Answer: 1 hour
Step-by-step explanation:
Let the distance they walked be
in time
and the distance they walked be
in time
,then
Walking speed=

Riding speed=

Also the total time taken=3 hours
∴
...(1)
Since the total distance= 20 km
∴![d_1+d_2=20\\\Rightarrow4t_1+12t_2=20\\\Rightarrow4t_1+12(3-t_1)=20..............[\text{from (1)}]\\\Rightarrow4t_1+36-12t_1=20\\\Rightarrow-8t_1=-8\\\Rightarrow\ t_1=1](https://tex.z-dn.net/?f=d_1%2Bd_2%3D20%5C%5C%5CRightarrow4t_1%2B12t_2%3D20%5C%5C%5CRightarrow4t_1%2B12%283-t_1%29%3D20..............%5B%5Ctext%7Bfrom%20%281%29%7D%5D%5C%5C%5CRightarrow4t_1%2B36-12t_1%3D20%5C%5C%5CRightarrow-8t_1%3D-8%5C%5C%5CRightarrow%5C%20t_1%3D1)
Thus, the they spent 1 hour time in walking.
The reading is 26°c as the small sections r of difference 2
We move all terms to the left:
+31x+32)-((+5)=0
So your answer is 0
Given :
The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by
.
To Find :
Find the time at which the concentration is a maximum. b. Find the maximum concentration.
Solution :
For maximum value of x, K'(x) = 0.

Since, time cannot be negative, so ignoring x = -3 .
Putting value of x = 3, we get, K(3) = 15/( 9 + 9) = 5/6
Therefore, maximum value drug in bloodstream is 5/6 at time x = 3 units.
Hence, this is the required solution.
Answer:
$2.7
Step-by-step explanation:
Let the price of one apple be x
and the price of one orange y
Hence ;
One apple and 3 oranges would cost:
x + 3×y = $5.10----------------------------(1)
One apple and 5 oranges would cost;
x + 5×y = $7.50--------------------------(2)
Subtracting eqn (1) from (2), we have :
x-x + 5y -3y = 7.5- 5.1
2y = 2.4
y = $1.2
From eqn(1)
x + 3y = $5.10
x= $5.10-3($1.2) = $5.10-$3.6 = $1.5
Hence x=$1.5 and y= $1.2
We are required to find the cost of one apple and one orange, hence :
x+y = $1.5+$1.2= $2.7