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vlabodo [156]
3 years ago
7

Which expression is equivalent to g−m ÷ gn?  g−m + n g−m ⋅ n  g−m − n g−m ÷ n

Mathematics
2 answers:
Elan Coil [88]3 years ago
6 0

The answer would be G=m-n

Sauron [17]3 years ago
4 0
<h2>Answer:</h2>

The expression that is equivalent to the given expression is:

                       g^{-m-n}

<h2>Step-by-step explanation:</h2>

Two expression are said to be equivalent if by some simplification i.e. by some multiplication or division it could be represented in the same manner.

Here we are given a expression as:

                    g^{-m} ÷ g^n i.e.

             \dfrac{g^{-m}}{g^n}

We know that:

\dfrac{a^m}{a^n}=a^{m-n}

                 Hence, we get:

\dfrac{g^{-m}}{g^n}=g^{-m-n}

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In basicallity if a question such as this pops up, the correct way to go would first be to graph this. y=x-4 would be equivalent to x=4.

My paper work is below:

y = x - 4

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Hope this helped..

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3 years ago
Write an equation for an equation with a slope of 1/2 and goes through the point C(−4,3).
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Answer:

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Idali owns 24 packs of gum which is only two-thirds (2/3) of the amount Noah owns. How many packs of gum does Noah own ?
elena-14-01-66 [18.8K]

First, divide 24 by 2 to find out what 1/3 of the gum is. The answer is 12. Then, multiply 12 by 3 to find the rest of the gum.

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4 0
3 years ago
Really need help guys!<br> Love you if you can solve them wiz steps!!!!
Marta_Voda [28]

Answer:

Step-by-step explanation:

x^2+y^2-8x+4y+4=0\\\\a)\\\\x^2-8x+y^2+4y+4=0\\\\x^2-2*x*4+(y^2+2*y*2+2^2)=0\\\\x^2-2*x*4+4^2+(y+2)^2=4^2\\\\(x-4)^2+(y+2)^2=4^2\\

Hence,

The radius of the circle is 4 units, coordinates of its centre are (4,-2).

b)\\\\y=0\\x^2+0^2-8x+4*0+4=0\\x^2-8x+4=0\\a=1\ \ \ \ b=-8\ \ \ \ c=4\\D=(-8)^2-4*1*4\\D=64-14\\D=48\\\sqrt{D}=\sqrt{48} \\ \sqrt{D}=\sqrt{16*3} \\\sqrt{D}=\sqrt{4^2*3}  \\\sqrt{D}=4\sqrt{3}  \\\displaystyle\\x=\frac{-(-8)б4\sqrt{3} }{2*1} \\\\x=\frac{8б4\sqrt{3} }{2} \\x_1=4-2\sqrt{3} \\x_2=4+2\sqrt{3}

c)\\\\A(6,2\sqrt{3}-2)\\ (x-4)^2+(y+2)^2=4^2\\(6-4)^2+(2\sqrt{3} -2+2)^2=16\\2^2+(2\sqrt{3} )^2=16\\4+2^2*(\sqrt{3})^2=16\\ 4+4*3=16\\4+12=16\\16\equiv16

d)\\A(6,2\sqrt{3}-2)}\\\sqrt{3} x+3y=\\\sqrt{3}(6)+3(2\sqrt{3} -2)=\\ 6\sqrt{3}+6\sqrt{3} -6=\\ 12\sqrt{3}-6

5 0
1 year ago
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