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wlad13 [49]
3 years ago
7

HELP TIMED Which is equivalent to ?

Mathematics
1 answer:
Nataliya [291]3 years ago
3 0

Answer:

C the one with the 12

so the 3 turn into 1/3 now 1/3 times 1/4 = 1/12

now turn the 1/12 into the outside of the sign

Step-by-step explanation:

You might be interested in
What is the solution to this equation y=4x<br> (1,3)<br> (-1,-4)<br> (-4,-1)<br> (1,-4)<br> !ASAP!
RideAnS [48]
(-1,-4) this is saying that when x=-1, y=-4. Plugging it into the line equation y=4x gives
(- 4) = 4( - 1)
Which is correct.
3 0
3 years ago
A closed cylindrical vessel contains a fluid at a 5MPa pressure. The cylinder, which has an outside diameter of 2500mm and a wal
Julli [10]

Answer:

1) Increase in the diameter equals 3.5 mm

2) Increase in the length equals 0.0003724L_{i} where L_{i} is the initial length of the vessel.

Step-by-step explanation:

The diametric strain in the vessel is given by

\epsilon_{D} =\epsilon_{diam}-\nu \epsilon _{axial}

We have

\epsilon _{diam}=\frac{\sigma _{hoop}}{E}\\\\\sigma _{hoop}=\frac{\Delta P\times D}{2t}\\\\\therefore \epsilon _{diam}=\frac{\Delta P\times D}{2t\times E}

Applying values we get

\therefore \epsilon _{diam}=\frac{5\times 10^{6}\times 2.5}{2\times 20\times 10^{-3}\times 193\times 10^{9}}\\\\\therefore \epsilon _{diam}=\frac{5}{3088}

Similarly axial strain is given by

\epsilon _{diam}=\frac{\sigma _{axial}}{E}

\sigma _{axial}=\frac{\Delta P\times D}{4t}\\\\\therefore \epsilon _{axial}=\frac{\Delta P\times D}{4t\times E}

Applying values we get

\therefore \epsilon _{axial}=\frac{5\times 10^{6}\times 2.5}{4\times 20\times 10^{-3}\times 193\times 10^{9}}\\\\\therefore \epsilon _{diam}=\frac{2.5}{3088}

Hence The effect of axial strain along the diameter is given by

-\nu \epsilon _{axial}

Applying values we get

-\nu \epsilon _{axial}=-0.27\times \frac{2.5}{3088}=-0.0002185

hence

\epsilon _{D} =\frac{5}{3088}-0.0002185\\\\\epsilon =0.00140

Now by definition of strain we have

\epsilon _{D} =\frac{D_{f}-D_{i}}{D_{i}}\\\\\therefore D_{f}=D_{i}+\epsilon D_{i}\\\\D_{f}=2.5+0.0014\times 2.5\\\\\therefore D_{f}=2503.5mm

Increase in the diameter is thus 3.5 mm

Using the same procedure for axial strain we have

\epsilon_{axial} =\epsilon_{axial}-\nu \epsilon _{diam}

Applying values we get

\epsilon_{axial} =\frac{2.5}{3088}-0.27\times \frac{5}{3088}

\epsilon_{axial} =0.0003724

Now by definition of strain we have

\epsilon _{axial} =\frac{L_{f}-L_{i}}{L_{i}}\\\\\therefore \Delta L=0.0003724L_{i}

where L_{i} is the initial length of the cylinder.

6 0
4 years ago
Find the average rate of change of f(x)=4x^2-6 on interval [1,a]​
stiv31 [10]

9514 1404 393

Answer:

  4a +4

Step-by-step explanation:

The average rate of change of f(x) on the interval is ...

  (f(a) -f(1))/(a -1)

  = ((4a² -6) -(4(1)² -6))/(a -1) = (4a² -4)/(a -1)

  = 4(a² -1)/(a -1) = 4(a +1) = 4a +4

The average rate of change on the interval is 4a +4.

4 0
3 years ago
Is 1.75 greater than 5.50
lord [1]

Answer:

No

Step-by-step explanation:

1.75 < 5.50

6 0
4 years ago
Read 2 more answers
Can somebody solve this?
lisabon 2012 [21]

Answer:

x= 2

Step-by-step explanation:

Isolate the radical

Raise each side of the equation to the power of 3

5 0
3 years ago
Read 2 more answers
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