So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:
Next, square both sides:
Next, subtract x and add 2 to both sides of the equation:
Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:
Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:
Now you can rewrite the equation as 
Now, apply the Zero Product Property and solve for x as such:
Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:
Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>
<span>(2.2 × 1012) + (1.7 × 109) = 2411.7</span>
Answer:
3π square units.
Step-by-step explanation:
We can use the disk method.
Since we are revolving around AB, we have a vertical axis of revolution.
So, our representative rectangle will be horizontal.
R₁ is bounded by y = 9x.
So, x = y/9.
Our radius since our axis is AB will be 1 - x or 1 - y/9.
And we are integrating from y = 0 to y = 9.
By the disk method (for a vertical axis of revolution):
![\displaystyle V=\pi \int_a^b [R(y)]^2\, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%3D%5Cpi%20%5Cint_a%5Eb%20%5BR%28y%29%5D%5E2%5C%2C%20dy)
So:
Simplify:
Integrate:
![\displaystyle V=\pi\Big[y-\frac{1}{9}y^2+\frac{1}{243}y^3\Big|_0^9\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%3D%5Cpi%5CBig%5By-%5Cfrac%7B1%7D%7B9%7Dy%5E2%2B%5Cfrac%7B1%7D%7B243%7Dy%5E3%5CBig%7C_0%5E9%5CBig%5D)
Evaluate (I ignored the 0):
![\displaystyle V=\pi[9-\frac{1}{9}(9)^2+\frac{1}{243}(9^3)]=3\pi](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%3D%5Cpi%5B9-%5Cfrac%7B1%7D%7B9%7D%289%29%5E2%2B%5Cfrac%7B1%7D%7B243%7D%289%5E3%29%5D%3D3%5Cpi)
The volume of the solid is 3π square units.
Note:
You can do this without calculus. Notice that R₁ revolved around AB is simply a right cone with radius 1 and height 9. Then by the volume for a cone formula:
We acquire the exact same answer.
Answer:
24x + 41
Step-by-step explanation:
Distribute the 6.
6(4x + 5) + 11 = 24x + 30 + 11
Now we just add the 30 and the 11 which are at the back of the expression.
24x + 30 + 11 = 24x + 41
Therefore the answer is 24x + 41. Please mark brainliest if possible and i hope you get a good grade on your homework
Area: If you multiply 2 1/3 by 5 2/5, you get an answer of about: 12.6, which means the area is about: 12.6in. squared.
Perimeter: If you add 2 1/3+ 2 1/3+ 5 2/5+ 5 2/5 you get an answer of about: 15.47. So, the perimeter is about 15.47in.
Note: If you aren't allowed to use a calculator on this test, you should probably multiply the numbers yourself so you get show the work and get the answer in a fraction, which is probably what they want. If you are allowed to use calculators, all the information you need is right here.
