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borishaifa [10]
2 years ago
12

?ABC has the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices. The measure of the longest side of ?ABC is units. ?ABC is tr

iangle. If ?ABD is formed with the point D(1, 2) as its third vertex, then ?ABD is triangle. The length of side AD is units.
Mathematics
1 answer:
Doss [256]2 years ago
4 0

Answer:

i. BC

ii. |AD|=|7-2|=5

Step-by-step explanation:

ABC has the points A(1, 7), B(-2, 2), and C(4, 2).

We use the distance formula to obtain

|AC|=\sqrt{(4-1)^2+(2-7)^2}

|AC|=\sqrt{9+25}

|AC|=\sqrt{34}=5.8

|AB|=\sqrt{(-2-1)^2+(2-7)^2}

|AB|=\sqrt{9+25}

|AB|=\sqrt{34}=5.8

Using the absolute value method

|BC|=|4--2|=6

The longest side is BC

We use the absolute value method to find the length of AD

|AD|=|7-2|=5

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We know the following relationship:

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The domain of a function are the inputs of the function, that is, a function f is a relation that assigns to each element x in the set A exactly one element in the set B. The set A is the domain (or set of inputs) of the function and the set B contains the range (or set of outputs).Then applying this concept to our function csc(\theta) we can write its domain as follows:

1. D<span>omain of validity for csc(\theta):
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D: \{\theta \in R/ sin(\theta) \neq 0 \} \\ In words: All \ \theta \ that \ are \ real \ values \ except \ those \ that \ makes \ sin(\theta)=0 
 
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</span>
2. which identity is not used in the proof of the identity 1+cot^{2}(\theta)=csc^{2}(\theta):

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The identity that is not used is as established in the statement above:

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</span>\frac{1+cos^{2}(\theta)}{sin^{2}(\theta)}=csc^{2}(\theta)<span>


</span>
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