Avery uses a blend of dark chocolate and milk chocolate to make the ice cream topping at her restaurant. she wants to buy 10kg more of dark chocolate than milk chocolate, and she needs 150kg of chocolate in total for her next order. let d be the number of kilograms of dark chocolate he buys and m be the number of kilograms of milk chocolate she buys. which system of equations represents this situation?

Step-by-step explanation:

Let d be the number of kilograms of dark chocolate and m be the number of kilograms of milk chocolate Avery buys.

She wants to buy 10 kg more of dark chocolate than milk chocolate, then

She needs 150 kg of chocolate in total for her next order, so

Hence, the following system of two equations represents this situation:

D

Step-by-step explanation:

The equations are similar and so if we compute the vertex we would see they are the same..

Let's try computing the maximum or minimum point of y= -3(x+2)^2 - 4 and y= 3(x+2)^2 - 4 ;

For y= 3(x+2)^2 - 4 ;

to compute the maximum or minimum point we find dy/dx of the expression equating it to 0; the value of x is determined and we substitute to the original expression for y.

If y is -ve we know it's a minimum graph and if y is +be it's a maximum graph.

From the foregoing;

For y= 3(x+2)^2 - 4 ;

dy/dx = 2× 3 ( x+2) ×1 = 6(x+2) = 6x + 12= 0

6x= -12=>x= -12/6= -2;

We substitute x= -2 in y =3(x+2)^2 - 4; y = 3(-2+2)^2-4 = -4

Since y = -4 is a hence y= 3(x+2)^2 - 4 is a minimum graph.

For y= -3(x+2)^2 - 4 ;

dy/dx = 2× -3 ( x+2) ×1 = -6(x+2) = -6x - 12= 0

-6x= 12=>x= -12/6= -2;

Substituting in the y= -3(x+2)^2 - 4 ;

We have;

We substitute x= -2 in y =-3(x+2)^2 - 4; y = -3(-2+2)^2-4 = -4

y= -4 ;

Since both graphs look like they are minimum that is the vertex is y= -4;

Let's explore a further step

The derivative of the previous derivative;

For y= 3(x+2)^2 - 4;

We means d/dy × [dx/dy] = d/dy [ 6x + 12 ] = 6

Hence y= 3(x+2)^2 - 4 is a maximum graph;

Similarly for y= -3(x+2)^2 - 4 ;

d/dy × [dx/dy]

d/dy [-6x-12 ] = -6

Hence y= -3(x+2)^2 - 4 is a minimum graph;

Conclusion: y= 3(x+2)^2 - 4 is a maximum graph and y= -3(x+2)^2 - 4 is a minimum graph;