How can I simplify this equation “(5x^8) • (7x^9)”?
Since multiplication is commutative, you can rearrange the four factors of your multiplication. (Note that there are also invisible multiplication signs between 5 and [math]x^8[/math] and between 7 and [math]x^9[/math], so it’s indeed just 4 things to be multiplied together:)[math](5x^8)(7x^9) = (5)(7)(x^8)(x^9) = 35 x^{8+9} = 35x^{17}[/math](Btw, [math](5x^8)\cdot(7x^9)[/math] is not called an equation since nothing is being compared to or equalled to anything else. It’s rather an “expression”, or you could also call it a “function of x”.)
Algebra 2 factor/simplify equations?
1) 8r^2 - 7r^2 + 3r + 2 = 7r^2 - 7r^2 r^2 + 3r + 2 = 0 (r + 1)(r + 2) = 0 2) 3x^2 - 8x - 16 = 0 x = (8 +/- sqrt(64 - 4(3)(-16)) / 2(3) x = (8 +/- sqrt(-128)) / 6 x = (8 +/- 8isqrt(2)) / 6 x = (4 +/- 4isqrt(2)) / 3 3) 28n^2 + 184n + 96 = 0 n = (-184 +/- sqrt(33,856 - 4(28)(96)) / 2(28) n = (-184 +/- sqrt(23,104)) / 56 n = (-184 +/- 152) / 56 n = (-23 +/- 19) / 7 n = -4/7 and n = -6 (n + 6)(n + 4/7) (n + 6)(7n + 4) I hope this information was very helpful.
(20+40x)/(20x) simplify the equation - algebra 2?
Divide top and bottom by 20, giving (1 + 2x)/x ... x ≠ 0 In some situations, rejigging it as 2 + 1/x might give a clearer indication of what happens as x approaches plus/minus infinity. –––––––––––––––––––––––––
Simplify alg. 2 equations?
1.) (2-3i)(1+2i) (a) -4-i (b) 8+i (c) -4+i (d) 8-i 2.) 8-4i / 2+2i (a) 1-3i (b) -2+i (c) -3+4i (d) 4-2i
Can someone simplify this equation quickly?!?
4/(X+3)=8 Multiply both sides by (x+3) 4 = 8(x+3) Divide both sides by 8 4/8 = x+3 Subtract 3 from both sides .5 - 3 = x -2.5 = x
Can you simplify x-1 = x+1 ?
Yes. Subtract X-1 from both sides to get(x-1) - (x-1) = (x+1) - (x-1) this gives you0 = 2 ??? OOPS something is wrong! Lets look at this another way:Let x = 2 Then, form the equation,x² = 2x True! Then, subtract 4 from both sides:x² - 4 = 2x - 4 True! Now, factor each side without changing any values:(x + 2) (x - 2) = 2 (x - 2) Done properly! Now, divide by both sides by (x - 2) to get(x + 2) = 2 Simplified! Therefore,x = 2 - 2 = 0 !!!!.!.!….!……..???? didn’t we start with x = 2??? we must then be saying:x = 2 = 0 So therefore,2 = 0 !!!!……????? what went wrong??? Or did we just prove that2 always equals 0?This tells me that I can eat 2 hamburgers at every meal and still lose weight! (Oh, how I wish!!! but probably that won’t happen!)Why does this not work? All of the algebra seems to be in order. For the answer, look at the line where we divided both sides by the term (x - 2). Since we had originally defined x = 2, then the term we divided by can be evaluated this way…IF x = 2, then x - 2 = 2 - 2 = 0Thus we divided both sides by zero, a process that is undefined. Although, lets try sneaking up on the answer: The limit of (x - 2) as x approaches 2 is zero. And any non-zero number divided by the term (x - 2) as x approaches 2 is infinite, also undefined. Thus all steps beyond this one are invalid. Although…I…..I am still wishing I could eat two hamburgers at every meal and still lose weight!QED
Algebra equation! 5s(-6t) Simplify! 10 points!!?
The answer is -30st! Five times a negative six is thirty and only in multiplication can you combine variables like that! You seem like you have it down, but I would brush up on your multiplication tables because you are going to need to know things like five times six! Sorry, I don't mean to be harsh, but I am just preparing you! Good Luck!
Simplifying expressions? (Algebra 2)?
(13m^4 + 2) + (m^4 n^2 + 2 - 2m^4) - (-13m^2 n^3 + 5m^4) The first and second parentheses are redundant and can be removed since they have a plus sign (implied for the first, explicit for the second) in front of them. Since the third parentheses has a minus sign, apply the minis sign to all of the terms in it by reversing the signs: 13m^4 + 2 + m^4n^2 + 2 - 2m^4 + 13m^2n^3 - 5m^4 Rearrange the terms in descending order and combine like terms: 13m^4 - 2m^4 - 5m^4 + m^4n^2 + 13m^2n^3 + 2 + 2 6m^4 + m^4n^2 + 13m^2n^3 + 4 Follow the same process for problem 2. (a+b) (c+d) Multiply using FOIL: First: a * c = ac Outside: a * d = ad Inside: b * c = bc Last: b * d = bd Combine into: ac + ad + bc + bd