Vertex coordinates: (h, k)
Vertex form: y = a(x-h)^2 + k
y = 2(x+3) + 2(x+4)
Use distributive property:
y = 2((x+3)+(x+4))
Simplify:
y = 2(2x+7)
y = 4x+14
This is slope - intercept form, not vertex form. Vertex form is for quadratic equations - this is a linear equation.
Answer (in slope - intercept form):
y = 4x+14
Answer:
The proportion is 2/7 = 12/42
x = 12
The answer is 12
Answer:
44.7
Step-by-step explanation:
I guessed based on what the side looks like so if it's wrong I"m so so sorry
Answer:
The rate of change is 0.75 gallons per minute
Step-by-step explanation:
The rate of change of the linear relation is represented by the slope of the line which represents this relation
The rule of the slope of a line is
m = Δy/Δx, where
- Δy is the vertical change
- Δx is the horizontal change
From the given graph
∵ The line passes through points (0, 0) and (4, 3)
∴ Δx = 4 - 0 = 4
∴ Δy = 3 - 0 = 3
→ Use the rule of the slope above to find the slope of the line
∴ m = 
= 0.75
∵ The x-axis represents the time in minutes
∵ The y-axis represents the amount in gallons
∵ m represents the rate of change
∴ The rate of change = 0.75 gallons per minute
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
