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Let's rewrite your problem in mathematical terms:

\((n)(n+1)=42\)

We can multiply the left side to get:

\(n^2+n=42\)

We can move 42 to the left side to get:

\(n^2+n-42=0\)

From here, we just solve for \(n\).

\((n+7)(n-6)=0\)

In order for this to be true, we need to either satisfy \(n-6=0\), meaning that \(n=6\), or \(n+7=0\), meaning that \(n=-7\). We need to find the positive solution for n, so we pick \(n=6\) . This does indeed work, because 6*7=42, so \(\boxed{n=6}\)

2x² – 13x + 21  will factor to (2x – 7)(x – 3)

2x² + 9x – 56    will factor to (2x – 7)(x + 8) 

The factor they have in common is (2x – 7) thus the common solution is x = 3.5  

.

Let's call the width "x."

The width is x and the length is 3x - 1

The area of the width and length is 30.

(x)(3x-1) = 30

3x^2-x = 30

3x^2-x-30 = 0

(3x+3)(x-10) = 0

Solve 3x+3=0 and x-10=0 and see which value of "x" is feasible. (hint: widths cannot be negative.)

You are very welcome!

:P

Six

.

If their solution is in common, we can set the equations equal to one another

So 2x^2-13x+21 = 2x^2+9x-56

-22x = -77

Solve x from here and round to the nearest tenth.

You are very welcome!

:P

I assume you mean \(3x^2-10x-8\)

You can factor it out as (3x+2)(x-4) = 0

From there you can find 3x+2 = 0 and x-4 = 0, solve for x

The positive x value is your solution.

You are very welcome!

:P

Area of the triangle = b*h/2 = (2x+4)(x+3)/2 = 2x^2+10x+12 all divided by 2 = x^2+5x+6.

\(x^2+5x+6 = 56, x^2+5x-50 = 0, (x-5)(x+10) = 0\)

can you solve from here?

You are very welcome!

:P

Much appreciated! 

[ 5 * 2^-3t ] / 40  =

(5/40) * (2)^-3t

(1/8) * ( 2)^-3t  =

(2)^-3  * (2)^-3t  =

2 ^{-3 - 3t}  =

2 ^{- (3 + 3t)}  =      

     1                    1                        1

_______  =      _______    =      ______ 

   2^{3 + 3t}             (2³) ^{(1 + t)}              8 ^{1 + t}

Let angle BAE  = 1  Let angle DAE  = 2   Let angle ABE = 3    Let angle ADE = 4  Let angle EBC = 5

Let angle ECB  = 6

So.....since triangle ABC  and triangle  ABD are isosceles....we have that

3 + 5  = 6       (1)

1 + 2  = 4       (2)

3  = 90° - 1     (3)

2 = 90° - 4     (4)

5 = 90° - 6      (5)

Sub (3), (4) and , (5)     into   (1) and (2)

(90° - 1) + (90° - 6)  = 6      ⇒  180° - 1  = 2*6      (7)

1  +  (90° -4)  = 4  ⇒            90° + 1  = 2*4         (8)

Add (7) and (8)  and we have that

270°  = 2*4 + 2*6

270° = 2 (4 + 6)  divide both sides by  2

135°  = 4 + 6  =  Angle ADE + angle ECB =  angle C + angle D

Combos that multiply to 20....sum to 12

Both must be positive

(1, 20)

(2, 10)

Notice that the last is what we want....

Combos that multiply to -15   are

(-1,15)

(1, -15)

(-5,3)

(5, -3)

Note that the last pair sums to 2

So...those are what we need

Look at   the possible combos that multiply to  4

(1, 4)

(2, 2)

(-2, -2)

(-1, - 4)

Note that only the last sums to -5....so....these are what we need....- 1  and - 4

Let a be the larger integer and b be the smaller....so...we have

a - b  =  12   and   ab  = 45

Rewrite the first equation  as  a = 12 + b

Sub this into the second for a

(12 + b) b  = 45      simplify

12b + b^2 =  45     rearrange as

b^2 + 12b  - 45  = 0

We need two factors that multiply to -45   and sum to 12

These are  15  and -3

So we write

(b +15) (b -3)  = 0  set each factor to 0  and solve for b and we get that b = -15  (reject)  and b = 3

And .... a   =  12 + b  =   12 + 3   = 15

And their  sum  is  18

1/17   has  a cycle  of  16 digits....this cycle is : 0.0588235294117647

So.....  4037 mod 16  = 5

So....this is the 5th digit in the cycle   = 2

The graph here shows that the functions have no common intersection points: https://www.desmos.com/calculator/qcbswxcjik

Thus....there are no real solutions

alternatively, you can derive the following equation from heureka's diagram-

label top left corner A, center of square B, and the bottom (or top, it's completely symmetrical) point where the circle intersects the diagonal of the square C. the length of AB is \(\frac{x}{\sqrt{2}}\) , the length of BC is 1 and the length of AC is x-7. angle ABC is a right angle so we can use pythagoras and solve for x

Thank you Heureka! That is a very elegant solution, this problem was so hard all my friends were stumped.

With the solution formula you always get a solution.

GA, please no flattery.
Tell me what is wrong!

  !

Congrats Mr. BB, your answer is correct; however, the identifier (G) gravitational constant in this equation should be (g) acceleration of gravity (earth average).

GA

Asinus, Are you nibbling on pot-laced candy?   

...Save some for me  

GA

tommarvoloriddle, Before your edit, your post read:

Ok, rethinking this: for A)

try plugging number to find a pattern... I am not just going to tell the answer in case of a hw prob, this web is not a hw website where you get your answers free. I am sorry. 

Deep regrets

-tommarvoloriddle

Your comment ...

“... I am not just going to tell the answer in case of a hw prob, this web is not a hw website where you get your answers free. ...”

...implies a payment is necessary.   

It also had this comment: 

Putting 2 problems in a post like that suggests to me like you don't have enough energy to make two.

You have on more than one occasion presented two questions in the same post.   Did you not have enough energy?

PE = m x G x h, where m=mass, G =Gravitational constant, h =Height above the ground.

4 = 0.112 x 9.81 x h

4 = 1.09872 x h 

h = 4 / 1.09872

h = 3.64 meters above the ground.

Find the number of real solutions to \(x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb\).

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb} \\ \\ x &=& \underbrace{ 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb }_{\text{Infinite Geometric Series}} \\ && a=1,\ r= \dfrac{-x}{1}=\dfrac{x^2}{-x}=\dfrac{-x^3}{x^2}=\dotsb=-x \\ \\ && \boxed{sum = \dfrac{a}{1-r} \text{ as long as } -1 < r < 1 } \\ \\ && sum = \dfrac{1}{1-(-x)} \\ && sum = \dfrac{1}{1+x } \\\\ x &=& \dfrac{1}{1+x } \\ x(1+x) &=& 1 \\ x+x^2 &=& 1 \\ x^2+x-1 &=& 0 \\\\ x &=& \dfrac{-1\pm\sqrt{1^2-4\cdot (-1)}} {2} \\ &=& \dfrac{-1\pm\sqrt{5}} {2} \\\\ \mathbf{x_1} &=& \mathbf{ \dfrac{-1+\sqrt{5}} {2}} \\ r &=& -x_1 \\ &=& -\dfrac{-1+\sqrt{5}} {2} \\ &=& \dfrac{1-\sqrt{5}} {2} \\ r &=& -0.61803398875 \quad | \quad -1 < -0.61803398875 < 1 \checkmark \\\\ \mathbf{x_2} &=& \mathbf{ \dfrac{-1-\sqrt{5}} {2}} \\ r &=& -x_2 \\ &=& -\dfrac{-1-\sqrt{5}} {2} \\ &=& \dfrac{ 1+\sqrt{5}} {2} \\ r &=& 1.61803398875 \quad | \quad -1 < 1.61803398875 < 1 \text{ false, no solution!} \\ \hline \end{array}\)

Real solution is \(x=\mathbf{ \dfrac{-1+\sqrt{5}} {2}}\)