Answer:
58.1 degrees
Step-by-step explanation:
Given the following
JK = 9.4miles (towards south) negative y axis
If the move 15.1 miles towards east (that will be towards the positive x axis)
Using the SOH CAH TOA identity
opposite= 15.1 miles(side facing m<J)
adjacent= JK = 9.4miles
tan theta = opposite/adjacent
tan m<J = 15.1/9.4
tan m<J = 1.6063
m<J = arctan (1.6063)
m<J = 58.09 degrees
Hence the measure of m<J to the nearest tenth is 58.1 degrees
Answer:
31-3n
Step-by-step explanation:
d=-3
a1=28
an= 28+(n-1)(-3)=28-3n+3
an=31-3n
Answer:
Total number of cards in a standard deck of card=52
As deck of card contain 13 club +13 heart +13 spade+13 Diamond.
P(Diamond)=13/52 =1/4
If the card is not diamond then
P(not diamond)=39/52=3/4
Since after drawing card is not replaced
P ( second Diamond)=12/51
now if second card is not Diamond.then
P(non diamond)=39/51
Chances of your winning i.e both cards are diamonds=1/4×12/51=1/17=0.0588..(approx)
chances of your friend winning=1/4×39/51=[1/4]×[13/17]=13/68=0.1911..(approx)
Your friend has got better chance to win because there are more chances that both the cards are not Diamonds.
Answer:
Step-by-step explanation:
b.
an=an-1+3
C.
when x=15
y=3×15-2=45-2=43
a15=43
Answer:
(A) Set A is linearly independent and spans 
. Set is a basis for .
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)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D1%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%201%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%201%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20)
, 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)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D%281%29%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%20%281%29%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%20%281%29%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20)
, 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. has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans .
Therefore Set A is linearly independent and spans . Thus it is basis for .
