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The diameter of a quarter is about 6 cm.  Find the area of the quarter. Use 3.14 for π  

If the diameter is 6 cm then the radius is 3 cm.  Area of a circle is  A = πr²   

substituting values                 A  =  (3.14) x (3 cm)²  =  3.14 x 9 cm²  =  28.26 cm²  

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Hi Melody,

looking at those you gave, the y value will be y+2

and f(n) will be f(n+5).....right?

g ( f(n) )   means  that we are putting the function f  into  g   (for n)

So we have

2 ( n^2 - 2)  +  2  =

2n^2  - 4  +  2   =

2n^2  - 2

Call the number of dimes  = D  = number of pennies    and the number of nickels = N 

So....we have these equations

Value of a  dime  * number of dimes  +  Value of a nickel * number of nickles =  6.10

.10D  +  .05N  =   6.10            (1)

And

Value of penny  *  number +  Value of a nickel * number = 3.49

.01 (D )  +  .05N  = 3.49             (2)

Subtract  (2)  from (1)

.09 D   = 2.61

D  = 2.61 / .09   =    29   = number of dimes = number of pennies

And to  find the number of nickels, use ( 1)

.10 (29) + .05N = 6.10

2.90 +.05N  = 6.10

.05N  =  6.10 - 2.90

.05N  = 3.20

N =  3.20 / .05 =   64 = number of nickels 

h (h (t) )   means   that we are putting the function itself  in for t

So we have

h(h(t)) =

-2 ( -2t - 1)  - 1   =

4t + 2  - 1     =

4t + 1

We have 3 equations that give us possible values for  x

!st....assume  that  these two sides are equal

3x + 62 = 7x + 30

   62 - 30  = 7x - 3x

       32  = 4x

         x =  8                        and the perimeter  =  3(8)+ 62 + 7(8)+ 30 + 5(8) + 60    = 272

Next assume that these two sides are equal

3x + 62  = 5x + 60

62 - 60  = 5x - 3x

2  = 2x

x = 1                               The perimeter  =  3(1) + 62  + 7(1)  + 30  + 5(1)  + 60 =   167

Finally, assume  that

7x + 30 = 5x + 60

7 - 5x  = 60 -30

2x = 30

x = 15....    it's obvious that this  value of x will lead to the greatest perimeter

So....the perimeter  is the least  when  3x + 62 = 5x + 60  and the perimeter  = 167

Hmm!  Check the resukt when a = -1 and b = 3.

ok    

I'm not seeing any quadrilateral here.

If f(x) = x^3 + 1, find f^{-1}(2010).

Let

\(y=x^3+1\\ y-1=x^3\\ x=\sqrt[3]{y-1}\\ so\; if\\ f(x)=x^3+1\\ f^{-1}(x)=\sqrt[3]{x-1}\\ f^{-1}(2010)=\sqrt[3]{2010-1}\\ f^{-1}(2010)=\sqrt[3]{2009}\\ \)

-8/9 if b=3 and a=-1

i am correct because b=3 and a=-3 then it is 0

hm..... not sure why but i think it is wrong....

i think there is something smaller..... maybe i am wrong?

thank you!

yup!!

\(\frac{1}{1+\frac{1}{\sqrt{3}+2}}\)

\(\frac{1}{\frac{\sqrt{3}+3}{\sqrt{3}+2}}\)     

\(\frac{1}{\frac{b}{c}}=\frac{c}{b}\)  \(\frac{\sqrt{3}+2}{\sqrt{3}+3}\)

\(\frac{3+\sqrt{3}}{6}\)

I guess... all 4 moderators gave him/her ++++ 

Draw this on a graph, and you will see that triangle ADE takes up half of \(1 \over 3\) of the rectangle, meaning that the ratio is \(\color{brown}\boxed{1 \over6}\)

Here is the image. The zone in black is triangle ADE, and the blue lines represent the rectangle. 

Simplify and rationalize the denominator.

Hello Guest!

\(\frac{1}{1+\frac{1}{\sqrt{3}+2}}\\ =\frac{1}{\frac{\sqrt{3}+2+1}{\sqrt{3}+2}}\\ =\frac{\sqrt{3}+2}{\sqrt{3}+3}\ |\ \times\frac{\sqrt{3}-3}{\sqrt{3}-3}\\ =\frac{3-3\cdot \sqrt{3} +2\cdot\sqrt{3} -6}{3-9}\\ \color{blue}=\frac{3+\sqrt{3}}{6}\)

  !

Thank You!!!

It would be the mass of 2.3 × 10²⁴ Ag atoms is about 410 g. I'm not the brightest at chemistry and I don't think this website is for chemistry but I recall some of this stuff.

This problem asks you to write in the form of "Egyptian Fractions".

If you go to this site:  https://goodcalculators.com/egyptian-fraction-calculator/

you can get 72 different answers (if I counted correctly!).