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1 answer:
Answer:
• multiplied by 4p: (x -h)² +4pk = 0
• zeros for k > 0: none
• zeros for k = 0: one
• zeros for k < 0: two
Step-by-step explanation:
a) Multiplying by 4p removes the 1/(4p) factor from the squared term, but adds a factor of 4p to the k term. (It has no effect on the subsequent questions or answers, so we wonder why we're doing this.) The result is ...
(x -h)² +4pk = 0
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b) The value of k is the vertical location of the vertex of the parabola with respect to the x-axis. The parabola opens upward, so for k > 0, the parabola does not cross the x-axis, and the number of real zeros is zero. (There are two complex zeros.)
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c) As in part b, the value of k defines the vertex location. When it is zero, the vertex of the parabola is on the x-axis, so there is one real zero (It is considered to have multiplicity 2.)
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d) As in part b, the value of k defines the vertex location. When it is negative, the vertex of the parabola is below the x-axis. Since the parabola opens upward, both branches will cross the x-axis, resulting in two real zeros.
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The attached graph shows a parabola with p=1/4 and h=2. The values shown for k are +1, 0, and -1. The coordinates of the real zeros are shown.
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Answer:
U₂₁ and U₂₂ fits in the line of upper triangular
Step-by-step explanation:
As explain in the question
An upper triangular matric is square matrix with all zeros below its daignal. A lower triangular matrix is a square matrix with all zeros above its diagonal.
(as shown in the attached image)
(a)for all 3×3 upper triangular matirices U, verify that U2 is also upper triangular
for a matrices U (having a 3*3 matrix), assuming for example, U is given below:
U =
the element in the upper triangular are 1,2,3,5,6,9 (U₁₁, U₁₂, U₁₃; , U₂₂, U₂₃, U₂₂)
while lower triangular are 0,0,0, the element in the lower triangular are (U₂₁, U₃₁,U₃₂)
U₂₁ and U₂₂ fits in the line of upper triangular
