Answer:
Area covered by the fences will be 16.1 unit²
Step-by-step explanation:
Let the first parabola is represented by the function f(x) = 6x²
and second parabola by g(x) = x² + 9
point of intersection of the graphs will be determined when f(x) = g(x)
6x² = x² + 9
5x² = 9
x² = 1.8
x = ± 1.34
Now we will find the area between these curves drawn on the graph.
Area = ![\int_{-1.34}^{1.34}[f(x)-g(x)]dx=\int_{-1.34}^{1.34}[6x^{2}-(x^{2}+9)]dx](https://tex.z-dn.net/?f=%5Cint_%7B-1.34%7D%5E%7B1.34%7D%5Bf%28x%29-g%28x%29%5Ddx%3D%5Cint_%7B-1.34%7D%5E%7B1.34%7D%5B6x%5E%7B2%7D-%28x%5E%7B2%7D%2B9%29%5Ddx)
= 
= ![[\frac{5}{3}x^{3}-9x]_{-1.34}^{1.34}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B5%7D%7B3%7Dx%5E%7B3%7D-9x%5D_%7B-1.34%7D%5E%7B1.34%7D)
= ![[\frac{5}{3}(-1.34)^{3}-9(-1.34)-\frac{5}{3}(1.34)^{3}+9(1.34)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B5%7D%7B3%7D%28-1.34%29%5E%7B3%7D-9%28-1.34%29-%5Cfrac%7B5%7D%7B3%7D%281.34%29%5E%7B3%7D%2B9%281.34%29%5D)
= ![[-4.01+12.06-4.01+12.06]](https://tex.z-dn.net/?f=%5B-4.01%2B12.06-4.01%2B12.06%5D)
= 16.1 unit²
Answer:
StartFraction negative 1 Over k cubed EndFraction
Step-by-step explanation:
3k / (k + 1) × (k²- 1) / 3k³
= 3k(k² - 1) / (k + 1)(3k³)
= 3k³ - 3k / 3k⁴ + 3k³
= -3k / 3k⁴
= -1/k³
StartFraction k + 1 Over k squared EndFraction
(k + 1) / k²
StartFraction k minus 1 Over k squared EndFraction
(k - 1)/k²
StartFraction negative 1 Over k cubed EndFraction
= -1/k³
StartFraction 1 Over k EndFraction
= 1/k
Answer
9/10=<u>90%</u> and 2/5=<u>40%</u>
Explanation
9÷10=0.9 0.9x100=90%
2÷5=0.4 0.4x100=40%
<em>hope this helps!</em>
<em>have a wonderful day :)</em>
Answer:
f(x)=6
Step-by-step explanation:
By Substitution
f(x or coefficient of x)
=6 since X doesn't exist in question above
