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Licemer1 [7]
4 years ago
11

IM DOIGN A MATH TEST AND I NEED HELP RN

Mathematics
1 answer:
Vladimir79 [104]4 years ago
6 0

Answer:

its 43

Step-by-step explanation:

i had a test with the same question, i promise its right :)

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Domain of f(x)= -x^2 -4 +2
garri49 [273]

Answer:

Domain : all real numbers

Step-by-step explanation:

The domain is the input values

What values can x be?

X can be any real number

Domain : all real numbers

6 0
3 years ago
Which property do i use to solve x-10=15
Stells [14]
Add both sides by 10

x=25
3 0
4 years ago
Explain how the sum and difference formula cos(A+/- B) = cosA cosB -/+ sinA sinB could be useful?
serg [7]
One way it's useful is in the application of phasors in physics. Phasors require you to add up vectors of two different angles, so the cosine angle addition formula can be used if you already know the cosine and sine of the original vector angles.
6 0
3 years ago
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Use the intersect method to solve the equation. 14x^3-53x^2+41x-4=-4x^3-x^2+1x+4
UNO [17]

Answer:

x = (68 2^(1/3) + (27 i sqrt(591) + 445)^(2/3))/(27 (1/2 (27 i sqrt(591) + 445))^(1/3)) + 26/27 or x = (68 (-2)^(2/3) - (-2)^(1/3) (27 i sqrt(591) + 445)^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) + 26/27 or x = 1/27 ((-2)/(27 i sqrt(591) + 445))^(1/3) ((-1)^(1/3) (27 i sqrt(591) + 445)^(2/3) - 68 2^(1/3)) + 26/27

Step-by-step explanation:

Solve for x over the real numbers:

14 x^3 - 53 x^2 + 41 x - 4 = -4 x^3 - x^2 + x + 4

Subtract -4 x^3 - x^2 + x + 4 from both sides:

18 x^3 - 52 x^2 + 40 x - 8 = 0

Factor constant terms from the left hand side:

2 (9 x^3 - 26 x^2 + 20 x - 4) = 0

Divide both sides by 2:

9 x^3 - 26 x^2 + 20 x - 4 = 0

Eliminate the quadratic term by substituting y = x - 26/27:

-4 + 20 (y + 26/27) - 26 (y + 26/27)^2 + 9 (y + 26/27)^3 = 0

Expand out terms of the left hand side:

9 y^3 - (136 y)/27 - 1780/2187 = 0

Divide both sides by 9:

y^3 - (136 y)/243 - 1780/19683 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

-1780/19683 - 136/243 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

z^6 + z^4 (3 λ - 136/243) - (1780 z^3)/19683 + z^2 (3 λ^2 - (136 λ)/243) + λ^3 = 0

Substitute λ = 136/729 and then u = z^3, yielding a quadratic equation in the variable u:

u^2 - (1780 u)/19683 + 2515456/387420489 = 0

Find the positive solution to the quadratic equation:

u = (2 (445 + 27 i sqrt(591)))/19683

Substitute back for u = z^3:

z^3 = (2 (445 + 27 i sqrt(591)))/19683

Taking cube roots gives 1/27 2^(1/3) (445 + 27 i sqrt(591))^(1/3) times the third roots of unity:

z = 1/27 2^(1/3) (445 + 27 i sqrt(591))^(1/3) or z = -1/27 (-2)^(1/3) (445 + 27 i sqrt(591))^(1/3) or z = 1/27 (-1)^(2/3) 2^(1/3) (445 + 27 i sqrt(591))^(1/3)

Substitute each value of z into y = z + 136/(729 z):

y = (68 2^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) + 1/27 (2 (27 i sqrt(591) + 445))^(1/3) or y = (68 (-2)^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) - 1/27 (-2)^(1/3) (27 i sqrt(591) + 445)^(1/3) or y = 1/27 (-1)^(2/3) (2 (27 i sqrt(591) + 445))^(1/3) - (68 (-1)^(1/3) 2^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3))

Bring each solution to a common denominator and simplify:

y = (2^(1/3) ((27 i sqrt(591) + 445)^(2/3) + 68 2^(1/3)))/(27 (445 + 27 i sqrt(591))^(1/3)) or y = (68 (-2)^(2/3) - (-2)^(1/3) (27 i sqrt(591) + 445)^(2/3))/(27 (445 + 27 i sqrt(591))^(1/3)) or y = 1/27 2^(1/3) (-1/(445 + 27 i sqrt(591)))^(1/3) ((-1)^(1/3) (27 i sqrt(591) + 445)^(2/3) - 68 2^(1/3))

Substitute back for x = y + 26/27:

Answer:  x = (68 2^(1/3) + (27 i sqrt(591) + 445)^(2/3))/(27 (1/2 (27 i sqrt(591) + 445))^(1/3)) + 26/27 or x = (68 (-2)^(2/3) - (-2)^(1/3) (27 i sqrt(591) + 445)^(2/3))/(27 (27 i sqrt(591) + 445)^(1/3)) + 26/27 or x = 1/27 ((-2)/(27 i sqrt(591) + 445))^(1/3) ((-1)^(1/3) (27 i sqrt(591) + 445)^(2/3) - 68 2^(1/3)) + 26/27

5 0
3 years ago
Please help with this
Zolol [24]
The value of X is B. 13
8 0
4 years ago
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