Let x be the number.
Then:
(x)+(x+1)+(x+2)>102
(x)+(x+1)+(x+2)<109
x+x+1+x+2
3x+3>102
3x>102-3
3x>99
x>22
3x+3<109
3x<109-3
3x<106
x<35.33
Therefore, the number can be bigger than 22 but less than 35.333 to meet the requirements.
Hope I helped :)
Answer:
x = -44/13
y = -65/13
Step-by-step explanation:
Using matrix form means using the crammers rule
The matrix form of the expression is written as;
![\left[\begin{array}{ccc}8&5\\-1&1\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}9\\7\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%265%5C%5C-1%261%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%5C%5C7%5C%5C%5Cend%7Barray%7D%5Cright%5D)
AX = B
taking the determinant of A;
|A| = 8(1) - 5(-1)
|A| = 8 + 5
|A| = 13
After replacing the first row with the column matrix;
![A_x =\left[\begin{array}{ccc}9&5\\7&-1\\\end{array}\right]](https://tex.z-dn.net/?f=A_x%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%265%5C%5C7%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
|Ax| = 9(-1)-5(7)
||Ax| = -9 - 35
|Ax| = -44
x = |Ax|/|A|
x = -44/13
similarly for y
![A_x =\left[\begin{array}{ccc}8&9\\-1&7\\\end{array}\right]](https://tex.z-dn.net/?f=A_x%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%269%5C%5C-1%267%5C%5C%5Cend%7Barray%7D%5Cright%5D)
|Ay| = 8(7)+9
|Ay| = 56+9
|Ay| = 65
y = |Ay|/|A|
y = -65/13
Answer: Sometimes
Explanation:
Quotient greater than 1:
3/4 divide by 2/3
= 3/4 x 3/2 = 9/8 > 1
4/5 divide by 2/5
= 4/5 x 5/2 = 2 > 1
Quotient less than 1
1/100 divide by 2/10
= 1/100 x 10/2 = 1/20 < 1
1/50 divide by 1/25
= 1/50 x 25/1 = 1/2 < 1
X = -6y - 12
4x + 5y =-39
4x + 5y = -39
-4(-6y - 12) + 5y = -39
-4(-6y) + 4(12) + 5y = -39
24y + 48 + 5y = -39
29y + 48 = -39
29y = -87
y = -3
x = -6y - 12
x = -6(-3) - 12
x = 28 - 12
x = 6
(x, y) = (6, -3)
