There are two bags: a bag with 4 blue marbles and 6 black marbles and another one containing 6 blue marbles and <span>6 black marbles. the probability of getting a blue marble in the first bag is 6/10 or 3/5 while that on the second bag is 6/12 or 1/2. Hence the total probability of getting two marbles is c. 3/10. </span>
Answer:
a
Step-by-step explanation:i did it
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:
Step-by-step explanation:
a. 16b^2c^12-0.25 can be rewritten as follows because it is the difference of two squares:
(4bc^6 - .5)(4bc^6 + .5)
b. 81x^6y^2-0.36a^2
is the difference of two squares, just like (a) (above); its factors are:
(9x^3y - 0.6a)(9x^3y + 0.6a)
Answer:
Option A. 65 trees
Step-by-step explanation:
From the question given:
50 trees was in the forest. Then 23 of them were cut down.
The remaining trees = 50 — 23 = 27 trees.
Now 38 new trees were plant.
Total trees after cutting and planting = 27 + 38 = 65 trees
