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Sergio [31]
3 years ago
13

In the diagram shown, XYZ is a dilation of TUV. The center of the dilation is point W.

Mathematics
2 answers:
saw5 [17]3 years ago
6 0

There ain't no diagram here, DAWGG :/

polet [3.4K]3 years ago
6 0

Answer:

No diargrammmmmmmmmmmm

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Answer with Step-by-step explanation:

We are given that

f(x)=\left\{\begin{matrix}\dfrac{x^2-2x}{x^2-4}&,if\ \ x\neq 2 \\ 1&,if\ \ x=2\end{matrix}\right.

We have to explain that why the function is discontinuous at x=2

We know that if function is continuous at x=a then LHL=RHL=f(a).

f(x)=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}

LHL=Left hand limit when x <2

Substitute x=2-h

where h is small positive value >0

\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2-h}{2-h+2}

\lim_{h\rightarrow 0}\frac{2-h}{4-h}=\frac{2}{4}=\frac{1}{2}

Right hand limit =RHL when x> 2

Substitute

x=2+h

\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2+h}{2+h+2}=\lim_{h\rightarrow 0}\frac{2+h}{4+h}

=\frac{2}{4}=\frac{1}{2}

LHL=RHL=\frac{1}{2}

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LHL=RHL\neq f(2)

Hence, function is discontinuous at x=2

4 0
4 years ago
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