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mrs_skeptik [129]
3 years ago
9

“There shall be a firm and perpetual Peace between his Brittanic Majesty and the said States… all Hostilities both by Sea and La

nd shall from henceforth cease: All prisoners on both Sides shall be set at Liberty.”
4. This excerpt would be found in which historical document?
Mathematics
1 answer:
icang [17]3 years ago
4 0

Answer:

Treaty of Paris (1783

Step-by-step explanation:

Some of the treaty key points are;

The aim of the parties to "forget" previous disputes and controversies and to gain "perpetual peace and harmony"

1) The acknowledgement of the United States as a sovereign state

2) The United States boundary establishment

3) The right to fish granted to the United States fishermen

4) Debt settlement of lawful contracts on either side

5) Prisoners of war are to be released

6) Protection of property of Loyalists by the United States.

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What is the equation when y=5x+4 is reflected over the x-axis, y-axis &amp; y=x?
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Answer:

A) y=-5x-4

B) y=-5x+4

C) y=\frac{x-4}{5}

Step-by-step explanation:

So we have the equation:

y=5x+4

Let's write this in function notation. Thus:

y=f(x)=5x+4

A)

To flip a function over the x-axis, multiply the function by -1. Thus:

f(x)=5x+4\\-(f(x))=-(5x+4)

Simplify:

-f(x)=-5x-4

B) To flip a function over the y-axis, change the variable x to -x. Thus:

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Simplify:

f(-x)=-5x+4

C) A reflection over the line y=x is synonymous with finding the inverse of the function.

To find the inverse, switch x and f(x) and solve for f(x):

f(x)=5x+4

Switch:

x=5f^{-1}(x)+4

Subtract 4 from both sides:

x-4=5f^{-1}(x)

Divide both sides by 5:

f^{-1}(x)=\frac{x-4}{5}

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2 years ago
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The similar circles P and Q can be made equal by dilation and translation

  • The horizontal distance between the center of circles P and Q is 11.70 units
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<h3>The horizontal distance between their centers?</h3>

From the figure, we have the centers to be:

P = (-5,4)

Q = (6,8)

The distance is then calculated using:

d = √(x2 - x1)^2 + (y2 - y1)^2

So, we have:

d = √(6 + 5)^2 + (8 - 4)^2

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d = 11.70

Hence, the horizontal distance between the center of circles P and Q is 11.70 units

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P = 2

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