If p, q are zeroes of polynomial f(x) =2x^2-7x+3,find the value of p^2+q^2

p^2 + q^2 = 22/9.

Step-by-step explanation:

For a quadratic function ax^2 + bx + c if the zeroes are A and B then

A + B = -b/a and AB =  c/a.

2x^2 - 7x + 3

Now p and q are the zeroes of the above function so

p + q =  7/3 and pq = 3/2.

Now p^2 + q^2 = (p + q)^2 - 2pq

= (7/3)^2 - 2* 3/2

=  49/9 - 3

= 49/4 -  27/9

= 22/9.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

correct answer is b. 27/8

step-by-step explanation: