Answer:
the solution set is r>7
step by step:
Subtract 3 from both sides
3−3r−3<−2r−4−3
Simplify
−3r<−2r−7
Add 2r to both sides
−3r+2r<−2r−7+2r
Simplify
−r<−7
Multiply both sides by −1 (reverse the inequality)
(−r)(−1)>(−7)(−1)
Simplify
r>7
I will do the first one only. The second question is done the same way.
Replace every x you see in the function with 2b and do the math.
f(x) = x^2 + 2x + 1
Let x = 2b
f(2b) = (2b)^2 + 2(2b) + 1
f(2b) = 4b^2 + 4b + 1
Did you follow? That's all there is to it.
You can tell that the equal share is more than 1 whole when the numerator is greater than the denominator.How? As we all know that in order to convert a fraction to a decimal number, we need to divide the numerator by the denominator.Now, if the numerator is less than the denominator, expect that the digit is now less than 1 whole. But if the numerator is greater than the denominator, for sure, it is more than 1 whole for example.=> 5/4 , where 5 is the numerator and 4 is the denominator<span>=> 1 and 1/4 or 1.25</span>
