Since he completes 10 minutes each day for five days, he completes 50 minutes.

The percentage of his total requirement would be 50/100 = 0.5 = 50%.

5 0

3 years ago

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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

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

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

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

6 0

2 years ago

3. A zookeeper is monitoring the supply of grain and hay that she has to feed the

4vir4ik [10]

Answer:

Day 6

Step-by-step explanation:

Every 3 days 1 pound of hay is used every day 1 pound of grain is used. There are 10 pounds of grain and 8 pounds of hay. On the 6th day there are 6 pounds of hay left and 6 pounds of grain left.

3 0

2 years ago