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djverab [1.8K]
1 year ago
14

Ellen said she spent half her money for lunch and half of what was left for a movie. She now has $1.20. How much did she spend f

or lunch?
Devise a plan

Carry out the plan

Look back (is it reasonable? Did we answer the question?)
Mathematics
1 answer:
goldenfox [79]1 year ago
8 0

Answer: She spend $1.20 for lunch.

Step-by-step explanation:

Let the total amount be 'x'.

Half of her money spend for lunch be \dfrac{x}{2}

Half of her money left for a movie be \dfrac{x}{2}

Amount she has now = $1.20

So, According to question, it becomes ,

\dfrac{x}{2}=1.20\\\\x=1.20\times 2\\\\x=\$2.40

Hence, Amount she spend for lunch is \dfrac{x}{2}=\dfrac{2.40}{2}=\$1.20

Therefore, she spend $1.20 for lunch.

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Step-by-step explanation:

If the linear scale factor of two solids is k, then the volume scale factor is k^3.

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se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y
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f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0    exists.

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f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}   exists.

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\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}  where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

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