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2 years ago

Determine if 0.515115111511115111115... is rational or irrational and give

1 answer:

7 0

It’s rational, for a number to be rational you have to be able to write it as a fraction and you can write 0.515115111511115111115... as a fraction.

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Han made some hot chocolate by mixing 4 cups of milk with 6 tablespoons of cocoa. How many tablespoons of cocoa per cup of milk

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Answer:

Step-by-step explanation:

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2 years ago

A phase modulator produces a maximum phase shift of 45 degrees. The modulation frequency range is 300 to 4000 Hz. What is the ma

sergejj [24]

Answer:

Maximum frequency deviation is 2616.29509 Hz

Maximum frequency deviation=(4000-3000)cos45

Max frequency deviation=2616.29509 Hz

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2 years ago

Tim explained a function in words and Paul wrote an equation. Tim: The amount of money in a savings account increases at a rate

masya89 [10]

D .........is the answer

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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:

$\frac{2}{3}$

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.

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

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

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

8 0

2 years ago

MODELING REAL LIFE In 60 seconds, a car in a parade travels 0.2 mile. The car traveled the last 0.05 mile in 12 seconds. How lon

schepotkina [342]

Well, if we put everything together we have gone .25 of a mile in 72 seconds.

Since .25 is a quarter of a mile (1/4) we can estimate that in 288 seconds the car has traveled one mile. We got 288 by multiplying .25 by 4.

8 0

2 years ago

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