Guide :

# is 1.6 cm exact value or approximate value?

the thickness of the math book is 1.6cm

## Research, Knowledge and Information :

### 1.calculer air of a sphere of radius 7 cm, giving an exact ...

1.calculer air of a sphere of radius 7 cm, giving an exact value and an ... 1.calculer air of a sphere of radius 7 cm, giving an exact value and an approximate value ...

### The Mathematics 11 Competency Test Exact and Approximate Numbers

... Exact and Approximate Numbers Page 1 of 2 The ... height is 196 cm, we imply that the “6 ... between approximate and exact values may seem to ...

### Section 2.2 Arc Length and Sector Area - mrsk.ca

A central angle in a circle of radius 3 cm cuts off an arc of length 6 cm. ... exact value and an approximate value ... intercepted by a central angle AOB of 1 ...

### 2.6 - Approximate Numbers and Significant Figures

... Approximate Numbers and Significant Figures An exact number is one that ... is 8.6 cm then we actually mean that the value is closer to 8.6 cm than it is to 8 ...

### What is the approximate value of x? Round to the nearest ...

Answer to What is the approximate value of ... Round to the nearest tenth. 3.1 cm 3.9 cm 4.6 cm 5.4 cm http://media.edgenuity.com/evresources/8101/8101-07/8101-07-04 ...

### Exact vs Approximate - CEEMRR.COM

Exact vs Approximate. ... This is the exact value of the side length. When circles are involved, the formulas C&nbsp;=&nbsp;2πr and A&nbsp; ...

### What is the approximate value of x? Round to the nearest ...

Round to the nearest tenth. 3.1 cm 3.9 cm 4.6 cm 5.4 cm - 4145552. 1. Log in Join now Katie; a few seconds ago; Hi there! ... What is the approximate value of x?

### What is the approximate value of x? Round to the nearest ...

Round to the nearest tenth 3.1 cm 3.9 cm 4.6 cm 5.4 cm. Log In with Facebook or. Keep me logged in. Forgot your password? ... What is the approximate value of x?

### Section 8.1 Finding the exact value and approximate ... - YouTube

Jul 24, 2013 · Section 8.1 Finding the exact value and approximate value for ... Find Exact Values of Trig ... Using Differentials to Approximate the Value of a ...

### trig study guide Flashcards | Quizlet

trig study guide. STUDY. PLAY ... The exact value for cot −1(√3) is 75 degrees. ... approximate in meters the altitude of the monument. 188.39.

## Suggested Questions And Answer :

### What is the exact value of tan eleven and five degree

Tan(11.5)=0.2034523 approximately. It cannot be represented exactly because the decimal goes on forever and it can't be expressed as a fraction. Tan(11.25) (eleven and a quarter degrees) can be found exactly in terms of irrational numbers because it's 45/4 and tan(45)=1 exactly, and there are formulas for finding the tangents of half angles, but there is still no exact decimal or fraction.

### tan-1(25)? without calculator

Simply speaking, you can't.  You cannot work out the exact value.  However, you can approximate it by expanding it using the taylor expansion which gives that  Although using this you would have to know what 25^5 is.  The further you expand the series, the closer the approximation is to the true value. Chances are that you will not learn about these types of expansion unless you go to university to study maths.    However, an additional way of working this out, is if this is looks something like this arctan(x) = pi/4 in which case you can take tan of both sides to give  x = tan(pi/4)

### 200= 250 - 120 (2.8)^-.5t

If p(t)=200, then 200=250-120*2.8^-0.5t. So 120*2.8^-0.5t=50, 2.8^-0.5t=50/120 and 2.8^0.5t=120/50=2.4. We need to take logs of both sides: 0.5t*log(2.8)=log2.4, so t=log2.4/0.5*log2.8=2log2.4/log2.8=1.7006 months. Let's put this value of t into the original equation: p(1.7006)=250-120*2.8^-.5*1.7006=200. Please note that because of the logs t=1.7006 is inevitably an approximation. However, because the number is so close to 1.7 months, there may in fact be approximations in other values in the function, so the solution 1.7 months may be the exact answer.

### is 1.6 cm exact value or approximate value?

1 milimeter is as good as yu kan get with a book

### Use different P and t, plot delta P/delta t against P n obtain the approximate value for r n k.

Inspection of the derivative function shows that it's a parabola​. When P=0 or k, dP/dt=0. And we can rewrite the function: P'=dP/dt=(rP/k)(k-P)=(r/k)(Pk-P^2)=(r/k)(k^2/4-P^2+Pk-k^2/4)=(r/k)((k/2)^2-(k-P)^2). The vertex is at (k/2,rk/2). So the distance between the point where the curve cuts the P (horizontal) axis gives us a measure for k; and, using this value for k we get r by inspecting the vertex height above the P axis and dividing by k/2. The table shows P against t, but we need P' against P; but we can roughly calculate dt and dP by subtracting pairs of numbers. When P=800 in the table we clearly have two values for t and the gradient P'=0. Therefore, k=800, implying that k/2=400. When P=400 in the table we can see that t is between 140 and 182 and P is between 371 and  534. So the gradient is roughly (534-371)/(182-140)=163/42=3.9. Therefore rk/2=3.9 so r=3.9/400=0.01. dP/P(1-P/k)=rdt. So int(dP/P(1-P/k))=rt+C, where C is to be determined from the table. Int(dP/P)-int(dP/(P-k))=rt+C; ln(P)-ln|P-k|=rt+C; ln(P/|P-k|)=rt+C; P/|P-k|=ae^rt where a replaces constant C (a=e^C). P=|P-k|ae^rt, so P=ake^rt/(1+ae^rt). From the table, when t=0, P=49, so 49=ak/(1+a); 49+49a=ak and a(k-49)=49, a=49/(k-49). Using the approximate value of k=800, a is approximately 49/(800-49)=0.065. dP/dt=rP-rP^2/k. Since r and k are constants the DE can be solved by separation of variables: So putting in our approximate values P=0.065*800*e^0.01t(1+0.065e^0.01t)⇒ P=52e^0.01t(1+0.065e^0.01t). The values in the table can be plotted on a graph of P against t, but to plot dP/dt against P would require us to know r and k, for which we have only the approximate values 0.01 and 800.

### use Euler’s method with the speciﬁed step size to determine the solution to the given initial-value problem at the speciﬁed point.

dy/dx=4y-1; separate the variables: dy/(4y-1)=dx, integrating: (1/4)ln(4y-1)=x. ln(4y-1)=4x. The general solution is 4y-1=Ae^4x, y=(Ae^4x + 1)/4 [CHECK: y'=Ae^4x=4y-1.] Initial condition, y(0)=1: 1=(A+1)/4, A+1=4, A=3 and y=(3e^4x + 1)/4. y(0.5)=(3e^2+1)/4=5.79. This is the value we hope to approximate using Euler's method of iterative increments. USING EULER'S METHOD This method uses the approximation that Dy/Dx = dy/dx where D is a small increment, Dx=h=0.05. Initially y=0 and x=1. This is our starting point The table shows the iterative procedure. By Euler's method y(0.50)=4.89, compared with the actual value of about 5.79. The value 5.82 is in fact closer and this is the value we should use for the following reason. The starting point is (0,1).  The next x-value is the first increment giving us the horizontal base of a triangle. The end points in this base are (0,1) and (0.05,1). So the vertical height of the triangle is found by multiplying the gradient dy/dx by the increment to give us the y increment. The table then has to be interpreted by adding the increment 0.05 to the the limiting value of x, which is 0.5, making it 0.55 and reading off the corresponding y value, 5.82 (columns 4 and 5). This is the solution, y(0.5)=5.82 using Euler's method. If the increment is made smaller y(0.5) gets closer to the actual value. The picture below shows the idea. Joining the hypotenuses of these little triangles gives us an approximation to the shape of the curve.

### Prove that mx(x^2+2x+3) = x^2-2x-3 has exactly 1 real root if m = 1 & exactly 3 real roots if m = -2/3

When m=1, x^3+2x^2+3x=x^2-2x-3; x^3+x^2+5x+3=0. When x=0 the cubic has a value of 3. Because all the signs are plus, when x>0 the cubic stays positive. When x=-1 the cubic has a value of -2, so there is a root between 0 and -1. When x<-1 it has negative values because x^3 is always more negative than x^2 and the combined negative values exceed -3 so the whole cubic remains negative. When m=-2/3, -2x^3/3-4x^2/3-2x=x^2-2x-3; 2x^3/3+7x^2/3-3=0; 2x^3+7x^2-9=0. This cubic has a root at x=1 because 2+7-9=0. We can use synthetic division to find the quadratic: 1 | 2 7 0  -9 .....2 2 9   9 .....2 9 9 | 0 = 2x^2+9x+9=0=(2x+3)(x+3). This shows there are three roots: x=1, -3/2, -3.

### What number can replace the blank space in the problem to find the square root of 115 by using the Babylonian method

The Babylonian method for square roots is based on the simple fact that sqrt(x)*sqrt(x)=sqrt(x^2)=x. So sqrt(x)=x/sqrt(x). If we approximate to sqrt(x) and call it x1, then we can write x2=x/x1 approximately. The actual square root of x will lie somewhere in between x1 and x2. The midpoint of these two values is the average (x1+x2)/2. Call this value x3. This will be a more accurate approximation than x1. So we start with x3 and x4=x/x3. The average of x3 and x4 is (x3+x4)/2=x5. And so on. Now the question. x=115. We could start with x1=10, so x2=11.5 and x3=10.75 (approx 10.8). Then x4=115/x3=10.7 (approx) and x5=10.72. So the progressive approximations to sqrt(115) are: 10,10.75,10.72380529. The question isn't clear as to where the blanks are, but it would appear that the first approximation as a result of starting with x1=10 is 10.8 (10.75 expressed to the nearest tenth). If we start with x1=11 we get: 11,10.7272...,10.72380586,10.72380529. So whatever starting value we use we end up with this number which, when squared, = 115. So if the answer is multi choice, then 10.8 is the first iteration to the nearest tenth.

### Use a growth rate of 7% to predict the population in 2074 of a country that in the year 2006 had a population of 100 million?

2074-2006=68 years. If the growth rate is 7% per year then the growth factor is 1.07^68=99.56275 approx. Multiply by 100,000,000=9,956,275,000. (1+r)^2=1+2r+r^2=1+2r approx for small r. This is the doubling formula. 68=64+4=2^6+2^2. If we apply the formula 6 times we get an approximate value for 1.07^64 (r=0.07): 1.07^2=1+2*0.07=1+0.14=1.14 (true value: 1.1449) Now put r=0.14 1.07^4=1.14^2=1.28 (1.3107...) 1.07^8=1.28^2=1.56 (1.7181...) 1.07^16=1.56^2=1+1.12=2.12 (2.9521...) 2.12=2*1.06=2(1+0.06); 1.07^32=2^2*1.06^2=4*1.12=4.48 (8.7152...) 4.48=4*1.12; 1.07^64=4^2(1.12)^2=16*1.24=16+3.84=19.84 (75.9559...) 1.07^68=1.07^64 * 1.07^4=19.84*1.28=25.3952 (99.56274...) So, using the doubling formula we have an underestimate of 100,000,000*25.4=2,540,000,000 instead of about 10,000,000,000. The doubling time formula should give a better approximation: when the growth is double (1+r)^T=2 where T is the time for growth to be doubled, so T=log(2)/log(1+0.07)=ln(2)/0.07 approx.=9.9 years. Call this 10 years. In 68 years then, the population should have doubled about 6.8 times. Let's work out doubling 6 times: 2^6=64 so after 6*10=60 years (2066) the population will be 6,400,000,000. Now, using our approximate doubling formula from earlier we can estimate for the remaining 8 years. We simply multiply by 1.56 (1.07^8) to give 9,984,000,000, which is very close to the true value of 1.07^68 * 100,000,000. If we use simple proportion to find the growth over the final 8 years, we would use the factor 8/10 * 2=1.6. This produces 1.6*64,000,000,000=102,400,000,000, which is an overestimate.