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algebra bool (What is the interest of simplifying a logic function


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Boolean Rules for Simplification | Boolean Algebra ...

Boolean Rules for Simplification ... algebra finds its most practical use in the simplification of logic circuits. If we translate a logic circuit’s function ...
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Any good boolean expression simplifiers out there? - Stack ...

Any good boolean expression simplifiers ... The website doesn't seem that good for simplifying boolean ... Simplifying a logic function using boolean algebra. 1.
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Boolean Algebra - Digital Electronics Course

Using Boolean Algebra to simplify or reduce Boolean expressions which represent circuits. ... Boolean Algebra. Boolean algebra, a logic algebra, ...
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What Is Boolean Algebra? - Quick and Dirty Tips

... What Is Boolean Algebra? play; pause; mute; unmute; max volume; Update Required To play the media you will need to either update your browser to a recent version ...
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Boolean Algebra and Reduction Techniques

Boolean Algebra and ... Simplification of Combinational Logic Circuits Using ... Reduction of a logic circuit means the same logic function with fewer gates ...
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Digital Logic - Using Minterms

Logic Functions In the ... That implies the following algorithm for generating a Boolean algebra function from a ... Getting A Function From a Truth Table & Simplifying.
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4 BOOLEAN ALGEBRA AND LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra.
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Boolean Algebra Calculator, Boolean Algebra Simplifier, Logic ...

Boolean Algebra Calculator (or Venn Diagram Calculator) ... Boolean Algebra Calculator (also known as Truth Table Generator or logic calculator) ...
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Boolean Algebra - University of Iowa

Boolean Algebra A + 0 = A A ... Synthesis of logic circuits ... Any function can be implemented using only NAND or only NOR gates. How can we prove this?
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Wolfram|Alpha Examples: Boolean Algebra

Boolean Algebra. A few examples of what you can ask Wolfram|Alpha about: ... compute a logic circuit for a Boolean function. logic circuit (p or ~q) and (r xor s)
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Suggested Questions And Answer :

algebra bool (What is the interest of simplifying a logic function

bool or not, simpel is alwaes gooder than komplex
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How to simplify/break down a factorial function?

(k+1)!=(k+1)k! Is the simple logical solution. Note that 0!=1.
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How do I find the simplified form of the difference quotient for the function: f(x) = ax^4

How do I find the simplified form of the difference quotient for the function: f(x) = ax^4  this is simple, remember that the a is a constant.  so there is NO PRODUCT. f'(x) = 4ax^3
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How to use "if" & "and" functions in formula bar for calculating exact value of particular cell

This the Excel formula bar? General format IF({test statement},{value if true},{value if false}). IF statements can be nested so that more IF statements can be inserted where the values are. The AND function has the format AND({condition1},{condition2}...) and returns TRUE or FALSE. Combining these we have: =IF(AND(A1>=3,A1<=6),0.3,IF(AND(A1>=6,A1<=12),0.5,IF(AND(A1>=12,A1<=24),1,0))) as the formula in B1. EXPLANATION The three closing brackets or parentheses (")")are necessary to satisfy the "grammar" or syntax. The last one belongs to the first IF statement, the middle one to the second IF statement and the first belongs to the third IF statement. Excel will show a syntax error if any brackets are missing. A quick way of checking is to count the number of opening brackets and the number of closing brackets; they must be the same. In this case, there are 6 of each. Because AND returns TRUE or FALSE it's not necessary to follow it with "=TRUE", although you may do if you wish, but if you do it must come before the comma. The formula bar must start with "=" otherwise, Excel will assume you are writing a string of text. Excel will then help you through construction of the formula. Excel encounters "=" in the formula bar and knows a formula follows. It sees the IF statement and expects 3 parameters separated by commas. The first parameter is the AND function; the second is 0.3 and the third is another IF statement. If the AND result is TRUE, 0.3 goes on B1. Otherwise it moves on to the third parameter, the second IF statement and applies the same logic. The third parameter of the second IF statement is activated if the second AND function returns FALSE, and Excel encounters the third IF statement, applying the same logic. The third parameter of the third IF statement is zero, so B1=0 if the last AND function returns FALSE.
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Explain the results of the following options: Option 1: 6% compound interest quarterly for 5 years. Option 2: 8% compound interest annually for 5 years. Option 3: 14.5% simple interest for 10 years.

MEMO TO CLIENT Simple interest applies the interest rate proportionately, so the amount of interest on a particular investment is directly proportional to the length of time invested. This means that, for example, if the investment period is 5 years, the interest is 5 times the interest earned in one year; for 10 years it is 10 times that earned in a year. It is also easy to calculate because of this simple proportion. Compound interest is more rewarding to the investor. After a period of time, for example, a year, the interest earned in the year is added to the original amount invested. So at this point, it is the same as simple interest. But what happens next is different. The investment plus the interest becomes the invested amount for the next period, the next year, for example. At the end of this period the process continues, and the interest is again added and becomes the investment amount for the next period. So it is clear that over a period of time more and more interest is earned. An important feature that investors need to be aware of is: how regularly is compound interest added? The shorter the period, the bigger the interest earned. Interest can be compounded annually, quarterly, monthly, daily or continuously. So, if the investor is quoted a particular annual rate of interest, then the largest amount of interest gained will correspond to the shortest compound interest period. As an example, take 6% per annum, or annual interest rate. After a year with interest compounded annually, 6% interest will be earned. If interest is compounded quarterly, then each quarter the interest will be added at a rate of 1.5% each quarter, but by the end of a year, the effective interest will be more than 6.1%. If interest is compounded monthly, the monthly rate would be 0.5% and after a year would be effectively closer to 6.2%. Interest compounded daily would be even closer to 6.2% and continuously would be slightly more. Growth is a convenient way of expressing the factor by which an investment increases over time, and companies will often publish tables to simplify calculations of expected returns on investments at fixed rates. The time periods will be typically 5, 10, 15, 25 years for a range of annual rates. So investors can quickly calculate the returns on varying amounts of money. As an example, take 15 years. The growth rates at 6% per annum would be: 1.9 (simple interest); 2.40 (compounded annually); 2.44 (compounded quarterly); 2.45 (compounded monthly); 2.46 (compounded daily or continuously). Option 1 6% annually is 1.5% quarterly, so growth is 1.015^20=1.3469, where 20 is 20*(1/4)=5 years. $500000*1.3469=$673,427.50 to best accuracy. Option 2 8% compounded annually: growth=1.08^5=1.4693 and amount is $500000*1.4693=$734,664.04. Option 3 14.5% simple interest for 10 years: 145% interest=1.45*500000=$725,000 interest+500000=$1.225 million. The first two options have the same investment time period, and option 2 is better. Option 3 has double the time period. If option 3 were to be applied over 5 years instead of 10, the interest would have been $362,500 (half of $725,000) and the total amount would have been $862,500. However, the investment is over ten years so the investor would need to wait 10 years before taking full advantage of the investment. Take option 2 over 10 years and we get a growth rate of 2.1589 making the investment worth $1,079,462.50, which is still smaller than option 3, which had a growth rate of 2.45 (1.45+1) because of the higher interest rate.  
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find a particular solution (e^x+y +ye^y)dx+(xe^y -1)dy=0 if y(0)=-1

This has the form of an exact DE. Let's see if it is.  First rewrite: (e^x+y+ye^y)+(xe^y-1)dy/dx=0=A(x,y)+B(x,y)dy/dx=0 where A=e^x+y+ye^y and B=xe^y-1. (p.d.=partially differentiate; wrt=with respect to)   Take A and p.d. wrt y=1+ye^y+e^y (a) Take B and p.d wrt x=e^y (b) Since (a)<>(b) we don't have an exact DE. (If it had been an exact DE, the two quantities would have been equal.) Can we find an integrating factor u such that p.d. wrt y of uA = p.d. wrt x of uB? That is, can we find u for (at this point u can be a function of x or y or both): D(ue^x)/Dy+D(uy)/Dy+D(uye^y)/Dy=D(uxe^y)/Dx-Du/Dx? (D is del or p.d.) e^xDu/Dy+u+yDu/Dy+u(ye^y+e^y)+ye^yDu/Dy=ue^y+xe^yDu/Dx-Du/Dx. This simplifies a little: e^xDu/Dy+u+yDu/Dy+uye^y+ye^yDu/Dy=xe^yDu/Dx-Du/Dx. u(1+ye^y)=(xe^y-1)Du/Dx-(e^x+y+ye^y)Du/Dy. If u is stipulated to be a function of x only then Du/Dy=0 and the above simplifies further: u(1+ye^y)=(xe^y-1)Du/Dx. Similarly, if u is a function of y only then Du/Dx=0 and: u(1+ye^y)=-(e^x+y+ye^y)Du/Dy. More to follow...
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Make a program on a TI-84 Plus Silver Edition Calculator?

There are some good programs that will do a lot of things at Look in the algebra section.
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I need help understanding a math question I am taking college Math

Simplify the algebraic expression 5xy - 3yz - 5xz + yz - 4xz + 7xy you find the like terms: 5xy and +7xy     - 3yz and +yz        -5xz and - 4xz there like terms if they have the same variables/exponents   THen you rewrite the problem with like terms next to eachother: 5xy  + 7xy  -3yz  +yz  -5xz  - 4xz Then you solve the like terms by themselfs: 5xy + 7 xy= 12 xy -3yz +yz= -2yz -5xz -4xz= -9xz so the answer is 12xy -2yz-9xz and you can't simplify it any more because none of them are like terms
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Basic functions

1. What is the range of k(x)=-|x| Asking for the range is another way of asking for valid Y values (or in this case k(x) values). So the range is (-infinity, 0]. 2.Is y=2 a constant function? Yes. Constant functions have no variation in the output for any input. 3. x=-3 a function? No it is not. In order to be a function, there are several properties which must be satisfied, one of which is that for any input, there must be a single output. Often we simplify this by asking "does the function pass a vertical line test?" So if you are to draw a vertical line, would it only cross the function in one place for all values of X? For X=-3, it is a vertical line, so it fails this property of functions because there are an infinite number of solutions (Y values) for this single X value (-3). 4. What is the minimum value of y=x^2-4 The minimum value of parabola's will occur either at the vertex (where its slope is 0) or at its domain boundaries. There are no stated domain restrictions for this function and the slope is 0 at X=0. The first derivative is y'=2x and y'=0 at x=0. The second derivative is y''=2, indicating it is concave for all values X. So the minimum is x=0 and y=-4. For a function with a higher power, there may be more than one minimum (each called local minimum) and any place the second derivative is concave up, both to the left and right segments where the first derivative is 0 will be one of these local minimums. You will take your inputs for each relative minimum and any domain limits (for functions which do not have all real numbers for their domain) and compare the Y values of the original function to determine the absolute minimum for that function.
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(1, -20), (7, -20), (2, -40), (6, -40), (5, -50), (8, 10), (9, 46), (0, 7), (-1, 47), (-2, 89)

First put all the points (x,y) in order of the x value: -2 89 -1 47 0 7 1 -20 2 -40 5 -50 6 -40 7 -20 8 10 9 46   When x=0 y=7, so the constant term in the function is 7 (y intercept). There's a minimum (turning point) near x=5 because the values of y for x=2 and 6 (-40) are more positive than -50. The gradient (dy/dx) gets positively steeper as x increases after x=5, and negatively steeper between x=-2 and +2. The degree of the polynomial is even because the function is positive for larger magnitude positive and negative values of x. The zeroes are between x=0 and 1 and between x=7 and 8. The function can be split into two parts: (1) y for -2 Read More: ...

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