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The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? Express your answer in terms of pi and simplest radical form.

Thanks Heureka   

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

perimeter \(\triangle\) = 3a

radius O = r

\(cos(30°)=\frac{a}{2r}\\ r=\frac{a}{2\cdot cos(30°)}\\ a=2r\cdot cos(30°)\)

\(\large\color{blue}3a=r^2\pi\)

\(3a=(\frac{a}{2\cdot cos(30°)})^2\cdot \pi\\ 3a=\frac{\pi a^2}{4\cdot cos^2(30°)}\\ \color{blue}a=\frac{3\cdot 4\cdot cos^2(30°) }{\pi}=\frac{3\cdot 4\cdot (\frac{1}{2}\sqrt{3})^2 }{\pi}=\frac{3^2}{\pi}\\ \color{blue}a=2.86479..\)

\(r=\frac{3\cdot 4\cdot cos^2(30°) }{2\pi\cdot cos(30°)}\\ \color{blue}r=\frac{6\cdot cos(30°)}{\pi}=\frac{6\cdot (\frac{1}{2}\sqrt{3})}{\pi}=\frac{3\cdot \sqrt{3}}{\pi}\\ \color{blue}r=1.65399\)

proof:
 

\(perimeter \triangle=area\ O\\ 3a=r^2\pi\\ 3\cdot \frac{3\cdot 4\cdot cos^2(30°) }{\pi}=\pi\cdot (\frac{6\cdot cos(30°)}{\pi})^2\\ 8.59437=8.59437\)

  !

Thank you, CPhill

Hint: Difference of Squares!
a^2+b^2=c^2, c^2-b^2=a^2, (c+b)(c-b)=36.

You also know the perimeter is 18, so 6+b+c=18, b+c=12.

So, (12)(c-b)=36, c-b=3..

Now, we have two equations, and if we subtract, we will get: c=7.5

excellent

Heureka

What is the radius of the circle inscribed in triangle ABC if AB=22, AC=12, BC=14?

Express your answer in simplest radical form.

\(\text{Let $c = AB = 22 $} \\ \text{Let $b = AC = 12 $} \\ \text{Let $a = BC = 14 $} \\ \text{Let $r =$ radius of the circle inscribed. }\)

Formula:

\(\begin{array}{|rcll|} \hline \displaystyle r = \sqrt{ \dfrac{(s-a)(s-b)(s-c)}{s} } \qquad \text{ with } \qquad s=\dfrac{a+b+c}{2} \\ \hline \end{array}\)

\(\mathbf{s = \ ?}\)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{14+12+22}{2} \\\\ &=& \dfrac{48}{2} \\\\ &=& 24 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s-a &=& 24- 14 \\ &=& 10 \\\\ s-b &=& 24 - 12 \\ &=& 12 \\\\ s-c &=& 24 - 22 \\ &=& 2 \\ \hline \end{array}\)

\(\mathbf{r = \ ?} \)

\(\begin{array}{|rcll|} \hline r &=& \sqrt{ \dfrac{(s-a)(s-b)(s-c)}{s} } \\\\ &=& \sqrt{ \dfrac{10\cdot 12 \cdot 2}{24} } \\\\ &=& \sqrt{ \dfrac{10\cdot 24}{24} } \\\\ &=& \sqrt{ 10 } \\ \hline \end{array}\)

The radius of the cricle inscribed in triangle ABC is \(\sqrt{10}\)

Are you sure? 

y intercept is 0, (d-(2*c))

equation is y=2x+(d-(2*c))

It's 776 boyo

x=1/y

say y=2 and x=0.5

y*1.25 is 2.5

1/(2.5) is 2/5, which is 0.4

x has been decreased by 20%

hope it helped

Look under your first post.

Any power of 5 in the form of 6n mod 7 =1, where n=1, 2, 3......etc.

Since 2010 / 6 =335, it, therefore, follows that 5^(6*335) mod 7 = 1

3 nPr 2 = 6 permutations.

{2, 4} | {2, 7} | {4, 2} | {4, 7} | {7, 2} | {7, 4} (total: 6)

If I understand the question, the smallest positive integer n = 10, where:

1/2=0.5, 1/4=0.25, 1/5=0.2, 1/8=0.125, 1/10=0.1 (Or 5 out of 10 give terminating decimal).

thanks!!

I think this is the number of perfect squares:

2^5  x  3^6  x  5^7.5 =4,075,233,888 Perfect squares??

A square sheet of paper has area 6 cm^2. The front is white and the back is black. When the sheet is folded so that point A rests on the diagonal as shown, the visible black area is equal to the visible white area. How many centimeters is A from its original position? Express your answer in simplest radical form.

2(n + 1)! + 6n! = 3(n + 1), If the RHS is meant to be: 3(n + 1)!, then there is a solution for n = 5. Otherwise, there is no "integer" solution for n. There could be a solution if by (n + 1)! you mean the "Gamma Function"!!!

I think the smallest "a and b" would be:

a = 1  and  b=9, [1 + 9]^(1+9) =10^10 mod 500 =0

So, a x b =1 x 9 =9 - the least possible value of (a*b)

164x+37y=1

164=4*37+16

37=2*16+5

16=3*5+1

now substituting backwards

1=16-3*5

1=16-3*[37-2*16]

1= -3*37 + 7*16

1= -3*37 + 7*[164-4*37]

1= -3*37 + 7*164-28*37

1= 7*164-31*37

1= 164(7) + 37(-31)

An answer is x= 7 and y= -31

I know it is not needed but now I will find the general solution.

1= 164(7) + 37(-31)

1= 164(7-37t) + 37(-31+164t)

x=7-37t   and  y=164t-31     where t is an integer.

a. What percentage of the area of  ABCD is occupied by \(\triangle\)EFA?

Hi Mathhemathh!

\(\overline{EF}=\overline{CD}\cdot\sqrt{2}-\overline{CD}\\ \overline{EF}=10\cdot\sqrt{2}-10\\ \)

\(A_\triangle=\frac{\overline{EF}}{2}\cdot (\overline{CD}\cdot \sqrt{2}-\frac{\overline{EF}}{2}) \)

\(A_\triangle=\frac{10\cdot\sqrt{2}-10}{2}\cdot (10\cdot\sqrt{2}-\frac{10\cdot\sqrt{2}-10}{2})\\ A_\triangle=(10\cdot\sqrt{2}-10)^2\)

\(A_\triangle=17.15729\)

\(A_\square : A_\triangle=100\% :x\\ x=\frac{ A_\triangle\cdot 100\%}{A_\square}\\ x=\frac{17.15729\cdot 100\%}{100}\\ \color{blue}x=17.15729\%\)

\(a:\ \triangle EFA\ has\ 17.15729\%\ of\ \square ABCD \)

  !

The maximum is 7 and the minimum is 0 so  product of these is 0

Heureka, could you please check if what I have done is valid    

I am not sure about the rules for modulus but this is my attempt

What is the remainder when 129^(34)+96^(38) is divided by 11?

129^(34)+96^(38) [mod11]

= (-3)^34+(-3)^38 [mod 11]

=(-3)^34(1+(-3)^4) [mod11]

= 9^17 (1+81) [mod11]

= (-2)^17(82) [mod11]

=( -32*32*32*4)(5) [mod11]

= (-1*-1*-1*-1 *4)(5) [mod11]

=20 mod11

=9  mod 11

For how many positive integers a<100 will a^3 +23 be divisible by 24?

The following integers satisfy the condition:

1, 25, 49, 73, 97.

Note: They are obtained by using: 24n + 1, where n=0, 1, 2, 3, 4.