C r = (n!)/(r!(n-r)!)
9 C 3 = (9!)/(3!*(9-3)!)
9 C 3 = (9!)/(3!*6!)
9 C 3 = (9*8*7*6!)/(3!*6!)
9 C 3 = (9*8*7)/(3!)
9 C 3 = (9*8*7)/(3*2*1)
9 C 3 = (504)/(6) 9 C 3 = 84

3 0

3 years ago

The set of months in the year that begins with letter B

Fynjy0 [20]

I don't really see how this pertains to Mathematics...

3 0

3 years ago

Please help me with the question please ASAP

erica [24]

Answer:

The ratio of perimeter of ABCD to perimeter of WXYZ = $\frac{2}{3}$

Step-by-step explanation:

First, we have to determine the multiplicative factor of the dimensions for both figures.

Considering sides AB and WX,

multiplicative factor = $\frac{12}{8}$

= 1.5

So that:

XY = 6 x 1.5 = 9

YZ = 7 x 1.5 = 10.5

ZW = 7 x 1.5 = 10.5

Perimeter of ABCD = 6 + 7 + 7 + 8

= 28

Perimeter of WXYZ = 9 + 10.5 + 10.5 + 12

= 42

The ratio of the perimeters of the two quadrilaterals can be determined as;

ratio = $\frac{perimeter of ABCD}{Perirmeter of WXYZ}$

= $\frac{28}{42}$

= $\frac{2}{3}$

The ratio of the perimeter of ABCD to perimeter of WXYZ is $\frac{2}{3}$ .

5 0

3 years ago