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RoseWind [281]
3 years ago
15

Encontrar la ecuacion paralela a y= -1/2x + 1, que pasa por el punto (2, -2). Graficar. Me ayudan con esto, POR FAVOR ❤️

Mathematics
1 answer:
Ymorist [56]3 years ago
7 0
Hola! Este línea hace no intersecarese con el par do puntos dada. ¿Ver?
¡Lo Siento! Dejar yo sabo si yo lata ayuda además.

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3 years ago
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Plz help ...........
bogdanovich [222]

Step-by-step explanation:

Given c = longest of triange = 15m

Given b = short side of triangle = 9m

By pythagoras' Theorem,

{c}^{2}  =  {a}^{2}   +  {b}^{2}  \\  {a}^{2}  +  {b}^{2} =  {c}^{2}   \\  {a}^{2}  +  {(9)}^{2}  =  {(15)}^{2}  \\  {a}^{2}  + 81 = 225 \\  {a}^{2}  = 225 - 81 \\  {a}^{2}  = 144 \\ a =  \sqrt{144}  \\  = 12m

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3 years ago
What is the side length of a 121 square centimeter launch pad
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11 centimeters is the length of the launch pad

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3 years ago
Point G is the centroid of the right △ABC with m∠C=90° and m∠B=30°. Find AG if CG=4 ft.
o-na [289]

Answer: \text{Length of AG=}\frac{2\sqrt{63}}{3}

Explanation:  

Please follow the diagram in attachment.  

As we know median from vertex C to hypotenuse is CM  

\therefore CM=\frac{1}{2}AB

We are given length of CG=4  

Median divide by centroid 2:1  

CG:GM=2:1  

Where, CG=4

\therefore GM=2 ft

Length of CM=4+2= 6 ft  

\therefore CM=\frac{1}{2}AB\Rightarrow AB=12

In \triangle ABC, \angle C=90^0

Using trigonometry ratio identities  

AC=AB\sin 30^0\Rightarrow AC=6 ft

BC=AB\cos 30^0\Rightarrow BC=6\sqrt{3} ft  

CN=\frac{1}{2}BC\Rightarrow CN=3\sqrt{3} ft

In \triangle CAN, \angle C=90^0  

Using pythagoreous theorem  

AN=\sqrt{6^2+(3\sqrt{3})^2\Rightarrow \sqrt{63}

Length of AG=2/3 AN

\text{Length of AG=}\frac{2\sqrt{63}}{3} ft


5 0
3 years ago
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