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Vitek1552 [10]
1 year ago
5

i need ideas on a poster. the topic is being a tactile learner (hands on learner). how should it look. what should go on it. im

using a website called canva so i,dk if that helps but i really need help
Advanced Placement (AP)
1 answer:
Fittoniya [83]1 year ago
4 0

If the purpose of the poster is to explain what a tactile learner is, use a picture of students doing an activity using their hands and an explanatory text.

<h3>Ideas for a poster about tactile learners</h3>

Tactile learners, also called kinesthetic learners, are those who need to touch or move in order to better learn new information. To help this type of learners to learn better in class, it is important to include, for example:

  • Activities that require them to stand up and move around.
  • Activities that require them to use their hands.
  • Activities that involve touching different objects.

Having that in mind, we can come up with some ideas for a poster with the purpose of explaining what a tactile learner is:

  • Find and a picture of students who are doing an activity using their hands, such as sculpting or drawing.
  • Add the title "Tactile learners".
  • Below the title, in smaller letters, provide a brief text describing what a tactile learner is. For example, "Learn better by touching or moving around."

With the information above in mind, you can create a poster about tactile learners.

Learn more about tactile or kinesthetic learners here:

brainly.com/question/3274282

#SPJ1

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The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded
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The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

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∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx

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A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

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A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

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A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526

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The sum of the area of the shaded regions, ∑A = A_{1 shaded} + A_{2 shaded}

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

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