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a_sh-v [17]
2 years ago
15

Q#1: Suppose that you have two different algorithms for solving a problem. To solve a problem of size n, the first algorithm use

s exactly n*exp1-[n+2n+3n+4n]/nlne(10) operations and the second algorithm uses exactly n! operations. As n grows, which algorithm uses fewer operations?
Mathematics
1 answer:
svetlana [45]2 years ago
7 0

Answer:

The one of -[n+2n+3n+4n]/nlne(10) has fewer operations because the value of n is <u>static</u><u>.</u>

The one of n! ( factorial ) is factorised up to n∞ hence has infinity operations.

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What is the true solution to the logarithmic equation below?<br> log₂ (6x)-log₂ (√x)=2
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Answer:

The true solution is x=4/9

EXPLANATION

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A horticulturist wants to improve the soil quality in a garden by adding 120 pounds of nitrogen and 54 pounds of potassium. A pa
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6 0
3 years ago
2x + 3 + 4x + 8 = 3x - 10
vodomira [7]

\text {Hello! Let's solve this Equation!}

\text {Your First Step is to Simplify the Equation:}

\text {(2x+4x)+(8+3)}

\text {6x+11=3x-10}

\text {The Next Step is to Subtract 3x:}

\text {6x+11-3x=3x-10-3x}\\\text {3x+11=-10}

\text {Next Step is to Subtract 11:}

\text {Q: Why Subtract 11?}

\text {A: So they cancel out. We cross them out once we subtract.}

\text {3x+11-11=-10-11}\\\text {3x=-21}

\text {The Final Step is to Divide 3:}

\text {3x/3=-21/3}

\text {Your Answer will be}

\fbox {x=-7}

\text {Best of Luck!}

\text {-LimitedX}

8 0
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