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Novosadov [1.4K]
2 years ago
14

Please help!!! Which expression is equivalent to (3m^-1n^2)^4/(2m^-2n)^3

Mathematics
2 answers:
Alik [6]2 years ago
8 0
<h3>The answer on edg is B!</h3>

Hope this helps!

Elza [17]2 years ago
7 0
\text{Use:}\\\\(a\cdot b)^n=a^nb^n\\\\(a^n)^m=a^{n\cdot m}\\\\\dfrac{a^n}{a^m}=a^{n-m}

\dfrac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}=\dfrac{3^4(m^{-1})^4(n^2)^4}{2^3(m^{-2})^3n^3}=\dfrac{81m^{-4}n^8}{8m^{-6}n^3}\\\\=10.125m^{-4-(-6)}n^{8-3}=10.125m^2n^5
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B) 8x10^(-8)

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A bridge in the shape of an arch connects two cities separated by a river. The two ends of the bridge are located at (–7, –13) a
sdas [7]

Answer:

y=-\dfrac{13}{49}x^2

Step-by-step explanation:

The shape of an arch corresponds to a parabola.

the general equation for a parabola is:

y=ax^2+bx+c

we're given three coordinates: (-7,-13),(7,-13) and (0,0)

so we can plug these values in the general equation to make 3 separate equations:

(x,y) = (-7,-13)

-13=a(-7)^2+b(-7)+c

49a-7b+c=-13

(x,y) = (7,-13)

-13=a(7)^2+b(7)+c

49a+7b+c=-13

(x,y) = (0,0)

0=a(0)^2+b(7)+c

c=0

so we have three equations. and we can solve them simultaneously to find the values of a,b, and c.

we've already found c = 0, let's use substitute it to other equations.

49a-7b+c=-13\quad\Rightarrow\quad49a-7b=-13

49a+7b+c=-13\quad\Rightarrow\quad49a+7b=-13

we can solve these two equation using the elimination method, by simply adding the two equations

\quad\quad49a-7b=-13\\+\quad49a+7b=-13

------------------------------

\quad\quad 98a=-26

\quad\quad a=-\dfrac{13}{49}

Now we can plug this value of a in any of the two equations.

49a-7b=-13

49\left(-\dfrac{13}{49}\right)-7b=-13

-13-7b=-13

-7b=0

b=0

We have the values of a,b, and c. We can plug them in the general equation to find the equation of the arch.

y=\left(-\dfrac{13}{49}\right)x^2+0x+0

y=-\dfrac{13}{49}x^2

49y=-13x^2

This our equation of the arch!

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2 years ago
Can you regroup 43 and 28
avanturin [10]

Answer:

no

Step-by-step explanation:


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