Answer:
a. when x=1 so substitute 1 to any X in the given equation.
y=-3(1)+2.5
y=-3+2.5
y=-0.5
b.when x is -1.5 so substitute -1.5 to any x in the equation
y=-3(-1.5)+2.5
y=-4.5+2.5
y=-2
W + 2w = 60
this equation is wrong
1) m = y2 - y1/x2 - x1
y = mx + b
m = 7 - 0/3 - 8
y = -7/5x + 11.2
2) y= -7/5x + 12
(the slope or m stays the same because the two equations are parallel)
Answer:
So, the first three nonzero terms in the Maclaurin series are:

Step-by-step explanation:
From exercise we have the next function y = 6 sec(3x).
We know that Maclaurin series for the function sec x, have the following form:

So, for the given function y = 6 sec(3x), we get:

So, the first three nonzero terms in the Maclaurin series are:

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