The answer to this problem is y= 1/10x + 60, the y-intercept of the function is $60, and the range is {y| y (less than or equal to) 60}
Answer:
= 49
Answer.
Step-by-step explanation:
Solution:-
Length of the road = 5 miles
Distance between every two streetlights = 1/10 miles
Total number of streetlights placed on the new road = 5 ÷ 1/10
= 5*10
= 50 streetlights
As there is already a streetlight at the beginning of the new road, so the number streetlights will be = 50 - 1
You would have to do 1/3x2/5 which is 2/15. So 2/15 of the girls in the school play sports.
Answer:
41 years old.
Step-by-step explanation:
Let x represent age of younger child.
We have been given that a mother has two children whose ages differ by 5 years. So the age of older child would be 
.
The sum of the squares of their ages is 97. We can represent this information in an equation as:
Let us solve for x.
Divide both sides by 2:
Since age cannot be negative, therefore, age of younger child is 4 years.
Age of older child would be 
Therefore, the age of older child would be 9 years.
We have been given that the square of the mother's age can be found by writing the squares of the children's ages one after the other as a four-digit number.
Square of 4: 
Square of 9: 
.
Square of mother's age: 
To find mother's age, we need to take positive square root of 1681 as:
Therefore, the mother is 41 years old.
Answer:
6 hours
Step-by-step explanation:
Answer:
6 hours
Step-by-step explanation:
Firstly, it is important to know where the train that leaves Washington will get after one hour since they didn't start at the same time between noon and 1 is 1 hour.
Speed = distance / time
Distance = speed * time
The distance S = 44 * 1
S= 44m
To determine the time they will pass each other
500(total distance)- 44 ( distance gotten to in 1 hour) - d(the distance it willl pass each other =44* t(time the train will pass each other.
456-d =44t
Second train
d = 32t
Substituting to the equation
456-32t=44t
456=76t
t= 6 hours.
The time they will pass each other is six hours.
