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# There are 7 members in a family. The average age of first 3 members is 25 years and the average age of last 3 members is 33 years. The sum of ages of all the members of a family is 210. Then find the age of 4th member of the family.
1. 32
2. 36
3. 40
4. 44
Option 2 : 36
## Detailed Solution
Given:
Number of members in the family = 7
Average age of first 3 members = 25 years
Average age of last 3 members = 33 years
Sum of ages of all the members of a family = 210
Formula used:
Average age = Sum of age of family members/Number of family members
Calculations:
Let the age of 4th member be x
Average age of first 3 members
⇒ Sum of age of first 3 members = 25 × 3
⇒ Sum of age of first 3 members = 75
Average age of last 3 members
⇒ Sum of age of last 3 members = 33 × 3
⇒ Sum of age of last 3 members = 99
Sum of ages of all the members of a family = Sum of age of first 3 members + x + Sum of age of last 3 members
⇒ 75 + x + 99 = 210
⇒ 174 + x = 210
⇒ x = 36 years
The age of 4th member of the family is 36 years |
The notation 32 and 23 is known as index form. The small digit is called the index number or power.
We have already seen that 32 = 3·3 = 9, and that 23 = 2·2·2 = 8.
In each case, the index number tells us how many times we should multiply the numbers together.
When the index number is two, we say "squared"
When the index number is three, we say "cubed"
When the index number is grater than three, we say "to the power of"
All scientific calculators have a "power" botton [xy]. This is particulary useful when tyhe index number is large. (After all, you're very likely not to make a mistake if you try to calculate 410 by typing 4·4·4·4·4·4·4·4·4·4 into your calculator!)
4 shift xy 10 =
INDEX LAWS FOR MULTIPLICATION AND DIVISION
Multiplication
23= 2·2·2
so, 23·25= 2·2·2·2·2·2·2·2= 28
25=2·2·2·2·2
Can you see what happened? We have 3 twos from 23 and 5 twos from 25, so alltogether we had 8 twos.
In general,
2m·2n= 2(m+n)
#### How Do You Multiply Two Numbers With Exponents?
Division
If we divide 25 by 23:
So 25: 23= 22
In general 2m: 2n= 2(m-n)
#### How Do You Divide Two Numbers With Exponents?
SQUARE ROOT
We already know that 3·3 can be written as 32 (3 squared)
The opposite of squaring a number is called finding the square root.
Examples:
The square rootof 16 is 4 (because 4·4 = 16)
The square root of 25 is 5 (because 5·5 = 25)
The square root of 100 is 10 (because 10·10 =100)
SCIENTIFIC NOTATION
For very large numbers, it is sometimes simpler to use "scientific notation" (so called, because scientifics often deal with very large numbers).
The format for writing a number in scientific notation is fairly simple:
First digit of the number followed by the decimal point and then all the rest of the digits of the number, times 10 to an appropiate power. The conversion is fairly simple.
Example: Write 12400 in scientific notation.
This is not a very large number, but it will work nicely for an example. To convert this to scientific notation, I first write "1.24". This is not the same number, but (1.24)·(10000)= 12400 and 10000= 104. Then, in scientific notation, 12400 is written as 1.24·104
Subpáginas (1): |
# A Lightweight Example and Tutorial on Genetic Algorithms
Genetic algorithms are a fun and sometimes almost magical problem solving tool. This write-up is a quick and dirty tutorial implementing a genetic algorithm to solve a particular geometry problem. I’ve often found it instructive to learn a new method, algorithm, technique on a problem I can already solve
First, let’s talk high-level. We’re going to first identify our problem mathematically. Then we’ll sample a region where we believe the solution should exist. And from there we’ll continue to refine our search. Now keep in mind that this problem will be easy to solve by hand, but our purpose here is to see how an evolutionary algorithm / method would find the solution.
Here’s a pretty picture that I’ll explain and we’ll build off of!
Genetic Algorithm teaser
### The Problem
I have a unit square whose bottom left corner is anchored at $$(0,0)$$. A circle of unknown radius intersects my square at the following points: $$(0,0), (0.5, 0), \mbox{ and }, (1,1)$$. What is the radius of the circle? This was a puzzle from Ed Southall and @srcav blogged about it here.
It might help if we could draw this out.
Our problem. What is the radius of the circle if the square is a unit square?
This problem can be viewed in a number of different ways, but a coordinate geometry approach seems to be simplest, at least for me.
We want to know the circle’s radius given that we have a unit square and a number of intersection points.
This yields the following set of equations if $$h$$ and $$k$$ are the $$x$$ and $$y$$ coordinates of the center of the circle:
$$\begin{eqnarray} h^{2} + k^{2} & = & r^{2}\\ (0.5-h)^{2} + k^{2} & = & r^{2}\\ (1-h)^{2} + (1-k)^{2} & = & r^{2} \end{eqnarray}$$
Remember that the equation of a circle with center $$(h,k)$$ is $$(x-h)^{2} + (y-k)^{2} = r^{2}$$
We have three equations, three unknowns. The algebra is pretty straightforward to go through as the square terms drop out. But! You may say, “Damn it, Shah, it’s straightforward for you, it’s not easy for me!” However, you’ve got programming chops that are beyond mine. So let’s put that to work. There’s more than one way to skin a math problem!
### An Approach
What we want are $$h$$ and $$k$$ from above and those values together should give the same value for $$r^{2}$$ for each of the equations.
I’ll introduce the jargon of genetic algorithms as we need it. The first thing we want to do is create an initial “population”. Our population is a collection of points $$(x,y)$$ within some region where we believe the solution exists. To be conservative, let’s sample from a large enough space, say, something like $$[-1,2]\times[-1,2]$$. We know that we can definitely contain the circle in an area of 9 square units in that range.
#### How should we sample this region?
Uniform random sampling is a good place. And we can pick, say, 10,000 points at random.
#### What should we do we these points?
The first thing we want to do is assign a “fitness score” that tells us how good these points are. In this case, I want to see the largest absolute difference in $$r^{2}$$ from all pairwise comparisons of the three equations. The larger the difference, the worse the fit. The closer the absolute difference is to zero, the better the fit. Here’s python code.
In [83]:
def opt(h,k):
f1 = h**2 + k**2
f2 = (0.5-h)**2 + k**2
f3 = (1-h)**2 + (1-k)**2
return max(abs(f1-f2), abs(f2-f3), abs(f1-f3)),((f1 + f2 + f3)/3)**0.5
So, if I create an initial population of 10,000 random points in $$[-1,2]\times[-1,2]$$ and apply a fitness score to each of those points, I now have the following visualization of our sample space. Notice that the bluer the region, the better the fit.
The blue region is where good candidates exist for our solution
#### Great, now what?
Ok! That was just an initial set up. Now the genetic part is going to come in. Each “individual” in the population is really a coordinate $$(x,y)$$. We can think of those dimensions as the “genetic code” for each individual. So, sticking with the analogy of evolutionary theory, we want to have a sort of “survival of the fittest” take place. That is, only the strong get to reproduce. Everyone else dies and their genetics do not get to passed on.
There are a lot of ways to do this and you can, for the most part, make up your own methodology, though there are some generally accepted principles and standards.
What we now have to do with this initial population is to pick two individuals and have them “mate”!
#### How do mathematical objects “mate”?
Glad you asked! There are a lot of ways to do this mating and, in general, the term is called “crossover” whereby the two selected individuals create some “offspring” which is composed of the two “parents” genetic code. Here is one example of a crossover in action.
Suppose we have an individual A with “DNA” $$(0.3, 0.8)$$ and another individual, B, with DNA $$0.1, 0.5$$. We could produce offspring M as $$(0.1, 0.8)$$ and offspring N as $$(0.3, 0.5)$$. Notice that we just swapped the $$x$$ and $$y$$ coordinates!
Another way is “alpha-averaging”. Pick a real number $$\alpha \in (0,1)$$ and produce offspring as the weighted average of the parents. Something like $$M = \alpha A + (1-\alpha) B$$ and $$N = (1-\alpha) A + \alpha B$$.
In [210]:
def arithmetic_crossover(pair,optfunc):
alpha = random.random()
offspring1 = [pair.iloc[0]['x']*alpha + pair.iloc[1]['x']*(1-alpha),pair.iloc[0]['y']*alpha + pair.iloc[1]['y']*(1-alpha)]
offspring2 = [pair.iloc[0]['x']*(1-alpha) + pair.iloc[1]['x']*(alpha),pair.iloc[0]['y']*(1-alpha) + pair.iloc[1]['y']*(alpha)]
score1 = optfunc(offspring1[0],offspring1[1])
score2 = optfunc(offspring2[0],offspring2[1])
df1 = [offspring1[0],offspring1[1],score1[0],score1[1]]
df2 = [offspring2[0],offspring2[1],score2[0],score2[1]]
return [df1,df2]
#### Who gets to mate?
It’s survival of the fittest! That’s why we computed a fitness score! There are several ways to decide on who gets to mate. Our goal is to pass on good individuals’ genes (in our case low fitness scores) to the next generation of offspring. This type of method is called “elitism”. You can choose how elitist you want to be. I chose to have the next population drawn from the top 1% of individuals per their fitness scores.
But there’s a catch! In this world of genetic reproduction, there is the possibility of a “mutation”. Think of mutation as a hedge against getting stuck in a local extremum when we are trying to optimize a function.
You might have caught on that there is a bit of ad-hocness to some of these parameters. How often should be mutate? It’s up to you! The higher the mutation rate, the closer the genetic algorithm is to a pure random search.
For this problem, I chose to reproduce from the top 1% of the population with 99% probability. With 1% probability I chose to draw a random individual from the bounding box of the current elite population (I could have chosen a wider range and doing so would be more inline with the hedge against getting stuck in local extrema).
Here’s code for how the next generation is created.
In [213]:
def next_pop(curr_pop, optfunc):
n = curr_pop.shape[0]
upperp = numpy.percentile(curr_pop['fitness'],1)
q = curr_pop.query('fitness < {z}'.format(z = upperp))
mins = q.min()
maxs = q.max()
pop = q.values.tolist()
sz = len(pop)
while len(pop) < n:
if random.random() < .99:
mates = q.sample(2)
mates.reset_index(inplace = True,drop = True)
offsprings = arithmetic_crossover(mates,optfunc)
pop.append(offsprings[0])
pop.append(offsprings[1])
else:
x = random.random()*(maxs['x']-mins['x']) + mins['x']
y = random.random()*(maxs['y']-mins['y']) + mins['y']
score = optfunc(x,y)
fitness = score[0]
And that’s it! Now we just rinse and repeat! Here’s how the solution evolved. If you solved the equations by hand you should have gotten a center of $$(0.25, 0.75)$$ resulting in a radius of $$\frac{\sqrt{10}}{4}$$. Notice in the title of the graphs, the best point found in each generation. I put the graph of the initial population in here for the third time in this post so that you can see how the initial random sample evolves.
Initial population
First generation, notice how our elitist sampling has reduced the space for our search.
Our approximate solutions are getting better
At this point we can feel that we have a fairly accurate solution. But when we stop is up to us! Often we will have an error tolerance set up to act as a stopping time. Other times, we may measure how much the solution has changed from one generation to the next to determine if there is still further room for improvement or if we’ve reached convergence.
By the fourth generation we are pretty well locked into a small region and indeed we see to have convergence.
### One piece of criticism!
It can feel like we are cheating because we keep trying “random” numbers and we’re just getting lucky. This is not so! One of the things I track is how many unique points I try over all generations. I then want to do a pure random search on my initial [larger, low-informed] search space. If I obtain comparable results [you can define your own tolerance here] with the genetic algorithm, then I probably can just get away with pure random sampling. It turns out that, at least for this problem, our genetic algorithm is quite superior. In each generation I had a population of 10000 and in the aggregate over the five generations, I generated 49600 unique points (elitism kept about 100 per generation). Here’s what a pure random search with 49600 points produced.
A pure random search
As you can see, while we are in the ballpark with a purely random search, the genetic algorithm with just as many points produced a far more accurate number. And remember we wanted to find the radius of the circle. Since we have the center, it’s a matter of picking one of the points on the circle ($$(0,0)$$ as an easy example) and computing the distance. That’s our radius!
### To Sum Up
If you want to continue to learn, DuckDuckGo is your friend, but now you have a bunch of terms (crossover, mutation, population, elitism) and a sample problem to work off of! Remember that the crossover functionality can be highly varied. We were working in a continuous framework. When we have to work in a discrete problem or a permutation problem, crossover works differently.
See if you can write your own genetic algorithm to maximize $$\sin(2\pi x) + \sin(2\pi y)$$ on $$[0,1]^{2}$$ subject to the constraint that $$x^{y} + y^{x} > 1.35$$. You can probably solve this by hand given that the objective function is neatly separated. The only thing you’d have to do is to check that the constraint is met. The tricky part from the genetic algorithm’s standpoint is making sure to generate points that abide by the constraint. A hit-or-miss implementation will work fine.
Let me know if this helps!
Do you enjoy this blog?
Consider supporting with a contribution!
## 2 thoughts on “A Lightweight Example and Tutorial on Genetic Algorithms”
1. Andy Novocin
So I’m interested in how we might, with a genetic algorithm, move from a numeric guess to a symbolic/exact solution. After a few generations you’ve got pretty darn close numerical answers. Now suppose we think that the answer might be a integer combination of square roots of integers. We could probably see the problem in a similar way for the exact version.
1. Manan Shah Post author
Yeah, there’s another line of “genetics” and this would be genetic programming where we have to endow our methodology with the primitive operations we want permissible — addition, subtraction, multiplication, division, exponentiation, logarithmiation (word?), etc along with an objective function (maximize / minimize something, eg) as well as the number of parameters to consider. From there, it’s a matter of constructing different functions and mating them. A typical approach here is to think of the genetic function as a tree and the crossover operation is a swapping of one part of a tree from parent A with one part of a tree from parent B. |
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# Pc12 sol c08_8-6
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### Pc12 sol c08_8-6
1. 1. 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:23 PM Page 25 Home Quit Lesson 8.6 Exercises, pages 743–749 A 3. Expand using Pascal’s triangle. a) (x + 1)5 The exponent is 5, so use the terms in row 6 of Pascal’s triangle as coefficients: 1, 5, 10, 10, 5, 1 (x ؉ 1)5 (1 ؍x)5 ؉ 5(x)4(1) ؉ 10(x)3(1)2 ؉ 10(x)2(1)3 ؉ 5(x)1(1)4 ؉ 1(1)5 ؍x5 ؉ 5x4 ؉ 10x3 ؉ 10x2 ؉ 5x ؉ 1 b) (x - 1)6 The exponent is 6, so use the terms in row 7 of Pascal’s triangle as coefficients: 1, 6, 15, 20, 15, 6, 1 (x ؊ 1)6 (1 ؍x)6 ؉ 6(x)5(؊1) ؉ 15(x)4(؊1)2 ؉ 20(x)3(؊1)3 ؉ 15(x)2(؊1)4 ؉ 6(x)(؊1)5 ؉ 1(؊1)6 ؍x6 ؊ 6x5 ؉ 15x4 ؊ 20x3 ؉ 15x2 ؊ 6x ؉ 1 c) (x + y)4 The exponent is 4, so use the terms in row 5 of Pascal’s triangle as coefficients: 1, 4, 6, 4, 1 (x ؉ y)4 (1 ؍x)4 ؉ 4(x)3(y) ؉ 6(x)2(y)2 ؉ 4(x)(y)3 ؉ 1(y)4 ؍x4 ؉ 4x3y ؉ 6x2y2 ؉ 4xy3 ؉ y4 d) (x - y)8 The exponent is 8, so use the terms in row 9 of Pascal’s triangle as coefficients: 1, 8, 28, 56, 70, 56, 28, 8, 1 (x ؊ y)8 (1 ؍x)8 ؉ 8(x)7(؊y) ؉ 28(x)6(؊y)2 ؉ 56(x)5(؊y)3 ؉ 70(x)4(؊y)4 ؉ 56(x)3(؊y)5 ؉ 28(x)2(؊y)6 ؉ 8(x)(؊y)7 ؉ 1(؊y)8 ؍x8 ؊ 8x7y ؉ 28x6y2 ؊ 56x5y3 ؉ 70x4y4 ؊ 56x3y5 ؉ 28x2y6 ؊ 8xy7 ؉ y8 4. Determine each missing number in the expansion of (x + y)7. x7 + nx6y + 21x5y2 + 35x ny3 + nx3y4 + 21x ny n + 7xy6 + y n The exponent is 7, so the coefficients of the terms in the expansion are the terms in row 8 of Pascal’s triangle: 1, 7, 21, 35, 35, 21, 7, 1 The exponents in each term must add to 7. The exponents of the powers of x start at 7 and decrease by 1 each time. The exponents of the powers of y start at 0 and increase by 1 each time. So, the missing numbers are: 7, 4, 35, 2, 5, 7 5. Determine the indicated term in each expansion. a) the last term in (x + 1)9 The last term in the expansion of (x ؉ y)n is yn. So, the last term in the expansion of (x ؉ 1)9 is 19, or 1. b) the 1st term in (x - 1)12 The first term in the expansion of (x ؉ y)n is xn. So, the first term in the expansion of (x ؊ 1)12 is x12. ©P DO NOT COPY. 8.6 The Binomial Theorem—Solutions 25
2. 2. 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:23 PM Page 26 Home Quit B 6. a) Multiply 4 factors of (x - 5). (x ؊ 5)4 ؍ (x ؊ 5)(x ؊ 5)(x ؊ 5)(x ؊ 5) ؍ (x2 ؊ 10x ؉ 25)(x2 ؊ 10x ؉ 25) ؍ x4 ؊ 10x3 ؉ 25x2 ؊ 10x3 ؉ 100x2 ؊ 250x ؉ 25x2 ؊ 250x ؉ 625 ؍ x4 ؊ 20x3 ؉ 150x2 ؊ 500x ؉ 625 b) Use the binomial theorem to expand (x - 5)4. (x ؉ y)n ؍nC0xn ؉ nC1xn ؊ 1y ؉ nC2xn ؊ 2y2 ؉ . . . ؉ nCnyn Substitute: n ,4 ؍y 5؊ ؍ (x ؊ 5)4 4 ؍C0(x)4 ؉ 4C1(x)4 ؊ 1(؊5) ؉ 4C2(x)4 ؊ 2(؊5)2 ؉ 4C3(x)4 ؊ 3(؊5)3 ؉ 4C4(؊5)4 (1 ؍x)4 ؉ 4(x)3(؊5) ؉ 6(x)2(25) ؉ 4(x)1(؊125) ؉ 1(625) ؍x4 ؊ 20x3 ؉ 150x2 ؊ 500x ؉ 625 c) Compare the two methods. What conclusions can you make? I find it easier to use the binomial theorem; it saves time and it is less cumbersome than multiplying 4 factors. 7. Expand using the binomial theorem. a) (x + 2)6 (x ؉ y)n ؍nC0xn ؉ nC1xn ؊ 1y ؉ nC2x n ؊ 2y2 ؉ . . . ؉ nCnyn Substitute: n ,6 ؍y 2 ؍ (x ؉ 2)6 6 ؍C0(x)6 ؉ 6C1(x)6 ؊ 1(2) ؉ 6C2(x)6 ؊ 2(2)2 ؉ 6C3(x)6 ؊ 3(2)3 ؉ 6C4(x)6 ؊ 4(2)4 ؉ 6C5(x)6 ؊ 5(2)5 ؉ 6C6(2)6 (1 ؍x)6 ؉ 6(x)5(2) ؉ 15(x)4(4) ؉ 20(x)3(8) ؉ 15(x)2(16) ؉ 6(x)1(32) ؉ 1(64) ؍x6 ؉ 12x5 ؉ 60x4 ؉ 160x3 ؉ 240x2 ؉ 192x ؉ 64 b) (x2 - 3)5 (x ؉ y)n ؍nC0xn ؉ nC1xn ؊ 1y ؉ nC2xn ؊ 2y2 ؉ . . . ؉ nCnyn Substitute: n ,5 ؍x ؍x2, y 3؊ ؍ (x2 ؊ 3)5 5 ؍C0(x2)5 ؉ 5C1(x2)4(؊3) ؉ 5C2(x2)3(؊3)2 ؉ 5C3(x2)2(؊3)3 ؉ 5C4(x2)1(؊3)4 ؉ 5C5(؊3)5 (1 ؍x10) ؉ 5(x2)4(؊3) ؉ 10(x2)3(9) ؉ 10(x2)2(؊27) ؉ 5(x2)1(81) ؉ 1(؊243) ؍x10 ؊ 15x8 ؉ 90x6 ؊ 270x4 ؉ 405x2 ؊ 243 26 8.6 The Binomial Theorem—Solutions DO NOT COPY. ©P
3. 3. 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:23 PM Page 27 Home Quit c) (3x - 2)4 (x ؉ y)n ؍nC0xn ؉ nC1xn ؊ 1y ؉ nC2xn ؊ 2y2 ؉ . . . ؉ nCnyn Substitute: n ,4 ؍x 3 ؍x, y 2؊ ؍ (3x ؊ 2)4 4 ؍C0(3x)4 ؉ 4C1(3x)3(؊2) ؉ 4C2(3x)2(؊2)2 ؉ 4C3(3x)(؊2)3 ؉ 4C4(؊2)4 18(1 ؍x4) ؉ 4(27x3)(؊2) ؉ 6(9x2)(4) ؉ 4(3x)(؊8) ؉ 1(16) 18 ؍x4 ؊ 216x3 ؉ 216x2 ؊ 96x ؉ 16 d) (-2 + 2x)4 (x ؉ y)n ؍nC0xn ؉ nC1xn ؊ 1y ؉ nC2xn ؊ 2y2 ؉ . . . ؉ nCnyn Substitute: n ,4 ؍x ,2؊ ؍y 2 ؍x (؊2 ؉ 2x)4 4 ؍C0(؊2)4 ؉ 4C1(؊2)3(2x) ؉ 4C2(؊2)2(2x)2 ؉ 4C3(؊2)(2x)3 ؉ 4C4(2x)4 2()8؊(4 ؉ )61(1 ؍x) ؉ 6(4)(4x2) ؉ 4(؊2)(8x3) ؉ 1(16x4) 46 ؊ 61 ؍x ؉ 96x2 ؊ 64x3 ؉ 16x4 e) (-4 + 3x4)5 (x ؉ y)n ؍nC0xn ؉ nC1xn ؊ 1y ؉ nC2xn ؊ 2y2 ؉ . . . ؉ nCnyn Substitute: n ,5 ؍x ,4؊ ؍y 3 ؍x4 (؊4 ؉ 3x4)5 5 ؍C0(؊4)5 ؉ 5C1(؊4)4(3x4) ؉ 5C2(؊4)3(3x4)2 ؉ 5C3(؊4)2(3x4)3 ؉ 5C4(؊4)(3x4)4 ؉ 5C5(3x4)5 3()652(5 ؉ )4201؊(1 ؍x4) ؉ 10(؊64)(9x8) ؉ 10(16)(27x12) ؉ 5(؊4)(81x16) ؉ 1(243x20) 0483 ؉ 4201؊ ؍x4 ؊ 5760x8 ؉ 4320x12 ؊ 1620x16 ؉ 243x20 8. a) Write the terms in row 7 of Pascal’s triangle. 1, 6, 15, 20, 15, 6, 1 b) Use your answer to part a to write the first 3 terms in each expansion. i) (x - 3)6 The exponent is 6, so the terms in row 7 of Pascal’s triangle are the coefficients of the terms in the expansion of the binomial. So, (x ؉ y)6 1 ؍x6 ؉ 6x5y ؉ 15x4y2 ؉ . . . Substitute: y 3؊ ؍ (x ؊ 3)6 1 ؍x6 ؉ 6x5(؊3) ؉ 15x4(؊3)2 ؉ . . . ؍x6 ؊ 18x5 ؉ 135x4 ؉ . . . So, the first 3 terms in the expansion are: x6 ؊ 18x5 ؉ 135x4 ©P DO NOT COPY. 8.6 The Binomial Theorem—Solutions 27
4. 4. 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:23 PM Page 28 Home Quit ii) (a + 4b)6 (x ؉ y)6 1 ؍x6 ؉ 6x5y ؉ 15x4y2 ؉ . . . Substitute: x ؍a, y 4 ؍b (a ؉ 4b)6 1 ؍a6 ؉ 6a5(4b) ؉ 15a4(4b)2 ؉ . . . ؍a6 ؉ 24a5b ؉ 240a4b2 ؉ . . . So, the first 3 terms in the expansion are: a6 ؉ 24a5b ؉ 240a4b2 iii) (-2a + 1)6 (x ؉ y)6 1 ؍x6 ؉ 6x5y ؉ 15x4y2 ؉ . . . Substitute: x 2؊ ؍a, y 1 ؍ (؊2a ؉ 1)6 2؊(1 ؍a)6 ؉ 6(؊2a)5(1) ؉ 15(؊2a)4(1)2 ؉ . . . 46 ؍a6 ؊ 192a5 ؉ 240a4 ؉ . . . So, the first 3 terms in the expansion are: 64a6 ؊ 192a5 ؉ 240a4 iv) (2x + 5y2)6 (x ؉ y)6 1 ؍x6 ؉ 6x5y ؉ 15x4y2 ؉ . . . Substitute: x 2 ؍x, y 5 ؍y2 (2x ؉ 5y2)6 2(1 ؍x)6 ؉ 6(2x)5(5y2) ؉ 15(2x)4(5y2)2 ؉ . . . 46 ؍x6 ؉ 960x5y2 ؉ 6000x4y4 ؉ . . . So, the first 3 terms in the expansion are: 64x6 ؉ 960x5y2 ؉ 6000x4y4 9. Determine the coefficient of each term. a) x5 in (x + 1)8 b) x9 in (x + y)9 x5 is the 4th term in (x ؉ 1)8. x9 is the first term in the expansion The coefficient of the 4th term is: of (x ؉ y)9. 8C4؊ 1 or 8C3 So, the coefficient of x9 is 1. 8C3 65 ؍ So, the coefficient of x5 is 56. c) x2y in (x + y)3 d) x2y3 in (x + y)5 The coefficients of the terms The coefficients of the terms will be the terms in row 4 of will be the terms in row 6 of Pascal’s triangle: 1, 3, 3, 1 Pascal’s triangle: 1, 5, 10, 10, 5, 1 x2y is the 2nd term in the x2y3 is the 4th term in the expansion of (x ؉ y)3. expansion of (x ؉ y)5. So, the coefficient of x2y is 3. So, the coefficient of x2y3 is 10. 28 8.6 The Binomial Theorem—Solutions DO NOT COPY. ©P
5. 5. 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:24 PM Page 29 Home Quit 10. Explain why the coefficients of the 3rd term and the 3rd-last term in the expansion of (x + y)n, n ≥ 2, are the same. Since each of x and y has coefficient 1, the coefficients of the terms in the expansion of (x ؉ y)n correspond to the terms in row (n ؉ 1) of Pascal’s triangle. In any row of Pascal’s triangle, the 3rd term and 3rd-last term are the same. 11. Determine the indicated term in each expansion. a) the last term in (3x + 2)5 b) the 1st term in (-2x + 5)7 The last term in the expansion The first term in the expansion of (x ؉ y)n is yn. of (x ؉ y)n is xn. Substitute: y ,2 ؍n 5 ؍ Substitute: x 2؊ ؍x, n 7 ؍ 25 23 ؍ (؊2x)7 821؊ ؍x7 So, the last term is 32. So, the 1st term is ؊128x7. c) the 2nd term in (3x - 3)4 d) the 6th term in (4x + 1)8 The second term in the expansion The kth term is: nCk ؊ 1xn ؊ (k ؊ 1)yk ؊ 1 of (x ؉ y)n is nxn ؊ 1y. Substitute: n ,8 ؍k ,6 ؍x 4 ؍x, Substitute: x 3 ؍x, y ,3؊ ؍ y1؍ n4؍ C (4x)3(1)5 46(65 ؍x3)(1) 8 5 4(3x)3(؊3) 423؊ ؍x3 4853 ؍x3 So, the 2nd term is ؊324x3. So, the 6th term is 3584x3. 12. When will the coefficients of the terms in the expansion of (ax + b)n be the same as the terms in row (n + 1) of Pascal’s triangle? The coefficients of the terms in the expansion of (x ؉ y)n correspond to the terms in row (n ؉ 1) of Pascal’s triangle. So, both a and b must equal 1. 13. Expand and simplify (x + 1)8 + (x - 1)8. What strategy did you use? The coefficients of the terms in the expansion of (x ؉ 1)8 are the terms in row 9 of Pascal’s triangle: 1, 8, 28, 56, 70, 56, 28, 8, 1 The coefficients of the terms in the expansion of (x ؊ 1)8 depend on whether the term number is odd or even. When the term numbers are odd, the coefficients are the terms in row 9 of Pascal’s triangle. When the term numbers are even, the coefficients are the opposites of the terms in row 9. So, the coefficients of the terms in the expansion of (x ؊ 1)8 are: 1, ؊8, 28, ؊56, 70, ؊56, 28, ؊8, 1 When the terms in the two expansions are added, every second term is eliminated as the sum of the terms is 0. (x ؉ 1)8 ؉ (x ؊ 1)8 2 ؍x8 ؉ 56x6 ؉ 140x4 ؉ 56x2 ؉ 2 ©P DO NOT COPY. 8.6 The Binomial Theorem—Solutions 29
6. 6. 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:24 PM Page 30 Home Quit 14. a) Show that the expansion of (-2x + 1)6 is the same as the expansion of (2x - 1)6. Use row 7 of Pascal’s triangle: 1, 6, 15, 20, 15, 6, 1 (؊2x ؉ 1)6 2؊(1 ؍x)6 ؉ 6(؊2x)5(1) ؉ 15( ؊2x)4(1)2 ؉ 20( ؊2x)3(1)3 ؉ 15(؊2x)2(1)4 ؉ 6(؊2x)1(1)5 ؉ (1)6 46 ؍x6 ؉ 6(؊32x5) ؉ 15(16x4) ؉ 20(؊8x3) ؉ 15(4x2) ؉ 6(؊2x) ؉ 1 46 ؍x6 ؊ 192x5 ؉ 240x4 ؊ 160x3 ؉ 60x2 ؊ 12x ؉ 1 (2x ؊ 1)6 2(1 ؍x)6 ؉ 6(2x)5(؊1) ؉ 15(2x)4(؊1)2 ؉ 20(2x)3(؊1)3 ؉ 15(2x)2(؊1)4 ؉ 6(2x)1(؊1)5 ؉ (؊1)6 46 ؍x6 ؉ 6(32x5)(؊1) ؉ 15(16x4) ؉ 20(8x3)(؊1) ؉ 15(4x2) ؉ 6(2x)(؊1) ؉ 1 46 ؍x6 ؊ 192x5 ؉ 240x4 ؊ 160x3 ؉ 60x2 ؊ 12x ؉ 1 b) Will (-ax + b)n always have the same expansion as (ax - b)n? Explain. No, when n is odd, the expansions will not be the same. For example, look at the first terms in the expansions of (؊2x ؉ 1)3 and (2x ؊ 1)3. For (؊2x ؉ 1)3: the 1st term in the expansion is: 1(؊2x)3 8؊ ؍x3 For (2x ؊ 1)3: the 1st term in the expansion is: 1(2x)3 8 ؍x3 Since the 1st terms are different, the expansions are not the same. 15. Which binomial power when expanded results in 16x4 - 32x3 + 24x2 - 8x + 1? What strategy did you use to find out? The first term is 16x 4. The exponent of x is 4, so the binomial has the form (ax ؉ b)4. √ The coefficient a must be 4 16: a ,2 ؍or a 2؊ ؍ The last term is 1, so b is either 1 or ؊1 because (؊1)4 ,41 ؍or 1. The terms alternate in sign, so the coefficients a and b must have different signs. The binomial power is either (؊2x ؉ 1)4 or (2x ؊ 1)4. 16. Expand using the binomial theorem. a) (0.2x - 1.2y)5 (x ؉ y)n ؍nC0xn ؉ nC1xn ؊ 1y ؉ nC2xn ؊ 2y2 ؉ . . . ؉ nCnyn Substitute: n ,5 ؍x 2.0 ؍x, y 2.1؊ ؍y (0.2x ؊ 1.2y)5 5 ؍C0(0.2x)5 ؉ 5C1(0.2x)4(؊1.2y) ؉ 5C2(0.2x)3(؊1.2y)2 ؉ 5C3(0.2x)2(؊1.2y)3 ؉ 5C4(0.2x)(؊1.2y)4 ؉ 5C5(؊1.2y)5 23 000.0(1 ؍x5) ؉ 5(0.0016x4)(؊1.2y) ؉ 10(0.008x3)(1.44y2) ؉ 10(0.04x2)(؊1.728y3) ؉ 5(0.2x)(2.0736y4) ؉ 1(؊2.488 32y5) 23 000.0 ؍x5 ؊ 0.0096x4y ؉ 0.1152x3y2 ؊ 0.6912x2y3 ؉ 2.0736xy4 ؊ 2.488 32y5 30 8.6 The Binomial Theorem—Solutions DO NOT COPY. ©P
7. 7. 10_ch08_pre-calculas12_wncp_solution.qxd 5/23/12 4:24 PM Page 31 Home Quit 4 b) a8a + 6bb 3 1 (x ؉ y)n ؍nC0xn ؉ nC1xn ؊ 1y ؉ nC2xn ؊ 2y2 ؉ . . . ؉ nCnyn 3 1 Substitute: n ,4 ؍x ؍a, y ؍b 8 6 4 4 3 2 2 a a ؉ bb 4 ؍C0 a ab ؉ 4C1 a ab a bb ؉ 4C2 a ab a bb 3 1 3 3 1 3 1 8 6 8 8 6 8 6 3 4 ؉ 4C3 a ab a bb ؉ 4C4 a bb 3 1 1 8 6 6 1 ؍a a b ؉ 4 a a3b a bb ؉ 6 a a2b a b2b 81 4 27 1 9 1 4096 512 6 64 36 ؉ 4 a ab a b b ؉ 1a b4b 3 1 3 1 8 216 1296 81 4 27 3 9 2 2 3 1 4 ؍ a ؉ ab؉ ab ؉ ab3 ؉ b 4096 768 384 432 1296 81 4 9 3 3 2 2 1 1 4 ؍ a ؉ ab؉ ab ؉ ab3 ؉ b 4096 256 128 144 1296 C 17. Determine the 3rd term in the expansion of (x2 + 2x + 1)6. (x2 ؉ 2x ؉ 1) ( ؍x ؉ 1)2 So, (x2 ؉ 2x ؉ 1)6 (( ؍x ؉ 1)2)6 ( ؍x ؉ 1)12 The coefficients in the expansion of (x ؉ 1)12 are the terms in row 13 of Pascal’s triangle. The 3rd term in row 13 is 66. So, the 3rd term in the expansion of (x2 ؉ 2x ؉ 1)6 is: 66(x)10(1)2, or 66x10 18. a) Show that nC0 + nC1 + nC2 + . . . + nCn - 1 + nCn = 2n Express 2n as the binomial (1 ؉ 1)n and expand: (1 ؉ 1)n ؍nC0(1)n ؉ nC1(1)n ؊ 1(1) ؉ nC2(1)n ؊ 2(1)2 ؉ . . . ؉ nCn ؊ 1(1)(1)n ؊ 1 ؉ nCn(1)n So, 2 ؍nC0 ؉ nC1 ؉ nC2 ؉ . . . ؉ nCn ؊ 1 ؉ nCn n b) What does the relationship in part a indicate about the sum of the terms in any row of Pascal’s triangle? nC0 , nC1, nC2, . . . nCn ؊ 1, nCn are the terms in row (n ؉ 1) of the triangle. From part a, the sum of these terms is 2n. So, the sum of the terms in any row of Pascal’s triangle is a power of 2. ©P DO NOT COPY. 8.6 The Binomial Theorem—Solutions 31 |
# Math 3070-1, Fall 2009 Solutions to Homework 7
```Math 3070-1, Fall 2009
Solutions to Homework 7
6.2. (a) Remember that independent normals add to a normal random variable; their means add; and their variances add. Therefore,
D
T = N (500 , 121).
(b) We are asked to compute P {T > 520}. So we transform 520 grams
into standard units:
520 − 500
√
= 1.81818181818;
121
The answer is the area to the right of 1.81818181818 under the
standard-normal curve; i.e.,
P {T > 520} ' 1 − 0.9656 = 0.0344.
(c) We expect each toy to weigh 500 grams; therefore, 24 toys weigh
a total of 24 × 500 = 12, 000 grams [or 12 kilograms]. This and
the 200 grams [for the box] together yield a total weight of 12, 200
grams [or 12.2 kilos]. The variances add: Each toy has variance 121
grams; therefore the total variance—or give and take squared—is
24 × 121 = 2, 904 grams squared [or 2.904 kilogram squared].
(d) The total weight of the box is 200 + the sum of 24 independent
normals. Therefore, its distribution is N (12200 , 2904). In particular,
the probability that the total weight is more than 12300 grams is the
probability that a standard normal is at least
12300 − 12200
√
' 1.86
2904
that probability is 1 − 0.9686 = 0.0314 [the answer!!].
6.4. (a) Since X̄ is N (2 , 0.04),
2.1 − 2
1.9 − 2
√
< N (0 , 1) < √
0.04
0.04
= P {−0.1 < N (0 , 1) < 0.1}
' 0.08.
P 1.9 < X̄ < 2.1 = P
(b) Since X̄ is distributed as N (2 , 4/n),
)
(
1.9 − 2
2.1 − 2
P 1.9 < X̄ < 2.1 = P p
< N (0 , 1) < p
4/n
4/n
√
√ = P −0.05 n < N (0 , 1) < 0.05 n .
1
√
In order for this probability to be 0.9, the area to the left of 0.05 n—
under the N√
(0 , 1) distribution—must be 0.95. The normal table tells
us that 0.05 n = 1.645. Solve to obtain:
n=
1.645
0.05
2
' 1, 082.41.
Therefore, n has to be around 1082 or 1083.
6.12. You need to draw a picture [as discussed in the lectures] in order to follow
this.
(a) We want to use a χ210 table; the area to the left of a is 0.05; and the
area to the left of b is 0.95. Therefore, a = 3.94 and b = 18.307.
(b) We want to use a χ215 table; the area to the left of a is 0.05; and the
area to the left of b is 0.95. Therefore, a = 7.261 and b = 24.996.
(c) We want to use a χ210 table; the area to the left of a is 0.025; and the
area to the left of b is 0.975. Therefore, a = 3.247 and b = 20.483.
(d) We want to use a χ220 table; the area to the left of a is 0.025; and the
area to the left of b is 0.975. Therefore, a = 9.591 and b = 34.170.
6.16. According to Theorem 6.4, (n − 1)S 2 /σ 2 is distributed according to χ2n−1 .
(a) Therefore,
13 × 0.00272484
(n − 1)S 2
<
σ2
0.001764
2
= P χ12 < 20.081 ' 0.95.
13 × 0.00309136
(n − 1)S 2
>
2
σ
0.001764
= P χ212 > 22.782 ,
P {S < 0.0522} ' P S 2 < 0.003 = P
(b) Therefore,
2
P {S > 0.0556} ' P S > 0.003 = P
and this is between 0.05 and 0.025.
6.20. (a) F3,20 (0.05) = 3.10.
(b) F3,20 (0.01) = 4.94.
(c) F4,30 (0.05) = 2.69.
(d) F4,30 (0.01) = 4.02.
2
``` |
# MATHS PROJECT-COMPLEX NUMBERS
Document Sample
``` Maths project
SWATHI.L.B
11 E
K.V.PATTOM
COMPLEX
NUMBERS AND
EQUATIONS
Introduction
We know that the equation x2 + 1 = 0 has no real
solution as x2 + 1 = 0 gives x2 = – 1 and square of every
real number is non-negative. So, we need to extend the
real number system to a larger system so that we can
find the solution of the equation x2 = – 1. In fact, the main
objective is to solve the equation ax2 + bx + c = 0, where
D = b2 – 4ac < 0, which is not possible in the system of
real numbers.
Complex Numbers
Let us denote −1 by the symbol i. Then, we have i2 = −1 . This means
that i is a
solution of the equation x2 + 1 = 0.
A number of the form a + ib, where a and b are real numbers, is defined
to be a
complex number. For example, 2 + i3, (– 1) + i 3 ,
Is a complex numbers.
For the complex number z = a + ib, a is called the real part, denoted by
Re z and
b is called the imaginary part denoted by Im z of the complex number z.
For example,
if z = 2 + i5, then Re z = 2 and Im z = 5.
Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and
b = d.
Algebra of Complex Numbers
In this Section, we shall develop the algebra of complex numbers.
Addition of two complex numbers Let z1 = a + ib and z2 = c + id
be any two
complex numbers. Then, the sum z1 + z2 is defined as follows:
z1 + z2 = (a + c) + i (b + d), which is again a complex number.
For example, (2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8
The addition of complex numbers satisfy the following properties:
(i) The closure law The sum of two complex numbers is a complex
number, i.e., z1 + z2 is a complex number for all complex numbers
z1 and z2.
(ii) The commutative law For any two complex numbers z1 and z2,
z1 + z2 = z2
+ z1
Difference of two complex numbers Given any two complex
numbers z1 and
z2, the difference z1 – z2 is defined as follows:
z1 – z2 = z1 + (– z2).
For example, (6 + 3i) – (2 – i) = (6 + 3i) + (– 2 + i ) = 4 + 4i
and (2 – i) – (6 + 3i) = (2 – i) + ( – 6 – 3i) = – 4 – 4i
COMPLEX NUMBERS AND QUADRATIC EQUATIONS 99
Multiplication of two complex numbers Let z1 = a + ib and z2 = c
+ id be any
two complex numbers. Then, the product z1 z2 is defined as
follows:
z1 z2 = (ac – bd) + i(ad + bc)
For example, (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = –
24 + i28
The multiplication of complex numbers possesses the following
properties, which
we state without proofs.
(i) The closure law The product of two complex numbers is a complex
number,
the product z1 z2 is a complex number for all complex numbers z1 and
z2.
(ii) The commutative law For any two complex numbers z1 and z2,
z1 z2 = z2 z1
.
(iii) The associative law For any three complex numbers z1, z2, z3,
(z1 z2) z3 = z1 (z2 z3).
(iv) The existence of multiplicative identity There exists the complex
number
1 + i 0 (denoted as 1), called the multiplicative identity such that z.1 = z,
for every complex number z.
(v) The existence of multiplicative inverse For every non-zero complex
number z = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number
The square roots of a negative real number
Note that i2 = –1 and ( – i)2 = i2 = – 1
Therefore, the square roots of – 1 are i, – i. However, by the symbol
− , we would
mean i only.
Now, we can see that i and –i both are the solutions of the equation
x2 + 1 = 0 or
x2 = –1.
Similarly ( ) ( ) 2 2
3i = 3 i2 = 3 (– 1) = – 3
( )2
− 3i = ( )2
− 3 i2 = – 3
Therefore, the square roots of –3 are 3 i and − 3i .
Again, the symbol −3 is meant to represent 3 i only, i.e., −3 = 3 i .
Generally, if a is a positive real number, −a = a −1 = a i ,
We already know that a× b = ab for all positive real number a and b.
This
result also holds true when either a > 0, b < 0 or a < 0, b > 0. What if
a < 0, b < 0?
Let us examine.
Note that
COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101
i2 = −1 −1= (−1) (−1) (by assuming a× b = ab for all real numbers)
= 1 = 1, which is a contradiction to the fact that i = − .
Therefore, a× b≠ ab if both a and b are negative real numbers.
Further, if any of a and b is zero, then, clearly, a× b= ab= 0.
5.3.7 Identities We prove the following identity
( )2 2 2
z1+z2 =z1+z2+2z1z2, for all complex numbers z1 and z2.
The Modulus and the Conjugate of a Complex Number
Let z = a + ib be a complex number. Then, the modulus of z,
denoted by | z |, is defined
to be the non-negative real number a2+b2, i.e., | z | = a2+b2
and the conjugate
of z, denoted as z , is the complex number a – ib, i.e., z = a –
ib.
For example, 3+i =32+12=10, 2−5i = 22+(−5)2= 29 ,
and 3+i=3−i, 2−5i=2+5i, −3i −5 = 3i – 5
Summary
���� A number of the form a + ib, where a and b are real numbers, is
called a
complex number, a is called the real part and b is called the
imaginary part
of the complex number.
���� Let z1 = a + ib and z2 = c + id. Then
(i) z1 + z2 = (a + c) + i (b + d)
(ii) z1 z2 = (ac – bd) + i (ad + bc)
���� For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there
exists the
complex number 2 2 2 2
���� For any integer k, i4k = 1, i4k + 1 = i, i4k + 2 = – 1, i4k + 3 = – i
���� The conjugate of the complex number z = a + ib, denoted by z , is given b
z = a – ib.
���� The polar form of the complex number z = x + iy is r (cosè + i sinè), wher
r = x2+ y2 (the modulus of z) and cosè =
x
r , sinè =
y
r . (è is known as the
argument of z. The value of è, such that – ð < è ≤ ð, is called the principal
argument of z.
SUMS
• Convert the given complex number in
polar form: –3
• Discussion
• –3
• Let r cos θ = –3 and r sin θ = 0
• On squaring and adding, we obtain
Convert the given complex number in polar form: –3
Discussion
3Convert the given complex number in polar form: –3
–
Discussion
–3
Let r cos θ = –3 and r sin θ = 0
On squaring and adding, we obtain
Let r cos θ = –3 and r sin θ = 0
On squaring and adding, we obtain
Historical Note
The fact that square root of a negative number
does not exist in the real number
system was recognised by the Greeks. But the
credit goes to the Indian
mathematician Mahavira (850 A.D.) who first
stated this difficulty clearly. “He
mentions in his work ‘Ganitasara Sangraha’ as in
the nature of things a negative
(quantity) is not a square (quantity)’, it has,
another I
Indian mathematician, also writes in his work Bijaganita,
written in
1150. A.D. “There is no square root of a negative quantity,
for it is not a
square.” Cardan (1545 A.D.) considered the problem of
solving
x + y = 10, xy = 40.
He obtained x = 5 + −15 and y = 5 – −15 as the solution of
it, which
was discarded by him by saying that these numbers are
‘useless’. Albert Girard
(about 1625 A.D.) accepted square root of negative
numbers and said that this
will enable us to get as many roots as the
degree of the polynomial equation.
Euler was the first to introduce the
symbol i for −1 and W.R. Hamilton
(about 1830 A.D.) regarded the complex
number a + ib as an ordered pair of
real numbers (a, b) thus giving it a purely
mathematical definition and avoiding
use of the so called ‘imaginary numbers’.
```
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1.06 Multiplication and division by 10 and 100
Lesson
Can you break a number into its parts , using place value?
Examples
Example 1
Complete the number expander for 27\,692.
a
Worked Solution
Create a strategy
Put the given number in a place value table.
Apply the idea
The complete number expander is:
b
Write 27\,692 as a number sentence.
⬚+⬚+⬚+90+2
Worked Solution
Create a strategy
Use the number expander from part (a).
Apply the idea
2 ten thousands is 20\,000, 7 thousands is 7000, and 6 hundreds is 600.
20\,000+7000+600+90+2
Idea summary
We can break a number into its parts by putting it in a place value table.
Multiply or divide by 10 and 100
The digits in our number tell us how many units, tens, hundreds or thousands (or more) we have. Remembering that each place is ten times larger than the place to its right can help us when we need to multiply or divide by 10.
Examples
Example 2
Fill in the boxes with the missing numbers.
6600\div⬚=66
Worked Solution
Create a strategy
Put the numbers in a place value table and see how many places the digits have moved.
Apply the idea
Put 6600 and 66 in a place value table:
The 6s in 6600 have moved 2 places to the right to get 66.
This is the same as dividing by 100.
6600 \div 100 = 66
Idea summary
When we multiply by 10, the digits move to the left one place, and our number gets bigger. We need to put a zero placeholder in the units place when we multiply by 10. When we divide by 10, our digits move to the right one place, and our number gets smaller.
When we multiply by 100, the digits move to the left two places, and our number gets bigger. We need to put zeros as placeholders in the tens and units places when we multiply by 100. When we divide by 100, our digits move to the right two places, and our number gets smaller. |
## Solving An Equation
### 1. The Question
“Take a look at the paper in front of you,” they said.
I looked down and got my pencil out of my pocket. (In retrospect, I wonder if any of them noticed that instinctive reach for the pencil. There were generations of us raised with pencils as necessary appendages to our brains, something that I suspect is not true today.)
It was a simple linear equation with one unknown variable.
“Can you tell us how you would teach a freshman with a seventh grade understanding how to solve for the variable?”
I started with, “The first thing I would do is ask them if they understand what it means to say there is an unknown variable in this equation.”
I explained how in my experience for many students (not only those with only seventh grade understanding) this thing about letters in math is baffling. They just haven’t fully internalized what is going on with the letters when math is after all about numbers.
They seemed to like this, because they all wrote down some notes at that point.
My answer that followed was probably good. But of course it could have been better. Here is that better answer.
### 2. Understanding the Equation
Let’s give the student a name: Pat.
I sit down at a table with Pat and write the equation out on a piece of paper:
10a + 1 = 46.
“Here’s an equation,” I say. “Pat, do you understand what I mean when I call this an equation?” We talk.
“And look at this darned thing. This a. Do you understand what that a is? It’s called an unknown variable. Do you understand what that means?” Pat says no.
“Right,” I say. “You are not alone. It almost never gets explained well. And it’s important to understand this, believe it or not. So let’s talk about that before we go further.”
“Let’s think of a as a box. It has something inside it, but we don’t know what that something is, because the lid of the box is closed. Our purpose in life at this point is to discover what’s in the box. Is a holding a 5? Is a holding a 43? We don’t know. But we’re on a mission to figure that out. We’re on a mission to open the box (to discover what a holds) using … math.”
“By the way, a is just a name of the box,” I point out. Someone working the same problem, looking at the same equation, trying to figure out what’s in the box might write the problem using x.”
10x + 1 = 46.
“So don’t get hung up on the specific letter. It’s a name for the box which could be x or b or z or … Fred.”
With that in place — explaining the problem we’re trying to solve in terms of this black-box thing we’re calling a, I point back at the equation.
“So let’s come back to this darned equation on the paper, here,” I say. “What does this whole thing mean?”
I look at Pat. Of course math is taught in a way that virtually no one knows how to answer that question meaningfully. I tell Pat that.
“Here’s what that equation means,” I say. “Someone has told us something about the box. They’ve given us a hint. The equation is the hint. It doesn’t tell us what’s in the box, but they promise us that the hint is a true statement about the box, something about a, something that we can rely on with 100% accuracy.”
I look at Pat.
“I know you don’t know why I’m telling you that.” I smile. “But do you hear what I’m saying. Our friend has told us that even though they won’t tell us what’s in the box, they promise that the following statement is true.”
10a + 1 = 46
I think Pat would nod at that point. And I would say, “Great! Let’s get to work!”
### 3. Matter and Anti-Matter
“So,” I say, kind of like we’re starting over. “What we know is this…” And I circle the equation dramatically.
“What we know with 100% certainty is that this equation is true. We know without a doubt that the left side of that equation equals the right side.”
“And here is the thing — the key to solving this problem. We already know enough about math to force this true statement that looks kinda useless into telling us what’s in the box.”
So I circle the equation again a few times. I point out that this true statement isn’t doing us any good. Things are jumbled up. Sure, a is in there, but it’s mixed up with other numbers. It’s multiplied by 10. 1 is added to it. It’s all a mess.
A mess? I ask, rhetorically. So let’s clean up the mess!
I ask if Pat knows the opposite of +1. If not, I guide Pat into understanding that if you add -1 to +1 you get zero.
“It’s like matter and anti-matter!” I say. “You add -1 to +1 and they both evaporate into zero… into nothingness!”
“But remember, this true equation has two sides. Whatever we do to the left side, we must do to the right side to keep everything in balance.”
And I look at Pat. “Because if you don’t, you turn the equation (the only thing we know about a) into something bogus.”
### 4. Making Things Less Messy
“Ok. So we’ve got this messy +1 sitting there on the left side of the equation. Let’s add -1 to it. And to the other side of the equation so that things stay balanced.”
I underline the left side and write -1 below it. And I underline the right side (looking at Pat and emphasizing that we have to do it on both sides), and I write -1 under it.
Here’s what we get:
10a + 1 – 1 = 46 – 1
I get Pat to help me do the math. I use a red pencil to draw parentheses around the places where we’re doing our cleanup.
10a + (1 – 1) = (46 – 1)
“Since we did the same thing to both sides, we know this is still a 100% true statement,” I say.
I get Pat’s help simplifying just that red stuff, first on the left and then on the right. Here’s what we get.
10a + (0) = (45)
“And all we did was clean things up,” I say. “So this is also a 100% true statement.”
And in a dramatic flourish, I say “Poof! Matter/antimatter! Zero! Nothingness!” And I rewrite it like this.
10a = 45
“And this is also a 100% true statement, but look at it. Now we’re talkin’!” I say. “The mess looks … less messy, doesn’t it?”
### 5. Even Less Messy
Then I run through the matter/antimatter thing again with the 10. But I point out that 10 is multiplying a, so we can’t just add some antimatter to get rid of it.
Here, I ask a leading question. “What do you get if you divide 10 by 10?” I’m expecting that a seventh grader can tell me that — that Pat can give me the answer I’m looking for.
“Right. 10 divided by 10 is 1. It’s a different way to clean things up. You’ll see. Let’s divide by 10.”
Wait! I look at Pat. If we do something to the left, we have to do it to the right, too. So I ask what we need to do to the right. (Right. Divide by 10.)
I underline the left and write a divide-by-10 annotation. I look over at Pat, “And over here?” I ask, pointing to the right side. “Right. Divide-by-10 on both sides.”
Here’s what we get:
10a / 10 = 45 / 10
I make the this-is-100%-true point yet again.
“Let’s rearrange this a bit,” I suggest.”We’re just rearranging, so this will still be 100% true.”
(10/10)a = (45/10)
With Pat’s help, we do the math.
(1)a = (4.5)
“And since we just cleaned things up, this is a 100% true statement. But look: Poof!” I say. A different kind of matter/antimatter.” And so we get this.
a = 4.5
“Is this statement still 100% true?” I ask. “Yes. As long as we kept things in balance.”
“So now you tell me: what is the value of a?”
We did it! We found what’s in the box!
### 6. Practice
At this point, I hope Pat understands the general idea. But in my opinion, general ideas like this need to be hammered in with reinforcement a few times with tiny (Tiny I say!) variations on the same problem.
So I would have Pat work thru some similar problems. To be honest, at this point it would work better if there were two students, so that they could work on these problems together. But Pat’s here alone.
“What if we rename the box?” I ask. “Show me how to solve this problem,” I say, writing down the same equation on a fresh piece of paper with a different name for the unknown variable:
10b + 1 = 46
“What if it’s 45 instead of 46?” I ask, writing down a different equation on a fresh piece of paper. “Show me how to solve this problem.”
10b + 1 = 45
“What if it’s 2 instead of 1?”
10c + 2 = 45
“What if it’s 3 instead of 10?”
3d + 2 = 45
And in that way, I hope to have taught a seventh grader how to solve that kind of linear equation.
For evidence of mastery, I write this equation down and say to Pat, “If you can take what we’ve talked about here and solve this equation, then you have completely mastered this. Try it. Come get me if you need help. I’ll be right over here.”
2x + 1 = -3x + 11 |
# Advice 1: How to find the area of the diagonal section
If both sides of a plane there are points belonging to three-dimensional figures (e.g., polyhedra), this plane can be called a secant. A two-dimensional figure formed by the General points in the plane and a polyhedron in this case is called a cross section. This section will be diagonal, if one of the diagonals of the base belongs to the section plane.
Instruction
1
The diagonal section of the cube has the shape of a rectangle, the area (S) it is easy to calculate, knowing the length of any edge (a) three-dimensional figures. In this rectangle one of the sides is the height, which coincides with the length of the edge. The length of the other diagonal - calculate the Pythagorean theorem for the triangle in which it is the hypotenuse, and the two edges of the base - the legs. In General it can be written as: a*√2. The area of the diagonal section find by multiplying the two sides, the length of which you found out: S = a*a*√2 = a2*√2. For example, if the edge length of 20 cm the area of the diagonal section of the cube should be approximately equal 202*√2 ≈ 565,686 cm2.
2
To compute the area of the diagonal cross section of the parallelepiped (S) proceed the same way, but note that the Pythagorean theorem in this case involved the legs different lengths - length (l) and width (w) volumetric shapes. The length of the diagonal in this case is equal to √(l2+w2). Height (h) may also vary from the edge lengths of the bases, so in General the formula for the area of the cross section can be written as: S = h*√(l2+w2). For example, if the length, width and height of a cuboid are equal, respectively, 10, 20 and 30 cm square of its diagonal section approximately 30*√(102+202) = 30*√500 ≈ 670,82 cm2.
3
The diagonal cross-section of the quadrangular pyramid has a triangular shape. If the height (H) of the polytope are known, and at its base lies a rectangle, the lengths of connected edges (a and b) which is also given in the conditions, the calculation of the area of the cross section (S) begin with computing the length of a diagonal of the base. As in previous steps, use the triangle of the two edges of the base and diagonal, where the Pythagorean theorem the length of the hypotenuse is equal to √(a2+b2). The height of the pyramid in this polyhedron coincides with the height of the triangle diagonal cross-section, is lowered to the side whose length you just determined. So to find the area of a triangle, find half of the product height on the length of the diagonal: S = ½*H*√(a2+b2). For example, at a height of 30 cm and lengths of adjacent sides of the base 40 and 50 cm the area of the diagonal section should be approximately equal to ½ *30*√(402+502) = 15*√4100 ≈ 960,47 cm2.
# Advice 2: How to find the area of a pyramid
Pyramid - complex geometrical body. It is formed of a flat polygon (the base of the pyramid), a point not lying in the plane of the polygon (the top of the pyramid) and all segments that connect the base of the pyramid to the vertex. How to find the area of the pyramid?
You will need
• ruler, pencil and paper
Instruction
1
The lateral surface area of any pyramid is equal to the sum of the areas of the lateral faces.
Because all the lateral faces of a pyramid are triangles, it is necessary to find the sum of the areas of all these triangles. The area of a triangle is calculated by multiplying the length of the base of the triangle to the length of its height.
2
The base of the pyramid is a polygon. If this polygon be divided into triangles, the area of the polygon simple to compute as the sum of the squares poluchivshimsya when dividing triangles by the already known formula.
3
Finding the sum of the lateral surface area of a pyramid and the base of the pyramid, you can find the total surface area of the pyramid.
4
For calculations of area of regular pyramid using a special formula.
Example:
Before us is a pyramid. At the base is a regular n-gon with side a. The height of the side faces h (by the way, is the name apofema pyramid). The area of each lateral face is 1/2ah. The entire side surface of the pyramid has an area of n/2ha, calculated by summing the squares of the side faces. na is the perimeter of the base of the pyramid. The area of this pyramid we find: a work of apogamy of the pyramid and one half of the perimeter of its base equal to the area of lateral surface of regular pyramid.
5
With regard to the area of the entire surface, then just add to the side area of the base, according to the principle discussed above.
# Advice 3: How to find the edge length of a pyramid
The pyramid is a figure that has a base in the form of a polygon and the lateral faces converging at the top with peaks. The boundaries of the lateral faces are called edges. And how to find the length of the edges of the pyramid?
Instruction
1
Find the boundary points of the ribs, the length of which are looking for. Let it be points A and B.
2
Set the coordinates of the points A and B. They need to ask three-dimensional, because the pyramid three – dimensional figure. Get A(x1, Y1, z1) and B(x2, y2, z2).
3
Calculate the needed length, using the General formula: edge length of a pyramid is equal to square root of the sum of squares of differences of corresponding coordinates of boundary points. Substitute the numbers for your coordinates into the formula and find the length of the edges of the pyramid. In the same way, find the length of the ribs is not only the right of the pyramid, but rectangular, and truncated and arbitrary.
4
Find the length of the edges of the pyramid, in which all edges are equal, set the base figure and known height. Determine the location of the base height, i.e. the lower point. Since edges are equal, then it is possible to draw a circle whose center is the intersection point of the diagonals of the base.
5
Draw straight lines connecting the opposite corners of the base of the pyramid. Mark the point where they intersect. This point will be the lower bound of the height of the pyramid.
6
Find the length of diagonal of a rectangle using the Pythagorean theorem, where the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Get A2+b2=c2, where a and b are the legs and C is the hypotenuse. The hypotenuse will then be equal to the square root of the sum of the squares of the legs.
7
Find the length of the edges of the pyramid. First divide the length of the diagonal in half. All the data, substitute values in the formula of Pythagoras, are described above. Similar to the previous example, find the square root of the sum of the squares of the height of the pyramid and the half-diagonal.
# Advice 4: How to find the area of the bases of the pyramid
Two grounds can only be a truncated pyramid. In this case, the second base is formed with a cross-section parallel to the larger base of the pyramid. Find one reason in that case, if you know the area or the linear elements of the second.
You will need
• properties of the pyramid;
• - trigonometric functions;
• the similarity of figures;
• - finding the areas of polygons.
Instruction
1
The area of the larger base of the pyramid is the area of the polygon that represents it. If this is the right pyramid, its base lies a regular polygon. To know his areaenough to know the only one of its sides.
2
If the large base is a right triangle, find its areaby multiplying the square of side square root of 3 divided by 4. If the base is a square, erect his side in the second degree. In General, for any regular polygon, use the formula S=(n/4)•a2•ctg(180º/n), where n is the number of sides of a regular polygon, a is the length of its side.
3
The side of the smaller base of the find, according to the formula b=2•(a/(2•tg(180º/n))-h/tg(α))•tg(180 ° /n). Here the a – side of the larger base, h – height of the truncated pyramid, α is the dihedral angle at its base, n is the number of sides of the bases (it is the same). The area of the second base look similar to the first, using the formula the length of its sides S=(n/4)• b2•ctg(180º/n).
4
If the reasons are other types of polygons, all known one reason, and one of the sides of the other, the other hand, calculate as such. For example, the side of larger base 4, 6, 8 see the Big side of the smaller base of the wound is 4 cm. Calculate the coefficient of proportionality, 4/8=2 (take the large side in each of the bases), and calculate the other sides 6/2=3 cm, 4/2=2, see Receive side 2, 3, 4 cm smaller base side. Then, calculate their area, as the areas of triangles.
5
If you know the ratio of the corresponding elements of the truncated pyramid, the ratio of the areas of the bases is equal to the ratio of the squares of these elements. For example, if you know the corresponding sides of the bases a and A1, A2/A12=S/S1.
# Advice 5: How to build a cross-section of a parallelepiped
In many textbooks there are jobs associated with the construction of sections of various geometric shapes including parallelepipeds. To cope with this task, you should be armed with some knowledge.
You will need
• paper;
• - handle;
• - the range.
Instruction
1
On a sheet of paper, draw a box. If your task says that the box needs to be rectangular, make it square corners. Remember that opposite edges must be parallel to each other. Name its vertices, for example, S1, T1, T, R, P, R1, P1 (as shown in the figure).
2
On the verge SS1TT1 put 2 points: A and C, let point a be on the segment S1T1 and a point on the segment S1S. If your task doesn't say where it should be these points, and do not specify the distance from the vertices, put them arbitrarily. Draw a straight line through points A and C. Continue this line to its intersection with the line segment ST. Label the point of intersection, let it be a point of M.
3
Put a point on the segment RT, label it as point B. draw a straight line through the points M and B. the intersection of this line with the edge of the label SP as point K.
4
Connect the points K and C. They must lie on the same face PP1SS1. Then through point B draw a straight line, parallel to the segment of the COP, will continue the line to the intersection with an edge of R1T1. The intersection of the label as point E.
5
Connect the points A and E. then select the resulting polygon ACKBE a different color – this will be a cross section of a given parallelepiped.
Note
Remember that when you build a cross-section of the parallelepiped is possible to connect only those points that lie in the same plane, if you had enough points for the build section, add them by extending the line segments to the intersection with the face on which the desired point.
Only the parallelepiped can be constructed of 4 sections: 2 diagonal and 2 transverse. For clarity, select the resulting polygon cross section, this can just stroke or stroke it in a different color.
# Advice 6: How to find a life-size section
Properties of figures in space involved in this section of geometry, as geometry of space. The basic method for solving problems in solid geometry is a method of cross-section polyhedra. It allows you to correctly build the cross-section of the polyhedra and to determine the species of these sections.
Instruction
1
The definition of the cross-section of any shape, that is, the actual size of this cross-section, often assumed in formulating the tasks of building a sloping section. The oblique section should be called front-secant projective plane. And build it life-size enough to perform several actions.
2
With a ruler and pencil, draw in figure 3 the projections – front view, top view and side view. The main projection in the front view show the path that is front-projecting section plane, draw a sloping line.
3
On an incline direct select main points: points of entry section and exit section. If a figure is a rectangle, the points of entry and exit will be one. If a figure is a prism, then the number of points is doubled. Two points define the entry into and figure out. The other two identify points on the sides of the prism.
4
At an arbitrary distance guide line parallel to the front projecting section plane. Then, from the points located on the axis of the main view, swipe to the auxiliary line perpendicular to the inclined straight line until they intersect with the parallel axis. Thus you will get the projection of the points of the shape in the new coordinate system.
5
To determine the width of the figure, lower straight from the main point of view the figure of top view. Indicate the appropriate indices of the projection points at each intersection of a line and a shape. For example, if point a belongs to the mind of the figure the points A’ and A” belong to projective planes.
6
Put in the new coordinate system the distance that is formed between the vertical projections of the main points. The figure, which is obtained as the result of development, and is a natural value of the inclined section.
# Advice 7: How to find the area of a regular quadrangular pyramid
Pyramid - a polyhedron made up of a certain number having one common vertex flat lateral surfaces and one base. The Foundation, in turn, has on each side a common edge, and therefore its form determines the total number of faces of the shape. In a regular quadrangular pyramid such faces five, but to determine the total surface area is sufficient to calculate areas of only two of them.
Instruction
1
The total surface area of any polyhedron is the sum of the areas of its faces. In a regular quadrangular pyramid there are two forms of the polygons in the base has a square, the sides have a triangular configuration. Start calculations, e.g. calculation of the area of the quadrangular base of the pyramid (Sₒ). By definition, the right of the pyramid at its base must lie a regular polygon, in this case a square. If in terms of the length of an edge of (a), just erect it in the second degree: Sₒ = a2. If you know only the length of the diagonal of the base (l), to compute the area find the half of its square: Sₒ = l2/2.
2
Determine the area of the triangular lateral faces of the pyramid Sₐ. If you know the length of its shared base edge (a) and apofema (h), calculate half of the product of these two quantities: Sₐ = a*h/2. Under these conditions, the lengths of a side edge (b) and fin base (a) find half of a work the length of its base at the root of the difference between the squared length of a side edge and a quarter of square base length: Sₐ = ½*a*√(b2-a2/4). If in addition the length of the common base edge (a) given a plane angle at the vertex of the pyramid (α), calculate the ratio of the squared length of the edge to twice the cosine of half the plane angle: Sₐ = a2/(2*cos(α/2)).
3
Calculating the area of one side face (Sₐ), increase the value four times to calculate the lateral surface area of the regular quadrangular pyramid. At a certain apofema (h) and the perimeter of the base (P) is the action along with all the previous step can be replaced by calculating half the product of these two parameters: 4*Sₐ = ½*h*P. In any case, the lateral surface area will add up to calculated in the first step, with an area of square base figures - this will be the total surface area of a pyramid: S = Sₒ+4*Sₐ.
Search |
A triangle is a three-sided shape whose three inner angles must sum to 180 degrees. The largest angle will be across from the longest side while the smallest angle will be across from the shortest side of the triangle. If and only if two sides of a triangle are equal, the angles opposite them will be equal as well.
You name triangles by their vertices, so a triangle with vertices A, B, and C is designated as ΔABC.
### Properties of Triangles
1. The sum of three interior angles in any triangle is 180 degrees.
2. The sum of any two sides of a triangle must be greater than the third side of a triangle.
3. The longest side of a triangle is opposite the largest angle of a triangle. Conversely, the smallest side of a triangle is opposite the smallest angle of a triangle.
4. Exterior angle of a triangle is equal to sum of two remote interior angles.
### Types of Triangles
You can identify triangle types by the measurements of their sides and angles.
1. Scalene: A scalene triangle has no equal sides and no equal angles.
2. Isosceles: An isosceles triangle has at least two equal sides, and the measures of the angles opposite those two sides are also equal to each other.
3. Equilateral: An equilateral triangle has three sides of equal lengths and three 60-degree angles.
4. Right Triangle: A right triangle has one angle that measures 90 degrees. The side opposite the right angle is called the hypotenuse.
A triangle with two equal sides is called isosceles. The angles opposite the equal sides are called the base angles, and they are congruent (equal). A triangle with all three sides equal is called equilateral, and each angle is 60 degrees. A triangle with no equal sides (and therefore no equal angles) is called scalene.
### Area of a Triangle
Area of a triangle is given by:
A = ½bh
A stands for area, b is the length of the base or bottom of the triangle, and h stands for the height (or altitude), which is the distance that a perpendicular line runs from the base to the angle opposite the base.
### Similar Triangles
Triangles are similar when they have exactly the same angle measures. Similar triangles have the same shape, even though their sides may have different lengths. The corresponding sides of similar triangles are in proportion to each other. The heights of the two triangles are also in proportion.
Two triangles are similar if:
• Corresponding angles are of the same measurement.
• The perimeter of each triangle is in the same ratio as the sides.
• Corresponding sides are all in the same proportion.
To check similarity of triangles, AAA, SAS or SSS rule can be used. |
# What can we learn as we reduce the fraction at the end of the quadratic formula process?
In the final throes of the quadratic formula, you reduce a fraction. Consider the following two examples.
1. $y = 6x^2 + 11x + 3$; the quadratic formula reveals the roots $x = -4/12$ or $x = -18/12$. The parts of the first fraction (4 and 12) have a common factor of 4, so the fraction reduces to -1/3. The parts of the second fraction (18 and 12) have a common factor of 6, so the fraction reduces to -3/2. The common factor of all three parts (4, 12 and 18) is 2.
2. $y = 42x^2 + 77x + 21$; the quadratic formula reveals the roots $x = -28/84$or $x = -126/84$. The parts of the first fraction (28 and 84) have a common factor of 28, so the fraction reduces to -1/3. The parts of the second fraction (126 and 84) have a common factor of 42, so the fraction reduces to -3/2. The common factor of all three parts (28, 84 and 126) is 14.
Now. Compare the two solutions.
Equation 2 is the product of Equation 1 times 7. Apparently as a result, the common factors of the equations were all multiplied by 7, and after reducing the fractions we discover the same roots.
My question is whether that step, reducing the fraction, reveals anything about the numerical factors of the equations.
It seems to me that any common numerical factor of a, b and c (like 7 in Equation 2) will also be a common factor of the parts of the fractions, and cancel when we reduce the fraction. But the parts of the fraction can have other common factors (like 2 in both Equations).
So when we come to that step, of reducing the fraction, does the common factor reveal anything about the numerical common factors of a, b and c?
• In most cases the square root cannot be evaluated to rational values. – user202729 Dec 12 '17 at 13:07
• The common factor in the first example is $2$, not $4$. – B. Goddard Dec 12 '17 at 13:10
• Your question will be easier to read, particularly the exponents, if you use mathjax math.meta.stackexchange.com/questions/5020/… – Lee Mosher Dec 12 '17 at 13:14
Suppose that we have some quadratic $q(x) = Ax^2 + Bx + C$, where $A, B, C$ are integers with a common factor $k$. Since they have a common factor $k$, there exist integers $a, b, c$ such that $A = ak$, $B = bk$, $C = ck$. Applying the quadratic formula to $q(x)$ gives $$\frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{-bk \pm \sqrt{k^2b^2 - 4k^2ac}}{2ka} = \frac{-bk \pm k\sqrt{b^2 - 4ac}}{2ka}$$ and so we can see that the common factor $k$ will definitely show up in both the numerator and denominator of the roots given by the quadratic formula.
However, you may get "false positives" this way. For example, consider $q(x) = x^2 + 2x + 1$. Then, the quadratic formula gives the roots as $$\frac{-2 \pm \sqrt{4 - 4}}{2} = -\frac{2}{2}$$ but $2$ is not a common divisor of $a, b, c$. |
85 Times Table
Greetings, fellow math enthusiasts! Today, we embark on an exciting journey into the world of multiplication, as we delve into the wonders of the 85 times table.
Get ready to uncover captivating patterns, embrace the power of mental math, and appreciate the beauty of numbers in a conversational and engaging manner.
Chapter 1: Introducing the Dynamic 85
Let's begin by acquainting ourselves with the basics. The 85 times table involves multiplying any number by 85.
But why 85, you may ask? Well, each number holds its own intrigue, and 85 is no exception. Let's set forth on our exploration of this multiplication table and discover its hidden secrets.
Chapter 2: Observing the Units Digit Patterns
Let's examine the initial multiples of 85:
• 1 x 85 = 85
• 2 x 85 = 170
• 3 x 85 = 255
At first glance, the pattern may not be immediately apparent. However, if we focus on the units digit of each result, we'll discover something fascinating.
It follows a dynamic pattern that cycles through 5, 0, 5, 0, and so on. Let's take a closer look:
• 85
• 170
• 255
• 340
• 425
• 510
The units digit oscillates between 5 and 0, creating a rhythmic and intriguing sequence.
Chapter 3: Understanding the Dynamic Nature
Now that we've identified the pattern in the units digit, let's explore its significance.
The alternating pattern of 5 and 0 in the units digit reveals that every multiple of 85 will always end with either 5 or 0.
It's an interesting observation that allows us to predict the units digit of any product involving 85.
Chapter 4: Mental Math Techniques
Understanding the patterns in the 85 times table can greatly enhance your mental math skills. Let's consider an example to illustrate this:
Suppose you're faced with the task of calculating 85 multiplied by 6. Instead of following the traditional multiplication steps, you can utilize the pattern we discovered earlier.
Since 6 is a single-digit number, we know that the units digit of the product must be either 5 or 0. In this case, the units digit is 0. Therefore, the answer is 510.
By employing this mental math technique, you can swiftly compute products involving 85, saving time and effort.
Chapter 5: Exploring Greater Multiples
Now, let's explore larger multiples of 85 to see if the pattern persists:
• 14 x 85 = 1,190 (units digit: 0)
• 27 x 85 = 2,295 (units digit: 5)
• 36 x 85 = 3,060 (units digit: 0)
Even with larger numbers, the alternating pattern in the units digit remains consistent. It's truly fascinating to witness the dynamic nature of the 85 times table.
Eighty-five Multiplication Table
Read, Repeat and Learn eighty-five times table and Check yourself by giving a test below
Table of 85
85 Times table Test
How much is 85 multiplied by other numbers?
@2024 PrintableMultiplicationTable.net |
# Beginning Algebra Tutorial 29
Beginning Algebra
Tutorial 29: Negative Exponents and Scientific Notation
Learning Objectives
After completing this tutorial, you should be able to:
1. Simplify exponential expressions involving negative exponents.
2. Write a number in scientific notation.
3. Write a number in standard notation, without exponents.
Introduction
This tutorial picks up where Tutorial 26: Exponents left off. It finishes the rules of exponents with negative exponents. Also we will go over scientific notation. Like it or not, the best way to master these exponents is to work through exponent problems. So I guess we better get to it.
Tutorial
Negative Exponents
or
Be careful with negative exponents. The temptation is to negate the base, which would not be a correct thing to do. Since exponents are another way to write multiplication and the negative is in the exponent, to write it as a positive exponent we do the multiplicative inverse which is to take the reciprocal of the base.
Example 1: Simplify .
*Rewrite with a pos. exp. by taking recip. of base
*Use def. of exponents to evaluate
Example 2: Simplify .
*Rewrite with a pos. exp. by taking recip. of base
*Use def. of exponents to evaluate
Simplifying an Exponential Expression
When simplifying an exponential expression, write it so that each base is written one time with one POSITIVE exponent
In other words, write it in the most condense form you can making sure that all your exponents are positive.
A lot of times you are having to use more than one rule to get the job done. As long as you are using the rule appropriately, you should be fine.
Review of Exponent Rules
Except for the negative exponent rule, examples of the following rules can be found in Tutorial 26: Exponents.
Product Rule:
Power Rule for Exponents:
Power of a Product:
Power of a Quotient:
Quotient Rule for Exponents:
Zero Exponent:
Negative Exponent:
Example 3: Simplify. Write answer with positive exponents.
Example 4: Simplify. Use positive exponents to write the answer.
Example 5: Simplify. Use positive exponents to write the answer.
Example 6: Simplify. Write answer with positive exponents.
Be careful going into the last line. Note that you do not see an exponent written with the number 5. This means that the exponent on 5 is understood to be 1. Since it doesn't have a negative exponent, we DO NOT take the reciprocal of 5. The only base that has a negative exponent is a, so a is the only base we take the reciprocal of.
Scientific Notation
A positive number is written in scientific notation if it is written in the form:
where 1 < a < 10 and r is an integer power of 10.
Writing a Number in Scientific Notation
Step 1: Move the decimal point so that you have a number that is between 1 and 10.
In other words, you will put your decimal after the first non zero number.
Step 2: Count the number of decimal places moved in Step 1 .
If the decimal point was moved to the left, the count is positive.
If the decimal point is moved to the right, the count is negative.
Step 3: Write as a product of the number (found in Step 1) and 10 raised to the power of the count (found in Step 2).
Example 7: Write the number in scientific notation: 483,000,000.
Step 1: Move the decimal point so that you have a number that is between 1 and 10.
*Decimal is at the end of the number
*Move decimal to create a number between 1 and 10
Step 2: Count the number of decimal places moved in Step 1 .
How many decimal places did we end up moving?
We started at the end of the number 483000000 and moved it between the 4 and 8. That looks like a move of 8 places.
What direction did it move?
Looks like we moved it to the left.
So, our count is +8.
Step 3: Write as a product of the number (found in Step 1) and 10 raised to the power of the count (found in Step 2).
Note how the number we started with is a bigger number than the one we are multiplying by in the scientific notation. When that is the case, we will end up with a positive exponent
Example 8: Write the number in scientific notation: .00054.
Step 1: Move the decimal point so that you have a number that is between 1 and 10.
*Decimal is at the beginning of the number
*Move decimal to create a number between 1 and 10
Step 2: Count the number of decimal places moved in Step 1 .
How many decimal places did we end up moving?
We started at the beginning of the number .00054 moved it between the 5 and 4. That looks like a move of 4 places.
What direction did it move?
Looks like we moved it to the right.
So, our count is - 4.
Step 3: Write as a product of the number (found in Step 1) and 10 raised to the power of the count (found in Step 2).
Note how the number we started with is a smaller number than the one we are multiplying by in the scientific notation. When that is the case we will end up with a negative exponent.
Write a Scientific Number in Standard Form
Basically, you just multiply the first number times the power of 10.
Whenever you multiply by a power of 10, in essence what you are doing is moving your decimal place.
If the power on 10 is positive, you move the decimal place that many units to the right.
If the power on 10 is negative, you move the decimal place that many units to the left.
Make sure you add in any zeros that are needed
Example 9: Write the number in standard notation, without exponents.
*Move the decimal 6 to the right
Example 10: Write the number in standard notation, without exponents.
*Move the decimal 5 to the left
Practice Problems
These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice.
To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problems 1a - 1b: Simplify, use positive exponents to write each answer.
Practice Problem 2a: Write the number in scientific notation.
2a. .00000146
Practice Problem 3a: Write the number in standard notation, without exponents.
Need Extra Help on these Topics?
The following are webpages that can assist you in the topics that were covered on this page:
http://www.sosmath.com/algebra/logs/log3/log32/log32.html
This webpage helps with the quotient rule for exponents.
http://www.purplemath.com/modules/exponent.htm
This webpage gives an overall review of exponents. It contains rules from both this tutorial and Tutorial 26: Exponents.
Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.
Last revised on August 2, 2011 by Kim Seward. |
# Finding the limit question
Struggled with this question, looked over examples still couldn't figure it out.
Question: Find the limit
$$\lim\limits_{x \to 1} \frac{2-x}{(x-1)^2}$$
What i got to: $$\lim\limits_{x \to 1} {2-x} \frac {1}{(x-1)(x+1)}$$
Understanding of infinite limits is "limited." For this question, am i supposed to do the right side first of the limit and then left side and then see if the outputs match, if so the limit exists?
• What is the issue here ? You get $\frac 10$ so limit does not exists (it is infinity with undefined sign). What is bothering you ? Also $(x-1)^2=(x-1)(x-1)\neq x^2-1$
– zwim
May 30, 2020 at 2:58
• The "naive" way is $\frac {2-x}{(x-1)^2} = \frac 10$. As $(x-1)^2 > 0$ we have that the answer should be $\lim \frac {2-x}{(x-1)^2} =+\infty$ so we need to set up our delta $N$ prove that for any $N$ there is a $\delta> 0$ so that $0< |x-1| < \delta$ then $\frac {2-x}{(x-1)^2} > N$ May 30, 2020 at 3:37
When we put $$x=1$$ in the limit, we get $$\frac{1}{0}$$ This is not an indeterminate form. Therefore we consider the right-hand and left-hand limits respectively. For the right-hand limit, the limit approaches positive infinity as both the top and bottom of the fraction are positive. For the left-hand limit, the limit also approaches positive infinity because both the top and bottom of the fraction are positive due to the square on the bottom. Therefore we conclude $$\frac{2-x}{(x-1)^2}\to+\infty$$ as $$x\to 1$$
A limit
$$\lim_{x\to a} f(x)$$
exists if and only if it is equal to a number. Note that $$\infty$$ is not a number.
The function is not defined for $$x = 1$$. In addition, no sign changes around $$x = +1$$ and $$x = -1$$ so the denominator grows quickly. So when $$x \rightarrow 1$$ the limit does not exist.
So we can calculate separately
$$\lim _{x\to \:-1} \frac{2-x}{\left(x-1\right)^2} = \lim _{x\to \:-1}\left(\left(2-x\right)\frac{1}{\left(x-1\right)^2}\right) \lim _{x\to \:-1}\left(2-x\right)\cdot \lim _{x\to \:-1}\left(\frac{1}{\left(x-1\right)^2}\right)$$
And the denominator grows without quota . You can use the same reasoning for growth from the right.
$$\lim _{x\to \:+1}\left(\frac{2-x}{\left(x-1\right)^2}\right) = \lim _{x\to \:+1}\left(\left(2-x\right)\frac{1}{\left(x-1\right)^2}\right) = \:\lim _{x\to \:+1-+}\left(2-x\right)\:\: \cdot \:\: \lim _{x\to \:+1}\left(\frac{1}{\left(x-1\right)^2}\right)$$
In the right part of the multiplication we get $$2-1 = 1$$. And on the othe for x approaching 1 from the right, denominator is a positive value that approaches 0 and $$\lim _{x\to \:1+}\left(\frac{1}{\left(x-1\right)^2}\right) =\infty \:$$
Let's find both the Right-hand limit and Left-hand limit separately: First, the left hand limit: $$\lim_{x\to 1^-} \frac{2-x}{(x-1)^2}$$ $$=\lim_{h\to 0^+}\frac{2-(1-h)}{[(1-h)-1]^2}$$ $$=\lim_{h\to 0^+}\frac{1-h}{h^2}$$ $$\implies \text{Left-hand limit}\to +\infty$$ Now the right hand limit: $$\lim_{x\to 1^+}\frac{2-x}{(x-1)^2}$$ $$=\lim_{h\to 0^+}\frac{2-(1+h)}{(1+h-1)^2}$$ $$=\lim_{h\to 0^+}\frac{1-h}{h^2}$$ $$\implies \text{Right-hand limit} \to +\infty$$ Since both the limits are non existent, and tend to $$+\infty$$ you can conclude that the limit does not exist at $$x=1$$.
P.S. : You have wrongly written $$(x-1)^2= (x-1)(x+1)$$ ( as pointed out by @zwim)
The "naive" way is $$\frac {2-x}{(x-1)^2} = \frac 10$$. As $$(x-1)^2 > 0$$ we have that the answer should be $$\lim \frac {2-x}{(x-1)^2} =+\infty$$ so we need to set up our delta $$N$$ prove that for any $$N$$ there is a $$\delta> 0$$ so that $$0< |x-1| < \delta$$ then $$\frac {2-x}{(x-1)^2} > N$$
So we want $$0 < x-1< \delta \implies$$
$$1 -\delta < x < 1+ \delta$$ and if we assume $$\delta < 1$$ then that implies
$$1 -\delta < 2-x < 1+\delta$$ and $$0 < (x-1)^2 < \delta^2$$ so
$$\frac {2-x}{(x-1)^2} > \frac {2-x}{\delta^2} >\frac {1-\delta}{\delta^2} > \frac {1-\delta}{\delta} =\frac 1\delta - 1$$(remember we are assuming $$0 < \delta < 1$$)
So so if we choose $$\delta$$ so that $$\frac 1\delta - 1 > N$$ or in other words if $$\delta < \frac 1{N+1}$$ we are done. Such a delta so that $$\delta < \frac 1{N+1}$$ and $$0 < \delta < 1$$ can always be found. (Well, it can if $$N > -1$$ but as we are trying to prove this for large $$N$$ we substitute $$N' = \max(N,0)$$ and do it.) |
# Difference between revisions of "2004 AMC 12A Problems/Problem 8"
The following problem is from both the 2004 AMC 12A #8 and 2004 AMC 10A #9, so both problems redirect to this page.
## Problem
In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$, $\angle{EAB}$ and $\angle{ABC}$ are right angles, $AB=4$, $BC=6$, $AE=8$, and $\overline{AC}$ and $\overline{BE}$ intersect at $D$. What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$?
$[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,3N); label("E",E,NW); //Side labels label("4",A--B,S); label("8",A--E,W); label("6",B--C,ENE); [/asy]$
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$
## Solution 1
Since $AE \perp AB$ and $BC \perp AB$, $AE \parallel BC$. By alternate interior angles and $AA\sim$, we find that $\triangle ADE \sim \triangle CDB$, with side length ratio $\frac{4}{3}$. Their heights also have the same ratio, and since the two heights add up to $4$, we have that $h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}$ and $h_{CDB} = 3 \cdot \frac 47 = \frac {12}7$. Subtracting the areas, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{\mathrm{(B)}\ 4}$.
## Solution 2
Let $[X]$ represent the area of figure $X$. Note that $[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$ and $[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$.
$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}$
## Solution 3
Put figure $ABCDE$ on graph. $\overline{AC}$ goes from (0, 0) to (4, 6) and $\overline{BE}$ goes from (4, 0) to (0, 8). $\overline{AC}$ is on line $y = 1.5x$. $\overline{BE}$ is on line $y = -2x + 8$. Finding intersection between these points,
$1.5x = -2x + 8$.
$3.5x = 8$
$x = 8 \times \frac{2}{7}$
$= \frac{16}{7}$
This gives us the x-coordinate of D. So, $\frac{16}{7}$ is the height of $\triangle ADE$, then area of $\triangle ADE$ is $\frac{16}{7} \times 8 \times \frac{1}{2}$ $= \frac{64}{7}$
Now, the height of $\triangle BDC$ is $4-\frac{16}{7} = \frac{12}{7}$ And the area of $\triangle BDC$ is $6 \times \frac{12}{7} \times \frac{1}{2} = \frac{36}{7}$
This gives us $\frac{64}{7} - \frac{36}{7} = 4$
\dot{.\hspace{.095in}.}\hspace{.5in} The difference is $4$ |
How to Find Vertical Angles? (+FREE Worksheet!)
In this article, you will learn how to Find Vertical angles in a few simple steps.
Step by step guide to Finding Vertical angles
The intersecting lines form an $$X$$-shape, and the angles on the two opposite sides of this $$X$$ are called vertical angles. The two vertical angles are always the same size and they have the same vertex. The bisector of two vertical angles makes a straight angle.
In the diagram at the right, lines and are straight:
• Angle$$1$$+ Angle$$3$$=$$180$$(because it is a straight angle)
• Angle$$2$$+ Angle$$3$$=$$180$$(because it is a straight angle)
Infer from the above two relations that angle $$1$$ and $$2$$ angle are equal; So, the vertical angles are equal.
In this diagram:
• Angle $$1$$ and angle $$2$$ are vertical angles.
• Angle $$3$$ and angle $$4$$ are vertical angles.
• Angle $$1$$ and angle $$3$$ are NOT vertical angles.
Finding Vertical Angles Example 1:
Find the number of degrees.
Solution: $$(3x+50)^{\circ}$$ and $$(5x+10)^{\circ}$$ are vertical angles.
$$3x+50=5x+10→3x+50-50=5x+10-50→3x=5x-40→3x-5x=-40→-2x=-40→x=20$$
$$(3x+50)^{\circ}=(3(20)+50)=110^{\circ}$$
$$(5x+10)^{\circ}=(5(20)+10=110^{\circ}$$
Finding Vertical Angles Example 2:
Find the number of degrees.
Solution: $$(4x)^{\circ}$$ and $$(6x-22)^{\circ}$$ are vertical angles.
$$4x=6x-22→4x-6x=-22→-2x=-22→x=11$$
$$(4x)^{\circ}=(4(11))=44^{\circ}$$
$$(6x-22)^{\circ}=(6(11)-22=44^{\circ}$$
Exercises for Finding Vertical Angles
Find the value of $$x$$.
1)
2)
1. $$x=96$$
2. $$x=30$$
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# Illustrative Mathematics Unit 6.5, Lesson 10: Using Long Division
Learning Targets:
• I can use long division to find a quotient of two whole numbers when the quotient is a whole number.
Related Pages
Illustrative Math
#### Lesson 10: Using Long Division
Let’s use long division.
Illustrative Math Unit 6.5, Lesson 10 (printable worksheets)
#### Lesson 10 Summary
Long division is another method for calculating quotients. It relies on place value to perform and record the division.
When we use long division, we work from left to right and with one digit at a time, starting with the leftmost digit of the dividend. We remove the largest group possible each time, using the placement of the digit to indicate the size of each group.
The following diagram shows how to use long division.
#### Lesson 10.1 Number Talk: Estimating Quotients
Estimate these quotients mentally.
500 ÷ 3.
1,394 ÷ 9
#### Lesson 10.2 Lin Uses Long Division
Lin has a method of calculating quotients that is different from Elena’s method and Andre’s method. Here is how she found the quotient of 657 ÷ 3:
1. Discuss with your partner how Lin’s method is similar to and different from drawing base-ten diagrams or using the partial quotients method.
• Lin subtracted 3 · 2 then 3 · 1, and lastly 3 · 9. Earlier, Andre subtracted 3 · 200 then 3 · 10, and lastly 3 · 9. Why did they have the same quotient?
• In the third step, why do you think Lin wrote the 7 next to the remainder of 2 rather than adding 7 and 2 to get 9?
1. Lin’s method is called long division. Use this method to find the following quotients. Check your answer by multiplying it by the divisor.
a. 846 ÷ 3
b. 1,816 ÷ 4
c. 768 ÷ 12
#### Lesson 10.3 Dividing Whole Numbers
1. Find each quotient. a. 633 ÷ 3
b. 1001 ÷ 7
2. Here is Priya’s calculation of 906 ÷ 3
a. Priya wrote 320 for the value of 906 ÷ 3. Check her answer by multiplying it by 3. What product do you get and what does it tell you about Priya’s answer? b. Describe Priya’s mistake, then show the correct calculation and answer.
#### Lesson 10 Practice Problems
1. Kiran is using long division to find 623 ÷ 7
He starts by dividing 62 by 7. In which decimal place should Kiran place the first digit of the quotient (8)?
A. Hundreds
B. Tens
C. Ones
D. Tenths
2. Here is a long-division calculation of 917 ÷ 7.
a. There is a 7 under the 9 of 917. What does this 7 represent?
b. What does the subtraction of 7 from 9 mean?
c. Why is a 1 written next to the 2 from 9 - 7?
3. Han’s calculation of 972 ÷ 9 is shown here.
a. Find 180 · 9.
b. Use your calculation of 180 · 9 to explain how you know Han has made a mistake.
c. Identify and correct Han’s mistake.
4. Find each quotient.
5. One ounce of a yogurt contains of 1.2 grams of sugar. How many grams of sugar are in 14.25 ounces of yogurt?
A. 0.171 grams
B. 1.71 grams
C. 17.1 grams
D. 171 grams
6. 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.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Symmetric Matrix
## Definition of a Symmetric Matrix
A square matrix A is symmetric if and only if A = AT where AT is the transpose of matrix A .
A symmetric matrix may be recognized visually: The entries that are symmetrically positioned with respect to the main diagonal are equal as shown in the example below of a symmetric matrix.
These are examples of symmetric matrices.
a)
By simple inspection, the entries that are symmetrically positioned with respect to the main diagonal are both equal to -2 and therefore matrix A is symmetric.
Also, the transpose of matrix A is obtained by interchanging the row of the matrix into a column. The transpose of matrix A is
Note that AT = A and therefore matrix A is symmetric.
b)
The transpose of matrix B is obtained by interchanging the column of the matrix into a row. The transpose of matrix B is
Note that BT = B and therefore matrix B is symmetric.
## The Product of Any Matrix and it Transpose is Symmetric
For any matrix $$A$$ of size $$m \times n$$ , the matrices $$A A^T$$ and $$A^T A$$ are symmetric and have sizes $$m \times m$$ and $$n \times n$$ respectively.
Let $$A = \begin{bmatrix} -3 & -3 & 9\\ 2 & -7 & 1 \end{bmatrix}$$ is a $$2 \times 3$$ matrix
$$A^T = \begin{bmatrix} -3 & 2\\ -3 & -7 \\ 9 & 1 \end{bmatrix}$$
$$A A^T = \begin{bmatrix} -3 & -3 & 9\\ 2 & -7 & 1 \end{bmatrix} \begin{bmatrix} -3 & 2\\ -3 & -7 \\ 9 & 1 \end{bmatrix}$$
$$\quad \quad = \begin{bmatrix} 99 & 24\\ 24 & 54 \end{bmatrix}$$
Hence $$A A^T$$ is a $$2 \times 2$$ symmetric matrix.
$$A^T A = \begin{bmatrix} -3 & 2\\ -3 & -7 \\ 9 & 1 \end{bmatrix} \begin{bmatrix} -3 & -3 & 9\\ 2 & -7 & 1 \end{bmatrix}$$
$$\quad \quad = \begin{bmatrix} 13 & -5 & -25\\ -5 & 58 & -34\\ -25 & -34 & 82 \end{bmatrix}$$ is a $$3 \times 3$$
Hence $$A^T A$$ is a $$3 \times 3$$ symmetric matrix.
## Properties of Symmetric Matrices
Some of the most important properties of the symmetric matrices are given below.
1. If $$A$$ is a symmetric matrix, then $$A^T = A$$, where$$A^T$$ is the transpose of matrix $$A$$.
2. If $$A$$ and $$B$$ are symmetric matrices, then $$A \pm B$$ are also symmetric.
3. If $$A$$ and $$B$$ are symmetric matrices of the same size, then $$AB + BA$$ is also symmetric.
4. If $$A$$ is a symmetric matrix, then its inverse matrix $$A^{-1}$$ ,if it exists ,is also symmetric.
5. If $$A$$ is a symmetric matrix, then $$k A$$ is also symmetric; where $$k$$ is any real number.
6. If $$A$$ is a symmetric matrix, then $$A^n$$ is also symmetric; $$n$$ is any positive integer.
7. For any matrix $$B$$, the matrices $$B B^T$$ and $$B^T B$$ are symmetric.
## Examples with Solutions
Example 1
Which if the following matrices are symmetric?
a) $$A = \begin{bmatrix} 0 & -1 & 1 \\ -1 & 0 & 0 \end{bmatrix}$$ b) $$B = \begin{bmatrix} -2 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix}$$ c) $$C = \begin{bmatrix} 5 & -7 & 0 & 0\\ 5 & -7 & 0 & 0\\ 5 & -7 & 0 & 0\\ 5 & -7 & 0 & 0 \end{bmatrix}$$ d) $$D = \begin{bmatrix} 1 & 4 & 0 & 9\\ 4 & 2 & -4 & 0\\ 0 & -4 & 4 & 0\\ 9 & 0 & 0 & -5 \end{bmatrix}$$
Solution
Matrices $$B$$ and $$D$$ are symmetric.
Example 2
Symmetric matrices $$A$$ and $$B$$ are given by $$A = \begin{bmatrix} -1 & 1 \\ 1 & 2 \end{bmatrix}$$ , $$B = \begin{bmatrix} 4 & 2 \\ 2 & -3 \end{bmatrix}$$.
Show that $$A+B$$ and $$A-B$$ are symmetric. (Verify property 2)
Solution
Calculate $$A + B$$
$$A + B = \begin{bmatrix} -1 & 1 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 2 \\ 2 & -3 \end{bmatrix}$$
$$\quad = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$$
Calculate $$A - B$$
$$A - B = \begin{bmatrix} -1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 4 & 2 \\ 2 & -3 \end{bmatrix}$$
$$\quad = \begin{bmatrix} -5 & -1 \\ -1 & 5 \end{bmatrix}$$
Hence $$A + B$$ and $$A - B$$ are also symmetric matrices.
Example 3
Find the inverse of the symmetric matrix $$A = \begin{bmatrix} -1 & -2\\ -2 & 0 \end{bmatrix}$$ and show that this inverse matrix is symmetric.(Verify properties 4)
Solution
Use the formula of the inverse of a $$2 \times 2$$ matrix $$\begin{bmatrix} x & y \\ z & w \\ \end{bmatrix}^{-1} = \dfrac{1}{xw - yz} \begin{bmatrix} w & - y \\ - z & x \\ \end{bmatrix}$$ to find $$A^{-1}$$
$$A^{-1} = \dfrac{1}{-4} \begin{bmatrix} 0 & 2\\ 2 & -1 \end{bmatrix}$$
$$\quad \quad = \begin{bmatrix} 0 & -\dfrac{1}{2}\\ -\dfrac{1}{2} & \dfrac{1}{4} \end{bmatrix}$$
Hence $$A^{-1}$$ is also symmetric.
Example 4
Let matrix $$A = \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix}$$
Calculate $$A^3$$ and verify that it is symmetric. (Verify properties 6)
Solution
$$A^3 = \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix} \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix} \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix}$$
$$\quad \quad = \begin{bmatrix} 10 & 4\\ 4 & 50 \end{bmatrix} \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix}$$
$$\quad \quad = \begin{bmatrix} -26 & 38\\ 38 & 354 \end{bmatrix}$$
Hence $$A^3$$ is a symmetric matrix.
Example 5
Find the real numbers $$a$$, $$b$$ and $$c$$ so that the matrix $$A = \begin{bmatrix} 0 & a+b & c+2 \\ a & 2 & c \\ 4 & a+b & 4 \end{bmatrix}$$ is symmetric.
Solution
For the given matrix to be symmetric, we have to have the following equations satisfied simultaneously:
$$a = a + b$$ , $$c + 2 = 4$$ , $$c = a + b$$
Solve the system of equations in $$a,\; b,\; c$$ above to obtain
$$b = 0 , \; c = 2 , \; a = 2$$ which are the values so that the given matrix is symmetric.
Example 6
Given the symmetric matrices $$A = \begin{bmatrix} -1 & 3 \\ 3 & 4 \end{bmatrix}$$ and $$B = \begin{bmatrix} 4 & 8 \\ 8 & 9 \end{bmatrix}$$, verify that $$AB + BA$$ is also symmetric (Property 3 above)
Solution
Calculate $$AB$$
$$AB = \begin{bmatrix} -1 & 3 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 4 & 8 \\ 8 & 9 \end{bmatrix}$$
$$\quad \quad = \begin{bmatrix} 20 & 19 \\ 44 & 60 \end{bmatrix}$$
Calculate $$BA$$
$$BA = \begin{bmatrix} 4 & 8 \\ 8 & 9 \end{bmatrix} \begin{bmatrix} -1 & 3 \\ 3 & 4 \end{bmatrix}$$
$$\quad \quad = \begin{bmatrix} 20 & 44\\ 19 & 60 \end{bmatrix}$$
Calculate $$AB + BA$$
$$AB + BA = \begin{bmatrix} 20 & 19 \\ 44 & 60 \end{bmatrix} + \begin{bmatrix} 20 & 44\\ 19 & 60 \end{bmatrix}$$
$$\quad \quad = \begin{bmatrix} 40 & 63\\ 63 & 120 \end{bmatrix}$$
Hence $$AB + BA$$ is a symmetric matrix.
## Questions (with solutions given below)
• Part 1
Matrices $$A$$ and $$B$$ are such that $$A B = B A = I$$ where $$I$$ is the identity matrix . Use any of the properties above to explain that if one of the matrices is symmetric, then the other one is also symmetric.
• Part 2
Given the matrices $$A = \begin{bmatrix} -5 & -2 \\ - 1 & 2 \end{bmatrix}$$ and $$B = \begin{bmatrix} 3 & 1 \\ 0 & 4 \end{bmatrix}$$.
Calculate $$A + B$$ and explain why $$(A + B)^T = A + B$$ without any further computations.
• Part 3
Let matrix $$A = \begin{bmatrix} - 1 & - b\\ c & c^2 \end{bmatrix}$$. Find the real numbers $$b$$ and $$c$$ so that matrix $$A$$ is symmetric and $$Det (A) = - 1$$.
• Part 4
Show that any symmetric matrix of the form $$A = \begin{bmatrix} a & \sqrt{1-a^2} \\ \\ \sqrt{1-a^2} & - a \\ \end{bmatrix}$$ , with $$|a| \lt 1$$, is such that $$A = A^T = A^{-1}$$ and calculate $$A^n$$ where $$n$$ is a positive integer.
### Solutions to the Above Questions
• Part 1
Given that $$A B = B A = I$$ where $$I$$ is the identity matrix, and according to the definition of the inverse of a matrix $$A$$ and $$B$$ are inverses of each other. Hence according to property 4, if $$A$$ is symmetric, then $$B$$ its inverse is also symmetric and if $$B$$ is symmetric, then its inverse $$A$$ is also symmetric.
• Part 2
$$A + B = \begin{bmatrix} -5 & -2 \\ - 1 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ -1 & 6 \end{bmatrix}$$
Note that $$A + B$$ is symmetric and therefore according to the definition (or property 1)
$$(A + B)^T = A + B$$
• Part 3
For matrix $$A$$ to be symmetric, we have to have $$- b = c$$ or $$b = - c$$.
$$Det(A) = - c^2 + b c = - 1$$
Hence the equation
$$- c^2 + b c = - 1$$
Substitute $$b$$ by $$- c$$ in the equation above
$$-c^2 - c^2 = - 1$$
Solve for $$C$$ to obtain two solutions
$$c_1 = \dfrac{\sqrt 2}{2}$$ and $$c_1 = - \dfrac{\sqrt 2}{2}$$
Hence two pairs of solutions
$$c = \dfrac{\sqrt 2}{2}$$ and $$b = - \dfrac{\sqrt 2}{2}$$
or
$$c = - \dfrac{\sqrt 2}{2}$$ and $$b = \dfrac{\sqrt 2}{2}$$
• Part 4
Given $$A = \begin{bmatrix} a & \sqrt{1-a^2} \\ \\ \sqrt{1-a^2} & - a \\ \end{bmatrix}$$
Determine $$A^T$$
$$A^T = \begin{bmatrix} a & \sqrt{1-a^2} \\ \\ \sqrt{1-a^2} & - a \\ \end{bmatrix}$$
Use the formula of the inverse of a $$2 \times 2$$ matrix $$\begin{bmatrix} x & y \\ z & w \\ \end{bmatrix}^{-1} = \dfrac{1}{xw - yz} \begin{bmatrix} w & - y \\ - z & x \\ \end{bmatrix}$$ to find $$A^{-1}$$
$$A^{-1} = \dfrac{1}{- a^2 - (\sqrt{1-a^2})^2} \begin{bmatrix} - a & - \sqrt{1-a^2} \\ \\ - \sqrt{1-a^2} & a \\ \end{bmatrix}$$
Simplify
$$A^{-1} = (-1) \begin{bmatrix} - a & - \sqrt{1-a^2} \\ \\ - \sqrt{1-a^2} & a \\ \end{bmatrix}$$
$$\quad \quad = \begin{bmatrix} a & \sqrt{1-a^2} \\ \\ \sqrt{1-a^2} & - a \\ \end{bmatrix}$$
and we note that $$A = A^T = A^{-1}$$
$$A^2 = A A = A A^{-1} = I$$
$$A^3 = A^2 A = I A = A$$
$$A^4 = A^3 A = A A = I$$
$$A^5 = A^4 A = I A = A$$
We may easily conclude that
$$A^n = I$$ if $$n$$ is even
and
$$A^n = A$$ if $$n$$ is odd |
# Order Of Operations With Exponents
Order of operations is determining the sequence of calculations in expressions that have more than one $$(1)$$ type of computation.
This lesson explains the Order of Operations when we deal with exponents & square roots.
1. Simplify.
$$(10 - 8)^{2} - (7 -5)^{3}$$
A.
B.
C.
D.
Question 1 of 2
2. Simplify:
$$-4 + 5(-4)^{3}$$
A.
B.
C.
D.
Question 2 of 2
This lesson is provided by Onsego GED Prep.
Next Lesson: Language of Algebra
This lesson is a part of our GED Math Study Guide.
### Video Transcription
The rules of “Order of Operations” allow us to simplify problems that include multiplication, division, addition, and subtraction, or grouping symbols.
What will happens if a problem includes exponents or square roots? We have to expand our rules regarding “Order of Operations” to include exponents and square roots.
First, perform all operations within grouping symbols. Grouping symbols are including $$brackets\, [ \,]$$, $$parentheses\, (\,)$$, $$braces \left \{ \right \}$$, and $$fraction\, bars$$.
Then, evaluate exponents and roots of numbers (for example, square roots).
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Then, apply multiplication and division, from left to right.
Then add and subtract, also from left to right.
If our expression includes exponents or square roots, they may be performed only after parentheses or other grouping symbols are simplified yet before multiplication, division, subtraction, or addition is done that’s outside of the parentheses or some other grouping symbol set.
Example: simplify $$14 + 28 \div 2^{2}$$
This expression includes addition, division, and has an exponent. Use our order of operations.
1. First, simplify $$2^{2}$$
2. Then, perform division before addition: $$14 + 28 \div 4$$
3. Then, add, to end up with $$21$$.
In summary
Order of Operations provides us with a consistent sequence to be used in computations. Without our order of operations, we could end up with totally different answers to exactly identical computation problems.
The Order of Operations and the use of a calculator: Well, there are some cheaper calculators that are NOT using our order of operations. When you use these calculators, you must input your numbers in exactly the correct order. Modern-day calculators will be implementing the correct order of operations, also without your input.
Last Updated on February 15, 2024. |
# An Introduction to Genetic Algorithms
## Putting It All Together
Now that we have learnt the basic theory behind the GA, lets put it all together. Imagine a simple scenario of 3 “wheels” on the combination lock each with 8 possible positions from 0 to 7. This would give us the problem space:
Stage 1 – generate 4 random guesses – g1, g2, g3, g4
g1 = 1,6,3 = 001110011
g2 = 6,5,1 = 110101001
g3 = 0,4,3 = 000001011
g4 = 1,5,2 = 001101010
Stage 2 – evaluate how good they are – their fitness
In a basic introduction such as this, it is not necessary to go into great deal with the evaluation stage, aka – the fitness function, however it is necessary to understand that this function will provide each guess with a “score” depending on how good they are.
Stage 3 – pick the parents
There are many ways to choose which guesses become parents, perhaps the easiest to imagine is to create a pie chart which represents each guesses share of the total points. First add up the points of each guess – g1 score = 2.12, g2 score = 1.61, g3 score = 4, g4 score = 3.26
Total score = 10.99
g1 – total share of pie = 20%
g2 – total share of pie = 15%
g3 – total share of pie = 36%
g4 – total share of pie = 29%
Then pick a random point on the pie and whichever guess you are pointing at becomes a parent, this is repeated 4 times.
Random parents = g1, g3, g4, g3
Stage 4 – breed the parents
Crossover 1 between g1 and g3
Before –
g1 = 001110011
g3 = 000001011
After –
Child1 = 001111011
Child2 = 000000011
Crossover 2 between g4 and g3
Before -
g4 = 001101010
g3 = 000001011
After –
Child3 = 001101011
Child4 = 000001010
The children now become the next generation:
g1 = c1 = 001111011
g2 = c2 = 000000011
g3 = c3 = 001101011
g4 = c4 = 000001010
Stage 5 - Random mutation is applied to a random 1% of the population:
g1 = 001111011
g2 = 010000011
g3 = 001101011
g4 = 000001010
Return to stage 2 – This process is repeated a predefined number of times. In a simple example like this the solution will not take long to converge upon (the solutions will all become the same) so a relatively low number of generations are necessary.
Feel free to download the samples.zip file which contains the C++ source code for a simple Genetic Algorithm as well as the source code for a Real-Coded GA, which is fundamentaly the same except the population consists of real numbers instead of binary digits.
Also included in the zip file is an exe which demonstrates the fundamentals of various search and optimisation techniques.
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 2/15 + 1/3 = 7/15 ≅ 0.4666667
Spelled result in words is seven fifteenths.
### How do you solve fractions step by step?
1. Add: 2/15 + 1/3 = 2/15 + 1 · 5/3 · 5 = 2/15 + 5/15 = 2 + 5/15 = 7/15
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(15, 3) = 15. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 15 × 3 = 45. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - two fifteenths plus one third = seven fifteenths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Frank
Frank will be riding his bike to school this year. The distance from his house to the end of the street is ⅜ mile. The distance from the end of the street to the school is ⅚ mile. How far is Frank's house from school?
• Mac painter
Mac is mixing red and white paint to make pink for his candy painting. He uses 1/4 ounce of red paint and 2/3 ounce of white paint. How many ounces of paint did he use?
• Evaluate expression
Evaluate expression using BODMAS rule: 1 1/4+1 1/5÷3/5-5/8
• Length subtracting
Express in mm: 5 3/10 cm - 2/5 mm
• Martin
Martin is making a model of a Native American canoe. He has 5 1/2 feet of wood. He uses 2 3/4 feet for the hull and 1 1/4 feet for a paddle. How much wood does he have left? Martin has feet of wood left.
• Fraction
Fraction ? write as fraction a/b, a, b is integers numerator/denominator.
• Team run
The first team member in a 926-person relay race must run 2 1/4 laps, the second team member must run 1 1/2 laps, and the third team member must run 3 1/4 laps. How many laps in all must each team run?
• Sum of 18
Sum of two fractions is 4 3/7. If one of the fractions is 2 1/5 find the other one .
• A dump
A dump truck bought 1/3 of a ton of rock on the first trip, 1/2 of a ton on the second trip, and 4/5 of a ton on the third trip. What was the total weight of the rock?
• The pet
Ananya has a bunny. She bought 4 7/8 pounds of carrots. She fed her bunny 1 1/4 pounds of carrots the first week. She fed her bunny 5/6 pounds of carrots the second week. All together, how many pounds of carrots did she feed her bunny? 1. Draw a tape diag
• Tartlets
After a special event, a caterer examined the leftovers on the serving table. She saw 10/11 of a tartlet with apples, 2/3 of a tartlet with strawberries, and 10/11 of a tartlet with raspberries. How many leftover tartlets did the caterer have?
• Food weight
Stacie is a resident at the medical facility where you work. You are asked to chart the amount of solid food that she consumes. For the noon meal, today she ate 1/2 of a 3-ounce serving of meatloaf, 3/4 of her 3-ounce serving of mashed potatoes, and 1/3 o
• Sum three fractions
Work out the sum of 1/4, 1/5 and 3/10. |
# How do you simplify (4 - sqrt 7) /(3+sqrt7)?
Jul 25, 2018
$\frac{19}{2} - \frac{7}{2} \sqrt{7}$
#### Explanation:
$\text{multiply the numerator/denominator by the conjugate of}$
$\text{the denominator}$
$\text{the conjugate of "3+sqrt7" is } 3 \textcolor{red}{-} \sqrt{7}$
$= \frac{\left(4 - \sqrt{7}\right) \left(3 - \sqrt{7}\right)}{\left(3 + \sqrt{7}\right) \left(3 - \sqrt{7}\right)}$
$= \frac{12 - 7 \sqrt{7} + 7}{9 - 7}$
$= \frac{19 - 7 \sqrt{7}}{2} = \frac{19}{2} - \frac{7}{2} \sqrt{7}$
$\frac{19 - 7 \setminus \sqrt{7}}{2}$
#### Explanation:
$\setminus \frac{4 - \setminus \sqrt{7}}{3 + \setminus \sqrt{7}}$
$= \setminus \frac{\left(4 - \setminus \sqrt{7}\right) \left(3 - \setminus \sqrt{7}\right)}{\left(3 + \setminus \sqrt{7}\right) \left(3 - \setminus \sqrt{7}\right)}$
$= \setminus \frac{12 - 3 \setminus \sqrt{7} - 4 \setminus \sqrt{7} + 7}{{3}^{2} - {\left(\setminus \sqrt{7}\right)}^{2}}$
$= \setminus \frac{19 - 7 \setminus \sqrt{7}}{9 - 7}$
$= \setminus \frac{19 - 7 \setminus \sqrt{7}}{2}$ |
# Question #715a8
Jun 5, 2017
$\left(f \circ g\right) \left(x\right) = 3 {x}^{2} - 27$
#### Explanation:
This is a composition of functions. It is essentially a function within a function. The notation states that you are evaluating the inner function, g(x), and then inserting the result into f(x) and re-evaluating.
$\left(f \circ g\right) \left(x\right) = f \left(g \left(x\right)\right)$
Read it as "f of g". I prefer the second notation as you can see which one is inside the other. $f \left(x\right)$ is the outer function and $g \left(x\right)$ is the inner function.
To evaluate, you replace all the $x$'s of the outer function with the inner function.
$f \left(x\right) = x - 4$
$g \left(x\right) = 3 {x}^{2} - 23$
$\left(f \circ g\right) \left(x\right) = f \left(g \left(x\right)\right) = f \left(3 {x}^{2} - 23\right)$
$= \left(3 {x}^{2} - 23\right) - 4$
$= 3 {x}^{2} - 27$
As an example, let's evaluate $\left(f \circ g\right) \left(x\right)$ when $x = 2$. You can do this by step by step:
i) Evaluate the inner function
$g \left(2\right) = 3 {\left(2\right)}^{2} - 23 = - 11$
ii) Insert this into the outer function and evaluate
$f \left(- 11\right) = - 11 - 4 = - 15$
But now that we have the composition function, we can insert $x = 2$ directly into it instead to get the same answer:
$\left(f \circ g\right) \left(2\right) = 3 {\left(2\right)}^{2} - 27 = - 15$
*As extra but vital information, an important rule for the composition to be defined is this:
$\text{range of inner function " sube " domain of outer function}$
I.e.
$\text{ran " g(x) sube "dom } f \left(x\right)$
This says that the range of $g \left(x\right)$ must be a subset or equal to the domain of $f \left(x\right)$. If this is not true, you trying to input values into $f \left(x\right)$ that are not part of its domain and, thus, cannot be evaluated.
In this case,
$\text{ran } g \left(x\right) = \left[- 23 , \infty\right)$
$\text{dom } f \left(x\right) = \mathbb{R}$
The below statement is true, so the composition is defined.
$\left[- 23 , \infty\right) \subseteq \mathbb{R}$ |
# Direct Variations Using Method of Proportion
Now we will learn how to solve direct variations using method of proportion.
We know, the two quantities may be linked in such a way that if one increases, other also increases. If one decreases, the other also decreases.
Some situations of direct variations:
● More articles, more money required to purchase.
● More men at work, more work will be done.
● More speed, more distance covered in fixed time.
● More money borrowed, more interest to be paid.
● More working hours, more work will be done.
Solved examples on direct variations using method of proportion:
1. The cost of 5 kg of rice is $30. What will the cost of 12 kg of sugar be? Solution: This is a situation of direct variation, now we solve using method of proportion. More quantity of rice results in more cost. Here, the two quantities vary directly (Quantity of rice and cost of rice) Weight of rice (kg) 5 12 Cost 30 x Since, they vary directly Therefore, 5/30 = 12/x (cross multiply) ⇒ 5x = 30 × 12 ⇒ x = (30 × 12)/5 = 72 Therefore, cost of 12 kg rice =$ 72
2. If 9 drawing books cost 171, what do 22 books cost?
Solution:
This is a situation of direct variation, now we solve using method of proportion.
More number of drawing books results in more cost.
Here, the two quantities vary directly (Number of drawing books and cost of drawing books)
Number of drawing books 9 22 Cost 171 x
Since, they vary directly
Therefore, 9/171 = 22/x (cross multiply)
⇒ 9x = 171 × 22
⇒ x = (171 × 22)/9 = 418
Therefore, cost of 22 drawing books = $418 3. A worker gets$ 504 for 7 days of work. How many days should he work to get $792? Solution: This is a situation of direct variation, now we solve using method of proportion. More money, more days of work Here, the two quantities vary directly. (Amount and days of work) Number of working days 7 x Amount Obtained ($) 504 792
Since, they vary directly
Therefore, 7/504 = x/792
⇒ 504x = 792 × 7
⇒ x = (792 × 7)/504
Therefore, 792 earned by the workers in = 11 days
`
Problems Using Unitary Method
Situations of Direct Variation
Direct Variations Using Unitary Method
Direct Variations Using Method of Proportion
Inverse Variation Using Unitary Method
Inverse Variation Using Method of Proportion
Problems on Unitary Method using Direct Variation
Problems on Unitary Method Using Inverse Variation |
# How do you factor the polynomials 48tu-90t+32u-60?
Apr 28, 2017
$48 t u - 90 t + 32 u - 60 = 2 \left(3 t + 2\right) \left(8 u - 15\right)$
#### Explanation:
Given:
$48 t u - 90 t + 32 u - 60$
Note that the ratio of the first and second terms is the same as the ratio of the third and fourth terms. So this quadrinomial will factor by grouping.
First separate out the common scalar factor $2$, since that is the greatest common factor of the coefficients.
$48 t u - 90 t + 32 u - 60 = 2 \left(24 t u - 45 t + 16 u - 30\right)$
$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(\left(24 t u - 45 t\right) + \left(16 u - 30\right)\right)$
$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(3 t \left(8 u - 15\right) + 2 \left(8 u - 15\right)\right)$
$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(3 t + 2\right) \left(8 u - 15\right)$
As an alternative, we could swap the middle two terms before grouping, which may make the arithmetic seem a little easier:
$48 t u - 90 t + 32 u - 60 = 48 t u + 32 u - 90 t - 60$
$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(24 t u + 16 u - 45 t - 30\right)$
$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(\left(24 t u + 16 u\right) - \left(45 t + 30\right)\right)$
$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(8 u \left(3 t + 2\right) - 15 \left(3 t + 2\right)\right)$
$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(8 u - 15\right) \left(3 t + 2\right)$ |
You will need
• A sheet of paper, ruler, pencil, calculator with function of calculate roots.
Instruction
1
A rectangle is a quadrilateral, all the angles which are straight. The diagonal of the rectangle is the line segment connecting two opposite vertices.
2
On a sheet of paper with a ruler and pencil draw an arbitrary rectangle ABCD. Better to do it on notebook sheet in a cage – it will be easier to draw right angles. Connect the cut vertices of the rectangle A and C. the resulting segment AC is a diagonalth rectangle ABCD.
3
Please note, the diagonal AC has divided the rectangle ABCD into the triangles ABC and АСD. The resulting triangles ABC and АСD straight triangles, because the angles ABC and АDС is 90 degrees (by definition of a rectangle). Remember the Pythagorean theorem – the square of the hypotenuse is equal to the sum of the squares of the other two sides.
4
The hypotenuse is the side of the triangle opposite the right angle. Sides – the sides of the triangle adjacent to the right angle. With respect to the triangles ABC and АСD: AB and BC, AD and DC sides of the AC common hypotenuse for both triangles (the desired diagonal). Therefore, as in a square = square AB + square BC or AC square = AD square + square DC. Input the side lengths of the rectangle in the above formula and calculate the length of the hypotenuse (diagonal of rectangle).
5
For example, the sides of the rectangle ABCD is equal to the following values: AB = 5 cm and BC = 7cm. the square of the diagonal AC of this rectangle is calculated by the Pythagorean theorem: AC square = AB square + a square sun = 52+72 = 25 + 49 = 74 sq cm using the calculator calculate the value of square root of 74. You should have 8.6 cm (rounded value). Keep in mind that one of the properties of a rectangle, its diagonals are equal. So the length of the second diagonal BD of the rectangle ABCD is equal to the length of the diagonal AC. For the above example this value is 8.6 cm |
# Study notes on Differential Equation 1 For Electrical Engineering Students
Updated : Nov 14, 2020, 10:00
By : Yash Bansal
## 1.1 Definitions
### 1.1.1 Order of a differential equation
Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation. Consider the following differential equations:
The first, second and third equations involve the highest derivative of first, second and third order respectively. Therefore, the order of these equations are 1, 2 and 3 respectively.
### 1.1.2 Degree of a differential equation
To study the degree of a differential equation, the key point is that the differential equation must be a polynomial equation in derivatives, i.e., y′, y″, y″′ etc. Consider the following differential equations:
We observe that first equation is a polynomial equation in y″′, y″ and y′ with degree 1, second equation is a polynomial equation in y′ (not a polynomial in y though) with degree of 2. Degree of such differential equations can be defined. But third equation is not a polynomial equation in y′ and degree of such a differential equation can not be defined.
By the degree of a differential equation, when it is a polynomial equation in derivatives, we mean the highest power (positive integral index) of the highest order derivative involved in the given differential equation.
NOTE: Order and degree (if defined) of a differential equation are always positive integers.
Example Find the order and degree, if defined, of each of the following differential equations:
Solution
(i) The highest order derivative present in the differential equation is , so its order is one. It is a polynomial equation in y′ and the highest power raised to is one, so its degree is one.
(ii) The highest order derivative present in the given differential equation is , so its order is two. It is a polynomial equation in and and the highest power raised to is one, so its degree is one.
(iii) The highest order derivative present in the differential equation is y′′′ , so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.
## 1.2 Separable Equations
### 1.2.1 Separable differential equation:
Example:
RC circuits: Charging:
Discharging:
### 1.2.2 Linear Differential Equations:
, integrating factor:
Example: , .
### 1.2.3 Exact Differential Equations
• Potential function: For , we can find a such that and is the potential function; is exact.
• Exact differential equation: a potential function exists; general solution: .
Example: .
• Theorem: Test for exactness:
Example: .
### 1.2.4 IntegratingFactors
• Integrating factor: such that is exact.
Example: .
• How to find integrating factor:
Example:
• Separable equations and integrating factors:
• Linear equations and integrating factors:
### 1.2.5 Homogeneous and Bernoulli Equations
• Homogeneous differential equation: ; let separable.
Example: .
• Bernoulli equation: ; linear; separable; otherwise, let linear
Example: .
# 2 HIGHER ORDER LINEAR DIFFERENTIAL EUQATION
## 2.1 Homogeneous Linear ODEs
An linear ordinary differential equation of order is said to be homogeneous if it is of the form
where
### , i.e., if all the terms are proportional to a derivative of (or itself) and there is no term that contains a function of alone.
However, there is also another entirely different meaning for a first-order ordinary differential equation. Such an equation is said to be homogeneous if it can be written in the form
1. a nth order ODE if the nth derivative of the unknown function is the highest occurring derivative.
2. Linear ODE: .
3. Homogeneous linear ODE: .
Theorem: Fundamental Theorem for the Homogeneous Linear ODE: For a homogeneous linear ODE, sums and constant multiples of solutions on some open interval I are again solutions on I. (This does not hold for a Non-Homogeneous or nonlinear ODE!).
General solution: , where is a basis (or fundamental system) of solutions on I; that is, these solutions are linearly independent on I.
1. Linear independence and dependence: n functions are called linearly independent on some interval I where they are defined if the equation on I implies that all are zero. These functions are called linearly dependent on I if this equation also holds on I for some not all zero.
Example: . Sol.: .
Theorem: Let the homogeneous linear ODE have continuous coefficients , on an open interval I. Then n solutions on I are linearly dependent on I if and only if their Wronskian/Determinant is zero for some in I. Furthermore, if W is zero for , then W is identically zero on I. Hence if there is an in I at which W is not zero, then are linearly independent on I, so that they form a basis of solutions of the homogeneous linear ODE on I.
Wronskian:
1. Initial value problem: An ODE with n initial conditions , , .
## 2.2 Homogeneous Linear ODEs with Constant Coefficients
1. : Substituting , we obtain the characteristic equation .
(i) Distinct real roots: The general solution is
Example: . Sol. .
(ii) Simple complex roots: , ,.
Example: . Sol..
(iii) Multiple real roots: Ifis a real root of order m, then m corresponding linearly independent solutions are: , , , .
Example: . Sol..
(iv) Multiple complex roots: If are complex double roots, the corresponding linearly independent solutions are: , , , .
1. Convert the higher-order differential equation to a system of first-order equations.
Example: .
## 2.3 Nonhomogeneous Linear ODEs
1. , the general solution is of the form: , where is the homogeneous solution and is a particular solution.
2. Method of undermined coefficients
Example: . Sol. .
1. Method of variation of parameters: , where , .
Example: . Sol. .
Thanks
Nov 14ESE & GATE EE
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Member since Nov 2018 |
### NCERT Solutions For Class 9 Math Chapter – 13 Exercise – 13.6
1. The circumference of the base of cylindrical vessel is 132cm and its height is 25cm.
How many litres of water can it hold? (1000 cm3= 1L) (Assume π = 22/7)
Solution:
Circumference of the base of cylindrical vessel = 132 cm
Height of vessel, h = 25 cm
Let r be the radius of the cylindrical vessel.
Step 1: Find the radius of vessel
We know that, circumference of base = 2πr, so
2πr = 132 (given)
r = (132/(2 π))
r = 66×7/22 = 21
Radius is 21 cm
Step 2: Find the volume of vessel
Formula: Volume of cylindrical vessel = πr2h
= (22/7)×212×25
= 34650
Therefore, volume is 34650 cm3
Since, 1000 cm= 1L
So,
Volume = 34650/1000 L= 34.65L
Therefore, vessel can hold 34.65 litres of water.
2. The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28 cm. The length of the pipe is 35cm.Find the mass of the pipe, if 1cm3 of wood has a mass of 0.6g. (Assume π = 22/7)
Solution:
Inner radius of cylindrical pipe, say r1 = diameter1/ 2 = 24/2 cm = 12cm
Outer radius of cylindrical pipe, say r2 = diameter2/ 2 = 28/2 cm = 14 cm
Height of pipe, h = Length of pipe = 35cm
Now, the Volume of pipe = π(r22-r12)h cm3
Substitute the values.
Volume of pipe = 110×52 cm= 5720 cm3
Since, Mass of 1 cm3 wood = 0.6 g
Mass of 5720 cm3 wood = (5720×0.6) g = 3432 g or 3.432 kg. Answer!
3. A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7cm and height 10cm. Which container has greater capacity and by how much? (Assume π=22/7)
Solution:
1. tin can will be cuboidal in shape
Dimensions of tin can are
Length, l = 5 cm
Breadth, b = 4 cm
Height, h = 15 cm
Capacity of tin can = l×b×h= (5×4×15) cm= 300 cm3
1. Plastic cylinder will be cylindrical in shape.
Dimensions of plastic can are:
Radius of circular end of plastic cylinder, r = 3.5cm
Height , H = 10 cm
Capacity of plastic cylinder = πr2H
Capacity of plastic cylinder = (22/7)×(3.5)2×10 = 385
Capacity of plastic cylinder is 385 cm3
From results of (i) and (ii), plastic cylinder has more capacity.
Difference in capacity = (385-300) cm3 = 85cm3
4. If the lateral surface of a cylinder is 94.2cm2 and its height is 5cm, then find
(i) radius of its base (ii) its volume.[Use π= 3.14]
Solution:
CSA of cylinder = 94.2 cm2
Height of cylinder, h = 5cm
(i) Let radius of cylinder be r.
Using CSA of cylinder, we get
2πrh = 94.2
2×3.14×r×5 = 94.2
r = 3
radius is 3 cm
(ii) Volume of cylinder
Formula for volume of cylinder = πr2h
Now, πr2h = (3.14×(3)2×5) (using value of r from (i))
= 141.3
Volume is 141.3 cm3
5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of Rs 20 per m2, find
(i) inner curved surface area of the vessel
(ii) radius of the base
(iii) capacity of the vessel
(Assume π = 22/7)
Solution:
(i) Rs 20 is the cost of painting 1 marea.
Rs 1 is the cost to paint 1/20 marea
So, Rs 2200 is the cost of painting = (1/20×2200) m2
= 110 m2 area
The inner surface area of the vessel is 110m2.
(ii) Radius of the base of the vessel, let us say r.
Height (h) = 10 m and
Surface area formula = 2πrh
Using result of (i)
2πrh = 110 m2
2×22/7×r×10 = 110
r = 1.75
Radius is 1.75 m.
(iii) Volume of vessel formula = πr2h
Here r = 1.75 and h = 10
Volume = (22/7)×(1.75)2× 10 = 96.25
Volume of vessel is 96.25 m3
Therefore, the capacity of the vessel is 96.25 m3 or 96250 litres.
6. The capacity of a closed cylindrical vessel of height 1m is15.4 liters. How many square meters of metal sheet would be needed to make it? (Assume π = 22/7)
Solution:
Height of cylindrical vessel, h = 1 m
Capacity of cylindrical vessel = 15.4 litres = 0.0154 m3
Let r be the radius of the circular end.
Now,
Capacity of cylindrical vessel = (22/7)×r2×1 = 0.0154
After simplifying, we get, r = 0.07 m
Again, total surface area of vessel = 2πr(r+h)
= 2×22/7×0.07×(0.07+1)
= 0.44×1.07
= 0.4708
Total surface area of vessel is 0.4708 m2
Therefore, 0.4708 m2 of the metal sheet would be required to make the cylindrical vessel.
7. A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite. (Assume π = 22/7)
Solution:
Radius of pencil, r= 7/2 mm = 0.7/2 cm = 0.35 cm
Radius of graphite, r= 1/2 mm = 0.1/2 cm = 0.05 cm
Height of pencil, h = 14 cm
Formula to find, volume of wood in pencil = (r12-r22)h cubic units
Substitute the values, we have
= [(22/7)×(0.352-0.052)×14]
= 44×0.12
= 5.28
This implies, volume of wood in pencil = 5.28 cm3
Again,
Volume of graphite = r22h cubic units
Substitute the values, we have
= (22/7)×0.052×14
= 44×0.0025
= 0.11
So, the volume of graphite is 0.11 cm3.
8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup does the hospital have to prepare daily to serve 250 patients? (Assume π = 22/7)
Solution:
Diameter of cylindrical bowl = 7 cm
Radius of cylindrical bowl, r = 7/2 cm = 3.5 cm
Bowl is filled with soup to a height of4cm, so h = 4 cm
Volume of soup in one bowl= πr2h
(22/7)×3.52×4 = 154
Volume of soup in one bowl is 154 cm3
Therefore,
Volume of soup given to 250 patients = (250×154) cm3= 38500 cm3
= 38.5litres. |
# Instantaneous acceleration: Definition, Formula, Examples
On the basis of the time interval considered for calculation, the acceleration is classified as average acceleration and instantaneous acceleration. In this article, we are discussing instantaneous acceleration in detail.
Contents:
## What is Instantaneous acceleration?
Instantaneous acceleration is the acceleration of the object at a specific instant during the motion. It indicates the change in velocity per unit time measured for a very small interval ‘dt’.
The object moving in a straight line may undergo an increase or decrease in acceleration or it may move with a uniform acceleration or zero acceleration. Thus in such cases, the average acceleration does not describe the motion of the object at every instant.
The average acceleration only provides the mean value of the acceleration instead of the actual acceleration of the object during the motion. While the instantaneous acceleration gives the exact acceleration at every instant during the motion.
In the above acceleration-time graph, the a_{\text{average}} indicates the average acceleration of the object throughout the motion while the a_{1} and a_{2} indicates the instantaneous acceleration of the object at specific instant t_{1} and t_{2} respectively.
## Equations:
In the form of limits, the instantaneous acceleration of the object is given by,
In the form of differentiation, the instantaneous acceleration of the object is given by,
## How to find the instantaneous acceleration?
It can be found by use of the following two methods:-
1] Analytical method:
This method is used to find the instantaneous acceleration when the equation of velocity in terms of time is given. It can be solved by using the method of limits or by using differentiation.
By using limits, the instantaneous acceleration is calculated as,
a_{\text{instantaneous}} = \lim_{\Delta t \to 0} \frac{V_{(t + \Delta t)}-V_{t}}{\Delta t}
By using differentiation, the instantaneous velocity can be calculated as,
a_{\text{instantaneous}} = \frac{dV}{dt}
2] Graphical method:-
This method is used for calculating instantaneous acceleration from the velocity-time graph.
## How to find instantaneous acceleration from a velocity-time graph?
The instantaneous acceleration from the different profiles of a velocity-time graph is calculated as follows:-
Case 1] When a velocity-time graph is linear:-
If the velocity-time plot has a linear nature, then it means that the object is moving with constant acceleration throughout its motion.
Thus in this case, the instantaneous acceleration is equal to the average acceleration of the object.
V_{\text{Instantaneous}} = V_{\text{Average}}
V_{\text{Instantaneous}} = \frac{\Delta V}{\Delta t}
V_{\text{Instantaneous}} = \frac{V_{2}-V_{1}}{t_{2}-t_{1}}
Case 2] When velocity-time profile is non-linear:-
If the velocity time profile is non-linear thus it means that the acceleration of the objeat is changing throughout the motion.
Thus in this case, the instantaneous acceleration of an object is different from the average acceleration. From such a graph, the instantaneous acceleration can be calculated by the use of the following steps:-
Step 1] Locate a point of a curve at a required time:-
Step 2] At this point draw a tangent to the curve:-
Step 3] Find the slope of the tangent:- The instantaneous acceleration is the slope of the tangent drawn to the velocity-time curve at that instant.
∴ The instantaneous acceleration is given by,
a_{\text{Instantaneous}} = Slope(1-2) = \frac{V_{2}-V_{1}}{t_{2}-t_{1}}
## Solved examples:
Given:
V = (5t^{2} +3t) m/s
t = 45 seconds
Solution:-
Method 1:- Using limits
The instantaneous acceleration of the object is given by,
a_{t} = \lim_{\Delta t \to 0} \frac{V_{(t + \Delta t)}-V_{t}}{\Delta t}
a_{t} = \lim_{\Delta t \to 0} \frac{[5(t + \Delta t)^{2} + 3(t + \Delta t)] – [5t^{2} +3t]}{\Delta t}
a_{t} = \lim_{\Delta t \to 0} \frac{5t^{2} + 10t\Delta t + 5\Delta t^{2}+3t+3\Delta t-5t^{2}-3t}{\Delta t}
a_{t} = \lim_{\Delta t \to 0} \frac{5\Delta t^{2}+10t\Delta t+3\Delta t}{\Delta t}
a_{t} = \lim_{\Delta t \to 0} 5\Delta t + 10t + 3
a_{t} = 10t + 3
Now the acceleration of the object at t = 45 seconds is given by,
a_{(t=45)} = 10(45) + 3
Method 2:- Using differentiation
a_{t} = \frac{dV}{dt}
a_{t} = \frac{d}{dt}(5t^{2} + 3t)
a_{t} = 10t + 3
The acceleration at t = 45 seconds is given by,
a_{(t=45)} = 10(45) + 3
Given:
t = 3.5 seconds
Solution:-
Step 1] Locate point on curve at time t = 3.5 seconds:
Step 2] Draw a tangent to the curve at point x:
Step 3] Find the slope of the tangent:
∴ The instantaneous acceleration at t = 3.5 second is given by,
a_{(t=3.5)} = Slope(1-2) = \frac{V_{2}-V_{1}}{t_{2}-t_{1}}
From above figure,
V_{2} = 0.5 m/s, V_{1}= 3 m/s,
t_{2} = 6 seconds, t_{1} = 1 second.
a_{(t=3.5)} = \frac{0.5-3}{6-1}
a_{(t=3.5)} = – 0.5 m/s²
Given:
V = (6t^{2} + 2t) m/s
a_{\text{average}} = 6 m/s²
Solution:-
The instantaneous acceleration of the particle is given by,
a_{\text{instantaneous}} = \frac{dV}{dt}
a_{\text{instantaneous}} = \frac{d}{dt}(6t^{2} + 2t)
a_{\text{instantaneous}} = 12t + 2
The time at which the V_{\text{average}} is equals to the V_{\text{instantaneous}} is given by,
a_{\text{average}} = a_{\text{instantaneous}}
6 = 12t + 2
## FAQs:
1. Why instantaneous acceleration is important?
It gives the acceleration of the object at a specific instant during the motion.
2. Is instantaneous acceleration always changing?
For the object moving with constant acceleration or constant retardation, the instantaneous acceleration never changes while for the object moving with continuously changing its acceleration, the instantaneous acceleration of the object always changes. |
# RELATIVELY PRIME NUMBERS
Two numbers are considered to be relatively prime, if there are no common factors other than 1. This means, no other integer can divide both the numbers exactly.
In other words, if the greatest common factor (GCF) of two numbers is 1, then the numbers are relatively prime.
or
If two numbers do not have common divisor other than 1, they are said to be relatively prime.
Here, the two numbers can both be primes as (3, 11) or both can be composites as (16, 35) or one can be a prime and other a composite as (7, 14).
We can extend this concept for than two numbers also.
Questions 1-9 : Find which of the following pairs of numbers are relatively prime.
Question 1 :
4 and 12
The above two numbers have the common divisors other than 1. They are 2 and 4.
So, 4 and 12 are not relatively prime.
Question 2 :
7 and 43
The above two numbers have no common divisor other than 1.
So, 7 and 43 are relatively prime.
Question 3 :
5 and 3
The above two numbers have no common divisor other than 1.
So, 5 and 3 are co-primes.
Question 4 :
8 and 17
The above two numbers have no common divisor other than 1.
So, 8 and 17 are co-primes.
Question 5 :
8 and 15
The above two numbers have no common divisor other than 1.
So, 8 and 15 are co-primes.
Question 6 :
14 and 21
The above two numbers have common divisor other than 1. That is 7.
So, 14 and 21 are not co-primes.
Question 7 :
2 and 4
The above two numbers have common divisor other than 1. That is 2.
So, 2 and 4 are not co-primes.
Question 8 :
1 and 2
The above two numbers have no common divisor other than 1.
So, 1 and 2 are co-primes.
Question 9 :
8, 15 and 49
Resolve 8, 15 and 49 into their prime factors.
8 = 2 x 2 x 2
15 = 3 x 5
49 = 7 x 7
There is no common factor or divisor for the numbers 8, 15 and 49.
So, the numbers 8, 15 and 49 are relatively prime.
Question 10 :
If two numbers are relatively prime, then, which is of the following must be true about the two numbers? Explain.
(A) Both the numbers must be prime
(B) One must be prime and other must be composite
(C) Both the numbers must be composite
(D) Both of them can be any numbers
The correct answer choice is (D).
That is, both of them can be any numbers.
Consider the following pairs of relatively prime numbers.
(2, 3) ----> both of them are prime
(4, 5) ----> one is composite and other one is prime
(8, 15) ----> both of them are composite
From the above examples, it is clear that if two two numbers are relatively prime, they can be any numbers.
Kindly mail your feedback to v4formath@gmail.com
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2. ### Implicit Differentiation
Mar 01, 24 08:48 PM
Implicit Differentiation - Concept - Examples |
Advice 1: How to find the side of a triangle if two sides are known
The solution of the problem developed by the ancient mathematician Pythagoras. All of the plurality of triangles select rectangular. In one of the angles is equal to 90 degrees. Sides that are adjacent to this angle are called the legs. And a third side connecting the other two sides is called the hypotenuse. Let one of the legs is equal to 15 centimeters, and the second was 9 inches. By the Pythagorean theorem we find the length of the hypotenuse.
Instruction
1
Find the square to the 1st side. Let's build the number 15 in the square, get the 225.
2
Find square 2nd side. Let's build the number 9 in the square, get 81.
3
Add the results of the 1st and 2nd step. Add 225 to 81, get 306.
4
Calculate the square root of the result of the 3rd step. Root of the number 306 is approximately equal to 17.49 inches. This is the length of the hypotenuse.
Note
If an unknown variable is one of the shorter sides, then at the 3rd step are different. From the square of the hypotenuse subtract the square of the leg. The rest is not changing. For example, was known to the hypotenuse is 17.49 inches. Also known side is 9 cm. Find the length of the other leg.
The number of 17.49 squared is equal to 305.9. The number 9 squared is 81. Subtract from the number to 305.9 81 received 224,9. Calculated from this number the root is received 14.99 inches - the length of the second leg. It turned out a little less than 15 centimeters because of 17.49 - we initially have a rough, rounded value.
To confidently solve problems by using the Pythagorean theorem, work out a few times. Decide on 50 different tasks with different rectangular triangles. You won't forget this theorem ever.
Advice 2 : How to find the side of a triangle
Side of the triangle is a direct, limited its vertices. All of them have figures of three, this number determines the number of almost all graphics characteristics: angle, midpoint, bisectors, etc. to find the side of the triangle, you should carefully examine the initial conditions of the problem and determine which of them may be basic or intermediate values to calculate.
Instruction
1
The sides of the triangle, like other polygons have their own names: the sides, base, and hypotenuse and legs of the figure with a right angle. This facilitates the calculations and formulas, making them more obvious even if the triangle is arbitrary. The figure of graphics, so it is always possible to arrange so as to make the solution of the problem more visible.
2
Sides of any triangle are connected and the other characteristics of the various ratios that help calculate the required value in one or more actions. Thus the more complex the task, the longer the sequence of steps.
3
The solution is simplified if the triangle is standard: the words "rectangular", "isosceles", "equilateral" immediately allocate a certain relationship between its sides and angles.
4
The lengths of the sides in a right triangle are connected by Pythagorean theorem: the sum of the squares of the legs equals the square of the hypotenuse. And the angles, in turn, are associated with the parties to the theorem of sines. It affirms the equality relations between the lengths of the sides and the trigonometric function sine of the opposite angle. However, this is true for any triangle.
5
Two sides of an isosceles triangle are equal. If their length is known, it is enough only one value to find the third. For example, suppose we know the height held to it. This cut divides the third side into two equal parts, and allocates two rectangular triangleH. Considered one of them, by the Pythagorean theorem find the leg and multiply it by 2. This will be the length of an unknown side.
6
Side of the triangle can be found through other sides, corners, length, altitude, median, bisector, perimeter size, area, inradius, etc. If you can't apply the same formula to produce a series of intermediate calculations.
7
Consider an example: find side of an arbitrary triangle, knowing the median ma=5, held for her, and the lengths of the other two medians mb=7 and mc=8.
8
Resented involves the use of formulas for the median. You need to find the way . Obviously, there should be three equations with three unknowns.
9
Write down the formulae for the medians:ma = 1/2•√(2•(b2 + c2) – a2) = 5;mb = 1/2•√(2•(a2 + c2) – b2) = 7;mc = 1/2•√(2•(a2 + b2) – c2) = 8.
10
Express c2 from the third equation and substitute it into the second:c2 = 256 – 2•a2 – 2•b2 b2 = 20 → c2 = 216 – a2.
11
Lift both sides of the first equation in the square and find a by entering explicit values:25 = 1/4•(2•20 + 2•(216 – a2) – a2) → a ≈ 11,1.
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# Algebra 1 : How to factor a trinomial
## Example Questions
### Example Question #11 : How To Factor A Trinomial
Factor completely:
The polynomial cannot be factored further.
Explanation:
Rewrite this as
Use the -method by splitting the middle term into two terms, finding two integers whose sum is 1 and whose product is ; these integers are , so rewrite this trinomial as follows:
Now, use grouping to factor this:
### Example Question #4241 : Algebra 1
For what value of allows one to factor a perfect square trinomial out of the following equation:
Explanation:
Factor out the 7:
Take the 8 from the x-term, cut it in half to get 4, then square it to get 16. Make this 16 equal to C/7:
Solve for C:
### Example Question #1 : Factoring Polynomials
Factor the trinomial .
Explanation:
We can factor this trinomial using the FOIL method backwards. This method allows us to immediately infer that our answer will be two binomials, one of which begins with and the other of which begins with . This is the only way the binomials will multiply to give us .
The next part, however, is slightly more difficult. The last part of the trinomial is , which could only happen through the multiplication of 1 and 2; since the 2 is negative, the binomials must also have opposite signs.
Finally, we look at the trinomial's middle term. For the final product to be , the 1 must be multiplied with the and be negative, and the 2 must be multiplied with the and be positive. This would give us , or the that we are looking for.
In other words, our answer must be
to properly multiply out to the trinomial given in this question.
### Example Question #11 : Polynomials
Factor:
Explanation:
One way to factor a trinomial like this one is to put the terms of the polynomial into a box/grid:
Notice that there are 4 boxes but only 3 terms. To fix this, we find two numbers that add to 22 [the middle coefficient] and multiply to -48 [the product of the first and last coefficients]. By examining the factors of -48, we discover that these numbers must be -2 and 24.
Now we can put the terms into the box, by splitting the 22x into -2x and 24x:
To finish factoring, determine the greatest common factor of each of the rows and columns. For instance, and have a greatest common factor of .
Our final answer just combines the factors on the top and side into binomials. In this case, .
### Example Question #11 : Polynomials
Factor:
Explanation:
One way to factor a trinomial like this one is to put the terms of the polynomial into a box/grid:
Notice that there are 4 boxes but only 3 terms. To fix this, we find two numbers that add to -7 [the middle coefficient] and multiply to -30 [the product of the first and last coefficients]. By examining the factors of -30, we discover that these numbers must be -10 and 3.
Now we can put the terms into the box, by splitting the -7x into -10x and 3x:
To finish factoring, determine the greatest common factor of each of the rows and columns. For instance, and have a greatest common factor of , and and -15 have a greatest common factor of 3.
Our final answer just combines the factors on the top and side into binomials. In this case, .
### Example Question #15 : Trinomials
Factor the trinomial:
Explanation:
The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of are :
Next, find the factors of . In this case, we could have or . Either combination can potentially produce , so the signage is important here.
Note that since the last term in the ordered trinomial is positive, both factors must have the same sign. Further, since the middle term in the ordered triniomial is negative, we know the signs must both be negative.
Therefore, we have two possibilities, or .
Let's solve for both, and check against the original triniomial.
Thus, our factors are and .
### Example Question #11 : Variables
Factor the trinomial:
Explanation:
The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of are :
Next, find the factors of . In this case, we could have or . Since the factor on our leading term is , and no additive combination of and can create , we know that our factors must be .
Note that since the last term in the ordered trinomial is negative, the factors must have different signs.
Therefore, we have two possibilities, or .
Let's solve for both, and check against the original triniomial.
Thus, our factors are and .
### Example Question #17 : Trinomials
Factor the trinomial:
Explanation:
The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of are :
Next, find the factors of . In this case, we could have or . Since the factor on our leading term is , and no additive combination of and can create , we know that our factors must be .
Note that since the last term in the ordered trinomial is positive, the factors must have different signs. Since the middle term is also positive, the signs must both be positive.
Therefore, we have only one possibility, .
Let's solve, and check against the original triniomial.
Thus, our factors are and .
### Example Question #18 : Trinomials
Factor this trinomial, then solve for
and
and
and
and
and
Explanation:
The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of are :
Next, find the factors of . In this case, we could have or . Since the factor on our leading term is , and no additive combination of and or of and can create our middle term of , we know that our factors must be .
Note that since the last term in the ordered trinomial is positive, the factors must have the same sign. Since the middle term is negative, both factors must have a negative sign.
Therefore, we have one possibility, .
Let's solve, and check against the original triniomial, before solving for .
Thus, our factors are and .
Now, let's solve for . Simply plug and play:
Therefore, and .
Note that, in algebra, we can represent this by showing . However, writing the word "and" is perfectly acceptable.
### Example Question #19 : Trinomials
Factor the trinomial below,
Explanation:
A factored trinomial is in the form , where the second term and the third term.
To factor a trinomial you first need to find the factors of the third term. In this case the third term is
Factors of are:
The factors you choose not only must multiply to equal the third term, they must also add together to equal the second term.
In this case they must equal .
To check your answer substitute the factors of into the binomials and use FOIL.
First terms:
Outside terms:
Inside terms:
Last terms:
Simplify from here by combining like terms
Using FOIL returned it to the original trinomial, therefore the answer is: |
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Q. 32
Expert-verified
Found in: Page 1
Calculus
Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861
In Exercises 24-34 sketch the parametric curve by eliminating the parameter.$x=\mathrm{sec}t,y=\mathrm{tan}t,t\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
The graphical representation by using the points$\left(-1,0\right)\left(1,0\right)\left(2,-\sqrt{3}\right)\left(2,\sqrt{3}\right)$ is as follows,
Therefore, the equation after elimination of the parameter is ${x}^{2}-{y}^{2}=1$
See the step by step solution
Step 1: Given information
$x=\mathrm{sec}t,y=\mathrm{tan}t,t\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
Step 2: Calculation
Consider the parametric equations $x=\mathrm{sec}t,y=\mathrm{tan}t,t\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right).$
The objective is to sketch the parametric curve by eliminating the parameter.
Take the equation $x=\mathrm{sec}t.$
Square the equation on both sides.
${x}^{2}={\mathrm{sec}}^{2}t$
Take the equation $y=\mathrm{tan}t$.
Square the equation on both sides. Then.
${y}^{2}={\mathrm{tan}}^{2}t$
Now subtract the equation ${y}^{2}={\mathrm{tan}}^{2}t$ from ${x}^{2}={\mathrm{sec}}^{2}t$.
Thus.
${x}^{2}-{y}^{2}={\mathrm{sec}}^{2}t-{\mathrm{tan}}^{2}t\phantom{\rule{0ex}{0ex}}{x}^{2}-{y}^{2}=1\left[\text{since}{\mathrm{sec}}^{2}t-{\mathrm{tan}}^{2}t=1\right]$
In order to draw the graph of the equation assume $x=-1,1,2$.
Substitute $x=-1$ in the equation ${x}^{2}-{y}^{2}=1$.
Then,
${1}^{2}-{y}^{2}=1\phantom{\rule{0ex}{0ex}}1-{y}^{2}=1\phantom{\rule{0ex}{0ex}}y=0\text{simplify}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(-1,0\right)$
Substitute $x=1$ in the equation ${x}^{2}-{y}^{2}=1$
Then.
$\left(-1{\right)}^{2}-{y}^{2}=1\phantom{\rule{0ex}{0ex}}1-{y}^{2}=1\phantom{\rule{0ex}{0ex}}y=0\text{simplify}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(-1,0\right)$
Substitute $x=2$ in the equation ${x}^{2}-{y}^{2}=1$.
Then,
${2}^{2}-{y}^{2}=1\phantom{\rule{0ex}{0ex}}4-{y}^{2}=1$
Add $-4$ on both sides of the equation.
$4-{y}^{2}-4=1-4\phantom{\rule{0ex}{0ex}}4-{y}^{2}-4--3\phantom{\rule{0ex}{0ex}}-{y}^{2}=-3\phantom{\rule{0ex}{0ex}}y=\sqrt{3}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(2,-\sqrt{3}\right)\left(2,\sqrt{3}\right)$
The graphical representation by using the points $\left(-1,0\right)\left(1,0\right)\left(2,-\sqrt{3}\right)\left(2,\sqrt{3}\right)$ is as follows,
Therefore, the equation after elimination of the parameter is ${x}^{2}-{y}^{2}=1$ |
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# wikiHow to Use Distributive Property to Solve an Equation
The distributive property is a rule in mathematics to help simplify an equation with parentheses. You learned early that you perform the operations inside parentheses first, but with algebraic expressions, that isn’t always possible. The distributive property allows you to multiply the term outside the parentheses by the terms inside. You need to make sure that you do it properly so you don’t lose any information and solve the equation correctly. You can also use the distributive property to simplify equations involving fractions.
### Method 1 Using the Basic Distributive Property
1. 1
Multiply the term outside of the parentheses by each term in the parentheses. To do this, you are essentially distributing the outer term into the inner terms. Multiply the term outside the parentheses by the first term in the parentheses. Then multiply it by the second term. If there are more than two terms, keep distributing the term until there are no terms left. Keep whatever operation (plus or minus) is in the parentheses.[1]
• ${\displaystyle 2(x-3)=10}$
• ${\displaystyle 2(x)-(2)(3)=10}$
• ${\displaystyle 2x-6=10}$
2. 2
Combine like terms. Before you can solve the equation, you will have to combine like terms. Combine all numerical terms with each other. Separately, combine any variable terms. To simplify the equation, arrange the terms so the variables are on one side of the equals sign and the constants (numbers only) are on the other.[2]
• ${\displaystyle 2x-6=10}$…..(original problem)
• ${\displaystyle 2x-6(+6)=10(+6)}$….. (Add 6 to both sides)
• ${\displaystyle 2x=16}$….. (Variable on left; constant on right)
3. 3
Solve the equation. Solve for ${\displaystyle x}$ by dividing both sides of the equation by the coefficient in front of the variable.[3]
• ${\displaystyle 2x=16}$…..(original problem)
• ${\displaystyle 2x/2=16/2}$…..(divide both sides by 2)
• ${\displaystyle x=8}$…..(solution)
### Method 2 Distributing Negative Coefficients
1. 1
Distribute a negative number together with its negative sign. If you have a negative number multiplying a term or terms within parentheses, be sure to distribute the negative to each term inside the parentheses.[4]
• Remember the basic rules of multiplying negatives:
• Neg. x Neg. = Pos.
• Neg. x Pos. = Neg.
• Consider the following example:
• ${\displaystyle -4(9-3x)=48}$….. (original problem)
• ${\displaystyle -4(9)-(-4)(3x)=48}$…..(distribute (-4) to each term)
• ${\displaystyle -36-(-12x)=48}$…..(simplify the multiplication)
• ${\displaystyle -36+12x=48}$…..(notice that ‘minus -12’ becomes +12)
2. 2
Combine like terms. After you complete the distribution, you then need to simplify the equation by moving all the variable terms to one side of the equals sign, and all the numbers without variables to the other. Do this by a combination of addition or subtraction.[5]
• ${\displaystyle -36+12x=48}$…..(original problem)
• ${\displaystyle -36(+36)+12x=48+36}$…..(add 36 to each side)
• ${\displaystyle 12x=84}$…..(simplify the addition to isolate the variable)
3. 3
Divide to find the final solution. Solve the equation by dividing both sides of the equation by whatever the coefficient of the variable is. This should result in a single variable on one side of the equation, with the result on the other.[6]
• ${\displaystyle 12x=84}$…..(original problem)
• ${\displaystyle 12x/12=84/12}$…..(divide both sides by 12)
• ${\displaystyle x=7}$…..(solution)
4. 4
Treat subtraction as adding (-1). Whenever you see a minus sign in an algebra problem, particularly if it comes before a parenthesis, you should imagine that it says + (-1). This will help you correctly distribute the negative to all the terms within the parentheses. Then solve the problem as before.[7]
• For example, consider the problem, ${\displaystyle 4x-(x+2)=4}$. To be certain that you distribute the negative properly, rewrite the problem to read:
• ${\displaystyle 4x+(-1)(x+2)=4}$
• Then distribute the (-1) to the terms inside the parentheses as follows:
• ${\displaystyle 4x+(-1)(x+2)=4}$…..(revised problem)
• ${\displaystyle 4x-x-2=4}$…..(multiply (-1) times x and times 2)
• ${\displaystyle 3x-2=4}$…..(combine terms)
• ${\displaystyle 3x-2+2=4+2}$…..(add 2 to both sides)
• ${\displaystyle 3x=6}$…..(simplify terms)
• ${\displaystyle 3x/3=6/3}$…..(divide both sides by 3)
• ${\displaystyle x=2}$…..(solution)
### Method 3 Using the Distributive Property to Simplify Fractions
1. 1
Identify any fractional coefficients or constants. Sometimes, you may have a problem that contains fractions as coefficients or constants. You are allowed to leave them as they are and apply the basic rules of algebra to solve the problem. However, using the distributive property can often simplify the solution by turning the fractions into integers.[8]
• Consider the example ${\displaystyle x-3={\frac {x}{3}}+{\frac {1}{6}}}$. The fractions in this problem are ${\displaystyle {\frac {x}{3}}}$ and ${\displaystyle {\frac {1}{6}}}$.
2. 2
Find the lowest common multiple (LCM) for all denominators. For this step, you can ignore all the integers. Look only at the fractions, and find the LCM for all the denominators. To find the LCM, you need the smallest number that is evenly divisible by the denominators of the fractions in the equation. In this example, the denominators are 3 and 6, so the LCM is 6.[9]
3. 3
Multiply all terms of the equation by the LCM. Remember that you can perform any operation that you want to an algebra equation, as long as you do it equally to both sides. Multiply all terms of the equation by the LCM, and the fractions will cancel out and “become” integers. Place parentheses around the entire left and right sides of the equation and then perform the distribution:[10]
• ${\displaystyle x-3={\frac {x}{3}}+{\frac {1}{6}}}$…..(original equation)
• ${\displaystyle (x-3)=({\frac {x}{3}}+{\frac {1}{6}})}$…..(insert parentheses)
• ${\displaystyle 6(x-3)=6({\frac {x}{3}}+{\frac {1}{6}})}$…..(multiply both sides by LCM)
• ${\displaystyle 6x-6(3)=6({\frac {x}{3}})+6({\frac {1}{6}})}$…..(distribute multiplication)
• ${\displaystyle 6x-18=2x+1}$…..(simplify multiplication)
4. 4
Combine like terms. Combine all of the terms so all the variables appear on one side of the equation, and all constants appear on the other. Use basic operations of addition and subtraction to move terms from one side to the other.[11]
• ${\displaystyle 6x-18=2x+1}$…..(simplified problem)
• ${\displaystyle 6x-2x-18=2x-2x+1}$…..(subtract 2x from both sides)
• ${\displaystyle 4x-18=1}$…..(simplify subtraction)
• ${\displaystyle 4x-18+18=1+18}$…..(add 18 to both sides)
• ${\displaystyle 4x=19}$…..(simplify addition)
5. 5
Solve the equation. Find the final solution by dividing both sides of the equation by the coefficient of the variable. This should leave a single x term on one side of the equation, and the numerical solution on the other.[12]
• ${\displaystyle 4x=19}$…..(revised problem)
• ${\displaystyle 4x/4=19/4}$…..(divide both sides by 4)
• ${\displaystyle x={\frac {19}{4}}{\text{ or }}4{\frac {3}{4}}}$…..(final solution)
### Method 4 Distributing a Long Fraction
1. 1
Interpret a long fraction as distributed division. You may occasionally see a problem that contains multiple terms in the numerator of a fraction, over a single denominator. You need to treat this as a distributive problem and apply the denominator to each term of the numerator. You can rewrite the fraction to show the distribution, as follows:
• ${\displaystyle {\frac {4x+8}{2}}=4}$.....(original problem)
• ${\displaystyle {\frac {4x}{2}}+{\frac {8}{2}}=4}$.....(distribute the denominator to each term of the numerator)
2. 2
Simplify each numerator as a separate fraction. After distributing the denominator to each term, you can then simplify each term individually.
• ${\displaystyle {\frac {4x}{2}}+{\frac {8}{2}}=4}$.....(revised problem)
• ${\displaystyle 2x+4=4}$.....(simplify the fractions)
3. 3
Isolate the variable. Proceed to solve the problem by isolating the variable on one side of the equation and moving the constant terms to the other side. Do this with a combination of addition and subtraction steps, as needed.
• ${\displaystyle 2x+4=4}$.....(revised problem)
• ${\displaystyle 2x+4-4=4-4}$.....(subtract 4 from both sides)
• ${\displaystyle 2x=0}$.....(isolated x on one side)
4. 4
Divide by the coefficient to solve the problem. In the final step, divide by the coefficient of the variable. This should lead you to the final solution, with the single variable on one side of the equation and the numerical solution on the other.
• ${\displaystyle 2x=0}$.....(revised problem)
• ${\displaystyle 2x/2=0/2}$.....(divide both sides by 2)
• ${\displaystyle x=0}$.....(solution)
5. 5
Avoid the common trap of dividing only one term. It is tempting (but incorrect) to divide the first numerator term by the denominator and cancel out the fraction. A mistake like this, for the problem above, would look like the following:
• ${\displaystyle {\frac {4x+8}{2}}=4}$.....(original problem)
• ${\displaystyle 2x+8=4}$.....(divide only 4x by 2 instead of the full numerator)
• ${\displaystyle 2x+8-8=4-8}$
• ${\displaystyle 2x=-4}$
• ${\displaystyle x=-2}$..... (incorrect solution)
6. 6
Check the correctness of your solution. You can always check your work by inserting your solution into the original problem. When you simplify, you should reach a true statement. If you simplify and get an incorrect statement, then your solution was incorrect. For this example, test the two solutions of x=0 and x=-2 to see which is correct.
• Begin with the solution x=0:
• ${\displaystyle {\frac {4x+8}{2}}=4}$.....(original problem)
• ${\displaystyle {\frac {4(0)+8}{2}}=4}$.....(insert 0 for x)
• ${\displaystyle {\frac {0+8}{2}}=4}$
• ${\displaystyle {\frac {8}{2}}=4}$
• ${\displaystyle 4=4}$.....(true statement. This is the correct solution.)
• Try the "false" solution of x=-2:
• ${\displaystyle {\frac {4x+8}{2}}=4}$.....(original problem)
• ${\displaystyle {\frac {4(-2)+8}{2}}=4}$.....(insert -2 for x)
• ${\displaystyle {\frac {-8+8}{2}}=4}$
• ${\displaystyle {\frac {0}{2}}=4}$
• ${\displaystyle 0=4}$.....(incorrect statement. Therefore, x=-2 is false.)
## Community Q&A
Search
• How would I do -6(a+8)?
wikiHow Contributor
-6(a+8) = -6a-48, the number outside the bracket multiplies everything inside the bracket. i.e. (-6 x a) + (-6 x 8) = -6a - 48.
• Where do I put the parentheses in this equation: 6x+2-2x=4x+12?
On the left side of the equation, enclose "x+2" inside parentheses. It will look like 6(x+2)-2x=4x+12.
• How would I solve 9x15=(9x9)+(9x )=?
wikiHow Contributor
This is an effort to show you that you can rewrite a multiplication problem, using the distributive property. Instead of performing 9x15, you can break the 15 into smaller parts. In this case, someone has chosen 9 and the first part and you have to fill in the rest. Since 9+6=15, the blank space in your problem will be filled in with a 6. You will then have 9x15=(9x9)+(9x6). If you can resolve those two parentheses in your head and add them, you can do the solution. 9x9=81, and 9+6=54, so 81+54=135. Therefore the original problem 9x15=135.
• What do I do with a problem evolving two numbers being subtracted outside the parentheses like: Ten minus five Parenthesesnine n minus nine?
wikiHow Contributor
If the equation looks like this: 10-5(9n-9), then distribute the 5 to the equation inside the parenthesis it will look like this 10-45n-45. You would solve for N like a regular equation and the answer you will later get it:-35-45n.
• How do I solve 8x-3 (4-5)?
wikiHow Contributor
I suspect you copied this problem incorrectly, but I'll work with what you wrote. First, you cannot "solve" this because there is no equals sign. All you can do is simplify this. Perform the step inside the parentheses first. You will get 8x-3(-1). Then multiply the -3 times -1 to get +3. The simplified problem is 8x+3. [However, as I said, I think you copied this wrong. Check your source.]
• How do I solve 8(22) as a distributive problem?
wikiHow Contributor
Find the "difficult" number in the problem and rewrite it as the sum of two "easy" numbers. For example, 22 can be rewritten as (20+2). Then you can write your original problem as 8(20+2). Use the distributive property to get (8)(20)+(8)(2)=160+16=176.
• If I had the expression -8a(3b), then would the answer be -24a + 3ab?
wikiHow Contributor
Since you can't multiply a's and b's together, the answer would just be -8a x 3b.
• How do I use the distribution property to find the product of 7 and 16?
wikiHow Contributor
The distributive property can help you do some mental math and perform multiplication in your head. You can rewrite 7 x 16 as 7(10+6). Then the distributive property makes this fairly easy, as it becomes 7*10=70, and 7*6=42. Just add those two parts together, 70+42=112. Therefore the answer to 7x16=112. If you practice this you can quickly do multiplications in your head.
• How can I change a decimal into a fraction?
Make the decimal the numerator, and use 100 as the denominator, then reduce the fraction if possible.
• How do you use the distributive property to solve 14x6?
wikiHow Contributor
When multiplying a number in the "teens," you can break it into 10 and whatever is left. This creates two simple multiplications that you can do in your head. Thus, 14x6=(10+4)(6). Then use the distributive property to rewrite this as (10x6)+(4x6). This is (60)+(24)=84.
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## Tips
• You can also use the distributive property to simplify some multiplication problems. You can "break" numbers into groups of 10 and whatever is left, to create easy mental math. For example, you can rewrite 8x16 into 8(10+6). This is then just 80+48=128. Another example, 7*24=7(20+4)=7(20)+7(4)=140+28=168. Practice doing these in your head and mental math becomes pretty easy.
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# Exponential derivative – Derivation, Explanation, and Example
In differential calculus, we’ll need to also establish a rule for exponential derivative. Our discussion will revolve around the formula for $\dfrac{d}{dx} a^x$ and $\dfrac{d}{dx} e^x$. Exponential functions have a wide range of applications in different STEM fields, so it’s essential to understand how its derivative behaves.
The derivative of an exponential function will be the function itself and a constant factor. A special case occurs for $\boldsymbol{e^x}$ since the derivative is $\boldsymbol{e^x}$ as well.
In this article, we’ll understand how we could come up with the exponential functions’ derivative rules. We’ll also see how we can apply them to differentiate a wide range of functions. This is why it’s important to take a refresher on the following topics:
Let’s begin by understanding the factors that we can find in the exponential function’s derivative. We’ll eventually learn how to apply this rule to solve problems involving exponential functions.
## What is the derivative of an exponential function?
Recall that an exponential function has a general form of $y = a^x$, where $a > 0$ but $a \neq 1$. We call $a$ the base, and $x$ is in the exponent part of the expression. For functions like this, the derivative will be the function itself times the base’s natural logarithm.
\begin{aligned}\dfrac{d}{dx} a^x &= a^x \ln a\end{aligned}
Let’s go ahead and observe how an exponential function and its derivative would look like when graphed on an $xy$-plane.
The graph of $f(x) = 2^x$ and $f’(x) = 2^x \cdot \ln2$ confirm the fact that the derivative of an exponential function is an exponential function itself. The only difference is that the derivative will have a constant factor.
There’s a special case for the exponential derivative rules, which occurs when $a = e$. Recall that $\ln e = 1$, so we can use this logarithmic property to simplify the formula for $\dfrac{d}{dx} e^x$ as shown below.
\begin{aligned}\dfrac{d}{dx} e^x &= e^x\ln e\\&= e^x \cdot 1\\&= e^x\end{aligned}
These two derivative rules will also apply to composite functions. We have to account for the inner function’s derivative through chain rule.
Here’s a table summarizing the derivative rules for exponential functions:
Derivative Rules Base of $\boldsymbol{a}$ Derivative RulesBase of $\boldsymbol{e}$ \begin{aligned}\dfrac{d}{dx} a^x &= a^x \ln a\end{aligned} \begin{aligned}\dfrac{d}{dx} e^x &= e^x\end{aligned} \begin{aligned}\dfrac{d}{dx} a^{[g(x)]} &= a^{[g(x)]} \ln [g(x)] \cdot g’(x)\end{aligned} \begin{aligned}\dfrac{d}{dx} e^{[g(x)]} &= e^{[g(x)]} \cdot g’(x)\end{aligned}
Keep these four variations in mind when working on the sample problems we’ve provided. For now, let’s understand the reason behind these derivative rules by differentiating $a^x$ using the formal definition of derivatives. We’ll eventually learn how to apply this rule to solve problems involving exponential functions.
### Proof of the derivative rule for exponential functions
Recall that $\dfrac{d}{dx} f(x) = \lim_{h\rightarrow 0}\dfrac{f(x + h) – f(x)}{h}$, so we can use this to confirm the derivative that we’ve just learned for $y = a^x$.
• Use the product rule for exponents,$a^{m} \cdot a^n = a^{m+n}$, to factor $a^x$ from the numerator.
• Since $a^x$ is considered a constant for this expression, we can factor it out of $\lim_{h \rightarrow 0}$.
• Evaluate the limit by setting $h$ to $0$.
\begin{aligned}y’ &= \lim_{h\rightarrow 0} \dfrac{a^{x + h} – a^x}{h}\\&= \lim_{h\rightarrow 0} \dfrac{a^x(a^h – 1)}{h}\\&= a^x\lim_{h\rightarrow 0}\dfrac{a^h – 1}{h}\end{aligned}
The limit of exponential functions of the form $\dfrac{a^h – 1}{h}$ as $h \rightarrow 0$ is said to be $\ln a$, so we’ll use this property to rewrite $y’$.
\begin{aligned}y’ &=a^x \ln a\end{aligned}
This confirms the exponential derivative rule – $\dfrac{d}{dx} a^x = a^x \ln a$.
## How to find the derivative of an exponential function?
The most important step in differentiating exponential functions is to make sure that we’re actually working with exponential functions. Make sure that the function has a constant base and $\boldsymbol{x}$ is found at the exponent.
Once we’ve confirmed that the function (or the composite function’s outer layer) has a form of either $y= a^x$ or $y = e^x$, we can then apply the derivative rule we’ve just learned.
• To find the function’s derivative, copy the original function.
• Multiply this with the natural logarithm of the base. (Skip this when working with $y= e^x$.)
• If working with composite functions, multiply the result with the derivative of the inner function.
We’ll show you four examples in which the derivative rule for exponential functions were applied. We’ll include the other rules used whenever possible.
\begin{aligned}\boldsymbol{\dfrac{d}{dx} a^x = a^x \ln a}\end{aligned} \begin{aligned}\boldsymbol{\dfrac{d}{dx} e^x = e^x}\end{aligned} \begin{aligned}\dfrac{d}{dx} (5^x) &= 5^x \cdot \ln 5,\phantom{x}\color{DarkOrange}\text{Derivative of }a^x \end{aligned} \begin{aligned}\dfrac{d}{dx} (6e^x) &= 6\cdot \dfrac{d}{dx}e^x,\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= 6\cdot{\color{DarkOrange}e^x},\phantom{x}\color{DarkOrange}\text{Derivative of }e^x \\&= 6e^x\end{aligned} \begin{aligned}\dfrac{d}{dx} (2^{3x}) &= {\color{DarkOrange}2^{3x} \cdot \ln 2}\cdot {\color{Green}\dfrac{d}{dx} 3x},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (2^{3x}\ln 2)\cdot\left({\color{DarkOrange}3\cdot \dfrac{d}{dx}x} \right ),\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= (2^{3x}\ln 2)\cdot3({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&=(3\ln 2)2^{3x} \end{aligned} \begin{aligned}\dfrac{d}{dx} (e^{5x}) &= {\color{DarkOrange}e^{5x}}\cdot {\color{Green}\dfrac{d}{dx} 5x},\phantom{x}{\color{DarkOrange}\text{Derivative of }e^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (e^{5x})\cdot\left({\color{DarkOrange}5\cdot \dfrac{d}{dx}x} \right ),\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= (e^{5x})\cdot5({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&=5e^{5x} \end{aligned}
These four examples show us how functions and composite functions can be differentiated using the exponential functions’ derivative rules. Review these four functions and when you’re ready, try out the practice problems below!
Example 1
Find the derivative of the following exponential functions.
a. $f(x) = 3^{4x}$
b. $g(x) = 2^x – 6^x$
c. $h(x) = 5^x + e^x – 4^x$
Solution
For $f(x)$, we can apply the derivative rule for exponential function and the chain rule to differentiate it.
\begin{aligned}\dfrac{d}{dx} (3^{4x}) &= {\color{DarkOrange}3^{4x} \cdot \ln 3}\cdot {\color{Green}\dfrac{d}{dx} 4x},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (3^{4x}\ln 3)\cdot\left({\color{DarkOrange}4\cdot \dfrac{d}{dx}x} \right ),\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= (3^{4x}\ln 3)\cdot4({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&=(4\ln 3)3^{4x} \end{aligned}
The second function is straightforward. We’ll have to apply the difference rule and the exponential function’s derivative rule to determine $g’(x)$.
\begin{aligned}\dfrac{d}{dx} (2^x – 6^x) &= \dfrac{d}{dx} 2^x – \dfrac{d}{dx} 6^x,\phantom{x}\color{DarkOrange}\text{Difference Rule}\\&= {\color{DarkOrange}(2^x\ln 2)} – {\color{DarkOrange}(6^x\ln 6)},\phantom{x}\color{DarkOrange}\text{Derivative of }a^x\end{aligned}
Hence, we have $g’(x) = (2^x\ln 2) – 6^x\ln 6$.
We’ll apply a similar process for $h(x)$ as shown below.
\begin{aligned}\dfrac{d}{dx} (5^x +e^x -4^x) &= \dfrac{d}{dx} 5^x+ \dfrac{d}{dx} e^x – \dfrac{d}{dx}4^x,\phantom{x}\color{DarkOrange}\text{Sum & Difference Rules}\\&= \dfrac{d}{dx} 5^x- \dfrac{d}{dx}4^x+ \dfrac{d}{dx} e^x \\&= {\color{DarkOrange}(5^x \ln 5 -4^x \ln 4)} + {\color{Green}e^x},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Derivative of }e^x\\&= 5^x \ln 5 -4^x \ln 4 +e^x \end{aligned}
This shows that $h’(x) = 5^x \ln 5 -4^x \ln 4 +e^x$.
Example 2
Find the derivative of the following exponential functions.
a. $f(x) = 2^{4x + 2}$
b. $g(x) = 3^{3x^2 – 4}$
c. $h(x) = e^{6^x}$
Solution
These three sets of functions are composite functions with $y = 2^x$, $y = 3^x$, and $y = e^x$, respectively. We’ll have to differentiate the inner functions for each item through chain rule. Let’s begin with $f(x) = 2^{4x + 2}$.
\begin{aligned}\dfrac{d}{dx} (2^{4x + 2}) &= {\color{DarkOrange}2^{4x + 2} \cdot \ln 2}\cdot {\color{Green}\dfrac{d}{dx} (4x + 2)},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (2^{4x + 2}\ln 2)\cdot\left({\color{DarkOrange}\dfrac{d}{dx}4x + \dfrac{d}{dx} 2 } \right ),\phantom{x}\color{DarkOrange}\text{Sum Rule}\\&= (2^{4x + 2}\ln 2)\cdot\left({\color{DarkOrange}4\dfrac{d}{dx}x } + {\color{Green}0}\right ),\phantom{x}{\color{DarkOrange}\text{Constant Multiple Rule}}\text{ & }\color{Green}\text{Constant Rule} \\&= (2^{4x + 2}\ln 2)\cdot4({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&= (4\ln 2)(2^{4x + 2})\end{aligned}
With the help of fundamental derivative rules, we have $f’(x) = (4\ln 2)(2^{4x + 2})$.We’ll apply a similar process to differentiate $g(x)$.
\begin{aligned}\dfrac{d}{dx} (3^{3x^2 – 4}) &= {\color{DarkOrange}3^{3x^2 – 4} \cdot \ln 3}\cdot {\color{Green}\dfrac{d}{dx} (3x^2 – 4)},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (3^{3x^2 – 4} \ln 3)\cdot\left({\color{DarkOrange}\dfrac{d}{dx}3x^2 – \dfrac{d}{dx} 4 } \right ),\phantom{x}\color{DarkOrange}\text{Difference Rule}\\&= (3^{3x^2 – 4} \ln 3)\cdot\left({\color{DarkOrange}3\dfrac{d}{dx}x^2 } + {\color{Green}0}\right ),\phantom{x}{\color{DarkOrange}\text{Constant Multiple Rule}}\text{ & }\color{Green}\text{Constant Rule} \\&= (3^{3x^2 – 4} \ln 3)\cdot3({\color{DarkOrange}2x}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&= (6x\ln 3)(3^{3x^2 – 4} )\end{aligned}
Hence, we have $g’(x) = (6x\ln 3)(3^{3x^2 – 4} )$. Let’s now work on $h(x)$ use the following rules to begin finding $h’(x)$.
• Derivative rule for $e^x$: $\dfrac{d}{dx} e^x = e^x$.
• Apply the chain rule and use $\dfrac{d}{dx} a^x = a^x \ln x$.
\begin{aligned}\dfrac{d}{dx} (e^{6^x}) &= {\color{DarkOrange}e^{6^x}}\cdot {\color{Green}\dfrac{d}{dx} 6^x},\phantom{x}{\color{DarkOrange}\text{Derivative of }e^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (e^{6^x})\cdot\left({\color{DarkOrange}} 6^x \ln 6\right ),\phantom{x}\color{DarkOrange}\text{Derivative of }a^x\\&= (6^x \ln 6)(e^{6x})\end{aligned}
This shows that $h’(x) = (6^x \ln 6)(e^{6x})$.
We’ve been working on differentiating different functions. Why don’t we try using this derivative rule to answer a word problem this time?
Example 3
An organism that is being studied has an initial population of $500$. After $x$ days, the population can be modeled by the function, $p(x)=500e^{0.4x}$. Jack tells his colleagues that they should expect the ratio of $p’(x)$ and $p(x)$ to be a constant. Is Jack correct?
Solution
Use the natural exponential function’s derivative rule and the chain rule to find the expression for $p’(x)$.
\begin{aligned}\dfrac{d}{dx} (500e^{0.4x}) &= 500 \dfrac{d}{dx}e^{0.4x},\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= 500({\color{DarkOrange}e^{0.4x}})\cdot {\color{Green}\dfrac{d}{dx} (0.4x)},\phantom{x}{\color{DarkOrange}\text{Derivative of }e^x}\text{ & }\color{Green}\text{Chain Rule}\\&= 500e^{0.4x}\cdot\left({\color{DarkOrange}0.4\dfrac{d}{dx}x } \right ),\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= 500e^{0.4x}\cdot (0.4)({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&=200e^{0.4x}\end{aligned}
Now that we have $p’(x)$, we can check whether $\dfrac{p’(x)}{p(x)}$ is indeed a constant. You can use fundamental derivative rules to further simplify the expression for $p’(x)$.
\begin{aligned} \dfrac{p’(x)}{p(x)} &= \dfrac{200e^{0.4x}}{500e^{0.4x}}\\&= \dfrac{200}{500}\\&= 0.4\end{aligned}
This shows that the ratio of the rate of change of the population and the actual population size is constant, so Jack is correct.
### Practice Questions
1. Find the derivative of the following exponential functions.
a. $f(x) = 6^{2x}$
b. $g(x) = 4^x + 5^x$
c. $h(x) = 8^x – 2e^x + 6^x$
2. Find the derivative of the following exponential functions.
a. $f(x) = 8^{3x – 5}$
b. $g(x) = 6^{\sqrt{x – 1}}$
c. $h(x) = 5^{e^x}$
3. An organism that is being studied has an initial population of $800$. After $x$ days, the population can be modeled by the function, $p(x)=800e^{0.6x}$.
a. What is the ratio of $p’(x)$ and $p(x)$?
b. What is the rate of change of $p’(x)$ after $2$ days?
1.
a. $f’(x) = 9^x2^{2x + 1}\ln 6$
b. $g’(x) = 4^x\ln 4 – 5^x \ln 5$
c. $h’(x) = 8^x\ln 8 – 2e^x + 6^x \ln 6$
2.
a. $f’(x) = (3x\ln8)(8^{3x – 5})$
b. $g’(x) = \dfrac{(\ln 6)(6^{\sqrt{x – 1}})}{2\sqrt{x – 1}}$
c. $h’(x) = 5^{e^x}(e^x \ln 5)$
3.
a. $0.60$
b. $480e^{1.2} \approx 1593.67$
Images/mathematical drawings are created with GeoGebra. |
# Relations and Functions
### Popular Tutorials in Relations and Functions
#### How Can You Tell if a Relation is Not a Function?
Every function is a relation, but not every relation is a function! Watch this video to learn how to tell which relations are functions and which are not.
#### How is a Function Defined?
You can't go through algebra without learning about functions. This tutorial shows you the definition of a function and gives you an example of a function. Take a look!
#### How Do You Find the Domain and Range of a Relation?
Finding the domain and range of a relation? No problem! Watch this tutorial and learn how to find the domain and range of a relation.
#### How Do You Graph a Linear Equation by Making a Table?
Graphing a function? It would be really helpful if you had a table of values that fit your equation. You could plot those values on a coordinate plane and connect the point to make your graph. See it all in this tutorial!
#### How Do You Figure Out If a Relation is a Function?
How do you figure out if a relation is a function? You could set up the relation as a table of ordered pairs. Then, test to see if each element in the domain is matched with exactly one element in the range. If so, you have a function! Watch this tutorial to see how you can determine if a relation is a function.
#### How Do You Use the Vertical Line Test to Figure Out if a Graph is a Function?
Trying to figure out if an equation is a function? Graph it and perform the vertical line test. If it passes, then it's a function! Get some practice by watching this tutorial!
#### How Do You Find f(x) If You Have a Value For x?
To solve a function for a given value, plug that value into the function and simplify. See this first-hand by watching this tutorial!
#### What is a Relation?
Got a set of ordered pairs? Then you have a relation! This tutorial takes a look at relations!
#### What is the Domain of a Relation?
Did you know that a relation has a domain? The domain of a relation is the set of the first coordinates from the ordered pairs. This tutorial defines the domain of a relation!
#### What is the Range of a Relation?
Did you know that a relation has a range? The range of a relation is the set of the second coordinates from the ordered pairs. This tutorial defines the range of a relation!
#### What's a Function?
You can't go through algebra without learning about functions. This tutorial shows you a great approach to thinking about functions! Learn the definition of a function and see the different ways functions can be represented. Take a look!
#### What's the Vertical Line Test?
Even graphs need to worry about tests! Using the vertical line test, you can figure out if a graph is a function or not. Watch this tutorial and learn about the vertical line test. Then, put your graphs to the test!
#### What is Function Notation?
Every see 'f(x)' in your math? That's function notation! It's a way to indicate that an equation is a function. Learn about function notation by watching this tutorial.
#### How Do You Find f(x) When the Value for x Contains Other Variables?
When you're evaluating a function, you're usually given a number to plug in for the variable, but what if the expression you're plugging in contains other variables? See how to plug an expression with variables into a function! This tutorial will show you!
#### What is a One-to-One Function?
What is a one-to-one function? What qualities make a function one-to-one? This tutorial is a great introduction and explanation for one-to-one functions!
#### What is a Continuous Function?
You've probably seen a continuous function before and not even known it! In this tutorial, you'll learn what a continuous function is and what a graph needs to have in order to be continuous.
#### How Do You Graph a Relation From a Table?
When you have a relation given as a table of x-values and y-values, it can sometimes be helpful to graph those points in order to get a visual representation of the relation. This tutorial will show you how to take values from a table and plot them on the coordinate plane! |
# What can you multiply to 120?
120 can be multiplied by any number, but some noteworthy examples are 12, 10, 8, 6, 5, 4, 3, 2 and 1. All of the above multiples of 120 equal 120 when multiplied together. Other than these numbers, any other number multiplied by 120 will result in a larger product.
For example, 4 x 120 = 480, 10 x 120 = 1,200, and 20 x 120 = 2,400.
## What times does 6 give you 90?
6 gives you 90 when multiplied by 15, x6 = 90. Additionally, 6 gives you 90 when multiplied by 18, x6 = 90. Another example of 6 giving you 90 is when it is multiplied by 30, x6 = 90. It can also be achieved when 6 is multiplied by 5, x6 = 30, then multiplied by 3, x30 = 90.
You can also get 90 when 6 is multiplied by 10, x6 = 60, and then multiplied by 3, x60 = 90. Lastly, 6 gives you 90 when multiplied by 3, x6 = 18, and then multiplied by 5, x18 = 90.
## How many times will 6 be contained in 90?
6 is contained in 90 15 times. This is because when you divide 90 by 6, the result is 15. Additionally, when you multiply 6 by 15, the result is 90.
## What times table equals 90?
The times table that equals 90 is the 9 times table. The 9 times table is all multiples of 9 from 1 to 10. The multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, and 90. Therefore, the 9 times table equals 90.
## What add up to 90?
90 can be added up in all sorts of combinations, depending on the context. Many of these combinations involve adding together two or more numbers.
For example, the numbers 9 and 81 can be added together to make 90. Likewise, 8 and 82, 7 and 83, and 6 and 84, can all add up to 90.
In addition, combinations of numbers can also add up to 90. For example, 1, 2, 3, 12, 27, and 45 are all different ways to make 90.
Finally, in mathematical settings, different operations can be used to add up to 90. For example, 3 x 30, 10 x 9 and 15 x 6 all equal 90. Similarly, 6 squared (6 x 6) or 10 squared (10 x 10) can both be used to make 90.
In general, if you’re looking to add up to 90, many combinations of numbers and mathematical operations can be used to do so.
## Which is the next multiple of 6 after 90?
The next multiple of 6 after 90 is 96. To find the next multiple of 6, we add 6 to 90, which is 96. In general, to find any multiple of any number, we just multiply the number by an increasing whole number.
For example, to find the next multiple of 8 after 20, we just multiply 8 by 3 (8 x 3 = 24).
## What are the times of 6?
The time of 6 can refer to a number of different things, depending on the context. Generally, 6 o’clock (or 06:00) is the official start of the day and is often thought of as early morning; it is also the middle of the night in 24-hour time.
Likewise, 18:00 (or 6 pm) is the official end of the day and is often thought of as nightfall. Additionally, 6 can refer to specific time increments in each hour, such as six minutes past the hour (06:06), six hours past the hour (06:06pm), six seconds past the hour (06:00:06), and so on.
Six can also refer to time zones, such as the Central Standard Time zone which is 6 hours behind Coordinated Universal Time (UTC).
## What all numbers make 6?
The numbers that make 6 when added together are 1, 2 and 3 – which can all be added up to make six. This is because 1 + 2 + 3 = 6. Additionally, the number 6 can be found by multiplying different numbers together.
For example, 2 x 3 = 6, or 3 x 2 = 6. Furthermore, when subtracting numbers, 6 can be found in multiple ways. For example, 8 – 2 = 6, or 7 – 1 = 6.
## How many tens are there in 90?
There are nine tens in 90. The number 90 can be broken down into two parts: the tens part, which is 9, and the ones part, which is 0. When you multiply 9 (the tens part) by 10 (the base number in the base 10 number system that we use), you get 90.
So, there are nine tens in the number 90.
## What is the 6 in math?
In math, 6 is a whole number, which is an integer that can be written without a fractional or decimal component. It is one of the first numbers we learn, along with 0, 1, 2, 3, 4, and 5. It has several properties that make it an important number in mathematics.
It is an even number, which means it is divisible by 2. It is also divisible by 3, which makes it a multiple of 3. It is also a composite number, since it can be divided more than just by 1 and itself.
Additionally, 6 is a perfect number, which means the sum of its divisors is equal to itself. For example, 1 + 2 + 3 = 6. This makes it a special number, and it can help in understanding advanced concepts in mathematics.
## What is a 6 out of 24?
6 out of 24 is a fraction that can also be expressed as a decimal. It is equal to 0.25, which is the same as 25%. It is the same as saying 6 divided by 24 or 6 ÷ 24.
## Is 24 a multiple of 6 yes or no?
Yes, 24 is a multiple of 6. This can be seen by divding 24 by 6, which gives 4. This means that 6 goes into 24 a total of 4 times, which makes 24 a multiple of 6.
## How do you get the answer 24?
The answer 24 can be obtained in many different ways depending on the context. If you are solving a math problem, 24 can be obtained by adding together numbers like 10+7+7 or 8+9+7. It can also be obtained by subtracting numbers like 28-4 or 30-6.
If you are working with fractions, 24 can be obtained by adding fractions of the same denominator or by multiplying fractions together. If you are multipliying or dividing numbers, 24 can be obtained by multiplying 3*8 or 6*4 or dividing 48/2.
Another way to get 24 is by using a combination of additions, subtractions, multiplications, and divisions. An example could be 17+6/3*4.
## How many ways can 24 be divided?
24 can be divided in a variety of ways, depending on how you want to divide it. Generally speaking, it can be divided into two groups (two groups of 12), three groups (three groups of 8), four groups (four groups of 6), or six groups (six groups of 4).
Additionally, it can be divided into two halves (12 and 12), four quarters (6 and 6), eight eighths (3 and 3), or even twenty-four twelfths (2 and 2). Additionally, you can divide it further within each of the above combinations.
For example, you could divide 12 into four and eight (4, 4, 4, and 8), or you could divide six into two and four (2, 2, and 4). Ultimately, there are several ways to divide 24, depending on the specific scenario and desired outcome.
## What is the answer for 6x 24?
The answer for 6x 24 is 144. Six multiplied by 24 is equal to 144. If you are using multiplication notation, 6×24 is the same as 6*24, both of which are equal to 144.
Categories FAQ |
# Equivalent Fraction Patterns
In this worksheet, students work with patterns when finding equivalent fractions.
Key stage: KS 2
Curriculum topic: Maths and Numerical Reasoning
Curriculum subtopic: Fractions
Difficulty level:
### QUESTION 1 of 10
In this worksheet, we will learn how to find equivalent fractions.
There are 20 small pieces in this bar of chocolate.
There are 5 strips.
The big chunk is 4 strips out of 5.
It is also 16 small pieces out of 20.
So...
4/5 of the bar is equivalent to 16/20.
Notice that:
4 x 4 = 16
4 x 5 = 20
We get equivalent fractions by multiplying the numerator (top) and the denominator (bottom) of a fraction by the same number.
Example
Find the values of A, B, C and D to form equivalent fractions.
3 9
=
6 18
=
A 27
=
12 B
=
C D
3/9 is the starting fraction.
In the second fraction, we get 6/18 by multiplying top and bottom of 3/9 by 2.
In the third fraction, we multiply top and bottom by 3. So A is 3 × 3 = 9.
In the fourth fraction, we multiply top and bottom by 4. So B is 4 × 9 = 36.
And in the last fraction, we multiply top and bottom by 5. So C is 5 × 3 = 15 and D is 5 × 9 = 45.
Notice the patterns.
The tops go 3, 6, 9, 12, 15.
The bottoms go 9, 18, 27, 36, 45.
In this worksheet, look for patterns like this.
Find the values of A, B, C and D to form equivalent fractions.
1 5
=
2 10
=
A 15
=
4 B
=
C D
## Column B
A
5
B
3
C
20
D
25
Find the values of A, B, C and D to form equivalent fractions.
2 6
=
4 12
=
A 18
=
8 B
=
C D
## Column B
A
24
B
30
C
6
D
10
Find the values of A, B, C and D to form equivalent fractions.
4 6
=
8 12
=
A 18
=
16 B
=
C D
## Column B
A
12
B
24
C
20
D
30
Find the values of A, B, C and D to form equivalent fractions.
7 8
=
14 16
=
A 24
=
28 B
=
C D
## Column B
A
32
B
35
C
40
D
21
Find the values of A, B, C and D to form equivalent fractions.
2 3
=
4 6
=
A 9
=
8 B
=
C D
## Column B
A
6
B
15
C
12
D
10
Find the values of A, B, C and D to form equivalent fractions.
1 4
=
2 8
=
A 12
=
4 B
=
C D
## Column B
A
5
B
16
C
20
D
3
Find the values of A, B, C and D to form equivalent fractions.
4 6
=
8 12
=
A 18
=
16 B
=
C D
## Column B
A
30
B
24
C
20
D
12
Find the values of A, B, C and D to form equivalent fractions.
3 4
=
6 8
=
A 12
=
12 B
=
C D
## Column B
A
20
B
9
C
15
D
16
Find the values of A, B, C and D to form equivalent fractions.
3 9
=
6 18
=
A 27
=
12 B
=
C D
## Column B
A
45
B
36
C
15
D
9
Find the values of A, B, C and D to form equivalent fractions.
1 2
=
2 4
=
A 6
=
4 B
=
C D
## Column B
A
5
B
8
C
3
D
10
• Question 1
Find the values of A, B, C and D to form equivalent fractions.
1 5
=
2 10
=
A 15
=
4 B
=
C D
## Column B
A
3
B
20
C
5
D
25
EDDIE SAYS
The tops go 1, 2, 3, 4, 5.
The bottoms go 5, 10, 15, 20, 25.
• Question 2
Find the values of A, B, C and D to form equivalent fractions.
2 6
=
4 12
=
A 18
=
8 B
=
C D
## Column B
A
6
B
24
C
10
D
30
EDDIE SAYS
The tops go 2, 4, 6, 8, 10.
The bottoms go 6, 12, 18, 24, 30.
• Question 3
Find the values of A, B, C and D to form equivalent fractions.
4 6
=
8 12
=
A 18
=
16 B
=
C D
## Column B
A
12
B
24
C
20
D
30
EDDIE SAYS
The tops go 4, 8, 12, 16, 20.
The bottoms go 6, 12, 18, 24, 30.
• Question 4
Find the values of A, B, C and D to form equivalent fractions.
7 8
=
14 16
=
A 24
=
28 B
=
C D
## Column B
A
21
B
32
C
35
D
40
EDDIE SAYS
The tops go 7, 14, 21, 28, 35.
The bottoms go 8, 16, 24, 32, 40.
• Question 5
Find the values of A, B, C and D to form equivalent fractions.
2 3
=
4 6
=
A 9
=
8 B
=
C D
## Column B
A
6
B
12
C
10
D
15
EDDIE SAYS
The tops go 2, 4, 6, 8, 10.
The bottoms go 3, 6, 9, 12, 15.
• Question 6
Find the values of A, B, C and D to form equivalent fractions.
1 4
=
2 8
=
A 12
=
4 B
=
C D
## Column B
A
3
B
16
C
5
D
20
EDDIE SAYS
The tops go 1, 2, 3, 4, 5.
The bottoms go 4, 8, 12, 16, 20.
• Question 7
Find the values of A, B, C and D to form equivalent fractions.
4 6
=
8 12
=
A 18
=
16 B
=
C D
## Column B
A
12
B
24
C
20
D
30
EDDIE SAYS
The tops go 4, 8, 12, 16, 20.
The bottoms go 6, 12, 18, 24, 30.
• Question 8
Find the values of A, B, C and D to form equivalent fractions.
3 4
=
6 8
=
A 12
=
12 B
=
C D
## Column B
A
9
B
16
C
15
D
20
EDDIE SAYS
The tops go 3, 6, 9, 12, 15.
The bottoms go 4, 8, 12, 16, 20.
• Question 9
Find the values of A, B, C and D to form equivalent fractions.
3 9
=
6 18
=
A 27
=
12 B
=
C D
## Column B
A
9
B
36
C
15
D
45
EDDIE SAYS
The tops go 3, 6, 9, 12, 15.
The bottoms go 9, 18, 27, 36, 45.
• Question 10
Find the values of A, B, C and D to form equivalent fractions.
1 2
=
2 4
=
A 6
=
4 B
=
C D
## Column B
A
3
B
8
C
5
D
10
EDDIE SAYS
The tops go 1, 2, 3, 4, 5.
The bottoms go 2, 4, 6, 8, 10.
---- OR ----
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# What is 1/4 of 450?
In this article, we'll show you exactly how to calculate 1/4 of 450 so you can work out the fraction of any number quickly and easily! Let's get to the math!
Want to quickly learn or show students how to convert 1/4 of 450? Play this very quick and fun video now!
You probably know that the number above the fraction line is called the numerator and the number below it is called the denominator. To work out the fraction of any number, we first need to convert that whole number into a fraction as well.
Here's a little tip for you. Any number can be converted to fraction if you use 1 as the denominator:
450 / 1
So now that we've converted 450 into a fraction, to work out the answer, we put the fraction 1/4 side by side with our new fraction, 450/1 so that we can multiply those two fractions.
That's right, all you need to do is convert the whole number to a fraction and then multiply the numerators and denominators. Let's take a look:
1 x 450 / 4 x 1 = 450 / 4
In this case, our new fraction can actually be simplified down further. To do that, we need to find the greatest common factor of both numbers.
You can use our handy GCF calculator to work this out yourself if you want to. We already did that, and the GCF of 450 and 4 is 2.
We can now divide both the new numerator and the denominator by 2 to simplify this fraction down to its lowest terms.
450/2 = 225
4/2 = 2
When we put that together, we can see that our complete answer is:
225 / 2
The complete and simplified answer to the question what is 1/4 of 450 is:
112 1/2
Hopefully this tutorial has helped you to understand how to find the fraction of any whole number. You can now go give it a go with more numbers to practice your newfound fraction skills.
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 1/4 of 450?". VisualFractions.com. Accessed on December 5, 2021. http://visualfractions.com/calculator/fraction-of-number/what-is-1-4-of-450/.
• "What is 1/4 of 450?". VisualFractions.com, http://visualfractions.com/calculator/fraction-of-number/what-is-1-4-of-450/. Accessed 5 December, 2021.
## Fraction of a Number
Enter a numerator, denominator and whole number |
# A Complete Overview On Addition of Fractions
We learn how to add two or more fractions with the same or different denominators through the addition of fractions. Fraction addition is dependent on two main factors:
• Same denominators
• Different denominators
If the denominators of two fractions are the same, the fractions can be combined directly since they are considered to be similar fractions. However, if the denominators differ (these fractions are known as unlike fractions), we must first make the denominators equal before adding the fractions. Learn more about like and unlike fractions on here.
## Addition of Fractions With Same Denominators
If the denominators of two or more fractions are identical, we can simply add the numerators while maintaining the equality of the denominators.
To add fractions with the same denominator, use these steps:
• Add the numerators together while maintaining the common denominator.
• Create the fraction in a more straightforward form.
For example: Add the fractions: 5/6 and 7/6.
Since the denominators are identical, therefore we can add the numerators directly.
(5/6) + (7/6) = (5 + 7)/6 = 12/6
Simplify the fraction
12/6 = 2
Hence, the sum of ⅚ and 7/6 is 2.
## Fraction Addition With Different Numerators
We cannot determine the numerators directly when two or more fractions with various denominators are combined.
• Check the denominators of the fractions.
• By calculating the LCM of the denominators and rationalizing them, make the fraction denominators the same.
• Keeping the denominator constant, multiply the fractions’ numerators.
• To obtain the total, simplify the fraction.
Just an example: Add 3/12 + 5/2
Solution: Both the fractions 3/12 and 5/2 have different denominators.
We can write 3/12 = ¼, in a simplified fraction.
Now, ¼ and 5/2 are two fractions.
LCM of 2 and 4 = 4
Multiply 5/2 by 2/2.
5/2 x 2/2 = 10/4
¼ + 10/4 = (1+10)/4 = 11/4
Hence, the sum of 3/12 and 5/2 is 11/4.
## Adding Fractions With Whole Numbers
Three easy steps will add the fraction and the whole number:
• In the form of a fraction, write the given full number (for instance, 3/1)
• Add the fractions after matching the denominators.
• Simplify the fraction
Just an example: Add 7/2 + 4
Here, 7/2 is a fraction and 4 is a whole number.
We can write 4 as 4/1.
Now making the denominators same, we get;
7/2 and 4/1 x (2/2) = 8/2
7/2 + 8/2 = 15/2
Hence, the sum of 7/2 and 4 is 15/2.
Co-prime denominators: The denominators that share only one other common factor are those.
Let’s use the following steps to learn how to add fractions with co-prime denominators:
• Verify whether the denominators are co-prime.
• Multiply the first fraction’s numerator and denominator by the other fraction’s denominator, then multiply the second fraction’s numerator and denominator by the first fraction’s denominator.
• Add the resulting fractions and simplify
Just an example, the addition of fractions 9/7 and 3/4 can be done as follows.
The denominators 7 and 4 are coprime since they have only one highest common factor 1.
So, (9/7) + (3/4) = [(9 × 4) + (3 × 7)]/ (7 × 4)
= (36 + 21)/28
= 57/28
Combining a whole number and a fraction results in a mixed fraction. Two mixed fractions must first be transformed into improper fractions before being added together.
• Create incorrect fractions from the provided mixed fraction.
• Verify whether the denominators are identical or different.
• If there are distinct denominators, then explain them.
• Fractions are added, then simplified.
Let’s use the following example to clarify how to combine mixed fractions:
Example: Add : 3 ⅓ + 1 ¾
Solution:
Step1: Convert the given mixed fractions to improper fractions.
3 ⅓ = 10/3
1 ¾ = 7/4
Step 2: Make the denominators same by taking the LCM and multiplying the suitables fractions for both.
LCM of 3 and 4 is 12.
So, 10/3 = (10/3) × (4/4) = 40/12
7/4 = (7/4) × (3/3) = 21/12
Step 3: Take the denominator as common and add numerators. Then, write the final answer.
(40/12) + (21/12) = (40 + 21)/12 = 61/12
Therefore, 3 ⅓ + 1 ¾ = 61/12 = 5 1/12
## Subtraction of Fractions
Mathematical procedures like addition and subtraction are comparable, as we all know. Additionally, addition involves adding two or more numbers, and subtraction involves taking a number away from another. As a result, fraction subtraction likewise adheres to the same rule as fraction addition.
For given fractions, if the denominators are the same, we can simply remove the numerator while maintaining the original denominator.
If a fraction has a different denominator, we must first rationalize it before we may subtract.
Some examples are:
Example 1: Subtract ⅓ from 8/3.
Solution: We need to find,
8/3 – ⅓ = ?
We may immediately subtract the two fractions 13 and 8/3 because their common denominator is 13.
8/3 – ⅓ = (8-1)/3 = 7/3
Example 2: Subtract ½ from ¾.
Solution: We need to subtract ½ from ¾, i.e.,
¾ – ½ = ?
We must rationalize the two fractions by using the LCM because the denominators of the two fractions differ.
LCM (4,2) = 4
Now multiply the ½ by 2/2, to get 2/4
Therefore,
¾ – 2/4 = (3-2)/4 = ¼
Hence, ¾ – ½ = ¼
## Solved Examples
Let’s work through some fraction addition-based challenges.
Q. 1: Add 1/2 and 7/2.
Solution: Given fractions: 1/2 and 7/2
Since the denominators are identical, we can simply add the numerators in this case while leaving the denominator alone.
Therefore,
1/2 + 7/2
= (1+7)/2
= 8/2
=4
Q. 2: Add 3/5 and 4.
Solution: We can write 4 as 4/1
Now, 3/5 and 4/1 are the two fractions to be added.
We must first simplify the denominators because they are different in this case before adding the fractions.
Hence,
3/5 + 4/1
Taking LCM of 5 and 1, we get;
LCM(5,1) = 5
In order to find the second fraction, 4/1, we must multiply it by 5 in both the numerator and denominator.
(4×5)/((1×5) = 20/5
Now 3/5 and 20/5 have a common denominator, i.e. 5, therefore, adding the fractions now;
3/5 + 20/5
= 23/5
One of the key concepts in grades 6, 7, and 8 are fraction addition. Here, a worksheet for adding fractions is available. You’ll be able to quickly and simply solve fraction addition sums after practicing the questions on this worksheet. Practice using the fraction addition worksheet provided on this site to succeed in tests.
### Practice Questions
1. ⅜ + ⅝ =
2. 1(⅓) + 3(5/2) =
3. 2(¾) + ___ = 7
4. ⅖ + ⅔ =
5. 3/7 + 2 + 4/3 = ?
Visit Knowledge Glow to learn much more mathematical topics in an interesting way.
## Frequently Asked Questions – FAQs
### How Do You Combine Two Fractions With Various Denominators?
To add two fractions with distinct denominators, we must remove the LCM from the denominators and change them to the same value. The fractions’ numerators should then be added while maintaining the common denominator. For instance, ½ + ⅗
LCM of 2 and 5 = 10
= (5/5) x (½) + (⅗) x (2/2)
= (5+6)/10
= 11/10
### What Guidelines Apply While Adding and Subtracting Fractions?
To add and subtract fractions, use two straightforward rules. We can add and subtract fractions directly if the denominators match. However, if the denominators differ, we must rationalize the denominators by determining the LCM of the two, and then we must add the fractions.
### How Do You Add Fractions and Whole Numbers?
If we add a whole number and a fraction, then we need to first write the whole number in the form of a fraction. For example, by adding 3 and ½ we get,
3+½ = 3x(2/2) + ½ = 6/2 + ½ = 7/2
### How Do You Add Big Fractions?
Let us take two fractions: 11/24 and 9/60
LCM of 24 and 60 = 120
Therefore,
= (11/24)x(5/5) + (9/60)x(2/2)
= (55+18)/120
= 73/120
### How May Fractions With Similar Denominators Be Added?
Let us say, ⅗ and 7/5 are two fractions. Since ⅗ and 7/5 are two like fractions, having the same denominator, we can add the numerators directly.
Therefore,
⅗ + 7/5 = (3+7)/5 = 10/5 = 2
### Knowledge Glow
I am Komal Gupta, the founder of Knowledge Glow, and my team and I aim to fuel dreams and help the readers achieve success. While you prepare for your competitive exams, we will be right here to assist you in improving your general knowledge and gaining maximum numbers from objective questions. We started this website in 2021 to help students prepare for upcoming competitive exams. Whether you are preparing for civil services or any other exam, our resources will be valuable in the process. |
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## 9.3 Quadratic Formula
### Learning Objective
1. Solve quadratic equations with real solutions using the quadratic formula.
## The Quadratic Formula
In this section, we will develop a formula that gives the solutions to any quadratic equation in standard form. To do this, we begin with a general quadratic equation in standard form and solve for x by completing the square. Here a, b, and c are real numbers and $a≠0$:
Determine the constant that completes the square: take the coefficient of x, divide it by 2, and then square it.
Add this to both sides of the equation and factor.
Solve by extracting roots.
This derivation gives us a formula that solves any quadratic equation in standard form. Given $ax2+bx+c=0$, where a, b, and c are real numbers and $a≠0$, then the solutions can be calculated using the quadratic formulaThe formula $x=−b±b2−4ac2a$, which gives the solutions to any quadratic equation in the form $ax2+bx+c=0$, where a, b, and c are real numbers and $a≠0$.:
Consider the quadratic equation $2x2−7x+3=0$. It can be solved by factoring as follows:
The solutions are 1/2 and 3. The following example shows that we can obtain the same results using the quadratic formula.
Example 1: Solve using the quadratic formula: $2x2−7x+3=0$.
Solution: Begin by identifying a, b, and c as the coefficients of each term.
Substitute these values into the quadratic formula and then simplify.
Separate the “plus or minus” into two equations and simplify each individually.
Answer: The solutions are 1/2 and 3.
Of course, if the quadratic factors, then it is a best practice to solve it by factoring. However, not all quadratic polynomials factor; nevertheless, the quadratic formula provides us with a means to solve such equations.
Example 2: Solve using the quadratic formula: $5x2+2x−1=0$.
Solution: Begin by identifying a, b, and c.
Substitute these values into the quadratic formula.
Answer: The solutions are $−1±65$.
Often terms are missing. When this is the case, use 0 as the coefficient.
Example 3: Solve using the quadratic formula: $x2−18=0$.
Solution: Think of this equation with the following coefficients:
Here
Substitute these values into the quadratic formula.
Answer: The solutions are $±32$.
Since the coefficient of x was 0, we could have solved the equation by extracting the roots. As an exercise, solve the previous example using this method and verify that the results are the same.
Example 4: Solve using the quadratic formula: $9x2−12x+4=0$.
Solution: In this case,
Substitute these values into the quadratic formula and then simplify.
In this example, notice that the radicand of the square root is 0. This results in only one solution to this quadratic equation. Normally, we expect two solutions. When we find only one solution, the solution is called a double root. If we solve this equation by factoring, then the solution appears twice.
Answer: 2/3, double root
Example 5: Solve using the quadratic formula: $x2+x+1=0$.
Solution: In this case,
Substitute these values into the quadratic formula.
The solution involves the square root of a negative number; hence the solutions are not real. This quadratic equation has two nonreal solutions and will be discussed in further detail as we continue in our study of algebra. For now, simply state that the equation does not have real solutions.
Answer: No real solutions
Try this! Solve: $x2−2x−2=0$.
Answer: $1±3$
### Video Solution
(click to see video)
It is important to place the quadratic equation in standard form before using the quadratic formula.
Example 6: Solve using the quadratic formula: $(2x+1)(2x−1)=24x+8$.
Solution: Begin by using the distributive property to expand the left side and combining like terms to obtain an equation in standard form, equal to 0.
Once the equation is in standard form, identify a, b, and c. Here
Substitute these values into the quadratic formula and then simplify.
Answer: The solutions are $6±352$.
Try this! Solve: $3x(x−2)=1$.
Answer: $3±233$
### Video Solution
(click to see video)
### Key Takeaways
• Use the quadratic formula to solve any quadratic equation in standard form.
• To solve any quadratic equation, first rewrite in standard form, $ax2+bx+c=0$, substitute the appropriate coefficients into the quadratic formula, $x=−b±b2−4ac2a$, and then simplify.
### Topic Exercises
Part A: Quadratic Formula
Identify the coefficients a, b, and c used in the quadratic formula. Do not solve.
1. $x2−x+5=0$
2. $x2−3x−1=0$
3. $3x2−10=0$
4. $−y2+5=0$
5. $5t2−7t=0$
6. $−y2+y=0$
7. $−x2+x=−6$
8. $−2x2−x=−15$
9. $(3x+1)(2x+5)=19x+4$
10. $(4x+1)(2x+1)=16x+4$
Solve by factoring and then solve using the quadratic formula. Check answers.
11. $x2−10x+24=0$
12. $x2−3x−18=0$
13. $t2+6t+5=0$
14. $t2+9t+14=0$
15. $2x2−7x−4=0$
16. $3x2−x−2=0$
17. $−2x2−x+3=0$
18. $−6x2+x+1=0$
19. $y2−2y+1=0$
20. $y2−1=0$
Use the quadratic formula to solve the following.
21. $x2−6x+4=0$
22. $x2−4x+1=0$
23. $x2+2x−5=0$
24. $x2+4x−6=0$
25. $t2−4t−1=0$
26. $t2−8t−2=0$
27. $−y2+y+1=0$
28. $−y2−3y+2=0$
29. $−x2+16x−62=0$
30. $−x2+14x−46=0$
31. $2t2−4t−3=0$
32. $4t2−8t−1=0$
33. $−4y2+12y−9=0$
34. $−25x2+10x−1=0$
35. $3x2+6x+2=0$
36. $5x2+10x+2=0$
37. $9t2+6t−11=0$
38. $8t2+8t+1=0$
39. $x2−2=0$
40. $x2−18=0$
41. $9x2−3=0$
42. $2x2−5=0$
43. $y2+9=0$
44. $y2+1=0$
45. $2x2=0$
46. $x2=0$
47. $−2y2+5y=0$
48. $−3y2+7y=0$
49. $t2−t=0$
50. $t2+2t=0$
51. $x2−0.6x−0.27=0$
52. $x2−1.6x−0.8=0$
53. $y2−1.4y−0.15=0$
54. $y2−3.6y+2.03=0$
55. $12t2+5t+32=0$
56. $−t2+3t−34=0$
57. $3y2+12y−13=0$
58. $−2y2+13y+12=0$
59. $2x2−10x+3=4$
60. $3x2+6x+1=8$
61. $−2y2=3(y−1)$
62. $3y2=5(2y−1)$
63. $(t+1)2=2t+7$
64. $(2t−1)2=73−4t$
65. $(x+5)(x−1)=2x+1$
66. $(x+7)(x−2)=3(x+1)$
67. $x(x+5)=3(x−1)$
68. $x(x+4)=−7$
69. $(5x+3)(5x−3)−10(x−1)=0$
70. $(3x+4)(3x−1)−33x=−20$
71. $27y(y+1)+2(3y−2)=0$
72. $8(4y2+3)−3(28y−1)=0$
73. $(x+2)2−2(x+7)=4(x+1)$
74. $(x+3)2−10(x+5)=−2(x+1)$
Part B: Discussion Board
75. When talking about a quadratic equation in standard form, $ax2+bx+c=0$, why is it necessary to state that $a≠0$? What happens if a is equal to 0?
76. Research and discuss the history of the quadratic formula and solutions to quadratic equations.
1: $a=1$, $b=−1$, and $c=5$
3: $a=3$, $b=0$, and $c=−10$
5: $a=5$, $b=−7$, and $c=0$
7: $a=−1$, $b=1$, and $c=6$
9: $a=6$, $b=−2$, and $c=1$
11: 4, 6
13: −5, −1
15: −1/2, 4
17: −3/2, 1
19: 1, double root
21: $3±5$
23: $−1±6$
25: $2±5$
27: $1±52$
29: $8±2$
31: $2±102$
33: 3/2, double root
35: $−3±33$
37: $−1±233$
39: $±2$
41: $±33$
43: No real solutions
45: 0, double root
47: 0, 5/2
49: 0, 1
51: −0.3, 0.9
53: −0.1, 1.5
55: $−5±22$
57: $−1±1712$
59: $5±332$
61: $−3±334$
63: $±6$
65: $−1±7$
67: No real solutions
69: 1/5, double root
71: −4/3, 1/9
73: $1±15$ |
# How To Find Standard Deviation In Statcrunch?
## How do you find the standard deviation from a frequency table?
Lesson 19 – Standard Deviation Of Data In A Frequency Table
## How do you find the standard deviation of the sample mean difference?
Work out the Standard Deviation.
• Work out the mean.
• Then for each number: subtract the Mean and square the result.
• Then work out the mean of those squared differences.
• Take the square root of that:
• Work out the mean.
• Then for each number: subtract the Mean and square the result.
## How do you find population standard deviation?
First, let’s review how to calculate the population standard deviation:
1. Calculate the mean (simple average of the numbers).
2. For each number: Subtract the mean. Square the result.
3. Calculate the mean of those squared differences.
4. Take the square root of that to obtain the population standard deviation.
## How do you find the mean and standard deviation of grouped data?
Standard Deviation of Grouped Data –
## What is the formula for variance?
To calculate variance, start by calculating the mean, or average, of your sample. Then, subtract the mean from each data point, and square the differences. Next, add up all of the squared differences. Finally, divide the sum by n minus 1, where n equals the total number of data points in your sample.
## What is a good standard deviation?
For an approximate answer, please estimate your coefficient of variation (CV=standard deviation / mean). As a rule of thumb, a CV >= 1 indicates a relatively high variation, while a CV < 1 can be considered low. A “good” SD depends if you expect your distribution to be centered or spread out around the mean.
## How do you find the percentage of data in one standard deviation of the mean?
Finding the area under the curve from x = 9 to x = 13. The Empirical Rule or 68-95-99.7% Rule gives the approximate percentage of data that fall within one standard deviation (68%), two standard deviations (95%), and three standard deviations (99.7%) of the mean.
## How do I find the sample variance?
To calculate variance, start by calculating the mean, or average, of your sample. Then, subtract the mean from each data point, and square the differences. Next, add up all of the squared differences. Finally, divide the sum by n minus 1, where n equals the total number of data points in your sample.
## What is Excel formula for standard deviation?
Use the Excel Formula =STDEV( ) and select the range of values which contain the data. This calculates the sample standard deviation (n-1). Use the web Standard Deviation calculator and paste your data, one per line.
## Why is it N 1 for standard deviation?
The reason n-1 is used is because that is the number of degrees of freedom in the sample. The sum of each value in a sample minus the mean must equal 0, so if you know what all the values except one are, you can calculate the value of the final one.
## What does M and SD mean in a study?
Updated May 7, 2019. The standard deviation (SD) measures the amount of variability, or dispersion, for a subject set of data from the mean, while the standard error of the mean (SEM) measures how far the sample mean of the data is likely to be from the true population mean.
We recommend reading: How To Find Perimeter Of A Trapezoid?
## How do you find the mean and standard deviation of grouped data in SPSS?
Find the Mean and Standard Deviation in SPSS for Two Groups
## How do you find variance in grouped data?
variance for grouped data –
## How do you find the standard deviation of grouped data on a calculator?
Calculate Mean and Standard Deviation of Grouped Data on TI |
MAM2000 (Essays/Dimension)
# Higher-Dimensional Spaces
The intuitions that students accumulate in dealing with coordinate pairs in the plane and coordinate triples in three-dimensional space lead naturally to coordinate geometry in higher dimensions. A thorough understanding of two and three dimensions provides an important foundation for the powerful generalizations of vector and matrix algebra in science and engineering, in economics and social science, and especially computer science and graphics. We illustrate this progression with two examples.
The vertices of a square can be given by four points (0,0), (1,0), (1, 1), and (0, 1). To obtain the vertices of a cube, we can take the points of a zero in the third coordinate and then move the square one unit in the third direction to obtain four more vertices, with a 1 in the last coordinate:
(0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 0, 1), (1, 0, 1), (1, 1, 1), (0, 1, 1).
Thus we can describe either the square or the cube as having vertices that are either 0 or 1 in each coordinate.
Figure 32. Generalizing the Pythagorean theorem to three dimensions by applying it to two different triangles found in a rectangular box.
The procedure generalizes automatically: to obtain the vertices of a hypercube, we start with the eight vertices of a cube and put 0 in the final coordinate and then "move the cube in a fourth direction" to obtain eight more points with 1 the last coordinate:
(0,0,0,0), (1,0,0,0), (1,1,0,0), (0,1,0,0), (0,0,1,0), (1,0,1,0), (1,1,1,0), (0,1,1,0), (0,0,0,1), (1,0,0,1), (1,1,0,1), (0,1,0,1), (0,0,1,1), (1,0,1,1), (1,1,1,1), (0,1,1,1).
We thus obtain the sixteen vertices of a hypercube, with 0 or 1 in each of four coordinates. It is this sort of representation that is ideal for communicating with a computer.
A second topic that generalizes in a very nice way is the Pythagorean theorem. If we think of this theorem as a way of calculating the length of the diagonal of a rectangle with given sides, then the extension to three dimensions is immediate: given a solid bounded by rectangular sides, we first apply the theorem to one side and then apply it to a rectangle built over the first diagonal (Figure 32). We easily get e2 = c2 + d2 = c2 + (a2 + b2), so the length of the diagonal of a rectangular prism with sides a, b, and c is (a2 + b2 + c2)1/2. The pattern is established, and the distance formula in four-dimensional space follows almost immediately. Students can then calculate the lengths of diagonals of the hypercube with the 0-1 coordinates. It turns out that the length of the major diagonal of a four-dimensional cube — say from (0,0,0,0) to (1,1,1,1) — is 41/2 = 2which is twice the length of a side. [an error occurred while processing this directive] |
Question Video: Writing a Relation between Two Sets given Its Rule | Nagwa Question Video: Writing a Relation between Two Sets given Its Rule | Nagwa
# Question Video: Writing a Relation between Two Sets given Its Rule Mathematics • Third Year of Preparatory School
## Join Nagwa Classes
For two sets π and π, π = {1, 5, 6, 7} and π = {1, 2, 5, 8, 9, 14}. Determine the relation π
from π to π, where ππ
π means π β π is a prime number, given π β π and π β π.
03:32
### Video Transcript
For two sets π and π, set π contains the numbers one, five, six, and seven and set π contains the numbers one, two, five, eight, nine, and 14. Determine the relation π
from π to π, where ππ
π means π minus π is a prime number, given π is an element of π and π is an element of π.
We recall that a relation is a set of ordered pairs π, π. In this question, π is an element of set π and π is an element of set π. We are told in this question that our ordered pairs must follow the rule that π minus π is a prime number. A prime number has exactly two factors. The first five of these are two, three, five, seven, and 11. As these are all positive and we are told that π minus π is a prime number, then π must be greater than π. The number from set π must be greater than the number from set π. This means that our value of π cannot be one as there are no numbers in set π that are less than one.
Letβs consider what can happen when π is equal to five. Both one and two are less than five, which suggests our ordered pair could be five, one or five, two. Five minus one is equal to four, and five minus two is equal to three. Three is one of our prime numbers, whereas four is not. This means that the ordered pair five, one is not in the relation π
. Letβs now consider what can happen when π is equal to six. The numbers in set π, one, two, and five, are all less than six, which suggests we could have the ordered pairs six, one; six, two; or six, five.
Six minus one is equal to five, which is a prime number. Six minus two is equal to four. This is not a prime number as it is divisible by two. The ordered pair six, two is therefore not in the relation π
. As six minus five is equal to one, which is not a prime number, the ordered pair six, five is not in the relation π
.
The last element of set π is seven, and the numbers one, two, and five in set π are less than this. This means that our potential ordered pairs are seven, one; seven, two; and seven, five. Seven minus one is equal to six, which is not a prime number. Seven minus two is equal to five, and seven minus five is equal to two. Both of these answers are prime numbers.
We can, therefore, conclude that the relation π
has four ordered pairs: five, two; six, one; seven, two; and seven, five. In all four of these cases, π minus π is a prime number, where π is an element of set π and π is an element of set π.
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# Common Core: 7th Grade Math : The Number System
## Example Questions
← Previous 1 3 4 5 6 7 8 9 33 34
### Example Question #1 : Describe Situations In Which Opposite Quantities Combine To Make 0: Ccss.Math.Content.7.Ns.A.1a
Compute the following:
Explanation:
Convert all the double signs to a single sign before solving. Remember, two minus (negative) signs combine to form a plus (positive) sign, and a plus (positive) sign and a minus (negative) sign combine to form a minus (negative) sign.
### Example Question #1 : Describe Situations In Which Opposite Quantities Combine To Make 0: Ccss.Math.Content.7.Ns.A.1a
For the equation provided, what value when substituted for , will equal
Explanation:
In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.
We can subtract to both sides in order to isolate the variable, .
### Example Question #3 : Describe Situations In Which Opposite Quantities Combine To Make 0: Ccss.Math.Content.7.Ns.A.1a
For the equation provided, what value when substituted for , will equal
Explanation:
In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.
We can subtract to both sides in order to isolate the variable, .
### Example Question #1 : Describe Situations In Which Opposite Quantities Combine To Make 0: Ccss.Math.Content.7.Ns.A.1a
For the equation provided, what value when substituted for , will equal
Explanation:
In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.
We can subtract to both sides in order to isolate the variable, .
### Example Question #191 : Grade 7
For the equation provided, what value when substituted for , will equal
Explanation:
In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.
We can subtract to both sides in order to isolate the variable, .
### Example Question #2 : Describe Situations In Which Opposite Quantities Combine To Make 0: Ccss.Math.Content.7.Ns.A.1a
For the equation provided, what value when substituted for , will equal
Explanation:
In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.
We can subtract to both sides in order to isolate the variable, .
### Example Question #1 : Describe Situations In Which Opposite Quantities Combine To Make 0: Ccss.Math.Content.7.Ns.A.1a
For the equation provided, what value when substituted for , will equal
Explanation:
In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.
We can subtract to both sides in order to isolate the variable, .
### Example Question #1 : Describe Situations In Which Opposite Quantities Combine To Make 0: Ccss.Math.Content.7.Ns.A.1a
For the equation provided, what value when substituted for , will equal
Explanation:
In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.
We can subtract to both sides in order to isolate the variable, .
### Example Question #1 : Describe Situations In Which Opposite Quantities Combine To Make 0: Ccss.Math.Content.7.Ns.A.1a
For the equation provided, what value when substituted for , will equal
Explanation:
In order to answer this question, we can solve for . When solving for we need to isolate the variable on one side of the equation.
We can subtract to both sides in order to isolate the variable, .
### Example Question #2 : Describe Situations In Which Opposite Quantities Combine To Make 0: Ccss.Math.Content.7.Ns.A.1a
For the equation provided, what value when substituted for , will equal |
# Tangent of an ellipse to an outside point
Let $C$ be a curve that is given by the equation:
$$2x^2 + y^2 = 1$$
and let P be a point $(1,1)$, which lies outside of the curve.
We want to find all lines that are tangent to $C$ and intersect $P$, and have found $y=1$, but are not sure how to find the other line.
• Write the general equation of a line that contains $P$. It should have one parameter $m$. Substitute in the equation of $C$ to get the intersection equation. Tangency means that this intersection equation (quadratic) has a double solution. This means its discriminant is $0$. This leads to a quadratic equation in the parameter $m$. You should find two solutions which is consistent with the geometric intuition. Aug 17, 2011 at 13:43
This is really a question in projective geometry, and has a constructive solution involving only straightedge. To see why it works, you have to know the theory, but here’s how to do it. From your outside point, call it $O$, draw two lines $\ell_1$ and $\ell_2$ each intersecting the ellipse in two points, say $\ell_1$ intersects in $A$ and $A'$, $\ell_2$ intersecting in $B$ and $B'$. Then draw the lines $\overline{AB}$ and $\overline{A'B'}$, intersecting at the point $P_1$ and the lines $\overline{AB'}$ and $\overline{A'B}$ intersecting at the point $P_2$. Then the line $m$ from $P_1$ to $P_2$ intersects your ellipse at the two points of tangency.
Let the equation of the line be $y = mx+c$. Since the line passes through $(1,1)$, we have $1 = m + c$ i.e. $c = 1 - m$. Hence, the equation of the line is $y = mx + 1 - m$.
Now we want the above line to be a tangent to $2x^2 + y^2 = 1$. This means the line should intersect the ellipse at exactly one point. Plug in $y = mx + 1 - m$ and the condition that the line intersects the ellipse at only one point means that the quadratic in $x$ must have only one root which means that the discriminant of the quadratic must be zero.
The quadratic we get is $2x^2 + (mx + 1 - m)^2 = 1$. Rearranging we get $$(m^2+2)x^2 + 2m(1-m)x + m^2 - 2m = 0$$ The discriminant is $4m^2(1-m)^2 - 4(m^2-2m)(m^2+2) = -4m(m-4)$
Setting this to zero, we get the two tangent from $(1,1)$ to the ellipse $2x^2 + y^2 = 1$ are $$y= 1$$ $$y = 4x - 3$$
• Thanks, that's a great answer. Does this technique work for all curves in $R^2$? Aug 17, 2011 at 15:37
• This idea holds good for any conic. This need not be true for arbitrary curves since for instance, the tangent to sin(x) from the point $(1,1)$ intersects the the sine curve at infinite points.
– user17762
Aug 17, 2011 at 15:41
• You're right even if for nice curves (say $C^1$) this is valid locally which means that in a small enough interval... for algebraic curves the contact equation is a polynomial and whenever this polynomial has a root of a multiplicity higher than 1 we have tangency, the other roots are other intersection points. Aug 18, 2011 at 6:06
General equation of a line that goes through $P$ is $y-1=m(x-1)$; substituting we get $(m^2+2)x^2-2m(m-1)x+(m-1)^2-1=0$. This quadratic equation in $x$ has a double solution if and only if its discriminant is $0$. This is equivalent to $m(m-4)=0$ (do the computation).
Take the derivative of the equation for the ellipse: $$4x+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0$$ to get the slope, $\frac{\mathrm{d}y}{\mathrm{d}x}$, of the ellipse at $(x,y)$. The slope of a line going through $(x,y)$ and $(1,1)$ is $\frac{y-1}{x-1}$. Therefore, the points of tangency on the ellipse must satisfy $$4x+2y\frac{y-1}{x-1}=0$$
• One of the points on this tangent is (1,1) - which does not satisfy your equation? Aug 17, 2011 at 16:02
• The answer says "points of tangency." The point of tangency of the line $y=1$ is $(0,1)$. Aug 17, 2011 at 16:21 |
# What is a vector?
What is a vector? A vector is a quantity which has a magnitude and a direction. A vector will always give you the following two pieces of information.
• How much?
• Which way?
If a quantity has only a magnitude, it is called a scalar, not a vector. Mass, volume, and temperature are all good examples of scalars.
A vehicle with a mass of 3000 kg is an example of magnitude or scalar.
A good example of vector is an airplane heading west with a speed of 150 miles per hour. This situation describes the airplane's velocity.
Therefore, the velocity is an example of vector.
In this case, the magnitude is 150 miles per hour and the direction is west.
You could also say that the airplane's velocity is 150 m/h due west.
Vectors are important because in physics, it is often useful to know not just the magnitude of things, but also the direction that those things are traveling as well.
## What is a vector? Mathematical representation of vectors
There are more than one way to represent a vector mathematically.
Generally speaking, a letter (a capital letter or a lower case letter) in bold can be used to represent a vector.
For example, s, a, F, and S can all represent vectors.
If the letters are not in bold, then it refers to the magnitude only.
If the letters are not in bold and we put an arrow on top of the letters, then it refers to a vector. Below, we show 2 vectors.
s
F
You could also use two capital letters if you like.
AB
## Graphical representation of vectors
We use big arrows to graph vectors. To graph vectors accurately, you must have the following things on your paper or graph.
• A scale
• A reference direction
Above, we see two vectors. The black line is the scale. Three of the black lines is equal to the length of the blue arrow.
Therefore, the blue arrow represents a velocity of 30 miles per hour due north.
Six of the black lines is equal to the length of the red arrow.
Therefore, the red arrow represents a velocity of 60 miles per hour due east.
## Vector quiz
This scalar and vector quiz will help you determine if you know the difference between scalar and vector. You will not need to use a paper and pencil to complete this quiz.
First, read carefully this lesson about what a vector is and then take this quiz.
Objective of the quiz:
• know the difference between vector and scalar.
• See real life examples of scalar and vector
• Understand how to graph vectors
• Understand some mathematical representation of vectors
• Understand the meaning of the words magnitude, scalar, and vector.
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
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# If $\tan x = \dfrac{{12}}{{13}}$ , then evaluate the value of $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}$ .
Last updated date: 20th May 2024
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We know the half angle formula for sin and cosine. These are $\sin 2a = 2\sin a\cos a$ and $\cos 2a = {\cos ^2}a - {\sin ^2}a$ . By using these identities, $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \dfrac{{\sin 2x}}{{\cos 2x}} = \tan 2x$ . Now, we have given $\tan x = \dfrac{{12}}{{13}}$ and we need to get the value of $\tan 2x$ . We can use the half angle formula again to write $\tan 2x$ in terms of $\tan x$ . We know that, $\tan 2a = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}$ . Putting the values, we’ll get,
$\tan 2a = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}} \\ \Rightarrow \tan 2a = \dfrac{{2\dfrac{{12}}{{13}}}}{{1 - {{(\dfrac{{12}}{{13}})}^2}}}{\text{ }}[{\text{Using}},\tan a = \dfrac{{12}}{{13}}] \\ \Rightarrow \tan 2a = \dfrac{{\dfrac{{24}}{{13}}}}{{1 - \dfrac{{144}}{{169}}}} \\ \Rightarrow \tan 2a = \dfrac{{\dfrac{{24}}{{13}}}}{{\dfrac{{169 - 144}}{{169}}}} \\ \Rightarrow \tan 2a = \dfrac{{\dfrac{{24}}{{13}}}}{{\dfrac{{25}}{{169}}}} \\ \Rightarrow \tan 2a = \dfrac{{24}}{{13}} \times \dfrac{{169}}{{25}} \\ \Rightarrow \tan 2a = \dfrac{{24}}{1} \times \dfrac{{13}}{{25}} \\ \Rightarrow \tan 2a = \dfrac{{312}}{{25}} \\$
Hence, the required value of $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}} = \dfrac{{312}}{{25}}$ .
Note: There is more process to solve this question. We have given$\tan$ratio so we can use it to get the ratios of$\sin {\text{ and cos}}$. Then just putting the value in $\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}$ will give us the answer. |
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The height of an isosceles triangle is the perpendicular distance from the base of the triangle to the opposite vertex. The height of a triangle is one of its important dimensions because it allows us to calculate the area of the triangle.
To find the length of the height of an isosceles triangle, we have to use the Pythagoras theorem to derive a formula.
##### GEOMETRY
Relevant for
Learning about the height of an isosceles triangle with examples.
See examples
##### GEOMETRY
Relevant for
Learning about the height of an isosceles triangle with examples.
See examples
## Formula for the height of an isosceles triangle
The height of an isosceles triangle is calculated using the length of its base and the length of one of the congruent sides. We can calculate the height using the following formula:
where a is the length of the congruent sides of the triangle and b is the length of the base of the triangle.
### Derivation of the height formula
To derive this formula, we can consider the following isosceles triangle:
By drawing a line representing the height, we can see that we divide the isosceles triangle into two congruent right triangles. We can use one of the obtained triangles and apply the Pythagorean theorem to calculate the height.
Recall that the Pythagorean theorem does not say that the square of the hypotenuse is equal to the sum of the squares of the legs. Therefore, we have:
We have obtained an expression for the height.
## Height of an isosceles triangle – Examples with answers
The following examples use the seen formula to find the height of isosceles triangles. Try to solve the exercises yourself before looking at the solution.
### EXAMPLE 1
What is the height of an isosceles triangle that has a base of 8 m and congruent sides of length 6 m?
From the question, we have the following data:
• Base, m
• Sides, m
Therefore, we use the height formula with these values:
The height of the triangle is 4.47 m.
### EXAMPLE 2
An isosceles triangle has a base of 10 m and congruent sides of length 12 m. What is the length of its height?
We can identify the following information:
• Base, m
• Sides, m
Substituting these values in the formula, we have:
The height of the triangle is 10.91 m.
### EXAMPLE 3
An isosceles triangle has a base of length 8 m and congruent sides of length 9 m. What is the length of the height?
From the question, we have the following values:
• Base, m
• Sides, m
Substituting these values in the height formula, we have:
The height of the triangle is 8.06 m.
### EXAMPLE 4
What is the height of a triangle that has a base of length 14 m and congruent sides of length 11 m?
We have the following information:
• Base, m
• Sides, m
Therefore, we use the height formula with these values:
The height of the triangle is 8.49 m.
## Height of an isosceles triangle – Practice problems
Use the formula for the height of isosceles triangles to solve the following problems. If you need help, you can look at the solved examples above. |
### Author Topic: Problem 3 (noon) (Read 1993 times)
#### Victor Ivrii
• Elder Member
• Posts: 2563
• Karma: 0
##### Problem 3 (noon)
« on: November 19, 2019, 04:22:43 AM »
(a) Find the general solution of
$$\mathbf{x}'=\begin{pmatrix} 1 &2\\ 1 &0\end{pmatrix}\mathbf{x}$$
and sketch trajectories.
(b) Find the general solution
$$\mathbf{x}'=\begin{pmatrix} 1 &2\\ 1 &0\end{pmatrix}\mathbf{x}+ \begin{pmatrix} 0 \\[1pt] \dfrac{6 e^{3t }}{e^{2t}+1}\end{pmatrix}.$$
#### Changhao Jiang
• Jr. Member
• Posts: 8
• Karma: 0
##### Re: Problem 3 (noon)
« Reply #1 on: November 19, 2019, 05:15:16 AM »
(a)
To find eigenvalues, $(1-\lambda x)(-\lambda)-2=0$,we can get $\lambda = 2$ or $\lambda = -1$
To find eigenvectors, when $\lambda=2$,
$\begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix} ~ \begin{pmatrix} 1 & -2 \\ 0 & 0 \end{pmatrix}$
let $x_2=t, x_1=2t$, so the eigenvector is \begin{bmatrix}1 \\ 2\end{bmatrix}
when $\lambda = -1$
$\begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix}$
~
$\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$
let $x_2=-t, x_1=t$, so the eigenvector is \begin{bmatrix}1 \\ -1\end{bmatrix}
Therefore, the general solution is y=$c_1 \begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t}$ + $c_2 \begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$
(b)
from (a), we can know $\phi(t)= \begin{pmatrix} e^{2t} & e^{-t} \\ 2e^{2t} & -e^{-t} \end{pmatrix}$
then
$\begin{pmatrix} e^{2t} & e^{-t} \\ 2e^{2t} & -e^{-t} \end{pmatrix} \begin{bmatrix}u_1'\\ u_2'\end{bmatrix} = \begin{bmatrix} 0 \\ \frac{6e^{3t}}{e^{2t}+1}\end{bmatrix}$
By ref form, we can know $u_1'= \frac{6e^t}{e^2t+1}, u_2' = 0$, then by integrating, we can get $u_1=6arctan(e^t)+c_1, u_2=c_2$
then the general solution is $x(t)=(6arctan(e^t)+c_1)\begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t} + c_2\begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$
« Last Edit: November 19, 2019, 06:00:07 AM by Changhao Jiang »
#### xuanzhong
• Jr. Member
• Posts: 12
• Karma: 1
##### Re: Problem 3 (noon)
« Reply #2 on: November 19, 2019, 06:00:45 AM »
Here's the solution for sketching.
#### Jingjing Cui
• Full Member
• Posts: 16
• Karma: 5
##### Re: Problem 3 (noon)
« Reply #3 on: November 19, 2019, 06:59:22 AM »
$$a)det(A-\lambda I)=0\\ det\begin{vmatrix} 1-\lambda&2\\ 1&-\lambda\\ \end{vmatrix}=0\\ (1-\lambda)(-\lambda)-2=0\\ \lambda_1=-1 \;\; \lambda_2=2\\ (A-\lambda I)x=0\\ \\ when\; \lambda=-1\\ \begin{pmatrix} 2&2\\ 1&1\\ \end{pmatrix}-> \begin{pmatrix} 2&2\\ 0&0\\ \end{pmatrix}\\ so\; 2x_1+2x_2=0\\ let\; x_2=t\;\;\;then\;x_1=-t\\ so\;the\;corresponding\;eigenvector\;is\; (\begin{array}{cc} -1 \\ 1 \end{array})\\ when\; \lambda=2\\ \begin{pmatrix} -1&2\\ 1&-2\\ \end{pmatrix}-> \begin{pmatrix} -1&2\\ 0&0\\ \end{pmatrix}\\ so\; -x_1+2x_2=0\\ let\; x_2=t\;\;\;then\;x_1=2t\\ so\;the\;corresponding\;eigenvector\;is\; (\begin{array}{cc} 2 \\ 1 \end{array})\\ so\;the\;general\;solution\;is\;x=c_1e^{-t}(\begin{array}{cc} -1 \\ 1 \end{array})+c_2e^{2t}(\begin{array}{cc} 2 \\ 1 \end{array})\\$$
#### Jingjing Cui
• Full Member
• Posts: 16
• Karma: 5
##### Re: Problem 3 (noon)
« Reply #4 on: November 19, 2019, 07:17:44 AM »
b)
$$\phi(t) U'(t)=g(t)\\ \begin{pmatrix} -e^{-t}&2e^{2t}\\ e^{-t}&e^{2t}\\ \end{pmatrix}(\begin{array}{cc} U_1' \\ U_2' \end{array})=(\begin{array}{cc} 0 \\\frac{6e^{3t}}{e^{2t}+1} \end{array})\\ -U_1'e^{-t}+2U_2'e^{2t}=0\\ U_1'e^{-t}+U_2'e^{2t}=\frac{6e^{3t}}{e^{2t}+1}\\ U_1'=\frac{4e^{4t}}{e^{2t}+1}\\ so\;U_1=2e^{2t}-2ln|e^{2t}+1|+C_1\\ U_2'=\frac{2e^{t}}{e^{2t}+1}\\ so\;U_2=2arctan(e^t)+C_2\\ x=\phi(t)U(t)\\ so\;the\;solution\;is\;:\\ x=(2e^{2t}-2ln|e^{2t}+1|+C_1)(\begin{array}{cc} -e^{-t} \\ e^{-t} \end{array})+(2arctan(e^t)+C_2)(\begin{array}{cc} 2e^{2t} \\ e^{2t} \end{array})\\$$
OK, except LaTeX sucks:
1) text should not be a part of math formulae or included like \text{blah blah}
2) "operators" should be escaped: \cos, \sin, \tan, \ln
« Last Edit: November 24, 2019, 09:59:17 AM by Victor Ivrii »
#### NANAC
• Jr. Member
• Posts: 10
• Karma: 2
##### Re: Problem 3 (noon)
« Reply #5 on: November 19, 2019, 09:15:02 AM »
Please see the attachment for the answer
#### anntara khan
• Jr. Member
• Posts: 6
• Karma: 0
##### Re: Problem 3 (noon)
« Reply #6 on: November 19, 2019, 02:20:39 PM »
Please see attached file, Thanks!
#### baixiaox
• Jr. Member
• Posts: 10
• Karma: 0
##### Re: Problem 3 (noon)
« Reply #7 on: November 19, 2019, 05:41:10 PM »
answer for tt2 question3 a)
#### baixiaox
• Jr. Member
• Posts: 10
• Karma: 0
##### Re: Problem 3 (noon)
« Reply #8 on: November 19, 2019, 05:42:18 PM »
answer for tt2 question3 b)
#### Mingdi Xie
• Jr. Member
• Posts: 14
• Karma: 0
##### Re: Problem 3 (noon)
« Reply #9 on: November 20, 2019, 02:45:02 PM »
this is my solution for part a
#### Mingdi Xie
• Jr. Member
• Posts: 14
• Karma: 0
##### Re: Problem 3 (noon)
« Reply #10 on: November 20, 2019, 02:54:05 PM »
Solution for part b
#### Victor Ivrii
In problem got lost "classify point $(0,0)$" |
# The Formula of a Cube Plus b Cube: Understanding the Basics
When it comes to mathematics, there are numerous formulas and equations that play a crucial role in solving problems and understanding various concepts. One such formula that often arises in algebraic expressions is the formula of a cube plus b cube. In this article, we will delve into the details of this formula, its applications, and how it can be used to solve mathematical problems.
## What is the Formula of a Cube Plus b Cube?
The formula of a cube plus b cube is an algebraic expression that represents the sum of two cubes. It can be written as:
a^3 + b^3 = (a + b)(a^2 – ab + b^2)
This formula is derived from the concept of factoring, where we break down a polynomial expression into its factors. In the case of the formula of a cube plus b cube, we factorize the sum of two cubes into a binomial multiplied by a trinomial.
## Understanding the Derivation of the Formula
To understand how the formula of a cube plus b cube is derived, let’s consider the following steps:
2. Recognize that this expression can be written as (a + b)(a^2 – ab + b^2) by factoring.
3. Expand the expression (a + b)(a^2 – ab + b^2) using the distributive property.
4. Simplify the expanded expression to obtain a^3 + b^3.
By following these steps, we can see that the formula of a cube plus b cube is indeed valid and can be used to simplify algebraic expressions.
## Applications of the Formula of a Cube Plus b Cube
The formula of a cube plus b cube finds its applications in various mathematical problems and real-life scenarios. Let’s explore some of these applications:
### 1. Algebraic Simplification
One of the primary applications of the formula of a cube plus b cube is in simplifying algebraic expressions. By using this formula, we can factorize expressions and make them easier to work with. For example, consider the expression 8x^3 + 27y^3. By applying the formula of a cube plus b cube, we can rewrite it as:
8x^3 + 27y^3 = (2x)^3 + (3y)^3 = (2x + 3y)((2x)^2 – (2x)(3y) + (3y)^2)
This simplification allows us to break down complex expressions into more manageable forms, making further calculations or analysis more straightforward.
### 2. Number Patterns
The formula of a cube plus b cube can also be used to identify number patterns and relationships. By substituting different values for a and b, we can observe the resulting sums and analyze any patterns that emerge. For instance, let’s consider the following examples:
• When a = 1 and b = 1, the formula becomes 1^3 + 1^3 = 2^3.
• When a = 2 and b = 1, the formula becomes 2^3 + 1^3 = 3^3.
• When a = 3 and b = 1, the formula becomes 3^3 + 1^3 = 4^3.
By observing these examples, we can notice a pattern where the sum of two cubes on the left side of the equation is equal to the cube of the next consecutive number on the right side. This pattern holds true for various values of a and b, allowing us to make generalizations and predictions.
## Examples and Case Studies
To further illustrate the applications of the formula of a cube plus b cube, let’s consider a few examples and case studies:
### Example 1: Factoring an Expression
Suppose we have the expression 64x^3 – 125y^3. To factorize this expression, we can use the formula of a cube plus b cube. By recognizing that 64x^3 is equal to (4x)^3 and 125y^3 is equal to (5y)^3, we can rewrite the expression as:
64x^3 – 125y^3 = (4x)^3 – (5y)^3 = (4x – 5y)((4x)^2 + (4x)(5y) + (5y)^2)
By applying the formula, we have successfully factored the expression into two factors, making it easier to work with or solve for specific values.
### Case Study: Volume of a Cube
The formula of a cube plus b cube also has practical applications in geometry, particularly when calculating the volume of a cube. The volume of a cube can be expressed as a^3, where a represents the length of one side of the cube. However, if we want to find the volume of a cube with an additional cube attached to one of its sides, we can use the formula of a cube plus b cube.
Let’s consider a cube with side length a and an additional cube with side length b attached to one of its sides. The total volume of this shape can be calculated using the formula a^3 + b^3. By substituting the appropriate values for a and b, we can determine the volume of the shape.
## Q&A
### Q1: Can the formula of a cube plus b cube be used for negative values of a and b?
A1: Yes, the formula of a cube plus b cube can be used for negative values of a and b. The formula holds true regardless of the sign of the variables. However, it is important to consider the signs when simplifying or solving equations involving negative values.
A2: Yes, there are other formulas related to cubes. Some notable examples include the formula for the difference of two cubes (a^3 – b^3 = (a – b)(a^2 + ab + b^2)) and the formula for the sum of cubes (a^3 + b^3 = (a + b)(a^2 – ab + b^2)). These formulas, along with the formula of a cube plus b cube, are essential tools in algebraic manipulations.
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# Maths on a Mug #16
Perhaps a “simple” #mathsonthemug this time. Simple, but very important. This is Bayes’ theorem: $P(A|B) = \frac{P(B|A)P(A)}{P(B)}.$ This relates the probability of observing $$A$$ given that $$B$$ is true where $$A$$ and $$B$$ are events, we also require that the probability of $$B$$ does not equal 0 (i.e. $$P(B)\neq 0$$). It allows us to find the probability of a cause, given an effect.
Bayes’ can lead to some surprising, count-intuitive, results. Dr. House never thinks his patients have Lupus. Lupus is a condition where antibodies that are supposed to attack foreign cells to prevent infections instead attack plasma proteins which can lead to blood clots. It is believed that 2% of the population have lupus. Suppose that a test is 98% accurate if a person has the disease and 74% accurate if they do not. Given the 98% accuracy of the test, should House really assume his patient with a positive result doesn’t have Lupus?
We can check the probability using Bayes’ theorem. Let $$L$$ represent Lupus, $$T$$ the test and $$1$$ and $$0$$ positive/negative respectively (I’ll leave as an exercise to the reader to calculate the individual probabilities):
$\begin{eqnarray*} P(L=1|T=1) &=& \frac{P(T=1|L=1)P(L=1)}{P(P=1)},\\[1em] &=& \frac{(0.02)(0.98)}{0.0196+0.2548},\\[1em] &=& \frac{0.0196}{0.2744},\\[1em] &=& 0.0714. \end{eqnarray*}$
So House’s insight is sensible, even given a positive result there is only a 7.14% chance of the person actually having Lupus!
How we interpret Bayes’ theorem depends on how we interpret probability, a Bayesian or Frequentist interpretation. This is a long running fight in mathematics, and it is worth reading about it to see where you come down on the argument. I’ll leave you with a quote from Savage (1954)
“It is unanimously agreed that statistics depends somehow on probability. But, as to what probability is and how it is connected with statistics, there has seldom been such complete disagreement and breakdown of communication since the Tower of Babel.”
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Question Video: Finding the Integration of a Function Using Integration by Substitution | Nagwa Question Video: Finding the Integration of a Function Using Integration by Substitution | Nagwa
# Question Video: Finding the Integration of a Function Using Integration by Substitution Mathematics • Third Year of Secondary School
## Join Nagwa Classes
Determine ∫ 8𝑥(8𝑥 + 9)² d𝑥 by using the substitution method.
02:28
### Video Transcript
Determine the integral of eight 𝑥 times eight 𝑥 plus nine squared with respect to 𝑥 by using the substitution method.
In this example, we’ve been very explicitly told to use the substitution method to evaluate this integral. Usually, we will look to choose our substitution 𝑢 to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it. Here though, it’s not instantly obvious what that might be. Instead, then, we try choosing 𝑢 to be some more complicated part of the function, perhaps the inner function in a composite function.
Let’s try 𝑢 equals eight 𝑥 plus nine. This means that d𝑢 by d𝑥 is equal to eight. And we can treat d𝑢 and d𝑥 as differentials. Remember, d𝑢 by d𝑥 is not a fraction but we certainly treat it like one when performing integration by substitution. We can say that d𝑢 is equal to eight d𝑥. Or, equivalently, an eighth d𝑢 is equal to d𝑥. Now, this isn’t instantly helpful. As if we replace d𝑥 with an eighth d𝑢 and eight 𝑥 plus nine with 𝑢, we’ll still have part of our function, that’s the eight 𝑥, which is in terms of 𝑥.
But if we look back to our substitution, we see that we can rearrange this. We subtract nine from both sides, and we see that eight 𝑥 is equal to 𝑢 minus nine. Then, our integral becomes 𝑢 minus nine times 𝑢 squared multiplied by an eighth d𝑢. Let’s take this eighth outside of the integral and then distribute the parentheses, and we see we have a simple polynomial that we can integrate.
The integral of 𝑢 cubed is 𝑢 to the fourth power divided by four. The integral of negative nine 𝑢 squared is negative nine 𝑢 cubed divided by three. And we mustn’t forget 𝐶, our constant of integration. We can simplify nine 𝑢 cubed divided by three to three 𝑢 cubed. But we mustn’t forget to replace 𝑢 with eight 𝑥 plus nine in our final step.
When we do, we see that our integral is equal to an eighth times eight 𝑥 plus nine to the fourth power divided by four minus three times eight 𝑥 plus nine cubed plus 𝐶. When we distribute our parentheses, we see we have our solution. It’s one over 32 times eight 𝑥 plus nine to the fourth power minus three-eighths times eight 𝑥 plus nine cubed plus 𝐶.
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# Thread: Simplify as far as possible, sovle the equation, sovle the quadratic equations
1. ## Simplify as far as possible, sovle the equation, sovle the quadratic equations
Simlify as far as possible
(i) 10y-5x+7x-3y-x-2y
(ii) 4(6x-7)-5(2x-3)
Sovle the equation
(i) 8x-9 = 4x-21
(ii) 2(2x-5)=4(2x-1)
(iii) 5-3x
----------- = 5 - x
4
(i) 3x² - x - 2 = 0
(ii) x² - 32 - 4x
I will appreciate if someone can help
thank you
2. Originally Posted by Steven777
Simlify as far as possible
(i) 10y-5x+7x-3y-x-2y
(ii) 4(6x-7)-5(2x-3)
To learn how to add polynomials and combine "like" terms, try here and here. To learn how to work with parentheses, try here.
Originally Posted by Steven777
Sovle the equation
(i) 8x-9 = 4x-21
(ii) 2(2x-5)=4(2x-1)
(iii) 5-3x
----------- = 5 - x
4
To learn how to solve linear equations, try here.
Originally Posted by Steven777
(i) 3x² - x - 2 = 0
(ii) x² - 32 - 4x
To learn how to solve quadratic equations, try here.
Once you have learned the basic terms and techniques, please attempt at least one exercise from each of the three sets above. If you get stuck, you will then be able to reply with a clear listing of your work and reasoning so far.
Thank you!
3. Originally Posted by Steven777
Simplify as far as possible
(i) 10y-5x+7x-3y-x-2y
(ii) 4(6x-7)-5(2x-3)
(i)
$= 10y - 3y - 2y - 5x + 7x - x$
$= 5y + x$
(ii)
$= 4 \times 6x - 4 \times 7 - 5 \times 2x - 5 \times -3$
$= 24x - 28 - 10x +15$
$= 24x - 10x - 28 + 15$
$= 14x - 13$
Solve the equation
(i) 8x-9 = 4x-21
(ii) 2(2x-5)=4(2x-1)
(iii) 5-3x
----------- = 5 - x
4
(i)
$8x - 4x = -21 + 9$
$4x = -12$
$x = -3$
(ii)
$2 \times 2x + 2 \times -5 = 4 \times 2x + 4 \times -1$
$4x -10 = 8x -4$
$-10 + 4 = 8x - 4x$
$-6 = 4x$
$x = -1.5$
(iii) dont understand what you have written for this question
(i) 3x² - x - 2 = 0
(ii) x² - 32 - 4x
(i)ok the way i was taught to solve these quadratic equations...
first you write it in the form $ax^2 + bx + c$, thus for this question, $a = 3, b = -1, c = -2$.
Then you work out $ac$, which is $ac = 3 \times -2 = -6$.
Now you have to find two numbers that multiply to give an answer of $-6$ and add to give an answer of $-1$ (which is $b$ in this question).
These two numbers are $-3$ and $2$ ( $-3 \times 2 = -6$ and $-3 + 2 = -1$).
So $3x^2 - x - 2 = 0$ can be rewritten as:
$3x^2 - 3x + 2x - 2 = 0$
$3x(x - 1) + 2(x - 1) = 0$
$(3x + 2) (x - 1) = 0$
This means that either $3x + 2$ or $x - 1$ have to be equal to 0 (or both)
Therefore,
$3x + 2 = 0$
$3x = -2$
$x = -\frac{2}{3}$
OR
$x - 1 = 0$
$x = 1$
(ii) have a go at this one by yourself now that I have shown you how to solve quadratic equations |
# Functional relationship between variables
### Functional Relationship Examples: Distance-Time Graphs and Temperature-Precipitation Graphs
A real-life example of a functional relationship is the relationship between of functional relationships helps us figure out how different variables work together. Functional relationships. • We frequently need to represent economic quantities as a mathematical function of one or more variables, e.g.: • A firm's costs as a. Determining if a Relationship is a Functional Relationship To determine whether or not a graph is a function, you can use the vertical line test. . A function exists when each x-value (input, independent variable) is paired with exactly one.
The following are examples of monomials: Polynomial comes from the Greek word, poly, which means many. A polynomial has two or more terms i. If there are only two terms in the polynomial, the polynomial is called a binomial. These terms are 4x3y2, - 2xy2, and 3. The coefficients of the terms are 4, -2, and 3. The degree of a term or monomial is the sum of the exponents of the variables. The degree of a polynomial is the degree of the term of highest degree.
In the above example the degrees of the terms are 5, 3, and 0. The degree of the polynomial is 5. Remember that variables are items which can assume different values.
## Variables, Functions and Equations
A function tries to explain one variable in terms of another. Consider the above example where the amount you choose to spend depends on your salary. Here there are two variables: Independent variables are those which do not depend on other variables. Dependent variables are those which are changed by the independent variables. The change is caused by the independent variable. In our example salary is the independent variable and the amount you spend is the dependent variable.
13-1 Relationships Between Variables
To continue with the same example what if the amount you choose to spend depends not only on your salary but also on the income you receive from investments in the stock market. Now there are three variables: A function is a mathematical relationship in which the values of a single dependent variable are determined by the values of one or more independent variables. Function means the dependent variable is determined by the independent variable s. A goal of economic analysis is to determine the independent variable s which explain certain dependent variables.
For example what explains changes in employment, in consumer spending, in business investment etc.? Functions with a single independent variable are called univariate functions.
There is a one to one correspondence.
Functions with more than one independent variable are called multivariate functions. The independent variable is often designated by x. The dependent variable is often designated by y. We say y is a function of x. This means y depends on or is determined by x. If we know the value of x, then we can find the value of y.
In pronunciation we say " y is f of x. In other words the parenthesis does not mean that f is multiplied by x. It is not necessary to use the letter f.
We may look at functions algebraically or graphically. If we use algebra we look at equations. If we use geometry we use graphs.
In other words a variable is something whose magnitude can change. It assumes different values at different times or places. Variables that are used in economics are income, expenditure, saving, interest, profit, investment, consumption, imports, exports, cost and so on.
It is represented by a symbol. Variables can be endogenous and exogenous. An endogenous variable is a variable that is explained within a theory. An exogenous variable influences endogenous variables, but the exogenous variable itself is determined by factors outside the theory.
### Basic Tools in Economic Analysis - WikiEducator
Ceteris Paribus is an assumption which we are compelled to make due to complexities in the reality. It is necessary for the sake of convenience. The limitations of human intelligence and capacity compel us to make this assumption.
Besides, without the assumption we cannot reach on economic relations, sequences and conclusions. In fact, there are large number of variables interacting simultaneously at a given time. If our analysis has to be accurate we may have to examine two variables at a time which makes it inevitable to assume other variables to remain unchanged.
For instance, if we try to establish the relationship between demand and price, there may be other variables which may also influence demand besides price. The influence of other factors may invalidate the hypothesis that quantity demanded of a commodity is inversely related to its price. If rise in price takes place along with an increasing in income or a change technology, then the effect of price change may not be the same.
However, we try to eliminate the interrupting influences of other variables by assuming them to remain unchanged. The assumption of Ceteris Paribus thus eliminates the influence of other factors which may get in the way of establishing a scientific statement regarding the behavior of economic variables.
A simple technical term is used to analyze and symbolizes a relationship between variables. It is called a function. It indicates how the value of dependent variable depends on the value of independent or other variables. It also explains how the value of one variable can be found by specifying the value of other variable. For instance, economist generally links demand for good depends upon its price. Functions are classifieds into two type namely explicit function and implicit function.
Explicit function is one in which the value of one variable depends on the other in a definite form. For instance, the relationships between demand and price Implicit function is one in which the variables are interdependent.
## Functional relations between variables
When the verbal expressions are transformed into algebraic form we get Equations. The term equation is a statement of equality of two expressions or variables.
The two expressions of an equation are called the sides of the equation. Equations are used to calculate the value of an unknown variable. An equation specifies the relationship between the dependent and independent variables.
### secondary mathematics functional relations between variables | Nuffield Foundation
Each equation is a concise statement of a particular relation. For example, the functional relationship between consumption C and income Y can take different forms. It says nothing about the form that this relation takes. |
## Rotation of Axes
### Learning Outcomes
• Identify nondegenerate conic sections given their general form equations.
• Write equations of rotated conics in standard form.
• Identify conics without rotating axes.
As we have seen, conic sections are formed when a plane intersects two right circular cones aligned tip to tip and extending infinitely far in opposite directions, which we also call a cone. The way in which we slice the cone will determine the type of conic section formed at the intersection. A circle is formed by slicing a cone with a plane perpendicular to the axis of symmetry of the cone. An ellipse is formed by slicing a single cone with a slanted plane not perpendicular to the axis of symmetry. A parabola is formed by slicing the plane through the top or bottom of the double-cone, whereas a hyperbola is formed when the plane slices both the top and bottom of the cone.
Figure 1. The nondegenerate conic sections
Ellipses, circles, hyperbolas, and parabolas are sometimes called the nondegenerate conic sections, in contrast to the degenerate conic sections, which are shown in Figure 2. A degenerate conic results when a plane intersects the double cone and passes through the apex. Depending on the angle of the plane, three types of degenerate conic sections are possible: a point, a line, or two intersecting lines.
Figure 2. Degenerate conic sections
## Identifying Nondegenerate Conics in General Form
In previous sections of this chapter, we have focused on the standard form equations for nondegenerate conic sections. In this section, we will shift our focus to the general form equation, which can be used for any conic. The general form is set equal to zero, and the terms and coefficients are given in a particular order, as shown below.
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$
where $A,B$, and $C$ are not all zero. We can use the values of the coefficients to identify which type conic is represented by a given equation.
You may notice that the general form equation has an $xy$ term that we have not seen in any of the standard form equations. As we will discuss later, the $xy$ term rotates the conic whenever $\text{ }B\text{ }$ is not equal to zero.
Conic Sections Example
ellipse $4{x}^{2}+9{y}^{2}=1$
circle $4{x}^{2}+4{y}^{2}=1$
hyperbola $4{x}^{2}-9{y}^{2}=1$
parabola $4{x}^{2}=9y\text{ or }4{y}^{2}=9x$
one line $4x+9y=1$
intersecting lines $\left(x - 4\right)\left(y+4\right)=0$
parallel lines $\left(x - 4\right)\left(x - 9\right)=0$
a point $4{x}^{2}+4{y}^{2}=0$
no graph $4{x}^{2}+4{y}^{2}=-1$
### A General Note: General Form of Conic Sections
A nondegenerate conic section has the general form
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$
where $A,B$, and $C$ are not all zero.
The table below summarizes the different conic sections where $B=0$, and $A$ and $C$ are nonzero real numbers. This indicates that the conic has not been rotated.
ellipse $A{x}^{2}+C{y}^{2}+Dx+Ey+F=0,\text{ }A\ne C\text{ and }AC>0$ circle $A{x}^{2}+C{y}^{2}+Dx+Ey+F=0,\text{ }A=C$ hyperbola $A{x}^{2}-C{y}^{2}+Dx+Ey+F=0\text{ or }-A{x}^{2}+C{y}^{2}+Dx+Ey+F=0$, where $A$ and $C$ are positive parabola $A{x}^{2}+Dx+Ey+F=0\text{ or }C{y}^{2}+Dx+Ey+F=0$
### How To: Given the equation of a conic, identify the type of conic.
1. Rewrite the equation in the general form, $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$.
2. Identify the values of $A$ and $C$ from the general form.
1. If $A$ and $C$ are nonzero, have the same sign, and are not equal to each other, then the graph is an ellipse.
2. If $A$ and $C$ are equal and nonzero and have the same sign, then the graph is a circle.
3. If $A$ and $C$ are nonzero and have opposite signs, then the graph is a hyperbola.
4. If either $A$ or $C$ is zero, then the graph is a parabola.
### Example 1: Identifying a Conic from Its General Form
Identify the graph of each of the following nondegenerate conic sections.
1. $4{x}^{2}-9{y}^{2}+36x+36y - 125=0$
2. $9{y}^{2}+16x+36y - 10=0$
3. $3{x}^{2}+3{y}^{2}-2x - 6y - 4=0$
4. $-25{x}^{2}-4{y}^{2}+100x+16y+20=0$
### Try It
Identify the graph of each of the following nondegenerate conic sections.
1. $16{y}^{2}-{x}^{2}+x - 4y - 9=0$
2. $16{x}^{2}+4{y}^{2}+16x+49y - 81=0$
## Finding a New Representation of the Given Equation after Rotating through a Given Angle
Until now, we have looked at equations of conic sections without an $xy$ term, which aligns the graphs with the x– and y-axes. When we add an $xy$ term, we are rotating the conic about the origin. If the x– and y-axes are rotated through an angle, say $\theta$, then every point on the plane may be thought of as having two representations: $\left(x,y\right)$ on the Cartesian plane with the original x-axis and y-axis, and \begin{align}\left({x}^{\prime },{y}^{\prime }\right)\end{align} on the new plane defined by the new, rotated axes, called the x’-axis and y’-axis.
Figure 3. The graph of the rotated ellipse ${x}^{2}+{y}^{2}-xy – 15=0$
We will find the relationships between $x$ and $y$ on the Cartesian plane with \begin{align}{x}^{\prime }\end{align} and \begin{align}{y}^{\prime }\end{align} on the new rotated plane.
Figure 4. The Cartesian plane with x- and y-axes and the resulting x′− and y′−axes formed by a rotation by an angle $\theta$.
The original coordinate x– and y-axes have unit vectors $i$ and $j$. The rotated coordinate axes have unit vectors \begin{align}{i}^{\prime }\end{align} and \begin{align}{j}^{\prime }\end{align}. The angle $\theta$ is known as the angle of rotation. We may write the new unit vectors in terms of the original ones.
\begin{align}&{i}^{\prime }=i\cos \theta +j\sin \theta \\ &{j}^{\prime }=-i\sin \theta +j\cos \theta \end{align}
Figure 5. Relationship between the old and new coordinate planes.
Consider a vector $u$ in the new coordinate plane. It may be represented in terms of its coordinate axes.
\begin{align}&u={x}^{\prime }{i}^{\prime }+{y}^{\prime }{j}^{\prime } \\ &u={x}^{\prime }\left(i\cos \theta +j\sin \theta \right)+{y}^{\prime }\left(-i\sin \theta +j\cos \theta \right) && \text{Substitute}. \\ &u=ix^{\prime}\cos \theta +jx^{\prime}\sin \theta -iy^{\prime}\sin \theta +jy^{\prime}\cos \theta && \text{Distribute}. \\ &u=ix^{\prime}\cos \theta -iy^{\prime}\sin \theta +jx^{\prime}\sin \theta +jy^{\prime}\cos \theta && \text{Apply commutative property}. \\ &u=\left(x^{\prime}\cos \theta -y^{\prime}\sin \theta \right)i+\left(x^{\prime}\sin \theta +y^{\prime}\cos \theta \right)j && \text{Factor by grouping}. \end{align}
Because \begin{align}u={x}^{\prime }{i}^{\prime }+{y}^{\prime }{j}^{\prime }\end{align}, we have representations of $x$ and $y$ in terms of the new coordinate system.
$\begin{gathered}x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \\ \text{and}\\ y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \end{gathered}$
### A General Note: Equations of Rotation
If a point $\left(x,y\right)$ on the Cartesian plane is represented on a new coordinate plane where the axes of rotation are formed by rotating an angle $\theta$ from the positive x-axis, then the coordinates of the point with respect to the new axes are \begin{align}\left({x}^{\prime },{y}^{\prime }\right)\end{align}. We can use the following equations of rotation to define the relationship between \begin{align}\left(x,y\right)\end{align} and \begin{align}\left({x}^{\prime },{y}^{\prime }\right):\end{align}
$\begin{gathered}x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \\ \text{and}\\ y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \end{gathered}$
### How To: Given the equation of a conic, find a new representation after rotating through an angle.
1. Find $x$ and $y$ where \begin{align}x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \end{align} and \begin{align}y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \end{align}.
2. Substitute the expression for $x$ and $y$ into in the given equation, then simplify.
3. Write the equations with \begin{align}{x}^{\prime }\end{align} and \begin{align}{y}^{\prime }\end{align} in standard form.
### Example 2: Finding a New Representation of an Equation after Rotating through a Given Angle
Find a new representation of the equation $2{x}^{2}-xy+2{y}^{2}-30=0$ after rotating through an angle of $\theta =45^\circ$.
## Writing Equations of Rotated Conics in Standard Form
Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the ${x}^{\prime }$ and ${y}^{\prime }$ coordinate system without the ${x}^{\prime }{y}^{\prime }$ term, by rotating the axes by a measure of $\theta$ that satisfies
$\cot \left(2\theta \right)=\frac{A-C}{B}$
We have learned already that any conic may be represented by the second degree equation
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$
where $A,B$, and $C$ are not all zero. However, if $B\ne 0$, then we have an $xy$ term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle $\theta$ where $\cot \left(2\theta \right)=\frac{A-C}{B}$.
• If $\cot \left(2\theta \right)>0$, then $2\theta$ is in the first quadrant, and $\theta$ is between $\left(0^\circ ,45^\circ \right)$.
• If $\cot \left(2\theta \right)<0$, then $2\theta$ is in the second quadrant, and $\theta$ is between $\left(45^\circ ,90^\circ \right)$.
• If $A=C$, then $\theta =45^\circ$.
### How To: Given an equation for a conic in the \begin{align}{x}^{\prime }{y}^{\prime }\end{align} system, rewrite the equation without the \begin{align}{x}^{\prime }{y}^{\prime }\end{align} term in terms of \begin{align}{x}^{\prime }\end{align} and \begin{align}{y}^{\prime }\end{align}, where the \begin{align}{x}^{\prime }\end{align} and \begin{align}{y}^{\prime }\end{align} axes are rotations of the standard axes by $\theta$ degrees.
1. Find $\cot \left(2\theta \right)$.
2. Find $\sin \theta$ and $\cos \theta$.
3. Substitute $\sin \theta$ and $\cos \theta$ into \begin{align}x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \end{align} and \begin{align} y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \end{align}.
4. Substitute the expression for $x$ and $y$ into in the given equation, and then simplify.
5. Write the equations with \begin{align}{x}^{\prime }\end{align} and \begin{align}{y}^{\prime }\end{align} in the standard form with respect to the rotated axes.
### Example 3: Rewriting an Equation with respect to the x′ and y′ axes without the x′y′ Term
Rewrite the equation $8{x}^{2}-12xy+17{y}^{2}=20$ in the \begin{align}{x}^{\prime }{y}^{\prime }\end{align} system without an \begin{align}{x}^{\prime }{y}^{\prime }\end{align} term.
### Try It
Rewrite the $13{x}^{2}-6\sqrt{3}xy+7{y}^{2}=16$ in the \begin{align}{x}^{\prime }{y}^{\prime }\end{align} system without the \begin{align}{x}^{\prime }{y}^{\prime }\end{align} term.
### Example 4: Graphing an Equation That Has No x′y′ Terms
Graph the following equation relative to the \begin{align}{x}^{\prime }{y}^{\prime }\end{align} system:
${x}^{2}+12xy - 4{y}^{2}=30$
## Identifying Conics without Rotating Axes
Now we have come full circle. How do we identify the type of conic described by an equation? What happens when the axes are rotated? Recall, the general form of a conic is
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$
If we apply the rotation formulas to this equation we get the form
\begin{align}{A}^{\prime }{{x}^{\prime }}^{2}+{B}^{\prime }{x}^{\prime }{y}^{\prime }+{C}^{\prime }{{y}^{\prime }}^{2}+{D}^{\prime }{x}^{\prime }+{E}^{\prime }{y}^{\prime }+{F}^{\prime }=0\end{align}
It may be shown that \begin{align}{B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime }\end{align}. The expression does not vary after rotation, so we call the expression invariant. The discriminant, ${B}^{2}-4AC$, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section.
### A General Note: Using the Discriminant to Identify a Conic
If the equation $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ is transformed by rotating axes into the equation \begin{align}{A}^{\prime }{{x}^{\prime }}^{2}+{B}^{\prime }{x}^{\prime }{y}^{\prime }+{C}^{\prime }{{y}^{\prime }}^{2}+{D}^{\prime }{x}^{\prime }+{E}^{\prime }{y}^{\prime }+{F}^{\prime }=0\end{align}, then \begin{align}{B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime }\end{align}.
The equation $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these.
If the discriminant, ${B}^{2}-4AC$, is
• $<0$, the conic section is an ellipse
• $=0$, the conic section is a parabola
• $>0$, the conic section is a hyperbola
### Example 5: Identifying the Conic without Rotating Axes
Identify the conic for each of the following without rotating axes.
1. $5{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0$
2. $5{x}^{2}+2\sqrt{3}xy+12{y}^{2}-5=0$
### Try It
Identify the conic for each of the following without rotating axes.
1. ${x}^{2}-9xy+3{y}^{2}-12=0$
2. $10{x}^{2}-9xy+4{y}^{2}-4=0$
## Key Equations
General Form equation of a conic section $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ Rotation of a conic section \begin{align}&x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \\ &y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \end{align} Angle of rotation $\theta ,\text{ where }\cot \left(2\theta \right)=\frac{A-C}{B}$
## Key Concepts
• Four basic shapes can result from the intersection of a plane with a pair of right circular cones connected tail to tail. They include an ellipse, a circle, a hyperbola, and a parabola.
• A nondegenerate conic section has the general form $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ where $A,B$ and $C$ are not all zero. The values of $A,B$, and $C$ determine the type of conic.
• Equations of conic sections with an $xy$ term have been rotated about the origin.
• The general form can be transformed into an equation in the \begin{align}{x}^{\prime }\end{align} and \begin{align}{y}^{\prime }\end{align} coordinate system without the \begin{align}{x}^{\prime }{y}^{\prime }\end{align} term.
• An expression is described as invariant if it remains unchanged after rotating. Because the discriminant is invariant, observing it enables us to identify the conic section.
## Glossary
angle of rotation
an acute angle formed by a set of axes rotated from the Cartesian plane where, if $\cot \left(2\theta \right)>0$, then $\theta$ is between $\left(0^\circ ,45^\circ \right)$; if $\cot \left(2\theta \right)<0$, then $\theta$ is between $\left(45^\circ ,90^\circ \right)$; and if $\cot \left(2\theta \right)=0$, then $\theta =45^\circ$
degenerate conic sections
any of the possible shapes formed when a plane intersects a double cone through the apex. Types of degenerate conic sections include a point, a line, and intersecting lines.
nondegenerate conic section
a shape formed by the intersection of a plane with a double right cone such that the plane does not pass through the apex; nondegenerate conics include circles, ellipses, hyperbolas, and parabolas |
Browse Questions
Home >> CBSE XII >> Math >> Matrices
# Using elementary transformations, find the inverse of the matrix if it exists - $\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix}$
$\begin{array}{1 1}A^{-1}=\begin{bmatrix} 1& -10/25 & -3/25 \\ -10/25 & 4/25 & 1\\ -3/25 & 1/25 & 0 \end{bmatrix} \\ A^{-1}=\begin{bmatrix} 1& -10/25 & 0 \\ -10/25 & 4/25 & 11/25\\ -3/25 & 1/25 & 9/25 \end{bmatrix} \\ A^{-1}=\begin{bmatrix} 1& -2/5 & -3/5 \\ -2/25 & 4/25 & 11/25\\ -3/5 & 1/25 & 9/25 \end{bmatrix} \\ \text{Does not exist} \end{array}$
Toolbox:
• There are six operations (transformations) on a matrix,three of which are due to rows and three due to columns which are known as elementary operations or transformations.
• Row/Column Switching: Interchange of any two rows or two columns, i.e, $R_i\leftrightarrow R_j$ or $\;C_i\leftrightarrow C_j$
• Row/Column Multiplication: The multiplication of the elements of any row or column by a non zero number: i.e, i.e $R_i\rightarrow kR_i$ where $k\neq 0$ or $\;C_j\rightarrow kC_j$ where $k\neq 0$
• Row/Column Addition:The addition to the element of any row or column ,the corresponding elements of any other row or column multiplied by any non zero number: i.e $R_i\rightarrow R_i+kR_j$ or $\;C_i\rightarrow C_i+kC_j$, where $i \neq j$.
• If A is a matrix such that A$^{-1}$ exists, then to find A$^{-1}$ using elementary row operations, write A = IA and apply a sequence of row operation on A = IA till we get, I = BA. The matrix B will be the inverse of A. Similarly, if we wish to find A$^{-1}$ using column operations, then, write A = AI and apply a sequence of column operations on A = AI till we get, I = AB.
Given $\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix}$
Step :1 In order to use row elementary transformation we write as$A=I_3A$
$\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} A$
Step 2: Applying $R_1\rightarrow R_2+ 3R_1$
$\begin{bmatrix} 1 & 3 & -2 \\ -3+3(1) & 0+3(3) & -5+3(-2) \\ 2 & 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0+3(0) & 1+3(0) & 0+3(0)\\ 0 & 0 & 1 \end{bmatrix} A$
$\begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 2 & 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} A$
Step 3: Applying $R_3\rightarrow R_3- 2R_1$
$\begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 2-2(1) & 5-2(3) & 0-2(-2) \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0\\ 0-2(1) & 0-0 & 1-2(0) \end{bmatrix} A$
$\begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0\\ -2 & 0 & 1 \end{bmatrix} A$
Step :4 Applying $R_2\rightarrow R_2+ 8R_3$
$\begin{bmatrix} 1 & 3 & -2 \\ 0+0 & 9+8(-1) & -11+8(4) \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3+8(-2) & 1+0 & 0+8(1)\\ -2 & 0 & 1 \end{bmatrix} A$
$\begin{bmatrix} 1 & 3 & -2 \\ 0 & 1 & 21 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -13 & 1 & 8\\ -2 & 0 & 1 \end{bmatrix} A$
Step:5 Applying $R_1\rightarrow R_1- 3R_2$
$\begin{bmatrix} 1-3(0) & 3-3(1 & -2-3(21) \\ 0 & 1 & 21 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} \\ 1-3(-13) & 0-3(1) & 0-3(8)\\ -13 & 1 & 8\\-2 \end{bmatrix} A$
$\begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} 40 & -3 & -24 \\ -13 & 1 & 8\\ -2 & 0 & 1 \end{bmatrix} A$
Step :6 Applying $R_3\rightarrow R_3+R_2$
$\begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0+0 & -1+1 & 4+21 \end{bmatrix} = \begin{bmatrix} 40 & -3 & -24 \\ -13 & 1 & 8\\ -2-13 & 0+1 & 1+8 \end{bmatrix} A$
$\begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & 0 & 25\end{bmatrix} = \begin{bmatrix} 40 & -3 & -24 \\ -13 & 1 & 8\\ -15 & 1 & 9 \end{bmatrix} A$
Step :7 Applying $R_3\rightarrow \frac{1}{25} R_3$
$\begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 40 & -3 & -24 \\ -12 & 1 & 8\\ -15/25 & 1/25 & 9/25 \end{bmatrix} A$
$\begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{bmatrix} = \frac{1}{25}\begin{bmatrix} 1000 & -75 & -600 \\ -325 & 25 & 200\\ -15 & 1 & 9 \end{bmatrix} A$
Step 8: Applying $R_1\rightarrow R_1+65 R_3$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{bmatrix} = \frac{1}{25}\begin{bmatrix} 25& -10 & -15 \\ -325 & 25 & 200\\ -15 & 1 & 9 \end{bmatrix} A$
Step :9 Applying $R_2\rightarrow R_2-21 R_3$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \frac{1}{25}\begin{bmatrix} 25& -10 & -15 \\ -10 & 4 & 11\\ -15 & 1 & 9 \end{bmatrix} A$
$A^{-1}=\begin{bmatrix} 1& -10/25 & -15/25 \\ -10/25 & 4/25 & 11/25\\ -15/25 & 1/25 & 9/25 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1& -2/5 & -3/5 \\ -2/5 & 4/25 & 11/25\\ -3/5 & 1/25 & 9/25 \end{bmatrix}$
edited Mar 18, 2013 |
9
Q:
# A man invests Rs.5000 for 3 years at 5% p.a. compound interest reckoned yearly. Income tax at the rate of 20% on the interest earned is deducted at the end of each year. Find the amount at the end of the third year
A) Rs.5624.32 B) Rs.5423 C) Rs.5634 D) Rs.5976
Explanation:
5% is the rate of interest. 20% of the interest amount is paid as tax.
i.e 80% of the interest amount stays back.
$\inline \therefore$ if we compute the rate of interest as 80% of 5% = 4% p.a., we will get the same value.
The interest accrued for 3 years in compound interest = 3 x simple interest on principal + 3 x interest on simple interest + 1 x interest on interest on interest.
= 3 x (200) + 3 x (8) + 1 x 0.32 =600 + 24 + 0.32 = 624.32
The amount at the end of 3 years = 5000 + 624.32 = 5624.32
Q:
A certain sum is invested for 2 years in scheme M at 20% p.a. compound interest (compounded annually), Same sum is also invested for the same period in scheme N at k% p.a. simple interest. The interest earned from scheme M is twice of that earned from scheme N. What is the value of k?
A) 7 B) 11 C) 9 D) 13
Explanation:
Interest earned in scheme M =
Interest earned in scheme N =
Now, from the given data,
k = 11
0 42
Q:
A sum is equally invested in two different schemes on CI at the rate of 15% and 20% for two years. If interest gained from the sum invested at 20% is Rs. 528.75 more than the sum invested at 15%, find the total sum?
A) Rs. 7000 B) Rs. 4500 C) Rs. 9000 D) Rs. 8200
Explanation:
Let Rs. K invested in each scheme
Two years C.I on 20% = 20 + 20 + 20x20/100 = 44%
Two years C.I on 15% = 15 + 15 + 15x15/100 = 32.25%
Now,
(P x 44/100) - (P x 32.25/100) = 528.75
=> 11.75 P = 52875
=> P = Rs. 4500
Hence, total invested money = P + P = 4500 + 4500 = Rs. 9000.
6 1088
Q:
What is the interest rate per annum, if a sum of money invested at compound interest amount to Rs. 2400 in 3 years and in 4 years to Rs. 2,520?
A) 3.5% B) 4% C) 5% D) 6.5%
Explanation:
Let 'R%' be the rate of interest
From the given data,
Hence, the rate of interest R = 5% per annum.
3 808
Q:
A sum of Rs. 8,000 is deposited for 3 years at 5% per annum compound interest (compounded annually). The difference of interest for 3 years and 2 years will be
A) Rs. 387 B) Rs. 441 C) Rs. 469 D) Rs. 503
Explanation:
Given principal amount = Rs. 8000
Time = 3yrs
Rate = 5%
C.I for 3 yrs =
Now, C.I for 2 yrs =
Hence, the required difference in C.I is 1261 - 820 = Rs. 441
4 781
Q:
Simple interest on a certain sum at 7% per annum for 4 years is Rs. 2415. What will be the compound interest on the same principal at 4% per annum in two years?
A) Rs. 704 B) Rs. 854 C) Rs. 893 D) Rs. 914
Explanation:
We know that,
From given data, P = Rs. 8625
Now, C.I =
3 1110
Q:
Find the compound interest on Rs. 6,500 for 4 years if the rate of interest is 10% p.a. for the first 2 years and 20% per annum for the next 2 years?
A) Rs. 3845 B) Rs. 4826 C) Rs. 5142 D) Rs. 4415
Explanation:
We know the formula for calculating
The compound interest where P = amount, r = rate of interest, n = time
Here P = 5000, r1 = 10, r2 = 20
Then
C = Rs. 4826.
10 1037
Q:
What is the difference between the compound interests on Rs. 5000 for 11⁄2 years at 4% per annum compounded yearly and half-yearly?
A) Rs. 1.80 B) Rs. 2.04 C) Rs. 3.18 D) Rs. 4.15
Explanation:
Compound Interest for 1 12 years when interest is compounded yearly = Rs.(5304 - 5000)
Amount after 112 years when interest is compounded half-yearly
Compound Interest for 1 12 years when interest is compounded half-yearly = Rs.(5306.04 - 5000)
Difference in the compound interests = (5306.04 - 5000) - (5304 - 5000)= 5306.04 - 5304 = Rs. 2.04
6 1107
Q:
The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is Rs. 56. Then the sum is :
A) Rs. 3680 B) Rs. 2650 C) Rs. 1400 D) Rs. 1170 |
# How do you integrate int 1/sqrt(2-5x^2) by trigonometric substitution?
Oct 18, 2017
$\int \setminus \frac{1}{\sqrt{2 - 5 {x}^{2}}} \setminus \mathrm{dx} = \frac{1}{\sqrt{5}} \setminus \arcsin \left(\sqrt{\frac{5}{2}} x\right) + C$
#### Explanation:
We seek:
$I = \int \setminus \frac{1}{\sqrt{2 - 5 {x}^{2}}} \setminus \mathrm{dx}$
Which, we can write as:
$I = \int \setminus \frac{1}{\sqrt{2 \left(1 - \frac{5}{2} {x}^{2}\right)}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{\sqrt{2} \sqrt{1 - {\left(\frac{\sqrt{5}}{\sqrt{2}} x\right)}^{2}}} \setminus \mathrm{dx}$
$\setminus \setminus = \frac{1}{\sqrt{2}} \setminus \int \setminus \frac{1}{\sqrt{1 - {\left(\sqrt{\frac{5}{2}} x\right)}^{2}}} \setminus \mathrm{dx}$
We can now perform a substitution, Let:
$u = \sqrt{\frac{5}{2}} x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \sqrt{\frac{5}{2}}$
$\therefore \sqrt{\frac{2}{5}} \frac{\mathrm{du}}{\mathrm{dx}} = 1$
Substituting into the integral, we get:
$I = \frac{1}{\sqrt{2}} \setminus \int \setminus \frac{1}{\sqrt{1 - {u}^{2}}} \setminus \left(\sqrt{\frac{2}{5}}\right) \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{\sqrt{5}} \setminus \int \setminus \frac{1}{\sqrt{1 - {u}^{2}}} \setminus \mathrm{du}$
Which is a standard integral, so we have:
$I = \frac{1}{\sqrt{5}} \setminus \arcsin u + C$
Restoring the substitution:
$I = \frac{1}{\sqrt{5}} \setminus \arcsin \left(\sqrt{\frac{5}{2}} x\right) + C$ |
# Do consecutive angles equal 180?
The consecutive interior angles theorem states that when the two lines are parallel, then the consecutive interior angles are supplementary to each other. Supplementary means that the two angles add up to 180 degrees.
## What is the sum of consecutive angles?
The ‘consecutive interior angle theorem’ states that if a transversal intersects two parallel lines, each pair of consecutive interior angles is supplementary, that is, the sum of the consecutive interior angles is 180°.
## What does consecutive mean in geometry?
Consecutive angles are the angles that formed when a transversal intersects two parallel lines. Each angle from a pair of consecutive angles lies on each of the parallel lines on any one side of the transversal (either interior or exterior).
## What is the difference between consecutive and corresponding angles?
Corresponding angles are at the same location on points of intersection. The last angle relationship is consecutive interior angles. These angles are located on the same side of the transversal and inside of the two lines. In the diagram above, angles 2 and 3 are consecutive interior angles, and so are angles 6 and 7.
## What are consecutive numbers?
Consecutive numbers meaning is “The numbers which continuously follow each other in the order from smallest to largest.” Consecutive numbers example are. Consecutive numbers from 1 to 8 are 1, 2, 3, 4, 5, 6, 7, 8. Here the difference between each number is 1.
## What is a meaning of consecutive?
Definition of consecutive : following one after the other in order : successive served four consecutive terms in office.
## What is a consecutive angle in a parallelogram?
Explanation: In a parallelogram, consecutive angles are supplementary (i.e. add to ) and opposite angles are congruent (i.e. equal).
## What are Cointerior angles?
Co-interior angles lie between two lines and on the same side of a transversal. In each diagram the two marked angles are called co-interior angles. If the two lines are parallel, then co-interior angles add to give 180o and so are supplementary.
## What is the sum of consecutive angles of parallelogram?
The consecutive angles of a parallelogram are supplementary. The sum of all the angles of a parallelogram is equal to 360°.
## What is consecutive sides and consecutive angles?
The angles that would be supplementary (next to each other, in this case) on one side of the transversal are called consecutive. A set of consecutive angles is always made up of one angle from each set: in the image above, the angles with arcs of the same color are consecutive to each other.
## Which Quadrilaterals always have consecutive angles are supplementary?
If a quadrilateral is a parallelogram, then its diagonals bisect each other. If a quadrilateral is a parallelogram, then consecutive angles are supplementary.
## Which is the corresponding angle to ∠ 1?
∠2 ≅ ∠60° since they are corresponding angles, and m and n are parallel. ∠1 and ∠2 form a straight angle, so∠1=120°.
## What are consecutive angles in a rectangle?
A rectangle is a parallelogram, so its opposite angles are congruent and its consecutive angles are supplementary.
## How do you name consecutive vertices?
When naming a polygon, its vertices are named in consecutive order either clockwise or counterclockwise. Consecutive sides are two sides that have an endpoint in common. The four‐sided polygon in Figure could have been named ABCD, BCDA, or ADCB, for example.
## What are consecutive vertices in a quadrilateral?
to create a random quadrilateral! Sides, angles, and vertices that are next to each other in a polygon are called consecutive. (Note that consecutive sides intersect at a single point.)
## What are 3 consecutive numbers?
Explanation: Three consecutive even integers can be represented by x, x+2, x+4. The sum is 3x+6, which is equal to 108. Thus, 3x+6=108.
## What are consecutive zeros?
If n=1 sequence will be {01} so number of pairs of consecutive zeros are 0, If n = 2 sequence will be {1001} so number of pairs of consecutive zeros are 1, If n=3 sequence will be {01101001} so number of pairs of consecutive zeros are 1, And in a segment of length 12, there are total 2 pairs of consecutive zeros.
## What is a consecutive multiple?
Numbers which follow each other in order, without gaps, from smallest to largest. 12, 13, 14 and 15 are consecutive numbers. 22, 24, 26, 28 and 30 are consecutive even numbers. 40, 45, 50 and 55 are consecutive multiples of 5. Number Sequences – Square, Cube and Fibonacci.
## What is consecutive and example?
Consecutive comes from the Latin consecutus, meaning “following closely” with no gap. Just like those snowstorms — one storm happened each day, back to back, for five days in a row. Consecutive numbers also follow each other, or advance in the right order. For example, 5, 6, 7, 8, 9, 10 are consecutive numbers.
## What are the consecutive days?
More Definitions of Consecutive days Consecutive days means days following one after the other without an interruption based on discharge. Means days occurring one after the other with no intervening days and does not mean sequential days or cyclical days.
## Where are consecutive angles in a parallelogram?
If one angle of a parallelogram is a right angle, then all the angles are right angles. Opposite angles of a parallelogram are equal (or congruent) Consecutive angles are supplementary angles to each other (that means they add up to 180 degrees) |
## The Reflective Educator
### Education ∪ Math ∪ Technology
Close
A binary number is a number written in base 2 format, like 101010101111. The binary number system is handy because it can be easily related to logical operators used in circuitry, and so almost all modern computers use this format for communication.
We use the decimal system for communication in our day to day lives because it is related to our original numbering systems, and this entire system was developed because of we have ten fingers between our two hands.
To convert a decimal number to a binary number, we want to rewrite the decimal number as a sum of powers of 2. For example, the number 5 is equal to 4 + 1 or 22 + 20 which is the same as 1×22 + 0×21 + 1×20. In binary, we write 5 as 101, since those are the coefficients of the powers of 2 (Try out this application which lets you switch between binary and decimal numbers).
Here is the basic multiplication table for binary, which only includes 0 and 1, since those are the only digits you have to multiply in binary (in a decimal system, you need a much larger multiplication table, since you need to be able to multiply each of 10 different digits, 0 – 9 by each of 10 different digits).
× 0 1 0 0 0 1 0 1
Compare this to the traditional 10 by 10 multiplication table for decimal numbers.
(Image credit: valilouve)
If you want to multiply numbers in binary, you could use some similar strategies to regular decimal multiplication. For example, 10101 times 101 looks like this:
If you want to double check, 10101 is the same as 1×24 + 0×23 + 1×22 + 0×21 + 1×20 = 16 + 0 + 4 + 0 + 1 = 21 and 101 is 5 (as we noted before) so this multiplication in decimal is 21×5, which is 105. 1101001 = 1×26 + 1×25 + 0×24 + 1×23 + 0×22 + 0×21 + 1×20 = 64 + 32 + 8 + 1 = 105. See this website for a more detailed example of binary multiplication.
The point of this activity is that you have taken something which is hard to do (memorizing a 10 by 10 times table) and switched it to something which is conceptually more difficult, but easier to memorize. For smaller numbers, it is faster to multiply directly in decimal, but for larger numbers, it will actually take less time to convert them to binary, do the multiplication, and convert back. You may notice that the multiplication step itself is much easier than decimal multiplication, since it’s just a matter of remembering 2 facts (0×0 = 0 and 0×1 = 1) and lining up the numbers correctly so that the place value matches. Check this page out for more information on binary number operations.
If all of this feels arbitrary and bizarre to you, now you know what many 3rd graders feel like when they are first introduced to multiplication. |
# Exponent Rules and Dividing Polynomials 5.6 1.Divide exponential forms with the same base. 2.Divide numbers in scientific notation. 3. Divide monomials.
## Presentation on theme: "Exponent Rules and Dividing Polynomials 5.6 1.Divide exponential forms with the same base. 2.Divide numbers in scientific notation. 3. Divide monomials."— Presentation transcript:
Exponent Rules and Dividing Polynomials 5.6 1.Divide exponential forms with the same base. 2.Divide numbers in scientific notation. 3. Divide monomials. 4.Divide a polynomial by a monomial. 6.Simplify expressions using rules of exponents. 5.Use long division to divide polynomials.
Objective 1 Divide exponential forms with the same base.
Quotient Rule for Exponents If m and n are integers and a is a real number, where a 0, then 11 1 1 11
When an equation in one variable is solved the answer is a point on a line. Divide:
Objective 2 Divide numbers in scientific notation.
Divide and write the result in scientific notation. Note: This is not in scientific notation.
When an equation in one variable is solved the answer is a point on a line. Divide and write the result in scientific notation:
Objective 3 Divide monomials.
3 4 1 x y3y3 1
Dividing Monomials 1. Divide the coefficients or simplify them to fractions in lowest terms. 2. Use the quotient rule for the exponents with like bases. 3. Write the final expression so that all exponents are positive.
When an equation in one variable is solved the answer is a point on a line. Divide:
Objective 4 Divide a polynomial by a monomial.
From arithmetic… or
Divide. Divide each term in the polynomial by the monomial.
If a, b, and c are real numbers, variables, or expressions with c 0, then Dividing a Polynomial by a Monomial Divide each term in the polynomial by the monomial.
When an equation in one variable is solved the answer is a point on a line. Divide:
Objective 6 Simplify expressions using rules of exponents.
Exponents Summary Zero as an exponent:a 0 = 1, where a 0. Negative exponents: Product rule for exponents: Quotient rule for exponents: Raising a power to a power: Raising a product to a power: Raising a quotient to a power:
Use the quotient rule for exponents. Use the power rule for exponents. 2 1 1 m 14 Evaluate coefficients. Reduce coefficients.
When an equation in one variable is solved the answer is a point on a line. Simplify:
Slide 5- 24 Copyright © 2011 Pearson Education, Inc. Simplify. a) b) c) d) 5.6
Slide 5- 25 Copyright © 2011 Pearson Education, Inc. Simplify. a) b) c) d) 5.6
Slide 5- 26 Copyright © 2011 Pearson Education, Inc. Divide. a)b) c)d) 5.6
Slide 5- 27 Copyright © 2011 Pearson Education, Inc. Divide. a)b) c)d) 5.6
Download ppt "Exponent Rules and Dividing Polynomials 5.6 1.Divide exponential forms with the same base. 2.Divide numbers in scientific notation. 3. Divide monomials."
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# Question #bb3f7
Jan 17, 2016
$x = \frac{\pi}{2} , \pi$
#### Explanation:
The first step is to rewrite $\cos \left(2 x\right)$ in terms of $\sin \left(x\right)$.
• $\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$
• ${\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$
• $\cos \left(2 x\right) = 1 - {\sin}^{2} \left(x\right) - {\sin}^{2} \left(x\right) = 1 - 2 {\sin}^{2} \left(x\right)$
Thus, the function can be rewritten as
$1 - 2 {\sin}^{2} \left(x\right) + 2 \sin \left(x\right) - 3 = - 2$
Move all the terms to the same side.
$0 = 2 {\sin}^{2} \left(x\right) - 2 \sin \left(x\right)$
Divide both sides by $2$.
${\sin}^{2} \left(x\right) - \sin \left(x\right) = 0$
Factor a $\sin \left(x\right)$ term.
$\sin \left(x\right) \left(\sin \left(x\right) - 1\right) = 0$
Here, we have a product of two terms that equals $0$. This means that either one of the two terms could be equal to $0$:
$\sin x = 0 \textcolor{w h i t e}{s s s s s s} \text{or} \textcolor{w h i t e}{s s s s s s} \sin \left(x\right) - 1 = 0 \implies \sin \left(x\right) = 1$
$x = \pi \textcolor{w h i t e}{s s s s s s s k l} \text{or} \textcolor{w h i t e}{s s s s s s s s s s s s s s s s s s s s s s s s} x = \frac{\pi}{2}$
The times when $\sin \left(x\right) = 0 , 1$ for $0 < x < 2 \pi$ are $x = \frac{\pi}{2} , \pi$. |
# What Are Twin Prime Numbers?
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• 0:03 Introducing Twin Primes
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#### Recommended Lessons and Courses for You
Lesson Transcript
Instructor: Mark Boster
There are all kinds of numbers: even, odd, positive, negative, and prime numbers, just to name a few. Did you know that there are even twin prime numbers? In this lesson, you'll learn what twin prime numbers are and some interesting facts about them.
## Introducing Twin Primes
Can you answer this riddle? There were 2 babies born on the same day, in the same hospital, by the same parents, but they were not twins. How did this happen? The answer may surprise you: They were 2 babies from a set of triplets!
You know what a twin is, but do you know what a twin prime number is? In this lesson, we're going to explore twin prime numbers and learn how to identify them. But first, let's quickly review prime numbers.
## Prime Numbers
A prime number is a whole number greater than 1 that only has whole number factors of 1 and itself. When we say that it only has whole number factors of 1 and itself, we mean the only two whole numbers you can multiply to get that number are 1 and the number itself.
For example, if you want to multiply two numbers to get the number 7, the only way to do it is 7 x 1 = 7. Another example of a prime number is 5. The only way to multiply whole numbers to get 5 is 5 x 1 = 5. There's no other way.
The first 9 prime numbers are 2 , 3 , 5 , 7, 11, 13, 17, 19, and 23.
If you do the math, you'll see that the only factors of these prime numbers are 1 and the number itself.
## Twin Prime Numbers
Now that we've refreshed our memory of prime numbers, let's dig into twin prime numbers. Twin prime numbers are two consecutive prime numbers that differ by 2. This is easy to remember when you remember that twins are a set of 2, and twin primes differ by 2.
For example, from our set of 9 prime numbers, there are 4 sets of twin prime numbers: 3 and 5, 5 and 7, 11 and 13, and 17 and 19. The numbers in each of these pairs differs by exactly 2.
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1.1111111111111113e+50 as a fraction
1.1111111111111113e+50 as a fraction - solution and the full explanation with calculations.
Below you can find the full step by step solution for you problem. We hope it will be very helpful for you and it will help you to understand the solving process.
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What is 1.1111111111111113e+50 as a fraction?
To write 1.1111111111111113e+50 as a fraction you have to write 1.1111111111111113e+50 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.
1.1111111111111113e+50 = 1.1111111111111113e+50/1 = 1.1111111111111E+51/10 = 1.1111111111111E+52/100 = 1.1111111111111E+53/1000 = 1.1111111111111E+54/10000 = 1.1111111111111E+55/100000 = 1.1111111111111E+56/1000000 = 1.1111111111111E+57/10000000 = 1.1111111111111E+58/100000000 = 1.1111111111111E+59/1000000000 = 1.1111111111111E+60/10000000000 = 1.1111111111111E+61/100000000000 = 1.1111111111111E+62/1000000000000 = 1.1111111111111E+63/10000000000000 = 1.1111111111111E+64/100000000000000 = 1.1111111111111E+65/1000000000000000 = 1.1111111111111E+66/10000000000000000 = 1.1111111111111E+67/100000000000000000 = 1.1111111111111E+68/1000000000000000000 = 1.1111111111111E+69/1.0E+19 = 1.1111111111111E+70/1.0E+20 = 1.1111111111111E+71/1.0E+21 = 1.1111111111111E+72/1.0E+22 = 1.1111111111111E+73/1.0E+23 = 1.1111111111111E+74/1.0E+24 = 1.1111111111111E+75/1.0E+25 = 1.1111111111111E+76/1.0E+26 = 1.1111111111111E+77/1.0E+27 = 1.1111111111111E+78/1.0E+28 = 1.1111111111111E+79/1.0E+29 = 1.1111111111111E+80/1.0E+30 = 1.1111111111111E+81/1.0E+31 = 1.1111111111111E+82/1.0E+32 = 1.1111111111111E+83/1.0E+33 = 1.1111111111111E+84/1.0E+34 = 1.1111111111111E+85/1.0E+35 = 1.1111111111111E+86/1.0E+36 = 1.1111111111111E+87/1.0E+37 = 1.1111111111111E+88/1.0E+38 = 1.1111111111111E+89/1.0E+39 = 1.1111111111111E+90/1.0E+40 = 1.1111111111111E+91/1.0E+41 = 1.1111111111111E+92/1.0E+42 = 1.1111111111111E+93/1.0E+43 = 1.1111111111111E+94/1.0E+44 = 1.1111111111111E+95/1.0E+45 = 1.1111111111111E+96/1.0E+46 = 1.1111111111111E+97/1.0E+47 = 1.1111111111111E+98/1.0E+48 = 1.1111111111111E+99/1.0E+49 = 1.1111111111111E+100/1.0E+50 = 1.1111111111111E+101/1.0E+51 = 1.1111111111111E+102/1.0E+52 = 1.1111111111111E+103/1.0E+53 = 1.1111111111111E+104/1.0E+54 = 1.1111111111111E+105/1.0E+55 = 1.1111111111111E+106/1.0E+56 = 1.1111111111111E+107/1.0E+57 = 1.1111111111111E+108/1.0E+58 = 1.1111111111111E+109/1.0E+59 = 1.1111111111111E+110/1.0E+60 = 1.1111111111111E+111/1.0E+61 = 1.1111111111111E+112/1.0E+62 = 1.1111111111111E+113/1.0E+63 = 1.1111111111111E+114/1.0E+64 = 1.1111111111111E+115/1.0E+65 = 1.1111111111111E+116/1.0E+66 = 1.1111111111111E+117/1.0E+67 = 1.1111111111111E+118/1.0E+68 = 1.1111111111111E+119/1.0E+69 = 1.1111111111111E+120/1.0E+70 = 1.1111111111111E+121/1.0E+71 = 1.1111111111111E+122/1.0E+72 = 1.1111111111111E+123/1.0E+73 = 1.1111111111111E+124/1.0E+74 = 1.1111111111111E+125/1.0E+75 = 1.1111111111111E+126/1.0E+76 = 1.1111111111111E+127/1.0E+77 = 1.1111111111111E+128/1.0E+78 = 1.1111111111111E+129/1.0E+79 = 1.1111111111111E+130/1.0E+80 = 1.1111111111111E+131/1.0E+81 = 1.1111111111111E+132/1.0E+82 = 1.1111111111111E+133/1.0E+83 = 1.1111111111111E+134/1.0E+84 = 1.1111111111111E+135/1.0E+85 = 1.1111111111111E+136/1.0E+86 = 1.1111111111111E+137/1.0E+87 = 1.1111111111111E+138/1.0E+88 = 1.1111111111111E+139/1.0E+89 = 1.1111111111111E+140/1.0E+90 = 1.1111111111111E+141/1.0E+91 = 1.1111111111111E+142/1.0E+92 = 1.1111111111111E+143/1.0E+93 = 1.1111111111111E+144/1.0E+94 = 1.1111111111111E+145/1.0E+95 = 1.1111111111111E+146/1.0E+96 = 1.1111111111111E+147/1.0E+97 = 1.1111111111111E+148/1.0E+98 = 1.1111111111111E+149/1.0E+99
And finally we have:
1.1111111111111113e+50 as a fraction equals 1.1111111111111E+149/1.0E+99 |
## Want to keep learning?
This content is taken from the University of Padova's online course, Advanced Precalculus: Geometry, Trigonometry and Exponentials. Join the course to learn more.
3.5
Skip to 0 minutes and 10 seconds Now, let’s go back to our plane, namely our plane equipped with our Cartesian coordinate system. And let’s consider two points in the plane. We’re now going to introduce the crucial concept of distance, which has been lacking so far. How do we define the distance between these two points? Well, the easiest way is to consider the segment PQ and imagine how we could calculate its length. The horizontal distance between P and Q, just the absolute value of the difference of the coordinates, is easily seen to be the base of a certain triangle whose vertical height is the absolute value of the difference of the y-coordinates.
Skip to 0 minutes and 53 seconds If we now invoke the classical theorem of Pythagoras, you know, hypotenuse squared equals the sum of the squares of the other two sides, we then see that the distance P, Q, the length of that hypotenuse, will be the square root of x1 minus x2 squared plus y1 minus y2 squared. That’s our definition. It’s the Euclidean distance, as it’s called, between the two points P and Q. It has a lot of applications. Let’s consider a point C in the plane. And can we imagine a point that would be a positive distance, let’s say, r from the point C? So r is a fixed number here. And I want one point that is distance r from C. OK? There it is.
Skip to 1 minute and 39 seconds But there are others, obviously. I could go up a bit and find another point of distance r. Or I could move more to the left and find a third point of distance r. In fact, if I look at all the points I can find that are distance r from the point C, the locus of those points, the set of those points, defines what is called a circle, the circle of radius r and centered at the point C. Now, consider a point P on that circle. It has coordinates. Let’s call them x, y. I also need a name for the coordinates of the center of the circle. Let’s call it x0, y0. Why nought?
Skip to 2 minutes and 23 seconds Well, actually nought, N-O-U-G-H-T is just a synonym for zero. So sometimes you would say x0, y0, same thing. What is the distance between P and C? Answer according to the formula we’ve have just determined, it’s what’s written here. And therefore, we can see that the point P, whose coordinates are x, y, lies on our circle if and only if it satisfies or its coordinates satisfy a certain equation for x and y. The equation that you see. This is called the general equation of a circle. It’s a circle centered at the point x0, y0 and of radius r greater than 0.
Skip to 3 minutes and 9 seconds Now, if you were to expand the two squares of parentheses that you see in the general equation, and of course, you’d have a term in x squared and you’d have a term in y squared. And you notice that these terms will appear. And you’ll also have linear terms– x multiplied by a certain coefficient in front. And a y multiplied by a certain number in front of that and then another number. But you will have no x y. There’ll be no term in which the product x y occurs. We’ll see them later on in another connection. So when you see an equation of this type, it might be the expanded equation of a circle. You would expect it to be so.
Skip to 3 minutes and 55 seconds Here’s an example in which we have to reverse the process of the expansion we’ve just seen and get back to the general equation. Suppose someone gives you this equation for the coordinates x, y of points in the plane. What does it describe? Well, you would anticipate a circle. You would anticipate a circle because the x squared is multiplied by 4, which is the same as what the y squared is multiplied by. And there’s no x y in the equation. How are you going to find the equation? Well, first, I divide it across by 4 just to have x squared and y squared without the coefficient 4. Next, you look at the terms involving x.
Skip to 4 minutes and 35 seconds There are two of them– the x squared and the minus 4 x. And you’re going to do the process we have called in the past completing the square. That is, you’re going to write those two terms as x minus 2 squared minus 4, which is the same thing. This doesn’t really change anything. They’re just written in a way that will be more convenient. Similarly, you look at the two terms involving y– y squared plus 2 y. You complete the square with those two terms. And they can be written as y plus 1 squared minus 1. Now, you keep the 29/4. And you simplify a little bit.
Skip to 5 minutes and 13 seconds And you’ll see that this equation, then, can be rewritten in the form x minus 2 squared, a positive term, plus y plus 1 squared, another positive term, equals something strictly negative. Well, of course, that’s crazy. You can’t have the sum of two positive things being negative. And in other words, what you have shown in this analysis, because the result here is negative on the right, is that this is not the equation of a circle. There are no points x, y that satisfy this equation. It’s what we will call the degenerate case of the equation of a circle. Didn’t really lead to a circle. Let’s look at another example, which may turn out differently.
Skip to 5 minutes and 57 seconds Again, we see that in this equation there’s no x y term. And the coefficient of x squared, 3, is the same as the coefficient of y squared 3, so we expect a circle or maybe a degenerate case. How do we analyze? The same way. We group the terms with x and complete the square. We look at the two terms involving y. We complete the square with them. And we simplify the equation. And it becomes this. And now, we recognize the general equation of a circle, mainly because at the end on the right hand side you have something that can be r squared. It’s a circle. And what is the centre of the circle? The x0, y0.
Skip to 6 minutes and 44 seconds You can read it off. It’s minus 1/3 minus 1. And what is the radius of the circle? Well, on the right, you have r squared. So the radius of the circle is root 2 over 3.
Skip to 7 minutes and 0 seconds General equation of a circle. Remark– the circle is the level curve of a certain function. The function x minus x0 squared plus y minus y0 squared minus r squared. It’s a level curve of that function. But contrary to the graph of a non-vertical line, this does not correspond to the graph of a function y equals f of x. However, if you were to just restrict attention to those y’s that are greater than the value y0, then it does look like the graph of a function. And locally, we could solve for y in terms of x as indicated.
Skip to 7 minutes and 44 seconds Graphically, it means that if you just look at the upper semicircle, then the x will go from x0 minus r to x0 plus r. And you know what the y is that corresponds for any x in that interval. The y can be solved for to get this. So a piece of our level curve, in this case the upper semicircle, is the graph of a function.
Skip to 8 minutes and 10 seconds We’ll be seeing a certain number of other famous level curves in the following segments.
# Distance and circles
In this video, Francis introduces the equation of a circle in the plane.
You can access a copy of the slides used in the video in the PDF file at the bottom of this step. |
## Tuesday, November 15, 2011
### 15/11/10 Math Homework
Homework for math: What is 22 % of 80? And to do pages 36, 37, 38, and 39 in your homework booklet.
What is 22% of 80?
There are a lot of ways you can get to your answer, but..
one way you can find the answer is to...
Step 1 : Find out what 10 % of 80 is.......... the answer is 8.
Step 2 : Do that again, 10% of 80 = 8.
Step 3: Find out what 1% of 80 is .......... it is 0.8
Step 4 : Do that again, 1 % of 80 = 0.8
Step 5 : Add the percentages together ( 10% + 10% + 1% + 1% = 22%)
The way I first found out my answer was by doing this ( I don't know if this is a correct way of doing it, but I'll just tell you how ) :
Step 1 : Find 25% of 80......You get 20
Step 2 : Find what 1% of 80........ It's 0.8
Step 3 : Again, find what 1% of 80 = 0.8
Step 4 : And again, find what 1% of 80 = 0.8
Step 5 : Add the Percentages, located in the purple "thing". Equals to 3%
Step 6 : Add the decimal numbers in the purple "thing". The answer is 2.4
Step 7 : Do 20 - 2.4 = 17.6, your answer! You're probably thinking where I got 20 from, that was when we did step one.
Step 8 : Lastly, Do 25% - 3% = 22%
Homework Booklet, Page 36, number 2a.
Shade hundred grids to represent each percent ( 3 % ).
Homework Booklet, Page 37, number 3a.
She diagram as a fraction, a decimal, and a percent.
decimal : 0.25
percent : 25 %
Homework Booklet, Page 38, number 1-4.
Match each sentence beginning in column A to an ending in column B.
1) To represent a percent great than 100%, shade more than one hundred grid.
2) To represent a fractional percent greater than 1, shade squares from a hundred grid to show the whole number and part of one square to show the fraction.
3) To represent a whole percent, shade squares on a grid of 100 squares called a hundred grid.
Homework Booklet, Page 39, number 9 a, b, c, and d.
How many hundred grids are needed to show each of the following percents? Explain your thinking.
9a) For 230%, you would need (3) hundred grids, because (2) hundred grids would be fully colored, and you would still need to color 30%, so 3.
9b) 680% would need (7) hundred grids, you always want to go one grid higher than how much hundreds you have in your number.
9d) 1420% Hmm.. Now let's think about that, you would need (15) hundred grids because, like I said in question 9b, " you always want to go one grid higher than how much hundreds you have in your number "
So, in this case, there are 14 hundreds in 1420, so you would go one grid higher, which is 15!
(When I publish this post, none of my colored fonts are showing, so sorry if it looks plain *sigh* )
Harbeck Pen
brought to you by Livescribe
1. Thanks Mary, I was a bit confused about the 22% of 80 but your example helped me. You explained all the notes clearly. Good Job, Mary. (The colored fonts thing happens to my posts to.)
2. Thanks Mary! Good job on explaining step by step on how to find the percent and the grid thingy! Haha.
3. Hi Mary. Great scribe post! The "22% of 80" is very well explained. Everything is clear. I think you should have added the different percents we did in class: " 50% of 60, 25% of 60, 10% of 60 and etc. "
Here is a link you could have added to you scribe post : http://www.mathgoodies.com/lessons/vol4/meaning_percent.html
Good Job Mary!
4. Hi mary! You did a good job on your scribe post. Everything was good but when you explained the second way to get 22% of 80, it was kind of confusing. Either than that you did great job.
Here is a video you should have added though:
5. Hi Mary! You did a great job on the scribe post! The "22% of 80" is a bit confusing. But overall, everything was clear.
Good Job Mary!
6. Great job Mary! I liked how you thoroughly explained how you got the answers for each of the questions. You explained the '22% of 80' very well(I used the same method; "25% - 3%") One thing I would suggest is that you include a link or a video or bold some key words.
So, I also found this great game/link that will give others a practice on percentages:
7. Fantastic Job Mary! I think you did an excellent job on finishing 2 posts in one day and how the results looked when you finished them. Your answers were very clear and easy to follow, it even helped me! :P. I also like how you gave pictures and bolded some words but i thought you could of bolded key words in you sentences just what like Angela said up there. I also thought you coluld of added a link or a video to help us understand the concept or converting percents and such sort.
8. Good job ! I like how you explained everything step by step. The only thing i got confused is when you were explaining 22% Of 80, but overall Everything was clear. Nice job!
9. Thanks I know what to do with the 22% of 80.
I didn't get confused. I thought the First explanation was more clear to me, when you had to just add 10% of 80.
10% of 80
1% of 80
1% of 80
10. I like how you explained everything step by step.
good job :D
11. Hi Mary! Good job on your scribe post. I liked how the questions were clearly explained, and how it was organized. I suggest that you put a video up. Here is a video link that might help. |
# Texas Go Math Grade 4 Module 7 Assessment Answer Key
Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Module 7 Assessment Answer Key.
## Texas Go Math Grade 4 Module 7 Assessment Answer Key
Vocabulary
• Distributive Property
• factor
• partial products
Choose the best term from the box to complete the sentence.
Question 1.
To find the product of a two-digit number and a 1-digit number, you can multiply the tens, multiply the ones, and find the sum of each ______________.
To find the product of a two-digit number and a 1-digit number, you can multiply the tens, multiply the ones, and find the sum of each factor.
Module 7 Test Answer Key Go Math Grade 4 Question 2.
The _____________ states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products.
The Distributive Property states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products.
Concepts and Skills
Find the product.
Question 3.
6 × 10 = ___________
6 × 10 = 60.
Explanation:
The product of 6 × 10 is 60.
Question 4.
7 × 100 = ___________
7 × 100 = 700.
Explanation:
The product of 7 × 100 is 700.
Question 5.
9 × 300 = ___________
9 × 300 = 2,700.
Explanation:
The product of 9 × 300 is 2,700.
Go Math Answer Key Grade 4 Module 7 Question 6.
8 × 7,000 = ___________
8 × 7,000 = 56,000.
Explanation:
The product of 8 × 7,000 is 56,000.
Use grid paper or base-ten blocks to model the product. Then record the product.
Question 7.
6 × 15 = ___________
6 × 15 = 90.
Explanation:
Given that to construct a block of 6 × 15 which is 90.
Question 8.
3 × 18 = ___________
3 × 18 = 54.
Explanation:
Given that to construct a block of 3 × 18 which is 54.
Question 9.
9 × 15 = ___________
9 × 15 = 135.
Explanation:
The product of 9 × 15 is 135.
Question 10.
8 × 17 = ___________
8 × 17 = 136.
Explanation:
The product of 8 × 17 is 136.
Record the product. Use expanded form to help.
Question 11.
5 × 64 = ___________
5 × 64 = 320.
Explanation:
The expanded form of 5 × 64 is
5 × 64 = 5 × (60+4)
= (5 × 60) +(5 × 4)
= 300+20
= 320.
Go Math Grade 4 Module 7 Answer Key Question 12.
3 × 272 = ___________
3 × 272 = 1,116.
Explanation:
The expanded form of 3 × 272 is
3 × 272 = 3 ×(300+70+2)
= (3 × 300)+(3 × 70)+(3 × 2)
= 900+210+6
= 1,116.
Estimate. Then record the product.
Question 13.
Estimate: __________
Estimated product is 400,
Actual product is 375.
Explanation:
The product of 75×5 is 375 and the estimated product is 80×5 which is 400.
Question 14.
Estimate: __________
Estimated product is 3,900,
Actual product is 3,882.
Explanation:
The product of 647×6 is 3,882 and the estimated product is 650×6 which is 3,900.
Question 15.
Estimate: __________
Estimated product is 17,400,
Actual product is 17,289.
Explanation:
The product of 5,763×3 is 17,289 and the estimated product is 5,800×3 which is 17,400.
Go Math Grade 4 Module 4 Assessment Answer Key Question 16.
Estimate: __________
Estimated product is 26,400,
Actual product is 26,172.
Explanation:
The product of 4,362×6 is 26,172 and the estimated product is 4,400×6 which is 26,400.
Use a strategy to find the product.
Question 17.
3 × 6 × 10 __________
3 × 6 × 10 = 180.
Explanation:
Here, Associative Property states that the product of three or more numbers remains the same regardless of how the numbers are grouped. So
3 × 6 × 10 = (3 × 6) × 10
= 18×10
= 180.
Question 18.
12 × 50 __________
12 × 50 = 600.
Explanation:
Here, Distributive property means if we multiply a sum by a number is the same as multiplying each addend by the number and adding the products. So
12 × 50 = (6+6)× 50
= (6× 50)+(6× 50)
= 300+300
= 600.
Question 19.
5 × 15 __________
5 × 15 = 75.
Explanation:
Here, Distributive property means if we multiply a sum by a number is the same as multiplying each addend by the number and adding the products. So
5 × 15 = 5 ×(10+5)
= (5 × 10)+(5 × 5)
= 50+25
= 75.
Question 20.
8 × 120 __________
8 × 120 = 960.
Explanation:
Here, Distributive property means if we multiply a sum by a number is the same as multiplying each addend by the number and adding the products. So
8 × 120 = 8 ×(100+20)
= (8 × 100)+(8 × 20)
= 800+160
= 960.
Fill in the bubble completely to show your answer.
Question 21.
Sunset School had 4 performances of their school play during the first week. The auditorium can seat 2,518 people and all of the seats at each performance were filled. During the second week, 7,450 people attended the play. How many people saw the play during the two weeks?
(A) 10,072
(B) 4,932
(C) 9,968
(D) 17,522
C.
Explanation:
Given that the auditorium can seat 2,518 people and all of the seats at each performance were filled for the first week and during the second week, 7,450 people attended the play, the number of people who saw the play during the two weeks is 2,518+7,450 which is 9,968 people.
Go Math Grade 4 Module 7 Assessment Question 22.
The table below shows the number of calories in 1 cup of different kinds of berries. How many more calories are in 4 cups of blackberries than in 4 cups of strawberries?
(A) 16
(B) 184
(C) 64
(D) 248
C.
Explanation:
Given that 1 cup of blackberries has 62 calories, for 4 cups it will be 62×4 which is 248 calories. And 1 cup of strawberries has 46 calories, so for 4 cups it will be 46×4 which is 184 calories. So the number of more calories in 4 cups of blackberries than in 4 cups of strawberries is 248-184 which is 64 calories.
Question 23.
The skating rink rents 184 pairs of skates in a month. About how many pairs of skates do the rink rent in 4 months?
Record your answer and fill in the bubbles on the grid. Be sure to use the correct place value. |
# Math Snap
## The volume of a rectangular box is given by $121500 \mathrm{~cm}^{3}$. The width, $(w)$, is $45 \mathrm{~cm}$ and the height, $(h)$, is twice the width. Calculate the length, $(l)$, of the rectangular box. Select one: a. $90 \mathrm{~cm}$ b. $120 \mathrm{~cm}$ c. $30 \mathrm{~cm}$ d. $60 \mathrm{~cm}$
#### STEP 1
Assumptions1. The volume of the rectangular box is $121500 \mathrm{~cm}^{3}$ . The width, $(w)$, of the box is $45 \mathrm{~cm}$
3. The height, $(h)$, of the box is twice the width4. The volume of a rectangular box is given by the formula $Volume = length \times width \times height$
#### STEP 2
First, we need to find the height of the box. We can do this by multiplying the width by2.
$Height =2 \times Width$
#### STEP 3
Now, plug in the given value for the width to calculate the height.
$Height =2 \times45 \mathrm{~cm}$
#### STEP 4
Calculate the height of the box.
$Height =2 \times45 \mathrm{~cm} =90 \mathrm{~cm}$
#### STEP 5
Now that we have the height, we can find the length of the box using the volume formula. We rearrange the formula to solve for length.
$Length = \frac{Volume}{Width \times Height}$
#### STEP 6
Plug in the values for the volume, width, and height to calculate the length.
$Length = \frac{121500 \mathrm{~cm}^{3}}{45 \mathrm{~cm} \times90 \mathrm{~cm}}$
##### SOLUTION
Calculate the length of the box.
$Length = \frac{121500 \mathrm{~cm}^{3}}{45 \mathrm{~cm} \times90 \mathrm{~cm}} =30 \mathrm{~cm}$The length of the rectangular box is $30 \mathrm{~cm}$. |
# CLEAR Calculus
## Lab 1: Dirac's belt trick
Carry out the following experiment:
Attach a belt to the back of a chair (or other upright object). Twist the belt through 2 full turns.
The problem now is to untwist the belt without rotating the end of the belt (the pencil) or moving the chair.
Try the same experiment with more or less than 2 full turns.
The mathematical object that captures the above behavior is called
the space of all rotations in 3-dimensions.
We now try to picture this object. A rotation in 3-dimensions is determined by two pieces of information: the axis about which to rotate and an angle through which we rotate. For example, rotate $$30^\circ$$ about the vertical axis. (We need to pick a "sense" through which to rotate. Here, we choose a "right-hand" rule.) Consider a single die oriented in 3-space as shown below:
1. In the following four pictures the die has been rotated about some axis by some angle. For each picture, draw the axis of rotation through the die and indicate the angle of rotation.
What you have done is identify exactly which rotation in "the space of all rotations in 3-dimensions" moved the die from the orientation in the original picture to each of those above.
2. Draw a die which has been moved from its original orientation as given at the beinning of Question 1 through each of the following rotations:
1. angle: $$\pi/2$$ axis: coming straight out of the face with 2 dots
2. angle: $$\pi/2$$ axis: coming straight out of the face with 3 dots
3. angle: $$\pi$$ axis: coming straight out of the edge between the 2 and 3
4. angle: $$\pi$$ axis: going straight into the the edge between the 2 and 3
Are rotations c and d really different?
3. Now we construct an actual picture of "the space of all rotations in 3-dimensions":
Consider a solid ball of radius $$\pi$$. Think of the center point as "zero rotation" or the rotation of 3-space by zero angle (through any axis).
Any other rotation will also be represented by a point in this solid ball. Specifically, a rotation through an angle $$\theta$$ about an axis $$L$$ will be represented by a point a distance $$\theta$$ out from the center along the axis $$L$$.
Locate a point in the solid ball that corresponds to each of the eight rotations from Questions 1 and 2.
Do some of these rotations correspond to more than one point in the ball?
Notice that a rotation through $$\pi$$ ($$180^\circ$$) about $$L$$ has the same outcome as a rotation through $$\pi$$ about the axis pointing in the opposite direction to $$L$$:
Therefore in our picture, we need to consider a point on the surface of the ball (at distance $$\pi$$ from the center) along an axis $$L$$ as the same rotation which is represented by the point on the opposite side of the surface (the antipodal point).
We are led to conclude that "the space of all rotations in 3-dimensions" can be pictured as a solid ball with antipodal points on the boundary identified as the same rotation. Every rotation is represented by a point on this ball, and every point represents a rotation.
With the above picture in mind, we can now explain the Dirac belt:
When we have rotated the belt through 2 turns, the belt contains a graphical illustration of a path through our picture. The path makes two "laps" through the space: beginning at the center, travelling out to point "a," back to the center, out to point "b" then back to the center again.
The sequence of moves which dissolves the rotations while keeping $$0$$ and $$4\pi$$ fixed is shown below. Here, we only show the disk where
all of the action is: |
Question #33e3f
2 Answers
Jun 14, 2017
See a solution process below:
Explanation:
$98.25$ is 98 and 25 hundredths, or:
$98 + \frac{25}{100}$
We can factor the fraction as:
$98 + \frac{25 \times 1}{25 \times 4} \implies 98 + \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{25}}} \times 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{25}}} \times 4} \implies 98 + \frac{1}{4} \implies$
$98 \frac{1}{4}$
Jun 14, 2017
$98.25 = 98 \frac{1}{4}$
Explanation:
A mixed number is comprised of a whole number and a fraction, like so:
$\text{Mixed number"="A whole number"+"A fraction}$.
Example: $1 \frac{1}{2} = 1 + \frac{1}{2}$
We know that we can turn decimals into fractions. Using the above definition of mixed numbers, we notice that we can turn a decimal into a mixed number by splitting the decimal up into its whole number and decimal parts. Then we keep the whole number as is while we turn the decimal into a fraction. We can combine the two to obtain the mixed number that is the value of the decimal. Applying these steps onto our problem, we get:
$98.25 = 98 + 0.25 = 98 + \frac{1}{4} = 98 \frac{1}{4}$
Hope this helped! |
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# Polynomials - PowerPoint PPT Presentation
Polynomials. Lesson 3.3 Factoring. Polynomials. A math equation consisting of one to many terms. Examples: 6, x , 6x, -1/2xy , 2y + x, x 2 – 5x - 9 Polynomials cannot have a variable as a denominator nor negative exponents. Are the following polynomials? 7/a ¼ xy – 10
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### Polynomials
Lesson 3.3 Factoring
• A math equation consisting of one to many terms.
• Examples:
• 6, x, 6x, -1/2xy, 2y + x, x2 – 5x - 9
• Polynomials cannot have a variable as a denominator nor negative exponents.
• Polynomials with
• one term are called monomials
• 5x3, 8, x2, etc
• two terms are called binomials
• 3x – 1, 2x2 + 8, etc
• three terms are called trinomials
• 2x2 – 4x + 9
• Variables – a letter that represents one or more numbers
• 4y = y is the variable
• Coefficient – number in front of a variable
• 4y = coefficient is 4
• The degree of a polynomial is the degree of the term with the highest exponent.
• Constant term: term without a variable.
• State the degree, coefficient’s and constant term of the polynomial.
• 5x3 + x2 – 7x + 9
• State the degree, coefficient and constant term of the polynomial.
• 6a – 4a2 - 3
• Find like terms and combine them in order to simplify polynomials.
• 4x – 2x2 + 3 – 6x2 + 5 – x
• a2b – ab2 + 4a3b – 7ab2 + 5a2b
• (3a – 4b + c) + (3b – 5c – 3a)
• (4x2 – 9x + 6) – (2x2 – 3x – 1)
• Just as natural numbers can be factored so can polynomials.
• Find the GCF in each term and then factor.
• 4m + 12
• GCF = 4
• = 4 (m + 3)
• 6n + 9 =
• 6c + 4c2 =
• 3g + 6 =
• 8d + 12d2 =
• ax2 + bx + c
• 5 – 10z – 5z2
• Find the GCF of all three terms.
• In this example the GCF is 5.
• Factor out a 5 from each and write as a product.
• 5 ( 1 – 2z – z2)
• 18a2 – 12a + 6
• 9 + 27x – 45x2
• Find all GCF’s, numbers and letters.
• -12 x3y – 20xy2 – 16x2y2
• GCF for numbers = 4
• GCF for letters = 1x and 1y
• 4xy (-3x2 – 5y – 4xy)
• 5ab a 2 + 10a2b3 – 15a2b4
• - a 20c4d - 30c3d2 – 25cd |
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# A deer is running at constant speed. It watches a lion and increases its speed by $3m/s$ .If after watching the lion it travels $400m$ in $10s$.Find its velocity before it watches the lion.
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Hint: The rate at which an object's position changes in relation to a frame of reference is called velocity, and it is a function of time. The term "velocity" refers to a description of an object's speed and direction of motion.
Complete step-by-step solution:
Speed is the rate at which an object moves along a path in terms of time, while velocity is the rate and direction of movement. In other words, while speed is a scalar value, velocity is a vector. Average velocity is determined by dividing the change in position $\left( {\Delta r} \right)$ by the change in time $\left( {\Delta t} \right)$ in its simplest form. Both terms are interchangeable because they mean the same thing, though when you're asked for a velocity, you'll usually be asked for both the speed and the direction.
Explanation:
Speed=distance/time
$\begin{array}{*{20}{l}} { = 400/10} \\ { = 40m/s} \end{array}$
$= 40m/s$
:.speed after watching lion $= 40m/s$
It increases speed 3m/s per second.
:.Total speed increased= $3 \times 10 = 30m/s$
Speed before watching lion = $40 - 30 = 10m/s$
Note:The coordinate system used to describe the position determines the sign of the velocity. A positive velocity indicates that the object is moving in the positive coordinate system direction, while a negative velocity indicates that the object is moving in the opposite direction. |
# 2.4: The Integrated Rate Law: The Dependence of Concentration on Time
Learning Objectives
• To apply rate laws to zeroth, first and second order reactions.
Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, the exponents in the rate law are the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data. The learning objective of this Module is to know how to determine the reaction order from experimental data.
## Zeroth-Order Reactions
A zeroth-order reaction is one whose rate is independent of concentration; its differential rate law is
$\text{rate} = k.$
We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \label{14.4.1}$
Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of $$−k$$. The value of $$k$$ is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of $$k$$, a positive value.
The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form
$[A] = [A]_0 − kt \label{14.4.2}$
where $$[A]_0$$ is the initial concentration of reactant $$A$$. Equation $$\ref{14.4.2}$$ has the form of the algebraic equation for a straight line,
$y = mx + b,$
with $$y = [A]$$, $$mx = −kt$$, and $$b = [A]_0$$.)
Units
In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.
Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N2O on a platinum (Pt) surface to produce N2 and O2, which occurs at temperatures ranging from 200°C to 400°C:
$\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \label{14.4.3}$
Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N2O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N2O to react with the entire Pt surface, doubling or quadrupling the N2O concentration will have no effect on the reaction rate. At very low concentrations of N2O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N2O concentration. The reaction rate is as follows:
$\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \label{14.4.4}$
Thus the rate at which N2O is consumed and the rates at which N2 and O2 are produced are independent of concentration. As shown in Figure $$\PageIndex{1}$$, the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N2O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.
A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzyme alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is
where NAD+ (nicotinamide adenine dinucleotide) and NADH (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (part (a) in Figure $$\PageIndex{3}$$). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in Figure $$\PageIndex{3}$$).
These examples illustrate two important points:
1. In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
2. A linear change in concentration with time is a clear indication of a zeroth-order reaction.
## First-Order Reactions
In a first-order reaction, the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \label{14.4.5}$
If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s−1).
The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
$[A] = [A]_0e^{−kt} \label{14.4.6}$
where [A]0 is the initial concentration of reactant A at t = 0; k is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation $$\ref{14.4.6}$$ predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation $$\ref{14.4.6}$$ and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and t:
$\ln[A] = \ln[A]_0 − kt \label{14.4.7}$
Because Equation $$\ref{14.4.7}$$ has the form of the algebraic equation for a straight line,
$y = mx + b,$
with $$y = \ln[A]$$ and $$b = \ln[A]_0$$, a plot of $$\ln[A]$$ versus $$t$$ for a first-order reaction should give a straight line with a slope of $$−k$$ and an intercept of $$\ln[A]_0$$. Either the differential rate law (Equation $$\ref{14.4.5}$$) or the integrated rate law (Equation $$\ref{14.4.7}$$) can be used to determine whether a particular reaction is first order.
First-order reactions are very common. We have already encountered two examples of first-order reactions: the hydrolysis of aspirin (Figure $$\PageIndex{4}$$) and the reaction of t-butyl bromide with water to give t-butanol (Equation $$\ref{14.10}$$). Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin.
Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:
Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure $$\PageIndex{5}$$ is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure $$\PageIndex{4}$$ have been studied extensively to find ways of maximizing the concentration of the active species.
If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.
The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Table $$\PageIndex{1}$$. The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin.
Table $$\PageIndex{1}$$: Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C
Experiment [Cisplatin]0 (M) Initial Rate (M/min)
1 0.0060 9.0 × 10−6
2 0.012 1.8 × 10−5
3 0.024 3.6 × 10−5
4 0.030 4.5 × 10−5
Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table $$\PageIndex{1}$$ shows that the reaction rate doubles [(1.8 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k[cisplatin]1. Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table $$\PageIndex{1}$$. For example, substituting the values for Experiment 3 into Equation $$\ref{14.4.5}$$,
3.6 × 10−5 M/min = k(0.024 M)
1.5 × 10−3 min−1 = k
Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.
Example $$\PageIndex{1}$$
At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:
$\ce{CH_3CH_2Cl(g) ->[\Delta] HCl(g) + C_2H_4(g)} \nonumber$
Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.
Experiment [CH3CH2Cl]0 (M) Initial Rate (M/s)
1 0.010 1.6 × 10−8
2 0.015 2.4 × 10−8
3 0.030 4.8 × 10−8
4 0.040 6.4 × 10−8
Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction
Asked for: reaction order and rate constant
Strategy:
1. Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.
2. Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.
C Use measured concentrations and rate data from any of the experiments to find the rate constant.
Solution
The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.
A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH3CH2Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH3CH2Cl].
B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k[CH3CH2Cl].
C We can calculate the rate constant (k) using any row in the table. Selecting Experiment 1 gives the following:
1.60 × 10−8 M/s = k(0.010 M)
1.6 × 10−6 s−1 = k
Exercise $$\PageIndex{1}$$
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 by the following reaction:
$SO_2Cl_2(g) → SO_2(g) + Cl_2(g) \nonumber$
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
Experiment [SO2Cl2]0 (M) Initial Rate (M/s)
1 0.0050 1.10 × 10−7
2 0.0075 1.65 × 10−7
3 0.0100 2.20 × 10−7
4 0.0125 2.75 × 10−7
first order; k = 2.2 × 10−5 s−1
We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Figure $$\PageIndex{6a}$$ shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C.
The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure $$\PageIndex{6}$$. The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure $$\PageIndex{6}$$ for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),
\begin{align*}\textrm{slope}&=\dfrac{\ln [\textrm{cisplatin}]_{1000}-\ln [\textrm{cisplatin}]_{100}}{\mathrm{1000\;min-100\;min}} \\[4pt] -k&=\dfrac{\ln 0.0022-\ln 0.0086}{\mathrm{1000\;min-100\;min}}=\dfrac{-6.12-(-4.76)}{\mathrm{900\;min}}=-1.51\times10^{-3}\;\mathrm{min^{-1}} \\[4pt] k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align*}
The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min−1 because the times plotted on the horizontal axes in parts (a) and (b) in Figure $$\PageIndex{6}$$ are in minutes rather than seconds.
The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.
Example $$\PageIndex{2}$$
If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M (k = 1.6 × 10−6 s−1) ?
Given: initial concentration, rate constant, and time interval
Asked for: concentration at specified time and time required to obtain particular concentration
Strategy:
1. Substitute values for the initial concentration ([A]0) and the calculated rate constant for the reaction (k) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time t.
2. Given a concentration [A], solve the integrated rate law for time t.
Solution
The exponential form of the integrated rate law for a first-order reaction (Equation $$\ref{14.4.6}$$) is [A] = [A]0ekt.
A Having been given the initial concentration of ethyl chloride ([A]0) and having the rate constant of k = 1.6 × 10−6 s−1, we can use the rate law to calculate the concentration of the reactant at a given time t. Substituting the known values into the integrated rate law,
\begin{align*}[\mathrm{CH_3CH_2Cl}]_{\mathrm{10\;h}}&=[\mathrm{CH_3CH_2Cl}]_0e^{-kt} \\[4pt] &=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}}) \\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*}
We could also have used the logarithmic form of the integrated rate law (Equation $$\ref{14.4.7}$$):
\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=\ln [\mathrm{CH_3CH_2Cl}]_0-kt \\[4pt] &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})] \\[4pt] &=-3.912-0.0576=-3.970 \nonumber \\[4pt] [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \nonumber \\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*}
B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for $$t$$. Equation $$\ref{14.4.7}$$ gives the following:
\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_t &=\ln[\mathrm{CH_3CH_2Cl}]_0-kt \\[4pt] kt &=\ln[\mathrm{CH_3CH_2Cl}]_0-\ln[\mathrm{CH_3CH_2Cl}]_t=\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \\[4pt] t &=\dfrac{1}{k}\left (\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \right )=\dfrac{1}{1.6\times10^{-6}\textrm{ s}^{-1}}\left(\ln \dfrac{0.0200\textrm{ M}}{0.0050\textrm{ M}}\right) \\[4pt] &=\dfrac{\ln 4.0}{1.6\times10^{-6}\textrm{ s}^{-1}}=8.7\times10^5\textrm{ s}=240\textrm{ h}=2.4\times10^2\textrm{ h} \nonumber \end{align*}
Exercise $$\PageIndex{2}$$
In the exercise in Example $$\PageIndex{1}$$, you found that the decomposition of sulfuryl chloride ($$\ce{SO2Cl2}$$) is first order, and you calculated the rate constant at 320°C.
1. Use the form(s) of the integrated rate law to find the amount of $$\ce{SO2Cl2}$$ that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C.
2. How long would it take for 90% of the SO2Cl2 to decompose?
0.0252 M
29 h
## Second-Order Reactions
The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form
$\ce{2A → products.}\nonumber$
A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).
The differential rate law for the simplest second-order reaction in which 2A → products is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8}$
Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).
For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:
$\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9}$
Because Equation $$\ref{14.4.9}$$ has the form of an algebraic equation for a straight line, y = mx + b, with y = 1/[A] and b = 1/[A]0, a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.
Second-order reactions generally have the form 2A → products or A + B → products.
Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO2 to NO and O2 and the decomposition of HI to I2 and H2. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.
Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:
Figure $$\PageIndex{7}$$
For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law (Equation $$\ref{14.4.8}$$) or the integrated rate law (Equation $$\ref{14.4.9}$$).
Table $$\PageIndex{2}$$: Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M
Time (min) [Monomer] (M) Instantaneous Rate (M/min)
10 0.0044 8.0 × 10−5
26 0.0034 5.0 × 10−5
44 0.0027 3.1 × 10−5
70 0.0020 1.8 × 10−5
120 0.0014 8.0 × 10−6
To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table $$\PageIndex{2}$$. From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:
$\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7$
Because (1.7)2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.
rate ∝ [monomer]2
This means that the reaction is second order in the monomer. Using Equation $$\ref{14.4.8}$$ and the data from any row in Table $$\PageIndex{2}$$, we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:
\begin{align}\textrm{rate}&=k[\textrm A]^2 \\8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \\4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align}
We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in Figure $$\PageIndex{8a}$$. The measurements show that the concentration of the monomer (initially 5.4 × 10−3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in Figure $$\PageIndex{8b}$$. Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M−1·min−1, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.
For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders.
Example $$\PageIndex{3}$$
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
$\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \nonumber$
Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:
Experiment [NO2]0 (M) Initial Rate (M/s)
1 0.015 1.22 × 10−4
2 0.010 5.40 × 10−5
3 0.0080 3.46 × 10−5
4 0.0050 1.35 × 10−5
Determine the reaction order and the rate constant.
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: reaction order and rate constant
Strategy:
1. From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
2. Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (k).
Solution
A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10−5) ÷ (1.35 × 10−5) = 4.0], which means that the reaction rate is proportional to [NO2]2. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO2]2. This behavior is characteristic of a second-order reaction.
B We have rate = k[NO2]2. We can calculate the rate constant (k) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:
\begin{align*}\textrm{rate}&=k[\mathrm{NO_2}]^2 \\5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2 \\0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align*}
Exercise $$\PageIndex{3}$$
When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:
$2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber$
The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:
Experiment [HO2]0 (M) Initial Rate (M/s)
1 1.1 × 10−8 1.7 × 10−7
2 2.5 × 10−8 8.8 × 10−7
3 3.4 × 10−8 1.6 × 10−6
4 5.0 × 10−8 3.5 × 10−6
Determine the reaction order and the rate constant.
second order in HO2; k = 1.4 × 109 M−1·s−1
If a plot of reactant concentration versus time is not linear, but a plot of 1/reaction concentration versus time is linear, then the reaction is second order.
Example $$\PageIndex{4}$$
If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will be the concentration of NO2 after 1.0 h? How long will it take for the concentration of NO2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction (Equation \ref{14.4.9}) and the rate constant calculated above.
Given: balanced chemical equation, rate constant, time interval, and initial concentration
Asked for: final concentration and time required to reach specified concentration
Strategy:
1. Given k, t, and [A]0, use the integrated rate law for a second-order reaction to calculate [A].
2. Setting [A] equal to 1/10 of [A]0, use the same equation to solve for $$t$$.
Solution
A We know k and [NO2]0, and we are asked to determine [NO2] at t = 1 h (3600 s). Substituting the appropriate values into Equation \ref{14.4.9},
\begin{align*}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt \\[4pt] &=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \\[4pt] &=2.0\times10^3\textrm{ M}^{-1}\end{align*}
Thus [NO2]3600 = 5.1 × 10−4 M.
B In this case, we know k and [NO2]0, and we are asked to calculate at what time [NO2] = 0.1[NO2]0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation $$\ref{14.4.9}$$ for t, using the concentrations given.
\begin{align*} t &=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k} \\[4pt] &=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}} \\[4pt] &=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \end{align*}
NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.
Exercise $$\PageIndex{4}$$
In the previous exercise, you calculated the rate constant for the decomposition of HO2 as k = 1.4 × 109 M−1·s−1. This high rate constant means that HO2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO2, calculate the concentration of HO2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO2 to decompose? Use the integrated rate law for a second-order reaction (Equation $$\ref{14.4.9}$$) and the rate constant calculated in the exercise in Example $$\PageIndex{3}$$.
2.0 × 10−13 M; 6.4 × 10−6 s
In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form $$A + B \rightarrow products$$, in which the reaction is first order in $$A$$ and first order in $$B$$. The differential rate law for this reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A][\textrm B] \label{14.4.10}$
Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant.
## Summary
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
• zeroth-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k \nonumber$ $[A] = [A]_0 − kt \nonumber$
• first-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \nonumber$ $[A] = [A]_0e^{−kt} \nonumber$ $\ln[A] = \ln[A]_0 − kt \nonumber$
• second-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^2 \nonumber$ $\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \nonumber$ |
# If the value of y varies directly with x and y = -8 when x = 20, find y if x = -4.
## Question:
If the value of y varies directly with x and y = -8 when x = 20, find y if x = -4.
## Proportions and Variation:
We can establish a relation between two numbers that are proportional by means of a ratio {eq}x {/eq} and {eq}y {/eq}. That is to say, a proportion shows the similarity between two ratios. When two variables are dependent, variations in the magnitude of one variable will have a proportional effect on the other. When there is an increase or decrease of a variable {eq}x {/eq} with respect to another {eq}y {/eq}, for a ratio or constant K, variations are present. In the case that we have a direct variation, it happens that when one variable increases the other increases, which can also be written as: {eq}\frac{{{y_1}}}{{{x_1}}} = \frac{{{y_2}}}{{{x_2}}} {/eq}.
## Answer and Explanation:
{eq}\eqalign{ & {\text{In this specific case we have two values }}x\,{\text{ and }}y\,{\text{ that have a }} \cr & {\text{variation in directly proportional form}}{\text{. So we have:}} \cr & \,\,\,\,{y_1} = - 8 \cr & \,\,\,\,{x_1} = 20 \cr & \,\,\,\,{x_2} = - 4 \cr & \,\,\,\,{y_2} = ? \cr & {\text{Since}}{\text{, }}x{\text{ and }}y{\text{ vary directly}}{\text{, then}}{\text{, when }}x{\text{ increases it also }} \cr & {\text{increases }}y{\text{. For this reason}}{\text{, it must be satisfied that:}} \cr & \,\,\,\,\frac{{{y_1}}}{{{x_1}}} = \frac{{{y_2}}}{{{x_2}}} \cr & {\text{Now}}{\text{, solving for }}\,{y_2}{\text{:}} \cr & \,\,\,\,{y_2} = \frac{{{y_1} \cdot {x_2}}}{{{x_1}}} \cr & {\text{So}}{\text{, substituting the given values:}} \cr & \,\,\,\,{y_2} = \frac{{ - 8 \cdot \left( { - 4} \right)}}{{20}} = \boxed{1.6}{\text{.}} \cr} {/eq} |
# Solving Triangles
In the 1991 film Shadows and Fog, the eerie shadow of a larger-than-life figure appears against the wall as the shady figure lurks around the corner. How tall is the ominous character really? Filmmakers use the geometry of shadows and triangles to make this special effect.
The shadow problem is a standard type of problem for teaching trigonometry and the geometry of triangles. In the standard shadow problem, several elements of a triangle will be given. The process by which the rest of the elements are found is referred to as solving a triangle.
# Basic Description
A triangle has six total elements: three sides and three angles. Sides are valued by length, and angles are valued by degree or radian measure. According to postulates for congruent triangles, given three elements, other elements can always be determined as long as at least one side length is given. Math problems that involve solving triangles, like shadow problems, typically provide certain information about just a few of the elements of a triangle, so that a variety of methods can be used to solve the triangle.
Shadow problems normally have a particular format. Some light source, often the sun, shines down at a given angle of elevation. The angle of elevation is the smallest—always acute—numerical angle measure that can be measured by swinging from the horizon from which the light source shines. Assuming that the horizon is parallel to the surface on which the light is shining, the angle of elevation is always equal to the angle of depression. The angle of depression is the angle at which the light shines down, compared to the angle of elevation which is the angle at which someone or something must look up to see the light source. Knowing the angle of elevation or depression can be helpful because trigonometry can be used to relate angle and side lengths.
In the typical shadow problem, the light shines down on an object or person of a given height. It casts a shadow on the ground below, so that the farthest tip of the shadow makes a direct line with the tallest point of the person or object and the light source. The line that directly connects the tip of the shadow and the tallest point of the object that casts the shadow can be viewed as the hypotenuse of a triangle. The length from the tip of the shadow to the point on the surface where the object stands can be viewed as the first leg, or base, of the triangle, and the height of the object can be viewed as the second leg of the triangle. In the most simple shadow problems, the triangle is a right triangle because the object stands perpendicular to the ground.
In the picture below, the sun casts a shadow on the man. The length of the shadow is the base of the triangle, the height of the man is the height of the triangle, and the length from the tip of the shadow to top of the man's head is the hypotenuse. The resulting triangle is a right triangle.
In another version of the shadow problem, the light source shines from the same surface on which the object or person stands. In this case the shadow is projected onto some wall or vertical surface, which is typically perpendicular to the first surface. In this situation, the line that connects the light source, the top of the object and the tip of the shadow on the wall is the hypotenuse. The height of the triangle is the length of the shadow on the wall, and the distance from the light source to the base of the wall can be viewed as the other leg other leg of the triangle. The picture below diagrams this type of shadow problem, and this page's main picture is an example of one of these types of shadows.
More difficult shadow problems will often involve a surface that is not level, like a hill. The person standing on the hill does not stand perpendicular to the surface of the ground, so the resulting triangle is not a right triangle. Other shadow problems may fix the light source, like a street lamp, at a given height. This scenario creates a set of two similar triangles.
Ultimately, a shadow problem asks you to solve a triangle given only a few elements of the possible six total. In the case of some shadow problems, like the one that involves two similar triangles, information about one triangle may be given and the question may ask to find elements of another.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Trigonometry, Geometry
Shadows are useful in the set-up of a triangle pro [...]
Shadows are useful in the set-up of a triangle problems because of the way light works. A shadow is cast when light cannot shine through a solid surface. Light shines in a linear fashion, that is to say it does not bend. Light waves travel forward in the same direction in which the light was shined.
In addition to the linear fashion in which light shines, light has certain angular properties. When light shines on an object that reflects light, it reflects back at the same angle at which it shined. Say a light shines onto a mirror. The angle between the beam of light and the wall that the mirror is the angle of approach. The angle from the wall at which the light reflects off of the mirror is the angle of departure. The angle of approach is equal to the angle of departure.
Light behaves the same way a cue ball does when it is bounced off of the wall of a pool table at a certain angle. Just like the way that the cue ball bounces off the wall, light reflects off of the mirror at exactly the same angle at which it shines. The beam of light has the same properties as the cue ball in this case: the angle of departure is the same as the angle of approach. This property will help with certain types of triangle problems, particularly those that involve mirrors.
Shadow problems are just one type of problem that involves solving triangles. There are numerous other formats and set ups for unsolved triangle problems. Most of these problems are formatted as word problems; they set up a triangle problem in terms of some real life scenario.
There are, however, many problems that simply provide numbers that represent angles and side lengths. In this type of problem, angles are denoted with capital letters, ${A, B, C,...}$, and the sides are denoted by lower-case letters,${a,b,c,...}$, where $a$ is the side opposite the angle $A$.
## Ways to Solve Triangles
In all cases, a triangle problem will only give a few elements of a triangle and will ask to find one or more of the lengths or angle measures that is not given. There are numerous formulas, methods, and operations that can help to solve a triangle depending on the information given in the problem.
The first step in any triangle problem is drawing a diagram. A picture can help to show which elements of the triangle are given and which elements are adjacent or opposite one another. By knowing where the elements are in relation to one another, we can use the trigonometric functions to relate angle and side lengths.
There are many techniques which can be implemented in solving triangles:
• Trigonometry: The basic trigonometric functions relate side lengths to angles. By substituting the appropriate values into the formulas for sine, cosine, or tangent, trigonometry can help to solve for a particular side length or angle measure of a right triangle. This is useful when given a side length and an angle measure.
• Inverse Trigonometry: Provided two side lengths, the inverse trig functions use the ratio of the two lengths and output an angle measure in right triangle trigonometry. Inverse trig is particularly useful in finding an angle measure when two side lengths are given in a right triangle.
• Special Right Triangles: Special right triangles are right triangles whose side lengths produce a particular ratio in trigonometry. A 30°− 60°− 90° triangle has a hypotenuse that is twice as long as one of its legs. A 45°− 45°− 90° is called an isosceles right triangle since both of its legs are the same length. These special cases can help to quicken the process of solving triangles.
• Pythagorean Theorem: The Pythagorean Theorem relates the squares of all three side lengths to one another in right triangles. This is useful when a triangle problem provides two side lengths and a third is needed.
$a^{2}+b^{2} = c^{2}$
• Pythagorean Triples: A Pythagorean triple is a set of three positive integers that satisfy the Pythagorean Theorem. The set {3,4,5} is one of the most commonly seen triples. Given a right triangle with legs of length 3 and 4, for example, the hypotenuse is known to be 5 by Pythagorean triples.
• Law of Cosines: The law of cosines is a generalization of the Pythagorean Theorem which can be used for solving non-right triangles. The law of cosines relates the squares of the side lengths to the cosine of one of the angle measures. This is particularly useful given a SAS configuration, or when three side lengths are known and no angles for non-right triangles.
$c^{2} = a^{2} + b^{2} - 2ab \cos C$
• Law of Sines: The law of sines is a formula that relates the sine of a given angle to the length of its opposite side. The law of sines is useful in any configuration when an angle measure and the length of its opposite side are given. It is also useful given an ASA configuration, and often the ASS configuration . The ASS configuration is known as the ambiguous case since it does not always provide one definite solution to the triangle.
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
When solving a triangle, one side length must always be given in the problem. Given an AAA configuration, there is no way to prove congruency. According to postulates for congruent triangles, the AAA configuration proves similarity in triangles, but there is no way to find the side lengths of a triangle. Knowing just angle measures is not helpful in solving triangles.
## Example Triangle Problems
Example 1: Using Trigonometry
A damsel in distress awaits her rescue from the tallest tower of the castle. A brave knight is on the way. He can see the castle in the distance and starts to plan his rescue, but he needs to know the height of the tower so he can plan properly. The knight sits on his horse 500 feet away from the castle. He uses his handy protractor to find the measure of the angle at which he looks up to see the princess in the tower, which is 15°. Sitting on the horse, the knight's eye level is 8 feet above the ground. What is the height of the tower?
We can use tangent to solve this problem. For a more in depth look at tangent, see Basic Trigonometric Functions.
Use the definition of tangent.
$\tan =\frac{\text{opposite}}{\text{adjacent}}$
Plug in the angle and the known side length.
$\tan 15^\circ =\frac{x \text{ft}}{500 \text{ft}}$
Clearing the fraction gives us
$\tan 15^\circ (500) =x$
Simplify for
$(.26795)(500) =x$
Round to get
$134 \text{ft} \approx x$
But this is only the height of the triangle and not the height of the tower. We need to add 8 ft to account for the height between the ground and the knight's eye-level which served as the base of the triangle.
$134 \text{ft} + 8 \text{ft} = h$
simplifying gives us
$142 \text{ft} = h$
The tower is approximately 142 feet tall.
Example 2: Using Law of Sines
A man stands 100 feet above the sea on top of a cliff. The captain of a white-sailed ship looks up at a 45° angle to see the man, and the captain of a black-sailed ship looks up at a 30° angle to see him. How far apart are the two ships?
To solve this problem, we can use the law of sines to solve for the bases of the two triangles since we have an AAS configuration with a known right angle. To find the distance between the two ships, we can take the difference in length between the bases of the two triangles.
First, we need to find the third angle for both of the triangles. Then we can use the law of sines.
For the white-sailed ship,
$180^\circ - 90^\circ - 45^\circ = 45^\circ$
Let the distance between this ship and the cliff be denoted by $a$.
By the law of sines,
$\frac{100}{\sin 45^\circ} = \frac{a}{\sin 45^\circ}$
Multiplying both sides by $\sin 45^\circ$ gives us
$(\sin 45^\circ)\frac{100}{\sin 45^\circ} = a$
Simplify for
$a = 100 \text{ft}$
For the black-sailed ship,
$180^\circ - 90^\circ - 30^\circ =60^\circ$
Let the distance between this ship and the cliff be denoted by $b$.
By the law of sines,
$\frac{100}{\sin 30^\circ} = \frac{b}{\sin 60^\circ}$
Clear the fractions to get,
$100(\sin 60^\circ) = b(\sin 30^\circ)$
Compute the sines of the angle to give us
$100\frac{\sqrt{3}}{2} = b\frac{1}{2}$
Simplify for
$100(\sqrt{3}) = b$
Multiply and round for
$b =173 \text{ft}$
The distance between the two ships, $x$, is the positive difference between the lengths of the bases of the triangle.
$b-a=x$
$173-100 = 73 \text{ft}$
The ships are about 73 feet apart from one another.
Example 3: Using Multiple Methods
At the park one afternoon, a tree casts a shadow on the lawn. A man stands at the edge of the shadow and wants to know the angle at which the sun shines down on the tree. If the tree is 51 feet tall and if he stands 68 feet away from the tree, what is the angle of elevation?
There are several ways to solve this problem. The following solution uses a combination of the methods described above.
First, we can use Pythagorean Theorem to find the length of the hypotenuse of the triangle, from the tip of the shadow to the top of the tree.
$a^{2}+b^{2} = c^{2}$
Substitute the length of the legs of the triangle for $a, b$
$51^{2}+68^{2} = c^{2}$
Simplifying gives us
$2601+4624 = c^{2}$
$7225 = c^{2}$
Take the square root of both sides for
$\sqrt{7225} = c$
$85 = c$
Next, we can use the law of cosines to find the measure of the angle of elevation.
$a^{2}=b^{2}+c^{2} - 2bc \cos A$
Plugging in the appropriate values gives us
$51^{2}=68^{2}+85^{2} - 2(85)(68) \cos A$
Computing the squares gives us
$2601= 4624+7225 - 11560 \cos A$
Simplify for
$2601= 11849 - 11560 \cos A$
Subtract $11849$ from both sides for
$-9248= -11560 \cos A$
Simplify to get
$.8 = \cos A$
Use inverse trigonometry to find the angle of elevation.
$A = 37^\circ$
# Why It's Interesting
Shadow Problems are one of the most common types of problems used in teaching trigonometry. A shadow problem sets up a scenario that is simple, visual, and easy to remember. Shadow problems are commonly used and highly applicable.
Shadows, while an effective paradigm in a word problem, can even be useful in real life applications. In this section, we can use real life examples of using shadows and triangles to calculate heights and distances.
## Example: Sizing Up Swarthmore
The Clothier Bell Tower is the tallest building on Swarthmore College's campus, yet few people know exactly how tall the tower stands. We can use shadows to determine the height of the tower. Here's how:
Step 1) Mark the shadow of the of the tower. Make sure to mark the time of day. The sun is at different heights throughout the day. The shadows are longest earlier in the morning and later in the afternoon. At around midday, the shadows aren't very long, so it might be harder to find a good shadow. When we marked the shadow of the bell tower, it was around 3:40 pm in mid-June.
Step 2) After marking the shadow, we can measure the distance from our mark to the bottom of the tower. This length will serve as the base of our triangle. In this case, the length of the shadow was 111 feet.
Step 3) Measure the angle of the sun at that time of day.
Use a yardstick to make a smaller, more manageable triangle. Because the sun shines down at the same angle as it does on the bell tower, the small triangle and the bell tower's triangle are similar and therefore have the same trigonometric ratios.
• Stand the yardstick so it's perpendicular to the ground so that it forms a right angle. The sun will cast a shadow. Mark the end of the shadow with a piece of chalk.
• Measure the length of the shadow. This will be considered the length of the base of the triangle.
Draw a diagram of the triangle made by connecting the top of the yardstick to the marked tip of the shadow. Use inverse trigonometry to determine the angle of elevation. $\tan X = \frac{36 \text{in}}{27 \text{in}}$ $\arctan \frac{36}{27} = X$ $\arctan \frac{4}{3} = X$ $X = 53^\circ$ Step 4) Now, we can use trigonometry to solve the triangle for the height of the bell tower. $\tan 53^\circ = \frac{h}{111 \text{ft}}$ Clearing the fractions, $111 (\tan 53^\circ) = h$ Plugging in the value of $\tan 53^\circ$ gives us $111 \frac{4}{3} = h$ Simplify for $148 \text{ft} = h$ According to our calculations, the height of the Clothier Bell Tower is 148 feet.
## History: Eratosthenes and the Earth
In ancient Greece, mathematician Eratosthenes made a name for himself in the history books by calculating the circumference of the Earth by using shadows. Many other mathematicians had attempted the problem before, but Eratosthenes was the first one to actually have any success. His rate of error was less than 2%.
Eratosthenes used shadows to calculate the distance around the Earth. As an astronomer, he determined the time of the summer solstice when the sun would be directly over the town of Syene in Egypt (now Aswan). On this day, with the sun directly above, there were no shadows, but in Alexandria, which is about 500 miles north of Syene, Eratosthenes saw shadows. He calculated based on the length of the shadow that the angle at which the sun hit the Earth was 7 °. He used this calculation, along with his knowledge of geometry, to determine the circumference of the Earth.
# References
All of the images on this page, unless otherwise stated on their own image page, were made or photographed by the author Richard Scott, Swarthmore College.
The information on Eratosthenes can be cited to http://www.math.twsu.edu/history/men/eratosthenes.html.
The main image and details about it were found at http://www.imdb.com/title/tt0105378/.
Some of the ideas for problems/pictures on this page are based from ideas or concepts in the Interactive Mathematics Program Textbooks by Fendel, Resek, Alper and Fraser. |
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1. Electronic Devices If I see an electronic device other than a calculator (including a phone being used as a calculator) I will pick it up and your parents can come an get it.
2. Sequences and Series 4.7 & 8 Standard: MM2A3d Students will explore arithmetic sequences and various ways of computing their sums. Standard: MM2A3e Students will explore sequences of Partial sums of arithmetic series as examples of quadratic functions.
3. Arithmetic Sequence • An arithmetic sequence is nothing more than alinear function with the specific domain of the natural numbers. The outputs of the function create the terms of the sequence. • The difference between any two terms of an arithmetic sequence is a constant, and is called the “common difference”
4. Practice • Page 140, # 1, 3, 5
5. Arithmetic Sequence • Look at the graph of the sequence: 2, 4, 6, 8, 10
6. Arithmetic Sequence • Let’s take the point-slope linear form (y – y1) = m(x –x1) Solving for y , and calling it f(x) gives: f(x) = m(x –x1) + y1 • The terms of a sequence are the outputs of some function , so f(x) = an an = m(x –x1) + y1
7. an = m(x –x1) + y1 • The domain of a sequence is usually the natural numbers. Let's use n for them. So, x = n in our formula. an = m(n –x1) + y1
8. an = m(n –x1) + y1 • The value m is the slope in a linear function. In the sequence world as we go from term to term, we find that the change in input is always 1 while the change in output never changes. It is common to all consecutive pairs of terms. In the sequence world the slope is exactly the same as the common difference, d. Then m = d. an = d(n –x1) + y1
9. an = d(n –x1) + y1 • The first term is always labeled a1. It is the ordered pair (1, a1). We'll use it for the (x1, y1) point in the point-slope form. • Putting them all together we have a rule for creating nth term formula: an = d(n – 1) + a1
10. Rule for nth term formula: an = a1 + d(n – 1) Where: an is value of the nth term d is the “common difference” n is the number of terms a1 is the first term NOTE: Be sure to simplify NOTE: Look at this on a graph
11. Practice – page 140 an = a1 + d(n – 1) • # 7 • # 9 an = 6n – 10; 50 an = 1/2 - 1/4n; 2
12. Problem 11 – 15 is like finding the linear equation given two points an = a1 + d(n – 1) • Find the common difference – d (slope) • Substitute a point and solve for a1 • Plug common difference and a1 into the general equation and simplify • # 11 • # 13 • # 15 an = 14n – 40 an = -5 - n an = n/4 + 2
13. Homework • Page 140, # 2 – 16 even
14. Finding the Sum of an Arithmetic Sequence • The expression formed by adding the terms of an arithmetic sequence is called an arithmetic series. • The sum of the first n terms of an arithmetic series is: (Determine the equation via a spreadsheet):
15. Practice – page 140 • # 17 • # 19 • # 21 • # 23 100 -210 an = n - 2 an = -n/2 + 5
16. Homework • Page 140, # 2 – 24 even |
• # How to reduce the fraction?
Oksana Logunova
March 28, 2013
Fraction reduction is used not only for regular ordinary fractions, but also for those fractions that are represented by the particular two polynomials, including the variable. A regular fraction is called the correct one, in which the number in the denominator is greater than the number in the numerator. Fractions are reduced for easier handling with lower numbers. From the materials of this article you will learn how to reduce the fraction of an ordinary fraction.
## Fractions reduction
• First, the estimation method is used. You need to look at what factors you can decompose the numerical value of the numerator and denominator. If they have a common factor, then divide them into it. For example, you want to reduce the fraction: 30/60. Looks at what factors the number 30 is expanded (these are numbers 5 and 6). Analyze the number 60, it can be decomposed into 5, 6 and 12. Take the common factor 5. Fraction divided by it, it turns out 6/12. The fraction is reduced again, by 6. It turns out ½.
• Another way to reduce the fraction. It is necessary to find a common factor for the numerator and denominator. To do this, each of these numbers is decomposed into prime factors. For example, the fraction 25/125.The numerator is decomposed like this: 25: 5: 5 = 1. The denominator is 125: 5: 5: 5 = 1.
• We are looking for the greatest common factor. For this purpose, all the factors are written out, which are repeated both in the first and in the second number once. It will be 5; 5. They multiply together. 5x5 = 25. This is the greatest common divisor by which the fraction will be reduced. As a result, we get 1/5.
• Finding the common factors of certain numbers will be faster if you know the signs of divisibility.
Now you know the rules for reducing fractions. This will help you in life as a general development or to explain this topic to your child at school. We wish you good luck!
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# How Do You Solve Time Problems?
## What is the 4 step method?
The “Four-Step Problem Solving” plan helps elementary math students to employ sound reasoning and to develop mathematical language while they complete a four-step problem-solving process.
This problem-solving plan consists of four steps: details, main idea, strategy, and how..
## What are the problems in life?
Life issues are common problems, issues and/or crises that happen to normal people living normal lives. Examples include managing one’s relationships so that they are healthy and functional, surviving disabilities, coping with grief, loss and self-esteem issues.
## What is the formula for elapsed time?
To calculate elapsed days is so easy, you just need to apply this formula = B2-A2, A2 is the start date, B2 is the end date. Tip: To calculate elapsed month, you can use this formula =DATEDIF(A2,B2,”m”), A2 is the start date, B2 is the end date.
## What is formula of mass?
The mass of an object can be calculated in a number of different ways: mass=density×volume (m=ρV). Density is a measure of mass per unit of volume, so the mass of an object can be determined by multiplying density by volume. mass=force÷acceleration (m=F/a).
## What’s the formula for distance?
Distance formula. Learn how to find the distance between two points by using the distance formula, which is an application of the Pythagorean theorem. We can rewrite the Pythagorean theorem as d=√((x_2-x_1)²+(y_2-y_1)²) to find the distance between any two points.
## What is a rate example?
A rate is a special ratio in which the two terms are in different units. For example, if a 12-ounce can of corn costs 69¢, the rate is 69¢ for 12 ounces. … When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates.
## What are the 7 steps to problem solving?
Here are seven-steps for an effective problem-solving process.Identify the issues. Be clear about what the problem is. … Understand everyone’s interests. … List the possible solutions (options) … Evaluate the options. … Select an option or options. … Document the agreement(s). … Agree on contingencies, monitoring, and evaluation.
## How do you manage problems?
5 Ways to Solve All Your ProblemsSolve the problem. Sometimes it’s as easy as that. … Avoid the problem. There just may be some things on that to-do list that will go away if you wait long enough. … Cut the problem down to size. Sometimes the best way to manage a problem is to figure out a way to do it in stages. … Address an underlying issue.
## What are some problems that need to be solved?
Top Problems that can be SolvedArmed Conflict.Chronic Disease.Education.Infectious Disease.Population Growth.Biodiversity.Climate Change.Hunger and Malnutrition.More items…•
## What problem do you solve for your customers?
What Problems Do You Solve for Your Customers?Finding new customers.Keeping existing customers.Selling more to existing customers.Improving customer service.Reducing personnel costs.Reducing customer complaints.Decreasing time to market.Improving market share (or mind share)More items…•
## What are the steps in solving real life problems?
The 5 Steps of Problem SolvingA “Real World” Math Drama. … Step #1: Stop and Think Before Doing Anything. … Step #2: English-to-Equation Translation. … Step #3: Solve for Whatever You’re Interested In. … Step #4: Make Sure You Understand the Result. … Step #5: Use Your Result to Solve Other Problems. … Wrap Up.
## How do you calculate problems?
If so, maybe these six quick tips can help you to solve it a little bit easier.First, ask yourself: is there really a problem here? … Accept it. … Ask for help. … Use 80 percent of your time to find solutions. … Break the problem down into smaller pieces. … Find the opportunity and/or lesson within the problem.
## What is the formula of calculating time?
To solve for time use the formula for time, t = d/s which means time equals distance divided by speed.
## What are the four steps for solving an equation?
Different ways to solve equations. We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal. If we subtract the same number from both sides of an equation, both sides will remain equal. |
Categories
How To Convert Months, Days And Seconds
To learn the skill of conversion of months, days and seconds is a beneficial practice. Information about history will be clearer to you when the time is converted into months or years.
You can easily calculate your monthly saving that at the last of the year how much you would be able to save. You can also calculate that if you sell the product of 100\$ daily then how much will you earn in one month.
Converting days months and seconds requires a bit skill of multiplication and division. If you are good at that then you can effortlessly transmute it. In this article, we will discuss how to convert months, days and seconds easily.
Number Of Days In Seconds
First of all, we need to know that how many seconds are there in a day so we could know the logic behind the conversion.
We know that
1 day = 24 hrs
1 hr = 60 min
1 min = 60 sec
So
1 day = 24 hrs x 60 minute x 60 seconds
1 day = 86400 seconds
With the help of this rule, we can easily convert months into seconds. Let’s discuss an example for better understanding of this concept.
Suppose we want to convert 8 days into seconds then we will simply multiply it with 86400.
8 days = 8 x 86400
8 days = 691200 seconds
Number Of Seconds In Days
Now will discuss the conversion of seconds in days
86400 seconds = 1 day
1 second = 1/86400 days
1 second = 1.157 x 10-5 days
If you are asked to convert 18400 seconds in days then either you can divide it by 86400 or simply multiply it by 1.157 x 10-5
18400 seconds = 18400/86400 days
18400 = 2 days
It is approximately equal to 2 days.
Number Of Months In Seconds
First, we need to know the logic behind this conversion
1 month = 30 days
1 day = 86400 seconds
1 month = 86400 x 30
1 month = 2.628 x 106 seconds
Using this we can easily convert months into seconds. If you are asked to convert 8 months into seconds then simply multiply it with 2.628 x 106
8 months = 8 x 2.628 x 106
8 months = 2. 102 x 107 seconds
Number Of Seconds In Months
To convert the seconds in months
2.628 x 106 seconds = 1 month
1 second = 1/2.628 x 106 month
1 second = 3.8052 x 10-7 month
Suppose we want to convert 8484844 seconds in months then we will multiply it with 3.8052 x 10-7
8484844 seconds = 8484844 x 3.8052 x 10-7
8484844 seconds = 3 months
It is approximately equal to 3 months.
Number Of Months In Days
Usually, we take 30 days to a month but some months also have 31 days. To convert days in month first we will take average numbers of days in a month as 30.5 days.
30.5 days = 1 month
So to calculate the number of days we will multiply the numbers of months by 30.5, for example, to calculate 4 months in days
4 months = 30.5 x 4 days
4 months = 122 days
4 months are approximately equal to 122 days.
Number Of Days In Months
Just like other reverse methods, to convert days in month
30.5 days = 1 month
1 day = 1/30.5 month
1 day = 0.0328 month
To convert 48 days in month we will multiply with 0.0328 or divide it by 30.5
48 days = 0.0328 x 48
48 days = 4.6 months
Remembering above rules we can convert time in different units so keep practicing. I hope you liked my post about “How to convert months, days and seconds” if you have any query then write it down in the comments section.
By VEQUILL
By profession, I'm a software engineer. Everyone has one strong driving force in self that let one evolve above boundaries, my passion is content creation. Following my ambition, I am founder and CEO at TapeDaily with aim of providing high-quality content and the ultimate goal of reader satisfaction. Consistent improvement has always been my priority. The spark with time ignites more and more and recognized me as one of the leading SEO experts in UAE. I've successfully delivered vast improvements in search engine rankings across a variety of clients and sectors, including property and real estate.
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# How do you find the zeros, real and imaginary, of y=3x^2+31x+9 using the quadratic formula?
Jan 6, 2016
Identify an substitute the values of $a$, $b$ and $c$ into the quadratic formula to find:
$x = \frac{- 31 \pm \sqrt{853}}{6}$
#### Explanation:
$y = 3 {x}^{2} + 31 x + 9$ is in the form $a {x}^{2} + b x + c$ with $a = 3$, $b = 31$ and $c = 9$.
This has zeros given by the quadratic formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$= \frac{- 31 \pm \sqrt{{31}^{2} - \left(4 \cdot 3 \cdot 9\right)}}{2 \cdot 3}$
$= \frac{- 31 \pm \sqrt{961 - 108}}{6}$
$= \frac{- 31 \pm \sqrt{853}}{6}$ |
## Actions to Develop Fractions Understanding
The seven mathematics processes – problem solving, reasoning and proving, reflecting, selecting tools and computational strategies, connecting, representing, and communicating – are integral to meaningful learning of fractions.
In this section, we will focus on learning and teaching fractions through the following three processes:
• representing
• reasoning and proving
• selecting tools and computational strategies
### Representing
Students of all ages need to represent fraction ideas and relationships by using concrete materials, pictures,diagrams, words and symbols. Often, symbols are privileged at the junior and intermediate/senior grades (students may use mathematical notation without knowing what it means). However, it is essential that all students develop the ability to represent mathematics in a variety of ways, as this skill allows them to make sense of the initial task, predict a reasonable answer, determine one or more possible strategies, check to see that their answer makes sense and communicate their thinking. Sometimes, models can be rough sketches that are approximate; at other times, they need to be precise. Students who have significant experience with constructing diagrams or models will be better able to discern when an approximate representation is sufficient and when an precise one is necessary. Constructing models will initially involve composing and decomposing fractions by using unit fractions. As students learn about equivalence, comparing and operations, it is essential that they can construct appropriate diagrams or models. This ability supports students in senior mathematics as they construct models or diagrams to solve for an unknown value (e.g., in a trigonometry question).
Into the Classroom: Consider asking students to share a pan of brownies equally among four people. When students solve this task by using concrete materials, such as paper folded into four equal regions, they are able to connect the physical solution with a symbolic notation indicating that each person would receive one one-fourth of the pan. By asking students how they might share this same pan of brownies with an additional four people, a number of different paper-folding strategies will be used, which generate different-looking yet equivalent solutions. Students justify that the solutions are equivalent by comparing the sizes of the regions. In being able to do so, students understand much more than how to generate a correct answer: they are able to see the difference between fourths and eighths, able to connect the symbolic notation to the concrete representation and able to consider how to generalize this strategy to other situations. Students quickly notice, for example, that other friendly numbers for sharing are 16 and 32 but that 10 and 12 would be more difficult. By using the concrete representation and the symbolic together, students can see that one-eighth is half of one-fourth and experience fractions across constructs in a meaningful and interconnected context.
### Reasoning and Proving
In the tasks described below, students are constantly engaged in reasoning about and proving both the strategies being used and the solutions being generated. Students who are given opportunities to make conjectures about fractions and explore the conjectures to refine or refute them will have a more solid understanding of fractions. Such opportunities frequently arise from an unexpected strategy or solution in a lesson.
In the Classroom: In one classroom, Grade 4 students were asked to identify examples and non-examples of fractions. Although the task seemed straightforward initially, it resulted in a number of questions being generated by the students, such as ”Can any number be a fraction?” and “Can we put fractions in a number sentence?” which the students explored further in a subsequent class. In a Grade 6 classroom, where students were using pattern blocks to create part-whole and part-part fractions by using set and area models, one student suggested that the fraction 16 could be used to represent one vertex on the hexagonal pattern block but wondered if that was an example of a set model. The teacher, rather than answering the question, asked students to discuss this in small groups and then engaged in a whole class discussion to share ideas. In this way, students were required to reason about the new question based on their understanding of set and area models, allowing them to extend their knowledge to more general situations.
### Selecting Tools and Computational Strategies
The learning of fractions is rife with opportunities to build students’ ability to thoughtfully select tools and computational strategies. A range of tasks allow for students to make, discuss and reflect on decisions for tool and calculation strategies. In some intermediate classrooms, students are hesitant to use a manipulative to solve a mathematical problem. Sometimes, a model can provide a simpler solution. Also, engaging students in selecting manipulatives and tools to solve a task requires them to consider all aspects of the question, including what the whole is and what the fractional units are.
In the Classroom: For example, if students are selecting a tool to represent a fraction task that includes fractional units of fourths and fifths, a number line or a rectangle area model will be much friendlier than a hand-drawn circular model (it is difficult to accurately partition circles into fifths).
A student who has experience with selecting from various tools will be able to quickly dismiss those that are less appropriate for given fractional units. Secondary students are frequently required to complete calculations involving fractions. Sometimes an estimate is sufficient, so a student who has a strong sense of fractions could correctly estimate the sum of 115 + 1617 to be approximately equal to 1. This knowledge could aid in checking an algebraic solution, such as 115 x + 1617 x as being close to 1x.
Secondary students benefit from using a range of manipulatives, including algebra tiles to represent the algebraic action of “completing the square,” which allows students to extend their understanding to expressions involving fractions. It is helpful for students to recognize that algebraically completing the square correlates to completing the physical model in which the tiles are arranged to create a square.
For example, to complete the square for x² + 6x, students might
• construct x² + 6x by using algebraic tiles
What would fit here to make this square complete?
• identify that the square is missing 9 ones tiles and add them in
Add in 9 negative ones tiles
• recognize that they need to keep the new image equivalent to the original expression, so add in 9 negative ones tiles
• write the expression by using the dimensions of the square (x + 3)² – 9
Note that for fractional situations, such as completing the square for x² + 5x, the diagram would be as follows (since the 5x has to be split to create a square):
And the expression would be (x + 2 12 )² – 6 14. |
# 6.11: Percent Equation to Find Part a
Difficulty Level: At Grade Created by: CK-12
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Do you like jelly beans? Take a look at this dilemma.
Taylor’s younger brother Max decided to visit her at the candy store. Max is only seven and can be a handful sometimes, so while Taylor loves to see him, she was a little hesitant to have him in the shop. Plus, what seven year old doesn’t love candy.
Taylor gave Max a small bag to put some candy in. She figured he would take a few pieces, but ended up with a whole bunch of candy.
“How many did you take?” Taylor asked him looking in the bag.
“I took 40 pieces,” Max said grinning. “I won’t eat it all now. I will save some for later.”
Taylor looked into the bag. There were candy canes, peanut butter cups and a whole bunch of jelly beans.
She gave Max the bag and watched him walk away chewing.
“I hope I don’t get into trouble for this,” Taylor murmured to herself.
In looking in the bag, Taylor discovered that 15% of Max's bag was peanut butter cups. If he put 40 pieces of candy in his bag, how many pieces of candy were peanut butter cups?
We can say this another way using a percent equation.
What is 15% of 40?
This Concept will teach you how to use the percent equation to find part a. Then we will come back to this dilemma at the end of the Concept.
### Guidance
Think about the proportion that you just learned to find the percent of a number.
ab=p100\begin{align*}\frac{a}{b}=\frac{p}{100}\end{align*}
When we used this proportion in problem solving, we multiplied b\begin{align*}b\end{align*} and p\begin{align*}p\end{align*} and a\begin{align*}a\end{align*} and 100. Then we divided the product of b\begin{align*}b\end{align*} and p\begin{align*}p\end{align*} by 100.
Let's look at a statement that uses this proportion.
What is 35% of 6?
First, we fill in the proportion.
a6=35100\begin{align*}\frac{a}{6}=\frac{35}{100}\end{align*}
Next, we multiply and solve for a\begin{align*}a\end{align*}.
100a100aa=35(6)=210=2.1\begin{align*}100a & = 35(6)\\ 100a & = 210\\ a & = 2.1\end{align*}
Notice that by dividing by 100, we moved the decimal place two places.
Hmmmm. This is the same two places that the percent is represented by. This means that if we changed the percent to a decimal FIRST, that we could skip a step and use an equation to find the missing value.
Take a look at the same problem.
What is 35% of 6?
First, change 35% to a decimal.
35%=.35\begin{align*}35\% = .35\end{align*}
Now we multiply it times 6, the base and find a\begin{align*}a\end{align*} the amount. Look at this equation.
aaa=p%(b)=.35(6)=2.1\begin{align*}a & = p\%(b)\\ a & = .35(6) \\ a & = 2.1 \end{align*}
Notice that we got the same answer as when we used the proportion. It just simplifies the process.
Let’s look at another one.
What is 25% of 50?
First, change 25% to a decimal.
25%=.25\begin{align*}25\% = .25\end{align*}
Now use the equation.
aaa=p%(b)=.25(50)=12.5\begin{align*}a & = p\%(b)\\ a & =.25(50)\\ a & =12.5 \end{align*}
Use the equation to find each amount. Include decimals in your answer.
#### Example A
What is 20% of 16?
Solution: 3.2\begin{align*}3.2\end{align*}
#### Example B
What is 5% of 40?
Solution:2\begin{align*}2\end{align*}
#### Example C
What is 15% of 65?
Solution:9.75\begin{align*}9.75\end{align*}
Here is the original problem once again.
Taylor’s younger brother Max decided to visit her at the candy store. Max is only seven and can be a handful sometimes, so while Taylor loves to see him, she was a little hesitant to have him in the shop. Plus, what seven year old doesn’t love candy.
Taylor gave Max a small bag to put some candy in. She figured he would take a few pieces, but ended up with a whole bunch of candy.
“How many did you take?” Taylor asked him looking in the bag.
“I took 40 pieces,” Max said grinning. “I won’t eat it all now. I will save some for later.”
Taylor looked into the bag. There were candy canes, peanut butter cups and a whole bunch of jelly beans.
She gave Max the bag and watched him walk away chewing.
“I hope I don’t get into trouble for this,” Taylor murmured to herself.
In looking in the bag, Taylor discovered that 15% of Max's bag was peanut butter cups. If he put 40 pieces of candy in his bag, how many pieces of candy were peanut butter cups?
We can say this another way using a percent equation.
What is 15% of 40?
To figure this out, we can convert 15% into a decimal.
Missing \end{align*}\begin{align*}15% = .15\end{align*}
Next, we multiply .15 by 40.
40×.15=6\begin{align*}40 \times .15 = 6\end{align*}
Six pieces of candy in the bag were peanut butter cups.
### Vocabulary
Here are the vocabulary words in this Concept.
Inverse Operation
the opposite operation.
Percent
a part of a whole calculated out of 100.
Amount
the part of the whole that “is” out of a base. “Is” is a key word showing amount.
Base
the part of the whole that the amount is out of. The word “Of what number” let you know that you are looking for the base.
### Guided Practice
Here is one for you to try on your own.
What is 32% of 200?
To figure this out, first we change the percent to a decimal.
Missing \end{align*}\begin{align*}32% = .32\end{align*}
Next, we multiply this decimal by 200.
200×.32=64\begin{align*}200 \times .32 = 64\end{align*}
32% of 200 is 64.
### Video Review
Here is a video for review.
### Practice
Directions: Use the percent equation to find each amount.
1. What is 20% of 18?
2. What is 10% of 30?
3. What is 5% of 90?
4. What is 12% of 27?
5. What is 18% of 30?
6. What is 50% of 88?
7. What is 75% of 12?
8. What is 75% of 90?
9. What is 22% of 40?
10. What is 25% of 60?
11. What is 8% of 15?
12. What is 99% of 200?
13. What is 90% of 12?
14. What is 18.5% of 230?
15. What is 20.5% of 160?
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# Show that: $\tan 3x - \tan 2x - \tan x = \tan 3x\tan 2x\tan x$
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Hint: - Here we go through by letting $\tan 3x$as $\tan (2x + x)$ . Because in question it is in terms of 2x and x. By applying this we can easily prove our question.
Let us take, $\tan 3x = \tan (2x + x)$
Now we can apply the formula of $\tan ({\text{A + B)}}$ i.e. $\tan ({\text{A + B) = }}\frac{{\tan {\text{A + }}\tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}}$
Now we can write, $\tan 3x = \tan (2x + x) = \frac{{\tan 2x + \tan x}}{{1 - \tan 2x\tan x}}$
Now we cross multiply it to get,
$\Rightarrow \tan 3x(1 - \tan 2x\tan x) = \tan 2x + \tan x \\ \Rightarrow \tan 3x - \tan 3x\tan 2x\tan x = \tan 2x + \tan x \\$
Now by rearranging the above equation we get
$\tan 3x - \tan 2x - \tan x = \tan 3x\tan 2x\tan x$ Hence, proved.
Note: - Whenever we face such a type of question the key concept for solving the question is that we
always try to make the bigger angle in sum of two smaller angles that are given in a question to apply
the formula to prove the question. |
## Tuesday, September 17, 2013
### Interactive Math Notebook: Factors and Multiples
Welcome to our Math Corner! Please come back often and leave a comment below to let me know your thoughts or any other ideas you have to add! Also, I would love a follow! :) Thanks!
We have been busy working on determining whether one number is a factor or multiple of another number. We started by building arrays both with tiles and on centimeter grid paper to determine a number's factors. Becoming proficient at building arrays has really helped some of my struggling students feel successful in finding factors even when they aren't proficient in their facts or have many strong strategies to use to determine a fact. After practicing many many times, I gave each student a pair of die (some were six sided, some were more for my kids who are proficient in their facts and wanted a challenge!) and grid paper and instructed them to roll the die and create an array with dimensions that match the numbers on the die. We then added this page to our notebooks. Below is an example.
Next, we used our Math Handbooks and its Table of Contents (oh hello, Language Arts skills! See....I'm learning how to incorporate it all!) to find the definition and examples of factor, prime number, composite number, and square number. We used this information to create a foldable to glue into the right hand side of the page. An example is below!
We are also learning about multiples. After many days of discovering, discussing, and applying this knew knowledge, we finally were able to put the information into our notebooks. I always try to wait until I feel they understand it to put it into the notebooks. We used a hundreds chart to choose a factor, then highlighted all of its multiples. We also used this factor to create a basic real world problem to show how to apply it to multiples.
This student chose the factor two, and wrote the problem: A store has CDs for \$2 each. She then drew a picture to show that one CD would cost \$2, two CDs would cost \$4, three would cost \$6, and so on. This shows that the price of buying CDs are multiples of 2
This student chose the factor four. His problem is about video games costing \$4 each, so two would cost \$8, three would be \$12, and so on. He goes on to begin to write a question associated with it!
Lastly, I really wanted to make sure we understood the difference between factors and multiples, as it can get very confusing. We used markers to circle all the factors in a list and all the multiples associated with it. Students were allowed to pick their own factor, or for those who are still not feeling comfortable with multiplication, were permitted to use the factor four as I did in my example. Then, they showed an example of an array that shows one of their listed factors and multiples, as well as a non array (just to make sure they understood that an array is a rectangle and cannot have any pieces sticking off the end!!). These last two pages might be my favorite!
This student wrote all the problems for the number four. He circled all of the fours to show that is the factor. He wrote: 4 is a factor of any whole number that it divides evenly. He also circled all the multiples to show that you can multiply any number by four to get a multiple of four.
This student did an excellent job of showing the difference between an array that four is a factor of one of its multiples, as well as a non example saying "21 is not a multiple of 4" and using his array as proof!
## Tuesday, September 10, 2013
### Arrays: The Gateway to Multiplication
Our first unit is all about factors, multiples, and arrays. I know from teaching 5th grade that if students don't understand what multiplication truly shows, they are missing the foundation to a deeper understanding of numbers, as well as the key to success with many many other parts of math. That is why I plan on really spending time this year developing a deep understanding of multiplication, how numbers are related, and strategies students can use to help them if they get stuck.
We first began, as always, using manipulatives. We have these wonderful one inch tiles that are perfect for creating arrays, but connecting cubes work as well. We discussed what the word "dimensions" mean when looking at the lengths of each side. We started by practicing arrays with smaller numbers and built our way up. Once I felt they understood, I put them in pairs and gave them each two numbers to find all the arrays for. An example of our work is shown below! (Please ignore the fact that the array she has showing shows 4x5 and is not an array for 18! She was actually building for their next number: 39)
This activity led into a discussion about why certain numbers had only one array, like 17. It was a "prime" time to talk about prime numbers! We displayed the posters throughout the room and refer to them often, especially when discussing multiples, factors, and special numbers!
## Monday, August 19, 2013
### New Beginnings
Is it really that time again?! SO SOON? August has truly flown by, and all of a sudden we are starting school. Except this year is going to be different. You see, last year, and the three years prior to that, I drove over an hour each day to my school (68 miles!!). While I loved my job, coworkers, students, and families, I needed to make a change. I couldn't continue using up two hours of my day (and all my extra cash for gas!) so I accepted a job closer to home. With this job is a different grade level (a 3-4 split....yowza) AND I get to teach all subjects (not just math as before)! I haven't figured out yet how this will affect my blog, but I'm sure that will come in due time. I look forward to doing math notebooks again, this year catering to 4th grade, while making many adjustments and improving it. I am also learning a new math curriculum, Investigations. Hopefully there will be lots of new and exciting ways to teach MATH! And perhaps, if I feel ambitious, I will broaden the blog to include all the wonderful things we will be doing in Language Arts. We shall see..... :)
## Friday, June 7, 2013
### Wrapping up Notebooks: A time for Reflection
Last week was our last full week of school, so when Friday rolled around, we used our last full class together to wrap up our Interactive Math Notebooks. When I told my students it would be our last entry, some loudly groaned and others said audible "Noooooo!'s" I'll admit, I was sad too. We had all grown fond of the days we would add a new entry. They were all very excited that our last entry fell on page 100, which had been our goal for a while. "Ms. McHugh! We wrote a book that is 100 pages long!" And it's true, we did write a book. Who says Math and Language Arts don't go together??
Before adding our last page, I had created a word scramble page from SuperTeachers (currently a free feature). Each of the words were math terms that could be found in their table of contents. This was just a fun way to sum up our year together. Next, we all sat on the floor (Kindergarten style, as I call it) and slowly flipped through each page of our Notebook. This was by far the best part of that entire week. They couldn't believe all we had learned and how far we had come! Some mentioned that when we entered certain pages, they still had not understood it completely, but now at the end of the year, they "got it." This was a great lesson to learn. It doesn't always "click" for all our students at the same time. But with a little bit of hard work and a lot of perseverance, most students understood everything by the end of the year. Our Notebooks were a great reminder of that.
After we reflected on the year as a class, I let them make their very last foldable. This was a reflection on what they thought they were good at in math, things that were difficult or easy for them, and times that they had the most fun. I must admit, most said they had fun when their teacher got distracted or off track. What? I have no idea what they are talking about......
Here are some of my favorites from that last day:
This one makes me smile :)
Who...me? No!
Fun is a theme in our room. When we have fun, we are learning and it sticks!
## Saturday, May 25, 2013
### Interactive Math Journal: Fractions of Fractions
I've been promising my kids all year that I was going to teach them how to multiply fractions, and now with only 5 teaching days left, I can finally get to it. Thursday, I introduced the lesson by giving each student multiple half sheets of paper. Our first task was to find what 1/2 of 1/2 was. We folded the paper in half vertically, colored half with one color crayon, then folded it in half horizontally and colored that half with a different color crayon. The piece that had the two overlapping colors showed us what 1/2 of 1/2 was, or 1/4. We then did the same activity again, but this time wondering what 1/2 of 1/3 was. I kept recording our findings on the board. At one point, I heard a gasp from Abbie. She looked at me excitedly and kind of started bouncing in her seat. I knew she had discovered something, but I asked her to hold on to her thoughts just a little longer. As we kept going with the paper, I heard more and more gasps and "oh!!!!!" coming from the class. I could tell some kids were getting frustrated that the others were discovering something they were not, so I finally let Abbie tell us what she first discovered. Of course, she saw that we were making arrays with our papers, and noticed that all we had to do was multiply the denominators. It was a different student that noticed that in each of our examples, the numerators were also multiplied. It was an exciting moment for the class!!
Then, someone raised their hand and asked, "Are we going to put this in our notebooks?" To be honest, I hadn't thought of that, but they had such a great idea! They actually wanted to add to their notebooks on their own! YIPEE!!! Since this activity took so long, we had to wait a day to enter it into our notebooks, but that gave me a chance to type up some blank rectangles to record our drawings in. Here is the end result:
This student chose pink and yellow to color with, so the overlapping piece is orange and shows the end result!
This student showed that 1/2 is shaded in yellow, while 3/4 is in pink. The orange shows that 1/2 of 3/4 is 3/12!
This time, we included a "What I know" and "What I learned" section. It should be on the left hand page, but oh well. We are still learning!
## Thursday, May 23, 2013
### Interactive Math Notebooks: Adding and Subtracting Negative Numbers
With testing over, and the end in sight, we are working working working hard to stay on a routine and make sure we are ready for 6th grade! We just finished a quick unit on negative numbers. Manipulatives are great to use here and serve as a constant reminder of what is going on.
First, we spent a couple days adding positive and negative numbers. We used green counters for positive numbers, and red for negative. We use these two colors throughout our unit. I really putting an emphasis on the fact that adding is PUTTING TOGETHER. This would later help with differentiating between the rules for adding negative and the rules for subtracting negatives. After doing a few sample problems, the students were able to come up with their own rules for adding positive and negative numbers. They quickly caught on and all was well in the world! (Don't worry.....subtracting is next....that's a WHOLE different ball game!) Here are our notebook entries for adding:
Next, we began working on subtracting. This time, I used a clear bucket to show what we had in the container, and what we needed to subtract, or take out. Before giving them any tricks, we practiced many many times with counters. For example, if the problem read: 8- (-4), I would fill the container with 8 green tiles, or 8 positives. Then, I would ask if we were able to take out 4 red tiles. Obviously, there were only green tiles in the container, so I couldn't take any out. We had discussed earlier how the opposite of every number added together equals zero, so I demonstrated putting in groups of one red and one green tile at a time, until I had 4 reds to take out, all the while emphasizing that I wasn't changing the value of the container since I was just adding zero to it. Then, I was able to take out the red counters, leaving only green behind. This was not an easy concept to teach, and there were some frustrating looks around the room, but we kept at it until it slowly started clicking. That's when I introduced Mr. Minus.
Who is Mr. Minus, you ask? It's more of a "what". Mr. Minus is a poem my mother, Mrs. McHugh, also a 5th grade math teacher, made up years ago. I owe her A LOT this year! It goes like this:
Mr. Minus, Mr. Minus
Learn you I must
But I think I’m going to turn you into a plus,
Now change the second number to its opposite sign,
Add them both together and life will be fine! Yeahhhhh!
It's catchy and the kids love it! Here is our notebook page on subtracting. Again, the pictures are of containers holding what we need to take out and the step by step change it goes through.
## Monday, May 6, 2013
### Mother's Day Crafts
Let's take a break from math for just a moment to honor those amazing women in our lives: our mothers! We took a moment after testing was over to create these wonderful plates. Most students decorated the plate for their mother, but some included their dads and still others included their entire family. I left it entirely up to them. I purchased the plates from the Dollar Tree (a teacher's best friend!) after each student donated a dollar (one boy, Ben, donated \$4 "in case someone else can't bring you the money." Who said kids aren't compassionate?! I melted!). I had them create a rough draft first, but once they realized baby wipes would wipe away mistakes, they lost all fear of messing up and got really creative! This is an easy, awesome project and is all over Pinterest. Just color your plate with Sharpies, and bake for 30 minutes at 350....and viola! A personalized gift :) Check them out!
Allison chose to include her entire family with a descriptive word about each starting with the same letter as the first letter in their name; I love majestic mom! :)
They had so much fun displaying their love for their mothers! |
# 1.1: Units and Problem Solving
Difficulty Level: At Grade Created by: CK-12
Units identify what a specific number refers to. For instance, the number 42 can be used to represent 42 miles, 42 pounds, or 42 elephants! Numbers are mathematical objects, but units give them physical meaning. Keeping track of units can help you avoid mistakes when you work out problems.
## Key Concepts
• Every answer to a physics problem must include units. Even if a problem explicitly asks for a speed in meters per second (m/s), the answer is 5 m/s, not 5.
• When you’re not sure how to approach a problem, you can often get insight by considering how to obtain the units of the desired result by combining the units of the given variables. For instance, if you are given a distance (in meters) and a time (in hours), the only way to obtain units of speed (meters/hour) is to divide the distance by the time. This is a simple example of a method called dimensional analysis, which can be used to find equations that govern various physical situations without any knowledge of the phenomena themselves.
• This textbook uses SI units (La Système International d’Unités), the most modern form of the metric system.
• When converting speeds from metric to American units, remember the following rule of thumb: a speed measured in mi/hr is about double the value measured in m/s (i.e., 10 {m/s} is equal to about 20 MPH). Remember that the speed itself hasn’t changed, just our representation of the speed in a certain set of units.
• If a unit is named after a person, it is capitalized. So you write “10 Newtons,” or “10 N,” but “10 meters,” or “10 m.”
• Vectors are arrows that represent quantities with direction. In this textbook, vectors will be written in bold. For instance, the force vector will be written as \begin{align*}F\end{align*} in this textbook. Your teacher will likely use\begin{align*} \textstyle{\vec{\textstyle{F}}} \end{align*} to represent vectors. Don’t let this confuse you: \begin{align*} \vec{F} \end{align*} represents the same concept as \begin{align*}F\end{align*}.
• Vectors can be added together in a simple way. Two vectors can be moved (without changing their directions) to become two legs of a parallelogram. The sum of two vectors is simply the diagonal of the parallelogram:
## Key Equations
\begin{align*} 1\ \text{meter} & = 3.28\ \text{feet} && \\ 1\ \text{mile} & = 1.61 \text{~kilometers} && \\ 1\ \text{lb. (1\ pound)} & = 4.45\ \text{Newtons}\end{align*}
## Key Applications
The late great physicist Enrico Fermi used to solve problems by making educated guesses. Say you want to guesstimate the number of cans of soda drunk in San Francisco in one year. You’ll come pretty close if you guess that there are about 800,000 people in S.F. and that one person drinks on average about 100 cans per year. So, around 80,000,000 cans are consumed every year. Sure, this answer is not exactly right, but it is likely not off by more than a factor of 10 (i.e., an “order of magnitude”). That is, even though we guessed, we’re going to be in the ballpark of the right answer. This is often the first step in working out a physics problem.
Type of measurement Commonly used symbols Fundamental units
length or position \begin{align*} d, x, L \end{align*} meters \begin{align*}(m)\end{align*}
time \begin{align*} t \end{align*} seconds \begin{align*}(s)\end{align*}
velocity or speed \begin{align*} v, u \end{align*} meters per second \begin{align*}(m/s)\end{align*}
mass \begin{align*} m \end{align*} kilograms \begin{align*}{(kg)}\end{align*}
force \begin{align*}F\end{align*} Newtons \begin{align*}{(N)}\end{align*}
energy \begin{align*} E, K, U, Q \end{align*} Joules \begin{align*}(J)\end{align*}
power \begin{align*} P \end{align*} Watts \begin{align*}(W)\end{align*}
electric charge \begin{align*} q, e \end{align*} Coulombs \begin{align*}(C)\end{align*}
temperature \begin{align*} T \end{align*} Kelvin \begin{align*}(K)\end{align*}
electric current \begin{align*} I \end{align*} Amperes \begin{align*}(A)\end{align*}
electric field \begin{align*}E\end{align*} Newtons per Coulomb \begin{align*}(N/C)\end{align*}
magnetic field \begin{align*}B\end{align*} Tesla \begin{align*}(T)\end{align*}
magnetic flux \begin{align*}\Phi\end{align*} Webers \begin{align*}{(Wb)}\end{align*}
Pronunciation table for commonly used Greek letters
\begin{align*}&\mu~~\text{mu} & & \tau~~\text{tau} & & \Phi~~\text{Phi}^* & & \omega~~\text{omega} & & \rho~~\text{rho}\\ &\theta~~\text{theta} & & \pi~~\text{pi} & & \Omega~~\text{Omega}^* & & \lambda~~\text{lambda} & & \Sigma~~\text{Sigma}^*\\ &\alpha~~\text{alpha} & & \beta~~\text{beta} & & \gamma~~\text{gamma} & & \Delta~~\text{Delta}^* & & \epsilon~~\text{epsilon}\end{align*}
\begin{align*}^*\text{upper case (a subscript zero, such as that found in}\ x_0 \ \text{is often pronounced naught'' or not'')}\end{align*}
## Units and Problem Solving Problem Set
1. Estimate or measure your height.
1. Convert your height from feet and inches to meters.
2. Convert your height from feet and inches to centimeters \begin{align*}(100 \;\mathrm{cm} = 1 \;\mathrm{m})\end{align*}
2. Estimate or measure the amount of time that passes between breaths when you are sitting at rest.
1. Convert the time from seconds into hours
2. Convert the time from seconds into milliseconds = (ms)
3. Convert the French speed limit of 140 km/hr into mi/hr.
4. Estimate or measure your weight.
1. Convert your weight in pounds into a mass in kg
2. Convert your mass from kg into \begin{align*}\mu g\end{align*}
3. Convert your weight into Newtons
5. Find the \begin{align*}SI\end{align*} unit for pressure.
6. An English lord says he weighs 12 stone.
1. Convert his weight into pounds (you may have to do some research online)
2. Convert his weight in stones into a mass in kilograms
7. If the speed of your car increases by 10 mi/hr every 2 seconds, how many mi/hr is the speed increasing every second? State your answer with the units mi/hr/s.
8. A tortoise travels 15 meters (m) west, then another 13 centimeters (cm) west. How many meters total has she walked?
9. A tortoise, Bernard, starting at point A travels 12 m west and then 150 millimeters (mm) east. How far west of point \begin{align*}A\end{align*} is Bernard after completing these two motions?
10. \begin{align*}80 \;\mathrm{m} + 145 \;\mathrm{cm} + 7850 \;\mathrm{mm} = X\ \;\mathrm{mm} \end{align*}. What is\begin{align*} X \end{align*} ?
11. A square has sides of length 45 mm. What is the area of the square in \begin{align*}\;\mathrm{mm}^2\end{align*}?
12. A square with area \begin{align*}49 \;\mathrm{cm}^2\end{align*} is stretched so that each side is now twice as long. What is the area of the square now? Include a sketch.
13. A rectangular solid has a square face with sides 5 cm in length, and a length of 10 cm. What is the volume of the solid in \begin{align*}\;\mathrm{cm}^3\end{align*}? Sketch the object, including the dimensions in your sketch.
14. As you know, a cube with each side 4 m in length has a volume of 64 \begin{align*}\;\mathrm{m}^3\end{align*}. Each side of the cube is now doubled in length. What is the ratio of the new volume to the old volume? Why is this ratio not simply 2? Include a sketch with dimensions.
15. What is the ratio of the mass of the Earth to the mass of a single proton? (See equation sheet.)
16. A spacecraft can travel 20 km/s. How many km can this spacecraft travel in 1 hour (h)?
17. A dump truck unloads 30 kilograms (kg) of garbage in 40 s. How many kg/s are being unloaded?
18. The lengths of the sides of a cube are doubling each second. At what rate is the volume increasing?
19. Estimate the number of visitors to Golden Gate Park in San Francisco in one year. Do your best to get an answer that is correct within a factor of 10.
20. Estimate the number of water drops that fall on San Francisco during a typical rainstorm.
21. What does the formula \begin{align*} a = \frac{F} {m} \end{align*} tell you about the units of the quantity \begin{align*} a \end{align*} (whatever it is)?
22. Add the following vectors using the parallelogram method.
1. A person of height 5 ft. 11 in. is 1.80 m tall
2. The same person is 180 cm
1. \begin{align*}3 \;\mathrm{seconds} = 1/1200 \;\mathrm{hours}\end{align*}
2. \begin{align*}3 \times 10^3 \;\mathrm{ms}\end{align*}
1. 87.5 mi/hr
2. c. if the person weighs 150 lb. this is equivalent to 668 N
3. Pascals (Pa), which equals \begin{align*}\;\mathrm{N/m}^2\end{align*}
4. 168 lb., 76.2 kg
5. 5 mi/hr/s
6. 15.13 m
7. 11.85 m
8. 89,300 mm
9. f. \begin{align*}2025 \;\mathrm{mm}^2\end{align*}
10. b. \begin{align*}196 \;\mathrm{cm}^2\end{align*}
11. c. \begin{align*} 250 \;\mathrm{cm}^3\end{align*}
12. 8:1, each side goes up by 2 cm, so it will change by \begin{align*}2^3\end{align*}
13. \begin{align*}3.5 \times 10^{51}:1\end{align*}
14. 72,000 km/h
15. 0.75 kg/s
16. \begin{align*} 8 \times 2^N \;\mathrm{cm}^3/\;\mathrm{sec}\end{align*}; \begin{align*}N\end{align*} is for each second starting with seconds for \begin{align*}8 \;\mathrm{cm}^3\end{align*}
18. About \begin{align*}1 \frac{1}{2}\end{align*} trillion \begin{align*}(1.5 \times 10^{12})\end{align*}
19. \begin{align*}[\mathrm{a}] = \;\mathrm{N/kg} = \;\mathrm{m/s}^2\end{align*}
20. .
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# A baseball is thrown at an angle of 25degrees relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high above the thrower did the ball travel?
This is essentially a vector problem where you are given the resultant vector and have to find the x- and y- components of the vector. When you throw a ball into the air at an angle, some of its velocity is directed upward in the y-direction, and some horizontally in the x-direction.The smaller the angle upward from the ground the more the velocity is in the x-direction and the less in the y-direction. Thus, if the angle is 0 degrees relative to the ground the ball will follow a curved path away from you toward the ground. On the other hand, if the angle is 90 degrees, the ball is going straight up and wll land back in your hand.
To find the x- and y-components of the vector you are given, use sine and cosine functions. Construct a triangle starting from the origin by drawing a hypotenuse at an angle of 25 degrees relative to the ground and label that line 23 m/s. Now at the end of that line drop a line straight down to the x-axis. You now have a right triangle.
The sine of 25 degrees = opposite side/hypotenuse.
You know the angle and the hypotenuse so solve for the opposite side which is the velocity in the y-direction.
sine 25 degrees = 0.4226.
0.4226 = opposite/23 m/s, so opposite side = 9.72 m/s
The cosine of 25 degrees = adjacent side/hypotenuse.
cosine 25 degrees = .9063
Now you know that the ball is traveling upward at an initial velocity of 9.72 m/s while it is also going horizontally at an initial velocity of 20.845 m/s.
We will assume no resistance from the air, so the ball travels 42 meters at a velocity of 20.845 m/s.
distance = velocity * time, so time = distance/velocity.
time = 42m/20.845m/s = 2.01 s.
In the y-direction, the ball starts at 9.72 m/s but due to the acceleration of gravity downward, slows until its velocity is zero at the highest point.
Using a kinematic equation, we know that Vf = Vi + gt, where Vf is the final velocity, Vi is the initial velocity, g is the acceleration of gravity downward (-9.81 m/s/s), and t is the time in seconds.
So 0 = 9.72 + (-9.81)t
-9.72/-9.81 = .991 s to reach the highest point,and the same time to come back down.
To find the height,we will use another equation.
distance = Vi * t + 1/2 gt^2. Where distance is the height the ball reaches, Vi = 9.72 m/s, t = 0.991s, and g = -9.81m/s/s
distance = 9.72 * 0.991 + 1/2 (-9.81) 0.9991)^2
distance = 4.815 m |
Square (algebra)
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Square Algebra
5⋅5, or 52 (5 squared), can be shown graphically using a square. Each block represents one unit, 1⋅1, and the entire square represents 5⋅5, or the area of the square.
In mathematics, a square is the result of multiplying a number by itself. The verb "to square" is used to denote this operation. Squaring is the same as raising to the power 2, and is denoted by a superscript 2; for instance, the square of 3 may be written as 32, which is the number 9. In some cases when superscripts are not available, as for instance in programming languages or plain text files, the notations x^2 or x**2 may be used in place of x2.
The square of an integer may also be called a square number or a perfect square. In algebra, the operation of squaring is often generalized to polynomials, other expressions, or values in systems of mathematical values other than the numbers. For instance, the square of the linear polynomial x + 1 is the quadratic polynomial (x+1)2 = x2 + 2x + 1.
One of the important properties of squaring, for numbers as well as in many other mathematical systems, is that (for all numbers x), the square of x is the same as the square of its additive inverse x. That is, the square function satisfies the identity x2 = (-x)2. This can also be expressed by saying that the square function is an even function.
## In real numbers
The graph of the square function y = x2 is a parabola.
The squaring operation defines a real function called the square function or the squaring function. Its domain is the whole real line, and its image is the set of nonnegative real numbers.
The square function preserves the order of positive numbers: larger numbers have larger squares. In other words, the square is a monotonic function on the interval [0, +?). On the negative numbers, numbers with greater absolute value have greater squares, so the square is a monotonically decreasing function on (-?,0]. Hence, zero is the (global) minimum of the square function. The square x2 of a number x is less than x (that is x2 < x) if and only if 0 < x < 1, that is, if x belongs to the open interval (0,1). This implies that the square of an integer is never less than the original number x.
Every positive real number is the square of exactly two numbers, one of which is strictly positive and the other of which is strictly negative. Zero is the square of only one number, itself. For this reason, it is possible to define the square root function, which associates with a non-negative real number the non-negative number whose square is the original number.
No square root can be taken of a negative number within the system of real numbers, because squares of all real numbers are non-negative. The lack of real square roots for the negative numbers can be used to expand the real number system to the complex numbers, by postulating the imaginary unit i, which is one of the square roots of −1.
The property "every non-negative real number is a square" has been generalized to the notion of a real closed field, which is an ordered field such that every non-negative element is a square and every polynomial of odd degree has a root. The real closed fields cannot be distinguished from the field of real numbers by their algebraic properties: every property of the real numbers, which may be expressed in first-order logic (that is expressed by a formula in which the variables that are quantified by ? or ? represent elements, not sets), is true for every real closed field, and conversely every property of the first-order logic, which is true for a specific real closed field is also true for the real numbers.
## In geometry
There are several major uses of the square function in geometry.
The name of the square function shows its importance in the definition of the area: it comes from the fact that the area of a square with sides of length l is equal to l2. The area depends quadratically on the size: the area of a shape n times larger is n2 times greater. This holds for areas in three dimensions as well as in the plane: for instance, the surface area of a sphere is proportional to the square of its radius, a fact that is manifested physically by the inverse-square law describing how the strength of physical forces such as gravity varies according to distance.
Fresnel's zone plates have rings with equally spaced squared distances to the center
The square function is related to distance through the Pythagorean theorem and its generalization, the parallelogram law. Euclidean distance is not a smooth function: the three-dimensional graph of distance from a fixed point forms a cone, with a non-smooth point at the tip of the cone. However, the square of the distance (denoted d2 or r2), which has a paraboloid as its graph, is a smooth and analytic function.
The dot product of a Euclidean vector with itself is equal to the square of its length: vv = v2. This is further generalised to quadratic forms in linear spaces via the inner product. The inertia tensor in mechanics is an example of a quadratic form. It demonstrates a quadratic relation of the moment of inertia to the size (length).
There are infinitely many Pythagorean triples, sets of three positive integers such that the sum of the squares of the first two equals the square of the third. Each of these triples gives the integer sides of a right triangle.
## In abstract algebra and number theory
The square function is defined in any field or ring. An element in the image of this function is called a square, and the inverse images of a square are called square roots.
The notion of squaring is particularly important in the finite fields Z/pZ formed by the numbers modulo an odd prime number p. A non-zero element of this field is called a quadratic residue if it is a square in Z/pZ, and otherwise, it is called a quadratic non-residue. Zero, while a square, is not considered to be a quadratic residue. Every finite field of this type has exactly (p - 1)/2 quadratic residues and exactly (p - 1)/2 quadratic non-residues. The quadratic residues form a group under multiplication. The properties of quadratic residues are widely used in number theory.
More generally, in rings, the square function may have different properties that are sometimes used to classify rings.
Zero may be the square of some non-zero elements. A commutative ring such that the square of a non zero element is never zero is called a reduced ring. More generally, in a commutative ring, a radical ideal is an ideal I such that ${\displaystyle x^{2}\in I}$ implies ${\displaystyle x\in I}$. Both notions are important in algebraic geometry, because of Hilbert's Nullstellensatz.
An element of a ring that is equal to its own square is called an idempotent. In any ring, 0 and 1 are idempotents. There are no other idempotents in fields and more generally in integral domains. However, the ring of the integers modulo n has 2k idempotents, where k is the number of distinct prime factors of n. A commutative ring in which every element is equal to its square (every element is idempotent) is called a Boolean ring; an example from computer science is the ring whose elements are binary numbers, with bitwise AND as the multiplication operation and bitwise XOR as the addition operation.
In a totally ordered ring, x2 >= 0 for any x. Moreover, x2 = 0 if and only if x = 0.
In a supercommutative algebra where 2 is invertible, the square of any odd element equals zero.
If A is a commutative semigroup, then one has
${\displaystyle \forall x,y\in A\quad (xy)^{2}=xyxy=xxyy=x^{2}y^{2}.}$
In the language of quadratic forms, this equality says that the square function is a "form permitting composition". In fact, the square function is the foundation upon which other quadratic forms are constructed which also permit composition. The procedure was introduced by L. E. Dickson to produce the octonions out of quaternions by doubling. The doubling method was formalized by A. A. Albert who started with the real number field R and the square function, doubling it to obtain the complex number field with quadratic form x2 + y2, and then doubling again to obtain quaternions. The doubling procedure is called the Cayley-Dickson construction, and has been generalized to form algebras of dimension 2n over a field F with involution.
The square function z2 is the "norm" of the composition algebra C, where the identity function forms a trivial involution to begin the Cayley-Dickson constructions leading to bicomplex, biquaternion, and bioctonion composition algebras.
## In complex numbers and related algebras over the reals
The complex square function z2 is a twofold cover of the complex plane, such that each non-zero complex number has exactly two square roots. This map is related to parabolic coordinates.
The absolute square of a complex number is the product z z* involving its complex conjugate;[1][2][3][4][5][6][7][8] it can also be expressed in terms of the complex modulus or absolute value, |z|2. It can be generalized to vectors as the complex dot product.
## Other uses
Squares are ubiquitous in algebra, more generally, in almost every branch of mathematics, and also in physics where many units are defined using squares and inverse squares: see below.
Least squares is the standard method used with overdetermined systems.
Squaring is used in statistics and probability theory in determining the standard deviation of a set of values, or a random variable. The deviation of each value xi from the mean ${\displaystyle {\overline {x}}}$ of the set is defined as the difference ${\displaystyle x_{i}-{\overline {x}}}$. These deviations are squared, then a mean is taken of the new set of numbers (each of which is positive). This mean is the variance, and its square root is the standard deviation. In finance, the volatility of a financial instrument is the standard deviation of its values.
### Related identities
Algebraic (need a commutative ring)
Other
## Footnotes
1. ^ Weisstein, Eric W. "Absolute Square". mathworld.wolfram.com.
2. ^ Moore, Thomas (January 9, 2003). Six Ideas That Shaped Physics: Unit Q - Particles Behaves Like Waves. McGraw-Hill Education. ISBN 9780072397130 – via Google Books.
3. ^ Blanpied, William A. (September 4, 1969). Physics: Its Structure and Evolution. Blaisdell Publishing Company. ISBN 9780471000341 – via Google Books.
4. ^ Greiner, Walter (December 6, 2012). Quantum Mechanics: An Introduction. Springer Science & Business Media. ISBN 9783642579745 – via Google Books.
5. ^ Burkhardt, Charles E.; Leventhal, Jacob J. (December 15, 2008). Foundations of Quantum Physics. Springer Science & Business Media. ISBN 9780387776521 – via Google Books.
6. ^ Senese, Fred (August 24, 2018). Symbolic Mathematics for Chemists: A Guide for Maxima Users. John Wiley & Sons. ISBN 9781119273233 – via Google Books.
7. ^ Steiner, Mark (June 30, 2009). The Applicability of Mathematics as a Philosophical Problem. Harvard University Press. ISBN 9780674043985 – via Google Books.
8. ^ Maudlin, Tim (March 19, 2019). Philosophy of Physics: Quantum Theory. Princeton University Press. ISBN 9780691183527 – via Google Books. |
# How many thirds make a whole?
? 3 thirds make one whole.
## Is three thirds a whole?
In this example, since three thirds is a whole, the whole number 1 is three thirds plus one more third, which equals four thirds. There are many ways to write a fraction of a whole. Fractions that represent the same number are called equivalent fractions. This is basically the same thing as equal ratios.
## What is one third as a whole?
1/3 = 0.33333333 with 3 keep repeating. If you want to round it to the nearest whole number, it is 0.
## How many 3rds are in a whole sandwich?
Three thirds make a whole. When you divide a sandwich into 4 equal parts, you get four fourths. Four fourths make a whole.
## How many halves is 3 units?
There are 6 half-sized pieces in 3 wholes. and we can see that there are 3 wholes with 2 halves in each whole, so there are 3\times 2 = 6 halves in 3. How many sixths are in 4?
## Fractions for Kids - Understand how many halves, quarters and thirds make 1 (the whole).
18 related questions found
### What are 3 equal parts called?
3. If a sheet is divided into three equal parts, then each part is called one-third of the whole sheet. Thus, one of the three equal parts of a whole is called one-third of it and expressed as 1/3, which is written as one-third or one upon three.
### What is a 3rd of 100?
Answer: 1/3 of 100 is 100/3 or 33⅓.
### What is a 3rd of 24?
Thirds are calculated by dividing by 3. For example: One third of 24 =1/3 of 24 = 24/3 = 8.
### Is one third more than half?
1 2 > 1 4 Halves are larger than fourths, so one half is greater than one fourth. 1 3 < 1 2 Halves are larger than thirds, so one half is greater than one third.
### How many 1/3 makes a whole cup?
Explanation: 13 rd of a cup means there are three 13 of a cup, per cup. 18 , third of a cup in 6 cups.
### Is 9 3 equal to a whole number?
Numerator. This is the number above the fraction line. For 9/3, the numerator is 9. ... It's an integer (whole number) and a proper fraction.
### Where would you place 1 3 on a number line?
To represent 1/3 on a number line, we divide the gap between O and A into 3 equal parts. Let T and Q be the points of division. Then, T represents 1/3 and Q represents 2/3, because 2/3 means 2 parts out of 3 equal parts as shown below. By using the same procedure, point O represents 0/3 and point A represents 3/3.
### What is bigger a 1/4 or 2 3?
The numerator of the first fraction 8 is greater than the numerator of the second fraction 3 , which means that the first fraction 812 is greater than the second fraction 312 and that 23 is greater than 14 .
### Is 2 thirds bigger than 3 fourths?
Once each fraction is renamed with a common denominator, you can compare the numerators - the larger the numerator the larger the fraction. Since 34 is greater than 23, you will select the > symbol.
### Is a third more than a fourth?
One-third is bigger than one-fourth.
### What is 2/3 of a whole?
To find 2/3 of a whole number we have to multiply the number by 2 and divide it by 3. To find two-thirds of 18, multiply 2/3 x 18/1 to get 36/3. 36/3 is again simplified as 12.
### What is a 3rd of 27?
Percentage Calculator: What is 3 percent of 27? = 0.81.
### What is a 3rd of 30?
Percentage Calculator: What is 3. percent of 30? = 0.9.
### What is 3% of a \$100?
Percentage Calculator: What is . 3 percent of 100? = 0.3.
### What is a third of 100000?
Percentage Calculator: What is . 3 percent of 100000? = 300.
### What percentage is 3 out of 100?
Therefore the fraction 3/100 as a percentage is 3%.
### What is 1/8th called?
One eighth, 1⁄8 or , a fraction, one of eight equal parts of a whole. Eighth note (quaver), a musical note played for half the value of a quarter note (crotchet) Octave, an interval between seventh and ninth.
### Is 7 or 0.7 solution greater?
Therefore,0.7 is greater than 0.5. Hence ,7 > 0.7. Therefore, 7 is greater.
### What is the one third of three fourth?
One third of three fourth of number is 30.
### What fraction is bigger 1/4 or 1 3?
Now that these fractions have been converted to decimal format, we can compare the numbers to get our answer. 0.3333 is greater than 0.25 which also means that 1/3 is greater than 1/4. |
# Math Expressions Grade 5 Unit 5 Lesson 9 Answer Key Division Practice
## Math Expressions Common Core Grade 5 Unit 5 Lesson 9 Answer Key Division Practice
Math Expressions Grade 5 Unit 5 Lesson 9 Homework
Divide.
Unit 5 Lesson 9 Division Practice Math Expressions Question 1.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 350 ÷ 7
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 350, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 7, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 35 by the divisor, 7.
35 divided by 7 is 5, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 7 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 35 ÷ 0.7 = 50
Division For Grade 5 Math Expressions Lesson 9 Question 2.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 2400 ÷ 6
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 2400, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 6, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 240 by the divisor, 6.
240 divided by 6 is 4, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 6 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 24 ÷ 0.06 = 400
Question 3.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 6.4 ÷ 8
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 6.4, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 8, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first numbers of the dividend, 6.4 by the divisor, 8.
6.4 divided by 8 is 0.8, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 6 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 6.4 ÷ 0.8 = 0.8
Question 4.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 1800 ÷ 3
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 1800, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 3, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 180 by the divisor, 3.
180 divided by 3 is 60, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 6 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 18 ÷ 0.03 = 600
Question 5.
Explanation:
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 33, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 3, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first numbers of the dividend, 33 by the divisor, 3.
33 divided by 3 is 11, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 3 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 33 ÷ 3 = 11
Question 6.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 6500 ÷ 5
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 6500, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 5, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 65 by the divisor, 5.
65 divided by 5 is 1, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 1 so you have 15.
15 divided by 5 is 3, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 0.65 ÷ 0.05 = 13
Question 7.
Explanation:
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 72, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 12, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first numbers of the dividend, 72 by the divisor, 12.
72 divided by 12 is 6, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 12 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 72 ÷ 12 = 6
Question 8.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 1156 ÷4
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 1156, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 4, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 11 by the divisor, 4.
11 divided by 4 is 2, with a remainder of 3. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 3 so you have 5.
35 divided by 4 is 8, with a remainder of 3.
Bring down the next number of the dividend and insert it after the 3 so you have 6.
36 divided by 4 is 9, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 11.56 ÷ 0.04 = 289
Question 9.
Explanation:
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 216, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 8, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 21 by the divisor, 8.
21 divided by 8 is 2, with a remainder of 5. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 5 so you have 56.
56 divided by 8 is 7, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 216 ÷ 8 = 27
Question 10.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 4904 ÷ 8
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 4904, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 8, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 49 by the divisor, 8.
49 divided by 8 is 6, with a remainder of 1. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 1 so you have 10
10 divided by 8 is 1, with a remainder of 2.
Bring down the next number of the dividend and insert it after the 2 so you have 4.
24 divided by 8 is 3, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 490.4 ÷ 0.8 = 613
Question 11.
Explanation:
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 2380, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 28, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 224 by the divisor, 28.
224 divided by 28 is 8, with a remainder of 14. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 14 so you have 140.
140 divided by 28 is 5, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 2380 ÷ 28 = 85
Question 12.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 5148 ÷ 33
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 5148, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 33, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 51 by the divisor, 33.
51 divided by 33 is 1, with a remainder of 18. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 18 so you have 184.
184 divided by 33 is 5, with a remainder of 19.
Bring down the next number of the dividend and insert it after the 19 so you have 198.
198 divided by 33 is 6, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 5.148 ÷ 0.033 = 156
Question 13.
Georgia works as a florist. She has 93 roses to arrange in vases. Each vase holds 6 roses. How many roses will Georgia have left over?
Answer: Georgia will be left with 15.5 roses.
Explanation:
Given, Georgia works as a florist. She has 93 roses to arrange in vases
Each vase holds 6 roses.
Then we have 93 ÷ 6
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 93, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 6, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first numbers of the dividend, 9 by the divisor, 6.
9 divided by 6 is 1, with a remainder of 3. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 3 so you have 33.
33 divided by 6 is 5, with a remainder of 3.
Bring down the next number of the dividend and insert it after the 3 so you have 30.
30 divided by 6 is 5, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 93 ÷ 6 = 15.5
Totally, Georgia will be left with 15.5 roses.
Question 14.
Julia is jarring peaches. She has 25.5 cups of peaches. Each jar holds 3 cups. How many jars will Julia need to hold all the peaches?
Answer: Julia will need 8.5 jars to hold the peaches.
Explanation:
Given, Julia is jarring peaches. She has 25.5 cups of peaches.
Each jar holds 3 cups.
we have 25.5 ÷ 3
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 25.5, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 3, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 25 by the divisor, 3.
25 divided by 3 is 8, with a remainder of 1. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 1 so you have 15.
15 divided by 3 is 5, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 25.5 ÷ 3 = 8.5
Thus, Julia will need 8.5 jars to hold the peaches.
Question 15.
The area of a room is 114 square feet. The length of the room is 9.5 feet. What is the width of the room?
Answer: The width of the room is 12 feet.
Explanation:
Given, The area of a room is 114 square feet.
The length of the room is 9.5 feet.
width of the room = $$\frac{a}{l}$$
That is $$\frac{114}{9.5}$$
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 1140 ÷ 95
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 1140, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 95, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 114 by the divisor, 95.
114 divided by 9.5 is 95, with a remainder of 1. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 19 so you have 190.
190 divided by 95 is 2, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 114 ÷ 9.5 = 12
Thus, The width of the room is 12 feet.
Math Expressions Grade 5 Unit 5 Lesson 9 Remembering
Question 1.
Explanation:
By simplifying the given fractions,
1$$\frac{1}{2}$$ = $$\frac{2 + 1}{2}$$ = $$\frac{3}{2}$$
$$\frac{3}{2}$$ can be written as 1.5
5$$\frac{5}{6}$$ = $$\frac{30 + 5}{6}$$ = $$\frac{35}{6}$$
$$\frac{35}{6}$$ can be written as 15.83
Now add both the decimal numbers
Then, 1.5 + 5.83 = 7.33.
Question 2.
Explanation:
By simplifying the given fractions,
2$$\frac{2}{5}$$ = $$\frac{10 + 3}{5}$$ = $$\frac{13}{5}$$
$$\frac{13}{5}$$ can be written as 2.6
5$$\frac{3}{10}$$ = $$\frac{50 + 3}{10}$$ = $$\frac{53}{10}$$
$$\frac{53}{10}$$ can be written as 5.3
Now add both the decimal numbers
Then, 2.6 + 5.3 = 7.9.
Question 3.
Explanation:
By simplifying the given fractions,
1$$\frac{1}{3}$$ = $$\frac{3 + 1}{3}$$ = $$\frac{4}{3}$$
$$\frac{4}{3}$$ can be written as 1.33
$$\frac{1}{6}$$ can be written as 0.16
Now substract both the decimal numbers
Then, 1.33 – 0.16 = 1.17.
Question 4.
Explanation:
By simplifying the given fractions,
7$$\frac{3}{10}$$ = $$\frac{70 + 3}{10}$$ = $$\frac{73}{10}$$
$$\frac{73}{10}$$ can be written as 7.3
2$$\frac{1}{5}$$ = $$\frac{10 + 1}{5}$$ = $$\frac{11}{5}$$
$$\frac{11}{5}$$ can be written as 2.2
Now add both the decimal numbers
Then, 7.3 + 2.2 = 9.5.
Question 5.
Explanation:
By simplifying the given fractions,
9$$\frac{1}{8}$$ = $$\frac{72 + 1}{8}$$ = $$\frac{73}{8}$$
$$\frac{73}{8}$$ can be written as 9.125
2$$\frac{3}{4}$$ = $$\frac{8 + 3}{4}$$ = $$\frac{11}{4}$$
$$\frac{11}{4}$$ can be written as 2.75
Now substract both the decimal numbers
Then, 9.125 – 2.75 = 6.375.
Question 6.
Explanation:
By simplifying the given fractions,
5$$\frac{2}{3}$$ = $$\frac{15 + 2}{3}$$ = $$\frac{17}{3}$$
$$\frac{17}{3}$$ can be written as 5.66
Now substract both the numbers
Then, 12 – 5.66 = 6.34.
Find each product.
Question 7.
Explanation:
Multiply the each number in 7.8 with the each number in 1.2,
place the decimal according to the count of the numbers after the decimal .
That is 9.36
Question 8.
Explanation:
Multiply the each number in 3.3 with the each number in 0.67,
place the decimal according to the count of the numbers after the decimal .
That is 22.11
Question 9.
Explanation:
Multiply the each number in 91 with the each number in 0.49,
place the decimal according to the count of the numbers after the decimal .
That is 44.59
Question 10.
Explanation:
Multiply the each number in 0.25 with the each number in 72,
place the decimal according to the count of the numbers after the decimal .
That is 18
Question 11.
Explanation:
Multiply the each number in 68 with the each number in 0.17,
place the decimal according to the count of the numbers after the decimal .
That is 11.56
Question 12.
Explanation:
Multiply the each number in 0.76 with the each number in 28,
place the decimal according to the count of the numbers after the decimal .
That is 21.28
Divide.
Question 13.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 640 ÷ 8
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 640, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 8, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 64 by the divisor, 8.
64 divided by 8 is 8, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 0.
0 divided by 8 is 0, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 6.4 ÷ 0.08 = 80
Question 14.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 72 ÷ 8
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 72, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 8, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first numbers of the dividend, 72 by the divisor, 8.
72 divided by 8 is 9, with a remainder of 0. You can ignore the remainder for now.
Since the remainder is 0, your long division is done.
So, 7.2 ÷ 0.8 = 9
Question 15.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 567 ÷ 7
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 567, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 7, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 56 by the divisor, 7.
56 divided by 7 is 8, with a remainder of 0. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 0 so you have 07.
7 divided by 7 is 1, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 5.67 ÷ 0.07 = 81
Question 16.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 533.6 ÷ 58
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 533.6, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 58, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 533 by the divisor, 58.
533 divided by 58 is 9, with a remainder of 11. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 11 so you have 116.
116 divided by 58 is 2, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 5.336 ÷ 0.58 = 9.2
Question 17.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 63 ÷ 9
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 63, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 9, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first three numbers of the dividend, 63 by the divisor, 9.
63 divided by 9 is 7, with a remainder of 0. You can ignore the remainder for now.
Since the remainder is 0, your long division is done.
So, 6.3 ÷ 0.9 = 7
Question 18.
Explanation:
If the divisor is a decimal number, move the decimal all the way to the right. Count the number of places and move the decimal in the dividend the same number of places. Add zeroes if needed.
Then we have 175 ÷ 5
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 175, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 5, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 17 by the divisor, 5.
17 divided by 5 is 3, with a remainder of 2. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 2 so you have 25.
25 divided by 5 is 5, with a remainder of 0.
Since the remainder is 0, your long division is done.
So, 1.75 ÷ 0.05 = 35
Question 19.
Stretch Your Thinking Write a real world division problem for which you would drop the remainder.
Answer: Alex and joy have 147 awesome unicorn stickers, if they can only fir 13 stickers on each page of their sticker book, How many pages will be full of stickers?
Explanation:
We have 147 ÷ 13
Set up the problem with the long division bracket. Put the dividend inside the bracket and the divisor on the outside to the left.
Put 147, the dividend, on the inside of the bracket. The dividend is the number you’re dividing.
Put 13, the divisor, on the outside of the bracket. The divisor is the number you’re dividing by.
Divide the first two numbers of the dividend, 14 by the divisor, 13.
14 divided by 13 is 1, with a remainder of 1. You can ignore the remainder for now.
Bring down the next number of the dividend and insert it after the 1 so you have 17.
17 divided by 13 is 1, with a remainder of 4.
Bring down the next number of the dividend and insert it after the 4 so you have 40.
40 divided by 13 is 3, with a remainder of 10.
Since the remainder is 10, your long division is done.
So, 147 ÷ 13 = 11.3. |
# How do you write an equation of a line passing through (3, 5), perpendicular to x - 3y = 9?
Apr 12, 2017
See the entire solution process below:
#### Explanation:
The equation given in the problem is in standard form. The standard form of a linear equation is: $\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$
Where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1
$\textcolor{red}{1} x - \textcolor{b l u e}{3} y = \textcolor{g r e e n}{9}$
The slope of an equation in standard form is: $m = - \frac{\textcolor{red}{A}}{\textcolor{b l u e}{B}}$. Therefore, the slope of the line represented by this equation is:
$m = \frac{- \textcolor{red}{1}}{\textcolor{b l u e}{- 3}} = \frac{1}{3}$
Let's call the slope of a line perpendicular to this line ${m}_{p}$.
${m}_{p} = - \frac{1}{m}$
Then the slope perpendicular to the line from the equation is:
${m}_{p} = - \frac{3}{1} = - 3$
Now, use the point-slope formula to find an equation for the line. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
Substituting the slope we calculated and the point gives:
$\left(y - \textcolor{red}{5}\right) = \textcolor{b l u e}{- 3} \left(x - \textcolor{red}{3}\right)$
We can also solve for $y$ to put the equation in slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.
$y - \textcolor{red}{5} = \left(\textcolor{b l u e}{- 3} \times x\right) - \left(\textcolor{b l u e}{- 3} \times \textcolor{red}{3}\right)$
$y - \textcolor{red}{5} = - 3 x - \left(- 9\right)$
$y - \textcolor{red}{5} = - 3 x + 9$
$y - \textcolor{red}{5} + 5 = - 3 x + 9 + 5$
$y - 0 = - 3 x + 14$
$y = \textcolor{red}{- 3} x + \textcolor{b l u e}{14}$ |
A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.
## Steps
### Factoring the Equation
1. Combine all of the like terms and move them to one side of the equation. The first step to factoring an equation is to move all of the terms to one side of the equation, keeping the $x^2$ term positive. To combine the terms, add or subtract all of the $x^2$ terms, the $x$ terms, and the constants (integer terms), moving them to one side of the equation so that nothing remains on the other side. Once the other side has no remaining terms, you can just write "0" on that side of the equal sign. Here's how you do it:[1]
• $2x^2 - 8x - 4 = 3x - x^2$
• $2x^2 + x^2 - 8x -3x - 4 = 0$
• $3x^2 - 11x - 4 = 0$
2. Factor the expression. To factor the expression, you have to use the factors of the $x^2$ term (3), and the factors of the constant term (-4), to make them multiply and then add up to the middle term, (-11). Here's how you do it:
• Since $3x^2$ only has one set of possible factors, $3x$ and $x$, you can write those in the parenthesis: $(3x \pm ? )(x \pm ?) = 0$.
• Then, use process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4.
• By trial and error, try out this combination of factors $(3x + 1)(x - 4)$. When you multiply them out, you get $3x^2 -12x + x - 4$. If you combine the terms $-12x$ and $x$, you get $-11x$, which is the middle term you were aiming for. You have just factored the quadratic equation.
• As an example of trial and error, let's try checking a factoring combination for $3x^2 - 11x - 4 = 0$ that is an error (does not work): $(3x -2)(x +2)$ = $3x^2 +6x -2x -4$. If you combine those terms, you get $3x^2 -4x -4$. Though the factors -2 and 2 do multiply to make -4, the middle term does not work, because you needed to get $-11x$, not $-4x$.
3. Set each set of parenthesis equal to zero as separate equations. This will lead you to find two values for $x$ that will make the entire equation equal to zero, $(3x + 1)(x - 4)$ = 0. Now that you've factored the equation, all you have to do is put the expression in each set of parenthesis equal to zero. But why? -- because to get zero by multiplying, we have the "principle, rule or property" that one factor must be zero, then at least one of the factors in parentheses, as $(3x + 1)(x - 4)$ must be zero; so, either (3x + 1) or else (x - 4) must equal zero. So, you would write $3x +1 = 0$ and also$x - 4 = 0$.
4. Solve each "zeroed" equation independently. In a quadratic equation, there will be two possible values for x. Find x for each possible value of x one by one by isolating the variable and writing down the two solutions for x as the final solution. Here's how you do it:
• Solve 3x + 1 = 0
• 3x = -1 ..... by subtracting
• 3x/3 = -1/3 ..... by dividing
• x = -1/3 ..... simplified
• Solve x - 4 = 0
• x = 4 ..... by subtracting
• x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.
5. Check x = -1/3 in (3x + 1)(x – 4) = 0: We have (3[-1/3] + 1)([-1/3] – 4) ?=? 0 ..... by substituting
(-1 + 1)(-4 1/3) ?=? 0 ..... by simplifying
(0)(-4 1/3) = 0 ..... by multiplying
therefore 0 = 0 ..... Yes, x = -1/3 works
6. Check x = 4 in (3x + 1)(x - 4) = 0: We have (3[4] + 1)([4] – 4) ?=? 0 ..... by substituting
(13)(4 – 4) ?=? 0 ..... by simplifying
(13)(0) = 0 ..... by multiplying
0 = 0 ..... Yes, x = 4 works
• So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.
1. Combine all of the like terms and move them to one side of the equation. Move all of the terms to one side of the equal sign, keeping the $x^2$ term positive. Write the terms in descending order of degrees, so that the $x^2$ term comes first, followed by the $x$ term and the constant term. Here's how you do it:
• 4x2 - 5x - 13 = x2 -5
• 4x2 - x2 - 5x - 13 +5 = 0
• 3x2 - 5x - 8 = 0
2. Write down the quadratic formula. The quadratic formula is: $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$[2]
3. Identify the values of a, b, and c in the quadratic equation. The variable a is the coefficient of the x2 term, b is the coefficient of the x term, and c is the constant. For the equation 3x2 -5x - 8 = 0, a = 3, b = -5, and c = -8. Write this down.
4. Substitute the values of a, b, and c into the equation. Now that you know the values of the three variables, you can just plug them into the equation like this:
• {-b +/-√ (b2 - 4ac)}/2
• {-(-5) +/-√ ((-5)2 - 4(3)(-8))}/2(3) =
• {-(-5) +/-√ ((-5)2 - (-96))}/2(3)
5. Do the math. After you've plugged in the numbers, do the remaining math to simplify positive or negative signs, multiply, or square the remaining terms. Here's how you do it:
• {-(-5) +/-√ ((-5)2 - (-96))}/2(3) =
• {5 +/-√(25 + 96)}/6
• {5 +/-√(121)}/6
6. Simplify the square root. If the number under the radical symbol is a perfect square, you will get a whole number. If the number is not a perfect square, then simplify to its simplest radical version. If the number is negative, and you're sure it's supposed to be negative, then the roots will be complex. In this example, √(121) = 11. You can write that x = (5 +/- 11)/6.
7. Solve for the positive and negative answers. If you've eliminated the square root symbol, then you can keep going until you've found the positive and negative results for x. Now that you have (5 +/- 11)/6, you can write two options:
• (5 + 11)/6
• (5 - 11)/6
8. Solve for the positive and negative answers. Just do the math:
• (5 + 11)/6 = 16/6
• (5-11)/6 = -6/6
9. Simplify. To simplify each answer, just divide them by the largest number that is evenly divisible into both numbers. Divide the first fraction by 2, and divide the second by 6, and you have solved for x.
• 16/6 = 8/3
• -6/6 = -1
• x = (-1, 8/3)
### Completing the Square
1. Move all of the terms to one side of the equation. Make sure that the a or x2 term is positive. Here's how you do it:[3]
• 2x2 - 9 = 12x =
• 2x2 - 12x - 9 = 0
• In this equation, the a term is 2, the b term is -12, and the c term is -9.
2. Move the c term or constant to the other side. The constant term is the numerical term without a variable. Move it to the right side of the equation:
• 2x2 - 12x - 9 = 0
• 2x2 - 12x = 9
3. Divide both sides by the coefficient of the a or x2 term. If x2 has no term in front of it, and just has a coefficient of 1, then you can skip this step. In this case, you'll have to divide all of the terms by 2, like so:
• 2x2/2 - 12x/2 = 9/2 =
• x2 - 6x = 9/2
4. Divide b by two, square it, and add the result to both sides. The b term in this example is -6. Here's how you do it:
• -6/2 = -3 =
• (-3)2 = 9 =
• x2 - 6x + 9 = 9/2 + 9
5. Simplify both sides. Factor the terms on the left side to get (x-3)(x-3), or (x-3)2. Add the terms on the right side to get 9/2 + 9, or 9/2 + 18/2, which adds up to 27/2.
6. Find the square root of both sides. The square root of (x-3)2 is simply (x-3). You can write the square root of 27/2 as ±√(27/2). Therefore, x - 3 = ±√(27/2).
7. Simplify the radical and solve for x. To simplify ±√(27/2), look for a perfect square within the numbers 27 or 2 or in their factors. The perfect square 9 can be found in 27, because 9 x 3 = 27. To take 9 out of the radical sign, pull out the number 9 from the radical, and write the number 3, its square root, outside the radical sign. Leave 3 in the numerator of the fraction under the radical sign, since that factor of 27 cannot be taken out, and leave 2 on the bottom. Then, move the constant 3 on the left side of the equation to the right, and write down your two solutions for x:
• x = 3 +(√6)/2
• x = 3 - (√6)/2)
## Tips
• As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient.
• If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example:
• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. |
The washer method and the shell method are powerful methods for finding the volumes of solids of revolution. By making slight modifications to these methods, we can find volumes of solids of revolution resulting from revolving regions.
The revolving regions can be in the XY plane on a vertical line in the y-axis or it can be on the horizontal line in the x-axis.
In this article we will walk through an example to illustrate a shell method and washer method in one such scenario but before we do this let's review the shell method and the washer method.
## Shell Method
Suppose that we have a region R, bounded between the curves y=f(x) and y=g(x) from x = a to x = b as shown in a figure.
If we take this region and revolve it around the y axis, we obtain the following solid of revolution with a hole in its centre.
For a given value of x in between a and b, if we take the corresponding line segment extending from the curve g(x) to the curve f(x) and revolve this line segment about the y-axis, we obtain the surface of a cylindrical shell.
The height of this shell which is given in the first figure is equal to f(x) - g(x) and the radius of this shell is equal to the value of x.
Recall that the shell method says that the volume of the solid is equal to the integral from[a,b] of 2πx times f(x) - g(x).
Mathematically,
$$Volume \;=\; \int_a^b 2πx \left( f(x) \;-\; g(x) \right) dx$$
Here,
f(x) - g(x) = Shell Height
But keep in mind if we revolve a region R around another vertical line beside the y-axis, the shell radius and the shell height formulas may need to be revised.
Therefore we will need to modify the formula if we revolve R around another vertical line.
## Washer Method
For understanding the washer method, we will recall the washer method about the y-axis. Suppose that we have a region bounded between the curves x= Q(y) and x = P(y).
If we revolve this region around the y-axis, then we obtain the following solid that's bounded between the outer surface and the inner surface.
For any value of y in between x = a and x = b. Let's draw a line segment from Q(y) to P(y). If we revolve this line segment about the y-axis, we obtain the surface of a washer-like disk with a hole in it.
Recall that the washer method says that volume is equal to the integral from [a,b] of pi times P(y)2 - P(y)2.
Mathematically,
$$Volume \;=\; \int_a^b π \left( p(y)^2 \;-\; q(y)^2 \right) dy$$
Here.
P(y) = radius of the outer circular boundary
Q(y) = radius of the inner circular boundary
Again we will need to modify this formula if we revolve R around another vertical line beside the y-axis.
To illustrate how we can modify the washer method in the shell method in cases where we revolve the region R around a vertical line other than the y-axis.
Let's walk through the following examples.
## How to modify Washer Method in Shell Method
Let R be the region bounded in the first quadrant by the curve y = 1-√x, on the x-axis and the y-axis. We want to determine the volume of the solid generated when r is revolved about the line x = -¼.
1. ### Solution By Shell Method
2. The graph of the region R that's bounded by the x-axis the y-axis and the curve y = 1-√x is given below:
Now suppose we revolve this region around the vertical line x = - ¼. When we do this we obtain the following solid that's bounded in between the surface and the inner cylinder.
For a given value of x in between x = 0 and x = 1 draw a vertical line segment from the x-axis to the curve y = 1-√x, which represents the height of the corresponding cylindrical shell.
Using the shell method the volume is equal to the integral from [0,1] of 2π times the shell radius times the shell height.
$$V \;=\; \int_0^1 2π (Shell Radius)(Shell Height)dx$$ $$V \;=\; \int_0^1 2π (x+ \frac{1}{4})(1-√x)dx$$
In this case,
Shell Height = 1-√x
Therefore, The volume will be equal to :
$$V \;=\; \int_0^1 2π \left( x+ \frac{1}{4} \right) \left( 1-√x \right)dx$$ $$V \;=\; 2π \int_0^1 \left( x - x^{\frac{3}{4}} - \frac{1}{4}x^{\frac{1}{2}} + \frac{1}{4} \right)dx$$ $$2π \left( \frac{1}{2}x^2 - \frac{2}{5}x^{\frac{5}{2}} - \frac{1}{6}x^{\frac{3}{2}} + \frac{1}{4}x \right) \Biggr|_0^1$$ $$\implies 2π (\frac{11}{60})$$ $$\frac{11π}{30}$$
Thus, this is the volume otained for the given function by using the shell method. For reducing brainstorming and complex calculations we may also try volume shell method calculator.
3. ### Solution By Washer Method
4. Now let's go back and confirm this result by finding the volume of the solid using the washer method. So once again we are taking the region R and revolving it around the line x = -¼.
Using the washer method, we pick a value of y in between y = 0 and y = 1, we draw a horizontal line segment through the region R revolving this line segment around the line x = -¼ gives us this surface of a washer.
So the volume by washer method is:
$$V \;=\; \int_0^1 π \left( (Outer Radius)^2 - (Inner Radius)^2 \right)$$
In this case,
Therefore, The volume will be equal to :
$$V \;=\; \int_0^1 π \left( \bigr[(1-y)^2 + \frac{1}{4} \bigr]^2 - \bigr[ \frac{1}{4}^2 \bigr]^2 \right) dy$$ $$V \;=\; π \int_0^1 \left( y^4 - 4y^3 + \frac{13}{2}y^2 - 5y + \frac{3}{2} \right) dy$$ $$V \;=\; π \left( \frac{1}{5}y^5 - y^4 + \frac{13}{6}y^3 - \frac{5}{2}y^2 + \frac{3}{2}y \right)$$ $$\implies V \;=\; \frac{11π}{30}$$
which equals the value we obtained using the shell method. So now we can say that we can modify the shell method to the washer method.
Volume by solid of revolution is somehow tricky techniques to do. So we can also try the online tools like washer method formula calculator and also volume of a disc calculator because the disc method is also the one of the valueable method for finding the volume of solid of revolution.
## Conclusion
As we know the washer method and shell method both apply in the calculations. But the uses of both methods are vital and beneficial method of integration. Sometimes, it is best to use the washer method in case of finding volume of solid revolutions and sometimes the shell method works more efficiently. Either way, both methods are very useful and widely used in finding the volume of solid revolutions. We hope you liked this article, do find other articles in the blog section. |
# Algebra: Solving Rational Equations
## Solving Rational Equations
Have you ever noticed that many mathematicians don't have great social skills? (If you haven't noticed, it must be because you don't know many mathematicians.) There's an old joke that goes "How can you spot an extroverted mathematician? He talks to your shoes instead of his."
Why is it that some math people can't deal with the outside world? Is it the bright daylight that sears their pupils behind their thick glasses, or the wedgies they endured in school? I don't think either is to blame. I think it's because they're used to a world completely and totally under their control. You see, in the math world, if you don't like something, you can manipulate the rules of the universe and make any unpleasantness disappear.
##### Kelley's Cautions
If you multiply an equation by something containing an x, you may be adding incorrect solutions. Always check to make sure any answer you get can be plugged into the original equation. (In other words, make sure none of the denominators become 0 when you plug your answers in for x.)
Case in point: Most mathematicians dislike fractions. It's not because they don't understand them, it's just that the constant need of common denominators is annoying. Therefore, whenever the opportunity arises, math people will completely eliminate fractions from their landscape. For instance, you can very easily eliminate every fraction from an equation just by multiplying everything in that equation by its least common denominator.
Example 1: Solve the equation.
Solution: This equation contains three rational expressions; your goal will be to eliminate all of those fractions to make the equation much simpler to solve. Start by factoring any expression you can in the equation. (In this case, the quadratic can be factored.)
The least common denominator of all three fractions is (x + 5)(x + 2). If you multiply the entire equation by that expression, the fractions will disappear. (Technically speaking, you'll multiply the equation by (x + 5)(x + 2)1, which is the exact same expressionit just reminds you to multiply the least common denominator by each individual fraction's numerator while leaving the denominators unchanged.)
You can simplify all of those fractions.
Notice that each of those denominators have been eliminated, and are technically now all equal to 1. However, there's no need to write a denominator of 1, so you can rewrite the equation using only the numerators.
• (x + 2) + (x + 5)x = 2x - 1
Distribute the x in the second term and combine all like terms (by setting the equation equal to 0).
• x + 2 + x2 + 5x = 2x - 1
• x2 + 6x + 2 = 2x - 1
• x2 + 4x + 3 = 0
Hey! There's a plain old quadratic equation left over that you can solve by factoring.
• (x + 3)(x + 1) = 0
• x = -3 or x = -1
If you plug both of those solutions into the original equation, you get true statements, so they are both valid answers.
Check x = -3
Check x = -1
##### You've Got Problems
Problem 1: Solve the equation x + 3x - 8 + xx2 - 6x - 16 = 1.
Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc. |
What is 2.9 as a Fraction?
2.9 as a fraction is 29/10. It can be reduced to 2 9/10 or to a mixed number of 2 3/10.
Checkout this video:
Introduction
When we talk about fractions, we are usually referring to a part of a whole, such as 1/2 or 3/4. A fraction is made up of two numbers: a numerator and a denominator. The numerator is the number on top of the fraction and the denominator is the number on the bottom of the fraction. In the fraction 1/2, for example, 1 is the numerator and 2 is the denominator.
What is 2.9 as a Fraction?
2.9 as a fraction is 2 9/100. In other words, it is two and nine-hundredths. When we convert a decimal like 2.9 to a fraction, we place the decimal point in between the 2 and the 9, and then count how many digits there are to the right of the decimal point. In this case, there are two digits. So, we put a 2 over a hundred (10^2). The final answer is 2/100, or 2%.
Decimal to Fraction
2.9 as a fraction is 29/10. To convert 2.9 to a fraction, divide 2.9 by 1 and then simplify the resulting fraction.
Fraction to Decimal
Converting a decimal to a fraction is a three-step process. The first step is to determine if the decimal is a terminating or repeating decimal. A terminating decimal is one where the digits after the decimal point all stop, such as 0.5, 0.75, 1.25 and so forth. A repeating decimal, on the other hand, has digits that go on forever after the decimal point, such as 0.33 or 1.66. You can tell if a decimal is repeating by seeing if any digit after the decimal point repeats itself over and over in a pattern. For example, in 0.333333333…, the “3” repeats itself indefinitely. In 1/3, on the other hand, there is no number that repeats indefinitely after the decimal point (although 1/3 = 0.333333333…).
To convert a terminating decimal to a fraction, count the number of digits that are to the right of the decimal point and put that many zeroes after the one in your fraction (or write out as many zeroes as there are digits). So 0.5 would be written as 1/2 (one-half), and 1.25 would be written as 1/4 (one-quarter).
To convert a repeating decimal to a fraction, you have to use algebraic methods (multiply both sides of an equation by some number so that the pattern stops repeating). For example:
0.333333333… = 3/9 because 3 x 0.333333333… = 1
1/3 = 3/9 because 1 x 3/9 = 3/9
Conclusion
To sum it up, 2.9 as a fraction is 29/10. This can be reduced to 2 9/10 or even just 2 9, which is read as “two point nine.” When writing fractions, it is industry practice to put the fraction in subscript after the whole number, like so: 2.9 = 2 9/10. |
# Thread: Changing the subject of the formulae
1. ## Changing the subject of the formulae
Please may someone show me the steps to changing the subject for the following questions, the subject needs to be y for each one.
1) h-y
h+y = k
2) y+3 d
y-2 = e
2. ## Re: Changing the subject of the formulae
Sorry everything has been moved to the left a little.
3. ## Re: Changing the subject of the formulae
Hello, joe345!
$\text{(1) Solve for }y\!:\;\frac{h-y}{h+y} \:=\:k$
Multiply by $(h+y)\!:\;h-y \:=\:k(h+y)$
Expand: . . . . . . . . $h-y \:=\: kh + ky$
Get all y-terms to one side:
. . . . . . . . . . . $\text{-}ky - y \:=\:kh - h$
Factor: . . . . $\text{-}(k+1)y \:=\:h(k-1)$
Divide by $\text{-}(k+1)\!:\;\:y \:=\:\frac{h(k-1)}{\text{-}(k+1)}$
Multiply by $\tfrac{\text{-}1}{\text{-}1}\!:\qquad\; y \:=\:\frac{h(1-k)}{1+k}$
[The last step is not necessary, but the answer is neater.]
$\text{(2) Solve for }y\!:\;\frac{y+3}{y-2} \:=\:\frac{d}{e}$
Cross-multiply: . $e(y+3) \:=\:d)y-2)$
Expand: . . . . . . $ey + 3e \:=\:dy - 2d$
Get y-terms on one side:
. . . . . . . . . $ey - dy \:=\:\text{-}3e -2d$
Factor: . . . $(e-d)y \:=\:\text{-}(3e+2d)$
Divide by $(e-d)\!:\; y \:=\:\frac{\text{-}(3e+2d)}{e-d}$
Multiply by $\tfrac{-1}{-1}\!:\quad\; y \:=\:\frac{2d+3e}{d-e}$
4. ## Re: Changing the subject of the formulae
Thankyou very much |
# How do you simplify sqrt(-9)?
Mar 25, 2018
The expression is equal to $3 i$.
#### Explanation:
$\sqrt{\textcolor{red}{a} \textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}}$
$\sqrt{{\textcolor{red}{a}}^{2}} = \textcolor{red}{a}$
And also the definition of the imaginary number:
$\sqrt{- 1} = i$
Here are these properties applied to our expression.
$\textcolor{w h i t e}{=} \sqrt{- 9}$
$= \sqrt{- 3 \cdot 3}$
$= \sqrt{- 1 \cdot 3 \cdot 3}$
$= \sqrt{- 1 \cdot {3}^{2}}$
$= \sqrt{- 1} \cdot \sqrt{{3}^{2}}$
$= \sqrt{- 1} \cdot 3$
$= i \cdot 3$
$= 3 i$
This is the result. Hope this helped!
Mar 25, 2018
$\sqrt{- 9} = 3 i$
#### Explanation:
Given:
$\sqrt{- 9}$
$\sqrt{{3}^{2} \times - 1}$
Apply rule: $\sqrt{{a}^{2}} = a$, and $\sqrt{- 1} = i$.
$3 i$ |
# 13.2. Properties of Covariance¶
Let’s examine how covariance behaves. In the next two sections we will use our observations to calculate variances of sample sums.
Establishing properties of covariance involves simple observations and routine algebra. We have done some of it below, and we expect that you can fill in the rest.
Recall that the covariance of $$X$$ and $$Y$$ is
$Cov(X, Y) ~ = ~ E(D_XD_Y) ~ = ~ E[(X - \mu_X)(Y - \mu_Y)]$
## 13.2.1. Constants Don’t Vary¶
That title has a “duh” quality. But it’s still worth noting that for any constant $$c$$,
$Cov(X, c) = 0$
## 13.2.2. Variance is a Covariance¶
Covariance is an extension of the concept of variance, because
$Var(X) = E(D_X^2) = E(D_XD_X) = Cov(X, X)$
The variance of $$X$$ is the covariance of $$X$$ and itself.
## 13.2.3. Covariance is Symmetric¶
Clearly $$Cov(Y, X) = Cov(X, Y)$$. It follows that
$Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y) = Var(X) + Var(Y) + Cov(X, Y) + Cov(Y, X)$
This way of thinking about the variance of a sum will be useful later.
## 13.2.4. Covariance and Expected Products¶
Covariance is an expected product: it is the expected product of deviations. It can also be written in terms of the expected product of $$X$$ and $$Y$$, as follows.
\begin{split} \begin{align*} Cov(X, Y) &= E[(X - \mu_X)(Y - \mu_Y)] \\ &= E(XY) - E(X)\mu_Y - \mu_XE(Y) + \mu_X\mu_Y \\ &= E(XY) - \mu_X\mu_Y \end{align*} \end{split}
So covariance is the mean of the product minus the product of the means.
Set $$X = Y$$ in this result to get the “computational” formula for the variance as the mean of the square minus the square of the mean.
This result simplifies proofs of facts about covariance, as you will see below. But as a computational tool, it is only useful when the distributions of $$X$$ and $$Y$$ are very simple – for example, when each has just a few possible values. In other calculations of covariance it is rarely a good idea to try to use this result. Rather, we will use the bilinearity property described at the end of this section.
Quick Check
Let $$(X, Y)$$ be one point picked at random from the four points $$(0, 0)$$, $$(1, 0)$$, $$(0.5, 1)$$, $$(1.5, 1)$$.
(a) Find $$E(XY)$$.
(b) Find $$Cov(X, Y)$$.
## 13.2.5. Independent Implies Uncorrelated¶
Let $$X$$ and $$Y$$ be independent. Then
\begin{split} \begin{align*} E(XY) &= \sum_x\sum_y xyP(X=x, Y=y) ~~~~~~ \text{(expectation of a function)} \\ &= \sum_x\sum_y xyP(X=x)P(Y=y) ~~~~ \text{(independence)} \\ &= \sum_x xP(X=x) \sum_y yP(Y=y) \\ &= E(X)E(Y) \end{align*} \end{split}
Therefore if $$X$$ and $$Y$$ are independent, then $$Cov(X, Y) = 0$$. We say that $$X$$ and $$Y$$ are uncorrelated.
To summarize, independent random variables are uncorrelated. But it is not true that uncorrelated random variables have to be independent.
A routine application of the calculation of covariance using the expected product shows that for any random variables $$X$$, $$Y$$, and $$Z$$,
$Cov(X+Y, Z) ~ = ~ Cov(X, Z) + Cov(Y, Z)$
Just write $$Cov(X+Y, Z) = E[(X+Y)Z] - E(X+Y)E(Z)$$, expand both products, and collect terms.
Quick Check
Let $$X$$ and $$Y$$ be independent and suppose $$Var(X) = 10$$. Find $$Cov(X, X+Y)$$.
## 13.2.7. The Main Property: Bilinearity¶
This property is the key to calculating covariance. First, easy algebra shows that for constants $$a$$ and $$b$$,
$Cov(aX, bY) = abCov(X, Y)$
Put this together with the addition rule to get
$Cov(aX + bY, cZ) = acCov(X, Z) + bcCov(Y, Z)$
You can see that covariance behaves like products. By induction,
$Cov(\sum_{i=1}^n a_iX_i, \sum_{j=1}^m b_jY_j) ~ = ~ \sum_{i=1}^n\sum_{j=1}^m a_ib_jCov(X_i, Y_j)$
That might look intimidating, but in fact this property greatly simplifies calculation. It says that you can expand covariance like the product of two sums. For example,
$Cov(10X - Y, 3Y + Z) = 30Cov(X, Y) + 10Cov(X, Z) - 3Cov(Y, Y) - Cov(Y, Z)$
You can replace $$Cov(Y, Y)$$ by $$Var(Y)$$.
Quick Check
Let $$Var(X) = 4$$, $$Var(Y) = 5$$, and $$Cov(X, Y) = 3$$. Find $$Cov(2X - Y + 7, X + 3Y - 10)$$. |
# Fraction calculator
This calculator adds two fractions. First, all fractions are converted to a common denominator when fractions have different denominators. Find the Least Common Denominator (LCD) or multiply all denominators to find a common denominator. When all denominators are the same, subtract the numerators and place the result over the common denominator. Then, simplify the result to the lowest terms or a mixed number.
## The result:
### 3/8 + 3/7 = 45/56 ≅ 0.8035714
The spelled result in words is forty-five fifty-sixths.
### How do we solve fractions step by step?
1. Add: 3/8 + 3/7 = 3 · 7/8 · 7 + 3 · 8/7 · 8 = 21/56 + 24/56 = 21 + 24/56 = 45/56
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(8, 7) = 56. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 8 × 7 = 56. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - three eighths plus three sevenths is forty-five fifty-sixths.
### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right. |
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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# Probability of Dependent Events - Concept
Carl Horowitz
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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When calculating the probability of multiple events, we must determine if the events are dependent or independent of one another. When calculating the probability of dependent events we must take into account the effect of one event on the other. An example of calculating the probability of dependent events is the probability of drawing two specific cards in a row with the second card being drawn from a smaller deck.
Often times we're finding the probability of events sometimes they are independent events where one event doesn't affect another one and often times they will also be dependent where one event does affect the other one. So we're going to look at this problem and sort of compare and contrast the difference between independent and dependent. Okay so what we're doing is we are grabbing marbles out of a bag we have 6 green and 4 blue and we're trying to find the probability you draw a green and then a blue.
Okay so the first thing we're going to do is what's called with replacement so we're taking out a marble and then we're putting it back in. Okay so the first draw we want to grab a green, we have 6 green marbles out of a 10 total. So the probability of drawing a green is going to be 6 out of 10 okay, we put that marble back because we're dealing with replacement and then we want to grab a blue. So when we grab our blue we have 4 blues out of 10 because we put it back. So now we just have a 4 tenth chance of that blue, to find the probability of both occurring we just multiply and end up with 24 over 100, that could be simplified but I'm not terribly concerned with the actual numeric value, just want the concept of what's going on. Okay so that is with replacement and those are independent events, it didn't matter that I drew this green first the probability of blue is still the same thing no matter what.
Okay without replacement so now I am taking a marble and I'm not putting it back in. Okay so we still need to grab that green first, there's still 10 marbles we're grabbing 1 there's 6 greens so the probability is still six tenth that we grab that green. We don't put that back so now instead of dealing with 10 marbles we're now dealing with 9. We want to grab a blue therefore a blue's probability is now just four ninths that we grab that blue multiply these probabilities together we end up with 24 out of 90. So our probability has increased because we didn't put that marble back. Okay, this is a dependent situation, this without replacement because where probability changes for the second marble depending on what happened with the first okay.
There are some formulas for dependent probability I tend to find them confusing, I tend to just sort of think about them logically and sort of look at the pool you're choosing from, first the outcome that you want. Okay, you can do a lot of these with t tree diagrams sort of diagrams okay what's going to happen here, here probability of each branch set off to another in general that will almost always give you the right answer it is just going to create a little bit more work than just thinking about it logically.
Okay so dependent events when one event varies the outcome of another just think about how they impact each other and how the probabilities change given that first event. |
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