Let the first number be 'x' and the second number is 'y'
Equation 1: x + y = 52
Equation 2: x - y = 38
Rearranging equation 2 to make either x or y the subject
x = 38 + y
Substituting x = 38 + y into equation 1
x + y = 52
(38+y) + y = 52
38 + 2y = 52
2y = 52 - 38
2y = 14
y = 7
Substitute y = 7 into either equation 1 or equation 2 to find x
x + y = 52
x + 7 = 52
x = 52 - 7
x = 45
x = 45
y = 7
Should be <span>10.75, 11.5, 12.25, 13, 13.75, 14.5</span>
Answer:
A
Step-by-step explanation:
It is linear because it is a pattern of adding 1 and not quadratic for there is no squaring. It is no exponential because the ratio stays the same.
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)
=8x + (-3) (5) + (-3) (9x)
=8x + -15 + -27x
=8x + -15 + -27x
(8x + -27x) + (-15)
19x +-15
19x -15