Answer:
![\boxed{4 \sqrt[8]{ {d}^{3} } }](https://tex.z-dn.net/?f=%20%5Cboxed%7B4%20%5Csqrt%5B8%5D%7B%20%7Bd%7D%5E%7B3%7D%20%7D%20%7D%20)
Step-by-step explanation:
![= > 4 {d}^{ \frac{3}{8} } \\ \\ = > 4({d}^{3 \times \frac{1}{8} }) \\ \\ = > 4( {d}^{3} \times {d}^{ \frac{1}{8} } ) \\ \\ = > 4( {d}^{3} \times \sqrt[8]{d} ) \\ \\ = > 4 \sqrt[8]{ {d}^{3} }](https://tex.z-dn.net/?f=%20%3D%20%20%3E%204%20%7Bd%7D%5E%7B%20%5Cfrac%7B3%7D%7B8%7D%20%7D%20%20%20%5C%5C%20%20%5C%5C%20%3D%20%20%20%3E%204%28%7Bd%7D%5E%7B3%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B8%7D%20%7D%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%3E%204%28%20%7Bd%7D%5E%7B3%7D%20%20%5Ctimes%20%20%20%7Bd%7D%5E%7B%20%5Cfrac%7B1%7D%7B8%7D%20%7D%20%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%3E%204%28%20%7Bd%7D%5E%7B3%7D%20%20%5Ctimes%20%20%5Csqrt%5B8%5D%7Bd%7D%20%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%3E%204%20%20%5Csqrt%5B8%5D%7B%20%7Bd%7D%5E%7B3%7D%20%7D%20)
Given that PQ and RS are drawn with KL as tranversal intersecting PQ at M and RS at point N. Angle QMN is congruent to angle LNS because they are alternate to each other. The theorem that Kari can use to show that the meansure of QML is supplementary to the measure of angle SNK is Alternate Exterior Angles Theorem.
This is because angle KNR is equal to QML by alternate exterior angles theorem so is angle MLP and SNK
Answer:
4.5
Step-by-step explanation:
Hello!
The find the volume of a cone you divide the product of the base and height by 3. First we find the area using

r².
A=49(3.14)=153.86
Now we multiply this by the height.
153.86(18)=2769.48
Now we divide by 3.
2769.48/3=923.16
Now we divide this number by 21 to see how many days it will take for the container to become empty.
923.16/21=43.96≈44
Therefore, our answer is
A) 44 days.
I hope this helps!
We will use the right Riemann sum. We can break this integral in two parts.

We take the interval and we divide it n times:

The area of the i-th rectangle in the right Riemann sum is:

For the first part of our integral we have:

For the second part we have:

We can now put it all together:
![\sum_{i=1}^{i=n} [(\Delta x)^4 i^3-6(\Delta x)^2i]\\\sum_{i=1}^{i=n}[ (\frac{3}{n})^4 i^3-6(\frac{3}{n})^2i]\\ \sum_{i=1}^{i=n}(\frac{3}{n})^2i[(\frac{3}{n})^2 i^2-6]](https://tex.z-dn.net/?f=%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%20%5B%28%5CDelta%20x%29%5E4%20i%5E3-6%28%5CDelta%20x%29%5E2i%5D%5C%5C%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%5B%20%28%5Cfrac%7B3%7D%7Bn%7D%29%5E4%20i%5E3-6%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2i%5D%5C%5C%0A%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2i%5B%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2%20i%5E2-6%5D)
We can also write n-th partial sum: