Answer:
i think its letter a)Zero pieces of candy split equally among 4 people
Answer: 
We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.
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Explanation:
It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.
I'm going to use these three log rules, which apply to any base.
- log(A) + log(B) = log(A*B)
- log(A) - log(B) = log(A/B)
- B*log(A) = log(A^B)
From there, we can then say the following:
The value of the derivative of functions h'(6) as requested in the task content is; 55.
<h3>What is the value of h'(6)?</h3>
Since it follows from the task content that the function h(x)=4f(x)+5g(x)+1.
Hence, the derivative of h(x) can be evaluated as;
h'(x)=4f'(x)+5g'(x)
On this note, by substitution, it follows that;
h'(6)=4(5)+5(7)
h'(6) = 55.
Read more on functions;
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S·S=500. S²=500. S=√500. S=22.36. 22.4 inches for 1 side, 22.4(4)=89.6, 89.6 for all 4 :)
<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.
