Answer:
a) there is s such that <u>r>s</u> and s is <u>positive</u>
b) For any <u>r>0</u> , <u>there exists s>0</u> such that s<r
Step-by-step explanation:
a) We are given a positive real number r. We need to wite that there is a positive real number that is smaller. Call that number s. Then r>s (this is equivalent to s<r, s is smaller than r) and s is positive (or s>0 if you prefer). We fill in the blanks using the bold words.
b) The last part claims that s<r, that is, s is smaller than r. We know that this must happen for all posirive real numbers r, that is, for any r>0, there is some positive s such that s<r. In other words, there exists s>0 such that s<r.
Answer:
4x+12/x^2+9
Step-by-step explanation:
Perimeter
x+3+x+3+x+3+x+3
Area
(x+3)(x+3)
9 x 6 = 54 is this right I'll do distributive property next time
The first thing you do is to find the lowest common multiple of 5 and 7.What you do on the bottom, you must do on the top 25/35 / 28/35. The answer is 25/28. Sara ran 25/28 of the total distance around the track.
