Consider the digit expansion of one of the numbers, say,
676₉ = 600₉ + 70₉ + 6₉
then distribute 874₉ over this sum.
874₉ • 6₉ = (8•6)(7•6)(4•6)₉ = (48)(42)(24)₉
• 48 = 45 + 3 = 5•9¹ + 3•9⁰ = 53₉
• 42 = 36 + 6 = 4•9¹ + 6•9⁰ = 46₉
• 24 = 18 + 6 = 2•9¹ + 6•9⁰ = 26₉
874₉ • 6₉ = 5(3 + 4)(6 + 2)6₉ = 5786₉
874₉ • 70₉ = (8•7)(7•7)(4•7)0₉ = (56)(49)(28)0₉
• 56 = 54 + 2 = 6•9¹ + 2•9⁰ = 62₉
• 49 = 45 + 4 = 5•9¹ + 4•9⁰ = 54₉
• 28 = 27 + 1 = 3•9¹ + 1•9⁰ = 31₉
874₉ • 70₉ = 6(2 + 5)(4 + 3)10₉ = 67710₉
874₉ • 600₉ = (874•6)00₉ = 578600₉
Then
874₉ • 676₉ = 578600₉ + 67710₉ + 5786₉
= 5(7 + 6)(8 + 7 + 5)(6 + 7 + 7)(0 + 1 + 8)(0 + 0 + 6)₉
= 5(13)(20)(20)(1•9)6₉
= 5(13)(20)(20 + 1)06₉
= 5(13)(20)(2•9 + 3)06₉
= 5(13)(20 + 2)306₉
= 5(13)(2•9 + 4)306₉
= 5(13 + 2)4306₉
= 5(1•9 + 6)4306₉
= (5 + 1)64306₉
= 664306₉
The law of cosines for this particular situation is b^2 = a^2 + c^2 - 2ac cosB.
Filling in what you know, you have b^2 = 25 + 49 - [2(5)(7)-.1908], which simplifies to b^2 = 74 - 69.8092 which gives you a b^2 value of 4.1908, but you have to take the square root of that so you get a side value for b of 2.05.
Answer:
6 i don't know what the unit is, but its 6 for time
Step-by-step explanation:
It is a two step problem
54=9t
divide both sides by 9 to get the value of t
6=t
I hope that helps you :)
Answer:
y= 1x
Step-by-step explanation:
Answer: D
Step-by-step explanation:
