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d1i1m1o1n [39]
2 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]2 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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Consider f(x) = 1.8x – 10 and g(x) = −4. A 2 column table with 6 rows. The first column, x, has the entries, negative 4, 0, 2, 4
viktelen [127]

Answer:1.8x – 10 = –4; x = 1.8 x minus 10 equals negative 4; x equals StartFraction 10 Over 2 EndFraction.

1.8 - 10= -4 x= -10/9

Step-by-step explanation:

5 0
2 years ago
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The tables represent two linear functions in a system.<br><br> What is the solution to this system?
Darya [45]

The solution to this system is (x, y) = (8, -22).

The y-values get closer together by 2 units for each 2-unit increase in x. The difference at x=2 is 6, so we expect the difference in y-values to be zero when we increase x by 6 (from 2 to 8).

You can extend each table after the same pattern.

In table 1, x-values increase by 2 and y-values decrease by 8.

In table 2, x-values increase by 2 and y-values decrease by 6.

The attachment shows the tables extended to x=10. We note that the y-values are the same (-22) for x=8 (as we predicted above). That means the solution is ...

... (x, y) = (8, -22)

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2 years ago
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Find the range of y=3/2cos4x-1
MrRissso [65]

Answer:

Range = [- 2.5, 0.5] = [ - 5/2, 1/2]

Step-by-step explanation:

Smallest value of cos α = - 1,

largest value of cos α = 1.

When cos 4x = - 1,  y=3/2cos4x-1 = 3/2*(-1) - 1 = - 5/2 = - 2 1/2 = - 2.5

When cos 4x = 1,  y=3/2cos4x-1 = 3/2*1 - 1 = 1/2 = 0.5

Range = [- 2.5, 0.5] = [ - 5/2, 1/2]

5 0
2 years ago
What is the quotient (x3 – 3x2 + 3x – 2) ÷ (x2 – x + 1)?
timofeeve [1]

We need to find the quotient of the given division problem.

\frac{x^{3}-3x^{2}+3x-2}{x^{2}-x+1}

In order to find its quotient, we will use long division.

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First of all, we put x in the quotient as x^{2} goes into x^{3}, x times.

So, we get:

x^{2}-x+1)x^{3}-3x^{2}+3x-2(x

\text{ ..................}x^{3}-x^{2}+x

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We can see that x^{2} goes into -2x^{2}, -2 times, therefore, the next term in the quotient will be -2. This makes our quotient as (x-2).


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