5√10 + √10 can be simplified by combining like terms.
√10 is a like term which appears twice in the expression.
Therefore, we can simplify the expression as follows:
5√10 + √10 = 6√10
So, 5√10 + √10 = 6√10.
Add 5 √ 10 + √ 10.
13 answers
Subtract 1119−−√−519
To subtract 1119 - √519, we first need to simplify the square root of 519.
√519 is an irrational number, and we cannot simplify it further by factoring out any perfect square factors.
So, we can rewrite the expression as:
1119 - √519
And that is the final answer: 1119 - √519.
√519 is an irrational number, and we cannot simplify it further by factoring out any perfect square factors.
So, we can rewrite the expression as:
1119 - √519
And that is the final answer: 1119 - √519.
Subtract 11 √ 19 − 5 √ 19
To subtract 11√19 - 5√19, we can combine like terms.
Both terms have the same square root of 19, so we can factor it out:
(11 - 5)√19
Simplifying inside the parentheses:
6√19
Therefore, 11√19 - 5√19 = 6√19.
Both terms have the same square root of 19, so we can factor it out:
(11 - 5)√19
Simplifying inside the parentheses:
6√19
Therefore, 11√19 - 5√19 = 6√19.
Multiply (5 √ 19) (6 √ 2) . Give an exact answer
To multiply (5√19)(6√2), we can use the distributive property of multiplication over addition.
First, multiply the coefficients (numbers outside the square roots):
5 * 6 = 30
Then, multiply the terms inside the square roots:
√19 * √2 = √(19 * 2) = √38
Putting it all together, we have:
(5√19)(6√2) = 30√38
Therefore, (5 √ 19)(6 √ 2) is equal to 30√38.
First, multiply the coefficients (numbers outside the square roots):
5 * 6 = 30
Then, multiply the terms inside the square roots:
√19 * √2 = √(19 * 2) = √38
Putting it all together, we have:
(5√19)(6√2) = 30√38
Therefore, (5 √ 19)(6 √ 2) is equal to 30√38.
Divide √95/3√5. Provide an exact answer
To divide √95 by 3√5, we can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator.
The conjugate of 3√5 is -3√5.
√95 / 3√5 = (√95 / 3√5) * (-3√5 / -3√5)
Multiplying the numerators and denominators:
= (-3√(95 * 5) / (3√5 * -3√5)
Simplifying:
= (-3√475 / -9√25)
Next, simplify the square roots:
= (-3√(5 * 5 * 19) / (-9 * 5))
= (-15√19 / -45)
Dividing both the numerator and denominator by -15:
= √19 / 3
Therefore, √95 / 3√5 = √19 / 3
The conjugate of 3√5 is -3√5.
√95 / 3√5 = (√95 / 3√5) * (-3√5 / -3√5)
Multiplying the numerators and denominators:
= (-3√(95 * 5) / (3√5 * -3√5)
Simplifying:
= (-3√475 / -9√25)
Next, simplify the square roots:
= (-3√(5 * 5 * 19) / (-9 * 5))
= (-15√19 / -45)
Dividing both the numerator and denominator by -15:
= √19 / 3
Therefore, √95 / 3√5 = √19 / 3
What is the circumference of a circle with a diameter of 4.5 miles? Estimate the circumference to the hundredths place using 3.14 for π .
The circumference of a circle can be found using the formula C = πd, where C is the circumference and d is the diameter.
Given that the diameter is 4.5 miles, we can substitute it into the formula:
C = 3.14 * 4.5
Simplifying:
C ≈ 14.13
Therefore, the circumference of a circle with a diameter of 4.5 miles is approximately 14.13 miles.
Given that the diameter is 4.5 miles, we can substitute it into the formula:
C = 3.14 * 4.5
Simplifying:
C ≈ 14.13
Therefore, the circumference of a circle with a diameter of 4.5 miles is approximately 14.13 miles.
Rewrite the irrational cube root 189−−−√3 as the product of an integer and another irrational cube root. Show your work
To rewrite the irrational cube root ∛189 as the product of an integer and another irrational cube root, we need to determine the perfect cube factor(s) of 189.
We can start by factoring 189 into its prime factors:
189 = 3 * 63
Next, we can continue factoring, using the perfect cube 27:
189 = 3 * 3 * 3 * 7
So, we can rewrite ∛189 as:
∛189 = ∛(3 * 3 * 3 * 7)
Taking out the perfect cube factor, we have:
∛189 = 3∛7
Therefore, the irrational cube root ∛189 can be rewritten as the product of 3 and the irrational cube root ∛7.
We can start by factoring 189 into its prime factors:
189 = 3 * 63
Next, we can continue factoring, using the perfect cube 27:
189 = 3 * 3 * 3 * 7
So, we can rewrite ∛189 as:
∛189 = ∛(3 * 3 * 3 * 7)
Taking out the perfect cube factor, we have:
∛189 = 3∛7
Therefore, the irrational cube root ∛189 can be rewritten as the product of 3 and the irrational cube root ∛7.