Simplify : (x^-2y^-4x^3)^-2

1 answer:

3 0

$\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (x^{-2}y^{-4}x^3)^{-2}\implies (x^{-2}x^3y^{-4})^{-2}\implies (x^{-2+3}y^{-4})^{-2}\implies (x^{1}y^{-4})^{-2} \\\\\\ \stackrel{\textit{distributing the exponent}~\hfill }{(x^{-2\cdot 1}y^{-2\cdot -4})\implies x^{-2}y^8}\implies \cfrac{1}{x^2}\cdot y^8\implies \cfrac{y^8}{x^2}$

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4x + 3y = 6 3x + 5y = -1 Solve the simultaneous equations. y= z=

Kisachek [45]

Answer:

x=3, y= -2

Step-by-step explanation:

Multiplying :

(4x+3y =6)*-3

(3x+5y =-1)*4

-12x -9y= -18

12x+20y= -4

Add the equations

20y-9y = -4-18

11y = -22

y= -22/11 = -2

3x +5* -2= -1

3x -10 = -1

3x = 10-1

x= 9/3 =3

4 0

3 years ago

prove that the quadrilateral whose vertices are the points A(-1,1), B(-3,4), C(1,5) and D(3,2) is a parallelogram.

Vsevolod [243]

Answer:

Since $\overrightarrow{AB} = \overrightarrow{DC}$ and $\overrightarrow{AD} = \overrightarrow{BC}$ , then the quadrilateral ABCD is a parallelogram.

Step-by-step explanation:

First, we label each point of the quadrilateral with the help of a graphing tool. If the quadrilateral ABCD is a parallelogram, then $\overrightarrow{AB} = \overrightarrow{DC}$ and $\overrightarrow{AD} = \overrightarrow{BC}$ . If we know that $A(x,y) =(-1, 1)$ , $B(x,y) =(-3,4)$ , $C(x,y) = (1,5)$ and $D(x,y) = (3,2)$ , then the measure of each vector is, respectively: