The numerator and denominator of a fraction are in the ratio of 3 to 5. if the numerator and denominator are both increased by 2, the fraction is now equal to 2/3 . if n = the numerator and d = the denominator, which of the following systems of equations could be used to solve the problem?

The two equation which can be used to solve the problem are:

5n - 3d = 0 and 3n - 2d = -2

The numerator of the fraction is "n" and the denominator of the fractions is "d"

The numerator and denominator of a fraction are in the ratio of 3 to 5

numerator : denominator = 3 : 5

On cross-multiplication we get,

5n = 3d

This is our first equation

The numerator and denominator are both increased by 2, the fraction is now equal to 2/3

The new numerator will be n + 2 and denominator will be d + 2

On cross multiplication we get,

3(n + 2) = 2(d + 2)

3n + 6 = 2d + 4

This is our second equation

Thus the two equation which can be used to solve the problem are:

5n - 3d = 0 and 3n - 2d = -2

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

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