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anonymous
 4 years ago
y=(sinx)(sqrt{25x^2}), what is the range? how do find that?
anonymous
 4 years ago
y=(sinx)(sqrt{25x^2}), what is the range? how do find that?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=(sinx)(\sqrt{25x^2}) \]

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1let's find out the domain first sin x x can be any real number sqrt (25x^2) 25x^2>=0 or x^2<=25 x will be in the [5,+5] so the domain is [5, 5] max value of square root will occur when x=0, it'll be 5 now sin x' s range is [1,1] so the range of the function is [5,5]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh okay, thank you. but it says estimate the range to two decimal places, explain why only an estimate is possible...

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1sorry i made a mistake, for the max value when x=0 root 25=5 but sin 0 =0 sin is max when x=pi/2 i.e when x=1.57 at x=1.57 the value of y is 4.74711 sin x is minimum at x=pi/2 or 1.57 therefore the minimum value is 4.74711 range is from [4.74711, 4.74711]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh okay, so you basically plug in 1.57 and 1.57 into the function to get the yvalue?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh okay makes sense :) thank you :D

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1welcome, sorry i made a mistake earlier
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