QUESTION 1
The dimensions of the rectangular blanket are;
[tex]l = (3x + 7)cm[/tex]
and
[tex]w = (2x - 3)cm[/tex]
The perimeter is given by,
[tex] p= 2w + 2l[/tex]
We substitute the dimensions to obtain,
[tex]p = 2(2x - 3) + 2(3x + 7)[/tex]
Expand bracket to get,
[tex]p = 4x - 6 + 6x + 14[/tex]
This simplifies to
[tex]p =( 10x + 8 )cm[/tex]
QUESTION 2
When
[tex]x = 4[/tex]
The perimeter becomes
[tex]p =( 10(4) + 8 )cm[/tex]
[tex]p =( 40 + 8 )cm[/tex]
[tex]p =48cm[/tex]
QUESTION 3
Area is given by
[tex]A=l \times w[/tex]
[tex]A=(3x + 7)(2x - 3)[/tex]
We expand to get,
[tex]A=6 {x}^{2} - 9x + 14x - 21[/tex]
This gives us,
[tex]A=(6 {x}^{2} + 5x - 21) {cm}^{2} [/tex]
QUESTION 4
If
[tex]x = 4[/tex]
Then the area becomes,
[tex]A=6 {(4)}^{2} - 9(4)+ 14(4) - 21[/tex]
[tex]A=6 \times 16 - 36+ 56- 21[/tex]
[tex]A=96 - 36+ 56- 21[/tex]
[tex]A=95 {cm}^{2} [/tex]
QUESTION 5.
When the length of the blanket is 5cm longer, then the length of the new blanket becomes
[tex]l = (3x + 7 + 5)[/tex]
[tex]l = (3x + 12)cm[/tex]
The width is still
[tex]w = (2x - 3)cm[/tex]
The perimeter of the new blanket is
[tex]p = 2(3x + 12) + 2(2x - 3)[/tex]
This implies that,
[tex]p = 6x + 24+ 4x - 6[/tex]
[tex]p = 10x +18[/tex]
Comparing to the old perimeter which is
[tex]p = 10x +8[/tex],
The perimeter changes by 10 units
