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3 years ago

8

2560 divided by 11 equals what

2 answers:

Dovator [93]3 years ago

6 0

Answer:

232.727

Step-by-step explanation:

babunello [35]3 years ago

5 0

Answer:

232.727272727

Step-by-step explanation:

just calculate it

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The area of the region in the first quadrant between the graph of y = k*sin(x) and x-axis (where k > 0 ) on the closed interv

astraxan [27]

Note that $0\le k\sin x\le k$ in the given closed interval, so the area is exactly given by the definite integral

$\displaystyle\int_0^\pi k\sin x\,\mathrm dx$

(no absolute values needed!)

Integrating gives

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3 years ago

Evaluate the line integral, where C is the given curve. (x + 6y) dx + x2 dy, C C consists of line segments from (0, 0) to (6, 1)

Dima020 [189]

Split $C$ into two component segments, $C_1$ and $C_2$ , parameterized by

$\mathbf r_1(t)=(1-t)(0,0)+t(6,1)=(6t,t)$

$\mathbf r_2(t)=(1-t)(6,1)+t(7,0)=(6+t,1-t)$

respectively, with $0\le t\le1$ , where $\mathbf r_i(t)=(x(t),y(t))$ .

We have

$\mathrm d\mathbf r_1=(6,1)\,\mathrm dt$

$\mathrm d\mathbf r_2=(1,-1)\,\mathrm dt$

where $\mathrm d\mathbf r_i=\left(\dfrac{\mathrm dx}{\mathrm dt},\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt$

so the line integral becomes

$\displaystyle\int_C(x+6y)\,\mathrm dx+x^2\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}(x+6y,x^2)\cdot(\mathrm dx,\mathrm dy)$

$=\displaystyle\int_0^1(6t+6t,(6t)^2)\cdot(6,1)\,\mathrm dt+\int_0^1((6+t)+6(1-t),(6+t)^2)\cdot(1,-1)\,\mathrm dt$

$=\displaystyle\int_0^1(35t^2+55t-24)\,\mathrm dt=\frac{91}6$

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3 years ago

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dybincka [34]

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Answer:

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Step-by-step explanation:

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