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Mars2501 [29]
3 years ago
13

Y= − 2 ( x − 2 ) ( x + 8 ) −2(x−2)(x+8) in standard form

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
7 0

Answer:

y=-4x^{2} -24x+64

Step-by-step explanation:

y=-2(x-2)(x+8)-2(x-2)(x+8)

y=-2x^{2} +12x+32-2(x-2)(x+8)

y=-2x^{2} -12x+32-2x^{2} -12x+32

y=-4x^{2} -24x+64

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What expression is not equivalent to 2/3 *4.
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Answer:

Step-by-step explanation:

Any expression that says (2 * 4)/3 is correct, even though there is a shift in the brackets. That means that A, B, C are all the same thing, and all are correct.

Only D is lying.

2*(1/3) = 2/3

4*(1/3) = 4/3     Now we have to add them

2/3 + 4/3 = 6/3 = 2

To show what the real answer is, try A

2*4/3

8/3  is the actual answer.

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2 years ago
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2 years ago
What are the fourth roots of 6+6√(3i) ?
Helen [10]

Answer:

Step-by-step explanation:

The genral form of a complex number in rectangular plane is expressed as z = x+iy

In polar coordinate, z =rcos ∅+irsin∅ where;

r is the modulus = √x²+y²

∅ is teh argument = arctan y/x

Given thr complex number z = 6+6√(3)i

r = √6²+(6√3)²

r = √36+108

r = √144

r = 12

∅ = arctan 6√3/6

∅ = arctan √3

∅ = 60°

In polar form, z = 12(cos60°+isin60°)

z = 12(cosπ/3+isinπ/3)

To get the fourth root of the equation, we will use the de moivres theorem; zⁿ = rⁿ(cosn∅+isinn∅)

z^1/4  = 12^1/4(cosπ/12+isinπ/12)

When n = 1;

z1 =  12^1/4(cosπ/3+isinn/3)

z1 = 12^1/4cis(π/3)

when n = 2;

z2 = 12^1/4(cos2π/3+isin2π/3)

z2 = 12^1/4cis(2π/3)

when n = 3;

z2 = 12^1/4(cosπ+isinπ)

z2 = 12^1/4cis(π)

when n = 4;

z2 = 12^1/4(cos4π/3+isin4π/3)

z2 = 12^1/4cis(4π/3)

8 0
3 years ago
Is (2,4) a solution of 3x - y = 2?
guajiro [1.7K]

yes because when filling in the unknown (3*2-4=2) gets you the answer of 2.

7 0
3 years ago
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Answer:

The answer is C. 0

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2 years ago
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