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Anvisha [2.4K]
3 years ago
12

The vertices of a triangle are given. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triang

le. (-4, 0, 0), (0, 0, 0), (7, 2, 6)
Mathematics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

Obtuse triangle

Step-by-step explanation:

Given are the vertices of a triangle

Let A (-4, 0, 0),B (0, 0, 0),C(7, 2, 6)

Let us find angles between AB, BC and CA

AB = (4, 0,0): BC = (7,2,6) : CA = (11, 2,6)

Cos B = \frac{AB.BC}{|AB||BC|} \\

B = arc cos \frac{AB.BC}{|AB||BC|} \\=137 deg 54 min 7 sec

Similarly

A=29 deg 53 min 53 seconds

C = 12 deg 12 min 3 sec

Obtuse triangle since one angle > 90 degrees

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The graph below shows the function f(x)=x-3/x^2-2x-3 which statement is true
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Answer:

The correct option is A.

Step-by-step explanation:

Domain:

The expression in the denominator is x^2-2x-3

x² - 2x-3 ≠0

-3 = +1 -4

(x²-2x+1)-4 ≠0

(x²-2x+1)=(x-1)²

(x-1)² - (2)² ≠0

∴a²-b² =(a-b)(a+b)

(x-1-2)(x-1+2) ≠0

(x-3)(x+1) ≠0

x≠3 for all x≠ -1

So there is a hole at x=3 and an asymptote at x= -1, so Option B is wrong

Asymptote:

x-3/x^2-2x-3

We know that denominator is equal to (x-3)(x+1)

x-3/(x-3)(x+1)

x-3 will be cancelled out by x-3

1/x+1

We have asymptote at x=-1 and hole at x=3, therefore the correct option is A....

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3 years ago
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Step-by-step explanation:

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