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3 years ago

Can you please help me on 6(1+n)=8n-3(n-7)

Mathematics

1 answer:

Nana76 [90]3 years ago

3 0

Answer:

n = 15

Step-by-step explanation:

6(1+n)=8n-3(n-7)

Distribute

6 +6n = 8n -3n +21

Combine like terms

6+6n = 5n +21

Subtract 5n from each side

6+6n-5n = 5n-5n +21

6 +n = 21

Subtract 6 from each side

6-6+n = 21-6

n = 15

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3 years ago

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Step-by-step explanation:

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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