Yes they will be equal at some point. Although Marc has more miles on his car at the current moment, Elena is adding miles at a higher rate.
Elena's car: 7000 + 21000t
Marc's car: 20000 + 11000t
To find at what time t they are equal, set these two expressions equal to each other and solve for t.
7000 + 21000t = 20000 + 11000t
13000 = 10000t
t = 1.3 years from now
Answer:
9u is a term
4 is a constant
Step-by-step explanation:
Coefficient is the number in front of the variable
The number of basketball that will fill up the entire office is <u>approximately 16,615.</u>
<em><u>Recall:</u></em>
Volume of a spherical shape = 
Volume of a rectangular prism = 
<em><u>Given:</u></em>
Diameter of basketball = 9.5 in.
Radius of the ball = 1/2 of 9.5 = 4.75 in.
Radius of the ball in ft = 0.4 ft (12 inches = 1 ft)
Dimension of the office (rectangular prism) = 20 ft by 18 ft by 12 ft
- First, find the volume of the basketball:
Volume of ball = 
Volume of basketball = 
- Convert to


<em>Therefore,</em>
- Volume of basketball =

- Find the volume of the office (rectangular prism):
Volume of the office = 
- Number of basket ball that will fill the office = Volume of office / volume of basketball
Number of basket ball that will fill the office = 
Therefore, it will take approximately <u>16,615 balls</u><u> to fill up the entire office</u>.
Learn more here:
brainly.com/question/16098833
Count the number of multiples of 3, 4, and 12 in the range 1-2005:
⌊2005/3⌋ ≈ ⌊668.333⌋ = 668
⌊2005/4⌋ = ⌊501.25⌋ = 501
⌊2005/12⌋ ≈ ⌊167.083⌋ = 167
(⌊<em>x</em>⌋ means the "floor" of <em>x</em>, i.e. the largest integer smaller than <em>x</em>, so ⌊<em>a</em>/<em>b</em>⌋ is what you get when you divide <em>a</em> by <em>b</em> and ignore the remainder)
Then using the inclusion/exclusion principle, there are
668 + 501 - 2•167 = 835
numbers that are multiples of 3 or 4 but not 12. We subtract the number multiples of 12 twice because the sets of multiples of 3 and 4 both contain multiples of 12. Subtracting once removes the multiples of 3 <em>and</em> 4 that occur twice. Subtracting again removes them altogether.