Answer:
it would be 1 million, ( 1,000,000 )
Step-by-step explanation:
1.) look to the right of the nine , (7)
2.) 7 is greater than five therefore rounding it up to 1,000,000
Answer:
4x^2 + 21x - 18
Step-by-step explanation:
(9x^2+24x-8)-(5x^2+3x+10)
9x^2+24x-8-5x^2-3x-10
9x^2 - 5x^2 = 4x^2
24x - 3x = 21x
-8 - 10 = -18
Answer: 4x^2 + 21x - 18
Let's compare apples to apples and oranges to oranges: convert each given proper fraction into its decimal counterpart:
3 63/80 => 3 .788 approx.
3 1/5 => 3.2
3 11/20 => 3.55
It's now an easy matter to arrange these numbers from least to greatest:
3.2, 3.55, 3.79, or
3 1/5, 3 11/20, 3 63/80
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 :
