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2+7+9=a+b+a+c+b+c=2a=2b+2c so we then finally divide by 2 to get a+b+c=9

so we have that the radius of the circle is 1 so the side of the square is sqrt2 and the area=2

We try 24608/9 and get that it is not divisible by 9. We keep on doing 24618/9 and it is the same 24628/9,24638/9,246148/9,24658/9,24668/9, and finally get that it is 7 . Since 24678/9=2742.

hi guest! 

i honestly don't know how to do this, but if you need quick answers, the answer is \(\boxed{x=\frac{17}{30}}\) (according to wolfram alpha)

ps how did you make the text colored? it looks really cool :)

there's 2 ways to solve this.

1. you can just experiment, and you can see that a=0, b=2, and c=7

2. we see that a+b+a+c=b+c

2a+b+c=b+c

2a=0

a=0

from there, you can find the other values :)

I reworked it but nothing was wrong...?

a+b=2

a+c=7

b+c=9

Add all three equations together to get 2a+2b+2c=18. We divide this equation by 2 to get a+b+c=9.

a+b+c=9

\(\sqrt[3]{x\sqrt{x}}=x^k\)

\(\sqrt[3]{x\cdot x^\frac{1}{2}}=x^k\)

\(\sqrt[3]{x^\frac{3}{2}}=x^k\)

\(x^{\frac{3}{2}\cdot\frac{1}{3}}=x^k\)

\(x^\frac{1}{2}=x^k\)

\(\boxed{k=\frac{1}{2}}\)

\(\)

ok, sorry.

Can you write one question at a time, please? 

10^10 has 121 divisors, 15^7 has 64 divisors, 18^11 has 276 divisors.  All three numbers have a common divisors of 1, which is triple-counted, so the answer is 121 + 64 + 276 - 2 = 459.

The green square has area 4, that means its sides are 2 

Therefore, its diagonal length is 2•sqrt(2) 

The entire dotted line is 7•sqrt(2) and the piece of it in the green square is 2•sqrt(2)

Therefore, the piece of dotted line in the blue square is 5•sqrt(2) 

The section of dotted line in the green square is its diagonal 

and the diagonal of a square runs at a 45^o angle 

Since it's a straight line, it is also running at a 45^o angle once it enters the blue square 

The 45^o line in the blue square creates an isoceles triangle out of the corner  

The hypotenuse of that triangle (the dotted line) is 5•sqrt(2) 

Therefore the legs of the triangle are 5 

The side of the green triangle is 2, and add the leg of the blue triangle 5

Therefore the entire length of the side of the blue square is 7 

Which makes the area of the blue square A = 7² = 49  

_.

Ok, Thank you for your efort, but it says that its wrong.

\(\sqrt{7+\sqrt{4-\sqrt{3+x}}}=3\)

\(7+\sqrt{4-\sqrt{3+x}}=9\)     (squared both sides)

\(\sqrt{4-\sqrt{3+x}}=2\)             (subtracted 7 from both sides)

\(4-\sqrt{3+x}=4\)                 (square both sides again)

\(-\sqrt{3+x}=0\)                     (subtracted 4 from both sides)

\(\sqrt{3+x}=0\)                        (divided both sides by -1)

\(3+x=0\)                            (squared both sides again)

\(\boxed{x=-3}\)

I find ONLY one value for "k" = 531,441 =3^12 =9^6

Equation of line: 2y = x + 2

Solution: x - 2y + 2 <= 0

First one: (-3,4)

Second: (2,-7)

1.

 \(2x-5\le-x+12\\ =3x\le17\\ =x\ge\frac{17}{3}\\ \boxed{x\in(-\infty, \frac{17}{3}]} \)

2. \(-4(x+4)>x+7\\ =-4x-16>x+7\\ =5x<-23\\ x<-\frac{23}{5}\\ \boxed{x\in(-\infty, -\frac{23}{5})}\)

It's just two, not a big deal in my opinion. Any more than that would be an overload.

plz post a question at a time 

the answer is 2