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sergey [27]
3 years ago
15

Can a translation and a reflection map QRS to TUV? Explain why or why not. No, the triangles are not congruent. Yes, a translati

on mapping vertex Q to vertex T and a reflection across the line containing QS will map △QRS to △TUV. No, the triangles are obtuse. Yes, a translation mapping vertex S to vertex T and a reflection across the line containing RS will map △QRS to △TUV.
Mathematics
2 answers:
N76 [4]3 years ago
5 0

Answer:

The answer is B

Step-by-step explanation:

just took the answer on e2020

ololo11 [35]3 years ago
3 0

Answer:

B:

<u><em>Yes, a translation mapping vertex Q to vertex T and a reflection across the line containing QS will map  △QRS to △TUV.</em></u>

<u><em /></u>

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3. For the polynomial: ()=−2(+19)3(−14)(+3)2, do the following:A. Create a table of values that have the x-intercepts of p(x) in
Pepsi [2]

Part A. We are given the following polynomial:

\mleft(\mright)=-2\mleft(+19\mright)^3\mleft(-14\mright)\mleft(+3\mright)^2

This is a polynomial of the form:

p=k(x-a)^b(x-c)^d\ldots(x-e)^f

The x-intercepts are the numbers that make the polynomial zero, that is:

\begin{gathered} p=0 \\ (x-a)^b(x-c)^d\ldots(x-e)^f=0 \end{gathered}

The values of x are then found by setting each factor to zero:

\begin{gathered} (x-a)=0 \\ (x-c)=0 \\ \text{.} \\ \text{.} \\ (x-e)=0 \end{gathered}

Therefore, this values are:

\begin{gathered} x=a \\ x=c \\ \text{.} \\ \text{.} \\ x=e \end{gathered}

In this case, the x-intercepts are:

\begin{gathered} x=-19 \\ x=14 \\ x=-3 \end{gathered}

The multiplicity are the exponents of the factor where we got the x-intercept, therefore, the multiplicities are:

Part B. The degree of a polynomial is the sum of its multiplicities, therefore, the degree in this case is:

\begin{gathered} n=3+1+2 \\ n=6 \end{gathered}

To determine the end behavior of the polynomial we need to know the sign of the leading coefficient that is, the sign of the coefficient of the term with the highest power. In this case, the leading coefficient is -2, since the degree of the polynomial is an even number this means that both ends are down. If the leading coefficient were a positive number then both ends would go up. In the case that the leading coefficient was positive and the degree and odd number then the left end would be down and the right end would be up, and if the leading coefficient were a negative number and the degree an odd number then the left end would be up and the right end would be down.

Part C. A sketch of the graph is the following:

If the multiplicity is an odd number the graph will cross the x-axis at that x-intercept and if the multiplicity is an even number it will tangent to the x-axis at that x-intercept.

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