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Arturiano [62]
2 years ago
13

In your own view, how can social media be used in teaching and learning in higher education?

Social Studies
2 answers:
kumpel [21]2 years ago
5 0

Answer:

Social media promotes self-directed learning, which prepares students to search for answers and make decisions independently. When reinforced in a classroom setting, these social media skills can be guided and refined to produce better learning outcomes and critical awareness.

Explanation:

ycow [4]2 years ago
5 0
If used with intention, social media can positively influence the way each individual learns and absorbs information in the classroom. Incorporating social media into a more traditional learning environment can expand students' creative freedom and encourage them to work harder and engage more.
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Features of liberalism theory​
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Answer:

Liberals espouse a wide array of views depending on their understanding of these principles, but they generally support free markets, free trade, limited government, individual rights (including civil rights and human rights), capitalism, democracy, secularism, gender equality, racial equality, internationalism.

Explanation:

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To ensure accurate measurement of cardiovascular activity, researchers typically:________
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For an accurate measurement of cardiovascular activity, researchers typically: D) measure the number of heartbeats per minute.

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Why did the professional dog walker go out of business math worksheet answers?
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Why Did the Professional Dog Walker ao Out of business, math worksheet Answers:

Q1. sin27°   = x/8

Solution:

We have to solve for x, therefore, we will rearrange the given equation for x.

We get,
x = 8 × sin27°

Using the calculator,

sin27° = 0.45

Now substitute the value of sin27° into the main equation.

we get,
x = 8 × 0.45
x = 3.63 (rounded to the nearest hundredth)


Q2. tan 18°  = n / 75

Solution:
We have to solve for n, therefore, we will rearrange the given equation for n.
We get,
n = 75 × tan 18°
Using the calculator,
tan 18° = 0.32
Now substitute the value of tan 18° into the main equation.
we get,
x = 75 × 0.32
x = 24.37 (rounded to the nearest hundredth)

Q3. sin40°  = 4 / a

Solution: We have to solve for a, therefore, we will rearrange the given equation for a.
We get,
a = 4 ÷ sin40°
Using the calculator,
sin40° = 0.64
Now substitute the value of sin40° into the main equation.
we get,
a = 4 ÷ 0.64
a = 6.25 (rounded to the nearest hundredth)

Q4. cos5°   = 92 / y

Solution: We have to solve for y, therefore, we will rearrange the given equation for y.
We get,
y = 92 ÷ cos5°
Using the calculator,
Cos5° = 0.99
Now substitute the value of cos5° into the main equation.
we get,
y = 92 ÷ 0.99
y = 92.92 (rounded to the nearest hundredth)

Q5:
Given the shape attached, therefore, using the triangle given, we have:
Angle of elevation = 35°
length of Opposite side to the angle = x
Length of Hypoteneus = 12
Calculations:
Using the SOH CAH TOA rules:
SOH stands for SineФ = Opposite ÷ Hypotenuse.

CAH stands for CosineФ = Adjacent ÷ Hypotenuse.

TOA stands for TangentФ = Opposite ÷ Adjacent.

Hence,

               SineФ = Opposite ÷ Hypotenuse

Substituting the values:

               Sine35° = x ÷ 12

               0.5735  = x ÷ 12

                          x = 0.5735 × 12

                          x = 6.88 (rounded to the nearest hundredth)

Q6: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 54°
length of the adjacent side to the angle = x
Length of Hypoteneus = 30
Calculations:
Using the SOH CAH TOA rules:

Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Cos54° = x ÷ 30

               0.5877  = x ÷ 30

                          x = 0.5877 × 30

                          x = 17.63 (rounded to the nearest hundredth)

Q7: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 22°
length of the adjacent side to the angle = 85
length of the opposite side to the angle = x
Calculations:

Using the SOH CAH TOA rules:

Hence,

               TangentФ = Opposite ÷ Adjacent

Substituting the values:

               tan22° = x ÷ 85

              0.4040 = x ÷ 85

                          x = 0.4040 × 85

                          x = 34.34 (rounded to the nearest hundredth)

Q8: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 16°
length of the opposite side to the angle = x
Length of Hypoteneus = 14
Calculations:
Using the SOH CAH TOA rules:
Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Sine16° = x ÷ 14

               0.2756  = x ÷ 14

                          x = 0.2756 × 14

                          x = 3.86 (rounded to the nearest hundredth)

Q9: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 65°
length of the adjacent side to the angle = 9
length of the opposite side to the angle = x
Calculations:
Using the SOH CAH TOA rules:
Hence,

               TangentФ = Opposite ÷ Adjacent

Substituting the values:

               tan65° = x ÷ 9

               2.1445 = x ÷ 9

                          x = 2.1445 × 9

                          x = 19.30 (rounded to the nearest hundredth)

Q10: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 51°
length of the adjacent side to the angle = x
Length of Hypoteneus = 70
Calculations:
Using the SOH CAH TOA rules:
Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Cos51° = x ÷ 70

              0.6293  = x ÷ 70

                          x = 0.6293 × 70

                          x = 44.05 (rounded to the nearest hundredth)

Q11: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 36°
length of the opposite side to the angle = 15
Length of Hypoteneus = x
Calculations:
Using the SOH CAH TOA rules:
Hence,

               CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

               Sine36° = 15 ÷ x

               0.5877  = 15 ÷ x

                          x = 15 ÷ 0.5877

                          x = 25.52 (rounded to the nearest hundredth)

Q12: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 60°
length of the adjacent side to the angle = x
length of the opposite side to the angle = 100

Calculations:
Using the SOH CAH TOA rules:

Hence,

               TangentФ = Opposite ÷ Adjacent

Substituting the values:

               tan65° = 100 ÷ x

               2.1445 = 100 ÷ x

                          x = 100 ÷ 2.1445

                          x = 46.63 (rounded to the nearest hundredth)

Q13: When a 25-ft ladder is leaned against a wall, it makes a 72° with the ground. How high up on wall does the ladder reach?

Solution: Given the shape attached, therefore, using the triangle given, we have:

The angle of elevation from the ground = 72°
length of the wall opposite to the angle = X
Length of ladder (Hypoteneus) = 25 feet

Calculations:
Using the SOH CAH TOA rules:
Hence,

               SineФ = Opposite ÷ Hypotenuse

Substituting the values:

               Sine72° = x ÷ 25

               0.9510  = x ÷ 25

                          x = 25 ÷ 0.9510

                          x = 23.77 (rounded to the nearest hundredth)


ANSWERS TO QUESTION 14 AND  15 ARE ATTACHED

7 0
3 years ago
A formal statement of the relation among the variables or factors of interest in a particular research area is a(n)
alexdok [17]

Answer:

It's a Theory.

Explanation:

8 0
2 years ago
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