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Elden [556K]
3 years ago
15

For the rational expression 2x^2-7x-15/x^2-25, do both of the following,

Mathematics
1 answer:
Mrrafil [7]3 years ago
3 0

Answer:

a. \frac{2x + 3}{x+5}

b. x ≠ -5 (Vertical asymptote) and x ≠ 5 (Hole)

Step-by-step explanation:

Factor the numerator (Grouping):

2x^2 - 7x - 15              

Two numbers that multiply to -30 and add to -7 = -3 and 10

[2x^2 - 10x] + [3x - 15]

(2x + 3)(x - 5)

Factor the denominator (Difference of Two Squares):

x^2 - 25 = (x +5)(x-5)

Factored Expression:

(x - 5) can be factored out of top and bottom as a hole-

\frac{(2x + 3)(x - 5)}{(x + 5)(x - 5)} = \frac{2x + 3}{x+5}

Variable Restrictions:

Denominator ≠ 0

x + 5 = 0\\x = -5

Vertical asymptote at x = -5 ⇒ x ≠ -5

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In order to answer this question, the figure in the first picture will be helpful to understand what a right triangle is. Here, a right angle refers to 90\°.


However, if we want to solve the problem we have to know certain things before:


In the second figure is shown a general right triangle with its three sides and another given angle, we will name it \alpha:


  • The side <u>opposite to the right angle</u> is called The Hypotenuse (h)
  • The side <u>opposite to the angle \alpha</u> is called the Opposite (O)
  • The side <u>next to the angle \alpha</u> is called the Adjacent (A)

So, going back to the triangle of our question (first figure):


  • The Hypotenuse is AB
  • The Opposite is BC
  • The Adjacent is AC

Now, if we want to find the length of each side of a right triangle, we have to use the <u>Pythagorean Theorem</u> and T<u>rigonometric Functions:</u>


Pythagorean Theorem


h^{2}=A^{2} +O^{2}    (1)  


Trigonometric Functions (here are shown three of them):


Sine: sin(\alpha)=\frac{O}{h}    (2)


Cosine: cos(\alpha)=\frac{A}{h}    (3)


Tangent: tan(\alpha)=\frac{O}{A}   (4)



In this case the function that works for this problem is cosine (3), let’s apply it here:


cos(40\°)=\frac{AC}{h}    


cos(40\°)=\frac{15}{h}    (5)


And we will use the Pythagorean Theorem to find the hypotenuse, as well:



h^{2}=AC^{2}+BC^{2}    


h^{2}=15^{2}+BC^{2}    (6)


h=\sqrt{225+BC^2}   (7)



Substitute (7) in (5):


cos(40\°)=\frac{15}{\sqrt{225+BC^2}}    


Then clear BC, which is the side we want:


{\sqrt{225+BC^2}}=\frac{15}{cos(40\°)}


{{\sqrt{225+BC^2}}^2={(\frac{15}{cos(40\°)})}^2


225+BC^{2}=\frac{225}{{(cos(40\°))}^2}


BC^2=\frac{225}{{(cos(40\°))}^2}-225


BC=\sqrt{158,41}


BC=12.58


Finally BC is approximately 13 cm



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