Answer:
Question 6
[tex]\left(5a+2\right)\left(a+4\right)=\:5a^2+22a+8[/tex]
Question 7
The width is 2x.
Step-by-step explanation:
Question 6)
Expand
[tex](5a+2)(a+4)[/tex]
solving to the expand
[tex](5a+2)(a+4)[/tex]
- [tex]\mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd[/tex]
[tex]a=5a,\:b=2,\:c=a,\:d=4[/tex]
so the expression becomes
[tex]=5aa+5a\cdot \:4+2a+2\cdot \:4[/tex]
[tex]=5aa+5\cdot \:4a+2a+2\cdot \:4[/tex]
[tex]=5a^2+20a+2a+8[/tex]
adding similar elements: 20a+2a=22a
[tex]=5a^2+22a+8[/tex]
Hence,
[tex]\left(5a+2\right)\left(a+4\right)=\:5a^2+22a+8[/tex]
Question 7)
Given
Using the formula
width = Area/Length
[tex]=\:\frac{8x^2}{4x}[/tex]
[tex]=2x[/tex] ∵ [tex]\mathrm{Divide\:the\:numbers:}\:\frac{8}{4}=2[/tex]
Therefore, the width is 2x.
