24 days,
12 & 8 have a least common multiple of 24 so that is when they would both travel
Answer:
no one can do it
Step-by-step explanation:
Answer:
There is one solution
Step-by-step explanation:
2x + y = -1
8x + 3y = -2
Multiply the first equation by -4
-8x -4y = 4
Then add the equations together to eliminate x
-8x -4y = 4
8x + 3y = -2
--------------------
-y = 2
Multiply by -1
y =-2
Now find x
2x+y =-1
2x+-2 =-1
Add 2 to each side
2x-2+2=-1+2
2x=1
Divide by 2
x = 1/2
Answer:
- (0, 4)
- (-1, 16/ 3)
- (-9/ 2, 10)
Step-by-step explanation:
In an ordered pair the term before the comma is the x variable and the term after the comma is the y variable.
1.
4x + 3y = 12 . . . . . .(0, _)
here were given the x variable
finding the other term
4× 0 + 3y = 12
3y = 12
y = 4
therefore the ordered pair is <u>(0, 4) </u>
2.
. . . . . . . (-1, _)
here were given the x variable
4 × -1 + 3y = 12
3y = 12 + 4
3y = 16
y = 16/ 3
ordered pair is <u>(-1, 16/ 3)</u>
3.
. . . . . . . . (_, 10)
here were given the y variable
finding the other term
4x + 3 × 10 = 12
4x + 30 = 12
4x = 12 - 30
4x = -18
dividing both sides by 2
2x = -9
x = -9/ 2
ordered pair is<u> (-9/ 2, 10)</u>
Problem 1
<h3>Answer: False</h3>
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Explanation:
The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.
So,
f(x) = x+1
f( g(x) ) = g(x) + 1 .... replace every x with g(x)
f( g(x) ) = 6x+1 ... plug in g(x) = 6x
(f o g)(x) = 6x+1
Now let's flip things around
g(x) = 6x
g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)
g( f(x) ) = 6(x+1) .... plug in f(x) = x+1
g( f(x) ) = 6x+6
(g o f)(x) = 6x+6
This shows that (f o g)(x) = (g o f)(x) is a false equation for the given f(x) and g(x) functions.
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Problem 2
<h3>Answer: True</h3>
---------------------------------
Explanation:
Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.
For example, let
f(x) = 1/(x+2)
g(x) = -2
The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.
So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).