Symbolab simplify

Excuse me????

2023.05.08 12:10 nico-ghost-king Excuse me????

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2023.04.19 10:22 Krypnicals i'm terribly sorry what

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2023.03.30 23:25 Cordobex29 Search for the criterion of the inverse of g(x) = - square root of x-1 plus 3.

I thought I was making everything good. I clearence the square root, so it looked like y-3 = square root of x-1. I took out the square root putting (y-3)² = x-1. Then, I tried to do (a+b)², I made something like y²+2y•-3+3² = -x-1. I did simplify that so it was looking like y²-6y+10 = -X, so I thought the result would be X=-y²-6y+10, but I looked for the result in Symbolab and it appeared that the result is X=y²-6y+10, with the expression positive so I don't understand how to get to that result. I apologize if I didn't use the correct words for explain myself.
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2023.03.23 18:04 Washingbloddysheets Finding the derivative of a function. How do I know when it consist of two functions? And how do I know which formula to use?

I have to find H'(x), where H(x)=2xsqrt{x}
The solution is 3x (written in my book)
I tried to solve it and got the right solution.
But when I plugged in the problem on “symbolab” and compared the way I solved it to the way symbolab did, it seems like I did it wrong.
Symbolab: So initially it simplified the expression first before finding the derivative. It took the constant 2 out. And found the derivative of x{\frac{2}{3}} - much simpler than what I did
The way I solved it was using the product rule, because I thought 2x could be seen as a function and \sqrt{x} could be seen as a function.
How can I learn from this next time - to know which kind of method to use?
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2023.02.17 14:36 your_friend_papu Bing Chat confused by a single 'fictional' maths page

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2022.12.31 07:15 WolfeTheMind Sqrt((x^3)-4)=x*Sqrt(x-2)

False input or output from Symbolab. Surely my bad. I thought I was going crazy trying to figure out how it could simplify to that because it's so obvious
Mods delete please. I probably should have just refreshed and retyped first. Still not sure, however, how I messed that up
Very embarrassed. Thanks for the assistance everyone
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2022.12.02 19:16 WolfeTheMind Do you have to factor first before solving a rational function?

So I'm trying to find a cross for my horizontal asymptote in my rational function. The asymptote is at 2 and my professor says I can solve using the original equation or the form that's already been factored and canceled out. Symbolab doesn't help as it just simplifies and cancels out factors but when I do the original equation I get a different answer than when I use the factored / cancelled form.
Let me show the example:
2x2-5x+2 / x2-4 = 2 ( the asymptote)
First I multiply 2 by the denominator to get 2x2-5x+2 = 2x2-8
Then I subtract 2x2 from both sides and am still left with the variable x in the form of -5x+2 = -8. Which then IS solvable.
However when I factor and cancel (x-2) first I get 2x-1=2x+4, which you can obviously see there is no solution for because the 2x's subtract out
Why am I getting different results for what the prof says is an identical equation (aside from the hole)? What am I getting wrong here
Update: This is the video and time I'm wondering about. He solves for the simplified version but explicitly says that can use either version, which I tried... With different results. That's I guess what I'm trying to figure out
FINAL UPDATE: I actually found a youtube comment just now that was asking the same exact thing. Comment:
"1:05:40 P.leonard said that it doesn’t matter whether or not you use the main or simplified version of the function when calculating intersection of the horizontal asymptote, but when i used the main function, i got x=2, meaning that there is an intersection. 2x^2-5x+2=2(x^2-4) 2x^2-5x+2=2x^2-8 -5x+2=-8 -5x=-10, so x=2….. what’s happening here?"
Top response was very clear, in case anybody is wondering:
"Great question, but yes, you're not going to get the same answer if its not a cross. In this case you're getting x=2 back showing where (if at all) you're going to have a H.A. cross. But because you know there's a hole at x=2, you know that there wont be a cross at that point. Using the simplified version of the fraction is just a different road to get to the same conclusion. Its just in this case you're getting that "non-sense answer" where two different numbers equal each other. Again great question, because yes, first time trying it you're expecting to get the same answer for both the original function and the simplified version. Two different "roads" that lead to two different "destinations" but the end and final conclusions are consistent "

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2022.11.04 10:28 Soggy_Leg_757 I can't fit my answers into the choices

Anyways, here are the two problems: 1. What is the first derivative of ln(ln(y)) + ln(y) = ln(x)? A. 2y/(x+y) B. y/(x-y) C. 2y/(x-y) D. y/(x+y)
I tried differentiating it and got y*ln(y)/(x*(1+ln(y))), which is close to choice D. I did it by first simplifying the ln using ln(xy) = ln(x) + ln(y) and then using the product rule. It's an implicit function, therefore, I had to isolate y' after differentiating them, which led me to what I have right now. The problem is, I can't find any way to remove ln to get to fit choice D. I even tried to evaluate each selection using Symbolab by letting it solve the problem initially but got everything wrong. Not a single one of those is correct according to Symbolab. And the answer I got is the same as the one that showed up using Symbolab. Therefore, how?!
  1. Differentiate y = 2e^x*8^x A. 6.16e^x*8^x B. 2e^x*8^x C. 3e^x*8^x D. 4.16e^x*8^x
I observed that I can simplify the given expression to (2^1+3x)(e^x) and I differentiated from there but ended up lost. I tried to study the solution I got from calculators but every single of them doesn't give the same result as the choices. I also tried evaluating each one of them and found out A was the real answer. But, the thing is, I have no idea where that decimal came from. Where do I even start?
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2022.09.03 16:36 BuilderBrilliant Can anybody tell me if this is right?

I am working on simplifying cos3x into cosx and sinx. The zn relation is used. I ended up using a Symbolab calculator for one of the steps, which is in the notebook. The final answer is the last row on the blackboard.
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2022.08.31 16:13 AnonymColonist My maths is just otherworldly

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2022.07.26 15:51 NarcolepticFlarp Why Did I Get This Integral Wrong?

I was trying to integrate 1/(x*sqrt(1-x^4)). I used the substitution x^2 = sin(theta). This transformed the integrand into 1/2*csc(theta), which I know how to integrate. After undoing the substitution and simplifying I got ln(x) - 1/2*ln(sqrt(1-x^4) + 1) + c. However, this is not the answer Wolfram Alpha and Symbolab give. What did I do wrong?

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2022.05.14 10:22 oobeing Multivariable limit, stuck at last step of evaluation

Limit is
lim (x,y) (0,0) x2y2/(x2+y4)
I've tried looking if it DNE by going along x and y-axis etc, get 0 on all of them.Changed to polar coordinates and then simplified I get
lim r 0 r2*cos2x*sin2x/(cos2x+r2sin4x)
Then if you plug in you get
and symbolab says the answer here is 0, because 0/a = 0 when a =/= 0.But how can one know the angle x is not pi/2, making the denominator 0 as well?
Edit: Since limit must be independent upon which path you take for it to exist, meaning it must be independent of the angle x here, wouldn't this mean it doesn't exist since there's one angle/path for which the limit is not defined?
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2022.05.14 00:49 Brown-Banannerz A question involving logs

Simple question
130 = 10log(x/10^-12)
Based on the log rules I'm familiar with, i simplify the equation like so (these are base 1`0 logs)
However, when I type the original question into an equation solver?or=input), the answer I get is x =10. In their step-by-step demonstration, they use a different method of simplifying logs than what I just did. However, as far as I'm aware, my way is also valid and so the answer should be the same. Anyone able to help me understand this discrepancy?
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2022.02.07 23:09 H4mSandw1ch Is there a way to get it to simplify like Symbolab does?

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2021.11.22 23:04 awendero Is there some kind of trick to solve this algebraic expression?

I am currently doing preparations for the University entrance exam here in Serbia and I encountered a problem with solving this expression.
I tried all possible ways to simplify it and solve it, but I didn't succeed in that. I tried to evaluate them in Photomath & Symbolab as well, but no luck there either.
I thought it was maybe a printing error, but instead of x I substituted it with -1 (which is the solution) and it really was equal, so I don't know if there is some kind of trick to apply here in order to get a normal solution.
Here is the picture:
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2021.10.17 21:15 Lukaeaeap0 Integral of ln(x)/x

I want to calculate the surface between two graphs and I have now simplified this to ∫ ln(x)/x dx. And I need to calculate the surface of the formula between x=1 and x=e.
When searching for an answer I came across this on symbolab%7D%7Bx%7D?or=input) where they substitute u=ln(x), which gives [u^2/2] between x=1 and x=0.
Can someone please explain how this works and how I could apply it myself in other integrals?
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2021.10.15 21:16 kingtun567 Modulo Algebra

I have an assignment that involves me coding using modulo. In one of my functions, I have the equation x = (a * b) mod n. I have to write another function that does the exact same equation, except solved/rewritten for a.
a = ?
How would you go about solving this? I never did modulo algebra in calculus or high school math, so I look at symbolab to see how they solved for a. You can find the solution here. What I found is very strange.
a = (94 mod^23 bx) / (8836 mod^25 b^2); b != 0. I have never seen that notation where the exponent is applied to the modulo or to the left of 'b'. What does it mean? How do they simplify at step 4? How do you write this solution in expanded form so that it can be legally written in code? Thanks for the help.
Edit: I'm also researching into solving for a using modulo identities rather than algebra. So far I've only looked at this:, but I cannot determine anything yet.
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2021.09.20 05:43 dcfan105 Confused about how to simplify complicated square root expression
I'm so confused about how they simplify the first circled expression to get the second circled expression. I've been playing around with this for hours and I just don't get it. I know it has to be right because this is a part of a larger problem involving computing the gradient of a voltage and verifying that it matches the previously derived expression for the electric field. It has to be equal to the second circled expression as that's the expression for the electric field and the problem started with the expression for the voltage, so it's not a mistake on Quizlet's part. But I can't figure out the algebra of going from the first expression to the second. The only thing I can think of is that they somehow factored the denominator of the denominator and canceled out common factors, but if that's what they did I have no idea how.
Whoever wrote this solution apparently thought the step was obvious enough to omit, but it's not at all obvious to me. I even tried putting it into symbolab, hoping it would show me the steps, but it couldn't figure out how to simplify it either, as it claimed it couldn't be simplified.
I'm not even sure what topics to Google to try to find similar problems elsewhere because this is just a very specific type of expression that I don't think has its own name. I don't normally have trouble with algebraic simplification -- heck I'm actually an algebra tutor, but this problem has me utterly stumped and quite frustrated. Can someone please help me understand?
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2021.08.09 04:23 jjrreett turning full worksheets into functions/blocks

I've been using desmos and an engineering ide for a bit now. I just wish it had a bit more functionality. I wish I could turn an entire worksheet into a function or block. Especially so I can use the regression tool as a root finder. Or a million other useful things.
I also wish symbolab was built-in. I often copy and paste equations into symbolab to simplify and solve for among other things.
Finally, a variable editor would be nice. Renaming variables is a hastle.
Desmos is great, wish it was the backbone to a real engineering ide tho (units handling, sig figs handling, error propagation, some spreadsheet things would be nice...
Im interested in Atlas ( Can't wait to see how that turns out.

Currently working on building richard nakka's rocket equations into a tool that you tell it a few inputs and it computes grain geometry.
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2021.06.22 19:28 lilpalp Help understanding how (x)(x) equals x^1 here?

Had to use the Symbolab Simplification calculator to understand where I was making a mistake with my working of this question, I'm confused with the workings shown in one of the steps.
Simplify x(2x^-1/3)^4
= x . 2^4x^-4/3
= 2^4x^1-(4/3)
The calculators method implies that xx^-(4/3) = x^1-(4/3).
Why is it that xx does not equals x^2 here?
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2021.04.25 03:33 hamzaff44 I am having a hard time with the following question concerning logarithms.

How should we simplify a^(log_a (8) + log_a (2))? Thanks for the help.
(solved by finding the answer on symbolab btw)
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2021.03.15 20:00 urquhartloch How do I solve this equation?

Ok, so I am studying for the FE next week and I am trying to solve a bunch of FE problems every day. This is from the Math section of the 2015 FE study problems I could find.
Problem: Find the center of the ellipse
Center: ((x-1)^2)/9+((y-1)^2)/16=1
Therefore the center is at (1,1)

This is all Im given. My problem is that I cant figure out how they got from the initial equation to the solution. I even put it in symbolab and it couldnt output any solution for simplifying it.
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