Which option best describes your approach to taking notes as you read.

1 answer:

8 0

The best option that describes one's approach to taking notes as one reads is: <em>you take good notes that helps you to recall </em><em>important information.</em>

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<h3>Taking Notes When Reading</h3>

When reading a book, textbook, or any other book, a technique for easily recalling to mind important details after reading is taking notes.
Taking of notes helps you to remember important information.

In summary, the best option that describes one's approach to taking notes as one reads is: <em>you take good notes that helps you to recall </em><em>important information.</em>

<em />

Learn more about taking notes on:

brainly.com/question/785675

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If the number of giraffes at congtin were reduced by 63.

zhenek [66]

Based on the information given, it can be noted that the computation shows that the ratio will be 3:7.

From the complete question, the data given was:

Giraffes = 124
Other = 108
New giraffe = 124 - (63% × 124) = 45

Therefore, the ratio will then be:

= 44:88 = 3:7

In conclusion, the ratio is 3:7.

Learn more about ratio on:

brainly.com/question/2328454

6 0

3 years ago

A point charge q1 = -4. 00 nc is at the point x = 0. 60 m, y = 0. 80 m , and a second point charge q2 = +6. 00 nc is at the poin

Alekssandra [29.7K]

The net electric field is the vector sum of the components of the electric

field produced by the two charges.

The values of the magnitude and direction of the net electric field at the origin (approximate values) are;

131.6 N/C

12.6 ° above the negative x–axis

<h3>How are the net electric field magnitude and direction calculated?</h3>

The possible questions based on a similar question posted online are;

(a) The net electric field at the origin.

The electric field due to charge q₁ is given as follows;

$\vec E_{1x} = \mathbf{ \dfrac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \dfrac{q_1}{\vec{r}^2_2}}$

Which gives;

$\vec{E}_{1x} =\mathbf{ \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot cos\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) =-21.6 \, N/C$

$\vec{E}_{1y} = \mathbf{\dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot sin\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) = 28.8 \, N/C$

Which gives;

$\vec{E}_1 = \mathbf{21.6 \, N/C \cdot \hat x + 28.8 \, N/C \hat y}$

$\vec{E}_{2x} = \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(6.00 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2} = 150 \, N/C$