F(t)= $100 x h + 300 A reasonable domain is (1,2,3) and the range is ($400,$500,$600)
F(x) = x^(1/3)
f(8000) = 8000^(1/3) = 20
f'(x) = 1/3 * x^(-2/3)
f'(x) = 1/(3x^(2/3))
f'(8000) = 1/(3*8000^(2/3)) = 1/(3*400) = 1/1200
Linear approximation for f(x) about x = 8000
L(x) = f(8000) + f'(8000) (x−8000)
L(x) = 20 + 1/1200 (x−8000)
8030^(1/3) ≈ L(8030)
= 20 + 1/1200 (8030−8000)
= 20 + 30/1200
= 20 + 1/40
= 20.025
Answer:
Subtractionproperty of inequality.
Step-by-step explanation:
From April to December =9 months
the depreciation expense
3,600×(9÷12)=2,700
Answer:
y intercept is (0,5)
x intercept is (5,0)
Step-by-step explanation:
re arange the equation to y = -x + 5.
To find the y intercept, you plug in 0 for x and find out what the value of y is. To find the x intercept, you plug in 0 for y and find the value of x.