Answer:
No solutions
Step-by-step explanation:
4(2x-3)=2(4x-6)

Same components are present on both sides of equal sign, therefore NO SOLUTIONS are possible.
Answer: Choice B
(-2, 5)
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Explanation:
The original system is

Multiply both sides of the second equation by 3. Doing so leads to this updated system of equations

Now add straight down
The x terms add to -4x+3x = -1x = -x
The y terms add to 3y+(-3y) = 0y = 0
The terms on the right hand sides add to 23+(-21) = 2
We end up with the equation -x = 2 which solves to x = -2
Now use this to find y. You can pick any equation with x,y in it
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-4x+3y = 23
-4(-2)+3y = 23
8+3y = 23
3y = 23-8
3y = 15
y = 15/3
y = 5
Or
x-y = -7
-2-y = -7
-y = -7+2
y = -5
y = 5
Either way, we get the same y value.
So that's why the solution is (x,y) = (-2, 5)
Answer:
c
Step-by-step explanation:
Given that:

since cos (kπ) = 
Then, the series can be expressed as:

In the sum of an alternating series, the best bound on the remainder for the approximation is related to its
term.
∴




Answer:
Least to greatest: 2, 3, 1
Step-by-step explanation:
Square root of 8 over 2 is about 1.41 which makes it the least value
2 is obviously equal to 2 which puts it in the middle
Square root of 7 is about 2.64 which makes it the greatest value