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Vladimir [108]
4 years ago
3

The quotation above is from an 1896 speech delivered by William Jennings Bryan, in which he was protesting the adoption of a str

ict gold standard in the United States, which could severely limit the national money supply. Bryan’s reference to a “cross of gold” is best referred to as which of the following selections?
Select one:

A. symbolism

B. bias

C. literal language

D. propaganda
Social Studies
2 answers:
Kruka [31]4 years ago
7 0
Bryan's reference to a "Cross of Gold" is a symbolism of the fact that the adoption of the gold standard in the United States at that time was imposing a heavier economic burden on the majority of the population. Instead of using Free Silver as the main material in producing coinage that would increase the money supply and therefore make it more available to all, the US followed the global standard of using gold as coinage material for money supply.

algol [13]4 years ago
7 0

A symbolism i think ;))))

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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
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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
4 years ago
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