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Mashcka [7]
3 years ago
6

Need a little help with the math questions I displayed

Mathematics
2 answers:
andrew11 [14]3 years ago
7 0
( xy^2/ x^-3y^3)^-4

oee [108]3 years ago
3 0
Use:\\\dfrac{a^n}{a^m}=a^{n-m}\\\\(a^n)^m=a^{n\cdot m}\\\\(a\cdot b)^n=a^n\cdot b^n


\left(\dfrac{xy^2}{x^{-3}y^3}\right)^{-4}=\left(x^{1-(-3)}y^{2-3}\right)^{-4}=\left(x^4y^{-1}\right)^{-4}\\\\x^{4\cdot(-4)}y^{-1\cdot(-4)}=\boxed{x^{-16}y^4}=\boxed{\frac{y^4}{x^{16}}}
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The sum of three numbers is 8. The third is 9
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→<u> </u><u>First</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>1</u>

→<u> </u><u>Second</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>2</u>

→<u> </u><u>Third</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>5</u>

Step-by-step explanation:

• let numbers be x, y and z

{ \tt{z = 8y - 9 -  -  - (eqn \: 1)}} \\  \\ { \tt{10x = 8y - 7 -  -  - (eqn \: 2)}} \\  \\ { \tt{x + y + z = 8 -  -  - (eqn \: 3)}}

• from eqn 2, make x the subject:

{ \tt{x =  \frac{8y - 7}{10} }} \\

• substitute all variables in eqn 3:

{ \tt{ \frac{8y - 7}{10}  + y + 8y - 9 = 8}} \\  \\ { \tt{8y - 7 + 10y + 80y - 90 = 80}} \\  \\ { \tt{98y = 177}} \\  \\ { \boxed{ \tt{ \: y = 1.8}}}

• find z

{ \tt{z = 8y - 9}} \\  \\ { \tt{z = 8(1.8) - 9}} \\  \\ { \tt{z = 14.4 - 9}} \\  \\ { \boxed{ \tt{ \: z = 5.4 \: }}}

• find x:

{ \tt{x =  \frac{8(1.8) - 7}{10} }} \\  \\ { \tt{x =  \frac{7.4}{10} }} \\  \\ { \boxed{ \tt{ \: x = 0.74 \: }}}

Rounding to nearest value:

{ \boxed{ \rm{x = 1}}} \\ { \boxed{ \rm{y =2 }}} \\ { \boxed{  \rm{z = 5}}}

7 0
2 years ago
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