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2 years ago

Indicate the method you would use to prove the two 's . If no method applies, enter "none". SSS SAS ASA AAS None

Mathematics

2 answers:

Goryan [66]2 years ago

6 0

Answer: ASA method must be applied to the given triangles to prove then congruent.

In two triangles, if two angles and their included side are congruent then the two triangles are said to be congruent .This method is known as Angle-Side -Angle postulate(ASA).

In the given triangles one side with their corresponding angles are equal. Therefore by ASA postulate they are congruent.

There we can see no other sides are given equal therefore all the other options cannot be applied.

Lorico [155]2 years ago

5 0

ASA
is your andswer i just got this correct.

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Mrs. Williams estimates that on average her family recycles 12.6 pounds of plastic each month. In a year's time, how much landfi

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Step-by-step explanation:

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2 years ago

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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

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2 years ago

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PSYCHO15rus [73]

We can start solving this problem by first identifying what the elements of the sets really are.

R is composed of real numbers. This means that all numbers, whether rational or not, are included in this set.

Z is composed of integers. Integers include all negative and positive numbers as well as zero (it is essentially a set of whole numbers as well as their negated values).

W on the other hand has 0,1,2, and onward as its elements. These numbers are known as whole numbers.

W ⊂ Z: TRUE. As mentioned earlier, Z includes all whole numbers thus W is a subset of it.

R ⊂ W: FALSE. Not all real numbers are whole numbers. Whole numbers must be rational and expressed without fractions. Some real numbers do not meet this criteria.

0 ∈ Z: TRUE. Zero is indeed an integer thus it is an element of Z.

∅ ⊂ R: TRUE. A null set is a subset of R, and in fact every set in general. There are no elements in a null set thus making it automatically a subset of any non-empty set by definition (since NONE of its elements are <u>not</u> an element of R).

{0,1,2,...} ⊆ W: TRUE. The set on the left is exactly what is defined on the problem statement for W. (The bar below the subset symbol just means that the subset is not strict, therefore the set on the left can be <u>equal</u> to the set on the right. Without it, the statement would be false since a strict subset requires that the two sets should not be equal).

-2 ∈ W: FALSE. W is just composed of whole numbers and not of its negated counterparts.

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This can be solved by Venn-diagram

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