Why do you think a removable discontinuity (hole) doesn't produce an asymptote on the graph of a polynomial function, even though it is excluded from the domain of the function?

Peter measures the angles in a triangle. he finds that the angles are 95, 10 and 75. is he correct? explain your answer

A+b= 17, ab=70, find the value of |a-b|

Evaluate 8x2 + 9x − 1 2x3 + 3x2 − 2x dx. solution since the degree of the numerator is less than the degree of the denominator, we don't need to divide. we factor the denominator as 2x3 + 3x2 − 2x = x(2x2 + 3x − 2) = x(2x − 1)(x + 2). since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the form† 8x2 + 9x − 1 x(2x − 1)(x + 2) = correct: your answer is correct. to determine the values of a, b, and c, we multiply both sides of this equation by the product of the denominators, x(2x − 1)(x + 2), obtaining 8x2 + 9x − 1 = a correct: your answer is correct. (x + 2) + bx(x + 2) + cx(2x − 1).