Answer:
The value at the end of year 2 is $4400.
Step-by-step explanation:
The best approach here is to determine the expression for the line depreciation and then calculate the depreciation value at x = 2 years.
A line is given by
where m is the slope and b the bias (aka y-intercept). You can determine both directly from what is given. The slope is change in y divided by change in x. We know that over 5 years the car loses (500-7000)=-6500 in value. So, the slope is m=-6500/5 (note the negative sign). At time 0, the y-intercept is 7000, since that is the initial value (at year 0). So our line function is fully identified:
and gives you the value of the car in any given year. To answer the question, we now plug in 2 as value of x:
Answer:
D. - 1/6
Step-by-step explanation:
M + 2/3 = 1/2
M = 1/2 - 2/3
The answer is c = .6x + 5
In order to find the equation, note that the price per pound is contingent upon the weight in x. Therefore, we can multiply the two together.
We also need to add the constant, which is 5.
<span>et us assume that the origin is the floor right below the 30 ft. fence
To work this one out, we'll start with acceleration and integrate our way up to position.
At the time that the player hits the ball, the only force in action is gravity where: a = g (vector)
ax = 0
ay = -g (let's assume that g = 32.8 ft/s^2. If you use a different value for gravity, change the numbers.
To get the velocity of the ball, we integrate the acceleration
vx = v0x = v0cos30 = 103.92
vy = -gt + v0y = -32.8t + v0sin40 = -32.8t + 60
To get the positioning, we integrate the speed.
x = v0cos30t + x0 = 103.92t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + 60t + 4
If the ball clears the fence, it means x = 0, y > 30
x = 0 -> 103.92 t - 350 = 0 -> t = 3.36 seconds
for t = 3.36s,
y = -16.4(3.36)^2 + 60*(3.36) + 4
= 20.45 ft
which is less than 30ft, so it means that the ball will NOT clear the fence.
Just for fun, let's check what the speed should have been :)
x = v0cos30t + x0 = v0cos30t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + v0sin30t + 4
x = 0 -> v0t = 350/cos30
y = 30 ->
-16.4t^2 + v0t(sin30) + 4 = 30
-16.4t^2 + 350sin30/cos30 = 26
t^2 = (26 - 350tan30)/-16.4
t = 3.2s
v0t = 350/cos30 -> v0 = 350/tcos30 = 123.34 ft/s
So he needed to hit the ball at at least 123.34 ft/s to clear the fence.
You're welcome, Thanks please :)
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