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rusak2 [61]
3 years ago
12

Una avioneta despega con un ángulo de elevación de 12° respecto al suelo si ha recorrido 48 metros en la misma dirección ¿a que

altura respecto al suelo se encuentra la avioneta en ese momento
Mathematics
1 answer:
victus00 [196]3 years ago
7 0

Answer: 9.98 meters

Step-by-step explanation:

Angle = 12°

Displacement = 48 m

The height, or displacement in the y-axis will be equal to the opposite cathetus of a triangle rectangle where the hypotenuse is the displacement, then we can find that the displacement in the y-axis is:

h = sin(12°)*48m = 9.98m

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<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Csqrt%7Bx%7D%20" id="TexFormula1" title="f(x) = \sqrt{x} " alt="f(x) =
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<u>Step-by-step explanation:</u>

\lim_{h \to 0} f(x)=\dfrac{f(x+h)-f(x)}{h}

f(x) = \sqrt x\\

f(x+h) = \sqrt{x+h}

\lim_{h \to 0} f(x)=\dfrac{\sqrt{x+h}-\sqrt x}{h}

                   =\dfrac{\sqrt{x+h}-\sqrt x}{h}\bigg(\dfrac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x}\bigg)

                   =\dfrac{(x + h)-(x)}{h(\sqrt{x+h}+\sqrt x)}

                   =\dfrac{h}{h(\sqrt{x+h}+\sqrt x)}

                   =\dfrac{1}{\sqrt{x+h}+\sqrt x}

                  =\dfrac{1}{\sqrt{x+0}+\sqrt x}

                  =\dfrac{1}{2\sqrt x}

4 0
3 years ago
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