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Elden [556K]
3 years ago
6

Evaluate the line integral 2 + x2y ds where c is the upper half of the circle x2 + y2 = 1.

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
8 0

Parameterize <em>C</em> by

r(t) = 〈x(t), y(t)〉 = 〈cos(t), sin(t)〉

with 0 ≤ t ≤ <em>π</em>. Then the line integral is

\displaystyle \int_C (2+x^2y)\,\mathrm ds = \int_0^\pi (2+\cos^2(t)\sin(t))\left\|\mathbf r'(t)\right\|\,\mathrm dt \\\\ = \int_0^\pi (2+\cos^2(t)\sin(t)) \,\mathrm dt = \boxed{\frac23+2\pi}

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Can you please help me with this
Vitek1552 [10]

Answer:

log9

Step-by-step explanation:

Using the rules of logarithms

logx + logy = log(xy)

logx - logy = log(\frac{x}{y})

logx^{n} ⇔ nlogx

Given

2(log18- log3) + \frac{1}{2}log\frac{1}{16}

= 2(log(\frac{18}{3} ) ) + log(\frac{1}{16}) ^{\frac{1}{2} }

= 2log6 + log\frac{1}{4}

= log6² + log\frac{1}{4}

= log36 + log\frac{1}{4}

= log( 36 × \frac{1}{4} )

= log9

5 0
3 years ago
Which point on the number line represents the quotient 12 divided by -2
Afina-wow [57]
Point C is the answer I believe
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3 years ago
The vertex form of the equation of a parabola y=(x-3)^+36
MrMuchimi

Answer:

ewrwewqetqwrtweq

Step-by-step explanation:

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Divide 30 cm into ther ratio 2:4
pantera1 [17]
In the ratio 2:4

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6 0
3 years ago
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What is the midpoint of the line segment with endpoints ( 3.5,2.2) and (1.5,-4.8)?
Fantom [35]

Answer:

Option A) (2.5,-1.3) is correct

The midpoint of the given line segment is M=(2.5,-1.3)

Step-by-step explanation:

Given that the line segment with end points (3.5, 2.2) and (1.5, -4.8)

To find the mid point of these endpoints midpoint formula is M=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)

Let (x_1,y_1) be the point (3.5, 2.2) and (x_2, y_2) be the point (1.5, -4.8)

substituting the points in the formula

M=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)

M=\left(\frac{3.5+1.5}{2}, \frac{2.2-4.8}{2}\right)

=\left(\frac{5}{2}, \frac{-2.6}{2}\right)

=\left(2.5,-1.3\right)

Therefore M=(2.5,-1.3)

The midpoint of the given line segment is M=(2.5,-1.3)

8 0
3 years ago
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