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example how to simplify integer expressions
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Question:I can not figure out how to help my daughter. The problem is to assign a variable and write a equation for teh following  a football and a basketball cost 65.00. a football costs $5.00 more than 1/2 of the basketball. How much does the football cost?
Answers:The easiest way to work it is to assign two variables. Let F = the cost of a football Let B = the cost of a basketball "A football and a basketball cost $65.00" means: F + B = 65 "A football costs $5.00 more than 1/2 of the basketball" means: F = 5 + B/2 We can rewrite the first equation by subtracting F from both sides: B = 65  F This lets us substitute "65  F" in place of B in the second equation: F = 5 + (65  F)/2 Multiply both sides by 2: 2F = 10 + (65  F) Simplify: 2F = 75  F Add F to both sides: 3F = 75 Divide by 3: F = 25 So a football costs $25 Sanity check: If a football is $25, then a basketball should be $65.00  $25.00 = $40.00 Using 25 for F and 40 for B, does our second equation still work? Yes. 25 does equal 40/2 + 5.
Answers:The easiest way to work it is to assign two variables. Let F = the cost of a football Let B = the cost of a basketball "A football and a basketball cost $65.00" means: F + B = 65 "A football costs $5.00 more than 1/2 of the basketball" means: F = 5 + B/2 We can rewrite the first equation by subtracting F from both sides: B = 65  F This lets us substitute "65  F" in place of B in the second equation: F = 5 + (65  F)/2 Multiply both sides by 2: 2F = 10 + (65  F) Simplify: 2F = 75  F Add F to both sides: 3F = 75 Divide by 3: F = 25 So a football costs $25 Sanity check: If a football is $25, then a basketball should be $65.00  $25.00 = $40.00 Using 25 for F and 40 for B, does our second equation still work? Yes. 25 does equal 40/2 + 5.
Question:i have a quiz tomorrow on simplifying algebraic expressions. &, im going to failhonestly. i dont know how to do these, so; what im asking is for the steps on how to do these. for example, we have questions like 12z + 4(z9)+30+z. and others like, 2a5(a+1).
please, if you know how to do them, can you please help me. i am in 7th grade advanced math and i really dont get this. thank you.
Answers:When simplifying any equation remember this: Please Excuse My Dear Aunt Sally Parentheses, Exponents, Multiplication, Division, Addition, Subtraction This is the order of simplification so for you first example the first thing to do is to get rid of the parentheses by distributing the 4 to both numbers inside. 12z + 4z 36 + 30 + z there are no exponents so you can skip that step, also there is no mult. or div. so the next thing to simplify is add/ sub. Remember, you can only combine like terms so no mixing up numbers with variables. All items with a "z" we can combine together. 12z +4z +z = 7z now all you have left is the +30 and 36 which you then combine together to get 6. now we cannot combine the last two tems so just put them together and get 7z 6. some teachers do not like the first term to be negative so to change that it would be 7z+6 all right now given these rules see if you can figure the next one on your own
Answers:When simplifying any equation remember this: Please Excuse My Dear Aunt Sally Parentheses, Exponents, Multiplication, Division, Addition, Subtraction This is the order of simplification so for you first example the first thing to do is to get rid of the parentheses by distributing the 4 to both numbers inside. 12z + 4z 36 + 30 + z there are no exponents so you can skip that step, also there is no mult. or div. so the next thing to simplify is add/ sub. Remember, you can only combine like terms so no mixing up numbers with variables. All items with a "z" we can combine together. 12z +4z +z = 7z now all you have left is the +30 and 36 which you then combine together to get 6. now we cannot combine the last two tems so just put them together and get 7z 6. some teachers do not like the first term to be negative so to change that it would be 7z+6 all right now given these rules see if you can figure the next one on your own
Question:How do I simplify
4x / 3xy^2
Also what are some rules of simplifiying these kind of problems? Im having trouble doing it when there are fractions and exponents, and multiple variables, etc/
Answers:4x / 3xy^2 = 4 / 3y^2 [Because the x in the denominator (bottom bit) cancells with the x in the numerator (top bit) But that's all that can be done with it. As for the rules if you google for them you'll find them Here's a powerpoint version thats pretty good: http://www.ltscotland.org.uk/Images/therulesofindices_tcm4123386.ppt It doesn't give the rule for fractional exponents because thats in another presentation which wasn't listed and I'm too lazy to try and find it but this rule might help: (a^m)^n = a^(mn) is used to help evaluate terms with fractional exponents a^(p/q) means the qth root of a^p So a^(p/q) means (a^p)^(1/q) but to simplify stuff its probably better to think of it as meaning [a^(1/q)]^p because this is easier to evaluate without having to use a calculator. When you have a fraction raised to an exponent the exponent refers to both the numerator and the denominator of the fraction For example (8/27)^(2/3) = [8^(2/3)] / [27^(2/3)] = [8^(1/3)]^2 / [27^(1/3)]^2 = 2^2 / 3^2 = 4 / 9 Not sure if that'll help ... hope it does.
Answers:4x / 3xy^2 = 4 / 3y^2 [Because the x in the denominator (bottom bit) cancells with the x in the numerator (top bit) But that's all that can be done with it. As for the rules if you google for them you'll find them Here's a powerpoint version thats pretty good: http://www.ltscotland.org.uk/Images/therulesofindices_tcm4123386.ppt It doesn't give the rule for fractional exponents because thats in another presentation which wasn't listed and I'm too lazy to try and find it but this rule might help: (a^m)^n = a^(mn) is used to help evaluate terms with fractional exponents a^(p/q) means the qth root of a^p So a^(p/q) means (a^p)^(1/q) but to simplify stuff its probably better to think of it as meaning [a^(1/q)]^p because this is easier to evaluate without having to use a calculator. When you have a fraction raised to an exponent the exponent refers to both the numerator and the denominator of the fraction For example (8/27)^(2/3) = [8^(2/3)] / [27^(2/3)] = [8^(1/3)]^2 / [27^(1/3)]^2 = 2^2 / 3^2 = 4 / 9 Not sure if that'll help ... hope it does.
Question:Ok, here is the problem..
I've a simple algebra expression like
A&B&C&D  A&B&D&E  B&E&F
by simple I mean, there are only 2 operations possible AND denoted by '&' and OR by ''.. the expression could be complex with brackets and involve like 50 terms. There is no NOT operator or any other fancy operators in my expression. Also AND has more precedence than OR.
I need to simplify this expression like so...
pulling B out of all the 3 terms
B & ( A&C&D  A&D&E  E&F )
again, pulling A out of the first 2 terms
B & ( A & (C&DD&E)  E&F)
again pulling D out of the innermost expression
B & ( A & D & (CE)  E&F)
This can be simplified any further as there are no more common elements in the terms. By simplification Im trying to minimize the number of operations in the final expression. Basically I'm applying the following axioms/rules to the expression again & again until I can find no more such patterns
A&A = A
AA = A
A&1=A
A1=1
A&B  A&C = A (BC)
(AB) & (AC) = A  B&C
I think there must be an algorithm in the literature somewhere that does this, only Im not able to find it. I thought I'll give it a try myself and I got to the point I had a binary tree equivalent of the input expression and transform the tree recursively whenever there are common terms to the left & right of the node .. but it created problems as the tree was not balanced and it refused to find common subtrees.
So here I am. Need an algorithm to simplify an algebraic expression.
more example....
A&(BC)  A&D&(BC)  A&E&(BC) algo should find both A and the subexpression (BC) as being common I ran into Karnaugh Map the moment I googled for a solution but I dont think it solves the same problem as mine. If they are the same problem, can you explain it how it solves with the example expression in the question?
Answers:You can use Boolean Algebra for this and what is called Karnaugh Mapping. http://en.wikipedia.org/wiki/Karnaugh_map The purpose of the karnaugh map is to reduce or simply the amount of logic gates needed in a circuit.
Answers:You can use Boolean Algebra for this and what is called Karnaugh Mapping. http://en.wikipedia.org/wiki/Karnaugh_map The purpose of the karnaugh map is to reduce or simply the amount of logic gates needed in a circuit.
From Youtube
simplifying algebraic expressions :WEBSITE: www.teachertube.com quick example on how to simplify an algebraic expression
Evaluating Algebraic Expressions: Division of Integers Problem 1 :www.greenemath.com In this video we explain how to evaluate an algebraic expression for a given value of the variable. This example involves division of integers.