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3 years ago

Pls4sssssssssssssssssssssssssssssssss

Mathematics

1 answer:

stellarik [79]3 years ago

6 0

Answer:

9.67 lbs

Step-by-step explanation:

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Find the quotient of n(x) = 6x^5y^7 and m(x) = 2x^3y

likoan [24]

Answer:

a. Yes

b. Yes

c. Yes

d. Degree 8

Step-by-step explanation:

a. Yes, n(x) is a polynomial of one single term (also called monomial) because it contains variables raised to positive integers.

b. Yes, m(x) is a polynomial of also one single term (also called monomial) because it contains variables raised to positive integers.

c. The quotient of n(x) / m(x) can be reduced to a polynomial of one single term as follows:

$\frac{n(x)}{m(x)} =\frac{6\,x^5\,y^7}{2\,x^3\,y} =3\,x^2\,y^6$

which as can be seen, also contains variables raised to positive integers.

d. The degree of the polynomial resultant is the addition of the powers of all variables present (x and y) which results in: 2 + 6 = 8

Therefore the degree of this polynomial is 8.

6 0

3 years ago

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

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3 years ago

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d1i1m1o1n [39]

Answer:

C: closer to one

D: much smaller than one

E: much greater than one

F: much smaller than one

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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Answer:

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