Using Laplace transform we have:L(x')+7L(x) = 5L(cos(2t))sL(x)-x(0) + 7L(x) = 5s/(s^2+4)(s+7)L(x)- 4 = 5s/(s^2+4)(s+7)L(x) = (5s - 4s^2 -16)/(s^2+4)
=> L(x) = -(4s^2 - 5s +16)/(s^2+4)(s+7)
now the boring part, using partial fractions we separate 1/(s^2+4)(s+7) that is:(7-s)/[53(s^2+4)] + 1/53(s+7). So:
L(x)= (1/53)[(-28s^2+4s^3-4s^2+35s-5s^2+5s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]L(x)= (1/53)[(4s^3 -37s^2 +40s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]
denoting T:= L^(-1)and x= (4/53) T(s^3/(s^2+4)) - (37/53)T(s^2/(s^2+4)) +(40/53) T(s^2+4)-(4/53) T(s^2/s+7) +(5/53)T(s/s+7) - (16/53) T(1/s+7)
Answer:
a
Step-by-step explanation:
easy
Isoceleese means then that 2 sides are equal
right triangle so the legs are equal
a^2+b^2=c^2
the legs aer a and b
in isocoleese, a=b
a^2+a^2=c^2
2a^2=c^2
if isocoleese=a then c=?
solve for c
sqrt both sides
a√2=c
if the legs are x feet, then the hyppotonuse is x√2 feet
Answer:
slope is undefined
Step-by-step explanation:
For the first question, simply plug in the value of 3 into the polynomial question. The final answer is your volume of the hot water tub.
For the second question, you can figure out that x = -5 is not a real root because it does not touch or cross the x axis.
If it does it would be an solution.
