3 years ago

Find the sum of the reciprocals of two consecutive even integers if the larger integer is x. please answer;-;

1 answer:

4 0

Step-by-step explanation:

Let the larger integer be x

Hence, the smaller integer = x-2

Hence,

Reciprocal of the x= 1/x

Reciprocal of x-2= 1/ x-2

Hence. their sum=

$\frac{1}{x} + \frac{1}{x - 2} \\ = \frac{x - 2 + x}{x(x - 2)} \\ = \frac{2x - 2}{ {x}^{2} - 2x}$

You might be interested in

spin [16.1K]

Well one week she'd have 23.50$. So per week you multiply that number by the number of weeks.

I hope this is what you were looking for, the question was a bit confusing.

8 0

3 years ago

horsena [70]

I believe the answer would be 25 + 0.35m = 42.50

8 0

2 years ago

Rom4ik [11]

let log2 32 = x

then 2^x = 32

but we know 2^5 = 32

therefore, x = 5

6 0

3 years ago

Step2247 [10]

Yeah The Answer Would Be A=299

6 0

3 years ago

kirill115 [55]

$S= \pi rh+ \pi r^2\ \ \ \ | subtract\ \pi r^2\\\\
s-\pi r^2= \pi rh\ \ \ \ | divide\ by\ \pi r \\\\
\frac{s-\pi r^2}{ \pi r}=h$

3 0

3 years ago

Find the sum of the reciprocals of two consecutive even integers if the larger integer is x. please answer;-;​