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goblinko [34]
3 years ago
8

Verify the Cauchy-Schwarz Inequality and the triangle inequality for the given vectors and inner product.

Mathematics
1 answer:
lions [1.4K]3 years ago
5 0

Answer:

To verify the Cauchy-Bunyakovsky-Schwarz Inequality, (p,q) must be less than (or equal to) ||p|| • ||q||

(1,1,1) is not equal to (-10,5)

Step-by-step explanation:

a°b° + a^1b^1 + a^2b^2 < 5x (-2x^2 + 1)

Any algebra raised to the power of zero is equal to 1.

a°b° = 1 × 1 = 1

1 + ab + a^2b^2 < -10x^3 + 5x

The vectors:

(1,1,1) < (-10,5)

This verifies the Cauchy-Schwarz Inequality

Triangle Inequality states that for any triangle, the sum of the lengths of two sides must be greater than or equal to the length of the third side.

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25 Points! Please answer asap! Carly stated “All pairs of rectangles are dilations”. Which pair of rectangles would prove that C
elena55 [62]

Answer:

C

Step-by-step explanation:

A. First two rectangles are dilations because

\dfrac{2}{4}=\dfrac{4}{8}=0.5

B. Second two rectangles are dilations because

\dfrac{2}{4}=\dfrac{3}{6}=0.5

C. Third two rectangles are not dilations because

\dfrac{3}{4}\neq \dfrac{2}{3}

D. Fourth two rectangles are dilations because

\dfrac{3}{4}=\dfrac{1.5}{2}=0.75

8 0
3 years ago
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Can sum1 helpp meee pweasee
wel

Answer:

105

Step-by-step explanation:

5 0
3 years ago
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M∠LON=77 ∘ m, angle, L, O, N, equals, 77, degrees \qquad m \angle LOM = 9x + 44^\circm∠LOM=9x+44 ∘ m, angle, L, O, M, equals, 9,
timama [110]

Answer:

Step-by-step explanation:

Given

<LON = 77°

<LOM = (9x+44)°

<MON = (6x+3)°

The addition postulate is true for the given angles since tey have a common point O:

<LON = <LOM+<MON

Since we are not told what to find we can as well look for the value of x, <LOM and <MON

Substitute the given parameters and get x

77 = 9x+44+6x+3

77 = 15x+47

77-47 = 15x

30 = 15x

x = 30/15

x = 2

Get <LOM:

<LOM = 9x+44

<LOM = 9(2)+44

<LOM = 18+44

<LOM = 62°

Get <MON:

<MON = 6x+3

<MON = 6(2)+3

<MON = 12+3

<MON = 15°

6 0
4 years ago
Find two consecutive positive intergers such that the square of the smaller interger is nineteen more than five times the larger
Naddika [18.5K]

Answer:

8 and 9

Step-by-step explanation:

Let the two consecutive integers be x - 1 and x

Smaller = x-1

Largere = x

If the square of the smaller integer is nineteen more than five times the larger integer, then;

(x-1)² = 19 + 5x

Expand

x²-2x+1= 19+5x

x²-2x-5x+1-19 -0

x²-7x-18 =0

Factorize

x²-9x+2x-18 =0

x(x-9)+2(x-9) = 0

(x+2) = 0 and x - 9 = 0

x = -2 and 9

SInce x cannot be negative;

x = 9

Smaller number = 9-1

Smaller number = 8

Hence the two consecutive integers are 8 and 9

6 0
3 years ago
Commutative property for 70*6000
Crazy boy [7]
(70*6000 = 420000) (6000*70 = 42000)

Bolded is the answer. 
5 0
4 years ago
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