In each of the these question two equation (I) and (II) are given. You have to solve both the equtions and give answer accordingly
I `2x^3=sqrt(256)`
II `2y^2-9y+10=0`

Options

A. `xgty`
B. `xgey`
C. `xley`
D. x=y

Answer

To solve the given equations step by step, we will start with each equation separately.

### Step 1: Solve Equation I
The first equation is:
\[ 2x^3 = \sqrt{256} \]

**Step 1.1:** Calculate \(\sqrt{256}\).
\[ \sqrt{256} = 16 \]

**Step 1.2:** Substitute this value back into the equation.
\[ 2x^3 = 16 \]

**Step 1.3:** Divide both sides by 2 to isolate \(x^3\).
\[ x^3 = \frac{16}{2} = 8 \]

**Step 1.4:** Take the cube root of both sides to solve for \(x\).
\[ x = \sqrt[3]{8} = 2 \]

### Step 2: Solve Equation II
The second equation is:
\[ 2y^2 - 9y + 10 = 0 \]

**Step 2.1:** Factor the quadratic equation.
We need to find two numbers that multiply to \(2 \times 10 = 20\) and sum to \(-9\). The numbers are \(-4\) and \(-5\).

**Step 2.2:** Rewrite the equation using these numbers.
\[ 2y^2 - 4y - 5y + 10 = 0 \]

**Step 2.3:** Group the terms.
\[ (2y^2 - 4y) + (-5y + 10) = 0 \]

**Step 2.4:** Factor by grouping.
\[ 2y(y - 2) - 5(y - 2) = 0 \]
\[ (y - 2)(2y - 5) = 0 \]

**Step 2.5:** Set each factor to zero.
1. \(y - 2 = 0 \Rightarrow y = 2\)
2. \(2y - 5 = 0 \Rightarrow y = \frac{5}{2} = 2.5\)

### Step 3: Summary of Solutions
From the solutions obtained:
- \(x = 2\)
- \(y = 2\) or \(y = 2.5\)

### Step 4: Compare Values
Now we compare the values of \(x\) and \(y\):
- When \(y = 2\), \(x = 2\) (equal)
- When \(y = 2.5\), \(x = 2\) (less than)

Thus, we conclude that:
\[ x \leq y \]

### Final Answer
The correct option is that \(x\) is less than or equal to \(y\).

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