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Mandarinka [93]
2 years ago
10

How do you solve the inequality 12 < x + 5 so that the variable "x" is on the left side?

Mathematics
2 answers:
Eduardwww [97]2 years ago
7 0
The like terms are 12 and 5, so you have to subtract 5 from both sides.
12 < x + 5
-5. -5
7 < x
We can just switch it around so x is in the left side.
x > 7
Sophie [7]2 years ago
5 0
12<x+5
=12-5<x
=7<x
we can just switch it and it becomes x>7
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igomit [66]

Answer:

x = - 3 ± \sqrt{10}

Step-by-step explanation:

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DiKsa [7]

Answer:

The correct option is;

4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3  \left (\dfrac{50(51) }{2} \right )

Step-by-step explanation:

The given expression is presented as follows;

\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3  \right )

Which can be expanded into the following form;

\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3  \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left  n^2 + 3  \times\sum\limits _{n = 1}^{50}  n

From which we have;

\sum\limits _{k = 1}^{n} \left  k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}

\sum\limits _{k = 1}^{n} \left  k = \dfrac{n \times (n+1) }{2}

Therefore, substituting the value of n = 50 we have;

\sum\limits _{n = 1}^{50} \left  k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}

\sum\limits _{k = 1}^{50} \left  k = \dfrac{50 \times (50+1) }{2}

Which gives;

4 \times \sum\limits _{n = 1}^{50} \left  n^2 =  4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}

3  \times\sum\limits _{n = 1}^{50}  n = 3  \times \dfrac{n \times (n+1) }{2} = 3  \times \dfrac{50 \times (51) }{2}

\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3  \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3  \times \dfrac{50 \times (51) }{2}

Therefore, we have;

4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3  \left (\dfrac{50(51) }{2} \right ).

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Answer:

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Step-by-step explanation:

2 x 2 = 4, then just plop the x and y down, and there's your answer

8 0
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