The product of a number and nine increase by four is 58 find the number

What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that t

Inga [223]

Answer:

f(2n)-f(n)=log2

b.lg(lg2+lgn)-lglgn

c. f(2n)/f(n)=2

d.2nlg2+nlgn

e.f(2n)/(n)=4

f.f(2n)/f(n)=8

g. f(2n)/f(n)=2

Step-by-step explanation:

What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos- sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]

from a

f(n)=logn

f(2n)=lg(2n)

f(2n)-f(n)=log2n-logn

lo(2*n)=lg2+lgn-lgn

f(2n)-f(n)=lg2+lgn-lgn

f(2n)-f(n)=log2

2.f(n)=lglgn

F(2n)=lglg2n

f(2n)-f(n)=lglg2n-lglgn

lg2n=lg2+lgn

lg(lg2+lgn)-lglgn

3.f(n)=100n

f(2n)=100(2n)

f(2n)/f(n)=200n/100n

f(2n)/f(n)=2

the time will double

4.f(n)=nlgn

f(2n)=2nlg2n

f(2n)-f(n)=2nlg2n-nlgn

f(2n)-f(n)=2n(lg2+lgn)-nlgn

2nLg2+2nlgn-nlgn

2nlg2+nlgn

5.we shall look for the ratio

f(n)=n^2

f(2n)=2n^2

f(2n)/(n)=2n^2/n^2

f(2n)/(n)=4n^2/n^2

f(2n)/(n)=4

the time will be times 4 the initial tiote tat ratio are used because it will be easier to calculate and compare

6.n^3

f(n)=n^3

f(2n)=(2n)^3

f(2n)/f(n)=(2n)^3/n^3

f(2n)/f(n)=8

the ratio will be times 8 the initial

7.2n

f(n)=2n

f(2n)=2(2n)

f(2n)/f(n)=2(2n)/2n

f(2n)/f(n)=2

5 0

3 years ago

8x +144 < 2000 Solve the in equation and graph

avanturin [10]

Answer:

8x + 144 < 2000

-144

8x < 1856

- ------

8 8

x < 232

7 0

3 years ago

What is the sum (Two-fifths x + StartFraction 5 over 8 EndFraction) + (one-fifth x minus one-fourth)?

liubo4ka [24]

The sum of ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) is $\frac{3}{5}$ x + $\frac{3}{8}$

Step-by-step explanation:

To add or subtract 2 fractions, they must have same denominators, if they not do that

Find the Lowest common multiple of the two denominators (LCM)
Replace each denominator by it
Divide The LCM by each denominator and multiply the numerator of each fraction by its quotient

∵ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ )

- Let us start with adding the like terms

∴ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) = (

∵ ( $\frac{2}{5}$ x + $\frac{1}{5}$ x ) have same denominators, then we can add them

∴ ( $\frac{2}{5}$ x + $\frac{1}{5}$ x ) = $\frac{3}{5}$ x

∵ ( $\frac{5}{8}$ - $\frac{1}{4}$ ) do not have the same denominators, then we must find

LCM of 8 and 4

- The LCM of 8 and 4 is 8 because 8 is the first common multiple

of 8 and 4

∵ LCM of 8 and 4 is 8

- Divide 8 by the denominator 4

∵ 8 ÷ 4 = 2

- Multiply the numerator of the fraction by 2

∴ $\frac{1}{4}=\frac{2}{8}$

∴ ( $\frac{5}{8}$ - $\frac{1}{4}$ ) = ( $\frac{5}{8}$ - $\frac{2}{8}$ ) = $\frac{3}{8}$

∴ ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) = $\frac{3}{5}$ x + $\frac{3}{8}$

The sum of ( $\frac{2}{5}$ x + $\frac{5}{8}$ ) + ( $\frac{1}{5}$ x - $\frac{1}{4}$ ) is $\frac{3}{5}$ x + $\frac{3}{8}$

Learn more:

You can learn more about the fractions in brainly.com/question/2456302

#LearnwithBrainly

7 0

3 years ago

Read 2 more answers

Simplify: -|-15| please answer correctly i'll give brainliest and 5-star rating.

12345 [234]

Answer:

-15

Step-by-step explanation:

Definition of absolute value:

Distance from zero on a number line.

Imagine a number line with 0 in the middle of it. Now mark the number 3 on it. Also mark the opposite of 3, which is -3. What is the distance from zero to those 2 points? it's 3.

This is exactly like saying |3| = 3 and |-3| = 3

Even if the number inside absolute value is negative, it simplifies to the positive of the same number.

_______________________________

So let's talk about your problem.

-|-15|

> the absolute value simplifies to 15

-|-15| = - 15

So that ends up being your answer.

$-15$