Which number is composite in 23897993?
A composite number is a number that has multiple factors (9: 1,3,9)
Answer:
yp = -x/8
Step-by-step explanation:
Given the differential equation: y′′−8y′=7x+1,
The solution of the DE will be the sum of the complementary solution (yc) and the particular integral (yp)
First we will calculate the complimentary solution by solving the homogenous part of the DE first i.e by equating the DE to zero and solving to have;
y′′−8y′=0
The auxiliary equation will give us;
m²-8m = 0
m(m-8) = 0
m = 0 and m-8 = 0
m1 = 0 and m2 = 8
Since the value of the roots are real and different, the complementary solution (yc) will give us
yc = Ae^m1x + Be^m2x
yc = Ae^0+Be^8x
yc = A+Be^8x
To get yp we will differentiate yc twice and substitute the answers into the original DE
yp = Ax+B (using the method of undetermined coefficients
y'p = A
y"p = 0
Substituting the differentials into the general DE to get the constants we have;
0-8A = 7x+1
Comparing coefficients
-8A = 1
A = -1/8
B = 0
yp = -1/8x+0
yp = -x/8 (particular integral)
y = yc+yp
y = A+Be^8x-x/8
-1400 + 800 - 350 = - 950 meters
A=number of seats in section A
B=number of seats in section B
C=number of seats in section C
We can suggest this system of equations:
A+B+C=55,000
A=B+C ⇒A-B-C=0
28A+16B+12C=1,158,000
We solve this system of equations by Gauss Method.
1 1 1 55,000
1 -1 -1 0
28 16 12 1,158,000
1 1 1 55,000
0 -2 -2 -55,000 (R₂-R₁)
0 12 16 382,000 (28R₁-R₂)
1 1 1 55,000
0 -2 -2 -55,000
0 0 4 52,000 (6R₂+R₃)
Therefore:
4C=52,000
C=52,000/4
C=13,000
-2B-2(13,000)=-55,000
-2B-26,000=-55,000
-2B=-55,000+26,000
-2B=-29,000
B=-29,000 / -2
B=14,500.
A + 14,500+13,000=55,000
A+27,500=55,000
A=55,000-27,500
A=27,500.
Answer: there are 27,500 seats in section A, 14,500 seats in section B and 13,000 seats in section C.
The answer is II and III, which is the last choice from above. This is because the symbol "less than (or equal to)" indicated a conjunction, which means we have to take into account that the absolute value can be less than or equal to 15, and greater than or equal to -15. This solution set falls in between the two values, hence, the conjunction.