If c(x)=5/x-2 and d(x)=x+3, what is the domain of (cd)(x)

An aircraft carrier left Port 35 traveling north five hours before a container ship. The container ship traveled in the opposite

Molodets [167]

Answer:

Aircraft carrier speed = 4.97 km/h

Step-by-step explanation:

Given that the time taken by the aircraft = 5 hours

Let the distance covered = X

Speed = V

Time taken by the ship = 7 hours

Let the distance covered = Y

Speed = 8V

Since it is 8 km/h faster than the aircraft carrier.

Using the speed formula

Speed = distance/time

Distance = speed × time

For the aircraft;

X = 5V

For the ship:

Y = 7 × 8V = 56V

Since they are 303km distance apart,

X + Y = 303

5V + 56V = 303

61V = 303

V = 303/61

V = 4.97 km/h

Aircraft carrier speed = 4.97 km/h

5 0

3 years ago

Find the domain of the graphed function.

Tpy6a [65]

Dddddddddddddddddddddd

8 0

3 years ago

Given the equation (x−13)2+y2=64, identify the center and radius. Do not enter any spaces when typing your answers.

Darya [45]

Answer:

centre = (13,0)

radius = 8

Step-by-step explanation:

The standard equation of the circle is

(x-x0)^2 + (y-y0)^2 = r^2 ...............(1)

where

(x0,y0) is the centre,

r is the radius.

For

(x-13)^2 + y^2 = 64 ..............(2)

we rewrite (2)

(x-13)^2 + (y-0)^2 = 8^2 ...............(3)

and compare (3) with (1)

to identify

x0 = 13, y0 = 0, and r = 8

Therefore

centre = (13,0)

radius = 8

3 0

3 years ago

Last year, the world produced 3.5 billon tons of cement and that amount could increase by a billion tons by year 2050. Express t

Svetradugi [14.3K]

Answer:

$4.5 \times 10^{9}$ tons.

300 times.

Step-by-step explanation:

Last year, the world produced 3.5 billion tons of cement and that amount could increase by a billion tons by the year 2050.

So, the net production of cement by the year 2050 will be (3.5 + 1) = 4.5 billion tons i.e. $4.5 \times 10^{9}$ tons. (Answer)

Now, this amount is larger than $1.5 \times 10^{7}$ by $\frac{4.5 \times 10^{9} }{1.5 \times 10^{7}} = 300$ times. (Answer)

5 0

3 years ago

A 63 foot long cable is anchored in the ground and

myrzilka [38]

Answer:

The height of the light pole to the nearest foot is approximately 50 feet

Step-by-step explanation:

The parameters given in the question are;

The length of the cable = 63 feet

The angle of elevation of the cable to the top of the light pole = 52°

With the assumption that the light pole is perpendicular to the ground, the figure formed by the light pole, the distance of the of the cable from the base of the light pole and the length of the cable form a right triangle

The question can be answered by using trigonometric relations for the sine of the given angle as follows;

$sin ( \theta) = \dfrac{Opposite \ leg \ length}{Hypotenuse \ length}$

The opposite leg length of the formed right triangle = The height of the light pole

The hypotenuse length = The length of the cable = 63 feet

The angle, θ = The angle of elevation = 52°

Plugging in the values, gives;

$sin (52 ^ {\circ}) = \dfrac{The \ height \ of \ the \ light \ pole }{63 \ feet}$

∴ The height of the light pole = 63 feet × sin(52°) ≈ 49.645 feet

The height of the light pole to the nearest foot ≈ 50 feet.

8 0

3 years ago