In a sequence, each number after the first number is obtained by adding 4 to the previous number. If the first number in the seq

Answer:

y

y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction y = StartFraction 3 menos 6 StartRoot 2 EndRoot Over 4 EndFraction

Explicación paso a paso:

La ecuación cuadrática que tenemos es (4y - 3)² = 72

Debemos encontrar el valor de y.

Ahora, 4y - 3 = ± 6√2

⇒ 4y = 3 ± 6√2

⇒ y

Por lo tanto, las soluciones son y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction y y = StartFraction 3 menos 6 StartRoot 2 EndRoot Over 4 EndFraction (Respuesta)

Answer:

x≥(500-87)/40

Step-by-step explanation:

You require 500 lbs and you're starting with 87, so you must acquire at least 413 bags. Since they are sold in 40 pound bags, you divide by 40. You must round up to ensure you have enough bags to meet the 500 lbs needed.

1 = 25 times 1 + 8 = 33 2 = 25 times 2 + 8 = 58 3 = 25 times 3 + 8 = 83 4 = 25 times 4 + 8 = 108 5 = 25 times 5 + 8 = 133 B 10 = 25 times 10 + 8 = 258 C 233 = 25 times x + 8 233 = 25x + 8 233 - 8 = 25x 225 = 25x 225 / 25 = 25x / 25 9 = x there are 9 classes.

Answer:

We have defined functions:

f(x) = IxI + 1

g(x) = 1/x^3.

Currently, it is evident that the composite functions are not commutative.

How can we demonstrate this?

To determine if two composite functions are commutative, the following must hold true:

f(g(x)) = g(f(x))

One could apply brute force (simply substituting values to see if the composite functions commute),

but I will opt for a more sophisticated approach.

There are two notable observations:

g(x) has a point of discontinuity at x = 0.

Thus:

f(g(x)) = I 1/x^3 I + 1

remains discontinuous at x = 0, whereas:

g(f(x)) = 1/(IxI + 1)^3

shows that the denominator IxI + 1 can never reach zero.

At this point, there is no discontinuity.

Consequently, the composite functions cannot be commutative.