What is the common difference or common ratio of the sequence 3,7,11,15...?

Whitepunk [10]

Answer:

0.206897

Step-by-step explanation:

it is 0.206897 because you have take the whole make uneven fraction an add it to 60 in 60/72 to get 348 divided equals 0.206897

3 0

3 years ago

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The sum of 3 consecutive odd integers is 18 less than five times the middle number. Find the three integers

Andreyy89

n, n + 2, n + 4 - three consecutive odd ntegers

n + (n + 2) + (n + 4) = 5(n + 2) - 18 |use distributive property

n + n + 2 + n + 4 = (5)(n) + (5)(2) - 18

3n + 6 = 5n + 10 - 18

3n + 6 = 5n - 8 |subtract 6 from both sides

3n = 5n - 14 |subtract 5n from both sides

-2n = -14 |divide both sides by (-2)

n = 7

n + 2 = 7 + 2 = 9

n + 4 = 7 + 4 = 11

Answer: 7, 9, 11.

8 0

3 years ago

Find LCM of: x2 - 9, 3x3 + 81

VikaD [51]

Answer:

$\boxed{\pink{\sf (x + 3) \ is\ the \ LCM . }}$

Step-by-step explanation:

Given two expressions ,

$\qquad (1)\: x^2 - 9 \\\\ \qquad (2)\:3x^3 + 81$

And , we need to find the LCM , that is lowest common factor . So , let's factorise them seperately .

Factorising x² - 9 :- 

$= x^2 - 9 \\\\= x^2 - 3^2 \\\\ \red{= (x+3)(x-3)} \qquad\bf [Using \ a^2-b^2 = (a+b)(a-b) ]$

Factorising 3x³ + 81 

$= 3x^3 + 81\\\\= 3(x^3+27) \\\\= 3( x^3+27) \\\\ = 3(x^3+3^3) \\\\ \red{= 3 [ (x+3)(x^2+9-3x) ] } \qquad \bf{[ Using \ a^3+b^3=(a+b)(a^2+b^2-ab) ] }$

Hence we can see that (x+3) is common factor in both expressions.

Hence the LCM is ( x+3 ) .

8 0

3 years ago

Dana surveyed students in her class. She found that 8 earned an A, 6 earned a B, 4 earned a C, and 2 earned a D on the last scie

lozanna [386]

Answer:

0.4 or 2/5

Step-by-step explanation:

Dana surveyed 20 students in total and out of those 20 only 8 got an A

so:

$\frac{8}{20}$ now if you simplify it $\frac{4}{10}$ and if you simplify it again:

$\frac{2}{5}$ or 0.4 in decimal

7 0

3 years ago

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An open box is to be made from a rectangular piece of material 9 inches by 12 inches by cutting equal squares from each corner a

grigory [225]

The questions for this problem would be:
1. What is the dimensions of the box that has the maximum volume?
2. What is the maximum volume of the box?

Volume of a rectangular box = length x width x height

From the problem statement,
length = 12 - 2x
width = 9 - 2x
height = x

where x is the height of the box or the side of the equal squares from each corner and turning up the sides

V = (12-2x) (9-2x) (x)
V = (12 - 2x) (9x - 2x^2)
V = 108x - 24x^2 -18x^2 + 4x^3
V = 4x^3 - 42x^2 + 108x

To maximize the volume, we differentiate the expression of the volume and equate it to zero.

V = 4x^3 - 42x^2 + 108x
dV/dx = 12x^2 - 84x + 108
12x^2 - 84x + 108 = 0x^2 - 7x + 9 = 0

Solving for x,
x1 = 5.30 ; Volume = -11.872 (cannot be negative)
x2 = 1.70 ; Volume = 81.872

So, the answers are as follows:

1. What is the dimensions of the box that has the maximum volume?
length = 12 - 2x = 8.60
width = 9 - 2x = 5.60
height = x = 1.70

2. What is the maximum volume of the box?

Volume = 81.872

7 0

4 years ago