The answer is three significant figures. The 1 and the 7 are both significant, because they are non-zero quantities.
This is where significant figures gets a little more complicated, because if a zero is used as a placeholder (i.e. 0.00027 cm) then it is insignificant.
But in the case above, the zero isn't being used as a placeholder, and thus, is significant.
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:
4
explanation:
well 18 minus 12 would equal 6. correct? so we would add x to 6 so the new equation would be 6 + x = 10 or to make it easier just do 10 - x = 6
Answer:
B.
Step-by-step explanation:
C=3.14D
Circumference is equal to pi multiplied by the diameter, so 50 times 3.14 is equal to 157. 157 is closest to 160.
