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
Evaluate them
A.
f(5), x>1 so square it, f(5)=25
A is wrong
B.f(2), x>1 so square it, f(5)=4
B is correct
C. f(1), x=1 so the result is 5, f(1)=5
C is correct
D. f(-2), x<1 so double it, f(-2)=-4
D is wrong
answers are B and C
The Answer to this Question is:
(2x^2+3y)(4x^4-6x^2y+9y^2)
This symbol ^ means that the number after is an exponent.
The greatest number of copies you can et for three dollars is 24.
Answer:
6 + (15) : 3 x (2) = 6 + 5 x 2 = 16
Step-by-step explanation:
sixteen with Jaywhyypeeeee
