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3^{11} + 5^7 + 1 ==255,273 == 3 * 85091

3 is the smallest prime factor.

no that is wrong the answer is 64

\(6tan^2(5x-3)=2\\ tan(5x-3)=\frac{1}{\sqrt3}\\ \quad \text { you should recognise that } \quad tan \frac{\pi}{6} =\frac{1}{\sqrt3}\\ 5x-3=\frac{\pi}{6}\\ \)

And you can finish it.

Wu rolls two fair, six-sided dice.  You are not told what the rolls were, but you are told that the sum of the two rolls is a prime.  What is the probability that the sum of the two rolls is 2?

Below is a table of all the possible combinations of two dice. 

There is only 1 way to roll a 2, out of 36 possible rolls.  

So the probability is 1/36.  

    1,1   1,2   1,3   1,4   1,5   1,6  

    2,1   2,2   2,3   2,4   2,5   2,6  

    3,1   3,2   3,3   3,4   3,5   3,6  

    4,1   4,2   4,3   4,4   4,5   4,6  

    5,1   5,2   5,3   5,4   5,5   5,6  

    6,1   6,2   6,3   6,4   6,5   6,6

_.

Let x be the number of chocolate cakes bought by Jeff

The total number of cakes including chocolate cakes and vanilla cakes is 288.

∴ Number of vanilla cakes = 288 - x.

Out of total cakes 4/9 are chocolate & the rest were vanilla.

∴ Number of chocolate cakes is

            x = 4/9 x (total number of cakes)

               = 4/9 * 288 = 1152/9 = 128

∴ Number of vanilla cakes = 288 - 128 = 160

∴ Jeff bought 128 chocolate cakes & 160 vanilla cakes

Jeff ate some chocolate cakes, the number of chocolates left was 5/8 of the total number of chocolate at first

Let y be the number of chocolate left.

∴ y = 5/8 x (total number of chocolate at first)

      = 5/8 * 128 = 640/8 = 80

∴ (chocolate cake ate by Jeff) + y = Total number of chocolate cakes

    ∴ (Chocolate cake ate by Jeff) + 80 = 128

      => Choclate cake ate by Jeff = 128 - 80 = 48.

∴ Jeff eat 48 chocolate cakes.

Hello Guest!

\(r_B=8cm\\\ d_A=x\\ (\pi \cdot64cm^2-\pi \cdot \dfrac{x^2}{4} ):(\pi \cdot \dfrac{x^2}{4})=8:1\\ \pi \cdot64cm^2-\pi \cdot \dfrac{x^2}{4} =2\pi x^2\\ \pi \cdot64cm^2=\dfrac{9\pi x^2}{4}\\ 8cm=\dfrac{3x}{2}\)

\(x=5\dfrac{1}{3}cm\)

The diameter of circle A is 5.33cm.

  !

This question is really hard. A math question with mixed numbers with this is really hard so it is easy to convert them to improper fractions.

Box A                Box B

W = x                      G

B = 8/5x                  BL

 
     White  marbles = x

    Black marbles B = 8/5x
 

\({X \over G + BL } = {8 \over 5} —- (1)\)

\({B\over G + BL + 138} = {4\over 3} —- (2)\)
Find B/G+BL = ?

B/G+BL = 8/5x/(G+BL) = 8/5(X/G+BL)

                                     = 8/5 * 8/5 (From (1)).

                                     = 64/25

So Ratio of Black marbles to the number of Blue and Green marbles = 64:25

X/G+BL = 8/5 ----(1)

So G+BL = 5x/8 --> Substitute in (2)

       B/G+BL+138 = 4/3

       B/5x/8 + 138 = 4/3

ie.   8/5x/8 + 138 = 4/3

         24/5x = 20x/8 + 552

       \( {192 - 100 \over 40}x = 552\)

             So x = 552 * 40/92 + 240

when x = 240

         B = 8/5x = 8/5 * 240

                       = 384

         G + BL = 5x/8 = 150

Ans:  Box A has 240 white marbles and 384 blue marbles So 384 - 240 = 144 more black marbles than white marbles.

The area of the overlap is 4*pi - 8.

nevermind, i think i got it;

heres how i did it

first draw quadrilateral ABCD, then draw the circle inside. start labeling the equal tangents from vertices. also label the points of tangency. I labeled mine

like this: W on AB, X on BC, Y on DC, and Z on AD. AW = AZ = x (variables), BX = WB = y, XC = YC = z, and ZD = DY = w. 

the equation is x + y + w + z = x + w + y + z

which is true

proved

this solution made no sense i know 

$\mathbf{A} \mathbf{u} = \begin{pmatrix} 4 \\ -4 \\ 8 \end{pmatrix}$

$\mathbf{A} \mathbf{v} = \begin{pmatrix} -8 \\ 12 \\ -16 \end{pmatrix}$

1. There are 9 dots, and to make triangles, you need 3.

C(9, 3) is 84. However, out of these 84 ways, there are 3 horizontal lines, 3 vertical, and two diagonal (these aren't considered triangles).

Now, we can use complementary counting to find the remaining number of true cases.

84 - (3 + 3 + 2) = 84 - 8 = 76.

Answer: 76 triangles

2. If a right isosceles triangle were to have a hypotenuse of 1, we could find the values of the base and height using the Pythagorean Theorem. 

\(x^2 + y^2 = 1^2 = 1\), \(x^2 + y^2 = 1\).

Now because the sides are the same length, we can simplify this equation to 

\(2x^2 = 1\)

\(x^2 = \frac{1}{2}\), \(x = \sqrt{\frac{1}{2}}\).

To find the area, we use the area equation of a triangle (\(\frac{bh}{2}\)). Because the base and height are the same, the equation becomes \(\frac{b^2}{2}\) or \(\frac{h^2}{2}\). 

Plugging the value we got into the equation, we get

\(\frac{\sqrt{\frac{1}{2}}^2}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4}.\)

Answer: \(\mathbf{\frac{1}{4}}\)

Hello Guest,

I understood and I am sorry for saying something nonsense... Yes, now I am not sure what to say, better I am not answering. But please speak normal English and not that weird. I don't know what to say, better to shut the mouth now.  Thank you for listening.

Hello Guest,

I know it's a joke, but please ask somewhere and not here.

Straight

Hello Guest,

it's really not that hard, still I am gonna to say the answer:

\(4^2+(1*5+7^2)+8\),

\(16+(5+7^2)+8\),

\(16+(5+49)+8\),

\(16+54+8\),

\(78\),

easy!

Straight

Hello Guest,

may you could something for it? It would at least be understandable. Here is what I did:

\(ab^2+3a^3b-8ab^2+4a^3b+2a^3b\),

\(-7ab^2+9a^3b\),

\(-ab*(7b-9a^2)\),

so finally the solution is: \(-ab*(7b-9a^2)\) . Hope this was helpful.

Straight

Hello Guest,

I think it's not possible to do, except you are doing: \(\mbox {if you are writing 2:3, it turns into }\frac{2}{3} \mbox{ and I think that's the onliest way.}\) Hope this was helpful.

Straight

Thanks, that's right!