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Genrish500 [490]
3 years ago
11

. When the process of solving a linear inequality results in the variable being eliminated with a false statement remaining, wha

t does this mean about the solution set of the inequality?
Mathematics
1 answer:
tino4ka555 [31]3 years ago
6 0

Answer:

No Solution

Step-by-step explanation:

When we solve linear inequalities using any one the methods such as

1) Elimination  or

2) Substitution

And we get  the variable eliminated with a false statement remaining, this means there is no solution.

In other words the solution does not exist or the solution is not possible in the given range or the values given are false.

But when the values are true we get a solution for the variable solved.

In the elimination method we eliminate the same terms by addition or subtraction after making them equal.

And the substitution method requires putting the values of one variable in place of another.

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aleksklad [387]
Question 1: 6 3/4 so option b
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3 years ago
Consider the function f(x)=9-x^2/x^2-4 For which intervals is f(x) positive? Check ALL that apply.
solong [7]

Answer:

a. f (x) < 0 for x ∈ (-∞ ,-3)

b. f (x) > 0 for x ∈ (-3,-2)

c. f (x) < 0 for x ∈ (-2,2)

d. f (x) > 0 for x ∈ (2,3)

e.f (x) < 0 for x ∈ (3,∞)

Step-by-step explanation:

Here, the given function is:f(x)=   \frac{9-x^2}{x^2-4}

Now, to check for the sign of f(x) at x = k, put the value of x from the given interval.

We get:

<u>a. (-infinity, -3) </u>

put k = -4 from the given interval

We get f(-4)=   \frac{9-(-4)^2}{(-4)^2-4}  = \frac{9-16}{16-4}  = \frac{-7}{12}

⇒ f (x) < 0 for x ∈ (-∞ ,-3)

b. (-3, -2)

put k = -2.5 from the given interval

We get f(-2.5)=   \frac{9-(-2.5)^2}{(-2.5)^2-4}  = \frac{9-6.25}{6.25-4}  = \frac{2.75}{2.25}  > 0

⇒ f (x) > 0 for x ∈ (-3,-2)

c. (-2, 2)

put k = 0 from the given interval

We get f(0)=   \frac{9-(0)^2}{(0)^2-4}  = \frac{9}{-4}  = -\frac{9}{4}  < 0

⇒ f (x) < 0 for x ∈ (-2,2)

d. (2, 3)

put k =2.5 from the given interval

We get f(2.5)=   \frac{9-(2.5)^2}{(2.5)^2-4}  = \frac{9-6.25}{6.25-4}  = \frac{2.75}{2.25}  > 0

⇒ f (x) > 0 for x ∈ (2,3)

e. (infinity, 3)

put k = 4 from the given interval

We get f(4)=   \frac{9-(4)^2}{(4)^2-4}  = \frac{9-16}{16-4}  = \frac{-7}{12}

⇒ f (x) < 0 for x ∈ (3,∞)

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4 years ago
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postnew [5]
Answer:
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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