Write a rational function that has the specified characteristics.

  a) f(x) = (x-5)/((x-3)(x-10))

  b) f(x) = (x-4)/((x+4)(x^2+1))

  c) f(x) = 2(x-1)(x+1)/((x+3)(x-4))

  d) f(x) = -2(x+5)(x-3)/((x+2)(x-5))

  e) f(x) = -3(x^2-1)(x-2)/(x(x^2-9))

Step-by-step explanation:

Ordinarily, we think of a horizontal (or slant) asymptote as a line that the function nears, but does not reach. Some of these questions ask for the horizontal asymptote to be zero and for a function zero at a specific place. That is, the actual value of the function must be the same as the asymptotic value, at least at one location.

There are several ways this can happen:

To make the horizontal asymptote be zero, the degree of the denominator must be greater than the degree of the numerator. That is, there must be additional real or complex zeros in the denominator beyond those for the required vertical asymptotes.

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a) f(x) = (x-5)/((x-3)(x-10)) . . . . vertical asymptote added at x=10 to make the horizontal asymptote be zero

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b) f(x) = (x-4)/((x+4)(x^2+1)) . . . . complex zero added to the denominator to make the horizontal asymptote be zero

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c) f(x) = 2(x-1)(x+1)/((x+3)(x-4)) . . . . factor of 2 added to the numerator to make the horizontal asymptote be 2. Numerator and denominator degrees are the same. (See the second attachment.)

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d) f(x) = -2(x+5)(x-3)/((x+2)(x-5)) . . . . similar to problem (c)

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e) f(x) = -3(x^2-1)(x-2)/(x(x^2-9)) . . . . similar to the previous two problems (See the third attachment.)

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You remember that the difference of squares factors as ...

  a² -b² = (a-b)(a+b)

so the factor that gives zeros at x=±3 can be written (x²-9).

i leave the calculator work to you.

7. diameter 4 miles, circumference 4π miles

8. radius 0.2 miles, circumference 0.4π miles

[this is a bad problem because the actual radius of a car on the track is less than 0.2 miles so the answer is less than 0.4π miles]

9. radius 6 feet, circumference 12π feet

10. diameter 14 cm, circumference 14π cm

11. diameter 4.8 feet, circumference 4.8π feet

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.