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# solve the expersion using the quadrtic formula

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

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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

### 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

### 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

### 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

### 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

### 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.

### 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.

### -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.

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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