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Solve the equation for x

x3 = 343/64

17 answers

To solve the equation \( x^3 = \frac{343}{64} \), we need to isolate \( x \).

First, take the cube root of both sides:

\[ x = \sqrt[3]{\frac{343}{64}} \]

Next, we can simplify the cube root of the fraction:

\[ x = \frac{\sqrt[3]{343}}{\sqrt[3]{64}} \]

Now, we calculate the cube roots:

  • The cube root of \( 343 \) is \( 7 \) because \( 7^3 = 343 \).
  • The cube root of \( 64 \) is \( 4 \) because \( 4^3 = 64 \).

Thus, we have:

\[ x = \frac{7}{4} \]

Therefore, the solution is:

\[ \boxed{\frac{7}{4}} \]

Determine which of the following are the solutions to the equation below.

x2 = 10

To solve the equation \( x^2 = 10 \), we need to isolate \( x \) by taking the square root of both sides.

Doing this gives us:

\[ x = \pm \sqrt{10} \]

This means there are two solutions:

  1. \( x = \sqrt{10} \)
  2. \( x = -\sqrt{10} \)

Thus, the solutions to the equation \( x^2 = 10 \) are:

\[ x = \sqrt{10} \quad \text{and} \quad x = -\sqrt{10}. \]

If you want numerical approximations, \( \sqrt{10} \approx 3.162 \) and \( -\sqrt{10} \approx -3.162 \).

Therefore, the final solutions are:

\[ \boxed{\sqrt{10}, -\sqrt{10}} \]

Determine which of the following are solutions to the equation below.

x2 = 16

To solve the equation \( x^2 = 16 \), we will isolate \( x \) by taking the square root of both sides.

Taking the square root gives:

\[ x = \pm \sqrt{16} \]

Since the square root of \( 16 \) is \( 4 \), we have:

\[ x = 4 \quad \text{or} \quad x = -4 \]

Thus, the solutions to the equation \( x^2 = 16 \) are:

\[ x = 4 \quad \text{and} \quad x = -4 \]

Therefore, the final solutions are:

\[ \boxed{4 \text{ and } -4} \]

Which of the following describes the positive solution to the equation below?

A: The solution is greater than two but less than three.
B:The solution is a rational number.
C:The solution is greater than one but less than two.
D: The solution is a repeating decimal.
E: The solution is greater than zero but less than one.
F:The solution is an irrational number.

Let's first solve the equation given by \( x^2 = 16 \):

Taking the square root of both sides, we find:

\[ x = \pm \sqrt{16} \]

This leads to the solutions:

\[ x = 4 \quad \text{and} \quad x = -4 \]

Since you are asking for the positive solution, we take:

\[ x = 4 \]

Now, let's evaluate each of the options in relation to this positive solution:

  • A: The solution is greater than two but less than three. (False, as \( 4 \) is greater than \( 3 \).)
  • B: The solution is a rational number. (True, as \( 4 \) can be expressed as \( \frac{4}{1} \).)
  • C: The solution is greater than one but less than two. (False, as \( 4 \) is greater than \( 2 \).)
  • D: The solution is a repeating decimal. (False, as \( 4 \) is a whole number, not a decimal.)
  • E: The solution is greater than zero but less than one. (False, as \( 4 \) is greater than \( 1 \).)
  • F: The solution is an irrational number. (False, as \( 4 \) is a rational number.)

Therefore, the only statement that correctly describes the positive solution \( 4 \) is:

\[ \boxed{B} \]

Select all the correct answers

Which of the following describes the positive solution to the equation below?

x2 = 5
A: The solution is greater than two but less than three.
B:The solution is a rational number.
C:The solution is greater than one but less than two.
D: The solution is a repeating decimal.
E: The solution is greater than zero but less than one.
F:The solution is an irrational number

To solve the equation \( x^2 = 5 \), we find the positive solution by taking the square root:

\[ x = \sqrt{5} \]

Now, let's determine the value of \( \sqrt{5} \) approximately. Since \( \sqrt{4} = 2 \) and \( \sqrt{9} = 3 \), we know:

\[ 2 < \sqrt{5} < 3 \]

Calculating further, \( \sqrt{5} \) is approximately \( 2.236 \).

Now let's evaluate each of the provided options regarding \( \sqrt{5} \):

  • A: The solution is greater than two but less than three.
    (True, as \( 2 < \sqrt{5} < 3 \).)

  • B: The solution is a rational number.
    (False, since \( \sqrt{5} \) is irrational.)

  • C: The solution is greater than one but less than two.
    (False, as \( \sqrt{5} \approx 2.236 \).)

  • D: The solution is a repeating decimal.
    (False, \( \sqrt{5} \) is an irrational number and does not terminate or repeat. It is non-repeating.)

  • E: The solution is greater than zero but less than one.
    (False, as \( \sqrt{5} \) is greater than \( 2 \).)

  • F: The solution is an irrational number.
    (True, since \( \sqrt{5} \) cannot be expressed as a fraction of two integers.)

From this evaluation, the correct answers are A and F. Thus, the final selection is:

\[ \boxed{A \text{ and } F} \]

Determine which of the following is the solution to the equation below.

x3 = 6

To solve the equation \( x^3 = 6 \), we need to isolate \( x \) by taking the cube root of both sides.

This gives us:

\[ x = \sqrt[3]{6} \]

We can express the cube root of 6 in its approximate decimal form, but its exact value is simply \( \sqrt[3]{6} \).

Calculating approximately:

\[ \sqrt[3]{6} \approx 1.817 \]

Thus, the solution to the equation \( x^3 = 6 \) is:

\[ \boxed{\sqrt[3]{6}} \]

This represents the exact solution. If you need a decimal approximation, it is approximately \( 1.817 \).

Simplify the expression

3 square root of 125

To simplify the expression \( 3 \sqrt{125} \), we first simplify \( \sqrt{125} \).

Start by factoring \( 125 \):

\[ 125 = 25 \times 5 = 5^2 \times 5 \]

Now, we can take the square root:

\[ \sqrt{125} = \sqrt{25 \times 5} = \sqrt{25} \times \sqrt{5} = 5\sqrt{5} \]

Now we can substitute this back into the original expression:

\[ 3 \sqrt{125} = 3 \times 5 \sqrt{5} = 15 \sqrt{5} \]

Therefore, the simplified expression is:

\[ \boxed{15 \sqrt{5}} \]

Solve the equation given below for x.

x2 = 67/25

To solve the equation \( x^2 = \frac{67}{25} \), we will isolate \( x \) by taking the square root of both sides.

Taking the square root gives us:

\[ x = \pm \sqrt{\frac{67}{25}} \]

This can be simplified as follows:

\[ x = \pm \frac{\sqrt{67}}{\sqrt{25}} = \pm \frac{\sqrt{67}}{5} \]

Thus, the solutions to the equation \( x^2 = \frac{67}{25} \) are:

\[ x = \frac{\sqrt{67}}{5} \quad \text{and} \quad x = -\frac{\sqrt{67}}{5} \]

Therefore, the final solutions are:

\[ \boxed{\frac{\sqrt{67}}{5} \text{ and } -\frac{\sqrt{67}}{5}} \]

Simplify.

3 sqrt 64

To simplify the expression \( 3 \sqrt{64} \), we need to first simplify \( \sqrt{64} \).

Since \( 64 \) is a perfect square:

\[ \sqrt{64} = 8 \]

Now, we can substitute this back into the original expression:

\[ 3 \sqrt{64} = 3 \times 8 = 24 \]

Thus, the simplified expression is:

\[ \boxed{24} \]