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2 years ago

I appreciate any help. Trying to get the hang of these.

Mathematics

2 answers:

aliya0001 [1]2 years ago

7 0

Answer:

x = 12

Step-by-step explanation:

Since the top line is equal to the bottom, you make the equation -

x + x + 4 = x + 16

You subtract an x on both sides because there is one on each side -

x + 4 = 16

Then you subtract the 4 from both sides to get x by its self, which makes

x = 12

jenyasd209 [6]2 years ago

6 0

Answer:

X=12

Step-by-step explanation:

This question can be answered in two ways:

Method 1

The length of the first line = x +x +4

The length of the second line = x +16

But both lines are equal, thus,

x +x +4 = x +16

2x +4 = x+16

2x - x =16-4

x= 12

Method 2

The mark for the first X in the two lines matches meaning they are equal.

Thus, we are left with

x + 4 = 16

x = 16 - 4

x =12

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2 years ago

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 \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} }=$
$\frac{(16)^{ \frac{3}{4} }}{(18)^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } \\ \\ 
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\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } = \frac{ 2^{4* \frac{3}{4} } }{ 3^{4* \frac{3}{4} } } * x^{4* \frac{3}{4} } * y^{8*\frac{3}{4}} = \frac{ 2^{3} }{ 3^{3} } * x^{3} *y^{6} =$
$= \frac{8x^{3}y^{6} }{27}$

Q2. The answer is 1/16

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