Use the three conditions to show that function f(x)=2x^2 +3x - 1 is continuous at x=0.

f(x)  = 2x^2 +3x - 1

f'(x) = 4x + 3

f''(x) = 4

f(0) = 2(0)^2 + 3(0) - 1 = -1 (valid)

f'(0) = 4(0) + 3 = 3 (valid)

f''(0) = 4 (valid)

=> f(x) is continous at x = 0

the answer is 175.4379

step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: