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Mekhanik [1.2K]
3 years ago
14

Find the roots of 2x^2 - 2x - 3= 0

Mathematics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

the roots are 0.823 and 1.823

Step-by-step explanation:

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PLEASE HELP ASAP What is the m and b for this graph??
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M&lt;2=4x+12 <br> Someone help
Katarina [22]
X=12. So this one says that it has 3 congruent sides so therefore it has to have 3 congruent angles. Here we have angle 2 and we have angle 60. Given the definition of triangle congruence, we can say that 60 is angle 2 and that the third angle is also 60 but it’s not relevant in this problem. So knowing that angle 2 is 4x+12 we solve for x by seeding that equal to 60 and then solving for x we get x=12. Hope this helps!!!!

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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.

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3 years ago
-3(x + 2)2 + 4 = -8<br> what’s the answer
podryga [215]
X=0

-3(0+2)2+4 distribute -3 to the parentheses-PEMDAS
(0-6)2+4 multiply 2 by the parentheses
(0-12)+4 solve the equation
= -8
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3 years ago
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