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Probabilities must be between 0 and 1

If something cannot happen it has a probability of 0

If something definitely will happen it has a probability of 1

6.02 x 10²³ (rounded) is avegadro's constant (it is a chemistry thing)

6.02 x 10²³ = 602000000000000000000000

To do 10 x 10^-2, move the decimal place along two (same as dividing by 10² which is 100)

i.e. 10/100 = 0.10

10 x 10^-3, move the decimal place along two (same as dividing by 10³ which is 1000)

i.e. 10/1000 = 0.01

\({1x}^{-1} = 1\left ( \frac{1}{x}\right ) = \frac{1}{x}\)

\({2x}^{-2} = 2\left ( \frac{1}{{x}^{2}} \right ) = \frac{2}{{x}^{2}}\)

\(\frac{2}{3}{x}^{-2} = \frac{2}{3}\left ( \frac{1}{{x}^{2}} \right ) = \frac{2}{3{x}^{2}}\)

\(\times {10}^{-1} = \times \frac{1}{{10}^{1}}\)which is the same as dividing by 10

\(35\times{10}^{-2}\)     \(=\)     \(35\times\frac{1}{{10}^{2}}\)     \(=\)     \(35\times\frac{1}{100}\)     \(=\)     \(\frac{35}{100} = 0.35\)

Find the value(s) of y such that the triangle with the given vertices has an area of 2 square units. (Enter your answers as a comma-separated list.)

(−5, 1), (0, 5), (−4, y)

y= ?

\(\begin{array}{lcll} \vec{A} = \binom{a_x}{a_y} = \binom{-5}{1} \qquad \vec{B} = \binom{b_x}{b_y} = \binom{0}{5} \qquad \vec{C} = \binom{c_x}{c_y} = \binom{-4}{y} \\\\ \vec{B}-\vec{A} = \binom{b_x}{b_y}-\binom{a_x}{a_y} = \binom{b_x-a_x}{b_y-a_y} = \binom{0-(-5)}{5-1} = \binom{5}{4} \\ \vec{C}-\vec{A} = \binom{c_x}{c_y}-\binom{a_x}{a_y} = \binom{c_x-a_x}{c_y-a_y} = \binom{-4-(-5)}{y-1} = \binom{1}{y-1} \\ \\ \end{array}\\ \begin{array}{rcll} 2\cdot A_{\text{triangle}} &=& | (\vec{B}-\vec{A}) \times (\vec{C}-\vec{A}) | \\ &=& | \binom{5}{4} \times \binom{1}{y-1} | \\ &=& | 5\cdot(y-1) - 4\cdot 1 | \\ &=& | 5\cdot y -5 - 4 | \\ &=& | 5\cdot y -9 | \\ \\ \hline \\ 2\cdot A_{\text{triangle}} &=& + ( 5\cdot y_2 -9 ) \\ 5\cdot y_2 &=& 9+ 2\cdot A_{\text{triangle}} \\ y_2 &=& \frac{9+ 2\cdot A_{\text{triangle}}}{5} \qquad &| \qquad A_{\text{triangle}} = 2 \\ y_2 &=& \frac{9+ 2\cdot 2}{5}\\ y_2 &=& \frac{13}{5}\\ \mathbf{y_2} & \mathbf{=} & \mathbf{2.6} \\ \\ \hline \\ 2\cdot A_{\text{triangle}} &=& - ( 5\cdot y_1 -9 ) \\ 2\cdot A_{\text{triangle}} &=& - 5\cdot y_1 +9 \\ 5\cdot y_1 &=& 9 - 2\cdot A_{\text{triangle}} \\ y_1 &=& \frac{9 - 2\cdot A_{\text{triangle}}} {5} \qquad &| \qquad A_{\text{triangle}} = 2 \\ y_1 &=& \frac{9 - 2\cdot2} {5} \\ y_1 &=& \frac{5} {5} \\ \mathbf{y_1} & \mathbf{=} & \mathbf{1} \\ \end{array} \)

Go online to this page and learn there how to do it:

https://www.mathsisfun.com/converting-decimals-fractions.html

4+4+4+1=13      the remainder is 1

(4x^3+2x^2-5)/2x-3    let's write this as

[4x^3 + 2x^2 + 0x  - 5 ]  /  [2x - 3]        now.....let's perform polynomial division

                2x^2   + 4x  + 6

2x - 3   [  4x^3 + 2x^2 + 0x  - 5 ]

               4x^3 -  6x^2

             ------------------------------

                           8x^2  + 0x

                           8x^2  - 12x

                           -----------------

                                       12x - 5

                                       12x - 18

                                        -----------

                                                13

So our answer is :   2x^2 + 4x + 6  +  13 / [2x - 3]

9.66 / 1.15=8.40 pounds before 15% tax.

-12+-22

= -12-22

= -34

That's pretty crafty, heureka.........!!!........thanks for showing me how you derived it.....

Thanks for that answer, heureka.........

could you explain how you derived your final formula?

...so far:

\(\begin{array}{rcll} a_{n}= \left\{ \begin{array}{ll} 2\cdot n-2 = (2\cdot n - 2) + 0\qquad & (\text{n odd})\\ 2\cdot n-3 = (2\cdot n - 2) -1 \qquad & (\text{n even}) \end{array} \right. \end{array} \)

...composite:

\(\begin{array}{rcll} a_n = (2\cdot n - 2) -1 \cdot s \end{array}\)

s is a switch. We need a function for a switch.

if n is even, s is on or equal 1,

and if n is odd, s is off or equal 0.

As the next we see we can switch with \((-1)^n\)

This function does switch beween -1 and 1 if n is odd or if n is even.

\(\begin{array}{lcr} (-1)^{\text{odd number}} &=& -1 \\ (-1)^{\text{even number}} &=& 1 \end{array} \)

We must modify this "switch" function to get our function.

But it is easy to do this.

First step we add 1.

\(\begin{array}{lcr} 1 + (-1)^{\text{odd number}} &=& -1+1 = 0\\ 1 + (-1)^{\text{even number}}&=& 1+1 = 2\\ \end{array} \)

Second step, the number 2 must be number 1, so we divide by 2
\(\begin{array}{lcr} \dfrac{1 + (-1)^{\text{odd number}} } {2} &=& \frac{0}{2} = 0\\\\ \dfrac{1 + (-1)^{\text{even number}} } {2}&=& \frac{2}{2} = 1\\ \end{array}\)

So our switch s is ready:

\(\begin{array}{lrcr} & s &=& \dfrac{1 + (-1)^n } {2} \\ & s &=& \frac12 [~ 1+(-1)^n ~]\\ n & \text{ is odd} & s = 0 \\ n & \text{ is even} & s = 1 \end{array}\)

5x-8=13     add 8 to both sides

5x = 21      divide both sides by 5

x = 21 / 5

Heureka probably knows how to do this better with vectors.......and maybe even come up with more values for y than I did......but, here's my attempt  :

The first value for y is pretty easily derived.......let the base be defined the two points (-5,1) and (-4,y).........if we let y = 1.........the length of the base = 1 and the height of the oblique triangle [ with  (0,5) as the "apex" point ] will be 4.....so......the area of this triangle =  (1/2)(b)(h)  = (1/2)(1)(4)  = 2 sq units

To get a second point....let the base be defined by the two points  (-5,1) and (0,5).....and the length of this base = sqrt (41)

And the equation of the line going through these two points is

y   = (4/5)x + 5   which we can write as    (4/5)x - y + 5  = 0

Using the area formula for a triangle, we can calculate the needed height  of our second triangle

2 = sqrt(41)/2 *h    →  h  = 4/sqrt(41)

So........we need to find a point with an x coordinate of -4 that is 4/sqrt(41)   units from the line

y = (4/5)x + 5

Using the equation for the perpendicular distance from a point to a given line, we have

abs [ (4/5)(-4)  -1y + 5 ] / ( sqrt [ 16/25 + 1 ] )  =  4/sqrt(41)

abs [ -16/5 -1y + 5 ] / sqrt (41)/ 5  = 4/sqrt(41)

abs [ 9/5 - y] * 5  = 4

abs [9/5 -y] = 4/5

And we have two equations here

9/5 - y  = 4/5                 or                  9/5 - y   = - 4/5

y = 9/5  - 4/5  = 1                                 y =  9/5 + 4/5  =  13/5   = 2.6

Notice that this gave us not only a second point, but also my original "guess" as well

So.......the two values of y are :     1, 2.6

Here's an  image of both triangles that we found......the first is triangle ABC   and the second is triangle ABD

In one hour, Nick does 1/12 of the job, Adam does 1/6 of the job and Mike does 1/8 of the job

So.......  1/12 + 1/6 + 1/8  =  3/8 of the job is done after one hr  and 6/8 = 3/4 of the job is done after two hours.....so  only 1/4 remains = 6/24

When Adam leaves......1/12 + 1/8  = 5/24 of the job gets done in one hr.....so....it will take the two other guys working together another  [ 6/24] / [5/24]  = 6/5 of an hour to finish = 72 more minutes

So.....the whole job is completed in 3 hrs and 12 miniutes.........therefore......Nick needn't worry

Here you go :  https://www.desmos.com/calculator/eq4xqzaeww

I think you might mean  ax^2 + bx + c   form

y = ( x -3)^2 - 5         expand

y = (x-3) (x -3)  - 5

y = x^2 - 6x + 9 - 5     simplify

y = x^2 - 6x + 4  

Note....in "standard form," y is taken to  be 0......so......the equation becomes

x^2 - 6x + 4  = 0

He had (3/5) x 130  =  78 red apples before purchasing some more.......let x be the  number of additional red ones that he pruchased

So......we have

total red apples / total apples = 5/6

[ 78 + x ]  / [130 + x ] =  5/6        cross- multiply

6[78 + x]  = 5(130 + x]     simplify

468 + 6x = 650 + 5x       subtract 5x, 468 from both sides

182 = x

Mr Soo has 130 green and red apples. 3/5 of the apples are red. After buying some more red apples,5/6 of them are now red.How many more red apple did he buy?

130 x 3/5 =78 red apples

78 + x= 5/6(130+x)

78+x=108 1/3 + 5/6x

x - 5/6x =108 1/3 - 78

1/6x=30 1/3 multiply both sides by 6

x=182 red apples that he purchased, so that:

182 + 78=260 total red apple

182+130=312 total green and red apples.

260/312=5/6 are red apples.

0.8 multiplied by 360 degrees

0.8*360 = 288

Yes, you most certainly can. These calculations are very common and simple for somebody who has a very good understanding of financial problems, such as yours. Good luck to you and wish you and your Charity the best.

Thank you very much Sir or Madam. So, we can rely on these numbers for our Budget plans?

biggest number