Evaluate f(2/3)-f(1/5)= <br><br> f(x)=6x-2 g(x)=2x^2+7

The point at which the slope of the tangent line is -3

AlexFokin [52]

F(x) = 6x^2 + 3x - 6

The derivative, and thus the slope of the T. L. to the graph, is f '(x) = 12x + 3. Set this result = to -3 and solve for x:

12x + 3 = -3 => 12x = - 6 and x = -1/2
6 + 6 - 24
If x = -1/2, y = f(-1/2) = 6(1/4) - 3(-1/2) - 6 = ----------------- = -6
4

The slope of the T. L. to the graph will be -3 at the point (-1/2, -6).

4 0

4 years ago

What is the next number to follow 17, 18, 22, 31, 47, ?

Minchanka [31]

Answer:

72

Step-by-step explanation:

First differences are ...

18 -17 = 1

22 -18 = 4

31 -22 = 9

47 -31 = 16

We observe that these are square numbers. The next square number is 25, so we expect the next number in sequence to be ...

47 +25 = 72

3 0

3 years ago

I need help, please help me this is a quiz and I don't know the answer.

k0ka [10]

Typically, you plot independent (x) vs dependent (y). Also, we list the variables on a table as x on top and y below. In this situation the miles travelled would be the variable the researcher controls (the independent variable) and number the of gallons used would be what was measured (the dependent variable). Answer J

4 0

3 years ago

Write 4.029 in expanded form and with number names.

irinina [24]

4 + .029 | Four and nine thousandth two hundredth tenths

5 0

3 years ago

Read 2 more answers

If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt

MatroZZZ [7]

Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

$\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t$
$\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$

with $0\le t\le1$ . Then we find the differential:

$\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$
$\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$

with $0\le t\le1$ .

Here, the line integral is

$\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle$
$=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$
$=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt$
$=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt$
$=x_1y_2-x_2y_1$

as required.

4 0

4 years ago

Evaluate f(2/3)-f(1/5)= f(x)=6x-2 g(x)=2x^2+7