
Actually Welcome to the Concept of the Modulus functions.
Since here, the value of h and k are zero,hence
the value of the function f(x) = 3|x| is 3
====> answer is 3
What is it?
The IQR describes the middle 50% of values when ordered from lowest to highest. To find the interquartile range (IQR), first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.
How do you find IQR?
<em>Step 1: Put the numbers in order. ...</em>
<em>Step 2: Find the median. ...</em>
<em>Step 3: Place parentheses around the numbers above and below the median. Not necessary statistically, but it makes Q1 and Q3 easier to spot. ...</em>
<em>Step 4: Find Q1 and Q3. ...</em>
<em>Step 5: Subtract Q1 from Q3 to find the interquartile range.</em>
Answer:
A , B, D
Step-by-step explanation:
YX is the segment joining points Y and X and is defined (A)
∠ XYZ is the angle between XY and YZ and is defined (B)
∠ YXZ is the angle between YX and XZ
However there is no segment joining XZ ← not defined
YW is the segment joining points Y and W and is defined (D)
First we have to assume that each quarter touched each other. Hence the area of the table not covered by the coins (A) is equal to the total area of the table (At) minus the total area of the coins (Ac). Coins are circle, so

and r =24.26mm. The area of one coin is then 1848.98mm^2. Hence the equation is A = At - xAc where x is the number of coins.
Total amount of the mixture =
ounces
Proportion of formula = 
Proportion of water = 
Amount of water needed to make the mixture can be given as:
Total amount of mixture x Proportion of water
Using the values, we get:
Amount of water needed =
= 7 ounces
Similarly the amount of formula needed will be:
Total amount of mixture x Proportion of formula
Using the values, we get:
Amount of formula needed =
= 3.5 ounces
Therefore, the home health care assistant needs 3.5 ounces of formula and 7 ounces of water to prepare 10 and a half ounces of formula.