Hello,
Let's be
![z=2x^4+8x \\ dz=(8x^3+8)\ dx=8(x^3+1)\ dx \\ \int\limits^0_{-1} {(2x^4+8x)^3*(4x^3+4)} \, dx =4* \int\limits^0_{-6} { \dfrac{z^3}{2}} \, dz \\ = 2* \left[\begin{array}{ccc} \dfrac{z^4}{4}\end{array}\right] ^0_{-6}= \left[\begin{array}{ccc} \dfrac{z^4}{2}\end{array}\right] ^0_{-6}=0-648=-648](https://tex.z-dn.net/?f=%20z%3D2x%5E4%2B8x%20%5C%5C%0Adz%3D%288x%5E3%2B8%29%5C%20dx%3D8%28x%5E3%2B1%29%5C%20dx%20%5C%5C%0A%0A%20%5Cint%5Climits%5E0_%7B-1%7D%20%7B%282x%5E4%2B8x%29%5E3%2A%284x%5E3%2B4%29%7D%20%5C%2C%20dx%20%3D4%2A%20%5Cint%5Climits%5E0_%7B-6%7D%20%7B%20%5Cdfrac%7Bz%5E3%7D%7B2%7D%7D%20%5C%2C%20dz%20%5C%5C%0A%3D%202%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cdfrac%7Bz%5E4%7D%7B4%7D%5Cend%7Barray%7D%5Cright%5D%20%5E0_%7B-6%7D%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cdfrac%7Bz%5E4%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%5E0_%7B-6%7D%3D0-648%3D-648)
Let's be
What is the exact angle between the clock hands at 6:45:00 PM? Plz show work
2 answers:
Answer:
67.5°
Step-by-step explanation:
You know the long hand would be at number 9 and you also know that from the number 6 to 9 it would be 90°. Between each number, there are also 5 smaller marks on the clock. Thus, for 90°, there would be 5×3=15 smaller marks. (I used 3 because between 6 and 9 there are 3 big numbers)
Now here's the trickier part: There are 60 minutes within 5 small marks. We can find out at which mark does the shorter hand move to at 45 minutes.
×5 = 3.75
Out of the 15 smaller marks of 90°, the 45th minute stands at 3.75. The angle of 3.75 would then be between the 15th mark and the 3.75th mark. The gap would be 15-3.75 = 11.25 marks.
×90° = 67.5°
You can refer to the picture for better visualisation of the clock. Hope this helps.
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