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2 years ago

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Find (f-g)(x). What is the domain of f/g f(x)=2x-5 g(x)=2-x

1 answer:

Elden [556K]2 years ago

6 0

$f(x)=2x-5 , \ \ \ \ g(x)=2-x \\\\(f-g)(x)=f(x)-g(x) = 2x-5 -(2-x)=2x-5-2+x = 3x+5 \\\\\\
 domain : \frac{f}{g}=\frac{2x-5}{2-x}\\\\2-x\neq 0 \\ \\x\neq 2 \\ \\D=R\setminus \left \{ 2 \right \}$

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Sergeeva-Olga [200]

Answer:

Standard form of the equation is:

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Step-by-step explanation:

Given equation:

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To convert the given equation to standard form of equation:

$A(x)+B(y)=c$

Using distribution.

$y-2=(-3x)+(-3\times-7)$

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Adding 2 both sides.

$y-2+2=-3x+21+2$

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Adding $3x$ to both sides.

$3x+y=3x-3x+23$

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Answer:

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Step-by-step explanation:

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3 years ago

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natita [175]

Answer:

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Step-by-step explanation:

Given

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This is determined when Demand = Supply

So:

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3 years ago

The quadratic formula gives which roots for the equation 2x^2 + x - 6 = 0?

Jet001 [13]

Answer:

The roots are $x = (\frac{3}{2}, -2)$ , which is given by option C.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question, we have that:

$2x^2 + x - 6 = 0$

Which is a quadratic equation with $a = 2, b = 1, c = -6$ . So

$\Delta = 1^{2} - 4*2(-6) = 1 + 48 = 49$

$x_{1} = \frac{-1 + \sqrt{49}}{2*2} = \frac{6}{4} = \frac{3}{2}$

$x_{2} = \frac{-1 - \sqrt{49}}{2*2} = \frac{-8}{4} = -2$

So the roots are $x = (\frac{3}{2}, -2)$ , which is given by option C.

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3 years ago