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A = A(0) * 1/2^(20,000/4,500)

A = 100 * 1/2 ^4.44.....

A =100 x 4.59....

A =~4.59g remaining after 20,000 years.

k = -(ln2) / half-life

k =- (0.693147) / 4,500

k = -0.000154 - which is the constant of decay.

Check: A =100 x e^-(0.000154*20,000)

             A =100 x e^ -3.08

             A =100 x 0.0459.....

             A =4.59 grams.

Thank you, I finally get it now  

Thank you so much! could you help me with more?

Look at the first table, Single.

Column number 6 is in the range, so: 0.25E-3190: 8300-3190=\(\boxed{5110}\), or (B).

Hi, RP!

1. Subtracting -3n^2 from -7n^2, we get: \(-7n^2-(-3n^2)=-7n^2+3n^2=4n^2.\)

2. \((-8x^4y^3)*(2x^5y^2)+7x^9y^5\) , we can remove the parenthesis, then multiply, then add, to attain: \(-16x^9y^5+7x^9y^5=-9x^9y^5\).

3. Again: \((-4a^3b^2)^2*(3a^2b)=16a^6b^4*3a^2b=48a^8b^5.\)

4. \(5c^{-7}d^2*(-cd^2)^4\) , simplify, apply the exponent rule, and finally: \(\frac{5d^{10}}{c^3}\) . This might be wrong!!!!

5. \(\frac{18w^4x^9}{14w^5x^5}\), division, subtract powers! So, \(\frac{18x^4}{14w}\). Reduce it, to \(\frac{9x^4}{7w}.\)

6. \(\frac{15x^2y*-6x^7y}{(3xy)^2}\), remove parenthesis and simplify to get: \(-10x^7\) .

Answers:

1. \(4n^2\)

2. \(-9x^9y^5\)

3. \(48a^8b^5\)

4. \(\frac{5d^{10}}{c^3}\)

5. \(\frac{9x^4}{7w}\)

6. \(-10x^7\)

deleted.

3/8" =0.375 arcseconds.

0.375 / 3,600 =1.04166667E-4=0.00010416667 degrees.

The red ones are above and the blue ones are below and you can finish it.

Thanks Mellie,

I hope we see a lot of you and I know that I am not the only one who hopes so.   :)

You can calculate this limit without l'hopital.

Hint: a³-1=(a-1)*(a²+a+1)

a²-1=(a-1)*(a+1)

When x is negative x^2/3 =1 will have three possible solutions, two of them complex.  If we are to stick with the reals, then we must do the squaring before the cube rooting.  i.e. we must express x^2/3 as (x²)^1/3.

When x approaches -1 from above then  (x²)^1/3 -1 is negative, x - 1 is negative, so f(x) is positive, and approaches the limit +infinity.

When x approaches -1 from below then  (x²)^1/3 -1 is positive, x - 1 is negative, so f(x) is negative, and approaches the limit -infinity. 

For x = +1, use l'Hopital's rule to show the limit is 3/2.

\(|A \cup B| = |A|+|B| - |A \cap B|\)

let A be brown haired students, and B be right handed students

\(|A|=\dfrac 3 4 \cdot 28 = 21 \\ |B| = \dfrac 6 7 \cdot 28 = 24 \\ |A \cup B| \leq 28 \\ 21 + 24 - |A \cap B| \leq 28 \\ 17 \leq |A \cap B|\)

so the smallest number of students that have both brown hair and are right handed is 17

3 distinct letters so 3!=6 arrangements

V=IR; I=40A; R=3Ω
V=40*3=120V

120V=I*R; this time, R=20Ω
120V=I*20Ω
I=120/20=6A

Since  ∠ABC  and  ∠CBE  form a straight line...

m∠ABC   =   180° - m∠CBE   =   180° - 125°   =   55°

The sum of the angles in a triangle is always  180° , and each angle is greater than  0° , so the sum of any two angles has to be less than 180° . So...

55° + (15x - 92x - 25)°   <   180°

55 + 15x - 92x - 25   <   180

30 - 77x   <   180

-77x   <   150

x   >   - \(\frac{150}{77}\)

And...

0°  <  (15x - 92x - 25)°

0   <   -77x - 25

25   <   -77x

-\(\frac{25}{77}\)  >  x

- \(\frac{25}{77}\)   >   x   >   - \(\frac{150}{77}\)

All possible values of  x  are in the interval  (-\(\frac{150}{77}\), -\(\frac{25}{77}\))

Let the amount of oranges that Kim could buy if the price rose to  4x  cents per orange be  n .

the amount of money Kim has   =   40 * 3x

the amount of money Kim has   =   n * 4x

n * 4x   =   40 * 3x

4nx   =   120x

                         Divide both sides of the equation by  x .

4n   =   120

                         Divide both sides of the equation by  4 .

n   =   30

The sum of the angles in every triangle is  180° , so..

∠A + ∠B + ∠C   =   180°

21° + 90° + ∠C   =   180°

∠C   =   180° - 21° - 90°

∠C   =  69°

∠D  corresponds to  ∠A  so...

∠D   =   ∠A   =   21°

∠F  corresponds to  ∠C  so...

∠F  =  ∠C  =  69°

△EDC  is similar to  △ABC ,  so side  DC  corresponds to side  BC .

Let the scale factor from  △EDC  to  △ABC  be  s .

DC * s   =   BC

4 * s   =   6

s   =   6/4

s   =   1.5

So each side of  △ABC  is  1.5  times the corresponding side of  △EDC.

△EDC  is similar to  △ABC ,  so side  EC  corresponds to side  AC .

EC * 1.5   =   AC

And we can see from the picture that   EC  =  x   and   AC  =  12.5 - x  , so....

x * 1.5   =   12.5 - x

1.5x   =   12.5 - x

                                 Add  x  to both sides of the equation.

1.5x + x   =   12.5

2.5x   =   12.5

                                 Divide both sides by  2.5 .

x   =   5

△GHI  is congruent to  △JKL .  So....

GH   =   JK   =   25 in

GI   =   JL   =   14 in

KL   =   HI   =   26 in

∠J  =  78°

∠K  =  31°

∠I  =  71°

idk since 1 in 1000 is the first sig fig, the final answer would therefore round to the thousands, so 831.136 should round off to 1000.
but, using online sig fig calculators (first few that pops up when googling "sig fig calculators"), the answer comes off as 831.
not sure if the calculators are correct or incorrect in this case. 

The smaller triangle has a hypotenuse length of \(\sqrt{4^2+6^2} = \sqrt{52}=2\sqrt{13}\)

as the triangles are similar the zip line will have the same ratio to that hypotenuse as the 140 ft leg does to the 4 ft leg.

\(\dfrac{140}{4} = 35 \cdot 2\sqrt{13} = 70\sqrt{13}\)

and thus the zip line has length \(70\sqrt{13} \approx 252 ft\)

The easiest way for me to think of these is to find the scale factor.

Let the scale factor from  △DEF  to  △HJK  be   s  .  So....

5s  =  8.75

s  =  8.75/5

s  =  1.75

So each side of  △HJK  is  1.75  times the corresponding side of  △DEF .

x   =   1.75 * 4   =   7

y   =   1.75 * 8   =   14

Thanks! I was a bti confused about the \(\bowtie\) sign.

Pick a letter, there are 26 choices, that letter is the first and last letter in the word.

We then have 2 slots left and can choose any letters to fill them.  There are 26² ways of choosing them

So there are 26x26² = 26³ =17576 words with this property