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tatiyna
3 years ago
12

You have a nightlight plugged into an outlet in the hallway, which uses 3.5 watts when plugged in. If the house circuit provides

120.0 volts, what is the current through this bulb?
Mathematics
1 answer:
Bess [88]3 years ago
5 0
This problem can be solved by manipulating the formula for power, which is: P = VI. Where:

P = Power in watts
V= Voltage in volts
I = Current in amperes

Since we are already given the power and voltage, simple substitution can be done.

P = VI
3.5 watts = 120 volts * I
I = 3.5 / 120
I = 0.0292 Amperes or 29.2 milliAmperes

Therefore, the current through the bulb is 29.2mA
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4:2

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What value of B makes the expression. x² + Bx + 81 a perfect square trinomial?
Oxana [17]

Answer:

B.) 18

Step-by-step explanation:

To find the correct b value we use this formula

b² - 4ac = 0

b = What we are solving for

a = 1

c = 81

b² - 4(1)(81) = 0

b² - 324 = 0

b² = 324

b = \sqrt{324}

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And 18 is one of the answer choices

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4 0
3 years ago
Read 2 more answers
F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?
SVEN [57.7K]

Answer: -4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)

Step-by-step explanation:

Given

Function is F(t)=t^2+19+60

Zeroes of the function are

t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15

Using completing the square method

y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)

4 0
3 years ago
2. The original cost of a clock is $60 but you
Cerrena [4.2K]

Answer:

The tax will be 3.60

Step-by-step explanation:

60 x 6% (0.06)

5 0
2 years ago
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