3

Mathematics

1 answer:

Alecsey [184]3 years ago

5 0

Answer:

Step-by-step explanation:

$y = 5x - 1 \\ 4 = 5(1) - 1 \\ 4 = 5 - 1 \\ 4 = 4 \\ \\ y = - x + 5 \\ 4 = - (1) + 5 \\ 4 = - 1 + 5 \\ 4 = 4$

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Given the following vector field and oriented curve C, evaluate ModifyingBelow Integral from nothing to nothing With Upper C Bo

sladkih [1.3K]

Answer:

The value of the line integral is $\frac{157}{2}$ .

Step-by-step explanation:

Given a path C, with parametrization r(t) for $t_0\leq t \leq t_1$ , we have that

$\int_{C}F dr = \int_{t_0}^{t_1} F(r(t)) \cdot r'(t) dt$ where $\cdot$ is the dot product between two vectors.

In our case, we have $r(t) = (6t,11t^2), 0\leq t\leq 1$ . Then $r'(t) = (6,22t)$ . In this case we have that F(x,y) = (x,y). Then, $F(r(t)) = (6t, 11t^2)$

$\int_{C}F dr = \int_{0}^{1} (6t,11t^2)\cdot(6,22t) dt = \int_{0}^{1} 36t+11\cdot 22 t^3= \left.(36\frac{t^2}{2}+11\cdot 22 \frac{t^4}{4})\right|_{0}^{1} = \frac{36}{2}+\frac{11\cdot 22}{4}= \frac{157}{2}$

4 0

4 years ago

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require