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:
−7r^(2)+12r+10x−18
Step-by-step explanation:
Grab the original equation: 10x+7r−r^(2)−6r^(2)−18+5r
For subtraction bits, treat them as negatives: 10x+7r+−r^(2)+−6r^(2)+−18+5r
Combine like terms: (−r^2+−6r^2)+(7r+5r)+(10x)+(−18)
Simplify that, and you get your final answer: −7r^(2)+12r+10x+−18
Answer:
-21, -20, -19, --18, -17, -16, -15, -14, -13, -12, -11
Step-by-step explanation:
There is no step by step explanation.
Use the formula 1/2(base1+base2)height. Substitute the values.
