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3 years ago

Isi drilled a hole that is 5/9 inch wide.she has a screw that is 5/6 inch wide.is the hole wide enough to fit the screw

Mathematics

1 answer:

koban [17]3 years ago

5 0

NO the screw is to big to fit

The variable in non simplified form will be

5/9 is 10/18

5/6 is 15/18

Since 5/6 is larger the screw will not be able to fit in a smaller hole.

Hope this helped!

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Help me solve it and please go fast

MariettaO [177]

Answer:

x/3 -3= 18 (add three to both sides)

x/3= 21

x/3(*3) =21(*3) (multiply by 3 on each side to cancel out the division)

x= 63

63/3-3=18

21-3=18

18= 18 (checking answers)

Step-by-step explanation:

5 0

3 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

I need help draw a array to represent the problem 5 x 8

xz_007 [3.2K]

Well draw a box with 10 squares going up and 20 square is going sideways and then shade and 5/8 of it

4 0

3 years ago

Lee wants to buy two pairs of shoes. One pair costs $37.07 and the other pair costs $42.48. A pair of socks cost $8.50. If Lee b

daser333 [38]

Answer:

$96.55

Step-by-step explanation:

The computation of the money he would require is shown below:

Since lee wants to buy both pairs of shoes and the two pairs of socks

So the amount of money he needed is

= $37.07 + $42.48 + 2 × $8.50

= $37.07 + $42.48 + $17

= $96.55

5 0

3 years ago

⎧

d1i1m1o1n [39]

Given:

The given expression to find the nth term of the sequence is $d(n)=d(n-1) \cdot (-5)$

The first term of the sequence is $d(1)=8$

We need to determine the third term of the sequence.

Second term:

The second term of the sequence can be determined by substituting n = 2 in the nth term of the sequence.

Thus, we have;

$d(2)=d(2-1) \cdot (-5)$

$d(2)=d(1) \cdot (-5)$

$d(2)=8 \cdot (-5)$

$d(2)=-40$

Thus, the second term of the sequence is -40.

Third term:

The third term of the sequence can be determined by substituting n = 3 in the nth term of the sequence.

Thus, we have;

$d(3)=d(3-1) \cdot (-5)$

$d(3)=-40 \cdot (-5)$

$d(3)=120$

Thus, the third term of the sequence is 120.

7 0

3 years ago

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