To add 5√10 + √10, you can combine the like terms:
5√10 + √10 = (5 + 1)√10 = 6√10
Therefore, 5√10 + √10 = 6√10.
add 5√10 + √10
37 answers
subtract 11√19 - 5√19
To subtract 11√19 - 5√19, we can follow a similar process as in addition.
Step 1: Recognize that the radical term is the same (√19). Let's label 11√19 as "a" and 5√19 as "b" to make it easier to write the equation:
a = 11√19
b = 5√19
Now, we substitute the values of a and b into the equation:
a - b = 11√19 - 5√19
Replacing a with 11√19 and b with 5√19, we have:
11√19 - 5√19 = 11√19 - 5√19
To simplify further, subtract the coefficients (numbers outside the radical):
11 - 5 = 6
Combine the radical term (√19):
6√19
Therefore, 11√19 - 5√19 simplifies to 6√19.
Step 1: Recognize that the radical term is the same (√19). Let's label 11√19 as "a" and 5√19 as "b" to make it easier to write the equation:
a = 11√19
b = 5√19
Now, we substitute the values of a and b into the equation:
a - b = 11√19 - 5√19
Replacing a with 11√19 and b with 5√19, we have:
11√19 - 5√19 = 11√19 - 5√19
To simplify further, subtract the coefficients (numbers outside the radical):
11 - 5 = 6
Combine the radical term (√19):
6√19
Therefore, 11√19 - 5√19 simplifies to 6√19.
multiply (5√19) (6√2). give an exact answer
To multiply (5√19) (6√2), we can use the distributive property.
First, multiply the coefficients (numbers outside the radicals):
5 * 6 = 30.
Next, multiply the terms inside the radicals:
√19 * √2 = √(19 * 2) = √38.
Combining the coefficients and radical terms, we have:
(5√19) (6√2) = 30√38.
Therefore, the exact answer is 30√38.
First, multiply the coefficients (numbers outside the radicals):
5 * 6 = 30.
Next, multiply the terms inside the radicals:
√19 * √2 = √(19 * 2) = √38.
Combining the coefficients and radical terms, we have:
(5√19) (6√2) = 30√38.
Therefore, the exact answer is 30√38.
divide √95/3√5. provide an exact answer
To divide √95/3√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. So, multiplying the numerator and denominator by -3√5:
(√95 * -3√5)/ (3√5 * -3√5)
Simplifying the expression:
-3√(95*5)/ 3 * √(5*5)
-3√(475)/ 3 * √25
-3√(475)/ 3 * 5
-√475/5
Therefore, √95/3√5 simplifies to -√475/5.
The conjugate of 3√5 is -3√5. So, multiplying the numerator and denominator by -3√5:
(√95 * -3√5)/ (3√5 * -3√5)
Simplifying the expression:
-3√(95*5)/ 3 * √(5*5)
-3√(475)/ 3 * √25
-3√(475)/ 3 * 5
-√475/5
Therefore, √95/3√5 simplifies to -√475/5.
divide 36√39/9√13. provide an exact answer
To divide 36√39 by 9√13, we can simplify by dividing the coefficients and then dividing the radicands.
First, divide the coefficients: 36 ÷ 9 = 4.
Next, divide the radicands: √39 ÷ √13 = √(39/13) = √3.
Combining the simplified coefficients and radicands, we have:
4√3
Therefore, 36√39 divided by 9√13 simplifies to 4√3.
First, divide the coefficients: 36 ÷ 9 = 4.
Next, divide the radicands: √39 ÷ √13 = √(39/13) = √3.
Combining the simplified coefficients and radicands, we have:
4√3
Therefore, 36√39 divided by 9√13 simplifies to 4√3.
simplify √48
To simplify √48, we look for factors of 48 that are perfect squares and can be taken out of the square root.
We notice that 16 is a perfect square that divides evenly into 48. Therefore, we can write:
√48 = √(16 * 3)
Next, using the product property of square roots, we can split the square root:
√(16 * 3) = √16 * √3 = 4√3
So, √48 simplifies to 4√3.
We notice that 16 is a perfect square that divides evenly into 48. Therefore, we can write:
√48 = √(16 * 3)
Next, using the product property of square roots, we can split the square root:
√(16 * 3) = √16 * √3 = 4√3
So, √48 simplifies to 4√3.
simplify ^3√54
To simplify ³√54, we can look for perfect cubes that divide evenly into 54.
We notice that 27 is a perfect cube and divides evenly into 54. Therefore, we can write:
³√54 = ³√(27 * 2)
Next, using the product property of cube roots, we can split the cube root:
³√(27 * 2) = ³√27 * ³√2 = 3 * ³√2
So, ³√54 simplifies to 3³√2.
We notice that 27 is a perfect cube and divides evenly into 54. Therefore, we can write:
³√54 = ³√(27 * 2)
Next, using the product property of cube roots, we can split the cube root:
³√(27 * 2) = ³√27 * ³√2 = 3 * ³√2
So, ³√54 simplifies to 3³√2.
what is the irrational square root of √63 Rewritten as the product of an integer and another irrational square root in simplest form
The irrational square root of √63 can be rewritten as:
√63 = √(9 * 7)
Using the product property of square roots, we can split the square root:
√(9 * 7) = √9 * √7
Simplifying, we find:
√9 = 3
Therefore, the simplified form of √63 can be written as:
√63 = 3√7
So, the irrational square root of √63 can be rewritten as the product of the integer 3 and another irrational square root √7 in its simplest form.
√63 = √(9 * 7)
Using the product property of square roots, we can split the square root:
√(9 * 7) = √9 * √7
Simplifying, we find:
√9 = 3
Therefore, the simplified form of √63 can be written as:
√63 = 3√7
So, the irrational square root of √63 can be rewritten as the product of the integer 3 and another irrational square root √7 in its simplest form.
and how would you write that down as?
To write the simplified form of √63 as the product of an integer and another irrational square root, it would be written as:
√63 = 3√7
√63 = 3√7
what is the irrational cube root of ^3√81 Rewritten as a product of an integer and another irrational cube root
The irrational cube root of ^3√81 can be rewritten as:
^3√81 = ^3√(27 * 3)
Using the product property of cube roots, we can split the cube root:
^3√(27 * 3) = ^3√27 * ^3√3
Simplifying, we find:
^3√27 = 3
Therefore, the simplified form of ^3√81 can be written as:
^3√81 = 3^3√3
So, the irrational cube root of ^3√81 can be rewritten as the product of the integer 3 and another irrational cube root ^3√3.
^3√81 = ^3√(27 * 3)
Using the product property of cube roots, we can split the cube root:
^3√(27 * 3) = ^3√27 * ^3√3
Simplifying, we find:
^3√27 = 3
Therefore, the simplified form of ^3√81 can be written as:
^3√81 = 3^3√3
So, the irrational cube root of ^3√81 can be rewritten as the product of the integer 3 and another irrational cube root ^3√3.
and how would you write that down?
To write the simplified form of ^3√81 as the product of an integer and another irrational cube root, it would be written as:
^3√81 = 3^3√3
^3√81 = 3^3√3
find the square root of 169: √169
The square root of 169 (√169) is 13.
find the cube root of -729:^3√-729
The cube root of -729 (^3√-729) is -9.
find the cube root of -8:^3√-8
The cube root of -8 (^3√-8) is -2.
find the cubic root of 135
The cube root of 135 (^3√135) is approximately 5.92.
find the cubic root of 135
^3√5
3√5
5^3√27
3^3√5
^3√5
3√5
5^3√27
3^3√5
The cubic root of 135 is ^3√135. It cannot be simplified further, so ^3√135 is the most accurate representation of the cubic root of 135.
just answer out of the options i gave you
Out of the options you provided, the cubic root of 135 would be:
^3√5
^3√5
what is the circumference of a circle with a diameter 4.5 MI. estimate the circumference to the hundredths place using 3.14 for pi
The circumference of a circle is given by the formula C = πd, where C represents the circumference and d represents the diameter.
Given that the diameter is 4.5 miles and using π ≈ 3.14, we can calculate the circumference:
C = 3.14 * 4.5
C ≈ 14.13
Therefore, the estimated circumference of the circle is approximately 14.13 miles to the hundredths place.
Given that the diameter is 4.5 miles and using π ≈ 3.14, we can calculate the circumference:
C = 3.14 * 4.5
C ≈ 14.13
Therefore, the estimated circumference of the circle is approximately 14.13 miles to the hundredths place.
rewrite the irrational cube root ^3√189 as the product of an integer and another irrational cube root. show ya work
To rewrite the irrational cube root ^3√189 as the product of an integer and another irrational cube root, we can look for perfect cubes that divide evenly into 189.
We notice that 27 is a perfect cube and divides evenly into 189. Therefore, we can write:
^3√189 = ^3√(27 * 7)
Using the product property of cube roots, we can split the cube root:
^3√(27 * 7) = ^3√27 * ^3√7
Simplifying, we find:
^3√27 = 3
Therefore, the simplified form of ^3√189 can be written as:
^3√189 = 3^3√7
So, the irrational cube root ^3√189 can be rewritten as the product of the integer 3 and another irrational cube root ^3√7.
We notice that 27 is a perfect cube and divides evenly into 189. Therefore, we can write:
^3√189 = ^3√(27 * 7)
Using the product property of cube roots, we can split the cube root:
^3√(27 * 7) = ^3√27 * ^3√7
Simplifying, we find:
^3√27 = 3
Therefore, the simplified form of ^3√189 can be written as:
^3√189 = 3^3√7
So, the irrational cube root ^3√189 can be rewritten as the product of the integer 3 and another irrational cube root ^3√7.