What is the answer to 5−2x^2=37
2 answers:
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Answer:
Do you want to be extremely boring?
Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?
is a valid solution.
Want something more fun? Why not a parabola? .
At this point you have three parameters to play with, and from the fact that we can already fix one of them, in particular
. At this point I would recommend picking an easy value for one of the two, let's say
(or even
, it will just flip everything upside down) and find out b accordingly:
Our function becomes
Notice that it works even by switching sign in the first two terms:
Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2:
Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need , and at that point the first condition is guaranteed; using the second to find k we get
Or how about a sine wave that oscillates around 2? with a similar reasoning you get
Sky is the limit.
Answer:
Step-by-step explanation:
LHS = Sin² ∅ + Sin² ∅*tan² ∅
= Sin² ∅ [1 + tan² ∅ ] {Sec² ∅ - tan² ∅ = 1}
= Sin² ∅ *Sec² ∅
= Sin² ∅ *
= Sin² ∅ /Cos² ∅
= tan² ∅ = RHS