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Although the answer is simple and you can guess it ver fast, here is an equation you can set up

EDIT: LOL this is incorrect now I am going to ask the question: Why doesn't my strategy work?

x = bees

y = flowers

Equation:

x = y + 1

x = 2y - 1

This is because for statement 1, notice how there are one more bees than the number of flowers

and for statement 2, the flowers would have to be doubled and have one flower subtracted from that in order to equal the amount of bees

So the answer is 2 flowers and 3 bees.

The sum of three numbers a, b, and c is 99. If we increase a by 6, decrease b by 6 and multiply c by 5, the three resulting numbers are equal. What is the value of b?

Die Summe der drei Zahlen a, b und c ist 99. Wenn wir a um 6 erhöhen, b um 6 verringern und c mit 5 multiplizieren, sind die drei resultierenden Zahlen gleich. Was ist der Wert von b?

\(a+b+c=99 \)

\(a+6=b-6\\ 5c=b-6\\ c=\frac{b-6}{5}\)

\(a+b+\frac{b-6}{5}=99\\ a-b=-12\\ 2b+ \frac{b-6}{5}=111\\ 10b+b-6=561\\ 11b=549\)
\(b=51\)

\(a+6=b-6\\ a+6=51-6\)

\(a=39\)

\(5c=b-6\\ 5c=51-6\)

\(c=9\)

  !

x^2 - 12x + k  = 0

By Vieta

The sum of the roots  =  -(-12) /1  =  12 / 1  =  12

The only two positive primes that sum to  12   are   5  and 7

So  we have that

(x - 7) ( x - 5)  =  0

x^2 - 7x - 5x  + 35  = 0

x^2 - 12x  +  35  = 0

So.....k = 35

a + b + c  = 99

And if

a + 6  = b - 6

Then  b = a + 12

And  a + 6  = 5c   ⇒   [a + 6] / 5  = c  

So

a + b + c  =  99

a  + ( a + 12)  +  [ a  + 6 ] / 5  = 99      multiply through by 5

5a + 5a  + 60 +  a + 6  =  495      simplify

11a  + 66 = 495       subtract 66 from both sides

11a  = 429     divide both sides by 11

a  =  39

b = a+ 12  =  51

8x^2 + 15x + c  = 0

Using  the discriminant  we will have two real roots when

15^2  - 4*8*C  >  0

225  - 32C  > 0

225 > 32C

225/32 > C

C <  225/32

C < ≈ 7.03

So....  

1 ≤ C ≤ 7 

So the product of the possible C's   =  7!   =   5040

actually, I eliminated all the integers divisible by 1000.

I think they have used 5% simple (non-compound) increase:

[260 x 4] + [5% x 260 x 4] = 1040 + 52 =1092 m.

length of one side of park   =   √[ 4225 sq m ]  =   65 m

perimeter of park   =   4 * 65 m   =   260 m

length of each circuit  =   5% greater than 260 m   =   105% * 260 m   =   273 m

total length needed  =  4 * 273 m   =   1092 m

Mmmm....don't know......I was assuming that each new "wrap" of wire would create a "new" perimeter

I see that hectictar is answering......she probably has the correct interpretation!!!!

Ah, I did not think about using floor(log(x)) as the upper limit!! That's even better!

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Click that and fill out the info there.

wait CPhill the answer you provided didn't match the multiple-choice, here are the four answer available. "1092,  1082.31,  1094.22, and 1089.21" did you do something wrong or is there something wrong with the answer choices?

sqrt (4225)  =   65  = side of the square

So....the perimeter of the park  =  4 * 65  =  260 ft

So......the length of wire needed  is

First perimeter + Second perimeter + Third perimeter + Fourth perimeter  = 

260 + 260(1.05) + 260(1.05)^2  + 260(1.05)^3  ≈  1120.6 ft

I'm glad to hear that it's been good!! 

and LOL it feels like summer where i am but it is fall 

GOOD....!!!

It's almost FALL in Louisiana.....cooler weather.....YEAH!!!!

idk, im confused too ngl :( sorry

Let A be the altitude, AD, of ABC

Let one base angle  = x

Then the other base angle is (180 - 45 - x)  =  135 - x

So.....

tan x  =  A/2      (1)

 tan (135 -x) = A/3    (2)

Using a trig identity to simplify (2)   and subbing in (1)

tan 135  - tan x              A

_____________  =     ____

1 +  tan135 tan x            3

-1  - A/2                 A

__________  =     ____

1 + (-1) A/2              3

-1 - A/2             A

_______  =  _____      cross-multiply

1 - A/2            3

-3 ( 1 + A/2)  =  A  (1 - A/2)

- 3 -(3/2)A  = A - (1/2)A^2

(1/2)A^2  - (5/2)A - 3  = 0

A^2   - 5A  - 6   =  0      factor

(A - 6) ( A + 1)  = 0

Assuming that A  is positive.....then A  = 6

So....the area of ABC  = 

(1/2)(BD + DC) (A) = 

(1/2)(2 + 3) (6)  =

(1/2) (5) (6)  =   

15

In acute triangle ABC, \(\angle \alpha=45^\circ \) Let D be the foot of the altitude from A to BC if BD=2 and CD=3 then find the area of triangle ABC.

(Im spitzen Dreieck ABC,\(\angle \alpha=45^\circ \) sei D der Fuß der Höhe von A nach BC, wenn BD = 2 und CD = 3, dann finde die Fläche des Dreiecks ABC.)

\(tan (\alpha_C)=\frac{3}{h} \ (Part\ of\ \alpha \ on\ page\ C)\\ tan (\alpha_B)=\frac{2}{h} \ (Part\ of\ \alpha \ on\ page\ B)\\ arc\ tan \frac{3}{h}+arc\ tan \frac{3}{h}-45°=0 \)

     \( \alpha_ {\small B}\)      +         \(\alpha_ {\small C}\)         \(-45°=0\) 

\(h= 6\)

\(A=\frac{1}{2}\cdot a\cdot h\\ A=\frac{1}{2}\cdot (3+2)\cdot6\)

\(A=15\)

  !

\(f(x)=\sum^{\lfloor\log_{10}x\rfloor}_{n=0}\Bigl\lfloor\dfrac{x}{10^n}\Bigr\rfloor\operatorname{mod}10\)

\(4\cos^2(2\theta-\pi)=3\\ \cos(2\theta-\pi) = \pm \dfrac{\sqrt{3}}{2}\\ \text{We need the 3 smallest positive values of $\theta$ so we have}\\ 2\theta - \pi = -\dfrac {5\pi}{6}, -\dfrac{\pi}{6}, \dfrac{\pi}{6}\\ 2\theta = \dfrac \pi 6,~\dfrac{5\pi}{6},~\dfrac{7\pi}{6}\\ \theta = \dfrac{\pi}{12},~\dfrac{5\pi}{12},~\dfrac{7\pi}{12}\)

Maybe not as difficult as it seems.....

Note that angles XNC  and ANC  are equal

And angles ACN and XCN  are equal

And NC = NC

Therefore, by ASA, triangles ANC and XNC are congruent

Then  AN  = XN  =  5     ....so.... AX  = 10

So ....N is the midpoint of  AX

Then.......if we connect MN.......this segment is paralell  to  BX because  MN splits the sides of triangle  ABX proportionally....that is   AM/BM  = AN/XN 

So  because MN is parallel to BX, then  triangle  AMN  is similar to triangle ABX  which implies that

MN / BX  =  AN / AX

MN  / 14   =  5 / 10

MN /14  = 1/2

MN  = (1/2) (14)    = 7   as found by CU!!!

I should have been more careful.

My statement is true if for all x such that f(x) is defined, then f(-x) is defined as well.

Draw  CM   perpendicular to AB  such that M lies on AB

Then triangle CBM  is a 30 - 60 - 90 right triangle

Then  BM is opposite the 30° angle MCB

So.....MB  = 1/2 CB  =  (1/2) (1)  = 1/2

And by symmetry

DC + 2(MB)  = AB

DC + 1  =  10

DC   =  9

This is made easier IF we can find the real roots right off the bat

So

-3x^3 -6x^2  + 3x + 6  =  0        multiply through by -1

3x^3  + 6x^2  - 3x - 6  =  0        factor

3x^2 ( x - 2)  -  3 ( x - 2)  = 0

(3x^2  - 3) (x - 2)  = 0     so

3x^2 - 3 = 0        x - 2  =  0

3x^2  = 3              x  = 2

x^2  = 1

x  = ± 1  

The roots are -1, 1 , 2

So....if we can  perform  synthetic division and get  alternating signs on the bottom row....then we will have found the lower bound

Test  -1                          Test  - 2                         Test  - 3

-1  [ 3  6   -3    -6  ]      -2  [ 3   6   - 3  - 6 ]       - 3  [ 3  6    - 3  - 6 ]

           3                                  -6                                 -9      9  -18

      _____________        ______________          _____________

       3  9   NO                      3  0  NO                       3   -3     6   -24

So  x  = -3   is the lower bound

Similarly.....if we perfrorm synthetic division and get all postives on the bottom row, then we will have found the upper bound

Test 2

2  [ 3   6    - 3  - 6 ]

           6   24    42

    _____________

     3   12   21  36

So....x  = 2 is the upper bound