MATHEMATICS GRADE 9 ASSIGNMENT 1 FEBRUARY Question 11.1. Given: $\sqrt{9+25}$ ; $\pi-4$ ; $\sqrt[3]{-27}$ ; $\frac{2}{3}$ ; $\frac{18}{2}$ ; $\sqrt{-27}$From The List Given Above, Write

MATHEMATICS GRADE 9 ASSIGNMENT 1 FEBRUARY QUESTION 11.1

Understanding the Given Expressions

In this assignment, we are given a list of mathematical expressions that we need to evaluate and simplify. The expressions include square roots, cube roots, and fractions. Our task is to carefully analyze each expression and determine its value.

Expression 1: $\sqrt{9+25}$

To evaluate this expression, we need to start by simplifying the expression inside the square root. We have $9+25=34$. Now, we need to find the square root of 34. However, we can simplify this expression further by finding the square root of the individual numbers first. We know that $\sqrt{9}=3$ and $\sqrt{25}=5$. Therefore, we can rewrite the expression as $\sqrt{9+25}=\sqrt{3^2+5^2}$. Using the Pythagorean theorem, we can simplify this expression to $\sqrt{3^2+5^2}=\sqrt{9+25}=\sqrt{34}$.

Expression 2: $\pi-4$

This expression involves the value of pi, which is approximately 3.14. To evaluate this expression, we need to subtract 4 from the value of pi. Therefore, $\pi-4=3.14-4=-0.86$.

Expression 3: $\sqrt[3]{-27}$

To evaluate this expression, we need to find the cube root of -27. We know that $-27=(-3)^3$. Therefore, we can rewrite the expression as $\sqrt[3]{-27}=\sqrt[3]{(-3)^3}$. Using the property of cube roots, we can simplify this expression to $\sqrt[3]{-27}=-3$.

Expression 4: $\frac{2}{3}$

This expression involves a fraction. To evaluate this expression, we need to simplify the fraction by dividing the numerator by the denominator. Therefore, $\frac{2}{3}$ is already in its simplest form.

Expression 5: $\frac{18}{2}$

This expression also involves a fraction. To evaluate this expression, we need to simplify the fraction by dividing the numerator by the denominator. Therefore, $\frac{18}{2}=9$.

Expression 6: $\sqrt{-27}$

To evaluate this expression, we need to find the square root of -27. However, we know that the square of any real number is always non-negative. Therefore, the square root of -27 is not a real number. In mathematics, we often use the term "imaginary number" to describe a number that is not real. Therefore, $\sqrt{-27}$ is an imaginary number.

Conclusion

In this assignment, we were given a list of mathematical expressions that we needed to evaluate and simplify. We carefully analyzed each expression and determined its value. We found that $\sqrt{9+25}=\sqrt{34}$, $\pi-4=-0.86$, $\sqrt[3]{-27}=-3$, $\frac{2}{3}$ is already in its simplest form, $\frac{18}{2}=9$, and $\sqrt{-27}$ is an imaginary number.

Key Takeaways

• We need to carefully analyze each expression and determine its value.
• We can simplify expressions by finding the square root of individual numbers first.
• We can use the Pythagorean theorem to simplify expressions involving square roots.
• We need to be careful when working with fractions and simplify them by dividing the numerator by the denominator.
• We need to be aware of the properties of cube roots and simplify expressions accordingly.
• We need to be aware of the properties of square roots and simplify expressions accordingly.

Practice Problems

• Evaluate the expression $\sqrt{16+9}$.
• Evaluate the expression $\pi-3$.
• Evaluate the expression $\sqrt[3]{-8}$.
• Evaluate the expression $\frac{4}{5}$.
• Evaluate the expression $\frac{20}{4}$.
• Evaluate the expression $\sqrt{-16}$.

Answer Key

• $\sqrt{16+9}=\sqrt{25}=5$
• $\pi-3=3.14-3=0.14$
• $\sqrt[3]{-8}=-2$
• $\frac{4}{5}$ is already in its simplest form.
• $\frac{20}{4}=5$
• $\sqrt{-16}$ is an imaginary number.
  MATHEMATICS GRADE 9 ASSIGNMENT 1 FEBRUARY QUESTION 11.1 Q&A

Frequently Asked Questions

In this Q&A section, we will address some of the most common questions related to the assignment. We will provide detailed answers to help you better understand the concepts and improve your problem-solving skills.

Q: What is the value of $\sqrt{9+25}$?

A: The value of $\sqrt{9+25}$ is $\sqrt{34}$. We can simplify this expression further by finding the square root of the individual numbers first. We know that $\sqrt{9}=3$ and $\sqrt{25}=5$. Therefore, we can rewrite the expression as $\sqrt{9+25}=\sqrt{3^2+5^2}$. Using the Pythagorean theorem, we can simplify this expression to $\sqrt{3^2+5^2}=\sqrt{9+25}=\sqrt{34}$.

Q: What is the value of $\pi-4$?

A: The value of $\pi-4$ is $-0.86$. We know that $\pi$ is approximately $3.14$. To evaluate this expression, we need to subtract $4$ from the value of $\pi$. Therefore, $\pi-4=3.14-4=-0.86$.

Q: What is the value of $\sqrt[3]{-27}$?

A: The value of $\sqrt[3]{-27}$ is $-3$. We know that $-27=(-3)^3$. Therefore, we can rewrite the expression as $\sqrt[3]{-27}=\sqrt[3]{(-3)^3}$. Using the property of cube roots, we can simplify this expression to $\sqrt[3]{-27}=-3$.

Q: What is the value of $\frac{2}{3}$?

A: The value of $\frac{2}{3}$ is already in its simplest form. We do not need to simplify this expression further.

Q: What is the value of $\frac{18}{2}$?

A: The value of $\frac{18}{2}$ is $9$. We can simplify this expression by dividing the numerator by the denominator. Therefore, $\frac{18}{2}=9$.

Q: What is the value of $\sqrt{-27}$?

A: The value of $\sqrt{-27}$ is an imaginary number. We know that the square of any real number is always non-negative. Therefore, the square root of $-27$ is not a real number.

Q: How do I simplify expressions involving square roots?

A: To simplify expressions involving square roots, we need to find the square root of the individual numbers first. We can use the Pythagorean theorem to simplify expressions involving square roots.

Q: How do I simplify expressions involving cube roots?

A: To simplify expressions involving cube roots, we need to use the property of cube roots. We can rewrite the expression as $\sqrt[3]{a^3}=a$.

Q: How do I simplify fractions?

A: To simplify fractions, we need to divide the numerator by the denominator. We can simplify fractions by finding the greatest common divisor (GCD) of the numerator and the denominator.

Q: What is the greatest common divisor (GCD) of two numbers?

A: The greatest common divisor (GCD) of two numbers is the largest number that divides both numbers without leaving a remainder.

Q: How do I find the GCD of two numbers?

A: To find the GCD of two numbers, we can use the Euclidean algorithm. We can also use the prime factorization method to find the GCD of two numbers.

Practice Problems

• Evaluate the expression $\sqrt{16+9}$.
• Evaluate the expression $\pi-3$.
• Evaluate the expression $\sqrt[3]{-8}$.
• Evaluate the expression $\frac{4}{5}$.
• Evaluate the expression $\frac{20}{4}$.
• Evaluate the expression $\sqrt{-16}$.

Answer Key

• $\sqrt{16+9}=\sqrt{25}=5$
• $\pi-3=3.14-3=0.14$
• $\sqrt[3]{-8}=-2$
• $\frac{4}{5}$ is already in its simplest form.
• $\frac{20}{4}=5$
• $\sqrt{-16}$ is an imaginary number.

Key Takeaways

• We need to carefully analyze each expression and determine its value.
• We can simplify expressions by finding the square root of individual numbers first.
• We can use the Pythagorean theorem to simplify expressions involving square roots.
• We need to be careful when working with fractions and simplify them by dividing the numerator by the denominator.
• We need to be aware of the properties of cube roots and simplify expressions accordingly.
• We need to be aware of the properties of square roots and simplify expressions accordingly.

Additional Resources

• For more practice problems and answers, please visit our website.
• For more information on mathematics, please visit our website.
• For more resources on mathematics, please visit our website.