The antecedent for both the possessive adjective 'their' and the objective personal pronoun 'them' is children<span>.</span>
Answer:

Step-by-step explanation:
The given function is

We want to find all values of x for which:
f(x) is greater than or equal to 14.
This implies;

Subtract 13 from both sides;


By definition of the absolute value function,

Divide through the first inequality and and reverse the inequality sign:




Here is your answer mate ,
10.7142860
Or
11
Hope it helps
Pls——-mark———-as————-brainliest
Answer:
<u>The correct answer is that 4√- 2 is the simplest radical form of this mathematical expression and 8i (i = imaginary unit) is the solution if we want to find out the square root of a negative signed number.</u>
Step-by-step explanation:
1. Let's write in a mathematical expression the information given:
Sum square root of -2 and the square root of - 18
√- 2 + √ - 18
√- 2 + √ 9 . - 2 ( 9 * -2 = - 18)
√- 2 + 3 √ - 2 (√ 9 = 3)
4 √- 2
This is the simplest radical form of this mathematical expression. If we want a concrete number, we should remember that if we refer to the imaginary numbers we can find the solution for √-2 = 2i, or for any other negative number, where i is the imaginary unit. This unit can be used for the development of the square root of negative signed numbers.
Then,
4 √- 2 = 4 (2i) = 8i
<u>The correct answer is that 4√- 2 is the simplest radical form of this mathematical expression and 8i (i = imaginary unit) is the solution if we want to find out the square root of a negative signed number.</u>
Answer:
1. 9 bounds of carbon dioxide
2. 27 bounds of carbon dioxide
Step-by-step explanation:
So let's find a unit rate: 1 to 9
"Why is it 1:9?"
If I can find divide 72 pounds into 8 hours? What would I find? You would find the pounds for each other
Example: 72/8=9. So If 9 carbon dioxide per hour. I can do 3 hours is 27 pounds of carbon dioxide.
So to sum it up, 72/8=9 so 1:9 can be used to figure out any amount of pounds used in a certain number of hours. So If 3 hours, then 27 pounds.