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jarptica [38.1K]
3 years ago
13

#54 Simplify and write the answers using positive exponents only.

Mathematics
1 answer:
kramer3 years ago
8 0
Gg easy

remember
(x^m)/(x^n)=x^(m-n)
and
(x^m)^n=x^(mn)
and
(xyz)^m=(x^m)(y^m)(z^m)
and
(m/n)^a=(m^a)/(n^a)
and
x^-n=1/(x^n)

so
( \frac{6mn^{-2}}{3m^{-1}n^2} )^{-3}=
( \frac{6}{3} )^{-3}( \frac{m}{m^{-1}} )^{-3}( \frac{n^{-2}}{n^2} )^{-3}=
(2^{-3})((m^2)^{-3})((n^{-4})^{-3})=
( \frac{1}{2^3})(m^{-6})(n^{12})=
( \frac{1}{8} )( \frac{1}{m^6})(n^{12})=
\frac{n^{12}}{8m^6}

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Ms. Carr's Class Is Selling Magazines To Raise Money For A Field Trip. The Students In The Class Device They Wanted To Make 5.50
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Here you are giving only the amount they want to raise (namely profit times number of magazines sold), and here you are also giving Money they want to raise... So clarifying, the money they want to raise, should include the money they will spend on buying the magazines (there is no statement saying they found them, or were given the magazines, so a cost should be involved) 

Now if they are only making the count of "Field trip costs X amount of money, and given we have to make a profit of $5.5, How many must we sell?" then the equation should be n=X/5.5 

Should the story be, how much money must they raise to have a profit of 5.5 on each magazine and still have enough for the field trip, then you have a different equation which varies only in adding the cost of each magazine, either case, M should be defined not as money they need to raise (cause here they will be short on their goal) but Money they must earn. And again, you should rewrite your equation to be: 

M=Amount they must raise 
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C+p=M/n 

And rewriting the previous they should make: 

n(C+p)=M -----> n(C+5.5)=M <span>m/n = 5.50 </span>
<span>m/n x n = 5.50 x n //// multiply each side by n </span>

<span>m = 5.5n</span>
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X-y=6<br> 2x-3z=16<br> 2y+z=4<br><br> Solve the systems of equations.
Neko [114]

Answer:

Given System of equation:

x-y =6                                  .....,[1]

2x-3z = 16                           ......[2]  

2y+z = 4                              .......[3]

Rewrite the equation [1] as

y = x - 6                                .......[4]

Substitute the value of [4] in [3], we get

2(x-6)+z = 4

Using distributive property on LHS ( i.e,  a \cdot (b+c) =a \cdot b+ b \cdot c )

then, we have

2x - 12 +z =4

Add 12 to both sides of an equation:

2x-12+z+12=4+12

Simplify:

2x +z = 16                         .......[5]

On substituting equation [2] in [5] we get;

2x+z=2x -3z

or

z = -3z

Add 3z both sides of an equation:

z+3z = -3z+3z

4z = 0

Simplify:

z = 0

Substitute the value of z = 0 in [2] to solve for x;

2x-3(0) = 16

or

2x = 16

Divide by 2 both sides of an equation:

\frac{2x}{2} =\frac{16}{2}

Simplify:

x= 8

Substitute the value of x =8 in equation [4] to solve for y;

y = 8-6 = 2

or

y = 2

Therefore, the solution for the given system of equation is;  x = 8 , y = 2 and z =0

5 0
3 years ago
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