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spayn [35]
4 years ago
15

How do you write 17 25 as a percentage?

Mathematics
2 answers:
joja [24]4 years ago
6 0
To write 17/25 as a percentage, you first multiply 17/25 by 100 which is 1700/25. then you divide it and the percentage would be 68%.
Sonja [21]4 years ago
5 0

is that supposed to be written as a fraction and you are looking for a percentage?


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-0.5,1.25,-1/3,0.5,-5/3 ordered from least to greatest
vampirchik [111]

Answer:

Step-by-step explanation:

-5/3,-0.5,-1/3,0.5,1.25

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3 years ago
What is 1080 miles on 15 gallons
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72 miles per gallons

Step-by-step explanation:

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Given vectors A (2,-1,5), B (4,3,-2) & C (5,4,0), find:?
MrRissso [65]
A) 4a - 3b + 2c

4(2, -1, 5) - 3(4, 3 , -2) + 2(5, 4, 0) = (8, -4, 20) - (12, 9, - 6) + (10, 8, 0) =

= (8 - 12 + 10 , -4 - 9 + 8 , 20 + 6 + 0) = (6, - 5, 26)

Answer: (6, - 5, 26)

b) magnitude of vector b

\sqrt{4^2+3^2+(-2)^2} = \sqrt{16+9+4} = \sqrt{29} ~ 5.4

c) vector of length 7 parallel to vector c

=> m(5,4,0) = (5m,4m,0)

=>    \sqrt{(5m)^2+(4m)^2+0}= \sqrt{25m^2+16m^2}= \sqrt{41m^2}=m \sqrt{41}=7

=> m = 7 / √41 ≈ 1.093

=> 1.093 (5, 4, 0) = (5.465 , 4.372, 0)

Answer: (5.465 , 4.372 , 0)
7 0
3 years ago
Arrange the entries of matrix A in increasing order of their cofactors values
givi [52]

To find the cofactor of

A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]

We cross out the Row and columns of the respective entries and find the determinant of the remaining 2\times 2 matrix with the alternating signs.


Ac_{11}=\left|\begin{array}{ccc}4&-1\\2&1\end{array}\right|


Ac_{11}=4\times 1- -1\times 2


Ac_{11}=4+ 2

Ac_{11}=6




Ac_{12}=-\left|\begin{array}{ccc}-7&-1\\-8&1\end{array}\right|


Ac_{12}=-(-7\times 1- -1\times -8)


Ac_{12}=-(-7- 8)

Ac_{12}=15




Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|


Ac_{21}=-(5\times 1- 3\times 2)


Ac_{21}=-(5-6)


Ac_{21}=1







A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|


Ac_{23}=-(7\times 2 -8\times 5)


Ac_{23}=-(14-40)


Ac_{23}=26




A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|


Ac_{31}=5\times -1 -4\times 3


Ac_{31}=-5-12


Ac_{31}=-17


A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|


Ac_{33}=7\times 4- -7\times 5


Ac_{33}=28+35


Ac_{33}=63


Therefore in increasing order, we have;

Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63



7 0
3 years ago
Read 2 more answers
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zheka24 [161]

Answer:

the chances are 5/11 that Jones will get an orange and 5/11 that Beth will get an orange also

Step-by-step explanation:


4 0
3 years ago
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