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∠POQ = 56 degrees

AB = sqrt(BC² + AC²)

CH = (BC * AC)/AB

x - (4x/5 + 3x/20 + 2x/60) = 50000

x = 3,000,000

uh..... do you mean \(\frac{3}{4x}?\) or \(\frac{3}{4}x\)?

EF = 10.43554858 m

\((1 + x)^{10}*(1 - x)^{10}\\ =[(1 + x)(1 - x)]^{10}\\ =[1-x^2]^{10} \)

This will have 11 terms

\(1-10x^2+ ..... -10x^{18}+x^{20}\)

Which integer is not in the range of f(x)?

Hello Guest!

-5 is not in the range of f(x).

  !

\(3x+7\equiv 2+x+5\;\;mod 16)\\ 2x\equiv 0 (mod 16)\\ x\equiv 8N\\ ~\\ 2x+11+x+3\;\; (mod16) \\ \equiv 0+x+14\;\; (mod16) \\ \equiv 8N-2 \;\; (mod16) \\ \)

So I think  \(2x+11+x+3(mod 16) \equiv 6\;(mod 16) \quad or \quad 14\;(mod 16) \)

I am not 100% certain that this is correct.     Please give feedback.

LaTex

3x+7\equiv  2+x+5\;\;mod 16)\\
2x\equiv  0 (mod 16)\\
x\equiv  8N\\
~\\
2x+11+x+3\;\; (mod16) \\
\equiv  0+x+14\;\; (mod16) \\
\equiv  8N-2 \;\; (mod16) \\

\(S_n=\frac{n}{2}(a+L) = \frac{20002}{2}(1+20002)=10001*20003 = 200\;050\;003\)

CQ = 2(9 - √33)

Both n and m ==15!

n  -  m ==0

Please resubmit your question and input your LaTex  properly, via the LaTex box (which can be found in the ribbon.)

Note: everything between $ signs is LaTex.

If you are too lazy to do this then why should anyone be bothered to answer you.

I have closed this question.

To find which two rational expressions sum to  (3x - 2) / (x² - x - 12):

1)  Factor the denominator:  (3x - 2) / [ (x - 4)(x + 3) ]

2)  This means that the two rational expressions are:  A / (x - 4)  and  B / (x + 3)

3)  Start with:  A / (x - 4)  +  B / (x + 3) 

     write as one fraction:  [ A(x + 3) + B(x - 4) ]  /  [ (x - 4)(x + 3) ]

4)  Simplify this:  =  ( Ax + 3A + Bx - 4B ) / [ (x - 4)(x + 3) ]

5)  When you compare this result with  (3x - 2) / [ (x - 4)(x + 3) ]

     notice that the numerators must be equal:  ( Ax + 3A + Bx - 4B )  =  (3x - 2)

6)  Rearranging the numerator:  [ Ax + Bx + 3A - 4B ]  =  (3x - 2)

7)  This means that   Ax + Bx  =  3x   and   3A - 4B  =  -2

     --->     Ax + Bx  =  3x     --->     A + B  =  3    

     --->                                        3A - 4B  =  -2 

8)  Solving for A and B:  A = 10/7   and   B = 11/7

9)  Which means that the two rational expressions are:  ( 10/7 ) / (x - 4)   and   ( 11/7 ) / (x + 3)

BE = 9 units

There are 100 numbers, that is the n(S)   the number in the sample space.

Prime actors of 18 are 2,3 .  The product of prime factors is 6

there are  50 factors of 2, all of those are reducible

There are  33 factors of 3 , all are reducible

There are  16 factors of 6, these have been counted twice

so number of desirable events is    50+33-16 = 67  

Prob = 67/100

Between 5000 and 8000 inclusive, there are [0 to 9]:

(814, 813, 813, 813, 813, 1542, 1542, 1542, 814, 813) >>>Total  w/o  repeats = 10319

814 integers that have at least 1 zero in them!

[20,002  x  20,003] / 2 ==200,050,003

There are 288 ways!

PLease like my question.

wait wait wait, why do have a different answer...?

*0**

The first digit can be 5, 6, 7           3 values

The second digit must be 0           1 value

The third digit can be 0-9             10 values

The fourth digit can be 0-9           10 values

3 * 1 * 10 * 10 = 300.

8000 is also a possible value.       +1

300 + 1 = 301 values.

uhm

500_               10 values (0 - 10)

510_               10 values (0 - 10)

520_               10 values (0 - 10)

530_               10 values (0 - 10)

540_               10 values (0 - 10)

550_               10 values (0 - 10)

560_               10 values (0 - 10)

570_               10 values (0 - 10)

580_               10 values (0 - 10)

590_               10 values (0 - 10)

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

                        10 * 10 = 100 values.

so when the first digit is 6 or 7, then there are (100 * 3 =) 300 values + 8000 works. So there are (300 + 1 =) 301 values.

@Straight I got a different answer...

*0**

the first digit can be 5,6,7

the third digit can be 1-9

the fourth digit can be 1-9

3*9*9=243...

so 243+243+243=729?

tell me how i am wrong because i think i am

I believe it is 3....

because 1/3/4 is equal to 4/3, which is greater than one....

I'm pretty shure I am wrong so @guest can you give an explanation?

My best shot:

Okay so i don't know how to explain this, but the hypotenuse of this triangle is the diameter of the circle, so the circle area is 

\(\pi\times r^2\)

\(169\pi\)

the area of the triangle is 

\(\frac{10*24}{2}\)

\(120\)

and $s = (10+24+26)/2 = 30$, so $r = [ABC]/s = 120/30 = 4$.

When $[ABC] = rs$, where $r$ is the inradius and $s$ is the semiperimeter of the triangle.

Therefore, the area of the incircle is $4^2\pi = 16\pi$.Finally, the area of the circumcircle is $169\pi - 16\pi = \(\boxed{153\pi}$\)

(2⁰ + 2)²              =    9 with 2022 in sequence   

    20   

—————           =  10 with 2022 in sequence

    sqrt(2²)  

.