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d1i1m1o1n [39]
4 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]4 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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patriot [66]

Answer:

m∠B = 110°

Step-by-step explanation:

We know that,

The sum of the measures of the angles in a pentagon is 540°.

So, we get,

130 + (x-5) + (x+30) + 75 + (x-35) = 540

i.e. 3x + (130+30+75) - (5+35) = 540

i.e. 3x + 235 - 40 = 540

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i.e. 3x = 345

i.e. x= 115°

Now, as m∠B = (x-5)° = (115-5)° = 110°

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4 0
3 years ago
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spayn [35]

Answer with explanation:

Let us assume that the 2 functions are:

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\frac{dy}{dx}

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Part 2)

The product of the 2 functions is shown below

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Diffrentiating both sides with respect to 'x' we get

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Now we can see the sign of the terms on the right hand side depend on the signs of the function's themselves hence we remain inconclusive about the sign of the product as a whole. Thus the product can be concave or convex.

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Step-by-step explanation:

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7 0
3 years ago
This better? If u seen my last one
ololo11 [35]
Ya I like the first one better than the second one but both are good
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