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3 years ago

HELP PLEASE WILL GIVE BRAINLIEST

Mathematics

2 answers:

abruzzese [7]3 years ago

7 0

Answer:
$1.21

Explanation:
What we know:
- $5.88=3 pounds of apples
- price went up by $0.75 per pound

First, we divide $5.88 by 3 to find out how much one pound of apples cost.

5.88/3
=1.96

Therefore, the new price of apples per pound is $1.96.

Because the old price of apples per pound was $0.75 cheaper, we subtract 0.75 from 1.96.

1.96-0.75
=1.21

Therefore, the old price of apples per pound was $1.21.

I hope this helps! Please comment if you have any questions.

Fynjy0 [20]3 years ago

5 0

$1.21 hope this helps

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Answer:

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Step-by-step explanation:

Andre calculated his variable and total expenses for last month.

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Answer:

<h2>The option B and D are the equations.</h2>

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At a conference, there are 121 math teachers and 11 science teachers. Write the

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3 years ago

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:

<em>First lamp rating</em>

Power (P) = 100W

Voltage (V) = 110V

<em>Second lamp rating</em>

Power (P) = 100W

Voltage (V) = 220V

<em>Source</em>

Voltage = 220V

i. <u>Get the resistance of each lamp</u>.

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

<em>Make R subject of the formula</em>

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

<em />

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

R₁ = $\frac{110^2}{100}$

R₁ = 121Ω

<em />

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

R₂ = $\frac{220^2}{100}$

R₂ = 484Ω

<em />

ii.<u> Get the equivalent resistance of the resistances of the lamps.</u>

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. <u>Get the current flowing through each of the lamps. </u>

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. <u>Get the power consumed by each lamp.</u>

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

P = I²R

Where;

I = current flowing through the lamp

R = resistance of the lamp.

<em>For the first lamp, power consumed is given by;</em>

P = I²R [Where I = 0.36 and R = 121Ω]

P = (0.36)² x 121

P = 15.68W

<em>For the second lamp, power consumed is given by;</em>

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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3 years ago