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2 years ago

The peak current passing through an inductor is 2. 0 aa. part a what is the peak current if the peak emf e0e0 is doubled?

Mathematics

1 answer:

kumpel [21]2 years ago

3 0

Answer:

I remember this question in school

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For every 1 litre of water used to make a medicine, 269ml of sucrose and 47ml of saline solution are used. Express the amount of

leonid [27]

Answer:

1000 : 269 : 47

Step-by-step explanation:

A ratio is a way of comparing the value or amount of one quantity to another.

We are to find the ration of water to sucrose to saline solution, hence, it must remain in that order.

Water : Sucrose : Saline solution

The amount of water is 1 litre (1 * 1000 ml = 1000 ml)

The amount of sucrose is 269 ml

The amount of saline solution is 47 ml

Therefore, the ratio of water to sucrose to saline solution is:

W : S : SS = 1000 : 269 : 47

Since 1000, 269 and 47 have no common factors, that is the simplest form of the ratio.

3 0

3 years ago

Lim x-1 x2 - 1/ sin(x-2)

balu736 [363]

Answer:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0$

Explanation:

Assuming the correct expression is to find the following limit:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}$

Use the property the limit of the quotient is the quotient of the limits:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}$

Evaluate the numerator:

$\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}$

Evaluate the denominator:

Since $\lim_{x \to1}sin(x-2)\neq 0$

$\frac{0}{\lim_{x \to1}sin(x-2)}=0$

4 0

3 years ago

Solve using long division <br> Please

madreJ [45]

1. $Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}$

$Steps:$

$\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}$

$\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}$

$\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}$

$\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}$

$\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}$

2. $Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}$

$Steps:$

$\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}$

$\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}$

$\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

4 0

3 years ago

Simplify the expression. (3.5x - 5) + (7.2x + 10)

Andru [333]

Answer: 10.7x + 5

Step-by-step explanation:

(3.5 + 7.2)x + (-5 + 10) = 10.7x + 5

8 0

3 years ago

How can you tell if a graph is proportional

marissa [1.9K]

Answer:

The fractions you make from the coordinates on the graph is equal. If it is then the graph is proportional

Hope this helped

5 0

3 years ago

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