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3 years ago

Help!!!! please!!!! Ill give brainliest!!!

Mathematics

2 answers:

Pani-rosa [81]3 years ago

8 0

Answer:

A) x < 9

Step-by-step explanation:

Here we have an open circle. Open circles are used for greater than signs (>) and less than signs (<).

Since we have an open circle, we can eliminate option C and option D.

Now, we look at the number line.

The point is at 9, and the line from the point is going towards the left (where the smaller numbers are.)

9 is greater than every number is going toward.

**Let x represent every number is going toward.**

Therefore, x < 9.

Fofino [41]3 years ago

4 0

Answer:

Step-by-step explanation:

hope this help u and hope u get a good grasde :}

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Answer:

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Step-by-step explanation:

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What is the range of the following function?

Lelechka [254]

Answer:

Option d

Step-by-step explanation:

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3 years ago

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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