No, The letter math

We have : \(x^2-180x+8102=\left(x^2-180x+8100\right)+2=\left(x-90\right)^2+2\ge2\forall x\left(1\right)\)

Applying the Bunhiacopxki inequality , we have : 

\(\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le\left(1+1\right)\left(x-89+91-x\right)\)

\(\Rightarrow\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le2.2=4\)

\(\Rightarrow\sqrt{x-89}+\sqrt{91-x}\le2\left(2\right)\)

Because \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102\)

So ( 1 ) ; ( 2 ) ; we have : \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102=2\)

Equal sign occurs \(\Leftrightarrow\sqrt{x-89}=\sqrt{91-x};x-90=0\left(3\right)\)

We have : \(\sqrt{x-89}=\sqrt{91-x}\)  \(\Leftrightarrow x-89=91-x\Leftrightarrow x=90\left(4\right)\)

 ( 3 ) ; ( 4 ) \(\Rightarrow x=90\) is the result of the equation

Đk : \(89\le x\le91\)

Applying the Bunhiacopxki inequality

We have : VT \(=1.\sqrt{x-89}+1.\sqrt{91-x}\le\sqrt{\left(1+1\right)\left(x-89+91-x\right)=2}\)

=> VT \(\le2\)

And VP = \(x^2-2.x.90+90^2+2=\left(x-90\right)^2+2\ge2\)

=> VP  \(\ge2\ge\) VT

The sign "=" occurs when and only when: 

\(\left\{{}\begin{matrix}\sqrt{x-89}=\sqrt{91-x}\\x-90=0\end{matrix}\right.\) => \(x=90\)(satisfy)

The answer is 90

Tran Anh ??? Le Quoc Tran Anh ??? 

El matodora : The Spam question

You should write the real question in English

The speed limit in this zone in feet per second is:
    88:60 x 15==22(feet)

22 feet

C1:  We have : 3x + 4x + 5x = x (3 + 4 + 5)

Replace x = 10 <=> 10 ( 3 + 4 + 5) = 10 . 12 = 120

C2 : 3x + 4x + 5x <=> 3 . 10 + 4 . 10 + 5 . 10 = 30 + 40 + 50 = 120

We have 

\(\dfrac{2}{8}.\dfrac{5}{7}=256.78125=20000000\)

\(=>1000000000:\left(\dfrac{2}{8}.\dfrac{5}{7}\right)=1000000000:20000000=50\)

The answer is 50

1 000 000 000 =10⁹=2⁹.5⁹

\(\Rightarrow\)1000000000 / (2⁸x5⁷)=2*5²=50

The answer ís 50

This is two