Ax+3x=bx+5 solve for x please helpill give brainliest please show steps
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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
Difference between the area of the triangle and square is 25
Step-by-step explanation:
- Step 1: Find the area of the triangle given its 3 sides using the Heron's formula.
Area of the triangle = where s =
⇒ s = (6 + 8 + 10)/2 = 24/2 = 12
=
= =
= 24 sq. units
- Step 2: Find the area of the square with perimeter = 28 units.
Perimeter of the square = 4 × side = 28
⇒ Side of the square = 28/4 = 7 units
⇒ Area of the square = (side)² = 7² = 49 sq. units
- Step 3: Find the difference between the area of the square and triangle.
Difference = 49 - 24 = 25