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JulijaS [17]
2 years ago
9

What is the area of this polygon?

Mathematics
2 answers:
kolezko [41]2 years ago
7 0
192cm squared. 17x10=170, 4x11= 44, 44/2= 22, 170+22= 192 cm squared
elena55 [62]2 years ago
3 0
A= 192 cm squared
Why?
17 x 10= 170
4 x 11= 44
44 divided by 2 =22
170+ 22= 192 cm squared
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Premium steak costs $18 per pound; how much steak cost per ounce ?
Angelina_Jolie [31]

Answer: $1.13 or $1.125

Step-by-step explanation:

divide 18 by 16

8 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:

<u>Region inside the quarter disc:</u>

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

<u>Region at the limit of the disc:</u>

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

<u>Analyse in g2:</u>

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

<u>Analyse in g3:</u>

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

<u>Analyse in g1:</u>

(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

<u>Points limits between g1, g2 y g3</u>

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.

<u>Conclusion:</u>

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)

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Whats the value of 8:3 ?
umka2103 [35]

Answer:

As a mixed number, 8/3 is equal to 2-2/3.

Step-by-step explanation:

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Answer:

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Cal's score was a third of Able's score, so that means c = a/3. And since a = 6b, that means c = 6b / 3 = 2b.

Thus, the ratio of Able's score to Ben's score to Cal's score, a:b:c, is 6:1:2, because c is twice as much as b and a is 6 times as much as b.

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Musya8 [376]

Answer: 1,4 is the answer

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2 years ago
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