3 years ago

Simplify 7 square root of 3 end root minus 4 square root of 6 end root plus square root of 48 end root minus square root of 54

1 answer:

3 0

$7\sqrt3-4\sqrt6+\sqrt{48}-\sqrt{54}\\\\=7\sqrt3-4\sqrt6+\sqrt{16\cdot3}-\sqrt{9\cdot6}\\\\=7\sqrt3-4\sqrt6+\sqrt{16}\cdot\sqrt3-\sqrt9\cdot\sqrt6\\\\=7\sqrt3-4\sqrt6+4\sqrt3-3\sqrt6\\\\=\boxed{11\sqrt3-7\sqrt6}$

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Answer:40% weight increase

Step-by-step explanation:

3 0

3 years ago

If f(x)=x^3−x, what is the average rate of change of f(x) over the interval [1, 5]? a)29.5 b)30 c)43 d)120

avanturin [10]

$\bf slope = m = \cfrac{rise}{run} \implies 
\cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby 
\begin{array}{llll}
average~rate\\
of~change
\end{array}\\\\
-------------------------------\\\\
f(x)= x^3-x \qquad 
\begin{cases}
x_1=1\\
x_2=5
\end{cases}\implies \cfrac{f(5)-f(1)}{5-1}
\\\\\\
\cfrac{[5^3-5]~-~[1^3-1]}{4}\implies \cfrac{125-5~~-~~0}{4}\implies \cfrac{120}{4}\implies 30$