Solving for a variable 3-2(2x-3)=17

I really need help on this pls :)

Rom4ik [11]

Answer:

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3 0

3 years ago

Read 2 more answers

Trey wants to plant grass in his backyard. His backyard is in the shape of a rectangle. Its length is 31

salantis [7]

Find the total area of the yard in feet.

Area of a rectangle is length x width.

31 x 24 = 744

Divide the total area of the yard by the coverage of the seeds.

744/12 = 62

He will need 62 packs of seed.

8 0

3 years ago

Should the quotient of an integer divided by a nonzero integer always be a rational number? Why or Why not?

lubasha [3.4K]

Yes, it will always be a rational number. I'll expound on this by defining what a rational number is. It is any number that can be expressed as a fraction. Otherwise, it is called an irrational number with a non-terminating decimal expansion. So, although 1/3 has a non-terminating decimal expansion because it is equal to 0.33333333...., it is still a rational number because it can be expressed into a fraction.

8 0

3 years ago

There are 8 trailmix bags in one box. Clarissa buys 5 boxes. She gives an equal number of bags of trailmix to 4 friends. How man

JulsSmile [24]

Each friend receives 2 bags because 4 divided by 8 is 2

3 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