Answer:
x = -7/40
, y = -11/20
Step-by-step explanation:
Solve the following system:
{12 x - 2 y = -1 | (equation 1)
4 x + 6 y = -4 | (equation 2)
Subtract 1/3 × (equation 1) from equation 2:
{12 x - 2 y = -1 | (equation 1)
0 x+(20 y)/3 = (-11)/3 | (equation 2)
Multiply equation 2 by 3:
{12 x - 2 y = -1 | (equation 1)
0 x+20 y = -11 | (equation 2)
Divide equation 2 by 20:
{12 x - 2 y = -1 | (equation 1)
0 x+y = (-11)/20 | (equation 2)
Add 2 × (equation 2) to equation 1:
{12 x+0 y = (-21)/10 | (equation 1)
0 x+y = -11/20 | (equation 2)
Divide equation 1 by 12:
{x+0 y = (-7)/40 | (equation 1)
0 x+y = -11/20 | (equation 2)
Collect results:
Answer: {x = -7/40
, y = -11/20
The answer would be C+A= $85 or A+C=$85
Explanation: Candaces total money + Amar’s total money equals out to $85, it can be wrote either way and still equal $85.
Answer:
P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Step-by-step explanation:
Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3
Basis: P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:
2(k+1) - 1 = 2k + 2 - 1
≤ 2 + k! (by the inductive hypothesis)
= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.
X = adult tickets and x+58 = student tickets
Create the equation from the problem.
x + x + 58 = 708
Combine like terms.
2x +58 = 708
Subtract 58.
2x = 650
Divide by 2.
x = 325
Therefore, your answer is:
there were 325 adult tickets sold.
(Adding 58 to that number will get you the student tickets sold which is 383 if they ask.)
