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3 years ago

Brainliest is waiting to be given to the correct answer!

Mathematics

1 answer:

Andrew [12]3 years ago

6 0

Answer:

Step-by-step explanation:

Let cost of member be "m" and cost of non-member be "n".

We know:

3 member and 4 non-member pay $79.

So we can say:

$3m+4n=79$

Also, from the problem, we know that:

non-member fee is 11 more than member fee

Thus, we can write:

n = 11 + m

Putting this in 1st equation, we can solve for "m":

$3m+4n=79\\3m+4(11+m)=79\\3m+44+4m=79\\7m+44=79\\7m=35\\m=5$

Members have to pay $5, so non-members will pay "11" more, that will be:

Non-Members = 11 + 5 = 16

<u>$16, D is the correct answer</u>

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Step-by-step explanation:

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3 years ago

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2 years ago

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2 years ago

What is 2.20 rounded to the nearest tenth?

Musya8 [376]

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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;

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

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

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

Therefore, we have;

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

4 0

3 years ago