3k - 1 = 7+2  ..... simplify the right side

3k -1  =   9       now....add 1 to both sides  so that we can have just 3k on the left side

     +1    +1

3k = 10     divide both sides by 3

k  = 10 / 3       and that's it  !!!

We have5 grams remaining after 1590 yeats.....so.....

5 = 10 e^[-k(1590)]     divide both sides by 10

(.5)  = e^[-k(1590]     take the ln of both sides

ln (.5)  = ln e^[-k(1590]    and  we can write

ln (.5)  = [-k(1590] *ln e      and ln e = 1....so we have

ln(.5) = -k(1590)    divide both sides by 1590

ln(.5)/1590  = -k

-0.0004359416229937 = -k   .....so.....k = 0.0004359416229937

And we need to find  t for 8 grams remaining....so.....

8 = 10 e^[-0.0004359416229937 (t)]       divide both sides by 10

.8  = e^[-0.0004359416229937 (t)]    take the ln of both sides

ln(.8)  =  ln  e^[-0.0004359416229937 (t)]     and we can write

ln(.8)  = [-0.0004359416229937 (t)] ln e     and ln e = 1   .....so we have...

ln(.8) = -0.0004359416229937 (t)    divide both sides by  -0.0004359416229937

ln(.8) / -0.0004359416229937  = t = about 511.87 years

that didnt help me understand my question. anyone else can answer my question?

3k - 1 = 7+2=3k=7+2+1=3k=10=k=10/3=3 1/3

So, k=3 1/3

6/7 (35 a + 14)

30a + 12

20log(400) - 26 = about 26 db  for the singing bird

For the second,  100Pa = 100,000,000 μPa...so we have

20log(100,000,000) - 26  = about 134 db

Prove that Log 5/36 - log 5/9 = -log 4......... IS FALSE!!!!!!!!!!!!!

Log 5/36-Log 5/9=Log 1/4

Let's assume that the cost per kg = c

Then, the total profit made was (the whole cost) x ( 1.10)  ......and the whole cost = 100c ....so....the profit was just (100c)(1.10)  = 110c

Now....let the amount that the 7% profit was made on = x.......and the cost of this was x*c......so.....the total profit made on the 7% part was just (cost)* (1.07) = (xc)(1.07)

And let the amount that the 17% profit was made on = (100 - x)....and the cost of this was (100 - x)c.... ....so....the total profit made on the 17% part was just  (cost)*(1.17)  = [(100 - x)c] (1.17)

Then.......the profit made on both parts = total profit.......so

(xc)(1.07) + [(100 - x)c](1.17)  = 110c  ..... simplify

1.07xc + 117c - 1.17xc  = 110c       divide through by c

1.07x + 117  - 1,17x = 110   

-.1x  = -7      divide both sides by -.1

x= 70     and this is the number of kg on which the 7% profit was made

Given do sides of a right triangle how would you go about finding one of the angle measures

Well, it depends on which two sides!!

If the two sides are the "OPPOSITE" & the "ADJACENT", then you divide the OPPOSITE/ADJACENT=tangent od the angle,

Once you have the tangent, then you can easily find the arctangent of that angle, or the angle you want in degrees.

If the two sides are the Hypotenuse & the Opposite, then you divide Opp/Hypo=sine of the angle. From that you can easily get the arcsine in degrees as above.

If the two sides are the Hypo & the Adj., then you divide the Adj/Hypo=cosine of the angle. From that you can easily get the arccosine of the angle in degrees.

solve(17000=X^2*37.1*1.12/1000000)

YOUR EQUATION IS FALSE!!!!!!!!!!!!!!!!

what is the square root for 10000

LOL! How about 100!!!!!!!!!!!!!!

$${\frac{{\frac{{\mathtt{7}}}{{\mathtt{8}}}}}{{\mathtt{7}}}} = {\frac{{\mathtt{1}}}{{\mathtt{8}}}} = {\mathtt{0.125}} = ABC$$

Thanks Badinage, I thought it looked like concrete too ://

Rosala please go take a proper look and tell us if it is concrete or just hard dirt :)   !!

I wonder why they would tar the middle of the road and concrete the edges :/

Oh sorry I misread.  You think the concrete is under the bitumen too.  Yes that is what it looks like.  

Would they do that to make the road less slippery?  

how many nanoseconds are in 1.5 years?

Just multiply that big number by 1.5 or

4.73364 x 10^16 nanoseconds

what is 7/8 divided by 7

You have 7/8 / 7 =7/8 x1/7=7/56=1/8

how many nanoseconds are in 1.5 years?

yr=365.25days

Then you have 365.25x 24 x 60 x60 x 10^9

which equals to:3.15576 x 10^16 nanoseconds

Since you earning 16% compounded monthly, then you have a monthly interest rate of 16/12 or 1 1/3%. We take this rate & divide by 100 & we get .013333333.....     (1)

Then take (1) above and add 1 to it or 1.013333333                                (2)

Then we have the future value of $40,000 which must be discounted........(3)

Then PV(present value)= FV(future value)/ (1.01333333^120) or

$40,000/4.900940.... =$8,161.70, which is the amount you must have today.

I'd say the "ground" is concrete; it's level, not pot-holed, and is neatly kerbed! Looks like the bitumen has been layed over a base of concrete, to me.

I'm assuming that the train travels at its normal rate for the first 120 km and then reduces its rate by 4km/hr for the rest of the trip.......

There is no specific answer for this.......to see why...let the normal rate be = R, the normal time = T, and the distance = D...and we have....

RT = D  →  T = D/R

So the time for the first part of the trip = D/R  = 120/R

And the fime for the second part of the trip = [D - 120] / (R - 4]

And the sum of these  equal to the normal time plus 1 hour.....so we have.....

120/R  + [D - 120] / [R - 4]  = T + 1 

120/R  + [D - 120] / [R - 4] =  D/R + 1       simplify

120[R - 4] + R[D - 120]  = (R^2 - 4R] [D/R + 1]

120R - 480  +  RD - 120R  = RD + R^2 - 4D - 4R     rearrange

R^2 - 4R - [4D - 480] = 0  ......   now let D = 480km     and we have

R^2 - 4R - [4(480) - 480] = 0

R^2 - 4R  - 1440  = 0    factor

(R + 36) (R - 40)  = 0     .......  and R = 40km/hr

So...when D = 480km, R = 40km/hr and the total time = 480/40 + 1 = 13 hrs   

Now,  let D = 960km   and we have

R^2 - 4R - [4(960) - 480] = 0

R^2 - 4R - 3360 = 0      factor

(R -60)(R + 56) = 0      and R = 60km/hr 

So, when D = 960km, R = 60km/hr, and the total time = 960/60 + 1 = 17 hrs

Thus......there are infinite possibilities........

√[4 - 12x^2] = 2√[√3 -3x]*√[ √3 + 3x] / √3   simplify

√[4 (1 - 3x^2] = 2 √ [ 3 - 9x^2] / √3

2 √[1 - 3x^2] = 2 √ [ 3 - 9x^2] / √3    divide both sides by 2

√[1 - 3x^2] = √ [ 3 - 9x^2] / √3  multiply both sides by √3

√3 * √[1 - 3x^2]  = √ [ 3 - 9x^2]

√[3[1 - 3x^2] ]  = √ [ 3 - 9x^2]

√[3 - 9x^2]  = √[ 3 - 9x^2]

And one side = the other side....just as asinus has said.....!!!!!

Hello friends!

The whole backwards.

(2 sqrt(sqrt(3)-3 x) sqrt(sqrt(3)+3 x))/sqrt(3)

= (2 * √ ( √ (3) - 3x ) * √ ( √ (3) + 3x )) / √ (3)

= (2 √ (3 - 9x²)) / √ (3)

= √ ( 12 - 36 x²) / √ (3)

= ( √ (3) * √ (4 - 12 x²)) / √ (3)

= √ ( 4 - 12 x² )

√ ( 4 - 12 x² ) = (2 sqrt(sqrt(3)-3 x) sqrt(sqrt(3)+3 x))/sqrt(3)

Q. E. D.

Greetings    :- )

Hallo anonymous!

Calculate the average speed if a golf cart runs 140 meters in 10 seconds .

v = s / t

v = 140 m / 10 sec = 14 m / sec * (3600 sec / 1 h) * (1 km / 1000m) = 50.4 km / h

The average speed is 14 m / sec or 50.4 km / h.

Greetings  :- )

Very nice problem and good answers from heureka and Melody.....!!!