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Fynjy0 [20]
3 years ago
6

The values in the table below represent Function B, which is a linear function.

Mathematics
1 answer:
yan [13]3 years ago
6 0
<h3>1. Compare Functions B and L by determining which one has the greater rate of change.</h3>

The rate of change is expressed as the ratio between a change in one variable relative to a corresponding change in another. On the other hand, a linear function is given as the form:

y=mx+b

And m is the rate of change we are looking for. For Function B we have a Table and the slope can be found by choosing two points, therefore:

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ Choosing \ (1,5) \ and \ (3,11) \\ \\ m=\frac{11-5}{3-1}=3

As you can see y=6x+4 has a rate of change of 6 while the function of the table has a rate of change of 3.

<em> In conclusion,</em> y=6x+4 <em>has the greater rate of change</em>

<h3>2. which one has a greater y-intercept?</h3>

We need to get the equation that rules the points given on the table. Two-Point Form of the Equation of a Line is:

y-y_{1}=m(x-x_{1}) \\ \\ We \ know \ m=3 \\ \\ y-5=3(x-1) \therefore y-5=3x-3 \therefore y=3x+2

As you can see y=6x+4 has a y-intercept of b=4 while the function of the table has a y-intercept of b=2

<em> In conclusion,</em> y=6x+4 <em>also has the greater y-intercept</em>

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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:

<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 />

<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

8 0
3 years ago
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