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3 years ago

7

Someone help ill mark brainliest

1 answer:

liq [111]3 years ago

7 0

Answer:

Step-by-step explanation: My variables are

d=big number q=small number

My equations are 6=3q2d+3=qI

will solve for q in first equation 6/3 = 3q/32=q

Now I will pluck back in to get d2d+3=(2) -3 -32d=-12d/2=-½d=-1/4 So the solution is(-1/4, .2)

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4 years ago

I have <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bdx%7D%7Bdt%7D%20%3D%20%20%5Cfrac%7B5%7D%7Bx%7D%20" id="TexFormula1" title=

mojhsa [17]

$\bf \cfrac{dx}{dt}=\cfrac{5}{t}\quad ,\quad x(0)=7\to 
\begin{cases}
f(t)=0\\
t=7
\end{cases}\\\\
-----------------------------\\\\
\displaystyle \int \cfrac{5}{t}\cdot dt\implies 5\int \cfrac{1}{t}\cdot dt\implies 5ln|t|+C= f(t)
\\\\\\
\textit{now, we know when f(t)=0, t=7}\implies 5ln|7|+C=0
\\\\\\
C=-5ln(7)
\\\\\\
thus\implies \boxed{5ln|t|-5ln(7)=f(t)=x}$

notice |7| 7 is positive, so we can simply remove the bars and use 7 by itself

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3 years ago