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gregori [183]
3 years ago
11

Suppose A is a matrix and k is a positive integer. What does the notation Ak mean?

Mathematics
1 answer:
Butoxors [25]3 years ago
7 0

Answer:

Scalar product

Step-by-step explanation:

The scalar product of a matrix refers to the product of a matrix and a number. The scalar is used to multiply each element in the matrix to get the final vector. In the question the scalar is k which is a constant and the matrix is A.

in the example below 2 is the scalar

2*\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]

the final matrix will be

\left[\begin{array}{ccc}2&4&6\\8&10&12\\14&16&18\end{array}\right]

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If the conditional statement, “If today is Friday, then yesterday was Thursday” is true, which other statement must also be true
myrzilka [38]

Answer:

If today is not Thursday, then tomorrow is not Friday.

Step-by-step explanation:

I presume this is the answer off other peoples responses on the same question elsewhere.... next time include the whole question.

4 0
3 years ago
Juan rents a car for $29.00 a week. He pays $17.00 per week for the insurance and spends $13.50 weekly on gas. What will his cos
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Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
3 years ago
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Answer: 6/10

Step-by-step explanation:

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