Mathematics 9 - Variation & Equations

A relationship between two variables where one variable increases as the other decreases, and their product remains constant.

A relationship between two variables where one variable increases as the other increases, and the ratio between them remains constant.

A relationship between variables where one variable changes based on the combined effect of multiple variables, either directly or inversely.

A special type of combined variation where one variable varies directly with more than one variable.

A fixed number which represents the relationship between variables in direct or inverse variation.

The force needed to move an object is directly proportional to its mass and acceleration.

As the value of one variable increases, the other variable decreases in a way that their product is constant.

If the force is doubled, while the mass remains the same, the acceleration also doubles.

Any non-zero number raised to the power of zero equals one.

A negative exponent indicates the reciprocal of the base raised to the positive value of the exponent.

When raising a power to another power, multiply the exponents.

The expression simplifies to 1. Any non-zero number raised to the power of zero is 1. The whole expression simplifies to 1 raised to -1, which is still 1.

In a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.

Use the Pythagorean theorem to find the diagonal distance between Yor's house and Anya's school. The distance between the house and the grocery store and the distance between the grocery store and the school form the two sides of a right triangle.

Exponents that are fractions. The numerator represents the power, and the denominator represents the root.

Using the laws of exponents, the expression simplifies to 2x^(4/3).

Combine the terms with the same base by adding their exponents and simplify the result.

Apply the power of a quotient rule: (a/b)^n = a^n / b^n. Then simplify each term by distributing the exponent and reducing the resulting exponents.

Apply the rule of negative exponents: a^-n = 1/a^n. Then simplify the fractions and combine like terms with common base.

Find the cube root of the coefficient, and simplify the variables by dividing their exponents by 3.

Combine the terms under one square root by using the rule: a / b = (a/b). Simplify the resulting expression.

No, because the 8 has a perfect square factor (4) that can be simplified further.

Find the fourth root of the coefficient. Simplify the variables by dividing their exponents by four.

Jimin's approach is more efficient because grouping terms into fourth powers minimizes steps, allowing for direct simplification.

To simplify this expression, first simplify the terms inside the parenthesis, then multiply by [$\sqrt{3}$]{.math.inline}. [$\sqrt{27} = 3\sqrt{3}$]{.math.inline}, so the expression becomes [$\sqrt{3}(3\sqrt{3} - \sqrt{3})$]{.math.inline}, which simplifies to [$\sqrt{3}(2\sqrt{3}) = 6$]{.math.inline}.

Nezuko's first step in simplifying [$\sqrt{3}(\sqrt{6} - 2\sqrt{6})$]{.math.inline} should be to subtract the terms inside the parentheses. [$\sqrt{6} - 2\sqrt{6} = -\sqrt{6}$]{.math.inline}, so the expression becomes [$\sqrt{3}(-\sqrt{6}) = -\sqrt{18}$]{.math.inline}.

The correct order of steps is: 1. Simplify the first term. 2. Simplify the second term. 3. Subtract the simplified terms.

To simplify, break down the radicals into their prime factors. [$\sqrt{18} = 3\sqrt{2}$]{.math.inline}, [$\sqrt{8} = 2\sqrt{2}$]{.math.inline}, and [$\sqrt{24} = 2\sqrt{6}$]{.math.inline}. Substitute these back into the original expression, then combine like terms: [$-9\sqrt{2} + 6\sqrt{2} - 2\sqrt{6} = -3\sqrt{2} - 2\sqrt{6}$]{.math.inline}.

Tanjiro is incorrect. While multiplying the numerator and denominator by [$\sqrt{3} + 2$]{.math.inline} rationalizes the denominator, it leads to a negative result. He should have multiplied by [$\sqrt{3} + 2$]{.math.inline} (the conjugate of the denominator).

The architect correctly calculated the area of the circular garden, using the correct formula and the radius from the diameter. The radius of the garden is [$\frac{\sqrt{50}}{2} = \frac{5\sqrt{2}}{2}$]{.math.inline}, and the area is calculated as [$\pi (\frac{5\sqrt{2}}{2})^2 = \frac{25\pi}{2}$]{.math.inline}.

The first step is to find the length of one side of the square playground since the area of a square is side * side. To find the length of a side, you need to find the square root of the area. The square root of 64 is 8, so the length of one side of the square playground is 8 meters.

The mesh net length needed to enclose the 8 meter by 8 meter square playground is 32 meters. 8 meters for each side of the square, times 4 sides, equals 32 meters.