Answer:
96 cm^2
Step-by-step explanation:
Total Surface Area of a square pyramid:
A = L + B = a2 + a√(a2 + 4h2))
A = a(a + √(a2 + 4h2))
<u>Answer:</u>
x = 4 (extraneous solution)
<u>Step-by-step explanation:</u>

This solution is extraneous. Reason being that even if it can be solved algebraically, it is still not a valid solution because if we substitute back
, we will get two fractions with zero denominator which would be undefined.
<u>Find fraction of snappers and catfish:</u>
1/3 + 2/9 = 3/9 + 2/9 = 5/9
<u>Find fraction of goldfish:</u>
1 - 5/9 = 4/9
4/9 = 40 goldfish
1/9 = 40 ÷ 4 = 10
9/9 = 10 x 9 = 90
Answer: 90 fish in the tank
El volumen <em>remanente</em> entre la esfera y el cubo es igual a 30.4897 centímetros cúbicos.
<h3>¿Cuál es el volumen remanente entre una caja cúbica vacía y una pelota?</h3>
En esta pregunta debemos encontrar el volumen <em>remanente</em> entre el espacio de una caja <em>cúbica</em> y una esfera introducida en el elemento anterior. El volumen <em>remanente</em> es igual a sustraer el volumen de la pelota del volumen de la caja.
Primero, se calcula los volúmenes del cubo y la esfera mediante las ecuaciones geométricas correspondientes:
Cubo
V = l³
V = (4 cm)³
V = 64 cm³
Esfera
V' = (4π / 3) · R³
V' = (4π / 3) · (2 cm)³
V' ≈ 33.5103 cm³
Segundo, determinamos la diferencia de volumen entre los dos elementos:
V'' = V - V'
V'' = 64 cm³ - 33.5103 cm³
V'' = 30.4897 cm³
El volumen <em>remanente</em> entre la esfera y el cubo es igual a 30.4897 centímetros cúbicos.
Para aprender más sobre volúmenes: brainly.com/question/23940577
#SPJ1

Employ a standard trick used in proving the chain rule:

The limit of a product is the product of limits, i.e. we can write

The rightmost limit is an exercise in differentiating

using the definition, which you probably already know is

.
For the leftmost limit, we make a substitution

. Now, if we make a slight change to

by adding a small number

, this propagates a similar small change in

that we'll call

, so that we can set

. Then as

, we see that it's also the case that

(since we fix

). So we can write the remaining limit as

which in turn is the derivative of

, another limit you probably already know how to compute. We'd end up with

, or

.
So we find that