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kondor19780726 [428]
3 years ago
12

if one of the sections labled 4 makes three complete turns counterclockwise,how many degrees will it have traveled

Mathematics
1 answer:
nadezda [96]3 years ago
7 0
It would have traveled 1080 degrees
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Jane has $5 more than three times as much money as Lonnie has.Bob has $1 less than four times as much money as Lonnie has. All t
kkurt [141]
Starting equation:
92=J+L+B

To find how much money Jane has, we use the equation:
J=5+3L
To find how much money Bob has, we use the equation:
B=4L-1

Enter those equations in the problem

92=(5+3L)+L+(4L-1)
Combine like terms
92=4+8L
Subtract 4 to isolate variable
88=8L
Divide by 8 to isolate variable
11=L

Answer:
Lonnie has $11
Jane has $38
Bob has $43

side note: Lonnie is that one poor friend we all have
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3 years ago
Whats the hcf of 108 and 308​
inna [77]

Answer:

i think this should help

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3 years ago
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aleksklad [387]
SAS i believe :) sorry if im wrong!
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3 years ago
If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.
Sophie [7]
There's no if about it, 

f(x)=x^3+3x^2-x-3


has a zero f(1)=0 so x-1 is a factor.   That's the special case of the Remainder Theorem; since f(1)=0 we'll get a remainder of zero when we divide f(x) by x-1.

At this point we can just divide or we can try more little numbers in the function.  It doesn't take too long to discover f(-1)=0 too, so  x+1 is a factor too by the remainder theorem.  I can find the third zero as well; but let's say that's out of range for most folks.

So far we have 

x^3+3x^2-x-3 = (x-1)(x+1)(x-r)

where r is the zero we haven't guessed yet.  Again we could divide f(x) by (x-1)(x+1)=x^2-1 but just looking at the constant term we must have

-3 = -1 (1)(-r) = r

so

x^3+3x^2-x-3 = (x-1)(x+1)(x+3)

We check f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark

We usually talk about the zeros of a function and the roots of an equation; here we have a function f(x) whose zeros are

x=1, x=-1, x=-3

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2 years ago
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