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3 years ago

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Who proposed that Missouri be allowed to enter the union as a slave state if no more slaves were brought into Missouri after tha

t?
A.)John C. Calhoun
B.)Henry Clay
C.)James Tallmadge
D.)Thomas Jefferson

2 answers:

tia_tia [17]3 years ago

7 0

Answer:

B: henry clay

Explanation:

Sliva [168]3 years ago

4 0

The answer is Henry Clay

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1 year ago

What are some ways that the French and Indian War may have started Colonists thinking about Independence?

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Answer:

Explanation:

The French and Indian War was the North American conflict in a larger imperial war between Great Britain and France known as the Seven Years’ War. The French and Indian War began in 1754 and ended with the Treaty of Paris in 1763. The war provided Great Britain enormous territorial gains in North America, but disputes over subsequent frontier policy and paying the war’s expenses led to colonial discontent, and ultimately to the American Revolution.

Map from the French and Indian War

The French and Indian War resulted from ongoing frontier tensions in North America as both French and British imperial officials and colonists sought to extend each country’s sphere of influence in frontier regions. In North America, the war pitted France, French colonists, and their Native allies against Great Britain, the Anglo-American colonists, and the Iroquois Confederacy, which controlled most of upstate New York and parts of northern Pennsylvania. In 1753, prior to the outbreak of hostilities, Great Britain controlled the 13 colonies up to the Appalachian Mountains, but beyond lay New France, a very large, sparsely settled colony that stretched from Louisiana through the Mississippi Valley and Great Lakes to Canada. (See Incidents Leading up to the French and Indian War and Albany Plan)

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3 years ago

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

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3 years ago