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Semenov [28]
3 years ago
8

Solve a Mixture Word Problem

Mathematics
2 answers:
saw5 [17]3 years ago
6 0

Answer: Orlando mixed five pounds of cereal squares and 25 pounds of nuts.

stich3 [128]3 years ago
3 0

Answer:

5 lbs cereal squares.

25 Ibs nuts.

Step-by-step explanation:

Let x be the weight of nuts in pounds and y be the weight of cereal squares in pounds.

Total weight of the party mix is 30 pounds.

x+y=30              ..... (1)

It is given that,

Cost of nuts = $7 per pound

Cost of cereal = $4 per pound

The cost of mix = $6.50 per pound.

7x+4y=6.50\times 30

7x+4y=195          ..... (2)

Substitute the value of x in equation (2) from equation (1).

7(30-y)+4y=195

210-7y+4y=195

-3y=195-210

-3y=-15

Divide both sides by -3.

y=5

Substitute y=5 in equation (1).

x+5=30

Subtract both sides by 5.

x=30-5

x=25

Therefore, he use 25 lbs of nuts and 5 lbs cereal squares for party mix.

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Step-by-step explanation:

\underline{ \underline{ \text{Given}}}  :

  • \tt{ {A}^{T}  = \begin{bmatrix} 2 &  - 4 \\ 4 & 3 \\ \end{bmatrix}}

\underline{ \underline { \text{To \: Find}}} :

  • \sf{ {A}^{2}  - 5A+ 22I= 0}

\underline{ \underline{ \text{Solution}}} :

The new matrix obtained from a given matrix by interchanging it's rows and columns is called the transposition of matrix. It is denoted by \sf{ {A}^{T}}. Again , Interchange it's rows and columns in order to find ' A '.

\tt{A = \begin{bmatrix} 2 &  4 \\  - 4 & 3 \\ \end{bmatrix}}

Now , LEFT HAND SIDE ( L.H.S )

\tt{ {A}^{2}  - 5A+ 22I}

Here, I refers to identity matrix. A diagonal matrix in which all the elements of leading diagonal are 1 ( unit ) is called unit or identity matrix.

⟼ \begin{bmatrix} 2 &   4 \\  - 4 & 3 \\ \end{bmatrix} \times \begin{bmatrix} 2 &  4 \\ -  4 & 3 \\ \end{bmatrix} - 5 \times \begin{bmatrix} 2 &   4 \\  - 4 & 3 \\ \end{bmatrix} + 22 \times \begin{bmatrix} 1 &   0 \\  0 & 1\\ \end{bmatrix}

⟼ \begin{bmatrix} 2  \times 2 + 4 \times ( - 4)&   2  \times 4 + 4 \times 3 \\  - 4 \times 2 + 3 \times ( - 4) &  - 4  \times 4 + 3 \times 3 \\ \end{bmatrix} - \begin{bmatrix} 10 &   20 \\   - 20& 15 \\ \end{bmatrix} + \begin{bmatrix} 22 &   0 \\  0 & 22 \\ \end{bmatrix}

⟼ \begin{bmatrix} 4 + ( - 16) &   8 + 12 \\   - 8 + ( - 12) &  - 16 + 9 \\ \end{bmatrix} - \begin{bmatrix} 10 &   20 \\   - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 &   0 \\  0 & 22 \\ \end{bmatrix}

⟼ \begin{bmatrix} - 12 &   20\\  - 20&  - 7 \\ \end{bmatrix} - \begin{bmatrix} 10 &   20 \\   - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 &   0 \\  0 & 22 \\ \end{bmatrix}

⟼ \begin{bmatrix}  - 22 &   0 \\  0&  - 22 \\ \end{bmatrix}  + \begin{bmatrix} 22 &   0 \\  0 & 22 \\ \end{bmatrix}

⟼ \begin{bmatrix}  - 22 + 22 &   0 + 0 \\  0 + 0 &  - 22  + 22 \\ \end{bmatrix}

⟼ \begin{bmatrix} 0 &   0\\  0 & 0 \\ \end{bmatrix}

⟼ \sf{0}

RIGHT HAND SIDE ( R.H.S ) : 0

L.H.S = R.H.S [ Hence , proved ! ]

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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4 0
3 years ago
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