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Very nice, mobro   !!!!

Note that  this function is ever-decreasing from 0  to  infinity

At 0, (1/3)^0  =  1

As x increases without bound, the function approaches  0   [ but never quite gets there ]

So....the range  on   [0, inf) =   (0, 1 ]

For the first row, we want to choose any 4 of the 9 pots  and arrange them in 4! ways

So  this is   C(9,4) * 4!   = 3024

For the second row, we  can arrange  any of the 5 remaining pots in 5!  ways  = 120 ways

So.....the total arrangements  are

3024 * 120  =

362880

vf = vo  +   a t

10 = 35 + (-5.35) t

t = 4.673 s

xf = xo + vo t  + 1/2 a t^2

xf = 0  + 35 (4.673) + 1/2 (-5.35)(4.673)^2

xf = 105.14  m

There are 4120 ways.

Here is my attempt at your question:

(7C2 * 7C3 * 9C4 * 7C5) ==1,944,810 - Number of ways the committee can be chosen.

Triangle ABC is an isosceles triangle with a base of 8 units.

angle (B) = 2(arctan3/4)

angle (A/2) = 90 - angle (B)

height AD = BD / tan(A/2)

area = BD * AD = 54.85714289 units squared

a mod 37 = 16
b mod 37 = 21
(a - b) mod 37 = n, solve for n

a = 16  and  b = 21
(a - b) mod 37 = n

(16 - 21) mod 37 = - 5

n =37 + (-5) = 32

124 , 132 , 134 , 142 , 152 , 154 , 214 , 234 , 254 , 312 , 314 , 324 , 342 , 352 , 354 , 412 , 432 , 452>>Total =  18 such numbers

See https://web2.0calc.com/questions/help-is-needed_5.

tysm

(x+7)(x-1) =0

x^2 +6x-7=0

So we know that xy=20, xz=35, and yz=14. If we multiply them together to form an equation, we get:

\(xy \cdot xz \cdot yz = 20 \cdot 35 \cdot 14\)

Finding the product, we have: 

\(x^2y^2z^2 = 9800\)

This is the same as 

\((xyz)^2 = 9800\)

which finding the square root gives us 

\(xyz=\sqrt{9800}\)

Simplifying, we find: 

\(\sqrt{9800} \implies \sqrt{4900 \cdot 2} \implies \sqrt{4900} \cdot \sqrt2 \implies \boxed{70\sqrt2}\)

First you can shuffle the order of the races.....   for intance :    vp  sec   pres     there are   6 ways to do this

then  pres can have    6 P 6 = 720 ways

VP can have                4 P 4 =  24 ways

Sec can have               4 P 4 = 24 ways

6 x 720 x 24 x 24 = ..............   ways to list the ballot

x = original price

   increase 50 %  =     1.5 x

      now decrease 40%  (there will be 60% left)          .6(1.5x)= .9x         .9 is 90%  of original x

We have the form  P(x) = ax^3 + bx^2 + cx^2  + d

Since f(0) =  0  .....then d   =  0

So we actually  have this system of equations

a(-1)^3  + b(-1)^2  + c(-1) = 32

a(1)^3  + b(1)^2   +  c(1)  = -24

a(2)^3  + b(2)^2  + c(2)  = -28           simplify these

-a + b  - c  =  32   (1)

 a  + b + c  = -24   (2)

8a  + 4b  + 2c  =  -28    (3)

Add (1)  and (2)

2b = 8

b = 4

Sub this into  (2) and (3)

a + 4 + c  =-24 →  a + c  = -28   (4)

8a + 4(4) + 2c  = -28  →   8a + 2c = -44→  4a + c = -22  →  -4a - c = 22   (5)

Add (4)  and (5)

-3a = -6

a = 2

Using (4)

2 + c  = -28

c  = -30

Our polynomial is

P(x)  =  2x^3 + 4x^2 - 30x

Setting this to 0  and dividing through by 2, we have that

x^3 + 2x^2  -15x  = 0           factor

x (x^2 + 2x - 15)  =  0

x ( x -3) (x + 5)  = 0

Setting  each factor to 0  and  solving for x, the intercepts are

x = 0    x = 3  and  x  = -5

See the graph here : https://www.desmos.com/calculator/zrdafdxpea

1 - N =13D  +  11, where D =0, 1, 2, 3........etc.

The smallest N = 11

5,000 - 11 =4,989

4,989 / 13 =383 - these are multiples of 13 beween 11 and 5000.

So you have a total of 383 + 1 =384 integers between 1 and 5000. They begin as follows:

[13 * 0 + 11] =11.  [13 *1 + 11] =24.  [13 * 2 + 11]=37.  [13 * 3 + 11] =50.....and so on until  [13 * 383  + 11] =4,990.

2 - N =156D  +  89, where D=0, 1, 2, 3......etc.

The smallest N = 89

5,000  -  89 =4,911

4,911 / 156 =31 - these are multiples of 156 between 89 and 5,000.

So, you have: 31 + 1 =32 integers between 1 and 5,000.

Follow exactly the same procedure as above.

5C 2  =  10

So we have ten possible sums

1 + 2  = 3       2 + 3  = 5      3  + 4  =  7      4 + 5  =  9

1+ 3  =  4       2 + 4  = 6      3 +  5  =  8

1 + 4 =  5       2 + 5 =  7

1 + 5  = 6

Expected value  =

3 *  Prob of occurence  + 4 * Prob of occurence + 5*Prob of occurence + 6 * Prob of occuence +

7 * Prob of occurence   + 8 * Prob of occurence  + 8 * Prob of occurence

3 (1/10)  + 4(1/10) + 5(2/10) + 6(2/10)  + 7 (2/10) + 8 (1/10) + 9(1/10)  =

3/10  + 4/10  + 10/10  + 12/10 + 14/10  + 8/10 + 9/10 =  

60/10

6

Consider triangle ABC...... the interior angle of a regular octagon = (6/8) 180  = 135°

So angles BCA and BAC are supplemental to this =  45°

Thus sides BA = BC 

And ABC is a right angle...so BA and BC are legs

And calling the length of these  sides, S......we  have  that

area of ABC =  (1/2) (product of legs)  =   (1/2) S^2

1  = (1/2) S^2

2 = S^2

S = sqrt (2)

Then AC =  sqrt [ BA^2  + BC^2 ]  =  sqrt  [  (sqrt (2))^2  + (sqrt (2))^2 ]  = sqrt [ 2 + 2]  = sqrt 4  =  2

Then the side of the octagon =  2

A formula for the octagon's area =  2 ( 1 + sqrt 2 ) * S^2 =

2 ( 1 + sqrt 2) * 2^2  =

8 ( 1 + sqrt 2)  units ^2  ≈   19.3 units^2

My mistake, I thought that AC was 25 for a moment (from a similar problem). Anyhow, Jugoslav's answer has a wonderful explaination.

Using the triangle inequality which says that the sum of any two sides of a triangle is greater than the remaining side, we have two triangles to consider

Top triangle....

22 + 18   = 40

So the greatest integer value that we can have for the dashed line is 39

18 + x  must be greater than 22

So the least integer value that we can have for the dashed line is 5

So the interval  that  solves  the dashed line in the top riangle is  [ 5, 39]     (1)

Bottom triangle

21 + 25 =  46

So the greatest integer value that we can have for the dashed line is  45

21 + x  = must be greater than 25

So the least integer value that we can have for the dashed line is 5

So the interval that solves the dashed line in the bottom triangle is  [5, 45 ]    (2)

Taking the intersection of (1)  and (2)  we have  that the   interval that solves the problem is

[ 5, 39 ]

So....the number of whole numbers that satisfies  the dashed line length  is

39  - 5  +  1   =

34  +  1  =

35

How many sides would there be in a convex polygon if the sum of all its angles is greater than 1070 degrees, but less than 1250 degrees?   

Assuming you mean the sum of its interior angles. 

The formula for the sum of the interior angles of a polygon with n sides is (n – 2)(180)  

Divide 180 into 1070 and you get 5.944444  .....  that quotient would be (n – 2) in the formula  

But you can't have 0.944444 of a side, so you have to raise that quotient to 6  

Now (n – 2) would be 6 and so n would be 8 .....  your polygon has 8 sides  

Check the answer:     (8 – 2)(180) = 1080  .....  is this between 1070 and 1250?  Yes. 

_.

1.)    60% of the boys    .6 (100) = 60 boys in Honors

         60% of girls          .6 ( 80)   = 48 girls in Honors           total 108 Honors students    and   72 nonHonors in the class

      Total number of students who passed = 75+65 = 140  passed

          80 % of this 140 are Honors    .8(140) = 112       

                       the rest are non-Honors who passed  = 28

                     28 nonHonors passed  out of 72 total nonHonors

                                                                        28/72 x 100% = 38.9% of nonHonors passed the test 

Too many Q's for one post ! 

(x)1/3 + (x)1/2 = 12 =

2/6 x  + 3/6 x   = 12

5/6 x = 12

x = 12 * 6/5 = .........