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Ksju [112]
3 years ago
14

Algebraically, find all remaining zeros of f(x) = 8x^3 - 6x^2-36x +27 in simplest form.

Mathematics
1 answer:
viva [34]3 years ago
7 0

Answer:

x=-\frac{3\sqrt{2}}{2},\frac{3}{4},\frac{3\sqrt{2} }{2}

Step-by-step explanation:

Given expression is,

f(x) = 8x³ - 6x² - 36x + 27

For zeros of the function,

f(x) = 0

8x³ - 6x² - 36x + 27 = 0

2x²(4x - 3) - 9(4x - 3) = 0

(2x² - 9)(4x - 3) = 0

(2x² - 9) = 0

x² = \frac{9}{2}

x = \pm\frac{3}{\sqrt{2}}

x = \pm \frac{3\sqrt{2}}{2}

Or (4x - 3) = 0

x = \frac{3}{4}

Therefore, all zeros of the given functions are  x=-\frac{3\sqrt{2}}{2},\frac{3}{4},\frac{3\sqrt{2} }{2}.

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Ronch [10]

Answer:

x = -1/5

Step-by-step explanation:

2+3 (x+2x) = -2x(2-4) +1

Combine like terms

2+3(3x) = -2x (-2) +1

Distribute

2+9x = 4x +1

Subtract 4x from each side

2+9x-4x = 4x+1-4x

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4 years ago
{20,1,6,10,11}<br> Find the subset and the proper subset.
liubo4ka [24]
With 5 elements in A={20,1,6,10,11}, there are 2^5=32 possible subsets, including
the null set, and A itself.
Any subset that is identical to A is NOT a proper subset.
Therefore there are 31 proper subsets, plus the subset {20,1,6,10,11}.

The subsets are:
null set  {} (has no elements) ........total 1
{20},{1},{6},{10},{11}.......................total 5
{20,1},{20,6}...{10,11}.....................total 10
{20,1,6},{20,1,10},...{6,10,11}.........total 10
{20,1,6,10}...{1,6,10,11}.................total 5
{20,1,6,10,11}.................................total 1
Altogether 32 subsets.


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Answer:

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Step-by-step explanation:

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