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nadezda [96]
3 years ago
13

8 (a+4) + 4 = 36 + 8a

Mathematics
2 answers:
CaHeK987 [17]3 years ago
6 0

Answer:

0a = 0

Step-by-step explanation:

Expand

8a + 32 + 4 = 36 + 8a

Rearrange equation

8a - 8a = 36 - 32 - 4

0a = 0

ololo11 [35]3 years ago
4 0

Answer:

a= 0

Step-by-step explanation:

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:

<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
Find the missing length to the nearest ten
Allushta [10]
It is 9

..............................................................................................................................

6 0
3 years ago
Plz help!!!! Need quick. 65 points
DanielleElmas [232]

Answer:

The area of the mat is 2/3 m^2

Step-by-step explanation:

Area of a rectangle is length times width or

A = l x w

8/9 x 3/4 =

24/36 =

2/3

4 0
3 years ago
Read 2 more answers
Which of the following shows of the slope of the line and one point on the line?
uysha [10]
Answer is choice D

This is the only answer choice with a negative slope, so we don't even need to look at the point really. Though if you wanted, you can confirm that the point (1,-3) is on the diagonal line. This point is shown in red in the attached image.

To find that the slope is -6/5, you pick two points on the diagonal line and use the slope formula. Two points you can use are (1,-3) and (-4,3).
The slope formula is m = (y2-y1)/(x2-x1)

Note: The negative slope is because we move downhill as we move from left to right along the diagonal line.

3 0
3 years ago
The average temperature of the week is 80 degrees. The first 3 days have an average of 78 and the average of the last 3 days is
Tju [1.3M]

Answer:

80

Step-by-step explanation:Solution:

step 1 Address the formula, input parameters & values.

Input parameters & values:

The given numbers are 78 & 80

step 2 Find the sum of the given two numbers.

sum = 78 + 80 = 158

step 3 Divide the sum by 2 to get the average.

average = 158/2

= 79

Thus, 79 is an average of positive integers 78 and 80.

3 0
3 years ago
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