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i to solve word problems

ok they said i have to divide it

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Solving Math Word Problems

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Techniques and strategies for solving math word problems
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Translating Word Problems: Keywords | Purplemath

Lists some of the keywords that are useful in translating word problems from English ... The first step to effectively translating and solving word problems is to ...
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Math Forum: Ask Dr. Math FAQ: Word Problems

While there is no ONE BEST way to solve word problems, we can work through an example to illustrate the typical steps involved.
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SOLVING WORD PROBLEMS - El Paso Community College

SOLVING WORD PROBLEMS . Word problems can be classified into different categories. Understanding each category will give be an advantage when trying to solve word ...
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WebMath - Solve Your Math Problem

WebMath is designed to help you solve your math problems. Composed of forms to fill-in and then returns analysis of a problem and, when possible, provides a step-by ...
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Solving Word Problems - Metropolitan Community College

SOLVING WORD PROBLEMS 1. Read the problem all the way through quickly, to see what kind of word problem it is and what it is about. 2. Look for a question at the end ...
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Free Math Problem Solver -

Free math problem solver The free math problem solver below is a sophisticated tool that will solve any math problems you enter quickly and then show you the answer.
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Is there a website that solves mathematical problems? - Quora

I noticed Mathway | Math Problem Solver does not and I was interested to see if any startup did. This would focus mostly on finance word problems. ie Compound ...
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The One Best Way To Solve Word Problems

Learn The One Best Way to solve word problems. A single approach to solve word problems the way they were meant to be solved, using a repeatable process.
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How to Solve Algebra Word Problems | Universal Class

Before you start solving word problems in algebra, you should first already know about real numbers, how to manipulate algebraic expressions, and how to solve math ...
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Suggested Questions And Answer :

how do you know what formulas to use when solving for two unknowns

Problem: how do you know what formulas to use when solving for two unknowns I am trying to solve a word problem and still have two unknowns and don't know where to go from there. the problem states: A 10kg object on a 26 degree inclined plane. What are the objects weight and the normal force exerted on the object by the inclined plane? I don't understand what I'm missing. :( An object sitting on a flat surface exerts force straight down, vertical. An object on an inclined plane exerts force in both a vertical and a horizontal direction. You multiply the weight by the cosine of the angle to determine the vertical force, and multiply the weight by the sine of the angle to determine the horizontal force.
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solve word problem using polynomials

Problem: solve word problem using polynomials Joe has a collection of nickles and dimes worth $2.15 If the number of dimes was doubled and the number of nickles was increased by 28, the value of the coins would be $4.65. How many nickles and dimes does he have? Let's say he has n nickles and d dimes. 0.05n + 0.10d = 2.15 The problem proposes to increase d to 2d and increase n to n + 28. 0.05(n + 28) + 0.10(2d) = 4.65 We have two equation to work with. Let's multiply both equations by 100 to eliminate the decimals. 100(0.05n + 0.10d) = 2.15 * 100 5n + 10d = 215 100(0.05(n + 28) + 0.10(2d)) = 4.65 * 100 5(n + 28) + 10(2d) = 465 We need to expand this second equation. 5(n + 28) + 10(2d) = 465 5n + 140 + 20d = 465 5n + 140 + 20d - 140 = 465 - 140 5n + 20d = 325 Here's what we have: 1) 5n + 10d = 215 2) 5n + 20d = 325 We can immediately eliminate the n by subtracting equatioin 1 from equation 2.    5n + 20d = 325 -(5n + 10d = 215) ------------------------           10d = 110 10d = 110 10d/10 = 110/10 d = 11 This tells us that Joe currently has 11 dimes. Let's use equation 1 directly above to solve for n. 5n + 10d = 215 5n + 10(11) = 215 5n + 110 = 215 5n + 110 - 110 = 215 - 110 5n = 105 5n/5 = 105/5 n = 21 This one tells us that Joe currently has 21 nickels. We need to check the two numbers against what the problem originally stated. 0.05n + 0.10d = 2.15 0.05(21) + 0.10(11) = 2.15 1.05 + 1.10 = 2.15 2.15 = 2.15 Verified! Answer: Joe has 21 nickels and 11 dimes.  
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i need to know how to figure out word problems dealing with solving mph and minutes

i need to know how to figure out word problems dealing with solving mph and minutes the question is 50mph between two towns and they arrived in 25 minutes. im not looking for an answer just the formula to solve so i can study it   (mph) is a mile per hour → this a velocity (speed) so if talking about two towns mean that there will be distance to travel in mile(s). 25 min is the time that used to travel this two town.   as you are going to examen the velocity (mph or mile / hour) this mean Velocity = mile(distance) / hour (Time)   If you will be asked for Distance then Distance = Velocity x Time   If you are asked for time then Time = Distance / Velocity     This is whar you are going to need in solving problem like this.   Remember that Jesus loves you.  Know Him in the Bible God Bless  
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Write a word problem requiring either factorials or binomial expansions to solve, and then solve it.

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what kind of viva question can be asked by a teacher about the course of algebra by a masters students

Type of question:  "How can your subjects (algebra and calculus) best be explained progressively to children and young people to enable them to develop their understanding gradually to grasp the principles involved in these abstract subjects, so that they can demonstrate their understanding in applying it in a practical way to everyday problems?" "How can such learning be communicated so as to be a fun thing, rather than cold, formal and analytical, as is often taught?" (Children will often attempt to understand the subjects by learning by rote formulas and the like without any comprehension of how their understanding can be applied. This makes the subjects boring for them when, with the right teaching techniques, it could be fun! If children can be taught at every step of the way in terms of what they already fully understand, they will enthusiastically and gradually develop their understanding of more and more complex, abstract topics.) If you're actually looking for a problem to set students, there are many examples, but the ideal problem is one that gets students to apply their understanding of mathematics in a practical way. Combining geometry with algebra, is one way. Choosing a problem that tests students' understanding of the formula for solving quadratic equations, or trigonometric identities, or rules like the sine and cosine rules, or simultaneous equations, etc., without specifically stating what methods to apply in solving the problem, is a good way of discovering students' mathematical ability in practical application. Such problems could be presented as word problems and could be expressed concisely to appear simple, but nevertheless demanding on the intellect and comprehension. 
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Which of the following word problems can be solved using the equation 48 ÷ 12 = y?

explain the how following experiment to random urn experiments: (a) flip a fair coin twice (b) toss a pair of fair dice (c) draw two cards from a deck of 52 distinct cards with replacement after the first draw; without replacement after the first draw.
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solving word problems

Problem: solving word problems Staples recently charged $17.99 per box of Pilot roller ball pens and $7.49 per box for Bic matic mechanical pencils. If Kelling Community College purchased 120 boxes for a total of $1234.80 how many boxes of each did they purchase? x = number of boxes of Pilot pens y = number of boxes of Bic pencils x + y = 120 17.99x + 7.49y = 1234.80 x + y = 120 y = -x + 120 17.99x + 7.49y = 1234.80 17.99x + 7.49(-x + 120) = 1234.80 17.99x - 7.49x + 898.80 = 1234.80 10.5x + 898.80 = 1234.80 10.5x + 898.80 - 898.80 = 1234.80 - 898.80 10.5x = 336 10.5x/10.5 = 336/10.5 x = 32 y = -x + 120 y = -32 + 120 y = 88 17.99x + 7.49y = 1234.80 17.99(32) + 7.49(88) = 1234.80 575.68 + 659.12 = 1234.80 1234.80 = 1234.80 Answer: 32 boxes of Pilot pens, 88 boxes of Bic pencils  
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if 2 1/7 gallons of water fit into 1 box, how many boxes are needed to pack 8 3/4 gallons of water? boxes are needed

if 2 1/7 gallons of water fit into 1 box, how many boxes are needed to pack 8 3/4 gallons of water? boxes are needed Iplease solve this word problem please Divide 8 3/4 by 2 1/7 8.75 / 2.1429 = 4.083 It is impossible to squeeze the extra water into one of the boxes, and it should be understood that all the boxes are the same size, so..... 5 boxes are needed
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how to solve word problems

Problem: how to solve word problems To buy both a new car and a new house, Tina sought two loans totaling $318,338. The simple interst rate loan on the first loan was2.4%, while the simple intrest rate on the second loan was4.7%. At the end of the first year Tina paid a combined intrest payment of $14,761.81. What were the amounts of the two loans? The two loans are x and (318338 - x) The interest on the first loan was 0.024x The interest on the second loan was 0.047(318338 - x) The total interest was 0.024x + 0.047(318338 - x) = 14761.81 0.024x + 0.047(318338 - x) = 14761.81 0.024x + 14961.89 - 0.047x = 14761.81 -0.023x + 14961.89 = 14761.81 -0.023x + 14961.89 - 14961.89 = 14761.81 - 14961.89 -0.023x = -200.08 -0.023x/-0.023 = -200.08/-0.023 x = 8699.13 The first loan, for the car, was $8,699.13 The second loan, for the house, was $309,638.87  
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how do you solve x-19/x=3/8

x-(19/x)=3/8 x^2-19=(3/8)x 8x^2-3x -19*8=0 or 8x^2 -3x -152=0 x=4.55042978 & x=-4.17542978
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