Answer:
f(-2) = 4
f(0) = 6
f(5) = 11
Step-by-step explanation:
You just need to plug in the different values of x into the equation
f(x)=x+6
When x = -2, then f(-2) = -2 + 6 = 4
When x = 0, then f(0) = 0 + 6 = 6
When x = 5, then f(5) = 5 + 6 = 11
Answer:

Step-by-step explanation:
The expression to transform is:
![(\sqrt[6]{x^5})^7](https://tex.z-dn.net/?f=%28%5Csqrt%5B6%5D%7Bx%5E5%7D%29%5E7)
Let's work first on the inside of the parenthesis.
Recall that the n-root of an expression can be written as a fractional exponent of the expression as follows:
![\sqrt[n]{a} = a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
Therefore ![\sqrt[6]{a} = a^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now let's replace
with
which is the algebraic form we are given inside the 6th root:
![\sqrt[6]{x^5} = (x^5)^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Bx%5E5%7D%20%3D%20%28x%5E5%29%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now use the property that tells us how to proceed when we have "exponent of an exponent":

Therefore we get: 
Finally remember that this expression was raised to the power 7, therefore:
[/tex]
An use again the property for the exponent of a exponent:
Answer:
t ≈ -2.014 or 3.647
Step-by-step explanation:
Add the opposite of the expression on the right side of the equal sign to put the equation into standard form.
4.9t² -8t -36 = 0
You can divide by 4.9 to make this a little easier to solve.
t² -(8/4.9)t -36/4.9 = 0
Now, add and subtract the square of half the x-coefficient to "complete the square."
t² -(8/4.9)t +(4/4.9)² -36/4.9 -(4/4.9)² = 0
(t -4/4.9)² -192.4/4.9² = 0 . . . . simplify
Add the constant term, then take the square root.
(t -4/4.9)² = 192.4/4.9²
t -4/4.9 = ±(√192.4)/4.9
t = (4 ± √192.4)/4.9
t ≈ {-2.014, 3.647}
What is the interquartile range for the data set 4,7,7,3,5,2,6,7,9 the answers are 3.5, 5.6, 7,9
polet [3.4K]
5.6 is the answer for your question