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# What is the geometric mean of the pair of numbers 576 and 4?
Nov 8, 2017
$48$
#### Explanation:
the geometric mean of two numbers $a \text{ &" b" }$is
$\sqrt{a b}$
we have $4 \text{ & } 576$
$G M = \sqrt{4 \times 576}$
$= \sqrt{4} \times \sqrt{576}$
$= 2 \times 24 = 48$ | HuggingFaceTB/finemath | |
# Arcs and Angles of Circles
Hello, and welcome to this video about arcs and angles of circles! In this video, we will explore the different parts of circles and how to use them to solve problems. Let’s learn about arcs and angles of circles!
When we look around us, after polygons, circles are the next most common shape that we are surrounded by in our environment. A circle is created by a 360° rotation.
An arc of the circle is a part of the circle, and we identify arcs using points on the circle. Arc $$RS$$ is the curved part of circle $$A$$ shown with the purple mark. Because arc $$RS$$ is less than 180°, we call this arc a minor arc. Arc $$RTS$$, or $$STR$$, is a major arc because it is greater than 180°. We can find the measure of an arc using the fact that a circle is 360°. If we know that the measure of arc $$RS$$ is 130°, we can find the measure of arc $$RTS$$ by subtracting the measure of arc $$RS$$ from 360°, which would make the measure of arc $$RTS$$ 230°.
Some important lines and segments associated with circles are chords, secants, and tangents.
A chord is a line segment with either end touching the circle. Line segment $$LM$$ is a chord on circle $$A$$. A secant line is similar to a chord, except it is a line that passes through two points on the circle instead of being a line segment. And a tangent line is a very special kind of line that only touches the circle at one point, called the point of tangency. Line $$PQ$$ is a tangent and point $$Q$$ is the point of tangency. Knowing these vocabulary words related to circles is important in helping understand the circle theorems which will help us solve problems.
The radius of the circle is always perpendicular to the point of tangency.
When two chords share a point on the circle, an inscribed angle is formed. In circle $$A$$, the chords $$RS$$ and $$TS$$ form the inscribed angle $$RST$$, which has its vertex on the circle. The arc that is formed by the legs of angle $$RST$$ is called an intercepted arc. An angle that has the center of the circle as its vertex is naturally called a central angle. In circle $$C$$, angle $$MCN$$ is a central angle and arc $$MN$$ is its intercepted arc.
If we are given the measure of the inscribed angle or the intercepted arc, we will be able to find the measure of the other using a circle theorem that tells us that the measure of an inscribed angle is one-half the measure of its intercepted arc. Or we could say that the intercepted arc is twice as long as the inscribed angle. For example, in circle $$A$$, the measure of inscribed angle $$RST$$ is given to us as 60°. According to the inscribed angle theorem, the measure of arc $$RT$$ is $$60° \times 2$$, which is 120°.
Here is a special case of an inscribed angle.
In circle $$A$$, inscribed angle $$PQR$$ encompasses the diameter of the circle, line segment $$PR$$ Remember, the diameter of the circle divides the circle into two equal parts, called semicircles. A circle has a total of 360°, so a semicircle has 180°, or half the measure of a full circle. So this must mean that the intercepted arc of angle $$PQR$$ is 180°. If we use the inscribed angle theorem, we find out that angle $$PQR$$ is 90° because it is half the degree measure of the intercepted arc.
Another important theorem we are going to take a look at is the central angle theorem. This theorem says that the measure of a central angle is equal to the measure of its intercepted arc. In circle $$C$$, since the measure of angle $$MCN$$, the central angle, is 110°, then arc $$MN$$, its intercepted arc, is also 110°.
Let’s take a look at an example problem. Line $$LM$$ is tangent to circle $$T$$ at point $$L$$. The measure of the central angle $$T$$ is 35°. What is the measure of angle $$LMT$$?
Since we know that the sum of the interior angles of any triangle is 180° and that the radius is always perpendicular to the point of tangency, we know that angle $$TLM$$ is 90°. We can subtract $$(90°+35°)$$ from 180° to find the measure of angle $$LMT$$. $$180 – (90+35) = 55$$, therefore the measure of angle $$LMT$$ is 55°.
When secants intersect inside a circle, we can find the measure of the vertical angles created by using the measure of the arcs formed by the angles. It also works the other way around, if we need to find the measure of the arcs using the measure of the angles.
Let’s take a look at circle $$K$$.
The secant lines $$DE$$ and $$FG$$ intersect at point $$H$$ and create the arcs $$DF$$, $$FE$$, $$EG$$, and $$GD$$. Find the measure of any of the vertical angles (notice that there are two sets) by adding together the measures of the intercepted arcs and dividing by 2.
We will use the formulas $$m\angle DHF=\frac{1}{2}$$ (arc $$DF$$ + arc $$GE$$) and $$m\angle FHE=\frac{1}{2}$$ (arc $$FE$$ + arc $$DG$$). Then we will use the vertical rule theorem to find $$m\angle EHG$$ and $$m\angle GHD$$.
Let’s practice. The measure of arc $$GE$$ is 170°, arc $$GD$$ is 80°, arc $$DF$$ is 60°, and arc $$FE$$ is 50°. To find the $$m\angle DHF$$, we will add the intercepted arcs, $$DF$$ and $$GE$$, then multiply by $$\frac{1}{2}$$. Therefore, $$m\angle DHF=\frac{1}{2}(60° +170 °)$$. So $$\angle DHF=115°$$. We will use the equation, $$m\angle FHE=\frac{1}{2}(50° +80°)$$ to find the measure of $$\angle FHE$$, which is $$m\angle FHE=65°$$.
When two secant lines intersect outside the circle, we use a different method to find the measure of the angles. When this happens, you look at the two intercepted arcs created by the angle and subtract the measure of the smaller arc from the measure of the larger arc, then multiply by $$\frac{1}{2}$$.
We want to find the $$m\angle WYU$$ that is created by the intersection of secants $$WX$$ and $$UV$$ on circle $$Z$$. $$\angle WYU$$ is intercepted by arcs $$XV$$ and $$WU$$. The measure of arc $$XV$$ is 30°; the measure of arc $$WU$$ is 80°. To find the measure of angle $$WYU$$, we use the formula: $$m\angle WYU= \frac{1}{2}$$ (arc $$WU$$ – arc $$XV$$), therefore, $$m\angle WYU=\frac{1}{2}(80°-30°) \text{, }m\angle WYU=25°$$.
A sector of a circle is a section of a circle between two radii. The red area in the diagram is an example of a sector.
We can find the length of an arc if we are given the length of the radius and the measure of the central angle using the formula $$s=\frac{\pi r\theta }{180°}$$, where $$s$$ is the arc length, $$r$$ is the length of the radius and $$\theta$$ is the measure of the central angle. In circle A, the length of the radius is 5 cm, and the measure of angle $$MAN$$ is 70°. We want to find the length of arc $$MN$$. We will start by substituting what we have into the formula, $$s=\frac{\pi (5)(70)}{180°}$$. We will simplify to get $$s=6.11\text{ cm}$$ , which is the length of arc $$MN$$.
Thanks for watching this video, and happy studying!
## Arcs and Angles of Circles Practice Questions
Question #1:
If angle $$B$$ is $$21°$$, what is the measure of arc $$AC$$?
$$82°$$
$$42°$$
$$165°$$
$$21°$$
According to the inscribed angle theorem, arc $$AC$$ will be twice as large as the inscribed angle $$B$$. This means that arc $$AC$$ will be $$2×21°=42°$$.
Question #2:
Line $$LM$$ is tangent to circle $$J$$ at point $$L$$. If angle $$J$$ is $$27°$$, what is the measure of angle $$M$$?
$$65°$$
$$83°$$
$$44°$$
$$63°$$
Since line $$LM$$ is tangent to circle $$J$$ at point $$L$$, the line $$JL$$ and $$LM$$ form a right angle. This means that angle $$L$$ is $$90°$$. Angle $$J$$ is $$27°$$, so add angle $$J$$ and angle $$L$$: $$90+27=117$$. Since there are $$180°$$ in a triangle, subtract $$117$$ from $$180$$: $$180-117=63$$. The measure of angle $$M$$ is $$63°$$.
Question #3:
Arc $$BD$$ is $$15°$$. Arc $$AE$$ is $$48°$$. Find the measure of angle $$ACE$$.
$$19.5°$$
$$16.5°$$
$$22.5°$$
$$19°$$
Look at the two intercepted arcs. The angle $$ACE$$ will be $$\frac{1}{2}$$ the difference between the arc lengths of these intercepted arcs. Subtract the smaller arc from the larger arc. In this case, subtract arc $$BD$$ from arc $$AE$$: $$48-15=33$$. Now multiply this by $$\frac{1}{2}$$. $$33×\frac{1}{2}=16.5°$$. Angle $$ACE$$ is $$16.5°$$.
Question #4:
You are walking around a circular pond from one star to the other. If the radius of the pond is $$8\text{ ft}$$, what distance have you walked? Use $$3.14$$ for $$π$$.
$$22.8\text{ ft}$$
$$19\text{ ft}$$
$$16.78\text{ ft}$$
$$34.5\text{ ft}$$
Find the measure of the arc by using the formula: $$s=\frac{πrϴ}{180°}$$
$$s$$ represents the arc length.
$$π$$ is approximated to $$3.14$$.
$$r$$ is the radius.
$$ϴ$$ is the central angle measure.
$$s=\frac{πrϴ}{180°}$$ becomes $$s=\frac{(3.14)(8)(120°)}{180°}=16.746$$, which simplifies to $$s=16.7\text{ ft}$$.
Question #5:
A flashlight shines across a circular field. What is the angle measure of the intercepted arc that will be illuminated?
$$104°$$
$$120°$$
$$106°$$
$$130°$$
The inscribed angle is $$52°$$. The intercepted arc will be twice as much as the inscribed angle. $$52×2=104$$. The angle measure that is illuminated across the field is $$104°$$. | HuggingFaceTB/finemath | |
• # question_answer A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1. The ratio of their nuclear sizes will be: A) ${{2}^{\frac{1}{3}}}:1$ B) $1:{{3}^{\frac{1}{2}}}$ C) ${{3}^{\frac{1}{3}}}:1$ D) $1:{{2}^{\frac{1}{3}}}$
According to the law of conservation of momentum $\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{v}_{2}}}{{{v}_{1}}}=\frac{1}{2}$ Since, masses are directly proportional to volume as densities are roughly same, so it follows that: $\frac{{{m}_{1}}}{{{m}_{2}}}\propto \frac{r_{1}^{3}}{r_{2}^{3}}$or $\frac{{{r}_{1}}}{{{r}_{2}}}\propto {{\left( \frac{{{m}_{1}}}{{{m}_{2}}} \right)}^{\frac{1}{3}}}$ $\therefore$ratio of nuclear sizes$=\frac{1}{{{2}^{1/3}}}.$ Hence, the correction option is (d). | HuggingFaceTB/finemath | |
# Phoenix Payback Method IRR and NPV In Evaluating Project Cash Flows Memo – professionalessaybuddy.com
Phoenix Payback Method IRR and NPV In Evaluating Project Cash Flows Memo – professionalessaybuddy.com
###### Purpose of Assignment
The purpose of this assignment is to allow the student to calculate the project cash flow using net present value (NPV), internal rate of return (IRR), and the payback methods.
###### Assignment Steps
Create a 350-word memo to management including the following:
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• Describe the use of internal rate of return (IRR), net present value (NPV), and the payback method in evaluating project cash flows.
• Describe the break-even point and its importance.
Calculate the following time value of money problems:
1. If you want to accumulate \$500,000 in 20 years, how much do you need to deposit today that pays an interest rate of 15%?
2. What is the future value if you plan to invest \$200,000 for 5 years and the interest rate is 5%?
3. What is the interest rate for an initial investment of \$100,000 to grow to \$300,000 in 10 years?
4. If your company purchases an annuity that will pay \$50,000/year for 10 years at a 11% discount rate, what is the value of the annuity on the purchase date if the first annuity payment is made on the date of purchase?
5. What is the rate of return required to accumulate \$400,000 if you invest \$10,000 per year for 20 years. Assume all payments are made at the end of the period.
Calculate the project cash flow generated for Project A and Project B using the NPV method.
• Which project would you select, and why?
• Which project would you select under the payback method? The discount rate is 10% for both projects.
• Note that a similar problem is in the textbook in Section 5.1.
Sample Template for Project A and Project B:
“Table showing investments and returns for Project A and Project B. Project A has \$10,000 initial investment with \$5,000 returns in each of the first 3 years. Project B has \$55,000 initial investment with \$20,000 in each of the first 3 years.”
Show all work.
Submit the memo and all calculations. | HuggingFaceTB/finemath | |
Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## Topic review (newest first)
mathsyperson
2005-12-12 09:39:21
True. That method wouldn't work if the range had a number of the form (n+0.5)π in it, because the sign changes at every point with that form, but they are not roots.
Ricky
2005-12-11 11:23:54
Just don't forget to also state the function is continuous from [1.1, 1.2].
mathsyperson
2005-12-10 23:18:36
To show that there is a root between 1.1 and 1.2, just find the values of the functions at each of those points and show that there is a change of sign.
f(1.1) ≈ 0.23
f(1.2) ≈ -0.17
There is a change of sign, so a root must exist between the two.
Flowers4Carlos
2005-12-10 16:29:54
i was working on problem one in degree mode and my calculator kept spitting out positiive results even for very large values of x. then i switched over to radian mode and it worked! sometimes a lil trial and error will get you on the right track!
2)
f(x) = x² + sin(x/2) f(0) = 0
f'(x) = 2x + (1/2)cos(x/2) f'(0) = 1/2
f''(x) = 2 - (1/4)sin(x/2) f''(0) = 0
f'''(x) = -(1/8)cos(x/2) f'''(0) = -1/8
f''''(x) = (1/16)sin(x/2) f''''(0) = 0
f'''''(x) = (1/32)cos(x/2) f'''''(0) = 1/32
0 + [(1/2)x]/1! + 0x²/2! - [(1/8)x³]/3! + [0x^(4)]/4! + [(1/32)x^5]/5!
irspow
2005-12-10 12:33:52
The first one is 1.165561185 and then my little solar calculator couldn't go any further. It took six iterations using Newton's Method.
x - f(x)/f '(x)
Oh, and be sure to use radians because degrees will not give a solution.
You can use degrees but then your equation will be: y = 2x - tan(180x/pi)
Mike
2005-12-09 23:55:16
I have this mathematics question that is really giving me a headache.......
Show that the equation 2x-tanx=0 has a root in the interval [1.1, 1.2]. hence find this root correct to 2 decimal places.
b) Find the Macluarin series expansion for f(x) = x2 + sin (x/2) up to and including the term in x5. | HuggingFaceTB/finemath | |
ArtooDetoo
Joined: Jan 9, 2012
• Threads: 8
• Posts: 33
January 9th, 2012 at 11:20:01 PM permalink
Hey guys, first of all, a big thank you to the Wizard for his amazing sites. Whenever I have questions about casino games, I always refer to his websites first.
I've been trying to learn Pai Gow recently, and have been using the advanced strategy with exceptions (http://wizardofodds.com/games/pai-gow-tiles/strategy/jb/advanced/), but have ran into a lot of problems when trying them out with the Pai Gow game provided. I'm not sure if these are because the game is using more optimal strategy than the JB strategy, or because I'm actually misinterpreting a lot of these rules. Sorry for this relatively newbie question, but these are some examples of my incompetence:
1) H4, L10, 9, H8
The possibilities are 9/2, 8/3, and 7/4. I interpret that 9/2 would be the choice because there is a high 9 (play high 9 whenever possible), and it does not follow the exception (with 11 point hands where best low hand is 4, play 7/4 if any tile is a Gee) since there is no Gee. Yet, the correct play is 7/4 according to the calculator.
2) H7, 9, GJ, L10
The possibilities are 9/3, 7/5, and 6/6. Again, I interpret that 9/3 would be the choice because of the high 9, but 6/6 is the correct choice.
3) 11, GJ, H8, 2
The possibilities are Gong/7 and 9/8, and once again, I interpret 9/8 to be the choice since there is a high 9 and not a low 8 gong, since the gong is with a high 8. Yet, gong/7 is correct.
4) 5, GJ, L10, H10
This is more of a situation that I didn't find in the strategy. The correct play is clearly 6/5. GJ and 5 are the lowest ranking tiles here, so the rank of the high and low hands depends on which 10 is played in each hand. I thought that the high 10 should be played in the high hand, but the optimal strategy states that the low 10 is the one that should be played in the high hand with the Gee.
5) H7, L10, GJ, L7
The possibilities are 7/3 and 6/4. I use the rule that since the point total is not 5 or less, and the best low hand is not 5, 6, 7, or 9, the best hand hand should be played, which is 7/3. Again, I am mistaken, as 6/4 is the correct play.
6) 9, 5, H10, L8
The possibilities are 9/3, 8/4, and 7/5. I once again use the rule to play a high 9, and the exception is not used (play 7/5, 7/6, or 7/7 instead of high 9 if tiles include 2 with any 4, 5, or 8) since there is no 2, so I play 9/3. Yet, the correct play is 7/5.
I'm sorry for what seems to be a long and unnecessary thread about a specific case, but these errors for me are happening at an alarmingly high rate. Please let me know if it's me making the errors, or if it's just the strategy that doesn't cover all of these situations. Thanks!
FinsRule
Joined: Dec 23, 2009
• Threads: 121
• Posts: 3722
January 10th, 2012 at 4:48:27 AM permalink
Ok, I am not even close to being the Pai Gow Tiles expert on this board, but let me give it a shot, and have the Wizard or someone else make a fool of me when I'm wrong.
When he refers to High 9, I believe he refers to a 9 that is made with the teen and day tiles only.
And in case #4, you want your low hand to be as high as possible, so you want the high tile in it.
Not sure about case #5, I would think 6-4 is right too...
dlevinelaw
Joined: Dec 3, 2009
• Threads: 15
• Posts: 230
January 10th, 2012 at 6:11:02 AM permalink
I believe, and Wiz or JB can correct me on this, that when you play the Pai Gow game on this site or WoO, and it offers to "correct" your plays, that this correction is based on the house way, and NOT optimal strategy. Therefore, when you make a play that may be optimal, but not the house way, it still offers to "correct" it to the house way, thus the conflicts you are seeing.
I'd like to know if I'm correct on this, as I'd love to be able to play with an optimal-strategy-based correction system as opposed to a house-way correction system.
kaysirtap
Joined: Nov 1, 2011
• Threads: 3
• Posts: 55
January 10th, 2012 at 7:44:23 AM permalink
1*) "High nine" (or any "high" number) must contain a tile from one of the top six individual rankings. That would be Teen (12), Day (2), Yun (H8), Gor (H4), Mooy (H10), and Chong (H6). It has nothing to do with whether it is the high hand or low hand. The tiles 9 and L10 do not make a high nine in your example (highest tile is Ping, which does not qualify). Tiles H6 and GJ make the lowest high nine possible.
2) Again, the tiles 9 and L10 do not make a high nine.
3) One of the first rules for the house way is to play a Teen (12) or Day (2) with 7, 8, or 9. I agree that 8/9 is a strong hand here, and as a player you may have a tough choice, but this is not the house way.
4) The house is trying to make the hands as close as possible in this case (the low hand as high as possible, if you want to look at it that way), so they will want to play the 5 as high as they can. To do that, you need the tiles H10 and 5 in the low hand, and L10 and GJ in the high hand (read together has H5/6).
5) I'm not sure I understand your rule for figuring out how to set the tiles here. 4/6 looks good to me here.
6) I'm not sure I understand your rule for figuring out how to set the tiles here. 5/7 looks good to me here as well.
*EDIT: Sadly, the explanation of high nine is incorrect. Oops.
ArtooDetoo
Joined: Jan 9, 2012
• Threads: 8
• Posts: 33
January 10th, 2012 at 10:42:27 AM permalink
Thanks for your explanation of high nine! It makes things a lot clearer now. And thanks to everyone who helped. As for the hands that you don't understand the rule I'm using, it's from JB Advanced Strategy with exceptions, where there are four main rules after the pairs and the obvious stuff:
1. Play 0/Wong, 1/Gong, and 1/Wong whenever possible.
2. Play High Nine whenever possible. If given a choice between 2 and 12, play the 12 in the high hand. If given a choice between Low 7 and High 7, play the High 7 in the low hand if it will improve it.
3. Play a Low 8 Gong whenever possible, except when you also have a High 8 and cannot use it to make a low hand of 7 or better. If given a choice between 2 and 12, play the 12 in the high hand.
4. Determine the best low hand and best high hand by comparing only those ways to play which have the highest point total. Play the best low hand with point totals of 5 or less, and when the best low hand is a 5, 6, 7, or 9. Play the best high hand in all other circumstances.
So going back to my situations:
3) 11, GJ, H8, 2
The possibilities are 7/Gong and 8/9, and rule 2 says that I should play high nine whenever possible. There is also no exception telling me to play a high 8 Gong instead of a high nine.
5) H7, L10, GJ, L7
The possibilities are 3/7 and 4/6. I follow rule 4, which tells me that since the point total is greater than 5 and the best low hand is not 5, 6, 7, or 9, I don't play the best low. Instead, I play the best high, which is 3/7.
6) 9, 5, H10, L8
The possibilities are 3/9, 4/8, and 5/7. I follow rule 2, telling me to play the high nine whenever possible. The exception of playing 5/7 instead of high nine only takes effect when the tiles include a 2 with any 4, 5, or 8, and there is no 2 in this hand.
Thanks again for helping me understand this wonderful game!
--EDIT--
Okay, I think I understand that a high 9 means a 2 with a 7 only, so that clears situations 3 and 6, and all that's left is why 4/6 is better than 3/7 in situation 5.
Tiltpoul
Joined: May 5, 2010
• Threads: 32
• Posts: 1573
January 10th, 2012 at 5:14:59 PM permalink
Quote: kaysirtap
1) "High nine" (or any "high" number) must contain a tile from one of the top six individual rankings. That would be Teen (12), Day (2), Yun (H8), Gor (H4), Mooy (H10), and Chong (H6). It has nothing to do with whether it is the high hand or low hand. The tiles 9 and L10 do not make a high nine in your example (highest tile is Ping, which does not qualify). Tiles H6 and GJ make the lowest high nine possible.
Most casinos I've played tiles at consider high nine to ONLY be a seven tile and Teen or Day. For example, a Yun +11, Teen + Tit (giggity), is declared as 9/HIGH 9. Now for advanced strategy play, it might mean the Chong 9 or better. Horseshoe Southern Indiana has a strange house rule where you set 7,8,9 with a Chong 3 or better, meaning the ranking of the 3 must be Chong or better. When I kept asking about this in Atlantic City, I got some very weird looks.
EDIT: I realized in my original explanation of HSI's house way, I didn't clarify. They play 7,8,9 first if they can't make a chong 3 or better... if they can, the split the hand and balance the values. Here's an example. Mixed 7, L10, Gor 4, and Look 6. The house way most places would be to go 3/4. However, the 3 is Look 6 an Mixed 7, neither which is higher than Chong. So HSI's house way sets the hand 0/7. It would also be the difference if there was a 2/9 or 3/8, but the three is chong 3, they would play the 3/8.
"One out of every four people are [morons]"- Kyle, South Park
kaysirtap
Joined: Nov 1, 2011
• Threads: 3
• Posts: 55
January 10th, 2012 at 7:30:44 PM permalink
Quote: Tiltpoul
Most casinos I've played tiles at consider high nine to ONLY be a seven tile and Teen or Day.
I'll admit that it's been a while since I've been on this game, so this may be more accurate than my explanation. I do remember that Chong factors into at least the High 3 rule... beyond that, I'm just not sure anymore. Sadness.
JB
Administrator
Joined: Oct 14, 2009
• Threads: 334
• Posts: 2089
January 10th, 2012 at 8:52:14 PM permalink
As others have mentioned, a High Nine is only [Teen + 7] or [Day + 7].
In some situations, like H4-L4-5-GJ, you can play 8/8, 7/9, or 7/9. In this case, we would distinguish between the two 7/9's as "high 7/9" and "7/high 9" based on which hand has the H4 tile in it, but "7/high 9" does not contain a High Nine.
With H7-L7-GJ-L0 you have four lousy low-ranking tiles, so making the best low hand (4/6) ends up being the best play.
The Pai Gow game on here and on WizardOfOdds.com offers optimal strategy advice, not the house way. (The Pai Gow Poker game offers the house way as advice.) But note that there are three types of optimal player strategy: optimal strategy when the dealer is banking, optimal strategy when you are banking, and an optimal single strategy regardless of who is banking. If you check the box "Use a single strategy" then it will offer the single-strategy advice. If this box is not checked, it will offer optimal dealer-banker or player-banker strategy, depending on whether or not you are banking.
ArtooDetoo
Joined: Jan 9, 2012
• Threads: 8
• Posts: 33
January 10th, 2012 at 9:28:46 PM permalink
Thanks a lot to JB and everyone else for helping me! JB, can you direct me to where in the strategy it tells me to play 6/4 over 7/3, or if it's not in there? I've just been making a lot of mistakes compared to the game on the Wizard of Odds site, and it makes me think that I'm not reducing the house edge to 1.78%/-0.08%. I came across a ton more mistakes today:
1) L10, H8, L6, 2
Correct play is 8/8, but I played 6/Gong since the last rule states that I play the best low hand if it's 5, 6, 7, or 9, and the best high hand otherwise, and since the best low hand is an 8, I played the best high hand.
2) H8, L8, L6, H6
Correct play is 4/4, but I play 2/6 since, again, I play best low hand if it's a 5, 6, 7, or 9, and the best high hand otherwise, and since the best low hand is a 4, I played the best high hand. An exception is that I play 4/4 if the tiles include two 7s without a 10, but there are no 7s.
3) 12, 2, H6, 11
Correctly play is 4/7, but I play 3/8 under the same exact rule. An exception is that I play 4/7 if the best low hand is a 4 and any tile is a Gee, but no tiles are Gee.
4) L7, H8, 9, L8
The correct play is 5/7, but i play 6/6 under the same rule again. Exceptions only talk about play 5/7 over High Nine.
Maybe I'm misinterpreting the rules/exceptions but it seems like these are very simple decisions that I am messing up on.
JB
Administrator
Joined: Oct 14, 2009
• Threads: 334
• Posts: 2089
January 11th, 2012 at 1:54:20 PM permalink
Quote: ArtooDetoo
JB, can you direct me to where in the strategy it tells me to play 6/4 over 7/3, or if it's not in there?
It's probably not in there. That strategy is just a guide, it is not perfect strategy. There are lots of differences between that and perfect strategy. Perfect strategy is difficult to put in simple terms.
I should probably ask the Wizard about removing those old JB strategies. You would be better off if you went here, click the Analyze button, (wait for it to analyze), click the Show Strategy button, and follow that strategy.
It will give you a computer-generated strategy which was generated by trying various rules, and settling for those which produce the fewest exceptions. It is better than the human-generated versions I made (the "JB" strategies); in fact, the base rules plus the exceptions represent perfect strategy.
The key rule for hands with no Pair/Wong/Gong/High Nine is: play the best low hand if it ranks between 4-with-H8 and 8-with-L10 inclusive, or if the best high hand ranks lower than 6-with-H6; otherwise play the best high hand. But of course, there are exceptions to it.
Quote: ArtooDetoo
L7, H8, 9, L8
7-8-8-9 is an interesting combination, where you always play 5/7, but sometimes you will play the H8 in the low hand and sometimes you will play it in the high hand. The way to remember when is to look at the 7 tile: if it's a L7, then play the lower-ranking 7-point high hand. If it's a H7, play the higher-ranking 7-point high hand. In other words: with L7-L8-H8-9, play H8+L7 in the low hand and L8+9 in the high hand (high 5/low 7). With H7-L8-H8-9 play H7+L8 in the low hand and H8+9 in the high hand (low 5/high 7). So if you have a L7 tile, play a low 7 in your high hand, and if you have a H7 tile, play the high 7 in your high hand.
• Jump to: | HuggingFaceTB/finemath | |
# Part II: Mass of the Death Star in Episode IV
This is the second in a two part post where I calculate the size and mass respectively of the Death Star in Episode IV (DS1). Estimating the mass will inform discussion about the power source of the station and other energy considerations.
Part II: Mass of DS1
As argued in Part I, I assert that the diameter of DS1 is approximately 60 km based on a self-consistent scale analysis of the station plan schematics as shown during the briefing prior to the Battle of Yavin.
A “realistic” upper limit for the mass is set if the 60 km volume of DS1 was filled with the densest (real, stable) element currently known. This is osmium with a mass density of 2.2E4 kilograms per cubic meter. This places the mass at 2.5E18 kg with a surface gravity of 0.05g. A filling fraction of 10% would then place a “realistic” estimate of the upper limit at 2.5E17 kg. Other analyses have made similar assessments using futuristic materials with some volume filling-fraction, also putting the mass somewhere around 10^18 kg assuming a radius of 160 km.
In this mass analysis, using information from the available footage from the Battle of Yavin, I find a DS1 mass of roughly 2.8E23 kg, about million times the mass of a “realistic” approximation Any supporting superstructure would be a small perturbation on this number. This implies a surface gravity of an astounding 448g. To account for this, my conclusion is that DS1 has a 40 m radius sphere of (contained) quark-gluon plasma or a 55 m radius quantity of neutronium at its core. Such materials, if converted to useful energy with an efficiency of 0.1%, would be ample to 1) provide the 2.21E32 J/shot of energy required to destroy a planet as well as 2) serve as a power source for sub-light propulsion.
Details
The approach here uses the information available in the schematics shown during the briefing. The briefing displays a simulation of the battle along the trench to the exhaust port. Again, as shown in Part I of this post, the simulation scale is self-consistent with other scales in both the schematic and the actual battle footage. As shown in Figure 1, the proton torpedo is launched into projectile motion only under the influence of gravity. It appears to be at rest with respect to the x-wing as it climbs at an angle of about 25 degrees.
Figure 1
Figure 2
From the previous scale analysis in Part I, the distance from the port, d, and height, h, above the the port can be estimated. They are approximately equal, h = d = 21 meters. The length of the x-wing is L = 12.5 m. After deployment, the trajectory slightly rises and then falls into the exhaust port as shown in Figure 2. A straightforward projectile motion calculation gives the formula for the necessary downward acceleration to follow the trajectory of an object under these conditions
$a=\frac{2 V_{0}^2}{d}(\frac{h}{d}+\tan{\theta})\cos^2{\theta}\ \ \ \ (1)$
Where t is the launch angle and Vo is the initial horizontal velocity of the projectile. If we assume for simplicity that the angle $\theta$ = 0 degrees and h = d, the formula simplifies to
$a=\frac{2 V_{0}^2}{d}\ \ \ \ (2)$.
From the surface gravity, the mass of can be obtained, assuming Newtonian gravity,
$M=\frac{a R^2}{G}\ \ \ \ (3)$.
Here G = 1.67E-11 Nm/kg, the gravitational constant. For a bombing run, let’s assume the initial speed of the projectile to be the speed of the x-wing coming down the trench. To estimate the speed, v, of the x-wing, information from the on-board battle computers is used. In Part I, the length of the trench leading to the exhaust port was estimated to be about x = 4.7 kilometers. On the battle computers, the number display coincidentally starts counting down from the range of about 47000 (units not displayed). However, from this connection I will assume that the battle computers are measuring the distance to the launch point in decimeters. From three battle computer approach edits, shown in Clip 1 below, and using the real time length of the different edits, the speed of an x-wing along the trench is estimated to be about 214 meters/second (481 miles/hour). This is close to the cruising speed of a typical airliner — exceptionally fast given the operating conditions, but not unphysical. This gives a realistic 22 seconds for an x-wing to travel down the trench on a bombing run.
Using this speed and the other information, this places the surface gravity of DS1 at about 448 g (where g is the acceleration due to gravity on the surface of the earth). DS1 would have to have a corresponding mass of 2.4E23 kg to be consistent with this.
However, it is clear that considerable liberty was taken in the above analysis and perhaps too much credibility was given to the battle simulation alone, which does not entirely match the dynamics show in the footage of the battle. Upon inspection of the footage, the proton torpedoes are clearly launched with thrust of their own at a speed greater than that of the x-wing. A reasonable estimate might put v (torpedo) to be roughly twice the cruising speed of the x-wing. Moreover, the torpedoes are obviously not launched a mere d = 21 meters from the port (although h = 21 is plausible), rather sufficiently far such that the port is just out of sight in the clip. Finally, the torpedoes enter the port at an awkward angle and appear to be “sucked in.” One might argue that there could be a heat seeking capability in the torpedo. However, this seems unlikely. If this were the case, then it greatly dilutes the narrative of the battle, which strongly indicates not only that the shot was very difficult but that it required the power of the Force to really be successful. Clearly, “heat seeking missiles along with the power of the Force” is a less satisfying message. Indeed, some have speculated that the shot could only have been made by Space Wizards. These scenarios, and other realistic permutations, are in tension with the simulation shown in the briefing. Based on different adjustments of the parameters v (torpedo), h, d, and th, one can tune the value of the surface gravity and mass to be just about anything.
However, if we attempt to be consistent with the battle footage, we might assume again that t=0 degrees while d = 210 m, and v (torpedo) = 2 v (x-wing) for propulsion. The speed of the x-wing can remain the same as before at 214 m/s. Even with this, the surface gravity will be 18g. This still leads to a mass over 10000 times larger than the mass of a realistic superstructure. In this case, a ball of neutronium 18 m in radius could still be contained in the center to account for this mass.
Nevertheless, my analysis is based on the following premise: the simulation indicates that the rebel analysts at least believed, based on the best information available, that a dead drop of a proton torpedo into the port, only under the influence of DS1’s gravity, was at least possible at d = h = 21 meters at the cruising speed of an x-wing flying along the trench under fire nap-of-the-earth. Any dynamics that occurred in real time under battle conditions would ultimately need to be consistent with this.
The large intrinsic surface acceleration may seem problematic (consider tidal forces or other substantial technological complications). However, as demonstrated repeatedly in the Star Wars universe, there already exists exquisite technology to manipulate gravity and create the appropriate artificial gravity conditions to accommodate human activities (e.g. within DS1, the x-wings, etc.) under a very wide range of activities (e.g. acceleration to hyperspace, rapid maneuvering of spacecraft, artificial gravity within spacecraft at arbitrary angles, etc.).
Implications for such a large mass.
One hypothesis that would explain such a large mass would be to assume DS1 had, at its core, a substantial quantity of localized neutrinoium or quark-gluon plasma contained as an energy source. Such a source with high energy density could be used for the purposes of powering a weapon capable of destroying a planet, as an energy source for propulsion, and other support activities. For example, the destiny of neutronium is about 4E17 kilograms per cubic meter and a quark-gluon plasma is about 1E18 kilograms per cubic meter. Specifically, a contained sphere of neutronium at the center of the death star of radius 55 meters would account for the calculated mass and surface gravity of DS1.
It has been estimated that approximately 2.4E32 joules of energy would be required to destroy an earth-sized planet. If 6.7 cubic meters of neutronium (e.g. a sphere of radius 1.88 m) could be converted to useful energy with an efficiency of 0.1%, this would be sufficient to destroy a planet (assuming the supporting technology was in place). This is using the formula
$\Delta E=\epsilon\Delta m c^2\ \ \ \ (4)$
where $\Delta E$ is the useful energy extracted from a mass $\Delta m$ with efficiency $\epsilon$. The mass is converted to a volume using the density of the material.
By using the work-energy theorem, the energy required to accelerate DS1 to an arbitrary speed can be estimated. Assuming the possibility for relativistic motion, it can be shown (left as an exerise for the reader) that the volume V of fuel of density $\rho$ required to accelerate an object of mass M to a light-speed fraction $\beta$ at efficiency $\epsilon$ is given by
$V=\frac{1}{\sqrt{1-\beta^2}}\left(\frac{M}{\epsilon\rho}\right)\ \ \ \ (5)$.
This does not account for the loss of mass as the fuel is used, so represents an upper limit. For example, to accelerate DS1 with M = 2.4E23 kg from rest to 0.1% the speed of light (0.001 c) would require about 296 cubic meters of neutronium (a sphere of radius 4.1 m).
From this, one concludes that the propulsion system may be the largest energy consideration rather than the primary weapon. For example, consider DS1 enters our solar system from hyperspace (whose energetics are not considered here) and found itself near the orbit of mars. It would take two days for it to travel to earth at 0.001 c.
# Part I: Size of the Death Star in Episode IV
This is the first in a two part post where I calculate the size and mass respectively of the Death Star in Episode IV (DS1). At the end of Part II I will discuss thoughts about the energy source of DS1.
Part I: Size of DS1
Conventional wisdom from multiple sources places the size of DS1 to about 100-160 km in diameter. Based on an analysis of the station’s plans acquired by the Rebels, I estimate that the diameter of DS1 is 60 kilometers, not 100 km to 160 km. To bolster the case, this scale is compared to other scales for self-consistency, such as the width of the trench leading to the exhaust port in the Battle of Yavin. Part II of the post will focus on the mass of DS1 using related methods.
To estimate the size of DS1, I will begin with the given length scale of the exhaust port w = 2 m. This information was provided in the briefing prior to the Battle of Yavin where the battle strategy and DS1 schematics are presented. This scale, when applied to Figure 1, is consistent with the accepted length of an x-wing L = 12.5 m. I assume that the x-wing has an equal wingspan (there does not seem to be consistent values available). I am also assuming that the “small, one-man fighter” referred to in the briefing is an x-wing, not a y-wing. The x-wing is a smaller, newer model than the y-wing and it is natural to take that as the template. The self-consistent length scales of w and L will establish the length calibration for the rest of the analysis.
Figure 1: A close up view of the exhaust port chamber during final phase of the bombing run. The port width is given as w = 2 m. The length of the x-wing is L = 12.5 m. The forward hole, of length l, is then determined to be about 10 m.
From this, I extract the length of the smaller forward hole in Figure 1 to be approximately l = 10 m.
Figure 2: As the plans zoom out, a larger view of the exhaust port chamber of width t = 186 m. The first hole is shown with width l = 10 m. The scale of width l was determined based on information in Figure 1. The width of t was determined based on the scale of l.
Using l as a calibration, this establishes the exhaust port chamber in Figure 2 to be approximately t = 186 m.
In Figure 3a and Figure 3b, circles of different radii were overlaid on the battle plans until a good match for the radius was established. Care was taken to have the sphere’s osculate the given curvature and to center the radial line down the exhaust conduit. From here, the size of the exhaust port chamber, of width t, was used as a calibration to approximate the diameter of DS1 as D = 60 km (red). Several other circles are show in Figure 3 to demonstrate that this estimation is sensible: 160 km (purple), 100 km (black), and 30 km (blue). It is clear that a diameter of 160 km is definitely not consistent with station’s schematics. A diameter of 100 km is not wildly off, but is clearly systematically large across the range over the given arc length. 30 km is clearly too small.
While a diameter of 60 km may seem modest in comparison to the previously estimated 100 km to 160 km range, an appropriately scaled image of New York City is overlaid in Figure 4 to illustrate the magnitude of this systems in real-world terms; even a sphere of 60 km (red) is an obscenely large space station, considering this is only the diameter — more than adequate to remain consistentwith existing canon. The size of the main ring of the LHC (8.6 km) is overlaid in light blue, also for scale.
Figure 3a (to the right of the exhaust port chamber): As the plans zoom out further, the exhaust port chamber of width t = 186 m is shown with the curvature of DS1 (the square blob is the proton torpedo that has entered the port). The scale of t was determined based on information in Figure 2. Several circles with calibrated diameters based on the scales set in Figures 1 and 2 are shown. The 60 km diameter circle in red is arguably the best match to the curvature. Care was taken to match the point of contact of the circles to a common central location along the radial port.
Figure 3b (to the left of the exhaust port chamber): The same idea as Figure 3a. The 60 km diameter is still arguably the best match, although is a little shy on this side. The 100 km diameter, the next best candidate, is shooting higher than the 60 km is shooting low. Since an exact mathematical fit wasn’t performed, the expected radius is probably a bit higher than 60 km, but significantly lower than 100 km.
Figure 4: A 60 km diameter circle in red (with yellow diameter indicator) shown overlaid on a Google Earth image of the greater New York City region. The blue ring is an overlay of the scale of the Large Hadron Collider at CERN (about 8.5 km in diameter) — note the blue ring is not a scaled representation of the main weapon! The main message here is that a 60 km station, although smaller than the accepted 100-150 km, is still freakin’ HUGE. At this scale, there is only a rather modest indication of the massive urban infrastructure associated with New York City.
As another check on self-consistency, the diameter D is then used to calibrate the successive zooms on the station schematics, as shown in Figures 5 and 6. The length B = 10 km is the width of the zoom patch from Figure 5, X = 4.7 km is the length of the trench run, and b = 134 m is the width of one trench sector. From Figure 6, the width of the trench is estimated to be b’ = 60 m, able to accommodate roughly five x-wing fighters lined wingtip-to-wingtip. This indicates that the zoom factor is about 1000x in the briefing.
Figure 7 is a busy plot. It overlays several accurately scaled images over the 60 m trench, shown with two parallel red lines, to reinforce plausibility. Starting from the top: an airport runway with a 737 ready for takeoff (wingspan 34 m); a 100 m-wide yellow calibration line; a 60 m-wide yellow calibration line; the widths of an x-wing (green, Wx = 12.5 m, where I’ve assumed the wingspan is about the same as the length — there does not seem to be a consensus online; I’ve seen the value quoted to be 10.9 m, but it isn’t well-sourced) and tie fighter (red, 6.34 m); and a scaled image from footage of two x-wings flying in formation, with a yellow 60 m calibration line as well as a calibrated green arrow placed over the nearer one to indicate 12.5 m. As predicted, about five x-wings could fit across based on the still image. Also from this, the depth of the trench is estimated to also be 60 m. The scales are all quite reasonable and consistent. It is worth noting that if the station were 100 km, the next possible sensible fit to the arc length in Figure 3, the width of the trench would be about 100 m, twice the current scale. This would not be consistent with either the visuals from the battle footage or the airport runway scales.
In short, while there is certainly worthy critique of this work, I argue that, after a reasonably careful analysis of the stolen plans for DS1, all scales paint a self-consistent picture that the diameter of DS1 is very close to 60 km.
Figure 5: A zoom-out of DS1 in the briefing based on the stolen battle plans. D = 60 km is the diameter and B = 10 km is the width of the patch in the region of interest near the exhaust port.
Figure 6: A zoom in in the region of interest patch near the exhaust port channel (see Figure 5) with B = 10 km. the channel itself is about X = 4.7 km long. The width of the channel is about b = 134 m. Inset is a further zoom of the insertion point along the channel. Width of the channel itself is about b’ = 60 m.
Figure 7: A zoom of the insertion point along the channel for the bombing run. Several elements are overplayed for a sense of scale and for consistency comparisons. The red parallel lines represent the left and right edges of the channel. From the top of the figure is a 737 with a wing span of 34 m. The 737 is on a runway (at SFO). Down from the 737 is a yellow line that represents 100 m. This would be the width of the channel if D = 100 km, which is clearly much too large based on the battle plans. The next horizontal yellow arrow is the 60 m width based on the scales assumed with D = 60 m. Next down, embedded in the vertical lines of the runway: a green block representing the width of an x-wing and a red block representing the width of a tie fighter. Finally, at the bottom is a shot from the battle footage. It has been scaled so the edges of the walls match the width of the channel (shown as a horizontal yellow arrow). The width of the near x-wing is shown with a green horizontal arrow, which matches the expected scale of an x-wing.
# The Best Nest
The Best Nest by P.D. Eastman
The classic children’s book The Best Nest by children’s author P.D. Eastman, published in 1968, is one of the books that really sticks with me from my childhood.
The church in the town featured in The Best Nest
One of the big turning points in the story is when the birds find this wonderful space for their nest. It is huge. It has all sorts of great views of the area. The mother bird thinks it is the best place. However, we, the reader, know that something will go terribly wrong: the space is really a bell tower for a church. The papa bird goes out to find new materials for their nest while the mama sets up shop. Well, sure enough, a funky beatnik proto-hippy guy named Mr. Parker, comes to the church and rings the hell out of that bell like he has no other outlet for his life’s frustrations. The guy clearly loves his job. The papa bird comes back to find the place littered with bird feathers and no mama bird. He fears the worst and goes on a quest to find her.
Oberlander, R.D. #1, Waldoboro, Ma…
Before they find the bell tower, they look in other places for a new nest. One of the potential nests is a mailbox. Now, as I mentioned, as a kid I had particular fixations in details I would never had seen as an adult; conversely, in reading it to my children, I also found details I would never have found as a kid. For example, one of the reasons they decided not to pick the mailbox is that, while they were checking it out, a mailman comes by and puts some mail into the mailbox. Definitely not an ideal space for a pair of birds.
However, the piece of mail has an address on it (upside down in the text of the book):
…Oberlander
R.D. #1
Waldoboro, Ma…
Circa 2016, there is indeed an [Old] Road 1 in Waldoboro, Maine. There is also an Oberlander family name that appears in that town’s older records. That’s sort of neat. Naturally, using Google Streeview, I wandered around to see if I could find the church where the bell tower was. While not definitive, I have two candidates. Sure, these churches are pretty generic shapes for the area. Nevertheless, with a specific town to focus on, you can be pretty sure it must be one of two churches, or a composite, that P.D. used as a template. He could have also just made something up from memory or imagination.
The first one, Broad Bay Congregational Church, has the correct weathervane, the correct three-window structure, a circular region in the middle, and an obvious bell tower. It also has a front that is roughly consistent with the drawing, although obviously updated (e.g. it has two windows on each side of the door).
Waldoboro Broad Bay Congregational Church 941 Main St, Waldoboro, Maine
The second one, Waldoboro United Methodist Church, also has the three window configuration in the side, has similar slats near the bell tower as the drawing in the story (the slats were one of the weirdly specific things I fixated on as a child), and a pointy tower that resembles the one in the drawing. But it does not have the right window configuration, the weathervane, nor the circular slats.
Waldoboro United Methodist Church (side view) 85 Friendship Street (Route 220), Waldoboro, Maine
Waldoboro United Methodist Church (front view), 85 Friendship Street (Route 220), Waldoboro, Maine.
My hunch is that the first one, Broad Bay Congregational Church, is the one in the story. I suspect that during the time since P.D. Eastman wrote the story (circa 1968), it has had a few upgrades.
But, as I said earlier, these are very common generic “Protestant-style” East Coast churches. The story might have nothing to do with these specific churches.
Anyway, I had fun with this little distraction. If anyone knows more about this Easter egg planted by P.D. Eastman, about any connection he may have had to the Waldoboro region, or the reason he might have picked “Oberlander” for the recipient of the letter on R.D #1, I’d love to hear about it.
I just listened to David Brin’s excellent Uplift War on audiobook, and unapologetically declare to the world that I “read the book.” For full disclosure, I’ve also read Martian Chronicles by Ray Bradbury, all three Hunger Games novels by Suzanne Collins, Sara Gruen’s Water for Elephants, Neverwhere by Neil Gaiman, Packing for Mars by Mary Roach, and Letters to a Young Contrarian by Christopher Hitchens, amongst others. All on audiobook. From a social point of view, if you and I got together to discuss these works, my experience of them would be such that you would not be able to determine by our conversation if I listened to it or physically read the words on a page. In this sense, I can responsibly claim to have “read the book” even if my eyes never looked at the words. That is, using Brin’s work as an example, unless you asked me to spell the names “Uthacalthing”, “Tymbrimi”, or “Athaclena” (which I did not know how to spell until I looked them up just now) — but then I’d ask you to pronounce them and we’d be even.
Do all books lend themselves to this audio mode of reading? No. Obviously not. Exceptions include works that rely directly on the shapes of words or encoding extra information in the precise layout of the text, font, or presentation. If the work involves lots of pictures, illustrations, data, or equations, audiobooks are not going to work very well. But the bulk of modern fiction lends itself wonderfully to audiobooks as does much non-fiction. Like so many other things in life, one needs to account for individual cases. Also, this equivocation is not appropriate for people (e.g. children) learning to read symbols on the page. An audio experience is not an adequate substitute for that kind of information processing during those fragile formative years. This argument is directed at people who have mastered both reading and listening and are educated adults.
A couple tangential examples, that inform the discussion. A formally trained and competent musician can look at a piece of written music and, for all practical purposes, “listen” to it by reading it with their eyes. The audio performance itself, of course, also has aesthetic value for that musician. But it would probably be appropriate for someone in that position to say, in either context, that they “listened” or “heard” the piece even if it merely involved reading the sheet music. Indeed, musicians who can read written music like that do refer to reading sheet music as having “heard” or having “listened” to the piece. In contrast, many bands we worship refer to “writing” music for their albums. However, rarely are any notes or music written down in any formal sense. Many rock/pop bands “write” music by playing it and piecing together sections into things than sound nice after editing (if they are lucky). Later, some music grad student, desperate to eat and pay rent, will be hired by a company to transcribe the sounds on the album into written notes, so other people without ear training can also play the songs; but that isn’t the way the band itself usually “writes” music — unless you are Yes or Dream Theater. If the Rolling Stones speak of “writing” music for a new album, they almost certainly mean a wanton, drug-infused geriatric orgy in the Caribbean that might have involved Keith Richards bringing his guitar. But the term “writing music” is still used. We can also reverse the situation and look at words on a page that were meant to be spoken out loud, such as plays. Take Shakespeare. Certainly the stage play is considered a respectable form of literary art and Shakespeare is arguably the greatest writer of the English language. But the plays he wrote were meant, designed, crafted to be read aloud and listened to. Yet we read them. Can you still read Shakespeare and claim to have experienced the work in an intellectually satisfying way and be conversational about it? Obviously. Does the stage work bring the work to life in a different way? Clearly.
Also, reading words on a page is not itself a magic recipe for intellectual absorption. Reading text can be pathologically passive if one is not actively engaged, and does not imply extra profound and deep understanding. Let me give an example from my own experience in the classroom. I tell students to “read chapter 10” from the text. And, indeed some do look at it with their eyes and the words are streamed through their thinking in some fashion. But in many cases no cognitive engagement has occurred. By speaking to them, I can tell that they did not, in fact, “read” the text as I meant the term “read.” In this context, “read” did not necessarily literally mean merely looking at the words, although it might conveniently involve that biomechanical process. I really just wanted them to come to class having processed and understood the material provided in the book by whatever means necessary. If that involves listening to the audiobook, it just doesn’t matter to me (although, good luck learning quantum mechanics from an audiobook).
Does watching a movie adaptation of a book count as “reading the book?” Not in my opinion. Putting audiobooks in the same category as movie interpretation of books is missing the point. I claim the unabridged audiobook is not, fundamentally, a different medium than the original work — not any different than the braille modes of reading that are considered “legitimate” reading. When we read books using the written word we are, in fact, “speaking” the words to ourselves in our head anyway exactly in the way the book is being read in an audiobook. A movie, even one adapted to be nearly identical to the book, is usually abridged and has been altered from the original work in fundamentally different ways. Moreover, one is not required to visualize the plot and characters in the same way as one does in reading text or listening to a reading of text.
I am not judging all these different modes or ranking them. They each serve their purpose and can give pleasure and intellectual stimulation in their own way. But I argue that, under many common situations, listening to audiobooks accomplishes the same social and intellectual function as reading text and can thus be responsibly declared a form of “reading the book.”
# The Universe: A Computer Simulation?
An unpublished paper on the arXiv is claiming to have formulated a suite of experiments, as informed by a particular kind of computer approximation (called “lattice QCD” or L-QCD), to determine if the universe we perceive is really just an elaborate computer simulation. It is creating a buzz (e.g. covered by the Skeptics Guide to the Universe, Technology Review, io9, and probably elsewhere).
I have some problems with the paper’s line of argument. But let me make it clear that I have no fundamental problem with the speculation itself. I think it is a fun and interesting to ponder the possibility of living in a simulation and to try and formulate experiments to demonstrate it. It is certainly an amusing intellectual exercise and, at least in my own experience, this was an occasional topic of my undergraduate years. More recently than my undergraduate years, Yale philosopher Nick Bostrom put forth this famous arguments in more quasiformal terms, but the idea had been hovering there (probably with a Pink Floyd soundtrack) for a long time.
The paper is not “crackpot”, but is highly speculative. It uses a legitimate argumentation technique, if used properly (and the authors basically do), called reductio ad absurdum: reduction to the absurd. Their argument goes like this:
1. Computer simulations of spacetime dynamics, as known to humans, always involve space and time lattices as a stage to perform dynamical approximations (e.g. finite difference methods etc.);
2. Lattice QCD (L-QCD) is a profound example of how (mere) humans have successfully simulated, on a lattice, arguably the most complex and pure sector of the Standard Model: SU(3) color, a.k.a. quantum chromodynamics, the gauge theory that governs the strong nuclear force as experienced by quarks and gluons;
3. L-QCD is not perfect, and is still quite crude in its absolute modern capabilities (I think most people reading these articles, given the hype imparted to L-QCD, would be shocked at how underwhelming L-QCD output actually is, given the extreme amount of computing effort and physics that goes into it). But it is, under the hood, the most physically complete of all computer simulations and should be taken as a proof-of-principle for the hypothetical possibility of bigger and better simulations — if we can do it, even at our humble scale, certainly an übersimulation should be possible with sufficient computing resources;
4. Extrapolating (this is the reductio ad absurdum part), L-QCD for us today implies L-Reality for some other beyond-our-imagination hypercreatures: for we are not to be taken as a special case for what is possible and we got quite a late start into the game as far as this sentience thing goes.
5. Nevertheless, nuanced flaws in the simulation that arise because of the intrinsic latticeworks required by the approximations might be experimentally detectable.
Cute.
Firstly, there is an amusing recursive metacognative aspect to this discussion that has its own strangeness; it essentially causes the discussion to implode. It is a goddamn hall of mirrors from a hypothesis testing point of view. This was, I believe, the point Steve Novella was getting at in the SGU discussion. So, let’s set aside the question of whether a simulation could
1. accurately reconstruct a simulation of itself and then
2. proceed to simulate and predict its own real errors and then
3. simulate the actual detection and accurate measurement of the unsimulated real errors.
Follow that? For the byproduct of a simulation to detect that it is part of an ongoing simulation via the artifacts of the main simulation, I think you have to have something like that. I’m not saying it’s not possible, but it is pretty unintuitive and recursive.
My main problem with the argument is this: a discrete or lattice-like character to spacetime, with all of its strange implications, is neither a necessary nor sufficient condition to conclude we live in a simulation. What it would tell us, if it were to be identified experimentally, is that: spacetime has a discrete or lattice-like character. Given the remarkably creative and far-seeing imaginative spirit of the project, it seems strangely naive to use such an immature, vague “simulation = discrete” connection to form a serious hypothesis. There very well may be some way to demonstrate we live in a simulation (or, phrased more responsibly, falsify the hypothesis that we don’t live in a simulation), but identifying a lattice-like spacetime structure is not the way. What would be the difference between a simulation and the “real” thing. Basically, a simulation would make error or have inexplicable quirks that “reality” would not contain. The “lattice approximation errors” approach is pressing along these lines, but is disappointingly shallow.
The evidence for living in a simulation would have to be much more profound and unsubtle to be convincing than mere latticworks. Something like, somewhat in a tongue-and-cheek tone:
1. Identifying the equivalent of commented out lines of code or documentation. This might be a steganographic exercise where one looks for messages buried in the noise floor of fundamental constants, or perhaps the laws of physics itself. For example, finding patterns in π sounds like a good lead, a la Contact, but literally everything is in π an infinite number of times, so one needs another strategy like perhaps π lacking certain statistical patterns. If the string 1111 didn’t appear in π at any point we could calculate, this would be stranger than finding “to be or not to be” from Hamlet in ASCII binary;
2. Finding software bugs (not just approximation errors); this might appear as inconsistencies in the laws of physics at different periods of time;
3. Finding dead pixels or places where the hardware just stopped working locally; this might look like a place where the laws of physics spontaneously changed or failed (e.g. not a black hole where there is a known mechanism for the breakdown, but something like “psychics are real”, “prayer works as advertised”, etc.);
I’m just making stuff up, and don’t really believe these efforts would bear fruit, but those kinds of thing, if demonstrated in a convincing way, would be an indication to me that something just wasn’t right. That said, the laws of physics are remarkably robust: there are no known violations of them (or nothing that hasn’t been able to be incorporated into them) despite vigorous testing and active efforts to find flaws.
I would also like to set a concept straight that I heard come up in the SGU discussion: the quantum theoretical notion of the Planck length does not imply any intrinsic clumpiness or discreteness to spacetime, although it is sometimes framed this way in casual physics discussions. The Planck length is the spatial scale where quantum mechanics encounters general relativity in an unavoidable way. In some sense, current formulations of quantum theory and general relativity “predict” the breakdown of spacetime itself at this scale. But, in the usual interpretation, this is just telling us that both theories as they are currently formulated cannot be correct at that scale, which we already hypothesized decades ago — indeed this is the point of the entire project of M-theory/Loop quantum gravity and its derivatives.
Moreover, even working within known quantum theory and general relativity, to consider the Planck length a “clump” or “smallest unit” of spacetime is not the correct visualization. The Planck length sets a scale of uncertainty. The word “scale” in physics does not imply a hard, discrete boundary, but rather a very, very soft one. It is the opposite of a clump of spacetime. The Planck length is then interpreted as the geometric scale at which spacetime is infinitely fuzzy and statistically uncertain. It does not imply a hard little impenetrable region embedded in some abstract spacetime latticeworks. This breakdown of spacetime occurs at each continuous point in space. That is, one could zoom into any arbitrarily chosen point and observe the uncertainty emerge at the same scale. Again, no latticeworks or lumpiness is implied.
# Forwards Backwards Songs
As a musical exercise, I’ve been experimenting with taking standard, famous songs and arranging a backwards version as a forward song (“backwards forwards” for short). I don’t use the original recording in any way, only part of the musical arrangement. The result is usually something that (naturally) has weird overtones of the original song, but also is a unique song in its own right.
My first effort was at tune called My Sweet Satan, an instrumental backwards forwards version of Stairway to Heaven by Led Zeppelin. The title is a play off of the famous backmasking fiasco that followed the song through its heyday and beyond. The haunting refrain of “my sweet Satan” can apparently be heard in verse five (somewhere around “There’s still time to change the road you’re on”). However, if you listen, it is clearly a combination of audio pareidolia and straightforward phonetic reversal rather than active backmasking. One can, of course, carefully craft forwards lyrics that do have phonetic reversals that sound like actual messages when played backwards. But Robert Plant’s lyric clearly isn’t that. I’ve tried doing such constructions myself and, with a specific backwards message in mind, you certainly don’t get anything nearly as coherent as the lyrics to Stairway (and that’s saying something).
Another effort is called But You Can Never Leave. Can you guess which backwards forward song it might be? A hint is that it is a song known (apocryphally) for having a backmasked message. The biggest clue is in the title.
If you like what you hear, take a look at my latest album called Pretty Blue Glow by Agapanthus and consider purchasing it (or parts you like). You can also find many of my tunes on Sutros under Agapanthus for free.
# Twilight Zone Revisited
Thanks to this Cracked article, I decided to revisit all the old Twilight Zone episodes streamed on Netflix. Wonderful!
# Mathematica One-Liner Competition 2012
Decided to enter Wolfram’s Mathematica One-Liner Competition 2012: “What can you do with one line of code?” That is, in under 140 characters (making it tweetable). Why, a Particle Zoo Calliope, of course! My entry (only slightly modified from that submitted):
SectorChart[ Button[{1, p[#, s]}, EmitSound@Sound@SoundNote@{2 p[#, s], Floor@p[#, "Mass"]^.3}] /.s -> "Spin" & /@ ParticleData[] /. p -> ParticleData]
W00t! Received an Honorable Mention! (the competition was fierce, lots of good one-liners). Give it a try below. You will need the free Mathematica CDF plugin installed. A figure will be generated. It is a musical instrument. Click on different locations on the figure to play different intervals. The first click is sometimes a bit awkward/slow, but after that it should play in real time.
Description:
A sector plot is generated based on the spin of all the known elementary particles (quarks, leptons, and gauge bosons) and the hadronic bound states (bayrons and mesons). The length of the tine on the sector plot is proportional to the particle’s intrinsic spin. There are around 1000 particles in the database. When you click on one of the sectors, representing a particle, two tones are played based on the spin and the mass of that particle. The mapping from values to notes is arbitrary, but selected to be “listenable.” I take two times the particle’s spin as one note and the integer part of the particle’s mass to the 0.3 as the second (this was selected by trial and error to give a reasonable range of tones for the full particle mass spectrum). A value of “0” is considered middle C and each integer above and below is a half-step. | open-web-math/open-web-math | |
# Please guide me : i have 0.5 grams of solid waste sample. It is acid digested and then diluted to 500 ml. This is tested for chromium by atomic absorption and the conc came out to be .243 mg/l. what will be the actual concentration in the solid sample?
By atomic absorption it is determined that the conc of chromium in 500 mL solution is $0.243 \text{ mg/L}$ .
So the mass of chromium present in 500mL or 0.5L solution will be $0.243 \frac{\text{mg}}{L} \times 0.5 L = 0.1215 m g$
And this amount of chromium has come from $0.5 g = 500 m g$ of solid waste sample.
=0.1215/500xx100%=0.0243% | HuggingFaceTB/finemath | |
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Z-test is a statistical method to determine whether the distribution of the test statistics can be approximated by a normal distribution. It is the method to determine whether two sample means are approximately the same or different when their variance is known and the sample size is large (should be >= 30).
When to Use Z-test:
• The sample size should be greater than 30. Otherwise, we should use the t-test.
• Samples should be drawn at random from the population.
• The standard deviation of the population should be known.
• Samples that are drawn from the population should be independent of each other.
• The data should be normally distributed, however for large sample size, it is assumed to have a normal distribution.
Hypothesis Testing
A hypothesis is an educated guess/claim about a particular property of an object. Hypothesis testing is a way to validate the claim of an experiment.
• Null Hypothesis: The null hypothesis is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value. We either reject or fail to reject the null hypothesis. Null Hypothesis is denoted by H0.
• Alternate Hypothesis: The alternative hypothesis is the statement that the parameter has a value that is different from the claimed value. It is denoted by HA.
Level of significance: It means the degree of significance in which we accept or reject the null-hypothesis. Since in most of the experiments 100% accuracy is not possible for accepting or rejecting a hypothesis, so we, therefore, select a level of significance. It is denoted by alpha (∝).
Steps to perform Z-test:
• First, identify the null and alternate hypotheses.
• Determine the level of significance (∝).
• Find the critical value of z in the z-test using
• Calculate the z-test statistics. Below is the formula for calculating the z-test statistics. • where,
• X¯: mean of the sample.
• Mu: mean of the population.
• Sd: Standard deviation of the population.
• n: sample size.
• Now compare with the hypothesis and decide whether to reject or not to reject the null hypothesis
Type of Z-test
• Left-tailed Test: In this test, our region of rejection is located to the extreme left of the distribution. Here our null hypothesis is that the claimed value is less than or equal to the mean population value. • Right-tailed Test: In this test, our region of rejection is located to the extreme right of the distribution. Here our null hypothesis is that the claimed value is less than or equal to the mean population value. • Two-tailed test: In this test, our region of rejection is located to both extremes of the distribution. Here our null hypothesis is that the claimed value is equal to the mean population value. Below is the example of performing the z-test:
Problem: A school claimed that the students’ study that is more intelligent than the average school. On calculating the IQ scores of 50 students, the average turns out to be 11. The mean of the population IQ is 100 and the standard deviation is 15. State whether the claim of principal is right or not at a 5% significance level.
• First, we define the null hypothesis and the alternate hypothesis. Our null hypothesis will be: and our alternate hypothesis. • State the level of significance. Here, our level of significance given in this question (∝ =0.05), if not given then we take ∝=0.05.
• Now, we look up to the z-table. For the value of ∝=0.05, the z-score for the right-tailed test is 1.645.
• Now, we perform the Z-test on the problem: • Where:
• X = 110
• Mean (mu) = 100
• Standard deviation (sigma) = 15
• Significance level (alpha) = 0.05
• n = 50 • Here 4.71 >1.645, so we reject the null hypothesis. If z-test statistics is less than z-score, then we will not reject the null hypothesis.
## Python3
# imports import math import numpy as np from numpy.random import randn from statsmodels.stats.weightstats import ztest # Generate a random array of 50 numbers having mean 110 and sd 15 # similar to the IQ scores data we assume above mean_iq = 110 sd_iq = 15/math.sqrt(50) alpha =0.05 null_mean =100 data = sd_iq*randn(50)+mean_iq # print mean and sd print('mean=%.2f stdv=%.2f' % (np.mean(data), np.std(data))) # now we perform the test. In this function, we passed data, in the value parameter # we passed mean value in the null hypothesis, in alternative hypothesis we check whether the # mean is larger ztest_Score, p_value= ztest(data,value = null_mean, alternative='larger') # the function outputs a p_value and z-score corresponding to that value, we compare the # p-value with alpha, if it is greater than alpha then we do not null hypothesis # else we reject it. if(p_value < alpha): print("Reject Null Hypothesis") else: print("Fail to Reject NUll Hypothesis")
Reject Null Hypothesis
Two-sampled z-test:
In this test, we have provided 2 normally distributed and independent populations, and we have drawn samples at random from both populations. Here, we consider u1 and u2 be the population mean X1 and X2 are the observed sample mean. Here, our null hypothesis could be like: and alternative hypothesis and the formula for calculating the z-test score: where sigma1 and sigma2 are the standard deviation and n1 and n2 are the sample size of population corresponding to u1 and u2 .
Type 1 error and Type II error:
• Type I error: Type 1 error has occurred when we reject the null hypothesis, even when the hypothesis is true. This error is denoted by alpha.
• Type II error: Type II error has occurred when we didn’t reject the null hypothesis, even when the hypothesis is false. This error is denoted by beta.
My Personal Notes arrow_drop_up | HuggingFaceTB/finemath | |
PowerPC assembly language beginners guide.
## Chapter 3
Logical Operations
AND
Logical operations are quite simple to understand and form a useful tool within the programmers armory. An "and" operation simply says if both A and B are equal to a logical 1, then set the result to a 1. Logical operations work at a bit level - each bit of both operands is logically tested and the same numbered bit in the destination is set or cleared as a result of the logical operation as applied to each bit. We humans can work them out serially; by starting at the first bit of each operand and making a note of the result of each operation. The CPU operates on all bits in parallel, hence all logical operations (in common with most PowerPC instructions) have an effective processing time of one cycle. The exception to this timing are integer multiplies and divides - avoid these like the plague if you can or use the FPU which is a multiply-add unit and can perform single sized (32 bit) multiplies in one clock cycle effectively and double sized (64 bit) multiplies in two.
Back to the logic...
Suppose we "and" 1 and 9. If we look at the number in binary, 1 is 0001 and 9 is 1001. When these two numbers are anded, the processor looks at the numbers like this:
```3210 <- Bit number 0001 <- 1 in binary 1001 <- 9 in binary 0001 <- result```
First it will look at both bit 3's. It says I have a 1 and a 0, so I don't have two 1's. Therefore the result is zero. Then it looks at bit 2's, which are both zero, so the result is zero. Bits 1 are both zero, so the result is zero. Bits 0 are both 1. It says if I have a 1 "and" a 1 then the result is 1, so bit zero result is 1.
The result of 9 anded with 1 is 1.
The and operation can be summarised by saying "only if both bits are set will the result be set"
Examine the following examples, to see if you can spot the main use of the "and" instruction.
What is the result of 0xFA anded with 0x0F?
```0xFA = 11111010
0x0F = 00001111
AND =
00001010 = 0x0A```
What is the result of 0xF1 anded with 0x0F?
```0xF1 = 11110001
0x0F = 00001111
AND = 00000001 =
0x01```
What is the result of 0x1220 anded with 0x00FF
```0x1220 = 0001001000100000
0x00FF = 0000000011111111
AND =
0000000000100000 = 0x0020```
The and instruction is mostly used to mask off wanted data in a register. By setting bits in the mask that identifies the bits you want to keep, and then anding this mask with the data, the bits you are not interested in will be set to zero.
For example if you had a routine that returns the ASCII value of a key pressed on the keyboard, and it returned the key in r3. The key can be specified in a byte, but there may be data from earlier processing in the upper three bytes of r3 - so to ensure you don't create errors further in the program, the byte can be masked off with the and instruction as follows:
andi. r3,r3,0xff
Irrespective of how much garbage is in the upper 24 bits of r3, after this instruction all that will be left in r3 is the byte defining the key press, the lower 8 bits. Note in particular the trinary operands and the dot at the end of the instruction.
If we wanted we could leave the contents of r3 intact and place the results of the AND operation in another register, say r4, with an instruction such as:
andi. r4,r3,0xff
The dot means that this instruction always affects the condition code register; a note is made of the result of the operation. If for example, the result of the AND was that all bits were cleared, then the Z (or zero) flag inthe condition code register was set. We will examine the condition code register in more detail later, but for now just note that there are 8 integer condition register "fields". An instruction followed by a dot means that the condition code register field 0 is affected - it notes the result of the operation. NOTE also that immediate type instructions such as andi. always work with 16 bit unsigned immediate data. There are shifted forms of the instructions which will affect the upper 16 bits of a 32 bit register, as an example see the movei macro from chapter 2.
#### where instruction is the mnemonic, [.] if present means you can use either the dotted or undotted form where the dotted form will affect the condition flags (cr0), rx,ry,rz are general purpose registers, fn means a floating point register, ui is an unsigned integer quantity (normally 16 bit, noted if different) and si is a signed integer quantity.
Examples of possible and instructions:
AND - and[.] rx,ry,rz
The contents of register rz is anded with register ry and the resultant 32 bit pattern is stored in rx.
AND with complement - andc[.] rx,ry,rz
The contents of register ry are anded with the complement of rz and the resultant 32 bit pattern is stored in rx. Complement means the inverse of - for example %1010 when complemented becomes %0101. Not to be confused with two's complement which is how computers subtract via addition. Two complement means to invert the data and then add 1.
AND Immediate - andi. rx,ry,ui
The contents of register ry are anded with the ui and the result stored in rx. Note that in this case, the upper 16 bits of the result will be cleared because ui is a 16 bit quantity! Note that this instruction is always dotted.
AND Immediate shifted - andis. rx,ry,ui
This is basically the same as andi, except the ui is shifted left 16 bits before being anded with ry, so the lower 16 bits of rx will always be cleared after this instruction.
OR
The OR instruction works like this:
If either or both of the bits are 1, then the result bit is a 1. The other way of looking at it is "If both bits are a zero then the result is a zero, otherwise its a 1".
Example - OR 1 with 2
```1 = 0001 2 = 0010 OR= 0011 = 0x03```
In PowerPC, the possible instructions are basically the same as and as follows:
OR immediate - ori rx,ry,ui (note no dot allowed unlike andi. which must have one)
OR immediate shifted - oris rx,ry,ui (note no dot allowed unlike andis. which must have one)
OR - or[.] rx,ry,rz
OR with complement - orc[.] rx,ry,rz
Fantasm uses the ori instruction to provide the "extended mnemonic" nop which stands for NO Operation:
nop is the same as ori 0,0,0
The or instruction is used to provide the useful extended mnemonic mr, which means move from one register to another:
mr rx,ry - move the contents of ry into rx and is the same as or rx,ry,ry
XOR
The eXclusive OR instruction works like:
"If one bit is a 1 and one bit is 0 then the result is 1, otherwise the result is 0".
Example - XOR 1 with 15
```1 = 0001
15 = 1111
XOR= 1110 = decimal 14 or
0x0e```
XOR 0 with 1
```1 = 0000
0 =
0001
XOR= 0001, so the result is 1.```
If we XOR the result with 1 again we get a 0 because both bits are now a 1. This is a neat way of toggling a bit, every time a loop executes for example. Initially the bit is set to 1. Each time round the loop, the bit is XOR'd with 1. Every time the loop executes. if the bit is a 1 its set to a 0, and if its a 0 its set to a 1.
What'sthe use of this? Suppose you want to flash something, say an alien spaceship on the screen between red and yellow. You simply xor the colour control bit with 1 and if it was a 1, use the colour red or if it was a zero use the colour yellow.
Possible XOR instructions:
xor immediate - xori rx,ry,ui
xor immediate shifted - xoris rx,ry,ui
xor - xor[.] rx,ry,rz
Miscellaneous logical instructions
NOR - NOT OR
NOT means to invert, so NOR performs an OR operation on the two operands, then inverts the result and stores it in the destination register:
nor[.] rx,ry,rz - the contents of ry are ORed with the contents of rz, then the result is inverted and stored in rx.
Fantasm uses the NOR instruction to provide the "extended instruction" NOT:
not rx,ry is the same as nor rx,ry,ry
NAND - NOT AND
NAND performs and AND operation on the two operands, inverts the result and stores it in the destination register.
nand[.] rx,ry,rz
EQV - Equivalent
eqv[.] rx,ry,rz - the contents of ry are XORed with the contents of rz, the result is inverted and stored in rx.
Sign extension instructions
Extend sign byte, Extend sign halfword, Extend sign word
extsb[.] rx,ry
extsh[.] rx,ry
extsw[.] rx,ry
These instructions copy the sized data (byte, half or word) to another or the same register and sign extend the result, so
extsb rx,ry copies the byte out of ry into rx and copies bit 7 to bits 8 through 31 of rx.
extsh rx,ry copies the lower 16 bits out of ry into rx and copies bit 15 to bits 16 through 31 of rx.
extsw rx,ry is a 64 bit instruction that is illegal on 32 processors such as 601/3/4. It copies the lower 32 bits out of ry into rx and copies bit 31 to bits 32 through 63 of rx.
NOTE that Fantasm 5 will assemble most 64 bit instructions (i.e. PowerPC 620 processor) just fine - you can turn on warnings about their use from Fantasm's preferences pane.
Count Leading Zeros (Word or Double)
cntlzw[.] rx,ry
cntlzd[.] rx,ry
These two instructions, one for 32 bit architectures and one for 64 bit (the double form) counts how many zeros from the leftmost bit position until the first binary 1 is encountered. The count is stored in the destination register.
We mentioned about the dotted form of some instructions and that if the dot was used the condition flags would be set. Your first question may be why not use them all the time? The answer is simply that to set the condition register flags the processor needs to do more work. Sometimes you may not care if the result of a logical operation was zero or not - you just want the data. In this case you would not use the dotted form. Some times you do need to know so you can make a decision based on the outcome of the operation; in this case you would use the dotted form.
This will form the basis of the next chapter, where we'll be looking at the processor in more detail and examining the condition register and branch processor in general along with a closer look at the whole architecture and introducing the branch processor instructions. We will then have enough information "under our belts" to be able to produce some rudimentary programmes to introduce more integer instructions.
In the mean-time, it would be worthwhile revising the first three chapters and if you're not subscribed to the Fantasm list server maybe you'd like to consider it. Some very useful discussions crop up from time to time.
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Learning Library
# Measuring Sugar in Soda
### Objective:
This experiment will examine and precisely measure the amount of sugar in various sodas.
### Research Questions:
-What is the percentage of sugar in different kinds of soda?
-Do all sodas have the same sugar content?
### Introduction:
We all know that sugar is filled with soda, but it’s hard to see just how much is in our drink by simply reading the label. Using a scientific instrument called a hydrometer, you can find out by precisely measuring the sugar content. Technically, a hydrometer measures any solids in a liquid, but since nearly all the dissolved solids in soda is sugar, it will work for this experiment.
### Materials:
• Floating hydrometer
• Several empty 2-liter bottles
• Several sodas of various brands and flavors
• Bag of pure sugar
• Funnel
• Kitchen scale
### Experimental Procedure:
1. Remove the carbon dioxide from each soda by dumping it back and forth between two clean containers until it stops bubbling.
2. Weigh out 100 grams of sugar with the kitchen scale, and use the funnel to pour the sugar into one of the bottles. Fill the bottle with 2 liters of water. Shake until all the sugar is dissolved. This is your 5% sugar solution.
3. Repeat steps 2 and 3 in two other bottles, with 200 grams and 500 grams of sugar. These are your 10% and 25% sugar solutions, respectively.
4. Fill one bottle with just water.
5. Place the hydrometer in the only water solution, and record the measurement. Wash and dry the hydrometer.
6. Repeat step 6 for bottles with the sugar solutions, and then the sodas.
7. Now it’s time to analyze your data. Using the sugar solutions as guides, calculate how many grams of sugar are in each soda and what the percentage of sugar is for each soda. Does it match up with the label? Are certain flavors or brands sweeter than others?
Concepts: solid measurement, sugar crystals, density
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Chapters
In the previous section on histograms and cumulative frequency polygons, we walked you through row and column frequency and how you could plot these frequencies by building dot plots, histograms and frequency polygons. In addition, we also introduced the basics of interpreting these plots. In this section, we’ll dive deeper into how to interpret histograms as well as provide you with some practice problems.
## Frequency Overview
Recall that there are three main types of frequencies: absolute, row and column. In the table below, you’ll find a brief summary of each as well as a description of when they should be used.
Calculation Description Uses Absolute The amount of times a variable occurs out of the total sample size When you want to compare variables between each other compared to the total Row The frequency of a variable out of the row total For comparison of one factor across the row total Column The frequency of a variable out of the column total For comparison of one factor across the column total
As you can see from the table above, frequencies can be used on a number of occasions. In fact, the application of frequencies can be seen in data visualizations such as dot plots, histograms and frequency polygons. Take a look at the three examples given below illustrating the differences between the three.
Group A Group B Row Total Song 1 136 195 331 Song 2 220 124 344 Song 3 100 204 304 Column Total 456 523 979
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## Absolute Frequency
As you can see in the image above, absolute frequency is found by dividing every individual value by the total sample size.
## Row Frequency
From the image above, observe the row frequency is simply the individual value over the row total.
## Column Frequency
From the image above, you can see the column frequency is simply the individual value over the column total.
## Interpreting Histograms and Frequency Polygons
If you recall, histograms and frequency polygons give us information about the distribution of the data set. The distribution of a data set is how the data is spread. The distribution of a data set in descriptive statistics has three main characteristics:
• Centre
• Skew
The first two characteristics deal with the main tools of descriptive statistics: measures of central tendency and of spread. These include measures such as the mean, standard deviation, mode and more. Skew, as we’ve discussed in previous sections, has to do with the position of the data points. In our section on absolute cumulative frequency distribution, we discussed these characteristics in depth, whose properties can be summarized in the table below.
Characteristic Interpretation Centre The centre describes the centre point of the data. If there is one, you should describe: How many there are Where it is located Use measures of central tendency Spread The spread of the data is how the data is distributed around the centre. You can use things like: The shape of the distribution If it is widely spread or clustered around the mean Use measures of variability Skew The skew of the data is when a large portion of the data is located to one side while there are a few extreme values to the other side. You can describe skew as: Asymmetrical Having outliers
The interpretation of both boxplots and frequency polygons can be done through these three characteristics.
## Problem 1
You want to display data on the different amounts of soda that are bought each day of the week. Given the data table below, build a frequency polygon.
Monday Tuesday Wednesday Thursday Friday Soda Bought 34 69 45 74 96
## Problem 2
You’re interested in making your data as clear and understandable as possible. Based on the data below, what do you think is the appropriate number of intervals, or “bins”, to group the data in?
Value Frequency 10 3 12 2 14 6 16 9 18 12 20 38 22 3 24 37 26 4 28 9 30 10 32 5 34 7 36 5 38 8 40 3 42 79 44 24 46 54 48 56 50 77 52 34 54 5 56 56 58 78 60 7 62 4 64 24 66 52 68 7 70 43 72 8 74 6 76 85 78 7 80 67
## Problem 3
Interpret the characteristics of the distribution of the following histogram. State at least one aspect of the spread, centre and skew. A data table has been provided in order to ease the interpretation.
Shoe Size Frequency 20-23 31 24-27 107 28-31 253 32-35 175 36-39 103 40-43 54 44-47 34 48-51 20
## Solution Problem 1
In this problem, you were asked to construct a frequency polygon from the data table provided. You should have come up with a chart similar to the one in the image below.
Where we can see the number of sodas bought increases throughout the week, with a slight dip in sales on Wednesday, and the highest number of sales made on Friday.
## Solution Problem 2
In this problem, you were asked to decide the appropriate number of bins to display the data. This is an important part of displaying data because, when choosing too few or too many bins, important information about your data can get lost.
Here, a good width for the interval would be 8. This would give us 9 bins, which can be illustrated by the histogram below.
## Solution Problem 3
For this problem, you were asked to interpret the characteristics of the distribution below.
As we can see from the table, this distribution charts the frequency of shoe sizes. First, we’ll tackle the centre of the distribution. There is one centre and it is located somewhere between 28 and 35. While we cannot say for sure what the mean, mode and median are without calculating them for the grouped data, we can say that the modal group is 28-31.
The spread is distributed unevenly around the mean, with more of the data set located to the right of the centre than to the left. While the picture suggests that the data are approximately normal, there seems to also be a slight right skew.
The platform that connects tutors and students | HuggingFaceTB/finemath | |
## What quadrilateral has 4 congruent sides but no congruent angles?
A rhombus is a parallelogram with four congruent sides. The plural of rhombus is rhombi .
What quadrilateral has no 4 right angles?
Other Types of Quadrilaterals Different from a rectangle, a parallelogram does not have to have four right angles. A rhombus is a quadrilateral where all four sides are equal in length. Different from a square, a rhombus does not have to have four right angles.
### What shapes have all 4 sides congruent?
1 Expert Answer Correct answer is a rhombus! A rhombus is a parallelogram with 4 congruent sides and it always has four congruent sides.
What is a shape with 4 congruent sides called?
Any figure with exactly 4 sides is called a quadrilateral. Every type of figure on here is a type of quadrilateral. Congruent sides have the same length.
## What parallelogram has no right angles?
Parallelogram
This parallelogram is a rhomboid as it has no right angles and unequal sides.
Edges and vertices 4
Symmetry group C2, [2]+,
What is a quadrilateral with 4 congruent sides?
Rhombus: A quadrilateral with four congruent sides; a rhombus is both a kite and a parallelogram.
### Which polygon has no right angles?
rhombus
The third special parallelogram is called a rhombus. Like a square, a rhombus has four congruent sides. It does not have right angles, but it does still have pairs of congruent angles opposite each other.
What is a parallelogram with no right angles?
This parallelogram is a rhomboid as it has no right angles and unequal sides.
## What quadrilateral always has 4 congruent sides?
Square: A quadrilateral with four congruent sides and four right angles; a square is both a rhombus and a rectangle.
What quadrilateral has 4 congruent sides? | HuggingFaceTB/finemath | |
Simplified Lorenz Cipher Toolkit
To make best use of this page, please refer to the Codes & Ciphers resources available
from Bletchley Park National Codes Centre. You can view these resources here.
NOTE: You can use standard copy and paste functions to move sequences of letters between the sections of this page.
Simplified Cipher Wheels
The original Lorenz Cipher Machine had twelve wheels, but this version only has two. The green area on each wheel indicates the current wheel position and cipher letter. You can use the two buttons on the left to set the initial positions of the two wheels.
K Position K Letter S Position S Letter
Method for "Adding" Letters
The Lorenz Cipher method requires cipher letters to be "added" to the letters in the plaintext message.
This is done by firstly converting both letters to a five-digit binary number, then combining them using a XOR operation.
You could just use the teleprinter addition table which can be found here.
Finally, you could just use the form below. Enter the two letters you want to add, then press the "add" button.
First letter Second letter Result
Enciphering/Deciphering Method
With the Lorenz Cipher, deciphering and enciphering use the same method.
The process below is described in terms of enciphering a plaintext message, but works identically for deciphering.
Once a starting position is chosen for each wheel, the message is enciphered as follows:
(i) Add the the cipher letter on the K-wheel to the letter from the plaintext message.
(ii) Add the cipher letter on the S-wheel to the resulting letter to get the first enciphered letter.
(ii) Rotate both wheels one place and move on to the next letter to be enciphered.
This carries on until all letters have been enciphered.
Example
Encipher the word "HELLO" with the starting position K=7, S=2.
K=7 S=2 K-Letter=G S-Letter=A H+G=C C+A=F The H in HELLO enciphers to F K=8 S=3 K-Letter=H S-Letter=B E+H=Y Y+B=N The E in HELLO enciphers to N K=9 S=4 K-Letter=I S-Letter=B L+I=H H+B=F The first L in HELLO enciphers to F K=10 S=1 K-Letter=J S-Letter=A L+J=B B+A=G The second L in HELLO enciphers to G K=11 S=2 K-Letter=K S-Letter=A O+K=Q Q+A=H The O in HELLO enciphers to H
So starting at K=7 S=2, the word "HELLO" enciphers to "FNFGH".
Automatic Enciphering and Deciphering Facility
We are interested in attempting to encipher and decipher Simplified Lorenz Cipher messages.
For this reason a facility for enciphering and deciphering messages quickly is provided here.
The same facility is used for both enciphering and deciphering, because the method is the same.
Simply put in the starting positions of the K and S wheels and the plaintext or ciphertext message, then click "Encipher/Decipher".
K-wheel start position = S-wheel start position = Plaintext message / ciphertext message (max 60 characters) Enciphered message / deciphered message
NOTE: You can only use "A" to "Z" and "3", "4", "8", "9", "+" or "/". Spaces are invariably represented by "99".
Tools To Help Break The Simplified Lorenz Cipher
In the pupil notes you will find Exercise 6 which relates to breaking this simplified Lorenz cipher.
This section is based around a coded message Z.
In the main example, Z = UDZDMR+JMSDC+TXUVQMYEDE8LWOKUD3TMK+G4UDC3NXWKOBYEFURWH
You could of course try to decipher this message with each possible setting of K and S.
There would be 56 combinations (14×4) but in a real Lorenz Cipher machine there would be several billion.
Producing a ΔZ sequence
The process for breaking the code described in Exercise 6 requires the production of a "ΔZ sequence".
ΔZ is found by "adding" each pair of values in Z (i.e. first and second, second and third and so on).
You can create your own ΔZ sequences from a message Z using the form below.
Z = ΔZ =
Producing a ΔK sequence
We also need to produce a "ΔK sequence".
ΔK is produced from K in the same way as ΔZ is produced from Z.
The K sequence must be the same length as the Z sequence, and is formed by cycling through the K-wheel letters.
e.g. If the K-wheel start-position is 7 i.e. letter "G", K = GHIJKLMNABCDEFGHIJKLMNABCDEFGHIJKLMNABCDEFGHIJKLMNABCD
You can create your own K and ΔK sequences using the form below. Just enter the start position and number of letters.
K-wheel start position = Number of letters in Z/K sequences = K = ΔK =
Adding the ΔZ and ΔK sequences and counting the "/" symbols.
To help us break the code, we also need to add the ΔZ to the ΔK sequence for each of the 14 K-wheel start positions.
We then need to count up the number of times the character "/" appears in these ΔZ+ΔK sequences.
Theoretically, the ΔZ+ΔK sequence where "/" appears the most often corresponds to the correct K-wheel start position.
You might like to try the main example, Z = UDZDMR+JMSDC+TXUVQMYEDE8LWOKUD3TMK+G4UDC3NXWKOBYEFURWH
You can create the 14 ΔZ+ΔK sequences using the form below.
You will only need to enter the Z sequence (the ciphertext message) and press the "Find ΔZ+ΔK" button.
The right-hand column shows the number of "/" characters in each sequence.
Z = No. of "/" K=1 ΔZ+ΔK= K=2 ΔZ+ΔK= K=3 ΔZ+ΔK= K=4 ΔZ+ΔK= K=5 ΔZ+ΔK= K=6 ΔZ+ΔK= K=7 ΔZ+ΔK= K=8 ΔZ+ΔK= K=9 ΔZ+ΔK= K=10 ΔZ+ΔK= K=11 ΔZ+ΔK= K=12 ΔZ+ΔK= K=13 ΔZ+ΔK= K=14 ΔZ+ΔK=
Produced by A.J.Reynolds March 2005 | HuggingFaceTB/finemath | |
## Mass percentage
The solubility of sodium chloride, NaCl, at 25 ºC is 36.0 g in 100 g water. If 60.0 g of sodium chloride is dissolved in 150 g of water at 100 ºC and the resulting solution is cooled to 25 ºC, what is the maximum amount of sodium chloride that can be separated at 25 ºC by filtration, and what is the mass percentage of sodium chloride in the filtrate?
• filtered can be = 60 - 36*1.5 = 6g
• amount of Na Cl that get dissolved in 150 mL = 36/100] * 150 = 54 g
therefore amount of NaCl that can be filtered = 60-54 = 6 grams.
mass percentage = 60/150] *100 = 40%
• sodium chloride = NaCl.
NaCl = 58.5 g/mol
a) 5.8 mol x 58.5 g/mol = 339.3 g
339.3 g of NaCl in 1 L water
339.3g / 10 = 33.93 g in 100 mL water
Unsaturated
b)3.25 mol x 58.5g/mol = 190.125g
190.125 g of NaCl in 500 mL water
190.125/5 = 38.025 g in 100 mL water
saturated
c)1.85 mol x 58.5 = 108.225
108.225 g of NaCl in 300 mL water
108.225/3 = 36.075 g in 100 mL water
saturated
• NaCl = 58.5 g/mol a) 5.8 mol x 58.5 g/mol = 339.3 g 339.3 g of NaCl in 1 L water 339.3g / 10 = 33.93 g in 100 mL water Unsaturated b)3.25 mol x 58.5g/mol = 190.125g 190.125 g of NaCl in 500 mL water 190.125/5 = 38.025 g in 100 mL water saturated c)1.85 mol x 58.5 = 108.225 108.225 g of NaCl in 300 mL water 108.225/3 = 36.075 g in 100 mL water saturated
Get homework help | HuggingFaceTB/finemath | |
# The vertical asymptotes of ##y=(x^2+1)/(3x-2x^2)## are ##x=0## and ##x=3/2##.
The vertical asymptotes of ##y=(x^2+1)/(3x-2x^2)## are ##x=0## and ##x=3/2##.
To find the vertical asymptotes we set the denominator equal to zero.
## 3x-2x^2=x(3-2x)=0rArrx=0 or x=3/2
Look at the graph below.
Sometimes, the denominator is equal to zero at the same x-value that makes the numerator zero. This will produce a instead of a vertical asymptote. | HuggingFaceTB/finemath | |
# Apps for Common Core Math Standards, Grades 6-8
12/1/2011 12:00:00 AM
The sixth grade standard includes five components:
• Ratios and Proportional Relationships - Understand ratio concepts and use ratio reasoning to solve problems.
• The Number System - Extend previous understanding of fractions including multiplying and dividing fractions, compute fluently with multi-digit numbers and find common factors and multiples.
• Expressions and Equations - Reason about and solve one-variable equations and inequalities.
• Geometry - Solve real-world and mathematical problems involving area, surface area, and volume.
• Statistics and Probability - Develop understanding of statistical variability.
6th Grade Math Testing Prep \$2.99; Teacher upgrade \$4.99 - This app covers all of the core standards, allowing students to move at their own pace with quizzes to evaluate student understanding. Topics include: ratios, probability, statistics, algebra, geometry, and multiple step procedures. The upgrade for teachers allows teachers to enter as many students as needed, set the number of questions, and review scores and track student improvement.
Algebra Touch \$2.99 - Current material covers: Simplification, Like Terms, Commutativity, Order of Operations, Factorization, Prime Numbers, Elimination, Isolation, Variables, Basic Equations, Distribution, Factoring Out, Substitution, and 'More Advanced' mode.
Greatest Common Factor \$.99 - Calculates the GCF or LCM of two numbers.
Elevated Math Free - Lessons include: Numbers and Operations, Measurement, Geometry Algebra and Data Analysis and Probability (grades 6-8).
Factor Race- Algebra Free - A game in which the player must identify the binomial factors of trinomial equations.
Geometry-Volume-Solids-lite Free - The lite version of this app provides a quick way to learn and calculate volume of solids. There are six computer-animated videos on the volume of the cube, rectangular solid, cylinder, sphere, cone and pyramid. The paid addition is \$1.99. It includes videos on the cube, cylinder, sphere, cone and pyramid.
The seventh grade standard includes five components:
• Ratios and Proportional Relationships - Analyze proportional relationships and use them to solve real-world and mathematical problems.
• The Number System - extend previous knowledge using the four operations with fractions and the ability to divide and multiply rational numbers.
• Expressions and Equations - Use properties of operations to generate equivalent expressions.
• Geometry - Draw, construct and describe geometrical figures and describe the relationships between them.
• Statistics and Probability - Ability to use random samplings and make inferences about populations, investigate chance processes and develop, use, and evaluate probability models.
7th grade math testing prep \$2.99; Teacher upgrade \$4.99- Covers the Core Standards from advanced problem solving, fractions, probability, and statistics to logical calculations.
Middle School Math 7th Grade Free - Three levels for each topic: Negative numbers, absolute value and order of operations.
Middle School Math Pro 7th Grade \$.99 - Three levels, 10 topics, including adding and subtracting fractions, multiplying and dividing fractions, factors and multiples, decimals, and more. Student guides a monkey down the ladder to get to the next level.
Algebra Concepts for the iPad \$1.99 - Geared for 7th grade students. The app includes Variables and Expressions, Properties of Real Numbers, Solving Equations and Polynomials.
Sketchpad Explorer Free - This app was mentioned in an earlier blog. Grades 7-9 specifically are geared to early algebra.
The eighth grade standard also has five components:
• The Number System - Know that there are numbers that are not rational, and approximate them by rational numbers.
• Expressions and Equations - Work with radicals and integer exponents, understand the connections between proportional relationships, lines, and linear equations.
• Functions - Define, evaluate, and compare functions.
• Geometry - Understand congruence and similarity using physical models, transparencies, or geometry; understand and apply the Pythagorean Theorem.
• Statistics and Probability - Investigate patterns of association in bivariate data
8th Grade Math- \$2.99 - Helps students understand major math terms and concepts: Sequences and Series, Polynomials, Square Roots, Introduction to Geometry, Triangles and Other Polygons, Pythagorean Theorem, and Trigonometry.
Algebra Prep \$3.99 - Review and practice equations, graphing, systems, exponents, factoring, rationals and more. Includes videos, practices tests and mini-tests.
Pythagorean Theorem \$4.99 - Includes five video examples, eight interactive practice problems, one challenge problem, one worksheet of extra problems and one notes page.
Symmetry-Shuffle \$1.99 - This mathematical puzzle allows users to explore line and rotational symmetry, while developing their spatial sense to create strategies to help them solve problems. The app is a fun way to understand the congruence, similarity, and line or rotational symmetry of objects using transformations.
ALinearEqn Linear Equations \$.99 - A combined interactive Coaching Calculator and Guide that helps students master the solving of linear equations.
Vicki Windman is a special education teacher at Clarkstown High School South.
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# Evaluate the triple integral where D is the region inside the cylinder $x^2 + y^2 = 1$ which is bounded…?
Evaluate the triple integral where D is the region inside the cylinder $x^2 + y^2 = 1$ which is bounded below by the plane z=0 and bounded above by the plane 2x+4y+z=11.
I started this problem by changing into cylindrical coordinates and got this triple integral and was wondering if you can give me an okay before I start solving it. Thank you!
$$\int_{-1}^1 \int_0^{2\pi} \int_0^{11-2rcos\theta -4rsin\theta} r(rcos\theta)^2 dzd\theta dr$$
• $r$ should run from $0$ to $1$. Use 2-dimensional polar-coordinates. – Frieder Mar 19 '15 at 23:48
The Jacobian of the cylindrical coordinates should be $r$.
$$\begin{gathered} \int_{ - 1}^1 {\int_{ - \sqrt {1 - {x^2}} }^{\sqrt {1 - {x^2}} } {\int_0^{11 - 2x - 4y} {dzdydx = } } } \int_{ - 1}^1 {\int_{ - \sqrt {1 - {x^2}} }^{\sqrt {1 - {x^2}} } {(11 - 2x - 4y} )dydx} \hfill \\ = \int_0^1 {\int_0^{2\pi } {11 - 2\cos (\phi ) - 4sin(\phi )} rd\varphi dr} = 11\pi \hfill \\ \end{gathered}$$
• @Ayoshna: The function in triple integral is the constant function $1$, so $$\int\limits_o^{11 - 2x - 4y} {dz} = 11 - 2x - 4y$$ – Frieder Mar 20 '15 at 0:00 | HuggingFaceTB/finemath | |
# GR and my journey to the centre of the Earth
That is awesome! And it makes complete sense too! (other than a possible misusage of the word "distance"). Let's have a look at the equations of motion of you in Earth's curved spacetime, assuming that your feet are not touching the ground:
$$\frac{\mathrm d^{2}x^{\mu}}{\mathrm ds^{2}}+\Gamma^{\mu}_{\nu\sigma}(x(s))\ \frac{\mathrm dx^{\nu}}{\mathrm ds}\frac{\mathrm dx^{\sigma}}{\mathrm ds}=0$$ where $x^{\mu}(s)$ is your world line, $s$ is some parameter,
$$\Gamma^{\mu}_{\nu\sigma}=\frac{1}{2}\ g^{\mu\tau}(\partial_{\nu}g_{\sigma\tau}+\partial_{\sigma}g_{\nu\tau}-\partial_{\tau}g_{\sigma\nu})$$ with $g^{\mu\tau}$ the inverse of the metric and $$g=\left( 1 - \frac{r_{s} r}{\rho^{2}} \right) c^{2}\, \mathrm dt^{2} - \frac{\rho^{2}}{\Delta} \mathrm dr^{2} - \rho^{2} \,\mathrm d\theta^{2}+ \\ - \left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta \,\mathrm d\phi^{2} + \frac{2r_{s} r\alpha \sin^{2} \theta }{\rho^{2}} \, c \,\mathrm dt \, \mathrm d\phi$$ where $$r_{s}=\frac{2GM}{c^{2}}\ ,\quad\alpha=\frac{J}{Mc} \ ,\quad \rho^{2}=r^{2}+\alpha^{2}\cos^{2}\theta\ ,\quad \Delta=r^{2}-r_{s}r+\alpha^{2}$$ with $M$ and $J$ Earth's mass and angular momentum.
The equations of motion can be derived from the action functional
$$S[x(s)]=-mc\int_{a}^{b}\sqrt{g_{\mu\nu}(x(s))\,\frac{\mathrm dx^{\mu}}{\mathrm ds}\frac{\mathrm dx^{\nu}}{\mathrm ds}}\ \mathrm ds$$ where $m$ is your mass and, as gravity goes, it plays no role at all in how you fall to the ground. You find the equations of motion by minimizing S with respect to the curve $x(s)$, which amounts to minimizing the (proper) time you spend on your worldline, times $-mc^{2}$ (this is why you are minimizing rather than maximizing): \begin{align} S[x(\tau)]&=-mc^{2}\int_\textrm{today}^\textrm{tomorrow}\sqrt{g_{\mu\nu}(x(\tau))\,\frac{\mathrm dx^{\mu}}{\mathrm d\tau}\frac{\mathrm dx^{\nu}}{\mathrm d\tau}}\,\mathrm d\tau\\ &= \text{the distance between today and tomorrow}\,. \end{align} As you'll fall in the direction that connects you to the center of the Earth, the shortest distance between today and tomorrow is indeed through the center of the Earth. The reason why you are sticking to the floor right now is really that the ground is preventing you from taking the shortest path from today to tomorrow, which passes through the center of the Earth.
What GR says is correct: the straight line between, say, London today and London tomorrow is not the curve that spends all the time between in London: whether it actually passes through the centre of the Earth I'm not sure, and it depends on how fast you are moving as well as where you are.
The caveat is that the straight line (geodesic) not the shortest path, it's the longest (there is no shortest path) and the length is proper time.
This not inconsistent with you being able to take other paths: you can, but they are not extrema of length and therefore you experience acceleration on the path: the acceleration which is currently sticking you to the ground, for instance.
It makes sense as a "visual" description.
In GR, free particles with mass move on time-like geodesics. A common description of geodesics are such curves that locally minimalize path length, but this desciption comes from Riemannian geometry, not Lorentzian geometry, which GR is. In Lorentzian geometry, timelike geodesics are those that locally maximalize proper time.
The reason the quote sounds so nonsensical, is that in GR time is also curved, and the geodesics move through space-time, not just space. If the ground was not beneath your feet, you'd fall through the center of the earth, as time would pass, hence you could say that the "path with greatest proper time between today and tomorrow leads through the center of the earth".
But there is a ground beneath your feet, the ground exerts EM force on you that makes you deviate from this geodesic, since you are no longer a "free particle". | open-web-math/open-web-math | |
# Mr.B Project
More Options: Make a Folding Card
#### Storyboard Description
This storyboard does not have a description.
#### Storyboard Text
• Learning the Chain Rule Story
• Pre-Calculus Class
• If f(u) is a differentiable function of u and u=g(x) is a differentiable function of x, then y =f(g(x)) is a differentiable function of x and
• Good Job class! We'll have a quiz on this tomorrow.
• The Chain Rule
• dy/dx = dy/du × du/dx
• Lunch
• Don't worry Chloe, I'll tutor you during Study Hall!
• The Chain Rule
• dy/dx = dy/du × du/dx
• Omg, we have a quiz on the Chain Rule ! I don't understand it, can you help me?
• Maybe I can help too!
• This is the story of a student from Riverside High school, named Chloe, who struggles with the Chain Rule. Enjoy!
• Study Hall
• The Power Rule Before solving the problem it looks like this: f(x)=xn When you find the derivative it looks like this : f'(x)=n*x(n-1)
• After Mr. C teaches the Chain Rule, Chloe doesn't understand the lesson. Even though Mr. C asks for any questions, she is ashamed to reveal her struggle.
• Study Hall
• f(x)=3x5
• You don't write f'(x) for your second step because you're only showing your work, you didn't find the derivative yet!
• Example
• f(x)=3*5(5-1)
• f'(x)=15x4
• During Lunch, Chloe asks her best friend, Essence, for help and Tj offered to help as well.
• Study Hall
• Keep the inside of the parenthesis the same. Think of it as 'x' it doesn't change
• f'(x)=4(x3+3x)3(3x2+3)
• f(x)=(x3+3x)4
• Now you take the derivative of what's inside the parenthesis and multiply it
• Essence proceeds to explain that you multiply the exponent (n) by the coefficient in front of 'x'. Remember: if you don't see no coefficient in front of x, then it's 1. Then, essence explains that 'n' in the equation is the exponent and when finding the derivative you always subtract the exponent (n) by 1, so (n-1).
• In order to understand the Chain Rule better, you'd have to be familiar with the Power Rule.
• definitely.
• Tj then elaborates on the Power Rule, by providing an example. He multiplied the coefficient (3) by the exponent (5) and then subtracted the exponent by 1 so, (5-1).
• Ohhh, I get it !
• Here's an example!
• Essence thoroughly explains an example of the Chain Rule. She says that you treat the problem similarly to a Power Rule one. Keep what's inside the parenthesis and simply multiply the coefficient by the exponent and then subtract that exponent by one (4-1).
• Now that you understand the Power Rule, you can do the Chain Rule! They're similar.
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# Dot (Scalar) Product of Two Vectors
The following physics revision questions are provided in support of the physics tutorial on Dot (Scalar) Product of Two Vectors. In addition to this tutorial, we also provide revision notes, a video tutorial, revision questions on this page (which allow you to check your understanding of the topic) and calculators which provide full, step by step calculations for each of the formula in the Dot (Scalar) Product of Two Vectors tutorials. The Dot (Scalar) Product of Two Vectors calculators are particularly useful for ensuring your stet-by-step calculations are correct as well as ensuring your final result is accurate.
Not sure on some or part of the Dot (Scalar) Product of Two Vectors questions? Review the tutorials and learning material for Dot (Scalar) Product of Two Vectors
Vectors and Scalars Learning Material
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Tutorial
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Notes
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2.4Dot (Scalar) Product of Two Vectors
## Dot (Scalar) Product of Two Vectors Revision Questions
1. Two forces F1 = 50N and F2 = 100N are pulling the same object as shown in the figure.
What is the size of the angle θ between the two forces?
Remark! Pay attention to the sign of components.
1. 900
2. 170
3. 880
4. 609
2. A 300N force acts on an object at a certain angle to the horizontal direction and as a result, the object moves horizontally at 2m/s. The output power delivered is 480W. What is the angle of the force to the horizontal direction?
1. 300
2. 370
3. 530
4. 600
3. Two forces are acting on the same object at 530 to each other (cos 530 = 0.6, sin 530 = 0.8). Their dot product is 120 units and the magnitude of the first force is 20N. What is the magnitude of the second force in newtons?
1. 6N
2. 8N
3. 10N
4. 12N
## Whats next?
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2. Vectors and Scalars Physics tutorial: Dot (Scalar) Product of Two Vectors. Read the Dot (Scalar) Product of Two Vectors physics tutorial and build your physics knowledge of Vectors and Scalars
3. Vectors and Scalars Video tutorial: Dot (Scalar) Product of Two Vectors. Watch or listen to the Dot (Scalar) Product of Two Vectors video tutorial, a useful way to help you revise when travelling to and from school/college
4. Vectors and Scalars Revision Notes: Dot (Scalar) Product of Two Vectors. Print the notes so you can revise the key points covered in the physics tutorial for Dot (Scalar) Product of Two Vectors
5. Check your calculations for Vectors and Scalars questions with our excellent Vectors and Scalars calculators which contain full equations and calculations clearly displayed line by line. See the Vectors and Scalars Calculators by iCalculator™ below.
6. Continuing learning vectors and scalars - read our next physics tutorial: Rounding and Significant Figures | HuggingFaceTB/finemath | |
# A 23.9 kg mass has a 3i-2jm/s velocity and a 21.3 kg mass has a 3i-10jm/s velocity. What is the magnitude of the total momentum of the two masses?
Feb 15, 2018
$\text{please have a look at the fallowing details.}$
#### Explanation:
${\vec{P}}_{1} = {m}_{1} \cdot {\vec{v}}_{1}$
${\vec{P}}_{1} = 23.9 \cdot \left(3 i - 2 j\right) = 71.7 i - 47.8 j$
${\vec{P}}_{2} = {m}_{2} \cdot {\vec{v}}_{2}$
${\vec{P}}_{2} = 21.3 \left(3 i - 10 j\right) = 63.9 i - 213 j$
$\Sigma \vec{P} = {\vec{P}}_{1} + {\vec{P}}_{2}$
$\Sigma \vec{P} = 71.7 i + 63.9 i + \left(- 47.8 j - 213 j\right)$
$\Sigma \vec{P} = 135.6 i - 260.8 j \text{ total momentum vector}$
$P = \sqrt{{\left(135.6\right)}^{2} + {\left(- 260.8\right)}^{2}} \text{ magnitude}$
$P = \sqrt{18387.36 + 68016.64}$
$P = 293.95 k g \text{ } \cdot m {s}^{-} 1$ | HuggingFaceTB/finemath | |
# Super-efficient heat engines and refrigerators
Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 4.16 – 4.17.
A heat engine has a maximum efficiency of
$\displaystyle e_{max}=1-\frac{T_{c}}{T_{h}} \ \ \ \ \ (1)$
where ${T_{h}}$ is the temperature of the hot reservoir from which heat ${Q_{h}}$ is extracted and ${T_{c}}$ is the temperature of the cold reservoir to which heat ${Q_{c}=Q_{h}-W}$ is expelled after work ${W}$ is extracted. From the second law of thermodynamics, the heat and temperature ratios satisfy (at the optimum efficiency):
$\displaystyle \frac{Q_{c}}{Q_{h}}=\frac{T_{c}}{T_{h}} \ \ \ \ \ (2)$
Thus the work done by an optimal heat engine is
$\displaystyle W=Q_{h}-Q_{c}=Q_{h}\left(1-\frac{T_{c}}{T_{h}}\right) \ \ \ \ \ (3)$
For an optimal refrigerator, the work required to extract ${Q_{c}}$ from the cold reservoir and expel ${Q_{h}=Q_{c}+W}$ to the hot reservoir is exactly the same as 3. Therefore, if we use the work from a Carnot engine (which operates at the maximum efficiency) to power a Carnot refrigerator, the refrigerator will exactly cancel the effect of the heat engine, since it removes exactly the same amount ${Q_{c}}$ of heat from the cold reservoir that the Carnot engine deposited there, and expels exactly the same amount ${Q_{h}}$ of heat back into the hot reservoir that the Carnot engine extracted in the first place.
If, however, we could make an engine that was more efficient than a Carnot engine, that engine could produce the same amount of work by extracting a smaller amount of heat ${Q_{he}}$ and expelling a smaller amount of waste heat ${Q_{ce}}$, provided that ${Q_{he}-Q_{ce}=W}$ as before. In this case, the heat engine still provides the amount of work required to power the refrigerator, but it produces less waste heat than the refrigerator extracts. The net effect of combining the two is to produce a system that requires no input of work, but still extracts a net amount ${Q_{c}-Q_{ce}}$ of heat from the cold reservoir. That is, we’ve produced a refrigerator that requires no input work.
Conversely, if we started with a refrigerator that had a coefficient of performance COP that is greater than the Carnot value of
$\displaystyle \mbox{COP}_{max}=\frac{T_{c}}{T_{h}-T_{c}} \ \ \ \ \ (4)$
this means that it requires an amount of work ${W_{r} to extract the heat ${Q_{c}}$ from the cold reservoir. Hooking this up to a Carnot engine that extracts ${Q_{h}}$ from the hot reservoir, produces work ${W}$ and expels heat ${Q_{c}}$ to the cold reservoir means that the compound system produces a net excess of work ${W-W_{r}}$ while producing no waste heat, since the refrigerator extracts exactly the same amount of heat that the Carnot engine expels.
Both of these systems satisfy the first law of thermodynamics (conservation of energy) but violate the second law (entropy cannot decrease), which is why we can’t make them in real life. | HuggingFaceTB/finemath | |
# Rigid Body Dynamics
Game engines are concerned with a ubset of physics called dynamics, which is the study of how forces affect the movement of objects.
• Newtonian mechanics: the objects in simulation obey Newton's laws of motions. There are no quantum effects, and speeds are low enough to not have relativistic effects.
• Rigid bodies: all objects are perfectly solid and cannot be deformed. Shape is constant. This greatly simplifies the mathematics required to simulate the dynamics of solid objects.
Physics engines are also intended to have realistic effects for the following:
• No interpenetrating objects (collisions look real)
• Hinges
• Wheels
• Ball joints
• Rag dolls
• Prismatic joints (sliders)
There is usually a one-to-one mapping of rigid bodies to collidables.
Every frame, we need to query the physics engine for the transform of every rigid body and apply it in some way to the transform of the corresponding game object.
See "Grant R. Fowles and George L. Cassiday. Analytical Mechanics, Seventh Edition. Pacific Grove, CA: Brooks Cole, 2005." for foundations in dynamics. See also: https://chrishecker.com/Rigid_Body_Dynamics
### Units
meters (m) for distance/length.
kilograms (kg) for mass.
seconds (s) for time.
### Linear and Angular Dynamics
An unconstrained rigid body translates and rotates freely along all three Cartesian axes. Such a body has six degrees of freedom (DOF).
• Linear dynamics: the description of the motion of a rigid body, ignoring rotational effects.
• Angular dynamics: the description of the rotational motion of a rigid body.
### Center of dynamics
In linear dynamics, an unconstrained rigid body acts as though all of its mass were concentrated on its center of mass.
A body with uniform density has its center of mass at the centroid of the body.
Convex bodies always have its center of mass inside the body. Concave bodies can have its center of mass outside the body.
### Linear Dynamics
A rigid body is described by a position vector identifying the location of its center of mass, $$\vec{r}$$. Position is measure in meters.
#### Linear Velocity and Acceleration
The derivative of position with respect to time is velocity.
$\vec{v}(t) = \frac{d \vec{r}(t)}{dt}$
Differentiating a vector is the same as differentiating each component independently.
$v_x(t) = \frac{d r_x(t)}{dt} = r_x(t)$
... and so on for y- and z- components.
The second derivative of position with respect to time is acceleration.
$\vec{a}(t) = \frac{d \vec{v}(t)}{dt} = \frac{d^2 \vec{r}(t)}{dt^2}$
### Force and Momentum
A force is anything that causes an object with mass to accelerate or decelerate. Force has magnitude and direction in space and are therefore represented as vectors.
Net effect of $$N$$ forces on a body:
$F_{\text{net}} = \sum_{i=1}^N F_i$
Newton's Second Law states that force is proportional to acceleration and mass:
$F(t) = m \vec{a}(t)$
Forces are measured in terms of $$\frac{kg-m}{s^2}$$.
A rigid body's linear momentum is its linear velocity multiplied by its mass, denoted with symbol $$p$$. (This is confusingly NOT position!)
$\vec{p}(t) = m \vec{v}(t)$
### Ordinary Differential Equations
An ordinary differential equation (ODE) is an equation involvin ga function of one independent variable and various derivatives of that function. If independent variable is time, and the function is $$x(t)$$, then an ODE is:
$\frac{d^n x}{dt^n} = f \left( t, x(t), \frac{dx(t)}{dt}, \frac{d^2 x(t)}{dt^2}, \cdots, \frac{d^{n-1} x(t)}{dt^{n-1}} \right)$
### Analytical Solutions
Differential equations of motion can rarely be solved analytically, which is the process of finding a simple, closed-form function that describes the rigid body's position for all possible values of time $$t$$.
In games this is usually impossible to find because closed-form solutions to some differential equations are not known. Games are also interactive and so you have no idea a priori how forces will interact over time.
### Numerical Integration
Numerical integration solves differential equations using a time-stepped approach. We use the solution from a previous timestep to arrive at the solution for the next time step.
The duration of the timestep is usually fixed, $$\Delta t$$.
#### Explicit Euler
A simple numerical solution to an ODE, and the most intuitive.
Assuming we know the current velocity, $$\vec{v}(t)$$ and we want to solve the following ODE to find the rigid body's position on the next frame:
$\vec{v}(t) = \vec{r}'(t)$
We convert velocity from $$\frac{m}{s}$$ to $$\frac{m}{\text{frame}}$$ by multiplying velocity by the time delta, then add "one frame's worth" of velocity onto the current position.
$\vec{r}(t_1) = \vec{r}(t_0) + \vec{v}(t_0) \Delta t$
This assumes that velocity is constant during the timestep.
By definition any derivative is the quotient of two infinitesimally small differences, $$\frac{d\vec{r}}{dt}$$. Explicit Euler approximates this using the quotient of two finite differences.
\begin{align} \frac{d\vec{r}}{dt} &\approx \frac{\Delta \vec{r}}{\Delta t'} \\ \vec{v}(t_0) &\approx \frac{\vec{r}(t_1) - \vec{r}(t_0)}{t_1 - t_0} \end{align}
Numerical solutions to ordinary differential equations have three properties:
• Convergence: as $$\Delta t$$ trends toward zero, does the approximate solution trend towards the real solution?
• Order: How bad is the error between the real solution and the approximated solution? It is usually reflected in big O notation, scaling with the size of $$\Delta t$$.
• Stability: does the numerical solution tend to find a stable equilibrium over time?
• If a numerical method adds energy into the system, velocities will eventually "explode", and the system will become unstable. If a numerical method removes energy from the system, it will have an overall damping effect, and the system will become stable.
#### Verlet Integration
Explicit Euler is simple but has high error and poor stability.
Alternatives include:
• Backward Euler
• Midpoint Euler
• Runge-Kutta methods, e.g., RK4
The most widely used method is called Verlet Integration. There is regular Verlet and velocity Verlet.
##### Regular Verlet
Regular Verlet is great because it offers a low error, is simple, and inexpensive to calculate. It works by adding two Taylor series expansions, one going forward in time and another going backward in time.
$\vec{r}(t_0 + \Delta t) = \vec{r}(t_0) + \vec{r}(t_0) \Delta t + \vec{r}'(t_0) \Delta t + \frac{1}{2} \vec{r}''(t_0) \Delta t^2 + \frac{1}{6} \vec{r}^{(3)}(t_0) \Delta t^3 + O(\Delta t^4)$
$\vec{r}(t_0 - \Delta t) = \vec{r}(t_0) + \vec{r}(t_0) \Delta t - \vec{r}'(t_0) \Delta t + \frac{1}{2} \vec{r}''(t_0) \Delta t^2 - \frac{1}{6} \vec{r}^{(3)}(t_0) \Delta t^3 + O(\Delta t^4)$
Adding these two together causes negative terms to cancel, which results in the regular Verlet method:
$\vec{r}(t_0 + \Delta t) = 2 \vec{r}(t_0) - \vec{r}(t_0 - \Delta t) + \vec{a}(t_0) \Delta t^2 + O(\Delta t^4)$
If you want to express $$\vec{a}$$ in terms of net force, since $$F = ma$$, you can replace it with $$\frac{F_{\text{net}}(t_0)}{m}$$.
##### Velocity Verlet
This is even more common than regular Verlet, it is a four-step process.
Given $$\vec{a}(t_0) = \frac{1}{m} \vec{F}\left(t_0, \vec{r}(t_0), \vec{v}(t_0)\right)$$:
1. Calculate $$\vec{r}(t_0 + \Delta t) = \vec{r}(t_0) + \vec{v}(t_0) \Delta t + \frac{1}{2} \vec{a}(t_0) \Delta t^2$$.
2. Calculate $$\vec{v}(t_0 + \frac{1}{2} \Delta t) = \vec{v}(t_0) + \frac{1}{2} \vec{a}(t_0) \Delta t$$.
3. Determine $$\vec{a}(t_0 + \Delta t) = \vec{a}(t_1) = \frac{1}{m} \vec{F} \left( t_1, \vec{r}(t_1), \vec{v}(t_1) \right)$$.
4. Calculate $$\vec{v}(t_0 + \Delta t) = \vec{v}(t_0 + \frac{1}{2} \Delta t) + \frac{1}{2}\vec{a}(t_0 + \Delta t) \Delta t$$
### Angular Dynamics
See:
Rigid bodies have different moments of inertia about different axes, since they have different distribution of mass about these axes.
The intertia tensor is a 3x3 matrix that represents a rigid body's rotational mass.
$I = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{bmatrix}$
The non-diagonal elements are usually 0 in games because they produce unintuitive motions. As a result, the inertia tensor is usually simplified down to $$I = \begin{bmatrix} I_{xx} & I_{yy} & I_{zz} \end{bmatrix}$$.
Orientation is defined in 3D using unit quaternions. Alternatives:
• A 3x3 rotational matrix (uses more space than needed)
• Euler angles, $$\begin{bmatrix} \theta_x & \theta_y & \theta_z \end{bmatrix}$$ (suffers from gimbal lock and more difficult mathematically).
\begin{align} q &= \begin{bmatrix} q_x & q_y & q_z & q_w \end{bmatrix} \\ &= \begin{bmatrix} \vec{q} & q_w \end{bmatrix} \\ &= \begin{bmatrix} \vec{u} \sin \frac{\theta}{2} & \cos \frac{\theta}{2} \end{bmatrix} \end{align}
If no forces are acting on a rigid body, then linear acceleration is zero and linear velocity is constant.
If no forces are acting on a rigid body, then angular acceleration is zero, but angular velocity is NOT constant, because axis of rotation can continually change direction.
Physics systems do not consider angular velocity as a primary quantity in simulations since it is not constant. Since angular momentum is constant (law of conservation of momentum), it is the primary quantity and velocity can be derived from it.
Angular momentum and Linear momentum are 3-element vectors.
Linear Momentum $\vec{p}(t) = m \vec{v}(t)$
Where:
• $$\vec{p}$$ is linear momentum.
• $$m$$ is mass.
• $$\vec{v}$$ is linear velocity.
Angular Momentum $\vec{L}(t) = I \vec{\omega}(t)$
Where:
• $$\vec{L}$$ is angular momentum.
• $$I$$ is moment of inertia (the inertia tensor).
• $$\vec{\omega}$$ is angular velocity.
Torque can be calculated as the cross product between the "radial position vector of the point of force application" and the force vector itself. In other words, a position vector extending out from the fulcrum to the position where a force is being applied to a lever cross multiplied against the size of the force.
However! Since angular velocity is not conserved, physics simulations typically express this in terms of angular momentum.
\begin{align} \vec{N} &= \vec{r} \cross \vec{F} \\ &= I \alpha(t) \\ &= I \frac{d \omega(t)}{dt} \\ &= \frac{d}{dt} \left(I \omega(t)\right) \\ &= \frac{d \vec{L}(t)}{dt} \end{align}
Where:
• $$\vec{N}$$ is torque.
• $$\vec{r}$$ is the position vector along a lever indicating where a force is being applied.
• $$\vec{F}$$ is force applied to a lever to create torque.
• $$\alpha(t)$$ is angular acceleration. Torque is proportional to moment of inertia and angular acceleration, just as Force is proportional to mass and linear acceleration.
### Solving equations of angular motion
Solving equations of angular motion are meant to find an object's orientation given some time $$t$$.
You cannot solve equations of angular motion the same way as linear motion: using the first- and second-order derivatives to find velocity and acceleration, respectively and using verlet integration or explicit euler to find the position at some given time $$t$$.
Instead, you'll want to solve directly for angular momentum $$\vec{L}$$ instead of angular velocity $$\vec{\omega}$$.
• Once you have this quantity, you can then calculate angular velocity using $$I$$ and $$\vec{L}$$.
In addition, angular velocity has three elements whereas orientation is a quaternion with four elements. You have to convert angular velocity into quaternion form, then apply a special euqation that relates the orientation quaternion to the angular velocity quaternion.
See section 13.4.6.5 of the Game Engine Architecture for more details. | HuggingFaceTB/finemath | |
In the last section, we set a couple of problems, and asked you to devise a program to come up with the answer.
A. Write a program to add up the integers from 1 to 10, and put the result in to the memory location at offset zero.
B. Alter the program to add up the integers from 1 to 15 and put the result in to the memory location at offset zero.
C. Alter the program again, to add up the numbers from 5 to 15, and put the result into the memory location at offset zero.
If you wrote a program for problem (A) which ended up with the number 55 in memory location zero, then congratulations - you have not only written your first program solo, but you've also devised your first algorithm: a description of a process required to take some input and produce a desired result.
In this section, we'll look at some programs I've devised to solve these 3 related problems. Your program may not look exactly like any of mine, and if you got the right number in the right place, then that's great. There's no 100% right answer - you're always making compromises and improvements. And even if you did get the right answer, it is always interesting to look at other people's solutions, and see if there's something you can learn from them.
We're going to evolve our solution from a very simple-minded approach, to a more sophisticated algorithm, and look at the implications and compromises we're making along the way.
So, here's a first stab at an algorithm for solving problem (A). It is not a very complicated one: we could do the addition one line at a time, long-hand.
`load r1 0`
`add r1 1`
`add r1 2`
`add r1 3`
`add r1 4`
`add r1 5`
`add r1 6`
`add r1 7`
`add r1 8`
`add r1 9`
`add r1 10`
`write 0 r1`
`exit`
(Notice that we remember to set the accumulated total to zero before we start - we don't know what might have been in the register before we began.)
That's a pretty decent stab at it. It gives us the right answer, and only takes 12 instructions to do it.
When we go to problem (B), it starts to look a bit less great. I got a bit bored typing this in, even with the help of copy and paste.
`load r1 0`
`add r1 1`
`add r1 2`
`add r1 3`
`add r1 4`
`add r1 5`
`add r1 6`
`add r1 7`
`add r1 8`
`add r1 9`
`add r1 10`
`add r1 11`
`add r1 12`
`add r1 13`
`add r1 14`
`add r1 15`
`write 0 r1`
`exit`
Imagine we wanted to add the whole sequence for 1…10,000! That'd be a lot of instructions. 10,003 to be precise! In fact, I don't think we've got enough program instruction memory in this toy computer to do that. Also, don't forget that every instruction we execute takes a little bit of time. If we're running in the public cloud that directly costs us a little bit of money (or indirectly via our electricity bill if we're running at home or in the office, in the form of the little bit of energy we expended in the execution of each instruction). So, ideally we'd also like to execute as few as possible.
It turns out that these are often competing constraints.
Let's look at the first part of this challenge. First - can we make our program more compact, such that we could reasonably sum integers up to 10,000 or more? To do that, we need to understand both the problem, and the algorithm we are using to solve it. Let's start by describing it more precisely.
## Describing our algorithm - flowcharts
There are lots of ways of describing an algorithm - we can do it in code, in everyday language, mathematically, or graphically, for example.
Instead of using more words, let's look at a graphical representation of our first algorithm, called a Flow Chart.
Long, isn't it, but quite expressive?
We're using four symbols in this diagram. The start or end of the algorithm A process or operation to be executed Some data The flow of control through the algorithm (hence 'flow chart')
It gives a nice, clear, step-by-step view of what we are doing. Notice that many of these steps are the same. If we were trying to explain this to someone in ordinary language, we might say
To start, set the result to zero.
Then, add 1 to the result.
Then, add 2 to the result.
(… and keep doing that sort of thing until …)
Then, add 9 to the result.
Then, add 10 to the result.
Then, you are done.
Anybody with some really basic maths would understand what you meant. But, unfortunately, computers are (in general) a bit slow on the uptake, and need everything explained to them in precise detail. How could we leave out those steps where we've used a hand-wavy description to avoid repeating ourselves, and yet still describe precisely what needs to be done?
One possible way is to use a loop.
## Loops (power and pitfalls)
Let's re-describe that process in ordinary language, and find some way of explaining the boring repetitive bit, such that we don't just wave our hands and skip over it. Here's my attempt.
To start, set the result to zero.
Then set the current number to 1
Then add the current number to the result
Then add 1 to the current number to give the next number
If the current number is less than or equal to 10 then go back and repeat the previous two steps, otherwise you are done.
If we want to be really precise, we could write this down again in language somewhere between an actual program and ordinary language. We call it pseudocode. Given the low-level language you've been dealing with up to now, this should be fairly easy for you to interpret - and that's the idea. It isn't any particular programming language, but it is just as precise as any programming language, so more or less any programmer should be able to understand it. We're also being meticulous about describing what the algorithm is supposed to do, what input it needs, and how to interpret the result.
Description: Calculate the sum of a particular sequence of consecutive positive integers
Input: The first number of the sequence, and the last number of the sequence an
Output: The sum of the integers between a1 and an(inclusive) Notice we're using a mixture of 'ordinary language' and mathematical symbols to describe precisely what our algorithm does.
There are a couple of terms that you might need help with: as "result changes to 0". The arrow is just a shorthand for a phrase like "changes to" or "becomes".
Then there's: You can read this as "while the current value is less than or equal to the last number in the sequence, keep doing the steps in the list below". Notice that we've used indentation to make it easy to see where that list of steps starts and ends. Those steps can be read as "the result changes to the value of result plus the current value", and "the current value change to the current value plus 1".
This pattern, where you go back and repeat a previous step based on some condition is very common in imperative programming, and we call it a loop. (This specific example is called a while loop for obvious reasons.) You can see why we call it a loop if we express this program as a flow chart. The first thing to notice is that we've added an extra symbol to our set, the diamond for a decision: The start or end of the algorithm A process or operation to be executed Some data The flow of control through the algorithm (hence 'flow chart') Decision
The diamond has 2 arrows coming out of it, one labelled 'yes' for the path to take if the statement in it is true. The other labelled 'no' for the path to take if the statement is false. Notice how the 'yes' arrow 'loops back' up the diagram from the decision point, to repeat a part of the process.
Let's restructure the program to implement this new, looping version of the algorithm. We're going to make use of registers to remember the minimum and maximum values of our sum, and take advantage of our `COMPARE `and `JUMPLTE `instructions to minimize the amount of repetitive code we have to write.
This program uses `r0 `to store the current number to be added, `r2 `to store the maximum number to be added, and `r1 `to accumulate the result. I've numbered the lines to make it easier to calculate the offset when we jump, and added some details comments. (You should leave both of these bits out if you edit your program and run it in our model computer.)
1. `load r2 10 ; We want to add up the numbers up to 10`
2. `load r0 1 ; We want to start with 1`
3. `add r1 r0 ; add the current number in r0 to the total in r1`
4. `add r0 1 ; increment the current number by 1`
5. `compare r0 r2 ; compare the current number with the maximum number `
6. `jumplte -3 ; if the current number is less than or equal to the maximum number, jump up 3`
7. `write 0 r1 ; otherwise, write the accumulated total from r1 into memory location at offset 0`
8. `exit`
The critical bit of code in this program that implements the loop is this:
`add r1 r0`
`add r0 1`
`compare r0 r2`
`jumplte -3`
`write 0 r1`
Here's how it works.
Once we've added the current number stored in `r0` to our accumulated total in `r1`, we increment ("add 1 to") `r0` so that it contains the next number to add. `r2` still contains the largest number we are interested in.
So, if the current number (`r0`) is less-than-or-equal-to the maximum number (`r2`), then we go back up 3 instructions. This takes us back to the line that adds the current number to the total. We keep going round and round like this, each time jumping back to the `add `instruction.
What happens when the current number is greater than the maximum number? We've seen before that in that case the `JUMPLTE `instruction does nothing, so we will step on and write the result into memory and exit.
Let's redraw the flowchart, with our actual instructions in the boxes so you can see how precisely similar they are. (We've added a hexagonal symbol on the diagram, which we use to represent steps that are purely a preparation for a comparison.)
Now that we've re-written the algorithm like this, it is easy to adapt it to solve problems (B) and (C) from the previous exercise.
For problem (B), to change the range of numbers to add from 1 to 15, we simply increase the value stored in `r2`
`load r2 15 ; We want to add up the numbers up to 15`
`load r1 0 ; Set the accumulated total to zero`
`load r0 1 ; We want to start with 1`
`add r1 r0 ; add the current number in r0 to the total in r1`
`add r0 1 ; increment the current number by 1`
`compare r0 r2 ; compare the current number with the maximum number `
`jumplte -3 ; if the current number is less than or equal to the maximum number, jump up 3`
`write 0 r1 ; otherwise, write the accumulated total from r1 into memory location at offset 0`
`exit`
For problem (C), to change the start of the range, we change the value stored in `r0`:
`load r2 15 ; We want to add up the numbers up to 15`
`load r1 0 ; Set the accumulated total to zero`
`load r0 5 ; We want to start with 5`
`add r1 r0 ; add the current number in r0 to the total in r1`
`add r0 1 ; increment the current number by 1`
`compare r0 r2 ; compare the current number with the maximum number `
`jumplte -3 ; if the current number is less than or equal to the maximum number, jump up 3`
`write 0 r1 ; otherwise, write the accumulated total from r1 into memory location at offset 0`
`exit`
That's great - we could imagine increasing the total of numbers in the sequence to any arbitrarily large number, and we wouldn't run out of program memory.
Let's remind ourselves of the original program:
`load r1 0`
`add r1 1`
`add r1 2`
`add r1 3`
`add r1 4`
`add r1 5`
`add r1 6`
`add r1 7`
`add r1 8`
`add r1 9`
`add r1 10`
`write 0 r1`
`exit`
This needed as many instructions as there were numbers to add, plus a few instructions to set up and finish. We say that the number of instructions in the program instructions memory scales linearly with the number of items to process, or is of order n (where 'n' is the number of items to be processed). We have a shorthand notation for this called big O notation: In general, we estimate the order of an algorithm by finding the bits that grow the fastest as we add more items into the input.
We could represent the storage cost of our algorithm diagrammatically, drawing a box scaled to the number of instructions required to implement each part of the algorithm. Now, imagine that we double the number of items in the sequence and sketch it again (reducing the scale by half so it fits on the page) And double the number of items again, reducing the scale by half once more. Each time, the central "add" section becomes a more and more significant part of the storage cost of the algorithm.
Now, imagine a huge number of items. The contribution of the start and finish become vanishingly small, and the budget is taken up (to all intents and purposes) entirely by the central piece of the work, so we can ignore them when we estimate the order of our algorithm.
So, back to our program. With the new version, no matter how big the number, we only need a program 9 instructions long. So, we've really optimized our program storage. The storage requirement of this program is constant, or of order 1: O(1).
(We will look at how to calculate this more formally later on.)
This is clearly a good thing. But, if you look closely, you'll realise we're actually executing a lot more instructions than we were before. To add each number in the original program, here's the code we add to execute:
`add r1 3`
But to add each number in the new program, here's the code we have to execute:
`add r1 r0`
`add r0 1`
`compare r0 r2`
`jumplte -3`
Four instructions in the new program for each instruction in the old program. So the new program is more computationally expensive. About four times more, in fact, if you ignore the few instructions they both have for setup and completion, and you assume that all those instructions take the same amount of time to execute (which they almost certainly don't – that `jumplte` is probably more effort than an add).
This may (or may not) be significant to us.
If we think about the computational cost of both implementations, they scale linearly with the number of items in the collection, so both the old and the new algorithm are O(n), but the new one is more expensive - especially for just a few items in the sequence. It could be cheaper to write the instructions out individually if we have less than, say, 4 numbers in the sequence, than it is to use the loop.
We call this process of trying to find a better, cheaper way to write our algorithms optimization. In this case, we've optimized our program for program storage, at the expense of some speed. We've made a significant improvement in the former (going from O(n) to O(1)), while still maintaining our O(n) computation cost – albeit that the new algorithm is potentially 4 times slower.
You will often find that you are making these trade offs - space for time, or vice versa. Sometimes, you discover that an optimization for space actually improves the time as well - if a significant portion of the time was taken allocating or deallocating resources.
## Optimization (premature & otherwise)
There's another point to make about optimization, though. I'm using a lot of contingent words like 'could' and 'may or may not' in terms of the computational cost or benefit of one algorithm or another. This is because we haven't measured it. And while it is almost certainly the case that an order-of-magnitude improvement in the efficiency of an algorithm in either storage or computation (or both) will improve the overall performance of your software, that is by no means guaranteed, especially where there are complex interactions with other parts of the system, and different usage patterns. We could expend a week's effort trying to improve this algorithm, get it as good as it could possibly be, and yet it would still not give us any real benefit, because the end users only cause the code to be executed once a month, it completes in 50ms, and that happens overnight as part of a job that takes 20 hours to run. Worse still, we didn't measure how long it took in a variety of real-life situations to provide a baseline, and then repeat those measurements to see if our so-called "optimized" version has actually had a positive effect
One of the things programmers, in general, are very bad at doing, is optimizing irrelevant things. We call this premature optimization. We also tend to focus on the optimization of little details. We call this micro-optimization. It is very tempting to get drawn into this as we have to be so focused on the minutiae of the code we are writing, that we can lose the big picture. Our optimizations can also make the code harder to read, and more difficult to maintain - and perhaps also more prone to bugs.
You could argue that both the programs we've written so far implement the same basic algorithm - they step through each item in the sequence in turn, adding it to the total. The second version uses a loop to do that. The first version just expanded it so that there is an instruction that represents each iteration around the loop (we call that loop unrolling).
If we're going to significantly improve both the storage requirements and the computational cost of this program, we're going to need an algorithm that is O(1) for both storage and computation. Algorithm selection almost always has a bigger impact on performance than micro-optimizations. So, can we come up with a better one?
## Looking for a better algorithm
Let's look at the sum of the numbers from 1 to 15 again.
We can write this as (leaving out some of the terms in the middle to avoid getting too bored…) Equally we could start at the other end, and write it as We get exactly the same answer.
Now we're going to do a bit of simple algebra - we're going to add these two equations together. In this case, the equations are so simple that we can do that by adding up all the terms on the left hand sides, and all the terms on the right hand sides. I've coloured them according to the original equations so you can see where the terms came from. Let's look at the right hand side of equation 3 more closely. For each of the 15 terms in our original sum, we've still got exactly one term - and it's an addition. Moreover, that sum is always the same - in this case 16. In fact, we've got 15 lots of 16.
Rewriting that again, then And dividing through by 2: Which looks very much like the right answer.
Before we delve into this further, let's try that again for the case when we're adding up the numbers from 5 to 15.
There are 11 numbers between 5 and 15 (inclusive). First, we'll write them out from low to high And then from high to low Again, we have one term on the right hand side of equation 3a for each of the 11 terms we had in the original sum, and this time each of those terms adds up to 20.
So, Right again.
Now, let's think about the general case of some starting integer m to some general maximum integer n.
First, how do we calculate the number of terms in this case?
Well, in the case of the sequence 1…15, there are 15 terms, which is (15 - 1) + 1.
For the sequence 5…15 there are 11 terms, which is (15 - 5) + 1.
In the general case, there are (n - m) + 1 terms, i.e. Another way of looking at the last term 'n' is to say that it is the starting term, plus the number of terms, less one (to account for the fact that we've already included the first term.) i.e. From our previous equation, we can see that So, what does our sum look like if we write it out in these general terms?
Here's the version starting with the first number in the sequence, m. Notice how we've expanded the last few terms to calculate them using our expression for the number of items in the sequence. This looks a bit redundant: after all, m + (n-m-1) = n-1 and m + (n-m) = n, but you'll see why it is important we express it that way in a moment.
As before, we'll write that the other way around, high to low, expressing it in terms of the highest number in the sequence, and our expression for the number of terms. And then add them up again Here, you can quickly see that each term adds up to (m + n) - which is suggestive of a solution. (Note that this isn't a proper proof! Anything could be happening in those bits in the middle. But, as it happens, this solution is correct, and generalizable.)
So, just as before, we can write our equation as
2S = (number of terms) * (value of each term) , or If we divide both sides by 2, we have a general formula for calculating the sum of all the integers from m to n: Or we could use the mathematical sigma notation - it means the same thing: "the sum of each i from i = m to i = n". Armed with this new algorithm, we can write a much better program to calculate the sum, in terms of both the space it needs, and the number of instructions it has to execute.
### Exercise 1
Can you devise a program to calculate the sum of any sequence of consecutive, positive integers?
Hint: There are instructions in our computer called `mul` and `div` which perform multiplication and division just like our `add `and `sub `instructions.
### Exercise 2
A geometric progression is one in which each term is a constant multiple of the previous term: For example 3, 6, 12, 24, 48 is a geometric progression where The formula for the sum of a geometric progression (which we call a geometric series) where a is the value of the first term in the sequence, and r is the constant multiplier is: Implement a program that can calculate the result of the geometric series for any arbitrary values of a, r, m and n.
Can you estimate the compute cost of your implementation, using big O notation? What about the storage cost?
## Summary
We've learned several important things in this section:
1. Several ways to represent algorithms, in the forms of code, flowcharts, pseudo-code and just ordinary language.
2. How to estimate the order of an algorithm, in terms of both its storage requirements and its computational cost, and how to communicate that using big O notation (e.g. O(1), O(n), O(n^2 ) etc.)
3. We've seen how to go about optimizing our algorithms for either space, storage or both - including the importance of measuring both before and after any change to see if it has had any significant positive effect for the effort we've put in to doing it
4. We've seen the importance of picking the best algorithm for the job, given the kind of input you are working with, and that the selection of a good algorithm can have far more impact on the cost of execution than micro-optimizations of the implementation.
In the next section, we're going to look at the data memory in more detail, and learn how we can represent numbers in a computer.
Learning To Program – A Beginners Guide – Part One - Introduction
Learning To Program – A Beginners Guide – Part Two - Setting Up
Learning To Program – A Beginners Guide – Part Three - What is a computer?
Learning To Program – A Beginners Guide – Part Four - A simple model of a computer
Learning To Program – A Beginners Guide – Part Five - Running a program
Learning To Program – A Beginners Guide – Part Six - A First Look at Algorithms
Learning To Program – A Beginners Guide – Part Seven - Representing Numbers
Learning To Program – A Beginners Guide – Part Eight - Working With Logic
Learning To Program – A Beginners Guide – Part Nine - Introducing Functions
Learning To Program – A Beginners Guide – Part Ten - Getting Started With Operators in F#
Learning to Program – A Beginners Guide – Part Eleven – More With Functions and Logic in F#: Minimizing Boolean Expressions
Learning to Program – A Beginners Guide – Part Twelve – Dealing with Repetitive Tasks - Recursion in F# | HuggingFaceTB/finemath | |
# difference between implicit and explicit solutions?
What is difference between implicit and explicit solution of an initial value problem? Please explain with example both solutions(implicit and explicit)of same initial value problem? Or without example but in some way that is understandable.
thanks
• Before anything else: from where did you encounter these terms? (I'm trying to gauge how to properly answer your question to suit you.) – J. M. is a poor mathematician Oct 29 '11 at 11:03
• during my assignment, I was making solution and I encounter these terms, I saw that overall these two are similar then where is difference? Is it just matter of writting the final solution in different way or actually procedure for finding implicit or explicit solution is different? – Hafiz Oct 29 '11 at 11:21
• these are common terms in differential equations – Hafiz Oct 29 '11 at 11:22
• Okay. Take for instance the differential equation $y^\prime y=-x$ with initial condition $y(0)=r$. The implicit solution of this differential equation is $x^2+y(x)^2=r^2$; here $y(x)$ is implicitly defined. The explicit solutions look like $y(x)=\pm\sqrt{r^2-x^2}$; the solution is "explicit" in that the expression for $y(x)$ can entirely be expressed in terms of $x$. Here, we are lucky to get an explicit solution since we know how to solve quadratics; it often happens that we can only be content with an implicitly expressed $y(x)$, like in the case of $y(x)-\varepsilon\sin(y(x))=x$... – J. M. is a poor mathematician Oct 29 '11 at 11:30
• Sometimes you solve an differential equation, and the answer is something of the type : $y+x=\sin(xy)$. While you still don't know exactly (i.e. explicitly) what $y$ is, this relation usually yields enough information to answer to many questions about $y$. If you can find the solution as $y=f(x)$, that is always best, but if you cannot, an equation between $x,y$ is still much better than nothing. – N. S. Oct 29 '11 at 17:20
## 3 Answers
As requested:
Let's use the example initial-value problem
$$y^\prime y=-x,\qquad y(0)=r, \qquad r\text{ constant}$$
One can derive both an implicit and explicit solution for this DE. The implicit solution to this DE is
$$x^2+y(x)^2=r^2$$
This solution implicitly defines $y(x)$; all we have here is an equation involving $y(x)$. On the other hand, the explicit solution looks like
$$y(x)=\pm\sqrt{r^2-x^2}$$
and in this case, $y(x)$ is explicitly defined: $y(x)$ is expressed here as an explicit function with $x$ as the only independent variable.
We aren't always this lucky when we solve differential equations that show up in practice. It often happens that we can only be content with an implicit solution (or a parametric solution, which is a somewhat better state of affairs than having just an implicit solution). One famous example is the differential equation that pops up in the brachistochrone problem:
$$(1+(y^\prime)^2)y=r^2$$
• I am wondering what the benefit of an implicit function is... If we know that the solution of an ODE has to satisfy some equation $f(x,y)=0$, what is then the advantage? – asd Oct 28 '15 at 10:23
• @asd, because in some ways normal equations are easier to analyze than differential equations. – J. M. is a poor mathematician Oct 22 '17 at 13:12
Explicit solution is a solution where the dependent variable can be separated. For example, $x+2y=0$ is explicit because if y is dependent, I can rewrite it as $y=-\frac{x}{2}$ and my y has been separated.
Implicit is when the dependent variable cannot be separated like $\sin(x+e^y)=3y$.
• Strictly speaking (even as mentioned in the accepted answer) $y+2y=0$ is an implicit solution, which can be converted into the explicit form $y=-\frac{x}{2}$ – Anubis Oct 31 '17 at 21:53
Let us consider a differential equation $$x +yy' =0 \label{1}\tag{1}$$ Now, the relation $$x²+y² -25 =0$$ is an implicit solution of the above equation for all $$x\in(-5,5)$$. This relation define two real functions $$f_1(x) =\sqrt{25 -x^2}$$ and $$f_2(x) = -\sqrt{25 -x²}$$ in the interval $$x\in(-5,5)$$: here $$f_1,f_2$$ are explicit solutions of the differential equation.
Now is it quite easy to understand that what are implicit and explicit solutions?
Implicit solution means a solution in which dependent variable is not separated and explicit means dependent variable is separated.
Now consider the relation $$x² +y² +25 =0$$ Is it also an implicit solution of the differential equation \eqref{1}?
The answer is 'No': formally it is solution of the equation \eqref{1} but not implicitly because it won't be identically zero for any real values of $$x,y$$. Precisely If you write it in the form $$y = -\sqrt{-25 -x^2}\:\text{}.$$ you will get always an imaginary $$y$$ for all real $$x$$. | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
1st number: 1 5/15, 2nd number: 6/90
20/15 + 6/90 is 7/5.
1. Find the least common denominator or LCM of the two denominators:
LCM of 15 and 90 is 90
2. For the 1st fraction, since 15 × 6 = 90,
20/15 = 20 × 6/15 × 6 = 120/90
3. Likewise, for the 2nd fraction, since 90 × 1 = 90,
6/90 = 6 × 1/90 × 1 = 6/90
120/90 + 6/90 = 120 + 6/90 = 126/90
5. After reducing the fraction, the answer is 7/5
6. In mixed form: 12/5
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time: | HuggingFaceTB/finemath | |
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# integral of secx
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1. ....obviously ( I was told this was the first step)
2. (Original post by wcp100)
....obviously ( I was told this was the first step)
Immediately there's a problem as your limits do not make sense. They must either be in radians or as a fraction of 360.
What you need to note is that your integrand can be rewritten as:
What is the derivative of ? Does it look familiar? What do you do with integrals with integrands of the form ? *cough* natural logarithm *cough*.
3. Recall the integral of f'(x)/f(x).
4. I don't like that way of finding the integral of the secant.
Why the hell would we multiply by that otherwise?
5. this needs to be moved to the health and relationships forum.
6. (Original post by a²+b² = c²)
I don't like that way of finding the integral of the secant.
Why the hell would we multiply by that otherwise?
You were thinking exactly what I was. Is there another way of doing it ?
7. (Original post by a²+b² = c²)
I don't like that way of finding the integral of the secant.
Why the hell would we multiply by that otherwise?
(Original post by wcp100)
You were thinking exactly what I was. Is there another way of doing it ?
Yes, there is.
Note that .
Then use a substitution of and use partial fractions.
8. (Original post by wcp100)
....obviously ( I was told this was the first step)
well you could just substitute in du for
and u for
to get
9. (Original post by wcp100)
You were thinking exactly what I was. Is there another way of doing it ?
You can also use the well known t substitution.
t=tan(x/2)
cos x = (1-t^2)/(1+t^2)
dx/dt=2/(1+t^2)
It's all very straightforward and requires no cleverness at all.
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# Question be392
Mar 15, 2017
We need the quotient, $\text{moles of solute"/"volume of solution}$ to solve for $\text{normality}$ (i.e. $\text{molarity}$).
#### Explanation:
We need the quotient, $\text{moles of solute"/"volume of solution}$
In $100 \cdot g$ of SOLUTION there are $17 \cdot g$ of $H C l$ solute; the volume of this solution is "Mass"/("density"(rho))=(100*g)/(1.21*g*mL^-1) $=$ $82.6 \cdot m L$.
Given this we can calculation the concentration with respect to molarity:
$\text{Concentration} = \frac{\frac{17 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1}}{82.6 \cdot m L \times {10}^{-} 3 \cdot m L \cdot {L}^{-} 1} \cong 6.0 \cdot m o l \cdot {L}^{-} 1$
The most concentrated hydrochloric acid you can buy("mass"/"mass")xx100% is approx. 36%. This corresponds to a concentration of approx. $12.0 \cdot m o l \cdot {L}^{-} 1$. Note that all I have done here is approach the problem dimensionally, and most of the problems you encounter as an undergrad may be solved by this method.
Also note that pure $H C l$ would be a room temperature gas ("boiling point"=-85.5 ""^@C)#. | HuggingFaceTB/finemath | |
# Significance of Jordan Canonical Form
1. Dec 19, 2014
### thelema418
I just finished a course on linear algebra which ended with Jordan Canonical Forms. There were many statements like "Jordan canonical forms are extremely useful," etc. However, we only learned a process to put things into Jordan canonical form, and that was it.
What makes Jordan canonical forms significant? Why would they be useful?
2. Dec 19, 2014
### coquelicot
Indeed, Jordan canonical form is very useful because
1) you cannot always diagonalize a matrix. This acts as a replacement for it.
For example, if you need to solve a system of linear differential equations, and cannot diagonalize the matrix of the system, then you put it into Jordan canonical form, and this leads to a solution (see textbooks).
Similarly, if you have a system of linear recurrence equations, and you cannot diagonalize the system, you put it into Jordan canonical form and solve the system. In brief, each time you would use diagonalization and you cannot diagonalize, use Jordan canonical form.
2) Theoretical aspects : this is used in several arguments in abstract mathematics
3. Dec 19, 2014
### Stephen Tashi
Do you accept that doing calculations (theoretcial or practical) with matrices is useful? If you accept that idea then one explanation is that Jordan canonical form makes doing calculations easier.
For example, suppose we have a matrix $A$ and we want to compute $A^{10}$. Suppose you find a matrix $B$ such that $B^{-1} A B = J$ where $J$ is in Jordan cannonical form. To compute $J^{10}$ you only have to raise the Jordan blocks to the 10th power. They individual blocks are usually smaller matrices than the matrix $A$. From $J^{10}$ you can compute $A^{10}$ from $A^{10} = B J^{10} B^{-1}$
You can think of Jordan cannonical form as a generalization of the the concept of a "diagonal matrix". It's easy to to multiplications and find inverses (if they exist) for diagonal matrices. If you can find a invertible matrix $B$ that makes matrix $B^{-1} A B = D$ where D is diagonal matrix then you can find $A^{10}$ by the procedure described above. To compute $D^{10}$ you only have to take the 10th power of each diagonal entry.
Why not use diagonal matrices and not worry about Jordan cannonical form? - it's because you can't always find an invertible matrix $B$ such that $B^{-1} A B$ is a diagonal matrix. The Jordan Cannonical form is important because it is the most nearly diagonal format that you can get reliably..
In applications of matrix algebra, expressions of the form $B^{-1}$ (...stuff...) $B$ often have a physical interpretation. The indicate "Change the coordinate system. Do the "stuff" operation in the new coordinate system.. Then change the coordinates of the result back to the original coordinate system. (It's analogous to reading a problem posed in cartesian coordinates, changing to polar coordinates to solve it and then changing back to cartesian coordinates to report the answer.)
If you want to compute the matrix product $A^2 C^3$ then you can daydream about finding an invertible matrix $B$ such that $B^{-1} A B$ and $B^{-1} C B$ are both diagonal. You use $B$ to change coordinates, you do the computation in the coordinate system where the matrices are diagonal. Then use $B^{-1}$ to change back to the original coordinate system.
You can't always find a matrix $B$ that simultaneously diagonalizes two matrices. You can't always find a matrix $B$ that changes coordinates so two matrices have the same Jordan Block structure. It's important to know special cases when you can. So it's of interest to have theorems that say "If ...so-and-so then the two matrices can be simultaneously diagonalized" or "put in compatible Jordan block form".
An even more general concept of expressing matrices in a convenient format is the "singular value decomposition" - but's that's a different topic.
4. Dec 19, 2014
### thelema418
Just for clarification, this part of the class did not revolve around a textbook. There was just a handout about how to put a matrix into Jordan canonical form using an algorithm for finding the dot representation of the cycles. So, I don't know how this connects to textbook problems or to real world problems.
I do work with partial and ordinary differential equations that often involves using matrix calculations, but I do not see how this would connect. I understand that if I did not have a calculator, that the process of computing A raised to a power on a similar matrix B could be easier. Yet, calculators and computers can calculate large matrixes raised to a power quickly using the standard left multiplication procedure.
Is this procedure efficient? Or, is it just worthy as a trick to handle situations where you have to calculate by hand?
5. Dec 19, 2014
### Stephen Tashi
Then perhaps you have encountered the exponential function $e^{tA}$ where $A$ is a matrix.
If you put a matrix in diagonal or Jordan cannonical form, you can often see what the result of long calculation will be without doing it - for example, taking the limit of a sequence of matrix operations. The function $e^{tA}$ is defined as a limit of an infinite series involving powers of the matrix $A$.
There are many numerical algorithms for matrices. If ordinary matrix multiplication met the needs of the world and its computers then many of them would be unnecessary.
6. Dec 19, 2014
### coquelicot
The problem is not always to compute something, but to obtain an analytic closed expression of something; For example, you may compute the Fibonacci sequence for a given initial condition up to a certain step n, but this does not provide you a general formula for the sequence, and hence you can not obtain approximations, general theorems etc. This is a general philosophical question that is important to understand in mathematics : the fact that you have an algorithm to obtain the solutions of a problem does not mean at all that you have entirely understood it, nor that this provides the tools to deal with questions related to the problem. For example, Lagrange was the first to give the algorithmic solution of the binary quadratic equations in numbers, but this gives nothing about question like "in how many way is it possible to divide a prime numbers in two squares" and an infinity of other questions. This is why Gauss wrote his famous Disquisitiones, where he studied the theoretical aspects of the binary quadratic equations in numbers.
7. Dec 22, 2014
### mathwonk
To see the connection between differential equations and Jordan forms, recall that a linear constant coefficient differential operator is a product of powers like (D-a)^n, where D stands for differentiation, and a is multiplication by the constant a. Then consider the kernel of such a basic operator acting on the space of smooth functions, e.g. for n = 5, and look at the matrix for D in an essentially standard basis for this space. Note it is already in jordan form:
E.g.: The matrix of D on ker(D-a)^5, with basis {(e^at)(t^r/r!)},0≤r≤4, is:
| a 1 0 0 0 |
| 0 a 1 0 0 |
| 0 0 a 1 0 |
| 0 0 0 a 1 |
| 0 0 0 0 a |, a classic example of a Jordan block, i.e. a “stretch plus a shift”. | HuggingFaceTB/finemath | |
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quiz1-key
# quiz1-key - 7(two points Using only U E L N and the three...
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Quiz 1. January 19, 2011 Seat # _________ Name: ___ < KEY > ___ Closed book and notes. No calculators. Set theory. For all questions below, consider the universe U = {1, 2, 3, 4, 5, 6}. Let E = {2, 4, 6}, the set of even integers. Let L = {4, 5, 6}, the set of large integers. Let N = {1, 4, 6}, the set of non-prime numbers. Recall that denotes the empty set, denotes union, denotes intersection, and denotes complement. 1. (one point) T F E ′= {1, 3, 4}. 2. (one point) T F U ′= . 3. (one point) T F L U = L . 4. (one point) T F ( L N ) ′= L N . 5. (one point) T F E E ′= U . 6. (one point) Sketch a Venn diagram showing U , E , and E . ____________________________________________________________ Many correct solutions. Maybe sketch a rectangle and label it U . Maybe partition the area U into six sets labeled one through six. Shade the sets two, four, and six and mark the shaded set as E and the rest as E . Or simply partition U into E and E
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Unformatted text preview: . ____________________________________________________________ 7. (two points) Using only U , E , L , N and the three set operations, write the set composed of odd integers that are also large. ____________________________________________________________ E ′ ∩ L ← ____________________________________________________________ 8. (two points) Using only U , E , L , N and the three set operations, write the set composed of integers that are neither even nor prime. ____________________________________________________________ E ′ ∩ N or ( E ∪ N ′ ) ′ ← ____________________________________________________________ ______________________________________________________________________ Write on the back any concerns about the weekly quizzes or the course in general. IE 230. – Page 1 of 1 – Schmeiser...
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### Grade 5 - Mathematics5.37 Fractions Review
A fraction is another way of expressing division. The expression x/y is also written as x ÷ y. x is known as the numerator and y is known as denominator. A fraction written as a combination of a whole number and a proper fraction is called a mixed fraction or mixed number. To generate equivalent fractions of any given fraction, we proceed as follows: Multiply the numerator and denominator by the same number (other than 0) or Divide the numerator and denominator by their common factor (other than 1), if any. The two fractions are equivalent if the product of numerator of the first and denominator of the second is equal to the product of the denominator of the first and numerator of the second. Reducing fractions: To reduce a fraction, divide both the numerator and denominator by the largest factor of both. A fraction is in its lowest term, if the numerator and denominator have no common factor other than 1. Comparing fractions: Of the two fractions having the same denominator, the one with greater numerator is greater. Of the two fractions having the same numerator, the one with greater denominator is smaller. Operations with Fractions: To add or subtract unlike fractions, first we convert them into like fractions and then perform the required operation on like fractions so obtained. The product of two or more fractions is a fraction whose numerator is the product of their numerators and whose denominator is the product of their denominators. When the product of two fractions or a fraction and a whole number is 1, then either of them is called the reciprocal of the other. The number zero (0) has no reciprocal. Dividing a fraction or a whole number by a fraction or a whole number (other than zero) is the same as multiplying the first by the reciprocal of the second. Directions: Answer the following questions.
Q 1: Which is the greatest and the least among these fractions - 1, 12/5, 2/3Answer: Q 2: To the quotient of 3/4 and 1/8, add 1/4.Answer: Q 3: In the fraction, 4/3, __ is the numerator.Answer: Q 4: Convert 1.3 to fraction.Answer: Q 5: What is one-third of 24?Answer: Q 6: Find the value of y in y/3 + 4/3 = 2Answer: Q 7: Which among these is an improper fraction - 12/5, 3/5, 6/5, 1 2/3Answer: Q 8: To the difference of 1 1/2 and 1/2, add 3/4.Answer: Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | HuggingFaceTB/finemath | |
### GTranslate Chapter 4
For a moment, think about what resistance looks like from the electrons’ point of view. Imagine you are an electron being pushed from atom to atom from the pressure of all your fellow electrons pushing on you. Think about your electron self, moving through the mix of glue binder to carbon particles, where you could either collide with an atom and be given off as heat in the form of atomic vibrations, or at other times, you are allowed to pass through the resistor to continue your path through the circuit.
It would stand to reason that if you had a certain pressure of electron friends pushing on you, lets say 9 volts (V or E) worth, and a certain amperage (I) working its way through the resistor of a given resistance, that your movement as an electron would be based on the relationship of all your pushy electron friends.
Georg Ohm (1789-- 1854) defined an important relationship between voltage, current, and resistance, now called Ohm’s law which states: a potential difference of one volt will force a current of one ampere through a resistance of one ohm.
Mathematically, Ohm's law can be written as: I = E/R, where I refers to the current, E is applied voltage, and R is the resistance in ohms. It can also be written as E = I ´ R or R = E/I.
The diagram below allows you to easily memorize this important law. By putting your finger on the value you are trying to solve, you will see the solution. So by putting your finger on the E, you solve for I and R or E = I x R or solving for I would be I = E divided by R. Ohms Law
To understand Ohm’s law and the relationship between voltage, resistance, and current flow, the water analogy is one way to think about this relationship. (directly below). Visualization of electron flow with a resistor as faucet analogy. More resistance means less electron flow.
The liquid flow decreases when the faucet is closed. This is analogous to high-value resistance constricting the width of the electron pipe, so fewer electrons pass in a given period. Therefore, high resistance decreases the flow of electrons, but the water analogy also allows identification of the idea of electron pressure or electron quantity as it relates to the possible flow.
With Ohm’s law, you can now calculate what size resistor you will need to protect a sensitive electronic part that is not able to withstand large quantities of electrons in a given period of time. Remember how we defined a fuse as a thin wire that would burn up and melt if too many electrons are being pushed through in a given period of time? Well, resistors can prevent sensitive parts from burning up by absorbing and giving off as heat some of that electron energy.
With Ohms law, if you are given any two values in a circuit, you can always solve for the third value using simple algebra. Often when you purchase parts the data sheet on the back of the package will say that the electronic device will work with a certain voltage like 5-volts DC. The package will also say that the electronic part only can withstand a certain amount of amperage or milliamperage.
With Ohm’s law, you can now calculate that a resistor of a particular size will allow a 5-volt source to deliver a certain amount of amperage in one second of time. Fortunately, Ohm’s law works for AC, DC, and radio frequencies (RF).
When working with electronics, it is important that we have a language to communicate electronic designs. The excitement of being able to read the maps will become apparent as you learn electronics since maps of possible circuits are freely available. A schematic is a map that tells us how to build circuits. They are constructed of symbols, which refer to different electronic parts.
When I first started studying electronics it was intimidating until I realized that artists and engineers are visual thinkers and can see things three dimensionally, however, engineers have also constructed a marvelous symbolic language to describe circuits and their connections, and this language also makes a lot of sense to artists.
For example, (image below) shows the symbol for the resistor and a 3D model of a resistor to the right. The crooked line communicates that as the electrons enter the resistor they are reduced in quantity or slowed down, as in a crooked road that may slow your car down. You can also think of the resistor symbols, crooked line, as a wave of heat which is the resistor getting hot with all the quantum vibrations of electrons striking the nuclei of the carbon atoms within the resistor. 470-ohm resistor: part on right and schematic drawing on left.
Following this paragraph is an example of a schematic that has a resistor, a lamp and a battery in a complete circuit. We will use this circuit schematic and some changing values to do some Ohm’s law calculations. | HuggingFaceTB/finemath | |
# 504 UK tablespoons in US quarts
## Conversion
504 UK tablespoons is equivalent to 7.98856286331041 US quarts.[1]
## Conversion formula How to convert 504 UK tablespoons to US quarts?
We know (by definition) that: $1\mathrm{brtablespoon}\approx 0.0158503231414889\mathrm{usquart}$
We can set up a proportion to solve for the number of US quarts.
$1 brtablespoon 504 brtablespoon ≈ 0.0158503231414889 usquart x usquart$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{usquart}\approx \frac{504\mathrm{brtablespoon}}{1\mathrm{brtablespoon}}*0.0158503231414889\mathrm{usquart}\to x\mathrm{usquart}\approx 7.988562863310405\mathrm{usquart}$
Conclusion: $504 brtablespoon ≈ 7.988562863310405 usquart$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 US quart is equal to 0.125178961111111 times 504 UK tablespoons.
It can also be expressed as: 504 UK tablespoons is equal to $\frac{1}{\mathrm{0.125178961111111}}$ US quarts.
## Approximation
An approximate numerical result would be: five hundred and four UK tablespoons is about seven point nine nine US quarts, or alternatively, a US quart is about zero point one three times five hundred and four UK tablespoons.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | HuggingFaceTB/finemath | |
# In the following figure, if AC = BD, then prove that AB = CD.
Solution:
From the figure, it can be observed that
$\mathrm{AC}=\mathrm{AB}+\mathrm{BC}$
$B D=B C+C D$
It is given that $A C=B D$
$\mathrm{AB}+\mathrm{BC}=\mathrm{BC}+\mathrm{CD}$(1)
According to Euclid’s axiom, when equals are subtracted from equals, the remainders are also equal.
Subtracting BC from equation (1), we obtain
$\mathrm{AB}+\mathrm{BC}-\mathrm{BC}=\mathrm{BC}+\mathrm{CD}-\mathrm{BC}$
$\mathrm{AB}=\mathrm{CD}$ | open-web-math/open-web-math | |
## Conversion formula
The conversion factor from centimeters to inches is 0.39370078740157, which means that 1 centimeter is equal to 0.39370078740157 inches:
1 cm = 0.39370078740157 in
To convert 1103 centimeters into inches we have to multiply 1103 by the conversion factor in order to get the length amount from centimeters to inches. We can also form a simple proportion to calculate the result:
1 cm → 0.39370078740157 in
1103 cm → L(in)
Solve the above proportion to obtain the length L in inches:
L(in) = 1103 cm × 0.39370078740157 in
L(in) = 434.25196850394 in
The final result is:
1103 cm → 434.25196850394 in
We conclude that 1103 centimeters is equivalent to 434.25196850394 inches:
1103 centimeters = 434.25196850394 inches
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 inch is equal to 0.0023028105167724 × 1103 centimeters.
Another way is saying that 1103 centimeters is equal to 1 ÷ 0.0023028105167724 inches.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand one hundred three centimeters is approximately four hundred thirty-four point two five two inches:
1103 cm ≅ 434.252 in
An alternative is also that one inch is approximately zero point zero zero two times one thousand one hundred three centimeters.
## Conversion table
### centimeters to inches chart
For quick reference purposes, below is the conversion table you can use to convert from centimeters to inches
centimeters (cm) inches (in)
1104 centimeters 434.646 inches
1105 centimeters 435.039 inches
1106 centimeters 435.433 inches
1107 centimeters 435.827 inches
1108 centimeters 436.22 inches
1109 centimeters 436.614 inches
1110 centimeters 437.008 inches
1111 centimeters 437.402 inches
1112 centimeters 437.795 inches
1113 centimeters 438.189 inches | HuggingFaceTB/finemath | |
Find the value of x if $\begin{bmatrix} x & 2 \\ 18 & x \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 18 & 6 \end{bmatrix}$
Toolbox:
• If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
Given
$\begin{bmatrix} x & 2 \\ 18 & x \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 18 & 6 \end{bmatrix}$
The above given two matrices are equal,hence their corresponding elements should be equal.
$\Rightarrow x=6.$ | HuggingFaceTB/finemath | |
# [Video] “Introduction to Cryptography” Writeup – Tryhackme “Security Engineer” Learning path
The purpose of this room is to introduce users to basic cryptography concepts such as:
• Symmetric encryption, such as AES
• Asymmetric encryption, such as RSA
• Diffie-Hellman Key Exchange
• Hashing
• PKI
Suppose you want to send a message that no one can understand except the intended recipient. How would you do that?
One of the simplest ciphers is the Caesar cipher, used more than 2000 years ago. Caesar Cipher shifts the letter by a fixed number of places to the left or to the right. Consider the case of shifting by 3 to the right to encrypt, as shown in the figure below.
The recipient needs to know that the text was shifted by 3 to the right to recover the original message.
Using the same key to encrypt “TRY HACK ME”, we get “WUB KDFN PH”.
The Caesar Cipher that we have described above can use a key between 1 and 25. With a key of 1, each letter is shifted by one position, where A becomes B, and Z becomes A. With a key of 25, each letter is shifted by 25 positions, where A becomes Z, and B becomes A. A key of 0 means no change; moreover, a key of 26 will also lead to no change as it would lead to a full rotation. Consequently, we conclude that Caesar Cipher has a keyspace of 25; there are 25 different keys that the user can choose from.
Consider the case where you have intercepted a message encrypted using Caesar Cipher: “YMNX NX FQUMF GWFAT HTSYFHYNSL YFSLT MTYJQ RNPJ”. We are asked to decrypt it without knowledge of the key. We can attempt this by using brute force, i.e., we can try all the possible keys and see which one makes the most sense. In the following figure, we noticed that key being 5 makes the most sense, “THIS IS ALPHA BRAVO CONTACTING TANGO HOTEL MIKE.”
Caesar cipher is considered a substitution cipher because each letter in the alphabet is substituted with another.
Another type of cipher is called transposition cipher, which encrypts the message by changing the order of the letters. Let’s consider a simple transposition cipher in the figure below. We start with the message, “THIS IS ALPHA BRAVO CONTACTING TANGO HOTEL MIKE”, and the key `42351`. After we write the letters of our message by filling one column after the other, we rearrange the columns based on the key and then read the rows. In other words, we write by columns and we read by rows. Also notice that we ignored all the space in the plaintext in this example. The resulting ciphertext “NPCOTGHOTH…” is read one row after the other. In other words, a transposition cipher simply rearranges the order of the letters, unlike the substitution cipher, which substitutes the letters without changing their order.
This task introduced simple substitution and transposition ciphers and applied them to messages made of alphabetic characters. For an encryption algorithm to be considered secure, it should be infeasible to recover the original message, i.e., plaintext. (In mathematical terms, we need a hard problem, i.e., a problem that cannot be solved in polynomial time. A problem that we can solve in polynomial time is a problem that’s feasible to solve even for large input, although it might take the computer quite some time to finish.)
If the encrypted message can be broken in one week, the encryption used would be considered insecure. However, if the encrypted message can be broken in 1 million years, the encryption would be considered practically secure.
Consider the mono-alphabetic substitution cipher, where each letter is mapped to a new letter. For example, in English, you would map “a” to one of the 26 English letters, then you would map “b” to one of the remaining 25 English letters, and then map “c” to one of the remaining 24 English letters, and so on.
For example, we might choose the letters in the alphabet “abcdefghijklmnopqrstuvwxyz” to be mapped to “xpatvrzyjhecsdikbfwunqgmol” respectively. In other words, “a” becomes “x”, “b” becomes “p”, and so on. The recipient needs to know the key, “xpatvrzyjhecsdikbfwunqgmol”, to decrypt the encrypted messages successfully.
This algorithm might look very secure, especially since trying all the possible keys is not feasible. However, different techniques can be used to break a ciphertext using such an encryption algorithm. One weakness of such an algorithm is letter frequency. In English texts, the most common letters are ‘e’, ‘t’, and ‘a’, as they appear at a frequency of 13%, 9.1%, and 8.2%, respectively. Moreover, in English texts, the most common first letters are ‘t’, ‘a’, and ‘o’, as they appear at 16%, 11.7% and 7.6%, respectively. Add to this the fact that most of the message words are dictionary words, and you will be able to break an encrypted text with the alphabetic substitution cipher in no time.
We don’t really need to use the encryption key to decrypt the received ciphertext, “Uyv sxd gyi siqvw x sinduxjd pvzjdw po axffojdz xgxo wsxcc wuidvw.” As shown in the figure below, using a website such as quipqiup, it will take a moment to discover that the original text was “The man who moves a mountain begins by carrying away small stones.” This example clearly indicates that this algorithm is broken and should not be used for confidential communication.
Answer the questions below
You have received the following encrypted message:
“Xjnvw lc sluxjmw jsqm wjpmcqbg jg wqcxqmnvw; xjzjmmjd lc wjpm sluxjmw jsqm bqccqm zqy.” Zlwvzjxj Zpcvcol
You can guess that it is a quote. Who said it?
Let’s review some terminology:
• Cryptographic Algorithm or Cipher: This algorithm defines the encryption and decryption processes.
• Key: The cryptographic algorithm needs a key to convert the plaintext into ciphertext and vice versa.
• plaintext is the original message that we want to encrypt
• ciphertext is the message in its encrypted form
A symmetric encryption algorithm uses the same key for encryption and decryption. Consequently, the communicating parties need to agree on a secret key before being able to exchange any messages.
In the following figure, the sender provides the encrypt process with the plaintext and the key to get the ciphertext. The ciphertext is usually sent over some communication channel.
On the other end, the recipient provides the decrypt process with the same key used by the sender to recover the original plaintext from the received ciphertext. Without knowledge of the key, the recipient won’t be able to recover the plaintext.
National Institute of Standard and Technology (NIST) published the Data Encryption Standard (DES) in 1977. DES is a symmetric encryption algorithm that uses a key size of 56 bits. In 1997, a challenge to break a message encrypted using DES was solved. Consequently, it was demonstrated that it had become feasible to use a brute-force search to find the key and break a message encrypted using DES. In 1998, a DES key was broken in 56 hours. These cases indicated that DES could no longer be considered secure.
NIST published the Advanced Encryption Standard (AES) in 2001. Like DES, it is a symmetric encryption algorithm; however, it uses a key size of 128, 192, or 256 bits, and it is still considered secure and in use today. AES repeats the following four transformations multiple times:
1. `SubBytes(state)`: This transformation looks up each byte in a given substitution table (S-box) and substitutes it with the respective value. The `state` is 16 bytes, i.e., 128 bits, saved in a 4 by 4 array.
2. `ShiftRows(state)`: The second row is shifted by one place, the third row is shifted by two places, and the fourth row is shifted by three places. This is shown in the figure below.
3. `MixColumns(state)`: Each column is multiplied by a fixed matrix (4 by 4 array).
4. `AddRoundKey(state)`: A round key is added to the state using the XOR operation.
The total number of transformation rounds depends on the key size.
Don’t worry if you find this cryptic because it is! Our purpose is not to learn the details of how AES works nor to implement it as a programming library; the purpose is to appreciate the difference in complexity between ancient encryption algorithms and modern ones. If you are curious to dive into details, you can check the AES specifications, including pseudocode and examples in its published standard, FIPS PUB 197.
In addition to AES, many other symmetric encryption algorithms are considered secure. Here is a list of symmetric encryption algorithms supported by GPG (GnuPG) 2.37.7, for example:
Encryption Algorithm Notes
AES, AES192, and AES256 AES with a key size of 128, 192, and 256 bits
IDEA International Data Encryption Algorithm (IDEA)
3DES Triple DES (Data Encryption Standard) and is based on DES. We should note that 3DES will be deprecated in 2023 and disallowed in 2024.
CAST5 Also known as CAST-128. Some sources state that CASE stands for the names of its authors: Carlisle Adams and Stafford Tavares.
BLOWFISH Designed by Bruce Schneier
TWOFISH Designed by Bruce Schneier and derived from Blowfish
CAMELLIA128, CAMELLIA192, and CAMELLIA256 Designed by Mitsubishi Electric and NTT in Japan. Its name is derived from the flower camellia japonica.
All the algorithms mentioned so far are block cipher symmetric encryption algorithms. A block cipher algorithm converts the input (plaintext) into blocks and encrypts each block. A block is usually 128 bits. In the figure below, we want to encrypt the plaintext “TANGO HOTEL MIKE”, a total of 16 characters. The first step is to represent it in binary. If we use ASCII, “T” is `0x54` in hexadecimal format, “A” is `0x41`, and so on. Every two hexadecimal digits constitute 8 bits and represent one byte. A block of 128 bits is practically 16 bytes and is represented in a 4 by 4 array. The 128-bit block is fed as one unit to the encryption method.
The other type of symmetric encryption algorithm is stream ciphers, which encrypt the plaintext byte by byte. Consider the case where we want to encrypt the message “TANGO HOTEL MIKE”; each character needs to be converted to its binary representation. If we use ASCII, “T” is `0x54` in hexadecimal, while “A” is `0x41`, and so on. The encryption method will process one byte at a time. This is represented in the figure below.
Symmetric encryption solves many security problems discussed in the Security Principles room. Let’s say that Alice and Bob met and chose an encryption algorithm and agreed on a specific key. We assume that the selected encryption algorithm is secure and that the secret key is kept safe. Let’s take a look at what we can achieve:
• Confidentiality: If Eve intercepted the encrypted message, she wouldn’t be able to recover the plaintext. Consequently, all messages exchanged between Alice and Bob are confidential as long as they are sent encrypted.
• Integrity: When Bob receives an encrypted message and decrypts it successfully using the key he agreed upon with Alice, Bob can be sure that no one could tamper with the message across the channel. When using secure modern encryption algorithms, any minor modification to the ciphertext would prevent successful decryption or would lead to gibberish as plaintext.
• Authenticity: Being able to decrypt the ciphertext using the secret key also proves the authenticity of the message because only Alice and Bob know the secret key.
We are just getting started, and we know how to maintain confidentiality, check the integrity and ensure the authenticity of the exchanged messages. More practical and efficient approaches will be presented in later tasks. The question, for now, is whether this is scalable.
With Alice and Bob, we needed one key. If we have Alice, Bob, and Charlie, we need three keys: one for Alice and Bob, another for Alice and Charlie, and a third for Bob and Charlie. However, the number of keys grows quickly; communication between 100 users requires almost 5000 different secret keys. (If you are curious about the mathematics behind it, that’s 99 + 98 + 97 + … + 1 = 4950).
Moreover, if one system gets compromised, they need to create new keys to be used with the other 99 users. Another problem would be finding a secure channel to exchange the keys with all the other users. Obviously, this quickly grows out of hand.
In the next task, we will cover asymmetric encryption. One of the problems solved with asymmetric encryption is when 100 users only need to share a total of 100 keys to communicate securely. (As explained earlier, symmetric encryption would require around 5000 keys to secure the communications for 100 users.)
There are many programs available for symmetric encryption. We will focus on two, which are widely used for asymmetric encryption as well:
• GNU Privacy Guard
• OpenSSL Project
### GNU Privacy Guard
The GNU Privacy Guard, also known as GnuPG or GPG, implements the OpenPGP standard.
We can encrypt a file using GnuPG (GPG) using the following command:
`gpg --symmetric --cipher-algo CIPHER message.txt`, where CIPHER is the name of the encryption algorithm. You can check supported ciphers using the command `gpg --version`. The encrypted file will be saved as `message.txt.gpg`.
The default output is in the binary OpenPGP format; however, if you prefer to create an ASCII armoured output, which can be opened in any text editor, you should add the option `--armor`. For example, `gpg --armor --symmetric --cipher-algo CIPHER message.txt`.
You can decrypt using the following command:
`gpg --output original_message.txt --decrypt message.gpg`
### OpenSSL Project
The OpenSSL Project maintains the OpenSSL software.
We can encrypt a file using OpenSSL using the following command:
`openssl aes-256-cbc -e -in message.txt -out encrypted_message`
We can decrypt the resulting file using the following command:
`openssl aes-256-cbc -d -in encrypted_message -out original_message.txt`
To make the encryption more secure and resilient against brute-force attacks, we can add `-pbkdf2` to use the Password-Based Key Derivation Function 2 (PBKDF2); moreover, we can specify the number of iterations on the password to derive the encryption key using `-iter NUMBER`. To iterate 10,000 times, the previous command would become:
`openssl aes-256-cbc -pbkdf2 -iter 10000 -e -in message.txt -out encrypted_message`
Consequently, the decryption command becomes:
`openssl aes-256-cbc -pbkdf2 -iter 10000 -d -in encrypted_message -out original_message.txt`
In the following questions, we will use `gpg` and `openssl` on the AttackBox to carry out symmetric encryption.
The necessary files for this task are located under `/root/Rooms/cryptographyintro/task02`The zip file attached to this task can be used to tackle the questions of tasks 2, 3, 4, 5, and 6.
Answer the questions below
Decrypt the file `quote01` encrypted (using AES256) with the key `s!kR3T55` using `gpg`. What is the third word in the file?
Decrypt the file `quote02` encrypted (using AES256-CBC) with the key `s!kR3T55` using `openssl`. What is the third word in the file?
Decrypt the file `quote03` encrypted (using CAMELLIA256) with the key `s!kR3T55` using `gpg`. What is the third word in the file?
Symmetric encryption requires the users to find a secure channel to exchange keys. By secure channel, we are mainly concerned with confidentiality and integrity. In other words, we need a channel where no third party can eavesdrop and read the traffic; moreover, no one can change the sent messages and data.
Asymmetric encryption makes it possible to exchange encrypted messages without a secure channel; we just need a reliable channel. By reliable channel, we mean that we are mainly concerned with the channel’s integrity and not confidentiality.
When using an asymmetric encryption algorithm, we would generate a key pair: a public key and a private key. The public key is shared with the world, or more specifically, with the people who want to communicate with us securely. The private key must be saved securely, and we must never let anyone access it. Moreover, it is not feasible to derive the private key despite the knowledge of the public key.
How does this key pair work?
If a message is encrypted with one key, it can be decrypted with the other. In other words:
• If Alice encrypts a message using Bob’s public key, it can be decrypted only using Bob’s private key.
• Reversely, if Bob encrypts a message using his private key, it can only be decrypted using Bob’s public key.
### Confidentiality
We can use asymmetric encryption to achieve confidentiality by encrypting the messages using the recipient’s public key. In the following two figures, we can see that:
Alice wants to ensure confidentiality in her communication with Bob. She encrypts the message using Bob’s public key, and Bob decrypts them using his private key. Bob’s public key is expected to be published on a public database or on his website, for instance.
When Bob wants to reply to Alice, he encrypts his messages using Alice’s public key, and Alice can decrypt them using her private key.
In other words, it becomes easy to communicate with Alice and Bob while ensuring the confidentiality of the messages. The only requirement is that all parties have their public keys available for interested senders.
Note: In practice, symmetric encryption algorithms allow faster operations than asymmetric encryption; therefore, we will cover later how we can use the best of both worlds.
### Integrity, Authenticity, and Nonrepudiation
Beyond confidentiality, asymmetric encryption can solve integrity, authenticity and nonrepudiation. Let’s say that Bob wants to make a statement and wants everyone to be able to confirm that this statement indeed came from him. Bob needs to encrypt the message using his private key; the recipients can decrypt it using Bob’s public key. If the message decrypts successfully with Bob’s public key, it means that the message was encrypted using Bob’s private key. (In practice, he would encrypt a hash of the original message. We will elaborate on this later.)
Being decrypted successfully using Bob’s public key leads to a few interesting conclusions.
• First, the message was not altered across the way (communication channel); this proves the message integrity.
• Second, knowing that no one has access to Bob’s private key, we can be sure that this message did indeed come from Bob; this proves the message authenticity.
• Finally, because no one other than Bob has access to Bob’s private key, Bob cannot deny sending this message; this establishes nonrepudiation.
We have seen how asymmetric encryption can help establish confidentiality, integrity, authenticity, and nonrepudiation. In real-life scenarios, asymmetric encryption can be relatively slow to encrypt large files and vast amounts of data. In another task, we will see how we can use asymmetric encryption in conjunction with symmetric encryption to achieve these security objectives relatively faster.
### RSA
RSA got its name from its inventors, Rivest, Shamir, and Adleman. It works as follows:
1. Choose two random prime numbers, p and q. Calculate N = p × q.
2. Choose two integers e and d such that e × d = 1 mod ϕ(N), where ϕ(N) = N − p − q + 1. This step will let us generate the public key (N,e) and the private key (N,d).
3. The sender can encrypt a value x by calculating y = xe mod N. (Modulus)
4. The recipient can decrypt y by calculating x = yd mod N. Note that yd = xed = xkϕ(N) + 1 = (xϕ(N))k × x = x. This step explains why we put a restriction on the choice of e and d.
Don’t worry if the above mathematical equations looked too complicated; you don’t need mathematics to be able to use RSA, as it is readily available via programs and programming libraries.
RSA security relies on factorization being a hard problem. It is easy to multiply p by q; however, it is time-consuming to find p and q given N. Moreover, for this to be secure, p and q should be pretty large numbers, for example, each being 1024 bits (that’s a number with more than 300 digits). It is important to note that RSA relies on secure random number generation, as with other asymmetric encryption algorithms. If an adversary can guess p and q, the whole system would be considered insecure.
Let’s consider the following practical example.
1. Bob chooses two prime numbers: p = 157 and q = 199. He calculates N = 31243.
2. With ϕ(N) = N − p − q + 1 = 31243 − 157 − 199 + 1 = 30888, Bob selects e = 163 and d = 379 where e × d = 163 × 379 = 61777 and 61777 mod 30888 = 1. The public key is (31243,163) and the private key is (31243,379).
3. Let’s say that the value to encrypt is x = 13, then Alice would calculate and send y = xe mod N = 13163 mod 31243 = 16342.
4. Bob will decrypt the received value by calculating x = yd mod N = 16341379 mod 31243 = 13.
The previous example was to understand the mathematics behind it better. To see real values for p and q, let’s create a real keypair using a tool such as `openssl`.
Terminal
``````user@TryHackMe\$ openssl genrsa -out private-key.pem 2048
user@TryHackMe\$ openssl rsa -in private-key.pem -pubout -out public-key.pem
writing RSA key
user@TryHackMe\$ cat public-key.pem
-----BEGIN PUBLIC KEY-----
MIIBIjANBgkqhkiG9w0BAQEFAAOCAQ8AMIIBCgKCAQEAymcAeYg1ohPQLHu7u9l1
UutN8bCP7r6czRX2zrQrpElYrm5mHERi1xweWEhTJ/0Q13FJcHLGtLbdQc0rGpOd
QmEctfucopZx5AVJ1vTn+qMv/0D6QU7Mm65MTSYg1SCRA0D0N9NLMj4rYlLOIr5q
5g3iunAE4tCROMcHf7fxWMuWdJTdtxTv7+4P5XGkWrWriO22JFHp9N22Fm96V9jH
dwIDAQAB
-----END PUBLIC KEY-----
user@TryHackMe\$ openssl rsa -in private-key.pem -text -noout
Private-Key: (2048 bit, 2 primes)
modulus:
00:ca:67:00:79:88:35:a2:13:d0:2c:7b:bb:bb:d9:
75:52:eb:4d:f1:b0:8f:ee:be:9c:cd:15:f6:ce:b4:
2b:a4:49:58:ae:6e:66:1c:44:62:d7:1c:1e:58:48:
53:27:fd:10:d7:71:49:70:72:c6:b4:b6:dd:41:cd:
2b:1a:93:9d:0e:76:09:06:ec:eb:a9:75:36:84:2e:
c4:ed:d9:6a:2e:c8:fa:dc:d3:c0:0e:a9:4e:19:80:
86:09:69:e6:fc:71:8c:d9:6b:95:b4:35:ea:46:2b:
4d:e3:37:8d:28:45:88:42:61:1c:b5:fb:9c:a2:96:
71:e4:05:49:d6:f4:e7:fa:a3:2f:ff:40:fa:41:4e:
cc:9b:ae:4c:4d:26:20:d5:20:91:03:40:f4:37:d3:
4b:32:3e:2b:62:52:ce:22:be:6a:e6:0d:e2:ba:70:
04:e2:d0:91:38:c7:07:7f:b7:f1:58:cb:96:74:94:
dd:b7:14:ef:ef:ee:0f:e5:71:a4:5a:b5:ab:88:ed:
b6:24:51:e9:f4:dd:b6:16:6f:7a:57:d8:c7:ed:a0:
12:46:42:19:bd:01:66:c7:ed:5d:97:b6:ed:64:38:
6c:9b:67:b3:53:4e:cb:05:a6:ef:f5:e7:e3:d2:02:
30:cf:a3:f7:99:32:55:9b:ec:31:68:31:fa:8e:25:
c1:77
publicExponent: 65537 (0x10001)
privateExponent:
10:fe:00:be:33:3f:3d:72:28:61:f3:a9:59:25:f2:
81:99:9b:9b:94:d5:20:98:04:15:fb:a8:12:c6:71:
7b:83:64:dc:90:0c:26:87:5f:3c:eb:f1:68:3b:fa:
2f:3b:41:b4:b4:a0:13:be:af:0b:f0:e6:36:66:01:
1e:64:12:25:6a:a7:6b:5b:6c:95:77:6f:b2:3d:32:
ef:3c:f7:7b:22:08:5d:8d:b1:6c:09:ae:b2:d9:65:
67:58:ea:b9:7a:d6:f6:51:df:e9:97:35:29:da:ec:
d9:0c:8a:df:3c:a7:29:db:79:4b:95:ea:1a:84:42:
df:7f:ca:29:2f:ba:62:02:37:05:c0:b0:c2:ff:42:
6b:fb:e1:36:40:10:ae:11:0f:d8:87:2f:fe:10:2e:
a4:60:de:ff:fe:c8:ab:0b:29:fa:6c:20:ec:87:33:
46:c0:cd:96:36:cb:9b:ca:81:17:e5:c3:eb:34:b2:
83:0f:52:cc:e9:68:bd:cb:d2:85:2f:fe:c4:47:76:
df:94:69:ce:7b:8a:50:71:36:96:e6:35:fb:fb:b4:
4a:ac:63:9b:9d:1b:bb:32:71:31:45:a2:25:33:cc:
f7:a5:fb:9f:66:b1:4e:30:ce:9d:71:e8:fa:7d:5f:
51
prime1:
00:e0:3d:87:b3:d3:1f:d2:c6:66:23:83:a5:95:d5:
20:35:f8:d8:c0:94:cf:cc:d2:04:d4:e4:ef:cf:c2:
94:00:10:cd:d1:4a:df:09:4e:7e:95:f8:70:08:b1:
20:98:8a:e3:88:f7:cc:a8:32:62:32:68:f6:1f:c0:
fb:c1:71:41:8c:21:a3:ff:20:e6:96:d0:6e:4b:66:
61:08:d0:b7:26:48:27:62:a7:d3:ff:36:55:c8:e1:
ab:91:48:90:fb:b5:b1:92:be:90:06:a8:40:1b:2a:
2d:53:1e:87:fc:a7:8a:57:72:0b:e5:35:71:7b:dd:
8c:e5:b5:ab:64:7c:37:c5:0d
prime2:
00:e7:11:ac:50:f5:dc:16:cf:20:46:77:5d:ca:16:
29:36:35:89:95:c0:f8:4b:42:ef:03:a0:f1:ce:2e:
1b:da:55:a9:ff:5a:28:4d:78:c5:8a:e2:55:9b:94:
b4:56:ec:ab:1b:dd:b8:07:be:dd:d5:0f:49:90:b3:
ed:a2:d7:78:38:24:d5:9e:7d:a2:e8:8c:e0:2a:33:
32:21:1f:0e:6b:aa:0b:b4:11:6a:bd:8f:d9:86:3f:
0a:8e:de:fb:1a:3c:51:9d:f2:dc:0a:59:80:d6:a4:
47:5c:02:a3:d0:30:1d:47:93
[...]``````
We executed three commands:
• `openssl genrsa -out private-key.pem 2048`: With `openssl`, we used `genrsa` to generate an RSA private key. Using `-out`, we specified that the resulting private key is saved as `private-key.pem`. We added `2048` to specify a key size of 2048 bits.
• `openssl rsa -in private-key.pem -pubout -out public-key.pem`: Using `openssl`, we specified that we are using the RSA algorithm with the `rsa` option. We specified that we wanted to get the public key using `-pubout`. Finally, we set the private key as input using `-in private-key.pem` and saved the output using `-out public-key.pem`.
• `openssl rsa -in private-key.pem -text -noout`: We are curious to see real RSA variables, so we used `-text -noout`. The values of p, q, N, e, and d are `prime1`, `prime2`, `modulus`, `publicExponent`, and `privateExponent`, respectively.
If we already have the recipient’s public key, we can encrypt it with the command `openssl pkeyutl -encrypt -in plaintext.txt -out ciphertext -inkey public-key.pem -pubin`
The recipient can decrypt it using the command `openssl pkeyutl -decrypt -in ciphertext -inkey private-key.pem -out decrypted.txt`
Answer the questions below
On the AttackBox, you can find the directory for this task located at `/root/Rooms/cryptographyintro/task03`; alternatively, you can use the task file from Task 2 to work on your own machine.
Bob has received the file `ciphertext_message` sent to him from Alice. You can find the key you need in the same folder. What is the first word of the original plaintext?
Take a look at Bob’s private RSA key. What is the last byte of p?
Take a look at Bob’s private RSA key. What is the last byte of q?
Alice and Bob can communicate over an insecure channel. By insecure, we mean that there are eavesdroppers who can read the messages exchanged on this channel. How can Alice and Bob agree on a secret key in such a setting? One way would be to use the Diffie-Hellman key exchange.
Diffie-Hellman is an asymmetric encryption algorithm. It allows the exchange of a secret over a public channel. We will skip the modular arithmetic background and provide a simple numeric example. We will need two mathematical operations: power and modulus. xp, i.e., x raised to the power p, is x multiplied by itself p times. Furthermore, x mod m, i.e., x modulus m, is the remainder of the division of x by m.
1. Alice and Bob agree on q and g. For this to work, q should be a prime number, and g is a number smaller than q that satisfies certain conditions. (In modular arithmetic, g is a generator.) In this example, we take q = 29 and g = 3.
2. Alice chooses a random number a smaller than q. She calculates A = (ga) mod q. The number a must be kept a secret; however, A is sent to Bob. Let’s say that Alice picks the number a = 13 and calculates A = 313%29 = 19 and sends it to Bob.
3. Bob picks a random number b smaller than q. He calculates B = (gb) mod q. Bob must keep b a secret; however, he sends B to Alice. Let’s consider the case where Bob chooses the number b = 15 and calculates B = 315%29 = 26. He proceeds to send it to Alice.
4. Alice receives B and calculates key = Ba mod q. Numeric example key = 2613 mod 29 = 10.
5. Bob receives A and calculates key = Ab mod q. Numeric example key = 1915 mod 29 = 10.
We can see that Alice and Bob reached the same key.
Although an eavesdropper has learned the values of q, g, A, and B, they won’t be able to calculate the secret key that Alice and Bob have exchanged. The above steps are summarized in the figure below.
Although the numbers we have chosen make it easy to find a and b, even without using a computer, real-world examples would select a q of 256 bits in length. In decimal numbers, that’s 115 with 75 zeroes to its right (I don’t know how to read that either, but I was told it is read as 115 quattuorvigintillion). Such a large q will make it infeasible to find a or b despite knowledge of q, g, A, and B.
Let’s take a look at actual Diffie-Hellman parameters. We can use `openssl` to generate them; we need to specify the option `dhparam` to indicate that we want to generate Diffie-Hellman parameters along with the specified size in bits, such as `2048` or `4096`.
In the console output below, we can view the prime number `P` and the generator `G` using the command `openssl dhparam -in dhparams.pem -text -noout`. (This is similar to what we did with the RSA private key.)
Terminal
``````user@TryHackMe\$ openssl dhparam -out dhparams.pem 2048
Generating DH parameters, 2048 bit long safe prime
[...]
\$ openssl dhparam -in dhparams.pem -text -noout
DH Parameters: (2048 bit)
P:
00:82:3b:9d:b5:29:31:f8:12:fe:21:e1:90:30:37:
ac:d2:48:41:f7:d7:55:e5:d2:5d:dd:87:67:9e:bd:
b3:97:df:05:a9:d2:d9:56:4f:66:b5:d9:d8:65:06:
58:c3:8f:b3:0e:30:d2:9a:0b:c3:0a:56:8d:fc:0f:
f2:e2:9e:4f:16:16:93:4e:b9:a4:c3:9c:09:2d:48:
a2:ec:b6:97:92:63:a3:b4:75:36:3f:51:77:ca:ac:
44:6d:99:eb:4d:4a:97:d5:4b:52:c8:07:f8:16:30:
37:d3:b2:47:30:e6:4e:bc:6a:53:d1:9b:6a:4d:91:
7a:4b:4f:af:3b:f0:ce:b9:ed:91:4d:8b:52:5a:3f:
bb:6b:06:ae:32:95:7d:53:da:9b:ce:b0:ec:7d:81:
25:05:d8:ce:ca:76:e7:d1:5a:31:13:d2:9f:62:b4:
be:61:49:cc:47:21:d8:e0:2c:e8:c6:35:4b:2f:ba:
35:36:8f:bb:41:c6:89:b2:60:3c:62:bb:fe:bf:59:
d3:7f:05:69:55:dc:61:1b:b4:bb:68:fa:65:1e:2e:
46:2f:2d:21:62:d1:9f:a0:2b:aa:81:df:3a:f9:7d:
0b:9d:0e:47:68:01:4f:6e:81:cc:4c:2a:91:fc:8c:
f4:6f
G: 2 (0x2)``````
Diffie-Hellman key exchange algorithm allows two parties to agree on a secret over an insecure channel. However, the discussed key exchange is prone to a Man-in-the-Middle (MitM) attack; an attacker might reply to Alice pretending to be Bob and reply to Bob pretending to be Alice. We discuss a solution to this problem in Task 6.
Answer the questions below
On the AttackBox, you can find the directory for this task located at `/root/Rooms/cryptographyintro/task04`; alternatively, you can use the task file from Task 2 to work on your own machine.
A set of Diffie-Hellman parameters can be found in the file `dhparam.pem`. What is the size of the prime number in bits?
What is the prime number’s last byte (least significant byte)?
A cryptographic hash function is an algorithm that takes data of arbitrary size as its input and returns a fixed size value, called message digest or checksum, as its output. For example, `sha256sum` calculates the SHA256 (Secure Hash Algorithm 256) message digest. SHA256, as the name indicates, returns a checksum of size 256 bits (32 bytes). This checksum is usually written using hexadecimal digits. Knowing that a hexadecimal digit represents 4 bits, the 256 bits checksum can be represented as 64 hexadecimal digits.
In the terminal output below, we calculate the SHA256 hash values for three files of varying sizes: 4 bytes, 275 MB, and 5.2 GB. Using `sha256sum` to calculate the message digest for each of the three files, we get three completely different values that appear random. It is worth stressing that the length of the resulting message digest or checksum is the same, no matter how small or big the file is. In particular, the four-byte file `abc.txt` and the 5.2 GB file resulted in message digests of equal length independent of the file size.
Terminal
``````user@TryHackMe\$ ls -lh
total 5.5G
-rw-r--r--. 1 strategos strategos 4 7月 21 12:46 abc.txt
-rw-r--r--. 1 strategos strategos 275M 2月 12 19:08 debian-hurd.img.tar.xz
-rw-r--r--. 1 strategos strategos 5.2G 4月 26 16:55 Win11_English_x64v1.iso
\$ sha256sum *
c38bb113c89d8fec6475a9936411007c45563ecb7ce8acd5db7fb58c0872bda0 abc.txt
0317ff0150e0d64b70284b28c97bb788310585ea7ac46cc8139d5a3c850dea55 debian-hurd.img.tar.xz
4bc6c7e7c61af4b5d1b086c5d279947357cff45c2f82021bb58628c2503eb64e Win11_English_x64v1.iso``````
But why would we need such a function? There are many uses, in particular:
• Storing passwords: Instead of storing passwords in plaintext, a hash of the password is stored instead. Consequently, if a data breach occurs, the attacker will get a list of password hashes instead of the original passwords. (In practice, passwords are also “salted”, as discussed in a later task.)
• Detecting modifications: Any minor modification to the original file would lead to a drastic change in hash value, i.e. checksum.
In the following terminal output, we have two files, `text1.txt` and `text2.txt`, which are almost identical except for (literally) one bit being different; the letters `T` and `t` are different in one bit in their ASCII representation. Even though we have flipped only a single bit, it is evident that the SHA256 checksums are entirely different. Consequently, if we use a secure hash function algorithm, we can easily confirm whether any modifications have taken place. This can help protect against both intentional tampering and file transfer errors.
Terminal
``````user@TryHackMe\$ hexdump text1.txt -C
00000000 54 72 79 48 61 63 6b 4d 65 0a |TryHackMe.|
0000000a
\$ hexdump text2.txt -C
00000000 74 72 79 48 61 63 6b 4d 65 0a |tryHackMe.|
0000000a
\$ sha256sum text1.txt
f4616fd825a10ded9af58fbaee09f3e31751d15591f9323ea68b03a0e8ac3783 text1.txt
\$ sha256sum text2.txt
9ffa3533ee33998aeb1df76026f8031c8af6ccabd8393eca002d5b7471a0b536 text2.txt``````
Some of the hashing algorithms in use and still considered secure are:
• SHA224, SHA256, SHA384, SHA512
• RIPEMD160
Some older hash functions, such as MD5 (Message Digest 5) and SHA-1, are cryptographically broken. By broken, we mean that it is possible to generate a different file with the same checksum as a given file. This means that we can create a hash collision. In other words, an attacker can create a new message with a given checksum, and detecting file or message tampering won’t be possible.
### HMAC
Hash-based message authentication code (HMAC) is a message authentication code (MAC) that uses a cryptographic key in addition to a hash function.
According to RFC2104, HMAC needs:
• secret key
• inner pad (ipad) a constant string. (RFC2104 uses the byte `0x36` repeated B times. The value of B depends on the chosen hash function.)
• outer pad (opad) a constant string. (RFC2104 uses the byte `0x5C` repeated B times.)
Calculating the HMAC follows the following steps as shown in the figure:
1. Append zeroes to the key to make it of length B, i.e., to make its length match that of the ipad.
2. Using bitwise exclusive-OR (XOR), represented by , calculate key ⊕ ipad.
3. Append the message to the XOR output from step 2.
4. Apply the hash function to the resulting stream of bytes (in step 3).
5. Using XOR, calculate key ⊕ opad.
6. Append the hash function output from step 4 to the XOR output from step 5.
7. Apply the hash function to the resulting stream of bytes (in step 6) to get the HMAC.
The figure above represents the steps expressed in the following formula: H(Kopad,H(Kipad,text)).
To calculate the HMAC on a Linux system, you can use any of the available tools such as `hmac256` (or `sha224hmac`, `sha256hmac`, `sha384hmac`, and `sha512hmac`, where the secret key is added after the option `--key`). Below we show an example of calculating the HMAC using `hmac256` and `sha256hmac` with two different keys.
Terminal
``````user@TryHackMe\$ hmac256 s!Kr37 message.txt
3ec65b7e80c5bf2e623e52e0528f1c6a74f605b10616621ba1c22a89fb244e65 message.txt
user@TryHackMe\$ hmac256 1234 message.txt
user@TryHackMe\$ sha256hmac message.txt --key s!Kr37
3ec65b7e80c5bf2e623e52e0528f1c6a74f605b10616621ba1c22a89fb244e65 message.txt
user@TryHackMe\$ sha256hmac message.txt --key 1234
Answer the questions below
On the AttackBox, you can find the directory for this task located at `/root/Rooms/cryptographyintro/task05`; alternatively, you can use the task file from Task 2 to work on your own machine.
What is the SHA256 checksum of the file `order.json`?
Open the file `order.json` and change the amount from `1000` to `9000`. What is the new SHA256 checksum?
Using SHA256 and the key `3RfDFz82`, what is the HMAC of `order.txt`?
Using a key exchange such as the Diffie-Hellman key exchange allows us to agree on a secret key under the eyes and ears of eavesdroppers. This key can be used with a symmetric encryption algorithm to ensure confidential communication. However, the key exchange we described earlier is not immune to Man-in-the-Middle (MITM) attack. The reason is that Alice has no way of ensuring that she is communicating with Bob, and Bob has no way of ensuring that he is communicating with Alice when exchanging the secret key.
Consider the figure below. It is an attack against the key exchange explained in the Diffie-Hellman Key Exchange task. The steps are as follows:
1. Alice and Bob agree on q and g. Anyone listening on the communication channel can read these two values, including the attacker, Mallory.
2. As she would normally do, Alice chooses a random variable a, calculates A ( A = (ga) mod q) and sends A to Bob. Mallory has been waiting for this step, and she has selected a random variable m and calculated the respective M. As soon as Mallory receives A, she sends M to Bob, pretending she is Alice.
3. Bob receives M thinking that Alice sent it. Bob has already picked a random variable b and calculated the respective B; he sends B to Alice. Similarly, Mallory intercepts the message, reads B and sends M to Alice instead.
4. Alice receives M and calculates key = Ma mod q.
5. Bob receives M and calculates key = Mb mod q.
Alice and Bob continue to communicate, thinking that they are communicating directly, unaware that they are communicating with Mallory, who can read and modify the messages before sending them to the intended recipient.
This susceptibility necessitates some mechanism that would allow us to confirm the other party’s identity. This brings us to Public Key Infrastructure (PKI).
Consider the case where you are browsing the website example.org over HTTPS. How can you be confident that you are indeed communicating with the `example.org` server(s)? In other words, how can you be sure that no man-in-the-middle intercepted the packets and altered them before they reached you? The answer lies in the website certificate.
The figure below shows the page we get when browsing example.org. Most browsers represent the encrypted connection with some kind of a lock icon. This lock icon indicates that the connection is secured over HTTPS with a valid certificate.
At the time of writing, example.org uses a certificate signed by DigiCert Inc., as shown in the figure below. In other words, DigiCert confirms that this certificate is valid (till a certain date).
For a certificate to get signed by a certificate authority, we need to:
1. Generate Certificate Signing Request (CSR): You create a certificate and send your public key to be signed by a third party.
2. Send your CSR to a Certificate Authority (CA): The purpose is for the CA to sign your certificate. The alternative and usually insecure solution would be to self-sign your certificate.
For this to work, the recipient should recognize and trust the CA that signed the certificate. And as we would expect, our browser trusts DigiCert Inc as a signing authority; otherwise, it would have issued a security warning instead of proceeding to the requested website.
You can use `openssl` to generate a certificate signing request using the command `openssl req -new -nodes -newkey rsa:4096 -keyout key.pem -out cert.csr`. We used the following options:
• `req -new` create a new certificate signing request
• `-nodes` save private key without a passphrase
• `-newkey` generate a new private key
• `rsa:4096` generate an RSA key of size 4096 bits
• `-keyout` specify where to save the key
• `-out` save the certificate signing request
Then you will be asked to answer a series of questions, as shown in the console output below.
Terminal
``````user@TryHackMe\$ openssl req -new -nodes -newkey rsa:4096 -keyout key.pem -out cert.csr
[...]
-----
You are about to be asked to enter information that will be incorporated
into your certificate request.
What you are about to enter is what is called a Distinguished Name or a DN.
There are quite a few fields but you can leave some blank
For some fields there will be a default value,
If you enter '.', the field will be left blank.
-----
Country Name (2 letter code) [XX]:UK
State or Province Name (full name) []:London
Locality Name (eg, city) [Default City]:London
[...]``````
Once the CSR file is ready, you can send it to a CA of your choice to get it signed and ready to use on your server.
Once the client, i.e., the browser, receives a signed certificate it trusts, the SSL/TLS handshake takes place. The purpose would be to agree on the ciphers and the secret key.
We have just described how PKI applies to the web and SSL/TLS certificates. A trusted third party is necessary for the system to be scalable.
For testing purposes, we have created a self-signed certificate. For example, the following command will generate a self-signed certificate.
`openssl req -x509 -newkey -nodes rsa:4096 -keyout key.pem -out cert.pem -sha256 -days 365`
The `-x509` indicates that we want to generate a self-signed certificate instead of a certificate request. The `-sha256` specifies the use of the SHA-256 digest. It will be valid for one year as we added `-days 365`.
To answer the questions below, you need to inspect the certificate file `cert.pem` in the `task06` directory. You can use the following command to view your certificate:
`openssl x509 -in cert.pem -text`
Answer the questions below
On the AttackBox, you can find the directory for this task located at `/root/Rooms/cryptographyintro/task06`; alternatively, you can use the task file from Task 2 to work on your own machine.
What is the size of the public key in bits?
Till which year is this certificate valid?
Let’s see how cryptography can help increase password security. With PKI and SSL/TLS, we can communicate with any server and provide our login credentials while ensuring that no one can read our passwords as they move across the network. This is an example of protecting data in transit. Let’s explore how we can safeguard passwords as they are saved in a database, i.e., data at rest.
The least secure method would be to save the username and the password in a database. This way, any data breach would expose the users’ passwords. No effort is required beyond reading the database containing the passwords.
`alice` `qwerty`
`bob` `dragon`
`charlie` `princess`
The improved approach would be to save the username and a hashed version of the password in a database. This way, a data breach will expose the hashed versions of the passwords. Since a hash function is irreversible, the attacker needs to keep trying different passwords to find the one that would result in the same hash. The table below shows the MD5 sum of the passwords. (We chose MD5 just to keep the password field small for the example; otherwise, we would have used SHA256 or something more secure.)
`alice` `d8578edf8458ce06fbc5bb76a58c5ca4`
`bob` `8621ffdbc5698829397d97767ac13db3`
`charlie` `8afa847f50a716e64932d995c8e7435a`
The previous approach looks secure; however, the availability of rainbow tables has made this approach insecure. A rainbow table contains a list of passwords along with their hash value. Hence, the attacker only needs to look up the hash to recover the password. For example, it would be easy to look up `d8578edf8458ce06fbc5bb76a58c5ca4` to discover the original password of `alice`. Consequently, we need to find more secure approaches to save passwords securely; we can add salt. A salt is a random value we can append to the password before hashing it. An example is shown below.
`alice` `8a43db01d06107fcad32f0bcfa651f2f` `12742`
`bob` `aab2b680e6a1cb43c79180b3d1a38beb` `22861`
`charlie` `3a40d108a068cdc8e7951b82d312129b` `16056`
The table above used `hash(password + salt)`; another approach would be to use `hash(hash(password) + salt)`. Note that we used a relatively small salt along with the MD5 hash function. We should switch to a (more) secure hash function and a large salt for better security if this were an actual setup.
Another improvement we can make before saving the password is to use a key derivation function such as PBKDF2 (Password-Based Key Derivation Function 2). PBKDF2 takes the password and the salt and submits it through a certain number of iterations, usually hundreds of thousands.
We recommend you check the Password Storage Cheat Sheet if you like to learn about other techniques related to password storage.
Answer the questions below
You were auditing a system when you discovered that the MD5 hash of the admin password is `3fc0a7acf087f549ac2b266baf94b8b1`. What is the original password?
In this task, we would like to explore what happens when we log into a website over HTTPS.
1. Client requests server’s SSL/TLS certificate
2. Server sends SSL/TLS certificate to the client
3. Client confirms that the certificate is valid
Cryptography’s role starts with checking the certificate. For a certificate to be considered valid, it means it is signed. Signing means that a hash of the certificate is encrypted with the private key of a trusted third party; the encrypted hash is appended to the certificate.
If the third party is trusted, the client will use the third party’s public key to decrypt the encrypted hash and compare it with the certificate’s hash. However, if the third party is not recognized, the connection will not proceed automatically.
Once the client confirms that the certificate is valid, an SSL/TLS handshake is started. This handshake allows the client and the server to agree on the secret key and the symmetric encryption algorithm, among other things. From this point onward, all the related session communication will be encrypted using symmetric encryption.
The final step would be to provide login credentials. The client uses the encrypted SSL/TLS session to send them to the server. The server receives the username and password and needs to confirm that they match.
Following security guidelines, we expect the server to save a hashed version of the password after appending a random salt to it. This way, if the database were breached, the passwords would be challenging to recover.
Answer the questions below
Make sure you read and understand the above scenario. The purpose is to see how symmetric and asymmetric encryption are used along with hashing in many secure communications.
Cryptography is a vast topic. In this room, we tried to focus on the core concepts that would help you understand the commonly used terms in cryptography. This knowledge is vital to understanding the configuration options of systems that use encryption and hashing.
Answer the questions below
Make sure you have taken notes of all the concepts and commands covered in this room. | HuggingFaceTB/finemath | |
# Commutator of Quotient Group Elements
## Theorem
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:
$\sqbrk {x, y} = x^{-1} y^{-1} x y$
Then:
$\forall x, y \in G: \sqbrk {x N, y N} = \sqbrk {x, y} N$
where $x N$ and $y N$ are left cosets of $N$, and so elements of the quotient group $G / N$ of $G$ by $N$.
## Proof
$\displaystyle \sqbrk {x N, y N}$ $=$ $\displaystyle \paren {x N}^{-1} \paren {y N}^{-1} \paren {x N} \paren {y N}$ Definition of Commutator of Group Elements $\displaystyle$ $=$ $\displaystyle \paren {x^{-1} N} \paren {y^{-1} N} \paren {x N} \paren {y N}$ Quotient Group is Group: inverse of $x N$ is $x^{-1} N$ $\displaystyle$ $=$ $\displaystyle \paren {x^{-1} y^{-1} N} \paren {x y N}$ Definition of Coset Product $\displaystyle$ $=$ $\displaystyle x^{-1} y^{-1} x y N$ Definition of Coset Product $\displaystyle$ $=$ $\displaystyle \sqbrk {x, y} N$ Definition of Commutator of Group Elements
$\blacksquare$ | HuggingFaceTB/finemath | |
## 2.8 Autocorrelation
Just as correlation measures the extent of a linear relationship between two variables, autocorrelation measures the linear relationship between lagged values of a time series.
There are several autocorrelation coefficients, corresponding to each panel in the lag plot. For example, $$r_{1}$$ measures the relationship between $$y_{t}$$ and $$y_{t-1}$$, $$r_{2}$$ measures the relationship between $$y_{t}$$ and $$y_{t-2}$$, and so on.
The value of $$r_{k}$$ can be written as $r_{k} = \frac{\sum\limits_{t=k+1}^T (y_{t}-\bar{y})(y_{t-k}-\bar{y})} {\sum\limits_{t=1}^T (y_{t}-\bar{y})^2},$ where $$T$$ is the length of the time series. The autocorrelation coefficients make up the autocorrelation function or ACF.
The autocorrelation coefficients for the beer production data can be computed using the ACF() function.
recent_production %>% ACF(Beer, lag_max = 9)
#> # A tsibble: 9 x 2 [1Q]
#> lag acf
#> <lag> <dbl>
#> 1 1Q -0.102
#> 2 2Q -0.657
#> 3 3Q -0.0603
#> 4 4Q 0.869
#> 5 5Q -0.0892
#> 6 6Q -0.635
#> 7 7Q -0.0542
#> 8 8Q 0.832
#> 9 9Q -0.108
The values in the acf column are $$r_1,\dots,r_9$$, corresponding to the nine scatterplots in Figure 2.16. We usually plot the ACF to see how the correlations change with the lag $$k$$. The plot is sometimes known as a correlogram.
recent_production %>% ACF(Beer) %>% autoplot() Figure 2.17: Autocorrelation function of quarterly beer production.
In this graph:
• $$r_{4}$$ is higher than for the other lags. This is due to the seasonal pattern in the data: the peaks tend to be four quarters apart and the troughs tend to be four quarters apart.
• $$r_{2}$$ is more negative than for the other lags because troughs tend to be two quarters behind peaks.
• The dashed blue lines indicate whether the correlations are significantly different from zero. These are explained in Section 2.9.
### Trend and seasonality in ACF plots
When data have a trend, the autocorrelations for small lags tend to be large and positive because observations nearby in time are also nearby in size. So the ACF of trended time series tend to have positive values that slowly decrease as the lags increase.
When data are seasonal, the autocorrelations will be larger for the seasonal lags (at multiples of the seasonal frequency) than for other lags.
When data are both trended and seasonal, you see a combination of these effects. The a10 data plotted in Figure 2.2 shows both trend and seasonality. Its ACF is shown in Figure 2.18.
a10 %>% ACF(Cost, lag_max = 48) %>% autoplot() Figure 2.18: ACF of monthly Australian electricity demand.
The slow decrease in the ACF as the lags increase is due to the trend, while the “scalloped” shape is due the seasonality. | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## What is the GCF of 192 and 972?
The gcf of 192 and 972 is 12.
#### Steps to find GCF
1. Find the prime factorization of 192
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
2. Find the prime factorization of 972
972 = 2 × 2 × 3 × 3 × 3 × 3 × 3
3. To find the gcf, multiply all the prime factors common to both numbers:
Therefore, GCF = 2 × 2 × 3
4. GCF = 12
MathStep (Works offline) Download our mobile app and learn how to find GCF of upto four numbers in your own time: | HuggingFaceTB/finemath | |
## Use Symbols to Compare_Worksheet.docx - Section 3: Independent Practice
Use Symbols to Compare_Worksheet.docx
Use Symbols to Compare_Worksheet.docx
# Use Symbols to Compare
Unit 6: Numbers and Place Value
Lesson 7 of 11
## Big Idea: Greedy Gator at work! This lesson has students comparing greater than, less than and equal to.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Math, greater than, less than, counting, tens
60 minutes
### Lisa Murdock
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Article: Bank Angle for Standard Rate Turn luizmonteiro.com is your source for: Online Aviation Instrument Simulators + E6b, CR3 and other Flight Computer Calculators
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Bank Angle and the Physics of Standard Rate Turns
I - Introduction II - Forces Acting on Aircraft III - Obtaining an Exact Equation for the Bank Angle required for a Standard Rate Turn IV - Obtaining Approximate Equations for Mentally Estimating the Bank Angle Required for Standard Rate Turn V - Comparing the Approximations to the Exact Formula VI - Half Standard Rate Turns VII - Standard Rate with TAS in Kilometers Per Hour (km/hr) VIII - Units and Variables Used IX - Other Useful Information that We Can Use From Our Study
I
- Introduction
In this article we are going to look deep into standard rate turns and examine the math and some of physics behind them. We are also going to look at the "rule of thumb" (mental calculation) methods that pilots use to estimate the bank angle required to achieve a standard rate turn. We will verify how precise are they in comparison to an exact formula that we will construct. A basic knowledge of algebra and trigonometry is required to fully understand the rest of this article.
Defining Standard Rate Turn
Definition: A standard rate turn is maneuver in which an aircraft turns at a rate 3o per second (3o/s) .
If this turn is held for exactly two minutes (120 seconds) the aircraft will complete a 360o turn since:
3o/s · 120s = 360o
Note: the middle dot (·) in the formula below means multiply. We will be using this notation throughout this text.
The bank required to achieve standard rate changes with your true airspeed (TAS). The higher that speed, the greater the bank angle will need to be in order to achieve a standard rate turn.
(a) (b)
Fig 1-1 Instruments used to measure rate of turn in aircraft.
Figure 1-1 a shows an analog turn coordinator instrument that currently indicates a standard rate turn to the right (tip of the miniature aircraft wing on right marking). Figure 1-1 b shows the same situation on a digital display (magenta arc on second right marking). Note that these instruments are showing how fast the heading of the aircraft is changing (rate of turn). They do not show bank directly.
Estimating the Bank Angle to Achieve a Standard Rate Turn
(a) (b)
Fig 1-2 Attitude indicator instruments used to bank the aircraft.
When banking the aircraft to execute a standard rate turn, it is useful for pilots to have an approximate idea of how much that bank should be for standard rate. They will be using the attitude indicator (Figure 1-2 a & b) first to bank the aircraft and once they are at the approximate bank, they will then rely on the turn coordinator instrument (Figure 1-1 a & b) to adjust the initial bank so that a standard rate turn is established.
The following was taken directly from a paragraph in the FAA-H-8083-15A Instrument Flying Handbook, Page 5-19 and 5-20, 2008 Edition, published by United States Department of transportation, Federal Aviation Administration Air Men Testing Standards Branch:
"A rule of thumb to determine the approximate angle of bank required for a standard rate turn is to use 15 percent of the true airspeed. A simple way to determine this amount is to divide the airspeed by 10 and add one-half the result. For example, at 100 knots, approximately 15o of bank is required (100 ÷ 10 + half of result = 10 + 5 = 15); at 120 knots, approximately 18o of bank is needed for a standard rate turn."
in other words:
0.15 · TAS
Where: is the bank angle in degrees and TAS is the true airspeed in knots. This equation will later be referred to as equation .
Note: the () in the formula above means approximately. It will be used throughout this text to distinguish approximate formulas from the exact (=) formulas.
There are also some variations to this approximation method. One of them is:
0.10 · TAS + 5
Basically meaning: take 10% of the true airspeed then add five to get the approximate bank angle in degrees for standard rate turn. This equation will later be referred to as equation .
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Precalc0108to0109-page20
# Precalc0108to0109-page20 - (Section 1.9 Inverses of...
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(Section 1.9: Inverses of One-to-One Functions) 1.9.10 PART E: FINDING FORMULAS FOR INVERSE FUNCTIONS If f is a one-to-one function , and if its formula can be expressed algebraically , then we should be able to find a formula for f ± 1 . Conceptual Approach to Finding the Inverse of a One-to-One Function f To determine f ± 1 , we need to invert (or “ undo ”) the steps applied by f in reverse order (WARNING 2 ) . (See the Exercises.) • Also, if necessary, impose restrictions and ensure that Dom f () = Range f ± 1 , and Dom f ± 1 = Range f . (See Example 10.) TIP 1 : Similarly, when dressing, you put on your socks before your shoes, but, when undressing, you remove your shoes before your socks. Example 6 (Conceptual Approach to Finding an Inverse)
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# How do you simplify 45/3(-5) using PEMDAS?
Jul 30, 2017
See a solution process below:
#### Explanation:
This is Division and Multiplication which are done in the same step, left to right.
Therefore, the first step is to do the division on the left:
$\frac{\textcolor{red}{45}}{\textcolor{b l u e}{3}} \left(- 5\right) \implies 15 \left(- 5\right)$
Next we can Multiply the result of the Division by the term within the parenthesis:
$\textcolor{red}{15} \left(\textcolor{b l u e}{- 5}\right) \implies - 75$ | HuggingFaceTB/finemath | |
# Notes 13 3 Set Venn Diagrams Organizing a
• Slides: 31
Notes #13 3 Set Venn Diagrams
Organizing a 3 set Venn Diagram
Example #1 • A survey of STAR WARS fans revealed the following information: • 51 admire Yoda • 49 admire Darth Vader • 60 admire Chewbacca “Chewy”
SOLUTION (10 Steps) Step 1: Draw & Create a Key for the Venn Diagram (Y): 51 (DV): 49 (C): 60 KEY Y: Yoda DV: Darth Vader C: Chewy
Step 2: Start with Y DV C “ 24 admire all three STAR WARS Icons” 51 49 24 60
Step 3: (Y DV C)’: “ 1 admires none of the STAR WARS Icons” Y DV 24 C 1
Step 4: “ 36 admire Yoda & Chewy Y DV 36 - 24 = 12 36 12 24 C 1
Step 5: “ 32 admire DV & Chewy” Y DV 12 24 C 8 32 32 - 24 = 8 1
Step 6: “ 34 admire Y & DV” Y DV 34 10 12 24 C 8 34 - 24 = 10 1
Step 7: Fans who only admire Chewy 51 49 10 12 60 -12 -24 -8 = 16 24 8 16 1 60
Step 8: Fans who ONLY admire Darth Vader 51 7 10 12 7 = 49 -10 -24 -8 49 24 8 16 60 1
Step 9: Fans who ONLY admire Yoda 51 51 -10 -24 -12= 5 5 49 7 10 12 24 8 16 60 1
Step 10: Use the Completed Venn Diagram to answer the given questions 51 5 49 7 10 12 24 8 16 60 1
Q 1: How many people were surveyed? Total People surveyed = 5 + 10 + 7+ 8 +16 +12 + 24 + 1 = 83 ppl were surveyed. 51 49 5 7 10 12 24 8 16 60 1
Q 2: How many admire Chewy, but not Yoda nor Darth Vader? ANS: 16 ppl admire Chewy, but not Yoda nor Darth Vader. 51 5 49 7 10 12 24 8 16 60 1
Q 3: How many admire Darth Vader or Chewy? Or “admire Darth Vader, or admire Chewy, or admire both? ” 10 + 7 + 8 + 16 + 12 + 24 = 77 51 49 5 7 10 12 24 8 16 60 1 ANS: 77 ppl admire Darth Vader or Chewy.
Q 4: How many admire exactly one of the Star Wars Icon? HINT: 3 Possibilities 5 + 7 + 16 = 28 51 5 49 7 10 12 24 8 16 60 1 ANS: 28 ppl admire exactly one of the Star Wars Icon.
Q 5: How many admire exactly 2 of the Star Wars Icons? 10 + 8 +12 = 30 51 5 49 7 10 12 24 8 16 60 1 ANS: 30 people admire exactly two of the Star Wars Icons.
Classwork #1 A group of 82 students were surveyed. The survey says that all students liked eating at least one or more of the following fruits (apricot, banana, and cantaloupes) 39 liked apricots. 50 liked bananas. 39 liked cantaloupes. 21 liked apricots and bananas. 18 liked bananas and cantaloupes. 19 liked apricots and cantaloupes. 22 liked exactly two of the following fruits: apricots, bananas, and cantaloupes
KEY A: Apricot B: Banana C: Cantaloupe (B): 50 (A): 39 21 STEP #1: 21 liked apricots and bananas. 18 liked bananas and cantaloupes. 19 18 19 liked apricots and cantaloupes. (C): 39
KEY A: Apricot B: Banana C: Cantaloupe n(B): 50 n(A): 39 21 21 - X STEP 2 Let X: A B X C 19 - X 18 - X 19 18 STEP 3 Set up equations to solve for Missing info. N(C): 39
KEY A: Apricot B: Banana C: Cantaloupe : “ 22 liked exactly two of the following fruits: apricots, bananas, and Cantaloupes” n(B): 50 n(A): 39 21 - X X 19 - X 18 - X Step 4: Solve for X 22= 21 - X + 18 - X + 19 - X 22 =21+18+19 -3 x 3 x = 21+18+19 -22 x = 36/3 = 12 N(C): 39
KEY A: Apricot B: Banana C: Cantaloupe n(B): 50 n(A): 39 21 – 12= 12 Step 5 Solve for the missing data 19 – 12= 18 – 12= N(C): 39
KEY A: Apricot B: Banana C: Cantaloupe n(B): 50 n(A): 39 11 23 9 12 6 7 14 N(C): 39
a. How many students liked apricots, but not bananas or cantaloupes? KEY A: Apricot B: Banana C: Cantaloupe ANS: 11 students liked apricots, but not bananas or cantaloupes.
b. How many students liked cantaloupes, but not bananas or apricots? ANS: 14 students liked cantaloupes, but not bananas or apricots. .
c. How many students liked all of the following three fruits: apricots, bananas, and cantaloupes? ANS: 12 students all 3 fruits.
d. How many students liked apricots and cantaloupes, but not bananas? ANS: 7 students liked apricots & cantaloupes, but not bananas.
Just for Fun…. At a restaurant, Harry, Ron, Hermonie, Snape, and Albus were asked if they had ordered (not eaten ) any of the following: chicken marsala, lasagna, or coke. 3 people ordered chicken marsala (Hermonie, Albus, Ron). 2 people ordered lasagna (Snape and Albus). 3 people ordered coke (Harry, Ron, and Snape). 1 person ordered both chicken marsala and lasagna (Albus).
Harry (coke) Hermonie (chicken marsala) chicken marsala Hermonie Albus Ron Snape lasagna Ron (chicken marsala, coke) Snape (lasagna, coke) Harry coke Albus (chicken marsala, lasagna) | HuggingFaceTB/finemath | |
# MATH
posted by on .
The cumulative number of car accidents from 2000 to 2010 can be modeled by the quadratic expression, C=-42x^2+924x+14,112, where x=1 corresponds to 2001, and so on until x=11 corresponds to 2010. Find the number of cumulative car accidents in 2003. Use the table of values to estimate the year in which the cumulative number of car accidents reached 18,100.
• MATH - ,
C = -42x^2 + 924x + 14112 = 18,100.
-42x^2 + 924x + 14,112 - 18,100 = 0.
-42x^2 + 924x - 3988 = 0. | HuggingFaceTB/finemath | |
# How can I prove that for every real number $a$ there exists a sequence $r_n$ of rational numbers such that $r_n$ approaches $a$.
How to prove that for every real number $a$ there exists a sequence $r_n$ of rational numbers such that $r_n \rightarrow a$.
-
The answer depends greatly on how you define your real numbers. If it is the completion of the rationals then this is trivial. – Ragib Zaman Oct 8 '12 at 0:59
Without loss of generality we may assume the real number $a$ is $\gt 0$. (If $a \lt 0$, we can apply the argument below to $|a|$ and then switch signs.) We sketch a fairly formal proof, based on the fact that the reals are a complete ordered field. In one of the remarks at the end, we give an easy informal but incomplete "proof."
Let $n$ be a natural number. Let $m=m(n)$ be the largest positive integer such that $\frac{m}{n}\lt a$. Then $\frac{m+1}{n}\ge a$, and therefore $|a-m/n|\lt 1/n$.
Let $r_n=m/n$. It is easy to show from the definition of limit that the sequence $(r_n)$ has limit $a$.
Remarks: $1.$ One really requires proof that there is a positive integer $m$ such that $\frac{m}{n}\ge a$. It is enough to show that there is a positive integer $k$ such that $k \ge a$, for then we can take $m=kn$. The fact that there is always an integer $\gt a$ is called the Archimedean property of the reals. We proceed to prove that the reals do have this property.
Suppose to the contrary that all positive integers are $\lt a$. Then the set $\mathbb{N}$ of positive integers is bounded, so has a least upper bound $b$. That means that for any $\epsilon \gt 0$ there is an integer $k$ such that $0\lt k\lt b$ and $b-k\lt \epsilon$. Pick $\epsilon=1/2$. Then $k+1\gt b$, contradicting the assumption that $b$ is an upper bound for $\mathbb{N}$.
$2.$ One can also give a very quick but not fully persuasive "proof" of the approximation result. Assume as before that $a\gt 0$. The numbers obtained by truncating the decimal expansion of $a$ at the $n$-th place are rational, and clearly have limit $a$. The problem is that we are then assuming that every real number has a decimal expansion.
-
Why is it problematic to assume that real numbers have decimal expansion? – leo Oct 8 '12 at 2:04
@leo: To prove that the real numbers have certain properties, we have to start with a formal definition of the reals. Then we can prove representability by an infinite series $a_0+\frac{a_1}{10}+\frac{a_2}{100}+\cdots$ where $a_0$ is an integer and the rest of the $a_i$ are digits, that is, integers between $0$ and $9$. But before a course in "analysis" one takes these facts for granted, and then for example $3, 3.1,3.14,3.141,3.1415, 3.14159,\dots$ is a sequence of rationals with limit $\pi$. To answer your question, one has to know at what level the question is being asked. – André Nicolas Oct 8 '12 at 2:13
Hint: The rational numbers are dense in the reals. If there is a real number $a$ for which there lacks such a sequence, what can we say about the neighborhood of $a$?
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This statement is equivalent to the statement that the rationals are dense in the reals. – Qiaochu Yuan Oct 8 '12 at 0:34
The point of the exercise is to see one such sequence. Isn't? – leo Oct 8 '12 at 2:03
I'm surprised that no one's talked about decimal notation yet, but here's an informal proof. (For familiarity, we'll use the base-10 system.)
If $x$ is rational, just use the sequence $\left(r_n\right)=\left(x, x, x, x, \dots\right)\to x$. If it's irrational, we'll have to do some work: Represent $x$ using decimal notation. It will be a non-repeating, non-terminating decimal. For example, let
$$x = \sqrt{2} = 1.414213562...$$
Then just make every term of your sequence a terminating decimal, a more refined approximation of $\sqrt{2}$.
$$\left(r_n\right)=\left(1, 1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414213, \dots\right)$$
Obviously, since every term in the sequence is a terminating decimal, it's a rational number. By the Cauchy criterion, the rational sequence converges, and it converges to the real number $\sqrt{2}$.
It's not a formal proof, but you can see how it works.
-
Certainly not a formal proof -- it's borderline circular! That any real number can be approximated by a decimal requires proof. But you're absolutely right that one can find a rational number with a fixed power denominator (i.e., a number of the form p/q^n for some fixed q) to approximate any real number. PS -- according to the answer right above yours, people did talk about decimals in October. – user54535 Jan 6 '13 at 4:20 | open-web-math/open-web-math | |
Binomial coefficient
# Binomial coefficient
The binomial coefficient is a mathematical function, which can be solved one of the fundamental tasks of combinatorics. It specifies on how many different types you can select objects from a set of different objects ( without replacement, without regard to the order). The binomial coefficient is therefore the number of subsets of a - element set.
" 49 6 " (or " 45 6 " in Austria and Switzerland ), for example, the number of possible contractions in the lottery (excluding the bonus number ).
A binomial coefficient depends on two numbers and down. He is the symbol
Written and spoken as "n over k ", " k of n " or "n low k". The abbreviation for nCr from n choose r is given as a label on pocket calculators.
The name given these numbers, as they occur as coefficients in the powers of the binomial; whichever is the so-called binomial theorem:
An extension of that coming from the combinatorics binomial coefficient represents the general binomial coefficient, which is used in the analysis.
• 5.1 Example
• 5.2 Combinatorial proofs
• 6.1 sums with binomial coefficients
• 6.2 sums with alternating binomial coefficients
• 6.3 sum of shifted binomial coefficients
• 6.4 Vandermonde identity
• 7.1 generalization
• 7.2 Binomial series
• 7.3 sum expression for the beta function
• 7.4 Gaussian product representation for the gamma function
• 7.5 Digammafunktion and Euler - Mascheroni constant
## Definition
Of a complex number and a non-negative integer the binomial coefficient "n over k " is defined as follows:
The Faculty of designated. The empty product () is.
If it is in a non-negative integer, we can use the well-known from the combinatorial definition:
## Properties
If besides also restricted to non-negative integers, then:
• Is always a non-negative integer.
• Is it so applies
• ( for the right summand ).
In the general case, real or complex values for some of the expressions set forth herein may be undefined in the above sense, that is, if should no longer be entirely non-negative; which relates to the statements, and. It appears, however, that these statements are correct if you allow according to the below analytical generalization of the beta function for complex values .
### Symmetry of the binomial coefficients
Integer binomial coefficients are symmetric in the sense of
For all non-negative and.
## Recursive representation and Pascal 's Triangle
For the binomial coefficients of non-negative integers and one has the following recursive representation:
This formula can also be used to determine all the binomial coefficients up to a predetermined limit for a scheme to is Pascal's Triangle: where it meets the design specification that each number is the sum of two standing on their numbers (in the formula above by replace ):
Or the other way around ( by replacing ):
Proof:
The coefficients can be found here in the -th row of the - th position (both from zero counting! )
The same triangle shown in the Binomialsymbolen:
## Algorithm for efficient calculation
For integer, an efficient algorithm which product formula exists
The binomial coefficient applies. Due to the constant change between multiplication and division, the intermediate results do not grow on unnecessarily. In addition, all intermediate results are natural numbers.
To avoid unnecessary computational cost, is calculated in the case of the binomial coefficients:
The following pseudo code illustrates the calculation of:
Binomial coefficient ( n, k) 1 if k = 0 then Return 1 2 if 2k > n 3 then do the results from binomial coefficient (n, nk ) 4 otherwise result from lead n -k 1 5 of i from 2 to k 6 leads from earnings earnings ( n - k i) 7 result result: i 8 Attributable profit The calculation method also use calculators if they offer the feature. Otherwise, the computational capacity of n = 69 would be exhausted. The labeling of the function key with nCr describes the sequence of input values in infix notation; first n number of elements, then the function key combinations, then the number of selected objects r ( in the article abbreviated k).
The calculation nPr (English permutation ) takes into account the permutations of r elements, the division by r! omitted:
## The binomial coefficient in combinatorics
The combinatorial definition of a number of subsets of a - element set plays a central role in enumerative combinatorics.
It can be clearly interpreted like this: First, you count all tuples with pairwise different elements that can be put together from the -element output quantity. There are ways of choosing the first tuple element. After any election of this first, there are only choices for the second element, at its option, only for the third, etc., to candidates for the -th and last tuple element. The total number of such tuples is compiled so the digit product, with the help of the faculty as can be noted. Now, however, the counted -tuple permutations are exactly each other and thus corresponds to einundderselben -element subset. After division by this counting multiplicities thus results in fact than the required number of groups.
Another, more symmetrical illustration emphasizes not and elements of the act of selecting of elements, but the aspect of the decomposition into two subsets. Suppose a -element Ausgangstupel consists of red and white somehow strung elements. By forming all permutations of this alignment, as are each of them indistinguishable in color, because the permutations of the red elements to each other do not change the color sequence, just as each of them independent permutations within the white. So there are only different color sequences of length with all possible different assignments by each red elements. Each sequence can now be unambiguously but one of the subsets of a - element set to assign. The same is true because of the symmetry of red and white or and also for the complementary subsets. The total number of the subsets is therefore ever.
The set of all subsets of a set is sometimes referred to because of their thickness with. This is true for any finite set:
### Example
The number of possible contractions or lottery tickets at the German Lotto 6 from 49 (without bonus number or super number)
Obviously there is only one chance to hit 6. one of the possibilities for 0 right thing, namely, to select all 6 Tips from the 43 wrong. The number of different tips with 5 correct results very easy to 6 ⋅ 43 = 258, because there are 6 ways to tap only 5 of the 6 numbers drawn (or omit it), and then 43 each = 49 - 6 options, the to put boisterous Tip on one of the 43 wrong numbers. In general, the number of different tips with r correct results in 6 out of 49 for the same consideration. At 6, 0 and 5 Right hardly noticeable that the factors used, and really simple binomial coefficients. The sum of all numbers called Tip yields of course the total number 13983816 all possible tips - this follows from the Vandermonde 's identity given below.
The probability of 6 scored with a tip right is therefore 1/13983816 which is for 5 correct 258/13983816. 0 for right result with 6096454/13983816 already about 44%. The overall probability for r right is a special case of the hypergeometric distribution, which combines just three binomial coefficients such.
### Combinatorial proofs
The combinatorial interpretation also allows simple proofs of relations between binomial coefficients, such as double counting. Example: For the following applies:
Proof: Let a - element set and a fixed element. Then decompose the subsets of two classes:
• The subsets that include; So they are made together with an -element subset of the - element set,
• The subsets that are not included; they are - element subsets of the - element set.
## Expressions with binomial coefficients
### Sums with binomial coefficients
This formula is a combinatorial facts as a basis. Since the number of all subsets of an amount corresponding with elements given by the summation of the number of all its subsets, ie. The formula can be derived also from the binomial theorem by setting.
### Sums with alternating binomial coefficients
This formula follows from the odd symmetry of the binomial coefficients. For arbitrary they can be derived from the binomial theorem, by and (or and ) is set.
### Vandermonde identity
There is also a combinatorial argument: The right side corresponds to the number of subsets of a - element set of balls. One can now imagine that the balls have two different colors: balls are red and green balls. A - element subset is then composed of a certain number of red balls and lots of green. For each possible are the corresponding term on the left side of the number of possibilities for such a split between red and green balls. The sum gives the total number. An often perceived as a simple proof uses the binomial theorem in the form
And approach
And comparing coefficients.
In the special case follows from the Vandermonde 's identity following formula for the sums of squares
## Binomial coefficients in the Analysis
### Generalization
A generalization, which plays a role in the analysis, obtained when one allows for an arbitrary complex number, but still requires as an integer. In this case,
The binomial coefficient " on" ( in the case of the blank the product is defined as 1). This definition is for non-negative integer with the combinatorial definition (ie the definition of as the number of all subsets of a fixed - element set ), respectively, and for non-negative with the algebraic definition (ie the definition of as the product ).
For example, is
And
The second parameter can be generalized to arbitrary complex assignment when defined using the beta function for:
Where the gamma function called. Is it or is a negative integer, then the value of the right-hand side is 0, because the non- positive integers, the ( single ) pole are.
Obviously still applies the symmetry relationship
In particular:
Whole and in nichtnegativem
To extract the sign of the first parameter, provided that it is an integer, the relation can be
Specify.
In general, for complex, the relationship
Another generalization offer the multinomial, which are required in the generalization of the binomial to multinomial theorem.
### Binomial series
For and with one obtains the relationship
Which is a generalization of the geometric series and is one of the binomial series.
Is, as well as the following series converges according to
( For more precise conditions and see Article binomial series. )
### Sum expression for the beta function
Another relationship can be relatively easy for all prove by induction,
Resulting directly the symmetry
Followed. A generalization for with and is
### Gaussian product representation for the gamma function
With the latest formula from the previous section is
Considering the case that breaks replaced in the sum by integrals according to
And summarizes the sum of the powers of the binomial corresponding together, one obtains
Where the last integral the substitution was applied. Finally, one has the equation
Resulting in through the border crossing directly the Gaussian product representation of the gamma function,
Results.
### Digammafunktion and Euler - Mascheroni constant
For with
Which can also be proved via induction on. For the special case of this equation simplifies to
The sequence of harmonic numbers, so the partial sums of the harmonic series. The transformation of the left-hand sum in a row ( instead of limit ) is allowed due to
On the other hand represented as
With the Digammafunktion and the Euler - Mascheroni constant
Can on complex values - to be continued - except on the negative integers. One gets then the series
As complex interpolation of the result of the harmonic numbers.
126304
de | HuggingFaceTB/finemath | |
# Number of non-abelian Groups of order 196
First I factorize $196=2^2\cdot 7^2$ and therefore the number of distinct Abelian groups is $P(2)\cdot P(2)=4$, where $P(\cdot)$ is the number of partitions of a natural number. So there are $4$ abelian groups. The isomorphic classes are:
The orders are not coprime so they're not cyclic? $$G_1=C_2\times C_2 \times C_7\times C_7$$ $$G_3=C_2\times C_2 \times C_{49}$$ $$G_4=C_4 \times C_7\times C_7$$ The orderes are coprime so this group is cyclic $$G=C_4 \times C_{49}$$
Now I know there is $1$ non-abelian group for sure, namely the dihedral group of the regular $98$ gon, is there anyway to tell how many there are? $$D_{196}=C_2\cdot C_{98}$$ I know the number of total groups is bounded by $n^{n^2}$ and the table with basic facts doesn't include $n=p^2q^2$. Not homework. The table T'm talking about is here https://groupprops.subwiki.org/wiki/Number_of_groups_of_given_order
• If you know dihedral groups, you can use the dihedral groups of order $14$ and $28$ and $98$ in direct products. Feb 17, 2018 at 8:26
• If you look on GAP, you find 8 nonabelian groups order 196. Feb 17, 2018 at 19:57
Using Sylow's third theorem, there is a normal Sylow $7$-subgroup, so the group is a semidirect product of a group of order $4$ acting on a group of order $49$. You need to classify the actions of $C_2\times C_2$ or $C_4$ on $C_7\times C_7$ or $C_{49}$.
Some interesting examples (not an exhaustive list). $C_4$ acts on $C_7\times C_7$ via the matrix $\pmatrix{0&1\\-1&0}$. This group has presentation $a^4=b^7=c^7=e$, $bc=cb$, $aba^{-1}=c$, $aca^{-1}=b^{-1}$.
Also $C_4$ acts on $C_{49}$ by the generator acting as inversion. Presentation: $a^4=b^{49}=e$, $aba^{-1}=b^{-1}$. | HuggingFaceTB/finemath | |
## Precalculus: Mathematics for Calculus, 7th Edition
a) $[-3,5]$ b) $(-3,5]$
a) For interval notation, the closed points at $-3$ and $5$ mean that those points are included, and therefore each end is enclosed with a square bracket instead of a parenthesis. b) For interval notation, the open point at $-3$ means $x$ cannot be -3 so this is enclosed by a parenthesis, however the closed point at $5$ means that $x$ can be $5$ and therefore $5$ is enclosed with a bracket. | HuggingFaceTB/finemath | |
# Origin in Math – Definition With Examples
## Definition of Origin in Math
In mathematics, especially in the field of coordinate geometry; origin is said to be the initial point or the starting point from where we begin our calculations or measurements.
On a ruler, the 0 is from where we start our measurements; hence it is said to be the origin of the scale.
On a weighing machine, we measure our weight when the scale shows zero.
In terms of temperature, when we go below 0 degrees Celsius, it starts getting colder whereas above 0 it starts getting hotter. Hence, it is the point from where we start our measurements.
## Origin on the Number Line
The origin on the number line is defined as 0. It divides the number line into two parts as given below:
On the left side of origin, we have the negative numbers and on the right side, we have the positive numbers. Whenever we say that we have started from a point, we always say that we have started from 0 or origin.
The origin, i.e., 0, is greater than the negative numbers and smaller than the positive numbers.
## Origin in the Cartesian Plane
The Cartesian Plane is a plane formed by the two number lines—vertical and horizontal. The point at which these two number lines intersect each other is known as the point of intersection which is also known as the origin. It is denoted by the letter O, which is used as a fixed point of reference for the geometry of the surrounding plane. The coordinates of the origin are denoted by (0, 0). It means that at origin, x=0 and y=0. The origin divides each of these axes into two halves—positive and negative.
## Distance of a point from the origin
1. Point lies on the x-axis.
Any point on the x axis is of the form (x, 0). Its distance from the origin will just be x units.
Consider two points, A(3, 0) and B(–3, 0). Now let’s find their distance from the origin.
Distance of A(3, 0) from the origin = |3| units = 3 units
Distance of B(–3, 0) from the origin = |–3| units = 3 units
This means that the distance of both the points is 3 units from the origin.
The reason for applying the absolute value sign is because distance cannot be negative.
1. Point lies on the y-axis.
Any point on the y axis is of the form (0, y). Its distance from the origin will just be y units.
Any point on the y-axis is of the form (0, y). Its distance from the origin will just be y units.
Consider two points, P(0, 4) and Q(0, –4). Now let’s find their distance from the origin.
Distance of P(0, 4) from the origin = |4| units = 4 units
Distance of Q(0, –4) from the origin = |–4| units = 4 units
So, these points are also equidistant from the origin.
## Equation of Line Passing through the Origin
The lines that pass through the origin are of the form y = mx where m is a constant. Here, “m” can be rational or irrational.
Let’s consider the equation: y = 2x. To plot this equation on the graph, we substitute a value of x in the equation to get the corresponding value of y. These values of x and y form the ordered pair (x, y). We repeat this for different values of x to get at least 3 ordered pairs. We may record our values in tabular form.
On plotting the graph, we will get a straight line passing through the origin.
## Origin Facts
We can check if the graph of an equation will pass through the origin by substituting x = 0 in the equation. If you get y = 0, then the line of the equation will pass through the origin.
## Solved Examples
Example 1: Find the distance of the point from the origin on a number line.
Solution: The distance of the point -10 from the origin, i.e., 0 on a number line =10 units
2. Find the distance of the point (0, -7) from the origin.
Solution: The distance of (0, -7) from (0, 0)=-7=7 units
3. What is the origin of the Cartesian Plane? Solution: In the Cartesian Plane, the point at which the two axes meet is known as origin. The coordinates of the origin are denoted by (0, 0).
## Practice Problems
1
### Which of the following is not an equation of line passing through the origin?
$y=-3x$
$y=-3x+1$
$y=$$\frac{x}{2}$
$y=2x$
CorrectIncorrect
Correct answer is: $y=-3x+1$
$y=-3x+1$ is not an equation passing through the origin as when we substitute
$x=0$, we don’t get $y=0$.
2
### What are the coordinates of the origin on the Cartesian Plane?
(0, 0, 0)
(10, 0)
(0, 0)
0
CorrectIncorrect
On the Cartesian Plane, the origin is denoted by (0, 0), which implies $x = 0$ and $y = 0$.
3
### Which of the following is a graph of a line passing through the origin?
B C D D CorrectIncorrect
D is the graph that has a line that passes through the origin, i.e., if $x=0$ then $y=0$. The rule for a rotation by 180about the origin is given by $(x,y) → (-x,−y)$. | HuggingFaceTB/finemath | |
# math
Simplify:
sqrt(49a^6)
Thanks.
1. 👍 0
2. 👎 0
3. 👁 68
1. 7a^3
1. 👍 0
2. 👎 0
posted by Jen
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More Similar Questions | HuggingFaceTB/finemath | |
# how can I show these two differential equation are the same?
the equations I need to show they are identical are..
$$\frac{dy}{dt}+r\int_{0}^{t}dt'\alpha e^{-\alpha (t-t')}y(t')=0$$ $$\frac{d^2y}{dt^2}+\alpha\frac{dy}{dt}+r \alpha y=0$$
I think differentiate the first term and used the condition $$\frac{dy}{dt}=-r\int_{0}^{t}dt'\alpha e^{-\alpha (t-t')}y(t')$$ can get the second equation.
my first attempt was changing the integral term of first equation with integration by parts, but it seems not working.
Since you have a mixture of derivatives and integrals (and you want a second order derivative in your ODE), you should think about differentiating your integro-differential equation. First we rewrite it slightly to make it easier for us:
$$\frac{dy}{dt} + r\alpha e^{-\alpha t}\int_0^t e^{\alpha t'}y(t')\,dt'.$$
Differentiating, we have (by using product rule and the fundamental theorem of calculus)
$$0 = \frac{d^2y}{dt^2} - r\alpha^2 e^{-\alpha t}\int_0^t e^{\alpha t'}y(t')\,dt' + r\alpha e^{-\alpha t} e^{\alpha t}y(t).$$
This isn't quite what you have but if you use your integro-differential equation, you can get the ODE.
• wow thanks a lot
– eric
Apr 8, 2015 at 1:20
• You're very welcome. Apr 8, 2015 at 1:25 | HuggingFaceTB/finemath | |
# Roman Numberals
Roman Numerals, although very old, are still in use for distinct applications. For example, when a list is enumerated especially in old mathematics books, you can see the roman numerals there. Or they are still being use in clocks hanging on a wall. It’s certainly useful to know a few facts about the Roman numerals. Here are a few of the Roman numerals:
1. – I
2. – II
3. – III
4. – IV
5. – V
6. – VI
7. – VII
8. – VIII
9. – IX
10. – X
After 10, number 11 would be for example would be XI or number 12 would be XII. Number 20 would be XX.
Here’s an overview of some of the larger Roman numerals:
I -> 1; V -> 5; X -> 10; L -> 50; C -> 100; D -> 500; M -> 1000.
There are a few rules regarding the Roman numerals. Knowing these rules, you’ll be able to read them if they get more complicated than the cases described above. You are usually not required to know the following rules but it does not hurt to know them:
Rule 1: If you repeat a numeral, its value will be added relative to the number of times you repeated it. Observe the following examples:
• To write the number 2 in Roman, you have to write the Roman 1 twice: II
• To write the number 20 in Roman, you have to write the Roman 10 Twice: XX
Rule 2: Although we said in Rule 1 that you can repeat “I” for example once to get to “II,” which is the number 2 in Roman, you are not supposed to repeat any numeral more than twice. In other words, no numeral may be written more than thrice in a row.
• For example, to write 30 in Roman, you repeat X twice, meaning that you write it once and repeat it twice so in total, you have three X’s. The number would be, “XXX.”But to write 40 in Roman you cannot write, “XXXX.” To do that, you’d have to write 50 and subtract 10 from it which would be, “XL.” We’ll talk about this subtraction in other rules.
One more part of Rule 2 is that the numerals V, L and D are not supposed to be repeated at all which stand for 5, 50 and 500 respectively.
Rule 3: This rule is somehow related to the “subtraction” we pointed to in the previous rule. To create the number 12 for example in Roman, you have to add 2 to 10. You do that by putting a 2 on the right hand side of 10. And you already know that a 2 in Roman is nothing but two 1’s. So the number becomes: XII. So from the example, you have hopefully understood that when you put a numeral of lower value to the right hand side of a numeral of higher value, the value of the numeral with the lower value will be added to the value of the numeral of the higher value.
Rule 4: This rule is the same thing as the previous rule but the other way around. To write the number 4 in Roman, you have to subtract 1 from 5. To do that, you write a Roman 1 on the “left” hand side of a Roman 5. That way the value of the numeral with a lower value on the left will be subtracted from the value of the numeral with a higher value on the right.
Also related to this subtraction rule, you never subtract the Roman 5, 50 and 500 (V, L and D respectively) from a numeral of greater value. Meaning that the three numeral will never be written on the left hand side of a numeral with a higher value than them.
Another related topic is the subtraction of Roman 1 which is “I.” The Roman 1 can be subtracted only from the Roman 5 and 10.
Another related topic is the subtraction of the Roman 10 which is “X.” The Roman 10 can be subtracted only from the Roman 50, 100 and 1000 (L, C and M respectively).
For your convenience, here’s a list of some Roman numbers:
• 1 -> I
• 2 -> II
• 3 -> III
• 4 -> IV
• 5 -> V
• 6 -> VI
• 7 -> VII
• 8 -> VIII
• 9 -> IX
• 10 -> X
• 20 -> XX
• 30 -> XXX
• 40 -> XL
• 50 -> L
• 60 -> LX
• 70 -> LXX
• 80 -> LXXX
• 90 -> XC
• 100 -> C
These articles are used by the author in a series of mathematics courses that will teach you mathematics from the sixth standard all the way up to 12th standard. The purpose of these courses are to help us understand mathematics so that you can use them professionally well. There is a road map that we have put together about the all the courses included, where to starts, etc. To know more about that and have access to those courses, please visit mathematics page on Great IT Courses. Thank you.
Web Developer | HuggingFaceTB/finemath | |
### Number letter counts
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
Solution:
```void Solution()
{
Dictionary<int, string> numbers = new Dictionary<int, string>();
int sum = 0;
for (int i = 1; i <= 20; i++)
{
sum += numbers[i].Length;
Console.WriteLine(numbers[i] + " " + numbers[i].Length);
}
for (int i = 21; i < 100; i++)
{
int key = i / 10;
if (i % 10 != 0)
sum += numbers[key * 10].Length + numbers[i % 10].Length;
else
sum += numbers[key * 10].Length;
}
int sumTill1To99 = sum;
for (int i = 101; i < 999; i = i + 100)
{
int key = i / 100;
sum += numbers[key].Length + numbers.Length;
sum += (numbers[key].Length + numbers.Length + "and".Length) * 99;
sum = sum + sumTill1To99;
}
sum = sum + numbers.Length + numbers.Length;
}
``` | HuggingFaceTB/finemath | |
greetings and a question!
• To: mathgroup at smc.vnet.net
• Subject: [mg83116] greetings and a question!
• From: dimitris <dimmechan at yahoo.com>
• Date: Sun, 11 Nov 2007 03:02:51 -0500 (EST)
```Hello to all of you!
Unfortunately, family, working and research issues
prevent me from participating to the forum as frequent
as I used to.
Anyway...
It is my first post since a long time so everybody be patient!
A student of mine came across the following
equation in a mathematical contest:
In[1]:=
eq = x^2 + y^2 + (a + b)*x - (a - b)*y + a^2 + b^2 - a - b + 1==0;
(all variables are assumed real)
Of course for Mathematica the solution is rather trivial.
In[1]:=
\$Version
Out[1]=
"5.2 for Microsoft Windows (June 20, 2005)"
In[2]:=
eq = x^2 + y^2 + (a + b)*x - (a - b)*y + a^2 + b^2 - a - b + 1==0;
In[3]:=
Reduce[eq, {a, b, x, y}, Reals]
ToRules[%]
eq /. %
Out[3]=
a == 1 && b == 1 && x == -1 && y == 0
Out[4]=
{a -> 1, b -> 1, x -> -1, y -> 0}
Out[5]=
True
Can somebody explain concisely the mathematica concept
behind this solution? In fact I would be much obliged if somebody
pointed me out how to obtain the result by hand. Also, by curiosity,
how Mathematica reaches the result?
Dimitris
```
• Prev by Date: Mathematica 6.0.0 crashes on OSX 10.4.10
• Next by Date: Re: AutoItalicWords
• Previous by thread: Re: Mathematica 6.0.0 crashes on OSX 10.4.10
• Next by thread: Re: greetings and a question! | HuggingFaceTB/finemath | |
# How Do You Solve 4+3x-6=3x+2-x?
Start by simplifying the question:
4 + 3x - 6 could be rewritten as 3x + 4 - 6 , or 3x - 2
3x + 2 - x could be rewritten as 3x - x + 2 , or 2x + 2
so then, 3x - 2 = 2x + 2. Subtract 2x from both sides:
3x - 2x - 2 = 2x - 2x +2, or x - 2 = 2
x = 4
thanked the writer.
michael finnegan commented
I guess it would've been easier to subtract 3x from both sides, then you'd have
4 - 6 = 2 - x , or -2 = 2 - x. Subtract 2 from both sides: -4 = -x , or 4 = x | HuggingFaceTB/finemath | |
Feb 10th, 2015
DreamIt
Category:
Mathematics
Price: \$50 USD
Question description
A type-Z option has the following payoff function:
{ K[1+sin(2*pi*St/(K))] if St<K
F(St)=
{K*exp^1-St/(k) if St>=K
1. Write a Python function with the following definition: def f(ST,K):
which calculates the payoff of a type-Z option.
Write a Python function with the following definition:
def plotTypeZPayoff (K,STMax):
which plots the payoff function of the type-Z option as a function of ST for given K. The range of the horizontal axis should be [0,STMax] and vertical axis [0,2.1K]. Include this graph in your written work for the case K = 20,STMax = 100.
2. Write a loop based binomial options pricing function with the definition: def binomOption(payoff ,steps ,T,sigma,S0,r): where payoff is a function of one variable. The variable sigma (σ) is the volatility, √√ related to u and d by u=exp(σ sqrt(δt)) and d=exp(−σ sqrt(δt)). Write a factory function with the definition:
def makeZPayoff(K):
which returns a type-Z payoff function with fixed K.
3. Fix T = 1,r = 0.05,K = 20,S0 = 15,steps = 200. Calculate the value of the type-Z option for σ ∈ {0.01, 0.1, 0.2, 0.3}. Produce a two column table of σ values and the corresponding option values in your written work. Explain why increasing σ increases the value of the option in this example. What do you expect to happen as σ → ∞ and why? If S0 = K, what would be the effect of increasing σ and why?
1. Fix T = 1,r = 0.05,K = 20,S0 = 15,steps = 200. Produce a graph of the type-Z option’s value at t = 0 as a function of S0 for S0 ∈ [0,100] for each of the following values of σ : {0.01, 0.1, 0.2, 0.3}. Show all four graphs on the same axes and include it in your written work. What effect does increasing volatility have on the price graph and why?
2. Fix T = 1,r = 0.05,K = 20,S0 = 15,steps = 200 and suppose the option’s value is V0 = 2.0. Write a root finding function to calculate the volatility σ correct to 6 significant figures. Include the result in your written work.
(Top Tutor) Daniel C.
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School: Cornell University
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# Find measure of angles word problem
Solve this word problem to find the measure of angles. In this example you'll split up a pie (don't forget to share!). Created by Sal Khan.
## Want to join the conversation?
• It seems like an equally valid answer would be that you cut yourself a 20 degree slice, and cut your brother a 10 degree slice. Where did the idea come from that these 3 slices would consume the entire remains of the pie?
• In fact, you're right. For it to be an exam question, it must be clearer.
• Couldn't you just do 180-30=150, then divide 150 by 3 which equals 50, and then multiply 50 times 2 to get 100? Wouldn't that be a lot simpler?
• yes but this strategy would only be valid with this specified problem
• I don't get any of this -_-
• Unfortunately, that is why we are here. to learn things like (9-4x) -2 (4y+8) +83y= -18x (14-13) -9x
• I feel like bad for the mom in this situation. She gets such a tiny amount. couldn't you just split it into 3 equal slices? Your slice isn't even a slice, it's like 2 slices. lol
• Two angles are supplementary. The measure of the larger angle is
10 degrees more than the product of 4 and the measure of the smaller angle.
What is the measure of the larger angle in degrees?
• Let's call larger angle "x" and the smaller angle "y".
We can use the given information to create a system of equations:
x + y = 180 (this is because they are supplementary)
x = 4y + 10 (The measure of the larger angle is 10 degrees more than the product of 4 and the measure of the smaller angle.)
Solving this system:
4y + y + 10 = 180
5y + 10 = 180
5y = 170
y = 34˚
x + 34 = 180
x = 146˚
• Use an algebraic equation to find the measures of the two angles described below. Begin by letting x represent the degree measure of the angle's supplement.
The measure of the angle is nineteen times greater than its supplement.
what is the measure of the supplement?
can someone help me with this.
• can you help me with angle addition postulate
• The angle addition postulate states two angles that share a ray add up to make up a larger angle. This is useful to know because if you know two of the angles, you can figure out the third.
Example:
<ABC + <ABD= <CBD
<ABC=42
<CBD=87
87-42=45
You subtract 42 from 87 because you are removing one angle from the larger angle to find the other smaller angle.
I hope this helped!
• Why do you need that much pie?!?!?!?!?!?!
• Is that question necessary
(1 vote)
• ok....this is a 4 min 33 sec video. when i found it out in less than a minut. half pie = 180 deg. mum slice so remaining 150. it will be 3x because little bro slice = x your slice equal to double so 2x. then x=150/3. x= 50. little bro slice = x so 50 ,yours 2x so 50*2 = 100. done | HuggingFaceTB/finemath | |
# A uniform sphere of weight $W$ and radius $5 \mathrm{~cm}$ is being held by a string as shown in the figure. The tension in the string will be:
Question:
A uniform sphere of weight $W$ and radius $5 \mathrm{~cm}$ is being held by a string as shown in the figure. The tension in the string will be:
1. $12 \frac{W}{5}$
2. $5 \frac{W}{12}$
3. $13 \frac{W}{5}$
4. $13 \frac{W}{12}$
JEE Main Previous Year Single Correct Question of JEE Main from Physics Laws of Motion chapter.
JEE Main Previous Year April 9, 2013
Correct Option: 4
Solution:
### Related Questions
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View Solution
• A metal sample carrying a current along $X$-axis with density $J_{x}$ is subjected to a magnetic field $\mathrm{B}_{\mathrm{z}}$ (along z-axis). The electric field $E_{y}$ developed along Y-axis is directly proportional to $J_{x}$ as well as $\mathrm{B}_{Z}$. The constant of proportionality has SI unit
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• The quantities $x=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}, y=\frac{E}{B}$ and $z=\frac{1}{C R}$ are defined where $C$-capacitance, $R$-Resistance, $l$-length, $E$-Electric field, $B$-magnetic field and $\varepsilon_{0}, \mu_{0},-$ free space permittivity and permeability respectively. Then :
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• Dimensional formula for thermal conductivity is (here $K$ denotes the temperature:
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View Solution
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View Solution
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View Solution
error: Content is protected !! | HuggingFaceTB/finemath | |
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A106729 Sum of two consecutive squares of Lucas numbers (A001254). 8
5, 10, 25, 65, 170, 445, 1165, 3050, 7985, 20905, 54730, 143285, 375125, 982090, 2571145, 6731345, 17622890, 46137325, 120789085, 316229930, 827900705, 2167472185, 5674515850, 14856075365, 38893710245, 101825055370, 266581455865 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Positive values of x (or y) satisfying x^2 - 3xy + y^2 + 25 = 0. - Colin Barker, Feb 08 2014 Positive values of x (or y) satisfying x^2 - 7xy + y^2 + 225 = 0. - Colin Barker, Feb 09 2014 Positive values of x (or y) satisfying x^2 - 18xy + y^2 + 1600 = 0. - Colin Barker, Feb 26 2014 LINKS Bruno Berselli, Table of n, a(n) for n = 0..300 Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (3,-1). FORMULA a(n) = L(n)^2 + L(n+1)^2 = 5*(F(n)^2 + F(n+1)^2) = 5*A001519(n+1). a(n) = 3*a(n-1) - a(n-2). - T. D. Noe, Dec 11 2006 G.f.: 5*(1-x)/(1-3*x+x^2). - Philippe Deléham, Nov 16 2008 a(n) = (5/2)*((3/2)+(1/2)*sqrt(5))^n+(1/2)*((3/2)+(1/2)*sqrt(5))^n*sqrt(5)-(1/2)*((3/2)-(1/2)*sqrt(5))^n *sqrt(5)+(5/2)*((3/2)-(1/2)*sqrt(5))^n, with n>=0. - Paolo P. Lava, Nov 19 2008 a(n) = Fibonacci(n-2)^2 + Fibonacci(n+3)^2. - Gary Detlefs, Dec 28 2010 For n>=3, a(n)=[1,1;1,2]^(n-2).{3,4}.{3,4}. - John M. Campbell, Jul 09 2011 a(n) = L(2n) + L(2n+2). - Richard R. Forberg, Nov 23 2014 From Robert Israel, Nov 23 2014: (Start) a(n) = 5*A000045(2*n+1). E.g.f.: (5+sqrt(5))/2 * exp((3+sqrt(5))*x/2) + (5-sqrt(5))/2 * exp((3-sqrt(5))*x/2). (End) MAPLE seq(combinat:-fibonacci(n-2)^2 + combinat:-fibonacci(n+3)^2, n=0..100); # Robert Israel, Nov 23 2014 MATHEMATICA Table[LucasL[n]^2 + LucasL[n + 1]^2, {n, 0, 30}] (* Wesley Ivan Hurt, Nov 23 2014 *) PROG (MAGMA) [Fibonacci(n-2)^2+Fibonacci(n+3)^2: n in [0..30]]; // Vincenzo Librandi, Jul 09 2011 (PARI) for(n=0, 30, print1(fibonacci(n-2)^2 + fibonacci(n+3)^2, ", ")) \\ G. C. Greubel, Dec 17 2017 CROSSREFS Cf. A000204. Sequence in context: A025625 A112024 A245415 * A212950 A038252 A211865 Adjacent sequences: A106726 A106727 A106728 * A106730 A106731 A106732 KEYWORD nonn,easy,changed AUTHOR Lekraj Beedassy, May 14 2005 EXTENSIONS Corrected by T. D. Noe, Dec 11 2006 More terms by Bruno Berselli, Jul 17 2011 STATUS approved
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# Which Of The Following Equations Applies To Parallel Circuits
By | January 7, 2023
Electricity is a basic part of life, and understanding the fundamentals of how electricity works is essential in order to make the most of it. One of those fundamental concepts is parallel circuits, and the equations that apply to them.
A parallel circuit is made up of two or more electrical devices connected in such a way that the current divides into multiple paths before it returns to the source. This means that each device receives the same voltage, but the current passing through each device can vary. Because of this, parallel circuits are often used for lighting and other applications where the intensity can be adjusted.
The equation that applies to parallel circuits is called Ohm's Law. This equation states that the total current passing through any point in a circuit is equal to the sum of the individual currents passing through each branch. This equation is written as I=I1+I2+I3+ ...etc., where I is the total current, and I1, I2, I3, etc. are the individual currents in each branch of the circuit.
In addition to Ohm's Law, there is another equation known as the "Maximum Power Transfer Theorem" which is used to determine the optimum load resistance in order to transfer the maximum amount of power in a given circuit. This equation states that the load resistance should be equal to the internal resistance of the source in order to achieve maximum power transfer.
Finally, Kirchhoff's Current Law states that for a closed loop circuit, the total current entering a node is equal to the total current leaving that node. This law is very useful for analyzing complex circuits with many nodes.
These equations are all essential for understanding how parallel circuits work and for designing circuits with optimal performance. Knowing these equations can help you make the most of your electrical systems and ensure that you get the most out of your power sources. | HuggingFaceTB/finemath | |
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# Scientific notation word problem: red blood cells
Vampires and math students want to know: How many red blood cells are in the a human body? We can find the answer using scientific notation. Created by Sal Khan.
## Want to join the conversation?
• Why wouldn't you just divide the 9 into 2 and have the quotient of 0.22 there instead of ''borrowing'' the 10?
• You could do that, and have 0.222 * 10^14. But then, to get it into official scientific notation, you would need change it to 2.22 * 10^13. So Sal is just doing the conversion into scientific notation in the earlier step. In both cases you would end up with the same answer,
• Why was 1/10^-14 the same as 10^14? Not quite sure.
• Well, you can think it as (1/10)^-14. Then, according to the rule (a/b)^c = ((a^c)/(b^c)), you can use the distributive property:
((1^-14)/(10^-14)
When a fraction has either a numerator or denominator or both with a negative exponent, you need to switch the position. Meaning the numerator with a negative exponent would switch to the denominator for a positive exponent, and vice versa.
((1^-14)/(10^-14)) = ((10^14)/(1^14))
And since we know that 1 with a positive exponent is still one, it becomes:
(10^14)/1
And since division by one means the quotient is the dividend, it becomes:
10^14
• At in the video, you start to mention that 90 isn't in scientific notation because it isn't less than ten, which I agree with, but under that understanding, wouldn't 10^-15 technically not be in sci.n. because it isn't less than ten?
• 90 isn't in scientific notation because the coefficient isn't less than 10. Since 90 is just a number, 90 is the coefficient which isn't less than 10.
With 10^-15, the coefficient is 1 which is less than 10.
However, it should be noted that to be in scientific notation, the coefficient must also be greater than or equal to 1.
I hope this clarifies what Sal meant!
• I realize I am the only person having trouble understanding this, but why did we divide by volume of 1 red blood cell rather than multiplying? total volume of 1 red cell/ volume of 1 cell = # of blood cells??
I understand the calculation itself just confused on the formula
• At , Sal wrote "5 (liters) x 40%". Can someone show me how to work this out? ( You don't have to include the liters)
• this is old
also how did he get the ten to turn into a nine that is some weird math right there. Does he have anymore things involving this kind of math because i still need to understand more.
• At in the video, couldn't Sal have done 2/9 first, then made it into scientific notation, rather then multiplying it by ten? I'm kind of confused, whether I'm doing it right or wrong. Does both ways always work?
• Is anyone else confused
• Probably | HuggingFaceTB/finemath | |
## Introductory Algebra for College Students (7th Edition)
$$7 + (-10) = -3$$
In this problem, we are adding a $-10$, meaning that we are going more left on the number line. If we are at $+7$ on the number line and go left $10$ spots, then we will be at $-3$. Therefore, $$7 + (-10) = -3$$ | HuggingFaceTB/finemath | |
# What mass of sodium metal is required to make a mass of 936*g sodium chloride?
Sep 29, 2017
Approx. $400 \cdot g$ of metal are required........
#### Explanation:
We interrogate the redox reaction....
$N a \left(s\right) + \frac{1}{2} C {l}_{2} \left(g\right) \rightarrow N a C l \left(s\right)$
We gots $\frac{0.568 \cdot k g}{70.90 \cdot g \cdot m o {l}^{-} 1} = 8.01 \cdot m o l$ with respect to dichlorine....
And we gots $\frac{936 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1} = 16.01 \cdot m o l$ with respect to sodium chloride....
And given the stoichiometry, CLEARLY, we require $16.01 \cdot m o l$ sodium metal.....
...which represents a mass of.....
$16.01 \cdot m o l \times 22.99 \cdot g \cdot m o {l}^{-} 1 \cong 400 \cdot g$ | HuggingFaceTB/finemath | |
## What is the value of sin 45 in trigonometry?
= 1 / 2
Sine 0° 0
Sine 30° or Sine π/6 1/2
Sine 45° or Sine π/4 1 / 2
Sine 60°or Sine π/3 3 / 2
Sine 90° or Sine π/2 1
What is Sohcatoa?
SOHCAHTOA is a mnemonic device used to remember the ratios of sine, cosine, and tangent in trigonometry.
### What does tan 45 mean?
Explanation: For tan 45 degrees, the angle 45° lies between 0° and 90° (First Quadrant). Since tangent function is positive in the first quadrant, thus tan 45° value = 1. Since the tangent function is a periodic function, we can represent tan 45° as, tan 45 degrees = tan(45° + n × 180°), n ∈ Z.
What is the exact value of tan 45?
The exact value of tan(45°) tan ( 45 ° ) is 1 .
#### How do you solve cos 45 degrees?
How can you evaluate the value of cos 45? We can use Pythagoras theorem to find the value of cos 45. Since, for a right angle, if the adjacent angles are 45 degrees, then the adjacent side and opposite side will be equal. Thus, we can find the value of cos 45 equal to 1/√2.
What is the exact value of sin 45 enter your answer as a simplified fraction?
The exact value of sin(45) is √22 .
## What is cosec in math?
Cosecant Formula. Cosecant is one of the six trigonometric ratios which is also denoted as cosec or csc. The cosecant formula is given by the length of the hypotenuse divided by the length of the opposite side in a right triangle.
What is csc on calculator?
The cosecant formula is: csc(α) = hypotenuse c opposite a. Thus, the cosecant of angle α in a right triangle is equal to the length of the hypotenuse c divided by the opposite side a. To solve csc, simply enter the length of the hypotenuse and opposite side, then solve.
### What is the value of cos0?
The value of cos 0 is 1. Here, we will discuss the value for cos 0 degrees and how the values are derived using the quadrants of a unit circle. The trigonometric functions are also known as an angle function that relates the angles of a triangle to the length of the triangle sides.
How do you find the value of sin 45?
How to Find the Value of Sin 45 Degrees? The value of sin 45 degrees can be calculated by constructing an angle of 45° with the x-axis, and then finding the coordinates of the corresponding point (0.7071, 0.7071) on the unit circle. The value of sin 45° is equal to the y-coordinate (0.7071). ∴ sin 45° = 0.7071.
#### How do you solve tan 45?
To find the value of tan 45 degrees using the unit circle:
1. Rotate ‘r’ anticlockwise to form 45° angle with the positive x-axis.
2. The tan of 45 degrees equals the y-coordinate(0.7071) divided by x-coordinate(0.7071) of the point of intersection (0.7071, 0.7071) of unit circle and r.
How do you write tan 45 degrees?
The value of tan 45 degrees is 1. Tan 45 degrees in radians is written as tan (45° × π/180°), i.e., tan (π/4) or tan (0.785398. . .). In this article, we will discuss the methods to find the value of tan 45 degrees with examples. Tan 45° in radians: tan (π/4) or tan (0.7853981 . . .) | HuggingFaceTB/finemath | |
## SQLite: Can Using Unicode Collations be Improved? Attempt 2
In my quest to improve using Unicode collations in SQLite, I figured I needed to accomplish two things first. First, persist the locale for the DB between closing and opening it again. Second, change the default collation from BINARY (ASCII) to the a collator for the locale.
Here is the logic that I used for my implementation:
1. try to load the ICU collation for the given locale or use the system’s default locale
2. check if any tables other than the “locale table” exist. If they exist, exit
3. update the “locale table” with the name of the locale. (This allows reloading the collation when opening the DB)
4. using the ICU collation, create an sqlite3 collation and make it the default collation for the DB
This is some of the source code for implementing the logic above:
int setDBDefaultLocale(sqlite3 * db, const char * const zLocale)
{
int rc;
sqlite3_stmt* stmt;
UErrorCode status = U_ZERO_ERROR;
UCollator *pUCollator= ucol_open(zLocale ? zLocale : uloc_getDefault(), &status);
if(!U_SUCCESS(status))
{
return SQLITE_ERROR;
}
// check if any tables other than the "locale table" have been created.
// if they exist, just quit, can not change default locale if a another table has already been created
if ((rc = sqlite3_prepare_v2(db, "select * from sqlite_master where name != 'locale_table' and type = 'table'", 76, &stmt, 0)) != SQLITE_OK ||
(rc = sqlite3_step(stmt)) != SQLITE_ROW)
{
sqlite3_finalize(stmt);
return rc;
}
if (sqlite3_column_int(stmt, 0))
{
sqlite3_finalize(stmt);
localeErrMsg(vdbe);
return SQLITE_MISUSE;
}
// creates a locale table if one doesn't exist and updates it with the value of zLocale
updateLocaleTable(zLocale ? zLocale : uloc_getDefault());
// create the collation
if ((rc = sqlite3_create_collation_v2(db, "LOCALE", SQLITE_UTF16,
(void *)pUCollator, icuCollationColl, icuCollationDel)) != SQLITE_OK)
{
ucol_close(pUCollator);
return rc;
}
// set the default collating sequence to the locale collation
db->pDfltColl = sqlite3FindCollSeq(db, SQLITE_UTF16, "LOCALE", 0);
return db->pDfltColl ? SQLITE_OK : SQLITE_ERROR;
}
With the implementation above, I was able to accomplish the first task i.e. persist the locale between connections. But when I tried to change the locale, it would throw an unexpected exception. After much digging, I figured out that the call to sqlite3FindCollseq was returning a buggy collating sequence. There was a mismatch between the DB encoding and the collating sequence encoding. That was solved by querying the encoding of the db object and passing that into the function call:
...
db->pDfltColl = sqlite3FindCollSeq(db, ENC(db), "LOCALE", 0);
...
But as I thought about it, because this implementation always creates a table, this breaks SQLite’s functionality that depends on having an empty database. (e.g. setting the database encoding, pragma encoding = ‘UTF-16’)
As such, I am now investigating other ways to persist the locale. I am thinking of actually appending the locale to the header of the database file. One of my colleagues suggested persisting the locale outside of the database, using either another DB instance or a file. I am now pursuing the file header option. So far I have found out that I may need to modify the Pager (pager.c) implementation, to add more space to store the locale and the Btree (btree.c) implementation, specifically the sqlite3BtreeOpen method to both read in the locale. From what I found out so far, seems like nothing has to change to persist the locale to the disk.
So now the fun begins to implement the above. Here goes…
## SQLite: Can using Unicode collations be improved? Attempt 1
Suppose you are working with a string data other than English. How well will SQLite handle that data? Turns out that SQLite support for languages other than English is not turned on by default. Your non-English data will most probably produce some unexpected results.
Last week I began on extending SQLite to provide a more “user friendly” Unicode support. The motivation behind this is to get SQLite into the NexJ Express framework. For this to happen, it has to have support full Unicode suport.
Out of the box, SQLite provides the following
• Unicode encoding of characters (UTF-8 & UTF-16)
• Support for custom collations
• Support for locales through the ICU library
• Open source code (** my favourite)
The default string collation that SQLite uses is BINARY. This collation just implements ASCII character manipulations. It “understands” only the 26 letters of the English alphabet. Feed the string manipulation functions unicode encoded characters for other languages and it breaks. The SQLite way to fix this is to
1. Get a build of SQLite with ICU enabled. It is easiest to just recompile. The ICU enabled library I found was built on an old version of SQLite.
2. Create an icu collation using “select icu_load_collation([locale], [collation name])
3. Attach the collation to columns using the keyword collate. E.g. create table foo (bar COLLATE [collation name])
4. If using full-text search then, use “create virtual table foo (bar tokenizer=icu [locale])
This works fine to an extent. The problem I immediately noticed is that SQLite does not remember custom collations. Every time a database is open, the “icu_load_collation” statements needs to be run. In the case of the NexJ Express framework, the locale is not available to the code that opens connections. So the idea I had was to enable persisting of custom collations. A colleague of mine implemented this using a table that is updated with the locale and collation name. When a connection is opened, the library loads the collation from the information in the “locale table”.
I thought of taking this further by modifying the database open code to instead set the system’s locale as SQLite’s default collation. I have an implementation of this but it breaks when the locale is changed.
This week I will spend time on finding a fix for my implementation otherwise, will just use the one my colleague came up with.
Tagged , , , ,
## SQLite Adapter for NexJ Express – Overcoming the limitations of SQLite part 2
This is a continuation of my previous post: SQLite Adapter for NexJ Express – Overcoming the limitations of SQLite part 1.
I made mention of the error codes SQLITE_LOCKED and SQLITE_BUSY the last time I blogged. These errors show up when new connections try to either read or write to an SQLite database instance that already has a connection with an active write operation. Question 5 from the SQLite FAQ provides a quick description of the overall problem. The gist of it is that when interacting with an SQLite database using multiple concurrent connections, it only takes one writer connection to lock the database file and lock out all the rest. NexJ Express makes use of threading on the persistence layer and as such ran into this issue.
For several weeks we tried various solutions ranging from setting statement query time outs, creating an sqlite3_busy_handler to creating a singleton connection for all threads. The final solution was found after reading about transactions in chapter 5 of “The Definitive Guide to SQLite, Second Edition”. The chapter explains the transaction life cycle and the transition between the different lock states. Write operations cycle through the following states
UNLOCKED -> PENDING (1) -> SHARED -> RESERVED -> PENDING (2) -> EXCLUSIVE -> COMMIT
Of note are PENDING states; PENDING locks are gateway locks.
Before a writer dumps to the disk, it needs to get an EXCLUSIVE lock. From RESERVED state when the writer gets PENDING(2), it affects the database in two ways:
1. It will hold onto the PENDING(2) lock no matter what
2. In PENDING(2) lock, no other connection can go from UNLOCKED to SHARED. This creates a deadlock when a previous writer has committed a transaction and it tries to get a SHARED lock. The PENDING(2) lock prevents that previous writer from entering PENDING(1) hence a deadlock.
Because of the second effect noted above, all the previous trial solutions failed. The application would retry infinitely but not get an EXCLUSIVE lock. The book suggested the use of the BEGIN IMMEDIATE command instead of the BEGIN command to start a write transaction. This prevents the deadlock situation noted in 2 above. Here is an extract from The Definitive Guide to SQLite to give more detail:
Since you know you want to write to the database, then you need to start by issuing begin IMMEDIATE. If you get a SQLITE_BUSY, then at least you know what state you’re in. You know you can safely keep trying without holding up another connection. And once you finally do succeed, you know what state you are in then as well—RESERVED. Now you can use brute force if you have to because you are the one in the right. If you start with a begin exclusive, on the other hand, then you are assured that you won’t have any busy conditions to deal with at all. Just remember that in this case you are doing your work in EXCLUSIVE, which is not as good for concurrency as doing the work in RESERVED"
As such, the code for setAutoCommit in the Conn class was changed to use “begin immediate“. The application then needed to keep retrying if it received an SQLITE_LOCKED or SQLITE_BUSY errors. A busy handler was created that returns 1 (i.e. try again). This wasn’t enough since the application still received the SQLITE_LOCKED and SQLITE_BUSY exceptions even when trying to set auto commit to false. Seems it has to do with SQLite internal detecting a deadlock and just throwing an exception. Changes needed to be made to the org.sqlite.PrepStmt and org.sqlite.Conn classes. The methods: execute, executeBatch, executeQuery, executeUpdate, setAutoCommit, rollback and commit, were wrapped in try-catch blocks to keep retrying if either an SQLITE_LOCKED or SQLITE_BUSY exception were thrown. The source code for this is in nexj.core.persistence.sql.SQLiteDataSourceWrapper.
With that solution implemented, the server now knows how to handle a locked database file. Using begin immediate has some side effects though and as such some unit tests in the JDBC driver fail.
The source code for the SQLite adapter is shared on bitbucket.org/cwdesautels/nexj-express-sqlite-adapter in the ‘sqlite’ branch. Check out the SQLite* classes in the package nexj.core.persistence.sql.
The source code to build a customized Xerial SQLite JDBC driver and SQLite shell can be found at bitbucket.org/gbatumbya/nexj-express-sqlite. (This repo use subrepos to track changes to gbatumbya/sqlite and gbatumbya/sqlite-jdbc)
## SQLite Adapter for NexJ Express – Overcoming the limitations of SQLite part 1
For the last two semesters I have been working on an SQLite persistence adapter for NexJ Model Server Express. Today it is finally ready for the first code review.
Getting to this stage has taken much longer than I first anticipated. Turns out that SQLite’s “liteness” was turning into a blocker and at one point I even thought we would abandon the project altogether. I should have known this though. The SQLite website says to “Think of SQLite not as a replacement for Oracle but as a replacement for fopen()” and provides a list of situations where other RDBMS work better. In developing this adapter, we were actually pitting SQLite up against the likes of Oracle, PostgreSQL, MySQL, SQL server and other enterprise SQL servers.
First to be tackled was SQLite’s implementation of SQL92 or rather lack thereof; granted though, the documentation for supported features/syntax is very good. We needed to come up with work-arounds to do things that are taken for granted when developing for an enterprise level RDBMSs. For example it didn’t support altering a column’s definition or table constraint, dropping a column or constraint and there is no procedures. Most of the work-arounds required renaming the old table, creating the new table and copying data from the old to the new and then finally deleting the old table.
There was also the case where it was required to convert columns containing hex string values into BLOBs. SQLite provides the x'ABC123' syntax for this, but it is limited. It does not take columns as input. For this we developed an extension function binary(). It pretty much does what the x'' does only that it is exposed as a function call.
It was also required to provide a ranking of results for FTS queries. Again an extension function rank() was developed. The rank for a returned row is calculated as: number of local hits / sum of global hits. The source of the extension functions is can be found here sqliteNexjExtension.c.
Next up was selecting a JDBC driver. SQLite does not provide an offical JDBC driver but their wiki provides a list of available drivers. The decision was taken to go with Xerial SQLite JDBC Driver (a fork of SQLiteJDBC). We found this driver to be the most up-to-date and its easy of use of native SQLite libraries was a big plus. The driver did not provide implementations of XADatasource and XAConnection. The team at NexJ Systems ended up providing us with implementations (nexj.core.persistence.sql.pseudoxa) that wraps DataSource and Connection and exposes them as XAResource‘s. (But just last week, I came across SQLiteConnectionPoolDataSource which I provides XA implementations for the SQLite JDBC driver. We may later on switch to using this.)
We found the driver to be buggy with setting: an encoding on a SQLiteDataSource and query time-out on a statement. Turns out that calling SQLiteDataSource.setEncoding("UTF8") causes an exception. The driver tries to execute the statement, pragma encoding = UTF-8. Note the hypen (-), SQLite expects that to be quoted. The javadoc on Statement.setQueryTimeout(int seconds) reads “Sets the number of seconds the driver will wait for a Statement object to execute to the given number of seconds“. Internal though, the driver calls sqlite3_busy_timeout which “sets a busy handler that sleeps for a specified amount of time when a table is locked”. This ends up having the effect that if a database is locked when first executed, the statement will sleep for the given number of seconds then try a second time instead of trying several times until the given number of seconds have past. The solution for this was to use a TimerTask (nexj.core.util.sql.StatementCancelationTask) that was already implemented.
Next week I will have a post about how the dreaded SQLITE_LOCKED and SQLITE_BUSY were overcome so that SQLite could be used in an JavaEE application server. The source code for the adapter is shared on bitbucket.org/cwdesautels/nexj-express-sqlite-adapter in the ‘sqlite’ branch. Check out the SQLite* classes in the package nexj.core.persistence.sql.
Tagged , , , ,
## NexJ Express Integration Layer Now Supporting JSON
I am glad to write that another project, JSON Integration adapter, to which I contributed passed its final code review and was merged into NexJ Express, at the beginning of February 2012.
It brings back lots of memories of my first project at CDOT during which I was first introduced to JSON as I developed the JSON RPC adapter. Some of my earlier work on classes from the RPC adapter in particular: JSONWriter and JSONParser were reused in building the integration adapter.
The integration layer in NexJ Express allows a Model to interact with external applications asynchronously. With the addition of this adapter, the integration layer can now produce and consume JSON messages. The most common use case where the adapter will be most utilized is when NexJ Express has to interact with a web services that serves JSON. JSON (JavaScript Object Notation) is a lightweight data-interchange format. It is easy for humans to read and write. It is easy for machines to parse and generate.
Thanks to Brian Lim who worked on the project at its very beginning during the summer of 2011 and helped get the ball rolling.
Seneca’s Centre for Development of Open Technology (CDOT) provides a physical and virtual environment for the development and research of open source software through collaboration with Seneca, the open source community, business, and other institutions. The centre is an integration point for knowledge, education, and relationships within the open source world.
NexJ Systems is a leading provider of enterprise private cloud software, delivering enterprise customer relationship management (CRM) solutions to the financial services, insurance, and healthcare industries.
Tagged , ,
## PostgreSQL Support Added to NexJ Express
Wow, been a while since I last blogged. But such an occasion calls for putting things aside and taking the time to blog.
After about 1 year of development, the PostgreSQL adapter code passed its final code review (there were 6 in total) and the code was merged into the core repository at mercurial.nexj.com/express. Thanks to Minoo and Anastasia who both contributed to the project in the past semesters.
The experience of developing this adapter involved lots of new learning and lots of refreshers. I learnt that there is more to writing code that interfaces with another system than just making it work. There was an emphasis placed on optimizing the code and making use of the things that PostgreSQL is “good at”. And while I got to appreciate PostgreSQL I also got to know that there were some small annoyances with it (I think that when on windows, creating a table space should understand windows folder paths syntax i.e. it shouldn’t be necessary to change C:\path\to\ts to c:/path/to/ts). I also got to appreciate the notation of open source community. It was neat that when I asked questions on the PostgreSQL forums or IRC, I got some very helpfully pointers from the community members. Thanks to RhodiumToad who answered alot of my questions on IRC and also Hiroshi who helped me out with UUID for PostgreSQL x64 on Windows.
To NexJ Express users out there, you now have a choice between two of the leading open source databases: MySQL and PostgreSQL, to use as the data source for your models.
## PostgreSQL Windows x64: Enabling UUID-OSSP.sql
The background to this is that for that for over 6 months now, I have been working on developing a PostgreSQL Adapter for NexJ Express, a part of Open Health Tools. Though most of the development is done in Java, I have also learnt a lot about PostgreSQL.
The Problem:
Although PostgreSQL has a UUID type, the UUID generation is dependent on the UUID-OSSP.sql module which in turn depends on the OSSP-UUID library.
The PostgreSQL installer for Windows includes the UUID-OSSP.sql module for x86 builds but not for x64 builds (lastest version that I checked was PostgreSQL 9.0.4).
When I asked on #postgresql irc channel, one of the things I was told was that there was not OSSP-UUID library build for windows x64.
The Solution: (tested with PG 9.0.4 x64 build)
Turns out that there was a build of OSSP-UUID for x64 windows in the wild. Thanks to Hiroshi Saito, who replied to my message on the pg_general mailing list.
2. Unzip the contents of the zip to path/to/PostgreSQL/9.0/
3. Open psql and install uuid-ossp.sql using the command:
\i 'path/to/PostgeSQL/9.0/share/contrib/uuid-ossp.sql'
Also, Hiroshi provides the source if you interested on compiling OSSP-UUID on x64 windows for yourself.
## Tying Shoelaces And What is Valid JSON
Up until this week, I thought I knew enough about tying shoelaces and JSON. Turns out that for both, I fell short.
First to the tying of shoelaces. You would have thought that I should have been a pro by now since I have done it from childhood and you may be laughing at me but before you do that, check out this very informative video from a talk from TED 2005 Terry Moore: How to tie your shoes.
So to the techy stuff.
I got introduced to JSON at the beginning of last summer as I worked on creating a JSON RPC adapter for NexJ Express.
At the moment I am also involved in building a JSON Integration Adapter, and as part of the design phase we had to figure out what vaild JSON was.
To my astonishment, I found out that Javascript “primitives”, that is String, Number and Boolean are not valid JSON. My misunderstanding of this was from missing reading the railway diagrams on json.org, I just thought that every thing that was a valid value, was also valid JSON.
So how did I know I was wrong, my colleague Brian, showed me the RFC for JSON, and right there it says
A JSON text is a serialized object or array. JSON-text = object / array
I also asked another colleague, Dave, who works a lot with Javascript and he started off by saying basically the same thing, JSON is “Object Notation”.
So what does this mean for me? Well, I have authored some web services which serve what I thought was “JSON”, so I may have to go back and fix them.
## PostgreSQL: Create a Large Object (lo) from binary data (bytea) on server side
Again, I found something that NexJ Express uses that PostgreSQL does not have support for, that is creating large objects when given bytea data.
Like with Comparing bytea data to a Large Object on server side, I created a CAST (bytea AS oid) and a procedure to solve this problem.
The algorithm is pretty simple, get the binary data, if it is null, return null. Else create a large object and in the lowrite function, pass it the binary value, instead of a path to a file.
The code for the procedure is below. Note that the lo_manage package should be installed for this to work.
create or replace function blob_write(lbytea bytea)
returns oid
volatile
language plpgsql as
$f$
declare
loid oid;
lfd integer;
lsize integer;
begin
if(lbytea is null) then
return null;
end if;
loid := lo_create(0);
lfd := lo_open(loid,131072);
lsize := lowrite(lfd,lbytea);
perform lo_close(lfd);
return loid;
end;
$f$;
CREATE CAST (bytea AS oid) WITH FUNCTION blob_write(bytea) AS ASSIGNMENT;
So now the following code works:
CREATE TABLE bytea_to_lo (
lo largeObj
);
INSERT INTO bytea_to_lo VALUES ( DECODE('00AB','hex'));
## PostgreSQL Compare bytea data to a Large Object on server side
While working on the PostgreSQL Adapter for NexJ Express, I found out that I needed to compare large objects to bytea data on the server side.
for example:
select * from Table_With_LOs where lo_value = bytea_value;
A little background:
PostgreSQL stores large object data in a separate table (pg_largeobject), and in the data column for the table it uses an oid (an integer value) as a “pointer” to this data.
So, doing "where lo_value = bytea_value", actually translates to
"where integer_value = bytea_value"
Solution:
Based on this blog post, Reading XML Files into Database ,by rhodiumtoad (he is very active in the postgresql irc channel), I created a cast function to go from oid to bytea data.
create or replace function blob_read(loid oid)
returns bytea
volatile
language plpgsql as
$f$
declare
content bytea;
lfd integer;
lsize integer;
begin
if(loid is null) then
return null;
end if;
lfd := lo_open(loid,262144);
lsize := lo_lseek(lfd,0,2);
perform lo_lseek(lfd,0,0);
$f$; | open-web-math/open-web-math | |
# Questions on Cross Product
1. Jan 17, 2013
### doctorjuice
First question:
a x b = -b x a
Why is this so?
As I understand, a major purpose of the cross product (if not, the purpose) is to find a third vector that is perpendicular to two other vectors simultaneously. Let's say a x b = c. Shouldn't the answer really be, a x b = +/- c? Since, of course, both c and -c are perpendicular to a and b simultaneously.
The situation of the sqrt(4) = +/- 2 is analogous to this.
Second question:
The cross product is said to be something that only works in 3 dimensional space. In 2D, it is said to be not applicable. Take two vectors in 2D that are just the negative of each other. Obviously a line can be drawn straight up relative to these two vectors that is perpendicular to both (visualize an upside down T). Since the cross product's main purpose is to find a third vector that is perpendicular to two other vectors, can it be said that the cross product fails in 2 dimensional space?
2. Jan 17, 2013
### Fredrik
Staff Emeritus
This follows easily from the definition. What definition have you been taught?
You may have to define the word "should".
Torque is a good example of the use of the cross product in physics. When you use a screwdriver on a screw, you're applying torque that's equal to a cross product. The direction of that vector contains the information about which way you're turning the screw. Angular momentum is another good example. In this case, the direction of the vector corresponds to the direction of the rotation. So that direction isn't irrelevant.
You may also have to define the word "fail".
I can't say that the cross product fails in 2-dimensional space, since there is no cross product on $\mathbb R^2$. But you are right that there are no vectors that are perpendicular to two linearly independent vectors.
3. Jan 17, 2013
### lurflurf
0=(a+b)x(a+b)=a x a+a x b+b x a+b x b=a x b+b x a
if
a x b = c
t c is perpendicular to both a and b for any t
The cross product is chosen to preserve distance and to be right handed.
There are generalizations of the cross product to other spaces, none of them have all the properties of the n=3 case, one reason is that several function are the same only for this case
for example consider functions
$$V \times V \rightarrow V \\ V \times V \rightarrow V^{n-2} \\ V \times V^{n-2} \rightarrow V \\$$
only when n=3 can we hope these functions are the same
There is the http://en.wikipedia.org/wiki/Seven-dimensional[/PLAIN] [Broken] cross_product]Seven dimensional cross product which is the closest.
Last edited by a moderator: May 6, 2017
4. Jan 17, 2013
### micromass
The definition is that the square root is always nonnegative. Thus by definition, we have $\sqrt{4}=2$. We do not define $\sqrt{4}=-2$ or $\sqrt{4}=\pm 2$. We only give one value to the square root because we want it to be a function.
5. Jan 17, 2013
### CompuChip
Similarly, for the cross product, the definition is such that the cross product is "right-handed". That is, if you take a right handed coordinate system (turning the fingers of your right hand from the x to the y-axis your thumb points along the z-axis) then ex x ey = ez (or, in alternative notation, $\hat i \times \hat j = \hat k$) rather than -ez.
6. Jan 17, 2013
### doctorjuice
I've been taught the same definition, I was just asking why it was defined this way, I didn't see the reason for it at first. There are actually a couple reasons, I believe (and I would like to hear confirmation or refutal of these reasons):
1) If a x b = +/-c (instead of +c), then the expression a x b would be ambiguous. By itself, this is not a major problem but once complicated calculations with cross products occur this problem becomes very large.
Analogous to sqrt(4), when it is part of complex calculations, if it were to ambiguously mean +/-2, then the whole mathematical expression of a complex operation involving multiple sqrt(4) terms would be very ambiguous and you could come up with many different answers for the same problem.
2) Even if the sign of c does not affect the end result in mathematics, in physics it does, and the information is very important in physics (as you said in other parts of your post).
I think you misunderstood this part of my post. I was referring to two vectors in 2D, that go in opposite directions (so a and -a would be a pair of such vectors). If a vector is drawn straight up in 2D relative to these two vectors (think an upside-down T, with the left part being vector a, right part being vector b (or -a) and the up part being a vector perpendicular to both, vector c) then that vector is perpendicular to both a and b in 2D space simultaneously.
Since a major purpose of the cross-product is to find a vector that is perpendicular to two vectors simultaneously, I was making the assertion that the cross product fails at fulfilling its purpose in 2D space, for this specific situation.
I would love to hear yours and anybody else's thoughts on what I have said here.
7. Jan 17, 2013
### CompuChip
1) is definitely true. It's just a matter of convention, and just like we chose sqrt(4) to give 2 and not -2 because that's often easier, we also choose (1,0,0) x (0,1,0) to be (0,0,1) and not (0,0,-1) because it's usually easier. You could view the definition of sqrt as just some function, which happens to have the property that it squares to its argument without claiming that there are no other such numbers, you could see the definition of the cross product in a similar way.
I'm not sure about 2) - I think quite the opposite is true. It does not matter which way you define it, but if you get it the other way around you might also want to change other conventions so you don't end up writing minus signs all the time.
8. Jan 17, 2013
### THSMathWhiz
A formal cross product is defined for $\mathbb{R}^3$ and $\mathbb{R}^7$. There is a projective cross product for $\mathbb{R}^2$, which returns a directed scalar. Given $\mathbf{u}=\langle u_1,u_2\rangle$ and $\mathbf{v}=\langle v_1,v_2\rangle$, the "cross product" is $\mathbf{u}\times_2\mathbf{v}=(\langle u_1,u_2,0\rangle\times_3\langle v_1,v_2,0\rangle)\cdot\langle0,0,1\rangle$, where $\times_n$ is the cross product in $\mathbb{R}^n$.
9. Jan 17, 2013
### Fredrik
Staff Emeritus
Sorry about that. I just saw that you were asking about cross products in 2D and jumped to the wrong conclusion about what you wanted to ask. I guess I should have actually read the question.
You could certainly define a "product" of two linearly dependent vectors in ℝ2 that has a vector perpendicular to both as the result. For example, you could define
$$(a_1,a_2)(b_1,b_2)=\operatorname{sgn}(a_1b_1)(a_2,-a_1)$$ where sgn is the sign function. But this seems rather pointless. This product also has very little in common with the cross product on ℝ3.
10. Jan 18, 2013
### lurflurf
1&2)The convention does not matter, but using multiple conventions would lead to errors. This is the same in mathematics and physics.
As for finding a vector that is perpendicular, as I mentioned above the cross product serves several purposes simultaneously in 3-space this cannot be done in other spaces. One generalization gives the (unique up to scalar multiplication) vector perpendicular to n-1 given vectors. In 2-space this means we take one vector and return a vector perpendicular to it. In 11-space this means we take ten vectors and return a vector perpendicular to it. Only in 3-space can we take two vectors and return one vector perpendicular to it. | HuggingFaceTB/finemath | |
# True or False. The domain and the range of the
True or False. The domain and the range of the reciprocal function are the set of all real numbers.
You can still ask an expert for help
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Step 1
The reciprocal function is $f\left(x\right)=\frac{1}{x}$
The domain of f(x) is the set
$\left\{x\mid x\ne 0\right\}$
Therefore, the statement is false.
vicki331g8
We went through this in class, but I forgot. I will wait for an answer. | open-web-math/open-web-math | |
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People often said, “This is important for your development.” Advertisements on television often try to help us understand what is necessary for our growth. We are often confused about growth and development. We are taught in schools that…
## Difference between Arithmetic and Geometric Sequence
What is the difference between arithmetic and geometric sequence? Is this the question in your mind right now? You have landed at the right place. Arithmetic sequences have a constant difference between each term. Example:…
## Difference between an Outcome and an Event
Do you want to know what is the difference between an outcome and an event? You have landed at the right place. Although the event and outcome of the words may look identical, there are…
## Difference between the Bohr Model of the Atom and the Rutherford Model
The Rutherford model, and the Bohr model, is models that describe the structure of Anatom. Ernest Rutherford proposed the Rutherford model in 1911. Niels Bohr proposed the Bohr model in 1915. Bohr model can be…
## Difference between Observation and Inference
You have probably heard the terms observation and inference used several times in statistics. The act of watching, or monitoring, something is called observation. Using senses, objects, units, people, or any other thing, is called… | HuggingFaceTB/finemath | |
Statement Questions
Chapter 6 Class 8 Squares and Square Roots
Concept wise
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### Transcript
Example 14 Area of a square plot is 2304 m2. Find the side of the square plot. Given area of the square plot = 2304 π^2 Let side of the square = π₯ We know that, Area of square = Side2 β΄ Area of square = π₯^2 2304 = π₯^2 π₯^2 = 2304 π₯ = β2304 β΄ Side = βππππ Finding square root of 2304 by Long Division :- β΄ Square root of 2304 = 48 Thus, Side of the square plot = 48 m Rough 86 Γ 6 = 516 87 Γ 7 = 609 88 Γ 8 = 704 | HuggingFaceTB/finemath | |
## 3171 Days Before November 21, 2023
Want to figure out the date that is exactly three thousand one hundred seventy one days before Nov 21, 2023 without counting?
Your starting date is November 21, 2023 so that means that 3171 days earlier would be March 17, 2015.
You can check this by using the date difference calculator to measure the number of days before Mar 17, 2015 to Nov 21, 2023.
March 2015
• Sunday
• Monday
• Tuesday
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• Friday
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March 17, 2015 is a Tuesday. It is the 76th day of the year, and in the 12nd week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2015 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 03/17/2015, and almost everywhere else in the world it's 17/03/2015.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 3171 weekdays before Nov 21, 2023, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Nov 21, 2023, which falls on a Tuesday. Counting forward, the next day would be a Wednesday.
To get exactly three thousand one hundred seventy one weekdays before Nov 21, 2023, you actually need to count 4439 total days (including weekend days). That means that 3171 weekdays before Nov 21, 2023 would be September 26, 2011.
If you're counting business days, don't forget to adjust this date for any holidays.
September 2011
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
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September 26, 2011 is a Monday. It is the 269th day of the year, and in the 269th week of the year (assuming each week starts on a Sunday), or the 3rd quarter of the year. There are 30 days in this month. 2011 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 09/26/2011, and almost everywhere else in the world it's 26/09/2011.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | HuggingFaceTB/finemath | |
Popular
# What are the closure properties of regular expression?
## What are the closure properties of regular expression?
Closure properties of Regular languages
• Kleen Closure: RS is a regular expression whose language is L, M.
• Positive closure: RS is a regular expression whose language is L, M.
• Complement:
• Reverse Operator:
• Complement:
• Union:
• Intersection:
• Set Difference operator:
## Are regular expressions commutative?
regular expressions as well, define exactly the same set of languages: the regular languages. Union and concatenation behave sort of like addition and multiplication. + is commutative and associative; concatenation is associative.
How do you concatenate in regular expressions?
Concatenation: If R1 and R2 are regular expressions, then R1R2 (also written as R1. R2) is also a regular expression. L(R1R2) = L(R1) concatenated with L(R2). Kleene closure: If R1 is a regular expression, then R1* (the Kleene closure of R1) is also a regular expression.
What is L * and L+?
L* denotes Kleene closure and is given by L* = ЄU Li i=0 example : 0* ={Є ,0,00,000,…………………………………} Language includes empty words also. Page 9. ? L+ denotes Positive closure and is given by L+= Li i=1 example:0+={0,00,000,……………………………………..}
### How do you prove closure property?
The Property of Closure
1. A set has the closure property under a particular operation if the result of the operation is always an element in the set.
2. a) The set of integers is closed under the operation of addition because the sum of any two integers is always another integer and is therefore in the set of integers.
### What do you mean by closure property?
The closure property means that a set is closed for some mathematical operation. For example, the set of even natural numbers, [2, 4, 6, 8, . . .], is closed with respect to addition because the sum of any two of them is another even natural number, which is also a member of the set. …
Can an infinite language be regular?
In the end, you can create infinite languages using finite descriptions (a regular expression). A finite language is a language containing a finite number of words. The simplest cases are those containing no words at all, the empty string, and a single string consisting of a single symbol (e.g. a in your example).
How do you prove a regular expression?
Proof: Since L and M are regular, they have regular expressions, say: Let L = L(E) and M = L(F). Then L ∪ M = L(E + F) by the definition of the + operator. If L is a regular language over alphabet Σ then L = Σ∗ \ L is also regular.
#### What is the regular expression?
A regular expression (shortened as regex or regexp; also referred to as rational expression) is a sequence of characters that specifies a search pattern. Usually such patterns are used by string-searching algorithms for “find” or “find and replace” operations on strings, or for input validation.
#### Which operation can be applied on regular expressions?
A regular expression describes a language using three operations. A regular expression (RE) describes a language. It uses the three regular operations. These are called union/or, concatenation and star.
How do you prove L is regular?
Which Cannot be accepted by a regular grammar?
Which among the following cannot be accepted by a regular grammar? Explanation: There exists no finite automata to accept the given language i.e. 0n1n. For other options, it is possible to make a dfa or nfa representing the language set. 6.
## Which is an example of a Kleene closure?
For example, in regular languages the representation of simpler versions such as the classings in regular expressions is common , allowing a close interrelation between the different types of forms of recognition and representation of regular languages, in finite automata and regular grammars .
## Which is a closure of a regular language?
Closure refers to some operation on a language, resulting in a new language that is of same “type” as originally operated on i.e., regular. Regular languages are closed under following operations. RS is a regular expression whose language is L, M. R* is a regular expression whose language is L*.
How to find the language of a regular expression?
Let L and M be the languages of regular expressions R and S, respectively.Then R+S is a regular expression whose language is (L U M). proof: Let A and B be DFA’s whose languages are L and M, respectively. | HuggingFaceTB/finemath | |
Preserving a propositional formula
I know I must be getting stuck on notation. However, I'm having trouble following the logic in Example 1.2 in https://arxiv.org/pdf/cs/0611018.pdf. They define what preserving a propositional formula means.
So, I'm trying to understand a basic example. Take X = A or (Not B) or (Not C) as an example Horn Clause. Define the assignments f1 and f2 by f1(A) = f1(B) = f1(C) = true and f2(A) = f2(B) = f2(C) = false. Both f1 and f2 satisfy X.
What does it mean to say AND preserves X with respect f1 and f2?
1 Answer
In your example the assignment $f_1\wedge f_2$ is $(f_1\wedge f_2)(A)=f_1(A)\wedge f_2(A)=\text{True} \land \text{False} = \text{False}$. similarly: $(f_1\wedge f_2)(B)=(f_1\wedge f_2)(C)=\text{False}$ so actually $f_1\wedge f_2 \equiv f_2$.
As they define, AND is preserving $X$ if for every two assignments $f_1, f_2$ that satisfy $X$, the assignment $f_1 \wedge f_2$ (defined by $(f_1\wedge f_2)(v)=f_1(v)\wedge f_2(v)$) also satisfies $X$.
• Thank you. I see now that $f_1\land f_2$ is defined componentwise. Aug 14, 2018 at 13:52 | HuggingFaceTB/finemath | |
# Lesson 18
Using Long Division
### Problem 1
Andre and Jada both found $$657 \div 3$$ using the partial quotients method, but they did the calculations differently, as shown here.
1. How is Jada's work the same as Andre’s work? How is it different?
2. Explain why they have the same answer.
### Problem 2
Here is a long-division calculation of $$917 \div 7$$.
1. There is a 7 under the 9 of 917. What does this 7 represent?
2. What does the subtraction of 7 from 9 mean?
3. Why is a 1 written next to the 2 from $$9-7$$?
### Problem 3
Han's calculation of $$972 \div 9$$ is shown here.
1. Find $$180 \boldcdot 9$$.
2. Use your calculation of $$180 \boldcdot 9$$ to explain how you know Han has made a mistake.
3. Identify and correct Han’s mistake.
### Problem 4
Find each quotient.
### Problem 5
The mass of one coin is 16.718 grams. The mass of a second coin is 27.22 grams. How much greater is the mass of the second coin than the first? Show your reasoning.
### Solution
(From Unit 3, Lesson 15.)
### Problem 6
One micrometer is a millionth of a meter. A certain spider web is 4 micrometers thick. A fiber in a shirt is 1 hundred-thousandth of a meter thick.
1. Which is wider, the spider web or the fiber? Explain your reasoning.
2. How many meters wider? | HuggingFaceTB/finemath | |
# 18.9 percent as a fraction
Here you will see step by step solution to convert 18.9 percent into fraction. 18.9 percent as a fraction is 189/1000. Please check the explanation that how to write 18.9% as a fraction.
## Answer: 18.9% as a fraction is
= 189/1000
### How to convert 18.9 percent to fraction?
To convert the 18.9% as a fraction form simply divide the number by 100 and simplify the fraction by finding GCF. If given number is in decimal form then multiply numerator and denominator by 10^n and simplify it as much as possible.
#### How to write 18.9% as a fraction?
Follow these easy steps to convert 18.9% into fraction-
Given number is => 18.9
• Write down the 18.9 in a percentage form like this:
• 18.9% = 18.9/100
• Since, 18.9 is not a whole number, now we need multiply numerator and denominator by 10^n, n = decimal points in number. [n=1]
• => 18.9 × 10/100 × 10 = 189/1000
• We also need to check to simplify the fraction.
• Greatest common factor [GCF] of 189 and 1000 is 1, so this is the simplest form is 189/1000.
• Conclusion: 18.9% = 189/1000
Therefore, the 18.9 percent as a fraction, final answer is 189/1000. | HuggingFaceTB/finemath | |
Anda di halaman 1dari 5
°c Copyright 2008. W. Marshall Leach, Jr.
, Professor, Georgia Institute of Technology, School of Electrical
and Computer Engineering.
The FET Bias Equation
Basic Bias Equation
(a) Look out of the 3 MOSFET terminals and replace the circuits with Thévenin equivalent circuits as showin
in Fig. 1.
(b) Solve the FET drain current equation for VGS .
r
ID
VGS = + VT O
K
(c) Write the gate-source loop equation in the gate-source loop and let IS = ID .
VGG − VSS = VGS + IS RSS = VGS + ID RSS
(d) Solve the loop equation for VGS .
VGS = VGG − VSS − ID RSS
(e) Equate the two expressions for VGS and rearrange the terms to obtain a quadratic equation in ID .
r
ID
ID RSS + − (VGG − VSS − VT O ) = 0
K
(f) Let a = RSS , b = 1/ K, and c = − (VGG − VSS − VT O ). In this case, the bias equation becomes
p
aID + b IC + c = 0
Use the quadratic equation to solve for ID , then square the result to obtain
à √ !2
−b + b2 − 4ac
ID =
2a
Note that there is a second solution using the minus sign for the radical. This solution results in VGS < VT O ,
which is a non realizable solution. The desired solution is the one which givesp the smaller value of ID .
(e) Check for the active mode. For the active mode, VDS > VGS − VT O = ID /K.
VDS = VD − VS = (VDD − ID RDD ) − (VSS + IS RSS ) = VDD − VSS − ID (RDD + RSS )
1
Example 1
Figure 2: Circuit for Example 1.
V + R2 + V − R1
VGG = RGG = R1 kR2
R1 + R2
VSS = V − RSS = RS VDD = V + RDD = RD
Example 2
Figure 3: Circuit for Example 2.
R2 RD
VGG = V + − ID R2 RGG = (R1 + RD ) kR2
RD + R1 + R2 RD + R1 + R2
R1 + R2
VDD = V + RDD = RD k (R1 + R2 )
RD + R1 + R2
VSS = 0 RSS = RS
The gate-source loop equation is
R2 RD
V+ − ID R2 = VGS + ID RS
RD + R1 + R2 RD + R1 + R2
2
p
This can be solved for VGS and equated to
ID /K + VT O to obtain
µ ¶ r µ ¶
RD R2 ID V + R2
ID RS + + − − VT O = 0
RD + R1 + R2 K RD + R1 + R2
The a, b, and c in the bias equation are given by
µ ¶
RD R2 1 V + R2
a = RS + b= √ c=− − VT O
RD + R1 + R2 K RD + R1 + R2
Example 3
Figure 4: Circuit for Example 3.
V + R2
VGG = RGG = R1 kR2
R1 + R2
RS RD
VSS = V + − ID RS RSS = RS k (RD + R3 )
RD + R3 + RS RD + R3 + RS
R3 + RS RS
VDD = V + + IS RD RDD = RD k (R3 + RS )
RD + R3 + RS RD + R3 + RS
Let IS = ID . The bias equation for ID is
µ ¶ r
V + R2 + RS RD ID
− V − ID RS = + VT O + ID [RS k (RD + R3 )]
R1 + R2 RD + R3 + RS RD + R3 + RS K
which gives
µ ¶ r µ ¶
RD RS ID V + R2 V + RS
ID RS k (RD + R3 ) − + − − − VT O =0
RD + R3 + RS K R1 + R2 RD + R3 + RS
The a, b, and c in the bias equation are given by
r
RD RS 1
a = RS k (RD + R3 ) − b=
RD + R3 + RS K
µ + ¶
V R2 V + RS
c=− − − VT O
R1 + R2 RD + R3 + RS
3
Figure 5: Circuit for Example 4.
Example 4
For M1
R2
VGG1 = V + RGG1 = R1 kR2 VSS1 = 0 RSS1 = RS1
R1 + R2
VDD1 = V + RDD1 = RD1
The loop equation for ID1 is
R2
V+ = VGS1 + ID1 RS
R1 + R2
This and the equation for VGS1 can be solved for ID1 .
For M2
VGG2 = V + − ID1 RD1 RGG2 = RD1
VSS2 = 0 RSS2 = RS2 VDD2 = V + RDD2 = RD2
The loop equation for ID2 is
V + − ID1 RD1 = VGS2 + ID2 RS2
This and the equation for VGS2 can be solve for ID2 .
Given ID1 and ID2 , it can be determined if the two MOSFETs are in the active mode.
Example 5
R2
VGG1 = V + RGG1 = R1 kR2 VSS1 = 0 RSS1 = RS1
R1 + R2
VGG2 = IS1 RS1 RGG2 = RS1 VSS2 = 0 RSS2 = RS2 VDD2 = V + RDD2 = RD2
Let the currents to be solved for be ID1 and ID2 and let IS1 = ID1 and IS2 = ID2 .
The gate-source loop equation for ID1 is
R2
V+ = VGS1 + ID1 RS1
R1 + R2
This and the equation for VGS1 can be solved for ID1 .
The gate-source loop equation for ID2 is
ID1 RS1 = VGS2 + ID2 RS2
Given ID1 and ID2 , it can be determined if the two MOSFETs are in the active mode.
4
Figure 6: Circuit for Example 5. | HuggingFaceTB/finemath | |
# math 117
posted by .
A launched rocket has an altitude in meters, given by the polynomial h+vt-4.9^2, h is the height in meters v is the velocity in meters per second and t is the number of seconds for which it takes the rocket to become airborne. If the rocket is launched from the top of a tower that is 100 meters and the initial speed is 60 meters per second, what will its height be after 4 seconds rounded to the nearest tenth?
• math 117 -
h = 100 + Vo*t + 0.5*gt^2.
h = 100 + 60*4 + 0.5*(-9.8)4^2,
h = 100 + 240 - 78.4 = 261.6m above ground.
h = 261.6 - 100 = 161.6m above the tower. | HuggingFaceTB/finemath | |
# Complex number
can someone help me solve this question as fasd as possible??
thx~~
determine the 3 cube roots of 3-i over 3+i giving the result in modulus argument form,express the principal root in the form a+Jb
hello..anyone can help me out??
berkeman
Mentor
can someone help me solve this question as fasd as possible??
thx~~
determine the 3 cube roots of 3-i over 3+i giving the result in modulus argument form,express the principal root in the form a+Jb
Do not bump your post after only 16 minutes. It is unreasonable to expect fast help all the time here on the PF.
Also, you must show your attempt at a solution before we can be of help. Show us how you would go about simplifying the complex fraction that you are describing....
Mark44
Mentor
First off, you're going to want to calculate (3 - i)/(3 + i), to get it in the form a + bi. After that, you should calculate the polar form of this complex number, r(cos$\theta$ + i sin$\theta$). After that, you can use DeMoivre's Formula to find a cube root, which says that
$$(r(cos\theta + i sin\theta))^{1/n} = r^{1/n}(cos(\theta/n) + i sin(\theta/n))$$. | HuggingFaceTB/finemath | |
# Answer to Question #10018 in Algebra for Sonny
Question #10018
a vehicle travels from A to B in 8 sec. at a constant speed of 25 mph. A second vehicle travels the distance in 8 sec. What is the distance from A to B and how fast is the second vehicle traveling ?
v=25, t=8/3600=1/450 => s=vt=25/450=1/9=0.111111111.....
So, speed of
second vehicle v'=s/(8 sec)=25 mph
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25.05.12, 18:51
v=25, t=8/3600=1/450 => s=vt=25/450=1/9
So, speed of the&
second vehicle v'=s/(3 sec)=1/27 mps=1/27*3600 mph=133.3 mph
Sonny
25.05.12, 17:58
I made a mistake in my problem. I am very sorry and appreciate your help. I need to know the speed of the second vehicle. It travels the distance in 3 sec. instead of the time I gave you for car #1. | HuggingFaceTB/finemath | |
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# Find, to 1 decimal place, the values of θ in the interval 0 ⩽ θ ⩽ 180° for which 4root3 sin (3theta + 20degrees) = 4 cos (3theta+ 20degrees)?
Make 3thetha + 20 = y So you’ll have: 4root3 siny = 4 cos y Then divide both sides by cos y So 4 root 3siny/cos y = 4cosy/cosy Siny/cosy = tan y and cosy/cosy = 1 Therefore: 4root 3 tan y = 4 Divide both sides by 4 So: root3 tan y = 1 So tan y = 1/root3 Change your intervals: 0<theta<180 => 0<3theta<540 => 0<3theta +20<560 y= 30,210,390,570 Let y = 3 theta+20 Theta = (y-20)/3 So plug each value into (y-20)/3 theta = 3.3, 63.3, 123.3, 183.3 Take any value under 180 degrees. So reject 183.3 degrees
Hiral Haria
·
1.2k students helped
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# Question: What Are The Importance Of Statistics In Mathematics?
## What is the importance of statistics?
Statistical knowledge helps you use the proper methods to collect the data, employ the correct analyses, and effectively present the results. Statistics is a crucial process behind how we make discoveries in science, make decisions based on data, and make predictions.
## How important is mathematics and statistics as a student?
By learning mathematics and statistics, students develop other important thinking skills. They learn to estimate with reasonableness, calculate with precision, and understand when results are precise and when they must be interpreted with uncertainty.
## What is the importance of statistics in students?
As a branch of knowledge it refers to collection, presentation analysis and interpretation of data. Statistics is very important in education as it helps in collecting, presenting analysis and interpreting data. It also helps in drawing general conclusion.
You might be interested: Question: What Is The Meaning Of Angles In Mathematics?
## What are the functions and importance of statistics?
(1) Statistics helps in providing a better understanding and accurate description of nature’s phenomena. (2) Statistics helps in the proper and efficient planning of a statistical inquiry in any field of study. (3) Statistics helps in collecting appropriate quantitative data.
## What is statistics in your own words?
Statistics is a branch of applied mathematics that deals with collecting, organising, analysing, reading and presenting data. Descriptive statistics make summaries of data. In addition to being the name of a field of study, the word ” statistics ” can also mean numbers that are used to describe data or relationships.
## What is the importance of statistics in our daily life?
It keeps us informed about, what is happening in the world around us. Statistics are important because today we live in the information world and much of this information’s are determined mathematically by Statistics Help. It means to be informed correct data and statics concepts are necessary.
## What is the study of mathematics called?
The short words are often used for arithmetic, geometry or simple algebra by students and their schools. Mathematics includes the study of: This subfield is usually called geometry. Change: how things become different. This subfield is usually called analysis.
## Why do we need to learn mathematics?
Math helps us have better problem-solving skills Math helps us think analytically and have better reasoning abilities. Analytical and reasoning skills are essential because they help us solve problems and look for solutions.
## Why do we study mathematics?
As a student of mathematics, you’ll develop better systems for problem-solving, learning how applied mathematics solves real-world issues. Knowledge of mathematics, and its complexities, can help in almost every career.
You might be interested: FAQ: What Is Mathematics Of Investment All About?
## What is the example of statistics?
A statistic is a number that represents a property of the sample. For example, if we consider one math class to be a sample of the population of all math classes, then the average number of points earned by students in that one math class at the end of the term is an example of a statistic.
## Why do we need to study statistics in education in everyday life?
Statistics is the study that deals with the collection and analysis of data. It is mostly used to keep records, calculate probabilities, and provide knowledge. Basically it helps us understand the world a little bit better through numbers and other quantitative information.
## What is the meaning of statistics in education?
Statistics education is the practice of teaching and learning of statistics, along with the associated scholarly research. Statistics is both a formal science and a practical theory of scientific inquiry, and both aspects are considered in statistics education.
## What are the applications of statistics?
Statistics help in providing data as well as tools to analyze the data. Some powerful techniques are index numbers, time series analysis, and also forecasting. These are immensely useful in the analysis of data in economic planning. Further, statistical techniques help in framing planning models too.
## What are the main features of statistics?
7 Main Characteristics of Statistics – Explained!
• It consists of aggregates of facts:
• It is effected by many causes:
• It should be numerically expressed:
• It must be enumerated or estimated accurately:
• It should be collected in a systematic manner:
• It should be collected for a predetermined purpose:
• It should be capable of being placed in relation to each other:
You might be interested: Often asked: What Are The Different Branches Of Mathematics?
## What is the importance of statistics in different fields?
Statistics plays a vital role in every field of human activity. Statistics helps in determining the existing position of per capita income, unemployment, population growth rates, housing, schooling medical facilities, etc., in a country. | HuggingFaceTB/finemath | |
# Week10homework (PDF)
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BMATH 127
Week 10
Week 10 Homework: Review for the Final
Directions. Work on the following problems on a separate sheet of paper. If you finish this worksheet
and upload your solution as a single PDF document into Canvas under Homework 10 by December
13th , you will receive 20 extra credit points. The entire worksheet must be completed neatly with work
shown on every problem to receive credit.
1. Suppose cos
and suppose
a. Calculate the values of the other five trig functions.
b. You may have talked in class about two identities that are called the double angle
identities. Here they are if you have not talked about them:
sin
cos 2
cos
sin 2
2 sin
cos
Use these identities and your answers to part (a) to calculate cos 2 and sin 2 .
2. Consider the functions
a.
b.
c.
d.
2
1
4 and
Sketch the graphs of
and
.
Calculate 2 and solve
5.
Calculate the average rate of change of
Calculate
and
. Is
4
1.
on the interval 1,1
? Explain.
.
3. The graph at right shows the amount of a
minutes
a. Find a formula for the amount, , of
the sample that remains after
minutes.
b. Find the half-life of the substance.
c. What percent of the substance
decays every minute?
4. Calculate the following without using a calculator:
1
log 81
2 log √1800
2
1
log 8
3
5. A Ferris wheel has a diameter of 100 meters and riders get on it from a platform that is 2 meters
above the ground. It takes 10 minutes for the Ferris wheel to make one full revolution.
a. Sketch a graph of the height of a rider above the ground minutes after the rider boards
the Ferris wheel.
b. Find a formula for the height of a rider above the ground minutes after the rider boards
the Ferris wheel.
c. Once the rider reaches a height of 27 meters, the rider can see the ocean. For how
many minutes can the rider see the ocean during each revolution?
Week10homework.pdf (PDF, 47.22 KB)
#### HTML Code
Copy the following HTML code to share your document on a Website or Blog | HuggingFaceTB/finemath | |
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# The force that pulls dogs toward the groins of strangers
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re: The force that pulls dogs toward the groins of strangers
7/8/2011 9:30:43 PM
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Put a ping pong ball on the edge of a turntable that plays LPs (if you have one). Set it up so that as the turntable turns the ball stays in the same place. Then give the ball a little poke to push it into the center. Watch it go in a circle. Figure out why and I think you'll have your answer.
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re: The force that pulls dogs toward the groins of strangers
6/16/2011 8:50:38 PM
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F=mv**2/r =100kg * ((447m/s)**2) / 6149045m =3.25 Newtons =.73 lbs less "weight" at equator for 100kg object
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re: The force that pulls dogs toward the groins of strangers
6/3/2011 4:38:52 PM
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@corndodger: "There is no "horizontal" force..." Yes, the word "horizontal" was misused -- in each case, the intended word was "tangential". @corndodger: "Cut the string and the ball flies away from you in the direction it was pointing when the string was cut." If you're going to be pedantic about it, the ball will fly away in a direction perpendicular to the "direction it was pointing when the string was cut." The ball continues in the direction it was moving, not pointing, at the instant the centripetal force was removed. @corndodger: "Sit in the center of a spinning merry-go-round, ball in hand. Let go of the ball. The ball leaves, rolling straight out from the center where you're sitting..." Actually, barring friction with the surface of the merry-go-round, in your rotating reference frame, the ball will appear to slide outwards in a *trailing spiral* since it's velocity remains constant but it's angular velocity decreases as the radius increases. Add friction, and the ball's outward spiral gets deformed due to the rotational inertia of the ball as it comes up to speed with the surface of the merry-go-round. As for your word "centrifical", the word for the stipulated force allowing you to treat a rotating frame as an inertial frame is "centrifugal". Perhaps you should consult something a little more advanced than _Sandbox Science_ before you go correcting others. Lance ==)-------------
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re: The force that pulls dogs toward the groins of strangers
6/3/2011 3:34:43 PM
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I was taken aback by you discussion of a "horizontal" component to the ball being swung around your head and feet. The ball must have hit your head a couple of times too many. There is no "horizontal" force, only the centrifical force that pulls the ball away from you, stretching the string. Cut the string and the ball flies away from you in the direction it was pointing when the string was cut. It does not follow the circular path it was in prior to release. This information can be found in any pre-school physics book. (This is too basic for Kindergarten students.) Want further confirmation? Sit in the center of a spinning merry-go-round, ball in hand. Let go of the ball. The ball leaves, rolling straight out from the center where you're sitting, and continues rolling straight away once it is on the ground. corndodger
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re: The force that pulls dogs toward the groins of strangers
6/3/2011 11:21:11 AM
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Air pressure, more specifically, air density and thus buoyancy, would have an effect, but it would be immeasurably small. The rate at which you lose mass due to sweating would likely have a greater effect. I once posed a puzzle along the lines of "Which weighs more at STP: a kilogram of gold or a kilogram of aluminum"; I was looking for the buoyancy effect, but found that the oxidation of the outer surface of the aluminum increased its weight by more than the decrease due to buoyancy. Lance ==)-------------
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re: The force that pulls dogs toward the groins of strangers
6/2/2011 6:47:03 PM
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@ck_02: Well, it's just so crazy it may work. Did you read my Discworld blog (http://bit.ly/mRLJN5). Your comment reminded me of the scene in Guards! Guards! Where our three heroes are on the roof with one about to shoot an arrow at a dragon to hit its "vulnerable spot" and one says something like "it's a million to one chance but it just might work" and another says "it has to work because a million to one chance always works in stories." But then one remembers that the guy who is going to fire the arrow won a competition in archery and is also going to use his lucky arrow .. and after some debate they decide that it's more of a thousand to one shot, and whoever heard on that working in a story? So they start to handicap the archer by having him stand on one leg and close one eye and ... until they recon they've managed to get the odds back up to a million to one ... I love this stuff :-)
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re: The force that pulls dogs toward the groins of strangers
6/2/2011 6:40:24 PM
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However, if a ventilation system were in place that kept constant pressure and air volume into account while providing said subject with a measured amount of breathable air that would have to be taken at the time of each weigh-in. Well, it's just so crazy it may work.
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re: The force that pulls dogs toward the groins of strangers
6/2/2011 6:32:46 PM
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Relative to any difference that might occur and be measurable in the effective weight due to your position on the globe, the donut factor, as alluded to by ck_02, would be an order or two of magnitude greater in impact. Certainly, sealing yourself into a plastic bag would prevent the donut factor from influencing your measurements. However, the plastic bag would in a short span of time, lead to your inability to continue with your measurements. Unfortunately, it would not make you a closed system as plastic does, at very small rates, allow the permeation of moisture and various gasses. In conclusion, I would not recommend sealing yourself into a plastic bag. While it would prevent some influences, it would ultimately fail in creating a completely closed system and therefore would call your data into question.
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re: The force that pulls dogs toward the groins of strangers
6/2/2011 6:30:41 PM
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That option screams terrible B-rated horror flick all over it. sorry. ;)
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re: The force that pulls dogs toward the groins of strangers
6/2/2011 5:07:00 PM
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How about if I seal myself in a plastic bag so as to become a closed system? :-)
Page 1 / 4 > >> | HuggingFaceTB/finemath | |
Does $\sum_{n = 2}^{\infty} [\zeta(n) - 1]$ converge?
Question in the title. Does $\sum_{n = 2}^{\infty} [\zeta(n) - 1]$ converge? If not, how about $\sum_{n = 1}^{\infty} [\zeta(2n) - 1]$?
$$S=\sum_{n\geq 2}\left(\zeta(n)-1\right)=\int_{0}^{+\infty}\sum_{m\geq 1}\frac{x^m}{m!}\left(\frac{1}{e^x-1}-\frac{1}{e^x}\right)\,dx$$ leads to: $$S = \int_{0}^{+\infty}(e^x-1)\left(\frac{1}{e^x-1}-\frac{1}{e^x}\right)\,dx = \int_{0}^{+\infty}e^{-x}\,dx = \color{red}{1}.$$
$$S= \sum_{n\geq 1}\left(\zeta(2n)-1\right) = \int_{0}^{+\infty}\sum_{m\geq 0}\frac{x^{2m+1}}{(2m+1)!}\left(\frac{1}{e^x-1}-\frac{1}{e^x}\right)\,dx$$ hence $$S = \int_{0}^{+\infty}\frac{\sinh(x)}{e^x(e^x-1)}\,dx = \frac{1}{2}\int_{1}^{+\infty}\frac{u-\frac{1}{u}}{u^2(u-1)}\,du=\frac{1}{2}\int_{1}^{+\infty}\frac{u+1}{u^3}\,du=\color{red}{\frac{3}{4}}.$$ The same technique (i.e. to exploit the integral representation for the $\zeta$ function) works pretty well for similar problems, too.
• Very cool, thank you. Can the same technique be used to find $\sum_{n = 1}^{\infty}[\zeta(mn) - 1]$ for an arbitrary positive integer $m$? I think it's interesting that these series of irrational numbers without any obvious pattern converge to such simple rational limits. – Vik78 Sep 17 '16 at 15:22
• @Vik78: you may use the same technique to compute the generating function $$f(x)=\sum_{n\geq 2}(\zeta(n)-1)x^{n}$$ in terms of the dilogarithm. If you apply a discrete Fourier transform to the above generating function (i.e. compute its values at the $m$-th roots of unity) you may compute such generalized series, too. They won't have rational values for every $m\in\mathbb{N}$, however. – Jack D'Aurizio Sep 17 '16 at 15:30
The family $(n^{-k})$ with $n, k \ge 2$ is summable, because the geometric series gives us the estimation
$$\sum_{k=2}^N \sum_{n=2}^N \frac{1}{n^k} < \sum_{n=2}^N \sum_{k=2}^\infty \frac{1}{n^k} = \sum_{n=2}^N \frac{1}{n(n-1)} = 1 - \frac{1}{N} < 1.$$
Therefore we can exchange the order of summation and write
$$\sum_{k=2}^\infty (\zeta(k) - 1) = \sum_{n=2}^\infty \sum_{k=2}^\infty \frac{1}{n^k} = \sum_{n=2}^\infty \frac{1}{n(n-1)} = 1.$$
Figured out the solution:$\sum_{n=2}^{\infty} [\zeta(n) - 1] = \sum_{n=2}^{\infty}\sum_{k=2}^{\infty} n^{-k} = \sum_{n=2}^{\infty}\frac{1}{1-\frac{1}{n}} - 1 - \frac{1}{n} = \sum_{n=2}^{\infty}\frac{1}{n(n-1)}$, which converges by comparison with $\zeta(2)$.
• The final sum converges to exactly $1$ (telescoping sum). – Erick Wong Sep 18 '16 at 19:36
$$0\le \zeta(n)-1=\frac{1}{2^n}+\sum_{k\ge3}\frac{1}{k^n}\le\frac{1}{2^n}+\int_2^\infty\frac{dx}{x^n}=\frac{1}{2^n}+\frac{1}{(n-1)2^{n-1}}=\frac{n+1}{n-1}\frac{1}{2^n}$$
which $\le\frac{3}{2^n}$ for $n\ge2$. Thus, the sum converges by comparison with the geometric series $\sum \frac{1}{2^n}$. | open-web-math/open-web-math | |
# Find asymptotically equivalent function for $\ln \binom{n^2}{2n}$
How can I find an asymptotically equivalent function for $$\ln \binom{n^2}{2n}$$?
I think I need to use the definition of the binomial coefficient and Stirling's approximation, but I'm not sure what to do after this.
• Where did you get to after using Stirling and the binomial coefficient? That expression may actually be fully simplified even though it may look complicated. Oct 2 '19 at 9:03
In the same spirit an in answer and comment, we can even get a very good approximation.
Writing $$\binom{n^2}{2 n}=\frac{(n^2)!}{(2n)! \, (n^2-2n)!}$$ $$\log \left(\binom{n^2}{2 n}\right)=\log\left((n^2)!\right)-\log\left((2n)!\right)-\log\left((n^2-2n)!\right)$$ Using Stirling approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({p}\right)\right)+\frac{1}{12 p}-\frac{1}{360 p^3}+O\left(\frac{1}{p^5}\right)$$ apply to each term and continue with Taylor series for large values of $$n$$. This should give $$\log \left(\binom{n^2}{2 n}\right)=2 n \left(\log (n)+\log \left(\frac{e}{2}\right)\right)-\log \left(2 e^2 \sqrt{\pi }\right)-\log{(\sqrt n)}-\frac 3 {8n}+O\left(\frac{1}{n^2}\right)$$
As shown below, this is even quite good for very small values of $$n$$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 3 & 4.49297 & 4.43082 \\ 4 & 9.49277 & 9.46265 \\ 5 & 15.0177 & 14.9999 \\ 6 & 20.9595 & 20.9478 \\ 7 & 27.2466 & 27.2383 \\ 8 & 33.8286 & 33.8224 \\ 9 & 40.6676 & 40.6628 \\ 10 & 47.7345 & 47.7306 \end{array} \right)$$
By Stirling approximation $$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$ we have that
$$\binom{n^2}{2n}=\frac{n^2!}{(2n)!(n^2-2n)!}\sim \frac{\sqrt{2 \pi n^2}\left(\frac{n^2}{e}\right)^{n^2}}{\sqrt{2 \pi 2n}\left(\frac{2n}{e}\right)^{2n}\sqrt{2 \pi (n^2-2n)}\left(\frac{n^2-2n}{e}\right)^{n^2-2n}}$$
$$\sim\frac{\sqrt{2 \pi }n^{2n^2+1}e^{-n^2}}{2\sqrt \pi n^\frac12 2^{2n}n^{2n}e^{-2n}\cdot \sqrt{2 \pi}nn^{2n^2-4n}e^{-n^2+2n}}=$$
$$=\frac1{2\sqrt \pi} \cdot {n^{2n-\frac12}} \cdot 2^{-2n}$$
and therefore
$$\ln \binom{n^2}{2n}\sim n\ln n$$ | HuggingFaceTB/finemath | |
### Home > CC2MN > Chapter 7 > Lesson 7.1.3 > Problem7-33
7-33.
A fair number cube labeled $1$, $2$, $3$, $4$, $5$, and $6$ is rolled $100$ times. About how many times would you expect the number $3$ to appear?
You would expect the number $3$ to appear about one sixth of the total rolls, $100$, since there are $6$ numbers on a cube.
What is one sixth of $100$? | HuggingFaceTB/finemath | |
# Search by Topic
#### Resources tagged with Combinatorics similar to Knights Moving:
Filter by: Content type:
Stage:
Challenge level:
### There are 41 results
Broad Topics > Decision Mathematics and Combinatorics > Combinatorics
### Magic Caterpillars
##### Stage: 4 and 5 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
### W Mates
##### Stage: 5 Challenge Level:
Show there are exactly 12 magic labellings of the Magic W using the numbers 1 to 9. Prove that for every labelling with a magic total T there is a corresponding labelling with a magic total 30-T.
### Plum Tree
##### Stage: 4 and 5 Challenge Level:
Label this plum tree graph to make it totally magic!
### Olympic Magic
##### Stage: 4 Challenge Level:
in how many ways can you place the numbers 1, 2, 3 … 9 in the nine regions of the Olympic Emblem (5 overlapping circles) so that the amount in each ring is the same?
### Magic W
##### Stage: 4 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
### The Eternity Puzzle
##### Stage: 5
A big prize was offered for solving The Eternity Puzzle, a jigsaw with no picture and every piece is the same on both sides. The finished result forms a regular dodecagon (12 sided polygon).
### Lost in Space
##### Stage: 4 Challenge Level:
How many ways are there to count 1 - 2 - 3 in the array of triangular numbers? What happens with larger arrays? Can you predict for any size array?
### Deep Roots
##### Stage: 4 Challenge Level:
Find integer solutions to: $\sqrt{a+b\sqrt{x}} + \sqrt{c+d.\sqrt{x}}=1$
### Jluuis or Even Asutguus?
##### Stage: 5 Challenge Level:
Sixth challenge cipher
### Stage 5 Cipher Challenge
##### Stage: 5 Challenge Level:
Can you crack these very difficult challenge ciphers? How might you systematise the cracking of unknown ciphers?
### Semicircle
##### Stage: 5 Challenge Level:
Fourth challenge cipher
### A Fine Thing?
##### Stage: 5 Challenge Level:
Second challenge cipher
### Vital?
##### Stage: 5 Challenge Level:
Third challenge cipher
### Ip?
##### Stage: 5 Challenge Level:
Seventh challenge cipher
### Up a Semitone?
##### Stage: 5 Challenge Level:
Fifth challenge cipher
### Counting Binary Ops
##### Stage: 4 Challenge Level:
How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms.
##### Stage: 5 Challenge Level:
With red and blue beads on a circular wire; 'put a red bead between any two of the same colour and a blue between different colours then remove the original beads'. Keep repeating this. What happens?
### Penta Colour
##### Stage: 4 Challenge Level:
In how many different ways can I colour the five edges of a pentagon red, blue and green so that no two adjacent edges are the same colour?
### One Basket or Group Photo
##### Stage: 2, 3, 4 and 5 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.
### Modular Knights
##### Stage: 5 Challenge Level:
Try to move the knight to visit each square once and return to the starting point on this unusual chessboard.
### Tangles
##### Stage: 3 and 4
A personal investigation of Conway's Rational Tangles. What were the interesting questions that needed to be asked, and where did they lead?
### Snooker Frames
##### Stage: 5 Challenge Level:
It is believed that weaker snooker players have a better chance of winning matches over eleven frames (i.e. first to win 6 frames) than they do over fifteen frames. Is this true?
##### Stage: 4 Challenge Level:
A walk is made up of diagonal steps from left to right, starting at the origin and ending on the x-axis. How many paths are there for 4 steps, for 6 steps, for 8 steps?
### Snowman
##### Stage: 4 Challenge Level:
All the words in the Snowman language consist of exactly seven letters formed from the letters {s, no, wm, an). How many words are there in the Snowman language?
### Factorial Fun
##### Stage: 5 Challenge Level:
How many divisors does factorial n (n!) have?
##### Stage: 4 and 5
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots, prime knots, crossing numbers and knot arithmetic.
### Knight Defeated
##### Stage: 4 Challenge Level:
The knight's move on a chess board is 2 steps in one direction and one step in the other direction. Prove that a knight cannot visit every square on the board once and only (a tour) on a 2 by n board. . . .
### Euler's Officers
##### Stage: 4 Challenge Level:
How many different solutions can you find to this problem? Arrange 25 officers, each having one of five different ranks a, b, c, d and e, and belonging to one of five different regiments p, q, r, s. . . .
### N000ughty Thoughts
##### Stage: 4 Challenge Level:
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the. . . .
### Symmetric Tangles
##### Stage: 4
The tangles created by the twists and turns of the Conway rope trick are surprisingly symmetrical. Here's why!
### Cube Net
##### Stage: 5 Challenge Level:
How many tours visit each vertex of a cube once and only once? How many return to the starting point?
### Magic W Wrap Up
##### Stage: 5 Challenge Level:
Prove that you cannot form a Magic W with a total of 12 or less or with a with a total of 18 or more.
### Doodles
##### Stage: 4 Challenge Level:
A 'doodle' is a closed intersecting curve drawn without taking pencil from paper. Only two lines cross at each intersection or vertex (never 3), that is the vertex points must be 'double points' not. . . .
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Transitivity
##### Stage: 5
Suppose A always beats B and B always beats C, then would you expect A to beat C? Not always! What seems obvious is not always true. Results always need to be proved in mathematics.
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### An Introduction to Computer Programming and Mathematics
##### Stage: 5
This article explains the concepts involved in scientific mathematical computing. It will be very useful and interesting to anyone interested in computer programming or mathematics.
### Russian Cubes
##### Stage: 4 Challenge Level:
How many different cubes can be painted with three blue faces and three red faces? A boy (using blue) and a girl (using red) paint the faces of a cube in turn so that the six faces are painted. . . .
### Molecular Sequencer
##### Stage: 4 and 5 Challenge Level:
Investigate the molecular masses in this sequence of molecules and deduce which molecule has been analysed in the mass spectrometer.
### Scratch Cards
##### Stage: 4 Challenge Level:
To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize?
### Snooker
##### Stage: 5 Challenge Level:
A player has probability 0.4 of winning a single game. What is his probability of winning a 'best of 15 games' tournament? | HuggingFaceTB/finemath | |
# Mensa IQ riddle - lines inside circles
I know that the correct answer is F, but I don't completely understand why.
The solution guide says the logic is to rotate around 180 and duplicate bars. Problem: in which direction are the bars duplicated? To say clockwise or counterclockwise fails in both cases.
Try "folding" them in half, and duplicating there. They're just exactly opposite and duplicated.
I.E. The bottom left of b is 12 because the 4+2(6) duplicates are overlapping at that point in the circle. Top of b is 18 as the duplicates of both 3 and 6 (9).
As I'm trying to edit this and clean it up I realize I think I misunderstood your question... But why does it have to be clockwise/counterclockwise? If you duplicate outward from the midpoint of each set, maybe that helps clarify?
• this idea with duplicate outward from the midpoint of each set went also through my head. Problems: how we duplicate if we have a group with odd number of bars. Futhermore for example the two bars group on the left seem to be duplicated only in one direction. Up to now the most promising idea was to draw a vertical line with splits the circle perfectly and deside to which of the both intersetions of the vertical line with the circle is closer to a group of bars, the upper or the lower. Feb 21, 2020 at 1:57
• And then - after mirroring horizontally - doplicate in the direction with shorter distance between intersection and the group of bars. That works but seems to circumstaneous for me so I was curious if it can be done in more simple way. Feb 21, 2020 at 1:57
I think the procedure is like so:
1. Visit every group of bars
2. For each group, replicate it and add it clockwise from the original group
3. Rotate the whole circle by 180 degrees
You could probably swap steps 2 and 3, but I find it harder to understand that way.
• hehe, maybe I need magnifying glass but I think there is a problem: let's look at the two group of bars in A at the bottom: so the 3 and the 6 group. We duplicate each of them and add them clockwise (2.) and rotate (3.). The new group on the top in B has no gap, on the other hand 6 bars seem not to be enough to fill the gap between considered 3 and 6 groups in A below. Indeed, by your logic the 3 new bars are added not between 3 and 6 groups, but on the other side, since clockwise. Or did I misunderstood your procedure? Feb 21, 2020 at 4:47
• You understood correctly. I also feel like there should be a gap. Perhaps the drawings are not supposed to be to scale. Otherwise I don't see how it can work. Well at least the counts make sense, each group always doubles in size. Feb 21, 2020 at 5:18
I think you want more from this "riddle" that it actually is. First of all the goal of this puzzle is to evaluate your ability to see patterns. To do that it's enough for one answer to fit better that all the rest. And this is what F does.
It doesn't have to be ideal both:
1. because it's a test-like question, you have discrete amount of options
2. because all patterns in the real life are approximate and it may test your ability to find patterns despite their distortions.
If you wish one can formulate the relation A->B and C->F like this:
1. Rotatate around 180 degree.
2. Duplucate amount of bars in each group.
3. Shift each group arbitrary by any angle in between -30 and 30 degree | HuggingFaceTB/finemath | |
m To inch Coversion
Conversion Rule
M(in) = M(m) x 39.37
[Where, M is a variable in inch/m]
Examples
Example 1. Convert 2 meter to inch.
Solution!
• 1 m = 39.37 inch.
• 2 m = 2x39.37 inch.
• 2 m = 78.74 inch
Example 2. Convert 21 meter to inch.
Solution!
• 1 m = 39.37 inch.
• 21 m = 21x39.37 inch.
• 21 m = 826.77 inch.
Example 3. Convert 19.7 meter to inch.
Solution!
• 1 m = 2.54 cm
• 19.7 m = 19.7x39.37 inch.
• 19.7 m = 775.589 inch.
Useful Information
10 millimeter (mm) = 1 centimeter (cm) 10 centimeter = 1 decimeter (dm) 10 centimeter = 100 millimeter 10 decimeter = 1 meter (m) 10 decimeter = 1000 millimeter 10 meters = 1 dekameter (dam) 10 dekameters = 1 hectometer (hm) 10 dekameters = 100 meters) 10 hectometers = 1 kilometer (km) 10 hectometers = 1000 meters | HuggingFaceTB/finemath | |
# Search by Topic
#### Resources tagged with Right angled triangles similar to Just Opposite:
Filter by: Content type:
Stage:
Challenge level:
### There are 17 results
Broad Topics > 2D Geometry, Shape and Space > Right angled triangles
### Just Opposite
##### Stage: 4 Challenge Level:
A and C are the opposite vertices of a square ABCD, and have coordinates (a,b) and (c,d), respectively. What are the coordinates of the vertices B and D? What is the area of the square?
### Coke Machine
##### Stage: 4 Challenge Level:
The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design. Coins inserted into the machine slide down a chute into the machine and a drink is duly. . . .
### Tied Up
##### Stage: 3 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
### Tennis
##### Stage: 3 Challenge Level:
A tennis ball is served from directly above the baseline (assume the ball travels in a straight line). What is the minimum height that the ball can be hit at to ensure it lands in the service area?
### Circle Box
##### Stage: 4 Challenge Level:
It is obvious that we can fit four circles of diameter 1 unit in a square of side 2 without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?
### How Big?
##### Stage: 3 Challenge Level:
If the sides of the triangle in the diagram are 3, 4 and 5, what is the area of the shaded square?
### Right Angles
##### Stage: 3 and 4 Challenge Level:
Can you make a right-angled triangle on this peg-board by joining up three points round the edge?
### In a Spin
##### Stage: 4 Challenge Level:
What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?
### Circle Scaling
##### Stage: 4 Challenge Level:
You are given a circle with centre O. Describe how to construct with a straight edge and a pair of compasses, two other circles centre O so that the three circles have areas in the ratio 1:2:3.
### Crescents and Triangles
##### Stage: 4 Challenge Level:
Triangle ABC is right angled at A and semi circles are drawn on all three sides producing two 'crescents'. Show that the sum of the areas of the two crescents equals the area of triangle ABC.
### Wedge on Wedge
##### Stage: 4 Challenge Level:
Two right-angled triangles are connected together as part of a structure. An object is dropped from the top of the green triangle where does it pass the base of the blue triangle?
### Parallel Universe
##### Stage: 4 Challenge Level:
An equilateral triangle is constructed on BC. A line QD is drawn, where Q is the midpoint of AC. Prove that AB // QD.
### Three Four Five
##### Stage: 4 Challenge Level:
Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.
### Matter of Scale
##### Stage: 4 Challenge Level:
Prove Pythagoras' Theorem using enlargements and scale factors.
### Lying and Cheating
##### Stage: 3 Challenge Level:
Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!
### Tricircle
##### Stage: 4 Challenge Level:
The centre of the larger circle is at the midpoint of one side of an equilateral triangle and the circle touches the other two sides of the triangle. A smaller circle touches the larger circle and. . . .
### Get Cross
##### Stage: 4 Challenge Level:
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing? | HuggingFaceTB/finemath | |
## PERT
Exclusive forum for iZenBirdge PMP® clients
KaurP
Participant
Posts: 94
Joined: Wed Apr 04, 2018 4:08 am
### PERT
One of the questions with no explanation for the answer-
Together with your team, you applied three-point estimation on a critical path which consists of two activities. The following duration uncertainties are all calculated assuming a ±3sigma confidence interval. The duration uncertainty—defined as pessimistic minus optimistic estimate—of the first activity is 18 days; the second estimate has an uncertainty of 24 days. Applying the PERT formula for paths, what is the duration uncertainty of the entire path?
o 21 days
o 30 days. Marked as correct Answer
o 42 days
o No statement is possible from the information given.
Thanks!
manishpn
Expert
Posts: 2603
Joined: Sat Jan 04, 2014 3:55 pm
### Re: PERT
This question from Oliver is quiet controversial, I am not an expert in statistics but us has been debated a lot.
Since almost all Math questions in exam are nowadays very simple you should not spend a lot of time on such questions
Data given is P = O for Activity 1 is 18 & Activity 2 is 24.
lets calculate P - O for project.
If we take information as per Rita, duration of uncertainity (P -O) is already provided for activities, so just adding up the two durations would give total duration of uncertainity for entire path.
Duration uncertainity of entire path = duration uncertainity of activity 1 + duration uncertainity of activity 2 = 18+24 = 42
however if we go with other approach
1. Calculate standard deviation of activities; SD = (P - O) / 6
2. Calculate variance of activities; variance = SD²
3. Calculate Total variance
4. Calculate Uncertainity; U = P-O
Activity A and Bs uncertianity is 18 and 24 respectively
SD of A = 18/6 = 3. of B = 24/6 = 4
Variance of A = 9 and of B =16
total variance = 20
Total variance = total SD²
0r 25= SD²
0r total SD = 5
total U = SD*6 =30
I
Br,
Manish P
PMP, PMI - ACP, SAFe Agilist
http://www.izenbridge.com/blog/7-effect ... ification/ | HuggingFaceTB/finemath | |
# energy signal and power signals | what are energy and power signals in signals and systems example
what are energy and power signals in signals and systems example , energy signal and power signals ? examples definition and meaning solved examples ?
Signals and Systems
Signals
Signal A signal is a single valued function of one or more variables that conveys some information.
Signals are represented mathematically as functions of one or more independent variables.
For example – A speech signal can be represented mathematically by acoustic pressure as a function of time and a picture can be represented by brightness as a function of two spatial variables.
Types of Signals
Signals may be classified in several ways
(a) Continuous Time Signals
A continuous time signal is one which is defined for all values of time.
A continuous time signal needs not be continuous (in the mathematical sense) at all points in time.
e.g., a rectangular wave is discontinuous at several points but it is continuous time signal.
(b) Discrete Time Signals
Discrete time signals are those signals which are defined only at a discrete set of points in time.
e.g., suppose we note down the temperature at a particular place at a fixed time say 5 AM everyday. we get a discrete time signal.
Here, between 1 and 2 we cannot say that value of signal is zero because we recorded temperature only at discrete value. generally, x[nT] is written as x[n] means discrete time signal is denoted by x[n].
(c) Periodic Signals
Periodic signals are those signals which repeat its value at a fixed particular time.
. A continuous time signal x(t) is said to be periodic in time, if
x(t) = x(t + mT)
where, m is any integer.
The smallest value of the positive constant T satisfying the above relation is known as fundamental periode of the periodic signal.
. A discrete time signal x[n] is periodic with period N, where N is a positive integer, if it is unchanged by a time shif of N, i.e., if
x[n] = x[n + N]
for all values of n, above equation holds, then x[n] is periodic with periods N, 2N, 3N,…..Fundamental period NO is the smallest positive value of N.
Example 1. Check whether the given signal is periodic or not.
x(t) = {cos t, if t < 0/sin t, if t > 0
Sol.
1. sin at and cos at has fundamental period (2/a).
2. DC term does not affect the fundamental period.
Example 2. is the signal x(t) = 10 cos2 (10 t) a periodic size if it is, what is its fundamental period?
Sol. x(t) = 10 cos2 (10 t)
since, cos 2t = 2 cos2 t – 1
cos2 t = 1 + cos 2t/2
x(t) = 5 + 5 cos 20 t
since, DC term does not affect fundamental period, so let we take
x(t) = 5 + 5y(t)
y(t) = cos 20 t
fundamental period of this signal is 20/20 {2/a = 1/10
Let z(t) = x(t) + y(t) with x(t) has fundamental period TX and y(t) has fundamental period TY, then fundamental period of z(t) is
T = LCM (TX, TY): if TX/TY is a rational number.
Even Signals
In continuous time, a signal is even if
x(-t) = x(t)
And in discrete time, signal is even if
x[-n] = x[n]
e.g., x(t) = cos t is even signal.
any signal which is symmetris about y-axis is even signal.
Odd Signals
A signal is referred to odd, if
x(-t) = – x(t) and x[-n] = -x[n]
An odd signal must necessarily be 0 at t = 0 or n = 0.
e.g., sin t is an odd signal.
Another Examples
in discrete time, example of even and odd signals are given below.
1. every signal needs not be even or odd.
2. any signal which is neither even nor odd can be represented as sum of even and odd signals.
if x(t) is a signal which is neither even nor odd. then x(t) can be written as
X(t) = XE (t) + XO(t)
XE(t) = x(t) + x(-t)/2
xo(t) = x(t) – x (-t)/2
Example 3. Find the even and odd components of x(t) = ejt
Sol. let, xe(t) and xo(t) be the even and odd components of ejt respectively.
ejt = xe (t) + xo(t)
xe(t) = 1/2 [ejt + e-jt] = cos t
x0(t) = 1/2 [ejt – e-jt] = j sin t
Energy Signals and Power Signals
for an arbitrary continuous time signal x(t), the normalized energy content E of x(t) is defined as
E = |X(t)2 dt
and normalized average power P of x(t) is defined as
P = lim 1/T |x(t)2| dt
similarly, for a discrete-time signal x[n], the normalized energy content E of x[n] is defined as
E = |x[n]|2
the normalized average power P of x[n] is defined as
P = lim 1/2N + 1n = – N |x[n]|2
1. x(t) (or x[n]) is said to be an energy signal (or sequence) if and only if 0 < E OO and P = 0
2. x(t) (or x[n]) is said to be a power signal (or sequence ) if and only if 0 < p < oo, thus implying that E = OO
3. Signals that satisfy neither property are preferred to as neither energy signals nor power signals.
Deterministic and Random Signals
Deterministic signals are those signals whose values ar completely specified for any given time. thus, a deterministic signal can be modelled by a known function of time t.
e.g., x(t) = 10 cos 100 t
At any instant of time one can calculate the value of x(t) from the given expression of x(t).
Random signals are those signals that take random values at any given time and must characterized statistically.
e.g., noise signals and EEG signals.
Random signals, if stationary may be described in terms of certain average values only.
One-dimensional and Multi-dimensional Signals
A signal which is a function of only one variable is referred to as a one-dimensional signal.
x(t) = a cos 0o t.
A signal which is a function of more than one variables is known as multi-dimensional signal.
e.g., image of any object, intensity of light reflected from any point.
Representation of Signals
(a) Continuous Time Signals
Continuous time signals are specified for all values of time by a function of time like, x(t) = e-|t|; –oo < t < oo or by a look up table which specifies its values for all time range. for example,
x(t) = {2, – oo < t < 0/1, 0 < t < oo
they are represented diagrammatically by their waveforms, i.e., graphs depicting their varition with time.
(b) Discrete Time Signals
A discrete time signals is a sequence of numbers. these number are assumed to occurring at regular intervals of T second. the nth sample of a discrete time signal is generally denoted by x[nT] or simply x[n].
A discrete time signal x[n] may be specified or described in one of the following ways
(i) by means of a look-up table
(ii) by means of an equation which specifies the nth and sample values x[n] as a function of n.
for example, x[n] = n2oo < n < oo
(iii) by means of a recursive formula such as
x[n] = 2x [n – 1] + [n] with x(-1) = 0
where, [n] = {z, for n = 0/0, otherwise
Some Commonly Used Signals for Continuous Time Signals
1. The unit impulse function (t)
Unit impulse function is defined as
(t) = {0 t = 0/ 1 t = 0
(t) dt = 1
Properties of unit impulse function
x(t) (t) dt = x(0)
the area under unit impulse function is equal to 1.
(t) dt = 1
the width of (t) along the time axis is zero.
Sampling property
if x(t) is continuous at t
x(t) (t) dt = x(t) t x(t)
for any t1 and t2 such that the interval t1 to t2 includes t
x(t) dt
= x(t) dt = x
from eqs. (1) and (2),
x(t) (t-) dt
this is called sampling property of the impulse function.
x(t) (t – to) = x(to) (t – to)
same as sampling property.
1. Unit step function u(t)
the unit step function is defined by
u(t) = {1, for t > 0/0, for t < 0
u(t – to) = {1, for t < to/0, for t < to
Relationship between unit impulse and unit step functions
u(t) = d
(t) = du(t)/dt
1. Unit ramp function
Unit ramp function is defined by
r(t) = {t, for t > 0/0, otherwise
Relationship between unit ramp and unit step functions
r(t) u(t) dt and u(t) = d/dt r(t)
1. The exponential signal
the continuous time exponential signal is defined by
x(t) = Ceat
where, c and a are complex. numbers.
if we take c and a as real, then it becomes real exponential signal.
x(t) = CRet
CR = Real part of C
a = real part of a
C = CR + jCI
a = 0 + joo
1. the rectangular pulse
A rectangular pulse, symmetrically located with respect to the time origin having an amplitude A and duration T is described by
x(t) = {A, – T/2 < t < T/2 0, otherwise
it is given a special symbol
area (t/T) or A (t/T)
A represents amplitude.
t represents function of time.
1. Sinusoidal signal
A continuous time sinusoidal signal can be expressed as
x(t) = A sin (0o t + 0)
A = amplitude
00 = angular frequency
0 = phase angle in radian
it is periodic function with fundamental period 2/00 = To
Discrete Time Signals
1. The unit sample sequence
it is denoted by [n] and is defined by
[n] = {1, for n = 0/0, otherwise
1. The unit step sequence
the unit step sequence is defined by
u[n] = (1, for n > 0/0, for n > 0
1. The unit ramp sequence
A unit ramp sequence is denoted by and is defined by
r[n] = [n, n > 0/0, n < 0
Relation between unit step and unit impulse sequences
[n] = u[n] = – u[ n – 1]
u[n] = [k]
x[n] = x[k] [n – k]
Properties of unit sample sequence
x[n] [n] = x[0] [n]
x[n] [n – n0] = x[n0] [n – no]
1. The exponential sequence
The exponential sequence is defined by x[n] = an u[n] where, a is constant.
1. Cosinusoidal sequence
A discrete time cosinusoidal function is defined by
x[n] = A cos n
Operations on Signals (Including Transformation of Independent Variables)
(a) Continuous Time Signals
let x(t) and y(t) be two signals, then their sum z(t) is defined by
z(t) = x(t) + y(t)
1. Multiplication of signal by a constant
z(t) = a x(t)
if |a|>1; then it is known as amplitude scaling.
if|a| <1; then it is n attenuated system.
1. Multiplication of two signals
the operation is defined by
z(t) = y(t) x(t)
it is very useful in communication system.
1. Differentiation and integration
for example , the voltage V(t) across an inductance L when a current I(t) flows through it, is given by
V(t) = L dV(t)/dt
voltage across capacitor is given by
V(t) = 1/C
1. Shifting in time
Consider a signal x(t). then, the signal x(t – to) represents a delayed version of x(t) delayed by 2 s, is shown as
1. Compressing and expanding a signal in time
let there is a signal x(t).
then x(at) shows expanding of a < 1 and x(at) shows compressing of a > 1.
in case of time shifting and scaling both then, first shift the signal and then scaling is done.
Example 4. plot x(2t – 3) if x(t) is given below,
Sol. first do time shift.
if time shifting scaling and time reversal have to be done, then
perform time shifting first
then time scaling
then time reversal
Example 5. x(t) is shown below, plot x(-2t + 1).
(b) Discrete Time Signals
1. Time scaling
Consider x[n] as shown below, then find x[2n].
1. Time Shifting
same as in CT signals.
z[n] = x[n] + y[n]. same as in CT.
1. Multiplying the signal by a constant
when a discrete time signal x[n] is to be multiplied by a constant, say a, we multiply every sample of x[n] by that constant.
Intro Exercise – 1
1. The period of signal x(t) = 24 + 50 cos 60t is
(a) 1/30s
(b) 60s
(c) 1/60s
(d) not periodic
1. The period of signal x(t) = 10 sin 5t – 4 cos 9t is
(a) 24/35
(b) 4/35
(c) 2
(d) not periodic
1. The period of signal x(t) = 5t – 2 cos 6000 t is
(a) 0.96 ms
(b) 1.4 ms
(c) 0.4 ms
(d) not periodic
1. The period of signal x(t) = 4 sin 6t + 3sin 3t is
(a) 2/3s
(b) 2/3s
(c) 2
(d) not periodic
1. Consider the following signals
x(t) = cos t + 2 cos 3 t + 3 cos 5 t
y(t) = sin 2t + 6 cos 2 t
z(t) = sin 3t cos 4t
periodic signals are
(a) x(t) and y(t)
(b) y(t) and z(t)
(c) x(t) and z(t)
(d) All of these
1. The signal x(t) = e-4t u(t) is a
(a) power signal with poo = 1/4
(b) power signal with Poo = 0
(c) energy signal with Eoo = 1/4
(d) energy signal with Eoo = 0
1. The signal x(t) = ej(2t + 6) is a
(a) power signal with Poo = 1
(b) power signal with Poo = 2
(c) energy signal with Eoo = 2
(d) energy signal with Eoo = 1
1. The raised cosine pulse x(t) is defined as x(t) = {1/2 cos ot + 1), – <t < otherwise
the total energy of x(t) is
(a) 3/4
(b) 3/8
(c) 3
(d) 3/2
1. Consider the voltage waveform shown below:
the equation for V(t) is
(a) u(t – 1) + u(t – 2) + u(t – 3)
(b) u(t – 1) + 2u(t – 2) + 3u(t – 3)
(c) u(t – 1) + u(t – 2) + u(t – 4)
(d) u(t – 1) + u(t – 2) + u(t – 3) -3u(t – 4)
1. Consider the following function for the rectangular voltage pulse shown below
2. V(t) = u(a – t) . u(t – b)
3. V(t) = u(b – t). u(t – a)
4. V(t) = u(t – a) -u(t -b)
the functions that describe the pulse are
(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) all of these
1. A signal is described by x(t) = r(t – 4) -r(t – 6), where r(t) is a unit ramp function starting at t = 0. the signal x(t) is represented as
2. The trapezoidal pulse y(t) is related to the x(t) as y(t) = x(10t – 5)
the sketch of y(t) is
1. The trapezoidal pulse x(t) is time scaled producting y(t) = x(5t). the sketch for y(t) is
2. The trapezoidal pulse x(t) is time scaled producting y(t) = x(t/5). the sketch for y(t) is
3. The trapezoidl pulse x(t) is applied to a differentiator, defined by y(t) = dx(t)/dt. the total energy of y(t) is
(a) 0
(b) 1
(c) 2
(d) 3
1. The total energy of x(t) is
(a) 0
(b) 13
(c) 13/3
(d) 26/3
1. u[n] + u[-n] is equal to
(a) 2
(b) 1 + [n]
(c) 2 + [n]
(d) 1
1. A discrete time signal is given as below
x[n] = cos 9 + sin (n/7 + 1/2)
the period of signal is
(a) periodic with period N = 126
(b) periodic with period N = 32
(c) periodic with period N = 252
(d) not periodic
1. A discrete time signal is given as below
x[n] = cos (n/8) cos (n/8)
(a) periodic with period 16
(b) periodic with period 16 + 1
(c) periodic with period 8
(d) not periodic
1. A discrete time signal is given as below
x[n] = cos (n/2) – sin (n/8) + 3cos (n/4 + 3)
the period of signal is
(a) periodic with period 16
(b) periodic with period 4
(c) periodic with period 2
(d) not periodic
1. (a)
2/t = 60 T = 1/30
1. (c)
2/t1 = 5 T1 = 2/5 and 2/T2 = 9
T2 = 2/9
LCM (2/5, 2/9) = 2
1. (d)
not periodic because of t.
1. (d)
not periodic because least common multiple is infinite.
1. (c)
y(t) is not periodic although sin t and 6 cos 2t are independently periodic. the fundamental frequency cannot be determined.
1. (c)
this is energy signal because
Eoo = |x(t)|2 dt < oo = e-4t u(t) dt = [e-4t dt = 1/4
1. (a)
|x(t)| = 1, EOO = |x(t)|2 dt = oo
so, this is a power signal not a energy.
P00 = lim 1/2t x(t)2 dt = 1
1. (a)
E = 1/4 cos (ot + 1)2 dt
= 1/2 (1/2 cos 2ot + 1/2 + c cos ot + 1) dt
= 1/2 (3/2) = 3/4
1. (d)
V(t) is sum of three unit step signals starting from 1, 2, and 3, all signals-end at 4.
1. (b)
The function 1 does not describe the given pulse. it can be shown as follows
1. (b)
the figure is as shown below.
1. (c)
the figure is as shown below
1. (d)
multiplication by 5 will bring contraction on time scale. it may be checked by x(5 x 0.8) = x(4).
1. (a)
division by 5 will bring expansion on time scale. it may be checked by
y(t) = x(20/5) = x(4)
1. (c)
y(t) = {1, for -5 < t <-4/-1, for 4 < t < 5/0, otherwise
E = (1)2 dt + (-1)2 dt = 2
1. (d)
E = 2 x2 (t) dt
= 2 (1)1 dt + 2 (5 – t)2 dt = 8 + 2/3 = 26/3
1. (b)
the pole are as follows :
1. (a)
both signals are periodic N1 = 18 N2 = 14
N = LCM (18, 14) = 126
1. (d)
cos (n/8) is not periodic. so, x[n] is not periodic.
1. (a)
N1 = 4, N2 = 16, N3 = 8, N = LCM (4, 16, 8) = 16 | HuggingFaceTB/finemath | |
# Resources tagged with: Comparing and ordering numbers
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### There are 15 results
Broad Topics > The Number System and Place Value > Comparing and ordering numbers
### Euromaths
##### Age 11 to 14 Challenge Level:
How many ways can you write the word EUROMATHS by starting at the top left hand corner and taking the next letter by stepping one step down or one step to the right in a 5x5 array?
### Snowman
##### Age 14 to 16 Challenge Level:
All the words in the Snowman language consist of exactly seven letters formed from the letters {s, no, wm, an). How many words are there in the Snowman language?
### Greetings
##### Age 11 to 14 Challenge Level:
From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . .
### Consecutive Seven
##### Age 11 to 14 Challenge Level:
Can you arrange these numbers into 7 subsets, each of three numbers, so that when the numbers in each are added together, they make seven consecutive numbers?
### Nice or Nasty
##### Age 7 to 14 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
### Dicey Operations
##### Age 11 to 14 Challenge Level:
Who said that adding, subtracting, multiplying and dividing couldn't be fun?
### More Carroll Diagrams
##### Age 7 to 14 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Mathland Election
##### Age 11 to 14 Challenge Level:
A political commentator summed up an election result. Given that there were just four candidates and that the figures quoted were exact find the number of votes polled for each candidate.
### Largest Number
##### Age 11 to 14 Challenge Level:
What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.
### Farey Sequences
##### Age 11 to 14 Challenge Level:
There are lots of ideas to explore in these sequences of ordered fractions.
### Nice or Nasty for Two
##### Age 7 to 14 Challenge Level:
Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your opponent.
### Writ Large
##### Age 11 to 14 Challenge Level:
Suppose you had to begin the never ending task of writing out the natural numbers: 1, 2, 3, 4, 5.... and so on. What would be the 1000th digit you would write down.
### Rachel's Problem
##### Age 14 to 16 Challenge Level:
Is it true that $99^n$ has 2n digits and $999^n$ has 3n digits? Investigate!
### Rod Fractions
##### Age 7 to 14 Challenge Level:
Pick two rods of different colours. Given an unlimited supply of rods of each of the two colours, how can we work out what fraction the shorter rod is of the longer one?
### Number Lines in Disguise
##### Age 7 to 14 Challenge Level:
Some of the numbers have fallen off Becky's number line. Can you figure out what they were? | HuggingFaceTB/finemath | |
# 9 Pint to Oz (9 pt to ounce conversion)
How many oz in 9 pint? - Are you a person trying to convert 9 pint to oz? You have landed at the right place! Our tool can convert 9 pt to oz at ease without any time delays! We consider the traditional calculation where 9 pint equals 144 oz.
9 pint = 144 ounce
pint
oz
9 pint = 144 ounce
(9 pt = 144 oz)
Hence to convert it to oz, to calculate the corresponding ounce of 9 pint, just multiplying 9 into 16 times is equal to 144 oz.
How to Calculate 9 pint to oz?
The base formula for this pint to oz converter is
ounce = pint * 16.
Applying the same formula to 9 pint, it can be shown in the form of
oz = 9 pint * 16 = 144 oz
How to Convert 9 pint to oz?
• Step 1: To convert 9 pint to oz, you should remember that oz equals a pint multiply value of 16.
• Step 9: 9 pint multiply the value by 16.
• Step 3: 9 * 16 is equal to 144.
• Step 9: Hence, the answer is 9 pint = 144 oz.
Some pint to oz Conversion Chart for your reference:
1 pint equal to 16 oz 2 pint equal to 32 oz 3 pint equal to 48 oz 4 pint equal to 64 oz 5 pint equal to 80 oz 6 pint equal to 96 oz 7 pint equal to 112 oz 8 pint equal to 128 oz 9 pint equal to 144 oz 10 pint equal to 160 oz
Reverse Calculation: Convert 9 oz to pint
Let us try to convert 9 oz to pint
From the oz to pint measurement converter, 9 oz = 0.5625 pt
9 oz = 9 / 16 = 0.5625 pt
Thus, 9 oz is the corresponding value of 0.5625 pint.
pint to oz Conversion Chart Links:
1.5 pint to oz conversion 2 pint to oz conversion 2.5 pint to oz conversion 3 pint to oz conversion 4 pint to oz conversion 5 pint to oz conversion 6 pint to oz conversion 7 pint to oz conversion 8 pint to oz conversion 9 pint to oz conversion 10 pint to oz conversion
This formula provides an instant answer for all your questions / People also search:
• How many oz in a 9 pint?
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• Formula to convert 9 pint to oz online
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• How much oz is equal to 9 pint?
• How to convert 9 pint to oz?
• How many oz in 9 pt?
• How many ounce in 9 pint?
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• 9 pint equals how many ounce? | HuggingFaceTB/finemath | |
Edmund Ford, teacher and City Council member, wonders if similar provisions will be made for students who miss getting an A or B." Intervention should be impartially accessible. How can we be fair to all our students when we are giving those not meeting standards more than enough chances?"
LOL WHAT DID HE JUST SAY! Its another one of those wrong kind of teachers talking again!!!!!!!! Please. Intervention is for failing students!!!! All great teachers give every student a chance to make any grade anyway- it's called either #1 make up work or #2 bonus points. That has not change in the last 75 years!
Mr. Ford this is for students who are f.a.i.l.i.n.g. Can you repeat the word after me…… f.a.i.l.i.n.g.
His words are like saying lets give all students pencils not just the students that have no pencils!!!!!! Why isn’t Mr. Keith Williams upset over the words of Mr. Ford? You talk about double working teachers- lets try Mr. Ford’s suggestion.
“ Hello kids, I know you already have an A but for those of you that want an A plus, let me re-teach the One dimensional motion. Now all you kids that are failing take a nap while I help the kids with an A make an A plus.
By one dimension we mean that the body is moving only in one plane and in a straight line. Like if we roll a marble on a flat table, and if we roll it in a straight line (not easy!), then it would be undergoing one-dimensional motion.
There are four variables which put together in an equation can describe this motion. These are Initial Velocity (u); Final Velocity (v), Acceleration (a), Distance Traveled (s) and Time elapsed (t). The equations which tell us the relationship between these variables are as given below.
v = u + at
v2 = u2 + 2as
s = ut + 1/2 at2
average velocity = (v + u)/2
Armed with these equations you can do wonderful things like calculating a cars acceleration from zero to whatever in 60 seconds.
Mr. Ford please do not make yourself look silly like you just did again.
Stay away from the reporters! Because of your last name you may have a future in politics.
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### Ppt
1. 1. LAW OF MOTION
2. 2. Who figured out most of what we know about ?how things move
3. 3. “If I have ever made any valuable discoveries, it has been owing more to patient attention, than to any other talent” Isaac Newton
4. 4. ?What is the law of motion• Newtons laws of motion are three physical laws which provide relationships between the forces acting on a body and the motion of the body.
5. 5. Newton’s first law
6. 6. An object at rest tends to stay at rest and object in motion tends to stay in motion unless acted upon by an unbalanced force.
7. 7. ?!What does this meanBasically, an object will keep doing what it was doing, unless acted on by an unbalanced force.If the object was sitting still, it will remain stationary. If it was moving at a constant velocity, it will keep moving.It takes force to change the motion of an object.
8. 8. What is meant byunbalanced force?
9. 9. If the forces on an object are equal and opposite, they are said to be balanced, and the object experiences no change in motion. If they are not equal and opposite, then the forces are unbalanced and the motion of the object changes.
10. 10. These are some examples from real lifeA soccer ball is sitting at rest.It takes an unbalanced force ofa kick to change its motion.
11. 11. Newton’s First Law is also called the Law of InertiaInertia: the tendency of an object to resist changes in its state of motionInertia: the tendency of an object to resist changes in its state of motionThe First Law states that all objects have inertia. The more mass an object has, the more inertia it has (and the harder it is to change its motion).
12. 12. More Examples from Real LifeA powerful locomotive begins to pull a long line ofboxcars that were sitting at rest. Since the boxcars areso massive,they have a great deal of inertia and it takes a largeforce to change their motion. Once they are moving, ittakes a large force to stop them.
13. 13. If objects in motion tend to stay in motion, why don’tmoving objects keep moving forever?
14. 14. Things don’t keep moving forever because there’s almost always an unbalanced force acting upon it.A book sliding across a table slows down and stops because of the force of friction.
15. 15. If you throw a ball upwards it will eventually slow down and fall because of the force of gravity. If you throw a ball upwards it willeventually slow down and fall because ofthe force of gravity.
16. 16. Newton’s Second Law
17. 17. Force equals mass times acceleration. F = ma Acceleration: a measurement of how quickly Acceleration an object is changing speed.Acceleration: a measurement of how quickly an object is changing speed
18. 18. ?What does F = ma meanForce is directly proportional to mass and acceleration. Imagine a ball of a certain mass moving at a certain acceleration. This ball has a certain force.
19. 19. Now imagine we make the ball twice as big (double the mass) but keep the acceleration constant. F = ma says that this new ball has twice the force of the old ball.Now imagine the original ball moving at twice the original acceleration. F = ma says that the ball will again have twice the force of the ball at the original acceleration.
20. 20. More about F = maIf you double the mass, you double the force. If you .double the acceleration, you double the force?What if you double the mass and the acceleration 2m)(2a) = 4F(Doubling the mass and the acceleration quadruples .the forceSo . . . what if you decrease the mass by half? How ?much force would the object have now
21. 21. ?What does F = ma say F = ma basically means that the force of an object comes from its mass and its acceleration. Something very massive (high mass) that’s changingspeed very slowly (low acceleration), like a glacier, can still have great force.
22. 22. Something very small (low mass) that’s changing speed very quickly (highacceleration), like a bullet, can still have a great force. Something very small changing speed very slowly will have a very weak force.
23. 23. Newton’s Third Law
24. 24. For every action there is an equal and opposite reaction For every action there is an equal and opposite reaction
25. 25. ?What does this meanFor every force acting on an object, there is an equal force acting in the opposite direction. Right now, gravity is pulling you down in your seat, but Newton’s Third Law says your seat is pushing up against you with equal force. This is why you are not moving. There is a balanced force acting on you– gravity pulling down, your seat pushing up.
26. 26. . . .Think about itWhat happens if you are standing on a skateboard or a slippery floor and push against a wall? You slide in the opposite direction (away from the wall), because you pushed on the wall but the wall pushed back on you with equal and opposite force.
27. 27. Why does it hurt so much when you stub your toe? When your toe exerts a force on a rock, the rock exerts an equal force back on your toe. The harder you hit your toe against it, the more force the rock exerts back on your toe (and the more your toe hurts).
28. 28. ReviewNewton’s First Law: Objects in motion tend to stay in motion and objects at rest tend to stay at rest unless acted upon by an unbalanced force.Newton’s Second Law: Force equals mass times acceleration (F = ma).Newton’s Third Law: For every action there is an equal and opposite reaction.
29. 29. :Done by | HuggingFaceTB/finemath | |
The neper (symbol: Np) is a logarithmic unit for ratios of measurements of physical field and power quantities, such as gain and loss of electronic signals. The unit's name is derived from the name of John Napier, the inventor of logarithms. As is the case for the decibel and bel, the neper is a unit defined in the international standard ISO 80000. It is not part of the International System of Units (SI), but is accepted for use alongside the SI.[1]
## Definition
Like the decibel, the neper is a unit in a logarithmic scale. While the bel uses the decadic (base-10) logarithm to compute ratios, the neper uses the natural logarithm, based on Euler's number (e ≈ 2.71828). The level of a ratio of two signal amplitudes or root-power quantities, with the unit neper, is given by[2]
${\displaystyle L=\ln {\frac {x_{1)){x_{2))}\mathrm {~Np} ,}$
where ${\displaystyle x_{1))$ and ${\displaystyle x_{2))$ are the signal amplitudes, and ln is the natural logarithm. The level of a ratio of two power quantities, with the unit neper, is given by[2]
${\displaystyle L={\frac {1}{2))\ln {\frac {p_{1)){p_{2))}\mathrm {~Np} ,}$
where ${\displaystyle p_{1))$ and ${\displaystyle p_{2))$ are the signal powers.
In the International System of Quantities, the neper is defined as 1 Np = 1.[3]
## Units
Main article: Level (logarithmic quantity)
The neper is defined in terms of ratios of field quantities — also called root-power quantities — (for example, voltage or current amplitudes in electrical circuits, or pressure in acoustics), whereas the decibel was originally defined in terms of power ratios. A power ratio 10 log r dB is equivalent to a field-quantity ratio 20 log r dB, since power in a linear system is proportional to the square (Joule's laws) of the amplitude. Hence the decibel and the neper have a fixed ratio to each other:[4]
${\displaystyle 1\ {\text{Np))=20\log _{10}e\ {\text{dB))\approx {\text{8.685889638 dB))}$
and
${\displaystyle 1\ \mathrm {dB} ={\frac {1}{20))\ln(10)\ \mathrm {Np} \approx {\text{0.115129255 Np)).}$
The (voltage) level ratio is
{\displaystyle {\begin{aligned}L&=10\log _{10}{\frac {x_{1}^{2)){x_{2}^{2))}&{\text{dB))\\&=10\log _{10}{\left({\frac {x_{1)){x_{2))}\right)}^{2}&{\text{dB))\\&=20\log _{10}{\frac {x_{1)){x_{2))}&{\text{dB))\\&=\ln {\frac {x_{1)){x_{2))}&{\text{Np)).\\\end{aligned))}
Like the decibel, the neper is a dimensionless unit. The International Telecommunication Union (ITU) recognizes both units. Only the neper is coherent with the SI.[5]
## Applications
The neper is a natural linear unit of relative difference, meaning in nepers (logarithmic units) relative differences add rather than multiply. This property is shared with logarithmic units in other bases, such as the bel.
The derived units decineper (1 dNp = 0.1 neper) and centineper (1 cNp = 0.01 neper) are also used.[6] The centineper for root-power quantities corresponds to a log point or log percentage, see Relative change and difference § Logarithmic scale.[7]
## References
1. ^ The International System of Units (SI) (9 ed.). International Bureau of Weights and Measures. 2019. pp. 145–146. Archived from the original on 2022-10-09.
2. ^ a b Letter symbols to be used in electrical technology – Part 3: Logarithmic and related quantities, and their units (International standard). International Electrotechnical Commission. 2002-07-19. IEC 60027-3:2002.
3. ^ Thor, A J (1994-01-01). "New International Standards for Quantities and Units". Metrologia. 30 (5): 517–522. doi:10.1088/0026-1394/30/5/010. ISSN 0026-1394.
4. ^ Ainslie, Michael A.; Halvorsen, Michele B.; Robinson, Stephen P. (January 2022) [2021-11-09]. "A terminology standard for underwater acoustics and the benefits of international standardization". IEEE Journal of Oceanic Engineering. 47 (1). IEEE: 179-200 [Appendix B Decibel: Past, Present, and Future – Section D]. doi:10.1109/JOE.2021.3085947. eISSN 1558-1691. ISSN 0364-9059. S2CID 243948953. Retrieved 2022-12-20. [1] (22 pages)
5. ^ ISO 80000-3:2007 §0.5
6. ^ Glossary of Telecommunication Terms. General Services Administration, Federal Supply Service. 1980. p. 73.
7. ^ Karjus, Andres; Blythe, Richard A.; Kirby, Simon; Smith, Kenny (10 February 2020). "Quantifying the dynamics of topical fluctuations in language". Language Dynamics and Change. 10 (1): 86–125. arXiv:1806.00699. doi:10.1163/22105832-01001200. S2CID 46928080. | HuggingFaceTB/finemath |
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