Answer:
300
Step-by-step explanation:
I believe the answer is -16x+35
Coefficient is the number before your variable. :)
You have to estimate the slope of the tangent line to the graph at <em>t</em> = 10 s. To do that, you can use points on the graph very close to <em>t</em> = 10 s, essentially applying the mean value theorem.
The MVT says that for some time <em>t</em> between two fixed instances <em>a</em> and <em>b</em>, one can guarantee that the slope of the secant line through (<em>a</em>, <em>v(a)</em> ) and (<em>b</em>, <em>v(b)</em> ) is equal to the slope of the tangent line through <em>t</em>. In this case, this would be saying that the <em>instantaneous</em> acceleration at <em>t</em> = 10 s is approximately equal to the <em>average</em> acceleration over some interval surrounding <em>t</em> = 10 s. The smaller the interval, the better the approximation.
For instance, the plot suggests that the velocity at <em>t</em> = 9 s is nearly 45 m/s, while the velocity at <em>t</em> = 11 s is nearly 47 m/s. Then the average acceleration over this interval is
(47 m/s - 45 m/s) / (11 s - 9 s) = (2 m/s) / (2 s) = 1 m/s²
Answer:
(x+4)(x+9)=0. Your answer would be x=-4 and x=-9. Explanation:: (x+4)(x+9) x times x=x2 x(9)+x(4)=13x 4 times 9=36 = x2+13x+36.(x+4)(x+9)=0. Your answer would be x=-4 and x=-9. Explanation:: (x+4)(x+9) x times x=x2 x(9)+x(4)=13x 4 times 9=36 = x2+13x+36.
