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2 years ago

Explain why the Leg-Leg (LL) theorem is the same as the Side-Angle-Side (SAS) postulate.

2 answers:

8 0

Both claim that if the two legs (the sides of the right triangle which meet to form 90°) and one of the corresponding acute angles of two right triangles are congruent then the two right triangles are also congruent :) They say the same but in a slightly different way!

Marysya12 [62]2 years ago

3 0

Answer:

SAS postulate of congruence refers to the congruence between two pairs of corresponding sides and the congruence between one pair of corresponding angles which are formed by the two sides.

Notice that to use this postulate we just need three congruent elements of those triangles.

Similarly we have the Leg-Leg theorem which is used to demonstrate congruence between right triangles. Using this theorem would give the same result than using SAS postulate, because Leg-Leg theorem is actually based on that.

In other words, we can use LL theorem for congruence because all right triangles has a congruent angle, the right angle. So, basically, when you use the LL theorem, you are actually use the SAS postulate indirectly.

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Answer:

Answers below

Step-by-step explanation:

1) 4 (x+4)

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4) 2x+8

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3 years ago

Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.

Romashka [77]

By definition, the average change of rate is given by:
$AVR = \frac{f(x2)-f(x1)}{x2-x1}$
We will calculate AVR for each of the functions.
We have then:

f(x) = x^2 + 3x interval: [-2, 3]:
$f(-2) = x^2 + 3x = (-2)^2 + 3(-2) = 4 - 6 = -2

f(3) = x^2 + 3x = (3)^2 + 3(3) = 9 + 9 = 18$
$AVR = \frac{-2-18}{-2-3}$
$AVR = \frac{-20}{-5}$
$AVR = 4$

f(x) = 3x - 8 interval: [4, 5]:
$f(4) = 3(4) - 8 = 12 - 8 = 4 f(5) = 3(5) - 8 = 15 - 8 = 7$
$AVR = \frac{7-4}{5-4}$
$AVR = \frac{3}{1}$
$AVR = 3$

f(x) = x^2 - 2x interval: [-3, 4]
$f(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15

f(4) = (4)^2 - 2(4) = 16 - 8 = 8$
$AVR = \frac{8-15}{4+3}$
$AVR = \frac{-7}{7}$
$AVR = -1$

f(x) = x^2 - 5 interval: [-1, 1]
$f(-1) = (-1)^2 - 5 = 1 - 5 = -4

f(1) = (1)^2 - 5 = 1 - 5 = -4$
$AVR = \frac{-4+4}{1+1}$
$AVR = \frac{0}{2}$
$AVR = 0$

Answer:
these functions from the greatest to the least value based on the average rate of change are:
f(x) = x^2 + 3x
f(x) = 3x - 8
f(x) = x^2 - 5
f(x) = x^2 - 2x

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3 years ago

What are the solutions for x when y is equal to 0 in the following quadratic

Scrat [10]

Answer: Option C

Step-by-step explanation:

Given the quadratic equation $y = x^2 - 13x$ , substitute $y=0$ :