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# Find the value of $k$, for which $f(x)=\begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, \text{if} \ -1\le x < 0 \\ \frac{2x+1}{x-1}, \text{if} \ 0\le x < 1 \end{cases}$ is continuous at $x=0$
## Example:
Find the value of $k$, for which $f(x)=\begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, \text{if} \ -1\le x<0 \\ \frac{2x+1}{x-1}, \text{if} \ 0\le x<1 \end{cases}$ is continuous at $x=0$.
### Solution:
Given that,
$f(x)=\begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, \text{if} \ -1\le x<0 \\ \frac{2x+1}{x-1}, \text{if} \ 0\le x<1 \end{cases}$
Since, f(x) is continuous at $x=0$,
$\displaystyle\lim_{x \to 0^-} f(x)=f(0)=\displaystyle\lim_{x \to 0^+} f(x)$ ............ (1)
Here, $f(0)=\frac{2(0)+1}{0-1}$
$=\frac{1}{-1}$
$f(0)=-1$ ..................... (2)
Now, $\displaystyle\lim_{x \to 0^-} f(x)=\displaystyle\lim_{x \to 0}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$
$=\displaystyle\lim_{x \to 0}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\times \frac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}}$
$=\displaystyle\lim_{x \to 0}\frac{\{1+kx\}-\{1-kx\}}{x(\sqrt{1+kx}+\sqrt{1-kx})}$
$=\displaystyle\lim_{x \to 0}\frac{2kx}{x(\sqrt{1+kx}+\sqrt{1-kx})}$
$=\displaystyle\lim_{x \to 0}\frac{2k}{\sqrt{1+kx}+\sqrt{1-kx}}$
$=\frac{2k}{\sqrt{1+0}+\sqrt{1-0}}$
$=\frac{2k}{2}$
$\therefore \displaystyle\lim_{x \to 0^-} f(x)=k$ ..................... (3)
From (1), (2) and (3), we have
$k=-1$
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# Inscribed Similar Triangles
## Division of a right triangle into two similar triangles.
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Inscribed Similar Triangles
### Inscribed Similar Triangles Theorem
Remember that if two objects are similar, their corresponding angles are congruent and their sides are proportional in length. The altitude of a right triangle creates similar triangles.
Inscribed Similar Triangles Theorem: If an altitude is drawn from the right angle of any right triangle, then the two triangles formed are similar to the original triangle and all three triangles are similar to each other.
In \begin{align*}\triangle ADB, m \angle A = 90^{\circ}\end{align*} and \begin{align*}\overline{AC} \bot \overline{DB}\end{align*}:
So, \begin{align*}\triangle ADB \sim \triangle CDA \sim \triangle CAB\end{align*}:
This means that all of the corresponding sides are proportional. You can use this fact to find missing lengths in right triangles.
What if you drew a line from the right angle of a right triangle perpendicular to the side that is opposite that angle? How could you determine the length of that line?
### Examples
#### Example 1
Find the value of \begin{align*}x\end{align*}.
Set up a proportion.
\begin{align*}\frac{\text{shorter leg in} \ \triangle SVT}{\text{shorter leg in} \ \triangle RST} &= \frac{\text{hypotenuse in} \ \triangle SVT}{\text{hypotenuse in} \ \triangle RST}\\ \frac{4}{x} &= \frac{x}{20}\\ x^2 &= 80\\ x &= \sqrt{80} = 4 \sqrt{5}\end{align*}
#### Example 2
Now find the value of \begin{align*}y\end{align*} in \begin{align*}\triangle RST\end{align*} above.
Use the Pythagorean Theorem.
\begin{align*}y^2 + \left( 4 \sqrt{5} \right )^2 &= 20^2\\ y^2 + 80 &= 400\\ y^2 &= 320\\ y &= \sqrt{320} = 8 \sqrt{5}\end{align*}
#### Example 3
Find the value of \begin{align*}x\end{align*}.
Separate the triangles to find the corresponding sides.
Set up a proportion.
\begin{align*}\frac{\text{shorter leg in} \ \triangle EDG}{\text{shorter leg in} \ \triangle DFG} &= \frac{\text{hypotenuse in} \ \triangle EDG}{\text{hypotenuse in} \ \triangle DFG}\\ \frac{6}{x} &= \frac{10}{8}\\ 48 &= 10x\\ 4.8 &= x\end{align*}
#### Example 4
Find the value of \begin{align*}x\end{align*}.
Set up a proportion.
\begin{align*}\frac{shorter \ leg \ of \ smallest \ \triangle}{shorter \ leg \ of \ middle \ \triangle} &= \frac{longer \ leg \ of \ smallest \ \triangle}{longer \ leg \ of \ middle \ \triangle}\\ \frac{9}{x} &= \frac{x}{27}\\ x^2 &= 243\\ x &= \sqrt{243} = 9 \sqrt{3}\end{align*}
#### Example 5
Find the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
Separate the triangles. Write a proportion for \begin{align*}x\end{align*}.
\begin{align*}\frac{20}{x} &= \frac{x}{35}\\ x^2 &= 20 \cdot 35\\ x &= \sqrt{20 \cdot 35}\\ x &= 10 \sqrt{7}\end{align*}
Set up a proportion for \begin{align*}y\end{align*}. Or, now that you know the value of \begin{align*}x\end{align*} you can use the Pythagorean Theorem to solve for \begin{align*}y\end{align*}. Use the method you feel most comfortable with.
\begin{align*}\frac{15}{y} &= \frac{y}{35} && (10 \sqrt{7})^2 + y^2 = 35^2\\ y^2 &= 15 \cdot 35 && \qquad 700 + y^2 =1225\\ y &= \sqrt{15 \cdot 35} && \qquad \qquad \quad \ y = \sqrt{525} = 5 \sqrt{21}\\ y &= 5 \sqrt{21}\end{align*}
### Review
Fill in the blanks.
1. \begin{align*}\triangle BAD \sim \triangle \underline{\;\;\;\;\;\;\;\;\;} \sim \triangle \underline{\;\;\;\;\;\;\;\;\;}\end{align*}
2. \begin{align*}\frac{BC}{?} = \frac{?}{CD}\end{align*}
3. \begin{align*}\frac{BC}{AB} = \frac{AB}{?}\end{align*}
4. \begin{align*}\frac{?}{AD} = \frac{AD}{BD}\end{align*}
Write the similarity statement for the right triangles in each diagram.
Use the diagram to answer questions 7-10.
1. Write the similarity statement for the three triangles in the diagram.
2. If \begin{align*}JM = 12\end{align*} and \begin{align*}ML = 9\end{align*}, find \begin{align*}KM\end{align*}.
3. Find \begin{align*}JK\end{align*}.
4. Find \begin{align*}KL\end{align*}.
Find the length of the missing variable(s). Simplify all radicals.
1. Fill in the blanks of the proof for the Inscribed Similar Triangles Theorem.
Given: \begin{align*}\triangle ABD\end{align*} with \begin{align*}\overline{AC} \perp \overline {DB}\end{align*} and \begin{align*}\angle DAB\end{align*} is a right angle.
Prove: \begin{align*}\triangle ABD \sim \triangle CBA \sim \triangle CAD\end{align*}
Statement Reason
1. 1. Given
2. \begin{align*}\angle DCA\end{align*} and \begin{align*}\angle ACB\end{align*} are right angles 2.
3. \begin{align*}\angle DAB \cong \angle DCA \cong \angle ACB\end{align*} 3.
4. 4. Reflexive PoC
5. 5. AA Similarity Postulate
6. \begin{align*}B \cong \angle B\end{align*} 6.
7. \begin{align*}\triangle CBA \cong \triangle ABD\end{align*} 7.
8. \begin{align*}\triangle CAD \cong \triangle CBA\end{align*} 8.
To see the Review answers, open this PDF file and look for section 8.4.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Inscribed Similar Triangles Theorem The Inscribed Similar Triangles Theorem states that if an altitude is drawn from the right angle of any right triangle, then the two triangles formed are similar to the original triangle and all three triangles are similar to each other.
Perpendicular Perpendicular lines are lines that intersect at a $90^{\circ}$ angle. The product of the slopes of two perpendicular lines is -1.
Proportion A proportion is an equation that shows two equivalent ratios.
Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.
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# Problems from Compound rule of three
In July, a family of $$4$$ people spends $$10$$ days in a hotel on the beach for $$2.500$$ €. How much it will be for a group of $$6$$ persons to spend the second two weeks of July in the same hotel?
See development and solution
### Development:
The first step to solve the exercise is to outline the compound rule of three and to analyze the relation of the unknown with the other two parameters:
$$\begin{eqnarray} & d & & d & \\\\ 4\ \mbox{people} & \rightarrow & 10 \ \mbox{days} & \rightarrow & 2500 \ \mbox{€} \\\\ 6\ \mbox{people} & \rightarrow & 15 \ \mbox{days} & \rightarrow & x \ \mbox{€} \\\\ & d & & d & \end{eqnarray}$$
So that the relation between the amount of euros and the number of persons is direct since: the more people there are, the more money it will cost. In the same way, the relation between the days and the price is also direct since the more days of accommodation the more expensive it will be.
As soon as the analysis is done, the transformation in fractions is immediate:
$$\dfrac{4}{6}\cdot\dfrac{10}{15}=\dfrac{2500}{x} \Rightarrow \dfrac{40}{90}=\dfrac{2500}{x} \Rightarrow 40x=90\cdot2500 \Rightarrow$$ $$40x=225.000 \Rightarrow x=\dfrac{225.000}{40}=5625$$€
### Solution:
$$5625$$€
Hide solution and development
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Conversion of Radians into Degrees and Vice Versa
We know that the circumference of a circle of radius $r$ is $2\pi r = \left( l \right)$ and the angle formed by one complete revolution is $\theta$ radian; therefore,
Thus we have the relationship
Further
Example:
Convert the following angles into degrees:
(i) $\frac{{2\pi }}{3}$ radians
(ii) $3 radians$
Solution:
(i)
(ii)
Example:
Convert $54^\circ \,\,45'$ into radians.
Solution:
Most calculators automatically would convert degrees into radians and radians into degrees.
Example:
An arc subtends an angle of $70^\circ$ at the center of a circle and its length is $132mm$. Find the radius of the circle.
Solution:
$70^\circ \,\,\, = 70 \times \frac{\pi }{{180}} = \frac{{70}}{{180}}\left( {\frac{{22}}{7}} \right) = \frac{{11}}{9}\,\,\,\,\,\,\,\,\,\,\left( {\because \pi = \frac{{22}}{7}} \right)$
$\therefore \theta = \frac{{11}}{9}\,radian$ and $l = 132m.m$
$\therefore \theta = \frac{1}{r} \Rightarrow r = \frac{l}{\theta } = 132 \times \frac{9}{{11}} = 108m.m$
Example:
Find the length of the equatorial arc subtending an angle of $1^\circ$ at the center of the Earth, taking the radius of the Earth as 5400km.
Solution:
$1^\circ = \frac{\pi }{{180}} = \frac{{3.1416}}{{180}}\,radians$
$\therefore \theta = \frac{{3.1416}}{{180}}$ and $r = 6400\,km$
Now
$\theta = \frac{l}{r}$
$\Rightarrow l = r\theta = 6400 \times \frac{{31416}}{{1800000}} = 111.7\,km$
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# How do you simplify cosx/(1+sinx) + (1+sinx)/cosx?
May 23, 2018
$\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 2 \sec x$
#### Explanation:
$\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = \frac{{\cos}^{2} x + {\left(1 + \sin x\right)}^{2}}{\left(1 + \sin x\right) \cos x}$
$\textcolor{w h i t e}{\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}} = \frac{{\cos}^{2} x + {\sin}^{2} x + 1 + 2 \sin x}{\left(1 + \sin x\right) \cos x}$
$\textcolor{w h i t e}{\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}} = \frac{2 + 2 \sin x}{\left(1 + \sin x\right) \cos x}$
$\textcolor{w h i t e}{\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}} = \frac{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(1 + \sin x\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(1 + \sin x\right)}}} \cos x}$
$\textcolor{w h i t e}{\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}} = \frac{2}{\cos} x$
$\textcolor{w h i t e}{\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}} = 2 \sec x$
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# Normal Approximation to solve a Binomial Problem
Binomial Theorem > Normal Approximation to solve a binomial problem
## Normal Approximation: Overview
A normal distribution curve, sometimes called a bell curve.
When n * p and n * q are greater than 5, you can use the normal approximation to solve a binomial distribution problem. This article shows you how to solve those types of problem using the continuity correction factor.
## Normal Approximation: Example#2
Sixty two percent of 12th graders attend school in a particular urban school district. If a sample of 500 12th grade children are selected, find the probability that at least 290 are actually enrolled in school.
Step 1: Determine p,q, and n:
p is defined in the question as 62%, or 0.62
To find q, subtract p from 1: 1 – 0.62 = 0.38
n is defined in the question as 500
Step 2: Determine if you can use the normal distribution:
n * p = 310 and n * q = 190. These are both larger than 5.
Step 3: Find the mean, μ by multiplying n and p:
n * p = 310
Step 4: Multiply step 3 by q :
310 * 0.38 = 117.8.
Step 5: Take the square root of step 4 to get the standard deviation, σ:
sqrt(117.8)=10.85
Note: The formula for the standard deviation for a binomial is &sqrt;(n*p*q).
Step 6: Write the problem using correct notation:
P(X≥290)
Step 7: Rewrite the problem using the continuity correction factor:
P (X ≥ 290-0.5)= P (X ≥ 289.5)
Step 8: Draw a diagram with the mean in the center. Shade the area that corresponds to the probability you are looking for. We’re looking for X ≥ 289.5, so…
Step 9: Find the z-score.
You can find this by subtracting the mean (μ) from the probability you found in step 7, then dividing by the standard deviation (σ): (289.5 – 310) / 10.85 = -1.89
Step 10: Look up the z-value in the z-table:
The area for -1.819 is 0.4706.
0.4706+ 0.5 = 0.9706.
That’s it! The probability is .9706, or 97.06%.
Check out our YouTube channel for hundreds more statistics help videos!
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Normal Approximation to solve a Binomial Problem was last modified: October 25th, 2017 by
# 24 thoughts on “Normal Approximation to solve a Binomial Problem”
1. jamie
oh my goodness, i feel like kissing you right now! thankyou thankyou thankyou! i will be buying the book for my end semester exam!!!
God bless you!
2. rc
Professor, there is something I don’t get in this example – how can p*q be >5? since they are both probabilities, this means they are maximum 1 each (and not in the same time :), so p*q can’t be >1 ever (in fact, I guess the max p*q =.25)
3. Zeshan
If n*p is less than 30 or in this case <5, then instead of normal distribution, you would be using t-distribution as the central limit theorem does not apply. t-distribution is similar but you would use the t-table chart and find the t-value instead of the z-value and work it the same way after
4. hunain turab
prof plz tell me the procedure to solve this problem by using normal distribution
if p(x>a)=0.4995 then find the value of a
5. Eddie
Wouldn’t it just be easier to find the z-score of -1.89
And then subtract it. So it would be 1-.0294
The last two steps confuse me greatly.
6. Andale
Eddie,
There’s more than one way to get the answer in stats. This is the way I taught my students, but do whatever works for you :)
Stephanie
7. Meliza
Well I do have a question how would we find the t-table and will the answer be correct since my z table stops at 3.49
8. Prasant Kumar
z- score of -1.819 is 0.0294 and why did you add 0.5 in that. we can simply subtract this z-value from 1 and we will get the answer as z-score is always calculate CDF (area to the left of the curve). Since we are interested in knowing area to the right of the curve.
9. Andale
The z-score isn’t always the whole area to the left of the curve. Sometimes it’s the area between the mean and the score (as in this case). You have to add .5 to account for the area to the left of the mean.
10. bilal
hi prof.plz guide me when should we use correction factor and when not we need to use this?how can v differentiate?…….
11. Andale
You use it when you have a discrete function and you want to use a continuous function (usually a binomial to a normal or, more rarely, a Poisson with a large λ to a normal). These are the only two cases I can think of where you would want to use it…
12. Lim Souchhy
Excuse me, I want to ask you something.
a. P=77.29% by binomial
b. P=76.94% by normal approximation
The question is how do I comment on answer (b) to exact percentage found in answer (a)?
13. Andale
Lim,
I’m not sure what you mean by “The question is how do I comment on answer (b) to exact percentage found in answer (a)?”. Could you rephrase your question?
Thanks.
14. Lim Souchhy
I mean “how to comment on (b) to exact percentage found in answer (a)”?
This exercise requires to calculate the percentage of probability by two ways. Question (a) calculate by binomial, Question (b) calculate by normal approximation plus compare and explain and comment why it is similar or equal to (a). But I can only calculate and cannot comment on it.
15. Andale
OK, if I have this right, you have found (a), your binomial, and (b) the normal approximation. What are your results? I’m assuming they are different: the question is probably asking you to say what the difference is and why (i.e. because you’re using the two different methods).
16. Salmi Tchitumba
OMG….. at first it was kind of difficult…… But i got it now.. thanks a lot
17. Eric Ndofor
Dear Professor,
Both examples are very clear up to the point of z tables.
I noticed you used the A-3 Table for both examples on this page. What is not clear to me is what criteria you use to determine what column on the Z table can be used. It is not clear to me why you used Column D for question 1 and column A for example 2. Thanks for explaining why.
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We think you are located in South Africa. Is this correct?
# Evaluating Internal Resistance In Circuits
Exercise 10.2
Describe what is meant by the internal resistance of a real battery.
Real batteries are made from materials which have resistance. This means that real batteries are not just sources of potential difference (voltage), but they also possess internal resistance.
So internal resistance is a measure of the resistance of the material that the battery is made of.
Explain why there is a difference between the emf and terminal voltage of a battery if the load (external resistance in the circuit) is comparable in size to the battery's internal resistance
The emf of a battery is essentially constant because it only depends on the chemical reaction (that converts chemical energy into electrical energy) going on inside the battery. Therefore, we can see that the voltage across the terminals of the battery is dependent on the current drawn by the load. The higher the current, the lower the voltage across the terminals, because the emf is constant. By the same reasoning, the voltage only equals the emf when the current is very small.
What is the internal resistance of a battery if its emf is $$\text{6}$$ $$\text{V}$$ and the potential difference across its terminals is $$\text{5,8}$$ $$\text{V}$$ when a current of $$\text{0,5}$$ $$\text{A}$$ flows in the circuit when it is connected across a load?
The voltage drop across the internal resistance is the difference between the emf and the terminal voltage when the battery is connected across a load. The internal resistance we treat as an ohmic resistor in series with a perfect battery. We know the voltage with and without a load and we know the current so we can use Ohm's law to determine the internal resistance:
\begin{align*} V_r&= Ir \\ (6)-(5,8)&= (0,5)r \\ r &= \frac{0,2}{0,5} \\ & = \text{0,4}\text{ Ω} \end{align*} $$\text{0,4}$$ $$\text{Ω}$$
A $$\text{12,0}$$ $$\text{V}$$ battery has an internal resistance of $$\text{7,0}$$ $$\text{Ω}$$.
1. What is the maximum current this battery could supply?
2. What is the potential difference across its terminals when it is supplying a current of 150.0 mA?
3. Draw a sketch graph to show how the terminal potential difference varies with the current supplied if the internal resistance remains constant. How could the internal resistance be obtained from the graph?
What is the maximum current this battery could supply?
\begin{align*} I_{\text{max}} & =\frac{\mathcal{E}}{r} \\ & =\frac{\text{12,0}}{\text{7,0}} \\ & =\text{1,71428} \\ & =\text{1,71}\text{ A} \end{align*}
What is the potential difference across its terminals when it is supplying a current of 150.0 mA?
\begin{align*} V & =\mathcal{E} - Ir \\ & =(\text{12,0})- (\text{1,50} \times \text{10}^{-\text{1}})(\text{7,0}) \\ & =\text{10,95} \\ & =\text{10,95}\text{ V} \end{align*}
Draw a sketch graph to show how the terminal potential difference varies with the current supplied if the internal resistance remains constant. How could the internal resistance be obtained from the graph?
The slope of the graph is the internal resistance.
In a hearing aid a battery supplies a current of 25.0 mA through a resistance of 400 $$\Omega$$. When the volume is increased, the resistance is changed to 100 $$\Omega$$ and the current rises to 60 mA. What is the emf and internal resistance of the cell?
We have two unknowns but information for two different scenarios so we can solve simultaneously.
In the first case:
\begin{align*} V & =\mathcal{E} - Ir \\ (\text{2,50} \times \text{10}^{-\text{2}})(\text{400}) = & =\mathcal{E} - (\text{2,50} \times \text{10}^{-\text{2}})r \\ \mathcal{E} & = (\text{2,50} \times \text{10}^{-\text{2}})(\text{400}) + (\text{2,50} \times \text{10}^{-\text{2}})r \end{align*}
In the second case we will get:
\begin{align*} V & =\mathcal{E} - Ir \\ (\text{6,0} \times \text{10}^{-\text{2}})(\text{100}) = & =\mathcal{E} - (\text{6,0} \times \text{10}^{-\text{2}})r \\ \mathcal{E} & = (\text{6,0} \times \text{10}^{-\text{2}})(\text{100}) + (\text{6,0} \times \text{10}^{-\text{2}})r \end{align*}
Subtracting the equation from the second case from the equation from the first case:
\begin{align*} \mathcal{E} - \mathcal{E} & = ((\text{2,50} \times \text{10}^{-\text{2}})(\text{400}) + (\text{2,50} \times \text{10}^{-\text{2}})r) - ((\text{6,0} \times \text{10}^{-\text{2}})(\text{100}) + (\text{6,0} \times \text{10}^{-\text{2}})r)\\ 0 &= 10 + \text{2,50} \times \text{10}^{-\text{2}}r -6 -\text{6,0} \times \text{10}^{-\text{2}}r \\ -\text{3,5} \times \text{10}^{-\text{2}}r = -4 \\ r &= \frac{4}{\text{3,5} \times \text{10}^{-\text{2}}} \\ &=\text{1,1428} \times \text{10}^{-\text{2}} \end{align*}
We substitute this back into one of the equations to get $$\mathcal{E}$$:
\begin{align*} \mathcal{E} & = (\text{2,50} \times \text{10}^{-\text{2}})(\text{400}) + (\text{2,50} \times \text{10}^{-\text{2}})r \\ \mathcal{E} & = (\text{2,50} \times \text{10}^{-\text{2}})(\text{400}) + (\text{2,50} \times \text{10}^{-\text{2}})(\text{1,1428} \times \text{10}^{-\text{2}}) \\ \mathcal{E} & = \text{10} \end{align*}
The emf is $$\text{10,0}$$ $$\text{V}$$ and the internal resistance is $$\text{1,1428} \times \text{10}^{-\text{2}}$$ $$\text{Ω}$$.
A battery is connected in series with a rheostat and an ammeter. When the resistance of the resistor is 10 $$\Omega$$ the current is 2.0 A. When the resistance is 5 $$\Omega$$ the current is 3.8 A. Find the emf and the internal resistance of the battery.
We have two unknowns but information for two different scenarios so we can solve simultaneously.
In the first case:
\begin{align*} V & =\mathcal{E} - Ir \\ (\text{2})(\text{10}) = & =\mathcal{E} - (\text{2})r \\ \mathcal{E} & = (\text{2})(\text{10}) + (\text{2})r \end{align*}
In the second case we will get:
\begin{align*} V & =\mathcal{E} - Ir \\ (\text{3,8})(\text{5}) = & =\mathcal{E} - (\text{3,8})r \\ \mathcal{E} & = (\text{3,8})(\text{5}) + (\text{3,8})r \end{align*}
Subtracting the equation from the second case from the equation from the first case:
\begin{align*} \mathcal{E} - \mathcal{E} & = ((\text{2})(\text{10}) + (\text{2})r) - ((\text{3,8})(\text{5}) + (\text{3,8})r)\\ 0 &= 20 + \text{2}r -19 - \text{3,8}r \\ r &= \frac{1}{\text{1,8}} \\ &=\text{0,55555} \end{align*}
We substitute this back into one of the equations to get $$\mathcal{E}$$:
\begin{align*} \mathcal{E} & = (\text{2})(\text{10}) + (\text{2})r \\ \mathcal{E} & = (\text{2})(\text{10}) + (\text{2})(\text{0,55555}) \\ \mathcal{E} & = \text{21,11111} \end{align*}
The emf is $$\text{21,11}$$ $$\text{V}$$ and the internal resistance is $$\text{0,55}$$ $$\text{Ω}$$.
When a cell is connected directly across a high resistance voltmeter the reading is $$\text{1,5}$$ $$\text{V}$$. When the cell is shorted through a low resistance ammeter the current is $$\text{2,5}$$ $$\text{A}$$. What is the emf and internal resistance of the cell?
In the case of the very high resistance voltmeter we need to assume that the current is zero. This means that the measurement is actually the emf and the potential difference across the internal resistance obeys Ohm's law, $$V=Ir=(0)r=0$$.
In the case of a very low resistance ammeter, we can assume that the current that is flowing is the maximum current possible. We know the emf and the maxmimum curren so we can find $$r$$:
\begin{align*} {I}_{c}& = \frac{\mathcal{E}}{r} \\ (\text{2,5})&= \frac{\text{1,5}}{r} \\ r&= \frac{\text{1,5}}{\text{2,5}} \\ r & = \text{0,6}\text{ Ω} \end{align*}
The emf is $$\text{1,5}$$ $$\text{V}$$ and the internal resistance is $$\text{0,6}$$ $$\text{Ω}$$.
## 10.4 Evaluating internal resistance in circuits (ESCPW)
### Approach (ESCPX)
The approach to solving problems that involve the internal resistance of batteries is straightforward, you just need to understand that each battery in previous examples was a source of emf, $$\mathcal{E}$$, and a small resistor, $$r$$, and then solve as before but include $$r$$ in your calculations.
An important thing to realise is that the potential difference you calculated or were given in previous examples is not the emf, it is the emf less that potential difference across the internal resistance.
To emphasise that internal resistance is an extension to what you have already done we are going to take previous worked examples and consider the internal resistance of the battery. If the internal resistance did not behave like an ohmic resistor this wouldn't be possible but we won't deal with cases like that.
## Worked example 7: Internal resistance in circuit with resistors in series
For the following circuit, calculate:
1. the potential differences $$V_\text{1}$$, $$V_\text{2}$$ and $$V_\text{3}$$ across the resistors $$R_\text{1}$$, $$R_\text{2}$$, and $$R_\text{3}$$
.
2. the resistance of $$R_\text{3}$$.
3. the resistance of $$R_\text{3}$$.
If the internal resistance is $$\text{0,1}$$ $$\text{Ω}$$, what is the emf of the battery and what power is dissipated by the internal resistance of the battery?
### Note
This is a very similar question to what you have seen earlier. This is to highlight the fact that the approach when dealing with internal resistance is built on all the same principles you have already been working with.
### Determine how to approach the problem
We are given the potential difference across the cell and the current in the circuit, as well as the resistances of two of the three resistors. We can use Ohm's Law to calculate the potential difference across the known resistors. Since the resistors are in a series circuit the potential difference is $$V = V_\text{1} + V_\text{2} + V_\text{3}$$ and we can calculate $$V_\text{3}$$. Now we can use this information to find the potential difference across the unknown resistor $$R_\text{3}$$.
### Calculate potential difference across $$R_\text{1}$$
Using Ohm's Law: \begin{align*} R_\text{1} &= \frac{V_\text{1}}{I} \\ I \cdot R_\text{1} &= I \cdot \frac{V_\text{1}}{I} \\ V_\text{1} &= {I}\cdot{R_\text{1}}\\ &= 2 \cdot 1 \\ V_\text{1} &= \text{2}\text{ V} \end{align*}
### Calculate potential difference across $$R_\text{2}$$
Again using Ohm's Law: \begin{align*} R_\text{2} &= \frac{V_\text{2}}{I} \\ I \cdot R_\text{2} &= I \cdot \frac{V_\text{2}}{I} \\ V_\text{2} &= {I}\cdot{R_\text{2}}\\ &= 2 \cdot 3 \\ V_\text{2} &= \text{6}\text{ V} \end{align*}
### Calculate potential difference across $$R_\text{3}$$
Since the potential difference across all the resistors combined must be the same as the potential difference across the cell in a series circuit, we can find $$V_\text{3}$$ using: \begin{align*} V &= V_\text{1} + V_\text{2} + V_\text{3}\\ V_\text{3} &= V - V_\text{1} - V_\text{2} \\ &= 23 - 2 - 6 \\ V_\text{3}&= \text{15}\text{ V} \end{align*}
### Find the resistance of $$R_\text{3}$$
We know the potential difference across $$R_\text{3}$$ and the current through it, so we can use Ohm's Law to calculate the value for the resistance: \begin{align*} R_\text{3} &= \frac{V_\text{3}}{I}\\ &= \frac{\text{15}}{\text{2}} \\ R_\text{3}&= \text{7,5}~\Omega \end{align*}
### Potential difference across the internal resistance of the battery
The value of the emf can be calculated from the potential difference of the load and the potential difference across the internal resistance.
\begin{align*} \mathcal{E}& = V+Ir\\ & = \text{23}+(\text{2})(\text{0,1})\\ & = \text{23,2}\text{ V} \end{align*}
### Power dissipated in the battery
We know that the power dissipated in a resistor is given by $$P=VI=I^2R=\dfrac{V^2}{R}$$ and we know the current in the circuit, the internal resistance and the potential difference across it so we can use any form of the equation for power:
\begin{align*} P_r &= V_rI_r\\ & = (\text{0,2})(\text{2})\\ & = \text{0,4}\text{ W} \end{align*}
• $$V_\text{1} = \text{2,0}\text{ V}$$
• $$V_\text{2} = \text{6,0}\text{ V}$$
• $$V_\text{3} = \text{10,0}\text{ V}$$
• $$R_\text{3} = \text{7,5} \Omega$$
• $$\mathcal{E} = \text{23,2}\text{ V}$$
• $$P_r = \text{0,4}\text{ W}$$
## Worked example 8: Internal resistance and resistors in parallel
The potential difference across a battery measures 18 V when it is connected to two parallel resistors of $$\text{4,00}$$ $$\Omega$$ and $$\text{12,00}$$ $$\Omega$$ respectively. Calculate the current through the cell and through each of the resistors. If the internal resistance of the battery is $$\text{0,375}$$ $$\text{Ω}$$ what is the emf of the battery?
### Determine how to approach the problem
We need to determine the current through the cell and each of the parallel resistors. We have been given the potential difference across the cell and the resistances of the resistors, so we can use Ohm's Law to calculate the current.
### Calculate the current through the cell
To calculate the current through the cell we first need to determine the equivalent resistance of the rest of the circuit. The resistors are in parallel and therefore: \begin{align*} \frac{\text{1}}{R} &= \frac{\text{1}}{R_\text{1}} + \frac{\text{1}}{R_\text{2}} \\ &= \frac{\text{1}}{\text{4}} + \frac{\text{1}}{\text{12}} \\ &= \frac{3+1}{\text{12}} \\ &= \frac{\text{4}}{\text{12}} \\ R &= \frac{\text{12}}{\text{4}} = \text{3,00} \ \Omega \end{align*} Now using Ohm's Law to find the current through the cell: \begin{align*} R &= \frac{V}{I} \\ I &= \frac{V}{R} \\ &= \frac{\text{18}}{\text{3}} \\ I &= \text{6,00}\text{ A} \end{align*}
### Now determine the current through one of the parallel resistors
We know that for a purely parallel resistor configuration, the potential difference across the cell is the same as the potential difference across each of the parallel resistors. For this circuit: \begin{align*} V &= V_\text{1} = V_\text{2} = \text{18}\text{ V} \end{align*} Let's start with calculating the current through $$R_\text{1}$$ using Ohm's Law: \begin{align*} R_\text{1} &= \frac{V_\text{1}}{I_\text{1}} \\ I_\text{1} &= \frac{V_\text{1}}{R_\text{1}} \\ &= \frac{\text{18}}{\text{4}} \\ I_\text{1} &= \text{4,50}\text{ A} \end{align*}
### Calculate the current through the other parallel resistor
We can use Ohm's Law again to find the current in $$R_\text{2}$$: \begin{align*} R_\text{2} &= \frac{V_\text{2}}{I_\text{2}} \\ I_\text{2} &= \frac{V_\text{2}}{R_\text{2}} \\ &= \frac{\text{18}}{\text{12}} \\ I_\text{2} &= \text{1,50}\text{ A} \end{align*} An alternative method of calculating $$I_\text{2}$$ would have been to use the fact that the currents through each of the parallel resistors must add up to the total current through the cell: \begin{align*} I &= I_\text{1} + I_\text{2} \\ I_\text{2} &= I - I_\text{1} \\ &= 6 - 4.5 \\ I_\text{2} &= \text{1,5}\text{ A} \end{align*}
### Determine the emf
This total current through the battery is the current through the internal resistance of the battery. Knowing the current and resistance allows us to use Ohm's law to determine the potential difference across the internal resistance and therefore the emf of the battery.
Using Ohm's law we can determine the potential difference across the internal resistance:
\begin{align*} V &=I \cdot r \\ &=\text{6} \cdot \text{0,375} \\ &= \text{2,25}\text{ V} \end{align*}
We know that the emf of the battery is the potential difference across the terminal summed with the potential difference across the internal resistance so:
\begin{align*} \mathcal{E}& = V+Ir \\ & = \text{18}+\text{2,25} \\ & = \text{20,25}\text{ V} \end{align*}
The current through the cell is $$\text{6,00}$$ $$\text{A}$$.
The current through the $$\text{4,00}$$ $$\Omega$$ resistor is $$\text{4,50}$$ $$\text{A}$$.
The current through the $$\text{12,00}$$ $$\Omega$$ resistor is $$\text{1,50}$$ $$\text{A}$$.
The emf of the battery is $$\text{20,25}$$ $$\text{V}$$.
## Worked example 9: Power in series and parallel networks of resistors
Given the following circuit:
The current leaving the battery is $$\text{1,07}$$ $$\text{A}$$, the total power dissipated in the external circuit is $$\text{6,42}$$ $$\text{W}$$, the ratio of the total resistances of the two parallel networks $$R_{P\text{1}} : R_{P\text{2}}$$ is 1:2, the ratio $$R_\text{1} : R_\text{2}$$ is 3:5 and $$R_\text{3}=\text{7,00}\text{ Ω}$$.
Determine the:
1. potential difference of the battery,
2. the power dissipated in $$R_{P\text{1}}$$ and $$R_{P\text{2}}$$, and
3. if the battery is labelled as having an emf of $$\text{6,50}$$ $$\text{V}$$ what is the value of the resistance of each resistor and the power dissipated in each of them.
### What is required
In this question you are given various pieces of information and asked to determine the power dissipated in each resistor and each combination of resistors. Notice that the information given is mostly for the overall circuit. This is a clue that you should start with the overall circuit and work downwards to more specific circuit elements.
### Calculating the potential difference of the battery
Firstly we focus on the battery. We are given the power for the overall circuit as well as the current leaving the battery. We know that the potential difference across the terminals of the battery is the potential difference across the circuit as a whole.
We can use the relationship $$P=VI$$ for the entire circuit because the potential difference is the same as the potential difference across the terminals of the battery: \begin{align*} P &=VI \\ V &= \frac{P}{I} \\ &=\frac{\text{6,42}}{\text{1,07}} \\ &= \text{6,00}\text{ V} \end{align*}
The potential difference across the battery is $$\text{6,00}$$ $$\text{V}$$.
### Power dissipated in $$R_{P\text{1}}$$ and $$R_{P\text{2}}$$
Remember that we are working from the overall circuit details down towards those for individual elements, this is opposite to how you treated this circuit earlier.
We can treat the parallel networks like the equivalent resistors so the circuit we are currently dealing with looks like:
We know that the current through the two circuit elements will be the same because it is a series circuit and that the resistance for the total circuit must be: $$R_{Ext}=R_{P\text{1}}+R_{P\text{2}}$$. We can determine the total resistance from Ohm's Law for the circuit as a whole: \begin{align*} V_{battery}&=IR_{Ext} \\ R_{Ext} &=\frac{V_{battery}}{I} \\ &=\frac{\text{6,00}}{\text{1,07}}\\ &=\text{5,61}\text{ Ω} \end{align*}
We know that the ratio between $$R_{P\text{1}} : R_{P\text{2}}$$ is 1:2 which means that we know: \begin{align*} R_{P\text{1}} &= \frac{\text{1}}{\text{2}}R_{P\text{2}} \ \ \text{and} \\ R_T &= R_{P\text{1}} + R_{P\text{2}} \\ & = \frac{\text{1}}{\text{2}}R_{P\text{2}} + R_{P\text{2}} \\ &=\frac{\text{3}}{\text{2}}R_{P\text{2}} \\ (\text{5,61}) &=\frac{\text{3}}{\text{2}}R_{P\text{2}} \\ R_{P\text{2}} &= \frac{\text{2}}{\text{3}}(\text{5,61}) \\ R_{P\text{2}} &= \text{3,74}\text{ Ω} \end{align*} and therefore: \begin{align*} R_{P\text{1}} &= \frac{\text{1}}{\text{2}}R_{P\text{2}} \\ &=\frac{\text{1}}{\text{2}}(3.74) \\ &= \text{1,87}\text{ Ω} \end{align*}
Now that we know the total resistance of each of the parallel networks we can calculate the power dissipated in each: \begin{align*} P_{P\text{1}} &= I^2R_{P\text{1}} \\ &= (\text{1,07})^2(\text{1,87}) \\ &= \text{2,14}\text{ W} \end{align*} and \begin{align*} P_{P\text{2}} &= I^2R_{P\text{2}} \\ &= (\text{1,07})^2(\text{3,74}) \\ &= \text{4,28}\text{ W} \end{align*} These values will add up to the original power value we had for the external circuit. If they didn't we would have made a calculation error.
### Parallel network 1 calculations
Now we can begin to do the detailed calculation for the first set of parallel resistors.
We know that the ratio between $$R_{\text{1}} : R_{\text{2}}$$ is 3:5 which means that we know $$R_{\text{1}}= \frac{\text{3}}{\text{5}}R_{\text{2}}$$. We also know the total resistance for the two parallel resistors in this network is $$\text{1,87}$$ $$\text{Ω}$$. We can use the relationship between the values of the two resistors as well as the formula for the total resistance ($$\frac{\text{1}}{R_PT}=\frac{\text{1}}{R_\text{1}}+\frac{\text{1}}{R_\text{2}}$$) to find the resistor values: \begin{align*} \frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{1}}{R_\text{1}}+\frac{\text{1}}{R_\text{2}} \\ \frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{5}}{3R_\text{2}}+\frac{\text{1}}{R_\text{2}} \\ \frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{1}}{R_\text{2}}(\frac{\text{5}}{\text{3}}+1) \\ \frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{1}}{R_\text{2}}(\frac{\text{5}}{\text{3}}+\frac{\text{3}}{\text{3}}) \\ \frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{1}}{R_\text{2}}\frac{\text{8}}{\text{3}} \\ R_\text{2}&=R_{P\text{1}}\frac{\text{8}}{\text{3}} \\ &=(\text{1,87})\frac{\text{8}}{\text{3}} \\ &=\text{4,99}\text{ Ω} \end{align*} We can also calculate $$R_{\text{1}}$$: \begin{align*} R_{\text{1}}&= \frac{\text{3}}{\text{5}}R_{\text{2}} \\ &= \frac{\text{3}}{\text{5}}(\text{4,99}) \\ &= \text{2,99}\text{ Ω} \end{align*}
To determine the power we need the resistance which we have calculated and either the potential difference or current. The two resistors are in parallel so the potential difference across them is the same as well as the same as the potential difference across the parallel network. We can use Ohm's Law to determine the potential difference across the network of parallel resistors as we know the total resistance and we know the current: \begin{align*} V &= I R \\ &=(\text{1,07})(\text{1,87}) \\ &=\text{2,00}\text{ V} \end{align*}
We now have the information we need to determine the power through each resistor: \begin{align*} P_\text{1}&=\frac{V^2}{R_\text{1}} \\ &=\frac{(\text{2,00})^2}{\text{2,99}} \\ &=\text{1,34}\text{ W} \end{align*} \begin{align*} P_\text{2}&=\frac{V^2}{R_\text{2}} \\ &=\frac{(\text{2,00})^2}{\text{4,99}} \\ &=\text{0,80}\text{ W} \end{align*}
### Parallel network 2 calculations
Now we can begin to do the detailed calculation for the second set of parallel resistors.
We are given $$R_\text{3}=\text{7,00}\text{ Ω}$$ and we know $$R_{P\text{2}}$$ so we can calculate $$R_\text{4}$$ from: \begin{align*} \frac{\text{1}}{R_{P\text{2}}} &= \frac{\text{1}}{R_\text{3}}+\frac{\text{1}}{R_\text{4}} \\ \frac{\text{1}}{\text{3,74}} &= \frac{\text{1}}{\text{7,00}}+\frac{\text{1}}{R_\text{4}} \\ R_\text{4}&=\text{8,03}\text{ Ω} \end{align*}
We can calculate the potential difference across the second parallel network by subtracting the potential difference of the first parallel network from the battery potential difference, $$V_{P\text{2}} = \text{6,00}-\text{2,00}=\text{4,00}\text{ V}$$.
We can now determine the power dissipated in each resistor: \begin{align*} P_\text{3}&=\frac{V^2}{R_\text{3}} \\ &=\frac{(\text{4,00})^2}{\text{7,00}} \\ &=\text{2,29}\text{ W} \end{align*} \begin{align*} P_\text{4}&=\frac{V^2}{R_\text{2}} \\ &=\frac{(\text{4,00})^2}{\text{8,03}} \\ &=\text{1,99}\text{ W} \end{align*}
### Internal resistance
We know that the emf of the battery is $$\text{6,5}$$ $$\text{V}$$ but that the potential difference measured across the terminals is only $$\text{6}$$ $$\text{V}$$. The difference is the potential difference across the internal resistance of the battery and we can use the known current and Ohm's law to determine the internal resistance:
\begin{align*} V&=I \cdot R \\ R&=\frac{V}{I} \\ & = \frac{\text{0,5}}{\text{1,07}} \\ & = \text{0,4672897}\\ & = \text{0,47}\text{ Ω} \end{align*}
The power dissipated by the internal resistance of the battery is:
\begin{align*} P &=VI \\ & = \text{0,5}\cdot\text{1,07} \\ &=\text{0,535}\text{ W} \end{align*}
## Worked example 10: Internal resistance and headlamps [NSC 2011 Paper 1]
The headlamp and two IDENTICAL tail lamps of a scooter are connected in parallel to a battery with unknown internal resistance as shown in the simplified circuit diagram below. The headlamp has a resistance of $$\text{2,4}$$ $$\text{Ω}$$ and is controlled by switch $$\textbf{S}_1$$. The tail lamps are controlled by switch $$\textbf{S}_2$$. The resistance of the connecting wires may be ignored.
The graph alongside shows the potential difference across the terminals of the battery before and after switch $$\textbf{S}_1$$ is closed (whilst switch $$\textbf{S}_2$$ is open). Switch $$\textbf{S}_1$$ is closed at time $$\textbf{t}_1$$.
1. Use the graph to determine the emf of the battery.
(1 mark)
2. WITH ONLY SWITCH $$\textbf{S}_1$$ CLOSED, calculate the following:
(3 marks)
2. Internal resistance, $$r$$, of the battery
(3 marks)
3. BOTH SWITCHES $$\textbf{S}_1$$ AND $$\textbf{S}_2$$ ARE NOW CLOSED. The battery delivers a current of $$\text{6}$$ $$\text{A}$$ during this period.
Calculate the resistance of each tail lamp.
(5 marks)
4. How will the reading on the voltmeter be affected if the headlamp burns out? (Both switches $$\textbf{S}_1$$ and $$\textbf{S}_2$$ are still closed.)
Write down only INCREASES, DECREASES or REMAINS THE SAME.
Give an explanation.
(3 marks)
Question 1
$$\text{12}$$ $$\text{V}$$
(1 mark)
Question 2.1
Option 1:
\begin{align*} I & = \frac{V}{R} \\ & = \frac{\text{9,6}}{\text{2,4}} \\ & = \text{4 A} \end{align*}
Option 2:
\begin{align*} \text{emf} & = IR + Ir \\ 12 & = I(\text{2,4}) + \text{2,4} \\ \therefore I & = \text{4 A} \end{align*}
(3 marks)
Question 2.2
Option 1:
\begin{align*} \text{emf} & = IR +Ir \\ 12 & = \text{9,4} + 4r \\ r & = \text{0,6}\ \Omega \end{align*}
Option 2:
\begin{align*} V_{lost} & = Ir \\ \text{2,4} & = \text{4}r \\ \therefore r & = \text{0,6}\ \Omega \end{align*}
Option 3:
\begin{align*} \text{emf} & = I(R + r) \\ \text{12} & = \text{4}(\text{2,4} + r)\\ \therefore r & = \text{0,6}\ \Omega \end{align*}
(3 marks)
Question 3
Option 1:
\begin{align*} \text{emf} & = IR +Ir \\ \text{12} & = \text{6}(R + \text{0,6}) \\ R_{\text{ext}} & = \text{1,4}\ \Omega \end{align*}\begin{align*} \frac{1}{R} & = \frac{1}{R_{1}} + \frac{1}{R_{2}} \\ \frac{1}{\text{1,4}} & = \frac{1}{\text{2,4}} + \frac{1}{R} \\ R & = \text{3,36}\ \Omega \end{align*}
Each tail lamp: $$R = \text{1,68}\ \Omega$$
Option 2:
\begin{align*} \text{Emf} & = V_{\text{terminal}} + Ir \\ 12 & = V_{\text{terminal}} + 6(\text{0,6}) \\ \therefore V_{\text{terminal}} & = \text{8,4}\text{ V} \end{align*}\begin{align*} I_{\text{2,4}\ \Omega} &= \frac{V}{R} \\ &= \frac{\text{8,4}}{\text{2,4}} \\ &= \text{3,5 A} \end{align*}\begin{align*} I_{\text{tail lamps}} & = 6 - \text{3,5} \\ & = \text{2,5}\text{ A} \\ R_{\text{tail lamps}} & = \frac{V}{I} \\ & = \frac{\text{8,4}}{\text{2,5}} \\ & = \text{3,36}\ \Omega \\ R_{\text{tail lamp}} & = \text{1,68}\ \Omega \end{align*}
Option 3:
\begin{align*} V & = IR \\ \text{12} & = \text{6}(R)\\ R_{\text{ext}} & = 2\ \Omega \end{align*}\begin{align*} R_{\text{parallel}} & = 2 - \text{0,6} \\ & = \text{1,4}\ \Omega \\ \frac{1}{R} & = \frac{1}{R_{1}} + \frac{1}{R_{2}} \\ \frac{1}{\text{1,4}} & = \frac{1}{\text{2,4}} + \frac{1}{R} \\ R & = \text{3,36}\ \Omega \end{align*}
Each tail lamp: $$R = \text{1,68}\ \Omega$$
Option 4:
For parallel combination: $$I_{1} + I_{2} = 6\text{ A}$$
\begin{align*} \therefore \frac{V}{\text{2,4}} + \frac{V}{R_{\text{tail lamps}}} & = \text{6} \\ \text{8,4}\left(\frac{1}{\text{2,4}} + \frac{1}{R_{\text{tail lamps}}} \right) & = \text{6}\\ \therefore R_{\text{tail lamps}} & = \text{3,36}\ \Omega \\ R_{\text{tail lamp}} & = \text{1,68}\ \Omega \end{align*}
(5 marks)
Question 4
Increases
The resistance increases and the current decreases. So $$Ir$$ (lost volts) must decrease which leads to an increase in the voltage.
(3 marks)
[TOTAL: 15 marks]
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Mathematics » Introducing Graphs » Understand Slope of a Line
# Solving Slope Applications
## Solving Slope Applications
At the beginning of this section, we said there are many applications of slope in the real world. Let’s look at a few now.
## Example
The pitch of a building’s roof is the slope of the roof. Knowing the pitch is important in climates where there is heavy snowfall. If the roof is too flat, the weight of the snow may cause it to collapse. What is the slope of the roof shown?
### Solution
Use the slope formula. $$m=\frac{\text{rise}}{\text{run}}$$ Substitute the values for rise and run. $$m=\frac{\text{9 ft}}{\text{18 ft}}$$ Simplify. $$m=\frac{1}{2}$$ The slope of the roof is $$\frac{1}{2}$$.
Have you ever thought about the sewage pipes going from your house to the street? Their slope is an important factor in how they take waste away from your house.
## Example
Sewage pipes must slope down $$\frac{1}{4}$$ inch per foot in order to drain properly. What is the required slope?
### Solution
Use the slope formula. $$m=\frac{\text{rise}}{\text{run}}$$ $$m=\frac{-\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\text{in}\text{.}}{1\phantom{\rule{0.2em}{0ex}}\text{ft}}$$ $$m=\frac{-\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\text{in}\text{.}}{1\phantom{\rule{0.2em}{0ex}}\text{ft}}$$ Convert 1 foot to 12 inches. $$m=\frac{-\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\text{in}\text{.}}{12\phantom{\rule{0.2em}{0ex}}\text{in.}}$$ Simplify. $$m=-\frac{1}{48}$$ The slope of the pipe is $$-\frac{1}{48}.$$
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# The Area of a Cyclic Quadrilateral
A quadrilateral inscribed is a circle is known as a cyclic quadrilateral. The proof is beyond the scope of this tutorial and will be discussed in an advanced tutorial, so only the formula is given here for application.
If $a$, $b$, $c$ and $d$ are the sides of a cyclic quadrilateral and if $s = \frac{{a + b + c + d}}{2}$, then
The area of a cyclic quadrilateral $= \sqrt {(s – a)(s – b)(s – c)(s – d)}$
Example:
In a circular grassy plot, a quadrilateral shape with its corners touching the boundary of the plot is to be paved with bricks. Find the area of the quadrilateral when the sides of the quadrilateral are $36$ m, $77$ m, $75$ m and $40$ m.
Solution:
Given that the sides of the quadrilateral are $a = 36$m, $b = 77$m, $c = 75$m and $d = 40$m
$s = \frac{{a + b + c + d}}{2} = \frac{{36 + 77 + 75 + 40}}{2} = \frac{{228}}{2} = 114m$
The area of the cyclic quadrilateral $= \sqrt {(s – a)(s – b)(s – c)(s – d)}$
The area of the cyclic quadrilateral $= \sqrt {(114 – 36)(114 – 77)(114 – 75)(114 – 40)}$
The area of the cyclic quadrilateral $= \sqrt {78 \times 37 \times 39 \times 74} = 39 \times 37 \times 2 = 2886$ square meters.
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# Viewing a Square
By Jonathan Osters @callmejosters & Chris Robinson @isomorphic2crob, The Blake School
This week, we were taking a look through How to Solve It, and came across this problem that intrigued us. The problem is as follows:
Point P is a point outside a square. Define the “viewing angle” as an angle that has P as a vertex and vertices of the square as points on its rays, as if P was on the side of a sculpture and viewing it by walking around it (Figure 1). What is the locus of points for which the viewing angle is 90˚? What is the locus for which the viewing angle is 45˚? (Paraphrased from p. 234 of How to Solve It)
This problem is really fun! Give it a try before you continue reading.
Solution for 90º viewing angle:
We began by creating convenient points in the locus. We reasoned that a point “half of a side length” from the midpoint of a side could create a 45-45-90 triangle.
Then we remembered Thales’ Theorem, that says that any point on a circle forms a right angle with endpoints of a diameter. Therefore, any point “half of a side length” from the midpoint of a side would create a right angle.
Thanks, Thales! Repeat that with all four sides, and you will get the following locus:
Solution for 45º viewing angle:
We knew a few things more about this one, having solved the 90˚ version of the problem. We knew that in this scenario, we would have to take into account that you might be able to see one or two sides, rather than just one, like in the 90˚ version. Not being able to see the solution with our mind’s eye, we began by calculating a few special points in the locus.
Special point 1: On a side (extended)
On the extension of one side, you will only be able to see one side, but you will be on the cusp of seeing two sides. This gave us the impression that this was an important point to investigate.
The point that forms a 45º viewing angle on the extended side forms an isosceles triangle which means that the point is a side length away from the vertex of the square. So the following 8 points are included in the locus.
Special Point 2: On an diagonal (extended)
The points along the extended diagonal would include two sides in their viewing angle. By symmetry we know that the diagonal is the bisector of the 45º angle and that the diagonal also bisects the 270º outside of the square.
Nicely, this leaves 22.5º, which means this a set of two isosceles triangles and that the point is a side length away from the vertex of the square. At that point we realized that to keep a 45˚ angle at the viewing angle, we could have the angle intercept a 90˚ arc on a circle. This circle:
We can see that any point on this quarter circle creates a 45˚ viewing angle.
There is such a quarter circle surrounding each vertex of the square.
This leaves us with the task of discovering what the locus of points is when only one side is visible. After playing around with it some more, and trying more things, we saw that the endpoints of the arcs and the closest vertices of the square, form a square.
If we create a circle at the center of that “side square,” then points on the outer edge will intercept a quarter of the circle at the viewing angle, thus making the viewing angle a 45˚ inscribed angle again!
There are these “bulbous bumpouts” on each of the sides, so the complete locus gives this “cloud” shape:
After completing this problem the other day, Chris and I started discussing the “cosmic significance” of the problem. This problem could be a great problem for an honors geometry class working on loci or circle properties, but more importantly, this problem is a reminder of what Chris and I like so much about working at Blake. We will often find ourselves on flights of mathematical fancy where we toss out some random problem like this, and will work to solve it together. We are all eager and excited mathematicians, and are always looking for problems that energize us and help us “recharge our mathematical batteries.” This is one such problem.
Next week, we will post the general solution to the problem. How do the loci change if the viewing angle is very small? What if it’s very large? See if you can sketch out a guess of what the loci look like, and snap a picture of it and send it to us on twitter! Our handles are @callmejosters and @Isomorphic2CRob. Have fun!
# Stars on the Flag – Solution
written by Alex Fisher, The Blake School
Hey folks! If you’ve been eagerly awaiting a solution to the flag problem, perhaps while stitching your own replacements and making plans to kick Texas out of the Union, wait no more!
The easiest way to think about this grid is to actually separate it into two overlapping grids:
From this picture, we can see two sub-grids: one in blue that has m rows and n columns (in this case, 5 and 6 respectively) and another in pink, that measures m-1 by n-1. Add them together and I have T stars total. It’s tempting to try to multiply (2m-1) by (2n-1), but that overcounts the total number of stars. However, what if we split these two grids up?
Now we have those same two rectangular grids, the blue m x n and the pink (m-1) x (n-1). Together they almost make a larger grid, but not quite: there’s a column missing. It’s tempting to just add those stars to the pink grid… However, if I add a column, I’m changing the total number of stars by m or n, making it harder to determine whether a given number of stars T can be arranged in this pattern. I want to find a way to make sure that the number of stars in my overall picture can be represented in terms of T and constants.
Now we’re getting somewhere! By doubling the previous arrangement, we actually end up with that previously mentioned (2n-1) x (2m-1) grid! Well, almost. There’s one star of overlap. Even so, this means that if I double T and take away that one overlapping star, I have a perfect grid. Thus, this equation: 2T-1 = (2m-1)(2n-1).
This seems like an odd conclusion to draw, but it is actually the key to answering all of our questions. Consider the situation where Puerto Rico, Guam, and Washington DC all are given statehood. We now have 53 states, so T = 53, and 2T-1 = 105. If I can factor 105, that means I can find numbers m and n to make that grid. In this case, since 105 factors into 3*5*7, I can actually do this in three different non-trivial ways: 5*21, 3*35, and 7*15. Solving for m and n in these scenarios gives us the following diagrams:
And while the last one is a little silly, the 4 x 8 case is relatively nice looking. Also, we can see that if only one state joined the Union, 2T-1 = 101, which isn’t factorable, so this arrangement wouldn’t be possible at all. Likewise, if we lost a state, we’d be left at 97, also not a number with nice factors. So, if you’re a big fan of this star arrangement, then bad news: you’re stuck with Texas. At least, unless you convinced them to take Florida on their way out…
Also, note that while I did this work geometrically, there is also an algebraic approach: in the original set-up, T = the blue rectangle + the pink rectangle, or T = m*n + (m-1)(n-1) = 2mn-m-n+1. I will leave the algebraic manipulation to get this to match our answer of 2T-1 = (2m-1)(2n-1) as an exercise for the reader.
# Stars on the Flag
Written by Alex Fisher, The Blake School
One problem that caught my interest recently involves the US flag. This is actually one I rediscovered from back in 2012 when Puerto Rico was considering becoming a state and there was much discussion online of how you add a star to the flag in such a scenario. If you look at the stars on the current flag, you’ll see that there are 5 rows of 6 stars, with another 4 rows of 5 stars between them.
If you were to keep this pattern, where a larger grid contains a smaller grid nestled between its rows, what numbers can you represent? Can you make one that has 51 stars? How about 49? And which numbers can be represented in more than one arrangement?
# Noticing and Wondering– The Beginning of a Mathematical Journey
Written by Jonathan Osters @CallMeJosters, The Blake School
A few weeks ago, we began the journey of parallel lines in geometry. I have taught this course for several years now, and the first day of this unit is typically excruciating for me – a lot of bland definitions that don’t really have any connection to anything interesting. This year, I decided to change things up and have students create the definitions and conjectures on their own, but I would need a means to do that.
I have seen a number of posts from the wonderfully knowledgeable teachers of #MTBoS (Math Twitter Blogosphere) on Twitter, and many of them had been beginning lessons asking students the questions, “What do you notice?” and “What do you wonder?” I have been skeptical about trying this in a high school level class, but I decided to give it a try anyway.
I used the following figure from Sketchpad and projected it on the board. Then I gave students a few minutes to notice and wonder (My only directions were that the “Notice” statements had to be statements, and the “Wonder” statements had to be questions).
The following were some of the “Notice” and “Wonder” statements I got throughout the day in my three sections of Geometry:
Then I asked if there was anything among the things we noticed that allowed us to answer any of the questions that we wonder. Some, like the distances, were directly measurable in Sketchpad, while “are the lines parallel?” required a different set of skills. These students have taken Algebra 1, and several pointed out that if these lines had been parallel, we would have been able to determine if they were parallel. I obliged them, despite the fact that discussions of slope weren’t an official part of the lesson.
So from here, we were able to re-familiarize ourselves with slope as a number that measures steepness, and that if two different lines have the same steepness, they will be parallel. At this point, many students were at least able to answer the question, as 6.63-1.131.87-(-14.08)=0.345, while 0.78-(-5.40)5.19-(-13.51)=0.330, so the lines are not parallel.
I asked if there was anything they could notice and wonder now that there is grid on the board. Several students wondered if the lines were making the same angle with the y-axis. I removed the grid and just placed a vertical line where the y-axis had been, and measured the angles.
Many students then made the argument that had the two angles been equal in measure, the lines would have been parallel. They also argued that they could find the measures of the other six angles, and that if they could determine if “the angle at any corner” were congruent “to the angle at the same corner on the other line,” then the lines would be parallel. Some noticed that it would be enough to know that the “two angles on the inside but the other side of that axis” were congruent. At this point, it was they who decided we would benefit from some terms for such angles. And voila, the definitions for Alternate Interior Angles, Corresponding Angles, and the like. We were able to modify their original conjecture to: If Corresponding Angles on two lines are congruent, then those lines are parallel. This is exactly what I had hoped they would see, and while we ran out of time to prove the conjecture, we would begin there the next day, and the lesson was so much the better for it.
Noticing and wondering was a very powerful technique, especially when introducing something brand new. I will be utilizing this strategy frequently. Thanks to the teachers of #MTBoS for helping me learn something new!
# A Comprehensive System for Student Assessment: Part 2
By Chris Robinson @Isomorphic2CRob & Jonathan Osters @callmejosters, The Blake School
Most of us think about two types of assessment on a daily basis. Formative Assessment should give students feedback on how close they are to meeting class expectations of knowledge and skill. Summative Assessment is a measure of what they know at the end of the learning process.
Another two pieces of the assessment picture that we often think about is the balance between a student knowing the skills of mathematics and students being able to solve novel problems. In the past, Mathematics education movements have swung far in each direction, at times over emphasizing skills and at times sacrificing skills practice.
As the Blake Mathematics Department renewed its discussion on teaching problem solving in 2009, we began thinking about how we could assess more effectively. We wanted to utilize frequent formative assessment but were unsure how to walk the line of making them worth enough that students took them seriously, but not worth an unfair amount as students are still in the learning process. We also wanted to assess authentic problem solving. The last post discussed our Skills Quiz System, while this post will discuss our Problem Solving Assessments.
Building a Culture of Problem Solving BEFORE Assessment
Being able to assess problem solving effectively requires that students are “practicing” solving novel problems on a daily basis. There is a great deal that can be said about buliding a culture in the classroom focused on problem solving and collaboration and every teacher’s experience will be a little different: see our blog and Carmel Schettino’s blog is a great resource as well. Carmel is a teacher at Deerfield and major proponent of problem based instruction. Suffice it to say that building a culture is more than selecting good problems but as we are trying to talk about assessment here, we shall move on.
My colleagues and I each take a different approach here but in all cases a majority of our lessons are centered around problems that explore new topics and are either scaffolded by printed leading questions or are scaffolded by interjected questions in small group discussions. Some problems are taken from the Phillips Exeter Academy Curriculum, some from Carmel Schettino, some from our current hodgepodge of curricular resources and many written by us. The goal in general is for students to encounter something approachable but new and for them to conjecture and test out approaches to solving a problem. Then students share their approaches in small group and then large group verbally or visually through doc cams or writing on the boards. There is a summary process of the approach and sometimes an urging of “if you want to remember the steps you just went through, you may want to take notes now.”
Problem Solving Under Pressure
The threat of assessment is real for all students, and although our skills quiz systems alleviates some pressure, solving novel or near-novel problems can be a hairy experience for most students and teachers. The constant practice of problem solving in the classroom should help prepare students. Our Problem Solving Assessments (PSA) come in a few varieties: Exploration Labs, Reflection Journals, In-Class PSAs and Take Home PSAs. Not all classes use all types but it is safe to say that all classes use at least two of the four.
Component 1: Exploration Labs
Exploration Labs come in many forms. A majority of them are guided questions utilizing a technology like Geometer’s Sketchpad, Fathom, or Desmos. In Geometry, students may be tasked with creating a diagram, manipulating the diagram and observing how things change. In algebra, students may use Desmos sliders to see how adjusting one part of an equation affects the graph of the equation. In statistics, we may create some statistic “from the ground up” that measures some particular facet of a distribution. Each of these activities puts students in new situations, where building something from scratch is required of them, but the stakes are low, so they can try different things until a “good,” “best,” or “most efficient” way to approach the problem comes up.
Pros: +Students have opportunities to conjecture in low risk situations (often using dynamic software).
+ Students often work together and develop math communication skills.
Con: – They can take a while to grade, depending on depth of expectation.
Component 2: Reflection Journals
Inspired by the work of Carmel Schettino and her frequent metacognitive journal assignments, we have experimented with a variety of writing assignments. My most recent Honors Algebra II written assignment was the following:
Which is your favorite representation for a line? (standard form, slope-intercept form, point-slope form) Be sure to compare and contrast your favorite to each other type regarding
1. a) ease of graphing.
2. b) ease of writing an equation given two points.
3. c) ease in finding the intersection of two lines.
Students get a few nights to compose their first response. A grade and a written commentary is given to each student, and they can then revise and resubmit to regain half the points they lost.
Pro: + Reflection and metacognition are powerful learning tools.
Con: – They can take a while to grade.
Component 3: In-Class Problem Solving Assessments
A traditional test has a mixture of skills and problem solving. Often coming in the form of 15 skill problems and 2-4 “word problems.” Those word problems were often the same as problems previously encountered with numbers changed; this is by necessity of students having so little time to approach them on a test filled with skills. And because there is so much to do, it was relatively common for students to skip or provide only a minimal attempt of the “word problems.”
It is important to have summative assessment on skills, but now that our skills quiz system is accomplishing both formative and summative assessment , we don’t need to do that on our tests. And thus was born the Problem Solving Assessment, or PSA. A PSA is essentially a traditional test with all the skill portions removed. It is a set of 2-6 problems that require synthesis of a number of skills.
Our PSAs have novel problems in the sense that they may have the same theme as a previously seen problems, but there is a major twist. In an Algebra 2 class, for example, we may have a two-variable systems word problem, but instead of giving them the problem and they find the solution, we might give them a solution and a framework like a paint mixing problem, and ask them to write the question. In order to write the problem they will have to create a system of equations with the correct solution, and then create the sentences in the word problem. Another example might be that of a race between several racers, where speeds, head starts, and starting points all vary, leading to different equations of position for the different racers. We may ask them a straightforward problem like “who won the race?”, but we also could ask them more thought-provoking questions like “ which racers were in 2nd place at any point in the race? How long were they in 2nd place? Use the graph to justify.”
Like any exam, these in-class PSAs have a time element to them. Students must complete the problems during the period. At times we don’t predict properly how long students will take to “solve a problem,” and so we at times let them take them home to revise and at times allow them to revise in the class after teacher comments. This has worked well for us so far, since the novelty of the problems make them such that students can’t find a similar problem in their book or online.
Pros: + Students have the extra time to problem solve, compared to a traditional test.
+ They are faster to grade than a journal, students gain comforter in a timed situation.
Con: – It’s a timed situation, and problem solving is tough with limited time.
Component 4: Take Home Problem Solving Assessments
Take-home PSAs are essentially the same as in-class PSAs, the only difference being the location. The advantage is students have more time to work through the problem if needed. The disadvantage is an increased risk of academic dishonesty. We might give students only one in depth problem rather than a couple of shorter problems as in the in-class PSAs.
Pro: + Students have more time to complete it.
Con: – Managing academic honesty becomes much trickier.
A teacher can use all or some of these components for a successful assessment of problem-solving. But what happens if a student has trouble even getting started? Or what if their work is haphazard and difficult to follow? Are these PSA’s graded differently than traditional exams? We will discuss those questions in next week’s post!
# A Comprehensive System for Student Assessment (Part 1)
by Jonathan Osters @callmejosters & Chris Robinson @Isomorphic2CRob, The Blake School
Most of us think about two types of assessment on a daily basis. Formative Assessment should give students feedback on how close they are to meeting class expectations of knowledge and skill. Summative Assessment is a measure of what they know at the end of the learning process.
Another two pieces of the assessment picture that we often think about is the balance between a student knowing the skills of mathematics and students being able to solve novel problems. In the past, Mathematics education movements have swung far in each direction, at times over emphasizing skills and at times sacrificing skills practice.
As the Blake Mathematics Department renewed its discussion on teaching problem solving in 2009, we began thinking about how we could assess more effectively. We wanted to utilize frequent formative assessment but were unsure how to walk the line of making them worth enough that students took them seriously, but not worth an unfair amount as students are still in the learning process. We also wanted to assess authentic problem solving. The remainder of this post will discuss our Skills Quiz System, while the following post will discuss our Problem Solving Assessments.
### Background
If you are reading this blog then it is fairly safe to assume that you have heard of Dan Meyer, and his blog dy/dan. He has made a name for himself by deconstructing problems into 3-acts – peaking students interest in solving real math problems. Dan Meyer is also a regular presenter at math teacher conferences like NCTM (as well as this fantastic TED talk) and is now the Chief Academic Officer at Desmos. In 2008, he wrote a blog about how he helps students develop and perfect their skills using a retakeable quiz system (http://blog.mrmeyer.com/2008/this-new-school-year/). Our department began discussing this idea in 2010 and invited Anna Maria Gaylord (A.G.A.P.E. High School) to a department meeting to share her skill system. Finding someone who had experimented with developing a skills quiz system was instrumental in our process of developing our system.
Both Dan’s and Anna Maria’s system began with a list of skills students need to master for success in their course. The teacher then develops a set of questions that assess that particular skill, with as little utilization of other skills to reach a solution. Students are then assessed a number of times on each skill, and allowed to retake each individually to show improvement and hopefully show mastery. Dan and Anna Maria would then modify student grades based on retake scores. After we reviewed and discussed these two approaches a number of teams of teachers at Blake decided to implement a similar system in 2010-2011 school year.
### Our Skills Quiz System
Our system takes and refines some of the details of Meyer’s and Gaylord’s systems. Under our system, there are several components. There is an in-class quizzing component, a remediation component, and a redemption component.
#### Component 1: Assessment
Students take a quiz each week. That quiz will contain some newer skills they have learned that week, but also older skills they have learned in previous weeks. That skill will also appear on the following week’s quiz, and the quiz the week after that, for a total of three times the student will have seen a question on that skill. Each time the student sees the question, it will be assessed on a scale from 0-4, inspired by the same scale used in the AP Statistics reading. And, like the AP Statistics reading, the scores themselves have descriptors that we can use to decide on a student’s score (4 = Complete, 3 = Substantial, 2 = Developing, 1 = Minimal). And rather than scoring the quiz as a whole, we follow the scores earned by the students on each skill.
Figure 1: This is a set of quizzes a student might take. On Skill #2, this student received a 3 the first attempt, a 2 the second attempt, and a 3 the third attempt. This student would receive an 8 on Skill #2.
The students receive scores for each attempt they make at a skill, and at the end of the three times they have attempted that skill, they receive a score, which is the sum of the three attempts (for example, if a student scores a 3 the first time a skill is on a quiz, a 2 the second time, and a 3 the third time, then the student receives a score of 3 + 2 + 3 = 8 for that skill). The highest score a student can receive in this system is 12. These scores are recorded in a google sheet to which each student has viewing access to only their own scores.
Figure 2: This student’s scores for Skill #2 have been entered into a google sheet.
#### Component 2: Remediation
If a student is not satisfied with their mastery (score) of a particular skill, they have the opportunity to take what we call “Redemption Quizzes” in order to show an increased level of mastery. But first, they need to to remediate their skill in order to show improvement on another assessment. Students are encouraged to review their past attempts on that skill with their teacher. In addition, on our course websites, we post “Redemption Assignments,” extra work that a student must complete in order to take a Redemption Quiz. They then check their answers against keys we post along with the assignment. We make it plain to them that if they don’t put in extra work to master a skill they have yet to master, then it is quite likely they will continue to earn the same scores they previously earned.
#### Component 3: Redemption
Once a student completes the Redemption Assignment, they can take a one-question Redemption Quiz, which covers the same skill as they have just remediated. This question is scored on the same 0-4 scale as the original quiz questions, and if it is an improvement over the previous scores, then the Redemption Quiz score replaces the lowest quiz score for that skill (for example, if the student who scored a 3 + 2 + 3 = 8 takes a redemption quiz and score a 4 on it, their score will improve to 3 + 3 + 4 = 10). These Redemption Quizzes are generally taken outside of class; our school has a few tutorial periods built into the schedule, but students can also take Redemption Quizzes during Study Hall, free periods in their schedule, or before or after school, contingent on there being someone available those times to proctor the student. We also allow for periodic “Days of Redemption” about once every six weeks, where students can use the class period to take Redemption Quizzes. We allow up to three Redemption Quizzes per skill, allowing for a student who scored poorly on all three in-class attempts to completely redeem themselves. We do have a rule that a student can take only one Redemption Quiz per skill, per day. We want to see sustained excellence rather than a one-time performance. This improvement gets recorded in the google sheet, and often we make a big fuss over entering a good score, as the student can actually see their score improving in real time. This incentive to improve and celebration of improvement is what made us want to create this system in the first place.
Figure 3: This student has taken a redemption quiz on Skill #1. His score has improved from 9 to 12.
To make sure students stay current with their remediation and redemption, Redemption Quizzes for a particular skill are only available for about 6 weeks after the last in-class attempt. This is about two months after a student first learns the skill.
### Reflection: Assessing the Assessment System
##### Benefits of using this system over a traditional quiz system
• Students have a clear message about what they know and don’t know, and a clear path to improvement.
• Students take ownership of their learning more willingly when they know what they need to study.
• The system produces increased feedback among students, parents, and teachers (and makes for rather easy conferences, since the path to improvement is so clear!).
• Not only can we pinpoint the skills with which a particular student struggles, but we also gain a clear understanding for which skills the class as a whole is struggling.
##### Challenges/Costs of this system and how we address them
• Enumerating the Skills – this takes some time upfront, but once it is done, then the only remaining big job is to write the questions.
• Writing good questions – This can be tricky, too, since we want to only test one skill, not multiple skills at once. It’s also important that each question for each skill be of similar difficulty without being nearly identical.
• Security of the Redemption Quizzes – we have tried a few different strategies for this over the years. The current way we try to keep the Redemption Quizzes secure is to have students turn in the Redemption Quiz, then place the graded Redemption Quiz in a file folder for each student, so they can request their folder from their teacher in order review any of their previously taken redemption quizzes. This has worked better than any other system we have used in the past.
• To be honest, it’s kind of a lot of grading and bookkeeping- but the results are impressive. Our students do significantly better on retakes, and it’s way more fun to grade if they can show mastery as a result of their hard work. One can reduce the grading by reducing the number of skills assessed or in finding another way to summarize mastery beyond the sum of the three best attempts. This is simply a system that has worked well for us.
In summary, skills quizzes are both formative and summative assessments. They provide effective feedback to the students, while encouraging remediation and redemption. A student’s skills quiz average is 40% of their grade, 5-10% is homework, while a variety of problem solving assessments complete the other half of their grade. Please read our next post to read about how we measure synthesis of the individual skills into a deeper understanding of mathematics.
Thanks for reading until the end!
# A couple quick thoughts for the weekend (A promise fulfilled)
A short one this week folks, here are two statements I have been thinking about:
“99.9% of all people have more than the average number of ears.” -Paul Vetscher
(I am not sure if he got it from somewhere else.)
“Most people don’t recognize opportunity when it comes, because it is usually dressed in overalls and looks a lot like work.” -Thomas Edison
# Chessboard (General Solution)
By Chris Robinson, The Blake School, @Isomorphic2CRob
Let’s begin with our assumptions:
• A square board with even side length of two squares or greater.
• Chessboard has the typical alternating color pattern.
If we remove a square of each color, there is only one rectangle, which contains those two squares as its corners.
The key concept is that this red rectangle is always even by odd or odd by even.
Imagine for a moment that the rectangle was of even width, then the top left and top right-most squares would be different colors, so the height would have to be 1, or 3, or 2n+1 to maintain opposite colored corners.
Imagine for a moment that the rectangle was of odd width, then the top corners would be the same color, so the height of the rectangle would need to be 2, 4, or 2n to ensure a different color in opposite corners.
We shall approach the tiling of the exterior and interior of the rectangle separately.
Exterior:
Once we remove the red rectangle we can divide the frame into up to 4 rectangles. Without loss of generality we shall assume that the odd-length of the red rectangle is the height. We take full vertical strips of the board on either side of the inner rectangle; these vertical strips have even height. They also leave even-width rectangles above and below the red rectangle as shown.
The fact that each of the blue border rectangles has at least one even side allows each rectangle to be easily tiled in an array.
Thus the exterior can always be tiled.
Interior:
Keeping the orientation the same, we can cut vertical strips of even length and width 1 above or below the two removed squares. This leaves a rectangle with an even width as we removed one square from each side of the original square. This remaining rectangle can be tiled with an array again.
Thus the interior can always be tiled.
So with both the interior and the exterior of our red rectangle tiled, we have shown it to always be possible to tile the chessboard when one tile of each color is removed.
A sample application of this technique is shown below.
# Indirect Reasoning and the Chessboard
By Jonathan Osters, The Blake School
I have always been a fan of the movie The Prestige (https://www.youtube.com/watch?v=LV-cXixgrho). The opening scene of the movie shows how to do a good magic trick. I think that sometimes, good teaching is similar to the outline described in the movie. Dan Meyer uses three acts frequently, and this problem uses three variations of the same problem.
In class this week, we have been studying indirect proof, which is a very challenging topic for beginning Geometry students. We show that a situation must be true by eliminating the chance that it is false. To illustrate this idea, we explored a classic problem in mathematics involving a checkerboard. The framework of the problem is very simple: you have a standard 8×8 checkerboard, and dominoes that cover two squares each.
The Pledge
The first question I ask them is, can you tile the checkerboard with dominoes? It does not take long for the first students to show a solution, and I ask if there are other solutions. Below are several such solutions.
You can see that there are a variety of different solutions to this problem. They all come to the conclusion that there are 32 dominoes to cover 64 squares, and that for the dominoes to cover the board, there must be an even number of squares.
The Turn
The second question I ask is: suppose the opposite corners of the checkerboard were cut away. Would it still be possible? Many emphatically express that it is, since there are still an even number of squares (62). I turn them loose to explore.
Some students insist, after failing the first time, that it can’t be done. I encourage them to continue trying; maybe they just haven’t found the solution yet! After attempting and failing numerous times, I call the class back together and ask if anyone came up with a solution. I contrast their early, frequent, and easy success with the current problem, and give them two choices: (1) A room full of intelligent, capable students failed to solve a solvable problem when they were able to solve a similar problem minutes ago, or (2) the problem is not actually solvable. Students become inclined to believe the latter, and I insist that if this is the case, then we should be able to explain why the problem is not solvable.
I have students generate a list of given information. There are only two main givens: first, is that each domino will cover two squares, and the other is that there are 62 squares. I remind them, only at this point, that the checkerboard has different colored squares, and ask if they can refine the givens with this new information. The given information is finally refined to: (1) each domino covers a white square and a black square, and (2) there are 32 white squares and 30 black squares on this board.
I then have them prove this checkerboard cannot be covered with dominoes.
Proof:
Assume that the board can be covered with dominoes. Since dominoes cover one black and one white square, n dominoes will cover n black and n white squares. This statement is a contradiction to the given information that there are more white squares than black. Therefore, this board cannot be covered with dominoes. QED
The Prestige
The key to the follow-up is to get the students to drive it with lots of “what if” questions. In class this week, I got the following “what if” questions as follow-up.
“What if you tried this on a 7×7 board, where the opposite corners are different colors?”
“What if, instead of removing a square from two corners, you added a square to two corners?”
“What if you removed a square from each corner instead of just two?”
“What if the squares you removed were different colors but not corner pieces?”
“Will you be able to cover the board if you remove ANY two squares, so long as they are different colors?”
These are absolutely wonderful questions, and get at the heart of what a true mathematician does. They are able to see that the first question, about the 7×7 board, does not work based on the odd number of squares. They come to realize that the adding of two squares creates a problem that reduces to the one we just solved, which is a mathematical technique that requires a very sharp mind.
The last question is the trickiest. It is the most general, to be sure. For instance, the scenario below is one such instance of removing two squares and being able to tile with dominoes.
As to whether it is always possible to do? Well, I will tell you that my friend and I sat down and we have a proof as to whether or not it is always possible… but a magician never reveals all his secrets. [We will post the proof next week 🙂 ]
# A Profound Respect for 1 Million
By Chris Robinson (@isomorphic2crob), The Blake School
How long will it take for your heart to beat ONE MILLION times?
Before you read on, imagine that the ball just dropped, January 1st, 2015, when will your one millionth heart beat occur? Will it be later that day? Some point in January? In July? In 2016? 2017? How long does it really take to reach ONE MILLION heartbeats? Before doing any figuring, take a guess, then after you have committed to that conjecture, go ahead and calculate it.
This question was posed by Marjory Baruch (Syracuse University) to a group of high school math teachers attending PROMYS for Teachers at Boston University this summer. The initial conjectures for this group of math teachers ranged from a couple months to a couple years. All of us were astounded when we calculated the result to be slightly less than two week’s time.
The questions stuck with me as I began to prepare for my year at Blake, and so I decided to ask my class of Algebra 1B ninth graders, my Honors Geometry ninth graders, and my Honors Algebra II tenth graders. Their answers were a much wider range, including a student or two in each class that guessed less than a month. But not too many predicted it to take more than a couple of years. There was very little difference in the guessing from one class to the next leading me to believe from a very small sample size that students ability to estimate the size of large numbers was not necessarily connected with their current success in mathematics. There was however a difference between the teachers and the students.
Why do math teachers have such a profound respect for the size of 1 million? Maybe its related to our salaries…. But I digress.
Here’s Hank Green discussing a very similar idea (How long was it a million seconds ago?) on his YouTube channel.
I also had my students place the number 1 million on a number line from 0 to 1 billion, like the one shown below.
Students struggled to see how much larger 1 billion was compared to 1 million, with one student commenting that they always felt like millionaires and billionaires were the essentially the same.
OK, so what’s the big deal (get it? Big.)? I am not sure I have a punch line here, so I will leave you with some observations, and the hope that you find this intriguing as well.
• For most students, and perhaps most teachers, our initial impression of numbers beyond 10,000 take on the same amount of “bigness” until they are compared to other large numbers.
• Asking students to reflect on why they guessed what they did is a useful exercise, and having them write about it for a few minutes can help them develop habits of explaining their mathematical reasoning.
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# What Do The Stars Say About Scott Speedman? (01/14/2020)
How will Scott Speedman do on 01/14/2020 and the days ahead? Let’s use astrology to complete a simple analysis. Note this is not at all guaranteed – take it with a grain of salt. I will first find the destiny number for Scott Speedman, and then something similar to the life path number, which we will calculate for today (01/14/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology enthusiasts.
PATH NUMBER FOR 01/14/2020: We will analyze the month (01), the day (14) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 01 and add the digits together: 0 + 1 = 1 (super simple). Then do the day: from 14 we do 1 + 4 = 5. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 1 + 5 + 4 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the path number for 01/14/2020.
DESTINY NUMBER FOR Scott Speedman: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Scott Speedman we have the letters S (1), c (3), o (6), t (2), t (2), S (1), p (7), e (5), e (5), d (4), m (4), a (1) and n (5). Adding all of that up (yes, this can get tedious) gives 46. This still isn’t a single-digit number, so we will add its digits together again: 4 + 6 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the destiny number for Scott Speedman.
CONCLUSION: The difference between the path number for today (1) and destiny number for Scott Speedman (1) is 0. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But this is just a shallow analysis! As mentioned earlier, this is not scientifically verified. If you want a reading that people really swear by, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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Four congruent equilateral triangles are placed in a row. A line connects the bottom left vertex of the first triangle to the top vertex of the last triangle. If each equilateral triangle has an area equal to 6, what is the area above the line contained within the triangles, as shaded in yellow?
As usual, watch the video for a solution.
Triangles In A Row
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Answer To Triangles In A Row Puzzle
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
Connect the top vertices and construct an upside down equilateral triangle to the left of the first triangle. Suppose each equilateral triangle has a side length s.
Corresponding sides of the congruent equilateral triangles are parallel. Thus we have two sets of similar triangles formed by the top line, the lines slanting to the right (or left), and the halfway dividing line.
The lines slanting to down to the left are proportions of 3 equilateral triangles, so they are in proportions of 1, 2/3, and 1/3. Similarly the lines slanting down to the right are in proportions of 4 equilateral triangles and are in proportions of 1, 3/4, 2/4, 1/4.
The three yellow triangles are similar. The largest yellow triangle has two consecutive sides of s and (3/4)s, and the angle between those sides is 60°, so its area is:
(1/2)s(3/4)s(sin 60°)
= (1/2)s(3/4)s(√3)/2
= (3/4)(s2/4)√3
The area of a single equilateral triangle is 6 = s2(√3)/4. Thus the largest triangle has an area equal to:
(3/4)(s2/4)√3
= (3/4)(6)
= 4.5
The middle yellow triangle has sides that are 2/3 the length of the largest, so its area is (2/3)2 the largest yellow triangle’s area:
= (2/3)2(4.5)
= 2
The smallest yellow triangle has sides that are 1/3 the length of the largest, so its area is (1/3)2 the largest yellow triangle’s area:
= (1/3)2(4.5)
= 0.5
The total area in yellow is thus:
4.5 + 2 + 0.5 = 7
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# McGraw Hill Math Grade 4 Chapter 10 Lesson 7 Answer Key Problem Solving: Draw a Picture and Write an Equation
Practice the questions of McGraw Hill Math Grade 4 Answer Key PDF Chapter 10 Lesson 7 Problem Solving: Draw a Picture and Write an Equation to secure good marks & knowledge in the exams.
## McGraw-Hill Math Grade 4 Answer Key Chapter 10 Lesson 7 Problem Solving: Draw a Picture and Write an Equation
Solve
Question 1.
Elsa is knitting a blanket. She used 4$$\frac{3}{8}$$ meters of red yarn and 2$$\frac{4}{8}$$ meters of gold yarn. How many meters of yarn did she use altogether?
6$$\frac{7}{8}$$ meters
Explanation:
Elsa is knitting a blanket
She used 4$$\frac{3}{8}$$ meters of red yarn and
2$$\frac{4}{8}$$ meters of gold yarn
Add to find the total number of meters of yarn
4$$\frac{3}{8}$$ + 2$$\frac{4}{8}$$ = 6$$\frac{7}{8}$$ meters
So, Elsa used 6$$\frac{7}{8}$$ meters of yarn altogether.
Question 2.
Katia is building a tree house. She needs 52 boards. Each board has to be 4 meters long. How much wood does Katia need?
208 meters of wood
Explanation:
Katia is building a tree house
She needs 52 boards
Each board has to be 4 meters long
Multiply to find
52 x 4 = 208
So, Katia used 208 meters of wood.
Question 3.
Ms. Morton has 167$$\frac{3}{5}$$ cm of string. She cuts the string into two pieces. One piece is 72$$\frac{1}{5}$$ cm long. How long is the other piece?
Explanation:
Ms. Morton has 167$$\frac{3}{5}$$ cm of string
She cuts the string into two pieces
One piece is 72$$\frac{1}{5}$$ cm long
Subtract to find the length of the other piece
167$$\frac{3}{5}$$ – 72$$\frac{1}{5}$$ = 95$$\frac{2}{5}$$
So, the other piece is 95$$\frac{2}{5}$$.
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# *1. Understand the concept of a constant number like pi. Know the formula for the circumference and area of a circle.
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1 Students: 1. Students deepen their understanding of measurement of plane and solid shapes and use this understanding to solve problems. *1. Understand the concept of a constant number like pi. Know the formula for the circumference and area of a circle. Develop the concept of a constant number like pi Graph the relationship of circumference (y) to diameter(x). x y Ordered Pair C/d 1 cm 3.14cm (1, 3.14) 2 cm 6.28 cm (2, 6.28) 3 cm cm (3, 10.42) If you divide the circumference by the diameter what do you discover? What is the ratio of circumference to diameter? Why are all the points you graphed approximately on a straight line? How many segments x will fit on the circumference of the circle? ( FW) x 26
2 Know the formula for the circumference and area of a circle 2. Know common estimates of pi (3.14 or 22/7) and use these values to estimate and calculate the circumference and the area of circles; compare with actual measurements. Compare actual measurements with estimated value of pi Given the following information, calculate the circumference and area. r = 15 d = 5 r = 5 d= 10 Choose which approximation (3.14 or 22/7) is easier to use to find circumference. r = 7m d = 12m d = 21m To find area. r = 7 inches r = 16m d = 6cm Measure the diameter and circumference of several circular objects. Use these measurements to calculate pi. How do your measurements compare to the known approximate value of pi? If there is a difference, explain. 3. Know and use the formulas for the volume of triangular prisms and cylinders (area of base x height); compare and explain the similarity between these formulas and the formula for the volume of a rectangular solid. 27
3 Calculate the volume of a rectangular solid Calculate the volume of a triangular prism Given a length of 4 cm, width of 5 cm, and height of 6 cm, calculate the volume. What would happen if the width doubled? If the length doubled? The area of the triangle in a triangular prism is 35 cm 2. Find the volume if the height is 15 cm. Find the volume of a triangular prism with a height of 8 inches and a triangular base with these measurements Find the volume of the triangular prism. 9m 6 7m 5m 6m 28
4 Find the volumes (dimensions are cm). ( FW) Calculate the volume of a cylinder Given a radius of 2.5 cm and a height of 6 cm, find the volume of the cylinder. Given a diameter of 10 meters and a height of 8 meters, find the volume of the cylinder. Which increases the volume more, doubling its height or doubling its radius? Compare and explain the similarity between the volume of a cylinder, triangular prism, and a rectangular solid Using the formula V = Bh, compare and explain the similarities and differences in calculating the volume of a rectangular prism, triangular prism, and a cylinder. 29
5 Students: 2. Students identify and describe the properties of two-dimensional figures. 1. Identify angles as vertical, adjacent, complementary and/or supplementary and provide descriptions of these terms. Adjacent Are 1 and 2 adjacent? ABC and 1? A B 1 2 C Are angle 1 and angle 2 adjacent? 1 2 Vertical Name the vertical angles. Explain why they are called vertical. Name the adjacent angles
6 Supplementary Are these supplementary angles? Complementary Are these complementary angles? Line L is parallel to line M. Line P is perpendicular to L and M. Name the following. If none can be named, leave blank. ( FW) P a b c d L e g f M Complementary Supplementary Vertical Acute Right Obtuse 31
7 *2. Use the properties of complementary and supplementary angles and of the angles of a triangle to solve problems involving an unknown angle. Find the value of 1 in the following figures Line L is parallel to line M. Find the missing angles. ( FW) 65 b L d a c 100 M Find the missing angles. ( FW) a b d 3. Draw quadrilaterals and triangles given information about them (e.g., a quadrilateral having equal sides but no right angles, a right isosceles triangle). Name the triangle and draw a picture given the following information. A triangle with all sides congruent. A triangle with no sides congruent. A triangle that contains 1 right angle. 32
8 A triangle with two congruent sides and at least one right angle. A triangle that contains 1 obtuse angle. A triangle that has all acute angles. Match the name of the quadrilateral with its description. Draw each quadrilateral. A quadrilateral having equal sides but not right angles. A quadrilateral with exactly one pair of parallel sides. A quadrilateral with 90 degree angles and congruent sides. A quadrilateral with two pairs of opposite parallel sides. A quadrilateral with four 90 degree angles and two pairs of opposite congruent sides. Square Rectangle Trapezoid Rhombus Parallelogram 33
9 34
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# Mensuration – 2D : : Download Mathematics Study Notes Free PDF For REET/UTET Exam
## Study Notes on 2D Mensuration
Mathematics is an equally important section for REETMPTET & DSSSB Exams and has even more abundant importance in some other exams conducted by central or state govt. Generally, there are questions asked related to basic concepts and formulas of the Mensuration.
To let you make the most of Mathematics section, we are providing important facts related to the 2D Mensuration. At least 3-4 questions are asked from mensuration topic in most of the teaching exams. We wish you all the best of luck to come over the fear of the Mathematics section.
How to Overcome Exam Fever, Especially When You Fear Maths
### Rectangle
(a) Its diagonals are equal and bisect each other
(b) Area = L×B
(c) Diagonal d=
(d) (i) Area of a path inside a rectangular field:
Area of Path= 2x(l+b-2x)
(ii) Perimeter (p)= inner p+ outer p
= 2(l+b)+2(l+b-4x)
= 4(l+b-2x)
(e) Room as a Rectangular figure:
Area of 4 walls of a room= perimeter × height
= 2(l+b)×h
(f) Area of roof and 4 walls = 2H (L+B)+LB
Example: A rectangular piece is 40cm long and 30m wide from its four corners, quadrants of radii 3.5m have been cut. The area of remaining part is:
Sol. Area of Rectangle – 4 × area of quarter circle
### Square
(a) Area = a² or ×(diagonal)²
(b) Perimeter = 4a
(c) Diagonal =
(d) Area of path inside square = 4d(x-d)
X= length of square, d= length of path
(e) Area of path outside square = 4d (x+d)
(f) Area of path midway square = d(2x-d)
Example: If each side of square park is increased by 35%. Find the percentage change in its area?
Practice REET Previous Year Maths Quiz For REET Exam Here
### Parallelogram
(a) Area = base × height
(b) Perimeter= 2(a + b)
(c) d₁²+ d₂²= 2(a²+b²)
Example: A parallelogram PQRS has side PQ = 36cm and PS = 24cm. The distance between the sides PQ and RS is 16cm. Find the distance between the side PS and RQ.
Mathematics Study Notes For All Teaching Exams
### Rhombus:
`It is a quadrilateral whose all four sides are equal. Diagonal bisect each other at 90°`
Example: The perimeter of Rhombus is 60cm and the measure of an angle is 60° then the area is:
### Trapezium
It is a quadrilateral, whose any two opposite sides are parallel.
Practice Mensuration 2D Quiz Here for CTET Exam
### Triangle
A triangle is a polygon with three edges and three vertices.
Types of triangles:
• Scalene Triangle: A scalene triangle has all its sides of different lengths.
Example: The ratio of sides of a triangle is 4:5:6. If perimeter of triangle is 90 cm then the Area of triangle is
• Isosceles Triangle: An Isosceles triangle is a triangle with two equal sides also their opposite angles are equal.
• Right angle triangle: It is a triangle with an angle of 90° The sides a, b and c of such a triangle satisfy the pythagoras theorem.
• Equilateral triangle: It is a triangle whose all sides and angle are equal.
• ∠A=∠B=∠C= 60°
• If P₁, P₂, and P₃ are perpendicular lengths from any interior point (O) of an equilateral ∆ ABC to all its three sides respectively, then:
Example: Each side of an equilateral triangle is 16 cm. the area of triangle is:
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# First Condition of Equilibrium
We may say that an object at rest is in equilibrium or in static equilibrium. An object at rest is described by Newton's First Law of Motion. An object in static equilibrium has zero net force acting upon it.
The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as
F = F1+ F2+ F3+ F4+. . . = 0
where , the Greek letter sigma, again means the summation of whatever follows -- the summation of the forces, in this case.
That's all there is!
However, remember the following
=>> Ensure that you have included all the forces. This means carefully draw a free body diagram. Include gravity (the weight) and all contact forces.
=>> Remember that forces are vectors.
That means that the first condition of equilibrium,
F = 0
really means
Fx = 0 Fy = 0
Example: Consider the 98 newton weight (or 10 kg mass) supported by a rope.
The tension in the rope attached to the 98 newton weight is just 98 newtons.
But this rope is now tied together with two other ropes as shown here.
What forces are exerted by the other two ropes?
To answer this, look at the forces exerted on the knot where the three ropes are joined. The knot at rest so the sum of the forces acting on it must be zero.
Draw a free-body diagram showing just those forces.
T is the magnitude of the force on the knot from the rope attached directly to the weight. TL is the magnitude of the force on the knot from the rope on the left that makes an angle of 45° with the horizontal. And TR is the magnitude of the force on the knot from the rope on the right that makes an angle of 30° with the horizontal. These three forces must add to zero. Graphically, they can be added as shown here; the force vectors form a closed triangle.
Now we apply the first condition of equilibrium,
F = 0
which really means
Fx = 0 Fy = 0
First, resolve all the forces into their x- and y-components.
Fx = 0 Fx = - TL cos 45o + TR cos 30o = 0 - 0.707 TL + 0.866 TR = 0 0.866 T R = 0.707 TL Fy = 0 Fy = TL sin 45o + TR sin 30o - 98 N = 0 0.707 TL + 0.5 TR - 98 N = 0 0.707 TL + 0.5 TR = 98 N Now we can substitute, 0.707 TL + 0.5 TR = 98 N 0.866 T R + 0.5 TR = 98 N ( 0.866 + 0.5 ) TR = 98 N 1.366 TR = 98 N TR = 98 N / 1.366 TR = 71.7 N 0.866 T R = 0.707 TL TL = (0.866 / 0.707) TR TL = 1.22 TR TL = 1.22 ( 71.7 N ) TL = 87.8 N
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# RS Aggarwal Solutions for Class 8 Maths Chapter 11 - Compound Interest Exercise 11C
RS Aggarwal Solutions for the Exercise 11C of Class 8 Maths Chapter 11, Compound Interest is provided here. BYJU’S expert team have solved in such a way that learners can understand easily and comfortably. This exercise includes ten questions based on finding the compound interest quarterly and half-yearly using the formulae. By exercising the RS Aggarwal Solutions for class 8, students will be able to grasp the concepts entirely. And, to acquire good scores, they have to go through RS Aggarwal Solutions are given here.
## Download PDF of RS Aggarwal Solutions for Class 8 Maths Chapter 11 – Compound Interest Exercise 11C
### Access answers to Maths RS Aggarwal Solutions for Class 8 Chapter 11 – Compound Interest Exercise 11C
1. Find the amount and the compound interest on ₹ 8000 for 1 year at 10% per annum, compounded half-yearly.
Solution:
Given:
Present value= ₹ 8000
Interest rate= 10 % per annum
Time=1 year and compounded half yearly
To find the amount we have the formula,
Amount (A) = P (1+(R/100)2n
Where P is present value, r is rate of interest, n is time in years.
Now substituting the values in above formula we get,
∴ A = 8000 (1 + (10/2)/100)2
⇒ A = 8000 (1+5/100)2
⇒ A = 8000 (1+1/20)2
⇒ A = 8000 (21/20)2
⇒ A = 20 × 441
⇒ A = ₹ 8820
∴ Compound interest = A – P
= 8820 – 8000 = ₹ 820
2. Find the amount and the compound interest on ₹ 31250 for 1 ½ year at 8% per annum, compounded half-yearly.
Solution:
Given:
Present value= ₹ 31250
Interest rate= 8 % per annum
Time=1 ½ year= 3/2 year and compounded half yearly
To find the amount we have the formula,
Amount (A) = P (1+(R/100)2n
Where P is present value, r is rate of interest, n is time in years.
Now substituting the values in above formula we get,
∴ A = 31250 (1 + (8/2) /100)3
⇒ A = 31250 (1+4/100)3
⇒ A = 31250 (1+1/25)3
⇒ A = 31250 (26/25)3
⇒ A = 31250 × 17576/15625
⇒ A = ₹ 35152
∴ Compound interest = A – P
= 35452 – 31250 = ₹ 3902
3. Find the amount and the compound interest on ₹ 12800 for 1 year at 7 ½ % per annum, compounded half-yearly.
Solution:
Given:
Present value= ₹ 12800
Interest rate= 7 ½ % per annum
Time=1 year and compounded half yearly
To find the amount we have the formula,
Amount (A) = P (1+(R/100)2n
Where P is present value, r is rate of interest, n is time in years.
Now substituting the values in above formula we get,
∴ A = 12800 (1 + (15/4)/100)2
⇒ A = 12800 (1+3/80)2
⇒ A = 12800 (83/80)2
⇒ A = 128 × 6889/64
⇒ A = ₹ 13778
∴ Compound interest = A – P
= 13778 – 12800 = ₹ 978
4. Find the amount and the compound interest on ₹ 160000 for 2 years at 10 % per annum, compounded half-yearly.
Solution:
Given:
Present value= ₹ 160000
Interest rate= 10 % per annum
Time=2 years and compounded half yearly
To find the amount we have the formula,
Amount (A) = P (1+(R/100)2n
Where P is present value, r is rate of interest, n is time in years.
Now substituting the values in above formula we get,
∴ A = 160000 (1 + (10/2)/100)4
⇒ A = 160000 (1+5/100)4
⇒ A = 160000 (21/20)4
⇒ A = 160000 × 194481/160000
⇒ A = ₹ 194481
∴ Compound interest = A – P
= 194481 – 160000 = ₹ 34481
5. Swati borrowed ₹ 40960 from a bank to buy a piece of land. If the bank charges 12 ½ % per annum, compounded half yearly. What amount will she have to pay after 1 ½ years? Also, find the interest paid by her.
Solution:
Given: Present value= ₹ 40960
Interest rate= 12 ½ % per annum
Time=1 ½ = 3/2 years and compounded half yearly
To find the amount we have the formula,
Amount (A) = P (1+(R/100)2n
Where P is present value, r is rate of interest, n is time in years.
Now substituting the values in above formula we get,
∴ A = 40960 (1 + (25/4)/100)3
⇒ A = 40960 (1+1/16)3
⇒ A = 40960 (17/16)3
⇒ A = 40960 × 4913/4096
⇒ A = ₹ 49130
∴ Compound interest = A – P
= 49130 – 40960 = ₹ 8170
## RS Aggarwal Solutions for Class 8 Maths Chapter 11 – Compound Interest Exercise 11C
Exercise 11C of RS Aggarwal Solutions for Chapter 11, Compound Interest covers all the topics related to compound interest and methods of solving compound interest using the formulae. Some of the topics focused prior to exercise 11C include the following.
• Definition of compound interest
• To find compound interest when interest is compounded annually
• To find compound interest when interest is compounded half-yearly
• Calculating the compound interest by using formulae
• Applications of compound interest formula
The RS Aggarwal Solutions can help the students in practising and learning each and every concept as it provides solutions to all questions asked in the RS Aggarwal textbook.
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# Sums and Differences of Independent Random Variables
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Sums and Differences of Independent Random Variables
In this Concept, you will learn how to construct probability distributions of independent random variables, as well as find the mean and standard deviation for sums and differences of independent random variables.
### Watch This
For more information on linear transformations, or linear combinations, see (5.0) , see mrjaffesclass, Linear Combinations of Random Variables (6:41).
### Guidance
A probability distribution is the set of values that a random variable can take on. At this time, there are three ways that you can create probability distributions from data. Sometimes previously collected data, relative to the random variable that you are studying, can help to create a probability distribution. In addition to this method, a simulation is also a good way to create an approximate probability distribution. A probability distribution can also be constructed from the basic principles, assumptions, and rules of theoretical probability. The examples in this lesson will lead you to a better understanding of these rules of theoretical probability.
Sums and Differences of Independent Random Variables
#### Example A
Create a table that shows all the possible outcomes when two dice are rolled simultaneously. (Hint: There are 36 possible outcomes.)
$2^{nd}$ Die
1 2 3 4 5 6
1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 $1^{st}$ Die
4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
This table of possible outcomes when two dice are rolled simultaneously that is shown above can now be used to construct various probability distributions. The first table below displays the probabilities for all the possible sums of the two dice, and the second table shows the probabilities for each of the possible results for the larger of the two numbers produced by the dice.
Sum of Two Dice, $x$ Probability, $p(x)$
2 $\frac{1}{36}$
3 $\frac{2}{36}$
4 $\frac{3}{36}$
5 $\frac{4}{36}$
6 $\frac{5}{36}$
7 $\frac{6}{36}$
8 $\frac{5}{36}$
9 $\frac{4}{36}$
10 $\frac{3}{36}$
11 $\frac{2}{36}$
12 $\frac{1}{36}$
Total 1
Larger Number, $x$ Probability, $p(x)$
1 $\frac{1}{36}$
2 $\frac{3}{36}$
3 $\frac{5}{36}$
4 $\frac{7}{36}$
5 $\frac{9}{36}$
6 $\frac{11}{36}$
Total 1
When you roll the two dice, what is the probability that the sum is 4? By looking at the first table above, you can see that the probability is $\frac{3}{36}$ .
What is the probability that the larger number is 4? By looking at the second table above, you can see that the probability is $\frac{7}{36}$ .
#### Example B
The Regional Hospital has recently opened a new pulmonary unit and has released the following data on the proportion of silicosis cases caused by working in the coal mines. Suppose two silicosis patients are randomly selected from a large population with the disease.
Silicosis Cases Proportion
Worked in the mine 0.80
Did not work in the mine 0.20
There are four possible outcomes for the two patients. With ‘yes’ representing “worked in the mines” and ‘no’ representing “did not work in the mines”, the possibilities are as follows:
First Patient Second Patient
1 No No
2 Yes No
3 No Yes
4 Yes Yes
As stated previously, the patients for this survey have been randomly selected from a large population, and therefore, the outcomes are independent. The probability for each outcome can be calculated by multiplying the appropriate proportions as shown:
$P(\text{no for} \ 1^{\text{st}}) \bullet P(\text{no for} \ 2^{\text{nd}}) &= (0.2)(0.2)=0.04\\P(\text{yes for} \ 1^{\text{st}}) \bullet P(\text{no for} \ 2^{\text{nd}}) &= (0.8)(0.2)=0.16\\P(\text{no for} \ 1^{\text{st}}) \bullet P(\text{yes for} \ 2^{\text{nd}}) &= (0.2)(0.8)=0.16\\P(\text{yes for} \ 1^{\text{st}}) \bullet P(\text{yes for} \ 2^{\text{nd}}) &= (0.8)(0.8)=0.64$
If $X$ represents the number silicosis patients who worked in the mines in this random sample, then the first of these outcomes results in $x = 0$ , the second and third each result in $x = 1$ , and the fourth results in $x = 2$ . Because the second and third outcomes are disjoint, their probabilities can be added. The probability distribution for $X$ is given in the table below:
$x$ Probability, $p(x)$
0 0.04
1 $0.16 + 0.16 = 0.32$
2 0.64
#### Example C
Suppose an individual plays a gambling game where it is possible to lose $2.00, break even, win$6.00, or win $20.00 each time he plays. The probability distribution for each outcome is provided by the following table: Winnings, $x$ Probability, $p(x)$ $-$$2 0.30
$0 0.40$6 0.20
$20 0.10 The table can be used to calculate the expected value and the variance of this distribution: $\mu_{} &= \sum x_{}p_{}(x)\\\mu_{} &= (-2 \cdot 0.30)+(0 \cdot 0.40)+(6 \cdot 0.20)+(20 \cdot 0.10)\\\mu_{} &= 2.6$ Thus, the player can expect to win$2.60 playing this game.
The variance of this distribution can be calculated as shown:
$\sigma{_{}}^2 &= \sum (x_{}-\mu_{})^2 p(x)\\\sigma{_{}}^2 &= (-2-2.6)^2 (0.30)+(0-2.6)^2 (0.40)+(6-2.6)^2 (0.20)+(20-2.6)^2 (0.10)\\\sigma{_{}}^2 & \approx 41.64\\\sigma_{} & \approx \sqrt{41.64} \approx \ 6.45$
#### Example D
The following probability distribution was constructed from the results of a survey at the local university. The random variable is the number of fast food meals purchased by a student during the preceding year (12 months). For this distribution, calculate the expected value and the standard deviation.
Number of Meals Purchased Within 12 Months, $x$ Probability, $p(x)$
0 0.04
$[1 - 6)$ 0.30
$[6 - 11)$ 0.29
$[11 - 21)$ 0.17
$[21 - 51)$ 0.15
$[51 - 60)$ 0.05
Total 1.00
You must begin by estimating a mean for each interval, and this can be done by finding the center of each interval. For the first interval of $[1 - 6)$ , 6 is not included in the interval, so a value of 3 would be the center. This same procedure can be used to estimate the mean of all the intervals. Therefore, the expected value can be calculated as follows:
$\mu_{} &= \sum x_{}p_{}(x)\\\mu_{} &= (0)(0.04)+(3)(0.30)+(8)(0.29)+(15.5)(0.17)+(35.5)(0.15)+(55)(0.05)\\\mu_{} &= 13.93$
Likewise, the standard deviation can be calculated:
$\sigma^2 &= \sum (x_{}-\mu_{})^2 p_{}(x)\\&= (0-13.93)^2 (0.04)+(3-13.93)^2 (0.30)\\& \quad +(8-13.93)^2(0.29)+(15.5-13.93)^2(0.17)\\& \quad +(35.5-13.93)^2(0.15)+(55-13.93)^2(0.05)\\& \approx 208.3451$
$\sigma_{} \approx 14.43$
Thus, the expected number of fast food meals purchased by a student at the local university is 13.93, and the standard deviation is 14.43. Note that the mean should not be rounded, since it does not have to be one of the values in the distribution. You should also notice that the standard deviation is very close to the expected value. This means that the distribution will be skewed to the right and have a long tail toward the larger numbers.
Technology Note: Calculating mean and variance for probability distribution on TI-83/84 Calculator
Notice that the mean, which is denoted by $\overline{x}$ in this case, is 13.93, and the standard deviation, which is denoted by $\sigma_x$ , is approximately 14.43.
Linear Transformations of $X$ on Mean of $X$ and Standard Deviation of $X$
If you add the same value to all the numbers of a data set, the shape and standard deviation of the data set remain the same, but the value is added to the mean. This is referred to as re-centering the data set. Likewise, if you rescale the data, or multiply all the data values by the same nonzero number, the basic shape will not change, but the mean and the standard deviation will each be a multiple of this number. (Note that the standard deviation must actually be multiplied by the absolute value of the number.) If you multiply the numbers of a data set by a constant $d$ and then add a constant $c$ , the mean and the standard deviation of the transformed values are expressed as follows:
$\mu_{c+dX} &= c+d \mu_{X}\\\sigma_{c+dX} &= |d| \sigma_{X}$
These are called linear transformations , and the implications of this can be better understood if you return to the casino example.
#### Example E
The casino has decided to triple the prizes for the game being played. What are the expected winnings for a person who plays one game? What is the standard deviation? Recall that the expected value was $2.60, and the standard deviation was$6.45.
Solution:
The simplest way to calculate the expected value of the tripled prize is (3)($2.60), or$7.80, with a standard deviation of (3)($6.45), or$19.35. Here, $c = 0$ and $d = 3$ . Another method of calculating the expected value and standard deviation would be to create a new table for the tripled prize:
Winnings, $x$ Probability, $p$
$-$ $6 0.30$0 0.40
$18 0.20$60 0.10
The calculations can be done using the formulas or by using a graphing calculator. Notice that the results are the same either way.
This same problem can be changed again in order to introduce the Addition Rule and the Subtraction Rule for random variables. Suppose the casino wants to encourage customers to play more, so it begins demanding that customers play the game in sets of three. What are the expected value (total winnings) and standard deviation now?
Let $X, Y$ and $Z$ represent the total winnings on each game played. If this is the case, then $\mu_{X+Y+Z}$ is the expected value of the total winnings when three games are played. The expected value of the total winnings for playing one game was $2.60, so for three games the expected value is: $\mu_{X+Y+Z} &= \mu_X+\mu_Y +\mu_Z\\\mu_{X+Y+Z} &= \ 2.60 + \ 2.60 + \ 2.60\\\mu_{X+Y+Z} &= \ 7.80$ Thus, the expected value is the same as that for the tripled prize. Since the winnings on the three games played are independent, the standard deviation of $X, Y$ and $Z$ can be calculated as shown below: $\sigma{^2}_{X+Y+Z} &= \sigma{^2}_X + \sigma{^2}_Y + \sigma{^2}_Z\\\sigma{^2}_{X+Y+Z} &= 6.45^2 + 6.45^2 + 6.45^2\\\sigma{^2}_{X+Y+Z} &\approx 124.8075\\\sigma_{X+Y+Z} &\approx \sqrt{124.8075}\\\sigma_{X+Y+Z} &\approx 11.17$ This means that the person playing the three games can expect to win$7.80 with a standard deviation of $11.17. Note that when the prize was tripled, there was a greater standard deviation ($19.36) than when the person played three games ($11.17). The Addition and Subtraction Rules for random variables are as follows: If $X$ and $Y$ are random variables, then: $\mu_{X+Y} &= \mu_X + \mu_Y\\\mu_{X-Y} &= \mu_X - \mu_Y$ If $X$ and $Y$ are independent, then: $\sigma{^2}_{X+Y} &= \sigma{^2}_X+\sigma{^2}_Y\\\sigma{^2}_{X-Y} &= \sigma{^2}_X+\sigma{^2}_Y$ Variances are added for both the sum and difference of two independent random variables, because the variation in each variable contributes to the overall variation in both cases. (Subtracting is the same as adding the opposite.) Suppose you have two dice, one die, $X$ , with the usual positive numbers 1 through 6, and another, $Y$ , with the negative numbers $-1$ through $-6$ . Next, suppose you perform two experiments. In the first, you roll the first die, $X$ , and then the second die, $Y$ , and you compute the difference of the two rolls. In the second experiment, you roll the first die and then the second die, and you calculate the sum of the two rolls. $\mu_X &= \sum x_{}p_{}(x) && \mu_Y = \sum y_{}p_{}(y)\\\mu_X &= 3.5 && \mu_Y=-3.5\\\sigma{^2}_X & \approx \sum (x_{}-\mu_X)^2 p_{}(x) && \sigma{^2}_Y \approx \sum (y_{}-\mu_Y)^2 p_{}(y)\\\sigma{^2}_X & \approx 2.917 && \sigma{^2}_Y \approx 2.917\\\mu_{X-Y}&=\mu_X - \mu_Y && \mu_{X+Y} = \mu_X+\mu_Y\\\mu_{X-Y} &= 3.5 - (-3.5)=7 && \mu_{X+Y} = 3.5 + (-3.5)=0\\\sigma{^2}_{X-Y}&=\sigma{^2}_X+\sigma{^2}_Y && \sigma{^2}_{X+Y} = \sigma{^2}_X+\sigma{^2}_Y\\\sigma{^2}_{X-Y} &\approx 2.917 + 2.917 = 5.834 && \sigma{^2}_{X+Y} \approx 2.917 + 2.917 = 5.834$ Notice how the expected values and the variances for the two dice combine in these two experiments. ### Vocabulary A chance process can be displayed as a probability distribution that describes all the possible outcomes, $x$ . You can also determine the probability of any set of possible outcomes. A probability distribution table for a random variable, $X$ , consists of a table with all the possible outcomes, along with the probability associated with each of the outcomes. The expected value and the variance of a probability distribution can be calculated using the following formulas: $E(x) &= \mu_X=\sum x_{}p_{}(x)\\\sigma{^2}_X &= \sum (x_{}-\mu _X)^2 p_{}(x)$ For the random variables $X$ and $Y$ and constants $c$ and $d$ , the mean and the standard deviation of a linear transformation are given by the following: $\mu_{c+dX} &= c+d \mu_X\\\sigma_{c+dX} &= |d| \sigma_X$ If the random variables $X$ and $Y$ are added or subtracted, the mean is calculated as shown below: $\mu_{X+Y} &= \mu_X+\mu_Y\\\mu_{X-Y} &= \mu_X-\mu_Y$ If $X$ and $Y$ are independent, then the following formulas can be used to compute the variance : $\sigma{^2}_{X+Y} &= \sigma{^2}_X+\sigma{^2}_Y\\\sigma{^2}_{X-Y} &= \sigma{^2}_X + \sigma{^2}_Y$ ### Guided Practice Beth earns$25.00 an hour for tutoring but spends $20.00 an hour for piano lessons. She saves the difference between her earnings for tutoring and the cost of the piano lessons. The numbers of hours she spends on each activity in one week vary independently according to the probability distributions shown below. Determine her expected weekly savings and the standard deviation of these savings. Hours of Piano Lessons, $x$ Probability, $p(x)$ 0 0.3 1 0.3 2 0.4 Hours of Tutoring, $y$ Probability, $p(y)$ 1 0.2 2 0.3 3 0.2 4 0.3 Solution: $X$ represents the number of hours per week taking piano lessons, and $Y$ represents the number of hours tutoring per week. The mean and standard deviation for each can be calculated as follows: $E(x) &= \mu_X = \sum x_{}p_{}(x) && \sigma{^2}_X = \sum (x_{}-\mu_X)^2 p{}(x)\\\mu_X &= (0)(0.3)+(1)(0.3)+(2)(0.4) && \sigma{^2}_X = (0-1.1)^2 (0.3)+(1-1.1)^2(0.3)+(2-1.1)^2(0.4)\\\mu_X &= 1.1 && \sigma{^2}_X = 0.69\\&&& \sigma_X = 0.831$ $E(y) &= \mu_Y = \sum y_{}p_{}(y) && \sigma{^2}_Y = \sum (y_{}-\mu_Y)^2p_{}(y)\\\mu_Y &= (1)(0.2)+(2)(0.3)+(3)(0.2)+(4)(0.3) && \sigma{^2}_Y = (1-2.6)^2 (0.2)+(2-2.6)^2(0.3)+(3-2.6)^2(0.2)\\&&& +(4-2.6)^2(0.3)\\\mu_Y &= 2.6 && \sigma{^2}_Y = 1.24\\&&& \sigma_Y = 1.11$ The expected number of hours Beth spends on piano lessons is 1.1 with a standard deviation of 0.831 hours. Likewise, the expected number of hours Beth spends tutoring is 2.6 with a standard deviation of 1.11 hours. Beth spends$20 for each hour of piano lessons, so her mean weekly cost for piano lessons can be calculated with the Linear Transformation Rule as shown:
$\mu_{20 X}=(20)(\mu_X)=(20)(1.1)=\ 22$ by the Linear Transformation Rule.
Beth earns $25 for each hour of tutoring, so her mean weekly earnings from tutoring are as follows: $\mu_{25 Y}=(25)(\mu_Y)=(25)(2.6)=\ 65$ by the Linear Transformation Rule. Thus, Beth's expected weekly savings are: $\mu_{25 Y}-\mu_{20 X}=\ 65 - \ 22 = \ 43$ by the Subtraction Rule. The standard deviation of the cost of her piano lessons is: $\sigma_{20 X}=(20)(0.831)=\ 16.62$ by the Linear Transformation Rule. The standard deviation of her earnings from tutoring is: $\sigma_{25 Y}=(25)(1.11)=\ 27.75$ by the Linear Transformation Rule. Finally, the variance and standard deviation of her weekly savings is: $\sigma{^2}_{25Y-20X} &= \sigma{^2}_{25 Y}+\sigma{^2}_{20 X}=(27.75)^2+(16.62)^2=1046.2896\\\sigma_{25Y-20X} &\approx \ 32.35$ ### Practice 1. Find the expected value for the sum of two fair dice. 2. Find the standard deviation for the sum of two fair dice. 3. It is estimated that 70% of the students attending a school in a rural area take the bus to school. Suppose you randomly select three students from the population. Construct the probability distribution of the random variable, $X$ , defined as the number of students who take the bus to school. (Hint: Begin by listing all of the possible outcomes.) 4. The Safe Grad Committee at a high school is selling raffle tickets on a Christmas Basket filled with gifts and gift cards. The prize is valued at$1200, and the committee has decided to sell only 500 tickets. What is the expected value of a ticket? If the students decide to sell tickets on three monetary prizes – one valued at $1500 dollars and two valued at$500 each, what is the expected value of the ticket now?
5. A recent law has been passed banning the use of hand-held cell phones while driving, and a survey has revealed that 76% of drivers now refrain from using their cell phones while driving. Three drivers were randomly selected, and a probability distribution table was constructed to record the outcomes. Let $N$ represent those drivers who never use their cell phones while driving and $S$ represent those who do use their cell phones while driving. Calculate the expected value and the variance using your calculator.
6. True or False? If $X$ and $Y$ are random variables then $X^2+Y^3$ is a random variable.
7. Are these concepts applicable to real-life situations?
8. Will knowing these concepts allow you estimate information about a population?
9. Suppose you have a six-sided fair die. Let the random variable X be the number that shows when you roll the die one time. Suppose in addition you have a fair four sided die with the numbers 1, 2, 3, 3. Let the random variable Y be the number that appears when you roll this die one time. Define a third random variable Z = X + Y.
1. Write the probability distribution for X.
2. Write the probability distribution for Y.
3. Write the probability distribution for Z.
10. Suppose in a box there are 4 tickets. Each ticket has two numbers on it. Ticket one has the numbers 1 and 2; ticket two has the numbers 1 and 3; ticket three has the numbers 5 and 7; and ticket four has the numbers 4 and 2. Define the following two random variables: is the first number on the ticket and is the second number on the ticket.
1. What is the probability that if you draw a ticket from random from the box that the first number will be a 1?
2. Define a new random variable $2X+3Y$ . What is the probability that this new random variable with have a value of 11?
11. Consider the following distributions:
$& X && 1 && 2\\& P(X) && 1/3 && 2/3\\& Y && 0 && 1\\& P(Y) && 1/2 && 1/2$
Find the probability distribution for the new random variable $Z=3X+4Y$ .
1. Suppose there are six numbers in a box: 1, 2, 3, 4, 5, 6. You draw two numbers out of the box, without replacement. Find the distribution table for the random variable $S=X_1+X_2$ where $X_1$ represents the first number drawn and $X_2$ represents the second number drawn.
Keywords
Expected value
Linear Transformation Rule
Linear transformations
Standard deviation
Subtraction Rule
Variance
### Vocabulary Language: English
expected value
expected value
The expected value is the return or cost you can expect on average, given many trials.
mean
mean
The mean, often called the average, of a numerical set of data is simply the sum of the data values divided by the number of values.
mean of a linear transformation
mean of a linear transformation
the mean of a linear transformation is given by the following: uc+dx = c+dux
random variables
random variables
Random variables are quantities that take on different values depending on chance, or probability.
standard deviation of a linear transformation
standard deviation of a linear transformation
The standard deviation of a linear transformation has the form: $\sigma_{c+dX} = |d| \sigma_X$
Transformation
Transformation
The transformation of a random variable describes the effect of performing operations (+, -, *, /) on the random variable.
variance
variance
A measure of the spread of the data set equal to the mean of the squared variations of each data value from the mean of the data set.
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The Radius of Convergence of a Power Series Examples 1
# The Radius of Convergence of a Power Series Examples 1
Recall from The Radius of Convergence of a Power Series page that we can calculate the radius of convergence of a power series using the ratio test, that is if $\lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = L$, then the radius of convergence is $R = \frac{1}{L}$. If $L = 0$ then the radius of convergence $R = \infty$ and if $L = \infty$ then the radius of convergence $R = 0$.
We will now look at some more examples of determining the radius of convergence of a given power series.
## Example 1
Determine the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{x^n}{n!}$.
We first notice that $a_n = \frac{1}{n!}$ in our power series, and applying the rule above we have that
(1)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{n!}{(n+1)!} = \lim_{n \to \infty} \frac{1}{n+1} = 0 \end{align}
Therefore the radius of convergence of this power series is $R = \infty$.
## Example 2
Determine the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{(3x + 4)^n}{(n^3 + 2)3^n}$.
Let's first rewrite this power series as $\sum_{n=0}^{\infty} \frac{(3x + 4)^n}{(n^3 + 2)3^n} = \sum_{n=0}^{\infty} \frac{1}{n^3 + 2} \left ( \frac{3x + 4}{3} \right )^n = \sum_{n=0}^{\infty} \frac{1}{n^3 + 2} \left ( x + \frac{4}{3} \right )^n$.
Now we note that $a_n = \frac{1}{n^3 + 2}$, and using the ratio test:
(2)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_{n}} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{\frac{1}{(n+1)^3 + 2}}{\frac{1}{n^3 + 2}} \biggr \rvert = \lim_{n \to \infty} \frac{n^3 + 2}{(n+1)^3 + 2} = 1 \end{align}
So the radius of convergence is $R = \frac{1}{L} = 1$.
## Example 3
Determine the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n!(5x + 3)^n}{(n+1)^2 + 4n}$.
Let's first rewrite this series as $\sum_{n=0}^{\infty} \frac{n!(5x + 3)^n}{(n+1)^2 + 4n} = \sum_{n=0}^{\infty} \frac{n!}{(n+1)^2 + 4n} (5x + 3)^n = \sum_{n=0}^{\infty}\frac{n!}{(n+1)^2 + 4n} \left (5 \left (x + \frac{3}{5} \right ) \right)^n = \sum_{n=0}^{\infty}\frac{5^nn!}{(n+1)^2 + 4n} \left (x + \frac{3}{5} \right)^n$.
Now we can see that $a_n = \frac{5^nn!}{(n+1)^2 + 4n}$. Using the ratio test to find the radius of convergence we have:
(3)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{\frac{5^{n+1}(n+1)!}{(n+2)^2 + 4(n+1)}}{\frac{5^nn!}{(n+1)^2 + 4n}} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{5^{n+1}(n+1)!\left [ (n+1)^2 +4n \right ]}{\left [ (n+2)^2 + 4(n+1)\right] 5^n n!} \biggr \rvert = \lim_{n \to \infty} \frac{5(n+1)[(n+1)^2 + 4n]}{[(n+2)^2 + 4(n+1)]} = \infty \end{align}
Therefore the radius of convergence $R = 0$.
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## How do you explain Bodmas rule?
The BODMAS rule states we should calculate the Brackets first (2 + 4 = 6), then the Orders (52 = 25), then any Division or Multiplication (3 x 6 (the answer to the brackets) = 18), and finally any Addition or Subtraction (18 + 25 = 43). Children can get the wrong answer of 35 by working from left to right.
### When did schools start using Bodmas?
Order of operations eg BODMAS was introduced in 1800s.
What is the remainder when 235 is divided by 5?
Thus, the remainder is $1 \times 3 = 3$. Hence, the remainder for $\dfrac{{{2^{35}}}}{5} = 3$. Note:As you can see in solving this example, we have used the concept of negative remainder.
Can you use Bodmas without brackets?
Ans: Yes, we use the BODMAS rule to get the correct answer even if there are no brackets. If there are no brackets, start solving from ‘order’ or ‘of’ followed by Division or multiplication (whatever comes first from left to right) then by addition or subtraction (whatever comes first from left to right).
## Why is multiplication done before addition?
Students should have answered something in their own words that gets across the concept: Multiplication and division are done before addition and subtraction in order to convert groups of items into subtotals of like items that can be combined for the total.
### Is Bodmas taught in UK schools?
There are conventions about the order of operations to try to resolve this, sometimes called BODMAS in UK schools. “If it still seems unclear, it’s best to include brackets to remove any possible ambiguity.
Is Bodmas American?
Most common in the UK, Pakistan, India, Bangladesh and Australia and some other English-speaking countries is BODMAS meaning either Brackets, Order, Division/Multiplication, Addition/Subtraction or Brackets, Of/Division/Multiplication, Addition/Subtraction. Nigeria and some other West African countries also use BODMAS.
What is Bodmas in maths for kids?
BODMAS rule is an acronym to help children to remember the order of operations in calculations. Operations are simply the different things that we can do to numbers in maths. It stands for, ‘Brackets, Order, Division, Multiplication, Addition, Subtraction. This is also the same for addition and subtraction.
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# Does A Quadratic Equation Have Equal Roots?
## Are there two distinct real roots?
If the discriminant is greater than zero, this means that the quadratic equation has two real, distinct (different) roots.
If the discriminant is equal to zero, this means that the quadratic equation has two real, identical roots..
## Which of the following has no real roots?
Hence the equation has real roots. Hence, the equation has real roots. Hence x2-4x+3√2=0 has no real roots. …
## How do you know if an equation has no real roots?
Explanation: If a quadratic equation has no real roots then its discriminant is less than zero. The discriminant of general quadratic equation ax2+bx+c=0 is b2−4ac .
## How do you know if roots are equal or unequal?
If Δ=0, the roots are equal and we can say that there is only one root. If Δ>0, the roots are unequal and there are two further possibilities. Δ is the square of a rational number: the roots are rational. Δ is not the square of a rational number: the roots are irrational and can be expressed in decimal or surd form.
## What are two equal roots?
Equal or double roots. If the discriminant b2 – 4ac equals zero, the radical in the quadratic formula becomes zero. In this case the roots are equal; such roots are sometimes called double roots.
## What are real and equal roots?
When a, b and c are real numbers, a ≠ 0 and discriminant is zero (i.e., b2 – 4ac = 0), then the roots α and β of the quadratic equation ax2 + bx + c = 0 are real and equal.
## Which equation has equal roots?
quadratic equationb^{2} – 4ac is called the discriminant of the equation and a quadratic equation has equal roots if and only if discriminant is zero.
## How do you find the other root when one is given?
8. Hint: Here, one root is given for the quadratic equation x2−5x+6=0. Now, we can find the other root by the formula for sum and product of the roots. If α and β are the two roots of the quadratic equation ax2+bx+c=0 then the sum and product of the roots are given by the formula: α+β=−ba and αβ=ca.
## Are roots and zeros the same?
A zero is of a function. A root is of an equation. … You could ask what are the roots of f(x) = 10, which would be the same as asking to solve 10 = x + 5. So, the zeros are only a specific kind of root for when the function equals 0.
## Can a quadratic equation have more than two roots?
We will discuss here that a quadratic equation cannot have more than two roots. Proof: Let us assumed that α, β and γ be three distinct roots of the quadratic equation of the general form ax2 + bx + c = 0, where a, b, c are three real numbers and a ≠ 0. … Hence, every quadratic equation cannot have more than 2 roots.
## Can real roots be negative?
Negative numbers don’t have real square roots since a square is either positive or 0. The square roots of numbers that are not a perfect square are members of the irrational numbers. This means that they can’t be written as the quotient of two integers.
## What does it mean if a quadratic equation has equal roots?
A quadratic equation has equal roots iff its discriminant is zero. A quadratic equation has equal roots iff these roots are both equal to the root of the derivative.
## What is real roots in quadratic equation?
A real number x will be called a solution or a root if it satisfies the equation, meaning . It is easy to see that the roots are exactly the x-intercepts of the quadratic function. , that is the intersection between the graph of the quadratic function with the x-axis.
## Which of the following has two distinct roots?
A quadratic equation, ax2 + bx + c = 0; a ≠ 0 will have two distinct real roots if its discriminant, D = b2 – 4ac > 0. Hence, the equation x2 –3x + 4 = 0 has no real roots. Hence, the equation 2×2 + x – 1 = 0 has two distinct real roots.
## What are real and non real roots?
Sometimes the quadratic factors do not factor into linear factors of the reals, but if you extend the numbers to the complexes, then they do. In other words, non-real roots imply that your polynomial cannot be factored into linear factors over the reals…you will inevitably have quadratic factors.
## Is 0 a real number?
Real numbers consist of zero (0), the positive and negative integers (-3, -1, 2, 4), and all the fractional and decimal values in between (0.4, 3.1415927, 1/2). Real numbers are divided into rational and irrational numbers.
## Can a quadratic equation have one root?
Thus, a parabola has exactly one real root when the vertex of the parabola lies right on the x-axis. The simplest example of a quadratic function that has only one real root is, y = x2, where the real root is x = 0.
## How can you tell how many solutions a discriminant has?
The discriminant can be positive, zero, or negative, and this determines how many solutions there are to the given quadratic equation.A positive discriminant indicates that the quadratic has two distinct real number solutions.A discriminant of zero indicates that the quadratic has a repeated real number solution.More items…
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# Physics
### Projectiles
In this tutorial, you will learn the following:
• Projectiles
• Vertical and horizontal motion of projectiles
• Interactive simulation for you to see how the vertical and horizontal velocities change with time
• Four different scenarios of projectiles with worked examples
• Projectile motion involving unit vectors - i and j
• Projectile motion formulae - with proof
• How to maximize the horizontal range of a projectile
Projectile
An object that moves under gravity in the air is a projectile.
E.g.
• › A cricket ball that moves through the air, after being struck by a bat
• › A baseball that moves through the air, after being struck by a baseball bat
• › A football that moves in the air after being kicked
• › A dart that moves through the air
• › An arrow that moves through the air, having been fired by an archer
Suppose an object is projected at an angle of t to the horizontal at the speed of v. Since it moves both vertically and horizontally, the initial velocity has components both vertically and horizontally: the vertical component takes it up and the horizontal component takes it horizontally. The following image illustrates it:
Since there is no horizontal force at work in the absence of air resistance, according to F = ma, the acceleration is zero. So, the horizontal velocity remains constant throughout the motion. The vertical velocity, however, goes down, as the gravitational pull acts in the opposite direction - downwards. When the projectile reaches the peak, it becomes zero and then changes the direction and increases as it goes down.
You can practise the changes in velocity of a projectile interactively with the following applet; just move the slider - time - and see the changes:
Maximum Height
This is he maximum vertical height reached by the projectile.
Time of flight
This is the time during which the object has been in the air.
Maximum Horizontal Range
This is he maximum horizontal distance travelled by the projectile.
The following worked examples cover the above in detail.
#### Projectiles - Problem Solving
Throughout these examples, the acceleration due to gravity, g, is assumed to be 10ms-2 to make calculations smooth.
E.g.1
A ball is projected at 20√2 m/s, at an angle of 450 to the horizontal. Find the following, by assuming that g = 10ms-2:
1. The maximum height
2. The time of flight
3. The maximum horizontal range
1. For vertical Motion, until the objects comes to an instant halt
↑ v2 = u2 + 2as
u = v sin(45) = 20√2 &1/√2 = 20; a = -10; s = h
0 = 400 - 2x10xs
h = 20m.
2. When the object reaches the highest point,
↑ v = u + at
u = v sin(45) = 20√2 &1/√2 = 20; a = -10; v = 0
0 = 20 - 10t
t = 2s
The time of flight = 2x2 = 4s.
3. → s = ut + 1/2 at2 u = 20√2 cos(45) = 20 x 1/√2 = 20; a = 0; t = 4
s = 20x4 + 0
s = 80m.
E.g.2
A ball is projected at 20 m/s, horizontally from a tower of height 20m. Find the following, by assuming that g = 10ms-2:
1. The time of flight
2. The maximum horizontal range
3. The velocity after 1 second
4. The velocity at which it hits the ground
Since it was projected horizontally, there would be no vertical speed at the beginning.
1. For vertical Motion, until it hits the ground,
↓ s = ut + 1/2 at2
u = 0; a = 10; s = 20
20 = 0 + 5t2
t2 = 4 => t = ±2
t = 2s.
2. When the object reaches the ground,
→ s = ut + 1/2 at2
u = 20; a = 0; t = 2
s = 20x2 + 0
s = 40m
3. After 1s, it stats moving downward; it has a vertical velocity and a horizontal velocity, which is constant.
↓ v = u + at
u = 0; a = 10; t = 1
v = 0 + 1x10
v = 10ms-1.
→ v = u + at
u = 20; a = 0; t = 1
v = 20 + 0
v = 20ms-1.
So, the resultant velocity = √(102 + 202)
↘ v = √500 = 10√5 m/s.
4. When it hits the ground too, it has a vertical velocity and a horizontal velocity, which is constant.
↓ v = u + at
u = 0; a = 10; t = 2
v = 0 + 2x10
v = 20ms-1.
→ v = u + at
u = 20; a = 0; t = 2
v = 20 + 0
v = 20ms-1.
So, the resultant velocity = √(202 + 202)
↘ v = √800 = 20√2 m/s.
E.g.3
A ball is projected at 20√2, at an angle of 450 to the horizontal, from the top of a tower of height 60m, upwards. Find the following, by assuming that g = 10ms-2:
1. The time of flight
2. The maximum horizontal range
3. The maximum height reached by the object
4. The velocity at which it hits the ground
When the object reaches the level of the top of the tower again, its displacement is zero; below that level, it is negative.
1. For vertical Motion, until it reaches the ground,
↑ s = ut + 1/2 at2
u = 20√2 sin(45) = 20; a = -10; s = -60
-60 = 20t - 5t2
5t2 - 20t - 60 = 0 => t² -4t - 12 = 0 => t = 6 or t = -2
t = 6s.
2. When the object reaches the ground,
→ s = ut + 1/2 at2
u = 20√2 cos(45) = 20; a = 0; t = 6
s = 20x6 + 0
s = 120m
3. ↑ v2 = u2 - 2as
u = 20; a = -10; v = 0
0 = 400 - 2x10xs
s = 20m
Total height = 20 + 60 = 80m.
4. ↓ v = u + at
u = 20; a = -10; v = ? t = 6
v = 20 - 10x6 = -40m/s
→ since horizontal velocity remains the same, v = 20
So, the resultant velocity = √(402 + 202)
↘ v = √2000 = 20√5 m/s.
E.g.4
A ball is projected at 12 m/s at an angle, 20° to the horizontal from a tower of height 10m. Find the following, by assuming that g = 10ms-2:
1. The time of flight
2. The maximum horizontal range
3. The velocity at which it hits the ground
1. ↓ u = 4.1, s = 10, t = ?, a = 10
s = ut + 1/2 at²
10 = 4.1t + 1/2 x 10 x t²
10 = 4.1t + 5t²
5t² + 4.1t - 10 = 0
t = 1.06s or t = -1.88s
Time of flight = 1.06s
2. → u = 11.28, a = 0, t = 1.06, s = ?
s = ut + 1/2 at²
s = 11.28 x 1.06
s = 11.95m
3. ↓ u = 4.1, a = 10, t = 1.06 v = ?
v = 4.1 + 10 x 1.06
v = 14.7 m/s
→ the horizontal velocity remains the same.
↘ v = √(14.7² + 11.27²)
v = 18.52 m/s
E.g.5
A ball is projected at 5i + 20j from a tower of height 5m Find the following, by assuming that g = 10ms-2:
1. The time of flight
2. The maximum height
3. The maximum horizontal range
4. The velocity at which it hits the ground
5. The angle of between the ball and the horizontal, when it hits the ground
1. ↑ u = 20, s = -5 t = ?, a = -10
s = ut + 1/2 at²
-5= 20t - 1/2 x 10 x t²
-5 = 20t - 5t²
5t² - 20t - 5 = 0
t = 4.2s or t = -0.2s
Time of flight = 4.2s
2. ↑ u = 20, a = -10, v = 0, s = ?
v² = u² + 2as
0 = 400 - 20s
s = 20m
Maximum height = 5 + 20 = 25m
3. → u = 5, a = 0, t = 4.2 s = ?
s = 5 x 4.2
s = 21m
4. → Since there is no horizontal acceleration, the horizontal velocity remains the same - 5i
↑ u = 20, a = -10, s = -5, v = ?
v² = u² + 2as
v² = 400 + 2(-10)(-5)
v² = 500
v = 22.4 m/s
↘ v = √(22.4² + 5²)
v = 22.9 m/s
5. ↘ tan θ = 22.4/5 = 4.48
θ = 1.35°
#### Formulae of Projectile Motion
These are the main formulae involving the projectile motion. You, however, do not have to remember them! Just try to derive them using the SUVAT; it makes the learning process easier.
↑ u = vsinθ, v = 0, a=-g, s = H, t = T/2
v = u + at
0 = vsinθ + -gT/2
gT/2 = vsinθ
T = 2vsinθ/g
↑ u = vsinθ, v = 0, a=-g, s=H
v² = u² + 2as
0 = v²sin²θ - 2gH
2gH = v²sin²θ
H = v²sin²θ / 2g
→ u = vcosθ, a = 0, s = R, t = T = 2vsinθ/g
R = vcosθ(2vsinθ/g)
R = 2v²sinθcosθ / g
since sin2θ = 2sinθcosθ
v²sin2θ/g
↑u = vsinθ, a = -g, t = t, s = y
s = ut + 1/2 at²
y = vsinθt - 1/2 gt² ------1)
→ u = vcosθ, t = t, s = x
x = vtcosθ => t = x/vcosθ
Sub in 1),
y = x tanθ - 1/2 gx²/v²cos²θ
y = x tanθ - gx²sec²θ/2v²
Since sec²θ = 1 + tan²θ
y = x tanθ - gx²(1 + tan²θ)/2v²
#### Maximizing the Horizontal Range
Suppose an object is projected at an angle, t, to the horizontal at v.
When it reaches the peak,
↑ u = v sin(t); v = 0; t = t
o = v sin(t) - at
t = v sin(t)/a
The time of flight= 2v sin(t)/a
→ s = x; u = v cos(t); a = 0; t = 2vsin(t)
x = 2v2sin(t)cos(t)
Since 2 sin(t) cos(t) = sin(2t) - a trigonometric identity
x = v2sin(2t)
dx/dt = 2v2cos(2t)
When x is maximum, dx/dt =0
2v2cos(2t) = 0
cos(2t) = 0
2t = 900
t = 450
So, the horizontal range is maximum when the angle of projection from the horizontal is 450.
That means, if a cricketer or a footballer think of maximizing the horizontal range, here is the way!
The best book for both teachers and students to learn physics - exactly like in the good old days:concepts are clearly explained in detail;no meaningless cartoons to devour space;the author rendered a great service in his unique approach for generations of students, with this being the fourth edition.
### GCSE Physics - Flash Cards
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# How do you simplify (7y -x)^2?
Jan 25, 2017
$49 {y}^{2} + {x}^{2} - 14 x y$
#### Explanation:
It is a binomial squared:
${\left(a + b\right)}^{2} = {a}^{2} + {b}^{2} + 2 a b$
Jan 25, 2017
See below.
#### Explanation:
This is a binomial so we can make use of the standard formula
${\left(\textcolor{red}{\text{a"-color(blue)"b}}\right)}^{2}$ $=$ ${\textcolor{red}{\text{a}}}^{2}$ - $2 \textcolor{red}{\text{a"color(blue)"b}}$ + ${\textcolor{b l u e}{\text{b}}}^{2}$
For our purposes, color(red)("a"=7y and color(blue)("b"=x . Putting the values in the above given equation we obtain:
${\left(7 y - x\right)}^{2}$ = $49 {y}^{2} - 2 \cdot 7 y \cdot x + {x}^{2}$
$\implies$ ${\left(7 y - x\right)}^{2}$ = $49 {y}^{2} - 14 x y + {x}^{2}$
Jan 25, 2017
$49 {y}^{2} - 14 y x + {x}^{2}$
#### Explanation:
${\left(7 y - x\right)}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$7 y - x$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$7 y - x$
$\textcolor{w h i t e}{a a a a a a a a a a a}$$- - -$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$49 {y}^{2} - 7 y x$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}$$- 7 y x + {x}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$- - - - - -$
$\textcolor{w h i t e}{a a a a a a a a a a a a}$$49 {y}^{2} - 14 y x + {x}^{2}$
|
Basic Math | Basic-2 Math | Prealgebra | Workbooks | Glossary | Standards | Site Map | Help
# QuickQuiz: Single-Digit Multiplication
This activity was designed to help you practice simple multiplication. You're going to see a variety of problems that ask you to multiply numbers between one (1) and nine (9). All of the questions combine random values. This quiz works with values in a horizontal format. We have other quizzes that work with the stacked layout. Try to do these problems in your head.
Good luck and have fun.
## Directions
This is another NumberNut three-choice quiz. Once you start the activity you will see a math problem. To the right side or below that problem are three possible answers. It's your job to figure out the correct answer and click the answer. The next screen will show you the correct answer. Some of these quizzes will show you how to get the right answer.
You will get a happy or sad face for every question you finish. Once you finish ten (10) questions, the quiz will be over. Take the quiz again because all of the questions are random. Chances are, you'll get a new quiz every time. It's good practice to learn these basic arithmetic operations.
RELATED LINKS LESSONS: - NumberNut.com: Multiplication ACTIVITIES: - Memory Challenge: Multiply by One - Memory Challenge: Multiply by Two - Memory Challenge: Multiply by Three - Memory Challenge: Multiply by Four - Memory Challenge: Multiply by Five - Memory Challenge: Multiply by Six - Memory Challenge: Multiply by Seven - Memory Challenge: Multiply by Eight - Memory Challenge: Multiply by Nine - Memory Challenge: Multiply by Ten - QuickQuiz: Single-Digit Multiplication (H) - QuickQuiz: Single-Digit Multiplication (V) - Next in Series: 4, 6, and 8 Multiplication - Next in Series: 3, 7, and 9 Multiplication - Next in Series: 2, 5, and 10 Multiplication - Pick-a-Card: Single-Digit Word Problems - Pick-a-Card: Word Problems - More or Less: One-Digit Multiplication - QuickQuiz: One/Two-Digit Mult. (No Carry, H) - QuickQuiz: One/Two-Digit Mult. (No Carry, V) - More or Less: One/Two-Digit Mult. - QuickQuiz: One/Two-Digit (Carrying) - More or Less: Two-Digit Mult. (Carry) - QuickQuiz: One/Three-Digit Multiplication - QuickQuiz: One/Four-Digit Multiplication - QuickQuiz: Two-Digit Multiplication - QuickQuiz: Two/Three-Digit Multiplication - QuickQuiz: Two/Four-Digit Multiplication
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# Algebra II : Graphing Polynomial Functions
## Example Questions
← Previous 1 3
### Example Question #1 : Graphing Polynomial Functions
Where does the graph of cross the axis?
Explanation:
To find where the graph crosses the horizontal axis, we need to set the function equal to 0, since the value at any point along the axis is always zero.
To find the possible rational zeroes of a polynomial, use the rational zeroes theorem:
Our constant is 10, and our leading coefficient is 1. So here are our possible roots:
Let's try all of them and see if they work! We're going to substitute each value in for using synthetic substitution. We'll try -1 first.
Looks like that worked! We got 0 as our final answer after synthetic substitution. What's left in the bottom row helps us factor down a little farther:
We keep doing this process until is completely factored:
Thus, crosses the axis at .
### Example Question #2 : Graphing Polynomial Functions
Where does cross the axis?
5
3
-7
7
-3
7
Explanation:
crosses the axis when equals 0. So, substitute in 0 for :
### Example Question #3 : Graphing Polynomial Functions
Which equation best represents the following graph?
None of these
Explanation:
We have the following answer choices.
The first equation is a cubic function, which produces a function similar to the graph. The second equation is quadratic and thus, a parabola. The graph does not look like a prabola, so the 2nd equation will be incorrect. The third equation describes a line, but the graph is not linear; the third equation is incorrect. The fourth equation is incorrect because it is an exponential, and the graph is not an exponential. So that leaves the first equation as the best possible choice.
### Example Question #1 : Understand Linear And Nonlinear Functions: Ccss.Math.Content.8.F.A.3
Which of the graphs best represents the following function?
None of these
Explanation:
The highest exponent of the variable term is two (). This tells that this function is quadratic, meaning that it is a parabola.
The graph below will be the answer, as it shows a parabolic curve.
### Example Question #4 : Graphing Polynomial Functions
Turns on a polynomial graph.
What is the maximum number of turns the graph of the below polynomial function could have?
3 turns
7 turns
4 turns
8 turns
7 turns
Explanation:
When determining the maximum number of turns a polynomial function might have, one must remember:
Max Number of Turns for Polynomial Function = degree - 1
First, we must find the degree, in order to determine the degree we must put the polynomial in standard form, which means organize the exponents in decreasing order:
Now that f(x) is in standard form, the degree is the largest exponent, which is 8.
We now plug this into the above:
Max Number of Turns for Polynomial Function = degree - 1
Max Number of Turns for Polynomial Function = 8 - 1
which is 7.
The correct answer is 7.
### Example Question #5 : Graphing Polynomial Functions
End Behavior
Determine the end behavior for below:
Explanation:
In order to determine the end behavior of a polynomial function, it must first be rewritten in standard form. Standard form means that the function begins with the variable with the largest exponent and then ends with the constant or variable with the smallest exponent.
For f(x) in this case, it would be rewritten in this way:
When this is done, we can see that the function is an Even (degree, 4) Negative (leading coefficent, -3) which means that both sides of the graph go down infinitely.
In order to answer questions of this nature, one must remember the four ways that all polynomial graphs can look:
Even Positive:
Even Negative:
Odd Positive:
Odd Negative:
### Example Question #6 : Graphing Polynomial Functions
Which of the following is a graph for the following equation:
Cannot be determined
Explanation:
The way to figure out this problem is by understanding behavior of polynomials.
The sign that occurs before the is positive and therefore it is understood that the function will open upwards. the "8" on the function is an even number which means that the function is going to be u-shaped. The only answer choice that fits both these criteria is:
### Example Question #4 : Understand Linear And Nonlinear Functions: Ccss.Math.Content.8.F.A.3
None of the above
Explanation:
Starting with
moves the parabola by units to the right.
Similarly moves the parabola by units to the left.
Hence the correct answer is option .
### Example Question #7 : Graphing Polynomial Functions
Explanation:
When we look at the function we see that the highest power of the function is a 3 which means it is an "odd degree" function. This means that the right and left side of the function will approach opposite directions. *Remember O for Odd and O for opposite.
In this case we also have a negative sign associated with the highest power portion of the function - this means that the function is flipped.
Both of these combine to make this an "odd negative" function.
Odd negative functions always have the right side of the function approaching down and the left side approaching up.
We represent this mathematically by saying that as x approaches negative infinity (left side), the function will approach positive infinity:
...and as x approaches positive infinity (right side) the function will approach negative infinity:
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# Sta220 - Statistics Mr. Smith Room 310 Class #13.
## Presentation on theme: "Sta220 - Statistics Mr. Smith Room 310 Class #13."— Presentation transcript:
Sta220 - Statistics Mr. Smith Room 310 Class #13
Section 4.4-4.5
The graphical form of the probability distribution for a continuous random variable is a smooth curve.
Copyright © 2013 Pearson Education, Inc.. All rights reserved. Figure 4.11 A probability f(x) for a continuous random variable x
One of the most common observed continuous random variable has a bell-shaped probability distribution (or bell curve).
Copyright © 2013 Pearson Education, Inc.. All rights reserved. Figure 4.14 Several normal distributions with different means and standard deviations
The normal distribution plays a very important role in the science of statistical inference. You can determine the appropriateness of the normal approximation to an existing population of data by comparing the relative frequency distribution of a large sample of the data with the normal probability distribution.
Copyright © 2013 Pearson Education, Inc.. All rights reserved. Figure 4.15 Standard normal distribution: 0 1
Example 4.15A Find the probability that the standard normal random variable z falls between -1.33 and +1.33.
Copyright © 2013 Pearson Education, Inc.. All rights reserved. Figure 4.16 Areas under the standard normal curve for Example 4.15
Solution
Example 4.15B Find the probability that the standard normal random variable z falls between 1.00 and 2.50.
Solution
Example 4.16 Find the probability that a standard normal random variable exceeds 1.64; that is, find P(z > 1.64).
Copyright © 2013 Pearson Education, Inc.. All rights reserved. Figure 4.18 Areas under the standard normal curve for Example 4.16
Solution
Example 4.17 Find the probability that a standard normal random variable lies to the left of.67.
Copyright © 2013 Pearson Education, Inc.. All rights reserved. Figure 4.19 Areas under the standard normal curve for Example 4.17
Solution
Example 18
Solution
Example 19
Solution
Example 4.20
Solution
STOP
Sta220 - Statistics Mr. Smith Room 310 Class #14
Section 4.4-4.5(Part 2)
Problem 4. 21 Assume that the length of time, x, between charges of a cellular phone is normally distributed with a mean of 10 hours and a standard deviation of 1.5 hours. Find the probability that the cell phone will last between 8 and 12 between charges.
Solution
Problem 4. 22 Suppose an automobile manufacturer introduces a new model that has an advertised mean in-city mileage of 27 miles per gallon. Although such advertisements seldom report any measure of variability, suppose you write the manufacturer for details on the tests and you find that the standard deviation is 3 miles per gallon. This information leads you to formulate a probability model for the random variable x, the in-city mileage for this car model. You believe that the probability distribution of x can be approximated by a normal distribution with a mean of 27 and a standard deviation of 3.
a.If you were to buy this model of automobile, what is the probability that you would purchase one that averages less than 20 miles per gallon for in-city driving? b.Suppose you purchase one of these new models and it does get less than 20 miles per gallon for in-city driving. Should you conclude that you probability model is incorrect?
Solution Part a.
Copyright © 2013 Pearson Education, Inc.. All rights reserved. Figure 4.21 Area under the normal curve for Example 4.21
Solution Part B. There are two possibilities that could exists: 1.The probability model is correct. You simply were unfortunate to have purchased one of the cars in the 1% that get less than 20 miles per gallon in the city. 2.The probability model is incorrect. Perhaps the assumption of a normal distribution is unwarranted, or the mean of 27 is an overestimate, or the stand deviation of 3 is an underestimate, or some combination of these errors occurred.
Keep in mind we have no way of knowing with certainty which possibility is correct, but the evidence points to the second on. We are again relying on the rare-event approach to statistical inference that we introduced earlier. In applying the rare-event approach, the calculated probability must be small (say, less than or equal to 0.05) in order to infer that the observed event is, indeed, unlikely.
Problem 4.22 Suppose that the scores x on a college entrance examination are normally distributed with a mean of 550 and the standard deviation of 100. A certain prestigious university will consider for admission only those applicants whose scores exceed the 90 th percentile of the distribution. Find the minimum score an applicant must achieve in order to receive consideration for admission to the university.
Copyright © 2013 Pearson Education, Inc.. All rights reserved. Figure 4.22 Area under the normal curve for Example 4.22
Solution Thus, the 90 th percentile of the test score distribution is 678. That is to say, an applicant must score at least 678 on the entrance exam to receive consideration for admission by the university.
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# Intuition behind a method of adding two fractions
1. Jan 6, 2010
### Juwane
One way to add two fractions is to multiply the numerators of both fractions with each other's denominator, then adding the two products, gives us the numerator of the final result. Then we multiply together the denominators of each other--this gives us the denominator of the final result.
Consider this example:
$$\frac{2}{3} + \frac{4}{5} = \frac{(5\times2)+(3\times4)}{(3\times5)} = \frac{22}{15}$$
Why are we able to do that? I'm not looking for a rigorous proof, just the intuition behind it.
2. Jan 6, 2010
### Bohrok
To add fractions, they need the same denominator. We "change" the fractions to have a common denominator by the identity property x·1 = x and then simplify the result.
$$\frac{2}{3} + \frac{4}{5} = \frac{2}{3}\times\frac{5}{5} + \frac{4}{5}\times\frac{3}{3}$$
$$= \frac{2\times 5}{3\times 5} + \frac{4\times 3}{5\times 3} = \frac{(5\times 2) + (3\times 4)}{3\times 5}$$
I suppose the "intuition" is to make the denominators the same and then combine the fractions. What you have could be used as a formula for adding fractions, bypassing all the steps that lead up to that result.
3. Jan 7, 2010
### Tac-Tics
This reminds me of something I told one of my friends a long time ago while driving down the highway. We saw a sign saying 3/4 mile to our exit. I mentioned to her, "there's only three quarter-miles to go." She thought about it for a second and was amused that she never thought of it as three "quarter miles." In her mind, it was always "three quarters" of a whole mile.
The nice thing is the two are interchangeable. Any hungry child will tell you that half a pie isn't the same as a third of a pie. But which is bigger? How do you compare halves and thirds? You convert each to an equal amount of sixths-pies. Then it's clear. If you cut a pie into 6 even pieces, you need to take 3 to have taken half the pie, so 1/2 = 3/6. And in the same way, you need 2/6 of a pie to equal that third-pie.
It's obvious with pie and pictures. But once you distill it to numbers, it does appear mysterious.
4. Jan 8, 2010
### HallsofIvy
One way of thinking about it is that the denominators of fractions are really "units". That is, if we talk about "4 feet" or "3 yards", the "4" and "3" alone would be meaningless without specifying what units we are using. If we were to add those, we would NOT get "7" anything! We must have the same units. Here, it is simplest to change "3 yards" to 9 feet, since there are 3 yards to a foot. "4 feet plus 3 yards" is the same as "4 feet plus 9 feet= (4 + 9) feet= 13 feet.
Similarly, two add 2/3 plus 4/5, I need to get the same "units". One "third" is not any integer multiple of one "fifth" or vice versa but one "third" is 5 "fifteenths" so 2 "thirds" is 10 "fifteenths" and one "fifth" is 3 "fifteenths" so 4 "fifths" is 12 "fifteenths".
2 thirds plus 4 fifths= 10 fifteenths plust 12 fifteenths= 22 fifteenths.
2/3+ 4/5= 10/15+ 12/15= 22/15.
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# How to Teach Fractions: All You Need to Know
Learning fractions can be hard for many children because it goes beyond basic simple arithmetic calculations. To understand the concept, kids need to practice with examples every day before they can master fractions. This article provides tips on how to learn fractions in a fun and engaging way.
## What is a fraction?
A fraction is a number that indicates a part of a whole. One can write a fraction using two numbers separated by a line instead of a division symbol.
The number above the line is a numerator and represents the number of parts you have. Similarly, the number below the line is called a denominator and represents the total number of parts in the whole area or object.
A simple fraction shows one part that has been partitioned into equal parts. A proper fraction is the one where a numerator is strictly less than a denominator. An improper fraction has a numerator that is greater than or equal to its denominator.
When teaching fractions, you can show kids that 2/5 means that 2 is divided by 5. Each piece that we divide the whole into is called a part, and each part has an equal value. The number on top tells us how many parts we have.
## How to teach fractions to a child?
At what grade do you learn fractions? Kids meet fractions as early as in their first grade. At this stage, you can draw circles or squares with different colors and then divide those shapes into parts. Such pictures will be among the fun ways to teach fractions.
While learning fractions for beginners, you can use colors. For example, you can divide a circle into 2 equal parts and apply blue color for one part and red color for the other part of the circle.
Such an approach will help kids understand that half of the circle is blue and half of it is red, which will let children understand fractions without using any numbers or symbols.
Once children master this concept, you can introduce multiplication. For example, start with simple examples like 1/2 × 1/3 = 1/6 if you are not sure how to teach multiplying fractions.
1:1 Math Lessons
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### The Nominator
A nominator is a number that indicates how many parts of a whole you have. In other words, a nominator is simply a number that represents the value of the fraction.
A nominator is the first part of a fraction. The other part is called a denominator, which is the bottom number in a fraction.
The purpose of a nominator in fractions is to tell you how many parts you have or what number to count up to. The following is an example of how to teach simplifying fractions:
2/3 means 2 out of 3 parts. So, if you were to cut a cake into 3 equal slices, you would eat 2 of them. Fraction explained, right?
1/2 means 1 out of 2 parts. If you cut that same cake into 2 slices, you would eat 1 slice and leave 1 slice for someone else to eat.
Another example is the following: if you have a pizza and slice it into eight pieces, each piece would be one-eighth of the pizza. You could also say that each piece is “one over eight” or “one/eight” of the whole pizza.
### The Numerator
A numerator is a term used in mathematics that refers to the number above the line in a common fraction. You can think of a numerator as an indicator representing the number of parts you want to take from the whole.
For example, in the fraction 5/8, “5” is the numerator, while the number below the line (8) is called the denominator.
A numerator represents many equal parts, indicating how many of those parts are taken or considered. In the fraction 7/10, for instance, seven parts are taken out of ten equal parts.
### Fraction Notation
Fraction notation is a form of writing simple fractions, ratios, and other numbers. It is represented by two digits or numbers separated by a line and placed over each other. In other words, fractional notation is a way to express ratios using the division sign and slashes.
Fractions are usually written using the slash symbol (/) but can also be presented as a division equation with a division sign (÷).
The examples are 3/4, 12/16, 1/3, etc. If you write 3/4, it means that there are three-quarters of something. This thing can be any object or concept. 1/2 is one half, 1/3 is one-third, and so on.
### Fractions in Everyday Life
There are numerous cases in everyday life where we apply fractions. They not only show up in math class but also can help measure things such as distance or time. For example, when measuring distance, we use fractions of a mile or a kilometer.
When you go to the gas station, you will see pumps that measure gasoline by fractions of gallons. Gallons are divided into halves, quarters, and halves or quarter’s eighths. More importantly, such division makes it easier to fill up a vehicle’s gas tank because some tanks only hold certain amounts of fuel.
Fractions are useful in cooking recipes. We use them when we are measuring ingredients while cooking. Recipes require you to cut, halve, or double ingredients.
In addition, measurements of liquids may require you to use fractions. Fractions also show up on measuring tapes and rulers to measure parts of an inch and parts of a foot accurately.
Another application of fractions is measuring time. For example, if you have three-quarters of an hour to complete an exam, you will have 45 minutes to complete the exam since there are 60 minutes in one hour.
## Conclusion
The process of learning fractions does not have to be tedious. Numerous useful tips can help make math lessons more interesting and engaging for students. A truly effective math tutor will know how to explain fractions to convey the mathematical concepts in a creative way that children love.
Book 1 to 1 Math Lesson
• Specify your child’s math level
• Get practice worksheets for self-paced learning
• Your teacher sets up a personalized math learning plan for your child
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# Additional Question Answers: Number System Notes | Study Mathematics (Maths) Class 9 - Class 9
## Class 9: Additional Question Answers: Number System Notes | Study Mathematics (Maths) Class 9 - Class 9
The document Additional Question Answers: Number System Notes | Study Mathematics (Maths) Class 9 - Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
Q1. Write four rational numbers equivalent to 5/7.
We have,
Four rational numbers equivalent to 5/7 are 10/14, 15/21, 20/28, 25/35.
Q2. Find nine rational numbers between 0.1 and 0.2.
Let x = 0.1, y = 0.2 and n = 9.
The nine rational numbers between x and y are:
(x + d), (x + 2d), (x + 3d), (x + 4d), (x + 5d), (x + 6d), (x + 7d), (x + 8d) and (x + 9d).
Nine rational numbers between 0.1 and 0.2 are:
(0.1 + 0.01); (0.1 + 0.02); (0.1 + 0.03); (0.1 + 0.04); (0.1 + 0.05); (0.1 + 0.06); (0.1 + 0.07); (0.1 + 0.08) and (0.1 + 0.09)
The nine rational numbers between 0.1 and 0.2 are 0.11, 0.12, 0.13, 0.14, 0.15, 0.16, 0.17, 0.18 and 0.19.
Q3. Represent √2 on the number line (or on the real line).
• Draw XOX' and mark O as 0.
• Take OA = 1 unit.
• Draw AB ⊥ OX and cut off AB = 1 unit.
• Now, OAB is a right triangle.
∵ OA2 + AB2 = OB2
⇒ 12 + 12 = OB2
⇒ 2 = OB2
⇒ OB =√2
• With centre O and OB as radius, draw an arc intersecting OX at C.
• Thus, OC = √2 on the real line XOX'.
Q4. Represent √3 on the real line.
• Draw XOX ' and mark O as 0 on it.
• Take OA = 1 unit.
• Draw AB ⊥ OX and cut off AB = 1 unit.
• Join OB such that OAB is a right triangle.
∵ OA2 + AB2 = OB2
⇒ 12 + 12 = OB2
⇒ OB =√2
• Again, draw BC ⊥ OB and cut off BC = 1 unit. Join OC.
• Now, OBC is a right triangle.
∵ OB2 + BC2 = OC2
⇒ √22 + 12 = OC2
⇒ 2 + 1 = OB2
⇒ OC =√3
• With centre as O and radius as OC draw an arc to intersect OX at D.
• Thus, OD =√3 on the real line XOX'.
Q5. Express as a rational number in simplest form.
⇒ 100x = 100 x (0.3838…)
⇒ 100x = 38.3838...(2)
Subtracting (1) from (2),
We have, 100x – x = (38.3838…) – (0.3838…)
Q6. Express in the form of .
∴ 10x = 10 x (0.5353…)
or 10x = 5.333... ...(1)
Also, 100x = 53.333... ...(2)
Subtracting (1) from (2),
⇒ 100x – 10x = (53.333…) – (5.333…)
⇒ 90x = 48
Q7. Express is the form of p/q in the simplest form.
∴ 1000x = 1000 x (0.003003…)
or 1000x = 3.003003... …(2)
Subtracting (1) from (2),
We have 1000x – x = (3.003003…) – (0.003003…)
⇒ 999x = 3
⇒ x = 3/999 = 1/333
Thus,
Q8. Find the sum of (3√3 +7√2) and (√3 - 5√2)
We have (3√3 +7√2) + (√3 - 5√2)
⇒ √3 3+7√2 + √3 - 5√2
⇒ (3√3+√3) + 7√2 - 5√2)
⇒ √3(3+1) + √2(7-2)
⇒ √3(4) + √2(5) = (4√3 + 2√2)
Q9. Divide 15 √12 by 3√3.
Q10. Rationalize the denominator of
(Prime factorize the numbers under the root in the denominator)
Q11. Rationalize the denominator of
Q12. If ‘a’ and ‘b’ are rational numbers and find the values of ‘a’ and ‘b’.
Comparing
Q13. If , what is the value of x3 - 5x2 + 8x - 4?
(x - 2)3 = x3 - 6x2 + 12x - 8 ...(1)
(x - 2)2 = x2 - 4x + 4 ... (2)
(1) + (2) = x3 - 5x2 + 8x - 4
Q14. Find the value of when
∵
Q15. Rationalise the denominator of 1/[7+3√3].
1/(7 + 3√3)
By rationalizing the denominator,
= [1/(7 + 3√3)] [(7 – 3√3)/(7 – 3√3)]
= (7 – 3√3)/[(7)2 – (3√3)2]
= (7 – 3√3)/(49 – 27)
= (7 – 3√3)/22
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# Problems on Size Transformation
Here we will solve different types of problems on size transformation.
1. A map of a rectangular park is drawn to a scale of 1 : 5000.
(i) Find the actual length of the park if the length of the same in the map is 25 cm.
(ii) If the actual width of the park is 1 km, find its width in the map.
Solution:
(i) Given scale = 1 : 5000
Let the actual length of the park be p
Therefore, $$\frac{\textrm{Length in Map}}{\textrm{Actual Length of the Park}}$$ = $$\frac{1}{5000}$$
⟹ $$\frac{25 cm}{p}$$ = $$\frac{1}{5000}$$
⟹ p = 25 × 5000 cm
⟹ p = 125000 cm
⟹ p = 125000 × 10$$^{-5}$$ km
⟹ p = 1.25 km
(ii) Let the width in the map be q
Therefore, $$\frac{\textrm{Width in Map}}{\textrm{Actual Width}}$$ = $$\frac{1}{5000}$$
⟹ $$\frac{q}{1 km}$$ = $$\frac{1}{5000}$$
⟹ q = $$\frac{1 km}{5000}$$
⟹ q = $$\frac{1 × 10^{5}}{5000}$$ cm
⟹ q = 20 cm
2. The length of a building is 40 m and the length of its model is 50 cm. Find the area of the terrace of the building if the area of the terrace in the model is 1400 cm$$^{2}$$.
Solution:
Let the scale factor = k
Therefore, k = $$\frac{\textrm{Length of the Building}}{\textrm{Length of the Model}}$$
= $$\frac{40 m}{50 cm}$$
= $$\frac{40 × 100 cm}{50 cm}$$
= 80 cm
Let the area of the terrace of the building = a
$$\frac{\textrm{Area of the Terrace of the Building}}{\textrm{Area of the Terrace in the Model}}$$ = k$$^{2}$$
Therefore, $$\frac{a}{1400 cm^{2}}$$ = 80 × 80
⟹ a = 80 × 80 × 1400 cm$$^{2}$$
⟹ a = $$\frac{80 × 80 × 1400}{100 × 100}$$ m$$^{2}$$
⟹ a = 896 m$$^{2}$$
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# IMAT 2011 Q80 [Semicircular]
A square piece of metal has a semicircular piece cut out of it as shown. The area of the remaining metal is {100cm^2}.
Which of the following is a correct expression of the length of the side of the square in cm?
A. 10 \sqrt{\frac {1}{8-π}}
B. 10 \sqrt{\frac {2}{4-π}}
C. 20 \sqrt{\frac{2}{8+π}}
D. 20 \sqrt{\frac{2}{8-π}}
E. 20 \sqrt{\frac{1}{4-π}}
When solving this types of questions the first thing you do is calculate the area of the main object (in this case, the square piece of metal)
In solving the question i will use b as a variable.
The area of the square is P_s=b^2.
A circle is defined by its radius. In this particular question, we can see that the radius is half of the length of the square edge, r=\frac{b}{2}.
The area of the circle is calculated by the formula: P_c=\pi r^2 Having seen previously what the value of r is we can get the final form of the area of the circle P_c=\pi \frac{b}{2}^2 \to P_c=\pi \frac{b^2}{4}.
In the area of the task, the semicircular piece cut out is actually half of the area of the circle we previously calculated.
So, in order to get the equation of the area of the remaining metal, we will subtract the area of half a circle from the square metal area: P = P_s - \frac{P_c}{2} \to P = b^2 - \frac{\pi\frac{b^2}{4}}{2} = b^2 - \frac{\pi b^2}{8} = b^2(1-\frac{\pi}{8}) = \frac{b^2}{8}(8-\pi) .
Using the value of the area given in the task we get an equation: P=\frac{b^2}{8}(8-\pi) =100cm^2. Multiplying the whole equation by 8 we get: b^2(8-\pi) =800cm^2 \to b^2 = \frac{800cm^2}{8-\pi} \to b=\sqrt{\frac{800cm^2}{8-\pi}} = 20\sqrt{\frac{2}{8-\pi}}cm.
Which is the solution to this question.
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# What is a section formula?
## What is a section formula?
The section formula gives the coordinates of a point which divides the line joining two points in a ratio, internally or externally. P ( x , y ) = ( c ⋅ m + a ⋅ n m + n , d ⋅ m + b ⋅ n m + n ) .
## How do you find the midpoint of a section?
The midpoint is the point on the segment halfway between the endpoints. It may be the case that the midpoint of a segment can be found simply by counting. If the segment is horizontal or vertical, you can find the midpoint by dividing the length of the segment by 2 and counting that value from either of the endpoints.
11.
## Where is section formula used?
In coordinate geometry, Section formula is used to find the ratio in which a line segment is divided by a point internally or externally. It is used to find out the centroid, incenter and excenters of a triangle. In physics, it is used to find the center of mass of systems, equilibrium points, etc.
## What is internal section formula?
Internal Section Formula When the line segment is divided internally in the ration m:n, we use this formula. That is when the point C lies somewhere between the points A and B. Understand the concept of Coordinate here. The Coordinates of point C will be, {[(mx2+nx1)/(m+n)],[(my2+ny1)/(m+n)]}
## What is Section Formula internally?
So, the coordinates of the point M(x,y) which divides the line segment joining points P(x_1, y_1) and Q(x_2, y_2) internally in the ratio m:n are. \LARGE\left(\frac{mx_2~+~nx_1}{m~+~n},\frac{my_2~+~ny_1}{m~+~n}\right) This is known as section formula.
## What is centroid formula?
Now, let us learn the centroid formula by considering a triangle. ... Then, we can calculate the centroid of the triangle by taking the average of the x coordinates and the y coordinates of all the three vertices. So, the centroid formula can be mathematically expressed as G(x, y) = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3).
## What is the midpoint formula for elasticity?
Let's calculate the arc elasticity following the example presented above: Midpoint Qd = (Qd1 + Qd2) / 2 = (40 + 60) / 2 = 50. Midpoint Price = (P1 + P2) / 2 = (10 + 8) / 2 = 9. % change in qty demanded = (60 – 40) / 50 = 0.
## What is midpoint formula used for?
The midpoint formula lets you find the exact center between two defined points. You might encounter this formula in your economics or geometry class or while prepping for a college entrance exam like the SAT or ACT.
## What does midpoint formula mean?
Definition: Midpoint formula is a mathematically equation used to measure the halfway point between two data points. The study of economics uses this calculation to find the coefficient of elasticity, either demand or supply, by measuring the average of the two points.
## What is midpoint of a segment?
In geometry, the midpoint is the middle point of a line segment. It is equidistant from both endpoints, and it is the centroid both of the segment and of the endpoints. It bisects the segment.
## What is the midpoint symbol?
Think of the midpoint as the “halfway” or middle point of a line segment. This so-called center point divides the line segment into two equal or congruent parts. NOTE: The midpoint of line segment A C AC AC denoted by the symbol A C ‾ \overline {AC} AC is located at point B.
## What does midpoint mean?
: a point at or near the center or middle.
## What is a midpoint in statistics?
Class Mark (Midpoint) The number in the middle of the class. It is found by adding the upper and lower limits and dividing by two. It can also be found by adding the upper and lower boundaries and dividing by two. Cumulative Frequency.
## How do you find the midpoint of a histogram?
The heights of the points represent the frequencies. A frequency polygon can be created from the histogram or by calculating the midpoints of the bins from the frequency distribution table. The midpoint of a bin is calculated by adding the upper and lower boundary values of the bin and dividing the sum by 2.
## What is the midpoint of a confidence interval?
Suppose a 95% confidence interval for the difference between two means is -1.
## Is midpoint the same as average?
It can be established in a meaningful manner that for any two points on the number line, the midpoint has the same value as the average of the two corresponding numbers.
## Why is average called mean?
A statistician or mathematician would use the terms mean and average to refer to the sum of all values divided by the total number of values, what you have called the average. This especially true if you have a list of numbers. ... There are times however that this is called the mean.
## How do you find the average point?
To calculate the overall average, the total points earned is divided by the total points possible. This is the default method. Each assignment is weighted equally, so a 20 point activity counts the same as a 10 point activity.
## How do we calculate?
1. How to calculate percentage of a number. Use the percentage formula: P% * X = Y
1. Convert the problem to an equation using the percentage formula: P% * X = Y.
2. P is 10%, X is 150, so the equation is 10% * 150 = Y.
3. Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.
## How do you find the percentage of a point?
Points System To do this, take the total number of points you earned on the assignment and divide by the number of points the assignment was worth. If, for example, you earned 38 points out of a total possible 50 points, then your percentage is 76, as shown here: 38 / 50 = . 76 or 76 percent.
## How do I know my CGPA?
To calculate a CGPA, you simply divide your total score of grade points for all subjects throughout your semesters by the total number of credit hours attended. GPA and CGPA are indicated by a number as opposed to the percentages or grades that are assigned under the Indian grading system.
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# Formulas for Area of a Triangle.
There are a lot of formulas and techniques to find the area of a triangle. We can use many different formulas to calculate area of a triangle according to the given conditions. Here we shall derive some of the main formulas used to calculate area of a triangle.
Formulas for Area of a Triangle:
The area of a triangle is denoted by the symbol delta ( $\Delta$ )
We shall appeal to the formula:
$\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B = \frac{1}{2} ab \sin C$
And the half angle formula:
$\sin \frac{1}{2} A = \sqrt{\dfrac{(s-b)(s-c)}{bc}} \, \, \, , \, \, \, \cos \frac{1}{2} A = \sqrt{\dfrac{s(s-a)}{bc}}$ etc.
Where “s” is the semi circumference of the triangle or, $s = \frac{a+b+c}{2}$
We shall now derive different formula for the area of triangle:
First formula for the area of a triangle:
$\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} bc . 2 \sin \frac{A}{2} \cos \frac{A}{2} \\ \\ \\ or , \Delta = bc . \sqrt{\dfrac{s(s-a)(s-b)(s-c)}{bc . bc}} \\ \\ \\ or , \Delta = \sqrt{s(s-a)(s-b)(s-c)}$
Second formula for the area of a triangle:
$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$
Now , as 2s = (a+b+c)
$\Delta = \frac{1}{4} \sqrt{(a+b-c)(b+c-a)(c+a-b)(a+b-c)} \\ \\ So , \Delta = \frac{1}{4} \sqrt{2b^2 c^2 + 2c^2 a^2 + 2a^2 b^2 - a^4 - b^4 - c^4}$
Third formula for the area of a triangle:
$\Delta = \frac{1}{2} bc \sin A$
Now using sine law:
$\Delta = \frac{1}{2} bc \frac{a}{2R} \\ \\ So , \Delta = \dfrac{abc}{4R}$
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## Engage NY Eureka Math 4th Grade Module 1 Lesson 11 Answer Key
### Eureka Math Grade 4 Module 1 Lesson 11 Problem Set Answer Key
Question 1.
Solve the addition problems below using the standard algorithm.
a.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 6,311 and 268
first adding the ones place = 1 + 8 = 9.
ten place : 1 + 6 = 7
hundreds place = 3 + 2 = 5
thousands place = 6
Sum = 6,579
b.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 6,311 and 1,268
first adding the ones place = 1 + 8 = 9.
ten place : 1 + 6 = 7
hundreds place = 3 + 2 = 5
thousands place = 6 + 1 = 7
Sum = 7,579
c.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 6,314 and 1,268
first adding the ones place = 4 + 8 = 12. 12 is represented as 10 ones + 2 ones that’s 10 ones make 1 ten and 2 ones. So 1 is carried to the tens place and added to the tens place values.
ten place : 1 + 6 + 1 = 8
hundreds place = 3+ 2 = 5
thousands place = 6 + 1 = 7
Sum = 7,582.
d.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 6,314 and 2,493
first adding the ones place = 4 + 3 = 7.
ten place : 1 + 9 = 10. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 0 is represented in tens place
hundreds place = 3+ 4 + 1 = 8
thousands place = 6 + 2 = 8
Sum = 8,807
e.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 8,314 and 2,493
first adding the ones place = 4 + 3 = 7.
ten place : 1 + 9 = 10. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 0 is represented in tens place
hundreds place = 3 + 4 + 1 = 8
thousands place = 8 + 2 = 10. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 0 is represented in thousands place
ten thousands = 1
Sum = 10,807.
f.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 12,378 and 5,463
first adding the ones place = 8 + 3 = 11. 10 ones + 1 ones = 11. 10 ones make 1 ten . So 1 is carried to the tens place and added to the tens place values and 1 is represented in ones place
ten place : 7 + 6 + 1(carry on) = 14. 10 ten and 4 tens = 14 .10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 4 is represented in tens place
hundreds place = 3 + 4 + 1(carry on) = 8
thousands place = 2 + 5 = 7
ten thousands = 1
Sum = 17,841
g.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 52,098 and 6,048
first adding the ones place = 8 + 8 = 16. 10 ones + 6 ones = 16. 10 ones make 1 ten . So 1 is carried to the tens place and added to the tens place values and 6 is represented in ones place
ten place : 9 + 4 + 1(carry on) = 14. 10 ten and 4 tens = 14 .10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 4 is represented in tens place
hundreds place = 0 + 0 + 1(carry on) = 1
thousands place = 2 + 6 = 8
ten thousands = 5
Sum = 58,146
h.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 34,698 and 71,840
first adding the ones place = 8 + 0 = 8.
ten place : 9 + 4 = 13. 10 tens + 3 tens = 13 .10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 3 is represented in tens place
hundreds place = 6 +8 + 1(carried number) = 15 = 10 hundred + 5 hundred . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 5 is represented in hundreds place
thousands place = 4 + 1 + 1(carried number) = 6.
ten thousands = 3 + 7 = 10 .10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 0 is represented in ten thousands place
Hundred thousands = 1(carried number) = 1
Sum = 106,538.
i.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 34,698 and 71,840
first adding the ones place = 1 + 5 = 6.
ten place : 1 + 4 = 5.
hundreds place = 8 + 4 = 12 = 10 hundred + 2 hundred . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 2 is represented in hundreds place
thousands place = 4 + 6 + 1(carried number) = 11 =10 thousand + 1 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 1 is represented in thousands place
ten thousands = 4 + 5 + 1(carried number) = 10 .10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 0 is represented in ten thousands place
Hundred thousands = 1(carried number) = 5 + 3 + 1(carried number) = 9
Sum = 901,256.
j. 527 + 275 + 752
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 527 , 275 and 752
first adding the ones place = 7 + 5 + 2 = 14 = 10 ones + 4 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 4 is represented in ones place
ten place : 2 + 7 + 5 + 1(carried number) = 15 = 10 tens + 5 tens. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 5 is represented in tens place
hundreds place = 5 +2 + 7 + 1(carried number)= 15 = 10 hundred + 5 hundred . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 5 is represented in hundreds place
thousands place = 4 + 6 + 1(carried number) = 11 =10 thousand + 1 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 1 is represented in thousands place
ten thousands = 4 + 5 + 1(carried number) = 10 .10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 0 is represented in ten thousands place
Hundred thousands = 1(carried number) = 5 + 3 + 1(carried number) = 9
Sum = 1,554
k. 38,193 + 6,376 + 241,457
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 38,193 + 6,376 + 241,457
first adding the ones place = 3 + 6 + 7= 16 = 10 ones + 6 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 6 is represented in ones place
ten place : 9 + 7 + 5 + 1(carried number) = 22 = 20 tens + 2 tens. 20 tens make 2 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 2 is represented in tens place
hundreds place = 1 + 3 + 4 + 1(carried number)= 10. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 0 is represented in hundreds place
thousands place = 8 + 6 + 1 +1(carried number) = 16 =10 thousand + 6 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 6 is represented in thousands place
ten thousands =3 + 4 + 1(carried number) = 8 .
Hundred thousands = 2
Sum = 286,026.
Draw a tape diagram to represent each problem. Use numbers to solve, and write your answer as a statement.
Question 2.
In September, Liberty Elementary School collected 32,537 cans for a fundraiser. In October, they collected 207,492 cans. How many cans were collected during September and October?
Number of cans collected by the Liberty Elementary School for a fundraiser in September are = 32,537
Number of cans collected by the Liberty Elementary School for a fundraiser in October are = 207,492
Total number of cans collected during September and October are = 32,537 + 207,492 =
Question 3.
A baseball stadium sold some burgers. 2,806 were cheeseburgers. 1,679 burgers didn’t have cheese. How many burgers did they sell in all?
Number of cheeseburgers sold at baseball stadium = 2,806
Number of burgers without cheese sold at baseball stadium are = 1,679
Total number of burgers sold in all are = 2,806 + 1,679
Question 4.
On Saturday night, 23,748 people attended the concert. On Sunday, 7,570 more people attended the concert than on Saturday. How many people attended the concert on Sunday?
Number of people attended the concert on Saturday night = 23,748
Number of people attended the concert on Sunday are = 7,570 more than on Saturday
Total people attended the concert on Sunday = 23,748 + 7, 570
### Eureka Math Grade 4 Module 1 Lesson 11 Exit Ticket Answer Key
Question 1.
Solve the addition problems below using the standard algorithm.
a.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 23,607 and 2,307
first adding the ones place = 7 + 7 = 14. 14 is represented as 10 + 4 that’s 1 ten and 4 ones. So 1 is carried to the tens place and added to the tens place values.
ten place : 0 + 0 + 1 = 1
hundreds place = 6 + 3 = 9
thousands place = 3 + 2 = 5
ten thousands place = 2
Sum = 25,914.
b.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 3,948 and 278
ones place = 8 + 8 = 16. 16 is represented as 10 + 6 that’s 1 ten and 6 ones. So 1 is carried to the tens place and added to the tens place values and 6 is represented in ones place.
ten place : 4 + 7 + 1(carried number) = 12 = 10 + 2 that’s 1 hundred and 2 tens. So 1 is carried to the hundreds place and added to the hundreds place values and 2 is represented in tens place.
hundreds place = 9 + 2 + 1(carried number) = 12 = 10 + 2 that’s 1 thousand and 2 hundred. So 1 is carried to the thousands place and added to the thousands place values and 2 is represented in hundreds place.
thousands place = 3 + 1(carried number) = 4
Sum = 4,226.
c. 5,983 + 2,097
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 5,983 and 2,097
ones place = 3 + 7 = 10. 10 is represented as 10 + 0 that’s 1 ten and 0 ones. So 1 is carried to the tens place and added to the tens place values and 0 is represented in ones place.
ten place : 8 + 9 + 1(carried number) = 18 = 10 + 8 that’s 1 hundred and 8 tens. So 1 is carried to the hundreds place and added to the hundreds place values and 8 is represented in tens place.
hundreds place = 9 + 0 + 1(carried number) = 10 = 10 + 0 that’s 1 thousand and 0 hundred. So 1 is carried to the thousands place and added to the thousands place values and 0 is represented in hundreds place.
thousands place = 5 + 2 + 1(carried number) = 8
Sum = 8,080
Question 2.
The office supply closet had 25,473 large paper clips, 13,648 medium paper clips, and 15,306 small paper clips. How many paper clips were in the closet?
Number of large paper clips = 25,473
Number of medium paper clips = 13,648
Number of small paper clips = 15,306
Total number of paper clips = 25,473 + 13,648 + 15,306
### Eureka Math Grade 4 Module 1 Lesson 11 Homework Answer Key
Question 1.
Solve the addition problems below using the standard algorithm.
a.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 7,909 and 1,044
first adding the ones place = 9 + 4 = 13 = 10 ones + 3 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 3 is represented in ones place
ten place : 0 + 4 + 1(carried number) = 5
hundreds place = 9 + 0 = 9.
thousands place = 7 + 1 = 8.
Sum = 8,953.
b.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 27,909 and 9,740
first adding the ones place = 9 + 0 = 9
ten place : 0 + 4 = 4.
hundreds place = 9 + 7 = 16. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 6 is represented in hundreds place
thousands place = 7 + 9 +1(carried number) = 17 =10 thousand + 7 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 7 is represented in thousands place.
ten thousands =2 + 1(carried number) = 3 .
Sum = 37,649.
c.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 827,909 and 42,989
first adding the ones place = 9 + 9= 18 = 10 ones + 8 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 8 is represented in ones place
ten place : 0 + 8 + 1(carried number) = 8.
hundreds place = 9 + 9 = 18. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 8 is represented in hundreds place
thousands place = 7 + 2 +1(carried number) = 10 =10 thousand + 0 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 0 is represented in thousands place
ten thousands =2+ 4 + 1(carried number) = 7 .
Hundred thousands = 8
Sum = 870,898.
d.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 289,205 and 11,845
first adding the ones place = 5 + 5 = 10 = 10 ones + 0 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 0 is represented in ones place
ten place : 0 + 4 + 1(carried number) = 5.
hundreds place = 2 + 8 = 10. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 0 is represented in hundreds place
thousands place = 9 + 1 +1(carried number) = 11 =10 thousand + 1 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 1 is represented in thousands place
ten thousands =8 + 1 + 1(carried number) =10. 10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 1 is represented in ten thousands place
Hundred thousands = 2 + 1(carried over number) = 3
Sum = 301,050.
e.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 547,982 and 114,849
first adding the ones place = 2 + 9 = 11 = 10 ones + 1 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 1 is represented in ones place
ten place : 8 + 4 + 1(carried number) = 13. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 3 is represented in tens place
hundreds place = 9 + 8 + 1(carried number)= 18. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 8 is represented in hundreds place
thousands place = 7 + 4 +1(carried number) = 12 =10 thousand + 2 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 2 is represented in thousands place
ten thousands =4 + 1 + 1(carried number) = 6.
Hundred thousands = 5 + 1 = 6.
Sum = 662,831.
f.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 258,983 and 121,897.
first adding the ones place = 3 + 7 = 10 = 10 ones + 0 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 0 is represented in ones place
ten place : 8 + 9+ 1(carried number) = 18. 10 tens make 1 hundred . So 1 is carried to the hundreds place and added to the hundreds place values and 8 is represented in tens place
hundreds place = 9 + 8 + 1(carried number)= 18. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 8 is represented in hundreds place
thousands place = 8 + 1 +1(carried number) = 10 =10 thousand + 0 thousand. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 0 is represented in thousands place
ten thousands =5 + 2 + 1(carried number) = 8.
Hundred thousands = 2 + 1 = 3.
Sum = 380,880.
g.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 83,906 and 35,808
first adding the ones place = 6 + 8 = 14 = 10 ones + 4 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 4 is represented in ones place
ten place : 0 + 0 + 1(carried number) = 1.
hundreds place = 9 + 8 = 17. 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 7 is represented in hundreds place
thousands place = 3 + 5 +1(carried number) = 9.
ten thousands =8 + 3 = 11. 10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 1 is represented in ten thousands place
Hundred thousands = 1(carried number) = 1.
Sum = 119,714.
h.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 289,999 and 91,849
first adding the ones place = 9 + 9 = 18 = 10 ones + 8 ones . 10 ones make 1 tens . So 1 is carried to the tens place and added to the tens place values and 8 is represented in ones place
ten place : 9 + 4 + 1(carried number) = 14. 10 tens make 1 hundreds . So 1 is carried to the hundreds place and added to the hundreds place values and 4 is represented in tens place
hundreds place = 9 + 8 +1(carried number)= 18 . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 8 is represented in hundreds place
thousands place = 9 + 1 +1(carried number) = 11. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 1 is represented in thousands place.
ten thousands =8 + 9 +1(carried number) = 18. 10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 8 is represented in ten thousands place
Hundred thousands = 2 +1(carried number) = 3.
Sum = 381,848
i.
Explanation:
While adding two numbers we add the numbers according to the place values individually. We add number from ones place to the largest place value.
given number 289,999 and 91,849
first adding the ones place = 0 + 0 = 0.
ten place : 0 + 0 = 0.
hundreds place = 9+1= 10 . 10 hundreds make 1 thousands . So 1 is carried to the thousands place and added to the thousands place values and 0 is represented in hundreds place
thousands place = 4 + 5 +1(carried number) = 10. 10 thousands make 1 ten thousands . So 1 is carried to the ten thousands place and added to the ten thousands place values and 0 is represented in thousands place.
ten thousands =5 + 4 +1(carried number) = 10. 10 ten thousands make 1 hundred thousands . So 1 is carried to the hundred thousands place and added to the hundred thousands place values and 0 is represented in ten thousands place
Hundred thousands = 7 + 2 +1(carried number) = 1. 10 hundred thousands make 1 million . So 1 is carried to the millions place and added to the millions place values and 0 is represented in hundred thousands place.
Millions = 1 (carried over number) = 1.
Sum = 1,000,000.
Draw a tape diagram to represent each problem. Use numbers to solve, and write your answer as a statement.
Question 2.
At the zoo, Brooke learned that one of the rhinos weighs 4,897 pounds, one of the giraffes weighs 2,667 pounds, one of the African elephants weighs 12,456 pounds, and one of the Komodo dragons weighs 123 pounds.
Given
One of the rhinos weights at the zoo = 4,897 pounds.
One of the giraffes weights at the zoo = 2,667 pounds
One of the African elephants weights at the zoo = 12, 456 pounds.
One of the Komodo dragons at the zoo weighs = 123 pounds.
a. What is the combined weight of the zoo’s African elephant and the giraffe?
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# Matrices I
• ### Introduction
A matrix (plural : matrices) is a rectangular arrangement of $m \times n$ numbers in $m$ rows and $n$ columns. It is written in between a pair of rectangular brackets. The numbers belonging to a matrix are known as its elements. Matrices are represented by capital Roman letters. For example, consider the following matrix $M$.
${M = \begin{bmatrix} 2& 3 & 0 \\[0.3em] 1 & -3 & \sqrt 2 \end{bmatrix}}$
It has 2 rows and 3 columns. Hence it is a $2 \times 3$ matrix.
If a matrix has $m$ rows and $n$ columns, it is said to be of the order $m$ by $n$. So, $M$ is of the order 2 by 3 and has 6 elements.
• ### Representation of an Element
$a_{ij}$ corresponds to that element of a matrix, which is present in the $i$th row and $j$th column of it. So, for matrix $M$,
${a_{21} = 1, a_{13}= 0, a_{23}= \sqrt 2}$
• ### Transpose of a matrix
A matrix obtained by interchanging the rows and columns of a matrix is known as the transpose of the matrix. So,
$M^T = \begin{bmatrix} 2& 1 \\[0.3em] 3 & -3 \\[0.3em] 0 & \sqrt 2 \end{bmatrix}$
• ### Equality of 2 matrices
2 matrices A and B are equal iff (if and only if)
i) their order is same and
ii) $\forall$ $a$ in $A$ and $b$ in $B$, $a_{ij}= b_{ij}$
• ### Row Matrix
A row matrix has only 1 row and more than 1 column. e.g.
${P = [1 \ 3 \ 2]}$
• ### Column Matrix
A column matrix has only 1 column and more than 1 row. e.g.
${Q = \begin {bmatrix} 2 \\ 3 \\ 5 \\ 6 \end {bmatrix}}$
• ### Null Matrix
A null matrix is the one, whose all elements are $0$. e.g.
${R = \begin {bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \end {bmatrix}}$
• ### Square Matrix
A square matrix has equal number of rows and columns. So, the total number of elements in a square matrix is always a perfect square. e.g.
${S = \begin {bmatrix} 1 & 4 & \sqrt 5 & -6 \\ 77 & 64 & \frac 1 2 & 0 \\ e & \pi & 5 & 3 \\ 0 & 1 & 1 & 4 \end {bmatrix}}$
So, $S$ has 4 rows as well as 4 columns, so 16 elements.
The elements $a_{ij}$ of a square matrix, for which $i=j$ are known as the diagonal elements. Thus, $1,64, 5$ and $4$ are diagonal elements of $S$.
• ### Diagonal Matrix
A square matrix, whose all elements are zero, except the diagonal elements, is known as the diagonal matrix.e.g.
${D = \begin {bmatrix} 3 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & 13 \end {bmatrix}}$
• ### Identity Matrix
An identity matrix is a square matrix, whose diagonal elements are equal to 1 and all other elements are 0. This matrix is of special importance in the matrix theory. Similar to the number $1$ in the theory of numbers, it acts as a unit entity. It is generally denoted by $I$. So, for any square matrix $A$, $AI = A$. Note that $A$ and $I$ must have the same order.
${I_{2} = \begin {bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}}$
${I_{3} = \begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {bmatrix}}$
• ### Upper- and Lower-Triangular Matrices
An upper-triangular matrix is the one, whose elements below the diagonal are $0$. Similarly, a lower-triangular matrix is the one, whose elements above the diagonal are $0$.
${U = \begin {bmatrix} 1 & 5 & 6 \\ 0 & 3 & 9 \\ 0 & 0 & 34 \end {bmatrix}}$
${L = \begin {bmatrix} 6 & 0 & 0 \\ 2 & 2 & 0 \\ 3 & 5 & 5 \end {bmatrix}}$
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# Mathematical Preliminaries (Advanced Methods in Computer Graphics) Part 3
## Curves
In Sect. 2.3, we came across the equation of a straight line expressed in terms of linear polynomials of a single parameter t (Eq. 2.12). Polynomials of a higher degree in t can be used to define curves in three-dimensional space. In the most general form, a curve can be represented as P(t) = (x(t), y(t), z(t)), where x(t), y(t), z(t) are continuous and differentiable functions of the parameter t. Polynomials of degree n have the property that their derivatives up to order n— 1 exist and are continuous over any finite interval in the parameter space. We can use the derivatives of the functions to define the tangential and normal directions to the curve at any point, and also to construct an orthonormal basis at any point on the curve.
The tangent vector at P(t) is given by the first derivative with respect to t, i.e., P’(t) = (x(t), y(t), z(t)). The unit tangent vector is denoted as
The tangent vector represents the local orientation of the curve at a point. If the parameter t denotes time, then P (t) represents the instantaneous velocity of the moving point P(t). The distance travelled from a starting point A = P(t0) to the current point, or in other words the arc length measured from A, is given by
Using the above equation we can express t as a function of arc length s, and re-parameterize the curve as P(s) = (x(s), y(s), z(s)). The chain rule for differentiation gives
from which we find that P’(s) is equivalent to the unit tangent vector T(t). For convenience, we denote P’(s) by T(s). Since T(s>T(s) = 1, it immediately follows that T(s>T’(s) = 0. Thus the instantaneous rate of change of the tangent direction is parallel to the normal vector at that point. If the unit normal direction at P(s) is denoted as N(s), we have
The proportionality factor k(s) is called the curvature of the curve at P(s). The curvature is a measure of the deviation of the curve from a straight line. For a straight line, k(s) = 0 at all points. The magnitude of the curvature is easily obtained as |k(s)| = T(s)|, and the unit normal direction at P(s) is given by
Fig. 2.8 Frenet frame attached to a curve at the point P
The plane containing the tangent vector and the normal vector is known as the osculating plane. The cross-product of the two unit vectors T(s) and N(s) gives the direction of the unit bi-normal vector denoted by B(s):
The three unit vectors T, N, B form an orthonormal basis as shown in Fig. 2.8. This local reference system is called the Frenet frame. The derivative of the binormal vector B’(s) is perpendicular to both B(s) and T(s), and hence parallel to N(s):
The term r (s) is called the torsion of the curve at s. Torsion is a measure of how much the curve deviates from the osculating plane.
The plane containing the tangent and binormal vectors is called the rectifying plane (Fig. 2.8). The plane formed by the normal and binormal vectors is called the normal plane.
The Frenet frame is useful for defining the local orientation of objects that move along a curved path. It can also be used for defining the eye-coordinate system for a camera that undergoes a curvilinear motion.
## Affine Transformations
In this section, we consider linear transformations of three-dimensional points and vectors. The homogeneous coordinate system (Sect. 2.1) allows all transformations including translations to be represented using 4 x 4 matrices. We denote a translation by a vector v = (xv, yv, zv), by Tv, a rotation about the x-axis by an angle by 9, by Re(x), and a scaling by a vector k = (xk, yk, zk), by Sk (Box 2.3).
Box 2.3 Fundamental 3D Transformations (Fig. 2.9)
Fig. 2.9 Examples showing transformations of (a) a translation by an offset vector v (b) a rotation about the x-axis by an angle d and (c) scaling by factors kx, ky, kz
A linear transformation followed by a translation is called an affine transform. A general transformation can be given in matrix form as follows:
In the above equation, the matrix elements aij’s are all constants. (a03, ai3, a23) denote the translation components, and (xp, yp, zp, 1) the point on which the transformation is applied. The translation parameters do not have any effect on a vector (xv, yv, zv, 0). Under an affine transformation, line segments transform into line segments, and parallel lines transform into parallel lines. A fixed point of a transformation is a point that remains invariant under that transformation. For example, every point along the x-axis is a fixed point for the transformation R0(x). Similarly, the origin is a fixed point for the scale transformation. The most general rotation of an object with the origin as a fixed point, is the rotation by an angle 6 about an arbitrary vector v = (xv, yv, zv, 0) passing through the origin. The matrix for this transformation is given below.
where A = (1—cos0), B = sin0, and C = cos0. A rotation about an axis parallel to the x-axis, with an arbitrary fixed point P, can be obtained by first applying a translation T_p from P to the origin, a rotation R0 (x) with origin as the fixed point, and finally a translation Tp back to the original position P. In matrix form, we write the composite transformation as TpR0(x)Tp_1. Here T_1 denotes the inverse of the transformation T. For a translation, the inverse of Tp is T_p ; and for a rotation, the inverse of R0 (v) is R_0(v). A transformation of the form TRT-1 is called the conjugate of R.
We have just seen a few examples of affine transformations that are commonly used for generating new points by transforming existing ones. We could also combine the coordinates of a set of points using a linear equation to obtain a new point. Such interpolation methods are discussed in the next section.
## Affine Combinations
A linear combination of a set of points Pi (i = 1,2,… n) produces a new point Q as shown below:
Fig. 2.10 (a) Linear interpolation and (b) trigonometric interpolation between two points where the coefficients (weights) wi are constants. If the weights satisfy the condition
then Eq. 2.41 gives an affine combination of points. Additionally, if wi > 0, for all i, then wi ’s form a partition of unity, and Eq. 2.41 is said to give a convex combination of points. As a special case, when n = 2, we get the formula for linear interpolation between two points Pi and P2:
An interesting variation of the above equation can be derived by expressing the parameter t as a function of an angle a, given by t = cos2a. Then the coefficient (1— t) becomes sin2a, and Eq. 2.43 takes the form Q = sin2a P1 + cos2a P2. However, this trigonometric interpolation formula gives a non-uniform distribution of points on the line when a is varied from 00 to 90° in equal steps. A comparison of linear and trigonometric interpolations is given in Fig. 2.10. In Fig. 2.10a, the parameter t is varied uniformly in the range [0-1] in steps of 0.1, and in Fig. 2.10b, the angle a is varied uniformly in the range [0-90] in steps of 91. Higher order interpolation between points is discussed in Chap. 7 (Box 2.4).
### Box 2.4 Bernstein Polynomials
Given a positive integer value n, we can construct n + 1 polynomials of degree n of a parameter t as follows:
These polynomials form a partition of unity, i.e.,
Therefore, they can be used to generate convex combinations of points. Given n + 1 points Pi, i = 0,… ,n, we define a point Q(t) as
As the parameter t is varied from 0 to 1, we get a continuous parametric curve called the Bezier curve. The equations for n = 1, 2, 3 are given below.
First degree (linear): Q(t) = (1—t) Po + tP1
Second degree (quadratic) : Q(t) = (1—t)2P0 + 2(1—t)t P1 + t2P2
Third degree (cubic) : Q(t) = (1—t)3P0 + 3(1— t)21P1 + 3(1— t)tP + t3P3
Fig. 2.11 A bilinear interpolation scheme first interpolates along the edges to get the values at A and B, and then uses another linear interpolation along the line AB to get the value at Q
Given a triangle with vertices P1, P2 and P3, we can perform a bilinear interpolation between the values defined at the vertices to get the interpolated value at an interior point Q (Fig. 2.11). Using this scheme, we can compute the colour value at any point inside a triangle, given the colour values at the vertices. A scan-line parallel to the base of the triangle sweeps the plane and generates the values of A and B using the linear interpolation equation in Eq. 2.43 with the same parameter t. Another linear interpolation between of A and B with a parameter s gives the value of Q. Thus we get
The above equation could be simplified into a simple convex combination of vertex points as
where k1 = s(1—t) and k2 = t. The bilinear interpolation of vertex coordinates shown above can be generalized to interpolate any quantity or attribute inside a triangle, given its values at the vertices. Examples of such vertex attributes are colour, texture coordinates and normal vectors. In the next section, we will consider another closely related interpolation method for triangles.
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660 views
### C1 - Simplifying a fraction that has a root on the denominator
(7+√5)/(√5-1)
In this example we are given (√5-1) as our denominator, and we want to simplify this complex fraction into the form a + b√5.
In order to do this our first step is to elimate the denominator. To do this, we use the rule associated with the Difference of Two Squares. For example, if you were told to factorise and simplify: x 2 - 16, you would get (x-4)(x+4) since when you expand the '-4x' and '+4x' would cancel each other out to leave x2 - 16.
To return to our question therefore, we need to multiply our fraction by something to eliminate the √5 on the denominator. This something is (√5 +1) as the opposite signs will mean the middle part of our expanded equation +√5-√5 will cancel each other out, and multiplying √5 by itself will eliminate the square root. But we must remember, that whatever we do to the bottom of the fraction we must also do to the top, therefore we must multiply both the numerator and the denominator by (√5 + 1)
So: (7+√5)/(√5-1) x (√5+1)/(√5+1)
Multiplying the numerators (top of the fractions) together, we get: (7x√5) + (7x1) + (√5x√5) + (√5x1), which equals: 7√5 + 7 + 5 + √5
By collecting like terms this simplifies to: 8√5 + 12
Multiplying the denominators (bottom of the fractions) together, we get: (√5x√5) + (√5x1) - (1x√5) - (1x1) which equals: 5 + √5 - √5 - 1
Again, by collecting like terms this simplifies to: 4
Therefore, we can also write (7+√5)/(√5-1) as:
(8√5 + 12) / 4
But this can be simplified further as both 8, and 12 are divisible by 4, and therefore this can finally be written as:
2√5 + 3 or 3 + 2√5
SOLUTION: a = 3 and b = 2
2 years ago
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#### Other A Level Maths questions
Differentiate y = (x^2 + 3)^2
How do I find dy/dx for the following equation: (x^2) + 2y = 4(y^3) + lnx?
The radius of a circular disc is increasing at a constant rate of 0.003cm/s. Find the rate at which the area is increasing when the radius is 20cm.
How do you add or subtract complex numbers?
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## How do u find the volume of a cylinder?
The formula for finding the Volume of a right circular cylinder is: V = πr2h, where r is the radius of the circle at one base of the cylinder, and h is the height of the cylinder (the distance between the bases.) Example: The radius of the base of a right circular cylinder is 8 cm.
## What is the formula for volume?
Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height.
## Why is the formula for volume of a cylinder?
Thus, the volume of the cylinder can be given by the product of the area of base and height. For any cylinder with base radius ‘r’, and height ‘h’, the volume will be base times the height. Therefore, the volume of a cylinder = πr2h cubic units.
## What is the formula for cubic volume?
Volume of a cube = side times side times side. Since each side of a square is the same, it can simply be the length of one side cubed. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches.
## What is a formula of cylinder?
Solution. The formula for the volume of a cylinder is V=Bh or V=πr2h . The radius of the cylinder is 8 cm and the height is 15 cm. Substitute 8 for r and 15 for h in the formula V=πr2h .
## What is the area and volume of cylinder?
So the total area = 2(π r2) + h(2π r2). VOLUME OF A CYLINDER is the area of the top times the height. The area of the top is given by the formula for the area of a circle (π r2). So the total volume = h(π r2).
## What is TSA of cylinder?
Consider a cylinder of radius r and height h. The total surface area (TSA) includes the area of the circular top and base, as well as the curved surface area (CSA).
## Which cylinder has the greatest volume?
Cylinder 1
### Releated
#### Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […]
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# 1709 Sometimes “Guess and Check” Is a Good Strategy
Contents
### Today’s Puzzle:
Most of the puzzles I publish are logic puzzles, and I encourage you to find the logic of the puzzle and not guess and check. However, guess and check is a legitimate strategy in mathematics, and it is a legitimate strategy to solve this particular puzzle.
Since one of the clues is -9, we know that the two boxes under it must be [1, 10], [2, 11], or [3, 12].
Suppose you assume it’s 1 – 10 = -9. If you fill out the rest of the boxes you would get:
You know that isn’t right because zero is not a number from 1 to 12. No problem. Simply add one to each of the numbers you wrote in, and the puzzle will be solved with only numbers from 1 to 12.
Suppose you assumed it’s 3 -12 = -9. The rest of the boxes would look like this:
Again, 13 is not included in the numbers from 1 to 12, but you can fix it by subtracting 1 from each of the numbers you wrote in. Easy Peasy.
### Factors of 1709:
• 1709 is a prime number.
• Prime factorization: 1709 is prime.
• 1709 has no exponents greater than 1 in its prime factorization, so √1709 cannot be simplified.
• The exponent in the prime factorization is 1. Adding one to that exponent we get (1 + 1) = 2. Therefore 1709 has exactly 2 factors.
• The factors of 1709 are outlined with their factor pair partners in the graphic below.
How do we know that 1709 is a prime number? If 1709 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1709. Since 1709 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, or 41, we know that 1709 is a prime number.
### More About the Number 1709:
1709 is the sum of two squares:
35² + 22² = 1709.
1709 is the hypotenuse of a Pythagorean triple:
741-1540-1709, calculated from 35² – 22², 2(35)(22), 35² + 22².
Here’s another way we know that 1709 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 35² + 22² = 1709 with 35 and 22 having no common prime factors, 1709 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √1709. Since 1709 is not divisible by 5, 13, 17, 29, 37, or 41, we know that 1709 is a prime number.
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# Goldbach's Conjecture Proof
$\color{#D61F06}{\text{This proof does not demonstrate Goldbach's Conjecture.}}$
$\color{#D61F06}{\text{ Can you figure out what this proof actually shows?}}$
Goldbach's Conjecture says that all even integers greater than $2$ can be written as a sum of $2$ prime numbers (may or maynot be distinct).
$\therefore 2n = p_1 + p_2$
PROOF
$\textbf{CASE 1}$ - $p_1$ & $p_2$ are not equal to $2$
We know that $p_1 + p_2$ is always divisible by $2$. It can easily be proven -
For all $p_i$ not equal to $2$,
$p_1 = 1 \pmod{2}$
$p_2 = 1 \pmod{2}$
$\therefore p_1 + p_2 = 0 \pmod{2}$
Let $p_2 \geq p_1$, so let $p_2 = p_1 + a$ for some integer $a$.
$\therefore 2n = p_1 + p_1 + a$
$\therefore n = \dfrac{2p_1 + a}{2}$
We know $2 \mid (p_1+ p_2)$ and $2 \mid 2p_1 \implies 2 \mid a$
Thus, let $a = 2x$
$\therefore n = \dfrac{2(p_1 + x)}{2}$
$\therefore n = p_1 + x$
Taking $\pmod{p_1}$ on both sides,
$\therefore n \pmod{p_1} = (p_1 + x) \pmod{p_1 }$
For $n=p_1$,
$\therefore x$ would be $0$ .
For $n\neq p1$,
$n \equiv p_1+x\equiv x \pmod{p_1}$
Now ($n \pmod{p_1}$) can be any number less than $p_1$ and we can assume that $(x\pmod{p_1})$ can also be any number less than $p_1$ keeping in mind that there are infinite primes and there is no sequence in the gap between any prime.
To keep in mind- Also, $x$ cannot be equal to $p_1$ because then, $p_2$ will have a prime factor.
One more fact is that for $n$ being a multiple of $p_1$, it cannot be proven.
Hence, it is proven for $p_1$ and $p_2$ not equal to $2$.
$\textbf{CASE 2}$- $p_1$ or/and $p_2$ equal to $2$
$2n = p_1 + p_2$
where one or both maybe 2. It is obvious that for $p_1$ being $2$ and $p_2$ any other prime, it will be an odd integer.
So, the only possibility is that $p_1$ and $p_2$ are equal to $2$.
So, $2n = 2 + 2 \implies n=2$ (See, $4$ can be written as sum of 2 primes as $2+2$ )
Hence, proved that both the primes have to be $2$ simultaneously, or both have to be $\neq 2$.
Thank You for reading it. If you want to know anything, you can e-mail me. Please tell if my proof is correct or not!!
Note by Kartik Sharma
6 years, 10 months ago
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I think you've misread the Goldbach Conjecture? If any two prime numbers $>2$ are added, the result will always be even. Goldbach's Conjecture proposes that for any integer $n>2$, there exists $2$ prime numbers that add up to $2n$. It' s trivial to show that at least one of them can be a prime number, but to show that both can be is a bit harder.
- 6 years, 10 months ago
A 'bit'?
- 6 years, 10 months ago
Yes, a bit.
- 6 years, 10 months ago
That's what was bugging me! @Kartik Sharma
- 6 years, 10 months ago
If only I could format the LaTeX for this.
- 6 years, 10 months ago
I will do it, I can do that as a moderator. @Sharky Kesa .
@Kartik Sharma, this is not the proof of Goldbach Conjecture, this doesn't prove the existence of 2 primes that sum up to give the number, for any even integer. What you proved is, if sum of 2 primes is $2n$, then either both primes are $2$ or none of them is $2$. Isn't this quite obvious ? If you want sum to be even , then both the numbers have to be odd, or both even. BTW, if it was this easy, why would it be set as a challenge in mathematics ?
- 6 years, 10 months ago
Thanks, @Krishna Ar for letting me know about this conjecture! And you too @Sharky Kesa for giving me some reviews! You can share your view on it here!
- 6 years, 10 months ago
Dear Kartik, Goldbach Conjecture is $NOT$ that
$\bullet$ If an even integer can be written as sum of 2 prime numbers, then both of them have to be odd primes or both have to be equal to $2$ $\implies \Huge{\color{#D61F06}{\times}}$
Goldbach Conjecture is that
$\bullet$ Every even integer greater than $2$ can be written as sum of $2$ primes. $\implies \Huge{\color{#20A900}{\surd}}$
Please note this. I have edited the note, do not re-edit it to remove the changes I have made.
- 6 years, 10 months ago
Hey, I think I have proved it only. The Case 2 should not have been there, it is misleading my 'proof'. Why do you think I haven't proved it?
I took the equation as 2n = p1 + p2 which is correct, right?
And I proved it equal, but it is wrong, why? Which step is wrong?
- 6 years, 10 months ago
Now $n \pmod{p_1}$ can be any $\ldots \ldots$ between any prime.
Could you explain what you did in that paragraph?
- 6 years, 10 months ago
Well, I have just assumed one thing. It says that $x mod {p}_{1}$ may vary as we don't know for sure what x is with respect to any prime ${p}_{1}$ and it is quite obvious that $n mod {p}_{1}$ can also vary.
- 6 years, 10 months ago
Nope. Can't understand a bit. You need to define a lot of things first. What is $p_1$? What is $p_2$? What is $n$? Are they fixed?
- 6 years, 10 months ago
Well, your proof just deals with that $p_1$ and $p_2$ will be odd primes, (started with that assumption only) , but nowhere in the proof you can assume what is to be proved. And btw, your proof also leads to that $18$ can be written as $3+15$ or $9+9$, which is not wanted (we want primes)
Your proof doesn't prove that there will be 2 primes for every even positive integer. Because you assumed only that $2n=p_1+p_2$ , and then you proved that $p_1\equiv p_2 \pmod{2}$. That's it.
- 6 years, 10 months ago
I still didn't understand what's wrong with it... Well, 18 can be written as 3 + 15 is not shown by my proof. What are you saying? I have already said ${p}_{1} + x$ should lead to another $prime$ ${p}_{2}$ and as that gap may vary, I have taken x as any number.
I have never proven the latter. Well, I am not convinced by your review. Actually, you are neglecting many parts of my proof- mainly the one Siddhartha Srivastva has asked.
- 6 years, 10 months ago
Typo, 13+5, not 3+5.
- 6 years, 10 months ago
Well, actually 3 + 15 another typo!
- 6 years, 10 months ago
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# Which number replaces the question mark?$a.{\text{ 10}} \\ b.{\text{ 12}} \\ c.{\text{ 14}} \\ d.{\text{ 16}} \\$
Last updated date: 16th Sep 2024
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Hint: - Addition of the corner numbers and divided by three is equal to the middle number.
As we see that the first figure follows the pattern which is addition of the corner numbers and divided by three is equal to the middle number.
$\therefore 5 = \dfrac{{2 + 12 + 1}}{3} \\ \therefore 5 = \dfrac{{15}}{3} \\ \therefore 5 = 5 \\$
Which is true.
Second figure also follows the pattern which is addition of the corner numbers and divided by three is equal to the middle number.
$\therefore 9 = \dfrac{{9 + 11 + 7}}{3} \\ \therefore 9 = \dfrac{{27}}{3} \\ \therefore 9 = 9 \\$
Which is true.
Therefore third figure also follow the same pattern
$\therefore ? = \dfrac{{4 + 15 + 17}}{3} = \dfrac{{36}}{3} = 12$
So, 12 replaces the question mark.
Hence, option (b) is correct.
Note: - Whenever we face such types of questions first determine which pattern is followed by the two figures, so the third figure also follows the same pattern, and then we will easily calculate which number is replaced by question mark.
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# Quick Tip: Smoothly Move an Entity to the Position of the Mouse
Difficulty:IntermediateLength:ShortLanguages:
Here's a simple snippet of code that comes in handy all the time: how to move an object from one point to another, in a smooth, flowing, continuous motion. We'll use the Pythagorean distance and a bit of easing to stop things getting jittery.
Note: Although this tutorial is written using JavaScript, you should be able to use the same techniques and concepts in almost any game development environment.
## 1. Explanation
In some games you may wish to move a character to the position of the mouse. Sometimes you'll wish to do this when the player clicks, other times you will want the character constantly moving toward the mouse position and coming to rest once it reaches the same position as the mouse. We will be doing the latter in this tutorial, but adjusting this to work with mouse clicks will be trivial.
## 2. Distance Formula
To move the entity to the position of the mouse, we first need to know how far away the entity is from the mouse.
To do this we will use the distance formula. This uses Pythagoras's theorem and is calculated as follows, for coordinates (x1, y1) and (x2, y2):
$d=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
In other words, you square the difference between the x-coordinates, square the difference between the y-coordinates, add the two squares together, and take the square root of the sum.
To help understand how this works I have created the following image.
In the above image the x-distance is 7 and the y-distance is 6. Working through the steps we arrive at a distance of approximately 9.21.
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ d=\sqrt{(10-3)^2+(3-9)^2}\\ d=\sqrt{(7)^2+(-6)^2}\\ d=\sqrt{49+36}\\ d=\sqrt{85}\\ d\approx9.21$
## 2. Code Implementation
To implement this in code we will take the x- and y-coordinates of both our mouse cursor and our entity (which in this case is a spaceship). Then, we will apply the distance formula, and increase or decrease the entity's x and y positions, as long as the distance is greater than 1.
Below is some code to illustrate this:
I have coded up a demo that shows all this in action:
## 3. What is Easing?
You may have noticed that in the step above we were multiplying the xDistance and yDistance by an easingAmount.
This makes our entity slow down as it approaches its target, rather than moving the same distance on each tick of the game loop. This is known as easing out.
Try adjusting the value of easingAmount in the jsFiddle of the demo above, and see what effect it has. You could also experiment with the code inside the tick() function.
## 4. Demo Game
I have coded a very simple demo to show you how this might be applied to a real game. Experiment with the code and see what you can come up with!
(Graphics from Everaldo Coelho, Sneh Roy, and our own Jacob Zinman-Jeanes.)
## Conclusion
In this short Quick Tip you learned how to move an entity to the mouse position. See if you can adjust the code so that the entity moves to the last position where the player clicked, rather than continuously following the mouse. I hope you found this useful - thanks for reading!
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# Class 12 RD Sharma Solutions – Chapter 17 Increasing and Decreasing Functions – Exercise 17.1
• Last Updated : 03 Jan, 2021
### Question 1: Prove that the function f(x) = loge x is increasing on (0,∞).
Solution:
Let x1, x2 ∈ (0, ∞)
We have, x1<x2
⇒ loge x1 < loge x
⇒ f(x1) < f(x2)
Therefore, f(x) is increasing in (0, ∞).
### Question 2: Prove that the function f(x) = loga (x) is increasing on (0,∞) if a>1 and decreasing on (0,∞) if 0<a<1.
Solution:
Case 1:
When a>1
Let x1, x2 ∈ (0, ∞)
We have, x1<x2
⇒ loge x1 < loge x2
⇒ f(x1) < f(x2)
Therefore, f(x) is increasing in (0, ∞).
Case 2:
When 0<a<1
f(x) = loga x = logx/loga
When a<1 ⇒ log a< 0
let x1<x2
⇒ log x1<log x2
⇒ ( log x1/log a) > (log x2/log a) [log a<0]
⇒ f(x1) > f(x2)
Therefore, f(x) is decreasing in (0, ∞).
### Question 3: Prove that f(x) = ax + b, where a, b are constants and a>0 is an increasing function on R.
Solution:
We have,
f(x) = ax + b, a > 0
Let x1, x2 ∈ R and x1 >x2
⇒ ax1 > ax2 for some a>0
⇒ ax1 + b > ax2 + b for some b
⇒ f(x1) > f(x2)
Hence, x1 > x2 ⇒ f(x1) > f(x2)
Therefore, f(x) is increasing function of R.
### Question 4: Prove that f(x) = ax + b, where a, b are constants and a<0 is a decreasing function on R.
Solution:
We have,
f(x) = ax + b, a < 0
Let x1, x2 ∈ R and x1 >x2
⇒ ax1 < ax2 for some a>0
⇒ ax1 + b <ax2 + b for some b
⇒ f(x1) <f(x2)
Hence, x1 > x2 ⇒ f(x1) <f(x2)
Therefore, f(x) is decreasing function of R.
### Question 5: Show that f(x) = 1/x is a decreasing function on (0,∞).
We have,
f(x) = 1/x
Let x1, x2 ∈ (0,∞) and x1 > x2
⇒ 1/x1 < 1/x2
⇒ f(x1) < f(x2)
Thus, x1 > x2 ⇒ f(x1) < f(x2)
Therefore, f(x) is decreasing function.
### Question 6: Show that f(x) = 1/(1+x2) decreases in the interval [0, ∞] and increases in the interval [-∞,0].
Solution:
We have,
f(x) = 1/1+ x
Case 1:
when x ∈ [0, ∞]
Let x1, x2 ∈ [0,∞] and x1 > x2
⇒ x12 > x22
⇒ 1+x12 < 1+x22
⇒ 1/(1+ x12 )> 1/(1+ x2 )
⇒ f(x1) < f(x2)
Therefore, f(x) is decreasing in [0, ∞].
Case 2:
when x ∈ [-∞, 0]
Let x1 > x2
⇒ x12 < x2 [-2>-3 ⇒ 4<9]
⇒ 1+x12 < 1+x22
⇒ 1/(1+ x12)> 1/(1+ x22 )
⇒ f(x1) > f(x2)
Therefore, f(x) is increasing in [-∞,0].
### Question 7: Show that f(x) = 1/(1+x2) is neither increasing nor decreasing on R.
Solution:
We have,
(x) = 1/1+ x2
R can be divided into two intervals [0, ∞] and [-∞,0]
Case 1:
when x ∈ [0, ∞]
Let x1 > x2
⇒ x12 > x22
⇒ 1+x12 < 1+x22
⇒ 1/(1+ x12 )> 1/(1+ x22 )
⇒ f(x1) < f(x2)
Therefore, f(x) is decreasing in [0, ∞].
Case 2:
when x ∈ [-∞, 0]
Let x1 > x2
⇒ x12 < x22 [-2>-3 ⇒ 4<9]
⇒ 1+x12 < 1+x22
⇒ 1/(1+ x12)> 1/(1+ x22 )
⇒ f(x1) > f(x2)
Therefore, f(x) is increasing in [-∞,0].
Here, f(x) is decreasing in [0, ∞] and f(x) is increasing in [-∞,0].
Thus, f(x) neither increases nor decreases on R.
### (i) strictly increasing in (0,∞) (ii) strictly decreasing in (-∞,0)
Solution:
(i). Let x1, x2 ∈ [0,∞] and x1 > x2
⇒ f(x1) > f(x2)
Thus, f(x) is strictly increasing in [0,∞].
(ii). Let x1, x2 ∈ [-∞, 0] and x1 > x2
⇒ -x1<-x2
⇒ f(x1) < f(x2)
Thus, f(x) is strictly decreasing in [-∞,0].
### Question 9: Without using the derivative show that the function f(x) = 7x – 3 is strictly increasing function on R.
Solution:
f(x) = 7x-3
Let x1, x2 ∈ R and x1 >x2
⇒ 7x1 > 7x2
⇒ 7x1 – 3 > 7x2 – 3
⇒ f(x1) > f(x2)
Thus, f(x) is strictly increasing on R
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## Apply these shortcuts to impress your interviewer with outstanding mental math skills
The best way to prepare for the quantitative portion of case interviews is to practice the types of calculations you will need to perform with numbers representative of actual Case Numbers and to be familiar with the types of problems you are most likely to encounter and how to solve them. As a reminder, candidates cannot use calculators or spreadsheets in the case interviews but may use pen and paper. All the Case Calculations take this into account, so Case Numbers and Case Calculations usually have certain characteristics you can use to simplify the calculations (if you are familiar with certain calculation methods).
For example, Case Numbers are usually round numbers with only a few significant digits, such as 300,000 or 4,000,000. While some people may be intimidated by having to do calculations with large numbers like this without a calculator, once you learn how to efficiently manage the zeroes, calculations with numbers like these become straightforward.
When Case Numbers have more than one significant digit, they usually have additional properties that can make certain calculations like multiplication and division much easier if you are aware of the appropriate methods. For example, let’s say you are told a company sells a product for \$32 per unit and sells 25,000 units per year, and you are asked to calculate the revenue. You would then need to do the calculation: (\$32 × 25,000). Most people would use pen and paper for this calculation. You can make the pen-and-paper calculation more efficient by not writing out the three zeroes from “25,000” and then adding them to the end of the result of (32 × 25).
These numbers are representative of actual Case Numbers, and here is an efficient method for this calculation (32 × 25). First decompose 25 as: (25 = ¼ × 100). Then we perform the calculation using this decomposition of 25.
32 × ¼ = 32 ÷ 4 = 8
8 × 100 = 800
Therefore (32 × 25 = 800) - you can check this with a calculator or pen-and-paper calculation. You can think of this new calculation method as follows: let’s say you are given 32 quarters (U.S. coins); how much money would you have? You might think to yourself: “Well, 4 quarters make 1 dollar, and 32 ÷ 4 = 8, so I would have \$8,” which is correct. Now think about how many cents are in \$8. The answer is clearly 800: (8 × 100 = 800), as there are 100 cents in 1 dollar. So 32 quarters make \$8, which is 800 cents. Now a quarter is also worth 25 cents, and since we have 32 quarters, then 32 × 25 = 800. So to multiply any number by 25, you divide the number by 4 and then multiply by 100. This method works because (4 × 25 = 100); so, when you divide 32 by 4, the result is the “number of groups of 100.” Now the answer to our original problem of (\$32 × 25,000) is then \$800,000.
Using this method relies on the fact that 25 = ¼ × 100. These numbers are representative of actual Case Numbers, and most actual multiplication calculations in case interviews involve numbers like 25,000 or 25 Million, for which there is a way to dramatically simplify the calculation.
Many case interview preparation resources have example calculations that you are highly unlikely to encounter in an actual case interview, such as multiplying by a number like 23,487, which has five significant digits and no clear way to simplify the calculation. Practicing quantitative problems that are not representative of actual Case Numbers of Case Calculations is not an effective way to prepare, as you won’t be practicing for what you will actually face in an interview. If the practice problems involve numbers with many significant digits, this becomes over-kill as you just practice brute-force mechanical calculations, and don’t practice the logical problem-solving aspects, and don’t get experience looking for and finding efficient calculation methods.
## Effective Quantitative Preparation Resources
This article has numerous examples of calculation methods and includes efficient solution methods to the example problems given in the previous article
## Mental Math Skills Are Extremely Valuable in Case Interviews
Because you cannot use calculators in case interviews — and because you may need to perform a large number of math operations — being able to do calculations mentally (without relying on pen and paper for the mechanics of calculations) is a very helpful skill in case interviews.
Consider the following example, where you are given Price, Quantity and Profit Margin for three different products and asked to calculate the percentage of the firm’s overall revenue and profit contributed by each product:
Let’s say your approach to answering this question is to first calculate the revenue and profit for each product. If you can do these operations mentally (and write down the answers) rather than writing out all the multiplication operations in longhand, you will have a large advantage in calculation speed. Longhand multiplication refers to doing a multiplication calculation like 7 × 25 as shown below:
If you need to do each multiplication operation from the above problem in longhand format, you would first need to copy the multiplication terms from the table onto pen and paper, and then do the calculation. Since calculating the revenue and profit for each product requires a large number of multiplication operations, a person who needs to do each multiplication operation in longhand format will be much slower than a person who can do the calculations mentally. It is much faster and more efficient to simply do each calculation mentally (without writing the original numbers out in longhand format) and then write down the result. The speed improvement of mental calculations is magnified if you need to multiply several numbers together when calculating a value.
For example: Profit = (Price) × (Quantity) × (Profit Margin)
For mental calculations, it is very helpful to “mentally picture” the numbers with which you are calculating —which I may refer to as “thinking” of a certain number or numbers. After doing a calculation mentally, you should always write down the result so you don’t forget it.
Given that the Case Numbers have few significant digits—and often have trailing significant digits of “5,” “25” or “75”—performing addition and subtraction with Case Numbers is relatively straightforward for most people. You may want to practice adding and subtracting numbers if you are rusty or don’t feel confident in being able to do this smoothly in an interview, which has added stress.
Example: Add the following numbers: 250 Million, 300 Million, and 150 Million.
### Methods for adding large numbers:
A helpful method for adding large numbers is to calculate with Units like Thousand (K), Million (M), Billion (B), and Trillion (T):
Units
K: Thousand
M: Million
B: Billion
T: Trillion
When adding large numbers, don’t write out (or mentally picture the number with) all the zeroes; simply write (or mentally picture) the leading digits and an abbreviation for the unit. For example, to represent “250 Million”—which, in long-form, would be written as “250,000,000”—write/think “250 M,” or mentally remember the Unit and do the calculations and, when you write down the result, include the appropriate Unit.
In case interviews, if you are given printed sheets with numerical values or tables of numbers, the printout will usually not include all the zeroes for a number like 25 Million; it will use a Unit like “Million” or the table will indicate the values are in “Millions” without writing out all the zeroes.
Using these Units, the prior example becomes:
## Multiplication methods
Multiplication is the most important quant skill to practice for case interviews because you will frequently have to perform numerous multiplication operations - sometimes involving three or more terms. You can improve your speed and efficiency with multiplication while reducing errors by learning and practicing a few efficient calculation methods. Below are a few examples of how to apply efficient calculation methods to case interview multiplication calculations.
Example: Calculate 160 × 350 M
Writing this calculation out in longhand with all the zeroes is a slow, error-prone process, as many people will make a mistake with the number of zeroes. It is also difficult to do this calculation mentally while keeping track of all the zeroes, and people frequently make a similar error regarding the number of zeroes with a mental calculation, which causes the answer to be off by a factor of 10 or more. This is clearly a very bad mistake to make in a job interview.
FastMath Solution Method
The FastMath Solution Method is different from how most people would approach this, but each step is very simple:
Step 1: What’s 160 ÷ 2
Answer: 160 ÷ 2 = 80
Step 2: What’s 350 × 2
Answer: 350 × 2 = 700
Step 3: What’s 80 × 700
Answer: 80 × 700 = 56,000
Guess what? That’s the same answer as 160 × 350 — you can check it with a calculator!
We get the answer by cutting 160 in half, which is 80, and doubling 350, which is 700, and multiplying these results (80 × 700). This is called the Halve and Double Method.
So the answer to the original problem is 56,000 Million, which is 56 Billion.
### Bullet Operator for Multiplication
To improve the readability of equations, we may use the “bullet” (also called the “dot”) symbol “⋅” to indicate multiplication. Therefore, the equation (20 × 4 = 80) could be written as (20 ⋅ 4 = 80) with the bullet/dot symbol. This can improve the readability of equations where you are multiplying the number of terms as the dot symbol is smaller than the cross symbol “×” and so it’s easier to read variable names and numbers.
Example: Calculate 125 ⋅ 2.5 ⋅ 4
Another method useful for multiplying numbers is to reorder the multiplication operations into a different sequence that is easier to calculate. We can multiply these numbers in this sequence:
125 ⋅ 4 = 500
500 ⋅ 2.5 = 1,250
We’ll call this the “Reordering” method, and it uses the mathematical property that, when multiplying a set of numbers, the order in which you multiply terms doesn’t matter; we’ll call this property the Reordering Property of multiplication. There are many potential ways in which you can Reorder the multiplication terms, and you can choose the order that is easiest for the particular numbers involved. There may be multiple efficient/effective ways to Reorder multiplication terms, and there is no single “correct” sequence. We could also calculate this as follows:
2.5 ⋅ 4 = 10
125 ⋅ 10 = 1,250
Most people find these new orders are much easier to execute than multiplying the first pair of numbers in the original calculation (125 ⋅ 2.5).
These are a few efficient multiplication methods covered—you can preview a video lesson that explains these and other multiplication methods in further detail here: Multiplication Lesson
## Division Methods
Some Case Calculations require division, and you frequently need to express the result as a percentage. There is usually a way to simplify the division Case Calculations, so you frequently don’t need to use Long Division. Candidates should still be familiar with and be prepared to use Long Division if they cannot find a simplification method, and candidates may encounter some Case Calculations where traditional Long Division is appropriate.
Below are a few examples of how to apply efficient calculation methods to division Case Calculations.
Example: Calculate 42 Billion ÷ 500
Doing the calculation using Long Division and writing out all the zeroes is a slow, error-prone process. It is also difficult to do this calculation purely mentally because of all the zeroes.
First, we write 42 Billion using our units, which becomes 42 B.
Second, we treat 500 as: (500 = ½ ⋅ 1,000 = ½ ⋅ K)
Then we calculate using this “decomposition” of 500.
42 B ÷ K = 42 M (since B ÷ K = M)
Next, we need to divide by ½, which is the same as multiplying by 2.
42 M ⋅ 2 = 84 M
To divide by 1,000 (K), we simply do the inverse of multiplying by 1,000 (K) and decrease the Unit size. Hence: (B ÷ K = M). We could do these operations in any order, so we could multiply by 2 first, and then divide by K.
## Multiplying and Dividing with “Clean” Numbers
Many of the Case Numbers involved in multiplication and/or division calculations in case interviews will have special properties and are what we will call “Clean Numbers,” defined as being a “Clean Fraction” times an Integer power of 10. In this context, a “Clean Fraction” means both the numerator and denominator (what we will call the “elements” of the fraction) are non-zero integers that are each smaller than 10 - and often smaller than 5.
For example, the number 25 can be represented as ¼ ⋅ 100; hence, 25 is a Clean Number, as the elements of the fraction are “1” and “4”, which are both smaller than 5. Below are some examples of how to express certain numbers as Clean Fractions times an Integer power of 10, showing that they are Clean Numbers:
A few things to note about Clean Numbers:
• The Numerator can be larger than 1, as shown with 75 = ¾ ⋅ 100.
• The Numerator can be larger than the Denominator, as shown with 150.
• The Integer exponent of 10 can be positive or negative, as shown with 0.05—which has a negative exponent of 10.
• You can represent percentages as a Clean Fraction, as shown with 2.5%.
• 33% and 67% are approximately equal to ⅓ and ⅔, respectively (the symbol “≈” means “approximately equal to.” These fractions are actually repeating decimals, but, in Case Interviews, you can often round these percentage values to these Clean Fractions.
There are specific efficient calculation methods you can learn and apply to make multiplying and dividing with Clean Numbers much easier than traditional calculation methods. The key concept of these efficient calculation methods is to identify Clean Numbers and multiply/divide with them while expressing them as a Clean Fraction times a power of 10. Keep in mind that, when multiplying two numbers, only one number needs to be a Clean Number to use these efficient calculation methods. With the example of 250 ⋅ 120 M, only 250 is a Clean Number.
## Percentage Growth Methods
Example: A company had annual revenue of \$400 Million in the past calendar year, and their revenue is projected to grow at 4% per year. Approximately, what will the company’s revenue be six years in the future?
The most efficient way to approximate the answer is to ignore compounding, which we will call the “non-compounding approximation”. Since the revenue grows 4% per year, for six years, if we ignore compounding, total revenue would grow by (4% ⋅ 6 = 24%). Since compounding will contribute slightly more growth, it is appropriate to round the total growth to 25%, which is ¼. Since the original revenue is \$400 M, the net change in revenue is 25% ⋅ \$400 M = \$100 M. Therefore, the revenue in six years is \$500 M = (\$400 M + \$100 M).
It is appropriate to ignore compounding when the percentage change and the number of years of growth are relatively small. The FastMath Ace the Case online course explains in further detail when you can use the non-compounding approximation and provides additional methods for performing percentage-growth calculations when it’s not appropriate to use this approximation.
## Net Present Value (NPV) Methods
NPV Example 1: How much would your company be willing to pay for another company that generates \$20 Million in profit annually, if your firm requires an annual return on investment of 10%?
There is a formula to determine the net NPV for a perpetuity, which is a fixed payment that occurs annually (\$C), for a given required annual return or interest rate (r), which is
Given that r =10%,
Therefore, NPV = 10 ⋅ \$C = 10 ⋅ \$20 M = \$200 M.
Your company would therefore be willing to pay up to \$200 Million to acquire this firm.
NPV Example 2: A real estate development firm is evaluating a project that involves buying a parcel of land and building condominiums on that parcel. The company forecasts they can sell the condominiums for a total of \$250 Million six years in the future.
What is the maximum the real estate company would be willing to spend now to buy the land and develop the condominiums, if all the associated costs for the project would be incurred today, and they require a 12% annual return on invested capital?
This is a trickier NPV calculation than the prior one, as this is not a perpetuity, and we have to determine the NPV of a payment of \$250 M in six years. There is a concept called the Rule of 72, which tells us that something growing at 12% annually will double in approximately six years, as: (6 ⋅ 12 = 72). In general, the doubling time for a quantity will be approximately 72 divided by the percentage growth rate (expressed as an integer) - in this example, 72 ÷ 12 = 6. This approximation is accurate for percentage growth in the range of 0%–20%.
Therefore, the NPV of a payment of \$250 M in six years, with a required return of 12%, is exactly half the future payment value of \$250 M, which is \$125 M.
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# Find the rate of change of the area of a circle with respect to its radius r, when r = 3 cm.
This question was previously asked in
AWES TGT 2015: Mathematics
View all AWES Army Public School Papers >
1. 2π cm/s
2. 6π cm/s
3. 16π cm/s
4. 12π cm/s
Option 2 : 6π cm/s
## Detailed Solution
Concept;
• Area of circle is given by: π ⋅ r2 where r is the radius of the circle.
• If y = f(x), then dy/dx denotes the rate of change of y with respect to x its value at x = a is denoted by: $${\left[ {\frac{{dy}}{{dx}}} \right]_{x = a}}$$
• Decreasing rate is represented by negative sign whereas the increasing rate is represented by the positive sign.
Calculation:
Here we have to find the rate of change of the area of a circle with respect to its radius r when r = 3 cm.
As we know that, area of circle is given by: πr2 where r is the radius of the circle.
Let A = πr2
As we know that, if y = f(x), then dy/dx denotes the rate of change of y with respect to x.
So, by differentiating A with respect to r we get,
$$\Rightarrow \frac{{dA}}{{dr}} = \frac{{d\left( {π \cdot {r^2}} \right)}}{{dr}} = 2π r$$
Now we have to find the value of dA/dr at r = 3 cm i.e $${\left[ {\frac{{dA}}{{dr}}} \right]_{r\ =\ 3}}$$
$$\Rightarrow {\left[ {\frac{{dA}}{{dr}}} \right]_{r\; =\ 3}} = 2π \cdot 3 = 6π \;cm$$
Hence, the rate of change of the area of a circle with respect to its radius r when r = 3 cm is 6π cm/s
|
Wheel Factorization
May 8, 2009
Having discussed prime numbers in several previous exercises, we are now interested in the problem of factoring an integer n; for instance, the prime factors of 42 are 2, 3, and 7. A simple factoring method is to perform trial division by all the integers counting from 2 to the square root of n. Your first task is to write that function.
An easy optimization is to divide only by 2 and then by odd integers greater than 2, which saves half the work. A better optimization is to divide by 2, then 3, then 5, and thereafter to alternately add 2 and 4 to the trial divisors — 7, 11, 13, 17, 19, 23, and so on — since all prime numbers greater than 3 are of the form 6k±1 for some integer k.
It turns out that both those optimizations are special cases of a technique called wheel factorization. Consider a 2-wheel of circumference 2 rolling along a number line with a “spoke” at the number 1; if you start with the spoke at 3 on the number line, it will strike the number line at 5, then 7, and then every odd number after that. Or consider a 2,3-wheel of circumference 2×3=6 with spokes at the number 1 and 5; if you start with the 5-spoke at 5 on the number line, it will strike the number line at 7, 11, 13, 17, 19, 23, and so on. Or consider a 2,3,5-wheel of circumference 2×3×5=30 with spokes at 1, 7, 11, 13, 17, 19, 23 and 29 starting with the 29-spoke at 7. And so on: next is a 2,3,5,7-wheel, then a 2,3,5,7,11-wheel, and the sequence continues infinitely.
Wheel factorization works by performing trial division at each place where a spoke touches the number line. As the wheels grow larger, more and more of the trial divisors are prime, so less and less unnecessary work is done. Of course, there is a point of diminishing returns; when the wheel gets too large, it is just as much work to compute the wheel as to compute the list of primes, and costs just as much to store. But a small wheel is easy to compute, and not too big, and provides a simple optimization over naive trial division.
The spokes of the wheel are computed by looking for co-primes, which are those numbers for which the spoke has no factors in common with the circumference of the wheel; in other words, where the greatest common divisor of the spoke and the circumference is 1. For instance, a 2,3,5-wheel has a spoke at 17 because the greatest common divisor of 17 and 30 is 1, but no spoke at 18 because the greatest common divisor of 18 and 30 is 6. These numbers are called totatives; if you’re curious about the math behind them, ask your favorite search engine for information about Euler’s totient function.
It is easy to see this visually. Here is a list of the positive integers to 42, with primes highlighted:
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42```
After the first row, all the primes are in two columns, which correspond to the two spokes of a 2,3-wheel. If 853 were input to the 2,3-wheel factorization function, we would trial divide by 2, 3, 5, 7, 11, 13, 17, 19, 23, 25,and 29 before concluding that 853 was prime; note that 25 is not prime, but is relatively prime to the circumference of the wheel.
Your second task is to write a function that finds the factors of a given number using wheel factorization; you should compute and use a 2,3,5,7-wheel. What are the factors of 600851475143? When you are finished, you are welcome to read or run a suggested solution, or post your own solution or discuss the exercise in the comments below.
Pages: 1 2
6 Responses to “Wheel Factorization”
1. […] Praxis – Wheel Factorization By Remco Niemeijer In today’s Programming Praxis problem we’re supposed to factor numbers both the naive way and using […]
2. Remco Niemeijer said
```import Data.List
tdFactors :: Integer -> [Integer] -> [Integer]
tdFactors n = nub . factor n . takeWhile (\x -> x * x < n)
where factor _ [] = []
factor r (x:xs) | mod r x == 0 = x : factor (div r x) (x:xs)
| otherwise = factor r xs
factorNaive :: Integer -> [Integer]
factorNaive n = tdFactors n [2..]
spokes :: Integer -> [Bool]
spokes c = map ((== 1) . gcd c) [1..c]
wheel :: [Integer] -> [Integer]
wheel ps = (ps ++) . drop 1 . map snd . filter fst \$
zip (cycle . spokes \$ product ps) [1..]
factorWheel :: Integer -> [Integer]
factorWheel n = tdFactors n \$ wheel [2,3,5,7]
main :: IO ()
main = do print \$ factorNaive 600851475143
print \$ factorWheel 600851475143
```
3. Remco Niemeijer said
Thanks to a suggestion from programmingpraxis I was able to speed up my program. Since tdFactors already more or less assumes a sorted list of integers as input we can stop looking as soon as x * x > r. The new code becomes:
```tdFactors :: Integer -> [Integer] -> [Integer]
tdFactors n = factor n
where factor _ [] = []
factor r (x:xs) | x * x > r = [r]
| mod r x == 0 = x : factor (div r x) (x:xs)
| otherwise = factor r xs
factorNaive :: Integer -> [Integer]
factorNaive n = tdFactors n [2..]
spokes :: Integer -> [Bool]
spokes c = map ((== 1) . gcd c) [1..c]
wheel :: [Integer] -> [Integer]
wheel ps = (ps ++) . drop 1 . map snd . filter fst \$
zip (cycle . spokes \$ product ps) [1..]
factorWheel :: Integer -> [Integer]
factorWheel n = tdFactors n \$ wheel [2,3,5,7]
main :: IO ()
main = do print \$ factorNaive 600851475143
print \$ factorWheel 600851475143
```
4. There isn’t much point in showing the code to compute the wheel; once it is calculated, it can be converted to a static (i.e., compile-time) value:
```
let wheel_factors =
let wheel2357 =
let rec w =
2 :: 4 :: 2 :: 4 :: 6 :: 2 :: 6 :: 4
:: 2 :: 4 :: 6 :: 6 :: 2 :: 6 :: 4 :: 2
:: 6 :: 4 :: 6 :: 8 :: 4 :: 2 :: 4 :: 2
:: 4 :: 8 :: 6 :: 4 :: 6 :: 2 :: 4 :: 6
:: 2 :: 6 :: 6 :: 4 :: 2 :: 4 :: 6 :: 2
:: 6 :: 4 :: 2 :: 4 :: 2 ::10 :: 2 ::10
:: w
in 1 :: 2 :: 2 :: 4 :: w
in
let rec go (w :: ws as wheel) l p n =
if p * p > n then List.rev (n :: l) else
let d = n / p in
if d * p == n
then go wheel (p :: l) p d
else go ws l (p + w) n
in go wheel2357 [] 2
```
5. David said
My FORTH solution; modified brute force factorization we did in (I think a future) exercise to use the factor wheel. Cool thing is the defining word “cycle” to create a new data type — the cyclic constant array.
```\ Quick word to create (non-cyclic, non-checked) constant array
: const-array create does> swap cells + @ ;
\ Create a cyclic data structure - constant array with
\ automatic wrap-around.
: cycle ( -- )
create here 0 ,
does>
@+ >r swap r> mod cells + @ ;
: end-cycle ( -- )
here over - cell / 1- swap ! ;
cycle wheel2357
2 , 4 , 2 , 4 , 6 , 2 , 6 , 4 ,
2 , 4 , 6 , 6 , 2 , 6 , 4 , 2 ,
6 , 4 , 6 , 8 , 4 , 2 , 4 , 2 ,
4 , 8 , 6 , 4 , 6 , 2 , 4 , 6 ,
2 , 6 , 6 , 4 , 2 , 4 , 6 , 2 ,
6 , 4 , 2 , 4 , 2 , 10 , 2 , 10 ,
end-cycle
const-array start2357
1 , 2 , 2 , 4 ,
variable wheel_ndx
: init_wheel ( -- )
-4 wheel_ndx ! ;
: wheel_step ( -- n )
wheel_ndx @ dup 0< IF
4 + start2357
ELSE
wheel2357
THEN 1 wheel_ndx +! ;
: d/mod ( d n -- d%n d/n )
locals| n |
\ FORTH does not have remainder function for double division
\ so we multiply the quotient by the divisor and subtract...
\
2dup 1 n M*/ 2dup 2>r \ get quotient & save a copy
n 1 M*/ d- drop \ compute remainder
2r> ; \ restore quotient
: .factors ( d -- )
2 locals| f | init_wheel
BEGIN 2dup f dup m* d< not WHILE
2dup f d/mod rot 0= IF
f .
2nip \ drop n; quotient on stack replaces it
ELSE
2drop \ drop quotient
f wheel_step + TO f
THEN
REPEAT
d. ;
```
6. Ethan Lee said
``` public class PrimeCalc {```
``` public static class PrimeWheel { // http://primes.utm.edu/glossary/xpage/WheelFactorization.html static final long[] PRIMES = { 2, 3, 5, 7 }; static final int WHEEL_FACTOR = 210; // 2 * 3 * 5 * 7; static final long[] SIEVES = { 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 209, 211 }; public long calcFactor(long n) { for (long p : PRIMES) { if (n % p == 0) { return p; } } long lim = (long) StrictMath.sqrt(n); for (long s = 0; s <= lim; s+= WHEEL_FACTOR) { for (long sieve : SIEVES) { long m = s + sieve; if (n % m == 0) { return m; } } } return n; } public boolean isPrime(long n) { if (n < 2) { return false; } return calcFactor(n) == n; } } public static void main(String[] args) { PrimeWheel pw = new PrimeWheel(); long num = 600851475143L; java.util.List factors = new java.util.ArrayList(); while (true) { long f = pw.calcFactor(num); factors.add(f); if (f == num) { break; } else { num /= f; } } System.out.println(factors); // [71, 839, 1471, 6857] } ```
```} ```
|
# How do you solve 2/3 = 2 - (5x-3)/(x-1)?
Jul 12, 2016
$x = \frac{5}{11}$
#### Explanation:
$\frac{2}{3} = 2 - \frac{5 x - 3}{x - 1}$
In an equation with fractions, we can get rid of denominators completely by multiplying by the LCM of the denominators. Then the denominators can cancel.
In this case the LCM = $\textcolor{red}{3 \left(x - 1\right)}$
$\frac{\textcolor{red}{3 \left(x - 1\right)} \times 2}{3} = \textcolor{red}{3 \left(x - 1\right)} \times 2 - \frac{\textcolor{red}{3 \left(x - 1\right)} \times \left(5 x - 3\right)}{x - 1}$
$\frac{\cancel{3} \left(x - 1\right) \times 2}{\cancel{3}} = 6 \left(x - 1\right) - \frac{3 \cancel{x - 1} \times \left(5 x - 3\right)}{\cancel{x - 1}}$
$2 \left(x - 1\right) = 6 \left(x - 1\right) - 3 \left(5 x - 3\right)$
$2 x - 2 = 6 x - 6 - 15 x + 9$
$2 x - 6 x + 15 x = - 6 + 9 + 2$
$11 x = 5$
$x = \frac{5}{11}$
|
# The graph of a linear equation contains the points (3.11) and (-2,1). Which point also lies on the graph?
May 12, 2018
(0, 5) [y-intercept], or any point on the graph below
#### Explanation:
First, find the slope with two points by using this equation:
$\frac{{Y}_{2} - {Y}_{1}}{{X}_{2} - {X}_{1}}$ = $m$, the slope
(3, 11) $\left({X}_{1} , {Y}_{1}\right)$
(-2, 1) $\left({X}_{2} , {Y}_{2}\right)$
$\frac{1 - 11}{- 2 - 3}$ = $m$
Simplify.
$\frac{- 10}{- 5}$ = $m$
Because two negatives divide to make a positive, your answer will be:
$2$ = $m$
Part Two
Now, use point-slope formula to figure out what your equation in y = mx + b form is:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - 11 = 2 \left(x - 3\right)$
Distribute and simplify.
$y - 11 = 2 x - 6$
Solve for each variable. To solve for the y = mx + b equation, add 11 to both sides to negate -11.
$y = 2 x + 5$
Now, plot this on a graph:
graph{2x + 5 [-10, 10, -5, 5]}
|
# Lesson Worksheet: Scale Drawings Mathematics
In this worksheet, we will practice interpreting simple scale drawings and finding real dimensions given a scale model.
Q1:
In a plan of a residential compound, if the buildings are 5 cm in height, the plan uses a 1 cm : 3 m scale. What is the height of the buildings in real life?
Q2:
In a museum design, the height of one of the statues is 4 cm. Use the scale to find the statue’s height in real life.
Q3:
An architect drew a scaled drawing of a school. The scale he used was . The school is 9 centimeters wide in drawing. How wide is the school in real life?
Q4:
Jackson made a drawing for his residence including a villa and a garden, as in the given figure. Using the scale drawing, what is the perimeter of the villa in real life?
Q5:
The two given plans, A and B, show two different scales of trees in the street. Which of the two plans has the taller tree in real life?
• APlan B
• BPlan A
Q6:
Consider this pool design. What is the perimeter of the pool in real life?
Q7:
Consider a map with scale . If the distance between two neighborhoods on the map is 11 cm, what is the distance in real life?
Q8:
Noah and Benjamin are brothers. They are at two different places and want to go home. Each of them has a differently scaled map. Noah’s map’s scale is and Benjamin’s map’s scale is . Noah’s map shows that he should drive 5 cm, and Benjamin’s map shows that he should drive 8 cm. Who has the shortest journey home?
• ANoah
• BBenjamin
Q9:
Consider this library design drawn on a centimeter paper. What is the area of the library in real life?
Q10:
A parking lot has been designed on a centimeter grid paper. What is the length of the parking lot in real life?
|
Hong Kong
Stage 4 - Stage 5
# Recurrence relationships for GP's
Lesson
The formula for the $n$nth term of a geometric sequence is given by $t_n=ar^{n-1}$tn=arn1, but there is another way to express the geometric relationship between terms.
It is generally known as a recurrence relationship and for geometric sequences, the recurrence formula is given by:
$t_{n+1}=r\times t_n,t_1=a$tn+1=r×tn,t1=a
The equation states that the $\left(n+1\right)$(n+1)th term is $r$r times the $n$nth term with the first term equal to $a$a.
Thus the second term, $t_2$t2 is $r$r times the first term $t_1$t1, or $ar$ar
The third term $t_3$t3 is $r$r times $t_2$t2 or $ar^2$ar2
The fourth term $t_4$t4 is $r$r times $t_3$t3, or $ar^3$ar3, and so on.
Hence, step by step, the sequence is revealed as $a$a, $ar$ar$ar^2$ar2$ar^3...$ar3... , $ar^{n-1}$arn1
Take for example the recursive relationship given as $t_{n+1}=\frac{t_n}{2}$tn+1=tn2 with $t_1=64$t1=64. From this formula, we see that
$t_2=\frac{t_1}{2}=32$t2=t12=32 and
$t_3=\frac{t_2}{2}=16$t3=t22=16, and so on.
This means that the sequence becomes $64,32,16,8,...$64,32,16,8,... which is clearly geometric with $a=64$a=64 and $r=\frac{1}{2}$r=12
Consider the recurrence relationship given as $t_{n+1}=3t_n+2$tn+1=3tn+2 with $t_1=5$t1=5.
To test whether or not the relationship is geometric, we can evaluate the first three terms.
$t_1=5$t1=5,
$t_2=3\times5+2=17$t2=3×5+2=17
$t_3=3\times17+2=53$t3=3×17+2=53.
Thus, the sequence begins $5,17,53,...$5,17,53,... and we immediately see that $\frac{53}{17}$5317 is not the same fraction as $\frac{17}{5}$175, and thus the recursive relationship is not geometric. In fact the only way the relationship given by $t_{n+1}=rt_n+k$tn+1=rtn+k is geometric is when the constant term $k$k is zero.
#### Worked Examples
##### Question 1
Consider the first-order recurrence relationship defined by $T_n=2T_{n-1},T_1=2$Tn=2Tn1,T1=2.
1. Determine the next three terms of the sequence from $T_2$T2 to $T_4$T4.
Write all three terms on the same line, separated by commas.
2. Plot the first four terms on the graph below.
3. Is the sequence generated from this definition arithmetic or geometric?
Arithmetic
A
Geometric
B
Neither
C
##### Question 2
The first term of a geometric sequence is $5$5. The third term is $80$80.
1. Solve for the possible values of the common ratio, $r$r, of this sequence.
2. State the recursive rule, $T_n$Tn, that defines the sequence with a positive common ratio.
Write both parts of the relationship on the same line, separated by a comma.
3. State the recursive rule, $T_n$Tn, that defines the sequence with a negative common ratio.
Write both parts of the relationship on the same line, separated by a comma.
##### Question 3
The average rate of depreciation of the value of a Ferrari is $14%$14% per year. A new Ferrari is bought for $\$9000090000.
1. What is the car worth after $1$1 year?
2. What is the car worth after $3$3 years?
3. Write a recursive rule for $V_n$Vn, defining the value of the car after $n$n years.
Write both parts of the rule on the same line, separated by a comma.
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# Topic 2 number system
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MTE3101 Knowing Numbers
Topic 2Elementary Number Theory
2.1 Synopsis This topic covers different number systems and focusses on the definition of number systems, classifications within the set of real numbers and number representation. The number systems referred to in this topic concentrates on Real Numbers that includes the set of Natural Numbers, Whole Numbers, Integers, Rational Numbers and Irrational Numbers.2.2 Learning Outcomes
1. Generate one set of numbers to another set of numbers.
2. Characterise natural, rational, irrational and real numbers.
2.3 Conceptual Framework
2.4 Number SystemsNumber theory is one of the oldest branches of pure mathematics and focusses on the study of natural numbers. Arithmetic is taught in schools where children begin with learning numbers and number operations. The first set of numbers encountered by children is the set of counting numbers or natural numbers.In mathematics, a number system is a set of numbers. As mentioned earlier, children begin by studying the natural numbers: 1,2,3, ... with the four basic operations of addition, subtraction, multiplication and division. Later, whole numbers 0,1,2, .... are introduced, followed by integers including the negative numbers. The next step will include rational numbers and irrational numbers. In short, examples of number systems covered in this topic include natural numbers, whole numbers, integers, rational numbers, irrational numbers and real numbers By studying the Number Systems it will help you to understand better the Elementary Number Theory in the next topic about Prime Numbers. Questions on divisibility, the use of the Euclidean algorithm to compute greatest common divisors, integer factorizations into prime numbers, and number recreations such as Fibonacci numbers are included in the next topic which will be delivered to you face-to-face.2.4.1 DefinitionTo be good mathematics teachers, we need to possess a sound knowledge on different number systems at our fingertips. Knowing how to define the various sets of numbers within the real number system is therefore an absolute MUST!( Real Numbers
Lets begin by defining real numbers. Try and answer the following question. What is a real number?
A real number refers to any number that you would expect to find on the number line. It is a number whose name will be the "address" of a point on the number line. Its absolute value will name the distance of that point from 0. Real numbers contain all the rational numbers (which are the infinite repeating decimals, positive, negative and zero) together with a new set of numbers called the irrational numbers. In other words, the set of real numbers is the set of all numbers that have an infinite decimal representation.In schools, counting numbers are taught first, followed by whole numbers, fractions and integers. The relationship among these sets is illustrated below.
Each arrow represents is a subset of, for example, the set of counting numbers is a subset of the set of whole numbers, and so on. Subsequently, both the fractions and integers extend the system of whole numbers.The diagram in the previous page can be extended to include the set of rational numbers as follows:
Lets revise by considering the following definitions for the different sets of numbers summarised in the table below. The definitions are written using set notation. The { } symbols, called braces indicate the closing and opening of a set or collection of numbers. The three dots after the three indicate that the pattern continues. Definition of sets of numbersNamesSetsNotes and examples
Natural numbers{1, 2, 3, . . .} Represents all the counting numbers beginning with 1
Whole numbers{0, 1, 2 , 3, . . .} Starts with zero plus all natural numbers
Integers {0, 1, 2, 3,. . .} Includes negative, 0 and positive whole numbers
Rationalnumbers { | p and q are integers, q 0 } Read as a fraction p over q, where p and q are both integers, q 0 .Rational numbers can be written in decimal form, but they always either end or repeat (recur). Examples include:
Irrational numbers
{x | x is a nonrepeating and nonterminating decimal}Examples include:
pi () 3.14159. . , ; e 2.71828 ; 2 , etc.
Real numbers{x | x can be written as a decimal}
Read as all numbers x, such that x can be written as a decimal
2.4.2 Classifications within the set of real numbersIn mathematics, different types of numbers are grouped together and given names. It is important to understand this organization of sets of numbers. Real numbers can be classified under different sets of numbers. Look at the list of numbers in the table given in the previous page. What do you notice? Note that as you go down the list, a new set will contain the set of numbers directly above it. For example, the whole numbers contain the natural numbers. In fact, the set of whole numbers consists of all the natural numbers together with one new number, zero. As you go down the list, the numbers get more "complicated." Theprogression of numbers is much the way we learn about numbers as we grow up. As small children, we start with the natural numbers when counting our fingers and toys. We then make an intellectual leap and learn about the idea of "all gone" or no more left and the concept of zero, which takes us towhole numbers. Fractions were introduced because of the need to deal with parts of a whole. At some time in our development, we learn about debts and negative numbers, and we start using integers. The same sort of progression happens in mathematics classes. You start doing mathematics with whole numbers, then fractions as well as decimals followed by operations with negatives and positives. Notice that the integers are all members of the rational numbers. Any integer can be written as a rational number by writng a one under it. The only exception to this progression is the irrational numbers. They are by their own. The set of irrational numbers is the set of numbers that have infinite nonrepeating decimal representations. Thus, the rationals and the irrationals are disjoint sets. These two sets together make up the real numbers, that is, when we put the irrational numbers together with the rational numbers, we finally have the complete set of real numbers. Any number that represents an amount of something, such as a weight, a volume, or the distance between two points, will always be a real number.From the above explanations on various sets within the real number system, you can now see how sets of numbers are related to one another and classified progressively. Now, can you describe the relationship between these sets?
Remember your Venn diagrams. The relationship between sets of numbers can be clearly shown with the help of Venn diagrams. The following diagram illustrates the relationships of the sets that make up the real numbers.
Next, something for you to think about.Testing your Understanding!1. Determine if the following statements are true or false. Give reasons for your answers.
i. Every integer is a rational number.
ii. Every rational number is an irrational number.
iii. Every natural number is an integer.
iv. Every integer is a natural number.2. Consider the following set of numbers:
{ - 81, - 0.315, 1, 3 , , 23, 6, 27, 3, 89.4, 100 000 }
Classify and list the numbers given according to the following sets. i natural numbers
ii whole numbers
iii integers
iv rational numbers
v irrational numbers
Besides using the set notation to represent various types of real numbers, we can also use other symbols such as alphabets to represent the set of real numbers. This is shown in the table below.Name of sets of numbersSymbols denoting the sets of numbers
Natural numbersN
Whole numbersW
IntegersZ
Rational NumbersQ
Irrational NumbersQ'
Real NumbersR
Apart from this, real numbers can also be represented on number lines. Writing numbers down on a number line makes it easy to tell which numbers are bigger or smaller. The ordered nature of the real numbers allow us to arrange them along a line (imagine that the line is made up of an infinite number of points all packed so closely together that they form a solid line). The points are ordered so that points to the right are greater than points to the left, as shown in the diagram below.The Number Line
Negative Numbers (-)Positive Numbers (+)
(The line continues left and right forever.)
Numbers on the right are bigger than numbers on the left:
8 is greater than 5
1 is greater than -1
But notice that -8 is smaller than -5
The number line above shows that
Every real number corresponds to a distance on the number line, starting at the centre (zero).
Negative numbers represent distances to the left of zero, and positive numbers are distances to the right.
The arrows on the end indicate that it keeps going forever in both directions.
For example, the number line below shows the set of Natural numbers:
Try representing the other sets of numbers discussed above using number lines. Have fun!
In conclusion, Real Numbers comprise the following: Rationals+Irrationals
All points on the number line
All possible distances on the number line.The discussion above serves to help you to recognise and cha
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# Solving For X Practice Problems
Given that the first time he took the test Brian had answered 150 questions correctly, how many correct answers did he answer in the second test?
If we use the method of addition in solving these two equations, we can see that what we get is a simplified equation in one variable, as shown below.There will be no change in the equation solving strategy and once you have learnt the above method, you do not need to bother about the coefficients at all.Next we present and try to solve the examples in a more detailed step-by-step approach. C 5x 2(x 7) = 14x – 7 5x 2x 14 = 14x – 7 7x 14 = 14x – 7 7x – 14x = -14 – 7 -7x = -21 x = 3 3. D 5x 3 = 7x – 1 now collect like terms 3 1 = 7x – 5x every time you move something it changes signs 4 = 2x anything multiplied is divided on the other side and vice versa 4/2 = x 2 = x 2. D The price increased from to () so the question is 5 is what percent of 20. You can review your answers and change them by checking the desired letter.Once you have finished, press "finish" and you get a table with your answers and the right answers to compare with. And that value is put into the second equation to solve for the two unknown values.The solution below will make the idea of Substitution clear. x y = 15 -----(2) (10 y) y = 15 10 2y = 15 2y = 15 – 10 = 5 y = 5/2 Putting this value of y into any of the two equations will give us the value of x.Let's try θ = 30°: sin(−30°) = −0.5 and −sin(30°) = −0.5 So it is true for θ = 30° Let's try θ = 90°: sin(−90°) = −1 and −sin(90°) = −1 So it is also true for θ = 90° Is it true for all values of θ? Click "Show Answer" underneath the problem to see the answer.
## One thought on “Solving For X Practice Problems”
1. Heroism at command, senseless brutality, deplorable love-of-country stance and all the loathsome nonsense that goes by the name of patriotism, how violently I hate all this, how despicable and ignoble war is; I would rather be torn to shreds than be part of so base an action!
2. Would it be useful to compare and contrast source C with source B?
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Pvillage.org
What are sequences ks3?
What are sequences ks3?
A number pattern which increases (or decreases) by the same amount each time is called a linear sequence.
What do you mean by limit of a sequence?
In mathematics, the limit of a sequence is the value that the terms of a sequence “tend to”, and is often denoted using the symbol (e.g., ). If such a limit exists, the sequence is called convergent. A sequence that does not converge is said to be divergent.
Is the limit of a sequence unique?
Theorem 3.1 If a sequence of real numbers {an}n∈N has a limit, then this limit is unique. We hope to prove “For all convergent sequences the limit is unique”. The negation of this is “There exists at least one convergent sequence which does not have a unique limit”.
What are the types of sequences?
Types of Sequence
• Arithmetic Sequences.
• Geometric Sequence.
• Fibonacci Sequence.
Which sequence is linear?
Linear sequences of numbers are characterized by the fact that to get from one term to the next we always add the same amount. The amount we add is known as the difference, frequently called the common difference. For example, the sequences: 3,7,11,15,19,23,…
What is a unique limit?
Theorem 3.1 If a sequence of real numbers {an}n∈N has a limit, then this limit is unique. We hope to prove “For all convergent sequences the limit is unique”. The negation of this is “There exists at least one convergent sequence which does not have a unique limit”. This is what we assume.
Can a sequence converge to two different limits?
A sequence {xn} converges to L if and only if every subsequence of {xn} converges to L. Therefore, if there exists two subsequences {xnk} and {xnl} converging to two different limits L′ and L″, then {xn} cannot be convergent.
What’s the limit of a sequence in math?
The limit of a sequence is the value the sequence approaches as the number of terms goes to infinity. Not every sequence has this behavior: those that do are called convergent, while those that don’t are called divergent.
Which is the limit of the sequence XNX _ NXN?
A real number LLL is the limit of the sequence xnx_nxn if the numbers in the sequence become closer and closer to LLL and not to any other number. In a general sense, the limit of a sequence is the value that it approaches with arbitrary closeness.
When does a sequence approach a specific value?
A sequence is “converging” if its terms approach a specific value at infinity. This video is a more formal definition of what it means for a sequence to converge.
How are the terms of a sequence accumulate?
Though the elements of the sequence oscillate, they “eventually approach” the single point 0. The common feature of these sequences is that the terms of each sequence “accumulate” at only one point. g ( n) = n − ⌊ n 2 ⌋ + ⌊ n 3 ⌋ − ⌊ n 4 ⌋ + ⋯ .
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IB Mathematics (HL)/Group Theory
The aims of this option are to provide the opportunity to study some important mathematical concepts, and introduce the principles of proof through abstract algebra. For this topic, you do not need any background knowledge of a core topic.
Naive Set TheoryEdit
Sets can be used as a foundation for constructing all the rest of mathematics. This is called axiomatic set theory. But axiomatic set theory is a very formal theory, too cumbersome for everyday mathematical use. On the other hand, sets are too useful a mathematical construct to be reserved for specialists investigating the foundations of mathematics. Naive set theory refers to using the framework of sets without formally defining them. This in what we will be doing.
A set is defined (informally) as any collection of thingsEdit
Ex: The set of primary colors, the set of my siblings, the set of positive integers.
Some sets can be described by listing the things in themEdit
We list the things separated by commas, and use curly braces to indicate that the belong to a set.
Ex: {red, yellow, blue}, {Donna, Linda, Bobbi, Julie, Steve}
If we can completely list (enumerate) all the things in a set, the set is said to be finite. The set of primary colors and the set of my siblings are finite sets. If a set isn’t finite, it is said to be infinite. The set of all positive integers is an infinite set.
Some sets can be described by a rule for enumerating themEdit
The set of positive even integers is infinite, so we can’t list them all. But, we can write it as {2, 4, 6, …}. The “…” says that there is a rule for constructing a list that contains every member of the set, even though the list can't be completed in a finite amount of time.
Problem:
1. Explicitly state a rule for constructing the set of positive even integers.
An interesting aside: some infinite sets are so “big” that it isn't possible to write a rule that constructs a list of all its members. The set of real numbers is one such set.
Some common mathematical setsEdit
Here are some common mathematical sets you are familiar with. You need to be able to recognize the symbols.
$\mathbb{N}$
the set of positive integers and zero, $\{0, 1, 2, \ldots\}$
$\mathbb{Z}$
the set of integers
$\mathbb{Z}^+$
the set of positive integers, $\{1, 2, 3, \ldots\}$
$\mathbb{Q}$
the set of rational numbers
$\mathbb{Q}^+$
the set of positive rational numbers
$\mathbb{R}$
the set of real numbers
$\mathbb{R}^+$
the set of positive real numbers
$\mathbb{C}$
the set of complex numbers
The “things” in a set are called elements of the setEdit
If something (say “x”) is in some particular set (say “S”), we say “x is an element of S.” To say this concisely, we write
$x \in S$
Mathematicians often use the word member as a synonym for element. For example, you’ll hear things like “2 is a member of the set of positive integers”.
Definition of equality for setsEdit
Two sets S and T are equal if every element of S is also an element of T and every element of T is also an element of S. Not surprisingly, we write this as
$S \mathbf= T$
Here are some implications of this definition.
The ordering of elements in a set is not important
The set {red, yellow, blue} equals (i.e. is the same as) the set {yellow, blue, red}. Why? Look at the definition of equality. Every element in the first set is an element of the second, and every element in the second set is an element of the first. So the two sets are equal.
Something is either an element of a set or not; it doesn’t make any difference if you list it multiple times
The set {red, red, yellow, blue, red} is the same as (i.e. is equal to) the set {red, yellow, blue}, even though “red” is listed multiple times in the first set. Don’t take my word for it; check the definition of equals.
Definition of subsetsEdit
What would happen if we broke up the definition of equal sets into its two parts? Suppose we required only that every element of S is also an element of T, without requiring the “vice versa”? For example, every element of the set {red, yellow, blue} is in the set {red, orange, yellow, green, blue, purple}, but not vice versa. This kind of thing happens often enough that we give it a name. We say that S is a subset of T if every element of S is also an element of T. To say this concisely, we write
$S \subseteq T$
Not surprisingly, we’ll also make up a definition for the “vice versa” part. That is, we say that S is a superset of T if every element of T is an element of S. We write this as
$S \supseteq T$
Notice the similarity between the symbols $\subseteq$ and $\supseteq$, which are defined for sets, and $\le$ and $\ge$, which are defined for numbers. This is not an accident. The subset relation between sets is not the same thing as the less-than-or-equal relation between the integers, but the two relations do have many similarities. (How many can you think of?) Good mathematical notation will often be suggestive of such similarities.
Problems:
1) Is $\mathbb N\supseteq\mathbb Z$? Is $\mathbb Z \supseteq\mathbb N$? Is $\mathbb Z \subseteq\mathbb Z^+$? Is $\mathbb Z\subseteq\mathbb Z$?
2) If S, T and U are sets with
$S \subseteq T$
and
$T \subseteq U$,
what can you say about the relationship between S and U? Show it using the definition of $\subseteq$. Do you remember what this property is called?
3) Put as many as you can of the common mathematical sets listed above into a single order of subsets, that is
? $\subseteq$ ? $\subseteq$ ? $\subseteq$
The relationship between subsets, supersets and equalityEdit
Reviewing the definitions, we see that for two sets $S$ and $T$,
$S \mathbf= T$
is true whenever both
$S \subseteq T$ and $S \supseteq T$
are true.
Problems:
1) What is the analogous statement for $\leq$ and $\geq$ with respect to real numbers? Is this statement true?
2) For x and y in $\mathbb N$, either x $\leq$ y or x $\geq$ y. Does the same hold true for $\subseteq$ and $\supseteq$ with respect to sets?
“Proper” subsetsEdit
Sometimes we have $S \subseteq T$ and we want to rule out the possibility that $S \mathbf= T$. To do this, we write
$S \subset T$
i.e. we omit the bar below the $\subset$. To say this in words, we say that S is a proper subset of T. The use of the word “proper” here is kind of funny. It doesn’t mean that there is something more accurate, or more polite, about being a proper subset. It is just the term that mathematicians have come to use to avoid having to say, “S is a subset of T but it isn’t equal to T.” Similarly,
$S \supset T$
is read “S is a proper superset of T ” and is a shorter way of writing
$S \supseteq T$ and $S \neq T$
Problems:
1. For any set S, which of the following are true?
$S \subseteq S$
$S \subset S$
$S \mathbf= S$
$S \supseteq S$
$S \supset S$
2. Suppose S = {2, 4, 6}. For each of the following sets T, list all the relationships that hold between S and T .
T = {6, 4, 2}
T = {2, 4, 4}
T = {2, 6, 10}
T = {2, 4, 6, 10}
T = {1, 3, 5}
The smallest possible setEdit
Is there a set X that is a subset of every possible set S? Yes. We can have a set with no elements at all. The definition of subset says that X is a subset of S if every element of X is also an element of S. If X has no elements, this statement is true regardless of what elements S contains.
We call the set containing no elements the null set. It sometimes is written as
{ }
but more often we write it as
$\empty$
Again, note the similarity between the notations for the set $\empty$ and the number 0. Just as $0 \le n$ for any natural number n, $\empty \subseteq S$ for any set S.
Combining sets: union and intersectionEdit
So far, we have defined various relations on pairs of sets $(=, \subset, \subseteq,$etc.) in terms of membership. It is also useful to define operations that take two sets and form a third set. Once again, we will define these operations in terms of membership.
We’ll start by defining the intersection of two sets S and T to be the set containing anything that is both an element of S and an element of T. We’ll write the intersection operation as
$S \cap T$
Similarly, we’ll define the union of two sets S and T to be the set containing anything that is either an element of S or an element of T. We’ll write the union operation as
$S \cup T$
If you have any trouble getting mixed up between $\cap$ and $\cup$, try remembering that $\cup$looks like U, which stands for Union.
Problems:
1. If S = {1, 2, 3} and T = {2, 3, 4}, what is $S \cap T$? What is $S \cup T$?
2. Are the following statements true for any sets S and T?
$S = (S\cap S)$
$S \subseteq (S\cup S)$
$S \subseteq (S\cup T)$
$S \subset (S\cap T)$
$S \supset (S\cup T)$
Is so, explain why. If not, give a counter-example.
Venn diagramsEdit
When doing math, it is often useful to draw a picture to help visualize a problem. For elementary set operations, there is a conventional method of drawing pictures called Venn diagrams, named after the British mathematician John Venn. To draw a set S, we simply draw a circle, with the name of the set inside the circle.
The intent of this drawing is that the inside of the circle represents all the elements in S. The outside of the circle represents everything that isn’t in the set S. There isn’t any significance to the fact that we use circles in Venn diagrams. We could just as well draw
or File:Trivial Pentagonal Venn Diagram
Now, to represent an operation on two sets, we draw two overlapping circles, like this:
File:Simple Venn Diagram
If you have colored pencils, you could color the inside of S red and the inside of T blue, like this:
File:Colored Simple Venn Diagram
We can now describe all the area that is colored as . The purple overlap between the circles represents . If you don’t have colored pencils, you can draw the circles and shade in the area you want to describe. For example, you can draw something like
to represent . To represent , you would shade in only the overlap of the two circles. (I would show this, but the word processor I am using to write this can’t do this easily.) If we never combined more than two sets, Venn diagrams wouldn’t be useful enough to bother learning about. But they are very useful when we want to illustrate relationships between three sets. In this case, we draw three overlapping circles, like this:
Problems: Illustrate the following set operations by drawing 3 overlapping circles for S, T and U and then shading in only the area described
Once we get more than three sets, Venn diagrams aren’t so useful. The problem is that we can’t simply draw four circles that show every possible combination of intersections. So Venn diagrams are mostly an introductory learning tool. Problems: Draw a Venn diagram (not limited to circles) that depicts every possible combination of intersections between four sets. What is the best you can do?
Set complement (first try)Edit
There is one more operation we want to define for sets – the complement. This operation is a unary operation, meaning it takes only one set as an argument. (Intersection and union are called binary operations, because they take two sets as arguments.) The idea behind the complement of a set S is that it should contain exactly those elements that the set S doesn’t contain. We’ll write the complement of S as S′ (Another notation that you’ll often see is .) Thus, we might try to define the complement of S as “the set of all elements that are not elements of S”. For example, if S were the set of odd integers, than the complement of S would include all the even integers. But according to our definition, the complement of S would also contain my sister Donna, the color red and Valentine’s Day, since they aren’t in S either. So the complement is going to have all kinds of useless junk in it. If this were an everyday discourse, we could just dismiss the junk as being irrelevant to our conversation and ignore it. But in mathematics, we like to be more precise than that. Our solution will be to introduce the notion of a “universal set”.
Universal setEdit
A set consisting of all elements under consideration is called the universal set.It is usually by the capital English Letter U. The universal set is simply the set of all things that we are currently talking about. The name “universal set” is misleading. It is not a set that contains everything in the universe. Nor is it universal in the sense that everyone has agreed to use it. Perhaps a better term would be “universe of discourse”. If we are making statements about the real numbers, then we say our universal set is the set of real numbers. If, on the other hand, we were discussing the natural numbers, our universe of discourse would be the set of all natural numbers. Another way to think of this is with a Venn diagram. We can draw the set S as
where the shaded area represents the elements in S. But how would we shade the diagram to represent the complement of S? We would want to shade the outside of the circle. But where would we stop? A “little ways” out? To the edge of the paper? Maybe the back of the paper as well? The universal set makes it clear where to stop:
Keep in mind that the only reason for defining a universal set is to remove ambiguity when we’re taking a set complement. If we’re not going to be using set complements, we don’t need to worry about what our universal set is.
Set complement (the real thing)Edit
Now that we have the concept of a universal set, we’ll give the real definition for set complement: The complement of a set T is the set of all elements that are in the universal set but not in T. Problems: Assume that R, S, and T are subsets of some universal set U. Draw Venn Diagrams to determine whether the following statements are always true.
(You had this last problem before, in the section titled Combining sets: union and intersection. Did you get the same answer this time? If not, did the Venn diagram mislead you? You have to be careful when interpreting Venn diagrams with proper subsets.)
DeMorgan’s lawsEdit
In mathematics, the term “law” describes a statement that can be proved from more fundamental mathematics. There are a number of synonyms: postulate, theorem, corollary, etc. The term “law” is often used in conjunction with statements that were discovered a long time ago, typically before that area of mathematics was formalized. DeMorgan’s laws, named after Augustus De Morgan, are
and
Note the symmetry between these two statements. If you take one and interchange the and , you get the other. DeMorgan’s laws are important because they provide a method for moving the complement from the “outside” to the “inside” of a complex expression. By applying DeMorgan’s laws repeatedly, we can transform any complex expression containing complements into an equivalent expression that has only complements of simple sets. Ex:
In the last step, we made use of the fact that taking the complement a second time returns us to the original set, i.e. for any set S,
Problems: Use Venn diagrams to show that DeMorgan’s laws are true. Use DeMorgan’s laws to transform these expressions into equivalent expressions that have complements only of basic sets.
Other useful lawsEdit
There are numerous laws relating various combinations of intersection, union and complement operations. Laws of the form X = Y are especially useful because we can use them for translating an expression into other, equivalent expressions. Here are some useful laws of this having this form. To the right of each law is a verbal description. union is commutative intersection is commutative When operations are commutative, we don’t have to be concerned about the order of the operands. union is associative intersection is associative When operations are associative, we can remove unnecessary parentheses. For example, we can write because there is no difference between and .
intersection distributes over union
This is analogous to the fact that for real numbers, multiplication distributes over addition. That is, . If we exchange for * and for +, we get the law above. Notice also that union does not distribute over intersection, such as addition does not distribute over multiplication for real numbers. The next two laws include U, the universal set. law of the excluded middle This is called the law of the excluded middle because any element in the universal set either is an element of S or isn’t an element of S. There is no middle ground. universal set is an identity for intersection This says that taking the intersection with the universal set makes no change, just as multiplying a real number by 1 makes no change. Problems: Verify each of the above laws using Venn diagrams.
Union-of-intersections formEdit
A good start in determining how two set expressions are related is to reduce them both to union-of-intersections form. This is an expression of the form (? ? … ?) (? ? … ?) (? ? … ?) … (? ? … ?) where all the question marks are either a named set or its complement. For example, the expression
is in union-of-intersections form. This is analogous to the sum-of-products form for expressions over R, such as
or, following the common conventions for omitting multiplication symbols and extra parentheses,
Question: What mathematical properties and/or notational conventions are needed to get the previous line from the one before it? Here’s a general technique for transforming any set expression into union-of-intersections form, Apply DeMorgan’s laws repeatedly to move all complements to named sets. Apply the distributive law repeatedly to change intersection of unions into unions of intersections. Make free use the associative laws to get rid of unnecessary parentheses and the commutative laws to put things in a nice order. Here’s a simple example.
(I had originally intended to take this topic farther, that is to develop a complete algorithm for deciding whether two set expressions were equivalent (or had a subset relationship). But I decided it was going to get into too many details, so I’m just going to stop here. Problems: Determine whether each of the following relationships is always true. Start by transform both sides of the expression into union-of-intersections form.
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# What is 122/109 as a decimal?
## Solution and how to convert 122 / 109 into a decimal
122 / 109 = 1.119
The basis of converting 122/109 to a decimal begins understanding why the fraction should be handled as a decimal. Both are used to handle numbers less than one or between whole numbers, known as integers. But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. Now, let's solve for how we convert 122/109 into a decimal.
## 122/109 is 122 divided by 109
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 122 is being divided into 109. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators and Denominators. This creates an equation. We must divide 122 into 109 to find out how many whole parts it will have plus representing the remainder in decimal form. This is how we look at our fraction as an equation:
### Numerator: 122
• Numerators represent the number of parts being taken from a denominator. Any value greater than fifty will be more difficult to covert to a decimal. But having an even numerator makes your mental math a bit easier. Large numerators make converting fractions more complex. So how does our denominator stack up?
### Denominator: 109
• Denominators represent the total number of parts, located below the vinculum or fraction bar. Larger values over fifty like 109 makes conversion to decimals tougher. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. Let's start converting!
## Converting 122/109 to 1.119
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 109 \enclose{longdiv}{ 122 }$$
We will be using the left-to-right method of calculation. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Solve for how many whole groups you can divide 109 into 122
$$\require{enclose} 00.1 \\ 109 \enclose{longdiv}{ 122.0 }$$
How many whole groups of 109 can you pull from 1220? 109 Multiple this number by our furthest left number, 109, (remember, left-to-right long division) to get our first number to our conversion.
### Step 3: Subtract the remainder
$$\require{enclose} 00.1 \\ 109 \enclose{longdiv}{ 122.0 } \\ \underline{ 109 \phantom{00} } \\ 1111 \phantom{0}$$
If there is no remainder, you’re done! If there is a remainder, extend 109 again and pull down the zero
### Step 4: Repeat step 3 until you have no remainder
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 122/109 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. They each bring clarity to numbers and values of every day life. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 122/109 and 1.119 bring clarity and value to numbers in every day life. Here are examples of when we should use each.
### When you should convert 122/109 into a decimal
Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.111 per hour and not$20 and 122/109.
### When to convert 1.119 to 122/109 as a fraction
Progress - If we were writing an essay and the teacher asked how close we are to done. We wouldn't say .5 of the way there. We'd say we're half-way there. A fraction here would be more clear and direct.
### Practice Decimal Conversion with your Classroom
• If 122/109 = 1.119 what would it be as a percentage?
• What is 1 + 122/109 in decimal form?
• What is 1 - 122/109 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 1.119 + 1/2?
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+
## Common Core Standards
In Grade 6, instructional time should focus on four critical areas: (1) connecting ratio and rate to whole number multiplication and division and using concepts of ratio and rate to solve problems; (2) completing understanding of division of fractions and extending the notion of number to the system of rational numbers, which includes negative numbers; (3) writing, interpreting, and using expressions and equations; and (4) developing understanding of statistical thinking.
Sophia and Common Core Standards
Our Many Ways to Learn method makes it easy to integrate the new standards into your teaching. By using multiple instructors and teaching styles for each tutorial, we make sure you're reaching your students, so they can reach their goals.
Note: Our Common Core-aligned content is middle and high school-focused. We're continually expanding our library, so stay tuned - more great content coming soon.
## 6.RP - Ratios and Proportional Relationships
##### Understand ratio concepts and use ratio reasoning to solve problems.
Code
Standard
Concepts
6.RP.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6
6.RP.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
5
6.RP.3a
Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
1
6.RP.3c
Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
5
6.RP.3d
Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
11
## 6.NS - The Number System
##### Apply and extend previous understandings of numbers to the system of rational numbers.
Code
Standard
Concepts
6.NS.5
Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, credits/debits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation.
1
6.NS.6b
Understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane; recognize that when two ordered pairs differ only by signs, the locations of the points are related by reflections across one or both axes.
5
6.NS.6c
Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane.
15
6.NS.7
Understand ordering and absolute value of rational numbers.
15
6.NS.7c
Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. For example, for an account balance of –30 dollars, write |–30| = 30 to describe the size of the debt in dollars.
7
6.NS.8
Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate.
5
## 6.EE - Expressions and Equations
##### Apply and extend previous understandings of arithmetic to algebraic expressions.
Code
Standard
Concepts
6.EE.1
Write and evaluate numerical expressions involving whole-number exponents.
3
6.EE.2a
Write expressions that record operations with numbers and with letters standing for numbers. For example, express the calculation “Subtract y from 5” as 5 – y.
4
6.EE.2b
Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity. For example, describe the expression 2 (8 + 7) as a product of two factors; view (8 + 7) as both a single entity and a sum of two terms.
1
6.EE.2c
Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole- number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). For example, use the formulas V = s3 and A = 6 s2 to find the volume and surface area of a cube with sides of length s = 1/2.
4
6.EE.3
Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6 (4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y.
12
## 6.SP - Statistics and Probability
##### Develop understanding of statistical variability.
Code
Standard
Concepts
6.SP.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.
2
##### Summarize and describe distributions.
Code
Standard
Concepts
6.SP.4
Display numerical data in plots on a number line, including dot plots, histograms, and box plots.
14
6.SP.5
Summarize numerical data sets in relation to their context, such as by:
20
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## Are A and B independent?
Events A and B are independent if the equation P(A∩B) = P(A) · P(B) holds true. You can use the equation to check if events are independent; multiply the probabilities of the two events together to see if they equal the probability of them both happening together.
## What does P A or B mean?
If A is an event and B is another event, then P(A or B) is the probability of either A occurring, orB occurring, or both occurring. ‘Or’ is commutative in the sense that P(A or B) = P(B or A).
## Is probability descriptive or inferential?
Probability distributions, hypothesis testing, correlation testing and regression analysis all fall under the category of inferential statistics.
## How do you find the probability of A or B?
The probability of two disjoint events A or B happening is: p(A or B) = p(A) + p(B).
## How do you calculate or probability?
Probability OR: Calculations The formula to calculate the “or” probability of two events A and B is this: P(A OR B) = P(A) + P(B) – P(A AND B).
## What does and mean in probability?
Or Probability. In probability, there’s a very important distinction between the words and and or. And means that the outcome has to satisfy both conditions at the same time. Or means that the outcome has to satisfy one condition, or the other condition, or both at the same time.
## Is probability an inferential statistic?
Inferential statistics is based on the probability of a certain outcome happening by chance. In probability theory, the word outcome refers to the result observed.
## What are two uses of probability?
Statistics Chapter 1
A B
two major branches of statistics descriptive and inferential
two uses of probability gambling (playing cards) and insurance industry
The group of subjects selected from the group of all subjects under study is called a(n) population
## How is probability used in everyday life?
Probability is the mathematical term for the likelihood that something will occur, such as drawing an ace from a deck of cards or picking a green piece of candy from a bag of assorted colors. You use probability in daily life to make decisions when you don’t know for sure what the outcome will be.
## How do probability problems work?
Divide the number of events by the number of possible outcomes.
1. Determine a single event with a single outcome.
2. Identify the total number of outcomes that can occur.
3. Divide the number of events by the number of possible outcomes.
4. Determine each event you will calculate.
5. Calculate the probability of each event.
## What is the role of probability in statistics?
The probability theory provides a means of getting an idea of the likelihood of occurrence of different events resulting from a random experiment in terms of quantitative measures ranging between zero and one. The probability is zero for an impossible event and one for an event which is certain to occur.
## What is the relationship between probability and inferential statistics?
Inferential statistics: The part of statistics that is concerned with drawing conclusions from data. Probability model: The mathematical assumptions relating to the likelihood of different data values.
## What is the difference between chance and probability?
Chance is the occurrence of events in the absence of any obvious intention or cause. It is, simply, the possibility of something happening. Probability is the extent to which an event is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible.
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# Geometric Sequences
A series of free, online lessons for Intermediate Algebra (Algebra II) with videos, examples and solutions.
In these lessons, we will learn
• how to find the common ratio of a geometric sequence
• how to find the formula for the nth term of an geometric sequence
• how to find the sum of an geometric series
The following figure gives the formula for the nth term of a geometric sequence. Scroll down the page for more examples and solutions.
Geometric Sequences
A geometric sequence is a sequence that has a pattern of multiplying by a constant to determine consecutive terms.
We say geometric sequences have a common ratio.
The formula is
an = an-1r
Examples:
1. A sequence is a function. What is the domain and range of the following sequence? What is r?
-12, 6, -3, 3/2, -3/4
2. Given the formula for the geometric sequence, determine the first 2 terms and then the 5th term. Also state the common ratio.
3. Given the geometric sequence, determine the formula. Then determine the 6th term.
1/3, 2/9, 4/27, 8/81, …
A Quick Introduction To Geometric Sequences
This video gives the definition of a geometric sequence and go through 4 examples, determining if each qualifies as a geometric sequence or not.
A geometric sequence is a sequence of numbers where each term after the first term is found by multiplying the previous one by a fixed non-zero number, called the common ratio.
Example: Determine which of the following sequences are geometric. If so, give the value of the common ratio, r.
1. 3,6,12,24,48,96, …
2. 3,3/2,3/4,3/8,3/16,3/32,3/62, …
3. 10,15,20,25,30, …
4. -1,.1,-.01,-.001,-.0001, …
Geometric Sequences
A list of numbers that follows a rule is called a sequence. Sequences whose rule is the multiplication of a constant are called geometric sequences, similar to arithmetic sequences that follow a rule of addition. Homework problems on geometric sequences often ask us to find the nth term of a sequence using a formula. Geometric sequences are important to understanding geometric series.
How To Find The General Term Of A Geometric Sequence?
Example:
Find the formula for the general term or nth term of a geometric sequence.
Geometric Sequences And Series
A short introduction to geometric sequences and series.
Math Skills & Equations: Solving Math Sequences
There are two kinds of math sequences that can be solved: arithmetic sequences and geometric sequences. An arithmetic sequence is solved by adding or subtracting the same number, while geometric sequences use division and multiplication. Learn more about solving math sequences.
Example:
Find the 9th term of 3,12,48,192, …
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# What is the limit of (x^2 + x- 30)/|x- 5| as x approaches infinity?
Mar 8, 2015
${\lim}_{x \rightarrow \infty} \frac{{x}^{2} + x - 30}{\left\mid x - 5 \right\mid} = \infty$
For positive values of $x$, $\left\mid x - 5 \right\mid = x - 5$. So,
${\lim}_{x \rightarrow \infty} \frac{{x}^{2} + x - 30}{\left\mid x - 5 \right\mid} = {\lim}_{x \rightarrow \infty} \frac{{x}^{2} + x - 30}{x - 5}$.
This limit evaluates to the indeterminate form $\frac{\infty}{\infty}$.
(If you have access to l'Hopital's rule, you could use it, but it is not necessary.)
Use algebra to make the denominator not go to $\infty$.
For all positive $x$ (which is all we're interested in as $x \rightarrow \infty$),
$\frac{{x}^{2} + x - 30}{\left\mid x - 5 \right\mid} = \frac{{x}^{2} + x - 30}{x - 5} = \frac{x \left(x + 1 - \frac{30}{x}\right)}{x \left(1 - \frac{5}{x}\right)} = \frac{x + 1 - \frac{30}{x}}{1 - \frac{5}{x}}$
As $x$ increases without bound, the numerator of this expression also increases without bound and the denominator approaches $1$.
Therefore:${\lim}_{x \rightarrow \infty} \frac{{x}^{2} + x - 30}{\left\mid x - 5 \right\mid} = {\lim}_{x \rightarrow \infty} \frac{x + 1 - \frac{30}{x}}{1 - \frac{5}{x}} = \infty$
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l2 (2)
# l2 (2) - Mathematical Foundations of Computer Science...
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Unformatted text preview: Mathematical Foundations of Computer Science Lecture Outline January 08, 2011 Permutations of Selected Elements. We looked at permutations of n elements out of the available n elements. Now we will consider permutations of r elements out of the available n elements. Such an arrangement is called an r-permutation. For example, ab,ba,ac,ca,bc,cb are all 2-permutations of the set { a,b,c } . Let P ( n,r ) denote the number of r-permutations of a set of n elements. What is the value of P ( n,r )? Forming an r-permutation of a set of n elements can be thought of as an r-step process such that in step i, 1 ≤ i ≤ r , we choose the i th element of the ordering. There are n- ( i- 1) = n- i + 1 ways of performing step i . By the multiplication rule, the number of r-permutations equals P ( n,r ) = n × n- 1 × n- 2 × ··· × n- ( r- 1) = n × n- 1 × n- 2 × ··· × n- r + 1 = n × ( n- 1) × ··· × ( n- r + 1) × ( n- r ) × ··· × 1 n- r × ( n- r- 1) × ( n- r- 2) × ··· × 1 = n ! ( n- r )! Example 1. How many ways are there to select a first-prize winner, a second-prize winner, and a third-prize winner from 100 different contestants? Solution. Selecting the winners can be done in 3 steps with each step i, 1 ≤ i ≤ 3 choosing the winner in the i th place. Step i can be performed in 100- ( i- 1) ways. By multiplication rule, the total number of possible ways in which the prizes can be given is 100 × 99 × 98 = 970200. Note that this is same as P (100 , 3). Example 2. In how many ways can we order 26 letters of the alphabet so that no two of the vowels a,e,i,o,u occur consecutively? Solution. The task of ordering the letters so that no two vowels appear consecutively can be performed in two steps. Step 1. Order the 21 consonants. 2 Lecture Outline January 08, 2011 Step 2. Choose locations for the 5 vowels. The vowels can be placed before the consonants, between the consonants and after the consonants. Step 1 can be performed in 21! ways. To count the number of ways of performing Step 2, observe that there is only one location for placing a vowel before and after the consonants, and 20 locations for placing the vowels between the consonants. This gives a total of 22 valid locations for placing 5 vowels. Thus the number of ways of placing the 5 vowels in 5 of the 22 locations is P (22 , 5). This is because there are 22 locations for a , 21 for e , 20 for i , 19 for o , and 18 for u . By multiplication rule, the total number of orderings in which no two vowels occur consecutively equals 21! × P (22 , 5) = 21! × 22! 17! The Inclusion-Exclusion Formula. If A,B , and C are any finite sets, then | A ∪ B | = | A | + | B | - | A ∩ B | | A ∪ B ∪ C | = | A | + | B | + | C | - | A ∩ B | - | A ∩ C | - | B ∩ C | + | A ∩ B ∩ C | Observe that if the sets A,B , and C are mutually disjoint, i.e., A ∩ B = A ∩ C = B ∩ C = ∅ then we get | A ∪ B | = | A | + | B | | A ∪ B ∪ C | = | A | + | B | + | C | This is often called the...
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## How do you calculate percentiles in business statistics?
Percentiles can be calculated using the formula n = (P/100) x N, where P = percentile, N = number of values in a data set (sorted from smallest to largest), and n = ordinal rank of a given value. Percentiles are frequently used to understand test scores and biometric measurements.
## What does 80th percentile mean?
Imagine that the height of a group of people is the set of data to study. If a height of 1.75 m is at P80 (80th percentile), it means that 80% of the people in the group are 1.75 or less.
How do you calculate 90th percentile?
What is the 90th percentile value? Multiply the number of samples by 0.9: 0.9 X 10 samples = 9 Therefore, the 9th highest ranked sample is the 90th percentile result to compare to the Action Level.
### What does 90th percentile mean in statistics?
n. the location of a score in a distribution expressed as the percentage of cases in the data set with scores equal to or below the score in question. Thus, if a score is said to be in the 90th percentile, this means that 90% of the scores in the distribution are equal to or lower than that score.
### How to calculate percentile?
How to Calculate Percentile 1 Arrange n number of data points in ascending order: x 1, x 2, x 3. 2 Calculate the rank r for the percentile p you want to find: r = (p/100) * (n – 1) + 1 3 If r is an integer then the data value at location r, x r, is the percentile p: p = x r
What is the p th percentile in statistics?
The p th percentile is the value in a set of data at which it can be split into two parts. The lower part contains p percent of the data, and the upper part consists of the remaining data; i.e., 100-p (the total data equates to 100%). Calculating the p th Percentile.
## How do I find the 5th percentile in Excel?
We use the same formula as the PERCENTILE () function in Excel, Google Sheets and Apple Numbers. The percentile calculator can create a table listing each 5th percentile, also showing quartiles and deciles. Click the check box before you click the Calculate button.
## How do I calculate the p-th percentile in Excel?
1) Input the numbers in the set separated by a comma (e.g., 1,9,18,12), space (e.g., 1 9 18 12), or line break. 2) Enter the percentile value you wish to determine. 3) Click on the “Calculate” button to generate the results. The p th percentile is the value in a set of data at which it can be split into two parts.
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## Calculus: Early Transcendentals (2nd Edition)
The solution is $$\lim_{x\to0^+}(\sin x)\sqrt{\frac{1-x}{x}}=0.$$
We will use L'Hopital's rule to calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to0^+}(\sin x)\sqrt{\frac{1-x}{x}}=\lim_{x\to0^+}\frac{\sin x}{\sqrt{\frac{x}{1-x}}}=\left[\frac{\sin 0^+}{\sqrt{\frac{0^+}{1-0^+}}}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0^+}\frac{(\sin x)'}{\left(\sqrt{\frac{x}{1-x}}\right)'}=\lim_{x\to0^+}\frac{\cos x}{\left(\sqrt{\frac{x}{1-x}}\right)'}.$$ Let us calculate the derivative $\left(\sqrt{\frac{x}{1-x}}\right)'$. We will use the chain rule and then the quotient rule: $$\left(\sqrt{\frac{x}{1-x}}\right)'=\frac{1}{2\sqrt{\frac{x}{1-x}}}\left(\frac{x}{1-x}\right)'=\frac{1}{2\sqrt{\frac{x}{1-x}}}\frac{(x)'(1-x)-x(1-x)'}{(1-x)^2}=\frac{1}{2\sqrt{\frac{x}{1-x}}}\frac{1-x+x}{(1-x)^2}=\frac{1}{2\sqrt{\frac{x}{1-x}}}\frac{1}{(1-x)^2}=\frac{1}{2\sqrt{x(1-x)^3}}.$$ Putting this into the limit we have $$\lim_{x\to0^+}(\sin x)\sqrt{\frac{1-x}{x}}=\lim_{x\to0^+}\frac{\cos x}{\frac{1}{2\sqrt{x(1-x)^3}}}=\lim_{x\to0^+}2\cos x\sqrt{x(1-x)^3}=2\cos 0^+\sqrt{0^+(1-0^+)^3}=0.$$
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Minuend and Subtrahend
# Minuend and Subtrahend
## Minuend and Subtrahend
Do you know that the three numbers involved in a subtraction equation are called using three different names? They are minuend, subtrahend and difference. We will discuss these terms one by one in this article.
We know that subtraction is the opposite of addition. Subtraction means taking away smaller number from a larger number. ‘–’ is the sign of subtraction.
Let us see the following example.
25 12 = 13
Minuend Subtrahend Difference
The number from which we subtract the other number is called the minuend.
The number which is subtracted from the minuend is called the subtrahend.
The result is called the difference.
## Minuend Definition
Minuend is the bigger number from which the smaller number is subtracted.
For example:
876 ← Minuend
642 ← Subtrahend
-------------
234 ← Difference
-------------
In the above example, 876 is the minuend, 642 is the subtrahend and 234 is the difference. If we change the position of the minuend and the subtrahend, the difference will change. It may be positive or negative. So, we cannot change the position the minuend and the subtrahend. Minuend is always bigger than the subtrahend.
## Subtrahend Definition
You have seen above that the number being subtracted from the minuend is called the subtrahend. The terms minuend, subtrahend and difference are used to differentiate among the three numbers used in a subtraction equation.
For example: Subtract 56 from 100.
Here, we have to subtract 56 from 100. So, 56 is the subtrahend and 100 is the minuend. To find the difference, let us subtract.
100 ← Minuend
56 ← Subtrahend
-------------
44 ← Difference
-------------
Remember this easy equation to under these three terms.
Minuend Subtrahend = Difference
## Solved Examples
Example 1: Find the minuend in the following subtraction equation.
18 – 12 = 6
Solution: We know that minuend is the number from which another number is subtracted. Thus, in this case minuend is 18.
Example 2: Find the subtrahend in the following subtraction equation.
248 – 135 = 113
Solution: We know that subtrahend is the number which is subtracted from the minuend. Thus, in this case subtrahend is 135.
Example 3: Subtract 3694 from 7329.
Solution: Here, 7329 is minuend and 3694 is subtrahend. So, the subtraction equation is: 7329 – 3694
7329 ← Minuend
3694 ← Subtrahend
-------------
3635 ← Difference
-------------
## Word Problems on Minuend and Subtrahend
Example 1: There were 578 seats in an auditorium. If 325 people attended a concert held in the auditorium, how many seats were empty?
Solution:
Total number of seats in auditorium = 578
Number of people in the concert = 325
------------
Number of empty seats = 253
------------
Example 2: In a cricket stadium, 48950 people can sit. If in a particular match, 25748 seats are occupied, how many seats are vacant?
Solution:
Total number of seats in stadium = 48950
Number of seats occupied = 25748
---------------
Number of vacant seats = 23202
---------------
Example 3: Mr. Sharma had Rs 76945 in his bank account. If he withdrew Rs 28538 from the account, how much money is left in his bank account?
Solution:
Amount in Mr. Sharma’s bank account = Rs 76945
Amount withdrawn by him = – Rs 28538
---------------
Amount left in his bank account = Rs 48407
---------------
Related Topics:
What is an Addend in Maths
Minuend and subtrahend
Multiplicand and multiplier
Dividend, divisor, quotient and remainder
Natural numbers
Whole numbers
Properties of rational numbers
Are all integers rational numbers?
Find five rational numbers between 3/5 and 4/5
Please do not enter any spam link in the comment box.
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# NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 – Linear Equations in two Variables
Download NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 – Linear Equations in two Variables. This Exercise contains 8 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 4 or other Chapters, you can click the link at the end of this Note.
NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Question 2
NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Question 3
NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Question 4
NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 Question 5
### NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 – Linear Equations in two Variables
NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 – Linear Equations in two Variables
1.Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x
(iv) 3 = 2x + y
Solution :
(i) x + y = 4
To draw the graph we need atleast two solutions. We can check that at x = 0 , y = 4 and at x = 4 , y = 0 .
Thus we can use the following solution table to draw the graph as follows :
(ii) x – y = 2
To draw the graph, we need atleast two solutions. We can check that at x = 0 , y = -2 and at x = 2 , y = 0 .
Thus we can use the following solution table to draw the graph as follows :
(iii) y = 3x
To draw the graph we need atleast two solutions. We can check that at x = 0 , y = 0 and at x = 1, y = 3 .
Thus we can use the following solution table to draw the graph as follows :
(iv) 3 = 2x + y
To draw the graph we need atleast two solutions. We can check that at x = 0 , y = 3 and at x = 1.5 , y = 0 .
Thus we can use the following solution table to draw the graph as follows :
2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solutions :
Following are the two lines passing through (2,14) :
y = 7x
y = x + 12
(2,14) is a point and we know that there exist infinite lines passing through a single point.
Therefore, there are infinitely many such lines.
3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
Given that the point lies on the graph of the given equation, this implies that the point satisfies the same equation.
Therefore, (3,4) must satisfy the equation 3y = ax + 7
Putting x = 3 and y = 4 in the equation we get,
3(4) = a(3) + 7
3a = 5
a = 5/3
4. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.
Solution :
Total distance covered = x km
Fare for 1 st kilometre = Rs 8
Fare for the subsequent distance = Rs (x − 1) 5
Total fare(y) = Rs [8 + (x − 1) 5]
y = 8 + 5x − 5
y = 5x + 3
5x − y + 3 = 0
To draw the graph we need atleast two solutions. We can check that the points (0, 3) and (-⅗ , 0) satisfies the above equation.
Thus we can use the following solution table to draw the graph as follows :
5. From the choices given below, choose the equation whose graphs are given in Fig.(1) and Fig. (2).
For Fig.(1)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x
For Fig.(2)
(i) y = x + 2
(ii) y = x – 2
(iii) y = –x + 2
(iv) x + 2y = 6
We know that there exists a unique line passing through two distinct points.
For Fig.(1)
Points on the given line are (−1, 1) and (1,−1).
We can check that both the coordinates satisfy the equation x + y = 0.
Therefore, x + y = 0 is the equation corresponding to the graph as shown in the first figure.
Hence (ii) alternative is the correct one.
For Fig.(2)
Points on the given line are (0,2) and (2,0).
We can check that both the coordinates satisfy the equation y = − x + 2.
Therefore, y = − x + 2 is the equation corresponding to the graph as shown in the first figure.
Hence (iii) alternative is the correct one.
6. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
(i) 2 units
(ii) 0 unit
Solution :
Let the distance travelled and the work done by the body be x and y respectively.
It is given that the work done is directly proportional to the distance travelled
i.e. y ∝ x
y = kx
Where, k = constant force
If constant force is 5 units, then work done is given as,
y = 5x
To draw the graph we need atleast two solutions. We can check that the points (1, 5) and (−1, −5) satisfies the above equation.
Thus we can use the following solution table to draw the graph as follows :
(i) From the graph, it can be seen that the value of y corresponding to x = 2 is 10. Hence, work done is 10 when distance travelled is 2.
(ii) From the graph, it can be seen that the value of y corresponding to x = 0 is 0. Hence, work done is 0 when distance travelled is 0.
7. Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the Same.
Solution :
Let the amount that Yamini and Fatima contributed be x and y respectively towards the Prime Minister’s Relief fund.
Amount contributed by Yamini + Amount contributed by Fatima = 100
i.e. x + y = 100
To draw the graph we need atleast two solutions. We can check that the points (0, 100) and (100 , 0) satisfies the above equation.
Thus we can use the following solution table to draw the graph as follows :
Here, it is important to note that variable x and y are representing the amount contributed by Yamini and Fatima respectively and these quantities cannot be negative. Hence, only those values of x and y which are lying in the 1st quadrant will be considered.
8. In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
F = (9/5)C + 32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Solution :
(i) Given equation
F = (9/5)C + 32 (1)
To draw the graph we need atleast two solutions. We can check that the points (0, 32) and (−40, −40) satisfies the above equation.
Thus we can use the following solution table to draw the graph as follows :
(ii) Putting C = 30 in equation (1) we get,
F = (9/5)30 + 32
F = 54 + 32
F = 86
Thus, If the temperature is 30°C, the temperature in Fahrenheit = 86oF
(iii) Putting F = 95 in equation (1) we get,
95 = (9/5)C + 32
(9/5)C = 63
C = 35
Thus, If the temperature is 95°F, temperature in Celsius = 35oC
(iv) Putting C = 0 in equation (1) we get,
F = (9/5)0 + 32
F = 0 + 32
F = 32
Thus, If the temperature is 0°C, the temperature in Fahrenheit = 32oF
Putting F = 0 in equation (1) we get,
0 = (9/5)C + 32
(9/5)C = -32
C = -160/9
C = -17.8
Thus, If the temperature is 0°F, temperature in Celsius = -17.5oC
(v) Putting C = F in equation (1) we get,
F = (9/5)F + 32
F(1 – 9/5) = 32
F(-⅘ ) = 32
F = -40
Thus, If the temperature is -40°C, the temperature in Fahrenheit = -40oF
NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 – Linear Equations in two Variables, has been designed by the NCERT to test the knowledge of the student on the topic – Graph of a Linear Equation in Two Variables
The next Exercise for NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.4 – Linear Equations in two Variables can be accessed by clicking here.
Download NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.3 – Linear Equations in two Variables
|
# Find length and width
• Oct 25th 2009, 06:47 PM
Nightasylum
Find length and width
of a rectangle that has the given perimeter and a maximum area.
Perimeter : 80 meters
Im still not sure how to set these up to prove with calculus.
2x+2y=80?
• Oct 25th 2009, 06:52 PM
Chris L T521
Quote:
Originally Posted by Nightasylum
of a rectangle that has the given perimeter and a maximum area.
Perimeter : 80 meters
Im still not sure how to set these up to prove with calculus.
Let a rectangle have length $l$ and width $w$.
Then by the first condition, we have $P=2l+2w=80$.
Let $A=lw$. It follows from above that $l=40-w$.
Substituting this into the area equation, we have $A=(40-w)w=40w-w^2$.
Since we want to maximize area, it follows that $\frac{\,dA}{\,dw}=40-2w$. Now, we find the critical point:
$40-2w=0\implies w=20$ (To verify its a max, we differentiate again to get $A^{\prime\prime}=-2<0$)
Since $w=20$, it follows that $l=40-(20)=20$.
Therefore, the dimensions of the rectangle that gives us the maximum area actually form a square.
Does this make sense?
• Oct 25th 2009, 06:57 PM
Nightasylum
Quote:
Originally Posted by Chris L T521
Let a rectangle have length $l$ and width $w$.
Then by the first condition, we have $P=2l+2w=80$.
Let $A=lw$. It follows from above that $l=40-w$.
Substituting this into the area equation, we have $A=(40-w)w=40w-w^2$.
Since we want to maximize area, it follows that $\frac{\,dA}{\,dw}=40-2w$. Now, we find the critical point:
$40-2w=0\implies w=20$ (To verify its a max, we differentiate again to get $A^{\prime\prime}=-2<0$)
Since $w=20$, it follows that $l=40-(20)=20$.
Therefore, the dimensions of the rectangle that gives us the maximum area actually form a square.
Does this make sense?
Yes perfect sense. Thanks alot
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# Graphical Representation of a Function that Represents Direct Proportionality
🏆Practice graphical representation
The graphical representation of a function that represents direct proportionality is actually the ability to express an algebraic expression through a graph. Since it's a direct proportionality, the graph will be that of a straight line.
The graphical representation of a function that represents direct proportionality is actually the ability to express an algebraic expression through a graph.
As it is a direct proportionality, the graph will be of a straight line.
A function that represents direct proportionality is a linear function of the family $y=ax+b$.
The graphical representation of this function is a straight line that is ascending, descending, or parallel to the $X$ axis but never parallel to the $Y$ axis.
Note: we observe the line from left to right.
We can now recognize in the equation of the line what the graphical representation of each function looks like:
(only when the equation is explicit $Y$ is isolated on one side and its coefficient is $1$)
## Test yourself on graphical representation!
Which statement is true according to the graph below?
## A -> the slope of the line
When $a > 0$ is positive: the line is ascending
When $a < 0$ is negative: the line is descending
When $a = 0$: the line is parallel to the $X$ axis
## B -> the point of intersection with the Y-axis
$b$ the y-intercept $Y$
$b$ indicates at which point the line crosses the $Y$ axis.
If $b$ has a positive coefficient, the line will intersect the positive part of the $Y$ axis at the point $b$.
If b has a negative coefficient, the line will intersect the negative part of the $Y$ axis at the point $b$.
If $b=0$, the line will cross the $Y$ axis at the origin where $Y=0$.
To know exactly what the graph of the line's equation looks like, we will have to examine both parameters at the same time, both a and $b$.
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## Examples of Graphical Representation of a Linear Function
### Example 1 (use of the graph)
$y=5x-4$
We will examine the linear equation.
$a=5$ The slope is positive, the line ascends
$b=-4$ The line crosses the $Y$ axis at the point where $Y=-4$
We will plot the graph based on the data:
Keep in mind that this is just a sketch.
If you want to draw the graph accurately, you can construct a table of values for $X$ and $Y$ and find out the points that form the line.
### Example 2 (using the table)
The function $y=2X$ represents a direct proportionality between the values of $X$ and $Y$. That is, for each value of $X$ that we input, the value of $Y$ will be double.
We will replace three different values and obtain:
Now let's plot the three points on the coordinate system and connect them. This is actually the graph of the function for $y=2X$.
Do you know what the answer is?
Related Subjects
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# 2017 AMC 8 Problems/Problem 21
## Problem
Suppose $a$, $b$, and $c$ are nonzero real numbers, and $a+b+c=0$. What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$?
$\text{(A) }0\qquad\text{(B) }1\text{ and }-1\qquad\text{(C) }2\text{ and }-2\qquad\text{(D) }0,2,\text{ and }-2\qquad\text{(E) }0,1,\text{ and }-1$
## Solution 1
There are $2$ cases to consider:
Case $1$: $2$ of $a$, $b$, and $c$ are positive and the other is negative. WLOG, we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that $$\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.$$
Case $2$: $2$ of $a$, $b$, and $c$ are negative and the other is positive. Without loss of generality, we can assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that $$\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.$$
In both cases, we get that the given expression equals $\boxed{\textbf{(A)}\ 0}$.
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# Tanisha is planning a backyard party. She will serve hamburgers, potato salad, strawberry shortcake, and lemonade. Including Tanisha, 28 people will be at the party. Use this information to help Tanisha plan her party. Tanisha expects that each person will drink two 8−ounce glasses of lemonade at the party, and makes 448 ounces of lemonade. Tanisha's mother buys 38 lemons. They know it takes 8 lemons to make 80 ounces of lemonade. Does Tanisha need more lemons?
2 months ago
## Solution 1
Guest #3805
2 months ago
For this case we can make the following rule of three:
8 lemons --------> 80 ounces of lemonade
38 lemons ------> x
Clearing x we have:
x = (38/8) * (80)
x = 380 ounces
We observed that:
380 <448
Therefore, more lemons are needed.
Yes, Tanisha needs more lemons
## 📚 Related Questions
Question
PLS HELP! Tanisha is planning a party. She will serve hamburgers, potato salad, strawberry shortcake, and lemonade. Including Tanisha, 28 people will be at the party. Tanisha expects that each person will drink two 8−ounce glasses of lemonade at the party, and makes 448 ounces of lemonade. Tanisha's mother buys 38 lemons. They know it takes 8 lemons to make 80 ounces of lemonade. Does Tanisha need more lemons?
Solution 1
Shee needs 448 ounces of lemonade.
448 ÷ 80 =5.8
⇒ There are 5.6 80-ounce of lemonade.
80 ounces of lemonade needed 8 lemons.
448 ounces of lemonade needed 8 x 5.6 = 44.8 lemons.
Tanisha has 38 lemons.
She does not have enough lemons.
She needs 44.8 - 38 = 7 more lemons.
Answer: Tanisha needs more lemons. She needs 7 more.
Solution 2
Step-by-step explanation:
Since she needs 448 ounces of lemonade you have to divide which is 5.6 . As a result there are 5.6 80 ounces of lemonade. 80 ounces needs 8 lemons. 448 ounces of lemonade are needed 8 x 5.6 = 44.8 lemons. Tanisha only has 38 lemons, which isn't enough. 44.8 - 38 = is about 7 more lemons.
Question
Solution 1
-3x - 3x = -24
Combine like terms:
- 6x = -24
Divide both side by -6:
x = 4
Question
Rafael has ridden 14 miles of a bike course. The course is 40 miles long. What percentage of the course has Rafael ridden so far
Solution 1
Percentage ridden = 14/40 x 100 = 35%
Question
Where can the numerator be found in a fraction? A. Above the fraction bar B. At the fraction bar C. Below the fraction bar D. Either A or C
Solution 1
The answer is A. above the fraction bar. The denominator lives below the fraction bar.
Question
What is the excluded value of -2x/17 a. x=0 b. x=2 c. x=17 d. no excluded values
Solution 1
Answer is d: no excluded values.
There is only excluded values if the variable x is in the denominator.
Question
A circle has a circumference of 10. It has an arc of length 9/2
Solution 1
Given that a circle has a circumference of 10 and an arc of length 9/2, thus the fraction of the arc length will be:
(9/2)/10
=0.45
thus the angle subtended by the arc will be:
0.45
×360°
=162°
Question
Anja collected data about the number of dogs 9th-grade students own. She created this histogram to represent the data and determined that it is skewed right. Which statement is true about Anja’s claim?
Solution 1
D) She is correct; the histogram is skewed to the right because there is less data on the right side.
Solution 2
Question
Mary has 7 1/4 yard of fabric. she cuts off 3/4 foot fabric to make the edge even. Mary wants to end up with 3 pieces of fabric that are the same size. how long will each piece be?
Solution 1
7 1/4 - 3/4 = 6 5/4 - 3/4 = 6 2/4
6 2/4 ÷ 3 = 26/4 ÷ 3 = 26/4 x 1/3 = 2 1/6
Question
Yun is making some investments in the stock market. Stock for Company A has a 30% chance of losing $12,000, a 45% chance of breaking even, and a 25% chance of earning Yun$11,000. Assuming Yun would like the odds to be in favor of her gaining money, should she invest in Company A? Explain. a. no, 7,700 > 3,600 b. yes, 7,700 > 3,600 c. yes, 3,600 > 2,750 d. no, 3,600 > 2,750
Solution 1
We have been given that company A has a 30% chance of losing $12,000. Let us calculate the amount that Yun can loose. Now, 45% chance of breaking even. It means neither loss nor profit. Then, 25% chance of earning Yun$11,000.
Let us calculate the amount of profit
Since, 3,600 > 2,750. It means that Yun will be in loss on investing in the company.
Hence, she will not invest in the company A.
d is the correct option.
d. no, 3,600 > 2,750
Solution 2
D. no, 3,600 > 2,750.
Question
A lawn roller is 1 m wide and 80 cm high. What area is covered in each revolution
Solution 1
The area for each revolution is given by:
A = 2 * pi * r * l
Where,
l: roller width
We have then:
A = 2 * 3.14 * (0.80 / 2) * 1
A = 2.512 m ^ 2
The area that is covered in each revolution is:
A = 2.512 m ^ 2
2647929
842281
748681
586256
406852
368373
348603
324927
199835
130075
112100
106146
77164
23213
22589
19607
17108
13966
10987
3389
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# MSC # 27 – Transpose and Apply
The Vedic Mathematics Sutra or word formula “Transpose and Apply” has a much wider application than the “transposition” which we know in Algebra as a short cut of the “golden rule’ of Algebra: do unto one side of the equation, what you do on the other side.
We first introduced this Sutra in MSC #12, Base Division. We will now apply this technique in polynomial division where the divisor is a quadratic expression.
In our example, the divisor is x2 – 2x + 3 and we transpose (reverse the signs of) the coefficients of all the terms of the divisor after the leading term. We also separate by a remainder bar the last two terms of the dividend. We write the transposed figures, 2 and -3, below the original coefficients.
We obtain the first term of the quotient by dividing the first term of the dividend by the first term of the divisor (the First by the First), 2x4/ x2 = 2x2.
We then multiply the coefficient of the first quotient term, 2, by the transposed figures to get 4 and -6, and place it under the 2nd and 3rd terms of the dividend as shown in the illustration.
By adding the coefficient of the x3 term of the dividend, -1, and 4, we get 3, the coefficient of the x term in the answer.
Again, we multiply this 3 by the transposed figures to give 6 and -9 which we will place in the 3rd and 4th columns of our solution.
We the add 1 which is the coefficient of the x2 term in the dividend, -6 and 6 to get 1 as the last answer figure. This is actually 1x2 but we have to divide it by xto get 1.
Since the last answer figure is 1, we need only to place the transposed figures in the fourth row as shown.
The first figure in the remainder is the sum of 9(x), – 9 and 2 which is 2x, while the last figure is 8 + – 3 = 5.
So we have (2x4 – x3 + x2 + 9x + 8) ÷ (x2 – 2x + 3) = 2x2 + 3x + 1 rem 2 x + 5
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# Find the Surface Area of a Cuboid
In this worksheet, students will learn the formula for finding the surface area of a cuboid and use it to solve simple problems.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR,
Curriculum topic: Geometry and Measures, Mensuration
Curriculum subtopic: Mensuration and Calculation Volume and Surface Area Calculations
Popular topics: Area worksheets
Difficulty level:
#### Worksheet Overview
The best part about a present when you were younger? Yes - the box.
All the care taken to wrap up the present when you might as well have just been given an empty box to play with.
The cuboid is a great shape for putting presents in - the shape makes it easier to wrap.
Have you ever wondered how much paper to buy to make sure the whole box is covered?
This is where surface area comes in.
The surface area of a shape is just the area of each face added together.
Cuboids can be found everywhere: house bricks, books, cameras, cookers - they all have to be wrapped to be transported to shops. Can you think of any more?
Retailers have to be able to calculate the surface area so that they can wrap goods before they are shipped off.
The surface area of a cuboid is simply the area of each face added together.
Find the surface area of this cuboid:
First, find the area of the face we are looking directly at: 5 x 4 = 20 cm²
There is another face exactly the same on the opposite side of the cuboid. No working out necessary we have already done it.
We now have: face 1 = 20 cm²
face 2 = 20 cm²
Now let's look at the face on the right-hand side and work out the area: 3 x 5 = 15 cm²
The opposite face is also the same.
We now have: face 1 = 20 cm²
face 2 = 20 cm²
face 3 = 15 cm²
face 4 = 15 cm²
Now find the remaining two faces in the same way: 4 x 3 = 12 cm²
We now have: face 1 = 20 cm²
face 2 = 20 cm²
face 3 = 15 cm²
face 4 = 15 cm²
face 5 = 12 cm²
face 6 = 12 cm²
Total surface area = 94 cm²
And remember that measurement to do with area is measured in units squared.
Happy wrapping.
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The MathBlog Percentage Calculator allows you to easily and quickly perform a variety of percentage calculations to meet all your needs. Whether you’re figuring out discounts, tips, or changes in values, use the calculators below to help you find the answers in just a few clicks.
## What is percentage of $$x$$?
What is
%
of
?
$$y$$ =
$$y = \frac {x \times p}{100}$$
## $$x$$ is what percentage of $$y$$?
is what percent
of
$$p =$$
$$p = \frac {x \times 100}{y}$$
## $$x$$ is $$p$$% of what?
is
% of what?
$$y =$$
$$y = \frac {x \times 100}{p}$$
## Percentage change calculator
from
to
CHANGE =
%
Percent change formula
$$\frac {(P2 - P1) \times 100}{|P1|}$$
## What Is a Percentage?
A percentage is a way to express a number as a fraction of 100. This method of representation helps us understand proportions and compare different quantities easily. The symbol “%” is used to denote percentages. For instance, 25% can be written as 0.25 in decimal form or $$\frac{25}{100}$$ as a fraction.
Percentages are a crucial tool in various real-life applications. Here are a few examples of how they work:
Imagine you are a teacher grading a test. If a student scores 18 out of 20, you can calculate their percentage score to understand their performance better. The calculation would be:
$$( \frac{18}{20} ) \times 100 = 90%$$
This means the student scored 90% on the test.
Percentages also come in handy when analyzing data, such as understanding population growth. Suppose a small town had 1,000 residents last year and now has 1,200 residents. To find the percentage increase in population, you would do the following calculation:
$$( \frac{1200 – 1000}{1000} ) \times 100 = 20%$$
Thus, the town’s population increased by 20%.
Another common use of percentages is in financial contexts, such as calculating interest rates. If you invest $500 in a savings account that offers a 5% annual interest rate, the interest earned in one year would be: $$( \frac{5}{100} ) \times 500 = 25$$ So, you would earn$25 in interest after one year.
Percentages make it simpler to understand and communicate changes, comparisons, and proportions. They are especially useful in everyday scenarios like shopping. For example, if a pair of shoes costs $80 and is on sale for 15% off, you can find the discount amount as follows: $$( \frac{15}{100} ) \times 80 = 12$$ This means you save$12, and the sale price of the shoes is:
$$80 – 12 = 68$$
Therefore, the shoes will cost \$68 after the discount.
## How to Calculate Percentages
Calculating percentages involves using simple formulas. Here’s a quick overview of what you can do with our calculators:
1. Find $$P \text{%}$$ of $$X$$: Determine what percentage of a number you need.
2. $$X$$ is what percent of $$Y$$: Figure out the percentage relationship between two numbers.
3. $$X$$ is $$P \text{%}$$ of what: Identify the base number when you know the percentage and the part.
4. Change Calculator: Calculate how much a value has increased or decreased in percentage terms.
### Basic Percentage Formulas
• Percentage of a number (formula):
$$P \text{%} \times X = Y$$
• What percent of $$X$$ is $$Y$$ (formula):
$$\frac{Y}{X} \times 100 = P \text{%}$$
• $$X$$ is $$P \text{%}$$ of what (formula):
$$\frac{X \times 100}{P} = Y$$
• Percentage change (formula):
$$\frac{\text{New Value} – \text{Old Value}}{\text{Old Value}} \times 100 = \text{% Change}$$
### Practical Applications
Percentages are everywhere! From calculating sales tax to determining body fat percentages, understanding how to work with percentages is crucial in many aspects of life. Here are some common uses:
• Shopping: Calculate discounts to know how much you’re saving.
• Finance: Understand interest rates, investment returns, and loan calculations.
• Health: Track fitness progress, such as body fat or caloric intake.
• Education: Calculate grades and test scores.
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#### Please solve RD Sharma Class 12 Chapter 21 Differential Equations Exercise Case Study Based Question (CSBQ) Question 1 Subquestion (v) maths textbook solution.
Answer: option(b) $r=(27+63 t)^{\frac{1}{3}}$
Hint: Here the case is given of inflate , so use formula $4 \pi r^{2} \frac{d r}{d t}=-\lambda$ and solve it.
Given: radius of balloon is 6 units.
Solution: as the case seems to be of radius increasing, after 3 seconds the differential equation gets modified as ,
\begin{aligned} &4 \pi r^{2} \frac{d r}{d t}=-\lambda \\ &4 \pi r^{2} d r=-\lambda d t \end{aligned}
Integrating both the sides.
\begin{aligned} &\int 4 \pi r^{2} d r=\int \lambda d t \\ &4 \pi \int r^{2} d r=\lambda \int d t \\ &4 \pi \frac{r^{3}}{3}=\lambda t+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(i) \end{aligned}
Now as stated before the radius was initially 3 units.
\begin{aligned} 4 \pi \frac{r^{3}}{3} &=\lambda t+c \\ 4 \pi \frac{(3)^{3}}{3} &=\lambda(0)+c \\ 36 \pi &=c\; \; \; \; \; \; \; \; \; \; \; \; \; .......(ii) \end{aligned}
Put (ii) in (i)
$\therefore 4 \pi \frac{r^{3}}{3}=\lambda t+36 \pi$
Now it is given that after 3 seconds radius is 6 times.
$\begin{gathered} 4 \pi \frac{(6)^{3}}{3}=\lambda(3)+c \\ 288 \pi=3 \lambda+c \\ 288 \pi-3 \lambda=c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ......(iii) \end{gathered}$
Put (ii) in (iii)
\begin{aligned} &288 \pi-3 \lambda=36 \pi \\ &288 \pi-36 \pi=3 \lambda \\ &\therefore \lambda=96 \pi-12 \pi \\ &\therefore \lambda=84 \pi \; \; \; \; \; \; \; \; \; \; \; \; ....(iv) \end{aligned}
Now put (ii) & (iv) in (i)
$\\\frac{4}{3} \pi r^{3}=84 \pi t+36 \pi\\ \begin{gathered} 4 \pi r^{3}=252 \pi t+108 \pi \\ r^{3}=63 t+\frac{108 \pi}{4 \pi} \\ r=(63 t+27)^{\frac{1}{3}} \\ \therefore r=(27+63 t)^{\frac{1}{3}} \end{gathered}$
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# Properties of Altitude
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Last updated date: 02nd Aug 2024
Total views: 170.1k
Views today: 4.70k
## Overview of Properties of Altitude
Have you ever wondered what that line which falls perpendicular to either of the sides of a triangle is called? Or ever wondered if there can only be one of those lines in a figure or if there can be even more? These perpendicular lines coming from the vertex of any triangle and then falling perpendicularly on the opposite side are called altitude in mathematics. In this article, we will look at these altitudes, will also learn the geometrical property of altitudes, and at last will also learn what is an orthocentre of a triangle. Sounds interesting right? So let's start learning.
Altitude of a Triangle
### What Is Altitude?
In a triangle, a particular Line segment which is drawn from its vertex and is connected to the opposite side of the triangle making a 90-degree inclination with that side is referred to as the altitude.
Triangles usually are of three kinds of obtuse, equilateral, and isosceles triangles.
In each type of triangle, the property of their altitude differs in various ways and usually is used to calculate the area of a triangle because the altitude is equivalent to the height of the triangle.
• In an obtuse-angled triangle the altitude is present outside the main triangle body for which we have to extend the base of the triangle and then draw a perpendicular line segment from the opposite vertex touching the extended base
• The altitude in an equilateral triangle interestingly divides the triangle into two equal parts.
• The isosceles triangle altitude bisects the angle of the vertex and bisects the base. It should be noted that an isosceles triangle is a triangle with two congruent sides and so, the altitude bisects the base and vertex.
Altitude of an Obtuse-Angled Triangle
## Properties of Altitude
Below are listed some properties of altitude:
• There can only be a maximum of three altitudes inside or outside a triangle.
• It is at a 90 degrees angle to the opposite side.
• The altitude of a triangle can either be inside it or can also be outside it depending upon the type of triangle being talked about.
• As we all know, there can only be a max of three altitudes inside a triangle. Therefore, the junction at which all those three altitudes meet is referred to as the orthocentre of a triangle.
## Properties of Orthocentre
In this section of the article, we will discuss about the various properties of orthocenter:
As discussed in the above section a triangle can only have a maximum of three altitudes and the portion at which those three altitudes meet with one another is called an Orthocentre. Now we will look at the orthocenter of a triangle properties:
The point indicated as ‘O’ is the orthocentre of a right-angled triangle
• For an acute angled Triangle, the orthocentre is drawn inside the triangle body.
• For an obtuse-angled triangle, the orthocentre is drawn outside the triangle body
• Whereas the orthocentre of a right-angled triangle is drawn at the vertices of the right angles.
Above were some unique orthocenters of a triangle properties.
## Formulas to Calculate the Altitudes of Various Triangles
Type of triangle Formula for calculating altitude Isosceles $h=\sqrt{a^2-\dfrac{b^2}{4}}$ where ‘a’ is the length of the two equal sides and ‘b’ is the other side. Equilateral $h=\dfrac{s \sqrt{3}}{2}$ where ‘s’ is the length of the side Right angle $h=\sqrt{x y}$ where ‘x’ and ‘y’ are the measures of the base as divided by the altitude falling on the base.
## Geometrical Property of Altitudes
Below are listed the geometrical properties of altitudes
1. All of the altitudes in a triangle are concurrent or congruent.
2. The altitude of a triangle lies inside or outside the triangle.
3. Orthocentre can either lie inside or outside the triangle
## Solved Examples
Below are some questions related to the altitude of a triangle:
Example 1: Calculate the altitude of a triangle having all equal sides of measure 6 cm.
Ans: According to the question the given triangle is an equilateral triangle
The formula for calculating the altitude of an equilateral triangle is $h=\dfrac{1}{2} \times \sqrt{3} \times s$ where $s$ is the length of the sides.
Therefore, $h=\dfrac{1}{2} \times \sqrt{3} \times 6$
$h=3 \sqrt{3} \mathrm{~cm}$
Example 2: Calculate the side of the equilateral triangle if the height or the altitude is 2cm.
Ans: For equilateral triangle
$h$ (height or altitude) $=\dfrac{1}{2} \times \sqrt{3} \times s$ where ' $s$ ' is the length of the side Given $h=2 \mathrm{~cm}$ noe putting the value in the equation we get
$(h \times 2) \div \sqrt{3}=s$
$(2 \times 2) \div \sqrt{3}=s$
Therefore, $\mathrm{s}=\dfrac{4}{\sqrt{3}} \mathrm{~cm}$
## Practice Questions
Q 1. Write down all the formulas for finding altitude of the following triangles:
1. Right angle triangle
2. Isosceles triangle
Ans: a. $h=\sqrt{x y}$
b. $h=\sqrt{a^2-\dfrac{b^2}{4}}$
Q 2. Find the measure of the altitude of a triangle having three equals sides of 8 cm
Ans: $4 \sqrt{3} \mathrm{~cm}$
Q 3. Calculate the height of a right angle triangle whose base values are $x=2 \mathrm{~cm}$ and $y=4 \mathrm{~cm}$
Ans: For right angle triangle $h=\sqrt{xy}$
Therefore, $h=\sqrt{2 \times 4}=2 \sqrt{2} \mathrm{~cm}$
## Summary
To complete all the learnings from this article we can say that altitudes are the heights of a triangle and are usually used to calculate the area of the triangle. Altitudes can help us find and relate various triangle properties with each other for example congruency. Through this article, we learned about the basic properties of altitudes of the triangle and also looked at the positions of the altitudes. With this, we would like to end this article and hope that we were clear and understandable enough to clear all your doubts yet if you still have doubts you can write them down in the comment section below.
## FAQs on Properties of Altitude
1. Are there any differences between the orthocentre and centroid of a triangle
The only main difference is that the orthocentre is the point where all the altitudes meet whereas the centroid is the point from which the line segments drawn from the center of each side of a triangle meet.
2. Are altitude and median the same?
No they are different median join the vertex to the center whereas altitude passes through vertex and join to opposite side.
3. Name all types of triangles.
Equilateral, isosceles, scalene, right angles, acute, obtuse are various types of triangle.
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### Home > MC1 > Chapter 1 > Lesson 1.1.3 > Problem1-22
1-22.
Study the dot patterns in parts (a) and (b) below. Assume that each pattern continues to increase by the same number of dots and in the same locations for each figure. For each pattern, sketch the $4$th and $5$th figures and then predict how many dots will be in the $100$th figure.
1. ${\large\begin{array}{c c} \;\;\bullet\\ \;\;\bullet &\bullet\\ \;\;\bullet\\ \end{array}\\ \text{Figure 1}} \qquad \qquad$
${\large\begin{array}{c c c} \bullet\\ \bullet\\ \bullet &\bullet & \bullet\\ \bullet\\ \bullet \end{array}\\ \\ \;\text{Figure 2}} \qquad \qquad$
${\large\begin{array}{c c c c} \bullet\\ \bullet\\ \bullet\\ \bullet &\bullet & \bullet & \bullet\\ \bullet\\ \bullet\\ \bullet\\ \end{array}\\ \quad \; \text{Figure 3}}$
Look at the vertical dots. How are they growing? What about the horizontal ones?
With each new figure, the vertical line gets one dot on the top and one dot on the bottom. The horizontal line gets one dot on the right side each time.
How can you find each figure number in the number of dots? Can you find groups of two dots in the second figure? Can you find groups of three dots in the third figure? Use this to help you with the $100$th figure.
${\large\begin{array}{c c c c c} \bullet\\ \bullet\\ \bullet\\ \bullet\\ \bullet &\bullet &\bullet &\bullet &\bullet\\ \bullet\\ \bullet\\ \bullet\\ \bullet\\ \end{array}\\ \qquad\text{Figure 4}}$
${\large\begin{array}{c c c c c} \bullet\\ \bullet\\ \bullet\\ \bullet\\ \bullet\\ \bullet &\bullet &\bullet &\bullet &\bullet &\bullet\\ \bullet\\ \bullet\\ \bullet\\ \bullet\\ \bullet \end{array}\\ \qquad \;\; \text{Figure 5}}$
The $100$th figure will have $301$ dots.
1. ${\large\begin{array}{c} \quad \; \bullet\\ \quad \; \bullet\\ \end{array}\\ \text{Figure 1}} \qquad \qquad$
${\large\begin{array}{c c} \; \; \bullet &\bullet\\ \; \; \bullet &\bullet\\ \end{array}\\ \text{Figure 2}}$
${\large\begin{array}{c c c} \bullet &\bullet &\bullet\\ \bullet &\bullet &\bullet\\ \end{array}\\ \;\; \text{Figure 3}}$
How many columns of two are in each figure? How many would you expect in figures $4$ and $5$? What about Figure $100$?
1. For each pattern, explain how you made your prediction for the $100$th figure.
Did you use numbers or diagrams? How do you know that your answer makes sense?
Use the eTool below to model the 4th and 5th figure for parts (a) and (b).
Click the link at right for the full version of the eTool: MC1 1-22 HW eTool
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# Trigonometry/For Enthusiasts/Regular Polygons
A regular polygon is a polygon with all its sides the same length and all its angles equal. A polygon can have any number (three or more) of sides so there are infinitely many different regular polygons.
### Interior Angles of Regular PolygonsEdit
To demonstrate that a square can be drawn so that each of its four corners lies on the circumference of a single circle: Draw a square and then draw its diagonals, calling the point at which they cross the center of the square. The center of the square is (by symmetry) the same distance from each corner. Consequently a circle whose center is co-incident with the center of the square can be drawn through the corners of the square.
A similar argument can be used to find the interior angles of any regular polygon. Consider a polygon of n sides. It will have n corners, through which a circle can be drawn. Draw a line from each corner to the center of the circle so that n equal apex angled triangles meet at the center, each such triangle must have an apex angle of 2π/n radians. Each such triangle is isoceles, so its other angles are equal and sum to π - 2π/n radians, that is each other angle is (π - 2π/n)/2 radians. Each corner angle of the polygon is split in two to form one of these other angles, so each corner of the polygon has 2*(π - 2π/n)/2 radians, that is π - 2π/n radians.
This formula predicts that a square, where the number of sides n is 4, will have interior angles of π - 2π/4 = π - π/2 = π/2 radians, which agrees with the calculation above.
Likewise, an equilateral triangle with 3 equal sides will have interior angles of π - 2π/3 = π/3 radians.
A hexagon will have interior angles of π - 2π/6 = 4π/6 = 2π/3 radians which is twice that of an equilateral triangle: thus the hexagon is divided into equilateral triangles by the splitting process described above.
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# Adding 3 Digit Numbers Using Standard Algorithm
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Adding 3 Digit Numbers Using Standard Algorithm
CCSS.MATH.CONTENT.4.NBT.B.4
### In this Video
Mr. Squeaks is taking off on the trip of a lifetime! He is going to visit a magical place that his beloved grandpa used to tell him stories about. He’s so excited about this trip and has never traveled this far, so he wants to record every moment in his journal! We can help figure out how far away his destination, is by totaling up the distance he travels each day. In this video, we are shown how to use standard algorithm to solve 3 digit addition without regrouping and 3 digit addition with regrouping. Mr. Squeaks' travels take him to the most beautiful place he has ever seen. His grandpa would be so proud!
## Adding 3 Digit Numbers Using Standard Algorithm Exercise
Would you like to practice what you’ve just learned? Practice problems for this video Adding 3 Digit Numbers Using Standard Algorithm help you practice and recap your knowledge.
• ### What is the place value of the underlined digit?
Hints
Put your number in the columns below.
Is the underlined digit in the hundreds, tens or ones column?
Solution
If your number has 3 digits it will have a hundreds, tens and ones place.
3 0 6 Tens
3 hundreds 0 tens 6 ones
2 7 1 Hundreds
2 hundreds 7 tens 1 one
9 8 1 Ones
9 hundreds 8 tens 1 one
• ### Can you find the mistake in the solution?
Hints
Make sure the answers have been placed in the correct columns.
Since the answer from the tens column is more than 2 digits, we need to carry the tens number over to the hundreds column.
Solution
For the ones place: 3 + 5 = 8
For the tens place: 4 + 7 = 11
We need to regroup the digit in the tens place to the hundreds column. So you are left with a 1 in the tens place.
For the hundreds place: 1 + 2 + 2 = 5
500 + 10 + 8 = 518
• ### Solve the equation by placing the digits into the correct boxes.
Hints
First, make sure you have put all of the digits for the number you're adding in the correct place.
Start by adding the ones, then the tens, and lastly the hundreds place.
Don't forget if your answer in the ones or tens column is greater than 9, you need to regroup the digit in the tens of the number to the next column.
Solution
Once the digits are in the correct column:
• Add the ones column first. 4 + 3 = 7.
• Then add the tens column. 4 + 6 = 10. We need to regroup the 1 to the hundreds column. So we're left with a 0 in the tens column.
• Finally add the hundreds column. 1 + 3 + 2 = 6.
• ### Solve the equations.
Hints
For this example, looking at the answer in the tens column, the 1 in 12 needs to be regrouped to the hundreds column. Then, the hundreds column will be 1 + 7 + 6.
Are there any that need to be regrouped for these equations above?
Start adding the ones place first.
Solution
473 + 326 = 799
306 + 291 = 597
651 + 367 = 1,018
• ### What's the solution?
Hints
Add the digits in the ones column first.
3 + 5 =
Add the digits in the tens column next.
4 + 2 =
Solution
Add the digits in the ones column first 3 + 5 = 8
Then add the digits in the tens column 4 + 2 = 6
Finally add the digits in the hundreds column 2 + 1 = 3
243 + 125 = 368
• ### Solve the answers to the equations.
Hints
Put the numbers on top of each other in the place value chart. Make sure all digits are in the correct column.
Solution
Put the numbers on top of each other in the place value chart.
Add the digits in each column, starting with the ones, then tens, and lastly, hundreds.
For 362 + 244:
Add the ones: 2 + 4 = 6
Then add the tens: 6 + 4 = 10 We need to regroup the 1 to the hundreds column.
Lastly, add the hundreds: 1 + 3 + 2 = 6
362 + 244 = 606
______________________________
For the rest of the equations:
271 + 149 = 420
745 + 639 = 1384
549 + 207 = 756
365 + 256 = 621
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RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions - GMS - Learning Simply
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# RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions
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# RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions
Class 12 is a crucial stage in a student’s life as it helps them achieve their career goals. We mainly focus on providing answers, which match the grasping abilities of students. To make this possible, our set of faculty has provided exercise-wise solutions for each chapter as per RD Sharma Solutions. The solutions are according to the latest CBSE syllabus in a stepwise manner as per the exam pattern and mark weightage. RD Sharma Solutions for Class 12 Chapter 4 Inverse Trigonometric Functions are given here. This chapter has fourteen exercises. Let us have a look at some of the important concepts that are discussed in this chapter.
• Definition and meaning of inverse trigonometric functions
• Inverse of sine function
• Inverse of cosine function
• Inverse of tangent function
• Inverse of secant function
• Inverse of cosecant function
• Inverse of cotangent function
• Properties of inverse trigonometric functions
### Exercise 4.1 Page No: 4.6
1. Find the principal value of the following:
Solution:
(iii) Given functions can be written as
(iv) The given question can be written as
(v) Let
(vi) Let
2.
(i)
(ii)
Solution:
(i) The given question can be written as,
(ii) Given question can be written as
### Exercise 4.2 Page No: 4.10
1. Find the domain of definition of f(x) = cos -1 (x2 – 4)
Solution:
Given f(x) = cos -1 (x2 – 4)
We know that domain of cos-1 (x2 – 4) lies in the interval [-1, 1]
Therefore, we can write as
-1 ≤ x2 – 4 ≤ 1
4 – 1 ≤ x2 ≤ 1 + 4
3 ≤ x2 ≤ 5
±√ 3 ≤ x ≤ ±√5
– √5 ≤ x ≤ – √3 and √3 ≤ x ≤ √5
Therefore domain of cos-1 (x2 – 4) is [- √5, – √3] ∪ [√3, √5]
2. Find the domain of f(x) = cos-1 2x + sin-1 x.
Solution:
Given that f(x) = cos-1 2x + sin-1 x.
Now we have to find the domain of f(x),
We know that domain of cos-1 x lies in the interval [-1, 1]
Also know that domain of sin-1 x lies in the interval [-1, 1]
Therefore, the domain of cos-1 (2x) lies in the interval [-1, 1]
Hence we can write as,
-1 ≤ 2x ≤ 1
– ½ ≤ x ≤ ½
Hence, domain of cos-1(2x) + sin-1 x lies in the interval [- ½, ½]
### Exercise 4.3 Page No: 4.14
1. Find the principal value of each of the following:
(i) tan-1 (1/√3)
(ii) tan-1 (-1/√3)
(iii) tan-1 (cos (π/2))
(iv) tan-1 (2 cos (2π/3))
Solution:
(i) Given tan-1 (1/√3)
We know that for any x ∈ R, tan-1 represents an angle in (-π/2, π/2) whose tangent is x.
So, tan-1 (1/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)
But we know that the value is equal to π/6
Therefore tan-1 (1/√3) = π/6
Hence the principal value of tan-1 (1/√3) = π/6
(ii) Given tan-1 (-1/√3)
We know that for any x ∈ R, tan-1 represents an angle in (-π/2, π/2) whose tangent is x.
So, tan-1 (-1/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)
But we know that the value is equal to -π/6
Therefore tan-1 (-1/√3) = -π/6
Hence the principal value of tan-1 (-1/√3) = – π/6
(iii) Given that tan-1 (cos (π/2))
But we know that cos (π/2) = 0
We know that for any x ∈ R, tan-1 represents an angle in (-π/2, π/2) whose tangent is x.
Therefore tan-1 (0) = 0
Hence the principal value of tan-1 (cos (π/2) is 0.
(iv) Given that tan-1 (2 cos (2π/3))
But we know that cos π/3 = 1/2
So, cos (2π/3) = -1/2
Therefore tan-1 (2 cos (2π/3)) = tan-1 (2 × – ½)
= tan-1(-1)
= – π/4
Hence, the principal value of tan-1 (2 cos (2π/3)) is – π/4
### Exercise 4.4 Page No: 4.18
1. Find the principal value of each of the following:
(i) sec-1 (-√2)
(ii) sec-1 (2)
(iii) sec-1 (2 sin (3π/4))
(iv) sec-1 (2 tan (3π/4))
Solution:
(i) Given sec-1 (-√2)
Now let y = sec-1 (-√2)
Sec y = -√2
We know that sec π/4 = √2
Therefore, -sec (π/4) = -√2
= sec (π – π/4)
= sec (3π/4)
Thus the range of principal value of sec-1 is [0, π] – {π/2}
And sec (3π/4) = – √2
Hence the principal value of sec-1 (-√2) is 3π/4
(ii) Given sec-1 (2)
Let y = sec-1 (2)
Sec y = 2
= Sec π/3
Therefore the range of principal value of sec-1 is [0, π] – {π/2} and sec π/3 = 2
Thus the principal value of sec-1 (2) is π/3
(iii) Given sec-1 (2 sin (3π/4))
But we know that sin (3π/4) = 1/√2
Therefore 2 sin (3π/4) = 2 × 1/√2
2 sin (3π/4) = √2
Therefore by substituting above values in sec-1 (2 sin (3π/4)), we get
Sec-1 (√2)
Let Sec-1 (√2) = y
Sec y = √2
Sec (π/4) = √2
Therefore range of principal value of sec-1 is [0, π] – {π/2} and sec (π/4) = √2
Thus the principal value of sec-1 (2 sin (3π/4)) is π/4.
(iv) Given sec-1 (2 tan (3π/4))
But we know that tan (3π/4) = -1
Therefore, 2 tan (3π/4) = 2 × -1
2 tan (3π/4) = -2
By substituting these values in sec-1 (2 tan (3π/4)), we get
Sec-1 (-2)
Now let y = Sec-1 (-2)
Sec y = – 2
– sec (π/3) = -2
= sec (π – π/3)
= sec (2π/3)
Therefore the range of principal value of sec-1 is [0, π] – {π/2} and sec (2π/3) = -2
Thus, the principal value of sec-1 (2 tan (3π/4)) is (2π/3).
### Exercise 4.5 Page No: 4.21
1. Find the principal values of each of the following:
(i) cosec-1 (-√2)
(ii) cosec-1 (-2)
(iii) cosec-1 (2/√3)
(iv) cosec-1 (2 cos (2π/3))
Solution:
(i) Given cosec-1 (-√2)
Let y = cosec-1 (-√2)
Cosec y = -√2
– Cosec y = √2
– Cosec (π/4) = √2
– Cosec (π/4) = cosec (-π/4) [since –cosec θ = cosec (-θ)]
The range of principal value of cosec-1 [-π/2, π/2] – {0} and cosec (-π/4) = – √2
Cosec (-π/4) = – √2
Therefore the principal value of cosec-1 (-√2) is – π/4
(ii) Given cosec-1 (-2)
Let y = cosec-1 (-2)
Cosec y = -2
– Cosec y = 2
– Cosec (π/6) = 2
– Cosec (π/6) = cosec (-π/6) [since –cosec θ = cosec (-θ)]
The range of principal value of cosec-1 [-π/2, π/2] – {0} and cosec (-π/6) = – 2
Cosec (-π/6) = – 2
Therefore the principal value of cosec-1 (-2) is – π/6
(iii) Given cosec-1 (2/√3)
Let y = cosec-1 (2/√3)
Cosec y = (2/√3)
Cosec (π/3) = (2/√3)
Therefore range of principal value of cosec-1 is [-π/2, π/2] – {0} and cosec (π/3) = (2/√3)
Thus, the principal value of cosec-1 (2/√3) is π/3
(iv) Given cosec-1 (2 cos (2π/3))
But we know that cos (2π/3) = – ½
Therefore 2 cos (2π/3) = 2 × – ½
2 cos (2π/3) = -1
By substituting these values in cosec-1 (2 cos (2π/3)) we get,
Cosec-1 (-1)
Let y = cosec-1 (-1)
– Cosec y = 1
– Cosec (π/2) = cosec (-π/2) [since –cosec θ = cosec (-θ)]
The range of principal value of cosec-1 [-π/2, π/2] – {0} and cosec (-π/2) = – 1
Cosec (-π/2) = – 1
Therefore the principal value of cosec-1 (2 cos (2π/3)) is – π/2
### Exercise 4.6 Page No: 4.24
1. Find the principal values of each of the following:
(i) cot-1(-√3)
(ii) Cot-1(√3)
(iii) cot-1(-1/√3)
(iv) cot-1(tan 3π/4)
Solution:
(i) Given cot-1(-√3)
Let y = cot-1(-√3)
– Cot (π/6) = √3
= Cot (π – π/6)
= cot (5π/6)
The range of principal value of cot-1 is (0, π) and cot (5 π/6) = – √3
Thus, the principal value of cot-1 (- √3) is 5π/6
(ii) Given Cot-1(√3)
Let y = cot-1(√3)
Cot (π/6) = √3
The range of principal value of cot-1 is (0, π) and
Thus, the principal value of cot-1 (√3) is π/6
(iii) Given cot-1(-1/√3)
Let y = cot-1(-1/√3)
Cot y = (-1/√3)
– Cot (π/3) = 1/√3
= Cot (π – π/3)
= cot (2π/3)
The range of principal value of cot-1(0, π) and cot (2π/3) = – 1/√3
Therefore the principal value of cot-1(-1/√3) is 2π/3
(iv) Given cot-1(tan 3π/4)
But we know that tan 3π/4 = -1
By substituting this value in cot-1(tan 3π/4) we get
Cot-1(-1)
Now, let y = cot-1(-1)
Cot y = (-1)
– Cot (π/4) = 1
= Cot (π – π/4)
= cot (3π/4)
The range of principal value of cot-1(0, π) and cot (3π/4) = – 1
Therefore the principal value of cot-1(tan 3π/4) is 3π/4
### Exercise 4.7 Page No: 4.42
1. Evaluate each of the following:
(i) sin-1(sin π/6)
(ii) sin-1(sin 7π/6)
(iii) sin-1(sin 5π/6)
(iv) sin-1(sin 13π/7)
(v) sin-1(sin 17π/8)
(vi) sin-1{(sin – 17π/8)}
(vii) sin-1(sin 3)
(viii) sin-1(sin 4)
(ix) sin-1(sin 12)
(x) sin-1(sin 2)
Solution:
(i) Given sin-1(sin π/6)
We know that the value of sin π/6 is ½
By substituting this value in sin-1(sin π/6)
We get, sin-1 (1/2)
Now let y = sin-1 (1/2)
Sin (π/6) = ½
The range of principal value of sin-1(-π/2, π/2) and sin (π/6) = ½
Therefore sin-1(sin π/6) = π/6
(ii) Given sin-1(sin 7π/6)
But we know that sin 7π/6 = – ½
By substituting this in sin-1(sin 7π/6) we get,
Sin-1 (-1/2)
Now let y = sin-1 (-1/2)
– Sin y = ½
– Sin (π/6) = ½
– Sin (π/6) = sin (- π/6)
The range of principal value of sin-1(-π/2, π/2) and sin (- π/6) = – ½
Therefore sin-1(sin 7π/6) = – π/6
(iii) Given sin-1(sin 5π/6)
We know that the value of sin 5π/6 is ½
By substituting this value in sin-1(sin 5π/6)
We get, sin-1 (1/2)
Now let y = sin-1 (1/2)
Sin (π/6) = ½
The range of principal value of sin-1(-π/2, π/2) and sin (π/6) = ½
Therefore sin-1(sin 5π/6) = π/6
(iv) Given sin-1(sin 13π/7)
Given question can be written as sin (2π – π/7)
Sin (2π – π/7) can be written as sin (-π/7) [since sin (2π – θ) = sin (-θ)]
By substituting these values in sin-1(sin 13π/7) we get sin-1(sin – π/7)
As sin-1(sin x) = x with x ∈ [-π/2, π/2]
Therefore sin-1(sin 13π/7) = – π/7
(v) Given sin-1(sin 17π/8)
Given question can be written as sin (2π + π/8)
Sin (2π + π/8) can be written as sin (π/8)
By substituting these values in sin-1(sin 17π/8) we get sin-1(sin π/8)
As sin-1(sin x) = x with x ∈ [-π/2, π/2]
Therefore sin-1(sin 17π/8) = π/8
(vi) Given sin-1{(sin – 17π/8)}
But we know that – sin θ = sin (-θ)
Therefore (sin -17π/8) = – sin 17π/8
– Sin 17π/8 = – sin (2π + π/8) [since sin (2π – θ) = -sin (θ)]
It can also be written as – sin (π/8)
– Sin (π/8) = sin (-π/8) [since – sin θ = sin (-θ)]
By substituting these values in sin-1{(sin – 17π/8)} we get,
Sin-1(sin – π/8)
As sin-1(sin x) = x with x ∈ [-π/2, π/2]
Therefore sin-1(sin -π/8) = – π/8
(vii) Given sin-1(sin 3)
We know that sin-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]
But here x = 3, which does not lie on the above range,
Therefore we know that sin (π – x) = sin (x)
Hence sin (π – 3) = sin (3) also π – 3 ∈ [-π/2, π/2]
Sin-1(sin 3) = π – 3
(viii) Given sin-1(sin 4)
We know that sin-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]
But here x = 4, which does not lie on the above range,
Therefore we know that sin (π – x) = sin (x)
Hence sin (π – 4) = sin (4) also π – 4 ∈ [-π/2, π/2]
Sin-1(sin 4) = π – 4
(ix) Given sin-1(sin 12)
We know that sin-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]
But here x = 12, which does not lie on the above range,
Therefore we know that sin (2nπ – x) = sin (-x)
Hence sin (2nπ – 12) = sin (-12)
Here n = 2 also 12 – 4π ∈ [-π/2, π/2]
Sin-1(sin 12) = 12 – 4π
(x) Given sin-1(sin 2)
We know that sin-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]
But here x = 2, which does not lie on the above range,
Therefore we know that sin (π – x) = sin (x)
Hence sin (π – 2) = sin (2) also π – 2 ∈ [-π/2, π/2]
Sin-1(sin 2) = π – 2
2. Evaluate each of the following:
(i) cos-1{cos (-π/4)}
(ii) cos-1(cos 5π/4)
(iii) cos-1(cos 4π/3)
(iv) cos-1(cos 13π/6)
(v) cos-1(cos 3)
(vi) cos-1(cos 4)
(vii) cos-1(cos 5)
(viii) cos-1(cos 12)
Solution:
(i) Given cos-1{cos (-π/4)}
We know that cos (-π/4) = cos (π/4) [since cos (-θ) = cos θ
Also know that cos (π/4) = 1/√2
By substituting these values in cos-1{cos (-π/4)} we get,
Cos-1(1/√2)
Now let y = cos-1(1/√2)
Therefore cos y = 1/√2
Hence range of principal value of cos-1 is [0, π] and cos (π/4) = 1/√2
Therefore cos-1{cos (-π/4)} = π/4
(ii) Given cos-1(cos 5π/4)
But we know that cos (5π/4) = -1/√2
By substituting these values in cos-1{cos (5π/4)} we get,
Cos-1(-1/√2)
Now let y = cos-1(-1/√2)
Therefore cos y = – 1/√2
– Cos (π/4) = 1/√2
Cos (π – π/4) = – 1/√2
Cos (3 π/4) = – 1/√2
Hence range of principal value of cos-1 is [0, π] and cos (3π/4) = -1/√2
Therefore cos-1{cos (5π/4)} = 3π/4
(iii) Given cos-1(cos 4π/3)
But we know that cos (4π/3) = -1/2
By substituting these values in cos-1{cos (4π/3)} we get,
Cos-1(-1/2)
Now let y = cos-1(-1/2)
Therefore cos y = – 1/2
– Cos (π/3) = 1/2
Cos (π – π/3) = – 1/2
Cos (2π/3) = – 1/2
Hence range of principal value of cos-1 is [0, π] and cos (2π/3) = -1/2
Therefore cos-1{cos (4π/3)} = 2π/3
(iv) Given cos-1(cos 13π/6)
But we know that cos (13π/6) = √3/2
By substituting these values in cos-1{cos (13π/6)} we get,
Cos-1(√3/2)
Now let y = cos-1(√3/2)
Therefore cos y = √3/2
Cos (π/6) = √3/2
Hence range of principal value of cos-1 is [0, π] and cos (π/6) = √3/2
Therefore cos-1{cos (13π/6)} = π/6
(v) Given cos-1(cos 3)
We know that cos-1(cos θ) = θ if 0 ≤ θ ≤ π
Therefore by applying this in given question we get,
Cos-1(cos 3) = 3, 3 ∈ [0, π]
(vi) Given cos-1(cos 4)
We have cos–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14]
And here x = 4 which does not lie in the above range.
We know that cos (2π – x) = cos(x)
Thus, cos (2π – 4) = cos (4) so 2π–4 belongs in [0, π]
Hence cos–1(cos 4) = 2π – 4
(vii) Given cos-1(cos 5)
We have cos–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14]
And here x = 5 which does not lie in the above range.
We know that cos (2π – x) = cos(x)
Thus, cos (2π – 5) = cos (5) so 2π–5 belongs in [0, π]
Hence cos–1(cos 5) = 2π – 5
(viii) Given cos-1(cos 12)
Cos–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14]
And here x = 12 which does not lie in the above range.
We know cos (2nπ – x) = cos (x)
Cos (2nπ – 12) = cos (12)
Here n = 2.
Also 4π – 12 belongs in [0, π]
∴ cos–1(cos 12) = 4π – 12
3. Evaluate each of the following:
(i) tan-1(tan π/3)
(ii) tan-1(tan 6π/7)
(iii) tan-1(tan 7π/6)
(iv) tan-1(tan 9π/4)
(v) tan-1(tan 1)
(vi) tan-1(tan 2)
(vii) tan-1(tan 4)
(viii) tan-1(tan 12)
Solution:
(i) Given tan-1(tan π/3)
As tan-1(tan x) = x if x ϵ [-π/2, π/2]
By applying this condition in the given question we get,
Tan-1(tan π/3) = π/3
(ii) Given tan-1(tan 6π/7)
We know that tan 6π/7 can be written as (π – π/7)
Tan (π – π/7) = – tan π/7
We know that tan-1(tan x) = x if x ϵ [-π/2, π/2]
Tan-1(tan 6π/7) = – π/7
(iii) Given tan-1(tan 7π/6)
We know that tan 7π/6 = 1/√3
By substituting this value in tan-1(tan 7π/6) we get,
Tan-1 (1/√3)
Now let tan-1 (1/√3) = y
Tan y = 1/√3
Tan (π/6) = 1/√3
The range of the principal value of tan-1 is (-π/2, π/2) and tan (π/6) = 1/√3
Therefore tan-1(tan 7π/6) = π/6
(iv) Given tan-1(tan 9π/4)
We know that tan 9π/4 = 1
By substituting this value in tan-1(tan 9π/4) we get,
Tan-1 (1)
Now let tan-1 (1) = y
Tan y = 1
Tan (π/4) = 1
The range of the principal value of tan-1 is (-π/2, π/2) and tan (π/4) = 1
Therefore tan-1(tan 9π/4) = π/4
(v) Given tan-1(tan 1)
But we have tan-1(tan x) = x if x ϵ [-π/2, π/2]
By substituting this condition in given question
Tan-1(tan 1) = 1
(vi) Given tan-1(tan 2)
As tan-1(tan x) = x if x ϵ [-π/2, π/2]
But here x = 2 which does not belongs to above range
We also have tan (π – θ) = –tan (θ)
Therefore tan (θ – π) = tan (θ)
Tan (2 – π) = tan (2)
Now 2 – π is in the given range
Hence tan–1 (tan 2) = 2 – π
(vii) Given tan-1(tan 4)
As tan-1(tan x) = x if x ϵ [-π/2, π/2]
But here x = 4 which does not belongs to above range
We also have tan (π – θ) = –tan (θ)
Therefore tan (θ – π) = tan (θ)
Tan (4 – π) = tan (4)
Now 4 – π is in the given range
Hence tan–1 (tan 2) = 4 – π
(viii) Given tan-1(tan 12)
As tan-1(tan x) = x if x ϵ [-π/2, π/2]
But here x = 12 which does not belongs to above range
We know that tan (2nπ – θ) = –tan (θ)
Tan (θ – 2nπ) = tan (θ)
Here n = 2
Tan (12 – 4π) = tan (12)
Now 12 – 4π is in the given range
∴ tan–1 (tan 12) = 12 – 4π.
### Exercise 4.8 Page No: 4.54
1. Evaluate each of the following:
(i) sin (sin-1 7/25)
(ii) Sin (cos-1 5/13)
(iv) Sin (sec-1 17/8)
(v) Cosec (cos-1 8/17)
(vi) Sec (sin-1 12/13)
(vii) Tan (cos-1 8/17)
(viii) cot (cos-1 3/5)
Solution:
(i) Given sin (sin-1 7/25)
Now let y = sin-1 7/25
Sin y = 7/25 where y ∈ [0, π/2]
Substituting these values in sin (sin-1 7/25) we get
Sin (sin-1 7/25) = 7/25
(ii) Given Sin (cos-1 5/13)
(iv) Given Sin (sec-1 17/8)
(v) Given Cosec (cos-1 8/17)
Let cos-1(8/17) = y
cos y = 8/17 where y ∈ [0, π/2]
Now, we have to find
Cosec (cos-1 8/17) = cosec y
We know that,
sin2 θ + cos2 θ = 1
sin2 θ = √ (1 – cos2 θ)
So,
sin y = √ (1 – cos2 y)
= √ (1 – (8/17)2)
= √ (1 – 64/289)
= √ (289 – 64/289)
= √ (225/289)
= 15/17
Hence,
Cosec y = 1/sin y = 1/ (15/17) = 17/15
Therefore,
Cosec (cos-1 8/17) = 17/15
(vi) Given Sec (sin-1 12/13)
(vii) Given Tan (cos-1 8/17)
(viii) Given cot (cos-1 3/5)
.
### Exercise 4.9 Page No: 4.58
1. Evaluate:
(i) Cos {sin-1 (-7/25)}
(ii) Sec {cot-1 (-5/12)}
(iii) Cot {sec-1 (-13/5)}
Solution:
(i) Given Cos {sin-1 (-7/25)}
(ii) Given Sec {cot-1 (-5/12)}
(iii) Given Cot {sec-1 (-13/5)}
### Exercise 4.10 Page No: 4.66
1. Evaluate:
(i) Cot (sin-1 (3/4) + sec-1 (4/3))
(ii) Sin (tan-1 x + tan-1 1/x) for x < 0
(iii) Sin (tan-1 x + tan-1 1/x) for x > 0
(iv) Cot (tan-1 a + cot-1 a)
(v) Cos (sec-1 x + cosec-1 x), |x| ≥ 1
Solution:
(i) Given Cot (sin-1 (3/4) + sec-1 (4/3))
(ii) Given Sin (tan-1 x + tan-1 1/x) for x < 0
(iii) Given Sin (tan-1 x + tan-1 1/x) for x > 0
(iv) Given Cot (tan-1 a + cot-1 a)
(v) Given Cos (sec-1 x + cosec-1 x), |x| ≥ 1
= 0
2. If cos-1 x + cos-1 y = π/4, find the value of sin-1 x + sin-1 y.
Solution:
Given cos-1 x + cos-1 y = π/4
3. If sin-1 x + sin-1 y = π/3 and cos-1 x – cos-1 y = π/6, find the values of x and y.
Solution:
Given sin-1 x + sin-1 y = π/3 ……. Equation (i)
And cos-1 x – cos-1 y = π/6 ……… Equation (ii)
4. If cot (cos-1 3/5 + sin-1 x) = 0, find the value of x.
Solution:
Given cot (cos-1 3/5 + sin-1 x) = 0
On rearranging we get,
(cos-1 3/5 + sin-1 x) = cot-1 (0)
(Cos-1 3/5 + sin-1 x) = π/2
We know that cos-1 x + sin-1 x = π/2
Then sin-1 x = π/2 – cos-1 x
Substituting the above in (cos-1 3/5 + sin-1 x) = π/2 we get,
(Cos-1 3/5 + π/2 – cos-1 x) = π/2
Now on rearranging we get,
(Cos-1 3/5 – cos-1 x) = π/2 – π/2
(Cos-1 3/5 – cos-1 x) = 0
Therefore Cos-1 3/5 = cos-1 x
On comparing the above equation we get,
x = 3/5
5. If (sin-1 x)2 + (cos-1 x)2 = 17 π2/36, find x.
Solution:
Given (sin-1 x)2 + (cos-1 x)2 = 17 π2/36
We know that cos-1 x + sin-1 x = π/2
Then cos-1 x = π/2 – sin-1 x
Substituting this in (sin-1 x)2 + (cos-1 x)2 = 17 π2/36 we get
(sin-1 x)2 + (π/2 – sin-1 x)2 = 17 π2/36
Let y = sin-1 x
y2 + ((π/2) – y)2 = 17 π2/36
y2 + π2/4 – y2 – 2y ((π/2) – y) = 17 π2/36
π2/4 – πy + 2 y= 17 π2/36
On rearranging and simplifying, we get
2y2 – πy + 2/9 π2 = 0
18y2 – 9 πy + 2 π2 = 0
18y2 – 12 πy + 3 πy + 2 π2 = 0
6y (3y – 2π) + π (3y – 2π) = 0
Now, (3y – 2π) = 0 and (6y + π) = 0
Therefore y = 2π/3 and y = – π/6
Now substituting y = – π/6 in y = sin-1 x we get
sin-1 x = – π/6
x = sin (- π/6)
x = -1/2
Now substituting y = -2π/3 in y = sin-1 x we get
x = sin (2π/3)
x = √3/2
Now substituting x = √3/2 in (sin-1 x)2 + (cos-1 x)2 = 17 π2/36 we get,
= π/3 + π/6
= π/2 which is not equal to 17 π2/36
So we have to neglect this root.
Now substituting x = -1/2 in (sin-1 x)2 + (cos-1 x)2 = 17 π2/36 we get,
= π2/36 + 4 π2/9
= 17 π2/36
Hence x = -1/2.
### Exercise 4.11 Page No: 4.82
1. Prove the following results:
(i) Tan-1 (1/7) + tan-1 (1/13) = tan-1 (2/9)
(ii) Sin-1 (12/13) + cos-1 (4/5) + tan-1 (63/16) = π
(iii) tan-1 (1/4) + tan-1 (2/9) = Sin-1 (1/ √5)
Solution:
(i) Given Tan-1 (1/7) + tan-1 (1/13) = tan-1 (2/9)
Hence, proved.
(ii) Given Sin-1 (12/13) + cos-1 (4/5) + tan-1 (63/16) = π
Consider LHS
Hence, proved.
(iii) Given tan-1 (1/4) + tan-1 (2/9) = Sin-1 (1/ √5)
2. Find the value of tan-1 (x/y) – tan-1 {(x-y)/(x + y)}
Solution:
Given tan-1 (x/y) – tan-1 {(x-y)/(x + y)}
### Exercise 4.12 Page No: 4.89
1. Evaluate: Cos (sin -1 3/5 + sin-1 5/13)
Solution:
Given Cos (sin -1 3/5 + sin-1 5/13)
We know that,
### Exercise 4.13 Page No: 4.92
1. If cos-1 (x/2) + cos-1 (y/3) = α, then prove that 9x2 – 12xy cos α + 4y2 = 36 sin2 α
Solution:
Given cos-1 (x/2) + cos-1 (y/3) = α
Hence, proved.
2. Solve the equation: cos-1 (a/x) – cos-1 (b/x) = cos-1 (1/b) – cos-1 (1/a)
Solution:
Given cos-1 (a/x) – cos-1 (b/x) = cos-1 (1/b) – cos-1 (1/a)
### Exercise 4.14 Page No: 4.115
1. Evaluate the following:
(i) tan {2 tan-1 (1/5) – π/4}
(ii) Tan {1/2 sin-1 (3/4)}
(iii) Sin {1/2 cos-1 (4/5)}
(iv) Sin (2 tan -1 2/3) + cos (tan-1 √3)
Solution:
(i) Given tan {2 tan-1 (1/5) – π/4}
(ii) Given tan {1/2 sin-1 (3/4)}
(iii) Given sin {1/2 cos-1 (4/5)}
(iv) Given Sin (2 tan -1 2/3) + cos (tan-1 √3)
2. Prove the following results:
(i) 2 sin-1 (3/5) = tan-1 (24/7)
(ii) tan-1 ¼ + tan-1 (2/9) = ½ cos-1 (3/5) = ½ sin-1 (4/5)
(iii) tan-1 (2/3) = ½ tan-1 (12/5)
(iv) tan-1 (1/7) + 2 tan-1 (1/3) = π/4
(v) sin-1 (4/5) + 2 tan-1 (1/3) = π/2
(vi) 2 sin-1 (3/5) – tan-1 (17/31) = π/4
(vii) 2 tan-1 (1/5) + tan-1 (1/8) = tan-1 (4/7)
(viii) 2 tan-1 (3/4) – tan-1 (17/31) = π/4
(ix) 2 tan-1 (1/2) + tan-1 (1/7) = tan-1 (31/17)
(x) 4 tan-1(1/5) – tan-1(1/239) = π/4
Solution:
(i) Given 2 sin-1 (3/5) = tan-1 (24/7)
Hence, proved.
(ii) Given tan-1 ¼ + tan-1 (2/9) = ½ cos-1 (3/5) = ½ sin-1 (4/5)
Hence, proved.
(iii) Given tan-1 (2/3) = ½ tan-1 (12/5)
Hence, proved.
(iv) Given tan-1 (1/7) + 2 tan-1 (1/3) = π/4
Hence, proved.
(v) Given sin-1 (4/5) + 2 tan-1 (1/3) = π/2
(vi) Given 2 sin-1 (3/5) – tan-1 (17/31) = π/4
(vii) Given 2 tan-1 (1/5) + tan-1 (1/8) = tan-1 (4/7)
Hence, proved.
(viii) Given 2 tan-1 (3/4) – tan-1 (17/31) = π/4
Hence, proved.
(ix) Given 2 tan-1 (1/2) + tan-1 (1/7) = tan-1 (31/17)
Hence, proved.
(x) Given 4 tan-1(1/5) – tan-1(1/239) = π/4
Hence, proved.
3. If sin-1 (2a/1 + a2) – cos-1(1 – b2/1 + b2) = tan-1(2x/1 – x2), then prove that x = (a – b)/ (1 + a b)
Solution:
Given sin-1 (2a/1 + a2) – cos-1(1 – b2/1 + b2) = tan-1(2x/1 – x2)
Hence, proved.
4. Prove that:
(i) tan-1{(1 – x2)/ 2x)} + cot-1{(1 – x2)/ 2x)} = π/2
(ii) sin {tan-1 (1 – x2)/ 2x) + cos-1 (1 – x2)/ (1 + x2)} = 1
Solution:
(i) Given tan-1{(1 – x2)/ 2x)} + cot-1{(1 – x2)/ 2x)} = π/2
Hence, proved.
(ii) Given sin {tan-1 (1 – x2)/ 2x) + cos-1 (1 – x2)/ (1 + x2)}
Hence, proved.
5. If sin-1 (2a/ 1+ a2) + sin-1 (2b/ 1+ b2) = 2 tan-1 x, prove that x = (a + b/ 1 – a b)
Solution:
Given sin-1 (2a/ 1+ a2) + sin-1 (2b/ 1+ b2) = 2 tan-1 x
Hence, proved.
At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio…
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Find the volume of an irregular 3D figure by dividing the figure into two rectangular prisms and finding the volume of each part.
# Volume word problems with cones, cylinders, and spheres
Volume word problems with cones, cylinders, and spheres
# 12.4 Volumes of Prisms and Cylinders (Lesson)
## By AutenMath
A lesson covering the formulas involved in finding the volume of prisms and cylinders
# Surface Area and Volume of Spheres - YouTube
## By Missy McCarthy
Surface Area and Volume of Spheres - YouTube
# Volume Word Problem (Intermediate)
## By mahalodotcom
YouTube presents Volume Word Problem (Intermediate) an educational video resources on math.
# Volume of Rectangular Prisms: Length Width and Height
## By Have Fun Teaching
The Volume Song and movie teaches students how to find the volume of a solid figure. Learn how to solve and find the volume of a 3D rectangular prism shape using length times width times height.
# Volume with unit cubes 2
Another way of finding the volume of a rectangular prism involves dividing it into fractional cubes, finding the volume of one, and then multiplying that area by the number of cubes that fit into our rectangular prism.
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# 7.2b Interquartile Range 1
Measures of Dispersion (Part 2)
7.2b Interquartile Range 1
(A) Interquartile Range of Ungrouped Data
Example 1:
Find the interquartile range of the following data.
(a) 7, 5, 1, 3, 6, 11, 8
(b) 12, 4, 6, 18, 9, 16, 2, 14
Solution:
(a) Rearrange the data according to ascending order.
Interquartile Range
= upper quartile – lower quartile
= 8 – 3
= 5
(b) Rearrange the data according to ascending order.
Interquartile Range
= upper quartile – lower quartile
$\frac{14+16}{2}-\frac{4+6}{2}$
= 15 – 5
= 10
(B) Interquartile Range of Grouped Data (without Class Interval)
Example 2:
The table below shows the marks obtained by a group of Form 4 students in school mid-term science test.
Marks 1 2 3 4 5 Number of students 4 7 5 2 6
Determine the interquartile range of the distribution.
Solution:
Lower quartile, Q1 = the $\frac{1}{4}{\left(24\right)}^{\text{th}}$ observation
= the 6thobservation
= 2
Upper quartile, Q3 = the $\frac{3}{4}{\left(24\right)}^{\text{th}}$ observation
= the 18thobservation
= 4
Marks Frequency Cumulative Frequency 1 4 4 2 7 11 (Q1 is located here) 3 5 16 4 2 18 (Q3 is located here) 5 6 24
Interquartile Range
= upper quartile – lower quartile
= Q3Q1
= 4 – 2
= 2
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We have already learned how to operate with integers (addition, subtraction, multiplication, and division) as well as its order of operation. Test your skills in addition of fraction in this quiz below. Fill in the blanks with the correct answer. Good luck!
A quiz on addition of integers.
How to Add Positive and Negative Integers
One of the topics in basic mathematics  that will likely be included in the the Philippine Civil Service Exam both professional and subprofessional are operations on integers. Although a few Civil Service test items may be given from this topic,  it is important that you master it because a lot of calculation in other topics will need knowledge of integers and its operations (addition, subtraction, multiplication, division). For example, solving some word problems in mathematics and solving equations will need knowledge on operations of integers.
Integers are whole numbers that are either positive or negative. Examples of integers are -5, 6, 0, and 10. If we place this on the number line, negative integers are the integers that are below 0 (left of 0), while the positive integers are the integers above 0 (right of 0).
Adding Integers that Are Both Positive
When you add integers that are both positive, it is just like adding whole numbers. Below are the examples.
Example 1:Â +2 + +4 = +6
Example 2:Â +9 + +41 + +6 + = +56
Example 3:Â +120 + +13 + +12 + = +145
Although we have created a small + before the number to indicate that it is positive, in reality, only negative numbers have signs. This means that +2 + +4 = +6 is just written as 2 + 4 = 6.
Adding Integers that Are Both Negative
Adding number that are both negative is just the same as adding numbers that are both positive. The only difference is that if you add two negative numbers, the result is negative.
Example 1:Â 5 + 8 = 13
Example 2:Â 10 + 18 + 32 + = 60
Example 3:Â 220 + 11 + 16 + = 247
How to Add Positive and Negative Integers
Before adding, you should always remember that +1 and 1 cancel out each other, or +1 + 1 is 0. So the strategy is to pair the positive and negative numbers and take out what’s left.
Example 1:  What is +13 + 8?
Solution:
We pair 8 positives and 8 negatives to cancel out. Then what’s left is of +13 is +5. In equation form, we have
 +13 + 8 = +5 + +8 + 8 = +5 + (+8 + 8) = +5 + (0) = +5
Example 2: What is +17 + 20?
Solution:
We pair 17 negatives and 17 positives. What’s left of 20 is 3. In equation form, we have
+17 +Â 20 = +17 + (17 +Â 3)Â =Â Â (+17 + 17) +Â 3 = 0 +Â 3 =Â 3
Example 3: What is +16 + +37 + 20 + 3 +9 ?
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# 4.2: Trig Proofs
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Trigonometry, Chapter 3, Lesson 2.
ID: 9776
Time required: 30 minutes
## Activity Overview
Students will perform trigonometric proofs and use the graphing capabilities of the calculator for verification.
Topic: Trigonometric Identities
• Use fundamental trigonometric identities to prove more complex trigonometric identities.
• Verify trigonometric identities by graphing.
Teacher Preparation and Notes
• Students should already be familiar with the Pythagorean trigonometric identities as well the fact that tangent=sinecosine,cosecant=1sine,secant=1cosine, and cotangent=1tangent.
• This activity is intended to be teacher-led. You may use the following pages to present the material to the class and encourage discussion. Students will follow along using their calculators, although the majority of the ideas and concepts are only presented in this document; be sure to cover all the material necessary for students’ total comprehension.
• The worksheet is intended to guide students through the main ideas of the activity, while providing more detailed instruction on how they are to perform specific actions using the tools of the calculator. It also serves as a place for students to record their answers. Alternatively, you may wish to have the class record their answers on separate sheets of paper, or just use the questions posed to engage a class discussion.
Associated Materials
## Problem 1 – Using the Calculator for verification
Prove: (1+cosx)(1cosx)=sin2x.
(1+cos(x))(1cos(x))=1cos2(x)=[sin2(x)+cos2(x)]cos2(x)=sin2x
To verify this proof graphically, you will determine if the graph of the expression on the left side of the equation coincides with the graph of the expression on the right side of the equation.
Enter the left side of the equation (1+cosx)(1cosx) in Y1.
Enter the right side of the equation (sinx)2 in Y2.
Arrow over to the icon to the left of Y2 and press ENTER to change it to a thick line. Now Y1 will be graphed as a thin line and Y2 will be graphed as a thick line. This will help you distinguish the two graphs (and make sure that they are equal).
Press ZOOM and select ZTrig to set the window size.
The two graphs coincide, so the two sides of the equation are equal, as we proved. Note that the calculator is only verifying what we have proven. The graph in no way constitutes a proof.
## Problems 2-5
For problems 2 through 5, prove the equation given and then verify it graphically. For cotx, type (1tanx). For secx, type (1cosx).
2. sinxcotxsecx=1
3. sec2x1sec2x=sin2x
4. tanx+cotx=secx(cscx)
5. sin2x49sin2x+14sinx+49=sinx7sinx+7
## Solutions
Problem 2
sin(x)cot(x)sec(x)=sin(x)cos(x)sin(x)1cos(x)=1
Problem 3
sec2(x)1sec2(x)=[1+tan2x]1sec2(x)=tan2(x)sec2(x)=(sin(x)cos(x))21cos2(x)=sin2(x)
Problem 4
tan(x)+cot(x)=sin(x)cos(x)+cos(x)sin(x)=sin2(x)cos(x)sin(x)+cos2(x)cos(x)sin(x)=sin2(x)+cos2(x)cos(x)sin(x)=1cos(x)sin(x)=sec(x)csc(x)
Problem 5
sin2(x)49sin2(x)+14sin(x)+49=(sin(x)7)(sin(x)+7)(sin(x)+7)(sin(x)+7)=sin(x)7sin(x)+7
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Date Created:
Feb 23, 2012
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Simplify Square Root of 12000
Here we will show you two methods that you can use to simplify the square root of 12000. In other words, we will show you how to find the square root of 12000 in its simplest radical form using two different methods.
To be more specific, we have created an illustration below showing what we want to calculate. Our goal is to make "A" outside the radical (√) as large as possible, and "B" inside the radical (√) as small as possible.
12000 = A√B
Greatest Perfect Square Factor Method
The Greatest Perfect Square Factor Method uses the greatest perfect square factor of 12000 to simplify the square root of 12000. This is how to calculate A and B using this method:
A = Calculate the square root of the greatest perfect square from the list of all factors of 12000. The factors of 12000 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 25, 30, 32, 40, 48, 50, 60, 75, 80, 96, 100, 120, 125, 150, 160, 200, 240, 250, 300, 375, 400, 480, 500, 600, 750, 800, 1000, 1200, 1500, 2000, 2400, 3000, 4000, 6000, and 12000. Furthermore, the greatest perfect square on this list is 400 and the square root of 400 is 20. Therefore, A equals 20.
B = Calculate 12000 divided by the greatest perfect square from the list of all factors of 12000. We determined above that the greatest perfect square from the list of all factors of 12000 is 400. Furthermore, 12000 divided by 400 is 30, therefore B equals 30.
Now we have A and B and can get our answer to 12000 in its simplest radical form as follows:
12000 = A√B
12000 = 20√30
Double Prime Factor Method
The Double Prime Factor Method uses the prime factors of 12000 to simplify the square root of 12000 to its simplest form possible. This is how to calculate A and B using this method:
A = Multiply all the double prime factors (pairs) of 12000 and then take the square root of that product. The prime factors that multiply together to make 12000 are 2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5. When we strip out the pairs only, we get 2 x 2 x 2 x 2 x 5 x 5 = 400 and the square root of 400 is 20. Therefore, A equals 20.
B = Divide 12000 by the number (A) squared. 20 squared is 400 and 12000 divided by 400 is 30. Therefore, B equals 30.
Once again we have A and B and can get our answer to 12000 in its simplest radical form as follows:
12000 = A√B
12000 = 20√30
Simplify Square Root
Please enter another square root in the box below for us to simplify.
Simplify Square Root of 12001
Here is the next square root on our list that we have simplifed for you.
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Smartick is a fun way to learn math!
Dec31
# Addition With and Without Carrying
In today’s post, we are going to learn how to solve addition problems with and without carrying. We will also see some examples of exercises that children complete in Smartick.
But first, let’s review what addition is!
Index
Addition is the act of adding, joining, or putting elements together. When we perform this action we are joining amounts or sets and as a result, there must always be a minimum of two elements.
Humans were performing addition before we even learned to write or had a language. It is believed that many species of animals know how to add small amounts.
## How to Explain Addition to a Primary School Student
Addition is a basic arithmetic operation that we learn at a very young age. Little ones use different strategies to add, such as counting all the elements, counting their fingers, or adding the smaller number to the larger one. We use addition often without realizing it. So if we have to help a child learn to add, it is important that we help them become aware of what they are doing. We can use counters (cubes, pasta, beans, etc.) to model quantities, represent them separately, and count them, count them all together again, repeat the process with other quantities… and then try to do so without representing the quantity.
What is described in the previous paragraph is not a quick process, you need to go slowly. The process comes together when the child sees that the numbers represent the quantities and the addition sign means that they need to put them all together and count them.
The first addition problems, those with small quantities, can be done horizontally as they are read. Later on, it will become more efficient to solve them vertically, which will be explained in our next point.
## How to Solve Addition Problems
### How to Solve Addition Without Carrying
• Place the numbers on top of one another so that the place values line up with one another in the same column.
• Add each column separately starting with the units.
• Write the result of each column below the line.
Let’s look at an example, 32 + 64.
We put 32 above 64 and line up the units column, 3 above 6.
We add 2 + 4 = 6 and we put the result below the line under the units column.
We add 3 + 6 = 9 and we put the result below the line under the tens column.
The sum is 96.
### Video to Learn to Add Without Carrying
If you would like, take a look at the following video to see more examples of addition without carrying.
This is a video of one of our interactive tutorials, and although it is not interactive you have the advantage of watching it as many times as you need to and can share it with your friends. If you would like to access our interactive tutorials, register with Smartick! The online method which helps children ages 4 to 14 learn and practice math.
### How to Solve Addition With Carrying
In this case, when you add the numbers in a column and get a two-digit number, you will carry the tens and add it to the next column.
Let’s take a look at an example. Amy and Max are playing a game and pretending to shop. Amy wants to buy a bottle of water that costs 27 cents and a package of juice boxes that costs 45 cents. We need to add 27 + 45.
First, we put 45 on top of 27 and line up the columns, units with units and tens with tens.
We add the units first, 5 + 7 = 12. We put the 2 below the units column and carry the 1 to the tens column and add it to the numbers there.
1 + 4 + 2 = 7, so we put the 7 below the tens column.
The sum is 72.
### Video to Learn to Add With Carrying
Once again, we invite you to take a look at the following video to see more examples of addition with carrying.
## Exercises
### Exercise 1
In this activity, you need to select which of the vertical addition problems is written correctly. Can you find the right answer?
Result: It is the second option.
### Exercise 2
In this activity, we are using a 100 Square to solve the addition problem. In this case, from the number 2, you can add ten units one by one or a ten, which would be the equivalent of jumping the row underneath to find the sum of the addition problem which is 12.
### Exercise 3
We’re using the 100 Square in this activity as well and if you look closely, in this case, we are adding a ten and subtracting a unit. This way, the operation 2 + 9, is done by adding 2 + 10, which is 12, and subtracting 1, to give us the result, 11. This exercise, and material, helps to develop strategies like mental calculation.
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# Let's Describe Algorithms Better
I was reading a cryptography textbook, and I came across an algorithm that reminded me why I often dislike descriptions of algorithms in textbooks. This is a short post going over this example, and thinking about how to improve these descriptions.
Specifically, I often dislike the imperative descriptions of algorithms, which do a great job of explaining the operational semantics, but don’t give any insight into why an algorithm actually works.
# The Example
This algorithm takes integers $N$, $g$, and $A$, and calculates:
$$g^A \mod N$$
As some background, the idea behind algorithms for doing this fast is to use the fact that if you write $A$ as powers of $2$, say $A = 2 + 4 + 16$ for the sake of example, then we have:
$$g^A = g^{(2 + 4 + 16)} = g^2 \cdot g^4 \cdot g^{16}$$
More generally, we can multiply all of the power of 2 exponents together.
An algorithm applying this technique is given an imperative description as:
1. Set $a = g$ and $b = 1$
2. Loop while $A > 0$
3. If $A \equiv 1 \mod 2$, set $b = b \cdot a \mod N$
4. Set $a = a^2 \mod N$, and $A = \lfloor A / 2 \rfloor$
5. End Loop
6. Return $b$, which is equal to $g^A \mod N$
At least for me, it’s not immediately obvious that this does the right thing, and I have an awfully hard time trying to remember this algorithm.
The advantage of an algorithm described imperatively like this, is that it translates immediately to an actual programming language. Let’s use Python, to illustrate this:
def pow(N, g, A):
a = g
b = 1
while A > 0:
if A % 2 == 1:
b = (b * a) % N
a = (a * a) % N
A = A // 2
return b
Even now, though the actual code makes it easier to trace through, it doesn’t immediately give you an understanding of what’s really going on.
## A better description
There’s a lot of essence in this algorithm to be distilled. One first insight is that our iteration is controlled entirely by A, so we should think about what values we’re using.
At each iteration, we care about A % 2, and then use A // 2 at the next iteration. What we’re really doing is iterating over all the bits of A, from least significant to most significant.
So, if A is 22, for example, the bits we’d see would be:
[0, 1, 1, 0, 1]
At each iteration, the value of a is the square of the previous iteration.
What we’re doing with b is multiplying it with some of these values, but only selectively. We only multiply it by the latest value of a if the bit for this iteration is 1.
A better description, in pseudo code, would be:
1. Let $G = [g, g^2, g^4, g^8, \ldots]$ be a list of successive squares (taken modulo $N$)
2. Let $B$ be a list of bits in $A$, from least to most significant
3. Pair up $G$ and $B$
4. Keep only those elements where the bit is $1$
5. Multiply all the remaining exponents, modulo $N$
6. This value is $g^A \mod N$
For me, this strategy is much easier to remember. Instead of describing things in the nitty-gritty details, we use a higher level description. This description does a better job, in my view, of distilling the essential details of the algorithm, and removing out less important details.
One goal of the original algorithm was efficiency, specificallly not storing values, and working only with a few accumulators. This declarative description talks about lists, albeit infinite ones, so it doesn’t do as good a job at describing the operational semantics.
Thankfully, in a language like Haskell, where you have a lazy lists, you can effectively implement this description with the described performance characteristics:
pow :: Int -> Int -> Int -> Int
pow n g a =
zip (bits a) (squares g)
|> filter (fst >>> (== 1))
|> map snd
|> prod
where
bits :: Int -> [Int]
bits 0 = []
bits n = mod n 2 : bits (div n 2)
squares :: Int -> [Int]
squares = iterate (\s -> s * s mod n)
prod :: [Int] -> Int
prod = foldr (\x acc -> x * acc mod n) 1
I find this algorithm much more satisfying, and much easier to remember:
“The fast squaring algorithm? Oh, it’s the one where you zip the bits of the exponent with the successive squares, and then multiply them!”
Now, this was a contrived example, in that I literally started this post just because I disliked the original presentation of the algorithm. Perhaps I’m biased, in that the true understanding comes from translating an imperative description into higher level insight.
That being said, maybe we should try and provide this kind of higher level insight alongside a lower level operational semantics.
One downside of the declarative description is that while the time complexity is still somewhat clear, the space complexity, and other performance characteristics are less clear.
It’s much easier to see why a given algorithm might have better performance characteristics when it’s described in terms of low level imperative operational details. This is because the performance ultimately does depend on all of these details.
It’s kind of hard to “forget” understanding the higher level insights though. Once you’ve understood why an imperative algorithm works, you conflate the higher intuition with the low level details, and it becomes hard to not read these details grouped with the lens of your intuition.
It’s because of this, I suspect, that so many algorithm descriptions in textbooks don’t that good a job of connecting imperative details to deeper understanding of the algorithm.
While it is ultimately good to develop your own framing and and intuition when working through a new piece of knowledge, it’s also great to provide some directions and insight as an author.
We can definitely do a lot better :)
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Trigonometric Functions
# Trigonometric Functions
Edited By Team Careers360 | Updated on Aug 14, 2023 04:44 PM IST
Introduction:
One of the most important concepts in mathematics is trigonometry. The measuring of angles and sides of a triangle is the subject of trigonometry. Normally, trigonometry is used to solve problems with right-angled triangles. Its functions are also used to calculate the length of a circle's arc, which is a portion of a circle with a radius and a centre point.
When we break down the word trigonometry, we find that 'Tri' is a Greek word that means three,' 'Gon' means length,' and metry' means trigonometric formulas measurement.' Trigonometry, in its simplest form, is the study of triangles with angles and lengths on their sides. The sin 0, sin 30, sin 45 , sin 60, sin 90, sin 180 and cosine formula such as cosine of 0, cosine of 30, cosine of 45, cosine of 60, cosine of 90, cosine of 180 tangent and trigonometric formulas class 10 are the foundations of trigonometry.
Trigonometry for class 10 and trigonometric formulas class 11 consists of trigonometric functions trigonometric identities class 10 that can be used to simplify the solution of complicated problems.
List of topics according to NCERT and JEE Main/NEET syllabus:
1. Angle.
2. Positive and Negative Angles
3. Measurement of Angles
1. Sexagesimal System/Degree Measure (English System)
2. Centesimal System (French System)
4. Relationship.
5. Trigonometric Ratios.
6. Trigonometric (or Circular) Functions.
7. Domain and Range
8. Range of Modulus Functions
9. Trigonometric Identities
10. Sign of Trigonometric Ratios
1. Trigonometric Ratios of Some Standard Angles
2. Trigonometric Ratios of Some Special Angles
3. Trigonometric Ratios of Allied Angles
11. Trigonometric Periodic Functions.
12. Maximum and Minimum Values of Trigonometric Expressions
13. Trigonometric Ratios of Compound Angles
14. Transformation Formulae
1. Trigonometric Ratios of Multiple Angles
2. Trigonometric Ratios of Some Useful Angles
15. Hyperbolic Functions
1. Domain and Range of Hyperbolic Function
16. Formulae for the Sum and Difference
17. Formulae to Transform the Product into Sum or Difference
18. Formulae for Multiples of x
Important concepts and Laws:
• The Trigonometry Formula List
• Basic Trigonometric Formulas
• identities of trigonometry
• Reciprocal Identities
• Pythagorean Identities
• Trigonometric Ratios Table
• Periodic Identities
• Cofunction Identities (in Degrees)
• all formulas of trigonometry
• Trigonometry Formulas of Sum and Difference of Identities.
• Half Angle Identities
• Double Angle Identities
• Triple Angle Identities
• Product Identities
• Sum of Product Identities
• inverse trigonometric functions
• Inverse Trigonometry Formula
• inverse sine and inverse Cosine Laws
Importance of trigonometric Functions class 11:
Maths 11 th grade Trigonometry is more than just a subject; it has several applications. In the exam, there will be a pure trigonometric question. Many physics courses need trigonometry, and it can make solving problems much easier and simpler. There will be some questions in math that need mastery of two or three courses. Most of the time, trigonometric equations were employed to confound you in such problems. In a nutshell, to succeed in tests, be thorough with trigonometry. For jee advanced, one should never rely on weightage.
Direct trigonometry queries, on the other hand, are less likely. It will be widely used in physics to create equations, solve problems, and even derive numerous formulae (for example, it will be widely utilised in simple harmonic motion, waves, sound waves, Newton's law, motion in two dimensions, and so on).... As a result, each and every formula, as well as their applications based on trigonometric values, such as where and how they should be applied to solve problems, should be thoroughly memorised.
For jee, trigonometry is a crucial subject.
Although, in the JEE exam, there are just two to three straight questions from trigonometry.
However, it is still very important.
If you are accepted into a good institute, they will begin with log and trigonometry.
The importance of trigonometry is demonstrated in the following subjects. Furthermore, you will see that trigonometry is employed in nearly every subject, including algebra, calculus, triangle properties, geometry, and so on. As a result, a solid understanding of trigonometry is required for JEE preparation. You can gain experience by solving books like SL Loney, which have introductory trigonometry questions. All trigonometric formulas must be memorised.
NCERT Exemplar Solutions Subject wise link:
1. What are some tricks for remembering trigonometric formulas in class 11?
In upper levels, we have a lot of formulas that are tough to remember, thus here are a few ways to remember them:
1. Learn how to read mathematical symbols.
2. The structure of the formulas and how they are generated follows.
3. Rehearse the formulas on a regular basis.
4. Review the formulas with flashcards, then revise and test yourself.
2. What is the trigonometry basic formula?
In trigonometry, there are six fundamental trigonometric ratios. Trigonometric functions are another name for these ratios. Sine, cosine, secant, cosecant, tangent, and cotangent are the six essential trigonometric functions. The right-angled triangle is used to generate trigonometric functions and identities. We may use trigonometric formulas to get the sine, cosine, tangent, secant, cosecant, and cotangent values when we know the height and base side of the right triangle.
1. sinθ=OppositeSide/Hypotenuse
5. cosec θ=Hypotenuse/OppositeSide
3. What are the eleven trigonometric identities?
The following are the eleven trigonometric identities in trigonometry:
1. Fundamental formulas
2. Identity Reciprocity
3. Ratio trigonometric Table
4. Recurring Identities
5. Identities of Co-functions
6. Identities of Sum and Difference
7. Identities with a Half-Angle
8. Identities with Two Angles
9. Identities with Triple Angles
10 Identities of Products
11. Product Identities as a Whole
4. What is the integral of cosine x?
The integral of cos x is sin x + C. i.e., ∫ cos x dx = sin x + C. Here, C is the integration constant.
Important Solutions of Triangle Formulas
Apr 27, 2022 - 12:42 p.m. IST ---STATIC
Escribed Circle of Triangle: Definition & Meaning
Apr 27, 2022 - 12:42 p.m. IST ---STATIC
In-Circle and In-Centre: Formula, Properties and Examples
Apr 27, 2022 - 12:42 p.m. IST ---STATIC
How to Find Area of Triangle: Formulas and Examples
Apr 27, 2022 - 12:42 p.m. IST ---STATIC
Semiperimeter And Half Angle Formulae
Apr 27, 2022 - 12:42 p.m. IST ---STATIC
Projection Formula & Overview
Apr 27, 2022 - 12:42 p.m. IST ---STATIC
Summation of Series in Trigonometry
Apr 27, 2022 - 12:42 p.m. IST ---STATIC
Simultaneous Trigonometric Equations
Apr 27, 2022 - 12:42 p.m. IST ---STATIC
Get answers from students and experts
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# 2.5: Factoring the GCF
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The distributive property of multiplication over addition/subtraction can be reversed.
$$a(b\pm c)=ab\pm ac$$ (right side equals left side) implies $$ab\pm ac=a(b\pm c)$$ (left side equals right side).
Factoring is the art of taking a sum (addition of terms) or difference (subtraction of terms) into a product (multiplication of factors).
Example $$\PageIndex{1}$$
Factor $$15x+20y$$.
Solution
$\begin{array}{rcl lll} 15x+20y&=&3\cdot 5x+4\cdot 5y\\ &=&5(3x+4y) \end{array}$
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Get Full Access to Elementary Statistics - 12 Edition - Chapter 3.3 - Problem 19bsc
Get Full Access to Elementary Statistics - 12 Edition - Chapter 3.3 - Problem 19bsc
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# In Exercises, find the range, variance, and | Ch 3.3 - 19BSC
ISBN: 9780321836960 18
## Solution for problem 19BSC Chapter 3.3
Elementary Statistics | 12th Edition
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Problem 19BSC
In Exercises, find the range, variance, and standard deviation for the given samph data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-2 where we found measures of center. Here we find measures of variation.) Then answer the given questions.
Saints in Super Bowl
Listed below are the numbers on the jerseys of the starting lineup for the New Orleans Saints when they recently won their first Super Bowl football game. What do the measures of variation tell us about the team? Does it make sense to compute the measures of variation for these data?
Step-by-Step Solution:
Step 1 of 3
Problem 19BSC
Step1 of 3:
We have Listed below are the numbers on the jerseys of the starting lineup for the New Orleans Saints when they recently won their first Super Bowl football game.
x 9 23 25 88 12 19 74 77 76 73 78
Step2 of 3:
We need to find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-2 where we found measures of center. Here we find measures of variation.) Then answer the given questions.
Step3 of 3:
1).Range can be calculated by using the formula Range = maximum value - minimum value
The given data in Ascending order
9 12 19 23 25 73 74 76 77 78 88
Here,
Maximum value = 88
Minimum value = 9
Now,
Range = maximum value - minimum value
= 88 - 9
= 79
Therefore, Range = 79.
2).consider,
...
x (x -)2 9 50.3636 1710.947 12 50.3636 1471.766 19 50.3636 983.6754 23 50.3636 748.7666 25 50.3636 643.3122 73 50.3636 512.4066 74 50.3636 558.6794 76 50.3636 657.225 77 50.3636 709.4978 78 50.3636 763.7706
Step 2 of 3
Step 3 of 3
##### ISBN: 9780321836960
This textbook survival guide was created for the textbook: Elementary Statistics, edition: 12. This full solution covers the following key subjects: measures, variation, data, Saints, super. This expansive textbook survival guide covers 121 chapters, and 3629 solutions. The full step-by-step solution to problem: 19BSC from chapter: 3.3 was answered by , our top Statistics solution expert on 03/15/17, 10:30PM. The answer to “In Exercises, find the range, variance, and standard deviation for the given samph data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-2 where we found measures of center. Here we find measures of variation.) Then answer the given questions.Saints in Super Bowl Listed below are the numbers on the jerseys of the starting lineup for the New Orleans Saints when they recently won their first Super Bowl football game. What do the measures of variation tell us about the team? Does it make sense to compute the measures of variation for these data?” is broken down into a number of easy to follow steps, and 102 words. Elementary Statistics was written by and is associated to the ISBN: 9780321836960. Since the solution to 19BSC from 3.3 chapter was answered, more than 325 students have viewed the full step-by-step answer.
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# Which ordered pair is a solution of the equation y=x-4?
a. (2,6)
b. (6,2)
c. (-2,6)
d. (3,1)
A)(-1,-5)
B) (0,-3)
C)(2,6)***??
D)(3,1)
## if (2,6) is right then
6 = 2 -4
but it is not so
try a
-5 = -1 -4
-5 = -5 that seems to work
## To determine which ordered pair is a solution of the equation y = x - 4, you need to substitute the values of x and y from each option into the equation and check if it holds true. Let's go through each option:
a. (2,6)
Substituting x = 2 and y = 6 into the equation, we get 6 = 2 - 4, which simplifies to 6 = -2. Since this is not true, option a is not a solution.
b. (6,2)
Substituting x = 6 and y = 2 into the equation, we get 2 = 6 - 4, which simplifies to 2 = 2. Since this is true, option b is a solution.
c. (-2,6)
Substituting x = -2 and y = 6 into the equation, we get 6 = -2 - 4, which simplifies to 6 = -6. Since this is not true, option c is not a solution.
d. (3,1)
Substituting x = 3 and y = 1 into the equation, we get 1 = 3 - 4, which simplifies to 1 = -1. Since this is not true, option d is not a solution.
Therefore, the ordered pair that is a solution of the equation y = x - 4 is option b, (6,2).
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# A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond
## 1 Answer
Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: $$\;P(E_i/A)$$=$$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$$
Let $E_1$ be the event that the lost card is a diamond. Given that there are 13 diamonds in the deck, P($E_1$) = $\large\frac{13}{52} = \frac{1}{4}$
Let $E_2$ be the event that the lost card is not a diamond. P($E_2$) =1 - $\large\frac{13}{52} = \frac{39}{52} = \frac{3}{4}$
Let A be the event that the two cards drawn are found to be both diamonds.
We need to calculate P (2 cards are both diamond w the lost card being a diamond) next.
Given 13 diamond cards, if one is lost, then we have 12 remaining diamonds out of 51 total cards now.
The two cards that are both diamond can be drawn in $^{12}C_2 = \large\frac{12 \times 11}{1 \times 2} = \frac{132}{2}$ = 66 ways.
Likewise, the two diamonds can be drawn from the pack of 51 remaining cards in $^{51}C_2 = \large\frac{51 \times 50}{1 \times 2} = \frac{2550}{2}$ = 1275 ways.
Therefore, the P (2 cards drwan are diamond given one is lost) = P (A|$E_1$) = $\large\frac{66}{1275}$
Now, consider the event where the two cards drawn are both diamonds but the lost card is not a diamond.
The two cards that are both diamond can be drawn in $^{13}C_2 = \large\frac{13 \times 12}{1 \times 2} = \frac{156}{2}$ = 78ways.
Likewise, the two diamonds can be drawn from the pack of 51 remaining cards in $^{51}C_2 = \large\frac{51 \times 50}{1 \times 2} = \frac{2550}{2}$ = 1275 ways.
Therefore, the P (2 cards drwan are diamond given one card which is not a diamond is lost) = P (A|$E_1$) = $\large\frac{78}{1275}$
We need to calcuate the probability that the lost card is a diamond. P ($E_1$|A).
We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
P($E_1$|A) = $\large\frac{\frac{66}{1275}.\frac{1}{4}}{\frac{66}{1275}.\frac{1}{4} + \frac{78}{1275}.\frac{3}{4}} = \frac{66}{66+78 \times 3} = \frac{66}{300} = \frac{11}{50}$
answered Jun 19, 2013
can i get the answer at 10th class level (cbse)
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# The Smallest and Greatest Number
## What is Greatest Number and the Smallest Number in Maths?
The smallest number in maths is 1. There is no end to numbers in mathematics. In mathematics, 1 is the smallest number and the largest number has no end. Apart from this, there are many other numbers in mathematics such as composite numbers, prime numbers, etc.
One-Digit Number: Smallest and Greatest Number
The Smallest 1 digit number is 1 and the Greatest one-digit number is 9. Based on this, we can find whatever number we want to find from the smallest number to the greatest number.
Name of the Chapter: The Smallest and Greatest Number
## How to Find the Smallest and the Greatest Number of any Digit?
The smallest 2-digit number will become 10 and the largest number will become 99 To make or write the smallest number we write a zero next to one becomes the smallest 2-digit number so on. To write the smallest number, write two zeros next to one, and then it will become the smallest number of three digits. Similarly, if you want to write the smallest number of 4 digits, 5 digits, 6 digits, and 9 digits, after writing one, write zero so many times that the number of digits becomes.
Like the smallest 8-digit number will be 10000000 as it is written 7 times zero after one.
Similarly, you can write the smallest number of any digit to write that number after writing one so much zero should be made that it becomes that number of digits.
Similarly, the Greatest number of any digit can also be written as the Greatest two-digit number is 99 in this you will see that 9 is written 2 times. That is, you can know that to write the largest number as many times as 9 is written then it will become the largest number of that digit e.g. the largest number of 8 digits will be 99999999, it will be the largest number of 8-digit numbers. Now that I understand how to write the smallest number and the largest number.
Smallest and Greatest Numbers Chart
## Solved Questions
1. Which is the smallest number?
Ans: 1 is the smallest number.
1. Which is the smallest whole number?
Ans: 0 is the smallest whole number.
1. What is the largest two-digit natural number(a part of the number system, which includes all positive integers from 1 to infinity)?
Ans: 99 is the largest two-digit number.
1. What is the smallest 3-digit number?
Ans: 100 is the smallest 3-digit number.
1. What is the smallest even number(Such natural numbers which are exactly divisible by 2)?
Ans: 2 is the smallest even number.
### Summary
First of all, we know which is the smallest number of one digit and which is the largest number, which is known, then based on this, we can find the smallest and largest number of any digit. The smallest one-digit number is 1 and the largest one-digit number is 9. Based on this, we can find whatever number we want to find from the smallest number to the greatest number.
## Learning by Doing
Write True or False.
1. The infinity of numbers exists.
2. 99999 is the largest 4-digit number.
3. 9 is the largest single-digit number.
4. 999999999 is the largest 7-digit number.
Last updated date: 30th Sep 2023
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Total views: 78k
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## FAQs on The Smallest and Greatest Number
1. What is the smallest negative number?
The smallest negative number is - infinity.
2. What is the largest 6-digit number?
999999 is the largest 6-digit number.
What is the largest two-digit prime number?
97 is the largest two-digit prime number.
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# 1999 AHSME Problems/Problem 23
## Problem
The equiangular convex hexagon $ABCDEF$ has $AB = 1, BC = 4, CD = 2,$ and $DE = 4.$ The area of the hexagon is $\mathrm{(A) \ } \frac {15}2\sqrt{3} \qquad \mathrm{(B) \ }9\sqrt{3} \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }\frac{39}4\sqrt{3} \qquad \mathrm{(E) \ } \frac{43}4\sqrt{3}$
# Solution
## Solution 1
Equiangularity means that each internal angle must be exactly $120^\circ$. The information given by the problem statement looks as follows:
$[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW)); dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label("A",O,SW); label("B",O+E,SE); label("C",O+E+4*NE,E); label("D",O+E+4*NE+2*NW,NE); label("E",O-3*E+4*NE+2*NW,NW); [/asy]$
We can now place this incomplete polygon onto a triangular grid, finish it, compute its area in unit triangles, and multiply the result by the area of the unit triangle.
$[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW) -- (O-3*E+3*NE+2*NW) -- cycle, 0.8red+4bp); for (int i=-5; i<=1; ++i) { draw( (O+i*E-1.5*NE)--(O+i*E+6.5*NE), dashed ); } for (int i=-2; i<=5; ++i) { draw( (O+i*E-1.5*NW)--(O+i*E+6.5*NW), dashed ); } for (int i=-1; i<=6; ++i) { draw( (O-2.5*E+i*NW)--(O+5.5*E+i*NW), dashed ); } dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label("A",O,SW); label("B",O+E,SE); label("C",O+E+4*NE,E); label("D",O+E+4*NE+2*NW,NE); label("E",O-3*E+4*NE+2*NW,NW); label("F",(O-3*E+3*NE+2*NW),W); [/asy]$
We see that the figure contains $43$ unit triangles, and therefore its area is $\boxed{\frac{43\sqrt{3}}4}$.
## Solution 2
$[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle); draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label("X",(7,13),S); label("Y",(14,0),S); label("Z",(0,0),S); label("A",(6,10.392),W); label("B",(8,10.392),E); label("C",(12,3.464),E); label("D",(10,0),S); label("E",(2,0),S); label("F",(1,1.732),W); label("1",(7,12.124)--(8,10.392),NE); label("1",(6,10.392)--(8,10.392),S); label("4",(8,10.392)--(12,3.464),NE); label("2",(10,0)--(12,3.464),NW); label("2",(14,0)--(12,3.464),NE); label("1",(1,1.732)--(2,0),NE); label("1",(0,0)--(2,0),S); label("1",(0,0)--(1,1.732),NW); label("1",(6,10.392)--(7,12.124),NW); label("2",(10,0)--(14,0),S); [/asy]$
An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle $X Y$ and $Z$ and drawing $AB$ of length one will remove one equilateral triangle of side length $1$, and drawing $CD$ will take out another equilateral triangle of side length $2$.Labeling the other sides of the smaller equilateral triangles, we can find that $XY$, or the side length of the equilateral triangle is $7$. Now, because we know what the side length of the triangle is, what $DY$ is, and it is given that $DE$ is $4$, we can find the length of $EZ$, $7-4-2=1$. Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are $\frac{1^2\sqrt{3}}{4}\implies\frac{1\sqrt{3}}{4}$, and $\frac{2^2\sqrt{3}}{4}\implies\frac{4\sqrt{3}}{4}\implies\sqrt{3}$ and the area of the large equilateral triangle is $\frac{7^2\sqrt{3}}{4}\implies\frac{49\sqrt{3}}{4}$ so the area of the hexagon would be $\frac{49\sqrt{3}}{4}-\frac{\sqrt {3}}{4}-\frac{\sqrt{3}}{4}-\sqrt{3}\implies\boxed{\frac{43\sqrt{3}}{4}}$
## Solution 2
$[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle); draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label("X",(7,13),S); label("Y",(14,0),S); label("Z",(0,0),S); label("A",(6,10.392),W); label("B",(8,10.392),E); label("C",(12,3.464),E); label("D",(10,0),S); label("E",(2,0),S); label("F",(1,1.732),W); label("1",(7,12.124)--(8,10.392),NE); label("1",(6,10.392)--(8,10.392),S); label("4",(8,10.392)--(12,3.464),NE); label("2",(10,0)--(12,3.464),NW); label("2",(14,0)--(12,3.464),NE); label("1",(1,1.732)--(2,0),NE); label("1",(0,0)--(2,0),S); label("1",(0,0)--(1,1.732),NW); label("1",(6,10.392)--(7,12.124),NW); label("2",(10,0)--(14,0),S); [/asy]$
An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle $X Y$ and $Z$ and drawing $AB$ of length one will remove one equilateral triangle of side length $1$, and drawing $CD$ will take out another equilateral triangle of side length $2$.Labeling the other sides of the smaller equilateral triangles, we can find that $XY$, or the side length of the equilateral triangle is $7$. Now, because we know what the side length of the triangle is, what $DY$ is, and it is given that $DE$ is $4$, we can find the length of $EZ$, $7-4-2=1$. Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are $\frac{1^2\sqrt{3}}{4}\implies\frac{1\sqrt{3}}{4}$, and $\frac{2^2\sqrt{3}}{4}\implies\frac{4\sqrt{3}}{4}\implies\sqrt{3}$ and the area of the large equilateral triangle is $\frac{7^2\sqrt{3}}{4}\implies\frac{49\sqrt{3}}{4}$ so the area of the hexagon would be $\frac{49\sqrt{3}}{4}-\frac{\sqrt {3}}{4}-\frac{\sqrt{3}}{4}-\sqrt{3}\implies\boxed{\frac{43\sqrt{3}}{4}}$
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+0
# geometric progression
0
35
1
When the same constant is added to the numbers \$60,\$ \$100,\$ and \$110,\$ a three-term geometric sequence arises. What is the common ratio of the resulting sequence?
Sep 8, 2023
#1
+850
+3
When the same constant is added to the numbers \$60,\$ \$100,\$ and \$110,\$ a three-term geometric sequence arises. What is the common ratio of the resulting sequence?
In a geometric progression the ratio between succeeding numbers is the same.
Therefore 100 + x 110 + x
–––––– = ––––––
60 + x 100 + x
Cross multiplying (x + 100)(x + 100) = (x + 60)(x + 110)
x2 + 200x +10,000 = x2 + 170x + 6600
Subtract x2 from both sides 200x + 10,000 = 170x + 6600
Subtract 170x from both sides 30x + 10,000 = 6600
Subtract 10,000 from both sides 30x = –3400
x = –113.333 • • •
Plugging –113.333
into the first ratio –13.333
––––––– = 0.24999 = 0.25
–53.333
Plugging –113.333
into the second ratio –3.333
––––––– = 0.24998 = 0.25
–13.333
Who'd a thunk it would really work.
The common ratio, after
.
Sep 8, 2023
#1
+850
+3
When the same constant is added to the numbers \$60,\$ \$100,\$ and \$110,\$ a three-term geometric sequence arises. What is the common ratio of the resulting sequence?
In a geometric progression the ratio between succeeding numbers is the same.
Therefore 100 + x 110 + x
–––––– = ––––––
60 + x 100 + x
Cross multiplying (x + 100)(x + 100) = (x + 60)(x + 110)
x2 + 200x +10,000 = x2 + 170x + 6600
Subtract x2 from both sides 200x + 10,000 = 170x + 6600
Subtract 170x from both sides 30x + 10,000 = 6600
Subtract 10,000 from both sides 30x = –3400
x = –113.333 • • •
Plugging –113.333
into the first ratio –13.333
––––––– = 0.24999 = 0.25
–53.333
Plugging –113.333
into the second ratio –3.333
––––––– = 0.24998 = 0.25
–13.333
Who'd a thunk it would really work.
The common ratio, after
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# Additive and Multiplicative Rules for Probability
## Apply tree diagrams to calculations.
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Additive and Multiplicative Rules for Probability
### The Additive and Multiplicative Rules
#### Venn Diagrams
When the probabilities of certain events are known, we can use these probabilities to calculate the probabilities of their respective unions and intersections. We use two rules, the Additive Rule and the Multiplicative Rule, to find these probabilities. The examples that follow will illustrate how we can do this.
#### Illustrating the Additive Rule of Probability
Suppose we have a loaded (unfair) die, and we toss it several times and record the outcomes. We will define the following events:
\begin{align*}& A: {\text{observe an even number}}\end{align*}
\begin{align*}& B: {\text{observe a number less than } 3}\end{align*}
Let us suppose that we have \begin{align*}P(A)=0.4, \ P(B)=0.3,\end{align*} and \begin{align*}P(A \cap B)=0.1\end{align*}. We want to find \begin{align*}P(A \cup B)\end{align*}.
It is probably best to draw a Venn diagram to illustrate this situation. As you can see, the probability of events \begin{align*}A\end{align*} or \begin{align*}B\end{align*} occurring is the union of the individual probabilities of each event.
Therefore, adding the probabilities together, we get the following:
\begin{align*}P(A \cup B) = P(1)+P(2)+P(4)+P(6)\end{align*}
We have also previously determined the probabilities below:
\begin{align*}P(A)& =P(2)+P(4)+P(6)=0.4\\ P(B)& =P(1)+P(2)=0.3\\ P(A \cap B)& =P(2)=0.1\end{align*}
If we add the probabilities \begin{align*}P(A)\end{align*} and \begin{align*}P(B)\end{align*}, we get:
\begin{align*}P(A)+P(B)=P(2)+P(4)+P(6)+P(1)+P(2)\end{align*}
Note that \begin{align*}P(2)\end{align*} is included twice. We need to be sure not to double-count this probability. Also note that 2 is in the intersection of \begin{align*}A\end{align*} and \begin{align*}B\end{align*}. It is where the two sets overlap. This leads us to the following:
\begin{align*}P(A \cup B)& =P(1)+P(2)+P(4)+P(6)\\ P(A)& =P(2)+P(4)+P(6)\\ P(B)& =P(1)+P(2)\\ P(A \cap B)& =P(2)\\ P(A \cup B)& =P(A)+P(B)-P(A \cap B)\end{align*}
This is the Additive Rule of Probability, which is demonstrated below:
\begin{align*}P(A \cup B)=0.4+0.3-0.1=0.6\end{align*}
What we have shown is that the probability of the union of two events, \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, can be obtained by adding the individual probabilities, \begin{align*}P(A)\end{align*} and \begin{align*}P(B),\end{align*} and subtracting the probability of their intersection (or overlap), \begin{align*}P(A \cap B)\end{align*}. The Venn diagram above illustrates this union.
#### The Additive Rule of Probability
The probability of the union of two events can be obtained by adding the individual probabilities and subtracting the probability of their intersection: \begin{align*}P(A \cup B)=P(A)+P(B)-P(A \cap B)\end{align*}.
We can rephrase the definition as follows: The probability that either event \begin{align*}A\end{align*} or event \begin{align*}B\end{align*} occurs is equal to the probability that event \begin{align*}A\end{align*} occurs plus the probability that event \begin{align*}B\end{align*} occurs minus the probability that both occur.
#### Using the Additive Rule of Probability
1. Consider the experiment of randomly selecting a card from a deck of 52 playing cards. What is the probability that the card selected is either a spade or a face card?
Our event is defined as follows:
\begin{align*}& E: {\text{card selected is either a spade or a face card}}\end{align*}
There are 13 spades and 12 face cards, and of the 12 face cards, 3 are spades. Therefore, the number of cards that are either a spade or a face card or both is \begin{align*}13 + 9 = 22.\end{align*} That is, event \begin{align*}E\end{align*} occurs when 1 of 22 cards is selected, the 22 cards being the 13 spade cards and the 9 face cards that are not spade. To find \begin{align*}P(E)\end{align*}, we use the Additive Rule of Probability. First, define two events as follows:
\begin{align*}& C: {\text{card selected is a spade}}\end{align*}
\begin{align*}& D: {\text{card selected is a face card}}\end{align*}
Note that \begin{align*}P(E)=P(C \cup D) = P(C) + P(D)-P(C \cap D).\end{align*} Remember, with event \begin{align*}C\end{align*}, 1 of 13 cards that are spades can be selected, and with event \begin{align*}D\end{align*}, 1 of 12 face cards can be selected. Event \begin{align*}C \cap D\end{align*} occurs when 1 of the 3 face card spades is selected. These cards are the king, jack, and queen of spades. Using the Additive Rule of Probability formula:
\begin{align*}P(A \cup B) & = P(A)+P(B)-P(A \cap B)\\ & =\frac{13}{52}+\frac{12}{52}-\frac{3}{52}\\ & =0.250+0.231-0.058\\ & =0.423\\ & =42.3\%\end{align*}
Recall that we are subtracting 0.058 because we do not want to double-count the cards that are at the same time spades and face cards.
2. If you know that 84.2% of the people arrested in the mid 1990’s were males, 18.3% of those arrested were under the age of 18, and 14.1% were males under the age of 18, what is the probability that a person selected at random from all those arrested is either male or under the age of 18?
First, define the events:
\begin{align*}& A: {\text{person selected is male}}\end{align*}
\begin{align*}& B: {\text{person selected in under 18}}\end{align*}
Also, keep in mind that the following probabilities have been given to us:
\begin{align*}P(A)=0.842 && P(B)=0.183 && P(A \cap B)=0.141\end{align*}
Therefore, the probability of the person selected being male or under 18 is \begin{align*}P(A \cup B)\end{align*} and is calculated as follows:
\begin{align*}P(A\cup B)& =P(A)+P(B)-P(A \cap B)\\ & = 0.842+0.183-0.141\\ & = 0.884\\ & = 88.4\%\end{align*}
This means that 88.4% of the people arrested in the mid 1990’s were either male or under 18.
#### Mutually Exclusive Events
If \begin{align*}A \cap B\end{align*} is empty \begin{align*}(A \cap B = \varnothing)\end{align*}, or, in other words, if there is not overlap between the two sets, we say that \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are mutually exclusive.
The figure below is a Venn diagram of mutually exclusive events. For example, set \begin{align*}A\end{align*} might represent all the outcomes of drawing a card, and set \begin{align*}B\end{align*} might represent all the outcomes of tossing three coins. These two sets have no elements in common.
If the events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are mutually exclusive, then the probability of the union of \begin{align*}A\end{align*} and \begin{align*}B\end{align*} is the sum of the probabilities of \begin{align*}A\end{align*} and \begin{align*}B\end{align*}: \begin{align*}P(A \cup B)=P(A)+P(B)\end{align*}.
Note that since the two events are mutually exclusive, there is no double-counting.
#### Calculating Probability Using the Additive Rule
If two coins are tossed, what is the probability of observing at least one head?
First, define the events as follows:
\begin{align*}& A: {\text{observe only one head}}\end{align*}
\begin{align*}& B: {\text{observe two heads}}\end{align*}
Now the probability of observing at least one head can be calculated as shown:
\begin{align*}P(A \cup B)=P(A)+P(B)=0.5+0.25=0.75=75\%\end{align*}
#### The Multiplicative Rule of Probability
Recall from the previous section that conditional probability is used to compute the probability of an event, given that another event has already occurred:
\begin{align*}P(A|B)=\frac{P(A\cap B)}{P(B)}\end{align*}
This can be rewritten as \begin{align*}P(A \cap B) = P(A|B) \bullet P(B)\end{align*} and is known as the Multiplicative Rule of Probability.
The Multiplicative Rule of Probability says that the probability that both \begin{align*}A\end{align*} and \begin{align*}B\end{align*} occur equals the probability that \begin{align*}B\end{align*} occurs times the conditional probability that \begin{align*}A\end{align*} occurs, given that \begin{align*}B\end{align*} has occurred.
#### Using the Multiplicative Rule of Probability
1. In a certain city in the USA some time ago, 30.7% of all employed female workers were white-collar workers. If 10.3% of all workers employed at the city government were female, what is the probability that a randomly selected employed worker would have been a female white-collar worker?
We first define the following events:
\begin{align*}& F: {\text{randomly selected worker who is female}}\end{align*}
\begin{align*}& W: {\text{randomly selected white-collar worker}}\end{align*}
We are trying to find the probability of randomly selecting a female worker who is also a white-collar worker. This can be expressed as \begin{align*}P(F \cap W)\end{align*}.
According to the given data, we have:
\begin{align*}P(F)& =10.3\%=0.103\\ P(W|F)& =30.7\%=0.307\end{align*}
Now, using the Multiplicative Rule of Probability, we get:
\begin{align*}P(F \cap W) = P(F)P(W|F)=(0.103)(0.307)=0.0316=3.16\%\end{align*}
Thus, 3.16% of all employed workers were white-collar female workers.
Suppose a coin was tossed twice, and the observed face was recorded on each toss. The following events are defined:
\begin{align*}& A: {\text{first toss is a head}}\end{align*}
\begin{align*}& B: {\text{second toss is a head}}\end{align*}
2. Does knowing that event \begin{align*}A\end{align*} has occurred affect the probability of the occurrence of \begin{align*}B\end{align*}?
The sample space of this experiment is \begin{align*}S=\left \{HH, HT, TH, TT\right \}\end{align*}, and each of these simple events has a probability of 0.25. So far we know the following information:
\begin{align*}P(A) & = P(HT) + P(HH) = \frac{1}{4} + \frac{1}{4} = 0.5\\ P(B) & = P(TH) + P(HH) = \frac{1}{4} + \frac{1}{4} = 0.5\\ A \cap B & = \left \{\text{HH}\right \}\\ P(A \cap B) & = 0.25\end{align*}
Now, what is the conditional probability? It is as follows:
\begin{align*}P(B|A)& =\frac{P(A \cap B)}{P(A)}\\ & =\frac{\frac{1}{4}}{\frac{1}{2}}\\ & =\frac{1}{2}\end{align*}
What does this tell us? It tells us that \begin{align*}P(B)=\frac{1}{2}\end{align*} and also that \begin{align*}P(B|A)=\frac{1}{2}\end{align*}. This means knowing that the first toss resulted in heads does not affect the probability of the second toss being heads. In other words, \begin{align*}P(B|A)=P(B)\end{align*}.
When this occurs, we say that events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent events.
Independence
If event \begin{align*}B\end{align*} is independent of event \begin{align*}A\end{align*}, then the occurrence of event \begin{align*}A\end{align*} does not affect the probability of the occurrence of event \begin{align*}B\end{align*}. Therefore, we can write \begin{align*}P(B)=P(B|A)\end{align*}.
Recall that \begin{align*}P(B|A)=\frac{P(B \cap A)}{P(A)}\end{align*}. Therefore, if \begin{align*}B\end{align*} and \begin{align*}A\end{align*} are independent, the following must be true:
\begin{align*}P(B|A)=\frac{P(A \cap B)}{P(A)}=P(B)\end{align*}
\begin{align*}P(A \cap B)=P(A) \bullet P(B)\end{align*}
That is, if two events are independent, \begin{align*}P(A \cap B)=P(A) \bullet P(B)\end{align*}.
#### Determining Independence
The table below gives the number of physicists (in thousands) in the US cross-classified by specialty \begin{align*}(P1, P2, P3, P4)\end{align*} and base of practice \begin{align*}(B1, B2, B3)\end{align*}. (Remark: The numbers are absolutely hypothetical and do not reflect the actual numbers in the three bases.) Suppose a physicist is selected at random. Is the event that the physicist selected is based in academia independent of the event that the physicist selected is a nuclear physicist? In other words, is event \begin{align*}B1\end{align*} independent of event \begin{align*}P3\end{align*}?
Academia \begin{align*}(B1)\end{align*} Industry \begin{align*}(B2)\end{align*} Government \begin{align*}(B3)\end{align*} Total
General Physics \begin{align*}(P1)\end{align*} 10.3 72.3 11.2 93.8
Semiconductors \begin{align*}(P2)\end{align*} 11.4 0.82 5.2 17.42
Nuclear Physics \begin{align*}(P3)\end{align*} 1.25 0.32 34.3 35.87
Astrophysics \begin{align*}(P4)\end{align*} 0.42 31.1 35.2 66.72
Total 23.37 104.54 85.9 213.81
Figure: A table showing the number of physicists in each specialty (thousands). These data are hypothetical.
We need to calculate \begin{align*}P(B1|P3)\end{align*} and \begin{align*}P(B1)\end{align*}. If these two probabilities are equal, then the two events \begin{align*}B1\end{align*} and \begin{align*}P3\end{align*} are indeed independent. From the table, we find the following:
\begin{align*}P(B1)=\frac{23.37}{213.81}=0.109\end{align*}
and
\begin{align*}P(B1|P3)=\frac{P(B1 \cap P3)}{P(P3)}=\frac{1.25}{35.87}=0.035\end{align*}
Thus, \begin{align*}P(B1|P3)\neq P(B1)\end{align*}, and so events \begin{align*}B1\end{align*} and \begin{align*}P3\end{align*} are not independent.
Caution! If two outcomes of one event are mutually exclusive (they have no overlap), they are not independent. If you know that outcomes \begin{align*}A\end{align*} and \begin{align*}B\end{align*} do not overlap, then knowing that \begin{align*}B\end{align*} has occurred gives you information about \begin{align*}A\end{align*} (specifically that \begin{align*}A\end{align*} has not occurred, since there is no overlap between the two events). Therefore, \begin{align*}P(A|B)\neq P(A)\end{align*}.
### Example
A college class has 42 students of which 17 are male and 25 are female. Suppose the teacher selects two students at random from the class. Assume that the first student who is selected is not returned to the class population.
#### Example1
What is the probability that the first student selected is female and the second is male?
Here we can define two events:
\begin{align*}& F1: {\text{first student selected is female}}\end{align*}
\begin{align*}& M2: {\text{second student selected is male}}\end{align*}
In this problem, we have a conditional probability situation. We want to determine the probability that the first student selected is female and the second student selected is male. To do so, we apply the Multiplicative Rule:
\begin{align*}P(F1 \cap M2) = P(F1)P(M2|F1)\end{align*}
Before we use this formula, we need to calculate the probability of randomly selecting a female student from the population. This can be done as follows:
\begin{align*}P(F1)=\frac{25}{42}=0.595\end{align*}
Now, given that the first student selected is not returned back to the population, the remaining number of students is 41, of which 24 are female and 17 are male.
Thus, the conditional probability that a male student is selected, given that the first student selected was a female, can be calculated as shown below:
\begin{align*}P(M2|F1)=P(M2)=\frac{17}{41}=0.415\end{align*}
Substituting these values into our equation, we get:
\begin{align*}P(F1 \cap M2)&=P(F1)P(M2|F1)\\ &=(0.595)(0.415)\\ &=0.247\\ &=24.7\%\end{align*}
We conclude that there is a probability of 24.7% that the first student selected is female and the second student selected is male.
### Review
For 1-4, you toss a coin and roll a die. Find each of the following probabilities:
1. P(a head and a 4)
2. P(a head and an odd number)
3. P(a tail and a number larger than 1)
4. P(a tail and a number less than 3)
5. Two fair dice are tossed, and the following events are identified: \begin{align*}& A: {\text{sum of the numbers is odd}}\\ & B: {\text{sum of the numbers is 9, 11, or 12}}\end{align*}
1. Are events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} independent? Why or why not?
2. Are events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} mutually exclusive? Why or why not?
6. The probability that a certain brand of television fails when first used is 0.1. If it does not fail immediately, the probability that it will work properly for 1 year is 0.99. What is the probability that a new television of the same brand will last 1 year?
7. A coin is tossed 3 times. Determine the probability of getting the following results:
8. Given that a couple decides to have 4 children, none of them adopted. What is the probability their children will be born in the order boy, girl, boy, girl?
9. Two archers, John and Mary, shoot at a target at the same time. John hits the bulls-eye 70% of the time and Mary hits the bulls-eye 90% of the time. Find the probability that
1. They both hit the bulls-eye
2. They both miss the bulls-eye
3. John hits the bulls-eye but Mary misses
4. Mary hits the bulls-eye but John misses
10. A box contains 8 red and 4 blue balls. Two balls are randomly selected from the box without replacement. Determine each of the following probabilities:
1. Both are red
2. The first is blue and the second is red
3. A blue and a red are obtained
11. A hat contains tickets with numbers 1, 2, 3, ……20 printed on them. If three tickets are drawn from the hat without replacement, determine the probability that none of them are primes.
12. Suppose you have a spinner with 4 sections: Black, black, yellow and red. You spin the spinner twice;
1. What is the probability that black appears on both spins?
2. What is the probability that red appears on both spins?
3. What is the probability that different colors appear on both spins?
4. What is the probability that black appears on either spin?
13. Bag A contains 4 red and 3 blue tickets. Bag B contains 3 red and 1 blue ticket. A bag is randomly selected by tossing a coin and one ticket is removed from it. Using a tree diagram, determine the probability that the ticket chosen is blue.
14. Matthew and Chris go out for dinner. They roll a die and if the number of dots comes up even Matthew will pay and if the number of dots comes up odd Chris will pay. They roll the die twice, once for the decision about who pays for dinner and the second roll for the decision about who leaves the tip. A possible outcome lists who pays for dinner and then who leaves the tip. For example a possible outcome could be Chris, Chris.
1. List all the possible simple events in this sample space.
2. Are these events equally likely?
3. What is the probability that Matthew will have to pay for lunch and leave the tip?
15. When a fair die is tossed each of the six sides is equally like to land face up. Suppose you toss two die, one red and the other blue. Explain if the following pairs of events are disjoint.
1. A = red die is 4 and B = blue die is 3
2. A = red die and blue die sum to 5 and B = blue die is 1
3. A = red die and blue die sum to 5 and B = red die is 5
16. Amy is taking a statistics class and a biology class. Suppose her probabilities of getting A’s are: P(grade of A in statistics) = .65 P( grade of A in biology) = .70 P(grade of A in statistics and a grade of A in biology) = .50
1. Are the events “a grade of A in statistics” and a grade of A in biology independent? Explain.
2. Find the probability that Amy will get at least one A between her statistics and biology classes.
To view the Review answers, open this PDF file and look for section 3.5.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Independent Events Two events are independent if the occurrence of one event does not impact the probability of the other event.
multiplicative rule of probability The multiplicative rule of probability then states that P(AB) = P(B) x P(A/B)
Mutually Exclusive Events Mutually exclusive events have no common outcomes.
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# Eureka Math Algebra 2 Module 3 Lesson 6 Answer Key
## Engage NY Eureka Math Algebra 2 Module 3 Lesson 6 Answer Key
### Eureka Math Algebra 2 Module 3 Lesson 6 Exercise Answer Key
Exercises 1 – 3:
Exercise 1.
Assume that there is initially 1 cm of water in the tank, and the height of the water doubles every 10 seconds. Write an equation that could be used to calculate the height H(t) of the water in the tank at any time t.
The height of the water at time t seconds con be modeled by H(t) = 2t/10
Exercise 2.
How would the equation in Exercise 1 change if…
a. the initial depth of water in the tank was 2 cm?
H(t) = 2 . 2t/10
b. the initial depth of water in the tank was cm?
H(t) = $$\frac{1}{2}$$ . 2t/10
c. the initial depth of water in the tank was 10 cm?
H(t) = 10 . 2t/10
d. the initial depth of water in the tank was A cm, for some positive real number A?
H(t) = A . 2t/10
Exercise 3.
How would the equation in Exercise 2, part (d), change if…
a. the height tripled every ten seconds?
H(t) = A . 3t/10
b. the height doubled every five seconds?
H(t) = A . 2t/5
c. the height quadrupled every second?
H(t) = A . 4t
d. the height halved every ten seconds?
H(t) = A . (0.5t/10)
Example 1.
Consider two identical water tanks, each of which begins with a height of water 1 cm and fills with water at a different rate. Which equations can be used to calculate the height of water in each tank at time t? Use H1 for tank 1 and H2 for tank 2.
a. If both tanks start filling at the same time, which one fills first?
Tank 2 fills first because the level is rising more quickly.
b. We want to know the average rate of change of the height of the water in these tanks over an interval that starts at a fixed time T as they are filling up. What is the formula for the average rate of change of a function f on an interval [a, b]?
$$\frac{f(b)-f(a)}{b-a}$$
c. What is the formula for the average rate of change of the function H1 on an interval [a, b]?
$$\frac{\boldsymbol{H}_{1}(\boldsymbol{b})-\boldsymbol{H}_{1}(\boldsymbol{a})}{\boldsymbol{b}-\boldsymbol{a}}$$
d. Let’s calculate the average rate of change of the function H1 on the interval [T, T + 0. 1], which is an interval one-tenth of a second long starting at an unknown time T.
Exercises 4 – 8:
Exercise 4.
For the second tank, calculate the average change in the height, H2, from time T seconds to T + 0. 1 second. Express the answer as a number times the value of the original function at time T. Explain the meaning of these findings.
On average, over the time interval [T, T + 0. 1], the water in tank 2 rises at a rate of approximately 1.16123H2(T) centimeters per second.
Exercise 5.
For each tank, calculate the average change in height from time T seconds to T + 0.001 second. Express the answer as a number times the value of the original function at time T. Explain the meaning of these findings.
Tank 1:
On overage, over the time interval [T, T + 0.01], the water in tank 1 rises at a rate of approximately 0.693387H1(T) centimeter per second.
Tank 2:
Over the time interval [T, T + 0.001], the water in tank 2 rises at an average rate of approximately 1. 09922H2(T) centimeters per second.
Exercise 6.
In Exercise 5, the average rate of change of the height of the water in tank 1 on the interval [T, T + 0.001] can be described by the expression c1 2T, and the average rate of change of the height of the water in tank 2 on the interval [T, T + 0. 001] can be described by the expression c2 3T• What are approximate values of c1 and c2?
c1 ≈ 0.69339 and c2 ≈ 1. 09922.
Exercise 7.
As an experiment, let’s look for a value of b so that if the height of the water can be described by H(t) = bt, then the expression for the average rate of change on the interval [T, T + 0. 001] is 1 . H(T).
a. Write out the expression for the average rate of change of H(t) = bt on the interval [T, T + 0.001].
$$\frac{\boldsymbol{H}_{b}(\boldsymbol{T}+\mathbf{0 . 0 0 1})-\boldsymbol{H}_{b}(\boldsymbol{T})}{\mathbf{0 . 0 0 1}}$$
b. Set your expression in part (a) equal to 1 H(T), and reduce to an expression involving a single b.
c. Now we want to find the value of b that satisfies the equation you found in part (b), but we do not have a way to explicitly solve this equation. Look back at Exercise 6; which two consecutive integers have b between them?
We are looking for the base of the exponent that produces a rate of change on a small interval near t that is 1. H(t). When that base is 2, the value of the rate is roughly 0.69H (t). When the base is 3, the value of the rate is roughly 1.1H. Since 0.69 < 1 < 1.1, the base we are looking for is somewhere between 2 and 3.
d. Use your calculator and a guess-and-check method to find an approximate value of b to 2 decimal places.
Students may choose to use a table such as a table shown below. Make sure that students are maintaining enough decimal places of b0.001 to determine which value is closest to 0.001.
Then b ≈ 2.72.
Exercise 8.
Verify that for the value of b found in Exercise 7, $$\frac{H_{b}(T+0.001)-H_{b}(T)}{0.001}$$ ≈ Hb(T), where Hb(T) = bT
When the height of the water increases by a factor of 2.72 units per second, the height at any time is equal to the rate of change of height at that time.
### Eureka Math Algebra 2 Module 3 Lesson 6 Problem Set Answer Key
Question 1.
The product 4 . 3 . 2 . 1 is called 4 factorial and is denoted by 4!. Then 10! = 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1, and for any positive integer n, n! = n(n – 1)(n – 2) ……… 3 . 2 . 1.
a. Complete the following table of factorial values:
b. Evaluate the sum $$1+\frac{1}{1 !}$$.
2
c. Evaluate the sum $$1+\frac{1}{1 !}+\frac{1}{2 !}$$
2.5
d. Use a calculator to approximate the sum $$1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}$$ to 7 decimal places. Do not round the fractions before evaluating the sum.
$$\frac{8}{3}$$ ≈ 2.6666667
e. Use a calculator to approximate the sum $$1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}$$ to 7 decimal places. Do not round the fractions before evaluating the sum.
$$\frac{65}{24}$$ ≈ 2.7083333
f. Use a calculator to approximate sums of the form $$1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}$$ to 7 decimal places for k = 5, 6, 7,8,9, 10. Do not round the fractions before evaluating the sums with a calculator.
If k = 5, the sum is $$\frac{163}{60}$$ ≈ 2.1766667.
If k = 6, the sum is $$\frac{1957}{720}$$ ≈ 2.7180556.
1f k = 7,the sum is $$\frac{685}{252}$$ ≈ 2. 7182540.
If k = 8, the sum is $$\frac{109 601}{40320}$$ ≈ 2.7182788.
If k = 9, the sum is $$\frac{98 461}{36 288}$$ ≈ 2.7182815.
If k = 10, the sum is $$\frac{9 864 101}{3 628 800}$$ ≈ 2.7182818.
g. Make a conjecture about the sums $$1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}$$ for positive integers k as k increases in size.
It seems that as k gets larger, the sums $$1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}$$ get closer to e.
h. Would calculating terms of this sequence ever yield an exact value of e? Why or why not?
No. The number e is irrational, so it cannot be written as a quotient of integers. Any finite sum $$1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}$$ can be expressed as a single rational number with denominator k!, so the sums are all rational numbers. However, the more terms that are calculated, the closer to e the sum becomes, so these sums provide better and better rational number approximations of e.
Question 2.
Consider the sequence given by an = (1 + $$\frac{1}{n}$$)n where n ≥ 1 is an integer.
a. Use your calculator to approximate the first 5 terms of this sequence to 7 decimal places.
a1 = (1 + $$\frac{1}{1}$$)1 = 2
a2 = (1 + $$\frac{1}{2}$$)2 = 2.25
a3 = (1 + $$\frac{1}{3}$$)3 ≈ 2.3703704
a4 = (1 + $$\frac{1}{4}$$)4 ≈ 2.4414063
a5 = (1 + $$\frac{1}{5}$$)5 = 2.4883200
b. Does it appear that this sequence settles near a particular value?
No, the numbers get bigger, but we cannot tell if it keeps getting bigger or settles on or near a particular value.
c. Use a calculator to approximate the following terms of this sequence to 7 decimal places.
i. a100 = 2.7081383
ii. a1000 = 2.7169239
iii. a10,000 = 2.7181459
iv. a100,000 = 2.7182682
v. a1,000,000 = 2.7182805
vi. a100,000 = 2.7182816
vii. a100,000,000 = 2.7182818
d. Does it appear that this sequence settles near a particular value?
Yes, it appears that as n gets really large (at least 100,000,000), the terms an of the sequence settle near the value of e.
e. Compare the results of this exercise with the results of Problem 1. What do you observe?
It took about 10 terms of the sum in Problem 1 to see that the sum settled at the value e, but it takes 100,000,000 terms of the sequence in this problem to see that the sum settles at the value e.
Question 3.
If x = 5a4 and a = 2e3, express x in terms of e, and approximate to the nearest whole number.
If x = 5a4 and a = 2e3, then x = 5(2e3)4. Rewriting the right side in an equivalent form gives x = 80e12 ≈ 13020383.
Question 4.
If a = 2b3 and b = –$$\frac{1}{2}$$e-2, express a in terms of e, and approximate to four decimal places.
If a = 2b3 and b = –$$\frac{1}{2}$$e-2, then a = 2 (-$$\frac{1}{2}$$ e-2)3. Rewriting the right side in an equivalent form gives a = –$$\frac{1}{4}$$e-6 ≈ -0.0006.
Question 5.
If x = 3e4 and e = $$\frac{s}{2 x^{3}}$$ show that s = 54e13, and approximate sto the nearest whole number.
Rewrite the equation e = $$\frac{s}{2 x^{3}}$$ to isolate the variable s.
e = $$\frac{s}{2 x^{3}}$$
2x3e = s
By the substitution property, if s = 2x3e and x = 3e4, then s = 2(3e4)3 e. Rewriting the right side in an equivalent form gives s = 2 . 27e12 . e = 54e13 ≈ 23890323.
Question 6.
The following graph shows the number of barrels of oil produced by the Glenn Pool well in Oklahoma from 1910 to 1916.
Source: Cutler, Willard W., Jr. Estimation of Underground Oil Reserves by Oil-Well Production Curves, U.S. Department of the Interior, 1924.
a. Estimate the average rate of change of the amount of oil produced by the well on the interval [1910, 1916], and explain what that number represents.
Student responses will vary based on how they read the points on the graph. Over the interval [1910, 1916], the average rate of change is roughly
$$\frac{300-3200}{1916-1910}=-\frac{2900}{6}$$ ≈ -483.33.
This says that the production of the well decreased by an average of about 483 barrels of oil each year between 1910 and 1916.
b. Estimate the average rate of change of the amount of oil produced by the well on the interval [1910, 1913], and explain what that number represents.
Student responses will vary based on how they read the points on the graph. Over the interval [1910, 19131], the average rate of change is roughly
$$\frac{800-3200}{1913-1910}=-\frac{2400}{3}$$ = -800.
This says that the production of the well decreased by an average of about 800 barrels of oil per year between 1910 and 1913.
c. Estimate the average rate of change of the amount of oil produced by the well on the interval [1913, 1916], and explain what that number represents.
Student responses will vary based on how they read the points on the graph. Over the interval [1913, 1916], the average rate of change is roughly
$$\frac{300-800}{1916-1913}=-\frac{500}{3}$$ ≈ -166.67
This says that the production of the well decreased by on an average of about 166.67 barrels of oil per year between 1913 and 1916.
d. Compare your results for the rates of change in oil production in the first half and the second half of the time period in question in parts (b) and (c). What do those numbers say about the production of oil from the well?
The production dropped much more rapidly in the first three years than it did in the second three years. Looking at the graph, it looks like the oil in the well might be running out, so less and less can be extracted each year.
e. Notice that the average rate of change of the amount of oil produced by the well on any interval starting and ending in two consecutive years is always negative. Explain what that means in the context of oil production.
Because the average rate of change of oil production over a one-year period is always negative, the well is producing less oil each year than it did the year before.
Question 7.
The following table lists the number of hybrid electric vehicles (HEVs) sold in the United States between 1999 and 2013.
Source: U.S. Department of Energy, Alternative Fuels and Advanced Vehicle Data Center, 2013.
a. During which one-year interval is the average rate of change of the number of HEVs sold the largest? Explain how you know.
The average rate of change of the number of HEVs sold is largest during [2011,2012] because of the number of HEVssold increases by the largest amount between those two years.
b. Calculate the average rate of change of the number of HEVs sold on the interval [2003,2004], and explain what that number represents.
On the interval [2003,2004], the average rate of change in sales of HEVs is $$\frac{84199-47600}{2004-2003}$$ which is 36,599. This means that during this one-year period, HEVs were selling at a rate of 36, 599 vehicles per year.
c. Calculate the average rate of change of the number of HEVs sold on the interval [2003, 2008], and explain what that number represents.
On the interval [2003, 2008], the average rate of change in sales of HEVs is $$\frac{312,386-47,600}{2008-2003}$$, which is 52,957.2. This means that during this five-year period, HEVs were selling at an average rate of 52,957 vehicles per year.
d. What does it mean if the average rate of change of the number of HEVs sold is negative?
If the average rate of change of the vehicles sold is negative, then the sales are declining. This means that fewer cards were sold than in the previous year.
Extension:
Question 8.
The formula for the area of a circle of radius r can be expressed as a function A(r) = πr2.
a. Find the average rate of change of the area of a circle on the interval [4, 5].
$$\frac{A(5)-A(4)}{5-4}=\frac{25 \pi-16 \pi}{1}$$ = 9π
b. Find the average rate of change of the area of a circle on the interval [4, 4.1].
$$\frac{A(4.1)-A(4)}{4.1-4}=\frac{16.81 \pi-16 \pi}{0.1}$$ = 8.1π
c. Find the average rate of change of the area of a circle on the interval [4, 4.01].
$$\frac{A(4.01)-A(4)}{4.01-4}=\frac{16.0801 \pi-16 \pi}{0.01}$$ = 8.01π
d. Find the average rate of change of the area of a circle on the interval [4, 4.001].
$$\frac{A(4.001)-A(4)}{4.001-4}=\frac{16.008001 \pi-16 \pi}{0.001}$$ = 8.001π
e. What is happening to the average rate of change of the area of the circle as the interval gets smaller and smaller?
The average rate of change of the area of the circle appears to be getting close to 8π.
f. Find the average rate of change of the area of a circle on the interval [4, 4 + h] for some small positive number h.
$$\frac{A(4+h)-A(4)}{(4+h)-4}$$ = $$\frac{(4+h)^{2} \pi-16 \pi}{h}$$
= $$\frac{\left(16+8 h+h^{2}\right) \pi-16 \pi}{h}$$
= $$\frac{1}{h}$$ (8h + h2
= $$\frac{1}{h}$$ . h(8 + h)π
= (8 + h)π
g. What happens to the average rate of change of the area of the circle on the interval [4, 4 + h] as h → 0? Does this agree with your answer to part (d)? Should it agree with your answer to part (e)?
As h → 0, 8 + h → 8, so as h gets smaller, the average rate of change approaches 8. This agrees with my response to part (e), and it should because as h → 0, the Interval [4, 4 + h] gets smaller.
h. Find the average rate of change of the area of a circle on the interval [r0, r0 + h] for some positive number r0 and some small positive number h.
i. What happens to the average rate of change of the area of the circle on the interval [r0, r0 + h] as h → 0? Do you recognize the resulting formula?
As h → 0, the expression for the average rate of change becomes 2πr0, which is the circumference of the circle with radius r0.
Question 9.
The formula for the volume of a sphere of radius r can be expressed as a function V(r) = $$\frac{4}{3}$$πr3. As you work through these questions, you will see the pattern develop more clearly if you leave your answers in the form of a coefficient times π. Approximate the coefficient to five decimal places.
a. Find the average rate of change of the volume of a sphere on the interval [2, 3].
$$\frac{V(3)-V(2)}{3-2}=\frac{\frac{4}{3} \cdot 27 \pi-\frac{4}{3} \cdot 8 \pi}{1}$$ = $$\frac{4}{3}$$ . 19π ≈ 25.33333π
b. Find the average rate of change of the volume of a sphere on the interval [2, 2. 1].
$$\frac{V(2.1)-V(2)}{2.1-2}=\frac{\frac{4}{3} \pi\left(2.1^{3}-8\right)}{0.1}$$ ≈ 16.81333π
c. Find the average rate of change of the volume of a sphere on the interval [2, 2. 01].
$$\frac{V(2.01)-V(2)}{2.01-2}=\frac{\frac{4}{3} \pi\left(2.01^{3}-8\right)}{0.01}$$ ≈ 16.08013π
d. Find the average rate of change of the volume of a sphere on the interval [2, 2.001].
$$\frac{V(2.001)-V(2)}{2.001-2}=\frac{\frac{4}{3} \pi\left(2.001^{3}-8\right)}{0.001}$$ ≈ 16.00800π
e. What is happening to the average rate of change of the volume of a sphere as the interval gets smaller and smaller?
The average rate of change of the volume of the sphere appears to be getting close to 16π.
f. Find the average rate of change of the volume of a sphere on the interval [2, 2 + h] for some small positive number h.
g. What happens to the average rate of change of the volume of a sphere on the interval [2, 2 + h] as h → 0? Does this agree with your answer to part (e)? Should it agree with your answer to part (e)?
As h → 0, the value of the polynomial 12 + 6h + h2 approaches 12. Then the average rate of change approaches $$\frac{4 \pi}{3}$$ . 12 = 16. This agrees with my response to part (e), and it should because as h → 0, the interval [2, 2 + h] gets smaller.
h. Find the average rate of change of the volume of a sphere on the interval [r0, r0 + h] for some positive number r and some small positive number h.
i. What happens to the average rate of change of the volume of a sphere on the interval [r0, r0 + h] as h → 0? Do you recognize the resulting formula?
As h → 0, the expression for the average rate of change becomes 4πr02, which is the surface area of the sphere with radius r0.
### Eureka Math Algebra 2 Module 3 Lesson 6 Exit Ticket Answer Key
Question 1.
Suppose that water is entering a cylindrical water tank so that the initial height of the water is 3 cm and the height of the water doubles every 30 seconds. Write an equation of the height of the water at time t seconds.
H(t) = 3($$2^{\frac{t}{30}}$$)
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# Basic Concepts Of Trigonometry
Trigonometry is a branch of mathematics. It is a central of mathematics. Trigonometry is all about the relation between the sides and angles of the triangles. The father of the trigonometry is “Hipparchus” introduce first trigonometry table. Trigonometry is the subject which is easy to understand if approach in a right manner. Go through the article to understand the basic concepts.
Right Angle Triangle:
Right angle triangle is a triangle in which one of its interior angle is 90 degree. The opposite side of the right angle triangle is hypothesis and the other two sides are adjacent and opposite. Basic sin, cos and tan functions are as follows,
a)Sin θ = Opposite/Hypothesis
Pythagoras Theorem:
Pythagoras theorem is used to find the sides of the right angle. Pythagoras theorem is,
Triangle Identities:
Triangle identities are the function. There are some identities which are true for right angle triangles and some are true for all triangles. Understand more about trigonometry identities.
Sine and cosine rule:
Sine and cosine rule are very important in trigonometry. This is useful to solve triangle which is not a right angle.
Sine rule:
a/sin A = b/sin B = c/sin C
Cosine rule:
a^2 = b^2+c^2-2ab cos A
b^2 = a^2+c^2-2ab cos B
c^2 = a^2+b^2-2ab cos C
Here a, b and c are the side of the triangle.
And A, B and C are the angle of the triangle.
Important points:
• Understand the Trigonometry table.
• To understand trigonometry you can have an online math tutor chat.
• To know more about trigonometry basic take online trig help.
Online Math Tutor Chat
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# The Commutative Property
Introducing terms such as the commutative property of arithmetic to kids in basic mathematics is in general the time we start to completely lose and confuse them! This is because we start to use language that they are not familiar with.
If we break these terms down to language they are familiar with understanding comes soon after. HOWEVER, once understanding is there, the proper terms must be used at the right times, as this is how mathematics is communicated!
## So what is the Commutative Property of numbers?
When we discuss this in basic arithmetic, or further on in Basic Algebra, we are simply referring to the order in which numbers appear and are operated upon.
If we operate on the numbers in one order, then change the order and operate again - if we get the same answer the operation is said to be commutative. And that is simply it!
## Let's look at each operation in turn.
Steps to be taken Action Result Addition problem Change Order Are the answers the same? Addition IS Commutative ie - the order in which you add numbers does not change the result. 3 + 5 5 + 3 8 = 8 8 8 Yes
### Subtraction
Steps to be taken Action Result Subtraction problem Change Order Are the answers the same? Subtraction is NOT Commutative ie - the order in which you add numbers does not change the result. 9 - 3 3 - 9 6 is not equal to -6 6 -6 NO
### Multiplication
Steps to be taken Action Result Multiplication problem Change Order Are the answers the same? Multiplication IS Commutative ie - the order in which you add numbers does not change the result. 3 x 5 5 x 3 15 = 15 15 15 YES
### Division
Steps to be taken Action Result Division problem Change Order Are the answers the same? Division Is NOT Commutative ie - the order in which you add numbers does not change the result. 9/3 3/9 3 is not equal to 1/3 3 1/3 NO
Your child must understand the commutative property and the other properties of arithmetic if they want to enjoy (yes enjoy!) Basic Algebra!
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# Combinations
## Using lists and tree diagrams.
%
Progress
Practice Combinations
Progress
%
Define and Apply Combinations
A lottery system consists of 40 balls numbered 1 through 40. In the Big 5, five balls are selected from these 40 balls. You need to match all five numbers to win the cash prize. How many ways can 5 balls from a selection of 40 numbers be drawn?
### Guidance
Combinations of a subset of a larger set of objects refer to the number of ways we can choose items in any order. For comparison, look at the table below to see when order matters and when order doesn’t matter.
Combinations Permutations
• Ways to select the members of a committee from a larger population
• Ways to select specific officers in a club-president, vice president, treasurer, etc.
• Ways to select a set number of pizza toppings from a larger list of choices
• Ways to select and arrange scoops of ice cream on a cone
• Ways to select books from a reading list
• Ways to select and order in which to read books selected from a reading list.
The simplest way to describe the difference between a combination and a permutation is to say that in a combination the order doesn’t matter. The members of a committee could be selected in any order but the officers in a club are assigned a specific position and therefore the order does matter. Be careful of the use of these words in the real world as they are sometimes misused. For example, a locker combination. The ways to select and order the three numbers for a locker combination is not actually a combination, but a permutation since the order does matter.
#### Example A
How many ways can we choose three different flavors of ice cream from a selection of 15 flavors to place in a bowl?
Solution: First, does order matter in the bowl? When we created an ice cream cone with three scoops in an earlier concept the order did matter but here it does not. Let’s work from the example of the ice cream cone. We determined the number of permutations of a subset of three flavors from the total 15 flavors using the formula: $\frac{n!}{(n-r)!}=\frac{15!}{(15-3)!}=\frac{15!}{12!}=\frac{15\times14\times13\times{\color{red}\cancel{12!}}}{{\color{red}\cancel{12!}}}=2730$ . Now that order doesn’t matter, this number includes the $3!$ ways to arrange each combination of 3 flavors. We can divide 2,730 by $3!$ to determine the number of combinations: $\frac{2730}{3\times2\times1}=\frac{2730}{6}=455$ .
The notation and formula for combinations can be written as: $\dbinom{n}{r}={_nC}_r=C^n_r=\frac{n!}{r!(n-r)!}$ , where $n$ represents the number of elements in the set and $r$ represents the number of elements in the subset.
#### Example B
Evaluate the following expressions:
1. $\dbinom{8}{5}$
2. $_8C_0$
3. $_8C_8$
4. $C^{10}_7$
5. Explain why the answers to 2 and 3 are the same.
Solution: All of the notations in problems 1-4 indicate that we should use the formula for a combination. We can use the graphing calculator to evaluate these as well. Problems 2 and 3 are set up in the form of the calculator notation so we will use the calculator to evaluate those two and the formula for the other two.
1. $\dbinom{8}{5}=\frac{8!}{5!(8-5)!}=\frac{8\times7\times{\color{red}6}\times{\color{red}\cancel{5!}}}{{\color{red}\cancel{5!}}\times{\color{red}3}\times{\color{red}2}\times1}=\frac{8\times7}{1}=56$ .
2. Type in 8 on the TI-83 Graphing calculator, then MATH $\rightarrow$ PRB, select 3: $_nC_r$ . Now type in 0 and your screen should read 8 $_nC_r$ 0 before your press ENTER to get the answer 1.
3. Type in 8 on the TI-83 Graphing calculator, then MATH $\rightarrow$ PRB, select 3: $_nC_r$ . Now type in 8 and your screen should read 8 $_nC_r$ 8 before your press ENTER to get the answer 1.
4. $C^{10}_7=\frac{10!}{7!(10-7)!}=\frac{10\times9\times8\times{\color{red}\cancel{7!}}}{{\color{red}\cancel{7!}}\times3!}=\frac{10\times{\color{red}\cancel{3}}\times3\times4\times{\color{red}\cancel{2}}}{{\color{red}\cancel{3}}\times{\color{red}\cancel{2}}}=120$
5. In problem 2, we are looking at the ways to choose 0 items from 8 choices. There is only one way to do this. In problem 3 we are looking at the ways to choose 8 items from 8 choices. Well, the only way to do that is to choose all 8 items. So, there is only 1 way to choose zero items or all the items from a set.
#### Example C
How many ways can a team of five players be selected from a class of 20 students?
Solution: We can express this problem using the notation $\dbinom{20}{5}$ and then use the formula to evaluate.
$\dbinom{20}{5}=\frac{20!}{5!\times15!}=\frac{{\color{red}\cancel{20}}\times19\times{\color{red}\cancel{6}}\times3\times17\times16\times{\color{red}\cancel{15!}}}{{\color{red}\cancel{5\times4}}\times{\color{red}\cancel{3\times2}}\times{\color{red}\cancel{15!}}}=15,504.$
Intro Problem Revisit First, we need to determine whether order matters. The winning numbers 5, 10, 15, 20, and 25 are the same as the winning numbers 25, 5, 20, 10, 15, so order does not matter.
Therefore, we use the combinations formula $\dbinom{n}{r}=\frac{n!}{r!(n-r)!}$
$\frac{40!}{5!(40-5)!}\\\frac{40!}{5!35!}\\\frac {40\cdot39\cdot38\cdot37\cdot36\cdot35!}{5\cdot4\cdot3\cdot2\cdot1\cdot 35!}\\\frac{78,960,960}{120} = 258,008$
Therefore, there are 258,008 ways five winning balls can be drawn from the 40 numbers.
### Guided Practice
1. Evaluate the following using the formula for combinations of the calculator.
a. $\dbinom{7}{5}$
b. $_{20}C_{12}$
c. $C^{15}_7$
2. How many ways can a committee of three students be formed from a club of fifteen members?
3. How many three-topping pizzas can be made if there are 10 topping choices?
1. Using the calculator for each of these we get:
a. $7 \ _nC_r \ 5 = 21$
b. $20 \ _nC_r \ 12 = 125,970$
c. $15 \ _nC_r \ 7 = 6,435$
2. $\dbinom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{{\color{red}\cancel{3}}\times5\times{\color{red}\cancel{2}}\times7\times13\times{\color{red}\cancel{12!}}}{{\color{red}\cancel{3\times2}}\times{\color{red}\cancel{12!}}}=455$ .
3. $C^{10}_3=10 \ _nC_r \ 3 = 120$ .
### Explore More
Evaluate the following combinations with or without a calculator.
1. $_{13}C_{10}$
2. $C^{10}_6$
3. $\dbinom{18}{10}$
4. Explain why $_9C_5={_9C}_4=126$ .
5. Decide whether the following situations are permutations and which are combinations.
1. Ways to arrange students in a row.
2. Ways to select a group of students.
3. Ways to organize books on a shelf.
4. Ways to select books to read from a larger collection.
5. Ways to select three different yogurt flavors from a collection of ten flavors.
In each scenario described below, use either a combination or permutation as appropriate to answer the question.
1. There are seven selections for appetizers on a caterer’s menu. How many ways can you select three of them?
2. You only have time for seven songs on your workout playlist. If you have 10 favorites, how many ways can you select seven of them for the list? Now, how many ways can you select them in a particular order?
3. How many ways can you select two teams of five players each from a group of ten players?
4. At the local frozen yogurt shop a sundae comes with your choice of three toppings. If there are 12 choices for toppings, how many combinations of toppings are possible?
5. How many ways can four people be selected from a group of 30 to serve on a committee? What if each of the four people was selected to fill a specific position on the committee?
6. A soccer team has 20 players, but only 11 play at any one time.
1. How many ways can the coach select a group of eleven players to start (disregard positions)?
2. Now, of the eleven players on the field, one is a goalie, four play defense, three play midfield and three play offense. How many ways are there to assign the eleven players to these positions?
3. Considering your answers to parts a and b, how many ways can the coach select eleven players and assign them positions on the field? Assume all players can play each position.
### Vocabulary Language: English
combination
combination
Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.
Permutation
Permutation
A permutation is an arrangement of objects where order is important.
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# Slope of a Line Between Two Points on a Function
We've already done this in the case where the function is a line, but what happens if the line isn't straight? What if it's snake-shaped or U-shaped? What if it's as curvy as a mudflap? Now we'll find the slope of a line between two points on any wonky function we like.
### Sample Problem
Find the slope of the line between the two points shown.
Since we're given a formula for the function f, we can say
Then we can find the rise and run from this picture:
A line between two points on a function is called a secant line.
Asking to find the slope of the "secant line" between two points on a function means the same thing as asking to find the slope of the "line" between those two points.
A secant line is a line between two points on a function. We've been thinking about a secant line as a line that starts at the point on f where x = a, and ends at the point on f where x = b. For what's coming, it will be helpful to think of starting at (af(a)) and ending at some "other point." The "other point" will be described by how far away its x-value is from a. Poor b is getting fired, to be replaced by a + h
h is whatever we need to make a + h equal the number formerly known as b, not to be confused with the artist formally known as Prince.
### Sample Problem
Before, we had a = 1 and b = 1.5. a and b were equally important:
Now we have a = 1 and h = 0.5. a is important, and the distance of the other point from a is important:
### Sample Problem
Before, we had a = 1 and b = -2:
Now we have a = 1 and h = -3:
The moral of the story is that we can think of a secant line on the function f as a line that starts at the point on f where x = a and ends at the point on f where xa + h:
Since h measures the distance (and direction) from our starting point to our ending point, h is conveniently equal to the "run" we need for the slope formula:
We find a lovely formula for the slope of the secant line on the function f from x = a to x = a + h:
If h is negative the formula will be the same, and the picture will look slightly different:
By letting h get smaller and smaller, we can get a better idea of what the slope of the function is at a particular point, x = a.
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Students can download Maths Chapter 8 Statistics Ex 8.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.
## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.3
Question 1.
The monthly salary of 10 employees in a factory are given below:
Rs 5000, Rs 7000, Rs 5000, Rs 7000, Rs 8000, Rs 7000, Rs 7000, Rs 8000, Rs 7000, Rs 5000
Find the mean, median and mode.
Solution:
Mean = 6600
Median:
Arrange in ascending order we get.
5000, 5000, 5000, 7000, 7000, 7000, 7000, 7000, 8000, 8000
The number of values = 10
Median = Average of ($$\frac{10}{2})^{th}$$ and ($$\frac{10}{2}+1)^{th}$$
= Average of 5th value and 6th value
= $$\frac{7000+7000}{2}$$
∴ Median = 7000
Mode: 7000 repeated 5 times
∴ Mode = 7000
Question 2.
Find the mode of the given data: 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1
Solution:
3.1 occuring two times
3.3 occuring two times
∴ 3.1 and 3.3 are the mode (bimodal)
Question 3.
For the data 11, 15, 17, x + 1, 19, x – 2, 3 if the mean is 14, find the value of x. Also find the mode of the data.
Solution:
Arithmetic mean
∴ 2x + 64 = 14 × 7
2x = 98 – 64
2x = 34
x = $$\frac{34}{2}$$
= 17
The given numbers are 11, 15, 17, 18, 19, 15 and 3
15 occuring two times
∴ Mode = 15
The value of x = 17 and mode = 15
Question 4.
The demand of track suit of different sizes as obtained by a survey is given below:
Which size is demanded more?
Solution:
The highest frequency is 37
The corresponding value is the mode
∴ Mode = 40
Size 40 is demanded more.
Question 5.
Find the mode of the following data:
Solution:
The highest frequency is 46
20 – 30 is the modal class
Here l = 20, f = 46, f1 = 38, f2 = 34 and c = 10
Mode
= 20 + 4
= 24
∴ Mode = 24
Question 6.
Find the mode of the following distribution
Solution:
In the given table the class intervals are in inclusive form; convert them into exclusive form.
The highest frequency is 14
Modal class is 54.5 – 64.5
Here l = 54.5, f = 14, f1 = 10, f2 = 8 and c = 10
mode
= 58.5
∴ Mode = 58.5
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# lim_(x->0)(1/x^2-1/sin^2x)= ?
Nov 26, 2016
$- \frac{1}{3}$
#### Explanation:
We have
$\frac{1}{x} ^ 2 - \frac{1}{\sin} ^ 2 x = \frac{{\sin}^{2} x - {x}^{2}}{{x}^{2} {\sin}^{2} x} = \frac{{\left(\sin \frac{x}{x}\right)}^{2} - 1}{1 - {\cos}^{2} x} =$
$= \left(\frac{\sin \frac{x}{x} + 1}{1 + \cos x}\right) \left(\frac{\sin \frac{x}{x} - 1}{1 - \cos x}\right)$
Now following with ${\lim}_{x \to 0} \left(\frac{\sin \frac{x}{x} - 1}{1 - \cos x}\right)$
At this point we introduce the series representation for $\sin x$ and $\cos x$
sinx = x-x^3/(3!)+x^5/(5!)+cdots
cosx = 1-x^2/(2!)+x^4/(4!)+cdots
so
sinx/x = 1-x^2/(3!)+x^4/(5!)+cdots
making substitutions
(sinx/x-1)/(1-cosx)=(-x^2/(!3)+x^4/(5!)-x^6/(7!)+cdots)/(x^2/(2!)-x^4/(4!)+x^6/(6!)+cdots)=(-1/(!3)+x^2/(5!)-x^4/(7!)+cdots)/(1/(2!)-x^2/(4!)+x^4/(6!)+cdots)
then
${\lim}_{x \to 0} \frac{\sin \frac{x}{x} - 1}{1 - \cos x} = - \frac{1}{3}$ and
${\lim}_{x \to 0} \frac{\sin \frac{x}{x} + 1}{1 + \cos x} = 1$
Finally
${\lim}_{x \to 0} \left(\frac{\sin \frac{x}{x} + 1}{1 + \cos x}\right) \left(\frac{\sin \frac{x}{x} - 1}{1 - \cos x}\right) = \left(1\right) \left(- \frac{1}{3}\right) = - \frac{1}{3}$
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