That would be C because there are two zeros right?
So you move two decimal places from 67.
And you get 0.67
1) Considering that those points (2,k) and (k,32) determines one same line, within the plane. Therefore we can say that
2) Considering that since they pass through the origin then their x -coordinate is equal to zero, and their y coordinate as well so
3) We can state that as a matter of fact (2, k) and (K, 32) can be rewritten as (2,0) and (0,32)
k=0
(b). → Simplify: [{5^(x-+1) + 5^x}/{6×5^x}]
= [{5^(x+1+x)}/{6×5^x}]
= [{5^(2x+1)}/{6×5^x}]
= [{5^(2x+1)}/{6×5^x}]
= [{5^(2x+1-x)}/6]
= [{5^(x+1)}/6]Ans.
(c). 4a³b²
= 4(3)³(-2)²
Since, a = 3 and b = -2.
so,
= 4(3*3*3)(-2*-2)
= 4(27)(4)
= 108(4)
= 432Ans.
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simplify the equation simplify x^5/8/x^1/6..
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