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what is the linear equation of 5.1x-2.8=2.5x-0.02

After 20 years of being out of school, I have decided to return and resume a higher education. Am taking a refresher math class and having trouble absorbing the material, need some guidance.Please help with the posted question.

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solve linear equation w/many operations


solve linear equation w/many operations ... Equation 10 is about as messy as a linear equation gets. ... After the equation "-x - 3 = 5x ...
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How Do I Solve This Equation - 5x 8 3 4x - Prijom


Solve 4x^2+5x+21=0 : 02.02 Solve : ... In Exercises 116, solve the equation. 1. 9x + x = 8. 2. 4x 5x = 3. 3. ... SOLVING LINEAR SYSTEMS Example 1 Example 2 : y = 5x ...
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Matrix Equations - University of Utah - UUMath


Matrix Equations This chapter ... Problem 2. Use that 0 @ 426 02 1 01 1 1 A 1 = 0 @ 1/4 27/2 011 0 12 1 A ... 0 @ 12 2 8 1 A The matrix equation above says that x =12 ...
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Solution - McGraw Hill Education


2.2 Solving General Linear Equations ... 5q 8(q 1) Original equation 2(2 3) 5(2) 8(2 1) Replace q by 2. 2( 1) ... 35. x 3.3 0.1x 3 36. y 2.4 0.2 y 2 Solve each equation.
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Solve the equation 3X+2(X+4)=5(X+1)+3? - Weknowtheanswer


... because 3X+2(X+4)=5(X+1)+3 Expand 3X + 2X + 8 = 5X + 5 + 3 5X + 8 = 5X + 8 Substitute literally any ... Answer #5 | 02/09 ... Equation Calculator Solve linear ...
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Linear Review - satecsite.org


y = 2.1x 5. x = 3y 6. y + 2 = 0.5x. ... Find the equation of the line through the points (5, 0) and (–2 ... SATEC/Algebra II/Linear Functions/2.02 Linear F'n Review ...
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2.1–2.4 DATE 2 Chapter Quiz NAME - Solution Manual Store


Write a linear equation giving the value V of ... 2.5-2.8 DATE 2 Chapter Quiz NAME 1. ... D1m22.2 1.1 0.7 1.0 1.7 2.5 3.3 4.0 4.4 3.8 2.8 1x - 422x + 2 Ú 0 f1x2= 5x2
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Problem 7 Solve Linear equation 3/4(7x-1) - (2x- (1-x)/2 ...


Apr 24, 2016 · Solve Linear equation 3/4(7x-1) - (2x- (1-x)/2) ... Solve 3x^2 + 5x + 2 = 0 by factoring - Duration: ... (1/2)n =5/8 - Duration: 3:02.

Suggested Questions And Answer :


how do you write an equation for the linear function f with the given values?

Standard linear function is f(x)=ax+b where a and b are constants. Substitute each point into the function: 21=-2a+b, -35=5a+b. We need go no further because we have two linear equations and two unknowns. f(-2)-f(5): 56=-7a so a=-8. Now we can work out b: b=21+2a=21-16=5 so f(x)=5-8x. Quick check shows that the function is correct: f(-2)=5+16=21; f(5)=5-40=-35. But we need to check all the other points: f(-6)=5+48=53; f(3)=5-24=-19; uh-oh, something wrong! The values don't fit. So either f(x) is not linear or it's piecewise. Closer inspection is needed. First plot the points. It's clear to see that the points are not colinear. Join the points with straight lines. We're not told that f is continuous. Let's assume it is. If it's piecewise, we need 5 different equations to define the function between the 6 points. In order these are (-9,-4), (-6,-2), (-2,21), (3,-5), (5,-35), (12,14). Call these points A, B, C, D, E, F. We can work out linear equations between A and B, B and C, C and D, etc. These equations will provide continuity for f in the domain -9 Read More: ...

Find the ordered triple of these equations 2x+5y-3z=-5, 3x+2y+7z=15, 5x-4y+6z=34

2x + 5y - 3z = -5 3x + 2y + 7z = 15 5x - 4y + 6z = 34 multiply the top equation by 2 on both sides 4x + 10y - 6z = -10 3x + 2y + 7z = 15 5x - 4y + 6z = 34 add the bottom equation to the top equation 9x + 6y + 0z = 24 3x + 2y + 7z = 15 5x - 4y + 6z = 34 multiply the middle equation by 6 on both sides and the bottom equation by -7 on both sides 9x + 6y + 0z = 24 18x + 12y + 42z = 90 -35x + 28y - 42z = -238 add the bottom equation to the middle equation. 9x + 6y + 0z = 24 -17x + 40y + 0z = -148 -35x + 28y - 42z = -238 mutiply the top equation by 20 on both sides and the middle equation by -3 on both sides 180x + 120y + 0z = 480 51x - 120y + 0z = 444 -35x + 28y - 42z = -238 add the middle equation to the top equation 231x + 0y + 0z = 924 51x - 120y + 0z = 444 -35x + 28y - 42z = -238 divide the top equation by 231 on both sides x + 0y + 0z = 924/231 51x - 120y + 0z = 444 -35x + 28y - 42z = -238 reduce 924/231 x + 0y + 0z = 4 51x - 120y + 0z = 444 -35x + 28y - 42z = -238 Now we know x = 4 Let's plug x = 4 into the middle equation: 204 - 120y + 0z = 444 -120y = 220 y = -220/120 y = -11/6 Now we know y = -11/6 Let's plug x = 4 and y = -11/6 into the bottom equation: -140 - 308/6 - 42z = -238 -140 - 154/3 - 42z = -238 -520 - 154 - 126z = -714 -674 - 126z = -714 126z = -40 z = -40/126 z = -20/63 Answer:  (4, -11/6, -20/63)
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what is the linear equation of $40 a month phone plan, and 20 cents per text messeage?

what is the linear equation of $40 a month phone plan, and 20 cents per text messeage? We have a system of linear equations project, we have to choose between two real life situations and then using system of lineat equations to decide what to buy. P = 0.20t + 40
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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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wich statement describes the graph 1.5x+0.2y=2.68 & 1.6 +0.3y=2.98

What you've been given is a pair of simultaneous equations, often called a system. Simultaneous just means the equations are both true at the same time. They're linear equations which means they're the sort that would be plotted as straight lines on a graph. There are two variables, x and y (and it looks like the second equation has a missing x after 1.6), and you need two independent equations to find out a solution for both variables. If you did plot their straight line graphs, where the lines cross is the solution to the equations as a point of intersection with coordinates x and y. So, back to the question, the linear graphs intersect at the solution to the equations. If the lines don't intersect because they're parallel there's no solution. These graphs do intersect because their slopes are different, the slope being derived from the coefficients of the x and y terms in each equation (-1.5/0.2 and -1.6/0.3). The standard form for a linear equation or the equation of a line is y=ax+b, where a and b are numbers called the slope and intercept  or y intercept (the point where the line cuts the y axis). Although the question doesn't ask for it, the point where the lines cross or intercept is (1.6,1.4) so x=1.6 and y=1.4 is the solution to the system.
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What equation models the data? What are the domain and range of the equation? Do you think your equation is a good fit for the data? Explain how you determined your answers.

Look first at the differences between the data. Call these D1, D2, etc. No need for $ sign. All figures in cents. D1: 31, 23, 45, 109, 43, 75, 27, -11, -38, -108 D2: -8, 22, -36, 34, 32, -48, -38, -27, -70 D3: 30, -58, 70, -2, -80, 10, 11, -43 D4: -88, 128, -72, -78, 90, 1, -54 No pattern, so perhaps statistical analysis will help. Linear regression, perhaps. Mean is $2.892 for money data (Y). Mean is 2010 for other data (X). A table will help: X Y XY X^2 2005 2.27 4551.35 4020025 2006 2.58 5175.48 4024036 2007 2.81 5639.67 4028049 2008 3.26 6546.08 4032064 2009 2.35 4721.15 4036081 2010 2.78 5587.80 4040100 2011 3.53 7098.83 4044121 2012 3.60 7243.20 4048144 2013 3.49 7025.37 4052169 2014 3.11 6263.54 4056196 2015 2.03 4090.45 4060225 TOTAL: 22110 31.81 63942.92 44441210 From these totals the slope=(11*(3rd column)-(1st col)*(2nd col))/(11*(4th col)-(1st col)^2)= (11*63942.92-22110*31.81)/(11*44441210-488852100)=0.04382. Intercept=((2nd col)-slope*(1st col))/11=(31.81-0.04382*22110)/11=-85.1827. The equation is Y=0.04382X-85.1827, which represents the best model to fit the data, assuming a linear relationship. Using this equation we get (2005,2.67), (2006,2.72), (2007,2.76), (2008,2.80), (2009,2.85), (2010,2.89), (2011,2.94), (2012,2.98), (2013,3.02), (2014,3.07), (2015,3.11) as the pairs of (X,Y) values. If the relation is non-linear, plotting the values may give us a clue. If we omit figures for 2009 and 2010, the graph resembles a parabola. We can find the average slope between consecutive pairs of points, ignoring 2009 and 2010: 2005-2006: 31; 2006-2007: 23; 2007-2008: 45; 2008-2011: 27/3=9; 2011-2012: 7; 2012-2013: -11; 2013-2014: -38; 2014-2015: -108. If this were a parabola, we would expect the negative and positive slopes to be opposites after the maximum at 2012, but this doesn't appear to be the case. However, this points to a maximum for the range at $3.60. The domain goes from 2005 to 2015 and we don't have a satisfactory model for the figures.
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Write a system of linear equations represented by the augmented matrix

QUESTION: I am struggling with keeps my steps in order, I need help to write a system of linear equations represented by the augmented matrix  2 0 5 -14  0 1 -5 12  5 2 0 4  I am to use x, y, and z for my variables. The above augmented matrix is a way of writing down, explicitly, the matrix equation AX = B, where A = | 2 0  5 |       | 0 1 -5 |       | 5 2  0 | X = | x |       | y |       | z | B = | -14 |        | 12 |        |   4 | So AX = B is  | 2 0  5 | | x | = | -14 |  | 0 1 -5 | | y |    | 12 |  | 5 2  0 | | z |    |   4 | where matrix multiplication now gives us the set of linear simultaneous equations, 2x       + 5z = -14          y - 5z = 12 5x + 2y       = 4      
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Math question about quadratic equation and linear equation

In a quadratic expression in x, c is the value of the expression when x=0; in a linear equation in x, b, the y intercept, is the value when x=0. So the quadratic and linear expressions both reduce to the constant term (b or c) when  x=0, and both are points along the y axis when the equations are plotted.
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independent variable linear equation

I don't have the image, but I guess ∑(Y-Yhat)^2=10591 is the variance and the standard error=√(10591/12) where 12 is the number of observations, n. SE=29.7083 approx. Since Y is measured in $K, SE=$K29.71 (approx) and the values are expected to be between 314.375-29.708=$K284.667 and 314.375+29.708=$K344.083. The given figure only gives the variance between the fitted and observed values, rather than the variance between the fitted values and the mean, and there is no linear regression equation provided. Are we to assume an equation from another question you've recently submitted? Y-Yhat=Y-Ybar-(Yhat-Ybar); 10591=∑(Y-Yhat)^2=∑(Y-Ybar-(Yhat-Ybar))^2 "=" ∑(Y-Ybar)^2+∑(Yhat-Ybar)^2-2∑(Y-Ybar)(Yhat-Ybar). We don't seem to have enough info to calculate ∑(Yhat-Ybar)^2.
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If the solution to a three variable system of linear equations is x+y=3 and z=1, is this a dependent linear system?

There are infinite solutions for x and y but z=1 consistently. There is linear dependence between x and y. z is independent of x and y.
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