Which ordered pairs are a solution are a solution to the equation?

Mathematics

2 answers:

notsponge [240]3 years ago

7 0

D is the answer because if you calculate the answer correctly. that will be so.

denis-greek [22]3 years ago

5 0

Answer:

I think the answer us

D no.

You might be interested in

Find the value of given expression<br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B5%20%2B%204%7D%20" id="TexFormula1" tit

mafiozo [28]

Answer:

Step-by-step explanation:

$\sqrt{5 + 4} \\ = \sqrt{9 } \\ = 3$

8 0

3 years ago

Read 2 more answers

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

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