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Definitely Aluminum. Then glass (unless much thicker than the plastic in question). Then plastic.
FV = PV(1 + i%)^(t) .... t would equal 4y in this case ... y as in years.
4500(1.07)^16 = 13284.73687
4000(1.07)^12 = 9008.766356
4500(1.07)^8 = 7731.837809
4000(1.07)^4 = 5243.18404
Oleg owes 35268.53
That is why no one should take a loan that is compounded quarterly!
The first one isn't bad.
P = ((pi r)/2 + 2h + w) ; w = 2r ; P = (pi r + 4h + 4r) / 2
A = (pi r^2)/2 + hw ; A = (pi r^2 + 4rh) / 2
Solve P for h and place in area function
h = 2P - 4r - pi r
A = (pi r^2 + 8Pr - 16 r^2 - 4pi r^2) / 2 = (-3 pi r^2 - 16 r^2 + 8Pr) / 2
dA/dr = (-6 pi r -32r + 8P) / 2
dA/dr = 0 when 8P = r(32 + 6 pi), r = 8P / (32 + 6 pi) ≈2.157
r = 2.157, w = 5.034, h = 16.595 (total height is 16.595 + 2.157)
That's not how I did it John. I just observed that if AB had an angle of arctan 14/5 above the horizontal then line C would have that same angle from the y axis or vertical because of the 90 degree angle between them. Constructing that triangle using C as the hypotenuse is how I used the trigonomic functions. Notice that the opposite side of this angle is horizontal and thus represents the change in x, hence the use of the sine function for the x position.
It is much worse than that! Because I flipped the x and y positions for the original points my slope was off also. In corrected form:
x = 15 + 5 sin arctan (14/5) = 19.708
y = 5 - 5 cos arctan (14/5) = 3.318
Thanks for catching the mistake though. (blush)
You can't learn physics without calculus. You would just be memorizing formulas otherwise. Then you would only be able to solve problems that have already been figured out. It would be like an "artist" who could only paint one picture again and again.
Math is a tool to communicate the logical progression of conceptual ideas. It is like any others science a figmant of the collective conscience. It is useless without a particular subjective viewpoint and set of assumptions. Gravity is no more of a true force than 1+1=2 unless we all agree to say it is so.
arctan slope of line AB = angle of the line AC from the vertical because of the 90 degree angle between AB and AC.
So x = 5 + 5 sin arctan (5/14) ≈ 6.68
y = 15 - 5 cos arctan (5/14) ≈ 10.29
Following your reasoning the height and radius are related to the original height and radius so no further manipulation is needed. However I came up with this:
A = square root of [(6561 V^2 / 676 pi r^3) + r^2]
Don't worry too much about the values I used, I kept it exact instead of rounding, but I think that you failed to square the term for h in the area formula.
Sadly, I doubt that this is what your teacher is looking for anyway.
This relationship between radius, height, and side(slant height) can not exist, at least not in a cone. If your height and radius were 1/3 the side would be the square root of 2/9 or ≈.471
I have tried to follow this thread and your previous posts and have still not figured out exactly what the original problem was. It seems that you have been going along for a while now on this problem.
I think that if you posted a more concise definition of the original problem that you would be pleasantly surprised by a more simple solution. It sounds like some type of basic related rate problem which ususally can be solved quite simply with all of the relevant information from the beginning.
Forgive me if I sound like some type of a-hole, but I think that you somehow made this problem more difficult than it truly needs to be. I, and I suspect others here, would love to help you, but your problem never seemed to be exactly clear from the beginning.
Sorry, my fingers are slower than my head.
.93968
This took four iterations using Newton's Method.
x - f(x)/f'(x)
The c should solve the equation at time zero. If your coordinate system is inverse to the data you should add pi to the previous c to change the sign of the position.
Perpendicular lines slopes always share a negative inverse relationship.
Mathematically this means that the product of the two slopes always equals negative one.
m1(m2) = -1
A linear equation can always be written as mx + b = y
You were on the right track to solving this problem.
Solve for y for the original equation: y= -2/3 x -2/3
Therefore the slope is -2/3
If the second line is perpendicular then -2/3 m2 = -1, so m2 = 3/2 or 1.5
The second equation must then be y = 1.5x + b
Plugging in the point specified by your question: 7 = 1.5(4) + b
Solving for b and keeping the slope give you your perpendicular line equation
y = 1.5x + 1
I would like to throw mathsyperson a bone here for being a good sport. I read an article proposing that the myth was indeed true. An experiment was conducted where hotter water was made to freeze faster than cold water, but the conclusion made by most was false. The final volumes were not the same. What in fact did occur was that a smaller amount of hot water froze faster than a larger amount of cool water. I'll pat myself on the back for suggesting this earlier.
So the myth is partly true, but it has nothing to do with temperature and everything to do with total energy.
I think that you meant:
y = (pi^2 r^6 + 4100625)^(1/2) / r - (pi^2 r^6 + 36905625)^(1/2) / 3r
There are no shortcuts here and the answer will not be pretty. It is easier if you treat each part of this equation as a separate function and then add them back together after differentiating.
We will call the first part A and the second part B. Also the constants have no effect upon the outcome so we will call them a and b respectively.
A = (pi^2 r^6 + a)^(1/2) / r = [(pi^2 r^6 + a) / r^2)]^(1/2)
dA/dr = {1/2 [(pi^2 r^6 + a) / r^2]^(-1/2)} {[(6 pi^2 r^5)r^2 - 2r(pi^2 r^6 + a)] / r^4}
dA/dr = (6 pi^2 r^7 - 2 pi^2 r^7 - 2ar) / [(4 pi^2 r^14 + 4a r^8) / r^2]^(1/2)
dA/dr = (4 pi^2 r^7 - 2ar) / [4 pi^2 r^12 + 4a r^6]^(1/2)
da/dr = (4 pi^2 r^7 - 2ar) / 2 r^3 [(pi^2 r^6 + a)^(1/2)]
da/dr = (2 pi^2 r^6 - a) / (pi^2 r^10 + a r^4)^(1/2)
B = (pi^2 r^6 + b)^(1/2) / 3r = [(pi^2 r^6 + b) / 9 r^2]^(1/2)
db/dr = {1/2[(pi^2 r^6 + b)/(9 r^2)]^(-1/2)}{[6 pi^2 r^5(9 r^2) - 18r(pi^2 r^6 + b)]/81 r^4}
db/dr = (54 pi^2 r^7 - 18 pi^2 r^7 + 18br)/[54 r^3(pi^2 r^6 + b)^(1/2)]
db/dr = (36 pi^2 r^7 + 18br) / [54 r^3(pi^2 r^6 + b)^(1/2)]
db/dr = (2 pi^2 r^6 + b) / [3 r^2 (pi^2 r^6 + b)^(1/2)]
whew....
dy/dr = {(2 pi^2 r^6 - a) / (pi^2 r^10 + a r^4)^(1/2)}
- (2 pi^2 r^6 + b) / [3 r^2 (pi^2 r^6 + b)^(1/2)]
This is why you have not gotten any answers. It is too long a computation just for the sake of argument. I have just found the rate of change for an unknown function. It is pretty dissatisfying.
Perhaps you could tell us what the function represented.
I think that you are just a little bit confused on this one. The derivative of e^x is e^x, but that is because you are differentiating a variable exponent. There are a lot of derivatives associated with logarithmic and exponential functions.
The derivative of e^2 would be 2e because the exponent was a constant and you could use the x^n = nx^(n-1) rule. This rule is used when a variable is raised to a constant power. Play close attention to what the variable is in the equation. Look around the internet for rules for differentiation of logarithmic functions and you will see where you went wrong.
I'd give it a stab also, but I too am unable to imagine this figure. Did you say an INSIDE angle of greater than 180 within a QUADRANGLE? And did you mean AB*BC = CD*DA? Sorry, but I am a little confused.
Technically speaking, setting the first derivative equal to zero only finds local extrema. To be sure that the minimum or maximum is truly the global extrema one needs to find the limit of the function at its end points.
If you do this in this case it indeed proves that the local minimum found is global also. There is no need to test in the negative direction because negative units do not exist, but....
lim 800/x + .04 + .0002x = + ∞
x⇒+∞
Many people forget to do this, but many functions continue to change after a local extrema and unless the value changes direction again the first derivative will not point it out.
If you look inside the brackets of the numerator the sum within them is raised to the one-half power which is the same thing as taking the square root but just different notation.
I couldn't resist this one. Heavier objects do fall faster than lighter ones in many examples. If two objects have the exact same shape and size but differ in density, the denser object will indeed fall faster. Drop a full bottle of water and an empty bottle of water from the second floor of your home and this will be quite evident. Only in a vaccuum do objects of different masses and shapes fall at the same rate.
Sadly, hot water does not freeze faster than cold water either. Temperature difference does greatly affect the rate of heat transfer because of equilibrium laws, but at some point the temperature differences of the hot versus cold water would be at comparable rates and thus would be cooling at similar rates. There could be an exception or rather "trick" in what you propose. I could easily freeze say a gallon of boiling water faster than five gallons of 40 degree cold water. This is because of the total energy within these samples. And unless someone shows you equal volumes of ice after such an experiment the experiment is faulty.
I came up with the same answer, but was a little puzzled at mathsyperson's approach. I am always amazed at the different approaches individuals will take. I did it like this.
3ln(-3x+5) = -9 divide by 3
ln(-3x+5) = -3 raise to e
-3x+5 = e^-3 subtract 5
-3x = e^-3 - 5 divide by -3
x = (e^-3 - 5) / -3
I realize that I am a little late on this one, but I did want to make a point. Mathyperson is absolutely correct in his approach, however the nature of this question suggests that the asker has limited understanding of fractions at this point. I would offer the additional explanation of:
"1 and 1/3" is a mixed fraction meaning that it is a composition of a whole number and a fraction.
To add or subtract with fractions all terms must be in fractional form and have identical denominators. The top number of the fraction is called the numerator and the bottom term is called the denominator. 1 1/3 is pronounced "one and one-third". To convert this mixed fraction into a regular fraction you need to multiply the whole number times the denominator then add the numerator and use this answer as the new numerator atop the previous denominator. Easier done than said.
1 1/3 = ( 1 X 3 + 1 ) / 3 = 4/3 or four-thirds
To subtract 7/8 from 4/3 you must make both denominators the same. As Mathsyperson stated the lowest common denominator is 24. There are various ways in which to determine this, but the simplest is simply to ask first if the smaller denominator goes into the larger one evenly. If it does, then you only have to alter one fraction by multiplying both the numerator and denominator of the smaller one by that amount. If it does not, then you will have to determine what the smallest number that both denominators go into evenly and then multiply each fraction appropriately.
The two denominators are 3 and 8. Since three does not go into eight evenly you have to try multiples of 8. 3 does not go into 16 either, but it does go into 24. So both denominators must be converted to 24 and whatever you do to the denominator you must also do to the numerator or you will change the ratio of what the fraction represents. Again, this is all much easier to do than to explain.
4/3 - 7/8 = 32/24 - 21/24 = 11/24
Sorry if this was a little long, but I just wanted to be clear in case you were unsure about other aspects of this problem.
I would personally use the quadratic equation for this problem:
-x^2 + 4x + 5
The quadratic equation would yield the same thing without trying different combinations.
[ -b +/- (b^2 - 4ac)^(1/2)] / 2a = [-4 +/- (16 + 20)^(1/2)] / -2, so x = 5 and -1
Maybe I am biased, but this method always seemed easier than figuring out values by trial and error.
Your question though was specifically about the "textbook" method of factoring. I would describe it as follows.
You want your answer to be like: (ax + b)(cx + d). This equals (ac)x^2 + (ad + bc)x + (bd).
So you are really concerned with the coefficients in the original equation.
In this case: ac = -1, ad + bx = 4, and bd = 5. b and d are the second terms, so just ask yourself what two numbers multiply to produce 5. The simple answer leaves only one and five. (and if you later find that you need fractional coefficients, God help you.)
( + 1)( + 5).......they both must be postitive otherwise the constant would be negative.
As ac = -1 and a and c represent the first terms just plug in a one into both brackets. Remember that one of these terms must be negative!
(1x +1)(1x +5)
The signs within the brackets come from the ad + bc term that produces the x^1 term in your original equation. So ad = 5 and bc = 1 and these two answers must add up to 4. The only way that a 5 and a 1 can equal 4 is if the 1 is negative. Therefore the bc term must be negative and b is already a positive one so c must be negative.
(x+1)(-x+5)
Remember the (ac)x^2 + (ad + bc)x + bd form of the equation to keep all of this straight in the harder factoralizations.
Now you see why I almost always use the quadratic equation to find the zeros of a 2nd degree equation. Just plug in the coefficients and your done.
1 inch stone reaches terminal velocity at about 23.93m/s (m=18.9g)
4 inch stone reaches terminal velocity at about 47.86m/s (m=1.2kg)
Our example would suggest that this stone would be traveling at 79.2m/s
A 10.9 inch rock with a mass of 24.74kg would be necessary to achieve this velocity!
I would not recommend throwing a 55 pound anything from that height.
I thought that I would drop a line to show how dramatic air resistance is in everyday situations.
My contribution for anyone that cares:
terminal velocity = {[8gr(density of object)] / 3(density of medium)}^(1/2)
where: g = -9.8m/s
r = radius of object
object must be smooth and round for this formula
The above stone calculations were made with these densities.
density of object = 2200kg/m^3 (based on the average density of the earth's crust)
density of air = 1.275kg/m^3 (sea level)