Answer:
x=0.5355 or x=-6.5355
First step is to: Isolate the constant term by adding 7 to both sides
Step-by-step explanation:
We want to solve this equation: 
On observation, the trinomial is not factorizable so we use the Completing the square method.
Step 1: Isolate the constant term by adding 7 to both sides
Step 2: Divide the equation all through by the coefficient of 
which is 2.
Step 3: Divide the coefficient of x by 2, square it and add it to both sides.
Coefficient of x=6
Divided by 2=3
Square of 3=
Therefore, we have:
Step 4: Write the Left Hand side in the form 
Step 5: Take the square root of both sides and solve for x
Answer:
uh, the answer is either (-6, 1) or (-1,6) because the when plugging them both into the equation, they're both true.
Remark
The trick here is to realize that you have to find the number both have in order to find the difference. That means you take (effectively) 3/8 and 5/8 to find out how much each of them has. The 8 stands for the 136
Step One
Find the number of stamps Alex has. Remember his ratio is the smaller so he will have 3/8
Alex = 3/8 * 136
Alex = 408/8 = 51
Step 2
Find the number that Freddy has.
The total is 136
<u>Formula</u>
Freddy = Total - Alex
Freddy = 136 - 51
Freddy = 85
<u>Check</u>
Let us see if 5/8 * 136 = 85
5*136/8 = 680 / 8 = 85 So your answers are correct.
Difference
Difference = 85 - 51 = 34 which is how many more stamps Freddy has than Alex.
5-4*9= 9
just use the properties
Answer:
Median.
Step-by-step explanation:
We have been given that in 2004, the mean net worth of families in a certain region was $470.2 thousand and the median net worth was $92.3 thousand.
We know that mean and median of a symmetric data set is equal.
We also know that when mean of a data set is greater than median, then the data set has a very large valued outlier.
Since mean net wroth of families is approximately 5 times more than median net wroth of families, this means that some of the families has very high net worth as outliers.
Since the net worth of families has very large outliers, therefore, I would prefer median as the appropriate measure of center as median is not affected by outliers.
