Answer:
B) x^2+6x+8
Explanation:
x-4 | x^3+2x^2-16x-32
- x^3-4x^2 <-- (x-4)(x^2)
_________________
6x^2-16x-32
- 6x^2-24x <-- (x-4)(6x)
_________________
8x-32
- 8x-32 <- (x-4)(8)
___________________________
0 | x^2+6x+8
This means the answer is B) x^2+6x+8
I think it’s 15cm
Might be 7cm
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as
Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as
where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then
Resolving for F,
PART B) We need here to apply the shear strength equation, then
In such a way that the material is more resistant to tensile strength than shear force.
Discrete systems are those systems in which are made up of finite component particles a which are non-homogeneously arranged such that no smooth variation exists. It is such that all constituent particles have properties which vary randomly. They are direct opposite to continuous systems, which are smooth arrangement of particles which cannot be individually taken into consideration.
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Answer:
First one, third one, and fourth one