Answer: 3H+4C<_ 30
The coefficients of the variables H and C represent the number of batteries each toy helicopter and each toy car uses.
Step-by-step explanation:
Each toy helicopter uses 3 batteries, and each toy car uses 4 batteries.
Now let's check whether Vugar has enough batteries for 5 toy helicopters and 4 toy cars. To do this, we substitute H=5 and C= 4 in the given inequality:
Does Vugar have enough batteries to play with 5 toy helicopters and 4 toy cars?
No, because if you plug in the value for H and C:
3H + 4C<_ 30
3(5) + 4(4) <_30
15 + 16 <_ 30
31 <_ 30; false
Since the inequality is false, Vugar does not have enough batteries to play with 555 toy helicopters and 444 toy cars.
Each toy helicopter uses 333 batteries, and each toy car uses 444 batteries.
No, Vugar does not have enough batteries to play with 555 toy helicopters and 444 toy cars.
The leading term of polynomial function is the the term contain highest degree so here in the given question leading term is 
and leading coefficient is the coefficient of the term with greatest exponent -3
RULES for End behaviour
we have following four cases
CASE1: Even degree and positive leading coefficient
CASE2: Even degree and negative leading coefficient
CASE3: Odd degree and positive leading coefficient
CASE4: Odd degree and negative leading coefficient
Here in the given case we have odd degree and negative leading coefficient
The answer would be 14 because 4x__=56 ? 4 x14 =56
Coefficient: 20 or 5
Variable: c or w
Constant: 95.50
Pt B: 20(5) + 5(20) + 95.50
100 + 100 + 95.50
295.50
Pt C:The coefficient of c would change to 10
Answer:
the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm
Step-by-step explanation:
since the volume of a cylinder is
V= π*R²*L → L =V/ (π*R²)
the cost function is
Cost = cost of side material * side area + cost of top and bottom material * top and bottom area
C = a* 2*π*R*L + b* 2*π*R²
replacing the value of L
C = a* 2*π*R* V/ (π*R²) + b* 2*π*R² = a* 2*V/R + b* 2*π*R²
then the optimal radius for minimum cost can be found when the derivative of the cost with respect to the radius equals 0 , then
dC/dR = -2*a*V/R² + 4*π*b*R = 0
4*π*b*R = 2*a*V/R²
R³ = a*V/(2*π*b)
R= ∛( a*V/(2*π*b))
replacing values
R= ∛( a*V/(2*π*b)) = ∛(0.03$/cm² * 600 cm³ /(2*π* 0.05$/cm²) )= 3.85 cm
then
L =V/ (π*R²) = 600 cm³/(π*(3.85 cm)²) = 12.88 cm
therefore the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm
