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well 9 being the highest number to yeild the best results, i would think you would divide,

1000000/9 this would be about 111111, so if you added 111,111 9's that would be 999,999.

then you need a 1 to make it a 1,000,000. put that in front to make it the smallest number possible.

so the answer would be 1999999999... basically 111,111 9's

just make a table where you plug in 1-5 for x

X|Y

___

1|12   (3(1)+9

2|15   (3(2)+9

3|18   (3(3)+9

4|21   (3(4)+9

5|24   (3(5)+9

Just divide 1998 / 2 =999 then just square it.

999 x 999 =998,001. I think that is pretty well it!!

By definition: 148/4 =37 - measure of one side.

By Pythagoras's Theorem:

24/2 =12 half of the given diagonal

37^2 - 12^2 =D^2, where D =1/2 of the 2nd diagonal

1,369 - 144 =D^2

1,225 =D^2

D =sqrt(1,225)

D =35 - This is 1/2 of the 2nd diagonal

35 x 2 = 70 - measure of the 2nd diagonal.

I could only think of 2 numbers that would do so in 6 steps:

79 +97=176 + 671 =847 + 748=1,595 + 5,951 =7,546 + 6,457 =14,003 + 30,041 =44,044

97 + 79=176.......and so on as before.

Now, 88 is a 2-digit palindrome. If you wanted it to be more than 2 digits, it will also take 6 steps to become 5-digit palindrome of 44,044.

There are 2 numbers that I don't think you can make a palindrome out of them unless I made a mistake. And they are: 89 and 98. Try and see if you can do it.

Let the size of Frank's yard = y Sq. feet
Then, Joe's yard =1.4y Sq. feet
2.4y / [y/45 + 1.4y/60 ] =m, solve for m
m =2,160/41 =52 + 28/41 minutes =52: 41 Time it takes both men to mow their combined yards.

I know how to code in Python and can easily code for your problem in a matter of minutes!! But, do you REALLY expect me to sit down and do your assignment for you, which you would just copy and paste and submit it as your own? How are you supposed to learn how to code, if you just let others do it for you? Put some effort into it man, and you will learn from your mistakes. Good luck.

Solution:

Subtract covered surface area from total surface area

Total Surface Area = 10*[(6)*(1²)] = 60 units²

Covered Surface Area = 14 units² (vertical) + 12 units² (horizontal) = 26 units²

Surface Area of resulting solid = 60 units² - 26 units² = 34 units²

Kudos to Hectictar for noticing my error.

GA

The  volume of a regualr tetrahedron with edge length " a "   is given  by

a^3 /  √72

So...the volume of  the tetrahedron  is

1^3  /√72   = 

1  / √72   [units^3]

So

V^2  =    1 /72   

Not as hard as it looks....!!!!

Note that  25^x  - 9^x   can be written as   (5^x)^2  - (3^x)^2

And this will factor as a difference of squares as

( 5^x - 3^x)  (5^x  + 3^x)  

And since this equals 18  and (5^x - 3^x)  = 3

Then  (5^x + 3^x)   must  = 6

If AC  = AB ....then triangle  ABC is isosceles....with angles ABC  and ACB  being equal

And the measure of angle ABC  = (180 - 42) / 2   =  69°

And since angle ABC  is an inscribed angle, then minor arc AC  will  have twice the measure of angle ABC  =  2 * 69  = 138°

And since ED is a tangent to the circle at  C, then, by Euclid, angle ACD will have half the measure of minor arc AC  = (1/2)138  =   69°

If we  unwrapped the label, we would have a rectangle  whose dimensions woud be :

6  *   circumference of the can  = area of the label

So....we have that

6 * 2*pi*r  =  60 pi    ....   divide out  pi

12r  =  60      divide both sides by 12

r  = 5 in =  radius of the can

Let   M  be the midpoint of   AQ  and N be the midpoint of BQ

Thus....M and N will be the centers of  both of the smaller semi circles

Connect  M and N

By Euclid, a segment connecting the centers of two tangent circles or semi-circles will pass through the point of tangency

So  MQ   will be the radius  of the semicircle with the diameter AQ  = 14

So MQ  = 7

And call the radius of the smallest semi-circle, r 

So....MQN  will form  right triangle with  leg  MQ  = 7, leg  QN  = 14 - r  and

hypotenuse MN = ( 7 + r )

So we have that, by the Pythagorean Theorem

7^2  +  (14 - r)^2 = (7 + r)^2

49 + r^2 - 28r + 196  = r^2 + 14r + 49

196 = 42r

196/42  = r

14/3   = r = the radius of the smaller semi-circle

Call the radius  of the semi-circle and the circle, r

Call the side of the  small square, s

So...using the Pythagorean Theorem, we have that

(1/2)s^2  + s^2  = r^2

(1/2)s^2 + s^2  = r^2

(5/4)s^2  = r^2

√5/2 s  = r

s  =   2r / √5

So...the area of the small square = s^2  =   (4/5)r^2

And..in the larger circle....the side of the square is √2r

So....the area of the square in the larger circle is just    2r^2

So....the  area of the smaller square's area to  the larger square's   area is

(4/5)r^2  / 2 r^2   =

4/10  = 

2/5  of the large square's area

Call the side of the equilateral triangle , S

Angle ABC  = 60°

Angle DBC  = 45°

And angle  ABD  =  ABC  - DBC  =  60 - 45  = 15°

The area of triangle  CDB  =

(1/2) * S * BD * sin 45

And the area of triangle ADB =

(1/2) * S * BD * sin 15

So...the area  of  triangle ADB  to the area of triangle CDB  =

(1/2) * S * BD * sin 15           sin  15

_________________   =     ______

(1/2 * S * BD  * sin 45          sin  45

sin 15   = sin (45 - 30)  = 

sin45 cos30  - sin 30 cos 45  =

√2/2 * √3/2    -  (1/2)(√2/2) =

[ √6  - √2 ] / 4

And sin 45  = √2/2

So    sin 15  / sin 45   =

[ √6  - √2 ] / 4   *  2 /  √2   =

[ √6  - √2 ]  / [ 2√2 ] =

√6 / [ 2√2 ]  - √2 / [ 2√2] = 

√3 / 2  -  1/ 2  =

(√3  - 1 )

______

     2

6S^2 =S^3+ 5S, solve for S
6 S^2 = S^3 + 5 S

Subtract S^3 + 5 S from both sides:
-S^3 + 6 S^2 - 5 S = 0

The left hand side factors into a product with four terms:
-S (S - 5) (S - 1) = 0

Multiply both sides by -1:
S (S - 5) (S - 1) = 0

Split into three equations:
S - 5 = 0 or S - 1 = 0 or S = 0

Add 5 to both sides:
S = 5 or S - 1 = 0 or S = 0

Add 1 to both sides:

 S = 5   or    S = 1     Sum= 1  + 5 = 6

Here's the third one

Orient the paper so that 22 is the width and 39 is the height

Note that  the number of retangles of (2 * 3)  that can  be cut from this sheet is

(39/3)  * ( 22/2)  = 

13  * 11  =

143

But each of the rectangles  can  be divided into two triangles with bases of 2 and heights of 3

So....the total number of triangles  is  143 * 2  =  286

This is similar to the previous linear question, I will use what I said last time.

First, we need acknowledge that 25 is the fixed number.

89 - 25 = 64

320/64= 5

This means that 5$ is charged every mile.

Thus, we obtain

C = 25 + 5x

Here's the first one....there might be an easier way to do it, however  !!!

The area of  ABC  is   (1/2) (16)^2 (√3/2)  =  64√3  cm

So...the area  of  equilateral triangle   PQR is   (1/4)  of this  = 16√3 cm^2

And we can find the   length of  PR  by using the area formula for PQR

(1/2) (PR)^2 (√3 /2 )  =  16√3 

(1/2) PR^2 (1/2)  = 16

PR^2  / 4  = 16

PR^2  = 64

PR  = 8  cm

And note that we have trapezoid BPRC  which has the same area as the small equilateral triangle

So   area  of trapezoid  BPRC  = area of triangle PQR

 also

(1/2) (height) ( sum of the bases)  = 16√3

height ( 16 + 8)  = 32√3

height (24)  = 32√3

height  =  32√3 / 24   =   (4/3)√3  cm

So...note  that if we draw a perpendicular  from P  to BC  this  will be the height of the trapezoid and will  also form one leg of a right triangle  with BP  as the hypotenuse

And the other leg length will be  ( BC  - PR)  / 2   = (16 - 8) /2  =  8/2  = 4

So....using the Pythagorean Theorem...we can find  BP  as

BP  =   √ [ 4^2  +  [ 4/3 * √3 ]^2  ]  =  √ [ 16  +  16/3  ]  =   4 √ [ 1 + 1/3] = 

4 √ [ 4/3 ]  = 8√ [ 1/3 ]  =  8 / √3  cm   =  8√3 / 3  cm

You are looking for the ways for partitioning a set with 66 elements in three subsets. If you allow empty subsets, the answer is given by

Credits to https://math.stackexchange.com/questions/2176796/how-many-ways-are-there-to-put-6-balls-in-3-boxes-if-the-balls-are-distinguishab for solution

Sorry, AWESOMEEE...I had the wrong value for the radius...see the corrected answer...!!!

Doesn't seem to work. I entered in 3.6pi, then 18pi/5.

I can't enter anymore, but I am still curious for the solution.

First, we need acknowledge that 25 is the fixed number.

67 - 25 = 42

210/42 = 5

This means that 5$ is charged every mile.

Thus, we obtain

C = 25 + 5x