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siniylev [52]
3 years ago
13

-6x+3y=9 -8x+4y=12 linear equation

Mathematics
1 answer:
vredina [299]3 years ago
5 0


System of Linear Equations entered :

[1] 4x + 3y = 9
[2] 3x + 4y = 12
Graphic Representation of the Equations :

3y + 4x = 9 4y + 3x = 12



Solve by Substitution :

// Solve equation [2] for the variable y


[2] 4y = -3x + 12

[2] y = -3x/4 + 3
// Plug this in for variable y in equation [1]

[1] 4x + 3•(-3x/4+3) = 9
[1] 7x/4 = 0
[1] 7x = 0
// Solve equation [1] for the variable x


[1] 7x = 0

[1] x = 0
// By now we know this much :

x = 0
y = -3x/4+3
// Use the x value to solve for y

y = -(3/4)(-0/17603848)+3 = 3
Solution :

{x,y} = {0/17603848,3}
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Answer:

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Step-by-step explanation:

1) Incomplete question. So completing the several terms:\left \{a_{n}\right \}_{n=1}^{\infty}=\left \{ 1,\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},... \right \}

We can realize this a Geometric sequence, with the ratio equal to:

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A) To find the next two terms of this sequence, simply follow multiplying the 5th term by the ratio (q):

\frac{1}{256}*\mathbf{\frac{1}{4}}=\frac{1}{1024}\\\\\frac{1}{1024}*\mathbf{\frac{1}{4}}=\frac{1}{4096}\\\\\left \{ 1,\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},\mathbf{\frac{1}{1024},\frac{1}{4096}}\right \}

B) To find a recurrence a relation, is to write it a function based on the last value. So that, the function relates to the last value.

\left\{\begin{matrix}a(1)=1 & \\ a(n)=a(n-1)*\frac{1}{4} &\:for\:n=1,2,3,4,... \end{matrix}\right.

C) The explicit formula, is one valid for any value since we have the first one to find any term of the Geometric Sequence, therefore:

\\a_{n}=nq^{n-1} \:for\:n=1,2,3,4,...

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Kirk wrapped 7 presents with those 23.8 cm of tape.

Step-by-step explanation:

Figure out the unit rate:

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Multiplying this by 23.8 cm results in the number of presents Kirk wrapped with that much tape:

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