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how do I do this problem 3A+7=49

I'm confused can you do it step by step so I can learn how to do equation

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40 problem 7 3a your answer is correct on terrell - Course Hero


40 Problem 7 3A Your answer is correct On May 31 2014 Terrell Company had a from ACC 557 at Strayer
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Problem Solving: If 3+2=7, 5+4=23, 7+6=47 and 9+8=79 ... - Quora


Problem Solving: If 3+2=7, ... first number with second one and adding a series of odd number starting from 1,3,5,7,9 ... the Logic Crash Course for only $49.
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Type Problem 7-3A On May 31, 2014, Terrell Company... | Chegg.com


Answer to Type Problem 7-3A On May 31, 2014, Terrell Company had a cash balance per books of $6,781.50. The bank statement from H...
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SOLUTION: how would i do this problem?? 3a^2 + 2a - 7 into ...


how would i do this problem?? 3a^2 + 2a - 7 into 27a^4 + 18a^2 - 2a + 42
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Financial Accounting Chapter 3 - Problem 3A Closing Entries ...


Oct 14, 2012 · Financial Accounting Chapter 3 - Problem 3A ... Financial Accounting Chapter 3 Problem 1A ... Unit 3 - Part 7: Closing Journal Entries - Duration: 13:49.

SOLUTION: how would i do this problem?? 3a^2 2a - 7 into 27a ...


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HOW DO YOU SOLVE THIS PROBLEM (simplify) -2(3a – 7) + 1/2 (4a ...


HOW DO YOU SOLVE THIS PROBLEM (simplify)-2(3a – 7) + 1/2 (4a -8) = Download it free from the WordPress Repository . Join Now .
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How do I work out this problem:5 2/3 - Wyzant Tutoring


How do I work out this problem:5 2/3 - 3 1/6 ? I know the answer is 2 1/2 but how do I arrive at that answer?
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Who Can Do the Math? Problem Solving - MathPlayground.com


Who Can Do The Math? is an activity designed to help students become more reflective in problem solving. Students begin by reading a math problem and thinking about ...
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Problem 7-3A - Instructure


Please watch the video and do Problem 7-3A. this is Part 1 video. Part 2. Blank Problem 7-3A 12th edition.xls
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problem: x/3 + 5 = 2x - when simplifying the fraction why isnt 3 x 1x 3x instead of simply x.

I believe I answered my own question.  In problem 1 we are canceling out because using the denominator 3. In problem 2 we are multiplying because using a common denominator.
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how do i add fractions

To add or subtract fractions, obtain a least common denominator. Subtract the numerators in the correct order and retain the same least common denominator for your answer. Simplify. To multiply fractions, multiple the numerators. The product will be the numerator of your answer. Repeat with denominators. Simplify. To divide fractions, take the reciprocal of what you are dividing by. Multiply the reciprocal with the initial number (see above for multiplication process). Simplify. Evaluate means to solve. You can solve fraction problems using the above processes. You can only simplify if both the numerator and denominator are divisible by the same number. If the denominator is odd, you can only simplify it if the numerator also is divisible by a same number. Ex. 88/33. Although the denominator is odd, both the numerator and denominator are divisible by 11 resulting in 8/3 as the simplified answer. To pace yourself during a test do the following. Find out how long you have for the test. Divide this by the total number of problems on the test. Example. 1 hour for 20 problems on your test. This means you have 3 minutes per problem. If you spend more than 3 minutes on a problem, skip it. Continue until you attempt all the problems. Go back with the remainder of the time to retry these problems you skipped. Most likely they are the most difficult, hence why you spent alot of time on them. This method of pacing allows you to skip the hard problems at first, attempt all problems, and finish the easier problems for sure.
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What fraction of the problems on the math test will be word problems?

????????? yu wanna no if me kan reed mind av yer teech ????????? first, giv me his name
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HCF of 81 and 162

Lets say the problem at hand is, we want to simplify this fraction by finding the HCF:  81 / 162.   You may have spotted that 81 is the HCF for this, but they want your method. The "Right" method, when followed, will give a correcct result for this problem, but also for any similar problem, even if harder than this particular problem at hand. The right method involves listing the prime number factors and number of occurrences in both numbers.  A prime factoring tool would make this easier, if you have one. 81 is 9 x 9  or        3 x 3 x 3 x 3 or written as 3^4 for factor of 3 with 4 occurrences. 162 is 9 x 9 x 2 or 3 x 3 x 3 x 3 x 2 or just 2 x 3^4 Then select only the primes and number of occurrences that are the same in both lists.   They do not have 2 in common, but do have 3 with 4 occurrences in common. 3^4 Multiplying these factors and occurrences will give the correct HCF. 81 Then divide the HCF into each number in the problem to simplify the fraction. 81 / 81 = 1 162 / 81 = 2 The fraction simplifies to 1 / 2. So much for the "right" method.   There may also be sometimes a shortcut for a particular problem at hand.  In this case, you may try dividing the larger number by the smaller one to see if it goes an exact number of times.  The answer is yes, it does go exactly 2 times.  Thus, even without a prime factoring tool, you know that the smaller number of 81 can serve as the Highest Common Factor to simplify this problem.
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Dierdre's work to solve a math problem is shown below. Problem: How many 1 1/2- foot pieces can be cut from a 12-foot length of a ribbon?

Dierdre's work to solve a math problem is shown below. Problem: How many 1 1/2- foot pieces can be cut from a 12-foot length of a ribbon? Step 1: 1 1/2 divide by 12 = n Step 2: 3/2 divide by 12 = n Step 3: 3/2 X 1/12 = n Step 4: 3/24 = n Answer: 1/8 = n What was Dierdre's first error? A. She switched the divisor and the dividend in step 1 when creating an equation to model the problem. B. She used an improper fraction in step 2 that is not equivalent to 1 1/2. C. She replaced the divisor by its reciprocal in step 3 instead of re[;acing the dividend by its reciprocal. D. She reduced the fraction in step 4 incorrectly. Her first error was A. She switched the divisor and the dividend in step 1 when creating an equation to model the problem.
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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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|3x-2|=2sqrtx+8

Four problems: Problem 1: 3x - 2 = 2sqrt(x) + 8 3x - 10 = 2sqrt(x) 9x^2 - 60x + 100 = 4x 9x^2 - 64x + 100 = 0 x = (64 +- sqrt(4096 - 3600))/18 x = (64 +- sqrt(496))/18 x = (64 +- 4sqrt(31))/18 x = (32 +- 2sqrt(31))/9 x = (32 + 2sqrt(31))/9, (32 - 2sqrt(31))/9 . Problem 2 -3x + 2 = 2sqrt(x) + 8 -3x - 6 = 2sqrt(x) 9x^2 + 36x  + 36 = 4x 9x^2 + 32x + 36 = 0 x = (-32 +- sqrt(1024 - 1296))/18 Can't do square root of a negative, so no solution. . Problem 3 3x - 2 = 2sqrt(x + 8) 9x^2 - 12x + 4 = 4(x+8) 9x^2 - 12x + 4 = 4x + 32 9x^2 - 16x - 28 = 0 x = (16 +- sqrt(256 + 36*28))/18 x = (16 +- sqrt(256 + 1008))/18 x = (16 +- sqrt(1264))/18 x = (16 +- 4sqrt(79))/18 x = (8 +- 2sqrt(79))/9 x = (8 + 2sqrt(79))/9, (8 - 2sqrt(79))/9 . Problem 4: -3x + 2 = 2sqrt(x+8) 9x^2 - 12x + 4 = 4(x+8) 9x^2 - 12x + 4 = 4x + 32 9x^2 - 16x - 28 = 0 same as Problem 3 x = (8 + 2sqrt(79))/9, (8 - 2sqrt(79))/9 . Answer: If you meant abs(3x-2) = 2sqrt(x) + 8 then the answer is:  x = (32 + 2sqrt(31))/9, (32 - 2sqrt(31))/9 If you meant abs(3x-2) = 2sqrt(x+8) then the answer is:  x = (8 + 2sqrt(79))/9, (8 - 2sqrt(79))/9
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 Eight is the quotient of a number and five.

A division problem 8=n÷5
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twice a number added to seven is equal to five less than five times the number

There's a problem. "Twice a number added to seven" could mean 2 * (x + 7) or it could mean 2x + 7 "Five less than five times the number" could mean 5x -5 or it could mean (5-5) * x Without knowing which one you mean, we have four problems: Problem 1: 2 * (x + 7) = 5x - 5 2x + 14 = 5x - 5 2x + 19 = 5x 19 = 3x x = 19/3 . Problem 2: 2x + 14 = (5-5) * x 2x + 14 = 0 * x 2x + 14 = 0 2x = -14 x = -7 . Problem 3: 2x + 7 = 5x - 5 2x + 12 = 5x 12 = 3x x = 4 . Problem 4: 2x + 7 = (5-5) * x 2x + 7 = 0 * x 2x + 7 = 0 2x = -7 x = -7/2
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How do I solve systems of equations using substitution?

solve for one variable(get it by itself on one side of the equal sign) ex. y= 2x-3 then plug it in (substitute it in where ever you see that variable) ex. x+y=2 becomes x+(2x-3)=2 example: -x+y=-2 2x+y=4 add x to both sides in the first problem y=x-2 substitute that into the second problem 2x+(x-2)=4 combine like terms 3x-2=4 add 2 to both sides 3x=6 and divide by 3 x=2 then plug 2 into the origional problem (either one) for x that will give you "y" in this case = 0
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