x + 2x + 8 − x + 6 = <u>2x + 14</u>
Answer:
45
Step-by-step explanation:
AEF is a similar triangle to ABC. that means it has the same angles, and the sides (and all other lines in the triangle) are scaled from the ABC length to the AEF length by the same factor f.
now, what is f ?
we know this from the relation of AC to FA.
FA = 12 mm
AC = 12 + 28 = 40 mm
so, going from AC to FA we multiply AC by f so that
AC × f = FA
40 × f = 12
f = 12/40 = 3/10
all other sides, heights, ... if ABC translate to their smaller counterparts in AEF by that multiplication with f (= 3/10).
the area of a triangle is
baseline × height / 2
aABC = 500
and because of the similarity we don't need to calculate the side and height in absolute numbers. we can use the relative sizes by referring to the original dimensions and the scaling factor f.
baseline small = baseline large × f
height small = height large × f
we know that
baseline large × height large / 2 = 500
baseline large × height large = 1000
aAEF = baseline small × height small / 2 =
= baseline large × f × height large × f / 2 =
= baseline large × height large × f² / 2 =
= 1000 × f² / 2 = 500 × f² = 500 ×(3/10)² =
= 500 × 9/100 = 5 × 9 = 45 mm²
The answer is C. 1 lb 14 oz.
Answer:
a)
b.) -12knots, 8 knots c) No e)
Step-by-step explanation:
We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.
a)
We know that at time t , the ship A has moved
and ship B has moved
. We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.
Using Pythagorean theorem, we can write the distance s as:

b)
We want to find
for t=0 and t=1

c)
We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.
Ships have seen each other = 
Since function
is quadratic, concave up and has no real roots, we know that
for every t. So, the ships haven't seen each other.
d)
Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.
e)
Function ds/dt has a horizontal asympote in the first quadrant if

So, lets check this limit:

Notice that:
=√(speed of ship A² + speed of ship B²)