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Lostsunrise [7]
3 years ago
15

Rachel types

Mathematics
1 answer:
Art [367]3 years ago
7 0

Answer:

4 im pretty sur

Step-by-step explanation:

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What property is this<br> 16+(c+17)=(16+6)+17
Art [367]
Answer: Associative Property

This equation would be the associative property because you would have the ability to add or multiply these numbers, regardless of how they were grouped together
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If l || m what is the value of y??
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b

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As the teams fundraiser,Agnes and Betty both sold candy bars at the end of the fundraiser Agnes determined that she sold 8 more
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356

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5 0
2 years ago
How can you represent all of the solutions to systems of inequalities?​
Dimas [21]

Answer:A closed, or shaded, circle is used to represent the inequalities greater than or equal to (≥) or less than or equal to (≤) . The point is part of the solution. An open circle is used for greater than (>) or less than (<). The point is not part of the solution. Solving" systems of linear inequalities means "graphing each individual inequality, and then finding the overlaps of the various solutions". So I graph each inequality, and then find the overlapping portions of the solution regions.Step 1: Line up the equations so that the variables are lined up vertically. Step 2: Choose the easiest variable to eliminate and multiply both equations by different numbers so that the coefficients of that variable are the same. Step 3: Subtract the two equations. Step 4: Solve the one variable system.

Step-by-step explanation:

hope this helps have a great night ❤️❤️❤️

6 0
3 years ago
Solve the following recurrence relation: <br> <img src="https://tex.z-dn.net/?f=A_%7Bn%7D%3Da_%7Bn-1%7D%2Bn%3B%20a_%7B1%7D%20%3D
-Dominant- [34]

By iteratively substituting, we have

a_n = a_{n-1} + n

a_{n-1} = a_{n-2} + (n - 1) \implies a_n = a_{n-2} + n + (n - 1)

a_{n-2} = a_{n-3} + (n - 2) \implies a_n = a_{n-3} + n + (n - 1) + (n - 2)

and the pattern continues down to the first term a_1=0,

a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))

\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)

\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k

Recall the formulas

\displaystyle \sum_{n=1}^N 1 = N

\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2

It follows that

a_n = n (n - 2) - \dfrac{(n-2)(n-1)}2

\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1

\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}

4 0
2 years ago
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