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djverab [1.8K]
3 years ago
11

See attached file. show all work.

Mathematics
1 answer:
andrew11 [14]3 years ago
3 0

The point-slope form of the equation for a line can be written as

... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)

Your function gives

... f'(h) = m

... f(h) = k

a) The tangent line is then

... y = 5(x -2) +3

b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.

... y = (-1/5)(x -2) +3

_____

You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.

In standard form, the tangent line's equation is 5x -y = 7. The normal line's equation is x +5y = 17.

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V=\pi \times r^{2}\times h\\\\\Longrightarrow V=r^{2}\times \left( \pi \times h\right)  \\\\\Longrightarrow r^{2}=\frac{V}{\pi \times h} \\\\\Longrightarrow r=\sqrt{\frac{V}{\pi \times h} }

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=\displaystyle\int_{-5}^{-5}\int_{-5}^5x^6e^y(25-y^2)\,\mathrm dy\,\mathrm dx

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please vote my answer brainliest. thanks!
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