The properties 1‚ 2‚ 4‚ and 5. are used
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: A
Step-by-step explanation:
One a normal a curve the mean or average always occurs in the middle/ top of the curve.
Answer:
16:9
Step-by-step explanation:
If the linear ratio is 4:3, the area ratio will be the square of that.
(4/3)² = 16/9
Answer:
b = ![\sqrt[3]{\frac{2d-c}{5} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B2d-c%7D%7B5%7D%20%7D)
Step-by-step explanation:
Given
2d = 5b³ + c ( subtract c from both sides )
2d - c = 5b³ ( divide both sides by 5 )
= b³ ( take the cube root of both sides )
= b
