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Marysya12 [62]
3 years ago
13

What is the midpoint of AB?

Mathematics
1 answer:
BabaBlast [244]3 years ago
7 0

Answer:

A picture of the problem might help

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DIA [1.3K]

Answer:

a. \  \dfrac{625 \cdot m}{27 \cdot n^{11}}

b. \  \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}

By expanding the expression, we get;

\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}

Collecting like terms gives;

\dfrac{m^{(3 + 4 - 6)}  \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}

\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}

(b) The given expression is presented as follows;

x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4

Therefore, we get;

x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times  x^{-4} \times y^{-4 \cdot n}

Collecting like terms gives;

x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )

x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}

x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times  x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}

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Look in the file, I've drew it there!

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