Well, since you have two variables in one equation, you would have to have at least one other equation to solve this problem. Their isn't enough information to solve this alone, sorry :/
Answer: la altura es 13.1 m
Step-by-step explanation:
El movimiento descripto es de tipo rectilineo uniformemente variado, más precisamente tiro vertical.
Para calcular la posición al cabo de 1 seg utilizaremos la ecuacionecuación del movimiento descrita como:
y = v0.t - 1/2 g t^2
Donde y es la altura para cualquier momento
v0 es la velocidad inicial 18 m/s
g la aceleración de la gravedad 9.81 m/s2
y t el tiempo medido en segundos
Entonces para calcular la altura después de un segundo:
y = 18 m/s x 1 seg - 1/2 9.81 m/s2 (1 seg)^2 = 18 m - 4.9 m = 13.1 m
<h3>
Answer: 125</h3>
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Explanation:
Ignore lines BC and FE. It might help to redraw the figure without those lines present. Refer to the diagram below. Doing this is optional if you can spot what's going on without having to erase those lines.
Notice how angles 1 and 3 are congruent corresponding angles, due to lines FJ and DC being parallel.
Since angle 3 is 55 degrees, angle 1 is also 55 degrees.
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Once we determine angle 1, we subtract from 180 because angle ABF is supplementary to angle 1 (aka angle ABJ)
In other words,
(angle ABJ) + (angle ABF) = 180
angle ABF = 180 - (angle ABJ)
angle ABF = 180 - (angle 1)
angle ABF = 180 - (55)
angle ABF = 125 degrees
Answer:
X is 20 :)
Step-by-step explanation:
this is what you would do
set (2+3x)= 62
3x=62-2
3x=60
60/3=x
x=20!
Answer:
-2/8is it's answer
Step-by-step explanation:
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