<h3>
Answer: False</h3>
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Explanation:
I'm assuming you meant to type out
(y-2)^2 = y^2-6y+4
This equation is not true for all real numbers because the left hand side expands out like so
(y-2)^2
(y-2)(y-2)
x(y-2) .... let x = y-2
xy-2x
y(x)-2(x)
y(y-2)-2(y-2) ... replace x with y-2
y^2-2y-2y+4
y^2-4y+4
So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4
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Another approach is to pick some y value such as y = 2 to find that
(y-2)^2 = y^2-6y+4
(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2
0^2 = 2^2 - 6(2) + 4
0 = 4 - 6(2) + 4
0 = 4 - 12 + 4
0 = -4
We get a false statement. This is one counterexample showing the given equation is not true for all values of y.
8/15 x 5/6
•multiply across the numerator and denominator.
= 40/90
• then find a common factor that will go into both the numerator and denominator equally.
Common factor: 5
= 8/18
•there is another common factor, which is 2
Simplified:
=4/9
Is this just free points or is there a question
Answer:
10
Step-by-step explanation:
Means back the numbers into multiples of several small numbers
Like:; 1. We take LCM of 40
Just break into multiples of small number
40= 2×2×2×5
2. We take LCM of 50
50= 5×5×2
So LCM for 100 is 2×2×5×5
after that see the pairs in the LCM like 2×2 or 3×3 or 4×4(same numbers)
Then write the the single number in place of two multipled numbers
Like:; 2×2 is written as 2 // 3×3 is written as 3
So we can write 100 into 2×2×5×5 and then after selecting pairs (2×2)×(5×5)
write pairs in single number 2×5
And so we get 2×5=10
So we find root of 100 that is 10
