All Answers 358150 Answers

|2t+5|< 3

To solve this kind of problem, let's take the expression out of the absolute value bars and solve two equations:

2t + 5 < 3         and   -(2t + 5) < 3

The reason for this is that if,   lal < b,  then  l-al is also < b........!!!!

Working with the frist one, we have....

2t < -2      divide by 2 on each side

t < -1

Working with the second one, we have

-2t - 5 < 3   add 5 to both sides

-2t < 8        divide by -2 on both sides and reverse the sign

t > -4

So the solution is given by

-4 < t < -1

This is always the Great Debate.......Was something always "out there" waiting to be "discovered"  - or - did the human mind  "invent" what was required???

I'm sure those with far more insight than me could present a good argument one way or the other......!!!

$${\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}} = {\mathtt{7}}$$

Becasue your brain is just like mine 

add 8 to both sides:4t-8(+8)=20(+8)

4t=28  negative and positive 8 cancel out

divide by 4 to get t by itself: 4t/4=28/4

t=7

Notice the diagram of the situation - I've increased everything by a factor of 10, but the idea is still the same.

Let's imagine that, for every 10 seconds he swims, the current is pushing him to the "right" (denoted by segmet CB) by 6.53m while he swims a total distance of 7.19m (denoted by segment AB). And the angle he is being "pushed" away from the "vertical," i.e., his intended direction of "straight across" the river, is given by

sinΘ = (6.53/7.19)   .....thus, to find "Θ" we can use the sine inverse .........

sin-1 (6.53/7.19)  = about 65.26°

So, to counter this, he needs to swm at the same angle on the opposite side of the vertical - i.e. to the "left" side of the vertical. And since you wanted the angle with respect to the shoreline,  then he must swim at an angle of (90 - 65.26)° = 24.74° with respect to the shoreline on his "left.".....Note that the width of the river doesn't matter since, no matter how wide the river is, we still have the same similar triangle !!!!

(I think i'm correct here, but I'm going to have one of our Physics pros look at this one just to see if I've made any false steps!!)

So, I'm assuming that the current exchange rate is being expressed as

1£ = $1.5

So £500 = $750

And if the old exchange rate was

1£ = $1.8,  then

500£ = $900

So your friend received   $900 - $750  = $150 more

All x's that count for 5/(x-2) < 3/(x-4) ?

Cross-multiplying, we have

5(x-4) < 3(x-2)     simplify

5x - 20 < 3x - 6     subtract 3x from both sides and add 20 to both sides

2x < 14                 divide by 2  on  both sides

x < 7

But we have to be careful with this answer......x cannot be equal to either 2  or 4  because it would make the denominator(s) in the original problem undefined.....also, when x < 2 or when  4 < x < 7 the original inequality is true, but it is not true when 2< x < 4

So, our answer is

(-∞, 2) U (4, 7)

This seems like you're tring to find the sine of an angle expressed in radian measure....just go to the calculator and choose radian mode at the bottom left...then key it in as you have it......

I don't know of any circumstance where a set is multiplied by a number, but naybe some other answerers on the site can think of something in set theory that I'm unaware of !!!

Elizabeth, just divide  2.50 by 50 to get the percentage.......

Remember to move the decimal in your answer two places to the right and add the "%" sign....

If you leave at 11:15AM British time, then that would be at 6:15AM New York time. And your arrival 8 hours later would be at 2:15PM New York time.

Just multiply 38.2 x 625 to get your answer.

We can find a generating polynomial by the sum-of-differences technique

0, 6, 16, 30, 48

 6  10   14  18

   4    4     4

      0     0

There are three differences required to get to the "0" row

This means that we have a possible 3rd degree polynomial with at least some of the coefficients ≠ 0

P(x) = ax³ + bx² + cx + d

 When x = 1, we have   a + b + c + d = 0             (1)

 When x = 2, we have  8a + 4b + 2c + d = 6         (2)             

 When x = 3, we have  27a + 9b + 3c + d = 16      (3)

 When x = 4, we have  64a + 16b + 4c + d = 30     (4)

Subtracting (1) from (2) we have       7a + 3b + c = 6      (5)

Subtracting (2) from (3) we have      19a + 5b + c = 10    (6)

Subtracting (3) from (4) we have      37a + 7b + c = 14    (7)

Subtracting (5) from (6) we have     12a + 2b = 4      (8)

Subtracting (6) from (7) we have     18a + 2b = 4      (9)      

And subtracting  (8) from (9)  we get 

6a = 0   →   a =  0        

And back-substituting  to find the other coefficients, we have 

b= 2   →   c = 0  →  d = -2

And our polynomial turns out to be the quadratic

P(x) = 2x² - 2

And the 40th term is given by

P(40 ) = 2(40)² - 2   =   3198

Uhm, im not positive what the answer is because, it varys. Depends on what the variable is representing.

Ok. I kind of don't get what you said in the trial thing.

The formula: Accumulated amount=p(1+i/m)^mn

p=principal amount

I=interest

M- if it's semiannual: m=2

     Quarterly: m=4

     Monthly: m=12

n: number of years.. In your case it's 3 

Glad to help

e^{(-1/2)x^2}= 1/2       take the natural log of both sides

ln e^{(-1/2)x^2}  = ln (1/2)    and by a property of logs - and remembering that ln e = 1 - we can write

(-1/2)x^2 =    ln(1/2)    multiply both sides by -2

x^2 = -2ln(1/2)

x^2 = about 1.38629        take the square root of both sides

x = about  ±1.1774

It depends opon what number the "percentage difference" is being calculated on

If it's on the 17.1, then 3.5 is about 20.4% of that

If it's on the 13.6, then 3.5 is about 25.9% of that

Thanks CPhill, but I actually meant pounds.... I get those confused all the time!!! So that would be in england... Anyone know how many school days there are in england? And how much I would pay?

9x + 3y = 12         subtract 9x from both sides

3y = -9x + 12       divide both sides by 3

y = -3x + 4         and, for "y,'  write "f(x)"

So

f(x) = -3x + 4

@@ End of Day Wrap:     Sun 14/9/14     Sydney, Australia      Time 2:05am   (Really Monday morning)   ♬

Hi all,

Great answers were provided by CPhill, Alan, Rosala, Genisous and Dragonslayer554.  Thanks everyone.   

I looked at the list and thought who is missing.  Aziz is missing!  That is one for the book on modern history! I hope you are of having fun Aziz.  

Now for the interest posts:

1) Break even point  (Thanks Chris)

Monday 15/9/14

1,2) Equations with absolute values  (Thanks Chris)

Sorry Dragon that is not right.   

$${\mathtt{\,-\,}}{\frac{{\mathtt{18}}{\mathtt{\,\times\,}}{\mathtt{4}}}{{\mathtt{3}}}} = -{\mathtt{24}}$$

$$\frac{\sqrt{27}}{3}=
\frac{\sqrt{9}\sqrt 3}{3}=
\frac{3\sqrt 3}{3}=\sqrt3$$