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# find the percent difference between 14.64 and 14.78

percent difference between 14.64 and 14.78

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### The percentage difference between 64.24% and 74.24% - Math ...

What is the percentage difference between 64 ... The short answer is 74.24% is 10 percentage points larger than 64 ... You can calculate the "percentage difference ...

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## Suggested Questions And Answer :

### Polynomial and Linear (?)

A. Domain is 0≤x≤100 because x is a percentage. However, if c(x)≥0 then there must be a minimum value for x. This implies there is no cost for removing less than the minimum percentage of pollutants. If c(x)=50x^2-100x-4900, and c(x)=0, x^2-2x-98=0, x^2-2x+1=99; (x-1)^2=99 and minimum x=1+√99=1+3√11=10.95 approx, making the domain 1+3√11≤x≤100 so that 0≤c(x)≤485100 (pesos). For removing less than 10.95% of pollutants there is no cost. B. If c(x)=50x^2-100x-4900, then putting x=50, 70, 90 and 99.5 into the function we get: c(50)=115100 pesos; c(70)=233100 pesos; c(90)=391100 pesos; c(99.5)=480162.5 pesos. c(x) would be negative if x=0 which suggests perhaps the function is meant to read differently or have the domain specified in A. If the formula for c(x) has not been correctly interpreted, please resubmit your question with a clearer representation of c(x). However, I hope the method shown is helpful to you.

### Interpolate the data set (1, 150), (3, 175), (4, 185), (6, 200), (8, 300) to estimate the amount of money Gracie may earn if she displays her items for 7 hours

Since 7 is halfway between 6 and 8, Gracie should earn an amount about halfway between 200 and 300, that is, 250 (triangular interpolation). This is the simplest interpolation, but see later. Interpolating for 2 and 5 we get 162.5 and 192.5, that is, respectively, a difference of 12.5 (162.5-150 or 175-162.5) and 7.5 (192.5-185 or 200-192.5), while 250 is a difference of 50 from 200 and 300. So the interpolated figures fluctuate. For a more sophisticated approach, we need to take the whole dataset and look for a formula that best fits. One way to do this is to fit a polynomial F(x)=ax^4+bx^3+cx^2+dx+e into the five given points. This polynomial has 5 unknown coefficients, so with 5 simultaneous equations we should be able to find them. The process can be simplified slightly by taking the lowest "x" coord and using that as the zero starting point. In this case the lowest coord is 1 (hour) so we subtract 1 from the first coord of each pair to get: (0,150), (2,175), etc. F(0)=e=150. So we have the constant 150. The next step is to subtract 150 from each of the other "y" coords so we arrive at the following set of equations: (1) F(2)=16a+8b+4c+2d=25 (2) F(3)=81a+27b+9c+3d=35 (3) F(5)=625a+125b+25c+5d=50 (4) F(7)=2401a+343b+49c+7d=150 and we already have F(0)=150=e. We can now eliminate d from (1) and (2): 2F(3)-3F(2): (162-48)a+(54-24)b+(18-12)c=70-75=-5; (5) 114a+30b+6c=-5. and we can eliminate d from (3) and (4): 5F(7)-7F(5): (12005-4375)a+(1715-875)b+(245-175)c=750-350; 7630a+840b+70c=400 which simplifies to (6) 763a+84b+7c=40 or 109a+12b+c=40/7 We can eliminate c between (5) and (6): 6(6)-(5): (654-114)a+(72-30)b=240/7+5=275/7; (7) 540a+42b=275/7 or 90a+7b=275/42. So b=(275/42-90a)/7. From (6) we have: 109a+12b+c=109a+12(275/42-90a)/7+c=40/7, so c=40/7-109a-12(275/42-90a)/7; c=40/7-550/7+(1080/7-109)a=-510/7+317a/7=(317a-510)/7. We now have b and c in terms of a. We can continue to find d in terms of a. From (1) d=(25-16a-8b-4c)/2=25-16a-8(275/42-90a)/7-4(317a-510)/7; d=25-1100/147+2040/7+(-16+720/7-1268/7)a= (3675-1100+42840)/147+(-112+720-1268)a/7; d=45415/147-660a/7. We have b, c and d in terms of a, so we can find a by substituting into an equation containing all four coefficients (but not (1), because we used it to find d). Let's pick (2) and hope we get a sensible result! 81a+27(275/42-90a)/7+9(317a-510)/7+3(45415/147-660a/7)=35. From this a=5143/2772=1.855. Therefore b=-22.919, c=78.510, d=-67.687, e=150. And F(x)=1.855x^4-22.919x^3+78.510x^2-67.687x+150. This results need to be checked before we use F to find an interpolated value. Unfortunately, this polynomial approach produces inconsistent results, and needs to be discarded. Lagrange's method seems the obvious choice, even if it is tedious to do. We have 5 x values which we'll symbolise as x0, x1, x2, x3, x4 and 5 function values f0, f1, f2, f3, f4. If the function we're looking for is f(x) then: f(x)=(x-x1)(x-x2)(x-x3)(x-x4)f0/((x0-x1)(x0-x2)(x0-x3)(x0-x4))+         (x-x0)(x-x2)(x-x3)(x-x4)f1/((x1-x0)(x1-x2)(x1-x3)(x1-x4))+         (x-x0)(x-x1)(x-x3)(x-x4)f2/((x2-x0)(x2-x1)(x2-x3)(x2-x4))+... x0=1, x1=3, x2=4, x3=6, x4=8; f0=150, f1=175, f2=185, f3=200, f4=300. We want x=7, so f(7) is given by: 4.3.1.-1.150/(-2.-3.-5.-7)+6.3.1.-1.175/(2.-1.-3.-5)+ 6.4.1.-1.185/(3.1.-2.-4)+6.4.3.-1.200/(5.3.2.-2)+ 6.4.3.1.300/(7.5.4.2) This comes to: -60/7+105-185+240+540/7=1600/7=228.57 (229 to the nearest whole number) compared with 250 from the simple interpolation.

### How to find equation of the circle passing through (9,1), (-3,-1) and (4,5)

The equation of a circle is more specific: (x-h)^2+(x-k)^2=a^2, where (h,k) is the centre and a the radius. Plug in the points: (9-h)^2+(1-k)^2=(-3-h)^2+(-1-k)^2=(4-h)^2+(5-k)^2=a^2 This can be written (9-h)^2+(1-k)^2=(3+h)^2+(1+k)^2=(4-h)^2+(5-k)^2=a^2 Leaving a out of it for the moment we can use pairs of equations, using difference of squares: 12(6-2h)+2(-2k)=0, 36-12h-2k=0, 18-6h-k=0, so k=18-6h. 7(-1-2h)+6(-4-2k)=0, -7-14h-24-12k=0, -31-14h-12k=0 or 31+14h+12k=0, 31+14h+12(18-6h)=0, 31+14h+216-72h=0. 247-58h=0 so h=247/58 This looks suspiciously ungainly. So rather than continuing, I'm going to call the three points: (Q,R), (S,T), (U,V) to provide a general answer to all questions of this sort. (Q-h)^2+(R-k)^2=(S-h)^2+(T-k)^2=(U-h)^2+(V-k)^2=a^2 [(equation 1)=(equation 2)=(equation 3)=a^2] Take the equations in pairs and temporarily ignore a^2. Equations 1 and 2: (Q-S)(Q+S-2h)+(R-T)(R+T-2k)=0 Q^2-S^2-2h(Q-S)+R^2-T^2-2k(R-T)=0 2k(R-T)=Q^2+R^2-(S^2+T^2)-2h(Q-S), so k=(Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T)) Equations 2 and 3: S^2-U^2-2h(S-U)+T^2-V^2-2k(T-V)=0, so k=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) At this point, we can substitute for k and end up with an equation involving the unknown h only: (Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T))=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) (T-V)(Q^2+R^2-(S^2+T^2)-2h(Q-S))=(R-T)(S^2+T^2-(U^2+V^2)-2h(S-U)) (T-V)(Q^2+R^2-(S^2+T^2))-2h(T-V)(Q-S)=(R-T)(S^2+T^2-(U^2+V^2))-2h(R-T)(S-U) 2h((R-T)(S-U)-(T-V)(Q-S))=(R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)) h=((R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)))/((R-T)(S-U)-(T-V)(Q-S)). Once h is found we can calculate k, and then we can substitute the values for h and k in any equation to find a^2 which is equal to each of the three equations. When we use Q=9, R=1, S=-3, T=-1, U=4, V=5 or -5, we appear to get very ungainly solutions. One way to find which values need to be changed may be to plot the values and work out where the circle should fit and where its centre has less complex values. If V=6 or -6, there is a simple solution: (x-3)^2+y^2=37 (centre at (3,0), a^2=37=6^2+1^2. (-3-3)^2=(9-3)^2=6^2; (-6)^2=6^2; (2-3)^2=(4-3)^2=1; (-1)^2=1^2 shows the combination of points that would lie on the circle: (9,1), (9,-1), (-3,1), (-3,-1), (4,6), (4,-6), (2,6), (2,-6) and this includes points A and B, but not C. Note that the equation k=18-6h is valid for h=3, k=0.

### Find the distance between circles centers

The general equation of a circle radius a is (x-b)^2+(y-c)^2=a^2, and the centre is at (b,c). Completing the square, y^2-2y+1=(y-1)^2=20. There is no x term so b=0 while c=1. The centre is at (0,1). The second circle equation can be written (x+1)^2+(y-2)^2-6=0 because x^2+2x+1+y^2-4y+4-6=x^2+y^2+2x-4y-1=0. So the centre of the second circle is at (-1,2). The y difference between the two centres is 2-1=1 while the difference between the x values is 1, so the difference between the centres is sqrt(1^2+1^2)=sqrt(2).

### how to find equation of elips?

We start with a line AB of length p. Now we imagine a length of string s>p. The string is attached to the line by fixing one end to A and the other to B. Now we attach a pointer or marker to the string using a very small loop so that the pointer can move along the length of the string, keeping it taut. The pointer starts at a point that is an extension of AB, and the pointer is moved out of the line AB to describe the ellipse on both sides of AB, until the pointer returns to its starting position. The geometry is that, if the pointer's position at any time is P, AP+BP=s. The midpoint of AB is O and represents the origin of rectangular coordinates x and y. The point A is (-p/2,0) and B is (p/2,0). P is the general point (x,y). By Pythagoras, AP^2=(x+p/2)^2+y^2; BP^2=(p/2-x)^2+y^2; s=sqrt((x+p/2)^2+y^2)+sqrt((x-p/2)^2+y^2). ((p/2-x)^2=(x-p/2)^2, so these may be used interchangeably. Note also that (x+p/2)^2+(x-p/2)^2=x^2+p^2/4 and (x+p/2)(x-p/2)=x^2-p^2/4.) s-sqrt((x+p/2)^2+y^2)=sqrt((x-p/2)^2+y^2) Squaring both sides: s^2+(x+p/2)^2+y^2-2ssqrt((x+p/2)^2+y^2)=(x-p/2)^2+y^2; but (x+p/2)^2=x^2+xp+p^2/4 and (x-p/2)^2=x^2-xp+p^2/4; so, s^2+2xp=2ssqrt((x+p/2)^2+y^2); Squaring again: s^4+4xps^2+4x^2p^2=4s^2(x+p/2)^2+y^2)=4s^2(x^2+xp+p^2/4+y^2); s^4+4x^2p^2=4s^2x^2+s^2p^2+4s^2y^2; 4x^2(s^2-p^2)+4s^2y^2=s^2(s^2-p^2); 4x^2+4s^2y^2/(s^2-p^2)=s^2; 4x^2/s^2+4y^2/(s^2-p^2)=1; Let a^2=s^2/4 and b^2=(s^2-p^2)/4: x^2/a^2+y^2/b^2=1 is the standard equation of an ellipse. Things to note: When p=0 the ellipse should be a circle: 4x^2/s^2+4y^2/s^2=1 is the equation of a circle radius s/2. When x=0, y=+sqrt(s^2-p^2)/2=+b and when y=0, x=+s/2=+a. The difference between the extreme points for x, (-a,0) and (a,0), is 2a=s, so s represents the major axis of the ellipse, and a the semimajor axis. Similarly the difference between extreme points for y, (0,-b) and (0,b), is 2b=sqrt(s^2-p^2), and b represents the semiminor axis. The points A and B of line AB are the foci of the ellipse. Since s=2a and b^2=(s^2-p^2)/4, we can find p. 4b^2=4a^2-p^2; p=2sqrt(a^2-b^2) and A(-p/2,0) and B(p/2,0)=A(-sqrt(a^2-b^2),0) and B(sqrt(a^2-b^2),0). For a circle, a=b and the foci converge to the centre of the circle.

### Prove, by contradiction, that if w,z∈C.... (elementary algebra)

Given the validity of the comment attached to the question, and it's resolution, we can take a look at the implication of the question. To be true for w and z generally, we can take a specific case to demonstrate the argument. Let w=0.8+0.6i, which satisfies the requirement |w|=1, a special case of |w|<1. For any value of n, the binomial expansion of (a+b)^n=1 only when a+b=1, i.e., b=1-a. The binomial expansion is (a+b)^n=a^n+nba^(n-1)+(n(n-1)/2)b^2a^(n-2)+...+b^n=1. The general term is (n!/(r!(n-r)!))a^(n-r)b^r, where 01/2. Let w=0.03+0.04i, |w|=0.05, z=1-(0.03+0.04i)=0.97-0.04i, |z|=0.9708 approx>1/2. As |w| gets smaller, |z| gets larger, so moving further away from 1/2 and towards 1. What is w+z? In all of the above cases w+z=1, by definition, and so the binomial expansion also =1. Just seen your comment. Let w=0.96+0.28i, |w|=1, z=0.04-0.28i, |z|=0.2828 approx, which is <1/2. So the proposition would not be true for the binomial expansion of (w+z)^n=1 for all n. That means we require a different expression than w+z. However, we do know that (...)^n must be 1 for all n and we need to find an alternative expression similar to the one given. Let w=a+isqrt(p^2-a^2), |w|=sqrt(a^2+p^2-a^2)=p<1. This is the general form for w. In the above, z=1-a-isqrt(p^2-a^2) and |z|=sqrt((1-a)^2+p^2-a^2)=sqrt(1-2a+a^2+p^2-a^2)=sqrt(1-2a+p^2). When p=1, this is sqrt(2(1-a)) which can be less than 1/2 (7/81/2) and wz/(w+z)=1 so ((w+z)/wz)^n=((1/z)+(1/w))^n=(wz/(w+z))^n=1. ((1/z)+(1/w))^n=(w+z)^n=1. Therefore, w and z are particular complex numbers for which the proposition is true, and the binomial expansion applies.

### seven times a two digit number

Short answer:  The two digit number is 36. Long answer: 7 * x1x2 = 4 * x2x1 "If the difference between the number is 3. . ." Last digits: x2: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 7*x2:  0, 7, 4, 1, 8, 5, 2, 9, 6, 1 x1:  0, 1, 2, 3, 4, 5, 6, 7, 8, 9 4*x1:  0, 4, 8, 2, 6, 0, 4, 8, 2, 6 The only way this works is when these last digits for 7*x2 and 4*x1 are the same.  That means the 7*x2 line can only be: 7*x2:  0, 4, 8, 2, 6 Which means the possible values for x2 are: x2:  0, 2, 4, 6, 8 Now let's look at x1.  Right now the possible values for x1 are: x1:  0, 1, 2, 3, 4, 5, 6, 7, 8, 9 But we have to end up with a choice from x1 and x2 having a difference of 3 (odd number).  There is no way to get an odd number by subtracting an even number from an even number.  That means x1 has to be odd.  Our possible values for x1 are now: x1:  1, 3, 5, 7, 9 And our possible values for x2 are: x2:  0, 2, 4, 6, 8 The possible combinations for x1x2 and x2x1 are: 10, 01 12, 21 14, 41 16, 61 18, 81 30, 03 32, 23 34, 43 36, 63 38, 83 50, 05 52, 25 54, 45 56, 65 58, 85 70, 07 72, 27 74, 47 76, 67 78, 87 90, 09 92, 29 94, 49 96, 69 98, 89 We want 7*x1x2 = 4*x2x1, so we can do 7 * the first column and 4 * the second column: 70, 4 84, 84 98, 164 112, 244 126, 324 210, 12 224, 92 238, 172 252, 252 266, 332 350, 20 364, 100 378, 180 392, 260 406, 340 490, 28 504, 108 518, 188 532, 268 546, 348 630, 36 644, 116 658, 196 672, 276 686, 356 But since we want 7*x1x2 to equal 4*x2x1, that list reduces to: 84, 84 252, 252 The corresponding x1 and x2 values are: 12, 21 36, 63 But the difference between x1 and x2 is 3, so we can't use x1 = 1, x2=2.  We have to use x1 = 3, x2 = 6. Answer:  The two digit number is 36. Check:  7 * 36 = 4 * 63 252 = 252 good. 6 - 3 = 3 good.

### I need help, I really need help pleeease.

I assume that (3) is a graph, because it doesn't display on my tablet. Question 1 a) 1 because the daily operating cost is stated specifically. b) 3 because the profit of \$500 will be a point on the graph corresponding to a value of t; 2 because if we put P=500 and solve for t we get t=1525/7.5=203.3 so the graph would show t=203-204. c) 2 because we set P=0 and solve for t=1025/7.5=136.7, so 137 tickets sold would give a profit of \$2.5, but 136 tickets would make a loss of \$5; alternatively, if (3) is a graph then P=0 is the t-axis so it's where the line cuts the axis between 136 and 137. d) The rate of change is 7.5 from (2).  e) 2, because the format shows it to be a linear relationship; a straight line graph for (3) also shows linearity. Question 2 a) profit=sales- operating costs so 1025 in (2) is the negative value representing these costs. If (3) is a graph, it's the intercept on the P axis at P=-1025; (4) is the P value when t=0. b) For (1) you would need to find out how many tickets at \$7.50 you would need to cover the operating costs of \$1025 plus the profit of \$500. That is, how many tickets make \$1525? Divide 1525 by 7.5; 2 and 3 have already been dealt with; to use (4) you would note that P=\$500 somewhere between t=200 and 250 in the table. c) For (1), work out how many tickets cover the operating costs. 1025/7.5=136.7, so pick 137 which gives the smallest profit to break even; 2 and 3 already given; in (4) it's where P goes from negative to positive, between 100 and 150. d) For (1) the only changing factor is the number of tickets sold. The rate is simply the price of the ticket, \$7.50; if (3) is a graph, the rate of change is the slope of the graph, to find it make a right-angled triangle using part of the line as the hypotenuse, then the ratio of the vertical side (P range) and the horizontal side (t range) is the slope=rate of change; for (4) take two P values and subtract the smallest from the biggest, then take the corresponding t values and subtract them, and finally divide the two differences to give the rate of change: example: (475-(-275))/(200-100)=750/100=7.5. e) For (1) it's clear that the profit increases (or loss decreases) with the sale of each ticket by the same amount as the price of a ticket, so there is a linear relationship; 2 and 3 already dealt with; in the table in (4) the profit changes by the fixed value of 50 tickets=\$375, showing that a linear relationship applies between P and t: for every 50 tickets we just add \$375 to the profit.

### what are the two tangent lines of x^2 and -x^2+2x-5

y=x^2, dy/dx=2x; y=-x^2+2x-5, dy/dx=-2x+2. These are the tangents at (x,y) for each equation, to find the tangent lines, you need to know at which point the tangent line is required, because it varies from point to point so I'm guessing that since you want two tangent lines, the two tangent lines touch each graph at different points. So that means we have two points on each tangent line to find. We know the equation of a straight line has the form y=mx+b where m=dy/dx. If one tangent line is to touch both curves, then the tangent of one must be equal to the tangent of the other. So if we call the point on the first curve (x1,y1) and on the other (x2,y2), m=2x1=-2x2+2, and x1=-x2+1=m/2. The equation of the tangent line is y=2x1x+b=(-2x2+2)x+b. But the points (x1,x1^2) and (x2,-x2^2+2x2-5) must both lie on this line. Therefore, x1^2=2x1^2+b and -x2^2+2x2-5=2x1x2+b. So b=-x1^2 from the first equation. Substituting for b: -x2^2+2x2-5=2x1x2-x1^2. And x1^2-2x1x2-x2^2+2x2-5=0. We know x2=1-x1 so we can substitute for x2 and find x1. x1^2-2x1(1-x1)-(1-x1)^2+2(1-x1)-5=0; x1^2-2x1+2x1^2-1+2x1-x1^2+2-2x1-5=0. 2x1^2-2x1-4=0=x1^2-x1-2=(x1-2)(x1+1), so x1=2 or -1. Therefore, x2=1-x1=-1 or 2. y1=4 or 1, y2=-8 or -5. The four points are (2,4) and (-1,-8), and (-1,1) and (2,-5), a pair of points on each line. Also, b=-4 or -1 and m=4 or -2. The two lines are y=4x-4 and y=-2x-1.