Expresa cada proposición mediante la gráfica, conjunto, desigualdad y la notación de
intervalo.
a. La estatura de los jugadores de un equipo de baloncesto es menor a 1,98 m y mayor o
igual a 1,82 m.
b. Los niveles normales de glucosa en ayunas en un ser humano deben ser mayores o
iguales a 70 mg/dl y menores que 100 mg/dl.
c. El tiempo que tarda una persona en llegar a su trabajo es mayor a 5/6 hora y menor o
igual a 3/2 hora.
d. La categoría juvenil es para personas mayores a 15 años y menores o iguales a 18
años.
e. La temperatura de la ciudad de Cali regularmente es menor a 31° C y mayor o igual a
19°C.

Answers

Answer 1

a. The height of the players of a basketball team is less than 1.98 m and greater than or equal to 1.82 m.

The Graph:

Draw a number line with 1.82 and 1.98 marked on it. Shade the region between the two points, including the point at 1.82, but not 1.98.

Set:

H = {h | 1.82 ≤ h < 1.98}

Inequality:

1.82 ≤ h < 1.98

Interval Notation:

[1.82, 1.98)

b. Normal fasting glucose levels in a human should be greater than or equal to 70 mg/dl and less than 100 mg/dl.

Graph:

Draw a number line with 70 and 100 marked on it. Shade the region between the two points, including both endpoints.

Set:

G = {g | 70 ≤ g < 100}

Inequality:

70 ≤ g < 100

Interval Notation:

[70, 100)

c. The time it takes a person to get to work is greater than 5/6 hours and less than or equal to 3/2 hour.

Graph:

Draw a number line with 5/6 and 3/2 marked on it. Shade the region between the two points, not including 5/6, but including 3/2.

Set:

T = {t | 5/6 < t ≤ 3/2}

Inequality:

5/6 < t ≤ 3/2

Interval Notation:

(5/6, 3/2]

d. The youth category is for people over 15 years of age and under or equal to 18 years.

Graph:

Draw a number line with 15 and 18 marked on it. Shade the region between the two points, not including 15, but including 18.

Set:

Y = {y | 15 < y ≤ 18}

Inequality:

15 < y ≤ 18

Interval Notation:

(15, 18]

e. The temperature in the city of Cali is regularly less than 31° C and greater than or equal to 19°C.

Graph:

Draw a number line with 19 and 31 marked on it. Shade the region between the two points, including 19, but not 31.

Set:

T_Cali = {t | 19 ≤ t < 31}

Inequality:

19 ≤ t < 31

Interval Notation:

[19, 31)

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The question in English is:

Express each statement using the graph, set, inequality, and the notation of

interval.

to. The height of the players of a basketball team is less than 1.98 m and greater than or

equal to 1.82 m.

b. Normal fasting glucose levels in a human should be greater than or

equal to 70 mg/dl and less than 100 mg/dl.

c. The time it takes a person to get to work is greater than 5/6 hours and less than or

equal to 3/2 hour.

d. The youth category is for people over 15 years of age and under or equal to 18

years.

and. The temperature in the city of Cali is regularly less than 31° C and greater than or equal to

19°C.


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A small block sits at one end of a flat board that is 4.00 m long. The coefficients of friction between the block and the board are μs = 0.550 and μk = 0.400. The end of the board where the block sits is slowly raised until the angle the board makes with the horizontal is α0 , and then the block starts to slide down the board.
Part A
If the angle is kept equal to α0 as the block slides, what is the speed of the block when it reaches the bottom of the board?

Answers

The speed of the block when it reaches the bottom of the board would be 3.19 m/s.

How to calculate the speed?

Let's start by finding the height of the board when the block starts to slide. The maximum angle α0 at which the block will remain stationary on the board is given by:

μs = tan(α0)

Solving for α0, we get:

α0 = tan^-1(μs) = tan^-1(0.550) = 29.59 degrees

The height of the board when the block starts to slide is given by:

h = 4.00 m sin(α0)

h = 4.00 m sin(29.59 degrees)

h = 2.09 m

Now let's find the work done by friction. The friction force is given by:

f= μk * N

where N is the normal force acting on the block. The normal force is equal in magnitude and opposite in direction to the component of the gravitational force acting perpendicular to the board, which is given by:

N = mg cos(α)

where m is the mass of the block, g is the acceleration due to gravity, and α is the angle of the board. The kinetic friction force does negative work, which is equal to:

Wf = -f * d

where d is the distance the block travels down the board. The work done by gravity is equal to the change in potential energy, which is given by:

Wg = mgh

where h is the height of the board. At the bottom of the board, all of the potential energy has been converted to kinetic energy, so we have:

1/2 * m * v^2 = mgh - f * d

Solving for v, we get:

v = sqrt(2gh - 2μk(N/m) * d)

We can now substitute the values we have found:

h = 2.09 m

μk = 0.400

N = mg cos(α0) = mg cos(29.59 degrees)

d = 4.00 m

Substituting these values and solving for v, we get:

v = sqrt(2gh - 2μk(N/m) * d)

v = sqrt(2 * 9.81 m/s^2 * 2.09 m - 2 * 0.400 * (m * g * cos(29.59 degrees)/m) * 4.00 m)

v = 3.19 m/s

Therefore, the speed of the block when it reaches the bottom of the board is 3.19 m/s.

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