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2 years ago

Why does (x-2)^2+9=0 have no real solutions?

Mathematics

1 answer:

alex41 [277]2 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Given

(x - 2)² + 9 = 0 ( subtract 9 from both sides )

(x - 2)² = - 9 ( take the square root of both sides )

x - 2 = ± $\sqrt{-9}$

$\sqrt{-9}$ yields no real solutions

There is no real number when multiplied by itself gives - 9

In fact the solutions to $\sqrt{-9}$ are said to be imaginary, that is ± 3i

where i = $\sqrt{-1}$

If interested investigate further.

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3 years ago

If f(x) = x^4-x^3+x^2 and g(x)= -x^2, where x does not = 0, what is (f/g)(x)?

Korvikt [17]

For this case we have the following functions:

$f (x) = x ^ 4-x ^ 3 x ^ 2\\g (x) = - x ^ 2$

By definition of composition of functions we have to: $(f \ g) (x) = \frac {f (x)} {g (x)}$

So:

$\frac {f (x)} {g (x)} = \frac {x ^ 4-x ^ 3 x ^ 2} {- x ^ 2} = \frac {x ^ 4} {- x ^ 2} +\frac {-x ^ 3} {- x ^ 2}+ \frac {x ^ 2} {- x ^ 2} =$

By definition of division of powers of the same base we have to place the same base and subtract the exponents:

$\frac {f (x)} {g (x)} = - x ^ 2+x-1$

ANswer:

$(\frac {f} {g}) (x) = - x ^ 2 +x-1$

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2 years ago

(8.4 - 4.5) × (7.04 - 0.04) =

iren [92.7K]

(8.4−4.5)(7.04−0.04)

=3.9(7.04−0.04)

=(3.9)(7)

=27.3

Ans : 27.3

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3 years ago

Read 2 more answers

Solve for f pleaseeeee

KonstantinChe [14]

Answer:

$\frac{3}{5}f = 4-2m\\ f= \frac{(4-2m) X 5}{3} \\f= \frac{20-10m}{3}$

P.S. If u want a numerical value for 'f' , now put the value of 'm'

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3 years ago

Which set of ordered pairs is not a functions?

Anit [1.1K]

Answer:

Step-by-step explanation:

X values cannot repeat in a function, 3 is used as an X value twice in A, and therefore cannot be a function.

If this is graphed, you can use the vertical line test, if a vertical line on any point goes through other points, it is not a function.

In this case, just check if X repeats.

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2 years ago