Hypotenuse of the right triangle: h=sqrt(4^2+6^2)
h=sqrt(16+36)
h=sqrt(52)
h=sqrt(4*13)
h=sqrt(4) sqrt(13)
h=2 sqrt(13)
Total Area: T.A.=2*(4*6/2)+[4+6+2 sqrt(13)] (8)
T.A.=2*(12)+[10+2 sqrt(13)](8)
T.A.=24+(10)(8)+[2 sqrt(13)](8)
T.A.=24+80+16 sqrt(13)
T.A.=104+16 sqrt(13)
Answer:
( 104+16 sqrt(13) )
Product is times so try that as it’s the product of a number
Just remember BODMAS, which B= brackets O= Order D= Division M= Multiply A=Addition and S=Subtract
So first you solve the brackets so 44+22 is 66
So 66-4+12 so 66-4 is 62
62+12 is 74
So now you have got 74 divided by 2 which is 37
Hope this helps xx
Answer:
f(9) =21
Step-by-step explanation:
I will assume you want to find f(9) when x=9
Let x=9
f(9) = 2(9) +3
= 18+3
=21
f(9) =21
