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)
I'm sorry, but there is not enough information for me to correctly answer this question, please provide more information.
I have summer school work too omg!!! it’s annoying
Answer:
B
Step-by-step explanation:
The differences in the terms of f(x) are + 3, + 5, + 7
Since the differences are not constant the relationship is not linear
Note the differences in the differences are + 2, + 2,
The second differences are constant indicating a quadratic relationship
Note the relationship between x and f(x)
x = 1 → 1² = 1 ← require to add 5, that is 1 + 5 = 6 ← value of f(x)
x = 2 → 2² = 4 ← require to add 5, that is 4 + 5 = 9 ← value of f(x)
x = 3 → 3² = 9 ← require to add 5, that is 9 + 5 = 14 ← value of f(x)
x = 4 → 4² = 16 ← require to add 5, that is 16 + 5 = 21 ← value of f(x)
Thus f(x) = x² + 5 → B
Answer:
C
Step-by-step explanation:
2,3,5,5,<u>6</u>,7,8,8,11
6 is the middle line, 8 is the end of the box, and 4 is the beginning of the bow.
Hope this helps.
