Check the picture below.
how do we know? well, notice h(t), starts off at 12, up up up reaches 47.84 then down down down, which is pretty much the trajectory of a flying object, by the time it gets to 44, is still going down.
now, let's look at g(t), starts off at 10, and goes up up up, never down, by the time it gets to 41, is still going up,
so at second 2, h(t) is 44 and going down, g(t) is 41 and going up, at 2.2 h(t) is 40.16, and g(t) is 44.1, between that lapse, h(t) became 44, 43, 42, 41, in the same lapse g(t) became 41, 42, 43, 44, so somewhere in those values h(t) = g(t).
what does the solution mean? It's the seconds or the instant lapse when the first cannon ball was at the same height as the second cannonball.
Answer:
Step-by-step explanation:
61%
Answer:
$8
Step-by-step explanation:
20% (= 0.20) is deducted from $40.40 * 0.20 = $8.
Answer:
y = 
Step-by-step explanation:
Let the equation of the line is,
y = mx + b
Here m = slope of the line
b = y-intercept
Slope of a line passing through two points
and
is,
m = 
From the graph attached,
Since, the given line passes through (0, -1) and (3, 4),
Slope 'm' = 
m = 
y - intercept 'b' = -1
Therefore, equation of the line will be,
y = 