Correct Answer: D. Federal Ministry of Finance

Explanation

The major categories of financial institutions include central banks, retail and commercial banks, internet banks, credit unions, savings, and loans associations, investment banks, investment companies, brokerage firms, insurance companies, and mortgage companies.

Correct Answer: B. steadfast love and knowledge of God

Correct Answer: D. prefers steadfast love to sacrifice

Explanation

\(y = \frac{x - 1}{2x + 1}, x \neq -\frac{1}{2}\)

Using the quotient rule, we have

\(\frac{\mathrm d y}{\mathrm d x} = \frac{(2x + 1). 1 - (x - 1). 2}{(2x - 1)^{2}}\)

= \(\frac{2x + 1 - 2x + 2}{(2x + 1)^{2}}\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{3}{(2x + 1)^{2}}\)

At (1, 0), \(\frac{\mathrm d y}{\mathrm d x} = \frac{3}{(2(1) + 1)^{2}} = \frac{3}{9} = \frac{1}{3}\)

Equation : \(\frac{y - 0}{x - 1} = \frac{1}{3}\)

\(y = \frac{1}{3}(x - 1)\)

Explanation

(a)(i) Hydrolysis

(ii) Dehydration

(iii) Bromination or substitution reactions.

(b) C\(_2\)H\(_4\) + H\(_2\) —› C\(_2\)H\(_6\)

Correct Answer: B. an aggregate of cells performing a similar function

Correct Answer: D. precipitation

Correct Answer: A. one hundred years

Correct Answer: C. 20mA

Explanation

Here's how to solve the problem:

1. Understand the Setup: When a galvanometer is converted into an ammeter, a shunt resistor (Rs) is connected in parallel with the galvanometer (Rg). The total current (I) of the ammeter splits between the galvanometer and the shunt.

2. Identify Given Values:

* Resistance of galvanometer (Rg) = 5 Ω

* Resistance of shunt wire (Rs) = 0.05 Ω

* Maximum current the ammeter can read (I) = 2 Amp

3. Principle of Parallel Connection: In a parallel connection, the voltage across each component is the same.

* Voltage across galvanometer (Vg) = Ig * Rg

* Voltage across shunt (Vs) = Is * Rs

* Since Vg = Vs, we have: Ig * Rg = Is * Rs

4. Current Relationship: The total current (I) is the sum of the current through the galvanometer (Ig) and the current through the shunt (Is):

* I = Ig + Is

* Therefore, Is = I - Ig

5. Substitute and Solve for Ig:

Substitute the expression for Is into the voltage equality:

Ig * Rg = (I - Ig) * Rs

Now, plug in the given numerical values:

Ig * 5 = (2 - Ig) * 0.05

Expand the right side:

5 * Ig = 2 * 0.05 - Ig * 0.05

5 * Ig = 0.1 - 0.05 * Ig

Move the Ig terms to one side:

5 * Ig + 0.05 * Ig = 0.1

(5 + 0.05) * Ig = 0.1

5.05 * Ig = 0.1

Solve for Ig:

Ig = 0.1 / 5.05 Amperes

6. Convert to Milliamperes (mA):

To convert Amperes to milliamperes, multiply by 1000:

Ig (mA) = (0.1 / 5.05) * 1000

Ig (mA) = 100 / 5.05

Ig (mA) = 10000 / 505

Simplify the fraction by dividing numerator and denominator by 5:

Ig (mA) = 2000 / 101

Now, perform the division:

Ig (mA) ≈ 19.80198... mA

7. Compare with Options:

The calculated value is approximately 19.8 mA. Looking at the options:

A. 2mA

B. 10mA

C. 20mA

D. 25mA

The closest option to 19.8 mA is 20 mA.

The final answer is 20mA.

Correct Answer: C. \(a = 9b^{3}\)

Explanation

\(\log_{3}a - 2 = 3\log_{3}b\)

Using the laws of logarithm, we know that \( 2 = 2\log_{3}3 = \log_{3}3^{2}\)

\(\therefore \log_{3}a - \log_{3}3^{2} = \log_{3}b^{3}\)

= \(\log_{3}(\frac{a}{3^{2}}) = \log_{3}b^{3} \implies \frac{a}{9} = b^{3}\)

\(\implies a = 9b^{3}\)