Answer: 3053.6 in^3
To find the volume of a sphere, you first must identify the formula: (4/3)pi(r)^3
You’re given the radius 9 inches
And you’re also given the pi approximation 3.14.
With that in mind, we can plug those variables in the formula to get:
(4/3)(3.14)(9in)^3 = 3053.63 in^3
And rounded to the tenths place:
3053.6 in^3
Answer:
4 cups
Step-by-step explanation:
Number of cups of milk for 5 servings =
cups
First find number of cups of milk for 1 serving.
Number of cups of milk for 1 serving =
cup
Now find umber of cups of milk for 8 servings.
Therefore,
Number of cups of milk for 8 servings =
cups
Answer:
![\left\{ \begin{array}{ll} a+s+r=350 \\ 4a+2.5s+2r=1095\\r=s-40 \end{array} \right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%20%20%20%20%20%20%20%20%5Cbegin%7Barray%7D%7Bll%7D%20a%2Bs%2Br%3D350%20%5C%5C%20%20%20%20%20%204a%2B2.5s%2B2r%3D1095%5C%5Cr%3Ds-40%20%20%20%20%20%20%20%20%5Cend%7Barray%7D%20%20%20%20%5Cright.)
Step-by-step explanation:
Let's define our variables. Let a represent the number of adult tickets sold, s represent the number of senior tickets sold, and r represent the number of senior tickets sold.
We know that a total of 350 tickets were sold. So, the number of adult, student, and senior tickets sold must total 350. Therefore, we can write the following equation:
![a+s+r=350](https://tex.z-dn.net/?f=a%2Bs%2Br%3D350)
We know that each adult ticket costs $4, each student ticket costs $2.50, and each senior ticket costs $2. We are given that a total of $1095 was collected. Therefore, the number of tickets multiplied by their respective price must equal $1095. So, we can write the following equation:
![4a+2.5s+2r=1095](https://tex.z-dn.net/?f=4a%2B2.5s%2B2r%3D1095)
Finally, we know that 40 fewer senior tickets were sold than student tickets. So, however many students tickets were sold, we can subtract 40 to get the number of senior tickets sold. Therefore:
![r=s-40](https://tex.z-dn.net/?f=r%3Ds-40)
So, our system of equations is:
![\left\{ \begin{array}{ll} a+s+r=350 \\ 4a+2.5s+2r=1095\\r=s-40 \end{array} \right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%20%20%20%20%20%20%20%20%5Cbegin%7Barray%7D%7Bll%7D%20a%2Bs%2Br%3D350%20%5C%5C%20%20%20%20%20%204a%2B2.5s%2B2r%3D1095%5C%5Cr%3Ds-40%20%20%20%20%20%20%20%20%5Cend%7Barray%7D%20%20%20%20%5Cright.)
And we're done!
Answer:
-30a⁵b + 12a⁴b² + 16a³b² - 4a²b⁴ - 2ab³
Step-by-step explanation:
( - 6a³b + 2ab²)( 5a² - 2ab² b)
multiply the second parantheses by each term from the first parantheses.
-6a³b×(5a²- 2ab²b -b)+2ab×(5a² -2ab²b-b)
-6a³b× (5a² - 2ab² b) +2ab × (5a² - 2ab² b)
-30 a⁵b + 12a⁴b² + 6a³b²+ 10a³b² - 4a²b⁴ - 2ab
combine like terms
-30a⁵b + 12a⁴b² + 16a³b² - 4a²b⁴ - 2ab³