All Answers 358468 Answers

Hallo

24^3=13824

28 ( 1 + .0725)  = $30.03

This is Fahrenheit

Celsius = (F - 32) (5/9)   = [70 - 32] (5/9)  = [ 38](5/9)  about 21.1° C

Kelvin =  [ F - 32 ] (5/9) + 273.15  =  [ 70 - 32] (5/9) + 273.15  = about 294.26° K

This is just a linear equation

-7h-5=4-4h      add 7h to both sides, subtract 4 from both sides

-9  = 3h       divide  both sides by 3

-3 = h

Look  here :  http://web2.0calc.com/questions/slope-intercept_2

Ex:  You are given two points...let's say  (1, 3)  and (4, 9)

First....find the slope,  (m)  =  [9 - 3 ] / [ 4 - 1]   =  6/3   = 2

Use the slope and one of the points to write the equation in slope-intercept form....I'll use the first point.....although, the second would work, too..........so we have, in point-slope form :

y - y₁  = m (x - x₁)       where y₁ = 3, m is the slope = 2, and x₁  = 1   so we have

y - 3 = 2(x - 1)   simplify the right side

y - 3 = 2x - 2      add 3 to both sides

y = 2x + 1      

And that's in slope-intercept form, i.e,  y = mx + b   

Hope that helps !!!

Deez Nutz

(2.2*10 to the -5th power)(2.4*10 to the 4th power  =

2.2 * 2.4  * 10^-5 * 10⁴ =

5.28 * 10^-1

i really need help

6(2x)^1/3   =  (216 *2x) ^1/3 =   (432 x)^1/3  

don't know bruv 

Press "2nd" key first and then "asin" key followed by = sign.

Go to this link and read the "Examples" section. it is all explained there. Here is the link:

https://en.wikipedia.org/wiki/Wien%27s_displacement_law

jeezzz why bother bruv 

the square root of -96 doesn't exist over the real numbers.

However over the complex numbers  if you have some number 

^{\(-a, ~~~~a>0 \\ \mbox{then }\sqrt{-a} = \sqrt{-1 \times a} = \sqrt{-1}\times \sqrt{a} = \imath \sqrt{a} \\ \mbox{The symbol } \imath \mbox{ also referred to as just }i \mbox{ or in electrical engineering as }j \\ \mbox{stands for }\sqrt{-1} \mbox{ and is a Complex number.}\)}

Just enter your fractions normally, e.g. 1/4. Type 1 then / key followed by 4. Look at very top dark area above the calculator and will see everything displayed there. Don't forget to press = sign to get your answer.

-2n+n+4=24

      -n+4=24    |-4

          -n=-20   |*(-1)

           n=20

This problem is a bit tricky.  They are mixing mass and weight and hoping you'll notice.

The gravitational potential is given by \(U=m g h\)

where g is the acceleration due to Earth's gravity and is 9.8 m/s²

But you have to be careful in that \(weight=m g\)

So in calculating the potential here we have to sum up all the weights and multiply them by the height.

\(TotalWeight = 4 \times 1000 \times g + 6 \times 150 =\\ 4000(9.8) + 900 = 40100 n \\ \mbox{and so the potential is}\\ U=112\times 40100 = 4491200 J ~~~~~\mbox{(J is for Joules.)}\)

3 5/6-2 1/2=3 5/6-2 3/6=1 2/6=1 1/3

What is 3 5/6 - 2 1/2?

First you turn them into improper fractions:

23/6 - 5/2, find the common denominator, which is 6,

23/6 - 15/6

8/6=1 1/3

343000000000000000000000000000000 in standard form please. I got 3.43*10^32

\(3\ 43\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000 = 3. 43 \cdot 10^{32}\)

This isn't the greatest question in the world.

What they are trying to get at is that the number of trainees that are still employed is a binomial random variable, in this case with parameters \(n=6, p=0.2\)

You should know or can look up that the variance of a binomial random variable is given by

\(\sigma_{binomial} = n p (1-p) \\ \mbox{so here we have}\\ \sigma = 6 (0.2)(0.8) = 0.96\)

The degrees apart in LONGITUDE depend on which direction you are measuring....the SMALLEST difference in LONGITUDE would be: (measuring Westward from Toronto to Melbourne)

(180 - 79.4E) + (180 - 144.96 W) = 135.64

Measuring the OTHER direction (Eastward from Toronro to Melbourne) would result in:

79.4E + 144.96W = 224.36

(note that the two answers add up to 360 !)

Find all zeros of f(x)=2x^3+3x^2-23x-12

Find all zeros of f(x)=2x^3+3x^2-23x-12

\(2x^3+3x^2-23x-12=0\)

1. Try all divider of 12: \(\pm 1, ~ \pm 2,~ \pm 3,~ \pm4, ~\pm 6,~\pm12\)

We find \(x_1 = 3\) and \(x_2 = -4\)

2. 

\((x-3)(x+4) = x^2+4x-3x-12 = x^2+x-12\)

\(2x^3+3x^2-23x-12 : x^2+x-12=2x+1\)

\(2x_3+1 = 0\\ 2x_3 = -1\\ x_3 = -\frac{1}{2}\)

We have zeros  \(\mathbf{x_1=3 \qquad x_2 = -4 \qquad x_3= -\frac{1}{2}}\)