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It's named Algebra Word Problem I posted it like 20 min ago

There are 50 multiples of 8 between 100 - 500

I know you are the same guy from like the problem two minutes ago so utilize these three steps. 

1. Find first multiple of 8 in 100-500  = 104

Now there are 396 numbers from 104-500 

2. Find how many multiples of 8 are in 396 = 49

3. Add 1 (first multiple) = 49+1

Thanks for your help! What problem do you need help with?

Can you help me with my problem labeled 'Algebra Word Problem'

Idea - find LCD of denominators 

6[\(\frac{x-1}{3}- \frac{x-1}{2} = -7 \)]

2x - 2 - 3x + 3 = -42

-x + 1 = -42                                        \(\left \{ 43 \right \}\)

      -1     -1

-x=-43

x = 43 

Crazy way of doing this, 

16*6=96

96+6=102

100/6= 16 (whole number wise) 

Now, add                                                     This represents ? multiples of 6

102                                                              1

96                                                                16

96                                                                16

96                                                                16

96                                                                16

6                                                                  1

6                                                                  1

=498

There are 67 multiples of 6 in 100-500

3) In rhombus ABCD, points E,F,G and H are the midpoints of AB, BC, CD and DA respectively. Quadrilateral EFGH has area 14 and perimeter 16. Find the side length for rhombus ABCD.

Let  the sides of the quadrilateral  be  W  and  L

And  W * L  =  14  ⇒   L  = 14 / W

And

2 ( W + L )  =  16

W + L   =  8    ⇒  W  +  14/ W   =  8   ⇒  W^2  + 14  =  8W   ⇒

W^2   -  8W  +  14   = 0

Solving this for W  we have that  W  =  4 + √2     and  L  =  4 - √2

The  side length  of  the rhombus  =   the length of a diagonal of the quadrilateral  = 

√  [   ( 4 + √2)²  +  (4  - √2)²  ]   =

√  [ 16 + 2  +  16 + 2  ]  =

√36   =

6 units

15   +  6n   =  363      subtract  15 from  both sides

6n  =  348                 divide both sides by 6

58 =  n     { plus  the first number, 15  } =      59 numbers

58 numbers.  Subtract fifteen from 363 and divide the difference by six.

5  +  3.2n  -  6  -  4.8n

=   5  +  3.2n  +  -6  +  -4.8n

                                                  We can add numbers in any order.

=   5  +  -6  +  3.2n  +  -4.8n

=   (5  +  -6)  +  (3.2n  +  -4.8n)

=   (5  -  6)  +  (3.2n  -  4.8n)

=   (  -1  )  +  (  -1.6n  )

=   -1  -  1.6n

2)A rectangle and a square have the same perimeter. One side-length of the rectangle is 25% longer than the other. What is the ratio between the areas of the rectangle and the square?
Write your answer as a common fraction (or an integer).

Perimeter of the square is  4s     where s  is the side length

And the area of the square  is   s²

Half the perimeter of the rectangle   = 2s

So....  W  +  L   =   2s

Let  W  = 1.25L

So.....  [ W  + L ]  = 2s    ⇒  [ 1.25L  + L ]   =  2s  ⇒  2.25L = 2s  ⇒

L =  (2/2.25)s ⇒  L  =  (4/5)s

So  W   =  1.25L  =   (5/4)L  =  (5/4)(4/5)s  =  s

And the area of the rectangle  is  L * W  =   (4/5)s * s  =  (4/5)s²

So...

area of rectangle                    (4/5)s²                      4

______________      =           ______        =          __     

area of square                            s²                         5

1)Square ABCD and rectangle ACEG have the same area. Find the ratio AC : CG

Let the side of the square  = s

Then  AC = √2s

The area of the square  is  s²

The  area  of the rectangle  is  also  s²  = AC * CE

So

s²  =  √2s *  CE    ⇒  CE  =   s / √2

So.........

CG   =   √  [ AC²  + CE² ]    =   √  [ 2s²  + s² / 2  ]   =  √  [ 5s² / 2 ]  =  s √ [ 5/2]

So    AC  : CE   =  

AC  / CE   =      √2s  /  [ s  √ [ 5/2 ]  ]    =  2 / √5

Thank you so much!

Thanks!

Thanks so much!

thank  you so much!

You meant to put  m  or you meant to put  x  ? 

Oh! I'm sorry, I meant to put m

7/18 %   =   (7/18) / 100   =   7/18  *  1/100   =   7 / 1800

7/18 ​%  of a quantity is equal to   7/1800  of the​ quantity.

Weekly commission  =

11%  *  7100  =

.11 *  7100  = 

$ 781

Commission  in 2011  =

52 weeks  *  781   =

$40,612

I think we want to find x, not m

Angle CDG  =  Angle JDE    { vertical angles}

Angle JDE  =  Angle HCD    {corresponding angles }

Thus

Angle CDG  = Angle HCD   { substitution }

So

5x  -  5   =   3x  +   13       add 5 to both sides, subtract 3x from both sides

2x   =  18            divide both sides by 2

x  =  9

Thank you so much!

Ninth one :

Definition of Bisector

Alternate Interior Angles Theorem

Substitution Property

Tenth one  :

[ x + 8 ]  / 10  =  [2x + 5 ] / 14      cross-multiply

14 [ x + 8 ]  =  10  [ 2x - 5 ]

14x + 112  =  20x - 50         add 50 to both sides, subtract 14x from both sides

162  =  6x              divide both sides by 6

27  =  x   

Last one :

Base  =  √ [ 18^2   - 9^2 ]   =  9√ [ 2^2 - 1^2 ]  =  9√3   ≈ 15.6 in

The question is...

The number of medium-sized jerseys out of the total number of jerseys is what out of 100 ?

number of medium-sized jerseys / total number of jerseys   =   x / 100

14 / 20   =   x / 100

100 * 14 / 20   =   x

70  =  x

70 percent of the jerseys are medium-sized jerseys. 

(a)  Yes. m∠ACB  =  m∠ECD since they are vertical angles, and  m∠ABC  =  m∠CDE . So by the AA congruency postulate,  △ABC ~ △EDC  .

(b)  Since  △ABC ~ △EDC  ....

DE / DC   =   BA / BC

 32 / 22    =   BA / 100     Multiply both sides of the equation by 100 .

145.45     ≈   BA

The width of the river is approximately  145.45 ft.