Need finding a constant k at which the function is continuous everywhere.

k = -6/35

Step-by-step explanation:

To make the function continuous

kx^2 = x+k

These must be equal where the function is defined for two different intervals

This is at the point x=-6  so let x=-6

k(-6)^2 = -6+k

36k = -6+k

Subtract k from each side

36k-k = -6+k-k

35k = -6

Divide by 35

35k/35 = -6/35

k = -6/35

b is the final answer! : )

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐