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3 years ago

PLEASE HELP I DONT KNOW

Mathematics

2 answers:

lubasha [3.4K]3 years ago

8 0

Answer:

Step-by-step explanation:

A line limited by the circumference going through the center is called the diameter.

riadik2000 [5.3K]3 years ago

5 0

Answer:

d, diameter

Step-by-step explanation:

radius is half of a diameter, or the distance from the center of the circle to the circumference

circle is a... circle

center is...

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Find the sum of the first four terms of the geometric sequence shown below. 4, 4/ 3, 4/9, ...

user100 [1]

It's evident that the first four terms are 4, 4/3, 4/9, and 4/27. So the fourth partial sum of the series is

$S_4=4+\dfrac43+\dfrac49+\dfrac4{27}$

It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.

Multiply the sum by 1/3 and you have

$\dfrac13S_4=\dfrac43+\dfrac49+\dfrac4{27}+\dfrac4{81}$

Now subtracting this from $S_4$ gives

$S_4-\dfrac13S_4=4-\dfrac4{81}$

That is, all the matching terms will cancel. Now solving for $S_4$ , you
have

$\dfrac23S_4=4\left(1-\dfrac1{81}\right)$
$S_4=6\left(1-\dfrac1{81}\right)$
$S_4=\dfrac{480}{81}=\dfrac{160}{27}$

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3 years ago

Simplify each expression. Assume that all variables are positive.

kozerog [31]

Q1. The answer is $\frac{8x^{3}y^{6} }{27}$

$( \frac{16 x^{5} y^{10}}{81x y^{2} } )^{ \frac{3}{4} }= ( \frac{16}{81}* \frac{ x^{5} }{x}* \frac{ y^{10} }{y^{2}} )^{ \frac{3}{4} } \\ \\ 
 \frac{ x^{a} }{ x^{b} }= x^{a-b} \\ \\ 
( \frac{16}{81}* \frac{ x^{5} }{x}*\frac{ y^{10} }{y^{2}} )^{ \frac{3}{4} }}=( \frac{16}{81 }* x^{5-1}* y^{10-2})^{ \frac{3}{4} }=( \frac{16}{81 }* x^{4}* y^{8})^{ \frac{3}{4} }= \\ \\ = (\frac{16}{18} )^{ \frac{3}{4} }*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=$
$\frac{(16)^{ \frac{3}{4} }}{(18)^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } \\ \\ 
 (x^{a} )^{b} = x^{a*b} \\ \\ 
\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } = \frac{ 2^{4* \frac{3}{4} } }{ 3^{4* \frac{3}{4} } } * x^{4* \frac{3}{4} } * y^{8*\frac{3}{4}} = \frac{ 2^{3} }{ 3^{3} } * x^{3} *y^{6} =$
$= \frac{8x^{3}y^{6} }{27}$

Q2. The answer is 1/16

$(-64) ^ \frac{-2}{3} =(-1* 2^{6} ) ^ \frac{-2}{3}=(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} \\\\x^{-a} = \frac{1}{ x^{a} } \\\\(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} = \frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{(2^{6})^ \frac{2}{3}} \\ \\ (x^{a} )^{b}=x^{a*b} \\\\x^{ \frac{a}{b} = \sqrt[b]{ x^{a} } } \\ \\ 
$
$\frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{2^{6*\frac{2}{3}}} = \frac{1}{ \sqrt[3]{(-1)^{2} } } * \frac{1}{ 2^{4} } = \frac{1}{ \sqrt[3]{1} } * \frac{1}{16} = \frac{1}{1} * \frac{1}{16}= \frac{1}{16}$

Q3. The answer is $a^{ \frac{7}{6} }$

$a^{ \frac{2}{3} } * a^{ \frac{1}{2} } \\ \\ 
 x^{a}* x^{b} =x^{a+b} \\ \\ 
a^{ \frac{2}{3} } * a^{ \frac{1}{2} }= a^{ \frac{2}{3} + \frac{1}{2} } =a^{ \frac{2*2}{3*2} + \frac{1*3}{2*3} }=a^{ \frac{4}{6} + \frac{3}{6} }=a^{ \frac{4+3}{6} }=a^{ \frac{7}{6} }$

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