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Thanks Alan :)

Both methods are correct. 

Melody has made a small numerical error in solving the last quadratic for reaching the rear of the truck. The value under the square root sign should be 601 rather than 761.

The other differences are due to small differences used for the numerical approximations used for 1/root2.  (I get values for velocity between 15.83 and 17.04 m/s).

Many of the people who answer questions here are retired, hence their eye sight has deteriorated.

You will increase your chance of an answer if a microscope is not needed to discern the question.

Simplify.

\(\begin{array}{|rcll|} \hline && \dfrac{ \dfrac{x^2+2x-3}{x-4} } { \dfrac{2x^2+5x-3}{x^2-16} } \\\\ &=& \dfrac{(x^2+2x-3)}{(x-4)} \cdot \dfrac{(x^2-16)}{(2x^2+5x-3)} \quad & | \quad x^2-16 = (x-4)(x+4) \\\\ &=& \dfrac{(x^2+2x-3)}{(x-4)} \cdot \dfrac{(x-4)(x+4)}{(2x^2+5x-3)} \\\\ &=& \dfrac{(x^2+2x-3)}{1} \cdot \dfrac{(x+4)}{(2x^2+5x-3)} \\\\ &=& \dfrac{(x^2+2x-3)(x+4)}{(2x^2+5x-3)} \quad & | \quad x^2+2x-3 = (x-1)(x+3) \\\\ &=& \dfrac{(x-1)(x+3)(x+4)}{(2x^2+5x-3)} \quad & | \quad 2x^2+5x-3 = (2x-1)(x+3) \\\\ &=& \dfrac{(x-1)(x+3)(x+4)}{(2x-1)(x+3)} \\\\ &=& \dfrac{(x-1)(x+4)}{(2x-1)} \\\\ &\mathbf{=}& \mathbf{ \dfrac{x^2+3x-4}{(2x-1)} } \\ \hline \end{array}\)

What is 2logx+logy−2logz2logx+logy−2logz written as a single logarithm?

\(\begin{array}{|rcll|} \hline && 2\log(x)+\log(y)-2\log(z) \\ &=& \log(x^2)+\log(y)-\log(z^2) \\ &=& \log\left(\dfrac{x^2y}{z^2} \right) \\ \hline \end{array}\)

3! 6! 5! 4!

Wow! That is more than 'a couple' of graphs...

1   Period is the same as wavelength....one full wave of this one is 4

2  Minimum is just that....the lowest point on the wave     1

3  Maximum is the hight point    5

4  wave goes from  -7 to +1  midline is exactly between these at y= -3

5  Ampltude is height from midline...or 1/2 the total height  this one is 6 tall amplitude = 3

Hey superman!

The goal is to count in base 3 using only binary digits.

The \(100^{\text{th}}\) smallest positive binary integer is \(100 = 1100100_2,\) 

so the \(100^{\text{th}}\) smallest positive integer that can be written with only the binary digits is

 \(1100100_3 = \boxed{981}\).

I hope this helped,

Gavin.

The probability that a girl is chosen is 2/3 the probability that a boy is chosen.

This means that if there are x boys, there are a total of 2/3 x girls. 

The total amount of kids in the class is 2/3x + 1x = 5/3x.

This is equal to 20. 

\(\frac53x=20\Rightarrow{x}=12\)

Therefore, there are \(12\cdot\frac23=8\) girls. 

We can verify this. The class has 20 people, 8 girls and 12 boys. 

The probabilty a girl is chosen is 8/20 and boy is 12/20. 

8/20 is also 2/3 of 12/20, so our answer is correct.

I hope this helped,

Gavin

First we need to figure out how many ways to draw two cards from a standard deck of 52 cards. 

To do this, we get: \(\binom{52}{2}=1326\)

The ways to chose them so their rank is the same is harder. 

There are 13 different ranks, and in each rank, their are 4 suits. 

To choose two cards in a rank, we have \(\binom42=6\) ways to do this. 

Since there are 13 ranks, we have: \(13\cdot6\) ways to chose the cards so their rank is the same. 

The probabilty is then \(\frac{78}{1326}=\frac1{17}\)

I hope this helped,

Gavin.

This is a reapeat question. Now ity has been answered twice.

One question per post thanks and not multiple questions posted in the same time frame.

$ charge = $20* #days + $.12 * miles

Hi EP

It is good that we have done it different ways. One (or both)  of us has probably made a little mistake as the answers should be closer. Still, the asker probably wanted one way or the other, now they have both.  :)

if you draw a 5 x 5 grid, you will see a 3 x 3 grid of cubes that have only one side painted

6 sides x 3 x 3 =54 cubes with only one side painted .

B   3 x 3 x 3 = 27 no paint.

C  125 cubes x 6 sides ea = 750 sides total

         25 x 6 sides are painted = 150 painted

     150/750 = 1/5

100 thru 999 is 900 numbers

110 is 11 x10

990 is 11=90

10 thru 90 is 81 numbers

81/900 = 9/100

1

x=3 y=-7   Hypotenuse = sqrt((-7)^2 +3^2)= sqrt(58) 7.616

sin =  -7/sqrt58    =-7sqrt58/58

2

In quadrant II    x is negative   and y is positive       they are equal magnitude because  x+y = 0

tan = sin/cos    =  -1

3

reference angle is terminal side referenced to x- axis   so  pi/4

Write ALL numbers from 1 to 342 in base 7 as 3 digits, such as the following:

000, 001, 002, 003, 004, 005, 006, 010, 011, 012, 013, 014, 015, 016.........and so on to 666, and  1000. This gives you a total of:
3 x 49 x 7 =1,029 digits in 342 numbers in base 7. Since each number from 0 to 6 is equally present in the above 1029 digits, it, therefore, follows that there are:1,029 / 7 =147 fours and 147 fives, in any combinations of the two.

8)

2 cos^2 = 1

cos^2 = 1/2

cos = +- sqrt (1/2) = +-.707       pi/4   3pi/4   5pi/4   7pi/4

The sample mean is the average of all the numbers in the sample. 

\(\frac{25+23+26+29+37+23+32+29+25+22+31+34}{12}=\frac{336}{12}=\boxed{28}\)

Therefore, 28 is the sample mean.

The population mean is the average of the whole set. 

This is already given in the problem, 27.

I hope this helped,

Gavin.

See first posting

I am going to assume it is 14 cm high (not 4 m as in your other posts)

Volume = pi r^2  h       h= 14 cm     volume = 2500 cm^3

2500 = pi r^2  (14)       Solve for r= 7.54 cm

1.

Yes 

No 

Yes

No

2.

Experiment

Observational study 

Survey 

hOpe i LE HeLp

Solve for all solutions 0-2pi

Sec^2x = 4tan^2x

I would add  11pi/6   (same as -pi/6 but in the 0-2pi range)