For this case we must solve each of the functions.
We have then:
f (x) = x2 - 9, and g (x) = x - 3
h (x) = (x2 - 9) / (x - 3)
h (x) = ((x-3) (x + 3)) / (x - 3)
h (x) = x + 3
f (x) = x2 - 4x + 3, and g (x) = x - 3
h (x) = (x2 - 4x + 3) / (x - 3)
h (x) = ((x-3) (x-1)) / (x - 3)
h (x) = x-1
f (x) = x2 + 4x - 5, and g (x) = x - 1
h (x) = (x2 + 4x - 5) / (x - 1)
h (x) = ((x + 5) (x-1)) / (x - 1)
h (x) = x + 5
f (x) = x2 - 16, and g (x) = x - 4
h (x) = (x2 - 16) / (x - 4)
h (x) = ((x-4) (x + 4)) / (x - 4)
h (x) = x + 4
1. $1.23 per pair
A dozen is 12 so you do 12x7=84
Then you do 103.32 divided by 84 which gives you 1.23
2. His error was doing 103.32 divided by 7 because he thought it was only 7 pairs but it’s 7 dozen.
Answer:
I think the ans is option 'a'
Final Answer:
Corresponding Angles Theorum; ∠AGF and ∠EHD are congruent
So long as the perimeters are the same, rectangles and squares share the same area. For example, a square that is 2m by 2m across is 4m squared. A rectangle of 4m by 1m across is still 4m squared.
Therefore all we want to do here is see how big we can make our “square” perimeter using the creek. We have three sides to spread 580ft across, therefore if we divide this by 3, we get 193.3ft of fencing per side. If we then square this figure, we will then get the maximum possible area, which comes to 37,377ft squared. (That’s a huge garden).
C (3rd Answer)
1st term is 900 and as n increases (number of terms increase), each time it goes down by 50 if n is greater than 1.
