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ASHA 777 [7]
3 years ago
10

999 less than 45 increased by the product of a number and 85

Mathematics
1 answer:
LekaFEV [45]3 years ago
8 0
45-999+Nx85 
(x) is times 
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How do I solve this?
KengaRu [80]

Answer with Step-by-step explanation:

The given differential euation is

\frac{dy}{dx}=(y-5)(y+5)\\\\\frac{dy}{(y-5)(y+5)}=dx\\\\(\frac{A}{y-5}+\frac{B}{y+5})dy=dx\\\\\frac{1}{100}\cdot (\frac{10}{y-5}-\frac{10}{y+5})dy=dx\\\\\frac{1}{100}\cdot \int (\frac{10}{y-5}-\frac{10}{y+5})dy=\int dx\\\\10[ln(y-5)-ln(y+5)]=100x+10c\\\\ln(\frac{y-5}{y+5})=10x+c\\\\\frac{y-5}{y+5}=ke^{10x}

where

'k' is constant of integration whose value is obtained by the given condition that y(2)=0\\

\frac{0-5}{0+5}=ke^{20}\\\\k=\frac{-1}{e^{20}}\\\\\therefore k=-e^{-20}

Thus the solution of the differential becomes

 \frac{y-5}{y+5}=e^{10x-20}

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3 years ago
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\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

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Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.

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