▶️ Ejercicio 163 álgebra - Solucionario Baldor ✅✅

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DESCARGA

CAPITULO XVIII

Fórmulas

Ejercicio 163

En la fórmula $e = v t$ , despejar v y t.
$\begin{matrix} e & = v t \\ v & = \frac{e}{t} \\ t & = \frac{e}{v} \end{matrix}$
En $A = h (\frac{b + b ‘}{2})$ hacer a h el sujeto de la fórmula.
$\begin{matrix} A & = h (\frac{b + b ‘}{2}) \\ 2 A & = h (b + b ‘) \\ h & = \frac{2 A}{b + b ‘} \end{matrix}$
En $e = \frac{1}{2} a t^{2}$ , despejar a.
$\begin{matrix} e & = \frac{1}{2} a t^{2} \\ 2 e & = a t^{2} \\ a & = \frac{2 e}{t^{2}} \end{matrix}$
En $A = \frac{1}{2} a l n$ , despejar a, l y n.
$\begin{matrix} A & = \frac{1}{2} a l n \\ 2 A & = a l n \\ a & = \frac{2 A}{l n} \\ 2 A & = a l n \\ l & = \frac{2 A}{a n} \\ 2 A & = a l n \\ n & = \frac{2 A}{l n} \end{matrix}$
En $A = π r^{2}$ , despejar r.
$\begin{matrix} A & = π r^{2} \\ \frac{A}{π} & = r^{2} \\ r & = \sqrt{\frac{A}{π}} \end{matrix}$
En $a^{2} = b^{2} + c^{2} - 2 b \times x$ , despejar x.
$\begin{matrix} a^{2} & = b^{2} + c^{2} - 2 b \times x \\ a^{2} - b^{2} - c^{2} & = - 2 b \times x \\ x & = - \frac{a^{2} - b^{2} - c^{2}}{2 b} \end{matrix}$
En $V = V_{o} + a t$ , despejar $V_{o}, a y t$ .
$\begin{matrix} V & = V_{o} + a t \\ V - a t & = V_{o} \\ V - V_{o} & = a t \\ a & = \frac{V - V_{o}}{t} \\ t & = \frac{V - V_{o}}{a} \end{matrix}$
En $V = V_{o} - a t$ , despejar $V_{o}, a y t$ .
$\begin{matrix} V & = V_{o} - a t \\ V + a t & = V_{o} \\ V + V_{o} & = a t \\ a & = \frac{V + V_{o}}{t} \\ t & = \frac{V + V_{o}}{a} \end{matrix}$
En $D = \frac{P}{V}$ , despejar V y P.
$\begin{matrix} D & = \frac{P}{V} \\ V & = \frac{P}{D} \\ P & = D V \end{matrix}$
En $a^{2} = b^{2} + c^{2}$ , despejar b y c.
$\begin{matrix} a^{2} & = b^{2} + c^{2} \\ a^{2} - c^{2} & = b^{2} \\ b & = \sqrt{a^{2} - c^{2}} \\ a^{2} - b^{2} & = c^{2} \\ c & = \sqrt{a^{2} - b^{2}} \end{matrix}$
En $V = a t$ , despejar a y t.
$\begin{matrix} V & = a t \\ V a & = t \\ V t & = a \end{matrix}$
En $\frac{1}{f} = \frac{1}{p ‘} - \frac{1}{p}$ , despejar p’ y p.
$\begin{matrix} \frac{1}{f} & = \frac{1}{p ‘} - \frac{1}{p} \\ \frac{1}{f} + \frac{1}{p} & = \frac{1}{p ‘} \\ \frac{p + f}{f p} & = \frac{1}{p ‘} \\ p ‘ (p + f) & = f p \\ p ‘ & = \frac{f p}{p + f} \\ \frac{1}{p} & = \frac{1}{p ‘} - \frac{1}{f} \\ \frac{1}{p} & = \frac{f - p ‘}{f p ‘} \\ f p ‘ & = (f - p ‘) p \\ p & = \frac{f p ‘}{f - p ‘} \end{matrix}$
En $v = \sqrt{\frac{e}{d}}$ , despejar d y e.
$\begin{matrix} v & = \sqrt{\frac{e}{d}} \\ v^{2} & = \frac{e}{d} \\ e & = v^{2} d \\ d & = \frac{e}{v^{2}} \end{matrix}$
En $e = V_{o} t + \frac{1}{2} a t^{2}$ , despejar $V_{o}$
$\begin{matrix} e & = V_{o} t + \frac{1}{2} a t^{2} \\ e - \frac{1}{2} a t^{2} & = V_{o} t \\ \frac{2 e - a t^{2}}{2} & = V_{o} t \\ V_{o} & = \frac{2 e - a t^{2}}{2 t} \end{matrix}$
En $e = V_{o} t - \frac{1}{2} a t^{2}$ , despejar $V_{o} y a$
$\begin{matrix} e & = V_{o} t - \frac{1}{2} a t^{2} \\ e + \frac{1}{2} a t^{2} & = V_{o} t \\ \frac{2 e + a t^{2}}{2} & = V_{o} t \\ V_{o} & = \frac{2 e + a t^{2}}{2 t} \\ e - V_{o} t & = - \frac{1}{2} a t^{2} \\ 2 V_{o} t - 2 e & = a t^{2} \\ a & = \frac{2 V_{o} t - 2 e}{t^{2}} \end{matrix}$
En $V = \frac{1}{3} h π r^{2}$ , despejar h y r.
$\begin{matrix} V & = \frac{1}{3} h π r^{2} \\ 3 V & = h π r^{2} \\ h & = \frac{3 v}{π r^{2}} \\ 3 V & = h π r^{2} \\ \frac{3 V}{h π} & = r^{2} \\ r & = \sqrt{\frac{3 V}{h π}} \end{matrix}$
En $I = \frac{c \times t \times r}{100}$ , despejar c, t y r.
$\begin{matrix} I & = \frac{c \times t \times r}{100} \\ 100 I & = c \times t \times r \\ c & = \frac{100 I}{t r} \\ 100 I & = c \times t \times r \\ t & = \frac{100 I}{c r} \\ 100 I & = c \times t \times r \\ r & = \frac{100 I}{c t} \end{matrix}$
En $E = I R$ , despejar R e I.
$\begin{matrix} E & = I R \\ R & = \frac{E}{I} \\ I & = \frac{E}{R} \end{matrix}$
En $e = \frac{v^{2}}{2 a}$ , despejar v.
$\begin{matrix} e & = \frac{v^{2}}{2 a} \\ 2 a e & = v^{2} \\ v & = \sqrt{2 a e} \end{matrix}$
En $u = a + (n - 1) r$ , despejar a, n y r.
$\begin{matrix} u & = a + (n - 1) r \\ u - (n - 1) r & = a \\ u - a & = (n - 1) r \\ u - a & = n r - r \\ u - a + r & = n r \\ n & = \frac{u - a + r}{r} \\ u - a & = (n - 1) r \\ r & = \frac{u - a}{n - 1} \end{matrix}$
En $u = a r^{n - 1}$ , despejar a y r.
$\begin{matrix} u & = a r^{n - 1} \\ a & = \frac{u}{r^{n - 1}} \\ \frac{u}{a} & = r^{n - 1} \\ r & = \sqrt[n - 1]{\frac{u}{a}} \end{matrix}$
En $I = \frac{Q}{t}$ , despejar Q y t.
$\begin{matrix} I & = \frac{Q}{t} \\ Q & = I t \\ t I & = Q \\ t & = \frac{Q}{I} \end{matrix}$