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wel
3 years ago
10

What postulate or theorem can be used to prove that these two triangles are congruent?

Mathematics
1 answer:
kati45 [8]3 years ago
5 0

Answer:

AAS postulate can be used to prove that these two triangles are congruent

Step-by-step explanation:

Let us revise the cases of congruence  

  1. SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
  2. SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
  3. ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
  4. AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
  5. HL ⇒ hypotenuse and leg of the 1st right Δ ≅ hypotenuse and leg of the 2nd right Δ  

In the given figure

∵ There is a pair of vertically opposite angles

∵ The vertically opposite angles are congruent ⇒ (1)

∵ There are two angles have the same mark

∴ These marked angles are congruent ⇒ (2)

∵ There are two sides have the same mark

∴ These two marked sides are congruent ⇒ (3)

→ From (1), (2), and (3)

∴ The two triangles have 2 angles and 1 side congruent

→ By using case 4 above

∴ The two triangles are congruent by the AAS postulate of congruency.

AAS postulate can be used to prove that these two triangles are congruent

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A. Use composition to prove whether or not the functions are inverses of each other.
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A. In a composition of two functions the first function is evaluated, and then the second function is evaluated on the result of the first function. In other word, you are going to evaluate the second function in the first function.

Remember that you can evaluate function at any number just replacing the variable in the function with the number. For example, let's evaluate our function f(x) at x=1:

f(x)=\frac{1}{x-3}

f(1)=\frac{1}{1-3}

f(1)=\frac{1}{-2}

Similarly, to find the composition of f(x) andg(x), we are going to evaluate f(x) at g(x). In other words, we are going to replace x in f(x) with \frac{3x+1}{x}:

f(x)=\frac{1}{x-3}

f(g(x) = f(\frac{3x+1}{x} ) = \frac{1}{\frac{3x+1}{x} -3}

Remember that two functions are inverse if after simplifying their composition, we end up with just x. Let's simplify and see what happens.

f(g(x)=\frac{1}{\frac{3x+1}{x} -3}

f(g(x)=\frac{1}{\frac{3x+1-3x}{x} }

f(g(x)=\frac{1}{\frac{1}{x} }

f(g(x)=x

Now let's do the same for g(f(x)):

g(\frac{1}{x-3} )=\frac{3(\frac{1}{x-3})+1}{x}

g(\frac{1}{x-3} )=\frac{\frac{3}{x-3}+1}{x}

g(\frac{1}{x-3} )=\frac{\frac{3+x-3}{x-3}}{x}

g(\frac{1}{x-3} )=\frac{\frac{x}{x-3}}{x}

g(\frac{1}{x-3} )=\frac{x}{x(x-3)}

g(f(x))=\frac{x}{x(x-3)}

We can conclude that g(x) is the inverse function of f(x), but f(x) is not the inverse function of g(x).

B. The domain of a function is the set of all the possible values the independent variable can have. In other words, the domain are all the possible x-values of function.

Now, interval notation is a way to represent and interval using an ordered pair of numbers called the end points; we use brackets [ ] to indicate that the end points are included in the interval and parenthesis ( ) to indicate that they are excluded.

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We can conclude that the domain of the composite function f(g(x)) in interval notation is (-∞,0)U(0,∞)

Now let's do the same for g(f(x)).

Notice that the composition is not defined when its denominator equals zero, so we are going to set its denominator equal to zero to find the values we should exclude from its domain:

x(x-3)=0

x=0 and x-3=0

x=0 and x=3

Know we know that we need to exclude x=0 and x=3 from the domain of g(f(x)).

We can conclude that the domain of the composition function g(f(x)) is (-∞,0)U(0,3)U(3,∞)

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