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3 years ago

10=16(x-2)^2+10 show work

2 answers:

8 0

$Cancel \ 10 \ on \ both \ sides.\\\\0 = 16(x - 2)^2\\\\\\Divide \ both \ sides \ by \ 16.\\0 = (x - 2)\\\\\\Now take the square root of both sides.\\\\0 = x - 2\\\\\\Add 2 to both sides.\\\\2 = x\\\\Switch sides.\\\\\fbox{x = 2}$

klemol [59]3 years ago

4 0

Answer:

x=2

Step-by-step explanation:

10=16(x-2)^2+10

Subtract 10 from each side

10-10=16(x-2)^2+10-10

0 =16(x-2)^2

Divide by 16

0/16 = 16/16 (x-2)^2

0 = (x-2)^2

Take the square root of each side

sqrt(0) = sqrt( (x-2)^2)

0 = x-2

Add 2 to each side

0+2 = x-2+2

2 =x

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3 years ago

dos aviones parten de una ciudad con direcciones S32°E y E57°N¿ cuál es el angulo que forma sus direcciones?

Elan Coil [88]

Answer:

El ángulo formado entre las dos direcciones es aproximadamente 115º.

Step-by-step explanation:

Las direcciones dadas en el enunciado significan lo siguiente:

S32ºE - <em>32º al este del sur.</em>

E57ºN - <em>57º al norte del este.</em>

Vectorialmente hablando, cada dirección es la siguiente:

S32ºE

$A(x,y) = (\sin 32^{\circ}, -\cos 32^{\circ})$

$A(x,y) = (0.530, -0.848)$

E57ºN

$B(x,y) = (\cos 57^{\circ},\sin 57^{\circ})$

$B(x,y) = (0.545, 0.839)$

El ángulo formado entre los dos vectores unitarios ( $\theta$ ), medido en grado sexagesimales, puede determinarse mediante la siguiente ecuación vectorial:

$\theta = \cos^{-1} \frac{A\,\bullet \,B}{\|A\|\cdot \|B\|}$

Donde:

$\|A\|$ - Norma de $A(x,y)$ .

$\|B\|$ - Norma de $B(x,y)$ .

Si sabemos que $A(x,y) = (0.530, -0.848)$ , $B(x,y) = (0.545, 0.839)$ , $\|A\| = 1$ y $\|B\| = 1$ , entonces el ángulo formado entre los dos vectores es: