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Finger [1]
3 years ago
8

Solve 2x - 1 < 7 and 5x + 3 < 3. {x | x < 0} {x | x < 4} {x | 0 < x < 4}

Mathematics
2 answers:
nataly862011 [7]3 years ago
7 0
2x -1 <7 ; then, 2x < 8; then x < 4;
5x + 3 < 3; then, 5x < 0; then x < 0;
Finally, {x / x < 0};
saul85 [17]3 years ago
3 0

Answer:

Solution for givrn system of inequalities: \{x ~| ~x < 0\}

Step-by-step explanation:

We are given a system of two inequalities:

2x-1 < 7\\5x+3 < 3

Solving the inequalities individually:

2x-1 < 7\\2x

The intersection of the two solution will give us the solution to the system of inequalities:

x \in (-\infty, 4) \cap (-\infty, 0)\\ x \in (-\infty, 0) \\\{x ~| ~x < 0\}

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First of all, you have to understand g<span> is a square-root function. 

</span>Square-root functions are continuous across their entire domain, and their domain is all real x-<span>values for which the expression within the square-root is non-negative. 
</span>
In other words, for any square-root function q and any input c in the domain of q (except for its endpoint), we know that this equality holds:  lim \ q(x)=q(c) 

Let's take \sqrt{x} <span>as an example. 
</span>
The domain of \sqrt x is all real numbers such that x \geq 0.  Since x=0  is the endpoint of the domain, the two-sided limit at that point doesn't exist (you can't approach 0 <span>from the left). 
</span>
<span>However, continuity at an endpoint only demands that the one-sided limit is equal to the function's value: 
</span>
lim \  \sqrt{x} =  \sqrt{0} =0 

In conclusion, the equality lim \ q(x)=q(c) holds for any square-root function q and any real number c  in the domain of q e<span>xcept for its endpoint, where the two-sided limit should be replaced with a one-sided limit. </span>

The input x=-3,  is within the domain of g<span>. 
</span>
Therefore, in order to find  lim \ g(x)  we can simply evaluate g at x-3<span>. 
</span>
g(x) 

\sqrt{7x+22} 

\sqrt{7(-3)+22} 

\sqrt{1} =1
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