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Tamiku [17]
3 years ago
5

6∛I= 9∴ = 2² - y3= Answer fast pls

Mathematics
1 answer:
3241004551 [841]3 years ago
6 0

Answer:

the last one is 4−y^3

the others i dont know sorry :(

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2x=5 what do we divide by?
larisa [96]
We divide by 2 on both sides.

We get the final answer as

x = 5/2 = 2.5
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3 years ago
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What is the LCD of 2/5 and 4/11
scoundrel [369]
To find the lowest common denominator you need to know the factors of 5 and 11.
So make two lists:
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11, 22, 33, 44, 55, 66, and so on.
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2 years ago
(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
An equation is given.<br><br> Solve the equation for r.
goblinko [34]
There is not an equation
3 0
3 years ago
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