Which of the choices best describes this arithmetic sequence?<br> -3, 1, 5, 9, 13,...

What is 7 1/7 - 3 5/7

Maru [420]

Answer:

Step-by-step explanation:

7 1/7- 3 5/7= 3 3/7

3 3/7 can also be 3.42 and 24/7

7 0

3 years ago

De un grupo de 22 estudiantes hay 13 que practican natación y 10 practican atletismo y 2 que no practican nada . ¿Cuántos practi

Goshia [24]

Usando conjuntos de Venn, se encuentra que 7 estudiantes practican solo atletismo.

Los conjuntos son:

Conjunto A: Estudiantes que practican natación.

Conjunto B: Estudiantes que practican atletismo.

Hay que:

13 que practican natación, o sea, $A = 13$ .

10 practican atletismo, o sea, $B = 10$ .

2 que no practican nada, o sea $A \cup B = 22 - 2 = 20$ .

La cantidad que practica ambos es:

$A \cap B = A + B - A \cup B$

$A \cap B = 13 + 10 - 20$

$A \cap B = 3$

Solo atletismo es:

$B - A \cap B = 10 - 3 = 7$

7 estudiantes practican solo atletismo.

Más informaciones a cerca de conjuntos de Venn pueden ser obtenidas en brainly.com/question/24388608

6 0

2 years ago

Find the absolute maximum and minimum values of the following function on the specified region R.

yan [13]

Answer:

Step-by-step explanation:

Since it is said that the region R is a semicircular disc, we asume that the boundaries of the region are given by $-2\leq x \leq 2, 0\leq y \leq \sqrt[]{4-x^2}$ . First, we must solve the optimization problem without any restrictions, and see if the points we get lay inside the region of interest. To do so, consider the given function F(x,y). We want to find the point for which it's gradient is equal to zero, that is

$\frac{dF}{dx}=9y=0$

$\frac{dF}{dy}=9x=0$

This implies that (x,y) = (0,0). This point lays inside the region R. We will use the Hessian criteria to check if its a minimum o r a maximum. To do so, we calculate the matrix of second derivates

$\frac{d^2F}{dx^2} = 0 = \frac{d^2F}{dy^2}$

$\frac{d^2F}{dxdy} = 9 = \frac{d^2F}{dydx}$

so we get the matrix

$\left[\begin{matrix} 0 & 9 \\ 9 & 0\end{matrix}\right]$

Note that the first determinant is 0, and the second determinant is -9. THis tell us that the point is a saddle point, hence not a minimum nor maximum.

Since the function is continous and the region R is closed and bound (hence compact) the maximum and minimum must be attained on the boundaries of R. REcall that when $-2\leq x \leq x$ and y=0 we have that F(x,0) = 0. So, we want to pay attention to the critical values over the circle, restricting that the values of y must be positive. To do so, consider the following function

$H(x,y, \lambda) = 9xy - \lambda(x^2+y^2-4)$ which consists of the original function and a function that describes the restriction (the circle x^2+y^2=4), we want that the gradient of H is 0.

Then,

$\frac{dH}{dx} = 9y-2\lambda x =0$

$\frac{dH}{dx} = 9x-2\lambda y =0$

$\frac{dH}{d\lambda} = x^2+y^2-4 =0$

From the first and second equation we get that

$\lambda = \frac{9y}{2x} = \frac{9x}{2y}$

which implies that $y^2=x^2$ . If we replace this in the restriction, we have that $x^2+x^2 = 2x^2 = 4$ which gives us that $x=\pm \sqrt[]{2}$ . Since we only care for the positive values of y, and that $y=\pm x$ , we have the following critical points $(\sqrt[]{2},\sqrt[]{2}), (-(\sqrt[]{2},\sqrt[]{2})$ . Note that for the first point, the value of the function is