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# in echelon form, tell how many solutions there are in nonnegative integers. x+3y+z=82 7y+2z=42

For the following system of equations in echelon form, tell how many solutions there are in nonnegative integers x+3y+z=82 7y+2z=42

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... For the following system of equations in echelon form, tell how many solutions there are in nonnegative integers. x+3y+z=82. 7y+2z=42. 1b) ... 7x-3y+z=8. 5y-9z=-4 ...

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Tutorials Hub. Menu. ... For the following system of equations in echelon form, tell how many solutions there are in nonnegative integers. x+3y+z=82. 7y+2z=42. 1b) ...

Tutorials Hub. Menu. ... For the following system of equations in echelon form, tell how many solutions there are in nonnegative integers. x+3y+z=82. 7y+2z=42. 1b) ...

### Algebra II Final Flashcards | Quizlet

Find the reduced row-echelon form of the augmented matrix that represents the ... {x + 2y + z = -17 {3x + 7y + 2z = -56 {x - y ... Round your answer to the nearest ...

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... in which case, there are inﬁnitely many solutions. ... reduced row-echelon form. Then there are ... 2 Since x, y, and z must all be integers ...

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5 Systems and Matrices. We found 20 results related to this asset. Document Information; Type: eBook; Total # of pages: 112. Avg Rating: Textbook Information.

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### in echelon form, tell how many solutions there are in nonnegative integers. x+3y+z=82 7y+2z=42

????????? "echelon form" ???????? ??? yu hav NE idea wot "echelon" meen ????

### Tell wether the system has one solution, infinetly many solutions or no solutions. 2x-5y=17 and 6x-15y=15

Tell whether the system has one solution, infinitely many solutions or no solutions. 2x-5y=17 and 6x-15y=51 The equations are, 2x – 5y = 17 6x – 15y = 51 Putting them into standard form, y = (2/5)x – 17/5 y = (6/15)x – 51/15 = (2/5)x – 17/5 The two equations are equations of straight lines with the same slope (m = 2/5) and with the same y-intercept (c = -17/5). This means that the two lines are co-linear. They lie on top of each other, so there are an infinite number of points of intersection. Answer: There are infinitely many solutions.

### Shawn took in \$69.15 in two hours work on Saturday selling Belts for \$8.05 and earrings for \$4.50. How many of each did Shawn sell?

Shawn took in \$69.15 in two hours work on Saturday selling Belts for \$8.05 and earrings for \$4.50. How many of each did Shawn sell? I'm not too sure how ​you are expected to solve this problem. You will end up with a Diophantine Equation (one involving integer coefficients and integer solutions only) If you have never heard of Diophantus, then the precise mathematical solution later on should be ignored. Instead ...   The (Diophatine) equation is 161B + 90E = 1363 You should be able to show that Shawn can only sell a maximum of 8 belts (i.e. if he sells no earrings), and a maximum of 15 earrings (i.e. if he sells no belts). So start with the smaller number (8). Shawn sells up to 8 belts. So put B = 1 in the (Diophantine ) equation and see if B is an integer. If not, then B=1 cannot be a solution, So now try again with B = 2, and so on ..     My precise mathematical solution now follows. Let B be the number of belts sold @ \$8.05 per belt. Let E be the number of earrings sold @ \$4.50 per pair. Total money made is P = \$69.15 Total money made is: B*\$8.05 + E*\$4.50 Equating the two expressions, 8.05B + 4.50E = 69.15 Making the coefficients as integers we end up with a Diophantine equation. 161B + 90E = 1383 We now manipulate the coefficients 161 and 90. We should end up with a relation between the two that should produce the value 1. This manipulation is in two parts. In the first line, we are writing the larger coefft, 161, as a multiple of the smaller coefft plus a remainder. 1st Part 161 = 1x90 + 71  90 = 1x71 + 19     now we write the 1st multiple as a multiple of the 1st remainder + remainder 71 = 3x19 + 14    we write the 2nd multiple as a multiple of the 2nd remainder + remainder 19 = 1x14 + 5 14 = 2x5 + 4 5 = 1x4 + 1 ============ 2nd Part We rewrite that last equation so that 1 is on the left hand side 1 = 5 – 1x4 We now use the remainder from the next equation up. 1 = 5 – 1x(14 – 2x5) 1 = 3x5 – 1x14 And again we use the remainder in the next equation up 1 = 3(19 – 1x14) – 1x14 1 = 3x19 – 4x14 1 = 3x19 – 4(71 – 3x19) 1 = 15x19 – 4x71 1 = 15(90 – 1x71) – 4x71 1 = 15x90 – 19x71 1 = 15x90 – 19(161 – 1x90) 1 = 34x90 – 19x161 ================ Therefore, 1383 = (-19x1383).161 + (34x1383).90 1383 = -26,277*161 + 57,022*90 Which implies B = -26,277 E = 47,022 which is obviously incorrect! But all is not lost. These two values for B and E are simply two values that happen to satisfy the original Diophantine equation. What we need is the general solution, which now follows. In general, B = -26,277 + 90k    (using the coefft of E) E = 47,022 – 161k    (using the coefft of B) If you substitute these expressions into the original Diophantine equation, the k-terms will cancel out, leaving you with the original eqn. We know that B and E must be smallish numbers and that neither can be negative. For example, if no earrings were sold, Shawn would need to sell over 8 belts to clear \$69, and if no belts were sold, then Shawn would need to sell over 15 earrings. So Shawn needs to sell between 0 and 8 belts and between 0 and 15 earrings. Setting k = 292, B = -26,277 + 90*292 = 3 E = 47,022 – 161*292 = 10 B = 3, E = 10 If k is greater than or less than 292 by 1, or more, then either B will be negative or E will be negative. So this is the answer. Check 3*8.05 + 10*4.50 = 25.15 + 45.00 = 69.15 – Correct! Answer: Shawn sells 3 belts and 10 earrings

### x=4 y=12, . Show how you built the equations using your integers.

x + y = 16 x - y = -8 Step 1) Choose two random numbers between -12 and 12 (4 and 12 in this case). Step 2) Let one of the numbers be x and the other be y. Step 3) Add both numbers together to form 1 equation. Step 4) Subtract one number from another to form the other equation. And with that, we are done.

### How can I know how many solutions has the trigonometric equation?

The trig functions sine, cosine, tangent, etc., are cyclic. For example, if sin(x)=1/2, x=30 degrees is only one solution out of an infinite number because x=150, 390, 510, 750, etc., are also solutions. If sin(30x)=1/2, then x=5, 13, 17, 25, etc. Unless there is a specific restraint in the trig equation, there will always be an infinite number of solutions. For example, x=sin(x) has only solution, x=0. Another example is x=2sin(x), which has only three solutions. Sometimes a question will state the limits of the answer. For example, find x between the limits 0 to 360, or 0 to 2(pi), depending on whether x is measured in radians or degrees. In this range there are usually two solutions. Graphically it is easier to tell how many solutions there will be. For example, x=2sin(x) can be represented by two functions plotted together on the same graph: y=x and y=2sin(x). The line y=x intersects the curve y=2sin(x) at 3 points, so there must be 3 solutions to x=2sin(x). You can see this, because the sine curve has humps, like the Loch Ness monster, that peak at 2 and -2 all the way along the x axis in both directions, positive and negative forever. The line joins the origin to (1,1), (2,2), etc. on the positive side and (-1,-1), (-2,-2), etc. on the negative side, and it cuts the sine curve through the first negative hump and the first positive hump including the origin, making three points of intersection only, because the line just rises over the humps after that. I hope this helps you to answer your trig questions.

### how many gallons of a 10% solution must you add to 30 gallons of a 40% solution to get a final solution that is at 25% level of strength

For the purpose of answering this type of question I consider a chemical solution to consist of two materials in liquid form. Then I think of the materials as each being represented by a quantity of balls of the same colour. One of the balls (say, a quantity of white balls of water) is the solvent and the other, the solute, is represented by a quantity of black balls. The strength of the solution is the percentage of black balls in the mixture. Call the quantity of each w and b. A 10% solution consists of 9v/10 white balls and v/10 black balls, where v is the volume, so we can write 9vw/10+bv/10=v(9w/10+b/10) as the 10% solution of v gallons. We can also write 30(60w/100+40b/100)=30(3w/5+2w/5) as 30 gallons of the 40% solution, so 30(3w/5+2b/5)+v(9w/10+b/10)=(v+30)(3w/4+b/4), which is the 25% strength solution, where there is 75% solvent and 25% solute. This equation simplifies to 18w+12b+v(9w/10+b/10)=(v+30)(3w/4+b/4) Multiply through by 20 to get rid of the fractions: 360w+240b+2v(9w+b)=5(v+30)(3w+b). This expands to 360w+240b+18vw+2vb=15vw+5vb+450w+150b. We're looking for v, so we get all the terms containing v together: 3vw-3vb=90w-90b, so v=90(w-b)/(3(w-b))=30. Therefore, 30 gallons of 10% solution are required to reduce the 30 gallons of 40% solution to 25% strength.

### What is a stem and leaf plot and when would I use one?

Stem and Leaf plots are just a method of ordering data in a dataset to produce a frequency chart. These plots are used in statistical analysis to draw conclusions about a dataset. The usual way this is done is to use part of each datum to create a data bin. Let's imagine a dataset where all the data consists of numbers between 1 and 99. It doesn't matter how big the dataset is or if there are duplicates. Now imagine 10 bins. The first bin is for numbers between 1 and 9; the second for numbers between 10 and 19, and so on. The numbers of the bins will be labelled 0 to 9. The bins are the stems. So we just go through all the data and put each datum into its appropriate bin. But we don't have to put the whole of the data into each bin, because the bin number is already numbered with the first digit of the data. So the contents of each bin just contain the second digit of the data. The bins (stems) are lined up in order 0 to 9 and we can also stack their contents so that the single digits are in order inside the bins. These are the leaves. Imagine the bins are made of glass. We can look at the bins and the heights of the stacks of contents. The heights of the contents form a shape as we run down the line of bins. These heights tell us how many data there are in each bin and indicate where the most data is and where the least data is. This is is a frequency distribution. It's the basis of the Stem and Leaf plot and can be represented by a table or chart. Each row of the table starts with the bin number (STEM) and along the row we have the contents of the bin (LEAVES). Turn the table on its side and we have a chart with the stem running along the bottom and the leaves forming towers over the stems. The chart resembles the row of bins with the stack, or column, of contents over them, but the bins are now invisible, and only their labels remain as regular horizontal divisions on the chart. But it doesn't stop there. This frequency chart tells us where most of the data can be found, where its middle is and the general shape of the data. These are important statistical observations. Not all the bins may have data in them, and some will have lots of data. Random data will produce no particular shape, but in many cases there will be a pattern. We've considered numbers from 1 to 99, but the data can have any range as long as the data is binned carefully to reflect the relative magnitude of the data. If the data were between 250 and 400, for example, we might take the first 2 digits as the bin label: 25 to 40 and the contents would be the third digit. So you need to make a decision based on the range of data values to decide how the data is going to be binned. I hope this helps you to understand Stem and Leaf plots.