(a) Here's a graph: https://www.desmos.com/calculator/fxto5gtfns
(b) P = (2, 8) Q = (x, x3)
Find the slope of the line PQ for the following values of x .
When x = 1.5, Q = (1.5, 1.53) = (1.5, 3.375)
We want to find the slope of the line passing through the points (2, 8) and (1.5, 3.375)
slope = \(\frac{8-f(1.5)}{2-1.5}\,=\,\frac{8-3.375}{2-1.5}\,=\,\frac{4.625}{0.5}\,=\,9.25\)
This is the first value for the slope of the secant on the table.
x | msec | |
1.5 | 9.25 | |
1.9 | \(\frac{8-f(1.9)}{2-1.9}\,=\,\frac{8-6.859}{2-1.9}\,=\,\frac{1.141}{0.1}\,=\,11.41\) | |
1.99 | \(\frac{8-f(1.99)}{2-1.99}\,=\,\frac{8-7.880599}{2-1.99}\,=\,\frac{0.119401}{0.01}\,=\,11.9401\) | |
2 | ? | |
2.01 | \(\frac{8-f(2.01)}{2-2.01}\,=\,\frac{8-8.120601}{2-2.01}\,=\,\frac{-0.120601}{-0.01}\,=\,12.0601\) | |
2.1 | \(\frac{8-f(2.1)}{2-2.1}\,=\,\frac{8-9.261}{2-2.1}\,=\,\) |
Do you see how I am doing these? See if you can finish the table...
(c)
The slope of the tangent line when x = 2 appears to be between 11.9401 and 12.0601
Let's just take the average of these two... (11.9401 + 12.0601) / 2 = 12.0001
So I'd guess that the slope of the tangent line at (2, 8) is 12
Solve for b . | ||
y = mx + b | Subtract mx from both sides of the equation. | |
y - mx = b | ||
Solve for b . | ||
A = h(b + c) | Divide both sides of the equation by h . | |
A / h = b + c | Subtract c from both sides of the equation. | |
A / h - c = b | ||
Solve for r2 . | ||
A = 4r2 | Divide both sides of the equation by 4 . | |
A / 4 = r2 | ||
Solve for x . | ||
7x - y = 14 | First add y to both sides. See if you can figure the rest out. | |
Solve for i . | ||
R = (E / i) | Multiply both sides of the equation by i . | |
i * R = E | Divide both sides of the equation by R. | |
i = E / R | ||
Solve for L . | Using this as the equation A = \(\frac{r}2\)L , | |
A = \(\frac{r}2\)L | multiply both sides by \(\frac2r\) . | |
\(\frac2r\) * A = \(\frac{2}{r}\) * \(\frac{r}{2}\) * L | ||
\(\frac{2A}{r}\) = L | If you meant for the equation to be A = \(\frac{r}{2L}\) , then it is different! |