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Nataly_w [17]
2 years ago
8

Find one value of x for which h(x)=4 and find h(4)

Mathematics
1 answer:
VARVARA [1.3K]2 years ago
7 0
Hp666799vvhnvrjnbbhhgvvvvvvvvvvvvgggvvvvvvvvvvvvv
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Which equation could be solved using the graph above?
nevsk [136]

Answer:

Option 3

Step-by-step explanation:

we are given the graph of a parabola

with vertex at (0,-1) and symmetrical about y axis and also open up.

Hence the equation of the graph would be transformation of y=x^2

by a vertical shift of 1 unit down.

So new equation after transformation would be

y+1=x^2

Ory=x^2-1

Hence option 3 is right choice.

Verify:

Put Solving the equation we get

x^2=1

x=±1

In the graph x intercepts are the same as -1 and 1

Hence our answer is right.

8 0
3 years ago
Read 2 more answers
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
Please please help!!
Olenka [21]

Answer:

The formula T= 10d +20

A) what does each term on the right side of the equation represent?

  • 10d⇒ 10 degrees increase per 1 km and 20 deg surface temperature

B) Estimate the depth where the temperature is 60 degrees C.

  • 60=10d+20
  • 10d=40
  • d=4 km

C) What is the approximate temperature at a depth of 4km?

  • T=10*4+20
  • T=60 deg
4 0
2 years ago
The sum of two consecutive even integers is -298. What are the integers?
Allushta [10]
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4 0
3 years ago
25 POINTS NEED QUICK What is the equation of the line of best fit given by a regression calculator for the data in the table?
Vaselesa [24]

Answer:

B

Step-by-step explanation:

3 0
3 years ago
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