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Grace [21]
3 years ago
13

Evaluate.     −64125‾‾‾‾‾√3

Mathematics
2 answers:
Alexeev081 [22]3 years ago
7 0

The given expression is :\sqrt[3]{\frac{-64}{125}}

64 can be written as 4x4x4=4^{3}

125 =5x5x5=5^{3}

The given expression can be written as :

\sqrt[3]{\frac{-64}{125}}  =\sqrt[3]{\frac{(-4)^{3}}{5^{3}}}     =[( \frac{-4}{5})^{3}  }]^\frac{1}{3}

Applying power rule of indices :

Powers are multiplied .

\frac{1}{3}  times 3 = 1.

= \frac{-4}{5}

The first option \frac{-4}{5} is the right asnwer

Kipish [7]3 years ago
6 0
-4/5 is correct, since (-4/5)³ = -4*-4*-4/(5*5*5) = -64/125 
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If m A. 27<br> B.63<br> C.90<br> D.17
bonufazy [111]

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Decimal: 5.567764)

Step-by-step explanation:

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Solve step by step solution then only i can do it plxx ​
Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So \triangle ABC and \triangle ABD Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
  • First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

  • We need to find base=b

According to Pythagoras thereon

{\boxed{\sf b^2=h^2-p^2}}

  • Substitutethe values

\longrightarrow\sf b^2=10^2-p^2

\longrightarrow\sf b={\sqrt {10^2-8^2}}

\longrightarrow\sf b={\sqrt{100-64}}

\longrightarrow\bf b={\sqrt {36}}

\longrightarrow\sf b=6

\therefore\overline{BC}=6cm

  • BD=BC+CD

\longrightarrowBD=9+6

\longrightarrowBD=15cm

  • Now in \triangle ABD

Perpendicular=p=8cm

Base =b=15cm

  • We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

{\boxed {\sf h^2=p^2+b^2}}

  • Substitute the values

\longrightarrow\sf h^2=8^2+15^2

\longrightarrow\sf h={\sqrt {8^2+15^2}}

\longrightarrow\sf h={\sqrt {64+225}}

\longrightarrow\sf h={\sqrt {289}}

\longrightarrow\sf h=17cm

\therefore{\underline{\boxed{\bf x=17cm}}}

3 0
2 years ago
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