so 
.![$A=\int\limits_0^k\dfrac{y^2}{4}\,dy=\frac{1}{4}\int\limits_0^ky^2\,dy=\frac{1}{4}\left[\frac{y^3}{3}\right]_0^k=\frac{1}{4}\left[\frac{k^3}{3}-\frac{0^3}{3}\right]=\frac{1}{4}\cdot\dfrac{k^3}{3}=\boxed{\frac{k^3}{12}}](/tpl/images/0264/3877/9b1e4.png)
![\dfrac{k^3}{12}=\dfrac{9}{4}\\\\\\k^3=\dfrac{9\cdot12}{4}\\\\\\k^3=9\cdot3\\\\k^3=3^3\quad|\sqrt[3]{(\ldots)}\\\\\boxed{k=3}](/tpl/images/0264/3877/f1458.png)
Anybody know the integration for the answer is k=3
...
Answers: 1
Another question on Mathematics
Mathematics, 21.06.2019 21:00
Oliver read for 450 minutes this month his goal was to read for 10% more minutes next month if all of her medicine go how many minutes will you read all during the next two months
Answers: 3
Mathematics, 21.06.2019 22:00
Write a description of each inequality. 1. -5a + 3 > 1 2. 27 - 2b < -6 (this < has a _ under it) 3. 1/2 (c + 1) > 5 (this > has a _ under it)
Answers: 3
You know the right answer?
Questions
English, 28.07.2020 09:01
Social Studies, 28.07.2020 09:01
Mathematics, 28.07.2020 09:01
History, 28.07.2020 09:01
Mathematics, 28.07.2020 09:01
Geography, 28.07.2020 09:01
Mathematics, 28.07.2020 09:01
Chemistry, 28.07.2020 09:01
Geography, 28.07.2020 09:01
Social Studies, 28.07.2020 09:01
