Guide :

# word problem

Cheryl is planning a garden that will be 8x7 feet. She wants it to be raised by 6 inches. How many cubic feet of topsoil will she need to add?

## Research, Knowledge and Information :

### Math Word Problems | MathPlayground.com

Solve math word problems with Thinking Blocks, Jake and Astro, IQ, and more. Model your word problems, draw a picture, and organize information!

### Word Problems - Dynamically Created Math Worksheets

These Word Problems Worksheets are perfect for practicing solving and working with different types of word problems.

Word Problems. Word problems are one of the first ways we see applied math in grade school, and also one of the first anxiety producing math challenges many grade ...

### IXL - Word problems

Welcome to IXL's word-problem page. We offer fun, unlimited practice in more than 200 different word-problem skills.

### Word problem (mathematics education) - Wikipedia

In science education, a word problem is a mathematical exercise where significant background information on the problem is presented as text rather than in ...

### Word Problem Worksheet Generator

Word Problem Worksheets. To create a set of free word problems, select the number of problems and difficulty level, then click the Generate button.

### explanation and exercises

Math series Solving math word problems. There are two steps to solving math word problems: Translate the wording into a numeric equation that combines smaller ...

### Word Problems - BrainPOP

In this educational animated movie about Math learn about problem-solving, mathematical, unknowns, and algebra.

### Translating Word Problems: Keywords | Purplemath

Purplemath. The hardest thing about doing word problems is using the part where you need to take the English words and translate them into mathematics.

## Suggested Questions And Answer :

### What fraction of the problems on the math test will be word problems?

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i need help wit this algebra problem  ac+xc+aw+xw²im trying to get the work and answer so i can check my work here are the vocabulary words  gcf, factors, grouping, prime factors, and perfect square

### what is word problem?

a word problem is a exersice of maths in words

### How do I figure this word problem out?

How do I figure this word problem out? Printing an article of 48000 words, using two sizes of type. The larger type contains 1800 words per page, the smaller type contains 2400 words.  The article is allowed 21 full pages. How many pages must be in the smaller type?   Large Type = L =1800 L Small Type = S = 2400S   1800S+ 2400L = 48000 <= eq1 L + S = 21 L = 21 - S <= eq2   1800 / (12 - S)+ 2400 / S = 48000 / (L+S) 1800S + 2400 L  = 48000 18S + 24L = 480 from eq2 18S + 24(21-S) = 480   18S + 24(12) - 24S)  = 480 - 6S + 504 = 480 6S = -(480 - 504) 6S = 24 Small Type (S) = 4 ◄ Ans

### 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.

### 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.

### Coin word problem help please

Coin word problem help please There is \$68.50. It is made up of 440 US coins. There are three types of coins. What are the coins? I will accept the premise that whoever created this problem believed that there is only one set of coins, and the combinations of those coins, that satisfy the stated requirements. It falls to me to prove that assumption false. There are five U.S. coins in circulation; the half-dollar (H), the quarter (Q), the dime (D), the nickel (N) and the penny (P). Taking any three of them, in certain combinations, we can have 440 coins that add up to \$68.50 as I will show. I will begin each list with the three letters repesenting the coins, as inidcated above. The largest coin is listed first, the mid-value coin second, and the smallest coin last. For each group, I will give the first combination and the last combination. Then, I will give the increments/decrements for each of the coins in order to progress to the next combination that satisfies the problem requirements. HQD There are 20 combinations.  1) 2/158/280 20) 59/6/375 To move to the next combination, H increases by 3, Q decreases by 8, D increases by 5. HQN There are 25 combinations.  1) 6/219/215 25) 102/3/335 For each step, H increases by 4, Q decreases by 9, N increases by 5. HQP There are 5 combinations.  1) 26/214/200  5) 122/18/300 For each step, H increases by 24, Q decreases by 49, P increases by 25. HDN There are 42 combinations.  1) 62/372/6 42) 103/3/334 For each step, H increases by 1, D decreases by 9, N increases by 8. HDP There are 7 combinations.  1) 68/342/30  7) 122/48/270 For each step, H increases by 9, D decreases by 49, P increases by 40. HNP There are 7 combinations.  1) 106/304/30  7) 130/10/300 For each step, H increases by 4, N decreases by 49, P increases by 45. QDN There are 69 combinations.  1) 164/274/2 69) 232/2/206 For each step, Q increases by 1, D decreases by 4, N increases by 3. QDP and QNP have no combinations that equal \$68.50, and for DNP there are no combinations that even reach \$44.00. As you can see, all five types of coins can be used, in the appropriate combinations, to have 440 coins add up to \$68.50, making the question, "What are the coins?" unanswerable, or at least, not unique.

### 1. What fraction of the problems on the math test will be word problems?

20 + 10 + 5 = 35 word problems = 5 / 35 or 1 /7 multiple choice = 20 / 35 = 4/7 you didnt ask but fill in the blank = 10 / 35 = 2 / 7 check 1 / 7 + 4 /7 + 2 / 7 = 7 / 7 or 100%

### 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