Guide :

use the given equation to complete each of the following tables ,,,,,y=3x-5

Grade 9 maths assignemnt graphs

Research, Knowledge and Information :


Linear Equation Table Of Values. Examples, how to, and Graph


Linear Equation Table. Algebra ... Equation: Y value : y = 3x ... you could check your answer by substituting the values from the table into your equation. Each and ...
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Use a table of values to graph each equation. State the ...


Use a table of values to graph each equation. State the domain and range. y = 2 x2 + 4 x í 6 62/87,21 Graph the ordered pairs, and connect them to create a smooth curve.
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Y=3x-5 for each equation,complete the given ordered pairs?


Y=3x-5 for each equation,complete the given ... Divide each polynomial by the factor given, then express each ... How do you solve the equation y= 3x+5. ... , solve ...
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Copy and Complete the following Tables of Values


Copy and Complete the following Tables of Values ... Determine whether each ordered pair is a solution of y = 3x ... some solutions to a given equation. Use x = 1 ...
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I need assistance Asap! Section 3.1: Exercise 8, 18, and 70 ...


Answer to I need assistance Asap! Section 3.1: ... and 70 Complete each ordered pair so that it satisfies the given equation. 8) y = 2x + 5: (8,?
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Solutions to 2-variable equations: substitution (old) (video ...


Solutions to 2-variable equations: substitution (old) ... Complete solutions to 2-variable equations. ... y equal negative 4, satisfy this equation, ...
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Using a Table of Values to Graph Linear Equations


Using a Table of Values to Graph Linear Equations. ... For any given linear equation, there are an infinite number of solutions or points on that line.
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1. Complete the function table. xy - Mrs. Obiedzenski's 4th ...


Using the variables x and y, write an equation that ... 8–5 Name Date 1. Complete the function table. 2. ... (4, 8), and (5, 10) Each y-coordinate is 2 times its ...
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HOW TO USE THE EQUATION TO COMPLETE THE TABLE. y=4x+3 x 1 2 5 ...


HOW TO USE THE EQUATION TO COMPLETE THE TABLE. y=4x+3 x 1 2 5 ... the equation 4x+y=9 The following given table is x y ... round tables with 4 people at each ...
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Suggested Questions And Answer :


With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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how do you solve T(r)=In(3.17)/12 In(1+0.0833r) where r>0

The equation for compound interest is usually given in the form P=A(1+r)^n, where A is the initial amount, P the amount after n time periods and r the compound interest rate for the time period. n=12T if the interest is compounded monthly and T is in years, with r as the monthly rate of interest. Example: if r=4.8% per annum (0.048) then r=0.4% per month (0.004). This equation can be rewritten: P/A (growth)=(1+r)^(12T), so ln(growth)=12Tln(1+r) and T=ln(growth)/(12ln(1+r)), which resembles the given equation, in which growth is 3.17 and r is the rate per annum, making 0.08333r (=r/12) the rate per month. T(r)=ln(3.17)/(12ln(1+0.08333r)).  Clearly, if r=0 there would be no growth at all and T would be infinite, hence r>0. This equation relates T and r and would be the basis of a table where the growth was fixed at 3.17 or 317%. Such a table appears below: r (rate % per annum) T years 1 115.42 2 57.73 3 38.51 4 28.89 5 22.12 6 19.28 7 16.53 8 14.47 9 12.87 10 11.59 A graph based on the table is also useful. The equation can't be "solved" for either T or r, since we would need either T or r to find r or T.
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use the given equation to complete each of the following tables ,,,,,y=3x-5

?????????????? tabels ????????? y=3x-5 be strate line with slope=3 & go thru point=(0,-5)
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Data Sets linear correlation coefficient

X Y XY X^2 Y^2 10 9.15 91.5 100 83.72 8 8.15 65.2 64 66.42 13 8.74 113.62 169 76.39 9 8.77 78.93 81 76.91 11 9.25 101.75 121 85.56 14 8.09 113.26 196 65.45 6 6.13 36.78 36 37.58 4 3.11 12.44 16 9.67 12 9.12 109.44 144 83.17 7 7.25 50.75 49 52.56 5 4.73 23.65 25 22.37 99 82.49 797.32 1001 659.81 The last row is the total of the figures in each column. I'll use the symbol S followed by the column number to show where the totals are used in the calculation. There are 11 XY pairs of data, shown as n below. r=(nS3-S1S2)/sqrt((nS4-S1^2)(nS5-S2^2))= (11*797.32-99*82.49)/sqrt((11*1001-9801)(11*659.81-6804.60))= 604.01/740.61=0.816 approx. This figure suggests a good linear correlation, since a perfect correlation is given by r=1. The critical value table for a=0.05 and 9 degrees of freedom (df=n-2=9), confirms this, because the minimum r value according to the table is 0.602 and we have calculated r to be 0.816 (to a confidence of 95%). To plot the scatter diagram or graph, use the X and Y values in the same way as you would plot the graph of a function. The range of X is 4 to 14 and of Y it's 3.11 to 9.25. To plot the linear regression line, the standard equation is A+BX, where B is the slope and A the Y intercept. B=(nS3-S1S2)/(nS4-S1^2)=0.50 approx., and A=(S2-BS1)/n=3.01 approx. Therefore the line Y=3+0.5X should be a fair approximation to the regression line and provide best fit for the points in the scatter diagram. Putting X=7 gives Y=6.5 (actual Y is 7.25).  
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how to find slope and y-intercept of y=3x-5

Any line equation which is in the form y=m*x + b is called slope-intercept form. What that gives you is the slope is the number which is multiplied by X (called the coefficient) while the slope intercept is the Y value when X=0. So if we take your first equation: y = 3x - 5 The slope (or m) = 3 The Y intercept = -5 Slope is defined as either "the change in y divided by the change in x" or "rise over run", so that 3, really can be considered as 3/1. Each change in X of 1 will change Y by 3. So slope = 3/1. Graphing any line can come from 2 methods. 1) create a table of values or 2) calculate a single point (x,y) and then apply the slope to find a second coordinate pair (x,y). For the equation y=3x - 5: y = 3x - 5 x y 0 -5 1 -2 Replacing the values for X into the original equation, will come out to the values for Y. So when X=0, y = 3*0 - 5, or simply -5.  When X = 1, Y = 3*1 -5 or simply -2. With these two coordinate pairs of points, you can plot a dot on your graph at each (0,-5) and (1, -2) then draw a straight line which goes through each point and continues straight in each direction, probably ending each end of this line with an arrow to show it continues. I do not have a way to include a picture here of a graph. The second way to graph a line is as follows. You need a starting point that will be on the line. Given the form of y=mx + b, you have a simple point which can be used at the y-intercept. The point is always in the form of (0, b), so in this case it is (0,-5). From that first point, you will apply your slope. The slope is 3 (or technically 3/1 which is a big help). From the initial point (0,-5) you will go UP 3 and RIGHT 1 and that will be the next point that is easy to find. Connect those two points and continue the line in each direction and that will be a graph of your line. Anytime your slope is positive, you will use it by going UP the top number (numerator of the slope) and going RIGHT the bottom number (denominator of the slope). But if your slope is negative (like your second problem is) you will use it by going DOWN the numerator and then RIGHT the denominator. The equation is y= -2/3x + 4 ( / = divided). I need to state the slope and y-intercept. I will not walk through the details on the second equation, but you should have enough information to get the answer from the above example.
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The following sample data show the average annual yield of wheat in bushels per acre

I can answer some of the question. To work out the regression equation we use the table to find the slope and intercept of the line y=a+bx, where a is the intercept and b the slope. b=(N(sum of)XY-(sum of)X*(sum of)Y)/(N(sum of)X^2-((sum of)X)^2). To abbreviate I'll use S() to mean (sum of). S(XY)=46861 (98*421=41258 not 5603). S(X)=196, S(Y)=842. N=9. S(X^2)=10910 (98*98=9604 not 1306). Therefore b=(9*46861-196*842)/(9*10910-196*196)=4.2948. And a=(S(Y)-bS(X))/N=(842-4.2948*196)/9=0.0245. The regression equation is y=0.0245+4.2948x. When x=9, y=38.68, close to actual 40. The mean rainfall is 196/9=21.778in and the mean yield is 842/9=93.556 bushels per acre. To calculate the correlation coefficient, we need to subtract the means from the data values. X-Mx Y-My (X-Mx)^2 (Y-My)^2 (X-Mx)(Y-My) -12.78 -53.56 163.27 2868.20 684.50 -11.78 -50.56 138.72 2555.86 595.60 -5.78 -24.56 33.38 602.98 41.96 -8.78 -41.56 77.05 1726.86 364.90 -8.78 -32.56 77.05 1059.86 285.88 -14.78 -66.56 218.38 4429.64 983.76 -10.78 -43.56 116.16 1897.09 469.58 -2.78 -14.56 7.72 211.86 40.48 76.22 327.44 5809.83 107219.86 24957.48 The Pearson correlation coefficient is the sum of the figures in the last column divided by the square root of the product of the separate sums of the third and fourth columns. The sums of the last three columns are respectively 6641.56, 122572.21 and 28424.14. The coefficient is 28424.14/sqrt(814070687)=28424.14/28531.92=0.9962.
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What is matricx

A matrix is a rectangular or tabular representation of quantities. Like a table, it has rows and columns and each "cell" is called an element. Matrices can be combined using addition, subtraction or multiplication, provided certain rules are followed about the sizes of the matrices involved. A matrix is a shorthand form of an expression that could consist of elements added together if written out in full. The advantage of using this shorthand form is that matrices can be used to solve problems, sometimes complex, that would be more tedious and difficult to solve using a "longhand" approach. There are rules that determine how matrices can be added, subtracted or multiplied, and matrices occupy a complete branch of mathematics of their own. The given matrix is a 1 by 4 matrix (one row and 4 columns) and it could represent the longhand: w+5x+4y+z (for example), where w, x, y, z are variables and the numbers are coefficients. But it could represent a 4-dimensional vector (a difficult concept to visualise) which is involved with other similar matrices representing vectors to solve a problem using vector analysis. Matrices are used in many branches of science to solve quite difficult physical problems.
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Prove the quadratic equation

A quadratic equation in the standard form is given by ax 2 + bx + c = 0 where a, b and c are constants with a not equal to zero. ...Solve the above equation to find the quadratic fomulas Given ax 2 + bx + c = 0 Divide all terms by a x 2 + (b / a) x + c / a = 0 Subtract c / a from both sides x 2 + (b / a) x + c / a - c / a = - c / a and simplify x 2 + (b / a) x = - c / a Add (b / 2a) 2 to both sides x 2 + (b / a) x + (b / 2a) 2 = - c / a + (b / 2a) 2 to complete the square [ x + (b / 2a) ] 2 = - c / a + (b / 2a) 2 Group the two terms on the right side of the equation [ x + (b / 2a) ] 2 = [ b 2 - 4a c ] / ( 4a2 ) Solve by taking the square root x + (b / 2a) = ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) } Solve for x to obtain two solutions x = - b / 2a ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) } The term sqrt { [ b 2 - 4a c ] / ( 4a2 ) } may be written sqrt { [ b 2 - 4a c ] / ( 4a2 ) } = sqrt(b 2 - 4a c) / 2 | a | Since 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written x = [ -b + sqrt( b 2 - 4a c ) ] / 2 a x = [ -b - sqrt( b 2 - 4a c ) ] / 2 a The term b 2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible case 1: If b 2 - 4a c > 0 , the equation has 2 solutions. case 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2. case 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions.
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The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m.

The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m. A) What is the temperature of the greenhouse at 1:00 a.m. to the nearest Fahrenheit? (B) Find the average temperature between 7:00 p.m. and 7:00 a.m. to the nearest tenth of a degree Fahrenheit. (C) When the temperature of the greenhouse drops below 80 degreeso Fahrenheit, a heating system will automatically be turned on a maintain the temperature at minimum of 80 degrees Fahrenheit. At what values of the to the nearest tenth is the heating system turned on? (D) The cost of heating the greenhouse is $0.25 per hour for each degree. What is the total cost to the nearest dollar to heat the greenhouse from 7:00 p.m. and 7:00 a.m.?   The equation is: f(t) = 96 – 20sin(t/4), 0 <= t <= 12 (A)  At 1.00 a.m. t = 6 f(6) = 96 – 20.sin(6/4) = 96 – 20*0.99749 = 96 – 19.9499 f(6) = 76 ⁰F (B)  The average temperature would need to be worked out by sampling the temperature at different times throughout the night. Divide the temperature range into N equal intervals, giving N+1 sampling points. We would then have T1 = f(δt), T2 = f(2δt), T3 = f(3δt), ... ,Tn = f(nδt) Where δt = range/N = 12/N, and n = 0..N Giving Tn = 96 – 20.sin((12n/N)/4) = 96 – 20.sin(3n/N) Then Tav = (1/N)*sum(Tn, n = 0 .. N) i.e. Tav = (1/N)*sum(96 – 20.sin(3n/N), n = 0 .. N Tav = 96 – 20. (1/N)*sum(sin(3n/N), n = 0 .. N I used Maple to evaluate the above summation. The results are tabulated as follows.                                  Average Temperature Num Intervals            3       4        6       10       20      50     100    200 Tav over the range 83.39 83.01 82.78 82.69 82.69 82.71 82.72 82.73 As can be seen from the table the temperature is averaging out at:  Tav = 82.7 ⁰F (C)  T = f(t) = 96 – 20sin(t/4), 0 <= t <= 12 At T = 80 ⁰F,            96 – 20sin(t/4) = 80 20.sin(t/4) = 96 – 80 = 16 sin(t/4) = 0.8 t/4 = 0.927295 t = 3.70918 t = 3.7 (to nearest tenth) (D)  The temperature will (normally) drop to 80 ⁰F after t = 3.7 hours and rise again to 80 ⁰F when t = 12 – 3.7 = 8.3 hours. Heating system is turned on for 8.3 – 3.7 = 4.6 hours Cost of heating is 4.6*80*0.25 = 4.6*20 = 92 Cost = $92  
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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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