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5^2 = 25 

 the answer is 25 because 5*5=25

Which of the following is equivalent to (n – 3)(n – 5)?

help help

5 

What you are describing is simply this:

10(x) / x    - 3     which is ALWAYS just   10-3 = 7     for ANY 'x' 

-2(x+9) = -2(x) +(-2)(9) = -2x-18

two hundredths       2/100      .02 

Easy to set this up as a RATIO

$10.38/666batteries =  x/888batteries

10.38/666  *  888 = $13.84

sec = 1/cos   = 5/4

1/(5/4) = cos = 4/5

cos^2 + sin^2 = 1   (Remember that?)

(4/5)^2 + sin^2 = 1

1-(4/5)^2 = sin^2

9/25 = sin^2

sin = 3/5

Thanks Heureka :)

Her TIME for the entire journey is given by   Distance/Rate = Time

45km/(75 km/hr) = .6 hr = 36 minutes

She spent 21 minutes going the first 2/3    2/3 of 45 = 30 km

SO she spent  36min-21min = 15 minutes travelling the remaining 15 km

Rate = distance/time = 15km/15min=15km/.25hr= 60km/hr for the final 1/3 (15 km)

You mean that there is a maximum of 8 steps if the element exists in the array,

but 9 steps if the element does not exist?

Hello Melody!

YES!

\(\begin{array}{|lrcll|} \hline & \text{We search number 1:} & & \text{set } \{1,\dots,193 \}\\\\ (1) & 1 > \lfloor \frac{1+193}{2} \rfloor = 97 & no & \text{new set } \{1,\dots,97 \}\\ (2) & 1 > \lfloor \frac{1+97}{2} \rfloor = 49 & no & \text{new set } \{1,\dots,49 \}\\ (3) & 1 > \lfloor \frac{1+49}{2} \rfloor = 25 & no & \text{new set } \{1,\dots,25 \}\\ (4) & 1 > \lfloor \frac{1+25}{2} \rfloor = 13 & no & \text{new set } \{1,\dots,13 \}\\ (5) & 1 > \lfloor \frac{1+13}{2} \rfloor = 7 & no & \text{new set } \{1,\dots,7 \}\\ (6) & 1 > \lfloor \frac{1+7}{2} \rfloor = 4 & no & \text{new set } \{1,\dots,4 \}\\ (7) & 1 > \lfloor \frac{1+4}{2} \rfloor = 2 & no & \text{new set } \{1,2 \}\\ (8) & 1 > \lfloor \frac{1+2}{2} \rfloor = 1 & no & \text{we have found number 1 in the last set}\\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline & \text{We search number 1.5:} & & \text{set } \{1,\dots,193 \}\\\\ (1) & 1.5 > \lfloor \frac{1+193}{2} \rfloor = 97 & no & \text{new set } \{1,\dots,97 \}\\ (2) & 1.5 > \lfloor \frac{1+97}{2} \rfloor = 49 & no & \text{new set } \{1,\dots,49 \}\\ (3) & 1.5 > \lfloor \frac{1+49}{2} \rfloor = 25 & no & \text{new set } \{1,\dots,25 \}\\ (4) & 1.5 > \lfloor \frac{1+25}{2} \rfloor = 13 & no & \text{new set } \{1,\dots,13 \}\\ (5) & 1.5 > \lfloor \frac{1+13}{2} \rfloor = 7 & no & \text{new set } \{1,\dots,7 \}\\ (6) & 1.5 > \lfloor \frac{1+7}{2} \rfloor = 4 & no & \text{new set } \{1,\dots,4 \}\\ (7) & 1.5 > \lfloor \frac{1+4}{2} \rfloor = 2 & no & \text{new set } \{1,2 \}\\ (8) & 1.5 > \lfloor \frac{1+2}{2} \rfloor = 1 & yes & \text{new set } \{ 2 \}\\ (9) & 1.5 = 2 & no & \text{we have not found number 1.5 in the last set}\\ \hline \hline \end{array}\)

Thank you!!!

.02 or 2 thousandths 

0.02

Thanks Heureka,

You mean that there is a maximum of 8 steps if the element exists in the array, but 9 steps if the element does not exist?

Thanks Heureka but can you explain to me why you have to add the 1 ?

Hello Melody!

The last compare on equality, if not equal, the name does not exist in the array.

what is 4.2857142857142857 as a mixed number?

\({\color{blue}4.\overline{285714}= \frac{30}{7}=4\frac{2}{7}}\)  !

Thanks Heureka but can you explain to me why you have to add the 1 ?

Hi EMP,

I did one of these for you the other day :)

Just assume that it is the very last one that is the one you are looking for and see how lon it would take.

193 in the array.

Let the first on be in postion 0 and the last one be in position 193

1.      (0+193)/2 =  96     rounded down

2.      (0+96)/2   =  48

3.      (0+48)/2   =  24

4.      (0+24)/2   = 12

5.      (0+12)/2=    6

6.      6/2=3

7.      3/2=1

8.       1/2=0

That is 8

So I think that 9 will do it.

Compute how many steps binary search would take to find an item in arrays of various sizes.

Let n = size of the array

Let c = comparisons in the worse case

\(\begin{array}{|rcll|} \hline c &=& 1 + \log_2(n) \qquad & | \qquad \log_2(n) = \dfrac{ \ln(n) }{\ln(2)} \\ c &=& 1 + \dfrac{ \ln(n) }{\ln(2)} \\\\ n &=& 193\\ c &=& 1 + \dfrac{ \ln(193) }{\ln(2)} \\\\ c &=& 1 + \dfrac{ 5.26269018890 }{0.69314718056} \\\\ c &=& 1 + 7.59245703727 \\ c &=& 8.59245703727 \\ c &\approx& 9 \\ \hline \end{array} \)

No more than 9.

At a Fourth of July celebration, a rocket is launched straight up with an initial velocity of 125 feet per second. The height 'h' of the rocket in feet above sea level is modeled by the formula h =125t - 16t^2, where 't' is the time in seconds after the rocket is launched. 

a. what is the height of the rocket when it returns to the ground?

b. let h=0 in the equation and solve for t

c. how many seconds will it take for the rover to return to the ground?

NOTamathematician
5 hours ago     

Gerrard found the slope and y-intercept of a line using equation for the line.Is his answer correct?If not,explain what he did wrong and provide the correct answer.

write the slope and y-intercept of the equation:-3y=3/4x+(-2)

answer:slope:3/4                  y-intercept:-2

find the and and y-intercept