(f o g)(x)
g(x) = x + 2
Since it’s telling you to use g(x), you use (x + 2) as your new x value for function f.
f(x + 2) = 5(x + 2) + 5
= 5x + 10 + 5
= 5x + 15
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:
see below
Step-by-step explanation:
List the coefficients and constant for an equation in one row of the matrix. The variables should be in the same order. Any missing terms are replaced by zero.
![\left[\begin{array}{ccc|c}9&-4&-5&9\\7&4&-4&-1\\6&-6&1&-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D9%26-4%26-5%269%5C%5C7%264%26-4%26-1%5C%5C6%26-6%261%26-5%5Cend%7Barray%7D%5Cright%5D)
Cubic centimetres since they are cubes. Did you add any picture of the question? There seems to be none.
Answer:
Slope = 4
Step-by-step explanation:
Use any two pairs from the table of values, say (0, 3) and (2, 11).
Slope = ∆y/∆x
Slope = 
Slope = 
Slope = 4