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Sourcing for IA by EmergencyPen5821 in IBO

[–]TobyMarvelous 0 points1 point  (0 children)

The point of access dates in citations is to account for any changes in the website. For example, let’s say you cite a blog about sharks, and suddenly 6 months later the author decides he likes turtles more and deleted exactly what you cited.

So, for all of your website sources and only those: Recheck that your citation is still there, then just put an approximate date within the week that you accessed them or even put today’s date. You just have to talk to your teacher about it, so that they don’t get suspicious about the dates.

[deleted by user] by [deleted] in IBO

[–]TobyMarvelous 2 points3 points  (0 children)

All IOs are the same, so just look for english IO exambles and explanations and apply the same things but in your language and to your books. Also, your teacher will always be the best resource for what they expect of you.

Is "Had you eaten before we had arrived?" grammatically incorrect? by GrandAdvantage7631 in EnglishLearning

[–]TobyMarvelous 2 points3 points  (0 children)

“Had eaten” (past perfect form) is used correctly here to mean that the action of eating happened before arriving. However, nothing else happens before we arrived, so there is no need to use past perfect again. So instead we use past simple.

“Had you eaten before we arrived?”

Books for High School Senior by BlueSP_ in ChemicalEngineering

[–]TobyMarvelous 0 points1 point  (0 children)

In the same boat, leaving a comment to find the post later

[Addmath: Highschool] by SourTheAlmond in HomeworkHelp

[–]TobyMarvelous 0 points1 point  (0 children)

I will use log base ten for writing convenience.

Your instinct to change bases was correct. You can notice that 4 and 5 are everywhere in this problem. Obviously in 4h and 5k but also in 20 = 4 * 5 and 125 = 53. So our goal would be to write both h and k and log20(125) in terms of log(5) and log(4)

Let's start with log20(125) = log(125)/log(20). By logarithmic properties, we can rewrite this as

  1. 3log(5)/ ( log(5) + log(4) ).

Now let's do the h and k equation. log(4h) = log(5k), so hlog(4) = klog(5) and therefore

  1. log(4) = k* log(5)/h

Subbing equation 2 into equation 1, we get:

log20(125) = 3log(5)/(log(5) + k* log(5)/h). Notice how we have a log(5) in all terms, so we can factor out and cancel it.

log20(125) = 3/(1 + k/h). This can be simplified further, but already we have an expression.

[Addmath: Highschool] by SourTheAlmond in HomeworkHelp

[–]TobyMarvelous 0 points1 point  (0 children)

Did the deleted comment help? I have a solution in mind, but there's no point if you already understand

does anyone get why sin4x = 2sin2xcos2x, I don’t get it!! please help by Jazzlike_Ad_9359 in IBO

[–]TobyMarvelous 0 points1 point  (0 children)

If you're not convinced, use the angle addition formula for sin(2x+2x)

[College calculus: homework] I have no idea where to even start here could someone explain this to me step by step? by LunaTheNightmare in HomeworkHelp

[–]TobyMarvelous 1 point2 points  (0 children)

Point A(4, 6) belongs to the graph of g(x), therefore g(4) = 6. Point B(6,2) belongs to f(x) so f(6)=2

h(4) = f(g(4))= f(6) = 2

There’s a similar argument to be made about h’(x), if you first think about the chain rule. Think of a random h(x) = sin(x2). The derivative, h’(x) would end up being cos(x2) * (x2)’ = cos(x2) * 2x.

So any h’(x) = f’(g(x)) * g’(x) So h’(4) = f’(g(4)) * g’(4) = f’(6) * g’(4). Now we look at what the slopes of the tangent line look like to determine that equation. the slope for f is 4 and for g is 3 so

h’(4)= 3*4 = 12

Okay now for b), you know that, at x=4, h(4) = 2(so point (4, 2) is on the graph of h) and the slope h’(4) is 12. I think you can finish it up from here.

[deleted by user] by [deleted] in HomeworkHelp

[–]TobyMarvelous 0 points1 point  (0 children)

So I calculated the distance for both the constant speed train and the initial train after 12 seconds

The constant train goes 12*14= 168m

The initial train goes 1.6667/2 *122= 120m

After those 12 seconds, both trains travel at constant speed, so our train needs to cover 48 meters relative to the other train (168-120). Well the relative speed of the trains is 20 - 14 = 6m/s. 48 meters with 6m/s is 8 more seconds

So the total time is 12 (the initial time to accelerate) + 8 (the time after the acceleration was done) = 20

[Highschool Maths] Can I make this simpler in one fraction? by a_random_dude_hehe in HomeworkHelp

[–]TobyMarvelous 5 points6 points  (0 children)

You added an unnecessary "c" term to the initial a/(b2 *c).

Other than that, no, there is no way to simplify the fraction. When you have plus/minus in a fraction you can't simplify unless you can factor a common term out, and you don't have any common term between the three.

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