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KengaRu [80]
2 years ago
14

Select the properties that equilateral triangles and isosceles triangles have in common.

Mathematics
1 answer:
defon2 years ago
7 0

The properties that equilateral triangles and isosceles triangles have in common is that they have two equal sides.

<h3>What is a triangle?</h3>

A triangle is a polygon that has three sides and three angles. Types of triangles are<em> isosceles, equilateral </em>and<em> scalene</em> triangle.

An equilateral triangle is a triangle that have three equal sides while isosceles triangle have two equal sides.

The properties that equilateral triangles and isosceles triangles have in common is that they have two equal sides.

Find out more on triangle at: brainly.com/question/17335144

#SPJ1

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find the value of x and the value of y. A. x = 15, y = 10 B. x = 20, y = 50 C. x = 50, y = 10 D. x = 50, y = 20
SIZIF [17.4K]
I think it will be b. x=20,y=50
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
Please just help !!!!!!
IceJOKER [234]

Answer:

what the hecker is that.

okay so your simplifying it, 12? i think.

3 0
2 years ago
Is this true or false?
marusya05 [52]

Answer:

TRUE

Step-by-step explanation:

4 0
3 years ago
Im going to give 10 points for this question please answer it
german
57 but it is an estimate because of the 4  so your answer is 57.4
8 0
3 years ago
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