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2 years ago

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Geoffrey is evaluating the expression StartFraction (negative 3) cubed (2 Superscript 6 Baseline) Over (Negative 3) Superscript

5 Baseline (2 squared) EndFraction as shown below. StartFraction (negative 3) cubed (2 Superscript 6 Baseline) Over (Negative 3) Superscript 5 Baseline (2 squared) EndFraction = StartFraction (2) Superscript a Baseline Over (negative 3) Superscript b Baseline EndFraction = StartFraction c Over d EndFraction What are the values of a, b, c, and d? a = 4, b = 2, c = 16, d = 9 a = 4, b = negative 2, c = 16, d = 9 a = 8, b = 8, c = 256, d = 6,561 a = 8, b = 8, c = 256, d = negative 6,561

2 answers:

vivado [14]2 years ago

5 0

Answer:

the answer is A

Step-by-step explanation:

I did the test and got it right for the explanation, the kind person above provided that.

makkiz [27]2 years ago

3 0

The mathematical expression does not seem clear but I have made an attempt to make sense of what is implied.

Answer:

<em>a</em> = 4, <em>b</em> = 2, <em>c</em> = 16, <em>d</em> = 9

Step-by-step explanation:

$\dfrac{(-3)^3(2^6)}{(-3)^5(2^2)} = \dfrac{(2)^a}{(-3)^b} = \dfrac{c}{d}$

Solving the first part of the question by indices,

$\dfrac{(-3)^3(2^6)}{(-3)^5(2^2)} = (-3)^{3-5}(2)^{6-2} = (-3)^{-2}(2)^{4} = \dfrac{(2)^4}{(-3)^2}$

Comparing the rightmost term with the second term in the question,

<em>a</em> = 4, <em>b</em> = 2

Solving on,

$\dfrac{(2)^4}{(-3)^2} = \dfrac{(2)\times(2)\times(2)\times(2)}{(-3)\times(-3)} = \dfrac{16}{9}$

Comparing with the final term in the question,

<em>c</em> = 16 and <em>d</em> = 9

Therefore,

<em>a</em> = 4, <em>b</em> = 2, <em>c</em> = 16, <em>d</em> = 9

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Answer:

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Step-by-step explanation:

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The distance between New York and Philadelphia by railroad is 90 miles; the distance between them by river and ocean is 240 mile

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2 years ago

If a is an arbitrary nonzero constant, what happens to a/b as b approaches 0

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2 years ago

Evaluate the line integral by the two following methods. xy dx + x2y3 dy C is counterclockwise around the triangle with vertices

nadezda [96]

Answer:

a)

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b)

$\frac{2}{3}$

Step-by-step explanation:

a) The first part requires that we use line integral to evaluate directly.

The line integral is

$\int_C xydx + {x}^{2} {y}^{3} dy$

where C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2)

The boundary of integration is shown in the attachment.

Our first line integral is

$L_1 = \int_ {(0,0)}^{(1,0)} xydx + {x}^{2} {y}^{3} dy$

The equation of this line is y=0, x varies from 0 to 1.

When we substitute y=0 every becomes zero.

$\therefore \: L_1 =0$

Our second line integral is

$L_2 = \int_ {(1,0)}^{(1,2)} xydx + {x}^{2} {y}^{3} dy$

The equation of this line is:

$x = 0 \implies \: dx = 0$

y varies from 1 to 2.

We substitute the boundary and the values to get:

$L_2 = \int_ {1}^{2}1 \cdot y(0) + {1}^{2} \cdot \: {y}^{3} dy$

$L_2 = \int_ {1}^2 {y}^{3} dy = \frac{8}{3}$

The 3rd line integral is:

$L_3 = \int_ {(1,2)}^{(0,0)} xydx + {x}^{2} {y}^{3} dy$

The equation of this line is

$y = 2x \implies \: dy = 2dx$

x varies from 0 to 1.

We substitute to get:

$L_3 = \int_ {1}^{0} x \cdot \: 2xdx + {x}^{2} {(2x)}^{3}(2 dx)$

$L_3 = \int_ {1}^{0} 8 {x}^{5} + 2 {x}^{2} dx = - 2$

The value of the line integral is

$L = L_1 + L_2 + L_3$

$L = 0 + \frac{8}{3} + - 2 = \frac{2}{3}$

b) The second part requires the use of Green's Theorem to evaluate:

$\int_C xydx + {x}^{2} {y}^{3} dy$

Since C is a closed curve with counterclockwise orientation, we can apply the Green's Theorem.

This is given by:

$\int_C \: Pdx +Q \: dy = \int \int_ R \: Q_y - P_x \: dA$

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int \int_ R \: 3 {x}^{2} {y}^{2} - y \: dA$

We choose our region of integration parallel to the y-axis.

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \int_ 0^{2x} \: 3 {x}^{2} {y}^{2} - y \: dydx$

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \: {x}^{2} {y}^{3} - \frac{1}{2} {y}^{2} |_ 0^{2x} dx$

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \: 8{x}^{5} - 2 {x}^{2} dx = \frac{2}{3}$

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2 years ago