i believe it's 22.5 out of 45 to get exactly .500
Answer:
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Step-by-step explanation:
Let
S = 2b/(b+a)^2 + 2a/(b^2-a^2) factor denominator
= 2b/(b+a)^2 + 2a/((b+a)(b-a)) factor denominators
= 1/(b+a) ( 2b/(b+a) + 2a/(b-a)) find common denominator
= 1/(b+a) ((2b*(b-a) + 2a*(b+a))/((b+a)(b-a)) expand
= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a)) simplify & factor
= 2/(b+a)(b^2+a^2)/((b+a)(b-a)) simplify & rearrange
= 2(b^2+a^2)/((b+a)^2(b-a))
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Answer:
Step-by-step explanation:
To find g(-4) , substitute the value of x that's - 4 into g(x). That is for every x in g(x) , replace it with - 4
We have
We have the final answer as
<h3>g(-4) = 14</h3>
Hope this helps you
Let the number of bike be x and the number of skates be y, then
21x + 20y ≥ 362 . . . (1)
2y = x . . . (2)
Putting (2) into (1), then
21(2y) + 20y ≥ 362
42y + 20y ≥ 362
62y ≥ 362
y ≥ 5.84
The least number of pairs of skates they need to rent each day to make their minimum is 6.
X=4 ........................................
