Pascals triangle to the 6th:
1 x^0
1 1 x^1
1 2 1 x^2
1 3 3 1 x^3
1 4 6 4 1 x^4
1 5 10 10 5 1 x^5
1 6 15 20 15 6 1 x^6<span>
</span>the problem is to the 6th power so your going to use the 6th row of pascals triangle (don't count the first row). these numbers represent the coefficients of the variables
1(d-5y)^6 + 6(d-5y)^5 + 15(d-5y)^4 + 20(d-5y)^3 + 15(d-5y)^2 + 6(d-5y) + 1
then simplify
Step-by-step explanation:
15 : 40
Both have table of 5 in common
5 : 8
Given the function :
![f(x)=\sqrt[]{x+2}+1](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B%5D%7Bx%2B2%7D%2B1)
We need to find each missing value
Given x = -3 , -2 , -1 , 2 , 7
So, substitute with each value of x to find the corresponding value of f(x)
![x=-3\rightarrow f(x)=\sqrt[]{-3+2}+1=\sqrt[]{-1}+1](https://tex.z-dn.net/?f=x%3D-3%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B-3%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B-1%7D%2B1)
So, there is no value for f(x) at x = -3 (the function undefined because the square root of -1)
![\begin{gathered} x=-2\rightarrow f(x)=\sqrt[]{-2+2}+1=\sqrt[]{0}+1=0+1=1 \\ \\ x=-1\rightarrow f(x)=\sqrt[]{-1+2}+1=\sqrt[]{1}+1=1+1=2 \\ \\ x=2\rightarrow f(x)=\sqrt[]{2+2}+1=\sqrt[]{4}+1=2+1=3 \\ \\ x=7\rightarrow f(x)=\sqrt[]{7+2}+1=\sqrt[]{9}+1=3+1=4 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D-2%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B-2%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B0%7D%2B1%3D0%2B1%3D1%20%5C%5C%20%20%5C%5C%20x%3D-1%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B-1%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B1%7D%2B1%3D1%2B1%3D2%20%5C%5C%20%20%5C%5C%20x%3D2%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B2%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B4%7D%2B1%3D2%2B1%3D3%20%5C%5C%20%20%5C%5C%20x%3D7%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B7%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B9%7D%2B1%3D3%2B1%3D4%20%5Cend%7Bgathered%7D)
the graph of the function and the points will be as shown in the following image :
Use the given functions to set up and simplify
h(8). The answer is −22. Hope this help! :)