Juan's time = x
Yumi's time = x + 2
Yumi's distance = 3 miles
Juan's distance = 2 miles
Rate = distance/time
Juan's rate = 2/x
Yumi's rate = 3/(x + 2)
The rates are equal so:
2/x = 3/(x + 2)
2x + 4 = 3x
x = 4 hours
Yumi's time = 4 + 2 = 6 hours
The answer is D.
Answer: The numbers are showing in descending order 3.6-1.8= 1.8 = 1.8 is half of 3.6 - therefore if 1200 is half again = 0.8 then what lies between 1.8 and 0.8? Answer to 1000 is 0.8 +5 = 1.3
Step-by-step explanation:
<u>Part (a)</u>
The variable y is the dependent variable and the variable x is the independent variable.
<u>Part (b)</u>
The cost of one ticket is $0.75. Therefore, the cost of 18 tickets will be:
dollars
Now, we know that Kendall spent her money only on ride tickets and fair admission and that she spent a total of $33.50.
Therefore, the price of the fair admission is: $33.50-$13.50=$20
If we use y to represent the total cost and x to represent the number of ride tickets, the linear equation that can be used to determine the cost for anyone who only pays for ride tickets and fair admission can be written as:
......Equation 1
<u>Part (c)</u>
The above equation is logical because, in general, the total cost of the rides will depend upon the number of ride tickets bought and that will be 0.75x. Now, even if one does not take any rides, that is when x=0, they still will have to pay for the fair admission, and thus their total cost, y=$20.
Likewise, any "additional" cost will depend upon the number of ride tickets bought as already suggested. Thus, the total cost will be the sum of the total ride ticket cost and the fixed fair admission cost. Thus, the above Equation 1 is the correct representative linear equation of the question given.
Answer:
We conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.
Step-by-step explanation:
We know that the perimeter of a rectangle = 2(l+w)
i.e.
P = 2(l+w)
Here
Given that the length and width of the playground by a scale factor of 2
A scale factor of 2 means we need to multiply both length and width by 2.
i.e
P = 2× 2(l+w)
P' = 2 (2(l+w))
= 2P ∵ P = 2(l+w)
Therefore, we conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.
Answer:
1.
2.543.6
Step-by-step explanation:
We are given that
y(0)=200
Let y be the number of bacteria at any time
=Number of bacteria per unit time


Where k=Proportionality constant
2.
,y'(0)=100
Integrating on both sides then, we get

We have y(0)=200
Substitute the values then , we get


Substitute the value of C then we get





Differentiate w.r.t

Substitute the given condition then, we get



Substitute t=2
Then, we get 

e=2.718
Hence, the number of bacteria after 2 hours=543.6