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andriy [413]
3 years ago
5

Explain how to subtract 247 from 538

Mathematics
2 answers:
Oliga [24]3 years ago
6 0
538
-247
-------
291

8-7=1
you cant do 3-4 so you borrow 10 numbers from 5 and got 13-4=9
4-2=2
Licemer1 [7]3 years ago
4 0
For this case we have the following operation:
 538 - 247
 To subtract both numbers we must follow the following steps:
 1) Subtraction is done from left to right:
 2) We subtract the first digits:
 8 - 7 = 1
 3) We subtract the second digits:
 3 - 4
 Since 4 is greater than 3 then we add a digit to 3 and accumulate it for the next subtraction:
 13 - 4 = 9
 4) We subtract the digit accumulated in the previous subtraction to the following number:
 5 - 2 - 1 = 2
 Finally the result is:
 538 - 247 = 291
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(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

4 0
3 years ago
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