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3 years ago

Hey I'm back anyone remembers me?

Mathematics

2 answers:

frozen [14]3 years ago

7 0

No I do not remember you but it’s nice to meet you I suppose.

-BARSIC- [3]3 years ago

7 0

Answer:

It's so sad no one answered saying that they do.

Step-by-step explanation:

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This algebra 1 solving inequalities right?

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3 years ago

The average of 5 numbers is 33. If the average of first 2 numbers is 35 and that of last 2 is 31, then find the 3rd number.

-BARSIC- [3]

The answer is c. 33.

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3 years ago

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nydimaria [60]

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Simplify<br> -4 1/5 (-13 1/10)<br> pls help

oksian1 [2.3K]

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55 1/50 or 2751/50

Step-by-step explanation:

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3 years ago

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

4 0

3 years ago

Hey I'm back anyone remembers me?​

Hey I'm back anyone remembers me?