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lana [24]
3 years ago
5

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:

A.

Step-by-step explanation:

4 0
2 years ago
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Please help asap<br> What is the least common denominator of the fractions?
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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

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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
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