Hypotenuse by Pythagoras theorem = 407.83.
Answer:
There is an asymptote at x = 0
There is an asymptote at y = 23
Step-by-step explanation:
Given the function:
(23x+14)/x
Vertical asymptote is gotten by equating the denominator to zero
Since the denominator is x, hence the vertical asymptote is at x = 0. This shows that there is an asymptote at x = 0
Also for the horizontal asymptote, we will take the ratio of the coefficient of the variables in the numerator and denominator
Coefficient of x at the numerator = 23
Coefficient of x at the denominator = 1
Ratio = 23/1 = 23
This means that there is an asymptote at y = 23
Resposta:
Primer rectangle:
Amplada = 11
Longitud = 14
Segon rectangle:
Amplada = 12
Longitud = 15
Tercer rectangle:
Amplada = 13
Longitud = 16
Explicació pas a pas:
Donat que:
Primer rectangle:
Amplada = x
Longitud = x + 3
2n rectangle:
Augment de la dimensió d'1 cm respecte al primer rectangle;
Amplada = x + 1
Longitud = x + 4
3r rectangle:
Augment de la dimensió de 2 cm respecte al primer rectangle;
Amplada = x + 2
Longitud = x + 5
Suma dels tres perímetres del rectangle:
Perímetre d'un rectangle: 2 (l + O)
Primer rectangle:
2 (x + x + 3) = 2 (2x + 3) = 4x + 6
2n:
2 (x + 1 + x + 4) = 2 (2x + 5) = 4x + 10
3r:
2 (x + 2 + x + 5) = 2 (2x + 7) = 4x + 14
Suma de perímetres = 162
(4x + 6 + 4x + 10 + 4x + 14) = 162
12x + 30 = 162
12x = 162 - 30
12x = 130
x = 11
Per tant,
Primer rectangle:
Amplada = 11
Longitud = 11 + 3 = 14
2n rectangle:
Amplada = 11 + 1 = 12
Longitud = 11 + 4 = 15
3r rectangle:
Amplada = 11 + 2 = 13
Longitud = 11 + 5 = 16
G(x) = (x+4)^4 is the answer. It's D. last choice
G(x) = (x+4)^4 shifted F(x) = x^4 4 units to the left
