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2 years ago

Answer the question provided in the picture below.

Mathematics

2 answers:

FromTheMoon [43]2 years ago

6 0

Answer:

Finding area: multiply length by width (multiply the top length by the side length)

Finding perimeter: Add the lengths of all of the sides together

Step-by-step explanation:

klio [65]2 years ago

4 0

Answer: see the explanation

Step-by-step explanation:The perimeter is the distance around the outside of a shape. Perimeter is measured in units (e.g., cm). The area is the amount of space the inside of the shape takes up. Area is measured in square units (e.g., cm2).

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Find the median, first quartile, third quartile, and interquartile range of the data.

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Answer:

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3 years ago

X²-10x+25=0 quadratic equation by factoring

lisov135 [29]

Answer:

x = 5

Step-by-step explanation:

${x}^{2} - 10x + 25 = 0 \\ {x}^{2} - 5x - 5x + 25 = 0 \\ x(x - 5) - 5(x - 5) = 0 \\ (x - 5)(x - 5) = 0 \\ ( x- 5) = 0 \: or \: (x - 5) = 0 \\ x = 5 \: or \: x = 5 \\ x = 5$

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2 years ago

Can someone please help me with my maths question

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;

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