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Leno4ka [110]
3 years ago
12

How to find variables x- and y-

Mathematics
1 answer:
Monica [59]3 years ago
5 0

It says that the triangles are congruent (same size). Also, a triangle's angles add up to 180 degrees. In both of these triangles, one angle is 90° and another angle is 52°. The other angle is 180°-90°-52°=38°.

Solve these equations to find x and y: 2x+18=38, 3y+11=38

2x+18=38\\2x=20\\x=10

3y+11=38\\3y=27\\y=9

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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;

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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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3 years ago
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