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3 years ago

Find the slope of the line.

Mathematics

2 answers:

Ainat [17]3 years ago

7 0

5/6 would be the slope of the line

rewona [7]3 years ago

3 0

Answer:

5/6

Step-by-step explanation:

Remember your rise over run!

2.5/3 = 5/6

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It has rained 3/5 inch in 1/4 hour. If it continues to rain at the same rate,how much rain would fall in an hour

Scorpion4ik [409]

Answer:

2 2/5 inches

Step-by-step explanation:

3/5 x 4

3/5 x 4/1

3/5 x 4/1 = 12/5

12/5 = 2 2/5

5 0

3 years ago

How do I simplify the n^-6*n^3 topic "Negative and Zero Exponents"

Alchen [17]

Since you’re multiplying n^-6 and n^3, you can add the exponents: -6+3 = -3

n^-6 * n^3 = n^-3

If you need to finish without having a negative exponent in your answer, then remember a negative exponent means that factor is on the wrong side of the fraction.

n^-3 = n^-3 / 1 = 1/n^3

when the factor moves to the other side of the fraction, the side of the exponent changes.

7 0

3 years ago

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;

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

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

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