Solving for y right?
2y = 3x + 4 - x
2y = 2x + 4
y = 2x + 4 over 2
y = 2 (x + 2) over 2
y = x + 2
y - 3 = 2x - 6 over 2
y - 3 = 2(x - 3) over 2
y - 3 = x - 3
y = x
x - y - 2 = 2(2x + 1)
-y - 2 = 2(2x + 1) - x
-y = 2(2x + 1) - x + 2
y = -2(2x + 1) + x - 2
Answer:
puedes resolver de dos maneras si diste teorema de Pitágoras lo aplicas
hipotenusa al cuadrado = cateto 2 +cateto 2 ( el 2 significa al cuadrado)
sustituyes
hipotenusa= 4m
cateto= 2,5
hay que hallar el otro cateto que nos daría la altura a la que está la escalera
despejamos y tenemos
cateto 2= hip2 -cat2
cat2=(4)2-(2,5)2=
= 16-6,25=9,75
luego hallamos la raíz cuadrada de 9,75= 3,12
la altura a la que se encuentra es 3,12m
-si aun no diste Pitágoras podes representarlo en una hoja utilizando cm en lugar de metros( a escala). trazas el triángulo rectángulo la base te la da la distancia a la cual se encuentra la escalera de la pared es decir 2,5cm trazas la hipotenusa de 4cm de manera que coincida con el cateto opuesto , y mides el valor de este ,será de 3,12cm no olvides que la respuesta la debes dar en metros ya que es la unidad de medida que te da.
Doesn't everyone? lol what can I help you with?
20% because you’d take 1/5 of 100 to find the percentage
Answer: Choice B. k(h(g(f(x))))
For choice B, the functions are k, h, g, f going from left to right.
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Explanation:
We have 4x involved, so we'll need f(x)
This 4x term is inside a cubic, so we'll need g(x) as well.
So far we have
g(x) = x^3
g( f(x) ) = ( f(x) )^3
g( f(x) ) = ( 4x )^3
Then note how we are dividing that result by 2. That's the same as applying the h(x) function
And finally, we subtract 1 from this, but that's the same as using k(x)
This leads to the answer choice B.
To be honest, this notation is a mess considering how many function compositions are going on. It's very easy to get lost. I recommend carefully stepping through the problem and building it up in the way I've done above, or in a similar fashion. The idea is to start from the inside and work your way out. Keep in mind that PEMDAS plays a role.