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arsen [322]
2 years ago
14

Which sequence explains a geometric method of determining 3 - 17i/2 +2i?

Mathematics
2 answers:
Katen [24]2 years ago
8 0

Answer:

D) Plot 3 – 17i. Scale by 2sqrt(2). Rotate 45° clockwise.

Step-by-step explanation:

never [62]2 years ago
7 0

Answer:

That would be D.

Step-by-step explanation:

first, turn both to polar form. Then you plot the first one and scale by the second. also, using a Rectangular to Polar calculator, we can see that 2.82 is 2 sqrt 2, so it cannot be A or B. It also can't be C since you graph 3-17i first.

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prohojiy [21]

Answer:

x=−5 and y=1

Step-by-step explanation:

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3 years ago
√6·√5·√2 what is the answer for this. Im very confused
miskamm [114]

Answer:

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2 years ago
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Harlamova29_29 [7]

the second one 6,6,6 it is a square all the others seem like a triangle

8 0
2 years ago
What is the sum? (1/x+2)+(1/x+3)+(1/X^2+5+6)
PolarNik [594]
For this case we have the following expression:
 (1 / x + 2) + (1 / x + 3) + (1 / X ^ 2 + 5 + 6)
 Rewriting we have:
 (1 / x + 2) + (1 / x + 3) + (1 / ((x + 2) * (x + 3)))
 By doing common factor we have:
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 Rewriting:
 (1 / ((x + 2) * (x + 3))) * (2x + 6)
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 Answer:
 
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7 0
2 years ago
Find all real zeros of 4x^3-20x+16
Pani-rosa [81]

Answer:

  {1, (-1±√17)/2}

Step-by-step explanation:

There are formulas for the real and/or complex roots of a cubic, but they are so complicated that they are rarely used. Instead, various other strategies are employed. My favorite is the simplest--let a graphing calculator show you the zeros.

___

Descartes observed that the sign changes in the coefficients can tell you the number of real roots. This expression has two sign changes (+-+), so has 0 or 2 positive real roots. If the odd-degree terms have their signs changed, there is only one sign change (-++), so one negative real root.

It can also be informative to add the coefficients in both cases--as is, and with the odd-degree term signs changed. Here, the sum is zero in the first case, so we know immediately that x=1 is a zero of the expression. That is sufficient to help us reduce the problem to finding the zeros of the remaining quadratic factor.

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Using synthetic division (or polynomial long division) to factor out x-1 (after removing the common factor of 4), we find the remaining quadratic factor to be x²+x-4.

The zeros of this quadratic factor can be found using the quadratic formula:

  a=1, b=1, c=-4

  x = (-b±√(b²-4ac))/(2a) = (-1±√1+16)/2

  x = (-1 ±√17)2

The zeros are 1 and (-1±√17)/2.

_____

The graph shows the zeros of the expression. It also shows the quadratic after dividing out the factor (x-1). The vertex of that quadratic can be used to find the remaining solutions exactly: -0.5 ± √4.25.

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The given expression factors as ...

  4(x -1)(x² +x -4)

5 0
2 years ago
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