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3 years ago

Order the values from least to greatest

2 answers:

8 0

1.) -4 , -2 , 1 , 3 , |-5|

-4 and -2 are first because they are the least, then 1 and 3 are next. The absolute value of -5 is last because it has the greatest value, 5. (The absolute value of a negative number is the same number but positive)

2.) |8| , -5 , 0 , |-9| , 11

The absolute value if 8 is first because it has the least value, -8. (The absolute value of a positive number is the same number but negative) Then -5 and 0 are next. The absolute value of -9 is next because it is actually 9, and 11 is last because it has the greatest value.

Natali5045456 [20]3 years ago

5 0

1. -4 -2 1 3 |-5|
2. -5 0 8 |-9| 11

Theses numbers |-9| are always positive but if it’s outside the box it would be negative for example -|4| is negative

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sleet_krkn [62]

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3 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

2 years ago

What is the basic ratio for 30:42

lianna [129]

30:42 is represented as 30/42
The lowest form for that is 5/7. Hope that helped. Good luck

6 0

2 years ago

Find the area of a triangle bounded by the x-axis the line f(x)=5-5/3x and the line perpendicular to f(x)

viktelen [127]

Answer:

answer fx

Step-by-step explanation:

its fx because it works

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1 year ago

The amount $0.8982 rounded to nearest cent is?

Butoxors [25]

Answer:

It would be rounded to $0.90

Step-by-step explanation:

The 9 would make the 8 go up to $0.90

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3 years ago