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A quadratic equation (t=s2+3). This kind of line will result in a parabola like curve.

Q: What is t equals s squared plus 3?

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s + 3 = 6s -s -s ------------- 3 = 5s /5 /5 s=3/5 or s=.6

Locate the turning point(s) for the following functions. (a) y=x3-x2 -3x + 5 3

3

102 = 100.I don't think it matters what 's' is.

s = -70

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2 squared is 4 3 cubed is 27 27 plus 4 is 31

s = p^2 - 5 *also, p = sqrt (s+5)

A = S squared is the formula for area of a square A = area S = lenght of side

This is a hyperbola. It is best approached using Fermat's factorisation method. Seefermat-s-factorization-methodor google wikepedia. I don't know of any faster approach.

-1

s + 3 = 6s -s -s ------------- 3 = 5s /5 /5 s=3/5 or s=.6

The solutions are: s = -3/2 and s = -3/2 they have equal roots

m/s(squared)

s = 7, t = -3

72 minus the root of squared minus equals 72 but you have to be smart to know that the minimum circumference s 56 there you go

Locate the turning point(s) for the following functions. (a) y=x3-x2 -3x + 5 3

6 or it equals butterfly with one of the 3's backwards and then the 0 and then the other 3