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anonymous
 5 years ago
can someone tell me how to graph this equation on my graphing calculator: 300(2+sin(4sqrt x+0.15)) for 0<x<20
anonymous
 5 years ago
can someone tell me how to graph this equation on my graphing calculator: 300(2+sin(4sqrt x+0.15)) for 0<x<20

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just do it... What type of calculator do you have?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Y= > Enter: 300(2 + sin(4 sqrt(x + 0.15)) Then, to see the region from 0<x<20: Window (at the top) > Xmin = 0; Xmax = 20; Ymin = 20 Ymax = 20

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Might have to change the Ymin / Ymax values  I'm just guessing :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0give me a sec and let me try it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what should mt table set be

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you need the table? If so > 2nd Window (for table settings) Tbl start = 0; triangleTable (change in table) = 1 or 0.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why's that  what did you enter on the calculator?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh yeah, our Ymax / Ymin values are not right at all. Make Ymax = 900 and Ymin 200

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do what I said above.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Btw is that sqrt(x) + 0.15 or sqrt(x + 0.15) I'm guessing it's the former.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey it worked good job but why did u choose 200 and 900

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cause on the table it starts out over 200 and then it goes up to a max of 900 ish.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have attached the graph. This is the accurate graph, and is drawn to scale

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you open up that file with?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0May be you couldn't open the file. Ok here it is

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey ty u for sending it over but i do not understand how to choose my max and min

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could anyone open the file?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah i did i saw it ty

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does anyone know how to choose 200 and 900

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why it has to be 200 and 900

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For that you will have to find the derivative, and equate it to zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah . do u know how to find the integral on the calculator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No I don't, but I can provide you that here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And I am attaching the graph in jpg format, as last time it was not openeable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did your problem got solved or it increased? :)
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