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Whoops i was wrong

Prime Factors of 2023 = 2023 is a prime number and cannot be factorize

The only factors of 2023 are 1 and 2023

2023 divided by 1, 2023, gives no remainder. They are integers and prime numbers of 2023, they are also called composite number.

Prime Factors of 2023 = 2023 is a prime number and cannot be factorize

The only factors of 2023 are 1 and 2023

64091 divided by 1, 2023, gives no remainder. They are integers and prime numbers of 2023, they are also called composite number.

i have same question

It’s easy to see why EP made a mistake. For standard formal English, this question is a poorly constructed if-then conditional question. ...It reads like a true or false question.

The conditional statement does not have an explicit question –it has an implied question.  (However, this form is common where English is an educational language.)

A much more clear form is: 

If it was -8° C sometime between 2 pm and 2:20 pm, then at what time interval was the temperature 9° C?

As you may see, the use of the word “what” and the phrase, “time interval,” and a question mark (?) makes this clearly a question, and clearly identifies what information the asker wants. 

This question is brain-dead, MathyGoo; you should be able to answer it yourself.  

If the temperature drops three (3) degrees per hour moving into the future, then it rises three (3) degrees per hour moving into the past. How many hours into the past are needed to elevate a temperature of -8 degrees to a temperature of 9 degrees? Then determine the time interval for this value. 

GA

--. .-

Repost of MathyGoo13's repost and comment:  [ https://web2.0calc.com/questions/weather-problem ]

The temperature changed throughout a certain day at a steady rate of
- 3 ° C every hour. If it was -8° C sometime between 2 pm and 2:20
pm, then the temperature was 9° C in the time interval

(Electric Pavlov tried to answer this, but he made the mistake of thinking it said "-9 degrees" meaning his solution was wrong.)

There are 100,000 integers
S==22.........6,000 integers.
S==23.........6,000 integers.
S==24.........5,875 integers.
______________________________
     Total ==17,875 integers.

The probability is: 17,875 / 100,000
==143 / 800

sorry, it's \((x^2-4x+2)^{x^2-5x+2}=1\)

bruh

no, it's 2/5

Since BC = 14 and BK = 7, CK = 7.

Since AK is an altitude, triangle(AKC) is a right triangle with hypotenuse AC.

Using the Pythagorean Theorem:  AK² + KC²  =  AC².

--->   AK² + 7²  =  10²

Find the value of AK.

The height of the triangle is AK and the base is BC.

Since you know these two values, you can find the area of the triangle.

The answer to a very similar problem was posted here:  

https://web2.0calc.com/questions/partial-fractions_8

See if you can follow that example.

25^o 33' 11"  =  25 + 33/60 + 11/3600  =  25.5530556^o

sine(25.5530556^o)  =  0.4313467008

6t² + 30  =  41t + 7t - 2t²

1)  Simplify the right side:  6t² + 30  =  48t - 2t²

2)  Get all the terms to the left side:   6t² + 30 - 48t + 2t²  =  0

3)  Simplify the left side:  8t² - 48t + 30  =  0   --->   4t² - 24t + 15  =  0

4)  Use the quadratic formula to get the two solutions:

     x  =  [ 24 + sqrt( (-24)² - 4(4)(15) ) ] / (2·4)  =  [ 24 + sqrt(336) ] / 8

                       =  [ 24 + 4sqrt(21) ] / 8  =  [ 6 + sqrt(21) ] / 2

    and  x =  [ 6 - sqrt(21) ] / 2

7)  Finish by subtracting the smaller solution from the larger solution.

(3x^2 + 6x + 5) is defined on the interval [2, infinity)

(2y^2 + 8y + 10) is defined on the interval [2, infinity)

Therefore, both 3x^2 + 6x + 5 = 2, and 2y^2 + 8y + 10 = 2.

Solve the equations and you have your answer :))

=^._.^=

Me too lol...

\(3.78^3 = 54.010152\)

\((10^3)^3 = 10^9\)

Move the decimal point to the left 1, and you have : \(5.4010152\)

Because we moved the decimal point back by 1, we add 1 to the power of 10, making the answer: \(\color{brown}\boxed{5.4010152 \times 10^{10}}\)

Thanks melody, I didn't realize the whole thing reduced to 3 haha

Also, if you were to go to WolframAlpha and try to minimize the function, it gives the correct minimum value but at a strangely specific point:

 g(x) = sqrt((x - 3)^2 - (x - 18)^2)

\( g(x) = \sqrt{(x - 3)^2 - (x - 18)^2}\\ g(x) = \sqrt{(x^2 - 6x+9) - (x^2 -36x+ 324)}\\ g(x) = \sqrt{ - 6x+9 +36x- 324}\\ g(x) = \sqrt{ 30x- 315}\\ 30x-315\ge 0\\ 30x\ge315\\ x\ge 10.5\)

Thanks guest.  You are totally correct.

I failed to recognise that all the even numbers have a factor of 2  

I'll start again

4 numbers are randomly selected from the subset (1,2,3,4,5,6,7,8,9) . What is the probability that the product of the 4 randomly selected integers is a multiple of 14?

One of the numbers has to be 7.  AND at least one has to be even.

Forget about the 7 and there are 4 even and 4 odd numbers from which to choose the other numbers

3 evens                                  4C3                       = 4

2 evens and an odd                4C2*4C1 = 6*4   = 24

1 even and 2 odds                 4C1* 4C2 = 4*6   = 24

  Total  4+24+24 = 52               (just like guest said)

Prob = 52 / 4C9 =  52 / 126

Now we are in agreement.

Melody: The computer lists 52 such numbers as follows:

1237 , 1247 , 1257 , 1267 , 1278 , 1279 , 1347 , 1367 , 1378 , 1457 , 1467 , 1478 , 1479 , 1567 , 1578 , 1678 , 1679 , 1789 , 2347 , 2357 , 2367 , 2378 , 2379 , 2457 , 2467 , 2478 , 2479 , 2567 , 2578 , 2579 , 2678 , 2679 , 2789 , 3457 , 3467 , 3478 , 3479 , 3567 , 3578 , 3678 , 3679 , 3789 , 4567 , 4578 , 4579 , 4678 , 4679 , 4789 , 5678 , 5679 , 5789 , 6789 , Total = 52 such numbers.

Probability ==52 / 126 ==26 / 63

\(\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}\\ =\frac{(y-x)^2(x-y)}{(y-z)(z-x)(x-y)} + \frac{(z-y)^2(y-z)}{(z-x)(x-y)(y-z)} + \frac{(x-z)^2(z-x)}{(x-y)(y-z)(z-x)}\\\\ =\frac{[(y-x)^2(x-y)]\;\;+\;\;[(z-y)^2(y-z)]\;\; +\;\; [(x-z)^2(z-x)]}{(y-z)(z-x)(x-y)} \\ =\frac{[-(y-x)^2(y-x)]\;\;-\;\;[(z-y)^2(z-y)]\;\; -\;\; [(x-z)^2(x-z)]}{(yz-yx-z^2+xz)(x-y)} \\ =\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\; +\;\; [(x-z)^3]\}}{(yz-yx-z^2+xz)(x-y)} \\ =\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\; +\;\; [(x-z)^3]\}}{(yzx-yzy-yxx+yxy-z^2x+z^2y+xzx-xzy)} \\ =\frac{-\{[y^3-3y^2x+3yx^2-x^3]\;\;+\;\;[z^3-3z^2y+3zy^2-y^3]\;\; +\;\; [x^3-3x^2z+3xz^2-z^3]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\ =\frac{-3\{[-y^2x+yx^2]\;\;+\;\;[-z^2y+zy^2]\;\; +\;\; [-x^2z+xz^2]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\ =\frac{-3(zy^2+yx^2 -y^2x+xz^2-z^2y-x^2z}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\ =3 \qquad \text{so long as the denominator isn't 0}\)

I'm sure Textot's formula is what you were supposed to use.

LaTex

\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}\\
=\frac{(y-x)^2(x-y)}{(y-z)(z-x)(x-y)} + \frac{(z-y)^2(y-z)}{(z-x)(x-y)(y-z)} + \frac{(x-z)^2(z-x)}{(x-y)(y-z)(z-x)}\\\\
=\frac{[(y-x)^2(x-y)]\;\;+\;\;[(z-y)^2(y-z)]\;\;  +\;\;  [(x-z)^2(z-x)]}{(y-z)(z-x)(x-y)} \\
=\frac{[-(y-x)^2(y-x)]\;\;-\;\;[(z-y)^2(z-y)]\;\;  -\;\;  [(x-z)^2(x-z)]}{(yz-yx-z^2+xz)(x-y)} \\
=\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\;  +\;\;  [(x-z)^3]\}}{(yz-yx-z^2+xz)(x-y)} \\
=\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\;  +\;\;  [(x-z)^3]\}}{(yzx-yzy-yxx+yxy-z^2x+z^2y+xzx-xzy)} \\
=\frac{-\{[y^3-3y^2x+3yx^2-x^3]\;\;+\;\;[z^3-3z^2y+3zy^2-y^3]\;\;  +\;\;  [x^3-3x^2z+3xz^2-z^3]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\
=\frac{-3\{[-y^2x+yx^2]\;\;+\;\;[-z^2y+zy^2]\;\;  +\;\;  [-x^2z+xz^2]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\
=\frac{-3(zy^2+yx^2 -y^2x+xz^2-z^2y-x^2z}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\
=3  \qquad \text{so long as the denominator isn't 0}

Guest, you made a small mistake.  I like seeing your answers though. :)

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I'm going to try and solve in pairs

m=-1 mod 6 and 1 mod 10

11  works

m=4 mod 7 and 1 mod 10

11 works

now I look, 11 works for all of them

---------

So the solutions will be of the form    m=(5*7*8*9)k + 11

the first 2 are 11 and  (5*7*8*9) + 11

the 3rd one is         2(5*7*8*9) + 11   and this is bigger than    6*7*8*9

so

There are just 2          they are     11 and  2531

The inverse of funtion x cannot be a function because for one segment of f(x)=-2  The inverse of that is  just -2

      I mean   for the function     For -3<=x<=-1     f(x)=-2

      the inverse would be when x=-2   \(f^{-1}(2)\)   can be any number between -3 and -1   

      Hence there is no unique value of \(f^{-1}(2) \)  Hence \(f^{-1}(x)\)    is not a function.

f(x)         domain [-3,5]    range  [-2,3]

\(f^{-1}(x)\)     domain [-2,3]         range  [-3,5] 

I assume for the last bit

 (x,y)  --> (y,x)     becasue the inverse of a function is its reflection about the line y=x

Here is a mapping