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3 years ago

Match with correct description

Mathematics

1 answer:

fgiga [73]3 years ago

8 0

First one: If the leading coeff. is negative, the graph begins in Quadrant III and ends in Quadrant IV. It's an even function. The fourth graph represents it.

If the leading coeff. is positive, but everything else remains the same, the graph opens upward, beginning in Q II and ending in Q I.

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Step-by-step explanation:

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Which expression is equivalent to *picture attached*

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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