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stiv31 [10]
2 years ago
13

To prove that ΔAED ˜ ΔACB by SAS, Jose shows that StartFraction A E Over A C EndFraction = StartFraction A D Over A B EndFractio

n.
Triangle A E D is shown. Line segment B C is drawn from side A D to A E to form triangle A C B.

Jose also has to state that
Mathematics
1 answer:
stich3 [128]2 years ago
7 0

The other equality which must be stated by Jose is that angles BAC and BAE are congruent and their measures are equal.

<h3>What other congruence statements must Jose state?</h3>

It follows from the task content that Jose is trying to prove the congruence of both triangles by means of the Side-Angle-Side congruence theorem.

It therefore follows that since, Jose has identified that the ratio of corresponding sides are equal as indicated in the task content, the equality which Jose has to state is the angle congruence equality.

Read more on congruence theorem;

brainly.com/question/3168048

#SPJ1

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Use Lagrange multipliers to find the volume of the largest rectangular box in the first octant with three faces in the coordinat
tensa zangetsu [6.8K]

Answer:

The volume of the largest rectangular box (V) = 81/4

Step-by-step explanation:

<u>Step 1</u>:-

Given volume of the largest rectangular box in the first octant

V = l b h

let (x ,y, z) be the one vertex in the given plane

V = f(x, y, z) = x y z

given plane.   Ф (x, y, z) = x + 9y + 4z = 27 ........(1)

<u>Step( ii):</u>

By using  Lagrange multipliers

Suppose it is required to find the extreme for the function f(x, y, z) subject to the condition Ф (x, y, z) =0

Form Lagrange function F(x, y, z) = f(x, y, z) + λ  Ф (x, y, z) where λ is called the Lagrange multipliers

F(x, y, z) = x y z+ λ ( x + 9y + 4z - 27) ......(2)

Obtain the equations are δ F / δ x = 0

                                  δ f / δ x +λ  δ Ф  / δ x =0

                           ⇒ y z + λ ( 1) = 0

                          ⇒ y z = - λ ......(a)

Obtain the equations are δ F / δ y = 0

                            δ f / δ x +λ  δ Ф  / δ x =0

                        ⇒ x z + λ ( 9) = 0

                        ⇒ \frac{xz}{9} = - λ .......(b)

Obtain the equations are δ F / δ z = 0

                                  δ f / δ x +λ  δ Ф  / δ z =0

                                   x y + λ ( 4) = 0

                                    ⇒ \frac{xy}{4} = - λ .......(c)

<u>Step (iii):-</u>

Equating (a) and (b) equations

we get         y z = \frac{xz}{9}

cancel 'z' value we get  x = 9y  .......(d)

Equating (b) and (c) equations

we get     \frac{xy}{4}  = \frac{xz}{9}

cancel 'x' value on both sides , we get

              4z =9y  ......(e)

substitute (d) (e) values in equation (1)  we get

x + 9y + 4z -27 = 0

9y + 9y + 9y -27 =0

27y -27 =0

<u> y = 1</u>

substitute y =1 in x = 9y

<u>x = 9</u>

substitute y =1 in 4z =9y

<u>z = 9 /4</u>

therefore the dimensions are x =9 , y=1 and z = 9 /4

<u>Conclusion</u>:-

The largest volume of the rectangular box V = x y z

substitute x =9 , y=1 and z = 9 /4

V = 9(1)(9/4) =81/4

The largest volume of the rectangular box (V) = 81/4

Verification :-

Given plane x + 9y + 4z = 27

substitute x =9 , y=1 and z = 9 /4

              9 + 9 + 4(9/4) = 27

           27 =27

so satisfied equation

     

3 0
3 years ago
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