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3 years ago

The quadratic formua can be used to solve an equation only if the highest degree in the quation is.....

1 answer:

4 0

2

By definition, a quadratic equation has 2 as the highest degree. If a variable is cubed, then the quadratic formula is not applicable.

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3 years ago

Find sin(a) (Picture)

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Answer:

Sine of A should be 5/13

Step-by-step explanation:

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2 years ago

Determine the domain of the function (fog)(x) where f(x)=3x-1/x-4 and g(x)=x+1/x

Katarina [22]

Answer: (-∞,-1) ∪ (0,+∞)

Step-by-step explanation: The representation fog(x) is a representation of composite function, meaning one depends on the other.

In this case, fog(x) means:

fog(x) = f(g(x))

fog(x) = $3(x+\frac{1}{x} )-\frac{1}{x+\frac{1}{x} } -4$

$fog(x)=3x+\frac{3}{x} -\frac{1}{\frac{x^{2}+x}{x} } -4$

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

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

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

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

This is the function fog(x).

The domain of a function is all the values the independent variable can assume.

For fog(x), denominator can be zero, so:

$x(x^{2}+x) \neq 0$

If x = 0, the function doesn't exist.

$x^{2}+x \neq0$

$x(x+1) \neq0$

$x+1\neq0$

$x\neq-1$

Therefore, the domain of this function is: -∞ < -1 or x > 0

5 0

3 years ago