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3 years ago

which set of number could represent the length of the sides of a right triangle? A 9 11 14 B 15 18 21 C 7 24 25 D 8 9 10

2 answers:

8 0

Common Pythagorean triples include
(3, 4, 5)
(5, 12, 13)
(7, 24, 25)
(9, 40, 41)
The only Pythagorean triple that is an arithmetic sequence is (3, 4, 5), so any arithmetic sequence that is a Pythagorean triple must be a multiple of that, such as (9, 12, 15) or (15, 20, 25).

The arithmetic sequences of selections B and D are unrelated to the (3, 4, 5) triple, so cannot be Pythagorean triples. For selection A, we know that 9² + 11² = 81 + 121 = 202 > 14², so that is not a right triangle.

The appropriate selection is ...
C. 7, 24, 25

Maslowich3 years ago

3 0

The correct answer is c 7 24 25
hope you did well

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zhenek [66]

Answer:

R(x) = -0.05x^2 +80x

Step-by-step explanation:

Given the two points (x, p), the equation for the price associated with a given demand quantity can be written using the 2-point form:

p = (p2 -p1)/(x2 -x1)(x -x1) +p1

p = (40 -50)/(800 -600)(x -600) +50

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3 years ago

Let Upper A equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 4 2nd Row 1st Column 1 2nd Column 3

Anastaziya [24]

Answer:

See explanation

Step-by-step explanation:

Given:

$A=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right]$

$B=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right]$

A. Find AB:

$AB=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right]\cdot \left[\begin{array}{cc}-2&1\\3&7\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+4\cdot 3&-2\cdot 1+4\cdot 7\\1\cdot (-2)+3\cdot 3&1\cdot 1+3\cdot 7\end{array}\right]=\left[\begin{array}{cc}16&26\\7&22\end{array}\right]$

B. Find BA:

$BA=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right]\cdot \left[\begin{array}{cc}-2&4\\1&3\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+1\cdot 1&-2\cdot 4+1\cdot 3\\3\cdot (-2)+7\cdot 1&3\cdot 4+7\cdot 3\end{array}\right]=\left[\begin{array}{cc}5&-5\\1&33\end{array}\right]$

C. Answers are not the same

D. Matrices multiplication is not commutastive in general, so

$AB\neq BA$

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3 years ago

The number of 5-digit numbers that can be formed if the last digit is odd and no

Softa [21]

Answer:

13,440

Step-by-step explanation:

10 possible digits 0 - 9

5 possible numbers for position 5 (odd digits only)

8 possible numbers for position 1 (cannot be zero or the odd number selected for position 5)

8 possible numbers remain for position 2 (10 minus 2 already selected)

7 possible numbers remain for position 3 (10 minus 3 already selected)

6 possible numbers remain for position 4 (10 minus 4 already selected)

5•8•8•7•6 = 13,440

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3 years ago

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bogdanovich [222]

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3 years ago

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3 years ago