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Serggg [28]
3 years ago
10

How do i do this ???

Mathematics
1 answer:
Westkost [7]3 years ago
4 0

a peculiar thing about the centroid of a triangle, is that it cuts all the medians in a 2:1 ratio, so JP : PM are on a 2:1 ratio, thus

\bf JP:PM\qquad 2:1\qquad \cfrac{JP}{PM}=\cfrac{2}{1}\implies \cfrac{27}{PM}=\cfrac{2}{1}\implies \cfrac{27}{2}=PM\\\\-------------------------------\\\\JM=JP+PM\implies JM=27+\cfrac{27}{2}\implies JM=\cfrac{54+27}{2}\\\\\\JM=\cfrac{81}{2}\implies JM=40\frac{1}{2}

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   ( x, y )  → ( x-3, y+4 )
A ( 1, 1) → A' (1-3, 1+4)      A' (-2, 5)
8 0
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an earthquake measures 3.6 on the richter scale. what is the amount of energy (ergs) released by the earthquake? give an exact a
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We have that 

<span>log10 E = 4.4 + 1.5<span>M
</span></span>E=<span>the released energy in Joules (J)
M= </span><span>earthquake measures---------- > 3.6
</span>log10 E = 4.4 + 1.5*3.6
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the answer is  </span>6.309573444 x10^9 J<span>

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5 0
3 years ago
Determine whether the set of vectors is a basis for ℛ3. Given the set of vectors , decide which of the following statements is t
schepotkina [342]

Answer:

(A) Set A is linearly independent and spans R^3. Set is a basis for R^3.

Step-by-Step Explanation

<u>Definition (Linear Independence)</u>

A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.

<u>Definition (Span of a Set of Vectors)</u>

The Span of a set of vectors is the set of all linear combinations of the vectors.

<u>Definition (A Basis of a Subspace).</u>

A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.

Given the set of vectors  A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right) , we are to decide which of the given statements is true:

In Matrix A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right) , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column. R^3 has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans R^3.

Therefore Set A is linearly independent and spans R^3. Thus it is basis for R^3.

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Yuliya22 [10]

Answer:

1/6

Step-by-step explanation:

favorable over total outcomes

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(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

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(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

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