I’m assuming you mean -1/4 -4, and since you’re subtracting a negative by a negative, it would be -4 1/4(mixed number)
8/12 and 4/6 and10/15 and20/30 and 40/60
Answer:
Step-by-step explanation:
From the given information:
The domain D of integration in polar coordinates can be represented by:
D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;
The partial derivates for z = xy can be expressed as:
Thus, the area of the surface is as follows:
![= 2 \pi \times \dfrac{1}{3} \Bigg [ (37)^{3/2} - 1 \Bigg]](https://tex.z-dn.net/?f=%3D%202%20%5Cpi%20%5Ctimes%20%5Cdfrac%7B1%7D%7B3%7D%20%20%5CBigg%20%5B%20%2837%29%5E%7B3%2F2%7D%20-%201%20%5CBigg%5D)
![= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B2%20%5Cpi%7D%7B3%7D%20%5CBigg%20%5B37%20%5Csqrt%7B37%7D%20-1%20%5CBigg%20%5D)
Answers:
- C) x = plus/minus 11
- B) No real solutions
- C) Two solutions
- A) One solution
- The value <u> 18 </u> goes in the first blank. The value <u> 17 </u> goes in the second blank.
========================================================
Explanations:
- Note how (11)^2 = (11)*(11) = 121 and also (-11)^2 = (-11)*(-11) = 121. The two negatives multiply to a positive. So that's why the solution is x = plus/minus 11. The plus minus breaks down into the two equations x = 11 or x = -11.
- There are no real solutions here because the left hand side can never be negative, no matter what real number you pick for x. As mentioned in problem 1, squaring -11 leads to a positive number 121. The same idea applies here as well.
- The two solutions are x = 0 and x = -2. We set each factor equal to zero through the zero product property. Then we solve each equation for x. The x+2 = 0 leads to x = -2.
- We use the zero product property here as well. We have a repeated factor, so we're only solving one equation and that is x-3 = 0 which leads to x = 3. The only root is x = 3.
- Apply the FOIL rule on (x+1)(x+17) to end up with x^2+17x+1x+17 which simplifies fully to x^2+18x+17. The middle x coefficient is 18, while the constant term is 17.
