The inequality equation that represents the interval − 7 , 2 and whether the interval is bounded or unbounded. The interval − 7 , 2 corresponds to − 7 < x ≤ 2 and the interval is bounded. Explanation: The solution of the inequality equation is the set of all real numbers. The real number line representing the solution of the inequality is the graph of the inequality. Graphs of the inequality are intervals on the real number line. Each interval can be classified as bounded or unbounded. The unbounded interval is, the interval where the end of the solution set or both go to infinity. The bounded interval is the interval where, both lower and upper are real numbers. The interval − 7 , 2 corresponds to − 7 < x ≤ 2 . The given interval is bounded since both the endpoints are real numbers.
The inequality equation that represents the interval − 7 , 2 and whether the interval is bounded or unbounded. The interval − 7 , 2 corresponds to − 7 < x ≤ 2 and the interval is bounded. Explanation: The solution of the inequality equation is the set of all real numbers. The real number line representing the solution of the inequality is the graph of the inequality. Graphs of the inequality are intervals on the real number line. Each interval can be classified as bounded or unbounded. The unbounded interval is, the interval where the end of the solution set or both go to infinity. The bounded interval is the interval where, both lower and upper are real numbers. The interval − 7 , 2 corresponds to − 7 < x ≤ 2 . The given interval is bounded since both the endpoints are real numbers.
Solution Summary: The author explains the inequality equation that represents the interval (-7,2right] and whether it is bounded. The real number line representing the solution is the graph of inequality.
The inequality equation that represents the interval −7,2 and whether the interval is bounded or unbounded.
The interval −7,2 corresponds to −7<x≤2 and the interval is bounded.
Explanation:
The solution of the inequality equation is the set of all real numbers. The real number line representing the solution of the inequality is the graph of the inequality. Graphs of the inequality are intervals on the real number line.
Each interval can be classified as bounded or unbounded.
The unbounded interval is, the interval where the end of the solution set or both go to infinity.
The bounded interval is the interval where, both lower and upper are real numbers.
The interval −7,2 corresponds to −7<x≤2.
The given interval is bounded since both the endpoints are real numbers.
Let
2
A =
4
3
-4
0
1
(a) Show that v =
eigenvalue.
()
is an eigenvector of A and find the corresponding
(b) Find the characteristic polynomial of A and factorise it. Hint: the answer to (a)
may be useful.
(c) Determine all eigenvalues of A and find bases for the corresponding eigenspaces.
(d) Find an invertible matrix P and a diagonal matrix D such that P-¹AP = D.
(c) Let
6
0 0
A =
-10 4 8
5 1 2
(i) Find the characteristic polynomial of A and factorise it.
(ii) Determine all eigenvalues of A and find bases for the corresponding
eigenspaces.
(iii) Is A diagonalisable? Give reasons for your answer.
most 2, and let
Let P2 denote the vector space of polynomials of degree at
D: P2➡ P2
be the transformation that sends a polynomial p(t) = at² + bt+c in P2 to its derivative
p'(t)
2at+b, that is,
D(p) = p'.
(a) Prove that D is a linear transformation.
(b) Find a basis for the kernel ker(D) of the linear transformation D and compute its
nullity.
(c) Find a basis for the image im(D) of the linear transformation D and compute its
rank.
(d) Verify that the Rank-Nullity Theorem holds for the linear transformation D.
(e) Find the matrix representation of D in the standard basis (1,t, t2) of P2.
Chapter 1 Solutions
Bundle: College Algebra, Loose-leaf Version, 10th + WebAssign Printed Access Card for Larson's College Algebra, 10th Edition, Single-Term
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