4. The quantities $$ A, B, C $$, and $$ D $$ all depend on time, and are related by the formula $$ A B=\log \left(C^{a}+D^{2}+1\right) $$ At time $$ t=10 $$, the following values are known: $$ A=0 $$ units and $$ A^{\prime}=2 $$ units $$ / \mathrm{sec} $$. What is $$ B $$ at $$ t=10 $$ ?

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To find $$ B $$ at $$ t=10 $$, we start with the given relationship:

$$
A B = \log \left(C^{a} + D^{2} + 1\right)
$$

We know that at $$ t=10 $$, $$ A = 0 $$ and $$ A' = 2 $$ units/sec.

1. **Substituting the known value of $$ A $$**:
   Since $$ A = 0 $$, we can substitute this into the equation:

$$
   0 \cdot B = \log \left(C^{a} + D^{2} + 1\right)
   $$

This simplifies to:

$$
   0 = \log \left(C^{a} + D^{2} + 1\right)
   $$

2. **Solving the logarithmic equation**:
   The equation $$ 0 = \log \left(C^{a} + D^{2} + 1\right) $$ implies that:

$$
   C^{a} + D^{2} + 1 = 1
   $$

Therefore:

$$
   C^{a} + D^{2} = 0
   $$

Since $$ C^{a} $$ and $$ D^{2} $$ are both non-negative (as they are raised to even powers), the only solution is:

$$
   C^{a} = 0 \quad \text{and} \quad D^{2} = 0
   $$

This means:

$$
   C = 0 \quad \text{and} \quad D = 0
   $$

3. **Finding $$ B $$**:
   Now, substituting back into the original equation, we have:

$$
   0 \cdot B = \log(0 + 0 + 1) = \log(1) = 0
   $$

This does not provide any information about $$ B $$ since $$ 0 \cdot B = 0 $$ is true for any value of $$ B $$.

Thus, we conclude that $$ B $$ can be any value at $$ t=10 $$ since the equation does not constrain it. Therefore, $$ B $$ is indeterminate based on the given information.
At $$ t=10 $$, $$ B $$ can be any value because the equation does not provide enough information to determine its specific value.

The quantities , and all depend on time, and are related by the formula

At time , the following values are known: units and units . What is at
?

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The quantities , and all depend on time, and are related by the formula At time , the following values are known: units and units . What is at ?