Jack's father drives to work in 30 minutes when driving at his usual speed. When traffic is bad, he drives 30 miles per hour slo
1 answer:
You might be interested in
Answer:
A number line goes from negative 10 to positive 10. An open circle appears at negative 9. The number line is shaded from negative 9 through positive 10.
Step-by-step explanation:
we have
Solve for x
Adds 7 both sides
Divide by -3 both sides
Remember that
When you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol
so
The solution is the interval (-9,∞)
therefore
A number line goes from negative 10 to positive 10. An open circle appears at negative 9. The number line is shaded from negative 9 through positive 10.
Answer:
1. Yes.
2. No.
3. Yes.
Step-by-step explanation:
Consider the following subsets of Pn given by
1.Let W1 be the set of all polynomials of the form , where a is in ℝ.
2.Let W2 be the set of all polynomials of the form , where a is in ℝ.
3. Let W3 be the set of all polynomials of the form , where a is in ℝ.
Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:
- The 0 vector of V is in W.
- Given v,w in W then v+w is in W.
- Given v in W and a a real number, then av is in W.
So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.
- First property:
Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.
For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.
- Second property:
W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials
.
We must check that p_1+p_2(t) is in W1.
Note that
Since a+b is another real number, we have that p1(t)+p2(t) is in W1.
W3. Consider two elements in W3. Say . Then
So, again, p1(t)+p2(t) is in W3.
- Third property.
W1. Consider an element in W1 and a real scalar b. Then
.
Since (ba) is another real scalar, we have that bp(t) is in W1.
W3. Consider an element in W3 and a real scalar b. Then
.
Since (ba) is another real scalar, we have that bp(t) is in W3.
After all,
W1 and W3 are subspaces of Pn for n= 2
and W2 is not a subspace of Pn.