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murzikaleks [220]
3 years ago
8

What is the range of the function y=x^2-5 where 0<x<10

Mathematics
1 answer:
abruzzese [7]3 years ago
3 0
The range is the bounds on y. If 0<x<10, since this is a function that increases as x increases, the maximum y value will be by x=10 and the minimum will be by x=0.

Plugging these values of x into the function gives y=(10)^2-5=100-5=95 and y=0-5=-5

However, since x can't actually equal 0 and 10, y cannot be exactly equal to - 5 and 95. Therefore, the range is -5<y<95
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sin² (A-B) = 1 - 9/25 = 16/25

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