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AfilCa [17]
3 years ago
5

Does changing the compound inequality x>-3 and x<3 from and to or change the solution set? Explain

Mathematics
2 answers:
Serggg [28]3 years ago
8 0

Answer:

So the answer is

Step-by-step explanation:

x)2943

aniked [119]3 years ago
3 0

Answer:The x values of the “and” inequality satisfy both inequalities.

The x values of the “or” inequality satisfy one or both inequalities.

Changing the compound inequality from “and” to “or” changes the solution set from the values between –3 and 3 for all real numbers.

Step-by-step explanation:

that's all u need

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When the steps are followed. They both pay the same amount

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Step-by-step explanation:

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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
Solve for x:<br> 3500 = 700.28x
jeka57 [31]
This isn’t an answer but you should use the app PhotoMath
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