Answer:
Yes.
Step-by-step explanation:
By adding like terms, -x - 2x = -3x. Therefore, they are equivalent.
The answer is B, at first the sum is 621 and then you divide by 7 to get your median (88), after you add the two #s of 86 and 87 to 621 and divide by 9. Resulting in the median being 88
We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.
Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.
(i)

![V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}](https://tex.z-dn.net/?f=V%3D%5Cpi%5Ccdot%5Cint%5Climits_0%5E9%28x%5E3%29%5E2%5C%2C%20dx%3D%5Cpi%5Ccdot%5Cint%5Climits_0%5E9x%5E6%5C%2C%20dx%3D%5Cpi%5Ccdot%5Cleft%5B%5Cdfrac%7Bx%5E7%7D%7B7%7D%5Cright%5D_0%5E9%3D%5Cpi%5Ccdot%5Cleft%28%5Cdfrac%7B9%5E7%7D%7B7%7D-%5Cdfrac%7B0%5E7%7D%7B7%7D%5Cright%29%3D%5Cdfrac%7B9%5E7%7D%7B7%7D%5Cpi%3D%5C%5C%5C%5C%5C%5C%3D%5Cboxed%7B%5Cdfrac%7B4782969%7D%7B7%7D%5Cpi%7D)
Answer B. or D.
(ii)

![V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx= 2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}](https://tex.z-dn.net/?f=V%3D2%5Cpi%5Ccdot%5Cint%5Climits_0%5E%7B9%7D%28x%5Ccdot%20x%5E3%29%5C%2C%20dx%3D2%5Cpi%5Ccdot%5Cint%5Climits_0%5E%7B9%7Dx%5E4%5C%2C%20dx%3D%0A2%5Cpi%5Ccdot%5Cleft%5B%5Cdfrac%7Bx%5E5%7D%7B5%7D%5Cright%5D_0%5E9%3D2%5Cpi%5Ccdot%5Cleft%28%5Cdfrac%7B9%5E5%7D%7B5%7D-%5Cdfrac%7B0%5E5%7D%7B5%7D%5Cright%29%3D%5C%5C%5C%5C%5C%5C%3D2%5Cpi%5Ccdot%5Cdfrac%7B9%5E5%7D%7B5%7D%3D%5Cboxed%7B%5Cdfrac%7B118098%7D%7B5%7D%5Cpi%7D)
So we know that the correct answer is D.
(iii)
Line x = h

![V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\](https://tex.z-dn.net/?f=V%3D2%5Cpi%5Ccdot%5Cint%5Climits_0%5E9%5Cbig%28%2818-x%29%5Ccdot%20x%5E3%5Cbig%29%5C%2C%20dx%3D2%5Cpi%5Ccdot%5Cint%5Climits_0%5E9%2818x%5E3-x%5E4%29%5C%2C%20dx%3D%5C%5C%5C%5C%5C%5C%3D2%5Cpi%5Ccdot%5Cleft%28%5Cint%5Climits_0%5E918x%5E3%5C%2C%20dx-%5Cint%5Climits_0%5E9x%5E4%5C%2C%20dx%5Cright%29%3D2%5Cpi%5Ccdot%5Cleft%2818%5Cint%5Climits_0%5E9x%5E3%5C%2C%20dx-%5Cint%5Climits_0%5E9x%5E4%5C%2C%20dx%5Cright%29%3D%5C%5C%5C%5C%5C%5C%3D2%5Cpi%5Ccdot%5Cleft%2818%5Cleft%5B%5Cdfrac%7Bx%5E4%7D%7B4%7D%5Cright%5D_0%5E9-%5Cleft%5B%5Cdfrac%7Bx%5E5%7D%7B5%7D%5Cright%5D_0%5E9%5Cright%29%3D2%5Cpi%5Ccdot%5CBiggl%2818%5Cbiggl%28%5Cdfrac%7B9%5E4%7D%7B4%7D-%5Cdfrac%7B0%5E4%7D%7B4%7D%5Cbiggr%29-%5Cbiggl%28%5Cdfrac%7B9%5E5%7D%7B5%7D-%5Cdfrac%7B0%5E5%7D%7B5%7D%5Cbiggr%29%5CBiggr%29%3D%5C%5C%5C%5C%5C%5C)

Answer D. just as before.
rounding off of 123.5201 is 123.5
rounding off of 424.9832 is 425
rounding off of 675.0608 is 675.1
rounding off of 1247.0057 is 1247
rounding off of 5653.5974 is 5653.6
here we have rounded off to first decimal place
one point to note that we can just approximate to get the rounded off value. We have to follow several rules, like we should increase the last digit to be retained by one when the digit following it is greater than 5 and we should not
change the last digit if the digit following the last digit to be retained is less than five
Answer: Angles: 42°, 114°, 24°, 30°, 150°.
Step-by-step explanation: