Step-by-step explanation:
Given that,
A meal plan lets students buy $20 meal cards.
Each meal card lasts about 8 days.
Let us assume that y denotes number of days for a 20 dollar meal card i.e. $20 every 8 days.
Let x = number of $ for another day.
So,
Hence, this is the required equation.
Sum/difference:
Let
This means that
Now, assume that 
is rational. The sum/difference of two rational numbers is still rational (so 5-x is rational), and the division by 3 doesn't change this. So, you have that the square root of 8 equals a rational number, which is false. The mistake must have been supposing that was rational, which proves that the sum/difference of the two given terms was irrational
Multiplication/division:
The logic is actually the same: if we multiply the two terms we get
if again we assume x to be rational, we have
But if x is rational, so is -x/15, and again we come to a contradiction: we have the square root of 8 on one side, which is irrational, and -x/15 on the other, which is rational. So, again, x must have been irrational. You can prove the same claim for the division in a totally similar fashion.
Answer:
Check the solution below
Step-by-step explanation:
2) Given the equation
x +y =5... 1 and
x-y =3 ... 2
Add both equations
x+x = 5+3
2x = 8
x = 8/2
x = 4
Substitute x = 4 into 1:
From 1: x+y = 5
4+y= 5
y = 5-4
y = 1
3) Given
x+3y =15 ... 1
2x+7y=19 .... 2
From 2: x = 15-3y
Substitute into 2
2(15-3y)+7y = 19
30-6y+7y = 19
30+y = 19
y = 19-30
y = -11
Substitute y=-11 into x = 15-3y
x =15-3(-11)
x = 15+33
x = 48
The solution set is (48, -11)
4) given
x/2 +y/3 =0 and x+2y=1
From 1
(3x+2y)/6 = 0
3x+2y = 0.. 3
x+2y= 1... 4
From 4: x = 1-2y
Substutute
3(1-2y) +2y = 0
3-6y+2y = 0
3 -4y = 0
4y = 3
y = 3/4
Since x = 1-2y
x = 1-2(3/4)
x = 1-3/2
x= -1/2
The solution set is (-1/2, 3/4)
5) Given
5.x=1/2 and y =x +1 then solution is
We already know the vkue of x
Get y
y= x+1
y = 1/2 + 1
y = 3/2
Hence the solution set is (1/2, 3/2)
6) Given
3x +y =5 and x -3y =5
From 3; x = 5+3y
Substitute into 1;
3(5+3y)+y = 5
15+9y+y = 5
10y = 5-15
10y =-10
y = -1
Get x;
x = 5+3y
x = 5+3(-1)
x = 5-3
x = 2
Hence two solution set is (2,-1)
Answer:
Tn = 64-4n
Step-by-step explanation:
The nth term of an AP is expressed as:
Tn = a+(n-1)d
a is the common difference
n is the number of terms
d is the common difference
Given the 6th term a6 = 40
T6 = a+(6-1)d
T6 = a+5d
40 = a+5d ... (1)
Given the 20th term a20 = -16
T20 = a+(20-1)d
T20 = a+19d
-16 = a+19d... (2)
Solving both equation simultaneously
40 = a+5d
-16 = a+19d
Subtracting both equation
40-(-16) = 5d-19d
56 = -14d
d = 56/-14
d = -4
Substituting d = -4 into equation
a+5d = 40
a+5(-4) = 40
a-20 = 40
a = 20+40
a = 60
Given a = 60, d = -4, to get the nth term of the sequence:
Tn = a+(n-1)d
Tn = 60+(n-1)(-4)
Tn = 60+(-4n+4)
Tn = 60-4n+4
Tn = 64-4n
