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inessss [21]
3 years ago
7

Someone help. Will give brainliest

Mathematics
2 answers:
Alexandra [31]3 years ago
5 0
3/4 = .75 and 1 1/2 = 1.5

The least number on that line is -1 and the greatest is 1.5

The order is -1, -.9, 3/4, 1.1, and 1 1/2
podryga [215]3 years ago
5 0

Answer:

Step-by-step explanation:

3/4 is 0.75 and 1 1/2 is 1.5

plot the numbers:

1.1, 3/4 , -0.9 , -1, 1 1/2 ( the line is not up to scale)

_|_______|__|_______|_____|___|__|_____|____|_

 -2           -1  -0.9            0       3/4    1  1.1      1 1/2       2

the number the least is -1

the number greatest  1 1/2

the correct order from least to greatest is  

-1  , -0.9 , 3/4  , 1.1 , 1 1/2  

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Complete #2 for me and show proof , will give brainliest to whose right .​
Stells [14]

Answer:

<u>X</u> <u>Y</u> <u>(</u><u>X,</u><u>Y)</u>

-4 14. (-4,14)

-3 12. (-3,12)

-2 10. (-2,10)

-1 8. (-1,8)

0 6 (0,6)

1 4. (1,4)

2 2. (2,2)

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2 years ago
If a Roadrunner ran 80 miles and 4 hours, how long will it take the Roadrunner to run 60 miles? Explain or show how you determin
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10 because 80 divided by 20 is 40 then that divided by 10
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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:

<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 answer is it for it to be simplified completely?
mamaluj [8]

Answer:

4

Step-by-step explanation:


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4 years ago
I just need help with this one problem
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Nicole is 36 years old
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3 years ago
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