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madreJ [45]
3 years ago
9

For the function F(x)= 1/x-2 whose graph is shown below, what is the relative value of F(x) when the value of x is close to 2?

Mathematics
2 answers:
Anton [14]3 years ago
8 0
It's B you can also see this on the graph.
kotegsom [21]3 years ago
8 0

Answer:

Either a very large positive or a very large negative number.

Step-by-step explanation:

There is a vertical asymptote at x=2.  From this graph, we can see that as the values of x approach 2, the graph goes up sharply and down sharply.  This means the values would be either very large positive numbers or very large negative numbers.

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Chase buys a bag of cookies that contains 6 chocolate chip cookies, 6 peanut butter cookies, 6 sugar cookies and 6 oatmeal cooki
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Answer:

0.0652173

Step-by-step explanation:

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6 chocolate chip cookies

6 peanut butter cookies

6 sugar cookies

6 oatmeal cookies

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4 0
3 years ago
Given that tan θ ≈ −0.087, where 3 2 π < θ < 2 , π find the values of sin θ and cos θ.
elena-s [515]

Answer:

  • sin θ ≈ -0.08667
  • cos θ ≈ 0.99624

Step-by-step explanation:

Straightforward use of the inverse tangent function of a calculator will tell you θ ≈ -0.08678 radians. This is an angle in the 4th quadrant, where your restriction on θ places it. (To comply with the restriction, you would need to consider the angle value to be 2π-0.08678 radians. The trig values for this angle are the same as the trig values for -0.08678 radians.)

Likewise, straightforward use of the calculator to find the other function values gives ...

  sin(-0.08678 radians) ≈ -0.08667

  cos(-0.08678 radians) ≈ 0.99624

_____

<em>Note on inverse tangent</em>

Depending on the mode setting of your calculator, the arctan or tan⁻¹ function may give you a value in degrees, not radians. That doesn't matter for this problem. sin(arctan(-0.087)) is the same whether the angle is degrees or radians, as long as you don't change the mode in the middle of the computation.

We have shown radians in the above answer because the restriction on the angle is written in terms of radians.

_____

<em>Alternate solution</em>

The relationship between tan and sin and cos in the 4th quadrant is ...

  \cos{\theta}=\dfrac{1}{\sqrt{1+\tan^2{\theta}}}\\\\\sin{\theta}=\dfrac{\tan{\theta}}{\sqrt{1+\tan^2{\theta}}}

That is, the cosine is positive, and the sign of the sine matches that of the tangent.

This more complicated computation gives the same result as above.

4 0
3 years ago
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