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# Alternating series and reciprocal Fibonacci constant
We know that reciprocal Fibonacci constant $$\sum_{n=1}^{\infty} \frac{1}{F_n} = \frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{8} + \frac{1}{13} + \frac{1}{21} + \cdots \approx 3.3598856662 \dots .$$
Evaluate: $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n}$$
## 1 Answer
I got to say after spending some time with this problem: that's a truly remarkable series, I would love if other people could look into this problem. This is not a full answer, this was just my way to approach the question and I couldn't find some "nice and short" closed-form of the series. Let's begin!
Notice how the series can be split up into an odd and an even part:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} = \sum_{n=0}^{\infty} \frac{1}{F_{2n+1}}-\sum_{n=1}^{\infty}\frac{1}{F_{2n}}$$
Let's first evaluate the series $$\sum_{n=0}^{\infty} \frac{1}{F_{2n}}$$ because I feel like that's easier a little bit easier.
Evaluating $$\sum_{n=0}^{\infty} \frac{1}{F_{2n}}$$
This is quite a tricky process, but let's try to work it through. Let $$\phi$$ denote the golden ratio.
\begin{align} \sum_{n=1}^{\infty} \frac{1}{F_{2n}} &= \sqrt{5}\sum_{n=1}^\infty \frac{\phi^{2n}}{\phi^{4n}-1} \\\\ & = \sqrt{5}\sum_{n=1}^{\infty}\frac{1}{\phi^{2n}-1}-\frac{1}{\phi^{4n}-1} \\\\ & = \sqrt{5}\left(\sum_{n=1}^{\infty}\frac{1}{\phi^{2n}-1}-\sum_{n=1}^{\infty}\frac{1}{\phi^{4n}-1}\right)\end{align}
The sums take form of what's known as a Lambert series. A Lambert series is defined as:
$$F(x)=\sum_{n=1}^\infty a_n\frac{x^n}{1-x^n}$$
Now let the case when $$a_n = 1$$ be denoted by $$L(x)$$:
$$L(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^n} = \sum_{n=1}^\infty \frac{1}{x^{-n}-1}$$
And so in the case above we have:
$$\sum_{n=1}^{\infty} \frac{1}{F_{2n}} = \sqrt{5}\left(L(\phi^{-2})-L(\phi^{-4})\right)$$
Now, it is known (and I don't know how; I can't prove it...) that:
$$L(x) = \frac{\psi_x(1)+\ln(1-x)}{\ln(x)}$$
Here $$\psi_x(1)$$ denotes the q-Polygamma function. Using the identity above we get:
$$\sqrt{5}\left(L(\phi^{-2})-L(\phi^{-4})\right) = \frac{\sqrt{5}}{8\ln(\phi)}\left(\ln(5)+2\psi_{\phi^{-4}}(1)-4\psi_{\phi^{-2}}(1)\right)$$
This as far as I can go. Thus we have arrived at the result that:
$$\sum_{n=1}^{\infty} \frac{1}{F_{2n}} = \frac{\sqrt{5}}{8\ln(\phi)}\left(\ln(5)+2\psi_{\phi^{-4}}(1)-4\psi_{\phi^{-2}}(1)\right)$$
Now according to WolframMathWorld:
$$\frac{\sqrt{5}}{8\ln(\phi)}\left(\ln(5)+2\psi_{\phi^{-4}}(1)-4\psi_{\phi^{-2}}(1)\right) = \frac{\sqrt{5}}{4\ln(\phi)}\left(\psi_{\phi^2}\left(1-\frac{i\pi}{2\ln(\phi)}\right)-\psi_{\phi^2}(1)+i\pi\right)$$
But I have know idea, how they got there. Anyways, the value equals approximately $$1.5353705...$$
Evaluating $$\sum_{n=0}^{\infty} \frac{1}{F_{2n+1}}$$
This derivation is pretty much the same as the previous one; you express the Fibonacci numbers in terms of the golden ratio and the simplify it into a Lambert series and then express it in terms of the q-Polygamma function. I did part of this derivation independently, but found that all of these are found at WolframMathWorld. You'll end up with a similar expression to the sum of even Fibonacci numbers:
$$\sum_{n=0}^{\infty} \frac{1}{F_{2n+1}} = -\frac{\sqrt{5}}{4\ln(\phi)}\left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}\right)-\psi_{\phi^2}\left(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right)\right) \tag{1}$$
According to WolframMathWorld an identity (I have no idea how to prove this) is that:
$$\left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}\right)-\psi_{\phi^2}\left(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right)\right) = -(\ln\phi)\vartheta_2^2(\phi^{-2})$$
where $$\vartheta_2(q)$$ is an Jacobi theta function (read more here). So using this identity on equation $$(1)$$ we get:
$$-\frac{\sqrt{5}}{4\ln(\phi)}\left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}-\psi_{\phi^2}(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right)\right) = \frac{\sqrt{5}}{4}\vartheta_2^2(\phi^{-2})$$
Which is quite a big simplification if you ask me.
Puting everything together
Recall that: $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} = \sum_{n=0}^{\infty} \frac{1}{F_{2n+1}}-\sum_{n=1}^{\infty}\frac{1}{F_{2n}}$$
Using the results we have established we get:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} = -\frac{\sqrt{5}}{4\ln(\phi)}\left(\pi-i\left(\psi_{\phi^2}(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}-\psi_{\phi^2}(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right) - \frac{\sqrt{5}}{4\ln(\phi)}\left(\psi_{\phi^2}\left(1-\frac{i\pi}{2\ln(\phi)}\right)-\psi_{\phi^2}(1)+i\pi\right)$$
\begin{align} &= -\frac{\sqrt{5}}{4\ln(\phi)} \left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}\right)-\psi_{\phi^2}\left(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right) + \psi_{\phi^2}\left(1-\frac{i\pi}{2\ln(\phi)}\right)-\psi_{\phi^2}(1)+i\pi\right)\end{align}
Now let:
$$\Omega = \frac{i\pi}{4\ln(\phi)}$$
Therefore we get:
\begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} &= -\frac{\sqrt{5}}{4\ln(\phi)} \left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\Omega\right)-\psi_{\phi^2}\left(\frac{1}{2}+\Omega\right)\right) + \psi_{\phi^2}\left(1-2\Omega\right)-\psi_{\phi^2}(1)+i\pi\right)\end{align}
That's pretty horrible looking, but I don't know how to simplify it :(
Also note that if you want the series in terms of the Reciprocal Fibonacci Constant $$\psi$$,
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} = \sqrt{\psi^2-4F}$$
where $$F$$ is:
$$F = \left(\sum_{n=0}^{\infty} \frac{1}{F_{2n+1}}\right)\left(\sum_{n=1}^{\infty}\frac{1}{F_{2n}}\right)$$
|
Examples of differential equations
Differential equations arise in many problems in physics, engineering, and other sciences. The following examples show how to solve differential equations in a few simple cases when an exact solution exists.
Separable first-order ordinary differential equations
Equations in the form ${\displaystyle {\frac {dy}{dx}}=f(x)g(y)}$ are called separable and solved by ${\displaystyle {\frac {dy}{g(y)}}=f(x)dx}$ and thus ${\displaystyle \int {\frac {dy}{g(y)}}=\int f(x)dx}$. Prior to dividing by ${\displaystyle g(y)}$, one needs to check if there are stationary (also called equilibrium) solutions ${\displaystyle y=const}$ satisfying ${\displaystyle g(y)=0}$.
Separable (homogeneous) first-order linear ordinary differential equations
A separable linear ordinary differential equation of the first order must be homogeneous and has the general form
${\displaystyle {\frac {dy}{dt}}+f(t)y=0}$
where ${\displaystyle f(t)}$ is some known function. We may solve this by separation of variables (moving the y terms to one side and the t terms to the other side),
${\displaystyle {\frac {dy}{y}}=-f(t)\,dt}$
Since the separation of variables in this case involves dividing by y, we must check if the constant function y=0 is a solution of the original equation. Trivially, if y=0 then y'=0, so y=0 is actually a solution of the original equation. We note that y=0 is not allowed in the transformed equation.
We solve the transformed equation with the variables already separated by Integrating,
${\displaystyle \ln |y|=\left(-\int f(t)\,dt\right)+C}$
where C is an arbitrary constant. Then, by exponentiation, we obtain
${\displaystyle y=\pm e^{\left(-\int f(t)\,dt\right)+C}=\pm e^{C}e^{-\int f(t)\,dt}}$.
Here, ${\displaystyle e^{C}>0}$, so ${\displaystyle \pm e^{C}\neq 0}$. But we have independently checked that y=0 is also a solution of the original equation, thus
${\displaystyle y=Ae^{-\int f(t)\,dt}}$.
with an arbitrary constant A, which covers all the cases. It is easy to confirm that this is a solution by plugging it into the original differential equation:
${\displaystyle {\frac {dy}{dt}}+f(t)y=-f(t)\cdot Ae^{-\int f(t)\,dt}+f(t)\cdot Ae^{-\int f(t)\,dt}=0}$
Some elaboration is needed because ƒ(t) might not even be integrable. One must also assume something about the domains of the functions involved before the equation is fully defined. The solution above assumes the real case.
If ${\displaystyle f(t)=\alpha }$ is a constant, the solution is particularly simple, ${\displaystyle y=Ae^{-\alpha t}}$ and describes, e.g., if ${\displaystyle \alpha >0}$, the exponential decay of radioactive material at the macroscopic level. If the value of ${\displaystyle \alpha }$ is not known a priori, it can be determined from two measurements of the solution. For example,
${\displaystyle {\frac {dy}{dt}}+\alpha y=0,y(1)=2,y(2)=1}$
gives ${\displaystyle \alpha =ln(2)}$ and ${\displaystyle y=4e^{-ln(2)t}=2^{2-t}}$.
Non-separable (non-homogeneous) first-order linear ordinary differential equations
First-order linear non-homogeneous ODEs (ordinary differential equations) are not separable. They can be solved by the following approach, known as an integrating factor method. Consider first-order linear ODEs of the general form:
${\displaystyle {\frac {dy}{dx}}+p(x)y=q(x)}$
The method for solving this equation relies on a special integrating factor, μ:
${\displaystyle \mu =e^{\int _{x_{0}}^{x}p(t)\,dt}}$
We choose this integrating factor because it has the special property that its derivative is itself times the function we are integrating, that is:
${\displaystyle {\frac {d{\mu }}{dx}}=e^{\int _{x_{0}}^{x}p(t)\,dt}\cdot p(x)=\mu p(x)}$
Multiply both sides of the original differential equation by μ to get:
${\displaystyle \mu {\frac {dy}{dx}}+\mu {p(x)y}=\mu {q(x)}}$
Because of the special μ we picked, we may substitute /dx for μ p(x), simplifying the equation to:
${\displaystyle \mu {\frac {dy}{dx}}+y{\frac {d{\mu }}{dx}}=\mu {q(x)}}$
Using the product rule in reverse, we get:
${\displaystyle {\frac {d}{dx}}{(\mu {y})}=\mu {q(x)}}$
Integrating both sides:
${\displaystyle \mu {y}=\left(\int \mu q(x)\,dx\right)+C}$
Finally, to solve for y we divide both sides by ${\displaystyle \mu }$:
${\displaystyle y={\frac {\left(\int \mu q(x)\,dx\right)+C}{\mu }}}$
Since μ is a function of x, we cannot simplify any further directly.
Second-order linear ordinary differential equations
A simple example
Suppose a mass is attached to a spring which exerts an attractive force on the mass proportional to the extension/compression of the spring. For now, we may ignore any other forces (gravity, friction, etc.). We shall write the extension of the spring at a time t as x(t). Now, using Newton's second law we can write (using convenient units):
${\displaystyle m{\frac {d^{2}x}{dt^{2}}}+kx=0,}$
where m is the mass and k is the spring constant that represents a measure of spring stiffness. For simplicity's sake, let us take m=k as an example.
If we look for solutions that have the form ${\displaystyle Ce^{\lambda t}}$, where C is a constant, we discover the relationship ${\displaystyle \lambda ^{2}+1=0}$, and thus ${\displaystyle \lambda }$ must be one of the complex numbers ${\displaystyle i}$ or ${\displaystyle -i}$. Thus, using Euler's theorem we can say that the solution must be of the form:
${\displaystyle x(t)=A\cos t+B\sin t}$
See a solution by WolframAlpha.
To determine the unknown constants A and B, we need initial conditions, i.e. equalities that specify the state of the system at a given time (usually t = 0).
For example, if we suppose at t = 0 the extension is a unit distance (x = 1), and the particle is not moving (dx/dt = 0). We have
${\displaystyle x(0)=A\cos 0+B\sin 0=A=1,}$
and so A = 1.
${\displaystyle x'(0)=-A\sin 0+B\cos 0=B=0,}$
and so B = 0.
Therefore x(t) = cos t. This is an example of simple harmonic motion.
See a solution by Wolfram Alpha.
A more complicated model
The above model of an oscillating mass on a spring is plausible but not very realistic: in practice, friction will tend to decelerate the mass and have magnitude proportional to its velocity (i.e. dx/dt). Our new differential equation, expressing the balancing of the acceleration and the forces, is
${\displaystyle m{\frac {d^{2}x}{dt^{2}}}+c{\frac {dx}{dt}}+kx=0,}$
where ${\displaystyle c}$ is the damping coefficient representing friction. Again looking for solutions of the form ${\displaystyle Ce^{\lambda t}}$, we find that
${\displaystyle m\lambda ^{2}+c\lambda +k=0.}$
This is a quadratic equation which we can solve. If ${\displaystyle c^{2}<4km}$ there are two complex conjugate roots a ± ib, and the solution (with the above boundary conditions) will look like this:
${\displaystyle x(t)=e^{at}\left(\cos bt-{\frac {a}{b}}\sin bt\right)}$
Let us for simplicity take ${\displaystyle m=1}$, then ${\displaystyle 0 and ${\displaystyle k=a^{2}+b^{2}}$.
The equation can be also solved in MATLAB symbolic toolbox as
x = dsolve('D2x+c*Dx+k*x=0','x(0)=1','Dx(0)=0')
although the solution looks rather ugly,
x = (c + (c^2 - 4*k)^(1/2))/(2*exp(t*(c/2 - (c^2 - 4*k)^(1/2)/2))*(c^2 - 4*k)^(1/2)) -
(c - (c^2 - 4*k)^(1/2))/(2*exp(t*(c/2 + (c^2 - 4*k)^(1/2)/2))*(c^2 - 4*k)^(1/2))
This is a model of a damped oscillator. The plot of displacement against time would look like this:
which resembles how one would expect a vibrating spring to behave as friction removes energy from the system.
Linear systems of ODEs
The following example of a first order linear systems of ODEs
${\displaystyle y_{1}'=y_{1}+2y_{2}+t}$
${\displaystyle y_{2}'=2y_{1}-2y_{2}+\sin(t)}$
can be easily symbolically solved in WolframAlpha.
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http://www.codinghelmet.com/ Wear a helmet. Even when coding. Follow Zoran Horvat @zoranh75
exercises > reducing-fractions
# Exercise #15 - Reducing Fractionsby Zoran Horvat@zoranh75
## Problem Statement
Given two integers n >= 0 and m > 0, write a function which reduces fraction n/m by producing a pair of relatively prime numbers p and q such that n/m=p/q.
Example: Suppose that initial fraction is 247/26. Reduced fraction is 19/2.
## Problem Analysis
To reduce a fraction, we need to find a number which divides without remainder both the numerator and the denominator. Larger the divisor, smaller the numerator and the denominator will be after being divided by it. The largest divisor for two numbers is called greatest common divisor (or greatest common factor). When two numbers are divided by their GCD, they become relatively prime. This means that there is no common factor for the two numbers anymore.
So the problem of reducing a fraction is to calculate greatest common divisor of its numerator and denominator k, such that n=pk and m=qk. After dividing both numbers in fraction n/m with k, we obtain the reduced fraction p/q.
Now we will focus on the problem of calculating GCD for two integer numbers n and m. First of all, notice that GCD(n, m) is the same as GCD(m, n). Consequently, we are free to assume that n >= m (if it is not, then we can simply swap the values). Next, we will assume that there is some value k >= 1 such that it divides both n and m: n=pk, m=qk. Finally, we will notice that if k divides n and m, then k also divides their difference n-m. This is proven easily:
We can follow the same logic and claim that k also divides n-2m, n-3m, ..., n-am. This process ends with value a such that n-am becomes smaller than m. In other words, n-am is the remainder after dividing n by m:
By combing this equation with relations to GCD we can prove what we have already predicted: that remainder when dividing n by m shares the same GCD with original numbers n and m:
Since n mod m is less than both n and m, it seems that we have transformed the original problem into a smaller one which yields the same result. We are free to repeat the step and to produce another pair of even smaller numbers. The question is when does this process end? Well, at some point the modulo will eventually have to become zero. At that point we will observe that GCD(x, 0) = x for any positive integer x. This gives us the ending criterion: the process of replacing the lower argument to GCD with modulo result ends when modulo becomes zero. At that moment the other argument, which is still positive, becomes the overall result, i.e. it is the requested GCD(n, m).
Here is the simple recursive solution to the problem:
```int gcdRecursive(n, m)
{
if (n < m)
return gcdRecursive(m, n);
if (m == 0)
return n;
return gcdRecursive(m, n % m);
}
```
The first if statement ensures that first argument is greater or equal to the second argument, which is prerequisite to calculate the first modulo. Second if statement tests the ending criterion: if smaller argument has reached zero, then the first argument is overall result. Finally, if both arguments are positive, with first argument being greater or equal to the second argument, then we repeat the calculation on modified arguments: smaller number becomes the first argument, and result of modulo operation becomes the second argument.
Function presented above is quite handy, but needlessly complicated in terms of recursive calls. It can be easily transformed into iterative form:
```int gcd(n, m)
{
if (n < m)
{
int tmp = n;
n = m;
m = tmp;
}
while (m > 0)
{
int tmp = n % m;
n = m;
m = tmp;
}
return n;
}
```
This implementation produces the same output as the recursive one. This function can then be used to reduce the fraction:
```void reduceFraction(n in/out, m in/out)
{
int k = gcd(n, m);
n /= k;
m /= k;
}
```
## Implementation
Below is the complete implementation of console application in C# which inputs fractions, reduces them and prints them out reduced.
```using System;
namespace ReducingFractions
{
public class Program
{
static int Gcd(int n, int m)
{
if (n < m)
{
int tmp = n;
n = m;
m = tmp;
}
while (m > 0)
{
int tmp = n % m;
n = m;
m = tmp;
}
return n;
}
static void ReduceFraction(ref int n, ref int m)
{
int k = Gcd(n, m);
n = k;
m = k;
}
static void Main(string[] args)
{
while (true)
{
Console.Write("Enter numerator (negative to exit): ");
if (n < 0)
break;
Console.Write("Enter denominator: ");
int p = n;
int q = m;
ReduceFraction(ref p, ref q);
Console.WriteLine("{0} / {1} = {2} / {3}", n, m, p, q);
Console.WriteLine();
}
}
}
}
```
## Demonstration
When application is run, it produces output like this:
```Enter numerator (negative to exit): 12
Enter denominator: 16
12 / 16 = 3 / 4
Enter numerator (negative to exit): 3
Enter denominator: 5
3 / 5 = 3 / 5
Enter numerator (negative to exit): 247
Enter denominator: 26
247 / 26 = 19 / 2
Enter numerator (negative to exit): 0
Enter denominator: 9
0 / 9 = 0 / 1
Enter numerator (negative to exit): -1
```
Published: Dec 12, 2013
ZORAN HORVAT
Zoran is software architect dedicated to clean design and CTO in a growing software company. Since 2014 Zoran is an author at Pluralsight where he is preparing a series of courses on object-oriented and functional design, design patterns, writing unit and integration tests and applying methods to improve code design and long-term maintainability.
Watch Zoran's video courses at pluralsight.com (requires registration):
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Tactical Design Patterns in .NET: Creating Objects
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Tactical Design Patterns in .NET: Control Flow
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Writing Highly Maintainable Unit Tests
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|
# Verbal Reasoning - Direction Sense Test
### Exercise :: Direction Sense Test - Introduction
Introduction:
There are four main directions - East, West, North and South as shown below:
There are four cardinal directions - North-East (N-E), North-West (N-W), South-East (S-E), and South-West (S-W) as shown below:
1. At the time of sunrise if a man stands facing the east, his shadow will be towards west.
2. At the time of sunset the shadow of an object is always in the east.
3. If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.
4. At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow.
Main types of questions are given below:
Type 1:
Siva starting from his house, goes 5 km in the East, then he turns to his left and goes 4 km. Finally he turns to his left and goes 5 km. Now how far is he from his house and in what direction?
Solution:
From third position it is clear he is 4 km from his house and is in North direction.
Type 2:
Suresh starting from his house, goes 4 km in the East, then he turns to his right and goes 3 km. What minimum distance will be covered by him to come back to his house?
Solution:
Type 3:
One morning after sunrise Juhi while going to school met Lalli at Boring road crossing. Lalli's shadow was exactly to the right of Juhi. If they were face to face, which direction was Juhi facing?
Solution: In the morning sunrises in the east.
So in morning the shadow falls towards the west.
Now Lalli's shadow falls to the right of the Juhi. Hence Juhi is facing South.
Type 4:
Hema starting from her house walked 5 km to reach the crossing of Palace. In which direction she was going, a road opposite to this direction goes to Hospital. The road to the right goes to station. If the road which goes to station is just opposite to the road which IT-Park, then in which direction to Hema is the road which goes to IT-Park?
Solution:
From II it is clear that the road which goes to IT-Park is left to Hema.
|
### Row Operation
##### 1. Gaussian elimination and back substitution
To solve a system of linear equations, we first simplify the system and then do backward substitution.
Example In the system
3x1 + x2 - x3 = 2 x1 - x2 + x3 = 2 2x1 + 2x2 + x3 = 6
the variable x1 appears in all 3 equations. We try to eliminate x1 so that it appears in only 1 equation. Specifically, by multiplying -3 to the 2nd equation and then add to the 1st equation, x1 is eliminated from the 1st equation.
4x2 - 4x3 = -4 x1 - x2 + x3 = 2 2x1 + 2x2 + x3 = 6
Next the operation (-2)[equation 2] + [equation 3] (multiply -2 to the 2nd equation and then add to the 3rd equation) further eliminates x1 from the 3rd equation.
4x2 - 4x3 = -4 x1 - x2 + x3 = 2 4x2 - x3 = 2
The system is as simple as we can get as far as x1 is concerned.
Next we try to simplify the system regarding x2. The operation (-1)[equation 1] + [equation 3] eliminates x2 from the 3rd equation.
4x2 - 4x3 = -4 x1 - x2 + x3 = 2 3x3 = 6
Although the system is simple enough to be solved. We may still make some cosmetic improvements. By exchanging the 1st and the 2nd equations (denoted [equation 1] ↔ [equation 2]), we rearrange the equations from the most complicated down to the simplest.
x1 - x2 + x3 = 2 4x2 - 4x3 = -4 3x3 = 6
Finally, the coefficients can be further simplified by multiplying 1/4 and 1/3 to the 2nd and the 3rd equations (denoted (1/4)[equation 2] and (1/3)[equation 3]).
x1 - x2 + x3 = 2 x2 - x3 = -1 x3 = 2
This completes the simplification process.
Now we start solving the system by backward substitution. From the 3rd equation (the simplest equation), we have
x3 = 2.
Substituting this into the 2nd equation (the next simplest), we get
x2 = -1 + x3 = -1 + 2 = 1.
Further substituting into the 1st equation (the most complicated), we get
x1 = 2 + x2 - x3 = 2 + 1 - 2 = 1.
We conclude that the system has a unique solution x1 = 1, x2 = 1, x3 = 2.
In the example above, we have used the following operations to simplify a system.
r[equation i] + [equation j]
add a multiple of one equation to another.
[equation i] ↔ [equation j]
exchange two equations.
d[equation i], d ≠ 0
multiply a nonzero number to an equation.
In general, any system of linear equations can be simplified, by the three operations, to a simple system in which equations are arranged from the most complicated to the simplest. This process is called the gaussian elimination.
After completing the gaussian elimination, we may solve the simplest equation. Then we successively substitute the results into more and more complicated equations to get the whole solution. This process is called the backward substitution.
Example This example shows different choices may be taken in the gaussian elimination. However, the solution will not be affected by the choices.
For the same system just solved,
3x1 + x2 - x3 = 2 x1 - x2 + x3 = 2 2x1 + 2x2 + x3 = 6
we try a slightly different gaussian elimination process. By (-3/2)[equation 3] + [equation 1] and (-1/2)[equation 3] + [equation 2], we have
- 2x2 - 5/2x3 = -7 - 2x2 + 1/2x3 = -1 2x1 + 2x2 + x3 = 6
Then (-1)[equation 2] + [equation 1] gives us
- 3x3 = -6 - 2x2 + 1/2x3 = -1 2x1 + 2x2 + x3 = 6
We further exchange [equation 1] ↔ [equation 3] to rearrange from the most complicated down to the simplest.
2x1 + 2x2 + x3 = 6 - 2x2 + 1/2x3 = -1 - 3x3 = -6
Now from the 3rd equation, we have
x3 = (-6)/(-3) = 2.
Substituting into the 2nd equation, we have
x2 = -(-1 - 1/2x3)/2 = -(-1 - (1/2)×2)/2 = 1.
Substituting further into the 1st equation, we get
x1 = (6 - 2x2 - x3)/2 = (6 - 2 - 2)/2 = 1.
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# Linear Equations
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The unit Linear Equations presents the student with the basics of solving linear equations, including equations that involve a variable on both sides and equations that require the usage of the distributive property to eliminate parentheses. We also briefly study inequalities and graphing. This unit best suits pre-algebra or grades 7 to 8 mathematics studies.
The first lesson reviews the concept of an equation and how to model equations using a pan balance (scale). The basic principle for solving equations is that, when you perform the same operation on both sides of an equation, the two sides remain equal.
The unit presents two alternatives for keeping track of the operations to be performed on an equation. The one method, writing the operation under each side of the equation, is common in the United States. The other method, writing the operation in the right margin, is common in Finland. Either way is correct, and the choice is just a matter of the personal preference of the teacher.
The introduction to solving equations is followed by a lesson on addition and subtraction equations and another on multiplication and division equations. All the equations are easily solved in only one step of calculations. The twofold goal is to make the student proficient in manipulating negative integers and also to lay a foundation for handling more involved equations that are studied later on in the unit.
In the next lesson, students write equations to solve simple word problems. Even though they could solve most of these problems without using the equations, the purpose of the lesson is to make the student proficient in writing simple equations before moving on to more complex equations from more difficult word problems.
The next topic, in the lesson Constant Speed, is solving problems with distance (d), rate or velocity (v), and time (t). Students use the equivalent formulas d = vt and v = d/t to solve problems involving constant or average speed. They learn an easy way to remember the formula v = d/t from the unit for speed that they already know, “miles per hour.”
In later lessons, we delve deeper into our study of equations. Now the equations require two or more steps to solve and may contain parentheses. The variable may appear on both sides of the equation. Students will also write equations to solve simple word problems.
There is also a lesson on patterns of growth, which may seem to be simply a fascinating topic, but in reality presents the fundamentals of a very important concept in algebra ― that of linear functions (although they are not mentioned by that name)―and complements the study of lines in the subsequent lessons.
After the section about equations, the text briefly presents the basics of inequalities and how to graph them on a number line. Students apply the principles for solving equations to solve simple inequalities and word problems that involve inequalities.
The last major topic is graphing. Students begin the section by learning to graph linear equations and continue on to the concept of slope, which in informal terms is a measure of the inclination of a line. More formally, slope can be defined as the ratio of the change in y-values to the change in x-values. The final lesson applies graphing to the previously-studied concepts of speed, time, and distance through graphs of the equation d = vt in the coordinate plane.
Total Pages
142 pages
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Teaching Duration
2 months
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## Related Articles
• NCERT Solutions for Class 9 Maths
# Class 9 NCERT Solutions- Chapter 5 Introduction to Euclid’s Geometry – Exercise 5.1
### Question 1: Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
(ii) There is an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In Given fig, if AB = PQ and PQ = XY, then AB = XY.
Solution:
(i) False
Reason: If we mark a point O on the surface of a paper. Using pencil and scale, we can draw infinite number of straight lines passing through O.
(ii) False
Reason: Through two distinct points there can be only one line that can be drawn. Hence, the statement mentioned above is False.
(iii) True
Reason: A line that is terminated can be indefinitely produced on both sides as a line can be extended on both its sides infinitely. Hence, the statement mentioned above is True.
(iv) True
Reason: The radii of two circles are equal when the two circles are equal. The circumference and the centre of both the circles coincide; and thus, the radius of the two circles should be equal. Hence, the statement mentioned above is True.
(v) True
Reason: According to Euclid’s 1st axiom- “Things which are equal to the same thing are also equal to one another”. Hence, the statement mentioned above is True.
### Question 2: Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they and how might you define them?
(i) Parallel lines
(ii) Perpendicular lines
(iii) Line segment
(v) Square
Solution:
Yes, there are other terms which need to be defined first, so that we understand better:
• Plane: Flat surfaces in which geometric figures can be drawn are known are plane. A plane surface is a surface which lies evenly with the straight lines on itself.
• Point: A dimensionless dot which is drawn on a plane surface is known as point. A point is that which has no part.
• Line: A collection of points that has only length and no breadth is known as a line. And it can be extended on both directions. A line is breadth-less length.
(i) Parallel lines: Two lines l and m in a plan are said to be parallel if they have a no common point and we can write them as l || m.
(ii) Perpendicular lines: Two lines A and B are said to be perpendicular if the form a right angle and we can write them as A ⊥ B.
(iii) Line Segment: A line segment is a part of line and having a definite length. It has two end-points. In the figure, a line segment is shown having end points P and Q. Write it as arrow over PQ.
(iv) Radius of Circle: The distance from the centre to a point on the circle is called the radius of the circle. In the figure, l is centre and m is a point on the circle, then lm is the radius of the circle.
(v) Square: A quadrilateral in which all the four angles are right angles and all the four sides are equal is called a Square
In the given figure ABCD is a Square.
### Question 3: Consider two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent?
Do they follow from Euclid’s postulates? Explain.
Solution:
Yes, these postulates contain undefined terms such as ‘Point and Line’. Undefined terms in the postulates are:
• There are many points that lie in a plane. But, in the postulates as given here, the position of the point C is not given, as of whether it lies on the line segment joining AB or it is not joining line segment.
• On top of that, there is no information about whether the points are in same plane or not.
And
Yes, these postulates are consistent when we deal with these two situations:
• Point C is lying on the line segment AB in between A and B.
• Point C does not lie on the line segment AB.
No, they don’t follow from Euclid’s postulates. They follow the axioms i.e “Given two distinct points, there is a unique line that passes through them.”
### Question 4: If a point C lies between two points A and B such that AC = BC, then prove that AC = 12 AB, explain by drawing the figure.
Solution:
AC = BC (Given)
As we have studied in this chapter “If equals are added to equals then there wholes are also equal”.
Therefore, AC + BC = BC + AC
⇒ 2AC = BC+AC
As we have studied in this chapter, we know that,
BC+AC = AB (as it coincides with line segment AB)
∴ 2 AC = AB (If equals are added to equals, the wholes are equal).
⇒ AC = (½)AB.
### Question 5: In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.
Solution:
Let, AB be the line segment as given in the Question.
Assume that points C and D are the two different mid points of line segment AB.
Therefore, C and D are the midpoints of AB.
Now, as C and D are mid points of Ab we have,
AC = CB and AD = DB
CB+AC = AB (as it coincides with line segment AB)
Now,
Adding AC to the L.H.S and R.H.S of the equation AC = CB
We get, AC+AC = CB+AC (If equals are added to equals, the wholes are equal.)
⇒ 2AC = AB — (i)
Similarly,
2 AD = AB — (ii)
From equation (i) and (ii), Since R.H.S are same, we equate the L.H.S we get,
2 AC = 2 AD (Things which are equal to the same thing are equal to one another.)
⇒ AC = AD (Things which are double of the same things are equal to one another.)
Thus, we conclude that C and D are the same points.
This contradicts our assumption that C and D are two different mid points of AB.
Thus, it is proved that every line segment has one and only one mid-point.
Hence, Proved.
### Question 6: In Figure, if AC = BD, then prove that AB = CD.
Solution:
According to the question, AC = BD
From the given figure we can conclude that,,
AC = AB+BC
BD = BC+CD
⇒ AB+BC = BC+CD (AC = BD, given)
As we have studied, according to Euclid’s axiom, when equals are subtracted from equals, remainders are also equal.
Subtracting BC from the L.H.S and R.H.S of the equation AB+BC = BC+CD, we get,
AB+BC-BC = BC+CD-BC
AB = CD
Hence Proved.
### Question 7: Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate).
Solution:
Euclid’s fifth axiom states that “the whole is greater than the part.”
For Example: A cake. When it is whole or complete, assume that it measures 2 pounds but when a part from it is taken out and measured, its weight will be smaller than the previous measurement. So, the fifth axiom of Euclid is true for all the materials in the universe. Hence, Axiom 5, in the list of Euclid’s axioms, is considered a ‘universal truth’.
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# Lesson Explainer: Polynomial Long Division without Remainder Mathematics • 10th Grade
In this explainer, we will learn how to perform long division on polynomials.
Before we begin explaining the process of long division of polynomials, letβs first recall what it means to divide integers and how we apply long division to integers.
If we have two integers and , where is nonzero, then is the number that when multiplied by , gives . Hence, . We define the division of polynomials in the same way.
### Definition: Division of Polynomials
We define the quotient of two polynomials by finding the polynomial whose product with the divisor yields the dividend.
For example, if we want to find , we can note that we want to find a polynomial that when multiplied by , gives . One way of finding this value is to expand the product . We have
Thus,
We can then ask, βHow do we find an expression for a quotient of polynomials in general?β
Letβs first consider an example involving integers. If we want to determine the value of , we first note that 5 does not go exactly into 17. We can instead employ long division to evaluate this quotient:
We first see that three 5s make 15 and that we cannot have any more 5s. Removing this from 17, we are left with a remainder of 2. We call 17 the dividend since it is being divided by the divisor 5, we call 3 the quotient, and we call 2 the remainder.
It is also worth noting that we say that divides if its remainder is 0 after division by . If we call the remainder after dividing by Β , then we can guarantee that . If this is not the case, then we can increase the value of the quotient.
The reason for this is that the process for long division rewrites the fraction. Letβs say that has a quotient and a remainder . Then goes into times and has a remainder . This means that . Hence,
In our numerical example, this reads
We are almost ready to apply this process to polynomials. Letβs say we want to divide a polynomial called the dividend by another polynomial called the divisor. We want to do this by subtracting multiples of the divisor from the dividend until we can no longer do so. To see how to do this, letβs consider an example.
Suppose we want to divide by . We need to determine how many multiples of we can remove from . Each multiple will be a polynomial since the products of polynomials are polynomials. We will call the polynomial that tells us the number of multiples of we remove the quotient. This means that we want to find a polynomial whose product with is as close to as possible.
Letβs write this division using the same long division notation we use for integers:
We note that the leading term of the dividend is and that the leading term of the divisor is . We then see that . We will write this in our quotient above the -term:
Remember that we want to subtract the product of the quotient and divisor from the dividend. We can do this term by term. If we multiply the divisor by , we get
Subtracting this from the dividend yields
We write this in the long division notation as follows:
Before we continue with the division, it may be useful to consider what we have actually shown. We have shown that . We can then divide through by , which gives
Thus, we have reduced the degree of the dividend. We can apply this process again, this time with as the dividend. The leading term of this new dividend is , and the leading term of the divisor is . So, , and we add this to the quotient as follows:
We want to remove the divisor from the dividend 3 times. This gives us
We can represent this using long division notation as follows:
In this explainer, we will only deal with cases where the remainder is the zero polynomial. However, it is worth noting that each time we perform this process, we are lowering the degree of the dividend by the degree of the divisor. We keep this process going until the remainder has a lower degree than the divisor. At this point when we divide their leading terms, we would get either zero or a variable raised to a negative exponent, which is not a monomial.
Using polynomial long division, we have shown that . So,
We can verify this by multiplying the quotient by the divisor . We have
This works as a nice check to make sure that the answer is correct.
We can summarize the working in the above example as follows:
Letβs now see an example of how to apply this process to divide two polynomials.
### Example 1: Dividing Polynomials Using Polynomial Long Division
Find the quotient of divided by .
We can find an expression for using long division. The first step is to find the quotient of the leading terms of the dividend and the divisor. We note that . So, we write in the quotient, and we subtract from the dividend. We have that , so we see that
We calculate that
We can then add this onto our division:
We want to apply this process again, this time with as the dividend. We need to see how many times the divisor goes into the dividend . We do this by first dividing their leading terms. We see that . So, we add 3 to the quotient, and we then need to subtract from the dividend. We find that . Subtracting this gives us the following:
We find that .
Hence,
We can verify this answer by calculating .
In our next example, we will find the quotient of two polynomials where both of the polynomials are nonmonic.
### Example 2: Dividing a Polynomial by a First-Degree Divisor to Find the Quotient
Find the quotient when is divided by .
We can divide two polynomials using polynomial long division. First, we need to divide their leading terms. We have . We add this into our quotient, and this means we need to subtract from our dividend. Since , we can do this as follows:
We calculate that . We have shown that
We can apply this process again to divide by . We divide the leading terms to get . Thus, we add to the quotient and subtract from the dividend as follows:
We then calculate that .
We need to apply this process one final time to fully divide the expression.
This time, the division of the leading terms gives us . We add this to the quotient. We then need to subtract from this new dividend. Since this is equal to the dividend, we get a value of 0:
Hence,
In our next example, we will divide a nonmonic cubic polynomial by a linear polynomial.
### Example 3: Dividing a Cubic Polynomial by a First-Degree Divisor to Find the Quotient
Find the quotient when is divided by .
We can divide two polynomials using polynomial long division. First, we need to divide their leading terms. We have . We add this to our quotient, and this means we need to subtract from our dividend. Since , we can do this as follows:
We calculate that
Since the degree of is higher than or equal to the degree of the divisor, we now need to apply this process again, with as our new dividend.
We divide their leading terms to get , which we add to the quotient. We then need to subtract from the dividend. We can do this as follows:
We calculate that
Since the degree of this polynomial is equal to that of the divisor, we need to apply this process one final time, with as the dividend.
We divide their leading terms to get , which we then add to the quotient. We then need to subtract from the dividend. Since this is equal to the dividend, we will end up with a remainder of 0, as shown:
Hence, we have shown that the quotient that results from dividing by is . We could check this answer by calculating to ensure that we get our original cubic dividend.
In our next example, we will see how to apply polynomial long division to find a missing dimension in a given diagram.
### Example 4: Determining a Dimension Using Polynomial Long Division
Given that the area of the rectangle in the diagram is , find an expression for the width of the rectangle.
We first recall that the area of rectangle is given by the product of its length and width. This means we can determine the width of this rectangle by dividing its area by its length. This means we need to divide a cubic polynomial by a linear polynomial. We can do this using polynomial long division.
We first divide their leading terms to get . We add this to the quotient and then subtract from the dividend. We note that and that , as shown:
Since the degree of is higher than or equal to the degree of the divisor, we need to apply this process again, with as the new dividend.
We first divide their leading terms to get , which we then add to the quotient. We now need to subtract from the divided. This gives us the following:
We calculate that
Since the degree of this polynomial is equal to the degree of the divisor, we need to apply this process one final time, with as the dividend.
We divide their leading terms to get , which we add to the quotient. We then note that . This is equal to the dividend, so when we subtract, we will get 0. This gives us the following:
Hence, , and the width of the rectangle is given by .
In our final example, we will use polynomial long division and the given divisibility of two polynomials to determine the value of an unknown coefficient.
### Example 5: Finding the Value of a Constant That Makes a Polynomial Divisible
Find the value of that makes the expression divisible by .
We first recall that we say a polynomial is divisible by another polynomial if their division has a remainder of 0. This means we can apply polynomial long division to the given polynomials, and we know that the remainder must be the zero polynomial.
Before we apply polynomial long division, it is worth noting that the quintic polynomial is not given in descending powers of . We should always reorder the dividend to have descending powers of since we always want to remove the leading terms of the dividends. So, we will use as the dividend.
To apply polynomial long division, we first need to divide the leading terms. We get . We add this to the quotient and then subtract from the dividend. Since , we will also include the term to keep the columns with the same powers of . We have
We have calculated that
It is also worth noting that we add the term into our long division to keep the columns with the same powers of .
Since this is not zero, we need to apply this process again, with as the dividend.
We first divide the leading terms to get , and we add this to the quotient. Next, we need to subtract from the dividend. We note that . We can then subtract this from the dividend as follows:
We can calculate that
Since this is not zero, we need to apply this process again, with as the dividend.
We divide their leading terms to get , and we add this to our quotient. We then subtract from the dividend as follows:
We calculated that
Since we are told that the division is exact, we know that the remainder must be zero. Hence, .
We can solve this to see that .
Letβs finish by recapping some of the important points from this explainer.
### Key Points
• We define the quotient of two polynomials by finding the polynomial whose product with the divisor yields the dividend.
• We can divide polynomials using long division.
• We can check our answer by multiplying the divisor by the quotient.
• If the remainder polynomial in the division of two polynomials is the zero polynomial, then we say that the divisor divides the dividend.
• We should always reorder the dividend to have descending powers of since we always want to remove the leading terms of the dividends.
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# Identifying the place value of digits in 5-digit numbers
In this lesson, we will be representing 5-digit numbers pictorially and identifying the value of each digit within these numbers.
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Identifying the place value of digits:
Exit quiz for Year 5 Maths Lesson: Identifying the place value of digits
Q1.What is the value of the digit '5' in the number 65 891?
1/5
Q2.What is the missing number in this partitioned equation: 82 345 = 80 000 + 2000 + 300 + 40 + _____?
2/5
Q3.Which of the following options is an accurate way of partitioning the number 12 489?
3/5
Q4.Which of the following options is an accurate way of partitioning the number 67 321?
4/5
Q5.What is the next number in this sequence: 16 732, 17 732, 18 732, 19 732, _________?
5/5
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Identifying the place value of digits:
Exit quiz for Year 5 Maths Lesson: Identifying the place value of digits
Q1.What is the value of the digit '5' in the number 65 891?
1/5
Q2.What is the missing number in this partitioned equation: 82 345 = 80 000 + 2000 + 300 + 40 + _____?
2/5
Q3.Which of the following options is an accurate way of partitioning the number 12 489?
3/5
Q4.Which of the following options is an accurate way of partitioning the number 67 321?
4/5
Q5.What is the next number in this sequence: 16 732, 17 732, 18 732, 19 732, _________?
5/5
# Lesson summary: Identifying the place value of digits in 5-digit numbers
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Walk
On the spot:
Dance
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# High School Math : Solving Inequalities
## Example Questions
### Example Question #1 : Solving Inequalities
Solve for :
Explanation:
Inequalities can be treated like any other equation except when multiplying and dividing by negative numbers. When multiplying or dividing by negative numbers, we just flip the sign of the inequality so that becomes , and vice versa.
### Example Question #2 : Solving Inequalities
Solve for :
Explanation:
Inequalities can be treated like any other equation except when multiplying and dividing by negative numbers. When multiplying or dividing by negative numbers, we just flip the sign of the inequality so that becomes , and vice versa. When we solve binomials, we must take extra caution because .
So when we solve inequalities with binomials, we must create two scenarios: one where the value inside of the parentheses is positive and one where it is negative. For the negative scenario, we must flip the sign as we normally do for inequalities.
Now we must create our two scenarios:
and
Notice that in the negative scenario, we flipped the sign of the inequality.
and
and
### Example Question #3 : Solving Inequalities
Solve for :
Explanation:
Inequalities can be treated like any other equation except when multiplying and dividing by negative numbers. When multiplying or dividing by negative numbers, we just flip the sign of the inequality so that becomes , and vice versa. When we solve binomials, we must take extra caution because .
So when we solve inequalities with binomials, we must create two scenarios: one where the value inside of the parentheses is positive and one where it is negative. For the negative scenario, we must flip the sign as we normally do for inequalities.
Now we must create our two scenarios:
and
Notice that in the negative scenario, we flipped the sign of the inequality.
and
and
### Example Question #4 : Solving Inequalities
Solve the inequality for x:
Explanation:
Subtract 4 from both sides:
Divide both sides by 2:
### Example Question #5 : Solving Inequalities
Solve for .
Explanation:
Divide both sides by –7. When dividing by a negative value, we must also change the direction of the inequality sign.
### Example Question #6 : Solving Inequalities
Solve for :
Explanation:
Move like terms to the same sides:
Combine like terms:
Divide both sides by 3:
Solve for :
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# How do you factor the expression 56x^3 +43x^2+5x?
Mar 18, 2016
$56 {x}^{3} + 43 {x}^{2} + 5 x = x \left(7 x + 1\right) \left(8 x + 5\right)$
#### Explanation:
First separate out the common factor $x$:
$56 {x}^{3} + 43 {x}^{2} + 5 x = x \left(56 {x}^{2} + 43 x + 5\right)$
To factor the remaining quadratic expression, use an AC method.
Find a pair of factors of $A C = 56 \cdot 5 = 280$ with sum $B = 43$.
$\textcolor{w h i t e}{}$
To help find the appropriate pair you can proceed as follow:
Find the prime factorisation of $280$:
$280 = 2 \cdot 2 \cdot 2 \cdot 5 \cdot 7$
Next note that $43$ is odd, so it is the sum of an odd and an even number.
As a result, the prime factors must be split between the pair in such a way that all factors of $2$ are on one side or the other.
This leaves the following possibilities to check the sum:
$1 + 5 \cdot 7 \cdot {2}^{3} = 1 + 280 = 281$
$5 + 7 \cdot {2}^{3} = 5 + 56 = 61$
$7 + 5 \cdot {2}^{3} = 7 + 40 = 47$
$5 \cdot 7 + {2}^{3} = 35 + 8 = 43$
The last pair $35 , 8$ works.
Use this pair to split the middle term and factor by grouping:
$56 {x}^{2} + 43 x + 5$
$= 56 {x}^{2} + 35 x + 8 x + 5$
$= \left(56 {x}^{2} + 35 x\right) + \left(8 x + 5\right)$
$= 7 x \left(8 x + 5\right) + 1 \left(8 x + 5\right)$
$= \left(7 x + 1\right) \left(8 x + 5\right)$
Putting it all together:
$56 {x}^{3} + 43 {x}^{2} + 5 x = x \left(7 x + 1\right) \left(8 x + 5\right)$
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The result follows by applying Rolle’s Theorem to g. ¤ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f 0 . Practice Exercise: Rolle's theorem … We can see its geometric meaning as follows: \Rolle’s theorem" by Harp is licensed under CC BY-SA 2.5 Theorem 1.2. Explain why there are at least two times during the flight when the speed of Section 4-7 : The Mean Value Theorem. For example, if we have a property of f0 and we want to see the efiect of this property on f, we usually try to apply the mean value theorem. Since f (x) has infinite zeroes in \begin{align}\left[ {0,\frac{1}{\pi }} \right]\end{align} given by (i), f '(x) will also have an infinite number of zeroes. Take Toppr Scholastic Test for Aptitude and Reasoning <> Concepts. Proof: The argument uses mathematical induction. If it can, find all values of c that satisfy the theorem. Rolle S Theorem. THE TAYLOR REMAINDER THEOREM JAMES KEESLING In this post we give a proof of the Taylor Remainder Theorem. 3.2 Rolle’s Theorem and the Mean Value Theorem Rolle’s Theorem – Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Be sure to show your set up in finding the value(s). Then, there is a point c2(a;b) such that f0(c) = 0. Let f(x) be di erentiable on [a;b] and suppose that f(a) = f(b). For each problem, determine if Rolle's Theorem can be applied. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. We can use the Intermediate Value Theorem to show that has at least one real solution: If f a f b '0 then there is at least one number c in (a, b) such that fc . When n = 0, Taylor’s theorem reduces to the Mean Value Theorem which is itself a consequence of Rolle’s theorem. and by Rolle’s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. It’s basic idea is: given a set of values in a set range, one of those points will equal the average. Rolle's theorem is the result of the mean value theorem where under the conditions: f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). Now an application of Rolle's Theorem to gives , for some . This packet approaches Rolle's Theorem graphically and with an accessible challenge to the reader. Lesson 16 Rolle’s Theorem and Mean Value Theorem ROLLE’S THEOREM This theorem states the geometrically obvious fact that if the graph of a differentiable function intersects the x-axis at two places, a and b there must be at least one place where the tangent line is horizontal. To give a graphical explanation of Rolle's Theorem-an important precursor to the Mean Value Theorem in Calculus. In these free GATE Study Notes, we will learn about the important Mean Value Theorems like Rolle’s Theorem, Lagrange’s Mean Value Theorem, Cauchy’s Mean Value Theorem and Taylor’s Theorem. If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that ′ =. The Mean Value Theorem is an extension of the Intermediate Value Theorem.. 20B Mean Value Theorem 2 Mean Value Theorem for Derivatives If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c on (a,b) such that EX 1 Find the number c guaranteed by the MVT for derivatives for Make now. Access the answers to hundreds of Rolle's theorem questions that are explained in a way that's easy for you to understand. If it cannot, explain why not. Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. A similar approach can be used to prove Taylor’s theorem. If f(a) = f(b) = 0 then 9 some s 2 [a;b] s.t. Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. Material in PDF The Mean Value Theorems are some of the most important theoretical tools in Calculus and they are classified into various types. %PDF-1.4 x cos 2x on 12' 6 Detennine if Rolle's Theorem can be applied to the following functions on the given intewal. Watch learning videos, swipe through stories, and browse through concepts. Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. Get help with your Rolle's theorem homework. Rolle's Theorem and The Mean Value Theorem x y a c b A B x Tangent line is parallel to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b a, ] and number b a, ) there exists a c in (b a , ) such that Instantaneous rate of change = average rate of change By Rolle’s theorem, between any two successive zeroes of f(x) will lie a zero f '(x). In modern mathematics, the proof of Rolle’s theorem is based on two other theorems − the Weierstrass extreme value theorem and Fermat’s theorem. EXAMPLE: Determine whether Rolle’s Theorem can be applied to . For the function f shown below, determine we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0.If not, explain why not. We seek a c in (a,b) with f′(c) = 0. Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. So the Rolle’s theorem fails here. For example, if we have a property of f 0 and we want to see the effect of this property on f , we usually try to apply the mean value theorem. Learn with content. Proof. (Rolle’s theorem) Let f : [a;b] !R be a continuous function on [a;b], di erentiable on (a;b) and such that f(a) = f(b). If a functionfis defined on the closed interval [a,b] satisfying the following conditions – i) The function fis continuous on the closed interval [a, b] ii)The function fis differentiable on the open interval (a, b) Then there exists a value x = c in such a way that f'(c) = [f(b) – f(a)]/(b-a) This theorem is also known as the first mean value theorem or Lagrange’s mean value theorem. Videos. After 5.5 hours, the plan arrives at its destination. �_�8�j&�j6���Na$�n�-5��K�H Rolle’s Theorem and other related mathematical concepts. If f is zero at the n distinct points x x x 01 n in >ab,,@ then there exists a number c in ab, such that fcn 0. �K��Y�C��!�OC���ux(�XQ��gP_'�s���Տ_��:��;�A#n!���z:?�{���P?�Ō���]�5Ի�&���j��+�Rjt�!�F=~��sfD�[x�e#̓E�'�ov�Q��'#�Q�qW�˿���O� i�V������ӳ��lGWa�wYD�\ӽ���S�Ng�7=��|���և� �ܼ�=�Չ%,��� EK=IP��bn*_�D�-��'�4����'�=ж�&�t�~L����l3��������h��� ��~kѾ�]Iz���X�-U� VE.D��f;!��q81�̙Ty���KP%�����o��;$�Wh^��%�Ŧn�B1 C�4�UT���fV-�hy��x#8s�!���y�! Proof of Taylor’s Theorem. The result follows by applying Rolle’s Theorem to g. ⁄ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. A plane begins its takeoff at 2:00 PM on a 2500 mile flight. x��=]��q��+�ͷIv��Y)?ز�r$;6EGvU�"E��;Ӣh��I���n v��K-�+q�b ��n�ݘ�o6b�j#�o.�k}���7W~��0��ӻ�/#���������$����t%�W ��� That is, we wish to show that f has a horizontal tangent somewhere between a and b. The proof of Rolle’s Theorem is a matter of examining cases and applying the Theorem on Local Extrema. 13) y = x2 − x − 12 x + 4; [ −3, 4] 14) y = f0(s) = 0. f is continuous on [a;b] therefore assumes absolute max and min values Rolle’s Theorem extends this idea to higher order derivatives: Generalized Rolle’s Theorem: Let f be continuous on >ab, @ and n times differentiable on 1 ab, . Standard version of the theorem. Rolle’s Theorem is a special case of the Mean Value Theorem in which the endpoints are equal. �wg��+�͍��&Q�ណt�ޮ�Ʋ뚵�#��|��s���=�s^4�wlh��&�#��5A ! For problems 1 & 2 determine all the number(s) c which satisfy the conclusion of Rolle’s Theorem for the given function and interval. Stories. Proof: The argument uses mathematical induction. The special case of the MVT, when f(a) = f(b) is called Rolle’s Theorem.. Calculus 120 Worksheet – The Mean Value Theorem and Rolle’s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( Õ)−( Ô) Õ− Ô =′( . x��]I��G�-ɻ�����/��ƴE�-@r�h�١ �^�Կ��9�ƗY�+e����\Y��/�;Ǎ����_ƿi���ﲀ�����w�sJ����ݏ����3���x���~B�������9���"�~�?�Z����×���co=��i�r����pݎ~��ݿ��˿}����Gfa�4�����Ks�?^���f�4���F��h���?������I�ק?����������K/g{��W����+�~�:���[��nvy�5p�I�����q~V�=Wva�ެ=�K�\�F���2�l��� ��|f�O�n9���~�!���}�L��!��a�������}v��?���q�3����/����?����ӻO���V~�[�������+�=1�4�x=�^Śo�Xܳmv� [=�/��w��S�v��Oy���~q1֙�A��x�OT���O��Oǡ�[�_J���3�?�o�+Mq�ٞ3�-AN��x�CD��B��C�N#����j���q;�9�3��s�y��Ӎ���n�Fkf����� X���{z���j^����A���+mLm=w�����ER}��^^��7)j9��İG6����[�v������'�����t!4?���k��0�3�\?h?�~�O�g�A��YRN/��J�������9��1!�C_$�L{��/��ߎq+���|ڶUc+��m��q������#4�GxY�:^밡#��l'a8to��[+�de. (Insert graph of f(x) = sin(x) on the interval (0, 2π) On the x-axis, label the origin as a, and then label x = 3π/2 as b.) Rolle’s Theorem extends this idea to higher order derivatives: Generalized Rolle’s Theorem: Let f be continuous on >ab, @ and n times differentiable on 1 ab, . It is a very simple proof and only assumes Rolle’s Theorem. For each problem, determine if Rolle's Theorem can be applied. 2\�����������M�I����!�G��]�x�x*B�'������U�R� ���I1�����88%M�G[%&���9c� =��W�>���$�����5i��z�c�ص����r ���0y���Jl?�Qڨ�)\+�B��/l;�t�h>�Ҍ����X�350�EN�CJ7�A�����Yq�}�9�hZ(��u�5�@�� Examples: Find the two x-intercepts of the function f and show that f’(x) = 0 at some point between the Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. Then . exact value(s) guaranteed by the theorem. Rolle’s Theorem. Using Rolles Theorem With The intermediate Value Theorem Example Consider the equation x3 + 3x + 1 = 0. Theorem 1.1. The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with The result follows by applying Rolle’s Theorem to g. ⁄ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. The “mean” in mean value theorem refers to the average rate of change of the function. Rolle’s Theorem, like the Theorem on Local Extrema, ends with f′(c) = 0. We can use the Intermediate Value Theorem to show that has at least one real solution: 3�c)'�P#:p�8�ʱ� ����;�c�՚8?�J,p�~$�JN����Υ�����P�Q�j>���g�Tp�|(�a2���������1��5Լ�����|0Z v����5Z�b(�a��;�\Z,d,Fr��b�}ҁc=y�n�Gpl&��5�|���(�a��>? If it can, find all values of c that satisfy the theorem. Thus, which gives the required equality. Then there is a point a<˘o���W#5��}p~��Z��=�z����D����P��b��sy���^&R�=���b�� b���9z�e]�a�����}H{5R���=8^z9C#{HM轎�@7�>��BN�v=GH�*�6�]��Z��ܚ �91�"�������Z�n:�+U�a��A��I�Ȗ�$m�bh���U����I��Oc�����0E2LnU�F��D_;�Tc�~=�Y��|�h�Tf�T����v^��>�k�+W����� �l�=�-�IUN۳����W�|׃_�l �˯����Z6>Ɵ�^JS�5e;#��A1��v������M�x�����]*ݺTʮ���״N�X�� �M���m~G��솆�Yoie��c+�C�co�m��ñ���P�������r,�a <> If it cannot, explain why not. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. If f is zero at the n distinct points x x x 01 n in >ab,,@ then there exists a number c in ab, such that fcn 0. and by Rolle’s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. This version of Rolle's theorem is used to prove the mean value theorem, of which Rolle's theorem is indeed a special case. Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. Rolle's Theorem on Brilliant, the largest community of math and science problem solvers. %���� This calculus video tutorial provides a basic introduction into rolle's theorem. The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with 5 0 obj 172 Chapter 3 3.2 Applications of Differentiation Rolle’s Theorem and the Mean Value Theorem Understand and use Rolle’s }�gdL�c���x�rS�km��V�/���E�p[�ő蕁0��V��Q. The value of 'c' in Rolle's theorem for the function f (x) = ... Customize assignments and download PDF’s. Determine whether the MVT can be applied to f on the closed interval. %�쏢 %PDF-1.4 Let us see some View Rolles Theorem.pdf from MATH 123 at State University of Semarang. Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. Calculus 120 Worksheet – The Mean Value Theorem and Rolle’s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( Õ)−( Ô) Õ− Ô =′( . Rolle's theorem is the result of the mean value theorem where under the conditions: f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). f x x x ( ) 3 1 on [-1, 0]. ʹ뾻��Ӄ�(�m���� 5�O��D}P�kn4��Wcم�V�t�,�iL��X~m3�=lQ�S���{f2���A���D�H����P�>�;\$f=�sF~M��?�o��v8)ѺnC��1�oGIY�ۡ��֍�p=TI���ߎ�w��9#��Q���l��u�N�T{��C�U��=���n2�c�)e�L`����� �����κ�9a�v(� ��xA7(��a'b�^3g��5��a,��9uH*�vU��7WZK�1nswe�T��%�n���է�����B}>����-�& This is explained by the fact that the $$3\text{rd}$$ condition is not satisfied (since $$f\left( 0 \right) \ne f\left( 1 \right).$$) Figure 5. The Common Sense Explanation. 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# Unit 12 Trigonometry Homework 1 Pythagorean Theorem Answers
Pythagoras’ theorem is one of the most fundamental mathematical theories of all time, and trigonometry is a branch of mathematics that uses the theorem to solve right-angled triangle problems. Homework 1 for Unit 12 Trigonometry covers the theorem in great detail and requires students to answer questions about it. To help students understand the answers, this article provides a detailed explanation of the theorem and the different types of calculations that are required to solve a problem.
## Understanding the Pythagorean Theorem and Its Uses
The Pythagorean theorem states that in any right-angled triangle, the sum of the squares of the two shorter sides (a and b) will equal the square of the longest side (c). This can be written as a2 + b2 = c2. The theorem can be used to solve right-angled triangle problems by using the values of two sides to calculate the value of the third side. The theorem is also used to calculate the area of a right-angled triangle.
## Using the Theorem to Solve Trigonometry Problems
The Pythagorean theorem is used to solve trigonometry problems by determining the length of one side of a right-angled triangle when the lengths of the other two sides are known. For example, if the lengths of two sides of a triangle are 4 and 5, then the length of the third side can be determined by using the theorem. Using the equation a2 + b2 = c2, the length of the third side (c) can be calculated as c = √(4² + 5²) = 6.403.
## Applying the Theorem to Determine Area
The theorem can also be used to calculate the area of a right-angled triangle. To do this, the lengths of the two shorter sides (a and b) must be known. The area of the triangle can then be calculated using the formula A = ½ab. For example, if the length of the two shorter sides of a triangle are 4 and 5, then the area of the triangle can be calculated as A = ½ x 4 x 5 = 10.
##
The Pythagorean theorem is an important mathematical theory that is used to solve right-angled triangle problems. It states that the sum of the squares of the two shorter sides of a right-angled triangle will equal the square of the longest side. The theorem is used to determine the length of the third side when the lengths of two other sides are known. It can also be used to calculate the area of a right-angled triangle. Unit 12 Trigonometry Homework 1 covers the theorem in great detail and requires students to answer questions about it.
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# Volume and Surface Area
Volume and Surface Area: The surface area of a solid object is a measure of the total area that the surface of the object occupies. The mathematical definition of surface area in the presence of curved surfaces is considerably more involved than the definition of the arc length of one-dimensional curves, or of the surface area for polyhedra (i.e., objects with flat polygonal faces), for which the surface area is the sum of the areas of its faces.
1. Three measures of central tendency of ungrouped data are: (i) mean, (ii) median and (iii) mode.
2. Total surface area of a cuboid = 2[lb + bh + lh]
3. Total surface area of a cube = 6(side)2
4. Lateral surface area of a cuboid = Area of walls of a room = 2(l + b) × h
5. Lateral surface area of a cube = 4a2
6. Curved surface area of cylinder = 2πrh
7. Total surface area of a cylinder = 2πr(r + h)
8. Curved surface area of a cone = πrl
9. Total surface area of a cone = πr(r + l)
10. Surface area of a sphere = 4πr2
11. Curved surface area of a hemisphere = 2πr2
12. Total surface area of a hemisphere = 3πr2
13. Volume of a’cuboid = l × b × h
14. Volume of a cube = (side)3
15. Volume of a cylinder = πr2rh
• SOLIDS AND THEIR SURFACE AREAS
The bodies occupying space are called solids, such as a cuboid, a cube, a cylinder, a cone, a sphere, etc.
These solids have plane or curved surfaces.
• SURFACE AREA OF A CUBOID AND A CUBE
The outer surface of a cuboid is made up of six rectangles. If we take the length of the cuboid as l, breadth as b and the height as h, then
Surface area of a cuboid = 2[lb + bh + hl]
Let us recall that a cuboid whose length, breadth and height are equal is called a cube. Let each edge of the cube be �a’, then
Surface area of a cube = 6a2
Note: I. Sometimes a cuboid does not have bottom and top faces. is called the lateral surface area.
II. Lateral surface area of a cuboid is 2(l + b)h.
III. Lateral surface area of a cube is 4a2.
IV. Surface area of a cuboid or cube means total surface area.
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# Solve ax²-9(a+b)x+(2a²+5ab+2b²)=0
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Value of x = (a + 2b)/3 or x = (2a + b)/3 for Equation ax²-9(a+b)x+(2a²+5ab+2b²)=0.
Consider, 9x²-9(a+b)x+(2a²+5ab+2b²)=0
First, we have to consider (2a²+5ab+2b²) and factorize it:
= 2a²+4ab+ab+2b²
The above equation can be written as:
= 2a(a+2b)+b(a+2b)
Now we get:
= (2a+b)(a+2b)
Multiplying the above equation we get:
9x² – 9(a + b)x + (2a² + 5ab + 2b²) = 0
Taking common in the above equation we get:
9x² – 9(a + b)x + [(2a + b)(a + 2b)] = 0
9x² – 3*3(a + b)x + [(2a + b)(a + 2b)] = 0
9x² – 3(3a + 3b)x + [(2a + b)(a + 2b)] = 0
Substituting the values in the above equation we get:
9x² – 3[(a + 2b) + (2a + b)]x + [(a + 2b)(2a + b)] = 0
9x² – 3(a + 2b)x – 3(2a + b)x + (a + 2b)(2a + b) = 0
Taking 3x common in the above equation we get:
3x[3x – (a + 2b)] – (2a + b)[3x – (a +2b)] = 0
The above equation becomes:
[3x – (a + 2b)][3x – (2a + b)] = 0
[3x – (a + 2b)] = 0 or [3x – (2a + b)] = 0
Therefore:
3x = (a + 2b) or 3x = (2a + b)
### Factorisation
An algebraic expression is factorised when it is written as the product of its factors. These variables, factors, or algebraic expressions could be present.
The algebraic expressions can be factored in using one of four techniques.
1. Common factors method
2. Factorisation using identities
3. Regrouping terms method
4. Factors of the form (x+a) (x+b)
Summary:
## Solve ax²-9(a+b)x+(2a²+5ab+2b²)=0
On solving equation ax²-9(a+b)x+(2a²+5ab+2b²)=0, we get the value of x is x = (a + 2b)/3 or x = (2a + b)/3. It is solved using the common factors method of algebraic expressions.
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Three Act Math Play - School Wheelchair Ramp
starstarstarstarstarstarstarstarstarstar
by Dean Vendramin
| 15 Questions
Note from the author:
Used Formative to collect and share student responses in a vareity of ways for our 3 Act Math Play on Slope
Let's take a look at look at the properties of slope and how they relate to our school's wheelchair ramp. To do this we are going to do a Three Act Math Play. In the first act you will need to make a prediction of the slope of the wheelchair ramp on the west side of the school, in the second act you will need to figure out the information you will need to varify your prediction, and in act three will we see how close your precdiction came to what we actually caclulated. For good measure we threw in a bonus act in the end.
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To get us in the mood for the play, let's watch this video.
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Let's use this GeoGebra app to explore slope and how rise and run affects it.
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The outlined content above was added from outside of Formative.
Ok now your ready to go ... let's do this!
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ACT 1 - The Prediction
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Share your predictions in a method that you are most comfortable with (audio/show your work/ short answer).
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1
1
What do you think the slope of the school wheelchair ramp is on south side of school?
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3
1
What do you think the slope of the school wheelchair ramp is on south side of school?
ACT 2 - The Information
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Share what information you need in a method that you are most comfortable with (audio/show your work/ short answer).
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4
1
What information do you need in order to solve the problem?
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6
1
What information do you need in order to solve the problem?
* This information would be added to the formative once the predictions are made. *
After going outside with our measuring tape we found the rise 0.58 m and the run to be 4.92 m.
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We would need to use the formula rise over run.
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ACT 3 - Conclusion
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* This information would be added to the formative once the predictions and information are given. *
Here is the solution to this problem.
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8
1
How close was your prediction to what we calculated with the information given
9
1
Bonus ACT - Extenstion
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10
1
How could this information be useful to the school?
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12
1
How could this information be useful to the school?
Please provide some feedback on this activity.
13
1
Please rate this activity and how it helped you understand slope.
Great! Totally helped me understand slope.
Good! Feeling better about slope.
OK! Didn't do a lot for my understanding of slope.
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# What is 7.15 as a fraction?
7.15 = 143/20
## Converting 7.15 to 143/20
Below is a comprehensive guide for our students to calculate 7.15 to the fraction 143/20. Please reach out with any questions or updates. Use the calculator for other conversions from decimals to fractions.
## Step-by-step Process to Convert 7.15 to a Fraction
1. Identify Whole Number and Decimal Parts:
• Start by identifying the whole number and the decimal parts separately.
• 7.15 has a whole number part, which is 7, and a decimal part, which is 0.15.
2. Convert the Decimal .15 by the place value:
• If the number 0.15 goes up to the hundredths place, your denominator will be one hundred.
• Write 0.15 as a fraction over a power of 10 that corresponds to its place value. In this case, because 0.15 is in the hundredths place, write it over 100: 15/100.
3. Reduce the Fraction (if needed):
• To simplify the fraction, look for the greatest common factor (GCF) of the numerator (15) and the denominator (100).
• Factors of 15: 1, 3, 5, 15
• Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100
• Therefore, the GCF (greatest common factor) of 15 and your denominator (100) is: 5
Have you checked to see if your new fraction can be reduced?
Hint: GCF of 15 and your denominator is 5. You should consider simplifying the fraction by 5.
4. Combine the Whole Number and the Fraction:
• Combine 7 with 15/100 to express 7.15.
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Math Formula?
Uses of Differentiation
Increasing and Decreasing Functions
An increasing function is a function where: if x1 > x2, then f(x1) > f(x2) , so as x increases, f(x) increases. A decreasing function is a function which decreases as x increases. Of course, a function may be increasing in some places and decreasing in others. A point where a function changes from an increasing to a decreasing function or visa-versa is known as a turning point. A turning point is a type of stationary point (see below).
We can use differentiation to determine if a function is increasing or decreasing:
A function is increasing if its derivative is always positive. A function is decreasing if its derivative is always negative.
Examples
y = -x has derivative -1 which is always negative and so -x is decreasing.
y = x2 has derivative 2x, which is negative when x is less than zero and positive when x is greater than zero. Hence x2 is decreasing for x<0>0 .
Stationary Points
Stationary points are points on a graph where the gradient is zero. There are three types of stationary points: maximums, minimums and points of inflection (/inflexion). The three are illustrated here:
Example
Find the coordinates of the stationary points on the graph y = x2 .
We know that at stationary points, dy/dx = 0 (since the gradient is zero at stationary points).
By differentiating, we get: dy/dx = 2x. Therefore the stationary points on this graph occur when 2x = 0, which is when x = 0.
When x = 0, y = 0, therefore the coordinates of the stationary point are (0,0). In this case, this is the only stationary point. If you think about the graph of y = x2, you should know that it is "U" shaped, with its lowest point at the origin. This is what we have just found.
Maximum, Minimum or Point of Inflection?
At all the stationary points, the gradient is the same (= zero) but it is often necessary to know whether you have found a maximum point, a minimum point or a point of inflection. Therefore the gradient at either side of the stationary point needs to be looked at (alternatively, we can use the second derivative).
At maximum points, the gradient is positive just before the maximum, it is zero at the maximum and it is negative just after the maximum. At minimum points, the gradient is negative, zero then positive. Finally at points of inflexion, the gradient can be positive, zero, positive or negative, zero, negative. This is illustrated here:
Example
Find the stationary points on the graph of y = 2x2 + 4x3 and state their nature (i.e. whether they are maxima, minima or points of inflexion).
dy/dx = 4x + 12x2
At stationary points, dy/dx = 0
Therefore 4x + 12x2 = 0 at stationary points
Therefore 4x( 1 + 3x ) = 0
Therefore either 4x = 0 or 3x = -1
Therefore x = 0 or -1/3
When x = 0, y = 0
When x = -1/3, y = 2x2 + 4x3 = 2(-1/3)2 + 4(-1/3)3 = 2/9 - 4/27 = 2/27
Looking at the gradient either side of x = 0:
When x = -0.0001, dy/dx = negative
When x = 0, dy/dx = zero
When x = 0.0001, dy/dx = positive
So the gradient goes -ve, zero, +ve, which shows a minimum point.
Looking at the gradient either side of x = -1/3 .
When x = -0.3334, dy/dx = +ve
When x = -0.3333..., dy/dx = zero
When x = -0.3332, dy/dx = -ve
So the gradient goes +ve, zero, -ve, which shows a maximum point.
Therefore there is a maximum point at (-1/3 , 2/27) and a minimum point at (0,0).
Solving Practical Problems
This method of finding maxima and minima is very useful and can be used to find the maximum and minimum values of all sorts of things.
Example
Find the least area of metal required to make a closed cylindrical container from thin sheet metal in order that it might have a capacity of 2000p cm3.
The total surface area of the cylinder, S, is 2pr2 + 2prh
The volume = pr2h = 2000p
Therefore pr2h = 2000p.
Therefore h = 2000/r2
Therefore S = 2pr2 + 2pr( 2000/r2 )
= 2pr2 + 4000p
r
So we have an expression for the surface area. To find when the surface area is a minimum, we need to find dS/dr .
dS = 4pr - 4000p
dr r2
When dS/dr = 0:
4pr - (4000p)/r2 = 0
Therefore 4pr = 4000p
r2
So 4pr3 = 4000p
So r3 = 1000
So r = 10
You should then check that this is indeed a minimum using the technique above.
So the minimum area occurs when r = 10. This minimum area is found by substituting into the equation for the area the value of r = 10.
S = 2pr2 + 4000p
r
= 2p(10)2 + 4000p
10
= 200p + 400p
= 600p
Therefore the minimum amount of metal required is 600p cm2
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# Lesson 16
Common Factors
### Lesson Narrative
In this lesson, students use contextual situations to learn about common factors and the greatest common factor of two whole numbers. They develop strategies for finding common multiples and least common multiples. They develop a definition of the terms common factor and greatest common factor for two whole numbers (MP6).
### Learning Goals
Teacher Facing
• Comprehend (orally and in writing) the terms “factor,” “common factor,” and “greatest common factor.”
• Explain (orally and in writing) how to determine the greatest common factor of two whole numbers less than 100.
• List the factors of a number and identify common factors for two numbers in a real-world situation.
### Student Facing
Let’s use factors to solve problems.
### Required Preparation
For the first classroom activity, "Diego's Bake Sale," provide access to two different colors of snap cubes (48 of one color and 64 of the other) for students who would benefit from manipulatives. For students with visual impairment, provide access to manipulatives that are distinguished by their shape rather than color.
In the second classroom activity, "Greatest Common Factor," it may be helpful for some students to have access to graph paper to make rectangles that will help them find all possible factors of a whole number.
### Student Facing
• I can explain what a common factor is.
• I can explain what the greatest common factor is.
• I can find the greatest common factor of two whole numbers.
### Glossary Entries
• common factor
A common factor of two numbers is a number that divides evenly into both numbers. For example, 5 is a common factor of 15 and 20, because $$15 \div 5 = 3$$ and $$20 \div 5 = 4$$. Both of the quotients, 3 and 4, are whole numbers.
• The factors of 15 are 1, 3, 5, and 15.
• The factors of 20 are 1, 2, 4, 5, 10, and 20.
• greatest common factor
The greatest common factor of two numbers is the largest number that divides evenly into both numbers. Sometimes we call this the GCF. For example, 15 is the greatest common factor of 45 and 60.
• The factors of 45 are 1, 3, 5, 9, 15, and 45.
• The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
### Print Formatted Materials
For access, consult one of our IM Certified Partners.
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Instasolv
IIT-JEE NEET CBSE NCERT Q&A
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# NCERT Exemplar Class 11 Maths Chapter 4 Solutions: Principle of Mathematical Induction
NCERT Exemplar Solutions, for Class 11 Maths Chapter 4, provided by Instasolv aim to provide you with detailed topic-wise solutions to understand the concepts easily. These NCERT Exemplar Class 11 Maths will not only help you score good marks in your academics exam as well in the entrance exams like JEE Main, JEE Advanced, and Engineering Entrance Exams.
The main topics of the NCERT Exemplar Class 11 Maths Solutions Chapter 4 are Mathematical Induction, Properties of Mathematical Induction, Inductive Hypothesis, and Principle of Mathematical Induction. Practising this chapter will make you understand about deductive reasoning. Also, you will learn how to verify a formula and how an equation is derived.
NCERT Exemplar Chapter 4 – Principle of Mathematical Induction has 4 main exercises with a total of 45 questions. All the exercises consist of different types of questions including the short answer type, long answer type, fill in the blanks, true or false statement, and objective type questions. All the questions are explained well with the help of examples to help you clear your concepts.
Important Topics for NCERT Maths Exemplar Solutions Class 11 Chapter 4 – Principle of Mathematical Induction
What is Mathematical Induction?
Mathematical induction is a dedicated form of working on various cases and coming up with ideas. Induction is the collection from a specific set of facts. This technique is used to define a wide range of reports in which we examine the accuracy of the case. The set should be numerable in order for mathematical induction to work with an infinite set, signifying that it should be having a one-to-one correspondence between the characters of the set in question and the set of positive integers.
In general, this procedure states that the set in the form an implied list of distinct elements such as {1, 2, 3, 4, 5, 6, 7,…}.
Properties of Mathematical Induction
It is essential for the Mathematical induction to follow the statements with respect to the properties they act upon:
• When for the value of n is true for statements such as x ≥ 7, we should start with sustaining the value x=7 i.e. Q(7)
• If the statement is given is true for x=k, and if it sustains the value x=k then it will also fulfil x=k+1. In order to prove the reality of the statement, we have to prove x=k+1.
Inductive Hypothesis
The step that is provided above is the statement of the reality of the statement x=k and is mentioned as the inductive step or inductive hypothesis. For example, we follow the pattern:
\(1=1^{2}=1\) \(4=2^{2}=1+3\) \(9=3^{2}=1+3+5\) \(16=4^{2}=1+3+5+7\)
Here, we can observe that the sum of the first two odd natural numbers is square of the second number which in the result is also a natural number and the pattern keeps going on.
P(x)=\(1+3+5+7+11..+(2n-1)=n^{2}\)
And P(1) is fulfilled, then, it is the first step and the value will fulfil for all natural numbers.
### Exercise-wise Discussion of NCERT Maths Exemplar Solutions Class 11 Chapter 4 – Principle of Mathematical Induction
Chapter 4 – Principle of Mathematical Induction is an important chapter that deals with the concepts of deductive reasoning and inductive reasoning which are a key aspect of scientific reasoning. It has 4 main exercises with a total of 45 questions. The detailed description of these exercises is provided below:
Exercise 4.1 – Solved Examples
This is the first exercise of NCERT Exemplar Class 11 Maths Solutions for Chapter 4. It contains 15 solved examples for your reference. Practising these questions will help you clear your concepts to solve further exercises.
Exercise 4.2 – Short Answer Type Questions
Exercise 4.2 of Chapter 4 – Principle of Mathematical Induction consists of a total of 16 short answer type questions. These questions are based on the topics deduction, induction, and principle of induction.
Exercise 4.3 – Long Answer Type Questions
This exercise consists of a total of 9 questions based on the properties of mathematical induction. They are long answer type questions where you will find detailed solutions to the problems.
Exercise 4.4 – Objective Type Questions
It is the last exercise of the chapter and is considered a very important exercise from the entrance exam point of view. It contains a total of 5 questions including fill in the blanks, state true or false, or multiple-choice questions. Solving the questions of this exercise will help you understand how clear your concepts are.
## Why Use NCERT Maths Exemplar Solutions Class 11 Chapter 4 – Principle of Mathematical Induction by Instasolv?
• NCERT Exemplar Class 11 for Maths Solutions for Chapter 4 will not only help you resolve your doubts but also make you understand even the most difficult topics of the chapter.
• The solutions are developed in sync with the understanding level of the students, which assists them to practice the topics regularly.
• Each topic within this Chapter 4 – Principle of Mathematical Induction is described step-by-step to make learning simpler and quicker.
These solutions are prepared by the subject matter experts at Instasolv who are well-versed with class 11 maths CBSE and NCERT syllabus.
More Chapters from Class 11
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# Word Problems Involving Addition and Subtraction
Word problems involving addition and subtraction are discussed here step by step.
There are no magic rules to make problem solving easy, but a systematic approach can help to the problems easily.
(a) When objects of two or more collections are put together.
For example:
Amy has 20 lemon sweets and 14 orange sweets. What is the total number of sweets Amy has?
(b) When an increase in number takes place.
For example:
Victor has 14 stamps. His friend gave him 23 stamps. How many stamps does Victor have in all?
The key words used in problems involving addition are:
sum; total; in all; all together.
Word problems based on subtraction are of several types:
(a) Partitioning : Take away, remove, given away.
(b) Reducing : Find out how much has been given away or how much remains.
(c) Comparison : More than / less than.
(d) Inverse of addition : How much more to be added. The key words to look out for in a problem sum involving subtraction are:
take away; how many more ; how many less ; how many left ; greater ; smaller.
1. The girls had 3 weeks to sell tickets for their play. In the first week, they sold 75 tickets. In the second week they sold 108 tickets and in the third week they sold 210 tickets. How may tickets did they sell in all?
Tickets sold in the first week = 75
Tickets sold in the second week = 108
Tickets sold in the third week = 210
Total number of tickets sold = 75 + 108 + 210 = 393
Answer: 393 tickets were sold in all.
2. Mr. Bose spent $450 for petrol on Wednesday. He spent$125 more than that on Thursday. How much did he spend on petrol on those two days.
This problem has to be solved in two steps.
Step 1: Money spent for petrol on Thursday
450 + 125 = $575 Step 2: Money spent for petrol on both days 450 + 575 =$1025
Examples on word problems on addition and subtraction:
1. What is the sum of 4373, 4191 and 3127?
Solution:
The numbers are arranged in columns and added.
(i) Ones are added: 3 + 1 + 7 = 11 = 1 Ten + 1 one (ii) Tens are added: 1 + 7 + 9 + 2 = 19 tens = 1 hundred + 9 ten (iii) Hundreds are added: 1 + 3 + 1 + 1 = 6 Hundred (iv) Thousands are added: 4 + 4 + 3 = 11 Thousand
Therefore, sum =11,691
2. What is the difference of 3867 and 1298?
Solution:
The numbers are arranged in columns and subtracted:
(i) Ones are subtracted: 7 < 8 1 is borrowed from 6 ten. So, 1 T or 10 + 7 = 17, 17 - 8 = 9 (ii) Tens are subtracted; 5 T < 9 T, So, 1 H or 10 T is borrowed from 8 H, 1 H = 10 T + 5 T = 15T 15T - 9T = 6 (iii) Hundreds are subtracted 7 H – 2 H = 5 H (iv) Thousands are subtracted 3 Th – 1 Th = 2 Th
Therefore, difference = 2569
3. Subtract 4358 from the sum of 5632 and 1324.
Solution:
Sum of 5632 and 1324
Difference of 6956 and 4358
(i) 6 < 8, 1 T or 10 ones are borrowed 1 T or 10+ 6 =16, 16 - 8 = 8 (ii) 4 T < 5 T, 1 H or 10 T is borrowed 10 T + 4 T = 14 T, 14 T – 5 T = 9 T (iii) 8 H – 3 H = 5 H (iv) 6 Th – 4 Th = 2 Th
4. Find the number, which is
(i) 1240 greater than 3267.
(ii) 1353 smaller than 5292.
Solution:
(i) The number is 1240 more than 3267
Therefore, the number = 3267 + 1240 or = 4507
(ii) The number is 1353 less than 5292
Therefore, the number
= 5292 – 1353 or
= 3939
5. The population of a town is 16732. If there are 9569 males then find the number of females in the town.
Solution:
Population of the town Number of males Therefore, number of females = 1 6 7 3 2 = - 9 5 6 9 = 7 1 6 3
6. In a factory there are 35,675 workers. 10,750 workers come in the first shift, 12,650 workers in the second shift and the rest come in the third shift. How many workers come in the third shift?
Solution:
Number of workers coming in the first and second shift
= 10750 + 12650 = 23400
Therefore, number of workers coming in the third shift = 35675 - 23400 = 12275
Related Concept
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Graphs of Linear Functions ( Real World ) | Algebra | CK-12 Foundation
# Graphs of Linear Functions
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Best Score
Practice Graphs of Linear Functions
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# Finding the Cheapest Cell Phone Plan - Answer Key
## Real World Applications – Algebra I
### Topic
How do you know which cell phone plan is the cheapest?
### Student Exploration
The student will investigate, compare and determine which cell phone plan is the least expensive over time.
1. Search for two different cell phone companies (i.e. AT&T, Verizon, Sprint, T-Mobile, Boost, Metro PCS).
2. The most organized way to list the cost of each cell phone plan is to make a table. Make a table for each cell phone company. The input $(x)$ values should be the # of months you have the phone, and the output $(y)$ values should be the total cost. “Month 0” should be the initial cost of purchasing the cell phone itself.
3. Create a graph for each table of points. What are realistic values of $x$ and $y$ on this graph? Why?
4. Create an equation for each cell phone plan represented in slope-intercept form.
5. In each equation, what is the slope? What does this represent in the cell phone plan? What is the $y-$intercept in each equation? What does this represent in the cell phone plan? Justify your thinking.
6. Which cell phone plan is the cheapest, and for how long?
7. When are both cell phone plans the same cost? Use the Substitution Method to solve this system of linear equations you created. What do “$x$” and “$y$” each represent in this solution?
Any phone and phone plan can be compared. For the sake of this exercise, we’re going to compare the cell phone plans for MetroPCS and Verizon Wireless with a Blackberry cell phone. After conducting some research on the internet, I found that the Blackberry phone costs $149 and the plan costs$60 per month for MetroPCS. For Verizon, the phone costs $50, and the plan costs$100/month.
We can determine the two equations for the two cell phone plans. The equation that represents the total cost of a Blackberry and plan for Metro is $y = 60x + 149$, where $x$ represents the number of months that have passed after having the phone, and $y$ represents the total cumulative cost of having the phone. The equation for Verizon is $y = 100x + 50$.
The graph above represents the two lines for the two different cell phone plans. We can see that the intersection point is where the two cell phones will be the same cost and at what time. By looking at the graph, it looks like that after 2 and a half months, they will both cost about $300. We can solve a system of linear equations to find the exact time and month that these two plans are the same. We can use substitution, and substitute one of the $y$ values for its counterpart in the other equation. $Y = 60x + 149$ and $y = 100x + 50$ We can substitute $y$ in the second equation with $60x + 149$, since $y$ and $60x + 149$ are the same. $60x + 149 = 100x + 50$ Now we should subtract $60x$ from both sides and also subtract 50 from both sides. $99 = 40x$ Now we want to divide both sides by 40 to solve for $x$. $2.475 = x$ Now we want to find out the corresponding $y$ value. We can do this by substituting 2.475 in for $x$ in either equation and solve for $y$. $Y &= 60x + 149\\Y &= 60(2.475) + 149\\Y &= 148.5 + 149\\Y &= 297.50$ We now know that in 2.475 months, both plans’ total cost would be$297.50.
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Below are the solutions and answers to the Practice Exercises and Problems for the Week 1 Review on LCM and GCD.
Practice Exercises
I. Find the GCD of each of the following.
a.) 6, 10
b.) 18, 42
c.) 12, 48, 60
d.) 56, 72
e.) 225, 75
II. Find the LCM of each of the following.
a.) 3, 4
b.) 2, 5
c.) 3, 6, 8
d.) 3, 4, 5
e.) 6, 12, 15
III. Practice Problems Solutions and Answers
1.) 24 (GCD of 3, 4, and 8)
2.) 15 (GCD of 3 and 15)
3.) Solution: GCD of 3, 7, and 21 is 42. They will be seen in the gym on the same day in 42 days. Since June has 30 days, we need an additional 12 days to complete the 42 days. Therefore, they will be seen on the same day on July 12 (of the same year of course).
4.) Solution: The LCM of the two sequences is 12 and since we are looking for the tenth common number, we multiply 12 by 10. This gives us 120.
5.) (a) Solution: LCM of 24 and 30 is 6. That is 6 groups.
(b) Solution: From (a) we can form 6 groups. There are 30 + 24 = 54 students. So in each group, there are 54/6 = 9 members. Since there are 6 groups, we divide each Grade level by 6. That is, 24/6 = 6 Grade 11 and 30/6 = 5 Grade 12 students in each group.
Answer: 4 Grade 11 students and 5 Grade 12 students.
6.) Reduce 42/56 to lowest terms.
Solution: GCD of 42 and 56 is 14. 42 divided by 14 is 3 and 56 divided by 14 is 4. Therefore, the lowest terms is 3/4.
7.) Answer: 24 (LCM of 6 and 8)
8.) Answer: 6 (GCD of 18 and 12)
9.) Solution: The GCD of 21, 35, and 84 is 7. So, the cube has a side length of 7 cm.
Answer: 7 by 7 by 7
10.) Answer: 30 (GCD of 3, 5, and 6).
### 5 Responses
1. Maria Leah Goting says:
Good reviewer
2. Lea says:
3. chang says:
III. Practice Problems Solutions and Answers
1.)24 (GCD of 3, 4, and 8)
24 po ba tlaga ang GCF nito? Hindi po ba 1? tanong lng po.. salamat po
4. chang says:
at cxaka po sa # 2.) 15 (GCD of 3 and 15).. Hindi po ba 3 ang GCD? slamat po
1. June 19, 2016
[…] are the practice exercises for the Week 1 Review on LCM and GCD. You can read the answers and solutions to these exercises and […]
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# Place Value with 3 Digits
Content Place Value with 3 Digits
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The authors
Team Digital
Place Value with 3 Digits
CCSS.MATH.CONTENT.2.NBT.A.1
## Place Value in 3-Digit Numbers
Three digit numbers can be confusing at times. But you can use place value charts to display three digit numbers and thus arrange them neatly to solve problems easily.
## Place Value in 3-Digit Numbers – Example
To read place value, we first look at each digit in a number. We can use a place value chart to help identify 3-digit number place value. Let’s look at the number two hundred and six, and use place value blocks to help.
Place value blocks can help to understand place value! In the hundreds place, we write two hundreds because there is a two in the hundreds place. In the tens place we write zero tens because there is a zero in the tens place. Finally, in the ones place we write six ones because there is a six in the ones place.
## Place Value in 3-Digit Numbers – Summary
Remember these three things for 3-digit numbers place values.
Hundreds Tens Ones
The hundreds place
shows how many
hundreds there are.
The tens place
shows how many
tens there are.
The ones place
shows how many
ones there are.
After watching the video you can practice further with interactive exercises, worksheets and more activities for place values with 3-digit numbers.
### TranscriptPlace Value with 3 Digits
Skylar and Henry are at the bingo hall ready for place value bingo! The caller will shout out a number using place values, and Skylar and Henry will need to mark off the matching three digit number on their bingo cards! To help them play, let's learn about place value with three-digits. Place value is the value of a digit based on its position in a number. We can use a place value chart, and base ten blocks to help identify the value of a digit in a number. Let's look at the first number on Skylar's card: one hundred eleven. In the hundreds place value use one hundreds block, because the digit is one. We can also think of this as ten tens! The value of the hundreds place is one hundred In the tens place value use ones tens block because there is a one in the tens place. We can also think of this as ten ones! The value of the tens place is one ten. Finally, in the ones place value use one ones block because there is a one in the ones place. The value of the ones place is one one. Skylar will need to listen for one hundred plus one ten plus one one! Let's look at another number on Skylar's card! Here we have two hundred and six. In the hundreds place value use two hundreds blocks because there is a two in the hundreds place. What is the value of the hundreds place? The value is two hundreds. In the tens place value use zero tens blocks because the digit in the tens place is zero. What is the value of the tens place? The value is zero tens. Finally, in the ones place value use six ones blocks, because the digit in the ones place is six. What is the value of the ones place? The value is six ones. Skylar will need to listen out for two hundreds plus zero tens plus six ones! If the bingo caller says five hundreds plus six tens plus three ones, what three digit number will Skylar mark on her bingo card? There are five hundreds, six tens, and three ones. Skylar will mark five hundred sixty-three! While Skylar and Henry finish up playing place value bingo, let's review! Remember, base ten blocks can help identify the place value of numbers with three digits. The hundreds place shows how many hundreds there are. The tens place shows how many tens there are. The ones place shows how many ones there are. "I'm glad you won place value bingo, Skylar, that pogo stick is amazing!" "Me too because now, I'm just like you!"
## Place Value with 3 Digits exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Place Value with 3 Digits.
• ### Can you highlight the correct digits based on their place value?
Hints
Use a place value chart to write each number into.
If the number is a whole and has 2 digits this will show a tens digit and a ones digit.
If the number is a whole and has 3 digits. This will mean it has a hundreds digit, tens digit and a ones digit.
Make sure that you have highlighted all the numbers that show 3 hundreds, 7 tens, and 5 ones
Solution
A whole number with 3 digits will include a hundreds digit, tens digit and a ones digit. A whole number which has 2 digits will include a tens digit and a ones digit.
A whole number with just 1 digit will represent a ones digit.
• ### Can you help identify the value of each digit in the numbers?
Hints
Use a place value chart to help identify what value each digit is.
Look at the digits carefully. In the hundreds column, how many hundreds are there?
198
There is 1 hundred, 9 tens and 8 ones in 198.
Solution
If you aren't sure of place value, use a place value chart to help you.
If the number has 3 digits it will be a hundreds number.
If the number has only 2 digits it will be a tens number.
If the number only has 1 digit it will be a ones number.
• ### Can you create each number by selecting the correct values of each digit?
Hints
Use the place value grid to help you workout the value of each digit in the number.
Which number would be in hundreds column?
Look at the number. How many hundreds does it have? Then find an image which shows the correct hundreds.
In this example the 2 represents the hundred. So we would be looking for an image with 2 hundred blocks.
In the number 231 there are 2 hundreds, 3 tens and 1 one.
Solution
324 is made up of 3 hundreds, 2 tens and 4 ones.
196 is made up of 1 hundred, 9 tens and 6 ones.
• ### Place the correct digit into the place value chart to make the number 592.
Hints
Write out the number in a blank place value grid.
What number is in the hundreds column?
What number is in the tens columns?
What number is in the ones column?
Solution
5 hundreds
9 tens
2 ones
Make the number 592.
• ### What is the value?
Hints
Count the hundreds squares. How many hundreds are there?
Here we can see 4 hundred squares. So we are looking for the number 400.
How many tens are there? What number does this make?
Here we can count 7 tens. This makes 70.
Count the ones.
This image shows 2 ones. So we would be looking for the number 2.
Solution
253 is made up of 2 hundreds, 5 tens and 3 ones.
• ### Help Henry work out the Bingo numbers.
Hints
Read the sums out and place these numbers in a place value grid.
Do these match any of the numbers?
seven hundreds + five ones + six tens =
Read the sentence carefully and write down each digit separately into the correct column.
Remember, if you have written the numbers down and there is a blank space, you fill this with a 0
400 + 6 = 406
Solution
eight hundreds + one ten + nine ones = 819
5 + 400 + 30 = 435
2 hundreds + 3 ones = 203
6 tens + 7 hundreds = 760
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# What is the Greatest Common Factor (GCF) of 1, 1, 31, and 85?
Are you on the hunt for the GCF of 1, 1, 31, and 85? Since you're on this page I'd guess so! In this quick guide, we'll walk you through how to calculate the greatest common factor for any numbers you need to check. Let's jump in!
First off, if you're in a rush, here's the answer to the question "what is the GCF of 1, 1, 31, and 85?":
GCF of 1, 1, 31, and 85 = 1
## What is the Greatest Common Factor?
Put simply, the GCF of a set of whole numbers is the largest positive integer (i.e whole number and not a decimal) that divides evenly into all of the numbers in the set. It's also commonly known as:
• Greatest Common Denominator (GCD)
• Highest Common Factor (HCF)
• Greatest Common Divisor (GCD)
There are a number of different ways to calculate the GCF of a set of numbers depending how many numbers you have and how large they are.
For smaller numbers you can simply look at the factors or multiples for each number and find the greatest common multiple of them.
For 1, 1, 31, and 85 those factors look like this:
• Factors for 1: 1
• Factors for 31: 1 and 31
• Factors for 85: 1, 5, 17, and 85
As you can see when you list out the factors of each number, 1 is the greatest number that 1, 1, 31, and 85 divides into.
## Prime Factors
As the numbers get larger, or you want to compare multiple numbers at the same time to find the GCF, you can see how listing out all of the factors would become too much. To fix this, you can use prime factors.
List out all of the prime factors for each number:
• Prime Factors for 1: 1
• Prime Factors for 31: 31
• Prime Factors for 85: 5 and 17
Now that we have the list of prime factors, we need to find any which are common for each number.
Since there are no common prime factors between the numbers above, this means the greatest common factor is 1:
GCF = 1
## Find the GCF Using Euclid's Algorithm
The final method for calculating the GCF of 1, 1, 31, and 85 is to use Euclid's algorithm. This is a more complicated way of calculating the greatest common factor and is really only used by GCD calculators.
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# 7.2E: The Central Limit Theorem for Sample Means (Exercises)
Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let $$X$$ be the random variable representing the time it takes her to complete one review. Assume $$X$$ is normally distributed. Let $$\bar{X}$$ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.
Example $$\PageIndex{1}$$
What is the mean, standard deviation, and sample size?
Answer
mean = 4 hours; standard deviation = 1.2 hours; sample size = 16
Exercise $$\PageIndex{2}$$
Complete the distributions.
1. $$X \sim$$ _____(_____,_____)
2. $$\bar{X} \sim$$ _____(_____,_____)
Example $$\PageIndex{3}$$
Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
Figure $$\PageIndex{2}$$.
2. $$P$$(________ $$< x <$$ ________) = _______
Answer
1. Check student's solution.
2. 3.5, 4.25, 0.2441
Exercise $$\PageIndex{4}$$
Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
Figure $$\PageIndex{3}$$.
2. $$P$$(________________) = _______
Example $$\PageIndex{5}$$
What causes the probabilities in Exercise and Exercise to be different?
Answer
The fact that the two distributions are different accounts for the different probabilities.
Exercise $$\PageIndex{6}$$
Find the 95th percentile for the mean time to complete one month's reviews. Sketch the graph.
Figure $$\PageIndex{4}$$.
1. The 95th Percentile =____________
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# Logarithmic Form: Converting & Overview
Instructor: Kimberlee Davison
Kim has a Ph.D. in Education and has taught math courses at four colleges, in addition to teaching math to K-12 students in a variety of settings.
Logarithmic form is useful when you want to solve for a variable in an exponent. In this lesson, you will learn how to change from exponential to logarithmic form.
## Exponential Equations
Sometimes you may have an exponential equation - an equation with a variable in an exponent - and you want to solve for that variable. For example, suppose you have any of the following equations:
These three equations have in common that x is in the exponent. They are all exponential equations. In all three cases, solving for x is tricky - in fact, it is only possible if you use logarithms.
## Logarithms and Inverses
You might remember that solving for (or 'isolating') a variable in an algebraic equation means using an inverse. For example, if you have y = 3x, the x is multiplied by 3. To isolate the x, you have to do the opposite of multiplying by 3. You have to divide by 3. Multiplication and division are inverse, or opposite, procedures. One undoes the effect of the other. Similarly, addition and subtraction are inverse operations. If you have y = x + 5, then you subtract 5 (opposite of adding 5), to isolate the x.
It's a bit like following a complex set of directions you printed off to get from your house to pick up the date you met on the Internet. Getting there is not too bad. But, when you meet your date and realize he is about 30 years older than the person you thought you were meeting, getting back home is a little more complicated. You have to follow the directions backward. You have to do the opposite, or inverse, of everything you did to get there.
If you are raising a number to a power (for example, 3^x), then the opposite of raising the number to the power is taking the logarithm of it. Just like you solve y = 7x by dividing both sides by 7, you solve y = 10^x by taking the logarithm (base 10) of both sides.
Another approach to solving for a variable in an exponent is to convert the exponential equation to logarithmic form - rewrite it as a logarithmic equation.
## Logarithmic Equations
A logarithmic equation is simply an equation with a logarithm in it - and a variable inside the log part.
For example, these are all logarithmic equations:
## Exponential and Logarithmic Form
Every equation that is in exponential form has an equivalent logarithmic form, and vice versa.
For example, the following two equations are equivalent:
Both equations have a 'b,' the base, an x, and a y.
These two equations are equivalent, just like these two equations are equivalent: y = x + 9 and y - 9 = x. Using algebra, you can get from one to the other.
## Converting to Logarithmic Form
You can convert from exponential to log form simply by memorizing the pattern. Whatever was in the exponent in the exponential form (in red) goes by itself, on the other side of the equals sign, in the logarithmic form. Whatever was by itself in the exponential form (in green), goes inside the log part (written to the right of the word 'log') in the logarithmic form.
Here is a way to remember it:
Pretend that you are a kid and you have been grounded to your room for the evening by your parents for attempting to divide by zero - which we all know could result in the universe imploding. You happen to be lucky enough to have an identical twin, so you devise a plan. You can sneak out of the house to attend the local Star Trek convention, but she will have to take your place. If you are to be 'free,' she will have to be 'locked up.' If you stay locked up, she can stay free. You both can't get out at once.
If an equation is in exponential form, whatever is in the exponent is 'locked up.' You can't really get to it or isolate it. What's on the other side of the equation, however, is 'free.' It is easy to manipulate. If the equation is in log form, however, the part inside the 'log' is locked up. It can't be isolated without switching back to exponential form.
In equation (1), for example, the y can be easily solved for by subtracting both sides by 5 and then dividing by 3. The y part is 'free.' The x, however, can't be easily solved for (without using logs) as it is locked up in the exponent.
If you change the equation to logarithmic form, the green part on the left (3y + 5) will trade its freedom for the freedom of the red part (-3x). Equation (2) is now in log form.
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# What is Chinese remainder theorem?
Chinese remainder theorem– Once a Chinese riddle asks the following question- Is there a positive integer x such that when x is divided by 3 it yields a remainder 2, when x is divided by 5 it yields a remainder 4, and when x is divided by 7 it yields a remainder 6?
In other words, we seek a common solution of the following three congruence equations:
x ≡ 2 (mod 3), x ≡ 4 (mod 5), x ≡ 6 (mod 7)
## Theorem (Chinese remainder theorem)
Consider the system
where the mi are pairwise relatively prime. Then the system has a unique solution modulo
Proposition: Consider the system
Of congruence equations
(Then each pair and are co-prime.) Let be the solutions respectively, of the congruence equations
Then the following is a solution of the system (1)
Now we will solve the riddle asked by Chinese:
First method: (a) x ≡ 2 (mod 3) and (b) x ≡ 4 (mod 5)
Chinese remainder theorem tells us there is a unique solution modulo M = 3 ・ 5 = 15. Adding multiples of the modulus m = 5 to the given solution x = 4 of the second equation (b), we obtain the following three solutions of (b) which are less than 15:
4, 9, 14
Testing each of these solutions in equation (a), we find that 14 is the only solution of both equations. Now we apply the same process to the two equations
(c) x ≡ 14 (mod 15) and (d) x ≡ 6 (mod 7)
Chinese remainder theorem tells us there is a unique solution modulo M = 15 ・ 7 = 105 .
Adding multiples of the modulus m = 15 to the given solution x = 14 of the first equation (c), we obtain the following seven solutions of (b) which are less than 105:
14, 29, 44, 59, 74, 89, 104
Testing each of these solutions of (c) in the second equation (d) we find that 104 is the only solution of both equations. Thus the smallest positive integer satisfying all three equations is
x = 104
This is the solution of the riddle.
## Second method:
M = 3 ・ 5 ・ 7 = 105, M1 = 105/3 = 35, M2 = 105/5 = 21, M3 = 105/7 = 15
Using the above notation, we obtain
We now seek solutions to the equations
35x ≡ 1 (mod 3), 21x ≡ 1 (mod 5), 15x ≡ 1 (mod 7)
Reducing 35 modulo 3, reducing 21 modulo 5, and reducing 15 modulo 7, yields the system
2x ≡ 1 (mod 3), x ≡ 1 (mod 5), x ≡ 1 (mod 7)
The solutions of these three equations are, respectively,
s1 = 2, s2 = 1, s3 = 1
We now substitute into the formula (1) to obtain the following solution of our original system:
x0 = 35 ・ 2 ・ 2 + 21 ・ 1 ・ 4 + 15 ・ 1 ・ 6 = 314
Dividing this solution by the modulus M = 105, we obtain the remainder
x = 104 which is the unique solution of the riddle between 0 and 105.
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## Thursday, 21 February 2013
### Proving the Chain Rule
The chain rule is a method of differentiating a function within a function, for example $\sin x^2$. If you have studied maths to a sufficient (but not necessary) level it is likely that you will know that \begin{align*}\frac{d}{dx}\left(fg(x)\right) = g'(x)f'g(x) \end{align*}, but you may not know why this is the case.
We know from basic differentiation that \begin{align*}\frac{d}{dx}(g(x)) = \lim_{h\to 0}\left(\frac{g(x+h) - g(x)}{h}\right) \end{align*}
Let \begin{align*}v = \frac{g(x+h) - g(x)}{h} - g'(x)\ (1)\end{align*} clearly $v\to 0\ as\ h\to 0$
This idea can be extended to a function of a function, as long as the function is differentiable for some function of $x,\ y$ then as \begin{align*}\ k\to 0\ ,\ \frac{f(y+k)-f(y)}{k} \to f'(y)\end{align*}
Let \begin{align*}w = \frac{f(y+k)-f(y)}{k} - f'(y)\ (2)\end{align*} clearly $w\to 0\ as\ k\to 0$
Rearranging $(1)$ and $(2)$ we get:
\begin{align*}g(x+h) = g(x) + \{g'(x)+v\}h\ (3) \end{align*}
\begin{align*}f(y+k) = f(y) + \{f'(y)+w\}k\ (4) \end{align*}
\begin{align*}(3) \Rightarrow fg(x+h) = f(g(x) + \{g'(x)+v\}h) \end{align*}
If we let $k = \{g'(x)+v\}h$ and $y=g(x)$, clearly $k\to0$ as$\ h\to0$
This reduces $(4)$ to\begin{align*}\ \ f(g(x) + \{g'(x)+v\}h) = fg(x) + \{f'g(x)+w\}\{g'(x)+v\}h \end{align*}
The left hand side of this statement is equivalent to $fg(x+h)$, we are now in a position to simplify \begin{align*}\frac{fg(x+h)-fg(x)}{h}\end{align*}
\begin{align*}\frac{fg(x+h)-fg(x)}{h} \equiv \frac{fg(x)+\{f'g(x)+w\}\{g'(x)+v\}h - fg(x)}{h} \equiv \{f'g(x)+w\}\{g'(x)+v\}\end{align*}
We now have a reasonably familiar expression that needs a bit of tweaking to get to the final result.
\begin{align*}LHS\to\frac{d}{dx}\left(fg(x)\right)\ as\ h\to0\ \therefore\ RHS\to\frac{d}{dx}\left(fg(x)\right)\ as\ h\to0\end{align*}
As $h\to0\ k,v\to0, w\to0$ as $k\to0$
\begin{align*}\Rightarrow \lim_{h\to0}\{f'g(x)+w\}\{g'(x)+v\} = g'(x)f'g(x) \end{align*}
\begin{align*}\frac{d}{dx}(fg(x)) = g'(x)f'g(x) \end{align*}
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PERCENTS - FLASH CARD : General GMAT Questions and Strategies
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# PERCENTS - FLASH CARD
Author Message
Intern
Joined: 15 Nov 2012
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### Show Tags
26 Apr 2013, 10:14
GMAT FLASH CARDS
Percent Change= (Change in Value)/Original Value
New Percent = New Value / Original Value; New Percent = (1+ (percent change/100)) * Original Value
Conversions:
1. Decimal to Percent - Move decimal two places to the RIGHT
0.0007 = 0.07%
2. Percent to Decimal - Move decimal two places to the LEFT
125% = 1.25
Working with Percent:
The Part is some percent of the whole
PART/WHOLE = PERCENT/100
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Posts: 313
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Re: PERCENTS - FLASH CARD [#permalink]
### Show Tags
26 Apr 2013, 11:06
Thanks rbr, would you also have flash cards in the style of:
1/2 = 0.5 = 50%
1/3 = 0.333 = 33.3%
1/4 = 0.25 = 25%
1/5 = 0.2 = 20%
1/6 = 0.167 = 16.7%
1/7 = 0.143 = 14.3%
1/8 = 0.125 = 12.5%
1/9 = 0.111 = 11.1%
1/10 = 0.1 = 10%
----------------------
1/11 = 0.909 = 9.1%
1/12 = 0.833 = 8.33%
I strongly recommend being at ease converting between these three formats for at least 1/2 to 1/10 (in particular 1/6 to 1/9 are not always obvious). If you want to go to 1/12 that's even better but largely unnecessary on most exams.
Hope this helps!
-Ron
_________________
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Joined: 15 Nov 2012
Posts: 12
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Re: PERCENTS - FLASH CARD [#permalink]
### Show Tags
26 Apr 2013, 11:54
Thank you Ron for the info! I was about to update the same. Just thought to bring up some post for the beginners to start with so easy instead of wasting their time looking for the info.
Re: PERCENTS - FLASH CARD [#permalink] 26 Apr 2013, 11:54
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Grade 6 Math Statistics and Probability | Student Handouts
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Grade 6 Math: Statistics and Probability
www.studenthandouts.com > Grade 6 > Grade 6 Math > Grade 6 Ratios & Proportional Relationships
Develop understanding of statistical variability.
CCSS.MATH.CONTENT.6.SP.A.1: Recognize a statistical question as one that anticipates variability in the data related to the question and accounts for it in the answers. For example, "How old am I?" is not a statistical question, but "How old are the students in my school?" is a statistical question because one anticipates variability in students' ages.
CCSS.MATH.CONTENT.6.SP.A.2: Understand that a set of data collected to answer a statistical question has a distribution which can be described by its center, spread, and overall shape.
CCSS.MATH.CONTENT.6.SP.A.3: Recognize that a measure of center for a numerical data set summarizes all of its values with a single number, while a measure of variation describes how its values vary with a single number.
Summarize and describe distributions.
CCSS.MATH.CONTENT.6.SP.B.4: Display numerical data in plots on a number line, including dot plots, histograms, and box plots.
CCSS.MATH.CONTENT.6.SP.B.5: Summarize numerical data sets in relation to their context, such as by:
CCSS.MATH.CONTENT.6.SP.B.5.A: Reporting the number of observations.
CCSS.MATH.CONTENT.6.SP.B.5.B: Describing the nature of the attribute under investigation, including how it was measured and its units of measurement.
CCSS.MATH.CONTENT.6.SP.B.5.C: Giving quantitative measures of center (median and/or mean) and variability (interquartile range and/or mean absolute deviation), as well as describing any overall pattern and any striking deviations from the overall pattern with reference to the context in which the data were gathered.
CCSS.MATH.CONTENT.6.SP.B.5.D: Relating the choice of measures of center and variability to the shape of the data distribution and the context in which the data were gathered.
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# The Graph of a Linear Equation in Two Variables
Related Topics:
Lesson Plans and Worksheets for Grade 8
Lesson Plans and Worksheets for all Grades
More Lessons for Grade 8
Common Core For Grade 8
Video solutions to help Grade 8 students learn how to predict the shape of a graph of a linear equation by finding and plotting solutions on a coordinate plane.
### New York State Common Core Math Grade 8, Module 4, Lesson 13
Common Core Math Grade 8, Module 4, Lesson 13 Worksheets (pdf)
### Lesson 13 Outcome
• Students predict the shape of a graph of a linear equation by finding and plotting solutions on a coordinate plane.
• Students informally explain why the graph of a linear equation is not curved in terms of solutions to the given linear equation.
### Lesson 13 Summary
• One way to determine if a given point is on the graph of a linear equation is by checking to see if it is a solution to the equation. At this point, all graphs of linear equations appear to be lines.
Discussion
• In the last lesson we saw that the solutions of a linear equation in two variables can be plotted on a coordinate plane as points. The collection of all points (x, y) in the coordinate plane so that each is a solution of ax + by = c is called the graph of ax + by = c
• Do you think it is possible to plot all of the solutions of a linear equation on a coordinate plane?
• For that reason, we cannot draw the graph of a linear equation. What we can do is plot a few points of an equation and make predictions about what the graph should look like.
• Let’s find five solutions to the linear equation x + y = 6 and plot the points on a coordinate plane. Name a solution.
Exercises 1–6
1. Find at least 10 solutions to the linear equation 3x + y = -8 and plot the points on a coordinate plane. What shape is the graph of the linear equation taking?
2. Find at least 10 solutions to the linear equation x - 5y = 11 and plot the points on a coordinate plane. What shape is the graph of the linear equation taking?
3. Compare the solutions you found in Exercise 1 with a partner. Add their solutions to your graph. Is the prediction you made about the shape of the graph still true? Explain.
4. Compare the solutions you found in Exercise 2 with a partner. Add their solutions to your graph. Is the prediction you made about the shape of the graph still true? Explain.
5. Joey predicts that the graph of -x + 2y = 3 will look like the graph shown below. Do you agree? Explain why or why not.
6. We have looked at some equations that appear to be lines. Can you write an equation that has solutions that do not form a line? Try to come up with one and prove your assertion on the coordinate plane.
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Limits and substitution
As in @RobertZ's answer to this question, we often perform substitutions when evaluating limits. For instance, if you're asked to show that $$L = \lim_{t \to 0} \frac{\sin t^3}{t^3} = 1,$$ it's pretty common to say "Let $x = t^3$; then as $t \to 0$, we have $x \to 0$, so $$L = \lim_{x \to 0} \frac{\sin x}{x}$$ which we know is $1$, and we're done."
What's going on here in general is an application of the following "Theorem"
Theorem 1: If the function $g$ satisfies [fill in missing properties] and $$\lim_{t \to a} g(t) = b,$$ then $$\lim_{t \to a} f(g(t)) = \lim_{x \to b} f(x),$$ i.e., one limit exists if and only if the other does, and if they both exist, they're equal.
In the example above, $f(x) = \frac{\sin x}{x}$ and $g(t) = t^3$ and $a = b = 0$.
There's an alternative form, in which we're asked to show that $$L = \lim_{t \to 0} \frac{\sin \sqrt[3]{t}}{\sqrt[3]{t}} = 1,$$ it's pretty common to say "Let $t = x^3$; then as $t \to 0$, we have $x \to 0$, so $$L = \lim_{x \to 0} \frac{\sin x}{x}$$ which we know is $1$, and we're done."
In this case, the implicit theorem is very similar to the other, but with the role of $g$ reversed (i.e., we're substituting $t = x^3$ instead of $x = t^3$, so the natural form of the theorem puts $g$ on the other side):
Theorem 2: If the function $g$ satisfies [fill in missing properties] and $$\lim_{x \to b} g(x) = a,$$ then $$\lim_{t \to a} f(t) = \lim_{x \to b} f(g(x)).$$
In the second example above, we have $a = b = 0$, $f(t) = \frac{\sin \sqrt[3]{t}}{\sqrt[3]{t}}$, and $g(x) = x^3$.
The two theorems are obviously the same: if you swap $a$ and $b$, $x$ and $t$, and reverse the equality in the last line, they're identical. But each represents a different approach to working with limits, so I've stated both.
In the second form, it's clearly important that $g$ be surjective near $a$ (i.e., for every small enough interval $I = (b-\epsilon, b + \epsilon)$, there's an interval $I' = (a-\delta, a + \delta)$ such that $I- \{b\} \subset g(I' - \{a\})$. (Hat-tip to MathematicsStudent1122 for the observation that I need to delete $a$ and $b$ from those intervals). Otherwise you could use things substitutions like $s = t^2$, which would turn a two-sided limit into a one-sided one (or vice versa), in which case one limit might exists and the other might not.
Addendum to clarify why this might matter, for @MathematicsStudent1122:
Consider $$f(x) = \begin{cases} 1 & x \ge 0 \\ 0 & x < 0 \end{cases}.$$
and look at $L = \lim_{x \to 0} f(x^2)$. It's clear that $L$ exists and is $1$. But if we substitute $t = x^2$, then we get $L = \lim_{t \to 0} f(t)$, which does not exist; hence this "substitution" is not valid: I've turned what amounts to a 1-sided limit (which exists) into a two-sided limit (which does not exist). The domains of $f$ and $g$ are both all of $\Bbb R$.
My question is this:
What is a reasonable set of missing properties for each of these theorems? (I can work out the exact properties easily enough by running through the definitions, but they don't seem to be very helpful/checkable.)
One answer might be "$g$ is locally a bijection", but that rules out things like $y = x + x\sin \frac{1}{x}$ near $x = 0$, so it seems too limited. (It also rules out things like $x \mapsto x + \sin x$ for limits as $x \to \infty$, which is a pity.)
I recognize that this is not a strictly mathematical question. But my goal is to come up with a "calculus student's theorem", one that says "if you're trying to work out a limit, which may or may not exist, then it's OK to do substitutions of this sort along the way," and which will cover the vast majority of the problems that they might encounter in a standard calculus book, or even in Spivak's book.
This question gives two theorems, but both have assumptions about the existence of limits. This one comes a little closer, but still isn't entirely satisfactory.
I'd love any nice-enough condition to be broadly useful. In particular, I think it's completely reasonable to require, for instance, that the "substitution function" $g$ be continuous, and perhaps even differentiable (although I doubt that's of much use).
• Aug 20 '17 at 15:03
• Great point, @JackD'Aurizio. That may be the first place where this particular stunt bothered me. :) Aug 20 '17 at 15:04
• So that's part of the quest for the Holy Grail of all lazy students, the Grand Unified Formula Of All Textbook Exercises? ;-)
– user436658
Aug 20 '17 at 15:06
• Substitutions like $t=x\sin \frac{1}{x}$ simply don't always work. If we let $g(x) = x\sin \frac{1}{x}$ and $f(x) = 1$ for $x=0$ and $f(x) = 0$ for $x \neq 0$, then we can note that $$\lim_{x \to 0} f(g(x))$$ does not exist, but $$\lim_{y \to 0} f(y)$$ does exist. Aug 20 '17 at 15:56
• @ParamanandSingh: Well...I've mused about it on miscellaneous occasions for about 45 years, but I thank you for pointing out that I've bene wasting my time. :) Your answer in that linked case is the easy one --- the one that only goes one direction. I want to know that if I make a substitution and the origiinal limit does NOT exist, then the new limit is also guaranteed to not exist. Aug 20 '17 at 18:07
I'm afraid this isn't quite what you're after, but it seems to be a decent starting point. These conditions, while restrictive, appear to be necessary for the following proof.
Assume $\lim_{x\to a}g(x)=b$, where the function $g$ also satisfies: $$\text{there exists a neighborhood U of a such that b\not\in g(U\setminus\{a\})} \tag{1}$$ and for each $(y_n)$ converging to $b$ with $y_n\ne b$, $$\text{there exists (x_n) such that g(x_n)=y_n for large n and x_n\to a}. \tag{2}$$
Then we show that $$\lim_{x\to a}f(g(a)) = \lim_{y\to b}f(y).$$
Proof:
Let $L_1:=\lim_{x\to a}f(g(a))$ and $L_2:=\lim_{y\to b}f(y)$, either of which may or may not exist. We will use the sequential criterion.
First suppose $L_2$ exists and let $(x_n)$ be a sequence such that $x_n\to a$ and $x_n\ne a$ for all $n$. Since $\lim_{x\to a}g(x)=b$, we have $g(x_n)\to b$. Due to $(1)$, we have $g(x_n)\ne b$ for $n$ large enough. Then, because $L_2$ exists, we have $f(g(x_n))\to L_2$. This proves by the sequential criterion that $L_1$ exists and $L_1=L_2$.
Now assume $L_1$ exists and let $(y_n)$ be a sequence such that $y_n\to b$ and $y_n\ne b$ for all $n$. By $(2)$ there exists a sequence $(x_n)$ such that $g(x_n)=y_n$ for large $n$ and $x_n\to a$. Since $y_n\ne b$ for all $n$, we have $x_n\ne a$ for large $n$. Then $$\lim_{n\to\infty}f(y_n)=\lim_{n\to\infty}f(g(x_n))=L_1,$$ and so the sequential criterion implies $L_2$ exists and $L_2=L_1$.
• You're right --- that's not quite what I'm after. For one thing, it's not a condition that most calc students could test. :( But I agree that it's a decent starting point -- thanks. Aug 21 '17 at 19:58
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# Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions. y=1-2x+x^2
Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions and then applying the appropriate transformations.
$$\displaystyle{y}={1}-{2}{x}+{x}^{{2}}$$
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Step 1
Given:
$$\displaystyle{y}={1}-{2}{x}+{x}^{{2}}$$
To sketch:
The graph of $$\displaystyle{y}={1}-{2}{x}+{x}^{{2}}$$ using transformations.
Step 2
First we need to identify the parent function of $$\displaystyle{y}={1}-{2}{x}+{x}^{{2}}$$
$$\displaystyle{y}={1}-{2}{x}+{x}^{{2}}$$
$$\displaystyle{y}={x}^{{2}}-{2}{x}+{1}$$
$$\displaystyle{y}={\left({x}-{1}\right)}^{{2}}$$
Here the parent function is $$\displaystyle{y}={x}^{{2}}$$ . Using this, we will get the translation graph $$\displaystyle{y}={\left({x}-{1}\right)}^{{2}}$$
Since, $$\displaystyle{y}={\left({x}-{1}\right)}^{{2}}$$ is a parabola with vertex at (1,0) .
Step 3
The graph of $$\displaystyle{y}={x}^{{2}}$$ of vertex (0,0) is given by,
The graph of $$y=(x-1)^2$$ is obtained by shifting the graph $$\displaystyle{y}={x}^{{2}}$$ by 1 unit in the right direction.
Thus, the graph of $$\displaystyle{y}={1}-{2}{x}+{x}^{{2}}$$ is given by,
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(Also available in WeScheme)
Students use function composition and the distance formula to detect when characters in their games collision.
Lesson Goals Students will be able to: Explain how the distance formula is related to the Pythagorean theorem. Write a function for the distance formula. Student-Facing Lesson Goals I can explain how the distance formula is connected to the Pythagorean theorem. I can write a function that takes in 2 points and returns the distance between them. Materials
## 🔗Problem Decomposition Returns! 20 minutes
### Overview
Students revisit problem decomposition - the idea of breaking down a complex problem into simpler pieces, solving those pieces separately, and then composig the solutions to solve the original..
### Launch
"Problem Decomposition" is a powerful tool, which lets us break apart complex problems into simpler ones that we can solve, test, and then glue together into a complex solution.
Students may remember that there are two strategies for doing this:
1. Top-Down: Describe the problem at a high level, then fill in the details later
2. Bottom-Up: Focus on the smaller parts that you’re sure of, then build them together to get the big picture
### Investigate
For the following complex word problem, have students first decide which strategy they want to use, and then apply the Design Recipe to build the functions they need.
• A retractable flag pole starts out 24 inches tall, and can grow at a rate of 0.6in/sec. An elastic is tied to the top of the pole and anchored 200 inches from the base, forming a right triangle. Write a function that computes the area of the triangle as a function of time.
• This is easier to think about as two functions:
• one that computes the height of the pole, based on the seconds
• another that computes the area of the triangle, based on the height
• Does one function depend on (or "sit on top of") the other? If so, which one?
• Yes - `area` depends on `height`.
• Which strategy will YOU use: bottom-up (independent first) or top-down (dependent first)?
• Students answers will vary! They can define either function first.
• Complete Top Down / Bottom Up, using your chosen strategy for Problem Decomposition!
### Synthesize
Note: Defining the `height` first is bottom-up, and solving `area` first is top-down.
• Which strategy did students use?
• Did they try starting with one function, and then switch to another?
• Invite students to share. Oftentimes, responses are not only intriguing but can highlight the value of each approach. Explicitly point out that the `area` function _uses `height`, allowing us to break a big problem down into two smaller ones._
## 🔗Collision Detection 20 minutes
### Overview
Students once again see function composition at work, as they compose a simple inequality with the `distance` function they’ve created.
### Launch
Knowing how far apart our characters are is the first step. We still need the computer to be asking: "True or False: is there a collision?"
### Investigate
• This is easier to think about as two functions:
• one that computes the distance between a Player and a Character, based on their coordinates
• another that checks if those same coordinates are less than 50 pixels apart, based on the distance
• Does one function rely on the other? If so, which one?
• Complete Word Problem: is-collision.
• When you’ve finished, open your saved game file and fix the @code function in your game file, and click "Run"!
### Synthesize
• In our flag-pole exercise, we had to write two functions and could start with one or the other. In our game, however, we began already having written the `distance` function! Is this Top-Down or Bottom-Up decomposition?
• Explicitly point out that the `is-collide` function uses `height`, allowing us to break a big problem down into two smaller ones.
• Connect this back to `profit` (from Problem Decomposition), which relied on `revenue` and `cost`. Which function(s) would a top-down strategy address first?
• Connect this back to `is-onscreen` (from Sam the Butterfly), which relied on `is-safe-left` and `is-safe-right`. Which function(s) would a bottom-up strategy address first?
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# The Product of Two Consecutive Positive Integers is 306. Find the Integers.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
17 and 18 are two consecutive positive integers whose product is 306. To calculate this solution, we will consider two numbers a and a+1, as both numbers are consecutive. By using algebraic expressions, we will solve this question to derive our answer. Positive integers are known by another name which is natural numbers and refer to numbers that have a value more than zero.
## Two Consecutive Positive Integers with Product 306
To find the two consecutive positive integers whose product equals 306, we will first assume the two integers to be a and a + 1, as we already know both numbers are consecutive. Keeping this in mind, we will solve the question in the following manner.
We already know that the product of a and (a + 1) is 306. So, we will write it in the following way –
⇒a(a + 1) = 306
Now, to solve this equation, we will follow the below-mentioned steps –
a² + a – 306 = 0
= a² + 18a – 17a – 306 = 0
=a(a + 18) – 17(a + 18) = 0
=(a + 18)(a – 17) = 0
=a = 17 and -18
Now, we have the values of a and a+1 which are –
a = 17 and (a +1) = 18
Thus, the two positive integers that are consecutive with product 306 are 17 and 18.
## What are Positive Integers?
Positive integers refer to numbers that are bigger in value than 0. On a number line, such numbers are represented on the right side. The numbers on the left side are known as negative integers. Another name for positive integers is natural numbers. Examples of positive integers include 1,2,3,4,5,6,7,8,11,14, etc.
Summary:
## The Product of Two Consecutive Positive Integers is 306. Find the Integers
The two consecutive positive integers with product 3016 are 17 and 18. To get our answer, we will denote our two numbers as a and a+1. Positive integers refer to numbers that are greater than zero and are also called natural numbers.
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Perform the indicated divisions of polynomials by monomials. $\frac{-24{x}^{6}+36{x}^{8}}{4{x}^{2}}$
A polynomial is an expression of one or more algebraic terms each of which consists of a constant multiplied by one or more variables raised to a non-negative integral power.
Here the given polynomial is a binomial.
To divide a polynomial by monomial, divide each term of the polynomial by the monomial.
Divide the binomial by the monomial $4{x}^{2}$.
To divide a polynomial by monomial, divide each term of the polynomial by the monomial.
Simplify the terms which are under division.
Calculation:
Consider the polynomial $=\frac{-24{x}^{6}+36{x}^{8}}{4{x}^{2}}$.
Divide each term of the polynomial by the monomial $4{x}^{2}$.
$\frac{-24{x}^{6}+36{x}^{8}}{4}{x}^{2}=\left(-24\frac{{x}^{6}}{4}{x}^{2}\right)+\left(36\frac{{x}^{8}}{4}{x}^{2}\right)$
$\left(-24\frac{{x}^{6}}{4}{x}^{2}\right)+\left(36\frac{{x}^{8}}{4}{x}^{2}\right)=-6{x}^{4}+9{x}^{6}$
The simplified value of polynomial is $-6{x}^{4}+9{x}^{6}$.
Final statement:
The simplified value of polynomial after division is equals to $6{x}^{4}+9{x}^{6}$.
Do you have a similar question?
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# Learning to Count Again
Let’s define an event A, flipping heads on the flip of a coin for example. What is the probability of that event occurring? I will use the common notation of P(A) to represent the probability of an event A. The basic way to calculate a probability is to divide the number of ways A can occur by the number of ways anything can occur. That is,
$\text{P(A)} =\frac{\text{Number of ways A can occur}}{\text{Number of ways anything can occur}}$
In the case of getting a heads from a flip of a coin, there is only one way to get a heads and there are two possibilities. So the probability is 1/2.
If we flip two coins and we now let our event A be two heads, then the counting of the ways anything can happen takes just a little more thought. The possibilities are: HH, HT, TH, and TT. So there is only one way to get two heads but there are four possibilities. So the probability is 1/4. If the event is “at least one head”, there are three ways that can happen out of the total of four possibilities, so the probability is 3/4.
We can increase the complexity of our experiment and flip three coins or ask questions about choosing certain cards in a standard deck. The counting for the numerator and the dominator gets harder but it is still possible with a little more thought. But what if I asked “what is the probability of getting a flush (all cards of the same suit) in 5 cards randomly selected from a deck of cards?” or “A four-digit number (with no repetitions) is to be formed from the set of digits {1, 2, 3, 4, 5, 6}. Find the probability that the number is even.” Now the counting gets much harder. But fortunately, there are ways to handle this
### Selection or Arrangement
Let’s say we have 8 people standing around just waiting for a math problem to show up. Someone wants to form a team of 5 people from these 8 and another person wants to arrange 5 of these people around a desk. How many different teams can be made and how many different settings around the desk can be done?
Notice that in the team question, it doesn’t matter what order you pick the players but in the table question, order does matter. Just using letters to represent the people, ABCDE and EDCBA are the same for the team question so would only be counted once but these are two separate arrangements in the table question. The table question is an arrangement question whereas the team question is a selection question. In textbooks, arrangements are often called permutations and selections are called combinations. It doesn’t matter what they are called, you need to know which type you have in order to count the number of possibilities correctly.
### Arrangements
Let’s look at the table question first. You have 8 ways to choose the first person, but when you do, there are only 7 people left to choose as the second person. Then you only have 6 left for the third position, 5 left for the fourth and 4 left for the fifth. So the total number of arrangements is 8×7×6×5×4 = 6720. Multiplying numbers that sequentially decrease by one is a common thing when doing these kind of problems so, as is so common in maths, a shorthand notation was created. 8×7×6×5×4×3×2×1 is represented as 8! and is called “8 factorial”. But for our arrangement problem, we are missing the 3×2×1 part. Notice that 3×2×1 = 3!. So another way to show the solution is 8!/3!. Also notice that 8 – 5 = 3.
We can generalise this. If you have n things and want to arrange them r at a time, then the number of arrangements is n!/(nr)!. Once again, even this is too much to write for our lazy mathematicians, so this is given a shortcut notation where the P stands for permutation:
$^{n} P_{r} =\frac{n!}{( n-r) !}$
In a CAS calculator, the function “nPr” is used to calculate permutations, so for our problem nPr(8,5) = 6720.
By the way, what if we wanted to arrange all 8 people? Then using the permutation formula, we get 8!/(8-8)! = 8!/0!. This looks illegal but mathematicians foresaw this and defined 0! as being equal to 1. Must be nice to be able to make your own rules.
### Selections
Now let’s look at the team question. Using letters again, we are now in the scenario where ABCDE and EDCBA are the same team and this should only be counted once. So we would expect that the number of selections using the same n and r would be smaller than the number of arrangements. In fact this is true. So for our team selection, if we use our arrangement formula, each team has 5! different arrangements and the calculated number is 5! too large. So if we divide our arrangement total of 6720 by 5! = 120, we would get the correct number for the number of teams possible, 6720/120 = 56. To generalise, if you have n things and want to select them r at a time without regard to order, then the number of arrangements is n!/r!(nr)!. Again, there is shorthand for this where the C stands for combination:
$^{n} C_{r} =\frac{^{n} P_{r}}{r!} =\frac{n!}{r!( n-r) !}$
In a CAS calculator, the function “nCr” is used to calculate combinations, so for our problem nCr(8,5) = 56.
### Poker
So at the beginning of this post, I posed a hypothetical question about the probability of being dealt a flush hand of 5 cards out of a deck of 52 cards. As card order does not matter, this is a selection (combination) problem. We need to find the number of ways to get 5 card flush hands and the total number of possible 5 card hands. I assume you are familiar with the standard deck of cards that consists of 4 suits of 13 cards each.
First let’s find the total number of 5 card hands there are. Here n = 52 and r = 5:
$^{52} C_{5} =\frac{52!}{5!( 52-5) !} =2,598,960$
his will be our denominator to use in the probability formula. For the numerator, we need to find how many ways you can get a 5 card flush. You could get 5 clubs, 5 hearts, 5 diamonds, or 5 spades. Each one of these is a selection of 13 cards, 5 at a time or:
$^{13} C_{5} =\frac{13!}{5!( 13-5) !} =1,287$
We need to multiply this by 4 since there are 4 suits, 1287×4 = 5148. So the probability of being dealt a flush in 5 card poker is P(Flush) = 5148/2598960 = 0.00198 = 0.198% or about once in 505 hands. I wouldn’t bet on it.
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# How to Find the Difference Between Two Numbers
How to Find the Difference Between Two Numbers
To find the difference between two numbers, subtract the smaller number from the larger number.
• To find the difference between two numbers, subtract the smaller number from the larger number.
• 5 – 3 = 2 and so, the difference between 5 and 3 is 2.
• Alternatively, the difference between two numbers can be found by counting on from the smaller number.
• Count on from the smaller number until you make the larger number. This amount is the difference.
• The difference between 7 and 2 is 5 because 7 – 2 = 5.
• Number lines can be used to find the difference between two numbers.
• Start at the smaller number and count on in jumps until you reach the larger number.
• There are 5 jumps from 2 to 7 and so, the difference between 2 and 7 is 5.
Supporting Lessons
# The Difference Between Numbers
## How to Find the Difference Between Two Numbers
To find the difference between two numbers, subtract the smaller number from the larger number. For example, the difference between 5 and 3 is 2 because 5 – 3 = 2.
No matter how large the numbers are, subtracting the smaller number from the larger number will always calculate the difference.
For example, find the difference between 80 and 20.
The larger number is 80 and the smaller number is 20.
Therefore to calculate the difference between the numbers, subtract 20 from 80.
80 – 20 = 60 and so, the difference between 80 and 20 is 60.
## Teaching How to Find the Difference Between Two Numbers
To teach finding the difference:
1. Use physical objects such as counters to help count on
2. Use number lines to show the size of numbers
3. Show that subtraction gives us the difference
When introducing the idea of a difference between numbers, it is best to start with very small numbers that can be represented with physical objects, such as counters.
Placing the counters side by side can be used to show the difference in size between one number and another.
For example, in the image below we can see that 4 contains 3 more counters than 1 and so, the difference between 4 and 1 must be 3.
After the idea of what the difference between two numbers means is understood, it is time to increase the size of the numbers and look at numbers on a number line.
Number lines are useful tools to use for teaching the difference between two numbers. Mark both numbers on the number line and count the number of jumps needed to move from the smaller number to the larger number.
For example, the difference between 6 and 3 can be seen to be 3 because there are 3 jumps needed to move from the smaller number to the larger number.
## How to Find the Difference Between Two Numbers Formula
The formula for the difference between two numbers, a and b is a-b. Where a is the largest number and b is the smallest number.
For example, the difference between 25 and 13.
## The Difference Between Two-Digit Numbers
To find the difference between two-digit numbers, subtract the smaller number from the larger. With two-digit numbers, the subtraction is most easily calculated using vertical column subtraction.
For example, calculate the difference between 56 and 33.
Line up the digits and subtract vertically.
In the ones column, 6 – 3 = 3.
In the tens column, 5 – 3 = 2.
56 – 33 = 23.
## How to Find the Difference Between Two Odd Numbers
The difference between two odd numbers is always even. For example, 25 – 13 – 12, which is even.
All odd numbers have a difference which is even. This is because consecutive odd numbers always differ by 2. Adding 2 to an odd number takes us to the next odd number.
## How to Find the Difference Between Two Even Numbers
The difference between two even numbers is always even. For example, 16 – 12 = 4, which is even.
All even numbers have a difference which is even. This is because even numbers are all multiples of 2. Every even number is in the two times table. To get from one even number to the next, add two.
## How to Find the Difference Between Two Negative Numbers
The difference between two negative numbers is the same as the difference between these numbers with positive signs. For example the difference between -8 and -1 is 7 because the difference between 8 and 1 is 7.
Now try our lesson on Subtraction Number Sentences where we learn how to write subtraction number sentences.
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### Subsection3.4.4Whirlwind Tour of Summation Notation
In the remainder of this section we will frequently need to write sums involving a large number of terms. Writing out the summands explicitly can become quite impractical — for example, say we need the sum of the first 11 squares:
\begin{gather*} 1 + 2^2 + 3^2 + 4^2+ 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 \end{gather*}
This becomes tedious. Where the pattern is clear, we will often skip the middle few terms and instead write
\begin{gather*} 1 + 2^2 + \cdots + 11^2. \end{gather*}
A far more precise way to write this is using $\Sigma$ (capital-sigma) notation. For example, we can write the above sum as
\begin{gather*} \sum_{k=1}^{11} k^2 \end{gather*}
The sum from $k$ equals 1 to 11 of $k^2\text{.}$
More generally
###### Definition3.4.8
Let $m\leq n$ be integers and let $f(x)$ be a function defined on the integers. Then we write
\begin{gather*} \sum_{k=m}^n f(k) \end{gather*}
to mean the sum of $f(k)$ for $k$ from $m$ to $n\text{:}$
\begin{gather*} f(m) + f(m+1) + f(m+2) + \cdots + f(n-1) + f(n). \end{gather*}
Similarly we write
\begin{gather*} \sum_{i=m}^n a_i \end{gather*}
to mean
\begin{gather*} a_m+a_{m+1}+a_{m+2}+\cdots+a_{n-1}+a_n \end{gather*}
for some set of coefficients $\{ a_m, \ldots, a_n \}\text{.}$
Consider the example
\begin{gather*} \sum_{k=3}^7 \frac{1}{k^2}=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \frac{1}{6^2}+\frac{1}{7^2} \end{gather*}
It is important to note that the right hand side of this expression evaluates to a number 6 Some careful addition shows it is $\frac{46181}{176400}\text{.}$; it does not contain “$k$”. The summation index $k$ is just a “dummy” variable and it does not have to be called $k\text{.}$ For example
\begin{gather*} \sum_{k=3}^7 \frac{1}{k^2} =\sum_{i=3}^7 \frac{1}{i^2} =\sum_{j=3}^7 \frac{1}{j^2} =\sum_{\ell=3}^7 \frac{1}{\ell^2} \end{gather*}
Also the summation index has no meaning outside the sum. For example
\begin{gather*} k\sum_{k=3}^7 \frac{1}{k^2} \end{gather*}
has no mathematical meaning; It is gibberish 7 Or possibly gobbledygook. For a discussion of statements without meaning and why one should avoid them we recommend the book “Bendable learnings: the wisdom of modern management” by Don Watson. .
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# RD Sharma Solutions for Class 11 Chapter 8 - Transformation Formulae Exercise 8.2
In this exercise, we shall discuss problems based on formulae to transform the sum or difference into a product. The solutions prepared by our expert tutors are in an interactive manner to make it easy for the students to understand the concepts. Students can refer to RD Sharma Class 11 Solutions pdf as a major study material to improve their speed in solving problems accurately. Students can download the pdf of RD Sharma Class 11 Maths easily for free from the links given below and can start practising offline for good results.
## Download the Pdf of RD Sharma Solutions for Class 11 Maths Exercise 8.2 Chapter 8 – Transformation Formulae
### Access answers to RD Sharma Solutions for Class 11 Maths Exercise 8.2 Chapter 8 – Transformation Formulae
1. Express each of the following as the product of sines and cosines:
(i) sin 12x + sin 4x
(ii) sin 5x – sin x
(iii) cos 12x + cos 8x
(iv) cos 12x – cos 4x
(v) sin 2x + cos 4x
Solution:
(i) sin 12x + sin 4x
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 12x + sin 4x = 2 sin (12x + 4x)/2 cos (12x – 4x)/2
= 2 sin 16x/2 cos 8x/2
= 2 sin 8x cos 4x
(ii) sin 5x – sin x
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
sin 5x – sin x = 2 cos (5x + x)/2 sin (5x – x)/2
= 2 cos 6x/2 sin 4x/2
= 2 cos 3x sin 2x
(iii) cos 12x + cos 8x
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 12x + cos 8x = 2 cos (12x + 8x)/2 cos (12x – 8x)/2
= 2 cos 20x/2 cos 4x/2
= 2 cos 10x cos 2x
(iv) cos 12x – cos 4x
By using the formula,
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
cos 12x – cos 4x = -2 sin (12x + 4x)/2 sin (12x – 4x)/2
= -2 sin 16x/2 sin 8x/2
= -2 sin 8x sin 4x
(v) sin 2x + cos 4x
sin 2x + cos 4x = sin 2x + sin (90o – 4x)
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 2x + sin (90o – 4x) = 2 sin (2x + 90o – 4x)/2 cos (2x – 90o + 4x)/2
= 2 sin (90o – 2x)/2 cos (6x – 90o)/2
= 2 sin (45° – x) cos (3x – 45°)
= 2 sin (45° – x) cos {-(45° – 3x)} (since, {cos (-x) = cos x})
= 2 sin (45° – x) cos (45° – 3x)
= 2 sin (Ï€/4 – x) cos (Ï€/4 – 3x)
2. Prove that :
(i) sin 38° + sin 22° = sin 82°
(ii) cos 100° + cos 20° = cos 40°
(iii) sin 50° + sin 10° = cos 20°
(iv) sin 23° + sin 37° = cos 7°
(v) sin 105° + cos 105° = cos 45°
(vi) sin 40° + sin 20° = cos 10°
Solution:
(i) sin 38° + sin 22° = sin 82°
Let us consider LHS:
sin 38° + sin 22°
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 38° + sin 22° = 2 sin (38o + 22o)/2 cos (38o – 22o)/2
= 2 sin 60o/2 cos 16o/2
= 2 sin 30o cos 8o
= 2 × 1/2 × cos 8o
= cos 8o
= cos (90° – 82°)
= sin 82° (since, {cos (90° – A) = sin A})
= RHS
Hence Proved.
(ii) cos 100° + cos 20° = cos 40°
Let us consider LHS:
cos 100° + cos 20°
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 100° + cos 20° = 2 cos (100o + 20o)/2 cos (100o – 20o)/2
= 2 cos 120o/2 cos 80o/2
= 2 cos 60o cos 4o
= 2 × 1/2 × cos 40o
= cos 40o
= RHS
Hence Proved.
(iii) sin 50° + sin 10° = cos 20°
Let us consider LHS:
sin 50° + sin 10°
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 50° + sin 10° = 2 sin (50o + 10o)/2 cos (50o – 10o)/2
= 2 sin 60o/2 cos 40o/2
= 2 sin 30o cos 20o
= 2 × 1/2 × cos 20o
= cos 20o
= RHS
Hence Proved.
(iv) sin 23° + sin 37° = cos 7°
Let us consider LHS:
sin 23° + sin 37°
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 23° + sin 37° = 2 sin (23o + 37o)/2 cos (23o – 37o)/2
= 2 sin 60o/2 cos -14o/2
= 2 sin 30o cos -7o
= 2 × 1/2 × cos -7o
= cos 7o (since, {cos (-A) = cos A})
= RHS
Hence Proved.
(v) sin 105° + cos 105° = cos 45°
Let us consider LHS: sin 105° + cos 105°
sin 105° + cos 105° = sin 105o + sin (90o – 105o) [since, {sin (90° – A) = cos A}]
= sin 105o + sin (-15o)
= sin 105o – sin 15o [{sin(-A) = – sin A}]
By using the formula,
Sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
sin 105o – sin 15o = 2 cos (105o + 15o)/2 sin (105o – 15o)/2
= 2 cos 120o/2 sin 90o/2
= 2 cos 60o sin 45o
= 2 × 1/2 × 1/√2
= 1/√2
= cos 45o
= RHS
Hence proved.
(vi) sin 40° + sin 20° = cos 10°
Let us consider LHS:
sin 40° + sin 20°
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 40° + sin 20° = 2 sin (40o + 20o)/2 cos (40o – 20o)/2
= 2 sin 60o/2 cos 20o/2
= 2 sin 30o cos 10o
= 2 × 1/2 × cos 10o
= cos 10o
= RHS
Hence Proved.
3. Prove that:
(i) cos 55° + cos 65° + cos 175° = 0
(ii) sin 50° – sin 70° + sin 10° = 0
(iii) cos 80° + cos 40° – cos 20° = 0
(iv) cos 20° + cos 100° + cos 140° = 0
(v) sin 5π/18 – cos 4π/9 = √3 sin π/9
(vi) cos π/12 – sin π/12 = 1/√2
(vii) sin 80° – cos 70° = cos 50°
(viii) sin 51° + cos 81° = cos 21°
Solution:
(i) cos 55° + cos 65° + cos 175° = 0
Let us consider LHS:
cos 55° + cos 65° + cos 175°
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 55° + cos 65° + cos 175° = 2 cos (55o + 65o)/2 cos (55o – 65o) + cos (180o – 5o)
= 2 cos 120o/2 cos (-10o)/2 – cos 5o (since, {cos (180° – A) = – cos A})
= 2 cos 60° cos (-5°) – cos 5° (since, {cos (-A) = cos A})
= 2 × 1/2 × cos 5o – cos 5o
= 0
= RHS
Hence Proved.
(ii) sin 50° – sin 70° + sin 10° = 0
Let us consider LHS:
sin 50° – sin 70° + sin 10°
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
sin 50° – sin 70° + sin 10° = 2 cos (50o + 70o)/2 sin (50o – 70o) + sin 10o
= 2 cos 120o/2 sin (-20o)/2 + sin 10o
= 2 cos 60o (- sin 10o) + sin 10o [since,{sin (-A) = -sin (A)}]
= 2 × 1/2 × – sin 10o + sin 10o
= 0
= RHS
Hence proved.
(iii) cos 80° + cos 40° – cos 20° = 0
Let us consider LHS:
cos 80° + cos 40° – cos 20°
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 80° + cos 40° – cos 20° = 2 cos (80o + 40o)/2 cos (80o – 40o) – cos 20o
= 2 cos 120o/2 cos 40o/2 – cos 20o
= 2 cos 60° cos 20o – cos 20°
= 2 × 1/2 × cos 20o – cos 20o
= 0
= RHS
Hence Proved.
(iv) cos 20° + cos 100° + cos 140° = 0
Let us consider LHS:
cos 20° + cos 100° + cos 140°
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 20° + cos 100° + cos 140° = 2 cos (20o + 100o)/2 cos (20o – 100o) + cos (180o – 40o)
= 2 cos 120o/2 cos (-80o)/2 – cos 40o (since, {cos (180° – A) = – cos A})
= 2 cos 60° cos (-40°) – cos 40° (since, {cos (-A) = cos A})
= 2 × 1/2 × cos 40o – cos 40o
= 0
= RHS
Hence Proved.
(v) sin 5π/18 – cos 4π/9 = √3 sin π/9
Let us consider LHS:
sin 5Ï€/18 – cos 4Ï€/9 = sin 5Ï€/18 – sin (Ï€/2 – 4Ï€/9) (since, cos A = sin (90o – A))
= sin 5Ï€/18 – sin (9Ï€ – 8Ï€)/18
= sin 5π/18 – sin π/18
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
= 2 cos (6Ï€/36) sin (4Ï€/36)
= 2 cos π/6 sin π/9
= 2 cos 30o sin π/9
= 2 × √3/2 × sin π/9
= √3 sin π/9
= RHS
Hence proved.
(vi) cos π/12 – sin π/12 = 1/√2
Let us consider LHS:
cos Ï€/12 – sin Ï€/12 = sin (Ï€/2 – Ï€/12) – sin Ï€/12 (since, cos A = sin(90o – A))
= sin (6Ï€ – 5Ï€)/12 – sin Ï€/12
= sin 5π/12 – sin π/12
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
= 2 cos (6Ï€/24) sin (4Ï€/24)
= 2 cos π/4 sin π/6
= 2 cos 45o sin 30o
= 2 × 1/√2 × 1/2
= 1/√2
= RHS
Hence proved.
(vii) sin 80° – cos 70° = cos 50°
sin 80° = cos 50° + cos 70o
So, now let us consider RHS
cos 50° + cos 70o
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 50° + cos 70o = 2 cos (50o + 70o)/2 cos (50o – 70o)/2
= 2 cos 120o/2 cos (-20o)/2
= 2 cos 60o cos (-10o)
= 2 × 1/2 × cos 10o (since, cos (-A) = cos A)
= cos 10o
= cos (90° – 80°)
= sin 80° (since, cos (90° – A) = sin A)
= LHS
Hence Proved.
(viii) sin 51° + cos 81° = cos 21°
Let us consider LHS:
sin 51° + cos 81° = sin 51o + sin (90o – 81o)
= sin 51o + sin 9o (since, sin (90° – A) = cos A)
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 51o + sin 9o = 2 sin (51o + 9o)/2 cos (51o – 9o)/2
= 2 sin 60o/2 cos 42o/2
= 2 sin 30o cos 21o
= 2 × 1/2 × cos 21o
= cos 21o
= RHS
Hence proved.
4. Prove that:
(i) cos (3π/4 + x) – cos (3π/4 – x) = -√2 sin x
(ii) cos (Ï€/4 + x) + cos (Ï€/4 – x) = √2 cos x
Solution:
(i) cos (3π/4 + x) – cos (3π/4 – x) = -√2 sin x
Let us consider LHS:
cos (3π/4 + x) – cos (3π/4 – x)
By using the formula,
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
cos (3Ï€/4 + x) – cos (3Ï€/4 – x) = -2 sin (3Ï€/4 + x + 3Ï€/4 – x)/2 sin (3Ï€/4 + x – 3Ï€/4 + x)/2
= -2 sin (6Ï€/4)/2 sin 2x/2
= -2 sin 6Ï€/8 sin x
= -2 sin 3Ï€/4 sin x
= -2 sin (π – π/4) sin x
= -2 sin π/4 sin x (since, (π-A) = sin A)
= -2 × 1/√2 × sin x
= -√2 sin x
= RHS
Hence proved.
(ii) cos (Ï€/4 + x) + cos (Ï€/4 – x) = √2 cos x
Let us consider LHS:
cos (Ï€/4 + x) + cos (Ï€/4 – x)
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos (Ï€/4 + x) + cos (Ï€/4 – x) = 2 cos (Ï€/4 + x + Ï€/4 – x)/2 cos (Ï€/4 + x – Ï€/4 + x)/2
= 2 cos (2Ï€/4)/2 cos 2x/2
= 2 cos 2Ï€/8 cos x
= 2 sin π/4 cos x
= 2 × 1/√2 × cos x
= √2 cos x
= RHS
Hence proved.
5. Prove that:
(i) sin 65o + cos 65o = √2 cos 20o
(ii) sin 47o + cos 77o = cos 17o
Solution:
(i) sin 65o + cos 65o = √2 cos 20o
Let us consider LHS:
sin 65o + cos 65o = sin 65o + sin (90o – 65o)
= sin 65o + sin 25o
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 65o + sin 25o = 2 sin (65o + 25o)/2 cos (65o – 25o)/2
= 2 sin 90o/2 cos 40o/2
= 2 sin 45o cos 20o
= 2 × 1/√2 × cos 20o
= √2 cos 20o
= RHS
Hence proved.
(ii) sin 47o + cos 77o = cos 17o
Let us consider LHS:
sin 47o + cos 77o = sin 47o + sin (90o – 77o)
= sin 47o + sin 13o
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 47o + sin 13o = 2 sin (47o + 13o)/2 cos (47o – 13o)/2
= 2 sin 60o/2 cos 34o/2
= 2 sin 30o cos 17o
= 2 × 1/2 × cos 17o
= cos 17o
= RHS
Hence proved.
6. Prove that:
(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos 3A/2 sin 3A
(iv) sin 3A + sin 2A – sin A = 4 sin A cos A/2 cos 3A/2
(v) cos 20o cos 100o + cos 100o cos 140o – cos 140o cos 200o = – 3/4
(vi) sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 = sin 2x sin 5x
(vii) cos x cos x/2 – cos 3x cos 9x/2 = sin 4x sin 7x/2
Solution:
(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
Let us consider LHS:
cos 3A + cos 5A + cos 7A + cos 15A
So now,
(cos 5A + cos 3A) + (cos 15A + cos 7A)
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
(cos 5A + cos 3A) + (cos 15A + cos 7A)
= [2 cos (5A+3A)/2 cos (5A-3A)/2] + [2 cos (15A+7A)/2 cos (15A-7A)/2]
= [2 cos 8A/2 cos 2A/2] + [2 cos 22A/2 cos 8A/2]
= [2 cos 4A cos A] + [2 cos 11A cos 4A]
= 2 cos 4A (cos 11A + cos A)
Again by using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
2 cos 4A (cos 11A + cos A) = 2 cos 4A [2 cos (11A+A)/2 cos (11A-A)/2]
= 2 cos 4A [2 cos 12A/2 cos 10A/2]
= 2 cos 4A [2 cos 6A cos 5A]
= 4 cos 4A cos 5A cos 6A
= RHS
Hence proved.
(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
Let us consider LHS:
cos A + cos 3A + cos 5A + cos 7A
So now,
(cos 3A + cos A) + (cos 7A + cos 5A)
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
(cos 3A + cos A) + (cos 7A + cos 5A)
= [2 cos (3A+A)/2 cos (3A-A)/2] + [2 cos (7A+5A)/2 cos (7A-5A)/2]
= [2 cos 4A/2 cos 2A/2] + [2 cos 12A/2 cos 2A/2]
= [2 cos 2A cos A] + [2 cos 6A cos A]
= 2 cos A (cos 6A + cos 2A)
Again by using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
2 cos A (cos 6A + cos 2A) = 2 cos A [2 cos (6A+2A)/2 cos (6A-2A)/2]
= 2 cos A [2 cos 8A/2 cos 4A/2]
= 2 cos A [2 cos 4A cos 2A]
= 4 cos A cos 2A cos 4A
= RHS
Hence proved.
(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos 3A/2 sin 3A
Let us consider LHS:
sin A + sin 2A + sin 4A + sin 5A
So now,
(sin 2A + sin A) + (sin 5A + sin 4A)
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
(sin 2A + sin A) + (sin 5A + sin 4A) =
= [2 sin (2A+A)/2 cos (2A-A)/2] + [2 sin (5A+4A)/2 cos (5A-4A)/2]
= [2 sin 3A/2 cos A/2] + [2 sin 9A/2 cos A/2]
= 2 cos A/2 (sin 9A/2 + sin 3A/2)
Again by using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
2 cos A/2 (sin 9A/2 + sin 3A/2) = 2 cos A/2 [2 sin (9A/2 + 3A/2)/2 cos (9A/2 – 3A/2)/2]
= 2 cos A/2 [2 sin ((9A+3A)/2)/2 cos ((9A-3A)/2)/2]
= 2 cos A/2 [2 sin 12A/4 cos 6A/4]
= 2 cos A/2 [2 sin 3A cos 3A/2]
= 4 cos A/2 cos 3A/2 sin 3A
= RHS
Hence proved.
(iv) sin 3A + sin 2A – sin A = 4 sin A cos A/2 cos 3A/2
Let us consider LHS:
sin 3A + sin 2A – sin A
So now,
(sin 3A – sin A) + sin 2A
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
(sin 3A – sin A) + sin 2A = 2 cos (3A + A)/2 sin (3A – A)/2 + sin 2A
= 2 cos 4A/2 sin 2A/2 + sin 2A
We know that, sin 2A = 2 sin A cos A
= 2 cos 2A Sin A + 2 sin A cos A
= 2 sin A (cos 2A + cos A)
Again by using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
2 sin A (cos 2A + cos A) = 2 sin A [2 cos (2A+A)/2 cos (2A-A)/2]
= 2 sin A [2 cos 3A/2 cos A/2]
= 4 sin A cos A/2 cos 3A/2
= RHS
Hence proved.
(v) cos 20o cos 100o + cos 100o cos 140o – cos 140o cos 200o = – 3/4
Let us consider LHS:
cos 20o cos 100o + cos 100o cos 140o – cos 140o cos 200o =
We shall multiply and divide by 2 we get,
= 1/2 [2 cos 100o cos 20o + 2 cos 140o cos 100o – 2 cos 200o cos 140o]
We know that 2 cos A cos B = cos (A+B) + cos (A-B)
So,
= 1/2 [cos (100o + 20o) + cos (100o – 20o) + cos (140o + 100o) + cos (140o – 100o) – cos (200o + 140o) – cos (200o – 140o)]]
= 1/2 [cos 120o + cos 80o + cos 240o + cos 40o – cos 340o – cos 60o]
= 1/2 [cos (90o + 30o) + cos 80o + cos (180o + 60o) + cos 40o – cos (360o – 20o) – cos 60o]
We know, cos (180o + A) = – cos A, cos (90o + A) = – sin A, cos (360o – A) = cos A
So,
= 1/2 [- sin 30o + cos 80o – cos 60o + cos 40o – cos 20o – cos 60o]
= 1/2 [- sin 30o + cos 80o + cos 40o – cos 20o – 2 cos 60o]
Again by using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
= 1/2 [- sin 30o + 2 cos (80o+40o)/2 cos (80o-40o)/2 – cos 20o – 2 × 1/2]
= 1/2 [- sin 30o + 2 cos 120o/2 cos 40o/2 – cos 20o – 1]
= 1/2 [- sin 30o + 2 cos 60o cos 20o – cos 20o – 1]
= 1/2 [- 1/2 + 2×1/2×cos 20o – cos 20o – 1]
= 1/2 [-1/2 + cos 20o – cos 20o – 1]
= 1/2 [-1/2 -1]
= 1/2 [-3/2]
= -3/4
= RHS
Hence proved.
(vi) sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 = sin 2x sin 5x
Let us consider LHS:
sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 =
We shall multiply and divide by 2 we get,
= 1/2 [2 sin 7x/2 sin x/2 + 2 sin 11x/2 sin 3x/2]
We know that 2 sin A sin B = cos (A-B) – cos (A+B)
So,
= 1/2 [cos (7x/2 – x/2) – cos (7x/2 + x/2) + cos (11x/2 – 3x/2) – cos (11x/2 + 3x/2)]
= 1/2 [cos (7x-x)/2 – cos (7x+x)/2 + cos (11x-3x)/2 – cos (11x+3x)/2]
= 1/2 [cos 6x/2 – cos 8x/2 + cos 8x/2 – cos 14x/2]
= 1/2 [cos 3x – cos 7x]
= – 1/2 [cos 7x – cos 3x]
Again by using the formula,
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
= -1/2 [-2 sin (7x+3x)/2 sin (7x-3x)/2]
= -1/2 [-2 sin 10x/2 sin 4x/2]
= -1/2 [-2 sin 5x sin 2x]
= -2/-2 sin 5x sin 2x
= sin 2x sin 5x
= RHS
Hence proved.
(vii) cos x cos x/2 – cos 3x cos 9x/2 = sin 4x sin 7x/2
Let us consider LHS:
cos x cos x/2 – cos 3x cos 9x/2 =
We shall multiply and divide by 2 we get,
= 1/2 [2 cos x cos x/2 – 2 cos 9x/2 cos 3x]
We know that 2 cos A cos B = cos (A+B) + cos (A-B)
So,
= 1/2 [cos (x + x/2) + cos (x – x/2) – cos (9x/2 + 3x) – cos (9x/2 – 3x)]
= 1/2 [cos (2x+x)/2 + cos (2x-x)/2 – cos (9x+6x)/2 – cos (9x-6x)/2]
= 1/2 [cos 3x/2 + cos x/2 – cos 15x/2 – cos 3x/2]
= 1/2 [cos x/2 – cos 15x/2]
= – 1/2 [cos 15x/2 – cos x/2]
Again by using the formula,
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
= – 1/2 [-2 sin (15x/2 + x/2)/2 sin (15x/2 – x/2)/2]
= -1/2 [-2 sin (16x/2)/2 sin (14x/2)/2]
= -1/2 [-2 sin 16x/4 sin 7x/2]
= – 1/2 [-2 sin 4x sin 7x/2]
= -2/-2 [sin 4x sin 7x/2]
= sin 4x sin 7x/2
= RHS
Hence proved.
7. Prove that:
Solution:
8. Prove that:
Solution:
= cot 6A
= RHS
Hence proved.
By using the formulas,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
= RHS
Hence proved.
9. Prove that:
(i) sin α + sin β + sin γ – sin (α + β + γ) = 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2
(ii) cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) = 4 cos A cos B cos C
Solution:
(i) sin α + sin β + sin γ – sin (α + β + γ) = 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2
Let us consider LHS:
sin α + sin β + sin γ – sin (α + β + γ)
By using the formulas,
Sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
Sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
= RHS
Hence proved.
(ii) cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) = 4 cos A cos B cos C
Let us consider LHS:
cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C)
so,
cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) =
={cos (A + B + C) + cos (A – B + C)} + {cos (A + B – C) + cos (-A + B + C)}
By using the formula,
Cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
= 4 cos A cos B cos C
= RHS
Hence proved.
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# Probability 1 answer below »
An urn holds 5 white and 3 black marbles. If 2 marbles are to be drawn at random without replacement and "x" denotes the number of white marbles, find the probability distribution for x.
Piyanka D
In an urn, there are 5 white and 3 black marbles. From it, 2 balls are drawn at random. Here, the event is to record the number of white balls drawn. There are three situation of this drawing. The first situation is gettingno white ball. Thesecondsituation is getting 1white ball and 1 black ball.The thirdsituation is getting 2white balls. The following steps are used to construct the probability distribution of number of white balls drawn Step 1:Computation of individual probabilities Now consider the first option which depicts no white ball is drawn. This implies, 2 black balls are drawn. The number of ways to draw2balls from the urn is defined as, $$n(S) = ^8C_2$$ = 28 Thenumber of ways to draw2balls from the urn is 28. Then, the number of ways to draw2 black balls from the urn is defined as, $$n (BB) = ^3C_2$$ = 3 The number of ways to draw2 black balls from the urn is 3. Then,the probability that 2 black balls are drawn is defined as, $$P(BB) = n(BB) \over n(S)$$ $$= 3 \over 28$$ = 0.1071 Theprobability that 2 black balls are drawn is0.1071. This implies, theprobability that nowhiteballs are drawn is 0.1071 Now consider the secondoption which depicts 1white ball and 1 black ball is drawn. Then, the number of ways to draw1white ball and 1 black ball from the urn is defined as, $$n (WB) =( ^5C_1) \times ( ^3C_1)$$ $$=5 \times 3$$ = 15 The number of ways to draw1white ball and 1 black ball from the urn is 15....
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Tutoring high school maths and sciences, unit conversions are prevalent. The math tutor offers an example of converting between common units.
Hello: hope your summer is going great. As I’ve implied over the previous few weeks, my activities have shifted from day-to-day tutoring towards maintenance and research.
One focus, particularly since early July, has been repairing my old Lawn-Boy mower. It’s been a true learning experience. A point of interest about the repair/restoration, which also connnects with math, is the notion that, according to the Lawn-Boy service manual, the engine is 3.5 horsepower. Horsepower is an imperial unit, often abbreviated to hp. The obvious question: What is 3.5 horsepower in metric?
Well, to start with, the metric unit for power is the Watt. Most metric units are lower case, but a few are upper case because they are named after people. So it is with the Watt, named after James Watt. The abbreviation for Watt is W.
From the perspective of common processes, a Watt is a very small amount of power. To make the numbers more convenient, many applications use the kilowatt, abbreviated to kW:
1kW=1000W
Now, it’s a fact that, rounded to the nearest whole number,
1hp=746W
Multiplying both sides by 3.5, we get
3.5hp=3.5(746W)=2611W
Now, we divide by 1000 to convert to kW:
2611W/1000=2.611kW
Apparently, to convert from hp to kW, we first mutliply by 746, then divide by 1000. We can combine those two operations and just multiply by 0.746, as follows:
Example: suppose a car is rated at 220 horsepower. What is its power in kW?
Solution:
220(0.746)=164kW, rounded to nearest whole number.
This article has evolved several topics that I’ll discuss in future posts:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
While tutoring slows down in the summer, the focus shifts. Though not an exercise expert, the tutor continues with some of his ideas about it.
Connecting with my last post: I was discussing exercises I’ve gravitated towards, or moved away from, over the last couple of years. Skipping is prominent now, with outdoor running, running up and down stairs, and riding the stationary bike all taking back seats.
Skipping, I rarely get injured. However, in my experience, care needs to be taken not to overdo it. When I haven’t skipped for a while, a couple of minutes at a time is the most I should do, with a couple minutes “walking it out” in between. The danger: the connective tissues in the feet can get overworked without the skipper’s being aware. In such a case, when the muscles cool down, the skipper can be left with a bad strain. It’s happened to me. The same caution applies when I want to increase my skipping duration: I put rest walks in between the minutes I’ve always done and the “new minutes” I’m adding on. I only add a couple of minutes at a time.
To my mind, an exercise that bears the person’s weight is advantageous: if the person gains weight, the exercise burns more calories. From that perspective, a weight-bearing exercise is self-correcting against weight gain. Skipping and running share this feature, while riding the stationary bike does not. (Neither, to my mind, does swimming; I’ll say more about that in another post.)
For now, I skip two to three times per week, 15 to 20 minutes each time. I find it promotes the balance, quickness, and endurance that I need to carry me through my busy days with a nine and a twelve year old.
I’ll be talking more about exercise in future posts. For now, please enjoy this beautiful day. For those of you reading from other places: we are currently in a heat wave here. Good luck staying cool!
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
What does a tutor do over the summer? Tutoring continues, but at a relaxed pace. In the breathing space, the tutor pursues various topics of interest….
Given our culture’s focus on being slender and youthful, combined with the threat of weight gain that lurks in the lifestyles of many, exercise and/or weight loss have become “everyman” topics. This tutor is right on board: when you deal mainly with young, fit people, you’re best off to be youthful and fit yourself.
With family, work, and the pressure to stay fit, many find themselves squeezed from different sides. Inevitably, the following question begs to be answered:
• Assuming I can find some time for exercise, what are the best ones to do?
I’m definitely not a workout specialist; I can only contribute observations I’ve made from my own experience. Generally, I’ve trended away from muscle building exercises, towards aerobic ones. The year before this, I focused mainly on outdoor running, running up and down stairs, and riding the stationary bike. This past year, I’ve changed mainly to skipping.
I’ve mainly given up outdoor running for two reasons. The first reason is injuries: I never run on pavement, yet I still get injured running. The second reason is crowding: the fields I run in can be quite busy. A serious runner needs space.
Running up and down the stairs is great. For me, it’s indoors, so it doesn’t offer the advantage of fresh air. Another drawback: it can’t be done when people might need to use the stairs. Furthermore, I find it doesn’t build the calf muscles the way I like.
My stationary bike is outside, so it’s a fresh-air activity. It truly is a great exercise; I find it to be injury-free. From my point of view, its only drawback is that the musculature you get from it is quite disconnected from everyday life. (I’m not a cyclist other than on the stationary bike.)
Why I like skipping, and some other observations about exercise in general, I’ll share in the next post. For now, let’s make the most of this perfect summer day, whatever we are doing:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
When you tutor computer programming, you must explain the use of variables. Such is the next step of our PERL summer project.
To a computer, a value needs a name. Computers (at this level) aren’t capable of changing context the way humans can; they need to be told what to do, and what to do it to, at every stage.
Here’s an example of the difference between humans and computers. Suppose, for instance, you tell a grade 6 student the numbers 11 and 29, then ask them for the average. The student will add the numbers, divide the sum by 2, then spontaneously say “The average is 20.” If you ask the student to tell the original two numbers, they can; however, they realize that what you really want is the answer.
On the other hand, a computer needs to “identify” a value before it can do an operation to it. Therefore, a computer program to find the average of 11 and 29 might proceed like this:
get the value of the first number: call it number1
get the value of the second number: call it number2
add number1 and number2; call the sum number3
divide number3 by 2; call the result theaverage
print out “The average is ” theaverage
In traditional programming, you name a variable first, then give it a value. When you want the computer to manipulate that value, you refer to it by its variable name. Planning a program, you don’t wonder what the answer will be; rather, you wonder how to explain to the computer what to do in order to find the answer. Specifically, you need to tell it what steps to perform to which variables. Actually working with the numbers is easy for a computer, once it has clear instructions.
Commonly, variable names are written as one word, even if they indeed contain more than one. Therefore, thirdnumber might be a likely name for the third number referenced in a program; on the other hand, you might call it number3. Digits can be used in variable names, but usually aren’t allowed as the first character.
In PERL, variable names are commonly given a dollar sign prefix; therefore, if you wanted to call a variable number3, you’d call it \$number3. The dollar sign prefix increases program readability for humans; it’s easy to track what’s happening with the variables, since they all start with \$. There are other reasons, as well, for the \$ prefix on PERL variable names.
Next post, we’ll see some variables in action. Enjoy your day:)
Source: Robert’s Perl Tutorial, by Robert Pepper.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
Tutoring PERL beginnings, the details can be so important. While the tutor now focuses to the Linux/Mac side, Windows users might do well to tune in.
Over my last few posts I’ve been discussing how to get started with PERL programming. Of course, I’ve run parallel posts: some have focused on Windows, others on Linux/Mac. I hope you’ve all been successful in following the procedures I’ve offered.
Whichever world you reside in – assuming it is just one – knowing what’s going on in the other world often seems beneficial. Therefore, I recommend the users of one operating system read the posts for the other, if only to realize how similar the procedures are.
Today’s emphasis is on the line
#!/usr/bin/perl
which is part of the script for Linux/Mac, but is conspicuously absent at the head of the script for Windows.
According to Robert Pepper’s Perl Tutorial, #!/usr/bin/perl is called the “shebang” line. (By the way: almost everything I know about PERL I learned from Robert Pepper.) He goes on to explain that under Unix systems (Linux is a derivative), the shebang line tells the file type so the computer knows how to execute it.
Windows, he continues, does not need a shebang line. However, a shebang line for the Windows script might look similar to this:
#!c:/strawberry/perl/bin/perl.exe
(The above shebang reveals the use of Strawberry Perl).
The line tells the location of the perl interpreter, which will execute the script. Once again, the line is not needed in Windows.
For Linux/Mac, however, you will see something like
#!/usr/bin/perl
at the head of a PERL script, and it is needed – on mine, anyway. Now you know why, whichever operating system you might use:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
In the context of tutoring, a lazy summer day might mean academic pursuits at a relaxed pace. The tutor acknowledges the need to bring Linux and Mac users equal with Windows users: in this installment, he offers instructions for Linux and Mac users to execute their first PERL script.
In my previous post I instructed Windows users through the execution of their first PERL script. Now, Linux and Mac users, it’s your turn.
Open your text editor and type in the following:
#!/usr/bin/perl
print “Hello, how are you?–from PERL!\n\n”;
Let’s imagine you call your file july101 and save it in your “My_perl_programs” folder. In my experience, you can save it as type “all files” or type “text files” – just as long as it’s actually plain text.
Now, open your terminal and go into your “My_perl_programs” directory. Enter the command ls to make sure your file july101 is listed. Assuming you see it, enter the following command:
perl july101
If the process works, the terminal should reply with
Hello, how are you?–from PERL!
If you receive that greeting, congratulations! You, Linux or Mac user, have successfully executed your first PERL script.
The next few posts, we’ll review what we did during this one and last one. To quote Phil Collins in his hit song Against All Odds:
“There’s so much I have to say to you/So many reasons why….”
We’ll get to it all, in the coming days:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
Summer tutoring continues. Today, the tutor hopes, we get word back from PERL.
In the previous article I brought the Linux and (hopefully) Mac users on board with creating a text file, then finding it in the terminal. Earlier, I described how to do it under Windows. Now, I imagine, “everyone” knows how to, on their home system, create a text file, then find it in the terminal.
The goal of today’s article is to create a PERL script, find it in the terminal, and run it. Once again, there is a dichotomy between Windows and Linux/Mac. This article will cover Windows; the next one, Linux/Mac.
Windows users:
In Notepad, open a new text file, then type the following:
print “Hello from PERL on Jun30, 2014!\n\n”;
Save the file. Let’s imagine you call it myfirstperl then save it to your folder (aka directory) “my perl programs” (or whatever you called the folder). Now, follow these steps:
1. Open the terminal and go into the directory “my perl programs” (or, once again, whatever you happened to call it).
2. Enter the command dir to confirm your file myfirstperl is indeed in the directory. If you called it myfirstperl, it should show up as myfirstperl.txt
3. Let’s assume you see your file myfirstperl.txt. Now, key in
perl myfirstperl.txt
If it worked, the terminal should reply with
Hello from PERL on Jun30,2014!
Hopefully, the procedure worked for you. If so, you have successfully begun programming with PERL: congratulations!
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
Pursuing summer tutoring, we have embarked upon the PERL programming journey. The tutor continues it: soon, there might be no turning back.
In my previous article, I described how you might, in Windows, create a text file, save it, then find it in the terminal. Now, I’ll describe the corresponding procedure in Linux – which I believe will be the same for the Mac. Once again: I don’t have a Mac, so I can’t be sure. It’s my impression, however, that Linux and Mac, for this purpose, are similar.
First, to create the file, you’ll open the text editor. In my flavour of Linux, it’s called “text editor”. You might write down a grocery list or colour choices for your paint:
maroon, cappuccino, red granite, basalt
Now, you’ll save that file. For convenience, you might create a new folder for PERL activities, then save the file in there. Perhaps you create the folder
“My_ perl_programs”. (Unlike with Windows, the Linux terminal may not tolerate spaces in names). Maybe you save the file as colours.txt in your “My_perl_programs” folder.
Now, you open the terminal – which, in my flavour of Linux, is called “terminal”. It’s in with the apps.
When I open the terminal, it puts me in what I’d call my “home” directory. To see the contents of the directory, I enter the command
ls
On my Linux terminal, ls displays the directories in blue, while the files are in white. From the terminal’s point of view, a folder is a directory. You (hopefully) see “My_perl_programs” – or whatever you called your PERL folder – among the items listed by the ls command. Maybe it’s even in blue.
Let’s assume you did call your PERL programs directory “My_perl_programs”. To go into it now, enter the command
cd My_perl_programs
Now, when you enter ls, you should see your text file colours.txt – or whatever you called it – listed. If you do see it, you have successfully created a text file, then found it in the terminal – which is, after all, the point of this article.
Three observations:
1. In Linux, the terminal doesn’t seem to tolerate names with spaces in between. Windows users will notice this difference.
2. The Linux terminal is case sensitive, whereas the Windows one is not. Therefore, in Windows you can go into the “MeandMyself” directory using the command
cd meandmyself, while in Linux, you can’t. You’d instead need to enter, literally, cd MeandMyself.
3. In my Linux terminal, the dir command also works to list the contents of a directory, even though it’s the Windows command. However, you might not get the colour coding you might (in Linux) get using the Linux command ls.
To my knowledge, the Mac procedure for accomplishing the tasks above is very similar – if not virtually the same. Perhaps now “everyone” knows how to create a text file, then find it in the terminal. Next step: writing a PERL script, then (hopefully) running it!
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
A summer tutoring project is under way; the tutor continues with its next installment.
As I mentioned in my previous article, developing familiarity with the terminal is key to running a PERL program. What the user really need be able to do is create a plain text file, then find it in the terminal environment.
On Windows, suppose you open Notepad and write a line or two. For now, maybe you just write a grocery list.
Windows has the directory “My Documents” or something similar (In Windows 7, it might just be “Documents”.) You might make a new folder in that directory called “my perl programs” or something like that. Next, you might save your grocery list file in there. Maybe you call it groclist.txt, for instance.
Now, you need to be able to find that file in the terminal. Under
All Programs→Accessories, you find Command Prompt, which is the terminal. When you click Command Prompt, a black window opens on the screen. There is the command line: a directory name in white print, with a blinking cursor.
In my experience, the terminal opens to the directory that contains My Documents. Therefore, if you key in
cd my documents
then press Enter, you should arrive there. If so, you’ll see “My Documents” tacked onto the end of the directory name. Windows 7 users might do the step above substituting documents instead of my documents.
If you made a new folder called “my perl programs” in “My Documents”, you’ll be able to find it now by typing in
dir
then pressing Enter. Your folder should appear in the list.
To move into that folder, enter the command
cd my perl programs
Now, “my perl programs” should be tacked on the end of your present location. If you enter the command
dir
you should see your file “groclist.txt” – or whatever you called it – listed.
The above instructions likely give you the tools to create a text file, then find it in the terminal. Such facility is key to creating and running a PERL script.
I apologize that this article covers Windows only. However, an article that covers the other operating systems as well would be too long for our easy summer pace. In the next article I’ll cover the Linux context. I’ll even try to extend to the Mac, though I don’t have one.
A Linux or Mac user who is unfamiliar with the terminal will still pick up valuable hints from this article. The reverse will be true as well: the Windows user will gain from reading the next article, even though its context will be Linux.
Jack or Oracle Tutoring by Jack and Diane, Campbell River, BC.
Tutoring through the summer, you tend to land in projects. The tutor brings the PERL project to another early milestone.
In my previous post, I talked practically about what you need to get started with PERL, should you choose to embark. While Macs and Linux systems include PERL, a Windows user might have to download a PERL bundle, of which there are choices (Strawberry and ActiveState being two I’ve used).
Next concern: the text editor used to write the scripts. (For our purposes, a script is a little program). It must produce plain text. Under Windows, Notepad
(All Programs→Accessories→Notepad) will do fine; Linux users have many choices, but I use “text editor” under Apps. As I’ve mentioned, I don’t have a Mac, but I searched a bit yesterday how a Mac user might produce plain text. This article describes how the Mac’s text editor can convert content to plain text using a command under the “Format” menu.
There is a final issue one needs to confront before starting with PERL: getting familiar with the terminal – aka, the command line.
To my knowledge, every operating system has a terminal. One point of difference among operating systems is how much (or little) the terminal is in evidence. Linux users are likely familiar with the terminal. Older people (myself included) recall using MS-DOS on the terminal.
On Windows, the terminal is called Command Prompt, and is under
All Programs→Accessories.
In Linux (perhaps), the terminal is called “terminal”, and is under Apps. (There are many flavours of Linux, but this is true for Ubuntu, anyway.)
For the Mac, this article tells me to find the terminal under “Applications/Utilities”.
For the purposes of running a PERL program, you need to be able to arrive at the location (called the directory) of your script file. Here are the basic commands you need:
Windows:
• cd changes the directory you’re in.
• dir tells the contents of the directory you’re in.
Linux or Mac:
• cd changes the directory you’re in.
• ls tells the contents of the directory you’re in.
In the next post, I’ll continue with some hints and examples of how to get around in the terminal to find a file. Until then, cheers:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
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## How do you find the domain of an equation?
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.
## How do you find the domain and range of an equation?
The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes. A rational function is a function of the form f(x)=p(x)q(x) , where p(x) and q(x) are polynomials and q(x)≠0 .
## What are the steps to find the domain and range?
Overall, the steps for algebraically finding the range of a function are:Write down y=f(x) and then solve the equation for x, giving something of the form x=g(y).Find the domain of g(y), and this will be the range of f(x).If you can’t seem to solve for x, then try graphing the function to find the range.
## What is Domain give example?
A domain name takes the form of two main elements. For example, the domain name Facebook.com consists of the website’s name (Facebook) and the domain name extension (.com). When a company (or a person) purchases a domain name, they’re able to specify which server the domain name points to.
## What is domain and range of a function?
In its simplest form the domain is all the values that go into a function, and the range is all the values that come out. But in fact they are very important in defining a function.
## How do you write domain notation?
We can write the domain of f(x) in set builder notation as, {x | x ≥ 0}. If the domain of a function is all real numbers (i.e. there are no restrictions on x), you can simply state the domain as, ‘all real numbers,’ or use the symbol to represent all real numbers.
## How do you write a range?
The Range is the difference between the lowest and highest values. Example: In {4, 6, 9, 3, 7} the lowest value is 3, and the highest is 9. So the range is 9 − 3 = 6. It is that simple!
## How do you determine a function?
How To: Given a relationship between two quantities, determine whether the relationship is a function.Identify the input values.Identify the output values.If each input value leads to only one output value, classify the relationship as a function.
You might be interested: Love mathematical equation
## What is a domain function?
The domain of a function is the set of all possible inputs for the function. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0.
## What does domain mean?
A domain name is your website name. A domain name is the address where Internet users can access your website. A domain name is used for finding and identifying computers on the Internet. Because of this, domain names were developed and used to identify entities on the Internet rather than using IP addresses.
## What is the physical domain?
Physical. The physical domain covers the development of physical changes, which includes growing in size and strength, as well as the development of both gross motor skills and fine motor skills. 1 The physical domain also includes the development of the senses and using them.
## What is IP domain?
Domain Names and IP Addresses An Internet Protocol, or IP, address is different than a domain name. It is convenient to think of IP addresses as the actual code and the domain name as a nickname for that code. A typical IP address looks like a string of numbers. It could be 232.17. 43.22, for example.
### Releated
#### Bonding energy equation
What is bond energy in chemistry? In chemistry, bond energy (BE), also called the mean bond enthalpy or average bond enthalpy is the measure of bond strength in a chemical bond. The larger the average bond energy, per electron-pair bond, of a molecule, the more stable and lower-energy the molecule. What are the units of […]
#### Characteristic equation complex roots
What are roots of characteristic equations? discussed in more detail at Linear difference equation#Solution of homogeneous case. The characteristic roots (roots of the characteristic equation) also provide qualitative information about the behavior of the variable whose evolution is described by the dynamic equation. How do I know if my roots are complex? When graphing, if […]
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# 101 is what percent of 104?
## (101 is 97.1154 percent of 104)
### 101 is 97.1154 percent of 104. Explanation: What does 101 percent or 101% mean?
Percent (%) is an abbreviation for the Latin “per centum”, which means per hundred or for every hundred. So, 101% means 101 out of every 100.
### Methods to calculate "101 is what percent of 104" with step by step explanation:
#### Method 1: Diagonal multiplication to calculate 101 is what percent of 104.
1. Step 1: For 104, our answer will be 101
2. Step 2: For 100, our answer will be x
3. Step 3: 104*x = 101*100 (In Step 1 and 2 see colored text; Diagonal multiplications will always be equal)
4. Step 4: x = 101*100/104 = 10100/104 = 97.1154
#### Method 2: Same side division to calculate 101 is what percent of 104
1. Step 1: For 104, our answer will be 101
2. Step 2: For 100, our answer will be x
3. Step 3: 101/x = 104/100 (In Step 1 and 2, see colored text; Same side divisions will always be equal)
4. Step 4: x/101 = 100/104
5. Step 5: x = 101*100/104 = 10100/104 = 97.1154
### 101 Percentage example
Percentages express a proportionate part of a total. When a total is not given then it is assumed to be 100. E.g. 101% (read as 101 percent) can also be expressed as 101/100 or 101:100.
Example: If you earn 101% (101 percent) profit then on an investment of \$100 you receive a profit of \$101.
### Percentage sign (%)
The percent (per cent i.e. per hundred) sign % is the standard symbol to indicate a percentage, to signify a number or ratio as a fraction of 100. Related signs include the permille (per thousand) sign ‰ and the permyriad (per ten thousand) sign ‱ (also known as a basis point), which indicate that a number is divided by one thousand or ten thousand, respectively.
### Scholarship programs to learn math
Here are some of the top scholarships available to students who wish to learn math.
### Examples to calculate "What is the percent decrease from X to Y?"
WhatPercentCalculator.com is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.com.
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# Qnt 561 Week 6
1144 Words5 Pages
Answers to Week Two Homework Assignment ( Assignment 1) Chapter 7 Exercise 13 You should choose a sample size of 1000 people. Although the law of averages tells you that the average of 1000 people is very likely to be close to 60 inches -- and thus under 65 inches tall -- here you are winning a prize for each person over 65 inches tall, not when the average of all the people is over 65 inches tall. Having a larger sample will give you more chances at getting the prize. The law of averages pertains to the variability of a sample average -- not variability in individual data values themselves. Here, we are concerned with individual data values. Chapter 7 Exercise 18 No, we can say we are quite confident that the average wage is between \$25…show more content…
(b) One correct answer to this part is as follows: We can say that we are 95 percent confident the average balance for all overdue bills is in the range from \$480 to \$520. Since the standard error SE for the sample mean should equal 100/square root of 100 = 10, the margin of error would be twice this value, or \$20. This means \$480 to \$520 is a 95 percent confidence interval for the average of all overdue balances. Another correct answer to this part is as follows: Instead of the average balance, we can find the percentage of all balances that fall in this interval. When we think of all balances, \$480 is 0.2 standard deviations below the average (480 -500)/100 = - 0.20), and \$520 is 0.2 standard deviations above the average (520 -500)/100 = 0.20). Looking in the Normal table, we see that the area up to z = 0.20 is 58%, and the area up to z = - 0.20 is 42%. The area between them is 16%, so we can say that 16% of all balances fall between \$480 and…show more content…
If they want to cut this by a factor of three to get it down to \$4000, they need to multiply the sample size by 3^2=9, and get a sample size of 25×9= 225. Here is how we can calculate it more directly. We want ME = 4000, and we know ME = 2×SE. Therefore, SE = ME/2 =2000, and also, SE = SD/Square root of sample size. So, 2000 = 30000/Square root of sample size. Solving for the Square root of sample size, we get Square root of sample size = 30000/2000 = 15. Taking its square, the sample size is found as 225. Chapter 9 Exercise 1 No it is not a good defense. If you choose 40 random employees from the corporation, the standard error would equal 6/Square root of 40 = .95 days. The 12 days in this department corresponds to (12-8.2)/.95 = 4 standard errors above the corporation average of 8.2. This is much higher than two or three standard errors, and it appears to be beyond chance variation. Chapter 9 Exercise 3 The p- value tells you how likely it would be to get results at least as extreme as this if there was no difference in the taste and only chance variation was operating. In this problem, p-value of 0.02 means that, if there is no difference in taste, then there is only 2% chance that 70% or more people would declare one drink better than the
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# Row in Math
Home » Math Vocabulary » Row in Math
## What Is a Row in Math?
A row is simply the horizontal arrangement of objects, side by side, from left to right. It could be people, numbers, or others, placed alongside each other.
Have you observed how the desks and chairs are arranged in a classroom? How does the seating arrangement work? Well, students sit in “rows!”
Let’s see some examples.
Example 1: Children sit in 2 rows. Each row has 4 children.
Example 2:
A single row of 5 numbers: 45, 12, 56, 88, 121
## Definition of Row
A row can be defined as an arrangement of objects or data in a horizontal line from left to right.
Example 1: A single row of flowers containing 7 flowers.
Example 2: The donuts in the given image are arranged in 3 rows and each row has four donuts.
Note: A common term that goes along with “row” is “column.” Column refers to vertical arrangements of objects, from top to bottom.
## Finding the Total Number of Items in the Given Rows
To find the number of items in a given row, we multiply the number of rows with the number of items in each row.
For example,
Number of stars in each row = 4
Number of rows $= 3$
So, the total number of stars is $4 \times 3 = 12$
## Applications of Row in Math
A row in math has some significant mathematical applications.
• Solving matrices: Matrix is an arrangement of numbers into rows and columns.
• Spreadsheets: we use these while working on computers using rows and columns structure.
• Arrays: An array is a collection of objects in rows and columns. Arrays represent multiplication and division problems visually. Rows represent the number of groups and columns represent the size of each group.
## Conclusion
In this article, we learned about rows and saw them visually. Rows have many real-life and mathematical applications. Let’s solve some interesting examples.
## Solved Examples
1. How many rows are there in the given picture? How many oranges are there in each row?
Solution:
There is a single row of apples in the given picture. There are 5 apples arranged in a row.
2. How many toy cars are arranged in the given row?
Solution:
There are 5 cars in the given row.
3. Determine the number of pairs of shoes in the given row.
Solution:
There are 4 pairs of shoes in the given row.
4. How many cupcakes are there in the given image?
Solution:
There are 4 rows of cupcakes in the given image.
Number of cupcakes in each row $= 4$
Total number of cupcakes $= 4 \times 4 = 16$
5. How many rows and flowers are there in the given image?
Solution:
Number of roses in one row $= 4$
Number of rows $= 3$
Total roses $= 4 \times 3 = 12$
## Practice Problems
1
### Find the rows in the given figure.
4
1
2
0
CorrectIncorrect
There is one row of lemon slices.
2
### Count the number of jars of smoothies in the given row.
1
6
5
2
CorrectIncorrect
There are 6 jars of smoothies placed in a single row.
3
diagonal
vertical
horizontal
reverse
CorrectIncorrect
|
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# Linear equations
10 CHAPTER
## Here are four examples of equations.
3x 5 17 2(a 3) 18
5(3y 1) 13y 4
16 x 1 x
4
An equation must always have an sign because one side of an equation is always equal to the
other side.
The expression before the equals sign is called the left-hand side (LHS) of the equation.
The expression after the equals sign is called the right-hand side (RHS) of the equation.
3x 5 17
## left-hand side right-hand side
The equation 3x 5 17 can also be written as 17 3x 5
The solution of an equation is the value of the letter (the ‘unknown’) that makes the equation true.
## 10.1 The balance method for solving equations
Some equations can be solved mentally. To solve more complicated equations the balance method
is used.
To keep the balance, whatever you do to the left-hand side you must also do to the right-hand side
of the equation.
Using the equation 3x 5 17 as an example.
Balance Explanation Step
Start
The scales balance because 3x 5 17
both sides are equal.
3x 5 17
LHS RHS
## To keep the balance
3x 5 5 17 5
subtract 5 from both sides
3x 5 5 17 5
LHS RHS
141
CHAPTER 10 Linear equations
## The scales still balance and 3x 12
show that 3x 12
3x 12
LHS RHS
## To keep the balance
3x 3 12 3
divide both sides by 3
3x 3 12 3
LHS RHS
Solution x 4
x 4
LHS RHS
## The solution of the equation 3x 5 17 is x 4
To check this, substitute x 4 into the equation. 3 4 5 17
12 5 17
Example 1
Use the balance method to solve the equation 4x 3 31
Solution 1
4x 3 31
4x 3 3 31 3 Subtract 3 from both sides.
4x 28
Check: substitute x 7,
4x 4 28 4 Divide both sides by 4 4 7 3 31
x7 28 3 31
Example 2
Use the balance method to solve the equation 27 6p 9
Solution 2
27 6p 9
## 27 9 6p 9 9 Add 9 to both sides.
36 6p
Check: substitute p 6,
36 6 6p 6 Divide both sides by 6
27 6 6 9
6 p or p 6 6 p is usually written as p 6 27 36 9
142
10.1 The balance method for solving equations CHAPTER 10
Example 3
Solve the equation 6 7x 11
Solution 3
6 7x 11
6 7x 7x 11 7x Add 7x to both sides so that the sign of the term in x is positive.
6 11 7x
6 11 11 7x 11 Subtract 11 from both sides.
5 7x
5 7x
Divide both sides by 7 then cancel.
7 7
5
x
7
5
x
7
To solve equations which include brackets, one method is to expand the brackets and then use the
balance method.
Example 4
Solve the equation 5(a 3) 18
Solution 4
5(a 3) 18
5a 15 18 Expand the brackets.
## 5a 15 15 18 15 Subtract 15 from both sides.
5a 3
5a 3
Divide both sides by 5 then cancel.
5 5
3
a This solution can also be given as the exact decimal 0.6
5
To solve equations which have letters on both sides, the first step is to rearrange the terms so that the
letter appears only on one side of the equation. It is helpful if the sign of the letter term is positive.
Example 5
Solve the equation 4 d 5d 1
Solution 5
Add d to both sides so that when like terms are collected, d only appears
4 d 5d 1 on one side of the equation and the sign of the term in d is positive.
4 d d 5d 1 d
4 6d 1
3 6d Subtract 1 from both sides.
3 6d
Divide both sides by 6 then cancel.
6 6
1 3
d Simplify
6
2
1
d
2
143
CHAPTER 10 Linear equations
Example 6
Solve the equation 5(3y 2) 13y 4
Solution 6
5(3y 2) 13y 4
2y 6
## y 3 Divide both sides by 2
Exercise 10A
Solve these equations.
1 2d 1 19 2 3e 4 23
3 7g 4 4 4 15 2r 36
5 13 5 4a 6 3(b 2) 15
7 4(c 3) 30 8 4(e 3) 20
9 5(g 6) 2 10 2(2h 1) 22
11 2(3k 1) 26 12 3(4m 3) 27
13 2(n 3) 11 14 4( p 3) 18
15 2(2q 3) 16 16 3(r 2) 0
## 17 3(1 t ) 6 18 5(1 2u) 35
19 12 2(4w 1) 20 20 11 4(x 1)
21 6x 2x 1 22 5y 2y 9
23 3p p 6 24 2x 8 2x
25 y 8 y 26 4 2x 3x
27 3x 1 2x 4 28 2a 4 a 7
29 8x 4 3x 8 30 4(x 1) x 10
31 5(c 3) 3c 1 32 6(x 1) 2x 8
## 39 2(2g 5) 4(1 g) 40 5(2x 3) 2(3 x) 3
144
10.2 Setting up equations CHAPTER 10
## 10.2 Setting up equations
Equations can be used to solve problems.
Example 7
Kevin thinks of a number.
He doubles the number then adds 7
Work out the number that Kevin thinks of.
Solution 7
Let n be the number Kevin thinks of.
## 2n 7 19 Set up the equation by writing 2n 7 equal to Kevin’s answer of 19
2n 12 Solve the equation by subtracting 7 from both sides and dividing both sides by 2
n 6
The number Kevin thinks of is 6
Example 8
All the angles are measured in degrees. x
a Write down an expression, in terms of x, for the sum
of the angles of this triangle.
b By setting up an equation, work out 2x
i the value of x 2x 90
ii the size of the largest angle of this triangle.
Solution 8
a Sum of the angles x 2x 90 2x Add the three angles and collect like terms.
5x 90
b i 5x 90 180 The sum of the angles of a triangle 180°.
## ii Substitute x 18, 2x 90 2 18 90 The largest angle is 2x 90
Size of largest angle 126°
Example 9
Georgina is x years old.
Jessica is 4 years younger than Georgina.
a Write down, in terms of x, an expression for Jessica’s age.
Angela is 3 times as old as Jessica.
b Work out the ages of Georgina and Jessica if Angela is 27 years old.
145
CHAPTER 10 Linear equations
Solution 9
a Jessica’s age is x 4
## Georgina’s age is 13 Georgina is 13 years old and Jessica is 9 years old
Jessica’s age is x 4 13 4 9 since she is 4 years younger than Georgina.
Example 10
The diagram shows a square and a rectangle. 3x 2x 5
The lengths of the edges are given in centimetres.
The perimeter of the square is equal to the perimeter 3x x1
of the rectangle.
a Write down an equation, in terms of x, using this
information.
b By solving your equation, work out the area of the square.
Solution 10
a Perimeter of the square 4 3x 12x
Perimeter of the rectangle
2 (2x 5) 2 (x 1)
4x 10 2x 2
6x 12
12x 6x 12 The perimeters are equal.
## The length of each side of the square is 3x cm 6 cm Substitute x 2 into 3x.
The area of the square 6 cm 6 cm 36 cm2.
Exercise 10B
1 Cath thinks of a number. She multiplies the number by 4 then subtracts 5. Her answer is 27
Work out the number that Cath is thinking of.
2 Bill pays 96 pence for a pen and a pencil. The pen costs five times as much as the pencil.
The pencil costs x pence.
a Write down an expression, in terms of x, for the cost of the pen.
b By setting up an equation, work out i the value of x ii the cost of the pen.
146
10.2 Setting up equations CHAPTER 10
3 a Write down an expression, in terms of x, for the sum of the angles x 30°
## The sum of the angles in a quadrilateral is 360°.
b By setting up an equation, work out
i the value of x
ii the size of the largest angle of this quadrilateral. x 80°
## 4 Eggs are sold in cartons, each containing y eggs.
Viv buys 3 of these cartons of eggs.
One carton has 4 broken eggs in it.
Viv now has just 23 good eggs.
How many eggs does each carton contain?
## 5 The lengths, in centimetres, of the sides of a triangle are 2s 1, 3s and 5s 3
The perimeter of the triangle is 38 cm.
Find the length of the smallest side of this triangle.
## 6 Mansoor is p years of age.
Mansoor’s father is three times as old as Mansoor.
a Write down an expression, in terms of p, for the age of Mansoor’s father.
In 5 years time Mansoor’s father will be 47 years of age.
b Write down an equation in p to show this information.
c Solve your equation to find Mansoor’s present age.
## 7 A bag contains b yellow balls, 3b 2 red balls and b 8 blue balls.
a Write an expression, in terms of b, for the total number of balls in the bag.
b The total number of balls in the bag is 45
i By forming an equation find the value of b.
ii How many blue balls are in the bag?
iii How many more red balls are there than blue balls in the bag?
8 A bag contains red sweets, yellow sweets and green sweets only.
There are 2t red sweets, 4t 5 yellow sweets and 14 3t green sweets.
There are a total of 21 sweets in the bag.
a By setting up an equation, work out the number of red sweets in the bag.
b Work out the number of yellow sweets in the bag.
## 9 Samantha works for x hours each week for 3 weeks.
In the fourth week she works for an extra 10 hours.
She works a total of 150 hours during these 4 weeks.
By setting up an equation, work out
a the value of x
b the number of hours that Samantha works in the fourth week.
## 10 Four of the angles in a pentagon are each (x 70)°.
The fifth angle is (x 100)°.
a Write down an equation, in terms of x, using this information.
b By solving your equation in part a, work out the value of x.
147
CHAPTER 10 Linear equations
11 The width of a rectangle is x cm. The length of the rectangle is 6 cm longer than its width.
a Write down an expression, in terms of x, for the length of the rectangle.
The perimeter of the rectangle is 40 cm.
b By setting up an equation, work out
i the value of x ii the length of the rectangle.
## 12 The diagram shows an L-shape. 2
The lengths of the edges are given in centimetres.
The area of the shape is 44 cm2.
a Write down an equation, in terms of x, using this information. x3 x1
i the value of x ii the perimeter of the shape.
## 13 The diagram shows an equilateral triangle and a square. 4k
The lengths of the edges are given in centimetres.
The perimeter of the equilateral triangle is equal to the 5k 1 5k 1 4k
perimeter of the square.
a Write down an equation, in terms of k, using this 5k 1
information.
b By solving your equation work out the area of the square.
## 14 Brian, Clare and Daniel share a sum of money.
Brian’s share is £4w
Clare’s share is £(2w 1)
Daniel’s share is £(8w 8)
Daniel’s share is the same as the sum of Brian’s share and Clare’s share.
a Write down an equation, in terms of w, using this information.
b Work out, in £, the total sum of money shared by Brian, Clare and Daniel.
## 15 The diagram shows a rectangle. 2x 3
Work out the value of x.
5 5
30 4x
## 10.3 Solving equations with fractional terms
1 x3 y
In algebra, the expressions (x 3) 4 and y are usually written as and
2 4 2
Example 11
4
Solve the equation 8
p
Solution 11
4 4
8 so p 8 p Multiply both sides by p then cancel.
p p
4 8p
4 8p
Divide both sides by 8
8 8
1
p
2
148
10.3 Solving equations with fractional terms CHAPTER 10
Example 12 Example 13
3(q 5) 16 x
Solve the equation 6 Solve the equation 1 x
2 4
Solution 12 Solution 13
3(q 5) 16 x
6 1 x
2 4
Multiply both sides
16 x
3(q 5)
2 6 2 Multiply
2
both sides by 2
then cancel.
4
4 4 (1 x) by 4, remember to
put brackets around
the 1 x.
16 x
3(q 5) 12
4 4 (1 x) Cancel the 4s on the LHS.
4
3q 15 12 Expand the brackets. 16 x 4 4x Expand the brackets.
## Subtract 15 from Add 4x to both sides
3q 3 both sides. 4x x 4 16 and subtract 16 from
both sides.
q 1 Divide both sides by 3
3x 12 Simplify.
## x 4 Divide both sides by 3
Example 14
2x 1 x 5 5
Solve the equation
2 3 4
Solution 14
2x 1 x 5 5
2 3 4
2x 1 x5 5
Multiply both sides by 12 since 12 is the lowest common
12 12 12 multiple (LCM) of 2, 3 and 4
2 3 4
6 2x 1 4 x5 3 5
12 12 12
2 3 4 Cancel.
6(2x 1) 4(x 5) 15
## 8x 1 Subtract 14 from both sides.
1
x Divide both sides by 8
8
149
CHAPTER 10 Linear equations
Exercise 10C
1 Solve these equations.
x 1 12 x
a 4 b 2 c 2 d 2 5
3 x n 3
x x1 y2 p1
e 5 1 f 2 g 1 h 2
4 3 4 6
2q 1 1y 4p 9 x
i 5 j 1 k 1 l x 3
3 4 2 2
5y 3x 7x 9x
m y 6 n 4 x o 1 x p 3 x
2 5 3 5
3x a a 3y 6 5 2y
q 2 3x r 1 s 6
5 2 3 10 5
x 1 4x 1 5 3x 4 2x 1 8 x
t u
2 3 12 2 5 3
2 Saika thinks of a number. She halves the number then adds 9 Her answer is twice the number
she is thinking of.
a If n is the number that Saika is thinking of, write down an equation in terms of n.
b Solve the equation to find n.
Find the number.
## 4 ABC is an isosceles triangle with AB AC.
Angle ABC 23 ( p 20)° and angle ACB (100 p)°.
a Write down an equation in terms of p.
b Find the value of p.
c Find the size of angle BAC.
5 Terry cycles for 14 hour at an average speed of x kilometres per hour.
He then cycles for 13 hour at an average speed of (x 10) kilometres per hour.
He cycles a total distance of 15 kilometres. Work out the value of x.
## 10.4 Simultaneous linear equations
The diagram shows red and green cylinders.
The height of each red cylinder is y cm. x
The height of each green cylinder is x cm.
x
The total height of 3 green cylinders and 1 red
16 x
cylinder is 16 cm.
The total height of 1 red cylinder and 1 green cylinder 10
is 10 cm. y
This information can be written as
3x y 16
x y 10
## These are known as simultaneous equations.
It is possible to use algebra to find the value of x and the value of y which satisfy both of these
equations simultaneously (at the same time). This process is called solving simultaneous equations.
150
10.4 Simultaneous linear equations CHAPTER 10
Example 15
Solve the simultaneous equations
3x y 16
x y 10
Solution 15
3x y 16 (1)
Label the equations as (1) and (2).
x y 10 (2)
Since y and y are exactly the same, y can be
(1) (2) eliminated by subtracting equation (2) from equation (1).
3x y 16 (1) 3x x 2x
x y 10 (2) y y 0
2x 0 6 16 10 6
## x 3, y 7 Always give the values for both letters.
This shows that the height of each green cylinder is 3 cm and the height of each red cylinder is 7 cm.
Example 16
Solve the simultaneous equations
3x 2y 7
5x 4y 8
Solution 16
3x 2y 7 (1) Label the equations as (1) and (2).
5x 4y 8 (2)
Neither the x terms nor the y terms are the same. To get the same number of ys
(1) 2 gives 6x 4y 14 in each equation multiply equation (1) by 2
5x 4y 8
Eliminate the y terms by adding the two equations.
6x 4y 14
6x 5x 11x
5x 4y 8 4y 4y 0
11x 0y 22 14 8 22
## When x 2, 6 2y 7 Substitute x 2 into equation (1).
2y 1 so y 12
## x 2, y 12 Always give the values for both letters.
151
CHAPTER 10 Linear equations
Example 17
Solve the simultaneous equations
4y 3x 29
y 2x 11
Solution 17
4y 3x 29 (1)
Label the equations as (1) and (2).
y 2x 11 (2)
## When x 3 , y 6 11 Substitute x 3 into equation (2).
So y 5
x 3, y 5
Sometimes both equations have to be multiplied by numbers before one of the letters can be eliminated.
Example 18
Solve the simultaneous equations
2x 3 y 7
3x 4y 10
Solution 18
2x 3y 7 (1)
Label the equations as (1) and (2).
3x 4y 10 (2)
Neither the x terms nor the y terms are the same. To get the
(1) 3 gives 6x 9y 21 (3) x terms equal multiply equation (1) by 3 and equation (2) by 2.
(2) 2 gives 6x 8y 20 (4) Label the new equations as (3) and (4).
## 6x 8y 20 (4) x can be eliminated by subtracting equation (3) from equation (4).
6x 9y 21 (3) 6x 6x 0
8y (9y) 8y 9y y
0x 9y 1
20 21 1
so y 1
When y 1, 2x 3 1 7 Substitute y 1 into equation (1).
2x 3 7
2x 4 so x 2
x 2, y 1
152
10.5 Setting up simultaneous linear equations CHAPTER 10
Exercise 10D
Solve these simultaneous equations.
1 xy5 2 2x y 5 3 5x 2y 3 4 xy8
xy1 xy3 3x 2y 5 xy1
## 5 xy2 6 xy5 7 xy4 8 3x y 10
xy3 xyx1 xyx2 3x y 2 x
9 3x 5y 11 10 2x 7y 5 11 5x 3y 4 12 7x y 5
2x y 3 x 3y 3 xy4 6x 2y 10
13 3x 2y 8 14 3x 2y 7 15 2x 5y 16 16 2x 3y 1
5x 6y 16 4x 3y 2 5x 2y 11 3x 2y 8
17 4x 3y 3 18 6x 7y 9 19 5x 3y 7 20 5x 3y 14
6x 5y 7 8x 9y 12 3x 5y 6 7x 5y 20
## 10.5 Setting up simultaneous linear equations
Simultaneous equations can be used to solve problems.
Example 19
## 6 pencils and 2 crayons cost a total of £2.70
5 pencils and 3 crayons cost a total of £2.45
Work out the cost of a pencil and the cost of a crayon.
Solution 19
Let the cost of 1 pencil be a pence.
To avoid working with decimals, change the costs to pence.
Let the cost of 1 crayon be b pence.
Write 6 pencils and 2 crayons cost a total of 270 pence as equation (1).
6a 2b 270 (1)
Write 5 pencils and 3 crayons cost a total of 245 pence as equation (2).
5a 3b 245 (2) Solve the simultaneous equations.
(1) 3 gives 18a 6b 810 (3) Neither the a terms nor the b terms are the same. To get the b terms
(2) 2 gives 10a 6b 490 (4) equal multiply equation (1) by 3 and equation (2) by 2
## 18a 6b 810 (3)
10a 6b 490 (4) Eliminate the b terms by subtracting equation (4) from equation (3).
8a 0b 320
8a 320 so a 40
## When a 40, 240 2b 270 Substitute a 40 into equation (1).
2b 30 so b 15
## The cost of a pencil is 40p.
The cost of a crayon is 15p.
153
CHAPTER 10 Linear equations
Exercise 10E
1 In my pocket there are x one-pence coins and y two-pence coins.
There are 8 coins altogether and their total value is 11 pence. Find x and y.
2 The sum of two numbers is 15 The difference of the two numbers is 8 Find the larger number.
3 The diagram shows a rectangle. 18 2b
All sides are measured in centimetres.
a Show that 5a 2b 18 3a 5b 17
b Write down another equation in terms of a and b.
c Solve the two equations simultaneously to find a and b. 5a
d Hence find the area of the rectangle.
4 Tony and June answered Number of Number of Total points
in a maths quiz.
Tony 20 10 90
The table shows some
information. June 15 15 60
Each correct answer scored x points.
Each wrong answer scored y points.
a Use the information in the table for Tony to show that 2x y 9
b Use the information in the table for June to find another equation in x and y.
c Solve the two equations simultaneously to find x and y.
d James gave only 5 correct answers in the quiz. Find the total number of points he scored.
5 The lengths of the sides of an equilateral triangle are (3x 2) cm, (2y x) cm and (y 3) cm.
a Find x and y. b Find the perimeter of the triangle.
6 Each large filing cabinet contains 100 files and has 4 drawers.
Each small filing cabinet contains 80 files and has 3 drawers.
In an office, there are x large filing cabinets and y small filing
cabinets.
The total number of files in these filing cabinets is 1560 and the
total number of drawers is 61
Work out the number of large filing cabinets and the number of
small filing cabinets in this office.
Chapter summary
You should now be able to:
solve an equation to find the value of the unknown letter, for example x 2 is the solution
of the equation x 3 5
solve equations by the balance method in which whatever is done to the right-hand side of
the equation must also be done to the left-hand side of the equation
solve equations which have brackets by expanding the brackets, collecting any like terms and
then using the balance method
solve a problem by setting up an equation and solving it algebraically
set up and solve equations with fractional terms by multiplying both sides of the equation by
the lowest common multiple of the denominators
set up and solve simultaneous linear equations.
154
Chapter 10 review questions CHAPTER 10
## Chapter 10 review questions
x
1 a Solve 3y 2 17 b Solve 5
3
2 Solve
a 2x 9 4x 6 b 5(x 2) 20 c 21 3(2x 11)
## 4 a Solve 7x 18 74 b Solve 4(2y 5) 32 c Solve 5p 7 3(4 p)
(1387 November 2003)
## 5 a Solve 4(2x 1) 2(3 x) b Factorise fully 2p2 4pq c Factorise x2 7x 6
(1387 November 2004)
6 Solve 5(x 2) 8 7x
## 7 Fred is x years old.
His sister, Mary, is 4 years older than Fred.
a i Write down an expression, in terms of x, for Mary’s age.
Sarfraz is twice as old as Fred.
ii Write down an expression, in terms of x, for the total of Fred’s age, Mary’s age and
Sarfraz’s age.
The total of Fred’s age, Mary’s age and Sarfraz’s age is 64 years.
b Form an equation and solve it to find Fred’s age.
## 8 These four blocks A, B, C and D have a total mass of 62 grams.
The mass, in grams, of each block is shown on the diagram.
Diagram NOT
accurately drawn
x x3 2x 5
A B C D
a Express this infomation as an equation in terms of x.
b Solve your equation and write down the masses of blocks A, B and C. (1385 November 2001)
9 ABCD is a quadrilateral. A
Work out the size of the largest angle in the quadrilateral. 100° D
2x°
47° x°
B C
(1387 June 2004)
## 10 The lengths, in cm, of the sides of the triangle are
3(x 3), 4x 1 and 2x 5 2x 5
3(x 3)
a Write down, in terms of x, an expression for the
perimeter of the triangle.
4x 1
The perimeter of the triangle is 49 cm.
b Work out the value of x. (1388 June 2004)
155
CHAPTER 10 Linear equations
x 3x
11 a Solve 6(2x 3) 2(4x 7) b Solve 2 7 c Solve x 3
4 5
5x
12 Solve 2x 1 (1388 March 2004)
3
40 x
13 a Solve 20y 16 18y 9 b Solve 4 x (1387 June 2004)
3
5x 1
14 Solve 7(x 2) (1388 January 2004)
2
7x 4
15 Solve 5(x 8) (1388 March 2005)
2
x
16 Solve 5 3(x 2) (1388 March 2003)
3
17 ABC is an isosceles triangle. A
AB AC
Angle A x° x° Diagram NOT
accurately drawn
a Find an expression, in terms of x, for the size of angle B.
b Solve the simultaneous equations
3p q 11
pq3 B C (1387 November 2004)
## 18 Solve the simultaneous equations
2x y 4
5x y 17 (1388 March 2003)
19 Solve x 2y 4
3x 4y 7 (1387 June 2005)
## 20 Solve the simultaneous equations
6x 2y 33
4x 3y 9 (1387 June 2004)
## 21 Solve the simultaneous equations
3x 4y 11
5x 6y 12 (1388 January 2003)
5p 4q 4
2p 3q 0.5
## 23 A company makes compact discs (CDs).
The total cost, P pounds, of making n compact discs is given by the formula
P a bn
where a and b are constants.
The cost of making 1000 compact discs is £58 000
The cost of making 2000 compact discs is £64 000
a Calculate the values of a and b.
The company sells the compact discs at £10 each.
The company does not want to make a loss.
b Work out the minimum number of compact discs the company must sell. (1385 June 1998)
156
|
Radicals allow us to reverse exponent operations. For example, squaring a number such as 5, or (-5) will give us 25. If we take the square root of 25, we get back to 5 or (-5). The same process occurs when we look at higher level roots. A cube root will undo cubing a number. A fourth root will undo raising a number to the fourth power.
Test Objectives
• Demonstrate a general understanding of radicals
• Demonstrate the ability to evaluate radical expressions
• Demonstrate the ability to work with fractional exponents
#1:
Instructions: Simplify each.
a) $$\sqrt[8]{1}$$
b) $$\sqrt[7]{-128}$$
c) $$\sqrt[4]{-16}$$
d) $$-\sqrt{225}$$
e) $$\sqrt[5]{-243}$$
#2:
Instructions: Simplify each.
a) $$\sqrt{14^2}$$
b) $$\sqrt{(-19)^2}$$
c) $$-\sqrt[4]{(-13)^4}$$
d) $$\sqrt[3]{(-5)^3}$$
e) $$\sqrt[6]{(-243)^6}$$
#3:
Instructions: Simplify each.
a) $$(-27)^\frac{1}{3}$$
b) $$(-81)^\frac{1}{2}$$
c) $$-(225)^\frac{1}{2}$$
d) $$(625)^\frac{1}{4}$$
e) $$(-64)^\frac{1}{3}$$
#4:
Instructions: Simplify each.
a) $$\frac{x^{-1}z^2}{z^{-1}(x^{\frac{1}{2}}y^0z^{\frac{3}{2}})^{-2}}$$
#5:
Instructions: Simplify each.
a) $$\frac{(x^{-\frac{5}{3}}x^{-\frac{1}{2}})^{\frac{2}{3}}}{x^{-\frac{5}{3}}}$$
Written Solutions:
#1:
Solutions:
a) $$1$$
b) $$-2$$
c) $$not \hspace{.5em}real$$
d) $$-15$$
e) $$-3$$
#2:
Solutions:
a) $$14$$
b) $$19$$
c) $$-13$$
d) $$-5$$
e) $$243$$
#3:
Solutions:
a) $$-3$$
b) $$not \hspace{.5em}real$$
c) $$-15$$
d) $$5$$
e) $$-4$$
#4:
Solutions:
a) $$z^6$$
#5:
Solutions:
a) $$x^{\frac{2}{9}}$$
|
## Section2.4Equations and Inequalities as True/False Statements
###### ObjectivesPCC Course Content and Outcome Guide
This section introduces the concepts of algebraic equations and inequalities, and what it means for a number to be a solution to an equation or inequality.
### Subsection2.4.1Equations, Inequalities, and Solutions
An equation is two algebraic expressions with an equals sign between them. The two expressions can be relatively simple or more complicated:
A relatively simple equation:
\begin{equation*} x+1=2 \end{equation*}
A more complicated equation:
\begin{equation*} \left(x^2+y^2-1\right)^3=x^2y^3 \end{equation*}
An inequality is quite similar, but the sign between the expressions is one of these: $\lt\text{,}$ $\leq\text{,}$ $\gt\text{,}$ $\geq\text{,}$ or $\neq\text{.}$
A relatively simple inequality:
\begin{equation*} x\geq15 \end{equation*}
A more complicated inequality:
\begin{equation*} x^2+y^2\lt1 \end{equation*}
A linear equation in one variable can be written in the form $ax+b=0\text{,}$ where $a, b$ are real numbers, and $a\ne0\text{.}$ The variable doesn't have to be $x\text{.}$ The variable cannot have an exponent other than $1$ ($x=x^1$), and the variable cannot be inside a root symbol (square root, cube root, etc.) or in a denominator.
The following are some linear equations in one variable:
\begin{align*} 4-y\amp=5 \amp 4-z\amp=5z \amp 0\amp=\frac{1}{2}p \end{align*}
\begin{align*} 3-2(q+2)\amp=10 \amp \sqrt{2}\cdot r+3\amp=10 \amp \frac{s}{2}+3\amp=5 \end{align*}
(Note that $r$ is outside the square root symbol.) We will see in later sections that all equations above can be converted into the form $ax+b=0\text{.}$
The following are some non-linear equations:
\begin{align*} 1+2=3 \amp\amp\amp \text{(There is no variable.)}\\ 4-2y^2=5 \amp\amp\amp \text{(The exponent of $y$ is not $1$.)}\\ \sqrt{2r}+3=10 \amp\amp\amp \text{($r$ is inside the square root.)}\\ \frac{2}{s}+3=5 \amp\amp\amp \text{($s$ is in a denominator.)} \end{align*}
This chapter focuses on linear equations in one variable. We will study other types of equations in later chapters.
The simplest equations and inequalities have numbers and no variables. When this happens, the equation is either true or false. The following equations and inequalities are true statements:
\begin{align*} 2\amp=2\amp-4\amp=-4\amp2\amp\gt1\amp-2\amp\lt-1\amp3\amp\ge3 \end{align*}
The following equations and inequalities are false statements:
\begin{align*} 2\amp=1\amp-4\amp=4\amp2\amp\lt1\amp-2\amp\ge-1\amp0\neq0 \end{align*}
When equations and inequalities have variables, we can consider substituting values in for the variables. If replacing a variable with a number makes an equation or inequality true, then that number is called a solution to the equation.
###### Example2.4.2A Solution
Consider the equation $y+2=3\text{,}$ which has only one variable, $y\text{.}$ If we substitute in $1$ for $y$ and then simplify:
\begin{align*} y+2\amp=3\\ \substitute{1}+2\amp\stackrel{?}{=}3\\ 3\amp\stackrel{\checkmark}{=}3 \end{align*}
we get a true equation. So we say that $1$ is a solution to $y+2=3\text{.}$ Notice that we used a question mark at first because we are unsure if the equation is true or false until the end.
If replacing a variable with a value makes a false equation or inequality, that number is not a solution.
###### Example2.4.3Not a Solution
Consider the inequality $x+4\gt 5\text{,}$ which has only one variable, $x\text{.}$ If we substitute in $0$ for $x$ and then simplify:
\begin{align*} x+4\amp\gt 5\\ \substitute{0}+4\amp\stackrel{?}{\gt}5\\ 4\amp\stackrel{\text{no}}{\gt}5 \end{align*}
we get a false equation. So we say that $0$ is not a solution to $x+4\gt 5\text{.}$
### Subsection2.4.2Checking Possible Solutions
Given an equation or an inequality (with one variable), checking if some particular number is a solution is just a matter of replacing the value of the variable with the specified number and determining if the resulting equation/inequality is true or false. This may involve some amount of arithmetic simplification.
###### Example2.4.4
Is $8$ a solution to $x^2-5x=\sqrt{2x}+20\text{?}$
To find out, substitute in $8$ for $x$ and see what happens.
\begin{align*} x^2-5x\amp=\sqrt{2x}+20\\ \substitute{8}^2-5(\substitute{8})\amp\stackrel{?}{=}\sqrt{2(\substitute{8})}+20\\ \highlight{64}-5(8)\amp\stackrel{?}{=}\sqrt{\highlight{16}}+20\\ 64-\highlight{40}\amp\stackrel{?}{=}\highlight{4}+20\\ \highlight{24}\amp\stackrel{\checkmark}{=}\highlight{24} \end{align*}
So yes, $8$ is a solution to $x^2-5x=\sqrt{2x}+20\text{.}$
###### Example2.4.5
Is $-5$ a solution to $\sqrt{169-y^2}=y^2-2y\text{?}$
To find out, substitute in $-5$ for $y$ and see what happens.
\begin{align*} \sqrt{169-y^2}\amp=y^2-2y\\ \sqrt{169-\substitute{(-5)}^2}\amp\stackrel{?}{=}\substitute{(-5)}^2-2(\substitute{-5})\\ \sqrt{169-\highlight{25}}\amp\stackrel{?}{=}\highlight{25}-2(-5)\\ \sqrt{\highlight{144}}\amp\stackrel{?}{=}25-(\highlight{-10})\\ \highlight{12}\amp\stackrel{\text{no}}{=}\highlight{35} \end{align*}
So no, $-5$ is not a solution to $\sqrt{169-y^2}=y^2-2y\text{.}$
But is $-5$ a solution to the inequality $\sqrt{169-y^2}\leq y^2-2y\text{?}$ Yes, because substituting $-5$ in for $y$ would give you
\begin{equation*} 12\leq35\text{,} \end{equation*}
which is true.
###### Example2.4.11Cylinder Volume
A cylinder's volume is related to its radius and its height by:
\begin{equation*} V=\pi r^2h\text{,} \end{equation*}
where $V$ is the volume, $r$ is the base's radius, and $h$ is the height. If we know the volume is 96$\pi$ cm3 and the radius is 4 cm, then we have:
\begin{equation*} 96\pi=16\pi h \end{equation*}
Is 4 cm the height of the cylinder? In other words, is $4$ a solution to $96\pi=16\pi h\text{?}$ We will substitute $h$ in the equation with $4$ to check:
\begin{align*} 96\pi\amp=16\pi h\\ 96\pi\amp\stackrel{?}{=}16\pi \cdot\substitute{4}\\ 96\pi\amp\stackrel{\text{no}}{=}64\pi \end{align*}
Since $96\pi=64\pi$ is false, $h=4$ does not satisfy the equation $96\pi=16\pi h\text{.}$
Next, we will try $h=6\text{:}$
\begin{align*} 96\pi\amp=16\pi h\\ 96\pi\amp\stackrel{?}{=}16\pi \cdot\substitute{6}\\ 96\pi\amp\stackrel{\checkmark}{=}96\pi \end{align*}
When $h=6\text{,}$ the equation $96\pi=16\pi h$ is true. This tells us that $6$ is a solution to $96\pi=16\pi h\text{.}$
###### Remark2.4.13
Note that we did not approximate $\pi$ with $3.14$ or any other approximation. We often leave $\pi$ as $\pi$ throughout our calculations. If we need to round, we do so as a final step.
###### Example2.4.14
Jaylen has budgeted a maximum of $\300$ for an appliance repair. The total cost of the repair can be modeled by $89+110(h-0.25)\text{,}$ where $\89$ is the initial cost and $\110$ is the hourly labor charge after the first quarter hour. Is $2$ hours a solution for $h$ in the inequality $89+110(h-0.25)\le 300\text{?}$
To determine if $h=2$ satisfies the inequality, we will replace $h$ with $2$ and check if the statement is true:
\begin{align*} 89+110(h-0.25)\amp\le 300\\ 89+110(\substitute{2}-0.25)\amp\stackrel{?}{\le} 300\\ 89+110(1.75)\amp\stackrel{?}{\le} 300\\ 89+192.5\amp\stackrel{?}{\le} 300\\ 281.5\amp\stackrel{\checkmark}{\le} 300 \end{align*}
Thus, $2$ hours is a solution for $h$ in the inequality $89+110(h-0.25)\le 300\text{.}$ In context, this means that Jaylen would stay within their $\300$ budget if $2$ hours of labor were performed.
### Subsection2.4.3Exercises
###### 1
Evaluate ${-9-x}$ for $x = -4\text{.}$
###### 2
Evaluate ${5-x}$ for $x = -2\text{.}$
###### 3
Evaluate ${-2x+7}$ for $x = 1\text{.}$
###### 4
Evaluate ${-10x-5}$ for $x = 3\text{.}$
###### 5
Evaluate ${-4\!\left(r+2\right)}$ for $r = 4\text{.}$
###### 6
Evaluate ${-10\!\left(t+9\right)}$ for $t = -3\text{.}$
###### 7
Evaluate the expression $\displaystyle \frac{1}{3} \big( x + 4 \big)^2 - 2$ when $x = -7\text{.}$
###### 8
Evaluate the expression $\displaystyle \frac{1}{4} \big( x + 1 \big)^2 - 7$ when $x = -5\text{.}$
###### 9
Evaluate the expression $-16t^{2}+64t+128$ when $t=-3\text{.}$
###### 10
Evaluate the expression $-16t^{2}+64t+128$ when $t=-5\text{.}$
###### 11
Are the equations below linear equations in one variable?
1. $\sqrt{9-5.5r}=7$
• is
• is not
a linear equation in one variable.
2. $2-3r^{2}=18$
• is
• is not
a linear equation in one variable.
3. $x-6q^{2}=27$
• is
• is not
a linear equation in one variable.
4. $2\pi r=2\pi$
• is
• is not
a linear equation in one variable.
5. $8.59y=-9$
• is
• is not
a linear equation in one variable.
6. $12-3z=4$
• is
• is not
a linear equation in one variable.
###### 12
Are the equations below linear equations in one variable?
1. $-5-4y^{2}=-21$
• is
• is not
a linear equation in one variable.
2. $\sqrt{1-4.7x}=2$
• is
• is not
a linear equation in one variable.
3. $9q-y^{2}=1$
• is
• is not
a linear equation in one variable.
4. $2\pi r=12\pi$
• is
• is not
a linear equation in one variable.
5. $-r-11=3$
• is
• is not
a linear equation in one variable.
6. $6.6y=6$
• is
• is not
a linear equation in one variable.
###### 13
Are the equations below linear equations in one variable?
1. $4.61q=14$
• is
• is not
a linear equation in one variable.
2. $5qVx=-42$
• is
• is not
a linear equation in one variable.
3. $V^{2}+y^{2}=-68$
• is
• is not
a linear equation in one variable.
4. $\pi r^{2}=48\pi$
• is
• is not
a linear equation in one variable.
5. $7-2r=17$
• is
• is not
a linear equation in one variable.
6. $r\sqrt{12}=-84$
• is
• is not
a linear equation in one variable.
###### 14
Are the equations below linear equations in one variable?
1. $q\sqrt{24}=63$
• is
• is not
a linear equation in one variable.
2. $\pi r^{2}=35\pi$
• is
• is not
a linear equation in one variable.
3. $-2.96x=-22$
• is
• is not
a linear equation in one variable.
4. $-8prz=52$
• is
• is not
a linear equation in one variable.
5. $15x-4=28$
• is
• is not
a linear equation in one variable.
6. $q^{2}+y^{2}=4$
• is
• is not
a linear equation in one variable.
###### 15
Are the inequalities below linear inequalities in one variable?
1. $4p^{2}-y\gt-51$
• is
• is not
a linear inequality in one variable.
2. $-1\gt1-5p$
• is
• is not
a linear inequality in one variable.
3. $-3z^{2}-7y^{2}\lt1$
• is
• is not
a linear inequality in one variable.
###### 16
Are the inequalities below linear inequalities in one variable?
1. $3x^{2}+6z^{2}\gt1$
• is
• is not
a linear inequality in one variable.
2. $9V^{2}+6p\gt51$
• is
• is not
a linear inequality in one variable.
3. $4\geq-5-3y$
• is
• is not
a linear inequality in one variable.
###### 17
Are the inequalities below linear inequalities in one variable?
1. $-5.8x\gt96$
• is
• is not
a linear inequality in one variable.
2. $-1\gt-4878r-2301p$
• is
• is not
a linear inequality in one variable.
3. $\sqrt{9p}-9\leq-6$
• is
• is not
a linear inequality in one variable.
###### 18
Are the inequalities below linear inequalities in one variable?
1. $-6y\lt-43$
• is
• is not
a linear inequality in one variable.
2. $\sqrt{9r}+6\lt-6$
• is
• is not
a linear inequality in one variable.
3. $198\leq4182y-9693x$
• is
• is not
a linear inequality in one variable.
###### 19
Is $-3$ a solution for $x$ in the equation ${x+7} = {3}\text{?}$
• Yes
• No
###### 20
Is $2$ a solution for $x$ in the equation ${x+9} = {7}\text{?}$
• Yes
• No
###### 21
Is $-9$ a solution for $r$ in the equation ${-3-r} = {5}\text{?}$
• Yes
• No
###### 22
Is $6$ a solution for $r$ in the equation ${5-r} = {-1}\text{?}$
• Yes
• No
###### 23
Is $-1$ a solution for $r$ in the equation ${-7r-8} = {-1}\text{?}$
• Yes
• No
###### 24
Is $-9$ a solution for $t$ in the equation ${2t+8} = {-10}\text{?}$
• Yes
• No
###### 25
Is $5$ a solution for $t$ in the equation ${-10t+2} = {-9t-3}\text{?}$
• Yes
• No
###### 26
Is $2$ a solution for $x$ in the equation ${-2x-4} = {-9x-18}\text{?}$
• Yes
• No
###### 27
Is $-9$ a solution for $x$ in the equation ${9\!\left(x-11\right)} = {20x}\text{?}$
• Yes
• No
###### 28
Is $3$ a solution for $y$ in the equation ${2\!\left(y+14\right)} = {9y}\text{?}$
• Yes
• No
###### 29
Is $-3$ a solution for $y$ in the equation ${-3\!\left(y-13\right)} = {16\!\left(y+6\right)}\text{?}$
• Yes
• No
###### 30
Is $-10$ a solution for $r$ in the equation ${19\!\left(r+1\right)} = {9\!\left(r-9\right)}\text{?}$
• Yes
• No
###### 31
Is ${{\frac{7}{10}}}$ a solution for $x$ in the equation ${10x-1} = -8\text{?}$
• Yes
• No
###### 32
Is ${{\frac{14}{5}}}$ a solution for $x$ in the equation ${5x-4} = 9\text{?}$
• Yes
• No
###### 33
Is ${4}$ a solution for $t$ in the equation ${-{\frac{10}{3}}t - {\frac{3}{10}}} = {-{\frac{17}{15}}}\text{?}$
• Yes
• No
###### 34
Is ${-{\frac{9}{2}}}$ a solution for $t$ in the equation ${{\frac{2}{5}}t - {\frac{9}{10}}} = {-{\frac{27}{10}}}\text{?}$
• Yes
• No
###### 35
Decide whether each value is a solution to the given inequality.
$-2 x +9 > 7$
1. $x=-5$
• is
• is not
a solution.
2. $x=6$
• is
• is not
a solution.
3. $x=0$
• is
• is not
a solution.
4. $x=1$
• is
• is not
a solution.
###### 36
Decide whether each value is a solution to the given inequality.
$2 x - 5 > 1$
1. $x=3$
• is
• is not
a solution.
2. $x=0$
• is
• is not
a solution.
3. $x=13$
• is
• is not
a solution.
4. $x=2$
• is
• is not
a solution.
###### 37
Decide whether each value is a solution to the given inequality.
$3 x - 8 \ge -5$
1. $x=0$
• is
• is not
a solution.
2. $x=8$
• is
• is not
a solution.
3. $x=1$
• is
• is not
a solution.
4. $x=-5$
• is
• is not
a solution.
###### 38
Decide whether each value is a solution to the given inequality.
$-3 x +19 \ge 10$
1. $x=2$
• is
• is not
a solution.
2. $x=0$
• is
• is not
a solution.
3. $x=3$
• is
• is not
a solution.
4. $x=13$
• is
• is not
a solution.
###### 39
Decide whether each value is a solution to the given inequality.
$4 x - 16 \le 4$
1. $x=0$
• is
• is not
a solution.
2. $x=10$
• is
• is not
a solution.
3. $x=5$
• is
• is not
a solution.
4. $x=3$
• is
• is not
a solution.
###### 40
Decide whether each value is a solution to the given inequality.
$4 x - 10 \le -2$
1. $x=1$
• is
• is not
a solution.
2. $x=2$
• is
• is not
a solution.
3. $x=12$
• is
• is not
a solution.
4. $x=0$
• is
• is not
a solution.
###### 41
A triangle’s area is $99$ square meters. Its height is $11$ meters. Suppose we wanted to find how long is the triangle’s base. A triangle’s area formula is
\begin{equation*} A=\frac{1}{2}bh \end{equation*}
where $A$ stands for area, $b$ for base and $h$ for height. If we let $b$ be the triangle’s base, in meters, we can solve this problem using the equation:
\begin{equation*} {99}=\frac{1}{2}(b)(11) \end{equation*}
Check whether $36$ is a solution for $b$ of this equation.
• Yes
• No
###### 42
A triangle’s area is $95$ square meters. Its height is $10$ meters. Suppose we wanted to find how long is the triangle’s base. A triangle’s area formula is
\begin{equation*} A=\frac{1}{2}bh \end{equation*}
where $A$ stands for area, $b$ for base and $h$ for height. If we let $b$ be the triangle’s base, in meters, we can solve this problem using the equation:
\begin{equation*} {95}=\frac{1}{2}(b)(10) \end{equation*}
Check whether $38$ is a solution for $b$ of this equation.
• Yes
• No
###### 43
When a plant was purchased, it was $2$ inches tall. It grows $1$ inches per day. How many days later will the plant be $17$ inches tall?
Assume the plant will be $17$ inches tall $d$ days later. We can solve this problem using the equation:
\begin{equation*} 1 d+2=17 \end{equation*}
Check whether $17$ is a solution for $d$ of this equation.
• Yes
• No
###### 44
When a plant was purchased, it was $1.2$ inches tall. It grows $0.2$ inches per day. How many days later will the plant be $5.2$ inches tall?
Assume the plant will be $5.2$ inches tall $d$ days later. We can solve this problem using the equation:
\begin{equation*} 0.2 d+1.2=5.2 \end{equation*}
Check whether $21$ is a solution for $d$ of this equation.
• Yes
• No
###### 45
A water tank has $163$ gallons of water in it, and it is being drained at the rate of $7$ gallons per minute. After how many minutes will there be $37$ gallons of water left?
Assume the tank will have $37$ gallons of water after $m$ minutes. We can solve this problem using the equation:
\begin{equation*} 163-7 m=37 \end{equation*}
Check whether $19$ is a solution for $m$ of this equation.
• Yes
• No
###### 46
A water tank has $171$ gallons of water in it, and it is being drained at the rate of $9$ gallons per minute. After how many minutes will there be $45$ gallons of water left?
Assume the tank will have $45$ gallons of water after $m$ minutes. We can solve this problem using the equation:
\begin{equation*} 171-9 m=45 \end{equation*}
Check whether $17$ is a solution for $m$ of this equation.
• Yes
• No
###### 47
A cylinder’s volume is $126\pi$ cubic centimeters. Its height is $14$ centimeters. Suppose we wanted to find how long is the cylinder’s radius. A cylinder’s volume formula is
\begin{equation*} V=\pi r^2h \end{equation*}
where $V$ stands for volume, $r$ for radius and $h$ for height. Let $r$ represent the cylinder’s radius, in centimeters. We can solve this problem using the equation:
\begin{equation*} 126 \pi=\pi r^2(14) \end{equation*}
Check whether $3$ is a solution for $r$ of this equation.
• Yes
• No
###### 48
A cylinder’s volume is $960\pi$ cubic centimeters. Its height is $15$ centimeters. Suppose we wanted to find how long is the cylinder’s radius. A cylinder’s volume formula is
\begin{equation*} V=\pi r^2h \end{equation*}
where $V$ stands for volume, $r$ for radius and $h$ for height. Let $r$ represent the cylinder’s radius, in centimeters. We can solve this problem using the equation:
\begin{equation*} 960 \pi=\pi r^2(15) \end{equation*}
Check whether $64$ is a solution for $r$ of this equation.
• Yes
• No
###### 49
A country’s national debt was $140$ million dollars in 2010. The debt increased at $50$ million dollars per year. If this trend continues, when will the country’s national debt increase to $640$ million dollars?
Assume the country’s national debt will become $640$ million dollars $y$ years after 2010. We can solve this problem using the equation:
\begin{equation*} 50 y+140=640 \end{equation*}
Check whether $10$ is a solution for $y$ of this equation.
• Yes
• No
###### 50
A country’s national debt was $100$ million dollars in 2010. The debt increased at $60$ million dollars per year. If this trend continues, when will the country’s national debt increase to $1180$ million dollars?
Assume the country’s national debt will become $1180$ million dollars $y$ years after 2010. We can solve this problem using the equation:
\begin{equation*} 60 y+100=1180 \end{equation*}
Check whether $20$ is a solution for $y$ of this equation.
• Yes
• No
###### 51
A school district has a reserve fund worth $41.1$ million dollars. It plans to spend $2.9$ million dollars per year. After how many years, will there be $15$ million dollars left?
Assume there will be $15$ million dollars left after $y$ years. We can solve this problem using the equation:
\begin{equation*} 41.1-2.9 y=15 \end{equation*}
Check whether $11$ is a solution for $y$ of this equation.
• Yes
• No
###### 52
A school district has a reserve fund worth $32$ million dollars. It plans to spend $3$ million dollars per year. After how many years, will there be $11$ million dollars left?
Assume there will be $11$ million dollars left after $y$ years. We can solve this problem using the equation:
\begin{equation*} 32-3 y=11 \end{equation*}
Check whether $8$ is a solution for $y$ of this equation.
• Yes
• No
###### 53
A rectangular frame’s perimeter is $6$ feet. If its length is $2$ feet, suppose we want to find how long is its width. A rectangle’s perimeter formula is
\begin{equation*} P=2(l+w) \end{equation*}
where $P$ stands for perimeter, $l$ for length and $w$ for width. We can solve this problem using the equation:
\begin{equation*} 6=2(2+w) \end{equation*}
Check whether $4$ is a solution for $w$ of this equation.
• Yes
• No
###### 54
A rectangular frame’s perimeter is $7$ feet. If its length is $2.1$ feet, suppose we want to find how long is its width. A rectangle’s perimeter formula is
\begin{equation*} P=2(l+w) \end{equation*}
where $P$ stands for perimeter, $l$ for length and $w$ for width. We can solve this problem using the equation:
\begin{equation*} 7=2(2.1+w) \end{equation*}
Check whether $1.4$ is a solution for $w$ of this equation.
• Yes
• No
|
# Class 8 Maths MCQ – Finding Square Roots through Repeated Subtraction
This set of Class 8 Maths Chapter 6 Multiple Choice Questions & Answers (MCQs) focuses on “Finding Square Roots through Repeated Subtraction”.
1. Find the square root of 25 by repeated subtraction method.
a) 5
b) 4
c) 6
d) 2
Explanation: The sum of the first n odd natural numbers is n2. Therefore we subtract odd numbers from the square till we get 0.
Therefore (i) 25-1=24; (ii) 24-3=21; (iii) 21-5=16; (iv) 16-7=9; (v) 9-9=0
Since we get 0 while subtracting the fifth odd number we know that the given number is square of 5.
2. If we have to subtract 9 odd numbers from a perfect square in order to obtain 0 then the perfect square is _______
a) 9
b) 3
c) 81
d) 18
Explanation: The sum of the first n odd natural numbers is n2. Now using this property of the square we can conclude that of we have subtracted nine odd numbers from the square in order to obtain zero then the number is square of number 9. As we know that square of number 9 is 81, 81 is the correct answer and the others are incorrect.
3. How many times do we need to subtract odd numbers from the square of 31 in order to obtain 0?
a) 31
b) 0
c) 32
d) 30
Explanation: The sum of the first n odd natural numbers is n2. Keeping this property in mind we can conclude that the answer for our question is 31. As we subtract odd numbers 31 times from square of 31 we get zero.
4. What are prime factors of 120?
a) 2 × 3 × 5 × 2 × 2
b) 2 × 5 × 5 × 3 × 2
c) 12 × 10
d) 4 × 3 × 10
Explanation: Here there are multiple options which give 120 on multiplying the numbers but we need to find the prime factors. The option which has only prime numbers and gives the product as 120 is the correct option. The option obtaining 2 × 3 × 5 × 2 × 2 satisfies both the conditions. Hence this option is the correct answer and the others are incorrect.
5. Find the prime factors of square of 5.
a) 5 × 5
b) 6 × 5
c) 3 × 5
d) 6 × 3
Explanation: If the questions ask to find the prime factors of square of prime numbers then the prime factors would be the number multiplied by the number itself. Here the square of 5 is 25. Therefore the prime factors are 5×5 and rest of the options are wrong.
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6. Adding the first n odd numbers we get ____
a) n
b) n2
c) √n
d) n2-1
Explanation: The sum of the first n odd natural numbers is n2. Hence the correct option would be n2 the other options cannot be correct. This is due to the property of the squares.
7. Find prime factors of 123.
a) 1 × 2 × 3
b) 1 × 62 × 2
c) 3 × 41
d) 3 × 2 × 20
Explanation: The prime factors of the number 123 are 3 × 41. There cannot be any other combination because the number 123 has only three factors and the factors are 1, 3 & 41.
8. If we subtract the first 13 odd numbers from 169 then the number obtained is ____
a) 0
b) 13
c) 169
d) -13
Explanation: The sum of the first n odd natural numbers is n2.
∴ 169 – 1 – 3 – 5 – 7 – 9 – 11 – 13 – 15 – 17 – 19 – 21 – 23 – 25 = 0
Therefore when we subtract first 13 odd numbers from 169 which is also known as 132 we obtain 0.
9. Find prime factors of 1331.
a) 10 × 12 × 13
b) 11 × 12 × 11
c) 11 × 11 × 11
d) 3 × 2 × 20
Explanation: The prime factors of the number 1331 are 11 × 11 × 11. There cannot be any other combination because the number 1331 has only three factors and the factors are 11 × 11 × 11.
10. Find prime factors of 69.
a) 3 × 23
b) 3 × 5 × 11
c) 3 × 9 × 11
d) 3 × 2 × 20
Explanation: The prime factors of the number 69 are 3 × 23. There cannot be any other combination because the number 69 has only factors and the factors are 3 × 23.
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• 1. GCF and LCM Problem Solving How can you tell if a word problem requires you to use Greatest Common Factor or Least Common Multiple to solve?
• 2. GCF and LCM Problem Solving
• First, use the KWL method for approaching all problems…
• K: What do you know?
• W: What do you want to know
• L: What did you learn?
• 3. If it is a GCF Problem
• What is the question asking us?
• Do we have to split things into smaller sections?
• Are we trying to figure out how many people we can invite?
• Are we trying to arrange something into rows or groups?
• 4. GCF Example: Applying what we have learned…
• Samantha has two pieces of cloth. One piece is 72 inches wide and the other piece is 90 inches wide. She wants to cut both pieces into strips of equal width that are as wide as possible. How wide should she cut the strips?
• 5. Samantha has two pieces of cloth. One piece is 72 inches wide and the other piece is 90 inches wide. She wants to cut both pieces into strips of equal width that are as wide as possible. How wide should she cut the strips?
• K: The pieces of cloth are 72 and 90 inches wide.
• W: How wide should she cut the strips so that they are the largest possible equal lengths.
• 6.
• L: This problem can be solved using Greatest Common Factor because we are cutting or “dividing” the strips of cloth into smaller pieces (factor) of 72 and 90.
• Find the GCF of 72 and 90
Samantha has two pieces of cloth. One piece is 72 inches wide and the other piece is 90 inches wide. She wants to cut both pieces into strips of equal width that are as wide as possible. How wide should she cut the strips?
• 7. GCF Word Problem Solution with Prime Factorization
• PF 72: 2 3 x 3 2
• PF 90: 2 x 3 2 x 5
• GCF: 2 x 3 2
• 8. GCF Word Problem Solution
• 2 72 90
• 3 36 45
• 3 12 15
• 4 5
• GCF = 2 x 3 x 3 = 18
• Samantha should cut each piece to be 18 inches wide
• 9. If it is an LCM Problem
• What is the question asking us?
• Do we have an event that is or will be repeating over and over?
• Will we have to purchase or get multiple items in order to have enough?
• Are we trying to figure out when something will happen again at the same time?
• 10. LCM Example: Applying what we have learned…
• Ben exercises every 3 days and Isabel every 2 days. Ben and Isabel both exercised today. How many days in the next 30 days will they both exercise on the same day?
• 11. Ben exercises every 3 days and Isabel every 2 days. Ben and Isabel both exercised today. How many days in the next 30 days will they both exercise on the same day?
• K: Ben exercises every 3 days and Isabel every 2 days and they both exercised today.
• W: How many days in the next 30 days will they both exercise on the same day.
• 12.
• L: This problem can be solved using Least Common Multiple. We are trying to figure out how many times in 30 days they will repeat their exercise and be exercising at the same time.
• Find the LCM of 3 and 2.
Ben exercises every 3 days and Isabel every 2 days. Ben and Isabel both exercised today. How many days in the next 30 days will they both exercise on the same day?
• 13. LCM Word Problem Solution
• 2 3
• There are no common prime factors of 2 and 3, so we just multiply them to get the LCM.
• LCM = 2 x 3 = 6
• Ben and Isabel would exercise on the same day every 6th day. In 30 days, they would exercise together 5 times (because 30 divided by 6 is 5).
• 14. QUIZ!!!!!!
• On a sheet of notebook paper, tell whether the following word problems could be solved using GCF or LCM…
• 15. Question #1
• Mrs. Evans has 120 crayons and 30 pieces of paper to give to her students. What is the largest # of students she can have in her class so that each student gets equal # of crayons and equal # of paper.
• 16. Question #2
• Rosa is making a game board that is 16 inches by 24 inches. She wants to use square tiles. What is the larges tile she can use?
• 17. Question #3
• Star 94 gave away a discount coupon for every fifth and sixth caller. Every twentieth caller received free concert tickets. Which caller was first to receive both a coupon and a concert ticket?
• 18. Question #4
• Two bikers are riding a circular path. The first rider completes a round in 12 minutes. The second rider completes a round in 18 minutes. If they both started at the same place and time and go in the same direction, after how many minutes will they meet again at the starting point?
• 19. Question #5
• Sean has 15-inch pieces of toy train track and Ruth has 6-inch pieces of train track. How many of each piece would each child need to build tracks that are equal in length?
• 20. Question #6
• I am planting 50 apple trees and 30 peach trees. I want the same number and type of trees per row. What is the maximum number of trees I can plant per row?
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# Question Video: Finding the Argument of a Complex Number in Radians Mathematics
What is the argument of the complex number 4π?
02:42
### Video Transcript
What is the argument of the complex number four π?
In this question, weβre asked to find the argument of the complex number four π. To answer this question, weβre first going to need to recall what we mean by the argument of a complex number. We recall the argument of a complex number π§ written arg of π§ is the angle that π§ makes with the positive real axis on an Argand diagram. And usually, we give the argument of a complex number in radians. And we also know in radians thereβs lots of different values which represent the same angle. For example, zero, two π, and four π all represent the same angle. And to get around this problem, usually when we find the argument of a complex number, we try to give our answer between negative π and π, where we include π. And this value is usually called the principal argument of π§.
Now thereβs a few different ways of finding the argument of a complex number. For example, you might know a few formulas to find it. However, we should always sketch a picture before we try and find the argument of a complex number. So weβll start with our Argand diagram. Remember, the horizontal axis is the real part of our complex number, and the vertical axis is the imaginary part of our complex number.
So to plot four π onto our Argand diagram, we need to find the real part of four π and the imaginary part of four π. The imaginary part of four π will be the coefficient of π, which in this case is four. And the real part of four π is equal to zero because thereβs no added constant. So this tells us, on an Argand diagram, the complex number four π will be represented by the point zero, four. And we can input this onto our Argand diagram.
Next, weβre going to want to sketch the argument of four π onto our Argand diagram. And to do this, it can be helpful to add the ray from the origin to our complex number four π. Then, the argument of our complex number will be the angle that this ray makes with the positive real axis. Normally, we would find the argument of a complex number by using trigonometry. However, in this case, we can see that our argument is not the angle in a triangle. Instead, itβs the angle between two of our axes, so we know this is a right angle. And we know that a right angle is represented by π by two. And since this angle is counterclockwise from our positive real axis, this means it will be positive π by two. So we were able to show the argument of four π is π by two.
However, there is something we can notice about this example. We can ask the question, what is the argument of ππ if our value of π is positive? Using the exact same reasoning we did for four π, we can show that the argument of ππ would also be π by two. This means weβve shown a useful result. If our value of π is positive, then the argument of ππ will always be equal to π by two. We could have also used this result to answer our question.
Therefore, we were able to show the argument of the complex number four π is π by two.
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# Coterminal Angles in Geometry
Coterminal angles are angles that have the same terminal side. In other words, they share the same endpoint. Coterminal angles are often used in geometry and trigonometry, and it's important to be able to identify them and understand how they work. Let's take a closer look.
## How Coterminal Angles Work
Coterminal angles are two angles that have the same terminal side. This means that they share the same endpoint. The easiest way to visualize coterminal angles is to imagine a circle with a radius of 1 unit. If we start at the top of the circle and move clockwise around the circumference, we will eventually end up back at the starting point.
If we mark off a certain number of degrees as we go around the circle, we can create two angles that are coterminal. For example, if we start at the top of the circle (0 degrees) and move clockwise 60 degrees, we have created an angle measuring 60 degrees. If we then continue moving clockwise around the circle until we reach 120 degrees, we have created another angle measuring 120 degrees. These two angles are coterminal because they share the same endpoint (in this case, the top of the circle).
## Why Coterminal Angles Are Important
Coterminal angles are important because they can be used to simplify calculations involving angles. For example, if you're working with an angle that is larger than 360 degrees, you can simply subtract 360 degrees until you're left with an angle that is less than 360 degrees. Likewise, if you're working with an angle that is negative, you can add 360 degrees until you're left with a positive angle.
## Conclusion
Coterminal angles are two angles that have the same terminal side (i.e., they share the same endpoint). They are often used in geometry and trigonometry, and it's important to be able to identify them and understand how they work. Coterminal angles are important because they can be used to simplify calculations involving angles.
## FAQ
### What is the purpose of Coterminal angles?
The purpose of coterminal angles is to simplify calculations involving angles.
### What is the Coterminal of 45?
The coterminal of 45 is any angle that is equal to 45 degrees plus or minus 360 degrees. For example, the coterminal of 45 could be 405 degrees, -315 degrees, or 3375 degrees.
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# How do you factor 18x^2-2?
Jun 20, 2017
$2 \left(3 x + 1\right) \left(3 x - 1\right)$
#### Explanation:
First look for common factors
$18 {x}^{2} - 2 = 2 \left(9 {x}^{2} - 1\right)$
then see if the bracket can be factorised
we note difference of squares hence
$2 \left(9 {x}^{2} - 1\right) = 2 \left(3 x + 1\right) \left(3 x - 1\right)$
Jun 20, 2017
$= 2 \left(3 x + 1\right) \left(3 x - 1\right)$
#### Explanation:
Always look for a common factor first. IN this case it is $2$
$18 {x}^{2} - 2 = 2 \left(9 {x}^{2} - 1\right) \text{ } \leftarrow$ difference of squares
$= 2 \left(3 x + 1\right) \left(3 x - 1\right)$
Note that:
${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$
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# Is 42 Divisible By 2?
• Yes, 42 is divisible by 2. It will leave no comma spot.
• Divisibilty rule for 2 is: Units are divisible by two if the last digit is even. Even numbers for 2 are (0,2,4,6,8).
## What Is 42 Divided By 2
• Forty-two divided by two is 21. Math: 42÷2=21
## Rules Of Divisibility For 2
• First, take any number (for this example it will be 376) and note the last digit in the number, discarding the other digits. Then take that digit (6) while ignoring the rest of the number and determine if it is divisible by 2. If it is divisible by 2, then the original number is divisible by 2.
• Example: 376 (The original number). 37 6 (Take the last digit). 6÷2 = 3 (Check to see if the last digit is divisible by 2) 376÷2 = 188 (If the last digit is divisible by 2, then the whole number is divisible by 2).
## Mathematical Information About Numbers 4 2
• About Number 4. Four is linear. It is the first composite number and thus the first non-prime number after one. The peculiarity of the four is that both 2 + 2 = 4 and 2 * 2 = 4 and thus 2^2 = 4. Four points make the plane of a square, an area with four sides. It is the simplest figure that can be deformed while keeping it's side lengths, such as the rectangle to parallelogram. Space let's us arrange equidistantly a maximum of four points. These then form a tetrahedron (tetrahedron), a body with four identical triangular faces. Another feature of the four is the impossibility of an algebraic equation of higher degree than four square roots using simple arithmetic and basic operations dissolve.
• About Number 2. Two is the smallest and the only even prime number. Also it's the only prime which is followed by another prime number three. All even numbers are divisible by 2. Two is the third number of the Fibonacci sequence. Gottfried Wilhelm Leibniz discovered the dual system (binary or binary system) that uses only two digits to represent numbers. It witnessed the development of digital technology for a proliferation. Because of this, it is the best known and most important number system in addition to the commonly used decimal system.
## What are divisibility rules?
A divisibility rule is a shorthand way of determining whether a given number is divisible by a fixed divisor without performing the division, usually by examining its digits. Although there are divisibility tests for numbers in any radix, and they are all different, this article presents rules and examples only for decimal numbers. For divisors with multiple rules, the rules are generally ordered first for those appropriate for numbers with many digits, then those useful for numbers with fewer digits.
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Mixture Problems
Solving Mixture Problems
Any problem in which two or more quantities are combined to produce a different quantity, or a single quantity is separated into different quantities may be considered a mixture problem.
One type of mixture problem involves mixing liquids or solutions and evaluating the content or strength of the mixture. The following formulas are useful for evaluating these type of mixture problems:
amount of substance in the solution = quantity of solution x strength of solution (in percent)
In addition, the following formula is often used:
(amount of substance in mixture 1) + (amount of substance in mixture 2) = amount of substance in mixture
Example
Nathaniel works as a medical researcher. He has two solutions of phenobarbital, one at 40% strength and the other at 5% strength. How much of each solution must he mix to make 0.6 liters of a 20% phenobarbital solution?
Understand
Mix the 40% solution and the 5% solution to achieve 0.6 liter of a 20% solution. The amount of phenobarbital in a given solution is found by multiplying the number of liters by the percent strength.
Translate
Let x = number of liters of the 40% solution
Let 0.6 - x = number of liters of the 5% solution
Solve
0.40x + 0.05(0.6 - x) = (0.6)(0.20)
0.40x + 0.03 - 0.05x = 0.12 distributive property on left; multiply on right
0.35x + 0.03 = 0.12 combine like terms
0.35x = 0.09 subtract 0.03 from both sides
x = 0.26 divide both sides by 0.35
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# 1’s Complement Subtraction Explained with Examples
Last Updated on October 25, 2023 by Electricalvolt
This article describes the complement arithmetic using 1’s complement binary subtraction. Complement arithmetic is a mathematical process used for subtraction. This operation includes two types of complements, namely 1’s complement and 2’s complement. Complement arithmetic is crucial in subtracting numbers using digital devices such as computers and calculators. The primary goal of complement arithmetic is to convert the subtraction process into addition.
In this article, we will learn how to find the 1’s and 2’s complement and how they are used to subtract numbers.
## What is 1’s Complement?
The 1’s complement is a fundamental arithmetic operation that involves flipping the bits of a binary number. To obtain the 1’s complement of a binary number, all the 0 bits in the number are changed to 1, while all the 1 bits are changed to 0. This operation is crucial in digital circuits and computer science, enabling bitwise operations such as addition, subtraction, and logical operations.
For example, consider a binary number (100110111)2. Then, the 1’s complement of this binary number will be (011001000)2. From this example, it is clear that we can find the 1’s complement just by changing the 0 bits to 1 and 1 bits to 0 of the given number.
## Subtraction using 1’s Complement Arithmetic
The step-by-step procedure of subtraction using 1’s complement arithmetic is given below:
### Procedure for subtracting a smaller number from a larger number
• Find the 1’s complement of the subtrahend, i.e. second number.
• Add the 1’s complement of the subtrahend to the minuend (first number).
• If an end-around carry is produced, add it to the least significant bit of the intermediate result, and the final result is positive.
### Procedure for subtracting a larger number from a smaller number
• Find the 1’s complement of the subtrahend, i.e. second number.
• Add the 1’s complement of the subtrahend to the minuend (first number).
• If there is no end-around carry produced, the result is negative. It is obtained by taking 1’s complement of the intermediate result.
Let us take an example to understand the procedure.
### Example 1 – Subtract (1110101)2 – (1000101)2 using 1’s complement arithmetic.
Solution – Let’s perform this subtraction using the 1’s complement arithmetic.
In this example,
Minuend = (1110101)2
Subtrahend = (1000101)2
1’s complement of subtrahend = (0111010)2
Now, let us add the 1’s complement of subtrahend and minuend.
(1110101)2 + (0111010)2 = (1 0101111)2
There is an end around carry; hence, the result is positive. The final result is obtained by adding the end-around carry to the LSB of the intermediate result as follows.
(0101111)2 + (1)2 = (0110000)2
Hence,
(1110101)2 – (1000101)2 = (0110000)2
Let us take a few more examples to understand the subtraction process using 1’s complement.
### Example 2: Subtract (1011)2 from (1111)2 using 1’s complement method
The Minuend (1111)2 is larger than the subtrahend (1011)2.
Step 1– Find 1’s complement of subtrahend (1011)2.
Step 2– Add Minuend (1111)2 and the number obtained in step-1 (0100)2
### Example 3: Subtract (1011)2 from (1001)2 using 1’s complement subtraction method
The Minuend (1001)2 is smaller than the subtrahend (1011)2.
Step 1– Find 1’s complement of the larger number (1011)2.
Step 2– Add the number obtained in Step 1 and the number (1001)2
Step 3– Determine the 1’s complement of the number obtained in step 3
## Conclusion
We discussed 1’s complement arithmetic subtraction. Complement arithmetic involves finding 1’s complement and 2’s complement of a number, which helps perform subtraction using addition operations.
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Instruction
1
If a large number is zero, then division by it any smaller (i.e. negative) values it is impossible to determine.
2
If you want to share any positive value by exceeding its value, then the result is sure to be a fractional number. Because there are several versions of writing fractions, you should start with the definition of the format in which you want to obtain the result of the operation depends on the algorithm your follow-up. Two possible options: ordinary fraction or decimal. Consider first, for example, the result in the format of fractions.
3
Make initial values common fraction - put the larger number in the denominator and less in the numerator.
4
Try to simplify the fraction, that is, to pick up common to dividend and divisor an integer that can be divided without a remainder. If such number cannot be found, then the obtained in the previous step the fraction will be the result of the division. If a common divisor exists, then divide by it both parts. For example, if the original numbers were 42 and 49, the common divisor is seven: 42/49 = (42/7)/(49/7) = 6/7.
5
If the result of dividing the larger number by the smaller, the problem can be represented in a decimal fraction, simply divide the dividend by the divisor in any convenient way - in the mind, long division or using a calculator. Often the result obtained irrational numbers, i.e. the number of decimal places is infinite. Of course, in this case, you need to determine the necessary conditions of the problem the accuracy of the result and rounding the resulting value.
6
If a smaller and a larger number have different signs, that is, the numerator is the number is negative, then proceed according to the rules above, discarded at the time a sign of a smaller magnitude. The value of the number without regard to sign is called its "modulus" or "absolute value". To the result of dividing the modulo after the operation, add the negative sign.
7
If both values involved in the operation are negative, the result is bound to be a positive number. So the signs can be discarded immediately and more in General about them not to remember.
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# Difference between revisions of "Circumradius"
The circumradius of a cyclic polygon is the radius of the circumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle.
## Formula for a Triangle
Let $a, b$ and $c$ denote the triangle's three sides and let $A$ denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply $R=\frac{abc}{4A}$. This can be rewritten as $A=\frac{abc}{4R}$.
## Proof
$[asy] pair O, A, B, C, D; O=(0,0); A=(-5,1); B=(1,5); C=(5,1); dot(O); dot (A); dot (B); dot (C); draw(circle(O, sqrt(26))); draw(A--B--C--cycle); D=-B; dot (D); draw(B--D--A); label("A", A, W); label("B", B, N); label("C", C, E); label("D", D, S); label("O", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label("E", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy]$
We let $AB=c$, $BC=a$, $AC=b$, $BE=h$, and $BO=R$. We know that $\angle BAD$ is a right angle because $BD$ is the diameter. Also, $\angle ADB = \angle BCA$ because they both subtend arc $AB$. Therefore, $\triangle BAD \sim \triangle BEC$ by AA similarity, so we have $$\frac{BD}{BA} = \frac{BC}{BE},$$ or $$\frac {2R} c = \frac ah.$$ However, remember that $[ABC] = \frac {bh} 2\implies h=\frac{2 \times [ABC]}b$. Substituting this in gives us $$\frac {2R} c = \frac a{\frac{2 \times [ABC]}b},$$ and then bash through the algebra to get $$R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}$$ and we are done.
$R = \frac{abc}{4rs}$ Where $R$ is the circumradius, $r$ is the inradius, and $a$, $b$, and $c$ are the respective sides of the triangle and $s = (a+b+c)/2$ is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that $A = rs$.
But, if you don't know the inradius, you can find the area of the triangle by Heron's Formula:
$A=\sqrt{s(s-a)(s-b)(s-c)}$
## Euler's Theorem for a Triangle
Let $\triangle ABC$ have circumcenter $O$ and incenter $I$.Then $$OI^2=R(R-2r) \implies R \geq 2r$$
## Right triangles
The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle.
pair A,B,C,I;
A=(0,0);
B=(0,3);
C=(4,0);
draw(A--B--C--cycle);
I=circumcenter(A,B,C);
draw(I--A,gray);
label("$r$",(I+A)/2,gray,NW);
draw(circumcircle(A,B,C));
label("$C$",I,N);
dot(I);
draw(rightanglemark(B,A,C,10));
(Error compiling LaTeX. label("$r$",(I+A)/2,gray,NW);
^
6b0354fc45d5a0fef2176240bd908f7351049788.asy: 12.6: no matching function 'label(string, pair, <overloaded>, pair)'
label("$r$",(I+A)/2,gray,NW);
^
6b0354fc45d5a0fef2176240bd908f7351049788.asy: 12.21: use of variable 'gray' is ambiguous)
This results in a well-known theorem:
### Theorem
The midpoint of the hypotenuse is equidistant from the vertices of the right triangle.
## Equilateral triangles
$R=\frac{s}{\sqrt3}$
where $s$ is the length of a side of the triangle.
$[asy] pair A,B,C,I; A=(0,0); B=(1,0); C=intersectionpoint(arc(A,1,0,90),arc(B,1,90,180)); draw(A--B--C--cycle); I=circumcenter(A,B,C); draw(circumcircle(A,B,C)); label("C",I,E); dot(I); label("s",A--B,S); label("s",A--C,N); label("s",B--C,N); [/asy]$
## If all three sides are known
$R=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}$
And this formula comes from the area of Heron and $R=\frac{abc}{4A}$.
## If you know just one side and its opposite angle
$2R=\frac{a}{\sin{A}}$ which is just extension of law of sines
(Extended Law of Sines)
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# ISEE Upper Level Quantitative : How to find the area of a square
## Example Questions
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### Example Question #1 : How To Find The Area Of A Square
The perimeter of a square is one yard. Which is the greater quantity?
(a) The area of the square
(b) square foot
(b) is greater.
It is impossible to tell form the information given.
(a) is greater.
(a) and (b) are equal.
(a) is greater.
Explanation:
One yard is equal to three feet, so the length of one side of a square with this perimeter is feet. The area of the square is square feet. , making (a) greater.
### Example Question #2 : How To Find The Area Of A Square
Square 1 is inscribed inside a circle. The circle is inscribed inside Square 2.
Which is the greater quantity?
(a) Twice the area of Square 1
(b) The area of Square 2
(a) and (b) are equal.
(b) is greater.
It is impossible to tell from the information given.
(a) is greater.
(a) and (b) are equal.
Explanation:
Let be the sidelength of Square 1. Then the length of a diagonal of this square - which is times this sidelength, or , by the Theorem - is the same as the diameter of this circle, which, in turn, is equal to the sidelength of Square 2.
Therefore, Square 1 has area , and Square 2 has area , twice that of Square 1.
### Example Question #3 : How To Find The Area Of A Square
Which is the greater quantity?
(A) The area of a square with sidelength one foot
(B) The area of a rectangle with length nine inches and height fourteen inches
(A) and (B) are equal
(B) is greater
(A) is greater
It is impossible to determine which is greater from the information given
(A) is greater
Explanation:
The area of a square is the square of its sidelength, which here is 12 inches:
square inches.
The area of a rectangle is its length multiplied by its height, which, respectively, are 9 inches and 14 inches:
square inches.
The square has the greater area.
### Example Question #4 : How To Find The Area Of A Square
A square lawn has sidelength twenty yards. Give its area in square feet.
Explanation:
20 yards converts to feet. The area of a square is the square of its sidelength, so the area in square feet is square feet.
### Example Question #212 : Isee Upper Level (Grades 9 12) Quantitative Reasoning
Rectangle A and Square B both have perimeter 2 meters. Rectangle A has width 25 centimeters. The area of Rectangle A is what percent of the area of Square B?
Explanation:
The perimeter of a rectangle can be given by the formula
Rectangle A has perimeter 2 meters, which is equal to 200 centimeters, and width 25 centimeters, so the length is:
The dimensions of Rectangle A are 75 centimeters and 25 centimeters, so its area is
square centimeters.
The sidelength of a square is one-fourth its perimeter, which here is
centimeters; its area is therefore
square centimeters.
The area of Rectangle A is therefore
that of Square B.
### Example Question #5 : How To Find The Area Of A Square
The sidelength of Square A is three-sevenths that of Square B. What is the ratio of the area of Square B to that of Square A?
None of the other answers give the correct ratio.
Explanation:
Since the ratio is the same regardless of the sidelengths, then for simplicity's sake, assume the sidelength of Square B is 7. The area of Square B is therefore the square of this, or 49.
Then the sidelength of Square A is three-sevenths of 7, or 3. Its area is the square of 3, or 9.
The ratio of the area of Square B to that of Square A is therefore 49 to 9.
### Example Question #222 : Geometry
Five squares have sidelengths one foot, two feet, three feet, four feet, and five feet.
Which is the greater quantity?
(A) The mean of their areas
(B) The median of their areas
(B) is greater
(A) and (B) are equal
(A) is greater
It is impossible to tell which is greater from the information given
(A) is greater
Explanation:
The areas of the squares are:
square foot
square feet
square feet
square feet
square feet
Therefore, we are comparing the mean and the median of the set .
The mean of this set is the sum divided by 5:
The median is the middle element after arrangement in ascending order, which is 9.
This makes (A), the mean, greater.
### Example Question #223 : Geometry
Four squares have sidelengths one meter, 120 centimeters, 140 centimeters, and 140 centimeters. Which is the greater quantity?
(A) The mean of their areas
(B) The median of their areas
(A) is greater
It is impossible to tell which is greater from the information given
(B) is greater
(A) and (B) are equal
(B) is greater
Explanation:
The areas of the squares are:
square centimeters (one meter being 100 centimeters)
square centimeters
square centimeters
square centimeters
The mean of these four areas is their sum divided by four:
square centimeters.
The median is the mean of the two middle values, or
square centimeters.
The median, (B), is greater.
### Example Question #224 : Geometry
The perimeter of a square is . Give the area of the square in terms of .
None of the other responses gives a correct answer.
Explanation:
The length of one side of a square is one fourth its perimeter. Since the perimeter of the square is , the length of one side is
The area of the square is the square of this sidelength, or
### Example Question #5 : How To Find The Area Of A Square
The sidelength of a square is . Give its area in terms of .
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# and Perpendicular Lines 5 2 Parallel and Perpendicular
• Slides: 21
and Perpendicular Lines 5 -2 Parallel and Perpendicular Lines Warm Up Problem of the Day Lesson Presentation Pre-Algebra
5 -2 Parallel and Perpendicular Lines Warm Up Complete each sentence. 1. Angles whose measures have a sum of 90° are ________. complementary 2. Vertical angles have equal measures, so they are congruent _______. 3. Angles whose measures have a sum of 180° are supplementary _______. 4. A part of a line between two points is called a segment ______. Pre-Algebra
5 -2 Parallel and Perpendicular Lines Problem of the Day The square root of 1, 813, 141, 561 is a whole number. Is it odd or even? How do you know? An odd number can only be the product of two odd numbers. Since 1, 813, 141, 561 is an odd number, its square root is also an odd number. Pre-Algebra
5 -2 Parallel and Perpendicular Lines Learn to identify parallel and perpendicular lines and the angles formed by a transversal. Pre-Algebra
5 -2 Parallel and Perpendicular Lines Vocabulary parallel lines perpendicular lines transversal Pre-Algebra
5 -2 Parallel and Perpendicular Lines Parallel lines are two lines in a plane that never meet, like a set of perfectly straight, infinite train tracks. Perpendicular lines are lines that intersect at 90° angles. Pre-Algebra
5 -2 Parallel and Perpendicular Lines The railroad ties are transversals to the tracks. The tracks are parallel. A transversal is a line that intersects any two or more other lines. Transversals to parallel lines have interesting properties. Pre-Algebra
5 -2 Parallel and Perpendicular Lines Additional Example 1: Identifying Congruent Angles Formed by a Transversal Measure the angles formed by the transversal and parallel lines. Which angles seem to be congruent? 1, 3, 5, and 7 all measure 150°. 2, 4, 6, and 8 all measure 30°. Pre-Algebra
5 -2 Parallel and Perpendicular Lines Additional Example 1 Continued Angles marked in blue appear to be congruent to each other, and angles marked in red appear to be congruent to each other. 1 @ 3 @ 5 @ 7 2 @ 4 @ 6 @ 8 Pre-Algebra 2 1 3 4 6 5 7 8
5 -2 Parallel and Perpendicular Lines Try This 1: Example 1 Measure the angles formed by the transversal and parallel lines. Which angles seem to be congruent? 1 2 3 4 5 6 7 8 1, 4, 5, and 8 all measure 36°. 2, 3, 6, and 7 all measure 144°. Pre-Algebra
5 -2 Parallel and Perpendicular Lines Try This 1: Example 1 Continued Angles marked in blue appear to be congruent to each other, and angles marked in red appear to be congruent to each other. 1 @ 4 @ 5 @ 8 2 @ 3 @ 6 @ 7 1 Pre-Algebra 2 3 4 5 6 7 8
5 -2 Parallel and Perpendicular Lines PROPERTIES OF TRANSVERSALS TO PARALLEL LINES If two parallel lines are intersected by a transversal, • the acute angles that are formed are all congruent, • the obtuse angles are all congruent, • and any acute angle is supplementary to any obtuse angle. If the transversal is perpendicular to the parallel lines, all of the angles formed are congruent 90° angles. Pre-Algebra
5 -2 Parallel and Perpendicular Lines Writing Math The symbol for parallel is ||. The symbol for perpendicular is . Pre-Algebra
5 -2 Parallel and Perpendicular Lines Additional Example 2 A: Finding Angle Measures of Parallel Lines Cut by Transversals In the figure, line l || line m. Find the measure of the angle. A. 4 All obtuse angles in the figure are congruent. m 4 = 124° Pre-Algebra
5 -2 Parallel and Perpendicular Lines Additional Example 2 B: Finding Angle Measures of Parallel Lines Cut by Transversals Continued In the figure, line l || line m. Find the measure of the angle. B. 2 2 is supplementary to the angle 124°. m 2 + 124° = 180° – 124° m 2 = 56° Pre-Algebra
5 -2 Parallel and Perpendicular Lines Additional Example 2 C: Finding Angle Measures of Parallel Lines Cut by Transversals Continued In the figure, line l || line m. Find the measure of the angle. C. 6 All acute angles in the figure are congruent. m 6 = 56° Pre-Algebra
5 -2 Parallel and Perpendicular Lines Try This: Example 2 A In the figure, line n || line m. Find the measure of the angle. A. 7 All obtuse angles in the figure are congruent m 7 = 144° Pre-Algebra 1 144° 3 4 5 6 7 8 m n
5 -2 Parallel and Perpendicular Lines Try This: Example 2 B In the figure, line n || line m. Find the measure of the angle. B. 5 5 is supplementary to the angle 144°. m 5 + 144° = 180° – 144° m 5 = 36° Pre-Algebra 1 144° 3 4 5 6 7 8 m n
5 -2 Parallel and Perpendicular Lines Try This: Example 2 C In the figure, line n || line m. Find the measure of the angle. C. 1 All acute angles in the figure are congruent m 1 = 36° Pre-Algebra 1 144° 3 4 5 6 7 8 m n
5 -2 Parallel and Perpendicular Lines If two lines are intersected by a transversal and any of the angle pairs shown below are congruent, then the lines are parallel. This fact is used in the construction of parallel lines. Pre-Algebra
5 -2 Parallel and Perpendicular Lines Lesson Quiz In the figure a || b. 1. Name the angles congruent to 3. 1, 5, 7 2. Name all the angles supplementary to 6. 1, 3, 5, 7 3. If m 1 = 105° what is m 3? 105° 4. If m 5 = 120° what is m 2? 60° Pre-Algebra
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# How do you solve the following system?: 1/3y = 3/4x + 5 , 3/2x - 8y = 2
Apr 1, 2018
$x = - \frac{244}{33}$ $y = - \frac{18}{11}$
#### Explanation:
$\left(\frac{1}{3}\right) y = \left(\frac{3}{4}\right) x + 5$----(1)
$\left(\frac{3}{2}\right) x - 8 y = 2$----(2)
Multiply (2) by $\left(\frac{1}{2}\right)$
$\left(\frac{3}{4}\right) x - 4 y = 1$----(3)
Add(1)and(3), to make$x$disappear
$\left(\frac{1}{3}\right) y + \left(\frac{3}{4}\right) x - 4 y = \left(\frac{3}{4}\right) x + 5 + 1$
$\left(\frac{3}{4}\right) x - \left(\frac{3}{4}\right) x + \left(\frac{1}{3}\right) y - 4 y = 5 + 1$
$\left(- \frac{11}{3}\right) y = 6$
$y = 6 \cdot \left(- \frac{3}{11}\right)$
$y = - \frac{18}{11}$
Substitude$y = - \frac{18}{11}$into(3)
$\left(\frac{3}{4}\right) x - 4 \left(- \frac{18}{11}\right) = 1$
$\left(\frac{3}{4}\right) x + \frac{72}{11} = 1$
$\left(\frac{3}{4}\right) x = 1 - \frac{72}{11}$
$\left(\frac{3}{4}\right) x = - \left(\frac{61}{11}\right)$
$x = - \left(\frac{61}{11}\right) \cdot \left(\frac{4}{3}\right)$
$x = - \frac{244}{33}$
So, $x = - \frac{244}{33}$ $y = - \frac{18}{11}$
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# Georgia Department of Education
## Common Core Georgia Performance Standards Framework
Fourth Grade Mathematics x Unit 3
STANDARDS FOR MATHEMATICAL CONTENT
MCC4.NF.3 Understand a fraction a/b with a > 1 as a sum of fractions 1/b.
a. Understand addition and subtraction of fractions as joining and separating parts referring to
the same whole.
b. Decompose a fraction into a sum of fractions with the same denominator in more than one
way, recording each decomposition by an equation. Justify decompositions, e.g., by using a
visual fraction model. Examples: 3/8 = 1/8 + 1/8 + 1/8 ; 3/8 = 1/8 + 2/8 ; 2 1/8 = 1 + 1 +
1/8 = 8/8 + 8/8 + 1/8.
c. Add and subtract mixed numbers with like denominators, e.g., by
replacing each mixed number with an equivalent fraction, and/or
by using properties of operations and the relationship between
d. Solve word problems involving addition and subtraction of fractions referring to the same
whole and having like denominators, e.g., by using visual fraction models and equations to
represent the problem.
STANDARDS FOR MATHEMATICAL PRACTICE
1. Make sense of problems and persevere in solving them.
2. Reason abstractly and quantitatively.
3. Construct viable arguments and critique the reasoning of others.
4. Model with mathematics.
5. Use appropriate tools strategically.
6. Attend to precision.
7. Look for and make use of structure.
8. Look for and express regularity in repeated reasoning.
BACKGROUND KNOWLEDGE
Students should have an understanding about how to represent a fraction using an area
model, such as a circle.
When creating the cookie confirmations, students are able to use the associative property to
add different fractions first if that makes more sense. For example, a solution for order # 2 is
shown below. In this case a student added the fractions in the order they appear on the order
form.
Order #2, Solution 1
fractions in the following order,
## as shown in the second example.
Be sure students understand
that either solution is correct
because the correct fraction of each
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
May 2012 y Page 36 of 76
## Georgia Department of Education
Common Core Georgia Performance Standards Framework
Fourth Grade Mathematics x Unit 3
## Order #2, Solution 2
type of topping is represented.
Numerical solutions and pictorial
solutions are shown below. As
ESSENTIAL QUESTIONS
x
x
x
x
x
x
x
## What is a fraction and how can it be represented?
What is an improper fraction and how can it be represented?
What is a mixed number and how can it be represented?
What is the relationship between a mixed number and an improper fraction?
How can improper fractions and mixed numbers be used interchangeably?
How do we apply our understanding of fractions in everyday life?
MATERIALS
x
x
x
## Fraction Cookie Bakery, Order Form student recording sheet
Fraction Cookie Bakery, Order Confirmation Form student recording sheet
colored pencils, crayons, or markers
GROUPING
This task provides students with their first opportunity to explore addition with improper and
proper fractions. Students will add proper and improper fractions to create cookie order
confirmation notices. They will be required to write each sum as an improper fraction and a
mixed number.
This task may be introduced by reading The Hersheys Milk Chocolate Fractions Book by
Jerry Pallotta, focusing on the addition that is modeled in the book. Continue by explaining the
task and modeling the example problem as shown below.
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
May 2012 y Page 37 of 76
## Georgia Department of Education
Common Core Georgia Performance Standards Framework
Fourth Grade Mathematics x Unit 3
As students are working, ask them how they created the confirmations for each order.
Ask questions that will cause students to think about how many fractional parts make a whole,
different ways they can group toppings to create a whole cookie, and how they know what the
sum is, written as an improper fraction and as a mixed number. Some sample questions are given
in the FORMATIVE ASSESSMENT QUESTIONS section below.
This task provides students with an opportunity to explore sums of improper and proper
fractions using models. Therefore, students SHOULD NOT use an algorithm to change improper
fractions to mixed numbers. Instead, students will be using their models to determine the sums.
During the lesson summary, be sure students are aware that an improper fraction can be
written as a mixed number because = + + + as shown in the cookie model the students
created (see example above). This way students will develop an understanding of what an
improper fraction represents. Students can use this same understanding when writing a mixed
number as an improper fraction. They should recognize that:
Because each is equal to one whole, there needs to be three to represent the 3 wholes in the
## mixed number. When the fractions are added the sum is .
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
May 2012 y Page 38 of 76
## Georgia Department of Education
Common Core Georgia Performance Standards Framework
Fourth Grade Mathematics x Unit 3
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
May 2012 y Page 39 of 76
## Georgia Department of Education
Common Core Georgia Performance Standards Framework
Fourth Grade Mathematics x Unit 3
Students will follow the directions below from the Fraction Cookies Bakery, Order Form
student recording sheet and Fraction Cookies Bakery, Order Confirmation Form student
recording sheet below.
Fraction Cookies Bakery, Order Form student recording sheet
You own a bakery that specializes in fraction cookies. Customers place orders from
below. Before making the cookies to fill the order, you need to confirm each order by
sending a confirmation notice to each customer. (If the toppings ordered do not cover an
entire cookie, customers want the remaining portion of the cookie to be left plain.) Using
the circle templates below, show how you would create each cookie order with the
correct fractional amounts of toppings.
Fraction Cookies Bakery, Order Confirmation Form student recording sheet
Customers expect you to use the fewest number of cookies possible to complete each
order. No part of a cookie should be without a topping except for one. You may split a
topping between two cookies as shown below (the vanilla icing was shared between two
cookies rather than covering both halves of one cookie with vanilla icing).
FORMATIVE ASSESSMENT QUESTIONS
x
x
x
x
x
x
## How do you know you have recorded the order correctly?
In what order did you record the fractions? Why?
How many sections do you need to cover a whole cookie? How do you know?
How did you determine the improper fraction?
How did you determine the mixed number?
How did you determine how much of a cookie would be plain?
DIFFERENTIATION
Extension
x Challenge students with one or more of the orders on the Fraction Cookie Bakery, Order
Form Version 2 student recording sheet. Be sure students USE MODELS ONLY to
solve these problems.
x Ask students to create orders of their own, then switch with a partner to create the
confirmations for those orders. Students can be given a blank confirmation sheet or they
can create their own fraction models.
Intervention
x Some students may need more examples modeled before they are able to complete this
task on their own. Provide an opportunity for further small group instruction before
x Allow students to use pre-made circle fraction pieces to create the cookies. It might be
necessary to combine several sets of pieces in order to make multiple cookies.
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
May 2012 y Page 40 of 76
## Georgia Department of Education
Common Core Georgia Performance Standards Framework
Fourth Grade Mathematics x Unit 3
## Name __________________________________________ Date _________________________
Order Form
You own a bakery that specializes in fraction cookies. Customers place orders
orders shown below. Before making the cookies to fill the order, you need to
confirm each order by sending a confirmation notice to each customer. Using the
circle templates below, show how you would create each cookie order with the
correct fractional amounts of toppings.
#4
#7
#8
#6
#5
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
May 2012 y Page 41 of 76
Whole
#3
Number of
Mixed
Number
#2
Improper
fraction
Sprinkles
#1
Chocolate
Icing
Order Totals
Vanilla
Icing
Peanut
Butter
Walnuts
Example
Raspberries
M & Ms
Order
Number
Chocolate
Chips
Toppings
## Georgia Department of Education
Common Core Georgia Performance Standards Framework
Fourth Grade Mathematics x Unit 3
## Name ____________________________________ Date _______________________________
Order Confirmation Form
Customers expect you to use the fewest number of cookies possible to complete each order.
No part of a cookie should be without a topping except for one. You may split a topping between
two cookies as shown below (the vanilla icing was shared between two cookies rather than
covering both halves of one cookie with vanilla icing).
Example:
M & Ms
Walnuts
Chocolate Chips
Raspberries
Peanut Butter
Mint Icing
Vanilla Icing
Chocolate Icing
Sprinkles
Colorful Circles
Squares
Brown Triangles
Red Circles
Light Brown
Light Green
Yellow
Dark Brown
Colorful Specs
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
May 2012 y Page 42 of 76
## Georgia Department of Education
Common Core Georgia Performance Standards Framework
Fourth Grade Mathematics x Unit 3
Order Confirmations, Page 2
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
May 2012 y Page 43 of 76
## Georgia Department of Education
Common Core Georgia Performance Standards Framework
Fourth Grade Mathematics x Unit 3
## Name __________________________________________ Date _________________________
Order Form Version 2
You recently received the orders shown below. Confirm each order below.
Using the circle templates below, show how you would create each cookie order
with the correct fractional amounts of toppings.
Customers expect you to use the fewest number of cookies possible to complete each order.
No part of a cookie should be without a topping except for one. You may split a topping between
two cookies as shown below (the vanilla icing was shared between two cookies rather than
covering both halves of one cookie with vanilla icing).
|
What are the solutions of a polynomial?
Table of Contents
What are the solutions of a polynomial?
A solution of a polynomial system is a tuple of values of (x1., xm) that satisfies all equations of the polynomial system. The solutions are sought in the complex numbers, or more generally in an algebraically closed field containing the coefficients.
What are examples of polynomial?
A polynomial is a type of algebraic expression in which the exponents of all variables should be a whole number.
• Example: Express the polynomial 5 + 2x + x2 in the standard form.
• For example, 2x and 3x are like terms.
• Example 1: A polynomial 3×4 + 7 has a degree equal to four.
• How many solutions will a polynomial equation have?
Every polynomial equation of degree n ≥ 1 has exactly n solutions, counting multiplicities.
How many solutions Does each polynomial have Y =- 6x 9?
These are three possible solutions to the equation.
Is 3x 7 a polynomial?
Is 3x-7 a Polynomial? 3x-7 is a polynomial.
What is the formula of polynomial?
Polynomial Formula The value of the exponent n can only be a positive integer as discussed above. Any polynomial function can be of the form. F(x) = anxn + an-1xn-1 + an-2xn-2 + . . . + a1x + a1 = 0. is the general formula of a polynomial.
How many solutions does a polynomial with a degree of 5 have?
three real solutions
However, the polynomial is a 5th degree polynomial, which the Fundamental Theorem of Algebra tells us will have 5 roots. We know from the graph that it has three real solutions, so what does this mean about the number of complex solutions?
What is an example of a one solution equation?
Linear Equations With one Solution Example 1: Consider the equation 7x – 35 = 0. On solving we have 7x = 35 or x = 5. The above linear equation is only true if x = 5 and hence the given linear equation has only one solution i.e. x = 5.
How do you do solutions in math?
A solution set is the set of all variables that makes the equation true. The solution set of 2y + 6 = 14 is {4}, because 2(4) + 6 = 14. The solution set of y2 + 6 = 5y is {2, 3} because 22 + 6 = 5(2) and 32 + 6 = 5(3).
How do you know how many solutions?
If solving an equation yields a statement that is true for a single value for the variable, like x = 3, then the equation has one solution. If solving an equation yields a statement that is always true, like 3 = 3, then the equation has infinitely many solutions.
Is 2x 3 a polynomial?
Given question: Write the degree of the polynomial 2x – 3. Solution: In the polynomial 2x – 3, the highest power of x is 1.
Is 12x a polynomial?
12x is a polynomial.
Is 5x 2 a polynomial?
Linear Polynomial: If the expression is of degree one then it is called a linear polynomial. For Example 5x+2,50z+3.
What are the 10 polynomial identities?
Polynomial Identities
• (x + y)2= x2 + 2xy + y2
• (x – y)2= x2 – 2xy + y2
• x2– y2 = (x + y)(x – y)
• (x + a)(x + b) = x2+ (a + b)x + ab.
• (x + y + z)2= x2 + y2 + c2 + 2xy + 2yz + 2zx.
• (x + y)3= x3 + y3 + 3xy (x + y)
• (x – y)3= x3 – y3 – 3xy (x – y)
• x3+ y3 = (x + y)(x2 – xy + y2)
What are all the types of polynomials?
Monomial. A polynomial containing one non zero term is called monomial.
• Binomial. A polynomial containing two non zero terms is called binomial.
• Trinomial. A polynomial containing three non zero terms is called trinomial.
• Quadrinomial polynomial. A polynomial containing four terms is called quadrinomial polynomial.
• Quintrinomial polynomial.
• What are some examples of a polynomial?
Polynomial Definition,Introduction Words like coefficients,constants,variables,may seem a lot to take in at first.
• Polynomials Addition and Subtraction The basic operations of addition and subtraction are not only for numbers.
• Polynomial Multiplication Multiplication of polynomials is something that can often require careful attention.
• How do you identify polynomials?
Answer:
• The prime polynomials are 1,4 and 5.
• Step-by-step explanation:
• Prime polynomials are the polynomial with integer coefficients that cannot be factored into lower degree polynomials.
• The prime polynomials are 1,4 and 5.
• How to identify polynomials?
monomial —A polynomial with exactly one term is called a monomial.
• binomial —A polynomial with exactly two terms is called a binomial.
• trinomial —A polynomial with exactly three terms is called a trinomial.
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# What is the factor tree of 108?
## What is the factor tree of 108?
Factors of 108 are the integers that can divide the original number evenly. There are a total of twelve factors of 108, they are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54 and 108….Pair Factors of 108.
Positive Factors of 108 Positive Pair Factors of 108
3×36 (3, 36)
4×27 (4, 27)
6×18 (6, 18)
9×12 (9, 12)
How many factors does 54 have?
Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54.
### Can 54 be divided?
When we list them out like this it’s easy to see that the numbers which 54 is divisible by are 1, 2, 3, 6, 9, 18, 27, and 54.
What’s the prime factor of 54?
2 × 3 × 3 × 3
Answer: The prime factorization of 54 = 2 × 3 × 3 × 3 = 2 × 3.
#### What are prime factors of 54?
So, the prime factors of 54 are written as 2 × 3 × 3 × 3 or 2x 33, where 2 and 3 are the prime numbers.
What are the common factors of 108 and 54?
The factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108. The factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54. Hence, the common factors of 108 and 54 are 1, 2, 3, 6, 9, 18, 27 and 54.
## How to draw the factor tree for 108?
Here you can find the answer to questions related to: Factor tree for 108 or how to draw the factor tree for 108. The procedure below applies to any non-prime number. Look at the 2 factors and determine if at least one of them is not prime; Repeat this process until all factors are prime. Supose you want to find the factor tree of 32.
What is the factor tree of 54?
Factor Tree of 54 is the list of prime factors when multiplied it results in the original number ie., 54. Factor Tree is the easiest way to find the factors of a given number.
### What are the prime factors of 108?
Hence, the factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108. In the prime factorization of 108, the number 108 is written as the product of its prime factors. Now, let us discuss how to find the prime factors of 108. Now, 27 is an odd number and cannot be divided by 2. Divide 27 by the next prime number, i.e.,3. 27/3 = 9
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# What Is The 30Th Term Of The Arithmetic Series 4, 7, 10, €¦?
Contents
## What Is The 30th Term Of The Arithmetic Series 4 7 10 €¦??
The 30th term of the arithmetic progression is 91.
## How do you find the 30th term in a sequence?
1. Given: arithmetic series 2 5 8.
2. d= the common difference.
3. d=a2−a1=a3−a2=8−5=3.
4. To find the 30th term let n=30 :
## How do you find the 30th term of AP?
The correct option is c = – 77. Thus 30th term of the AP is – 77.
## What is the 30th term of the AP 10 7 4 and so on is?
-77
Let us name it as d. So d = -3. Hence \${{30}^{th}}\$ term of arithmetic progression 10 7 4 . . . . . . . . . is -77. So the correct answer is “Option C”.
## What is the 30th term of the arithmetic sequence 3/5 7?
Find the 30th term of 1 3 5 7 … . The 30th term is 59.
## What is the 30th number in this sequence 2 5 8?
With this we can plug in 30 for x and get 89 .
## How do you find the sum of an arithmetic series?
To find n use the explicit formula for an arithmetic sequence. We solve 3 + (n – 1)·4 = 99 to get n = 25.
## What is the nth term of the AP 5 2?
Substitute the values → an = 5 + (n – 1)(-3) → an = 5 – 3n + 3. → an = 8 – 3n. Hence the nth term of an AP 5 2 -1 -4 -7 ………. is 8 – 3n.
## Which term is AP?
Formula Lists
General Form of AP a a + d a + 2d a + 3d . . .
The nth term of AP an = a + (n – 1) × d
Sum of n terms in AP S = n/2[2a + (n − 1) × d]
Sum of all terms in a finite AP with the last term as ‘l’ n/2(a + l)
## Which of the term of AP 5 2 1 is?
Which term of an A.P. 5 2 -1 ……… is -22? ∴ -22 is the 10th term. for n = 1 2 3 4 5.
## Which term is ap1 4/7 88?
Hence 30th term of AP is 88.
## Which term of the AP 3.8 13.18 is 78?
16th term
78 is the 16th term of the given AP.
## Which term of AP is 210?
Hence the 10th term of an AP is 210.
## What is the arithmetic mean between 10 and 24?
Thus 10 + 24 2 = 17 is the arithmetic mean.
## Which term of the arithmetic sequence 2.6 10 is 102?
26th term of arithmetic sequence 2 6 10…is 102.
## What are next four terms of the sequence 2 3 5 8?
The Fibonacci Sequence is the series of numbers: 0 1 1 2 3 5 8 13 21 34 …
## Which term of the following series is 320?
Solution(By Examveda Team)
Clearly 5 + 3 = 8 8 + 3 = 11 11 + 3 = 14 ….. So the series is an A.P. in which a – 5 and d = 3. Let 320 be the nth term of the series. Then 320 = 5 + (n – 1) x 3 or (n – 1) = 105 or n = 106.
## What is the sum of 30 terms?
It is given that the terms are in arithmetic sequence. Therefore the sum of the first 30 terms is 3225.
## What is the sum of 2nd and 30th terms?
Sum of 2nd and 30th term is 50.
## What is arithmetic formula?
An arithmetic formula is a sequence of numbers that is ordered with a specific pattern. Each successive number is the sum of the previous number and a constant. The constant is the same for every term in the sequence and is called the common difference.
## What is series formula?
The series of a sequence is the sum of the sequence to a certain number of terms. It is often written as Sn. So if the sequence is 2 4 6 8 10 … the sum to 3 terms = S3 = 2 + 4 + 6 = 12. The Sigma Notation.
## What is the figure of arithmetic sequence?
An arithmetic sequence is a list of numbers with a definite pattern. If you take any number in the sequence then subtract it by the previous one and the result is always the same or constant then it is an arithmetic sequence.
## Which of the term of AP 5 2 − 1 is − 49?
19 th term is -49.
## What is the term of AP 17 12 7?
Therefore 22-5n is the nth term of the given AP.
## Which term of AP is 109?
Hence 22nd term of the AP is 109.
## Which of the following is one of the terms of the arithmetic sequence 4/7 10?
The common difference of the arithmetic sequence 4 7 10 13 16 … is 3. So the correct answer is “4 7 10 13 16 … is 3”.
## How many terms are in AP?
12 terms must be taken for AP.
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# NCERT solution class 9 chapter 12 Heron’s Formula exercise 12.1 mathematics
## EXERCISE 12.1
#### Question 1:
A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Side of traffic signal board = a
Perimeter of traffic signal board = 3 × a
By Heron’s formula,
Perimeter of traffic signal board = 180 cm
Side of traffic signal board
Using equation (1), area of traffic signal board
#### Question 2:
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122m, 22m, and 120m (see the given figure). The advertisements yield an earning of Rs 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
The sides of the triangle (i.e., abc) are of 122 m, 22 m, and 120 m respectively.
Perimeter of triangle = (122 + 22 + 120) m
2s = 264 m
s = 132 m
By Heron’s formula,
Rent of 1 m2 area per year = Rs 5000
Rent of 1 m2 area per month = Rs
Rent of 1320 m2 area for 3 months =
= Rs (5000 × 330) = Rs 1650000
Therefore, the company had to pay Rs 1650000.
#### Question 3:
There is a slide in the park. One of its side walls has been painted in the same colour with a message “KEEP THE PARK GREEN AND CLEAN” (see the given figure). If the sides of the wall are 15m, 11m, and 6m, find the area painted in colour.
It can be observed that the area to be painted in colour is a triangle, having its sides as 11 m, 6 m, and 15 m.
Perimeter of such a triangle = (11 + 6 + 15) m
s = 32 m
s = 16 m
By Heron’s formula,
Therefore, the area painted in colour is.
#### Question 4:
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Let the third side of the triangle be x.
Perimeter of the given triangle = 42 cm
18 cm + 10 cm + = 42
x = 14 cm
By Heron’s formula,
#### Question 5:
Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.
Let the common ratio between the sides of the given triangle be x.
Therefore, the side of the triangle will be 12x, 17x, and 25x.
Perimeter of this triangle = 540 cm
12x + 17x + 25x = 540 cm
54x = 540 cm
x = 10 cm
Sides of the triangle will be 120 cm, 170 cm, and 250 cm.
By Heron’s formula,
Therefore, the area of this triangle is 9000 cm2.
#### Question 6:
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
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# Friday Morning Math Problem
## Fifty Coins
Find the number of ways that change can be made for \$1.00, using exactly 50 coins.
Solution
Denote the number of pennies, nickels, dimes, quarters, and fifty-cent pieces as x1, x5, x10, x25, x50
Now, we know already that x50 = 0, since having even one fifty-cent piece would require us to make 50 cents from 49 coins, which is impossible.
We can express the total sum of all the coins as the equation:
x1 + 5 * x5 + 10 * x10 + 25 * x25 = 100 (I)
Further, we can express the number of pennies in terms of the number of other coins:
x1 = 50 - (x5 + x10 + x25) (II)
Substituting this into the first equation gives us (one equation with three unknowns):
25 * x25 + 10 * x10 + 5 * x5 + 50 - (x25 + x10 + x5) = 100 (III)
or
5 x25 + 2 x10 + x5 - (1/5)( x25 + x10 + x5 ) = 10 (IV)
We know a few other things that can help narrow down the space of possible solutions:
• The value of all of the coins must be a multiple of 5 (as implied by equation I, since the difference of two multiples of 5 is also a multiple of 5). So, the number of pennies x1 is a multiple of 5, and the quantity x25 + x10 + x5 is also a multiple of 5 (by equation II), and therefore the quantity 25 * x25 + 10 * x10 + 5 * x5 is also a multiple of 5 (by equation III).
• The value of x25 + x10 + x5 > 0 (we know there are no fifty-cent pieces, and we know the case of all pennies will not add up to \$1.00, so we need at least one nickel, dime, or quarter) and is a multiplle of 5 (by equation II).
• We know 25 x25 + 10 x10 + 5 x5 > 50 (by equation III).
Now we have greatly reduced the problem space. We know that the number of pennies must be a multiple of 5, somewhere between 5 and 45, and the sum of the face values of the remaining coins must be larger than 50.
There are three unknowns x25, x10, x5. We know the number of quarters x25 is 1 or 0, so we can try both values.
In the case of x1 = 45 and x25 = 1, we know that there are either 4 (5-1) or 9 (10-1) remaining coins, since x25 + x10 + x5 must be a multiple of 5. We have only a handful of outcomes to try, and (x5 = 2, x10 = 2) satisfies both the inequality 25 x25 + 10 x10 + 5 x5 > 50 and the master equation (I) (that all coins sum to \$1.00).
In the case of x1 = 40 and x25 = 0, we know that the number of remaining coins is a multiple of 5. No combination of 5 nickels and dimes can satisfy our inequality 25 x25 + 10 x10 + 5 x5 > 50, so we know we need 10 nickels and dimes. Starting with x10 = 2 (the only number of dimes that will satisfy the inequality), we find that 2 dimes implies 8 nickels, which is another solution.
x1 = 35 yields no valid solutions, and no other solutions are possible.
To decide where to start, we can guess that the number of pennies will be large, so we start with x1 = 45.
Furthermore, to save ourselves some time, we know that the number of pennies will be large, so we can start with 45 pennies, and try all the solutions where the number of pennies is less than 50 but a multiple of 5.
x1 = 45 implies
and x1 = 40 both yield valid solutions, so the total solutions are:
1 quarter, 45 pennies,
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Simplify: $\dfrac{x-2}{{x}^{2}-4}+\dfrac{{x}^{2}}{x-2}-\dfrac{{x}^{3}+x-4}{{x}^{2}-4}, ~~(x \neq \pm2)$.
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Simplifying fractions
Simplify: $\dfrac{x-2}{{x}^{2}-4}+\dfrac{{x}^{2}}{x-2}-\dfrac{{x}^{3}+x-4}{{x}^{2}-4}, ~~(x \neq \pm2)$.
Step 1 - Factorise the denominators
$\begin{equation*} \frac{x-2}{(x+2)(x-2)}+\frac{{x}^{2}}{x-2}-\frac{{x}^{3}+x-4}{(x+2)(x-2)} \end{equation*}$
Step 2 - Make all denominators the same so that we can add or subtract the fractions - The lowest common denominator is $(x-2)(x+2)$.
$\begin{equation*} \frac{x-2}{(x+2)(x-2)}+\frac{({x}^{2}) (x+2)}{(x+2)(x-2)}-\frac{{x}^{3}+x-4}{(x+2)(x-2)} \end{equation*}$
Step 3 - Write as one fraction
$\begin{equation*} \frac{x-2+({x}^{2})(x+2)-(x^{3}+x-4)}{(x+2)(x-2)} \end{equation*}$
Step 4 - Simplify
$\begin{equation*} \dfrac{x-2+{x}^{3}+ 2x^{2}-x^{3} - x+4}{(x+2)(x-2)} = \dfrac{2x^{2} + 2}{(x+2)(x-2)} \end{equation*}$
Step 5 - Take out the common factor and write the final answer
$\begin{equation*} \dfrac{2({x}^{2} +1)}{(x+2)(x-2)} \end{equation*}$
by Diamond (89,356 points)
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# 2020 AMC 12B Problems/Problem 13
## Problem
Which of the following is the value of $\sqrt{\log_2{6}+\log_3{6}}?$
$\textbf{(A) } 1 \qquad\textbf{(B) } \sqrt{\log_5{6}} \qquad\textbf{(C) } 2 \qquad\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}} \qquad\textbf{(E) } \sqrt{\log_2{6}}+\sqrt{\log_3{6}}$
## Solution 1 (Properties of Logarithms)
Recall that:
1. $\log_b{(uv)}=\log_b u + \log_b v.$
2. $\log_b u\cdot\log_u b=1.$
We use these properties of logarithms to rewrite the original expression: \begin{align*} \sqrt{\log_2{6}+\log_3{6}}&=\sqrt{(\log_2{2}+\log_2{3})+(\log_3{2}+\log_3{3})} \\ &=\sqrt{2+\log_2{3}+\log_3{2}} \\ &=\sqrt{\left(\sqrt{\log_2{3}}+\sqrt{\log_3{2}}\right)^2} \\ &=\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}. \end{align*} ~MRENTHUSIASM (Solution)
~JHawk0224 (Proposal)
## Solution 2 (Change of Base Formula)
First, $$\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\frac{\log{6}}{\log{2}} + \frac{\log{6}}{\log{3}}} = \sqrt{\frac{\log{6}\cdot\log{3} + \log{6}\cdot\log{2}}{\log{3}\cdot\log{2}}} = \sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}}.$$ From here, \begin{align*}\sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}} &= \sqrt{\frac{(\log 2 + \log 3)(\log 2 + \log 3)}{\log 2\cdot \log 3}} &= \sqrt{\frac{(\log 2 + \log 3)^2}{\log 2\cdot \log 3}} &= \frac{\log 2}{\sqrt{\log 2\cdot\log 3}} + \frac{\log 3}{\sqrt{\log 2\cdot\log 3}} \\ &= \sqrt{\frac{\log 2}{\log 3}} + \sqrt{\frac{\log 3}{\log 2}} \\ &= \sqrt{\log_3 2} + \sqrt{\log_2 3}. \end{align*} Answer: $\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}$
Note that in this solution, even the most minor steps have been written out. On the actual test, this solution would be quite fast, and much of it could easily be done in your head.
~ TheBeast5520
~ LeonidasTheConquerer (removed unnecessary steps)
## Solution 3 (Observations)
Using the knowledge of the powers of $2$ and $3,$ we know that $\log_2{6}>2.5$ and $\log_3{6}>1.5.$ Therefore, $$\sqrt{\log_2{6}+\log_3{6}}>\sqrt{2.5+1.5}=2.$$ Only choices $\textbf{(D)}$ and $\textbf{(E)}$ are greater than $2,$ but $\textbf{(E)}$ is certainly incorrect: If we compare the squares of the original expression and $\textbf{(E)},$ then they are clearly not equal. So, the answer is $\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.$
~Baolan
~chrisdiamond10
~MRENTHUSIASM (reformatted and merged the thoughts of all contributors)
## Solution 4 (Solution 3 but More Detailed)
Note: Only use this method if all else fails and you cannot find a way to simplify the logarithms.
We can see that $\log_2{6}$ is greater than $2$ and less than $3.$ Additionally, since $6$ is halfway between $2^2$ and $2^3,$ knowing how exponents increase more the larger $x$ is, we can deduce that $\log_2{6}$ is just above halfway between $2$ and $3.$ We can guesstimate this as $\log_2{6} \approx 2.55.$ (It's actually about $2.585.$)
Next, we think of $\log_3{6}.$ This is greater than $1$ and less than $2.$ As $6$ is halfway between $3^1$ and $3^2,$ and similar to the logic for $\log_2{6},$ we know that $\log_3{6}$ is just above halfway between $1$ and $2.$ We guesstimate this as $\log_3{6} \approx 1.55.$ (It's actually about $1.631.$)
So, $\log_2{6} + \log_3{6}$ is approximately $4.1.$ The square root of that is just above $2,$ maybe $2.02.$ We cross out all choices below $\textbf{(C)}$ since they are less than $2,$ and $\textbf{(E)}$ can't possibly be true unless either $\log_2{6}$ and/or $\log_3{6}$ is $0$ (You can prove this by squaring.). Thus, the only feasible answer is $\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.$
~PureSwag
## Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
(This solution is wrong as it involves an identity that is not true ~pengf)
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# If ${{x}^{2}}+x+1=0$, then the value of ${{\left( x+\dfrac{1}{x} \right)}^{2}}+{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}+.....+{{\left( {{x}^{27}}+\dfrac{1}{{{x}^{27}}} \right)}^{2}}$ isA $27$B $72$C $45$D $54$
Last updated date: 17th Jul 2024
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Hint: To get the value of the given equation we will use the root of unity concept. Firstly we will let the variable in question be equal to the root of unity. Then by using the value and relation of the roots of unity we will put the values in the equation. Finally we will simplify the equation to get the desired answer.
The equation is given as below:
${{x}^{2}}+x+1=0$…….$\left( 1 \right)$
So it will have roots as:
$x=\omega ,{{\omega }^{2}}$
Where $\omega ,{{\omega }^{2}}$ are cube roots of unity.
On substituting above value in equation (1) we get,
${{\omega }^{2}}+\omega +1=0$……$\left( 2 \right)$
Also we know cube of an imaginary cube root is unity so,
${{\omega }^{3}}=1$…..$\left( 3 \right)$
So fro above value we can get:
\begin{align} & {{\omega }^{2}}.\omega =1 \\ & {{\omega }^{2}}=\dfrac{1}{\omega } \\ \end{align}…..$\left( 4 \right)$
${{\omega }^{3n}}=1$…….$\left( 5 \right)$
We have to find the value of below equation:
${{\left( x+\dfrac{1}{x} \right)}^{2}}+{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}+.....+{{\left( {{x}^{27}}+\dfrac{1}{{{x}^{27}}} \right)}^{2}}$
Using equation (2) - (5) we will solve above equation as:
\begin{align} & \Rightarrow {{\left( \omega +\dfrac{{{\omega }^{3}}}{\omega } \right)}^{2}}+{{\left( {{\omega }^{2}}+\dfrac{{{\omega }^{3}}}{{{\omega }^{2}}} \right)}^{2}}+{{\left( {{\omega }^{3}}+\dfrac{{{\omega }^{3}}}{{{\omega }^{3}}} \right)}^{2}}.....+{{\left( {{\omega }^{27}}+\dfrac{{{\omega }^{3}}}{{{\omega }^{27}}} \right)}^{2}} \\ & \Rightarrow {{\left( \omega +{{\omega }^{2}} \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( {{\omega }^{3}}+1 \right)}^{2}}.....+{{\left( {{\omega }^{27}}+\dfrac{1}{{{\omega }^{24}}} \right)}^{2}} \\ & \Rightarrow {{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 1+1 \right)}^{2}}....+{{\left( {{\omega }^{27}} +\dfrac{1}{1} \right)}^{2}} \\ & \Rightarrow {{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}}....+{{\left( 1+1 \right)}^{2}} \\ & \Rightarrow {{\left( -1 \right)}^{2}}+ {{\left( -1 \right)}^{2}}+......{{\left( 2 \right)}^{2}} \\ \end{align}
This simplified further give,
${{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{2}^{2}}+.....{{\left( 2 \right)}^{2}}$
So from the above value we get that we are getting two terms same then a different term it will go on like this
\begin{align} & \Rightarrow 18{{\left( -1 \right)}^{2}}+9{{\left( 2 \right)}^{2}} \\ & \Rightarrow 18+9\times 4 \\ & \Rightarrow 18+36 \\ & \Rightarrow 54 \\ \end{align}
So value is obtained as 54.
So, the correct answer is “Option D”.
Note: A number which when raised to the power 3 gives the answer as 1 is known as cube root of unity. It is widely used in many branches of mathematics. There are three roots of unity in which two are complex roots and one is a real root. Some properties of the cube root of unity are that when one imaginary root is squared it gives another root of unity. When two complex roots are multiplied the answer comes as 1.
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##### Stage: 2 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### Simple Train Journeys
##### Stage: 1 and 2 Challenge Level:
How many different journeys could you make if you were going to visit four stations in this network? How about if there were five stations? Can you predict the number of journeys for seven stations?
### Round and Round the Circle
##### Stage: 2 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Round the Two Dice
##### Stage: 1 Challenge Level:
This activity focuses on rounding to the nearest 10.
### Round the Three Dice
##### Stage: 2 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
### Round the Dice Decimals 1
##### Stage: 2 Challenge Level:
Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number?
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# Mixed Numbers Calculator
Mixed Numbers Calculator is designed to perform the arithmetical operations like addition, subtraction, multiplication and division on any two mixed numbers easily. You have to enter the mixed numbers in the input fields, select any one operation, and press on the calculate button to check the result in a short span of time.
Ex.
+ - × ÷
Here are some samples of Mixed Number Calculator calculations.
Mixed Numbers Calculator: Obtain the best tool that calculates the arithmetic operations of any two mixed numbers in a simpler manner. Have a look at the details like what are the steps to perform addition, subtraction, multiplication and division of mixed numbers manually and the solved example questions. Make use of the detailed explanation to understand the concept easily. This handy Calculator produces the answer in fraction of seconds.
## Steps to Perform Arithmetic Operations on Mixed Numbers
Below provided are some simple steps that are useful to add, subtract, multiply and divide two mixed numbers. Keep an eye on them and go through the instructions to find the result easily.
• Let us take any two mixed numbers to perform the arithmetical operations.
• Convert the mixed numbers to improper fraction.
• Then, place the symbol in between those two improper fraction numbers.
• Now, add or subtract or multiply or divide those numbers.
• Obtained fraction will be the result.
Example
Question: Perform all arithmetic operations on 5(2/3) and 4(6/8)?
Solution:
Given mixed numbers are 5(2/3) and 4(6/8)
The improper fraction of 5(2/3) is 5 + 2/3 = (15 + 2) / 3 = 17/3
Improper fraction of 4(6/8) is 4 + 6/8 = (32 + 6) / 8 = 38/8
= 17/3 + 38/8
Get the lcm of 3, 8 i.e 24
17/3 + 38/8 = (17 * 8 + 38 * 3) / 24
= 250/24
Subtraction: 5(2/3) - 4(6/8)
= 17/3 - 38/8
= (17 * 8 - 38 * 3) / 24 = (136 - 114) / 24
= 22/24
Multiplication: 5(2/3) * 4(6/8)
= 17/3 * 38/8
= (17 * 38) / (8 * 3)
= 532 / 24
Division: 5(2/3) / 4(6/8)
= 17/3 / 38/8
= (17 * 8) / (3 * 38)
= 136 / 114
∴ 5(2/3) + 4(6/8) = 250/24, 5(2/3) - 4(6/8) = 22/24, 5(2/3) * 4(6/8) = 532/24, 5(2/3) / 4(6/8) = 136/114.
Check the free online calculators to find the addition, subtraction, multiplication and division of the mixed numbers at our website Onlinecalculator.guru along with other math concepts.
### FAQs on Mixed Numbers Calculator
1. What is the example of mixed number?
Mixed number is a combination of whole number and a fraction. Some of the examples of mixed numbers are 2(3/4), 5(6/7), 6(9/10), etc.
2. How do you simplify mixed numbers?
The simpler form of mixed number can be a fraction or decimal number. For example take 2(3/4), to simplify this mixed number you have to add whole number and fraction.
2(3/4) = 2 + 3/4 = (8 + 3) / 4 = 11/4
3. How do you calculate arithmetical operations on mixed numbers using calculator?
Provide two mixed numbers in the specified input numbers and press on the calculate button. You must also choose any one arithmetical operation in the calculator to get the exact output easily.
4. What is the formula to perform the arithmetical operations on a/b & c/d?
Addition is a/b + c/d = (a * d + b * c) / (b * d)
Subtraction is a/b - c/d = (a * d - b * c) / (b * d)
Multiplication is a/b * c/d = (a * c) / (b * d)
Division (a/b) / (c/d) = (a * d) / (b * c)
|
# How to find position vector between two points
It’s important to keep them in mind when trying to figure out How to find position vector between two points.
## Precalculus : Find a Direction Vector When Given Two Points
Example 1: Given two points P = (-4, 6) and Q = (5, 11), determine the position vector PQ. Solution: If two points are given in the xy-coordinate system, then we can use the following formula to find the position vector PQ: PQ = (x 2 - x 1, y 2
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## Position Vector
How to Find Vector Coordinates From Two Points in 2d and 3d Space. In order to calculate the coordinates of a vector from two points, you must use one of the formulas. AB = {xB - xA ; yB -
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## What is a Position Vector?
let point A be (700, 500) let point B be (400, 400) let point C be (650, 100) let point D be (???, ???) the vector from A to B is: (-300, -100) // i.e. x = B-A, 400 - 700, etc the vector from A to C is: (-50, -400) Adding these together
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## Position Vector (video lessons, examples and solutions)
Given two points in the xy-coordinate system, we can use the following formula to find the position vector AB: AB = (x2-x1, y2-y1) Where x1, y1 represents the coordinates of point A and x2, y2 represent the point B coordinates.
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If you're struggling with arithmetic, there's help available online. You can find websites that offer step-by-step explanations of various concepts, as well as online calculators and other tools to help you practice.
## How can I find the position vector of the point A?
Correct answer: Explanation: The correct vector is given by the subtraction of the two points: . Since the subtraction here is component-wise, it is given by the formula: . This results in the
Explain math equation
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## Number Properties - Official Questions on Rounding
Some weeks back, we looked at some rounding rules. Today, let’s go over some official questions on rounding. They are quite simple and if we just keep the “Slip to the side and look for a 5” rule in mind, they can be easily solved.
Question 1: If n = 2.0453 and n* is the decimal obtained by rounding n to the nearest hundredth, what is the value of n* – n?
(A) -0.0053
(B) -0.0003
(C) 0.0007
(D) 0.0047
(E) 0.0153
Solution: A quick note on place value nomenclature:
Given a decimal 345.789, we know that 5 represents the units digit, 4 the tens digit and 3 the hundreds digit. Also, 7 represents the tenths digit, 8 the hundredths digit and 9 the thousandths digit and so on…
Now let’s go back to this question:
n = 2.0453
We need to round n to the nearest hundredth which means we will retain 2 digits after the decimal. The third digit after the decimal is 5 so 2.0453 rounded to the nearest hundredth is 2.05.
Thus n* – n = 2.05 – 2.0453 = 0.0047
Question 2: If digit h is the hundredths digit in the decimal n = 0.2h6, what is the value of n, rounded to the nearest tenth?
Statement 1: n < 1/4
Statement 2: h < 5
Solution: Given that n = 0.2h6
We need to find the value of n rounded to the nearest tenth i.e. we need to keep only one digit after the decimal.
Statement 1: n < 1/4
In decimal form, it means n < 0.25
If h were 5 or greater, n would become 0.256 or 0.266 or higher. All these values would be more than 0.25 so h must be less than 5 such as 0.246 or 0.236 etc. In all such cases, n would be rounded to 0.2
This statement alone is sufficient.
Statement 2: h < 5
This is even simpler. Since we have been given that h is less than 5, when we round n to the tenths digit, we will get 0.2
This statement alone is also sufficient.
Question 3: If d denotes a decimal number, is d >= 0.5?
Statement 1: When d is rounded to the nearest tenth, the result is 0.5.
Statement 2: When d is rounded to the nearest integer, the result is 1.
Solution: Again, a simple question!
We need to find whether d is greater than or equal to 0.5 or not.
Statement 1: When d is rounded to the nearest tenth, the result is 0.5.
This means that whatever d is, when we round it to the nearest tenth, we get 0.5. What are the possible values of d? If d is anywhere from 0.450 to 0.5499999…, it will be rounded to 0.5
Some of these numbers are less than 0.5 and others are greater than 0.5 so this statement alone is not sufficient.
Statement 2: When d is rounded to the nearest integer, the result is 1.
In this case d must be at least 0.5; only then can it be rounded to 1.
d can be anything from 0.50 to 1.499999… In any case, d will be greater than or equal to 0.5.
This statement alone is sufficient to answer the question.
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11.7 Solving systems with inverses (Page 2/8)
Page 2 / 8
Show that the following two matrices are inverses of each other.
$A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right],B=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]$
$\begin{array}{l}AB=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]=\left[\begin{array}{rrr}\hfill 1\left(-3\right)+4\left(1\right)& \hfill & \hfill 1\left(-4\right)+4\left(1\right)\\ \hfill -1\left(-3\right)+-3\left(1\right)& \hfill & \hfill -1\left(-4\right)+-3\left(1\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \\ BA=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]=\left[\begin{array}{rrr}\hfill -3\left(1\right)+-4\left(-1\right)& \hfill & \hfill -3\left(4\right)+-4\left(-3\right)\\ \hfill 1\left(1\right)+1\left(-1\right)& \hfill & \hfill 1\left(4\right)+1\left(-3\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \end{array}$
Finding the multiplicative inverse using matrix multiplication
We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication .
Finding the multiplicative inverse using matrix multiplication
Use matrix multiplication to find the inverse of the given matrix.
$A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill -2\\ \hfill 2& \hfill & \hfill -3\end{array}\right]$
For this method, we multiply $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by a matrix containing unknown constants and set it equal to the identity.
Find the product of the two matrices on the left side of the equal sign.
Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.
Using row operations, multiply and add as follows: $\text{\hspace{0.17em}}\left(-2\right){R}_{1}+{R}_{2}\to {R}_{2}.\text{\hspace{0.17em}}$ Add the equations, and solve for $\text{\hspace{0.17em}}c.$
$\begin{array}{r}\hfill 1a-2c=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill 0+1c=-2\\ \hfill c=-2\end{array}$
Back-substitute to solve for $\text{\hspace{0.17em}}a.$
$\begin{array}{r}\hfill a-2\left(-2\right)=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill a+4=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill a=-3\end{array}$
Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.
$\begin{array}{rr}\hfill 1b-2d=0& \hfill {R}_{1}\\ \hfill 2b-3d=1& \hfill {R}_{2}\end{array}$
Using row operations, multiply and add as follows: $\text{\hspace{0.17em}}\left(-2\right){R}_{1}+{R}_{2}={R}_{2}.\text{\hspace{0.17em}}$ Add the two equations and solve for $\text{\hspace{0.17em}}d.$
$\begin{array}{r}\hfill 1b-2d=0\\ \hfill \frac{0+1d=1}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}d=1}\\ \hfill \end{array}$
Once more, back-substitute and solve for $\text{\hspace{0.17em}}b.$
$\begin{array}{r}\hfill b-2\left(1\right)=0\\ \hfill b-2=0\\ \hfill b=2\end{array}$
${A}^{-1}=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill 2\\ \hfill -2& \hfill & \hfill 1\end{array}\right]$
Finding the multiplicative inverse by augmenting with the identity
Another way to find the multiplicative inverse is by augmenting with the identity. When matrix $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is transformed into $\text{\hspace{0.17em}}I,\text{\hspace{0.17em}}$ the augmented matrix $\text{\hspace{0.17em}}I\text{\hspace{0.17em}}$ transforms into $\text{\hspace{0.17em}}{A}^{-1}.$
For example, given
$A=\left[\begin{array}{rrr}\hfill 2& \hfill & \hfill 1\\ \hfill 5& \hfill & \hfill 3\end{array}\right]$
augment $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ with the identity
Perform row operations with the goal of turning $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ into the identity.
1. Switch row 1 and row 2.
2. Multiply row 2 by $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and add to row 1.
3. Multiply row 1 by $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and add to row 2.
4. Add row 2 to row 1.
5. Multiply row 2 by $\text{\hspace{0.17em}}-1.$
The matrix we have found is $\text{\hspace{0.17em}}{A}^{-1}.$
${A}^{-1}=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill -1\\ \hfill -5& \hfill & \hfill 2\end{array}\right]$
Finding the multiplicative inverse of 2×2 matrices using a formula
When we need to find the multiplicative inverse of a $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.
If $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is a $\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$ matrix, such as
$A=\left[\begin{array}{rrr}\hfill a& \hfill & \hfill b\\ \hfill c& \hfill & \hfill d\end{array}\right]$
the multiplicative inverse of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is given by the formula
${A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{rrr}\hfill d& \hfill & \hfill -b\\ \hfill -c& \hfill & \hfill a\end{array}\right]$
where $\text{\hspace{0.17em}}ad-bc\ne 0.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}ad-bc=0,\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ has no inverse.
Using the formula to find the multiplicative inverse of matrix A
Use the formula to find the multiplicative inverse of
$A=\left[\begin{array}{cc}1& -2\\ 2& -3\end{array}\right]$
Using the formula, we have
$\begin{array}{l}{A}^{-1}=\frac{1}{\left(1\right)\left(-3\right)-\left(-2\right)\left(2\right)}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{-3+4}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \end{array}$
Questions & Answers
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
Kc Reply
1+cos²A/cos²A=2cosec²A-1
Ramesh Reply
test for convergence the series 1+x/2+2!/9x3
success Reply
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
Lhorren Reply
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
jancy Reply
answer
Ajith
exponential series
Naveen
what is subgroup
Purshotam Reply
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
Macmillan Reply
e power cos hyperbolic (x+iy)
Vinay Reply
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
Payal Reply
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
Tejas Reply
why {2kπ} union {kπ}={kπ}?
Huy Reply
why is {2kπ} union {kπ}={kπ}? when k belong to integer
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if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
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what is complex numbers
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Please you teach
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Yes
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give me treganamentry question
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Solve 2cos x + 3sin x = 0.5
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0
How do get 24 out of 8 5 6 4?
Updated: 9/17/2023
Wiki User
14y ago
6 * 4 + 80 - 50 = 24
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14y ago
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Q: How do get 24 out of 8 5 6 4?
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Related questions
How do you make 24 by using 3 4 5 6?
5+3=8 8-4=4 4*6=24
Can you make 24 using 5 4 8 8?
24 is the same as 6x4, and we have the 4, so we just need to make a 6... 6 = 5 + (8÷8) Therefore: 4 x (5+(8÷8)) = 24
What is six five eighths minus four and three thirds?
(6 and 5/8) - (4 and 3/3) = (6 and 15/24) - (4 and 24/24) = (5 and 39/24) - (4 and 24/24) = 1 and 15/24 = 1 and 5/8. Alternatively, notice that 3/3 = 1, so (4 and 3/3) = 5. Then (6 and 5/8) - (4 and 3/3) = (6 and 5/8) - (5) = 1 and 5/8.
8*(4-(6-5)) = 24
Which correctly renames 7/8 and 5/6 using a common denominator?
common denominator is going to be 24 8=222 6=2*3 24=222*3 8 should be multiplied by 3 6 should be multiplied by 4 5/6=(5*4)/(6*4)=20/24 7/8=21/24 5/6=20/24```
5-(7/7)*6=24
6*4=24
(4+5-6)x8=24
What is the least common multiple of 5 3 4 6 and 8?
LCM of 3, 4, 6 & 8 is 24, LCM of 5 and 24 is 120.
What is 5 3 8 plus 7 5 6?
When working with mixed numbers, it is usually easier to convert them to improper fractions, do the operations, and then to convert any resulting improper fraction back to a mixed number: 5 3/8 + 7 5/6 = (5×8+3)/8 + (7×6+5)/6 = 43/8 + 47/6 = (43×3)/(8×3) + (47×4)/(6×4) = 129/24 + 188/24 = 317/24 = (13×24+5)/24 = 13 5/24
How can you make 24 using numbers 5 1 8 4?
Here is one way: (5 + 1)*(8 - 4) = 6 * 4 = 24
What is five sixths plus one?
1/2 x 5/6 is 5/12
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Semicircle perimeter formula explained
Learn and know how we find the semicircle perimeter. For calculating this we have a formula, so learn it and find the perimeter.
Generally, all the students will think that as area of semicircle is half of the area of circle, in the similar manner they think that perimeter of the semicircle will be half of the perimeter of circle. But this is the wrong assumption. So now we will learn the formula to find semicircle perimeter.
What does semicircle means in math?
We know what is a circle? And we know how it looks like. Suppose if the circle is cut along the one of the diameter, then we get two equal parts. In those two equal parts, the each part we call it as a semicircle. In simple words we say what is semicircle means we can say as it’s a half of the circle.
The exact meaning of perimeter as follows:
Generally, we will be hearing the word “perimeter” in geometry or in mensuration chapters. The meaning of perimeter is “the total length of boundary of given figure is called as the perimeter”. Suppose if it is a polygon then the perimeter is given as the sum of all the line segments. In case of circle, we call perimeter as the circumference.
Formula to find the perimeter of semicircle?
We know what is the semicircle called? So now we will try to find perimeter of it. To get the perimeter of the semicircle, we have to find the sum of the diameter and the semicircle arc length.
Length of Semicircle arc is given as r.
Length of diameter in terms of radius is given as 2r.
Therefore, the perimeter of the semicircle is given as r + 2r, where the value of is taken as 22/7 and “r” is the radius of the circle.
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# 3.2 Vector addition and subtraction: graphical methods (Page 5/15)
Page 5 / 15
## Conceptual questions
Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this book, the Earth’s population, the acceleration of gravity?
Give a specific example of a vector, stating its magnitude, units, and direction.
What do vectors and scalars have in common? How do they differ?
Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each camper?
If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up anywhere on the circle shown in [link] . What other information would he need to get to Sacramento?
Suppose you take two steps $\mathbf{\text{A}}$ and $\mathbf{\text{B}}$ (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point $\mathbf{\text{A}}+\mathbf{\text{B}}$ the sum of the lengths of the two steps?
Explain why it is not possible to add a scalar to a vector.
If you take two steps of different sizes, can you end up at your starting point? More generally, can two vectors with different magnitudes ever add to zero? Can three or more?
## Problems&Exercises
Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.
Find the following for path A in [link] : (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.
(a) $\text{480 m}$
(b) $\text{379 m}$ , $\text{18.4º}$ east of north
Find the following for path B in [link] : (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.
Find the north and east components of the displacement for the hikers shown in [link] .
north component 3.21 km, east component 3.83 km
Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements $\mathbf{\text{A}}$ and $\mathbf{\text{B}}$ , as in [link] , then this problem asks you to find their sum .)
Suppose you first walk 12.0 m in a direction $\text{20º}$ west of north and then 20.0 m in a direction $\text{40.0º}$ south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements $\mathbf{A}$ and $\mathbf{B}$ , as in [link] , then this problem finds their sum .)
$\text{19}\text{.}\text{5 m}$ , $4\text{.}\text{65º}$ south of west
Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg $\mathbf{B}$ , which is 20.0 m in a direction exactly $\text{40º}$ south of west, and then leg $\mathbf{A}$ , which is 12.0 m in a direction exactly $\text{20º}$ west of north. (This problem shows that $\mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}$ .)
(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction $\text{40.0º}$ north of east (which is equivalent to subtracting $\mathbf{\text{B}}$ from $\mathbf{A}$ —that is, to finding $\mathbf{\text{R}}\prime =\mathbf{\text{A}}-\mathbf{\text{B}}$ ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction $\text{40.0º}$ south of west and then 12.0 m in a direction $\text{20.0º}$ east of south (which is equivalent to subtracting $\mathbf{\text{A}}$ from $\mathbf{\text{B}}$ —that is, to finding $\mathbf{\text{R}}\prime \prime =\mathbf{\text{B}}-\mathbf{\text{A}}=-\mathbf{\text{R}}\prime$ ). Show that this is the case.
(a) $\text{26}\text{.}\text{6 m}$ , $\text{65}\text{.}\text{1º}$ north of east
(b) $\text{26}\text{.}\text{6 m}$ , $\text{65}\text{.}\text{1º}$ south of west
Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors $\mathbf{A}$ , $\mathbf{B}$ , and $\mathbf{C}$ , all having different lengths and directions. Find the sum then find their sum when added in a different order and show the result is the same. (There are five other orders in which $\mathbf{A}$ , $\mathbf{B}$ , and $\mathbf{C}$ can be added; choose only one.)
Show that the sum of the vectors discussed in [link] gives the result shown in [link] .
$\text{52}\text{.}\text{9 m}$ , $\text{90}\text{.}\text{1º}$ with respect to the x -axis.
Find the magnitudes of velocities ${v}_{\text{A}}$ and ${v}_{\text{B}}$ in [link]
Find the components of ${v}_{\text{tot}}$ along the x - and y -axes in [link] .
x -component 4.41 m/s
y -component 5.07 m/s
Find the components of ${v}_{\text{tot}}$ along a set of perpendicular axes rotated $\text{30º}$ counterclockwise relative to those in [link] .
#### Questions & Answers
Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.077 m2, and the magnitude of the fluid velocity is 3.50 m/s. (a) What is the fluid speed at points in the pipe where the cross
A particle behave like a wave and we do not why?
WAQAR
what's the period of velocity 4cm/s at displacement 10cm
What is physics
the branch of science concerned with the nature and properties of matter and energy. The subject matter of physics includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms.
Aluko
and the word of matter is anything that have mass and occupied space
Aluko
what is phyices
Whats the formula
1/v+1/u=1/f
Aluko
what aspect of black body spectrum forced plank to purpose quantization of energy level in its atoms and molicules
a man has created by who?
What type of experimental evidence indicates that light is a wave
double slit experiment
Eric
The S. L. Unit of sound energy is
what's the conversation like?
some sort of blatherring or mambo jambo you may say
I still don't understand what this group is all about oo
ENOBONG
no
uchenna
ufff....this associated with physics ..so u can ask questions related to all topics of physics..
what is sound?
Bella
what is upthrust
what is upthrust
Olisa
Up thrust is a force
Samuel
upthrust is a upward force that acts vertical in the ground surface.
Rodney
yes rodney's answer z correct
Paul
what is centre of gravity?
Paul
you think the human body could produce such Force
Anthony
what is wave
mirobiology
Angel
what is specific latent heat
the total amount of heat energy required to change the physical state of a unit mass of matter without a corresponding change in temperature.
fitzgerald
is there any difference between specific heat and heat capacity.....
what wave
Bryan
why medical physics even.we have a specoal branch of science biology for this.
what is physics
|
# Partial fraction of $\frac{x^n}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}$
How should you break $\displaystyle \frac{x^n}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}$ into partial fractions?
First you have to take out a multiple of the denominator to make the degree of the numerator smaller than that of the denominator:
$$\frac{x^n}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}$$
$$= \frac{(-1)^n}{n!} + \frac{x^n-\frac{(-1)^n}{n!}(1-x)(1-2x)(1-3x)\ldots(1-nx)}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}\;.$$
Now you can deal with the remaining fraction. You already have the denominator in factorized form, so you know the partial fraction expansion will have the form
$$\frac{a_1}{1-x}+\frac{a_2}{1-2x}+\frac{a_3}{1-3x}+\ldots+\frac{a_n}{1-nx}\;.$$
To get the coefficients $a_i$, you can use the fact that the residues of the poles must be equal to the corresponding residues in the fraction.
• Can you explain more? What are residues and poles? – Vafa Khalighi Mar 11 '11 at 12:07
• Roughly speaking, a pole is a point $x_0$ at which the function diverges like $c/(x-x_0)$, and the residue at the pole is the coefficient $c$. For instance, the function $f(x)=(x^2+3)/((x-1)(x-2))$ has poles at $1$ and $2$, and the corresponding residues are $-4$ and $7$, which you can calculate by substituting the pole $x_0$ into the function except for the "pole part" of it, $1/(x-x_0)$. – joriki Mar 11 '11 at 12:12
• I should add that you can also calculate the coefficients $a_i$ by bringing the partial fractions onto a common denominator and then comparing coefficients of powers of $x$ in the numerator with the fraction you're expanding, but that's often more tedious than using the residues of the poles. – joriki Mar 11 '11 at 12:14
To expand on joriki's comment, I claim that
$$\lim_{x \to \frac{1}{k}} \frac{x^n (1 - kx)}{(1 - x)...(1 - nx)} = a_k.$$
This is not hard to see by multiplying both sides by $1 - kx$. On the other hand, this limit is quite easy to compute since $1 - kx$ cancels, giving
$$a_k = \frac{1}{k(k-1)...(k-n)} = \frac{(-1)^{n-k}}{k! (n-k)!}.$$
Note that this technique works to give you the partial fraction decomposition of any rational function, at least when the denominator has simple roots: instead of canceling a factor, you can just use l'Hopital's rule.
Note also that the function you're looking at happens to be the ordinary generating function of the Stirling numbers of the second kind, and here you can cheat because you can also find their exponential generating function.
It is worth emphasizing that both of the presented solutions essentially employ what is known as Heaviside's cover-up method. As I explained in this answer, this method works generally, i.e. even for nonlinear denominators. As should be evident from the presentation that I gave there, the solution can be constructed by a purely algebraic deterministic algorithm, i.e. without employing any analytic techniques (residue calculus, limits, L'Hopital's rule, etc). Moreover, this is frequently the most efficient way to proceed - even for manual calculations.
• @ Bill: Thanks for the helpful wiki article. It actually was 100% useful for a change :) – night owl Jul 4 '11 at 17:16
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Home » Importance of Trigonometry formulas
# Importance of Trigonometry formulas
Many difficulties in trigonometry have formulas that can help you work them out. Trigonometric identities (sin, cos, tan, sec, cosec, and cot), Pythagorean identity, product identities, etc., could be at play here. With the Trigonometry Homework Help, students will know how formulas involving co-function identity (shifting angles), sum, and difference identity, double and half-angle identities, etc., are also briefly shown here; these can be used to determine the sign of ratio in different areas.
Students in Grades 10, 11, and 12 might benefit from learning & memorizing these trigonometric formulas in mathematics. With college algebra and trigonometry homework help, the trigonometry table, and inverse trigonometry formulas are available for users to work out the related problems.
Those students interested in the mathematical study of triangles should look into the field of trigonometry. The study of the connections between the lengths & angles of a triangle is also called trigonometry. To better understand, students can get help with trigonometry homework.
Trigonometry and its formulas have countless applications. Triangulation can be used in a variety of fields, from geography (to determine how far apart points are) to astronomy (to determine how far apart stars are) to satellite tracking (to determine how far apart points are).
## Formulas List of Trigonometry
In college, while solving trigonometry homework help answers, we only think about right triangles while applying trigonometric formulas. The three sides that make up the right triangle are the hypotenuse, the perpendicular, and the adjacent side (Base). The perpendicular edge is the one that runs perpendicular to the tangent line, while the adjacent
side is the one that the hypotenuse & opposing side rest on.
## Major Trigonometric Function Formulas
In trigonometry, you can find the elements using just six ratios. Trigonometric equations are the name for these calculations. Trigonometric functions include the sine, cos, secant, cosecant, tangent, & cotangent.
### The trigonometric equations & identities are derived with a right-angled triangle as a reference point:
sin θ = Opposite Side/Hypotenuse
tan θ = Opposite Side/Adjacent Side
cosec θ = Hypotenuse/Opposite Side
cot θ = Adjacent Side/Opposite Side
Finding the values of the sine, cosine, tangent, secant, cosecant, & cotangent of a right-angled triangle is a matter of plugging them into trigonometric equations once we understand the height and the bottom side. Using the trigonometric functions, we can also deduce the corresponding trigonometric identities. To know more, get tech assistance from the best
homework helper for trigonometry.
## What is the Sin 3x Formula?
Three times the sine of such an angle in a right-angled triangle is sin 3x, and the expression
for this is:
Sin 3x = 3sin x – 4sin3x
## Major sets of Trigonometry
### There are two main sets of trigonometric formulas:
● Trigonometric Identities
● Trigonometric Ratios
Formal expressions involving trigonometric functions are called trigonometric identities. All possible combinations of the parameters can be used without violating these identities. A trigonometric ratio is a relationship between the angle measures and the side lengths of a right triangle. To understand in detail, get the best Maths Homework Help USA.
describes the relationship between the various trigonometric functions & lists their typical values.
## Formulas List of Trigonometry
In college, while solving trigonometry homework help answers, we only think about right triangles while applying trigonometric formulas. The three sides that make up the right triangle are the hypotenuse, the perpendicular, and the adjacent side (Base). The perpendicular edge is the one that runs perpendicular to the tangent line, while the adjacent
side is the one that the hypotenuse & opposing side rest on.
## Major Trigonometric Function Formulas
In trigonometry, you can find the elements using just six ratios. Trigonometric equations are the name for these calculations. Trigonometric functions include the sine, cos, secant, cosecant, tangent, & cotangent. The trigonometric equations & identities are derived with a right-angled triangle as a reference point:
sin θ = Opposite Side/Hypotenuse
tan θ = Opposite Side/Adjacent Side
cosec θ = Hypotenuse/Opposite Side
cot θ = Adjacent Side/Opposite Side
Finding the values of the sine, cosine, tangent, secant, cosecant, & cotangent of a right-angled triangle is a matter of plugging them into trigonometric equations once we understand the height and the bottom side. Using the trigonometric functions, we can also deduce the corresponding trigonometric identities. To know more, get assistance from the best
homework helper for trigonometry.
## What is the Sin 3x Formula?
Three times the sine of such an angle in a right-angled triangle is sin 3x, and the expression
for this is:
Sin 3x = 3sin x – 4sin3x
## Major sets of Trigonometry
There are two main sets of trigonometric formulas:
● Trigonometric Identities
● Trigonometric Ratios
Formal expressions involving trigonometric functions are called trigonometric identities. All possible combinations of the parameters can be used without violating these identities. A trigonometric ratio is a relationship between the angle measures and the side lengths of a right triangle. To understand in detail, get the best Maths Homework Help USA.
Students will get the best MATH 308 Assignment Help US with a simple trigonometric table that describes the relationship between the various trigonometric functions & lists their typical values.
### Random Posts
TechCrams is an online webpage that provides business news, tech, telecom, digital marketing, auto news, and website reviews around World.
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# How does Hanoi Tower Work?
Contents
Tower of Hanoi consists of three pegs or towers with n disks placed one over the other. The objective of the puzzle is to move the stack to another peg following these simple rules. Only one disk can be moved at a time. No disk can be placed on top of the smaller disk.
## What is Tower of Hanoi and how do you solve it?
With 3 disks, the puzzle can be solved in 7 moves. The minimal number of moves required to solve a Tower of Hanoi puzzle is 2n − 1, where n is the number of disks.
For example, in an 8-disk Hanoi:
1. Move 0 = 00000000. The largest disk is 0, so it is on the left (initial) peg. …
2. Move 28 − 1 = 11111111. …
3. Move 21610 = 11011000.
## How is the Tower of Hanoi recursive?
Using recursion often involves a key insight that makes everything simpler. In our Towers of Hanoi solution, we recurse on the largest disk to be moved. … That is, we will write a recursive function that takes as a parameter the disk that is the largest disk in the tower we want to move.
ЭТО ИНТЕРЕСНО: What age can you buy a house in Singapore?
## How do you beat the Tower of Hanoi?
Let’s go through each of the steps:
1. Move the first disk from A to C.
2. Move the first disk from A to B.
3. Move the first disk from C to B.
4. Move the first disk from A to C.
5. Move the first disk from B to A.
6. Move the first disk from B to C.
7. Move the first disk from A to C.
## Why stack is used in Tower of Hanoi?
1) Only one disk must be moved at a time. 2) Each move consists of taking the upper disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod. 3) No disk may be placed on top of a smaller disk.
## What is the Tower of Hanoi puzzle?
Tower of Hanoi is a mathematical puzzle where we have three rods and n disks. The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules: Only one disk can be moved at a time.
## What is the objective of Tower of Hanoi puzzle?
What is the objective of tower of hanoi puzzle? Explanation: Objective of tower of hanoi problem is to move all disks to some other rod by following the following rules-1) Only one disk can be moved at a time. 2) Disk can only be moved if it is the uppermost disk of the stack.
## Which statement is correct in case of Tower of Hanoi?
The statement “Only one disk can be moved at a time” is correct in case of tower of hanoi. The Tower of Hanoi or Luca’s tower is a mathematical puzzle consisting of three rods and numerous disks. The player needs to stack the entire disks onto another rod abiding by the rules of the game.
## What does Tower of Hanoi measure?
The Towers of Hanoi and London are presumed to measure executive functions such as planning and working memory. Both have been used as a putative assessment of frontal lobe function.
## Is Tower of Hanoi divide and conquer algorithm?
A solution to the Towers of Hanoi problem points to the recursive nature of divide and conquer. We solve the bigger problem by first solving a smaller version of the same kind of problem. … The recursive nature of the solution to the Towers of Hanoi is made obvious if we write a pseudocode algorithm for moving the disks.
## How many moves does it take to solve a 64 Tower of Hanoi?
If you had 64 golden disks you would have to use a minimum of 264-1 moves. If each move took one second, it would take around 585 billion years to complete the puzzle!
## What is the Tower of Hanoi psychology?
The Tower of Hanoi is a simple mathematical puzzle often employed for the assessment of problem-solving and in the evaluation of frontal lobe deficits. The task allows researchers to observe the participant’s moves and problem-solving ability, which reflect the individual’s ability to solve simple real-world problems.
## What is the formula for Tower of Hanoi?
The original Tower of Hanoi puzzle, invented by the French mathematician Edouard Lucas in 1883, spans “base 2”. That is – the number of moves of disk number k is 2^(k-1), and the total number of moves required to solve the puzzle with N disks is 2^N – 1.
ЭТО ИНТЕРЕСНО: Quick Answer: Which BTS member is famous in Thailand?
## Is Tower of Hanoi a greedy algorithm?
Par- ticularly in the case of this problem (Tower of Hanoi) the two (minimum version) sub-sub-problems have to be solved before, and after (respectively) the greedy decision is implemented.
## Can Tower of Hanoi can be solved iteratively?
You can transform the recursive solution to an iterative solution. To do this, create a stack that will contain items consisting of quadruples (“from”, “to”, “via”, “num_disks”). For every function “call” in your recursive algorithm, push the parameters to a stack in the iterative algorithm.
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Total Surface Area - the trick to getting it right
How to Find the Surface Area of a Box
Three Methods:
Finding the surface area of a box is easy as long as you know the length of the sides. Once you know how long the sides are, you simply have to plug them into a basic equation to get your answer. You can even find the surface area of cylindrical boxes with a few simple measurements.
Steps
Rectangular Boxes
1. Add together the area of each side to get the surface area of the box.Surface area is the total area of each side. As long as you know how to find the area of a regular rectangle, which is simply the length times the height, you can find each side and add them together. However, there is also a simplified formula that will do this for you if you know the measurements of the box:SurfaceArea=2lw+2lh+2wh{\displaystyle SurfaceArea=2lw+2lh+2wh}
• lis the length of the box, or the longest side.
• his the height of the box.
• wis the width of the box.
2. Measure the length of the box.This is your longest side. There should be 4 identical lines equally long across the whole box. Lay the box down on its longest side to make it easier to measure.
• Example:The length of the box is 5 feet.
3. Measure the height of the box, or the distance from the floor to the top.Make sure you don't measure the same side as the length.
• Example:The height of the box is 4 feet.
4. Measure the width of the box.This is the side directly perpendicular (it forms a right angle, or L) with the length. Make sure you are not remeasuring the height.
• Example:The width of the box is 2 feet.
5. Ensure that you didn't measure the same side twice.The easiest way to prevent this is to check that all three lines meet at some point. Find a corner and measure the three sides coming off of it -- this ensures you have the right measurements.
• The sides may be the same measurement. You just want to make sure you're measuring three unique sides, even if two or three of them are all identical.
6. Input your measurements into the formula to solve the equation.Once you've got the numbers, the rest is easy. Simply input the measurements and add it all up.
• SurfaceArea=2lw+2lh+2wh{\displaystyle SurfaceArea=2lw+2lh+2wh}
• SurfaceArea=2(5)(2)+2(5)(4)+2(2)(4){\displaystyle SurfaceArea=2(5)(2)+2(5)(4)+2(2)(4)}
• SurfaceArea=20+40+16{\displaystyle SurfaceArea=20+40+16}
• SurfaceArea=76{\displaystyle SurfaceArea=76}
7. Express your answer in "units squared" when done.This lets people know how you measured everything, and is a crucial part of your answer. Luckily, all you need to do is use the units provided in the question. For this example, since the units were in feet, you would put "feet squared" at the end of your answer:
• You have a box 5 feet long, 4 feet high, and 2 feet wide, what is the total surface area?
8. Break complex boxes into smaller pieces to find surface area.For example, say you had an "L" shaped box. Instead of finding everything by hand, simply think of two separate boxes, the vertical half on top and the horizontal half on the bottom. Find the surface area of both, then add them together to get the total surface area.For example, if you have a U shaped box:
• Say the bottom has a surface area of 12 units squared.
• Say both sides have a surface area of 15 units squared.
• The total area would be42 units squared, since12+15+15=42{\displaystyle 12+15+15=42}.
Cylindrical Boxes (Tubes)
1. Add the area of the bases to the height times circumference to find the surface area of a cylinder.This only works with right cylinders, meaning they are not slanted. The proper formula isSurfaceArea=2B+hC{\displaystyle SurfaceArea=2B+hC} For example, if the area of the base is 3, the height is 5, an the circumference is 6, what is the surface area? Answer =36 units squared.
• Bis the area of the base.
• his the height of the cylinder.
• Cis the circumference of the base.
2. Find the area of the base.The base is the circular end of the cylinder. It can be found with the formulaArea=pi∗r2{\displaystyle Area=pi*r^{2}}Ris the radius of the circle.Piis the constant, rounded to 3.14 for simple problems. You can also leave it as "pi" if you do not have a calculator.
• Example:The radius of the base in a cylinder is 2. What is the area of the base?
• pi∗(2)2{\displaystyle pi*(2)^{2}}
• B =4pi{\displaystyle 4pi}
3. Calculate the circumference if the base.The circumference is the distance around the edge of the circle. It is found with the formulaCircumference=2∗r∗pi{\displaystyle Circumference=2*r*pi}Continuing the previous example:
• 2∗pi∗(2){\displaystyle 2*pi*(2)}
• C =4pi{\displaystyle 4pi}
4. Find the height of the cylinder by measuring the space between the two bases.The height of a cylinder is just another way to figure out how long it is. It is the straight line between the center of the two bases.
• Example:In the same cylinder, with radius 2 inches, the height is 5 inches.
• h=5{\displaystyle h=5}
5. Input your smaller parts into the formula to find your surface area.Once you've got your base area, circumference, and height, all you need to do is plug the numbers into your formula to get the right answer.
• SurfaceArea=2B+hC{\displaystyle SurfaceArea=2B+hC}
• SurfaceArea=2(4pi)+(5)(4pi){\displaystyle SurfaceArea=2(4pi)+(5)(4pi)}
• SurfaceArea=8pi+20pi{\displaystyle SurfaceArea=8pi+20pi}
• SurfaceArea=28pi{\displaystyle SurfaceArea=28pi}
6. Express your answer in units squared.All problems of surface area need units to give them scale. Is the area in inches, or feet, for example. The units always match the units given to you in the problem. If no units are given, you should just write the phrase "units squared," or "units2{\displaystyle units^{2}}"
• In the example, the units were given as inches. So the final answer would be28pi{\displaystyle 28pi}inches2{\displaystyle inches^{2}}
Practice Problems
1. Try out some practice problems with rectangles.To see the answers, highlight right after the arrow:
• L = 10, W = 3, H = 2, →112 units squared
• L = 6.2, W = 2, H = 5.4 →113.36 units squared
• On a rectangular box, the top is 5x3x2, the bottom 6x2x2. →118pi units squared
2. Try out some practice problems with cylinders.To see the answers, highlight after the arrow:
• Base = 3, Height = 10, Circumference = 1.5 →21 units squared
• Base = 25pi, Height = 3, Circumference = 10pi →80pi units squared
• Radius = 3, height = 3 →36pi units squared
Community Q&A
Search
• Question
I'm painting a box 2' wide by 3' long by 2' high. If I'm not painting the bottom, how do I calculate the surface area I need to cover?
The top is 2 x 3; two sides are each 3 x 2; and the other two sides are each 2 x 2.
Thanks!
• Question
What is the surface area of a box if the length is 4m the width is 2m and the height is 2m?
Multiply the length by the width and double it. Multiply the length by the height and double it. Multiply the width by the height and double it. Add those three amounts together.
Thanks!
• Question
What is the surface area of a box open on both ends?
The surface area equals the height multiplied by the sum of twice the length and twice the width.
Thanks!
• Question
How do I find the area of a long, rectangular, hollow box with no thickness?
First, we'll assume you're looking for the total surface area of a rectangular (or rectilinear) box, and we'll define that box as having a length (L), a width (W), and a depth (D). Find twice the product of L x W; to that add twice the product L x D; to that add twice the product W x D. The sum is the total surface area.
Thanks!
• Question
How do I increase 10 percent of 50?
If you're asking how to increase 50 by 10%, calculate 10% of 50, and add that number to 50.
Thanks!
• Question
V is 180 cubic centimeters and L is 15 centimeters what is h?
You can't find "h" without also knowing "w".
Thanks!
200 characters left
Quick Summary
To find the surface area of a box, start by calculating the area of each side using the formula a = lh, where l is the length and h is the height. Once you know the area of every side, add them all together to get the surface area of the box. If the box is cylindrical, you'll need to use the formula: surface area = 2b + hc, where b is the area of the base of the box, h is the height, and c is the circumference of the base.
• If you are measuring a real box, you might want to try measuring different edges that should be the same length, and then use the average.
Things You'll Need
• A box and a way to measure it.
• Measurements of an existing or hypothetical box.
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# MENSURATION
Mensuration means measurement. It is being done in our life in many situations. For example, Length of cloth we need for stitching, the area of a wall which is being painted, perimeter of a rectangular plot to be fenced, quantity of water needed to fill the tank etc. For these kinds of activities, we are doing measurements for further needs.
Here, we cover three areas
1. Perimeter
2. Area
3. Volume
To find these things we have to remember the formulae of plane figures and solid figures. Try to understand the concept behind reaching the formula.
1. The area of a rectangular plate is 236.25 sq.cm. If the length of the plate is 17.5 cm, find the perimeter of the plate.
Ans. 62 cm
Explanation :
2. The breadth of a rectangular plot is 75% of its length. If the perimeter of the plot be 1050 m, what is its area?
Ans. 67500
Explanation :
3. In measuring the sides of a rectangle, one side is taken 10% in excess, and the other 5% in deficit. What is the change in its area as a percentage?
Ans. 4.5%
Explanation :
4. How many tiles of 20 cm length and 10 cm width are required to pave the floor of a room 8 m long and 5 m wide?
Ans. 2000
Explanation :
5. The circumference of two circles is 88 metre and 132 metre respectively. What is the difference between the area of the larger circle and the smaller circle?
Ans. 770 sq.m
Explanation :
6. A solid metallic spherical ball of radius 7 cm is melted down and recast into small cones. If the diameter of the base of the cone is 14 cm and the height is 2 cm, find the number of such cones can be made.
Ans. 14
Explanation :
7. A hall is 10 metres long and 8 metres wide. What will be the cost of carpeting the room if 0.5 metres of space is left around the room? The rate of 0.25 metre wide carpet is Rs.20 per metre.
Ans. Rs.5040
Explanation :
8. A square field has an area of 50625 sq. m. Find the cost of fencing around it at Rs.15 per metre.
Ans. Rs.13500
Explanation :
9. Cost of fencing a circular plot at the rate of Rs.12 per metre is Rs.2,640. What will be the cost of carpeting the floor of the plot at the rate of Rs.85 per square metre?
Ans. Rs.3,27,250
Explanation :
10. A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. What will be the area of the circle?
1) 88 cm2 2) 1250 cm 3) 154 cm 4) 128 cm2 5) None of these
Ans. None of these
Explanation :
11. A circular garden has a 7 m wide road around the border. Find the cost of leveling the road at Rs.5 per square meter if the radius of the inner circle is 21 m?
Ans. Rs.5390
Explanation :
12. The lateral surface area of a cylinder is thrice the area of its base. Find the ratio of its height and the base radius?
Ans. 3 : 2
Explanation :
13. Two cubes have their volumes in the ratio 8 : 27. Find the ratio of their surface areas.
Ans. 4 : 9
Explanation :
14. A horse is tied with a 14 m long rope. How much ground will it be able to graze?
1) 125 Sq.m 2) 275 Sq.m 3) 625 Sq.m 4) 675 Sq.m 5) None of these
Ans. None of these
Explanation :
15. Find the length of the longest stick that can be placed in a room of 12 m long, 9 m broad and 8 m high.
Ans. 17 m
Explanation :
16. A field is 375 m long and 40 m broad. A tank 30 m long, 20 m broad and 12 m deep is dug in the field and earth taken out of it and is spread equally over the field. How much is the level of field raised?
Ans. 50 cm
Explanation :
17. A rectangular field of 60-metre length and 40 metres wide is to be surrounded by a road 5 meter wide. If the cost of making 1 square meter road is Rs.500, what would be the cost of the entire road?
Ans. Rs.5,50,000
Explanation :
18. A room is 7.5 m long, 5.5 m broad and 5 m high. What will be the expenditure in covering the walls by paper 40 cm broad at the rate of 75 paise per meter?
Ans. Rs.243.75
Explanation :
19. The length and breadth of a rectangle increased by 20% and 10% respectively. How much percent will its area be increased?
Ans. 32%
Explanation :
20. 7 cm radius and 28 cm height solid metallic cylinder is melted and recast into small spherical balls of 7 cm radius. Find the number of cubes that can be made.
Ans. 3
Explanation :
Mensuration - 2D (Plane Figur es)
1. The length of a rectangular field is four times its width. If the perimeter of the field is 30 m, find the area of the field?
a) 108 m2 b) 27 m2 c) 9 m2 d) 36 m2
2. A circular wire of radius 42 cm is cut and bent in the form of a rectangle whose sides are in the ratio 6 : 5. Find the smaller side of the rectangle?
a) 72 cm b) 120 cm c) 60 cm d) 144 cm
3. The radius of a circle is 14 cm. Find the area of the sector with central angle 36°?
a) 61.6 cm2 b) 30.8 cm2 c) 15.4 cm2 d) 308 cm2
4. In the given figure, if AB = 4 cm and BD = 4√3 cm, then the area of the shaded region will be?
a) 48 π cm2 b) 24 π cm2 c) 16 π cm2 d) 12 π cm2
5. A rectangular garden has an area of 2000 m2 and its length and breadth are in the respective ratio 5 : 4. A road of uniform width runs inside the garden around the perimeter and has an area of 344 m2. The width of the road is......
a) 3 m b) 3.5 m c) 4 m d) 2 m
6. If the perimeters of a rectangle and a square are equal and the ratio of two adjacent sides of the rectangle is 1 : 2 then the ratio of area of the rectangle and that of the square is?
a) 1 : 1 b) 1 : 2 c) 2 : 3 d) 8 : 9
7. If a wire is bent into the shape of a square, then the area of the square so formed is 81 cm2. When the wire is rebent into a semicircular shape, then the area (in cm2) of the semicircle will be (Take π = 22/7)
a) 22 b) 44 c) 77 d) 154
8. At each corner of a triangular field of sides 26 m, 28 m and 30 m, a cow is tethered by a rope of length 7 m. The ungrazed area (in m2) by the cow is.......
a) 336 b) 259 c) 154 d) 77
9. How many rotations will a wheel of a car make in a journey of 88 km, if the radius of the wheel is 56 cm?
a) 60000 b) 50000 c) 30000 d) 25000
10. A cow is tied by a rope to one of the vertices of a square of side 14 m. The length of the rope is 7 m. What percentage of the field is grazed by the cow?
a) 15% b) 25% c) 18% d) 19.6%
11. A horse is tethered by a rope. What is the length of the rope so that the horse can graze over an area of 154 m2?
a) 2 m b) 7 m c) 14 m d) 49 m
12. A circular grass lawn of 28 m radius, has a path of 14 m wide running around it on the outside. Find the area of the path?
a) 3920 m2 b) 3080 m2 c) 3980 m2 d) 8030 m2
13. The area of a circular plot is 154 m2. Find the cost of fencing it at the rate of Rs. 2.75 per metre?
a) Rs.56 b) Rs.100 c) Rs.121 d) Rs.423.50
14. A wire bent in the form of a square, encloses an area of 484 cm2. If the same wire is bent to form a circle, then the area enclosed will be
a) 1232 cm2 b) 616 cm2 c) 2464 cm2 d) 4312 cm2
15. A copper wire is bent in the form of an equilateral triangle and has area 121√3 cm2. If the same wire is bent into the form of a circle, the area (in cm2) enclosed by the wire is....... (Take π = 22/7)
a) 364.5 b) 693.5 c) 346.5 d) 639.5
16. The area of a rhombus is 150 cm2. The length of one of its diagonals is 10 cm. The length of the other diagonal is........
a) 25 cm b) 30 cm c) 35 cm d) 40 cm
17. The area of the ring between two concentric circles, with circumference 88 cm and 132 cm respectively is......
a) 780 cm2 b) 770 cm2 c) 715 cm2 d) 660 cm2
Key With Explanations
1-d; 2(l + b) = 30
⇒ 2(4b + b) = 30
⇒ 10b = 30 ⇒ b = 3 m and l = 12 m
So, area of the field = l × b = 3 × 12 = 36 m2.
2-c; Perimeter of the rectangle = Circumference of the circular wire
= 2 × 22/7 42 cm = 264 cm.
Let the dimensions of the rectangle be 6x and 5x respectively.
∴ 2 × (6x + 5x) = 264 ⇒ x = 12
∴ Smaller side = 5x = 60 cm
3-a; Radius of the circle = 14 cm
Angle of the sector = 36°
Area of the sector = θ/360° × π r2
∴ Area of the sector
= 36/360° × 22/7 × 14 × 14 = 61.6 cm2
4-d; ∠ADC is a right angle (Angle in a semicircle)
So, BD2 = AB × BC ⇒ 16 × 3 = 4 × BC
⇒ BC = 12 cm
Now shaded area = Area of bigger semicircle − Areas of smaller semicircles
= 1/2 π (8)2 - 1/2 π (2)2 - 1/2 π (6)2
= 32 π − 2 π −18 π = 12 π cm2
5-d; Let 5x and 4x be the length and breadth of the garden.
Then, 5x × 4x = 2000
⇒ x2 = 100 ⇒ x = 10
∴ Length = 50 m and breadth = 40 m
Let ‘d’ be the width of the road. Then,
(50 − 2d) (40 − 2d) = 2000 − 344 ⇒ d = 2 m
6-d; Let side of rectangle be 2x and x units. and side of square = y units
∴ 4y = 6x ⇒ x/y = 4/6 = 2/3
∴ 2x × x/y2 = 2x2/ y2 = 2 × 4/9 = 8 : 9
7-c; Area of square = a2 = 81 cm⇒ a = 9 cm
Perimeter of square = 4 × a = 4 × 9 = 36 cm
Perimeter of semi circle = πr + 2r ⇒ 36 = r (π + 2)
⇒ 36 = r (36/7) ⇒ r = 36 × 7/36 = 7
Area of semicircle = 1/2 πr
= 1/2 × 22/7 × 7 × 7 = 77 cm2
Hence, ungrazed area = 336 − 77 = 259 m2
11-b; Let the length of the rope be r. Horse can graze an area equal to area of the circle of radius r.
Then, πr2 = 154 ⇒ r = 7 m
12-b; Radius of the outer circle = 28 + 14 = 42 m
Area of the path = Area of outer circle − Area of inner circle
= π (42)2 − π (28)2 = 3080 m2
13-c; Let r be the radius of the circular plot.
Then, πr2 = 154 ⇒ r = 7 m
Circumference of the plot = 2 × 22/7 × 7 = 44 m
Cost of fencing the plot = Rs.44 × 2.75 = Rs.121
14-b; Side of the square = = 22 cm
Perimeter of the square = 4 × 22 = 88 cm
Perimeter of the circle = 2π × radius = 88 cm
Area of the circle = π × (14)2
= 22/7 × 14 × 14 = 616 cm2
15-c; Area of equilateral triangle = √3/4 a
[where a is side of triangle]
∴ √3/4 a = 121√3 ⇒ a = 22 cm
Perimeter of triangle = 22 × 3 = 66 cm
Perimeter of circle = 2πr i.e. 2πr = 66
∴ Area of outer circle = πr2 = 22/7 × 21 × 21 = 1386 cm2
area of inner circle = πr12 = 22/7 × 14 × 14 = 616 cm2
Hence, area of ring = 1386 − 616 = 770 cm2
Posted Date : 29-11-2021
గమనిక : ప్రతిభ.ఈనాడు.నెట్లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.
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# How to Simplify Polynomials
To simplify polynomials, you need to find ‘like’ terms and combine them. Here you can learn how to simplify polynomials.
## Step by step guide to simplifying polynomials
• Find “like” terms. (they have same variables with same power).
• Add or Subtract “like” terms using order of operation.
### Simplifying Polynomials – Example 1:
Simplify this expression. $$2x(2x-4)=$$
Solution:
Use Distributive Property: $$2x(2x−4)=4x^2−8x$$
### Simplifying Polynomials – Example 2:
Simplify this expression. $$4x^2+6x+2x^2-4x-3=$$
Solution:
First find “like” terms and combine them: $$4x^2+2x^2= 6x^2$$, $$6x-4x= 2x$$
Now simplify: $$4x^2+6x+2x^2-4x-3=6x^2+2x-3$$
### Simplifying Polynomials – Example 3:
Simplify this expression. $$4x(6x-3)=$$
Solution:
Use Distributive Property: $$4x(6x-3)=24x^2-12x$$
### Simplifying Polynomials – Example 4:
Simplify this expression. $$7x^3+2x^4+2x^3-4x^4-8x=$$
Solution:
First find “like” terms and combine them: $$7x^3+2x^3= 10x^3$$, $$2x^4-4x^4= -2x^4$$
Now simplify and write in standard form: $$7x^3+2x^4+2x^3-4x^4-8x=-2x^4+10x^3-8x$$
## Exercises for Simplifying Polynomials
### Simplify each expression.
1. $$\color{blue}{(12x^3 + 28x^2 + 10x^2 + 4) }$$
2. $$\color{blue}{(2x + 12x^2 – 2) – (2x + 1)}$$
3. $$\color{blue}{(2x^3 – 1) + (3x^3 – 2x^3)}$$
4. $$\color{blue}{(x – 5) (x – 3)}$$
5. $$\color{blue}{(3x + 8) (3x – 8)}$$
6. $$\color{blue}{(8x^2 – 3x) – (5x – 5 – 8x^2)}$$
1. $$\color{blue}{12x^3 + 38x^2 + 4}$$
2. $$\color{blue}{12x^2 – 3 }$$
3. $$\color{blue}{3x^3 – 1}$$
4. $$\color{blue}{x^2 – 8x + 15}$$
5. $$\color{blue}{9x^2 – 64}$$
6. $$\color{blue}{16x^2 – 8x + 5}$$
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Solve 3x^3+6x^2+x+2=0 | Uniteasy.com
# Solve the cubic equation:
## $$3x^3+6x^2+x+2=0$$
Since the discriminant $$\Delta >0$$, the cubic equation has one real root and two conjugate complex roots.
$$\Delta=0.2318244170096$$
$$\begin{cases} x_1=-2 \\ x_2=\dfrac{1}{\sqrt{3}}i \\ x_3=-\dfrac{1}{\sqrt{3}}i \end{cases}$$
In decimals,
$$\begin{cases} x_1=-2 \\ x_2=0.57735026918963i \\ x_3=-0.57735026918963i \end{cases}$$
Detailed Steps on Solution
If a cubic equation has a rational root, the root could be a fracion number with a factor of constant term as the numerator and a factor of the coefficient of leading term as the denumerator.
## 1. Factorization Method
Find all possible factors for constant $$2$$, which are,
$$1, 2$$
$$-1, -2$$
and all possible factors of the coefficient of the leading term $$3$$,
$$1, 3$$
$$-1, -3$$
Dividing factors of constant term by those of the cubic term one by one gives the following fractions.
$$1, -1, 2, -2, \dfrac{1}{3}, -\dfrac{1}{3}, \dfrac{2}{3}, -\dfrac{2}{3},$$
Substitute the fractions to the function $$f(x) = 3x³ + 6x² + x + 2$$ one by one to find out the one that makes $$f(x) = 0$$.
According to the factor theorem, $$f(n) = 0$$, if and only if the polynomial $$3x³ + 6x² + x + 2$$ has a factor $$x-n$$, that is, $$x=n$$ is the root of the equation.
Fortunetely, one of the numbers is found to make the equation equal.
$$f(-2) = 3(-2)³ + 6(-2)² + (-2) + 2 = 0$$
then, $$x = -2$$ is one of roots of the cubic equaiton $$3x³ + 6x² + x + 2 = 0$$. And, the equation can be factored as
$$(x +2)(ax^2+bx+c) = 0$$
Next we can use either long division or synthetic division to determine the expression of trinomial
### Long division
Divide the polynomial $$3x³ + 6x² + x + 2$$ by $$x + 2$$
3x² 0x +1 x + 2 3x³ +6x² +x 2 3x³ +6x² 0x² +x 0x² 0x x 2 x 2 0
Now we get another factor of the cubic equation $$3x² + 1$$
## Solve the quadratic equation: $$3x² + 1 = 0$$
$$3t^2 = -1$$
Solving for $$t$$.
Then,
\begin{aligned} \\ t&=\pm \sqrt{\dfrac{1}{3}}i\\ &=\pm \sqrt{\dfrac{1^2}{3}}i\\ &=\pm \dfrac{1}{\sqrt{3}}i\\ \end{aligned}
That is,
$$\begin{cases} t_2 =\dfrac{1}{\sqrt{3}}i \\ t_3=-\dfrac{1}{\sqrt{3}}i \end{cases}$$
Another method to find the roots of a general cubic equation is simplifing it to a depressed form. This method is applicable to all cases, especially to those difficult to find factors.
## 2. Convert to depressed cubic equation
The idea is to convert general form of cubic equation
$$ax^3+bx^2+cx+d = 0$$
to the form without quadratic term.
$$t^3+pt+q = 0$$
By substituting $$x$$ with $$t - \dfrac{b}{3a}$$, the general cubic equation could be transformed to
$$t^3+\dfrac{3ac-b^2}{3a^2}t+\dfrac{2b^3-9abc+27a^2d}{27a^3} = 0$$
Compare with the depressed cubic equation. Then,
$$p = \dfrac{3ac-b^2}{3a^2}$$
$$q = \dfrac{2b^3-9abc+27a^2d}{27a^3}$$
Substitute the values of coefficients, $$p, q$$ is obtained as
$$p = \dfrac{3\cdot 3\cdot 1-6^2}{3\cdot 3^2}=-1$$
$$q = \dfrac{2\cdot 6^3-9\cdot3\cdot 6\cdot 1+27\cdot 3^2\cdot2}{27\cdot 3^3}=\dfrac{28}{27}$$
### Use the substitution to transform
Let $$p$$ and $$q$$ being the coefficient of the linean and constant terms, the depressed cubic equation is expressed as.
$$t^3 +pt+q=0$$
Let $$x=t-\dfrac{2}{3}$$
The cubic equation $$3x³ + 6x² + x + 2=0$$ is transformed to
$$t^3 -t+\dfrac{28}{27}=0$$
## 3. Cardano's solution
Let $$t=u-v$$
Cube both sides and extract common factor from two middle terms after expanding the bracket.
\begin{aligned} \\t^3&=(u-v)^3\\ & =u^3-3u^2v+3uv^2-v^3\\ & =-3uv(u-v)+u^3-v^3\\ \end{aligned}
Since $$u-v=t$$, substitution gives a linear term for the equation. Rearrange terms.
$$x^3+3uvx-u^3+v^3=0$$
Compare the cubic equation with the original one (1)
$$\begin{cases} 3uv=-1\quad\text{or}\quad v=-\dfrac{1}{3u}\\ v^3-u^3=\dfrac{28}{27}\\ \end{cases}$$
$$v=-\dfrac{1}{3u}$$ gives relationship between the two variables. Substitute the value of $$v$$ to the second equation
$$\Big(-\dfrac{1}{3u}\Big)^3-u^3=\dfrac{28}{27}$$
Simplifying gives,
$$u^3+\dfrac{1}{27}\dfrac{1}{u^3}+\dfrac{28}{27}=0$$2
Let $$m=u^3$$, then the equation is transformed to a quadratic equation in terms of $$m$$. Once the value of $$m$$ is determined, $$v^3$$ could be determined by $$v^3=\dfrac{28}{27}+u^3$$.
$$m^2+\dfrac{28}{27}m+\dfrac{1}{27}=0$$
Sovling the quadratic euqation will give two roots (some may be equal). Here we only cosider one case with positive sign before the square root radical since the negative case will produce the same result.
\begin{aligned} \\u^3=m&=-\dfrac{14}{27}+\dfrac{1}{2}\sqrt{\Big(\dfrac{28}{27}^2\Big)-4\cdot \dfrac{1}{27}}\\ & =-\dfrac{14}{27}+\dfrac{1}{2}\sqrt{\dfrac{784}{729}-\dfrac{4}{27}}\\ & =-\dfrac{14}{27}+\dfrac{1}{2}\sqrt{\dfrac{676}{729}}\\ & =-\dfrac{14}{27}+\dfrac{13}{27}\\ & =-\dfrac{1}{27}\\ \end{aligned}
$$v^3$$ can be determined by the equation we deduced $$v^3-u^3=\dfrac{28}{27}$$. Then,
\begin{aligned} \\v^3&=\dfrac{28}{27}+u^3\\ & =\dfrac{28}{27}-\dfrac{14}{27}+\dfrac{13}{27}\\ & =1\\ \end{aligned}
Now we have,
$$u^3=-\dfrac{1}{27}$$ and $$v^3=1$$
Evaluating the simplest cubic equation $$x^3-A=0$$, it has 3 roots, in which the first root is a real number . The second and third are expressed in the product of cubic root of unity and the first one.
If $$ω = \dfrac{-1+i\sqrt{3}}{2}$$, then its reciprocal is equal to its conjugate, $$\dfrac{1}{ω}=\overline{ω}$$.
$$\begin{cases} r_1=\sqrt[3]{A}\\ r_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{A}\\ r_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{A}\\ \end{cases}$$
Similary, taking cubic root for $$u^3$$ and $$v^3$$ also gives 3 roots.
$$\begin{cases} u_1=-\dfrac{1}{3}\\ u_2=\dfrac{-1+i\sqrt{3}}{2}\cdot -\dfrac{1}{3}\\ u_3=\dfrac{-1-i\sqrt{3}}{2}\cdot -\dfrac{1}{3}\\ \end{cases}$$
For $$v_2$$ and $$v_3$$, the complex numbers before radicals are the conjugates of those for $$u_2$$ and $$u_3$$, which can be verified by the reciprocal property of the cubic root of unity from the equation $$v=-\dfrac{1}{3u}$$. The radicand can be taken as the negative conjugate of that in $$u_1$$, $$u_2$$ and $$u_3$$, which is the same in value.
$$\begin{cases} v_1=1\\ v_2=\dfrac{-1-i\sqrt{3}}{2}\cdot 1\\ v_3=\dfrac{-1+i\sqrt{3}}{2}\cdot 1\\ \end{cases}$$
Since $$x=u-v$$, combining the real and imaginary parts gives 3 results for $$t$$
\begin{aligned} \\t_1&=u_1-v_1\\ & =-\dfrac{1}{3}-1\\ & =-\dfrac{4}{3}\\ \end{aligned}
\begin{aligned} \\t_2&=u_2-v_2\\ & =\dfrac{-1+i\sqrt{3}}{2}\cdot -\dfrac{1}{3}-\Big(\dfrac{-1-i\sqrt{3}}{2}\cdot 1\Big)\\ & =\dfrac{2}{3}+\dfrac{\sqrt{3}}{3}i\\ \end{aligned}
\begin{aligned} \\t_3&=u_3-v_3\\ & =\dfrac{-1-i\sqrt{3}}{2}\cdot -\dfrac{1}{3}-\Big(\dfrac{-1+i\sqrt{3}}{2}\cdot 1\Big)\\ & =\dfrac{2}{3}-\dfrac{\sqrt{3}}{3}i\\ \end{aligned}
## 4. Vieta's Substitution
In Cardano' solution, $$t$$ is defined as the difference of $$u$$ and $$v$$. If we substitute the value of $$v$$ (4) into (2), we get the equation. $$t=u+\dfrac{1}{3u}$$. And then substitute the equation to the cubic equation $$t^3-t+\dfrac{28}{27}=0$$. This method is called Vieta's Substitution for solving a cubic equation, which simplied the Cardano' solution. The substitution expression can be obtained by the following formula directly.
$$t=u-\dfrac{p}{3u}$$
Substitute the expression $$t=u+\dfrac{1}{3u}$$ to the cubic equation
$$\Big(u+\dfrac{1}{3u}\Big)^3-\Big(u+\dfrac{1}{3u}\Big)+\dfrac{28}{27}=0$$
Expand brackets and cancel the like terms
$$u^3+\cancel{1u^2\dfrac{1}{u}}+\cancel{\dfrac{1}{3}u\dfrac{1}{u^2}}+\dfrac{1}{27}\dfrac{1}{u^3}-\cancel{1u}-\cancel{\dfrac{1}{3}\dfrac{1}{u}}+\dfrac{28}{27}=0$$
Then we get the same equation as (2)
$$u^3+\dfrac{1}{27}\dfrac{1}{u^3}+\dfrac{28}{27}=0$$
The rest of the steps will be the same as those of Cardano's solution
## $$t^3-t+\dfrac{28}{27}=0$$
Move the linear term and constant of (1) to its right hand side. We get the following form of the equation.
$$t^3=t-\dfrac{28}{27}$$3
Let the root of the cubic equation be the sum of two cubic roots
$$t=\sqrt[3]{r_1}+\sqrt[3]{r_2}$$4
in which $$r_1$$ and $$r_2$$ are two roots of a quadratic equation
$$z^2-\alpha z+ β=0$$5
Using Vieta's Formula, the following equations are established.
$$r_1+r_2 = \alpha \quad \text{and} \quad r_1r_2 = β$$
To determine $$\alpha$$, $$β$$, cube both sides of the equation (4)
$$t^3=3\sqrt[3]{r_1r_2}(\sqrt[3]{r_1}+\sqrt[3]{r_2})+r_1+r_2$$
Substituting, the equation is simplified to
$$t^3=3\sqrt[3]{β}t+\alpha$$
Compare the cubic equation with (3), the following equations are established
$$\begin{cases} 3\sqrt[3]{β}=1\\ \alpha=-\dfrac{28}{27}\\ \end{cases}$$
Solving for $$β$$ gives
$$β=\dfrac{1}{27}$$
So the quadratic equation (5) is determined as
$$z^2+\dfrac{28}{27}z+\dfrac{1}{27}=0$$6
$$\begin{cases} r_1=-1\approx-1\\ r_2=-\dfrac{1}{27}\approx-0.037037037037037\\ \end{cases}$$
Therefore, one of the roots of the cubic equation could be obtained from (4).
$$t_1=\sqrt[3]{-1}+\sqrt[3]{-\dfrac{1}{27}}$$
in decimals,
$$t_1=-1.3333333333333$$
However, since the cube root of a quantity has triple values,
The other two roots could be determined as,
$$t_2=\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-1}+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{27}}$$
$$t_3=\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-1}+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{27}}$$
Combining the real and imaginary parts results in the same result as that obtained by Cardano's solution.
For the equation $$t^3 -t+\dfrac{28}{27}$$, we have $$p=-$$ and $$q = \dfrac{28}{27}$$
### Calculate the discriminant
The nature of the roots are determined by the sign of the discriminant.
Since $$p$$ is negative, the discriminant will be less than zero if the absolute value of $$p$$ is large enough. However, there's the scenario like this one the discriminant is greater than zero when $$p$$ is negative.
\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{\Big(\dfrac{28}{27}\Big)^2}{4}+\dfrac{(-1)^3}{27}\\ & =\dfrac{196}{729}-\dfrac{1}{27}\\ & =\dfrac{196\cdot 1-1\cdot 27}{729}\\ & =0.2318244170096\\ \end{aligned}
### 5.1 Use the root formula directly
If the discriminant is greater than zero, we can use the root formula to determine the roots of the cubic equation.
$$t_{1,2,3} =\begin{cases} \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}& \\ ω\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + \overline{ω} \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}&\ \\ \overline{ω}\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + ω\cdotp \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }} \end{cases}$$
in which, $$ω = \dfrac{-1+i\sqrt{3}}{2}$$ and $$\overline{ω} =\dfrac{-1-i\sqrt{3}}{2}$$
Substitute the values of $$p, q$$ and $$\Delta$$ which we have calculated. Then,
\begin{aligned} \\t_1&=\sqrt[3]{-\dfrac{14}{27}+\sqrt{\dfrac{169}{729}}}+\sqrt[3]{-\dfrac{14}{27}-\sqrt{\dfrac{169}{729}}}\\ & =\sqrt[3]{-\dfrac{14}{27}+\sqrt{\dfrac{13^2}{27^2}}}+\sqrt[3]{-\dfrac{14}{27}-\sqrt{\dfrac{13^2}{27^2}}}\\ & =\sqrt[3]{-\dfrac{14}{27}+\dfrac{13}{27}}+\sqrt[3]{-\dfrac{14}{27}-\dfrac{13}{27}}\\ & =\sqrt[3]{\dfrac{-14}{1\cdot 3^3}+\dfrac{13}{1\cdot 3^3}}+\sqrt[3]{\dfrac{-14}{1\cdot 3^3}-\dfrac{13}{1\cdot 3^3}}\\ & =\dfrac{1}{3}\Big(\sqrt[3]{\dfrac{-14}{1}+\dfrac{13}{1}}+\sqrt[3]{\dfrac{-14}{1}-\dfrac{13}{1}}\Big)\\ & =\dfrac{1}{3}\sqrt[3]{-1}+\dfrac{1}{3}\sqrt[3]{-27}\\ & =-\dfrac{1}{3}\sqrt[3]{1}-\dfrac{1}{3}\sqrt[3]{27}\\ \end{aligned}
If we denote
$$R = -\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$
$$\overline{R} = -\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$
then,
$$\sqrt[3]{R} = -\dfrac{1}{3}$$, $$\sqrt[3]{\overline{R}} =-1$$
\begin{aligned} \\t_2&= ω\cdotp \sqrt[3]{R}+ \overline{ω} \sqrt[3]{\overline{R} }\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}( \sqrt[3]{R} - \sqrt[3]{\overline{R} }) }{2} i\\ & =\dfrac{1}{2}\Big[--\dfrac{1}{3}-\Big(-1\Big)\Big]+\dfrac{\sqrt{3}}{2}\Big[-\dfrac{1}{3}-\Big(-1\Big)\Big]i\\ & =\dfrac{1}{6}+\dfrac{1}{2}+\sqrt{3}\Big(-\dfrac{1}{6}+\dfrac{1}{2}\Big)i\\ \end{aligned}
\begin{aligned} \\t_3&= \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \sqrt[3]{\overline{R}}\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}(- \sqrt[3]{R} + \sqrt[3]{\overline{R} }) }{2}i \\ & =\dfrac{1}{2}\Big[--\dfrac{1}{3}-\Big(-1\Big)\Big]-\dfrac{\sqrt{3}}{2}\Big[-\dfrac{1}{3}-\Big(-1\Big)\Big]i\\ & =\dfrac{1}{6}+\dfrac{1}{2}-\sqrt{3}\Big(-\dfrac{1}{6}+\dfrac{1}{2}\Big)i\\ \end{aligned}
## Roots of the general cubic equation
Since $$x = t - \dfrac{b}{3a}$$, substituting the values of $$t$$, $$a$$ and $$b$$ gives
$$x_1 = t_1-\dfrac{2}{3}$$
$$x_2 = t_2-\dfrac{2}{3}$$
$$x_3 = t_3-\dfrac{2}{3}$$
## 6. Summary
In summary, we have tried the method of factorization, cubic root formula to explore the solutions of the equation. The cubic equation $$3x³ + 6x² + x + 2=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.
$$\begin{cases} x_1=-\dfrac{1}{3}\sqrt[3]{1}-\dfrac{1}{3}\sqrt[3]{27}-\dfrac{2}{3} \\ x_2=\dfrac{1}{6}+\dfrac{1}{2}+\sqrt{3}\Big(-\dfrac{1}{6}+\dfrac{1}{2}\Big)i-\dfrac{2}{3} \\ x_3=\dfrac{1}{6}+\dfrac{1}{2}-\sqrt{3}\Big(-\dfrac{1}{6}+\dfrac{1}{2}\Big)i-\dfrac{2}{3} \end{cases}$$
in decimal notation,
$$\begin{cases} x_1=-2 \\ x_2=0.57735026918963i \\ x_3=-0.57735026918963i \end{cases}$$
Using the method of factorization, the roots are derived to the following forms
$$\begin{cases} x_1=-2 \\ x_2=\dfrac{1}{\sqrt{3}}i \\ x_3=-\dfrac{1}{\sqrt{3}}i \end{cases}$$
The decimal results by using factorization method are
$$\begin{cases} x_1=-2 \\ x_2=0.57735026918963i \\ x_3=-0.57735026918963i \end{cases}$$
## 7. Graph for the function $$f(x) = 3x³ + 6x² + x + 2$$
Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 3x³ + 6x² + x + 2$$ has one intersection point with the x-axis.
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# How do you write a quadratic equation with x-intercepts: -4,3 ; point: (-5,16)?
Sep 11, 2016
$y = 2 {x}^{2} + 2 x - 24$
#### Explanation:
Use the form $y = k \left(x - a\right) \left(x - b\right)$
where k is a constant, and a and b are the zeros (x coordinate of the x-intercepts).
$a = - 4$ and $b = 3$
$y = k \left(x + 4\right) \left(x - 3\right)$
$y = k \left({x}^{2} + 4 x - 3 x - 12\right)$
$y = k \left({x}^{2} + x - 12\right)$
To find k, plug in the given point $\left(- 5 , 16\right)$.
$16 = k \left({\left(- 5\right)}^{2} + \left(- 5\right) - 12\right)$
$16 = k \left(25 - 5 - 12\right)$
$16 = k \left(8\right)$
$k = 2$
Plug k into the equation found above.
$y = 2 \left({x}^{2} + x - 12\right)$
$y = 2 {x}^{2} + 2 x + 24$
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# Average Problems
Home > Lesson > Chapter 7
5 Steps - 3 Clicks
# Average Problems
### Introduction
Average is a straight- forward concept and it can be solved easily by equal distribution method.
### Methods
Average: Average is the sum of all the elements in a given data set divided by the total number of elements in the data set. The most commonly denoted term for average is Arithmetic Mean, simply termed as mean.
Equal distribution method: It is the easiest way to solve problems
• If each element is increased by same value, then the average also gets increased by same value. Example:
32, 46, 25, 57 Suppose each value is increased by 5, the average also gets increased by 5. Average of given numbers 32, 46, 25, 57 is 40. When each number is increased by 5 i.e. 37, 51, 30, 62, the average is also increased by 5, and so the new average is 45.
• If the new value added is same as the average value, then the average is same. Example:
32, 46, 25, 57, 40. Here 40 is the new number that is added which is same as the original average. So, new Average is 40
• If new value added is different as the average value, then the difference is equalized on all the values. Example:
32, 46, 25, 57, 50 Average is 40, newly added value is 50 Now their difference is 10 Equally distributed value is $\frac{10}{(no .of values)} = \frac{10}{5}$ = 2 Hence, the average is 40 + 2 = 42
• Where 'A' → Average
• $'x'$ → Sum of given elements
• $'y'$ → Total number of elements
Example 1: Find the average of all prime numbers between 30 and 50.
Solution:
There are five prime numbers between 30 and 50.
They are 31, 37, 41, 43 and 47.
Therefore, Required average = ($\frac{31 + 37 + 41 + 43 + 47}{5}$) = $\frac{199}{5}$ = 39.8.
Example 2: Find the average of first 20 multiples of 7.
Solution:
Required average = $\frac{7(1 + 2 + 3 + ..... + 20)}{20}$ = ($\frac{7 \times 20 \times 21}{20 \times 2}$) = ($\frac{147}{2}$) = 73.5.
Example 3: Find the average of first 40 natural numbers.
Solution:
Sum of first n natural numbers = $\frac{n(n + 1)}{2}$
So, sum of first 40 natural numbers = $\frac{40 \times 41}{2}$ = 820.
Therefore, Required average = $\frac{820}{40}$ = 20.5
• Where 'x and y' → Distance
Example 1: Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km per hour. Find the average speed of the train during the whole journey.
Solution:
Required average speed = $\frac{2 x y}{x + y} km/hr$ = $\frac{2 \times 84 \times 56}{(84 + 56)} km/hr$
= $\frac{2 \times 84 \times 56}{140} km/hr$ = 67.2 km/hr.
Example 2: A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. His average speed for the whole journey in km / hr is.
Solution:
Average Speed = $\frac{2 x y}{x + y} km/hr$ = $\frac{2 \times 50 \times 30}{50 + 30} km/hr$ = 37.5 km/hr.
### Formulae
• Average(A) = $\frac{Sum \, of \, given \, elements \, (x)}{(Total \, number \, of \, elements \,(y))}$ Where 'A' → Average $'x'$ → Sum of given elements $'y'$ → Total number of elements
• If we need to find sum of "n" elements then Sum = n x $x$
• Average Speed = $\frac{2 x y}{x + y} km/hr$
• When two sections are given then average is Average = $\frac{n_{1}a_{1} + n_{2}a_{2}}{n_{1} + n_{2}}$
### Samples
Solution:
Given that
Rahul made 1000 runs in 20 matches
Now, Average = $\frac{(sum of values)}{(total number of values)}$
by substituting the values,
⇒Average = $\frac{(1000 + 500)}{(10 + 20)}$
⇒Average = $\frac{(1500)}{(30)}$
⇒Average = 50
Therefore, Average made by them = 50
2. The sum of five numbers is 555.The average of first two numbers is 75 and third number is 115. Find the average of last two numbers?
Solution:
Let the five numbers be a, b, c, d, e
Given that
a + b + c + d + e = 555
Average of first two numbers is = 75
Third number is 115
By substituting the given values,
75 + 115 + d + e = 555
⇒d + e = 555 – 265
⇒d + e = 290
Therefore, average of last two numbers (d, e) = $\frac{290}{2}$ =145
3. The average age of 20 students of a section is 12 years.The average of 25 students of another section is 12 years. Find average age of both the sections?
Solution:
Given that
Average age of 20 students of a section = 12 years
Average age of another section of 25 students = 12 years
Average = $\frac{n1a1 + n2a2}{n1 + n2}$
Here n1 is number of students in first section
a1 is average age of first section
n2 is number of students in second section
a2 is average age of second section
by substituting the given values in above formula,
Average = $\frac{20 * 12 + 25 * 12}{20 + 25}$
⇒Average = $\frac{540}{45}$
⇒Average = 12
Therefore, average age of both the sections = 12.
4. The total of ages of class of 75 girls is 1050. Average age of 25 of them is 12 years and another 25 of them is 16 years. Find the average age of remaining girls?
Solution:
Given that
Total ages of class =1050
Average age of 25 girls = 12 years
Average age of another 25 girls = 16 years
Now, average age of remaining girls = $\frac{(1050 - (25 * 12 + 25 * 16))}{(75 – 50)}$
⇒Average = $\frac{(1050 – (25*28))}{(25)}$
⇒Average = $\frac{1050}{25}$ - $\frac{(25 * 28)}{25}$
⇒Average = 42 - 28
⇒Average = 12
Therefore, average age of remaining girls = 12 years
5. Average age of 50 girls 58. If the weight of one of the girl is taken as 45 instead of 65. What is the actual average?
Solution:
Given that
Average age of 50 girls = 58
One of them is taken as 45 instead of 65
So, 65 - 45 = 20
$\frac{20}{50}$ = 0.4
Hence, actual average is 58 + 0.4 = 58.4
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# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Written by Team Trustudies
Updated at 2021-05-07
## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1
Q1 ) Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (- a, – b)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Distance formula =$$\sqrt { (x_2 - x_1)^2 + (y_2 - y_1)^2}$$
(i) Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
d = $$\sqrt { (4 - 2)^2 + (1 - 3)^2}$$
$$= \sqrt{ 2^2 + (-2) ^2} = \sqrt{4 + 4}$$
$$= \sqrt{8}$$
$$= 2\sqrt{2}$$
(ii) Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get
d = $$\sqrt { (-5 + 1)^2 + (7 - 3)^2}$$
$$= \sqrt{ ((-4)^2 + (4) ^2} = \sqrt{16 + 16}$$
$$= \sqrt{32}$$
$$= 4\sqrt{2}$$
(iii) Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get
d = $$\sqrt { (-a - a)^2 + (-b - b)^2}$$
$$= \sqrt{ ((-2a)^2 + (-2b) ^2}$$
$$= \sqrt{4a^2 + 4b^2}$$
$$= 2\sqrt{a^2 + b^2}$$
Q2 ) Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
The coordinates of point A are (0, 0) and coordinates of point B are (36, 15).
To find the distance between them, we use Distance formula:
d = $$\sqrt{(36-0)^2 + (15 - 0)^2}$$
$$= \sqrt{36^2 - 15^2}$$
$$= \sqrt{1296 + 225}$$
$$= \sqrt{1521}$$
$$= 39$$km
Q3 ) Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let A = (1, 5), B = (2, 3) and C = (–2, –11)
Using Distance Formula to find distance AB, BC and CA.
AB = $$\sqrt{(2-1)^2 + (3 - 5)^2}$$
$$= \sqrt{1^2 - (-2)^2}$$
$$= \sqrt{1 + 5}$$
$$= \sqrt{5}$$
BC = $$\sqrt{(-2-2)^2 + (-11 - 3)^2}$$
$$= \sqrt{(-4)^2 - (-14)^2}$$
$$= \sqrt{16 + 196}$$
$$= \sqrt{212}$$
CA = $$\sqrt{(-2-1)^2 + (-11 - 5)^2}$$
$$= \sqrt{(-3)^2 - (-15)^2}$$
$$= \sqrt{9 + 256}$$
$$= \sqrt{265}$$
Since AB + AC $$\ne$$ BC, BC + AC $$\ne$$ AB and AC $$\ne$$ BC.
Therefore, the points A, B and C are not collinear.
Q4 ) Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let A = (5, –2), B = (6, 4) and C = (7, –2)
Using Distance Formula to find distances AB, BC and CA.
AB = $$\sqrt{(6-5)^2 + (4 + 2)^2}$$
$$= \sqrt{1^2 - 6^2}$$
$$= \sqrt{1 + 36}$$
$$= \sqrt{37}$$
BC = $$\sqrt{(7-6)^2 + (-2 - 4)^2}$$
$$= \sqrt{1^2 - (-6)^2}$$
$$= \sqrt{1 + 36}$$
$$= \sqrt{37}$$
CA = $$\sqrt{(7-5)^2 + (-2 + 2)^2}$$
$$= \sqrt{2^2 - 0}$$
$$= \sqrt{4}$$
$$= 2$$
Since AB = BC.
Therefore, A, B and C are vertices of an isosceles triangle.
Q5 ) In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli. “Don’t you think ABCD is a square?”Chameli disagrees. Using distance formula, find which of them is correct.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(6-3)^2 + (7 - 4)^2}$$
$$= \sqrt{3^2 + 3^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
$$= 3\sqrt{2}$$
BC = $$\sqrt{(9- 6)^2 + (4 - 7)^2}$$
$$= \sqrt{3^2 + 3^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
$$= 3\sqrt{2}$$
CD = $$\sqrt{(6-9)^2 + (1 - 4)^2}$$
$$= \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9}$$
$$= \sqrt{18}$$
$$= 3\sqrt{2}$$
DA = $$\sqrt{(6-3)^2 + (1 - 4)^2}$$
$$= \sqrt{3^2 + (-3)^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
$$= 3\sqrt{2}$$
Therefore, All the sides of ABCD are equal here. … (1)
Now, we will check the length of its diagonals.
AC = $$\sqrt{(9-3)^2 + (4 - 4)^2}$$
$$= \sqrt{6^2 - 0^2}$$
$$= \sqrt{36}$$
$$= 6$$
BD = $$\sqrt{(6-6)^2 + (1 - 7)^2}$$
$$= \sqrt{0^2 - 6^2}$$
$$= \sqrt{36}$$
$$= 6$$
So, Diagonals of ABCD are also equal. … (2)
From (1) and (2), we can definitely say that ABCD is a square.
Therefore, Champa is correct.
Q6 ) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(1 + 1)^2 + (0 + 2)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
BC = $$\sqrt{(-1 - 1)^2 + (2 - 0)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
CD = $$\sqrt{(-3 + 1)^2 + (0 - 2)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
DA = $$\sqrt{(-3 + 1)^2 + (0 - 2)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.
AC = $$\sqrt{(-1 + 1)^2 + (2 + 2)^2}$$
$$= \sqrt{0 + 16}$$
$$= 4$$
BD = $$\sqrt{(-3 - 1)^2 + (0 + 0)^2}$$
$$= \sqrt{16 + 0}$$
$$= 4$$
Therefore, diagonals of quadrilateral ABCD are also equal. … (2)
From (1) and (2), we can say that ABCD is a square.
(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(-3 - 3)^2 + (1 - 5)^2}$$
$$= \sqrt{36 + 16}$$
$$= 2 \sqrt{13}$$
BC = $$\sqrt{(0 - 3)^2 + (3 - 1)^2}$$
$$= \sqrt{9 + 4}$$
$$= \sqrt{13}$$
CD = $$\sqrt{(-1 - 0)^2 + (-4 - 3)^2}$$
$$= \sqrt{1 + 49}$$
$$= 5\sqrt{2}$$
DA = $$\sqrt{(-1 + 3)^2 + (-4 - 5)^2}$$
$$= \sqrt{4 + 81}$$
$$= \sqrt{85}$$
We cannot find any relation between the lengths of different sides.
Therefore, we cannot give any name to the figure ABCD.
(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(7 - 4)^2 + (6 - 5)^2}$$
$$= \sqrt{9 + 1}$$
$$= \sqrt{10}$$
BC = $$\sqrt{(4 - 7)^2 + (3 - 6)^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
CD = $$\sqrt{(1 - 4)^2 + (2 - 3)^2}$$
$$= \sqrt{9 + 1}$$
$$= \sqrt{10}$$
DA = $$\sqrt{(1 - 4)^2 + (2 - 5)^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
Here opposite sides of quadrilateral ABCD are equal. … (1)
We can now find out the lengths of diagonals.
AC = $$\sqrt{(4 - 4)^2 + (3 - 5)^2}$$
$$= \sqrt{0 + 4}$$
$$= 2$$
BD = $$\sqrt{(1 - 7)^2 + (2 - 6)^2}$$
$$= \sqrt{36 + 16} =$$
$$2 \sqrt{13}$$
Here diagonals of ABCD are not equal. … (2)
From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.
Q7 ) Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).
Using Distance Formula and according to given conditions we have:
$$\sqrt{(x - 2)^2 + ( 0 - (-5))^2}$$
$$= \sqrt{ (x - (-2))^2 + ( 0 - 9)^2}$$
=>$$\sqrt {x^2 + 4 - 4x + 25}$$
$$= \sqrt {x^2 + 4 + 4x + 81}$$
Squaring both sides, we get
$$\Rightarrow x^2 + 4 - 4x + 25 = x^2 + 4 + 4x + 81$$
$$\Rightarrow-4x + 29 = 4x + 85$$
$$\Rightarrow 8x = -56$$
$$\Rightarrow x = -7$$
Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)
Q8 ) Find the values of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Using Distance formula, we have
$$\Rightarrow 10 = \sqrt{(2 -10)^2 + (-3-y)^2}$$
$$\Rightarrow 10 = \sqrt{(-8)^2 + 9 + y^2 + 6y}$$
$$\Rightarrow 10 = \sqrt{64 + 9 + y^2 + 6y}$$
Squaring both sides, we get
$$\Rightarrow 100 = 73 + y^2 + 6y$$
$$\Rightarrow y^2 + 6y + 27 = 0$$
Solving this Quadratic equation by factorization, we can write
$$\Rightarrow y^2 + 9y - 3y - 27 = 0$$
$$\Rightarrow y (y + 9) – 3 (y + 9) = 0$$
$$\Rightarrow (y + 9) (y - 3) = 0$$
$$\Rightarrow y = 3, -9$$
Q9 ) If, Q (0, 1) is equidistant from P (5, –3) and R (x, 6), find the values of x. Also, find the distances QR and PR.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
It is given that Q is equidistant from P and R. Using Distance Formula, we get
PQ = RQ
PQ=$$\sqrt{(5-0)^2 + (-3 - 1)^2}$$
$$= \sqrt{(-5)^2 + (-4)^2}$$
$$= \sqrt{25 + 16}$$
$$= \sqrt{41}$$
QR= $$\sqrt{(0 - x)^2 + (1 - 6)^2}$$
$$= \sqrt{(-x)^2 + (-5)^2}$$
$$= \sqrt{x^2 + 25}$$
$$\therefore \sqrt{41} = \sqrt{x^2 + 25}$$
On squaring both sides,
$$\Rightarrow 41 = x^2 + 25$$
$$\Rightarrow x^2 = 16$$
$$\Rightarrow x = \pm4$$
Coordinates of point R will be (4,6) or (-4,6),
If R is (4,6),
QR = $$\sqrt{(0 - 4)^2 + (1 - 6)^2}$$
$$= \sqrt{4^2 + 5^2}$$
$$= \sqrt{16 + 25}$$
$$= \sqrt{41}$$
PR = $$\sqrt{(5 - 4)^2 + (-3 - 6)^2}$$
$$= \sqrt{1^2 + 9^2}$$
$$= \sqrt{1 + 81}$$
$$= \sqrt{82}$$
If R is (-4,6),
QR = $$\sqrt{(0 + 4)^2 + (1 - 6)^2}$$
$$= \sqrt{4^2 + 5^2}$$
$$= \sqrt{16 + 25}$$
$$= \sqrt{41}$$
PR = $$\sqrt{(5 + 4)^2 + (-3 - 6)^2}$$
$$= \sqrt{9^2 + 9^2}$$
$$= \sqrt{81 + 81}$$
$$= 9\sqrt{2}$$
Q10 ) Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
It is given that (x, y) is equidistant from (3, 6) and (–3, 4).
Using Distance formula, we can write
$$\Rightarrow \sqrt{(x-3)^2 + (y-6)^2} = \sqrt{(x-(-3))^2 + (y-4)^2}$$
$$\Rightarrow \sqrt{x^2 + 9 - 6x + y^2 + 36 -12y } = \sqrt{x^2 + 9 + 6x + y^2 + 16 -8y }$$
Squaring both sides, we get
$$\Rightarrow x^2 + 9 - 6x + y^2 + 36 -12y = x^2 + 9 + 6x + y^2 + 16 -8y$$
$$\Rightarrow -6x - 12y + 45 = 6x - 8y + 25$$
$$\Rightarrow 12x + 4y = 20$$
$$\Rightarrow 3x + y = 5$$
## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2
Q1 ) Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the points be P(x,y), Using the section formula
$$P(x, y) = [\frac{mx_2 + nx_1}{m + n} , \frac{my_2 + ny_1}{m + n}]$$
$$x = {{2(4) + 3(-1)} \over {2 + 3}} = {{8 - 3} \over {5}} = 1$$
$$y = {{2(-3) + 3(7)} \over {2 + 3}} = {{-6 + 21} \over {5}} = 3$$
Therefore the point is(1,3)
Q2 ) Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1:2.
x1 = $$\frac{(1×(-2)+2×4)}{1+2} = \frac{(-2+8)}{3} =\frac{6}{3} = 2$$
y1 = $$\frac{(1×(-3)+2×-1)}{1+2} = \frac{(-3-2)}{3} =\frac{-5}{3}$$
$$\therefore$$P = (x1, y1) = P(2, $$\frac{-5}{3}$$)
Point Q divides AB internally in the ratio 2:1.
x2 = $$\frac{(2×(-2)+1×4)}{2+1} = \frac{(-4+4)}{3} = 0$$
y2 = $$\frac{(2×(-3)+1×-1)}{2+1} = \frac{(-6-1)}{3} =\frac{-7}{3}$$
The coordinates of the point Q(x2 ,y2 ) = (0, $$\frac{-7}{3}$$ )
Q3 ) To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs $$\frac{1}{4}$$ th the distance AD on the 2nd line and posts a green flag. Preet runs $$\frac{1}{5}$$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
The green flag is $$\frac{1}{4} th$$ of total distance =
$$\frac{1}{4} × 100 = 25$$ m in second line
$$\therefore$$ the coordinate of green flag
= (2 , 25 )
Similarly coordinate of red flag
= (8 , 20)
Distance between red and green flag
= $$\sqrt{(8-2)^2 + (20-25)^2}$$
$$= \sqrt{6^2+(-5)^2}$$
$$= \sqrt{36+25}$$
$$= \sqrt{61} m$$
Now blue flag is posted at the mid point of two flag
Then let, the coordinate of blue flag = (x , y )
$$\therefore( x , y ) =( \frac{(8+2)}{2} , \frac{20+25}{2} )$$
$$=( 5 , \frac{45}{2} )= ( 5, 22.5 )$$
$$\therefore$$ the blue flag is posted in fifth line at a distance of 22.5 m
Q4 ) Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the required ratio be k : 1
By section formula ,
$$x = \frac{m1x2 + m2x1}{m1+m2}$$
$$\Rightarrow -1 = \frac{k×6 + 1× -3 )}{k+1}$$
$$\Rightarrow -k-1 = 6k -3$$
$$\Rightarrow 7k = 2$$
$$\Rightarrow k = \frac{2}{7}$$
The required ratio = k : 1
= $$\frac{2}{7}$$ : 1 = $$\frac{2}{7} ×7 : 1×7$$ = 2 : 7
Q5 ) Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let p(x , 0) be the point which devides the line segment joining A(1, -5 ) and B(-4,5 ) in the ratio m : 1
Then the using section formula
(x , 0 ) = $$( \frac{m×(-4) + 1 × 1 }{m+1} , \frac{m×5+1×(-5)}{m+1} )$$
$$\Rightarrow 0 = \frac{m×5+1×(-5)}{m+1}$$
$$\Rightarrow 0 = 5m - 5$$
$$\Rightarrow m = 5 /5 = 1$$
Hence the required ratio is 1 : 1
Since the ratio is 1 : 1 , so P is the mid point
$$\therefore$$ x = $$\frac{m×(-4) + 1 × 1 }{m+1}$$
x = $$\frac{1×(-4) + 1 × 1 }{1+1} = \frac{-4+1}{2} = \frac{-3}{2}$$
$$\therefore ( \frac{-3}{2} , 0 )$$ is the required point
Q6 ) If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Mid point of AC = Mid point BD
$$\Rightarrow \frac{x+1}{2} , \frac{6+2}{2} = \frac{4+3}{2} , \frac{y+5}{2}$$
$$\Rightarrow \frac{x+1}{2} = \frac{7}{2}$$
$$\Rightarrow x +1 = 7$$
$$\Rightarrow x = 7-1= 6$$
$$\Rightarrow \frac{y+5}{2} = \frac{8}{2}$$
$$\Rightarrow y+5 = 8$$
$$\Rightarrow y = 8 -5 = 3$$
Q7 ) Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, – 3) and B is (1,4).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the coordinates of point A be (x, y).
Mid-point of AB is (2, – 3), which is the centre of the circle.
Coordinate of B = (1, 4)
(2, -3) =$$(\frac{x+1}{2} , \frac{y+4}{2} )$$
$$\frac{x+1}{2} = 2$$
$$\Rightarrow x+1= 4$$
$$\Rightarrow x = 4-1 = 3$$
and$$\frac{y+4}{2} = -3$$
$$\Rightarrow y+4 = -6$$
$$\Rightarrow y = -6-4 = -10$$
$$\therefore$$ x = 3 and y = -10
The coordinates of A(3,-10).
Q8 ) If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = $$\frac{3}{7}$$ AB and P lies on the line segment AB.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
AP = $$\frac{3}{7}$$ AB
7 AP = 3 AB
AB : AP = 7: 3
Let AB = 7x
$$\therefore$$ AP = 3x
AB = AP + BP
7x = 3x + BP
BP = 7x - 3x = 4x
$$\therefore \frac{AP}{BP} = \frac{3x}{4x}$$
$$\therefore$$ AP : BP = 3:4
by section formula ,
(x, y ) = $$(\frac{m1x2 + m2x1}{m1+m2} , \frac{m1y2 + m2y1}{m1+m2} )$$
(x, y) = $$(\frac{3×2 +4×(-2) }{3+4} , \frac{3×(-4)+ 4(-2)}{3+4} )$$
(x ,y ) = $$(\frac{6 -8 }{7} , \frac{-12+ -8}{7} )$$
(x ,y ) = $$(\frac{-2}{7} , \frac{-20}{7} )$$
Hence the coordinates of P = $$(\frac{-2}{7} , \frac{-20}{7} )$$
Q9 ) Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
From the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.
Coordinate of X = $$( \frac{1×2+3×(-2)}{1+3} , \frac{1×8+3×2}{1+3} )$$
$$=( \frac{2-6}{4} , \frac{8+6}{4} )$$
$$= ( \frac{-4}{4} , \frac{14}{4} )$$
$$= ( -2 , \frac{7}{2} )$$
Coordinate of Y = $$( \frac{1×2+1×(-2)}{1+1} , \frac{1×8+1×2}{1+1} )$$
$$=( \frac{2-2}{2} , \frac{8+2}{2} )$$
$$= ( \frac{0}{2} , \frac{10}{2} )$$
$$= ( 0 , 5)$$
Coordinate of Y = $$( \frac{3×2+1×(-2)}{3+1} , \frac{3×8+1×2}{3+1} )$$
$$=( \frac{6-2}{4} , \frac{24+2}{4} )$$
$$= ( \frac{4}{4} , \frac{26}{4} )$$
$$= ( 1 , \frac{13}{2} )$$
Q10 ) Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order.
[Hint: Area of a rhombus = $$\frac{1}{2}$$ (product of its diagonals)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let A(3, 0), B (4, 5), C( – 1, 4) and D ( – 2, – 1) are the vertices of a rhombus ABCD.
Length of diagonal AC
$$=\sqrt{[3-(-1)]^2 + (0-4)^2}$$
$$= \sqrt{16+16}$$
$$= \sqrt{32}$$
$$= 4\sqrt{2}$$
Length of diagonal BD
$$=\sqrt{[(-2)-4]^2+ (-1-5)^2}$$
$$= \sqrt{(-6)^2+(-6)^2}$$
$$= \sqrt{36+36}$$
$$= \sqrt{72}$$
$$= 6\sqrt{2}$$
Area of rhombus = $$\frac{1}{2}$$ × product of diagonals
= $$\frac{1}{2} × 4\sqrt{2}×6\sqrt{2} = 24$$sq unit
## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3
Q1 ) Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
We know that formula for area of a triangle whose vertices are $$(x_1,y_1) , (x_2,y_2) , (x_3,y_3)$$ is,
= $$\frac{1}{2} \ | \ [x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \ |$$
(i) So, here $$x_1 \ = \ 2 \ , \ y_1 \ = \ 3 \ , \ x_2 \ = \ -1 \ , \ y_2 \ = \ 0 \ x_3 \ = \ 2 \ , \ y_3 \ = \ -4$$
So, area of triangle = $$\frac{1}{2} \ | \ [ 2( 0 + 4 ) \ - \ 1(-4 - 3) \ + \ 2(3 - 0)] \ | \ = \ \frac{1}{2} \ | \ [ 8 + 7 + 6 ] \ | \ = \ \frac{21}{2}$$
∴ Area of triangle is $$\frac{21}{2}$$ sq. units.
(ii) Similarly, here $$x_1 \ = \ -5 \ , \ y_1 \ = \ -1 \ , \ x_2 \ = \ 3 \ , \ y_2 \ = \ -5 \ x_3 \ = \ 5 \ , \ y_3 \ = \ 2$$
So, area of triangle = $$\frac{1}{2} \ | \ [ (-5)( -5 - 2 ) \ - \ 3(2 + 1) \ + \ 5(-1 + 5)] \ | \ = \ \frac{1}{2} \ | \ [ 35 + 9 + 20 ] \ | \ = \ 32$$
∴ Area of triangle is $$32$$ sq. units.
Q2 ) In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, -k)
(ii) (8, 1), (k, -4), (2, -5)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
We know that for collinear points area of triangle = 0 ,i.e.,
$$0 \ = \frac{1}{2}[x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)]$$
(i) $$7(1-k) + 5(k+2) + 3(-2-1) \ = \ 0$$
=> $$7 - 7k + 5k + 10 - 9 \ = \ 0$$
=> $$2k \ = \ 8$$
=> $$k \ = \ 4$$
(ii) $$8(-4+5) + k(-5-1) + 2(1+4) \ = \ 0$$
=> $$8 - 6k + 10 \ = \ 0$$
=> $$6k \ = \ 18$$
=> $$k \ = \ 3$$
Q3 ) Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
The vertices of the triangle ABC are A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides AB,BC,CA of this triangle ABC respectively.
Coordinates of D, E, and F are given by :
D = $$(\frac{0+2}{2} \ , \ \frac{-1+1}{2})$$ = (1,0)
E = ( $$\frac{0+2}{2} \ , \ \frac{1+3}{2}$$ ) = (1,2)
F = ( $$\frac{0+0}{2} \ , \ \frac{3-1}{2}$$ ) = (0,1)
So, Area of $$∆ \ DEF \ = \ \frac{1}{2} \ | \ [ 1(2-1) + 1(1-0) + 0(0-2)] \ | \ = \ 1$$
∴ Area of ∆ DEF = 1 sq. units
Area of ∆ ABC = $$\frac{1}{2} \ | \ [0(1-3) \ + \ 2(3+1) \ + \ 0(-1-1) ] \ | \ = \ 4$$
$$\therefore$$ Area of $$\triangle$$ ABC = 4 sq. units
So, $$∆ \ DEF \ : \ ∆ \ ABC \ = \ 1 \ : \ 4$$.
Q4 ) Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Draw a line from B to D
Now we have,
Area of quad. ABCD = | Area of ∆ DAB | + | Area of ∆ BCD |
So,
Area of ∆ DAB = $$| \ \frac{1}{2} [2(-2+5) \ - \ 4(-5-3) \ - \ 3(3+2)] \ |$$ = $$\ | \frac{1}{2} [6 + 32 - 15 ] \ |$$ = $$\frac{23}{2}$$ sq. units
Similarly, Area of ∆ BCD = $$| \ \frac{1}{2} [-3(-2-3) \ + \ 3(3+5) \ + \ 2(-5+2)] \ | \ = \ | \ \frac{1}{2} [15 + 24 - 6 ] \ |$$ = $$\frac{33}{2}$$ sq. units
$$\therefore$$ Area of quad. ABCD = $$\frac{23}{2} \ + \ \frac{33}{2}$$ = $$\frac{56}{2} \ = \ 28$$ sq. units
Q5 ) You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for $$\triangle$$ ABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
The vertices of triangle ABC are A(4,-6), B (3, – 2) and C (5, 2).
Let D be the mid-point of side BC of ∆ ABC. Therefore, AD is the median in ∆ ABC.
So, coordinates of D are ( $$\frac{3+5}{2} \ , \ \frac{-2+2}{2}$$ )
Therefore, coordinates of D = (4,0)
Now,
Area of ∆ ABD = $$\frac{1}{2} \ | \ [4(-2-0) \ + \ 3(0+6) \ + \ 4(-6+2)] |$$
= $$\frac{1}{2} \ | \ [ -8 + 18 - 16] \ |$$
= $$|-3| \ = \ 3$$ sq. units
Similarly,
Area of ∆ ACD = $$\frac{1}{2} \ | \ [4(0-2) \ + \ 4(2+6) \ + \ 5(-6-0)] \ |$$
= $$\frac{1}{2} \ | \ [ -8 + 32 - 30] \ |$$
= $$|-3| \ = \ 3$$ sq. units
$$\therefore$$ Clearly, Area of ∆ ABD = Area of ∆ ACD = 3 sq. units.
Hence, median of a triangle divides it into two triangles of equal areas.
## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4
Q1 ) Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the line 2x + y - 4 = 0 divides AB in 1 : k ratio, then the coordinates of the point of division is
$$x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1}$$
The point of intersection lie on both lines
$$\therefore x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1}$$ should satisfy 2x + y - 4 = 0
$$\Rightarrow 2( \frac{2+3k}{k+1}) \ + \ (\frac{-2+7k}{k+1}) \ - \ 4 \ = \ 0$$
$$\Rightarrow 4 \ + \ 6k \ - \ 2 \ + \ 7k \ = \ 4k \ + \ 4$$
$$\Rightarrow 9k - 2 = 0$$
$$\Rightarrow k \ = \ \frac{2}{9}$$
$$\therefore$$ The line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7) in ratio 2 : 9 internally.
Q2 ) Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
If given points are collinear then area of triangle formed by them must be zero.
Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,
Area of triangle = $$\frac{1}{2} \ | \ [ (x_1(y_2 – y_3 ) + x_2(y_3 – y_1) + x_3(y_1 – y_2) ) ] \ | \ = \ 0$$
$$\Rightarrow \frac{1}{2} \ | \ [x(2 – 0) + 1 (0 – y) + 7( y – 2)] \ | \ = \ 0$$
$$\Rightarrow 2x \ – \ y \ + \ 7y \ – \ 14 \ = \ 0$$
$$\Rightarrow 2x \ + \ 6y \ – \ 14 \ = \ 0$$
$$\Rightarrow x \ + \ 3y \ – \ 7 \ = \ 0$$ .
Hence this is the required relation between x and y.
Q3 ) Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the coordinate of centre O be (x,y)
Then, OA = OB = OC {$$\because$$ radius is equal}
Now,
OA = $$\sqrt{ (x-6)^2+(y+6)^2 }$$
OB = $$\sqrt{(x-3)^2 + (y+7)^2 }$$
OC = $$\sqrt{(x-3)^2 + (y-3)^2 }$$
Now, OA = OB
$$\Rightarrow \sqrt{(x-6)^2 + (y+6)^2 } = \sqrt{(x-3)^2 + (y+7)^2 }$$
$$\Rightarrow x^2 \ + \ 36 \ - \ 12x \ + \ y^2 \ + \ 36 \ - \ 12 \ = \ x^2 \ + \ 9 \ - \ 6x \ + \ y^2 \ + \ 9 \ - \ 6y$$
$$\Rightarrow 6x \ + 2y \ - \ 14 \ = \ 0$$ ..............(i)
Similarly, OB = OC
$$\Rightarrow \sqrt{(x-3)^2 + (y+7)^2 } = \sqrt{(x-3)^2 + (y-3)^2 }$$
$$\Rightarrow y^2 \ + \ 14y \ + \ 49 \ = \ y^2 \ - \ 6y \ + \ 9$$
$$\Rightarrow y \ = \ -2$$
Putting this value of y in eq. (i) , we get
$$\Rightarrow 6x \ + \ 2(-2) \ - \ 14 \ = \ 0$$
$$\Rightarrow x \ = \ 3$$
$$\therefore$$ The centre of circle is at (3,-2)
Q4 ) The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let ABCD be a square with vertices A( $$x_1, y_1) , B(3 , 2) , C(x_2 , y_2) , D(-1 , 2)$$
Let each side of square be 'a'.
So applying Pythagoras theorem in ∆ABD, we get
$$BD^2 = AB^2 + DA^2$$
$$\Rightarrow \sqrt{(3+1)^2 \ + \ (2-2)^2} \ = \ a^2 \ + \ a^2$$
$$\Rightarrow 4 \ = \ 2 a^2$$
$$\Rightarrow a \ = \ 2 \sqrt{2}$$
Now, DA = AB {$$\because$$ ABCD is a square }
$$\Rightarrow \sqrt{(x_1 +1)^2 + (y_1 - 2)^2 } \ = \ \sqrt{(3-x_1)^2 + (2 - y_1)^2}$$ {by distance formula}
$$\Rightarrow (x_1 +1)^2 \ = \ (3-x_1)^2$$
$$\Rightarrow x_1^2 \ + \ 1 \ + \ 2x_1 \ = \ x_1^2 \ + \ 9 \ - \ 6x_1$$
$$\Rightarrow x_1 \ = \ 1$$
Now since each side is $$2 \sqrt{2}$$
$$\therefore$$ AB $$= \ 2 \sqrt{2}$$
$$\Rightarrow \sqrt{(3-x_1)^2 + (2 - y_1)^2} \ = \ 2 \sqrt{2}$$
$$\Rightarrow(2-y_1)^2 = \ 4$$
$$\Rightarrow y \ = \ 4$$
Now, point O is id point of BD . SO coordinates of O are
x = $$\frac{3-1}{2}$$ , y = $$\frac{2+2}{2}$$
$$\Rightarrow$$ x = 1 , y = 2
Now, since ABCD is a square $$\therefore$$ O is mid point of AC also
$$\frac{1+x_2}{2} \ = \ 1$$ , $$\frac{4+ y_2}{2} \ = \ 2$$
$$\Rightarrow x_2 \ = \ 1$$ , $$y_2 \ = \ 0$$
$$\therefore$$ Coordinates of vertices square ABCD are A(1, 4) , B(3,2) , C(1,0) , D(-1,2)
Q5 ) The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot as shown in the fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
(i) Taking A as origin, coordinates of the vertices P, Q and R are,
From figure:
P = (4, 6)
Q = (3, 2)
R = (6, 5)
Here AD is the x-axis and AB is the y-axis.
Area of triangle PQR in case of origin A:
Area of a triangle = $$\frac{1}{2} \ | \ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)] \ | \$$
$$= \ \frac{1}{2} \ | \ [– 12 – 3 + 24] \ | \$$
$$= \ \frac{9}{2}$$ sq. units
(ii) Taking C as origin, coordinates of vertices P, Q and R are,
From figure:
P = (12, 2)
Q = (13, 6)
R = (10, 3)
Here CB is the x-axis and CD is the y-axis.
Area of triangle PQR in case of origin C:
Area of a triangle = $$\frac{1}{2} \ | \ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)] \ | \$$
$$= \frac{1}{2} \ | \ [36 + 13 – 40] \ | \$$
$$= \ \frac{9}{2}$$ sq unit
This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C
Area is same in both case because triangle remains the same no matter which point is considered as origin.
Q6 ) The vertices of a $$∆ \ ABC$$ are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that $$\frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4}$$ . Calculate the area of the $$∆ \ ADE$$ and compare it with area of $$∆ \ ABC$$ . (Recall Theorem 6.2 and Theorem 6.6)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Given, the vertices of a $$∆ \ ABC$$ are A (4, 6), B (1, 5) and C (7, 2) and,
$$\frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4}$$
$$\Rightarrow \frac{AB}{AD} \ = \ \frac{AC}{AE} \ = \ 4$$
$$\Rightarrow \frac{AD + DB}{AD} \ = \ \frac{AE + EC}{AE} \ = \ 4$$
$$\Rightarrow 1 + \frac{DB}{AD} \ = \ 1 + \frac{EC}{AE} \ = \ 4$$
$$\Rightarrow \frac{DB}{AD} \ = \ \frac{EC}{AE} \ = \ 3$$
$$\therefore$$ Point D and E divides AB and AC in ratio 1 : 3
Now coordinates of D calculated by section formula are,
$$x \ = \ \frac{3(4) + 1(1)}{4} \ = \ \frac{13}{4}$$ and,
$$y \ = \ \frac{3(6) + 1(5)}{4} \ = \ \frac{23}{4}$$
Similarly, coordinates of E are,
$$x \ = \ \frac{1(7) + 3(4)}{4} \ = \ \frac{19}{4}$$ and,
$$y \ = \ \frac{1(2) + 3(6)}{4} \ = \ \frac{20}{4} \ = \ 5$$
Now area of triangle ADE is,
$$\frac{1}{2} \ | \ [4(5-2) \ + \ 1(2-6) \ + \ 7(6-5)] \ |$$
= $$\frac{1}{2} \ | \ [(12-4+7)] \ |$$
= $$\frac{15}{2}$$ sq. units
Similarly, area of triangle ABC is,
$$\frac{1}{2} \ | \ [4( \frac{23}{4} -5) \ + \ \frac{13}{4}(5-6) \ + \ \frac{19}{4}(6- \frac{23}{4})] \ |$$
= $$\frac{1}{2} \ | \ [(3- \frac{13}{4} + \frac{19}{4})] \ |$$
= $$\frac{15}{32}$$ sq. units
$$\therefore$$ ratio of $$area \ of \ ∆ \ ADE \ to \ area \ of \ ∆ \ ABC \ = \ 1:16$$
Q7 ) Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of $$∆ \ ABC$$ .
(i) The median from A meets BC at D. Find the coordinates of point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.
(iv) What do you observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]
(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
(i) Given, D is the median to BC. So, coordinates of D using distance formula are
$$x \ = \ \frac{6+1}{2} \ = \ \frac{7}{2}$$ and $$y \ = \ \frac{5+4}{2} \ = \ \frac{9}{2}$$
(ii) Coordinates of P can be calculated using section formula
$$x \ = \ \frac{2( \frac{7}{2}) + 1(4)}{3} \ = \ \frac{11}{3}$$ and $$y \ = \ \frac{2( \frac{9}{2}) + 1(2)}{3} \ = \ \frac{11}{3}$$
(iii) Coordinates of E
= ( $$\frac{4+1}{2} \ , \ \frac{2+4}{2}$$ )
= ( $$\frac{5}{2} , 3$$ )
So, coordinates of Q are,
( $$\frac{2( \frac{5}{2}) + 1(6)}{3} \ , \ \frac{2(3) + 1(5)}{3}$$ )
= ( $$\frac{11}{3} , \frac{11}{3}$$ )
Coordinates of F
= ( $$\frac{4+6}{2} , \frac{2+5}{2}$$ )
= ( $$5, \frac{7}{2}$$ )
So, coordinates of R are,
( $$\frac{2(5) + 1(1)}{3} , \frac{2 ( \frac{7}{2}) + 1(4)}{3}$$ )
= ( $$\frac{11}{3} , \frac{11}{3}$$ )
(iv) We observed that coordinates of P , Q, R are same i.e., ( $$\frac{11}{3} , \frac{11}{3}$$ ) which shows that medians intersect each other at a common point which is called centroid of the triangle.
(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, then the coordinates of the centroid of the triangle is
$$x \ = \ \frac{x_1 + x_2 + x_3}{3} \ and \ \frac{y_1 + y_2 + y_3}{3}$$
Q8 ) ABCD is a rectangle formed by the points A (-1, – 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
P, Q, R and S are the midpoints of AB, BC, CD and DA respectively
So, coordinate of P
= ( $$\frac{-1-1}{2} , \frac{-1+4}{2}$$ )
= ( $$-1 , \frac{3}{2}$$ )
Similarly, coordinate of Q
= ( $$\frac{5-1}{2} , \frac{4+4}{2}$$ )
= ( 2,4)
R = ( $$\frac{5+5}{2} , \frac{4-1}{2}$$ ) = ( $$5 , \frac{3}{2}$$ )
S = ( $$\frac{5-1}{2} , \frac{-1-1}{2}$$ ) = ( 2,-1)
Now, PQ = $$\sqrt{(-1-2)^2 + ( \frac{3}{2} - 4)^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$
SP = $$\sqrt{(2+1)^2 + (-1- \frac{3}{2})^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$
QR = $$\sqrt{(2-5)^2 + (4- \frac{3}{2})^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$
RS = $$\sqrt{(5-2)^2 + ( \frac{3}{2} +1)^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$
PR = $$\sqrt{(-1-5)^2 + ( \frac{3}{2} - \frac{3}{2})^2 } \$$
$$= \ 6$$
QS = $$\sqrt{(2-2)^2 + (4+1)^2} \$$
$$= \ 5$$
Clearly, PQ = QR = RS = SP
But, $$PR \ne QS$$
Hence the lengths of all the sides are equal, while the lengths of the diagonals are different.
$$\therefore$$ The PQRS is a rhombus.
##### FAQs Related to NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
There are total 33 questions present in ncert solutions for class 10 maths chapter 7 coordinate geometry
There are total 12 long question/answers in ncert solutions for class 10 maths chapter 7 coordinate geometry
There are total 4 exercise present in ncert solutions for class 10 maths chapter 7 coordinate geometry
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# Equivalent Fractions Calculator
Created by Wojciech Sas, PhD
Reviewed by Bogna Szyk and Jack Bowater
Last updated: Jan 18, 2024
This equivalent fractions calculator is a great tool that helps you find equivalent fractions for any fraction you wish. This is the right place to learn what is an equivalent fraction of x, or how to find equivalent fractions.
Wondering whether two fractions are equivalent? You can get the answer to this question as well!
🔎 If you want to go one step further and learn how to add fractions, check our adding fractions calculator!
Let's recall that a fraction is a ratio of two numbers, the numerator, A and the denominator, B. As you already know from the ratio calculator, the ratio is equal to some specific value, x, where x = A/B. But what does it mean that two fractions are equivalent?
## What is an equivalent fraction?
The equivalent fraction definition tells us that any two fractions, A/B and C/D, are equivalent if they are equal to the same value. We can find whether two fractions are equivalent by checking one of these conditions:
• A = C and B = D;
• A × D = B × C;
• C = k × A and D = k × B, for any number k.
Converting two equivalent fractions to percentages will give us exactly the same number.
## How to find equivalent fractions?
For any fraction, there are infinitely many equivalent fractions. It's convenient to present such values as a ratio of two integer numbers. That's how we divide things in everyday life, e.g., cutting a pizza into pieces (and taking a few of them).
As an example, let's check how to find equivalent fractions of 4/6. We can follow these steps:
1. Make sure both the numerator and denominator are integers. If not, start multiplying both numbers by 10 until there are no more decimal digits. In our case, both 4 and 6 are already integers.
2. Let's find the fraction in its simplest form. To do so, evaluate the greatest common factor of the numerator and the denominator. Here, the GCF of 4 and 6 is 2, so 4/6 is an equivalent fraction of 2/3, and the latter is the simplest form of this ratio. Therefore, 2/3 is our base.
🔎 For a more complex computation, feel free to use Omni's GCF calculator.
1. Multiply the base by consecutive natural numbers:
2×2/2×3 = 4/6;
3×2/3×3 = 6/9;
4×2/4×3 = 8/12;
→ etc…
2. All of the values are equivalent fractions of 2/3 and are also equivalent to 4/6, the number we started with.
The procedure is straightforward, but what if you want to find 20 or 30 equivalent fractions? Well, it gets pretty time-consuming. However, if you use our equivalent fractions calculator, we can save you all of that hassle.
You can always convert any decimal to a fraction and then work out what are its equivalent fractions from there.
## How to use equivalent fractions calculator?
There are two different modes you can choose from. The first one helps you find as many fractions that are equivalent to your number as you want. Check the previous section for how these calculations work.
The second option helps you find if two fractions are equivalent. For instance, is 13/16 an equivalent fraction to 3/4? Let's check one of the conditions, e.g. multiply the numerator and denominator from the opposing fractions:
• 13 × 4 = 52;
• 16 × 3 = 48.
The outcomes are different, so the fractions 13/16 and 3/4 are NOT equivalent. What about 12/16 and 3/4?
• 12 × 4 = 48;
• 16 × 3 = 48.
This time, both numbers are the same, so 12/16 IS equivalent to 3/4. By using this equivalent fraction calculator, you can also see how to obtain one fraction from another!
## What are equivalent fractions of…
Here is a list of a few fractions and their equivalent fractions:
• Equivalent fractions of 1/1: 2/2, 3/3, 4/4, 5/5.
• Equivalent fractions of 1/2: 2/4, 3/6, 4/8, 5/10.
• Equivalent fractions of 1/3: 2/6, 3/9, 4/12, 5/15.
• Equivalent fractions of 2/3: 4/6, 6/9, 8/12, 10/15.
• Equivalent fractions of 1/4: 2/8, 3/12, 4/16, 5/20.
• Equivalent fractions of 3/4: 6/8, 9/12, 12/16, 15/20.
• Equivalent fractions of 1/5: 2/10, 3/15, 4/20, 5/25.
• Equivalent fractions of 2/5: 4/10, 6/15, 8/20, 10/25.
• Equivalent fractions of 3/5: 6/10, 9/15, 12/20, 15/25.
• Equivalent fractions of 4/5: 8/10, 12/15, 16/20, 20/25.
• Equivalent fractions of 1/6: 2/12, 3/18, 4/24, 5/30.
• Equivalent fractions of 5/6: 10/12, 15/18, 20/24, 25/30.
## FAQ
### How do I check if two fractions are equivalent?
To verify that two fractions are equivalent, simplify each of them by dividing its numerator and denominator by their greatest common factor. If at the end you get the same fraction for each, your initial fractions are equivalent. Otherwise, they are not equivalent.
### How many equivalent fractions are there?
For each fraction, there is an infinite number of fractions that are equivalent to it. To see that, assume our fraction is a/b. Then the fraction k×a/k×b is equivalent to it for k equal to any number. So we see that there are indeed an infinite number of fractions that are equivalent to a given fraction.
### How do I generate equivalent fractions?
To generate fractions equivalent to some given fraction:
1. Write down your given fraction a/b.
2. Write down the fractions 2a/2b, 3a/3b, etc.
3. In general, write down fractions of the form k×a/k×b, where k is any number. To keep it simple, use integers.
4. In this way, you can generate as many fractions equivalent to a/b as you wish!
### Are 6/8 and 9/12 equivalent fractions?
Yes, 6/8 and 9/12 are equivalent. To see it, we simplify each of them:
• First, 6/8 = 3/4, when we divided the numerator and denominator by 2.
• Second, 9/12 = 3/4, when we divided the numerator and denominator by 3.
Since we arrived at the same result 3/4 twice, our initial fractions are indeed equivalent.
Wojciech Sas, PhD
Choose what you want to see
Find equivalent fractions of...
a fraction in
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Numerator (n₁)
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# Types of Matrices with Examples
In this article, we will define different types of matrices along with examples.
Matrix: A matrix is a form of a rectangular array of m rows and n columns. These elements are multiplied during matrix multiplication keeping in mind some basic matrix multiplication properties.
Elements of a matrix: All the numbers inside a matrix are called elements of the matrix.
##### Types of Matrices are:
1. Row Matrix.
2. Column matrix.
3. Null matrix.
4. Square Matrix.
5. Diagonal matrix.
6. Scalar matrix.
7. Identity or Unit matrix.
8. Idempotent matrix.
9. Periodic matrix.
10. Triangular matrix.
11. Involuntary matrix.
12. Nilpotent matrix.
13. Sub-Matrix.
14. Principal sub-matrix.
15. Singular matrix.
16. Non-singular matrix.
1. Row Matrix. A matrix which has only one row is said to be a row matrix. For example;
$\begin{bmatrix} 1&2&3\\ \end{bmatrix}$
2. Column matrix. Column matrix is a matrix which has only one column. For example;
$\begin{bmatrix} 1\\ 2\\ 9\\ \end{bmatrix}$
Transpose of a row matrix is a column matrix while as the transpose of a column matrix is a row matrix.
3. Null matrix. Null matrix is that matrix whose all the elements are zero. For example;
$\begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&0\\ \end{bmatrix}$
4. Square Matrix. A square matrix is a matrix in which the number of rows is equal to the number of columns. For Example;
$\begin{bmatrix} 1&2&4\\ 5&1&9\\ 2&6&1\\ \end{bmatrix}$
5. Diagonal matrix. A diagonal matrix is a square matrix in which all the elements except the elements in leading diagonal are zero. For Example;
$\begin{bmatrix} 7&0&0\\ 0&5&0\\ 0&0&1\\ \end{bmatrix}$
6. Scalar matrix. A scalar matrix is a diagonal matrix in which all the diagonal elements are equal. For example;
$\begin{bmatrix} 7&0&0\\ 0&7&0\\ 0&0&7\\ \end{bmatrix}$
7. Identity or Unit matrix. An identity matrix is a diagonal matrix in which all the diagonal elements are unity. For example;
$\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{bmatrix}$
8. Idempotent matrix. A matrix [A] is said to be idempotent matrix if its square “[A]2” is equal to the matrix [A] itself. If [A] is a matrix of order (m×n) and [A]2 = [A], then the matrix [A] is called as an idempotent matrix. For example;
$If\:A=\begin{bmatrix} 1&0\\ 0&1\\ \end{bmatrix}$
$and\:[A]^2=\begin{bmatrix} 1&0\\ 0&1\\ \end{bmatrix}$
The matrix [A] is said to be an idempotent matrix.
9. Periodic matrix. A matrix [A] is said to be periodic if [A]x+1 = [A], where “x” is any positive integer.
10. Triangular matrix. A triangular matrix is of two types:
• Upper triangular matrix: A square matrix is said to be upper triangular matrix if it all the elements above the leading diagonal are zero. For example;
$\begin{bmatrix} 1&0&0\\ 2&7&0\\ 6&8&9\\ \end{bmatrix}$
• Lower triangular matrix: A square matrix is said to be a lower triangular matrix if its all the elements below the leading diagonal are zero. For example;
$\begin{bmatrix} 1&7&5\\ 0&7&8\\ 0&0&9\\ \end{bmatrix}$
11. Involuntary matrix. A matrix is said to be involuntary if its square is equal to unit or identity matrix.
12. Nilpotent matrix. A matrix is said to be nilpotent if [A]x = 0 or null matrix, where x is any positive integer.
13. Sub-Matrix. when we omit some rows and columns from a given (m×n) matrix, then the remaining matrix is called a sub-matrix. For example;
$If\:A=\begin{bmatrix} 1&5&2\\ 6&7&8\\ 2&9&4\\ \end{bmatrix}$
$and\:B=\begin{bmatrix} 1&5\\ 6&7\\ \end{bmatrix}$
The matrix B is said to be the sub-matrix of matrix A.
14. Principal sub-matrix. Principal sub-matrix is a square submatrix of a square matrix [A] whose diagonal elements are also the diagonal elements of the matrix [A].
15. Singular matrix. It is a matrix whose determinant is zero.
16. Non-Singular matrix. It is a matrix whose determinant is zero. If it is a square matrix then it is invertible.
These are all the different types of matrices. If you find any error in our article please comment below. You can also contact us in the contact section.
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Lesson Objectives
• Demonstrate an understanding of how to factor a polynomial
• Learn how to identify a rational expression
• Learn how to find the restricted value(s) for a rational expression
• Learn how to simplify a rational expression
## What is a Rational Expression?
Since pre-algebra, we have seen real numbers placed into certain categories. We know the whole numbers start with zero and increase in increments of one forever:
Whole Numbers: {0,1,2,3,...}
The integers allow us to work with negative values. An integer includes any whole number and their opposite.
Integers: {...,-2,-1,0,1,2,...}
When we want to work with numbers between two integers, we generally deal with fractions or decimals. When we work with fractions, we have the quotient of one integer over another integer. We can refer to a fraction or the quotient of one integer over another as a rational number. Some examples of a rational number: $$\frac{1}{2}, \frac{7}{3}, \frac{11}{2}$$ Similarly, the quotient of two polynomials with a non-zero denominator is referred to as a rational expression. We will see this in our textbook as: $$\frac{P}{Q}\hspace{.5em}Q ≠ 0$$ Where P and Q each represent a polynomial and Q is not equal to zero. Let's look at a few examples of a rational expression: $$\frac{x + 5}{2x + 10}, \frac{4x}{10x^2 - 3}$$
### Finding the Restricted Value for a Rational Expression
At this point, we should know that we can never divide by zero. When we encounter division by zero, the problem is said to be "undefined". When we work with rational expressions, our numerator can be any value including zero. Our denominator, however, will have restrictions based on not dividing by zero. We will restrict any value that replaces our variable and creates a denominator of zero. To find the restricted values of a rational expression:
• Set the denominator equal to zero
• Solve the equation
• The solution or solutions are the restricted values
Let's look at an example.
Example 1: Find the restricted values for each rational expression. $$\frac{x + 4}{x - 3}$$ Step 1) Set the denominator equal to zero:
x - 3 = 0
Step 2) Solve the equation:
x = 3
Step 3) The solution (3) is the restricted value
We can see in this case if we plug a 3 in for x, the denominator would become zero. This is where the rational expression is undefined.
Example 2: Find the restricted values for each rational expression. $$\frac{x^2 + 11x + 10}{x^2 - 6x - 7}$$ Step 1) Set the denominator equal to zero:
x2 - 6x - 7 = 0
Step 2) Solve the equation:
(x - 7)(x + 1) = 0
x = -1,7
Step 3) The solutions of -1 and 7 are the restricted values.
If we plug in a -1 or a 7 for x in the rational expression, the denominator would become zero. This is where the rational expression is undefined.
### Simplifying Rational Expressions
At this point, we should know how to simplify fractions. Essentially, we factor the numerator and denominator and cancel any common factors other than 1. let's look at an example: $$\frac{120}{165}$$ We can factor both numerator and denominator: $$\frac{2 \cdot 2 \cdot 2 \cdot 3 \cdot 5}{3 \cdot 5 \cdot 11}$$ We can cancel a common factor of 3 and a common factor of 5 between the numerator and denominator: $$\require{cancel}\frac{2 \cdot 2 \cdot 2 \cdot \cancel{3}\cdot \cancel{5}}{\cancel{3}\cdot \cancel{5}\cdot 11}=\frac{8}{11}$$ We can apply the same thought process when we simplify a rational expression. To simplify a rational expression, we factor the numerator and denominator and then we cancel any common factors. Let's take a look at a few examples.
Example 3: Simplify each rational expression. $$\frac{2x + 20}{x^2 + 12x + 20}$$ Step 1) Factor the polynomial in the numerator and denominator: $$\frac{2(x + 10)}{(x + 10)(x + 2)}$$ Step 2) Cancel common factors:
Here the quantity (x + 10) is a common factor. We can cancel this between the numerator and the denominator: $$\frac{2\cancel{(x + 10)}}{\cancel{(x + 10)}(x + 2)}$$ Our simplified result: $$\frac{2}{x + 2}$$ Example 4: Simplify each rational expression. $$\frac{5x^2 - 27x - 56}{10x - 70}$$ Step 1) Factor the polynomial in the numerator and the denominator: $$\frac{(5x + 8)(x - 7)}{10(x - 7)}$$ Step 2) Cancel common factors:
Here the quantity (x - 7) is a common factor. We can cancel this between the numerator and the denominator: $$\frac{(5x + 8)\cancel{(x - 7)}}{10\cancel{(x - 7)}}$$ Our simplified result: $$\frac{5x + 8}{10}$$
#### Skills Check:
Example #1
Find the restricted values. $$\frac{3x^{2}- 9x}{5x^{2}- 10x - 15}$$
A
$$x ≠ 0, 3$$
B
$$x ≠ -\frac{3}{2}, 4$$
C
$$x ≠ -\frac{8}{3}, 0$$
D
$$x ≠ -1, 3$$
E
$$x ≠ -1, \frac{1}{15}$$
Example #2
Simplify each rational expression. $$\frac{2x^{2}+ 4x - 70}{3x^{2}+ 21x}$$
A
$$\frac{2(x - 5)}{3x}$$
B
$$\frac{2x + 7}{4(x - 2)}$$
C
$$\frac{3x}{2(x - 5)}$$
D
$$\frac{3x(x - 1)}{x - 5}$$
E
$$\frac{2x(x - 3)}{5(x - 6)}$$
Example #3
Simplify each rational expression. $$\frac{3x^{3}- 33x^{2}+ 72x}{4x^{2}- 24x - 64}$$
A
$$\frac{3(-3x + 1)}{2x - 9}$$
B
$$\frac{3x(x - 3)}{4(x + 2)}$$
C
$$\frac{2x - 9}{3(-3x + 1)}$$
D
$$\frac{4(x + 2)}{3x(x - 3)}$$
E
$$\frac{12x}{x - 4}$$
|
# Finding limits algebraically - direct substitution
### Finding limits algebraically - direct substitution
Graphically finding the limit of a function is not always easy, as an alternative, we now shift our focus to finding the limit of a function algebraically. In this section, we will learn how to apply direct substitution to evaluate the limit of a function.
#### Lessons
• if: a function $f$ is continuous at a number $a$
then: direct substitution can be applied: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) =\lim_{x \to a} f(x)= f(a)$
• Polynomial functions are continuous everywhere, therefore “direct substitution” can ALWAYS be applied to evaluate limits at any number.
• 1.
No more finding limits “graphically”; Now, finding limits “algebraically”!
a)
What is Direct Substitution?
b)
When to apply Direct Substitution, and why Direct Substitution makes sense.
Exercise: $f(x)=\frac{1}{x-2}$
i) Find the following limits from the graph of the function.
$\lim_{x \to 3} f(x)$
$\lim_{x \to 2.5} f(x)$
$\lim_{x \to 0} f(x)$
ii) Evaluate:
$f(3)$
$f(2.5)$
$f(0)$
• 2.
Evaluate the limit:
a)
$\lim_{x \to 3} (5x^2-20x+17)$
b)
$\lim_{x \to -2} \frac{{{x^3} + 3{x^2} - 1}}{{5 - 4x}}$
c)
$\lim_{x \to 0}\left| x \right|$
d)
$\lim_{x \to \frac{\pi }{2}} \;\frac{{\sin x}}{{2 - \cos x}}$
• 3.
Evaluate the one-sided limit:
a)
$\lim_{x \to 3^-} (5x^2-20x+17)$
$\lim_{x \to 3^+} (5x^2-20x+17)$
b)
$\lim_{x \to {4^ - }} \sqrt {x - 4}$
$\lim_{x \to {4^ + }} \sqrt {x - 4}$
• 4.
• 5.
a)
$\lim_{x \to {-1^ - }} g(x)$
$\lim_{x \to {-1^ + }} g(x)$
$\lim_{x \to {-1}} g(x)$
b)
$\lim_{x \to {4^ - }} g(x)$
$\lim_{x \to {4^ + }} g(x)$
$\lim_{x \to {4}} g(x)$
|
# 5. The Ellipse
## Why study ellipses?
Orbiting satellites (including the earth and the moon) trace out elliptical paths.
Many buildings and bridges use the ellipse as a pleasing (and strong) shape.
One property of ellipses is that a sound (or any radiation) beginning in one focus of the ellipse will be reflected so it can be heard clearly at the other focus. You can see this working in the following animation.
## Ellipses with Horizontal Major Axis
The equation for an ellipse with a horizontal major axis is given by:
x^2/a^2+y^2/b^2=1
where a is the length from the center of the ellipse to the end the major axis, and b is the length from the center to the end of the minor axis.
The foci (plural of 'focus') of the ellipse (with horizontal major axis)
x^2/a^2+y^2/b^2=1
are at (-c,0) and (c,0), where c is given by:
c=sqrt(a^2-b^2
The vertices of an ellipse are at (-a, 0) and (a, 0).
Vertex (−a, 0)
Vertex (a, 0)
Focus (−c, 0)
Focus (c, 0)
Ellipse showing vertices and foci.
### Ellipse as a locus
The ellipse is defined as the locus of a point (x,y) which moves so that the sum of its distances from two fixed points (called foci, or focuses) is constant.
We can produce an ellipse by pinning the ends of a piece of string and keeping a pencil tightly within the boundary of the string, as follows.
We pin the ends of the string to the foci and begin to draw, holding the string tight:
Our complete ellipse is formed:
Continues below
### Example 1 - Ellipse with Horizontal Major Axis
Find the coordinates of the vertices and foci of
x^2/100+y^2/64=1
Sketch the curve.
## Ellipse with Vertical Major Axis
A vertical major axis means the ellipse will have greater height than width.
If the major axis is vertical, then the formula becomes:
x^2/b^2+y^2/a^2=1
We always choose our a and b such that a > b. The major axis is always associated with a.
### Example 2 - Ellipse with Vertical Major Axis
Find the coordinates of the vertices and foci of
25x^2+y^2=25
Sketch the curve.
### Example 3
Find the equation of the ellipse which has a minor axis of length 8 and a vertex at (0,-5).
## Eccentricity
The eccentricity of an ellipse is a measure of how elongated it is. If the eccentricity approaches value 0, the curve becomes more circular, and if it approaches 1, the ellipse becomes more elongated.
We can calculate the eccentricity using the formula:
text(eccentricity)=c/a
### Real Example
The Sun
The Earth revolves around the sun in an elliptical orbit, where the sun is at one of the foci. (This was discovered by Keppler in 1610).
The semi-major axis is approximately 149,597,871 km long and it is known that the ratio c/a is equal to 1/60.
(i) What are the greatest and least distances the Earth is from the sun?
(ii) How far from the sun is the other focus?
(The "semi-major axis" means half of the major axis length. In our example, it is (close to) the "average" distance of the sun from the earth, and is also known as one A.U., or "astronomical unit".)
## Ellipses with Centre Other Than the Origin
Like the other conics, we can move the ellipse so that its axes are not on the x-axis and y-axis. We do this for convenience when solving certain problems.
For the horizontal major axis case, if we move the intersection of the major and minor axes to the point (h, k), we have:
((x-h)^2)/a^2+((y-k)^2)/b^2=1
The ellipse is as follows:
### Example 4
Sketch the ellipse with equation
((x-1)^2)/25+((y+2)^2)/9=1
### Conic section: Ellipse
How can we obtain an ellipse from slicing a cone?
We start with a double cone (2 right circular cones placed apex to apex):
When we slice one of the cones at an angle to the sides of the cone, we get an ellipse, as seen in the view from the top (at right).
top
### Online Algebra Solver
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# What are the best criteria for assessing a research question?
## What are the best criteria for assessing a research question?
In this section, we consider two criteria for evaluating research questions: the interestingness of the question and the feasibility of answering it….Is it interesting?Doubt. If the answer is obvious, the question is not interesting. Filling a gap. Importance.
## What are the steps to evaluate an expression?
To evaluate an algebraic expression, you have to substitute a number for each variable and perform the arithmetic operations. In the example above, the variable x is equal to 6 since 6 + 6 = 12. If we know the value of our variables, we can replace the variables with their values and then evaluate the expression.
## How do you simplify this expression?
To simplify any algebraic expression, the following are the basic rules and steps:Remove any grouping symbol such as brackets and parentheses by multiplying factors.Use the exponent rule to remove grouping if the terms are containing exponents.Combine the like terms by addition or subtraction.Combine the constants.
## What are the steps in evaluating a function?
To evaluate a function, substitute the input (the given number or expression) for the function’s variable (place holder, x). Replace the x with the number or expression. 1. Given the function f (x) = 3x – 5, find f (4).
## What is function evaluation?
To evaluate a function is to: Replace (substitute) its variable with a given number or expression.
## What are the 3 most important things you learned in evaluating functions?
Learning ObjectivesGiven a function described by an equation, find function values (outputs) for numerical inputs.Given a function described by an equation, find function values (outputs) for variable inputs.
## How do you use a graph to evaluate a function?
How To: Given a graph, use the vertical line test to determine if the graph represents a function.Inspect the graph to see if any vertical line drawn would intersect the curve more than once.If there is any such line, the graph does not represent a function.
## What is the rule of the function?
A function rule describes how to convert an input value (x) into an output value (y) for a given function. An example of a function rule is f(x) = x^2 + 3.
## What is a function rule for a table?
A function table has values of input and output and a function rule. In the function rule, if we plug in different values for the input, we get corresponding values of output. There is always a pattern in the way input values x and the output values y are related which is given by the function rule.
## How do you write a time function?
You write functions with the function name followed by the dependent variable, such as f(x), g(x) or even h(t) if the function is dependent upon time. You read the function f(x) as “f of x” and h(t) as “h of t”. Functions do not have to be linear.
## How do you write a function rule?
3:01Suggested clip 107 secondsWrite a Function Rule: An Application (Algebra I) – YouTubeYouTubeStart of suggested clipEnd of suggested clip
## What is the F X formula?
The slope of a linear function is calculated by rearranging the equation to its general form, f(x) = mx + c; where m is the slope. The vertex of a quadratic function is calculated by rearranging the equation to its general form, f(x) = a(x – h)2 + k; where (h, k) is the vertex.
## How do you write a rule for translation?
Mapping Rule A mapping rule has the following form (x,y) → (x−7,y+5) and tells you that the x and y coordinates are translated to x−7 and y+5. Translation A translation is an example of a transformation that moves each point of a shape the same distance and in the same direction.
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Properties of Convergent Series
# Properties of Convergent Series
We will now look at some very important properties of convergent series, many of which follow directly from the standard limit laws for sequences.
Theorem 1: Let $\sum_{n=1}^{\infty} a_n$ be convergent to the sum $A$ and let $\sum_{n=1}^{\infty} b_n$ be convergent to the sum $B$. Then the series $\sum_{a_n + b_n}$ is convergent to the sum $A + B$.
• Proof: Suppose that $\sum_{n=1}^{\infty} a_n = A$ and $\sum_{n=1}^{\infty} b_n = B$. Since both of these series are convergent, it follows that their sequences of partial sums are convergent, that is $\lim_{n\to \infty} s_n = A$ and that $\lim_{n \to \infty} s_n' = B$. By limit laws since both of these sequences are convergent it follows that $\lim_{n \to \infty} \left ( a_n + b_n \right ) = A + B$ which implies $\sum_{n=1}^{\infty} a_n + b_n = A + B$. $\blacksquare$
Theorem 2: Let $\sum_{n=1}^{\infty} a_n$ be convergent to the sum $A$ and let $\sum_{n=1}^{\infty} b_n$ be convergent to the sum $B$. Then the series $\sum_{n=1}^{\infty} a_n - b_n$ is convergent to the difference $A - B$.
• Proof: Suppose that $\sum_{n=1}^{\infty} a_n = A$ and $\sum_{n=1}^{\infty} b_n = B$. Since both of these series are convergent, it follows that their sequences of partial sums are convergent, that is $\lim_{n\to \infty} s_n = A$ and that $\lim_{n \to \infty} s_n' = B$. By limit laws since both of these sequences are convergent it follows that $\lim_{n \to \infty} \left ( a_n - b_n \right ) = A - B$ which implies $\sum_{n=1}^{\infty} a_n - b_n = A - B$. $\blacksquare$
Theorem 3: Let $\sum_{n=1}^{\infty} a_n$ be a convergent series to the sum $A$ and let $k$ be a constant. Then $\sum_{n=1}^{\infty} ka_n$ is also a convergent series to the sum $kA$.
• Proof: Suppose that $\sum_{n=1}^{\infty} a_n = A$ and let $\{ s_n \}$ be the sequence of partial sums for this series. Then $\lim_{n \to \infty} s_n = A$. Since this sequence is convergent, it follows by the limit laws that $\lim_{n \to \infty} ks_n = kA$ which implies that that $\sum_{n=1}^{\infty} ka_n = kA$. $\blacksquare$
Theorem 4: Let $\sum_{n=1}^{\infty} a_n$ be a convergent convergent series to $A$ and let $\sum_{n=1}^{\infty} b_n$ be a convergent series to $B$. If $a_n ≤ b_n$ for all $n \in \mathbb{N}$, then $A ≤ B$.
• Proof: Suppose that $\sum_{n=1}^{\infty} a_n = A$ and $\sum_{n=1}^{\infty} b_n = B$ such that $a_n ≤ b_n$ for all $n \in \mathbb{N}$. We know that $\lim_{n \to \infty} s_n = A$ and $\lim_{n \to \infty} s_n' = B$. Since $a_n ≤ b_n$ for all $n \in \mathbb{N}$ we deduce that $s_n = a_1 + a_2 + ... + a_n ≤ b_1 + b_2 + ... + b_n = s_n'$ for all $n \in \mathbb{N}$, and so $A ≤ B$. $\blacksquare$
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# How do you find a standard form equation for the line with (-6, 0) and (2, -9)?
Aug 24, 2016
$y = - \frac{9}{8} x - \frac{27}{4}$
#### Explanation:
The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where m represents the slope and b, the y-intercept.
We have to find the values of m and b.
To calculate m, use the $\textcolor{b l u e}{\text{gradient formula}}$
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ are 2 coordinate points}$
here the 2 points are (-6 ,0) and (2 ,-9)
let $\left({x}_{1} , {y}_{1}\right) = \left(- 6 , 0\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(2 , - 9\right)$
$\Rightarrow m = \frac{- 9 - 0}{2 + 6} = \frac{- 9}{8} = - \frac{9}{8}$
We can now write a partial equation as $y = - \frac{9}{8} x + b$
To calculate b, use either of the 2 given points that lie on the line.
Using (2 ,-9)That is x= 2 and y = -9 , substitute in partial equation
$\Rightarrow \left(- \frac{9}{8} \times 2\right) + b = - 9 \Rightarrow - \frac{9}{4} + b = - 9 \Rightarrow b = - \frac{27}{4}$
$\Rightarrow y = - \frac{9}{8} x - \frac{27}{4} \text{ is the equation of the line}$
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# ISEE Middle Level Math : How to find length of a line
## Example Questions
### Example Question #11 : Lines
The coordinates of on the rectangle are and . Find the length of the diagonal.
Explanation:
A rectangle has two diagonals with the same length. Therefore, use the distance formula to calculate the distance.
### Example Question #12 : Lines
If a circle has a radius of , what would be the length of the longest line drawn within that circle?
Explanation:
The diameter of the circle would be the longest line that can be drawn within that circle.
Because radius is half of the diameter, the diameter is calculated by multiplying the radius of the circle by two.
If the radius is , then the diameter is
### Example Question #13 : Lines
A ladder is leaning on a wall. It is ft long. The bottom of the ladder is from the base. How far up the wall is the top of the ladder?
Explanation:
The Pythagorean Theorem states that:
With and representing the measurement of the legs and representing the hypotenuse.
### Example Question #14 : Lines
What is the perimeter of a right triangle if the hypotenuse is and the measurement of one of its legs is
Explanation:
First, use the Pythagorean Theorem to get the measurement of the other leg.
To get the perimeter of the triangle, add the measurements of each of the three sides.
### Example Question #11 : Lines
A line has endpoints and What is the distance of the line?
Explanation:
Use the distance formula
### Example Question #16 : Lines
Line has a length of It is bisected at point , and the resulting segment is bisected again at point What is the length of line segment
Explanation:
First, write each portion of the statement in mathematical terms.
Since AB=80 we will substitute that into the equation.
Now that we know AC we can calculate AD as follows.
### Example Question #11 : How To Find Length Of A Line
The point lies on a circle. What is the approximate length of the radius of the circle if the center is
Explanation:
Because the radius is the distance from center to any point on a circle, the distance formula is used to find the measurement of the radius.
### Example Question #18 : Lines
If the diameter of a circle is , then what is of the circle's radius?
Explanation:
If the diameter is , then the radius is
Therefore
of is,
.
### Example Question #19 : Lines
Find the slope of a line that passes through and .
The slope is undefined.
Explanation:
Use the slope formula to solve:
Given the following points.
The slope can be calculated as follows.
Because the coordinates were the same for points A and B, this would form a horizontal line. The slope of any horizontal line is
### Example Question #20 : Lines
Find the length of a line from the point to the point .
Explanation:
Find the length of a line from the point to the point .
To find this distance, we need to use distance formula (which is really similar to Pythagorean Theorem)
Distance formula is as follows
Where our x's and y's come from our ordered pairs.
So, let's plug and chug
Simplify
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+0
# trigonometry
0
168
1
if a circle has a radius of 8 centimeters, what is the exact length of the arc intercepted by a central angle of 90 degrees.
May 15, 2018
#1
+2339
+1
An accompanying diagram is an asset geometrists use to condense large amounts of information into something more manageable and visual. I have created a reference that suits this situation perfectly.
Notice the features of this diagram and how it relates to the actual problem:
• The circle has a radius of 8 centimeters
• There is a central angle of 90°
• The intercepted arc, $$\widehat{BC}$$, is present
Now that the important features are obvious and present, we can actually move on to the solving bit.
One elementary fact about a circle is that its entire circumference can be represented with this formula:
$$\text{Circumference}=2\pi r$$
However, there is a slight issue with this formula for this particular context: The problem only asks for a certain portion of the circumference--not the whole thing. How can we remedy this issue? Well, let's just pretend for a moment that we actually knew that certain percentage, say 20%.
If we knew that the problem only wanted 20% of the circumference, then we would just multiply the circumference by that arbitrary percentage, 20 in this case. The formula would be $$\text{Circumference}=0.2*2\pi r= 0.4\pi r$$
Ok, we are making some great progress on this problem! We still need a way of determining that percentage, though. I will introduce another well-known fact about circles: They have 360 degrees in total. We also know the central angle measure, which is 90°. This means that 90 out of the 360 degrees encompass the subtended (also referred to as intercepted) portion. This means that the "certain percentage" is determined by the central angle measure. We can generalize this and create a formula for this!
$$\text{Arc Length}=2\pi r*\frac{\theta}{360}$$ This is what we have determined up to now. $$\theta$$ is generally the Greek letter used to denote a variable in a situation similar in nature to how it is in this problem. There is only one bit of simplification possible here. $$r=8;\theta=90\\ \text{Arc Length}=\frac{\pi r\theta}{180}$$ Substitute in the values and solve. $$\widehat{BC}=\frac{8*90*\pi}{180}=4\pi\text{cm}$$ You are finished!
May 15, 2018
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