I need help it’s geometry
2 answers:
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Answer:
x ∈ (-∞, -1) ∪ (1, ∞)
Step-by-step explanation:
To solve this problem we must factor the expression that is shown in the denominator of the inequality.
So, we have:
So the roots are:
Therefore we can write the expression in the following way:
Now the expression is as follows:
Now we use the study of signs to solve this inequality.
We have 3 roots for the polynomials that make up the expression:
We know that the first two are not allowed because they make the denominator zero.
Observe the attached image.
Note that:
when
when
and
is always
Finally after the study of signs we can reach the conclusion that:
x ∈ (-∞, -1) ∪ (1, 2] ∪ [2, ∞)
This is the same as
x ∈ (-∞, -1) ∪ (1, ∞)
Answer:
a) No
b) No
c) No
d) No
Step-by-step explanation:
Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:
1. u+v∈V
2. u+v=v+u
3. (u+v)+w=u+(v+w).
4. Exist 0∈V such that u+0=u
5. For each u∈V exist −u∈V such that u+(−u)=0.
6. if c is an escalar and u∈V, then cu∈V
7. c(u+v)=cu+cv
8. (c+d)u=cu+du
9. c(du)=(cd)u
10. 1u=u
let's check each of the properties for the respective operations:
Let
Observe that
1. u+v∈V
2. u+v=v+u, because the adittion of reals is conmutative
3. (u+v)+w=u+(v+w). because the adittion of reals is associative
4.
5.
then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.
a)
6.
7.
8.
Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product
b) 6.
7.
8.
9.
10
Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product
c) Observe that
Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product .
d) Observe that
Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product .