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photoshop1234 [79]
2 years ago
8

Calculate the potential difference across a 40 ohm resistor when a current of 5A flows through it​

Mathematics
1 answer:
MrRissso [65]2 years ago
4 0

Answer:

I don't know if this is what your looking for but if you had 40 ohms and 5 amps, you would get 200 volts / and 1000 watts.

Hope that helps :)

Step-by-step explanation:

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How much money will you make if you make 11.35 for 10 hours 4 days a week?
TEA [102]
Correct me if im wrong, if you multiply 11.35 by 10 you get $113.5. multiply by 4 and that equals to $454. 

-hope this helps
4 0
3 years ago
The scale factor of the blueprint of a garage door to the actual garage door is 0.05. the are of the actual garage door is 220 f
erastovalidia [21]
The scale factor 0.05 refers to length, so for area the scale factor is 0.05²=0.0025.

We then multiply: 220 ft²*0.0025=0.55 ft² (choice A)
8 0
3 years ago
Read 2 more answers
Which is equivalent?
Arlecino [84]

Steps to simplify:

(2x + 3)(x - 7)

~Use FOIL to multiply

(2x * x) + (2 * -7) + (3 * x) + (3 * -7)

~Simplify

=2x² - 14x + 3x - 21

~Combine Like Terms

2x² + (-14x + 3x) - 21

~Simplify

2x² - 11x - 21 (Option 2)

Best of Luck!

6 0
3 years ago
Can someone help ASAP please? X
Tju [1.3M]
I think the answer 5 over 2
6 0
3 years ago
An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click
Sergeeva-Olga [200]

Answer:

A.    \mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

B.    \mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

Step-by-step explanation:

Given that:

An investment of  Amount = $8000

earns  at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A)  Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

A = Pe^{rt}

where;

A = (8000) \ e ^{0.7t}

The instantaneous rate of change = \dfrac{dA}{dt}

\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )

= 8000 \dfrac{d}{dt}e^{0.07 \ t}

\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 2 years; the instantaneous rate of change is:

\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}

\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

12000 = (8000)e^{0.07 \ t}

\dfrac{12000 }{8000}= e^{0.07 \ t}

1.5= e^{0.07 \ t}

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 5.79

\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}

\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

7 0
3 years ago
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