Answer:
Total tax = $25.51
Step-by-step explanation:
Total collection = $1265.23
Tax-Free amount = $840
so
Taxable amount = $1265.23 - $840
= $425.23
Given that the taxable amount was taxed at 6%
Thus,
The total tax = 6% of 425.23
= 6/100 × 425.23
= 0.06 × 425.23
= $25.51
Therefore,
Answer:
C = -4
Step-by-step explanation:
First, combine like terms (terms with the same amount of variables):
14 = -2C - 3C - 6
14 = (-2C - 3C) - 6
14 = (-5C) - 6
Isolate the variable, C. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS. PEMDAS is the order of operation, and equals:
Parenthesis
Exponents & Roots
Multiplication
Division
Addition
Subtraction.
First, add 6 to both sides:
14 (+6) = -5C - 6 (+6)
14 + 6 = -5C
-5C = 20
Next, divide -5 from both sides:
(-5C)/-5 = (20)/-5
C = 20/-5
C = -4
-4 is your value for C.
~
Rewrite the fractions as improper fractions:
1. −4 2/5 ÷ (−5)
= -22/5 ÷ −5/1
= -22/5 x -1/5
=22/25
2. −4 4/5÷4
= -24/5 ÷ 4/1
= -24/5 x 1/4
= -6/5
= -1 1/5
3. −2 1/8 ÷ 1 1/4
= -17/8 ÷ 5/4
= -17/8 x 4/5
= -68/40
= -17/10
=-1 7/10
4. −6 7/8 ÷ (−3 3/4)
= -55/8 ÷ -15/4
= -55/8 x -4/15
= 220/40
= 11/6
= 1 5/6
5. −8 3/4÷ 2 1/6
= -35/4 ÷ 13/6
= -35/4 x 6/13
= -210/52
= - 105/26
= -4 1/26
Answer:
Explanation:
Given:
The equation describing the forest wood biomass per hectare as a function of plantation age t is:
y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4
The equation that describes the annual growth in wood biomass is:
y ′ (t) = 0.01t + 0.072t^2 - 0.018t^3
To find:
a) The year the annual growth achieved its highest possible value
b) when does y ′ (t) achieve its highest value?
a)
To determine the year the highest possible value was achieved, we will set the derivative y'(t) to zero. The values of t will be substituted into the second derivative to get the highest value


SInce t = 4.13, gives y ′' (t) = -0.316 (< 0). This makes it the maximum value of t
The year the annual growth achieved its highest possible value to the nearest whole number will be
year 4
b) y ′ (t) will achieve its highest value, when we substitute the value of t that gives into the initial function.
Initial function: y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4