Answer:
The answer is at the bottom!!
Step-by-step explanation:
And All of those Natural Numbers will be starting with 1 so we will be having other Numbers as well on which No Number will be mapped like 2 or 211 or 79 So This Means Set of Natural Numbers is Grater then Real Numbers Between 0 and 1. So Set of Real Numbers Between 0 and 1 is Countably Infinite.
Hope this helps!!
Answer:
Step-by-step explanation:
hola
Cuando dos numeros tienen el mismo signo la respuesta siempre es positiva. cuando un signo es positivo y e lotro es negativo la respuesta siempre sera negativa. Esta es la regla para multiplicar numeros con signos diferentes. el orden de los signos no importa en multiplicacon siempre va ha ser la misma respuesta.
Taking

and differentiating both sides with respect to

yields
![\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B3x%5E2%2By%5E2%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B7%5Cbigg%5D%5Cimplies%206x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D0)
Solving for the first derivative, we have

Differentiating again gives
![\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B6x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B0%5Cbigg%5D%5Cimplies%206%2B2%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%2B2y%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D0)
Solving for the second derivative, we have

Now, when

and

, we have
Answer:a=2.1 :)
Step-by-step explanation:a+(-1.6)=-3.7
+1.6 +1.6
a=-2.1
Answer:
Step-by-step explanation:
Given that:

Since the given function is equal to zero, then the function :

where;
and



Thus; f(1) < 0 and f(2) > 0