The summation of the considered expression in terms of n from n = 1 to 14 is given by: Option D: 343
<h3>How to find the sum of consecutive integers?</h3>
<h3>
What are the properties of summation?</h3>
![\sum_{i=r}^s (a \times f(i) + b) = a \times [\: \sum_{i=r}^s f(i)] + (s-r)b](https://tex.z-dn.net/?f=%5Csum_%7Bi%3Dr%7D%5Es%20%20%28a%20%5Ctimes%20f%28i%29%20%2B%20b%29%20%3D%20a%20%5Ctimes%20%5B%5C%3A%20%5Csum_%7Bi%3Dr%7D%5Es%20f%28i%29%5D%20%2B%20%28s-r%29b)
where a, b, r, and s are constants, f(i) is function of i, i ranging from r to s (integral assuming).
For the given case, the considered summation can be written symbolically as:
It is evaluated as;
![\sum_{n=1}^{14} (3n + 2) = 3 \times [ \: \sum_{n=1}^{14} n ] + \sum_{n=1}^{14} 2\\\\\sum_{n=1}^{14} (3n + 2) = 3 \times \dfrac{(14)(14 + 1)}{2} + (2 + 2 + .. + 2(\text{14 times}))\\\\\sum_{n=1}^{14} (3n + 2) = 3 \times 105 + 28 = 343\\](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7B14%7D%20%20%283n%20%2B%202%29%20%3D%203%20%5Ctimes%20%5B%20%5C%3A%20%5Csum_%7Bn%3D1%7D%5E%7B14%7D%20n%20%5D%20%2B%20%5Csum_%7Bn%3D1%7D%5E%7B14%7D%202%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B14%7D%20%20%283n%20%2B%202%29%20%3D%203%20%5Ctimes%20%5Cdfrac%7B%2814%29%2814%20%2B%201%29%7D%7B2%7D%20%2B%20%282%20%2B%202%20%2B%20..%20%2B%202%28%5Ctext%7B14%20times%7D%29%29%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B14%7D%20%20%283n%20%2B%202%29%20%3D%203%20%5Ctimes%20105%20%2B%2028%20%3D%20343%5C%5C)
Thus, the summation of the considered expression in terms of n from n = 1 to 14 is given by: Option D: 343
Learn more about summation here:
brainly.com/question/14322177
Answer: Use the coordinates (0,-7)
Step-by-step explanation:
First aquarium dimensions:
Length = 6 m.
Width = 4 m and
Height = 2 meter.
Second aquarium dimensions:
Length = 8 m.
Width = 9 m and
Height = 3 meter.
We know formula for volume of a cuboidal box = Length*Width*Height.
Plugging values of length, width and height of first aquarium in formula of volume. We get
V1 = 6*4*2 = 48 m^3.
Plugging values of length, width and height of second aquarium in formula of volume. We get
V2 = 8*9*3 = 216 m^3.
In order to find the total cubic meters of space do the sea turtles have in their habitat, we need to add both volumes.
Therefore, Total voulme of both aquarium = V1 +V2 = 48+216 = 264 m^3.
Therefore, total 264 m^3 cubic meters of space the sea turtles have in their habitat.
Area of chalkboard = 4 feet x 6 feet = 24 square feet.
Total cost = 24 square feet x $7.21 per square foot = $173.04
Answer: $173.04
