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how can inverse operations help you find things

Suppose you lost your wallet or purse. How can inverse operation or actions help you find it?

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Inverse Operations in Math: Definition & Examples - Video ...

How to Use Inverse Operations. Inverse operations can be ... There are other inverse operations that can help you ... Inverse Operations in Math: Definition ...
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Inverse Operation | BetterLesson

Inverse operations can help you find ... SWBAT use inverse operations to find the missing numbers ... Using the Inverse Operation for Addition and Subtraction;
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Algebra 1: Things to Know Flashcards | Quizlet

How can we help? You can also find ... Algebra 1: Things to Know. STUDY. PLAY. ... Step 3: Use inverse operations to undo operations and isolate the variable. Literal ...
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Inverse Functions - Cool math Algebra Help Lessons - How to ...

This algebra lesson explains how to find the inverse of a function
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Using Inverse Operations to Solve Equations: Overview

Using Inverse Operations to Solve ... Once students understand that using inverse operations to solve an equation requires performing the same operation on ...
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Inverse Operations Flashcards | Quizlet

This is a set that can be used to learn your inverse operations. ... How can we help? You can also find more resources in our Help ... is the inverse of 7 ...
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Inverse Functions - Video & Lesson Transcript |

Inverse functions are two ... and subtraction are inverse operations. We can then also undo ... and career path that can help you find the school ...
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Solving Equations - Monterey Institute

Solving Equations . Learning Objective(s) ... Multiplication and division are also inverse operations. Multiplication can be undone by division, ...
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Intro to inverse functions (video) | Khan Academy

Intro to inverse functions. ... So this is going to be all real, but we're making it a nice contained set here just to help you visualize it. Now, ...
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How do you find the inverse of f(x)=(2x+7)/(3x-1)? | Socratic

How do you find the inverse of #f(x)= ... The inverse function is found by performing the inverse operations of each operation done on #x# that led to #y# in the ...
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Suggested Questions And Answer :

how can inverse operations help you find things

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I need help, I really need help pleeease.

I assume that (3) is a graph, because it doesn't display on my tablet. Question 1 a) 1 because the daily operating cost is stated specifically. b) 3 because the profit of $500 will be a point on the graph corresponding to a value of t; 2 because if we put P=500 and solve for t we get t=1525/7.5=203.3 so the graph would show t=203-204. c) 2 because we set P=0 and solve for t=1025/7.5=136.7, so 137 tickets sold would give a profit of $2.5, but 136 tickets would make a loss of $5; alternatively, if (3) is a graph then P=0 is the t-axis so it's where the line cuts the axis between 136 and 137. d) The rate of change is 7.5 from (2).  e) 2, because the format shows it to be a linear relationship; a straight line graph for (3) also shows linearity. Question 2 a) profit=sales- operating costs so 1025 in (2) is the negative value representing these costs. If (3) is a graph, it's the intercept on the P axis at P=-1025; (4) is the P value when t=0. b) For (1) you would need to find out how many tickets at $7.50 you would need to cover the operating costs of $1025 plus the profit of $500. That is, how many tickets make $1525? Divide 1525 by 7.5; 2 and 3 have already been dealt with; to use (4) you would note that P=$500 somewhere between t=200 and 250 in the table. c) For (1), work out how many tickets cover the operating costs. 1025/7.5=136.7, so pick 137 which gives the smallest profit to break even; 2 and 3 already given; in (4) it's where P goes from negative to positive, between 100 and 150. d) For (1) the only changing factor is the number of tickets sold. The rate is simply the price of the ticket, $7.50; if (3) is a graph, the rate of change is the slope of the graph, to find it make a right-angled triangle using part of the line as the hypotenuse, then the ratio of the vertical side (P range) and the horizontal side (t range) is the slope=rate of change; for (4) take two P values and subtract the smallest from the biggest, then take the corresponding t values and subtract them, and finally divide the two differences to give the rate of change: example: (475-(-275))/(200-100)=750/100=7.5. e) For (1) it's clear that the profit increases (or loss decreases) with the sale of each ticket by the same amount as the price of a ticket, so there is a linear relationship; 2 and 3 already dealt with; in the table in (4) the profit changes by the fixed value of 50 tickets=$375, showing that a linear relationship applies between P and t: for every 50 tickets we just add $375 to the profit.
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math inverse operation of 500-143 or help with understand inverse

i think you use the operation such as, + to - or * to /
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Please help

You need to see through the problem and apply whatever is necessary to reduce the number of variables till eventually there's only one to find. Remember a simple fact: if you have two variables you always need two independent equations to find them; for three variables, three equations; four variables, four equations. You use the multiplication property if it helps you to eliminate a variable between two equations. Take some examples: x+y=10, x-y=3; simply adding these two equations will eliminate y and help you find x. 2x=13 so x=6.5 and y=3.5; 2x+y=10, x-2y=10; we could double one equation or the other so as to match the coefficients of one or other of the variables; but since it's easier to add two equations rather than to subtract them, where we have a minus in one equation and a plus in the other, we would prefer to use the multiplier for the relevant variable. So we double the first equation and add to the second: 4x+2y=20 PLUS x-2y=10: 5x=30, making x=6 and y=-2. The last pair of equations could have been written: 2x+y=x-2y=10, but it's still two equations. There is no one way to solve equations, and you can save yourself a lot of stress by not assuming you have to remember a rigid technique or formula as “The Way to do it”. You'll find mathematics is more fun when you intelligently try different methods and use your natural creativity to guide you. And here's another interesting thing. Those questions about finding a missing number in a series can be tackled in many cases as solving simultaneous equations. You only need n equations to find n variables, and a series can be seen as a set of terms generated by a function y=f(x) for different values of x (the position in the series), giving different values of y (the terms in the series). f(x) is a polynomial of the type ax^n+bx^(n-1)+cx^(n-2)+... If there are four given terms n=3 and the variables are a, b, c and d; if there are 3 given terms, n=2 and the variables are a, b and c. There is always a solution, we just have to work through and find it!
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Can anyone HELP with the indicated matrix operations, let B=[-144-3]. Find -2B.

if yu hav 2 numbers in 1 roe, that be vekter... size =root av sum av squares... 3^2=9, 144^2=20,736, sum=20,745, root=144.0312466
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In algebra, why are all odd powers reversible?

Strictly speaking "NO" power operation (raising to an exponent) is reversible, because the inverse operation is ALWAYS multi-valued, unless we restrict the domain. In the case of even powers, for example: (+3) ^2 = 9 (- 3) ^2 = 9 Therefore the reverse operation thus has two possible solutions: Sqrt(9) = +3 Sqrt(9) =  -3 Also, it is "NOT" true that odd powers are reversible, if we consider complex numbers: (+3                        ) ^3  = 27 (-1.5 + j 2.59808) ^3  = 27 (-1.5 -  j 2.59808) ^3  = 27 The cube root of three (above) therefore has a total of three solutions, one "real"  and two complex conjugates. In the general case, the inverse operation of: y = (x)^n        --->   x = nth root of (y) has "n" solution. Odd roots(y) have a single real solution and (n-1)/2 pairs of complex conjugate solutions. Even roots of (y) have two real solutions (+/-) and (n-2)/2 pairs of complex conjugate solutions, assuming that "y" is positive. In the general case, the "nth root" of any real or complex power can be plotted as a set of equally spaced vectors in the complex plane, each with a magnitude of:    (|y|  / n) separated by angles of:   (2 pi / n)   or    (360 degrees /n)   In summary:     Even powers are reversible if we restrict our domain to positive numbers.     Even powers are NOT reversible in the INTEGER or larger domains,     because the roots are multi valued.     Odd powers are reversible in the REAL domain,     but NOT in the COMPLEX or larger domains. Hope this helps, MJS
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10 + 2 (3 + 2x) = 0 PEMDAS dictates you solve what is in parentheses first, but you cannot add what is in parentheses since they are not the same "language;" one is an integer and the other is a variable being multiplied by 2 (a coefficient).  No exponents here, so we start with multiplying/dividing:  Here our multiplication is in the form of distributive property: 2(3 + 2x) becomes 2*3 + 2*2x which is 6+4x.  Your equation is now 10 + 6 + 4x = 0. There is nothing more to multiply or divide since we don't know what x is yet., so we can add/subtract same "language" numbers:  10 + 6 is 16. Your equation is now 16 + 4x = 0, a simple two-step linear equation.  We want to find what x equals, so we need to get rid of anything on the side of the equation with x. The first step here is to get rid of the integer by doing the inverse of its operation.  Our integer is 16 and it's positive so it's being added.  The inverse (opposite) would be to subtract:  16 (- 16) + 4x = 0 (-16).  You must subtract on both sides of the equation or it will no longer be equal.  The 16's cancel out to zero on one side of the equation while 0 - 16 becomes -16 on the other side and leaves us with 4x = -16.  We need to change our 4x to 1x by using the inverse operation.  4x is 4 times x and the inverse of multiplication is division, so we will divide by 4 on both sides of the equal mark:  4x/4 = -16/4.  4x/4 is now 1x or just x (a coefficient of 1 is understood and not written next to the variable) and -16/4 is -4 (a negative divided by a positive is a negative), so x = -4.
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What is 5/3 divide by 1/3?

If the question had been: what's 5 apples divided by one apple you would have said 5, wouldn't you? The question has thirds instead of apples, but the answer is still 5. Don't forget that to divide by a fraction you invert the fraction and multiply: 5/3 times 3 is 5, because 3 cancels out the denominator in 5/3. I'm sorry you didn't get your answer in 10 minutes as you wished, but I've only just seen it! There are thousands of questions coming in and questions cannot always be answered just as they come in, and you have to remember that the website is available all over the world, which is split into many time zones. Add to that the fact that the questions are answered by users, like me, who have a life outside answering and asking questions! There are meals to eat, shopping to be done, going to the office or factory, going to school, you know the sort of thing. And, of course, while your question is coming in, a user may be answering another question, perhaps a harder one. I'm sure everyone does their best to help people who have questions to answer. But, be assured that there are many users who will be only too pleased to help you. It's just a question of when. Here's a clue that may help you. Below your question you will usually find a list of similar questions, many of which will probably have been answered (shown in green with the number of answers), so if you look at them, you may well find out how to do your own question. Now that will save everyone's time!  
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How do you add and subtract negative numbers? Like -4 + -9?

You may need a number line to help you, or if you understand how to use a thermometer, that will help. Your number line is like a thermometer. Somewhere round the middle of the line mark 0. To the left are all the negative numbers, to the right all the positives. On a thermometer (Celsius) 0 is the freezing point. All to the left are below freezing; all to the right are above freezing. -4 and -9 are on the left of zero, and -9 is further to the left of zero than -4, which means -9 is to the left of -4. Now take a positive number, say, 5 to the right of zero. What is 5-(-4)? To find out you want to know the distance between them on the number line. The answer is 9, because the distance between zero and 5 is 5 and the distance between -4 and zero is 4; the distance between -4 and 5 is therefore 9. So 5-(-4)=5+4=9. Remember, negative is left and positive is right. What is -1-(-4)? What does it mean? We start at -1, which is 1 to the left of zero. How far away is -4? It's 3 further to the left. But we know from above that 5-(-4) is the same as 5+4, so -1-(-4) is -1+4=4-1=3. So when we write -1-(-4), we're really saying how far is -1 away from -4. First thing to note is that it's to the right of 4, so right means positive so we know the answer is that -1 and -4 are separated by 3, and -1 is bigger than -4 by (positive) 3. What about -4-(-1)? To get from -1 to -4 we have to move 3 to the left so the answer is -3, because left is negative.  -4+(-9)  means start at -4 and go another 9 places to the left, ending up at -13. -4+(-9) is the same as -4-9. Play around with these ideas, and think also about thermometers and temperatures. -4 is a lower temperature than -1, and they're both below freezing. The better you can visualise numbers and abstract concepts the easier you will find the math.
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I need help with some math. Do you mind helping me anyone?

1 4 5 6 7
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