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AlladinOne [14]
2 years ago
9

Plz help me fast urgent thank you :) 20 points

Mathematics
2 answers:
natima [27]2 years ago
6 0

Answer:

number 3 dual federalism

NeTakaya2 years ago
5 0

Answer:

I believe it is A. Cooperative Federalism

Step-by-step explanation:

The other ones don't seem applicable to the diagram

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How would you convince a fellow student that the number 0.57 is a rational number? (5+ sentences please) Please help me with thi
Gre4nikov [31]

Answer:  0.57 = 57/100 this means it has to be rational.

Explanation: You have to explain to them that any number that can be written in fraction form is a rational number. This includes integers, terminating decimals, and repeating decimals as well as fractions. An integer can be written as a fraction simply by giving it a denominator of one, so any integer is a rational number.

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<span>D. Zahra can make more than one triangle with these angle measures.
(Infinite similar triangles)</span>

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Determine the area of the coin to the nearest tenth.
Snowcat [4.5K]

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Step-by-step explanation:

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The formula κ​(x)= f′′(x) 1+f′(x)23/2 expresses the curvature of a​ twice-differentiable plane curve as a function of x. Use thi
Katyanochek1 [597]

Answer:

K(x) =  \frac{-10}{[1 + (-10x)^2]^{\frac{3}{2} } }    ( curvature function)

Step-by-step explanation:

considering the Given function

F(x) = -5x^2

first Determine the value of F'(x)

F'(x) = \frac{d(-5x^2)}{dy}

F'(x) = -10x

next we Determine the value of F"(x)

F"(x) = \frac{d(-10x)}{dy}

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To find the curvature function we have to insert the values above into the given formula

K(x)  = \frac{|f"(x)|}{[1 +( f'(x)^2)]^{\frac{3}{2} } }

 K(x) =  \frac{-10}{[1 + (-10x)^2]^{\frac{3}{2} } }    ( curvature function)

       

6 0
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75

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o find the area of a triangle, multiply the base by the height, and then divide by 2. The division by 2 comes from the fact that a parallelogram can be divided into 2 triangles. For example, in the diagram to the left, the area of each triangle is equal to one-half the area of the parallelogram.

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