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Rina8888 [55]
3 years ago
6

HELP ASAP PLEASE WILL MARK BRAINIEST

Mathematics
2 answers:
Rom4ik [11]3 years ago
8 0

Answer:

The correct answer is option A.

Step-by-step explanation:

Hi! Nice to meet you and have a great day.

Gekata [30.6K]3 years ago
3 0

Answer:

The answer is B.

Step-by-step explanation:

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Solve the inequality. 4 + 9x > 8 − 3x
mestny [16]

Answer:

x > 1/3

Step-by-step explanation:

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79,002 this is the answer
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Line segmentAB has vertices A(−3, 4) and B(1, −2) . A dilation, centered at the origin, is applied to AB⎯⎯⎯⎯⎯ . The image has ve
zysi [14]

Answer:

The scale factor is

\frac{1}{3}

Step-by-step explanation:

Line segmentAB has vertices A(−3, 4) and B(1, −2) .

A dilation, centered at the origin, is applied to AB. The image has vertices

A'(-1,4/3) and B′(1/3, −2/3) .

We know the rule for dilation by scale factor of k, is

(x,y)\to(kx,ky)

This means that;

A(-3,4)\to A'(-3k,4k)

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- 3k =  - 1

k =  \frac{1}{3}

We could also compare:

4k=4/3

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k=⅓

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3 years ago
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Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integ
liq [111]

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

  • Mx(x,y) = d/dx 8x²y = 16xy
  • Ly(x,y) = d/dy 7y²x = 14xy

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2

We conclude that the line integral is 1/2

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2 years ago
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