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Ash90
i need help solving this replace r and g with the following and find the time-out. r=91-7* sqrt(t) g=14*t
idk what to do with this?
Do we just have to find the time in this question? And is any other equation given? :)
yes we just need to find the time in this equation but I have to write it out by showing my work. let me check and see if anything else is given...
that's all nothing else is need right now.
what this is is a programm that when you do the calculations the colors are set up to do what you need them to.
That's cool! Well, if you have to find the value of t in terms of r or in terms of g, you can do that. But if you want to find the value of t, in terms of real numbers, [Although, I don't think it is possible] I suggest you tag someone else. The equations you have posted are called Parametric equations, you can also express r in terms of g here but I am not sure how to go about calculating the sole value of 't'. :)
could i show you what i have and maybe you can help from there?
Yes, I can help you from there! But is it given that r=g ? Because you have equated it over there! :D
well i think what i had to do was set them equal to each other if i'm not mistaking i think thats what the teacher wanted.
Okay. So if r=g \(91-7 \sqrt{t} = 14t\) \(91-14t = 7\sqrt{t}\) \(Squaring ~~both ~~sides ~~we~~ get:\) \((91-14t)^2 = 49t\) \(196t^2 + 8281 - 2548t = 49t\) \(196t^2 - 2597t + 8281\) \(~~~~~~~~~~~~Where~~[a = 196, ~b = -2597,~c = 8281]\) Use the quadratic equation below to solve for \(t\): \[\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] Got it? :)
\[t=\frac{ 53+\sqrt{105} }{ 8 }, t=\frac{ 53-\sqrt{105} }{ 8 }\]
Yes, if you got that by calculating the above equation, then yes. :)
for this one i thought using quadratic formula right of hand would be a good idea
You cannot always do that. The quadratic formula can only be used when f(x) = ax^2 + bx + c = 0 [And zero only] As it gives us the zeroes of the polynomial/expression. What you could do is write it like this: 7t^2 - 51t + 92 - r = 0 Now apply the quadratic formula, considering, [c = 92-r] Getting this? :) You may also try this by the completing-the-square method, although I am not a 100% sure if it'll work here.