How many even k-digit numbers can be constructed out of the digits 3, 4, 5, 6, and 7? a. Assume first that you may use the same digit again.
b. Assume that you cannot use a digit more than once.

A. 1562 numbers , B. 130 numbers

Step-by-step explanation:

A) Using same digit again

1 digit nos = 2 ( ie 4 , 6 )

2 digit nos = 5  (ie any no) x 2 ( ie 4 ,6 )

Similarly

3 digit nos = 5 x 5 x 2

4 digit nos = 5 x 5 x 5 x 2

5 digit nos = 5 x 5 x 5 x 5 x 2

Adding all 5 equations

Total = 2 + 10 + 50 + 250 + 1250  = 1562

B) Cant use a digit more than once

1 digit nos = 2 ( ie 4 , 6 )

2 digit nos = 4 ( ie any no other than 4 or 6) x 2  ( ie 4 , 6 )

Similarly

3 digit nos = 4 x 3 x 2

4 digit nos = 4 x 3 x 2 x 2

5 digit nos = 4 x 3 x 2 x 1 x 2

Adding all five equations

Total = 2 + 8 + 24 + 48 + 48 = 130

the answer would be d.

step-by-step explanation:

move all variables to the left side and all constants to the right side

(x−3)^2+(y−1)^2=16

this is the form of a circle. use this form to determine the center and radius of the circle.

( x − h ) 2 + ( y − k ) 2 = r 2

match the values in this circle to those of the standard form. the variable   r   represents the radius of the circle,   h   represents the x-offset from the origin, and   k   represents the y-offset from origin.

r = 4

h = 3

k = 1

the center of the circle is found at   ( h , k ) .

center:   ( 3 , 1 )

these values represent the important values for graphing and analyzing a circle.

center:   ( 3 , 1 ).

radius:   4

- 4, v72, 32/5

i think this is right.

hope it