Please help!!! Will mark Brainliest Want with explanation... Evaluate it...

Mathematics

2 answers:

attashe74 [19]3 years ago

8 0

$\sqrt{5 ^{ - 3 } } x \sqrt{2 ^{ - 3} }$

Multiply the terms with equal exponents by exponents by multiplaying their bases.

$( \sqrt{5 \sqrt{2) ^{ - 3} } }$

Calculate the product.

$\sqrt{10 ^{ - 3} }$

Express with a positive exponent using

a ^ {-n }}} = 1

a ^ {n }}}

$\large \frac{1}{ \sqrt{10 ^{3} } }$

Evaluate the Power.

$\large \frac{1}{10 \sqrt{10} }$

Rationaliz the denominator

$\frac{ \sqrt{10} }{100}$

$\color{IRON}{SOLUTION}$

$\frac{ \sqrt{10} }{100}$

Alternate form

$\color{iron}{≈0.0316228}$

vitfil [10]3 years ago

7 0

Answer:

0.0316227766 Is the answer

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Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess

Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2$