(1+1)=23
Thus, the limit of the function is 23_.
Hence the required proof is obtained.
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# Special Lines in Triangles - Medians
### New York State Common Core Math Geometry, Module 1, Lesson 30
Student Outcomes
• Students examine the relationships created by special lines in triangles, namely medians.
Special Lines in Triangles - Medians
Classwork
Opening Exercise
In β³ π΄π΅πΆ to the right, π· is the midpoint of π΄π΅, πΈ is the midpoint of π΅πΆ, and πΉ is the midpoint of π΄πΆ. Complete each statement below.
π·πΈ is parallel to ____ and measures ____ the length of ____. π·πΉ is parallel to ____ and measures ____ the length of ____. πΈπΉ is parallel to ____ and measures ____ the length of ____.
Discussion
In the previous two lessons, we proved that (a) the midsegment of a triangle is parallel to the third side and half the length of the third side and (b) diagonals of a parallelogram bisect each other. We use both of these facts to prove the following assertion:
All medians of a triangle are ____. That is, the three medians of a triangle (the segments connecting each vertex to the midpoint of the opposite side) meet at a single point. This point of concurrency is called the ____, or the center of gravity, of the triangle. The proof also shows a length relationship for each median: The length from the vertex to the centroid is ____ the length from the centroid to the midpoint of the side.
Example 1
Provide a valid reason for each step in the proof below.
Given: β³ π΄π΅πΆ with π·, πΈ, and πΉ the midpoints of sides Μ
π΄π΅Μ
Μ
Μ
, π΅πΆΜ
Μ
Μ
Μ
, and π΄πΆΜ
Μ
Μ
Μ
, respectively
Prove: The three medians of β³ π΄π΅πΆ meet at a single point.
(1) Draw midsegment π·πΈ. Draw π΄πΈ and π·πΆ; label their intersection as point πΊ.
(2) Construct and label the midpoint of π΄πΊ as point π» and the midpoint of πΊπΆ as point π½.
(3) π·πΈ β₯ π΄πΆ,
(4) π»π½ β₯ π΄πΆ,
(5) π·πΈ β₯ π»π½,
(6) π·πΈ = 1/2 π΄πΆ and π»π½ = 1/2π΄πΆ,
(7) π·πΈπ½π» is a parallelogram.
(8) π»πΊ = πΈπΊ and π½πΊ = π·πΊ,
(9) π΄π» = π»πΊ and πΆπ½ = π½πΊ,
(10) π΄π» = π»πΊ = πΊπΈ and πΆπ½ = π½πΊ = πΊπ·,
(11) π΄πΊ = 2πΊπΈ and πΆπΊ = 2πΊπ·,
(12) We can complete Steps (1)β(11) to include the median from π΅; the third median, π΅πΉ, passes through point πΊ, which divides it into two segments such that the longer part is twice the shorter.
(13) The intersection point of the medians divides each median into two parts with lengths in a ratio of 2:1; therefore, all medians are concurrent at that point.
The three medians of a triangle are concurrent at the , or the center of gravity. This point of concurrency divides the length of each median in a ratio of ; the length from the vertex to the centroid is the length from the centroid to the midpoint of the side.
Example 2
In β³ π΄π΅πΆ, the medians are concurrent at πΉ. π·πΉ = 4, π΅πΉ = 16, π΄πΊ = 30. Find each of the following measures.
a. πΉπΆ =
b. π·πΆ =
c. π΄πΉ =
d. π΅πΈ =
e. πΉπΊ =
f. πΈπΉ =
Example 3
In the figure to the right, β³ π΄π΅πΆ is reflected over π΄π΅ to create β³ π΄π΅π·. Points π, πΈ, and πΉ are midpoints of π΄π΅, π΅π·, and π΅πΆ, respectively. If π΄π» = π΄πΊ, prove that ππ» = πΊπ.
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. |
# Video: Writing an Exponential Equation from a Table of Values
Write an exponential equation in the form π¦ = π^π₯ for the numbers in the table.
02:46
### Video Transcript
Write an exponential equation in the form π¦ equals π to the power of π₯ for the numbers in the table. π₯ equals zero, π¦ equals one. π₯ equals one, π¦ equals five. π₯ equals two, π¦ equals 25. And π₯ equals three, π¦ equals 125.
There are lots of ways of approaching this question. One way would be to substitute the values of π₯ and π¦ into the equation π¦ equals π to the power of π₯. Letβs consider the first column when π₯ equals zero, π¦ equals one. Substituting in these values gives us one is equal to π to the power of zero. We know from our laws of exponents or indices that anything to the power of zero is equal to one. Therefore, this doesnβt help us work out the value of π.
In the next column, weβre told that π₯ is equal to one and π¦ is equal to five. This gives us the equation five is equal to π to the power of one. Once again, from our laws of exponents, we know that anything to the power of one is equal to itself. We can therefore say that if π to the power of one is equal to five, π must be equal to five. As we have now calculated the value of π, we can rewrite the exponential equation as π¦ equals five to the power of π₯. Whilst this appears to be the correct answer, it is worth substituting in our values in column three and four to check that we are correct.
Substituting in π₯ equals two gives us π¦ is equal to five squared. Squaring a number is the same as multiplying it by itself. And five multiplied by five is 25. This means that the numbers π₯ equals two and π¦ equals 25 do fit the equation. Substituting in π₯ equals three gives us π¦ is equal to five cubed or five to the power of three. This is the same as five multiplied by five multiplied by five. Five cubed is therefore equal to 125. So this pair of numbers also fits the equation.
We can therefore conclude that the exponential equation π¦ equals five to the power of π₯ is correct. |
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# A ladder of mass $M$ and length $L$ is supported in equilibrium against a smooth vertical wall and a rough horizontal surface, as shown in figure. If $\theta$ is the angle of the angle of inclination of the rod with the horizontal, then calculate(a). normal reaction of the wall on the ladder(b). normal reaction of the ground on the ladder(c). net force applied by the ground on the ladder
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Hint: The ladder is in equilibrium against a wall, this means it is at rest or in uniform motion and no external forces act on it. We can apply Newton’s second law for forces acting in different directions and make equations. We can also apply equilibrium in its rotational motion and make equations for torque acting on it and solve the equations to calculate the value of forces.
Formulas used:
$\tau =F\times r$
${{N}_{1}}-F=0$
${{N}_{2}}-mg=0$
The system of ladder is an isolated system, analyzing all the forces acting on it and resolving the forces into its components as follows
Since the ladder is on rest against the wall, the resultant of all forces acting on it will be 0.
Forces acting on the ladder in the x-direction
\begin{align} & {{N}_{1}}-F=0 \\ & \Rightarrow {{N}_{1}}=F \\ \end{align}
Forces acting in the y-direction
\begin{align} & {{N}_{2}}-mg=0 \\ & \Rightarrow {{N}_{2}}=mg \\ \end{align}
Now taking torque about point O, we have,
Since the ladder is in equilibrium, the net torque acting on it is 0.
Therefore, the torque is calculated as-
$\tau =F\times r$ - (2)
Here, $\tau$ is the torque
$F$ is the force acting on the ladder
$r$ is the distance from the axis
The axis is passing through point O
Distance of point of action of ${{N}_{1}}$ from the axis is calculated as-
From the triangle given in the above figure
\begin{align} & \dfrac{x}{L}=\sin \theta \\ & \Rightarrow x=L\sin \theta \\ \end{align}
Distance of point of action of ${{N}_{2}}$ from the axis is calculated as-
\begin{align} & \dfrac{y}{L}=\cos \theta \\ & \Rightarrow y=L\cos \theta \\ \end{align}
Distance of point of action of $mg$ from the axis is $\dfrac{L}{2}\cos \theta$
Therefore, from eq (2), the torque acting on the ladder is-
\begin{align} & {{N}_{1}}L\sin \theta +{{N}_{2}}L\cos \theta -mg\dfrac{L}{2}\cos \theta =0 \\ & \Rightarrow FL\sin \theta +mgL\cos \theta -mg\dfrac{L}{2}\cos \theta =0 \\ & \Rightarrow FL\sin \theta +mg\dfrac{L}{2}\cos \theta =0 \\ & \Rightarrow FL\sin \theta =-mg\dfrac{L}{2}\cos \theta \\ & \Rightarrow F=-\dfrac{mg}{2}\cot \theta \\ \end{align}
Substituting values for ladder in the above equation, we get,
$\therefore F=-\dfrac{Mg}{2}\cot \theta$
Therefore,
(a). The normal reaction of the wall on the ladder is $\dfrac{Mg}{2}\cot \theta$.
(b). The normal reaction of the ground on the ladder is $Mg$.
(c). the force applied by the ground on the ladder is $\dfrac{Mg}{2}\cot \theta$
Note:
The negative sign indicates that the force is opposite to the assumed direction. The normal reaction is perpendicular to the surfaces and by Newton’s third law, it acts equal and opposite on both the surfaces. The frictional force resists the motion of a body; it is applied by rough surfaces. When an object is in equilibrium, no net force or torque acts on it. |
## Small Group Activity: The Wire
Vector Calculus II 2021
Students compute a vector line integral, then investigate whether this integral is path independent.
What students learn
• Practice evaluating line integrals;
• Practice choosing appropriate coordinates and basis vectors;
• Introduction to the geometry behind conservative vector fields.
Consider the vector field given by ($\mu_0$ and $I$ are constants): $\boldsymbol{\vec{B}} = {\mu_0 I\over2\pi} \left({-y\,\boldsymbol{\hat{x}}+x\,\boldsymbol{\hat{y}}\over x^2+y^2}\right) = {\mu_0 I\over2\pi} \, {\boldsymbol{\hat{\phi}}\over r}$
$\boldsymbol{\vec{B}}$ is the magnetic field around a wire along the $z$-axis carrying a constant current $I$ in the $z$-direction.
• Determine $\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}$ on any radial line of the form $y=mx$, where $m$ is a constant.
• Determine $\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}$ on any circle of the form $x^2+y^2=a^2$, where $a$ is a constant.
You may wish to express the equations for these curves in polar coordinates.
Go: For each of the following curves $C_i$, evaluate the line integral $\int\limits_{C_i}\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}$.
• $C_1$, the top half of the circle $r=5$, traversed in a counterclockwise direction.
• $C_2$, the top half of the circle $r=2$, traversed in a counterclockwise direction.
• $C_3$, the top half of the circle $r=2$, traversed in a clockwise direction.
• $C_4$, the bottom half of the circle $r=2$, traversed in a clockwise direction.
• $C_5$, the radial line from $(2,0)$ to $(5,0)$.
• $C_6$, the radial line from $(-5,0)$ to $(-2,0)$.
FOOD FOR THOUGHT
• Construct closed curves $C_7$ and $C_8$ such that this integral $\int\limits_{C_i}\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}$ is nonzero over $C_7$ and zero over $C_8$.
It is enough to draw your curves; you do not need to parameterize them.
• Ampère's Law says that, for any closed curve $C$, this integral is ($\mu_0$ times) the current flowing through $C$ (in the $z$ direction). Can you use this fact to explain your results to part (a)?
• Is $\boldsymbol{\vec{B}}$ conservative?
#### Main ideas
• Calculating (vector) line integrals.
• Use what you know!
#### Prerequisites
• Familiarity with $d\boldsymbol{\vec{r}}$.
• Familiarity with “Use what you know” strategy.
#### Warmup
This activity should be preceded by a short lecture on (vector) line integrals, which emphasizes that $\int_C\boldsymbol{\vec{F}}\cdot d\boldsymbol{\vec{r}}$ represents chopping up the curve into small pieces. Integrals are sums; in this case, one is adding up the component of $\boldsymbol{\vec{B}}$ parallel to the curve times the length of each piece.
#### Props
• whiteboards and pens
#### Wrapup
Emphasize that students must express everything in terms of a single variable prior to integration.
Point out that in polar coordinates (and basis vectors) \begin{eqnarray*} \boldsymbol{\vec{B}}= {\mu_0 I\over2\pi} {\boldsymbol{\hat{\phi}}\over r} \end{eqnarray*} so that using $d\boldsymbol{\vec{r}} = dr\,\boldsymbol{\hat{r}} + r\,d\phi\,\boldsymbol{\hat{\phi}}$ quickly yields $\boldsymbol{\vec{B}}\cdot dd\boldsymbol{\vec{r}}$ along a circular arc (${\mu_0 I\over2\pi}\,d\phi$) or a radial line ($0$), respectively.
### Details
#### In the Classroom
• Sketching the vector field takes some students a long time. If time is short, have them do this before class, or consider using MATLAB or similar technology to plot the field. Still, it's important to plot a few vectors by hand.
• Students who have not had physics don't know which way the current goes; they may need to be told about the right-hand rule.
• Some students may confuse the wire with the paths of integration.
• Students working in rectangular coordinates often get lost in the algebra of Question 2b. Make sure that nobody gets stuck here.
• Students who calculate $\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}={dy\over x}$ on a circle need to be reminded that at the end of the day a line integral must be expressed in terms of a single variable.
• Some students will be surprised when they calculate $\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}=0$ for radial lines. They should be encouraged to think about the directions of $\boldsymbol{\vec{B}}$ and $d\boldsymbol{\vec{r}}$.
• Most students will either write everything in terms of $x$ or $y$ or switch to polar coordinates. We discuss each of these in turn.
• This problem cries out for polar coordinates. Along a circular arc, $r=a$ yields $x=a\cos\phi$, $y=a\sin\phi$, so that $d\boldsymbol{\vec{r}}=-a\sin\phi\,d\phi\,\boldsymbol{\hat{x}}+a\cos\phi\,d\phi\,\boldsymbol{\hat{y}}$, from which one gets $\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}} = {\mu_0 I\over2\pi}\,d\phi$.
• Students who fail to switch to polar coordinates can take the differential of both sides of the equation $x^2+y^2=a^2$, yielding $x\,dx+y\,dy=0$, which can be solved for $dx$ (or $dy$) and inserted into the fundamental formula $d\boldsymbol{\vec{r}}=dx\,\boldsymbol{\hat{x}}+dy\,\boldsymbol{\hat{y}}$. Taking the dot product then yields, $\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}= {\mu_0 I\over2\pi} {dy\over x}$. Students may get stuck here, not realizing that they need to write $x$ in terms of $y$. The resulting integral cries out for a trig substitution --- which is really just switching to polar coordinates.
In either case, sketching $\boldsymbol{\vec{B}}$ should convince students that $\boldsymbol{\vec{B}}$ is tangent to the circular arcs, hence orthogonal to radial lines. Thus, along such lines, $\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}=0$; no calculation is necessary. (This calculation is straightforward even in rectangular coordinates.)
• Watch out for folks who go from $r^2=x^2+y^2$ to $d\boldsymbol{\vec{r}} = 2x\,dx\,\boldsymbol{\hat{x}} + 2y\,dy\,\boldsymbol{\hat{y}}$.
• Working in rectangular coordinates leads to an integral of the form $\int-{dx\over y}$, with $y=\sqrt{r^2-x^2}$. Maple integrates this to $-\tan^{-1}\left({x\over y}\right)$, which many students will not recognize as the polar angle $\phi$. If $r=1$, Maple instead integrates this to $-\sin^{-1}x$; same problem. One calculator (the TI-89?) appears to use arcsin in both cases.
#### Subsidiary ideas
• Independence of path.
#### Homework
• Any vector line integral for which the path is given geometrically, that is, without an explicit parameterization.
#### Essay questions
• Discuss when $\oint\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}$ around a closed curve will or will not be zero.
#### Enrichment
• This activity leads naturally into a discussion of path independence.
• Point out that $\boldsymbol{\vec{B}}\sim\nabla\phi$ everywhere (except the origin), but that $\boldsymbol{\vec{B}}$ is only conservative on domains where $\phi$ is single-valued.
• Discuss winding number, perhaps pointing out that $\boldsymbol{\vec{B}}\cdot d\boldsymbol{\hat{r}}$ is proportional to $d\phi$ along any curve.
• Discuss Ampère's Law, which says that $\oint\!\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}$ is ($\mu_0$ times) the current flowing through $C$ (in the $z$ direction).
Keywords
Line integrals conservative vector fields Ampere's Law simply-connectedness
Learning Outcomes |
# What is the formula of loan calculation?
## What is the formula of loan calculation?
A = Payment amount per period. P = Initial principal or loan amount (in this example, \$10,000) r = Interest rate per period (in our example, that’s 7.5% divided by 12 months) n = Total number of payments or periods.
## What is the formula to calculate monthly payments?
If you want to do the monthly mortgage payment calculation by hand, you’ll need the monthly interest rate — just divide the annual interest rate by 12 (the number of months in a year). For example, if the annual interest rate is 4%, the monthly interest rate would be 0.33% (0.04/12 = 0.0033).
How do you calculate loan payments manually?
To figure your mortgage payment, start by converting your annual interest rate to a monthly interest rate by dividing by 12. Next, add 1 to the monthly rate. Third, multiply the number of years in the term of the mortgage by 12 to calculate the number of monthly payments you’ll make.
How do I calculate loan amount in Excel?
=PMT(17%/12,2*12,5400) The rate argument is the interest rate per period for the loan. For example, in this formula the 17% annual interest rate is divided by 12, the number of months in a year. The NPER argument of 2*12 is the total number of payment periods for the loan. The PV or present value argument is 5400.
### How do you find the original amount of a loan?
We can calculate an original loan amount by using the Present Value Function (PV) if we know the interest rate, periodic payment, and the given loan term. This function tells the present value of an investment….Explanation
1. 0.0125.
2. The cell containing the interest rate divided by 12.
3. 15%/12.
### How do you calculate initial loan amount?
How to Calculate Initial Mortgage Loan Amount With Known Loan Amount
1. Calculate the interest rate per month by dividing the interest rate by 12 months.
2. Add 1 to the interest rate per month.
3. Raise the number calculated in Step 2 to the negative power of the number of payments made.
How do you find the original amount borrowed?
Principal Amount Formulas We can rearrange the interest formula, I = PRT to calculate the principal amount. The new, rearranged formula would be P = I / (RT), which is principal amount equals interest divided by interest rate times the amount of time.
How do I calculate a loan amount in Excel?
Enter “=PMT(A2/12,A3*12,A1)” into cell B4. This will calculate the monthly payment on your loan. The interest rate is divided by 12 to find the monthly interest rate and the term is multiplied by 12 to determine how many monthly payments you will make.
#### How do banks calculate loans?
Calculation
1. Divide your interest rate by the number of payments you’ll make that year.
2. Multiply that number by your remaining loan balance to find out how much you’ll pay in interest that month.
3. Subtract that interest from your fixed monthly payment to see how much in principal you will pay in the first month.
#### What does PV stand for in Excel?
present value
Use the Excel Formula Coach to find the present value (loan amount) you can afford, based on a set monthly payment. At the same time, you’ll learn how to use the PV function in a formula. Or, use the Excel Formula Coach to find the present value of your financial investment goal.
How do I calculate principal and interest on a loan in Excel?
Excel PPMT Function
1. Summary.
2. Get principal payment in given period.
3. The principal payment.
4. =PPMT (rate, per, nper, pv, [fv], [type])
5. rate – The interest rate per period.
6. The Excel PPMT function is used to calculate the principal portion of a given loan payment.
What is the formula for calculating a loan?
The formula for calculating a loan payment is: Monthly payment = P [{r(1+r)^n}/{(1+r)^n-1}] An explanation of the symbols: ^ : This denotes an exponent; in the equation, it would read, “One plus r raised to the power of n.”.
## What is the formula for calculating monthly mortgage?
Formula to Calculate Mortgage Payments The Formula. Principal amount: This is the amount of the mortgage or amount you want to borrow. Determine Overall Interest. If you want to know how much interest you’ll pay over the life of the loan, multiply the amount of your monthly payment by the term of Calculating Your Payment.
## How do you calculate a bank loan?
Compute the total number of payments you have to make on your bank loan. Multiply the payments per year by the number of years you will take to repay the loan. For example, if you will repay your loan over three years, you would multiply three by 12 to get 36 payments.
How to calculate monthly loan payment formula?
a: 100,000,the amount of the loan
• r: 0.005 (6% annual rate—expressed as 0.06—divided by 12 monthly payments per year)
• n: 360 (12 monthly payments per year times 30 years)
• Calculation: 100,000/{[(1+0.005)^360]-1}/[0.005 (1+0.005)^360]=599.55,or 100,000/166.7916=599.55 |
Instruction
1
The easiest way to determine the diameter (D) circumference can be used in the case when you know the radius (R) of the circle. By definition, the radius is the segment connecting the center of the circle with any point lying on the circle. From this it follows that the diameter be two sections, the length of each of which is equal to the radius: D=2*R.
2
Use to calculate the diameter (D) ratio called PI, if you are aware of the length of the perimeter (L). Perimeter for a circle is called the circumference and the number PI expresses a constant ratio between the diameter and the length of a circle in Euclidean geometry, the division of the perimeter of a circle to its diameter always equals PI. Means for finding the diameter and circumference you need to divide this constant: D=L/π.
3
From the formula of finding the area of a circle (S) it follows that to find the diameter (D) you should find the square root of the result of dividing the area by PI and double the value obtained is: D=2*√(S/π).
4
If the circle described by the rectangle and the length of its side is known, there is nothing to calculate there is no need in such a rectangle can be a square, and the length of its side is equal to the diameter of the circle.
5
In the case of circle inscribed in a rectangle the length of the diameter is the same as the length of its diagonal. To identify it with the known width (H) and height (V) of the rectangle, you can use the Pythagorean theorem since the triangle formed by the diagonal, width and height will be rectangular. From theorem it follows that the length of the diagonal of the rectangle, and hence the diameter of a circle is equal to the square root of the sum of squares of width and height: D= √(H2+V2). |
# AP Statistics Curriculum 2007 Hypothesis Proportion
(Difference between revisions)
Revision as of 00:29, 17 February 2008 (view source)IvoDinov (Talk | contribs)m (→Hypothesis Testing about a Mean: Small Samples)← Older edit Revision as of 22:53, 1 March 2008 (view source) (→Hypothesis Testing about a Sinlge Sample Proportion)Newer edit → Line 10: Line 10: : $SE_{\hat{p}} = \sqrt{\hat{p}(1-\hat{p})\over n} \longrightarrow SE_{\tilde{p}} = \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.$ : $SE_{\hat{p}} = \sqrt{\hat{p}(1-\hat{p})\over n} \longrightarrow SE_{\tilde{p}} = \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.$ - === Hypothesis Testing about a Sinlge Sample Proportion=== + === Hypothesis Testing about a Single Sample Proportion=== * Null Hypothesis: $H_o: p=p_o$ (e.g., 0), where ''p'' is the population proportion of interest. * Null Hypothesis: $H_o: p=p_o$ (e.g., 0), where ''p'' is the population proportion of interest. * Alternative Research Hypotheses: * Alternative Research Hypotheses:
## General Advance-Placement (AP) Statistics Curriculum - Testing a Claim about Proportion
### Testing a Claim about Proportion
Recall that for large samples, the sampling distribution of the sample proportion $\hat{p}$ is approximately Normal, by CLT, as the sample proportion may be presented as a sample average or Bernoulli random variables. When the sample size is small, the normal approximation may be inadequate. To accommodate this, we will modify the sample-proportion $\hat{p}$ slightly and obtain the corrected-sample-proportion $\tilde{p}$:
$\hat{p}={y\over n} \longrightarrow \tilde{y}={y+0.5z_{\alpha \over 2}^2 \over n+z_{\alpha \over 2}^2},$
The standard error of $\hat{p}$ also needs a slight modification
$SE_{\hat{p}} = \sqrt{\hat{p}(1-\hat{p})\over n} \longrightarrow SE_{\tilde{p}} = \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.$
### Hypothesis Testing about a Single Sample Proportion
• Null Hypothesis: Ho:p = po (e.g., 0), where p is the population proportion of interest.
• Alternative Research Hypotheses:
• One sided (uni-directional): H1:p > po, or Ho:p < po
• Double sided: $H_1: p \not= p_o$
• Test Statistics: $Z_o={\tilde{p} -p_o \over SE_{\tilde{p}}}$
### Example
Suppose a researcher is interested in studying the effect of aspirin in reducing heart attacks. He randomly recruits 500 subjects with evidence of early heart disease and has them take one aspirin daily for two years. At the end of the two years, he finds that during the study only 17 subjects had a heart attack. Use α = 0.05 to formulate a test a research hypothesis that the proportion of subject on aspirin treatment that have heart attacks within 2 years of treatment is po = 0.04.
$\tilde{p} = {17+0.5z_{0.025}^2\over 500+z_{0.025}^2}== {17+1.92\over 500+3.84}=0.038$
$SE_{\tilde{p}}= \sqrt{0.038(1-0.038)\over 500+3.84}=0.0085$
And the corresponding test statistics is
$Z_o={\tilde{p} - 0.04 \over SE_{\tilde{p}}}={0.002 \over 0.0085}=0.2353$
The p-value corresponding to this test-statistics is clearly insignificant.
### Genders of Siblings Example
Is the gender of a second child influenced by the gender of the first child, in families with >1 kid? Research hypothesis needs to be formulated first before collecting/looking/interpreting the data that will be used to address it. Mothers whose 1st child is a girl are more likely to have a girl, as a second child, compared to mothers with boys as 1st children. Data: 20 yrs of birth records of 1 Hospital in Auckland, New Zealand.
Second Child Male Female Total First Child Male 3,202 2,776 5,978 Female 2,620 2,792 5,412 Total 5,822 5,568 11,390
Let p1=true proportion of girls in mothers with girl as first child, p2=true proportion of girls in mothers with boy as first child. The parameter of interest is p1p2.
• Hypotheses: Ho:p1p2 = 0 (skeptical reaction). H1:p1p2 > 0 (research hypothesis).
Second Child Number of births Number of girls Proportion Group 1 (Previous child was girl) n1 = 5412 2792 $\hat{p}_1=0.516$ 2 (Previous child was boy) n2 = 5978 2776 $\hat{p}_2=0.464$
• Test Statistics: $Z_o = {Estimate-HypothesizedValue\over SE(Estimate)} = {\hat{p}_1 - \hat{p}_2 - 0 \over SE(\hat{p}_1 - \hat{p}_2)} = {\hat{p}_1 - \hat{p}_2 - 0 \over \sqrt{{\hat{p}_1(1-\hat{p}_1)\over n_1} + {\hat{p}_2(1-\hat{p}_2)\over n_2}}} \sim N(0,1)$ and Zo = 5.4996.
• $P_value = P(Z>Z_o)< 1.9\times 10^{-8}$. This small p-values provides extremely strong evidence to reject the null hypothesis that there are no differences between the proportions of mothers that had a girl as a second child but had either boy or girl as their first child. Hence there is strong statistical evidence implying that genders of siblings are not independent.
• Practical significance: The practical significance of the effect (of the gender of the first child on the gender of the second child, in this case) can only be assessed using confidence intervals. A 95% CI(p1p2) = [0.033;0.070] is computed by $p_1-p_2 \pm 1.96 SE(p_1 - p_2)$. Clearly, this is a practically negligible effect and no reasonable person would make important prospective family decisions based on the gender of their (first) child. |
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Exercise 11.1 Page: 208
1. The Length and the breadth of a rectangular piece of land are 500 m and 300 m, respectively. Find
(i) Its area (ii) the cost of the land, if 1 m2 of the land costs Rs 10,000.
Solution:-
From the question it is given that,
Length of the rectangular piece of land = 500 m
Breadth of the rectangular piece of land = 300 m
Then,
(i) Area of rectangle = Length × Breadth
= 500 × 300
= 150000 m2
(ii) Cost of the land for 1 m2 = Rs 10000
Cost of the land for 150000 m2 = 10000 × 150000
= Rs 1500000000
2. Find the area of a square park whose perimeter is 320m.
Solution:-
From the question it is given that,
Perimeter of the square park = 320 m
4 × Length of the side of park = 320 m
Then,
Length of the side of park = 320/4
= 80 m
So, Area of the square park = (length of the side of park)2
= 802
= 6400 m2
3. Find the breadth of a rectangular plot of land, if its area is 440 m2Â and the length is 22 m. Also find its perimeter.
Solution:-
From the question it is given that,
Area of the rectangular plot = 440 m2
Length of the rectangular plot = 22 m
We know that,
Area of the rectangle = Length × Breadth
Then,
Perimeter of the rectangle = 2(Length + Breadth)
= 2 (22 + 20)
= 2(42)
= 84 m
∴Perimeter of the rectangular plot is 84 m.
4. The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth.
Also find the area.
Solution:-
From the question it is given that,
Perimeter of the a rectangular sheet = 100 cm
Length of the rectangular sheet = 35 cm
We know that,
Perimeter of the rectangle = 2 (Length + Breadth)
100 = 2 (35 + Breadth)
Then,
Area of the rectangle = Length × Breadth
= 35 × 15
= 525 cm2
∴Area of the rectangular sheet is 525 cm2
5. The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:-
From the question it is given that,
Area of a square park is the same as of a rectangular park.
Side of the square park = 60 m
Length of the rectangular park = 90 m
We know that,
Area of the square park = (one of the side of square)2
= 602
= 3600 m2
Area of the rectangular park = 3600 m2 … [∵ given]
6. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side?
Also find which shape encloses more area?
Solution:-
By reading the question we can conclude that, perimeter of the square is same as perimeter of rectangle.
From the question it is given that,
Length of the rectangle = 40 cm
Breadth of the square = 22 cm
Then,
Perimeter of the rectangle = Perimeter of the Square
2 (Length + Breadth) = 4 × side
2 (40 + 22) = 4 × side
2 (62) = 4 × side
124 = 4 × side
Side = 124/4
Side = 31 cm
So, Area of the rectangle = (Length × Breadth)
= 40 × 22
= 880 cm2
Area of square = side2
= 312
= 31 × 31
= 961 cm2
∴Square shaped wire encloses more area.
7. The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is
30 cm, find its length. Also find the area of the rectangle.
Solution:-
From the question it is given that.
Perimeter of the rectangle = 130 cm
Breadth of the rectangle = 30
We know that,
Perimeter of rectangle = 2 (Length + Breadth)
130 = 2 (length + 30)
130/2 = length + 30
Length + 30 = 65
Length = 65 – 30
Length = 35 cm
Then,
Area of the rectangle = Length × Breadth
= 35 × 30
= 1050 cm2
8. A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (Fig). Find the cost of white washing the wall, if the rate of white washing the wall is Rs 20 per m2.
Solution:-
From the question it is given that,
Length of the door = 2 m
Breadth of the door = 1 m
Length of the wall = 4.5 m
Breadth of the wall = 3.6 m
Then,
Area of the door = Length × Breadth
= 2 × 1
= 2 m2
Area of the wall = Length × Breadth
= 4.5 × 3.6
= 16.2 m2
So, Area to be white washed = 16.2 – 2 = 14.2 m2
Cost of white washing 1 m2 area = Rs 20
Hence cost of whit washing 14.2 m2 area = 14.2 × 20
= Rs 284 |
# 4.2 Transformation of graphs by modulus function (Page 4/4)
Page 4 / 4
$y=f\left(x\right)\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}x=|f\left(y\right)|$
The invertible function x= f(y) has its inverse function given by y=f⁻¹(x). Alternatively, if a function is defined as y=f⁻¹(x), then variables x and y are related to each other such that x=f(y). We conclude that graph of y=f⁻¹(x) is same as graph of x=f(y) with the same orientation of x and y axes. It is important to underline here that we transform (change) graph of inverse of given function i.e. y=f⁻¹(x) to get the transformation of graph of x=f(y). Further x and y coordinates on the graph correspond to x and y values.
We interpret assignment of |f(y)| to x in the given graph in accordance with the definition of modulus function. Consider x=|f(y)|. But, modulus can not be equated to negative value. Hence, x can not be negative. It means we need to discard left half of the graph of inverse function y=f⁻¹(x). On the other hand, modulus of negative or positive value is always positive. Hence, positive value of x=a correspond to two values of function in dependent variable, a=±f(y). Corresponding to these two function values in y, we have two values of y i.e. f⁻¹(a) and f⁻¹(-a). In order to plot two values, we need to take mirror image of the left half of the graph of y=f⁻¹(x) across y-axis. This is image in y-axis.
From the point of construction of the graph of x=|f(y)|, we need to modify the graph of y=f⁻¹(x) i.e. x=f(y) as :
1 : take mirror image of left half of the graph in y-axis
2 : remove left half of the graph
This completes the construction for x=|f(y)|.
Problem : Draw graph of $x=|\mathrm{cosec}y|;\phantom{\rule{1em}{0ex}}x\in \left\{-\pi /2,\pi /2\right\}$ .
Solution : The inverse of base function is cosec⁻¹x. We first draw the graph of inverse function. Then, we take mirror image of left half of the graph in y-axis and remove left half of the graph to complete the construction of graph of $x=|\mathrm{cosec}y|$ .
## Examples
Problem : Find domain of the function given by :
$f\left(x\right)=\frac{1}{\sqrt{|\mathrm{sin}x|+\mathrm{sin}x}}$
Solution : The square root gives the condition :
$⇒|\mathrm{sin}x|+\mathrm{sin}x\ge 0$
But denominator can not be zero. Hence,
$⇒|\mathrm{sin}x|+\mathrm{sin}x>0$
$⇒|\mathrm{sin}x|>-\mathrm{sin}x$
We shall make use of graphing technique to evaluate the interval of x. Since both functions are periodic. It would be indicative of the domain if we confine our consideration to 1 period of sine function (0, 2π) and then extend the result subsequently to other periodic intervals.
We first draw sine function. To draw |sinx|, we take image of lower half in x-axis and remove the lower half. To draw “–sinx”, we take image of y=sinx in x-axis.
From the graph, we see that |sinx| is greater than “-sinx” in (0,π). Note that end points are not included. The domain is written with general notation as :
$x\in \left(2n\pi ,\left(2n+1\right)\pi \right)$
Problem : Determine graphically the points where graphs of $|y|={\mathrm{log}}_{e}|x|$ and ${\left(x-1\right)}^{2}+{y}^{2}-4=0$ intersect each other.
Solution : The function $|y|={\mathrm{log}}_{e}|x|$ is obtained by transforming $y=\mathrm{log}{}_{e}x$ . To draw $y={\mathrm{log}}_{e}|x|$ , we need to remove left half (but here there is no left half) and take image of right half in y-axis. To draw $|y|=\mathrm{log}{}_{e}|x|$ , we transform the graph of $y=\mathrm{log}{}_{e}|x|$ . For this, we remove the lower half and take image of upper half in x-axis.
On the other hand, ${\left(x-1\right)}^{2}+{y}^{2}-4=0$ is a circle with center at 1,0 having radius of 2 units. Finally, superposing two graphs, we determine the intersection points.
Clearly, there are three intersection points as shown by solid circles.
## Exercises
Draw the graph of function given by :
$f\left(x\right)=\frac{1}{\left[x\right]-1}$
Hints : Draw 1/x, which is a hyperbola with center at (0,0). Then draw 1/x-1. It is a hyperbola shifted right by 1 unit. Its center is (1,0). Remove left half and take the image of right half in y-axis.
2. Draw the graph of function given by :
$f\left(x\right)=||\frac{1}{x}|-1|$
Hints : Draw 1/x, which is a hyperbola with center at (0,0). Then draw |1/x|. Take image of lower half in x-axis. Remove lower half. To draw |1/x|-1, shift down the graph of |1/x| by 1 unit. To draw ||1/x|-1|, Take image of lower half of the graph of |1/x|-1 in x-axis. Remove lower half.
#### Questions & Answers
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Period of sin^6 3x+ cos^6 3x
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Period of sin^6 3x+ cos^6 3x
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# Lesson video
In progress...
Hello, Ms. Seton here again.
Welcome back to lesson 12.
Where are we going to continue to explore the simplest form in equivalent fractions.
I left you with this practise activity where you had to find the equivalent fraction first and then identify which one was in its simplest form.
Did you get the same pairings as me? Excellent, well done.
Let's have a look at one of those together.
Let's take 1/10 and 10/100.
Did you say that 1/10 was in its simplest form? Fantastic, perhaps we can use that STEM sentence to explain why again did you say that 1/10 is in its simplest form because the numerator denominator have been made as small as possible whilst keeping the proportion the same.
And how have the numerator and denominator been made as small as possible? The numerator and denominator, that's right.
They've been divided by the same number, which is 10.
Excellent.
So the numerator is 1/10 of the denominator and the denominator is 10 times the numerator.
Did you find that with all of the fraction pass? Excellent.
What can you tell me about the fractions that we've identified as the simplest form? Do you notice one that's maybe an odd one out? Yes, super, 3/4.
Why? Why did you identify that as being a bit of an odd one out? Because it's not a unit fraction.
And we did say, didn't we in the previous lesson that very often the simplest form has been reflected by a unit fraction but it doesn't always have to be.
I'm going to look at that more in this session.
What could you find out about 2/10, I've left it in the middle there because I wasn't sure where to put it.
Why did you put it? Did you say it was not in its simplest form or in its simplest form ? It's not in its simplest form, fantastic.
And why not? Because the numerator and denominator can be divided by the same amount to make an equivalent fraction.
And what is the amount? They can be divided by two.
Fantastic.
And the equivalent fraction will be 1/5 which is a unit fraction again, isn't it? And that would be in its simplest form, really well done.
How did you get on with a challenge? Do you think my puppy the preferred 1/2 an hour or 30/60 of an hour for a walk? Yes, you're quite right.
T decided to choose 30/60, yes because he thought that would be a longer walk.
What do you think? No, you're quite right.
1/2 an hour and 30/60 of an hour are the same amount because our equivalent fractions.
What else do you notice about 30/60? Every 60 minutes in an hour, fantastic.
So the denominator represents the whole house being divided into 60 equal parts, which in the minutes, and 30 of those minutes is 1/2 an hour really well done.
I like you to have a look at this fraction wall.
So have a think back to what we did in the last session.
I'd like you to pause me now and come back and tell me everything that you notice.
Did you come back? How did he get on? What did you notice about this family of equivalent fractions.
Excellent, so you noticed haven't you? That they're equivalent because they're all the same proportion of the whole.
So for example, if we look at 9/12 here, that's the same proportion of the whole as is 3/4.
Brilliant.
Did you notice anything else? Excellent, yes.
So for 3/4, the numerator and denominator have been kept as small as possible.
But there's still the same proportion of the whole and that vertical relationship is the same proportion as well.
Did you notice what that was? Brilliant, yes.
So the numerator is 3/4 of the denominator, excellent.
Could you tell me anything else that you noticed, I bet you've written down haven't you? That, let's say it together, 3/4 is the simplest form in this family fractions.
And we met that STEM centres, didn't we, in the last session? Should we say it again together? The simplest form in this family of fractions, is 3/4.
Well done.
Let me look at this fraction wall in the last session.
We looked at 1/4.
The same and what's different between 1/4 and 3/4? When we look at them on this fraction wall.
Yes, that's right.
1/4 was a unit fraction.
And we did say the, on the whole a lot of the simplest farmer fractions are unit fractions.
Well, in this case, three quarters is not a unit fraction.
Well done.
We also looked at this multiplication square didn't we? In the last session and we found an equivalent fraction family with the simplest form was 1/7 really well done.
And we can see that here can't we? 1/7.
Have a look at this yellow shady bar.
And I wonder, could you pause me? And write the equivalent fraction family represented by these two bars pause me now and have a ago.
How did you get on? Did he get the same family of equivalent fractions as me? Fantastic.
Yes.
So we can say count with a 5/7 is equivalent to 10/14 which is also the same proportion of the whole is 20/28.
And that's the same proportion of the whole is 35/49.
Why? Why are they the same proportion of the whole? What else might you have noticed? Brilliant.
Excellent.
So you can say can't you? That the numerator denominator multiplied by the same amount and the value of the fraction remains the same.
It's some of you write as examples.
So for example, here, 5/7 is the same proportion of the whole as 10/14 because they've both been multiplied by two.
The numerator and denominator have been multiplied by two.
15/21, the numerator denominator has been divided or multiplied by three really well done What about 30/42 What if the numerator denominator being multiplied or divided by? They've been multiplied by six really well done.
So we can see here can't we, that this generalisations correct? Let's say it again together.
When the numerator and denominator are multiplied or divided by the same number the value of the fraction remains the same.
Excellent.
So yes, 5/7 is the simplest form.
Now, how is that different to what we saw in the previous lesson? Yes, we said the simplest form was 1/7.
Didn't we? So the simplest form in this family of fractions is 5/7 And we're beginning to see that, that a simplest form, doesn't always have to be a unit fraction.
Welcome.
Can you tell me what the simplest farm is in this family, the equivalent fractions? Yes.
That's right.
The simplest form is 2/3.
Fantastic.
Why is it 2/3? And you tell me using the STEM sentence that we looked at.
So 2/3 is the simplest form in this family of fractions because the numerator denominator are made as small as possible while keeping the proportion the same.
And we can see can't we? The 2/3 is the same proportion of the whole.
So these are equivalent fractions, well done Taken two of those equivalent fractions.
And I'd like you to tell me what we said in the last session is still true.
So we said in the simplest form, we have scaled.
Do you remember that language? The numerator and denominator to make them a smaller so we can.
Pause me, now see if that's true and see if you can tell me anything else that you notice.
We come back.
What did you find? Is this correct? Did you agree? Yes.
You did agree so So, what have the numerator and denominator been scaled by to keep them as small as we can.
So 6/9 the numerator and denominator have been divided by three.
Excellent, well done.
And so 2/3 is the simplest farm.
Now we also said in the simplest form we made the numerator one, have a little look at those again.
Is this true now? What do you think? No Not true.
Is it? So in the simplest part, we don't always have to make the numerator one.
So we finding out now aren't we that simply spam can look very different than a unit fraction.
And we also said in the simplest form we are making the fractions as small as possible.
And I hope you remember, we didn't agree with that.
Did we? Have a look at those and see, do you still disagree with that statement that we're making the fractions as small as possible? And tell me why as well.
What did you think? No, that's right.
We're not making the fractions any smaller are we? The numerator denominator are made as small as possible while keeping the proportion the same.
Shall we say that one again together? The numerator and denominator are made as small as possible while keeping the proportion the same.
Well, done.
What's the sequence of lessons? You've met the two generalisations at the top of this screen.
I'd like us to take a moment now just to check if these are the true and also for you to prove it to me.
So if we take this first generalisation let's say it together.
The numerator and denominator made as small as possible while keeping the proportion the same and have a look at 4/5.
I want you to pause me now, find out if this is true.
And could you draw a pictorial representation to prove it see you in a moment? How did you get on? Did you say that this is true? Yes, it is.
Isn't it? The numerator and denominator been kept as small as possible.
And in the case of 4/5 here, the numerator and denominator, are as small as possible but they're the same proportion of the whole as these other fractions.
And how did you represent that pictorially? Fantastic.
Did you do ordinary and model like me? So here we can say the whole has been divided into five equal parts and four of those parts is the same proportion of the whole as if I divide the whole into 20 equal parts.
And then 16 of those equal parts will be the same proportion of the whole.
So well done.
So that generalisation is true and you proved it You've also met this generalisation in several of the previous lessons, shall we made it together? When the numerator and denominator are multiplied or divided by the same number the value of the fraction remains the same.
Do you remember that one? Super.
Well, you just proved to me in the last slide that 8/10 was equivalent 4/5 because they're the same proportion of the whole when you do give me some grey area models to prove that do you think you can prove that the numerator and denominator have been divided by the same amount to keep the value in 4/5 the same? You might need you multiplication square now and have a look at those relationships as well.
So pause me now.
I find out what 1/2 the numerator and denominator both been divided by.
How did you get on? What did you find? What did you say? The numerator and denominator have been divided by? They've been divided by two.
Fantastic.
So they've been divided by two.
What did you notice looking at your multiplication squares? Did you notice that both eight and 10 are in the two times tables? Great.
So how else can we say that? We can also say the eight and 10 are multiples of two brilliant.
And that too, in that case, and two must be a factor of eight and 10.
Fantastic.
Were there any other ways that you could divide the numerator and denominator and keep the value of the same? No no.
You had to divide by two.
What I'd like you to do now, is pause me and see if you can find out what the numerator and denominator The 12/15 have been divided by to keep the value the same and then also do the same 16/20 and usual multiplication square to use that language of multiples and see which times tables you're using to help you do that.
How did you get on? Let's have a look at 12/15.
What did you find the numerator and dominator had been divided by? Did you say they'd been divided by three? Really well done.
And what did you notice about three? Yes.
It's a factor of 12 and 15.
Isn't it? So 12 and 15 are in the three times tables and multiples of three.
And actually we had a look at that language.
Didn't we? A few lessons ago.
And you said that a common factor is a factor that is shared by two or more numbers.
So three is a common factor of 12 and 15.
Well done.
What did you find out about 16/20? What did you divide the numerator denominator by to keep the value the same? Did you divide by four? Really well done.
Is four a common factor of 16 and 20.
Yes, it is.
Why, why is this a common factor? Because 16 and 20 are in the four times tables.
They're both multiples of four.
Excellent.
Really well done.
So have proven this generalisation? Yes.
So when the numerator and denominator are multiplied or divided in this case, by the same number the value of the fraction remains the same.
And we've used that language.
Haven't we? Have common factors.
So perhaps we could sharpen up our realisation and saying, instead Say it with me when the numerator and denominator are multiplied or divided by the same common factor, the value of the fraction remains the same.
Well done.
What about these equivalent fractions? I'd like you to pause me now and have a think about everything you notice about them.
What did you find? Yes.
You found that there's the same proportion of the whole because 4/12 and 1/3 are equivalent.
What else did you notice? Fantastic So the numerator and denominator have been divided by the same amount.
Can you tell me what that was? They'd be divided by four, really well done.
And what else can you tell me about four? So the numerator and denominator can both be divided by it because it is a common factor.
So it's a factor of the numerator four and it's also a factor of the denominator 12th.
And we've said, haven't we? That when the numerator and denominator are multiplied or divided in this case, by the same common factor, the value of the fraction remains the same.
Brilliant Are there any other common factors? So, four is a common factor of four and 12.
And you can probably see that on your multiplication square again, can't you? So four and 12 were in the four times tables four and 12, the multiples of four.
That's why four is a common factor of both the numerator and denominator.
Are there any other common factors? I'm just going to pause you for a moment.
I'd like you to find all the other common factors.
Did you find any? What did you find? Yes.
Fantastic.
There are other common factors and what are they? So four and 12 have a common factor of one.
Excellent.
And they also have a common factor of two.
Fantastic.
Let's have a look at what happens if we divide by the different common factors.
So let's divide by the common factor one and have a little think what happens when we divide any number by one.
So pause me and divide the numerator and denominator by one and tell me which equivalent fraction you get.
How did you get on, what did you get? Did you find that this was equivalent to 4/12? It's in the same form, isn't it? So we can say that the fraction stays in the same form when we divide it by one.
How about if we divide it by the other common factor two? Would you like to have a go with that? So pause me again and divide the numerator and denominator by two and tell me which equivalent fraction that you get What did you get? Okay.
So if we divide the numerator and denominator by two did you find that you got 2/6? So this is an equivalent fraction, is it in its simplest form? No, it's not in its simplest form.
How could I get you to be in its simplest form? I could divide it by two again, really well done.
If I divided by two again, what will my fraction be? It will be 1/3, well done.
Because, why is that? Because they are the same proportion of the whole, and actually, we can see again, can't we? The numerator denominator have been divided by the same common factor and the value of the fraction remains the same.
Do you notice anything about what's different between.
We divided 4/12 by four didn't we? In the previous slide and we found the equivalent fraction of 1/3.
We divided it by two.
And then by two again, do you notice anything about that? Excellent.
So divided by two and then divided by two again is the same as divided by four.
Isn't it? Well, done.
You quite rightly told me that the common factors of four and 12 are one, two and four.
Excellent.
And we could divide the numerator denominator by any of these common factors and equivalent fraction, actually which common factor expressed 4/12 in its simplest form which common factor? Brilliant, screaming at me now, aren't you? You screaming at me.
You saying it was fours Ms. Seton, it was four, you're quite right.
So when we divided by four, we expressed 4/12 in its simplest form.
Can you tell me anything about the common factor four? So have a look at one, two and four which are the only common factors of four and 12.
And what can you tell me about four? You remember the language? It's the highest common factor.
Isn't it? Let's say that together.
The highest common factor the numerator and denominator is four really well done.
One more time.
The highest common factor of the numerator and denominator is four.
We've made a lot of connections in this session.
We now know how to identify a fraction in its simplest form, because we know that the numerator and denominator have been kept as small as possible whilst keeping the value the same.
And we now know that we can find the highest common factor of both the numerator and denominator and divide the numerator and denominator by the highest common factor to find the simplest form.
So I find you to do that for your independent activity.
I put 6/15, and find all the common factors for the numerator and denominator.
And then I'd like you to just approve if you can that when you divide by the highest common factor it expresses it and its simply form and you might want to have a look at those STEM sentences because you'll be using them in the next session.
And see if you agree with what they say and a little challenge there.
Thinking back to my poppy do you think you would have wanted to go out for 45/60 of an hour? And how could we express that In it's simplest form? Those using the highest common factor help us.
I hope you really enjoyed our session today.
I have, I really look forward to seeing you again.
Bye.
Bye. |
How do you solve -4x^2+8x= -3?
Apr 18, 2018
$x = 1 \pm \frac{1}{2} \sqrt{7}$
Explanation:
$\text{rearrange the equation in "color(blue)"standard form}$
•color(white)(x)ax^2+bx+c=0
$\Rightarrow - 4 {x}^{2} + 8 x + 3 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$
$\text{consider the a-c method of factorising}$
$\text{there are no whole numbers of - 12 which sum to + 8}$
$\text{we can solve using the method of "color(blue)"completing the square}$
• " the coefficient of the "x^2" term must be 1"
$\text{factor out } - 4$
$\Rightarrow - 4 \left({x}^{2} - 2 x - \frac{3}{4}\right) = 0$
• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - 2 x$
$\Rightarrow - 4 \left({x}^{2} + 2 \left(- 1\right) x \textcolor{red}{+ 1} \textcolor{red}{- 1} - \frac{3}{4}\right) = 0$
$\Rightarrow - 4 {\left(x - 1\right)}^{2} - 4 \left(- 1 - \frac{3}{4}\right) = 0$
$\Rightarrow - 4 {\left(x - 1\right)}^{2} + 7 = 0$
$\text{subtract 7 and divide by } - 4$
$\Rightarrow {\left(x - 1\right)}^{2} = \frac{7}{4}$
$\textcolor{b l u e}{\text{take the square root of both sides}}$
$\Rightarrow x - 1 = \pm \sqrt{\frac{7}{4}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$
$\text{add 1 to both sides}$
$\Rightarrow x = 1 \pm \frac{1}{2} \sqrt{7} \leftarrow \textcolor{red}{\text{exact solutions}}$ |
# N 0 2 n 1 w n and 1 2 w 1 1 2 n 0 1 n w n therefore f
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Chapter 4 / Exercise 2
Calculus
Stewart
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= n = 0 2 ( n + 1 ) w n and - 1 2 ( w + 1 ) = - 1 2 n = 0 ( - 1 ) n w n . Therefore, f ( z ) = 3 w + n = 0 2 n + 9 2 - ( - 1 ) n 2 w n = 3 z + 1 + n = 0 2 n + 9 2 - ( - 1 ) n 2 ( z + 1 ) n .
##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
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Chapter 4 / Exercise 2
Calculus
Stewart
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82 Chapter 4. Series 9. Write the two Laurent series in powers of z that represent the function f ( z ) = 1 z ( 1 + z 2 ) in certain domains and specify these domains. Solution. Since f ( z ) is analytic at z 6 = 0 , ± i , it is analytic in 0 < | z | < 1 and 1 < | z | < . For 0 < | z | < 1, f ( z ) = 1 z ( 1 + z 2 ) = 1 z 1 1 - ( - z 2 ) = 1 z n = 0 ( - 1 ) n z 2 n = n = 0 ( - 1 ) n z 2 n - 1 and for 1 < | z | < , f ( z ) = 1 z ( 1 + z 2 ) = 1 z 3 1 1 - ( - z - 2 ) = 1 z 3 n = 0 ( - 1 ) n z - 2 n = n = 0 ( - 1 ) n z - 2 n - 3 10. Let f ( z ) = z 2 z 2 - 3 z + 2 Find the Laurent series of f ( z ) in each of the following domains: (a) 1 < | z | < 2 (b) 1 < | z - 3 | < 2 Solution. First, we write f ( z ) as a sum of partial fractions: z 2 z 2 - 3 z + 2 = 1 + 3 z - 2 ( z - 2 )( z - 1 ) = 1 + 4 z - 2 - 1 z - 1 In 1 < | z | < 2, z 2 z 2 - 3 z + 2 = 1 - 2 1 - z / 2 - 1 z 1 1 - 1 / z = 1 - 2 n = 0 2 - n z n - 1 z n = 0 z - n = - 1 - n = 1 2 1 - n z n - n = 1 z - n
4.1 Taylor and Laurent series 83 In 1 < | z - 3 | < 2, z 2 z 2 - 3 z + 2 = 1 + 4 ( z - 3 )+ 1 - 1 2 +( z - 3 ) = 1 + 4 z - 3 1 1 + 1 / ( z - 3 ) - 1 2 1 1 +( z - 3 ) / 2 = 1 + 4 z - 3 n = 0 ( - 1 ) n ( z - 3 ) - n - 1 2 n = 0 ( - 1 ) n 2 - n ( z - 3 ) n = 1 2 + 4 n = 1 ( - 1 ) n + 1 ( z - 3 ) - n - n = 1 ( - 1 ) n 2 - n - 1 ( z - 3 ) n 11. Let f ( z ) = z 2 z 2 - z - 2 Find the Laurent series of f ( z ) in each of the following domains: (a) 1 < | z | < 2 (b) 0 < | z - 2 | < 1 Solution. First, we write f ( z ) as a sum of partial fractions: z 2 z 2 - z - 2 = 1 + z + 2 ( z - 2 )( z + 1 ) = 1 + 4 3 ( z - 2 ) - 1 3 ( z + 1 ) In 1 < | z | < 2, z 2 z 2 - z - 2 = 1 - 2 3 1 1 - z / 2 - 1 3 z 1 1 + 1 / z = 1 - 2 3 n = 0 2 - n z n - 1 3 z n = 0 ( - 1 ) n z - n = 1 3 - 1 3 n = 1 2 1 - n z n + 1 3 n = 1 ( - 1 ) n z - n In 0 < | z - 2 | < 1, z 2 z 2 - z - 2 = 1 + 4 3 ( z - 2 ) - 1 3 1 3 +( z - 2 ) = 1 + 4 3 ( z - 2 ) - 1 9 1 1 +( z - 2 ) / 3 = 1 + 4 3 ( z - 2 ) - 1 9 n = 0 ( - 1 ) n 3 - n ( z - 2 ) n = 8 9 + 4 3 ( z - 2 ) + n = 1 ( - 1 ) n + 1 3 - n - 2 ( z - 2 ) n
84 Chapter 4. Series 12. Find the Laurent series of 1 e z 2 - 1 in z up to z 6 and show the series converges in 0 < | z | < 2 π . Solution. Let f ( z ) = 1 / ( e z - 1 ) . Since e z - 1 = n = 0 z n n ! - 1 = n = 1 z n n ! = z n = 0 z n ( n + 1 ) ! e z - 1 has a zero at 0 of multiplicity one and hence f ( z ) has pole at 0 of order 1 . So the Laurent series of f ( z ) is given by f ( z ) = n = - 1 a n z n = a - 1 z + a 0 + a 1 z + a 2 z 2 + a 3 z 3 + n 4 a n z n in 0 < | z | < r for some r > 0. Since ( e z - 1 ) f ( z ) = 1, we have 1 = a - 1 + a 0 z + a 1 z 2 + a 2 z 3 + a 3 z 4 + n 5 a n - 1 z n ! 1 + z 2 + z 2 6 + z 3 24 + z 4 120 + n 5 z n ( n + 1 ) ! ! . Comparing the coefficients of 1, z , z 2 , z 3 and z 4 on both sides, we obtain a - 1 = 1 a 0 + a - 1 2 = 0 a 1 + a 0 2 + a - 1 6 = 0 a 2 + a 1 2 + a 0 6 + a - 1 24 = 0 a 3 + a 2 2 + a 1 6 + a 0 24 + a - 1 120 = 0 Solving it, we have a - 1 = 1 , a 0 = - 1 / 2 , a 1 = 1 / 12 , a 2 = 0 and a 3 = - 1 / 720 . Hence f ( z ) = 1 z - 1 2 + z 12 - z 3 720 + n 4 a n z n and 1 e z 2 - 1 = f ( z 2 ) = 1 z 2 - 1 2 + z 2 12 - z 6 720 + n 4 a n z 2 n . Note that f ( z ) is analytic in { z : e z - 1 6 = 0 } = { z 6 = 2 n π i } . So it is analytic in 0 < | z | < 2 π . Therefore, f ( z 2 ) is analytic in 0 < | z 2 | < 2 π , i.e., 0 < | z | < 2 π . So the series converges in 0 < | z | < 2 π .
4.2 Classification of singularities 85 4.2 Classification of singularities 1. For each of the following complex functions, do the following: find all its singularities in C ; write the principal part of the function at each singularity; |
# 1st Grade CCSS Math Worksheets and Activities
Operations and Algebraic Thinking
1.OA.A
Represent and solve problems involving addition and subtraction
Add and subtract within 20 to solve contextual problems, with unknowns in all positions, involving situations of add to, take from, put together/take apart, and compare. Use objects, drawings, and equations with a symbol for the unknown number to represent the problem.
1.OA.A.2 Worksheets
Add three whole numbers whose sum is within 20 to solve contextual problems using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
1.OA.B
Understand and apply properties of operations and the relationship between addition and subtraction
1.OA.B.3 Worksheets
Apply properties of operations (additive identity, commutative, and associative) as strategies to add and subtract. (Students need not use formal terms for these properties.)
1.OA.B.4 Worksheets
Understand subtraction as an unknown-addend problem. For example, For example, to solve 10 – 8 = ___, a student can use 8 + ___ = 10.
1.OA.C
Add and subtract within 20
1.OA.C.5 Worksheets
Add and subtract within 20 using strategies such as counting on, counting back, making 10, using fact families and related known facts, and composing/ decomposing numbers with an emphasis on making ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1=9 or adding 6 + 7 by creating the known equivalent 6 + 4 + 3 = 10 + 3 = 13).
1.OA.C.6 Worksheets
Fluently add and subtract within 20 using mental strategies. By the end of 1st grade, know from memory all sums up to 10.
1.OA.D
Work with addition and subtraction equations
1.OA.D.7 Worksheets
Understand the meaning of the equal sign (e.g., 6 = 6; 5 + 2 = 4 + 3; 7 = 8 - 1). Determine if equations involving addition and subtraction are true or false.
1.OA.D.8 Worksheets
Determine the unknown whole number in an addition or subtraction equation, with the unknown in any position (e.g., 8 + ? = 11, 5 = ? - 3, 6 + 6 = ?).
Number and Operations in Base Ten
1.NBT.A
Extend the counting sequence
Count to 120, starting at any number. Read and write numerals to 120 and represent a number of objects with a written numeral. Count backward from 20.
1.NBT.B
Understand place value
1.NBT.B.2 Worksheets
Know that the digits of a two-digit number represent groups of tens and ones (e.g., 39 can be represented as 39 ones, 2 tens and 19 ones, or 3 tens and 9 ones).
Compare two two-digit numbers based on the meanings of the digits in each place and use the symbols >, =, and < to show the relationship.
1.NBT.C
Use place value understanding and properties of operations to add and subtract
Add a two-digit number to a one-digit number and a two-digit number to a multiple of ten (within 100). Use concrete models, drawings, strategies based on place value, properties of operations, and/or the relationship between addition and subtraction to explain the reasoning used.
1.NBT.C.5 Worksheets
Mentally find 10 more or 10 less than a given two-digit number without having to count by ones and explain the reasoning used.
1.NBT.C.6 Worksheets
Subtract multiples of 10 from multiples of 10 in the range 10-90 using concrete models, drawings, strategies based on place value, properties of operations, and/or the relationship between addition and subtraction.
Measurement and Data
1.MD.A
Measure lengths indirectly and by iterating length units
1.MD.A.1 Worksheets
Order three objects by length. Compare the lengths of two objects indirectly by using a third object. For example, to compare indirectly the heights of Bill and Susan: if Bill is taller than mother and mother is taller than Susan, then Bill is taller than Susan.
1.MD.A.2 Worksheets
Measure the length of an object using non-standard units and express this length as a whole number of units.
Tell and write time in hours and half-hours using analog and digital clocks.
1.MD.B.4 Worksheets
Count the value of a set of like coins less than one dollar using the ¢ symbol only.
1.MD.C.5 Worksheets
Organize, represent, and interpret data with up to three categories. Ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another.
Geometry
1.G.A
Reason about shapes and their attributes
1.G.A.1 Worksheets
Distinguish between attributes that define a shape (e.g., number of sides and vertices) versus attributes that do not define the shape (e.g., color, orientation, overall size); build and draw two-dimensional shapes to possess defining attributes.
1.G.A.2 Worksheets
Create a composite shape and use the composite shape to make new shapes by using two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, rectangular prisms, cones, and cylinders).
1.G.A.3 Worksheets
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that partitioning into more equal shares creates smaller shares.
1st Grade common core math worksheets with answers is available online for free in printable & downloadable (PDF) format to teach, practice or learn mathematics. The K-1 curriculum includes the above cluster topics under the CCSS domains operations and algebraic thinking 1.OA, number and operations in base ten 1.NBT, measurement and data 1.MD and geometry 1.G.A. Refer the cluster headings and content standards for 1.OA.A, 1.OA.B, 1.OA.C, 1.OA.D, 1.NBT.A, 1.NBT.B, 1.NBT.C, 1.MD.A and 1.G.A to select intended common core math worksheet for grade-1.
All common core math worksheets for 1st Grade provisioned with the corresponding answer key which contains step by step calculation or complete work with steps for each exercise in the worksheet. The key activities included in the 1st Grade common core math worksheets (questions and answers) to increase the student’s ability to apply mathematics in real world problems, conceptual understanding, procedural fluency, problem solving skills, critically evaluate the reasoning or prepare the students to learn 1st Grade common core mathematics in best ways is available in printable and downloadable (PDF & PNG) formats too.
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27 November 2023
No matter what level you are at learning algebra topics, exponents are part of math exercises. Therefore, it is vital to know what they are and whatHow important is it in a mathematical equation?. These usually appear from the perfect square binomials and trinomials, to indicate the power to which a high term is found.
Therefore, today you will have the opportunity to understand what its definition is, as well as the uses that can be given to exponents within a mathematical or algebraic formula. At the end, you will know the most important aspects and it will be easier for you to solve exercises with less difficulty.
## Definition and concept of an exponent in mathematics
It is a mathematical operation that derives from multiplication. So we can say that it is a slightly more complex mathematical form, but that, at the same time, it helps us to simplify an algebraic expression. In a simple way, it is a number which tells us the number of times that we are going to multiply a term by itself. It can be expressed as a whole number, positive or negative, in addition to the fact that fractions can be applied to identify a power or exponent.
In general, the exponents are usually expressed by the number two and three. In such a case, it is said that the term is raised to the square or to the cube respectively. However, this does not mean that a term or elements cannot be raised to another power greater or less than these. The detail of this is that it is not always the case and for practical purposes, it is not very useful in the workplace. Of course, it is very useful to identify which are the powers that have the similar terms, in order to be able toapply the associative property and solve a problem more accurately.
So, in summary, we can define the concept of algebraic exponent as follows:
• It is a number or index that is located at the top right of a number or term
• It can be applied to both real numbers and terms with independent variables.
• Indicates the number of times that term will multiply itself
• You can provide us with information regarding the degree of a mathematical equation.
• A term can be raised to a positive, negative, or fractional number.
## What is an exponent used for in algebra?
The exponent or the powers, in general, are usually used when in a multiplication with similar terms, it is too large or repetitive. For example, if we have the following multiplication: 20x*(-x)*4(xy). It is much more practical to accommodate this algebraic expression, in such a way that it is as follows: -40yx³. But why do we do it this way?
It is thanks to the propertytocommutative d in multiplication, that I can put the variable y and Y first, followed by the cubed X. Of course, always respecting the law of signs, which tells us that more for less is less. To make it easier to read, I put the negative sign at the beginning of everything, followed by the number, then the variable with the lowest exponent followed by the one with the highest power.
On the other hand, the exponents have a particular use, which is to indicate the degree of a term, a polynomial or an equation. For example, the quadratic equation is so called because its degree is of index two. However, there are equations and polynomials that have a degree more than 2. So you will find mathematical equations of degree three, four, five and even more. The advantage and benefits that algebra gives us, is that the quadratic equation, like many others in mathematics, can be solved. That is, they can be written in a certain way, which makes their use more practical.
## What are the characteristics of an exponent in mathematics?
There are two fundamental characteristics that define exponents in mathematics. The first of these is that there must be a basis for a power to exist. That is, a variable number or a term that combines both elements. Example of this can be: 8, X or 9X. Here we have three bases with different elements, so the first condition or characteristics is met. Now, there is the second, which is the power, index or exponent.
The latter indicates both the degree of the term and the number of times it will be repeated. For example, if we have: They are two different cases, but at the same time similar. That is to say, that, for the first case, only the independent variable Y is cubed, while, in the second case, everything that is inside the parentheses is being affected by the power. When there is a base, we can apply one of the different leyes of exponents in algebra. In case you do not know what these laws are, we will detail each one shortly. So you can make a custom form after finishing the article.
## What are the laws of exponents in algebra?
There are many laws of which, you have to learn each and every one of them, since you will use it in almost everything that has to do with mathematics. Let’s start from the simplest and we will increase the difficulty.
1. Any term that is raised to the zero power will be equal to 1.
2. That number that is raised to the power one, will be equivalent to the same term.
3. Any number or variable with a negative exponent will be the inverse of the term. That is, the base will divide. For example: X^-1 = 1/X. Salways keeping the base with its exponent.
4. When there is a multiplication of terms with the same base, their exponents are added and the same base is placed.
5. When it comes to a division of terms with the same base, the exponents are subtracted and the same base is placed.
6. If a term with a certain exponent is raised to a power, we proceed to multiply the exponents and leave the same base.
7. Everything that is inside a parenthesis and is raised to a power X, all the elements will have the same power. In other words, the power outside the parenthesis is distributed among all the internal elements.
8. If a term is raised to a rational power or fraction, this has its equivalent in a root. The denominator becomes the index of the root, while the numerator is the power of the term or radicand.
The good thing about these properties is that you will find many simple examples where you can better visualize each application according to the corresponding property. Keep in mind that it is very important that you master this perfectly, since it will serve you for almost everything and for any subject that is related to calculus.
## What are the parts of exponents in math and algebra?
Unlike other properties in mathematics, this one in particular has only two parts. The first one is the base, which can be expressed with a number, a variable, or a combination of both. The second part is the exponent or also known as power. may be represented by through a number a variable or the combination of both. In addition, the element that is considered as power can be both positive and negative.
## Examples of exponents in algebraic terms
The idea that you can learn to deal with exponents is by applying the standard formulas that exist for each case according to the property. For example, any number raised to 0 will give you one as a result. That is, if you have X^0 = 1; 1000⁰ = 1 ; (2X)^0 = 1. If you look closely, there are tons of ways to write the number, as long as you know how to apply the correct property.
Now, to complicate it a bit, imagine that you have the following algebraic expressions: (2X)^-4. According to one of the power properties, this is equivalent to writing: 1/ (2X)^4. This is due to one of the properties, which tells us that everything term raised to a negative power, can be written as a fraction, keeping the same power, but in a positive way. To understand this better, look at it as if a negative exponent is the inverse. By having such a term, in the numerator, its inverse would be a fraction. Now, if the exponent is in the denominator, its inverse would be a numerator. |
# Group theory: course and worksheet for beginners
Group theory is a big part of advanced algebra. The Group is a special set with a particular structure (the set of real numbers is a particular group). A set is called a group if it has a law of composition which allows calculations to be made. In this article, we make this concept clear with definitions, properties, and great exercises.
### A short course
In this section, we give a concise course on group theory. The most important properties of groups will be discussed.
Definition: Let $G$ be a no empty set. A composition law on $G$ is an application $\ast: G\times G\to G$ such that for any $(x,y)\in G$ we have $x\ast y\in G$.
Definition: A set $(G,\ast)$ is called a group is $\ast$ is a composition law with the following properties:
• associativity of $\ast$: for any $x,y$ and $z$ in $G$, $x\ast (y\ast z)=(x\ast y)\ast z$.
• there exists an element $e\in G$ such that $x\ast e=e\ast x=x$. This element $e$ is called neutral element,
• For any $x\in G$ there exists $y\in G$ such that $x\ast y=y\ast x=e$. The element $y$ is called the inverse of $x$ and will be denoted by $x^{-1}$.
A group $(G,\ast)$ is called commutative group if for any $x,y\in G$, $x\ast y=y\ast x$.
$(\mathbb{R},+)$ (here $\ast=+$ the usual addition operation) is a commutative group.
If $(H,\ast)$ and $(H,\star)$ are two groups, then we can define another group $(G\times H, \diamond )$ by introducing the following composition law \begin{align*} (x,y)\diamond (x’,y’)=(x\ast x’,y\star y’).\end{align*}
### Group theory worksheet
Let us now give a selection of exercises on group theory.
Exercise: Does the following groups are isomorph?
• $\mathbb{Z}/6\mathbb{Z}$ and $\mathfrak{S}_3,$ the group of permutations.
• $\mathbb{Z}/(nm)\mathbb{Z}$ and $\left(\mathbb{Z}/n\mathbb{Z}\right)\times \left(\mathbb{Z}/m\mathbb{Z}\right)$ for $n,m\in\mathbb{N}$.
• $(\mathbb{R},+)$ and $(\mathbb{Q},+)$.
• $(\mathbb{R},+)$ and $(\mathbb{R}^\ast_{+},\times)$.
• $(\mathbb{Q},+)$ and $(\mathbb{Q}^\ast_{+},\times)$.
Solution: 1) The groups $\mathbb{Z}/6\mathbb{Z}$ and $\mathfrak{S}_3$ are not isomorphic because one is commutative and not the other.
2) When $n$ and $m$ are not coprime (relatively prime), the groups $\mathbb{Z}/(nm)\mathbb{Z}$ and $\left(\mathbb{Z}/n\mathbb{Z}\right)\times \left(\mathbb{Z}/m\mathbb{Z}\right)$ are not isomorph. In fact if $p=gcd(n,m)$, then we must have $0 < p < nm$. Hence if we denote $[x]_r$ the elements of $\mathbb{Z}/r\mathbb{Z}$ (for $r\in \mathbb{N}$), we have $p[1]_n=[p]_n\neq [0]_{nm}$.
If we assume that $\mathbb{Z}/(nm)\mathbb{Z}$ and $\left(\mathbb{Z}/n\mathbb{Z}\right)\times \left(\mathbb{Z}/m\mathbb{Z}\right)$ are isomorph, then there would be an element $(x,y)\left(\mathbb{Z}/n\mathbb{Z}\right)\times \left(\mathbb{Z}/m\mathbb{Z}\right)$ such that $p(x,y)\neq ([0]_n,[0]_m)$, which is absurd since $px=[0]_n$ ($p$ is a multiple of $n$) and $py=[0]_m$ ($p$ is a multiple of $m$).
3) $(\mathbb{R},+)$ and $(\mathbb{Q},+)$ are not isomorph. In fact we that $\mathbb{R}$ is not countable while $\mathbb{Q}$ is.
4) It is well known that the application $f:(\mathbb{R},+)\to \mathbb{R}^\ast_+$ such that $f(x)=e^x$ is bijective and $f(x+y)=e^{x+y}=e^x e^y=f(x)f(y)$ for any $x,y\in\mathbb{R}$. This shows that $f$ is an isomorphism of groups. Hence the groups $(\mathbb{R},+)$ and $(\mathbb{R}^\ast,\times)$ are isomorph.
5) The group $(\mathbb{Q},+)$ satisfies the following property \begin{align*} \forall y\in \mathbb{Q},\qquad \exists\,x\in \mathbb{Q},\quad \text{s.t.}\;y=x+x. \end{align*} Now if there exists an isomorphism of groups between $(\mathbb{Q},+)$ and $(\mathbb{Q}^\ast_+,\times)$ then we will have \begin{align*} \forall y\in \mathbb{Q}^\ast_+,\qquad \exists\,x\in \mathbb{Q}^\ast_+,\quad \text{such that}\;y=x\times x. \end{align*} This is not possible as e.g. the number $2$ does not admit a square root in $\mathbb{Q}$.
Exercise: We denote by $GL_2(\mathbb{R})$ the group of invertible matrices. Let \begin{align*} \mathcal{H}=\left\{\begin{pmatrix} 3^n&n3^{n-1}\\ 0&3^n\end{pmatrix}:n\in \mathbb{Z}\right\} \end{align*} a subgroup of $GL_2(\mathbb{R})$. Prove that $\mathcal{H}$ is isomorph to $\mathbb{Z}$.
Solution: We shall use the fact that any infinite monogenic group is isomorph to $\mathbb{Z}$
We select \begin{align*} A=\begin{pmatrix} 3&1\\0&3\end{pmatrix}\in GL_2(\mathbb{R}). \end{align*} By using a reccurrence argument one can see that \begin{align*} A^n=\begin{pmatrix} 3^n&n3^{n-1}\\ 0&3^n\end{pmatrix},\quad n\in \mathbb{Z}. \end{align*} Then \begin{align*} \mathcal{H}=\{A^n:n\in\mathbb{Z}\}. \end{align*} Then $\mathcal{H}$ is a infinite monogenic group, so it is isomorph to $\mathbb{Z}$.
Exercise: Let $G$ be a group (supposed not Abelian). To any $a\in G,$ we associate an application \begin{align*} f_a:G\to G,\quad x\mapsto f_a(x)=axa^{-1}. \end{align*}
• Prove that $f_a$ is an isomorphism from $G$ to $G$.
• We denote \begin{align*} {\rm Int}(G)=\{f_a:a\in G\}. \end{align*} Prove that $({\rm Int}(G),\circ)$ is a group.
• Prove that \begin{align*} \psi: G\to {\rm Int}(G),\quad a\mapsto f_a \end{align*} is an homomorphism of groups. Determine its kernel.
Solution: 1) We denote by $e$ the identity element of $G$, we then have $f_a(e)=aea^{-1}=aa^{-1}=e$. For $x,y\in G$, we have \begin{align*} f_a(xy)&=axya^{-1}=axeya^{-1}=axa^{-1}aya^{-1}\cr &=(axa^{-1})(aya^{-1})\cr &= f_a(x)f_a(y). \end{align*} Then $f_a$ is an homomorphism of groups. Let us prove that $f_a$ is bijective. It suffices to show that for any $y\in G$ there is a unique $x\in G$ such that $f_a(x)=y$, which means that $axa^{-1}=y$. This is equivalent to $x=a^{-1}ya$. This element is unique, so $f_a$ is an isomorphism.
2) We denote by $(B(G),\circ)$ the group of all isomorphism from $G$ to $G$. Observe that ${\rm Int}(G)\subset B(G)$. It suffice then to show that ${\rm Int}(G)$ is a subgroup of $B(G)$. In fact, remark that for any $x\in G$, we have ${\rm id}_G(x)=x=exe^{-1}=f_e(x)$. This means that ${\rm id}_G\in {\rm Int}(G)$, and then ${\rm Int}(G)$ is not empty. Let $\ell_1,\ell_2\in {\rm Int}(G)$. Then, there exist $a,b\in G$ such that $\ell_1=f_a$ and $\ell_2=f_b$. Now for any $x\in G,$ we have \begin{align*} (\ell_1\circ\ell_2)(x)&=\ell_1(\ell_2(x))\cr &=f_a(f_b(x))\cr &= af_b(x)a^{-1}\cr &= abxb^{-1}a^{-1}\cr &=(ab)x(ab)^{-1}\cr &= f_{ab}(x). \end{align*} As $ab\in G,$ then $f_{ab}\in {\rm Int}(G)$. Hence $\ell_1\circ\ell_2\in {\rm Int}(G)$. According to the proof of the first question, for any $a\in G,$ we have $f^{-1}_a=f_{a^{-1}}\in {\rm Int}(G),$ because $a^{-1}\in G$. This ends the proof.
3) In the proof of the second question we have seen that for $a,b\in G$ we gave $f_a\circ f_b=f_{ab}$. Hence \begin{align*} \psi(ab)&=f_{ab}=f_a\circ f_b\cr &= \psi(a)\circ \psi(b). \end{align*} This shows that $\psi$ is an homomorphism of groups.
Let $a\in \ker(\psi)$, which means that $f_a={\rm id}_G$. For all $x\in G,$ $f_a(x)=x$, then $axa^{-1}=x$. This implies that $a$ satisfies $ax=xa$ for all $x\in G$. Hence \begin{align*} \ker(\psi)=\{a\in G\;|\; ax=xa,\;\forall x\in G\}. \end{align*}
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## A Multiplication Based Logic Puzzle
### What Makes 689 Amazing?
689 is an amazing number for several reasons. I decided to make graphics to illustrate many of those ways. (689’s factoring information is at the end of this post.)
689 is the sum of consecutive prime numbers 227, 229, and 233.
Also 689 is the sum of the primes from 83 to 109. Do you know what those 7 prime numbers are?
Stetson.edu informs us that 689 is the smallest number that can be expressed as the sum of three different square numbers NINE ways. I decided to figure out what those nine ways are and make this first graphic to share with you:
Note: 614 can also be expressed as the sum of 3 squares 9 different ways, but one of those ways is 17² + 17² + 6² = 614, and that duplicates 17² in the same sum.
689 is the same number when it is turned upside down. Numbers with that characteristic are called Strobogrammatic numbers.
689 BASE 10 isn’t a palindrome, but 373 BASE 14 is; note that 3(196) + 7(14) + 3(1) = 689
Both of 689’s prime factors have a remainder of 1 when divided by 4, so they are hypotenuses of Pythagorean triples. That fact also means 689 can be expressed as the sum of two square numbers TWO different ways, and it makes 689 the hypotenuse of FOUR Pythagorean triples. Can you tell by looking at the graphic which two are primitive and which two aren’t?
689 is the sum of consecutive numbers three different ways:
• 344 + 345 = 689; that’s 2 consecutive numbers.
• 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 = 689; that’s 13 consecutive numbers.
• 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 = 689; that’s 26 consecutive numbers.
Now you have a few reasons why 689 is an amazing number. 13 and 53 were part of some of those reasons so it shouldn’t surprise anyone to see 13 and 53 pop up in its factoring information, too:
——————————————————————————
• 689 is a composite number.
• Prime factorization: 689 = 13 x 53
• The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 689 has exactly 4 factors.
• Factors of 689: 1, 13, 53, 689
• Factor pairs: 689 = 1 x 689 or 13 x 53
• 689 has no square factors that allow its square root to be simplified. √689 ≈ 26.248809.
#### Comments on: "What Makes 689 Amazing?" (23)
1. What an interesting number! And here’s one fact you didn’t mention – 689 was the winning number in our jazz club raffle on Saturday. And I – very stupidly – said “Are you sure you’re reading it the right way up?”!
Liked by 3 people
• No, no, no. I think you were inspired. How many people learned about the Strobogrammatic property of that number because you asked that? Regardless of how it made you feel, I love your story!
Like
2. brandon said:
And 689 is also the number of votes the current Chief Executive of Hong Kong received out of 1200 selected voters.
Liked by 2 people
• That’s interesting. I wonder why only 1200 voters were selected to make such an important choice. 689 is about 57% of 1200.
Perhaps that is why so far today 628 people from Hong Kong have viewed this blog. That’s so close to 689. I will update today’s view count if it changes. Update: I’ve had 1385 Views from Hong Kong and 1109 views from Taiwan in a single day. Wow.
Liked by 1 person
You pointed out something extremely obvious but neither China nor Hong Kong government is willing to admit/understand
Like
3. kevin cheung said:
689 is the name of the most famous person in hk
Liked by 1 person
4. CY said:
I was brought here because of the “Chief Executive” 689 lol
Liked by 1 person
• These two comments made me google 689 Hong Kong. I found an article that helps me understand why my post has had 3277 views in Hong Kong and 2284 in Taiwan: Why are Hong Kong’s protesters rallying around the number 689? I had no idea this number had so much political meaning when I wrote this post. Now I am a little better informed.
Like
• haha said:
Actually there is another meaning for 689 at Taiwan, but I am unable to find the explanation in English. http://zh.pttpedia.wikia.com/wiki/689%E3%80%81609
Liked by 1 person
• Thank you. I clicked on the link and pasted it into google translate which doesn’t do that good of a job translating it. I will have to study it some more.
Like
• John Fung said:
This is a Taiwanese stuff, meaning 6.89 million votes for the incumbent Taiwanese president Ma Ying-jeou in the 2012 election.
(In Chinese language, 6.89 million = 689萬, in which 萬 means ten thousand)
Liked by 1 person
5. MaIngJiu said:
Magic number for Taiwan and Hong Kong
Magic!
Liked by 1 person
• I guess so!
Like
6. Thank you for clarifying it, John Fung!
Like
7. TW#1 said:
Again, 689 for Taiwan!!!
Liked by 1 person
• Wow! 689 x 10,000 votes for your new president as well. Thank you so much for letting me know!
Like
• It was 689 X 10,000 (689 萬）.
Like
• Thank you. I remembered it incorrectly, but now I’ve corrected my comment.
Like
8. F said:
689 for Taiwan, Twice!!!
Liked by 1 person
• Democracy has won! I’m very happy for you. Also, congratulations electing your first female president.
Like
9. […] 689 is also a magical number. As findthefactors.com explains, it is a strobogrammatic number, meaning it can be read upside […]
Like
10. In the above referenced article the Hong Kong Free Press does an excellent job explaining in English why 689 is causing such a stir in Hong Kong and Taiwan. I recommend reading it.
Like
11. […] 689 is also a magical number. As findthefactors.com explains, it is a strobogrammatic number, meaning it can be read upside […]
Like |
# Pillow-Problems: Problem #45
Math Lair Home > Source Material > Pillow-Problems > Problem #45
Problem #44 | Problem #47
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).
## Problem:
45.
If an infinite number of rods be broken: find the chance that one at least is broken in the middle.
[5/84
45.
.6321207 &c.
## Solution:
45.
Divide each rod into (n + 1) parts, where n is assumed to be odd, and the n points of division are assumed to be the only points where the rod will break, and to be equally frangible.
The chance of one failure is
(n − 1)/ (n)
;
∴ " " n failures is (
(n − 1)/ (n)
)n
= (1 −
(1)/ (n)
)n
Now, if m =
(1)/ (n)
; then, when n =
(1)/ (0)
, m = 0;
∴ the chance that no rod is broke in the middle = (1 − m)
(1)/ (m)
, when m = 0;
i. e. it approaches the limit (1 − 0)
(1)/ (0)
.
And Ans. = 1 − (1 − 0)
(1)/ (0)
.
Now (1 − 0)
(1)/ (0)
= e. Hence if, in the series for e, we call the sum of the odd terms 'a', and, of the even terms 'b'; then e = a + b; and (1 − 0)
(1)/ (0)
= ab = 2ae.
Q.E.F.
[N.B. What follows here was not thought out.]
Now a = 1 +
(1)/ (2!)
+
(1)/ (4!)
+ &c.
1 = 1
(1)/ (2!)
= .5
(1)/ (4!)
= .04166666 &c.
(1)/ (6!)
= .00138888 &c.
(1)/ (8!)
= .00002480 &c.
(1)/ (10!)
= .00000027 &c.
a = 1.5430806 & c.
∴ 2a = 3.0861612 & c.
e = 2.7182818 & c.
∴ (1 − 0)
(1)/ (10!)
= .3678793 &c. ∴ Ans = 1 − (1 − 0)
(1)/ (10!)
= .6321207 &c. |
# How do you simplify sqrt12 times sqrt 6?
Mar 4, 2016
$6 \sqrt{2}$
#### Explanation:
To simplify $\sqrt{12} \times \sqrt{6}$, first let us factorize each of the numbers
$\sqrt{12} \times \sqrt{6}$ = $\sqrt{2 \times 2 \times 3} \times \sqrt{2 \times 3}$
=$\sqrt{2 \times 2 \times 3 \times 2 \times 3} = \sqrt{2 \times 2 \times 3 \times 3 \times 2}$
=$2 \times 3 \times \sqrt{2}$
=$6 \sqrt{2}$
Mar 4, 2016
$\sqrt{12} \times \sqrt{6} = \textcolor{b l u e}{6 \sqrt{2}}$
#### Explanation:
$\sqrt{12} \times \sqrt{6}$
Simplify $\sqrt{12}$.
$\sqrt{12} = \left(2 \times 2 \times 3\right)$
$\sqrt{12} = \left({2}^{2} \times 3\right)$
Apply square root rule ${\sqrt{a}}^{2} = a$.
$\sqrt{12} = 2 \sqrt{3}$
Rewrite the expression.
$2 \sqrt{3} \times \sqrt{6} =$
$2 \sqrt{18}$
Simplify $2 \sqrt{18}$.
$2 \sqrt{18} = 2 \sqrt{2 \times 3 \times 3}$
$2 \sqrt{18} = 2 \sqrt{2 \times {3}^{2}}$
Apply square root rule ${\sqrt{a}}^{2} = a$.
$2 \sqrt{18} = 2 \times 3 \sqrt{2}$
$2 \sqrt{18} = 6 \sqrt{2}$ |
# Vectors Subtraction Quantity
This post categorized under Vector and posted on November 20th, 2019.
In Physics Vector Quangraphicies are quangraphicies that have a magnitude and direction. It is important to understand how operations like addition and subtraction are carried out on Vectors. Let us begin with the addition of vectors followed of subtraction. This graphic discusses the addition subtraction and scalar multiplication of vectors. In maths we have learned the different operations we perform on numbers. Let us learn here the vector operation such as Addition Subtraction Multiplication on vectors. Vector Addition. The two vectors a and b can be added giving the sum to be a b. This requires joining them head to tail.
Another quangraphicy represented by a vector is force since it has a magnitude and direction and follows the rules of vector addition. Vectors also describe many other physical quangraphicies such as linear displacement displacement linear acceleration angular acceleration linear momentum and angular momentum. A lot of mathematical quangraphicies are used in Physics to explain the concepts clearly. A few examples of these include force speed velocity and work. These quangraphicies are often described as being a scalar or a vector quangraphicy. Scalars and vectors are differentiated depending on their definition. A scalar quangraphicy is defined as the physical VECTORS - ADDITION AND SUBTRACTION. VECTORS AND SCALARS Certain quangraphicies can be described by a numerical value alone and certain quangraphicies need direction along with the numerical value to be defined.
How to Add or Subtract Vectors. Many common physical quangraphicies are often vectors or scalars. Vectors are akin to arrows and consist of a positive magnitude (graphicgth) and importantly a direction. on the other hand scalars are just numerical Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point. Add vectors end-to-end component-wise and by the parallelogram rule. Understand that the magnitude of a sum of two vectors is typically not the sum of the magnitudes. Adding Vector subtraction including boat example Introduction to head to tail vector subtraction in the geometric sense. This is then applied to an example of working out a boats velocity relative to water given the velocity of the current and the velocity of the boat relative to land are both known. This physics graphic tutorial provides a basic introduction into vectors. It explains the process of vector addition and subtraction using the head to tail met
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