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AVprozaik [17]
3 years ago
9

kimmys savings account has a balance of $76.23 in june.by september, her balance is 5 times as much as her june balance.between

september and december , kimmy deposits a total of 87.83 into her account if she does not withdraw any moneyfrom her account, what should kimmy's balance be in december
Mathematics
1 answer:
nalin [4]3 years ago
4 0
From the information you have given me, I would say Kimmy has $<span>468.98 </span>dollars in her bank account.

Does she get more money during the other months, just as she had gotten 5 times as much as she had in a 3 month span? (From june to september.) All I could tell was her money was multiplied by 5, then you add $87.83 more into her account.

Please check my math if you want to be sure.
$76.23 * 5 = $381.15
$381.15 + 87.83 = 468.98
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The given function is

f(x)=\frac{1}{x^2}                  .... (1)

According to the first principle of the derivative,

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

f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}

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

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

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

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

Cancel out common factors.

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

By applying limit, we get

f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}

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Therefore f'(x)=-\frac{2}{x^3}.

(b)

Put x=2, to find the y-coordinate of point of tangency.

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The slope of tangent at x=2 is

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Substitute x=2 in equation 2.

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The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

y-y_1=m(x-x_1)

y-0.25=-0.25(x-2)

y-0.25=-0.25x+0.5

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y=-0.25x+0.75

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

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