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2 years ago

Does the point (6, 2) satisfy the equation y = x - -4?

1 answer:

5 0

Answer:
No it does not.

Explanation:

If you plug in (6,2) into the equation it will become false

y = x - -4

2 = 6-(-4)
2 = 6 + 4
2 = 10

So the answer is no it does not.

You might be interested in

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

At the theater, tickets for adults cost $4.75

shutvik [7]

X+y= 26 and 4.75x + 2.25y= 83.50

y=26-x( substitute this into second equation for y and solve for x)

4.75x + 2.25(26-x)= 83.50

4.75x + 58.5-2.25x= 83.50

x= 10

Now solve for y by substituting your answer for x

10+ y= 26

y=16

Therefore, 16 tickets were purchased for kids and 10 for adults.

7 0

3 years ago

Which points are the vertices of the ellipse?

SIZIF [17.4K]

Answer

The points of intersection of the ellipse and its major axis are called its vertices. Here the vertices of the ellipse are A (a, 0) and A′ (− a, 0).

Step-by-step explanation:

4 0

2 years ago

What value of t makes (-2) +(-11) = t a true sentence?

Bas_tet [7]

Answer:

t = -13

Step-by-step explanation:

Given expression: