2. Rewrite ∛40 as a product of an integer and another irrational cube root
3. Find area of circle with diameter of 10.2 cm (round to tenth and put units)
4. Find circumference of circle with radius of 2.7 cm (round to tenth and put units)
5. Rewrite √45 in as a product of an integer and irrational square root.
(do simple work)
2. ∛40 = ∛(8×5)=2∛5
3. The diameter of the circle is 10.2 cm, so the radius is half of that, which is 5.1 cm. The formula for the area of a circle is A = πr², substituting the values we have:
A = π(5.1)² = 25.91 cm² (rounded to tenth)
4. The radius of the circle is 2.7 cm. The formula for the circumference of a circle is C = 2πr, substituting the value we have:
C = 2π(2.7) = 16.96 cm (rounded to tenth)
5. √45 can be written as √(9×5) = 3√5
2. To rewrite ∛40 as a product of an integer and another irrational cube root, we can start by finding the prime factorization of 40.
The prime factorization of 40 is 2 * 2 * 2 * 5.
Since we are looking for the cube root, we can group the factors in sets of 3.
So, we can rewrite ∛40 as ∛(2 * 2 * 2 * 5) as ∛(2^3 * 5).
Simplifying this, we have 2∛(5).
Therefore, ∛40 can be rewritten as 2∛5.
3. The formula to find the area of a circle is A = π * r^2, where A is the area and r is the radius. Given that the diameter of the circle is 10.2 cm, we can find the radius by dividing the diameter by 2.
Radius = Diameter / 2 = 10.2 cm / 2 = 5.1 cm.
Now, we can plug the radius into the formula to find the area:
A = π * r^2 = π * (5.1 cm)^2 ≈ 81.17 cm² (rounded to the tenth).
Therefore, the area of the circle with a diameter of 10.2 cm is approximately 81.17 cm².
4. The formula to find the circumference of a circle is C = 2 * π * r, where C is the circumference and r is the radius. Given that the radius of the circle is 2.7 cm, we can plug the radius into the formula to find the circumference:
C = 2 * π * r = 2 * π * 2.7 cm ≈ 16.98 cm (rounded to the tenth).
Therefore, the circumference of the circle with a radius of 2.7 cm is approximately 16.98 cm.
5. To rewrite √45 as a product of an integer and an irrational square root, we can find the prime factorization of 45.
The prime factorization of 45 is 3 * 3 * 5.
Since we are looking for the square root, we can group the factors in pairs.
So, we can rewrite √45 as √(3 * 3 * 5) as √(3^2 * 5).
Simplifying this, we have 3√(5).
Therefore, √45 can be rewritten as 3√5.
2. To rewrite ∛40 as a product of an integer and another irrational cube root, we can start by finding the prime factorization of 40, which is 2^3 * 5.
Next, we can rewrite ∛40 as ∛(2^3 * 5).
Taking the cube root of 2^3 gives us ∛(2) * ∛(2) * ∛(2) * ∛(5).
So, we can rewrite ∛40 as 2 * ∛8 * ∛5.
Simplifying further, we have 2 * 2∛5 or 4∛5, where ∛5 is an irrational cube root.
3. To find the area of a circle with a diameter of 10.2 cm, we need to use the formula for the area of a circle, which is A = πr^2.
First, we can find the radius by dividing the diameter by 2. So, the radius is 10.2 cm / 2 = 5.1 cm.
Now, we can substitute the radius into the formula to find the area:
A = π * (5.1 cm)^2
= π * 26.01 cm^2 (rounded to two decimal places)
Therefore, the area of the circle is approximately 81.29 cm^2.
4. To find the circumference of a circle with a radius of 2.7 cm, we can use the formula C = 2πr, where C represents the circumference and r represents the radius.
Substituting the given radius into the formula, we have:
C = 2π * 2.7 cm
= 5.4π cm (rounded to one decimal place)
Therefore, the circumference of the circle is approximately 16.97 cm.
5. To rewrite √45 as a product of an integer and an irrational square root, we can start by finding the prime factorization of 45, which is 3^2 * 5.
Next, we can rewrite √45 as √(3^2 * 5).
Taking the square root of 3^2 gives us 3, and the square root of 5 remains unchanged.
So, we can rewrite √45 as 3√5, where √5 is an irrational square root.