Due tomorrow i will like!

(x 3 +y 3 )(xy 4 +7)

KonstantinChe [14]

Answer:

question is not proper

Step-by-step explanation:

question is

5 0

2 years ago

Very important!! Solve the inequality 6r + 30 ≥ 12 or -r > 12 (the question is important for high school application c:)

Fofino [41]

First simplify the one inequality, 6r + 30 greater than or equal to 12 just divide by 6 then subtract 30 from both sides of the inequality. You should get r is greater than or equal to -3. For this one you need to reverse the inequality symbol then divide both sides by -1 to get a positive r, and you get r is less than -12. It's not A because there's two solutions, and it's not D for the same reason. It's not C because r isn't greater than -12. In conclusion it's B because it correctly represents the solutions that make the inequality true. I hope this helps you on your high school application :)

7 0

2 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately 2020.022

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$