3 years ago

What does 1.2h-2.8=2 in one step equation form ?

Mathematics

1 answer:

nekit [7.7K]3 years ago

8 0

Answer:

h=4

Step-by-step explanation:

1.2h-2.8=2

1.2h=2+2.8

1.2h=4.8

h=4.8/1.2

h=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, $F(r(t)) = (6t, 11t^2)$

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