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andreyandreev [35.5K]
3 years ago
12

What is the answer to the problem

rmula1" title="(x {3})^{3} = x^n" alt="(x {3})^{3} = x^n" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0
Solve for x:
x^9 = n x

Subtract n x from both sides:
x^9 - n x = 0

Factor x and constant terms from the left hand side:
-x (n - x^8) = 0

Multiply both sides by -1:
x (n - x^8) = 0

Split into two equations:
x = 0 or n - x^8 = 0

Subtract n from both sides:
x = 0 or -x^8 = -n

Multiply both sides by -1:
x = 0 or x^8 = n

Taking 8^th roots gives n^(1/8) times the 8^th roots of unity:
Answer: x = 0 or x = -n^(1/8) or x = -i n^(1/8) or x = i n^(1/8) or x = n^(1/8) or x = -(-1)^(1/4) n^(1/8) or x = (-1)^(1/4) n^(1/8) or x = -(-1)^(3/4) n^(1/8) or x = (-1)^(3/4) n^(1/8)
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Subtract (3x^3+4x^2)-(2x^3+3x^2-x)
insens350 [35]

Answer:

X^3 + x^2 + x

Step-by-step explanation:

First, distribute the negative sign to the numbers in the second parentheses. Then you’ll need to combine like terms.

4 0
3 years ago
****BRAINLIEST GOES TO THE FIRST CORRECT ANSWER****
erik [133]
Given
Segment Addition Postulate
Transitive Property of Equality
4 0
3 years ago
Which polynomial can be simplified to a difference of squares
Mrrafil [7]
<h2>Hello!</h2>

The answer is:

The polynomial that can be simplified to a difference of squares is the second polynomial:

16a^{2}-4a+4a-1=16a^{2}=(4a)^{2}-(1)^{2}=(4-1)(4+1)

<h2>Why?</h2>

To solve this problem, we need to look for which of the given quadratic terms given for the different polynomials can be a result of squaring (elevating by two).

So,

Discarding, we have:

The quadratic terms of the given polynomials are:

First=10a^{2}

Second=16a^{2}

Third=25a^{2}

Fourth=24a^{2}

We have that the coefficients of the quadratic terms that can be obtained by squaring are:

16a^{2} =(4a)^{2} \\\\25a^{2} =(5a)^{2}

The other two coefficients are not perfect squares since they can not be obtained by square rooting whole numbers.

So, the first and the fourth polynomial are discarded and cannot be simplified to a difference of squares at least using whole numbers.

Therefore, we need to work with the second and the third polynomial.

For the second polynomial, we have:

16a^{2} -4a+4a-1=16a^{2}=(4a)^{2}-(1)^{2} =(4-1)(4+1)

So, the second polynomial can be simplified to a difference of squares.

For the third polynomial, we have:

25a^{2} +6a-6a+36=16a^{2}+36=(5a)^{2}+(6)^{2}

So, the third polynomial cannot be simplified to a difference of squares since it's a sum of squares.

Hence, the polynomial that can be simplified to a difference of squares is the second polynomial:

16a^{2}-4a+4a-1=16a^{2}=(4a)^{2}-(1)^{2}

7 0
3 years ago
Read 2 more answers
The following function is probability Distribution function.
Rina8888 [55]

Answer:

9.60 ; - 60.96

Step-by-step explanation:

Given the function :

F(x)=6(x+1) /25, x=0, 1, 2, 3, 4.

x = 0

F(0)=6(0+1)/25 = 6/25 = 0.24

x = 1

F(1)=6(1+1)/25 = 12/25 = 0.48

x = 2

F(2)=6(2+1)/25 = 18/25 = 0.72

x = 3

F(2)=6(3+1)/25 = 24/25 = 0.96

x = 4

F(2)=6(4+1)/25 = 30/25 = 1.2

X ______0 _____ 1 ______ 2 ______ 3 ____ 4

P(x) ___ 0.24 __ 0.48 ___ 0.72 ____ 0.96 __ 1.2

Mean, μ = Σx*p(x) :

(0*0.24) + (1*0.48) + (2*0.72) + (3*0.96) + (4*1.2)

= 9.60

Variance : Σx²*p(x) - μ²

(0^2*0.24) + (1^2*0.48) + (2^2*0.72) + (3^2*0.96) + (4^2*1.2) - 9.6^2

= 31.2 - 92.16

= - 60.96

4 0
2 years ago
Is the following relation a function? (4,2), (1,1), (0,0), (1, -1), (4, -2). Give the domain and range.
arsen [322]
No it is not a function. tell me if you need explanation
4 0
3 years ago
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