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statuscvo [17]
3 years ago
12

Write the following in exponential form. 2��2��3��3��3��5

Mathematics
1 answer:
vampirchik [111]3 years ago
3 0
Good luck I have no freaking
 clue

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Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
Write the first five terms of the arithmetic sequence. a4 = 22, a10 = 64
dedylja [7]
Hello :
the general term is : 
an = a1+(n-1)×d  
or : an = ap +(n-p)×d......d is common diffrence
let : n=10 and p= 4
a10 =  a4 +(10-4)×d
64 = 22 + 6d
6d =  42
d= 7
conclusion :
a4 = 22
a3 = 22-7=15
a2=15-7=8
a1= 8-7=1
<span>the first five terms : 1 , 8 , 15 , 22</span>

4 0
3 years ago
An equivalent expression to a2/3
scoundrel [369]

Answer:

4/6  6/9  8/12  10/15 and so on, just multiply both the numerator and denominator by the same number and get your answer.

Step-by-step explanation:

6 0
3 years ago
Anybody need help with some work?
otez555 [7]

Answer:

nope. im done with school :D

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
A packet of vegetable seeds has a germination rate of
Doss [256]

N = number of seeds planted

P = probability of germination

K = number that germinated

Using probability calculation:
p(x = k) = (n k) x 0.90^K x 0.10^n-k


p(10) = (12 10) x 0.90^10 x 0.10^2

P(10) = 0.2310

8 0
2 years ago
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