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4 years ago

The graph of a quadratic function f(x) = x2 – 6x + 5 is shown here. What is its axis of symmetry?

1 answer:

5 0

If you put it in vertex form, it is easy to see the axis of symmetry.

f(x)-5=x^2-6x

f(x)-5+9=x^2-6x+9

f(x)+4=(x-3)^2

f(x)=(x-3)^2-4 so any value of x other than 3 increases the value of f(x), therefore the axis of symmetry is the vertical line x=3.

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Serjik [45]

Try Photomath; to see if it helps

7 0

3 years ago

Is 6200 ft greater or less than 1 mile 900 ft

Sphinxa [80]

1 mile is equal to 5280 feet

Add 900 to it:

5280 + 900 = 6180

6180 is less than 6200, so 1 mile and 900 feet is less than 6200 feet.

7 0

3 years ago

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Please help ill mark brainliest

anyanavicka [17]

The value of the missing length is 20

Let the missing length be x

From Parallel lines and transversal theorem, we have that

If three or more parallel lines intersect two transversals, then they cut off the transversals proportionally.

Then, we can write that

$\frac{x}{15} =\frac{16}{12}$

∴ $x = \frac{15\times 16}{12}$

x = 20

Hence, the value of the missing length is 20

Learn more here: brainly.com/question/24258397

6 0

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

Find the product. 4x - 3y(-2x 2y)

liubo4ka [24]

(4x + 3y)² =

(4x)² + 2 (4x) (3y) + (3y)² = 

16x² + 24xy + 9y²

4 0

3 years ago

Read 2 more answers