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3 years ago

What is the solution to the linear equation 4b + 6 = 2 - b+ 4

Mathematics

1 answer:

inessss [21]3 years ago

5 0

Final Answer: $b = 0$

Steps/Reasons/Explanation:

Question: What is the solution to the linear equation $4b + 6 = 2 - b + 4$ ?

Step 1: Simplify $2 - b + 4$ to $6 - b$ .

$4b + 6 = 6 - b$

Step 2: Cancel $6$ on both sides.

$4b = -b$

Step 3: Subtract $-b$ from both sides.

$4b + b = 0$

Step 4: Simplify $4b + b$ to $5b$ .

$5b = 0$

Step 5: Divide both sides by $5$ .

$b = 0$

~I hope I helped you :)~

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Two lamps marked 100 W - 110 V and 100 W - 220 V are connected i

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<h2>Answer:</h2>

The power consumed in the lamp marked 100W - 110V is 15.68W

The power consumed in the lamp marked 100W - 220V is 62.73W

<h2>Step-by-step explanation:</h2>

Given:

First lamp rating

Power (P) = 100W

Voltage (V) = 110V

Second lamp rating

Power (P) = 100W

Voltage (V) = 220V

Source

Voltage = 220V

i. Get the resistance of each lamp.

Remember that power (P) of each of the lamps is given by the quotient of the square of their voltage ratings (V) and their resistances (R). i.e

P = $\frac{V^2}{R}$

Make R subject of the formula

⇒ R = $\frac{V^2}{P}$ ------------------(i)

For first lamp, let the resistance be R₁. Now substitute R = R₁, V = 110V and P = 100W into equation (i)

R₁ = $\frac{110^2}{100}$

R₁ = 121Ω

For second lamp, let the resistance be R₂. Now substitute R = R₂, V = 220V and P = 100W into equation (i)

R₂ = $\frac{220^2}{100}$

R₂ = 484Ω

ii. Get the equivalent resistance of the resistances of the lamps.

Since the lamps are connected in series, their equivalent resistance (R) is the sum of their individual resistances. i.e

R = R₁ + R₂

R = 121 + 484

R = 605Ω

iii. Get the current flowing through each of the lamps.

Since the lamps are connected in series, then the same current flows through them. This current (I) is produced by the source voltage (V = 220V) of the line and their equivalent resistance (R = 605Ω). i.e

V = IR [From Ohm's law]

I = $\frac{V}{R}$

I = $\frac{220}{605}$

I = 0.36A

iv. Get the power consumed by each lamp.

From Ohm's law, the power consumed is given by;

P = I²R

Where;

I = current flowing through the lamp

R = resistance of the lamp.

For the first lamp, power consumed is given by;

P = I²R [Where I = 0.36 and R = 121Ω]

P = (0.36)² x 121

P = 15.68W

For the second lamp, power consumed is given by;

P = I²R [Where I = 0.36 and R = 484Ω]

P = (0.36)² x 484

P = 62.73W

Therefore;

The power consumed in the lamp marked 100W - 110V is 15.68W

The power consumed in the lamp marked 100W - 220V is 62.73W

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