Answer:The value of the bulldozer after 3 years is $121950
Step-by-step explanation:
We would apply the straight line depreciation method. In this method, the value of the asset(bulldozer) is reduced linearly over its useful life until it reaches its salvage value. The formula is expressed as
Annual depreciation expense =
(Cost of the asset - salvage value)/useful life of the asset.
From the given information,
Useful life = 23 years
Salvage value of the bulldozer = $14950
Cost of the new bulldozer is $138000
Therefore
Annual depreciation = (138000 - 14950)/ 23 = $5350
The value of the bulldozer at any point would be V. Therefore
5350 = (138000 - V)/ t
5350t = 138000 - V
V = 138000 - 5350t
The value of the bulldozer after 3 years would be
V = 138000 - 5350×3 = $121950
Answer:
the distance is 10
Step-by-step explanation:
-5,2 to 0,0 is 5
and the refection of -5,2 is 5,2
5,2 to 0,0. is also 5
5+5=10
Answer:
x = 42.5
Step-by-step explanation:
It is said:
If two angles are not corresponding, then they make up 180°. They are more known as supplementary angles.
So the equation is:
a + b = 180°
Substitute
x + 8 + 3 x + 2 = 180°
x + 3x + 8 + 2 = 180°
4x + 10° = 180°
4x = 180° - 10°
4x = 170°
x = 170° ÷ 4
x = 42.5
Answer: function 1
Rate of change of function 1:
Following the format of y=mx+c, the rate of change should be m, so, the rate of change for function 1 = 4
To find the gradient (rate of change):
The two points the line passes through are (x1, y1) and (x2, y2), which in this case is (1, 6) and (3, 10)
(Doesn't matter which is which but you need to make sure that once you decide which is which, you stick to it)
To calculate the gradient, you substitute these values following (y1 - y2)/(x1 - x2)
Gradient of function 2 = (10 - 6)/(3 - 1)
= 2
Therefore, since 4 > 2, rate of change of function 1 > rate of change of function 2.
Answer:
y≤−6. Everthing more or equal to -6 should be shaded. So like fro-6 to -10
Step-by-step explanation:
Multiply each term in 12y≤−3 by 2 to eliminate the fractions.
