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HECTICTAR!!!!..good to see you!!!!..where have you been keeping yourself????????

Very nice answer.....this one had me stumped   !!!!!

Hi Rom,

Thanks for your answer  

but...

I do not understand that initial plug in.  Why have you used the conjugates??

Could you explain a little more please. 

Note that area in question  =  area of △UYZ  +  area of trapezoid UYXW

m∠VUW + m∠WUY  =  90°

m∠ZUY + m∠WUY  =  90°

So

m∠VUW  =  m∠ZUY

By ASA congruence, △UVW is congruent to △UYZ.

area of △UVW  +  area of trapezoid UYXW  =  area of square UVXY  =  5 * 5  =  25 sq units

Since △UVW is congruent to △UYZ,

area of △UYZ  +  area of trapezoid UYXW  =  area of △UVW  +  area of trapezoid UYXW  =  25 sq units

Your reasoning looks good to me!

As follows:

1(n) + 2(n^2)   =  0.28

2n^2  + n  - 0.28  = 0      [ multiply through by 100 ]

200n^2 + 100n - 28  = 0

50n^2 + 25n - 7   = 0

(5n - 1)(10n + 7)  = 0

Only the second factor will produce a negative n

10n + 7  = 0

10n  = -7

n  = -7/10

We need to find the   x values (if any) that make  the inside function < 0

A graph seems best [ IMHO  !!! ]  

Look at the graph here : https://www.desmos.com/calculator/im64nbdu1v

The function under the radical is always > 0 for all x values....so...the domain is ( -inf, inf)

And the graph of the function itself, here, https://www.desmos.com/calculator/xytn5zb5ky

Shows the range to be   ( ≈ 111.504, inf )

Here's my best attempt....

4, a, b       b, c, 5                                                         

b + (c - b) + (c - b) = 5              

2c - b  = 5

b = 2c - 5

4b  = 8c - 20    (1)

4 * (a/4)^2  = b            

a^2/4 = b

a^2 = 4b    (2)       equate (1) and (2)

a^2  = 8c - 20               

c        a         b

9        x

8        x

7        6        9

6        x

5        x

4        x

3        2        1

The largest possible  "abc"  is 697

Proof

4, 6 , 9     is geometrc if  r  = 1.5

9, 7,  5      is  arithmetic if  d  = -2

c-b=5-c

2c+b=5

a*a/4=b

2c+a^2 /4=5

we try to make a as big as possible while still being an interger

a = 4 

c is not interger

a=2 is the biggest 

so c=2

and b=1

x + y + 3z  = 10     multiply  through by -2  ⇒   -2x - 2y - 6z  = -20   (1)

-4x +2y + 5z  = 7   (2)

kx + z  = 3            (3)

Add (1)  and (2)   and we have that       -6x - z  = -17   multiply through by -1  ⇒   6x + z  = 17    (4)

So....we have that....using (3)  qnd (4)

6x + z  = 17    and

kx + z  = 3

And this will have no solution if  k  = 6

\(\text{The polynomial evaluated at 1 is the sum of the coefficients. This must be a rational number}\\ \text{Thus each root that is not a rational number must have it's conjugate also as a root}\\ \text{This will be all the roots of the form }a + \sqrt{b} \text{ where }\sqrt{b} \not \in \{2,3,\dots,1001\}\)

\(\text{There are 30 numbers in 2-1001 that are perfect squares}\\ \text{So }1000-30=970 \text{ of the roots must also have their conjugate as a root}\\ \text{Thus we have a total of }\\ 30 + 2\cdot 970 = 1970 \text{ is the smallest possible degree of the polynomial}\)

I'm pretty sure this is correct but I might be missing something.  Someone else should take a look.

How many five-digit numbers have distinct digits which are decreasing from left to right? (For example, 96531 is such a number.)

1- ABCDE. I will give this a shot.

2 - For A, you have 9 choices.

3 - For B, you still have 9 choices out of 10, because we are excluding A

4 - For C, you have 8 choices because we are excluding A and B

5 - For D, you have 7 choices because we are excluding A, B, and C.

6 - For E, you have 6 choices because we are excluding A, B, C and D.

7 - Then we have: 9 x 9 x 8 x 7 x 6 =27,216 - numbers.

8 - The biggest such number would be: 98765.

9 - The smallest such number would be: 43210

10 - The difference:98765 - 43210 =55,555 from which 27,216 would be descending numbers.

Note: somebody should check this. Thanks.

1 - ABCDE:

1- For A you have 9 choices.

2 - For B, you have 5 choices if A is even and 4 choices if A is odd.

3- For C, you have 7 choices exlcluding A and B and either a 0 or a 5 which must be at the beginning of the number.

4- For D, you have 6 choices excluding A, B, C and either 0 or 5

5-For E, you ONLY have 2 choices, since the number is a multiple of 5, it must begin with either 0 or 5.

Conclusion: When A is EVEN, you have: 9 x 5 x 7 x 6 x 2 =3,780 - such numbers.

                    When A is ODD, you have  : 9 x 4 x 7 x 6 x 2 =3,024 - such numbers.

                    We add the 2 numbers together =3,780 + 3024 =6,804 - The Total.

Thanks Guest

Just reverse the process by going backwards:

1 - 243 This is after she pressed the x key. So,it must have been:

2 - 324 Since 2nd last step was squaring the number, then that means we must take its square root:

3 - Square root of 324 is 18. Then she pressed the x key and this becomes:

4 - 81 - Since the first step was to square the number, that means we do the reverse by taking its square root:

5- Square root of 81 is 9.

6- 9 is the number that Thea started with:

7 - Check: 9. square it =81. press the x key and you get: 18. Square it again and you get: 324. Press the x key and you get: 243.

I don't have time so I'm going to answer the last one

If you are to choose 4 out of x students it will be \(\frac{x!}{4!(x-4)!}=y\)

Then if the number of students in the group increases by 1, the committees double.

So \(\frac{(x+1)!}{4!((x+1)-4)!}=2y\)

So we have two equations with 2 variables.

You can try and solve now, (try substituting)

xxxxxxxxxxxx

You originally said 36, what was wrong with that answer?

2- See the solution here: https://web2.0calc.com/questions/homework-problem-and-need-help#r1

No problem...and the answer to the second question should be \(\frac{8!}{3!*5!}=\boxed{56}\) different sequences.

Thanks!

Yes, congrats to both of you!

1. We can use the handshakes formula here...n(n-1) / 2 =21 n(n-1)=42, n=7 people at the party.

2. Think of arranging three identical letters with five other identical letters. How many ways are there to do this?

1 - 66!_(12) =a162b4b289973a1a4b2336638b21ba13a710b618a5bb5a83343a2840,000,000,000,000,000,000,000,000,000,000_12

It has 31 zeros

Ok, when there is a geometric sequence, each TERM is multiplied by the common ratio to get the NEXT TERM.

So you can make the common ratio 5, for example.

Geometric sequence: She starts with 1 penny, and adds five times her current amount to her penny collection every day.

How many pennies she would have after 10 days?

Formula for sum of finite geometric sequence \(S_n=\frac{a(r^n-1)}{r-1}\)

Plugging in what we know - 

n = 10 (term)

r = 5 (common ratio)

a = 1 (first term)

9765624/4 = 2441406 pennies.