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NikAS [45]
2 years ago
9

What is one root of this equation? 2x^-4x+9=0

Mathematics
1 answer:
Nutka1998 [239]2 years ago
3 0

9514 1404 393

Answer:

  1 +i√3.5

Step-by-step explanation:

In vertex form, the equation is ...

  2(x² -2x +1) +7 = 0

  2(x -1)² +7 = 0

Then the solutions are ...

  (x -1)² = -7/2

  x = 1 ±i√3.5

One solution is 1+i√3.5.

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There is one integer for which there does not exist another integer with the same absolute value, What is that integer?
kotykmax [81]
Zero, it is not negative or positive.
7 0
3 years ago
Read 2 more answers
What is the solution of this equation? -1/2n2+18=0
SashulF [63]

Answer:

n=-6\text{ or } n=6

Step-by-step explanation:

So we have the equation:

-\frac{1}{2}n^2+18=0

Subtract 18 from both sides. The left cancels:

-\frac{1}{2}n^2=-18

Multiply both sides by -2. The left cancels again:

n^2=36

Take the square root of both sides:

n=\pm\sqrt{36}

Since we're taking the square root, we need to use plus-minus.

Evaluate:

n=\pm 6

So, the solutions are -6 or 6.

5 0
3 years ago
Solve the equation and show each step:<br> 25 - 4x = 15 - 3x + 10 - X
dimulka [17.4K]

Answer:

All real numbers.

Step-by-step explanation:

25-4x=15-3x+10-x

25-4x-(-3x)-(-x)=15+10

25-4x+3x+x=25

25-x+x=25

25=25

3 0
3 years ago
Simplify (xy)^-3/(x^-5y)^3
Nesterboy [21]
Use:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\a^n:a^m=a^{n-m}\\\\\left(a^n\right)^m=a^{n\cdot m}\\\\a^{-n}=\left(\frac{1}{a}\right)^{n}\\==================================\\\frac{(xy)^{-3}}{(x^{-5}y)^3}=\frac{x^{-3}y^{-3}}{(x^{-5})^3y^3}=\frac{x^{-3}y^{-3}}{x^{-15}y^3}=x^{-3-(-15)}y^{-3-3}=\boxed{x^{12}y^{-6}=\frac{x^{12}}{y^6}}
6 0
3 years ago
A linear relation is given by the function f (x) = −7x +18 . If (x, 46) is on the function, the value of x is
prohojiy [21]

Answer:

x=-4

Step-by-step explanation:

46 = -7x + 18

subtract 18 from each side to get:

28 = -7x

divide each side by -7 to get:

x = -4

4 0
3 years ago
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