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AVprozaik [17]
3 years ago
8

What is the missing constant term in the perfect square that starts with x^2-4x

Mathematics
2 answers:
Jet001 [13]3 years ago
7 0

4 will be added to make the given expression a perfect square

Further explanation:

Given term is:

x^2-4x

As there is a square term and a linear term involving x, the expression will be completed using the formula

a^2-2ab+b^2=(a-b)^2

We know that

a^2=x^2

so,

2ab=4x\\2xb=4x\\b=\frac{4x}{2x}\\b=2

Adding (2)^2

x^2-4x+(2)^2

x^2-4x+4

4 will be added to make the given expression a perfect square

Keywords: Perfect square, polynomials

Learn more about perfect squares at:

  • brainly.com/question/4703820
  • brainly.com/question/4706270

#LearnwithBrainly

Basile [38]3 years ago
5 0

Answer:4 pleb

Step-by-step explanation:

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plain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = x2 − 2x x2 − 4 if x ≠ 2 1 if x =
ElenaW [278]

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}

f(2)=1

LHL=RHL\neq f(2)

Hence, function is discontinuous at x=2

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4 years ago
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