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# solve the following a/b/c/or d

(6/y+5)-(9/y-5)=(15/y^2-25) a -30 b 90 c 64 d 30

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### Solve The Following: A = [a B C D] = [1 2 3 7] A^-... | Chegg.com

Question: Solve the following: A = [a b c d] = [1 2 3 7] ... Solve the following: Show transcribed image text Solve the following: A = [a b c d] = [1 2 3 7] ...

### solve the following matrix equation for a, b, c, ... - OpenStudy

solve the following matrix equation for a, b, c, d. [a - b b + c ] = [8 1] [3d + c 2a ... => c=6-2d Finally plug this into equation 3 and solve for d ...

### Solve the following equation: 6y – 20 = 2y – 4. A. y = 3 B. y ...

Solve the following equation: 6y – 20 = 2y – 4. ... Solve the equation 2$$log_{9}$$ (x+2) + $$log_{3}$$ x = 1 please helppp Answer

### Solve the following equation: 2x+3=-1 A. x=2 B. x=-2 C. x=-4 ...

Solve the following equation: - 2534518. 1. Log in Join now Katie; ... Solve the following equation: 2x+3=-1 A. x=2 B. x=-2 C. x=-4 D. x=4 2. Ask for details ; Follow;

### Solving Systems 2 1.)Solve The System Of ... - Chegg.com

Answer to Solving Systems 2 1.)Solve the system of equations: ... Solve the following problem using substitution or elimination: To skate at Roller Heaven, ...

### Solve the following equation: 6y – 20 = 2y – 4. A. y = 2 B. y ...

Solve the following equation: 6y – 20 = 2y – 4. ... User: Solve this equation: y/9 + 5 = 0. A. y = –5 B. y = –45 C. y = 45 D. y = 5 Weegy: ...

### Solving linear equations - A complete course in algebra

LINEAR EQUATIONS. The law of inverses ... ax − b + c = d. We must get a, b, c over to the other side, so that x is alone. ... Solve for x: ab(c + d)x − e + f = 0 ...

## Suggested Questions And Answer :

### solve following system of equations

Problem: solve following system of equations solve the following system of equations: x-2y+z=6 , 2x+y-3z=-3, x-3y+3z=10 1) x - 2y + z = 6 2) 2x + y - 3z = -3 3) x - 3y + 3z = 10 Multiply equation 1 by 3. 3(x - 2y + z) = 6 * 3 4) 3x - 6y + 3z = 18 Add equation 2 to equation 4.   3x - 6y + 3z = 18 +(2x +  y - 3z = -3) -----------------------   5x - 5y      = 15 6) 5x - 5y = 15 Add equation 3 to equation 2.   2x +  y - 3z = -3 +( x - 3y + 3z = 10) ----------------------   3x - 2y      = 7 7) 3x - 2y = 7 Multiply equation 6 by 2. 2(5x - 5y) = 15 * 2 8)10x - 10y = 30 Multiply equation 7 by 5. 5(3x - 2y) = 7 * 5 9) 15x - 10y = 35 Subtract equation 9 from equation 8.   10x - 10y = 30 -(15x - 10y = 35) -------------------   -5x       = -5 -5x = -5 -5x/-5 = -5/-5 x = 1 Use equation 6 to solve for y. 5x - 5y = 15 5(1) - 5y = 15 5 - 5y = 15 5 - 5y - 5 = 15 - 5 -5y = 10 -5y/-5 = 10/-5 y = -2 Use equation 3 to solve for z. x - 3y + 3z = 10 1 - 3(-2) + 3z = 10 1 + 6 + 3z = 10 7 + 3z = 10 3z = 3 3z/3 = 3/3 z = 1 Answer: x = 1, y = -2, z = 1

### 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

### 7x+2y=5 and 3y=16-2x

7x+2y=5 and 3y=16-2x  solve each of the following systems using the method of youre choice. However, you must show an organize work. We could solve both equations, but what would we have? We would still have two equations. Each would show y in terms of x, or x in terms of y. We still wouldn't know the values of x or y. E.g., y = 16/3 - 2/3 x (the second equation) When you are given two equations in the same problem, you can be sure that you are supposed to solve them simultaneously, to find the one unique pair of x,y values that will solve both equations. Let's proceed with that thought in mind. 7x+2y=5   3y=16-2x -> rewrite the second:  2x + 3y = 16 We'll eliminate the y terms by subtracting one equation from the other. To do that, the y terms must be the same. 3 * (7x+2y) = 3 * 5           21x + 6y = 15 2 * (2x + 3y) = 2 * 16      (  4x + 6y = 32)                                     ----------------------- Subtract and get---->       17x         = -17 17x = -17 x = -1 We will substitute that x value into the first equation. 7*(-1) +2y = 5 -7 + 2y = 5 2y = 12 y = 6 Finally, substitute both values, x and y, into the second equation. 3 * 6 = 16 - (2 * -1) 18 = 16 - (-2) 18 = 18 Now, you know that x=-1 and y=6 solves both equations at the same time. This tells you that if you plot the two equations on the same graph, you will have two lines that intersect at (-1, 6).

### solve the equations using elimination

Problem: solve the equations using elimination solve the following equations using elimination x-2y=-1 and 2x+y=4 1) x - 2y = -1 2) 2x + y = 4 Multiply equation 2 by 2. 2(2x + y) = 4 * 2 3) 4x + 2y = 8 Add equation 1 to equation 3, eliminating the y term.  4x + 2y =  8 +(x - 2y = -1) ------------------  5x        = 7 5x = 7 x = 7/5 Use equation 1 to solve for y. x - 2y = -1 7/5 - 2y = -1 7/5 - 2y - 7/5 = -1 - 7/5 -2y = -5/5 - 7/5 -2y = -12/5 -2y/-2 = (-12/5)/-2 y = (-12/5)(-1/2) y = 12/10 y = 6/5 Use equation 2 to check the values. 4x + 2y = 8 4(7/5) + 2(6/5) = 8 28/5 + 12/5 = 8 40/5 = 8 8 = 8 Answer: x = 7/5, y = 6/5

### How would I solve the following equation by completing the Square? x^2-18x+86=0

How would I solve the following equation by completing the Square? x^2-18x+86=0 I just need to know how to solve it, it's Algebra 2 math homework and its on completing the square! When completing the square, you move the constant to the right side of the equation; in this case you do that by subracting it from both sides. That leaves the x squared and x terms on the left side of the equation. Divide the co-efficient of the x term by the co-efficient of the x squared term. It's easy when the co-efficient of the x squared term is 1. You simply end up with the co-efficient of the x term. Take half of that and square it. Add that number to both sides of the equation. You now have to take the square root of both sides of the equation. On the left side, you will have  x plus one-half of the x term. On the right side perform the addition and take the square root of the result. In general terms, you will now have something like (x + d) = sqrt (h). d may be negative if the co-efficient of the x term was negative. Add or subtract d to/from both sides of the equation, leaving x all by itself of the left side.

### how do you solve the equation w=Cr^-2 for r

how do you solve the equation w=Cr^-2 for r the weight of an object follows this equation:  w=Cr^-2, where C is a constant, and r is the distance that the object is from the center of the earth. solve the equation w=Cr^-2 for r   w = Cr¯² w = C / r² r = √(C/w) suppose that an object is 100 pounds when it is at sea level.  Find the value of C that makes the equation true.  (sea level is 3,963 miles from the center of the earth) C = wr² C = 100 ( 3963)² C = 15.7 x 10^8 use the value of C you found in the previous question to determine how much the object would weigh in           i.  Death Valley (282 feet below sea level)   r = 3963 - 282 w = C / r² w = 15.7 x 10^8 / (3963 - 282)² w =  15.7 x 10^8/ (3681)² w =15.7 x 10^8/ 13.55 x 10^6 w = 115.88 pounds          ii.  The top of Mt McKinley (20,320 feet above sea level)'     w = C / r² w = 15.7 x 10^8 / (3963 + 20320)² w =  15.7 x 10^8/ (24283)² w =15.7 x 10^8/ 5.897 x 10^8 w = 2.66  pounds

### Solve the following: 25-32*5/8=

25-32*5/8 =25 -(32*5)/8 =25 -160/8 =25 -20 =5

### When the perimeter of a petagon is given how do you solve for missing side

The perimeter is 25x+8 of a pentagon the given sides are as follow: a= 4x, b= 6x-8, c= 0, d=5x+4, e=5x-4 How would you solve c the missing value? 25x + 8 = 4x + 6x - 8 + c + 5x + 4 + 5x - 4 put common terms together. 25x + 8 = (4x + 6x + 5x + 5x) + (-8 + 4 - 4) + c  solve for c the side that is missing. 25x + 8 = 20x -8 + c 5x + 16 = c

### Factoring by grouping?

Is there a typo here? Maybe it should be 0 = x^3 + 6x^2 -3x - 18 .  This can be solved by grouping like so:   0 = (x^3 + 6x^2) + (-3x - 18 )  This is the grouping step.  I groupred the first two terms, and the last two terms.   0 = x^2(x+6) - 3(x+6)   In this step, I factored out the greatest common factor from each set of parenthesis.   0 =  ( x+6 )(x^2 - 3)  In this step, I factored out the common (x+6) binomial factor from the right side of the equation. (a challenging notion to follow, but it goes something like this:  In  x^2(x+6) - 3(x+6) , think of x^2 as "a", 3 as "b", and (x+6) as "c",  Then x^2(x+6) - 3(x+6) is of the form  ac - bc.  I factored out the "c" to get c(a-b).  Substituting back, using  x^2 as "a", 3 as "b", and (x+6) as "c", get c(a-b) = (x+6)(x^2-3)   Continuing on to solve the equation, if 0 =  ( x+6 )(x^2 - 3) , then 0 = x+6   or    0 = x^2 -3 Solving these equations, get x = -6, x =sqrt(3), x = - sqrt(3)  (remember when taking the square root of each side you must include both the positive and negative results)

### How do I solve the equation z = 6y?

Problem: How do I solve the equation z = 6y? Solve the following equation for y, z = 6y z = 6y z/6 = 6y/6 z/6 = y