Answer:
B
Step-by-step explanation:
The given equations satisfy the given conditions. There are 2 equations and 2 unknowns, so a certain solution can be found.
This can be solved using substitution,
Substituting eqn 2 to eqn 1:
2(2y – 10) + 3y =1240
Simplifying,
y = 180
x = 350
Answer with explanation:
The given differential equation is
y" -y'+y=2 sin 3x------(1)
Let, y'=z
y"=z'

Substituting the value of , y, y' and y" in equation (1)
z'-z+zx=2 sin 3 x
z'+z(x-1)=2 sin 3 x-----------(1)
This is a type of linear differential equation.
Integrating factor

Multiplying both sides of equation (1) by integrating factor and integrating we get


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Answer:
a right cylinder with a radius of 7 inches
Answer:
1'3'5'4
Step-by-step explanation: