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valina [46]
3 years ago
13

What is equivalent to 2x^2+9x+9

Mathematics
1 answer:
melamori03 [73]3 years ago
7 0

(2x + 3) • (x + 3)

2x • x = 2x^2

2x • 3 = 6x

3 • x = 3x

3 • 3 = 9

6x and 3x combine to get 9x, which when all written out is equal to your original trinomial.

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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
Shonda goes to a restaurant with no more than $15 to spend. She buys a burger that costs
andrew11 [14]

Answer:

she has 4.5

and she should´ve got bacon egg and cheese than that burger

5 0
2 years ago
35 POINTS!!! PLEASE HELP !!!!!!!!!!!!!!
lyudmila [28]

Answer:

The function is increasing from x = 0 to x = 1.

Step-by-step explanation:

A <u>function</u> is increasing when the <u>y-value increases</u> as the <u>x-value increases</u>.

A <u>function</u> is decreasing when the <u>y-value decreases</u> as the <u>x-value increases</u>.

From x = -2 to x = -1 the function is decreasing as the y-value decreases as the x-value increases:  

  • x-value -2 to -1 → increase
  • y-value 2 to 0 → decrease

From x = -1 to x = 0 the function is increasing as the y-value increases as the x-value increase:  

  • x-value -1 to 0 → increase
  • y-value 0 to 2 → increase

From x = 0 to x = 1 the function is increasing as the y-value increases as the x-value increase:  

  • x-value 0 to 1 → increase
  • y-value 2 to 8 → increase
7 0
1 year ago
The question is below please help me
Cloud [144]

Answer:

103&(#!92($+#”39214849)?lm

7 0
1 year ago
Find the measure of the missing angle
lapo4ka [179]

Answer:

Approximately 35, or 34.67

Step-by-step explanation:

You can set up a cosine (adjacent over hypotenuse) equation.

cos38 = x/44

Move the 44 over and plug into the calcultaor.

x = 44cos38

(make sure the mode is on degree and not radian!)

8 0
3 years ago
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