We can do like we did before, distribute the denominator,

$\bf \cfrac{6x^4-5x^3+6x^2}{2x^2}\implies \cfrac{6x^4}{2x^2}-\cfrac{5x^3}{2x^2}+\cfrac{6x^2}{2x^2}\implies \cfrac{6}{2}\cdot \cfrac{x^4}{x^2}-\cfrac{5}{2}\cdot \cfrac{x^3}{x^2}+\cfrac{6}{2}\cdot \cfrac{x^2}{x^2}
\\\\\\
3 x^4x^{-2}-\cfrac{5}{2}x^3x^{-2}+3x^2x^{-2}\implies 3x^{4-2}-\cfrac{5}{2}x^{3-2}+3x^{2-2}
\\\\\\
3x^2-\cfrac{5}{2}x+3x^0\implies 3x^2-\cfrac{5}{2}x+3$

6 0

3 years ago

Evaluate the indicated function for f(x) = x2 − 1 and g(x) = x − 2 algebraically. If possible, use a graphing utility to verify

Maru [420]

(F-g)(-2) is the answer

5 0

3 years ago

Skip counting represents 8×4

Blizzard [7]

It represents that because multiplication is like adding continually so it represent 8+8+8+8 which is 8 x 4

5 0

4 years ago