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yKpoI14uk [10]
2 years ago
14

Why did Spain declare war on England?

History
1 answer:
alisha [4.7K]2 years ago
5 0

Answer: Spain did not want to be seen as directly assisting a British colony in revolting against its mother country. They feared the repercussions with their own colonies. In June 1779, Spain declared war on Britain under the terms of the treaty with France.

Explanation:

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As commander in chief the _______ often sends troops on mission around the world without declaration of war.
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As commander in chief, the President often sends troops on missions around the world without a declaration of war.

Explanation:

Article 2, Section 2, of the Executive Branch states, "The President shall be Commander in Chief of the Army and Navy of the United States, and of the Militia of the several States, when called into the actual Service of the United States.”

PLEASE MARK BRAINLIEST IF THIS HELPED! :)

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What determines the value of fiat currency?
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What was the Bush Doctrine?
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Generally speaking, the Bush Doctrine was "<span>b. a belief that the Unites States had the right to remove perceived threats," since this was a "strike before you're struck" type of foreign policy. </span>
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What does the Circular/ Economic Flow Model represent?
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this is what i looked up an got

7 0
2 years ago
(x+1/x-1 - x-1/x+1 - 4x/x^2 +1) + 4x/x^4-1​
Lunna [17]

Answer:

\frac{12x}{x^4-1}

Explanation:

Given

(\frac{x+1}{x-1} - \frac{x-1}{x+1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}

Required

Solve:

Take L.C.M of the first two fractions

(\frac{(x+1)(x+1)-(x-1)(x-1)}{(x-1)(x+1)} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}

(\frac{(x+1)(x+1)-(x-1)(x-1)}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}

(\frac{x^2+2x+1-(x^2-2x+1)}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}

(\frac{x^2+2x+1-x^2+2x-1}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}

Collect Like Terms

(\frac{x^2-x^2+2x+2x+1-1}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}

(\frac{4x}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}

Solve the expression in bracket

(\frac{4x(x^2+1) - 4x(x^2-1)}{(x^2-1)(x^2+1)} ) + \frac{4x}{x^4 - 1}

(\frac{4x(x^2+1) - 4x(x^2-1)}{x^4-1} ) + \frac{4x}{x^4 - 1}

(\frac{4x^3+4x - 4x^3+4x)}{x^4-1} ) + \frac{4x}{x^4 - 1}

(\frac{4x^3- 4x^3+4x +4x)}{x^4-1} ) + \frac{4x}{x^4 - 1}

\frac{8x}{x^4-1} + \frac{4x}{x^4 - 1}

Take LCM

\frac{8x+4x}{x^4-1}

\frac{12x}{x^4-1}

<em>The expression can not be further simplified</em>

7 0
2 years ago
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