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Svet_ta [14]
3 years ago
12

LOOK HERE IF YOU ANSWER CORRECTLY YOU PASS 8th grade

Mathematics
1 answer:
Helen [10]3 years ago
5 0

Answer:

$3,700

Step-by-step explanation:

16% of 2500=400

400×3(years)=1200

2500+1200=3700

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myrzilka [38]
What the regression is a conflict
3 0
3 years ago
Find all of the polar coordinates for (6, 31 degrees)
Jobisdone [24]
(6, 31+2pi degrees)
(6, 31+4pi degrees)
(-6, 31-pi degrees)
(6, 31-2pi degrees)

etc. etc.
8 0
2 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)

4 0
3 years ago
Which statements are true?
djyliett [7]

Answer:

All Triangles are closed figure polygons that have three straight sides

Step-by-step explanation:

All 2D shapes are plane figures so this one is false. Three right angles are greater than 180 so this is fake. Polygon and closed figures are shapes with three or more closed sides. And triangles have straight sides.

3 0
2 years ago
Express the following ratios as fractions in their lowest term
ANTONII [103]

Answer:

1/4 is A and 1/20 is B

Step-by-step explanation:

Not much of an explanation needed

7 0
2 years ago
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