Cuál es la receta para 4 persona

Maggie is training to run a 6 mile race. Her training program has her increase her weekly long run each week. This week she ran

Korvikt [17]

Answer:

Maggie will run 5.85 miles next week.

Step-by-step explanation:

Given:

Maggie is training to run a 6 mile race.

This week Maggie ran = 4.5 miles

Also Given:

Next week she will run 1.3 times as far as she did this week.

We need to distance she will run next week.

Now we know Next week she will run 1.3 times as far as she did this week.

It means Distance to be run next week will be equal to 1.3 times distance she ran this week.

Framing in equation form we get;

distance she will run next week = $1.3\times 4.5 = 5.85 \ miles$

Hence Maggie will run 5.85 miles next week.

6 0

3 years ago

Find the absolute maximum and minimum values of f(x, y) = x+y+ p 1 − x 2 − y 2 on the quarter disc {(x, y) | x ≥ 0, y ≥ 0, x2 +

Andreas93 [3]

Answer:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

Step-by-step explanation:

In order to find the absolute max and min, we need to analyse the region inside the quarter disc and the region at the limit of the disc:

Region inside the quarter disc:

There could be Minimums and Maximums, if:

∇f(x,y)=(0,0) (gradient)

we develop:

(1-2x, 1-2y)=(0,0)

x=1/2

y=1/2

Critic point P(1/2,1/2) is inside the quarter disc.

f(P)=1/2+1/2+p1-1/4-1/4=1/2+p1

f(0,0)=p1

We see that:

f(P)>f(0,0), then P(1/2,1/2) is a maximum relative

Region at the limit of the disc:

We use the Method of Lagrange Multipliers, when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant). In our case the constraint are the curves of the quarter disc:

g1(x, y)=x^2+y^2=1

g2(x, y)=x=0

g3(x, y)=y=0

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

Analyse in g2:

x=0;

1-2y=0;

y=1/2

Q(0,1/2) critical point

f(Q)=1/4+p1

We do the same reflexion as for P. Q is a maximum relative

Analyse in g3:

y=0;

1-2x=0;

x=1/2

R(1/2,0) critical point

f(R)=1/4+p1

We do the same reflexion as for P. R is a maximum relative

Analyse in g1:

(1-2x, 1-2y)=λ(2x,2y)

x^2+y^2=1

Developing:

x=1/(2λ+2)

y=1/(2λ+2)

x^2+y^2=1

So:

(1/(2λ+2))^2+(1/(2λ+2))^2=1

$\lambda_{1}=\sqrt{1/2}*-1 =-0.29$

$\lambda_{2}=-\sqrt{1/2}*-1 =-1.71$

$\lambda_{2}$ give us (x,y) values negatives, outside the region, so we do not take it in account

For $\lambda_{1}$ : S(x,y)=(0.70, 070)

and

f(S)=0.70+0.70+p1-0.70^2-0.70^2=0.42+p1

We do the same reflexion as for P. S is a maximum relative

Points limits between g1, g2 y g3

we need also to analyse the points limits between g1, g2 y g3, that means U(0,0), V(1,0), W(0,1)

f(U)=p1

f(V)=p1

f(W)=p1

We can see that this 3 points are minimums relatives.

Conclusion:

We compare all the critical points P,Q,R,S,T,U,V,W an their respective values f(x,y). We find that:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

4 0

3 years ago

You need to buy some chicken for dinner tonight. You found an ad showing that the store across town has it on sale for $2.79 a p

san4es73 [151]

Answer:

B) Going to the close store is cheaper.

Step-by-step explanation:

Cost of going to the next town:

(8.3mi/22mi) × $3.79 = $1.43

Cost of going to the close store:

(2.3mi/22mi) × $3.79 = $0.40

3 0

3 years ago

Find the volume of a cone with a height of 12 and a base diameter of 6.

vichka [17]

Answer: 466.56

Step-by-step explanation:

Volume = pi times radius^2 times height/3

Radius = 1/2 of the diameter

Volume = 3.24 times 6^2 times 12/3

Volume = 3.24 times 36 times 4

Volume = 116.64 times 4

Volume = 466.56

5 0

2 years ago

Answer this question if you wanna be my guy best friend

Charra [1.4K]

Answer:

b. 0.50x+15=25

so the answer is 20

7 0

2 years ago