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Ugo [173]
3 years ago
10

Kenneth is making chocolate cakes. For each cup of milk uses, he needs to use one and 3/4 cups of flour. For each cup of flour h

e uses, he needs to use 3/7 cup of cocoa powder.kenneth is making enough cakes that he needs to use 4 cups of milk. How many cups of cocoa powder does Kenneth need to use?
Mathematics
1 answer:
Klio2033 [76]3 years ago
8 0

Answer:

see below  PLEASE GIVE BRAINLIEST

Step-by-step explanation:

1 milk = 1 3/4 cup of flour

1 flour = 3/7 cup of cocoa powder


4 cups of milk = 1 3/4 x 4 = 7/4 x 4/1 = 7 cups of flour

7 cups of flour = 7 x 3/7 =  7/1 x 3/7 = 3 cups of cocoa powder

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Answer:

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Step-by-step explanation:

3x + y = 26

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2 years ago
Which of the following equations listed below has a radius of 3?
Evgesh-ka [11]

Answer:

C.

Step-by-step explanation:

WE need to convert the given equations  into the standard form

(x - a)^2 + (y - b)^2 = r^2 where r = the radius.

x^2 + y^2 + 12x - 20y+ 132 = 0

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2 years ago
Read 2 more answers
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

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<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)

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Equating (a) and (b) equations

we get         y z = \frac{xz}{9}

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Equating (b) and (c) equations

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cancel 'x' value on both sides , we get

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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 :-

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

              9 + 9 + 4(9/4) = 27

           27 =27

so satisfied equation

     

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