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3 years ago

Only answer if your 100% sure What equation is shown in the graph?

Mathematics

2 answers:

elena-s [515]3 years ago

6 0

(0,2) and (1,1)
slope = (2 - 1)/(0 -1) = -1
y intercept b = 2
equation
y = -x + 2

answer
equation
y = -x + 2

dedylja [7]3 years ago

6 0

Slope is -1 because from (0,2) to (1,1), that's m = slope = ( rise ) / ( run ) = ( 1 - 2 ) / (1 - 0) = -1 / 1 = -1

And y-intercept is y=2 at x=0, so b = 2

Putting it together:

y = mx+b

y = (-1)x + (2)

Final answer: y = -x + 2

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Answer:

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Step-by-step explanation:

9+n/4 = -7

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9+n= -28

Subtract 9 from both sides

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The probability that an archer hits the target when it is windy is 0.4; when it is not windy, her probability of hitting the tar

stiks02 [169]

Answer:

x = 6

Step-by-step explanation:

The probability that this archer hits her first shot is given by t the probability of being windy and that she hits the shot added to the probability of not being windy and that she hits the shot:

$P(H_{first}) = P(H_w)*P(W)+P(H_{nw})*(1-P(W))\\P(H_{first})=0.4*0.3+0.7*(1-0.3)\\P(H_{first})=0.12+0.49 = 0.61$

If the probability that she hits the target on her first shot is 0.x1, the value of x is:

$P(H_{first})=0.61=0.x1\\x=6$

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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)

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3 years ago