The dolphins football team gained 16 yards on their first play, then lost 11 yards on the next play. write an addition expression to represent this situation. find the sum and explain its meaning.

The Dolphins football team gained 16 yards on their first play, then lost 11 yards on the next play. Write an addition expression to represent this situation. Find the sum and explain its meaning.

On the first play, Dolphin's football team gained 16 yards.

On the 2nd play, Dolphin's football team lost 11 yards.

Therefore, the addition expression to represent this situation is:

The sum of this expression is:

Therefore, we can interpret it as: Dolphin's football team gained 5 yards over the 2 plays

Answer : The team Dolphins gained 5 yards over the 2 plays.

Step-by-step explanation:

Given information is -

The Dolphins football team gained 16 yards on their first play

And then lost 11 yards on the next play.

Means they had 2 plays.

When we have two conditions and have to apply addition expression, we put brackets around A and B as they are expressions, and not simply numbers.

Then the addition is supposed to apply to whatever A and B may contain.

So, the addition expression to represent this given situation is:

= 5

Answer : The team Dolphins gained 5 yards over the 2 plays.

  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.