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Natali5045456 [20]
3 years ago
6

Use the quadratic formula to find both solutions to the quadratic equation given below. 3x^2-x+4=0

Mathematics
1 answer:
Nuetrik [128]3 years ago
4 0

Answer:

B. and D.

Step-by-step explanation:

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(f o g)(-2)= 11

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Lastima Nelson, Inc. is recruiting a new Chief Financial Officer (CFO). The recruiting budget is $44,000. Lastima has spent $10,
Musya8 [376]
The budget is $44000

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Let the maximum price of the home be x
This value of x will be the 100% before the 6% commission is calculated of it.

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3 years ago
Consider the functions f and g defined by \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\qquad\qquad\text{and}\qquad\qquad g(x) = \dfrac{\sqrt
tino4ka555 [31]

Answer:

The given functions are not same because the domain of both functions are different.

Step-by-step explanation:

The given functions are

f(x)= \sqrt{\dfrac{x+1}{x-1}}

g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}

First find the domain of both functions. Radicand can not be negative.

Domain of f(x):

\dfrac{x+1}{x-1}>0

This is possible if both numerator or denominator are either positive or negative.

Case 1: Both numerator or denominator are positive.

x+1\geq 0\Rightarrow x\geq -1

x-1\geq 0\Rightarrow x\geq 1

So, the function is defined for x≥1.

Case 2: Both numerator or denominator are negative.

x+1\leq 0\Rightarrow x\leq -1

x-1\leq 0\Rightarrow x\leq 1

So, the function is defined for x≤-1.

From case 1 and 2 the domain of the function f(x) is (-∞,-1]∪[1,∞).

Domain of g(x):

x+1\geq 0\Rightarrow x\geq -1

x-1\geq 0\Rightarrow x\geq 1

So, the function is defined for x≥1.

So, domain of g(x) is [1,∞).

Therefore, the given functions are not same because the domain of both functions are different.

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