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# solve the expersion using the quadrtic formula

(6/x)+(6/x+9=1

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### Solving Quadratic Equations with the Quadratic Formula ...

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### Solve a Quadratic Equation Using the Quadratic Formula

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## Suggested Questions And Answer :

### solve the expersion using the quadrtic formula

me spekt that trash shood be (6/x) +6/ (x+9)=1 or 6 +6x/(x+9)=x 6(x+9)+6x=x*(x+9) x^2 +9x=6x+6x+54 x^2 +9x-12x-54=0 x^2-3x-54=0 x=9 or -6
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### i need the answer for these questions

Part 1 Newton’s Method for Vector-Valued Functions Our system of equations is, f1(x,y,z) = 0 f2(x,y,z) = 0 f3(x,y,z) = 0 with, f1(x,y,z) = xyz – x^2 + y^2 – 1.34 f2(x,y,z) = xy –z^2 – 0.09 f3(x,y,z) = e^x + e^y + z – 0.41 we can think of (x,y,z) as a vector x and (f1,f2,f3) as a vector-valued function f. With this notation, we can write the system of equations as, f(x) = 0 i.e. we wish to find a vector x that makes the vector function f equal to the zero vector. Linear Approximation for Vector Functions In the single variable case, Newton’s method was derived by considering the linear approximation of the function f at the initial guess x0. From Calculus, the following is the linear approximation of f at x0, for vectors and vector-valued functions: f(x) ≈ f(x0) + Df(x0)(x − x0). Here Df(x0) is a 3 × 3 matrix whose entries are the various partial derivative of the components of f. Speciﬁcally,     ∂f1/ ∂x (x0) ∂f1/ ∂y (x0) ∂f1/ ∂z (x0) Df(x0) = ∂f2/ ∂x (x0) ∂f2/ ∂y (x0) ∂f2/ ∂z (x0)     ∂f3/ ∂x (x0) ∂f3/ ∂y (x0) ∂f3/ ∂z (x0)   Newton’s Method We wish to find x that makes f equal to the zero vector, so let’s choose x = x1 so that f(x0) + Df(x0)(x1 − x0) = f(x) =  0. Since Df(x0)) is a square matrix, we can solve this equation by x1 = x0 − (Df(x0))^(−1)f(x0), provided that the inverse exists. The formula is the vector equivalent of the Newton’s method formula for single variable functions. However, in practice we never use the inverse of a matrix for computations, so we cannot use this formula directly. Rather, we can do the following. First solve the equation Df(x0)∆x = −f(x0) Since Df(x0) is a known matrix and −f(x0) is a known vector, this equation is just a system of linear equations, which can be solved efficiently and accurately. Once we have the solution vector ∆x, we can obtain our improved estimate x1 by x1 = x0 + ∆x. For subsequent steps, we have the following process: • Solve Df(xi)∆x = −f(xi). • Let xi+1 = xi + ∆x
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### solve for x when (.182+x)/(.106-2x)=3.24

solve for x when (.182+x)/(.106-2x)=3.24 Im trying to solve for x its for a chemistry equation but i need to solve for x using the quadratic equation and i dont know which equals a b and c to find x. ************************** You can't solve this with the quadratic formula, because there is no x^2 term. Observe: (.182+x)/(.106-2x)=3.24 Multiply both sides by (.106-2x) to eliminate the fraction on the left. (.106-2x) * (.182+x)/(.106-2x) = (.106-2x) * 3.24 0.182 + x = 0.34344 - 6.48x This is a linear equation. Even if you arranged it so that all the terms are on the left, leaving a zero on the right side, you still would not have an x^2 term. That means that the "a" coefficient would be zero. When you try to use that in the quadratic formula, the denominator would be 2a = 2 * 0 = 0. You cannot divide by zero, so you are now stuck, with no possible solution. If you solve the equation from where I have taken it, you will find the value of x. 0.182 + x = 0.34344 - 6.48x 6.48x + x = 0.34344 - 0.182 7.48x = 0.16144 (Note that you could now subtract 0.16144 from both sides to get 7.48x - 0.16144 = 0, but you still don't have an x^2 term, so a=0, b=7.48 and c=-0.16144. That pesky 0 for the value of a will give that 0 denominator. Impossible to divide.) x = 0.0215828877
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### Solve the quadratic function ax^2 + bx + c =0

What do you use to solve for x? Do you use the quadratic formula or do you complete the square? Actually, the quadratic formula is proven by completing the square with variables instead of numbers. So, just complete the square. If you already know how to solve for these functions then it stands to reason that you can for any arbitrary quadratic.  You start out with the quadratic:  ax² + bx + c = 0  You want to divide by 'a'  x² + (b/a)x + c/a = 0  Put the constant on the other side  x² + (b/a)x = −c/a  Complete the square...  x² + (b/a)x + (b/2a)² = −c/a + (b/2a)²  Factor the left side:  (x + (b/2a))² = −c/a + (b/2a)²  Simplify:  (x + (b/2a))² = (b/2a)² − c/a  (x + (b/2a))² = b²/(4a²) − c/a  (x + (b/2a))² = b²/(4a²) − (4ac)/(4a²)  (x + (b/2a))² = (b² − 4ac) / (4a²)  Square root to isolate 'x'  x + (b/2a) = ±√[ (b² − 4ac) / (4a²) ]  x + (b/2a) = ±√[ b² − 4ac ] / (2a)  x = −(b/2a) ±√[ b² − 4ac ] / (2a)  x = [ −b ±√( b² − 4ac ) ] / (2a)  And you are left with the quadratic formula
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### What is the diameter of a spiral coil of .65265 inch diameter pipe 100 feet long?

The equation of a spiral in polar coordinates has the general form r=A+Bø, where A is the starting radius of the spiral and B is a factor governing the growth of the spiral outwards. For example, if B=0, there is no outward growth and we just have a circle of radius A. A horizontal line length A represents the initial r, and the angle ø is the angle between r and this horizontal line. So r increases in length as ø increases (this angle is measured in radians where 2(pi) radians = 360 degrees, so 1 radian is 180/(pi)=57.3 degrees approximately.) If B=1/2 and A=5", for example, the minimum radius would be 5" when ø=0. When ø=2(pi) (360 degrees), r=5+(pi), or about 8.14". This angle would bring r back to the horizontal position, but it would be 8.14" instead of the initial 5". At ø=720 degrees, the horizontal line would increase by a further 3.14". Everywhere on the spiral the spiral arms would be 3.14" apart. What would B be if the spiral arms were 0.65625" apart? 2(pi)B=0.65625, so B=0.65625/(2(pi))=0.10445". The equation of the spiral is r=5+0.10445ø. To calculate the length of the spiral we have two possible ways: an approximate value based on the similarity between concentric circles and a spiral; or an accurate value obtainable through calculus. The approximate way is to add together the circumferences of the concentric circles: L=2(pi)(5+(5+0.65625)+...+(5+0.65625N)) where L=spiral length and N is the number of turns. L=2(pi)(5N+0.65625S) where S=0+1+2+3+...+(N-1)=N(N-1)/2. This formula arises from the fact that the first and last terms (0, N-1) the second and penultimate terms (1, N-2) and so on add up to N-1. So, for example, if N were 10 we would have (0+9)+(1+8)+(2+7)+(3+6)+(4+5)=5*9=45=10*9/2. If N were 5 we would have 0+1+2+3+4=10=(0+4)+(1+3)+2=5*4/2. L=12*100 inches. L=1200=2(pi)(5N+0.65625N(N-1)/2)=(pi)N(10+0.65625(N-1))=(pi)N(9.34375+0.65625N). If the external radius is r1 and the internal radius is r then the thickness of the spiral is r1-r and since 0.65625 is the gap between the spiral arms N=(r1-r)/0.65625. N is an integer, but, since it is unlikely that this equation would actually produce an integer we would settle for the nearest integer. If we solve this equation for N, we can deduce the external radius and diameter of the spiral: N(9.34375+0.65625N)=1200/(pi)=381.97; 0.65625N^2+9.34375N-381.97=0 and N=(-9.34375+sqrt(1089.98))/1.3125=18 (nearest integer). This means that there are 18 turns of the spiral to make the total length about 100 feet. If X is the final external diameter of the coiled pipe and the internal radius is 5" (the minimum allowable) then X/2 is the external radius, so N=((X/2)-5)/0.65625. We found N=18 so we can find X: X=2*(0.65625*18+5)=33.625in. Solution using calculus Using calculus, we can work out the relationship between the length of the spiral and other parameters. We start with any polar equation r(ø) and a picture: draw a line representing a general value of r. At a small angle dø to this line we draw another line a little bit longer, length r+dr. Now we join the ends together to make a narrow-angled triangle AOB where angle AOB=dø and AB=ds, the small section of the curve. In the triangle AO is length r and BO is length r+dr. If we mark the point C along BO so that CO is length r, the same as AO, we have an isosceles triangle COA. Because the apex angle is small, CA=rdø, the length of the arc of the sector. In triangle ABC, CB=dr, AB=ds and CA=rdø. By Pythagoras, AB^2=CB^2+CA^2, that is, ds^2=dr^2+r^2dø^2, because angle BCA is a right angle as dø tends to zero. The length of the curve is the result of adding the tiny ds values together between limits of r or ø. We can write ds=sqrt(dr^2+r^2dø^2). If we divide both sides by dr, we get ds/dr=sqrt(1+(rdø/dr)^2) so s=integral(sqrt(1+(rdø/dr)^2)dr, where s is the length of the curve. The integral is definite if we define the limits of r. For our spiral we have r=A+Bø, making ø=(r-A)/B and B=p/(2(pi)), where p is the diameter of the pipe=0.65625", so we can substitute for ø in the integral and the limits for r are A to X/2, where A is the inner radius (A=5") and X/2 is the outer radius. dø/dr=2(pi)/p, a constant=9.57 approx. s=integral(sqrt(1+(2(pi)r/p)^2)dr) between limits r=A to X/2. After the integral is calculated, we solve for X putting s=1200". The expression (2(pi)r/p)^2 is large compared to 1, so s=integral((2(pi)r/p)dr) approximately and s=[(pi)r^2/p] (r=A to X/2); therefore, since we know s=1200, we can write ((pi)/p)(X^2/4-A^2)=1200. Therefore X=2sqrt(1200p/(pi))+A^2)=33.21". Compare this answer with the one we got before and we can see they are close. [We could get a formal solution to the integral, using hyperbolic trigonometric or other logarithmic functions, but such a solution would make it very difficult or tedious to solve for X, since X would appear in logarithmic expressions and in other expressions making it difficult or impossible to isolate X. For example, the next term in the expansion of the integral would be (p/(4(pi))ln(X/2A), having a value of about 0.06. It is anticipated, therefore, that an approximation would be sufficient in this problem with the given figures.] We can feel justified in using the formula for finding the length of pipe, L, when X=6'=72": L=((pi)/p)(1296-25)=6084.52"=507' approximately. This length of pipe would hold 507/100*0.96 gallons=4.87 gallons.
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### how do you solve x^2-3x by using the quadratic formula

x^2 -3x=x*(x-3) quadratik dont apli kauz yu dont hav all 3 terms it assume +0...x^2 -3x +0=0 and giv (x-0)*(x-3) quadratik...ansers=-b/2a +-root(b*b -4ac)/2a
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### NEED HELP FAST!!!!!

A. The circumference is (pi)d where d is the diameter. So (pi) for circle A is 21.98/7=3.14 and for B is 18.84/6=3.14. B. Area is (pi)r^2 where r=radius or d/2. So for circle A (pi)=38.465/(3.5)^2=3.14 and circle B (pi)=28.26/(3)^2=3.14. C. To two places of decimals (pi) has a value of 3.14 in all cases. This indicates or implies that (pi) is a universal constant applying to all circles using the formulae.
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### use the continuous compound interest formula to find t a=96000 p=67785 r=8.7% t=?

Answer is: Given A=96000, P=67785  and r = 8.7%    We know that, continuous compound interest formula is A = Pe^(rt)   where P = principal amount  r = annual interest rate t =  years A = amount after t   Now Use A = Pe^(rt) with A=96000, P=67785  and r = 0.087   Therefore, 96000 = 67785 e^(0.087 t) 96000/ 67785 = e^(0.087 t) 1.42 = e^(0.087 t)   log (1.42) = 0.087 t 0.351 = 0.087 t   t = 0.351/0.087   t = 4.034   There are many methods to solve the simple interest, compound interest and continuous compound interest this was the one method which i solved for you.
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### -4.905x^2 +1.85x+69.3=0

You can simply apply the formula for solving a quadratic equation, but you'll need a calculator, judging by the awkwardness of the numerical coefficients. The standard quadratic is usually written ax^2+bx+c=0. In your question a=-4.905, b=1.85, c=69.3. The formula is (-b+/-sqrt(b^2-4ac))/2a, where I've used +/- to mean plus-or-minus so that two different solutions are possible, one using the positive square root and the other using the negative one. Substituting in the formula the two solutions are: (-1.85+sqrt(3.4225+4*4.905*69.3))/-9.81 = -3.5749 and  (-1.85-sqrt(3.4225+4*4.905*69.3))/-9.81=3.9521.
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### a quadratic equation

x^2-2x-2=0 by using the qudratic formula, x=(-(-2)+sqrt((-2)^2-4(1)(-2)))/2(1) or (-(-2)-sqrt((-2)^2-4(1)(-2)))/2(1) =(2+sqrt(4+8))/2 or (2-sqrt(4+8)/2 =1+sqrt(3) or 1-2sqrt(3)
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