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d1i1m1o1n [39]
3 years ago
10

Decide, without calculation, if each of the integrals below are positive, negative, or zero. Let D be the region inside the unit

circle centered at the origin. Let T, B, R, and L denote the regions enclosed by the top half, the bottom half, the right half, and the left half of unit circle, respectively.1. ∬B xe^xdA2. ∬R xe^xdA3. ∬T xe^xdA4. ∬D xe^xdA5. ∬L xe^xdA? Positive Negative Zero
Mathematics
1 answer:
Schach [20]3 years ago
4 0

The integrals over B and T will be positive. Keeping y fixed, xe^x is strictly increasing over D as x increases, so the integrals over x (i.e. the bottom/top left quadrants of D) is negative but the integrals over x>0 are *more* positive.

The integrals over R and L are zero. If we take f(x,y)=xe^x, then f(x,-y)=f(x,y), which is to say f is symmetric across the x-axis. For the same reason, the integral over all of D is also zero.

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In the diagram AB is parallel to CD. What is the value of X
aivan3 [116]

Answer:

B) 120

Step-by-step explanation:

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A lamina with constant density rho(x, y) = rho occupies the given region. Find the moments of inertia Ix and Iy and the radii of
jenyasd209 [6]

Answer:

Ix = Iy = \frac{ρπR^{4} }{16}

Radius of gyration x = y =  \frac{R}{4}

Step-by-step explanation:

Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.

Mass of disk = ρπR2

Moment of inertia about its perpendicular axis is \frac{MR^{2} }{2}. Moment of inertia of quarter disk about its perpendicular is \frac{MR^{2} }{8}.

Now using perpendicular axis theorem, Ix = Iy = \frac{MR^{2} }{16} = \frac{ρπR^{4} }{16}.

For Radius of gyration K, equate MK2 = MR2/16, K= R/4.

3 0
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Copy and complet the table compare the values of 2^x-2 with the values of 2^x-2
Vladimir79 [104]

#2^(x-2)

For 2

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For 3

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For 4

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For 5

  • 2^{5-2}=2³=8

For 6

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#2^x-2

For 2

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For 3

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For 4

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For 5

  • 2⁵-2=30

For 6

  • 2⁶-2=62
3 0
2 years ago
Read 2 more answers
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