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2 years ago

Somebody please help

Mathematics

1 answer:

DaniilM [7]2 years ago

6 0

Answer:

3/5x−3/7

Step-by-step explanation:

Simplify the expression.

3/5x−3/7

-hope it helps

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The table below represents Bree's trip from her home to the library and back. On her way to the library, she travels at 6 miles

user100 [1]

The distance in miles from Bree's house to the library is 8 miles.

Step-by-step explanation:

Time to the library from home is d/6 hours

Time spend to reach home from library is d/12 hours

Hence, the equation can be written as:

d/6 +d/12 = 2 hours ( 2 hours is the total time spend to and fro)

Solving the equation:

2d+d= 24 => 3d=24

Now d is d=24/3 =8

Hence the distance from the home to the library is 8 miles.

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4 years ago

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Solve for h. −110=13+3(4h−6)

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−110=13+3(4h−6)
−110=13+12h−18
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3 years ago

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A triangle has two angles with measures 43 and 51.

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In Euclidean geometry, the angles of a triangle add to 180, which means the remaining angle has a measure of 86 degrees.

In spherical geometry, it's impossible to know the measure of the remaining angle precisely without knowing any more info about the triangle. The sum of the interior angles is between 180 and 540 degrees, so the missing angle is somewhere between 87 and 445 degrees.

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3 years ago

Point G is located at (3, −1) and point H is located at (−2, 3). Find y value for the point that is the distance from point G to

-Dominant- [34]

B. 1.67
m : n = 2 : 1
( x, y ) = ( (1*3 + 2*(-2))/(2+1) ; (1 * (-1) + 2 * 3) /(2 + 1) ) =
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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)

4 0

3 years ago

Somebody please help​

Somebody please help