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I don't know about factoring primes but my calculator says 8.27677 in other words you omitted the "2" in red. 

This Q has been asked before.....do a search for it....

Impulse = change in momentum

100 N = m * change in v

100N = 20 kg * change in v

5 m/s = change in velocity  

if the impulse is applied in line with the direction of travel, the object will now be travelling 55 m/s

if the impulse is applied OPPOSITE to the direction of travel , the object is now traveling 45 m/s

7t + h = 67

5t +h  = 49      subtract bottom eq from top

2t = 18      t = 9

5(t) + h = 49

5(9) + h = 49

45 + h = 49

h = 4                                 Edited for mistake Guest pointed out to me...Thanx !

Person A does 1/5 of the job in an hour.

Person B does 1/x of the job in an hour.

A and B do 1/5+1/x of the job in an hour.

1/5+1/x = 1/2.

\(\frac{1}{5}+\frac{1}{x}=\frac{1}{2}\)

\(x=\frac{10}{3}\) hours.

You are very welcome!

:P

Thank you! Sorry for the late response.

I need help, I can't use the calculator of this page, the web says "access denied for adblock users" but I don't have any adblock, i don't know what is happening.

(a - 1/a)^2 = a^2 +1/a^2 - 2 = 25/36,

a^2 + 1/a^2 + 2 = 25/36 + 4 = 169/36,

(a + 1/a)^2 = 169/36,

a + 1/a  = 13/6.

Prime factorize 567

3^4*7^1

We can factor out a 3 to get

\(3\sqrt[3]{21}\)

or around 8.767

Sum, S, of the  first 20  terms  is given by

S   =   ( a₁  +  a₂₀) * (n /2)   

790  =  (-8 +  a₂₀)  * (20/2)

790  = (-8 + a₂₀) * 10               divide both sides by 10

79 =  -8 +  a₂₀       add 8 to both sides

87   = a₂₀

a -  1/a =  5/6                 multiply through by a

a^2  - 1  = (5/6)a           multiply through by 6

6a^2  - 6  = 5a              rearrange as  

6a^2  -  5a  -  6  =  0        factor

(3a  + 2) ( 2a - 3)  = 0

Set each factor to 0  and solve for a   and we have that

a = -2/3   (reject)     or     a = 3/2

So

a +  1/a  =

3/2  +  1 / (3/2)  =

3/2 +  2/3  =

13 / 6

If a > 0 and a - 1/a = 5/6, then find a + 1/a.

Danke

It takes the cyclist of 6 mph 10/3 minutes for 1 lap.
It takes the cyclist of 9 mph 20/9 minutes for 1 lap.
It takes the cyclist of 12 mph 5/3 minutes for 1 lap.
It takes the cyclist of 15 mph 4/3 minutes for 1 lap.

The LCM [10/3, 20/9, 5/3, 4/3] = 20/3 = 6 2/3 minutes when all four will meet at the start for the first time.
By the way, when they meet the first time, they will have done:2, 3, 4, 5 laps each respectively.
6 2/3 x 4 =26 2/3 minutes after noon hour when they will all meet for the 4th time.

1 3/4  x 3/4 = 21/16

1 3/4 - 21/16 =   28/16 - 21/16 = 7/16

2^8 is 256 , so this is equivalent to:

2^8(x-2) =2^{(-x-7)    Now equate the exponents}

8(x-2) = -x-7

8x-16 = -x-7

9x = 9

x=1

I'm not sure where you learned this, but there is a lot of extraneous math in your solution. All the "1^ -1" is completely unnecessary and confusing, although it does somehow give the correct solution. Here's a simpler way to do it:

256^x-2 = 2^-x-7 | 256 is equal to 2⁸, so you can make the bases the same

2^8x-16=2^-x-7 | Now, you can take away the bases and solve algebraically. If you perform log₂ on both sides, you're left with just the exponents, so I'm just not showing a step.

8x-16=-x-7 | Next, you solve algebraically.

8x+x-16+16=-x+x-7+16

9x=9 | Divide both sides by nine.

x=1 | You have your answer!

Now, let's plug it back in to check if it works.

256^x-2 = 2^-x-7 | Replace x with our answer, 1

256^1-2=2^-1-7 | Change 256 to 2⁸

2^8-16=2^-1-7 | Simplify

2^-8=2^-8

It works! It may look hard at first, but once you try it out it's much easier and saves you the extra steps.

TLDR; Match your bases, take the log and solve algebraically.

i agree, you should refix this question, so it would make a little more sense, or is it a trick question:)???!!!!!

Solve for x :
256^(x - 2) = 2^(-x - 7)

-256^(x - 2) = (-1)^1·2^(8 x - 16):
(-1)^1·2^(-x - 7) = (-1)^1·2^(8 x - 16)

Equate exponents of -1 and 2 on both sides:
1 = 1 and -x - 7 = 8 x - 16

All equations give x = 1 as the solution:
 
x = 1

i haven't been going through any troubles, but i feel like i will, with your technical difficulties!

Constant is a number that doesn't have a variable next to it.

All of them have x next to them. So, No constant in this equation also we could say the constant= 0.....

(Good question: Is \(\frac{4}{x} ,\frac{1}{x^4}\) considered to have variables?) 

Yes!

\(\frac{4}{x}\)=4x^-1 

\(\frac{1}{x^4}=\)x^-4

So The constant of this equation is 0 (I.e. there isn') 

Why do you think your rebuke is worth 3 points NewMember? 

I think it is worth -3 points. 

Behave yourself and say thanks for the person's interest before you suggest that an anwer is wrong.

AND

If you believe an answer is wrong then you are expected to say why you know this to be true. 

  e.g. the answer in the book is .....