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ELEN [110]
3 years ago
13

DF=78, DE=5x-9, and EF-2x+10, find EF

Mathematics
1 answer:
lbvjy [14]3 years ago
8 0

Answer:

As e is point between df. So we can write df as

df = de + ef

As given df = 78 , de = 5x - 9 , ef = 2x + 10.

So we can write

78 = 5x - 9 + 2x + 10

78 = 7x + 1

On subtracting 1 on booth side

78 - 1 = 7x +1 - 1

77 = 7x

On dividing both side by 7.

11 = x

So x = 11.

So ef = 2x + 10 = 2*11 + 10 = 22 + 10 = 32

Step-by-step explanation:

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OlgaM077 [116]

Answer: Los catetos medirán 12m y 9m.

Step-by-step explanation:

Bueno, sabemos que los lados de nuestro triangulo son 6m, 8m y 10m.

Podemos usar relaciones trigonométricas para encontrar los ángulos de este triangulo.

Cos(X) = cateto adyacente/hipotenusa

Si tomamos al cateto adyacente como 8m, tenemos:

Cos(X) = 8m/10m

X = Acos(8/10)  

Esto nos dice que para que nuestro otro triangulo rectángulo sea equivalente al primero, entonces los nuevos catetos C1  y C2 tienen que ser tal que los cocientes entre los catetos y la hipotenusa se mantengan constantes:

C1/15m = 8m/10m

C1 = (8/10)*15m = 12m

C2/15m = 6m/10m  

C2 = (6/10)*15m = 9m.

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4 years ago
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Salsk061 [2.6K]

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5 0
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Step-by-step explanation:

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