1. 6.73 x 10^6
2. 1.33 x 10^4
3.9.77 x 10^22
4. 3.84 x 10^5
The rest i didn’t understand what where you trying to say
To complete the square:
we take the coefficient ox "x" (which in this problem is -20)
we divide it by 2
square that number
then add it to both sides of the equation
-20 / 2 = -10
-10^2 = 100
then we add 100 to both sides of the equation:
x^2 -20x
x^2 -20x +100 = 100
******************************************************
To get the roots of the equation, we take the square root of both sides:
(x -10) * (x-10) = 10
(x-10) = square root (10)
x-10 =
<span>
<span>
<span>
3.1622776602
</span>
</span>
</span>
x1 =
<span>
<span>
<span>
13.1622776602
</span>
and don't forget that square root of 10 also equals </span></span><span><span><span> -3.1622776602
</span>
</span>
</span>
x2 = 10
-<span>
<span>
<span>
3.1622776602
x2 = </span></span></span>
<span>
<span>
<span>
6.8377223398
</span>
</span>
</span>
They are exactly the same thing
9514 1404 393
Answer:
False
Step-by-step explanation:
f(-2) = (1/2)(-2) -3 = -1 -3 = -4
g(-2) = -2(-2) +2 = 4 +2 = 6
The function values are not the same at x=-2, so the graphs do not intersect there.
__
The graphs intersect at x=2.
Piecewise Function is like multiple functions with a speific/given domain in one set, or three in one for easier understanding, perhaps.
To evaluate the function, we have to check which value to evalue and which domain is fit or perfect for the three functions.
Since we want to evaluate x = -8 and x = 4. That means x^2 cannot be used because the given domain is less than -8 and 4. For the cube root of x, the domain is given from -8 to 1. That meand we can substitute x = -8 in the cube root function because the cube root contains -8 in domain but can't substitute x = 4 in since it doesn't contain 4 in domain.
Last is the constant function where x ≥ 1. We can substitute x = 4 because it is contained in domain.
Therefore:
![\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 8} \\ f(4) = 3 \end{cases}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%20%5Cbegin%7Bcases%7D%20f%28%20-%208%20%29%20%3D%20%20%20%5Csqrt%5B3%5D%7B%20-%208%7D%20%20%5C%5C%20f%284%29%20%3D%203%20%5Cend%7Bcases%7D%7D)
The nth root of a can contain negative number only if n is an odd number.
![\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 2 \times - 2 \times - 2} \\ f(4) = 3 \end{cases}} \\ \large{ \begin{cases} f( - 8 ) = - 2\\ f(4) = 3 \end{cases}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%20%5Cbegin%7Bcases%7D%20f%28%20-%208%20%29%20%3D%20%20%20%5Csqrt%5B3%5D%7B%20-%202%20%5Ctimes%20-%20%202%20%5Ctimes%20%20%20-%202%7D%20%20%5C%5C%20f%284%29%20%3D%203%20%5Cend%7Bcases%7D%7D%20%5C%5C%20%20%5Clarge%7B%20%20%5Cbegin%7Bcases%7D%20f%28%20-%208%20%29%20%3D%20%20-%202%5C%5C%20f%284%29%20%3D%203%20%5Cend%7Bcases%7D%7D)
Answer