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Zarrin [17]
3 years ago
13

⦁ Evaluate ⦁ 6! ⦁ 8P5 ⦁ 12C4

Mathematics
2 answers:
lisabon 2012 [21]3 years ago
7 0
6! = 6 * 5 *4*3*2*1 =720

8p5 = 8*7*6*5*4 = 6720

12C4 = (12p4)/4! = (12*11*10*9)/4*3*2*1 = 495
Vesnalui [34]3 years ago
6 0
N! = n(n-1)(n-2)(n-3)....
∴6! = 6×5×4×3×2×1 = 720
⁸P₅ = \frac{8!}{(8-5)!} =  \frac{8*7*6*5*4*3*2*1}{3*2*1} = 6720
¹²C₄ = \frac{12!}{4!*(12-4)!} =  \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4!*8!}
       = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1*8*7*6*5*4*3*2*1} = 495

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Answer:

Dennis need 288.91 square inches of construction paper for his project.

Step-by-step explanation:

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So the area of 1st rectangle is

A_1 = 12\times \frac{61}{4} = 183 \: in^{2}

He needs to cut another rectangle that is 10 1/3"" x 10 1/4""

The symbol " means inches

First convert the mixed fraction to improper fraction

10\frac{1}{3} = \frac{(10\times3) + 1}{3} = \frac{31}{3}

10\frac{1}{4} = \frac{(10\times4) + 1}{4} = \frac{41}{4}

So the area of 2nd rectangle is

A_2 = \frac{31}{3}\times \frac{41}{4} = 105.91 \: in^{2}

The total construction paper needed for this project is

A = A_1 + A_2\\\\A = 183 + 105.91\\\\A = 288.91 \: in^{2}

Therefore, Dennis need 288.91 square inches of construction paper for his project.

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