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3 years ago

Math help please!! Ty

Mathematics

1 answer:

Ronch [10]3 years ago

8 0

The answer would be the third one.
twice the number of c would be 2c
four more than 2c is 2c +4

Hope I helped!!

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Jolene wants to invest $2000 into a money market account. Interest is to be 6.5% compounded annually, and she wants to take the

timama [110]

Answer:

Interest = 650$

Step-by-step explanation:

Given:

Invest Inv = 2000$, interest rate r = 6.5% compounded annually

and time n = 5 years

Formula for calculation is:

Int = (Inv · r · n) / 100

Int = (2000 · 6.5 · 5) / 100 = 650$

Int = 650$

God with you!!!

3 0

3 years ago

Idek the answer please help

VashaNatasha [74]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Please help asap<br> What is the least common denominator of the fractions?

Reil [10]

The answer is B. 6m^3n^5. Hope I actually helped!

6 0

3 years ago

What is an equation in point-slope form for the line that passes through the points (−3,5) and (2,−3)?

fredd [130]

(-3,5). (2,-3)
x1,y1. x2,y2

y2-y1. -3-5. -8
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x2-x1. 2-3. -1

plug the rest in

4 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;

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

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

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

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