2 years ago

Sixety five divided by fifteen

Mathematics

1 answer:

oee [108]2 years ago

7 0

Answer:

the answer is 4 with a remainder of three

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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

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

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

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

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Randomly picking a blue card from a bag containing all blue cards

scoundrel [369]

What's the question

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It’s question 14 I need help with please it’s due soon

belka [17]

if you are trying to work out -3/4 - (-1/4 - 1/2) then the answer is 0

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The probability of A is 7/20, the probability of A intersection B is 49/400 what is the probability of B?

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Answer:

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n(B)=(400+49-140)/400

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Which expression is a trinomial of degree 1?<br> A. x^3+x^2-x<br> B. xy^2<br> C. 6+x-y<br> D. 3x-y

ki77a [65]

C

Trinomial means three parts
A and C or trinomial's but the degree on a is at three and the degree of C is a one

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Sixety five divided by fifteen ​

Sixety five divided by fifteen