Answer:f(x) has three real roots
Step-by-step explanation:
The solution to a system of equations is the point where the 2 lines of a graph cross each other.
On the pictured graph, the lines cross each other at x = -6 and Y = -2.
The answer would be (-6,-2)
Answer:
Remember the property:
a^-1 = (1/a)^1
and:
(a/b)^n = (a^n)/(b^n)
A table for a function like:
![\left[\begin{array}{ccc}x&f(x)\\&\\&\\&\\&\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26f%28x%29%5C%5C%26%5C%5C%26%5C%5C%26%5C%5C%26%5Cend%7Barray%7D%5Cright%5D)
Is just completed as:
![\left[\begin{array}{ccc}x&f(x)\\x_1&f(x_1)\\x_2&f(x_2)\\x_3&f(x_3)\\x_4&f(x_4)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26f%28x%29%5C%5Cx_1%26f%28x_1%29%5C%5Cx_2%26f%28x_2%29%5C%5Cx_3%26f%28x_3%29%5C%5Cx_4%26f%28x_4%29%5Cend%7Barray%7D%5Cright%5D)
So, here we have:
y = f(x) = (1/6)^x
To complete the table, we need to find:
f(-1)
and
f(2)
So let's find these two values:
f(-1) = (1/6)^-1 = (6/1)^1 = 6
and the other value is:
f(2) = (1/6)^2 = 1/36
Then the complete table is:
![\left[\begin{array}{ccc}x&f(x)\\-2&36\\-1&6\\0&1\\1&1/6\\2&1/36\\1&1/216\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26f%28x%29%5C%5C-2%2636%5C%5C-1%266%5C%5C0%261%5C%5C1%261%2F6%5C%5C2%261%2F36%5C%5C1%261%2F216%5Cend%7Barray%7D%5Cright%5D)
Hi! So I got :
-3 —>20
-2 —>13
-1 —>8
0 —>5
1 —>4
2 —>5
3 —>8
I’m sorry if any of these are wrong but In order to find the answers I used the “graphing quadratics” one (don’t know the name so sorry) but yeah first I found the axis of symmetry which was 1 and from there I built a chart and substituted. Again, I didn’t see some answer choices above in the image so sorry if there wrong, but yeah I hope this helped! (This is from algebra 1 by the way)
633,600 ft. per hour I used a calculator converter so its a 100% right
