The best approximation to by elements of a subspace W is given by the vector −projW(). 2. If an ×p matrix has orthonormal columns, then T= for all in ℝ. 3. If W is a subspace of ℝ and if is in both W and W⊥, then must be the zero vector. 4. If =1+2, where 1 is in a subspace W and 2 is in W⊥, then 1 must be the orthogonal projection of onto W. 5. Let 1,2,…, be an orthonormal basis of a vector space . In the Orthogonal Decomposition Theorem, each term =⟨,1⟩1+…+⟨,⟩ is itself an orthogonal projection of onto a subspace of .

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

2 2/3 hours duhh silly

step-by-step explanation:

slope =   (y2 -y1) ÷ (x2 -x1)

slope = (0 -0) / (-5, -7)

slope = 0 / -12 = 0

the slope of the line is zero.   it is perfectly horizontal.

step-by-step explanation: