Matemáticas, pregunta formulada por lalisamanoban59, hace 4 meses

Calcular senx si: tg(3x-10") = ctg(2x-50°)

A) 1/2
B) 2
C) 4/5
D) 3/5
E) 1

Doy corona plis. ​

Respuestas a la pregunta

Contestado por rafaelhanibalpayenew
1

Explicación paso a paso:

a)

8c = 8(1) = 88c=8(1)=8

b)

\begin{gathered}2a + b = \\ 2(3) + 7 = \\ 6 + 7 = \\ 13\end{gathered}2a+b=2(3)+7=6+7=13

c)

\begin{gathered} {a}^{2} + b + c \\ 3 {}^{2} + 7 + 1 \\ 9 + 7 + 1 \\ 17\end{gathered}a2+b+c32+7+19+7+117

d)

\begin{gathered} {c}^{3} + 2b + \frac{a}{3} - 2 \\ 1 {}^{3} + 2(7) + \frac{3}{3} - 2 \\ 1 + 14 + 1 - 2 \\ 14\end{gathered}c3+2b+3a−213+2(7)+33−21+14+1−214

e)

\begin{gathered} {c}^{5} + {a}^{3} - b {}^{2} \\ 1 {}^{5} + 3 {}^{3} - 7 {}^{2} \\ 1 + 27 + 49 \\ 77\end{gathered}c5+a3−b215+33−721+27+4977

f)

\begin{gathered} - 2a {}^{2} b {c}^{3 } \\ - 2(3 {}^{2} )(7)(1 {}^{3} ) \\ - 2(9)(7)(1) \\ - 126\end{gathered}−2a2bc3−2(32)(7)(13)−2(9)(7)(1)−126

g)

\begin{gathered}b {}^{2} + 5ac + 2 \\ 7 {}^{2} + 5(3)(1) + 2 \\ 49 + 15 + 2 \\ 66\end{gathered}b2+5ac+272+5(3)(1)+249+15+266

h)

\begin{gathered} \frac{ab}{3} + 2.7c - \frac{3}{2} \\ \\ \frac{(3)(7)}{3} + 2.7(1) - \frac{3}{2} \\ \\ 7 + 2.7 - \frac{3}{2} \\ \\ 7 + \frac{27}{10} - \frac{3}{2} \\ \\ \frac{70 + 27 - 15}{10} \\ \\ \frac{82}{10} = \frac{41}{5} \end{gathered}3ab+2.7c−233(3)(7)+2.7(1)−237+2.7−237+1027−231070+27−151082=541

espero que sea de tu ayuda

a)

8c = 8(1) = 88c=8(1)=8

b)

\begin{gathered}2a + b = \\ 2(3) + 7 = \\ 6 + 7 = \\ 13\end{gathered}2a+b=2(3)+7=6+7=13

c)

\begin{gathered} {a}^{2} + b + c \\ 3 {}^{2} + 7 + 1 \\ 9 + 7 + 1 \\ 17\end{gathered}a2+b+c32+7+19+7+117

d)

\begin{gathered} {c}^{3} + 2b + \frac{a}{3} - 2 \\ 1 {}^{3} + 2(7) + \frac{3}{3} - 2 \\ 1 + 14 + 1 - 2 \\ 14\end{gathered}c3+2b+3a−213+2(7)+33−21+14+1−214

e)

\begin{gathered} {c}^{5} + {a}^{3} - b {}^{2} \\ 1 {}^{5} + 3 {}^{3} - 7 {}^{2} \\ 1 + 27 + 49 \\ 77\end{gathered}c5+a3−b215+33−721+27+4977

f)

\begin{gathered} - 2a {}^{2} b {c}^{3 } \\ - 2(3 {}^{2} )(7)(1 {}^{3} ) \\ - 2(9)(7)(1) \\ - 126\end{gathered}−2a2bc3−2(32)(7)(13)−2(9)(7)(1)−126

g)

\begin{gathered}b {}^{2} + 5ac + 2 \\ 7 {}^{2} + 5(3)(1) + 2 \\ 49 + 15 + 2 \\ 66\end{gathered}b2+5ac+272+5(3)(1)+249+15+266

h)

\begin{gathered} \frac{ab}{3} + 2.7c - \frac{3}{2} \\ \\ \frac{(3)(7)}{3} + 2.7(1) - \frac{3}{2} \\ \\ 7 + 2.7 - \frac{3}{2} \\ \\ 7 + \frac{27}{10} - \frac{3}{2} \\ \\ \frac{70 + 27 - 15}{10} \\ \\ \frac{82}{10} = \frac{41}{5} \end{gathered}3ab+2.7c−233(3)(7)+2.7(1)−237+2.7−237+1027−231070+27−151082=541

espero que sea de tu ayuda

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