the eagle trucking company must deliver 7/8 ton of cement blocks and 5/8 ton of bricks how much will this load weigh?
1 answer:
This load will weigh 1.5 TONS ALTOGETHER.
So, the truck is delivering a
ton of cement blocks, and a 
of bricks. To find how much this load weighs, we'd have to add the two fractions together, which proves to be quite easy, because the denominators are already the same, so we wouldn't have to change anything.
+ = 
And now, we would want to convert that fraction into a whole number, or at least a decimal, to find how many tons the truck is carrying. So we would divide. Once the work is done, here is what would happen:
12 ÷ 8 = 1.5
And there's your answer. The truck is carrying 1.5 TONS.
So, the truck is delivering a
And now, we would want to convert that fraction into a whole number, or at least a decimal, to find how many tons the truck is carrying. So we would divide. Once the work is done, here is what would happen:
12 ÷ 8 = 1.5
And there's your answer. The truck is carrying 1.5 TONS.
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h(x) = ((x + 2)² - 7x - 8) / (x + 2)
h(x) = ((x + 2)² - 7 (x + 2) - 14 - 8) / (x + 2)
h(x) = ((x + 2)² - 7 (x + 2) - 22) / (x + 2)
h(x) = (x + 2) - 7 - 22/(x + 2)
h(x) = x - 5 - 22/(x + 2)
An oblique asymptote of h(x) is a linear function p(x) = ax + b such that
In the simplified form of h(x), taking the limit as x gets arbitrarily large, we obviously have -22/(x + 2) converging to 0, while x - 5 approaches either +∞ or -∞. If we let p(x) = x - 5, however, we do have h(x) - p(x) approaching 0. So the oblique asymptote is the line y = x - 5.