Answer:
32.66 units
Step-by-step explanation:
We are given that
![y=6x^2+14x](https://tex.z-dn.net/?f=y%3D6x%5E2%2B14x)
Point A=(-2,-4) and point B=(1,20)
Differentiate w.r. t x
![\frac{dy}{dx}=12x+14](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%3D12x%2B14)
We know that length of curve
![s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx](https://tex.z-dn.net/?f=s%3D%5Cint_%7Ba%7D%5E%7Bb%7D%5Csqrt%7B1%2B%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E2%7Ddx)
We have a=-2 and b=1
Using the formula
Length of curve=![s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx](https://tex.z-dn.net/?f=s%3D%5Cint_%7B-2%7D%5E%7B1%7D%5Csqrt%7B1%2B%2812x%2B14%29%5E2%7Ddx)
Using substitution method
Substitute t=12x+14
Differentiate w.r t. x
![dt=12dx](https://tex.z-dn.net/?f=dt%3D12dx)
![dx=\frac{1}{12}dt](https://tex.z-dn.net/?f=dx%3D%5Cfrac%7B1%7D%7B12%7Ddt)
Length of curve=![s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B1%7D%7B12%7D%5Cint_%7B-2%7D%5E%7B1%7D%5Csqrt%7B1%2Bt%5E2%7Ddt)
We know that
![\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2%2Ba%5E2%7Ddx%3D%5Cfrac%7Bx%5Csqrt%20%7Bx%5E2%2Ba%5E2%7D%7D%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Cln%28x%2B%5Csqrt%20%7Bx%5E2%2Ba%5E2%7D%29%2BC)
By using the formula
Length of curve=![s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B1%7D%7B12%7D%5B%5Cfrac%7Bt%7D%7B2%7D%5Csqrt%7B1%2Bt%5E2%7D%2B%5Cfrac%7B1%7D%7B2%7Dln%28t%2B%5Csqrt%7B1%2Bt%5E2%7D%29%5D%5E%7B1%7D_%7B-2%7D)
Length of curve=![s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B1%7D%7B12%7D%5B%5Cfrac%7B12x%2B14%7D%7B2%7D%5Csqrt%7B1%2B%2812x%2B14%29%5E2%7D%2B%5Cfrac%7B1%7D%7B2%7Dln%2812x%2B14%2B%5Csqrt%7B1%2B%2812x%2B14%29%5E2%7D%29%5D%5E%7B1%7D_%7B-2%7D)
Length of curve=![s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B1%7D%7B12%7D%28%5Cfrac%7B%2812%2B14%29%5Csqrt%7B1%2B%2826%29%5E2%7D%7D%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7Dln%2826%2B%5Csqrt%7B1%2B%2826%29%5E2%7D%29-%5Cfrac%7B12%28-2%29%2B14%7D%7B2%7D%5Csqrt%7B1%2B%28-10%29%5E2%7D-%5Cfrac%7B1%7D%7B2%7Dln%28-10%2B%5Csqrt%7B1%2B%28-10%29%5E2%7D%29)
Length of curve=![s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B1%7D%7B12%7D%2813%5Csqrt%7B677%7D%2B%5Cfrac%7B1%7D%7B2%7Dln%2826%2B%5Csqrt%7B677%7D%29%2B5%5Csqrt%7B101%7D-%5Cfrac%7B1%7D%7B2%7Dln%28-10%2B%5Csqrt%7B101%7D%29)
Length of curve=![s=32.66](https://tex.z-dn.net/?f=s%3D32.66)
Multiply the height and base of the triangle, then divide by 2. To find the area of the semi-circle, you must do this formula. 3.14 x Diameter × 1/2
Answer : 96 + 25.12 = 121.12cm
The area of a parallelogram is:
A = b * h
Where,
b: base
h: height
Clearing the base we have:
b = A / h
Substituting values we have:
b = (6x2 + x + 3) / 3x
Rewriting we have:
b = 2x + 1 / x + 1/3
Answer:
the length of the base is:
b = 2x + 1 / x + 1/3
Answer: Factors are the numbers and/or variables between your transactions (division, multiplication, addition, and subtraction signs). So the answer would be x2, 16x, and 64.
Step-by-step explanation:
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