All categories

3 years ago

What is the difference between an irrational number and an integer

1 answer:

7 0

Learn to tell the difference between relational and a relational numbers ration on numbers are the numbers that can be written in like fractions with them one in together the numerator over the integrator the denominator original numbers on the other hand are all the numbers that cannot be written this way.

You might be interested in

Help me plss I have a f

Serjik [45]

Answer:

31/3 or 31 over 3

Step-by-step explanation:

If you do the whole number (10) times the denominator (bottom number of the fraction,in this case 3) the add the numerator (top number of the fraction,in this case 1) you will get your improper fraction.

3 0

2 years ago

Expand and Simplify 4(g + 5) + 3(g – 2)

Montano1993 [528]

The solution is 7g+14

8 0

2 years ago

A mass of 1 g is set in motion from its equilibrium position with an initial velocity of 6in/sec, with no damping and a spring c

yan [13]

a) $y(t)=0.0016 sin(94.9t)$ [m]

b) 0.033 s

c) -0.152 m/s

Step-by-step explanation:

The force acting on the mass-spring system is (restoring force)

$F=-ky$

where

k = 9 is the spring constant

y is the displacement

Also, from Newton's second law of motion, we know that

$F=my''$

where

m = 1 g = 0.001 kg is the mass

y'' is the acceleration

Combining the two equations,

$my''=-ky$

This is a second order differential equation; the solution for y(t) is

$y(t)=A sin(\omega t-\phi)$

where

A is the amplitude of motion

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{9}{0.001}}=94.9 rad/s$ is the angular frequency

The spring starts its motion from its equilibrium position, this means that y=0 when t=0; therefore, the phase shift must be

$\phi=0$

So the displacement is

$y(t)=A sin(\omega t)$

The velocity of the spring is equal to the derivative of the displacement:

$v(t)=y'(t)=\omega A cos(\omega t)$

We know that at t = 0, the initial velocity is 6 in/s; since 1 in = 2.54 cm = 0.0254 m,

$v_0=6(0.0254)=0.152 m/s$

And since at t = 0, $cos(\omega t)=1$

Then we have:

$v_0=\omega A$

From which we find the amplitude:

$A=\frac{v_0}{\omega}=\frac{0.152}{94.9}=0.0016 m$

So the solution for the displacement is

$y(t)=0.0016 sin(94.9t)$ [m]

Here we want to find the time t at which the mass returns to equilibrium, so the time t at which

$y=0$

This means that

$sin(\omega t)=0$

We know already that the first time at which this occurs is

t = 0

Which is the beginning of the motion.

The next occurence of y = 0 is instead when

$\omega t = \pi$

which means:

$t=\frac{\pi}{\omega}=\frac{\pi}{94.9}=0.033 s$

As said in part a), the velocity of the mass-spring system at time t is given by the derivative of the displacement, so

$v(t)=\omega A cos(\omega t)$

where we have

$\omega=94.9 rad/s$ is the angular frequency

$A=0.0016 m$ is the amplitude of motion

t is the time

Here we want to find the velocity of the mass when the time is that calculated in part b):

t = 0.033 s

Substituting into the equation, we find:

$v(0.033)=(94.9)(0.0016)cos(94.9\cdot 0.033)=-0.152 m/s$

4 0

3 years ago

A cab company charges a $3 boarding rate in addition to its meter which is $2 for every mile. What is the equation?

STALIN [3.7K]

3+2(x) x is for miles

8 0

2 years ago

Read 2 more answers

Which number line correctly shows 1. 5 + 2. 5? A number line going from 0 to 6. 5 in increments of 0. 5. An arrow goes from 0 to

Usimov [2.4K]

So we want to see which number line shows the sum 1.5 + 2.5. We will see that the correct option is "An arrow goes from 0 to 1. 5 and from 1. 5 to 4"

So the number line usually starts at 0 and goes to the first number, 1.5 in this case.

Then, it moves accordingly to the second number, 2.5 (so it will move 2.5 units to the right)

So the number line should start at 0 and move to the right until it meets 1.5 + 2.5 = 4.

From the options, the only that is correct is:

"An arrow goes from 0 to 1. 5 and from 1. 5 to 4"

If you want to learn more about number lines, you can read:

brainly.com/question/10851163

6 0

2 years ago